Masonry Designer's Guide.pdf

MASONRY DESIGNERS’ GUIDE

Based on

Building Code Requirements for Masonry Structures (AC1 530-92/ASCE 5-92/”MS 402-92) and Specifications for Masonry Structures (AC1 530.1-92/ASCE 6 - 9 2 m S 602-92) with Illustrated Design Applications

COPYRIGHT ACI International (American Concrete Institute) Licensed by Information Handling Services

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MASONRY DESIGNERS GUIDE

Based on

Building Code Requirements for Masonry Structures (AC1 530-92/ASCE 5-92/TMS 402-92) and Specifications for Masonry Structures (AC1 530.1-92/ASCE 6-92/TMS 602-92) with Illustrated Design Applications

John H. Matthys, editor


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The Masonry Society American Concrete Institute

The Masonrv Desimers’ Guide is not intended to teach a novice how to design or build masonry structures, or to replace sound engineering knowledge,experience, and judgment. The Guide should be used by professionals who are qualified to evaluate the significance, limitations, and applicability of the information reported, and whowill accept the responsibility for its proper use.

Direct all correspondence to: Masonry Designers’ Guide The Masonry Society 2619 Spruce Street Boulder, Colorado 80302-3808 (303) 939-9700


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FOREWORD

This Masonrv Desieners’ Guide (MDG) is intended to assist those involved in the design, construction, and regulation of masonry structures, The Guide was prepared to help users apply the provisions ofBuildine; Code Reauirements for Masonrv Structures (AC1 53092/ASCE 5-92DMS 402-92) and SDecificationsfor Masonry Structures (AC1 530.1-92/ASCE 6-92DMS 602-92). These two documents were developed by the Masonry Standards Joint Committee (MSJC) which includes membersof the American Concrete Institute, American Society of Civil Engineers, and The Masonry Society. Emphasis in the MDG is on application of the two documents. Background information on the development of the Code and Specifications provisions is not emphasized. For such information the reader is referred to Commentarv on Buildine: Code Reauirements for Masonrv Structures (AC1 530-92/ASCE 5-92DMS 402-92) and Commentarv on SDecifications for Masonrv Structures (AC1 530.1/ASCE 6-92/TMS 602-92). For ease in referencing the documents described above, an abbreviated notation has been used in the MDG. Building Code

Reauirements for Masonrv Structures is shortened to

MSJC Code or Code. Specifications or Specs. means SDecificationsfor Masonry Structures. Code C and Specs. C refer to the respective commentaries. The MDG is a first-of-its-kind document for the masonry industry and is a culmination

of

the efforts of The Masonry Society (TMS), the Councilfor Masonry Research (CMR), and theAmerican Concrete Institute(ACI).TMS, industry,providedthemasonryexpertise

the professionalsociety of the masonry

of their members to write the document. The

CMR, a consortiumof masonry industry associations, hadthe financial resourcesto fund the


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Guide’s development. AC1 had the qualifications and resources

to publish the MDG.

The MDG should be a valuable reference to engineers, contractors, architects, inspectors, building code authorities, and educators.The first chapters address materials, testing, quality assurance, quality control, and construction methods, with reference to specific provisions of the MSJC Code and Specifications. Following chapters on design illustrate applications of Code provisions to the structural design of masonry. There are more than 80 numerical example problems. A Code Reference Index and a SpecificationReference Index correlate each discussion and design application example to a particular MSJC Codeor Specification section. The Guide was developed under the auspices of The Masonry Society under the direction of John H.Matthys,Professor

ofCivil

Engineering and Director of the Construction

Research Center, The University of Texas at Arlington. Dr. Matthys provided guidance to individual authors in development ofall

chapters,servedasmanagingeditor,

and

coordinator for the production of the document. The production of Part I, General, was the direct responsibility of John H. Matthys. The production of Part II, Materials and Testing, and Part III, Construction, was the direct responsibility of the TMS Construction Practices Technical Committee chairedby Howard Droz. The production of Part IV, Design, was the direct responsibility of the TMS Design Practices Technical Committee chaired by John Tawresey. The voluntary contributions of allprimary authors and reviewers are recognized. Each section/chapter of the finished Guide is a meshing

of concepts of authors and numerous

reviewers. In addition manyof the example problems were developed and refined by several authors. iv


Authors/Sections/Chapters

Dr. Daniel P. Abrams - Professor of Civil Engineering, University of Illinois, ChampaignUrbana, Illinois - Sections 11.2 and 12.3 on Pilasters Dr.Subhash C. h a n d - Professor of Civil Engineering,ClemsonUniversity,Clemson, South Carolina

-

Section 9.0 Introduction and Section 9.3 on IntrawallLoad

Distribution Jefferson W. Asher - KPFF Consulting Engineering, Santa Monica, California Load Distribution ChristineBeall

-

- Lateral

- Computer Calculations for RCJ Hotel

Architect,Austin,Texas

- Section 5.1 on Submittals,Section 6.1 on

Preparation, Section 7.1 on Hot Weather Construction, and Section

7.2 on Cold

Weather Construction WilliamBretnall

-

Gensert BretnallAssociates,Cleveland,Ohio

-

Chapter 8 onDesign

Methodology and Philosophy Dr.RussellBrown

-

Chairman, Department ofCivil

Engineering,ClemsonUniversity,

Clemson, South Carolina - Section 12.1 on Columns and Section 12.2 on Walls Mario J. Catani - Dur-O-Wal,Inc.,ArlingtonHeights,Illinois

- Section 3.5 on Metal

Connectors and Reinforcement Howard Droz - Architect, Smith, Hinchman & Grylls, Detroit, Michigan - Section 6.3 on Tolerances

- Former ExecutiveDirector,MasonryInstitute Missouri - Section 5.2 on Sample Panels - Deceased 1992

Harry A. Fine

V


ofSt.

Louis,St.Louis,

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Richard Gensert

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Gensert Associates,Sewickley,Pennsylvania

-

Chapter 8 on Design

Methodology and Philosophy Clayford T. Grimm

-

Masonry Consultant, Austin, Texas

- Section 3.4 on Masonry and

Chapter 10 on Movements Dr. Ahmad A. Hamid

- Professor ofCivil

Engineering, Drexel University, Philadelphia,

Pennsylvania - Section 13.2 on Shear Walls and Chapter 16 on Provisionsfor Seismic Design

- Smith & Huston, Consulting Engineers, Seattle, Washington - Lateral Load Distribution - Hand Calculations for RCJ Hotel

Edwin T. Huston

Albert W. Isberner - Consultant, Portage, Wisconsin

- Chapter 3 on Materials,

Chapter 4

on Testing and Section 5.5 on Compliance Rochelle C. Jaffe

- Raths, Raths, and Johnson, Willowbrook, Illinois - Development and

BasicDesign

of theTMSShopping

Center, DPC Gymnasium andRCJ

Chapter 8 onDesignMethodologyandPhilosophy,

Hotel,

and Section 6.5 on Quality

Assurance/Quality Control Checklist Dr. Richard E. Klingner - Professor of Civil Engineering, The University of Texas at Austin, Austin, Texas

- Chapter 14 on Reinforcement and Connectors

Robert Kudder - Raths, Raths, and Johnson, Willowbrook, Illinois - Section 5.4 on Testing Dr. W. Mark McGinley

- Professor

of Architectural Engineering, North Carolina A&T

University, Greensboro, NorthCarolina

- Section 9.1

Section 9.2 on Intenvall Load Distribution

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- MunsellAssociates,Inc.,Southfield,Michigan

- Chapter 15 on

Empirical Design Dr.Max Porter

-

Professor of CivilEngineering,Iowa

State University,Ames,Iowa

-

Section 13.1 on Design for Shear in Masonry Components John G. Tawresey - VicePresident, KPFF ConsultingEngineers, Seattle, Washington Development of TMS ShoppingCenter, DPC Gymnasium, and RCJ Hotel - Section 11.0 Introduction and 11.1 on Walls,Section

12.0 Introduction and Section 13.0

Introduction Al Tomassetti - Masonry Consultant, Louisville, Kentucky

- Section 5.3 on Inspection and

Section 6.4 on Cleaning Terence A. Weigel - Professor of CivilEngineering,University

of Louisville, Louisville,

Kentucky - Section 11.1 on Walls

- Professor of CivilEngineering,University College Park, Maryland - Section 11.3 on Beams and Lintels

Dr.Amde

M.Wolde-Tinsae

Gary L Zwayer

- Wiss, Janney, Elstner Associates, Northbrook, Illinois

of Maryland,

- Section

6.2 on

Placement and Section 6.5 on Quality Assurance/Quality Control Checklist Reviewers

To encourage input and acceptance of the Guide by the design communityand themasonry industry as awhole,numerousreviewsduring planned.Appreciation

the development of this document were

is extended to allreviewers for their voluntarycontributionsin

production of this unique document.

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Review of the first draft of Part II, Materials and Testing, and Part III, Construction, was conducted by a Technical Review Committee of:

- National Concrete Masonry Association - Herndon, Virginia Mario Catani - Dur-O-Wal, Inc. - Arlington Heights, Illinois Howard Droz - Smith, Hinchman, & Grylls - Detroit, Michigan John Grogan - Brick Institute of America, Region 9 - Atlanta, Georgia Dr. John Matthys - University of Texas at Arlington - Arlington, Texas.

Kevin Callahan

Review of the first draft of Part IV, Design,wasconducted

by aTechnicalReview

Committee of: Dr. James Colville

- Professor ofCivil

Engineering

- University of

Maryland and

Chairman of Masonry Standards Joint Committee - College Park, Maryland Ed Huston - Smith & Huston - Seattle, Washington

- Raths, Raths & Johnson - Willowbrook, Illinois Dr. Richard E. Klingner - University of Texas at Austin - Austin, Texas Dr. John H. Matthys - University of Texas at Arlington - Arlington, Texas.

Rochelle Jaffe

A review of the revised first draft was conducted by a Combined Review Group of:

J. Gregg Borchelt - Brick Institute of America

- Reston, Virginia - Representing

CMR Dan Shapiro - SOHA - San Francisco, California

J. A. "Tony"Wintz,III

-

- Representing TMS

Wiss, Janney,ElstnerAssociates

- Washington, D.C.

-

Representing ACI. Afinaltechnicalreview

of thesubmittedproposeddocumentwasconducted

by the

Technical Activities Committeeof TMS, the Technical Activities Committee of ACI, and the Technical Committee of CMR. viii


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A final editorial review of the entire Guide was conducted by an editorial committee of:

Rochelle C. Jaffe - Raths, Raths & Johnson - Willowbrook, Illinois Dr. Richard E. Klingner - University of Texas at Austin - Austin, Texas Dr. John H. Matthys - University of Texas at Arlington - Arlington, Texas The staffof

theConstruction

Research Center (CRC) at theUniversity

of Texas at

Arlington was in charge of the production of the Guide in camera ready form, both hard copy and electronic disks. Special thanks go to: Barbara Wallace - CRC secretary for the word processing of the chapters’ text. Debra Roberts

-

CRC staffandcivilengineering

student forproduction

of the

design example problems and coordinating production of all figures. Finny Samuel, Titus Benny, and Asher Mahmood- students at UTA for production of the electronic graphics. Although this document has undergone numerous examinations, errors and inconsistencies are sure to exist. The Masonry Society would appreciate the findings of such discrepancies being brought to its attention. John H. Matthys Professor of Civil Engineering Director of Construction Research Center University of Texas at Arlington

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CONTENTS

-

PART I GENERAL

1. 2.

.

FOREWORD ................................................ CONTENTS ................................................... CODE REFERENCE INDEX ................................... SPECIFICATIONS REFERENCE INDEX ......................... INTRODUCTION ............................................. NOTATIONS, DEFINITIONS, AND ABBREVIATIONS ..............

iii xxv xxx

1-1 2-1

PART II - MATERIALS AND TESTING

3.

MATERIALS 3.0 INTRODUCTION ....................................... General Intent 3.0.1 3.0.2 Specifications-Preface and Checklists 3.0.3 Mandatory Specification Checklist 3.0.4 Optional Specification Checklist 3.0.5 Submittals 3.0.6 Material Specification References 3.0.7 Material Specification Requirements (ASTM) 3.1

3.2

UNITS 3.1.1 3.1.2 3.1.3 3.1.4 3.1.5 3.1.6

................................................

3-1

3-5

Product Specifications for Clay or Shale Masonry Units Product Specificationsfor Concrete Masonry Units Product Specifications for Stone Masonry Units Product Testing and Conformance Product Receipt and Storage Manufacturers’ Recommendations

MORTARS ............................................ 3.2.1 Selection of Mortar Type 3.2.2 ASTM C 270 Mortar Types 3.2.2.1 Proportion Specification 3.2.2.2 Property Specification 3.2.3 Cementitious Materials 3.2.4 Aggregates 3.2.5 Mortars Containing Hydrated Lime 3.2.6 Mortars Containing Masonry Cement 3.2.7 Admixtures

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3.3

GROUT

..............................................

3-17

3.4

MASONRY ............................................ Brick Masonry In Compression 3.4.1 3.4.2 Brick Masonry Elastic Modulus 3.4.3 Brick Masonry In Flexure 3.4.4 Brick Masonry In Shear 3.4.5 Concrete Masonry In Compression 3.4.6 Concrete Masonry Elastic Modulus 3.4.7 Concrete Masonry In Flexure 3.4.8 Concrete Masonry In Shear 3.4.9 Grout 3.4.10 Steel Reinforcement

3-17

3.5

METAL CONNECTORS AND REINFORCEMENT General 3.5.1 3.5.2 Steel Wire 3.5.3 Steel Sheet Metal 3.5.4 Reinforcing Bars 3.5.5 Connectors 3.5.6 Corrosion Protection 3.5.7 Deformed vs. Smooth Reinforcement

REFERENCES 4.

93

............. 3-23

..............................................

3-26

TESTING

4.0

INTRODUCTION

.......................................

4-1

4.1

MATERIALS TESTING .................................. 4.1.1 Preconstruction 4.1.2 Construction Testing

4-2

4.2

ASSEMBLAGE TESTING .PRISMS

........................

4-4

PART III - CONSTRUCTION 5.

QUALITY ASSURANCE

5.0

INTRODUCTION ....................................... 5.0.1 Quality Assurance 5.0.2 Quality Control

5-1

5.1

SUBMITTALS ..........................................

5-4

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SAMPLE PANELS ...................................... 5.2.1 Recommended Practices 5.2.2 Suggested Criteria for Construction

5-5

5.3

INSPECTION .......................................... 5.3.1 Purpose 5.3.2 Planning 5.3.3 Material Submittals 5.3.4 Inspection Files 5.3.5 Construction Inspection

5-7

5.4

TESTING ............................................. 5-10 Testing as Part of a ConstructionQualityAssurance Program 5.4.1 5.4.1.1 Initial Rate of Absorption Tests 5.4.1.2 Testing to Verify the PrismCompressive Strength of Masonry 5.4.1.3 Testing to Evaluate Mortar 5.4.1.4 Testing to Evaluate Grout 5.4.1.5 Testing to Determine the FlexuralModulus of Rupture 5.4.2 Procedures Useful for Inspection Programs 5.4.2.1 OmittingMasonry to Permit Inspection Within a Cavity Wall 5.4.2.2 Fiber-optic Borescope

5.5

COMPLIANCE

REFERENCES

6.

93

......................................... ..............................................

5-18 5-20

QUALITY CONTROL

6.0

INTRODUCTION

.......................................

6-1

6.1

PREPARATION ........................................ 6.1.1 Material Delivery, Storage, and Handling 6.1.2 InspectingSurfaces to ReceiveMasonry 6.1.3 Masonry Units 6.1.4 Reinforcement, Connectors, and Accessories 6.1.5 Mortar and Grout 6.1.5.1 Mortar 6.1.5.2 Grout 6.1.6 Protections

6-1

6.2

PUCEMENT .......................................... Mortar Placement 6.2.1

6-11

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Reinforcement Placement Tie and Anchor Placement Unit Placement Grout Placement Flashing and Weephole Placement Movement Joint Construction

6.3

TOLERANCE .......................................... 6.3.1 Introduction 6.3.2 AC1 530.1/ASCE 6/”S 602 Reference 6.3.3 Tolerance 6.3.4 Tolerance Examples 6.3.4.1 Mortar Joint 6.3.4.2 Masonry Openings 6.3.4.3 VerticalExpansionJoints

6-25

6.4

CLEANING

............................................

6-31

6.5

QUALITY ASSURANCE/QUALITY CONTROL CHECKLIST

REFERENCES

7.

93

. . . . 6-32

..............................................

6-33

HOT ANDCOLD WEATHER CONSTRUCTION

7.0

INTRODUCTION

.......................................

7-1

7.1

HOT WEATHER CONSTRUCTION ........................ Performance of Masonry and Mortar 7.1.1 7.1.2 Material Storage, Protection, and Preparation

7-1

7.2

COLD WEATHER CONSTRUCTION . . . . . . . . . . . . . . . . . . . . . . . 7-3 Performance of Masonry and Mortar 7.2.1 7.2.2 Material Storage, Protection, and Preparation

REFERENCES

..............................................

7-9

-

PART IV DESIGN 8.

DESIGN PHILOSOPHY AND METHODOLOGY

.......................................

8-1

...................................

8-3

8.0

INTRODUCTION

8.1

WHATIS MASONRY. 8.1.1 Masonry Units

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Mortars Grout Masonry Assemblages Volume Changes Details of Construction

...................

8.2

STRUCTURAL ANALYSISANDDESIGN 8.2.1 General Requirements 8.2.2 Analysis Considerations

8.3

LOADS AND LOAD COMBINATIONS

8.4

STRUCTURAL CONSIDERATIONS FOR MASONRY WALLS . . . 8-14 8.4.1 WallContinuity and Support Conditions 8.4.2 Vertical Loadbearing Walls 8.4.2.1 Types of Vertical Loads 8.4.2.2 Failure Mode 8.4.2.3 Effects of Openings 8.4.2.4 Gravity Stresses Resultingfrom Interaction ofWalls and Horizontal Diaphragms 8.4.2.5 Engineering Analysis 8.4.3 Shear Walls 8.4.3.1 Horizontal Diaphragm Stiffness 8.4.3.2 Effects ofWall Proportions 8.4.3.3 Effects ofAxial Loads 8.4.3.4 Effects of Openings 8.4.3.5 Effects ofWall Placements 8.4.3.6 Effects of Interconnection of Perpendicular Walls 8.4.3.7 Effects of Location of Plan Center of Resistance 8.4.3.8 Wall Reinforcing Patterns 8.4.3.9 Engineered Design of Masonry Shear Walls 8.4.4 Progressive Collapse

8.5

STRUCTURAL CONSIDERATIONS FOR MASONRY BEAMS 8.5.1 Beam Behavior 8.5.2 Engineered Design of Reinforced MasonryBeams

8.6

STRUCTURAL CONSIDERATIONS FOR MASONRY BEAMCOLUMNS ...................................... 8.6.1 Beam Column Behavior 8.6.2 Engineered Design of MasonryBeamColumns 8.6.3 Interaction Diagrams

REFERENCES

......................

.............................................. xv


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DISTRIBUTION OF LOADS

9.0

INTRODUCTION

.......................................

9-1

9.1

BUILDING EXAMPLES .................................. 9.1.1 TMS Shopping Center 9.1.1.1 Gravity Design Loads 9.1.1.2 Lateral Design Loads 9.1.2 DPC Gymnasium 9.1.2.1 Gravity Design Loads 9.1.2.2 Lateral DesignLoads 9.1.3 RCJ Hotel 9.1.3.1 Gravity Design Loads 9.1.3.2 Lateral Design Loads

9-2

9.2

INTERWALL LOADDISTRIBUTION ....................... 9-49 9.2.1 Global Lateral LoadDistributionon Shear WallsinBuildings with Flexible Diaphragms 9.2.2 Global Lateral LoadDistributionon Shear Walls in Buildings with Rigid Diaphragms

9.3

INTRAWALLLOADDISTRIBUTION ....................... 9-56 General 9.3.0 9.3.1 LocalDistribution UnderConcentrated Loads 9.3.2 LocalDistribution of Concentrated LoadsActingonBondBeams 9.3.2.1 Hollow Masonry Walls 9.3.2.2 SolidMasonryWalls 9.3.3 EffectiveBearing AreaUnder Concentrated Loads 9.3.4 LocalLoadDistributioninMultiwytheNoncomposite(Cavity)Walls 9.3.5 LocalLoadDistributioninMultiwytheCompositeMasonryWalls 9.3.6 Local Lateral and Axial Load Distribution Single in Wythe Loadbearing Wall Systems 9.3.7 LocalDistribution of Lateral LoadWithin PerforatedShear Walls

REFERENCES

..............................................

9-63

EXAMPLES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 - 6 5 9.2-1 TMSShopping Center - Lateral LoadDistribution 9.2-2 DPC Gymnasium - Lateral LoadDistribution 9.2-3 RCJHotel - Lateral LoadDistribution - Hand Calculations 9.2-4 RCJ Hotel - Lateral LoadDistribution - Computer Calculations 9.3-1 TMSShopping Center - LoadDistributionWithinSingleWythe Walls Under Concentrated Loads 9.3-2 T M S Shopping Center - Distribution of Concentrated Loads Acting xvi


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9.3-3 9.3-4 9.3-5 9.3-6 9.3-7 9.3-8 9.3-9 9.3-10 9.3-11 9.3-12

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on a Bond Beam TMS Shopping Center - Effective Bearing Area Under Concentrated Load TMS ShoppingCenter - Effective BearingArea Under Concentrated Load TMS ShoppingCenter - Effective BearingArea Under Concentrated Load Moment in DPC Gymnasium - Distribution of GravityLoad Multiwythe Noncomposite (Cavity) Walls DPC Gymnasium - In-Plane Lateral Load Distribution in Multiwythe Noncomposite (Cavity) Walls DPC Gymnasium - Distribution of Out-of-Plane Lateral Loads in Multiwythe Noncomposite (Cavity) Walls DPC Gymnasium - Shear Stress Distribution in the Collar Joint of a Multiwythe Composite Wall Due to Out-of-Plane Wind Load TMSShopping Center - Lateral and Axial Load Distribution in Single Wythe Loadbearing Wall Systems TMS Shopping Center - Distribution of Horizontal Load Within Reinforced Perforated Shear Walls TMS Shopping Center - Distribution of Horizontal Load Within Unreinforced Perforated Shear Walls

MOVEMENTS 10.1 CAUSES AND CONSEQUENCES OF MOVEMENTS

. . . . . . . . . . . 10-1

10.2 DETERMINATION OF STRUCTURAL MOVEMENTS . . . . . . . . . 10-1 Probabilistic Concepts 10.2.1 Short-Term Movements Due to External Forces 10.2.2 Long-Term Movement of Masonry 10.2.3 10.2.3.1 Creep of BrickMasonry 10.2.3.2 Creep of Concrete Masonry Thermal Movement 10.2.4 10.2.4.1 Temperature Change in Exterior Walls 10.2.4.2 Coefficient of Thermal Expansion Moisture Movements 10.2.5 10.2.5.1 Brick 10.2.5.2 Mortar 10.2.5.3 Concrete Masonry Freezing Expansion 10.2.6 Restraint of Masonry 10.2.7 10.3 STRUCTURAL MOVEMENT 10.3.1 Frame Movement xvii


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10.3.1.1 Concrete Column Shortening 10.3.1.2 Steel Column Shortening 10.3.1.3 Sidesway Deflections of Horizontal Members 10.3.2.1 Beam Deflection 10.3.2.2 ShelfAngleDeflection Foundation Movement Slab Movement Differential Structural Movement 10.3.5.1 Nonloadbearing Walls 10.3.5.2 Loadbearing Walls 10.3.5.3 Loadbearing/Nonloadbearing Wall Intersection

10.3.2 10.3.3 10.3.4 10.3.5

10.4 ACCOMMODATION OF MOVEMENTS ..................... 10.4.1 Design of Movement Joints 10.4.1.1 Sealants Used in Movement Joints 10.4.1.2 Control Joints 10.4.1.3 Expansion Joints 10.4.1.4 Construction Joints REFERENCES EXAMPLES 10.4-1 10.4-2 10.4-3 11.

.............................................

10-8

10-14

............................................... 10-22 TMS Shopping Center - Vertical Control Joint Location RCJ Hotel - VerticalExpansion Joint Size and SpacingDesign RCJ Hotel - DifferentialMovementin BrickBlock Exterior Wall

FLEXURE

11.0 INTRODUCTION ....................................... 11.0.1 Organization of Chapter 11 11.0.2 Flexural Masonry Design 11.0.3 Direction of Flexure 11.0.4 Effects of Bonding Pattern 11.0.5 Flexure: Working Stress Design 11.0.5.1 Unreinforced Masonry 11.0.5.2 Reinforced Masonry 11.1 WALLS .............................................. 11.1.1 FlexuralDesign of Unreinforced MasonryWalls 11.1.2 FlexuralDesign of Reinforced MasonryWalls 11.1.2.1 Initial Depthand Steel Estimate 11.1.2.2 Balanced Design

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11.2 PILASTERS .......................................... 11.2.1 General Description 11.2.2 Role of Pilaster in Resisting Loads 11.2.3 Coursing Layout 11.2.4 Effective Section 11.2.5 Flexural Design Considerations 11.2.5.1 Unreinforced Pilasters 11.2.5.2 Reinforced Pilasters 11.2.6 Shear Design Considerations

11-15

11.3 BEAMSAND LINTELS ................................. Introduction 11.3.1 11.3.2 Assumptions 11.3.3 Basic Equations - Singly Reinforced Sections 11.3.4 Basic Equations - Doubly Reinforced Sections 11.3.5 LoadDistributionsonLintels 11.3.6 Beam Depth Determination 11.3.7 Deflection 11.3.8 Deep Beams

11-25

.............................................

11-42

REFERENCES

EXAMPLES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1 - 4 4 11.1-1 TMS Shopping Center - Design of Unreinforced CMU Nonloadbearing Wall for Flexure Only 11.1-2 TMS ShoppingCenter - Design of Reinforced CMU Nonloadbearing Wall for Flexure Only 11.1-3 TMS Shopping Center - Unreinforced Wall Designfor Out-of-Plane Flexure 11.1-4 DPC Gymnasium - Design of an Unreinforced Multiwythe BrickBlock Noncomposite (Cavity) Wall for Flexure Only 11.1-5 DPC Gymnasium - Design of an Unreinforced Multiwythe Composite Wall for Flexure Only 11.1-6 DPC Gymnasium - Design of a Reinforced Multiwythe Composite Wall for Flexure Only 11.1-7 Wythe Reinforced DPC Gymnasium - Design of a Single Nonloadbearing Hollow Clay Masonry Wall for Flexure 11.1-8 RCJ Hotel - Design of a Reinforced Clay Brick Lintel 11.1-9 RCJ Hotel - Unreinforced Retaining Wall Design for Out-of-Plane Flexure 11.1-10 RCJ Hotel - Reinforced Retaining WallDesign for Out-of-Plane Flexure 11.1-11 RCJ Hotel - Design of an Unreinforced Multiwythe Noncomposite (Cavity) Brick-Block MasonryNonloadbearing Wall for Flexure Only XiX


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11.1-12 11.2-1 11.2-2 11.3-1 11.3-2 11.3-3 11.3-4 11.3-5 11.3-6 11.3-7 11.3-8 12.

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FLEXURE AND AXIAL LOAD

12.0 INTRODUCTION

.......................................

12.1 COLUMNS ............................................ 12.l.1 General 12.1.2 Development of Interaction Diagram 12.1.2.1 Compression Controls 12.1.2.2 Tension Controls

12-1 12-2

12.2 WALLS .............................................. Unreinforced Masonry Walls 12.2.1 12.2.1.1 Unity Inequality 12.2.1.2 Euler Buckling 12.2.1.3 Flexural Tensile Stress 12.2.2 Reinforced Masonry Walls 12.2.2.1 Interaction Diagram for Reinforced Walls 12.2.2.2 Typical Iterative Method

12-18

12.3 PILASTERS .......................................... 12.3.1 Critical Loading Cases 12.3.2 DesignConsiderations for Unreinforced Pilasters 12.3.3 DesignConsiderations for Reinforced Pilasters

12-42

.............................................

12-48

REFERENCES EXAMPLES 12.1-1 12.2-1 12.2-2 12.2-3

...............................................

12-49 RCJ Hotel - LobbyColumnDesign TMSShopping Center - Design of Reinforced Loadbearing Wall TMS Shopping Center - Design of Unreinforced Loadbearing Wall DPC Gymnasium - Design of Unreinforced Multiwythe



Noncomposite Masonry Wall DPC Gymnasium - Design of Unreinforced Composite Masonry Wall DPC Gymnasium - Design of Reinforced Hollow Clay Masonry Wall DPC Gymnasium - Unreinforced Pilaster Subject to Flexure and Axial Load DPC Gymnasium - Reinforced Pilaster Subject to Flexure and Axial Load

12.2-4 12.2-5 12.3-1 12.3-2 13.

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SHEAR

13.0 INTRODUCTION

.......................................

13.1 DESIGN FOR SHEAR INMASONRY COMPONENTS 13.1.1 OverallPhilosophy for Shear Design 13.1.2 Unreinforced Masonry Shear Design 13.1.3 Reinforced Masonry Shear Design 13.1.3.1 Shear Reinforcement Not Required 13.1.3.2 Shear Reinforcement Required 13.1.4 SpecialProvisions for Diaphragms 13.2 SHEAR 13.2.1 13.2.2 13.2.3 13.2.4 13.2.5 13.2.6

13-1

.......... 13-2

WALLS ....................................... Definition of a Shear Wall Function of Shear Walls Layout of Shear Walls Analysisof Shear Walls Flexural Design 13.2.5.1 Unreinforced Shear Walls 13.2.5.2 Reinforced Shear Walls Shear Design 13.2.6.1 Unreinforced Shear Walls 13.2.6.2 Reinforced Shear Walls

13-10

.............................................

13-22

REFERENCES

EXAMPLES . . . . . . . . . . . . . . . . , . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 3 - 2 3 13.1-1 RCJ Hotel - Shear Design of a Reinforced BrickCouplingBeam 13.1-2 RCJHotel - Shear Design for CanopyBeam 13.1-3 RCJ Hotel - Shear Design of Continuous MasonryBeam 13.1-4 RCJHotel - Shear Design of a Reinforced ClayBrickNonloadbearing Wall 13.1-5 DPC Gymnasium - Shear Design for a CompositeMasonryWall 13.1-6 TMS Shopping Center - Shear Design of an Unreinforced Wall Due to Out-of-Plane Bending



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DPC Gymnasium - Shear Design of a Reinforced Composite Wall RCJ Hotel - Shear Design of a Coupling Beam TMSShopping Center - Shear Design of a Reinforced CMU Nonloadkaring Wall 13.1-10 TMS Shopping Center - Shear Design for a Doubly Reinforced Masonry Lintel 13.2-1 TMS Shopping Center - Unreinforced Shear Wall Design 13.2-2 TMS Shopping Center - Reinforced Shear Wall Design 13.2-3 DPC Gymnasium - Shear Wall Design 13.2-4 RCJ Hotel - Design of Unreinforced Masonry Shear Wall for InPlane Lateral Loads RCJ Hotel- Design of Reinforced Masonry Shear Wall for In-Plane 13.2-5 Lateral Loads 13.2-6 RCJ Hotel - Reinforced Masonry Shear Wall Design

13.1-7 13.1-8 13.1-9

14.

REINFORCEMENT ANDCONNECTORS

14.1 GENERAL ............................................ 14.1.1 Steel Reinforcement 14.1.2 Connectors 14.1.3 ConnectionsBetween Intersecting Walls

14-1

14.2 STEELREINFORCEMENT ............................... 14-6 14.2.1 Requirements for Steel Reinforcement 14.2.1.1 Strength Requirements for Reinforcement 14.2.1.2 Corrosion Resistance and Protection Requirements for Reinforcernent 14.2.1.3 Embedment Requirements for Reinforcement 14.2.2 Design of Steel Reinforcement 14.3 CONNECTORS ......................................... 14-8 Requirements for Connectors 14.3.1 14.3.1.1 Strength Requirements for Connectors 14.3.1.2 CorrosionResistanceand Protection Requirements for Connectors 14.3.1.3 Embedment Requirements for Connectors 14.3.1.4 Stiffness Requirements for Connectors 14.3.2 Design of Connectors REFERENCES

.............................................

14-11

EXAMPLES . . . . . . . . . . . . . . . , . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 4 - 1 3 14.2-1 T M S Shopping Center - Design of a Straight Bar Anchorage 14.2-2 TMSShopping Center - Design of a Hooked Bar Anchorage xxii


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14.2-3 14.3-1 14.3-2 14.3-3 14.3-4 14.3-5 14.3-6 14.3-7 14.3-8 14.3-9 14.3-10 14.3-11 14.3-12 14.3-13 14.3-14 14.3-15 14.3-16 15.

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TMS ShoppingCenter - Design of Anchorage at a Simple Supported Lintel Design of Anchor Bolts of a Shear Wall to Roof Diaphragm DPC Gymnasium - Design ofWall Ties in an Unreinforced Multiwythe Noncomposite (Cavity) Masonry Wall DPC Gymnasium - Design of Shear Wall-Floor Connection for Composite Nonloadbearing Wall DPC Gymnasium - Design of Shear Wall-Floor Connection for Unreinforced Multiwythe Noncomposite (Cavity) Wall TMS Shopping Center - Joist Connection to Loadbearing Wall RCJ Hotel - Connection Between Canopy Beam and Column Typical Reinforcing Details RCJ Hotel - Connection of RigidRoof Diaphragm to Exterior Loadbearing Wall RCJ Hotel - Connection of Floor Diaphragm to Nonloadbearing Wall - Connection of Floor Diaphragm to Interior RCJ Hotel Loadbearing Wall TMS Shopping Center - Connection of Steel Beam Bearing Detail TMS ShoppingCenter - Roof Diaphragm Connection to ShearWall RCJHotel - Connection of Exterior Nonloadbearing Wall to Exterior Loadbearing Wall RCJ Hotel- Connection of Interior Nonloadbearing Wall to Interior Loadbearing Wall DPC Gymnasium - Roof Diaphragm Connection to Nonloadbearing Wall RCJ Hotel - Termination of FlexuralReinforcement for Continuous Masonry Beam

EMPIRICAL DESIGN

15.1 HISTORY

.............................................

15-1

15.2 GENERAL DESCRIPTION

15-2

15.3 LIMITATIONS

............................... .........................................

15-3

15.4 EMPIRICAL DESIGN REQUIREMENTS .................... 15.4.1 Materials and Specifications 15.4.2 Lateral Stability 15.4.3 Compressive Stress Requirements 15.4.4 Lateral Support 15.4.5 Thickness of Masonry 15.4.5.1 Minimum Thickness Criteria

15-4

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15.4.6 15.4.7

15.4.8

REFERENCES

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15.4.5.2 Foundation Walls Bond 15.4.6.1 Masonry Headers 15.4.6.2 Metal Ties Anchorage 15.4.7.1 Intersecting Walls 15.4.7.2 Floor and Roof Anchorage 15.4.7.3 WallsAdjoining Structural Framing Miscellaneous Requirements 15.4.8.1 Chases and Recesses 15.4.8.2 Lintels 15.4.8.3 Support on Wood 15.4.8.4 Corbelling

.............................................

15-13

EXAMPLES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . , . 1 5 - 1 4 15.4-1 TMSShopping Center - EmpiricalDesign of MasonryWalls 15.4-2 DPC Gymnasium - EmpiricalDesign of MasonryWalls 15.4-3 RCJ Hotel - EmpiricalDesign of MasonryWalls

16.

PROVISIONS FOR SEISMICDESIGN

16-1

16.2 MATERIALS

16-2

16.3 DESIGN OF MASONRY ELEMENTS

....................... ...........................................

16-3

16.4 DETAILING 16.4.1 Reinforcement 16.4.2 Anchorage 16.4.3 Minimum Dimensions

16-4

..............................................

16-8

................................................ ClayMasonrySection Properties ................................. Concrete MasonrySection Properties .............................. Conversion Factors .SI Units ....................................

A-1

REFERENCES A

....................................... ..........................................

16.1 INTRODUCTION

APPENDIX

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5.9.1.6(a) 5.10 5.10.1 5.10.2 5.11.2 5.12 5.12.1

5.12.2 5.12.3

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5.14.2.2 5.16 6.3 6.3.1 6.3.l(a) 6.3.l(c) 6.3.1.1

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7.3.2 7.3.2.1 7.3.3 7.3.3.1 7.3.3.3 7.3.3.4 7.3.3.5 7.5 7.5.1 7.5.2 7.5.2.1

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13-34, 14-56 13-57, 14-8, 14-14 14-56 13-57 13-57 13-34 14-57 8-8 15-3 15-14, 15-21, 15-,28, 15-31, 15-33 15-14, 15-21, 15-28, 15-31, 15-33 15-14, 15-23, 15-28, 15-31, 15-33 15-4, 15-15, 15-21, 15-29, 15-33 15-15, 15-17, 15-21, 15-22, 15-23, 15-29, 15-34 15-4, 15-15, 15-17, 15-22, 15-29, 15-34 15-4, 15-22, 15-23, 15-24, 15-33 15-18 15-5, 15-15, 15-20, 15-29, 15-31, 15-34 15-5 15-16, 15-18, 15-23, 15-24, 15-31 15-6 15-7 15-3, 15-8 15-6, 15-16, 15-18 15-9 15-6 15-10 15-10 15-6, 15-10 15-3

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2.1.4.4 2.2.1 2.2.1.1 2.2.1.2 22.1.3 2.2.2 2.2.2.1 2.2.2.2 2.2.2.3 2.2.6 2.2.7 2.3.1 2.3.1.l(a) 2.3.1.l(b) 2.3.1.l(d) 2.3.1.l(e) 2.3.1.2(b) 2.3.2.1 2.3.2.1.1 2.3.2.1.2 2.3.2.2 2.3.2.3 2.3.2.4 2.3.2.4(a) 2.3.2.4(b) 2.3.3.2 2.3.3.2(a)l 2.3.3.2(a)2 2.3.3.3 2.3.3.3(d)2 2.3.3.3(d)4 2.3.3.3(e) 2.3.3.3(f)1 2.3.3.6(f) 2.3.3.6(g)

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1.6.3.2 1.6.3.3 2.1.1.1 2.1.2 2.1.2.1 2.1.2.l(a) 2.1.2.1(b) 2.1.2.l(c) 2.1.2.l(d) 2.1.2.1(f) 2.1.24g) 2.1.2.1(h) 2.1.2.2 2.1.2.2(b) 2.1.2.3 2.1.3.2 2.1.3.2(a) 2.1.3.2(b) 2.1.3.2(c) 2.1.3.3 2.1.3.4 2.1.3.5 2.1.4

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1 INTRODUCTION

Early masonry codes were totally empiricalas evidenced by requirements of minimum wall thicknesses, maximum building heights, etc. The so-called modern empirical masonry code, ANSI A41.1, has for years been the basis for the empirical design provisions for masonry found in model building codes. In the early 1960's, masonry industry associations began development of a technologicaldata base of masonry materials and assemblage performance through internally sponsored research and testing programs.

or externally

The result of these efforts culminated in such

design standards as the Brick Institute of America's (BIA) Recommended Practice For EngineeredBrickMasonryin (NCMA)SDecifications

1966 andthe

National Concrete MasonryAssociation's

for Loadbearing;ConcreteMasonryin

addressed only selected masonry materials. In Committee 531 published a

1970. Each document

1970 American Concrete Institute (ACI)

report, "Concrete Masonry

Structures

-

Design and

Construction" and in 1976 published Specificationsfor Concrete Masonry Construction(AC1

531.1-76). Both of these documents servedas the basis for Building Code Requirements for Concrete Masonry Structures (AC1 531-79), which addressed only concrete masonry. In the mid-70's The MasonrySociety(TMS)begandevelopment masonry standard that addressed both clay and concretemasonry.

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of a single structural The TMS standard,

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completed in 1981, served as the source document for the major changes to Chapter 24 of the Uniform Building Code that first appeared in the 1985 edition of the UBC. The masonry industry associations recognized the need for a national design code covering

all masonry materials. The American Society of Civil Engineers (ASCE) and the American Concrete Institute undertook this activity in the late 1970's. An agreement resulted in the ACI/ASCE 530 MasonryStructuresJointCommittee,formed consensus standard formasonrydesign.

in 1978, to developa

The committeemembersconsisted

of building

officials,contractors,researchers,professors,consultants,andmaterialproducers. developeddocumenthad

The

to meet the rigid procedural and consensusacceptance

requirements of both organizations. A code, to address design, and specifications,to address construction,weredraftedforcommitteeballot

by 1984.

Final adoption of Code,

Specifications, and Commentaries by ASCE and AC1 occurred in October 1988. The 530 Building: Code Requirements for Masonry Structures is primarily directed

to the designer

and code enforcement officials. The 530.1 Soecifications for MasonrvStructures is primarily directed to the contractor and inspector. Significant aspects related to these documents are that: 1.

Brick,block,andcombination

ofbrick

andblock

are covered in asingle

document. 2.

Designisbasedonthepremise

3.

Acceptancehascomefrom

that allworkwill

be inspected.

the masonryindustry,engineeringorganizations

and model code groups. This MDG and all design examples hereinare based on the allowable stresses for insDected masonryconstruction.

Thereare no alternativeallowablestressesbecause

uninmected

workmanship is not permitted. Seminars on the 530 Codeand 530.1 Specificationshave been conductedthroughthe auspices of ASCE and the sponsorship of the Council for MasonryResearch (CMR) for the 1-2


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past several years. The original joint ACUASCE530 Masonry Structures Committeeis now under the auspices of TMS/ACI/ASCE and hasbeen renamed the Masonry Standards Joint Committee (MSJC). This committee oversees revisions and expansions to the original 530 Code (MSJC Code) and the original 530.1 Specifications (MSJC Specifications). The first revision of the original document passed public review of the sponsoring organizations in 1992. This document reflects those revisions. Based on these activities, it becameevident

to TMS,CMR,

and AC1 that amanual,

handbook, or guide type document that specifically addressed the application of the MSJC Code and Specifications with illustrative examples would be a tremendous benefit to the industry. An agreement was made among TMS, CMR, and AC1 to address this issue. The Masonry Society would writethe document, CMR would provide financial resources during its development, and AC1 would review and publish the product. The result of these efforts Guide. is the Masonrv Designers’

The Masonrv Desipners’ Guide (MDG) is composed of four major parts divided into 16 chapters. Part I, General, is administrativeandapplies

to all other parts.Background

information on development of MDG including author and reviewer contributions is given in the Foreword andChapter 1 on Introduction. The Code Reference Index and the Specifications Reference Index tie discussions and design example problem

procedures to

the appropriate MSJC Code/Specifications sections. Chapter 2 on Notations, Definitions, and Abbreviations presents the MSJC Code notations and definitions with modifications and abbreviations found in

the MDG. Where appropriate, notations are defined within the

MDG text. Part II, Materials and Testing, primarily addresses the Specifications provisions as related to materials and testing. The Code dictates compliance with the Specifications. Chapter 3 on Materials examinesthe provisions for clay or shale masonry units, concrete masonry units, stone masonry units, mortar, grout, masonry assemblages, reinforcement

and connectors.

Chapter 4 on Testing addresses Specificationsrequirements on testing frequencyand quality 1-3


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assurance provisions. The material provisions during preconstruction and construction are addressed along with assemblage testing. Part III, Construction, addresses quality assurance, quality control, and hot and cold weather construction. Qualityassurance includes the administrative policiesand requirements related to quality control measures that will provide the owner’squalityobjectives. addresses the items comprising quality assurance including

Chapter 5

organizational responsibilities,

materials control, inspection, testing and evaluating, noncomplying conditions, and records. Quality control is the systematic performance of construction, testing, and inspection. It consists of the operationsof the contractor at theconstruction site to obtain compliance with the contract documents. Chapter 6 addresses quality controlby examining the Specifications

provisions for masonryconstruction

preparation, suchas:

materials, placement of materials,includingunits,

storage and protection of

mortar, grout, reinforcement and

connector; and tolerances. The MDG Chapters 5 and 6 deal with masonry construction, in particular in light of quality assuranceand control provisionsas related to the Specifications. The Specifications contain some requirements that arealways mandatory and others that are optional. The latter become mandatory when required by the specifier. A compilation of these requirements in the form of a checklist is given in MDG Table 6.5.1. The extent of the quality assurance and quality control program

willvary with the sizeof the project.

Suggested applications of the Specifications QA/QC provisions to three typical types of masonry buildings (TMS Shopping Center, DPC Gymnasium, RCJ Hotel) are presentedin MDG Table 6.5.2. Chapter 7 addresses hot and cold weather construction. Part IV, Design, basically covers the application of the Code provisions to the structural design of different types of masonry assemblages (beams,

walls,columns, pilasters) for

different types of construction(multiwythecompositeandnoncomposite,singlewythe, unreinforced and reinforced) based on the structural analysis of three typicaltypes of masonry buildings (TMS Shopping Center, DPC Gymnasium, RCJ Hotel) for various load conditions. These are the same buildings for whichQualityAssurance/Quality recommendations are suggested in Chapter 6.

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Control

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Chapter 8 on DesignPhilosophy

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and Methodologygives

strengths, loads, masonry construction,

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the background on material

and performance that have produced a masonry

philosophy of structural design. Loads and load combinations are covered. Significant time is spent covering the structural behavior of walls under different load conditions; boundary conditions and wall configurations. Basic beam behavior is examined along with the design equations. Basic axial column behavior

is presented and the effect of the combination of

axial loads and bending is presented with Interaction Diagrams.

The design methodology

for each component type as found in the Code is discussed and referenced to appropriate Code sections in this chapter. With the basic design philosophy and methodology established, appropriate application of the conceptsfoundin

the Code is accomplished by conducting structural analyses of

structures and presenting design examples.These structural application aspects are covered in MDG Chapters 9 through 16. One of the unique features of the MDG is that the applications of the Code provisions are based on the same three typical masonry structuresa one-story strip shopping center, a one-story gymnasium, and a four-story hotel. Chapter 9 deals with structural analysis aspects

ofgravity and lateral load distributions. the three basic structures - TMS Shopping

These are evaluated first in global terms for

Center, the DPC Gymnasium, and the RCJ Hotel. Next the evaluation of the global loads into loadson orwithin individual components is considered. Example problems with respect

to globalgravityand

lateral load distribution,alongwithinterwall

and intrawallload

distribution, are presented at the end of the chapter. Chapter 10 on Movements covers causes and consequences masonry construction.

of movements as related to

Methods for determination of the magnitude of specifictypes of

movements are presented. The chapter deals with ways of accommodating the calculated movement in masonry construction. Example problemson determining size and location of control joints and expansion joints are given.

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Based on the information generated in Chapters 9 (Loads) and 10 (Movements), design of individualwalls,lintels,columns,

and pilasterscan be addressed inlight of the design

methodology in the Code. Chapter 11on Flexure addresses the structural design aspectsof elements where flexure may occur

-

walls,pilasters,

and beams.Bothunreinforced

considered. Design example problems

and reinforced elements are

detail the design procedure for elements such as

multiwythe composite and noncomposite walls, unreinforced and reinforced retaining walls, unreinforced and reinforced pilasters, lintels,and simple and continuous reinforced masonry beams. Chapter 12 expands Chapter 11 into the flexural and axial load structural design aspects of columns, walls,and pilasters. Columns are examined onlyas reinforced elements as required by the Code. Walls and pilasters are addressed for both the unreinforced and reinforced

state.Designexampleproblems

from the three masonry structures illustrate the Code

methodology. Chapter 13 on Shear presents the topic from the viewpoint of out-of-plane loads (Shear in Masonry Components) and in-planeloads(ShearWalls).Exampleproblems,some coordinated to previous problems consideredfor flexure only, show applicationof the Code shear provisions. Chapter 14 on Reinforcement and Connectors addresses strength requirements, corrosion resistance and protection provisions, embedment criteria,

and design aspects of

reinforcement and connectors.Numerousdesignexampleproblemsshow

not only

application of specific Code provisions for reinforcement and connectors but also typical design methodology for several typical connections. Chapter 15 on EmpiricalDesign

presents the background of the originalempirical

Hammurabi Code through the present empirical provisions found in the Code. The Code’s 1-6


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specific criteria on restrictions, strength requirements, support provisions, and minimum wall thickness are discussed. Aspects of bonding wythesand anchoring intersecting walls, roofs, and floor diagramsare presented. Design example problemsrelated specifically to thethree designated buildings show application of the empirical provisions of the Code. Chapter 16 presents Provisions For SeismicDesign as related to masonry construction. Seismic resistant design of masonry buildings requires provisions for ductility not generally required for wind or other lateral loads.This presents thosecriteria

chapter discusses these provisions and

of Code Appendix A that includesminimum

requirements for

differentseismic zones intended to provide proper performance of masonry structures subjected to earthquake shaking.

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2 NOTATIONS, DEFINITIONS, AND ABBREVIATIONS

2.1 NOTATIONS

A, = AA = Ast

=

A', = A, = A, =

db = nominal diameter of reinforcement, in. d, = actualdepth ofmasonry in direction of

cross-sectionalarea of element cross-sectionalarea of an anchor bolt, in2 net cross-sectional area of masonry, in.2 projected area, on themasonry surface, of a right circular cone for anchor bolt allowable shear andtension calculations, in.2 area of tension reinforcement, in.2 area of tension reinforcement for balanced condition, in.2 A, + A'*,in.? area of compression reinforcement, i n . 2 cross-sectionalarea of shear reinforcement, in.? pilaster cross-sectionalarea without flange,

shear considered, in. D = dead load or related internal moments and forces E = strain, in./in. E,,, = compressive strain in masonry, inJin. es = tensile strain in reinforcement, in&. e = eccentricity of axial load, in. E = load effects of earthquake, or relatedinternal moments and forces E, = modulus of elasticity of grout, psi E,,, = modulus ofelasticityofmasonry in compression, psi E, = modulus of elasticity of steel, psi E, = modulus of rigidity (shear modulus) of masonry, psi f = calculated stress, psi fa = calculated compressive stress inmasonry due to axial load only, psi fob = combined axial and flexure masonrycompressive stress, psi fb = calculated compressive stress in masonry due to flexure only, psi fb, = calculated bearing pressure, psi fb = clay brick compressive strength, psi fbt = calculated tensile stress in masonry due to flexure only, psi PC = specified compressive strength of concrete, psi = CMU net area compressive strength, psi fg = compressive strength of groutdetermined in accordance with AC1 530.1/ASCE 6TMS 602 Section 1.6.2.l(c) or 1.6.2.2(c), psi fm = masonrycompressive strength, psi f m = Specified compressive strength of masonry,

in.2

AI AI A2 b b,

= wall influence area per ASCE 7-88, ft2 = bearing area,

in?

= effective bearing area, in.2 = width of section, in. = total applied design axial force on an an-

chor bolt, lb the transformed section at the plane of interest, in. total applied design shear force on an anchor bolt, lb width of wall beam, in. allowable axial force on ananchor bolt, lb allowable shear force on an anchorbolt, lb distance from neutral axis to extreme fiber in bending, in. compression force, lb numeric coefficient in seismic load calculations, ASCE 7-88 distance from extreme compressionfiber to centroid of tension reinforcement, in. distance from extreme compression fiber to centroid of compression reinforcement

b, = widthof b, = b, = B, = B, = c

=

c = c = d

=

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psi fi

= modulus of rupture, psi

I

=

fs

= calculated tensile or compressive stress in

1,

=

f, = fsa = f, = fy = F, = Fb =

Fb, = Fbr = F, = F, = =

g

h

=

h'

=

H = Z = Z = Z, =

I, Z,

= =

Z, = j

k

=

=

kc = k, = ki = km =

k, =

K =

reinforcement, psi steel compressive stress, psi brickmasonry ultimate shear strength, psi calculated shear stress in masonry,psi specifiedyield stress of steel forreinforcement and anchors, psi allowablecompressive stress due to axial load only, psi allowable compressive stress due to flexure only, psi allowable bearing pressure, psi allowable tensile stress due to flexure only, psi allowable tensile or compressive stress in reinforcement, psi allowable shear stress in masonry,psi ratio of distance between tension steel and compression steel to the overallcolumn depth effectiveheight of column, wall or pilaster, in. height of column, wall, or pilaster, in. lateral pressure of soil or related internal moments and forces moment of inertia of masonry,in.4 importance factor,ASCE 7-88 moment of inertia of cracked transformed section, in.4 effective moment of inertia, in.4 gross section moment of inertia, neglecting reinforcement, in.4 moment of inertia of the transformed area about the neutralaxis, in.4 ratio of distance between centroid offlexural compressive forces and centroid of tensile forces to depth, d ratio of thedistance between theneutral axis and the extreme fiber in compression to the depth, d coefficient of creep of masonry, per psi coefficient of irreversible moisture expansion of clay masonry element stiffness, in." coefficient of shrinkage of concrete masonry coefficient of thermal expansion of masonry per degree fahrenheit horizontal force factor in seismic load

lk =

=

1,

=

1,

= =

L

LB = L, =

M = M. =

Mb = Mc*= Mm= M,= M" = M, =

M, = M? = n =

N, = P = P, = Pl = P = p' = Pt

=

Q =

r

=

R = R = Ri = S

2-2


1,

=

m

calculations, ASCE 7-88 clear span between supports, in. effective embedment length of plate, headed or bentanchor bolts, in. anchor bolt edge distance measured from the surfaceof an anchor bolt the to nearest free edge of masonry, in. embedment length of straight reinforcement, in. equivalent embedmentlength provided by standard hooks, in. horizontal length of wall, in. live load or related internal moments and forces bearing width, in. length of bearing plate, in. maximum moment occurringsimultaneously with design shear forœ V at the section under consideration,in.-lb maximum moment in member at stage deflection is computed moment at balanced condition without compression steel, in.-lb cracking moment, in.-lb moment as limited by allowable bending compression stress in masonry, in.-lb midspan bending moment of members member nominal moment strength moment of applied load with respect to the centroid of internal compressive forœ moment as limited by allowable tension stress in reinforcement, in.-lb moment due to compression steel, in.-lb modular ratio of elasticity force acting normal to shear surface,lb design axial load, lb Euler buckling load, lb lateral load, lb ratio of tension reinforcement = AJbd ratio of compression reinforcement = A 'Jbd AJbt fnst moment about the neutral axis of a section of that portion of the cross section lying between the plane under consideration and extremefiber, in.3 radius of gyration, in. radius of curvature, in. slenderness reduction factor relative rigidity spacing of reinforcement, in.

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S,

S = S = T = T =

r;,

=

v

=

v =

W = W = W

=

Y

=

y = YI

=

z = a,,, = a+,,,= a, = 0: =

A = A =

2.2

which a masonry unit is laid. Budding cy).icial - The officer or other designated authority chargedwith the administration and enforcement of this code, or his duly authorized representative. CoZZur joint - Vertical longitudinal joint between wythes of masonry or between masonIy wythe and back up constructionwhich is permitted to be filled with mortar or grout. Column - An isolated vertical member whose horizontal dimension measured at right angles to the thickness does not exceed 3 times its thickness and whose height is at least 3 times its thickness. Composite action - Transfer of stress between so that in componentsofamemberdesigned resisting loads, the combined componentsact together as a single member. Composite masonry - Multicomponent masonry members acting with composite action. Compressive snength of masonry, fm - Maximum compressive force resisted per unit of net crosssectional area ofmasonry, determined by the testing of masonry prismsor a function of individual masonry units, mortar and grout in accordance with the provisions of AC1 530.1/ASCE 6EMS 602. Connector - A mechanical device for securing two or more pieces, parts, or members together, including anchors, wall ties, and fasteners. Diaphragm - A roof or floor system designed to transmit lateral forces to shear walls or other vertical resisting elements. Eflective height- Clear height of a braced member between lateral supports and used for calculating the slenderness ratio of a member.Effective height for unbraced members shall be calculated. Head joint - Vertical mortar joint placed between masonryunitswithin the wythe at the time the masonry units are laid. Heder (Bonder) - A masonry unit that connects two or more adjacentwythes of masonry. Load, dead - Dead weight supported by a member, as defined by the general building code. Load, live - Live load specified by the general building code. Modulus of eZusr¿c¿ry - Ratio of normal stress to corresponding strain for tensile or compressive stresses below proportional limit of material. Modulus of r¿&¿t -yRatio of unit shear stress to unit shear stress for unit shear strain below the proportional limit of the material. Project Dawings - The drawings which accompany

= total linear drying shrinkage of concrete

masonry units determined in accordance with ASTM C 426 section modulus, in? soil factor, ASCE 7-88 tension force, lb fundamental elastic period of vibration of the building or structure, ASCE 7-88 nominal thickness of wall, or overall depth of member cross-section, in. thickness of pilaster, in. shear stress given in Code 6.5.2(c), psi design shear force, lb wind load or related internalmoments and forces total dead load, ASCE 7-88 uniform loading, plf distance from neutral axis to a fiber in cross-section, in. centroid of compression forces, in. distanœ from centroidal axis of gross section, neglecting reinforcement, to extreme fiber in tension seismic coefficient, ASCE 7-88 masonry compressive stress, psi stress in masonry at location of compression reinforcement, psi steel tension stress, psi steel compression stress, psi change in length or deflection, in. distance from the axial load to the centroid of the tension steel divided by d long term deflection factor

DEFINITIONS

Anchor - Metal rod, wire or strap that secures masonry to its structural support. Architect/Engineer - The architect, engineer, architectural firm, engineering firm, or architectural and engineering fiim, issuing project drawings and specifications, or administering the work under contract specifications and drawings, or both. Area, gross cross-sectional - The area delineated by the out-to-Out dimensions of masonry in the plane under consideration. Area, net cross-sectional - The area of masonry units, grout and mortarcrossed by the plane under consideration based on out-toautdimensions. Bed joint - The horizontal layer of mortar on

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Code or MSJC Code or 530 Code or ACIIASCE - Building Code Requirements for Masonry Structures (AC1 5301ASCE 5/TMS 402) Code C - Commentarv on Building Code Reauirements for Masonry Structures (AC1 530/ASCE 5 /TMS 402) CMR - Council For Masonry Research CM - center of mass CMU - Concrete Masonry Unit CR - center of rigidity IRA - Initial Rate of Absorption LLRF - Live Load Reduction Factor MDG - Masonrv Designers’ Guide MOR - Flexural Modulus of Rupture MSJC - Masonry Standards JointCommittee N.A. - Neutral &is NCMA - National Concrete Masonry Association PCA - Portland Cement Association PCL - Portland CementIHydrated Lime pg - pounds per squarefoot plf - pounds per linear foot QA - Quality Assurance QC - lluality Control SCF - Slenderness Correction Factor Specs. or 530.1 Specifications - SDecifications for Masonrv Structures (AC1 530.1lASCE 6/TMS602) Specs. C - Commentary on SDecifications for Masonry Structures (AC1 530.1/ASCE6DMS 602) STP r Special Technical Publication TMS - The Masonry Society UBC - Uniform Building Code

Project Specifications for the construction of the work and complete the descriptive information for construction work required or referred to in the Project Specifications. Running bond The placement of masonry units such thathead joints in successive mursesare horizontally offset at least onequarter the unit length. Specified compressive strength of masonry, f Minimum compressivestrength expressed as force per unit of net cross-sectionalarea required of the masonry used in construction by the project documents, and upon which the project design is based. Whenever the quantity f I,,, is under the radical sign, the square root of numerical value only is intended and the result has units of pounds per square inch. Stack bond - For the purpose of this code stack the bond is otherthan runningbond.Usually placement of units is such that the head joints in successive murses arevertically aligned. Stonemasonry - Masonrycomposed of field, quarried, or cast stone units bonded by mortar. Stone masonry, ashlar - Stone masonry composed of rectangular units havingsawed, dressed, or squared bed surfaces and bonded by mortar. Stone masonry, rubble - Stone masonry composed of irregular shaped units bonded by mortar. Tie, lateral - Loop of reinforcing baror wire enclosing longitudinal reinforcement. Tie, wall - Metal connector whichconnects wythes of masonry walls together. Wall - A vertical element with a horizontal length at least 3 times its thickness, used to enclose space. Wall, load bearing - Wall carrying vertical loads greater than 200 lblft in addition to its own weight. Wythe - Each continuous,vertical section of a wall, one masonry unit in thickness.

530

-

I,,,

2 3 ABBREVIATIONS ACI - American Concrete Institute AIE - ArchitecEngineer ANSI - American National Standards Institute ASCE - American Society of Civil Engineers ASCE 7 - AmericanSociety of Civil Engineers Minimum Loadsfor Buildings and OtherStructures ASTM - American Society of Testing and Materials B U - Brick Institute of America

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3 MATERIALS

3.0 INTRODUCTION Specification criteriafor structures designedunder the Code are written in conformance with AC1 530.1/ASCE 6TMS 602. The Specifications encompass provisions for commonly used masonry materials and the Specifications integrate provisions for construction

and quality

assurance common to these materials. By direct reference in Code 3.1.1, the Specifications become a part of the Code and have the same force of law as the Code when the Code is adopted by a local governing body. Code 3.1.1 states: "Composition,quality,storage,handling, materials, quality assurance for materials

preparation andplacement

of

and masonry and construction

of

masonry shall comply with AC1 530.1/ASCE6TMS 602." ThisMSJC

Code requirement dictates as aminimumcompliancewith

the MSJC

Specifications. The variety and availability of materials producedby the masonry industry provides designers with extensive options to satisfy structural, aesthetic, fire resistance, andother requirements

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for construction. 3.0.1 GeneralIntent

The general intent of the material provisions of the Code and Specifications is to ensure that products of acceptable and defined quality are used throughout the masonry construction. 3.0.2

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Specifications PrefaceandChecklists

To assist all members of the design and construction team, the Preface (Specs. P3) and the Specification Checklist (Specs.P5) identify those areas of the Specifications where decisions regardingadministration

of thejob,materials,andsubmittalshave

to be made. The

Specification Checklist consists of three parts:

* * *

Mandatory Checklist Optional Checklist Submittals

3.03 MandatorySpecificationChecklist

Mandatory items required by the A/E are significant since the A/E designates the desired level of quality and performance of the masonry.

A typical example of the Mandatory Specification Checklist in Section 2--Masonry,2.2.1 and 2.2.2, Masonry units and mortar--alerts the A/E to: "Specify the masonry units and mortar to be used for the variousparts of the project and the type of mortar to be used with each type of masonry unit."

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Optional SpecificationChecklist

The Optional Specification Checklist lists various items that should be considered by the A/E while preparing the ContractDocuments.

The decision to incorporate specificitems

depends on the level of quality assurance required or, where project

requirements may

necessitate, more specific information, suchas: bond pattern, mortar bedding, cold weather constructionprocedures,etc.When

no decisions are madewhilerequiringcertain

specifications, the provisions revert to default requirements. Caution and care are required to ensure the quality of the end product. Each project is different and decisionsfrom previous projects should not be applied indiscriminately to new projects. 3.0.5

Submittals

The Submittals portion of the Specification Checklist addresses Specs. 2.1.2, Submittals. When required by the Contract Documents, various submittals are to be made. These are optional requirements of the Specifications. Good industry practice requires submittals.

These submittals, after approval, become the

reference for quality control acceptance/rejection of materials and construction practices. Submittals should be maintained throughout the entire construction period and be handled in accordance with the Quality Assurance program. Masonry units should be submitted whenever aesthetics are a primary consideration. Units submitted should reflectthe full range of colors, textures,and finishes. Additionally, mortar samples shouldbe submitted when coloredmortars are to be used. These submittals should reflect the acceptable ranges. When required, mortar material proportions resulting from mortar test results obtained in accordancewith the property specificationrequirements 3-3


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should be submitted. 3.0.6

MaterialSpecificationReferences

Materials standards are referencedwithinSpecs.

1.3. It is important that the specific

standard be used for acceptance or rejection of a given material, basing such acceptanceor the standard. rejection onconformance/nonconformance with requirements contained within Product standards may contain provisions for Manufacturer’s Certification of Compliance of the product. Such certification by the manufacturer provides an alternate to extensive testing before product acceptance. 3.0.7 MaterialSpecificationRequirements (ASTM)

ASTM material specifications contain provisions for acceptance or rejection of a material. The provisions are either chemical or physical or both depending on the specific product. Product classifications are used to further identify the desired product from within a range of similar products with differing performance characteristics. Acceptance without concern as to product sub-classificationsresultsinconfusionunless

the materialsspecification

contains default. a Intheabsence

the defaultbecomes

requirement. For example,ASTM

of specificdirection

C 270 contains both aproportion

the

and a property

specification for mortars for unit masonry. The A/E may select either mortar specification but not both from within the standard. If the A/E fails to indicate the mortar specification that will apply, the standard dictates that the proportion specification shall apply. Material specifications are identified by designation number and the year given in Specs.

1.3. If a manufacturer supplies materialunder a standard with a different year, the designer needs to examine the standard for variations from the required Code version.

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3.1 UNITS

The A/E should select the desired unit based on compliance with

product specification,

including requirements such as aesthetics, strength, durability, availability, and such other attributes as may be important. Masonry units covered within the Specifications includeclay or shale masonry units,concrete masonryunits or stone. Providing the design criteria are met, individualunits selected during design may be of varied composition and possess differing chemical and physical properties. In addition to the design and serviceability criteria, the longevity and durability of the unit in service and under expected exposure conditions must be considered. 3.1.1 Product Specifications for Clay or Shale Masonry Units

Clay or shale masonry units

are available with varied coring

and performance

characteristics. These units are covered by several product specifications per Specs. 2.2.1.2:

34

Structural Clay Load-Bearing Wall Tile

ASTM C 56

Structural Clay Non-Load-Bearing Tile

ASTM C 62

BuildingBrick(Solid

ASTM C 126

Ceramic Glazed Structural ClayFacing Tile, FacingBrickand

ASTM C

Units fromClay or Shale)

Solid Masonry Units ASTM C 212

Structural Clay Facing

ASTMC216FacingBrick(Solid

Tile

Units from Clay or Shale)

ASTMC652HollowBrick(Hollow

Units fromClay or Shale)

As indicated by the title of the individual specifications, the specifications address clay or shale products consisting of brick, both solid and hollow, and tile, both loadbearing and nonloadbearing. Ceramic glazed units are available as facing brick and tile and as solid units.

3-5



93

Ob62949 0508536 3bL

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These specifications for the units containrequirements addressing durabilityand appearance when units are intended for use as facing components as well as structural components of the masonry. For examplemostbrickused requirements of ASTM C 216

for facing will be required to meet the

or ASTM C 652. The only difference between these two

specifications is the allowable percentage of void area. Under these specifications the Grade classification addresses durability and the Type classification addresses appearance. Color, texture, and size are not covered by the specification and must be specified by the purchaser. Brick strength requirements in ASTM standards are for durability consideration. Brick produced inthe US. typically have compressive strengthsthat substantially exceedthe ASTM minimum values. The average compressive strength of brick in the marketplace is in the order of 12,000 psi. Strengths needed above the required ASTM minimums mustbe specified by the designer. Acceptance/rejection of the product is based on conformance to the requirements of the specification. The requirements for Clay or Shale Masonry units are listed in MDG Table 3.1.1. 3.1.2 ProductSpecificationsforConcrete Masonry Units

Concrete masonry units

are available with varied compositions

and performance

characteristics. These units are covered by several product specificationsper Specs. 2.2.1.1:

ASTM C 55

Concrete Building Brick

ASTM C 73 Calcium Silicate

Face Brick

ASTM C 90 Load-Bearing

Concrete Masonry Units

ASTM C 129 Non-Load-Bearing ASTMC744Prefaced

Concrete Masonry Units Concrete and CalciumSilicateMasonryUnits

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Common to each of the product specifications are classifications, i.e., Types, Grades, and Physical Requirements. Additionally, the optional combination of materials, i.e., cement and normal weight or lightweight aggregate, lime and aggregate, are covered within the material specification. The limits on certain properties of the concrete masonry units vary so the individual specification should be consulted. Acceptance/rejection of the product is based on conformance to the requirements of the specification. The classification and physical requirements for concrete masonry units are listed in MDG Table 3.1.2. ASTM standards include provisions for limiting the moisture content of block, depending upon their shrinkage properties and the environmental moisture conditions (averageannual relativehumidity)

at the job site. These provisions are structured so that similar

performance (residual shrinkagepotential)can

be expectedregardless

of the inherent

shrinkage properties of the units. Units with higher shrinkage potentialare required to have lower moisture content than units with low shrinkage potential. Maximum linear shrinkage is limited to 0.065% for these units, which are classified as Type I, moisture controlled units.

Type II units are classified as non-moisturecontrolled.

These provisions for moisture

content were formulated to eliminate the need for specifying shrinkage limits. Concrete masonry units are typically manufactured to the minimum compressive strength as listed in the ASTM standards. Where design requires higher strength units,

the higher

strength must be specified. 3.13 ProductSpecifications for StoneMasonryUnits

ASTMspecificationscover

an array ofbuilding

stoneswithcompositionsvaryingfrom

marble to slate. These are covered by several specifications per Specs. 2.2.1.3: ASTM C 503 Marble Building

Stone (Exterior)

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m

0662947 0508538 L34

ASTM C 568

Limestone Building Stone

ASTM C 615

Granite Building Stone

ASTM C 616

Sandstone Building Stone

ASTM C 629

Slate Building Stone

m

These specifications classify the building stone by composition, density, and application. The specification requires conformance to physical requirements including one or more of the following physical properties: absorption, density, compressive strength, tensile strength (modulus of rupture and flexural strength), abrasion and acid resistance. Acceptance/rejection of the product is based on conformance to the requirements of the specification. The classifications and physical requirements for stone masonry unitsare listed in MDG Table 3.1.3. 3.1.4 ProductTestingandConformance

Individual product standards reference the applicable test methods used physical properties identifiedasrequirements.

Test methods of special interest to the

designer and the properties covered in each include:

Clay and Shale Masonry Units A S N C 67SamplingandTestingBrickandStructuralClayTile Modulus of Rupture Compressive Strength Absorption Saturation Coefficient Freezing & Thawing Efflorescence 3-8


to measure the

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Initial Rate of Absorption Weight Size Warpage Length Change Void Area Out of Square

Concrete Masonry Units ASTM C 140

Sampling and Testing Concrete Masonry Units Compressive Strength Absorption Weight Moisture Content Dimensions

ASTM C 426

Drying Shrinkage of Concrete Block Drying Shrinkage

Stone ASTM C 97

Absorption and Bulk Specific Gravity of Natural Building Stone Absorption Bulk Specific Gravity

ASTM C 99

Modulus of Rupture of Natural Building Stone Modulus of Rupture

ASTM C 120

Flexural Testing of Slate Modulus of Rupture Modulus of Elasticity

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Water Absorption of Slate Water Absorption

ASTM C 170

CompressiveStrength

of NaturalBuilding

Stone

Compressive Strength The test procedures are delineated for each of the individual physicalproperties of interest, pertaining to a specific product.

Test methods provide no acceptance/rejection criteria.

Acceptance/rejection criteria are contained in the product specification.

3.1.5ProductReceiptandStorage With acceptance of the product, handling and storage quality of the product. Basically, the units should

of the units on site

will affect the

be received and stored to prevent the

degradation of desired properties. See MDG 6.1.1. Unprotected units and stone can be degraded both by allowing water to contact the units, thereby altering their moisture content, and

by allowingground or soil to contaminate

surfaces, thereby altering the bonding characteristics and/or appearance of the units. The degree of protection to be provided a product from the time of its manufacture until the time of its use should be mutually agreed

to by the manufacturer and the purchaser.

Plasticwrapping that envelopspalletizedproductsprovide

an easysystemforproduct

protection.

3.1.6Manufacturers'Recommendations Productmanufacturers

are inkeypositions

to learn of successfulapplications

procedures.Unsuccessfulapplicationsandprocedures

and

are also wellknown to them. In

specifymg product applications and procedures, manufacturers' recommendations usually received in a product description, technicalsheet or reflected on a tag attached to palletized 3-10


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units, should be considered together with past performances of the products and personal experiences of the designer.

3.2 MORTARS Mortar is the bonding agent that holds the individual units and connectors together to act as a complete assembly.Mostmasonry

mortars are produced at the constructionsite,

though pre-batched mortars are available in certain geographical areas. The Specifications address mortars prepared at the constructionsite.

By exception to Specs. 2.2.2 and to

ASTM C 270, the Contract Documents may permit the use of ready-mixed mortars or pre-batched mortar mixes. Dry mortar ingredientsmayalso

be delivered to the job in

pre-batched silos whichare capable of producing highly controlled, on-siteautomatic mixing.

3.2.1

Selection of Mortar Type

The selection of the proper mortar coupled with a specified masonry unit to attain the desiredmasonry strength and other performance characteristicsshould be basedona knowledge of the units, the various mortar types available, and the compatibility of the two components. No single mortar type or composition of mortar type and unit type is considered appropriate for allapplications.Allowableflexuraltension

as afunction of

mortar type and mortar materials is given in Code Table 6.3.1.1. The compressive strength of clay masonry and concrete masonry as a function of mortar type is given in Specs.Tables

1.6.2.1 and 1.6.2.2, respectively. Specs. 2.2.2.1 requires the use of ASTM Specification C 270 on Mortars for UnitMasonry. The mortar specification consistsof a proportion specification and a property specification. When neither the proportion or property specifications are specified, the proportion specification is the governing requirement. Bothspecificationsrecognize

four types of

mortar. Cementitious materials that can be combined in the production of the mortar will involve portland cement, lime, and masonry cement. Ten individual mortar combinations

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are listed. To ease selection of the mortar type, ASTM C 270 contains an extensive appendix that lists and discusses many considerations. The A/E should recognize that the provisions of the proportion specifications of ASTM C 270 are based on performance of the indicated combinations of cementitious materials and aggregate ratios. The combinations have been time tested. The A/E should weigh the merits of the various mortar compositions and types with the materials available within the project area. The following considerations are considered to be good practice: 1.

Use the mortar type that provides the desiredphysicalcharacteristics; do not use a higher compressive strength mortar type than necessary.

2.

Unlessdictated by compressive strength requirement, avoid the selection of

Type M mortar over Type S mortar. However, when masonry contact withsoil,e.g.inretainingwalls,

isin direct

Type M mortar should be used

regardless of strength requirements. Mortar shall be specified either by proportion or by property, but not by both. 3.2.2 ASTM C 270 Mortar Qpes 3.2.2.1 Proportion Specification - Under the proportion specification within ASTMC 270,

four mortar types are recognized. Composition of the mortar is based on the selection of portland cement in combination with hydrated lime, portland cement in combination with masonry cement, or masonry cement alone. The proportions of cementitious materials(portland cement, limeand masonry cement) and the ratio of aggregate to the sum of the cementitious materials are prescribed.

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The proportion specificationgivesmorecontrol

5TL

m

to the specifier. It is not equivalent in

performance to the property specification, yielding higher compressive strength mortars than their property specification counterparts. 3.2.2.2 Property Specification - Under the optional property specification within ASTMC

270, combinations of cementitious materialsand aggregates are laboratory tested to establish the proportions of ingredients which yield a mortar which complies with the specifications. The property specification allows any combinationof permitted cementitious materials with 2-1/4 to 3-1/2 volume parts of damp, loose aggregate per volume part of cementitious materials. Additionally, water retention testing is mandatory. In-situ mortar compressive strength is not covered in ASTM C 270. 3.23

CementitiousMaterials

Cementitious materials used in the production of masonry mortars include portland cements, blended cements,slagcements,masonry

cements and hydratedlime.

Each of these

materials is covered by a material specification. Of the variouscementitiousmaterials, combinations of portland cement with lime, combinationsof portland cement with masonry cement, and masonry cements are the most common in masonry construction. Slagcements are allowed when used under the property specification of ASTM C 270. Individual materials possess certain desired characteristics that influence masonry mortar workability. Masonry cements are now available as Types M, S and N. These cements are special formulations addressing the needs of the masonry industry. In some geographical areas, mixtures of portland cement and hydrated lime identified variouslyas being suitable for the production of ASTM C 270 mortars, TypeM, S and N are available. These preblended cementitious materials, thoughnot covered by an existing product specification, provide the one-bag advantages common to masonry cements.

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For masonry structural design considerations involving flexural tension, the design allowable per Code Table 6.3.1.1 is influencedby the type of cementitious material used inthe mortar. Code 6.3.1.1 assigns lower allowable flexural tension values

for masonry that consists of

masonry cement or air entrained portland cement and lime mortar as compared to non-air entrained portland cement and lime mortars. Type N mortar and masonry cement are not allowed to be used in any part of the structural system in Seismic Zones 3 and 4. 3.2.4 Aggregates

Aggregates for masonry mortar are covered by ASTM C 144. This specification recognizes the need for certain chemical and physical characteristics of fine aggregate used in production of masonry mortars.

the

The specification recognizes natural and manufactured

sand, while requiring conformanceto grading, composition controls on deleterious materials and organic impurities, and soundness. The mostdesiredmasonrysand

is one that impartsgoodworkability

and strength

characteristics to the masonry mortar. Fine, round particles are desired, along with uniform gradation to allow support of masonryunits after placement.Masonrysandsgenerally possessfinegradationswhichwhenusedin

mortar facilitateshandling and placement.

However, with fine gradations, the ratio of paste to aggregate of mortar composed of such aggregates is low, resulting in

greater extensibility of the mortar and lower compressive

strengths. Acceptance of the sand, even if it does not comply with the grading requirements, should be based on conformance to the specification using the waiver clause of ASTM C 144 that requires testing in accordance with the property specification of ASTM C 270. 3.2.5 Mortars ContainingHydratedLime

Masonry mortars prepared using a combination of portland cement and hydrated lime have

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TITLErMDG 93 m 0662949 0508545 374 D

performance characteristics dependent on the amount of the hydrated lime added to the mixture. Low lime content mortars possess higher compressive strength; high lime content mortars possess more desirable workability characteristics. These properties are influenced by the composition of the hydratedlime.Normally,dolomiticlimesyield

more desired

workability characteristics than their high calcium counterparts. Portland cement-lime mortars generally require high water contents and have higher water retention characteristics,consideredbeneficial absorption and cement hydration.

for satisfymgwater

demand for unit

The paste resulting from the combination of cement,

hydrated lime and water promotes the intimate contact of mortar with unit, thus enhancing bond between the two components. Hydrated lime

hardens upon contact with air.

Thus

complete hardening of PCL mortar takes place over a long time period. This characteristic aids in recementing small hairline shrinkage cracks. During cold weather masonry construction, high water

contents of portland cement-lime

mortars makes them more susceptible to early freezing. For above normal temperatures, dry conditions or while using highly absorptive masonry units,portland cement-lime mortars perform better as their lime content increases. Unhydrated oxidesin

mortar will hydrateovertime,causingmasonryexpansion

and

consequent cracking. ASTMC 207 gives the limits on the unhydrated oxide content of Type

S , special hydrated lime, and Type N, normal hydrated lime. ASTM C 270 permits the use of ASTM C 207 Types S or SA lime. Type N or NA limes may be permitted provided tests

or performance records are acceptable. 3.2.6

Mortars Containing Masonry Cement

Manufacturing masonrycement involves blendingor intergrinding portland cement with filler materialssuch

as groundlimestone.Air-entrainingadditives

are added to improve

freeze-thaw durability, workabilityand water retention characteristics. The portland cement 3-15


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0662947 050854b 200

fraction of the masonry cement is a finer grind and is more activethan mixtures containing the same ingredients but produced under different processes. ASTM Specification C 91 recognizes three types of masonry cement, i.e., types M,

S, and

N. All of these masonry cements promote ease of mortar preparation at the project. The single bag concept tends to reduce on-site variables during mortar preparation. Masonry cement mortars, becauseof their intentionallyentrained air, require less water than portland cement-lime mortars. -The lower water content is beneficial during cold weather masonry construction. However the lower water content becomes a detriment when using very absorptive masonry units on hot,

low relative humidity days.

The information in the

Appendix of ASTM C 270 provides the designer guidance in determining the selection of mortar for a specific use.

3.2.7 Admixtures

As indicated in ASTM C 270, the use of admixtures is prohibited, unless specified by the A/E or the owner’s designated representative. This attitude is in keeping with the concepts

that admixtures should not be used indiscriminately, and

that proof of the suitability of

admixtures should be demonstrated by tests involving the materials under temperature and relativehumidityconditionswhichpresumablyrequiretheiruse.Admixturescontaining chlorides are disallowed by Specs. 2.2.2.3. When specified, mineral oxidesor carbon blackmay be used to impart color to the mortars. Specs. 2.2.2.2 delineates types and proportions of pigment in masonry mortars that may be used. Using excessive amounts of pigments may reduce compressive and bond strength of mortars.Providinglimits

in theSpecifications is an attempt to keep suchlosses to an

acceptable level. The Specificationsimply that only job sitepigments are considered.However,colored

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TITLE*MDG 9 3 W 0 6 6 2 9 4 90 5 0 8 5 4 7

cementitiousmaterials

are availablefromvariousmanufacturers.

147

If specified,colored

mortar must comply withthe property specification ASTMC 270 no matter what procedure is used for obtaining the colored mortar. Testing in accordance with the property specification of ASTM C 270 is recommended for any mortar containing an admixture.Acceptance

of themortar mixturecontaining

an

admixture shouldbe based on materials tested under temperature and relative humidity that will prevailduringuse.Considerationshouldbe

given to the effect of admixtures on

embedded materials, i.e., steel, aluminum, etc. 3 3 GROUT

Grout is a fluid cementitious mixture used either to bond adjacent masonry units, wythes, or tobond the steel reinforcement positioned in the collar joint between adjacent wythesor in cores of masonry units to the masonry. Grout material and proportionrequirements

are specified in ASTM

C 476. This

specification includes provisions for both fine and coarse grout. Selectionof the appropriate grout depends on thegrout space available inthe masonry; selectionof either fine or coarse grout is controlled by the Code 3.1.2. Fine grout is composed of one part of portland cement, O to 1/10 part of hydrated lime, and fine aggregate, as defined by product specification ASTM C 404. The aggregate to cement ratio is 2-1/4 to 3 parts per volume of cementitious material. Coarse grout is similar to fine grout but contains an additional 1 to 2 parts per volume of coarse aggregate as defined in ASTM C 404. Both fine and coarse grout should have a slump of 8 to 11 in. 3.4

MASONRY

The basic materials used in masonry assemblages include clay units, concrete units, mortar,

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TITLEtMDG 93 m Obb2949 0508548 O83 M

grout, and steel reinforcement. Masonry physical properties used by the designer in sizing an assemblage or evaluating performance include compressive strength, flexural strength, shear strength, and modulus of elasticity. Fire rating, accoustics, and unit size also must be considered. Masonry strength in compression, flexure, and shear is directly affected by many different factors. Its strength willvary

depending on the mortar type selected, by themortar

materials, by the units used, and by the workmanship. The unit size as well as the direction

of loading, parallel or perpendicular to the mortar bed joint, affects masonry strength. Specs. 1.5.1.3 requires that all masonry work be inspected. Bond strength results are affected by the initial rate of absorption, texture, and the cleanliness of the masonry units. Mortar water retentivity, flow, cement,and air content also affect the plastic properties and the bond relative to strength. 3.4.1 Brick Masonry inCompression

Specs. Table 1.6.2.1 permits the assumption that for Types M or S mortar f, (psi) = 0.25

fb

+ 400, and for Type N mortarf, (psi)

= 0.2fb

+ 400.

Typically, the mean compressive

strength of brick masonry is about 40% greater than the values assumed in Specs. Table

1.6.2.1. The mean expected compressive strength perpendicular to bed joints of standard modular brick masonryat 28 days, built with inspected workmanship and ASTM C 270 Type S mortar without air-entrainment, may be estimated (3.4.1) as:

f, (psi)

= 0.283

Cf'), + 8,380)

Eq. 3.4-1

That strength is reduced about 29% by use of Type N mortar (3.4.2), 27% by uninspected workmanship (3.4.2), and 10% to 20% by increasing mortar air content from 5% to 18%

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(3.4.3, 3.4.4, 3.4.5). Other sources indicate little or no effect of mortar air content on brick masonry compression strength (3.4.6, 3.4.7). Mortars without air entrainment typically have air contents of 5% or less. The specification for Type N masonry cement (ASTM C 91) permits air content of 22%, which may result in typical mortar air content of 24% (3.4.8,

3.4.9). Strength is increased about 6% by use of Type M mortar (3.4.2) and 22% by use of 3-5/8 in. high brick with

3/8 in. mortar joints rather than a 2-1/4 in. high brick

(3.4.10).

Other things being equal, compressive strengthof brick masonry is reduced a total of about

36% by use ofhigh

air-entrainment and uninspectedworkmanship

(3.4.1, 3.4.5). Brick

masonry in compression has a mean cracking strength of 47% of its ultimate strength (Jrn) with a coefficientof variation of 15% (3.4.11). Frequent application and withdrawal of load may cause fatigue and strength reduction, and may therefore increase cracking probability

(3.4.12). 3.4.2 BrickMasonry ElasticModulus

Masonry elastic modulus in compression is rather constant over the stress range from 5% to 33% of the masonry's ultimate strength. Over that stress range the mean chord modulus of elasticity of solid brick masonry with a prism aspect ratio of five is 464 fm (psi) with a

standard deviation of 185 fm (psi) (3.4.13). E, is reduced about 24% when stress is parallel to bed joints (3.4.14). E,,, and fm may be determined in accordance with Code 5.5.1.2 and Specs. 1.6.3. Code Table 5.5.1.2 states that with Type N mortar the elastic modulus of clay masonry can be taken as E,,,(psi x 106)= 0.20 [(f',/lOOO)

+ 21; with Type S mortar &(psi

[(f'b/1000)+ 1.61; and with Type M mortar E,,, (psi x 106) = 0.30 [(f',/~ooO)

x 106) = 0.25

+ 1.331.

3.43 Brick Masonry inFlexure

Flexural cracks in masonry form primarily at the unit-mortar interface. Resistance to such cracks depends on the tensile bond strength of the unit-mortar combination. Bond strength

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is an important physical property of masonry, because higher bond strength reduces cracking, leaking, staining, and spalling. Bond strength is a function

of:

l.

Initial rate of absorption,texture,andcleanliness

2.

Mortar waterretentivity,flow,

3.

Quality of workmanship.

In somecasestheallowablestress

of brick;

cement andaircontent;

and

inflexuraltensionwithnonair-entrainedportland

cement-lime mortars is 100% greater than for masonry cement or air-entrained portland cement-lime mortar, but averages about 56% higher. The higher bond strength achieved by the use of nonair-entrainedportlandcement-lime

mortar is indicated in Code Table

6.3.1.1. The 28-day flexural strength of brick masonry walls built with ASTM C 270, Type S , PCL, non air-entrained mortar, and inspected workmanship, with stress perpendicular joints, has a

mean value of 140 psi with a

accordancewithASTM

to bed

standard deviation of 31 psi, when tested in

E 72. Wallswith Type M mortar are about 10% stronger in

flexure. With Type N mortar, strength is reduced about 23% (3.4.2). Increasing air content reduces flexural bond strength

(3.4.15, 3.4.16). Uninspected workmanship reduces mean

strength by 23% (3.4.2). Other things being equal, the use of Type N rather than Type S mortar, of air content of 18%,and of uninspected workmanship, typically collectively reduce flexural strength about 60%. For 4 in. wythes of standard modular brick, flexural strength is about 3.7 times greater when stress is parallel rather than perpendicular to bed joints. The Code allowable stresses are based on inspected workmanship.

The Code makes no

provisions for allowable stresses for uninspected construction. The first crackin brick masonry in flexure occursat about 80% of ultimate strength. Out-ofplane cracking occursat a deflectionof about 0.05% of wall span (Z/2,000),with a coefficient

of variation of 26% (3.4.2). 3-20


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3.4.4 Brick Masonry inShear

(shear modulus) E, is about 40% of E,.

The modulus ofrigidity

In the absence of

compressive stress the ultimate shear strength of brick masonry can be approximated as follows with a coefficient of variation of about 30%: fsb

(psi)

=

40 esp

(A] 1,780

Eq. 3.4-2

ConcreteMasonryinCompression

3.4.5

Compressive strengthof concrete masonryis a functionof unit compressive strength,mortar type, stress direction, and workmanship. For ASTM C 270 Type M or S non air-entrained

mortars, stress perpendicular to bed joints, and inspected workmanship, the mean 28-day compressive strength on the net area of a standard prism of concrete masonry

(Jm)

has an

estimated mean value of:

-

fm

(psi) = pa - 576

Eq. 3.4-3

Strength is reduced about 25% for Type N mortar, 35% for uninspected workmanship,10% by increasing air content from 5% to 18%, and 25% for stress parallel to bed joints (3.4.17,

3.4.18,3.4.19,3.4.20,3.4.21,3.4.22,3.4.23). inspected workmanship.

The Codeallowablestresses

The Code makes

are based on

no provision for allowable stresses for

uninspected construction. Specs. Table 1.6.2.2 permits the assumption that for Types M or S mortar fm (psi)

fa

+ 391 and for Type

N mortar fm (psi)

-

+

-

0.555

0.508 fa( 396. Typically the compressive

strength of concrete masonry is about equal to the values assumed in Specs. Table 1.6.2.2.

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3.4.6ConcreteMasonryElastic

Modulus

The mean chord modulus of elasticity of ungrouted concrete masonry with a prism aspect ratio of two is 615

fm

(psi) with a standard deviation of 365

fm

(psi)based

on 374

measurements (3.4.13).

3.4.7ConcreteMasonryinFlexure Concretemasonrywalls

28 daysoldconstructedwithhollowConcreteMasonryUnits

(CMU) with ASTM Type S mortar, inspected workmanshipand stress perpendicular to bed joints has an estimated mean flexural tensile strengthof 76 psi with a standard deviation of about 12 psi (3.4.13). Strength is reduced 21% by use of Type N mortar (3.4.24). High air content associated withthe use of masonry cement and air-entrainment in portland cementlime mortars reduces flexural strength about 50% (3.4.15). Strength is increased by 105% by the use of solid units instead of hollow units for flexure normal to bed joints (3.4.25). Strength is more than doubled when bending stress acts parallel rather than perpendicular to bed joints. Deflection at flexural cracking is about 0.036% of wall span (1/2800).

3.4.8ConcreteMasonry

in Shear

The mean shear stress at first crack in concrete masonry

is about 64% of the ultimate

compressive strength, with a coefficient of variation of 25%.

3.4.9 Grout Code 5.5.1.4 gives the modulus of elasticity of grout as 500 fr

3.4.10 Steel Reinforcement The modulus of elasticity of steel reinforcement, per Code 5.5.1.1, is taken as 29,000 ksi.

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METALCONNECTORSANDREINFORCEMENT

General

Connectorsused

for modernmasonryconstructioninNorthAmerica

are allmetallic.

Connectors can be made from wire, sheet metal or structural steel shapes. MDG Table

3.5.1 indicates the majority of the metals used in masonry. 3.5.2

Steel Wire

Steel wire used for reinforcement or connectors is cold drawn wire. Wire must conform

to

the requirements of ASTMA 82. Thistype of wireusually does not have a verywell definedyieldpoint.Yieldstrengthisgenerallydefined

as the stress at an extension of

O.OOSin./in.of gage length. In addition, ASTM A 82 specifies a minimum amount of area reduction at the point of rupture to evaluateductility.

See MDG Table 3.5.2 for wire

properties. Stainless steel wire is nickel-chromiumsteel manufactured in accordance with ASTMA 580 and is annealed in the manufacturingprocess.Annealednickel-chromiumsteels austenitic and as a result are non-magnetic. This is applications reinforcement in masonry walls

are

an important property since in some

is undesirable due to its effect on magnetic

fields created by medical diagnostic equipment and some military equipment.

3.53 Steel Sheet Metal Sheet metal used for connectors is made from either cold-rolled carbon steel conforming to ASTM A 366 and galvanizedinaccordancewith

ASTM A 525, or from stainless steel

conforming to ASTM A 167 Type 304. See MDG Table 3.5.3 for sheet metal sizes and weights. The colddrawncarbon

steel is not as ductile as the annealed stainless steel

material.

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3.5.4 Reinforcing Bars

Reinforcing bars are the same as used in reinforced concrete work. The bars can conform to a number of ASTM Specifications (A 615, A 616, A 617, and A 706) depending on the materialstrength(grade)

orother

properties desired. A majordistinctionbetween

reinforcing bars used in masonry and concrete is the limitation on size of bars for masonry to #11 as stated in Code 8.2.1, and the limitation of the maximum yield strength that is obtained with Grade 60 reinforcement. ASTM A 706 should be specified when controlled tensile properties or controlled chemical composition for weldability, or both are required. See MDG Table 3.5.4 for reinforcing bar sizes. 3.5.5 Connectors

Structural steel connectors such as those used

to support stone work, or as strap ties at

intersecting walls are required to conform to ASTM A 36. This is the steel typically used throughout the industry. 3.5.6 CorrosionProtection

All steel except reinforcing bars and wire fabric is required to be corrosion protected. This can be accomplished by galvanizing (Specs. 3.2.1.4) or by the use of Type 304 stainless steel (Specs. 3.2.1.3). Galvanizing must be by hot dipping. Electro-galvanizing is not permitted. The amount of zinc coating required on the galvanized product varies in accordance with the exposure (exterior vs. interior) and the amount of protection afforded by the mortar. Three different conditions of galvanizing are specified. For metal in interior walls, the galvanizing results in0.1 oz of zinc per sq ft of surface area. For sheet metal it amounts to 0.6 oz of zinc per sq ft of area (0.3 oz on each side). For and hence there ASTM A 525 and A 641 the metal is normally galvanized before fabrication is no zinc coating on sheared ends, at holes or atwelds. Also since the metal is usually bent

3-24


A C 1 T I T L E S M D G 93

m

Obb2949 0508555 213

during fabrication, the thickness of zinc coating is limited to prevent flaking when metal forming takes place. For exterior walls, sheet metal ties and wire ties mustbe "hot dipped" galvanized per ASTM A 153. If galvanizing is performed after a part is fabricated, the sheared edges and welds are coated with zinc. Galvanic action can occur whenever dissimilarmetals are in contact with each other. Many timesconditionsexist

where an A/E may want to usedissimilar metals together. The

severity of thisgalvanic

action depends on the relativeposition

electrochemical series.

See MDG Table 3.5.5(3.5.1).

of the metals in the

In somecases

(for example,

aluminum and structural steel), the possibility of galvanicaction is serious enough to warrant use of an insulating separator between the two materials. In other cases (for example, steels withslightly

different alloy contents), galvanicaction

is theoreticallypossible

but not

sufficiently serious to warrant the use of insulating separators. 3.5.7

Deformed vs. SmoothReinforcement

All bar reinforcement is required to be deformed per Specs. 3.2.1.1.1. There can be no

confusion for reinforcing bars since the standards are clearly defined and virtuallyall reinforcing barsare

deformed. This is not the case for wire products. Thereare

specifications for deformations in wire used to make deformed mesh (ASTM A 496) but there are no standards for deformations of wire used to make joint reinforcement. As a result, Specs. 3.2.1.1.2 contains special requirements for wire deformations in joint reinforcement. It is also important to note that reinforcing rods are hot rolled to form protrusions or lugs, but wire is cold formed by rolling indentations into it. Wire fabric isallowed

tobe used for principal reinforcement of masonry walls. The

application of this wire fabric would be in multiwythe walls which have continuous grout space between masonry wythes. When this material is used it must conform to eitherASTM

3-25


A C 1 TITLE*MDG 93

0662949 0 5 0 8 5 5 6 I S T

A 185 or A 497. Welded wire fabric of plain wire is acceptable since wire fabric derives its bond from the grip of the cross-wires in the grout. REFERENCES 3.4.1 National Testing Program, Brick Institute of America, October 1964, pp. 5-16. 3.4.2

Gross, J. G., R. D. Dickers, andJ. C. Grogan, Recommended Practice for Engineered Brick Masonry, Brick Institute of America, McLean, VA, November 1969, pp. 252262.

3.4.3

Allen, M.H. and R.B. Taylor, "Compressive, Flexural, and Diagonal Tensile Testing of SmallScaleFour-inchBrickMasonrySpecimens,"Progress

Report 1, Brick

Institute of America, McLean, VA, October 1964. 3.4.4 Allen,M.H.and

C.B. MonkJr., Tompressive andTransverseStrengthTests

Eight-InchBrickWalls,"Research

Report No. 10, BrickInstitute

of

of America,

McLean, VA, October 1966. 3.4.5

Davison,J.I.,"EffectofAir

Content on Durability of Cement-LimeMortars,"

Durabilitv of Building Materials, Elsevier Scientific PublishingCo., Vol. 1, 1982, pp. 23-34. 3.4.6

Fishburn, C.C., "Effect of Mortar Properties on the Strength ofMasonry,"U.S. National Bureau Standards, Monograph 36, November 1961, pp. 1-45.

3.4.7

Matthys, J.H., Tonventional Masonry Mortar Investigation," National Lime Association, August 1988.

Current 3.4.8 Huizer, A., M.A. Ward, and H. Mustead, "Field and Laboratory Study Using and Proposed Procedure For Testing Masonry Mortar," Masonry: Past and Present, ASTM STP 589, American Society for Testing and Materials, Philadelphia, PA, 1975,

PP. 107-122. 3.4.9

Dubovoy, V.S. and J.W. Ribar, "Masonry Cements

- A Laboratory Investigation,"

Construction Technology Laboratories Inc., Skokie, IL, January 1989, pp 9 and 12. 3.4.10 Grimm, C.T. and J.T. Houston, "Effect ofBrick Height on Masonry Compressive Strength," Journal of Materials,AmericanSocietyforTesting 3-26


and Materials,

A C 1 TITLElrMDG 9 3

0662949 0508557 O96

m

Philadelphia, PA, September 1972, Vol. 7, No. 2, pp. 388-392 3.4.11 Grimm, C.T. and C. Fok, "BrickMasonryStrength

at FirstCrack,Masonry

International, University of Edinburgh, Edinburgh, Scotland, UK,Vol. 1, No. 2, July 1984, PP. 18-23. 3.4.12 Abrams, D. P., J. L. Noland, and R. H. Atkinson, "Response of Clay-Unit Masonry

To Repeated CompressiveForces,"Proceedings Masonry Conference, Brick Development

of the 7th International Brick

Research Institute, University

of

Melbourne, Australia, February, 1985, p. 565. 3.4.13 Atkinson,R.

H. and G. G. Yan,AStatisticalStudy

of MasonryDeformability,

Atkinson-Noland & Associates, Boulder, CO, February 1990, pp. 20-21. 3.4.14 "Testing forEngineered Brick MasonryDetermination of Allowable Design Stresses," Technical Notes on Brick Construction, #39A, Brick Institute

of America, Reston,

VA, July/August, 1975. 3.4.15 Johnson, W. V., "Effect of Mortar Properties on the Flexural Strength of Masonry," Master Thesis, Clemson University, May 1986. 3.4.16 Ritchie, T., and J.I. Davison, "Factors Affecting Bond Strength Moisture Penetration of Brick Masonry," Symposium

and Resistance to

on Masonry Testing, ASTM,

STP 320, June 1962, p. 16. 3.4.17 Commentary on SpecificationsforMasonryStructures(AC1530.1-88),American Concrete Institute, Detroit, MI, 1990. 3.4.18 Cheema, T. and R. E. Klingner, "Compressive Strengthof Concrete Masonry Prisms," AC1 Journal, Detroit, MI, January/February 1986, pp. 88-97. 3.4.19 Hamid, A. and R. G. Drysdale, Toncrete Masonry Under Combined Shear and Compression Along the Mortar Joints," AC1 Journal, Detroit, MI, September/October 1980, PP. 341-320. 3.4.20 Drysdale, R. G. and A. A. Hamid, "Behaviorof Concrete Block MasonryUnder Axial Compression," AC1 Journal, Detroit, MI, June 1979, pp. 707-721. 3.4.21 Drysdale, R. G.

and A. A. Hamid,"Capacity of Concrete BlockMasonryPrisms

Under Eccentric Compression Loading," AC1 Journal, Detroit, MI, March/Aprill983,

PP. 102-108. 3-27


3.4.22 Maurenbrecher, A.H.P., "CompressiveStrength of Hollow Concrete Blockwork," Proceedings of the 4th Canadian MasonrySvmposium, Universityof New Brunswick, N.B., Canada, Vol. 1, June 1986, pp. 553-568. 3.4.23 Compressive Strengthof Composite Masonry Prisms and Walls, NCMA/BIA, Reston, VA, May 1972. 3.4.24 Recommended BuildingCode Requirements forEngineered

Concrete Masonry,

"Research Data and Comments," NCMA, Herndon, VA, 1967. 3.4.25 Drysdale, R. G. and A. Hamid, "Effect of Grouting on the Flexural Tensile Strength

of Concrete Block Masonry," The Masonry Society Journal, Vol. 3, No. 2, 1984, pp. T10-Tl9. 3.5.1

"Curtain-Wall Connections," Architectural Technology, May/June 1986, p. 37.

3-28


A C 1 TITLE*MDG

93 M Obb2949 0 5 0 8 5 5 9 9b9

Table 3.1.1 Product Specifications and Requirements

I

Item

C34

C 56

3-29


C 62

m

- Clay or Shale Masonry Units C 126

C 212

C 216

C 652

AC1

TITLE*NDG 9 3 m 0662949 0508560 680

Table 3.1.1 Cont’d.

I

Item C

C 34

C56

1 Physical Requirements

E D

Specification Entry Default Specification

3-30


62

C 126

C 212

A C 1 TITLE*IDG

0662949 0508563 517

93

Table 3.1.2 Product Specifications and Requirements

E D I U

Specification Entry Default Specification Type I units, singly Base Unit Specification Applies 3-3 1


- Concrete Masonry Units

m

A C 1 T I T L E x M D G 9 3 H 0662949 0508562 4 5 3

Table 3.13 ProductSpecificationsandRequirements Item

C 503

I

C 568

- Stone C 615

C 616

I

E

Specification Entry

3-32


C 629

I

A C 1 T I T L E U M D G 93

m

o

P

e

c1

S S

o

I

F

e

G c O

.I

P 3-33


0662949 0508563 39T

A C 1 TITLE*flDG 93

Table 3.5.2

m

Nominal" Nominal Diameter, Area in. in. sq in.

1/4"

Nominal

Tensile Strength

O. 1205

0.379 0.0114

O. 1483

0.465 0.0173 1,210.

1,380.

O. 1620

0.5 09 0.0206 1,442.

1,648.

0.5 89

2,220.

O. 1875

**

m

Properties of Wire for Masonry'

Wire Size

*

Obb2949 0 5 0 8 5 6 4 22b

0.0277

0.2500

1,940.

0.785 0.0491

909. 797.

3,935. 3,430.

Basedon requirements ofASTM A 82forcolddrawn steel wire ASTM A 82 permits variation of plus or minus 0.003 in. from diameters shown

Table 3.53 Sheet MetalThicknesses

3-34


AC1

T I T L E t M D G 9 3 m 0662949 0508565 Lb2 m

Table 3.5.4 StandardReinforcing Bars Nominal

Bar Size

Diameter

in.

Nominal Area

Nominal Weight

sq in.

Plf

#3

0.375

0.11

0.376

4

0.500

0.20

0.668

5

0.625

0.3 1

1.043

6

0.750

0.44

1.502

7

O. 875

0.60

2.044

8

1.000

O. 79

2.670

9

1.128

1.00

3.400

10

1.270

1.27

4.303

11

1.410

1.56

5.313

3-35


AC1

TITLE*MDG 7 3 m Obb29Ll7 0508566 O T 7 m

Table 3.5.5 Galvanic Compatibility of Metals'

Lead

LI

LI

LI

o

o

Brass

LI

m

m

LI

m

LI

Bronze

LI

m

LI

m

LI

Monel

LI

o

m

LI

m

LI

Uncured Mortar or Cement

O

m

o

o

o

m

o

m

o

LI

H

o

m

0

LI

LI

m

o

Woods With Acid (Redwood and Red Cedar) Iron/Steel

Galvanic action will occur LI

Galvanicactionmayoccur

under certaincircumstances and/or overaperiod

of time

O

Galvanicactionisinsignificantundernormalcircumstances

*

This table compares galvanic compatibility of metals commonly used with some of the more common building materials. Galvanic reactions occur most readily when materials touch, but may also occur when water runs from one material onto another.

3-36


A C 1T I T L E w M D G

93

m

0bb2949 05085b7 T 3 5 W

4 TESTING

4.0

INTRODUCTION

Test report requirements are identified by the Specificationsandserve

as the Quality

Control portion of Quality Assurance Program. The A/E determines when the testing is to be included. Test reports are given to the A/E as required in Submittals: Specs. 2.1.2.1

(b)

Results of mortar testsperformed

in accordancewith

the property

specification requirements of ASTM C 270. (Frequency indicated - once, when required.) (h)

Results of tests of masonryunits and materialsattestingcompliance with the specified requirements. (Frequency indicated - once, when required.)

Testingdoes

not affect the quality of the masonry,itonly

determines compliance to

specifications. See MDG 5.0.1 for resolution of noncomplying conditions. Testingfrequencies are indicated in the Specifications for the followingmaterials assemblages: QualityAssurance:Specs. .2

Masonry units

2.1.3 are tested in accordance with indicated materials

specification.

4-1


and

AC1

.3

T I T L E m M D G 93 W Ob62949 0508568 971 m

Masonry prisms when required. (Preconstruction and one testper 5000 square feet of wall) per Specs.

1.6.3 .4

Mortar tests in accordance with ASTM C 270 preconstruction evaluation of property specifications (once for each mortar type)and ASTM C 780 (When required)

.5

Grout tests - Specs. 4.1.3 (When required)

Quality Assurance: Specs. 4.1.3 Grout tests ASTM C 1019 (When required - one test per each 5000 square feet of masonry.)

4.1 MATERIALS TESTING 4.1.1 Preconstruction Preconstruction testingis recommended to ensure that materials selectedby the A/E comply with the requirements of the product specification and provide the required performance. In many casesthe Manufacturer’s Certificationmay be acceptable in lieu of preconstruction testing. The selected masonry mortar must be in compliance with ASTM C 270. For this mortar selected physical tests from ASTM C 780 should be completed prior to construction, so a set of comparative values is produced. Test results obtained during construction are more easily interpreted when compared with preconstruction tests. It is not the intent of ASTM C 780 that tests are performed at some fixed frequency. The test methods are classed as quality control tests that allow rapid isolation of the reason for any loss of quality control

that might occur. The array of tests, except for compressive 4-2


AC1

TITLE*NDG 73 m 0662949 0508569 808 m

strength and tensile splitting tests, can easily be performed in a periodof one or two working days. Thus, immediate appraisal

of the quality control exercised at the projectmay be

documented. Variations from batch to batch and day to day can be easily evaluated. In performing preconstruction tests, materials

to be used during construction should

be

sampled, combined and prepared using equipment and procedures that will be used during actual construction. The merits in performing these tests in addition to establishing basic comparative values are that they allow all

parties to agree on the interpretation of the

project specification and to the execution of the specification requirements. 4.1.2

Construction Testing

Testing done during actual construction should be directed toward establishing compliance with the Specifications requirements and the quality control requirements delineated in the project specifications.

As indicated in ASTM C 270, testing of hardened masonry mortar samples removed from a structure is not addressed.Specializedchemical

and petrographic tests can be used;

however, the qualitycontrol tests ofASTM C 780 are more easilycompleted.Testing during construction is preferable to testing after construction has been completed. Strength characteristics of masonry mortar prepared at the project are not related to the property requirement strengths of ASTM C 270. This is due to theincreased water content of jobprepared

mortars. Mortar qualitycontrol

under C 270 is actuallyobtained

by

measurement of materials used to mix the mortar. Any testing of site prepared mortars should be referenced back to ASTM C 780, namely the preconstruction tests and the base test results. Testing of masonry mortars during actual construction should involve selected

tests from

ASTM C 780. If the intent is to track air content of mortars, the air content test should be 4-3


Obb2949 0508570 52T

A C 1 TITLExMDG 93

performed. If the intent is to establish the exercise of qualitycontrolduring

mortar

preparation, both water content and cement to aggregate ratio tests should be performed. The test results should be referenced back to the preconstruction test results. External factors, such as temperature, may influence the results so the external influences should be systematically recorded. The data sheet provided as a part of ASTM C 780 (Annex A8) is suggested as a general format. The selectedmasonrygroutmust provisionsfor

either finegrout

be incompliancewithASTM or coarsegrout

C 476 whichmakes

using acceptablematerialsbasedon

proportions by volume. This ASTM standard does not require any physical testing. Code

4.3.3.4, however, requires a minimum compressive strength

of 2,000 psi and 28 days. The

sampling and testing of grout is conducted in accordance with ASTMC 1019. This standard can be used to initially select the ASTM C 476 materials and proportionsto satisfy the Code minimum compressive strength requirement. ASTM

C 1019 can also be used as a quality

control test for uniformity of grout preparation during construction. 4.2

ASSEMBLAGE TESTING - PRISMS

Testing for compressive strength of masonry with the masonry units and mortar combined in an assemblage, will be as indicated by the M. The test method involves fabrication of brick or concrete masonry assemblages called masonry prisms, in accordance with ASTM

E 447 Method B as modified in Specs. 1.6.3.2. Because masonry units are of different sizes, the test method indicates the number of units to be used during the fabrication of the test specimen. The A/E may require that prism testing be a part of the design process and the materials selection process. Thereafter, the masonry prism test may be required during construction, serving as a quality control test. The combined materials forming the assemblage should reflect the individual materials, as used, and also the workmanship. The specimen may be

4-4


A C 1 TITLE*flDG

93

0662349 0508571 466

subject to temperature influences; so, similar to mortar testing in accordance with ASTM C 780, interpretation of test results mustbe tempered with knowledgeof external influences.

In performingprismcompressivestrengthtests,assemblages

are loaded concentrically.

Because test specimen size and shape influence indicated compressive strength, the tested strength must be corrected depending upon height to thickness ratio. Test specimens with a lower height to thickness ratio produce higher indicated strengths. Correction factors for clay brick masonry and concrete masonry units are listed in the Specs. Table 1.6.3.3(b) and Table 1.6.3.3(c), respectively.

4-5


A C 1 TITLExNDG 93

m

0662947 0 5 0 8 5 7 2 3T2

m

5 QUALITY ASSURANCE

5.0

INTRODUCTION

Both quality assurance and quality control requirements should be incorporated within the Project Specification. Quality assurance provides administrative policies related to the quality control measures expected

and requirements

to assure the Owner’s quality objectives.

Quality control is the systematic performance of construction testing and inspection. The extent of the quality assurance and the quality control program generally will vary with the size of the project. The quality objective of the Owner should be met when construction is completed in accordance with proper design concepts, acceptable construction practices, and materials complying with product specifications. Success is dependent on open, but direct, communicationsamongresponsiblepartieswithinthedesign

and constructionteam.

Records documenting the successful completion of the structure in accordance with

the

Owner’s objectives complement the Contract Documents.

5.0.1

Quality Assurance

The quality assurance program incorporated in the contract documents, includes both the project specifications and drawings, and should address 1)

organizational responsibilities,


the following:

AC1

T I T L E x M D G 9 3 W 0682949 0508573 2 3 9 m

2)

materials control,

3)

inspection,

4)

testing and evaluation,

5)

identification and resolution ofnoncomplyingconditions,

6)

records.

and

Quality assurance considerations are delineated in an AC1 Committee 121 report, entitled "Quality Assurance Systems for Concrete Construction" (5.1.1). According to Specs. C.1.5, it is necessary to delineate the responsibility, authority,and lines

of communication of the parties involvedinqualityassurance.

Procedures should be

established for the identification and resolution of nonconformances. Materials control verifies the chemical and physical characteristics of individual materials requiredin

the contractdocuments.

These controlsshould

be monitoredthroughout

construction for compliance with the contract documents. Inspectioncontrolshould

be established to assure thatthe

masonrymaterials

and

construction practices comply with the requirements of the contract documents. Inspection program, personnel, and records should be regarded as inspection control measures. Testing and evaluationshould

be describedin

the qualityassuranceprogram.

The

or noncompliance evaluation of test results by the testing agency should indicate compliance with a referenced standard. Procedures for identification and resolution of noncomplying conditions shouldbe described in the contractdocuments.In

agreement with others in the desigrdconstructteam,all

resolutions should be either accepted as is, or rejected and repaired or reworked. Repaired and reworked conditions should initiate reinspection.

5-2


I

AC1

TITLE*MDG 93 m 0662949 0 5 0 8 5 7 4 L75 m

Records control should be described in the contract documents. Distribution of documents during and after construction should be delineated. Review of documents should continue throughout the construction periodto ascertain that all partiesare informed and thatrecords for documenting construction occurrences are available and correct after construction has been completed.

As an example of quality assurance, the general contractor may require that the masonry subcontractor submit a written procedure for cold weather masonry construction practices. 5.0.2

Quality Control

The specific requirements of the projectspecification

related to design,materials

procurement and use should be implemented by quality control measures. Quality control measures may be dictated by the Owner's representative or self-imposed by a responsible party charged with the conductof a specific task associated withor contained in the project specification. The qualitycontrolprogramessentiallyfollows

the qualityassurance

requirements as

documented from a specific project specification section or allied sections. Members of the Design and Construction team affected by quality control requirements include the owner,

A/E, inspection agency, contractor and subcontractor. The person responsible for the quality assurance provisions contained within the Contract Documents should impose certain responsibilitieson those whowill perform quality control testing while granting them authority to access the construction site forthe performance of sampling, testing, and inspection. Those Inspection and Testing Agencies being considered for performance of the Quality Control tests should be evaluated for their conformance to the requirements of the "Standard Practice for The Accreditation of Testing Agencies for Unit Masonry," ASTM Designation C 1093.

5-3


A C 1 TITLESMDG 93

m

O b b 2 9 4 9 0508575 001

The Testing Agency should be required to produce its laboratory accreditation document whichisessentially

their qualityassuranceprogram.Thisdocumentshouldindicate

organization, testing capabilities, qualification of personnel, test procedures, presende and calibration of physical test equipment, and records control. The approved Testing Agency should document its representatives who will be a part of the quality control segment,hisher qualifications and procedures duringthe performance of the qualitycontrolphase.Knowledge

of the proper procedures for sampling and testing

masonry materials is essential.

As an example of quality control, the masonry subcontractor may be required to carry out the cold weather masonry construction practices described in the approved quality assurance program. 5.1

SUBMITTALS

Prior to construction, the A/E or the owner’s representative who is responsible for submittal acceptance/rejection and compliance can require establishment of conformance of a product either by sampling and testing prior to construction or acceptance of the Manufacturer’s Certification for Compliance. Tests allow establishing conformance with a product specification.The measured test result is compared with the specification limit for the chemical/physical property. The alternative to sampling and testing the product is to rely on the manufacturer’s quality control data obtained during production of the actual product to be used on the project. Certification by the manufacturer provides the consumer with documentation indicating that on a certain date the product was tested in accordance with indicated test methods and measuredtestresultsconfirmingcompliancewith

the requirements of the product

specification. The documentation is certified by the company representative. The consumer 5-4


A C 1T I T L E W N D G

93

m

0662949 0508576 T 4 8

m

using these data can reduce overall testing during the product selection phase. The specifiermustidentifyallsubmittalsrequired.(SeeSpecs.

Optional Checklist and

Specs. 2.1.2 and 4.1.2). Documentation of all submittals should follow the guidelines given in MDG 5.5. 5.2 SAMPLE PANELS 5.2.1 Recommended Practices

The Specs. 2.1.2.2 states that "When required, construct sample panels

of masonry walls

using materials and procedures conforming to the Project Specifications." See also Specs. 2.1.2.3. In accordance with Specs. 2.1.2.2(b), job-site sample panels should

be constructed for the

purpose of establishing an accepted standard of qualityfor the project. Sample panels should be used to evaluate the appearance and construction of the finished masonry work. Job-site sample panels should contain all aspects of the combined masonry materials and the specified construction procedures per MDG 5.2.2. The representative of the owner and/or

PJE may choose to use more than one type of bond and mortar joint finish on the sample panel to help make final choices as to the finished appearance. The mortar joint finishes greatly influence the appearance of the wall. All submittals should be approved before the sample panel is constructed.

The panel should be constructed by the mason contractor

selected for the project before the masonry work begins, and should not be removed or destroyed until al1,work has been accepted (Specs. 2.1.2.3). The construction of the sample panel is based on the contract documents and is approved when the owner or their authorized representative acceptsthe appearance and construction characteristics of the panel.Whenapproved,

the constructionproject

may proceed.

Usually, one panel is constructed forappearance, any required tests (e.g., ASTM C 780), and 5-5


A C 1 TITLExMDG 93

m

0662949 0508577 98Y

m

workmanship. The panel is then referred to as the approved project standard. It becomes an important device in the evaluation and acceptance of the masonry work during constructionperiod.Opinionsrelating

toappearanceare

much easierto

the

resolve if a

standard of quality, embodied by an approved sample, is available for evaluation at the construction site. Therefore, it is to everyone’s benefit that sample panels be made a part of the project requirements for masonry work. It is especially important where appearance, establishment of procedures, and construction are important issues. 5.2.2

Suggested Criteria for Construction

-

SAMPLE PANEL LEVEL 1 SCOPE OF COMPLIANCE SHOULD INCLUDE Masonry unit types, sizes, shapes, color range, texture, surface configuration, and other characteristics. Chippage dimensions, and warpage limits

of units controlled

by product specifications or more restrictive in wall criteria determined by the A/E. Mortar joint size, alignment, color, tooling configuration, and texture. Mortar should be evaluated for appearance after it is surface dry.If

colored mortar is used, the

color should be judged after the sample panel has had sufficient time to dry (5.2.1). Bond pattern and color pattern if masonry units are more than one color. The conformance of workmanship representative to that specified in the contract documents. Quality of appearance of approved cleaning material applied by the approved methods proposed for the finished work if required. Quality of appearance of approved water-repellent masonry surface treatment material applied by the approved methods proposed for the finished workif required. Workmanship with respect to mortar placement and dimensional tolerances.

-

SAMPLE PANEL LEVEL 2 SCOPE OF COMPLIANCE SHOULD INCLUDE: In addition to Level 1 items, characteristics should include: 5 -6



93

m

0 6 6 2 9 4 9 0 5 0 8 5 7 8 810

O

Weephole type and spacing, and finish material, if any.

O

Flashingmaterial,configuration,lapping

and sealing joint details, end dams,

back-dam height, and termination conditions. Anchors and tie types, material, spacing, and placement requirements. Reinforcement type, location, and securement methods. Expansion joint and control joint material and configuration. Joint sealant type, color, and configuration. Gasket and joint filler type and configuration. Port holes for observation of hidden construction components, clean-outs,and other items as specified in the contract documents. O

Full capability for testing the sample panel for tests as required by the code and/or the contract documents. Sample panels shouldbe sized accordinglyand verified with applicable industry standards for tests.

5.3 5.3.1

INSPECTION Purpose

Thepurpose

of inspection is to observe and record thatthe work performed during

construction is in

compliance with

the contract documents.

Thus the duties and

responsibilities of the inspector are to observe that the work generally complies with the drawings and specifications of the

A/E. The inspector can be the A/E or an independent

consultant. The intent of the Code and Specs. is to require inspection, at anextent and frequency to be determined by the A/E,based on theproject requirements. Possible inspection activities and procedures are presented in this section.

5 -7


A C 1 TITLEtMDG 93

53.2

m

Obb2949 0 5 0 8 5 7 9 757

m

Planning

The inspector represents the owner. The inspector should have a complete knowledge of the contract documents, construction practices, materials, and test procedures. Ideally the inspector should attend pre-bid meetings with the A/E and project bidders. The inspector should attend pre-construction meetings with the A/E and the successful contractor. 5 3 3 MaterialSubmittals

Products specified in the contract documents must meet applicable material specifications. All submittals shallbe approved by the A/Ei or the owner’s designatedrepresentative before

any construction begins as described in MDG 5.1 and 5.2. All changes must be approved by the A/E or owner’s designated representative in writing prior to installation. All shop drawings must be approved by the N E or owner’s designated representative in writing before beginning fabrication or production. 53.4 InspectionFiles The inspector should have all approved submittals, shop drawings, changes, job site and laboratory tests, as required by the contract documents, for all products, techniques, location, and procedures. In accordance with the procedures established in the quality assurance program, accurate quality control records should be kept ofall material deliveries, as to type of product, manufacturer, date of delivery, where it is placed on the job site, and how it is protected from the environment (Specs. 2.3.1).

5-8


AC1

TITLE*NDG 93 m 0 6 6 2 9 4 9 0508580 477 m

In accordance with the procedures established in the quality assurance program, a daily quality control log should be kept which includes weather conditions at the job site (e.g., temperature, relative humidity,

general conditions, both A.M. and P.M.) and any

observations not in compliance with the contract documents. 53.5 ConstructionInspection

Inspectors should observe such things as workmanship, masonry bond, mortar joint finish, and placement of connectors, reinforcement, and grout.

As designated by the owner, the inspector hasthe authority to judge materials, workmanship and procedures employedinconstruction,based

on contract document requirements.

Approval or disapproval should be reported to the owner and contractor. The inspector does not have the authority to direct the work of the contractor. The inspector should observe how materials are stored and handled and also observe the handling of equipment used to prepare, transport, and install masonry materials to be sure it is being used properly and will in no way adversely affect the finished wall. The inspector shouldunderstand the materials andthe effect that weather changes may have on them. For example, the effect of high and low temperatures and humidity on a mortar systemmay be extremely important.

The absorption of clayproducts and the moisture

content of concrete masonry products mustbe considered. Material properties and weather can have a direct effect on the quality and performance of the finished masonry. Where required, the use of reinforcingsteel,

joint reinforcement,ties,

anchors and

accessories shouldmeet the requirements of Specs. 3.2 and 3.3. Verify that the appropriate corrosion resistant material is used. Flashing,weepholes,anchors,ties,

wall vents,etc. are installedin accordance with the

5-9


contract documents. See MDG 6.2.6. Weepholes are constructed as detailed and at the specified spacings. Weepholes constructed as a void or with prefabricated units should be installed over all flashingat 24 in. intervals. Rope weepholes should be installed at 16 in. intervals. See MDG 6.2.6. Observe the constructiqn of movement joints installed as called for in the contract documents. See MDG 6.2.7. Observe and note preparation of jobsite test samples. 5.4

"ING

A well planned, thoroughly specified and properly implemented testing program during the construction of a project is an important component of an overall inspection and quality control program. Tests are performed to verify the consistency of materials, workmanship, protection and curing conditions,and the in-place performance and strengthcharacteristics achieved during construction. The tests discussed below are commonly applied to masonry construction. In most cases, the procedures are well established and governed by standards. The interpretation of test results is based on job requirements. 5.4.1 Testing as Part of a Construction Quality Assurance Program

Correct implementation of a testing program should result in productive and meaningful activity serving the interests of all parties involved in the project. The construction testing program must be rationally related to preconstruction testing requirements, and be based onthe

knowledge of masonrybehavior,

the required properties,andthe

performance of the completed masonry. 5-10


required

AC1

TITLE*HDG 9 3 m 0662949 0508582 241 m

Projectdesignassumptionsandmaterialacceptancecriteriacan

be based on the unit

strength method in Specs. 1.6.2, in which the required properties of the individual masonry materials are verified by testing. A rational quality control program for a project based on the unitstrengthmethodmightinclude

a plannedrepetition

of the unittestsforthe

materials in construction, accompanied by an inspection program. Sampling

and testing

should be in accordance with ASTM C 67 for clay masonry units and ASTM

C 140 for

concrete masonry units.

For grouted masonry, the grout may be

proportioning and materialrequirements

required to satisfy the

in ASTMC 476. For aprojectdesigned

and

specified on the basis of the unit strength method, it should not be necessary to test masonry assemblies fabricated during construction specifically for testing purposes.

As an alternative to the unit strength method, design assumptionsand materials acceptance criteria could include measured

properties of masonry assemblages. These criteria could

include: the prismstrengths,Specs.

1.6.3; measuredmodulus

permeabilityperformance;modulus

of rupture; and other physical properties of the

completed masonry. Testing

of elasticity,Code

is normally required before construction begins,

5.5.1;

to develop

acceptable and agreed upon values. Similar tests are repeated during construction as apart of a quality control program. Specifications for a properly implemented construction testing program must include more than a simple requirement for a certain number of tests. The program will only serve its intended purpose if the following issues are addressed in the project specifications: a.

Correspondingpreconstruction testingrequirements.Constructiontestsshould not be specifiedunless

an equivalentpreconstructiontesthasalso

been

specified. b.

A clear definition of an acceptance criterion, which should be based on careful consideration of actual project requirements. Minimum test results for

both

the preconstruction and construction tests must be specified, and they must be related to each other. It would not be unreasonable for required test results 5-11


AC1

T I T L E x M D G 9 3 m Ob62949 0508583 188 m

during construction to be less stringent than corresponding preconstruction laboratory test results.Initssimplest

form, a minimumaverage

can be

specified. For designs based on strength, a minimum average related to the measured coefficient of variation of the test results is more appropriate. c.

A cleardefinition of the requiredfrequency

of testingbasedontime,

the

maximum constructed area, or the maximum number of installed masonry units between test cycles. As a minimum, require the testing of prisms and thus the application of Specs. 2.1.3.3 which when specified, requiresone prism test for each 5,000 square feet of wall area or portion thereof. The sampling method, sample handling, and the test methods must also be specified. d.

A protocol for recording,reporting, and disseminating the test results, and a process for obtaining written review

of the results from

the design

professional.

e.

A planned response for test results that do not meet the specified acceptance criteria. The response could include retesting, re-evaluation of strength and performance requirements by the design professional specificallyfor the area affected, and finally, a rejection protocol.

f.

A cleardefinition

ofwhichpartyisresponsible

for scheduling the tests,

performing the tests, obtaining the samples, and paying for the tests. g.

A clear statement on the reconstruction of rejectedwork.

In thesections below, various tests, that areuseful for a construction quality control program are briefly discussed. Not all

of the tests are necessarily needed for every project.

design professional should formulate a construction testing program

The

to address masonry

properties that are necessary for compliance with the Code,and those critical to the success and durability of the project. 5.4.1.1

Initial Rate of Absorption Tests

- Both laboratory

and field investigations have

shown that high suction clay brick may cause excess loss of mixing water from the mortar, resulting in poor adhesion, incomplete bond and water-permeable joints of low strength. 5-12


A C 1 TITLE+MDG 93

m 0662949

0 5 0 8 5 8 4 014

m

Thus, some clay masonry products may require wetting so that the proper interaction with the mortar can be achieved, and proper bond and weather resistance can be realized

(5.4.1). The initial rate of absorption (IRA) test is used to determine if wetting is necessary. Preconstruction IRA testing will determine if wettingshould

be a general project

requirement. Construction IRA testing can be used for periodic quality control checks or as part of a response to extreme temperature and wind conditions which might adversely affect the curing of the mortar and

necessitate a temporary adjustment in construction

procedures. ASTM C 67 contains two methods for measuring the IRA. The brick is suspended in a pan of water. For laboratory tests, the amount of water absorbed by a dried unit is determined by weighing the brick before and after soaking, using an accurate balance. For field tests, the amount of water absorbed by the unit can be determined by measuring the volume of water in the pan before and after soaking, using a graduated flask called a pycnometer. Weighing the brick is best suitedas a laboratory procedure; the pycnometer method can be used on-site, butmay be difficult because of the precision necessary in controllingthe depth and levelness of the soaking process. Masons often use a more convenient approximate method to determine the I R A A circle, one inch in diameter, is marked with a wax pencil on the bedding surface of the brick, and

25 drops of water are deposited using an eyedropper. The outline of a quarter ($0.25) is the customary guide for this circle. If it takes more than 1%minutes for the brick to absorb the dropsof water, wetting is usuallynot required. If this procedure is allowed on a project, its acceptability should be determined by preconstruction laboratory tests for comparison with the more rigorous ASTM C 67 IRA test. 5.4.1.2

Testing to Verify thePrismCompressiveStrength

of Masonry

-

Testing to

determine the masonry compressive strength, fm, is a useful quality control procedure for comparison with preconstruction qualifymg tests, and is required when specifiedor when the properties of the masonry materials or construction do not qualify for the use of the unit 5-13



73

m

Obb27Y7 0 5 0 8 5 8 5 T 5 0

m

strength method by Specs. 1.6.. Code 5.5.1.3 permits designing on the basis of the actual modulus of elasticity of masonry usedfor a project. The modulus of elasticity is determined duringaprismcompressive

strength test. The standard procedure for prismtests is a

modification of ASTM E 447 as described in Specs. 1.6.3; for the modulus of elasticity, the secant method and theprocedures of ASTM E 111are permitted by Code 5.5.1.2 and Code 5.5.1.3. Using project materials and techniques, three prisms are fabricated in stack bond one unit wide and thick, with full a

mortar joint. Clay masonry prisms must have a

height-to-thickness ratio between 2.0 and 5.0. Concrete masonryprismsmusthavea height-to-thickness ratio between 1.33 and 5.0, and contain at least one mortar joint. If a prism is too large for the intended testing machine, half units can be used for the top and bottom as shown in Fig. 5.4-1. Solid Joints

e' Unit

'

/ h Full Unit

Minimum:

One M o r t a r

Clay Brick

Concrete Block

Fig. 5.4-1 PrismCompressiveStrengthTestSpecimens After fabrication, the prisms are stored on-site for 48 hours in an environment similar to the air exposure conditions of the masonry wall, and wherethey

will not be disturbed.

Temporary plywood top and bottom caps are then strapped onto the transported to a laboratory for additional curing. On-site storing

prism, and it is

and transporting of the

prisms must be done carefully to avoid damaging the bond between the mortar and the 5-14


AC1

TITLE*IDG 9 3 m 0662949 0508586 9 9 7 m

masonry units, which will cause invalid test results. The prisms are cured in the laboratory until theyare 28 days old,capped, and tested in compression to failure. Prism weight should be limited to approximately 150 pounds unless special

arrangements are made with the

testing laboratory for handling and capping. The physical size and load capacity of the testing machine should also be considered in determining the size of a prism. Test results are corrected for the aspect ratio of the prism using the factors in Specs. 1.6.3.3. 5.4.13

Testing to Evaluate Mortar

-

Mortars are classified by type using theletter

designations M, S, N and O. Two methods of specifymg mortars are given in ASTM C 270 (1)proportion specifications and (2) property specifications. One or the othermethod, but not both, should be used to specify the mortar. The procedures and criteria for testing in ASTM C 270 are for laboratory-prepared samples only,

and should be thought of as a

method for qualifymg materialsand the mix design for use in a project. The strength criteria of ASTM C 270 are therefore not the appropriate basis for a construction quality control

testing program unless specifically so stated in the project requirements, as suggested in Specs. 2.1.3.4. Once a mortar mix design qualifies in accordance withthe laboratory ASTM

C 270, the volume proportions of ingredients, as added to the mixer, serve as the quality control measure. Alternatively the procedures of ASTM C 780 can be used to establish a preconstruction datum and a construction quality control testing program. Seven test procedures are given in ASTM C 780: (1) consistency by cone penetration, (2) consistency retention, (3) mortar aggregate ratio, (4)mortar water content, (5) mortar air content, (6) compressive strength of molded mortar cubes and cylinders, and (7) splitting tensile strength of molded mortar cylinders. The results of one or more of these procedures, repeated over the course of construction, are compared to preconstruction results from the same procedure. Currently,

mortar compressive strength and mortar air content tests are

the two most common procedures specified for quality control testing. Perhaps the results of these procedures are more easily and intuitively related to projectperformance requirements. However, the other five tests shouldnot be ignored (5.4.1). Consistent results within acceptable bounds based on the preconstruction evaluation tests is a good indicator

5-15


A C 1 TITLExMDG 93

Obb2949 0508587 823

m

of consistent batch-to-batch mortar properties and reliable in-place mortar properties. The test procedures, and the appropriate interpretation of results for allseventests,

are

thoroughly discussed in ASTM C 780. The mortar compressive strength and mortar air content tests are briefly discussed below. Mortar compressive strength tests are performed on cubes or cylinders cast from samples of the project mortar.Sincemosttesting

laboratories havejigs for cappingcylindrical

specimens, it may be easier to achieve top and bottom caps that are flat and parallel for cylindrical specimensthan for cube specimens. To achieve consistency inthe cast specimens, complete instructions for filling the mold, spading the wet mortar, and curing, transporting, storing and capping the specimen are given. They are similar to the instructions for the more familiar concrete cylinder tests. There is also a prescribed protocol for rejecting data from specimens which are judged to be "manifestly faulty". Mortar air content is determined using special apparatus. Either a volumetric method or a pressure method can be used. The air in the mortar sample is replaced with water. The volume of water necessary for a thorough replacement of the air is a measure of the air content of the mortar. For the volumetric method, the sample is jarred and rolled to cause the air to migrate out of the mortar and be replaced by water. For the pressure method, air pressure is used. Mortar compressivestrength and mortar splittingtensiontests

require time for curing

specimens. Therefore, they cannot provide immediate information necessary for adjusting mortar batching procedures in a timely fashion. 5.4.1.4 Testing to Evaluate Grout - Specs. 1.6.2.1, Specs. 1.6.2.2, Specs. 4.1.3, and Specs.

4.2 require that grout conform to the proportion requirements of ASTM C 476 and thatits strength be determined in accordance with ASTM

C 1019. The procedures of ASTM C

1019 can be used both to verify compliance of a grout designmix and as thebasis of a grout quality control testing program. The objective of the test is to subject the grout specimen

5-16


AC1

TITLESUDG 9 3 U 0 6 6 2 9 4 9 0508588 7bT m

to curing and absorption conditions similar to those in the wall. To achieve this objective, masonry units, intended for use in the wall, are used as the form for grout specimen. As in prism testing, the storage, handling, and transport of the grout specimen must be done carefully to avoid damage which will invalidate the test results. The grout must be removed from the masonry form for testing, so it is necessary to prevent bonding by lining the form with a permeable sheet such as a paper towel. 5.4.1.5 Testing to Determine the Flexural Modulusof Rupture - If the flexural modulus of

rupture (MOR) is important for the performance of a project, preconstruction testing should be specified to demonstrate the compatibility of masonry materials and their ability to provide the required flexural bond strength. Corresponding MOR tests can be performed during construction as a quality control procedure, and should be specified if the reliability of a design depends on theMOR. However currently neither the Code or the Specifications

require such testing. Laboratory MOR tests are performed on a stack bond prism fabricated in a jig

to assure

flatness and proper alignment of the masonry. After curing, the sample can be tested either as a beam in bending, according to ASTM E 518, or by peeling the bricks from the prism using a special bond wrench device, according

to ASTM C 1072. Neither test method is

referenced in the Code nor in the Specs. Thus they would haveto beincluded in the project specifications. There is currently no correlation between the test results from these methods of test and the allowable flexural tension values in Code Table 6.3.1.1. 5.4.2 ProceduresUsefulforInspection

Programs

Specs. 1.5 discussesan inspection program duringthe construction process. Important items, such as anchor and tie spacing, filling of collar joints, grouting, flashing installation details, clearances and obstructions inthe cavity, clogging of the weep system and flashing, and joint filling are easier to observe while a wall is being built

than to evaluate after the wall is

completed. There may be occasions when it is necessaryto determine conditions which are 5-17


A C 1 TITLExMDG 93

m

0662947 0508589 bTb

concealed by the finished masonry, and several techniques are available for this purpose. 5.4.2.1 Omitting Masonry to Permit Inspection Withina Cavity Wall - The most direct way

to facilitate inspection of conditions within a wall after it is built is to anticipate the need, and to specify that masonry be periodicallyomitted to provide an inspectionopening. Inspection openings are particularly useful immediately above flashing and at the bottom of grout pours. They can also

be used to clean out debris which falls on the flashing during

construction. If inspection openings have not

been left during construction, they

can be

created by removing one or two units of masonry. If done carefully, adjacent construction will not be damaged, and only the removed units will require replacement. After inspection, the opening mustbe closed and properly pointed. Exactly matching the surrounding mortar color is virtually impossible, and some allowance must be made for color variations. Looking through an opening to observe the adjacent construction requires the use of a small inspection mirror and a flashlight. By shining the flashlight beam at the inspection mirror along the line of sight, the area within view in the mirror can be illuminated from outside the wall. With a little practice, a sense

of orientation and distances will develop.

5.4.2.2 Fiber-optic Borescope - A fiber-optic borescope can be used to inspect concealed

conditions. The viewing wand of the device is inserted in a small hole which can usually be drilled inthe mortar joints without damage to the adjacent masonry. Light to illuminate the view is transmitted from a light box to the end of the wand along optical fibers, imageistransmitted

to an eyepiecealong the samefibers.

and the

The resulting image can

be

recorded by replacing the eyepiece with a camera.

5.5 COMPLIANCE Documentscontainedwithin

the qualityassurancefileshouldshowcompliance

of all

construction activities withthe individual specification requirements.The documents should reflect the following:

5-18


A C 1 TITLE*RDG 93

m

O662949 0508590 318

m

1.

Documentation for approved submittals.

2.

Documentation for masonry materials certifiedor tested and in conformance with the product specification.

3.

Documentation for mortar selection and conformance of thatmortar

to the

requirements of ASTM C 270.

4.

Documentation of mortar quality control measures and tests.

5.

Documentation of masonry materials quality control tests.

6.

Documentation of masonry prism compressive strength tests.

7.

Documentation of special construction practices: a) b)Hot

Cold weather masonry protection. weather masonry protection.

8.

Documentation of grout compressive strength.

9.

Documentation of inspection and findings.

practice with Compliance of the construction

the Specifications requirements is

demonstrated by: identifylngthe Project Specification requirement; entries reflecting action taken as required, suchas sampling and testing; test reporting indicatingboth test results and imposedlimits;signatures

by tester and agency; and receipt and approval of the QA

manager. The supporting documents are appropriate for materials submittals and testing, as well as inspections required by the A/E to assure quality. Compliance isindicatedwhenaspecific

part of the QualityAssuranceplan

has been

completed in its entirety.Any nonconformance that occurs duringthe construction sequence should be documented to reflect: specification requirement,nature of the nonconformance, correctiveactiontaken,

and verification that correctiveaction

occurrence, and changes to prevent its recurrence.

taken has resolved the

Appropriate signatures and approvals

by the A/E or the owner’s designated representative should be a part of the documentation. Basically, what is sought is certification

that the requirement was completed and where

completed improperly, has been corrected to restore the structure to its designed condition.

5-19


A C 1 TITLE*flDG 93

m

0662949 0 5 0 8 5 9 1 254

m

For example, if quality control tests of a material, e.g., masonry units, are found to be in nonconformance with the materials specification, documentation should allow tracing the sample from manufacture to and through testing, duringwhich testing the nonconformance was detected. The same document should establish the people involved and the corrective action taken, and should verify that no part of the masonry or that part of the masonry structure containing questionable materials hasbeen impaired or degraded by the inclusion of the product. The corrective measures should also consider additional

steps to prevent

been sampled, tested, and recurrence, such as non-use of that product until on-site lots have approved. An additional part of the compliance should include record retentionand disposal. Records

should be provided to the Ownerfor

useby

maintenancepersonnel

and consultants

attempting to recreate past happenings.

REFERENCES 5.1.1 American Concrete Institute, "Quality Assurance Systems for Concrete Construction", AC1 121R-85, American Concrete Institute, Detroit, Michigan, 1985.

5.2.1 Beall, C., "Coloring Mortar (Mortar--How to Specify and Use Masonry Mortar)," Aberdeen's Magazine of Masonrv Construction, October 1989, p. 33. 5.2.2

Beall, C., Masonrv Design and Detailing for Architects, Engineers and Builders, 2nd ed., 1987, p. 361.

5.4.1 Brick Institute of America,"MortarsforBrickMasonry,"Technical

Note No. 8,

Revised, Brick Institute of America, Reston, Virginia, November 1989.

5.4.2 McGinley, W.M., "IRA and theFlexural Bond Strengthof Brick Masonry," Masonry: ComDonents to Assemblages, ASTM STP 1063, John H. Matthys, Ed., American Society for Testing and Materials, Philadelphia, 1990.

5.4.3 Sarker, A, R. H. Brown,"FlexuralStrength

of BrickMasonryUsing

the Bond

Wrench," Research Reuort Number 20, BrickInstitute of American, Reston, Virginia, November, 1987. 5 -20


AC1

TITLErMDG 9 3 m 0662947 0508572 170 m

6 QUALITY CONTROL

6.0

INTRODUCTION

Quality Control, as discussed in this section, consistsof the operations of the contractors and other members of the design team at the construction site to obtain compliance with the contractdocuments.

The CodeandSpecifications

operations may be desirableorhelpfultoachieve operations at the sitebeginwith

mandate certain operations; other the desiredlevel

of quality. The

the receipt,inspection,storageandhandling

of the

materials. Prior to beginning the work,the areas to receive the masonry should be inspected to ensure that they are ready to receive the workand that the specified tolerances havebeen met. The actual placement of the masonry begins with the preparation of the mortar and continues through the installation of a variety of differing materials and products that, when properly combined, forms the completed product. The partially completed and completed portions of the work should be protected during the work to prevent damage. 6.1 PREPARATION

Preparations for masonry construction include proper storage and protection of materials from the weather, inspectionof supporting elements for completion and accuracy, examining units and materials, and providing construction protections.

6-1


A C 1 TITLE*:MDG 93

m 0662949 0508573

027

6.1.1 Material Delivery,Storage,andHandling

The methods of material delivery, storageand handling prior to placement are critical to the performance and appearanceof the finished masonry. Improper procedures can easily result in physical damage to units and accessories, or contamination or degradation of mortar and grout ingredients. Materials should

always be stored off the ground and protected from

weather (Specs. 2.1.4, 3.1.3, and 4.1.4). Masonryunitsshould

be delivered and stored to preventmoisturemigration

from the

ground, and covered with water-repellenttarps or plastic coversto protect from the weather and from staining or discoloration during construction (Fig.6.1-1). Units should be handled to avoid chipping or breaking. Aggregates should also be protected against contamination from rain, ice and snow and from blowing dust and soil during construction (Fig. Different aggregates should always

be stored in separate stockpiles (Specs. 2.1.4.4).

Fig. 6.1-1 Protection of Units 6-2


6.1-2).

A C 1 TITLE*flDG

93 M 0bb2949 0508594 Tb3

m

Fig. 6.1-2 Protection of Aggregate Packaged mortar and grout ingredients should

be stored off the ground and covered to

prevent moisture penetration, deterioration and intrusion of foreign materials (Fig. 6.1-3).

Also, packaged mortar and groutingredientsshould manufacturer’s labels intact and legible. Broken packages,

be in originalcontainerswith open containers, or materials

with missing or illegible labels should be rejected. Reinforcement, tiesand metal accessories should beprotected against permanent distortion, and should be stored off the ground to prevent soiling or wetting that could inhibit bond or promote corrosion (Specs. 3.1.3.1). Weather conditionsshould not affect the performance of weather-protected materials. Masonry units properly stored and covered immediately after delivery will remain in good condition. Dry masonry units that are subjected to freezing temperatures may be used in construction without damage to the units or to the masonry under the proper procedures. Wet masonry units that have frozen, however, must be thawed prior to use. 6-3


A C 1T I T L E S N D G

93

m

Obb2949 0508595 9 T T

Fig. 6.1-3 Protection of Mortar and Grout Materials For special material storageand protection requirements duringhot and cold weather, refer to MDG 7.1 and 7.2. 6.1.2 Inspecting Surfaces toReceiveMasonry The Specifications state that, when required, foundations be inspected prior to the start of masonrywork(Specs.

1.5.1.2 and 2.3.2.1). Supporting elements must be levelled within

acceptable tolerances set by the Contract Documents (seeMDG 6.3). Concrete foundations and brick ledges should be inspected for correct conformanceto design, dimensions and for condition of surfaces. Other masonry supports above the foundationlevelshould

be

inspected for correct location. Deficiencies should be noted and corrected by the General Contractor before masonry construction begins. Specs. 2.3.2.1.2 requires that the masonry contractor remove laitance, loose aggregate and 6-4


A C 1 TITLE*flDG

93

0662943 0508596 836

m

other substances which would prevent mortar from bonding to the foundation. 6.13 Masonry Units

Masonry units mustbe examined and sometimes modifiedfor installation. The color, texture and size of units delivered to the job site should be compared with the approved sample panel. Both concrete

and clay units should be checked for cracks, chips, and warpage and

size tolerances as defined by the appropriate ASTM Standard. During construction the product should be maintained in the "as received" condition

or

upgraded to benefit the masonry. On receipt, protected products should be maintained in this condition until use. Nonprotected products may haveto be upgraded prior to use in the masonry. Drying and cleaning of the units may forestall their rejection. The IRA (initial rate of absorption, or suction) of clay brick units must be checked well in advance of construction. Specs. 2.3.2.4(b) requires

that clay brick with IRA's in excess of

one gram per minute per sq in., when measured in accordance with ASTM C 67 field test, be wetted to produce an IRA not to exceed one gram per minute per sq in. when the units are used.Brickshould

be wetted by spray,dip, or soaker hose.Brickmay

immediately before laying, but it is recommended

be wetted

that they be thoroughly wetted 3 to 24

hours prior to use to allow time for moisture to become distributed throughout the unit. Units should be saturated but surface dry when laid. They can be broken in half to check wetting conditions as shown in Fig. 6.1-4. Brick generally should

not be wetted in winter

because some higher suction units in cold weather could produce better bond strength than low IRA units (see MDG 7.2 on cold weather construction). Brick having an IRA of less than 0.10 gr/in.2/min should not

be wetted, and concrete masonry should not be wetted

before placement (refer to Specs. 2.3.2.4(a)

and ASTM C 90).

All masonry units, however, should be clean and free of contaminants such as dirt, oil or sand that might inhibit bond or proper suction. See MDG 3.1.5.

6-5


Dry

Moist

Saturated

Surface

Saturated but Surface

Dry

Fig. 6.1-4UnitMoistureState 6.1.4Reinforcement,ConnectorsandAccessories Per Specs, 3.1.2.l(a) reinforcement, ties and accessories should be checked for correct size and configuration. Before placing reinforcing steel or metal accessories in the wall, Specs. 3.3.2.1 requires that oil, dirt, iceand other contaminants be removed so that good bond with the mortar or grout can be achieved.

6.1.5MortarandGrout Mortar and grout ingredients should be checked for compliance with Contract Documents and with the material storage and protection requirements of Specs. 2.1.4, 3.1.3, and 4.1.4 (refer to MDG 6.1.1). Mortar and grout are usually mixed at the job site throughout the work day, but ready-mixed

and pre-batched mortars and grout are also available. These

newer mortar and grout systems prepared at central batching locations attempt to control field variables that often adversely affect the quality and consistency of mortar.

6.1.5.1Mortar

- Mortar mixturesshould be prepared so thatthe desiredproportions

(either selectedfrom

the proportionspecification

of ASTM C 270 or derivedfrom

laboratory tests in accordance withthe property specificationof ASTM C 270) are obtained. The method of measuring and batching dry ingredients should be either by weight or by

6-6


A C 1 T I T L E s M D G 93

m 0662949

0508598 609

m

volume, so that the specified proportions can be controlled and consistently maintained. Ingredient proportions and properties of the various types of conventional ASTM C 270 mortar are covered in MDG 3.2. Inspection should concentrate not on actual water content, but on assuring batch to batch consistency of the volume of cementitious ingredients and aggregate. The mortar should be prepared by mixing the ingredients in a mechanical drum or paddle

type mixer. A good mix generally results when about three-fourths of the required water, one-half the sand, and all of the cementitious materials are briefly mixed together.

The

balance of the sand is then added, along with the remaining water. Mixing time should usually be a minimum of three minutes and a maximum of five minutes after thelast mixing water has been added. Overmixing causes segregation of ingredients, and also entraps air in the mortar, thus reducing bond strength. After all other ingredients are mixed, specified admixtures and pigments should be added in the approved quantities. Pigments should always be added in pre-batched amounts. Retempering of non-pigmented mortars should be permitted, but onlyto replace water lost by evaporation. Retempering of pigmented mortar may cause changes

of mortar color.

Retempering slightly decreases mortar compressive strengthbut restores bond strength. The amount of strength loss increases with time after mixing. Mortar will begin to stiffen and lose workability as it loses its moisture. Moisture

loss is a function of cement hydration,

wind, temperature, humidity, and time. Mortar that has begun to set should be discarded. 6.1.5.2 Grout - Neither the Specifications nor ASTM C 476 address many of the specifics

regarding grout preparation. ASTM C 476 requires that themethod of measuring materials be such that specified proportions can be controlled and accurately maintained. Required mixing time is a minimum of five minutes. Air-entraining admixtures are not recommended when bond to reinforcing steel is required. The use of expansive admixtures in grouts is

6-7


AC1

T I T L E x M D G 9 3 m Obb2949 0508599 5 4 5 H

encouraged. Considerations should be given to grout preparation so costly surprises don’t occur at the job site such as low strengths or segregation when grout is pumped (6.1.1). ASTM C 476 listsgroutproportions

byvolume.However,testinglaboratories

suppliers typically batch by weight. Volume proportions can

and ready mix grout be changed to weight values

by using the specific gravitiesof the materials. The design criteria for grout should consider compressive strength, durability, grout space, and consistency. The Code requires grout to have compressive strength equal to or exceeding the specified compressive strength of the masonry, f m , but not less than 2,000 psi. The durability of grout is typically not a concern since the grout is usually protected from moisture saturation and not susceptible to freeze-thaw conditions. ASTM C 476 classifies grouts as fine or coarse depending on the maximum aggregate size used. If the maximum aggregate size is less than 3/8 in. the grout is classified as fine; 3/8 in. or larger is classified as coarse grout. Since grout must flow easily into confined spaces, the smaller the space the smaller the grout’s maximum aggregate size. However, coarse grout is preferred when possible because it shrinks less, requires a smaller

proportion of

cement, and is more economical. Specs. Table 4.3.3.4 gives recommended grout types for different grout spaces. For a project that requires grouting of both a collar joint and cells

of masonry units, two different grout mixes might possibly be used on the job. To flow around reinforcement and to completely fill cavities, grout,

whether pumped or

poured, needs to be very fluid. This necessitates a higher water content than for masonry mortars. Specs. 4.2.2.2requires a consistency that is achieved when the slump falls between

8 and 11 in. Use the lower slump for masonry units with low absorption

and the higher

slump for masonry units with high absorption. In addition consideration should be given to temperature and humidity conditions and sizes of cavities in determininggrout consistency. Cavity size affectsthe grout’s contact surface area. The greater the surface area the greater

6-8


AC1

TITLE*MDG 93 m 0662947 0508600 0 7 7 m

the amount of water absorbed. Although a contractor can use volume proportioningon the job, usually a contractor selects a grout producer to providegrout

that economically meets designcriteria.

The grout

producer or test lab will make trial batches using materials that conform to ASTM standards in developing a grout mix that meets design requirements. 6.1.6

Protections

Masonry should be covered at the end of each day and when work is not in progress per Specs. 2.3.3.11. Excess moisture entering

the masonry during construction

can cause the

masonry to become saturated. Such masonry may take weeks or months to dry out. Such prolonged wetting will dissolve even slightly soluble salts,

and may result in efflorescence.

Covers such as water-repellent tarps or heavy plastic sheets should extend at least two feet down each side of the masonry and be held securely in place (Fig. 6.1-5).

Fig. 6.1-5 Wall Protection 6-9



93

m

Obb2949 0508603 T23 W

Mortar boards and scaffold planks are not acceptable as covers. During construction, scaffold boards should be turned on edge whenever work

is not in

progress, so that rain will not splash mortar droppings or dirt onto the face of the masonry and causestaining(Fig.6.1-6).

The base of masonryshouldalso

be protected from

rain-splashed mud andmortar droppings by spreading plastic sheets3 to 4 ft. on the ground and 2 to 3 ft. up the masonry.

Fig. 6.1-6 Scaffold Board Storage

Partially completed work should be braced

per Specs. 2.3.3.9 during construction against

lateral loads from wind or other forces applied before full design strength is attained (Fig. 6.1-7). Bracing should remain in place until sufficient strength

is reached or elements of

permanent construction provide adequate support. Protection of work necessary during extremely hot or cold weather is outlined in MDG 7.1 and 7.2. 6-10



m

0662949 0508b02 9bT H

Adjacent work shouldbe protected from damage during masonry construction. In particular, door and window frames, sills, ledges

and exposed finish materials should be covered to

avoid mortar splatter. During cleaning operations, these elements should also be protected from stains caused by cleaning solutions that run down the masonry.

Fig.6.1-7 6.2

Wall Bracing

PLACEMENT

Specifications give requirements for the placement of mortar, reinforcement, ties, units, grout, flashing and weepholes, and movement joints. This section discussesthe purpose of, Code/Specs. mandated procedures for the placement of items. Fig. 6.2-1 diagrammatically shows many of the items that are discussed.

Joint Reinforcement

Fig.6.2-1

Wall Components 6-11


A C 1 T I T L E v f l D G 93

6.2.1

0662949 0 5 0 B b 0 3 BTb

m

Mortar Placement

The specific requirements of Specs. 2.3.3.3 are: l.

Use bed joint between 1/4 in. and 3/4 in. thick at foundations;

2.

Use 3/8 in. thick joints between units;

3.

Tool all joints with around jointer when the mortar is thumbprint hard, unless otherwise required by the Contract Documents;

4.

Place mortar on clean units while the mortar is soft and plastic;

5.

Do not disturb the unit after it is initially positioned;

6.

Place mortar so that all joints of solid units are fully filled with mortar;

7.

Fill the bed and head joints of hollow units with mortar, spread across the width of the face shells;

8.

Mortar cross webs in hollow unitsfor the following situations: (a) adjacent to cells to be grouted for partially grouted construction, (b) starting course on foundations, and (c) all piers, columns and pilasters that are to be fully filled with grout;

9.

Remove protrusions of mortar into collar joints, cavities and cells of hollow units if they project more than 1/2 in.;

10.

Do not slush mortar into head joints;

11.

Fill all holes in the mortar.

Furrowing of the mortar bed joints is discouraged, since it can reduce the contact area and create voids as shown in Fig. 6.2-2. Specs. 2.3.3.3(e) prohibits deeply furrowed

bed joints.

Slushing to fill the head joints after the units are placed is to be avoided, because slushed mortar will not be placed under compression and may not develop proper contact for bond. The slushing process could also disturb the unit and break the contact which occurs when the unit is initially placed. Disturbing the unit at any time after it has initiallyset breaks the initial bond of the mortar to the unit. Any unit that is disturbed should be removed and reset, using new mortar. 6-12


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m

0662747 0508604 732

m

Fig. 6.2-2 Furrowed Mortar Bed Joint In walls to be fully grouted, only the face shells of hollow units need to be fully mortared. Where only selected cells are to be grouted, crosswebs adjacent to those cells should also be fully mortared. Cells should align. Protrusions that exceed 1/2 in. should be removed in hollow cells and collar joints to be grouted. Protrusions in spaces for collar joints or cells that are to remain open or to be grouted can fall to the bottom and block weep holes, inhibit free flow of moisture along the flashing, or prevent the proper filling of the space by the grout. At levels abovethe flashing inthe wall, mortar that bridges across the collar joint can transfer moisture. Three methods assist in keeping the spaces clean (6.2.1, 6.2.2). Fig.

6.2-3 shows two of the methods that can be employed to keep the spaces clean and free of unwanted mortar protrusions. The properplacement of mortar along the bed jointincludes not placing excessive amounts of mortar and beveling the back sideof the mortar bedbefore placing the next unit. A wooden

strip should be used in the cavity to catch the mortar

droppings and facilitate their removal. A similar system usinga compressible material could be employed in cells to be grouted. A

third method is to use cleanout openings

6-13


at the

A C 1 TITLExMDG 93

I

m

Obb29Y9 0508b05 6 7 9

m

Beveled Bed Joints-,

nn

Fig. 6.2-3 Techniques for Maintaining a Clean Cavity 6.2.2 ReinforcementPlacement

Placement of reinforcing bars and joint reinforcement are quite different. Reinforcing bars must always be embedded in grout; joint reinforcement is embedded in mortar and grout. Individual ties and anchors are discussed in MDG 6.2.3. Code 1.2.1 requires that the size, grade, type, and location of reinforcement be shown on

the project drawings and typical

details. These requirements are to be followed during the placement. If this information is not provided in the contract documents, the contractor should obtain that information before proceeding with the work. It is generally acceptable to lay joint reinforcement directly on top of the masonry course. Mortar is then spread over the wire and face shell in one operation. Due to irregularities in the masonry

and the wire, mortar surrounds the wire and provides strength 6-14


A C 1T I T L E * f l D G

73

m

Obb2747 0508bOb 5 0 5

development.Fullscaletestshaveconsistentlyverifiedthisphenomenon.

m

It is not

recommended to place the wire between thin layers of bed-joint mortar, since this mortar has a tendency to dry out and lose bond. Cover requirements for joint reinforcement per Specs. 3.3.3.4(d) are 1/2 in. on the inside face and a minimum of 5/8 in. on the exterior face of a wall; these are clearly less than that recommended for other reinforcement. The placement of reinforcement has to be carried out carefully to ensure that there is enough grout around the bar, and to ensure that the bar is located where theA/E intended. In order to guarantee proper bond between reinforcing steel and grout, bars must have a clear distance from any face of masonry or formed surface of 1/4 in. for fine grout and 1/2 in. for coarse grout per Specs. 3.3.3.4. In order to ensure that bars are in the proper location to resist stresses, bars mustbe placed within specific tolerances as listed in Table 6.2.1 per Specs. 3.3.3.2. If interferences exist that require movement of the reinforcement greater than one bar diameter or the specified tolerances, the A/E or the owner’s designated representative should be notified as stated in Specs. 3.3.3.2(c).

Table 6.2.1 ReinforcementPlacementTolerances

I Distance From Centerline of Steel To the Opposite Face Element

of Masonry > 8 in. but S 24 in.

> 24 in. Walls and Flexural Elements 2 112 in. f 1 in. & 1 1/4 in. For vertical bars, 2 in. from location along length of wall Walls indicated on the Project Drawing S

8 in.

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m Obb2747

0508b07 4 4 1

Reinforcing bars must be held in position during grouting. This ofways;

one of themostreliableis

can be done in a number

to use reinforcing bar positioners. There are no

requirements for the spacing of these positioners, but they are generally located at the bottom and top of the wall and at 10 foot intervals. Typical reinforcing bar positioners are shownin the commentary inSpecs.C.Fig.3.3-1.

Galvaniccorrosion can occurwhen

dissimilar metals are placed in contact with each other. Therefore Specs. 3.3.3.1 states that contact of dissimilar metals is not to occur. Reinforcing bar positioners should either be of plastic, or of the same material as the reinforcement. Reinforcement is not to be bent after being embedded in the mortar or groutunless approved by the PJE or the owner’s designated representative per Specs. 3.3.3.4(c). Bending of the reinforcement can

break bond, create voids in mortar or grout, and weaken the

reinforcement.

To ensure the adequate transfer of the forces that the reinforcement is intended to resist, Specs 3.3.3.4(b) stipulates that all splices madeto connect reinforcing together should be as indicated in the contract documents or as approved by the A/E or the owner’s designated representative.

6.23

Tie and Anchor Placement

The Specifications requirements for placement of individual ties and anchors are the same as for reinforcement. Code 4.3 states that the type, size and location of connectors shall be shown or indicated in the contractdocuments, and that these requirements are to be followed during the placement. While it is not the intent of the Code that every connector must be shown, there should be enough information on the drawingsto make it clear to the contractor whatisrequired.

These requirements are necessary to achieve the design

strength and provide proper connection. If this information is not provided in the drawings and specifications, the contractor should obtain the information before proceeding with the work. 6- 16


AC1

TITLEaMDG 9 3 m Obb2949 0508608 388 m

Proper interactionbetween

the tiesandanchorsand

the masonry elements relieson

sufficient bond to the embedded portion of the tie or anchor in the mortar or grout. Per Specs. 3.3.2.1 the embedded portions per Specs. 3.3.2.1 should be kept free of laitance, dirt, sand, oil, or debris that will inhibit bond with the

mortar and grout. By Specs. 3.3.3.5(c)

portions of the ties and anchors are not to be bent after being embedded in the mortar or grout unless approved. Bending of these items can break the bond that exists, create voids in the mortar or grout and weaken the member through improper bending techniques. The Code and Specifications do not require the use of extra wall ties around the perimeter ofwall panels and around openings.Both

the BIA and NCMA recommend extra ties.

Using those recommendations, wall ties should be spaced at 12 in. around openings and at the edges of masonry walls. Code 5.14 coversstructuraldesignaspects

of anchor boltssolidlygroutedinmasonry.

Spacing and location are addressed in light of potential failure modes. Wall ties and anchors can be placed directly on top of masonry units and covered with the bed jointmortar; alternatively the embedded portion canbe placed in the fresh mortar prior to the placement of the next unit, so that the embedded portion is fully surrounded and the specified reinforcement minimum coveris achieved. Because joint reinforcement should be placed in fresh mortar, two-piece ties are better suited for multiwythe walls in which the wythes are not constructed simultaneously. to be embedded at least For solid units and hollow units with cells filled solid, wall ties need

1-1/2 in. per Specs. 3.3.3.5(a). Minimum wall tie cover requirements from the exterior face of the masonry are not given; however the requirements for joint reinforcement serve as a good guideline. Hence, for a 3-5/8 in. brick unit the wall ties can be embedded from 1-1/2 in. to 3 in. into the brick. This allows standard sizes of wall ties to be used for a variety of cavity widths.

6-17


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m

Obb2949 0508609 214

m

Specs. 3.3.3.5(a) also stipulates that for hollow units, ties must extend to the outer face shell and be embedded at least 1/2 in. into it. This is clearly less embedment than that typically obtained with solid units. For this reason, anchors extending into grouted cores should be considered. The placement of connectors on mortared crosswebs of hollow units is not recommended since crosswebs seldom line up. 6.2.4 UnitPlacement

The placement of the units, Specs. 2.3.3.3, is an operation that occurs in conjunction with the placement of the mortar. The requirements governing placementof mortar are equally important in the placement of the units. The plastic mortar serves as a separator to allow for dimensional tolerances in

the units. The hardened mortar is the bonding agent that

holds the individual units together

to act as acompleteassembly.Unless

the units are

properly placed, the mortar bond will be weakened or destroyed, creating paths for air and water infiltration. All units should be cut with a masonry sawor other means that will not damage the exposed manufactured faces or edges. Wet cutting of units will add moisture to the units. Failure to allow the moisture to evaporate before the unit is set in the mortar could inhibit bondby decreasing the unit’s ability to absorb water-cement matrix. Concrete masonry units must be allowed to dry before laying to reduce subsequent shrinkage. Dry cutting is preferred if the units are to be set immediately after they are cut. Bond strength beginsto develop the instant the unit comes in contact withthe mortar. This makes itessential that theunit is shoved tightlyinto position so that thejoint size is achieved without disturbing the unit. Disturbing the unit after it has been placed breaks the initial bond of the mortar to the unit and will reduce bond. Per Specs. 2.3.3.3(f)1 any unit that is moved after initial set will require: the unit be removed, mortar to be removed from the unit and masonry be reset after placement of new mortar. Fig. 6.2-4 shows the unit being properly shoved into position. To enable the appropriate amount of water cement matrix 6-18


to be absorbed from the mortar by the unit, the units should be free of laitance or other bond inhibitors, such as dirt, sand, oil, or debris that will inhibit suction by the units. The placement of the units should be soon after mortar placement to prevent water loss from the mortar by evaporation.

Fig. 6.2-4 UnitPlacement The performance of the units during the normal course of construction may require some modification. On dry, normal temperature days, absorptive clay masonry units may haveto be wetted to reduce the initial rate of absorption. See MDG 6.1.3 and Specs. 2.3.2.4. Water additions should be adequate to alter theabsorption characteristics to the desired range at the time of laying, i.e., 5 to 30 grams of water per minute per 30 sq in. of masonry unit. Bond strength will be increased, provided the units are essentially uniform in their rate of suction. Concrete masonry units normally should not be wetted before use in masonry per Specs. 2.3.2.4(a).

6-19


By altering normal constructionpractices, performance of the masonryassemblyin

the masoncontractorcanalsoinfluence

the

the completedconstruction.Duringdry,high

temperature periods, the mortar bed length laid in advanceof placing the subsequent course must be reduced(Specs. 2.3.2.3).

See MDG 7.1. The "pick and dip"method,whereby

mortar sufficient for one unit is laid and the upper unit is immediately laid, is one method for increasing the bond strength of the masonry under these conditions. Moist curing of masonry will promote further hydration of the cement and increase all performance characteristics, most significantlythe bond strength of the masonry. The field application of excess water should be avoided as soluble bases and salts may be dissolved and concentrate on the surface of the masonry as efflorescence. See MDG 6.1. Cells in hollow units to be grouted are to be aligned per Specs. 2.3.3.3(d)4. Consideration should be given to the coring pattern of the unit so that core alignment can be achieved. 6.2.5

GroutPlacement

The preparation of spaces to be grouted andthe

placement of the groutshould

be

performed in accordance with Specs. 4.3 and with the project drawings and specifications. Prior to placing the grout, the cells or voids to receive the grout should be inspected to ensure that they are clean and free of all debris that could inhibit the free flow of the grout and complete filling of the space. All reinforcement in place shouldbe inspected to ensure that it is properly located, supported to minimize movement, and has sufficient surrounding space for grout. Proper bond between the grout and the reinforcernent and the proper placement of the reinforcement will not occur if the reinforcement is forced into the grout. All required cleanouts should be in place and fillers ready to plug the cleanouts. Cleanouts

are required at vertical reinforcement bar locations and at a minimum horizontal spacingof 32in.when

the groutpourexceeds

5 ft inheight.

The minimumcleanoutopening

dimension shall be at least 3 in. and of sufficient size to permit removal of the debris. See reference (6.2.3). 6-20


Grout should be placedwithin 1 1/2 hours from the time water is introduced into the mixture to ensure that the proper flow and bond is achieved. Cold

weather procedures

discussed in MDG 7.2 apply to the placement of grout. The grout shall be confined to the spaces to be grouted without inhibiting the bond between the masonry unit and mortar. Grout is poured in lifts not more than 5 ft high. Lifts should

not be terminated at a bed

joint in the masonry. Stopping 1-1/2 in. below the bed joint is recommended. One or more lifts comprise a grout pour. A grout pour is the total height of grout placed in a masonry wall before constructing additional masonry; it usually should not be higher than 24 ft. The maximum height of a grout pour is limited by the type of grout (fine or coarse) and the size of the grout space. See Specs. Table 4.3.3.4. This table gives maximum grout pour height

for minimum width of the grout space or minimum grout space of cells in hollow units when either fine and coarse grouts are used. The maximum lift heightsand pour heights specified for each project must take into account the ability of the wall, with or without bracing, to resist the fluid grout pressure without damage to the wall. Each lift of grout should be properly consolidated and reconsolidated per Specs. 4.3.3.6 to ensure complete filling of the space, surrounding of the reinforcement, and the filling of voids created by water loss and settlement. The Specifications require mechanical vibration of the grout during the placement and reconsolidation after the initialwaterloss

and

settlement for all pours exceeding 12 in. high. Pours 12 in. or less in height need only be vibrated or puddled. The vibrator size and type, velocity, time in the grout and spacing of the points vibrated are a function of the type of grout and the size of the space being grouted. Generally, a low velocity vibrator placed in cells

12 to 16 in. apart for one to two

seconds is considered sufficient. Additional discussions on theconventional methods of low lift groutingand high lift grouting are found in reference (6.2.4). In some parts of the USA full-height grouting(24 foot height of wall in one lift at one time) has been successfully used and is acceptable by the UBC

(6.2.5). 6-21


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Obb2949 0 5 0 8 b L 3 745

m

6.2.6 Flashing and WeepholePlacement

Flashing and weepholes are required to be in accordance with the Contract Documents by Specs. 2.1.1.1 and are mentioned in Specs. 2.3.3.6(f). Proper placement of the flashing and weeps is required to allow them to perform their intended purpose. Thus suggestions are presented in the MDG.

The specified flashing and weep systems should be installed in a manner that will direct water to the exterior of the wall without allowing it to flow to the interior. Flashing should collect and contain the water; the weepholes providea conduit for the water to travel to the exterior. Typically flashing is installedat all interruptions in the vertical plane of a masonry wall, such as tops of the foundation, above shelf angles, over openings, above bond beams, etc. All lap joints should be sealed with adhesive to maintain continuity of the flashing and prevent intrusion of water into the exterior wall and the interior of the building. End dams should be used where required; i.e., at ends of flashing runs. The flashing should alsobe installed so that it channelsthe water to the exterior of the wall. This requires that the flashing extend to or beyond the exterior face of the masonry. Some flashing materials cannot maintain a permanent configurations such as a formed drip with

a hem. Weepholes transfer watercollected by flashings to the wall exterior. Theyshould be installed and protected during construction to maintain drainage. Types of weepholes used

in today’s construction are open head joints, open head joints filled with louvers or cellular material, cotton sash cord, and polyethylene tubes. Cotton sash cord and the polyethylene tubes do not provide a direct path for the water to follow. The base of the flashing should be keptclean to prevent accumulated debris from blockingthe open path. Small open holes and polyethylene tubes can be clogged by much smaller pieces of mortar or other debris than open head joints. Open head joints result in a larger opening for positivedrainage, but could also allow intrusion from bugs,

dirt and water. Louvers or cellular material should 6-22


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93

m

0662949 0508634 bBL

m

inhibit intrusion while still allowing drainage.Cotton sash cord weeps remove the water by absorption and evaporation and may become clogged by remaining salts. This is a slower means of removing the water than open paths.However,

the sashcord also prevents

infiltration. To allow for significant absorption it is best to leave an 8 in. minimum tail of sash cord in the cavity. Open weepholes are normally recommended to be spaced at a maximum of 24 in. on center, and sash cord or other wick type at a maximum of 16 in. on center. Care must be taken when sealants or paints are applied in the area of weepholes to prevent clogging or coating the weeps and inhibiting the flow or evaporation. 6.2.7 MovementJointConstruction

Code 5.2.4 requires considerations of the effects of forces and deformations involved in movement due to manyfactors.

MDG Chapter 10 specifically addresses suchfactors.

Control joint and expansion joint materials are specified in Specs. 2.2.6; the installation is required to be in accordance with

the Contract Documents by Specs.2.3.3.6(g).

Proper

placement of the joints is required to allow them to perform their intended purpose. Control joints and expansion joints should be installed in a manner that will allow for the expected movement and provide a watertight condition. These joints should be located and detailed on the drawings and described in the project specifications. If this information is not provided in

the contract documents, the contractor should request it before

construction. Control joints in concrete masonry create a weakened section forcing cracking to occur at that predetermined locationwheretensilestress

from restrained shrinkage

exceed the material strength. Expansion joints in clay brick masonry are intended to allow for the expansion from thermal effects, moisture effects,and freeze-thaw effects. Both joint types should have their exterior surface sealed to prevent water penetration.

Control joints in concrete masonry are typically constructedby aligning a vertical head joint, raking back the mortar in the joint to create a weak plane for the crack to occur and installing sealant to prevent migration of moisture. The ends between the units may be 6-23



93

m

Obb2949 0508635 5LB

laterally supported with an interlock, cross-shaped gasket,

m

or cementitious fill and bond

breaker such as building paper is installed to prevent bond to one side. Fig. 6.2-5 shows three methods of constructing control joints.

Preformed Gasket

Special

Joint Units Units

Rake Joint and Caulk

Fig. 6.2-5 Control Joint Wpes

Expansion joints are made by leaving an unobstructed void in clay brick masonry, thus allowing the two way movement. joints,keeping

These joints are constructed by aligning vertical head

the joints free of all mortar or other materials that couldrestrict

the

movement of the masonry units toward each other, and sealing the exterior with a sealant and backer rod or a manufactured joint cover to prevent moisture from entering the joint. Compressible materials such as those shown in Fig. 6.2-6 can be placed in the joint during construction and left in placeif they havethe ability to be compressed the necessary amount to prevent damage to the units.

6-24


AC1

T I T L E x M D G 93 W 0662949 0 5 0 8 6 3 6 4 5 4 m

/-2o -Sealant

OZ’

foam 7Premolded rubber or plastid

Copper

and backer rod

-Sealant

and backer rod

/-Extruded pad -Neoprene

-Sealant

plastic

and backer rod

-Sealant

and backer rod

Fig. 6.2-6 ExpansionJoint mpes 6 3 TOLERANCE

63.1 Introduction Construction is not an absolute and completely controllable process. As such, consideration for dimensional limitations of materials and systems and for workmanship is essential. Every building component and construction operation is subject to dimensional variations which must be understood and allowed for in the design and constructionprocess.Thismust include not only the masonry components and assemblages, but also systems such

the other building

as foundations, structural frame, floor slabs and finishes. The final

manufactured or fabricated product must be integrated with the ongoing construction. be detailed and identified as totolerances. During the design process, building systems must

This establishes the dimensional relationships among the various building systems. These dimensional tolerances accommodate the needed variations and allow the various systems and components to form an integrated building. It should be anticipated that design dimensions applied during constructionare not absolute but in fact vary. Some aspects in design are more critical than others and require more

6-25


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93

m 0662949 050BbL7 390 m

attention. These variations are the difference between the theoretical size and location (as designed) and the finished size and location (as-built). The design dimension (absolute) combined with the tolerance dimension (controlled variation) locates within the established limits. This anticipated dimensioned range than the design dimension. This method

the design elements

can be larger or smaller

of dimensioned work in design provides

for the

variation so materials and components will fit when constructed. The constructionprocessconsists

of a series of independent actions that must be

coordinated to result in a predictable product. example, windows must fit

All of the building systems must fit.

into openings, concrete foundation walls must align with

For the

construction above, steel supports must be level and align within the wall. These are real every day conditions that need to be considered. The different trades working together as

a team will coordinate with each other on how their work will interface with existing and

of anticipating tolerance requirements can

future construction. Only through this process

the completed construction meet the design requirements. 63.2

AC1 530.UASCE 6/TMS 602 Reference

Most successful construction work is based

on repeated successes involving professionals

working together in a timely manner. Design professionals maintain these successes from project to project by following the standards recommended by each industry. The masonry industry has provided recommended tolerances. Specs.

2.3.2.1, 2.3.3.2 and 3.3.3.2. include

the present masonry standards on tolerances. Specs. 2.3.2.1 addresses foundation tolerances, Specs. 2.3.3.2 presents tolerances on erecting masonry, and Specs. 3.3.3.2 covers tolerances on placing reinforcement. 633 Tolerance

Product placement or fabricated units shouldbe located at the specified location withinthe established deviations. The theoretical line or plane should be plumb, in alignment or level

6-26


93 W Ob62949 0508618 2 2 7


m

for that reference construction. These conditions are defined as follows: (1) alignment (plan or elevation oriented) is an in and out placement variation (tolerance) from the design dimension (absolute) measured from the control reference; (2) the level (elevationoriented) is an up and down placement variation (tolerance) from the design dimension (absolute) measured from the control reference in elevation- floor elevation; and (3) plumb (vertically oriented) is an in and out placementvariation

(tolerance) from the designdimension

(absolute) measured from the control reference in section, see Figs. 6.3-1, 6.3-2, and 6.3-3. In the three figures the actual element variation is not shown but would fall within the boundary represented by C. These dimensionaldeviationsshould

be annotated in the

contract documents where required.

Also, tolerance restricts the size of construction elements. For example, the thickness of the mortar joint, the collar joint width, the thickness of the masonry walls, the size of openings in the wall and other conditions are controlled to account for variations in size of masonry units and workmanship of the mason.However,

the specifications does not nowgive

dimensional tolerances for movement joints. Tolerance requirements set the limits for how workshould

be performed to fitwithinthedesign.

Remember, tolerances are not

cumulative.

...

?-

Plan View (Looking Down) Legend A

-

B-

C-

Control Reference;i.e., Grid Centerline Design Dimension (Absolute)

D

-

Fig. 63-1 Line Tolerance

6-27


Tolerance Range (Limited Variation) Theoretical Line (Absolute Location)

I

A

B

-

Elevation View Legend ControlReferenœ; i.e. Finish Hmr Elevation Top of Steel or Concrete Top of Foundation Design Dimension (Absolute)

C

-

D

-

Toleranœ Range (Limited Variation) Theoretical Line (Absolute Location)

Fig. 63-2 LevelTolerance

Vertical Section View

-

Control Reference; i.e. Grid Centerline Vertical D Orientation Dimension BDesign (Absolute) A

Legend

C-

Tolerance Range (Limited Variation) - Theoretical Line Location) (Absolute

Fig. 6.33 PlumbTolerance

6-28


A C 1 TITLESMDG 93

m

0 6 6 2 9 4 9 O508620 985

m

63.4 Tolerance Examples 63.4.1 Mortar Joint - The most basic tolerance contribution in the exterior masonry wall construction is the variation in the mortar joint thickness. This variation compensates for the nominal changes in actual size in the manufactured masonry units and for the dexterity and skill level of the mason in determining the amount of mortar that should be placed on each joint. The Specs. 2.3.3.2(a)2 requirement for the mortarbed joint states thattolerance should not vary more than +/- 1/8 in. If the design thickness is 3/8 in., the maximum bed joint thickness variation would be from a minimum of 1/4 in. to a maximum of 1/2 in. The same design thickness for the mortar head joint could vary from the design value by -1/4 in. to +3/8 in. as given in Specs. 2.3.3.2(a)2, equivalent to a minimum joint thickness of 1/8 in. to a maximum width thicknessof 3/4 inch. These variations in the mortarjoint thickness are in part caused by the unit variation and the workmanship of the individual mason.

63.4.2 Masonry Openings - From Specs.2.3.3.2(a)lamasonry elevation dimension from -1/4 in. to

opening can varyin

+ 1/2 in. If the actual masonry opening width is 4 ft-O

3/8 in., the actual permissible dimension could range from 4 ft-O 1/8 in. to 4 ft-O 7/8 in. If

the A/E designs the sealant joint width to be 1/4 in. as shown in Fig. 6.3-4, the sealant will fit only if the masonry openingis constructed accordingto the design dimension which isnot always possible. The window manufacturer will fabricate the window to 3 ft-11 7/8 in. A potential for the masonry opening being constructedat the smaller dimension exists, leaving the possibility for the joint width to be O in. on one side and 1/4 in. on the otherside, or 1/8 in. on both sides (see Fig. 6.3-5). Three out of the four combinations are not acceptable relative to the minimum sealant width requirement of 1/4 in. Therefore, wider perimeter sealant joints should be designed. To anticipate variations inthe construction of a masonry opening, the minimum perimeter sealant joint should be 1/2 in. wide instead of the 1/4 in. shown in the example.

6-29


A C 1 TITLE*NDG

W,

93

m 0662949

050Bb2L Bll

m

4'- O 318" M .O.

3"11 718"

W .o.

114" Sealant

Aluminum Frame

Glass

M.O.- Masonly Opening W.O.

Design Dimensions Legend

- Width Window

Fig. 63-4 Comer of Window as Designed (Elevation) M.O.constructed -1/4" from the design dimension

-

Unacceptable Sealant Width

AluminumFrame

r F -

M.O.- Masonry Opening

Constructed Condition Legend

Glass

W.O.- Width Window

Fig. 63-5 Comer of Window as Built (Elevation) 6-30


TITLE*HDG 9 3 m 0662949 0508622 758 m

AC1

63.43 Vertical Expansion Joints

- Vertical expansion

joints in brick masonry are often

undersized; and one such aspect not considered usuallyis tolerance. Other factors, such as aesthetics, have causedthe undersizing of this type of joint in cases, where narrow joints are desired to match the design width of the mortar joint. In fact no expansion joints at all are used in some projects. If, for example, the joint width were designed for 3/8 in., this width would be inadequate if construction tolerances were

not considered. See Fig. 6.3-6. The

tolerance for expansion joint width must be specified. The tolerances should be such that the minimum sealant joint width as recommended by different sealant manufacturers is met. This sealant joint would not perform under these circumstances.

Backer -

I

I Design f

Tolerance As Specified

Fig. 63-6 Undersized Expansion Joints 6.4

CLEANING

Even with construction protection in place,some mortar smears and splatters will inevitably occurduringconstruction.Whilefresh

mortar splatters can be cleanedfairlyeasily,

6-31


removing stains may not be as simple. Cleaning operations affect the finished appearance

of masonry. Improper cleaning materials or procedures can often be the source of difficult or permanent discolorations.Specs.2.2.7addressesmasonrycleaners.

For specific

information on identifylng unknown stains and determining optimum cleaning methods, refer to Grimm, Tleaning Masonry - A Review of the Literature" (6.4.1). 6.5

QUALITYASSURANCIYQUALITY CONTROL CHECKLIST

MDGChapters 5, 6, and 7 deal withmasonryconstructionmaterials,testing,quality assurance, and qualitycontrolprovisions

related to the Code and Specifications. The

Specifications provide either mandatory or optional quality assurance and quality control requirements compiled in MDG Table 6.5.1, Quality Assurance/Quality Control Checklist. This table references sections of the Specifications and MDG and states provisions are mandatory or optional.ThisChecklistshould

whether the

be applied to allmasonry

projects. Quality assurance and control requirements can vary from project to project. A requirement for one project can be different for another. The design dictates what aspects are tobe used and what importance it will have for that condition. Generic examples should not be applied without evaluating each aspect under consideration. These conditions have to be evaluated for each project. Application of the Checklist to an individual project should begin with the identification of the materials to be used in the construction. This willallow immediate identification of those itemsthat are eithermandatory or not applicable. For example, certificationand shop drawings for reinforcing would notrequire shop drawings sincejoint reinforcing is considered to be a tie and not reinforcement. To complete the Checklist for an individual project, the remaining items must be analyzed to determine their importance tothe

successful

completion and serviceability of the building. An example of this would be where a specific manufacturer's product is specified and no substitutions are allowed. In this case samples may not be necessary once certification is received that these products will be used.

6-32


A C 1 T I T L E W N D G 73

m

Obb2949 0508624 5 2 0

m

To illustrate the application of this Checklist, MDG Table 6.5.2 is a suggested application to the threedistinct masonry buildings: 1) T M S Shopping Center, 2) DPC Gymnasium, and 3) RCJ Hotel that arespecifically presented in MDG Chapter 9 and used throughout MDG

Chapters 9 through 16 for application of the structural design methodology and provisions found in the Code. Each building can have different wall construction types (options) as follows:

TMS Shopping Center Unreinforced Concrete Masonry (A) Reinforced Concrete Masonry (B) DPC Gymnasium Unreinforced Brick-Block Noncomposite (A) Unreinforced Brick-Block Composite (B) Reinforced Brick-Block Composite (C) Reinforced Hollow Clay Masonry (D) RCJ Hotel Unreinforced Brick-Block Noncomposite (A) Reinforced Hollow Clay Masonry (B) Based on thewall systemand the significance of the facilities, Table 6.5.2 indicates the most appropriate options: (1) Mandatory = M; (2) Not Applicable = -- (3) Recommended = R, (4) Not Recommended = N.

REFERENCES 6.1.1 "Designing Grout Mixes," Magazine of Masonry Construction, Addison, Illinois, June 1991, PP. 218-220.

6-33


A C 1 T I T L E * M D G 93

6.2.1

Ob62747 0 5 0 8 b 2 5 4b7

m

"Recommended Practice For Engineered Brick Masonry," BrickInstitute of America, Reston, Virginia, 1969, p.263.

6.2.2

Brick Technical Note 7B, "Water Resistance of

Masonry Construction and

Workmanship Part III of III," Brick Institute of America, Reston, Virginia, April 1985, PP. 3-4. 6.2.3 "How to Place Grout," Magazine of Masonry Construction, Addison, Illinois, June

1991, PP. 216-217. May 1988, 6.2.4 "High Lift Grouting," Magazine of Masonry Construction, Addison, Illinois,

PP. 60-62. 6.2.5

"Grout 24-Foot-High Wall inOne Lift," Magazineof Masonry Construction, Addison, Illinois, April 1989, pp. 137-139.

6.4.1

Grimm, C. T., "Cleaning Masonry - A Review of the Literature," Arlington, Texas, Construction Research Center of the University of Texas at Arlington, 1988.

6-34


AC1

TITLE+MDG 93

0662949 0508626 3T3

m

Table 6.5.1 Quality Assurance/Quality Control Checklist

O = Optional

M = Mandatory

Quality Assurance (Owner’s effort to measurequality and determine its acceptability)

Certification

Brick Units (Specs. 2.1.2.1 h) ......................................O CMU Units (Specs. 2.1.2.lh) ...................................... O Mortar Mix (Specs. 2.1.2.lh) ......................................O Grout Mix (Specs. 2.1.2.lh,4.1.2.2) ................................. O Reinforcing Steel (Specs. 3.1.2.lb) .................................. M Joint Reinforcing (Specs. 3.1.2.1~) .................................. M Anchor Bolts (Specs. 3.1.2.1~) ..................................... M Ties and Anchors (Specs 3.1.2.1~) .................................. M M Metal Accessories (Specs. 3.1.2.1~)..................................

Hot Weather Construction (Specs. 2.1.2.ld) ........................... Cold Weather Construction (Specs. 2.1.2.1~)........................... Cleaning Method and Materials (Specs. 2.2.7) .........................

O O O

Material Samples

Brick Units (Specs.2.1.2.la) ...................................... O CMU Units (Specs.2.1.2.la) ...................................... O Colored Mortar (Specs.2.1.2.la) ................................... O Sample Panel (Specs. 2.1.2.2) ...................................... O Joint Reinforcing (Specs. 3.1.2.1~) .................................. O Anchor Bolts(Specs. 3.1.2.1~) ..................................... O Ties and Anchors (Specs. 3.1.2.1~) .................................. O Metal Accessories (Specs. 3.1.2.1~).................................. O Flashing and Joints (Specs. 2.1.2.lg) ................................. O Shop Drawings

Reinforcing Steel (Specs. 2.1.2.lf, 3.1.2.la) ............................ M Lintels and Door Frames (Specs. 2.1.2.lf’ 3.1.2.la) ...................... M Shelf Angles and Lintels (Specs. 2.1.2.lf, 3.1.2.la) ...................... M

6-35



73

m 0662749

0508627 2 3 T D

Table 6.5.1 Cont’d. ~

I

Quality Control (ContractoWManufacturer’seffort to achieve a specified result)

Pre-Construction Testing

Brick Units .Compression ASTM C 67 (Specs. 2.1.3.2b) ................. M Brick Units .IRA C 67 (Specs. 2.3.2.4b) ............................. O CMU Units .ASTM C 140 (Specs. 2.1.32) .......................... M Structural Granite .Compression ASTM C 97 (Specs. 2.1.3.2~) ............M Building Sandstone.Compression ASTM C 120 (Specs. 2.1.3.2~) ........... M Structural Slate .Compression ASTM C 99 (Specs. 2.1.3.2~)............... M Exterior Marble Building Stone-CompressionASTM C 170 (Specs. 2.1.3.2~) ... M Limestone Building Stone .Compression ASTM C 121 (Specs. 2.1.3.2) ...... M Masonry Modulus of Rupture (MDG 5.4.1.5) .......................... O Freeze-Thaw Durability (MDG 3.1.4) ................................ O Prism Compression Strength (Specs. 2.1.3.3) ........................... O Mortar Compression Strength (Specs.2.1.2.lb,2.1.3.4) ................ O Grout Compression Strength (Specs.1.6.2.1,1.6.2.2,2.1.3.5,4.1.3) .......... O Grout Slump (Specs. 4.2.2.2) ...................................... M Construction Testing

Brick Unit Compression (Specs. 2.3.1.ld) ............................. CMU Unit Compression (Specs, 2.3.1.ld) ............................. Prism Compression Strength (Specs. 2.3.1.ld) .......................... Field IRA (MDG 5.4.1.1) ........................................ Mortar (Specs. 2.3.1.lb) .......................................... Grout (Specs.2.3.1.lb,4.1.3) ......................................

M M M O O O

Review Manufacturer‘s Reports

Brick(Specs.2.3.1.la) ........................................... CMU (Specs. 2.3.1.la) ........................................... Mortar (Specs. 2.3.1.lb) .......................................... Grout (Specs. 2.3.1.lb) ........................................... Cement Materials (Specs. 2.3.1.le) .................................. Aggregate(Specs.2.3.1.le) ....................................... Reinforcing Steel (Specs. 2.3.1.le) .................................. Ties and Anchors (Specs. 2.3.1.le) ..................................

I’

11

6-36


O O O O O O O O

A C 1 TITLExMDG 93

Obb2747 O508628 176 m

Table 6.5.1 Cont’d.

Inspection

Delivery, Storage, and Handling (Specs. 2.3.1.2b) ....................... O Brick Units (MDG 6.1.3) ......................................... O CMU Units (MDG 6.1.3) ......................................... O Mortar Mixing (MDG 5.3.5) ...................................... O Grout Mixing (MDG 5.3.5) ....................................... O Reinforcing Steel (MDG 6.1.4) .................................... O Joint Reinforcement (MDG 6.1.4) ..................................O Ties and Anchors (MDG 6.1.4) ....................................O Flashing and Weephole Materials (MDG 6.2.6) ........................ O Movement Joints (MDG 6.2.7) ..................................... O Foundation Tolerances (Specs. 2.3.2.1.1) ............................. M O Dimensional Tolerances (MDG 6.3) ................................. Foundations (Specs. 1.5.1.2) ....................................... M O Attend Pre-Bid Meeting (MDG 5.3.2) ............................... Attend Pre-Construction Meeting (MDG 5.3.2) ........................ O Procedures

Hot Weather Construction Procedure (Specs. 2.3.2.3) .................... M Cold Weather Construction Procedure (Specs. 2.3.2.2) ...................M O Unit Placement (MDG 6.2.4) ...................................... Steel Placement (MDG 6.2.2) ..................................... O Tie and Anchor Placement (MDG 6.2.3) ............................. O Mortar Placement (MDG 6.2.1) .................................... O Grouting (MDG 6.2.5) ........................................... O Flashing and Weephole Placement (MDG 6.2.6) ....................... O Movement Joint Placement (MDG 6.2.7) ............................. O Cleaning Procedure (MDG 6.4) .................................... O Maintain Daily Log (MDG 5.3.4) ................................... O


A C 1 TITLE*UDG

m

93

0662747 0508627 002

m

Table 6.5.2 Quality Assurance/Quality Control Checklist For Masonry Buildings

--

R = Recommended for this project N = Not recommended for this project

= Not applicable to this project

M = Mandatory per Specs. 530.1

q E E E Wall Construction Type

I

Quality Assurance Certification

Brick Units CMU Units

I

t

"

R

"

R R

R

R

R

R

--

R

Reinforcing Steel

--

M

Joint Reinforcing

M

M

Anchor Bolts

M

M

Ties and Anchors

M

M

M

M

Metal Accessories

M

M

M

M

Hot Weather Construction

R

R

R

R

R

R

R

Cold Weather Construction

R

R

R

R

R

R

R

Cleaning Method and Materials

R

R

R

R

R

R

R

Brick Units

R

R

R

R

CMU Units

R

Mortar Mix

I Grout Mix

II

R

R "

M

"

R M

M M

M

PIWedureS

-

Material Samples

Colored Mortar

"

R "

R

"

I

Sample Panel

R

R

R

R

Joint Reinforcing

R

R

R

R

6-38


R

A C 1 TITLE*MDG

93

m

0bb2747 0508b30 8 2 4

m

Table 6.5.2 Cont'd.

I

I

1

DPC Gymnasium

Building

RCJ Hotel

Anchor Bolts Ties and Anchors Metal Accessories Hashing and Joints

II

Shop Drawings

Reinforcing Steel

__

M

__

--

Lintels and Door Frames

M

M

M

M

M

Shelf Angles and Lintels

M

M

M

M

M

6-39


"

M

M

M

M

M

M

M

M M


Obb2949 0508b3L 760

Table 6.5.2 Cont'd.

Building

DPC Gymnasium RCJ

Wall Construction Type

A l B l C l D

I

Hotel

CMU Unit Compression

Prism Compression Strength Field IRA Mortar

--

- - I R I R I R

Grout

I

R

Review Manufacturer's Reports Brick Units CMU Units Mortar Grout Cement Material Aggregates Reinforcing Steel

RIR

Ties and Anchors Inspection

Delivery, Storage, & Handling

R

R

R

Brick Units

R

R

R

R

R

CMU Units Mortar Mixing Grout Mixing Reinforcing Steel

x-

Joint Reinforcement Ties and Anchors Flashing and Weephole Materials

"

Movement Joints Foundation Tolerances

M I M I M

>

6-40


" "

M


93 H 0 6 6 2 9 4 9 0508632 b T 7

m

Table 6.5.2 Cont'd.

Hot Weather Construction Procedure

M

M

Cold Weather Construction Procedure

M

M

Unit Placement

N

N

Steel Placement

"

R

Tie and Anchor Placement

N

N

Mortar Placement

N

N R

Grout Placement

"

Cleaning Procedure

R

R

Maintain Daily Log

R

R

Flashing and Weephole Placement

R

R

Movement Joint Placement

R

6-41


"

A C 1 TITLE*MDG

93

m

0 b b 2 9 4 9 0508633 5 3 3

m

7 HOT AND COLD WEATHER CONSTRUCTION

7.0

INTRODUCTION

Weather conditionsduringconstructionaffect

the performance of completedmasonry

structures. Special precautions must be taken during both hot and cold weather to assure the desiredquality,

and the masonrycontractorshouldfollow

Contract Document

requirements or submit proposed procedures for approval before beginning work

7.1 HOT WEATHER CONSTRUCTION Hot weather causesspecialproblems temperatures, lowhumidity

inmasonryconstruction.Combinations

of high

and wind create conditions that mayadverselyaffectthe

properties and performance of materials and components. 7.1.1 Performance of Masonry and Mortar

Rapid evaporation during dry weather reduces the water lowering both compressive and bondstrength. reduced, and set occursfaster.With

content of mortar and grout,

Mortar workability and grout flow are

high temperature, mortars shouldhavelowerair

contents and high water retention. Board life of mortar is shorter, and joints must be tooled sooner than normal. Rapid evaporation at the exterior face of mortar joints decreases 7-1



73

m 0662947 0508634

47T

m

durability and strength at the surface. When the water needed for mortar and grout curing is lost byevaporation and unit suction, the hydration process may actually

stop, and is reactivated only

by subsequent

re-introduction of moisture. This phenomenon, known as dry-out, canbe avoided by careful attention to material selection, storage and preparation and masonry protection. 7.1.2 Material Storage,Protection, and Preparation

Specs. 2.3.2.3 requires that approved hot weather procedures, as outlined in the Contract Documents or proposed by the masonrycontractor,

be implementedwhen

ambient

temperatures exceed 100" F, or 90" F with wind velocities greater than 8 mph. Materials should be stored in a cool, shaded location. Covering aggregate stockpiles with

a plastic

sheet will retard the evaporation of moisture. High-suction brick canbe wetted, and additional mixing water maybe needed in mortar and grout. Metal accessories, reinforcing steel, wheelbarrows, mixersand mortar boards should be kept cool by flushing with water. Increasing cement content accelerates early strength gain. Adding extra lime also increases water retentivity. Adding ice

or cooling the mixing

water significantly lowers the temperature of mortar and grout. Hoses stretched too great a distance from

the source to the mixer can allow the water to become hot, potentially

contributing to flash set of the mortar. Set-retarding or water-reducingadmixturesshould

be usedonlywhen

approved by the

Owner's representative, and should be tested in advance in accordance with ASTM C 780 and C 1019 (see MDG 5.4). Sun shades,wind screens and water fog sprays can reduce the effects of dry, windy weather. Mortar should not be mixed too far ahead and, when mixed, should

be stored in a cool,

shady location. Hot weather construction requirements, Specs. 2.3.2.3., prohibit spreading

7-2


A C 1 TITLElKMDG 9 3

m

0662949 0508635 306

mortar beds more than four feet ahead of the masonry and mandate that units be set within one minute of spreading mortar. A good rule of thumb to remember is that if the mason has to tap the unit down to the line with his trowel, the mortar bed has probably lost most

of its plasticity and ability to developitsfullbondpotential.Wetting

the constructed

masonry by water spray after the tooled joints have set provides moisture needed for curing, preventsdry-out,

and effectivelyincreases

the tensile bond strength of the masonry.

Covering masonry slowsthe natural rate of evaporation and creates a greenhouse effect that aids in moist curing. However, this could cause increased efflorescence. When temperatures are extremely high, consideration should also

be given to scheduling work to avoid the

hottest part of the day.

7.2 COLD WEATHER CONSTRUCTION Cold weather also poses special concerns

for masonry construction and for protection of

completed work. Cold weather construction is defined as any construction occurring when either the ambient temperature or the temperature of the masonry units is below 40" F. Temperatures below 40" F affect both materials and performance, and as temperatures drop, additional protective measures are required (Specs. 2.3.2.2). It is recommended that the suggestions in "Recommended Practices

and Guide Specifications for Cold Weather

Masonry Construction" by the International Masonry Industry All Weather Council (7.2.2) should be followed unless amended to more stringent requirements based on experience or preference. Some A/E firms require that, when the ambient temperature falls below 40"

F, the masonry construction shouldbe protected to maintain the temperatureof the masonry at or above 40" F for 24 hours.

7.2.1Performance

of Masonry and Mortar

Plastic properties of mortar and grout are changed significantlyat low temperatures. Water requirements for a given consistency are less, air entraining admixtures are more effective, and initial and final sets take longer. Early strength development is slower in cold weather, 7-3


A C 1 TITLE*MDG

93

m

Obb29Y9 0508636 2 4 2 M

but final strength is equal to or greater than that attained at normal temperatures. One particular concern in cold weather is the possibility of masonry freezing. The rate at which masonry freezes is influenced by the severity of ambient temperature and wind, the temperature andabsorption characteristics of the units, the temperature of reinforcing steel and metal accessories, and the temperatureof the mortar mixture when placed. When fresh mortar freezes, its performance characteristics are affected by several factors including water content, age at freezing and the amount of strength developed prior to freezing. Hydration of the mortar cannot take place when the temperature is below 40"

content (internal relativehumidity)isless

F and the moisture

than 75%. Frozen mortar takes on all the

outward appearances of hardened mortar. It develops some compressivestrength

as

indicated by its ability to carry loads, and some bond strength as evidenced by its ability to adhere to other materials. But it is NOT cured, and does not develop full design strength until thawed and water is again available to complete the cement hydration process. Spring rains and watering the masonry walls will help provide enough water to starthydration again

so that the mortar can reach minimum strength. Water in the mortar expands when it freezes.The higher the water content, the greater the expansion. If the moisture content is low, expansive forces willbe minimized, so low water content mortar and high suction units are desirable. Grout is not as weather sensitive as mortar because it is not exposed to wind, but because of its higher water content, freezing grout can cause significant disruptive expansion. Mortar dry-outs can occur during winter

as well as in summer. Heated walls must have

moisture to cure, and walls may have to be wetted to provide adequate water. Sublimation may also require the addition of water to assure hydration. Heating and protection of masonry materials prior

to construction are the best defense

against adverse cold weather effects, but mortar and grout mixtures may also be modified for better performance.

Type III portland cement provides higher early strength 7-4



93

m

Obb29q9 0508b37 189

m

development but can also cause color variations. Air entrainment increases workability and freeze-thaw resistance. Excessive entrapped air lowers the compressive and bond strength of mortar and grout, so mortar mixing time should be carefully controlled.

Dispersing agents used in some color pigments have a retarding and drying effect on cement hydration. This is particularly undesirable in cold weather. Accelerating agents increase the rate of early strength development by promoting more rapid cement hydration. Calcium chlorideaccelerators,however,should

not be usedinmasonryconstructionbecause

of

adverse side effects such as increased efflorescence and corrosion of embedded metals. Non-chloride accelerators such as soluble carbonates, silicates and fluorosilicates, calcium aluminate, and triethanolamine are available. These should be approved and tested before use since potentially they could cause

other problems. So called "anti-freeze" admixtures

that claim to lower the freezing point of mortar must be used in such large quantities to be effective that theylower both compressive and bond strengthbelow acceptable levels. Proposedadmixtures

for masonry mortar or groutmust

be approved by the Owner's

representative, and should be tested in advance in accordance withASTM C 780 or C 1019 at the lowest expected temperature at which they will be used in construction (see MDG 5.4). Admixture manufacturers shouldbe required to provide written certification that their

products do not contain chlorides, and do not have adverse effects on either the plastic or hardened properties of the mortar, grout, or the masonry. Any masonry that has been constructed during cold weather should be inspected to ensure that no frozen mortar is present, and that mortar strength development has begun and continues. Frozen mortar can be detected inseveral ways: (1) a''crow's

feet"pattern

appears on the surface of tooled joints; (2) flaking indicates freezing expansion; and (3) friable material scratched

from the joint surface indicates frozen

mortar dried through

sublimation. When frozen mortar is detected, the masonryshould be thawed and then sprayed with a water fog to reactivate hydration. Additional necessary to assurecontinuation

of the hydrationprocess

strengths. 7-5


heat should be provided as and development of design


7.2.2

93

m

Obb2949 0508638 015

m

MaterialStorage,Protection,andPreparation

The cold weather considerations given in the Specifications are intended to permit masonry construction to proceed during inclement weather. The protection required is considered adequate, but minimal for the temperature ranges listed. Proper storage and protection of materials, heating of materials and mortar ingredients before construction, and protection of completed work duringmortar and grout curing will prevent early freeze damage. MDG

Table 7.2.1 summarizes the minimum requirements of Specs. 2.3.2.2 for heating and protection.

Table 7.2.1 Summary of Specifications Requirements

I Mean Daily Air Temp., I

Construction Requirements

Deg=F

Heating of Materials

Protection

32 - 40

Heat mortar sand or mixing water to produce mortar between 40-120°F at time of mixing; maintain mortar above freezing until used in the masonry

Protect completed masonry from rain or snow with weather-resistant membrane for 24 hours after construction

25-32

In addition to the above, thawing wet, frozen masonry units; heat dry masonry units to not less than 20°F when laid in the masonry

Completely cover completed masonry with weather-resistive membrane for24 hours after construction

20-2s

In addition to the above, thawing wet, frozen masonry units; heat dry masonry units to not less than 20°F when laid in the masonry

Install wind breaks when wind velocity exceeds 15 mph; provide heat sources on both sides of masonry under construction; completely cover completed masonry with insulating blankets or equal protection for 24 hours after construction

Below 20

In addition to theabove, thawing wet, frozen masonry units; heat dry masonry units to not less than 20°F when laid ih the masonry

Provide enclosure for masonry under construction; use heat sources during construction to maintain temperatures above 32°F within the enclosure; maintain 32°F in the enclosure for 24 hours after construction

Heating of materials is intended to assure adequate cement hydration in mortar and grout by maintaining temperatures above 40" F and moisture content above 75%. Mixing water 7-6



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is the easiest material to heat, and it also stores more heat per pound than any other ingredient (Fig. 7.2-1). Water that is too hot (above 140" F) is a safety hazard.

Fig. 7.2-1 HeatingWater Because water hotter than 180" F can cause flashset when it comes incontact with cement, the sand and water should be combined in the mixer first to lower excessive temperatures, before the cement is added. The temperature of the mixed mortar or grout should be maintained between 40-70" F until it is used. Smaller batches will be used more quickly, and will stay warm until placed in the wall. Some mortar silos can be heated to maintain batch to batch mixtures at a constant temperature. Sand is generally heated to a minimum temperature of 45-50" F to assure that all frozen be mixedperiodically to assure uniform

lumps are thawed.Coveredstockpilesshould

heating, and to avoid any possibility of scorching near the heat source. The aggregate can be piled over a metal pipe containing a fire, or can be heated by steam coils or an ordinary water heater (Fig.7.2-2).

Cold or frozenmasonryunitscan 7-7


be heated withoil,gas

or

electric hot-air heaters. During verycold weather, frozen wallsmust be thawed before grouting, and heated enclosures can also be used for this purpose. When air temperatures are below 40" F, or have been below 32" F during the previous two hours, the air temperature in the bottom grout space should be raised above 32" before beginning the grout pour.

Fig. 7.2-2

Heating Aggregates

In addition to normal material storage and protection procedures, a temporary cover should beerected

over themortar

mixing area to provide a sheltered location for mixing

operations. Temporary enclosures may also be necessary at work areas. Heated enclosures of plastic sheeting or other materials attached to the scaffolding or supported on other framework permit year-round construction, provide protection for materials and workers, help maintain elevated mortar and grout temperatures, and assure sufficient heat for proper cement hydration (Fig. 7.2-3). Heat can be provided by natural gas, fuel oil, electricity, steam or bottled propane, but enclosures must also be well ventilated for safety.

7-8



m

0662947 0.508641 bOT

m

Fig. 7.2-3 Heated Enclosures

REFERENCES 7.2.1 Frohmader, D. L,"Cold Weather Checklist," The Magazine of Masonry Construction, Addison, Illinois, November 1990, pp. 503-504. 7.2.2

International Masonry Industry All-Weather Council, Recommended Practices and Guide Specifications for Cold Weather Masonry Construction, Washington, D.C., 1970.

7.2.3

International MasonryIndustryAll-WeatherCouncilTechnical

Task Committee,

"All-WeatherMasonryConstructionState-of-the-ArtReport,"Washington, 1968.

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DESIGN PHILOSOPHY & METHODOLOGY

8.0

INTRODUCTION

The design and analysis of masonry building structures is an art as well as a science, since both creative judgment and engineering principles are required. Decisions must about the type(s) of masonryunit(s) accessoriesmust

be selected.

tobe

used.

be made

Mortar, grout, reinforcement, and

The configuration of the masonryassemblymust

determined. Architectural considerations, such

be

as weatherproofing, fireproofing,

soundproofing and appearance, must be coordinated withstructural behavioral requirements. The nature and magnitude of loading that the structure will experience must be predicted. Although building structure self weight can be determined fairly accurately, the magnitude of live loads is more uncertain. Superimposed gravitydead andlive loads may vary overthe life of abuilding.

Lateral loads, most often

due to wind or earthquake, are probability

based. Just as loads can vary, so can the strength of materials that make up the structural system, due to material inconsistencies and manufacturing tolerances. The quality of workmanship used to assemble these materials is also variable, depending upon locale,the experience of the mason, and the extent of inspection. 8-1


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Previous masonry codes recognizedthe variability in workmanship quality by differentiating between "inspected" and "uninspected" construction. The designer wasrequired to use lower allowable masonrystresses for uninspected construction. The design provisionsof the Code, however, are predicated upon the fact that good workmanship is a

requirement of the

Specifications, withwhich compliance is required by Code 3.1.1. Good workmanship must be verified by a qualityassuranceprogram.

The Code does not permit"uninspected"

workmanship. Different design philosophies have

been developed to account for the load and material

variabilities and uncertainties. In the strength design method, applied loads are increased by load factors, with gravity live loads

and lateral loads increased by larger factors than

gravity dead loads. The requiredstrengthbased

on factoredloads is compared to the

strength capacity of the section, reducedby a materials variability factor. The UBC permits the strength design of shear walls and slender walls subjected to out-of-plane lateral loads. The MSJC is currently developing a limit

states design standard for masonry structures.

Limit states design is based on the theory of probability in which statistically determined "expected" valuesof materials, assembly,and system properties are used. Capacity reduction factors and load amplification factors are applied similarto the strength design method.The capacity reduction factor reflects not only the material variability, but also the reliability of the design equation and the potentiallyundesirableconsequencesassociatedwiththe occurrence of the limit state for which the capacity is being calculated. The Code is based upon the allowable working stress design philosophy(Code 5.1.). In this method, calculated stresses resulting from service or working loads (not increased by load factors) are compared to Code-specifiedallowablevalues.

The serviceloads are of a

magnitude that may be assumed to actually occur during the lifetimeof the structure. This design philosophy is discussed in more detail in MDG 8.2.

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8.1 WHAT IS MASONRY

Masonry consists of relatively small units bonded together with

mortar, and sometimes

grouted and reinforced with steel. Masonry units may be stone, fired clay units, cementitious concrete units, glass, or ceramics. The Code addresses only fired clay (brick) units, concrete masonry units, and stone. The strength of a masonry assemblage (wall, beam, etc.) depends upon the combination of units, mortar, grout, and reinforcement. Assemblage strength depends on individual material strengths, bond betweenmaterials,and

the dimensions of the components andthe

assemblage. For example, the strength of masonry construction depends on the strength of the units, the strength of the mortar, mortar-unit bond,and the thickness of the mortar joint. The use of grout in masonry in single

or multiplewythewalls

or in the collar joint in

multiwythe walls increases the masonry strength. 8.1.1 Masonry Units

The most commonly used masonry units

are made of clay brick and concrete block. See

MDG 3.1. Because of costconsiderations,concreteblockis structural applications than brick,althoughbrickmay

more commonlyusedin

be more economicalwherehigh

strengths are required and where appearance is important. For exposed exterior wythes, consideration mustbe given to appearance, weathering characteristics,and water permeance, as well as structural requirements. Various masonry unit properties are defined in the applicable ASTM specifications, listed in MDG 3.1. The ASTM standards reflect minimum requirements for any project. If the designer determines that a more stringent requirement is necessary, then this must be stated in the contract documents in addition to the governing ASTM standard. In this way, project’s special requirements will be communicated to the contractor.

8-3


the

Masonry units are generally defined as eithersolid or hollow. In general, solid units are not

100% solid unless so specified. Units are usually formed with core holes, to allow for more even curing or firing, and to decrease unit weight without significantly sacrificing crosssectional properties. Units having core holes that constitute 25% of the gross area or less are considered solid units. sectional area,the

If the core holes remove more than 25% of the gross cross-

units are hollow.Hollow

concrete masonry units are generally

approximately 50% solid. Hollow clay bricks are generally from 40% to 60% solid.

8.1.2 Mortars Mortar holds the individual masonry units, reinforcement, and connectors together so that the components act as a complete assemblage. Many mortar propertiescontribute to proper performance of a wall. Compressive strength is only one of the properties, and may not be the most important one. Mortar-masonry bond has a more significant impact on masonry flexural strength and moisture resistance. Flowability

and workability affect the mason’s

ability to place the mortar, and may affect the quality of workmanship. See MDG 3.2 for further discussion of these issues. For construction projects in which masonrystrength properties arecritical, pre-construction testing of prisms built with

the specified units and mortar may be warranted. Flexural

strength can be assessed by flexural prism tests or by bond wrench testing. See MDG 4.1,

4.2, 5.4.1.5. 8.13 Grout

Grout is made from portland cement and sand, withpea gravel sometimes added when large spaces are to be filled. However, an ASTM C 476 fine grout is more commonlyused. Slump ranges of 8 in. to 11in. allow the grout to flow properly into thecavities or cells. See MDG 3.3. Grout is used to create a solid wythe by filling the cores of hollow units, or to create a composite wall by filling the collar joint between wythes. Grout is also used to fill 8-4


A C 1T I T L E v M D G

reinforced bond beams.

93

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ObbP947 0 5 0 8 6 4 6 191

Mortar shouldnot

be usedfor

m

these purposes, because its

flowabilty is inadequate to completely fill voids. 8.1.4 MasonryAssemblages

Masonry units, mortar, and grout can be combined in several ways to construct different types of masonryassemblages.Mostmasonryassemblages

are walls.Wallsmay

be

classified by their construction type or by their primary load-resisting function. Using construction type as the criterion, walls may be classified as:

O

Singlewythe:

A singlethickness(wythe)

of masonry,whichmay be reinforced or

unreinforced. 0

Cavity: Two similar

or dissimilar wythes of masonry separated by an air space, and

tied together with metal anchors. The cavity serves as a drainage path for any water entering the wall and provides a space in which insulation can be placed. For this

type of assembly to function properly, the cavity must be kept free of mortar bridges and droppings, and flashing and weepholes must be correctlyinstalled.

There is

general agreement in the industry that a 2 in. air space, exclusive of insulation, is the minimum that can be kept clean during construction. Code 5.8.2 defines this system as a noncomposite, multiwythe wall. 0

Prefabricated: A plant-manufactured panelizedwallsystem.Connections

between

the panel and the supporting structure might haveto be designed to allow differential movement between the two. Sources of differential movement are volume changes in the panel, live load deflections, dead load creep of the supporting structure, and wind or seismic drift. See MDG Chapter 10 - Movements. Panel connections to the structure may be designedaccording

to the recommendations of the Precast

Prestressed Concrete Institute. O

Barrier: Multiwythe construction with the collar joint grouted or mortared solid. The filled collar joint acts as a barrier against water entering the inner wythe(s). Barrier

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walls require the same attention to flashing details as other walls. The wythes may be of similar or dissimilar material units. Either masonry units or metalties are used to mechanicallyconnect the wythes together. Code 5.8.1 defines thissystemas composite. Masonry Veneer: A single wythe of masonry mechanically tied to a backing with a cavity between them. The Code does not apply to this type of assembly. Masonry Bonded Hollow Wall: These walls are built of hollow or combined hollow and solid masonry units with multiple wythesbonded by masonry headers. Usingloadingas

the criterion, wallsmay

be classifiedas

veneer, nonloadbearing,

loadbearing, or shear wall. Veneer walls are not covered by the Code. The backing of the veneer wall resists the lateral(out-of-plane) loads on the wall. A loadbearing wall is defined by Code 2.2 as a wall carrying, in addition to its own weight, vertical loads greater than 200 plf. Walls which do not meet these criteria may be considered nonloadbearing. Shear walls resist lateral loads such as wind or seismic by in-plane shear stresses and overturning moments.

8.1.5 Volume Changes

As discussed in MDG 6.2.7 and Chapter 10,masonry units change volumedue to moisture, temperature and other effects. Masonry units also

creep under sustained load. If these

movements are restrained, theresulting additional loads introduced into the masonry must be considered. Unintended stresses may be avoided by introducing vertical and horizontal movementjoints.

The geometry of the building,wallcomposition,masonry

material

properties, and anticipated differential movement must be carefully considered in locating and sizing these joints. 8.1.6

Details of Construction

Proper performance of masonry walls requires attention toconstruction details as wellas to

8-6


A C 1 TITLE*NDG 9 3

structural analysis.Detailing

m

Ob62949 0508b48 Tb4

to accommodate water whichwill

m

penetrate single-wythe

exterior masonry walls is a critical design aspect affecting wall performance. The designer may choose to prevent water entry by providing a barrier wall, or to control the water path by providing proper flashing and weeping of a cavity wall. Proper cavity wall performance depends on anunimpeded drainage path and flashings to collect and direct all water to the exterior.Top-of-wallclosures

and sill materials/details that prevent water entryare

important to bothwall types.

To minimize water penetration, mortar joints should be completely filled and compacted against the sides of the adjacent masonry units by concave tooling. Joints which are raked, struck or not tooled are poor barriers to water entry. See MDG Chapter 6. Construction details must be consistent with the assumptions of the structuralanalysis, since each impacts the other. Forexample, if a joint is introduced between two intersecting walls, a monolithic flanged wall analysis is not appropriate. Special CMU control joints, such as those illustrated in Fig. 10.4-1, will transfer out-of-plane shear but not in-plane shear. Such joints effectivelydivide a shear wall into segments.Flashingalsoaffectswall

structural

behavior, since it creates a discontinuity in the wall, limiting shear and bending moment transfer across the flashed joint. 8.2

8.2.1

STRUCTURAL ANALYSIS AND DESIGN General Requirements

The design procedures of the Code are predicated upon allowable stress methods, in which the effect of service loads on structural members permits an elastic analysis, and computed stresses arecompared to specifiedallowablestresses.

See Code 5.1. Serviceloads, or

working loads, are those which the general building code determines may actually occur during the structure’s service life.

8-7


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9TO

m

In an elasticanalysis, it is assumed that the structuralmaterialsfollowHooke’s

Law:

deformations or strains are linearly proportional to the loads or stresses. It is also assumed that all materialsare homogeneous, and sectionsthat are plane before bending remain plane the specified compressive after bending. To ensure that the materials remain linearly elastic, strength’f,

(Code 5.4), is divided by a factor of safety to obtain the allowable stress. The

required factor of safety depends on the variability of material strength, the variability of construction quality, the accuracy with which the applied forces can be predicted, and the accuracy with whichthe actual stresses canbe calculated. Code stipulated allowable stresses incorporate adequate factors of safety. The Code requires that allstructuresandtheircomponentmembers

be designedin

accordance with Code Chapters 5 and 8. Additionally, structures and members must follow the provisions of Chapter 6 or Chapter 7. Alternatively, structures or members may be designed according to the provisions of Chapter 9 (Code 9.1). Chapters 6 and 7 are based upon rational design methods, whereasChapter 9 presents an empirical design philosophy. Empiricaldesign isonly

permittedforbuildingsthat

meet the sizeandconfiguration

limitations of the Code (see MDG Chapter 15)’ or for component members which are not part of the lateral force resisting system of a rationally designed building. Code 5.5 providesmaterial

properties of steelreinforcement,

claymasonry,concrete

masonry, and grout. See MDG 3.1, 3.2, 3.3, and 3.4. The specified compressive strength of masonry,

fm,

is notdefinedfor

the variousmasonryunitand

However, Code 5.4.2 requires the design designer to specify the required

fm

to be based upon fm.

mortar combinations. Code 5.4.1 compels the

onthedesigndrawings,butdoesnot

mandate

specification of unit strength or mortar type. It is apparent that the intent of the Code is for projects to be built under a performance specification, wherein the designer determines the required level of performance

Vm), and it is the responsibility of the contractor to meet

that level of performance. Specs. Tables 1.6.2.1and 1.6.2.2are intended for the contractor’s use, and provide proven means of producing a

8-8


fm

that will satisfy the specified

fm

with

AC1

TITLE*NDG 9 3 9 O b b 2 9 4 9 0508650 b L 2 m

different masonry unit strengths and mortars, bypassing the need for prism testing (Specs. 1.6.1 and 1.6.2). The Specifications require specifying not only f’, (Specs. 1.5.2), but also the masonry units and mortar (Specs. 2.2.1 and 2.2.2). The masonry units and mortar are specified when the structural designer needs modulus

of elasticityvalues for stiffnesscalculations.When

a

specified masonry unit and mortar combination cannot meet the specifiedf’, requirement according to Spec. Tables 1.6.2.1 and 1.6.2.2,prism testing must be conducted

to prove

compliance; i.e., f, equals or exceeds f’,. Depending upon the masonry unit, mortar, and

f’, combination specified,it may be difficult for thecontractor to achieve the desired results. This situation presents a condition which must be resolved. 8.2.2

Analysis Considerations

Masonrycomponentsmust

be analyzedforverticaland

lateral loads.Verticalloads

generally come from gravity loads or from wind acting on a roof. Lateral loads result from wind, earthquake, or soil pressure. Gravity loads are shared by all wythes in a composite multiwythe wall (Code 5.8.1). In a is resisted noncomposite wall (Code 5.8.2), the axial compression resulting from gravity loads onlyby the wythe nearest the center of span of the supported members. However, any bending about the weak axis of the wall, due to eccentric support of the gravity loadson one wythe, is considered to be resisted by all wythes. The amount of bending resisted by each wythe is determined by the flexural stiffness of that wythe relative to the others. Stress computations are based upon the minimum net cross-sectional area of the masonry (Code 5.13.1). Inmultiwythemasonryofdissimilarmaterials,

the section properties are

determined by transforming each minimum net cross-sectional area to that of asingle material, using the relative elastic moduli of the different materials. This method uses the

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elastic analysis concept required by Code 5.13.1.2 To determine the relative stiffness of different wythes, a similar approach is used, except that the average net cross-sectional member area may be used, per Code 5.13.2. Values for themodulus of elasticity of masonry are given in Code Tables 5.5.1.2 and 5.5.1.3.

As a loadbearing wallbecomeshigher

and supports more floors, the loads at its base

increase. These increased loads can be accommodated by increasing the thickness of the wall; grouting the wall solid; keeping the same thickness and increasing the strength of the materials; reducing the size of any openings in the wall; or by any combination of these options. Lateral loads acting perpendicular to the plane of a masonrywall create out-of-plane bending. All wythes of multiwythe masonry resistthese loads, in proportion to theirrelative stiffnesses. See Code 5.8.1.3,5.8.2.1(d), 5.13.1.2. In composite masonry, the resulting shear stresses between wythes is limited to the values specified inCode 5.8.1.2. In noncomposite masonry, when the width of the space between wythes exceeds 4 in., a detailed analysis of the wall ties is required per Code 5.8.2.l(f').

A wall without openings can be analyzed as an assemblage of crossing strips. Each strip is considered a beam of unit width, spanning either vertically or horizontally. A wall with openings canbe similarly analyzed, or can require a two way plate analysis to determine the stresses around the openings. The diaphragm action of floors and roof transfers lateral loads to shear walls that are parallel to thedirection of the lateralload, resulting in in-plane shear and bending on those walls. In performing the lateral load analysis on the structure, flanges of intersecting walls may be considered as adding to the stiffness ofwalls resisting in-plane loads. See Code

5.7.1.1. Parameters that limit the stiffening effect of flanges are given in Code 5.13.4.2. The stiffness of the horizontal diaphragms influences the distribution of the lateral loads to the shear walls, and is discussed in more detail in MDG 8.4.3.1 and Chapter 9.

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In composite multiwythe masonry, all wythesare considered to resist in-plane lateral loads, in proportion to their relative stiffnesses. In noncomposite masonry, onlythe wythe to which the in-plane lateral loads are applied resists the load. Any transfer of in-plane stresses between noncomposite wythes is neglected, per Code 5.8.2.1(c). Wallsegmentsin

the same plane may be coupled or uncoupled.

Coupled walls are

structurally linked together and, for resistance to in-plane lateral forces, must be analyzed together. The shearresisted by each segment of the wall will be in proportionto its relative stiffness, just asin uncoupledwalls. However, the link (or coupling beam) must be analyzed for axial, shear, andbending forces, which result from the action of the wall segments. See

MDG 8.4.3.8 and RCJ Hotel in MDG 9.1.3. In Building Construction Option I of the RCJ Hotel, the interior masonrywalls on Grid Line 2 are coupled shear walls. In Building Construction Option II, the same shear walls are uncoupled. Stresses are also introduced into masonry when it is restrained against thermal expansion and contraction, moisture expansion and contraction, and shrinkage and creep. These movements, and the extent to which the masonryis

restrained against them, must be

considered in structural design. Values for the magnitude of these effects are given in Code 5.5.2, 5.5.3, 5.5.4, 5.5.5, and MDG Chapter 10. Allowable stresses in masonry are influenced by masonry unit type; bond pattern; mortar type; unit compressive strength; specifiedcompressive strength of masonry; amount of grouting; and the ratio of unsupported height to thickness. Stronger units and/or higher mortar strengths permit higher allowable stresses. masonrylaidinrunning

Higher shear stresses are allowed for

bond than in stack bond (Code 6.5.2).

The Code permits all

allowable stresses in Chapter 6 and Chapter 7 to be increased by one-third when the load combination considered includes wind or earthquake togetherwith dead and/or live loads. See Code 5.3.2 and MDG 8.3. Quality assurance and quality control to ensure good workmanship are required by the Specifications, withwhich compliance isrequired by Code

3.1.1. Since "uninspected" constructionis not permitted, theCode does not have a provision 8-11


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for reduced allowable stress in "uninspected" construction, as in earlier codes. 8.3

LOADS AND LOAD COMBINATIONS

Structures and theircomponents are required toresist gravity loads,lateral loads, and other types of loads. Gravity loads act vertically; lateral loads act horizontally. The Code also requires the designer to consider the effects of prestressing, vibrations, impact, shrinkage, expansion, temperature changes, creep,and

differential movement.

See Code 5.2.

Restrained volume changes induce loads. Gravity loads are of two types:live

and dead.

Live loads aretransientinnature,

representing theassumed weightof building occupants, furnishings, equipment, and the like. Minimum live loads for different occupancies are mandatedby the governing building code, or by ASCE-7 (8.3.1). Reductions in live loads are permittedby the building code or ASCE7, depending upon the tributary areas supported (Code 5.2.2). Dead loads are generally permanent in nature, and can be subdivided into structure selfweight and superimposed loads.

The largest portion of thedeadload

isusually

the

structure's self-weight; the smaller portion is the superimposed dead load. Superimposed loads may include, but arenot

limited to roofingsystems;

concrete topping;ceilings;

mechanical equipment; and nonloadbearing masonry walls. Analysis of the structureduring construction would not include superimposed dead loads which are generally placed later. Lateral loads are typically inducedby wind, earthquake, andfluid or earthpressures. Other lateral load sources include blast,crane loads, and horizontal thrust from untied sloped roof framing. Wind pressures are applied over the surface of a structure. The magnitude and distribution of the wind pressures to the various surfaces are given in the appropriate building code or ASCE-7. A document designed to assist the professional in the understanding and use of

8-12



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the complex wind load provisions of ASCE-7 is available

m

(8.3.2). The surfaces, to which

wind is applied, usually span vertically between floors or horizontal diaphragms and are analyzed for out-of-plane flexure. Horizontal diaphragms

span from support to support;

these supports often are shear walls. Shear walls and diaphragms are analyzed for in-plane flexure, shear, and deflection. See Code

5.2.3.

Seismicloads are inertial forces, frequently expressed

as a fraction of the mass of the

building components. For purposes of design, these are often applied as lateral forces in accordance with magnitude and distribution formulas defined by the appropriate building code or byASCE-7.Likewindloads,seismic

forces are transferred by horizontal

diaphragms to the shear walls. Component analysis for seismic forces is similarto that used for wind pressures. See Code 5.2.3. Fluid and earth pressures are generally treated as linearly varying loads acting normal to the wall surface; these loads induce out-of-plane flexure and shear stresses in walls. Loads must generally be consideredin combination with each other. Loading combinations must be examined to identifywhich one causesthehigheststresses.Unless

mandated

otherwise by the governing building code, loading combinations given in Code 5.3.1 must be considered. These are: 1.

Dead load acting alone

2.

Dead load plus live load

3.

Dead loadplus live loadplus

either wind or seismicload.Windload

and

seismic loadare considered as non-concurrent events. While wind and seismic loadscanactfromanydirection,they

are usuallyappliedin

one oftwo

orthogonal directions coinciding with the major axis of the building, since these are typically the critical directions. oftheload

Per the UBC, one often considers 100%

in one orthogonaldirectionplus

perpendicular direction.

8-13


30% of theload

in the


93

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Ob629490508b55

IT4

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4.

Dead load plus wind load acting from any direction.

5.

0.9 times dead load, plus seismic load. In this combination the dead load is reduced because it can help in resisting seismic overturning forces.

The 0.9

reduction factor recognizes that: a) the magnitude of the structureself-weight may have been over-estimated, b) vertical acceleration can reduce dead load effects.

6.

Dead load plus live load plus lateral load from fluids or earth pressures.

7.

Dead load plus lateral load from fluids or earth pressures. The dead load component is not reduced when combined with fluidor earth pressures.

8.

Dead load pluslive

load plus the effects of restrained movementsfrom

temperature, shrinkage and creep, and other effects.

9.

Dead load plus the effects of restrained movements fromtemperature, shrinkage and creep, and other effects.

Code 5.3.2 permits an increase in allowable stresses by one-third when considering load combinations 3, 4, or 5 (8.3.3). When the structure is adequately provided with movement joints, effects from restrained volume change are mitigated. However, consideration must still be given to the effects of differential movement in noncomposite multiwythe walls of multi-story buildings. See MDG Example 10.4.3. 8.4

STRUCTURAL CONSIDERATIONS FOR MASONRY WALLS

Walls act as:

O

nonloadbearing facade elements subject to out-of-plane flexure

O

loadbearing elements subject to axial compression

O

shear wall elements subject to in-plane shear and flexure

O

or combinations of the above elements

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8.4.1WallContinuityandSupportConditions

A wall is described as having a simple span when it is laterally supported on two opposite edges only, and when rotational fixity is not provided at the supports. A two-way simple span wall is pin-supported on all four edges. The walls in most single story structures have a simple span from the top of the foundation to the roof diaphragm.

A wall is described as multi-span when it is continuous across supports. Continuity may be provided in only one direction (i.e., either vertically or horizontally), or it may be provided in both directions. Fig. 8.4-1 is a schematicrepresentation of deflected shape of a multispan wall subject to out-of-plane loading. Continuity, or rotationalfixity, between adjacent spans depends upon the mortadunit bond, unit interlock, or reinforcement at the joint over the support. Walls in multi-story and/or multi-bay

structures generally are continuous multi-

span.

Fig.8.4-1Multi-spanWall The joint between perpendicular vertical panels can transfer shear forces between flanges and the web of a shear wall as shown in Fig. 8.4-2. For this to happen, the connection at the joint must meet the requirements of Code 5.13.4.2.

Fig. 8.4-2

PerpendicularVerticalPanelConnection

8-15



93

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The joint between twoverticallyaligned

Obb2949 0508657 T77

m

elements and a horizontal panel can provide

buckling restraint for the vertical elements as shown in Fig. 8.4-3. The connection at the joint between thewall and horizontal diaphragm mustbe able to resist the forces developed when the diaphragm acts as a support for out-of-plane flexure of the wall.

Fig. 8.4-3 Horizontal to Vertical Panel Connection

- Out-of-PlaneLoads

The joint between verticallyaligned elements can resistin-plane shear resultingfrom diaphragm action, by providingclamping

due to vertical compression, or by positive

connection with reinforcing steel. See Fig. 8.4-4 and Code 6.5.2(c).

Fig. 8.4-4 Horizontal To Vertical Panel Connection

- In-Plane Loads

Pilasters, addressed in MDG 11.2 and 12.3, act as vertical beams to support wall panels. See Fig. 8.4-5. Panels, supported by pilasters, span horizontally between the pilasters, vertically between horizontal diaphragms, or both ways. Pilasters at joints between panels may provide continuity between panels and add stiffness. See Code 5.10.

8-16


A C 1 TITLErMDG 93 D 0662949 0508658 903

Fig. 8.4-5 8.4.2

m

IntegralPilasters

VerticalLoadbearingWalls

8.4.2.1 Types of Vertical Loads

- Uniform loads cause uniformly distributed stresses over

wall lengths as shown in Fig. 8.4-6. Concentrated loads create concentrations of stresswherethey

are applied.Stress

distribution becomes uniform as stressesspread out across the wall length, as shown in Fig. 8.4-7.

Fig. 8.4-6 Uniform Load Fig.

8.4-7 Concentrated Load

As a general rule, concentrated loads canbe considered to have the same effect as uniform loads when their spacing is less than or equal to the widthof bearing plusfour times the wall thickness, per Code 5.12.1. See Fig. 8.4-8. The effect of bond beams on the distribution of concentrated loads in masonry presented in MDG Chapter 9.

8-17


walls is

A C 1 T I T L E x M D G 93 m 0662949 0508659 8 4 T m

P I

D

Fig. 8.4-8 Concentrated Load Spacing for Uniform Wall 8.4.2.2 Failure Mode

Stress

- The capacity of loadbearing walls, shown in Fig. 8.4-9, is generally

controlled by compression failure of the material or by buckling of the panel. Since masonry walls usually have a relatively small thickness, material compression failure rarelyoccurs.Localizedcrushingmayoccur under concentrated loads with inadequate bearing area. See Fig. 8.4-10 and Code 5.12.2 and 5.12.3.

h TEICKNESS

CONPRESSION

Fig. 8.4-9Crushing 8.4-10 Bearing Mode Fig. Panel Where large concentrated vertical loads exceed the bearing capacity of the wall, and economic or architecturalconsiderations prevent increasing the wall thickness,pilasters may be added at the concentrated load locations. Wall panel buckling shown in Fig. 8.4-11, is a stability problem. Variables that influence stability are: 1) the number of wall panel edges supported; 2) the thickness to span ratios

8-18


A C 1 T I T L E * M D G 93

m 0662949 0508660

561

m

of the panels; and 3) the length to height ratios of the panels.

For means of comparison, assume that the panels shown in Figs. 8.4-12 through 8.4-14 all have length equal to height, as well as equal thickness. The buckling resistance factor relative to the support conditions is indicated for each figure.

Fig. 8.4-11 Buckling Mode Fig. 8.4-12 Supports

at Horizontal Edges

Only:

Buckling Resistance = 1.0

Fig. 8.4-13 Supports Horizontal at Fig. 8.4-14 All Edges Supported: Edges and One Vertical Edge: Buckling Buckling Resistance = 1.5

Resistance = 3.3

Increasing the number of supported edges increases the buckling resistance (8.4.1). For the support conditions shown in Figs. 8.4-13 and 8.4-14, the buckling resistance decreases as the wall length becomes increasingly greater than the height.

8.4.2.3 Effects

of Openings - The distribution of stresses and loads around an opening is

8-19



93

m

Obb2949 0508bb3 YTB

m

governed by the depth to span ratio of the masonry material above the opening. See Fig.

8.4-15. Load

1

rcompression

Dept orizontal Reaction

Fig. 8.4-15 PanelOpenings

- ArchAction

Where the deptwspan ratioof material above an opening is equal to 1/1.5 or more, vertical load is distributed by arch action. For masonry above an opening to act as an arch, there must be sufficient masonry masson eachside of the opening to resist the horizontal thrust. Lintels at these locations only need to be designed for the weight of the triangular area of masonry above the opening, and forany other loads applied within that triangle. See MDG

11.3. With lesser deptwspan ratios, masonry above openings acts as a beam rather than anarch. The resulting stresses are illustrated in Fig. 8.4-16. Lintels at these locations must support all of the masonry above and all loads applied above. See MDG 11.3. Load

-7-

Depth

Fig. 8.4-16 PanelOpenings

8-20


- Beam Action

A C 1 TITLE*flDG

93

m

0662947 0508662 334

m

It is preferable to locate openings so that vertical loads can be transferred directly to the foundation through continuous vertical elements. See Fig. 8.4-17(a) or @).When openings are not aligned, vertical loads must be transferred through lintels. See Fig. 8.4-17(c).

m Loads

Direct Load Paths

Indirect Load Paths

0 ) Shaded Areas Denote Beam

(c) Or Arch Action

Fig. 8.4-17 Vertical Load Paths

Arch action within a panel having non-aligned openings as shown in Fig. 8.4-18 is possible, provided that the following conditions are satisfied:

I

I

I

M 1S d

Fig. 8.4-18 ArchAction 1)

The minimum vertical distance, d, between openings must be greater than or equal to 2/3 maximum horizontal dimension of lower opening.

2)

Dimensions dl and da must be sufficient for the masonry to resist compressive 8-21


A C 1 T I T L E * M D G 93 M 0662947 0508663 270 W

stresses by arch action without exceeding the allowable stress.

3)

There is sufficient masonry mass each side of the lower opening to resist the horizontal thrust resulting from arch action.

As a building’s height and floor span increases, so do the vertical compressive stresses in bearing wall elements. In a wall panel with openings, the masonry piers between openings support the gravity loads, and must be designed accordingly. Required minimum net wall area at the base is greater than that required at the roof, due to the accumulation of loads as shown in Fig. 8.4-19. Since wall strength and thickness are often kept the same throughout the height of the building, over-capacities exist at higher levels in the building if the same opening pattern is used at all levels. For more uniform capacity to required strength ratios, use smaller openings materials at lower floors, or thinner walls at upper floors.

n 5th 4th

3rd

2nd ist

Fig. 8.4-19 BuildingWall

8-22


at lower floors, higher strength

A C 1 TITLEvHDG 93

8.4.2.4

m

0662949 0508664 107 H

Gravity Stresses Resulting from Interaction of Walls and Horizontal Diaphragms

The deformed shape of a gravity loaded horizontal diaphragm restrained at its edges by the supporting masonry walls is illustrated in Fig. 8.4-20. Each panel of masonry influences adjacent perpendicular panels as the floor slab deflects.The influence resultsfrom restraint at the joints produced by bonding of the

units.

Deformations

Stresses slabs

in Stresses

in walls

Fig. 8.4-20 Interaction of Walls and Slabs Under Gravity Loads The slab and wall flexural stresses resulting from their interaction under gravity loading is illustrated in Fig. 8.4-21. The flexural stresses shown should

be superimposed upon the

vertical compressive stress in the walls resulting from gravity load support.

8-23


E:

Top Surfrn h

It.

(Typical A t Middle Of Each Edge)

Clamping Moments

From Roof And Floor Slabs

Tieing Stresses And Shear Stresses Corners At

Fig. 8.4-21 Slab and Wall Stresses Under

8.4.2.5 Engineering Analysis

Of Walls

Gravity Loads

- Wall analysis depends upon the following assumptions:

1)

direction of span (horizontal or vertical)

2)

number of continuous spans

3)

degree of fjty at each end of each span

For example, a wall that is f i e d at its base by dowels embedded into the foundation, and is laterally unsupported at its top,is designed as a pure cantilever as indicated in Fig. 8.4-22. This situation occurs when an expansion joint is located at the top of a loadbearing wall.

8-24


~~

0662949 050Abbb T A T W

A C 1 T I T L E * N D G 93

P

P Fig. 8.4-22 Unsupported Wall A loadbearing wall that is rotationally unrestrained at the base, and is laterally supported

by a diaphragm at thetop, is analyzed with simplesupports at each end as shown in Fig. 8.4-

23. Walls in typical one story buildings are analyzed in this manner.

Note that the Euler

buckling capacity of the pinned-pinned wall is four times as great as thecantilever wall.

[ \

Fig. 8.4-23 Pinned-Pinned Wall The Eulerbuckling equation for a wall pinned at one endand fixed at the otheris presented

in Fig. 8.4-24. A one-story wall with lateral supports at the top and bottom of a deep roof system may be considered rotationallyfixed at its top providing that appropriate connection details are used at top and bottom of roof framing members and also provided that the framing system is braced against lateral sway. The point offixitymay

8-25


be conservatively

assumed at the top of the wall as shown. One might consider placing

the fixity point at a

wall location between the top and bottom of the roof framing system.

P

i P

Fig.8.4-24Pinned-FixedWall Wall panels between intermediate floors of a multistorywall

are often analyzedas

continuous at each end. The bottom panel may also be considered fixed when adequately attached to a rigid foundation. See Fig. 8.4-25. These assumptions are valid only when the wall construction is not interrupted by the floor construction. The Euler buckling capacity of a fiied-fixed wall isfour times as great as that of the pinned-pinned wall, or 16 times the capacity of the cantilever wall.

i F h’

I!

P

}3

= h’l2

+

}3

= h‘12

T P

Fig. 8.425 Multistory Walls (Fixed-Fixed)

F, is defined as the allowable compressive stressdue to axial load only, and is related to the

8-26


strength Sm. The value of F, is obtained by reducing

specifiedmasonrycompressive

specified masonry compressivestrength by a safety factor against compressive crushingand a slenderness or buckling factor. In previous codes,the slenderness factor was based on the ratio of masonry wall heightto thickness. The same wall thicknesses were usedwhether the units were hollow or solid. In this Code, height to the radiusofgyration

the slenderness factor is based on the ratio of

of the crosssection,where

r =

m. This results in

allowable stresses that vary with height, unit thickness, and grouting of the cells. F, is given as 0.25 Sm[l-(h/l40r)q when h/r does not exceed 99, and as 0.25 Sm(70r//~)~ when h/r is greater than

99.

See Code 6.3.1 and 7.3.1.1.

The slenderness reduction factors for h/r not greaterthan 99 are derivedfromtests. Slenderness values for h/r greater than 99 are based on elastic stability failure theory. The curve based on these values is shown in Fig. 8.4-26, which appears in Code C. Fig. 6.3.1.

I

O Test

Results

1.2 1.o

0.8

0.6 0.4

O O

0.2

o ! O

-

I

5

10

15

20

25

30

40

35

45

hlt I

O

25

50

75 hlr

99 150

125

Fig. 8.4-26 Effects of Slenderness On Axial Compressive Strength If the wall design allows tensilestresses in the masonry (Code Chapter 6), an additional limit

8-27


~~

~

A C 1 TITLESMDG 93

Obb29Ll7 0508669 7 9 9

m

is placed on the allowable axial load. The design axialload, P,must be equal to or less than

0.25 P,,where P, is the critical Euler buckling load. P, is sensitive to slight variations in eh. The basis and development of the basic equations for unreinforced masonry construction are given in reference (8.4.3).

Pe = Perthe

n2

(1 - 0.577

h2

4)'

Code the P, equation isused

for both hollow and solidcross-section.

The

mathematical relationship is developed by considering a solid section. The applicability of the above equation to hollow cross sections or partially grouted masonry construction has been examined (8.4.4). Allowable flexural tension due to out of plane loading in unreinforced masonry is given in Code Table 6.3.1.1. Actual flexural tensilestresses under service loadingare limited to these allowables.

Fb is defined as the allowable compressive stress due to flexure only, and is given as l/3f,. See Code 6.3.1 and 7.3.1.2. An unreinforced wall, subject to axial load and bending from sources other than wind or earthquake, isdesigned by the interaction equation (unity equation) f'/F,

+ fJFb S 1.00 per Code 6.3.1.

When the bending stress results from wind

or seismic loads,the allowable stresses are increased by 1/3 per Code 5.3.2. In this casef,/F,

+ fdFb <

1.33.

The interaction equation recognizes that compressive stresses may

simultaneously occurfrom two types of loading: axial and bending. The unity equation can be extended to biaxial bending by simply adding the ratio of the calculated bending stress to the allowable bending stress for the additional axis of bending. It should be noted that this equation ignores the secondary bendingstresses resulting from the axial load. Although this omission is not conservative, the Code committee felt that it was not significant. See Code C.6.3.1. Flexural shear for unreinforced walls due to out of plane loading is seldom a controlling design criterion. See Code 6.5 and Code C.6.5.

8-28


~~


93

~

m 0662949 0508670 400 m

+ fb is limited to 1/3 r,,,(or 1.33 times 113

In reinforced masonry design, the s u m of fa

when wind or seismic loads are considered); also fa must be less than Fa. See Code 7.3.1.2 The interaction formula Code Eq. 6-1is applicable only to unreinforced, uncracked masonry. The reinforced masonrysectionisassumed

to be cracked. Therefore,the interaction

equation (unity equation) is technically not valid; however it is conservative when used to check the adequacy of the compression portion of the wall, beam, or column. The designer

must also check stresses in the tension reinforcement. Flexural shear for reinforced walls due to out-of-plane loading is rarely a design problem. See Code 7.5.

MDG 11.1 discusses the provisionsof

the Code pertaining to structural design of

unreinforced and reinforced masonrywallssubject structural design of wall systems

to flexural loads.

MDG 12.2 covers

under combined flexural and axial loading. MDG 13.1

addresses shear design in walls due to out-of-plane loading. Shear due to in-plane loading, examined in MDG 8.4.3 and MDG 13, is often a controlling design condition.

8.43 Shear Walls - Transverse lateral loads, applied to walls spanning vertically between horizontal diaphragms, are transferred to shear walls located parallel to the applied load. This transfer takes place through the horizontal diaphragms whichspan between shear walls. The shearwalls stabilize a building by transferring lateral forces to the foundations, and by resisting overturning with the gravity loads on thewalls and/or with reinforcement. See Fig.

8.4-27. If the uplift due to lateral loads cannot be resisted by the gravity load of the

Flexural Tension Vertical Reaction

Flexural Compression

Vertical Reaction

Fig. 8.4-27 Schematic Representation of In-Plane Load Transfer by a Shear Wall 8-29


A C 1 T I T L E * H D G 93

Ob62949 0508671 3Y7

m

structure and foundation, special soil anchors may be required. See MDG Chapter 9 for masonry shear wall analysis methods. The following factors are known to affect wall resistance:

1.

Horizontal diaphragm stiffness (MDG8.4.3.1)

2.

Shear wall proportions (MDG8.4.3.2)

3.

Shear resistance and axial load in stacked walls (MDG8.4.3.3)

4.

Openings in walls (MDG8.4.3.4)

5.

Placement of walls (MDG 8.4.3.5)

6.

Interconnection of perpendicular walls (MDG 8.4.3.6)

7.

Location of center of resistance (MDG 8.4.3.7)

8.

Reinforcement (MDG 8.4.3.8)

8.43.1 Horizontal Diaphragm Stiffness- Horizontal diaphragms must be carefully analyzed and designed.

Diaphragm to shear wall and non-shear wall connection details must

correspond with the design assumptions. The stiffness of diaphragms is critical for several reasons. Excessive in-plane deflection under lateralloads may overstress the masonry walls perpendicular to the load. Nonstructural interior elements, such as drywall partitions, may experience distress due to the unintentional transfer of lateral loads through connections to the diaphragm. Because shear walls are usually stiff in their own planes, the drift of a masonry illustrated

shear

wall system,

in Fig.

ordinarily small. However, designer should verify

8.4-28,

is the

Fig. 8.4-28

thatthe

8-30


Shear Wall Drift Under Lateral Loading

AC1 TITLE*MDG

93

Obb2949 0 5 0 8 b 7 2 283 H

computed drift will not distress interior nonstructural components connected to the wall. The in-plane stiffness of the floor diaphragms, relativeto the rigidity of the supporting shear walls, determines the distribution of loads to the shear walls. A rigid floor diaphragm transfers loads to the walls in proportion to the walls’ relative stiffnesses. Flexible floor diaphragms transfer loads to the walls in proportion to the distance

of the walls fromthe applied loads. In reality, diaphragms purelyrigid

are not hear Wall

or flexible, butare

generally assumedto be one or the other for the purposes of analysis. The determination of rigid flexiblediaphragms based

on

or

is commonly

the type of

Horizontal Load

Fig.8.4-29

Span to Width Ratios For Floor Diaphragms

floor

construction and the diaphragm’s span to width ratio, as shown in Fig. 8.4-29. MDG Table 8.4.1 gives the critical span to width ratios, within which the diaphragms of various materials may be considered as rigid. When the diaphragm of the structure has a greater span to width ratio, it should be analyzed as a flexible diaphragm (8.4.2).

Table8.4.1MaximumSpan-to-WidthRatiosforRigidFloorDiaphragms Floor Construction

Span-&Width

Cast-in-place solid concrete slab

5: 1

Precast concrete interconnected

41

Metal deck with concrete fill

3:1

Metal deck with no fill

21

Cast-in-place gypsum deck (roof)

3: 1

8-31


Effects of Wall Proportions - Shear walls resist in-plane lateral loads through shear and flexure. Depending on the wall proportions, the shearing or the flexural response can

8.43.2

sometimes dominate the behavior with the other response being of secondary importance. Fig. 8.4-30 divides shear walls into three categories, based upon the height to length ratio.

P 4

n

.

Type I: h’llw I 0.25

Type II: 0.25 c h’/lw c 4.0

Type III: h’/$ >4.0

Fig. 8.4-30 Flexure vs. Shear Resistance in Shear Walls Type I -

wall drift dominated by shear deformation. Wall lateral load capacity is governed by shear resistance.

Type II -

wall drift depends on shear and flexure deformation. Wall lateral load capacity depends on both shear and flexural resistance.

Type III

-

wall drift dominated by flexural deformation. Wall lateral load capacity is governed by flexural resistance.

Thus wall drifts are sometimes computed considering only shear or flexure deformations. However, for stress design, one must consider both shear and flexure. In unreinforced masonry, Types I & II shear walls are more common, since their most efficient resistance is primarily in shear.

8-32


A C 1 TITLEsMDG 9 3

m

Obb29q9 0508674 056

m

The distribution of lateral loads to all shear walls from a rigid diaphragm, or to shear walls in the same line from

a flexible diaphragm, depends upon the relative stiffness of those

walls. Wall stiffness is a function of the material used and the wall proportions. For Type

I walls of the same materials and heights, lateral stiffness isproportional to wall lengths. For

Type III walls of the same materials and heights, lateral stiffness isproportional to the wall moments of inertia (the cube of the wall length). For example, in Fig. 8.4-31,suppose that

h '/Zw < 0.25 for both walls. Their lateral stiffness w l i be proportional to their areas (or lengths, in the case of uniform thickness). Therefore Wall 2 has 10 times the length of Wall 1 and will receive approximately 10 times the shear resulting from a lateral load.

AResistance Wall 1

Wall 2 L e n g t h = 101,

Length = I ,

Fig. 8.4-31 Lateral Force Distribution

If, however, h '/Zw > 4 for both walls, their lateral stiffnesses will be proportional to their moments of inertia (length cubed). Therefore, Wall 2 will receive approximately1,000 times as much lateral load, or flexure as Wall 1. Stiffness-based load distribution assumes that a continuous load path (usually a continuous diaphragm) exists to transfer the lateral load to allresistingwalls.

For example, if the

horizontal diaphragm in Fig. 8.4-31 were interrupted by flooropenings,

it would be

reasonable to provide a load path to transfer 90% of the load to Wall 2.

8.433 Effects of Axial Loads - As shown in Fig. 8.4-32,compressive axial loads may coexist with shear forces. Masonry in compression is better able to resist shear. This phenomenon is recognized by Code 6.5.2(c).

8-33


A C 1 TITLE*flDG

93

m

Obb29Y9 0508675 T92

m

Roof

Fig. 8.4-32 Action of Shear and Vertical Load Thus, shear walls in compressionhave

an increased shear capacity.Addingvertical

compression is best accomplished by making the shear walls loadbearing.

Effects of Openings - Fig. 8.4-33 illustrates the behavior of an unperforated shear wall under in-plane lateral force P,.Lateral forces are transferred to the shearwall through shear in the the horizontal diaphragm connections to the wall. As showninFig.8.4-33, 8.43.4

masonry is resisted primarily by a diagonal compressionstrut. In an unreinforced wall, shear capacity is usually governed

by diagonal tension capacity, where

the diagonal tension is

perpendicular to the compression strut. Floor Diaphragm

Connection To Wall Panel

P,

+

1

I

r-

" J

Floor Diaohtanm Connect& TÕ

d-,

ZPl (Reaction At Base Of Wall)

Wall Panel

Fig. 8.433 Shear Resistance Mechanism in Unperf'oratedWall Walls with openings resistlateral load in flexureas well as in shear. Fig. 8.4-34(a) shows the primary shear resistance mechanismin a wall with a large central opening. The stiff portions 8-34


~


93

m

Obb2949 0 5 0 8 6 7 b 929

of the panel above and below opening resist shear through diagonal compression strut either side of the panel opening deform in

action. The more flexiile pier portions on

flexure as well as shear as shown in Fig. 8.4-34@). Based on aspect ratio, their resistance may be governed by flexure as well as shear.

pI h‘lLw

I Pl h’ltw

(b) Pier Deformation

(a) Compression Strut

Shear Resistance Mechanism of Perforated Walls

Fig. 8.4-34

Fig. 8.4-35 illustrates how the size of the opening affects the mode of shear wall resistance.

If the opening is relatively small,as in Fig. 8.4-35(a), the diagonal shear stresses flow around the opening and the behavior is similar to a wall without openings.

/

Compression Strut

Fig. 8.4-35

Effect of Opening Size on Diagonal Strut

Large openings divide the panel into smaller shear resisting panels that are tied together by connecting portions subject to flexure and shear. This is illustrated in Fig. 8.4-35@), which shows a wall panel divided into two portions connected by coupling beams above and below the opening. 8-35


8.43.5

Effects of WallPlacements

-

Lateral loads are distributed amongwallsin

proportion to their lateral stiffness. Since the lateral stiffness of a wall about its weak axis (Fig. 8.4-36(a)) is relatively small,

it is often neglected in analysis. Only the walls parallel

to the applied load are considered (Fig. 8.4-36@)). Neutral Axis

Neutral Axis

Lateral Force

I

lb

'

FLateral o r e r f J

I

I Plan View

Plan View

a) negligible lateral stiffness Fig. 8.4-36 Effect

b) significant lateral stiffness of Wall Orientation on Stiffness

Since lateral loads act in any direction, as indicated in

Fig. 8.4-37, shear walls must be

oriented along both majoraxes of a building. If shear wallsprovideresistancein

one

direction only,other lateralforce resisting elements must be provided in the other direction.

Force L a t e r a l

I

Lateral Force

Plan View

Plan View

Fig. 8.4-37Multi-AxisLateralWalls 8.43.6 Effects of Interconnection of Perpendicular Walls - Connecting perpendicular walls can greatly enhance the structure's resistance to lateral loads. For example, without the 8-36


TITLEtMDG 93 m 0662949 0508678 7TL

AC1

interconnection of perpendicular walls in Fig. 8.4-38(a), the lateral flexural stiffness is only

12

. If the wall connections meet the requirements of Code 5.13.4, the flexural stiffness 2

increases to

t(1J3 ):+()G(.

12

. In a multistory building, the walls are usually tall enough

for flexural deformation to dominate lateral drifts and, the increased stiffness will significantly reduce lateral drift.

Wall

(a>

Plan View Must Resist SIhnetaerr f a c e (b) For Composite Action

Fig. 8.4-38 AdjacentWallInterconnection However, if movement joints are required at these locations, the designer cannot utilize the extra stiffness potential of the perpendicular wall. Thus the designer must carefully review volumechange

requirements in ordertoensurethattherestraint

provided by these

perpendicular wall connections will not adversely affect the structure's performance. (See

MDG Chapter 10). The requirements and limitations of Code 5.13.4.2, relative to the connections of flanged walls, are presentedin Figs. 8.4-39 through 8.4-42. The designer has the option of providing wall connections in accordance with Fig. 8.4-39 and Fig. 8.4-40, or Fig. 8.4-41 or Fig. 8.4-42

8-37


Flange Thickness

-

Flange

'Flange

Fig. 8.4-39 Flanged Wall

'Flange

u

-

Effective Flange = Six Times Flange Thickness

-

- T Intersection

Fig. 8.4-40 Flanged wall L Intersection (Code 5.13.4.2(e)l)

(Code 5.13.4.2(e)l)

Wall Flange Thickness Wall And Flange Interface Regularly Tooth With 8" Maximum Offsets

1Reinforcing (If Required)

Strap Anchors Maximum Spacing Of 4 Feet Vertically On Center. 4"x$" x 28Grout in Wall 24 "

2"

m2" Or

2"

Fig. 8.4-41 FlangedWall

- L Intersection(Code5.13.4.2(e)2) Intersecting Reinforced Bond Beams At 4"O" On Center .um

Continuous Reinforcing Bars Grout In Wall (If Required)

cWebBondBeam

Reinforcing

BondBeam Reinforcement = 0.1 i& er f t . Of Wall Minimum Develop Reinforcement EachSide !O Intersection.

Fig.8.4-42FlangedWall

- T Intersection(Code5.13.4.2(e)3) 8-38


A C 1 TITLE*MDG

8.43.7

73

m 0662749 0508680

35T 9

Effects of Location of Plan Center of Resistance - Ideally, the line of action of

lateral loads should coincide in plan with the line of action of resistance of the structural system, as shown in Fig. 8.4-43 (a). In that situation, the shear walls are subject to direct shear only.If

8.4-43(b), a

these two lines of action do not coincide, as showninFig.

torsional moment equal to PItimes e develops. In this situation, the walls parallel to the applied lateral force are subject to direct shear, and all wallsare subjected to torsional shear.

Line Of Action Of Load And Resistance

Line Of Action Of Resistance

Lateral Loa+ Twist of Structur

L L i n e Of Action Of Load

Plan View

Plan View

(a) Concentric Load

(b) Eccentric Load

Fig. 8.4-43 Lines of Action of Lateral Load and Lateral Resistance To determine the magnitude of torsional shear force applied to each wall, determine the ratio of each wall’s stiffness to the total stiffnesses for allwallsin

both directions. The

torsional shear force is equal to P, times e times the stiffness ratio divided by the wall’s distance from the applied load. See MDG 9.2.2.

8.43.8

Wall Reinforcing Patterns

-

In masonry construction,

placement of wall

reinforcement is influenced bywall configuration, wall openings, and the wall’s intended structural action. For example, consider the shear walls of Fig.

8.4-44. Because they are

intended to be uncoupled under lateral load, the walls are connected only by a shallow floor slab, and they are reinforced for individual cantilever action. In contrast, the shear walls of Fig. 8.4-45, which are intended to act as a unit, are coupled by deep, appropriately reinforced masonry lintel beams and are reinforced for coupled wall action.

8-39


A C 1 T I T L E x N D G 93

0662747 0508682 296

Continuous Tie Reinforcement At Roof Diaphragm (If Required) Wall To Diaphragm Connections Required To Activate All Shear Wall Panels

Continuous Tie Reinforcement A t Floor Diaphragm(1f Required) Panel

Joint Reinforcing Continuous Reinforcement At Ends Of Walls (If Required)

Fig. 8.4-44 Typical Reinforcement Pattern for Uncoupled Shear Walls

Continuous Tie Reinforcement At Roof Diaphragm (If Required) Continuous Tie Reinforcement At Floor Diaphragm (If Requited) Panel

Horizontal Reinforcement A t Coupling Beam Shear Reinforcement In Coupling Beam

Joint Reinforcing Continuous Reinforcement A t Ends Of Walls (If Required) Effective Length

Fig. 8.4-45 Typical Reinforcement Pattern for Coupled Shear Walls

'8-40


A C 1 TITLE*NDG

8.43.9

0 6 6 2 9 4 9 0508b82 1 2 2

93

EngineeredDesign of MasonryShearWalls

- The shear

m

design of shear walls

depends upon the magnitude of axial load in the wall. If the axial load is sufficient

to

overcome the flexural tension resulting from lateral loads, the section may be designed as unreinforced in accordance with Code6.5. Shear reinforcement is onlyrequired iff. exceeds

F, If flexural tension exists inthe wall, F, is first calculated fromCode 7.5.2.2 Iff. exceeds F, shear reinforcement must be provided, and F, is recalculated from Code 7.5.2.3. If f v exceeds the recalculated F,,,the wall must be increased in size. Fig. 8.4-46 illustrates wall sections without and withnet flexural tension (from Code C Fig. 7.5-1(a) and Fig. 7.5-l(b)). Fig. 8.4-47 presents a flow chart for shear design (adapted from Code C.Fig. 7.5-1).

%

Flexural

m

Flexural

Axial Axial Combined Flexural and Axial

Combined Flexural and Axial

Shear fv =

Ib

Fig. 8.4-46(a) Illustration Of Design Fig. 8.4-46(b) Illustration Section Without Tension NetSection With Tension Net

8-41


Of Design


93

m

0662949 0508683 Ob9

m

Determine Design Forces Campte Maximum Stresses from Combined Forces.

Designed per Chapter 7 Reinforced Masonry?

1 I

t

A to Flexural

\

Tension?/

No

I

Code Eq. 6-7. Calculate f, by

No

I See Fig. 8.4-4Ka).

II

:A;' \ \Cade

1

65.2?/

I

Calculate f,

Reinforce According to Code 7.5.1 and Redesign

by Ccde Eq. 7-3. See Fig. 8.4-4qb).

Shear Requirement Satisfied

rement 1 I

I

l

Fig. 8.447 FlowChartforShearDesign

8-42


No

Shear Requirement "~_.-*.->

A C 1T I T L E S M D G

93

m 0662749

O508684 T T 5

m

-

8.43.9.1 Shear Walls Without Net Flexural Tension If M/S is less than PIA,,, then theaxial

compressionexceeds the flexuraltension,and illustrated in Fig. 8.4-46(a).

no net flexuraltensionresults.Thisis

In this case, the actual shear stress is calculated from Code

6.5.1, and the allowable shear stress is calculated from Code 6.5.2.

F,,is the least value of

fv

=

V0

"g

a)

1.56

b)

120 psi

c)

U +

0.45

4 N Y

(see Code 6.5.2(c) for u)

d) 15 psi other than running bond with other than open end units grouted solid Iff,, is less than or equal to F,, the section is satisfactory. If not, the wall must be increased

in size, or reinforced to resist all calculatedshear. If the designer elects to reinforce the wall, the allowable shear stress is calculated from Code 7.5.2.3 and the amount of reinforcement is calculated from Code 7.5.3. See MDG 8.4.3.9.2. 8.43.9.2 Shear Walls With Net Flexural Tension

- If M/S exceeds PIA,,, then the wall has

net flexural tension, as illustrated in Fig. 8.4-46(b). In this case, the actual shear stress is calculated from Code 7.5.2.1. It isusually desirable to avoid the requirement for shear reinforcement. Therefore the allowable shear stress is first calculated from Code 7.5.2.2, where shear reinforcement is not provided to resist the shear.

M Vd

If-<

I f -M2 1

1

Vd

Iff,, is less than or equal to F,, the section is satisfactory. Iff,, exceeds F,, calculated above, 8-43



m

Ob62747 0 5 0 8 b 8 5 931

m

shear reinforcement is required, and F, is recalculated from Code 7.5.2.3.

M

If-<

Vd

M

1

If-r1

Vd

Fv = 1.5

and Fv

If f v stillexceeds F,, the wallmust

S

be increased in size.

fi

75 psi

Whenever reinforcement is

necessary, the required area of reinforcement is calculated by Code 7.5.3, where F, is the allowable stress in thereinforcement.

Steel tensilestressallowables

are giveninCode

7.2.1.1.

The shear reinforcement is placed parallel to the direction of the applied shear force. The reinforcement spacing is limited to the lesser of d/2 or 48 in. Additional reinforcement, perpendicular to the shear reinforcement, is required by Code 7.5.3.2. The total additional reinforcement area must be at least one-third of the shear reinforcement, and is spaced uniformly at a maximum spacing of 8 ft.

MDG 13.2 specifically addresses the provisions of the Code to the structural design aspects of unreinforced and reinforced masonry shear walls. 8.4.4

Progressive Collapse

Failure of an individual element that leads to failure of additional elements is referred to as progressive collapse. The initial localized failure maybe relatively insignificant; however, the progressive failure may be major and may include complete collapse of the structure. See Fig. 8.4-48.

8-44


A C 1 T I T L E * H D G 9 3 D 0bb2947 0508b8b 8 7 8

m

Impact Destroys Relined Boundq: DiaphragmAction of Floor Act1 I S a Beam Between Adjacent Undamaged W%.

Explosion Damages Unreinforced Panel

Continuity

Tie Reinforcement

Precast Flous: Explosion Removes Unreinforced or LlGhtly Reinforced Section of Wall. Reinforced Boundaries Remain And Carry Buildin Loads A s Masonry Columns. At h m r Line Develops Resulting A n h, Action

Fig. 8.4-48 Progressive Collapse

\ .

Vehicle Removes Wall Panel And Exterior Reinforced Vertical Boundary

Reinforced Boundaries Remain And Carry Buildiag Loads As A Masonry Column And As ATic

- Examplesin Masonry Structures

Although not required by the Code, it is prudent for an engineer to provide sufficient element continuity to reduce the risk of progressive collapse. The key to achieving masonry element continuity is to tie the structural elements together with reinforcing steel.

The

following items should be considered: vertical reinforcement at each end of a bearing or shear wall; horizontal reinforcement inwalls at the floor levels; reinforcement around openings in walls; anchorage between walls and floor/roof diaphragms; redundant systems; and alternate load paths. For example, verticalreinforcement at the endsof all walls serves as compression or tension reinforcement. Horizontal reinforcement in walls at the floor levels can serve as a tensile tie for a masonry arch if part of a wall is removed. 8.5

8.5.1

STRUCTURAL CONSIDERATIONS FOR MASONRY BEAMS Beam Behavior

A beam is a horizontal member that is subjected to vertical or horizontal loads, and that spans between points of support. Axial loads in beams are usuallyignored for design purposes except in shear wall coupling beams. 8-45


A C 1 TITLEvMDG 9 3

0662949 0508687 704

m

Beams, like other structural elements, must satisfy conditions of equilibrium, stress-strain relationships, and kinematics(deformations). supported beam isshowninFig.

The deformation pattern of a simply

8.5-1. The upper fibers shorten, and the lower fibers

lengthen. Fiber strain is proportional to eachfiber's distance from the beam's neutral axis. For linear elastic material, stress is proportional to strain; therefore, stresses also increase in proportion to distance from the neutral axis.

A

Radius of Curvature (R)

Fig. 8.5-1 BeamBending Axial equilibrium requires that the beam's neutral axis lie at the geometric centroid of the cross-section. The resultingdistribution

of stresses isshowninFig.8.5-2(a)

foran

uncracked, unreinforced rectangular section, and in Fig. 8.5-2(b) for a cracked, reinforced rectangular section. In the latter case, the section is often treated as being of a single material. For this purpose, the reinforcement is transformed into an equivalent area of masonry by multiplying the steel area by the modular ratio (steel modulus divided by the masonry modulus).

fbt

t

k"+ Ir

a) Unreinforced

b) Reinforced (Cracked)

(Uncracked)

Fig. 8.5-2 FlexuralStressDistribution 8-46


A C 1 TITLE+MDG 93

m 0662947 0508688 h40

W

The requirement that the internal resisting momentequal the external applied moment leads to therelationship between moment and flexural stress at a distance y from the neutralaxis.

f = M,Y The maximum internal flexure stress occurs inthe extreme fiber, located at a distance c from the neutral axis, and is given by

Mc - M f = ÏS - The quantity S iscall the sectionmodulus.Given

the same external moment, a larger

section modulus will result in lower stresses. Kinematics leads to thefollowing relationship between the beams radiusof curvature, R, and the change in the length A of a fiber at a distance y from the neutral axis, and the corresponding strain in the fiber.

The two previousexpressionscanbecombined

to obtain a singleexpressionrelating

moments and curvatures:

The quantity EI is sometime referred to as the flexural stiffness of the cross section. A

larger modulus of elasticity or moment of inertia will result in smaller deformations, given the same external moment. The compressive portion of a beam is subjected to flexural stresses that are similar to those in a beam column. Like a beam column (MDG 8.6), a beam can buckle sidewaysif it is not sufficiently supported laterally. Factors that affect transverse stability are illustrated in Fig.

8.5-3.

8-47


AC1

TITLE*NDG 93

Ob62949 0508b89 587

m

Lateral Buckling Is Proportional To The Unsup orted S an The Beam Depth, lnnd The$ak Width

Compression

Tension Reaction

Y Fig. 8.53

Lateral Buckling

To avoid the need to reduce the allowablecompressive stress inbeams,similar

to the

column design method, Code 7.3.3.4 requires bracing of the compression flange against lateral displacement. Supports must be provided at the level of the compression face at a distance not to exceed 32 times the width of the compression face. In addition to bending moments,beams are also subjectedto shear. Like bending moments, shear forces are required to maintain beam equilibrium. As illustrated in Fig. 8.5-4, for the case of a simple beam with uniform load,shear is equal to the rateof change (slope) of the bending moment diagram.

Fig. 8.5-4 Relation Between Shear And Bending Moment Shear forces acting on a beam produce shear stresses, which act horizontally and vertically.

8-48


A C 1T I T L E * U D G

93

m

Ob62949 0508690 2 T 9

m

As shown in Fig. 8.5-5, the magnitude of the horizontal stresses can be computed from the stresses (proportional to the shear), and the properties of the cross-

change in flexural

section. In general, horizontal shear stress is given by

a

Difference Of Flexural Stresses

I

1 1

Vertical Shear Stress

d

I

dx

a)

Derivation of Horizontal Shear Stresses

Equilibrium b) Vertical and

of Forces From Horizontal Shear Stresses

Fig. 8.5-5 Beam Shearing Stresses As shown in Fig. 8.5-5, to achieve equilibriumof a differential element, the horizontal shear stresses must be accompanied by equal and opposite vertical shear stresses. For the particular case of a crackedsection, the aboveexpression for horizontal (and vertical) shear stress leads to the following expression for maximum shear.

V

fv

=

bjd

Under combinations of moment and shear, a beam is subject to combined flexural and shearing stresses. The directions of the resulting principal tensile and compressive stresses can be shown in the form of stress contours, illustrated in Fig. 8.5-6 for a simple beam with uniformload.

Atthe

beam midspan,where

8-49


shear is zero, pure tension and pure


93

m

Obb2949 0508b93 II35

compression act on the bottom and top of the beam, andthe

m

principal stresses are

horizontal. Near the beam ends, principal stresses act at an angle.

\

\

Pure Tension

Tendan and Compression

Fig. 8.5-6 Stress Contours 8.5.2

Engineered Design of Reinforced Masonry Beams

Code 7.3.3 defines beam span length for the purposes of analysis. Simple span beams, or those not built integrally with supports, have a span length equal to the lesser of the clear span plus the member depth, or the distance between centers of supports. The span length of beams continuous over supports is the center to centerdistance between supports. Code

7.3.3.3 further specifies that the minimum length of masonry beam bearing over supports is 4 inches in the direction of the span. The span of a cantilever beam is not defined by the

Code but is usually measured to the face of the support. For the reinforced rectangular beam illustrated in Fig. 8.5-2, the equations for flexural and shear stresses in the beam (see MDG 11.1) are as given below.

C = T

f

b

=

c r --

-bkd 2

8-50


M 1

-bjkd2 2

A C 1T I T L E * I I D G

93 M Ob62949 0508692 0 7 2 D

The allowable masonry flexural compressive stress is 1/3rm. The allowable tensile stress for steel reinforcement is 20,000psi for Grade 40 or Grade 50 steel, and 24,000psi for Grade 60 steel. The resistance of steel reinforcement in the compression zone is neglected, unless lateral reinforcement that meets the requirements of Code 5.9.1.6 is provided. In that case, the steel is transformed to the equivalent masonry area as defined in Code 5.13.1.2. The design for allowable shear follows the flow chart presented in Fig. 8.4-47. Since beams are usually subject to flexural tension, the allowable shear stress F, is first calculated from Code 7.5.2.2(a): F,=@oPsi

If f, is less than or equal to F,,,no shear reinforcement is required. If f, calculated above exceeds F,,shear reinforcement is required, and the amount is calculated by Code 7.5.3.

Either the calculated shear stress must be less than the allowable stress F, recalculated by Code 7.5.2.3, or the cross section must be increased.

F,,= 3.0/$$ S 150 psi Code 7.3.3.5 and Code 5.6.1 limit the deflection of beams supporting unreinforced masonry to U600 or 0.3 inch. MDG 11.3 specially addresses the Code provisions for flexural structural design of masonry beams, while MDG 13.1 addresses beam shear considerations.

8-5 1


TITLE+MDG 93

AC1

8.6

8.6.1

Obb2949 0508b93 T08

STRUCTURAL CONSIDERATIONSFORMASONRY BEAM COLUMNS

BeamColumnBehavior

Columns resist compressionand flexure. As discussed inMDG 8.4.2.3 for loadbearing walls, columns may fail in one of two modes. 1. Material compression failure, illustrated in Fig. 8.6-1: Load

Strength is proportional to the ultimate compression or bearing capacity of the material, and the cross-sectional area of the column.

-.-. 0.

/*

/

Fig. 8.6-1 MaterialCompressionFailure 2. Buckling, illustrated in Fig. 8.6-2 Column II carries 4 times the buckling

IP

Fixed

load of Column I. See Fig. 8.4-22 and 8.4-

24 for other end fixity conditions. For any end condition: h'

', \

h '14

Fixed Column I

Column II

Fig. 8.6-2 CompressiveBucklingFailure 8-52


A C 1 TITLE+MDG 93

Ob62949 0508694 9 4 4

m

Column buckling load is proportional to the material's modulus of elasticity and the crosssectional shape, and isinversely

proportional to thesquare of the column'seffective

unbraced length. The shape of the cross section determines the moment of inertia and the area An index for thecross sectionalshape is expressedas@,

and is called the radius of gyration, r. The

slenderness of the column is expressed by hlr. 8.6.2

Engineered Design of Masonry Beam Columns

Code 5.9 gives general requirements for column dimensions, reinforcement, and analysis. Each nominal side dimension mustbe at least 8 in., and the ratioof effective heightto least nominal side dimension must not exceed 25. Columns must have a minimum of 4 vertical reinforcing bars, and the areaof reinforcing steel may not be less than 0.0025 A,, nor exceed

0.04An. Lateral tie reinforcement at least 0.25 in. in diameter must be provided at a spacing and configuration s-pecifiedby

Code 5.9.1.6.

For analysispurposes,columnsmust

be

designed assuminga minimum axial load eccentricity equal to0.1 times each side dimension, where each axis is considered independently. Beam columns are designed for axial load, axial plus flexure, and shear as needed. The allowable masonry compressive stress due to flexure plus axial load, is 1/3fm, provided that the calculated compressive stress due to the axial load component, fa, does not exceed the allowable stress F,, given in Code 7.3.1.1.

For hlr c 99, F,, =

1 -$, 4

8-53


93 M 0662949 0508695 880


For hlr r 99, Fa =

m

$r

id( 4

Previous codes and industry standards permitted the area for allowable axialstress to include some percentage of the transformed steel reinforcement area. However, the Code is silent

on this issue. The MSJC is currently considering a revision to address this subject. MDG 12.1 specifically addresses the provisions of the Code to the structuraldesign aspects of reinforced masonry columns. 8.63 Interaction Diagrams

Columns can be subject to both axialloadandflexure.

Depending on the relative

magnitudes of the axial load and bending moment, one of three cases will exist:

I.

The allowable axial load on the beam column (independent of moment) is based on the net area of masonry only, and is governed by the allowable axial compressive stress, F,, as defined in Code 7.3.1.1.

II.

Combinations of allowable beam column moment and axial force, computed using a cracked transformed section, are governed by the allowableflexuralcompressive stress, Fb, as defined in Code 7.3.1.2.

III.

Combinations of allowable beam column moment and axial force, computed using a cracked transformed section, are governed by the allowabletensile

stress in

reinforcement, F,, as defined in Code 7.2.1.1.

As discussedin detail in MDG 12.1, these three casescan be graphicallyexpressed as interaction diagrams. A simplified diagram is shown in Fig. 8.6-3.

8-54


A C 1 T I T L E * H D G 93 W Ob62949 0 5 0 8 b 9 b 717 W

Bending Moment

M

Fig. 8.63 InteractionDiagram Combinations of beam column axial load and bending moments lying within the shaded area are allowed; combinations outside the shaded area are not allowed.

REFERENCES 8.3.1 ASCE Standard, ASCE 7-88, "MinimumDesign Loads for Buildings and Other Structures," American Society of Civil Engineers, New York, NY, 1990.

8.3.2 Guide to the Use of the Wind Load Provisions of ASCE 7-88, American Society of Civil Engineers, 1992.

8.3.3 Ellifrit D. S., IlThe Mysterious 1/3 Stress Increase",AISC Journal, 4th Quarter, American Institute of Steel Construction, Chicago, IL, 1977.

8.4.1 L McGraw-Hill, 1965. 8.4.2 Building Code Requirements for Masonry Structures, American Concrete Institute, Detroit, Michigan, 1988.

8.4.3 Colville, James, "Service Load Design

Equation for Unreinforced Masonry

Construction," TMS Journal, Vol. II, No. 1, August 1992.

8.4.4 Colville, James, "Stability of Hollow Masonry Walls," August 1992.

8-55


TMS Journal, Vol. II, No. 1,

A C 1 T I T L E + M D G 93

m 0662949 0508697 653 m

DISTRIBUTION OF LOADS

9.0

INTRODUCTION

A major criterion that influences the design of any structure is the distribution within the structure of the effects of dead, live, and other superimposed loads. This chapter deals with the various load distribution concepts with respect

to the design and analysis of masonry

structures. Two types of load distribution in masonry structures are considered. Global distribution is concerned with the transfer of loads to the various components of the structure (i.e., slabs, beams,walls,pilasters,columns,

and footings).Thisdistributionmayhave

to take into

account the stiffness of structural components, and must consider the existence and location

of control and expansion joints. Local load distribution is concerned with how loads

are

distributed within individual structural components; i.e., the distribution of concentrated loads in single wythe, multiwythe

and composite walls, distribution

of loads under bond

beams, and load distribution within perforated shear walls. The concepts of loaddistribution,

as well as the application of the Code to the

comprehensive design of typical masonry buildings,are presented in this Guide using three typical, and realistic masonry structures as examples. The three structures and their design loads are describedin MDG 9.1. The global distribution of these loads to the various components of each structure is presented in MDG 9.2, Interwall Load Distribution. Typical component examples considering various construction options are examined. The local load

9-1


AC1 TITLExMDG 93 m 0662949 0508698 5 9 T m

distribution within a wall is presented in MDG 9.3, Intrawall Load Distribution. Numerous situations commonly found in structural design of masonry elements are examined. 9.1 BUILDING EXAMPLES

Three buildings arepresented

in thissection:

a singlestory

strip shopping center, a

gymnasium, and a four story hotel. These buildings and their masonry elements will be used in subsequent chapters to demonstrate masonry structural design procedures. The purpose ofusing three typical buildings is not

to provide the reader with final designs

for each

structure, but to illustrate the application of design methodology and philosophy presented

in MDG Chapter 8 to typical real life situations and to provide some insight into the practical masonry design requirements of the MSJC structural masonry design code. Each masonry system (suchas anexterior loadbearing wall) may havea number of different configurations(forexample,solidwall,noncompositewall,etc.).Both

reinforced and

unreinforced examples of different configurations will be investigated for each building. Since the design of masonry structural systems requires a detailed analysis of the building loads, a typicalgloballoadanalysis

will be performed and presented for each building

example. Code 5.1 requires that structures and their components be designed by elastic analysis under service load conditions. The following analyses are based on elastic material behavior ând will be restricted to the determination of gravity loads on walls, and global earthquake andwind loadings. Lateral load distribution to individual building elements can

-

involve the determination of masonry wall stiffness and will

be addressed in MDG 9.2

Interwall Load Distribution. Specific examples

loads are distributed within

ofhowwall

individual wall elements are presented within MDG 9.3 - Intrawall Load Distribution. 9.1.1 TMS ShoppingCenter

Figs. 9.1-1 and 9.1-2show the plan and elevations of a single-story, 16,000 square foot

9-2


A C 1 TITLE*MDG

93

m

Ob62949 0508679 426

shopping center. The north, east, west and central fire wall are constructed of concrete masonry. The south wallisprimarily

a glass curtain wallwith one masonry shear wall

element. The roof framing system consistsof a one-way steel joist and beam system supported on the concrete masonry wallsand steelcolumns, with a five-foot overhang on thesouth side. This roof framing system is typical of many low-rise masonry commercial buildings.

To illustrate the application of the Code to typical structural design considerations for this type of structure, 29 example problemsare presented throughout the MDG Design Chapters 9 through 16 for theT M S Shopping Center. A listing is included in thissection to assist the reader in correlating the illustrated example design issue to the plans and elevations of the structure.

T M S SHOPPING CENTER EXAMPLE PROBLEM INDEX All Wall Construction Options Examule #

Desim Issue

9.2-1

Lateral load distribution (all walls)

9.3-1

Concentrated load distribution without bond

9.3-2

Concentrated load distribution

9.3-3

Concentrated load bearing area(east

and west walls)

9.3-4

Concentrated load bearing

and west walls)

9.3-5

Concentrated load bearing area(ea@ and west walls)

9.3-10

Out-of-plane lateral and axialload

beam(north

wall)

with bond beam (north wall)

area(east

distribution within peforated wall

(east wall)

14.3-5

Joist bearing connection to wall (north wall)

14.3-11

Steel beam bearing connection to wall (eastand westwalls)

14.3-12

Roofdiaphragm to shear wall connection (interior wall)

9-3


A C 1 TITLExMDG 93

m

0662949 0508700 T78 M

va

W

l

o Fig. 9.1-1 TMS Shopping Center 9-4


~~

ACI TITLErMDG 93 m 0 b b 2 9 4 9 0508703 904 m

Fig. 9.1-2 T M S Shopping Center Elevations 9-5


A C 1 TITLE*UDG

93

m 0662949 0508702 840

Wall Construction Option A (Unreinforced CMU) Example #

Desim Issue

9.3-12

In-plane lateral load distribution within perforated wall (east wall)

10.4-1

Control joint locations (all walls)

11.1-1

Wall design

for out-of-plane flexure (east wall)

11.1-3 Wall design

for out-of-plane flexure (east wall)

12.2-2Walldesign

for axial and out-of-plane flexure (north wall)

13.1-6 Wall design

for out-of-plane shear(interior wall)

13.2-1Walldesign

for in-planeflexure and shear(east wall)

15.4-1

Empirical wall design (all walls)

Wall Construction Option B (Reinforced CMU) Examde # 9.3-11

Issue

Design

In-plane lateral load distribution within perforated wall (east wall)

11.1-2 Wall design 11.3-2Doubly

for out-of-plane flexure (east wall) reinforced masonrylinteldesign

for flexure (east wall)

12.2-1Walldesign

for axial and out-of-plane flexure (north wall)

13.1-9 Wall design

for out-of-plane shear(east wall)

13.1-10Doubly 13.2-2Walldesign

reinforced masonry lintel design forshear(east for in-planeflexureand

shear(east wall)

14.2-1

Straight bar foundation dowel anchorage (all walls)

14.2-2

Hooked bar foundation dowel anchorage (all walls)

14.2-3Doubly

wall)

reinforced masonrylintel reinforcement anchorage (east wall)

Miscellaneous Example # 14.3-1 Roof

Desim Issue diaphragm to shear

wall connection

9.1.1.1 Gravity Design Loads - The gravity load analysis is straightforward, requiring the application of the tributary area concept and simple statics. It is assumed that the dead load of the roofing system, including all framing members, 9-6


mechanical systems and ceiling, is a uniformly distributed 15 psf. The governing roof live load is a 30 psf snow load.

In addition, the height of the parapet wall is such that lateral

drift need not be considered. The steel joists spanning between Grid Lines A and B are spaced at 5 ft, so that each is uniformly loaded by: W =

(30psf x 5 ft) + ( 15 psf x 5 ft) = 150.0 plf

Live Load

+

75.0 plf = 225 plf

Dead Load

Assuming an average wall thickness of 8 in., the joist reactions on the north bearing wall (Grid Line A) and on the beams on Grid Line B are:

RA

=

RBI = (41

*-

h‘12 h*’ft) x 225 plf 2

=

4,538 lb;

use 4,540lb

For the cantilevered joist (steel joists with extended ends) spanning between Grid Lines B and C, it is conservatively assumed that full snow and dead load on the center span and dead load minus wind uplift on the cantilevered span will produce the greatest reaction at Grid Line B. Assuming a fascia dead load of 10 psf (or 70 plf for the 7 ft deep fascia) and a wind uplift of 1.5 x 20 psf = 30 psf, this reaction is:

RE=- 1 40.5 ft

X

[

-(70 plf

+ 225 plf x

RB2 = 4,536 lb;

X

5 ft

X

5 ft) + (30 psf- 15 psf)

X

5 ft

X

( 5 ftI2 2

2 use 4,540lb

(Note that thewind load is increasedby a factor of 1.5 to account for pressure build up under the canopy) The total load on Grid Line B from the joists is 4,540 lb

9-7


+ 4,540 lb = 9,080 lb.

A C 1 T I T L E * H D G 93

5 f t Typical

6,050 lb

Ob62949 05087011 b L 3

9,080 lb Typical

f l t t t - + t t t I

4

4 " 1"

t

t

T Ï 12 "9"

Suspended Span

B- 1 Cantilevered Span

B

24 '- 1"

Fig. 9.1-3 Cantilevered Beam System on Grid Line B - Section A-A (Fig. 9.1-1)

These reactions are spaced at 5 ft centers along the cantilevered beam system.

The joist

reaction located at 1ft - 6 in. from the outside face of the walls is only 6,050 lb due to the reduction in tributary area (see Fig. 9.1-3). The total beam reaction load on the masonry walls at B-1 and B-3 is 15,770 lb, of which 5,090 lb is dead load and 10,680 lb is live load. Assuming full dead and live load on the cantilevered joist between Grid LinesB and C, the typical load from each joist on Grid Line C (ignoring the wind uplift under the canopy) is:

Rc

=

5,830 lb

(2,000 lb dead load and 3,830 lb live load)

The corresponding load from the joist located 1 ft - 6 in. from the exterior face of the wall on Grid Line 1 onto the W 16 x 26 is 3,890lb (1,330 lb dead load and 2,560 lb live load).

9-8


A C 1 TITLESMDG 93

m

Obb2747 0508705 S S T

m

The W 16 X 26 beam reaction totthe masonrywalls at the junctions of Grid Line C and Grid Lines 1 or 3 is:

- 13,890lb R1-C,3-C

-

Rl-C,3-C

=

x 20

11,940lb

ft

+

5,830 lb(15 ft + 10 ft + 5 21.167 ft

(4,090 lbdeadloadand

fi)]

7,850 lb live load)

(assumes an 8 in. wall thickness) The reaction of the W 16 x 26 beam on the short wall at Grid Line C, between Grid Lines 1 and 2, is:

&z-z

= (5

&z-2

=

fi

10 fi + 15 ft + 20 fi) x

+

13,666lb

=

13,670lb

5,830 lb 21.33 fi

(4,690 lbdeadload

and 8,980lblive

load)

(assuming 6 in. of bearing of the steel beam onto the masonry wall fields a

21 ft - 4 in. span) The reaction of the beam and joist at the junction of Grid Lines C and 2 is:

-

Rc-2 =

12 h/ft

5ft 12,412lb;

use 12,410lb

x 5,830 lb +

5,830lb ( 5 ft

(4,260 lb deadloadand

+

19

10 ft + 15 ft)

ft

8,150lblive

load)

The above calculation used the ratio of joist tributary widths to calculate the joist reaction and assumed a beam span of 19 ft.

9.1.1.2

Lateral Design Loads - Both seismic and wind loads must be investigated for the

lateral loading on the strip shopping center. 9-9


A C 1 TITLE*NDG 93

Obb2949 050870b

496

m

Seismic Loads The TMS shopping center is located in Seismic Zone 1. According to ASCE 7-88(9.1.1), Section 9.4, the minimum total seismic force applied to a structure in the direction of each principal plan direction is V,given by

v = ZIKCSW For Seismic Zone 1, the various coefficients on the right hand side of this equation are obtained from ASCE 7-88 as follows: 2 = Seismic Coefficient from Table 21 = 3/16,

I = Importance Factor from Table 22 = 1.0,

K

= Horizontal Force Factor from Table 23 = 1.33,

C = Numeric Cotficient =

1 15n

in which

T =

0.05 h,

0

where h, is the height and D is the lateral dimension of the structurein the direction under consideration, and T is the fundamental elastic period of vibration of the building or structure in the direction under consideration, in seconds. S=

Soil factor from Table 24 in ASCE 7-88 = 2.5 assuming an S, soil profile due to lack of soil information.

W = the total dead load. For the east-west direction, the longest dimensionis approximately 204 ft and the height to the diaphragm is 16 ft. Therefore:

I

9-10


A C 1 T I T L E S M D G 73 W Obb2949 0508707 322 W

1

C = Numeric Coeficieni L

= 0.282

15 40.056 seconds but the maximum value of C = 0.12 Therefore use C = 0.12 By inspection, C in the north-south direction = 0.12

ASCE 7-88, Section 9.4.2 indicates that the product of C and S need not exceed 0.14.

C

x S =

V

=

0.12 x 1.5

=

:.

0.18,

Use C x S = 0.14

Thus,

3 X 1.0 16

X

1.33

X

0.14 X W

=

0.035 X W in both directions

Since the masonry walls in this example problem potentially may have many configurations, assume that the average constructed weight of these walls is 60 psf. Assume that theweight of the glass and the partitions is 10 psf each, and that 1/2 of the lateral inertia force from the partition mass is transferred to the roof. Ignoring the parapet and assuming an 8 in. thick wall, the total roof area =

(81.5ft

+

5 ft -

)

in*

12 in&

x (204.67

ft - 3 (8 12 in@

in*)

)

The weight of the roof and 1/2 of the partitions =

1”@ + 15 psf) A

x 17,400 f t 2 =

348,000 lb

The weight of the glass = 1 o p s f x - l2 ft x 182 ft

2

=

10,920 lb

The weight of the fascia =

10 psf x 7 ft x 204.67 ft

=

14,330 lb

The 1/2 weight of the short masonry wall on Grid Line C =

9-11


=

17,400 ft2

A C 1 TITLEtMDG 9 3 W Obb29Y9 0508708 269 W

The 1/2 weight of the short masonry wall on Grid Line C =

* x 21.33 ft

60 psf x-

2

=

11,520 lb

Assuming that the remaining masonry walls are simply supported at the roofline and foundation, and extend 2 ft above the roof support, theweight of these walls applied tothe roof =

( 6OPf

X

18ft

16 ft =

W

=

Y ) x 202.67 ft

-

(6Opsfx18ftx~) +

3 x

16 ft

x

82 ft

272,600 lb

348,000 lb

+

10,920 lb + 14,330 lb

+

11,520 lb

+

272,600 lb

=

657,400 lb

Therefore the total seismic force to be resisted (i.e., base shear) is,

Y

=

0.035 x 657,400 lb

=

23,000 lb in either direction

The previous calculations didnot include anyportion of the roof live load in the calculations. However, ASCE 7-88 requires the entire roof snow load to be included in the calculation of Wwhen the ground snow load exceeds30 psf. To simplify the design example,the added weight due to snow load was not included. In a real design situation this added weight must be addressed.

Wind Loads The wind loading on buildings can vary significantly.

Different building codes across the

USA use different wind load values and distributions. These vary from a simplistic value of

X psf over the wall surface, to the complexity of ASCE 7-88 with its internal and external coefficients. A document designed to assist the professional in the understanding and use of the complexwindloadprovisions

of ASCE 7-88 is available (9.1.2). Sincethisis

a

masonry design guide and not a wind manual, a simplistic approach to wind loads will be followed. A wind pressure of 20 psf, acting uniformly over the wall areas, will be used. If

9-12


A CT1I T L E * M D G

93

Obb2949 0 5 0 8 7 0 9 I T 5

m

a more complex analysis were used the wind pressure on the individual elements, such as the parapets and wall comers, would vary significantly. The walls are assumed to be simply supported between the foundations and the roof diaphragm, with a 2 ft cantilevered parapet. For wind from the north:

I.,

(20 psf x 18 ft x 18 ft

mrUiload=

L

j

16 ft

=

203 plf

For wind from the south, blowing against the glass and fascia:

(

Windload=2Opsfx 12* + 7 f t ) =260plf 2 (Governs for the north-south direction) For wind in the east-west direction:

Wind load

=

203 plf

Also, an additional wind load from the fascia of 4.5 ft x 7 ft x 20 psf

= 630 lb will be

applied

to the wall on Grid Line C. It should be noted that wind loading produces the largest total force in the north-south direction, and therefore governs. In east-west direction, the seismic loading produces the largest total force, and governs in this direction. The global distribution ofthese lateral design loads to individual masonry walls is discussed in MDG 9.2. 9.1.2

DPC Gymnasium

Figs. 9.1-4 and 9.1-5 show the planandelevation

9-13


of a 7,500 square foot, single-story

Obb2949 0508730 917

A C 1 T I T L E * N D G 93

4t

8

'OH .o-.81

1 .o-.zr 1 .o-.

.o-. PE P9

Fig. 9.1-4 DPC Gymnasium Plan

9-14


A C 1 TITLEaMDG 93

m 0662949 0508711

853

-

N

l .

Fig. 9.1-5 DPC Gymnasium Elevations

9-15


.

m

A C 1 TITLE*:MDG 7 3

D

0 6 6 2 7 4095 0 8 7 l 1729 T

gymnasiumbuilding.Thisbuildinghas

four wall construction options, including two

unreinforced and two reinforced masonry wall systems. The roof framing system consistsof gabled rooftrusses supporting a metal roof deck, insulation,a membrane, andmiscellaneous equipment. The north and south walls are loadbearing. The roof framing system is, again, a simple one-way system. Simple tributary area analysis of the loading can be performed. To illustrate the application of the Code to the typical structural design considerations for this type of structure, 27 example problems are presented throughout the MDG Design Chapters 9 through 16 for theDPC Gymnasium. A listing of theMDG Examples is included in this sectionto assist the readerin correlating the illustrated example design issue to the plans and elevations of the structure.

DPC GYMNASIUM EXAMPLE PROBLEM INDEX All Wall Construction Options Example #

9.2-2

Issue

Design


Wall Construction Option A (Unreinforced Noncomposite Brick and Block) Examde #

Desim Issue

9.3-6

Eccentric gravityloaddistribution to the wythes (northand south walls)

9.3-7

In-plane lateral load distribution to the wythes (east and westwalls)

9.3-8

Out-of-plane lateral load distribution tothe

11.1-4

Wall design

11.2-1

Unreinforced pilaster design for out-of-plane flexure (east and west

wythes (all walls)

for out-of-plane flexure (wast wall)

walls)

11.2-2

Reinforced pilaster design for out-of-plane flexure (eastand westwalls)

11.3-3

Steel lintel design

(south wall)

9-16


AC1

TITLE*HDG 9 3 M 0bb2949 0508733 626 m

12.2-3

Walldesignforaxial

12.3-1

Unreinforced pilasterdesign for axial and out-of-plane flexure (north

and out-of-plane flexure (north wall)

and south walls)

12.3-2

Reinforced pilaster design for axial and out-of-plane flexure (north and south walls)

14.3-2

Walltiedesign

14.3-4

Shear wall to foundation connection (eastand westwalls)

14.3-15

Roof diaphragm toshear wall connection (east and westwalls)

for out-of-plane loading(allwalls)

Wall Construction Option B (Unreinforced Composite Brick and Block) Examde #

Issue

Desien

9.3-9

Collar joint design for out-of-plane shear(eastand

11.1-5

Wall design

12.2-4

Walldesign for axial and out-of-plane flexure (north wall)

13.1-5

Wall design

13.2-3

Walldesignfor

14.3-3

Shear wall to foundation connection (east and westwalls)

15.4-2


westwalls)

for out-of-plane flexure (west wall) for out-of-plane shear(north

wall)

in-plane flexure and shear(east wall)

Wall Construction Option C (Reinforced Composite Brick and Block) Examde #

Issue

Design

11.1-6

Wall design

11.3-8

Steel lintel design (south wall)

13.1-7

Wall design

for out-of-plane flexure (west wall) for out-of-plane shear (west wall)

Wall Construction Option D (Reinforced Clay Masonry) ExamDle #

Issue

Desien

11.1-7

Wall design

12.2-5

Walldesign for axial and out-of-plane flexure (north wall)

for out-of-plane flexure (west wall)

9-17


Miscellaneous

14.3-7

Typical reinforcing details

Gravity Design Loads - It is assumed that the dead load of the roofing system,

9.1.2.1

including all framing members, mechanical systems and ceiling, is a uniformly distributed load of 20 psf. The governing roof live load

is a 40 psfsnow load. This load

is either

applied across the entire span, or 20 psf is applied on one slope and 40 psf on the other

slope. The building is64 ft wide; the masonry wall thicknesswill vary slightlydepending on the wall construction option selected. To determine the loads it is assumed that the wall width will be 12 in. This assumption leaves a net truss clear span of 62 ft.

The reactions of the fully loaded truss under dead load and live load are therefore:

R

= [8

ft

X

(20 psf + 40 M)] X

62 ft = 2

14,880 lb

The reactions of the truss loaded with the unbalanced snow load are:

[ 8 f t x ( 2 0 p s f +40psf)] x

=

(31 2

11,160 lb

[ 8 f t x ( 2 0 p s f + 2 0 p s f ) ] x (31

fiY

2

R-

+

[ 8 ft x (20 psf + 40 psf)] x 31 ft x (31 fi +

=

13,640 lb

9-18


=)} 2

AC1

TITLEsMDG 93 D 0662949 0508735 4 T 9 m

The exterior trusses and the first interior trusses have a reduced tributary area. Taking into account this reduction, reactions were calculated and are summarized in Table 9.1.1. This table summarizes all the truss reactions.

Table9.1.1

Truss Reactions

Truss

Reaction

Tributary Width

(lb)

(fi)

Full Load Unbalanced

Snow Maximum

Minimum Typical (Dead Load) Exterior (Dead Load) 1st Interior

8.00

3.50

(4,960) 6,5 10

11,160 14,88013,640 (4,960) (4,960) 5,970 4,880

(2,170)

(2,170)

7.17

(2,170)

10,000 13,33012,220 (4,440)

(Dead Load)

(4,440)

I

(4,440)

9.1.2.2 Lateral Design Loads - Both seismic and wind loads must be investigated for lateral loading on the DPC Gymnasium.

Seismic Loads Using an analysis similar to that described for the TMS Shopping Center in MDG 9.1.1.2 the seismic loading for the DPC Gymnasium can be calculated using the formula

V

= ZIKCSW.

Z = 3/8 (category II building ASCE 7-88)

I = 1.25 K = 1.33

9-19


To find C for the north-south direction: The mean roof height is:

h,

= 24.67

ft

+

5.33 ft = 26.44 ft -

u ~ 26.5 e

3

1

C = Numeric Coeflcient =

ft

= 0.16 > 0.12

15 40.17 seconds

.:

use C = 0.12 (maximum allowed)

To find C for the east-west direction:

T = 0.05 x 26.5 ft Jim? C

=

= o.12

1

Numeric Coeficient =

= 0.19

> 0.12

15 40.12 seconds 2.

use C = 0.12

Using a soil factor, S, of 1.5 as in the TMS Shopping Center,

C x S = 0.18 However, use maximum of C x S = 0.14 (maximum required) in both directions. The total seismic load is therefore the same in both directions, and has a value of:

V

=

3 8

X

1.25

X

1.33

X

0.14

X

W

= 0.0875

9-20


X

W

a variety of configurations, a 100 psf

Since the masonry walls in this example may have average constructed weightisassumed.

It is alsoassumed that the gymnasiumhasno

partitions or any other items that add to its seismic weight. The walls are assumed pinned at the top and bottom; thus, only half of that weight was included in the calculation of the seismic forces. The weight of the roof = 20 psf x 64 ft x 128 ft = 163,800 lb

The weight of north or south walls = 100 psf

X

24.67 ft

X

128 ft = 2

157,900 lb

The weight of the east or west walls = 24.67 ft +

W

=

)

ft 2

x

y

=

84,740 lb

163,800 lb + 2 x ( 157,900 lb + 84,740 lb )

=

649,100 lb

Note that the snow loads on the roof are not included in the weight calculation to simplify the problem (see discussion in MDG 9.1.1.2). The total seismic load is therefore:

V

= 0.0873 x 649,100

lb

=

56,700 lb in each direction

Wind Loads

The same assumptions used for the TMS Shopping Center apply to the DPC Gymnasium.

9-21



93

m

Obb29Y9 0508718 108 M

Wind load in the north and south direction =

Assuming the east and west walls span from the foundations to the roof, the average span =

Wind load in the east-west direction =

The global distribution of these lateral loads to individual masonry walls is discussed in MDG 9.2 9.13 RCJ Hotel Figs. 9.1-6 through 9.1-15 show the typical floor plans, elevationsand details of a multi-story hotel. The north and south walls are glass curtain walls and the masonry walls are shown hatched. General notes for the RCJ Hotel are given on pages 9-29 and 9-30. There are two wall construction options. Option A uses unreinforced single wythe concrete masonry units on the interior walls. The exterior walls are composed of 4 in. face brick, a

3 in. air cavity, 1 in. of insulation and an interior wythe of concrete masonry units. Option B uses reinforced single wythe hollow clay masonry for both exterior and interior walls.

9-22


A CT1I T L E a M D G

93

Obb2949 0 5 0 8 7 3 9 044

m

R

(-J;IÂ

-a

II

.I

N

11

Fig. 9.1-6 RCJ Hotel Floor Plan 9-23


A C 1T I T L E r U D G

9 3 W Obb29Y9 0508720 8bb

f "T I

a B

I

;;

ri

I

l

i?"-

I

! !

I

L

Fig. 9.1-7 RCJ Hotel Floor Plan 9-24


A C 1 T I T L E * H D G 93 W Ob62949 0508723 7 T 2 W

Fig. 9.1-8 RCJ HotelBuildingSection

9-25


A C 1 T I T L E x M D G 93 m Obb2949 0 5 0 8 7 2 2 639 m

Fig. 9.1-9 RCJ HotelElevations

9-26



93 W 0662749 0508723 575 W

N

L k L

r,

Fig. 9.1-10 RCJ HotelBuildingSection

9-27


A C 1 TITLESMDG 93

m

Obb2949 0508724 401

; I !

i

!

!

j

!

-!! I

! !

I !

I ! +

I ! ! !

+ ! !

1 j

I

i

! !

j

I

4;. ! I

I ! ! !

i

j

ll

! ! ! !

+j ! !

I !

I I

! ! ! ! ! ! i !

-

-8-. 6

"8-.8

I

!

Fig. 9.1-11 RCJ Hotel Elevations 9-28


m


93

m

0662949 0508725 348

m

RCJ HOTEL NOTES 1.

Design Dead Loads Roof (Hotel)

95 psf

Roof (Canopy)

50 psf

Floor

110 psf 10 psf

Glass curtainwall 2.

(includes partitions)

Design Live Loads Roof

20 psf

Dwelling Rooms

40 psf

Public Rooms

loo psf

1st Floor Corridor

100 psf

Corridors above 1st

100 psf

Stairways

loo psf

No Snow

Wind pressure or suction

i

3.

on vertical surfaces

25 psf

Wind uplift on open roofs

40 psf

Seismic Zone

See Item 4c

Conditions Soil

4000 psf

Allowable soil bearing pressure

30 pcf

Equivalent fluid pressure S

4.

Building Construction a.

Floor and Masonry Elevations Option I

Option II

10’ - 10” 9’ - 8’’

First floor to second floor

40‘ - 4”

Overall masonry wall height

8’

Floor to floor above second floor

- 10“

First floor to second floor

9-29



m

Ob62947 050872b 284

8’ - 8”

Floor to floor above second floor

35’ - O” b.

m

Overall masonry wall height

Roof and Floor Construction

8” precast hollowcore planks with 2“ thick normal-weight topping at Hotel 4” precast hollowcore planks with no topping at Canopy C.

Wall Construction Option A - Non-Reinforced, Seismic Zone 2 Interior loadbearing and nonloadbearing walls, and retaining wall, single wythe CMU Exterior loadbearing walls 4” face brick, 3” cavity with 1” rigid insulation interior wythe of CMU Option B - Reinforced, Seismic Zone 4 All walls - single wythe hollow clay masonry

d.

Canopy Construction Beams and columns - reinforced clay masonry

e.

Lintels and Beams Beam B1

-

W21 x 78

+ 2 - L 3-1/2” x 3-1/2” x 3/8” - bear each end on

stiffened grillage beam W8 x 21 x 2’-O’’ at Wall Construction Option A only Beam B2 -

W10 x 26 with 6” long bearing each end andbearing plate 3/4” X

Lintel L1 -

6“

X

O”8”

W8 x 18 with 8” long bearing each end at Wall Construction Option A or reinforced masonry beam at Wall Construction b

Option B Lintel L2 -

2 Angles 3-1/2” x 3-1/2” x 1/4” with 4” long bearing each end

Lintel L3 -

W8 x 15with 6” long bearing each end Add 5/16”’ thick bottom plate for Wall Construction Option A

f.

Masonry Openings All door openings are 3’-4” wide x 7’-O’’ high, unless noted otherwise

9-30



93 H 0 b b 2 9 4 9 0 5 0 8 7 2 7 110

Fig. 9.1-12 RCJ HotelSections 9-31



Obb2949 0508728 057

c -1

Fig. 9.1-13 RCJ Hotel Section

9-32


AC1

TITLE+MDG 93 m Obb2949 0508729 T93 m

Fig. 9.1-14 RCJ Hotel Sections

9-33



93

m

Obb2949 0508730 705

m

8 ."c

o"

c

Fig. 9.1-15 RCJ Hotel Sections 9-34



93

m Obb2949

0508732 b4L

m

To illustrate how to design both coupled and uncoupled shear walls the hotelwill have two possible elevation configurations, Building Option I (40 ft- 4 in.overall masonry wallheight) and Option II (35 ft- O in. overall masonry wall height), respectively. See Figs. 9.1-8, 9.1-9,

9.1-10, and 9.1-11.

As shown in Fig. 9.1-6 and in canopy section N 6 of Fig. 9.1-12, there is a canopy over the entrance to the lobby. This canopy issupported on reinforced masonry columnsand beams. The columns have an unsupported height of 12 ft.

To illustrate the application of the Code to typical structural design considerations for this

type of structure, 31 example problems are presented throughout the MDGDesign Chapters 9 through 16 for the RCJ Hotel. A listing of the MDG Examples is included in this section to assist the reader in correlating the illustrated example design issue to the plans and elevations of the structure.

RCJ HOTEL EXAMPLE PROBLEM INDEX All Wall Construction Options Example #

Issue

Desien

11.1-9

Unreinforced retaining wall design

11.1-10

Reinforced retaining wall design

11.3-1

Singly reinforcedmasonry beam design for flexure (canopy)

13.1-2

Singly reinforced masonry beam design forshear (canopy)

14.3-6

Beam to column connection (canopy)

Wall Construction Option A (Unreinforced Noncomposite Brick and Block) Example #

Issue

DesiFn

9.2-3


10.4-2

Expansion joint locations (all walls)

10.4-3

Differential movement between wythes (all walls)

9-35


AC1 TITLE*NDG 93 m 0662949 0508732 588 m

11.1-11

Wall design for out-of-plane flexure (stairwell wall)

11.3-6

Continuous masonry beam design for flexure (Grid E)

12.1-1

Column design for axial and flexure (Grid E-3)

13.1-3

Continuous masonry beam design for shear (Grid E)

13.2-4

Wall design for in-plane flexure and shear (Grid C)

14.3-8

Roof diaphragm to shear wall connection (Grid F)

14.3-16

Continuous masonry beam reinforcement termination (Grid E)

15.4-3


Wall Construction Option B (Reinforced Clay Masonry) Examde #

Desim Issue

9.2-4


11.1-18

Singly reinforced masonrylinteldesign

11.1-12

Walldesign

11.3-4

Wall beam design for flexure (Grid E)

11.3-5

Coupling beam design for flexure (Grid 2)

11.3-7

Coupling beam design for flexure (Grid B)

13.1-1

Coupling beam design forshear

13.1-4

Wall design

13.1-8

Coupling beam design for shear (Grid B)

13.2-5

Walldesign for in-planeflexure and shear (Grid C)

13.2-6

Walldesign for in-planeflexure andshear (Grid 2)

14.3-9

Floor diaphragm to wall connection (Grid 2)

14.3-10

Floor diaphragm to wall connection (Grid D)

14.3-13

Wall to wall connection (Grid 2)

14.3-14

Wall to wall connection (Grid 2-C)

for flexure (Grid B)

for out-of-plane flexure(stairwellwall)

(Grid 2)

for out-of-plane shear (stairwell wall)

9.13.1 Gravity Design Laads - The analysis of this structure is significantly more complex than the previous two building examples. The floor and roof framing systems are hollow core planks which are assumed to be simply supported on the masonry walls and beams on

9-36



m

Obb2949 0508733 4L4

m

the Grid Lines running in the north-south direction. Figs. 9.1-16 and 9.1-17 show the vertical load carrying elements on the first and second through fourth floors, respectively. The dead load on the roof is 95 psf and the dead load on the floor systems is 110 psf. The roof has a minimum live load of 20 psf and a wind uplift of 40 psf. Each floor has a live load of 40 psf and the corridor live load is 100 psf.

An average selfweight for the wall

systems was assumed to be 70 psf and the curtain walls have a selfweight of 10 psf. For live loads less than or equal to 100 psf, ASCE 7-88 allows a reduction in live loads on any member supporting influence areas in excess of 400 ft2. Influence areas are taken as twice the tributary areafor beams and fourtimes

the tributary area for columns. A

maximum reduction of 50 percent isallowed for members supporting one floor and a maximumof 60 percent for members supporting more than one floor. ASCE 7-88 also allows a different reduction inrooflive

loads for tributary areas inexcess of 200 ft2.

However, the maximum allowable reduction only changes the total loads on the top of the fourth floor walls by 14 percent. The change in the lower floor loads is significantly less than this value. Since the reduction in loading is small, no reduction of the roof live load was calculated for this building example. This simplifies the analysis and produces slightly conservative loadings. If so desired, the designer may reduce these live loads as defined in ASCE 7- 88, Section 4.8. Using the tributary areas shown in Figs. 9.1-16 and 9.1-17 the uniform loads on top of each wall at each floor level can be calculated. The total uniform roof load on top of the fourth floor wall on Grid Line C, between Grid Lines 1 and 2 is WC,l-Zp =

( 95 psf + 20 psf ) x 30 ft = 3,450 plf

(2,850 plf dead load) 9-37


U

O cv3

a

' m l

m

2

d "L

u

""_

m

o -

Fig. 9.1-16 RCJ Hotel Vertical Load Carrying Elements O n First Floor

9-38


93

A CT 1I T L E * M D G

0bb2949 0508735 297

m

L!

O

*

cy! W

2

P

c?

u

a d " "

"

"-F m

o -

Fig. 9.1-17 RCJ Hotel Vertical Load Carrying Elements On Second Through Fourth

Floor 9-39



93

m

0662949 0508736 1 2 3

Since roof uplift mayproduce a critical loadingfor shear evaluation on these walls, this load combination should be calculated, and is I Wc,l-w = (95 psf

- 40 psf)

X

30 ft

=

1,650 plf

The roof loads on the remainder of the building elements on the top of the fourth floor walls are summarized in Table 9.1.2. Similarly, the total uniform load on top of the third floor wall on Grid Line C, between Grids 1 and 2, is calculated as follows:

Tributary Area

29.67 ft x 30 ft

=

=

890

ft2

The above calculation ignores the tributary area of the beam in the corridor. However, since the beam has a higher live load and produces a concentrated load on the wall, it isnot included in the calculation. Assuming the wall influence area, AI is twice the tributary area

A,

=

2

X

890

fi2 =

1,780

fi2

From ASCE 7-88the live load reduction factor, LLRF,is

URF

15

= 0.25 + - = 0.25 +

6

l5

d

m

= 0.606

Note: The Code does not specifically address walls, and some designers may not apply a live load reduction for a wall. The wall load from the fourth floor is therefore,

F h r wall load

=

(110 psf + 40 psf x 0.606) x 30 ft

=

4,027 pl€

The total load on the top of the third floor wall on Grid Line C, between Grid Lines 1 and

2 is: -23 =

4,027 plf + 3,450 plf + 9.67 ft x 70 psf = 8,154 plf

(6,827 plf dead load)

9-40


A C 1 TITLE*MDG 93

m

Obb29Y9 0.508737 ObT

12

3-

4,8 4 Y

5 M"

4" ?$!

x

Po

d

P<

3

S

9-41


m

A C 1T I T L E l r M D G

m

93

0662747 0508738 T T 6

m

where 9.67 ft x 70 psf is the wall weight from the story above The wall loads should be reduced by 1 x 70 = 70 plf for the shorterwall construction option dimensions, Building Option II (8 ft - 8 in. floor to floor instead of 9 ft - 8 in.). The loads on the top of all the masonry elements from the third floor to the first floor are summarized in Tables 9.1.3 to 9.1.5.

Stair Well Landings The stringers on the stairs have a dead load of 129 psf due to theincreased concrete weight of a slab and treads, and the 3 ft

- 6 in. wide

landings have a dead load of 85 psf. The

landings transfer the stair load to the walls on Grid Lines 2 and 3. It is assumed that the stair load is applied to the walls over a 2 ft section near the ends of the landing slab. This portion of wall also supports the landing load over a 2 ft section. The 1.5 ft stairwell wall section on Grid Lines 2 and 3 nearest the doorways supports only the landing loads (see Fig. 9.1-18). For each section, the loads on these walls at each landing are:

W on Grid Lines 2 and 3, 2 ft section

=

(129psf+ lOOpsf) x **]x;+(SSpaf

2

W

=

2,630 plf

+ l O O p s f ) x -8.67 ft

( 1,400 plf dead load)

(assumes 1/2 the stair load goes to each side and is distributed over 2 ft)

W on Grid Lines 2 and 3, 1.5 ft section W = ( 8 5 p ~ f +lOOpsf)x-

W

=

802 plf

=

8.67 ft 2

(368 plfdead load) 9-42


2


93

m 0662949 0508737 932 m

8

E

Rm-

t

E

r4

c1

B

B E;

8

21

(*

f

8

21 f

(*

:

S

:

O

5

9

d

9-43


O

5

A C 1T I T L E s N D G

93

m

Obb2949 0508740 654 M

=

QS

8

E

CJ

CJ

9

9

5

5

H

H

O

5

9.

v)

r(

9-44


5

E

Ei - - O

5

O

5

5

H

A C 1 T I T L E * N D G 93

m

0662947 0508741 570

n

c v c

f

9-45


m

A C 1 TITLEwMDG 9 3

0662949 0508742 427

9-46


A CT 1I T L E t M D G

93

Obb2949 0508743 3 6 3

Fig. 9.1-18TypicalStairwellAma

No live load reduction is taken for the stairwell area, the loads on the walls immediately below each landing are summarized in Table 9.1.6. Table 9.1.6 also summarizes the loads on the canopy.

9.13.2 Lateral Design Loads - The lateral loads applied to the hotel are produced by wind or seismic forces. The determination of seismic forces on the four story hotel is relatively complex and will be discussed in MDG 9.2.2.

Wind Loads For the same reasons described in MDG 9.1.1, a simplistic approach will be taken for wind loading. It will be assumed that all vertical faces have a uniform 25 psf suction or pressure applied to them, and that the first floor wall transfers 1/2 the force applied to it directly to the foundation. This assumption produces a total wind load in each direction for Building Option I Dimensions of, Wind load in north-south direction = (9.67, x 3

+

2

*

+ 0.5

ft x 150 ft x 25 psf

9-47


A C 1T I T L E I M D G

93

m

0662747 0508744 2 T T

9-48


m

A C 1 TITLExMDG 93

m 0662747 0508745

L3b

m

Wind load in east-west direction =

1

(9.67 ft x 3 + 'OA3 ft + 0.5 ft x 67.67 ft x 25 psf 2

(neglecting differences in grade) The global distribution of the above loadsto eachstory, and to each resisting element within the story, is discussed in MDG 9.2.2.

9.2 INTERWALL LOAD DISTRIBUTION The global distribution of loads to walls withina masonry building is discussed in this section. This discussionwill be restricted to theglobal distributionof lateral forces to the top of wall elements since the global distributionof vertical loadsfollows commonengineering principles as discussed previouslyin MDG 9.1. The local distributionof both lateral andvertical forces within a wall will be discussed in MDG 9.3. The global distributionof lateral forces to the resisting masonry wall systems depends on the rigidity of the floor or roof diaphragm used to transfer the lateral loads to the top of the walls, and on the relative rigidity of the wall elements themselves.

9.2.1 Global Lateral Load Distribution

on Shear Walls in Buildings with Flexible

Diaphragms The in-plane flexibility of the roof diaphragm is evaluated relative to the flexibility of the supporting walls. If the diaphragm undergoes significantlateral deformation when compared to the deformation of the supporting walls, the diaphragm is considered flexible. When flexible diaphragms are used, the lateral loads are distributed to the resisting elements in proportion to each elements tributary area. MDG Examples 9.2-1 and 9.2-2 illustrate this concept Òn the TMS ShoppingCenter and the DPC Gymnasium, both of which have flexible 9-49


A C 1 TITLEvMDG 9 3

m Obb2949

0508746 O72

m

roof diaphragms. 9.2.2

Global Lateral Load Distribution on Shear Walls in Buildings with Rigid Diaphragms

If a diaphragm does not undergo significant in-plane

deformation relative to the lateral

deformations of supporting walls, it isconsidered rigid. When a rigid diaphragm is used, the lateral loads are distributed to resistingwall elements in proportion to each element's relativestiffness and distance from the plan center of rigidity. The eccentricity of the applied load (torsion effects) from the center of rigidity also affects the amount of load distributed to the shear wall elements. The type of construction used in the RCJ Hotel would classify the floor diaphragms as rigid. Several methods are currently used to determine the relative stiffness of the wall elements, and to subsequently distribute the lateral loads to the top of each wall. Schneider and Dickey (9.2.1) list three methods; Amrhein (9.2.2) shows two variations on another method. Other authors use similar methods. However, when certain configurations of wall openings are analyzed, some of these methods produce results that erroneously suggest that a wall with openings is stifferthan a wall without openings! In fact, the applicability of one method over another may depend on the building configuration. Only one "hand calculation method,"Schnieder and Dickey's Method I (9.2.1), will be briefly presented in this chapter. This method has the advantages of being simple to apply and generally avoids the erroneous stiffness results described above. For more information on these methods, detailed descriptions are included in the indicated references. It should be noted that this method does not accurately describe the lateral deformation of the building system and should not be used to calculate building drift. Using Method I, a hand analysis of the RCJ Hotel was made and is presented in MDG Example 9.2-3. With increasing micro-computer use and the development of relatively inexpensivestructural

9-50



93

m Obb2949

0508747 T 0 9

analysis programs, more and more structural analysis is performed electronically. These programs allow a more rational 3-dimensional vertical andlateral load analysis of masonry buildings and, presumably, a more accurate distribution of lateral loads. A computerized analysis of the RCJ Hotel was made, and the results are presented in MDG Example 9.2-4.

Hand Calculation Method By hand calculations Schneider and Dickey's Method I takes the total lateral load at each diaphragm location(Vm,al) and distributesthe load to eachresisting element according to the ratio of the element's stiffness (ki)and the sum of the stiffness of all the elements at this elevation (kmal).In equation form this is

vr

=

VAd x

(2)

Eq. 9.2-1

The stiffness of individual elements is determined based on the assumption that the walls act as beams that have significantshear deformations. These beams are assumed to be fixed at the base of the structure and either free atthe diaphragm location (a cantilevered beam -

Ac), or free to translatebut fully restrained against rotation (fixed end -Ap) (see Fig. 9.2-1). Boundary conditions determine which equation should be used.

Fig. 9.2-1 Deflection of a Shear Wall Element Assuming linear elastic behavior, the deflection (A,) at the top of a cantilevered, solid, 9-51 ,


A C 1 TITLELMDG 93

m Obb2949 0508748 945 m

rectangular shaped wall element is given by:

Ac =

Pl (h')3 3 E,,,Z

+

1.2 Pl h'

Eq. 9.2-2

Ev A

Since only relative stiffness is being considered, the value of P can be set to an arbitrary value and E, assumed to be approximately 0.4 x E,. Including the formulas for A and I,

Eq. 9.2-2 becomes Ac

=

[

3 * Em

4

+

3

(81

Eq. 9.2-3

A similar expression can be developed for the deflection at the top of a solid, rectangular shaped wall element, where both ends are fixed against rotation (Ap).

Eq. 9.2-4

Pl can be taken as any convenient constant value. If all walls are of the same material Emf

Eqs. 9.2-3 and 9.2-4 do not account for flanged walls or cracked section behavior.

The

designer can use Eq. 9.2-2 for Ac and a similar equation for A p to include these effects. However, in light of the assumptions used for Method I, it is questionable whether these refinements will produce significantly more accurate results. The element stiffness is calculated by taking the reciprocalof the deflection at the topof the wall under the applied load, P.

Ri

=

1 -

Eq. 9.2-5

A

For perforated shear

walls,Eqs.

9.2-3 and 9.2-4 cannot be useddirectly to determine

deflection. Schneider and Dickey (9.2.1) suggest a method for calculating the deflection at the top of perforated shear walls. In their Method I, this deflection is approximated using

9-5 2


A C 1 TITLE*MDG

93

m

Obb2949 0508749 881

m

the following procedures: 1.

The deflection atthetop

of the wallis

calculatedusingEq.

9.2-3. This

assumes that the wall acts as a cantilevered beam.

2.

The deflection of a section of the wall equal in height to the tallest opening is calculated using Eq. 9.2-3 (again assumes cantilever action). The total wall depth is used for I,.

3.

The deflection obtained in step 2 is subtracted from that obtained in step 1.

4.

The deflection ofall piers lying within the strip used for step 2 are calculated assuming these piers are fixed, top and bottom (Eq. 9.2-4).

5.

The stiffness of each pier is determined using Eq. 9.2-5 and the stiffness of all piers in the opening strip are summed. The total deflection of the opening strip is obtained by taking the reciprocal of the stiffness sum. If there are additional openings in the strip, the wall is separated into sections and the deflection at the top of each pier within the section is calculated using steps 1 through 5.

6.

The deflection of the piers in the opening strip is addedtothe

deflection

calculated in step 3.

7.

The stiffness of theperforated wallis then the reciprocal of the deflection obtained in step 5.

It should be noted that there are some significant inconsistencies in

the above method,

particularly with respect to compatibility and load distribution. However, the method has been found to give both reasonable and, in most cases, conservative results. When a rigid diaphragm analysis is used and the lateral resisting elements are not placed symmetrically, or if they have unequal length and cross-section, the center of rigidity of the systemmust be determined. The center ofrigidity

can be located by determining the

stiffness of the walls resisting the load in each of the coordinate directions (Xand Y) and then calculating the static moments of these stiffnesses about the center line of one of the

9-53


AC T I1T L E * U D G

0662949 0508750 5 T 3

93

wall systems (see Fig. 9.2-2). The formulas used are

-

c Ri xi c R,

Eq. 9.2-6

Y

'Y

Eq. 9.2-7

where x, = the distance to the center of rigidity, along the x axis

y, = the distance to the center of rigidity, along the y axis Rt, = the relative rigidity of each wall element resisting forces in the x direction

=

(21,

Rb= the relative rigidity of each wall element resisting forces in the y

xi = the distance to the center of the wall element, along the x axis yi = the distance to the center of the wall element, along the y axis

' ,1 Y

/Center

Fig. 9.2-2

of Rigidity

Center of Rigidity 9-54


A C 1T I T L E m M D G

73

m

0662749 0 5 0 8 7 5 1 4 3 T W

The distance fromthe centerof rigidity to the centerof mass for seismic force, or to the line

of action of resultant wind force, produces a torsional moment in the plane of each rigid diaphragm. This moment must be resisted by the lateral resisting elements as a shear that acts in addition to direct shear forces each receives. Schneider and Dickey (9.2-1) suggest distributing both direct shear and the torsional shear using the following equations. Eq. 9.2-8

Eq. 9.2-9 where

(K)i

is the total shear force applied to a particular wall element oriented to resist loads parallel to the y axis is the total shear force applied to a particular wall element orientedto resist loads parallel to the x axis

PY

is the y component of the resultant lateral force at the diaphragm elevation

Px

is the x component of the resultant lateral force at the diaphragm elevation

X'

the distancefrom the center of the wall to the center of rigidity, parallel to the x axis

Y'

the distancefrom the center of the wall to the center ofrigidity, parallel to the y axis the distance from the center of rigidity to the line of action of P, parallel to the y axis the distance from the center ofrigidity to the line of action of P,,, parallel to the x axis

J,

relative polar moment of inertia

E (R& Y'2

+

Riy X'2) 9-55


Eq. 9.2-10

AC1

TITLE*MDG 9 3 W Obb29490508752376

This method of analysis has been applied to the RCJ Hotel to illustrate hand calculation procedures. Refer to MDG Example 9.2-3.

9 3 INTRAWALLLOADDISTRIBUTION 93.0 General MDG 9.1 and 9.2 described the procedures for calculating global loadsthat act on masonry structures and theirdistribution to various componentsof a structure. MDG Examples 9.2-1 through 9.2-4 give illustrative detailed steps of this process. The second area of interest in the load analysis of masonry structures concerns the further distribution of the loads within each component element. The various phenomena which have been considered here include distribution of in-plane and out-of-plane vertical and horizontal loads on cavity walls and composite walls. In addition, distribution

of in-plane

loads on perforated shear walls is also calculated. The designs of some typical elements of the example buildingsare then presented in the subsequent chapters and examples of MDG.

93.1 Local DistributionUnderConcentratedLoads

To calculate the compressive stress

va)on a wall system subjected

to concentrated loads,

the wall system mustbe made. some assumptionof the concentrated load distribution within The Code C 5.12.1 suggests distributing the load using an angle of 45" from the vertical. Schneider and Dickey (9.2.1) suggest that a 30" angle might be used. The designer must determine what angle should be used with the knowledge that the smaller the angle, the smaller the effective length values. Whateverthe angle, Code 5.12.1 states thatfor walls laid in running bond, the effective wall length resulting from the assumed distribution is limited to the width of the bearing plate plus four times the thicknessof the wall, but not to exceed the center to center spacing between the concentrated loads.

9-56


A CT1I T L E * H D G

0662949 0 5 0 8 7 5 3 202

93

m

93.2 Local Distribution of Concentrated Loads Acting on Bond Beams 93.2.1 Hollow Masonry Walls

- Bond beams are often used in masonry construction

for

distribution of concentrated loads on hollow walls. Sincethe Code 5.13 requires that applied loads be resisted by net area only, the use of bond beams allows concentrated loads to be distributed over lengths greater than those directly under the bearing plates. The absence of bond beams could result in bearing stresses which exceed the maximum bearing stress allowed by Code 5.12.3. Neither the Code nor the Commentary provide any information regarding load distribution through bond beams. However, research carried out by Page and Shrive (9.3.1,9.3.2)

has shown that the angle of load distribution in a bond beam for a

hollow wall is smaller than the commonly assumed value of 45"

from the vertical. Their

research shows that this angle varies according to the number of courses used in a bond beam. For an 8 in. deep bond beam an angle of distribution equal to 30" from the vertical should be assumed. For a 16 in. deep bond beam, the suggested angle reduces to 25". The bearing area can then be calculated as if the load were applied as a patch load on the hollow wall. The effective length for calculation of compressive stresses is determined by assuming that the angle of distribution changes to 45" from the vertical once the load has been transferred into the hollow masonry section. According to Code 5.12.1 the effective length shall be limited to the bearing width (LB)plus four times the thickness of the wall, but not to exceed the center-to-centerdistance betweenthe concentratedloads. Code 5.12.1 applies to walls laidinrunning

bond; for wallslaid

in stack bond, the load transfer

terminates at the head joint nearest to the edge of the bearing area. See MDG Example 9.3-2 for distribution of concentrated load acting ona bond beam in a hollow masonry wall.

93.2.2 Solid Masonry Walls- The following statement from the paperby Page and Hendry (9.3.3) concerns loads on bond beams in solid masonry walls:

in this area,the

logical approachto

thisproblem

l'In the absence of research

is to assume a dispersion of the

concentrated load through the beam (say at 60" or 45" from the vertical) and then to evaluate the bearing strength enhancement for the masonry beneath the beam as though it was loaded by a patch loading of this size." 9-57


A C 1 T I T L E * N D G 93

0662949 0508754 L49

m

No example is presented for the case of solid masonry walls since the procedure of load distribution is verysimilar to the onegiven in MDG Example 9.3-2 for hollow masonry walls. 9 3 3 Effective Bearing Area Under Concentrated Loads

When calculating the effective bearing area of a concentrated load on a wall system, the Code allows for an increase in area if the supporting masonry is larger on all sides than the direct bearing area. This increase ispermissible because the confinement of the direct bearing area by the surrounding masonry increases the bearing capacity of the wall in the vicinity of the concentrated load. The Code 5.12.2 states that thebearing stresses in a wall shall be computed by distributing the load over an area determined as follows: direct bearing area AI, or

(a)The (b)

A,

\i 5

but not greater than U ,

A,

where A2is the supporting surface wider than AI on all sides, or A2 is the area of the lower base of the largest frustrum of a right pyramid or cone having AI as upper base, sloping at

45 o from the horizontal, and wholly contained within the support. Area A2 shall terminate at head joints in other than running bond. The Code 5.12.3 also states that the bearing stress shall not exceed the allowable value of

0.25 Pm. Typical situations related to effective bearing area under concentrated load are presented in MDG Examples 9.3-3, 9.3-4 and 9.3-5.

9-58


A C 1 TITLExMDG 93

9.3.4

m

Obb2949 0508755 0 8 5

m

Local Load Distribution in Multiwythe Noncomposite (Cavity) Walls

Multiwythe noncomposite masonry walls, also

known as cavitywalls, are a type of wall

construction in which an aircavity isprovided betweenthe wythes of a multiwythe wall. This

type of wall construction is widely usedand, if properly constructed, has excellent moisture, sound, and thermal resistance. The Code 5.8.21 states that each wythe of the noncomposite wall is to be designed to resist individually the effects of loads imposed on it. It continues by saying: "Unless a more detailed analysis is performed, the following requirements shall be satisfied... Gravity loads from supported horizontal members shall be resisted by the wythe nearest to the center of span of the supported member. Any resulting bending moment about the weak axis of the wall shall be distributed to each wythe in proportion to its relative stiffness. Loads acting parallel to the plane of a wall shall be carried only on the wythe on which they are applied. Transfer of stresses between wythes from such loads, shall be neglected. Loads actingtransverse to the plane of the wall shall be resisted by all wythes in proportion to their flexural stiffness..." The noncomposite wall in MDG Examples 9.3-6, 9.3-7, and 9.3-8 consists of an 8 in. hollow concrete block wythe, a 4 in. cavity, and a 4 in.claybrickwythe.

The three examples

presented cover the following load distributions: (a)

Eccentrically placed vertical gravity load from a roof truss transferred to the block wythe of the wall (MDG Example 9.3-6).

(b)

Lateral in-plane loads transferred to the wall through the roof system which acts as a flexible diaphragm (MDG Example 9.3-7).

(c)

Lateral out-of-plane loads,suchas 9-59


wind or seismicloads,which

act on an

AC1 TITLElwMDG

93

0662797 0508756 T l 1 1

individual wythe of the nonloadbearing multiwythe noncomposite wall(MDG Example 9.3-8).

To design wall systemsfor out-of-plane lateral loads, suchas wind and earthquake loads, the Code 5.8.2.2 requires that theload transverse to the plane of the wall be distributed to the individual wythes of the wall system in proportion to their relative flexural stiffnesses. In general, the stiffness of each wythe is affected by its height and boundary conditions. In the three examples, i.e., MDG Examples 9.3-6 to 9.3-8, the height and boundary conditions of each wythe of the noncomposite (cavity) wall are assumed to be the same. However, for morecomplicatedsituations,

as in the case of the RCJ Hotel building

described in MDG 9.1.3, the story height and support conditions of the two wythes are different. Shown below is the cross-section of the exterior wall system proposed in Option A of the RCJ Hotel. In that option, the outerwythe is verticallycontinuous for four stories, while the inner wytheissimply location and axialstiffnessaffect

supported at each floorlevel.Since

factors such as tie

the transverseload distribution between wythes, it is

suggested that a simple plane frame model be used to determine thestresses in each wythe of such a wall system. Since the wall system usually acts in one-way bending

between the

floor slabs, especially in the critical areas between wall openings, a simple two dimensional model can be used in conjunction with any availablecomputer analysis program to quickly provide a sufficiently accurate solution. The stresses and tie forces calculated from these analyses canthen be checked for acceptability usingthe methods described in MDG 8.4 and 11.1. In determining the member properties for theabove model, itis suggested that thetributary width of the inner and outer wythes be taken as the horizontal distance between ties. The stiffnesses of the wythes are based on this width andthe net area where the distributed load is applied. When the vertical masonry strip modelled is adjacent to a window opening, the load will include the load applied over the tributary width plus the load on one-half of the adjacent wall opening.

9-60


Typical Exterior Wall

Suggested Analytical Model 4-

4-

tinuous

e Out of Plane Lord

4-

e 4-

Fig. 93-1 Wall System Model Where opening edge stiffeners or building frame members are arranged to permit significant two-way action, other analysis methods that account forthe increase in load carrying capacity of the wall system may be used. The wind load on this system canbe determined using the procedures outlined in ASCE 788 or the governing building code. The critical sections forwind load will be near openings

or at the corners of the top level of the building. 93.5

Local Load Distribution in Multiwythe Composite Masonry Walls

The Code 5.8.1.1 states that a multiwythe wall designed

for composite action shall have

collar joints which are either (a)

crossed by connecting headers or

(b)

filledwith grout or mortar and connected bywall ties

The composite wall thus becomes a structural assemblage whose mechanical properties are dependent on those of its components (i.e., block, brick,and grout). The Code 5.8.1.2 states further that the average shear stress developed in the planes of the interface between the

9-61



93

m

Obb2949 0508758 894

m

wythes and collar joint or within headers shall not exceed 5 psi for mortared collar joints,

10 psi for grouted collar joints, and

8

for headers. If the shear stress at the wythe-collar

joint interface becomes too large, delamination, or header splitting can occur. Then loads can no longer be transferred between the wythes and the wallwill act as a multiwythe noncomposite wall. The above-cited average values of maximum allowable shear stresses in the collar joints of multiwythe composite masonry walls

relate strictly to shear stresses due to out-of-plane

loads. The Code has as yet not addressed the situation of collar jointshear stresses produced due to the applied in-plane vertical and horizontal loads acting onlyon oneof the wythes in a composite masonry wall. Research at Clemson University(9.3.5,9.3.6,9.3.7) has shown that the shearstresses in the collar joint for such in-plane loadingsituations are much larger than 10 psi. Nevertheless, due to theabsence at the time of this writing of any Code provisions for a detailed analysis of collar joint shearstresses due to in-plane loads, this load case is not considered in the MDG. The load transfer in a multiwythe composite masonry wall can be explained schematically with the help of the following figure. Applied Load

Brick Wythe

Block Wythe

L Grouted Collar Joint

Fig. 93-2 MultiwytheWall Load Transfer The composite wall in MDG Example 9.3-9 consists of an 8 in. hollow concrete block wythe, a 2 in. grouted collar joint and a 4 in. claybrick wythe. The example illustrates the development of shear stress in the collar joint due to out-of-plane wind loads acting on the

9-62


A C 1 T I T L E * N D G 93

m O662947 0508757 720 m

wall. The various symbols used in this example have the same meaning as those presented

in MDG 9.3.4. 93.6 Local Lateral and Axial Load Distribution in Single Wythe Loadbearing Wall Systems To design a masonry wall system

and its constituent elements, one must determine the

distribution of the axial and lateral loads throughout the wall. For axial and out-of-plane loads, critical wall sections are usually around openings. See MDG Example 9.3-10 for an investigation of the lateral and axial load distribution in a single-wythe loadbearing wall.

93.7 Local Distribution of Lateral Load Within Perforated Shear Walls See MDG Examples 9.3-11 and 9.3-12.

REFERENCES 9. l.1

ASCE Standard, ASCE 7-88, "Minimum Design Loads

for Buildings and Other

Structures," American Society of Civil Engineers, New York, NY, 1990. 9.1.2

Guide To The Use Of The WindLoadProvisionsOfASCE7-88,American Society of Civil Engineers, 1992.

9.2.1

Schneider, R., and W. Dickey, Reinforced MasonrvDesign,Second

Edition,

Prentice Hall, Englewood Cliffs, NJ, 1987. 9.2.2

Amrhein,J., Reinforced Masonrv Engineerine Handbook, Third Edition, The Masonry Institute of America, Los Angeles, CA, 1987.

9.3.1

Page, A. W., and N. G. Shrive, "Concentrated Loads on Hollow Masonry - Load Dispersion Through Bond Beams," The Masonry Society Journal, July-December 1987, PP. T45-T51.

9.3.2

Page, A. W., and N. G. Shrive, "Concentrated Loads on Hollow Concrete 9-63


AC1

TITLE*HDG 93 m 0662949 0508760 442 m

9.3.3

Masonry," AC1 Structural Journal, July-August 1990, pp. 436-444. Page, A. W.,and A. W. Hendry, "Design Rules for Concentrated Loads on

9.3.4

Masonry," The Structural Enpineer, Vol. 66, No. 17, September 1988, pp. 273-281. NCMA-TEK 141, "Concrete Masonry Section Properties for Design," National Concrete Masonry Association, Herndon, VA, 1984.

9.3.5

Anand, S. C., Young, D. T., and Stevens, D.J.,"AModel StressesBetweenWythesinCompositeMasonryWalls

to Predict Shearing Due to Differential

Movement," Proceedings, 2nd North American Masonry Conference (2 NAMC), University of Maryland, College Park, MD, August 1982, pp. 7.1 - 7.16.

9.3.6

Rahman, M. A., "AnalyticalInvestigations

of theBehavior

and Failure of

Composite Masonry Walls," a Dissertation submitted to the Graduate School of Clemson University in partial fulfillment of the. requirements for the degree of Doctor of Philosophy, August 1989.

9.3.7

Yalamanchili, K. IC,"Finite Element Computations on Super Computers and Their Applications to FailureAnalysis of CompositeMasonryWalls,"August

9.3.8

1990.

NCMA-TEK 81, "Lintels forConcrete Masonry Walls,"National Concrete Masonry Association, Herndon, VA, 1976.

9-64


A C 1 TITLEMMDG 93 W Obb2949 0508763 389 W

-

Example 9.2-1 T M S Shopping Center LateralLoadDistribution

The roof diaphragm in this example is considered flexible and is located at a height of 16 ft. All lateral loads will be assumed to be applied at this elevation. Determine the lateral load distribution to the east-west walls on Grid LinesA and C and to the north-south walls on Grid Lines 1, 2, and 3. Consider both seismic loads and wind loads.

N

o V 2

v1

v3

Lateral Loads on Masonry Walls ~~

Calculations and Discussion

Code Reference

Seismic Loads: East-West Direction Since the roof diaphragm is flexible andthe tributary areas of the roof loads to the walls on GridLines

A and C are equal, they sharethelateral

loadequally.Using

earthquake load computed in MDG. 9.1.1.2, each is subjected to a load OE

VA =

vc =

23'000 lb 2

=

11,500 lb

9-65


thetotal

m 0662949 0508762 215 m


Example 9.2-1 Cont'd. ~


Code Reference

North-South Direction The tributary areas of the roof to the walls on Grid Lines 1 and 3 are equal. As explained above, they will share the lateral load equally. The tributary area of the roof load for the wall on Grid Line 2 is twice that of the walls on Grid Lines 1 and 3. The lateral load on each wall is given as

VI =

v3 =

v2 =

23'000 lb = 11,500lb 2

23'000 lb 4

=

5,750lb

Wind Loads: East-West Direction The tributary areas of the masonry walls on Grid Lines A and C are equal. However, the wall on line C also carries thewind load applied to the 4.5 ft projection of the fascia. Using the distributed wind load calculated in MDG 9.1.1.2, wind loads resisted by these elements are:

VA = 202.5 plf

plf

X

X

82 ft = 8,300 lb 2

")

82 2

+ (4.5

ft

x 7

9-66


ft

x 20

psf) = 8,930 lb

A C 1 TITLExMDG 93

Obb2747 05087b3 L 5 L

m

Ejrample 9.2-1 Cont'd. Calculations and Discussion

Reference

Code

North-South Direction The governing wind load will be produced by winds from the south applied to the glass and fascia. The load is 260 plf, as given in MDG 9.1.1.2. Since the tributary areas of the walls on Grid Lines 1 and 3 are equal, they will share the lateral load equally. The tributary area of the wall on Grid Line 2 is twice that of the walls on Grid Lines 1 and 3. The lateral load on each wall is therefore:

VI =

v3 = 260 plf x

204*7 4

")

204*7 4

=

=

13,300 lb 26,600 lb

9-67


I

AC1

TITLE*IDG 9 3 m 0662947 0508764 098 m

Example 9.2-2

DPCGymnasium

- LateralLoad

Distribution

Determine the lateralload distribution to the walls in the north-south (Grid Lines 1 and 2) and east-west (Grid Lines A and B) directions consideringboth earthquakeloading and wind loading. This applies to all four Wall Construction Options A, B, C, and D. The influence

of the pilasters in Wall Construction Option A is neglected. TT

%S

128 '

\W *

B-

N

64 '

v,, U

B-

o

o

o Lateral Loads on Masonry Walls

~~~

and

Calculations

Code Reference

Seismic Loads: The seismic load on the DPC Gymnasium has been calculated in MDG 9.1.2.2 and is equal to 56,700 lb in each direction, applied horizontally. The roof diaphragm is considered to be flexible in its own plane. Wall shear is calculated assuming that the horizontal seismic load in

each direction (north-south or east-west) is

distributed equally between the walls oriented parallel to the direction of loading.

9-68


Example 9.2-2 Cont’d. and

Calculations

Conservatively, the shear acting on each wall can be applied at the level of the roof diaphragm. More realistically,the seismic loadsproduced by the acceleration of each wall’s mass can be applied at the wall’s centroid, and the roof load can be applied at the level of the roofdiaphragm.

For example,east-westseismicloadingwouldapply

the following

horizontal shear to the walls on Grid Lines A and B: a)

the product of the roof weight and the seismic coefficient, applied at the roof diaphragm midheight;

b)

the product of each wall’sself-weight and the seismiccoefficient, applied at each wall’s centroid, and

c)

the product of one-half of each north-south wall’s self-weight and the seismic coefficient, transferred at the roof diaphragm mid-height.

The remaining seismic forces from the north-south walls are transferred directly to the foundations. Both approaches result in the same wall shear. However, the second approach gives a lower (and more realistic) overturning momentand a lower shear force to be transferred from the roof diaphragm to the shear walls. Under the conservative approach, each wall is subjected to an equal seismic load of:

Vm

=

vNs=

569700 lb 2

=

28,350 lb

28,400 lb

This load is assumedto be applied at the average diaphragm height (top of metal deck) of

9-69


AC1 TITLEIMDG 9 3 W 066271.19 0508766 960 m

Example 9.2-2 Cont'd. ~~~~~

~

Code Reference


27.0 ft on the walls on Grid Lines 1 and 2, and at a 24.33 ft height on the walls on Grid

Lines A and B. Using the more realistic approach with the weights of the diaphragm and walls 'from MDG 9.1.2.2, the seismic loads are:

East-West Direction

'EW

D&p.= 0.0875 =

X

[ 163'800 lb 2

1.

+ 2 (2)(84,74O lb)]

14,580 lb

applied at the diaphragm height of 24.33 ft on the walls on Grid Lines A and B and VEw

= 0.0875 x (157,900 lb) x 2 = 27,630 lb

applied at the centroid of the walls on Grid Lines A and B North-South Direction VNS Dhp. = 0.0875 x =

163*800lb

1

+ (2) 2 (157,9OO lb)

20,983 lb

9-70


A C 1 TITLExMDG 9 3 M Obb2949 0 5 0 8 7 b 7 B T 7 M


Calculations

applied at the diaphragm height of 27.0 ft on the walls on Grid Lines 1 and 2 and

VNs

=

0.873

=

7,400 lb

x

(84,740 lb)

applied at the centroid of the walls on Grid Lines 1 and 2 Wind Loads: East-West Direction

As calculated in MDG 9.1.2.2, the tributary areas of the walls on Grid Lines A and B are equal, and each carries a load of

vew= 64 ft x

273 plf 2

=

8,750 lb

North-South Direction The tributary areas of the walls on Grid Lines1 and 2 are equal, and eachcarries a load of

VNs =

128 ft x 353 plf 2

=

22,590 lb

9-71


AC1

Example 9.2-3

TITLE*NDG 9 3

O662949 0 5 0 8 7 6 8 733

-

RCJ Hotel LateralLoadDistribution

- HandCalculations

In the following example, a load distribution to the resisting elements in the RCJ Hotel is obtained by hand calculation using Schneider and Dickey’s Method I (described in MDG 9.2.2). As directed by the Code 5.7.1, the method distributes the load with respect to each element’s relative stiffness. The configuration of the RCJ Hotel is described in MDG 9.1.3 Figs. 9.1-6 through 9.1-15. The unreinforced noncomposite masonry WallConstruction Option A with BuildingOption

II dimensions will be examined under Seismic Zone 2 loading.

Code Reference


The first step in the solution of any lateral load distribution is to determine whether wind or earthquake loadings govern. Seismic Loads: Using an analysis similar to that described for the T M S Shopping Center in MDG 9.1.12, the earthquakeloading for the RCJ Hotel can be calculated usingthe formula

V

= ZIKCSW, per ASCE 7, Chapter 9. The canopy is free standing, and will be

analyzed separately.

Z

= 318

I = 1.0 (category I building) K = 1.33

To find the coefficient C for the north-south direction:

9-72


A C 1 T I T L E * N D G 93 W 0662947 0508769 b 7 T

m

Example 9.2-3 Cont’d.


Code Reference

The mean roof height is

h, = 34.83 ft

T

= 0.05 x

T = 0.05

x

-

@

34.83

(ASCE 7-88)

fi

= 0.21 sec

@mi

C = NumericCoffkient =

.=

1

15J E K G

= 0.14

> 0.12

use C = 0.12 (required by ASCE 7, Section 9.4.2)

To find C for the east-west direction:

T = 0.05

x

34.83 ft

= 0.14 sec

Jm

C = NumericCofficient =

1

= 0.18

use C = 0.12

15JCiGG

Since no information on the soil is given, assume that the product, C x S = 0.14 permitted by ASCE 7) in both directions.

9-73


( m a x.

A C 1 TITLE+MDG 73 H Obb2949 0508770 391

m

Example 9.2-3 Cont'd. and

Calculations

Code Reference

The total seismic load is therefore the same in both directions and has a value of:

V

=

-3 X 8

1.0

X

1.33

X

0.14

X

W

=

0.07

X

W

Assuming each story height of wall mass is distributed equallyto the diaphragms above and below, the total weight at each diaphragm elevation can be calculated.

The dead load of

each floor is 110 psf, including partitions, and the roof dead load is 95 psf. The weight at each diaphragm level is calculated based on thetotal floor or roof area, the applicable dead loads and the total area of masonry walls. The approximate weight of the masonry walls is based on a 10 in. wall thickness, resulting in a 70 psf self weight. The weights at each floor are summarized in Table 1. Note that per ASCE-7, only dead loads contribute to the seismic lateral loads (except in storage or warehouse buildings). Table 1 Level

h'

W

(ft)

(kips)

W h '

Story Shear (kips) (kips) (ft-kips)

9-74


Floor Force

A C 1 T I T L E + M D G 93

m 0662949 0508771

226

m

ExamDle 9.2-3 Cont'd.

and

Calculations

Code Reference

The total building shear can therefore be calculated as

V

= 0.07 x 5,115

kip = 358 kips

The loads that are carried by the first floor diaphragm go directly to the foundations. Since the period of the building is less than 0.7 seconds, the concentrated load that ASCE 7-88 requires at the top of the building can be taken as zero. The remainder of the shear

is proportioned to each floor using Equation 13 of ASCE 7-88. For example,

Tire force at the roof =

Wh' J%'

c

x Total Shear

The force calculations for each floor and the story shears are shown in Table 1. The total wind load at the base from the north or south is

The total wind load at the base from the east or west is

9-75


A C 1 TITLE*flDG

93 W 0662749 0506772 L b 4

Example 9.2-3 Cont’d. Calculations and Discussion

Cade Reference

Since the seismic loads significantly exceed those produced by wind, seismic forces govern. The next step in the lateral load analysis is to determine the center of mass and center of rigidity at each diaphragm. Since the hotel is approximately symmetrical,the centerof mass,

CM,is assumed to be at the centerof each floor area. The location of these centersof mass are shown in Figs. 1 and 2. These mass centers can be more accurately determined. However, ASCE 7-88 requires that an accidental eccentricity of 5 percent of each of the horizontal building dimensions be added to the actual eccentricity between the center of mass and the center of rigidity. This

additional torsion will cover the inaccuracies in this

approximate hand calculation. To locate the center of rigidity at each diaphragm, the relative rigidity of each resisting element (walls and columns which support the diaphragm below the diaphragm elevation must be determined. Fig. 1 shows the lateral resisting elements below the second floor diaphragm, and Fig. 2 shows the lateralresisting elements below the third floor, fourth floor and roof diaphragms. Since some of these elements have control joints that create a break in the continuity of the wall, the wall segments between joints are analyzed as individual piers. Fig. 3 shows elevations of the walls on Grid Line2 and Grid Lines B and C. The wall on line 2 and the wall on the north side of the elevator shaft are assumed to resist shear in the east - west, or x, direction. The walls on Grid Lines B through F are assumed to resist shear in the north - south, or y, direction. The walls surrounding the stairwell on Grid Lines

3, A and G are not included in the analysis because holes in the diaphragms at the stair openings make it difficult to transmit lateral loads to these walls.

9-76


A C 1 TITLExMDG 93

m

Obb2949 0508773 O T O

m

Example 9.23 Cont'd. Code Reference


Wall

Wall

B,1-2

Wall

C,1-2

B.3-4

@

LC4 LC3

tCM"0& t

I

L

I

Y x

I

LC1

J Lobby D.3-4

I

-

F.1-2

Wall

F,3-4

I

I

Wall

Wall

J

J

0 ,

E.1-2

I I

m

I

Defines Lateral Load Carrying Walls

For Canopy

Fig. 1: Lateral Load Resisting Elements Below Second Floor Diaphragm

1

Wall E.l-2

Wall B.1-2

J

Wall F,l-2

J Wall C,1-2

J 2 3

CM0-l CR 1

Wall C,3-4 Wall

Y 4

I

I

83-4

Wall D,3-4

/

J

J

Wall

J

E.3-4

I Wall F.3-4

Lobby

X

-

Defines Lateral Load Carrying Walls

Fig. 2: L a t e r a l Load Resisting Elements Below Third Floor Through Roof

9-77


A C 1 TITLE*IDG 93

0662949 0508774 T37

Example 9.23 Cont’d. and

Calculations

Piers On Grid Line 2

8

9* 9.

9

Q

?. Q. Roof 4th Floor

3rd Floor 2nd Floor

X Indicates Opening

8 ?Y ? Piers On Grid Line B

Roof 4th Floor

o

8 7 Pier On Grid 3

3rd Floor 2nd Floor Piers On Grid Line

C Roof 4th Floor 3rd Floor

.MW

.

Piers On Grid Line D

2nd Floor

;:;;:h;

o

? Y Pier On Grid 3

m a ;;LI;; 2nd Floor

Piers On Grid Line E

2nd Floor

Piers On Grid Line F

Roof 4th Floor

.

3rd Floor 2nd Floor

Fig. 3: Piers on Grid Lines 2 and B through F Due to the presence of control joints at the intersection of most perpendicular walls, none

of the walls are considered flanged. 9-78



m 0662949 0508775 973 m

Example 9.2-3 Cont'd. Code Reference


To further simplify the analysis, the small piers formed in the walls on Grid Lines B and F, between Grid Lines 2 and 3, are neglected. The stair doors are offset between the second and third through fourth floors, creating small disjointed piers. As a result, these piers will resist very little shear and can be ignored. Following this

same reasoning, the first floor

columns on Grid Line E, between Grid Lines 3 and 4, are also ignored in the analysis. The stiffness of each pier is calculated using Eqs. 9.2-3 and 9.2-5. All piers are assumed to be cantilevered from the foundations. Since only the relative stiffness of the wall elements D

are important, and all wall thickness are assumed to be equal, the ratio of

'I may be Emf

taken as 0.10 in. The stiffness of the pier on Grid Line B between Grid Lines 1 and 2 (Wall B, 1-2) at the second floor is therefore

Ael

=

r

[

fi)]

0.10 X 4(8.83 ft (8.83 + 3 29.3 ft 29.3 fi

and

k = -1

MDG Eq. 9.2-5

A 1

k1

=

0.1012 in.

=

9.88 in." 9-79


=

0.1012 in.

A C 1 TITLESNDG 93

m

O b 6 2 9 4 90 5 0 8 7 7 6

BOT

m

Example 9.2-3 Cont'd. ~~

and

Calculations

The stiffnesses of the remaining piers below the second floor diaphragm are calculated in the same manner, and the results are summarized in Table 2.

The origin is selected as thejunction of Grid Lines A and 4. The x and y coordinates of the center of rigidity can be obtained using Eqs. 9.2-6 and 9.2-7. Since the relative rigidity, R, is found by dividing the stiffness by the total stiffness, these equations can be modified to incorporate stiffness values. Using the values in Table 2,

To determine the amount of shear carried by each element, the torsional moment, MT, must be calculated. This moment is center ofrigidity.

produced by the story shear applied eccentrically to the

As mentionedpreviously,

ASCE 7-88 requiresthatan

accidental

eccentricity of 5% of the horizontal building dimensionbe added to any actual eccentricity to determine thedesign torsional moment. Since this eccentricity can force the applied story shear to either side of the center of rigidity, four load cases must be investigated: 1.

Add 5% of the x buildingdimension

to the actual eccentricity in the x

direction 2.

Subtract 5% if the x buildingdimension to the actual eccentricityin the x

9-80


0662949 0508777 74b

A C 1 TITLE*MDG 93


Calculations

direction

3.

Add 5% of the y buildingdimension

to the actual eccentricityin

the y

direction 4.

Subtract 5% of the y buildingdimension tothe actual eccentricityin the y direction

The governing lateral load is produced by earthquake forces.

Therefore,the

actual

eccentricity is the distance between the center of mass and the center of rigidity. For the second floor level this distance is 3.54 ft and 3.30 ft in the x and y directions, respectively, assuming the center of mass at the geometric center of the building. Five percent of the building dimension in the x direction is 7.50 ft and five percent of the building dimension in the y direction is 3.38 ft. MT for each load case listed previously is shown inTables 2 to 5. The second floor shear can now be distributed to thewall elements using Eqs. 9.2-8 and 9.2-

P ey are replaced by the applicable MT and R is replaced by k,Eq. 9.2-8 for 9. If Pye, and ' the second story diaphragm, load case 2, becomes

t

P

Direct Shear

Torsional

Component

Component

The torsional component is added to the direct shear component when the wall under

9-81



93

m

Obb2949 0508778 b 8 2

m

Example 9 3 3 Cont'd. Code Reference


consideration is on the same side of the center of rigidity as the application of load, and is subtracted when the wall is on the opposite side of the center of rigidity as the application of load. Since the accidental eccentricity may reverse the sign of the torsional moment, the governing load case for a shear wall design may be the result of subtracting the accidental eccentricity from the actual eccentricity. Substituting relative stiffnesses for relative rigidities, the relative polar moment of inertia, J, is calculated using Eq. 9.2-10. For load cases 1 through 4, the calculations for shear applied to eachresisting wall element atthe second floor diaphragm level are summarized in Tables

2 to 5. Piers located above the first floor openingsare assumed to cantilever from the base of the structure.

Thisassumption

is madebecause

surrounding theseopenings.Thecalculations

of the restraint provided by the piers for shear applied to each resistingwall

element at levels 3 through the roof are summarized in Tables 6 through 17.

Canopy The canopy weight is 48.8 kips, including roof and masonry weight. Since the structure is symmetrical, the center of mass and the center of rigidity coincide and are located at the center of the canopy. This symmetry also allowsequal distribution of the direct shear force. Seismic loading governs the canopy design, and the total diaphragm shear is calculated as follows

9-82


A C 1 TITLExMDG 93

m

0662949 0508779 519


Calculations

9-83


m


93 W 0bb2949 0508780 230

m

Example 9.23 Cont'd. and

Calculations

Code Reference

9-84


1

A C 1 T I T L E x M D G 93 W Ob62949 0508783 177


Calculations

Discussion

Code Reference

9-85


AC1

TITLEvMDG 93 m Obb2949 0 5 0 8 7 8 2 003 m

Example 9.2-3 Cont'd. Calculations and Discussion

Cade Reference

9-86


A C 1 TITLESMDG 9 3

Obb2949 0508783 T 4 T

Example 9.23 Cont'd. Calculations and Discussion

Reference

9-87


Code

A C 1 TITLExMDG 93 D 0662949 0508784 98b


Calculations

9-88


m


93 D Ob62949 0508785 8L2

m


Reference

1111 9-89


Code

A C 1 TITLE*UDG 93

m O662949 0508786

759 D

Example 9.23 Cont’d. Calculations and Discussion

Code Reference

9-90


AC1

TITLExMDG 93

Obb2747 0 5 0 8 7 8 7 695


Calculations

t

9-91


m

A C 1 TITLE*HDG 93

Obb2949 0508788 521 W

Example 9 3 3 Cont’d. Calculations and Discussion


Code Reference

A C 1 TITLErflDG 93

m

0662949 0508787 4 b A

m


Reference

Code

r

9-93



93

m

Ob62949 0508790

LBT m


Cade Reference


9-94


A CT 1I T L E t M D G

93

Obb2949 0 5 0 8 7 9 1 016


Reference

9-95


Code


93

m 0662949


Calculations

9-96


0508792 T 5 2

~


93

m 0662949

~~~~~

0508793 9 9 9


Reference

9-97


Code

A C 1 T I T L E * M D G 93

m

Ob62949 0508794 825

m


Code Reference

9-98


A C 1 TITLE+MDG 93

Ob62949 0508795 761

m


~~

~

Code Reference


V

=

3 -

a

X

1

X

2

X

0.14

X

48.8 kip

=

5.12 kip in

both directians

Allowing for an accidental eccentricity of 5 % of 20 ft, equal to1 ft, each column must resist a maximum shear of

Shear to column in the &-west

The torsion of 5.12 kips

X

diredon =

'*O5 x 30 19 ft

*

=

kips

x

E)] = 1.55 kips 19

0.43 kips due to north-south seismic loads is

resisted by the east-west portal frame. These analyses assume that the columns are spaced at 19 ft on center. The columns form a resisting couple in the east-westdirectiononly

because beams are provided in that

direction only. This is a conservative assumption, since it neglects the contribution of the columns in the north-south direction, under north-south loading.

9-99


AC1

TITLExMDG 93 m Obb29Ll9 050879b bT8

Example 9.2-4

RCJHotel

- LateralLoadDistribution - ComputerCalculations

In the following example, a lateral load distribution to the resisting elements in the RCJ Hotel is obtained using a computer analysis. The configuration of the RCJ Hotel is described in MDG 9.1.3 Figs. 9.1-6 through 9.1-15.

ill be examined under The Wall Construction Option B, with BuildingOption I dimensions, w Seismic Zone 4 loading.

and

Calculations

Discussion

Reference

Code

Many PC-based structural analysis programs are available to the practicing engineer. Some

of the most common are ETABS, STAADIII, SAP90, and GTSTRUDL These programs are simple to use and provide a variety of pre- and postprocessors. If a computerized analysis program is used, a more detailed structural analysis of masonry structures is possible. Applying this method of analysis to the RCJ Hotel allows the effect of the coupling beams and openings in both the shearwalls and the diaphragm to be readily incorporated in the analysis. Most computer structural analysis programs model degrees of freedom. The designer must review

diaphragms as rigid elements with 3

the program manual to ensure that the

element properties and limitations are clearly understood. When determining the cross-sectional properties for use in the analysis, uncracked section properties are used. If a more refined analysis is desired, the individual elements can be subsequently checked to ensure thatthey remain uncracked. If any element exceeds the cracking loading, cracked

9-100



Code Reference

Calculations

section properties can be calculated based on analysis rerun.

the results of the initial loading and the

The analysis and crackedsectioncalculationscan

be repeated until

convergence is reached. However, unless the computed design forces in critical elements changesignificantly as a result of the cracking the additional computational effort of repeated iterations may not be justified.

If shear wall elements intersect, the designer is required by Code 5.7.1.1 to consider flanged wall' actions. These wall connections must comply with the requirements of Code 5.13.4.2. In many building configurations it is more conservative

to ignore flanged walls and Code

5.7.1.1 can be neglected. Code 5.7.1.1 limits the effective flange width on each side of the web to 6 times the flange thickness and requires minimum connection detail to ensure composite action. Since it is often difficult to model the building with the 6 times the flange thickness limitation, the designer must use sub-elements to determine the relative importance of these conflicting Code provisions.

The lateral load resisting elements of a typical floor of the RCJ Hotel is shown in Fig. 1. Thisconfiguration

isslightly

different from that presented for the previousexample,

incorporating the resistance of both North and South stairwell wallsand having removedthe center column on Grid Line E, between Grid Lines 3 and 4. Note that the building is taller with this building option as well. Fig. 2 shows the pier and coupling beam configurations on the major Grid Lines. Where there are perforations in the walls pier, identification labels are shown. 9-101


A C 1 TITLExMDG 93

m Ob62949 0508798

470 W

&le 9.2-4 Cont'd. Calculations and Discussion

@-

Reference

I

Code

" " " 1 1

Tr

a

@-

o-

m

@-

"

L

x

Fig. 1 Latewl Load Carrying Elements of RCJ Hotel at First Floor (Neglecting Canopy)

9-102


A C 1 TITLExMDG 9 3

m

m

Obb2949 0508799 307


Calculations

Code Reference

Piers On G r i d Line 2

...

i

Roo1 4 t h Floor

3rd Floor 2nd Floor m Indicates Beam

X

Piers On Grid Line B

??

?

Piers On Grid Line E

Floor Floor Floor

4th 3rd 2nd

C

Floor Floor Floor

Pl

PP P

0

PRoof 4th Floor 3rd Floor 2nd Floor

Piers On Grid Line

I n d i c a t eO s pening

Roof 4th Floor 3rd Floor 2nd Floor

B P

Piers On Grid Line

Roof 4th 3rd 2nd

Roof

P P

Piers On Grid Line D

Pl

F

?Piers On Grid Line 3

3rd Floor 2nd Floor

Pl

P2

P1

P2

Pl

P2

P1

P2

Fig. 2 L a t e r a l Load Resisting Piers Along the Major Grid Lines

Using the modelof the lateral load resisting elements shown in Figs.1 and 2, an analysis of 9-103


A C 1 TITLEdMDG 9 3

m

Obb2949 0508800 7 5 7

m


Calculations

Discussion

Code Reference

the RCJ Hotel was performed using the ETABS program. Seismic loads govern and the magnitude of the story shears were calculated usingthe procedures described in the previous example. These calculations and their results are summarized below. The three load cases shown in Fig. 1were analyzed to account for the accidental eccentricity described in MDG Example 9.2-3. A fourth ,loadcase, the east-west shear applied north of the center of mass was neglected, since by inspection it does not govern. Load Calculations

The total seismic load is the same in both directions and has a value of:

V

= 1.0 x 1.0 x

1.33 x 0.14 x W

=

The loads presented in Table 1 wereused

(for Seismic Zone 4)

0.186 x W

in the analysis and the critical wall shears

produced by the three loading cases are summarized in Table 2. The critical end moments on the coupling beams are summarized in Table 3. Table 1 Analysis Loads Level

h' (ft)

W

m'

(kips)

(ft-kips)

Floor Force (kips)

Roof

39.8

1,090323

43,200

323

4th Floor

30.2

1,320624

40,300

301

3rd Floor

20.5

1,320

27,400

205

2nd Floor

10.8

1,320938

14,700

110

Sum

939 5,050

125,600

,

9-104


Story Shear (kips)

829

A C 1 T I T L E + M D G 93

I

m 0662749 050880L 875 m


Calculations

Table 2 Critical Shear Force Results


Code Reference

A C 1 T I T L E S H D G 93

m

0662949 0508802 721

m


Calculations

Discussion

Code Reference

9-106


Example 9.2-4 Cont’d. ~~

~~


Code Reference

able 2 Critical Shear Force Results (Cont’d.)

3rd

P5

58

P4

58 66

P5 9-107


A C 1 T I T L E * N D G 93 D 0662949 0508804 5 T 4

m

Examde 9.2-4 Cont’d. and

Calculations

Table 3 Critical Beam Moment Reaction Results Wall Bay At Line B

I

I.D.

Moment At Left End (in.-kips)

Roof

B3 B3 4th B3 II B1 I B2 B1 I

303 450 446 65 -10 38

I

3rd 2nd

I

2nd

I

At Line C

Level

I I

E

I

I I

I

1

Line At

I

I

2nd

I

B2 B1

I m

I

40 Neglected

251 409 425 95 75 40 38 Neglected h.

9- 108


(in.-kips)

A C 1T I T L E * R D G

93

m 0662949 0508805 430 m


Calculations

9-109


A C 1 T I T L E J M D G 93

m

0 b b 2 9 4 9 0508806 377

m

Example 9.24 Cont’d. Calculations and Discussion

Code Reference

Table 3 Critical Beam Moment Reaction Results (Cont’d.) Moment At Left End Moment At Right End Level Bay I.D. Wall (in.-kips) (in.-kips)

461 533 -21

B7 B8 B9 B10

955 435 919

9-110


849

A C 1 TITLE*NDG 93

Example 93-1

m

Obb2949 0508807 203 W

-

T M S Shopping Center Load Distribution Within Single WytheWalls Under Concentrated Loads

Determine the effective area ofwallacting compressive stress

under concentrated loads for calculation

of

va).

The north loadbearing wall of the TMS Shopping Center, Grid Line A, is used to illustrate the distribution of load under typical roof joist supports.The portion of the loadbearing wall above the joist bearing elevation is ignored and the loading of a section of the wall is as shown below.

t- Effective -It; Length Effective

4

Length

a" CMU

Load

Spacing

and

Calculations The limiting effective length

per the Code

equals the bearing width

plus 4 times the wall thickness. See MDG Appendix A for concrete masonry unit dimensions.

9-111


5.12.1

A C 1 T I T L E * H D G 9 3 W 0662949 0508808 L4T W

Example 93-1Cont'd. ~~~~

~~

Code Reference


Eflective Length

=

3 in. + ( 4 x 7.63 in.)

=

33.5 in.

(thickness of 8 in. nominal concrete block is 7.63 in.) The Code states that the effective length shall not exceed the center-tocenter distance between concentrated loads.

33.5 in. < 5 ft x 12 in./ft The net area of the wallis

=

60 in.

5.12.1

.: OK

then calculated based on the limiting

effective length calculated above. If the wall was grouted solid, the net area would be: A,, = 33.5 in. x 7.63 in. = 255

in?

If the wall was hollow, the Code would require that the minimum net area be used; for face shell bedding thiswould equal 33.5 in. times the thickness of 2 face shells, or 2.5 in.

5.13.1.1

For walls laid in running bond the load distribution must be terminated at movement joints if the distribution intersects the joint. For walls laid inother thanrunning bond,the Code C.5.12.2 states that no stress shall be transferred across head joints. Therefore, the load distribution must stop at the head joint closest to each edge of the bearing plate. The Commentary suggests distributing the load over a 45' angle. Thus, if the previous wall system was laid instack bond, the

distribution of load shown by the shaded areas in the figure below should be assumed.Consequently,

the effectivelength

limited to one unit.Again,minimum

is

area (based on face shell

9-112


ofwall


9 3 W Ob62949 0508809 O8b W

Example 93-1Cont’d. Code Reference


bedding) should be used if wall is not grouted solid.

9-113



Example 93-2

TMS ShoppingCenter

0bb29Ll9 0508810 B T B

- Distribution of Concentrated Loads Acting on

a Bond Beam

The north wall on Grid Line A of TMS Shopping Center is a loadbearing wall supporting steel joists spaced on 5 ft centers. Assuming that the topcourse of the wall is replaced with a bond beam, calculate the net bearing area of the hollow wall section and the effective length over which the concentrated load can be distributed.


Code Reference

P 7.63"

1 8

I

"

o

It

r

The bearing plate dimensions are given in MDG Fig. 9.1-1 as 5 in. x 7 in. The nominal wall thickness and the bond beam depth are assumed to be 8 in. The angle of load dispersion through the bond beam is taken as 30" from the vertical as suggested in MDG 9.3.2.1. See MDG Appendix A for properties of concrete masonry units. The gross effective bearing area of the hollow wall that can be used to resist the loal given by:

Gross Bearing Area

= t x

LB 9-114


AC1 TITLESNDG 93 m 0662747 05088LL 734 m

Example 93-2 Cont'd. Calculations and Discussion

Reference

Code

in which t is equal to the nominal thickness of the wall and LBis calculated as

LB = [ L p + 2 ( 8 in.)(& 30°)] where Lp is the length of the bearing plate. For this example, t = 7.63 in. (8 in. nominal block), and Lp = 7 in., which leads to a value for LB OE LB =

7 h + 2 ( 8 in.)(tan30" )

16.2 in.

=

Combining above equations and substituting these values gives: Gross Bearing Area

= ( 7.63

in.) ( 16.2 in.) = 124.0 in?

The Code 5.13.1.1 requires that the load must be resisted by the net area. As the net area for an 8 in. CMU is equal to 41.5 in.2, the net effective bearing area below the bond beam can be calculated by proportionality as 41.5

Net Bearing Area =

The effective length over

I

ia2

(15.63 in.) (7.63 in.)

124.0 i a 2

which concentrated loads can

bearing width plus four times

=

43.1

ia2

be distributed is limited to the

the wall thickness, but not to exceed the center-to-center

spacing of the concentrated loads. For this example, the effective length is given by

Efective Length = LB + 4 (7.63 in.) 5.121

EflectiveLength

=

16.2 in. + 4 (7.63

in.)

9-115


=

46.7 in.

A C 1 TITLE*HDG 93

Obb29V9 0508832 670

m

Example 93-2 Cont’d. Calculations and Discussion

Code Reference

Thisdistanceislessthanthecenter-to-centerspacing

of theconcentratedloadsand,

A,, for calculating compressive stressin the therefore, governs. The net cross-sectional area,

wall can again be calculated by proportionality as

.

9-116


AC1 TITLEtMDG 93 m Obb2949 050BBL3 507

-

Example 93-3 T M S Shopping Center Effective Bearing Area Under Concentrated h d For thewalls on Grid Lines1 or 3 of the TMS Shopping Center, theW 16 x 31 steel beams running along Line B are supported on 5 in. x 10 in. x % in. bearing plates. Assume the bearing plate is placed at the centerof an 8 in. grouted block wall laidin running bond. The bearing plate is located at a distance of 1-5/16 in. from each face of the wall. Calculate the effective bearing area for this plate location.

r

7.63" Bearing Area Al

45

L

i

A

A

Bearing Area 2

Section A

-A


Code Reference

Referring to MDG 9.3.3,

A,

=

5 in. x 10 in.

=

50 h2

A, = [Sin. + 2( 1.31 in.)][ 10 in. + 2( 1.31 h)]= 96.3 The load can be distributed over an area given by

9-117


in?

A C 1 TITLExMDG 93

m 0662949 0508814

443


Calculations

Area

Discussion

= A,

5

Code Reference

but not greater than

5.12.2

2Al

Al

in?

Area

= 50

Area

= 69.4 in.2 but not greater than

50

in?

but not greater than 100 in.2

Therefore,

Effective Bearing Area = 69.4 in.2

9-118


2(50 in.2) = 100 in?

A CT 1I T L E x M D G

Obb2949 0 5 0 8 8 3 5 3 8 T

93

m

-

Example 93-4 TMS Shopping Center Effective Bearing Area Under Concentrated Load Assume the 5 in. x 10 in. x ?4in. bearing plate in MDG Example 9.3-3 is located at the edge of the masonry wall. Calculate the effective bearing area for this plate location.

T

Bearing Area Al

Area A2 Is Measured On

L A

t

1

I

This Plane

A

Bearing Area A2

and

Calculations

Reference

Discussion

Since the edge of the bearing plate, in this case, is in line with

Code

the face of the block, the

effective bearing area which can be considered is equal to the actual area of the bearing plate. Therefore,

Effective Bearing Area = 50 in.2 Note:Typically

the bearing plate is placed no closer

masonry unit.

9-119


than 0.5 in. from the face of the


Example 9.3-5

93 W Obb29Y9 OSOBBLb 2Lb

m

-

TMS Shopping Center Effective Bearing Area Under ConcentratedLoad

Assume the 5 in. x 10 in. x % in. bearing plate in MDG Example 9.3-3 is placed at the center of an 8 in. grouted block wall laid in stack bond

but at a distance of % in. from a

head joint. Calculate the effective bearing area for this plate location.

>

"""""""""""

7

Area A2 Is Measured On This Plane

Section B - B

ReferenceCalculations Cade and Discussion The Code states that A, shall terminate at head joints in other than running bond. As the bearing plate is assumed to end at a distance of % in. from the head joint of the stack bond masonry the value of A2

can be calculated as:

A2 =

[ 5 h. + 2(0.75 in.)][ 10 h. + 2(0.75 h.)]= 74.8 h2

The load can be distributed over an area given by

9-120


5.12.2

A C 1 T I T L E * H D G 93

m 0662949 050BBL7 L52 m

Example 93-5 Cont'd. and

Calculations

E

Area

= A,

Area

= 50

Area

= 61.1 in.2whichis less than 100 in.2

in?

but not greater than

74'8 i a 2 50 h2

2A1

but not greater than 100 in.2

:. In this case, effective bearing area = 61.1 in.2

9-121


A C 1 T I T L E * l D G 93

Example 93-6

m

0662949 0508838 O99

m

-

DPC Gymnasium Distribution of Gravity Load Moment in Multiwythe Noncomposite (Cavity) Walls

Determine the distribution of bending moment resulting from gravity loads that act on the block wythe of a two-wythe brick block noncomposite wall on Grid Lines A and B in the

DPC Gymnasium, Wall Construction Option k The following material properties have been assumed for the purpose of the load distriiution: Concrete Block Masonry Clay Brick Masonry (Hollow Block) 2,000

Unit Strength (psi) Mortar

6,000

Type N

Type S

f’m

(psi)

1,500

2,500

E m

(psi)

1.8 x 106

1.9 x 106


B

Reference

Code

O*@

Roof Truss With Sloped Top Chord With Continuous 6 x 4 To Support

Joint Reinforcement Brick

Continuous Reinforcement Bond Bearing Plate 6 x 12 With Headed

9-122


A C 1 TITLESMDG 93

m

Obb2949 0508819 T25

m

Example 93-6 Cont’d. ~~

~

Code Reference

Calculations and Reference

Loads for this wall are taken from the north wall (Grid Line A) of the

DPC Gymnasium given in

MDG 9.1-2. As the vertical gravity

loads

5.8.2.1(b)

acts only on the block wythe, these will be .resisted entirely by that wythe. At the point of load application, any bending moment caused by the eccentricity of the load with respect to the centroid of the block wythe will be resisted completely by the block wythe. Away from the point of load application, however, some of this momentis assumed to be transferred to thebrick wythe in proportion to the relative flexural

stiffness of the two wythes. One can conservatively assumea triangular distribution of gravity load bearing stress on the block wythe as shown.

The total moment, M,is then equal to load, P,multiplied by the eccentricity, e, where e is measured from the centroid of the load distribution to the centerline of the block wythe.

9- 123



93

m

0662949 0508820 747

m

Example 93-6 Cont'd. ~~~

Calculations and Reference

Code Reference

Thus, the total moment, M,is given as:

M

=

Pe

The critical truss reaction is calculated in MDG 9.1.2.1 as 14,880 lb. A bearing plate of size

6 in. x 12 in. placed one-half in. from the inner face of the block wythe, as shown, yields an eccentricity, e, of e = 7*63h* - 0.5 2

- 6

in.

(i) =

1.31 in.

P

7.63"

Therefore, the total moment, M,is given by

Although this may not always be true, the wall ties connecting the two wythes are assumed to have sufficient axial stiffness

so that the two

wythes must have equal curvature. This fact leads to the relationship

9- 124


5.8.1.5

AC1

TITLE*HDG 93 U 0662949 0508821 b 8 3 U

Examde 93-6 Cont'd. Calculations and Reference

Code Reference

that

In these equations, the subscripts 61 and br correspond to the block and brick wythe, respectively. Assuming only face shellmortar bedding for the block wythe, IN is given by MDG Appendix A as 309 in.4/ft. While Code 5.13.2 permits the use of the average net sections, the minimum section is used here for simplicity. The value of Ibr= (12)(3.625)3/12 = 47.6 in.ll/ft. Using the section and material properties above, each

wythe's share of the total moment acting in the wall can be computed as follows:

Mar

=

19.5 h - kip [ (1.8 X lo6 psi) (309 h.")] [ (1.8 x 106 psi) (309 in.")] + [ (1.9 x lob psi) (47.6 in.")]

M M = (19.5 h-kip) (556 X lob lb-h2) (646 X

lo6 lb+.*)

or

M M = 16.8 h - k i ~ 9-125


5.13.2

A C 1 T I T L E S f l D G 93

m

Obb29Y9 0508822 51T D


Calculations

and

Ma

=

19.5 h-kips [ (1.9 X 106 psi) (47.6 i n . ' ) ] [ (1.8 x 106 psi) (309 in.3 ] + [ (1.9 x lob psi) (47.6 in?) ]

Ma

=

(19.5 in.-kip) (90.4 X lo6 1b-h2) (646 X 106 lb-in?)

or

Ma = 2.7 i n . - k i p ~

Check

M M + Ma = 16.8 M

=

in.-kip~ + 2.7 h-kip

19.5 h.-kip

19.5 h-kips

Note: At the application point of the vertical load the block wythe is subjected to a total moment, due to the eccentricity of the load of 19.5 in.-kips. Accordingly, the block wythe must be designed for this moment magnitude at that level.

9-126



Example 93-7

DPCGymnasium

Ob62949 0508823 45b

m

- In-PlaneLateral Load DistributioninMultiwythe

Noncomposite (Cavity) Walls ~~~

~

Consider the noncomposite wall in the DPC Gymnasium on Grid Lines 1or 2 (Option A) subjected to in-plane lateral loads. Determine the distribution of in-plane lateral load to the individual wythes assuming Wall Construction Option A (unreinforced brick and block cavity wall).


Reference

Code

~~

Reinforced BondBeam

k TEK Screw ! ! ! !

Strap Anchor

!

! ! !

I

As in MDG Example 9.2-2, the roof system is assumed to be flexible in its own plane, and is attached only to the block wythe, as shown.

Thus, according to the Code, the in-plane lateral load is resisted only by the block wythe, and any transfer of loads between the brick and block wythes is neglected. 9-127


5.8.2.1(c)

A C 1 TITLEzMDG 9 3

Example 9.3-8

DPCGymnasium

m

0662949 0508824 392

- DistributionofOut-of-PlaneLateralLoads

in

Multiwythe Noncomposite (Cavity) Walls

Consider the noncomposite wall (Wall Construction Option A) in the DPC Gymnasium on GridLines 1, 2, A, and B. The wallis subjected to lateral loadsshownin

the figure.

Determine the out-of-plane lateral load distribution to the brick and block wythes. RD:$ragm Reaction

20 psf

and

Calculations

To design wall systems

for out-of-plane lateral loads, such as wind and

earthquake loads, the Code requires that the load transverse to the plane of the wall be distributed to the individual wythes of the wall system in proportion to their relative flexural stiffnesses. In general, the stiffness of each wytheisaffectedbyitsheight

and boundary

conditions. In this building, the height and boundary conditions of each wythe of the cavity wall are assumed to be the same.

9-128


5.8.2.2

A C 1 TITLE*MDG 9 3

m

Ob62949 0 5 0 8 8 2 5 229

m

Example 93-8 Cont’d. and

Calculations

Reference

Discussion

Code

Defining the distributed transverse wind load acting on the wall as equal to a force P per unit area, the resulting load on each wythe can be obtained from:

inwhich the subscripts bl and br refer to the block and brick wythe, respectively. This equation is valid regardless of the boundary conditions, as long as they are thesame for both wythes. Substituting the previously calculatedstiffnessvalues fromMDG Example 9.3-6 and the design wind pressure of 20 psf into the above equation yields:

PM

=

P

=

(E%atal

(20 Psf) (556

X

106 1b-h2)

(646

X

106 lb-in?)

9-129


=

1,.2


Example 93-9

DPCGymnasium

m 0662747

0 5 0 8 8 2 6 Lb5

m

- Shear Stress Distribution in the Collar Joint of a

Multiwythe Composite Wall Due to Out-of-Plane Wind Load Determine the maximum shear stress in the collarjoint of the multiwythecomposite masonry wall on Grid Line 1 or 2 of the DPC Gymnasium Wall Construction Option B, due to the out-of-plane wind loads. The following material properties are used in the analysis. Clay Brick

Concrete Block Masonry Masonry (Hollow) Unit Strength (psi) Mortar (psi) Em or Es (psi)

f ' m

and

5000

6,OOo

Type N

Types

1,500

2,5 O0

1.8 x 106

1.9 x 106

Calculations

2" Grouted Collar Joint 4"

Brick WytPe

I

18,' Block Wythe

30 ' max.

20 psf

9-130


AC1

TITLEtMDG 93 M 0662947 0508827 OTL M

Examde 93-9 Cont’d. and

Calculations

The wind load for the wall is given previously in MDG 9.1.2 as 20 psf. The highest point of the wall, which has a height of 30 ft, is utilized to compute the maximum interface shear stress in the collar joint. This shear stress due to out-of-plane loads is computed after the crosssection of the composite wall has been transformed into one material. The standard shearstress formula of the Code, Eq. 6.7, is modified to take into account the existence of more than one material in the cross

5.13.1.2

section as

The location of the neutral axis and the magnitude of the moment of inertia of the composite section are calculated as follows: According to the Code, in walls designed for composite

action, the

shear stress shall be computed using section properties based on the ~

5.13.1.2

minimum transformed net cross-sectional area of the composite wall. The generally accepted transformed area concept for elastic analysis,

in which areas of dissimilar materials are transformed according to their relative moduli, shall apply.

Using the material propertiesof various components given above, a one ft width of wall can be transformed to an equivalent concrete block as shown. The location of the neutral axis

from the centroid of the block is calculated by taking the first moment of area about the block centroid as: 9-131


TITLE*MDG 9 3 m Obb2949 0508828 T38 m

AC1

Example 93-9 Cont'd.

Code Reference


From MDG Appendix A

P -"l"l

r 1

Block

I=309 in4 A=41.5 in.

12"

J4 J

+

7.63"

I

2

Grouted Collar Joint

I

-" -

X = 4.11"

I

I

3.63"

/

///// Brick

-

1L-E

c Fvst moment of

x 12=12.67"

area

Y

r =

-x =

2

+ -2h)(2in. 2

x ,.,,in.)+( 7.63in. 2

+

2h

+

3*63in.k . 6 3in. x 12.67in.)

2 2in. x 6.67in. + 3.63in. x 12.67in. + 41.5in.2

414.41 h? = 4.11 in, 100.77

from the block centtoid

9-132



m

0662949 0508829 974

m

Example 93-9 Cont’d. Code Reference


:.

Zw

=

6.67in.(2hJ3 3 W h 4 + 41.5in?(4.11i~~)~ + 12

+ 2h(6.67 in.)

Zw

=

1,639

ia4

7.63h + -2in. r. , , -, . , 2

+

12.67in.(3.63ix~)~ 12

for a 12 in. wide strip

The value of the shear stress is critical at either the block-collar joint interface or brick-collar joint interface depending upon where the value of the first moment of area Q is larger. The value of Q at these interfaces is computed from the transformed area as follows:

Qbl = 41.5

Qbr = 161

x 4.1 1

in. = 171 h

3

in.’

Therefore, Qbl = 171 in.3 governs

9-133



93

m

Obb2949 0508830 6 9 6

m


Code Reference

The maximum shear stress at the block-collar joint interface is given by:

This value is less than (1.33)(10) = 13.3 psi

.:

OK

5.3.2 5.8.1.2

9-134


A C 1 TITLEsMDG 9 3 W 0662747 0 5 0 8 8 3 3 522 W

Example 93-10

-

TMS Shopping Center LateralandAxial Load Distribution in Single

Wythe Loadbearing Wall Systems Determine the axial and out-of-plane lateral load distribution to the elements in a single wythe wall system. The east wall section under consideration is located on Grid Line

3 of the T M S

Shopping Center layout shown in Fig. 9.1-1. This section

wall ofis

subject to a

concentrated reaction from the roof girder

Pier t 2

/

(W 16 x 31) on Grid Line 3, near Grid Line

B. Wind Load 20 psf positive pressure or suction pressure (includes interior pressure loading)

~~

~

Code Reference

Calculations and Discussion Loading on Wall Pier 1: Whendesigningmasonrysystems,

the Code requires a number of different load

combinations to be investigated to determine the critical loading conditions.

D

+ L, 0.9D + E, D + L + (W or E ) or D + W are the

govern the design of wall systems.

9-135


load combinations that usually

A C 1 TITLE*MDG

93

m 0662949 0508832

qb9

m


Calculations

Code Reference

For the purpose of this example, the following assumptions are made:

1.

When designing wall sections to resist out-of-plane loads in Seismic Zone 1, the D

L

+ W loadcombinationusuallygoverns

+

and will be assumed to governinthis

example.

2.

The wall system is constructed of 8 in. reinforced concrete block with a 88 psf dead load.

3.

The roofsystemprovidessimple

lateralsupporttothe

wall at thebase

of the

parapet. 4.

The foundation providesnegligible restraint to out-of-plane rotations, i.e., it can be assumed to act as a simple support.

It should be noted that when analyzing unreinforced wall systems, control joints will affect the continuity of the wall system and therefore the load distribution. In these cases, the designer can assume that the wall system is separated into discrete elements defined by the control joints. In the subsequent analysis, it is assumed that the control joints transfer outof-plane shear but not in-plane shear or axial stresses. Axial Loads: From the analysis presented in MDG 9.1.1.1, the W 16 x 31 girder reaction, P,to the wall at B3 is equal to 15,770 lb. It should be noted that the uplift effect of the dead load of the fascia suspended at the end of cantilevered joists betweenGrid B and C was ignored in the calculations. This uplift was ignored because

it simplified the calculations and the uplift

forces acting on the soffit would likely balance out a significant portion of the fascia dead load. This simplification results in a slightly conservative value for the reaction.

9-136


A C 1T I T L E l k M D G

93

m

Obb2949 0508833 3T5 M

Example 93-10Cont’d.

and

Code Reference

Calculations

The girder reaction is applied to the wall at the center of a door opening. This girder reaction is transferred symmetrically to thewall sections oneither side of the opening. Thus, Pier #1 will be subjected to 1/2 of the girder reaction, 7,890lb. Even though the 7,890lb axial load is applied to Pier #1 over the height of the wall above the opening, it can be conservativelyassumed that the entire

loadcomes onto Pier #1 at the girder bearing

elevation. The rotation of girder produces a varying bearing stress distribution on the wall. in the figurebelow, a simplified triangular stress distributionisassumed

As shown

to model the

conditions under the bearing plate. The axialload applied to Pier #l will act at an eccentricity fromthe centriodal axis of the wall cross section. For a 5 in. wide bearing plate, Pl2

e =

( L x 7.63 in.) 2

(i

x

5 in. + 0.5 in. 5 in.

0.5 in

Interior

Other than thegirder reaction, the wall section has no other applied axial loads. However, a portion of the dead loads from the sections above the door openings are also transferred to Pier #l. Half of this load will be assumed to go to each side of these openings. The additional axial load on Pier #1 (Po) is therefore:

Po = 88 psf

x

3.33 ft

x 12

ft

+

(y) ft] x

9

=

5,720 lb

This load is assumedto be applied to Pier #1 at the top of the tallest opening, i.e., at 10 ft 9-137


AC1

TITLESMDG 93 m 0662949 0508834 231


Reference

Code

from the floor level. If desired, the wall dead loads can be considered as two concentrated loads applied at heights of 10 ft and 7 ft. Lateral Loads due to Wind It is assumed that the interior and exterior wind pressures combine to produce a uniform wind loading of 20 psf on the wall system. Because of the wall system configuration,these wind loads are primarily distributed in oneway bending to the foundation and roof supports. Further, due to the flexibility from the openings, most of the load on sections of the wall system with openings is transferred first to the continuous wall sections on either side, and subsequently to the foundation and roof. Based on these assumptions, the uniform lateral load on Pier #1 is calculated as:

W

=

20 psf

X

2

2

It is important that the wall sections above the door openings be analyzed to determine whether they have sufficientstrength to transfer the loads to the supporting piers. A simply supported beam analysis usingthe opening length as the beam span is suggested for this wall section. The total loading applied to Pier #1 is summarized in the figure below. From this loading, critical moments and shears, and subsequently stresses, can be determined. These stresses can then be used to evaluate whether the maximum stresses in the proposed wall section design are within allowable limits. During this evaluation,the self weight of the wall system must be added to the axial stress to provide an accuratestress determination at these critical 9-138



Obb2949 0508835 L78

m

Example 93-10Cont'd.


Code Reference

sections. 1.65"

In most cases, the suction wind loading (or the outward Seismic loading), combined with eccentric axial load,will produce the critical design conditions

i%: I p2+tl/ I

for a wall system.

If the designer so desires the loading shown for Pier #1 can be further simplified by assuming that both axial loads are applied at the girder bearing elevation.

Loading on Wall Pier #2 Using procedures similar to those outlined above,the axial and lateral loads on Pier #2 can be calculated. To simplify calculations, since it is not subjected to an axial load from the roof, Pier #2 is assumed to have a design width of 1ft. Axial Loads at top of door:

Po

=

88 psf x

(y) x 12

ft

=

1,760 lb 53.3 plf

Lateral Loads:

w=2opsfx

2

The loading for Pier #2 is summarized in the figure shown.

9-139

. COPYRIGHT ACI International (American Concrete Institute) Licensed by Information Handling Services


m

0662949 0 5 0 8 8 3 b 004

m


Code Reference

Design of Lintels A and B: The analysis of the lintelsover

thedoor

openings issurprisingly

indeterminate. The

distribution of vertical load to the lintels will depend on the size of the opening, the depth of the masonryabove

the lintel, the restraint provided by the piers on each side, the

presence of control joints, and the type of vertical loading. As the ratio of the length of opening to the depth of masonry above the lintel decreases, more of the vertical load is transferred to the piers by arching action. This arching action will only occur if sufficient restraint is provided by a tension tie over the opening, or if sufficient masonry exists on either side of the opening to resist the arch thrusts. Tension ties can be provided by sizing the lintel reinforcement to resist the beam stress and to provide the tie for arch action over the opening. This reinforcement must be adequately anchored within the piers on either side if these bars are tobe expected to develop sufficient strength to resist the arch thrusts.

For uniformly distributed vertical loadsapplied to walls which haveheights of masonry above the lintel that are less than one-half the lintel support spacing plus 16 in., NCMA Tek-Note

81 (9.3.11) suggests that the lintel be designed for the entire vertical load, and be assumed simply supported over the wall opening. This same analysis can be used for cases where movement joints are present on one or both sides of the openings. Concentrated loads can be distributed to the lintel in a manner similar to that described in MDG 9.3.1.

For lintels which have heights of masonry above the lintel that exceed one-half the lintel span spacing plus 16 in., the designer can assume that all of the uniformly applied vertical

9-140


A C 1 T I T L E v N D G 93 W 0662949 0508837 T 4 0 W


Calculations

load above the opening, and above the apex of a 450 triangular area over the opening, is carried by arching action. If arching action is taken into account, then the lintel should be designed to support the sum of the dead weight of the masonry in a 450 triangular area above the lintel and the self weight of the lintel (9.3.12). If a concentrated load is applied to the wall away from the center of the lintel opening someof this load may be distributed to the lintel and must be accounted. The reader should refer to Schneider and Dickey

(9.2.1) for further information on this subject and a suggested analysis method.

If the designer is uncertain whether there is sufficient masonry aboveor oneach side of the opening to form the arch, arching action should be neglected. Lintel A: The location, dimensions and loads for lintel A are as shown.

A minimum bearing length of 4 in. is assumed on each side of the opening, so the span of the lintel is:

Span

=

3.33 ft

+

4*0

in*

12 in./fi

=

3.67 ft Apex

j

The height of the masonry above the lintel

=

19 ft

- 7.0 ft

12ft > -3+.67 ft 2

j

45O o+,

to the top of wall is: Height

Arch

=

12 ft

in. 12 in./ft

=

3.17 ft

H 3.67 ’

Therefore, arching action is present.

9-141


Loads Above Of Triangle, If Present. Are Carried By Arching Action

AC1

TITLE*NDG 93 W 0662949 0508838 987 m


Code Reference

Calculations

Since action is present, the lintel can be design for the weight of the lintel and the weight

of the wallwithin the triangular area defined by the arch. The loads on lintel A are summarized below. If a 45" angle of distribution is assumed,the arch thrusts can be calculated as the horizontal component of the arch forces. Therefore the thrust is [(19 ft

Thrust

=

"1'

- 7 ft) X 3.67 fi] - (3*67 2

x 88 psf x cos 45"

2,320 lb

- Assumed 90 plf Lintel

Lintel

J 3.67

'

and must be resisted with a tension tie or the in-plane shear capacity of the piers on either side of the opening. Lintel B: The location, dimensions and loads for lintel B are as shown. A minimum bearing length of 4 in. was assumed on each side of the opening, so the center

9-142


A C 1 T I T L E * H D G 93

m

0662949 0508837 813

m


Calculations

to center span of the lintel is: P

The height of the masonry above the lintel (to Girder bearing elevation) is:

Height 4.3 ft

=

14.6 ft - 10.3 ft 10.3 ft +

12 2

l6

=

4.3 ft

in'

in./ft

=

6.5 ft

Therefore, no arching action The loading for the lintel is shown below. This load includes

the weight of the lintel, the

weight of the wall immediately above the opening and the distributed concentrated load. Note that a distribution angle of 30" was used and the distributed length was limitedto the bearing plate width plus 4 times the wall thickness (see MDG Example 9.5). The effect of the bond beam was ignored since this results in a conservative loading

and simplifies the

analysis. If these loads producean uneconomical lintel designfurther refinement of the load analysis can be conducted

9-143


A C 1 TITLE*MDG

m

93

Obb2949 0508840 535

m

Examde 93-10 Cont'd. and

Calculations

Code Reference

o:['n:ght

Of

, .

,-, Bearing Plate

I

\

J. . "&L""""""""-A

15,770 1 (40.5I 12 ) = 4,670plf

10" + 4 X 7.63" = 40.5"

88 psf (9') = 792 plf

Lintel

Lintel B must be designed to resist the two distributed loads

9-144


A C 1 T I T L E * M D G 93

m

0662749 0508841 471

Example 93-11 T M S ShoppingCenter

-

m

DistributionofHorizontalLoadWithin

Reinforced Perforated Shear Walls Determine thehorizontal load distributionto the piers in a reinforced perforated shearwall. The wall section under consideration is Wall Construction Option B located in the TMS Shopping Center on Grid Line 3, between Grid Lines

A and C. This 8 in.reinforced,

partially grouted, concrete block wall has two door openings separated by a 2 ft - 8 in. pier

of masonry. The figure below shows the configuration of this wall and its openings. 82 '

1

I

~

J4-

4 1

and

13.300 lb

Calculations Using Method I described in MDG 9.2.2, the lateral load applied to the top of this wall can be distributed to each wall section.

Since the loads in this example are relatively small, it is appropriate to use a linear elastic, uncracked section analysis for the distribution of the horizontal shear loads. With higher loadlevels

and/or smaller wall piers, a crackedanalysis

may be more appropriate.

Furthermore, thesection analyzed is assumedto be reinforced and has no movement joints 9-145


A C 1 TITLEMMDG 9 3

Obb29Y9 05088q2 308

m

Example 93-11 Cont‘d. and

Calculations

within the wall. If movement joints are present, they will separate the wall into isolated piers and must be accounted for in the analysis. See MDG Example 9.3-12.

As shown in MDG Example 9.2-1, the largest shear load applied near the top of the wall on Grid Line 3 is produced by the wind loading and has a value of 13,300 lb. It is assumed that only the stiffness of the wall below the diaphragm will be effective in resisting the lateralloads. The totalcantilever deflectionof the wall can be calculated at the

16 ft elevation using MDG Eq. 9.2-3.

Since all walls have equal thickness and only relative stiffness is required, assume pl

- 1 in.

”-

Emf

10

The relative cantilever deflection of a solid 16 ft x 82 ft shear wall would be

-

Ae16x82

in. [4(=)’ -10 82 A

+

This produces a relative stiffness, k16x

3(=) 82

-

A

1

= 0.062

0.062 in.

=

in.

16.3 in.-’

The largest opening in the wall is the 10 ft by 10 doorway. Thus, the relative cantilever deflection of a 10 ft by 10 ft opening strip,

A,, -

[ (i: ir

1 in.410

+

3 ( 3 ]

is =

9-146


0.037 in.

A CT 1I T L E * f l D G

93

0662949 05088q3 244

m

Example 93-11Cont'd. and

Calculations

There are two sections of wall within the opening strip, a 10 ft x 36 ft section to the right of the door opening, pier 3, and a 10 ft x 36 ft section to the left of the 10 ft by 10 ft opening. The left section has a 3.33 ft x 7.0 ft opening and contains piers 1 and 2. For either 10 x 36 section, the fixed pier deflection from MDG Eq. 9.2-4 is

4 0

-

x

36

1in.

- 10

x 36

'P10

[ (gr

'

-

3(%)

+

=

11.7

0.085 in.

]

=

0.085 in.

in.-'

Since there is a 3.33 ft by 7 ft opening in the left section, the composite stiffness of the two piers, 1 and 2, must be determined before the stiffness of this section can be calculated. The fixed pier deflection of the 7 ft x 36 ft section, ignoring the opening, is AF7X36

1

- 10 in.

[ (Er ] +

3(%)

= 0.059

in.

The fixed pier deflection of pier 1 is F'

pier 1 -

therefore,

kph. 1 --

210 in.

' in.

[(ET

+ 3 ( 2 ) ] = 0.071 in.

=

14.0

in.-'

0.071

9-147


AC1

TITLEtMDG 93 W 0662949 0508844 L80 m

Example 93-11Cont'd. Calculations and Discussion

Code Reference I

The fixed pier deflection of pier 2 is ' F pkr 2

- 10 -Lin.[( 2.67 7 f tftT + 3 ( 2.67 ' * )ft] = 2 . 6 O h L

therefore,

Adding the stiffness of pier 1 and 2 results in a composite stiffness for this pier group of 14.03

+ 0.385

in.-1

in.-='14.4 in."

The deflection of this pier group is therefore, 'Fpier 1+2

--

1 14.4

in.-='0.069 in

The net deflection of the 10 x 36 section to the left of the 10 ft opening is

Al&+.nrOn

-- A c 1 0 x M -

A&# OectfON

= 0.085

A F 7 x 3 6 -b

A

P p k 1+2

in. - 0.059 in. + 0.069 in.

= 0.096

in.

therefore

-

1

kmwnlocl - 0.096 in.

= 10.4

in."

This left section stiffness, 10.4, is less than the solid right section stiffness,11.7, as expected. The composite Stiffness of the two 10 x 36 wall sections is kcaaiPlOw36 =

10.4 h-'+ 11.7 h-'= 22.1 h.-'

therefore producing a relative deflection of

9-148


A C 1 T I T L E r M D G 93 M 0662749 0508845 Olt7 D


Calculations

1

-

~co!mplOx36

=

0.045 in.

22.1 in."

The method discussed in MDG 9.2.2 defines the net stiffness of the perforated wall as the reciprocal of the solid wall cantilever deflection, minus the cantilever deflection of a solid wall with same heightas thelargest opening, plusthe f i e d end deflection of the piers in the opening strip. While this

net wall stiffness will not be used for proportioning the loads

within the wall in the TMS shopping Center, it would be used to obtain the load on the top of the wall if the building had a rigid diaphragm. To illustrate how this quantity would be determined, the required calculations are shown below.

-

'CO

'netpqforated 16 x 82 =

'c

An#parforcrrcd

0.062 in. - 0.037 in. + 0.045 in. = 0.070 in.

16 x 82 =

+

'CO,

10 x 36

therefore

This stiffness compares well to the solid wall stiffness of 16.3 in.". The loads are proportionedto each wall section by the ratio of their relative stiffness to the total stiffness of the section. Using MDG Eq. 9.2-1, the load carried by the left section is

9-149

j


93


m

Obb2949 0 5 0 6 8 4 b T 5 3

m

Example 9.3-11Cont'd. Calculations and Discussion

Code Reference

Similarly, the load carried by the right section is

v*.

3 =

13,300 lb x

[i:::

=

7,030 lb

The load in the left section is distributed to piers 1 and 2 in proportion to their relative stiffnesses to the total stiffness of the section. Therefore, the shearforce applied to the top of pier 1 is =

6,270 lb

x

14.4 in."

and the shear applied to the top of pier 2 is

Check 7,030 lb

+ 6,100 lb + 168 lb

= 13,300 lb

..

OK

It should be noted that pier 2 resists very little of the lateral load and could have been neglected in the analysis with very little effect, while greatly simplifylng the analysis. Further discussion and examples of shear wall load distribution are included in MDG 13.2.

9-150


A C 1 T I T L E j N D G 9 3 M 0662749 0508847 99T

Example 93-12 TMS ShoppingCenter

-

m

Distribution of Horizontal Load Within

Unreinforced Perforated Shear Walls Determine the horizontal load distributionto the piers in the unreinforced perforated shear wall on Grid Line 3 (East Wall) of the TMS Shopping Center for Wall ConstructionOption k

The analysis will considertheexistingcontroljointswhich

are requiredin

the

unreinforced concrete block masonry wall. The wall section under consideration 'is located on Grid Line 3, between Grid Lines A and

C. This wall has two door openings separated by a 2 ft - 8 in. pier of masonry. The figure below shows the configuration of this wall, the expansion joints and the openings. Control Joints 82 '

-

8

and

13,300 lb

Calculations Since the loads in this example are relatively small and the wall is unreinforced, it is again appropriate to use a linear elastic, uncracked section analysis for horizontal shear loads.

9-151


the distribution of the

A C 1 TITLE*flDG

93 9 Ob62949 0508848 8 2 b

m


CaIcuIations

Discussion

Reference

Code

As before, the largest shear load of 13,300 lb, applied near the top of the wall on grid line 3, is produced by the wind loading. It is assumed that the entire wall acts to resist the lateral loads even though the load is applied at the diaphragm location, 2 ft below the top. There are five sections ofwall separated by control joints. It is usually a good idea to assume that the control joints will not transfer in-plane shear so that each section of the wall acts as an independent pier, joined by the diaphragm. As shown in MDG Example 9.3-11, the small pier between the two doors will resist very little of the wall shear. Thus, the section with openings can be

ignored with little effect on the analysis. Therefore, four piers resist the in-plane load. The loadcan

be distributed toeach

pier in proportionto itsrelative

stiffness. The total

cantilever deflection of the wall can be calculated using MDG Eq. 9.2-3

Since all walls have equal thickness and only relative stiffness is required, assume pl - 1 in. "-

Emt

10

The deflection at the top of the 14 ft-8 in.wide piers, Aclox 'c

18 x 14-8

[(

-1 10 i n . 4 -

l:fir

+ 3(%)]

9-152


is

= 1.11 in.

93 W 0 b b 2 9 4 9 0508849 7 6 2 W



Calculations

This produces a stiffness, ki k18 x

=

=

1 A

0.90 in."

- 1.11 in.

Similarly the deflection of the two 17 ft-8 in. piers is 'c

18 x 17-8

-

I rif:yl

in. 4 (1 10

+

=

0.73 in.

Therefore -

k18 x 17-8

- 0.73 in.

=

1.37 in.-'

The total stiffness of the wall system is

ktotol = 2 x 1.37 in." + 2 x 0.90 in." The total load of 13,300 lb can be

=

4.55 in.-'

distributed to the piers by the ratio of pier stiffness to

total wall stiffness as summarized in the table below.

Piers

Pier Stiff., kPb

kpuAotai

1 2 5

0.90 0.90 1.37 1.37

0.20 0.20 0.30 O. 30

2,660 2,660 3,990 3,990

Total

4.5 5

1.00

13,300

4

9-153


Pier Shear (lb)


93

m

0 b b 2 9 4 9 0508850 484 D

10 MOVEMENTS

10.1 CAUSESANDCONSEQUENCES OF MOVEMENTS Wallmovements are caused by volumetricchange and support deflection. All building materials are subject to volumetric changeas a resultof internally generated strains, external forces, and material properties. Different materials have

different volume change

characteristics. Restraint of movements, either by adjacentdissimilarmaterials

or by

external supports, causes stress which can result in cracks. Cracks can be prevented by reducing volume change, and by accommodating differential movement between different materials and between different parts of the same material (10.1.1). By estimating the magnitudes of the different movementswhichmayoccurin masonry,itispossible

to rationallydesign

a system of movement joints and flexible

anchorage to eliminate or greatly reduce cracks and the problems they cause (10.1.2).

10.2DETERMINATION

OF STRUCTURAL MOVEMENTS

10.2.1ProbabilisticConcepts Because volume changes

are highly variable, precise deterministic prediction

of building

movements is not possible. However, they can be described statistically in terms of mean

10-1



93

m

Obb2949 050885L 310

m

values and standard deviations, which are measures of variability. This permits rational selection of designvalues

(characteristic values)with

a specifiedprobability

of being

exceeded or not(10.2.1). The Code commentary discussesmaterial properties andprovides the source of the Code material property values and indicates whether they are mean or maximum values.

10.2.2Short-TermMovementsDue

to External Forces

Short-term movements due toexternal forces depend on the masonry’s elastic modulus. See

MDG 3.4.2 and 3.4.7. Determination of structural movements based on theCode would use the assumed values of modulus in Code Table 5.5.1.2 and Code Table 5.5.1.3.

10.23 Long-TermMovement of Masonry Under sustained load, some materials exhibit creep (time dependent strain under load). The magnitude of creep in masonry and concrete depends on stress level, material age and strength when loaded, duration of stress, material quality, and exposure conditions. Creep in structural concrete isdiscussedin

the section onframe movement. Creep is often

described in terms of specific creep, i.e., additional strain per unit stress.

10.23.1

Creep of Brick Masonry

- About 80% of creep in brick masonry occurs

in the

mortar joints. Code 5.5.5.1 gives a specific creep value of kc = 0.7 x l t 7inJin. per psi which is reasonable for brick masonry walls. Generally, creep is not a major design concern with clay masonry (10.22).

10.23.2 Creep of Concrete Masonry - About 80% of creep in concrete masonry occurs in concrete masonryunits. Code 5.5.5.2 gives a specific creep value of kc = 2.5 x per psi,whichis

reasonable for concrete masonrywalls

lightweight aggregate.

10-2


in./in.

of both normal weight and

A C 1 T I T L E * N D G 93 W Obb2949 0508852 257 W

10.2.4ThermalMovement Unrestrained materials subjected to temperaturechanges undergo thermal strains which are the product of thetemperature

a coefficient of thermal expansion.

changetimes

Computation of these thermal movements therefore requires estimates of the temperature change and of the coefficient of thermal expansions.

10.2.4.1 Temperature Change masonrysubjected

in Exterior Walls

- The maximum surface temperature of

to solar radiation iswellabove

theairtemperature.

The upper

characteristic value in mean wall temperature rise is about 100' F, a common design value used in evaluating wall temperature movement. For more precise determination of such movements, temperature based on specific localities shouldbe used (10.2.3).

10.2.4.2 Coefficient of Thermal Expansion - For masonry and other materials the mean coefficient of thermal expansion, the standarddeviation, and the upper characteristic value are given in" D G Table 10.2.1. The coefficient of thermal expansion for clay masonry given in Code 5.5.2.1 is 4 x 10-6 inJin. deg F. There is about one chance in six that this value for clay brick masonry will be exceeded. The coefficient of thermal expansion for concrete masonry,givenin

Code 5.5.2.2 as 4.5 x N P inJin.deg F, may be compared with the

characteristicvalues for concrete masonry withdifferent types of aggregate asgiven inMDG Table 10.2.1.

10.2.5 Moisture Movements 10.2.5.1 Brick - Unlike mostmaterials, ceramics exhibit irreversible moisture expansion (net long-term growth). As indicated in

0.051%, orabout 3/16in.in

MDG Table 10.2.2, the upper characteristic value is

30 ft. The BTA recommends the value of 0.05% when

calculating maximum differential movement in brick walls in structural frames. There is about one chance in six that the value in Code 5.5.3, kc = 3 x 1 P in./in., will be exceeded.

10-3


AC1

TITLE*MDG 93 m Obb29Y9 0508853 193 m

A test method is in general use for measuring brick moisture expansion(10.2.4), but it has not been adopted by ASTM. 10.2.5.2

Mortar - Cementitious materials undergo time-dependent shrinkage due to drying

and carbonation. Such shrinkage depends on cement fineness, aggregate type, water content, mix proportions, presence of chemical admixture, member size and shape, and curing conditions. Drying shrinkage is partially reversible.

Carbonation shrinkage, which

is the result of the reaction of portland cement paste with carbon dioxide and moisture in the air, is irreversible. The ultimate (23-year) shrinkage of mortar in contact with masonry units is estimated at 1260 x 106 inJin. with a standard deviation of 600 x 106in./in. The 28day shrinkage is about half that value. Mortar shrinkage is increased by increase in water content, sand fineness, air-entrainment, and calcium chloride. 10.2.53 S,

Concrete Masonry

-

The total linear drying shrinkage of concrete masonry units,

(determined in accordance withASTM

C 426),varieswith

the units’ method of

manufacture and type of aggregate. The mean is about 330 x 1W in./in. with a standard deviation of about 140 x 106 inJin. Values for unrestrained shrinkage of concrete masonry are given in MDG Table 10.2.3. Horizontal joint reinforcement and friction at the wall foundation reduce shrinkage of the wall. Code 5.5.4 requires a design value for wall shrinkage of 0.15 S, for moisture controlled CMU and 0.5 S, for non-moisture controlled CMU. Moisture controlled units must be kept dry at the job site. 10.2.6

Freezing Expansion

The Code does not now address the freezing expansion of masonry. When water freezes, its volume increases about 9%. Three investigators (10.2.6, 10.2.7, 10.2.8) have

measured

residual expansion insaturated clay brick after a few freeze-thaw cycles. Measurements on 71 specimens gave a mean expansion of about 0.012% with a standard deviation of 0.01%. 10-4


AC T I1T L E * M D G

0 b b 2 9 4 9 0 5 0 8 8 5 4 02T

93

The characteristic valuewas 0.028% or about 3/32 in. in 30 ft. No published data are available for the freezing expansion of mortar or of concrete masonry units.

10.2.7Restraint

of Masonry

Horizontal contraction or expansion of masonry walls isgreater at the top than at the base, where it is restrained by bond or friction withsupports. Coefficients of friction betweenwall and base as suggested by the author are given in MDG Table 10.2.4. Masonry is typically anchored to a structuralframe movement.However,

by connectors which offer little restraintto

vertical

horizontal differential movement between the wall and frame is

restrained, first by bond and then by friction with the supporting structure. Coefficient of friction is influenced by support material, presence of mortar at interface, surface coatings on supports, type and location

of flashing(10.2.9,10.2.10).

Horizontal differential strain

between brick masonry and a concrete frame is virtually always sufficient to rupture the bond between masonry and concrete. Accordingly,frictionoffers further strain. Restraint istypicallycaused

byrigid

the onlyresistance to

connectors, by improperlyplaced,

designed, or built expansion joints, or by the complete absence of such joints.

103 STRUCTURAL MOVEMENT 103.1 FrameMovement 10.3.1.1ConcreteColumnShortening

-

The shortening of normal weight reinforced

concrete columns due to elastic deformation, shrinkage, and creep is estimated to have a mean of 0.1% with a standard deviation of 0.05%, i.e., about 1/8 in. in 10 ft.

103.1.2 Steel Column Shortening - The shortening of typical steel columns due to elastic strain at service load is estimated to be 0.06%with a standard deviation of 0.006%.

103.13

Sidesway - The Code does not address relative lateral deflection. The literature

10-5



93

m

Obb2949 0508855 Tbb W

suggests limiting this movement to one 1/10oO(0.1%) of the wall height not to exceed 0.15 in. (10.3.1).

The National Earthquake Hazards Reduction Program has set limits of story

drift from 0.01h to 0.0%. Sidesway is not a problem in shear wall buildings. However, in infilled frame structures, the frame movement must be examined.

103.2 Deflections of HorizontalMembers 103.2.1 Beam Deflection - When walls are built tightly under floors or beam soffits, vertical deflection of those flexural members will place a vertical compressive load on the wall. Underthe

rightcircumstances

that load may accumulate through successivefloors

to

produce very high compressive stress in walls not necessarily designed as bearing elements. Obviously, expansion joints are required between nonbearing walls and overlying flexural members. Code 5.6 limits deflectionof beams and lintels supporting non reinforced masonry

to 1/600 or 0.3 in. for dead loadplusliveload.This

Code provision does not apply to

supported reinforced masonry elements.

10.3.2.2 Shelf Angle Deflection

- The horizontal leg of a shelf angle supporting masonry on

a structural frame,will deflect due to bending overthe span between anchor bolts, torsional rotation, and bending of the cantilevered horizontal leg (10.3.2). Inadequate shimming of shelf angles permits additional deflection due to rotation of the angle. If these deflections are not accommodated, the toeof the horizontal legwill place a concentrated line load near the face of the underlying masonry below the angle, possibly causing spalling.

1 0 3 3 FoundationMovement Soils, like all materials, deform under load. While uniform, limited

foundation settlement

is not objectionable, excessive differential foundation movements can cause problems. Differential foundation movements are usually caused by volume changes inthe underlying soils. Foundations can also be lifted due to frost heave in frozen soils and swelling of clay.

10-6


A C 1T I T L E * P l D G

93

m

0 6 6 2 9 Y 9 0508856 7 T 2

m

103.4 Slab Movement

If a concrete slab is cool and/or dry on top and warm and/or moist on the bottom, the top contracts relative to the bottom, causing the slab to curl upward. Because

the diagonal

dimension in plan is longer than the sides, the comers tend to curl upward more than the sides. If a 4-in. thick, 15-ft square concrete slab has a strain gradient of 0.03% through its thickness, the unrestrained comers will rise about 5/8 in. If the slab is bolted tightly to a masonry loadbearing wall, the uplift may cause the masonry comers to crack. 103.5 DifferentialStructuralMovement 103.5.1

NonloadbearingWalls

- For masonry

supported on a structural frame the net

masonry expansive strain is added to the net frame contraction to give the total differential strain. The Code does not specifically address the net differential strain between masonry walls and frame systems, althoughit does provide movement coefficientsfor design purposes.

MDG Table 10.3.1 tabulates estimated differential verticalstrain between masonry wallsand steel or concrete structural frames (10.3.2). 103.5.2 Loadbearing Walls - When brick masonry and concrete masonry wythes are used

in a composite loadbearing wall with a filled collar joint, expansion of brick masonry and contraction of concrete masonry shifts mostof the load to thebrick masonry (10.3.3,10.3.4). Shear stress in the collar joint typically exceeds 80% of the ultimate shear strength, and vertical steel reinforcement is required to control tensile stress (10.3.4,10.3.5, 10.3.6,10.3.7). 103.53 LoadbearingMonloadbearing Wall Intersection

- Masonry loadbearing walls are

subject to elastic deformation, temperature, shrinkage, and creep. Nonloadbearing walls of concrete masonry are subject only to shrinkage and temperature deformation. The vertical joint between a loadbearing and a nonloadbearing wall is, therefore, subject to considerable shear which should be considered in light of Code 6.5.2.

10-7



m

Obb29Y9 0500057 839

m

If lateral support of a wall is not required at an intersection, a control joint may be located at the intersection. Otherwise, the walls must be bonded in accordance with Code 5.13.4.2.

10.4ACCOMMODATION 10.4.1Design

OF MOVEMENTS

of MovementJoints

Three types of movement joints are used for crack control in masonry: 1) control joints which open to accommodate shrinkage of concrete masonry; 2) expansionjoints which close to accommodate expansion of brick or stone masonry; and 3) construction joints to seal the crack between masonry and other materials, such as beams, columns, windows, and doors.

10.4.1.1

- Joint

Sealants UsedinMovementJoints

strain capacities ranging from 25% to 50%. Use

sealants typicallyhavecompressive

of sealants with greater strain capacity

permits the use of thinner joints for the same anticipated joint movement. Sealants are available in several colors and finish to match the appearanceof mortar joints. Sealant joints in walls designedfor fire resistance should be designed as fire stops. The water permeance integrity of a facade often depends heavilyonsuch

sealant, whichhave

a mean life

expectancy of seven years.

10.4.1.2 Control

Joints - The maximum horizontal spacing between vertical control joints

in concrete masonry walls is determined by: 1) the local average annual relative humidity; 2) the type of CMU (ASTM C 90, moisture controlled or non-moisture controlled); 3) the vertical spacing of bed joint reinforcement; and 4) exposure conditions. MDG Table 10.4.1 gives suggested spacingsfor control joints based on thosefour criteria (10.4.1). Additionally, control joints in concrete masonry should be placed at the following positions:

l. At all abrupt changes in wall height 2. At all changes in wall thickness, such

as those at pipe and duct chases and those

adjacent to columns or pilasters

3. Above joints in foundations and floors 10-8


A C 1 TITLE*UDG

93

O662949 0508858 775

4. Below joints in roofs and floors that bear in the wall

5. At a distance not over one-halfthe allowable joint spacing frombonded intersections or corners

6. At the end of a lintel on one side of wall openings six feet or less in width and at both sides of wall openings more than six feet wide unless bond beams or equivalent joint reinforcement is placed at top ofwindows and doors and at the bottom

of

windows. One lintel bearing should be built to slide on a slip plane. Control joints for concrete masonry as shown in Fig. 10.4-1 are typically 3/8 in. wide, and should be spaced at horizontal intervals as givenin MDG Table 10.4.1. Half-inch wide control joints separate CMU walls and concrete columns to accommodate sidesway of the frame (See Fig. 10.4-2). Special Control Joint CMU7

Building Paper Backer

Rod And Sealant G r o u t or Mortar

Fig. 10.4-1 Control Joints for Concrete Masonry

The bed joint reinforcement referredto in MDG Table 10.4.1 normally consists of two No. 9 colddrawn steel wires, one in each face shell bed. Bed

joint reinforcement may be

replaced by bond beams reinforced with two No. 4 continuous reinforcingrods in 8 in. wide bond beams and two No. 5 bars in 10 in. or 12 in. wide bond beams. Bond beams, if used, should be spaced at most four times the required vertical spacing of joint reinforcement.

10-9


A C 1 TITLErMDG 93

m Obb2949

1

Expansion Joint Wire Anchor

m

0508859 601

,-

p i r e Anchor

7

Masonry

2" Min.

MU Cell

routed t Anchor

v

1/2" Control Joint

Dove Tail Slot

Fig. 10.4-2 Flexible Wall Anchorage to Concrete Columns When bond beams are used, they should be placed at the top and the base of walls, and below windows. Fig. 10.4-3 provides approximate average annual relative humidities in the United States, but local weather records will provide better data.

AMRH > 75% O 7558 > AMRH > 50% II AMRH c 50%

Fig. 10.4-3 AnnualMeanRelativeHumidity

10-10


(AMRH), 96

A C 1T I T L E v M D G

9 3 W Obb291)9 05088b0 3 2 3 W

10.4.13 Expansion Joints - Vertical expansionjoints are not required in concrete masonry

walls, because initial drying shrinkage usually exceeds thermal expansion. Clay brick masonry walls expand both horizontally and vertically. Allowable movement in an expansion joint in clay brick isthe product of joint thickness and allowable sealant strain. For a 3/4 to 1 in. wide expansion joint with a sealant extensibility of 50%, the allowable joint movement is 3/8 in. to 1/2 in. Allowable joint movement divided by anticipated masonry strain equals maximum joint spacing. Vertical expansion joints should be placed in brick masonry at the following locations:

1. At approximately 25 ft intervals in enclosure walls 2. At approximately 12 ft intervals in parapet and fences 3. At offsets, junctions, and

corners such as those at pipe and duct chases and those

adjacent to columns or pilasters

4. Above joints in foundations and floors 5. Below joints in roofs and floors that bear in the wall

6. At a distance not over one-halfthe allowable joint spacing frombonded intersections or corners.

In addition, a horizontal expansion (softjoint) joint should be placed at the top of masonry walls infilled in a structural frame between the spandrel beam soffit and the wall. When masonry walls are support on shelf angles, a horizontal expansion joint should be placed immediately below the shelf angle. The thickness of horizontal expansion joints may be determined by the anticipated deflection of the spandrel beam and

shelf angle and the

characteristic value for differential movement given in MDG Table 10.3.1. See Figs. 10.4-4 and 10.4-5.

10-11


A C 1 TITLErMDG 9 3

m

Ob62949 05OBBbL 2bT

m

Fig. 10.4-4 Horizontal ExpansionJoint

Concrete Spandrel Beam

Clip Angle

Movement Joint Masonry Wall

m

Fig. 10.4-5 Expansion Joint Between Interior Partition and Beam Soffit The required movement in an expansion joint under a shelfangleis

the s u m of the

differential movement between wall and frame, the long-term spandrel deflection, and the shelf angle deflection. Design

of sealant joints under shelf angles is discussed in MDG

10.3.2. 10-12


AC1

TITLExMDG 9 3 M Obb2949 05088b2 1Tb

10.4.1.4ConstructionJoints

- Differentialmovement

m

between masonry and adjacent

materials is accommodatedby construction joints. Where masonry is anchored to structural frames, flexible anchorage should be provided to accommodate wall expansion, beam deflection, column sidesway, and column contraction.

10.4.1.4.1 WalVColumn Joints - Masonry walls infilledin structural frames may be designed as shearwalls to provide lateral support forthe structure, eliminating the need for moment resistant frame connections or frame bracing. To avoid loading masonry infillnot designed for shear,construction joints provide clearance between wall and frame asillustrated in Figs.

10.4-2 and 10.4-6. Because 2 inches is the narrowest cavity width a mason can be expected to keep clean of mortar droppings, masonry should haveat least a 2 in. clearance from the column face.

Movement Joint

Steel Column

L #2 Bar Welded To Column

Fig. 10.4-6 Flexible Wall Anchorage

To Steel Column

10.4.1.4.2 Wall/Floor Joints - Roofs and walls expand and contract, frequently in opposite directions. To avoidmasonrycracks,

break the bond between roofs and walls. If roof

anchorage is essential, provide flexible anchorage. The detail shown in Fig. 10.4-7 provides such flexible anchorage.

10-13


Notched Slab

Roof Slab

Sealant Joint

Fig. 10.4-7 Wall/Roof Anchor

REF'ERENCES

10.1.1

Grimm, C. T., "Masonry Cracks:A Review of the Literature," Masonry: Materials,

% STP 992, American Societyfor Testing and Materials, Philadelphia, PA, 1988, pp. 257-280.

10.1.2

Grimm, C T., "Water Permeance of Masonry Walls: A Review of the Literature," Masonry: Materials. Properties. and Performance, STP 778, American Societyfor Testing and Materials, Philadelphia, PA, 1982, pp. 178-199.

10.2.1

Grimm, C T., 'Statistical Primer for Brick Masonry," Masonrv. Materials. Desirm,

STP 992, AmericanSociety for Testing and Materials, Philadelphia, PA, 1988, pp. 169-192.

10.2.2

Lenczner, D., "Design of Brick Masonry for Elastic and Creep Movements," 2nd Canadian Masonry Symposium,Carlton University, Ottawa, Ontario, Canada, June 9, 1980.

10.2.3

Grimm, C. T., ''Thermal Strain in Brick Masonry," 2nd North American Masonry

Conference. University of Maryland, College Park, MD, August 1982, p. 34. 10-14


A C 1 T I T L E * H D G 93 D 0662747 O508864 T 7 9 W

10.2.4

Baker, L. R. and E. L Jessop,"MoistureMovements

in ClayBrickwork

-

a

Review," International Journal of Masonrv Construction, British Masonry Society, Stoke-on-Trent, England, Vol. 2, No. 3, 1982.

10.2.5

Russell, WilliamA. Shrinkage Characteristicsof Concrete Masonry Walls, Housing and Home Finance Agency, Housing Research Paper No. 34, Supt. of Doc., U.S. Govt. Printing Office, Washington D.C., April 1954, p. 23.

10.26

"Bricks,"ReDortof

the Tests of Metals and Other Materials for Industrial

Purnoses, Watertown Arsenal, Watertown, MA, June 30, 1896, pp. 344-372.

10.2.7

Davidson, J. I., "LinearExpansion Due to Freezing and Other Properties of Bricks,"2nd

Canadian MasonrvSymposium,

Carlton Univ., Ottawa, Ontario,

Canada, June, 1980, pp. 13 -24.

10.2.8

Palmer, L A., "Volume Changesin Brick Masonry Materials,"Journal of Research, National Bureau of Standards, Washington, D. C., June 1931, pp. 1003 - 1026.

10.2.9

McGinley, W.M., and J. G. Borchelt, "Friction at Supports of Clay Brick Walls," Proceedings of 5th North American Masonrv Conference, University of Illinois, Urbana-Champaign, June 3-6, 1990, pp. 1053-1066.

10.2.10 McGinley, W.M., and J. G. Borchelt, "Friction Between Brick and Its Supports," Proceedings of 5th Canadian Masonrv SvmDosium, University of British Columbia, Vancouver, B.C., June 1989, pp. 713-722.

10.3.1

"Structural Serviceability: A Critical Appraisal and Research Needs," Journal of Structural Engineering, American Societyof Civil Engineers, New York, NY, Vol.

112, No. 12, December 1986, p. 2649. 10.3.2

Grimm, C. T. and J. A. Yura, "ShelfAngles for Masonry Veneer," Journal of Structural Engineering, American Societyof Civil Engineers, New York, NY, Vol.

115, No. 3, March 1989, pp. 509-525. 10.3.3

Yi, F. L. and R. L Carrisquillo: "A Studyof the ThermalBehavior of Brick Under ServiceConditions,"Masonrv.Research.ADDlications.

and Problems, STP 871,

American Society for Testing and Materials, Philadelphia, PA, 1985, pp. 101-119.

10.3.4

Anand, S. C. and M. A. Rahman, "Temperature and CreepStresses in Composite Masonry Walls," Advances in Analvsis of Structural Masonry, American Societyof

10-15


AC1

TITLE*HDG 93 m Ob629490508865905

m

Civil Engineers, New York, N Y , 1986, pp. 94.

10.3.5

McCarthy, J. A., R. H. Brown, and Cousins, T. E., "An Experimental Study of the Shear Strength of Collar Joints in Grouted Slushed Composite Masonry Walls," Proceedings of the Third North American Masonry Conference, University of Texas, Arlington, TX, June 1985, pp. 39-1 through 39-16.

10.3.6

Williams, R., and L Geschwinder, "ShearStress Across Collar Joints in Composite Masonry," presented at Proceedings.Second

Canadian MasonrySymposium,

Carleton University, Ottawa, Canada, June 1980.

10.3.7

Colville, J., S. k Matty, and k M. Wolde-Tinsae, "Shear Capacity of Mortared Collar Joints," Proceedings of the Fourth North American Masonry Conference, University of California, Los Angeles, CA, Vol. 2, August 1987, pp. 60-1 through

60-15. 10.4.1

Masonry Structural Desipn for Buildings, Army TM 5-809-3, Departments of the Army,Navy,

and Air Force, August, 1982, pages 3-1 ff, National Technical

Information Services 5285 Port Royal Road, Springfield VA 22161.

10-16



Table 10.2.1

93

m

0662949 0508866 441 W

EstimatedCoefficients of ThermalExpansion inJin./ x lod deg F Mean

Standard Deviation

Characteristic Value (a)

1 Brick

3.2

0.8

4. S

2

Mortar

5.4

1.1

7.2

3

Brick Masonry

3.9

1.0

5.6

Material

Ja

Concrete 4

Quartz

6.6

O. 6

7.6

5

Sandstone

6.5

1.2

as

6

Gravel

6.O

0.6

7.0

l

Granite

5.3

1.4

7.6

a

Basalt

4.8

0.5

5.6

9

Limestone

3.8

1.5

6.3

Concrete Masonry 10

Dense Aggregate

5.2

O. 3

5.7

11

Cinders

3.1

O. 6

4.0

12

Expanded Shale

4.3

O. 6

5.2

13

Ewpanded Slag

4.6

1.1

6.3

14

Pumice

4.1

0.3

4.5

Stone 15

Sandstone

5.4

1.0

7.0

16

Marble

4.2

0.8

5.5

17

Granite

4.4

1.2

6.4

18

Basalt

3.1

O. 9

4.6

14

Limestone

4.2

1.5

6.7

u:

Slate

5.0

1.1

6.8

Steel 21

Mild

6.5

"_

2;

Stainless

9.9

"_

"_ "_

21

O. 3

26

2

Wood

Plaster

2r

Perlite

4.3 - 6.1

ND

ND

2!

Vermiculite

5.5

- 6.2

ND

ND

24

Gypsum

6.5

- 8.6

ND

ND

(a) 95% probability of not being

exceeded

ND - No data available Source:

10-17


C T. Grimm

A C 1 TITLE+MDG 93 W 0662949 0508867 788 W

Table10.2.2EstimatedMoistureMovement,

%

Reversible Moisture Movement Mean

1

Clay Brick

Characteristic Standard Deviation

Irreversible Moisture Movement Standard Characteristic Mean Deviation

Value

Value

0.0200

ND

ND

+ 0.020

0.W

+ 0.051

Concrete 2

Gravel

0.04oO

0.01u)

0.060

- 0.060

0.015

- 0.085

3

Stone Crushed

0.0700

0.0210

0.105

- 0.060

0.015

- 0.085

4

Limestone

0.0250

0.0030

0.030

- 0.035

0.003

- 0.040

5

Light Weight

0.0450

O. o090

0.060

- 0.060

0.018

- 0.090

Concrete Masonry 6

Sand & Gravel

O.Oo80

0.0010

0.010

- 0.023

0.003

7

Expanded Shale

0.0100

0.0030

0.015

- 0.031

0.009

- 0.028 - 0.046

O.Oo90

0.0010

0.011

- 0.027

0.004

- 0.034

Concrete Masonry Units 8

Sand & Gravel

9

Light Weight

10

Sand Lime

Stone

13

Granite

14

Limestone

15

Marble

16

Sandstone

ND - No dataavailable

Source: C T. Grimm

Table 10.23 EstimatedUnrestrainedShrinkage of Low-PressureSteam-CuredConcrete Masonry (10.2.5)

III

Cinders

425

Expanded Slag

340 I

hoanded Shale

I 10-18


310

I


m

Ob62949 0508868 b L 4

Table 10.2.4 Coefficient of Friction(Dimensionless)*

Source: C. T. Grimm

10-19


m


93

m 0662947

0508869 550

m

Table 103.1 Estimated Differential Vertical Strain Between Masonry Walls and Structural

Frames, 10" i n . / h a Upper Standard No. Mean Wall Material Deviation

Characteristicb Value

Concrete Frame

I'

Brick

1,150

1,750

2

Dense Aggregate 908 CMUc

3

Lt. Wt. Aggregate CMU

4

Granite, limestone, or marble 1,100

5

Sandstone

741

1,850

I 532

2,070

513

I I

1,940

623

I

2,880

517

1,590

Steel Frame

6

Brick

7

Dense Aggregate CMU

8

Aggregate Lt.Wt.

CMU

290 565 182

9

Granite, limestone, or marble

10

Sandstone

1,880

123

424

488

171

1,230

392

aExclusive of beam and shelf angle deflection b5% probability of being exceeded

'Low pressure steam cured Source: (10.3.2)

10-20


167

769

Masonry Walls, Ft Average Annual Relative Humidtity

Wall Location

Less Than 50%

Exterior

Vertical Spacing of Bed Joint Reinforcement, inches

Interior

Between 50 and 75%

Exterior

Interior

Greater Than 75 %

Exterior

II Non-moisture Moisture Controlled Controlled

12 6 18 24

None 16 8

16.5 24 31.6 19

9 14

18 24 30 20

12 16

* None

16

22.5 30 37.6 24 18 30 36 28.5 36 26 43.6

Interior

Source: (10.4.1)

10-21


I

None 16 8

None 16 8

I

Type of CMU, ASTM C 90

10 14

15 20 25 22 26 21 31


93 W 0662949 0508873 307

m

-

Example 10.4-1 T M S Shopping Center Vertical Control Joint Location Determine the control joint locations in the TMS Shopping Center for theconcrete masonry exterior and interior walls for Wall Construction Option A (unreinforced).

and

Calculations

Discussion

Reference

Code

This example illustrates the application of the recommended material properties found in Code 5.5 in comparison to the author's recommendations found in MDG Table 10.4.1, 'Maximum Horizontal Spacing of Vertical Control Joints in Concrete Masonry Walls", for determination of location of vertical control joints. For the TMS Shopping Center the exterior and interior walls are constructed of concrete masonry with assumedC 90 Type I units. The building is assumed to be in a region having, an average annual relative humidity between 50% and 75%. Bed joint reinforcement is assumed to be placed at 16 in. vertically. Code Considerations For Concrete Masonry

k, = 4.5 x 10" in./in. per deg F Assume AT = 100 deg F for exterior wall R,,,

= 0.15

sl

5.5.2.2

5.5.4.1

Assume total linear drying shrinkage of CMU is 0.00065 inJin. Temperature strain deformation 100 x 4.5 x 10" = 0.00045 inJin. Shrinkage strain deformation 0.15 x 0.00065 = 0.0000975 in./in. The Code does not provide means for direct consideration of relative humidity and reinforcement. 10-22


AC1

TITLExMDG 93

Obb29Y9 0508872 045



Code Reference

Total Strain = 0.000548 inJin. Assume 3/8 in. joint & sealant extensibility of 50%

Joint spacing

=

3/16 = 342 in. = 28.5 ft 0.000548

Assuming AT = 40 deg F for interior wall, joint spacing

=

55 ft.

Author’s Recommendation

MDG Table 10.4.1 indicates a maximum horizontal spacing between vertical control joints of 24 ft for exterior walls and 30 ft for interior walls. The author considers a deviation of 10% as acceptable to meet practical project conditions.

The actual control joints for theexterior walls are indicated in the elevation drawings in Fig. 9.1-2. The control joint locations in the North, East and West exterior wall satisfy the computed (28.5 ft) andsuggested (24ft) maximum spacing. For the Northexterior walls the 26 ft control joint spacing is within the author’s acceptable 10% deviation range. The interiorwall control joint locations are indicated in theT M S plan drawings in Fig. 9.1-1. The actualcontrol joint spacing of 26 ft-8 in.and 27 ft-4 in.are less than the computed (55 ft) and suggested (30 ft) maximum spacing.

10-23


A C 1T I T L E s M D G

93

m

0bb2949 0508873 T 8 1

m

-

Example 10.4-2 RCJ Hotel Vertical Expansion Joint Size and Spacing Design Determine theexpansion joint locations inthe RCJ Hotel forOption A (unreinforced) using Building Construction Option II (short dimensions). The exterior wall is a brickblock noncomposite (cavity) wall with the exterior brick wythe being non-loadbearing.

and

Calculations The RCJ Hotel plan dimensions are shown in Figs. 9.1-6 and 9.1-7. The locations of the brick exterior wythe expansion joints are determined as follows: Clay brick masonry exterior wythe: Thermal expansion:Assume

at 40 deg F and

wallbuilt

maximumservice

mean temperature of

140 deg F. Thus the temperature rise is 100 deg F. From the Code the coefficient of thermal expansion, k,, is 4 x 10" inJin.deg F.

5.5.2.1

Therefore, thermal expansion

strain is:

4 x 106 x 100 =

0.0004in./in. 0.0003 inJin.

Moisture expansion (kc):

in./in.

Subtotalmasonry clay expansion: 0.0007 Sealant joint: Assumed width, 3/8 in.

0.375 in.

Sealant compressibility

50%

Therefore, maximum joint movement: 0.5 x 0.375

0.188 in.

10-24


5.5.3

AC1

TITLExMDG 93 m 0662949 0508874 918 m

Example 10.4-2 Cont'd.

and

Calculations

Maximum joint spacing:

It is reasonable to exclude characteristic freezing expansion because it is unlikely that all three expansions will be simultaneously at their high characteristic values. Examining the RCJ Hotel floor plans in Fig. 9.1-6, and the exterior wall elevations in Fig. 9.1-11, the exterior brick wythe expansionjoint spacing of 14 ft - 8 in. is lessthan the 22 ft-4

in. maximum spacing.

10-25

i COPYRIGHT ACI International (American Concrete Institute) Licensed by Information Handling Services

l

AC1

TITLEaMDG 93 m Obb2949 0508875 854 m

-

Example 10.4-3 RCJ Hotel Differential Movement in BriclJBlock Exterior Wall The unreinforced exterior wall on Grid Line B between Grids 1 and 2 is composed of a noncomposite exterior brick wythe and an interior block wythe

(cavity wall), for Wall

Construction Option I. The block wythe is loadbearing. The exteriorbrick wythe is subject to thermal, moisture, and freezing expansions. The interiorblock wythe is subject to elastic deformation, shrinkage, and creep. To accommodate these differential movements an expansion joint should be provided at the top of the exterior brick wythe. This example estimates the magnitude of those differential movements and designs the necessary expansion joint.

and

Code Reference

Calculations

Clay Brick Masonry Exterior Wythe: wall built at 40 deg F and

Thermal expansion: Assume

maximum service mean temperature of 140 deg F. The temperature rise is 100 deg F

5.5.2.1

The coefficient of thermal expansion, k,,is

4 x 10" in./in.-deg F. Therefore, thermal expansion strain is: 4.x10" x loo =

0.0004 inJin.

Moisture expansion (kc):

0.0003 inJin.

Subtotal clay masonry expansion:

0.0007 inJin.

Total clay masonry height Total clay masonry expansion

5.5.3

484 in.

=

0.339 in.

=

Freezing expansion is excluded because it is not covered

by the Code and because it is

unlikely that all three types of expansion will be simultaneously at their high value

10-26


m


Obb2949 0508876790

m


Calculations

Discussion

Reference

Code

Concrete Masonry Interior Wythe: CMU elastic deformation: From MDG Example 15.4-9 the compressive strength of the CMU is 4,500 psi,and the CMU wythe is 10 in. nominal, fully grouted. Mortar is ASTM C 270 Type S. Elastic modulus of the concrete masonryis 3.05 x lo6 psi FromMDG

Appendix A:

Table 5.5.1.3

Area = 115.5 in.2

Story Height Elastic Deformation (ED):

Wall Load Dead Story MDG 9.13.1

Wt. inJin.

75

PIA Stress, psi

story

EDlstory

25

130

0.0033

Unit ED, in. P/!, lo6,

Height in.

1st 340

8,376

2nd

6,060

200

54

180.0021

116

3rd

3,742

130

34

11

116

0.0013

4th

1,425

60

13

4

116

0.0005 0.0072

Total wall elastic deformation

CMU Shrinkage CMU are assumed to be Type II (Non-moisture controlled). Totallinear 0.00065inJin.

dryingshrinkage

of CMU isassumed

to be

The coefficient of shrinkage of non-moisture

controlled concrete masonry is 0.5 S,, in./in. Shrinkage strain deformation: 10-27


5.5.4.2

AC1 TITLE*MDG 93 W 0662949 0508877 627 W


Reference

0.5 x 0.00065 =

0.000325

Wall height =

468 in.

Total shrinkage

Code

inJin.

0.15 in.

=

CMU Creep: Creep strain in concrete masonry is:

lo-' inJin. - psi

2.5 x

5.5.5.2

Creep deformation (CD):

story

Stress, psi

CD/story in.

inJin.

Story Height in.

Unit CD,

10'

1st

75

25

130

0.0033

2nd

54

18

116

0.0021

3rd

34

11

116

0.0013

4th

13

4

116

-.""""""0.0005

Subtotal Creep

0.0072

0.0072

Subtotal concrete masonrycontraction: Total differential movement = 0.339

+ 0.16 =

+ 0.15 + 0.0072 = 0.16 in.

0.50 in.

50%

Expansion joint sealant elasticity = Total expansion joint thickness: 0.50/0.5

1.00 in.

A top of wall detail that accommodates this calculated differential movement is shown on the next page. 10-28


AC1 TITLEjNDG

93

0662949 0508878 563


Calculations

10-29


m

A C 1 TITLE*MDG

93

0662949 0508879 4 T T

11 FLEXURE

11.0 INTRODUCTION 11.0.1 Organization of Chapter 11

The subject of Chapter 11is flexure. The chapter is divided into sections basedon the type of structural element to be designed and whether the element is unreinforced or reinforced. Walls are treated first, followed by pilasters, lintels and beams. The introduction describes the general basis for the analysis, and developsthe flexural working stress design equations. 11.0.2 Flexural MasonryDesign

Unreinforcedmasonrydesign

is based on allowingtensionin

masonry design is based on design neglecting tensile strength

the masonry.Reinforced of masonry and relying on

reinforcement to resist tension. Assumptions common to both are as follows: 1.

Planesections before bendingremainplane

2.

Masonrycomponents(units,mortar,etc.)combine member.

3.

Stress is proportional to strain.

11-1


after bending. to formahomogeneous

AC1

TITLE*MDG 93 W Ob62949 0508880 111

11.03 Direction of Flexure It is useful to classify flexure as either in-plane flexure (shear walls) or out-of-plane flexure (flexure normal to the plane of the wall). In-plane flexure is a result of lateral forces on the lateral load resisting system (shear walls). This flexure is often referred to as overturning moment. Out-of-plane flexure is a result of forces applied perpendicular to the surface of the wall, such as wind loading or seismic inertia loading.

In-Plane

Out-of -Plane

Fig. 11.0-1 WallLoading For in-plane flexure, the width used in design is the thickness of the wall, and the depth is approximately the length of the wall. For out-of-plane flexure, the design width is the length of the wall, and the design depth is all or part of the wall thickness.

11-2


A C 1 TITLEaMDG 93

m

Ob62949 0508883 058

m

11.0.4 Effects of BondingPattern For unreinforced masonry design (design allowing tension in

the masonry), the tension

allowable stresses per Code 6.3.1.1 vary with the direction of the tensile stress. Allowable tensile stresses are higherwhentensionisnormal masonry.Thisincreaseis

to the head joints inrunningbond

due to the interlocking of the masonryunits,whichprovides

tensile strength comparable to the unit tensile strength, rather than the lower tensile bond strength of the mortarunit interface. For stack bonded masonry the Code allowable tension across the head joints is zero. Other than theseexceptions,masonryisassumedisotropic.

Thus the design and the

allowable stresses are assumed independent of the unit orientation and pattern.

11.0.5Flexure:WorkingStressDesign Fig.11.0-2

presentscommonlyusedsymbols

for flexuralanalysis.Masonrywidthis

designated by the letter b. The depth to the centroid of the reinforcing steel is designated by the letter d. Location of the neutral axis from the compression face is designated

kd,

where k is the ratio of the distance from the extreme compression fiber to the neutral axis location divided by the depth d.

Fig. 11.0-2 Flexure-Tensile Reinforcement Only 11-3



m

Ob62949 O508882 T 9 4

m

The first assumption from MDG 11.0.2, results in the following equation:

-- Y

E,

Eq. 11.0-1

R

where G is the strain in the masonry at a distance y from the neutral axis and R is the radius of curvature of the flexural element. The relationship between the curvature and the applied moment can be expressed by the following:

"1 - M R

Eq. 11.0-2

E,I

Where E,,, is the modulus of elasticity of the masonry and I is the second moment of the bending area or moment of inertia. Substitution of Eq. 11.0-2 into Eq. 11.0-1 results in the following: Eq. 11.0-3

Using the assumption, that stress is proportional to strain, or:

The stress in the masonry may be determined as: fb =

11.0.5.1

MY 7 UnreinforcedMasonry

Eq. 11.0-5

-

For unreinforcedmasonry,

W.11.0-5isused

to

determine the allowable moment as limited by the allowable tensile stress in the masonry. Usually the allowable tensile stress controls. However, when axial load acts simultaneously with flexure, the allowable compression stress may control as discussed in MDG 12.2. 11.0.5.2

ReinforcedMasonry

- The commonbasicassumptions

of the working stress

method for flexural designof unreinforced and reinforced masonry werepresented in MDG 11.0.2. Additional assumptions pertaining only to reinforced masonry include:

1.

Modulus of elasticity of the reinforcement remains constant throughout the 11-4


A C 1 TITLE*llDG

93

0 b b 2 9 4 9 0508883 920

working load range. 2.

Tensileforces

are resisted onlyby

the tensilereinforcement.

strength of the masonryunits,mortar,

The tensile

and grout isneglectedinflexure

analysis and design.

3.

Reinforcement is completely surrounded by and bonded to masonry material, and full composite action between the two materials is assumed.

For reinforced masonry design tension is resisted by reinforcement. The tensile strength of masonry is neglected. Thus, the masonry is assumed cracked from the tension edge of the masonry to the neutral axis. The neutral axis is located a distance kd from the extreme compression fiber.

The depth of the section "d" is now defined as the distance from the

extreme compression fiber to the centroid of the tensile force in the reinforcement. Using assumption 1 in MDG 11.0.2 and thesimilartrianglesin

Fig.11.0-2, the strain in the

masonry and reinforcement are related as follows: E s . 11.0-6

where

E,,,

is the maximummasonrycompressive

strain and

is the steel tensionstrain.

When the applied external loading is only bending, the compression force in the masonry equals the tension force in the reinforcement. Summing forces on the section illustrated in Fig. 11.0-2 results in the following equation:

Eq. 11.0-7 where

fs

is the stress in the reinforcement and A, is the reinforcement steel area.

Substitution of Eq. 11.0-4 into Eq.ll.O-7 results in: Eq. 11.0-8

Defining two new terms(the first is the reinforcement ratio, P, and thesecond is the modular ratio, n ) as follows:

11-5


A C 1 TITLE*flDG

93

m

Obb2949 0508884 8 6 7

m

Eq. 11.0-9 and substituting into Eq. 11.0-8, results in the following: Eq. 11.0-10 and using Eq. 11.0-6 results in:

k - 2nP " (1-R) k

Eq. 11.0-11

which can be rearranged to:

k2

+ 2npk

- 2np

=

Eq. 11.0-12

O

solving the quadratic equation results in the following expressionfor k.

k

=

d

m -np

Eq. 11.0-13

Summing moments about the centroid of the tensile force results in an expression of the allowable applied moment as limited by the allowable bending compression stress,Fb, in the masonry. Eq. 11.0-14

(

9.

Wherej= 1--

Summing moments about the centroid of the compression force results in an expression for the allowable applied moment as limited by the allowable tensilestress in the reinforcement,

F,.

Mt = ASjdFs Equations 11.0-9,11.0-13,11.0-14,

Eq. 11.0-15 and 11.0-15 provide the basis for flexuraldesign of

11-6


A C 1 TITLE*MDG 9 3

rectangular sectionswithtension

m

0662949 0 5 0 8 8 8 57 T 3

m

reinforcement only. Additional equations for pilasters

where the section may not be rectangular are developed in MDG 11.2. Flexural elements may contain both compression steel andtension steel (doubly reinforced).

See Fig. 11.0-3. For the compression steel tobe considered for load carjing purposes, the compression reinforcement must meet all the applicable requirements of Code 5.9.1.6. The appropriate equations for doubly reinforced flexural elements are developed as for singly reinforced flexuralelements. Summing forces alongthe length of the beam as shown in Fig. 11.0-3, results in:

ES. 11.0-16 where A', and

$'f

are the area of compression reinforcement and the stress in the

compression reinforcement, and f; is the stress in the masonry at the location of the compression reinforcement. Notice that the compression reinforcement has replaixd masonry and a corresponding force has been subtracted. Substitution of the stress-strain equations of Eq. 11.0-4 results in: Eq. 11.0-17 The centroid of the compression reinforcement is located a distance d' from the extreme compression fiber as shown in Fig. 11.0-3.

Fig. 11.0-3 Flexure - Compression And Tension Reinforcement

11-7


A C 1 TITLE*MDG 9 3 W O b b 2 9 4 9 0 5 0 8 8 8 b b 3 T

m

By similar triangles: Eq. 11.0-18 Where

E'# = E',,,

is the strain atthe centroid of the compression reinforcement. Substituting

Eq. 11.0-18 into Eq. 11.0-17 results in the following: Eq. 11.0-19

Substitution of Eq. 11.0-9 and defining the compression steel ratioas p' = AIs/bd results in: E, - 2np 2np' ""ES k Æ

(Æ - d'/d) (1

-

- k)

2p' (Æ - d'/d) k (1 - k)

Eq. 11.0-20

This equation can be rearranged to the quadratic form: Æ* + [2np+2(n-l)p']Æ

- [2np+2(n -

l)p'dy4 =

o

Eq. 11.0-21

From which the following expression for k is obtained:

k

=

/ [ n p +(n - 1)p'l2 +[2np+2(n - l)p'd'/d) - [np + ( n - l ) p q

Eq. 11.0-22

As with the case of tension reinforcement only, once the location of the neutral axis is known, expressions for moment as a function of allowable stress can be determined. By summing moments about the centroid of the tension force an expression for moment as a function of masonry compression allowable can be found.By summing moments about the centroid of the compression force an expression for moment as a function of tension reinforcement allowable canbe found. Unfortunately, the centroid of the compression force is not as obvious in a doubly reinforced beam as in the previous singly reinforced beam derivation. Selecting the extreme compression f i k r as a reference location, the following equations for the force in the masonry and the force in the compression reinforcement and their first

11-8



m

0662949 0508887 576

m

moment about the reference location are obtained

(force) ( u m ) The compression force in the masonry and corresponding arm are:

Eq. 11.0-23

Eq. 11.0-24 The compression force in the reinforcement(subtracting the area of masonry displaced by the reinforcement) and corresponding arm are:

(A'+?#, - A'#,E'J *(d')

11.0-25

Eq.

or

[A',@, - E,) (1 - m?" (

11.0-26 Eq.

0

The centroid of the compression forces, y, is then obtained by dividing the first moment by the sum of the forces:

Y =

Wkd 2 3

+

5 bd (Es - Em)(l - d'/kd)E,d'/kd

kbd Afs E , , , E +~ ~ (Es- Em) (1 kbd

-

m.11.0-27

d'/khi) E,

Substitution of Eq. 11.0-9 results in: -

r116

+

P'(n -

r112

(1 -

-

1)(1 - &/W]

k

Y=kd +

&/mdf/m

'1

P'(n

Eq. 11.0-28

k

An expression for the moments can now be obtained. Summing

the moments about the

centroid of the compression force results in:

Eq. 11.0-29

11-9


AC1

TITLE*MDG 93 W 0662949 0508888 402 W

Summing the moments about the centroid of the tension force results in:

Mt = A,(d - RF,

Eq. 11.0-30

Equations 11.0-22 , 11.0-28,11.0-29, and 11.0-30 provide the basis for flexural design of rectangularmasonry

elements with both tension and compressionreinforcement.

equations are not applicable to sectionssubjected

The

to flexure and axial load acting

simultaneously (see MDG Chapter 12).

11.1 WALLS Most masonry elements are walls. A wall is defined per Code 2.2 as a "Vertical element with a horizontal length greater than three times its thickness." Elements with a horizontal lengthless than or equal to three times their thickness are usuallycolumns.They

are

columns if in addition to satisfying the length to thickness requirements, the height is also at least three times the thickness. The distinctionbetween

the definition ofwalls

and columnsis

important because of

restrictions placedon columns inthe Code. For example, columns are limited to a minimum thickness of 8 in. by Code 5.9.1.1, an effective heightto thickness ratio of 25 by Code 5.9.1.2 and must be reinforced to comply with Code5.9.1.4,5.9.1.6and A.4.6. Walls are not limited to aminimumthickness

or amaximumheight

to thicknessratio.Wallshave

different

reinforcement requirements which are generally less restrictive.

1.

Code 5.16 requireswalls of stackbondedmasonry

to be reinforcedwith a

minimum reinforcement of 0.0007 times the vertical cross section of the wall spaced not further than 48 inches on center.

2.

Code A.3.8 and A.4.5 contain special provisions for masonry in Seismic Zones

2, 3 and 4.

11-10



93 W 0662949 0508889 349 W

A literal reading of the Code will reveal that some elements do not fit the definition of either a wall or a column. Elements not meeting the definition of either a wall or a column are typically designed as walls. 11.1.1 Flexural Design of Unreinforced Masonry Walls

Code Chapter 6 contains provisions for unreinforced masonry wall design. Reinforcement may be present in the wall, but its effect is ignored for design purposes. Code EQ. 6-1 is the unity equation for flexurewithaxialload.Whenaxialloadis

not

present the equation reduces to:

-f b < l

Eq. 11.1-1

'b

Small letters are used for computed stresses and capital letters are used for allowable stresses. The computed bending stress is normally calculated by using MDG Eq. 11.0-5. For out-of-plane bending of solid unit masonry or hollow unit masonry fully grouted, the equation becomes:

Mc - M - 6M f b = I " - -

S

Eq. 11.1-2

bt2

Where t is the specified thickness of the wall and b is the width of the wall taken as the same width used to calculate the moment M. For in-plane bending ofsolid

unitmasonry

or fully grouted hollowunitmasonry,

the

equation becomes: fb

=

6M

-

Eq. 11.1-3

2;

11-11


A C 1 TITLE*MDG

93 D Ob62949 0508890 Ob0

m

Where ,Z is the length of the wall. The allowable bending compression stress is given in Code Eq. 6-5. A one-third increase for short duration loading (wind or seismic) applies to both the compression and tension allowable stresses, Code 5.3.2. Allowable flexural tension stresses due to out-of-plane bending are given in Code Table

6.3.1.1. The values depend on the type of mortar, type of masonry and the direction of the tensionstress.

M a l tension (not tensionresulting

from bending) is not allowedin

unreinforced masonry (Code 6.4). In running bond construction, values for tension parallel to the bed joint are higher than values for tensionperpendicular to the bed joint because of the interlocking of the masonry units. Walls are often supported on more than two sides. Under load they often exhiiit plate behavior (two-way bending) rather than beam behavior (one-way bending). The actual load distribution and resulting stresses can be very complex Such determination would require values for stiffness of masonry both parallel and perpendicular to the bed joint. The Code assumes a single value per Code 5.5.1. Code 5.13.4 containsspecific requirements for including the effects of and designing intersecting walls. Code 5.7.1.1 requires that intersecting walls be included in the stiffness determination for lateral load distribution unless shear transfer is prevented. The Code limits the effectiveflangewidth

to 6 timesthethickness

of the wall (Code 5.13.4.2).

Reference (11.1.1) discusses masonrylateral load resisting systemsand the structural model for distributing the forces. The hand calculation of lateral load distribution for the unreinforced masonrywall construction in the RCJ Hotel, MDG Example 9.2.3, neglected intersecting wall stiffnessdue

11-12


A C 1 TITLESVDG 93

0 b b 2 9 4 9 0508891 T T 7

to the presence of control joints at the intersection of walls. The computer calculation of the lateral load distribution using the ETABS program for the hotel example considering reinforced masonry wall construction included the effects ofwall intersections since the program considers all walls in its three dimensional analysis.

11.1.2

Flexural Design of Reinforced MasonryWalls

Chapter 7 of the Code contains provisions for reinforced masonry wall design. Using MDG Eqs. 11.0-14 and 11.0-15, or 11.0-29 and 11.0-30, and substituting the allowable stresses, the allowable applied moments can be obtained. The allowable tensile stresses in the reinforcement are given in reinforcement.

Code 7.2.

The allowable tension

Grade 60 is the mostcommonlyused

stress 24,000 is psi.

The allowable flexural

compression stress in masonry isone third the specified compressivestrength of masonry per Code 7.3.1.2. A one-third increase for short duration loading (wind or seismic) in this allowable stress is allowed for both the

reinforcement allowable stress and the masonryflexural or axial

compression allowable stress, Code 5.3.2. For out-of-plane bending, the effective width to use in the flexural equations is limited by Code 7.3.2 to the center-to-center spacing of the bars, six times the wall thickness or 72 in., whichever is least. The following iterative procedure isnormallyusedby

the structural engineer to design

reinforced masonry for flexure: 1.

Determine the

applied moment.

2.

Estimate the required reinforcement basedon an assumed j or k.

3.

Calculate k (neutral axis location). 11-13


AC1 TITLE*NDG 93 m Ob62949 0508892 9 3 3 m

11.1.2.1

4.

Calculate M, and M,.

5.

Compare M, and M, tothe applied moment.

Initial Depth and Steel Estimate

- The required depth of the masonry and the

amount of reinforcement can be estimated using the two flexure equations and assuming k equals to 0.3. Rearranging MDG Eq. 11.0-14 gives

Eq. 11.1-1

Rearranging MDG Eq. 11.0-15 gives Eq. 11.1-2

11.1.2.2

Balanced Design - While many textbooks present the concept of balanced design,

this concept is of limited practical use. Balanced design is a condition where the allowable tensile bending stress and the allowable masonry compressive stress occur in the bending element at the same time, Some designers use the balanced condition as a starting point indesignsincedesign allowable stresses. This

parameters such as k, j, and p are known based on the known results in easily generated design aids. Refer to MDG 11.3.3 for

balanced beam design. Other designers simply use balanced parameters, such as

Pb,

as a

check to indicate whether in their design the allowable moment is limited by tension or compression. If this condition (an additional equation) is added to our previous derived equations, all manner of design procedures result, none of which is of much value.

The problem with balanced design isthat it usually is not the minimum cost design,because the cost per unit of masonry strength is not the same as the cost per unit of steel strength.

11-14


A C 1 T I T L E L M D G 93

Obb2949 0508893 8 7 T

Moreover, the factor of safety for masonry in flexural compressionis 3.0 while the factor of safety for reinforcement varies between 2.0 and 2.5. Thus, the concept of balanced design is dependent upon the somewhat arbitrary selections of factors of safety made by the building codes. Thus balanced design using the UBC is not the same as balanced design using the Code. Other methods use various tables for easy reference; computer programs are also available for the design of wall elements. When using computer programs, the designer is cautioned to check the code provisions employed. In recent years, masonry codes havebeen changing rapidly and many programs quickly become outdated.

11.2 PILASTERS

11.2.1

General Description

Pilasters consist of a column section built integrally with a wall. Because of the modular nature of both clay and concrete masonry units, pilasters can be built within the coursing pattern of a wall. Units in alternate courses of a pilaster may be arranged such that they interlock in directions both normal and parallel to the plane of a wall. Unlike an isolatedcolumnmember reinforced per Code 5.10.

Theymay

(Code 5.9.1.4), pilasters are not required to be

even be ungrouted.However,

for hollowunit

construction pilasters typically should be grouted, since a beam or truss is often supported

in bearing at the top of a pilaster. Furthermore, grouting of at least the column section of a pilaster will enhance itsflexural strength, not only because of the increased section modulus, but more significantly because of the increased values of allowable tensile stress normal to bed joints that arepermitted for a fully grouted section per Code 6.3.1.1. Pilasters typically support wall panels, acting as flexural members spanning vertically while carrying little verticalload.

Thus they are commonly grouted and reinforced regardless of unit

construction. 11-15


AC1

TITLESHDG 93 m 0662949 0508894 706

Reinforced and grouted pilasters should have a minimum of four vertical reinforcing bars placed through the cores of hollow units, or placed within a cavity formed with solid units. Vertical reinforcement greatly increases flexural strength of a pilaster. Flexure is generally a controlling factor for pilasters; consequently reinforced pilasters can often be much smaller than unreinforced pilasters.

11.2.2 Role

of Pilaster in Resisting Loads

A pilaster serves the two basic functions depicted in Fig. 11.2-1. The first is to provide lateral support for walls subjected to wind or seismic forces normal to their plane (out-ofplane). The second is thesupport of gravity loads transferred from beams or trusses supporting roofs, floors, or overhead cranes. Gravity Beam or

L '. \ Midheight of Pilaster

Fig. 11.2-1 Forces on Pilaster-Wall System

11-16



m

Ob62949 0508895 b 4 2

m

Vertical reactions from beams or trusses are usually transferred to a piIaster through steeI bearing plates placed at the top of a pilaster. The reaction often does not coincide with the centroid of the pilaster. Since the pilaster spacing is usually greater than thebeam spacing, beams or truss members are often supported on top of masonry walls between pilasters. When allbeams or truss membersare detailed for identical spans, the reaction will be closer to the centroid of the wall, rather than the centroid of the pilaster sections. The eccentricity of the vertical force results in bending moment at the top of the pilaster. Walls are commonlydesigned

to span horizontally toeach pilaster. This iscommonly

assumed when the spacing of pilasters is less than one half the unsupported vertical span of an out-of-plane wall. The pilaster spans vertically and sometimestransfers half of its lateral force to the roof or floor diaphragm above or below. For unreinforced walls, span length is not the only criterion, since allowable stresses differ for vertical and horizontal spans. A system of pilasters and walls resisting lateral load acts like a system of beams and slabs resisting uniform floor loads. The design of pilasters used in low-rise buildings will rarely be governed by gravity forces, unless the centroid or resultant of the vertical force is outside the pilaster’s kern. The kern is that portion of the cross-sectional area within which the resultant vertical load must act to produce only compression throughout the cross section. For loads outside the kern, the resulting tensilestress may govern overthe compressive stress, since the vertical compressive stresses are usually low for this class of building (in the range of 20 to 50 psi). Furthermore, pilasters may be located in walls that run parallel to roof trusses or floor beams, and thus do not resist any gravity forces other than their own self weight. In

these cases, pilasters

may be thought of as simple flexural memberswith light amounts of axial compressive force which may be conservatively neglected. This MDG section presents pilaster design for the special case of flexure without axial forces. Design is governed by flexure resulting fromlateral forces applied normal to a wall containing a pilaster. Both unreinforced and reinforced pilasters are discussed. In MDG 11-17


A C 1 T I T L E * H D G 93

0662747 0508876 587

12.3, pilaster design is discussed with respect to combined axial force and flexure.

11.23 Coursing Layout In hollow-unit masonry, pilasters may be built within the thickness of a wall by grouting and reinforcing those cells assigned to a pilaster. However, the more common situation is when a deeper and stronger pilaster is needed. The pilaster may be centered in or through the wall,fully offset from the wall, or somewhere in between. The position of the pilaster relative to the wall is often dictated by nonstructural considerations. Sometimes a pilaster may be placed so that its centroid will coincide with a truss or beam reaction. Unfortunately this isnot the common case, and flexural strength must be provided to account for theeccentricity of the vertical reaction. See MDG 12.3 for design of pilasters subjected to axial load and flexure. To qualify as an integral system, at least 50% of the masonry units must interlock at the interface between the pilaster and the wall per Code5.13.4.2(e)l.Wallsmayalso

be

effective as pilaster flanges if steel connectors or intersecting bondbeams are used per Code 5.13.4.2(e)2 and 5.13.4.2(e)3 in lieuofmasonry

unitinterlock.However,

the later two

methods are impractical and unnecessary for pilasters. In hollow unit masonry, pilastersare commonly madeof units laidin a coursing pattern that complies with the running bond pattern of a wall, and provides good keyingaction with the pilaster. Sixteen-inch pilasters, shown in Fig. 11.2-2, are the most common for this reason. However, similar layouts lend themselves

well to 24 in. and 32 in. sections. For larger

pilasters, it may not be necessary for units to be placed in the interior. The cavity that is formed by the face units can be filled with grout and a reinforcing cage, if necessary. The effective section of the pilaster varies with the mortar bed configuration and extent of the grout-filled cores. See MDG 11.2.4.

11-18


~~


93

m

Obb2949 0508897 415

m.

8” CMU(Typica1)

Alternate Courses

16“

Courses Alternate

x 16” Pilaster

Fig.11.2-2Coursing

16” x 24” Pilaster

Layout for CMU Pilasters

Unreinforced pilasters with solid units are made up by laying units in a full pattern or about a grouted and reinforced interior cavity. Since bed joints are fully mortared, the effective

section is the grosssection.Solidunitpilasterscanbereinforced

by placingvertical

reinforcement in grouted cavities.

11.2.4EffectiveSection The cross section that may be assumed effective in resisting axial antrl flexural stresses is often much different from what

the architect conceives or the mason builds. Movement

joints are typically located along both sides of pilasters, thus limitingthe effective section to the area of the pilaster. If the pilaster is not isolated by joints and the requirements of Code 5.13.4.2(e) are met, then the effective width of flange on either side of the web shall equal

6 times the flange thickness per Code 5.13.4.2(c). If the section is unreinforced, then both the tensile and compressive masonry

areas are

considered. If the section is grouted, then the full gross area is considered. If the section

is ungrouted, only the mortar bedded area is considered.Effectivesections combinations of these classifications are shown in Figs. 11.2-3. 11-19


for three

A C 1 TITLElkMDG 9 3

III c-

m

0662949 0508898 351

f

6t

6t

(a) Pilaster

""'

'L

And Wall Fully Grouted

Compressive Zone

(b) Pilaster Grouted, Wall Ungrouted

Face Shell Bedding Compressive Zone 6t

(c) Pilaster And Wall Ungrouted

Fig. 11.2-3 Effective Sections for Unreinforced Pilasters-NoExpansion/Control Joints If the section is reinforced, the tensile

strength of masonry is neglected, and only the

transformed area of the tension steel is considered. It thenbecomes important to recognize

11-20


the direction of the moment, because the width of the compressionflange may vary substantially for different directions of bending. The reinforcement in compression may be considered effective(steel transformed to masonry by n) if lateral ties are provided meeting Code 5.9.1.6. Effective sections for three vastly different cases are shown in Figs. 11.2-4.

(a) Pilaster Reinforced, Wall Unreinforced, FlangeIn Tension

I

I

6t

HAS 6t

I

(b) Pilaster Reinforced, Wall Grouted, Flange In Compression

lkd

$1 II (c) Pilaster Reinforced, Wall Ungrouted, FlangeIn Compression

Fig. 11.2-4 EffectiveSectionsforReinforcedPilasters

11-21


A C 1 TITLEvMDG 93

W

0bb2949 0 5 0 8 9 0803 T

11.2.5 Flexural Design Considerations

As mentioned in the introduction to

this section, pilaster flexure is

a result of transverse

lateral loads such as wind or earthquake forces, and eccentricity of vertical gravity loads applied to the top of a pilaster. Design considerations depend on the relative participation of each of these effects.

The latter loading case involves flexure plus axial compression,

discussed in MDG 123. This section focuses on pilaster designs governed by the former loading case, where flexural stresses only result

from lateral forces. For these cases, the

critical states that typically must be examined are flexural tension normal to bed joints for an unreinforced pilaster, and tensile stress in vertical reinforcement and compression stress in masonry for a reinforced pilaster. In this section, light amounts of axial compression will only be considered if needed since they tend to reduce effects of flexural tension. If walls span horizontally with respect to lateralforces, then the full tributary width of wind or seismic forces must be applied to the pilaster. The triiutary width is simply half the distance to the adjacent pilaster or other support element on both sides of the pilaster in question. Portions of the wall wouldspan vertically if the height of a wall were less than the spacing of pilasters, or if an end wall were omitted in an open frontbuilding. For the case where the pilaster is loaded only witha portion of the lateral load, valuesof load distribution on walls and pilasters have been calculated based on plate theory (11.21J1.2.2). Pilasters are usuallyassumed

tospan vertically between storylevels.

For single-story

buildings, thismeans that half the lateral load is transferred to the foundation, and the other half is transferred to the roof diaphragm. It is common to assume a simple support at the top of a pilaster even though it translates with a flexiile diaphragm. Since the bottom of a pilaster is also considered to be a simple support, the vertical element will be statically determinate, and any translation at the top will be insignificant provided that the roof diaphragm is designed for the shear. This is done for simplicity, although some restraint against rotation may exist if the pilaster does not rock and the foundation is relatively stiff.

11-22


A C 1 TITLExMDG 93

0662949 0508903 776

m

Though the examples presented herein all consider pilastersto have simple supports at top and bottom, it should not be inferred that these boundary conditions will be universal. For example, a pilaster can act as a cantilever, fixed at the bottom and free at the top. Earthquakes produce inertial loads that must be assumed to act laterally in any direction. Wind forces are normally applied as direct pressures on the windward sideof a building, but should also be considered as applying a suction of equal value to the leeward side, unless a comprehensive wind analysis is done.

Loading standards such as ASCE-7 describe the

wind analysis procedure for both the Main Lateral Force Resisting Systems (forshear wall analysis),and

for Components andCladding

(for out-of-plane lateral loads on walls).

Because unreinforced pilasters are asymmetrical sections, flexural tensile stress should be checked for bothdirections of lateral loading since the subtractive load combinationmay be more critical than the additive combination depending on the section properties. Vertical bars in reinforced pilasters must also be checked for the critical loading direction.

11.2.5.1

Unreinforced Pilasters - Flexural capacity is limited by flexural tensile stress. For

designallowingtensile

stress in masonry, the values of allowableflexuraltension

are

summarized in Code Table 6.3.1.1. Pilasters are vertical elements rather than horizontal ones, and thus only the values for tension normal to bed joints are usually applicable. The allowablemasonrytensile

stresses are givenin Code Table 6.3.1.1. For ungrouted

hollow-unit masonry, the allowable flexural tensile

stress is dependent upon the type of

masonry unit and type of mortar. In addition to increasing the section modulus, grouting can more than triple the allowable flexural tensile stress, because tension across bed joints does not need to be relied on exclusively.

For an unreinforced pilaster subjected to pure flexure, there is no need to check compressive stress. The allowable flexural tensile stress will control the design. It is common practice to grout the column portion of a pilaster and leave the wall portion

11-23


(the flange of the pilaster) ungrouted. For this situation, allowable tensile stresses may be interpolated based on the relative proportions of grouted and ungrouted masonry areas. Bending moments resulting from lateral forces are largest near the midheight. Thus, flexural tensile stress should be checked at thislocation.

It isadmissible to subtract from the

computed flexural tensile stress, the compressive stress resulting from the self-weight of one half the pilaster height. It is conservative in this case to estimate the lower bound of the wall weight. If the pilaster supports roof or floor loads, a minimum value of gravity dead load may also be used to reduce the flexural tensile stress. However, this may not lead to thinner pilasters, because the gravity load may be accompanied by an additional moment resulting from the load through its eccentricity (see MDG 123).

11.2.5.2

Reinforced Pilasters

- Reinforced pilasters are much stronger in flexure than

unreinforced ones, particularly if axialcompressiveforces

are negligiile. For thiscase,

design is essentially the same as for reinforced beams, once an effective section has been defined.

As noted inFig.

11.2-4, severaldifferenteffectivesectionscan

be defined

depending on the direction of the bending moment, and on whether the flange is grouted and/or reinforced. Typically, movement joints will

be located adjacent to pilasters, thus

defining the cross section as rectangular. Once determined, flexural stresses are compared with Code allowable values. In Code 7.2.1, values of allowable reinforcement stress are 20 ksi for Grades 40 and 50 steel and24 ksi for Grade 60 steel. In Code 7.3.1.2 the limiting masonry compressivestress, F*,is 0.33 times the specified compressive strength. Allowablestress values for both steel and masonry may be increased by one third for the case of the wind or earthquake per Code 5.3.2.

11.2.6

Shear Design Considerations

Shear design of pilasters is no different than for masonry walls or beams. Shear stress for unreinforced pilasters is determined using Code Eq. 6-7. For simplicity, only the pilaster without wall flange needs to be considered in resisting shear. For a solid rectangular web

11-24


A C 1T I T L E M M D G

9 3 W 0662749 0508903 549

m

Code Eq. 6-7 reduces to:

3 v fy=zA,

Es. 11.2-1

Because pilasters are usually tall, slender elements, shear usually does not control, and this simplification is warranted. Allowable in-plane shear stresses are given in Code 6.5.2 for unreinforced members. For pilasters made with hollow units and face-shell mortar bedding, the shear area should only be the net bedded area. Shear design of reinforced pilasters follows Code 7.5. Shear stress is computed by dividing the shear forceby the width, b, and the internal lever arm,jd, of the cross-section (Code Eq.

7-3). No distinction is madebetween

a rectangular section and a T-section. If one

interprets thederivation of the flexural shear stress equation literally, the width b should be taken as thewidth of the compression zone, whether of the web or of the effective flange. However, for very large flange widths, this wouldnot be reasonable because of large shear lag effects. Therefore, it is recommended that the web width be used in all cases. Again, shear should not control the design and such a simplification should be warranted. 113 BEAMS AND LINTELS 113.1 Introduction The use of reinforcing steel in masonry constructionpermits the design of flexural members such as beams, lintels and deep wall beams. Lintels are horizontal members used to span openings in masonrywalls.Theymay

be reinforcedmasonry, precast or cast-in-place

concrete, or structural steel. Reinforced masonry lintels may be constructed of specially formed lintel units, bondbeam units or standard units with cut-out webs. Special lintel units arranged to form a channel for placement of reinforcement and grout are shown in Fig.

11.3-1. Typical reinforced brick masonry lintel sections anda temporary shoring detail are shown in Figs.11.3-2 and 11.3-3 respectively. Steel lintels may consist ofone ormore angles,

11-25


A C 1 TITLEtMDG 93

Obb2949 0508904 485 W

or a C or T section. For wider spans and heavier loads, a.shape with a suspended soffit

plate may be used (Fig. 11.3-4). Analysis and design of lintels subjected to vertical and lateral loads are discussed in this section.

A brief discussion of deep wall beams and an

example problem are also included. The Code gives general provisions for masonry beams in Code 7.3.3, and specific deflection criteria for beam and lintels supporting unreinforced masonry in Code 5.6. The deflection of steel or concrete beams that support unreinforced masonry above openings is limited to 1/600 or 0.3 in. under dead and live load. Beams and lintels supporting reinforced masonry are not required to meet these deflection limits.

Fig. 113-1 Lintel Blocks in Place (113.1)

11-26


Fig. 113-2 Reinforced Masonry Lintels (113.1)

Clvity Wall

Stop Brick

Fig. 113-3 Shoring for Reinforced Masonry Lintel (113.1)

11-27



93 M Obb2949 0 5 0 8 9 0 6 258 M

6” TTW Brick

Single Angle

Double Angle

(a>

(b)

Triple Angle

Rolled Shape And Suspended Plate

(c)

(a

Fig. 113-4 Steel Lintels

113.2 Assumptions The working stress method is used to design beams and lintels in accordance with Code Chapters 5 and 7. Structural elements aredesigned so that stresses from working or service loads computed under the assumption of linear elastic behavior do not exceed specified allowable stresses. The working or service loads represent the maximum expected loads under service conditions,and may include dead, live, snow, wind,and earthquake loads. The allowablestresses are givenintheCode strength of masonry,

y,,,,and

as a percentage of the specifiedcompressive

a fraction ofyield

reinforcement (see Code 7.3.1.2,

strength based on the grade of the

7.2.1.and 1, 7.21.2). Fig. 11.3-5 shows that the stress-strain 11-28


A C 1 TITLEgMDG 93

m

Ob62949 0508907 L94

m

behavior of masonry in a standard compression test is approximatelylinear up to about 50 per cent of Pm,and that of reinforcing steel in tension or compression is linear up to yield. In the working stress method, stresses under service loads are assumed to be within the linear and elastic range, so that a straight-line relationship is assumed between stress and strain. The basic assumptions of the working stress method were detailed in MDG 11.0.2 and MDG 11.0.5.2.

I

Strain

Elastic

1 Repon

Strain

Fig. 113-5 TypicalStress-StrainCurves 1133 Basic Equations

- Singly Reinforced Sections

The basic equations forgeneral reinforced masonry design concept are presented in MDG

11.0.5.2. This section examines these concepts specifically as related to singly reinforced beams. The transformed section for a reinforced beam with tension steel only is shown in Fig. 11.36. The neutral axis coincides with the centroidal axis of the cracked transformed section.

The neutral axis is generally located by setting the first moment of area about the neutral axis equal to zero.

T)-

@)(M)(

nA,(d - M ) = O

11-29


A C 1T I T L E v H D G

93

m

0662949 O508908 020

8

.+!

P

W

W

11-30


m

A C 1 T I T L E t f l D G 73

Substituting p "

=

m 0662747

0 5 0 8 9 0 7 Tb7

m

A,/bd, the steel ratio,

L

Dividing by bd2,

R* - pn(1 -k) 2

=

o

Solving,

R

Eq. 11.3-1

[ 2 p n + ( ~ n ) '-] pn ~

=

Equation 11.3-1 can be used to locate the neutral axis for a given beam cross-section. For the triangular masonry stress block (Fig. 11.3-6), the resultant compressive force is,

C

bkdf,@

=

The tensile force is,

T

I

I

=

AJs

The moment

M

=

M

=

M can be expressed as, 1 Cjd = -fbkjbd2 = M,,, 2

Tjd

=

A,& j d

=

pbdf, j d

=

M,

Solvingmasonry the for stress, fb =

2M

Eq. 11.3-2

-

b d 'jk

Solving for the steel stress,

&=-

M

Eq. 11.3-3

A,jd

11-31


~

AC T I1T L E * H D G

93

0662947 0 5 0 8 9 3 0 7 8 9

The lever arm, jd, may be expressed as,

from which we obtain, j = 1" k 3

Eq. 11.3-4

Balanced Beam Design Balanced condition is a state where both the steel andmasonry reach their prescribed allowablelimitssimultaneously.Thus,

fb

= F b = allowable masonry compressive stress in flexure per Code 7.3.1.2

fs = F,

= steel allowable tensile stress per Code 7.2.1

From similar triangles (see Fig. 11.3-6), 'bd

-

"

d

Fb

Fb + F J n

where kb is the k value at balanced condition. Thus

Or

Or

11-32



m

Obb2949 Q50891L bL5

m

Eq. 11.3-5 To obtain the balanced steel ratio, pbbdFs

Pb,

equate C and T,

= Fbkbdb/2

Substituting for kb and solving, ES. 11.3-6 If p < p b the beam is said to be under-reinforced and the steelwill reach its allowable value first.

If p >

Pb

the beam issaid

to be over-reinforced and the masonry will reach its

allowable value first.

-

113.4 Basic Equations Doubly Reinforced Sections

The basic equations for general reinforced masonry designconcept are presented in MDG

11.0.5.2. This section examines those concepts specifically as related to doubly reinforced beams. Compression reinforcement may be provided in a masonry beam to increase the compressive resistance of the section and/or to minimize creep and reduce deflections. The total resisting moment M for a doubly reinforced section may be expressed as,

M

=

Eq. 11.3-7

Mb+M2

where

M&,= moment capacity for the section at balanced condition without compression reinforcement. The tensile reinforcement needed to develop M b is denoted

Ash* 11-33


A C 1 T I T L E l r M D G 93

m

0662949 0508932 553

m

M 2= additional moment capacity developed by A', and Asa where, AS2= A, -Asb. The compressive steel stress can be derived from the stress diagram in Fig. 11.3-7,

Or

Eq.11.3-8

Or in terms of masonry stress,

Eq.11.3-9

Eq.11.3-10

where

Mb

Es. 11.3-11

= Fbjbkbbd2/2

In design the additional tensile steel required to resist M 2may be estimated using,

Eq.11.3-12

and the compressive reinforcement may be estimated using,

Eq.11.3-13

The location of the neutral axis for the doubly reinforced section is given by

11-34



93

m

Obb2949 0508913 498

I

m

v)

a

p

a

I

I

11-35


k

[

= [np + (n

- l)p’]*

+ 2[np + (n- 1) p’d’/d]]m- [np + (n - 1)p’I

.

Eq. 11.3-14

where p/ = A,’/bd

Check the adequacy of the section using k as per Eq. 11.3-14. Calculate

f’#

and fs and

determine M (see MDG Example 11.3-2). 113.5 Load Distributions on Lintels

Lintel beams spanning openings in masonry wallsmay be subjected to two types of vertical loading: (1) distributed loads from the dead weight of the lintel and the masonry above the lintel, and floor and roof dead and live loads, where the floor and roof construction is uniformly supported by the masonry; and (2) concentrated loads from floor girders, roof trusses, etc. framing into the wall. Because masonry can arch across openings, lintel beams usually do not need to be designed for the total tributary loading above the lintel. That is, if a lintel fails or is removed, only a triangular portion of the wall immediately above the opening would collapse, because the masonry will form an arch over the opening. However, for arching action to take place, there must be sufficient masonry mass on each side of the opening, or tension ties across the opening to resist the horizontal thrust resulting

from arch action. The lintel reinforcement can not be assumed to act asa tension tie unless it is extended beyond the lintel bearing and sufficientlyanchored.Arching

action also

requires that the height or depth of the masonry above point C (see Fig. 11.3-8) must be sufficient to provide resistance to arching thrusts. Assuming arching

action exists, design

loads for lintels may include: Dead weight of the masonry wall above the opening within a triangular area ABC as shown in Fig. 11.3-8. O

Uniform live and dead loads of floors and roofs which bear on the wall above the opening and below the apex of triangle ABC. Since the floor loads shown

11-36


A C 1 TITLE+MDG 93

m 0662949 0508915

2b0

m

in rig. 11.3-8 are above the apex, they can be neglected. 0

Concentrated loads from floorgirders,rooftrusses,etc.framing

into a wall

may be assumed to be distniuted at 45 degrees as shown in Fig. 11.3-8 (see Code 5.12 and Code C 5.12).

Only the portion of the distributed

concentrated load that is directly over the lintel is assumed to act upon the lintel.

i

i

! Effective Length c BearingWidth + 4t 2 Spacing of Concentrate5Load

7

! Ciacentrated

I

Load

m

!

1

Fig. 113-8 Load Distributions on Lintel

In Fig. 11.3-8, the portion of the concentrated load shown shaded above segment EF may be considered as a uniform load partially distributed over the lintel. Code 5.12.1 requires that for walls laid in running bond, the length of wall to be considered effective for each concentrated load shouldnot exceed the widthof bearing plus four times the wall thickness, nor the center-to-center distance between concentrated loads. The Commentary states that

11-37


A C 1 TITLEsflDG 93

M 0662947 0508916 1 T 7 M

when other than running bond is used, concentrated loads can only be spread across the

length of one unit unless a bond beam or other technique is used to distriiute the load. 113.6 Beam Depth Determination

Beams that are parts of a wall are normally designed to have the same width as the wall thickness. The effective depth assumed for the beam design will depend on the height of the wall above the opening. For walls with heightsup to 3 ft above the beam soffit, the full height could be considered as the effective depth (11.3.8). For wall heights greater than 3 ft above the beam sofit, the effective beam depth is commonly taken equal to the beam

depth required for the masonry to resist all the shear. MDG Example 11.3-4 on wall beam design illustrates these concepts.

113.7 Deflection Realistic predictions of masonry beam and lintel deflections require the use of reasonable estimates for moment of inertia, modulus of elasticity, modulusof rupture, creepfactors and, in the case of concrete masonry, shrinkage factors. Also, the procedure used to calculate deflections has to be based on a comprehensive understanding of the mechanics of shortterm and additional long-term deformation of masonry beams and lintels. The Code provides tables and expressions for the material properties of masonry, but does not provide guidelinesfor deflection calculations. An important consideration is the method to account for the effect of tension stiffening. Tension stiffening is the flexural resistance provided by undamaged tensile masonry between flexural cracksand in regions of lowtensile stress; i.e., between tips of tensile cracks and the neutral axis. Tension stiffeninghas

been accounted for inreinforced

concrete design byusing

an

empirical equation to define the effectivemoment of inertia, I, (11.3.9). The effective moment of inertia provides a transition between the well-defined upper bound of I' (moment

11-38


A CT1I T L E M M D G

93

Obb291)9 0508717 033

m

of inertia of gross section about centroidal axis, neglecting reinforcement), and the lower bound of I, (moment of inertia of the cracked transformed section). I, is defined as a function of the level of cracking, as represented by MJMa as follows: Eq. 11.3-15

The cracking moment is defined as: Eq. 11.3-16

Using these concepts, the designer would use I, along with standard elastic engineering deflection equations to predict initial (short-term) deflections of reinforced concrete members.Comparison

of suchcalculateddeflectionswith

experimental test results on

masonry members show that this method under-predicts masonry deflections (11.3.10). The UBC (11.3.14) shows that the method provides fair estimates for short-term deflection of masonry members although the method tends to underestimate deflections in some cases

(11.3.10). This method will be used for calculating the short-termdeflections of b e a m s and lintels in this section. The following formulasto predict mid-height out-of-plane deflections of uniformly loaded simply supported beams provide a fair estimate for short-term deflection of masonry members (11.3.14).

Eq. 11.3-17

Eq. 11.3-18

In some cases, this mathematical model underestimates the deflection (11.3.10). Equation 11.3-17is based onflexural stiffnessof the uncracked sectionand should give good estimates

of deflections of uniformly loaded, simply supported flexural members up to first cracking. 11-39


A C 1 TITLExMDG 93

m

Obb29Y9 0508938 T 7 T W

Equation 11.3-18 is based on a bilinear load-deflection relation.

It is assumed that the

member deflects as an uncracked section until the modulus of rupture is reached, and whereupon thereafter the member cracks and the cracked moment of inertia is used in calculating the additional deflection. Equation 11.3-18 alsoassumes a simple span and uniform loading. These equations will be used for calculating the short-term deflections of beams and lintels in the MDG example problems. Expressions for I, for singly and doubly reinforced rectangular sections are given in Eqs.

11.3-19 and 11.3-20. For the singly reinforced section,

I,, = bk3d3 + M,(d-M2 3 where kd =

(4-

-

Eq. 11.3-19

1)

B

For doubly reinforced section, ''r

--

bk3d3 + M,(d-kdy + (n- 1)A,'(kd-d')2 3

2dB where kd

=

+ (1 +

r)2

- (1

Eq. 11.3-20

+ r)

B

Long-term deflection due to creep from sustainedloadsand,in

11-40


the case of concrete

A C 1T I T L E x f l D G

9 3 W Ob62949 05089L9 906

m

masonry, deflection due to shrinkage need to be considered to estimate total long-term deflection of masonry beams and lintels. The Code does not give specific guidelines for calculating long-term deflectionof masonry beams and lintels. One technique (11.3.9) used to estimate long-term deflectionsof flexural members is to compute the additional long-term deflection resultingfrom creep andshrinkage by multiplyingthe immediate deflection caused by the sustained load considered, by a factor

A =

m.11.3-21

f 1 +50p‘

where

p’

=

A,’/bd at midspan for simple and continuousspans, and at supportfor cantilevers

=

2.0 for sustained loading of 5 years or longer duration, 1.4 for 12 months, 1.2 for 6 months, and 1.0 for 3 months duration.

In the absence of other recommendations, and unlessvalues are obtained by a more comprehensiveanalysis,

the multipliergivenabovemay

be considered satisfactory for

predicting long-term deflection of concrete masonry flexural members. Since Eq. combines creep and shrinkage, it is not applicable to clay brickmasonry.

11.3-21

Thecreep

coefficient given in Code 5.5.5 may be used to estimate long-term creep deflection of claybrick flexural members.

113.8 Deep Beams

A deep beam may be defined as one whose depth is equal to or exceeds the span length. Deep masonry beams are not addressed in the Code. However design considerations and provisions for deepbeams have been developed by various groups (11.3.15,11.3.16,11.3.17,

11.3.18).

11-41


A C 1 TITLErMDG 93

Ob62949 O508920 628

m

REFERENCES 11.1.1

"Concrete Masonry Load Bearing Walls-Lateral Load Distribution," NCMATek #61A, NCMA 1981.

11.2.1

"Engineered Concrete Masonry-Wind Loads," NCMA Tek #24, NCMA, 1970.

11.2.2

"Engineered Concrete Masonry Warehouse Walls," NCMA Tek #89, NCMA, 1977.

11.3.1

Smith, R.C., T.L Honkala, CK. Andres, Masonrv: Materials.

Desim,

Construction, Reston Publishing Company, A Prentice Hall Company, Reston, Virginia, 1979. 11.3.2

Hendry, kW., R.E. Bradshaw, D.J. Rutherford, 'Tests on Cavity Walls and the Effect of Concentrated Loads and Joint Thickness on the Strength of Brickwork," Clay Products Technical Bureau (England), Vol. 1, No. 2, July 1968.

11.3.3

"Building Code Requirements for Concrete Masonry Structures (AC1 531-79) and Commentary (AC1 531R-79)," American Concrete Institute, Detroit, Michigan, 1978.

11.3.4

"Structural Recommendations for Loadbearing Walls," British Standard Code of Practice CPIII: 1964, The Council of Codes of Practice, British Standards Institution, March 1964.

11.3.5

'Wall Masonry, design and Execution," German Standard DIN 1053, November 1962

11.3.6

"Standard for Calculation and Execution of Manufactured and Natural Bricks," Technical Standard 113, Swiss Engineers and Architects Society, 1965.

11.3.7

"Recommended Practice For Engineered BrickMasonry,"Brick

Institute of

America, McLean, Virginia, 1969. 11.3.8

Schneider, R.R., W.L Dickey, Reinforced Masonrv Desim, Prentice-Hall, Inc., Englewood Cliffs, New Jersey, 1987.

11.3.9

Building Code Requirementsfor

Reinforced Concrete (ACI 318-89) and

Commentary - AC1 318R-89, American Concrete Institute, Detroit, 1989. 11.3.10

Horton, R.T.,M.K. Tadros, "Deflection of Reinforced Masonry Members," AC1 11-42


m


0662947 0508923 564

m

Structural Journal, V.87, No. 4, July-August 1990. 11.3.11

Wolde-Tinsae, AM., M.L. Porter, M. Ahmed, "Shear Strength of Composite Brick-to-BlockPanels,"Proceedings

of the Third North AmericanMasonry

Conference, Arlington, Texas, June 1985. 11.3.12

Wolde-Tinsae, A.M., J. Colville, R.H. Najib, ltModulus of Elasticity of Clay Brick Masonry," Ninth International BrickBlock Masonry Conference, Berlin, October 1991.

11.3.13

Hamid, A, G. Ziab, O. El-Nawawy, "Modulus of Elasticity of

Concrete Block

Masonry," Proceedingsof the Fourth North American MasonryConference, Los Angeles, August 1987. 11.3.14

Commentary to Chapter24, Masonry ofthe Uniform Building Code 1988 Edition, TMS Codes and StandardsCommittee, The Masonry Society, Boulder,Colorado, 1990.

11.3.15

Portland Cement Association, "Designof Deep Girders," IS079D, Skokie, Illinois,

11.3.16

10 PPPortland Cement Association,"Tilt-UpLoad-BearingWalls,

A DesignAid,"

EB074D, Skokie, Illinois, 1980, 28 pp. 11.3.17

International Conference of Building OfficialsResearch Committee, "Reinforced Brick Wall Panels," Report No. 2727.

11.3.18

International Conference of Building Officials Research Committee, "Concrete Masonry Wall Panels," Report No. 2868.

11.3.19

"Structural Steel Lintels," BrickInstitute of America, TechnicalNotes 31B revised, McLean, Virginia, November/December 1981.

11.3.20

"Manual of Steel Construction," American Institute of Steel Construction, Inc., Chicago, Illinois, 9th edition, 1989.

11.3.21

Blodgett, Orner W., Desien - Of Welded Structures, The James F. Lincoln Arc Welding Foundation, Cleveland, Ohio, 1976.

11-43


A C 1 T I T L E S f l D G 93

Example 11.1-1

m

0662949 0508922 4 T 0

m

-

TMS ShoppingCenter Design of UnreinforcedCMUNonloadbearing Wall for Flexure Only

Design the East Wall on Grid Line 3 of the T M S Shopping Center for out-of-plane flexure using Wall Construction Option A (single-wythe unreinforced concrete masonry).

Loading:

Materials:

Wind: 20 psf

Unit Strength: 1,900 psi (net area)

Neglect Self Weight

Mortar: Type N (PCL)

Sm: 1,500 psi E,,,: 1.8 x 106 psi Calculations and Discussion

See MDG Example 14.3-15 For Wall Connection To Roof Diaphragm

2”O”

Reference

Code

o

8

I

g

‘

This Example Pertains To These Areas

Assume wall is pin

1

supported at the top of the foundation wall and at the joist bearing

elevation (+ 16 ft-O in.).

11-44


A C 1 TITLE*MDG 93

m

Obb2949 0508923 337



Part I - Calculate Imposed Bending Moment Per 1ft-O in. Tributary Width of Wall )40 f t lblft

18'-O"

Moment

Part II - Determine Wall Section Properties Alternate A 12 in. face shell bedded hollow CMU

Tributary Width

I = 929 i n !

See Table 7 in MDG Appendix.

Section Modulus

=

S

=

929 in? = 160 h3 0.5 (11.63 in.) 11-45


25 "

A C 1 TITLErflDG 93

m

Ob62949 0508924 273

m


Calculations

Discussion

Code Reference

Alternate B: 10 in. hollow CMU, grouted solid (grout per ASTM C 476)

Part III - Compare Actual Stresses to Allowables Alternate A 12 in. face shell bedded hollow CMU

M - 620 ft-Ib (12 f”=s160

=

46.5 psi > F& = 19 psi(1.33) = 25 psi .-.N.G. 6.3.1.1

5.3.2 Alternate B: 10 in. hollow CMU, grouted solid

..

Tension stresses acceptable for 10 in.solid grouted construction alternate only.

In a nonloadbearing, unreinforced masonry wall, tension capacity will govern over flexural compression capacity. Therefore, a 12 in.singlewythe

face shell bedded hollow CMU (1,900 psi compressive

strength) wall,with Type N Portland Cement Lime mortar is not an acceptable design solution. The 10 in. hollowsinglewythe

CMU, grouted solidis an acceptable solution.

Considering the compressive stress contribution of wall weight per MDG Chapter 12, the alternate A might be acceptable. Note that thisdesign depends upon providing lateral support for the wall at the joist bearing elevation. The connections must be capable of resisting 203 plf.

11-46


A C 1 TITLE*lDG

93

Obb2949 0508925 L O T

T M S ShoppingCenter

Example 11.1-2

m

- Design of Reinforced CMU Nonloadbearing

Wall for Flexure Only Design the East Wall on Grid Line 3 of the TMS Shopping Center for out-of-plane flexure. Use hollow concrete masonry, reinforced (Wall Construction Option B). Loading:

Materials: Unit Strength Mortar

E m

1,500 psi 1.8 x 106 psi

n

16.1

Reinforcement

Grade 60

P m

Wind 20 psf

2,000 psi Type N

Neglect Self-weight

~~

Code Reference


Roof 1

1

Part 2

Part 1

” ” ” ”

18‘ O”

\

Design For These Wall Elements

IlEllE

“I I1 3 I I EIIF

Assume pin support at the wall foundation. Assume pin support at the 16 ft O-in. roof level.

11-47


I

Obb2949 0 5 0 8 9 2 b 04b

A C 1 T I T L E * M D G 93


Discussion

Calculations

Reference

Code

Wall Element Part 1 Calculate Wall Reactions at Roof and Foundation:

m)= 20 psf x (18 f i ) '

%=

203 plf

16 ft

RF=20psfx(

=

16,fi

20 psf x 18 ft = 203 plf

157 plf

+ 157 plf

:. OK

Calculate Maximum Moment: If 3

203 plf

16'"''

575 @O:Id[head

20 psf 620 ft-lb/ft

7'-10" 157 plf

157 plf

Momant

M

=

157 plf

X

7.9 ft -2

620 fi-lb

ft 11-48


ft-lblft

Ob62999 0508927 T82

A C 1 TITLExMDG 9 3

m



Estimate Reinforcement: Try 6 in. CMU, assume steel at mid-depth

Assume j = 0.9 for initial estimate. 5.3.2

Use the 1/3 stress allowable increase for wind load -

-

620 ft -lb X 12 h/ft %,O00 psi x 1.33 x 0.9 x 2.8 in.

Try #4 @ 24in. o . ~ (. A

=

0.20

%mdfkd

=

0.093

IL x - = 0.10

24

Check Strength: Use 2 ft-O in. wide strip Design Moment = 620 ft-lb/ft x 2 = 1240 ft-lb/ft 0.20 24

=

k

0.265,

=

in. x 2.8 in.

=

0.003

k2 + 2pnR - 2pn=0

0.048

np

i n :

j

=

0.911

11-49


i n . ' / f t

id/fi.)

.

7.2.1.1

93

A C 1T I T L E S H D G

m

Obb2949 0508928 919

m



Allowable Tension Flexural Capacity

Mt = A, jdFs Mt = (0.20 i a 2 )

X

0.911 x 2.8 h. X 24,OOO psi x

1.33 12 in./ft

:. OK

Mt = 1,360 ft-lb/ft > 1,240 ft-lb/ft Allowable Compression Flexural Capacity

Fb = M m = 24

1 3

x 1.33 = -(1,500 x (2*8

in'

psi) 1.33 = 665 psi

x 0.265 x

0.911 x 665 psi 12 in/ft

2 M,,,= 1,260 ft -lb/ft > 1,240 ft -lb/ft Use #4 @ 24 in.

.+.OK

O.C.

Wall Element Part 2 Calculate Bending Moment: Assume door wind load is transferred to door head as a concentrated reaction.

11-50


7.3.1.2


93

m

0662949 0508929 855

m

Examde 11.1-2 Cont'd. and

Code Reference

Calculations

Reactions at roof and foundation due to door load

1

:. OK

Calculate Maximum Moment Maximum moment is located at the concentrated load location (by inspection at head location)

M

= 187 lb X 10 ft = 1,870 ft-lb

Assume wall area above door opening spans horizontally to jamb masonry strip.

Reactions Due to Loading From Wall Area Above Door Opening

RF

= 100

plf

2 16 ft

Total

=

800 lb

11-51


.: OK

A C 1 TITLESRDG 93

0662949 0508930 577


Code Reference

Moment at head location due to masonry loading above lintel.

M = 100 lb X 10 ft

= 1,000 Et-lb

Moment at head location due to uniform wind load on Wall Element Part 2

M

= 157

plf

X

10 fi - 20 psf

X

(lo *I2 2

=

575 ft-lb/ft

Assume 2 ft-8 in. effective strip (size of strip on opposite jamb). Moment:

M

= 575

M

=

ft-lb/&

X

2.67 ft

+ 1 O ,O O

ft-lb + 1,870 fi-lb

4,410 fi-lb

Estimate Reinforcement

Try 6 in. CMU

(Assume j’ = 0.9 for estimate)

As =

4,410 ft-lb X 12 h/ft 24,OOO psi x 1.33 x 0.9 x 2.8 in.

= 0.658

Try (2) #6, A, = 0.88 in2 11-52


h2

A C 1 TITLEStMDG 9 3 W Ob62949 0508933 403 W

and

Calculations

1-

2‘ 8”

I Effective Width Of Assumed Strip (Same Width As Isolated Pier)

Check Effective Width of Compression Area shall not exceed the least of:

W = 1/2 center to center bar spacing each side of bar W = 36 in. each side of bar side W = 3t each of bar W = 3 x 5.63 = 16.9 in. c 1 ft-8 in. assumed in design Move #6 Bar to next cell

I #6

I#6

I

Check Strength Design Moment 4,410 ft-lb 0.88 =

in?

32 in. x 2.8 in.

=

0.010

11-53


7.3.2.1

A C 1 TITLESMDG 9 3

m

Obb2949 0508932 3 4 T


Code Reference

Calculations

k

np = 0.158

j = 0.85

= 0.42

Appears compression limited because of low j value.

Mt = A, jdF, Mt = 0.88 in? X 0.85

X

2.8 in. x 24,OOO psi x

Mt = 5,570 ft-lb > 4,410 ft-lb

1.33 12 in./ft

.: OK

hd2 Mm = kjFb

2

Mm=

32

in*

x (2.8 h)2

2

x

0.42 x 0.85 x

Mm= 2,480 fi-lb < 4,410 fi-lb

1,500 psi 3

X

1.33 12 in./fi

Does not work

Options:

1.

Increase wallthickness to 8 in. CMU

2

Add pilaster

3.

Increase masonry strength

4.

Placedoublebars

to increase d

Since this is a local condition, Option 2 is probably the most cost effective. Option 4 may bedifficult to achieve in some areas where the ability Option 1 is probably the most expensive solution.

As an illustration, Options 3 and 4 will be investigated 11-54


of mason contractors islimited.


m

Ob62949 0508933 2 8 b

m

Examde 11.1-2 Cont'd. Calculations and Discussion

Reference

Code

Option 3: Assume newf, 49410*-lb 2,480 fi-lb

Estimate:

X

1,500 psi

=

2,670 psi

Tryfm = 3,000 psi Use Type S Mortar Unit Strength for 3,000 psi f, needs to be determined.

By Specs. prism test assume 4,800 psi unit strength required.

Specs. 1.6.3

New Modulus:

O '0

1

psi (3.2 - 2.9) 106 psi 1 O ,O O psi

n = - 29

=

=

3.14 x lo6 psi

9.17

3.14

np = 9.17 x 0.0098 = 0.090

k

=

j

0.34

=

0.88

Mt = A,FJd Mt

=

0.88

h.*

x 0.88

Mt = 5,770 ft -lb

X

2.8 h. X 24 O , OO psi

.: OK

11-55


X

1.33 12 k/ft

Table 5.5.1.3

A C 1 TITLE*NDG

93

m

Obb2949 0508934 L12

m


Code Reference


bd2 M m = -kjFb

2

M,,, = 4,160 ft-lb

4,410 ft-lb

Margin of Safety = -6% This does not satisfy the Code; however this mayor may not be acceptable depending on local structural design practice. One method to further increase compression capacity isto reduce the modulus of elasticity. Thus, by using Type N mortar instead of switching to Type S may make it work: assuming

Pmdoes not change. Em

=

I

2.6 +

k

=

1

psi (2.8 - 2.6) 106 psi

1,OOO psi

0.36 j

=

=

2.76 x lo6 psi, n

=

10.5, np

=

0.10

0.88

Mt 5,770 ft-lb

:. OK

M,,, = 4,400 ft -lb

.: OK

Another method to increase the compression capacity is to add more steel. This

is not

usually efficient, but to correct for the 6% deficiency may be the best approach. Note, the estimate did not work because of the change in the value of n. Try (4) #6.

11-56


A C 1 TITLExMDG 93

m

Obb29Y9 0508935 059

m


Code Reference

L Neglect

Increase In Effective Width

=

np

1.76 i a 2 32 in. x 2.8 in.

=

0.180

k

=

=

0.44

Mt = 1.76 in? X 0.85

X

Mm=

32

in*

j

=

0.85

2.8 in. x 2 4 O , OO psi x

1.33 12 h/f3

.-. OK

Mt = 11,100 fi-lb

Mm =

0.020

x (2.8 h>2

x

0.44 x 0.85

2

x 3,000 psi

3

X

1.33 12 h/ft

:. OK

fi-lb

Note: This will probably work on the other side of the door as well. 1 13/16 2 1 13116

,

~

2' 8"

Minimum bar spacing is 1 in. or the nominal bar diameter

11-57


8.3.1

,

A C 1 T I T L E * H D G 93

m

Ob62949 0508936 T95 D

Example 11.1-2 Confd.


Code Reference

1 1/2 in.

Minimum is bar cover

8.4.l(a) 1“ Min. use (2) t5 1 112” Min.cover o c .

1 13/16”

Minimum thicknessof grout between the masonry and thebar for coarse grout is 1/2 in. 1 13116

8.3.5

112”

Select (4) #5 spaced as above. Neglect bar closest to compression face

As = (2) x 0.31 h2= 0.62 h2 d

P =

0.62 h2 32 in. x 3.81 in.

np = 0.082,

R

=

=

0.33,

0.005,

= 3.81

in.

It = 16.1

j = 0.89

Mt = As jdFs

Mt = 0.62

in2

x 0.89 x 3.81 in. x %,O00 psi x

Mt = 5,580 ft-lb > 4,410 ft-lb

.: OK

11-58


1.33 12 in./fi


93

m

Obb2949 0508937 921

m


and

Code Reference

Calculations

bd2 M m = -kjFb 2

:.

M m = 3,780 ft-lb 4,410 ft-lb

Does not work

This approach was not successful. By increasing f, to 2,000 psi, it should work. Stay with Type N mortar. This keeps the modulus lower and favors compression stress. Table 1.6.2.2

Unit Strength 3,050 psi

1

50

n

=

np

(2.6 - 2.3) lo6 in./in.

2.3 x lo6 h/h.

12.6 =

0.064,

k

=

0.30,

j = 0.90

Mt = A, jdF, Mt = 0.62 in? X 0.90 X 3.81 in. X %,O00 psi Mt

=

5,660 ft-lb 4,410

ft-lb

:.

OK

11-59


x

1.33 12 in./ft

Table 5.5.1.3

A C 1 TITLE+MDG 93

m

Ob62949 0508938 8b8

m


Code Reference

bd2 M m = -kjFb 2

M,,,= 4,630 ft-lb > 4,410 ft-lb

:. OK

For shear considerations see MDG Example 13.1-9

11-60



73

m

Obb27470508937

7TY

For the interior wallon Grid Line 2 of the TMS Shopping Center (Wall Construction Option A) design a hollow unreinforced 8 in. concrete masonry wallfor out-of-plane flexure. fn= r 1,500 psi Unit Weight = 46.5 psf (see MDG Appendix A)

Type N Mortar (PCL)

and

Calculations Since a bar joist is placed alongeach side of this wall,no roof load istransmitted to the wall. The wall span for bending due to seismic effect is 16 ft. Although the wall extends two ft above the joist bearing elevation, this portion is not being included in calculations to be conservative. The axial force at midpoint between lateral supports is:

P

= (8 ft)(46.5 psf) = 372 plf

The seismic loading from ASCE 7-88 is:

Fp = ZICpWp

Z

= 3/16 for Seismic Zone 1

I = 1.0

See MDG 9.1.1.2

Cp = 0.3

(interior partition normal to flat surface)

Wp = 46.5psf

Fp = (3/16)(1)(0.3)(46.5 psf) = 2.62psf

11-61



m

0662949 0508940 4Lb


Code Reference

The associated bending moment and shear force are:

M,, = 1/8 Fp H1 = l/8 (2.62 p~f)(l6ft)2 = 83.8 ft-lb/ft V,, = 1/2 Fp H = 1/2 (2.62 psf)(16 ft) = 21.0 plf Assuming face shell bedding and using MDG Appendix A, the calculated stresses are:

M f"'s-

- (83.8 ft-lb/ft)(l2 h./ft)

81

=

12.4 psi

The allowable stresses assumingf', = 1,500 psi are:

Fb = 1/3f', = 1/3 (1,500 psi) = 500 psi

6.3.l(c)

The h/r ratio is: 5.1.3.3

Note that A is the average area of unit - not face shell area

h/r = (16 ft)(12 in./ft)/2.73 in. = 70.3 < 99 Hence,

11-62



m 0662947 0508943 352 m



Fa

Reference

'f:( 3 7 4

=

[l - 14Or

Code

Eq. (6.3)

Fb, = 19 psi for Type N mortar

6.3.1.1

For axial comparison for flexure, 6.3.1

:. OK Hence, thisrequirementissatisfied.Notethattherightside

of the

interaction equation could be increased to 1.33 since seismic loading is included in the load combination.

5.3.2

An additional requirement is

P+", 1

4

where P,

=

7t2 EJ

h2

e = O;

11-63



93

m 0662949 0508942

299 W


Calculations

Assuming 2,000 psi net compressive strength of units

E, = 1.8 x 106 psi

5.5.1.3

Then, 372 plf <

f (148,900 plf) 4

=

:. OK

37,230 plf

Although the above is technicallycorrect, a minimum elr is advisable in lightof construction imperfections not otherwise anticipated by analysis. Suggest e = 0.U = 0.763 in.

I = 309 in4

.: OK

:. P, = 87,800 plf > 372 plf Hence, this requirement is also satisfied: The net tensile stress at mid-height is f,,/net = 12.4 psi

- 12.4 psi

= 0.0 psi

11-64


A C 1 TITLExMDG 93

m

Obb2949 0508943 L25

m


Calculations

Fb, = (19 psi)l.33 = 24 psi

6.3.1.1

Hence fbbet

Fbt

Hence, this requirement is satisfied. For shear considerations see MDG Example 13.1-6 Therefore, theinterior nonloadbearingwall issatisfactorywith respect to out-of-plane flexure if constructed of 8 in. concrete block units with Type N or S mortar.

11-65


Example 11.1-4

-

DPC Gymnasium Design of an Unreinforced Multiwythe Brick-Block

Noncomposite (Cavity) Wall for Flexure Only

For thematerial properties assumed below, designthe west wall on Grid Line 1of the DPC Gymnasium as an unreinforced multiwythe brick-block noncomposite (cavity) wall, Wall Construction Option A, for flexure only. Concrete BlockMasonryClayBrickMasonry 2,ooo

6,OOo

Mortar (PCL)

Type S

Type S

2,500

1,500

Unit Strength (psi)

2.2 x 106

1.9 x 106

(Code Table 5.5.1.3)

(Code Table 5.5.1.2)

Design Wind Pressure = 20 psf

~

~~

Code Reference


The multiwythe wall will be constructed with a concrete block wythe of 8 in. nominal width, a clay brick of 4 in. nominal width, and a 3 in. cavity. In addition, only face shell bedding will be assumed for the block wythe.

The roof truss and wall arrangement are shown in MDG Fig. 9.1-4. As there is a roof truss located adjacent to the wall under consideration, it is treated as a nonloadbearing wall subjected to wind loads only. The dimensions of the wall along with the locations of the movement joints andpilasters are shown in MDG Fig. 9.1-5 as West Elevation. The expansion joints in the clay brick masonry

11-66


~~

A C 1 TITLEsMDG 9 3

m

Obb2949 0508945 T T 8 D


Code Reference

Calculations

and the control joints in the concrete block masonry are assumed to be located at the same points. The two wythes are assumed to be sufficiently well connected with wall ties so that the wind load is transferred directly from the brick wythe to the block wythe through the wall ties. The expansion joints in the brick wythe are unable to transfer any shear force. Therefore the total wind load is assumed to be resisted by the concrete block wythe, since the control joints have the capability to transfer shear. Considering pilasters to act as supports, a one foot strip of the wall just above the door spanning horizontally will be designed.

A'

c C' "

D' D

11-67


A C 1 TITLE*MDG 93

m

0662949 0508946 9 3 4

m


Calculations

Discussion

Reference

Code

Structurally, this horizontal strip of the wall can be treated as a continuous beam (with hinges) resting against pilasters. The actual beam structural system along with its division into subcomponents for analysis purposes are shown in the above sketches. The moments in the statically determinate Sections A’CC’ and C’D’, shown above, can be easilycalculatedfromsimplestatics.

The

moments in the statically indeterminate Section D’DEB’ have been computed by moment distribution. The maximum moment values in these sections are 4,610 in.-lb, 5,330 in.-lb,and 4,570 in.-lb, respectively. Thus, M = 5,330 in.-lb is utilized for the wall design. Allowable Tension Stress:

As the tension is parallel to the bed joints in hollow block masonry,the

6.3.1.1

maximum allowable tension from Code Table 6.3.1.1 is = 50 psi. This

5.3.2

value is increased by 33% due to wind moments to yield

Fa = 1.33(50pi)

=

66.5psi

Actual Maximum Tension: From MDG Appendix A, the value of I for a 8 x 8 x 16 concrete hollow block with only face shell mortar bedding = 309

11-68



Code Reference


fa

=

65.8psi < Fa

=

66.5psi

:. O.K.

The above computations assumed the total wind load carried by the block wythe. If found to be deficient one might then examine the wall to span vertically and distribute the wind load to each wythe based upon their respective flexural stiffness.

11-69


93


Example 11.1-5

DPCGymnasium

m

0662949 0508948 707

m

- Designof an UnreinforcedMultiwytheComposite

Wall for Flexure Only Given the following material properties,design an unreinforced multiwythe composite brick block wall (Wall Construction Option B) for Grid Line 1 of the DPC Gymnasium as given in MDG 9.1.2 The design isto consider flexureonly and is to consist of a 4 in. nominal face brick wythe, a 3 in. grouted collar joint and a CMU wythe. Material properties areassumed as follows:

Unit Strength (psi) Mortar (PCL)

Concrete Block

Clay Brick

Masonry (Hollow)

Masonry

Grout

2,000 Type S

6,000

N.A.

Type S

N.A. 2,000 cf2 L O X 106(Code

S m

(psi)

1,500

23O0

E m

(psi)

2.2 x 106 (Code

1.9 x 106(Code

Table 5.5.1.3)

Table 5.5.1.4) 5.5.1.2)

Calculations And Discussion

Reference

Code

In this design the wall contains no pilasters and is considered to span vertically. Although expansion/control joints are present in this design, they do not affect the wall flexurally. According to MDG 9.1.2, a uniformly distributed wind load of 20 psf is thegoverning lateral load in this design. The critical one ft section is taken at the center of the wall (see MDG Fig. 9.1-5) and is considered to be pinned at the top and bottom. The maximum moment at the midspan is given by

M = -wz2 - - (20Psf)(12~/ft)(29*33~)2 = 25.8 h-kipslft& 8

8

11-70


length


Calculations

and the maximum shear, V,near supports equals

The west wall is parallel to the roof trusses and does not resist any gravity loads other than its own weight. In thisdesign, the wallis considered as a simple flexural element and the minimal amount of axial compression is neglected.

The allowable compressive stress

in unreinforced masonryismuch

greater than the allowable tensile

stress; thus, for wall systems in simple flexure, tensile stress is the critical design composite walldesigns

theshear

the allowable flexural

parameter.

In addition, for

stress at the wythe-collar joint

5.8.1.2

interface must be checked to insure that its value does not exceed 10 psi for grouted collar joints. In walls designedfor composite action,the stresses are tobe computed using section properties based on the minimum transformed net crosssectional area of the compositemember.

The generally accepted

transformed area concept for elastic analysis, in which area of dissimilar materials are transformed in accordance with their relative elastic moduli ratios, shall apply. For the design example, trya multiwythe composite wall composedof a 4 in. facebrickwythe,

a 3 in. grouted collar joint, and a 12 in.

concrete block wythe (50% grouted). The position of the neutral axis from an axis through the center of block is found by equating the first

11-71


5.13.1.2

A C 1 TITLESMDG 73

m

Obb27470508750

365 W


Reference

Code

moment of the transformed areas about this axis as follows:

i

r

CMU 12"

I

1

1S ' '

8.63"

I 1.5"

I

I

-R

"

3.63"

Grouted Collar Joint

Clay Brick

Transformed Area for Composite Wall From MDG Appendix A, for hollow block withfull mortar bedding, the area and'I for one ft length are equal to 57.8 in2 and 1070in.4,respectively.Using

these values and the

dimensions of the transformed section shown above, the centroid distance

11-72


2 is given by

A CT1I T L E r M D G

0662949 0508953 2 T L

93

m


Calculations

5.45in. x 3 h x( 14*63h.) + 10.4h x3.63hx( 11.63h +3.63h+3h 2 2

-

r =

x =

or

519h3 120h3+399h3

=-

57.8in.2+ 18.6in.2+ 16.3id +37.6in?

-

180m.2

3.98in. belowC.G.ofblockwythe

X =

.;

Reference

I,

= 107Oin.'+57.8in?x(3.98h.)2

+2.16h. ~(8.63in.)'/12 +2.16in.~ 8 . 6 3 h ~ ( 3 . 9 8 h - y

(

+ 5.45 in.x (3 h.)3/12+ 5.45 m. x 3in.x 14'6;3h. -3.98h

r

+ 10.36in. x(3.63hJ3/12

+ 10.36in. x3.63h. ~ ( 1 0 . 6 3 h -3.98in.y

I,

or

I~

=

=

1O7Oin~+916in!+115h4+295in!+l2h4+182h.4+41in!+166oin!

4,280

in."

The distances of the extreme fibers from the neutral axis of concrete block and clay brick can be calculated as 9.79 and 8.46 in., respectively. The maximum tensile stresses normal to bed joints in concrete block and clay brick masonry, assuming compression and suction, can be computed as follows. For Concrete Block

For Clay Brick


equal values for wind

A C 1 TITLExMDG 93

m

Ob62949 0508952 L38

m


Discussion

Calculations

Reference

Code

The maximum allowable flexural tension for solid clay brick masonry and 50% grouted concrete block masonry (by

interpolation) are taken

6.3.1.1

from Code Table 6.3.1.1 for tension normal to the bed joints. These values are increased by 33% for wind

to yield the following allowable

5.3.2

tension stresses: For Concrete Block

For Clay Brick

F&

=

1.33(40pi) = 53.2psi > 44.Opsi

As these maximum allowable tension magnitudes are larger than actual tension, design is safe for flexure. Shear Stress at Collar Joint Interface For the composite action to be considered, the Code limits the shear stress at the wythe - collar joint interfaces to 10 psi for grouted collar joints. The shear stress is computed by the formula, Code Eq. 6-7, fv

=

VQ

where the standard meanings apply to the symbols used. The critical section is at the block - collar joint interface. Using the figure shown earlier in this example,

11-74


6.5.1

A C 1 TITLE*NDG

93

m

Ob62949 0508953 074

m


Calculations

Code Reference

Q = 10.4in. x3.63in.X

Q

=

.:

fv

11.63ia +3.63h +3h. -3.98h

25Oh3+54.5h3 = 304h'

=

2931bx304h3 = 3.82psi 4,280in.4 x 5.45 in.

lopsi, F,, :. OK for shear

Since the brick and block wythes are not bonded by headers, minimum 5.8.1.5

wall ties must be provided per Code.

11-75


Example 11.1-6

-

DPC Gymnasium Design of a Reinforced Multiwythe Composite Wall for Flexure Only

Design a reinforced composite brick-block wall (Wall Construction Option C) for thewall on Grid Line 1 of the DPC Gymnasium. Since this is

a nonloadbearing wall, the design

considers flexure only. The wall is composed of a nominal 4 in. clay brick wythe,

a 2 in.

grouted collar joint, and a nominal 8 in. concrete masonry wythe, to create anoverall 13.25 in. thick wall. The reinforcement is located in the collar joint. The material properties are assumed as follows: Concrete Block

Clay Brick

Masonry (Hollow)

Masonry

Grout

2,OOo

8,OOo

N.A.

Type N

Type N

N.A.

1,500 1.8 x 106 (Code

2,m

2,000 v g psi)

2.0 x 106 (Code

1.0 x 106 (Egpsi)

Table 5.5.1.3)

Table 5.5.1.2)

Unit Strength (psi) Mortar P m

(psi)

E m (psi)

16.1

It

14.5

Reinforcement - Grade 60, E = 29 x 106 psi


Reference

Code

The wall on Grid Line 1 of the DPC Gymnasium spans vertically between pin supports at the top of the foundation wall and at the roof diaphragm. A one footwide section located at the center

of the wallgoverns the design. The moment at midspanis25.8in.-kips,

considering a lateral wind pressure of 20 psf (see MDG Example 11.1-5). Initially, wind pressure is considered to act in a direction that causes compressionin the brick wythe. The wallwill also be checked for the wind suction case, where

11-76


the inside face shell of the

AC1

TITLExMDG 93 m 0 b b 2 9 4 9 0508955 947 m


Calculations

concrete block is in compression.Determination of wind pressure and suction in accordance with ASCE 7-88 islikely to result in design suction forces

greater than design pressure

forces. However, for simplification of calculations in this Guide, the design pressure and suction are considered to have the same value. The required steel area can be calculated from the internal couple shown below:

i

1.25”

7.63”

13.3“

3.63”

Wind Pressure Case Estimating the initial value o f j as 0.9, and taking the allowable tensile stress, F,, in steel (increased by 33% from 24,000 psi due to wind) as

11-77


5.3.2

..

AC1

.

~

TITLExMDG 93 9 Obb2949 0508956 883

9


and

Code Reference

Calculations

gives psi,32,000

the required steel area as

A,

=

7.2.1.1

25,800 h-lb = 0.194 (32,000 psi)(0.9)(4.63in.)

h2/ft

of wall

Check if maximum allowable masonry compression stress, Fb, controls:

Also, from the internal couple

for j = 0.9, k = 0.3

fb =

2(25,800 h -lb) 12 in.(0.3)(4.63 in.)(0.9)(4.63 in.)

=

Therefore steel stress governs. For A, req'd = 0.194

11-78


745 psi

Fb = 889 psi

7.3.1.2

A C 1 T I T L E r M D G 9 3 W 0662949 0508957 7 1 T

m


and

Calculations

Code Reference

25,800 ia-lb (0.207 in?)(32,000 psi)(4.63

j = - -

A,F,d

R

=

3 - 3j

=

3

-

3(0.842)

=

= 0.842

in.)

0.479

2(25,800 h-lb) = 504 psi - (12 in.)(0.479) (4.63 in.)(0.842)(4.63 in.)

2M fb =

f b = 504 psi < Fb

=

:.

889 psi

OK

Verify:

T = C

or (0.207 in?)(32,000 psi) 6,625 lb

Mt = Tjd

=

=

504 psi (12 in.)(0.479) (4.63 in.) -

(J

:. OK

= 6,630 lb

6,625Ibs(0.842)(4.63

in.)

=

11-79


25,800 in.-lb

.:

OK


93

m

Obb2949 0508958 b5b

Examde 11.1-6 Cont'd.

m

-

Code Reference


Wind Suction Case

fbl

f

t = 1.25"

f

b = 12"

-A

d = 8.63"

7.38"

L

n .As = .207(16.1) = 3.33 in2

The transformed sectionis considered with face shellmortar bedding only inCMU masonry.

For values o f j and k

n AJd - kd) = bt(kd ot

(16.1)(0.207)(8.63

- 8.63k) = 12(1.25)(8.63k

or 28.70 - 28.70R

from which, k

=

0.241 and j

i)

= 1 -

=

-A 3

2

129.0k - 9.38

=

0.241 = 0.920 1 -3

Assuming that CMU reaches the maximum allowable stress, Pb, first

11-80


7.3.1.2


Cade Reference


5.3.2

From the sketch on the previous page,

1.25

or fb2 =

(' - cd)

fbl =

- 0.601)fbl

fb2 = ( 1

.: c

:.

=

226 psi

+

2

=

]

1.25 in. - (0.241) (8.63 h) f b l 0.399fbl = (0.399)(667 psi) = 266 psi

['

+(1.25 in.)(12 in.) = 7,000 lb

Steel governs, and the assumption thatthe CMU reaches its maximumallowable stress first is incorrect.

Try letting the steel reach its allowable stress first:

Since the position of the neutral axisis unchanged, the masonry stresses can simply be determined by proportion:

f, = F, = 32,000 psi 11-81




33,800 psi

Reference

=

631 psi

fbz = 266 psi (33,800 32’000 psi) = 252 psi

667 psi

:.

OK

667 psi

:,

OK

T = A,

=

(32,000psi)(0.207 h.?= 6,630 lb

Mt = Tjd

=

6,630 lb(0.92)(8.63 h.)

Mt = 52,600 in.-lb > Me

..

<

=

25,800 in.-lb

:.

Code

OK

Provide #S @ 18 in. O.C.

NOTE:

As discussed in MDG Chapter 9, it is possible that composite brick-block walls may experience delamination at the collar joint, due to shear from differential movement, in-plane loads, and out-of-plane loads.

The calculationsofthis

example are valid only if such delamination does not occur.

11-82


A C 1 TITLE8KMDG 93

Example 11.1-7

0662949 0508961 L40

-

DPC Gymnasium

m

Design of a Single Wythe Reinforced

Nonloadbearing Hollow Clay Masonry Wall for Flexure Given the following material, design a single-wythe reinforced hollow clay masonry wall for the wall on Grid Line 1of the DPCGymnasium (Wall Construction Option D). The design is to consider flexure only. Reinforcement is placed in the cells of the hollow units. MasonryBrick

Clay

Grout

Unit Strength (psi)

6,000

NA

Mortar

Type S 2,500

NA

P m

(psi)

39800 G)

x 106

1.9 x 106 (Code Table 5.5.1.2) 1.9

E (Psi) n

15.3

Steel Reinforcement Grade 60, E = 29 x 106 psi

and

Calculations The wall spans vertically. A one-foot section located at the center

of the wall w ill be

designed. The wall is assumed to be pinned at the top and bottom. For a vertical span of 29 ft

-

4 in., the midspanbendingmoment

is25.8in.-kips/ft.,

considering a lateral

compressive wind load of 20 psf (see MDG Example 11.1-5). For simplicity, it is assumed that the grout has the same stiffness as the clay masonry, Sm.A 6 in. thick x 4 in. high 8 in. long nominal hollow clay unit will be used with Grade 60 reinforcement. From

the

geometry

d = 5*50 in. = 2.75 2

of

the

units, shown below,

the

effective

depth

is

in. The first step is to estimate the required steel area for design.

11-83


A C 1 TITLElKMDG 9 3 M 0662749 05089b2 O87

m

Example 11.1-7 Cont’d. ~

CalcuIations and Discussion

~~

Code Reference

From the preceding sketch of the internal couple,

Assuming an initial value for j = 0.9, and increasing the allowable

7.2. l. 1

tensile stress in steel (24,000 psi) by 33% for wind, the required A, can

5.3.2

be computed as:

A,

25,800 h - l b (24,OOO psi x 4/3)(0.9)(2.75 in.)

=

=

0.33

in2/ft

of wall

Check if allowable masonry stress controls: Because of the 33% increase due to wind, the allowable stress in brick,

Fb, is given as

7.3.1.2 (2,500 psi)

or

5.3.2

Fb = 1,110 psi 11-84


TITLE*NDG AC1

93

m

0 b b 2 9 4 9 0 5 0 8 9 b 3 T13

m


I

Calculations

For the assumed value of j = 0.9, the corresponding value of k is 3 - 3j = 3 - 3(0.9) = 3 -

2.7 = 0.3. For this value of k, the compressive stress in masonry is fb=”

2 (25,8OO in -lb) in.(0.9)(0.3) (2.75

2M bjkd212

=

2,110 psi

>

Fb =

1,110 psi

in.)2

Therefore, for the previously assumed values ofj and k, masonry compressive stress controls. Continue the problem assuming that

fb

= Fb = 1,100 psi.

Let

2M

- Fb

”

=

1,100 psi

bjkd’

1,100

k(1 -

2(25,800 in.-lb)

=

$) k

= 0.51

=

0.66

j = 1 - -

=

0.78

3

Bystraincompatibility,

I

f, =

(y).. 11-85


A C 1 T I T L E r f l D G 93 D 0662949 0508964 95T


-

0.66

= (

)

Code Reference

(15.3)(1,110 psi)

=

8,750 psi

Fs = %,O00 psi

Therefore, the assumption that compression controls is correct. 25,800 h.-lb = 1.37 h2/ft - " fJd (8,750 psi)(O.78)(2.75 h) -

"

Use 1 #6 bar in each cell of the hollow clay units.

A,

=

3 x 0.44

in? =

1.32 in."/ft

E

1.37 h2/ft

Note that this is a heavily reinforced wall. An alternative design would involve a thicker masonry unit and less reinforcement.

11-86


A C 1 TITLEsMDG 73

m

Obb2947 05089b5 89b

m

RCJ Hotel - Design of aReinforcedClay Brick Lintel

Example 11.1-8

For thematerial properties given below, designa single-wythe reinforced clay brick lintelfor the service entrance on Grid Line B of RCJ Hotel (Wall Construction Option B, Building Construction Option II; see MDG Fig. 9.1-11). The design is to consider flexure only. Masonry

Brick

Clay

Grout

Unit Strength (psi)

~,ooo

N.A.

Mortar

Type S

N.A.

T m (Psi)

2,500

3,800 cfg) 1.9 x 106

x 106 (Code Table 5.5.1.2)

E 1.9 (Psi) n

15.3


and

Calculations

Discussion

Reference

Code

The lintel will be built with clay bricks of dimensions 7 1/2 in. thick x 3 1/2 in. high x 11 1/2

in. long. The floor height in the Building Construction Option II is 8 ft - 10 in. Reduction of 7 ft - O in. for the height of the service entrance gives total lintel height of 1 ft

- 10 in.

The entrance is 6 ft - 8 in. wide. Allowing for 8 in. bearing at each end (theCode requires a minimal of 4 in., see Code 7.3.3.3), the center line span, I, to be used for calculation is

The dead load and live load for RCJ Hotel are given in MDG 9.1.3.1 as 110 psf and 40 psf, respectively. The corresponding self weight of the wall from the same MDG section is 70 psf. It is conservatively assumed that the floor above the kitchen (See MDG Fig. 9.1-6) is supported along Grid Lines B and C. In addition, the lintel only supports the wall load defined by 45" angles from its supports due to the arching action. This wall load will be

11-87


AC1

TITLE*IDG 93 m 0662949 0508966 7 2 2 m


and

Code Reference

Calculations

assumed to act at the midspan of the lintel.

:.

Floor & a d + live Load

(1 10 psf

=

+

40

psf)(y)

1 Self weight of 7- in. x 22 h lintel =

=

(140 Pd)

2

=

W =

I

-I

7.33' Lintel Span = 1

M - - -- +w12 8

Pl 4

- 2,410 plf(7.33 hJ2(12 hm)

940 Ib(7.33 in.)(12 4(1ow 194.0 & k i p s+ 20.7 h-kip = 215.0 h-kips +

wow

M-

=

V-(ìnsidè supprts)

V-

6.67 ft

=

2,410

=

8,040lb + 470 lb

+

940 lb -

=

8,510 lb

2

11-88


2,410plf

7

Self Weight of wall above lintel (see sketch)

P = i(7.33 f i ) (7.33 ft)(70 prf) 2 P = 940 lb

2,250 plf

hm)


0662949 0508967 669 W



Reference

For Grade 60 steel, allowable stress F# = 24,000psi. depth, d , can be conservatively assumed

The effective

as (22 in. - 2 in.) = 20 in. 7.2.1.1

Using the sketch of moment resisting couple shown in MDG Example

11.1-7, the required steel areasis given by

in which j will be initially estimated as 0.9

.: A,

=

215 h.-kips(loOO lb/ldp) (24,000 psi) (0.9)(20 in.)

=

h2

Recompute k for steel provided: First moments of area about NA are equal

11-89


Code

AC1

TITLESMDG 93 W 0662949 0508968 5T5 W


Code Reference

.. or (7,5)(2Ok)(lOk) = (15.3)(0.613)(20)(1 - k) or 1500k2 = 188 - 187k or k2 + 0.125k - 0.125 = O Solving the quadratic equation yields

k = 0.296 .: j = 1 - - =k1 - - = 1 -0.2% 0.099=0.901 3

3

CheckingSteelStress:vs.

F,)

Check Stresses in Masonry:

&, vs. F b )

fb="

2M 2(215 h.-kip~)(lOOO) bkjd' (7.5 in.)(0.296)(0.901)(20 in.)2

CheckforShearStress:

vs. F")

From Code Es. 7.3, shear stressf v is given by

11-90


7.5.2.1

AC1

TITLExMDG 93 m Ob62949 0 5 0 8 9 b 9 431 m



V

fv =

which in

bjd

for computing maximum

shear stress, V is computed at d/2

from the face of the support. The shear

d from face V8 2

.. *

fv

=

=

stress shall

8330 lb - 2,410 plf

Check for Bearing Stress:

('fb

not exceed

( ly:m)i

6,320 lb = 46.8 psi (7.5 in.)(0.901)(20 in.)

7.5.5

50 psi

=

7.5.2.2(a)

6,320 lb

:.

OK

vs. Fb)

5.12.3

The maximum allowable bearing stress is given as

0.25(f',)

= 0.25(2,500 psi) = 625 psi

End reaction

=

2,410 plf

"1 y

-+-

r3: R = 8,830 lb + 470lb

=

lb 9,303 lb

For the bearing length on each end of 8 in. assumed earlier,

Bearing Stress

=

9,303 lb

(7.5 in.)@ in.)

=

155 psi < 625 psi

11-91


:.

OK

AC1

Example 11.1-9

TITLEIMDG 93 m 0662949 0508970 L53 m

- UnreinforcedRetainingWallDesign

RCJHotel

for Outsf-Plane

Flexure For theretaining wall shown on Grid Line A of the RCJ Hotelin MDG Fig 9.1-6determine

if an unreinforced concrete masonry wall is sufficient.

S,,, = 1,500 psi

Weight of masonry (12-in. thick

- solid grouted) = 100 psf

Type N mortar (PCL)

Running bond Active earth pressure coefficient K =

Weight of soil y = 100 pcf

and

’I’S

Calculations 113’’

4

Elevation 104’-O’’

Unmeinforced Sidewalk Elevation

H

\

‘Reinforced Concrete

1.

Footing

Stem Height

For purposes of setting the stem height, assume that the footing will be 12 in. thick. The bottom of the footing should be below the frost line. In this example, it is assumed that 30

in. of cover above the bottom of the footing is sufficient to place it below the frost line.

Height of stem

= 104.0

ft - 99.5 ft + 30h’ - 1 f t = 6 f i 12 in./ft 11-92


A C 1 TITLE*MDG

93

m

0662949 0508973 09T

m


Calculations

2.

StemDesign

- for a 1ft - O in.long

strip ofwall

The allowable flexural tensile stress normal to the bed jointswith Type

N mortar is 58 psi. At

the base

of the wall, where the interface

6.3.1.1 is

masonry to concrete, this allowable value may not be applicable. However, no Code values are given for this condition and the Code allowable value for masonry to masonry interface will be used. Section modulus of 1 ft long strip of wall =

(12in.)(11.63 in.)* 6

=

270

ia3

Pressure at base of stem = 200 psf =

600 lb

Moment at base of stem =

Check shear stress in stem

f v‘Q=-x3‘ -forz

rectangular a

section

11-93


6.5.1

A CT 1I T L E l r M D G

0662949 0508972 T26 W

93


and

Calculations

Discussion

600 lb fv

=

3.

=

4.3 psi < 1.5

=

58.1 psi :. OK

(11.63 in.)(12 in.)

1.5 d m

Code Reference

g

=

6.5.2

Joint Reinforcement

Use joint reinforcement consisting of No. 9 longitudinal wires and No. 9 cross rods in each mortar bed for crack control.

4.

Stability and Soil Pressure Check

The dimensions of the reinforced concrete footing should be set to prevent overturning and/or sliding of the wall. In addition, the pressure under the toe of the footing must be smaller thanthe

allowable pressure for the type of soilinvolved.

Referto

textson

geotechnical engineering or concrete design for examples of these calculations. 5.

Footing Design

The reinforced concrete footing shouldbe designed in accordance with the provisions of the latest edition of ACI-318.

11-94


A C 1 T I T L E S M D G 9 3 9 Obb2949 0508773 7b2

Example 11.1-10

RCJHotel

- ReinforcedRetainingWallDesign

m

for Out-of-Plane

Flexure For the retaining wall shown on Grid Line A of the RCJ Hotel in MDG Fig 9.1-6, design the necessary wall size and reinforcement.

Sm = 2,500 psi

8 in. hollow clay brick (7.5 in. x 3.5 in. x 11.5 in.)

Type S mortar

Running bond Active earth pressure coefficient K = 0.45

Weight of soil y = 100 pcf

n = 15.3

Grade 60 steel (Fs= 24,000 psi)


Reference

A7*"&Elevation

Code

104'-O"

yK,h = (100)(0.45)(6.0) = 270 PSf

Reinforced Brick

H

Reinforced Concrete Footing

1.

Stem Height

The stem height for this problem will be the same as for MDG Example 11.1-9. 2.

StemDesign

- For a 1 ft-O in. long strip ofwall

Lateral pressure at base of stem = 270 psf

11-95


m

A C 1 TITLErMDG 93

Obb29Y9 050897Y & T 9


and

Code Reference

Calculations

Momentatbaseofstem

= (810

d =Assume 3.75 in.

3

)'

=

P =

np

k

=

=

19,400 h-lb/ft

j = 0.875

A , % -" 19,400 h -lb/ft Fsj d (%,O00 psi) (0.875) (3.75 in.)

Try No. bars 5

1,620 ft-lb/ft

@ 12 in.

0.31 in? (3.75 in.)(12 in.)

0.247 in?/ft

1 11.5"

A, = 0.31 in.2/ft

O.C.

=

=

0.007

(15.3)(0.007) = 0.105

-4 =

- It p

=

40.1052 + (2)(0.105)

- 0.105 = 0.365

Check steel stress

Provide matching dowels from wall to footing. Check masonry stress 7.3.1.2

11-96


AC1 TITLEtHDG 9 3 9 0662949 0508975 735 m


and

Calculations

Reference

-

"=bf

jkbd2

:. OK

(2)( 19,400 in.-lb) = 720 psi < 833 psi (0.878) (0.365)(12 in.)(3.75 i n . ) '

Check shear stress Allowable shear stress Fv fv =

v -

bjd -

=

6

=

@ÖÖ

810 lb (12 in.)(0.878) (3.75 in.)

=

=

7.5.2.2

50 psi

20.5 psi

50 psi

.: OK

7.5.2.1

Shear steel is not needed.

3.

Horizontal Steel

Although not required by the Code, it is advisable to provide somehorizontal steel. Provide a bond beam at the top of the wall,reinforcedwith

1 No. 5. Provide standard joint

reinforcement every 16 in. for crack control. 4.

Stability and Soil Pressure Check

The dimensions of the reinforced concrete footing should be set to prevent overturning and/or sliding of the wall. In addition, the pressure under the toe of the footing must be smaller than the allowable pressure for the type ofsoilinvolved.

Refer to textson

geotechnical engineering or concrete design for examples of these calculations.

5.

Footing Design

The reinforced concrete footing should be designed in accordance with the provisions of the latest edition of ACI-318.

11-97


A C 1 TITLE*MDG 93

- Designof

Example11.1-11RCJHotel

m

0662747 050897b b7L

an UnreinforcedMultiwytheNoncomposite

(Cavity) Brick-Block Masonry Nonloadbearing Wall for Flexure! Only For the material properties given below, designthe wall on Grid Line 3 between Grid Lines

F and G from MDG Fig.9.1-7

for RCJ Hotel. The wall is unreinforced multiwythe

noncomposite (cavity) brick-blockwall of Building Option II and Wall Construction Option

A, and is to be designed for wind loads only. Concrete Block Masonry Clay Brick Masonry Unit Strength (psi)6,000

2,000

Mortar

Type S (PCL)

S m

Type S (PCL) 2,5 O0

(Psi)

(psi) 2.2 x 106 (Code Table 5.5.1.3) Design Wind Pressure = 25 psf

1.9 x 106 (Code Table

Em


Reference

Code

The multiwythe wall willbe constructed of

8 in.wide nominal concrete block, 4 in. thick nominal clay brick with a 3 in. cavity separatingthe twowythes.

In addition,

f.

only face shell bedding will be considered

8"8"

Stair I < ,

for the block wythe. I

It is assumed that thewall rests against the stair and landings and accordingly spans in the vertical direction, with a span

11-98


5.5.1.2)

A C 1 TITLE+MDG 93 W Obb27g7 0508777 508 W


Calculations

length of 8 ft - 8 in. as shown in the figure. One foot strip of wall spanning in the vertical direction will be designed. For the design wind load of 25 psf, the maximum moment is

5.7.1

The totalmoment isdistributed to individual blockand brick wythes in proportion to their bending rigidities as shown in MDG Example 9.3-6. The moment of inertias of the block and brick wythes, respectively,can be taken from MDG Example 9.3-6 as

309 i n ! and Ibr = 47.6 in? The individual wvthe moments are

IM

=

MM = (2,820 in.-lb)

(1.9

106 in?-lb) 106 in?-lb)

M M = (2,820 h-lb) (679 (770

X

X

3 (2.2 x 106 psi)(309 h lo6 psi) (47.6 i a 4 ) + (2.2 x lob psi)(309 i n ! )

X

or Mbl

=

2,490 in.-lb

Ma

=

(2,820 h.-lb)

M&

=

(2,820 ia -1b)(9O.4 X 106 h2-1b) (770 X lob in?-lb)

or M ,

=

331

(1.9 x 106 psi)(47.6 i a 4 ) (770 X lob h?-lb)

in.-lb 11-99


1

A C 1 TITLESNDG 93

m

Obb2949 0508978 444

m


and

Code Reference

Calculations

Bending Stresses: The bending in the vertical direction produces tension perpendicular to the bed joints. In order to be conservative, the normal compressive

6.3.1.1

stresses due to the weight of the wall have been neglected. However,

5.3.2

a 33% increase due to wind loads yields an allowable flexural tension stress as:

Fa

=

(1.33)(25 psi)

=

33.3 psi for Concrete Masonry

Fa

=

(1.33)(40 psi)

=

53.3 psi for Clay Masonry

In Concrete Block f h MmC = Ï -

o (2,490 h-lb)

7.63 in.

=

30.7 psi

33.3 psi Fa .: OK

309 h4

In Clay Brick: f a Mbrc = Ï --

)

(331 h-lb)( 3.63 in.

=

12.6 psi < 53.3 psi

Fa .: OK

47.6 h4

=.

h unreinforced multiwythenoncompositebrick-blockmasonrywall

made of 8 in.

wide concrete blocks, 4 in. wide clay bricks with a 3 in. cavity is satisfactory for this design.

11-100


A C 1 TITLExMDG 9 3

Example 11.1-12

RCJ Hotel

I

Obb2949 0508979 380

- Design Reinforced of a Clay Brick Nonloadbearing Wall

I ~

for Flexure Only

For the material properties

Il

given below, design

the wallof MDG Example 11.1-11 for

Building Option I, Wall Construction Option B, as a hollow reinforced claybrick nonloadbearing wall for out-of-plane flexure only. Masonry Brick Clay Unit Strength (psi)

Grout 6,OOo

NA

Type S

NA

2,5O0

3,800 6 )

1.9 x 106 (Code Table 5.5.1.2)1.9

n

x 106

15.3


and

Calculations The stair and landing arrangements forthis wall are similar to those shown in the figure of MDG Example 11.1-11. except that the floor height in this building option = 9 ft - 8 in. The design wind load in this example is equal to 25 psf. The following assumptions are made in this reinforced wall example,

1.

The landings are connected tothe walls.

2.

The stairs are not connected to the wall.

3.

The wall above and belowlandings behaves as a strip and spans in the vertical direction between landings.

11-101



4.

Code Reference

The 7 ft wide horizontal portion of the wall spans horizontally between the vertical strips.

5.

The wallwill be designedfor flexure only.

Horizontal Steel Center to center of landing is considered as the span

Considering a one foot wide strip, the maximum moment is given as

M="

wz2

- 25

Theallowable

ft12(12 h*/fi)= 4,540 h-lb/ft 8

8

stress in Grade 60 steel is 24,000 which canpsi

increased by 33% for wind load, thus

'F

=

1.33(24,000 psi) = 32,000 psi

I

3.75"

brick wall, the effective depth, d as shown, is equal to 3.75 in. Assuming j = 0.9, the required steel area in the horizontal direction is calculated as

-" " 4,540 h.-lb/fi - 'F j d (32,000 psi)(0.9)(3.75 in.)

Use 1 - #4 @ 40 in., A, = 0.06 in.2/ft or

7.2.1.1 5.3.2

Considering a 71/2 in. thick hollow clay

As

be

2 - #3 @ 40 in., A, = 0.066 in.2/ft

11-102


=

0.042 in?/ft

AC1 TITLE*MDG 93 M Obb2949 0508981 T39 D


Calculations

or

Discussion

Code Reference

joint reinforcement Typically for seismic areas whole bars would be used with notched masonry units. In non-seismic areas joint reinforcement would be common.

Check k

7

or

(12)(3.75k)(

or

84.4k2 = 3.38 - 3.38k

or

k2 + 0.04k - 0.04 = O

k) = (15.3) (0.06)(3.75

- 3.75k)

from which

k

= 0.18

.. j = l - L = 1 - o*18 --- 1 3

3

:.

0.06 = 0.94

-

" 4,540 h.-lb Steel Stress, f, = A, j d (0.06i1~)~(0.94)(3.75 in.)

fs

=

PS = 32,000 psi

21,500 psi

Masonry Stress:

With a 33% increase due to wind

Fa

=

(i)

(1.33) - (2,500 psi)

=

1,110 psi 11-103


.-. OK

A C 1 TITLElrMDG 93

Ob62949 0508982 975

m


Code Reference

Calculations

fbh-"

2 " bkjd2

(4,540 in.-lb)(2)

(12 in.)(O.18) (O"(3.75

in.>"

Vertical Steel: Distance between landings = 9 ft - 8 in. It will be conservatively assumedthat thewind on half of 15 ft wall widthbetween Grid Line

F and G acts as a load on 4 ft wide wall strip spanning vertically between the landings. The strip will be assumed to be simply supported at its ends. Based on this assumption, Wind load on 4 ft wide strip = l5 *(Spsf) = 188 plf 2

Assuming j = 0.9,

AJreq'a

=

" 6,580 h.-lb Fsj d (32,000 psi) (0.9)(3.75 in.)

Use #4 @ 36 in., A, = 0.066 in.2/ft

11-104


=

0.061 h2/*

A C 1 TITLExMDG 9 3

m

0bb2949 0 5 0 8 9 8 3 B O L

m



Check k

y

k)

(15.3)(0.066)(3.75 - 3.75k)

or

(12) (3.75k)(

or

84.4k2 = 3.79 - 3.79k

01

k2

+

=

0.045k - 0.045

=

O

From which

k

= 0.19

j = 1 - -R= 1 - 3

=

3

1 - 0.063

=

0.937

Steel Stress:

=

<

28,400 psi

Fs = 32,000 psi

:. OK

Masonry Stress: "=bf

-

(2) (6,580 in.-lb) bkjd' (12 in.)(0.19)(0.937)(3.75 =

438 psi

c

in.>"

Fb = 1,110 psi

Provide steel as shown.

11-105


:.

OK

I


m

0662949 0508984 748

m


Calculations

Steel Areas Horizontal

Vertical

0.20

=

in2

(40in.)(7.5 in.)

0.00067

o20 i n 2 = 0.00074 (36 in.)(7.5 _ . in.) To& = 0.00141 k4.5

For Seismic Zone 3 or 4 the total minimumempiricalseismic reinforcement required by Appendix is 0.002. A minimum of 0.0007 reinforcement is required in each direction. Increase steel amount to that shown in sketch to meet seismic criteria.

Horizontal Steel 94 One At Level Of

In Between Landings.

Horizontal

Vertical

W."-"

29 in.(7.5 in.) 0.20

in?

24 in47.5 in.)

' Vertical Steel t4 @ 24" O C .

- 0.00111

0.0020

11-106


____

A C 1 TITLE*MDG 9 3

Example 11.2-1

DPCGymnasium

m

Obb2949 0508985 bB4

m

- Design of Unreinforced Pilaster for Flexure

The DPC Gymnasium walls (see MDG 9.1.2, Fig. 9.1-4) on Grid Lines 1 and 2 are to be braced against lateral loads with pilasters at 16 ft centers for Wall Construction Option A. Determine the size of an unreinforced concrete masonry pilaster that is needed to resist bending resulting from lateral forces acting on the east or west wall. Since the roof trusses run in the north-south direction, the east and west walls are not loadbearing walls. Pilasters are needed only to resist flexure resulting from lateral wind or earthquake forces applied to the wall in theeastor

westdirection.

The pilaster is

considered to be fully grouted to achieve the needed flexural capacity. The 8 in. CMU wall is considered to be ungrouted with face shell bedding. Since a control joint is to be placed immediately north of the column portion, the resulting sectioncontains a flange ononly one side as shown below. The block wythe is assumed

to carry all wind loading. Preliminary

estimates have indicated that a 32 in. square pilaster is needed to resist flexure under wind loadings. Alternate coursing is shown below.

.63"

OR

Coursing for 32 in. Square Pilaster The earthquake loading is equal to 0.0875 times the weight of the wall per MDG 9.1.2.2. Considering the weight of the exterior brick wythe with the weight of the 8 in. CMU wythe gives a total weight equal to 140 psf, or an equivalent static lateral seismic force equal to

11-107


A C 1 TITLEStMDG 9 3

m

Obb29Y9 0508986 510


Reference

Code

12.3 psf. Thus, the 20 psf wind loading governs. Each pilaster is assumed to resist a 16 ft tributary width of wind pressure, or 320 plf along its height. Assuming pin supports at the top andat the bottom of the pilaster, the maximum moment at midheight is:

Because flexural stresses may be high for the 30 ft tall unreinforced pilaster, the controlling parameter islikely

to be flexuraltension

normal to bed joints. Allowable code values for Fbcare 58 psi and 19 psi for fully grouted and ungrouted constructionrespectively

6.3.l.1

for

Portland Cement Type N mortar. Interpolating between these values based on percentages of grouted (74%) or ungrouted areas (26%) gives a value of 48 psi. For this interpolation, the grouted area was taken as 31.63 in. x 31.63 in. or 1,OOO in.2, and the ungrouted area was

5.3.2

taken as 45.78 in. x 7.63 in. or 349 in.2. Thus, 74% of the section may be considered as grouted. This allowable may be increased by a third, for comparison to wind induced stresses, giving a value equal to64 psi. The effective section of the pilaster isshownbelow.

Because the

column portion is fully grouted, all of its area isconsidered. overhanging flange wall, or 45.78 in.

portion is taken to Only thearea

The

be six times the width of the

5.10.1

of the mortared bed joints is

considered in the flange, because theiswall

ungrouted.

11-108


5.13.4.2

AC1

TITLESMDG 93 m 0662949 0508987 457 m

~~

1


Calculations

The location of the centroid of the sectionfrom the interior face,

7, is determined by

summing the first moments of the areas of each segment and dividing by the total area as 4

below. shown Exterior 15.41”

1.25

j !

” _

3 1.63”

7=16.22”

Interior

Item

Area

Arm

Moment 15,800

A

31.63 in. x 31.63 in. = 1,000 x

15.81 in.

-

B

1.25 in. x 45.78 in. = 57.2

x

23.00 in.

-

1,315

C

1.25 in. x 45.78 in. = 57.2

x

16.63 in.

-

95 1

1,114 in.2

Total

18,070 in.3

Y

= 18,070/1,114 = 16.22 in.

Effective Section of Pilaster The moment of inertia is taken about the centroid as follows:

I = (31.63 h)4 + 1,OOO in?(0.39 12 I = 86,220 h4

in.)2 +

57.2 h~~(6.80 in.)2 + 57.2 in?(0.43

11-109


ia)2

AC1

TITLESMDG 9 3

0662949 O508988 3 9 3


Flexural tension stress,

h,, is computed

Reference

Code

for the case of a wind pressure acting on the

windward face, or when tension is developed on the interior side of the pilaster.

-

Mwrnd Y -- (34.4k-fi X 12) (16.22 h) x 1,OOO = 77.6 psi f&= I 86,220 in? If

Fat = 64 psi

:. N.G.

were less than halfof the pilaster depth, then wind should be assumed as a suction

acting on the leeward face, and tension on the exterior side checked. The self weight of the pilaster at midheight creates an axial compressivestress that may be deducted from the flexural tension stress to determine the nettensile stress. Assuming fully grouted concrete masonry to weigh140pcf,

a height of 29.3 ft/2 would result in a

compressive stress equal to 14.2 psi. Thus the net tension on the unreinforced pilaster is 77.6 - 14.2 = 63.4 psi. This isjust less than the allowable value of 64.0 psi so the section has adequate flexural strength to resist the wind loading. For this problem, the effectiveness of the flange portion is very small since it is ungrouted.

If only the 32 in. square, grouted section is considered, the net flexural tensile stress is 64.1 psi which is very closeto the 63.4 psi calculated abovefor the"L" shaped pilaster. However, since 100% of the square section is grouted, the allowable flexural tension stress is 58 psi x 1.33 or 77.1 psi which is considerably larger than the value of 64.0 psi assumed for the "L" shaped section. Thus,

through this code interpretation, the smaller section has a higher

allowable flexural strength. Shear stress must also be checked using Code Eq. 6-7. Because the pilaster is fairlyslender, it is assumed that shear will not control. All of the shear is conservatively assumed to be 11-110


A C 1 TITLE*HDG 93 W Obb2949 0508989 2 2 T W

,


and

Calculations

resisted by the column portion. Thus, the maximum shear stress is equal to 1.5 times the average shear stress across the 32 in. square area. Themaximum shear due towind occurs at the top and thebottom of the pilaster, and is equal to 320 plf times halfthe height of 29 ft-4 in. or 4,693 lb.

The applied shear stress per MDG Eq. 11.2-1 is then: 6.5.1 Allowable shear stresses are given in the code to be equal to the lesser of: (a) Fv = 1 . 5 6 psi

(b) F, = 120 (c) Fv

=

N Y v + 0.45 ;where v = 60 psi for solid grouted masonry in running bond

4

No matter what the specified compressive strength is (i.e., 1,000 psi, 1,500 psi, etc.),

the

applied shear stress is much less than these values, and does not control, as was surmised.

11-111


A C 1 T I T L E x M D G 93 M Obb2949 0508990 T 4 1

Example.11.2-2

DPCGymnasium

- DesignofReinforcedPilasterforFlexure

The unreinforced pilasters for the walls on Grid Lines 1 and 2 of the DPC Gymnasium were felt to be too expensive because of their 32 in. width and depth. Thus, a reinforced concrete masonry pilaster is designed for Wall Construction Option A to resist the same wind loading as prescribed for the previous example, and costs are compared. A minimum compressive strength will be specified as a result of these calculations.

As noted for MDG Example 11.2-1, the walls on Grid Lines 1 and 2 are not loadbearing walls. Pilasters are needed only to resist bending resulting from lateral loads. A wind load of 20 psf governs over the seismic load, and the design moment at midheight of a pilaster is equal to 34.4 ft-kips. As in MDG Example 11.2-1 the block wythe is assumedto carry all the wind loading. Preliminary estimates have suggested the use of a 16-in. square pilaster with four vertical reinforcing bars, as shown below. Only the cells containingreinforcement are grouted. The 48-in. CMU wall is ungrouted with face shell mortar bedding.

Joint Reinforcement

i i d = 11.63"

Detail for 16 in. Square Reinforced CMU Pilaster

11-112


7.63"


93

m

Obb2949 0508973 988

m


and

Calculations

The controlling parameters are compressionin

tension of reinforcement, and

the masonry due to bending.Since

wind can be

applied either asa direct pressure or as a suction, flexuralstrength will be provided for bending in either direction. 5.13.4.2(c)

As in MDG Example 11.2-1, an effective flange width

equal to six

times the wall thickness can be considered. However, since bending is considered in either direction, the governing case is when the interior face is in compression.

For this case, the width of the compression

zone is the width of the column section (15.63 in.), as shown below. Thus, for this example it makesno difference what the effective flange width is assumed to be, or whether the wall is grouted or not.

Exterior Face

\

d

In

Effective Section Reinforcement in the compression zone could be relied on to enhance the 11-113


compressive

A C 1 TITLE*NDG 93 D Obb2949 0508992 814 D


Code Reference

resistance of the masonry provided it is properly tied. However,

the contribution of the

compression reinforcement is neglected here since the tensile reinforcement is expected to be light, and large compressive resistance should not be needed. Furthermore, it isexpected that the compressive reinforcement will be close to the neutral axis and thus ineffective. Based on the tension steel controlling, the area of reinforcement required is determined using EQ. 11.0-15:

A s = -Mwind Fsjd

-

(34.4 fi-kips x 12) = 1.24 (24 ksi x 1.33)(0.9 x 11.63 in-)

h 2

7.2.1(b)

The value of j is assumed as 0.9, and the allowable tensile stress is increased by a factor of 1.33 since wind isthe source of the stress. The steel requirement canbe satisfied withtwo No. 8 bars (As = 1.58 in.". The gravity stress due to the self-weight of the wall at midheight (12

5.3.2

psi) would slightly reduce the required area of tensile reinforcement, and can conservativelybe neglected. If considered one would probably use two No. 7 bars (A, = 1.20 in.2). For A, equal to 1.58 in.2, p becomes 0.00869using

the section

dimensions for b and d as shown in the figure. The modular ratio, n, is 10.5 if E, is assumed to be 2,760 psi (considering an approximate unit strength equal to 4,800 psi based on net area). The depth to the neutral axis, k, can then be found using MDG Eq. 11.0-13 to be equal to 0.346. Theterm j is thenequal to 0.88 per MDG Eq. 11.3-4. Therefore, the assumption o f j = 0.9 was valid. Compressive stress on

11-114


5.5.1.3

A C 1 TITLElrMDG 93

m

Ob62949 0508993 750

I

m



Code Reference

the interior face, fb, is then determined using MDG fb="

mwn fd jkbd2

-

Eq. 11.0-14.

2(34.4 ft-kips X 12,O) (0.88)(0.346)(15.63 inJ(11.63 in.)2

=

1,280 psi

Equating this compressive stress with the allowable value of 0.33 P,,, x 1.33 results in a required prism compressive strength equal to 2,923

7.3.1.2

psi. A contractor can comply with this requirement by using the unit strength method (Specs. 1.6.2.2) or the prism test method (Specs.

1.6.3). In either case, the block strength will need to considerably exceed the minimum ASTM C 90 strength of 1,900 psi (based on net area). In projects where high CMU strength is required, it must be so specified on project drawings and specifications.

Also; tensilestressin

the reinforcementcan be checkedwith the

allowable value:

Since more steel was provided than required (1.58 in.2 vs. 1.24 in.2) it is no surprise that the stress,f,, is less than the allowable. Shear stress is determined using Code Eq.(7-3) of the Code. From MDG Example

7.5.2.1

11.2-1, the shearforce of 4,690 lb withb, j , and d as defined previously gives a shear stress equal to 29 psi, less than the allowable valueof 50 psi according to Code Eq.(7-4).

11-115


7.5.2.2

A CT 1I T L E r M D G

93

O b 6 2 9 40 95 0 8 9 96 49 7

Ehmple 11.2-2 Cont'd. and

Code Reference

Calculations

In summary, the size of the pilaster can be reduced by half if four No.

8 bars are run vertically downthe cells of the units. It is likely that the costs of the four reinforcing bars will be less than the added cost of construction of a 32 in. pilaster. Since the vertical reinforcingbars are not relied upon

to resist compression, lateral ties are not required.

However, a nominal amount of ties will be specified to enhanceoverall integrity.

Pilaster Cross Section

11-116


5.10.2

A C 1 T I T L E * M D G 93 D Ob62949 0508995 523

Example 11.3-1

RCJ Hotel - Design of aSinglyReinforcedMasonryBeam

Consider the canopy beam that is part of a masonry frame in the RCJ Hotel. The canopy beam spans 31 ft center to centerin the east-west direction(see MDG Fig. 9.1-6). Assuming a tributary width of 10 ft and 50 psf dead load and 20 psf live load, the loads on the canopy beam are determined to be: dead load = 500 plf and live load = 200 plf. A canopy section is shown below. Given:ClayBrick

Unit Strength = 6,000 psi (from manufacturer)

Type S Mortar

f ' ,= 2,500 psi

fB

= 3,800psi

Grade 60 steel

E, = 1.9 x 106 psi (Code Table 5.5.1.3) 4"

Hollowcore Planks(Span NorthSouth). Grout Cores Solid At Bearing Ends.

Reinforced Masonry Beam

Bottom Of Masonry Beam Elevation = 12 '-0' '

- Reinforced Masonry Column Beyond h

4

Beam Width Equal to Column Width

Canopy Section N6 Calculations and Discussion

1.

Reference

Determine effective depth, modular ratio, andallowablestresses: 11-117


Code

A C 1 TITLExMDG 9 3

m

Obb2949 050899b YbT

m

Examde 113-1 Cont'd. Calculations and Discussion

Code Reference

Assume the total depth that is to be solid grouted is h = 24 in.

A structural analysis of the reinforced brick beam-column frame for Seismic Zone 4 yields a controlling load combination of D

+ L + E. The maximum positive moment = 49 ft-kips.

The maximum negative moment at the centerline of the beam-column frame intersection is 55.2 ft-kips.

Assume cover to centroid of tension steel = 3.5 in. then, d = 24 in. - 3.5 in. = 20.5 in. Assume beam width equal to column width (11.5 in.)

b = 11.5 in. E, = 29 x 106 psi

5.5.1.1

The modular ratio, n = EJE,,,

n = 15.3 The allowable compressive masonry stress, Fb

7.3.1.2

= (1/3)$m

Fb = 1/3 (2,500 psi) = 833 psi The allowable tensile steel stress,

F, = 24,000 psi

7.2.1.1

Since the load combination used allows for one-third increase in allowable stresses

11-118


A C 1 TITLEtMDG 93

0662949 0508997 3Tb H

Example 113-1 Cont’d. ~~

Code Reference


.:

Fb = (833 psi) 1.33 = 1,110 psi

5.3.2

and F’ = (24,000 psi) 1.33 = 32,000 psi

2.

Determine steel area required for negativemoment

M

= Mncg= 55.2 ft-kips

Assume j = 0.9

M As = V A,

=

- 55.2 fi-kip(12 in./ft) d - (32 ksi)(0.9)(20.5 in.)

1.12 in?

Use 2 #7 bars, A, = 1.20 i a 2

If the steelbecomes excessive the designer could correct the design moment to the moment at the column face (a reduction) resulting in smaller quantity of steel required. Check stresses: 1.20 =

in.2

(1 1.5 in.)(20.5 in.)

= 0.005

pn = 0.078

R R

=

[2pn + ( p r ~ ) ’ ] ~- pn

=

0.324

11-119



Obb29Y9 0508998 232

Example 113-1 Cont'd. ReferenceCaIcuIations Code and Discussion

0.89

55.2 ft-kips (12,000) (1.2 i n . ' ) (0.89)(20.5 in.)

fs

=

.-. OK

30,300 psi e Fs = 32,000 psi

-

fb="

bd2jk

2 (55.2 ft-kips) (12,000) (11.5 in.)(20.5 in.)2(0.89)(0.324)

Use 2 #7 bars for negative moment For possible cut off locationsof reinforcement, see MDG 14.2.2, Design of Reinforcement.

3.

Determinesteelarearequired

M

for positivemoment

= Mp = 49 ft-kips

Assume J' = 0.9 A , = - -"49 ft-kip (12 in./ft) F, j d (32 ksi)(O.9)(20.5 in.)

=

1.00 h2

Use 2 - #7 bars, A, = 1.20 i n 2 11-120


A C 1 T I T L E S M D G 93 m Obb2949 0508999 179 W


Calculations

Discussion

Reference

Code

No need to check stresses (see Section 2 above) 4.

Check Shear Stresses (see MDG Example 13.1-2 shear design)

A figure of the canopy beam construction detail is shown below. Cut In Grout

tcher

-Alg/

11-121


AC1

Example 113-2

TITLE*NDG 93 W Obb2949 0509000 5 9 9 W

- DoublyReinforced


Masonry LintelDesign

Consider the concrete masonry lintel above the10 ft x 10 ft opening in the east wall on Grid Line 3 of the TMS Shopping Center for Wall Construction Type A. The loading for the lintel from MDG Example 9.3-10, is shown below. The figure showsa uniformly distributed load,

W,

= 792 plf, and a uniform load at the center distributed over a span of 40.5 in.,

= 4,670 plf. Load

W,

W,

includes the weight of the lintel and the weight of the wall above the

lintel. Load W, is the effect of the concentrated load from a girder. No arching action is assumed. The concrete masonry lintel is to be designed assuming:

Type N mortar

Reinforcement = Grade 60

r,,,= 1,500 psi

E,,, = 1.8 x lo6 psi (Code Table 5.5.1.3)

Concrete masonry unit strength = 2,000 psi

10" Length Of

,", Bearing Plate

, . , ,

I(40.5 I 12 ) = 4,670 plf

&"$5,770

h 10"

;j

" 1 "

+4x

7.63" = 40.5"

- 88 psf

(9') = 792 plf

Lintel 10.3'


Code Reference

1. Determine effective depth, modular ratio, and allowable stresses: Assume the total depth that is to be solidly grouted and considered for lintel depth is limited to h = 24 in.

11-122



93

m 0662949 050900L 4 2 5 1

"


Calculations

Assume cover to centroid of tension steel = 3.0 in. then, d = 32 in. - 3.0in. = 29 in. Assume b = 7.63 in.

E, = 29 x 106 psi The modular ratio, n

5.5.1.1 29 x lob

Es = -

=

16.1

1.8 x lo6 psi

E,,,

The allowable compressive masonry stress, 7.3.1.2 The allowable tensile steel stress, 7.2.1.1

F, = 24,000 psi 2. Determine span length and maximum moment:

Assume minimum end bearing length of 4 in.

7.3.3.3

Then span length,

7.3.3.1

I=lOfi+ M

M

in. 12 in./ft

=

10.3 fi

1 (792 PU')(10.3 8

= -

+

[4,670 plf

=

10,560 fi-lb + 40,700 fi-lb - 6,650 fi-lb = 44,640 fi-lb

(40.5 in. )/2](10: 12 in./fi

fi)

4,670 plf

"

11-123


2

=

44.6 fi-kips

A C 1 TITLE*flDG

93

0662949 0509002 361


Calculations

3. Determine moment capacity of section for balanced condition:

Using MDG Eq. 11.3-5,

Using MDG Eq. 11.3-4, jb

=

1 -

'b = 0.916

3


Mb =

Fb 'b j b

M2

2

Mb =

Additional moment capacity needed,

M2 = M - Mb = 44.6 ft-kips - 30.7 ft-kips = 13.9 ft-kips Design as a doubly reinforced section Assume d ' = 2.0 in. 4. Determine tension steel area: Using MDG Eq. 11.3-10,

11-124


A C 1 T I T L E I M D G 93

m

0662949 0509003 2T8 D

Example 113-2Cont’d. and

Calculations

Code Reference

Discussion


Then, A, = Asb + An = 0.83 in.2 Use 2 #6 bars (A, = 0.88 in.2) 5. Determine compression steel area:

Assume k = 0.251 for the doubly reinforced section Using MDG Eqs. 11.3-9 and 11.3-8

fis

= (16.1)(500

fl,

=

fts

= Fs

f’,

= (~,OOopsi)

fis

=

psi)

5,838 psi

(

(0.251) (29 h.) - 2 h. (0.251)(29 in.)

7.2.1.2

Fs = 24,000 psi

Or

k-d Vd ( n)

5,833psi

0.25 1 - 2 h./29 in. 1 - 0.251

(

Fs

=

24,000 psi

11-125


7.2.1.2

m

A C 1 T I T L E * M D G 93

0bb2949 0509004 L34 W


Calculations

Use smaller f’J to determine A’,

Using MDG Eq. 11.3-13

Als

(13.9ft-kips)(12,OOO) 16.1 (5,833 psi) (29 in. - 2 h)( 16.1

=

=

1.12 i n 2

’)

Use 2 #7 bars A’, provided = 1.20 i n 2

6. Revise k and check adequacy of Section

R

=

[(np + ( n - l ) ~ ’ ) +~ 2(np As

P = - -

bd

-

0.88

A’

bd

(n-l)p‘d‘/d)]lP

=

0.00398

ia2

(7.63 in)(29 in.)

np = (16.1)(0.00398) p/=>=

+

=

0.064

1.20 h* = 0.0054 (7.63 in.)(29 in.)

(n - 1) p/ = (16.1 - 1) (0.0054)

=

0.082

11-126


- [np

+

(n-1)p’l


93

m

Ob62949 0509005 070

m

Examde 113-2 Cont'd. ReferenceCalculations Code and Discussion

np + (n- 1) pl = 0.686

1

:. R

=

E0.0212 + 0.1392]" - 0.1455

R

=

0.255

Rd

=

7.39 in.

0

2 5.4"

Assumed 0.251 21.6"

Assuming masonry compression stress = 500 psi and neglecting hole effect fis

fs

=

=

M =

(500 psi)( 5.4'")(16.1) 7.4 in.

=

5,874 psi

(

(16.1)(500 psi) 21*6in. = 23,500 psi < Fs 7.4 h.)

'O 0 psi (7.63 in.)(7.4 in.) 2

7.4 in. 3

+ (5,874 psi) (1.2 in?)(29 in. - 2 h.)

M = 31.2 ft-kips + 15.9 ft-kip

M

=

47.1 ft -kips > 44.6 ft -kips

.-. OK

.-. Section isadequate

11-127


:.

Masonry controls

AC1

TITLE*HDG 93 m Obb29Y9 0509006 T07 m


Code Reference


Check placement limits for 2 #6 bars for A, ; and 2 #7 bars for A’,

8. Check Placement of Reinforcement Clear distance between parallel bars shall not be less than

8.3.1

the nominal diameter of the bars, nor less than 1 in. Thickness of grout between the reinforcement and masonry

8.3.5

units shall not be less than 1/4 in. for fine grout and 1/2 in. for coarse grout. 0

Minimum masonry cover required for reinforcement:

-

8.4.1

If exposed to earth or weather, 2 in. for bars larger

than #5; 1 1/2 in. for # bars or smaller

-

If not exposed to earth or weather, 1 1/2 in.

(a) Check if 2 #6 bars for A, will fit Assume 1 row of 2 #6 bars grout and From MDG Appendix Table 4 face shell thicknessof 1-1/4 in., assume fine

#3 stirrups Minimum beam width required =

+ 2 (minimumcover required; or thickness of grout required plus face shell thickness) + 2 (diameter of 2 (diameter of #6 bar, 0.75 in.)

+ (1 in.

clear distance)

stirrup) = 2 (0.75 in.) = 6.25 in.

+ (1 in.) + 2 (1.5 in.) + 2 (0.375 in.) :. OK

7.63 in.

(b) Check if 2 #7 bars for A

will fit

11-128


A C 1 TITLExMDG 93

m Obb2949 0509007 943 m


Calculations

Minimum beam width required =

+ 1 in. + 2(1.5in.) + 2(0.375in.)

2 (0.875in.) 6.5 in.

< 7.63 in.

The final cross section and steel placement are as shown:

2 - i7

32

2

-

’I

#6

9. CheckDeflection: Using MDG Eq. 11.3-20

I,,

=

bk

3

+

nA,(d -

+

(n - l)A’,(kd -

11-129


::

OK

A C 1 TITLE*NDG

93

m

Obb2949 0509008 8 8 T

m


Calculations

k

Code Reference

0.255

=

I,, = 1,029 in4 I,,

=

+

6,610

in4 +

528

in?

8,167 i n !

Use MDG Eq. 11.3-18 to calculate short-term deflection f , = 2.5

Ma

=

=

2.5 (2,500 psi)'fl

=

125 psi

535 h.-kips

Then short-term deflection,

A =

5M,L2 5(Ma",>L2 + 48 E,,,Ig 48EmI,

A = 5(163 h.-kip~)(10.3 fix12in./ft)2 48 (2.6 x I d ksi) (20,835 in.4)

+ 5 (535 in.-kips - 163 in -kips) (10.3 ft X 12 48 ( 2 . 6 lo3 ~ ksi) (8,167 in.3 A = 0.0048 in. + 0.028 in.

=

0.033 in.

Determine additional long-term deflection for creep and shrinkage for total load. 11-130


AC1

TITLEsMDG 93 m 0bb2949 0509009 7Lb m


Calculations


x = pf =

I 1+ m p f

1.20 i n 2 7.63 in.(29 in.)

Use E = 2

I =

=

0.0054

then,

2 = 1.57 1 + 50(0.0054)

:. Total deflection = 0.033 in.

+ 1.57(0.033in.)

= 0.085 in.

Allowable deflection,

5.6

- (10.33 ft)(12 in./ft)

Aauonub& -

~auonwb& =

600

0.207 in. > 0.085 in.

:. OK

11-131


AC1

Example 113-3

TITLE*HDG 93 H Obb2949 0507010 438

DPCGymnasium

- DesignofaSteelLintel

Consider the 8 ft wide masonry opening on Gymnasium. Design a

the south wall on Grid Line B of the DPC

steel lintel to support the masonry in Wall Construction Option A,

unreinforced brick and block cavity wall.

Sm (brick) and

= 2,500 psi;

(block) = 1,500 psi; A-36 structural steel

Calculations

Partial Elevation

1.

Loading Conditions and Geometry Per MDG Example 9.3-2, the truss bearing length at the bottom of the bond beam is 16.2 in., and the maximum width of the load distribution is 46.7 in., or 3.89 ft. For a 30" spread of load, the truss reaction becomes a distributed load well above the apex of apotentialtriangulararchabovethelintel,

as showninabovesketch.

However, the location of control and expansion joints at both ends of the lintel, as shown aboveand in Fig. 9.1-5, prevents any possible arching action, since vertical and in-plane horizontal forces cannot be transferred across those joints. The lintel must 11-132


A C 1T I T L E * R D G

93

m

0 b b 2 9 4 9 05090LL 374

m

Example 113-3Cont'd. Code Reference


therefore be designed to support theweight of all the masonry above, as well as the truss gravity load reaction.

For lateral loads, the masonry will span horizontally to the pilasters, since out-ofplane shear can be transferred across properlydetailed control joints. (For instance, see the first detail in MDG Fig. 10.4-1). Brick expansion joints do not transfer shear forces acrossthe joint. However, the horizontal joint reinforcement continuously ties the brick to the CMU and therefore transfers the lateral loads to the CMU wythe. The only lateral load acting onthe lintel, therefore, is the wind pressure on thedoors. Loads - From figure below:

2.

Brick weight:

1

P, = 4.0 psf (16.7 fi) = 667 plf

p3

CMU weight:

P, = Brick Weight

Pz = CMU Weight

Pz

=

60 psf (16.7 fi) = 1O , OO plf

Lintelself-weight:Assume

P = Truss Reaction

40 plf

Trussreaction: (MDG 9.1.2.1)

Assume 6 in. bearing length each end.

11-133


A C 1 TITLEUMDG 93

m Ob62949

H

0509032 200

Example 113-3 Cont’d.

and

Code Reference

Calculations

Vertical Loading

9

= 3,825 plf

Self weight = 40 plf

pi

= 667 plf

Ej = 1,000 plf

14,700 lb

14,700 lb

0 -

Center To Center Of Bearing

M = 39.8 ft-kips

Lateral Loading

NOTE:

120 psf (8 ft)(

M =

i)]( 8.5

=

8 (1,OOo lb/kiP)

0.72 ft -kips

The horizontalloading may be neglected,since it is so small and a 1/3 stress increase is permitted when it is included.

3.

Selectlintelbased

on flexuralstresses.

A structural steel shape is required, due to themagnitude

of the verticalload

moment. It is desirable to select a steel section depth that will match the masonry coursing, so that masonry units need not be cut to fit around the lintel. A bottom plate is also required,both structurally to support the two wythes whichare separated by the nominal 3 in. (actual 3.75 in.) cavity, and architecturally to provide a seamless surface against which to install the door frame. Select a W8 x 24 with bottom plate 5/16 in. x 14 in.

11-134


AC1

TITLE*MDG 9 3 m 0 6 6 2 9 4 9 0509033 347 m


Code Reference

Calculations

Determine the composite section properties: Using the AISC Construction Manual (11.3.20) For the W8 x 24, area = 7.08 in2, depth = 7.93 in., 'I = 82.8 in.4 For the plate, area = 4.375 in.2, depth = 0.3125 in., I, = negligible Composite neutral axis: Using reference line at top of W shape

7.08in?(7.93 in.)(0.5) +4.375 in.217.93 in.+0.5(0.3125 in.)] -- 63.4 in? 11.455 i n 2 7.08 h2+4.375 in? =

5.54 in. h m the top of W8x24

Composite moment of inertia:

I

=

82.8 h4+ 7.08

in?[5.54 in. - 0.5(7.93 in.)I2

+ 4.375 h2[7.93

in? +

I

=

82.8

I

=

129 in."

17.6 in." + 28.4

in. + 0.5(0.3125 in.) - 5.54

in?

Composite section moduli:

Since the lintel is supporting all of the masonry above, and the masonry has joints at each end of the lintel, the lintel is not laterally supported by the masonry. Steel lintel allowable

11-135



93

m

Obb2949 0509014 083

m


Code Reference

Calculations

stresses are therefore based on an unsupported length of 8.5 ft. Following the procedures of Chapter F of the Manual of Steel Construction (11.3.20), the allowable flexural stress for this composite section is 21.2 ksi.

4.

Check Deflection

For a total moment = 39.8 ft-kips, the equivalent uniform load

W

is

For uniform loading, deflection is A = - -5w14 - S(4.41 klf)(8.5 ft14(l,728h3/ft3) = 0.138 384 E I 384 (29,000 ksi)(129 i n ! )

The maximum allowable deflection in unreinforced masonry is

5.6

~aenccJ=

0.138 h. < A&,,,

=

0.17 in.

11-136


:. OK


m

93

Obb2949 0509035 T L T H


Calculations

5.

Check shear and torsion - referto figure on page11-133

Under full dead and live load, the torsional loading at the center of the lintel is 3,825 plf(2.2 in) + 1,ooO plf (2.2 in.) - 667 plf(7.2 in.)

in.-lb/ft = 5,813 in.-lb/ft =

8,415

+ 2,200 h-lb/ft - 4,802 in.-lb/ft

Under dead load only, the torsional loading at the center of the lintel is

[ =

1/3(14’880 lb)](2.2 in.) + 1O , OO plf(2.2 in.) - 667 plf(7.2 in.) 3.89 ft

2,805 in.-lb/ft + 2,200 h.-lb/ft

- 4,802 in. -lb/ft

203 h.-lb/ft

The torsional loading near the ends of the lintel is 1,OOO plf(2.2 in.)

- 667 plf(7.2 h.) = -2,602 h.-lb/ft

Under full dead and live load, the torsional reaction at each end support is 5,813 h.-lb/ft(3.89 fi)(0.5)

- 2,602 h.-lb/ft(2.305

ft) = 5,309 in.-lb

and under dead load only it is 203 in. -lb/ft( 3.89 ft)(O.5)

- 2,602in.-lb/ft(2.305

ft) = -5,603in.-lb

governs

The lintel’s torsional resistanceis based on the aspect ratio of its component parts (11.3.21) 11-137


A C 1 T I T L E S M D G 9 3 W Obb2949 050901b 9 5 b W


Reference

Code

A W8 x 24 flange has 0.4 in. thickness and is 6.495 in. long. The portionof the web between flanges has 0.245 in. thickness and is 7.13 in. long. The torsional resistance of the composite section is

R

=

1/3[2(6.495 in.)(0.4 in.)3 + 7.13 h(0.245

R

=

0.455 h !

The maximum torsional shear stress

Web torsional shear stress r =

5.603 h.-kips(0.245 in.) 0.455

=

3.02 ksi

i n :

Direct shear stress of W beam is fw="

v dt,

14.7 kip 7.93 h(O.245 in.)

=

7.57 ksi

Combined shear stress f,, +

t =

7.57 ksi + 3.02 ksi

=

10.6 ksi

11-138


+ 14

in(0.3125


93

m

Obb2949 0509037 892

m

Example 113-3 Cont'd. ~~~~~


Code Reference

Allowable shear stress

Fv = 0.4FY = 0.4(36 ksi) ( P r o m11.3.20) = 14.4

6.

ksi > f,, = 10.6 ksi .-. OK

Check bearing

Check web crippling per Chapter K1.3 of reference (11.3.20).

R t,(N + 2.5k)

5

0.66Fy

For k, use the k value for W8 x 24 plus the bottom plate thickness

14.7 kips - 14'7 kips (0.245 in.)[ 6 in. + 2.5(0.875 ia + 0.313 in.)] 2.20 h 2 0.66(36 ksi)

=

24 ksi

=

6-69ksi

:. OK

Therefore, no bearing stiffeners are required. Check masonry bearing stresses. Per MDG Example 11.1-4, the brickf, = 2,500 psi; and the CMUf,

= 1,500 psi.

The allowable bearing stress Fb = 0.25 f m

5.12.3

Therefore, the allowable brick bearing is

0.25

(2,500 psi)

=

625 psi 11-139


A C 1 T I T L E r M D G 9 3 W 0662747 05090LB 727


Calculations

Code Reference

and the allowable CMU bearing is

0.25( 1,50psi) 0

=

375 psi

Under full dead and live load, the centroid of the end reaction is located at

8.5ft[(667plf)(1.8h.) + l,OOOplf( 11.2h.) +4Oplf(9 h.)] +3.89ft(3,825~lf)(11.2ia) 2( 14,7001b)

- 2759113in"1b 29,400 lb

9.36 in. from brick outside face

=

3.63" d

,

" "

I

138 psi

,

3.75"

"

7.63"

14,700 lb

313 psi

9.36"

5.64 " "

4

"

The centroid of the masonry bearing area is

(0.5)(3.63

-

in.)2

+

7.63 hi3.63 in. + 3.75 in. + 0.5(7.63 in.)] 3.63 in. + 7.63 in.

in. = 8.17 in. from brick outside face

11.3 in.

11-140


A C 1 TITLE*HDG

93

m

Obb2947 0509039 bb5

m


Calculations

The moment of inertia of the bearing area is

IE =

6 in.( 15 12

in.)3

-

6 in.(3.75 12

in.)3

- 6 h(3.75 h)[8.17 h - 3.63 in. - 0.5(3.75

IE

=

1,687.5

i n !- 26.4

in4

in.)I2

- 160.4 h' = 1,500 h'

The bearing area section moduli are S

=

1,500 h4 (15 in. - 8.17 in.)

=

220

ia3

The maximum CMU bearing pressure is

assuming grouted solid CMU at the lintel bearing fbr= 313 psi < Fb

=

375 psi .-. OK

However, the CMU must be grouted solid at the lintel bearing.

NOTE

By inspection,bearing stresses for the dead load onlycasewill

not govern

since the overall reaction is lower. Also, the eccentricity of the reaction will result in higher compression in the higher strength material, the brick, rather than the CMU.



93

m Ob62949

0509020 387

m


Calculations

7.

Weld of the plate to the W beam

Since this section was designed assuming composite action of the plate and W beam, they must be welded together to resist the shear flow between them.

q = ~t q =

Q

=

14.7 kips(ll.14 129 in."

in3)

=

kip/in.

12 inJft(1.27 kip/ in.) = 15.2 kîp/ft

(14 h.)(0.313 in.)[(7.93 in. + 0.313 h.) - 5.54 in. - O.S(O.313 in.)

Q = 11.14

For welding each side of the W beam flange, capacity required is

The capacity of 2-1/2 in. long, 1/4 in. fillet weld is

14.84 hî(0.25h.)(2.5 in.) = 9.28 kips > 7.6 kips

:.

OK

OK to use 1/4 in. fillet weld, 2-1/2 in. long at 12 in. on center each side.

11-142



93

m

0 6 6 2 9 4 9 0509021 213

I

m


Calculations

8.

Check Shoring

Since the lintel is designed to support all loads above it,

no shoring of the lintel during

masonry construction is required per (11.3.19). Other Considerations An open cavity, properly detailed flashings,and adequate weepholes are essential for the weather (water) resistance of cavitywalls.

See the MDG 6.2.1 and 6.2.6 for

further discussion. The designer may consider placing insulation in the cavity or in the CMU cells for thermal control of the building interior. Moisture and thermal movement

of the exterior masonry

walls has been

accommodated by the locations of expansion and control joints, illustrated in MDG Fig. 9.1-5. See MDG Chapter 10 for discussion. The severity of the climate is the basisfordetermining

the level of corrosion

protection that should be provided for the steel lintel. In an area of 40 psf design snow load, galvanizing is recommended.

See MDG 3.5.6 for further discussion.

Masonry anchors are required to tie the CMU to the steel lintel. Select triangular wire ties and weld on rods at 24 in. on center, placed into fully mortared head joints. The brick is tied to the CMU by the joint reinforcement.

11-143


A TC I1T L E * f l D G

93

Obb2949 0509022 &!äT

m

Example 11.33 Cont'd.

and

Calculations

Code Reference

The final lintel design is given below:

f

Horizontal Joint Reinforcement

Located In CMU

SteelLintel: W8x24 Plus Plate 5/16" x 1-1/2" _I- 5.5"

8.5"

"

15 "

Final Lintel Design

11-144


"

NOTE: Items Such As

Flashing, Weeps, And Insulation Are Not Shown For Clarity

A C 1 TITLE*HDG

Example 113-4 ~~~~

93

m Ob62949 0509023

096

RCJ Hotel - WallBeamDesign

~~

Design the wall beam on Grid Line E, between Grid Lines 3 and 4, for Wall Construction Option B, Building Construction Option I. The design is for in-plane bending. The wall is a loadbearing wall. The most desirable material is 8 in. hollow brick to match the other walls in the hotel.


Code Reference

Geometry & Loads: (See MDG 9.1.3.1)

p,fR: Cumulative

i k f T 2nd

'O'

-

Roof

3rd

lO"10'

41h

2,s o

600

3d

6,830

1,320

2"

10,800

1,820

1st

1"

-

Load at top of the floor Lintel load includes the Total

P d

floor D.L. and LL plus the lintel D.L.

D.L.

11,000 plf

L.L.

1,820 plf

By inspection, the 34 in. deep lintel (24 in.

+ 10 in.) cannot resist the totalloads.

By shoring

the wall during construction, part or all of the wall above will participate in resisting the

11-145


A C 1 T I T L E + M D G 9 3 W Ob62949 0507024 T 2 2 W


Code Reference


loads. However, the wall is a deep beam and the Code is silent about this type of element. A generally accepted procedure is to divide the wall into strips with depth to span ratio's less than 215. The first strip assumed is the 34 in. deep lintel. Span = 24 ft-8 in. 1 in.

+

7.3.3.1

+ 1 in. D.L. ofwall

Assume2ndfloorloadingplus

1

above. Neglect any Live

10"

D.L

T

'i

LL

Load reduction.

110 psf x 30 ft

=

psf 70

=

x 9.67 ft

3,300 plf

70 psf x 2 ft

-

680 plf 140 plf

40 psf

-

1.200 Dlf

x 30 ft

Total

plf

M = 5,320 plf (26*67 fil2 8

Estimate Reinforcement (Assume Grade 60 reinforcement)

5,320 =

473,000 fi-lb

7.2.1.1

Use 3 ft additional wall above. Shore until strength is reached. Vertical steel should not be spliced in this area. Total Loading 5,320 plf

M

+ 210plf

= 5,530 plf

= 491,000 ft-lb

11-146


TITLE*MDG 9 3

AC1

Obb2949 0509025 9b9

m

Examde 113-4Cont'd. Calculations and Discussion

Reference

Code

Estimate Reinforcement:

A,

=

4.80 in?

Specify masonqf,

= 4,000 psi, Type S mortar with a brick net area compressive strength

Use Type S mortar.

of 10,000 psi.

Table 5.5.1.2

E, = 2.9 x 106 psi 29 x lo6 psi

4.80 i a 2 7.5 in. x 50 in.

k = 0.38

=

0.128

j = 0.87

M, = 518,000 ft-lb > 491,000 ft-lb :. OK M,,, = 530,000 ft-lb > 491,000 ft-lb

OK

Check Shear: 7.5.5

v=- V

7.5.2.1

bjd 59,400 lb

v =

A,,

7.5 in. x 0.87 x 62 in. =

=

146 psi

- useeverycells=6in. Fs d 11-147


150 psi

7.5.2.3(a)

7.5.3

A C 1 TITLE*VDG

0 6 6 2 7 4 7 0507026 8 T 5 D

73


A,, =

Reference

lb x 24,000 psi x 62 in. in*

=

0.23 in?

#5 Q 6 in. 0.C.

Code

d -

7.5.3.1

2

Code requires reinforcement of AJ3 perpendicularshear to

7.5.3.2

reinforcement. The designed flexural steel satisfies this requirement. The next strip begins 3 ft above the 2nd floor. Assume the strip is 80 in. deep (to the top of the 3d floor). Include the weight of the wall and contribution weight and load from the 3d floor. Shoring on the first strip to remain until strength of this strip is achieved.

D.L. 30 x 110

=

3,300 plf

70 x 6.67

=

plf 470

=

1.200 plf

L.L. 30 x 40

4,970 plf

M = 441,000 ft-lb

n p = 0.058 k = 0.28

M,,, = 524,000 ft-lb

.: OK

M, = 679,000 ft-lb

.: OK

Shear

V

=

4,970 lb

11-148


j = a91

TITLE*MDG 93

AC1

0662949 0509027 731

m


Calculations

Reference

Code

try S = 12 in.

v = 91 psi

7.5.3

The Code requirement for steel in the amount of AJ3 to be provided is perpendicular to the shear reinforcementsatisfied

by the designed

7.5.3.1

flexural reinforcement. The floors above are basically the same. Note: Using the 3 ft of wall above the 2ndfloor requires a check to assure the 2nd floor is adequately attached to the beam. Bearing is not sufficient since the majority of the beam is above the floor. Assume all of the floor load must be transformed.

W = (110 + 40) x 30 = 4,500 plf (2) #5 fs = 7,300 psi Add standard hooks to #5 in bottom bond beam Check anchorage of flexural steel Id

= 0.0015 db

Id =

0.0015

X

8.5.3.1(f)

Fs 7 8

-

8.5.2 X

24,000

ld = 31.5 in. 11-149


0662747 0509028 b 7 8

A C 1 TITLESMDG 93


Reference

Code

Embed Flexural bars into support column

n t L

7

I

3rd

!

0 0 0 0 0 0

#5 x 12 "2" @, 12" O.C.

I

A

4

r

!

i

LAP 2 '-6"

(6) #7 x 3 2 '

#5 @ 6" O.C.

""". +

T

"O"

""_.

2 "10"

""_.

" " " " "

Stnd Hook

LShore Lintel For Entire Length Until T h e Full Height Of T h e Wall Is Complete.

Reinforcement Arrangement Note: Additional horizontal reinforcement is required for seismicdesign and minimum reinforcement (Code A.4.7) The design approach used in this example israther novel but might be considered by some to be rather conservative. For heights greater than 12 ft-4 in.above the opening the designer might consider arching of the masonry with the floors acting as tension ties.

11-150


Example 113-5

-

RCJ Hotel Coupling Beam for Flexure Only

Given the following material properties, design a single-wythe reinforced brick lintel for the

RCJ Hotel, Wall Construction Option B. The design is for the door openings along wall Grid Lide 2 (see MDG Fig. 9.1-6) and is to consider-in-plane flexure only. MasonryBrick

Clay

Grout

Unit Strength (psi)

6,000

Mortar

Type S

(Psi)

2,500

E m (psi)

1.9 x 15.3

S m

n

lo6

Steel Reinforcement - Grade 60

1"lO"

11-151


AC T I1T L E x M D G

Obb2949 0 5 0 9 0 3 0 226

93


and

Calculations

Code Reference

The lintels of wall Grid Line 2 of the RCJ Hotel have a clear span of 3 ft-4 in. They are considered fixed on both ends and resist flexure

7.3.3.3

due primarily to seismic loads. From MDG Example 9.2-4 Table 3, the

in beam B5 at the 3rd floor. 7.3.3.1

maximum coupling moments occur

Higher moments exist on the second floor but this beam is deeper. Additional moments exist due to the dead load of the lintel, but these are small relative to the coupling moments and are neglected. The maximum moment is taken as 759 in.-kips. Using working stress design, the value o f j is initially assumed as 0.9, and the area of steel required to resist the maximum moment is estimated using the expression:

where M = 759,000in.-lb,

5.3.2

F, = 24,000 psi x 1.33 = 32,000 psi

7.2.1.1

j = 0.9 (assumed), and d = 29in. Substitution yields A,

=

759,000 in.-lb (32,000 psi) (0.9)(29 in.)

:. use 2 - #6 P = -As- bd

=

0.91

in?

bars (A, = 0.88 in.2) 0.88 (7.5 in.)(29 in.)

=

0.00405

11-152


A C 1 TITLE*flDG

93

m

0662949 050903L L b 2

m


Calculations

np = 0.0619

k = 0.29

Mt = A, jdF,

j = 0.90 = (0.88

in?)(0.9)(29 in.)(32,000 psi)

=

735,000 h - l b

Margin of Safety = 4%, may be ok in some local jurisdictions. Net Moment capacity is higher because beam has been doubly reinforced.

bd2 M,,, = -kjFb 2

in.)' (0.29)(0.9)(25': M,,, = (7.5 in.)(29 2

p6i)(1.33)

=

912,000 h - l b

..

OK

Use (2) #6 top and bottom Shear Stress: See MDG Example 13.1-1 All requirements of Code 8.5.3 should be met concerning the flexural reinforcement and

development lengths.

11-153


A C 1 T I T L E z f l D G 93

Example 113-6

RCJHotel

Obb2949 0509032 O T 9

m

- Design of a Continuous Masonry Beam

Design a continuous masonry beam on Grid Line E spanning from column E-3 to E-3.5 to E-4 above the opening on the first floorof the RCJ Hotel (Wall Construction Option A with Building Construction Option I). The continuous beam spans over two openings each 11

ft wide and is supported by three columnsover a length of 2.67 ft each.Assuming

a

tributary widthof 30 ft, dead load of 110 psf and live load of 40 psf, the loads on the continuous beam are determined to be: dead load = 14,750plf and total beam load =

17,020 plf (includes live load reductions). Design

a continuous concrete masonry beam

assuming: Concrete Masonry - 12 in. CMU Concrete masonry unit strength = 2,000 psi (from manufacturer) Type N mortar fm

fg

= 1,800 psi

= 3,600 psi

Grade 60 steel

E,,, = 1.8 x lo6 psi (from tests) Note: This problem is worked as an illustrative alternative to the steel beam at this location.

and

Calculations 1.

Determine effective depth, modular ratio andallowablestresses: Assume the total depth that is to be solid grouted is h = 72 in. Assume cover to centroid of tension steel = 6 in.

then, d = 72 in. - 5 in. = 67 in.

11-154


AC1

TITLE*HDG 73 M 0662747 0509033 T35 m


Calculations

Assume b = 11.63in.

E, = 29 x 106 psi

5.5.1.1

The modular ratio, n = E,/E,,, = 29D.8 = 16.1 The allowable compressive masonry stress,

7.3.1.2

Fb

=

1

- (1,800psi) 3

=

600 psi

The allowable tensile steel stress, 7.2.1.1

F, = 24,000 psi

2.

Determine span length and maximumpositive and nega1tive moments: Assume end bearing length = 8 in.

7.3.3.3

Assume pin support at center of middle column Then span length for each span,

I = - 4 f t + l l f t + - 2.67 ft 12 2

=

7.3.3.1

12.7 ft

11-155


Obb2949 0509034 771 W

93

A CT 1I T L E * f l D G

Example 113-6 Cont'd. ReferenceCalculations Code and Discussion

Check the depth to span ratio: depth 67 in. span (12.7 ft)(12 in./ft)

"

=

0.44

Per MDG Example 11.3-4, it is desirable to limit the depth to span ratio 2/5, to since the Code is silent on the subject of deep beams. Although the actual ratio is slightly greater than 2/5, it is acceptable for this example.

MF - -W12 9

- 17*020 plf

(

128

128

- wZ2 -

17,020 plf (12.7

-

3.

=

8- (

1,OOO

)

Determinesteelarearequiredfor

M

=

MF

=

)

1,OOO

(12.7

fi)2

=

=

192 ft-kips

fi-kips

8

positive moment

192 ft-kip

Assumej = 0.9

M - 192 fi-kips (12 ialft) A, = F,jd - (24 ksi)(O.9)(67 in.) A, = 1.59 in?

11-156



m 0662947

0509035 B O B

m


and

Calculations

Code Reference

Try 4 #6 bars, A,

=

1.76 in.2

Check stresses: =

1.76 i n . ' = 0.00226 (1 1.63 in.)(67 in.)

pn = 0.0364

k

k

=

[2pn + (pn)'Iw - pn

=

[2(0.0364) + (O.O364)']'' 0.236

=

M

- 192 ft-kips (12 in./ft)(1,OOO lb/kips)

h=-@f,

=

M'jk fa

=

(1.76 i n . ' ) (0.921) (67

-

in.)

.: OK

21,200 psi < Fs = 24,000 psi

2M

fb="

- 0.0364

2 (192 ft -kips) (12,000) (11.63 in.)(67 in.)'(O.921)(0.236)

406 psi < Fb = 600 psi

:. OK

Use 4 #6 bars for positive moment For possible cut off locations of reinforcement, see MDG Example Problem 14.3-16 11-157


A C 1 TITLExMDG 93

m O662949 0509036 744 m


Calculations

4.

Determine steel arearequired for negativemoment

M

M,

=

341 ft-kips

=

Assume j' = 0.9

Use 5 #7 bars, A,

= 3.00 in.2

Check stresses: 3.00 (11.63 in.)(67 in.)

=

=

0.00385

pn

=

k k

[2pn + ( ~ n )- ~pn] ~ 0.296

= =

0.0620

&" =a-

341 fi-kip~(12,OoO) (3.00 in?)(O.902) (67 in.)

fs

=

-

fb="

b d 2 jk fb =

:. OK

22,570 psi < Fs = 24,000 psi

2 (341 ft -kips)(12,000) (1 1.63 in.)(67 in.)2 (0.902)(0.296)

587 psi < Fb

=

600 psi

:. OK

11-158


2

A C 1 T I T L E + M D G 93

m

Ob62949 0509037 b B 0


Calculations

Use 5 #7 bars for negative moment For possible cut off locations of reinforcement, see MDG Example 14.3-16.

5.

Checkplacement requirements for flexuralreinforcement. Assume #4 stirrups and coarse grout. Minimum clear distance between

bars is the bar diameter or 1in. 8.3.1

Minimum grout thickness between rebarand masonryunitin1/2in. is

8.3.5

cover masonry Minimum

A 12 in. CMU face shall is 1-1/2 in. thick; assume the bottom of the lintel block is 2 in. thick. Check width required for negative moment reinforcement (governs over positive moment reinforcement). 2(1.5 in.)shell + 2(0.5 in.)grout + 2(0.5 h.)stirrup + 3(0.87 in.)rebar + 2 (1 in.)clear = 9.62

11.62 in. provided .:

in. required

OK

Check depth of cover to negative moment reinforcernent (from the top) 1.5 in. grout + 0.5 in. stirrup 1 2

+ -( 1

+

0.87 in. rebar

5 in. assumed .:

in.)clear = 3.37 in. actual

OK

Check depth of cover to positive moment reinforcernent centroid (from bottom) 2 in. + 0.5 1 2

+ -( 1

in. grout

in.)clear

+ 0.5

= 4.25

in. stirrup

+

0.75

in. rebar

in. actual < 5 in. assumed

11-159


:. OK


m

0662949 0509038 517

m


Calculations ~

~~

The final beam cross section and steel arrangements is shown below.

72 "

I

at 8"

O.C.

4 #6 Bottom

L

6.

11.63"

c

Check shear stress (See MDG Example 13.1-3)

11-160


A C 1 TITLE*MDG

93

m

0 6 6 2 9 4 9 0509039 453

m

-

RCJ Hotel Designofa Masonry Coupling Beam

Example 113-7

Design the masonrycoupling

beam onLine

Construction Option B,Building

B, fourth floor of theRCJHotel

Construction Option I).

(Wall

The left end and right end

moments on the coupling beam are both clockwise with the magnitude of 450 k-in. and 409 k-in., respectively. The end moments wereobtained from an envelope considering four load cases. Assume the opening below the beam to be 3 ft - 4 in. wide and the totalbeam depth (top of opening to floor slab) to be 2 ft

-

8 in. Determine the flexural reinforcement

required assuming:

fg

Clay Brick Unit Strength = 6,000 psi Type S Mortar

Reinforcement Grade 60

Sm= 2,500 psi

Em = 1.6 x 106 psi (from tests)

Code Reference

Calculations and Discussion 1.

= 3,200psi

Determine effective depth, modular ratio andallowable stresses. Assume cover to centroid of tension steel = 4 in. then, d = 32 in. - 4 in. = 28 in. 5.5.1.1

E, = 29 x 106 psi The modular ratio, n = EJErn= 18.1 The allowable compressive masonry stress, (113) S m Fb = 1/3 (2,500 psi) = 833 psi Fb

=

11-161


7.3.1.2


93

m

0662949 0509040 175

m


Calculations'

Reference

Code

Allowable stresses are permitted to be increased by one-third whenconsideringload combination 3, 4, or 5 of Code 5.3.1.

.: Fb = psi) 1.33 (833 5.3.2 psi= 1,110 The allowable tensile steel stress,

F,psi)1.33 = (24,000 7.2.1.1 2.

psi = 32,000

Determinesteelarea

required atthe leftend:

M = 450 in.-kips Assume

b = 7.50 in.

Assume

j = 0.9

Use2 #5 bars, A, =: 0.62 in.2 Check stresses: P = -

0.62 h? (7.50 in.)(28 in.)

=

0.00295

pn = 0.053

k

= [2pn + =

R

( ~ n ) * ] -' ~pn

[2(0.053)

+

(0.053)2]'/r - 0.053

= 0.25'7

j = 1 - k;/3 = 0.91

11-162


A CT 1I T L E w M D G

93

Ob62949 0509043 001

m


Calculations

- (450 h.-kip)(1,OOO lb/kip) &M = q -(0.91) (28 ia) (0.62 ia2)

&

=

24,500 psi < Fs = 32,000 psi

.: OK

- 2 (450 h.-kip)(1,OOO lb/kip) bd2jÆ (7.5 in.)(28 i1~)~(0.91) (0.277) 607 psi Fa = 1,110 psi :. OK 2M

fb="

fb =

"q

7.5'

t."

Use 2 #5 bars at the left end (check placement tolerances) 3.

Check placement limits for reinforcement ..

Assume #4 stirrup, and fine grout

15'

Assume face shell thickness = 1 1/4 in. Minimum beam width required = 2(diam. required;

+ 1 ('1 in. cleardistance) + 2 (minimumcover or thickness of grout required plus face shell thickness) + 2 (diam. of #4

of #5 bar, 0.625 in.)

stirrup) =

-

+

+

2 (0.625in.) 1 in. 2 (1.5 in.) 6.25 in. < 7.5 in. :. OK

+ 2 (0.5 in.)

BY inspection the 4 in. distance assumed to centroid of tension steel is adequate. For possible reinforcement cut off locations see MDG 14.2.2

11-163


A C 1 T I T L E x M D G 93 W O662949 0509042 T 4 8 W


4.

Determinesteelarea

Reference

required at the rightend:

M = 409 in.-kips Assume j = 0.9

As

- " 409 in.-kips - F'# (32 ksi)(0.9)(28 in.)

A,

=

"

0.51

in?

Use 2 #5 bars, A, = 0.62 in.2 For possible reinforcement cut off locations see MDG 14.2.2

11-164


Code

A C 1 TITLESNDG 93

Example 113-8

DPCGymnasium

m

Obb2949 0509043 984

m

- Design of a SteelLintel

Consider the 8 ft wide masonry opening on the south wall of Grid Line B of the DPC Gymnasium (MDG Figs. 9.1-4 and 9.1-5). Design a steel lintel to support the masonry in Wall Construction Option C, reinforced composite brick and block wall.

~

Code Reference


Truss Reaction

Truss Bearing Elevation,

-

24" 8" 4" 2"

8'- O" OPG

-

Partial Elmtion

1.

Loading Conditions

and Geometry

Per MDG Example 9.3-2, the truss bearing length at the bottom of the bond beam is 16.2 in., and the maximum width of the load distribution is 46.7 in., or 3.89 ft. For a 30" spread of load, the truss reaction becomes a distributed load 3 ft - 4 1/2 in. downfromthepoint

of application,and 8 ft

-

5 1/2 in.above the apex of the

triangular arch above the lintel as shown in figure above. Since

there is adequate

masonry mass on each side of the lintel to resist the arch's horizontal thrust, arching action will occur. Therefore the only vertical loads on the lintel will be from self11-165


A C 1 TITLE*flDG 9 3

0 b b 2 9 4 9 0509044 810

Example 113-8 Cont'ld. and

Calculations

weight and weight of the masonry within the triangular arch area. The lintel is laterally braced by the masonry mass. The angles are not free to twist, and the compression "flange"cannot buckle laterallydue to the restraintof the grout and the bottom plate. Refer to Lintel Section figure. Lateral loads acting on the lintel derive from wind pressure within the triangular arch area and acting on the doors.

2.

Loads - Refer to LintelSectionfigure and MDG Example 11.1-6. Vertical Loading:

Masonry weight = 4 in. brick =

40 psf

=

125 psf

+

2 in. grout

2 12

+ "(150

+

8 in. CMU

psf) + 60 pSf 125 psf (0.5)(8.33 f t ) = 521 plf

&

Self Weight = 25 plf

1,190 l b

M = -w12 + --W1 8

8.33' C-C Of Bearing

6'

M = 25 plf(8.33

8 + - [521 plf (8.33 fi)(0.91 (8.33 fi)

6

M

=

1,190 lb

217 ft-lb + 3,013 ft-lb = 3,230 ft-lb

11-166



m

0 6 6 2 7 4 7 0507045 757

m


Calculations

Code Reference

Discussion

Lateral Loading:

20 psf (0.5) (8.33 f t ) = 83.3 plf

M = 80 plf(8.33 8

507 lb

+ [83.3 plf( 8.33 ft)(0.91 (8.33 ft)

t

t

L

20 psf (0.5) (8 f t ) = 80 plf 507 lb

i

8.33' C-COf Bearing

6

M

3.

=

694 ft-lb + 482 fi-lb

Selectlintelsize

=

1,176 ft-lb

based onflexuralstresses.

Try three angles 3-1/2 in. x 3-112 in. x 5/16 in. To provide a finished surface against which to erect the door frame, and to contain the grout in the collar joint, add a bottom plate. The plate is not designed compositely with the angles, the required welding to the angles is minimal, and the bottom plate need not be continuous. The plate thickness of 5/16 in. is selected for durability in exposed usage, and a 13 in. length is selected to support the 13-1/4 in. overall thick wall. For each angle, Area = 2.09 in.2, I, = 2.45 in.", S, = 0.976 in.3 For vertical loading, S = 3(0.976 in.3) = 2.93 in.3

11-167


I

A C 1 I’ITLE*HDG

93

m

0662949 0509046 693

m


Code Reference

For horizontal loading, the section modulus of the angles is greater than 2.93 in.3, since they are spaced apart. Since the lateral load moment is less than the vertical load moment, the horizontal section properties are greater, and theallowable stress is 1/3 greater when the wind load is considered, flexuralstresses are within allowable by inspection. 4.

Check Deflection

A = - + -w14 384EI

w13 60EI

A = 0.025 khF( 8.33

fi)4(

1,728

h3/ft3)

384 (29,000 ksi) [3 (2.45in?)]

A

=

+ [0.0833 klf (8.33 ft)(0.91(8.33 60(29,000 ksi) [3 (2.45 0.0025 h. + 0.0271 in = 0.0296 in.

( 1,728 h3/f ’)t ia4)]

The Code onhy limitsdeflection in beams supporting unreinforced masonry. However, A =: 0.0296 in. = 1/3,370 is very small deflection and OK 5.

Check shear and torsion

By inspection, shear is not a problem. Torsion is no problem since the angles are not free to rotate.

11-168


5.6

A C 1T I T L E I M D G

93

m 0662949

0509047 5 2 T

m

Example 113-8Cont’d. Reference Calculations Code and Discussion

6.

Check bearing Because the wall is composite, the load will be fairly uniform across the wall cross section. For 4 in. long bearing each end, and assuming solid brick and 50% solid CMU, the bearing area is

3.63 in.(4 in.) + 7.63 in.(4 î&)(OS)

=

29.8

in?

The bearing pressure is

The allowable bearing is Fbr = 0.25$, Per MDG Example 11.1-6, f m = 1,500 psi in the CMU, and$,

5.12.3 = 2,500 psi in the

brick. Therefore the minimum Fbr = 0.25(1,500 psi) = 375 psi. Sincefb, = 40 psi <

Fbr = 375 psi, bearing stress is OK, and the CMU need not be grouted at the lintel bearing.

7.

Weldof

the plate to the angles

Since the plate is not part of the structural member, the weld is only required to hold the pieces together. The minimum weld size for 5/16 in. thick members is 3/16 in. (11.3.20).Use 3/16 in. 11-169


A C 1 TITLE+MDG 93 M O662949 0509048 466 W

Example 113-8ConVd. Calculations and Discussion

Code Reference

welds 2 in. long at 12 in. on center at the toe of each angle.

8.

Check Shoring When arching action has been assumed to reduce the loads on the lintel, temporary shoring of the lintelmust be provideduntil the masonry has attained sufficient strength to accommodate arching action. Shoring must be provided for a minimum of 24 hours (11.3.19). The shoring period should be increased to three days when imposed loads are tobe supported, and should be even longer when masonry is built under cold weiather conditions (11.3.19). For this lintel design, require shoring for a minimum of three days.

9.

Other Considerations Although not required in a solid (barrier) wall, the designer may choose to place a flashing at the lintel location. While sizing the wall reinforcement, the designer must consider stresses due to moisture and thermal movement, in addition to thegravity and lateralload stresses, since no expansion or control joints are provided in the reinforced composite masonry wall. The severity of the climate and the level of maintenance that will be provided determine thequality of corrosion protection that the steellintel should have. For this lintel, require galvanizing of the steel. See MDG 3.5.6 for further

11-170


AC1

TITLE*HDG 9 3 m Ob62947 0507047 3 T 2 m

Example 113-8Cont'd.

and

Code Reference

Calculations

discussion. d)

Masonry anchors to the lintel are not required for this lintel configuration.

The final lintel design is shown below: y Rebar In Grouted Collar Joint

CMU-Notch Webs Around Vertical Legs Of Angles

Steel Lintel: 3 - L 3-112" X 3-112" X 5/16'' Plus Plate 5/16' x 13'

13.75" 5.5"

"

3.75" I-

I

13"

Lintel Section

11-171


-1


93

m

Obb2949 0509050 014 H

12 FLEXURE AND AXIAL LOAD

12.0 INTRODUCTION Most masonry elements support axial load, even if the axial load is onlythe dead weight of the element itself. When lateral loads such as wind or seismic are added, the element is designed for both flexure and axial load. For elements that primarily resist lateral forces (such as walls on a one-story building), the addition of axial load usually improves an element’s ability to resist flexure. This is true in both unreinforced and reinforced masonry wherethe allowable load is controlledby tension. In these situationsit is often acceptable to neglect the axial load effect since this assumption is conservative. For elements that primarily resist axial forces, such as columns, the addition of lateral forces usually reduces an element’s ability to resist axial load.

In these cases, it is generally not

acceptable to neglect the effects of lateral force. This chapter presents methodsfordesign

of masonrycolumns,walls

and pilastersfor

combined bending and axial loading. The methods presented are a sampling of the many approaches available. Many approaches exist because solutions to the combined loading case are more easily solved with the use of iterative or graphic methods.

12-1


A C 1 TITLE*:MDG 93

m

Obb2949 0509051 T50

m

12.1 COLUMNS

12.1.1

General

The Code 2.2 defines a column as an isolated vertical member whose horizontal dimension measured at right angles to the thickness does not exceed 3 times its thickness and whose height is at least 3 times its thickness. Most of the Code provisions are based on rational engineering, not rules of thumb. Because

of the structural importance of columns, special provisions have been imposed. These are listed in Code 5.9 and are summarized as follows:

1.

Least lateral dimension of 8 in. (nominal);

2.

Maximum height to thickness ratio of 2 5 ;

3.

Minimum eccentricity

of 0.1 x each side dimension

about each

axis

independently, for design; 4.

VerticaJ reinforcement with at least 4 bars having an area between O.OOZA, and O . O U , ;

5.

Lateral ties at least 1/4 in. in diameter, whichmay be placed either in mortar or grout.

Other special requirements which relate to columns are as follows: 1.

Code 5.11.2 requires that devicesusedfor

transferring lateralsupport from

longitudinal members to columnsmusthave

a capacity of at least 1,000

pounds.

2.

For ap~plicationsin Seismic Zone 2, Code A.3.7 requires structural members framing into or supportedby masonry columnsbe anchored to those columns. Anchor bolts located in the tops of columns are required to be set entirely

12-2


A C 1T I T L E * l D G

93

m

Obb29Y7 0509052 9 9 7

within the reinforcingcagecomposed

m

of column bars and lateral ties. A

minimum of two #4 lateral ties must be provided in the top 5

in. of

the

column. Welded or mechanical connections for reinforcing bars in tension are required to develop 125%of the yield strength of the bars in tension. In Seismic Zones 3 and 4, Code k4.6 requires #3 lateral ties embedded in

3.

grout at closer spacings. A least lateral dimension of 12 in. is also required per Code k4.8. Sincecolumnsmust

be reinforced, the applicabledesign

procedure of Chapter 7

-

Reinforced Masonry, which uses linear stress-strain assumptionsand linear bending theory, applies. Allowable tension

and compression stresses in

steel are given in Code 7.2, and

allowable compressive stresses in masonry are provided in Code 7.3.1.2. Tensile stresses in masonry are neglected. In addition

to the limit on the ratio of effective height to least

nominal dimension < 25 and the allowable compressive stress due to combined flexure and axial load

S S /’ fm,

the overall axial stress per Code 7.3.1.1 must not exceed the following:

Code Eq. 7-1

for h/r c 99 and

Fa =

for hlr

L

($r

(i) f’,

Code Eq. 7-2

99

Per Code 5.13.1.2 the cross-section to which F, isapplicablecouldcertainlyinclude transformed longitudinal column steel. Obviously neglecting

the longitudinal steel for the

axial load conditionas illustrated in the MDG Example problemsis conservative. The Code committee is currently addressing this issue.

12-3


A C 1 TII’LExMDG 93

m 0662947 0509053

823

m

The necessary equatiolnsfor allowable stress design of masonry columns are very difficultto apply manually. 12.1.2 Development of InteractionDiagram The axial load-bending moment interaction diagram is developed using equations and assumptions very similarto those used in analysisand design of reinforced masonry flexural members. The only difference lies in the fact that the compression force is not equal to the tension force, due to the additional net axial force on the structural element. This minor difference, however, results in much greater complexity in solving the equations directly. Interaction diagrams can be produced which permit a rapid graphical solution to the problem.

To illustrate the process of interaction diagram development, a rectangular column withtwo layers of steel will firsit be examined. This choice is based on the code requirement that a minimum of 4 bars must be placed in any column;hence, at least two layers are most likely.

As illustrated in Fig. 1:2.1-1,several states of stress (strain gradients) can occur in a column cross-section.

These stress statesare

shown as Cases 1-5. Case 1 illustrates pure

compression applied concentrically. Case 2 is the result of a load applied eccentrically to produce maximum stress at the top fiber and zero stress at the bottom fiber. In Case 3, there is compressive mess in the top fiber and zero stress in the bottom bars. Tension stress in the bottom fibers is neglected. Case 4 is the balanced condition, with allowable compression stress on thetop fiber, and allowable tension stress in the bottomlayer of steel. Case 5 is the pure bending case, with no axial load present. Finally, Case 6 (not shown on Fig. 12.1-1) is a required Code check for axial compressive stress. these cases will be derived below.

12-4


Equations for each of

A C 1 TITLE+MDG 93

m

Ob62949 0507054 7bT

m

Fb

(b) Tension Controls

(a) Compression

Controls

Fig. 12.1-1 Column States of Stress

Case

1

(correspondstostraingradient1inFig.12.1-1) Neglectinggroutdisplaced

e =O

by steel andtransformingcompressionstee i1 to

masonry using a transformation factor of n:

1 d'

-4

"-

e As e

12-5


C', = nF, A',

S

F, A',

A C 1 TITLEaMDG 93 W 0662949 0509055 bTb W

In this case and subsequent cases, steel in compression is assumed to have adequate lateral reinforcement to satisfy Code5.9.1.6. If lateral reinforcement is inadequate, the compression steel stress is zero. Summing internal for,ces axially

P

=

cm,+ cl,

+ Cs

Summing internal moments about the centroid

Case 2 (zero strain ai. bottom fiber)

/-

Cm =

Summing forces axia1:ly

P

=

cm+ cfs+ Cs

Summing moments about the centroid

M

=

($)

CER

+

C',

(f)- Cs (f)

12-6


1/2F, bt

A C 1 TITLErMDG 93

0662949 O509056 532

Case 3 (zero strain in bottom steel)

Fb

I

Cs = o

Summing forces axially

P

=

cm+ c',


Case 4 (Balanced Conditions)

T

l"-4 b 12-7


=

A,F,

m

A C 1 TITLE*MDG

93

m 0662949 0509057 479 m

The neutral axis can be defiied in terms of allowable stresses using similar triangles

-

'bd

"

d

4 Fs

Fb + -

n

from which 1

kb =

1 + - Fs nFb


*)

Pb =

CD,+ Cis - T


[i

M b = C,,, .- Rd

+ Cls($)

+

T(:)

Case 5 (Pure Bending)

1r

fb

f

_r b

11

12-8


AC1

TITLErMDG 73 m Ob62747 0507058

305 H


P

=

o,

T=

c,

From similar triangles,

+ C',

(Fjf,n

=

-)

nAs( d-kd kd

fb =

fbbkd

+

nfb

[ kd-d'y A's ]

Define A , = pbd, A', = p'bd, multiply by k d , and divide byfb 1

npbd(d - k d ) = - b ( W 2 2

k2 + ( n p l 2

+

+

n(kd - d')p'bd

np) k -

Check k against kb, If k

L

kb Use F b in above equations

If k

S

kb Use F, in above equations


M

=

;( - y)

cm

+

cfs(f)+ Ts(f)

Case 6 (Axial Load Only) Check P / !S F, 12-9


A C 1T I T L E : * H D G

93

m

Obb2949 0509059 241

m

To illustrate these equations, a specific example will be worked as follows:

-

A,, = 4 #7 bars

g = 0.6

symmetrically placed A, = A', = 1.20 in.2

EL

Adequate lateral reinforcement

is

provided to satisfy Code 5.9.1.6.

f',= 1,500 psi

15.63"

L

9.38

"

CMU Strength = 1,900 psi

e

Type S mortar Grouted Solid, fg = 4,160 psi

F, = 24000 psi (tension and compression for grade 60 reinforcing)

Fb = 1/3 f',= 500 psi E, = 2.08 x 106 psi (Code Table 5.5.1.3)

Eg = SOO& = 2.08 x lo6 psi

n = - -29 - 13.9, use n 2.08

Using the equations previously developed for Cases 1 through 5 Case 1 (Uniform strain over section)

cm=

P

=

O '0

psi

1 O ,OO lb/kip

(15.63 in.)2

=

122 kips

122 kips + 8.4 kips + 8.4 kips

=

139 kips (Plot on Fig. 12.1-2) 12-10


= 14

A CT 1I T L E * N D G

Obb2949 05090bO Tb3

93

Case 2 (Zero strain at bottom fiber) 1 Cm = -( 122 kips) 2

=

61 kips

8.4kips

in - 3.13 in ( 15.63 15.63 in.

Cs = 8.4 kips

15.63 h. - 12.5 h. 15.63 in.

C',

P

=

=

(

61 kips + 6.72 kips + 1.68 kips

=

) + 6.72 kips (9.38 )

M = 6 l k i p s ( 15'63

in'

M

(Plot on Fig.12.1-2)

183 in.-kips

=

69.4 kips (Plot on Fig. 12.1-2) in. - 1.68kips

Case 3 (Zero strain in bottom steel)

cm=

500 pi (15.63in.)(12.5 v2 (lo00 1bPa.p)

Cls = 8.4

P

M

=

=

kips(:) =

207 in.-kips

Cs =

6.3 kips

48.8 kips + 6.3 kip

=

in.) = 48.8 kips

o

55.1 kips

Plot

Fjg. 12.1

-2

Case 4 (Balanced Case)

kb =

.

I +

1 24 ksi 14(.5 ksi)

=

0.226

kbd

=

2.82 in.

12-11


(9.38 in.)

m

A C 1 TITLErMDG 93

dl

Ob62949 0509063 9 T T

m

3.13 in.

=

Cm = 1/2(500psi)(15.63inJ(2.82 in.) = 11.0 kips

T = 1.2 h2(24 ksi)

28.8 kips

=

P = 11.0kips - 0.9 kips - 28.8 kips

(

M = 11.0 kips 15.63 in. 3

2

2.82 h.)

-

=

-18.7kips (tension)

o.9 kip (9.382 h.)

~

(

28.8 kips 9.38

in.

)

M = 207 in.-kips (Plot on Fig 12.1 -2) Case 5 (Pure Bending)

I+

1.20 in? 1.20 in? + 14 15.63 in.(12.5 in.) (15.63 in.:)(12.5 in.)

4

[0.172a +

2(14)(1.2 in?)(3.13 in.) 15.63 h(12.5 in.) (12.5 in.)

+

2 (14) (1.2 in?) 15.63 in. (12.5 in.)

=

0.323

Since k > kb, the compressive allowable stress in the masonry is reached before the tensile allowable in the steel. Hence the steel stress is

d-M M

fs

=

n (113 f',)

T

=

1.2 h2(14.7 ksi)

cm=

=

=

14.7 ksi e Fs 17.6kips

O '0

psi (15.63 in.)(0.323)(12.5 in.) = 15.8 kips 1,Ooo Ib/ki*)

C/,= 8.4 kips 4.04 h. - 3.13 in. 4.04 in.

M

= 15.8

kips

(

(

15*63 h: + 1.90 kip 9.38 2 4.043 "')

12-12


in.

)+

9.38 in.

17.6 kips

(

)


M = 194 in.-kips

m 0662747 0507062 836 m

Plot on Fig. 12.1-2

Case 6 (Axial Compressive Stress)

P

=

(0.37 k~i)(15.63h.)2 = 90 kips

The allowable axial capacitycomputations have neglected the axial capacity contribution of the column steel.

See MDG 8.6.2 and 12.1.1.

compression stress due to axial

Note that in the development the

+ bending was limited to 1/3 Fm,satisfymg the Code.

One

would still have to check to ensure that theaxial stress satisfied Code 7.3.1.1, Case 6. These 6 cases are plotted on Fig. 12.1-2 to form the interaction diagram for this column only.

150

100

P KIPS

90

50

Fig. 12.1-2 Column Interaction Diagram It is possible to construct such diagrams ina non-dimensional formfor a family of columns. This will be illustrated by reworking the previous example using different values of b and t,

12-13


A C 1 T I T L E S U D G 93

m

0662743 0509063 772

m

while maintaining the same steel ratio pl, the same allowable compressive stress Fb, and same spacing ratio belween steel layers, g. This second example is summarized along with the previous example in the table below. Now, if the values of ,P and M are non-dimensionalized by dividing P by (F&) and M by (Fat2),each value of Examples 1 and 2 results in the same non-dimensionalized values for

load and moment for all 5 cases. These non-dimensionalized values are summarized in the table below as Example 3.

Table 12.1.1 Load Moment Interaction Diagram Data Case Number Example in. No.

b in.

A Pt

t

1

2

P 69 139 1

2

15.6

16

15.6

8

3

4

5

55

-19

O

2.4

M

194207 O 207183

P

-107329 36

O

1.26

M

O

49

56

55

52

1.14

0.57

0.45

-0.15

0

P Fbbt

3

non-dimensionalized

M Fbbt2 0.0

0.096 0.109 0.108

0.101

12.1.2.1 Compression Controls- If these non-dimensionalized values are plotted, they form an interaction diagramwhich has muchbroader application than those previously developed. 12-14


A C 1 T I T L E r H D G 93 D 0662949 0509064 609

m

Such a plot is illustrated in Fig. 12.1-4. If another set of examples were calculated using a if steel spacing ratio (g) is

larger steel percentage, another curve would result. Likewise,

changed, stillanother non-dimensionalized curve results. Two such sets of curves are shown in Figs. 12.1-3 and 12.1-5, g = 0.4 and g = 0.8. These curves are valid only for rectangular columns with two equal areasof top and bottom steel (As = A’J, symmetrically placed and

c,

I O

\9 d

U

O

o: .A

O

o?

d

O

9 d

O

O

09

P

O

O

O

oO:

Fig. 12.1-3 Non-Dimensional Column Interaction Diagram 12-15


O

o? O

O

9 O

- Compression Controls


9 3 W Obb2949 05090b5 5 4 5 D

N

-

Fig. 12.1-4 Non-Dimensional Column Interaction Diagram Compression Controls

12-16


A C 1 TITLE*IDG

9 3 M 0662949 0509066 481 M

r"'

9

09 O

Il bc

I

Fig. 12.1-5 Non-Dimensional Column Interaction Diagram - Compression Controls

12-17


A C 1 T I T L E * H D G 93

m

Obb2747 0507067 318 W

adequately laterally tied. The total steel area is A, (A,+ A’, = A,), and pr is the ratio of /

total steel area to gross column area (P, = AJbt). The curves in Fig. 12.1-3 through 12.1-5 are based on the assumption that allowable masonry compressivestress controls the limiting capacity.

12.1.2.2

Tension Controls

-

If the eccentricity of load is large enough, capacity will be

controlled by allowable steel tensile stressrather than allowable masonry compressive stress. Strain gradients for this case are shown in Fig. 12.1-1(b). Using basic mechanics as before, design aids can be developed for this case as well as for the cases illustrated previously. Three such aids are shown in Figs. 12.1-6, 12.1-7 and 12.1-8. These curves control when e

> eb or k < kb. Often the designer cannot readily tell which conditions control.In this case, calculate the required value of pI from both the compression and tension controls curvesand use the larger steel requirement.

12.2 WALLS 12.2.1 Unreinforced Masonry Walls Design of unreinforced masonry walls mustmeet several criteria required in Code Chapter

6 as follows:

1.

The unityinequality (Code Eq. 6-l), which hasbeen in masonry codesfor years, must be satisfied. This

fa

inequality assumes straight line a

Fa

interaction

between

axial

and

flexural compressive stresses.

12-18



93

m

Obb2949 0509068 254

m

IC

O

\9 O

O

‘J:

O

O

r? O

-

Fig. 12.1-6 Column Design Aid Tension Controls - g = 0.4 12-19


A C 1 TITLExMDG 9 3 D 0662949 0509069 L90

O

O

O

O

09

\9

O

O

O

O

7

m

W

Wqd/d

-

-

Fig. 12.1-7 Column Design Aid Tension Controls g = 0.6 12-20


A C 1 TITLE*MDG

93

m

Ob62949 0509070 902

-

-

m

Fig. 12.1-S Column Design Aid Tension Controls g = 0.8 12-21



2.

93

m

0662949 0509073 849

m

In lieu of limits on hlt, the Code uses limits on buckling load (Code Eq. 6-2) to more rationally restrict the slendernessratio.

This requirement does not apply to

reinforced masonry walls.

3.

Flexural tensile stresses caused by eccentricity ofaxial load or lateral load must be limited to allowable stresses from Code Table 6.3.1.1. h i a l tension forces are not permitted in unreinforced masonry walls in Code 6.4.

-

Unity Inequality The unity inequality for checking compression stress

12.2.1.1

f,

+

Fa

fb - S 1 Fb

can be restated in terms of P (allowable), by substituting for

fa

=

P

and fb =

Pe S

By definingthe kern eccentricity

e, astheeccentricity

at which extreme fiberstress

undergoes a transition from compression to tension, another substitution is for F

e, =

3 -

A

Replacing these terms in Code Eq. 6-1, results in Eq. 12.2-1

Code Eqs. 6-3 and 6-4 may be rewritten in the form

where R is a slenderness reduction factor given by

12-22


AC1 TITLE*HDG 93

m

Ob62949 0507072 785

m

Eq. 12.2-2

Eq. 12.2-3 Fig. 12.2-1 illustrates the variation of R over a wide range of hlr values. 1.0 O .8

O .6

R O .4 0.2

O

20

O

40

60 14080120 100

160

hlr

Fig. 12.2-1 Slenderness Reduction Factor Code Eq. 6-1 can now be written as Eq. 12.2-4

This equation can be non-dimensionalized as follows:

P

f/,A

R

S

4

Eq. 12.2-5

+ 3Rp ek

Note: For seismic or wind conditions the right side of Eq. 12.2-5 would be increased by ‘h. 12-23


A C 1 TITLExMDG 93

m

Obb2949 0509073 b l l

This equation in its non-dimensionalized form is plotted in Fig. 12.2-2 for values of R = 1,

R = 0.5, and R = 0.25. Fig. 12.2-2 is a general curve for any masonry strength and wall thickness, except that the ordinate can be increased by a factor of 4/3 when a 1/3 increase in allowable stress is permitted in accordance with Code 5.3.2. This curve may be used to determine the maximum value of P for a given e or the maximum value of e for a given P as limited by Eq. 6-1, i.e., compression allowable controlling. Note: C = 314 If The 1/3 Increase In Allowable Stress In Code 5.3.2 Is Audicable. Otherwise, C = 1.dS - Section Modulus el,=”Ä Area =For Solid Sections

0’3 o -20 P fhAC

O . 15

0.10

O .O5

O

O

0.25

O -75

0.50

1.o

1,25

ele,

Fig. 12.2-2 General Wall Design Aid For Allowable Compression More specific design aids can easilybe developed for specific wall thicknessesand masonry compressive strength. Fig. 12.2-3 illustrates Eq. 12.2-5 applied to an8 in. concrete masonry wall having a specified compressive strength of masonry of 1,500 psi. Four curves are given in Fig. 12.2-3. Load capacities for fully grouted, full mortar bedding and face shell bedding,

12-24



m

0662949 0509074 5 5 8

m

assuming ASTM C 90 (hollow) block, are given for R = 1. The curve for AST" C 90 (solid) CMU is not plotted. The face shell bedding case is also plotted for R = 1/2 For other values of R and hollow CMU face shell bedding,linear interpolationmay be used as an approximation. If more accuracy is needed, the value for load capacity obtained from Figs. 12.2-2or 12.2-3 usingR = 1can be multiplied by a slenderness correction factor (SCF), 4+3-

SCF=R

e

ek

Eq. 12.2-6

3Re 4+-

35

30

25

20

15

10

5

O

1 J

O

o .S

1 .o

1.S

2.o

3.0 2.S

Eccentricity e,in

'ASTM C 90 (Hollow) CMU

-

Fig. 12.2-3 Wall Design Curves 8 in. Concrete Masonry

12-25



m

Obb2949 0509075 494

m

The slenderness correction factor (SCF) is equal to R for zero eccentricity and gets larger as eccentricity increases. For example, for e/ek = 1and R = 0.5, SCF = 0.64. Figs. 12.2-4

and 12.2-5 show the variation of SCF with elek and R, respectively. R= 1

ele k Fig. 12.2-4 Slendernesss Correction Factors vs. elek

1.0

O .8

O .6

O .4

o .2 O Slenderness Reduction Factor, R

Fig. 12.2-5 Slenderness Correction Factors vs. Slenderness Reduction Factor 12-26


A C 1 TITLErMDG 93

12.2.1.2 Euler Buckling

m

0662747 0509076 320

m

- Limitations on Euler buckling as required by Code Eq. 6-2 using

Eq. 6-6 effectively limitsh/? for unreinforced walls. Code Eq. 6-6 divided by Af', to be nonon the approximation that E,,, = 1000 Sm.

dimensionalized is plotted in Fig. 12.2-6 based

Fig. 12.2-6 can be used in conjunction with Fig. 12.2-2, selecting the lower value of PTA.

If a curve such as Fig. 12.2-6 is used, then axial load must be calculated from the value of

Perdand compared to the P from Fig. 12.2-3. The smaller value governs. 1.o

O .6

P, Af 'm

O .4

O.,t

O

1 O

50

75

100

125

150

hlr

Fig. 12.2-6 Non-Dimensionalized Euler Effect vs. hlr CodeEq.(6-2)requires

that P be limited to amaximumvalueof

PJ4. This criteria

indirectly places a limit on the maximum slenderness that is acceptable. In the application of Code Eq. 6-6, the eccentricity shouldbe based only on the actual or estimated eccentricity of the gravity load P. In fact, e is defined in the Code as the eccentricity of axial load. It

is not necessary to include virtual eccentricity which results fromlateral load due to wind or seismic effects.

12.2.13 Flexural Tensile Stress allowablevaluesgiveninCodeTable

- Calculated flexural tensile 6.3.1.1.

stresses must not exceed the

Thistableapplies

to claymasonry and

concrete masonry and both portland cement lime and masonry cement mortars, Type M, S ,


AC1

TITLEwMDG 93 W Ob62747 0507077 2b7 m

or N. Values are also given for stress parallel to the bed joints (horizontal span) and perpendicular to bed joints (vertical span). For concrete masonry, allowable flexural tensile stresses are given for fully grouted construction. Forpartial grouting, interpolation is permitted. Applied flexural tensile stresses resulting from combined axial load and bending moments are calculated from the equation

-

-P + -Pe
Eq. 12.2-7

S

Using terms defined previously, this equation can be rewritten

Eq. 12.2-8

This expression for checking tension is valid only for e > ek and the equation is plotted in Fig. 12.2-7. The value of P obtained from this equation or Fig. 12.2-7 is then compared to the other values of P as controlled by the unity inequality where compression controlsand

by Euler buckling. The smallest value of the three will control. Obviously, flexural tension will not control unless eccentricities, e, become somewhat large, at least > ek. Note: This curve ensures compliance with Code 6.3.1.1 allowable flexural tension provisions.

To use this curve, determine the allowable Fb, from Code Table 6.3.1.1 (applying the 1/3 increase from Code 5.3.2 where applicable). Calculate the kern eccentricity of the wall, ek, from the expression e,

- S

--

Eq. 12.2-9

A

where: S = section modulus A = area

12-28


A C 1 T I T L E * M D G 93

m 0662949

0509078 I T 3

m

0.8

0.7

O .6 O5

O .4 O .3

f ; : : O

1

2

3

4

5

6

7

ele

Fig. 12.2-7 Wall Design Aid for Allowable Flexural Tension Design aids can easilybe constructed from Eq.12.2-8. Fig. 12.2-8 illustrates a typical design curve for 8 in. concrete masonry walls. Curves are shown for solid walls, walls with hollow units with fullmortar bedding, and walls with hollow units with face shell bedding. Different allowable stresses are shown to illustrate the effect. The allowable load is read directly from the ordinate. In cases where allowable stresses are increased by one-third, the designer can either increase Fb, or simply increase the calculated load, P,by one-third.

12.2.2 Reinforced Masonry Walls Design of Reinforced Masonry Walls for combined axial load and bending requires that several conditions be met: 1)

The average axial compressive stress cannot exceed the allowable values given by Code Eqs. 7-1 and 7-2,

12-29


A C 1 T I T L E * M D G 93

m

0662949 0509079 0 3 T

m

P , kips

S

e . in.

'ASTM C 90 (hollow) CMU

Fig. 12.2-8 Wall Design Aid - 8 in. Concrete Masonry 2)

The extreme fiber compressive stress due to combined axial load and bending and cannot exceed the allowable value of 1/3Sm,

3)Thesteel

tensile stress cannot exceeditsallowablevaluegiveninCode

7.2.1.1. An important code provision affecting the design of reinforced masonry walls is contained

in Code 7.2.1.2(a).

This requirement permitscompression

forces tobe

resisted by

compression reinforcement only if the lateral supportrequirements of Code 5.9.1.6 are met. Sinceitisvirtuallyimpossible

to meet these provisions in walls, the contribution of

reinforcing steel to compressive forces must be neglected for walls. This very conservative 12-30


A C 1T I T L E b f l D G

93

m

0662949 0509080 851

m

requirement does, however, result in greatly simplified design procedures for eccentricities

of e < t/3. Assuming the wall to be fully grouted with steel at the wall centerline for this region, the capacity of a wall is independent of the amount of reinforcement. Only when e

> t/3 is it necessary to account for reinforcement in the design. This concept will be

clearly illustrated with the development of the interaction diagram.

12.2.2.1 Interaction Diagram for Reinforced Walls - The interaction diagram for reinforced masonry walls withreinforcement located at the center of the wall can be divided into three distinctregions: e c t/6;t/6

S

e

S

t/3; and e > t/3. Each of these will be developed

separately, and a final diagram developed. The masonry is assumed to have zero tensile capacity. The developed equations are valid for a solidly grouted wall or for a section in which kd e

5

t/6

S

face shell thickness.

This region of the diagram corresponds to a strain gradient ranging between Case

1 and Case 2 as shown in Fig. 12.2-9. Since Case 2 coincides with the kern eccentricity, this case corresponds to e = t/6 for fully grouted sectionsor S/A for partially ungoruted sections.

T

Tension

Reinforced Wall

Compression

Strain Gradient

Fig. 12.2-9 Variable Compression Throughout Cross Section The resulting interaction diagram for any pointin this region is shown in the top portion of Fig.12.2-10.Since

the reinforcement is in compression and cannot be laterally tied, its

12-31


A CT1I T L E v M D G

Obb2949 0509083 7 9 8 M

93

effect is neglected in thisregion.

The shape of the curveconnecting

the points

corresponding to Case 1 and Case 2 is, very conveniently, a straight line. O
-1"

"

Fbbt

t/6

Eq. 12.2-10

Fbbt2'

Note: One would still have to check PIA versus Code Eq. 7-1 or 7-2.

\

O .E

P=Fbbt - 6P e/t

O.Í O .6

Fbbt

0.5 O .4

]

t16

e

t/3

0.3

o .2 0.1

c

O

0.02 0.04 0.06

0.08 0.10

Pe

"7

F,bt

-

Fig. 12.2-10 Interaction Diagram Reinforced Wall t/6

ic

e

ic

t/3

This region of the diagram corresponds to strain gradients between Case 2

and Case 3 on Fig.12.2-9.

Thisregionis

a curvedlinewhen

plotted using the normal

interaction diagram parameters as shown in the middle section of Fig. 12.2-10. It is linear, however, if expressed in terms of P/F& and e/t and can be expressed as Eq. 12.2-11

e > t/3

In this region the steel is in tension and resists all tensile stresses.

Points on the

interaction diagram in this region correspond to strain gradients between Case 3 and Case

12-32



93

m 0662949 0509082 624 m

4 in Fig. 12.2-9. Case 4 can represent any magnitude of tensile strain in steel. All values

obtained on the interaction diagram for this region assume that the allowable compressive stress in masonry, Fb,controls the capacity. This

further implies that k

2

kb, where kb is

defined such that c

Eq. 12.2-12

Points on the interaction diagram for various selections of the location of Case 4 are shown in the lower portion of Fig. 12.2-10 for a broad selection of values of n p , where n is the modular ratio EsErnand pI is the ratio of total steel area togross area of masonry. In order to facilitate more accurate use of this figure as a design aid, Fig. 12.2-11 is provided as an enlargement of the lower part of Fig. 12.2-10. Values of k are also shown in Fig. 12.2-11 in order to ensure that k > kb.

e > e,

For large eccentricities, steel allowable stress will control. Values of n p obtained

from Fig. 12.2-11 would not be correct for such cases.

Another interaction diagram must

be developed by limiting the steel tensile stressto the allowable value and varying the strain

gradient in the cross section to obtain curves which correspond to different values of np. Such a diagram is illustrated in Fig. 12.2-12. This diagram also indicates valuesof k which can be used to verify that k C kb is a necessary condition for using Fig. 12.2-12. For partially grouted reinforced walls the following procedure should be used.If the neutral

axis depth, k d , is less than the face shell thickness,

ffi

of partially grouted units, the values

obtained from the interaction curve need not be changed. Only the value in Code Eq. 7-1 or Code Eq. 7-2 would have to be adjusted to account for the change in area and radius of gyration. Since the interaction diagrams also include values confirm whether or not M

5

tf

of k, the designer can easily

If kd > tr, a basic mechanics approach should be employed.

12-33



93

m

Obb2949 0509083 560

m

ö

"i9 O

""-3

8'0 = Y

" "

L'O =

m

9

O

d

9 O

m 9 O

O

O

o? O

2 ö

2 o

-

Fig. 12.2-11 Interaction Diagram Reinforced Wall 12-34


O

AC1

TITLE*MDG 93 m 0662949 0509084 4T7 m

-

Fig. 12.2-12 Interaction Diagram Reinforced Wall

12-35


A C 1 TITLExMDG 93

12.2.2.2 Typical Iterative Method

Ob62949 0509085 3 3 3

m

- In the United States most masonry elements are walls.

Walls commonly have to resist the effects of combined axial and bending loads. Bending loads can be either in-plane or out-of-plane as shown in Fig. 12.2-13.

In-Plane

Out-of-Plane

Fig. 12.2-13 Wall Loading There are no easyclosed form solutions to the stress equations for walls subjected to bending combined with axial compression.

Structural engineers typically rely on iterative

methods or design charts. Designcharts for most masonry systemsare not readily available. See MDG 12.2.2.1for discussion concerning developmentof wall interaction diagrams. Thus iterative methods are most common. The following procedure is one example of an iterative method (12.2.1, 12.2.2).

Fig. 12.2-14 shows a diagram of a wall. The axial load is applied to the centroid of the

12-36


A CT 1I T L E x M D G

73

Obb2749 0509086 27T

masonry area with the applied moment adjusted as necessary. This diagram,although depicting in-plane bending, is applicable to both in-plane and out-of-plane bending of walls. There are three possible conditions for the wall. It is either uncracked, cracked with the steel in compression(the extent of the crack has not reached the steel), or cracked with the steelin tension. The first step in the analysis is to determine which condition applies to the wall. It will depend on both geometry and

Fig. 12.2-14 Flexure and Axial Wall Loading

loading.

Fig. 12.2-15 shows the typical bending and axial load diagram. Loading conditions beyond the limits of the diagram are beyond the allowable stresses. Loading conditions inside the limits od the diagram are within the allowable stresses. The condition of the wall canbe quickly established by using the non-dimensional parameter

MIPd. This parameter represents a straight line radiating from the diagram’sorigin as shown. By summing moments, it can be shown that certain values of the MIPd divide the diagram into the wall conditions. These values are given in MDG Table 12.2.1. The determination of the allowable moments of Regions 1 and 2 can be obtained in closed form with a simple equation. In Region 3 it is more complicated. Normally the process of evaluating a wall consists of guessing the thickness, length, etc.(or as given by the architect) and then determining the required area of steel. This is the basis for the following iterative solution for Region 3.

12-37


AC T I1T L E * H D G

93

Obb2949 0509087 LOb

P4

a

O H

12-38



m

93

0662749 0509088 042

In Region 3, the wall is either limited by the compression of the masonry or the tension in the reinforcement. Begin by assuming the tension in the steel controls. By making an initial guess about the location of the neutral axis, the MDG Eqs. 12.2-15 and 12.2-18 provide an iterative process that quickly converges. The equations are derived from the summation of forces and moments and using linear stresshtrain relationships.Typically, converges, but sometimes results in negative

the system

areas of reinforcement. Do not allow this

negative solution to discourage continuing with the process. Following convergence, a check is made of the initial assumption that tension in the steel controls. If valid, the analysis is complete. If not valid, then the compression controls and a new set of equations that do not require iteration result. MDG Eqs. 12.2-20 and 12.2-21. The following is a detailed step-by-step procedure of the analysis described above. Step 1

Determine the wall condition 1. Calculate A as the distance from the axial load to the centroid of the tension steel divided by d. 2. Calculate the quantity MIPd.

3. Use Table 12.2.1 to determine the region for analysis (See Fig. 12.2-15). Table 12.2.1 Flexure and Axial Loading Region

- Wall Analysis

Condition of Wall

Test

Wall is in compression and not cracked

Pd

2

Wall is cracked but steel is in compression.

Pd

5

(5

3

Wall is cracked but steel is in tension.

MPd

>

(-A) ;

1

I

12-39


-A)

A C 1 T I T L E x U D G 93 W 0662949 0509089 T89

Step 2

m

Calculate the allowable moment. REGION 1: The moment is limited by flexural compression in the masonry. Eq. 12.2-13

If M, is greater than M applied, the section is satisfactory. REGION 2 The moment is also limited by flexural compression in the masonry. Eq. 12.2-14

If M, is greater than M applied, the section is satisfactory. REGION 3: The moment may be limited by either the compression in the masonry or tension in the steel.

An iterative approach is required.

1. Assume a compression centroid location, a. 2. Perform the followingiteration that assumes the tension inthe steel controls:

Mp = P[+

-u]

Eq. 122-15 Eq. 122-16

c=

(P+AsFan

Eq. 12-2-17

Fsb Eq. 12.2-18

Where :

a = Estimate of internal compression load centroiddistancefromthe extreme compression fiber

P

= Applied axial load at the center of the wall

M = Appliedmoment 12-40



m 0662949

93

O509090 7 T 0

m

A, = Area of trim steel Mp = Moment of applied axial load with respect to the centroid of internal compression force 1, = Length of the wall for in-plane bending 1, = Thickness of the wall for out-of-plane bending

d

= Distance from the extreme compression fiber to the steel centroid

F, = Allowable steel tension stress b

= Width of the wall for out-of-plane bending

b = Thickness of the wall for in-plane bending Use a, for a, and repeat until the value of a converges.

3. If the iteration converges and the resulting a is less than the following value, the wall is limited by the tension reinforcement and the analysis is complete. Otherwise continue: d

[

31+-

“3

Eq. 12.2-19

4. If the value of u is larger than the above value, determine the required steel area using the following: a =

_2

4

d2

2

2(PAd + M ) 3Fbb

Eq. 12.2-20

If 3a is larger than d , then the masonry wall needs to be longer and/or thicker. If 3a is smaller than d, then the steel area is

Eq. 12.2-21

Often this area is smaller than minimum steel requirements oris a negative value. Minimum steel should then be used. It is normal

12-41



m

Obb29Y9 0509093 b37

m

practice in loadbearing masonry buildings to use at least (2) #5 minimumum bars for trim reinforcement.

12.3 PILASTERS The description of pilasters given in MDG 11.2.1 through 11.2.4 is sufficiently general to apply to nearly allpilasters whether they are subjected to flexure onlyor toflexure and axial load as described in this section.

It is recommended that MDG 11.2.1

-

11.2.4 be read

before proceeding with this section,so that the reader may become acquainted with loading conditions, coursing layouts, and effective sections.

123.1

CriticalLoadingCases

As noted in Fig. 11.2-1, pilasters may be subjected to vertical loads at roof or floor levels, and lateral loads along their height. If simple supports are assumed at the top and bottom (refer to MDG 11.2.5 for discussion of boundary conditions), moments from each load will be as shown in Fig. 12.3-1 for a single story pilaster. While

the sense of the gravity load

moment will be defined by the orientation of the eccentricity,wind load moment may reverseitself

if a suction is applied rather than direct pressure. If a seismicloadis

considered, both senses of sway should be considered as well. Thus, moments resulting from gravity and lateral force must be added as noted in Eq. 12.3-1. If the lateral force can act in either direction, then the effects of gravity and lateral loads should be summed algebraically, and the two cases (sum and differences of effects) should be checked. M = -W h 2 * -Pe Eq. 12.3-1 8 2 If the lateral moment is small relative to the gravity moment, then the critical case may be at the top of the pilaster with gravity loads alone, in which case, Eq. 12.3-2 would apply.

M

=

Pe

Eq. 12.3-2

A case involving zero moment may govern if the gravity load is relatively large and the

12-42


A C 1 TITLE*MDG

93

m

Obb2949 0509092 573 M

eccentricity small. In this case, the critical location for design would be at the base where the vertical compressive stress is maximum.

Fig. 123-1 MomentDiagrams for Pilaster 123.2 DesignConsiderations for UnreinforcedPilasters

Flexural tensile

stress usually governs

compressive stresses areanorder

for unreinforced pilasters. Since allowable

of magnitude more than allowabletensilestresses,

compression typically does not control except for the case of small eccentricity. However, both cases should be checked. For nearly all practical cases, design is controlled by flexure rather than axial force. It is likely that wind or seismic cases with a minimum

amount of

vertical compressive force will govern. However, if vertical load is applied at a relatively large eccentricity,then it may be conservative to consider the largest gravity forcestogether with the lateral forces, since the gravity moments will increase flexural tension stress. With large eccentricities and small lateral forces, gravity effects alone may govern, since the onethird increase in allowable stressesdoes not apply. In summary, the following loading cases should be considered when flexural tensile stress governs.

12-43


A C 1 T I T L ExM DG 93 W Ob62949 0509093 q O T W

1. full lateral

+ minimum gravity (for small eccentricities)

2. full lateral

+ maximum gravity (for large eccentricities)

3. maximum gravity (for large eccentricities, small lateral) According to Code 5.3.1, when earthquake forces are considered, deadload shall be multiplied by 0.9 for loading case (1). When effects of wind are considered, minimum dead loads shall be taken attheir full value. No load factor is applied to the wind or earthquake loads. Although compressive stress will probably not control for an unreinforced pilaster, it still should be checked (it could control for cases of high gravity force and small eccentricity as noted above).Compressive

stresses resulting from both axial and flexureshould

be

considered using Code Eq. 6-1: c

-+J~

c

J' S

1 for gravity loads only

Eq. 12.3-3

Fa

c

c

Fa

Fb

Ja Jb + <

1.33 for combined gravity and wind or earthquake

Axial compressive stress, fa, should be computed over the minimum net area of masonry. Flexural compressive stress,

fb, should be calculated based on the section modulus for the

minimum net section. The minimum net section shallbe based on the mortar bedded area for ungrouted construction. The allowable axial compressivestress is given by Code Eq. 6-3 and Eq. 6-4. c

Eq. 12.3-4

or Eq. 12.3-5

12-44


A C 1 TITLE*:MDG 93

m

Obb2949 0509094 3Yb

m

The allowable flexural compressive stress is given by Code Eq. 6-5.

Fb

=

(U31 f',

Es. 12.3-6

If wind or seismic is considered, then the "1"in Eq. 12.3-3 may,be replaced by a "1.33" per Code 5.3.2. Even though compressive stress may not govern the size of an unreinforced section, it will dictate what the required prism compressive strength should be. Shear strength of an unreinforced pilaster subject

to both axial loads and flexure is no

different than for the case of pure flexure (see MDG 11.2.5). 1233 DesignConsiderations for ReinforcedPilasters

All masonry subjected to tension is neglected in a reinforced pilaster. The unity equation (Eq. 12.3-3) does not apply for checking the compressive stress under combined flexure and axial stress since the section is considered to be cracked, and the axial force will not be distributed across the entire net area. Instead, the Code 7.3.1.2 states that the compressive stress due to flexure in combination with axial load shallnot exceedfJ3, provided that the axial stress alone is less than the allowable F, given by Eq. 12.3-4 or Eq. 12.3-5. When the compressive stress is due to combined lateral and gravity forces, the allowable stresses may be increased by 'h. As noted in MDG 12.1.2, an axial load-moment interaction diagram can bedeveloped

byassigningthislimitingcompressivestress

determining different combinations differentassumed

ofaxial

to theextremefiber,

and

load and bending moment that result from

depths to the neutral axis.Similarinteractioncurves

can also be

developed based on limiting the reinforcement to its allowable tensile stress as prescribed in Code 7.2. These tension-controlledcurves will govern the reinforcementdesignfor pilasters with relatively small axial force. The effectivesection will depend on whethertheweb

or flange isin

compression,as

discussed in MDG 11.2.4. If the bending moment.is primarily a result of lateral force rather 12-45


A C 1 TITLEsMDG 93

m

0662949 0509095 2 8 2

m

than eccentric vertical force, then moment reversal needs to be considered, and the critical direction of bending mustbe identified. If compression controlsthe design, then thebending that produces compression onthe web will control (Fig. 11.2-4 (a)), because the compressed area is much smaller than if the flange were in compression (Fig. 11.2-4 (b)). If the pilaster is reinforced symmetrically, and the effective flange width isunreinforced, the critical sense of moment will be when the webis in compression. This assumption

will

reduce the problem to one of a cracked rectangular beam for which standard load-moment interaction diagrams are available. Design may then be as simple as plotting the normalized load and moments on the standardcharts to check that the assumed section size is adequate and to estimate the needed reinforcement. An equivalent rectangular section can also be used for the case of the flange in compression, provided that the neutral axis is within the wall thickness for a fully grouted wall, and within the exterior face shell for an ungrouted wall. If reinforcement is not placed symmetrically across the pilaster section, then both senses of bending need to be considered. It may not be obvious which sense will control the design. For example, if the effective flange width contains reinforcement as shown in Fig. 12.3-2, then compression stresses will be highest when the flange is in tension. However,the critical case for tension will be when the web is in tension becausethe amount of reinforcement will be limited. For eitherof these two cases, an analysis based on a cracked rectangular section would suffice provided that the compressed zone remains a rectangle. In the rare case that the compressed area is not rectangular, an interaction diagram needs to be generated for theparticular T-shaped pilaster. No standard load-moment interaction diagramsexist

for suchsections,becausethey

cannot be generalized in terms of

non-dimensionalized parameters. To generate such a diagram, maximum compressive stress is set equal to a constant Fb (Fig. 12.3-3a), and the depth to the neutral axis, M ,is varied arbitrarily to result in different combinations of axial force and moment. This will define points on thel'compression-controls" interaction curve. Correspondingly, tensilestress in the

12-46


AC1

TITLE*NDG 93 M Obb29L19 0 5 0 9 0 9 b L19

reinforcement at the extreme layer is set equal to the allowable tensile stress (Fig. 12.3-3b), and the resulting combinations

of axialforce and moment define the tensioned-control

interaction curve.

(a) Wall Reinforced With Flange In Tension

6t

6t

(b) Wall Reinforced With Web In Tension

Fig. 123-2 Effective Sections for Pilaster with Reinforced Wall For sections with unsymmetrical reinforcementor geometry, axial loads mustbe referenced to theplastic centroid (uncracked transformed section) rather than to the geometric centroid (cracked transfromed section). The plastic centroid is located at the centroid of all internal forces when a uniform stress distribution, with an amplitudeequal to Fb is applied acrossthe transformed section. For this stress distributionthe plastic centroid is defined as the line of action ofthe resultant of all internal forces. This definitionis consistent with zero curvature of the section under pure axial load.

12-47


A C 1 TITLElltNDG 93 W Obb2949 0509077 055 W

L", -

d

Plastic Centroid

1"'

Fig. 12.33 Stress Distributions for Development of Load-Moment Interaction Diagrams for Non-rectangular Sections

REFERENCES 12.2.1 Tawresey, J. G., "Applied Stress Equations-Walls With Axial Load Combined With Bending Moment," TMS Journal, Vol. 5, No. 2, July-December 1986, pp.T16-T20. 12.2.2 Tawresey, J.G., "Masonry P-M Diagrams July-December 1989, T31-T35.

12-48


Made Easy," TMS Journal, Vol. 8, No. 2,


Example 12.1-1 RCJHotel

m

Obb27Lt7 0509078 T 7 1

m

- Lobby ColumnDesign

Design the column at Grid Line E and Grid Line 3on the first floor of the RCJ Hotel, Wall Construction Option A. See MDG Fig. 9.1-6. The column is rectangular in plan, having a

2 ft

- 8 in. dimension along

Grid Line E. The width b will be established by structural

requirements. From gravity load analysis in MDG 9.1.3.1, the loads are D = 181 kips, L = 143 kips. Lateral load analysis (Seismic Zone 4) results in a shear force of 31 kips and

strong axis moment of 137 ft-kips and axial compression of 50 kips. Clay Brick Construction

11

f L = 2,400 psi E,,, = 2.4 x lo6 psi (from test)

32"

and

It

= 12

Calculations

Try b = 16 in. 5.3.1

Loading Combinations

D + L

P

=

181 + 143 = 324 kips

Minimum eccentricity

e,,, = O.lt = 3.2in. D + L + E

P

=

181 + 143 + 50 = 374 kips

M = 137 ft-kips

12-49


5.9.1.3

A C 1 TITLEtMDG 93

0662949 0509099 928


Calculations

0.9D

+E

Fa = 0.33 f111

P

=

= 163

+ 50 = 213kips

0.33(2,400 psi)

=

800 psi

=

0.8 ksi

7.3.1.2

Using the interaction diagram (MDG Fig. 12.1-5) for g = 0.8

P -

324 kips 0.8ksi (16in.)(32 in.)

"

Fbbt

=

0.79 np, = 0.21

D+L =

Fbbt2

3.2 in. 0.79 -= 0.079 32 m.

For the other two loading cases a 1/3 increase in allowablestress F b is permitted.

12-5O


5.3.2

TITLE*RDG 93 m 0662947 0507100 4 7 T m

AC1


D+L+E

1

Code Reference

374kips = 0.69 1.33(0.8 ksi)(16in.)(32 in.) $ t

=

np, = 0.20

374kips (4.4 in.) 1.33(0.8 hi)(l6 in.)(32

“

Fbbt2

P -

=

0.094

in.)2

213 kips = 0.39 1.33(0.8 ksi)(l6 in.)(32 in.)

”

F,bt 0.9D + E

*

<

=

Fbbt2

0.39

7-72 32 in.

i n *

=

np, = 0.02

0.094

The larger amount of steel area results from loading case D

+L

np, = 0.21 0.21 pt = - = 0.0175 12 A , = p,bt = 8.96 in.2

Select 8 -#lo bars, 4 in each face A, = 10.16 i n 2 p, = 0.020 C 0.04

.-.OK

5.9.1.4 12-51


A C 1 TITLExMDG 93

m

Obb2949 0509LOL 30b


Calculations

Reference

Code

The Code imposes an upper bound on axial compressive stress

Fa=

(- 3A [

1-

EQ. 7-1

Using the trial size of 32 in. x 16 in. and an effective height, h, of 10 ft

- 10 in. minus

the

depth of the second floor slab (8 in.), and a radius of gyration of 0.289 x 16 in. about the weak axis, the allowable stress becomes

Therefore the maximum allowable load is

Pa = Fabt = 0.579 x 32 in. x 16 in. = 2% kips

324 kips

Note that the allowable axial load computations have neglected any contribution

of the

longitudinal column steel which is conservative. See MDG 8.6.2 and 12.1.1. Hence a 32 in.

x 16 in. column is too small to meet this requirement in spite of otherwise being capable of carrying the imposed loads. Actually, this calculation should

be carried out immediately

upon selecting a trial size, not at the time shown here. This sequence was used in order to illustrate the use of thedesignchartsforacolumnwithsignificantloadsrequiring appreciable amounts of reinforcing. If the column is laterally supported to resist bending about theweak axis, then the 32 in. x 16 in. columncan be retained. One couldalso consider increasing the design value off:.

I’)

Try a larger colu n size, 32 in. x 24 in. Fa = 2,400 psiT1 122 h’ 140(0.289) (24 in.) 4

(

=

591 psi

12-52



0662949 0509102 242


Calculations

Pa

=

0.591 ksi(32 in.)(24 in.) = 453 kips > 324 kips

.-.OK

Now the procedure used previously to determine steel reinforcing requirements for the 32

in. x 16 in. column will be repeated for the 32 in. x 24 in. column. Using the interaction diagram (Fig. 12.1-5) for g = 0.8

P -

324 kips

”

=

0.53

0.8 ksi ( 2 4 in.)(32 in.)

F,bt

np, = O

D + L =

0.53(0.1) = 0.053

Fbbt2

For the other two loading cases a 1/3 increase in allowable stress Fb is permitted.

I

$ t

374 kips = 0.46 = 1,33(0.8 ksi)(24 in.)(32 in.)

D + L + E 374 kips (4.4 in.) 1.33(0.8 ksi)(24 in.)(32

=

”

F,&*

12-5 3


in.)2

0.063

I

I

np, = O


73

m

Obb2747 0507L03 L87

m


and

Calculations

Code Reference

I1

$ t

0.9D

+E

213 kips = 0.26 = 1.33(0.8lcsi) (24 in.)(32 i n.)

I

npt = O

Since allthree loading cases resultin a zero calculated steel area, use minimum columnsteel area (Code 5.9.1.4), p, = 0.0025 A , = O.0O25An= 0.0025(24 in.)(32 in.) = 1.92 i n 2 Select 4 - #7 bars, 2 in each face, A , = 2.4 in.2

12-54


5.9.1.4

m

A C 1 TITLEvHDG 93

0662949 0509304 015

m

Example 12.1-1 Coned. and

Calculations

Repeat the design of the same column using CMU construction having pt

r,,,= 1,500 psi,

= 14.3.

Pa =

[ ( 1-

19500PSi

4

)'I

24 in. x 32 in. = 283 kips

122 in. 140(6.94 in.)

324 kips

The reduction in allowable stress results in having to increase the column width. Try a 32 in. square column.

[ (

Pa = 19500 psi 1 4

-

122

in*

r]

140(9.25 in.)

32 in. x 32 in. = 371 kips > 324 kips .: OK

5.3.1

Loading Combinations

D+L

- P-

Fa&

-

324 kips (0.5) (32in.)(32 in.)

=

0.633

e/t =

0.1, np,

=

0.02

p, = 0.0014

The other loading cases (D

+ L + E and 0.9D + E ) result in zero calculated steel area.

Minimum steel for columns is p, = 0.0025 (Code 5.9.1.4), hence A, = 0.0025 (32 in.) (32 in.) = 2.56 in.2

Use 4 - #8 bars

Provide lateral ties in accordance with Code 5.9.1.6. Ties shall be 1/4 in. diameter spaced at 12 in., with the first tie located not less

than 6 in. from the bottom or the top of the

column. Since Code 8.2.3 limits joint reinforcement to one-half joint thickness. Thus ? in.'i

12-55


AC T I1T L E a N D G

Obb2'74'7 050'7305 T53

'73

Example 12.1-1 Cont'd. ~~~~~~~

~


Code Reference

ties cannot be placed in a 3/8 in. mortar joint. The CMU must be notched out to provide sufficient clearance ties as shown in the figure.

32"

.

As an alternate the design might consider placing reinforcement inthe center solid grouted area since this would eliminate notchingthe masonry units. Designer would need to verify steel required based on new location of steel.

32"

12-56



93

m 0662747 0507306

998

m


and

Calculations

Code Reference

For Seismic Zones 3 or 4, use 3/8 in. ties spaced at 16 in. centers with the first tie located no more than 8 in. from the top or the bottom. In Zones 3 or 4 the tie must be embedded in grout, not in the mortar joint (Code Appendix k 4 . 6 ) . Check shear requirements for CMU column. From Code Eq. 7-3, and a design shear force due to D

+ L + E of 31 kips,

Fv

=

1.33@

=

1.33Jm

=

38.7 psi x 1.33 > 38.4 psi

Hence no shear reinforcing is required. Check shear requirements for the brick masonry column.

F" Since fv<

=

1.334m

=

65 psi > f v

F,,, shear reinforcement need not be provided. Actually, the assumed value

= 0.9 is conservative since for the loading case

of j

that produces the design shear load of 31

kips the entire column cross-section is in compression. Although not required, calculations for shear reinforcement are shown to illustrate to the reader the procedure for shear consideration for columns.

12-57


Obb2949 0509307 8 2 4 U

A C 1 TITLExMDG 93



Code Reference

Shear reinforcement is determined by Code Eq. 7-10, solved for S using #3 ties, Grade 60 steel. S="

A,F#

V

- (0.22 i~)~(24,000 psi)(28 in.) = 4.77

in.

31,000 lb

Using larger diameter ties or a double tie configuration can increase the spacing. Use a #4 tie, hence

S

= 8.67 in. Use #4 ties at 8 in. centers to meet shear requirements. This size

and spacing will also satisfy Code 5.9.1.6.

12-58



H Obb29Y9 0509308 7 6 0

93

-

Example 12.2-1 T M S Shopping Center Design of Reinforced Loadbearing Wall For the North Wall on Grid Line A of the TMS Shopping Center determine the size of wall needed for Wall Construction Option B (Reinforced Concrete Masonry).

P

J

f', = 1,500 psi Unit Compressive Strength = 1,900 psi Type S Mortar Grade 60 Reinforcing Steel.

and

Calculations

Reference

Code

Fb = 1/3f', = 500 psi

7.3.1.2

F, = 24,000 psi

7.2.1.1 Table 5.5.1.3

E, = 2.08 x 106 psi From analysis MDG 9.1.1, PD = 315plf,

PL =

12-59


630 plf,

A C 1 TITLEvMDG 93

m

Obb2949 0509309 bT7

m


Calculations

Eccentricity of dead and live load, e, = e,

Wind load

Note:

=

wZ2 -- 2Op~f(16ft)~ in 20 psf, MW= x 12 8 8 A

=

Parapet loading is neglected for simplicity.Moment

7,680 in.-lb at midheight

of7,680in.-lbisslightly

conservative. For purposes of analysis, the base of the wall is considered pinned.

Try 8 in. wall, solid grouted t = 7 5/8 in., wall weight = 78 psf, e,

-- eL -- 7.63 in. 2

1 in. - 4 - -

2

3

in.

=

1.98 in.

Wall weight= 156 plf (at joist bearing), 1,404 plf (at base) and 780 plf (at midheight) Loading Combinations D + L

P

= 315plf

M = (315 The eccentricity of D

D+L+W

+ 630plf + 156 plf

= 1,100 plf at joist bearing

+ 630) 1.98 = 1,870 in.-lb/ft at joist bearing plate

+ L including wall weight is maximum at the joist bearing elevation,

P = 1,730 plf @ 8 ft abovebase

The maximum moment at midheight results from eccentricity of D

12-60


+ L plus wind load,


m

93

Ob62949 0509LLO 3L9

m

ExamDIe 12.2-1 Cont'd. and

Calculations

M = 1'871 in"1b + 7,680 in.-lb 2

=

8,610 in.-lb/ft

Using the interaction diagram (MDG Fig. 12.2-11)

P -

Pe

v

1,looplf = 0.0241 5OOpsi (12 in.)(7.63h)

"

Fbbf

npt = O

D+L pe 1,870h.-lb F,bf 500psi (12in.)(7.63in.)2

"

P -

=

0.0054

l ,730plf 4/3 (500psi)(12in.) (7.63i.n.)

"

Fbbf

= 0.028

np, = O

D+L+W

P

pe 8,620h. -lb F,bf 4/3 (5OOpsi)(12in.)(7.63in.)2

"

12-61


=

k

0.0185 J

=

0.15

A C 1 TITLEnMDG 93

Obb29Y9 0509LLL 255

m


Calculations

Code Reference

kb

=

-

Fb

500 psi

24 O , oo psi n

=

0.225 > 0.15

14 ,

MDG Fig. 12.2-11 is not applicable (i.e., compressive allowable stress is not the controlling factor) since k < kb. Use MDG Fig. 12.2-12 (steel controls), with the same ordinate and abcissa and read pt

Fs -

=

0.012, k

=

0.16

kb :.

OK

to use Fig 12.2-12

'b

D+W

P

+ 780 = 1,100 plf @ 8 ft above base

= 315

e = -M- - 7,990 in.-lb/& = 7.3

P

plf 1,100

P -

1,100 plf = 0.018 4/3(500 psi)(l2 in.)(7.63 in.)

"

F,bt

in.

= 0.0180

12-62


Example 12.2-1 Cont’d. Code Reference


From MDG Fig. 12.2-11, np, = 0.0024, k = 0.13 < kb .: compression does not control and

MDG Fig. 12.2-12 must be used.

A,

=

p,bt

=

p,[$]?bt

=

in.)(7.63 in.) = 0.034 h2/ft

(0.0180) (12

Spacing of various bar sizes

Bar

Area

#3

0.11 i n 2

38in.use

4

0.20

69

5

0.3 1

&e*.

#3 @ 32 in.

108

The average axial compressive stress cannot exceed Code

Eq. 7-1, or 7-2 depending upon

hlr.

h = 16 x 12 = 192 in.;

r = -

m

= 0.29t =

12-63


2.12 in.

Code C.6.3.1

A C 1 TITLExMDG 93

m Obb2949 0509LL3 O28 m


hlr = 19212.21 = 86.9 c 99

Fa =

:A

(Lr]

[l - 140r

Max applied stress is

=

Reference

2.

Code

Code Eq. 7-1 controls.

i(1,500 4 psi)[ 1 -

2,350 plf (12 in. x 7.63 in.)

=

r]

( 140lg2 2.ia21 in.

26 psi,

x

=

231 psi

OK

Use 8 in. concrete masonry wall with #3 vertical bars spaced at 32 in. centers, fully grouted with$,

= 1,500 psi.

12-64



m

Ob62949 0509LL4 Tb4

m

Example 12.2-2 T M S Shopping Center - Design of Unreinforced Loadbearing Wall For theNorth Wall on Grid Line A of the TMS ShoppingCenter, determine the size of wall needed for Wall Construction Option A (Unreinforced Concrete Masonry).

and

Calculations The obvious difficulty in designing this wall without reinforcing willbe accommodating the very large eccentricity for the D

+ W combination of 7.3 in. calculated in

MDG Example

12.2-1 without exceedingthe allowable flexural tensile stresses inCode Table 6.3.1.1. Using MDG Eq. 12.2-8 with section properties from MDG Appendix A for a fully grouted 8 in. wall,

Note that a similar calculation using a trial8 in. ungrouted section results in failure to satisfy 12-65



m

0662949 0509LL5 9 T O W


Code Reference

MDG Eq. 12.2-8 using either full mortar bedding or faceshell mortar bedding.

A similar check of other load cases using the fully grouted 8 in. trial section indicates

(1.27 in.)

-

The trial 8 in. solid grouted section satisfies MDG Eq. 12.2-8 for all three loading conditions.

Now check the unity equation for the same loadings, by MDG Eq. 12.2-4.

D + L, * .

(1.27 h)(4)

(1.27 in.)(4)

D + W,

P S (91°5 h2)(229 Psi)(1*33) (7.3 in.)(3)(0.611) 1+ (1.27 jn.)(4)

12-66


=

7,670 lb > 1,100 lb

.. OK

A CT1I T L E * I D G

m

Obb2949 0507LLb 837

93


and

Calculations

All three loadings meet the requirements of Code Eq.6-1 using the trial section. Finally, check Euler Buckling (Code Eq. 6-2) for each loading case.

D + L, e

=

1.98 in.,

PC =

~ ~ ( 2 . 0 8 ~@(M3 1 0 ~ h4) 1-0.577(1.98 h )

(192

P

= 1,100

lb

- = 27’400 4 4

Euler buckling loads

lb

for load cases

lb

27,400 lb

(2.20 in.)

in.)2

= 6,850

I’

=

.: OK

D +L + W and D+ W are calculated using actual

eccentricities from gravity loads without including effects of lateral loads. Hence, the largest value of e will be 1.98 in. from the previously considered load case. Since the allowable load from this case exceeds the design loads of cases D + L + W and D+ W, no furthur check is necessary. In summary, use an 8 in. solid grouted unreinforced concrete masonry wall.

12-67


A C 1 T I T L E * M D G 93

Example 12.2-3

M 0662949 0509LL7 773 M

-

DPC Gymnasium Design of UnreinforcedMultiwytheNoncomposite Masonry Wall

Design the North Wall (Grid Line A) of the DPC Gymnasium using Wall Construction Option A. Assume a horizontal span of 16 ft to carry wind load to pilasters. The wall must still span vertically to resist eccentric bearing load from roof trusses. Wall

Construction

Option A consists of a 4 in. nominal brick wythe, 8 in. block, 3 in. cavity. Roof Truss

Type S, PCL Mortar Concrete Masonry fYm

Brickl

= 1,500 psi

E, = 2.08 x lo6 psi Clay Masonry L

L

L

3.63"13"1

Unit Strength = 8 ksi

L

7.63"

E, = 2.4 x lo6 psi Wind Load = 20 psf

and

Calculations

Discussion

Reference

Code

Since the roof truss acts only onthe block wythe, the block wythe must resist all

of its effect. However

both wythes may resist weak

axis

5.8.2.1(b)

bending from gravity load. The load from a typical roof truss bearing on the wall between pilasters is shown below:

12-68


A C 1 TITLExMDG 93

m

0 b b 2 9 4 9 0509LLB bOT

m


Calculations

8'

k

4

Wall Support

12" Long Bearing Plate

3 = 12" + 4(7.63)" = 42.5"

The angle of load distribution and resulting width of bearing would be modified if the masonry course immediatelyunder the bearing plate weremade of a continuous bond beam. See MDG Ex. 9.3-2. Although the trusses are 8 ft apart, only 42.5 in. of the wall is effective in carrying the gravity load from the trusses between the pilasters. From structural analysis in MDG 9.1.2 within the 42.5 in. or 3.54 ft length of wall resisting these loads: PD = PD

4,960 lb,

PD + L

= 1,400 plf,

= 14,900 lb

4,180 plf = 5,490 in.-lb/ft

PD + L =

MD = 1,840 in.-lb/ft,

MD + L

5.8.2.1

If the bearing plate is 1/2 in. from the interior face, and assuming the inner wythe only resists the eccentric gravity load -

-

t

eD-eL-

1 in. 2 -2 - -

- -3in.

- 1.31 in. I I

i"

E

MD

=

PDeD = (1,400 lb)(1.31 in.) = 1,840in-lb/ft, MD

12-69


+

L =

5,490in.-lb/ft

A C 1 T I T L E * M D G 93

m

Obb2949 0509119 5Yb

m


Code Reference

Calculations

Check Euler Buckling (Code Eqs. 6-2 and 6-6) Use section properties in MDG Appendix A.

Pa

=

*I

[l

h'

Eq. 6-6

- 0.577fr

lo6 psi)(309 (24 ft x 12 in./ft)2

PC = rr'(2.08

x

1.31 in. 3.21 in.

0.577

ia4)

x

PC = 34,100 plf

P

= 4,180 plf < 1/4 P, = 8,530 plf

.-. OK

Eq. 6-2

Check Code Unity Equation (6-1) Eq. 6-3

R

-

Ad

4 +

=

MDG

e 3R -

W.12.2-5.

ek

P

Ad

S

0.59 = 0.121 3 ( 0 . 5 9 ) (1.3 1i n . ) 4 +

P = 4,180plf

2.70 in.

S

0.12 (30 in.2)(1,500psi) = 5,450 plf

12-70


.-.

OK

A C 1 T I T L E * N D G 73

m 0662747 0507320

268


Calculations

Note: The above calculations didnot rely onthe exterior wythe of brick to share in resisting the out-of-planebendingresultingfrom

the eccentricity ofgravityloads.If

the

interior CMU wythe had not had adequate capacity, the moment could have been resisted by both wythesinproportion

to theirrelativestiffnesses.

5.8.2.1(b)

WIND CONDITION Withpilasters

at16 ftcenters,

the windcan

be resisted by the cavitywallspanning

horizontally between pilasters. Per Code 5.8.2.1(d) the load is distributed between wythes according to their relative stiffness. For the brick wythe, EI = (2.4 x 103 ksi)(l2 in.)(3.63 in.)3/12 = 114,000 in.2-kips in.4) block wythe, EI = (2.08 x 106 ksi)(309

= 643,000 in.2-kips

1 14,000kip -in?

Brick carries

0.151 or 15% of the wind

=

114,000 kip-h2 + 643,000 kip-h2

Block carries wind 85% of Calculate stresses using moment coefficient,

2

M

=

W4leU ~

11

.

This value is used for those

horizontal span areas between pilasters without control joints.For spans containing control joints the designer would need to modify the moment value. For Block Wythe,

12-71


A C 1 T I T L E + f l D G 93

Example 12.23 and

Obb2949 0509323 3 T 4

m

Cont'd.

Calculations

M - 348 ft -lb fk's81

X

12 h/ft

in.3

=

52 psi

Allowable stress for ungrouted CMU parallel to bed joint

Table 6.3.l. 1

:. OK

Fbc = 50 psi x 1.33 = 66 psi > 52 psi,

5.3.2

For Brick Wythe,

M = (0.15/0.85)348 ft-lb

= 61 ft-lb

12 h./ft f k M=-s6112-fi-lb h(3.63 in.)'/6 X

Allowable stress

=

Fb, = 80 psi x 1.33 >psi 28

27.9 psi

:. OK

Note: These walls must be tied in accordance with Code transfer across the cavity.

5.8.2.2 in order to ensure load

See MDG Example 14.3-2.

12-72


Table 6.3.1.1

Example 12.2-4

DPCGymnasium

- Design of UnreinforcedComposite

Masonry Wall

Redesign the DPC Gymnasium north wall on Grid Line A as anunreinforced composite wall (Wall Construction Option B) using 4 in. nominal face brick and concrete masonry units.

Try 12 in. CMU wall

P

2 in. grout, fg = 5,100 psi Other propertiessame as Option

A. (MDG Example 12.2-3)

2"

11.63"*

Wall Section

Code Reference


Ir

Transforming grout to block

14.7"13.9"

12

"

Transforming the brick to block kY-4

Y

Transformed Section

-

y = distance from block centroid to composite section centroid

The block wall is ungrouted.

12-73


A C 1 T I T L E x M D G 9 3 W 0662949 0509323 T 7 7 W

Example 12.2-4

and

Cont'd.

Code Reference

Calculations

-

CAY CA

y = - -

6.81 in. (2 in.) (14.7 in.) + 9.62 in. (13.9 in.)(3.63 in.) = 5.9 36 h2+ 14.7 in. x 2 in. + 13.9 in. x 3.63 in.

i n !+

I = 929

36

in? (5.89

in.)2

+ 14" in*(2 in.)3

12

+ 29.4 in?(0.91 in.)2

+ S&

=

I

-

2,960 h4 = 253 5.89 in. + 5.81 h.

"

-

t

Y+z

Check D

in.3

+ L + W condition first -

e, = e L = y -

2

+ 4 h =5.9oin. - 11.63 2

i n *

+

4 in,

=

4.09 in.

PD+ L is spread over 12 + 4t = 81 in., PD+L= 14,900 lb= 2,200 plf 81 in. "

12-74


"

in.

A C 1 T I T L E x M D G 93 m Obb2949 0509324 903 m


Calculations

Reference

Self weight of wall at midheight based on wall weight of 178.5 psf = 1,785 lb e = -" -

P

21,800 in.-lb 2,200 lb + 1,790 lb

=

5.46 in.

Unity Equation

P

R

ALI

e

=

116

in2

(1,500psi)

4+3R-

Maximum applied load at midheight is 2,204 plf

0.83 x 1.33 5.46 in. 4 + 3 (0.83) 2.18 in.

=

18,800 plf

+ 1,785 plf = 3,989 plf < 18,764 plf :. OK

Euler Buckling

PC =

* h'

PC = PC =

[l

- 0.577

51

3

P

lo6 psi)(2,963 in?) (24 ft x 12 b~./ft)~ 111 kips x' (2.08 x

4.09 in. 5.05 in.

(Using gravity load eccentricity only as explained in MDG 12.2.1.2)

P

=

4 kips

S

:. OK

'/4 P, = 27.8 kips,

Note that this value of %Pe exceedsallaxial

load combinations. HenceEuler

conditions are satisfied with this trial section for all loadings.

12-75


A C 1T I T L E l k M D G

93

m

Obb2947 0507325 B Y T

m


Calculations

Check D

PD = 49960 lb = 735plfplus1,785plfself 81 in. 12 in./ft

+ W,

e =

15,777 in.-lb 735lb + 1,785lb

=

weight at midheight

6.26

Unity Equation P

=

2,520 lb

S

116 in?(1,500 psi) 4

P, = 111 kipssinceeccentricity

Euler

+

0.83 (1.33) 6.26 in. 3(0.83) 2.18 in.

=

15,400 lb

:.

OK

is thesame as loading case D+L.

Check Flexural Tension (use Type S PCL Mortar) Thiscondition e, =

A

=

need be checkedonlyif

e > e,.

In this

case, e = 6.26 in. and

2.18 in. The allowable flexural stress is Fb, = 25 psi using the lower value for

hollow units (Table 6.3.1.1). From MDG Eq. 12.2-8 or MDG Fig. 12.2-7,

ek

2.18 in. 12-76


A C 1 TITLExMDG 73

m 0662947

0509326 786

m



Code Reference

Since the applied axial load (2,520lb) is greater than 2,061 lb, flexural tension requirements are not satisfied. If the concrete masonry is solidly grouted, the allowable flexural tensile stress increases from

25 psi to 68 psi. The section properties of the wall change as a result, but all of the previous calculations for the ungrouted wall are found tobe within code allowables whenthe wall is solidly grouted. Use a solidly grouted wall. Note: As discussed in MDG Chapter 9, it is possible that a composite brick-block wallmay experience delamination at the collar joint, due to shear from differential movement, inplane loads, and out-of-plane loads. The calculations of this example are valid only if such delamination does not occur.

12-77


AC T I1T L E r M D G

0662949 0 5 0 9 3 2 7 632

93

-

Example 12.2-5 DPC Gymnasium Design of Reinforced Hollow Clay Masonry Wall

Design the wall on Grid Line Aof the DPC Gymnasium using Wall ConstructionOption D (Single Wythe, Reinforced Hollow Clay Masonry).

Roof Truss Type S, PCL Mortar Hollow Clay Brick

Pm = 1,500 psi Wind Load = 20 psf n = 14


Reference

Code

Try nominal 8 in. Hollow Clay Brick Unit (Actual Width = 7 1/2 in.) The load

width is spread over 42.0 in.

(12 in. brg.

+ 4 x 7.5 in.)

per MDG Example 12.2-3.

PD + L

= 4,180 plf,

PD = 1,400 plf,

e = 1.25 in. (From MDG Example 12.2-3)

For the wall spanning vertically MW= *O

8

*I2

(12 in./ft)

=

17,300 in.-lb/*

Use MDG Fig. 12.2-11

12-78


5.12.1


73

m

Ob62949 0509128 559

Cont'd.

Example 12.2-5

Discussion Calculations and

Code Reference

The midheight moment is approximately the maximum. F* = 0.33 P,,,

P

-

"

Fbbt

7.3.2.1 4,180 lb

4 (500 psi) (12 ia) (7.5 in.)

= 0.071 "_.

3

-

pe

19,900 h.-lb

0.044

23 (500 psi) (12 in.) (7.5 in.)2

t2

'b

=

From MDG Fig. 12.2-11

np, = 0.02 k = 0.40 > kb .-. Compression controls and MDG Fig.12.2-11isvalid D + L P

4,180 lb = 0.095 (500 psi) (12 h.)(7.5 h.)

-

"

Fb b

2

0.093

=

Fb b t 2

( i")n

=

0.0155

7.5

From Fig. 12.2-11,

k = 0.12 P,-

C

kb

:. use MDG Fig.12.2-12 andread

Fs = o Fb

12-79


A C 1 TITLExMDG 9 3

Example 12.2-5

m

Obb23Ll9 0509329 4 9 5

Cont’d.

ReferenceCalculations Code and Discussion

D+W

From MDG Fig. 12.2-11

np, = 0.035

k = 0.35 > kb The largest requirement is np, = 0.035 from D A, =

0.11

pbt

=

-(12 in.) (7.5

in.) = 0.229 in?/ft

Area

S

14

in.2

#3 10.6 #4

+ W.

in.2

in.

5.8

0.20

#5

0.31 in.2

#6

0.44 in.2

#7

0.60 in.2

in. in. 16.2

Use #7 @ 32 in.

23.0 in. in.

31.4

J 7 32

12-80


”

AC1

Example 12.3-1

TITLE8NDG 93 m Obb2949 0509330 L07 m

DPCGymnasium

- UnreinforcedPilaster

Subjected to Flexureand

Axial Load Design the pilasters on the north and south walls on Grid Lines A and B, respectively, of the DPC Gymnasium for Wall Construction Option k

The brick veneerlateral load

resistance is assumed negligible. Determine the minimum size of an unreinforced pilaster, and the minimum specified compressive strength. Assume Type N PCL Mortar.

and

Code Reference

Calculations

Alternate roof trusses are supported by each pilaster. Vertical reactions are applied to the pilasters through 6 in. x 12 in. bearing plates that are located 112in. from inside edge of pilaster. From MDG 9.1.2 truss reactions are 4,960 lb for roof dead load and 9,920 lb for roof snow load. The governing lateral load is wind (20 psf) which produces a moment at midheight equal to 23 kip-ft for a 24 ft-O in. high pilaster.

A 32 in. square pilaster will be checked which was the size determined for pilasters on the eastand westwalls.

The 8 in. CMU walls are

The proposed section isshownbelow.

ungrouted with face shell bedding of mortar, and the 32 in. pilaster is fully grouted. Exterior Face

Plastic Centroid Of Section Centroid Of Vertical Load

7

f

48.00 31.63"

1

14.07'

W

3 1.63"

4

6t

=

6t = 45.78" 4

45.78"

Proposed Pilas ter 12-81


e

f Interi.or Face

AC1

TITLExMDG 9 3 m 0662949 0509L3L 043 m


Calculations

Section properties for the proposed sectionare as determined in MDG Example 11.2-1 with the exception that flanges will be on both sides of the web since there are no control joints. The centroid of the section is 15.06 in. from the exterior face and 16.57in. from the interior face. The area of the section is 1,230 in.2 and the moment of inertia is 88,700 in4. The centroid of the truss reaction is assumed to occur at one-third of the 6 in. plate width, or 2 1/2 in. from the interior face of the pilaster. This results in vertical load equal to 14.07 in. Because the location is outside of will resultfromtheeccentricverticalforceand

an eccentricity of the the kern, tensile stresses

the governingloadcombinationisnot

obvious. Allowable flexural tensile stress must be interpolated between values

6.3.1.1

of 58 psi for fully grouted sections and 19 psi for ungrouted sections (Table 6.3.1.1). The square columnportionconsists of 59% of the total gross area of the effective section which gives a flexural tensile stress equal to 42 psi. Combined axial and flexural compressive stresses are checked usingthe

unity equation (MDG Eq. 12.3-3). The allowableaxialcompressive

6.3.1

stress is obtained using MDG Eq. 12.3-4 noting that

Fa =

(1/4)

f”, [I

-

(Er]f’, =

0.235

12-82


6.3.l(a)


93

m

Ob62949 0 5 0 9 3 3 2

TBT m


Code Reference

Calculations

The value of 'Y, 8.50 in. used in MDG Eq. 12.3-3 isthe square root of the ratio of I (88,700 in4) over A (1,230 in.2)for the proposed pilaster

is 0.33 Pm.

bending stress section shown previously. The allowable

6.3.l(c)

Three loading cases are considered: (a) roof dead plus roof snow, (b) roof dead plus wind, and (c) roof dead plus roof snow plus wind. Case "all Roof Dead Load Plus Roof Snow Load Net Flexural Tension Stress at Top - Exterior Face

M = P e = (14,900 lb) (14.07 in.)

=

1,o00

fa =

PD

PS

+

Anet

fb* - fe

=

-

14,900 lb

209

=

in"kipS

14.9 psi

31.6 in. x 31.6 in.

35.5 psi - 14.9 psi

=

20.6 psi

Fb*= 42.0 psi OK

Compressive Stress at Top - Interior Face fb =

M (16.57 in.) -- (209 ia-kip~) (16.57 in.) (1O , OO lb/kip) - 39.0 psi Inet

88,700

12-83


i n !

9 3 M 0 6 6 2 9 4 90 5 0 9 3 3 39 3 b


m


Calculations

Code Reference

Discussion

14.9 0.235 f m

+

3900 = 1.0 0.333 fRI

fln required

=

180 psi

Case "b" Roof Dead Load Plus Wind Load Net Flexural Tension Stress at Midheight - Exterior Face Consider wind pressure to be a suction so that lateral moment will add with eccentric load moment.

MW

=

PD e/2

23.0 ft.-kips

=

(4,960 lb)

Total Moment

f&

=

=

276 in.-kips

( 14*02 ")

1 1,OOO lb/kip

=

34.9 in.-kips

31 1 in.-kips

M (15.06 in.) -- (311 h - k i p ~ )(15.06 in.) (1,OOO lb/kip) -Iner 88,700 in."

f = -P + D f

= . JDL

4,960 lb 1,Ooo

(140 pcf)

(y)

+

in.2

144 in.2/ftz 12-84


psi

ACL- T I T L E * M D G 9 3

m

Obb2949 0509334 852

m

Exarnule 123-1 Cont'd. and

Calculations

fa

=

5.0 psi + 11.7 psi = 16.7 psi

f a -fa = 52.9 psi - 16.7 psi = 36.2 psi < Fbt=42.0 psi x 1.33 =55.9 psi

.-.OK

Compressive Stress at Midheight - Interior Face fb =

M (16.6 in.)

- (311 in.-kips) (16.6 h)X 1,OOO lb/kip - 58.1 psi

88,700 in?

Inet

16.7 0.235 f

+

)II

58.1 = 1-33 0.333 f ,

f m rcqrrited =

185 psi

Case "ctt - Roof Dead Load Plus Roof Snow Load Plus Wind Net Flexural Tension Stress at Midheight - Exterior Face

Mm&

=

Mwind + (PD +Ps)e/2

Mrotor= 276 h-kips + 105 i n . - k i p ~ = 381 h - k i p ~

12-85


A C 1 TITLESMDG 93 W Obb2949 0509335 799 W


fk =

fa =

M (15.06 Inet

PD

PS

+

Anet

h.) - (381 h-kip)(15.06 h.) X 1,OOO lb/kip =

88,700

in4

+ fDL = 14.9 psi + 11.7 psi = 26.6 psi

f k -fa = 64.7 psi -26.6 psi =38.1 psi

Fa =42.0 psi x 1.33 =55.9 psi

Compressive Stress at Midheight - Interior Face

fb =

M (16.6 in.) - (381 h-kips) (16.6 Inet 88,700 i n 4

26.6 0.235 f ',,,

+

psi

71*2 = 1.33 ; 0.333 f',,,

psi

f',REQolRED -- 246 psi

12-86


h.) = 71.2

:. OK

AC T I1T L E r M D G

0bb2749 0509136 b25

93


and

Code Reference

Calculations

SUMMARY fa Case Condition Net

-fa =

Flexural Tensile Stress

Fa

(Psi)

(Psi)

fa -fa Fa

fm

REQulRED (Psi)

a

D+S

20.6

42.0

0.49

180

b

D + W

36.2

55.9

0.65

185

c

D+S+W

38.1

55.9

0.68

246

Thus, the critical loading case for both flexural tension and flexural compression is the combined effects of roof dead and snow load with the wind loading.

Case l'a'' does not

control becausethe truss reaction results in an eccentric-load moment that is relatively small when compared to the wind load moment of the other two cases. Flexural tensile stresses are quite close for Cases "b" and "c", and it is not obvious which case should control. Thus, a 32 in. pilaster is needed to resist flexural tension stress, and a minimum prism strength equal to 246psiis

necessary to resist combined axial and flexural compressive

stresses. Note that compression capacity does notgovern the design of the pilaster sincep, for minimum strength ASTM C 90 block and Type N mortar is approximately 1,300 psi. Because compressive stresses are so light, and flexural tensile stresses are well below allowable values, it may be possible to reduce the section size. However, for concrete block masonry, this must be done in increments of 8 in. A quick check using a 24 in. square section shows a net flexural tensile stress equal to 114 psi for Case "c". Therefore, the 32 12-87


A C 1 T I T L E a N D G 73

m

Obb2749 0509137 561

m



in. pilaster as designed will be used. Shear is checked by adding the wind shear force (3,840 lb) with a shear due to the eccentric load. The eccentric gravity moment is equal to 14,880 lb times 14.07 in. When divided by the pilaster height of 24.0 ft, the shear force dueto the eccentric gravity load is 727 lb. Thus the total design shear is4,567lb.

Shear stress is checked in the same

manner as in MDG Example 11.2-1. The applied shear stress is 6.8 psi which is much

smaller than the allowable value of 62 psi per Code

6.5.2(c).

12-88


6.5.1

m

A C 1 T I T L E * N D G 93

06629470507338

4T8

m

-

Example 123-2 DPC Gymnasium Reinforced Pilaster Subjectto Flexure and Axial Load

The 32 in. width ofthe unreinforced concrete masonry pilasters for the north and south walls (Grid Lines A and B) of the DPC Gymnasium (with Wall ConstructionOption A) was felt to be excessive. It is of interest to examine the feasibility of reinforcing the pilaster. The

adequacy of an assumed 16-in. pilaster is checked, and

the necessary amount of vertical

steel, and the required prism strength are determined. Exterior Face

7.63" Centroid Of Vertical Load Interior FaceA- 15.63" 6 t = 45.8" c ~

8 .OO" c

1 6t = 45.8"

-

Assumed Configuration for 16 in. Square Reinforced Pilaster

f',= 1,500 psi

F,, = 500 psi

Concrete Block ASTM C 90

E, = 1.8 x 106 psi

n = 16.1

Type N Mortar and

Calculations

Discussion

Reference

Code

As noted in MDG Example 12.3-1, truss reactions are 4,960 lb for roof dead load and 9,920 lb for roof snow load.The governing lateral load is wind (20 psf) which produces a moment at midheight of a pilaster equal to 23.0 kip-ft (276 kip-in.) as given inMDG Example 12.3-1. The 16-in. section usedfor the pilasters in nonloadbearingwalls on Grid Lines 1and 2 (see MDG Example 11.2-2) will be checked for combined effects of axial load and moment for the loadbearingwalls on Grid Lines A and B. As for the east and west pilasters, only those cells containing reinforcementare grouted. Control jointsare assumed absent at the design 12-89


m

A C 1 TITLESNDG 93

0662949 0509139 3 3 4 D


Calculations

section. The 8-in. CMU wall is considered to be ungrouted with face shell mortar bedding. The 6 in. x 12 in. bearing plate for the roof truss is assumed to be located 0.5 in. from the interior face of the pilaster. If a triangular distribution of bearing pressure is assumed, the centroid of the vertical truss reactions is 2.5 in. from the interior face. Because the web portion is narrower than the effective flange width,

the critical case for

compression stress underwind loading is when wind acts as a suction, resulting in compression on the interior face. This condition is true even when the depth of the effective flange is limited to the thickness of the face shell as is for this case of an ungrouted wall. This sense of wind moment is the same as the sense of the eccentric load moment, and should therefore control for all cases of web compressive stress. Since vertical

reinforcement is

assumed to be symmetrical about the column portion, the case of wind suction should also govern for design of reinforcement. The same three loading cases are considered as for MDG Example 12.3-1: (a) roof dead load plus roof snow load;

(b) roof dead load plus wind; and (c) roof

dead load plus roof

at the top of the pilaster if the eccentric

snowloadpluswind.Case"a"shouldgovern

vertical load ismuch larger than the wind load moment for cases "b"

and tictt. Case "b"

should result in the minimum axial force, and thus the lowest flexural strength, but the applied moment will not be a minimum. Case l'cl' should result in the largest moment, but the axial force (and thus the flexural strength) will not be a minimum. Theassumed16-in.section

ischeckedusing

rectangular section with twolayersof

load-momentinteractiondiagrams

for a

reinforcement (see MDG 12.1.2). Because only a

portion of the web is considered to be in compression, the rest of the T-section is assumed to be cracked, and thus ineffective. Flexural tension 12-90


is assumed only for the two vertical

A C 1 TITLExMDG 93

Ob62949 0509340 056

m


Calculations ~~

~

reinforcing bars. The overhanging flanges are neglected entirely. The first step is to select an interaction diagram based on the ratio of the distance between rebars to the overall thickness, g. If it is assumed that the bar will have 3 in. of cover (distance from edge to centroid of bar), the g value is (15.63 in. - 2 x 3.00 in.)/15.63 in. = 0.62. Interaction diagrams for a g value of 0.6 will be used (Fig. 12.1-4). Any number of design solutions are possible because there are three variables: the section size, the masonry strength and the amount of vertical reinforcement. Because the section size must be in increments of the block size, it will be fixed first at 16 in. Prism strength also is restricted within a narrow range if standard strength units are to be used. Therefore, the amount of reinforcement will be used as the primary variable for each of the three loading cases. The case resulting in the most required reinforcement will be the one that governs. Reinforcement requirements are determined as indicated in the following table for each of the three loading cases. The total axial compressive load is listed as well as the maximum bending moment foreach

case. The effective eccentricity issimply

the total bending

moment divided by the total axial load. By dividingit by the thickness (15.63in.),

the

eccentricity can be used with the normalized axial load P/Fb bt, as an alternative to moment for locating points on the interaction diagram. Combinations of axial load and eccentricity are plotted on theinteraction diagram to identify the required percentage of reinforcement, p. For cases of light axial loads, required amounts of reinforcement must be checked with

respect to curves based on allowable compressive stress (expressed in terms of pn) as well as with respect to curves based on allowable reinforcement tensile stress (expressedin terms of

PWFb).

12-91


AC1

TITLE*HDG 93

0662949 0509141 T92


Calculations

% l

2

a

J

Q fl

d a Y

L

V

d O tF;1

3 h

Y

(v

8V

U

.8

2 S

v)

+

a

12-92


A C 1 T I T L E x M D G 9 3 W Obb2947 0507142 7 2 9 W


Code Reference

Calculations

Based on compression controlling, the maximum value of pn is 0.26, which results in a p value equal to 0.0161 (Case c). Based on tension controlling, the maximumvalue of pF,./Fb is equal to 0.24 (Case b) which results in a p value equal to 0.0050 for F, = 24 ksi x 1.33 = 32ksi.

Compression governs, andthe

required totalamount

of

7.2.1.1

reinforcement is equal to 0.0161 times b and t, or 3.95 in2. This can be

5.3.2

satisfied with four No. 9 bars (As = 4.00 in2). Because the anchorage of a No. 9 bar in an 8 in. CMU may be difficult,it

isof

interest to see what reductions in steel areaare

possible if the masonrycompressive strength is increased. A prism strength of 2,500 psi will be tried. Since thiswill require a unit strength in excess of the minimum set forth by ASTM C 90, the higher prism strength must be specified on project drawings and in job specifications. A contractor may be able to attain such a strength according to the and Type N mortar (see

Specs. 1.6.2

Table 1.6.2.2). He may also verifysuch a strength using the prism test

Specs. 1.6.3

unit strength method using 4,000 psi units method.

The assumed modulus of elasticity, E, is2.6

x 106 psi

according to Table 5.5.1.3 for units with a strength of 4,000 psi Type N mortar. This gives a modular ratio, n, equal to

10.8.

and 5.5.1.3

With the masonry strength increased to 2,500 psi, the required amount of reinforcement is controlled by tension. For compression controlling, the maximum value of pn is 0.053 which results in a p value equal to 0.0049 (Case b). For tension controlling, the maximum value of pFs/Fbis equal to 0.15 (Case b) which results in a p value equal to 0.0069. The required total amount of reinforcement is equal to 0.0069 times b and t, or 1.70 in2. This can be 12-93


A C 1 TITLElwMDG 93

m

Obb2949 0509343 Ab5

m


Calculations

satisfied with four No. 6 bars (As = 1.76 in.2). In summary, the required area of reinforcement is 3.95 in.2 with$, equal to 1,500 psi, and 1.70 in.2 withf',

equal to 2,500 psi. With the stronger masonry, the controlling concern is

allowable reinforcement tensile stress

rather than allowable masonry compressive stress.

Roof dead load plus wind required slightly more reinforcement than roof dead plus roof snow plus wind. This was because the eccentricity the of vertical truss reaction was relatively

small. Added vertical forces tended to increase the flexural strength more than the applied bending moment. The final design is then:

r

Exterior Face>

1

",,"R,.

7.63" 8,OO" Interior Face

#

3 a t 24"

Type N mortar

The No. 3 ties at 24 in. spacing are nominal transverse reinforcement. The ties are needed to support the vertical reinforcement. Although ties are not essential since reinforcement was not considered to resist compressive stress, it is prescribed

here simply as an added

measure. Note that the ties should bend around the vertical reinforcement rather than being sized to fit within the face shell. This is

preferred for lateral support of the vertical bars. 12-94



93

m

Obb2949 0509344 7 T L

m

Example 123-2 Cont’d.

and

Code Reference

Calculations

This design is similar to that for the east and west walls (Grid Lines 1 and 2) of the DPC Gymnasiumwhich were the same size but reinforced with four No. 8 bars. The design requirements are less for the north and south pilasters because of their shorter height (24 ft-O in. rather than29 ft-4 in.), and the axial compressiveforce which increased their flexural capacity. The fact that a control joint cut off one flange of the east and west pilasters was of little concern since the tensile area of the flange wasneglected for thereinforced sections.

As was seen with the east and west pilasters, the addition of four reinforcing bars to the already grouted section can result in a required reduction of 50% in the required dimensions of the pilaster.

12-95



93

m

Obb2949 0509345 b 3 8

m

13 SHEAR

13.0 INTRODUCTION Different methods of analysis for shear design are employed for unreinforced and reinforced masonry. The computed shear stress for unreinforced masonry is obtained by the VQ/fi equation. For reinforced masonry the shear stress is obtained by using the V/bjd equation. The allowable stresses are also different for unreinforced and reinforced masonry. The designer should not combine the two types of masonrydesign techniques. One should consider them distinct and separate; use only one or the other. In unreinforced masonry, the allowable stresses depend on the type of masonry element I

(wall, beam, etc.), the bonding pattern, and the amount of axialload. In reinforced masonry, the allowable stresses depend onthe type of masonry element (shear wall, beam, etc.), the magnitude of MWd, and the amount of reinforcement. Chapter 13 first addresses the design of unreinforced and reinforced non-shear wall elements. Design for shear wall elements is then presented.

13-1


A C 1 T I T L E g N D G 73 W 0662949 0509346 574

13.1 DESIGN FOR SHEAR IN MASONRYCOMPONENTS

13.1.1 OverallPhilosophyforShearDesign Shear and flexure occurtogether in masonry components such as beams and beam-columns (when axial loadis present) (Fig. 13.1-1 [a]), in walls spanning vertically to resist out-of-plane loads (Fig. 13.1-1 [b]), in walls spanning horizontally to resist out-of-plane loads(Fig. 13.1-1 [c]), and in shear walls that resist in-plane forces (Fig. 13.1-1 [dl). Shear design of beams, beam columns, and walls subjected to out-of-plane loading are addressed in MDG 13.1. Shear design of walls subject to in-plane loading is addressed in MDG 13.2. Shear designin the Code is considered in two casesformemberssubjected

to flexure.

These two design cases are: 1) Code 6.5

- Unreinforced Masonry -

Members where flexural tension is allowed to be carried by the masonry, and

2) Code 7.5 - ReinforcedMasonry

-

Memberswhereflexuraltension

is not

allowed to be carried by the masonry. The first case (Code 6.5) is often referred to as "unreinforced masonry," even though some nominal reinforcementmay be present for reasonsother than load carrying purposes.Shear computations are then based upon "uncracked section" concepts. The secondcase

(Code 7.5) is further divided into the categories of membersbeing

subjected to flexural tension, and members not subjected to flexural tension. For members withflexuraltension,

the shear computationmustbebasedupon"crackedsection"

properties. As innormalreinforcedmasonryconcepts,

the masonry area subjected to

tension is neglected and the reinforcement carries all the tension forces.

For members

without flexural tension the shear computations may be considered in light of either Code 6.5 or 7.5. 13-2


AC1

TITLE*MDG 93 H 0662749 0509347 400 W

(b) Vertically Spanning Wall Out-of-Plane

(a) Masonry Beam Subjected to

Shear and Flexure

Shear and Flexure

Lateral Loads

u

(d) In-Plane Shear and Flexure

(c) Horizontally Spanning Wall Out-of-Plane Shear and Flexure

Fig. 13.1-1 Masonry Elements Subjected to Shear and Flexure Fig. 8.4-47 provides an overall flow diagram for the shear design of masonry. This chapter discusses the Code shear provisions, followed by examples for these various subdivisions of each case.

13-3


A C 1 T I T L E * M D G 93

m

Obb2949 0509L48 347

m

13.1.2 Unreinforced Masonry Shear Design Shear design allowing tensile stresses in masonryis used when the flexural masonry tensile stresses do not exceed the maximumallowablevaluesinCode

Table 6.3.1.1. For cases

involving axialloads and flexure, the combined stress is compared to the maximum allowable tension in Code Table 6.3.1.1. Fig. 13.1-2 depicts these combined stresses. Note that this case is based upon "uncracked section" concepts. Thus, the flexural stress in Fig. 13.1-2 is found from

the simple flexure

equation fb =

MC

Forces

Eq. 13.1-1

Axial Flexural

Combined Axial Shear Stresses Flexural and Stresses (Include Others if Present)

Stresses Stresses

Fig. 13.1-2 Combined Flexural Stress and Shear Stress Distribution in Uncracked Section, Unreinforced Masonry where the moment of inertia, I, is based upon the net uncracked section area. The axial stress in Fig. 13.1-2 is found from the simple axial equation

P fa = -

Eq.13.1-2

A"

where the A, is the net uncracked cross-sectionalarea. Any other axial and flexural stresses,

13-4


A C 1 T I T L E x M D G 93 W Obb2949 0509349 283 W

for,

due to factors such as restrained differential movement, temperature change, moisture

expansion, or shrinkagemust

be combinedwith

dead and live loadstresses.

The

combination must satisfythe provisions of Code Chapter 6; thus, the algebraic sumof fd fb

+ fo,

+

cannot exceed the values in Code Table 6.3.1.1. Stresses resulting from restraint

should be controlled by movement joints (see MDG

Chapter 10) or other construction

techniques to ensure that the combined stress does not exceed the allowable values. I

The

allowable stresses in Code Table 6.3.1.1 govern for out-of-plane bending.

If Code Table 6.3.1.1 is satisfied, then the shear calculation is based upon the uncracked section and the parabolic shear distribution found from (13.1.1): fv =

V0

Code Eq. (6-7)

Any reinforcement in the cross sectionis simply neglected. For a rectangular cross-section,

Code Eq. (6-7) gives fv =

3 v A,

Eq. 13.1-3

-

at midheight, the point ofmaximum

shear stress.SinceCodeEq.

(6-7) isused

for

calculations of both in-plane and out-of-plane shear in walls, the reader is also referred to MDG 13.2. Eq. 13.1-3 is sometimes used as a conservative approximation of the composite interlaminar shear which occurs at the wythe-to-collar joint interface under out-of-plane bending.

The

collar joint is usually only a short distance from the point of maximum shear stress which occurs at the cross sectionmid-depth.See

MDG 9.3.5 for further discussionofthis

composite section shear computation. Even though Code 6.5.2 specifies allowable stresses for "in-plane shear,"

these allowable

values are considered applicable to out-of-plane bending for Case (1) masonry elements utilizing Chapter 6 - Unreinforced Masonry. Thus, the maximum allowable shear stress is

13-5


A C 1 TITLEtMDG 93

0662949 0509350 TT5

the least of the following: 1 S K ;120 psi; u

+ 0.45 N, / A , inwhich

u = 37 psi for

masonry in running bondand not solid grouted, and for stack bond masonry withopen end units and grouted solid or 60 psi for masonry in running bond and solid grouted; 15 psi for masonry in other than running bond with other than open end units grouted solid. See Fig. 13.2-7 for a graphical depiction of these allowable values.

13.13 Reinforced Masonry ShearDesign

Although it is rare, if masonry is subjectedto axial tension, it must be reinforced. The bond between masonry units and mortar cannot be counted onto resist this type of loading. See Code 6.4. When net axial tension exists, or when the flexural tensile stress or shear stress values exceed the allowable stresses of Code Chapter 6, then Case (2) in MDG 13.1.1 applies, and the member is designed as reinforced by Code Chapter 7. Any tensile stress contribution of the masonry is neglected. Thus, when the combined tension stress exceeds Code Table 6.3.1.1 allowables, the stress distributions in Fig. 13.1-2 change to those shown in Fig. 13.1-3. Note that in Fig. 13.1-3, regions of tensile stresses in the masonry are ignored, and all of the tension force is carried by the reinforcement. For the reinforced masonry Case (2), the shear stress is found from

V

fv =

Code Eq.(7-3)

bjd

where cracked sectionproperties are (MDG 11.1) used to determine theid distance between the internal C and T forces for compression and tension, respectively.Fig. 13.1-3 shows this

jd distance and a derivation for Code Eq. (7-3). Generally, masonry members without net axial tension can be designed

to resist shear by

either Code 6.5 or Code 7.5, provided that the member (unreinforced or reinforced) is not subjected to flexuraltension.

For memberssubject to flexuraltension shear must be

considered according to Code7.5. The philosophy of Code 7.5 is to allow shear stress to a 13-6



m 0662949 0509151 931 m

limited maximum value. If shearing stresses exceed the maximum allowable value for masonry, theentireshearstress

must be resisted by horizontal and vertical shear

reinforcement. No credit is givenfor shear carriedby the masonry once shear reinforcement is required. The required amount of shear reinforcement, A,, is:

A,

=

vs -

Code Eq. (7-10)

Fsd The derivation of Code Eq. (7-10) is shown in Fig. 13.1-3.

2

i

fb

Forces on Section A-A

F N.A.

+T

Axial

Flexural Stresses Stresses

= Asfs

Combined Flexural and Axial Cases

a) Longitudinal Stress Distribution For a unit length or dx

T MqJ[ Lpi::M ldxl

Zforces = O (on top segment) f,(b)(dx) = C + dC - C

f,(b)(dx) = dC = @ (assuming jd = moment arm) ~d

jd

T - " T

' fv(b)(dx) v + d v -"

f,(b)(jd) =

f

+ dT

2= V Y*

L "-W 1

pdadx = Bond Force

b) Shear Stress Derivation (for Flexure and Shear in Combination)

c) Shear Reinforcement Derivation

Fig. 13.1-3 Combined Longitudinal Stress and Shear

Stress Derivations for Cracked

Concepts, Neglecting Tension in Masonry 13-7


9 3 M 0 b b 2 9 4 9 0509352 8 7 8 M


13.13.1

Shear Reinforcement Not Required - Where shear reinforcement is not required,

the calculated shear stress,f,,per Code 7.5.2 shall not exceedthe allowable shear stress, F,, determined as: a) Flexural members:

F,,=

fi

S

Code Eq. (7-4)

50 psi

b) Shear walls: 1) Where M/Vd c 1.0:

Code Eq. (7-5)

with (M/Vd) taken always as a positive numtber 2) Where M/Vd r 1.0:

F,,= 13.13.2

c

S

Code Eq. (7-6)

35 psi

ShearReinforcementRequired

- Where shear reinforcement is required,the

calculated shear, fv, is limited by the allowable shear stress F,, which is determined as: a) Flexural members:

F,,= 3.0

fi

S

150 psi

Eq.Code

(7-7)

b) Shear walls:

1) Where M/Vd c 1.0

Code Eq. (7-8)

13-8


A C 1 TITLExMDG 9 3

m 0662949

O509353 7 O Y

m

2) Where M/Vd r 1.0

F”

=

1.5

c

S

Code Eq. (7-9)

75 psi

with (M/Vd) taken always as a positive number Per Code 7.5.3.1, the shear reinforcement is to be provided parallel to the direction of the applied shear force, and the spacing, S, shall not exceed d/2 or 48 in. The amount of required shear reinforcement, A,, is: A,,

=

vs Fs

Code Eq. (7-10)

The derivation of Code Eq. (7-10) is shown in Fig. 13.1-3. Code 7.5.3.2 requires additional reinforcement of at least (AJ3) whichmust

be placed perpendicular totheshear

reinforcement at a uniform spacing not to exceed S ft. Thus, in Code Chapter 7, the masonry elements must be reinforced for the primary tension force in the member, but could be either reinforced or unreinforced for shear force. The details for Code Eqs. 7-5,7-6,7-7, and 7-S addressing shear walls are included in MDG 13.2.

13.1.4

Special Provisions for Diaphragms

MDG 13.1.1 - 13.1.3 discuss combined shear and flexure for masonry members subjected to beam bending, out-of-plane wall bending (vertically or horizontally) due to out-of-plane forces or due to in-plane forces (see MDG 13.2). However, some structural components such as floor or roof diaphragms serve as primary shear elements. These diaphragm elements are very important to the overall lateral stability of buildings subjected to wind or earthquake loads. The forcedistribution in these diaphragms is illustrated in MDG 9.2.1 for flexible diaphragms and MDG9.2.2. for rigid diaphragms. Actual as-built diaphragm stiffness varies between these two theoretical ideal cases.

13-9


A C 1 TITLE*flDG

0662949 0509354 b 4 0

93

m

Criteria for analyzing diaphragms in masonry buildings have been developed based on the diaphragmmaterial

type, thickness, configuration, number of sides connected, and the

number of fasteners (13.1.2, 13.1.3). The designer must assure that the connection between the horizontal diaphragm (floor or roof) and the vertical lateral load-resisting elements is capable of transferring the shear. The design of connectors is covered in MDG Chapter 14.

As shown by test results, the number of sides of a diaphragm that are connected to vertical lateral load-resisting elements significantly affects the load capacity of the diaphragm

(13.1.2). 13.2 SHEAR WALLS The design of shear walls is covered in Code 6.3 and 6.5 for unreinforced shear walls and Code 7.3 and 7.5 for reinforced shear walls.

13.2.1 Definition

of aShearWall

Code 2.1 defines a column as a member whose width to thickness ratio does not exceed 3. It is, therefore, assumed that a member with a ratio greater than 3 is a wall. A shear wall is a vertical member which resists lateral in-plane shear forces from wind or earthquakes (Fig. 13.2-1). Vertrcal Load If

Lateral

I

Y / & Shear

t ->3.0

Fig. 13.2-1 Illustration

13-10


of aShearWall


93

m

0662949 0509355 587

m

13.2.2 Function of ShearWalls

Shear wallsresist lateral shear forces due to wind or earthquake, and can also act as loadbearing elements to supportvertical loads from the floors and/or the roof. The stability of the masonry building depends predominantly on the in-plane capacity of the shear walls. The building’s lateral drift under wind or earthquake loads is a function of the in-plane stiffness of the shear walls. A shear wallsystemutilizesfloor

and roof diaphragms to

distribute lateral forces to the shear walls. Load distribution is based upon the relative lateral stiffnesses of the shear walls if the floor diaphragms are rigid, or upon the tributary floor widths if the floor diaphragms are flexible. Shear walls also serve as fire walls and as building enclosures; in such cases their thermal, moisture permeance and acousticalcharacteristics

are important, in addition to their

appearance and structural characteristics. 13.26 Layout of ShearWalls

Loadbearing building construction is used predominantly in layouts where the floor area is divided into arelatively large number of compartments and inwhich the floor plan is repeated at each level for the full height of the building. Stability in such

construction is

usuallyderived from gravityloads, and from a careful planning of the layoutwithfull utilization of elevator shafts and stairwells to provide lateral stiffness. The layout is mainly determined based on the type and function of the building. A large variety of wall arrangements is possible.

1)

Cellular wallsystemwithtwo-way

2) Single

slabaction(Fig.13.2-2a),

or double cross wall system with one-way

slab action (Fig. 13.2-2b and

Fig. 13.2-2c), 3) Longitudinal 4)

or spine wall system with one-way

slab action (Fig. 13.2-2d), and

Complexsystem combination of the abovesystems. 13-11


A C 1 T I T L E * U D G 93 D Ob62949 O509356 413 M

In single-story masonry buildings, shear walls are concentratedat the perimeter where they serve as enclosures; some also carry vertical and lateral loads. Intermediate shear walls can be added to provide stability and support of the roof system, or to reduce the diaphragm stresses.

4"

FF ""- F F "_ l"_-

4"

" "

" "

a) Cellular Wall Arrangement

b) SimpleCross-WallStructure

c) Double Cross-Weight Structure

d) Spine Wall Structure

Fig. 13.2-2 Arrangement of Shear Walls in Multistory Building

13.2.4 Analysis of ShearWalls

The global analysis of buildings is covered in Code 5.1, 5.2, 5.3, and 5.7. Determination of wall stiffness is required for lateral load distribution and for calculation of building drift. 13-12


A C 1 T I T L E g N D G 93 W Obb27470507357

35T

m

MDG 9.2.1 and 9.2.2 discusses global lateral load distribution on shear walls. Wall stiffness is primarily a function of:

1)

Wallgeometry - wall aspect ratio h/l, affects the contribution of shear and bending deformation to the wall rigidity.

2)Boundarycondition

- restraint at the top of the wallaffectswallstiffness.

Two conditions are commonly considered; cantilever and fixed-fixed wall. 3) Openings

- the

size, location and arrangement of openings dramatically affect

wall stiffness. The lateral stiffness of a solid cantilever shear wall is:

The lateral stiffness of a solid fixed-fixed shear wall is:

The effect of openings on wall stiffness depends on the size, shape, and distributionof those openings. Finite element methods can be used to accurately determine wall stiffness and a wall's response to lateral loading.However,approximatemethods estimate wallrigidity.

For acombination ofwalls

stiffness is expressed as:

k=k,+k,+k,

13-13


may beused

in parallel(Fig.13.2-3a)

to

the system

m

A C 1 TITLEsMDG 93

For a combinationofwallsin

Ob62949 0509358 29b

m

series (Fig. 13.2-3b) the system stiffness is approximately

expressed as:

k=(

A,

+

A2

+

A,

1

- 1+ - +1-

1

a) Walls in Parallel

b) Walls in Series

Fig. 13.2-3 Stiffness of Combinations of Walls

For single story buildings whose wall height to width ratio is less than1, three approximate methods based on the parallel and series models can be used per reference (13.2.1). The

MDG will present only methods I and III. Method I - In this method the wall deflection is first calculated assuming a solid cantilever wall, andthen modified to account for openings by subtracting deflection of cantilever strips containing openings, and then adding deflection due to deformation of piers (considered fixed top and bottom) within each strip. Using this method the rigidity of the wall shown in Fig. 13.2-4 can be calculated with the following formula:

k = -1 A A

=

A solid - A &ipA

+ AuA5

13-14


kz

kzSA1 =

+

%41

1

k3,4*5 =

A3.41

A3,41 A3,4

=

-

~ 0 x 4-, A~ stripB ~

A

+ A3,4

1

S

+

R4

This method has the advantage that it does not give results thatareerroneous

when

compared to solid- walls. Method III - In this method the wall is divided into wall elements stacked on top of each other in series; the lateral deflection of each element is added to obtain the total wall deflection. For each part, which is a horizontal strip, the deflection of piers connected in parallel is the reciprocal of the sum of rigidities. For the wall shown in Fig.

13.2-4, the

rigidity is expressed as:

k=

1 1

In-series model

1

In-parallel model In-series model

R3,4 = k3

+

In-parallel model

k4

Wall deflection is calculated based on fixed boundary at top and bottom. Therefore, the method ignores the rotation at the top of each strip which is more applicable for squat walls where shear deformation is the predominant mode compared to flexural deformation. This

13-15


AC1

TITLExMDG 93

0662949 0509360 944

method can alsogive erroneous results. In certain configurations a wall with an opening can appear to be stiffer than a wall without an opening. This can affect not only

the wall in

question but the distribution of shears to adjacent walls if the floor and roof diaphragmsare rigid or if the walls are in line with each other. The percent reduction of wall stiffness due to openings shown in Fig. 13.2-4 is calculated using the above two methods and the results are presented in Table 13.2.1 for the given geometry. As can be seen, Method III results

in a higherwallstiffness

than Method I.

Obviously engineering judgement must be used to select the mostappropriate method. For a further discussion of this method, see MDG 9.2.2.

o

. ......

-k

o

4’

D

Fig. 13.2-4SingleStoryWallwithOpenings Table 13.2.1 Reduction

of Wall Stiffness Due to Openings Stiffness of Wall with Openings

Method Wall Stiffness Solid of

MethodI Stiffness

0.143 E,t

0.073 Emt

III

% Reduction Stiffness 48.9%

13-16


% Reduction 0.097 Emt

32.3 %


93

m

Obb29Ll9 0509363 AA0

m

In multistory construction where corridor openings are common, coupling between adjacent piers may be ignored in preliminary design, and thepiers assumed to act ascantilevers. This would provide a lower bound value for wall stiffness. The continuum approach for coupled shear walls or the finite element method would provide a more accurate prediction of wall stiffness and stresses, as long as the coupling beams are stiff enough and are appropriately detailed to transfer the loads between wall elements. Lateral load distribution among walls depends on relative wall-diaphragm rigidities. When diaphragms are stiffer than shear walls,(e.g.,rigid

concrete floor slabs) lateral load is

distributed among walls in proportion to their relative lateral stiffness. In that case, the center of wall rigidities does not coincide with the line of action of the lateral load torsion results, and the additional wall shears from this torsional effect should be considered in the wall design. Most building codes also require the addition of an "accidental" torsion force when earthquake loads are considered. See MDG 9.2.2 and MDG Example 9.2-3. Cracking reduces thestiffness of reinforced masonry shear walls, and thereforeaffects lateral force distribution (in the case of rigid floor diaphragms), and also affects lateral force distribution among wall piers in all cases. 13.2.5 FlexuralDesign 13.2.5.1 Unreinforced Shear Walls - Shear walls are commonly exposed to combined axial

force and in-plane bending moment resulting from lateralshear forces. The resulting normal stresses, fob, can be calculated, assuming plane sections remain plane, as:

f& = fa * f b P

Mc

fob=-*A Z

The resulting stress distribution is shown in Fig. 13.2-5. The maximum extreme fiber stress 13-17


A C 1 TITLESHDG 73

0 6 6 2 7 4 7 O507362 7 3 7

has two components; one fromaxial load(PIA) and the other from bending(McII). Because the Code provides different compression allowable stresses

under axial load and under

flexure, the following unity equation is used to provide a more accurate margin of safety for compression:

t Fig. 13.2-5 Stress Distribution under Combined Axial and Flexure for Unreinforced ShearWalls

Code 6.3.1.1allows

in the transversedirection for unreinforced

flexuraltensiononly

masonry. It is therefore assumed that flexuraltensionisnotallowedin direction(i.e.,for

shear walls),and

the in-plane

that an unreinforced shear wallmusthave

net

compression in the extreme fiber. This condition will control the design in the case of low gravity loads and high lateral wind or earthquake loads. In this situation, walls should be carefully arranged so that enough gravity compression load exists

to fully negate flexural

tension from lateral loads. Flanges may be considered effective in resisting normal stresses from axial load and bending moment. Code 5.13.4 specifies the effective flange width. wallsmustconform

The connection of intersecting

to the requirements of Code5.13.4.2(e) to achieve adequate shear

transfer at the interface. Code C. Figs. 5.13.1 through 5.13.4 show recommended detailsof wall intersections.

13-18


A C 1 T I T L E * M D G 93

0662949 0509163

653

13.2.5.2 Reinforced ShearWalls - If flexural tension develops inan extreme fiberof a shear

wall, the wall must have flexural reinforcement by Code 7.5.2 so that vertical steel will carry all tension. In this case, Code 7.3.1.2 limits the masonry compressive stress to 1/3f m and Code 7.2 limits the steel stress to 20 ksi for Grade 40 or 24 ksi for Grade 60 steel. In areas of high seismicity, reinforcement of masonry shear walls is required by local building codes. Referring to Fig. 13.2-6, the following compatibility and equilibrium equations can be utilized to solve for the location of the neutral axis and the required amount of steel. See MDG 12.3.3. Compatibility Equation:

Equilibrium Equation: P = C - T P = - 1f a b t M - C & A s i 2

fsi

L

-

T

I-

T

C

Fig. 13.2-6 Stress Distribution Under Axial Load and Bending for Reinforced Masonry Shear Walls 13-19


‘

A C 1 TITLEtMDG 93

m

Ob62949 0509364 59T

m

This is a balanced design if the masonry and steel at extreme location reach theirmaximum stress allowable simultaneously. For less reinforcement than thebalanced amount,allowable tension reinforcement will control the design. This would be the most common case for reinforced masonry walls.

13.2.6Shear

Design

13.2.6.1 Unreinforced Shear Walls

- For unreinforced

shear walls, the maximum shear,f,,

is determined per Code 6.5:

where Q, I and b are calculated based on the uncracked net cross section of the wall. See Flowchart Shear Design, Fig. 8.4-47, and MDG 8.4.3.9. The calculated maximum shear stress is not allowed to exceed the least of four values given in Code 6.5.2. Graphical presentation of the code allowables for shear are shown in Fig.

13.2-7. The Code allows an increase in shear due tofriction from gravity load compression with an upper bound value as shown in the figure. The friction contribution shall be due to the assumed force (Nv)acting normal to shear surface. N,, is typically assumed to come

from dead load only.

50

100

f

200

,(PS11

Fig. 13.2-7 Unreinforced Walls 13-20


150

- Code Allowables

93


m

Obb2949 0509365 426

m

Effective flange width should be determined asshown in Code C. Fig. 5.13-5 for calculating section properties of T, I, or C Sections. Proper connection details should be provided at the flange-web interface to assure adequate shear transfercapabilityrequired

for full

interaction betweenthe flange and the web. It is common to grout adjacent cells and place horizontal reinforcement in hollow unit construction for this purpose.

See details in Code

C. Figs 5.13-2, 5.13-3, 5.13-4. 13.2.6.2 Reinforced Shear Walls

- Shear resistance of a wall is providedeither by masonry

or by reinforcement. According to the Code these two contributions are not additive. See

MDG 8.4.3.9. If the calculated shear stress exceedsthe allowable shear for the masonry, all shear must be resisted by reinforcement placed parallelto the applied shear direction per Code 7.5.3.1. In addition Code 7.5.3.2 requires an accompanying portion of perpendicular either case is given as a function of Mwd ratio and the

reinforcement. The allowable in square root off,.

Shear walls with loweraspect ratios (expressedby M/Vd) have higher allowableshear stress (Fig. 13.2-8).

Reinforcement Resists Entire Shea Masonry Shear Wall Fixed at Bottom Only

Masonry Resists Entire Shear 4

4

0.5

-

I I l l

o!s

o:

l!S

210

MlVd ratio

M

-w=k Masonry Shear Wall 2d Fixed Top and Bottom

"

VdVd

Fig. 13.2-8 Reinforced Shear Walls

13-21


- CodeAllowables


93 W 0662747 0509366 362

m

REFERENCES

13.l.1

Higdon, Ohlsen & Stiles, "Mechanics of Materials",John Wiley & Sons, Inc., New York, 1960.

13.1.2

Porter M. L, Sabri, AA., "PlankDiaphragmCharacteristics.Task Report, Task 5.1, TechnicalCoordinatingCommittee

5.1", Final

for MasonryResearch,

Submitted to National Science Foundation, Engineering Research Institute, Iowa State University, Ames, IA; July 1990. 13.1.3

Porter, M.L, Yeomans, F.S., Johnson, A. W., "Assemblv of Existing

Diaphragm

Datal' Final Report, Task 5.2, Technical Coordinating Committee for Masonry Research,Submitted

to NationalScienceFoundation,EngineeringResearch

Institute, Iowa State University, Ames, IA, July, 1990. 13.2.1

Schneider, R., and W. Dickey, "Reinforced Masonry Design", 2ndedition, Prentice Hall, Englewood Cliffs, NJ, 1987.

13-22



Example 13.1-1 RCJ Hotel

93 H Obb2949 0509167 2 T 9

m

- Shear Design of a Reinforced Brick Coupling Beam

Given the material properties shown below (whichare repeated from MDG Example 11.35 ) , perform the shear design for the corresponding lintel using Code Chapter 7. The lintel

is for a door opening along Grid Line 2 (see MDG Fig. 9.1-6). The lintel is a single wythe

reinforced clay brick lintel beam assembly with cross section and properties shown below. Unit strength of clay masonry = 6,000 psi Mortar

=

Type S

Sm= 2,500 psi fg = 3,800 psi E,,, of clay masonry = 1.9 x 106 psi

n = 15.3 Steel reinforcement = Grade 60 d = 32 - 3 = 29 in. j = 0.9 (from MDG Example 11.3-5) b = 7.5 in. (from MDG Example 11.3-5)

and

Calculations

The clear span of lintels on Grid Line 2 of the RCJ Hotel is 3 ft. 4 in. and subjected to a uniform load of 264 plf (from MDG Example 9.2-4 - Seismic Controls). Assumean effective bearing 4 in. on each end. Thus, the shear forces V, and V, are calculated from

VI = 759 i n . - k i p ~ + 621 in.-kips - 0.264 klf(3.33 ft)(1.67ft) (12 in./ft) 40 in.

VI = 34.1 kips

13-23



93

m

Obb2949 0509168 L35

m


and

Calculations

v. =

Reference

759 in.-kip~ + 621 h-kips + 17.6 h-kips

40 in.

=

Code

34.9 kips

According to Code 7.5.5, the calculated maximum shear for noncantilever beams can be taken at d/2 from the face of support. Thus design V

=

34.9 kips - 0.3 kips = 34.6 kips

Shear design of beams is governed by Code Chapter 7. Therefore, the shear stress is found from Code Eq. 7-3: V Using j

7.5.2.1

=

fv

=

0.9 from MDG Example 11.3-5 gives:

34.6 kips fv

= (7.5

in.)(O.9) (29 in.)

=

177 psi

The maximum shear stress allowed, F, , if shear reinforcement is not provided is

F,,= 66.5

< 50 psi x 1.33

=

66.5 psi

7.5.2.2

177 psi

Shear 1reinforcement is required.

F,,= 3.0

fi

X

1.33

200 psi > 177 psi

A,="

vs Fvd

=

200 psi

.+.OK

34,600 lb x 6 in. 24,OOO psi x 1.33 x 29 in.

Use DBL #3 @ 6 in. O.C. 13-24


=

0.22 in?

A C 1T I T L E l K M D G

93

m

Q b b 2 9 Y 9 0509369 071 W


Code Reference


Check steel placement b = 7.5in.

Face shell thickness = 1 1/4 in. Shear Reinforcing = #3 = 0.375in. Flexural Reinforcing = #6 = 0.75in.

7 .S "

i

Code requires 1/2in. thickness of grout between masonry unit and bar

8.3.1

and 1 in. minimum between bars.

8.3.5

Minimum beam width = 2 (0.375 in.) 2 (0.5 in.)

+ 1 in. =

6.75in. < 7.5in.

+ 2 (1.25 in.) + 2 (0.75 in.) + .: OK

Reinforcement is required perpendicular to shearreinforcement by the amount of 1/3A,. The longitudinal flexural steel may be used to satisfy this requirement.

13-25


7.5.3.2


Example 13.1-2

73

RCJHotel

m

0 b b 2 7 4 9 0509370 8 9 3

m

- Shear DesignforCanopyBeam

MDG Example 11.3-1 considered the flexural designof the canopy frame beam forthe RCJ Hotel. This beam spans 31 ft center to center in the east-west direction. The data used in MDG Example 11.3-1 are:

f',= 2,500 psi (from prism tests)

Dead Load = 500 plf LoadLive Brick Clay

fg = 3,800 psi (specified)

= 200 plf

Strength Unit

= 6,000 psi

Grade 60 steel

Type S Mortar

E, = 1.9 x 106 psi (Code Table 5.5.1.2)

h = 24 in., d = 20.5 in., b = 11.5 in., j = 0.89 Evaluate the canopy beam for shear considerations.

and

Calculations

1.

The structural analysisof the reinforced brick beam-column frame for Seismic Zone 4 yields a controlling load combination

of D

+ L + E; therefore, the allowable

stresses can be multiplied by 1.33 according to Code 5.3.2. The maximum negative moment at the beam end is 55.2 ft-kips, the maximum positive moment is 49 ft-kips, the axial load is 4.4 kips. The j from MDG Example 11.3-1 is 0.89 associated with the negative moment where the shear is maximum.

2.

The resulting shear diagram is: 9.6kips

Shear Section

13-26


A C 1 TITLExMDG 93

m

Obb2949 0 5 0 9 3 7 37 2 T

m


and

Code Reference

Calculations

3.

Find design shear V at d/2 from face of column:

distance

1 2

=

-(coIumn width)

=

(

7.5.5

d

+ -

2

11.: h) (20.; h) +

distance = 16.0 in.

V 4.

=

11,200 lb

Find shear stress f.:

Eq. (7-3) fv

5.

11,200 lb = (11.5 h.)(0.89) (20.5 in.)

=

53.4 psi

Check Code maximum shear stress, if no shear reinforcing: Neglect the benefit of the 4.4 kips axial compressive load in the canopy beam since Code does not provide for increased shear strength in compression.

F,,=

&

=

d

m

SI =

50 psi

S

50 psi

Thus F, (1.33) = 66.5 psi (for load combination) > 53.4 psi. Therefore, no shear reinforcing is needed.

13-27


7.5.2.2

5.3.2

93


Example 13.13 RCJ Hotel

m

m

Obb2949 0509372 bbb

- Shear Design of Continuous

Masonry Beam

Perform the shear design for a continuous masonry beam on Grid Line E spanning from column E-3 to E-3.5 to E-4 above the opening on the first floor of the RCJ Hotel (Wall Construction A withBuildingConstruction

I).

The continuousbeamspansover

two

openings each 11ft wide. The length overeach of the three support columns is 2.67ft. The total beam load is 17,020 plf with the following assumed data (see MDG Example 11.3-6): Concrete Masonry Unit Strength = 2,000 psi

b = 11.63in.

Type N Mortar

d = 67 in.

f', = 1,800 psi fg

j = 0.912 for

= 3,600 psi

+A4

j = 0.902 for -A4

Grade 60 steel

F, = 24 ksi

E,,, = 1.8 x lo6 psi and

Calculations

From MDG Example 11.3-6, the moment diagram gives a moment of 341.4 ft-kips at the center support using a span of 12.7 ft for flexure. 17,020 plf

l

t

Bekring 8''

i

i

i

i

i I

l

i

i

i

t

i

l I

I

11'-O'' Clear

11'-O'' Clear

-

Column

2'4" Column

Loadlna

341.4 ft-kips

MomentDlaaram

13-28


-

2/43" Column

192 ft-kips

192 ft-kips

l

_____

1

93 D 0662749 0 5 0 9 3 7 3 5 T 2 D



Code Reference

Calculations

From statics:

VR = 17,020 plf(12.67 ft) - 80,900 lb

t

'

t

+

=

134,700 lb

17,020plf

t

t

t

i

80,900lb 12'43''

4

($,J$,

Of Bearing

''

c 4

+

t

269,500 lb 12'4''

t

t

80,900 lb Q-q Of Bearing

CT

b4

Reactions 12'"' Shear Reinforcement

4

Reinforcement Not Required

t

c = 2 #4 @ 8" oc

6'4''

Reinforcement Not Required

?c?19 m o m - T

Kr u3

/ / /

/

W

2a -

r n

,3

!

S

! 2 ? d e

'.2

* "

I-

Face Of

l"5"

d/2 From Face Of Support'

Face Of Support

7"11" d12 From Face Of Support

1"4"

Shear Diagram

13-29


i

4"9"

AC1

TITLEaMDG 9 3 D 0662949 0509374 Y39 m

Example 13.13 Cont’d. Code Reference


From MDG Example 11.3-6, the length of the center column is 2 ft - 8 in. Thus, the shear, V, at the face of the center support is:

V

=

112,100 lb

At d/2 from the face, the shear, V, is:

V

=

112,100 lb -

V

=

64,600 lb

(17,020pH)

Neglecting flexural tension (consistent with

7.5.5

MDG Example 11.3-6), the shear stress from

Code Eq. 7-3 is as follows:

V fv

7.5.2.1

=

using j = 0.902 (from MDG Example 11.3-6): 64,600 lb =

(1 1.63 in.)(0.902)(67 in.)

=

91.9 psi

The maximum allowed shear stress, F, , is given by: 50 psi

Fv =

Fv =

c / -

=

7.5.2.2

42.4 psi

50 psi

For this beam, f, = 91.9 psi > 42.4 psi; therefore, shear reinforcement is required. Code ,

7.5.2.3 says that the maximum F, when shear reinforcement exists is:

13-30


TITLE*NDG 93

AC1

Obb29Y9 0509375 375 H

Example 13.13 Cont'd.

and

Code Reference

Calculations

Fv = 3.0

fi

S

150 psi

.: F,,= 3.0 4-

7.5.2.3

127 psi

=

S

150 psi

:.

OK

For this example: =

fv

91.9 psi < 127 psi

Thus, shearreinforcement

is acceptable.

NotethatCode

7.5.3says

thattheshear

reinforcement shall resist all of the calculated shear (i.e.? no contribution is counted by the masonry). Code Eq.(7-10) is used:

4

rd vs

=

7.5.3

Options: A"

S

8 in.

0.32 in.2

12 in.

0.48 in.2 (n.g. - need 8 in. multiples)

16 in.

0.64 in.2

Select 2 - #4 Bars @ 8 in.

O.C.

(vertically). A , = 2(0.20 in.2) = 0.40 in.2/8 in.

Check spacing:

7.5.3.1

Maximum S < d/2 < 48 in. 13-31


A C 1 TITLEvMDG 93

m

Obb2949 0509376 201

m


Reference

Code

"- 67 in. = 33.5 in. > 8 in. :. OK 2

2

Check shear reinforcement requirement throughout span. a)

Whereisshearreinforcementnotrequired? where

fv

=

E

=

42.4 psi

V f'=bjd V

= 29.8 =

b)

kips occuts at

V

=

42.4 psi( 11.63 in.)(0.902)(67 in.) = 29,800 lb

29.8 kips (12.67 f t) + 4.75 ft 80.9 kips + 134.7 kips

6.5 ft fiom center of end bearing

Wherecanshearreinforcement

A

vs Fsd

be decreasedto2 #4 at 16 in.

O.C.

S

=-

V = 0.40 h2(24 ksi) (67 h) = 40.2 kips 16 in.

V

= 40.2 =

kips occurs at

40.2 kips 80.9 kips + 134.7 kips

7.11 ft from center of end beating

13-32


(12.67 f t ) + 4.75 ft


93

m

m

Obb2949 0509377 L48


Calculations

Discussion

Reference

Code

Since the point at which shear reinforcement spacing can be doubled is so close to the point at which shear reinforcement is not required, provide the maximum shear reinforcement everywhere thatshear reinforcement is required. See the shear diagram for layout. Where shear reinforcement is not required, the designer may choose to use #3 stirrups at 24 in. on center, as a means of supporting the flexural reinforcement. Reinforcement is required perpendicular to the shear reinforcement, i.e., horizontal, in this case, by the amount of (1/3) A , = 0.032 in.2/3 = 0.11 in.2. Although not specifically stated in the Code or Commentary, itis the author’s understanding that for the bearing element under consideration the flexural steel can be used to satisfythis

Code

provision. From MDG Example 11.3-6, A, = 3.0 in.2 > 0.11 in.2

:. OK If the designer decides to satisfythis

Code provision by adding

additional reinforcement, then an acceptable solution is as follows: this reinforcement must be uniformly distributed, and the spacing must be less than 8 ft-Oin.

Prefabricated joint reinforcement with two W1.7

side rods and cross rods, with an effective area of2(0.0173in.2) 0.0345 in.2 can be placed 1 ft-6 in.

O.C.

=

horizontally. For four layers of

reinforcement, the horizontal reinforcing area is 4(0.0345 in.2) = 0.138 in.2 > 0.11 in.2 required. OK

13-33


7.5.3.2

A C 1 T I T L E * M D G 93 D Obb29Y9 0509178 084

m

Example13.1-3Cont’d. Code Reference


Laddertype

joint reinforcement is preferable to trusstype

joint

reinforcement when vertical bars are also used. In the truss type, the diagonal cross wires interfere with the bar placement. Provide U-shaped The shear reinforcement must be anchored.

stirrups

8.5.6.1(e)

with lap slice length = 1.7Zd 1.7 Id 1.7 Id

=

1.7(0.0015 d,F,)

=

1.7(0.0015) (0.5 in.)(24,000 psi)

=

30.6 in.

8.5.2

use 31 in. lap length

See MDG Example 11.3-6 for cover requirements. The horizontal joint reinforcement can be developed or anchored by extending into the neighboring masonry elements.

For a check on steel placement within the beam, see MDG Example 11.3-6. The required out-to-out dimension of the stirrup is 11.62 in.

- 2(1.5

in.) face shell

- 2(0.5 in.)

grout = 7.62 in.

The minimum bend radius is 6 bar diameters or 6(0.5 in.) = 3 in. Therefore the minimum out-to-out dimension of the stirrup is 3 in.

+ 3 in. + 0.5 in. + 0.5 in. = 7 in. < 7.62in. :.

13-34


OK

8.5.5.1

i

A C 1 TITLE*MDG

73

m

0662747 0507177 T10


Calculations

Code Reference

Discussion

The final beam cross section and steel arrangement are shown below.

.C. Where Indicated

On Shear Diagram; 2

1.7b = 31"

H 1163''

13-35


- #3 At

24" O.C. Elsewhere

A C 1 TITLE*UDG 93

Example 13.1-4

Obb2949 0509380 732

m

-

RCJ Hotel Shear Design of aReinforcedClayBrickNonloadbearing Wall

Consider the shear design of a reinforced hollow clay brickstair wall located between Grid Lines F and G along Grid Line 3 (See also MDG 9.1.3). This wall is designed for out-ofplane bending only (See MDG Example 11.1-12). The wall is an exterior stair wall in the

RCJ Hotel with Building Option I and Wall Construction Option B. The wall cross section and properties are shown below: Unit Strength of Clay Brick = 6,000 psi Mortar = Type S I

y,,,= 2,500 psi

1 /d = 3.75”

fg

= 3,800 psi

E,,, = 1.9 x 106 psi

5”

Eg = 1.9 x 106 psi

n = 15.3 Steel = Grade 60 j = 0.94 (MDG Example 11.1-12)


The cross section and materials are the same as those usedin MDG Example 11.1-12. The wall height, momentand forces are the same as those in MDG Examples 11.1-11and 11.112. The wall is subjected to a wind load of 25 psf. The resulting moment for the 11ft span was 4,540 in.-lb/ft (MDG Example 11.1-12). Thus,

v = (W)- (0 2

13-36


the shear, V , is

A C 1 TITLE*NDG 93

m

Obb2949 0509LBL b 7 9

m


Calculations

This wall is a single-wythe wall composed of 8 in. nominal clay units with No. 4 reinforcing steel bars placed in the center of a grouted cell at 40 in.

O.C.

From MDG Example 11.1-12:

d = 3.75 in.

j = 0.94 According to Code 7.55, the calculated maximum shear for"non-cantilever beams" is taken at d/2 from the face. The commentary says "Beam or WallLoading".

Even though the

shear at d/2 or 2 in., in this case, could be used, most designers simply use the end span reaction for walls for use in finding the shear stress in Code Eq. (7-3) as shown below:

V

fv =

bjd

fv =

138 lbs (12 in.)(0.94) (3.75 in.)

Eq. (7-3) 7.5.2.1 =

3.3 psi

The maximum allowed shear stress, F,,, is given by:

Fv

=

:. Fv

=

@<

4

50 psi

50 psi x 1.33

=

7.5.2.2 5.3.2

66.5 psi

For this wallf, = 3.3 psi C 66.5 psi; therefore, no shear reinforcing is needed.

13-37


A C 1 TITLESUDG 93

Example 13.1-5

DPCGymnasium

Perform the shear design for the wall

Obb29q7 0 5 0 9 3 8 2 505

m

- Shear Design for a Composite Masonry Wall on Grid Line A of the DPC Gymnasium assuming

Wall Construction Option B. The flexural design was given in MDG Example 12.2-4.

,

Grout

fg

=

5,100 psi

Eg = 2.55 x lo6 psi /+3642$

EBIoat= 2.08 x lo6 psi

4

11.63'

Em = 2.40 x 106 psi

Wall Section

Transformed grout to block

bg = 14.7 in.

I

y

13.9'

=

Transformed brick to block

H 2 - k 11.63'

4

bb = 13.05 in.

i

Transformed Section

Note: The listed values of

=

Calculations 1.

Reference

Out-of-planeshear due to induced 20 psfwindload:

13-38


5.89 in.

i are based on

the CMU wall not being grouted.

and

3.74 in.

Code


93

m

Obb2949 0509383 4 4 3

m

~

I


Calculations

The calculated shear stress is:

2.

fv

=

=

VQ

6.5.1

distance from unit centroid to composite section centroid

I = 2,960 in.4 (from MDG Example 12.2-4)

-

ybl = 5.89 in.,

5,

=

3.74 in. based on CMU wall not being grouted.

Q = statical moment of area outside interface in question In this problem two interfaces exist -- (1) between CMU and grout and (2) between brick and grout. Both calculations will be shown (For QIand Q2,respectively).

QI = 36 in? (5.89 in.) = 211 Q2

=

13.9 h ( 3 . 6 3 in.)(3.74 in.) = 188

in.3

f., controls 3.

Check the allowable interface shear

F,,= 10 psi I

Thusf,, < F,

x

1.33

=

5.3.2

13.3psi

.-. OK 13-39


5.8.1.2(b)

A C 1 TITLExMDG 9 3

Example 13.1-6

m

0bb2949 0509384 3 8 8

m

-

T M S Shopping Center Shear Design of an Unreinforced Wall Due to Out-of-Plane Bending

Consider the shear design of an unreinforced wall (Wall Construction Option A) subjected to out-of-plane bendingfor the interior wall on Grid Line 2 of the TMS Shopping Center. Initial data is: Use 8 in. CMU with face-shell bedding of A)

Appendix MDG(See Wall span

and

=

S,,, = 1,500 psi

mortar weight Wall

psf = 46.5

16 ft (See MDG Example 11.1-3)

Calculations

From MDG Example 11.1-3: 1

V-

= - ( F p

Seismic Controls

2

1

= -(2.6

V', =

p~Q(l6f t )

2 20.8 plf

The axial force at midheight due to wall weight is

P

=

(8 ft)(46.5 psf)

=

372 PE

6.5.1

The shear stress is: fv =

VQ

Eq. (6-7)

For a 1 ft width of wall, the shear stress is

13-40


1 i


Calculations

Reference

Discussion

Code

Note that the shearis assumed to beresisted by two face shells sincethe wall is unreinforced and uncracked. Some engineers conservatively use only a single face shell in their calculations to cover both the uncracked and cracked conditions. For this case f, = 2.0 psi. Code 6.5.2 provides allowable stresses for in-plane shear stresses in combination with axial load. Separate shear stresses for out-of-plane bending shear are not stated. However the Code C. suggests using Code 6.5.2 values for out-of-plane loading. For this example, Code 6.5.2 allowables will be used. Using Code 6.5.2 for combined axial load and shear provides the following allowable shear stress, F,, values: (u)

Fv

=

1.5&

= 1.54-

= 58.1

psi

(b) Fv = 120 psi (c)

Fv

=

6.5.2

Nv v + 0.45 An

=

37 psi + 0.45(372 lb) 30 i n 2

=

42.6 psi

Where v = 37 psi for running bond not solidly grouted

F, = 42.6 psi controls 5.3.2

For seismic, F, = 1.33(42.6 psi) = 56.6 psi For this example, f v = 2.0 psi c 56.6 psi, :. OK

13-41


A C 1 TITLE*flDG 93

Example 13.1-7

DPCGymnasium

0662949 0507186 150

- ShearDesignofaReinforcedCompositeWall

Consider the shear design of a reinforced composite brick-block wall (Wall Construction Option C) on Grid Line 1of the DPC Gymnasium. This composite wall is the same as that in MDG Example 11.1-6 and is composed of a 4 in. clay brick wythe, a 2 in. grouted collar joint,and an 8 in. concretemasonrywythe.

The material properties are (From MDG

Example 11.1-6): Concrete Block Clay Brick

Unit Strength (psi) Mortar P m

(Psi)

E m

(psi)

Masonry(Hollow)Masonry

Grout

2,000 Type N

8,000

N.A.

Type N

N.A.

1,500 1.8 x 106 (Code

2,500 2.0 x 106

2,000 (fg psi) 1.0 x 106 (Eg psi)

Table 5.5.1.3) 14.5

16.1

It

Reinforcement - Grade 60, E = 29 x lo6 psi Calculations and Discussion

Code Reference

The cross-section (from MDG Example 11.1-6) is:

T I

7.63"

8.63' '

13.3"

13-42



m

Ob62949 0509187 097

m


Code Reference

Calculations

This wall isconsidered to span vertically on a 30ft span andis subjected to lateralwind load of 20 psf. Initially the wall is bent under wind pressure producing compression in the brick wythe; however, the wall must also be checked for suction where the inside face shell of the CMU is in compression (see also MDG Example 11.1-6). For Commession in Brick Wythe: From MDG

d = 4.63 in.

Example 11.1-6

j = 0.932

From Code Chapter 7, the shear stress, fv, is:

Eq. (7-3)

V

fv =

bjd

7.5.2.1

The shear force, V , is: W1 v =-

2

According to Code 7.5.5, the shear at d/2 from the face can be used for design; however, as discussed in MDG Example 13.1-4, the span of I will be used, thus

.. *

= fv =

I

300 lb (12 in.)(0.932)(4.63 in.) 5.8 psi

The maximum allowable shear stress, F", is given by: 13-43


7.5.2.2

A C 1 T I T L E x M D G 9 3 W Obb29Y9 0509LBB T23 D


Code Reference

Calculations

Fv =

6

Fv = F,, =

d m = 50 psi d m = 38.7 psi

< 50 psi

Eq. (7-4)

For wind, F, = 1.33(50 psi) = 66.5 psi For this case, fy = 5.8 psi

C

66.5 psi,

5.3.2

.-. OK

For CornDression in the Concrete Block Wvthe: From MDG Example 11.1-6 (with the same wind pressure assumed in both directions) d = 8.63 in.

jd = 7.94 in. 7.5.2.1

V

fv =

= fv

=

bid 300 lb (12 in.)(0.92)(8.63 in.) 3.2 psi

The maximumF,

=

dm

=

For wind,Fv = 1.33(38.7 psi)

38.7 psi =

7.5.2.2 5.3.2

51.5 psi

This design is okay for shear.

13-44


A C 1 TITLEtMDG 73 D O662949 0 5 0 7 3 8 7 7bT

m


Code Reference

See Note in MDG Example 11.1-6 regarding delamination. Check for potential delamination 5.8.1.2

Maximum allowable collar joint shear = 10 psi

5.3.2

Increase 1/3 for wind = 13.3 psi Delamination computations are generallybased

on the resistance

provided by the contact area. However, since this exampleis based on a cracked section the transverse shear computed above will be conservatively used for this in plane shear check.

f v = 5.8 psi c 13.3 psi fy

= 3.2 psi

psi 13.3

.: OK .s.

OK

13-45


A C 1 TITLE*rMDG 9 3

Example 13.1-8

On GridLineB,

RCJHotel

m

m

0662949 0509390 b 8 L

- ShearDesign

of a CouplingBeam

fourth floor of the RCJ Hotel (Wall Construction Option B,Building

Option 1) perform a shear investigation of the coupling beams. From MDG Example 11.3-

7, the clockwise end moments are 450 and 409 in.-kips, respectively, the opening is 3 ft 4 in. wide, and the total beam depth is 2 ft 8 in. The properties are:

fg

Clay Brick Unit Strength = 6,000 psi

=

3,200 psi (from prism tests)

Type S Mortar

Reinforcement Grade 60 E, = 1.6 x 106 psi (from tests)

= 2,500 psi

n

=

Eg = 1.6 x 106 psi

18.1

d = 28 in.,

j = 0.914,

b = 7.5 in.


Reference

First find the end shears from the free-body diagram

of beam forces:

Code

for dead weight,

assume solidly grouted. Thus, weight, W = (88 psQ(2.67 ft) = 235 plf

W

450 in.-kips

fi

450 h-kips + 409 h-kips -

VL = VL = 21,100lb;

= 235 plf

235 plf (3.33

3.33 ft(12 in./ft)

Vk = 21,900 lb

13-46


12 in./ft)

l,O00 lb/kip

A C 1 TITLE+flDG 93

m

0662949 0509191 518

m


Calculations

7.5.5

Find the shear force, V,at d/2 from face

Y

=

21,600 lb

7.5.2.1

Find the shear stress, f,,from

Eq. (7-3)

V

bjd

fv

=

G

21,600 lb = (7.50 in.) (0.914) (28 in.)

&,

=

113 psi

The maximum allowable stress, F,, is

Fy =

@ S 50 psi

F,,=

Jm 50 psi

7.5.2.2

Eq. (7-4)

=

F, is permitted to be increased by 1/3

..

5.3.2

F, = (50 psi)(l.33) = 66.5 psi

For this example, f v = 113 psi > 66.5 psi Therefore, special shear reinforcement is needed. Check maximum shear stress (When shear reinforcement exists). Thus, maximum

13-47


7.5.2.3

A C 1 TITLEsMDG 93

m

0662747O509392454

W

Elxample 13.1-8 Cont’d. Calculations and Discussion

F,

= (1.33)3.0c

F,

=

S

Reference

Code

(150psi)(1.33) = 200 psi

Eq. (7-7)

( 1 . 3 3 ) 3 . 0 @ Ö = 200 psi

For this example, 113 psi < 200 psi, so shear reinforcement is possible. All of shear must be carried by reinforcement

7.5.3

Maximum F, = (24,000 psi) 1.33 = 32,000 psi

7.2.1.1 5.3.2

Thus, A,,

=

A,

=

if

(21,600 1b)s

32,000 lb(32 in.) 0.0211s

S =

16 in.,

A , = 0.338in.2

S =

12 in.,

A , = 0.25 in.2

S

..

A , = 0.51 in.2

S =in., 24

= 8 in.,

Use 1 - #4 bar at 8 in.

A , = 0.169 in.2 O.C.

+

1 - #4 each cell

for vertical shear reinforcing.

Check maximum spacing of d/2 < 48 in.

7.5.3.1

13-48



93

m

Ob62947 0507173 390

m


Calculations

Discussion

- 16 in. > 8 in. :.

"

Reference

Code

OK

2

7.5.3.2

must be provided perpendicular to shear reinforcement.

0.17 h*= 0.057

h 2

3 This can be provided by the flexural tension steel. Check steel placement

b

=

7.5in.

Face shell thickness = 1.25 in. Shear reinforcing = #4 = 0.5in. Flexural reinforcing = #5 = 0.625 in.

t"---+ 7S ' '

MDG Example 11.3.7 chose 2-#5 for flexural steel. A single #7 would also satisfy flexural requirements and actually be a better selection with regard to shear tie support. Use 1-#7 Code requires in. 0.5 and

bars minimum1 in.

thickness of grout between masonry units

and

between bars.

Minimum beam width = 2 (1.25 in.) =

8.3.1

+ 2 (0.5in.) + 0.875in. + 2 (0.5in.) :. OK

5.4 in. cin.7.5 13-49


8.3.5

A C 1 TITLExMDG 93 D 0662949 0509194 227

Example 13.1-9

TMS Shopping Center

-

Shear Design of

a Reinforced CMU

Nonloadbearing Wall

Design the East Wall (Grid Line 3) of the TMS Shopping Center for shear from out-ofplane bending assuming Wall Construction OptionB. The wall was designed for flexure in MDG Example 11.1-2. The material and other load properties (see MDG Example 11.1-2) are:

CMU Strength psi= 3,050

Also, j = 0.9

Mortar = Type N

in.

d

psi f, = 2,000

=

3.81

b = 32 in.

E, = 2.3 x 106 psi


n = 12.6

psf

Wind = 20

Assumed pinned at foundation and at 16 ft level (See Sketch) and

Calculations

MDG Example 11.1-2 has established the load distributions

for wall elements Part 1 and

Part 2. Part 2

i

Part 1

16' O"

Design For These Wall Elements

13-50


A C 1 T I T L E * M D G 93 W Obb2949 0509395 L b 3

m


and

Code Reference

Calculations

Shear and Moment Diapams Part 1 & Part 2

Part 2

Wind on Wall

Wind on Door

Wind on Wall above Door Lintel

(Width = 2.67’)

f

i-

Wall Element Part 2 will be examined for shear first. Maximum shear from above diagrams occurs just below the roof support reaction i.e.

V = (163 plf)(2.67 ft)

+ (312.5 lb) + (500 lb) = 1,250 lb

13-51


A C 1 TITLExMDG 93

0 b b 2 9 4 9 0509196 O T T D


Calculations

According to Code 7.5.5, the shear at d/2 from face of support could be used provided support reaction is in compression; however, the amount of change is small and therefore neglected. The shear stress is found from Code Eq. (7-3):

V

fv =

7.5.2.1

bjd

From MDG Example 11.1-2 fv

=

b = 32in.,

j = 0.90,

d = 3.81 in.

1,250 lb = 11.4 psi (32 in.)(0.9)(3.81 in.)

The allowable shear, F,, is:

1.33E

S

1.334m

7.5.2.2

1.33(50 psi) =

59.5 psi < 1.33(50.0 psi)

=

66.5 psi

For this problem 11.4 psi < 59.5 psi, thus shear is okay and no special shear reinforcing is needed. Typically out-of-plane shear for walls is small enough that no shear reinforcing is required. Since the actual shear stress is less than the Code allowable shear stress for Wall Element Part 2 (the critical element), Wall Element Part 1 need not be checked. A sliding shear failure at the base of the wall may occur if there is insufficient bond between

the floor slab and the wall. See MDG Chapter 14 for connections of walls to foundations design. 13-52


A C 1 TITLEtMDG 93

Example 13.1-10

m

Obb2949 0509197 T3b

-


ShearDesignfora

m

Doubly Reinforced

Masonry Lintel

The shear design is to be performed on the doubly reinforced masonry lintel beam in MDG Example 11.3-2. The lintel dimensions and properties are: Type N Mortar


= 1,500 psi

E,,, = 1.8 x 106 psi

fm

CMU Strength = 2,000 psi

fg

Eg = 1.8 x 106 psi

d = 29 in., b = 7.63 in. wI = 792 plf

= 3,600 psi

+ W, = 4,670 plf

(Distributed over central 40.4 in.)

j = 0.915

,A, J'llo:rngth Bearing

,,

Of

Plate

\\ lb I(40.5 in. I 12 in.lft 1 = 4.670 plf

7& $-01 '5,

"""""."""" 10"

+4X

7.63" = 40.5"

88 psf (9') = 792 plf

Lintel -

10.33'

and

Calculations

1.

Discussion

Reference

The shear force at theface of reaction is:

where a

=

length of load w2 13-53


Code

A C 1 T I T L E * N D G 93

a

0662949 0509398 972



Code Reference

V = 792 plf(10.33 fi)

4,670 plf( +

)

=

12,000 lb

According to Code 7.5.5, find shear @ d/2 fromface:

V 3.

l2 hlfi

2

2

2.

40.5 in.

=

(

"P)

12,000 lb - 792 plf :2)( -

=

11,043 lb

FromCode Q. (7-3), findthe shear stress:

V

fv

=

fv

=

fv

4.

bjd 7.5.2.1

11,043 lb (7.63 in.)(O.915) (29 h) = 54.5 psi

The maximum shear stress for no shear reinforcement is:

F,, = 1 . O E

5

50 psi

7.5.2.2

.: F,,= l.O\/m = 38.7 psi 54.5 psi > 38.7 psi :. N.G. 5.

Check to see if F, is exceededwhen shear reinforcement isused: 150 psi

F,,= 3.0@',,,

F,,= 3 . 0 4 m

=

7.5.2.3

116 psi > 54.5 psi

13-54


:. OK


93

m

0662949 0509399 8 0 9 H



Reference

Code

Thus, shear reinforcement is permissible. 6.

Design shear reinforcement: Code 7.5.3 requires that all of shear be carried by reinforcement, according to: A,

=

vs -

(7-10)

EQ.

Fs (i 7.2.1.1

The maximum F, = 24,000 psi thus,

if S = 16 in. A, = 0.25 in2 if S = 24 in. A, = 0.38 in.2 if S = 8 in.

Use 1 - #4 bar @ 8 in.

O.C.

A , = 0.13 i n 2

Use 1 - #4 each cell

for vertical shear reinforcing. Check maximum spacing of d/2

< 48 in. 7.5.3.1

7.

Av Code 7.5.3.2 requires be provided perpendicular to shear reinforcement. 3

13-55



93 D Ob62949 0509200 350

m


Code Reference

If this Code criteriais assumed to apply to designed flexural steel,then 2 - #6 are sufficient. If this Code criteria is assumedto be additional steel (above and beyondthe design flexural steel) then one way to satisfy this situation is to use joint reinforcement in mortar joints.

If place in both joints

=

i&* 2

=

0.m h

2

per joint

Place one standard truss joint reinforcement in mortar joint near beam midheight. Each joint effective A, = 0.048 in.2 per joint

A

Check steel placement

b = 7.63 in. Face shell thickness = 1.25 in.

29 "

Shear reinforcing = #4 = 0.5 in. Flexural reinforcing A, = 2 #6

I'

A,' = 2 #7

3

"

Minimum beam width required = 2 (minimumcoverrequired;orthickness

thickness)

+

2 (diameter of #7 bar, 0.875)

of groutrequiredplus

+

face shell

1 (1 in. cleardistancebetween

+ 1 (diameter of #4 stirrup) + 2 (0.875in.) + 1 (1 in.) + (0.5 in.) = 6.75in.

longitudinal bar) = 2 (1.75in.)

13-56


< 7.63in.

.-. OK

A TCI1T L E l k f l D G

93

0662749 0 5 0 9 2 0 1 2 7 7


Reference

Check development of shear reinforcement 1, = 0.0015 dd;, = 0.0015 (0.5) (24,000) = 18 in.

Code

8.5.6 8.5.2

For standard hook in tension an equivalent embedment length 8.5.5.2

= 11.25db = 5.6 in.

A singleleg stirrup canbeanchored i

by providingastandardhookplus

an effective embedment of 0.51, 0.51, = 9 in. < 14.5 in. - 3 in. = 11.5 in. :. OK

13-57


8.5.6.1

A C 1 TITLE*NDG

Example 13.2-1

93

m

Obb2949 0509202 L23

TMS ShoppingCenter

m

- UnreinforcedShearWallDesign

Design one of the piers in the perforated shear wall elements of the TMS Shopping Center, located on Grid Line 3. between Grid Lines A and C. Assume Wall Construction Option

A (unreinforced concrete masonry). The analysis shown inMDG Example 9.3-12 indicates that pier 1is subjected to a shear load fromthe diaphragm of 2.66 kips, applied at the mean roof height of 16 ft. Control Joints

-

\ I

\

17'"''

13,300 l b

17'""

4

36'

b

16/rl/83


k

14.7'

Code Reference

The loading of pier 1 isshownbelow.

It is

2.66 kips

assumed that there is no gravity roof load applied

to this wall pier.

TV M

Assume that 12in.,hollow,faceshell

CMU's are used with a weight of 46.5 psf. See

MDG Appendix A for unit and wall properties tables.

13-58


bedded


m

93

0 6 6 2 9 4 9 0509203 ObT

m


M

=

(2.66 kip~)(l6fi)

=

Reference

Code

42.6 &-kips

Check Normal Stresses: Assuming face shell bedding (t = 1.5 in. from MDG Appendix A)

:. A,

=

2 x (14.7 ft x 12 h./ft) x 1.5 in. = 529 h?

Max. tensile stress

=

P +M -S

An = -

( 12.3 x lo3) lb

529 =

+

in?

-23.3 psi + 32.9psi

(42.6 x lo3 x 12)in.-lb in?15,600 =

9.6psi

Tension stresses arenot allowed for unreinforced wallelements subjected to in-plane forces since values in

to out-of-plane loading.

Code Table 6.3.1.1 apply only

Try 12 in. solid grouted CMU’s (Assume 100 pcf)

A,

=

S =

(11.63 in.)(14.7 ft x 12 in./ft) = 2,050 (11.63 h.) X (14.7 ft 6

X

12 h./fQ2

=

13-59


60y300

in.2

in:

6.3.1.1

A C 1 TITLE*:MDG 73 W Obb2747 0507204 TTb

m


Calculations

Max. tensile stress

= =

:.

25,600 lb + (42.6 x lo3 x 12) in.-lb 2,050 in? 60,300 h3 - 12.5psi + 8.49psi = - 4.0 psi

-

No net tensile stresses.

Since the compressive stresses are so low the unity equation: c

c

6.3.1

is OK by inspection.

Shear Stress

=

VQ -

6.5.1

Ibw

for rectangular sections f v

3 v

= --

An

.:

fv

=

-[

3 2.66 x lo3 lb 2,050 in?

2

]

=

psi

The allowable shear stress (Fv) is the least of:

b)

120 psi

C)

V

+

N V 0.45 -

An

Assume fm = 1,000 psi from prism testing of 1,300 psi units and Type N mortar. Use f, = 1,000 psi 13-60


6.5.2

A C 1T I T L E + f l D G

93

m

0662949 0509205 932

m


~

Code Reference

c) ..

F, = 60 psi(solid

..

F, = 47.4 psi governs and ismuch greaterthan f v = 1.95 psi

F, = 60 psi

grouted units)

+ 0.45(12.5

psi) = 65.6 psi

.: OK

Use grouted 12 in. CMU's with a minimum compressive strength of 1,300 psi and Type N Mortar.

13-61


A C 1 TITLE*UDG 93

Example 13.2-2

m

Obb2949 0509206 8 7 9

TMS ShoppingCenter

m

- ReinforcedShearWallDesign

For the TMS Shopping Center, design the shear wall on Grid Line C for Seismic Zone 1. Use Wall Construction A or B as required.

8 in. Hollow CMU - 2,000 psi fm

Type S Mortar

= 1,800 psi

Grade 60 Steel


From MDG Example 9.2-1 the seismic analysis gives V, = 11,500lb at roof diaphragm location. The moment at wall base is V& = 11,500 lb (16 ft) = 184,000 ft-lb. Check Wall Construction Option A - Unreinforced Concrete Block From MDG 9.1.1.1, the joist reactions on the wall are 2,000 lb dead load and 3,830 lb live load, except for the joist next to Grid 2 where half these values apply. The left beam reaction (look fromthe outside) is 4,690 lb dead load and 8,980 lb live load. The right beam reaction is 4,260 lb dead load and 8,150 lb live load. Using the self weight of the wall at 46.3 psf and uniformally distributing the loads (by inspection a reasonable assumption) the dead load is:

D.L. = 4,690 lb + 3 D.L = 31.7 kips

x 2,000

lb + 1,OOO lb

+

4,260 lb + 46.3 psf x 16 ft x 21.3 ft

And the live load is:

L.L. L.L.

=

8,890 lb + 3 x 3,830 lb + 1,915 lb + 8,150 lb

= 30.5

kips

13-62


A C 1 TITLE*:MDG 93

m

Ob62949 0509207 705

m


~

Code Reference


At base of wall

P

Mc

f=-Äf-I 5.3.1

The Code requires 90% of dead load, thus:

P

=

0.9 x 31.7 kip = 28.5 kip

The net bedded area and moment of intertia are: Net &&d

areu = (2 x 1.25 in.)(21.3 ft)(l2 in./ft)

Moment of Inertia

f = -28,500

640

f

=

ft

=

lb

ia2

-44.5 psi + 36.4 psi

=

bd3 -

=

640 i a 2

2.5 h(256 h)3= 3,500,000

12

h4

12

( ")

184,OOO fi-lb (12 h./ft)256

2

3,500,000 h4 80.9 psi fc = -125.4 psi

- 125-4

No tension for in plane loadingis allowed for unreinforced shear walls in Code. Thus design for Wall Construction Option B - Reinforced Concrete Masonry. = psi) 1.33(600

= 798 psi

13-63


7.3.1.2

A C 1 TITLExMDG 93 M 0bb2949 0509208 b4L

m


Code Reference

1.25"

Assume

steel

is

I .-

I , = 21.3' = 256" P

located in the last cellofthewall

I '

I

on

M

bothends.Consider only tension steel.

d = 252"

I-

-I

Apply iterative method given in

MDG 122.2.2 IC

#1

Load Case

.9D

+E

5.3.1

M = 184 ft-kips

P

-2 - A

= 28.5 kips

3

= 0.17

d = 252 in.

M = 0.31 -

A

->"

Pd

Pd

3

Assume a = 75 in.

13-64


Thus Region 3 applies


93

m

0bb2949 0509209 5BB

m


Discussion Calculations and

M,,= P(

+

-

Code Reference

MDG

a)

m.12.2-15 /12 in./ft

Mp = 28.5 kips

=

As =

M - Mp F,(d - a)

As

(184 fi-kips - 126 ft-kips)( 12 in./ft) 24 ksi x 1.33 x (252 in. - 75 in.)

=

MDG

A, = 0.12 in?

(small

126 ft-kips

W.12.2-16

as expected)

MDG Eq. 12.2-17

c=

(28.5 kips

0.12 in? x 24 ksi x 1.33) x 16.1 24 ksi x 1.33 x 2.5 in.

+

=

6.52 in.

MDG Eq. 12.2-18 System obviously will not converge (too close to Region 2)

d

a r

311 + L

n >

Tl

MDG

m.12.2-19

4 .

252 in.

=

2S.9 in.

13-65


A C 1 T I T L E * H D G 93

m

O b b 2 9 Y 9 0509210 2 T T

m


CaIcuIations and Discussion

MDG (252 in.)'

a = x0.666 3

- 2(28.5

W.12.2-20

Itips ~0.492x252 in. + 184 ft-kips X 12 h./ft) x2.5 ksi x4/3

2

in.

a = 47.3 in.

Since 3a = 3(47.3 in.) = 141.9 in. < d , then steel area is:

MDG Eq. 12.2-21

A, =

F, 3 x 0.666 ksi x 4/3 x 47.3 in. x 2.5 in. 2 A, = * 24 ksi x 4/3

A,

= 4.03

- 28.5 kips-

.

in?

This is more steel area than required without axial load. Thus, neglect axial load.

Assume (1) #5

P =

A, = 0.31 in.2

0.31 h* = o.Ooo49 2.5 in. x 252 in.

np

=

0.0079

k

i(0.0(n9)2 + 2(0.0079) - 0.0079 = 0.12

13-66


A C 1 T I T L E * H D G 93 W 0662947 05092LL L 3 b W


and

Code Reference

Calculations

i = (1 -

y)

= 0.96

Mt = A,FJd Mt = 0.31 i a 2

x

24 ksi

Mt = 199 ft-kip

252 in. 12 in./ft

x 4/3 x 0.96 x

184 ft-kips

2.

OK

bd2 M,,, = -kjFb 2

M,,,= 2.5 in.

x 252

x o.12 x o,96 x 0.666 ksi x 4/3

2

M,,, = 677 ft-kips > 184 ft-kips

12 h./&

:. OK

Use (1) #5 at the end of each wall Note: Since this wall is key to the stability of the entire structure for lateral loads, a good design would be (2) #5 at each end, and grout the wall solid, the cost increase will

be justified by performance.

SHEAR For the shear wall the bedded thickness is

the face shell thickness

2(1.25 in.) = 2.5 in.

13-67


=

A.4.9.1

A C 1 TITLE+MDG 93

m

Obb29Y9 0509232 072 W

Example 13.2-2Cont'd. and

Calculations

Computing the actual shear stress:

Eq. (7.3)

V

fv

fv

=

bid

=

11,500 lb 2.5 in. x 0.97 x 252 in.

=

18.8 psi

Compare actual shear to Code allowable shear

M

- 184 ft-kips

"

Vd

12 h./ft 11.5 kips x 252 in.

Fv = '[4 3

Fv = S

13 [4

-

X

o.76

(31E

Eq. (7.5)

- (0.76)]-4

4 [80 - 45 x 0.761 x 3

F,, f y

=

x =

4 -

3

= 64

psi

61 psi

No shear reinforcement required

Load Case #2

D + L + E

5.3.1

M = 184 ft-kips

P = 62.2 kips

- - A 3

d = 252 in.

13-68


=

0.17

AC1

TITLExIDG 93 m Obb2949 0509233 T09 m


Calculations

Reference

M = 0.14 2 - A Pd

:.

3

=

<(L

- I,

<(l

-

3d

-.) 0.17

I

256 in. - 0.492 = 0.169 3 x 252 in.

Region #1 Applies

Use face shells only (canservative) 2.5 h. X (256 h)2

M,,, = ...

6

x 0.888 h i

-

62.2 kips x 256 in. 6

12 in./ft

Mm= 1,800 fi-kips > 184 fi-kip

:. OK

13-69


1

Code


Example 13.2-3 DPC Gymnasium

0 6 6 2 9 4 9 0509214 945

- ShearWallDesign

Design the East Wall of the DPC Gymnasium on Grid Line 2 for Seismic Zone 2. The East Wall is subject to a seismic in-plane shear load of 28,400 lb from MDG Example9.2-2. Use Wall Construction Option B, unreinforced composite wall.

fg

= 5,100 psi

Eg = 2.55 x 106 psi EBIoac = 2.08 x 106 psi EM = 2.40 x 106 psi = 1,500 psi rm)b,.j&

= 2,400 psi

The concrete block wythe is ungrouted.

and

Calculations

Average Wall Height

=

24.67 ft + 30 ft 2

Wall Weight = Concrete Masonry

Wall Weight

= 40

=

27.3 ft

+ Clay Masonry + Grout

psf + 36.25 psf + 23.33 psf = 99.6 psf

Check to see if wall is subject to in-plane flexural tension: Overturning Moment = 28.4 kips (27.3 ft) = 776 ft-kips = 9,320 in.-kips For nonloadbearing wall, P is due only to the dead weight

P = (99.6 psg(24.67 ft)

=

2,460 plf (using conservative minimum height)

The contribution of the wall flanges willbe neglected in this problem. Transform 12 a

in. wall length equivalent to concrete 5.13.1.2

block 13-70


A C 1 TITLElrMDG 93

m

Ob62949 0509235 881

m


Reference

Actual :

Code

Transformed:

3.63'

Mal Stress, fa = P --

A

2,460 lb = 22.3 p i (4.18 in. + 2.5 in. + 2.5 in.) (12 in.)

Bending Stress due to overturning in-plane moments:

fb =

(9,320 in.-kip)(6) (1,o00 lb/kip) (4.18 in. + 2.5 in. + 2.5 in.) (64 (144 in./ft)

f b = 10.3 psi

Walls subjected to flexural tension mustbe reinforced and designed for shear according to Code 7.5.2

E

-

:.

Wall is not subject to flexuraltension;

A

S

=

22.3 psi - 10.3 psi = 12.0 psi compression

unreinforced shear wall, Code Chapter 6.

13-71


thus, check the walldesign

as an

A C 1 TITLE*NDG

93

m

Obb29Y9 0 5 0 9 2 1 b 718

m


Code Reference

Calculations

Consider shear stresses: Note that a combined shear due to direct in-plane shear plus shear due to twisting exists as illustrated by the eccentricity distance, e, at the beginning of the problem. Therefore, a torsion stress, T, exists and can be computed from:

where c is the distance from the centerof gravity to location of torsion stress, T is the torsional moment, and J is the polar moment of inertia of the cross section. This torsion stress r can be computed and added to the direct shear stress. However, this torsional shear stress is usually small. For this problem, if the direct shear stress is close to the allowable, then the torsional shear stress would need to be computed, otherwise it can be neglected. Calculate direct shear stress:

VQ Ib

- 3V -

6.5

3V 2Lb 3 (28,400lb)

" " "

fv-

fv

=

&,

=

2A

2 (64 ft)(12 in./ft) (4.18 in. + 2.5 in. + 2.5 in.) 6.0 psi in the CMU

& , = 6 p s i 2*4 x 'O6 Psi 2.08 x 106 psi fv

= 6

x 'O6

2.08

x

"1

lo6 psi

=

6.9 psi

=

7.4 psi in grout

13-72


Eq. (6.7)

A C 1 T I T L E x M D G 93 W Ob62949 0509217 b5Y


Code Reference


Checking the CMU the allowable shear stress, F,, is the least of (a)

Fv

=

1.56

=

1.54-

=

6.5.2

58 psi

(Note: could split shear by the proportional amount carried by each wythe and useallowable

for each. Since stresses are small, the lower

fm

will

conservatively be used.)

(b)

F, = 120 psi

(c)

F"

=

2)

v + 0.45(

Use running bond and not solidly grouted, so v = 37 psi thus,

0.45(:)

Fv =

V +

Fv

47.0 psi

(d) does

=

=

37 psi

+

not apply sinceunitslaidin

0.45( 2,460 (4.18 in. + 2.5

running bond

The allowable shear stress, F, = 1.33(47.0 psi) = 62.6 psi

f, < F, i.e.6.0psi

C

62.6psi

lb

in. + 2.5 in.)(12 in.)

5.3.2

.: OK

It is obvious that the shear in the brick and grout are OK 3.

Interface stresses dueto differential volumechanges: Note that claybrick expansion coupled with concrete shrinkage may induce an interface shear stress that should be checked. These differential displacements are not part of this problem and are discuss elsewhere in this MDG (See Chapter 10). 13-73



93 M 0662949 0509218 590 M

Examde 13.2-3 Cont'd. Calculations and Discussion

4.

Code Reference

Interface shear stresses for multiwythewalls: The ties across the interface between wythes must be capable of taking the interface shear stress, if this stress is deemed to be beyond the usual small amount. Code 5.8.1.2 provides for an allowable of 10 psi. In this case

the proportional amount of

shear carried across the interfaces does not need to be computed since the f v = 4.0 psi < 10 psi 5.

Ties across

=$

already OK

the interface:

Code 5.8.1.1 requires wall ties across the grouted collar joint. Code 5.8.1.5 requires at least one #9 gage wall ties per 2.67 ft2 of wall with a horizontal spacing S 36 in. and vertical spacing S 24 in.

#9 gagewalltie(styles

Placea

manufacturer'scatalogs)

other than "2"wallties

can be selectedfrom

at 16 in. on center verticallyand

24 in. on center

horizontally. 2 wall ties are not acceptable for this wall as per Code 5.8.1.5.

6.

Check the unity equationfor the compressionside of the in-planeflexure: Thischeckwill

beillustrated(eventhough

the stresses are smallandcould

be

ignored.) From Code 6.3.1:

Eq. (6-1)

fa

= 22.3 psi (see above)

13-74



93

m 0662949 0509239 427 m

1 Example 13.2-3 Cont'd.

and

Calculations

Code Reference

r

140r

Eq. (6-4)

h. Based on -. r

r = 0.287t r = 0.287(4.18 in.

+ 2.5in. + 2.5 in.)

= 2.63 in.

Use peak height as conservative slenderness: h - (3O fi)(12 in*/ft) 2.63 r in. "

usingf,

=

136.9 > 99

= 1,500 psi

Thus,

Fa = 1 ( 1,5OO p i ) (

6.3.1

30 ft( 12 in./ft) 1.33(98 psi) = 130 psi 4

Fa =

5.3.2 Eq. (6-5)

(with the 1.33 factor from Code 5.3.2 since in-plane bending is due to seismic load) Fb = 1.33( $)(1,500 psi) Fb

=

665 psi

13-75


A C 1 T I T L E S H D G 9 3 D Obb2949 0509220 149

m


Code Reference

Unity Eq. :

Eq. (6-1)

thus, as stated previously, this check was not expected to be a problem, but is included for illustrative purposes. Since the shear wall is in Seismic Zone 2 both vertical and horizontal steel must be provided. Provide vertical reinforcement of 0.2 in.2 (#4 reinforcing bar) at the two wall ends. Provide horizontal reinforcement

of 0.2 in.2 (#4 reinforcing bar)attop

and bottom of wall and

intermediate locations with maximum vertical spacing of 10 ft. Place both vertical and horizontal reinforcement in grouted collar joint.

13-76


k3.8

A C 1 TITLE+flDG 93 W Obb2949 0509223 085

Example 13.2-4

m

-

RCJ Hotel Design of Unreinforced Masonry Shear Wall for In-Plane

Lateral Loads Design the shear wall on Grid Line C between Grid Lines 1 and 2 using Wall Construction Option A (Unreinforced)and

Use hollow concrete

Building Construction Option II.

masonry units. Seismic Zone 2.


Reference

Code

Consider 8 in. wall and design 1” floor wall section for different load combinations. For unreinforced wall two critical parameters need to be checked:

1)

No tension is developed under minimum dead load and maximum lateral load, i.e., load case of 0.9 D

2)

Compression stress attheotherend

+ E.

of the wall complies with theCode

allowables using the unity equation, i.e., load case D

+ L + ( E + W>.

The above two cases are considered below. Other cases may be checked.

Loads Lateral loads are calculated using hand 33 kip-

calculations in MDG Example 9.2-3. Gravity loads are given in Maximum shear and moment

MDG 9.1.3.

42 kip-

develop at

54 kips __t

the base section since wall acts as a freestanding cantilever.

13-77


5.3.1

93 W 0662949 0509222 TIIL W



and

Calculations

Code Reference

For the lSt floor wall section - assume ungrouted D = 14,781 plf x 29.34 ft

+ 46.5 psf(8.83

ft x 29.34 ft) = 442,000 lb

L = 2,239 plf x 29.34 ft = 65,700 lb Y = 146 kips M = 17.2 kips (34.8 ft) + 33 kips (26.2 ft) + 42 kips (17.5 ft) + 54 kips (8.83 ft) M = 599 ft-kips + 863.6 ft-kips + 735 ft-kips + 477 ft-kips = 2,680 ft-kips Section Properties - Consider face shell bedding, face shell thickness = 1.25 in.

Net area, A,, = (1.25 in. x 2) (29.34 ft x 12 in./ft) = 880 in.2

Section modulus, S = (1.25 in. x 2) (29.34 ft x 12 in./ft)2/6 = 52,000 in.3 Normal Stresses

+E

Load Case 0.9 D

P = 0.9 x 442,000 lb M = 2,675 ft-kips f = - -

P

= -

f

=

S

398,000 lb 880

f

M

f -

An f = -

= 398,000 lb

f

2,680 ft-kip

12,000 h-lb/ft-kip

52,000

ia2

452 psi

X

ia3

618 psi

- 1,070 psi and

+

166 psi

Tension develops- thus this 8 in. hollow concrete block wall mustbe reinforcedor modified. 13-78


A C 1 TITLE*PlDG

93

m

0662949 0509223 9 5 8

m


Calculations Assume that 2.67 ft (2 block units) on each end of wall will be grouted. 2.67’

2,67’

I -

l

I ,= 29.34’ = 352”

Area

=

7.63 in.(32 in.)(2 in.) + 24 ft(l2 inJfQ(1.25 in. x 2)

1 (7.63 in.)(352

I

12

I = 17,500,000

S =

=

f =

1 -(5.125 in.)(24 ft

X

in4

398,000 lb 1310 in?

f

2,680 ft-kips

- 329.6 psi * 322 psi - 652 psi and - 8 psi

Load Case D

12,000 in.-lb/ft-kiP 99,500 in.3 X

.: No Tension

+L +E

D = 442,000 lb

+ 65,700 lb = 508,000 lb

M = 2,680 ft-kips 13-79


1,210 in?

12 h ~ / f t ) ~

1C = 99,500 in?

f = -

f

in.)3

=

A C 1 T I T L E I M D G 93

m ObbZ949 0509224

894

m


and

Code Reference

Calculations

Use unity equation to determine required compressive strength of masonry Eq. (6-1) 5.3.2

[

Fa = 0.25 f i m 1

-(&r]

Eq. (6-3)

For axial capacity buckling inthe out-of-plane direction controls - Ignore grouted ends for

F*

-F

Consider a 12 in. strip

Ix = 2[ l2 I* A,

.

= =

309

12

29.34‘

I

h)’ + 12 h(1.25 in.)(3.19 h)2

ia4

2(1.25 in.)(12 h) = 30 h’

E -4

=

3-21

=

30 in?

[-

(8.83 140ftx x3.21 12 in.

F,, = 0.25 f i , 1

in,/ftr]

=

0.236 fi,,,

13-80


I

I

I

I

_____c(

A C 1 TITLExMDG 93

m

0bb2949 0509225 720


Calculations

M - 2,675ft-kips(12 in&) = 322 psi 99,500 in.3

fb'y-

Fb

=

:.

f',

6.3.1

0.33 f',

=

2,080psi

required

Shear Stress

V = 146 kips fv =

VQ

fv =

146 kips[7.63 in. x 32 in. x 160 in. + 144 in. x 2 x 1.25 in. x 72 in.] 17,500,000 in?(2 x 1.25 in.)

fv =

146 kips(64,960 in.3) (17,5OO,OOO in?)(2.5 in.)

fv

21.7 psi

Fv

=

=

(1.33)lSK

=

(1.33) 1.54-

=

(1.33)68.3 psi = 90.8 psi

or

5.3.2

F, = (1.33)120 psi = 160 psi or

13-81


6.5.2


0662749 0509226 667

m


Calculations

F,,= 1.33(v

N

+ 0.45 -)

A

=

1.33[37 psi + 0.45 (339 psi)] = 1.33(185 psi) = 246 psi

Thus F, = 90.8psi > 21.7psi

:. OK

No shear reinforcement is required. Provide vertical reinforcement of 0.2 in.2 (#4 reinforcing bar) at the two ends of the wall. Provide horizontal reinforcement of 0.2 in.2 (#4 reinforcing bar) at top

andbottom ofwall.

Steel to beplacedin

concrete block wythe.

13-82


k3.8

93


Example 13.2-5

m

Obb2949 0 5 0 9 2 2 7 5 T 3

-

m

Shear Wall for In-Plane

I

1 and 2 for Seismic Zone 4 using

I

RCJ Hotel DesignofReinforcedMasonry Lateral Loads

Design the shear wall on GridLine C betweenGridLines

Wall Construction Option B and Building Construction Option I. Use Hollow Clay Masonry Units.


Reference

Code

Loads, from MDG Table 9.1.3 to 9.1.5 and MDG Example 9.2-3

Shear

LL

D.L Level Floor Height

Moment

(Table 2)

(Top of Wall)

(plf)

(Pif)

4

2,850

600

32

3

6,830

1,330

64

(kips)

(ft)

(Bottom of Wall) (ft-kips)

9.67

309 928 9.67

2

1,800 10,80083

9.67

1,731

1

2,240 14,800

10.83 108

2,900

Geometry: Assume 6 in. brick wall; Type S mortar, grouted solid 1, = 29 ft - 8 in.

d

b = 7.5 in.

r,,,= 2,500 psi

=

344 in. ( assumed)

A = 0.17

"

3

n = 15.3

13-83


A

= 0.483 in.


93

m

0662949 0509228 4 3 T

m

Example 13.2-5 Cont’d. ~~


Code Reference

MIPd

Level

*

(0.9 D.L.

MIPd

+ E)

(D

+ L + E)

4

O. 14

o. 10

3

O. 18

O. 13

2

0.21

O. 16

1

0.26

o.20

(Approximate, neglectsweight ofwall

forone level)

Check shear for wall thickness assumption.

M values. Assume j = 0.9, conservative for low Pd The seismic shear is multiplied by 1.5

A.4.9.1 Eq. (7-3)

V

fv

=

bjd

M in allowable equations, V is not multiplied by 1.5 for Vd

13-84


AC1

TITLE*NDG 9 3 m Obb27Y7 O509229 3 7 6 m


M -

Reference

Code

2,900 ft-kips X 1,200 h-lb/ft-kip = o.93 108,000 lb x 344 in.

”

Vd

F,,= 314

-

$1

E x 43

=

-4

1 [ 4 - (0.93)]

3

4 = [80 - 45 (0.93)] x 3

x =

-4 3

= 73.3 psi

Eq. (7-5)

50.7 psi < 95.0 psi

Shear steel required, assume 24 in. O.C.horizontally.

Fv = 1 [ 4 - 0.931-4

X :

2

Fv

S

[ 120

A,,

=

vs F#

- 45

X

0.931

3 X

4 -

3

=

Eq. (7-8)

110 psi

104 psi

.+. OK

Eq. (7-10)

1.5 x 108,000 lb x 24 in. Av = 24,000 psi x 1.33 x 344 in.

A,

=

0.35 h2

Use (2) #4 @ 24 in. O.C.

TheCoderequiresthatreinforcementin

the amountof

provided perpendicular to the shear reinforcement

13-85


A V be

3

7.5.3.2

A C 1 TITLErMDG 93

m

Obb29Ll9 0509230 098



53 = 0.117 i a 2 This will be satisfied by the flexural reinforcement requirements and

A45

the minimum steel requirements for Seismic Zone 4. Trim or tension reinforcement: First floor

M = 0.26 -

P

Pd

= 395 kips

Try a = 103 in.

Mp = P(

i-

a)

kips(

'Wp =

395

As =

M - Mp F,(d - a)

MDG Eq. 12.2-15

356 in. - 103 in. 112 2

MDG

in.m = 2,470 ft-kips

m.12.2-16

As = (2,900 fi-kips - 2,470 ft-kip~)(12 ialft) 24 ksi X 1.33 X (344h. - 103 h) A, = 0.67

in? MDG

m.12.2-17

13-86


M = 2,900 ft -kips

A C 1 TITLE*MDG

93

m

Obb2949 0509231 T 2 4

m


Calculations

MDG a = (d(36.2

+

in.)2

m.12.2-18

2 x36.2 in. x344 in. - 36.2 in.) = 41.9 in. 3

2nditeration

395 kips(

M =

\

-

356

in*

- 41.9 in.)

L

=

12 in./í%

4,480 fi-kips > M applied

System will not converge (too close to Region 2)

d

MDG

m.12.2-19

1 344 in.

a r

(3

x

[l

+

24 ksi/0.833 15.3

ksi]]

=

39.7 in.

MDG

a = 344

2

in. -

4[

Eq. 12.2-20

- 2(395 kips x0.483 x 344 in. +2,900 fi-kips x 12 in./ft) 3 x0.833 ksi x4/3 x5.5 in. W2in2)

a = 35.5 in.-kips

13-87



Obb2947 0509232 960


Calculations

A,

Code Reference

Discussion

MDG Q. 12.2-21

=

- l]

A,

13

=

X

1

0.833 ksi x 4/3L n x 35.5 i n. x 5.5 in. - 395

15.3 x 0.833 ksii x 4/;1x A,

-1.84

(3 344 iain.) x 35.5

kips] 1

-11

h 2

This is negative steel area. Thus use minimum steel.

Thus use (2) #5 at each wall end. Similar calculations for the remainder of the wall. Since the wallis

in Seismic Zone 4, provide total verticalplus

horizontalreinforcement

of at least 0.002 times the gross cross-

sectional area of wall. The minimum ineach direction shallnot be less than 0.0007 times the gross cross-sectional area. Vertical

+ horizontal steel = 0.002t(Zw)= 0.002(5.5 in.)(356 in.) = 3.92 in.2

Minimum Steel in either direction = 0.00O7t(Zw)= 1.37 in.2 Provide 5-#5 horizontally at 2 ft centers. Provide 8-#5 vertically at 4 ft centers.

13-88


A.4.5


Example 13.2-6

Obb2949 0509233 8 T 7

93

-

RCJ Hotel Reinforced Masonry ShearWallDesign

Design the shear wall on Grid Line 2 between Grid Lines E and F using reinforced hollow clay masonry (Wall Construction Option B) and Building Construction Option I. Seismic Zone 4. and

Calculations

Code Reference

Discussion

The critical lateral load case on this wall is seismic load.The wall geometry and loading are obtained from MDG Fig. 9.1-10 and Table 2, MDG Example 9.2-4.

P

22'"'

61 kip?

(Cumulative)

125 k i E (Cumulative)

136 k i E

(Cumulative)

Shear Wall

13-89


P

A C 1 TITLExMDG 93

m 0662749 0509234 733

M


Reference Calculations Code and Discussion

LOADS

Moment

Shear

h

61 kips

- 100 kips - 125 kips -

590 ft-kips

136 kips

I 4,240 ft-kips

Lateral: Axial: Loads for this wall were not previously calculated in MDG 9.1.3.1. Dead Load: Roof

95 psf x 4 ft =

380 plf

Corridor

110 psf x 4 ft

440 plf

Wall:

70 psf x 9.67 ft = 677 plf 70 psf x 10.8 ft = 758 plf

=

Length for Corridors and Roof = 22 ft Length for Wall

=

22 ft

Neglect Coupling Beams 13-90


+ 3 ft - 4 in. = 25.3 ft


93

m

Ob62949 0509235 b 7 T

m


and

Calculations

Live Load:

Roof

20 psf

x 4ft

80 plf

=

Floors 100 psf x 4 ft =

400plf

Roof

Live Load (kips) 2.o

Cumulative Live Load (kips) 2.0

4Ih

10.1

12.1

3'(1

10.1

2"d

10.1

Element

22.1 32.1

TRIM STEEL DESIGN .9DL

+ E controls by inspection

5.3.1

1" Floor .9DL = .9 x 104.3kips = 93.9 kips

M = 4,240 ft-kips 13-91


A C 1 T I T L E * H D G 73

m

Obb2749 0509236 506


Calculations

(See MDG 12.2.2.2)

FirstEstimate:

P Assume 6 Bars

Assume 8 in. Wall b = 7.5 in.

d = 264 in. - 18 in. = 246 in.

[('Th)

- 18 in.]

A =

246 in. M - 4,240 &-kips X 12 h/ft Pd 93.9 kips x 246 in.

"

=

=

2.2

0.463 2 3

--A

[

> 1"

:. Region 3 Applies

Step 1 Assume a = 20 in.

MF = P[+ -

.)

MDG

m.12.2-15

13-92


:-A)

m

7

93


m

0 b b 2 9 4 9 0509237 442

m


M

As = A,

=

Mp F,(d - a) -

MDG

MDG

(93,900lb

h2

m.12.2-17

5.58 in? x 24O , OO psi x 1.33) x 15.3 = 17.4 ia 24,000 psi x 1.33 x 7.5 in.

+

MDG

a = a = 25.6

Code

m.12.2-16

(4,240 ft-kip - 876 ft-kip)X 12 h/ft = 24 ksi X 1.33 (246 ia - 20 h.)

t.=

c=

Reference

R. 12.2-18

in.

PdIteration

M’

=

A, =

t.

=

U =

832 ft-kips 6.8 in? 17.8 in. 25.8 in.

:. OK

Check For Compression d

246 in.

=

MDG Eq. 12.2-19

15.3 1 28.4 in. > a .-.Tension Controls

13-93


93

m


0 6 6 2 9 4 9 0509239 215

m

3

~ ~ _ _ _

A C 1 T I T L E * U D G 93 W Obb29Y9 0509238 389

m


Calculations

Use (6) #9 Bars A, = 6.0 in.2 Alternative method is to use the conservative unity equation: 1

1

Ja Jb + -5

1.0

Fb

a'

Eq. (7-1)

Fa = 679 psi

[assume grouted solid]

= f -(1.33) ',

=

3

2,500

(1.33)

=

1,110 psi

7.3.1.2 5.3.2

-47*4 +679

''íi

1,110

=

0.89 e 1.0

:. OK

Check Tension Steel For Unity Equation Estimate (8)#9 13-94



m

Ob62949 0509239 215

m


Code Reference

Calculations

d=264-24=240in. = 8.0 h2

A,

np = 15.3 x

K

=

.O044 = .O68

/(.o68)’ + 2(.068)

- .O68

=

.306

j = .89 Mt = 8.0 x 24 x 4/3 x .89 x 240/12 =

4557 kip-ft

4240 kip-ft

2.

OK

Check Shear:

V = lSO(136 kips)

=

A.4.9.1

204 kips

V f,, = bjd

7.5.2.1

204,000 lb fv = (7.5 in.)(0.91)(246 in.) = 122 psi

M -

7.5.2.3

Vd

F,, =

E -4

=

=

F,, S 35 psi(1.33)

=

50 psi (1.33)

=

66.5 psi

Eq. (7-6)

46.6 psi

Shear reinforcement must be provided F,,

=

1SK

Max. F,,

=

x

1.33

=

100 psi

75 psi x 1.33

=

122 psi

100 psi

(Does not help to raise f’J 13-95


(.-.

N.G.)

Eq. (7-9) 5.3.2

A C 1 TITLE*NDG

93

m

0662749 0509240 T37

m


Calculations Use 10 in. brick wall first floor then 8 in.

f,=bjdv -

204,000 lb

=

96.3 psi

100 psi

:.

OK

(9.5 in.)(0.91)(246in.) A,

=

vs Fsd

7.5.3

Use (2) #5 @ 24 in. O.C. horizontal. Since wall is in Seismic Zone 4 provide total vertical plus horizontal

k4.5

reinforcement of at least 0.002 times the gross cross-sectional area of wall. The minimumin each direction shall not be less than 0.0007 times the gross cross-sectional area. Vertical

+ Horizontal Steel = O.O02t(Zw) = 0.002(7.5 in.)(22 ft x

12 in./ft) = 3.96 i n 2

Minimum steel in either direction = 0.0007~(2,,,)= 1.39 in.2 Provide 5-#5 horizontally at 24 in. O.C. Provide 8-#5 vertically at 36 in. O.C.

The Code requires reinforcement perpendicular to provided shear reinforcement in theamount

AV of -. 3

7.5.3.2 This will be satisfied by the

computed flexural steel requirements and by the minimumwall reinforcement required for Seismic Zone 4. 13-96


A C 1 TITLE*UDG 93

m

Obb2949 0509243 973

m

14 REINFORCEMENT AND CONNECTORS

14.1 GENERAL Code Chapter 8, titled "Details of Reinforcement," Code 5.14, titled "Anchor Bolts Solidly Grouted inMasonry,"

and Specifications Section 3, titled "Reinforcement and Metal

Accessories," address most of the issues in this chapter of the MDG.

14.1.1 Steel Reinforcement Steel reinforcement for masonry consists of deformed reinforcing bars, deformed wire, or welded wire fabric, completely embedded in the masonry.

Materialrequirementsfor

reinforcement are specified as detailed in Specs. 3.2.1.1. Those items are also discussed in MDG 3.5, and will be briefly mentioned here for completeness. Specs. 3.2 addresses material requirements for reinforcing bars, joint reinforcement, wire, and welded wire fabric. Deformed reinforcing bars must conform to ASTM A 615, A 616, A 617, or A 706 (billet, rail, axle, and low alloy respectively). Deformed wire must conform

to ASTM A 496, and deformed welded wire fabric must conform to ASTM A 497. Typical uses of each type of reinforcement are shown in Figs. 14.1-1 through 14.1-3. Fig.

14.1-1 shows deformed reinforcing bars in a grouted masonry wall. Fig. 14.1-2 shows joint 14-1



93

m

Ob629990509292

BOT 9

reinforcement in bed joints of a concrete masonry wall.Fig. 14.1-3 shows welded wire fabric

in the topping of a floor slab connected to a masonrywall.

:

I

:

I " " " " "

Fig. 14.1-1 Typical Application of Deformed Reinforcing Bars in Grouted Masonry Wall

" " " " "

" " " "

...................................

Fig. 14.1-2 Typical Application of Bed Joint Reinforcement

14-2


.


m

Ob62747 0509243 74b

m

Welded Wire Fabric

Fig. 14.1-3 Typical Use of Welded Wire Fabric

14.1.2 Connectors

In Code 2.2, a connectoris defined as "a mechanical device for securing two or more pieces, parts, or members together, including anchors, wall ties, and fasteners." Anchors connect wythes of masonry to intersecting masonry wythes or

other structural elements; wall ties

interconnect wythes of multiple-wythemasonry;and

fasteners connectnon-structural

elements to masonry. Material requirements for connectors are detailed in Specs. 3.2.1.2.

Those items are also

discussed in MDG 3.5, and will be briefly mentioned here for completeness. Connectors must conform to ASTM A 36 (plate, headed and bent bar anchors), ASTM A 325 (high-strength bolt anchors),ASTM A 366 (sheet steel anchors and ties), ASTM A 185 (weldedwirefabricties),ASTM

A 82 (plain wireties and anchors), or AST" A 167

(stainless steel sheet anchors and ties). Optional requirementsfor stainless steel connectors 14-3


A C 1 TITLE*MDG 93

m

0bb2949 05092LlLl 682 W

are given in ASTM A 167. Optional requirementsfor galvanized steel connectors are given in ASTM A 641, ASTM A 153, or ASTM A 525 as appropriate. Typical uses of each type of connector are shown in Figs. 14.1-4 through 14.1-6. Fig. 14.14(a) showshow dovetail anchors are used to connect masonry wall panels

to steel and

concrete columns; in Fig. 14.1-4(b) anchor bolts are used to connect structural elements of floors and roofsto masonry walls;and Fig. 14.1-4(c) indicates howstrap connectors are used to connect wythes of intersecting masonrywalls. Fig. 14.1-5 shows the use of adjustable ties and wire tiesto interconnect masonry wythes. Fig. 14.1-6 gives an example of fasteners used to connect non-structural elements to masonry.

.

.

(a) Dovetail Anchors

(b) Anchor Bolts

(c) Strap Connectors

Fig. 14.1-4 Typical Anchors

14-4


A C 1 TITLE*MDG 93

0 b b 2 9 4 9 0509245 519 W

(a) Adjustable Ties

(b) Wire Ties

Fig. 14.1-5 'Qpical Wall Ties

Power

-

Actuated Fasteners

Holding Furring Strips

Fig. 14.1-6 Typical Fasteners

14.13 ConnectionsBetweenIntersectingWalls

According to Code 5.13.4.2(e),masonrywalls support shallbeanchored

dependingupon one another for lateral

or bonded at locationswherethey 14-5


meet or intersect, using

A CT1I T L E * H D G

0662949 0509246 455

93

m

interlocking units, reinforcement,or connectors. One example of the use of connectors for this application is shown in Fig. 14.1-4(c). Additional illustrations of these techniques

are

given in Code C. Fig. 5.13-2, Code C. Fig. 5.13-3, and Code C. Fig. 5.13-4. 14.2 STEEL REINFORCEMENT 14.2.1Requirementsfor

Steel Reinforcement

In general, steel reinforcement must be Sufficiently strongto resist the forcesto which it will be subjected; it must be sufficiently corrosion resistant

to last as long as the structure in

which it is used; and it must have enough embedment

at each end to transmit the forces

acting on it. Inunreinforcedmasonrydesign(Code

Chapter 6), reinforcementisnotused

to resist

calculated tensile stresses from applied loads, but rather to control the effects of movements from temperature changes, shrinkage, and other effects. In reinforced masonry design (Code Chapter

7), reinforcement resists calculated tensile

stresses and sometimes calculated compressive stresses from applied loads. For example, flexural stresses in reinforcement caused by wind forces actingperpendicular to the masonry wall are limited to Code allowable values, the reinforcement area must not be significantly reduced by corrosion; and the reinforcernent must be firmly anchored,at each point, against the stresses developed at that point. 14.2.1.1StrengthRequirementsforReinforcement

- The designer supplies

a sufficiently

large reinforcement area so that thetensile or compressive stress in the reinforcement does not exceed the corresponding allowable values (Code 7.2.1). 14.2.1.2 Corrosion Resistance

and ProtectionRequirementsforReinforcement

- In the

context of the Code, the corrosion resistance and protection of reinforcing bars and welded

14-6



93 W Obb2949 O509247 391

m

wire fabric are provided for by ensuring that the reinforcement has sufficient cover

as

specified in Code 8.4. Joint reinforcement requires both corrosion resistant materials and sufficient cover. 14.2.13

EmbedmentRequirementsforReinforcement

anchorageprimarilyform

- Weldedwire

fabric derivesits

the grip of thecrosswires.Deformedreinforcingbars

anchoredprimarily by interlock at the deformationsthemselves.

are

If the cover and the

distance between reinforcing bars are large, failure occurs by yielding and fracture of the reinforcement itself, or by pullout of the reinforcing bar. If the cover to the reinforcing bar or the spacing between adjacent bars is relatively small, failure can occurby splitting from the reinforcement to the free surface, or by splitting from one bar to another. Code 8.2.1 prohibits the use of reinforcing bars larger than bar size #ll. Code 8.5 specifies that at least the required embedment length be provided on each side of each critical section. The required embedment length corresponds to a uniform allowable bond stress of 160 psi. By treating anchorage requirements for deformed reinforcing bars in terms of a required development length, additional checks are required for some specific situations. In addition todesignagainst

bar-to-bar and bar-to-surfacesplitting,discussedpreviously,

another

significant check concernsthe required development lengthfor reinforcing bars terminating at a simple support or point of inflection. At first glance, since the reinforcing bars stress at the simple support is zero, no development length would apparentlybe required beyond the simple support, regardless of the bar diameter. However, if the reinforcing bar stress (and consequent reinforcing bar force) increases away from the simple support, at a rate faster than that force can be resisted

by bond, the reinforcing bar will be insufficiently

anchored at the simple support, and additional embedment must be provided, either by a standard hook or by extending the reinforcing bar past the simple support. A code for a different material (14.2.1), which also uses the development length approach, 14-7


A C 1 T I T L E * M D G 93

m

Obb2949 0509248 228

m

addresses this issue. The Code currently contains no provisions or recommendations. The author suggests a check for this situation in MDG Example 14.2-3. Code 8.5.3.2 requires that for flexural members that are part of a primary lateral load resisting system, no less than 25% of the positive moment reinforcement is extend into the support and anchored to develop the allowabletensilestress,

required to F,. This

anchorage attempts to provide a ductile response in the event of overstress. Code 8.5.3.3(b) requires that at least one third of the total reinforcement provided for moment at a support be extended beyond the point of inflection a certain limiting distance to provide for possible shifting of the moment diagram. 14.2.2 Design ofSteelReinforcement

Examples of design of reinforcement for strength and corrosion resistanceare given in the

MDG. Code 8.2 addresses limitations on sizesof reinforcing bars and joint reinforcement. Code 8.3 covers placement limits for reinforcement. 143 CONNECTORS 143.1 Requirements for Connectors

In general, connectors must be sufficiently strong to safely resistthe forces to which they will be subjected; they must be sufficiently corrosion resistantto last as long as the structure in

which they are used; they must have sufficient anchorage at each end to transmit the forces acting on them; and they must be sufficiently stiff so that the masonry they connect actually behaves as assumed in design.

two wythes of a masonry cavity wall must have an adequate For example, ties connecting the factor of safety against failure underthe loads imposedby wind or earthquake forces acting 14-8



93

m

0662949 0509249 L b 4 W

on the outer wythe; their area must not be significantly reduced by corrosion; they must remain firmly attached to both wythes; and their stiffness must be consistent with that assumed in computing the distribution of load between the wythes.Usually,

this last

requirement means that the ties, when loaded axially, must be rigid compared to the walls when loaded out of plane. The Code requires that wythes of multiwythe wallsbe connected using wire ties or cross wires of joint reinforcement. Code 5.8.1.5,5.8.2.2 specifies the number and size of the ties.

143.1.1 Strength Requirements for Connectors - The Code assumes that connectors with negligible shear stiffness (for example, dovetail anchors andwire ties) act primarily in tension and compression rather thanshear. Connectors with significant shear stiffness (for example, bolts) can act in shear aswell as tension or compression. The strength of a connector itself is provided for by ensuring that theconnector hassufficient cross-sectional area toresist the forces acting on it. This is accomplished either by specifymg maximum allowable loads on connectors (Code 5.14), or by specifylng maximumspacing requirements (Specs. 3.3.3.5(b)). Connectors acting in compression must have sufficient elastic buckling resistance. In the case of wire ties, this is ensured by limits on maximum cavity width and spacing (Code 5.8.2.1(f)).

143.1.2

Corrosion Resistance and Protection Requirements for Connectors - The Code

requires that a connector either be of stainless steel (Specs. 3.2.1.3), or have a sufficient thickness of corrosion-resistant galvanizing (Specs. 3.2.1.4). Embedded connectors must be I

protected by a sufficient thickness of alkaline cementitious material (Specs. 3.3.3.5 (a)).

143.13 EmbedmentRequirements for Connectors - Headed connectors(for example, bolts) transfer tensile forces by direct bearing against the surrounding mortar or grout. Typical bolt heads have sufficient area to preclude local bearing failure of the surrounding cementitious material. Headed connectorslocated far from a free edge and

loaded in

tension or shear eitherfail in the anchor itself, or by the breaking out of a roughly conical

14-9



m

m

Obb2949 0509250 9 8 b

volume of grouted masonry surroundingthe anchor (Fig. 14.3-1). A headed connector’s tensile or shear capacity as governed by steel failure depends on the connector steel embedment depth anditscross-sectional area; a headed connector’s tensile or shear capacity as governed by grout, mortar, or unitfailure

depends on the embedment

depth of the connector, the tensile strength material, and of the surrounding proximity of

the

the connector to other

connectors or to free surfaces.

Fig. 143-1

Connectors without heads (for example, flat anchors, bolts, forces by frictional resistance between

Typical Conical Failure Headed Anchor

of

or wall ties) transfer tensile

the connector and the surrounding cementitious

material,bond, and bearing. If the cover to the connector issmall and the connector transfers sufficient force to the surrounding material, failure can also occur by transverse splitting betweenthe connector andthe free surface. This type of failure is discussedfurther in MDG 14.2.1.3 dealing with reinforcing bars. The Code requires that embedment requirementsfor connectors be satisfied inthe following ways: headed anchors, as shown in Code C. Fig. 5.14-1, must be embedded as specified in Code 5.14; and walltiesmust

be embedded as specifiedinSpecs.

3.3.3.5. No specific

requirements apply to unbent anchors without heads. 143.1.4

StiffnessRequirements

for Connectors

- Althoughdesignistypicallymore

be sufficiently stiff concerned with strength rather than stiffness, in general, connectors must

so that the masonry they connect actually behaves as assumed in design. As noted above,

connectors(with

the exception of bolts) are usuallyassumed

to transfer tensile and

compressiveforces only. The axialstiffness of mostconnectors is simplygivenby 14-10


the

~~

~~

A C 1 T I T L E t M D G 93 D 0 6 6 2 9 4 9 0 5 0 9 2 5 1 B12

m

product of their cross-sectionalarea and modulusof elasticity, divided by their unsupported length. Even ties with small cross-sectional areas have a higher axial stiffness than the outof-plane stiffness of the masonry wythes they connect,and are therefore sufficiently stiff to satisfy the general requirement given above. However, adjustable ties, described in Code 5.8.22, are usually much more flexible than straight ties. Their flemiility is due to the fact that they deform in flexureand shear, as well as axially; also, adjustable ties have a gap between the two parts comprising them.

The

flexibility of adjustable ties is intended to be controlled by the maximum limitation (Code 5.8.2.2(c)) of 1-1/4 in. on misalignment of bed joints between wythes.

143.2 Design of' Connectors Design procedures for connectors are illustrated in MDG examples 14.3-1 through 14.3-16.

REFERENCES 14.2.1

Building Code Requirements for Reinforced

Concrete AC1 318-89 and

Commentary AC1 318R-89 (Revised 1992), American Concrete Institute, Detroit, 1992. 14.2.2

"BondandAnchorage

to Concrete," AC1 Committee 408, AmericanConcrete

Institute, 1988. 14.2.3

Cheema, T. S. and Klingner,R. E., 'Tensile AnchorageBehavior of Deformed Reinforcement in Grouted Concrete Masonry," AC1 Journal, V. 82, No. 3, MayJune 1985, pp. 372-380.

14.2.4

Cheema, T. S. and Klingner, R. E., "Failure Criteria for Deformed Reinforcement Anchored in Grouted Concrete Masonry," AC1 Journal, V. 82, No. 4, July-August 1985, PP. 434-442.

14.2.5

Cheema, T. S. andKlingner,

R. E., "Design Recommendations for Tensile

Anchorages of Deformed Reinforcement in Grouted Concrete Masonry," 14-11


m

AC1

TITLE*NDG 93 m 0662949 0509252 759 M

Journal, V. 82, No. 5, September-October 1985, pp. 616-621.

14.2.6

Soric, Zorislav andTulin,Leonard

G., "BondStress/DeformationinPull-Out

Masonry Specimens, Journal of Structural Engineering, ASCE, V. 115, No. 10, October 1989, pp. 2588-2602.

14-12


A C 1T I T L E g M D G

m

93

0 b b 2 9 4 9 0509253 695

m

-

Example 14.2-1 TMS Shopping Center Design of a Straight Bar Anchorage

A straight bar anchorage is to be designed. To make this example relevant to the overall building designs which are part of this MDG, the design will involve the anchorage of a #6 foundation dowel,embedded vertically intothe foundation of the TMS Shopping Center and one of its walls. Use of this example does not implythat such a foundation-wall connection wouldalwayshave

to bereinforced.MDGExamples14.3-3and14.3-4

deal with an

unreinforced connection.

and

Calculations

Use #6 dowel, Grade 60. The requireddevelopmentlength

is

given by Eq. (8-1)

ld

= (0.0015)(0.75 in.)(24,000 psi)

rd

= 27 in.

A s o check l d not less than 12 in. for bars. .: OK

14-13


8.5.2

A CT1I T L E t M D G

0 6 6 2 9 4 9 0509254 5 2 1

93

m

Example 14.2-2 TMS Shopping Center - Design of a Hooked Bar Anchorage

A hooked bar anchorage is to be designed. To make this example relevant to the overall building designs which are part of this MDG, the design will involve the anchorage of a #6 foundation dowel, embedded vertically into the foundation of the T M S Shopping Center and one of its walls. Use of this example does not implythat such a foundation-wall connection wouldalwayshave

to bereinforced.

MDG Examples 14.3-3 and 14.3-4 deal with an

unreinforced connection. Calculations and Discussion

Code Reference

Use #6 dowel, Grade 60.

As previouslycomputed in MDG Example 14.1-1, the total required embedmentlength is 27 in.

equivalent The

e m b e d m el ennt g t h provided by a standard hook is equal to

8.5.5.2

B""*d

1, = 11.25db or 1, = 11.25(0.75 in.)

1, = 8.44 in.

The remaining required embedment length is therefore

27 in. - 8.44 in. = 18.6 in. Use 19 in. additionalstraightembedmentmeasuredfrom standard 90" hook.

14-14


start of

A C 1 TITLE*NDG 93

Example 14.2-3

m

0662949 0509255 4bA

m

-

T M S ShoppingCenter Design of Anchorage at aSimpleSupported Lintel

An anchorage at a simple support is to be designed. The design will involve the anchorage

of the main reinforcement of the lintel shown below. The lintel, designed in MDG Example

11.3-2, is above the 10 ft x 10 ft opening in the wall on Grid Line 3 of the TMS Shopping Center. The lintel has bottom reinforcement consisting of 3-#6 bars. It has an allowable flexural capacity of 46.4 ft-kips and a design shear of 11,976 lb.

As pointed out in the text of this chapter, the current Code does not require a check for anchorage at a simple support. An approach to this problem, consistent with other codes which use a development length approach, would be to calculate the ratio of (allowable flexural capacity/ service-level shear), and compare itwith the required development length. That procedure will be followed in this example. This check is not a Code.However,it

is consistentwiththeCode's

part of the current

intent, andsimilarprovisionsmaybe

included in future editions of the Code.

and

Calculations See Above

I

3

i# 6

bars

I

Shear

Allowable Flexural Capacity at Simple Support = 46.4 f t - kips

14-15


A C 1T I T L E S H D G

93 W 0662949 0509256 3 T 4 W


Calculations

For a #6 bar,

2, = 0.0O15dfls

Eq. (8-1)

2, = 0.0015(0.75 in.)(24,000 psi) 2, = 27 in. The ratio of allowable flexural capacityto service level shear gives the available development length: (46.4 ft-kips x 12 in./ft / 11.98kips)

=

46.5 in.

This exceeds the required development length of 27 in. and therefore no bar extension or hook is needed at the simple support for this situation. However for flexural elements that are part of primary a

lateral load

resisting system, at least 25% of the positive moment reinforcement mustextendinto

the support and be properlyanchored.

For this

example to satisfy this Code provision 1-#6 would extend beyond the support adistance

of27

in. A shorter extensionispossible

standard hook is used.

14-16


if a

8.5.3.2

AC1

Example 143-1

1

TITLExMDG 73 m Obb2949 0507257 230

Design of AnchorBolts of a ShearWall to Roof Diaphragm

~

Bolts will be designed for the connection of a CMU nonloadbearing shearwall to metal roof deck. The bolts transfer the wind-induced shears from the metal deck roof diaphragm to the wall. Bolt heads will be anchored in a bond beam formed by grouting the top course of the wall. The free endsof the bolts will project through a steel plate. Angles at theedge

of the roof deck will be welded to the plate. Assume wall length is 81.5 ft. and

Calculations

Angles Welded To Plate Metal

,

Assume total shear fromthe metal deck

Deck

is 26,600 lb.

Welded to Angle

Half of this shear is

transferred in from each side. Try 1/2in. diameter A 307 bolts. Assume bolt threads will be outside shear plane, so effective tensile stress area equals gross area. Use the

B,

=

350

lesser of:

7 i-

Eq. (5-5)

Eq. (5-6) Assume f', = 2,000 psi and Ab = 0.20 in2

B,,

4

=

350 J2,OOOpsix 0.20in2

B, = 1,565 lb Assume bolt is in the center of the wall.

14-17


Example 143-1 Coned. ~

~~~

~~

~

Code Reference


However, the distance measured from the anchor bolt to the nearest free surface (Ih) is given by:

7.63 - db 2

Zbe -

7.63 h - 0.5 in.

2

lk = 3.56 in.

Because 1, is less than 12 db, the value of B, in Code Eq. (5-5) must be

reduced by linear interpolation to zero at an B,, (reduced) = B,, B,, (reduced)

=

x

B,, x

(Zk -

lacdistance of 1 in.:

1 in.)

12 db

(3.56 h - 1 in.) 12 x 0.5 in.

B,, (reduced) = 0.43 x B,, B,, (reduced) = 0.43 x 1,565 lb = 668 lb (governs)

B,

=

0.12 x 0.20 in2 x 36,000 psi

B"

=

864 lb (does not govern) 14-18


5.14.2.2


73 M 0 b b 2 7 4 9 0509259 O03 M

Example 143-1 Cont’d. ~~~~


Code Reference

The governing loadper bolt canbe increased by one-third, because the loading combination involves wind. The allowable load per bolt is therefore 668 lb (1.33) = 891 lb/bolt Number of bolts = (26,600 lb)/(891 Ib/bolt) = 30 bolts Wall length is 81.5 ft; mace bolts at 32 in. alone toD of wall

14-19


5.3.2


Example 143-2

73 W 0662747 0507260 825 W

DPCGymnasium

- Design of Wall Ties in an UnreinforcedMulti-

wythe Noncomposite (Cavity) Masonry Wall two wythes of the DPC Gymnasium for Wall

Wall ties are to be provided between the

Construction Option A. The ties will transfer out-of-plane loads between the two wythes.


Reference

multi-wythe of Wythes

5.8.2.2

wallsshallbe by wall

connected

ties consisting

either of #9 gage or 24 "

Code

3/165.8.1.5

in. diameter wire.Cross

w i r e osjfo i n t reinforcement may be used for this purpose.

1'24"1

Sheet metalties are not permitted.

Use 3/16-in. wall ties meetingASTh4 A 82, with hot-dippedgalvanizingSpecs.

3.3.3.5

meeting ASTM A 153 Class B2 (1.50 oz/ft2).

3.2.1.4

Embedties at leastSpecs.

0.5 in. in mortar beds of each wythe.

Spacingrequirements:

One wall tie p e r 4.5 ft2 ofwall;maximum spacing 36 in.horizontallyand vertically.

5.8.2.2 5.8.1.5

Space ties at 24 in. each direction.

14-20


24 in.

A C 1 TITLExMDG 93

Example 14.33

DPC Gymnasium

m

-

0 b 0b 52709b94l29 6 3

I

m

Design of Shear Wall-Floor Connection for

Composite Nonloadbearing Wall The connection to be designed for the walls on Grids 1 or 2 and the foundation or floor slab. The Gymnasium is located in Seismic Zone 2. Assume Wall Construction Option C composite wall with filled collar joint.

The connection will be designed for the in-plane

shear carried by the wall as a vertical diaphragm.

and

Calculations

The seismic shear carried by the wall on Grid Lines 1 or 2 is 22.6 kips from MDG Example 9.2-2. That shear will be carried by the interface between the wall and the foundation.

The allowable stress is the least of: 1-5

{Pnt

or

120 psi or v + 0.45

[2]

14-21


6.5.2

A C 1 TITLElrMDG 9 3

Obb2949 05092b2

bTB


Calculations

The allowablestresscan

be increased by 1/3forloadcombinationsinvolving5.3.2

earthquake. In evaluating the last of these, conservatively assume N, = O (neglect compressive axial stresses from self weight). Assume solidgrouted,and exceeding608psi,this

v therefore equals 37psi. 37 psivaluewill

the wall is not

For allvalues of

belowerthan

that givenby the

6.5.2(c)

other two equations, and will therefore govern. The Coderequiresfullmortaring

of foundationcourses.BeginningSpecs.

2.3.3.3(d)2

with the first bedjoint above the foundation assume face shell bedding only. Using an 8 in. wall with face-shell bedding, plus the filled collar joint (1 in.), plus the outer wythe (3.63 in.), the available area is: (81.5 ft)(12 in./ft)(1.25 in.

+ 1.25 in. + 1 in. + 3.63in.)

= 6,973 in2

The maximum shear stress for a rectangular cross sectionis fv

=

fv

=

V0

Eq. (6-7)

V

1.5 -

A

Therefore f v = 1.5 x (22,600 lb / 6,973 in.2) = 4.86 psi.

14-22


A C 1 T I T L E + M D G 73

m 0662749 0507263 534 m

Example 14.33 Cont’d Code Reference


This is much less than 37 psi increased by 1/3, and the design is acceptable withoutdowels. An analogous calculation would also have to be carried out forthe out-of-plane loads onthe

wall due to wind in the east-west direction. Although no allowableshear values for this case are specifically provided in the Code, the Code C recommends the shear allowable in Code 6.5.2 be used for limiting out-of-plane shear stress.

14-23


A C 1 T I T L E * N D G 93

Example 143-4

DPC Gymnasium

O662949 0509264 470

-

Design of Shear Wall-Floor Connection

for

Unreinforced Multiwythe Noncomposite (Cavity) Wall The connection to be designed is between the walls on Grids 1 or 2 and the foundation or floor slab.

It is assumed that because the gymnasium is located within Seismic

Zone 2,

minimum reinforcement is required in the wall. However, for purposes of this calculation, no dowels will be assumed between the floor andthe wall. The connection will be designed for the in-plane seismicshear carried by the wall as a vertical diaphragm, andfor the out-ofplane shear due to wind loads. Assume Wall Construction

and

Option A.

Calculations

.

Design of this connection for in-plane shear is very similar to that of MDG Example 14.3-3, with one exception: in MDG Example 14.3-4,

Shear Is Transferred On Inside Face Shell Only

Flashin

shear is transferred only on the inner face shell of the inner wythe. The reason for this is that

. &.

the flashing actsas a bond breaker through the

.*

entire bottom bedjoint of the outer wythe, and

*.

.

9 Q

.

..

.C

U

"

.

.. . .P

. -.

..

.. .

through the outer faceshell of the firstbed joint up from the bottom in the inner wythe.

Desim for In-Plane Shear: The governing shear, due to wind, is 22.6 kips per MDG Example 9.2-2.

14-24


Example 1 4 3 4 Cont’d. I

Code Reference


That shear will be carried by the interface between the wall and the foundation. Allowable stress, F,, is the least of:

6.5.2 or

120 psi or Y + 0.45

(21

The allowable stress can be increased by 1/3 for load combinations

5.3.2

involving seismic stress.

In evaluating the last of these, conservatively assume N,,= O (neglect As in MDG Example

compressiveaxialstressesfromselfweight).

14.3-4, the governing (lowest) value of v will be 37 psi. Using an 8 in. wall with face-shell bedding, the availablearea on the inside face shell only is:

(64ft) x (12 in./ft) x (1 face) x (1.25 in.)

= 960 in.2

The maximum shear stress for a rectangular cross section is

&=x

- 1.5

V -

Eq. (6-7)

A

14-25


AC1

TITLE*MDG 93 m Obb2949 0 5 0 9 2 b b 2 4 3 m


Calculations

Therefore fv = 1.5 x (22,600 lb / 960 in.2) = 35.3 psi This is less than 37 psi increased by 1/3, and the design is acceptable without dowels. Desiyn for Out-of-Plane Shear: The critical point on these walls for out-of-plane wind is at the peak of the roof, where the walls are 30 ft high.Windloadis

20 psf.Assuming

the wall to be simply supported

between the foundation and the roof diaphragm, the maximum base shear per foot of wall length is therefore

Again assuming that only the inner face shell of the inner wythe is effective in resisting shear, the available cross-sectional area per foot of wall length is therefore (12 in.)

X

(1.25 in.)

=

15.0

in?

The corresponding shear stress is 1.5 x 300 lb 15 i a 2

TheCode does notspecificallyprovideallowable

=

20 psi

shear stress for out-of-planeforces.

However Code C recommends using the Code 6.5.2 values for both in-plane and out-ofplane shear. Thus the actual shear stress is again less than 37 psi increased by 1/3, and the design is acceptable without dowels.

14-26


A C 1 T I T L E x M D G 9 3 9 O662949 0509267 1 8 T W

Example 143-5

TMS ShoppingCenter

- Joist ConnectiontoLoadbearingWall

A joist connection detail is to be designed for the wall on Grid Line A of the TMS Shopping

Center. The detail must transfer in-plane gravity loads, in-plane shear loads, and out-ofplane wind loads. ~~~

Code Reference

Calculations and Discussion Two details are shown below:

T"i-

4.75"

.Steel Joist

'

H

"-

Alternate Detail

Preferred Detail

W recommenned unless bond beam mechanically attached to wall by grout andlor reinforcement

a) The b)

gravity-load joist reaction is 4.54 kips per joist from MDG Example 9.2-1 The in-plane seismic shear on the walls on Grid Lines A and C is the same as on Grid Lines 1 and 2 11.5 kips from MDG Example 9.2-2. From MDG Example 9.1.12 the wall is 204.7 ft long, and the joist spacing is 5 ft, the in-plane shear per joist is 11.5 kips x (5 ft / 204.7 ft) = 281 lb /joist

14-27


A C 1 T I T L E * M D G 93 D 0662949 0509268 016

m


c)

Code Reference

The out-of-plane shear fromwindloadsisbasedonauniformpressureof20psf: 203 lb / joist x 5 ft spacing = 1,015 lb /joist

The connections between the joist and the wall must be designed for these loads, using the same procedures demonstrated in MDG Example 14.3-1. In-Plane Shear: Shear per joist (each end) is 281 lb. Try 1/2-in. diameter A 307 bolts. Assume bolt threads will be outside shear plane, so effective tensile stress area equals gross area. Use the least Of:

350

7-

(Shear inMasonry)

Eq. (5-5)

or B, = 0.12Abfy (Shear

in Bolt)

Eq. (5-6)

Assume f L = 2,000 psi and Ab = 0.20 in.2 For Shear in Masonry:

Bv = 350 V2,OOO psi x 0.20 B,, = 1,565 lb/bolt

i n . '

Checkallowableboltloadconsideringedgedistance.Thedistancemeasuredfrom anchor bolt to the nearest free surface (Ih) is given by:

14-28


the


73

m

0 b b 2 9 4 9 0509267 T52

m


Calculations

- 7.63 in. - db la 2 - 7.63 h - 0.5 h.

la -

lk

Because lb, isless

=

2

3.56 in.

than 12 db(6 in.), the value of B, in Code Eq. (5-5)

5.14.2.2

must be reduced by linear interpolation to zero at an Zbc distance of 1

in.: B, (reduced) = B,,

x

(Za -

1 in.)

12 db (3.56 in. - 1 in.) 12 x 0.5 in.

B, (reduced)

=

B,, x

B,, (reduced)

=

0.43 x By

B, (reduced)

=

0.43 x 1,565 lb = 668 lb

For shear in bolts : B, = 0.12 x 0.20 i n 2 x 36,000 psi

B, = 864 lb (does not govern) .m.

B, (reduced) = 668 lb governs

The governing allowable load per bolt (668 lb) can be increased by one-third because the loading condition involves earthquake.

14-29


5.3.2

m


0662949 0509270 774

m


Calculations

The allowable load per bolt is therefore

668 lb x (1.33) = 891 lb / bolt The applied loadof 281 lb per bolt is less than this value, thus the design using one bolt per joist is acceptable. Out-of-Plane Shear: The shear load on the end of the joist is 1,013 lb. This exceeds the allowable load per bolt (computed above) of 891 lb. Try 3/4-in. diameter A 307 bolts. Assume bolt threads will be outside shear plane, so effective tensile stress area equals gross area. Use the lesser of:

B,,

Eq. (5-5)

350

=

Assume f’, = 2,000 psi and A b = 0.44 in.2

B,,

4

=

350 J2,OOO psi x 0.44

in.’

B,, = 1,910 lb B,, = 0.12 x 0.44 B,,

=

in? x

1,900 lb

14-30


36,000 psi (go==)

I

1

T I TAL C E 1w M D G

93

m

Ob62949 0509273 6 0 0

m


Calculations

Code Reference

The distance measured from the anchor bolt to the nearest free surface (Ik) is given by:

- 7.63 in - db

-

la

2

- 7.63 in. - 0.5 in. lb# 2 lh

=

3.56 in.

Because lb, is less than 12 db, the value of B, in Code Eq. (5-5) must be reduced by linear interpolation to zero at an lb, distance of 1 in.:

By(reduced)

=

B,, x

By(reduced)

=

B,, x

(lk -

1)

12 db

(3.56 in. - 1 in.) 12 x 0.5 in

B,, (reduced) = 0.43 x Bv B, (reduced)

=

0.43 x 1,910lb

=

820 lb (governs)

The governing allowable load per bolt (820 lb) can be increased by

5.3.2

one-third because the loading condition involves wind. The allowable load per bolt is therefore 820 lb x (4/3) = 1,090 lb / bolt The applied load of 1,015 lb per joist is less than this value, and the connection design is acceptable.

14-31


AC1

Example 143-6

TITLEgMDG 73 m 0662749 0507272 547

RCJHotel

- ConnectionBetweenCanopyBeam

and Column

The connection between the canopy beam and columnof the RCJ Hotel is to be designed.

The canopy beams are 11.5 in. wide by 23.5 in. deep masonry beams; the column is 11.5 in. square, and is constructed of 6 in. hollow clay units.

~

and

~~~~~~~

Calculations

A connection detail is shown below. The beam must be designed as described in thisMDG Example 11.3-1. Column Reinforcement = 4 #7 With #2 Ties 12" O.C. Hook Bars A t Top Into Beams ---_ "

--

A Reinforcement Beam

#7 Bars With 12" Hooks TyDical (2 Bars ?Ôp. . 4Bottom)

-"

11.5" Actual 12" Nominal

-

Plan View Canopy Beams

14-32


TITLE*RDG 93 m Ob62949 0509273 4 8 3 m

AC1


and

Cade Reference

Calculations

Beam Reinforcement Hook At End With Perpendicular Beam

-

Column Ties

12“

1Column

Vertical Reinforcement

12” l

i

lf-

.-.

m 12” Nominal

-

Section View Beam/Cohunn Connection

14-33


A C 1 TITLE*NDG 93

Example 143-7

m 0662949 0509274

3LT

m

Typical Reinforcing Details

Typical reinforcing details are to be provided for the following situations: a)

wall corners

b)

between wythes - walls

pilasters c)

/ columns


Code Reference

Wall Corners: Unless wall intersections are designed to prevent shear transfer between the intersecting walls (isolate the walls from each other), they must be able to transfer shear, using one of the following means: e

50% of the units at the interfaceshallinterlock

e

Wallsshallbe

e

Intersecting,reinforcedbondbeamsshall

toothed andjoined by steel connectors

The approach of Code 5.13.4.2(e)2is

be provided

illustrated inFig.14.1-4(c).

Note that the intersection mustbe laid in running bondso that at least some of the units overlap. The steel connectors must be at least 1/4 in. thick by 1-1/2 in. wide by 28in. long, including 2 in. long,90" bends at each end. The connectors must be spaced no more than 4 ft apart.

14-34


5.13.4.1 (b)

A CT1I T L E S f l D G

0 b b 2 9 4 9 0 5 0 9 2 7 5 256 D

93


Calculations

Connections between Wythes: Wythes of multi-wythe walls shall be connected by wall ties consisting

5.8.1.5

either of #9 gage or 3/16-in. diameter wire. Sheet metal ties are not

5.8.2.2

permitted. Use #9 gagewirewalltiesmeeting

ASTM A 82,with hot-dipped

Specs. 3.3.3.5

galvanizing meeting ASTM A 153 Class B2 (1.50 oz/ft2). Embed ties at least 0.5in.

in mortar beds of the outer face shell of the outer

wythe.

Pilasters / Columns:

As shown in sectionat left, ties are used in pilasters and columns to resist shear and to resist bucklingof the longitudinal reinforcing bars. Ties

are more

effective if they are placeddirectlyin contact with the longitudinal reinforcing bars. Design of pilasters and columns is

I

discussed in detail in MDG Example 12.2-1.

14-35


5.9.1.6.(a)

A C 1T I T L E S f l D G

Example 143-8

m 0662949 0509276 L92

93

RCJ Hotel

-

D

to Exterior

Connection of Rigid Roof Diaphragm

Loadbearing Wall A connection is to be designed between the roof diaphragm and the exterior loadbearing

wall of the RCJ Hotel. The connection must transfer gravity loads, in-planeshear, and outof-plane shear from the roofdiaphragm to the exteriorloadbearing wall. Differential vertical movement must be accommodated between the interior and exterior wythes of the loadbearing wall. Consider the wall on Grid Line F, for Wall Construction Option A and Building Construction Option II.


Reference ~~

Roofing

f

Bellows Insulation,

Code

~~

Metal Coping 2" Foam Gap

Treated

"

Precast Hollowcofe Planks

connection Detail The detail is shown above. Refer to MDG Example 10.4.3

As in previous examples,the wall mustbe designed for gravity loads, in-planeshear and out14-36


A C 1 TITLEsMDG 9 3

m 0662949 0509277 O29 m


Code Reference

Calculations of-plane shear. Gravity Load: The critical gravity load (DL

+ LL) is 1,725 lb/ft

from MDG Table

9.1.2. The critical in-plane seismic shear is 14.5 kips, distributed over 29.3 ft. See MDG Example 9.2-3.

The critical out-of-planeshear comes from the20 psf wind load, actingon a total wall height of: 7.83 ft in.0.67 0.17 in.

story height parapet parapet extension

8.67 ft Assuming the wall to be simply supported at the base and at a height of 7.83 ft, the out-of-plane shear in lb/ft is then

20psf

x

(8.67ftI2 = gfj

2 x 7.83ft

Gravitv Load: The connection resists gravity load by bearing. If one assumed all bearing to take place on the inner face shell, the bearing stress is:

14-37


A C 1 T I T L E*M DG 9 3

m

0bb2949 0509278 Tb5

m


Reference

1,725 lb 1.25 in. x 12 in. This will be less than the allowable bearing stress

=

Code

115psi

of Fh = 0.25 f;.

5.12.3

Although the Code does not provide for a minimum bearing length, the author suggests that more than the face shell thickness of the masonry unit be used. A reasonable bearing length would be 4 in. In-Plane Shear: The in-plane shear can either be transferred by shear stresses between the inner face shell of the CMU and the plank, or by shear friction due to the tension developed in the dowel bars. The shear stress is 14.5 kips x 1,0oO lbs/ kip 1.25in x 29.3ft x 12in/ft

=

33 psi

This is less than the governing shear allowable in the masonry of 37

6.5.2(c)

psi, which can be increased by one-third for this load cases involving

5.3.2

earthquake forces. The design will be acceptable for in-plane shear. Out-of-Plane Shear: The out-of-plane shear in lb/ft is less than the in-plane shear, and will not govern. The desim is acceptable.

14-38



Example 143-9

m

Obb2949 0509279 9 T L

I

m

-

RCJ Hotel Connection of Floor DiaphragmtoNonloadbearing Wall

Design the connection between a floor diaphragm and a nonloadbearing wall in Hotel. Use wall on Grid Line

the RCJ

2 between Grid Lines C and D on second floor using Wall

Construction Option B with Building Construction OptionI. The connection must transfer shear loads only. and

Calculations

A connection detail is shown below. Refer to MDG Example 9.2-4.

The connection is designed for in-plane seismicshear loads. Because the precast planksare cambered, they cannot be attached directly to the wall. In-plane shear transfer is achieved through the topping. The total shear transferred through the floor to the wall is 139 kips, over a length of 29.3

14-39


Ob62949 0509280 bL3



Calculations

ft from MDG Example 9.2-4 Table 2. Because this shear must be transferred along 2 sides of the wall, through a topping 2 in. thick, the corresponding shear stress is 139 kips x l,OOOlb/kip 2sides x 29.3ft/side x 12in/ft x 2in

Using topping

f:= 3,000 psi,

the allowable

=

98.7psi

shear stress of topping

per AC1 Code is (increased by 1/3 for seismic) 1.1

E

=

80.1 Psi

6.5.2(c) 5.3.2

The shear can be transferred by shear friction. Assuming a coefficient of friction equal to unity, it is therefore necessary to generate a clamping stress perpendicular to the wall-floor interface

98.7 ofpsi. 7.2.1.l(b)

Assuming the reinforcing steel has an allowable stress (againincreased

by one-third), thenecessary

24,000 of psi 5.3.2

steel percentage is

therefore 98.7 psi 24,OOOpsi x (4/3)

=

0.31 %

The required :steelarea per linear foot of topping is therefore: % x 12in. x 2 in. 100

=

0.07 h2/ft

The required steel could be providedby welded wire fabricor by deformed reinforcing bars.

14-40


A C 1 TITLEJtMDG 93

Example 143-10

RCJHotel

m

m

Obb2949 0 5 0 9 2 8 1 5 5 T

- Connection of FloorDiaphragm to Interior badbearing

Wall Design the connection between a floor diaphragm and an interior loadbearing wall in the

RCJ Hotel. Use Grid Line D between Grid Lines 1 and 2 (Wall Construction Option B with Building Construction I). The connection must transfer gravity loads plus shear.

and

Calculations The connection detail is shown below, refer to MDG Example 9.2-4:

@ 5'-O'' Each Cell Break

S5

Reinforcement As Required

Out Planks To Set

I

Planks Set On Face Shells

Grout Dams

Prior To Grouting

!

The connection is designed for gravity loads plus in-plane seismic shear. Because the precast planks span perpendicular to the wall (in contrast to the previous example), they can bear directly on the face shells.

14-41


A CT 1I T L E l r M D G

93

Obb2949 0 5 0 9 2 8 2 4 9 b

m

Example 143-10Coned. ~

~~

Code Reference


Desien for Gravitv Loads: (See The maximum gravity load occurs

MDG Example 9.2-4, Table 2) at the 2nd floor level, and is17.2kips/ft.

This load is

transferred from the planks to the wall by bearing of the planks on the face shells of the units. For construction purposes, the planks should extend far enough over the edge of the face shells so that they bear securely, but notso far that they obstruct the flow of grout into the cells of the wall. The minimum grout space is given in Code Table 3.1.2. The bearing stress on the face shells of the 6 in. wall units is 17.2 kips x l,OOOlb/kip 12 in x 2face shells x 1 in

This bearing stress

less is

=

than the maximum allowable value of

717 psi

0.25

5.12.3

Pm.If needed, some of the plank gravity load could be transferred by bearing to the grout core of the wall. In-Plane Shear: (See

MDG Example 9.2-4, Table 2)

The maximum in-plane shear transferred from the floor to the wall (2 sides) is 91 kips over a length of 29.3 ft. This shear is transferred through the topping. The required amount of steel required for transfer by shear friction is calculated as in MDG Example 14.3-9.

14-42


1

A C 1 T I T L E * H D G 93

Example 143-11

m 0662947 0509283

322

m

-

T M S ShoppingCenter Connection of SteelBeamBearingDetail

In the TMSShopping Center, design a bearing detail for theW 16 x 31 beam onGrid Lines

1 and B.

and

Calculations

The bearing detail is shown below:

/

Plate

x S" x 8"

\ Plate, f " x 4" x 12" With 33"@x 4" Headed Stud Flush With Top Of Grout

I

l

As in previous examples, the wall must be designed for gravity loads, in-plane shear, and out-of-plane shear. Bearing under the plate must also be checked. These conditions are described in more detail below. Use f,' = 2,000 psi.

14-43


A C 1 TITLEMMDG 9 3

m

Obb2949 0509284 2b9

m


Calculations

GravityLoads:(SeeMDGExample

9.2-1)

The masonry bond beam must be designed for

the vertical reaction from the W 16 x 31.

The governing reaction is 15.8 kips. First check bearing stress directly under the plate: 15.800 lb 5 x 8

in.’

395 psi

This bearing stressis less than0.25 f L. If the stress distributionunder

5.12.3

the bearing plate is conservatively assumed as triangular rather than uniform, the maximum bearing stress would double, to 788 psi. This stress would require that f’i be increased to 3,150 psi. The bearing stress in the hollow wall under the the bond beam would also need to be evaluated. See MDG Example 9.3-2. Other Loads: If in- or out-of-plane shears are present, they would be designed for using the techniques illustrated in MDG Example 14.3-5.

14-44


AC1 TITLE*NDG 73 Obb27Y7 0507285 IT5

Example 143-12

TMS ShoppingCenter

m

- Roof DiaphragmConnection to Shear Wall

In the TMS Shopping Center, design the connection between the roof diaphragm and the shear wall on Grid Line 2. The connection must transfer lateral loads from the roof to the wall, but will not transfer vertical load.

and

Calculations The connection detail is shown. The "butterfly plate," as given in detail at the bottom of the figure, allows vertical flexibility. The lateral tie must be designed for the in-plane shear transmitted to the wall. In-Plane DiaDhraem Shear: The total shear is 26.6 kips from MDG Example 9.2-1. Half comes from each side of the wall. The wall is 82 ft long. The shear per ft is therefore

1-1/2 in. x 22 gage metal deck with nominal fastening will be satisfactory for transferring shear to open-web joists.

14-45



93 D 0bb2949 0509286 031 D


Calculations

1

Bond Beam With

2 4 5 Continuous

Open W e b S t e e l

$\

A

Detail Of "Butterfly Plate"

14-46


A C 1 TITLEvMDG 9 3

m

Obb29q9 0509287 T 7 8

m


Calculations

Discussion

Reference

Code

Butterflv Plates: Try 1/8-in. butterfly plates at 4 ft. Plate shear (horizontal) is 4 ft x 162plf = 648 lb

Dimensions of the legs of butterfly plate must be checked for horizontal shear and flexure.

14-47


A C 1 TITLEJMDG 9 3

Example 143-13

RCJHotel

0 6 6 2 9 4 9 0509288 904

- Connection of ExteriorNonloadbearingWall

to Exterior

Loadbearing Wall The stair tower wall (Grid Line2) between Grid LineF and G is to be connected to the rest of the RCJ Hotel through Grid Line F. Use Wall Construction Option B with Building Construction Option I. Drag struts will provide the connection

at each floor level. The

figure below shows the overallarrangement of the stair tower with respectto the rest of the

RCJ Hotel. Story shears in the stair tower are (from top to bottom) 6, 37,60, and 52 kips from MDG Example 9.2-4, Table 2. The drag struts must be designed for these forces.

Grid Line 2

""_

Grid Line F

Elevation

Plan


Code Reference

Shears are produced by the lateral forces transmitted to the tower at each floor level. For purposes of this example, it is conservatively assumed that all the lateral force transmitted to the tower at each level must be transmitted throughthe drag strut. The drag strut forces at each floor level are therefore given by the difference between the story shear above and below that floor level. From top to bottom, the drag strut forces are therefore 6,31,23, and

8 kips, as shown on the figure above.

14-48


A C 1 T I T L E + M D G 93

Obb2949 0509289 840

m


Reference

Code

The figure below shows how a typical drag strut will connect the stair tower to the rest of the RCJ Hotel. The required area of the drag strut is determined by the load it must resist.

In this case, the critical drag strut has a tensile forceof 31 kips. Using an allowable steel stress of24,000psi,

increased by 1/3 for loading

5.3.2

combinations involving earthquake, the required steel area is Grid! Line F

Stair Tower

Connection Detail

=

0.97

14-49


in?

7.2.1.l(b)

A C 1 T I T L E S N D G 93

m 0662949 0509290

562

m



This can be satisfied using 3 4 5 bars, providing a total steel area of 0.93 in.2 The required embedment length of the bars at each end (the stair tower bond beam and the topping of the hotel slab) is determined by development length requirements. The development length of the drag strut bars is measured from the near face of the wall in each direction.

Zr

=

0.0015 db Fs

Eq. (8-1) Zr

= 0.0015 (0.63in.) 24,000 psi

Id

=

22.5 in.

14-50



Example 143-14

RCJ Hotel

m

93

-

0 6 6 2 9 4 9 0509293 4 T 9

m

Connection of InteriorNonloadbearingWall to Interior

Loadbearing Wall

Design the connection between corridor wall (nonloadbearing)and cross wall (loadbearing on Grid Line C between Grid Lines 1 and 2) in the RCJ Hotel. Use Wall Construction Option B and Building Construction Option I. The corridor slab is composed of precast, prestressed concrete planks, spanning along the corridor. The planks are supported on W 10 structural steel beams which span across the corridor. Code Reference


Grid Line 2

Cross Wall

W 10

c. c.

Beam g V

Partial Plan

m I

#4 Bars

\

" "

" "

" "

" "

Section B-B

View

Grid Line 2

2 #4 Bars Plus

Room Slab

Beam

Section A-A

14-5 1


A C 1 T I T L E t M D G 93 H Obb2949 O509292 335 H


Code Reference

Calculations

The figures show plan and section views of the connection. TheW 10 beam sits in pockets in each corridor wall. Precast, prestressed concrete planks

(8 in. thick) sit on top of the

lower flanges of the W 10 beam. The W 10 beams are tied to the cross walls and the rest of the slab by the 2 in. slab topping (reinforced with welded wire fabric), and also by 2-#4 bars placed above the planks on either side of the W 10 web.

14-52


A CT 1I T L E * M D G

93

Obb29Y9 0509293

271

m

I Example 143-15

DPCGymnasium

- Roof DiaphragmConnection to Nonloadbearing

Wall

Design a connection between the roof diaphragm and a nonloadbearing wall of the

DPC

gymnasium.


Reference

Code

'V

Connection Detail

This detail concerns the wallof MDG Example 9.3-7. The bent plate is very similar to that of MDG Example 14.3-12. Design calculations are also similar.

14-53 ~


A C 1 T I T L E S N D G 93 W Ob62949 0509294 108

Example 143-16

RCJHotel

- Termination of FlexuralReinforcementforContinuous

Masonry Beam

For the continuous masonry beam designed in MDG Examples 11.3-6 and 13.1-3, determine the required lengths of the flexural reinforcement. This beam is located below the second floor of the RCJ Hotel (Wall Construction A with Building Construction I) on Grid Line E

between Grid Lines 3 and 4.

d = 67 in.

b = 11.63 in.

Fb = 600 psi

n = 16.1

References Calculations Codeand Discussions l. Shear Diagram and Shear Reinforcement per MDG Example 13.1-3.

6'-6"

12'""

Shear SteelShear Steel = 2#4 @ 8" Shear Steel O C . Not Required Not Required 134,7 kips 80.9 kips

80.9 kips

4"9" 4

12'""

-

134.7 kips

7'- 11"

D"

7'-11"

"

'4"9"

t

$-Eof Bearing

12'-8"

E-$of

Shear Diagram

14-54


Bearing

A C 1 TITLEaHDG 93

m

0662949 0509295 044 D

Example 143-16Cont'd. Calculations and Discussions

Reference

Code

2. Moment Diagram and Required Reinforcement per MDG Example 11.3-6.

-D

\

I -341 ft-kips

2'-8"

$-Eof Beari

'-8"

%-Eof

T

Bearing-

2#7 X 17"6"

Moment Diagram t

2#7 x 14'-4"' 4#6 -U

\2#7 x 17'"" 1#7 X 14'"''

12"

T'

'I I

kinforcement Layout

3. Positive Moment Reinforcement Anchorage and Cut-Off Lengths. At least 25% of the positive moment reinforcement must extend into the support and be anchored to develop 8.5.3.2 the stress. tensile allowable Choose to make 2 bars, of the total 4 required, continuous. This decision is based on the symmetry of the bar locations (see MDG Example 11.3-6) and the need to have bars to which to tie the stirrups. 14-55


~

~~

A C 1 T I T L E * M D G 93 M 0662949 050929b T B 0 M


Reference

Calculations

The two extended bars must be anchored for length, 8.5.2

For a #6 rebar, 364, = 36(0.75 in.) = 27 in. A standard hook develops equivalent an length

= 11.25db, or

8.5.5.2

11.25(0.75 in.) = 8.44 in. for a #6 rebar. A standard 90" hook for a #6 rebar consistsof a 6 bar diameter bend

bar diameters for plus an extension of 12 The bar must be extended 27 in.

a total length

- 8.44 in.

of 12 in.

8.5.5.1 8.5.4.1(b)

= 18.56 in. past the center of bearing before

hooking into the column. To fit into masonry cores use20 in. extension. It is 4 in. from the face of the support to the center of bearing. 20 in.

+ 4 in. = 24 in. C 32 in. column

.: OK

To determine where the remainingtwobarsmay

be cut off,

it isnecessary to find the

flexural capacity of the section with 2-#6 bars. P =

k

2(o*44 11.63 in.(67 in.)

= -pn +

=

0.00113 =

pn = 0.00182

-0.00182 + 42(0.00182)

+

(0.00182)2 = O.174

Mt = F,A,jd Mt = (24ksi)[2(0.44 in?)](0.942)(67 in.) 12 in./ft Mt = 111 ft-kip based on steel stress allowable OR 14-56



93

m

Obb2949 0509297 917

m


Calculations

Mm =

(600 psi)(ll.63 in.)(67 in.)2(0.942)(0.174) 12,000 h -1bIft -kips

M m = 428 ft-kips

based on masonry stress allowable

+M = 111 ft-kips occurs at 1 ft - 8 in. from the point of zero moment. These bars may not terminate at the point at which they are no longer required to resist flexure, but instead must be extended for a length equal to the greater of the member depth or 12db.

8.5.3.1(c)

The member depth of 67 in. or 5 ft - 7 in., exceeds the distance from the point at which

+M

= 111 ft-kips to thecenter

of bearing.

Therefore,no

positive moment flexural

reinforcement bars may be terminated short of the supports. Make all 4-#6 bottom bars continuous and hooked into the end supports for a distance of 32 in.

If the bars cannot bebe provided full length in one piece, provide lap splices of 8.5.7.1.1 0.002dps r 12 in. For #6 bars, lap length = 0.002(0.75 in.)(24,000 psi) = 36 in. 4. Negative Moment Anchorage and Cut-Off Lengths. At least 1/3 of the total negative moment reinforcement must be extended beyond the point of zero moment for a distance equal to the member depth or equal toone-sixteenth of the span.

8.5.3.3(b)

14-57



Code Reference

Calculations

Since the total negative moment reinforcement consists of 5-#7 bars, extend 2-#7 bars for

a distance of

12.7ft(12 in./ft) 16

Id

=

9.53 in.

The total length of these bars is 2(3.17 ft)

+ 2(67 in./12) = 17.5 ft.

It may be desirable to

extend these bars full length of the beam, to provide something to which to tie the stirrup bars.

To determine where the remaining bars may be cut off, it is necessary to find the flexural capacity of the section with the 2-#7 rebars.

2(o.60 in.2)

= 11.63 in.(67

-+ 42pn

k = -pn

Mm=

in.)

+

=

0.00154

bn

=

0.0248

= -0.0248 + 42(0.0248) + (0.0248)2 = 0.199

(600 psi)(ll.63 in.)(67 in.)2(0.934)(0.199)

12,000 in.-lb/& -kips

Mm= 485 ft-kips

based on masonry stress allowable

OR

14-58



93

m 0662947 0509299

79T

m


Calculations

Discussion

Reference

Code

M, = F,A,jd Mt =

(24 ksi)[2(0.60 h2)](0.934)(67 in.) 12 in./ft

based on steel stress allowable

Mt = 150 ft-kips GOVERNS

-M = 150 ft-kips occurs at 1 ft - 7 in. from the point of zero moment. The remaining three bars may not terminate at the point at whichthey are no longer required to resist flexure but instead must be extended for a length equal to the greaterof orthe member depth

lub.

8.5.3.1(c)

The member depth of 67 in. governs over lub= 12(0.875 in.) = 10.5 in. for #7 bars. The extension of 67 in. or 5 ft - 7 in. means that the bars terminate in the zone of positive moment, or in the compression zone. Therefore the requirements of Code 8.5.3.1(e) need not be met. By inspection, the continuing reinforcement has embedment length r Zd beyond the point of terminated reinforcement.

8.5.3.1(d)

The total length of the terminated bars is 2(1 ft - 7 in.)

+ 2(5 ft - 7 in) = 14 ft - 4 in.

Bond capacity to develop the bars should also be checked at the point of zero moment. See MDG 14.2.1.3 and MDG Example 14.2-3

14-59



93

m Obb2949 0509300 231 m

15 EMPIRICAL DESIGN

15.1

HISTORY

Beginning with the Code of Hammurabi (2123 - 2081 B.C.), man hasbeen developing design standards for structural masonry. Some of the first standards specified such requirements as minimum wall thickness and maximum building height. The structural design of historic

buildings was based upon requirements for mass, minimum wall thickness, maximum wall height, crosswalls, quality of materials, and workmanship. The modern empirical standard (ANSI A 41.1, AmericanStandard

Building Code

Requirements for Masonry) was originally issued by the National Bureau of Standards as Miscellaneous Publication 211 in1944.

The document has been the basis for empirical

design standards in model building codes, engineering reference standards, architectural graphic standards, and textbooks. These standardshave been based upon sometype of engineering rationale,including lateral stability, buckling of masonry walls, compressive strength, minimum wall thickness, and associated requirements. Masonry materials have substantial compressive strengthand durability but low resistance to tensile stresses. Therefore, masonry should be designed thick enough for the vertical compressive force resultant to act in the middle third of the

15-1


A C 1 T I T L E S N D G 93

m

Obb2949 0509301 178

m

structural element. Otherwise, flexural tension will develop in the nonreinforced masonry

which may exceed the allowable tensile stresses assigned to masonry in codes.

MONADNOCK BUILDING, CHICAGO, IL Sixteen Story Building,Six Foot Thick Walls. A Building Example of Empirical Design Requirements Before Rational Engineering Methods.

BRICK AND CONCRETE MASONRY CONSTRUCTION EXAMPLE Recommended Empirical Design According to the Code Requirements.

Fig.15.1-1EmpiricalDesignMethodExamples 15.2GENERALDESCRIPTION

Code Chapter 9, Empirical Design of Masonry, describes requirements for nonreinforced masonry only; reinforcementis not considered. Masonry designed by the empirical structural 15-2


A C 1 T I T L E * H D G 93

0662749 0509302 004

design method has a proven success record, and has exhibited economics of construction. If the proposed structural masonry systemis not permitted to be designed withthe empirical design requirements, then the designer is required to design according to the rationaldesign requirements presented in Code Chapter 6 for Unreinforced Masonry and Code Chapter

7 for Reinforced Masonry. Members not participating in the lateral force resisting system of a building may be empirically designed by Code Chapter 9 even though the lateral force resisting system is designed under Code Chapters 5, 6, 7, 8. The Code empirical design section has been improved over the previous empirical standard,

ANSI A 41.1. Arbitrary masonry design requirements have been removed which required very thick masonry walls,and unrealistic masonry sizeswhen compared to masonry designed using rational requirements. The empirical design procedure continues to be successful whenused

according to the criteria prescribed in the Code. The important difference

between this Code and previous codes is the restricted use of Empirical Design under certain loading criteria and building configurations. 1 5 3 LIMITATIONS

In accordance with Code 9.1.1 the empirical design procedure shall not be used for masonry structures with the following characteristics: Location inSeismic Zone 3 or 4. Empirical design procedures have been successfully used for centuries in Seismic Zones 0,1, and 2; Design Wind pressure in excess of 25 psf; Horizontal loads from sources other than permitted wind or seismic loads; Foundation walls not satisfying Code 9.6.2; Unreinforced masonry laid in other than running bond. (See Code 9.7.5.2); Masonry veneer walls (this design material isusually

included in model

building codes); Buildings taller than 35 ft that rely on masonry walls for lateral load resistance 15-3


~~

A C 1 T I T L E * M D G 93

m Obb2949

0509303 T 4 0

m

(except for masonry infill walls); and 8)

Cantilever retaining

walls.

Masonry infill walls not part of the lateral load resistance system can be designed by the empirical procedure. 15.4EMPIRICALDESIGNREQUIREMENTS 15.4.1MaterialsandSpecifications

Empirically designed masonry must be constructedin accordance with AC1 530.1/ ASCE6/ TMS 602 Specifications for Masonry Structures to assure compliance with the design. The Specifications require the designer to specify selected materials and methods,as well as the extent of quality control and inspection necessary for each project. The reader is referred to MDG Chapters 3, 4,5, 6, and 7 for additional information. 15.4.2 Lateral Stability

Shear wall spacing requirementsare an integral part of the Code requirementsfor Empirical Design. The structural integrityof an exterior masonry wall is based not only on flexural and axial strength, but also on lateral stability normally provided by shear walls. Out-of-plane forces caused by lateral wind or seismic loads, imposed on

an exterior wall surface, are

transferred to the roof or floor diaphragm system and then to the shear walls. Code 9.3.1.1 stipulates a minimum thickness of 8 in. for masonryshear walls. Code 9.3.1.3 specifies shear wallspacingrequirements,whichpreviouslyhadbeenlistedin

other documents as

recommended practices. 15.43 Compressive Stress Requirements

Code 9.4 specifies conservative valuesof allowable compressive stress for empirical design

15-4


A C 1 T I T L E * H D G 93 W Ob62949 0509304 9 8 7 W

ofmasonry.Theallowablestresses

in CodeTable

9.4.2, are baseduponthe

cross-sectional area of themasonrywall,usingactualdimensions

gross

rather than nominal

dimensions. 15.4.4

Lateral Support

Lateral supportis a primary requirement in the structural design of masonry walls usingthe empirical requirements. Lateral support elements brace the masonrypermit and the transfer of loads to theresistingelements.Decisionsaboutthetypesandlocations

of support

a wall mayspan vertically elements for the masonry affect the masonry design. For example,

between floors and roof. The floor to floor height then determines the span length of the wall and the Code limits the h/t ratio. Alternatively, walls may span horizontally between crosswallsandtheCodelimitsthenapply

to the

Z/t

ratio.Code

9.5 specifiesspan to

thickness ratios, which determine the required thickness for a given span (see Figs. 15.4-1 and 15.4-2). No matter what lateral support system is used, the masonry element must be adequately anchored to the support element in order fortoitfunction properly. See MDG 15.4.7.

Fig. 15.4-1 h/t Ratios

15-5


Obb2749 0509305 AL3

A C 1 T I T L E 3 M D G 93

15.4-11for Masonry Connection to Steel Column

See Figure

1 r 'L

Expansion Gap.Typical Steel

-Y

m

Column, Typical

1 = Horizontal Span

1 = Horizontal Span

"""""-

Typical

Simple Span Deflection

Fig. 15.4-2 Z/t Ratios

Lateral support design includes two distinct issues: 1)

Lateral support systems(crosswalls,

pilasters,buttresses, or other structural

frame members) based upon empirical and/or rational design methods, and 2)

Masonrywallthicknessbasedupon

h 'It or Z/t ratios and minimum thickness

requirements. Code Table9.5.1, Code 9.5.1.1, and Code9.6.3 provisions forlateral support are summarized in Table 15.4-1. In computingthe ratio (Ut or h 'It)in Table 15.4-1 for walls bonded in accordance with Code 9.7.2 or Code 9.7.3, although not included inthe Code, it is recommended that the value for

thickness (t) shall be the sum of the nominal thicknesses of the wythes.

15-6


A C 1 TITLELMDG 9 3

m

Obb2949 050930b 75T

m

Table 15.4-1 WallLateral Support Requirements ~~

MAXIMUM llt or h ‘/t

CONSTRUCTION

Bearing Walls Solid or Grouted All Other Non Bearing Walls Exterior Interior Cantilever Walls Solid Hollow Parapets (8 in. thick minimum)

20 18 18 36

6 4 3

The designer must not only determine the wall thickness but also design the lateral support system and the connections transferring the forces from the wall to the lateral support. As an example, roof diaphragms must be designed to transmit out of plane forces on walls to the in-plane lateral force resisting shear walls. A bond beam, although not required by the Code, is commonly installed at thelevel of the diaphragm support element to transfer forces from the wall to the diaphragm, and vice versa. 15.4.5 Thickness of Masonry

15.4.5.1 Minimum Thickness Criteria - Code 9.6.1 requires an arbitraryminimum nominal masonry thickness to limit inadvertent axial load eccentricity in masonrybearing

wall

structures. Unreinforced masonry wall structures may crack under eccentric loading

that

produces tensile stressesin the wall. The arbitraryminimum thickness requirement, together with appropriate details, will limit cracking. 15.4.5.2 Foundation Walls

- Masonry foundation walls as shown in Fig. 15.4-3, per Code 15-7


A C 1 TITLElwMDG 93

m

0bb2949 0509307 b 9 b

9.6.2, may be designed according to the empirical design chapter.

Drain

Fig. 15.4-3 Foundation Wall Section

The following empirical criteriaare intended for basement foundationwall design: l.

Maximumheight

of wallbetween

lateralsupports(usuallybetweenthe

basement floor and the first floor) = 8 ft- O in. Cantilever walls (no lateral support at top) are not permitted.

2.

Maximumequivalentfluidweightofbackfill

3.

Provisions for

= 30 pcf

adequate drainage:

a. clean granular backfill, b. an operational drain tile, and c. backfill sloping away from 4.

structure (15.4.1).

Maximumheightoffinishgradeabovebasementfloor

= 8 ft- O in.Masonry

thickness requirements vary with the heightof finish grade. See Code Table 9.6.2.2 below. 5.

Type Mor S Mortar 15-8


m

A C 1 T I T L E x M D G 93 W 0662749 0507308 5 2 2

Code Table 9.6.2.2 Thickness of Foundation Walls

Nominal Thickness, in.

Foundation Wall Construction

Unbalanced Fill, ft

Masonry of hollow units, ungrouted

Masonry of hollow or solid units, fully grouted

i Masonry of hollow units reinforced vertically with #4 bars and grout at 24 in. on centers. Bars located not less than 4% in. from pressure side of wall

15.4.6

Maximum Depth of

8 10 12

4 5 6

8 10 12

5 6 7

8 10 12

7

8

7

8

8

Bond

Code 9.7 addresses the required connections between the facing and backing of multiple wythe masonry walls. 15.4.6.1 Masonry Headers - Before the development of metal ties, masonry headers were

used to connect the wythes of multiwythe walls. The details shown in Code C. Fig. 9.7-1 describe the Coderequirements

for the lapping ofmasonryunit

monolithicaction ofmultiwythewalls.

headers to achieve

The Codeestablishesminimumrequirements

to

connect the two wythes for any masonry material types. There is the potential for water

15-9


A C 1 T I T L E x M D G 93 M 0662949 0509309 469

m

transmission across wythes. Differential movement between the inner and outer wythes due to unequal gravityloading

or due to differential thermal or moisturebehaviorcould

potentially result infracture of the masonry header. If masonry headers are used, then the same type of masonry unit should be used for both wythes, and movement joints must be provided. Masonry header details for both solid masonry units,per Code 9.7.21, and hollow masonry units, per Code 9.7.2.2, are shown in Code C. Fig. 9.7-1. If both hollow and solid masonry units are combined in a wall, then the stricter Code requirement governs. 15.4.6.2 Metal Ties - The alternate system for connecting multiwythe masonry walls metal is

ties. See Code 9.7.3. The metal tie system has several advantages compared to the masonry header system: 1)it accommodates vertical and horizontal adjustment during construction, resulting in straighter walls; 2) it allows minor differential movement between multiwythe walls which may be caused by different masonry materials, shrinkage or expansion due to moisture and temperature; and 3) it allows for a more watertight construction systemthan has been experienced with masonryheader construction. The empirical metal tie requirements shown in Code C. Fig. 5.8-2 have been successfully used in multiwythe masonry walls, with cavity widthsnot exceeding four inches, to transfer lateral loads to both wythes. A typical cavity wall system has a 4 in. brick masonry outer wythe, a 3 in. cavity which includes 1 in. rigid insulation, and an interior concrete masonry wythewhosethicknessis

based upon other empirical requirements described in this

Chapter. 15.4.7 Anchorage 15.4.7.1 Intersecting Walls - Intersecting masonry walls whichdepend upon each other for

lateral support must be anchored at the intersection to ensure structural integrity of the building. Code 9.8.2 requirements represent past, successful standards for masonry walls. 15-10


A C 1 T I T L E * N D G 93 9 Obb2949 0509330 360

Interior nonloadbearing walls may be mortared or mechanically tied to the floor below for lateral support.Withoutspecific

restraint at the top,suchwallsmust

be considered

cantilevered from the floor. It is undesirable to connect the interior nonloadbearing wall rigidly to the roof or floor above, becausethe temporary live load deflection of the structure will imposeunanticipatedstresses

whichmaycausecracking.

Connections that permit

vertical slip may be used (See MDG 15.4.7.2), or the intersecting walls may be anchored with masonry bonding, as described in Code 9.8.2.1 and shown in Code C. Fig. 5.13-2, or mechanical fastening as described in Code 9.8.22 through Code 9.8.2.5. and illustrated in Code C.5.13.3. 15.4.7.2

Floor and Roof Anchorage - Floors and roofs must be anchored to the exterior

masonry wall to provide lateral support for the wall and to transfer lateral loads to the horizontal diaphragm. Diaphragm connections are required not only at the bearing walls, but also at the nonloadbearing walls, where floor or roof framing is parallel to the wall. Examples of the fastening systemsdescriied in Code9.8.3 are shown in Fig. 15.4-4. Interior nonloadbearing walls maybe connected to the floors or roof abovefor lateral support, if the connection permits the diaphragm to freely deflect vertically.

Fill With CMU Between Joists Wood Floor On Wood Joist

CMU Header Block

Joint Attached to Bond Beam

Beam Solid CMU

6"O" Long Into ooked t o

STEEL ROOF JOIST

WOOD FLOOR JOIST

Fig. 15.4-4 Anchorage

15-11


A C 1 T I T L E x M D G 73 M 0 6 6 2 7 4 7 0507311 017 M

15.4.73

WallsAdjoiningStructuralFraming

- Code 9.8.4 presents a general minimum

requirement of metal anchor area and spacing that permits the masonry to be keyed to the structure. Where masonry walls are anchored to structural framing for lateral support, care must be taken to provide the lateral support while keeping the masonry isolated from structural deformations. Inadvertent loading of unreinforced masonry walls connected to ductile steel frames with rigid connections may cause masonry fracture. 15.4.8MiscellaneousRequirements 15.4.8.1 Chases and Recesses

- Lintels are required above openings wider than 12 inches,

per Code 9.9.1. See Fig. 15.4-5.

1 r

4"min.

--- ---

Section Through Lintel

Lintels Required for Openings > 12"

Fig.15.4-5OpeningSupport 15.4.8.2Lintels

- Lintels may be steel,concrete,

or masonry.Masonrylintelsmust

be

designed in accordance with Code 5.6 and Code 7.3.3. Code 9.9.2 requires a minimum lintel end bearing length of 4 in. as shown in Fig. 15.4-6. The minimum bearing length allows for minor imperfections within the several building materials involved. All masonry bearing conditions mustbe sized for width, length, andmaterial type using a bearing pressure stress analysis. The required end bearing lengths will be greater for heavier loads, or longer span lintels. Lintel bearings, in general, do not require bearing plates. If they do the edge of the bearing plate should be at least 1 in. back from the opening. Lintels rarely need anchor bolts. For longlintels care should be take to avoid edge loading of the masonryby

15-12


A C 1 TITLEtMDG 93

0bb2747 0507332

T53 W

deflection of the lintel.

ening

Grout Hollow Units At Lintel Bearing Or Use

Fig.15.4-6MinimumEndBearing

Length

15.4.83 Support on Wood - Code 9.9.3 prohibits the support of masonry by wooden girders

or other form of wood construction. The basis of the exclusion isthe concern of the wooden elements fire resistance as comparedto that of the masonry, and the potential implications to public health and safety.

15.4.8.4 Corbelling

- Code 9.9.4 states the provisions for corbelling in masonry walls

hollow and solid units.

of

Only solid units may be used for corbelling. The general design

requirements are shown in Code C.Fig9.9-1.

If the designerchoosesacorbelling

arrangement in excessof that permitted by Code 9.9.4, then the designer must verifygeneral structural stability and flexural tensile stress capacity using other sections of the Code. REFERENCES

15.4.1 "Guide

to Residential Cast-In-Place Concrete Construction",

American Concrete Institute.

15-13


AC1 332R-84,

A C 1 T I T L E S M D G 93 9 0bb2949 0509333 99T

m

-

Example 154-1 TMS Shopping Center Empirical Design of Masonry Walls

Design the masonry walls for the TMS Shopping Center Wall Construction Option k See MDG 9.1.1 for building plans and elevations. Roof Deck Metal Deck Load Live Roof

With No

Concrete Masonry ASTM C 90

Fill

= 30 psf

Load Dead Roof

(Hollow) Grade N

= 15 psf

Weight Unit

pcf = 120

Thickness to be determined Mortar ASTM C 270

Seismic Zone = 1 Wind Design

Type N

Pressure psf = 20

o

8

102"4"

12" 41"O"

41"O"

TMS Shopping Center Framing Plan

N


Code Reference

1. EMPIRICAL DESIGN CRITERIA CHECK

A.

Seismic Zone 1 < Zone 3 .: OK

9.1.1.1

B.

DesignWind Pressure = 20psf C 25 psf :. OK

9.1.1.2

C.

Building Height = 18 ft

C

35 ft .: OK

:. Empirical Design maybeused 15-14


9.2

A C 1 T I T L E * U D G 93

1

m 0662947 0507324 826 m


Reference

Code

2. DESIGN OF LOADBEARING WAILS, GRID LINES A & C Try 12 in. hollow, concrete masonry units,based upon lateral support requirements

A.

8 in. < 12 in. :. OK

Minimum Shear Wall Thickness:

9.3.1.1

B. Minimum Shear Wall Length 1.

Shear walls are required on Grid Lines A & C to resist wind in the east or west direction. The Code requires shear wall spacing and shear wall length considerations. Solving for the minimum shear wall length minimum length =

shear wall spacing

9.3.1.3

ratio

1 = - =824 1ftf t < L e n g t h W a l l G r i d L i n e A = 2 0 4 . 6 f t

2

> Length Wall Grid Line C

=

20 ft

For the TMS Shopping Center to comply withEmpirical Design,an additional shear wall length of approximately 20 ft must be added on Grid Line C 2.

MinimumCumulativeLength = 0.4 x longdimension minimum length

= 204.7 ft x 0.4 = 81.9 ft

actual length (deducting openings) = North

I

=

204.7 ft - 6 X 3.33 ft

9.3.1.2

+ South walls

+ 21.33 ft + 20 ft = 226 ft > 81.9 ft

C. Compressive Stress: 1.

Wall Grid Line A axial compressive stress, fa

=

P Fa = 70 psi -

9.4.2

As

15-15


A C 1 T I T L E r f l D G 93 W 0662949 0509315 7b2


Calculations

Compute compressive stressat base of wall for roof live

+ dead loads + wall

weight

2.

fa =

((30 psf + 15 psf) 1 ft x 40 ft x 0.5) + (18 ft x 55.5 psf x 1 ft) 12 in. x 11.63 in.

f,

13.6 psi

=

.-.OK

70 psi

WallGridLineC Overhang on wall = 253 plf

f‘

D.

1,900 plf + 253 plf

=

12 in./ft x 11.63 in.

=

2,153 plf = 15.4 psi 12 in./ft x 11.63 in.

Lateral Support:GridLines

70 psi

2.

OK

A&C

(a) 16 ft tall; assume lateral brace at roof with roof diaphragm 9.5.1 Use 12 in. hollow CMU (b) 2 ft parapet cantilever - Assume 8 in. units

h’ - 2ft x 12 in./ft = 3 = 3

”

t

8 in.

:.OK

Use 8 in. hollow CMU 3. DESIGN OF NONLOADBEARING WALLS, GRID LINES 1,2, & 3

A.

Minimum Shear WallThickness:

8 in.

15-16


9.6.3

A C 1 TITLE*UDG 93 D Obb271r7 O507336 bT7 D



Reference

Code

Shear walls are required on Grid Lines 1, 2, & 3 to resist wind in the north or

south direction.

1.

Minimum length = shear wall spacing

9.3.1.3

ratio

82 ft > 50.7 ft

GridLine 1 Walllength

=

Grid Line 2 Wall length

= 82 ft

> 50.7 ft

Grid Line 3 Wall Length

= 82 ft

- 10 ft - 3.33 ft

= 68.7 ft

2.

> 50.7 ft

2.

OK

Minimum cumulative length = 0.4 x long dimension Minimum length = 204.7 ft x 0.4 = 81.9 ft

+2+3 = 82 ft - 3.33 ft - 10 ft + 82 ft + 82 ft

Actuallength Actuallength

= 232.7 ft

C.

1

= GridLines

> 81.9 ft

2.

OK

Compressive Stress

1.

WallGridLines

1&3

Try 12 in. hollow CMU

2.

WallGridLine

2

Try 8 in. hollow CMU l8 fa

=

x 40 psf x 12 in. x 7.63 in.

=

7.87

15-17


70 psi

9.3.1.2

m

A C 1 TITLESMDG 93

Obb29V9 0509337 535


Calculations

Reference

fu

D.

fi

x 40

psf x fi 12 in. x 7.63 in.

=

=

Code

7.87 psi <' 70

Lateral Support 1.

GridLine 1 & 3, exteriornonloadbearing (a) 16 ft tall, assume lateral brace at roof with roof diaphragm

- l6

"

t

l2 in'/fi 12 in.

fi x

=

:. OK

16 < 18

9.5.1

Use 12 in. hollow CMU (b) 2 ft parapet cantilever

h' - 2 fi x 12 in./ft t 8 in.

"

=

= ,a.

OK

9.6.3

Use 8 in. hollow CMU 2.

GridLine 2, interiornonloadbearing wall 9.5.1

Use 8 in. hollow CMU 4. COLUMN OR PILASTERDESIGN

At B-1, B-3, C-1.1, C-1.2, C-2.2 Note:ColumnsandPilasters

are notspecificallydescribedinCode

Chapter 9,

Empirical Design; however, the designer may make certain assumptions of load distribution while utilizing the Code requirements.

A.

Column or Pilaster at B-1 and B-3

15-18


93


m

0662949 0509318 471

m



Control Joint

Code Reference

1

I

I

&Beam

k

Thicken Wall with 12" x 15.63"CMU,

15.63"

Running Bond.

Base of Wall Reaction = Roof Load

I

Reaction

=

P -

(ai *

+

S)10 ft x 45 psf + 2

20,800 lb 23.6 in. x 15.6

"

=

A,

B.

+ Wall Weight

in.

= 54.2

psi

18 ft x 111 psf x 1.33 ft

70 psi

Column or Pilaster C-1.1,C-1.2,C-2.2 Author's Recommendation:

P 15.63"

23.63"


Thicken Wall with r 1 2 " x 15.63" CMU, Running Bond.

=

20,800 lb


9 3 D Obb29Y9 0509339 308

m


*

Reaction =

Code Reference

+

5 ft o v e h g IO ft width x 45 psf

+ wallwgt 18 ft x 111 psf x 1.33 ft fa

=

14,100 lb 23.6 in. x 15.6

in.

=

=

36.7 psi < 70 psi

14,100 lb

:. OK

9.4.2

ShODDinP Center Desim Summarv Use single wythe 12 in. hollow concrete masonry units, ASTM C 90, Grade N with ASTM C 270 Type N mortar. For parapet use 8 in. hollow CMU. For pilaster add 12 in. x 16 in. hollow concretemasonryunit;all

other specificationsremainsame.Many

details are required forasuccessfulproject

other design

as well as to satisfy the EmpiricalDesign

requirements in Code Chapter 9. The roof structure must be properly fastened to the top of the walls to provide lateral bracing and to transfer diaphragm shear loads to the lateral load resisting shear walls.

*For Pilaster DesiEo See MDG 11.2

tN

TMS shoppins center Masonry wall Layout

15-20



m 0662949 0509320

93

02T

m

-

DPC Gymnasium Empirical Design of Masonry Walls

Example 15.4-2

Design the masonry walls for the DPC Gymnasium.See MDG 9.1.2 for building plans and elevations.

ASTM C 216

Roof Deck Metal Deck With Fill Brick NoMasonry Load Live Roof

Grade SW 4 in. thickness

= 40 psf

Roof Dead Load = 20 psf Seismic Zone

ASTM C 90

= 20Concrete psfMasonry

Design Wind Pressure =2

= 120 pcf

Grade N, Weight Unit

Mortar, ASTM C 270 and

Reference

Calculations

1. EMPIRICALDESIGNCRITFdUACHECK

A. Zone Seismic

2 < Zone 3

:. OK

9.1.1.1

However, minimum reinforcement is required for Seismic Zone B.

Design Wind Pressure

= 20 psf

< 25 psf

2

:. OK

k3.8 9.1.1.2

.: Empirical Design may be used 2. WALL CONSTRUCTION OPTION A: (4 in. brick, 3 in. cavity with rigid insulation, hollow CMU with pilasters at 16 ft o.c.): estimate clear spacing = 13 ft - 4 in. for interior wythe of CMU, try 6 in. hollow, concrete masonry units, A.

ShearWallDesign: 1.

Minimum shear wall thickness shall be 8 in. or more Assume 6 in. concrete masonry units shear wall t = 4 in. + 6 in.

2.

=

10 in. > 8 in.

.: OK

9.3.1.1

Minimum totalshearwall length a.GridLines

9.3.1.2

1 and 2

15-21


m

A C 1 TITLEaMDG 93 M Obb2949 0 5 0 9 3 2 3 Tbb


Calculations

Shear Walls: North/South direction minimum cumulative length = 0.4 x long dimension min. length = 128 ft x 0.4 = 51.2 ft actual length = 64 ft

+ 52 ft = 116 ft > 51.2 ft :.

OK

b. Grid Lines A and B

9.3.1.2

Shear Walls: East/West direction. minimum cumulative length = 51.2 ft actual length = (128 ft - 8 ft) x 2 = 240 ft > 51.2 ft 3.

:.

OK

Maximum spacing of shear walls a. Grid Line A and B

9.3.1.3

Shear Wall Spacing/Shear Wall Length

= 64 ft/128 ft

- 8 ft

:.

OK

=

0.50 < 2

b. Grid Line 1 and 2

9.3.1.3

Shear Wall SpacingBhear Wall Length

= 128 ft/64 ft =

B.

2.0

=

2

:. OK

Compressive Stress: Nonloadbearing Wall on Grid Lines 1and 2 (Assume gravity loads are resisted by pilasters alone)

f, < F,

=

70 psi, 6 in. hollow CMU

P = bottom of wall reaction

Use 4 in. brick and 6 in. CMU, Type N mortar C.

Lateral Support: 1.

13 ft - 4 in. horizontal span wall; 15-22


9.4

Ob62949 0507322 9 T 2

A C 1 TITLEvMDG 9 3

Examde 15.4-2

Cont'd. Code Reference


assume lateral support at pilaster

height = 30 ft < 35 ft :. D.

OK

9.2

Pilaster Design: 1.

Pilastersmustberationallydesigned loads. See MDG Chap.

for combined lateral androofgravity

11 and 12.

3. WALLCONSTRUCTIONOPTION

B:

(Compositewall

4 in.brick,

3/4 in.full

collar joint, and hollow CMU), determine the wall thickness. No pilasters. Grid Lines A and B, try 12 in. hollow CMU for interior wythe, Grid Lines 1 and 2, try 16 in. hollow CMU for interior wythe,

A.

Shear WallDesign: Same as for Wall Construction Option A

B.

Compressive Stress: Lines l. Grid fa

1 and 2, nonloadbearing

< F,

=

9.3.1.2

70 psi for Type N mortar (Governed by hollow CMU allowable per

Code 9.4.2.2) f a =

P -

P = wall wgt.

at bottomwall of

reaction

A, brick wgt. = 40 psf; CMU Wall wgt. = 80 psf(See MDG Appendix A)

Use 4 in. brick and 16 in. CMU, Type N mortar 2.

Grid Lines A and

B, loadbearing walls 15-23


9.4 9.3.1.2


m

m

Obb2949 0509323 839


Calculations

Discussion

fa

Reference

Code

C F, = 70 psi for Type N mortar (Governed by hollow CMU allowable per

Code 9.4.2.2)

P

fa = -

P = roof live and dead loads

+ wall wgt.

A*

brick wgt. = 40 psf; CMU wgt. = 80 psf

fa =

(24 ft ht x 80 Psf) 1 ft + (32 ft x 60 psf) 1 ft = 25.0psi 12 in. x 12 in.

Use 4 in. brick and 12 in. CMU, Type N mortar C.

7opsi

9.4

Lateral Support: 1.

GridLines 1 and2 9.5.1

2.

GridLines A and B 9.5.1

DESIGN SUMMARY WALL CONSTRUCTION OPTION A: Use double wythe, 4 in. brick, 3 in. insulated cavity, 6 in. hollow concrete masonry units,

ASTM C 90 with ASTM C 270 Type N Mortar. Pilasters at 16 ft. WALL CONSTRUCTION OPTION B Use composite wall:

15-24


A C 1 TITLELMDG 93

m

Ob62949 0509324 7 7 5

m


and

Code Reference

Calculations

loadbearing wall:

4 in. nominal brick, 3/4in. full collar jt.,

12 in. hollow

concrete masonry units mortar, ASTM C 270 Type N nonloadbearing wall:

4 in. brick, 3/4 in.full

collar jt., 2 wythes 8 in.hollow

concrete masonry units mortar, ASTM C 270 Type N Author’s Notes: The empirical design for this construction option may be an impractical and uneconomical solution depending on localmasonry construction experience. A rational approach should be considered in accordance with MDG Chapters 11and 12 to determine a reinforced wall design. Intermediate lateral braces, such as steel columns or buttresses may also be considered. SEISMIC ZONE 2 REQUIREMENTS The Code A.3.8 requires minimum steel reinforcement for structures in Seismic Zone 2. General steel reinforcement requirements are as follows:

l. Vertical Reinforcement (from support to support) a. at each wall corner, b. at each side of each opening,

c. at ends of walls, 2. Horizontal Reinforcement a. at bottom and top of wall openings, b. at structurally connected roof level, c. bottom of walls, and other areas, see A.3. d. uniformly distributed inwalls.

15-25


TITLEaNDG 73 m 0662747 0507325 601 m

AC1


Calculations

Control Joints (CMU) Exp. Joints(Brick)

T

#4 Typical

24 '-8''

I 28 "O"

24 '-0"

LIA

24 '-0"

T

1- 23 '-0"

128 "O"

4

h

South Elevation, North Elevation Opposite Hand

m

Control Joint, Typical,

f For CMU

I

24

t {"-

I L"_

L A ""_

II

I I

I

I.

64 "O"

""_

"_

64 "O"

DPC Gymnasium

Seismic Zone 2 Reinforcement Requirements

15-26


I

""-

East Elevation

West Elevation

-

4

A C 1 T I T L E * M D G 73

0662947 050732b 5 4 8

-

Example 15.4-3 RCJ Hotel Empirical Design of Masonry Walls

Design the masonrywalls

for the RCJ Hotel.

Use WallConstruction

Option "A"

(Unreinforced brickand block noncomposite exterior walls and unreinforced blockinterior walls), and Building Construction Option I dimensions. See MDG 9.1.3 for building plan dimensions and elevations. Roof Deck Masonry Precast Concrete Concrete

ASTM C 90, Grade N

Roof (Hotel) Dead Load psf = 95 Hollow Floor Dead psfLoad = 95

Units

Partition = 15psf

D. L = 10 psf

Curtainwall Glass Live Roof

Mortar, ASTM C 270

Load = 20 psf

Mortar Type to be specified

Dwelling Room L L = 40 psf Public Rooms L L = 100 psf Corridors L. L = 100 psf Stairs L L

=

100 psf

Seismic Zone = 2;

Pressure Wind Design

= 25 psf

Use modified framing plan shown on the next page and

Calculations ~~~

~

~

~~~

The building scheme presented in MDG Figs. 9.1-6 through 9.1-15 cannot be designed by the empirical sectionof the Code, since it does not satisfythe Code requirements for shear walls in the east-west direction. A revised scheme, which may be designed empirically is presented in the modified framing plan shown onthe next page. In this revised scheme,the interior nonloadbearing wall on Grid Line 2 is removed and exterior nonloadbearingwalls are added to Grid Lines 1and 4. All other aspects of the building scheme are as shown in

MDG Chapter 9.

15-27


A C 1 TITLExMDG 93

m

0662949 0509327 484

m


Calculations

Reference

30 ' I

29 "8"

,

30

Code

30

30

a

I

E

[c'

7 '-8"

28 "8" typical

I

E [c,

,

2 n d , 3rd, and 4th Floor Plan

RCJ Hotel - Modified Framing Plan l. TYPICALINTERIORLOADBEARINGWALL A. Empirical Design Criteria Check

1.

.-.

Seismic Zone 2 < Zone 3

OK

However, minimum reinforcement is required for Seismic Zone 2. 2.

Design WindPressure = 25 psf

3.

Maximum Height = 35 ft

S

S

25 psf mm. permitted .-. OK

35 ft mm. permitted .-. OK

B. ShearWall Design Try 8 in. fully grouted concrete masonry units, unit strength = 4,500 psi 15-28


9.1.1.1 A.3.8

9.l.1.2 9.2

A C 1 TITLE*HDG

m

93

Ob62949 0509328 310

m


Reference

Calculations

Code

8 in.

1.

Minimumthickness shear wall

2.

Minimum total shear walllength

9.3.1.1

Shear Walls: NortWSouth direction. Minimum cumulative length required = 0.4 x long dimension. Min. length

= 150 ft x 0.4 = 60 ft required

= 9 x 30 ft = 270 ft

Actual length

Note: stairwells and elevator shafts

> 60 ft :. OK

9.3.1.2

may be included, but the design is adequate

without them. 3.

Maximumspacingof

shear walls:

:. OK

ratio = 30 ft/30 ft = 1.0 < 4

9.3.1.3

with precast concrete slab diaphragms C.Compressivestress:typicalinteriorwall axial compressive stress,

fa =

P - < Fa A*

9.4.2

F, = 225 psi for fully grouted units, Type M or S Mortar, Code Table 9.4.2

wall wt.

=

140 pcf x

7.63 in. x 1 ft x 1 ft

=

89.0

12 in./ft

Check stresses at the base of the wall. Loading Case: Reduced Live Load

+ Dead Load (ASCE7-88)

Dwelling unit live load reduction: factor = 0.25

+

factor = 0.25

+

15 -

fi

15

= 0.441

46 x 30 x 34.33 15-29


A C 1 T I T L E v M D G 93

Obb2949 0509329 257


CaIcuIations

Discussion

Load Live Roof

Reference

Code

= 30 ft x 20 psf =

Roof Dead Load = 30psf ft x 95

plf 600 plf2,850

=

Roof Load over Corridor: Concentrated Load that spreads through wall LL = 4.0 ft x 30 ft X 20 psf =

lb2,400

DL = 4.0 ft x 30 ft x 95 psf =

11,400 lb

2nd, 3rd,

4th Floor L.L. = 30 ft (40 psf) (0.441) (3) =

1,590 plf

2nd, 3rd,

4th Floor D.L. = 30 ft (110 (3) psf)

9,900 plf

2nd, 3rd, 4th Floor Corridor:

LL. (30ft= 4 D.L. Wall

Concentrated Load that spreads through wall

ft) (100(0.441) psf)

D.L. (30 ft = 4

=

15,900 lb

(3) =

ft) (110 psf) (3) =

39,600 lb

= 35 ft (89.0 psf) =

plf 3,115 Total = 18,055 plf

+ 69,300 lb

The concentrated loads fromthe beams supportingthe roof and floors over the corridors may be spread over distance a thickness,say 12 in.

+ 4 (8 in.)

= bearing plus 4

5.12.1

= 44 in. (spreads over both legs of

flanged wall).

18,055 plf

lb fa = 12(7.63 in.) 44 69y300 h(7.63 in.) +

=

197 + 207

=

404 psi

225 psk N.G.

Reevaluate using nominal 12 in. fully grouted wall. =

19,700 plf 12 (11.63in.)

+

69y300lb = 141 + 99 60 in. (11.63in.)

15-30


=

240 psi > 225 psi: N.G.

A C 1 TITLE*:MDG 93

m 0662747

0507330T77

m


Calculations

The reader should comparethe allowable compressive stresses within MDG Example 13.2-4 to this example to evaluate the alternate design method. D. Lateral Support Design

1.

8 ft - 10 in. floor to floor dimension, 8 ft - O in. clear, assume lateral brace at each floor with floor diaphragm action. 9.5.1

2. TYPICAL EXTERIOR LOADBEARING WALL

A. Empirical Design Criteria Check 1.

Seismic Zone 2 < Zone 3 :.

OK

However, minimum reinforcement 2.

Design Wind

Pressure = 25 psf Height = 35 ft

Maximum 3.

S

9.1.1.1

is required for Seismic Zone 2. 25 psf :. OK

S

35 ft max. permitted

k3.8 9.1.1.2

:.

OK

9.2

B. Shear Wall Design Try 8 in. fully grouted concrete masonry units.Per Code k3.5 brick outside wythe is not used to resist shear loads. See Calculations under item l.B above C.Compressivestress:typicalexteriorwall axial compress stress,

fu

=

P -

S

Fu

9.4.2

A, wall wt.

=

140 pcf x 7.63 in. x 1 ft x 1 fi

15-31


=

89.0 psf


m

0662949 0509333 905

m


Calculations

+ Dead Load

Loading Case: Reduced Live Load Check stress at base of wall Dwelling unit live load reduction: 15 factor = 0.25 + -

J;I; factor

15

= 0.25 +

= 0.520

46 x 15 x 34.33

Load Live Roof

=

15psf ft x 20

ft x 95 Roof Dead Load = 15psf

plf 300

= =

plf1,425

=

lb 1,200

Roof Load Over Corridor

LL. = 4.0 ft (15psf) ft) (20

D.L. = 4.0 ft (15 ft) (95 psfj =

5,700 lb

2nd, 3rd, 4th Floor

L.L. = 15(40 psf) ft

Floor 2nd, 4th 3rd,

D.L. = 15 ft (110 (3) psf)

(0.520) (3) =

D.L. = 4.0 ft (15 ft) (110 (3)psf)

4,950 plf

=

2nd, 3rd, 4th Floor Corridor: Concentrated Load

LL. = 4.0 ft (15 ft) (100 psf) (0.520) (3)

936 plf

that spreads through wall. 9,360 lb

=

19,800 lb

=

Wall D.L. = 35 ft (89.0 psf) =

plf 3,115 Total = 10,726 plf

+ 36,060 lb

The concentrated loads from the beams supporting the roof and floors over the corridors maybe spread over a distance

15-32


= bearing plus 4

5.12.1

1


Obb2949 0509332 841

93

Examele 15.43 Cont’d. Code Reference


(thickness), say 12 in.

+ (4) (8 in.) = 44 in. (spreads out over bothlegs

of flanged wall. fa

=

10,726 pl€ 12(7.63 in.)

+

369060 lb = 224.7 psi 44 in. (7.63 in.)

225 psi:

: ,

OK

9.4

Use fully grouted 2,500 psi unit, Type M or S Mortar

D. Lateral Support Design See item 1.D above 3. EXTERIOR NONLOADBEARING WALL, COLUMN LINE 1AND 4

Between B-C, C-D, D-E, and E-F A. Empirical Design Criteria Check 1.

.: OK

Seismic Zone 2 < Zone 3

However, minimum reinforcement is required for 2.

Design Wind Pressure

3.

Maximum Height = 35 ft

=

25 psf S

9.1.1.1

Seismic Zone 2

k3.8

:. OK

9.1.1.2

25 psf m a . permitted

S

35 ft max. permitted

:. OK

9.2

B. Shear Wall Design Try 8 in. solid, concrete masonry units. Per Code k 3 . 5 , brick outside wythe is not used to resist shear loads.

1.

Minimum shear wall thickness 8 in.

2.

Minimumtotal shear wall length: Shear Walls to resist an east or west wind. minimum cumulative length = 0.4 x long dimension min. length = 150 ft x 0.4

=

60 ft 15-33


9.3.1.1


m

Obb29Y9 0509333 7 8 8 D


Calculations

actual length (see modified framing plan)

=8ftx8x2

= 128 ft > 60 ft

3.

=.OK

9.3.1.2

Minimum wall length by shear wall ratio:

2 wall l e n a spacing

64 ft 67.67 ft

= 0.95 > 0.25

.: OK

9.3.1.3

C. Compressive Stress: typical exterior or interior nonloadbearing wall axial compressive stress, f,

=

P - < F,

9.4.2

A,

estimate average wall wgt. = 89 psf Loading Case: Wall Weight only Wall D. L. = 35 ft x 89 psf = 3,115 plf fa

3,115 lb = 12 in. x 7.63 in.

=

34.0 psi

160 psi

.-. OK

The low axial compressive stress can be adequately supportedby many types of concrete masonry units and type of mortar. The same masonry unit and mortar as theadjoining wall described above is recommended, for ease of construction.

D. Lateral Support Design See item 1.D above

MINIMUM REINFORCEMENT Use single wythe 8 in. fully grouted concrete masonry units with special strengths noted 15-34


AC1

TITLE*NDG 93 W 0 6 6 2 9 4 9 0509334 bL4 W

Example 15.4-3 Cont’d. i

Code Reference


above, ASTM C 90 with ASTM C 270, Type M or S Mortar. The Code Appendix k3.8 requires minimum steel reinforcement for structures inSeismic Zone 2. General steel reinforcement requirements are as follows:

1. Vertical Reinforcement (from support to support) a. at each wall corner, b. at each side of each opening, c. at ends of walls,

2. Horizontal Reinforcement, a. at bottom and top of wall openings,

b. at structurally connected roof and floor levels, c. bottom of walls, see Code Appendix A 3 d. uniformly distributed inwalls.

The figure on the next page shows typical reinforcement locations. Author’s Notes: Code Appendix A.3.6 and A.3.7 specify additional connection requirements for the anchorage of masonry walls to floor and roof diaphragms and at tops of columns. Many architectural design details are required for a properly functioning masonry structure. Required masonry joints, including control joints and expansion joints, are described in MDG Chapter 10. The cavity must be adequately flashed, weeped and vented. See MDG Chapter 6. Another critical detail location is the top of the wall, where the coping must be capable of withstanding differential movement of the brick and block wythes. See MDG Chapter 10.

15-35



m

O662949 0509335 550

m


Calculations

Code Reference

?

QQ

'Typical Floor Thickness

Vertical #4 Reinforcement At: 1. End Of Each Wall, 2. Each Side Of Each Opening.

Horizontal # 4 Reinforcement At: 1. Bottom And Top Of Wall Openings, Extend 24" Or 40 d b Past Opening 2 . Roof And Floor Levels And Top Of Wall 3. And A t Bottom Of Wall.

SOUTHELEVATION Typical Reinforcement Locations

15-36


T IA T LCE1 v M D G

I

93 W 0662949 0509336 497 W

16 PROVISIONS FOR SEISMIC DESIGN

16.1 INTRODUCTION Masonry buildings, likethose of other buildings, respond to theseismic inertial forcescaused by ground shaking.Soil

characteristics and many other factors will influence the actual

ground motion to which a building will be subjected in an earthquake. Because the resisting forces are caused by an inertial response, thebuilding’s distribution of mass, stiffnesses, and strength are important characteristics which must be considered in assigning design forces to be carried to the resisting elements of the building. Unlike designing for wind forces, in which all structural elements are normally assumed to remain elastic, earthquake-resistant design assumes that structuralelements must retaintheir

integrity even when stressed

beyond the elastic limit. Therefore it is important to detail masonry elements for ductility, that is, the ability to resist cyclic inelastic deformation without significant loss of strength. Code Appendix A provides special provisions to be used in masonry design to improve the ductile performance characteristics of masonry elements consistent with earthquake

I

probabilities in the various seismic zones. This chapter of the MDG is intended to provide a general overview of the additional Code seismic requirements(Appendix A) pertaining to the configuration, materials, element design, and reinforcement detailing of masonry structures. Additional information regarding the seismic design of masonry is available in references such as (16.1.1), (16.1.2),and (16.1.3).

16-1



93

m

Obb29Ll9 0509337 323

An important part of seismic design is the conceptual planning of the building to determine

the proper configurationandlayout discontinuity,and

of shear walls.If

possible,largeplanasymmetry,

abrupt changeinbuildingstiffnessshouldbeavoided.Collapse

of

buildings in earthquakes throughout the world has been attributed to designs having these irregularities. Conceptual designis followed by the lateral force determinationand distribution andlateral force design. The inelastic deformation capacityof lateral force resistingelements (masonry shear walls) should beadequate. Design provisions and reinforcernent details given in Code Appendix Aare intended to provide the building with a lateral-force resisting system which should retain asubstantialportion

of itsstrengtheven

if it is subjected to reversed

displacements into the inelastic range. Code Appendix A 2 does not require special provisions for Seismic Zones O and 1 because of the low inelastic deformation demands on building structures in these zones. Special Code provisions(k3) are given for Zone 2, with additional Code provisions(A.4) for Zones 3 and 4.

Under severeseismicloads,masonryelementscanbesubjected

to reversedinelastic

deformations and high axial, flexural, and shear forces. Masonry elements in Zones 3 and 4 should be designed to withstand these conditions without significant loss of strength or

stiffness. 16.2 MATERIALS

Code Appendix k 4 . 4 does not allow the use of Type

N mortar nor masonry cement in

Seismic Zones 3 and 4. Grout must properly surround all the reinforcement and fill all the voids in the wall.

16-2


A C 1 T I T L E * f l D G 93

m O662949

0509338 2bT

m

163 DESIGN OF MASONRYELEMENTS

In Seismic Zone 2, Code Appendix A.3.1 permits the tensile strength of masonry to be considered; as such, masonry elements can be designed according to Code Chapter 6. In Seismic Zones 3 and 4, Code Appendix A.4.2 does not permit consideration of masonry tensile strength; therefore, reinforcement must be provided to carry all tensile forces, and masonry elements shall be designed according to Code Chapter 7. For the shear design of reinforced masonry shear walls in Seismic Zones 3 and 4, Code Appendix A.4.9.1 requires that thecalculated seismic shear force be increased by 50 percent. This provisionis intendedto

prevent brittle shear failure and assure desirable ductile

behavior. When compression stresses exceed 20 percent of the specified compressive strength, good designpractice utilizes boundary elements such as returns andflanges, and/or confinement for the compressive stress block toes of shear walls. Code Appendix A.3.4 requires that all masonry walls and columns which are not considered part of the lateral force resisting system be designed for vertical load and induced moment corresponding to 2.25 times the drift resulting from Code seismic forces. An example is a flexural wall that is perpendicular to the direction of the seismic shear force (Fig. 16.3-1). This wall will experience out-of-plane drift from the displacement of the diaphragm, and should be designed to resist internal shear and moments corresponding to 2.25 times the Code-level drift. This induced displacement would result in out-of-plane wall shears and moments, also influenced by P-A effects and boundary conditions (Fig. 16.3-1). If the building has more than one level, relative displacements of the levels at the top and bottom of this wall should be considered.

16-3



93

m

0 6 6 2 9 4 9 0509339 1 T b

r""""""""-

!

T

m

A

2.25 A

Applied Force Moment

Applied Moment Force

Wall A-Fixed At Top

Wall A-Free To Rotate At Top

Fig. 163-1 Effect of Out of Plane Building Drift on Flexural Walls

16.4 DETAILING 16.4.1 Reinforcement

In Seismic Zone 2, Code Appendix k 3 . 8 specifies that minimum vertical and horizontal reinforcement of 0.2 in.2 be provided at key locations, as shownin Fig. 16.4-1. For walls laid in other than running bond, Code Appendix k 3 . 9 and k 4 . 7 require minimum horizontal reinforcement of 0.0007 and 0.0015 times the gross cross-sectional area of the wallfor Seismic Zone 2 and for Zones 3 and 4, respectively. This can be satisfied either by joint reinforcement or by bond beams as shown in Fig. 16.4-2.

16-4


A C 1 TITLE*NDG 93

m 0662947 0507340

918

,"- On Each Side Of Opening

At TOP

T-T

At Top & Bottom Of Opening

- At Floor

Or Roof Level

I

T-.-~""-T""T"T""""

T-.-7""-7"""

r-t-t-

L I

I

l

" " "

I

I

"t+I"r""n

I

fi

I

A t Wall Ends

RC Frame A t Wall Perimeter

Infill Masonry Wall

Around

Opening

Fig. 16.4-1 Minimum Reinforcement Requirements (Seismic Zone

r

2)

h

A, = 0.0007 In Open Ended Bondbeams Otherwise A, = 0.0015 4

t

I ""_

i

1

" " " " " " "

Figure 16.4-2 Minimum

Horizontal Reinforcement in Walls Laid in Other Than Running Bond (Seismic Zone 2)

16-5


A C 1 TITLExMDG 9 3 M O b b 2 9 4 90 5 0 9 3 4 38 5 4

M

Additional reinforcement requirementsare specified for masonry walls in Seismic Zones 3 and 4. In accordance with Code Appendix k4.5, all walls must be reinforced with both having a minimum area of 0.002 timesthe gross

vertical and horizontal reinforcement

cross-sectional area of the wall. A minimum area of reinforcement equal to 0.0007 times the gross cross sectional area must be provided ineach direction, at a maximum spacing of 4 ft. Fig. 16.4-3 shows minimum reinforcement requirements for masonry walls in Seismic

Zones 3 and 4. Horizontal reinforcement must be adequately anchored around vertical reinforcement. Code Appendix k 4 . 6 specifies minimum requirementsfor column ties which are shown in Fig. 16.4-4. Lateral tiesshould be embedded in grout rather than in mortar joints for columns in Seismic Zones 3 and 4, per Code Appendix k4.6.1.

4'

max I

I

Horizontal Steel

-> 0.0007Ag

Adequate Anchorage

Vertical Steel >

0.0007Ag

Fig. 16.4-3 - Minimum Wall Reinforcement (Seismic Zones 3 and 4)

16-6


~

~~

A C 1 T I T L E + M D G 9 3 9 0662949 0509342 7 9 0 9

I

i!/i!i t"t t--t C"+ Columns Reslstlng Overturning Forces

Ties 8" OS!. Over Full Height

Ties 8" O C . 116 Col. Ht.Or 18" Or Maximum Column Side Dimension Whichever Is Greater

-1

Ties @ 16 Bar Diameters. - 48 Tie Diameters, The Least Column Ihmension Or 18" O.C.

It?

-+--r -I"--c -I--+

All Columns Other

Fig. 16.4-4 - Tie Requirements for Masonry Columns (Seismic a n e s 3 and 4)

16.4.2 Anchorage Code Appendix k 3 . 6 requires that all masonry walls be anchored to floors and roofs, with connections capable of resisting the horizontal forces required in Code 5.2 or a minimum of 200 lb per linear foot of the wall, whichever is greater. Anchors must be embedded in reinforced bond beams or reinforced vertical cells, as shown in Fig. 16.4-5.

.

16.43 MinimumDimensions

The nominal minimum wall thicknessrequired in SeismicZones 3 and 4 per CodeAppendix

k4.8.1 is 6 in., and the least minimum dimension of a masonry column per Code Appendix k4.8.2 is 12 in. There is an exception for hollowclaymasonry requirements are met (Code Appendix k4.8.1).

16-7


provided that specified

A CT 1I T L E M M D G

93

Obb2949 0 5 0 9 3 4 3 b 2 7

m

Fig. 16.4-5 - Anchorage of Masonry Walls to Roofs and Floors

REFERENCES

16.1.1

Schneider, R.R., W.L. Dickey, ReinforcedMasonryDesign,PrenticeHall,

Inc.,

Englewood Cliffs, New Jersey, 1987. 16.1.2Englekirk,

K.,Hart, G.C., Earthquake Design Of ConcreteMasonrv

Buildings,

Prentice Hall, Inc., Englewood Cliffs, New Jersey, 1984. 16.1.3

Pauley, T.and Priestley, M. Seismic Design Of Reinforced Concrete And Masonrv Buildings, Prentice Hall, Inc., Englewood Cliffs, New Jersey, 1992.

16-8


A C 1 TITLExMDG 93

m

Obb29490509344563

m

A APPENDIX

CLAY MASONRY SECTION PROPERTIES Table 1 Minimum Thickness* of Face-Shkll and Webs

- ASTM C 652 Units 7

Nominal Width (W) of Unit, in

Face Shell Thickness Interior (FST) min, in.

Web Thickness (WT) min, in.

End Web Thickness min, in

4

314

1/2

314

6

314

112

1

8

314

1/2

1

10

314

1/2

1 118

12

314

1/2

1 118

*Most manufacturers produce units which have greater face-shell thicknesses than the minimum values listed in Table 1.

Table 2 Section Properties* - Hollow Clay Unit Walls - ASTM C 652 Units Mortar Bedding

Solid or Fully Grouted

Property per ft of Wall Full Face-Shell

li

Actual Unit Thickness = 3 V2"

4" Walls I

kea

I A , in.2 I

18.0

1

22.0

1

42.0

42.934.0

35.7

S, in.3

19.4

20.4

24.5

e&, in.

1.07

0.93

0.58

1.27 r, in.

1.37

Moment of Inertia

I , in!

Section Modulus Kern Eccentricity Radius of Gyration 1.01

I

A-1


Table 2 Cont'd. Section

-

Properties Hollow Clay Unit Walls

- ASTM C 652 Units

Mortar Bedding


Property per ft of Wall

Full

Face-Shell

Actual Unit Thickness = 5 IJ2"

6" Walls

Area

A , in.*66.0

24.0

32.8

Moment of Inertia

I, in.4 166

122

132

Section Modulus

S,in.3

44.2

48.1

Kern Eccentricity

eb in. 0.92 *

1.84

1.47

r, in. 1.59

2.25

2.01

Radius of Gyration

I

li

60.5

Actual Unit Thickness = 7 42"

S"walls I

I

kea

A , in.2

30.0

Moment of Inertia

I, in.4

293

42.5 323

Section Modulus

S,in.3

78.1

86.1

Kern Eccentricity

e,, in.1.25

2.6

2.03

Radius of Gyration

r, in. 2.17

3.13

2.75

L

114

Moment of Inertia

51.6A , i n . 2

33.0

I, in?

857 544

Section Modulus

180.5

131 S,i n . 3

115

Kern Eccentricity

1.58

2.54e,, in.

3.47

3.47 r, in.

4.06

Radius of Gyration 2.75

Area Moment of Inertia Section Modulus 265

Radius of Gyration

113

621

Unit Thickness = 11 U2"

Actual 12"Wall

Kern Eccentricity

90.0 422

= 9 1/2"

Thickness Unit Actual 10" Walls

Area

I

1.92

59.4 A , in.2

36.0

I , in.4

1,520 900

182 S,in.3

157

3.06e,, in.

4.36

r, in.

3.32 5.00

138 1,048

4.20

*Table 2 values are based on the minimum thicknesses given in Table 1.

A-2


A C 1 T I T L E S N D G 9 3 D Obb2949 0509346 336

Table 3 Average Wall Weights

I

Nominal Width of Unit, in

- Hollow Clay Units With Grout

I None

4'

- O''

3'

2'

O"

O''

1'

- 6"

24.1 22.3 21.3 18.6

6"

27.6

32.1

8

35.7

42.0

44.1

49.7

- O"

Fully

29.7 26.0 40.7

39.5

45.5

63.3

48.3

52.5

61.0

85.8

76.3 65.3 107.8 59.8 54.3 70.4 63.5

60.0

A-3


1'

36.5 33.5

51.5 43.3

10" 12"

(psf)

Grout Spacing

4"

"

m

77.3

91.2

130.8

.5


m

m

Ob62747 0507347 2 7 2

CONCRETE MASONRY SECTION PROPERTIES Table 4 Minimum Thickness of Face - Shell and Webs

-

Nominal Width (W) of Units, in.

- ASTM C 90

Face-Shell Thickness Thickness Web (FST) Min., in.

min., in.

4

314

314

6

1

1

8

1 114

1

10

1 318

1 118

12

1 112

1 118

-

Table 5 Average Weight of Concrete Masonry Units (lb) ASTM C 90 Nominal Size

22.0

Density (pcf) 80

20.0 4 x18.0 8 ~ 116.5 6

14.5

23.5 6 21.5 x 8 ~ 119.0 6

17.0

8 x 8 ~ 39.0 16

90 110

36.0 225 33.525.030.5

100

120

130

.

140

25.5 25.5

27.5

30.0

50.5

54.5

28.0

48.0 1044.5 x 841.0 x 37.5 1634.531.027.5 12 x 8 47.0 x 16 43.0 39.0 35.0 31.0

A-4


A C 1 TITLExflDG 93

m

Ob62949 0509348 L O 9 D

Table 6 Average Weight of Ungrouted Completed Walls, plf7ft of height

Table 7 Section Properties

- CMU Walls - ASTM C 90 Hollow Units Mortar Bedding

I

Shell Face I


Full

I

I

Unit Designation 4 x 8 x 16 Average Unit Dimensions: Thickness = 3.63",Length = 15.63", Height = 7.63"

3.5

Area21.6

A , in.2

18.0

47.6

39.4 Moment of Inertia

I , in.4

38.0

6.3

21.7 Section Modulus

S, in.3

21.0

Kern Eccentricity

ek, in.

1.17

Radius 1.05of Gyration

1.35 r, in.

1.45

6 in. Walls

67.5

3.3

Area32.2

i

1.00

0.0

Unit Designation 6 x 8 x 16 Actual Unit Dimensions: Thickness = 5.63",Length = 15.63", Height = 7.63"

A, in.2

24.0

139 Moment of Inertia

I, i n . 4

130

49.5 Section Modulus

S, in.'

46.3

1.54 Kern Eccentricity

et, in.

1.93

Radius of Gyration

r, in.

2.33

0.94

208

A-5


1.62

9 3 M Obb2949 0509349 045 M


-

Table 7 Cont'd. Section Properties CMU Walls - ASTM C 90 Hollow Units Mortar Bedding

perProperty ft of Wall 8 in. Walls

Area

Full

Unit Designation 8 x 8 x 16 Average Unit Dimensions: Thickness = 7.63', Length = 15.63", Height = 7.63'

41.5

A , in.2

30.0

Moment of Inertia 334

I, in.'

309

116

Section Modulus 87.6

S) in.3

81.0

1.27

Kern Eccentricity 2.11

e,

2.20


91.5

43

10 in. Walls

I


I

Shell Face

in.

2.70

r, in.

3.21

Unit Designation 10x 8 x 16 Actual Unit Dimensions: Thickness = 9.63", Length = 15.63", Height = 7.63"

Area

A, in.*

33.0

2

Moment of Inertia 635

I ) in.'

567

5

Section Modulus 132

S) in.3

118

Kern Eccentricity

e,, in. 1.60

3.57

2.62

Radius of Gyration

r, in. 2.78

4.14

3.55

12 in. Walls

Area 1,571

0 1.94

36.0 140

I ) in.'

929

Section Modulus 183

S) in.3

160

Kern Eccentricity 3.17

e,)

in.

4.44

r, in.

5.08


116

Unit Designation 12x 8 x 16 Actual Unit Dimensions: Thickness = 11.63", Length = 15.63", Height = 7.63"

A, i n e 2

Moment of Inertia 1,065

50.4

3.36

A-6


57.8

A C 1 TITLE*HDG 93

0662949 0509350 867 H

Table 8 SI Units Conversion Factors'

I

To convert from

kilometer

to'

inch foot

Length millimeter (mm) meter (m)

Yard

meter (m)

mile (statute)

I

multiply by2

25.4E 0.3048E 0.9144E 1.609

(km) Area

foot

square inch square square yard

centimeter square meter

(cm2)

6.452

(m2) (m2)

square

0.09290

0.8361

Force

kilogram-force kip-force pound-force (N)

newton (N) kilonewton (kN) newton Pressure or Stress (Force perArea)

kilogram.force/square meter (Pa) (MPa) kip-forcehquare inch (ksi) newtonhquare meter (Pa) (N/m*) pound.force/square foot (Pa) pound.force/square (Pa) inch (psi)

pascal megapascal pascal pascal pascal

9.807 4.448 4.448 9.807 6.895 1. W E 47.88 6895

Bending Moment or Torque

inchpound-force foot-pound-force meter-kilogram.force

(N-m)

newton-meter (Nem) newton-meter newtonmmeter (Nm)

0.1130 1.356 9.807

Mass

ounce-mass (avoirdupois) pound-mass (avoirdupois) ton (metric) ton (short, 2000 lbm) pound.mass/cubic foot pound.mass/cubic yard pound.mass/gallon

gram (g) kilogram (kg) megagram (Mg) megagram (Mg) Mass per Volume kilogram/cubic meter (kg/m3) kilogram/cubic meter (kg/m3) kilogram/cubic meter (kg/m3)

28.35 0.4536 1.OOOE 0.9072 16.02 0.5933 119.8

Temperature

deg Farenheit (F) deg Celsius (C)

deg Celsius (C) deg Farenheit (F)

k = (fF - 32)/1.8 tF

= 1.8k

+ 32

This list gives practical conversion factors of units found in masonry technology. The reference source for information on SI units and more conversion factors is "Standard for Metric Practice" ASTM E 380. Symbols of metric units are given in parentheses. * E indicates that the factor given is exact.

A-7


Masonry Designer's Guide.pdf

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