UNISA Reinforced Concrete Design Study Guide

Reinforced Concrete Design IV Prepared by:

Greg Parrott Pr Tech (Eng) M Dip Tech (TN) T MSAICE

INTRODUCTION This text covers the design of reinforced concrete structures in accordance with SABS 0100-1:2000. The level of study is National Qualifications Framework (NQF) level eight, and covers the entire syllabus for the subject “Reinforced Concrete Design IV” as offered by the Department of Civil Engineering at the University of South Africa (UNISA) for students studying towards the Bachelor’s Degree in Technology in the specialised discipline of Structural Engineering. It is assumed that the t he reader has already passed the subject “Reinforced Concrete and Masonry Design III” which is offered as one of the subjects in the National Diploma in Civil Engineering offered by universities of technology (previously known known as technikons) in South Africa.

INTRODUCTION This text covers the design of reinforced concrete structures in accordance with SABS 0100-1:2000. The level of study is National Qualifications Framework (NQF) level eight, and covers the entire syllabus for the subject “Reinforced Concrete Design IV” as offered by the Department of Civil Engineering at the University of South Africa (UNISA) for students studying towards the Bachelor’s Degree in Technology in the specialised discipline of Structural Engineering. It is assumed that the t he reader has already passed the subject “Reinforced Concrete and Masonry Design III” which is offered as one of the subjects in the National Diploma in Civil Engineering offered by universities of technology (previously known known as technikons) in South Africa.

MODULE

TOPIC

1

I n tr o d u c ti o n

2

A n a l y s i s o f i n d e te r m i n a te s tr u c tu r e s

3

Beam s

4

S la b s

5

S ta i r s

6

C o lu m n s

7

F o u n d a ti o n s

8

R e ta in i n g w a l ls

9

S ilo s

10

W a te r - r e ta i n i n g s tr u c tu r e s

TABLE OF CONTENTS MODULE 1

INTRODUCTION

1.1

SYMBOLS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1

1.2 1.2.1 1.2.2 1.2.3 1.2.3.1 1.2.3.2 1.2.3.3

MATERIALS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Concrete . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Reinforcing steel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Cover . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Durability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Fire resistance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Practical considerations . . . . . . . . . . . . . . . . . . . . . . . .

1.3 1.3.1 1.3.1.1 1.3.2 1.3.2.1 1.3.2.2 1.3.3 1.3.3.1 1.3.3.2 1.3.3.3 1.3.4 1.3.5

LOADS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.9 Dead loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.9 1.9 Unit weight . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1. 1 .9 Live loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.10 Floor loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1. 1.10 Roof loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1. 1 .11 Loading patterns . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 1.11 Uniformly distributed . . . . . . . . . . . . . . . . . . . . . . . . . . 1. 1.11 Point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1. 1.12 Triangular and trapezoidal . . . . . . . . . . . . . . . . . . . . . . 1. 1.12 Unit conversions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 1.13 Continuity effe ffects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.13

1.4 1.4.1 1.4.1.1 1.4.1.2 1.4.2 1.4.3 1.4.3.1 1.4.3.2

LIMIT STATES DESIGN . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Limit states approach . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Ultimate limit state . . . . . . . . . . . . . . . . . . . . . . . . . . . . Serviceability limit state . . . . . . . . . . . . . . . . . . . . . . . . Limit state equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Partial factors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Load factors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Resistance factors . . . . . . . . . . . . . . . . . . . . . . . . . . . .

MODULE 2

1.5 1.6 1.6 1.6 1.6 1.7 1. 1 .7 1. 1.8 1. 1.8

1.14 1.1 1.14 1. 1.14 1. 1.15 1.15 1.15 1. 1.16 1. 1.17

ANALYSIS OF BEAMS AND FRAMES

2.1

ARRANGEMENT OF LOADS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1

2.2 2.2.1 2.2.1.1 2.2.2 2.2.2.1 2.2.3 2.2.3.1

CONTINUOUS BEAMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Moment distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 2.2 Example 2.1 (moment distribution) distribution) . . . . . . . . . . . . . . . . 2.4 Redistribution of moment . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6 2.6 Example 2.2 (redistribution) redistribution) . . . . . . . . . . . . . . . . . . . . . 2.7 Analysis coeffi fficients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 2.10 Example 2.3 (analysis coefficients) coefficients) . . . . . . . . . . . . . . . 2.11 i

2.3 2.3.1 2.3.1.1 2.3.1.2 2.3.1.3

RIGID FRAMES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Simplification into sub-frames . . . . . . . . . . . . . . . . . . . . . . . Vertical load only . . . . . . . . . . . . . . . . . . . . . . . . . . . . Vertical and lateral loads . . . . . . . . . . . . . . . . . . . . . . Example 2.4 (sub-frame analysis) . . . . . . . . . . . . . . .

MODULE 3

2.13 2.13 2.13 2.14 2.15

BEAMS

3.1 3.1.1 3.1.1.1 3.1.1.2 3.1.2 3.1.2.1 3.1.3 3.1.3.1 3.1.3.2 3.1.3.3 3.1.4 3.1.4.1 3.1.4.2 3.1.4.3 3.1.4.4 3.1.4.5 3.1.4.6 3.1.4.7 3.1.4.8 3.1.5 3.1.5.1 3.1.5.2 3.1.6

FLEXURE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 Rectangular beams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 Example 3.1 (moment of resistance) . . . . . . . . . . . . . . 3.7 Example 3.2 (area of tension reinforcement ) . . . . . . . 3.10 Rectangular beams doubly reinforced . . . . . . . . . . . . . . . . 3.11 Example 3.3 (area of compression reinforcement ) . . . 3.14 Flanged beams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.16 Effective flange width . . . . . . . . . . . . . . . . . . . . . . . . . 3.16 Flexural strength . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.17 Example 3.4 (area of tension reinforcement ) . . . . . . . 3.19 Nominal reinforcement and spacing of bars . . . . . . . . . . . . 3.20 Minimum area of tension reinforcement . . . . . . . . . . . 3.21 Minimum area of compression reinforcement . . . . . . 3.22 Maximum area of reinforcement . . . . . . . . . . . . . . . . 3.22 Additional reinforcement in deep beams . . . . . . . . . . 3.22 Minimum reinforcement in flange . . . . . . . . . . . . . . . . 3.23 Minimum spacing of all bars . . . . . . . . . . . . . . . . . . . . 3.23 Maximum spacing of bars in tension . . . . . . . . . . . . . 3.25 Choice of reinforcing bars . . . . . . . . . . . . . . . . . . . . . 3.26 Curtailment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.27 Anchorage length . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.28 Simplified rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.29 Design flowchart for flexure . . . . . . . . . . . . . . . . . . . . . . . . 3.31

3.2 3.2.1 3.2.2 3.2.2.1 3.2.2.2 3.2.3 3.2.3.1 3.2.3.2 3.2.3.3 3.2.4 3.2.4.1 3.2.4.2 3.2.4.3 3.2.4.4 3.2.4.5 3.2.5 3.2.6

SHEAR . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Sections un-reinforced for shear . . . . . . . . . . . . . . . . . . . . Sections reinforced for shear . . . . . . . . . . . . . . . . . . . . . . . Stirrups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Inclined bars . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Nominal Reinforcement and spacing of bars . . . . . . . . . . . Minimum area of stirrups . . . . . . . . . . . . . . . . . . . . . . Maximum spacing of stirrups . . . . . . . . . . . . . . . . . . . Stirrups for compression reinforcement . . . . . . . . . . . Other considerations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Concentrated load close to support . . . . . . . . . . . . . . Practical fixing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Anchorage of stirrups . . . . . . . . . . . . . . . . . . . . . . . . . Curtailment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Local bond . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Design flowchart for shear . . . . . . . . . . . . . . . . . . . . . . . . . Example 3.5 (shear reinforcement ) . . . . . . . . . . . . . . . . . .

ii

3.33 3.33 3.34 3.35 3.36 3.37 3.37 3.37 3.37 3.38 3.38 3.38 3.39 3.39 3.39 3.40 3.40

3.3 3.3.1 3.3.2 3.3.2.1 3.3.2.2

TORSION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Torsional shear stress . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Reinforcement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Links . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Longitudinal bars . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

3.4 3.4.1

COMBINED SHEAR AND TORSION . . . . . . . . . . . . . . . . . . . . . . . . 3.45 Example 3.6 (combined shear and torsion) . . . . . . . . . . . . . 3.46

3.5 3.5.1 3.5.2 3.5.3

DEFLECTION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Rectangular beams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Flanged beams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Example 3.7 (deflection) . . . . . . . . . . . . . . . . . . . . . . . . . . .

3.6

DESIGN EXAMPLE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.55

MODULE 4

3.43 3.43 3.44 3.44 3.45

3.49 3.49 3.51 3.52

SLABS

4.1

INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1

4.2 4.2.1 4.2.2 4.2.2.1 4.2.2.2 4.2.3 4.2.3.1 4.2.3.2 4.2.3.3

ANALYSIS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 One-way spanning . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3 Two-way spanning . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3 Simply supported unrestrained slabs . . . . . . . . . . . . . . 4.4 Restrained slabs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.5 Flat slabs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.6 General configuration . . . . . . . . . . . . . . . . . . . . . . . . . . 4.6 Division of panels . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.7 Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.8

4.3 4.3.1 4.3.2 4.3.2.1 4.3.2.2 4.3.2.3 4.3.3 4.3.3.1 4.3.3.2 4.3.3.3

DESIGN FOR FLEXURE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Reinforcement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Nominal reinforcement and spacing of bars . . . . . . . . . . . . Minimum area of tension reinforcement . . . . . . . . . . . Minimum spacing of bars . . . . . . . . . . . . . . . . . . . . . . Maximum spacing of bars . . . . . . . . . . . . . . . . . . . . . . Placement and curtailment of bars . . . . . . . . . . . . . . . . . . . One-way spanning slabs . . . . . . . . . . . . . . . . . . . . . . . Two-way spanning slabs . . . . . . . . . . . . . . . . . . . . . . . Flat slabs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

4.11 4.11 4.12 4.12 4.12 4.12 4.12 4.13 4.13 4.14

4.4 4.4.1 4.4.2 4.4.3

DEFLECTION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . One-way spanning slabs . . . . . . . . . . . . . . . . . . . . . . . . . . . Two-way spanning slabs . . . . . . . . . . . . . . . . . . . . . . . . . . . Flat slabs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

4.15 4.15 4.15 4.16

4.5 4.5.1 4.5.2 4.5.2.1 4.5.2.2

DESIGN FOR SHEAR . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Beam shear . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Punching shear . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Point load on solid slabs . . . . . . . . . . . . . . . . . . . . . . . Flat slabs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

4.16 4.16 4.17 4.17 4.19

iii

4.6 4.6.1 4.6.2 4.6.3

DESIGN EXAMPLES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Simply supported slab . . . . . . . . . . . . . . . . . . . . . . . . . . . . Restrained slab . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Flat slab . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

4.19 4.19 4.24 4.27

4.7 4.7.1 4.7.2 4.7.3 4.7.4 4.7.5

OTHER SLAB SYSTEMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Rib and hollow block . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Coffer or waffle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Voided construction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Permanent formwork . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Ribbed beam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

4.35 4.35 4.36 4.37 4.37 4.38

MODULE 5

STAIRS

5.1 5.1.1 5.1.2 5.1.3

LOADING . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Imposed load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Dead load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

5.2 5.2.1

ANALYSIS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Effective span . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2

5.3 5.3.1 5.3.1.1 5.3.2

DESIGN . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Longitudinally spanning . . . . . . . . . . . . . . . . . . . . . . . . . . . . Example 5.1 (longitudinally spanning stair ) . . . . . . . . . Laterally spanning . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

MODULE 6

5.1 5.1 5.1 5.2

5.3 5.3 5.4 5.7

COLUMNS AND ARCHES

6.1

INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1

6.2

BRACING . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1

6.3 6.3.1 6.3.2 6.3.3

EFFECTIVE HEIGHT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . General method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Rigorous method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Nomograph . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

6.4 6.4.1 6.4.2

SLENDERNESS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.5 Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.6 Example 6.1 (slenderness classification) . . . . . . . . . . . . . . . 6.7

6.5

DESIGN AXIAL LOAD . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.8

iv

6.2 6.2 6.3 6.4

6.6 6.6.1 6.6.2 6.6.2.1 6.6.2.2 6.6.2.3 6.6.3 6.6.4

DESIGN MOMENT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.9 Additional Moments in slender columns . . . . . . . . . . . . . . . . 6.9 Uni-axial bending . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.11 Braced slender columns . . . . . . . . . . . . . . . . . . . . . . . 6.11 Unbraced slender columns . . . . . . . . . . . . . . . . . . . . . 6.13 Short columns . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.14 Bi-axial bending . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.14 Example 6.2 (design moment ) . . . . . . . . . . . . . . . . . . . . . . . 6.15

6.7 6.7.1 6.7.1.1 6.7.1.2 6.7.2

LONGITUDINAL REINFORCEMENT . . . . . . . . . . . . . . . . . . . . . . . . Derivation of design equations . . . . . . . . . . . . . . . . . . . . . . Full compression . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Partial compression . . . . . . . . . . . . . . . . . . . . . . . . . . . Design charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

6.17 6.17 6.18 6.19 6.20

6.8 6.8.1 6.8.2 6.8.3

OTHER CONSIDERATIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Minimum area of main reinforcement . . . . . . . . . . . . . . . . . Maximum area of main reinforcement . . . . . . . . . . . . . . . . . Minimum requirement for links . . . . . . . . . . . . . . . . . . . . . . .

6.28 6.28 6.28 6.28

6.9

EXAMPLE 6.3 (reinforcement ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.29

MODULE 7

FOUNDATIONS

7.1

INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1

7.2 7.2.1 7.2.2 7.2.2.1 7.2.3 7.2.3.1 7.2.3.2 7.2.4 7.2.5

ISOLATED BASES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1 Base dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2 Bending . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2 Layout of reinforcement . . . . . . . . . . . . . . . . . . . . . . . . 7.3 Shear . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.3 Beam shear . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.3 Punching shear . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.4 Example 7.1 (centric axial load ) . . . . . . . . . . . . . . . . . . . . . . . 7.4 Example 7.2 (centric axial load and applied moment ) . . . . . . 7.8

7.3 7.3.1 7.3.2 7.3.3 7.3.4

COMBINED BASES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Base dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Bending . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Shear . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Example 7.3 (rectangular combined base) . . . . . . . . . . . . .

7.13 7.13 7.14 7.14 7.14

7.4 7.4.1 7.4.2 7.4.3 7.4.4

STRAPPED BASES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Base dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Bending . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Shear . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Example 7.4 (strapped base) . . . . . . . . . . . . . . . . . . . . . . . .

7.22 7.23 7.23 7.23 7.24

v

MODULE 8

RETAINING WALLS

8.1 8.1.1 8.1.2 8.1.3

TYPES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Gravity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Cantilever . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Counter-fort or buttress . . . . . . . . . . . . . . . . . . . . . . . . . . . .

8.2

COMPONENTS OF A RETAINING WALL . . . . . . . . . . . . . . . . . . . . 8.2

8.3 8.3.1 8.3.2 8.3.3 8.3.4

MODES OF FAILURE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Overturning . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Sliding . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Soil failure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Fracture of elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

8.2 8.2 8.3 8.3 8.3

8.4 8.4.1 8.4.2 8.4.3 8.4.4

SOIL PARAMETERS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Angle of internal friction . . . . . . . . . . . . . . . . . . . . . . . . . . . . Soil density . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Sliding friction coefficient . . . . . . . . . . . . . . . . . . . . . . . . . . . Maximum ground pressure . . . . . . . . . . . . . . . . . . . . . . . . . .

8.4 8.4 8.4 8.4 8.4

8.5 8.5.1 8.5.2 8.5.3 8.5.4 8.5.5

APPLIED PRESSURES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Active soil pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Passive soil pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Surcharge pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Water pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Resulting pressure distribution . . . . . . . . . . . . . . . . . . . . . . .

8.4 8.4 8.5 8.5 8.5 8.6

8.6

DESIGN APPROACH . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.8

8.7

EXAMPLE 8.1 (cantilever retaining wall ) . . . . . . . . . . . . . . . . . . . . . 8.10

MODULE 9

8.1 8.1 8.1 8.1

SILOS

9.1

INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.1

9.2 9.2.1 9.2.2

PARAMETERS REQUIRED FOR DESIGN . . . . . . . . . . . . . . . . . . . . 9.1 Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.1 Properties of retained material . . . . . . . . . . . . . . . . . . . . . . . 9.1

9.3 9.3.1 9.3.2 9.3.3 9.3.3.1 9.3.3.2

ANALYSIS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Horizontal pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Ring tension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Vertical pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Walls . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Floor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

9.4

EXAMPLE 9.1 (silo) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.5

vi

9.2 9.3 9.4 9.4 9.4 9.4

MODULE 10

WATER-RETAINING STRUCTURES

10.1

LOADS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.1

10.2 10.2.1 10.2.2 10.2.3

ANALYSIS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Slabs and floors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Walls of cylindrical tanks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Walls of rectangular tanks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

10.1 10.1 10.1 10.1

10.3 10.3.1 10.3.2 10.3.2.1 10.3.2.2

LIMIT STATE REQUIREMENTS . . . . . . . . . . . . . . . . . . . . . . . . . . . Ultimate limit state . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Serviceability limit state . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Deflection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Cracking . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

10.2 10.2 10.2 10.2 10.2

10.4 10.4.1 10.4.2 10.4.2.1 10.4.2.2 10.4.2.3 10.4.3 10.4.4

CRACK CONTROL . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.3 Direct tension in mature concrete . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.3 Direct tension in immature concrete . . . . . . . . . . . . . . . . . . . . . . . . . 10.4 Minimum reinforcement . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.4 Calculation of crack spacing . . . . . . . . . . . . . . . . . . . . . . . . 10.5 Calculation of crack width . . . . . . . . . . . . . . . . . . . . . . . . . . 10.5 Flexural tension in mature concrete . . . . . . . . . . . . . . . . . . . . . . . . . 10.6 Minimum reinforcement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.10

10.5 10.5.1 10.5.2 10.5.3 10.5.3.1 10.5.3.2 10.5.3.3 10.5.3.4

DESIGN AND DETAILING OF JOINTS . . . . . . . . . . . . . . . . . . . . . Construction joints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Sliding joints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Movement joints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Contraction joint . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Induced contraction joint . . . . . . . . . . . . . . . . . . . . . . . . . . Expansion joint . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Spacing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

10.10 10.10 10.10 10.11 10.11 10.12 10.12 10.12

10.6 10.6.1 10.6.2 10.6.2.1 10.6.2.2 10.6.3 10.6.4 10.6.4.1 10.6.4.2 10.6.4.3

SPECIFICATION OF MATERIALS . . . . . . . . . . . . . . . . . . . . . . . . . Blinding layer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Concrete . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Cement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Aggregate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Formwork . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Joining materials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Joint fillers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Water-stops . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Joint sealing compounds . . . . . . . . . . . . . . . . . . . . . . . . . .

10.14 10.14 10.14 10.14 10.14 10.14 10.14 10.14 10.15 10.15

10.7

DESIGN EXAMPLE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.16

ANNEXURE A

Bending moments in circular tanks . . . . . . . . . . . . . . . . . . 10.20

ANNEXURE B

Ring tension in circular tanks . . . . . . . . . . . . . . . . . . . . . . . 10.24

vii

MODULE 1

INTRODUCTION This section covers the basic introduction to reinforced concrete design and introduces you to the symbols, the materials used, loads that the elements will be subjected to, and the design philosophy adopted.

Reinforced Concrete Design IV

G.K. Parrott

Module 1 - Introduction

CONTENTS Page 1.1

SYMBOLS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1

1.2 MATERIALS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2.1 Concrete . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2.2 Reinforcing steel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2.3 Cover . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2.3.1 Durability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2.3.2 Fire resistance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2.3.3 Practical considerations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1.5 1.6 1.6 1.7 1.7 1.8 1.8

1.3 LOADS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.9 1.3.1 Dead loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.9 1.3.1.1 Unit weight . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.9 1.3.2 Live Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.10 1.3.2.1 Floor loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.10 1.3.2.2 Roof loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.11 1.3.3 Loading patterns . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.11 1.3.3.1 Uniformly distributed . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.11 1.3.3.2 Point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.12 1.3.3.3 Triangular and trapezoidal . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.12 1.3.4 Unit conversions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.13 1.3.5 Continuity effects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.13 1.4 LIMIT STATES DESIGN . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4.1 Limit states approach . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4.1.1 Ultimate limit state . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4.1.2 Serviceability limit state . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4.2 Limit state equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4.3 Partial factors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4.3.1 Load factors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4.3.2 Resistance factors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1.14 1.14 1.14 1.15 1.15 1.15 1.16 1.17


1.1

G.K. Parrott


SYMBOLS

The following list gives the symbols which are commonly used in reinforced concrete design in South Africa and in the notes that follow. area of tension reinforcement in beams and slabs

m m²

area of compression reinforcement in beams and slabs

m m²

area of longitudinal reinforcement in columns

m m²

area of bent-up bars for shear reinforcem ent

m m²

induced deflection in slender colum ns

mm

distance measured from the face of support (for shear enhancement)

mm

sm aller dim ension of rectangular foundation

mm

width of rectangular beam

mm

sm aller dim ension of rectangular colum n

mm

effective width of flange in beams

mm

width of section at the level of tension reinforcem ent

mm

web width of flanged beam

mm

com pression force (used for derivation of equations for flexure) nominal dead load

kN kN, kN/m, kN/m²

distance from compression face to tension reinforcement

mm

distance from compression face to compression reinforcement

mm

average distance from compression face to tension reinforcem ent

mm

Young's modulus

N/m m²

eccentricity of resultant force

mm

m inim um eccentricity of axial load on columns

mm

horizontal force on retaining walls

kN

ultim ate anchorage bond stress for reinforcem ent in concrete

N/mm²

characteristic compressive strength of concrete

N/mm ²

service stress in tension reinforcement

N/m m²

stress in reinforcement in the more highly compressed face of a column

N/m m²

stress in reinforcement in the less highly compressed face of a colum n

N/m m²

characteristic tensile strength of reinforcement

N/mm ²

design stress in compression reinforcem ent

N/mm ²

characteristic tensile strength of shear reinforcem ent

N/m m²

overall depth of beam

mm

larger dimension of rectangular column appropriate depth of backfill behind a retaining wall

mm mm

depth of soil offering passive resistance to sliding of a retaining wall

mm

effective diameter of column head for flat slabs

mm

thickness of beam flange

mm

1.1


G.K. Parrott

second moment of area (m oment of inertia)

Module 1 - Introduction 4

mm

bending stress ratio factor to reduce the additional moment in slender columns limit to bending stress ratio, above which compression reinforcem ent is required stiffness of mem ber active soil pressure coefficient passive soil pressure coefficient larger dimension of rectangular foundation

mm

length of retaining wall base

mm

nominal live load

kN, kN/m, kN/m²

distance between points of zero moment in continuous beams

mm

effective span of beam

mm

width of stair landing

mm

effective column height

mm

average panel dim ension of flat slab

mm

clear column height

mm

smaller effective span of two-way spanning slab

mm

larger effective span of two-way spanning slab

mm

span of flat slab panel in the direction under consideration

mm

span of flat slab panel perpendicular to the direction under consideration

mm

ultimate bending moment

kNm

additional mom ent in slender columns

kNm

initial m om ent near mid-height of colum n

kNm

ultimate moment on short span of two-way spanning slab

kNm

ultimate moment on long span of two-way spanning slab

kNm

ultimate moment of resistance

kNm

smaller end moment on colum n (taken as negative for double curvature)

kNm

larger end mom ent on colum n

kNm

ultimate axial column load design load on two-way spanning slabs

kN kN/m²

factor used to determ ine the active soil pressure coefficient soil pressure

kN/m²

perm issible soil-bearing pressure

kN/m²

m axim um resulting soil pressure below retaining walls

kN/m²

m inim um resulting soil pressure below retaining walls

kN/m²

nominal imposed load

kN, kN/m, kN/m²

non-dominant imposed load

kN, kN/m, kN/m²

dominant im posed load

kN, kN/m, kN/m²

resultant force on retaining wall base support reaction nominal resistance of element (limit state equation) prefix for mild steel reinforcement 1.2

kN kN varies


G.K. Parrott


spacing of reinforcing bars

mm

spacing of shear link reinforcement

mm

tension force (used for derivation of equations for flexure)

kN

total ultimate torsion mom ent

kNm

partial ultimate torsion moment on rectangular segment

kNm

effective perimeter for punching shear

mm

ultim ate shear force

kN

shear resistance of a single system of bent-up bars

kN

effective punching shear force

kN

full ultimate shear force from flat slab panel

kN

ultim ate shear stress

N/mm ²

shear resistance of section unreinforced for shear

N/m m²

limiting ultimate shear stress

N/mm ²

torsional shear stress

N/m m²

torsional resistance of section unreinforced for torsion

N/m m²

limiting ultimate stress to the sum of shear and torsional stresses

N/m m²

vertical force on retaining walls

kN

self-weight of a foundation

kN

nominal wind load

kN, kN/m, kN/m²

depth of concrete section in compression

mm

sm aller centre-to-centre dim ension of a link

mm

prefix for high-tensile reinforcement larger centre-to-centre dimension of a link

mm

lever-arm used to calculate internal moment of resistance

mm

inclination of bent-up bars for shear

degrees

ratio of sum of column stiffness to sum of in-plane beam stiffness coefficient for bending moment in short span of unrestrained slabs coefficient for bending moment in long short span of unrestrained slabs effective length factor for columns inclination of 'com pression strut' between bent-up bars for shear

degrees

factor used to calculate the induced deflection in slender column s ratio of redistributed moment to the elastic moment coefficient for design moment in columns subjected to bi-axial bending coefficient for bending moment in short span of restrained slabs coefficient for bending moment in long short span of restrained slabs ratio of longer to shorter dimension of a rectangu lar foundation unit weight of material

kN/m

3

partial load factor for limit state design partial material factor for limit state design partial load factor for d ead load at the serviceability limit state partial load factor for live load at the serviceability limit state

1.3


G.K. Parrott


partial load factor for dead load at the ultimate limit state partial load factor for live load at the ultimate limit state deflection

mm

deflection

mm

maximum strain in concrete in compression (0.0035) strain in compression reinforcement angle of internal friction in soil (angle of repose)

degrees

coefficient of friction m axim um ultim ate compressive stress in concrete slope of backfill behind a retaining wall

N/mm² degrees

resistance factor (limit state equation) load comb ination factor (limit state equation) ultimate uniform load on beams retaining wall surcharge

1.4

kN/m kN/m²


1.2

G.K. Parrott


MATERIALS

Reinforced concrete is a strong, durable building material that can be formed into almost any shape. Use is made of steel and concrete by combining the best features of each. The table below shows the good and bad properties of each material.

Strength in tension Strength in compression Strength in shear Durability Fire Resistance

CONCRETE

STEEL

POOR GOOD FAIR GOOD GOOD

GOOD GOOD (but buckles if slender) GOOD POOR POOR

Thus when the materials are combined, the reinforcement provides the tensile strength and some shear strength while the concrete, strong in compression, protects the steel to give durability and fire resistance. Nearly all reinforced concrete structures are designed on the assumption that concrete does not resist any tensile force. Reinforcement is designed to carry these tensile forces, which are transferred by the bond between the interface of the two materials.

Referring to the above figure, we can see that the concrete will resist the compressive force, and the reinforcement will resist the tensile force. Cracks will develop in the tension face of the member, but providing that the cracks are small enough to continue offering protection to the reinforcement and the bond is good, there will be no reduction in the strength of the member.

1.5


G.K. Parrott


Steel reinforcement may also be required in areas where the compressive strength exceeds the strength of the concrete. Stirrups are used to prevent the compression reinforcement from buckling and also to provide the required shear reinforcement. A basic cross-section of a reinforced concrete rectangular beam is shown below.

1.2.1

CONCRETE

The concrete strength is dependant on the mix proportions of sand, stone, cement and water. The strength is assessed by measuring the cube crushing strength f cu at 28 days according to standard procedures. Typical concrete strengths : Mass concrete General structures High loadings Prestressed concrete

1.2.2

15 N/mm² 25 N/mm² 30 N/mm² 50 N/mm²

(MPa)

REINFORCING STEEL

There are three basic types of reinforcement. a)

1.6

Mild steel

-

Characteristic strength

= 250 N/mm²

-

Usually has a smooth surface. Can be readily bent so is usually used where small radius bends are required or where re-bending on site is necessary.


b)

c)

1.2.3

G.K. Parrott

High tensile -

Drawn wire



= 450 N/mm²

-

Usually manufactured with a ribbed surface. More economical to use as the price is much the same as for mild steel.

-


-

Used in welded fabric and suitable for floors, shells, walls, etc.

= 485 N/mm²

COVER

The shortest distance measured from the surface of the concrete to the nearest reinforcing bar is known as the cover , which is determined by considering durability , fire resistance and practical considerations. The effective depth d is that dimension that will be used to determine the resistance of the beam, and is given by the distance measured from the centre of area of the tension reinforcement to the extreme compression fibre.

1.2.3.1

Durability

This is covered by part two of SABS 0100, but generally the following table may be used to choose suitable nominal cover according to the conditions of exposure. Exposure conditions

Concrete grade 20

25

30

40

50

Mild

25

25

25

25

25

Moderate

--

40

30

25

20

Severe

--

50

40

40

35

Very severe

--

--

--

60

50

1.7


1.2.3.2

G.K. Parrott


Fire resistance

[7]

A reinforced concrete element will undergo a gradual reduction in strength when subjected to fire. We therefore need to retain the structural strength by resisting the penetration of flames and heat transmission. Concretes made with siliceous aggregates have a tendency to spall when exposed to high temperature, but this tendency can be reduced by incorporating supplementary reinforcement (expanded metal).

The fire resistance of a reinforced concrete beam depends on the amount of protective cover provided to the main reinforcement. Table 43 in the code gives the minimum average concrete cover and minimum beam width for a fire resistance measured in hours.

1.2.3.3

Practical considerations

[4.11.2.4 & 5]

The nominal cover to all reinforcement should be at least equal to the maximum nominal size of the aggregate (typical nominal sizes being 19 and 26 mm). The cover to any particular bar should also be at least equal to the diameter of that bar, or in the case of bundles of more than two bars, equal to the diameter of a single bar of equivalent area.

= nominal size of aggregate = diameter of longitudinal bar

Reference should be made to clause 4.11.2 for further cover requirements. Note that the link diameter and the longitudinal bar diameter will have to be assumed at the beginning of our design calculations.

1.8


1.3

G.K. Parrott


LOADS

The two broad categories of loads are: 1) 2)

Permanent load Dn Imposed load Qn

Imposed loads will include the following main types: a) b) c) d) e)

Live Ln Wind Wn Overhead cranes Cn Temperature Tn Earthquake En

The subscript ‘n’ denotes that the load is ‘nominal’ (un-factored). Load calculations are usually the very first step in the actual design procedure for structural elements, and the importance of determining the load intensities accurately cannot be emphasised enough. This book will cover only dead and live loads. All loading and load calculations will be based on the code of practice for The general procedures and loadings to be adopted in the design of buildings SABS 0160-1989.

1.3.1

DEAD LOADS

The symbol for nominal dead load (permanent or self-weight load) is: This nominal self-weight includes the weight of the building itself and all finishes and materials which may be considered 'permanent'. Note that movable partitions, domestic appliances, etc. are treated as live loads.

1.3.1.1

Unit weight

Appendix B of SABS 0160 gives the nominal unit mass of a large variety of materials. A few of the more commonly used material weights are given below: MATERIAL UNIT WEIGHT (kN/m3) Structural steel 77 Reinforced concrete 24 Brickwork 20 Asphalt 23 Plaster (cement and sand) 23 Sand (normal MC) 18 SA PINE (up to grade 5) 5 SA PINE (up to grade 10) 7 1.9


1.3.2

G.K. Parrott


LIVE LOADS

The symbol for nominal live load (that produced by people, merchandise, etc.) is: Live loads are separated into two areas:

1.3.2.1

a) b)

floors roofs

Floor Loads

Table 4 of SABS 0160-1989 gives the appropriate loading for diff erent descriptions of floor loading. It is important to consider the most severe condition of either a uniformly distributed load or a concentrated load. A summary of this table is given below for quick reference. Minimum uniformly distributed imposed floor load KN/m2

Minimum concentrated load (applied over the area given adjacent kN

All rooms i n a dwelling unit Private rooms in institutional buildings

1.5

1.5

Classrooms, lecture theatres X-ray rooms, operating theatres

2.0

5.0

Garages and parking areas for vehicles of gross weight less than 25 kN

2.0

10.0

Offices for general use

2.5

Occupancy class of building or floor zone

Offices with data-processing equipment

9.0

Area over which concentrated load is applied (m)

0.1 x 0.1

0.75 x 0.75

3.0

Cafes, restaurants

3.0

5.0

Areas of public assembl y with fixed individual seating

4.0

3.0

Light laboratories Sales and display areas in departmental stores

4.0

5.0

Areas of public assembl y without fixed individual seating

5.0

3.0

0.1 x 0.1

Note that loads as high as 7 kN/m2 can be experienced at points of congestion (passageways, etc.) when people are forced by panic to crowd together. Load reduction:

There is little chance that all floors in a multi-storey building will be fully loaded over their entire area at any one time. For t his reason the code allows for a reduction in the imposed floor load for calculating the loads on the primary supporting structural elements (columns, walls and foundations). Where the total tributary area of a floor A (considering all levels), used for the assembly of persons or for storage, manufacturing or garaging exceeds 80 m², the load on the supporting elements may be mult iplied by the following factor, but with a minimum value of 0.7: 0.7



1.10


1.3.2.2

G.K. Parrott


Roof loads

The intensity of roof loads depend’s generally on the accessibility of the roof, and these loads are primarily maintenance or construction loads. These loads will, however, also cater for limited accumulations of snow (250 mm), hail (60 mm) and rain (50 mm) An accessible flat roof should generally be designed for a load of 2 kN/m², but consider the appropriate loading if the roof has an intended use as a floor. Where no public access is provided to a roof, allow for the most severe of the following: a)

A concentrated load of 0.9 kN applied over a solitary area of plan dimension 0.1 m x 0.1 m

b)

A uniformly distributed load of

where A is the tributary area. In the adjacent diagram the shaded area represents the tributary area for the first interior rafter of the roof. The intensity of this load should, however, not be taken as less than 0.3 kN/m² or greater than 0.5 kN/m².

c)

A distributed load corresponding to the expected depth of snow, if it is known that the depth of snow could exceed 250 mm

1.3.3 1.3.3.1

LOADING PATTERNS Uniformly distributed

A uniformly distributed load (UDL) is that which is constant over any given length or area of a structure. The units are kN/m² (applied to slab elements) or kN/m (applied to beam elements). The UDL applied to a beam is usually indicated by one of the two sketches shown below.

1.11


1.3.3.2

G.K. Parrott


Point

A point load is that applied by some intense load over a small enough area for it to be considered as acting at a point. The unit is kN.

1.3.3.3

Triangular and trapezoidal

These loads are common to beams supporting a two-way spanning slab. The units will be kN/m but the intensity of the load will vary along the length of the beam. The load from the slab is assumed to disperse at 450 from the corners of the slab if the two edges of the slab meeting at that corner have the same restraint (support) condition. This results in the beams sharing the load f rom the slab as shown in the following diagram.

These trapezoidal and triangular loads can become awkward to work with, and for this reason we shall equate these loads to an equivalent UDL for the purpose of calculating the bending moment. That is, we will convert these trapezoidal or triangular loads to an equivalent UDL which will produce approximately the same maximum bending moment. Note: The use of this equivalent UDL produces a conservatively high value for the reactions and hence shear forces. The correct shear force should therefore be calculated using the actual load area.

Trapezoidal load:

Triangular load:

1.12

N/m

kN/m


where:

G.K. Parrott

= = = =


equivalent UDL (kN/m) load on slab (kN/m²) length of the longer side of slab length of the shorter side of slab

Where the two edges of a slab meeting at a corner have different restraint conditions, the distribution of load onto the adjacent beams is as shown:

1.3.4

UNIT CONVERSIONS

The load intensity needs to be converted to the appropriate units (depending on the type of element being considered) and the f ollowing table may prove useful.

1.3.5

Original units

Required units

kg

kN

Multiply by:

kN/m3

kN/m2

(slabs)

thickness

kN/m2

kN/m

(beams)

effective width perpendicular to the direction of the span

kN/m

kN

(columns)

effective length parallel to the direction of the span

CONTINUITY EFFECTS

The continuity of an element usually renders the element indeterminate to some degree. Consider the beams shown below where it can be seen how the support reaction at ‘A’ varies due to the continuity of the beam at ‘B’.

1.13


G.K. Parrott


The third beam shown here is statically indeterminate and requires a special analysis to solve for the reactions. Coefficients to determine the reactions can be obtained for continuous beams having equal spans and uniform load, and these will be discussed in the section on analysis in module 2.

1.4

LIMIT STATES DESIGN

1.4.1

LIMIT STATES APPROACH

A structure, or part of a structure, is considered unfit for use or to have failed when it exceeds a particular state, called the limit state, beyond which its performance or use is impaired. The limit states are classified into two categories: a) Ultimate b) Serviceability

1.4.1.1

Ultimate limit state

The ultimate limit states are those concerning safety and they correspond to the maximum load-carrying capacity. Typical ultimate limit states are:

1.14

Overturning or uplift Fracture of members Deformations resulting in failure


1.4.1.2

G.K. Parrott


Serviceability limit state

The serviceability limit states are those which restrict the normal use and occupancy, or which effect durability. Typical serviceability limit states are:

1.4.2

Excessive deflection Cracking or spalling Excessive vibration

LIMIT STATE EQUATION

The basic equation for checking the ultimate li mit state is as follows:

where:

= = = = = = =

nominal resistance of element (moment, shear, etc.) partial resistance factor partial load factor dead load effect dominant imposed load effect non-dominant imposed load effect load combination factor

A load effect is the magnitude of the moment, shear, axial stress, etc. resulting from the applied load.

1.4.3

PARTIAL FACTORS

It is not possible to precisely define either the loads acting or the resistance of the members. It therefore becomes necessary to apply a factor to each - an amplifying factor for the applied loads, and a reducing factor for the members’ resistance. A load factor is applied to the nominal loads to allow for the fact that loads higher than anticipated may exist and also to allow for approximations made in the analysis of the load effects. A resistance factor is applied to the nominal strength of the member to allow for the strength being less than anticipated because of variability of material properties, dimensions and workmanship, and uncertainties in modelling the as-built structure.

1.15


G.K. Parrott


Consider the graph shown:

The first curve on the left represents the amplified load effect with the frequency of occurrence plotted against the magnitude. The far right end of this curve will therefore indicate cases where the actual load effect exceeds the amplified load effect. The second curve represents the factored resistance with the frequency of occurrence plotted against the magnitude. The far left end of this curve will therefore indicate cases where the actual strength of a member is for some reason less than the calculated factored resistance. Thus the shaded area where the two curves overlap will indicate the rare case where the amplified applied load effect exceeds the members’ factored resistance and failure could occur.

1.4.3.1

Load factors

The factor that is applied to the nominal load varies according to the nature of the load and consideration of the other loads acting simultaneously. It is thus possible to adjust the factor according to the precision with which the load can be predicted. Load combination factors The total imposed load may consist of many components such as floor or roof loads, wind, snow, temperature, etc. It is highly unlikely that all the components of imposed load would be applied simultaneously at their maximum values. Consider the extreme case where a maximum high temperature load would not apply simultaneously with snow loading. Nor would the maximum roof load apply simultaneously with maximum wind intensity. The approach to overcome this 'over-conservative' (and sometimes impossible) combination of loading is to consider the dominant imposed loads alone, and then to reduce all other imposed loads (non-dominant) by a combination factor .

1.16


G.K. Parrott


The partial load factors and load combination factors are given in Table 2 of SABS 01601991, an extract from which is shown below. Partial load factor Type of load

Load combination factor



1.5 1.2 0.9

1.1 1.1 1.0

1.0 1.0

1.6 1.6 1.6 1.6 1.3 1.3

1.0 1.0 1.0 1.0 0.6 1.0

0.6 0.3 0.3 0 0 1.0

Permanent In isolation In combination with imposed loads When considering uplift or overturning Imposed Garages, filing or storage floor areas All other floor areas Accessible roof Inaccessible roof Wind Fluids

1.4.3.2

Resistance factors

For reinforced concrete elements designed in accordance with SABS 0100-1 1992, the resistance factor cannot simply be applied since we are dealing with a combination of two different materials, viz. concrete and steel reinforcement. The strength/resistance of each material is therefore divided by a partial material f actor relating to each particular material, which takes account of the following: Differences between actual and laboratory values of strength Local weakness Inaccuracies in the assessment of the resistance of the section The importance of the limit state being considered Values for ultimate limit state

[3.3.3.2]

Steel reinforcement

=

1.15

Concrete (flexure and axial load)

=

1.5

Concrete (shear and bond)

=

1.4

Values for serviceability limit state

[3.3.4.2]

Steel reinforcement

=

1.0

Concrete

=

1.0 1.17

MODULE 2

ANALYSIS OF BEAMS AND FRAMES The analysis of a structure produces the load effects of ‘bending’, ‘shear’, ‘torsion’, ‘axial tension’, ‘axial compression’ and ‘deflection’ which the structure is to be designed to resist. It is therefore a critical component of the overall design process of ensuring a safe and serviceable structure.


G.K. Parrott

Module 2 - Analysis of beams and frames

CONTENTS Page 2.1

ARRANGEMENT OF LOADS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1

2.2 CONTINUOUS BEAMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 2.2.1 Moment distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 2.2.1.1 Example 2.1 (moment distribution) . . . . . . . . . . . . . . . . . . . . . . . 2.4 2.2.2 Redistribution of moment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6 2.2.2.1 Example 2.2 (redistribution) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.7 2.2.3 Analysis coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.10 2.2.3.1 Example 2.3 (analysis coefficients) . . . . . . . . . . . . . . . . . . . . . . 2.11 2.3 RIGID FRAMES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.1 Simplification into sub-frames . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.1.1 Vertical load only . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.1.2 Vertical and lateral loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.1.3 Example 2.4 (sub-frame analysis) . . . . . . . . . . . . . . . . . . . . . .

2.13 2.13 2.13 2.14 2.15


2.1

G.K. Parrott


ARRANGEM ENT OF LOAD S

The loads on a beam must be arranged (positioned) in such a way as to obtain the worst load effect. It will normally be sufficient to consider the following critical arrangements of vertical load: -

All spans loaded with total ultimate load (

-

All spans loaded with ultimate self-weight ( spans loaded with ultimate imposed load (

) ) and alternate )

No provision need be made for the arrangement of permanent loads as the ultimate load factor of 1.2 includes allowance for modelling uncertainties. Consider the continuous beam shown below which is loaded firstly with the total ultimate load on all spans. The exaggerated deflected shape is indicated by the solid line.

If the live load component is now removed from alternate spans, it can be seen that those spans still containing the ultimate live load will be permitted to deflect further as indicated by the broken line.

The effect of this will be an increase in the bending moment on those spans with increased deflection. To obtain the worst possible positive bending moment on any particular span, we therefore load that span and alternate spans with the total ulti mate load, all other spans being loaded with the ultimate dead load. It has been found to be too conservative to load any two adjacent spans to obtain the higher negative moments and shear forces over internal supports caused by this pattern of loading.

2.1


2.2

G.K. Parrott


CONTINUOUS BEAM S

A continuous beam is that which is continuous over three or more supports. The ends of the beam may be either fixed, pinned or free (cantilever). It is assumed that the supports offer no rotational restraint, i.e. the column ends are considered as ‘pinned ’ (Note that these columns will need to be analysed according to clause 4.2.2.4 of SABS 0100.)

There are various suitable methods available to analyse continuous beams, only two of these being considered in this module (viz. moment distribution and the use of coefficient tables). Various computer programs are availabl e which may prove convenient and time-efficient. The use of such software is certainly encouraged but it is extremely important that the designer have a thorough understanding of both the input and the output and never simply accept the answers without confirming their validity.

2.2.1

MOMENT DISTRIBUTION

This is an accurate method for calculating the bending moments and shear forces on a beam. It is therefore more suitable for accurate curtai lment of reinforcement which will yield a more economical design. This method is, however, quite time-consuming. The process of moment distribution will by itself produce all support bending moments and enable us to calculate the reactions at each support. The negative moments at the supports resulting from the applied loads will be distributed according to the stiffness of each span. The span with the greatest stiffness will attract the greatest load. The stiffness (k) of a member is given by where:

2.2

k E I L

= = = =

stiffness Young's modulus moment of inertia effective span


G.K. Parrott


Distribution factors are obtained either side of each support, which are used to calculate the distribution of moment about that support. The distribution factor for any member meeting at a joint (or support) is given by the stiffness of that member divided by the sum of the stiffness of all members meeting at that joint. For the continuous beams analysed here, we will consider all internal supports as ‘pinned’ and hence no moment will be transferred into the support, i.e. stiffness of the support = 0. Consider span ABC shown below:

The distribution factor for span AB at B would be given by:

and for span BC at B would be given by:

Both E and I are usually constant throughout the beam and the stiffness k in this case is then given by:

The distribution factors for certain beam end conditions are given below.

The method of applying moment distribution is as follows: 1)

Calculate the distribution factors to either side of each support.

2)

Calculate the fixed-end moments for each span independently. Write these down as negative on the LHS and positive on the RHS (fixed-end moments for various load types are available in many handbooks). The most common load type is the UDL, for which the fixed-end moment = ²/12

2.3


G.K. Parrott


3)

Calculate the out-of-balance moment at each support and multiply this by the distribution factor. (Out-of-balance moment = opp. sign of the algebraic sum of the moments either side of the support.)

4)

Carry half of this value across to the opposite side of the span.

5)

Repeat steps 3) to 4) until the distribution becomes small.

6)

Sum each column of values to obtain the support moments. (End on a distribution, not a carry-over, before adding up the columns.)

2.2.1.1

Example 2.1 (moment distribution)

Consider the 3-span beam shown below that is fixed at one end and pinned at the other. The load case will be considered where the end spans are loaded with 1.2 Dn + 1.6 Ln and the centre span loaded with 1.2 Dn.

The reactions at any support are then obtained by calculating the sum of the simply supported reactions and the reaction induced by the difference in moment at either end of each span. The reaction induced at the one end of a span is obtained by dividing that end moment, less the end moment at the other end of the span, by the length of the span.

2.4


G.K. Parrott


There will be an increase in the reaction on the end of a span having the highest negati ve end moment on that span. For example, at support C:

Similarly :

=

56.5 kN

=

101.8 kN

=

96.5 kN

From these reactions, the shear force diagram may then be drawn.

Once we have the shear force diagram, the position of the maximum span moments is easily obtained, (.ie. the position where the shear force is equal to zero). The values for the positive bending moments at these points are obtained by taking moments to one side of that point or by simply using the ‘area of shear force diagram’ method.

2.5


G.K. Parrott


By taking moments:

By using the area of shear force:

It will now be necessary to repeat the above process for all other load combinations. The results obtained from all load combinations are combined to produce a shear force and bending moment envelope. The maximum values from this envelope are the values to be used in our design.

2.2.2

REDISTRIBUTION OF MOMENT

Reinforced concrete beams behave in a manner that shares the elastic-plastic behaviour of the steel and the limited plastic deformation of the concrete. Basically, the continuous reinforced concrete beam may be considered elastic until the steel yields and then plastic until the concrete fails in compression. The concrete failure limits the plastic behaviour by li miting the rotation of the plastic hinge. As the bending moments develop in an elastic-plastic member, the beam will behave elastically until the plastic moment is reached. Further loading causes these hinges to rotate without changing the moments. The extra moment is then carried by other parts of the member. From a design point of view, this conditi on may be obtained by calculating the elastic BMD (moment distribution) and then reducing the support moments while increasing the midspan moments by a corresponding amount to maintain equilibrium. This operation is known as redistribution of moments.

NOTE

2.6

To maintain equilibrium, the magnitude of the reactions is recalculated to suit the adjusted negative moment. The corresponding new values for the positive mid-span moments are then calculated as before by taking moments about the new points of zero shear.


G.K. Parrott


The following restrictions are, however, given by SABS 0100: a)

Equilibrium between internal and external forces must be maintained. (This is covered by applying the method of redistribution as explained above.)

b)

Where the ultimate moment of resistance of a section is reduced, ensure that the neutral axis depth does not exceed

(The symbols

,

and

will be explained in the next section.)

c)

The ultimate moment of resistance of any section must be at least 75% of that required before redistribution for beams of constant cross-section and 80% for non-uniform members.

d)

Limit the redistribution to 10% for structures exceeding four storeys in height where the frame provides the stability.

2.2.2.1

Example 2.2 (redistribution)

A reinforced concrete beam is continuous over two spans of 7 m and 9 m respectively, and subjected to a nominal dead load of 25 kN/m and a nominal live load of 20 kN/m.

Draw the ultimate bending moment envelope considering all required arrangements of load and then redistribute the maximum ultimate negative moment obtained by 15% and redraw the final bending moment envelope.

2.7


G.K. Parrott


SABS Load Case #1

All spans loaded with maximum ultimate load

Note: For a 2-span continuous beam, the method of moment distribution produces an exact solution after only two ‘distributions’. The shear forces above are determined by firstl calculating the simply supported reactions, and then adjusting those values by the shear induced by moment, which is taken as the difference in support moments divided by the span.

2.8


G.K. Parrott


SABS Considering the other two required arrangements of load at the ultimate limit state, the following bending moment envelope is obtained.

The maximum negative bending moment of 519.2 kNm is obtained from load case #1. This moment is now reduced by 15% and the shear forces are re-calculated to maintain equilibrium as shown below:

Recalculating the positive bending moments, we get 191.2 kNm on the 7 m span and 426.5 kNm on the 9 m span, which are both less than the maximum positive moments obtained in the design envelope considering all three arrangements of load. The final bending moment envelope considering 15% redistribution is:

2.9


2.2.3

G.K. Parrott


ANALYSIS COEFFICIENTS

The coefficients given in Table 4 of SABS 0100 may be used to calculate the values f or the maximum moments and shear force on a continuous beam provided that the following conditions are met. a)

The characteristic imposed load does not exceed the characteristic dead load by more than 25%

b)

The loading is substantially uniformly distributed over at least 3 spans

c)

The lengths of the spans do not differ by more than 15%

Do not make any redistribution of the moments obtained from the table.

Table 4 - Ultimate bending moments *) and shear forces 1

2

Position

Moment

At outer support Near middle of end span

At first interior support

At middle of interior spans

At interior supports *)

3 Shear 0.45 F -

0.6 F

-

0.55 F

Do not redistribute the moments obtained from the table.

NOTE:

is the total ultimate load (1.2 Dn + 1.6 L n) and L is the effective span.

Further tables are available which give a range of coefficients more specific than those obtained from table 4. Reference should be made to the following: -

2.10

South African steel construction handbook pp. 5.84 ,5.85 and 5.86 Steel designers’ manual (Constrado, fourth edition) pp. 57, 58 and 59 Reinforced concrete designers’ handbook (Reynolds and Steedman, tenth edition), tables 33, 34 and 35


2.2.3.1

G.K. Parrott


Example 2.3 (analysis coefficients)

Draw the shear force and bending moment diagrams for a continuous beam with three equal spans of 6 m. The beam carries a dead load of 12 kN/m and a live load of 10 kN/m. The table below gives the shear force and bending moment coefficients for a 3-span continuous beam where the loading is uniformly distributed and all spans are equal.

Bending moment and shear force coefficients (uniform load) LOAD PATTERN

@

V

M

A

+0.400

0.000

B

-0.600 +0.500

-0.100

C

-0.500 +0.600

-0.100

D

-0.400

0.000

A-B

-

0.080

B-C

-

0.025

C-D

-

0.080

A

+0.450

0.000

B

-0.550 0.000

-0.050

C

0.000 +0.550

-0.050

D

-0.450

0.000

A-B

-

+0.101

B-C

-

-

C-D

-

+0.101

A

-0.050

0.000

B

-0.050 +0.500

-0.050

C

-0.500 +0.050

-0.050

D

+0.050

0.000

A-B

-

-

B-C

-

+0.075

C-D

-

-

The coefficients for the shear force V are in terms of moments M are in terms of

, where

, and the coefficients for the

is the UDL in kN/m on that span of length

.

2.11


2.12

G.K. Parrott



2.3

G.K. Parrott


RIGID FRAMES

2.3.1 2.3.1.1

SIMPLIFICATION INTO SUB-FRAMES Vertical load only

Consider the following structural frame:

The structure may be divided into a series of sub-frames, each consisting of the beams at one level together with the columns above and below. The columns may be assumed to be ‘fixed ’ at their remote ends unless a pinned end is more appropriate (as in the case of a column ‘pinned ’ to a pad foundation). If the second floor beam were to be analysed using this assumption, then the simplified structure would be as shown below:

2.13


G.K. Parrott


Alternatively, the structure may be divided into a series of sub-frames, each consisting of a single beam span together with the columns and beams at each end. Both the column and beam remote ends may be assumed to be ‘fixed ’ unless a pinned end is more appropriate. Note that if the beams on either side have been assumed to have ‘fixed ’ ends, then only half of their actual stiffness should be taken. If the second floor beam were to be analysed using this assumption, then the simplified structure would be as shown below:

Note that in both of these cases, the arrangements of load as discussed in 1.1 should be considered.

2.3.1.2

Vertical and lateral loads

Lateral loads need to be considered where the frame provides the lateral stability to the structure as a whole (i.e. the structure is not braced by shear walls or other bracing systems). The complete frame needs to be analysed using only the design wind load (self-weight and live load ignored). The moments obtained from this analysis are then added to those obtained using the simplified sub-frames.

2.14


2.3.1.3

G.K. Parrott


Example 2.4 (sub-frame analysis)

Draw the bending moment diagram for the sub-frame below.

The moment of inertia of the beam may be taken as 6.144 x109 mm4 . Internal columns are 350 x 350 and external columns are 450 x 350. Consider an ultimate load of 75.85 kN/m on both spans.

SABS (given)

2.15


SABS Moment distribution :

Beam reactions:

2.16

G.K. Parrott



G.K. Parrott


SABS

Bending moment diagram

2.17

MODULE 3

BEAMS Beams are the structural elements which generally support the slabs and transfer the load to the columns. Beams are designed for ‘flexure’ (bending), ‘shear’, ‘torsion’ and ‘deflecti on’.


G.K. Parrott

Module 3 - Beams

CONTENTS Page 3.1 FLEXURE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 3.1.1 Rectangular beams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 3.1.1.1 Example 3.1 (moment of resistance) . . . . . . . . . . . . . . . . 3.7 3.1.1.2 Example 3.2 (area of tension reinforcement ) . . . . . . . . . 3.10 3.1.2 Rectangular beams doubly reinforced . . . . . . . . . . . . . . . . . . . . . . . . 3.11 3.1.2.1 Example 3.3 (area of compression reinforcement ) . . . . 3.14 3.1.3 Flanged beams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.16 3.1.3.1 Effective flange width . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.16 3.1.3.2 Flexural strength . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.17 3.1.3.3 Example 3.4 (area of tension reinforcement ) . . . . . . . . . 3.19 3.1.4 Nominal reinforcement and spacing of bars . . . . . . . . . . . . . . . . . . . . 3.20 3.1.4.1 Minimum area of tension reinforcement . . . . . . . . . . . . . 3.21 3.1.4.2 Minimum area of compression reinforcement . . . . . . . . 3.22 3.1.4.3 Maximum area of reinforcement . . . . . . . . . . . . . . . . . . 3.22 3.1.4.4 Additional reinforcement in deep beams . . . . . . . . . . . . 3.22 3.1.4.5 Minimum reinforcement in flange . . . . . . . . . . . . . . . . . . 3.23 3.1.4.6 Minimum spacing of all bars . . . . . . . . . . . . . . . . . . . . . 3.23 3.1.4.7 Maximum spacing of bars in tension . . . . . . . . . . . . . . . 3.25 3.1.4.8 Choice of reinforcing bars . . . . . . . . . . . . . . . . . . . . . . . 3.26 3.1.5 Curtailment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.27 3.1.5.1 Anchorage length . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.28 3.1.5.2 Simplified rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.29 3.1.6 Design flowchart for flexure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.31 3.2 SHEAR . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.1 Sections un-reinforced for shear . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.2 Sections reinforced for shear . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.2.1 Stirrups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.2.2 Inclined bars . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.3 Nominal Reinforcement and spacing of bars . . . . . . . . . . . . . . . . . . . 3.2.3.1 Minimum area of stirrups . . . . . . . . . . . . . . . . . . . . . . . . 3.2.3.2 Maximum spacing of stirrups . . . . . . . . . . . . . . . . . . . . . 3.2.3.3 Stirrups for compression reinforcement . . . . . . . . . . . . . 3.2.4 Other considerations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.4.1 Concentrated load close to support . . . . . . . . . . . . . . . . 3.2.4.2 Practical fixing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.4.3 Anchorage of stirrups . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.4.4 Curtailment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.4.5 Local bond . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.5 Design flowchart for shear . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.6 Example 3.5 (shear reinforcement ) . . . . . . . . . . . . . . . . . . . . . . . . . . .

3.33 3.33 3.34 3.35 3.36 3.37 3.37 3.37 3.37 3.38 3.38 3.38 3.39 3.39 3.39 3.40 3.40

3.3 TORSION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3.1 Torsional shear stress . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3.2 Reinforcement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3.2.1 Links . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3.2.2 Longitudinal bars . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

3.43 3.43 3.44 3.44 3.45


G.K. Parrott

Module 3 - Beams

3.4 COMBINED SHEAR AND TORSION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.45 3.4.1 Example 3.6 (combined shear and torsion) . . . . . . . . . . . . . . . . . . . . . 3.46 3.5 DEFLECTION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5.1 Rectangular beams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5.2 Flanged beams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5.3 Example 3.7 (deflection) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6

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

3.49 3.49 3.51 3.52

DESIGN EXAMPLE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.55


3.1

G.K. Parrott

Module 3 - Beams

FLEXURE

This section takes a look at deriving the necessary equations for determining the amount of reinforcement required to resist bending in concrete elements.

3.1.1

RECTANGULAR BEAMS

It will be assumed in the derivations that foll ow that the concrete does not resist any tensile force. Consider the stress distribution diagram shown below.

The maximum stress in the concrete will be given by the 28-day cube crushing strength multiplied by a factor of 0.67, which takes into account the difference between the laboratory and site strength of concrete. This stress must then be divided by the partial safety factor for materials , which is taken as 1.5 for concrete.

The maximum stress in the concrete is therefore given by:

The depth to the neutral axis is the depth of the actual parabolic compression stress, which is simplified to the equivalent rectangular stress block having a maximum depth in compression of . 3.1


G.K. Parrott

Module 3 - Beams

The maximum depth to the neutral axis is limited to where the applied maximum ultimate elastic bending moment is not reduced by more than 10% as a result of redistribution. Where the applied maximum ultimate elastic bending moment is reduced by more than 10% as a result of redistribution, the maximum depth to the neutral axis x is limited further to

where:

Referring to the stress diagram on the previous page: The compressive force is denoted by ' The tensile force is denoted by ' '

'.

For equilibrium within the cross-section, the forces acting towards the ri ght must be equal to the forces acting towards the left. Therefore = The distance between these two forces is the lever arm which will be denoted by ' '. For a perfectly balanced design, the applied ultimate design moment internal ultimate moment of resistance of the section .

will be equal to the

i.e. or:

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . eq. (a)

and:

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . eq. (b)

; therefore the force is equal to the maximum stress in the concrete multiplied by the area of concrete in compression.

where:

3.2

= = =

characteristic strength of concrete width of section depth to N.A.


G.K. Parrott

Module 3 - Beams

We shall firstly consider the case where the redistribution is limited to 10%, in which case the maximum depth to the neutral axis is given by . We then have or

.....(i)

The lever arm (measured from the centroid of the full compression area to the centroid of the tension reinforcement) is given as: .....(ii) Substituting (i) and (ii) into eq. (a)

CONCRETE RESISTANCE (for redistribution not exceeding 10%) If

then the concrete will fail in compression.

In a similar way we can derive an equation for the ultimate moment of resistance considering the concrete strength where redistribution exceeds 10% (i.e. by setting ).

CONCRETE RESISTANCE (for redistribution exceeding 10%)

If

then the concrete will fail in compression.

We shall refer to the term

as

, the limiting value of

being

.

when redistribution does not exceed 10% when redistribution exceeds 10% A more general expression may now be given for the moment of resistance of a rectangular section considering the strength of the concrete.

3.3


3.4

G.K. Parrott

Module 3 - Beams


The force steel.

G.K. Parrott

Module 3 - Beams

is equal to the maximum stress in the reinforcement multiplied by the area of

where:

=

characteristic strength of steel

=

area of tension reinforcement

=

partial safety factor for materials, which is taken as 1.15 for steel

Substituting into eq. (b) gives:

STEEL RESISTANCE When a beam section is subjected to an applied ultimate moment less than the maximum moment of resistance , the force in the concrete required to resist this reduced moment will also reduce. The depth to the neutral axis reduces such that the stress in the concrete remains at the full value of . This implies that the lever arm will increase as the applied moment is reduced as shown below.

We therefore need to obtain an expression to enable us to calculate the magnitude of the lever arm depending on the magnitude of the applied moment.

. . . . . . . . . . . . . . . . . . . . . eq. (c)

3.5


G.K. Parrott

Module 3 - Beams

We have shown that: Substituting eq. (c) into this expression above gives:

Now substituting this expression for

into eq. (a) gives:

or rewriting:

Note: Do not take the value of

(LEVER ARM EQUATION)

as exceeding 0.95.

We now have the necessary equations to enable us to calculate the ultimate moment of resistance of a singly reinforced rectangular concrete beam for a given cross-section and area of reinforcement.

3.6


3.1.1.1

G.K. Parrott

Module 3 - Beams

Example 3.1 (moment of resistance)

Calculate the ultimate moment of resistance of the beam section shown below if redistribution of the maximum moments will be limited to 10%.

=

450 N/mm²

=

25 MPa

SABS

4.3.3.4

The moment of resistance is therefore 82.23 kNm.

3.7


G.K. Parrott

Module 3 - Beams

The equation for the ultimate moment of resistance of a concrete section, considering the strength of the reinforcement ( ), is that which will be used to calculate the required area of reinforcement in the beam by setting . This equation is therefore best written in the form:

In attempting to solve for the lever arm Z from the lever arm equation we will see that it contains the term , but to solve for we need the value of . To remove the term from the lever arm equation we can substitute the above equation into the lever-arm equation as follows:

Multiplying through by Z we get:

therefore

Solving the quadratic equation produces:

which can be simplified to that shown below.

Note: Do not take the value of

as exceeding 0.95.

This equation may be given in the form of a graph known as the LEVER ARM CURVE shown on the following page.

3.8


G.K. Parrott

Module 3 - Beams

3.9


G.K. Parrott

Module 3 - Beams

We now have the necessary equations to enable us to calculate the area of tension reinforcement required to resist a given applied ultimate moment to a rectangular section using the lever arm curve and the equation for steel resistance. It is important to note that we must always ensure that the equation giving the moment of resistance with respect to the concrete strength is greater than that for the steel strength (i.e. the section must be under-reinforced) to ensure that a failure (say to an excessive overload) will result from the gradual yielding of the reinforcement rather than as a sudden compression failure in the concrete.

3.1.1.2

Example 3.2 (area of tension reinforcement )

Calculate the area of reinforcement required to resist an appli ed ultimate bending moment of 200 kNm to a 750 X 300 beam. Given:

SABS 4.3.3.4

LA curve

4.3.3.4

3.10

=

450 N/mm²

=

30 N/mm²

=

675 mm


G.K. Parrott

Module 3 - Beams

3.1.2 RECTANGULAR BEAMS DOUBLY REINFORCED

[4.3.3.4]

We have shown that if , the applied force in the concrete will exceed the maximum permissible and the concrete will fail in compression. Reinforcement may, however, be placed in the compression zone to resist the resulting additional compression force. In deriving the formulae for compression reinforcement, the concrete is assumed to resist its maximum possible compressive force. Hence the moment of resistance of the concrete alone will be that given by: for redistribution  10% for redistribution > 10%

Both of these equations can be expressed as

.

The required increase in resistance moment would be the internal moment given by the force in the compression reinforcement multiplied by the lever arm to the tension reinforcement .

The increase in resistance moment due to the compression reinforcement alone will be:

Adding this to the equation for the maximum moment of resistance due to the concrete alone, we get an equation to calculate the maximum total moment of resistance of a concrete beam reinforced with compression steel.

3.11


G.K. Parrott

Module 3 - Beams

Changing the subject of this formula we can show that:

Now:

Substituting this into the above equation for

gives:

or:

where:

when redistribution does not exceed 10% when redistribution exceeds 10%

Figure 2 in the code shows that the maximum compressive stress

It therefore follows that:

.

for high tensile steel for mild steel

Note: This maximum compressive stress will only be able to develop if sufficient strain occurs at the level of the compression reinforcement.

3.12


G.K. Parrott

Module 3 - Beams

Figure 1 in SABS 0100 shows that the maximum strain in the concrete in compression is 0.0035. Consider the strain diagram shown below:

but not exceeding or

327.3 N/mm² for HT reinf. 196.1 N/mm² for MS reinf.

We now need to derive an equation to calculate the required area of tension reinforcement to balance the section by maintaining equilibrium of the internal forces.

Tension force in reinforcement

=

Compression force in concrete

=

Compression force in reinforcement

=

3.13


G.K. Parrott

Module 3 - Beams

Balancing these three forces we get:

or re-writing:

where:

We now have the necessary equations for the design of a rectangular reinf orced concrete beam in bending when it requires compression reinforcement.

3.1.2.1

Example 3.3 (area of compression reinforcement )

Calculate the reinforcement required for an 8.6 m long 600 X 300 R.C. beam simply supported on supports that are 300 mm wide and have a clear distance of 8 m between them. The beam carries its self-weight plus a live load of 30 kN/m. The exposure conditions may be considered as mild. = 450 MPa = 25 N/mm² SABS 4.3.2.1.2

3.14


G.K. Parrott

Module 3 - Beams

SABS 3.3.1.1

4.3.3.4

4.3.3.4.1

3.15


G.K. Parrott

Module 3 - Beams

3.1.3 FLANGED BEAMS When a concrete beam is cast integral with a concrete slab,a portion of the slab (beam flange) may be used to assist in resisting the compressive force providing the applied bending moment results in the flange being in compression - i.e. a positive moment for the flange on the top of the beam.

3.1.3.1

Effective flange width

The width of the flange which can be used to assist with the resistance of the compressive force must be limited to allow for full transferral of the compressive stress. The compressive stress may be assumed to be fully effective within a zone which extends from the edge of the web at the points of contraflexure to a maximum of one-tenth of the distance between the points of contraflexure (LZ ) to either side of the web midway between the points of contraflexure as shown below:

Note: No common area of flange may be shared between any two adjacent beams, so the effective width of the flange is therefore limited to the actual available flange width.

3.16

For a 'T' beam

-

the web width +

For an 'L' beam

-

the web width +


G.K. Parrott

Module 3 - Beams

For a simply supported beam the distance between the points of contraflexure is obviously equal to the effective span. For a continuous span of a beam the distance between the points of contraflexure can be determined from first principles by setting the appropri ate bending moment equation equal to zero and solving for the unknown distance/s. For continuous beams with approximately equal spans that support uniform loading the distances between the points of contraflexure may be taken as 0.7L for int erior spans and as 0.85L for end spans as shown below:

3.1.3.2

Flexural strength

Two cases of compressive stress distribution may occur: firstly when the neutral axis (above which the concrete is in compression) lies in the flange, and secondly when it lies in the web, as shown in figures (a) and (b) respectively.

In the first case:

The section may be designed as a rectangular beam by substituting the breadth with the full effective flange width. Most flanged beams will fall into this category.

In the second case: An exact solution would require special analysis, but if we ignore the portion of the web in compression the following equations may be derived, which will give slightly conservative values for the ultimate moment of resistance. 3.17


G.K. Parrott

Module 3 - Beams

Here the full depth of the flange is taken as being in compression and the lever arm will be given by: Substituting this into the equation derived earli er for the ultimate moment of resistance for rectangular beams, viz. , we get:

(STEEL RESISTANCE)

To calculate the maximum moment of resistance for a flanged beam considering the maximum compressive stress in the concrete, consider the diagram below:

The maximum compressive force and the lever arm

(CONC. RESISTANCE)

SABS 0100-1:2000 does, however, give the following equation which may be used to calculate the required area of reinforcement for a fl anged beam where the neutral axis falls below the flange, this equation being subject to three requirements: a)

3.18

.

b)

The design ultimate moment is less than

c)

( is obtained from table 5 in the code). Not more than 10% of redistribution has been carried out.


G.K. Parrott

Module 3 - Beams

How would we know whether or not the N.A. falls within the flange ? 1)

Calculate the lever arm Z using our lever arm curve as for a rectangular beam, assuming that the breadth of our beam = the full effective flange width.

2)

Calculate the depth to the neutral axis:

3)

If

: then the N.A. falls in the flange.

It is important to remember that over a support where negative moments arise, the effect of a top flange can NOT be used since the flange will be in tension, and as we know, the tensile resistance of the concrete is ignored in our design equations.

3.1.3.3

Example 3.4 (area of tension reinforcement )

Calculate the area of tension reinforcement required to resist a positive ultimate bending moment of 550 kNm for a 550 X 250 beam which is cast integral with a 170 thick slab. There are 4 beams spaced at 3 m centres which are continuous over 5 equal spans of 6 m. Design one of the internal beams only. Use high tensile reinforcement and grade 25/26 concrete. Exposure conditions are moderate.

3.19


G.K. Parrott

Module 3 - Beams

SABS 4.3.1.5

LA curve

3.1.4 NOMINAL REINFORCEMENT AND SPACING OF BARS Once the required area of reinforcement has been calculated and we have decided on what arrangement of bars to try, there are certain checks which must be carried out to comply with the requirements of the code which are deemed to satisfy 'normal' requirements for the serviceability limit state of cracking. Note that for special structures such as water-retaining structures, further serviceabili ty limit state checks are required to control the crack width. 3.20


3.1.4.1

G.K. Parrott

Module 3 - Beams

Minimum area of tension reinforcement

[4.11.4]

This check should generally satisfy the serviceabil ity limit state for cracking in most beams less than 750 mm deep.

For rectangular beams and slabs, the minimum percentage is expressed in terms of the gross cross-section, i.e. .

For flanged beams the minimum percentage is expressed in terms of the gross cross-section excluding the flange, .ie. .

The following table gives the minimum percentages of tension reinforcement in flexural members (from table 23 in the code).

Situation

Definition of percentage

Minimum percentage Mild steel

High tensile

0.24

0.13

1)

0.32

0.18

2)

0.24

0.13

1) T-beam

0.48

0.26

2) L-beam

0.36

0.20

Rectangular section (including secondary reinforcement in slabs) Flanged beams, web in tension:

Flanged beams, flange in tension:

3.21


3.1.4.2

G.K. Parrott

Module 3 - Beams

Minimum area of compression reinforcement

Where this reinforcement is required for the ultimate limit state, the following minimum values shall apply.

Situation

3.1.4.3

Definition of percentage

Minimum percentage Mild steel

High tensile

Rectangular section

0.2

0.2

Flanged beams, flange in compression

0.4

0.4

Flanged beams, flange in tension

0.2

0.2

Maximum area of reinforcement

[4.11.5.1]

Ensure that the area of tension or compression reinforcement does not exceed 4% of the gross cross-sectional area of the concrete. (The beam dimensions will need to be increased if this requirement is not met.)

3.1.4.4

Additional reinforcement in deep beams

[4.11.4.4]

Where the overall depth of a beam exceeds 750 mm, additional reinforcement is required on the sides of the beam to control cracking. These bars should be positioned inside the links up the side of the beam at a spacing not exceeding 250 mm, and must extend to a distance of two-thirds the depth of the beam measured from the tension face.

The diameter of these bars should be at least

where:

is the spacing of the bars b

3.22

is the width of the beam (or 500 mm, whichever is the smaller)


3.1.4.5

G.K. Parrott

Minimum reinforcement in flange

Module 3 - Beams

[4.11.4]

Transverse reinforcement: Where a beam is cast monolithically with the slab and the slab provides the flange to the beam, the amount of reinforcement required in the top of the slab, at right angles to the span of the beam, will be that necessary to resist the negative moment in the slab that occurs over the beam. For an isolated 'T' beam, the amount of reinforcement required in the top of the slab, at right angles to the span of the beam, will be that necessary to resist the negative moment caused by the flange cantilevering from the beam. In both of the above cases, we must however also ensure that the amount of reinforcement provided across the full effective width of the flange, expressed as a percentage of the longitudinal cross-sectional area of the flange, is at least 0.15%. Longitudinal reinforcement: We shall see in the next section that in a solid suspended slab we need to ensure that the amount of reinforcement provided at right angl es to the main reinforcement, expressed as a percentage of the gross cross-section, is at least 0.13% for high yield reinforcement or, alternatively, at least 0.24% for mild steel reinforcement. In either case, ensure that the distance between bars does not exceed five times the effective depth of the slab.

3.1.4.6

Minimum spacing of all bars

[4.11.8.1]

It is essential that the spacing of any reinforcement bars allows for the proper placement and compaction of the concrete. Generally, the gap between the bars should be no less than the dia meter of the bar for bars placed individually, and no less than the equivalent diameter of the sum of the area of all the bars in the group when they are placed in pairs or in bundles of three or four bars. The following minimum spacings are however also required:

3.23


G.K. Parrott

-

Allow a horizontal gap between individual bars of at least 5 mm greater than the maximum stone size.

-

If more than one layer of individual bars is used, the bars must be aligned vertically with a vertical gap between rows of at least 2/3 of the maximum stone size. The spacer bars used should be spaced sufficiently close to prevent ‘draping’ of the longitudinal bars between the spacers.

-

For bars placed in pairs, the same spacing requirements given above apply.

-

For bars placed in bundles of three or four bars, the gap between the bars both vertically and horizontally must be at least 15 mm greater than the maximum stone size.

-

Ensure also that the reinforcement in the top of a beam allows for a vibrator needle to be inserted.

3.24

Module 3 - Beams


3.1.4.7

G.K. Parrott

Module 3 - Beams

Maximum spacing of bars in tension

[4.11.8.2.1.4]

Ensure that the clear horizontal distance between adjacent bars or groups, near the tension face of a beam, depending on the percentage of redistribution, is not greater than the values given in table 25 of SABS 0100. Characteristic strength of reinforcement f y MPa

Clear distance between bars mm Percentage redistribution to or from section considered -30

-25

-20

-15

-10

0

+10

+15

+20

+25

+30

250

215

230

245

260

275

300

300

300

300

300

300

450

120

130

135

145

155

170

185

195

205

210

220

485

110

120

125

135

140

155

170

180

190

195

205

Ensure that the clear distance from the corner of a beam to the surface of the nearest longitudinal bar is not greater than half the values given above.

Alternatively, the maximum clear distance between bars may be given as: (but not exceeding 300 mm)

where

is the design service stress and may be estimated as follows:

where:

=

self-weight factor for SLS (serviceability)

=

imposed load factor for SLS

=

self-weight factor for ULS (ultimate)

=

imposed load factor for ULS

Where all applied loads are of a similar nature (e.g. all UDL), it is possibly more accurate to calculate the service stress using the following equation, which allows for varying proportions of dead and live loads.

The load effect can be taken as the load itself if both dead and live loads are of the same nature (e.g. uniformly distributed), but must be taken as the bending moment or some other load effect if the loading units are not consistent.

3.25


3.1.4.8

G.K. Parrott

Module 3 - Beams

Choice of reinforcing bars

Considering all of the above requirements, we may now choose a suitable arrangement of reinforcement bars that will give an area at least equal to that required. Total area for different diameters and number of bars ( No.

)

BAR DIAMETER 8

10

12

16

20

25

32

1

50

78

113

201

314

491

804

2

101

157

226

402

628

982

1608

3

151

236

339

603

942

1473

2413

4

201

314

452

804

1257

1963

3217

5

251

393

565

1005

1571

2454

4021

6

302

471

679

1206

1885

2945

4825

Mild steel bars are denoted by: High tensile bars are denoted by:

R Y

The maximum available length of any bar is 13 m. Note that high tensile bars are not available in 8 mm diameter. Larger diameter high tensile bars (40 and 50 mm diameter) are sometimes available, but are not preferable.

When using a combination of bar sizes to obtain the required area, the diameter of those bars should not differ by more than two sizes.

Bars should be chosen such that the arrangement is symmetrical about the vertical axis through the beam.

3.26


G.K. Parrott

Module 3 - Beams

Bars may be placed in pairs or in bundles of up to four bars, but the further spacing requirements given in clause 4.11.8.1 must be considered. If more than one full layer of reinforcement is required, the layers should be separated by a spacer bar equal to at least the size of the largest bar in either layer.

3.1.5 CURTAILMENT

[4.11.7.1]

Up until this stage, we have only considered the reinforcement required to resist the maximum bending moment. However, as the bending moment reduces along the span of a beam, so too will the required area of reinforcement also reduce. To determine at what point (theoretical cut-off point) on a beam some of the reinforcement may be curtailed, the following procedure may be adopted: a)

Decide which bars you wish to curtail (ensuring that all other considerations are still satisfied for the remaining bars).

b)

Calculate the moment of resistance of the section considering the remaining bars only (see the example given on page 3.7).

c)

Calculate the theoretical point at which the curtailed bars are no longer required, i.e. the distance from one of the supports to the point at which the moment of resistance is equal to the applied moment. Alternatively, if the bending moment diagram has been plotted to scale, this distance may be scaled off.

d)

Ensure that the curtailed bars are effectively anchored a full effective anchorage length beyond this point to enable the development of the full design strength (0.87 fy) of the bars (see 3.1.5.1). However:

If the shear or bending resistance is greater than twice the shear force or bending moment actually present at a point 12 times the bar diameter or d, whichever is the greater, beyond the theoretical cut-off point, then the bars may be extended 12 times the bar diameter or d, whichever is the greater, beyond the theoretical cut-off point.

3.27


e)

3.1.5.1

G.K. Parrott

Module 3 - Beams

At the simply supported end of a beam, anchor each remaining tension bar by one of the following: -

An effective anchorage length equivalent to 12 times the bar diameter beyond the centre-line of the support; no bend or hook should begin before the centre of the support.

-

An effective anchorage length equivalent to 12 times the bar diameter plus d/2 from the face of the support; no bend or hook should begin before d/2 from the face of the support.

Anchorage length

[4.11.6.2]

To prevent anchorage bond failure, ensure that the tension in any bar at any section due to ultimate loads will be developed on either side of the section by an appropriate embedment length or other end anchorage. To achieve this, ensure that the anchorage bond stress values given in table 24 of SABS 0100.

does not exceed the allowable

Let us consider the required anchorage length for a high tensile bar in tension using 25 MPa concrete: From table 24,

= 2.5 N/mm2

Note that for the top bars in members exceeding 300 mm in depth, the ultimate anchorage bond stress should be reduced by 30% for deformed bars and by 50% for plain bars. 3.28


G.K. Parrott

Module 3 - Beams

The following table gives the effective anchorage length in terms of the bar diameter for different grades of concrete, for both high tensile and mild steel bars:

Concrete grade

Force

20

25

30

40

Tension

46

40

37

30

Compression

37

32

29

24

For high tensile bars:

The effective anchorage length may be reduced by 6.5 D for every standard 900 bend within the anchorage zone.

For mild steel bars:

The effective anchorage length may be reduced by 4 D for every standard 900 bend within the anchorage zone.

3.1.5.2

Simplified rules

[4.11.7.2]

a) Simply supported beams:

Extend at least 50% of the tension reinforcement provided at mid-span to the supports and give it an effective anchorage of 12 times the bar diameter beyond the centre-line of the support. Extend the remaining 50% to within 0.08 times the effective span from the centre of the support.

b) Cantilever beams:

Extend at least 50% of the tension reinforcement provided at the support to the end of the cantilever. Extend the remaining 50% a distance of half the span or 45 times the bar diameter, whichever is the greater, from the face of the support.

c) Continuous beams:

These simplified rules for curtailment are given on the next page, but it is important to note that they apply only to continuous beams having approximately equal spans and uniform loading.

3.29


G.K. Parrott

Module 3 - Beams

top reinforcement: No more than 40% of the tension reinforcement over the supports may be curtailed at 0.15 L from the face of the support, and a further 40% may be curtailed at 0.25 L from the face of the support. The remaining reinforcement is to continue through the spans.

Bottom reinforcement: No more than 70% of the tension reinforcement may be curtailed at 0.15 L from the centre of interior supports and at 0.1 L from the centre of exterior supports. The remaining reinforcement is to extend at least to the centre of interior supports and extend 12 dia. past the centre of exterior supports.

3.30


G.K. Parrott

Module 3 - Beams

3.1.6 DESIGN FLOWCHART FOR FLEXURE 1.

Carry out an appropriate analysis of the structure considering all required combinations of load to obtain the maximum ultimate design bending moment M at all critical points along the length of the beam.

2.

Determine the required cover to reinforcement.

3.

Estimate stirrup and longitudinal bars sizes to obtain the effective depth d.

4.

Considering the nature of the bending moment at the point under consideration (sagging or hogging), determine the effective width of concrete in compression b.

5.

Decide on a suitable grade of concrete f cu to be used (25 MPa for most general structures, but consider durability).

6.

Calculate the ratio K

7.

Determine the limit K’ considering any possible reduction in the design mom ent due to redistribution (K’ = 0.156 if redistribution  10%).

8.

If K > K’ then compression reinforcement is required and proceed to step 12.

9.

Determine the lever arm ratio Z/d

.

and the lever arm Z

(remember that Z/d may not exceed 0.95) 10.

For flanged beams, confirm that the compression zone falls within the flange depth by ensuring that

11.

.

Calculate the required area of tension reinforcement As

.

Proceed to step 17. 12.

Estimate the size of the compression bars and obtain the depth from the extreme compression fibre to the centroid of these bars d’.

13.

Determine the design stress in the compression reinforcement f yc (considerati on m ust be given to strain incompatibility in slabs and shallow beams).

14.

Calculate the required area of compression reinforcement A’s

.

15.

Determ ine the lev er arm Z (remember that Z/d may not exceed 0.95).

16.

Calculate the corresponding required area of tension reinforcement As

.

17.

Calculate the minimum and maximum allowable areas of steel.

18.

Choose an appropriate and practical arrangement of steel and sketch a cross-section to show the placement of these bars.

3.31


G.K. Parrott

Module 3 - Beams

19.

Check the minimum and maximum allowable spacing of bars and adjust the arrangement if necessary.

20.

Choose which bars may be curtailed such that all serviceability requirements are still met.

21.

Calculate the moment of resistance of the cross-section considering the area and effective depth of the remaining bars only.

(not exceeding 0.95)

22.

Determine the theoretical cut-off points by setting the general bending moment equation equal to the moment of resistance and solving.

23.

Provide a full anchorage of the curtailed bars beyond these points.

24.

Provide additional bars in the sides of the web if the overall depth of the beam exceeds 750 mm.

3.32


3.2

G.K. Parrott

Module 3 - Beams

SHEAR

3.2.1

SECTIONS UNREINFORCED FOR SHEAR

Shear forces give rise to diagonal tension in the concrete and bond stresses between the reinforcement and the concrete. The shear force in a concrete beam without shear reinforcement is resisted by the following:

1)

Uniform shear stresses in the compression zone.

2)

Aggregate interlock along the shear crack.

3)

Dowel action in the bars where the concrete between cracks transmits shear force to the bars.

The actual behaviour of shear forces in a reinforced concrete section is complex and very difficult to analyse, but results from experimental investigations give reasonable design procedures. We know that:

Therefore: where :

= = = =

shear stress shear force due to ultimate loads effective depth width of the section which, for a flanged beam, shall be taken as the average width of web below the flange.

Where a section has a longitudinal hole through it, the area bd in the above equation must be reduced by that area of the hole. i.e.

3.33


G.K. Parrott

Module 3 - Beams

The ultimate allowable shear stress in beams is the maximum shear stress that the beam can resist without providing any form of shear reinforcement. The value for is obtained from experimental investigations and may be determined from the following equation.

where:

is the partial material factor = 1.4 (see cl 3.3.3.2) is the percentage of properly anchored tension reinforcement at the point being considered, and should not be taken as greater than 3% shall not be taken to exceed 40 MPa

The read er should note that uncertainty exists regarding the va lue of the constant in the above equation, which is given in SABS 0100:1 (2000) as 0.75, and as 0.79 by numerous other sources. 0.75 has been adopted in this text as it gives a more co nservative result.

Ensure that in no case, even with shear reinforcement, will the shear stress value the lesser of

3.2.2

or

exceed a

.

SECTIONS REINFORCED FOR SHEAR

Where the actual ultimate shear stress in a beam

exceeds the allowable value

given

by the above equation, it will be required to design shear reinforcement. Shear reinforcement may be in the form of either stirrups alone, or inclined bars used in conjunction with the stirrups.

The use of inclined bars has become less favourable over the years and is generally only used in exceptional circumstances where the use of stirrups alone prooves inadequate. 3.34


3.2.2.1

Let

G.K. Parrott

Module 3 - Beams

Stirrups

[4.3.4.1.3]

be the spacing of the links.

The shear crack is assumed to be at 450. Horizontal length of crack = d (approx.). =

resistance to shear force of the section without shear reinforcement.

=

ultimate shear force

The shear force to be taken by the links = ultimate shear force - shear resistance of section = ie. shear force taken by links

=

The number of links crossing the crack =

The strength of the links where:

= is characteristic strength of the links is the area of the total number of legs to each link (usually 2 legs per link)

Equating the strength of the links to the shear force to be taken by the links, we get:

or rewriting:

NOTE :

may not exceed 450 N/mm² in this equation.

3.35


3.2.2.2

G.K. Parrott

Module 3 - Beams

Inclined bars

[4.3.4.1.4]

Where longitudinal bars are no longer required to resist moment, they can be bent up to assist as shear reinforcement. However, bent-up bars alone are not satisfactory as shear reinf orcement. Links must also be provided, and they must resist at least 50% of the ultimate shear force. Because the beam would already have been designed for bending, the longitudinal reinforcement available to be used as bent-up bars is predetermined. We therefore calculate the shear force that the bent up bars can resist, subtract this from the applied shear force and design links to resist the greater of the remaining shear force, or 50% of the maximum shear force. These bent-up bars are assumed to form the tension members of a ‘lattice girder’ in which the concrete forms the compression members.

The shear resistance of a single system of bent-up bars with the bars inclined at 45o or more, may be calculated from the following equation:

where:

=

area of bent-up bars across shear failure plane

=

the angles as shown in the figure above (must be at least 45o) horizontal spacing of bent-up bars

= Note that

must be at least

Note also that if

3.36

=

and not more than

.

, the above equation may be simplified to:


3.2.3

G.K. Parrott

Module 3 - Beams

NOMINAL REINFORCEMENT AND SPACING OF BARS

3.2.3.1

Minimum area of stirrups

[4.11.4.5.3]

In all beams, except those of minor structural importance, where the maximum shear stress, calculated in accordance with 4.3.4, is less than half the recommended value, provide nominal links throughout the span such that:

where :

=

the cross-sectional area of the legs of the link.

=

the breadth of the beam at the level of the tension reinforcement. the spacing of the link.

=

3.2.3.2

Maximum spacing of stirrups

[4.11.4.5.4]

To ensure that any potential shear crack will be intercepted by at least one link, the spacing of links must not exceed 0.75 times the effective depth of the beam, and the lateral spacing of the individual legs of the links must not exceed this figure. Links must enclose all tension reinforcement.

3.2.3.3

Stirrups for compression reinforcement

[4.11.4.5.1]

When in a beam, part or all of the main reinforcement is required to resist compression, provide stirrups of at least one-quarter of the size of the largest compression bar at a maximum spacing of twelve times the size of the smallest compression bar. Ensure that all other bars within a compression zone are within 150 mm of a restrained bar. Arrange links so that every corner and alternate bar or group in an outer layer of reinforcement is supported by a link passing round the bar and having an inclined angle of not more than 135o.

3.37


3.2.4

G.K. Parrott

Module 3 - Beams

OTHER CONSIDERATIONS

3.2.4.1

Concentrated load close to support

[4.3.4.2]

The ultimate allowable shear stress in beams may be increased if a concentrated load is less than twice the effective depth away from the face of the support. This enhancement is particularly useful for the design of corbels (see 6.2.4.2).

Consider the adjacent diagram where it can be seen that any load applied to the left of section x-x will have a reduced effect in producing a shear failure in the plane as indicated.

The ultimate allowable shear stress stress

may be increased to

provided that the shear

at the face of the support remains less than the smaller of

and

.

3.2.4.2

Practical fixing

Consider the space confinement for positioning the longitudinal reinforcement in the beam. It is very often practical to use shape code 72 links for shear reinforcement in preference to the closed link (shape code 60 or 74) to allow for the longitudinal bars to be dropped into position.

3.38


3.2.4.3

G.K. Parrott

Anchorage of stirrups

Module 3 - Beams

[4.11.6.4]

A link may be considered fully anchored if it complies with the following: a)

It passes round another bar of at least its own diameter, through an angle of 90o, and continues for a minimum length of eight times its own diameter.

b)

It passes round another bar of at least its own diameter, through an angle of 180o, and continues for a minimum length of four times its own diameter.

3.2.4.4

Curtailment

As we curtailed the tension reinforcement where it was no longer required, so too can we curtail the shear reinforcement in the areas where the shear forces become sufficiently lower than that used to design the links. To curtail the links we decide on a suitable i ncreased spacing (or reduced diameter) of the stirrups, calculate the shear resistance of these stirrups and then calculate f rom the shear force diagram at what positions this increased spacing may occur. It is important to remember that the reduced shear reinforcement must still satisfy the requirements of nominal area and maximum spacing.

3.2.4.5

Local bond

[4.11.6.1 ]

Local bond stresses occur at sections of beams subjected t o shear where the force in the tension bar is changing along its length. If, however, at both sides of any cross-section of a beam, the force in each bar is permitted to develop as a result of an appropriate embedment length or end anchorage, local bond stress may be ignored.

3.39


G.K. Parrott

Module 3 - Beams

3.2.5 DESIGN FLOWCHART FOR SHEAR 1.

Carry out an appropriate analysis of the structure considering all required combinations of load to obtain the maximum ultimate design shear force V at all critical points along the length of the bea m.

2.

O btain the area of reinfo rcem ent A s and effective depth d of the tension steel that has bee n provided and fully anchored beyond the point being considered .

3.

C alcu la te th e ultim ate s hea r stre ss v

4.

Ensure that this stress does not exceed the absolute ultimate allowable shear stress which is the lesser of 4.75 or 0.75

5.

(where b is the web width).

.

Calculate the ultimate allowable shear stress for the section unreinforced for shear: vc

(the term

may not be taken as exceeding 3).

6.

I f v < vc th en on ly n om in al s tir ru ps ar e r eq uir ed an d p ro ce ed to ste p.

7.

C alc ula te th e re qu ire d s tir ru p a re a / s pa cin g ra tio

8.

D ete rm in e th e m in im u m re qu ir ed are a / s pa cin g ra tio .

9.

N ote th e m ax im u m a llo we d s tir ru p s pa cin g o f 0 .7 5d .

1 0.

C ho os e a su ita ble st irr up arr an ge me nt.

3.2.6

.

EXAMPLE 3.5 (shear reinforcement )

Calculate the shear reinforcement required for a rectangular 600 X 300 RC beam to resist a maximum ultimate shear force of 267 kN. The beam is simply supported over a single span and reinforced with 4 Y20 longitudinal bars, 2 of which extend to the supports. The cover is 25 mm and the concrete is grade 30/26

3.40

1)

Use HT stirrups only

2)

Use a combination of MS stirrups and pairs of Y20 bent-up bars.


G.K. Parrott

Module 3 - Beams

SABS 1) Using HT stirrups only Assume Y10 stirrups: 4.3.4.1.1

4.11.4.5.3

4.11.4.5.4

Adopt

3.41


G.K. Parrott

S.A.B.S 2) Using MS stirrups with bend-up bars

4.3.4.1.4

4.3.4.1.1

4.11.4.5.3 4.11.4.5.4

Adopt

3.42

Module 3 - Beams


3.3

G.K. Parrott

Module 3 - Beams

TORSION

Shear reinforcement is usually adequate to control mi nor torsional cracking under normal conditions, but when the design of an element relies on t he torsional resistance of a beam, torsional reinforcement will need to be provided where required.

3.3.1

TORSIONAL SHEAR STRESS

The torsional shear stress

(N/mm²) results from the applied ultimate torsional moment

(kNm) and is calculated from the following equation:

where:

and

are the smaller and larger dimensions of a rectangular section respectively.

For flanged beams, the cross-section can be divided into a number of rectangles with each rectangle being subjected to a partial torsional moment

where:

Divide these rectangles in such way that the widest rectangle is made as long as possible.

The torsional shear stress

is then calculated for each rectangle.

3.43


G.K. Parrott

Module 3 - Beams

Table 8 in SABS 0100:1 gives the minimum torsional shear stress , which is the maximum torsional stress that the rectangle section can resist without providing any additional reinforcement for torsion. Table 8 - Minimum and ultimate torsional shear stress Values in mega pascals 1

2

3

Concrete grade

Minimum torsional shear stress v t min

Ultimate torsional shear stress v t u

0.27

3.18

25

0.30

3.56

30

0.33

4.00

0.36

4.50 < vt u < 4.75

20



*)

40

*)

Grade not recommended

NOTES 1.

Allowance has been made in these figures for a

2.

Values of v t u are derived from the equation

 m of

1.40.

but not exceeding 4.75 .

Ensure that in no case will the sum of the shear stress exceed the ultimate torsional shear stress When the larger dimension of a link ensure that 3.3.2

and the torsional shear stress

(also given in table 8).

(measured centre-to-centre) is less than 550 mm,

. REINFORCEMENT

Where the actual ultimate torsional shear stress

exceeds the allowable value

, it

shall be required to design torsional reinforcement. Torsional reinforcement is in the form of both links and longitudinal bars and is provided in addition to that required for bending and shear.

3.3.2.1

Links

Provide stirrups to satisfy the following area / spacing ratio:

3.44


The spacing

G.K. Parrott

Module 3 - Beams

should not exceed the least of -

= smaller centre-to-centre link dimension. = larger centre-to-centre link dimension. The link should be of the closed type (shape code 74).

Note that for thin rectangular sections (such as beam flanges), the maximum spacing of the links will be very small and it may be more economical to consider the larger rectangle (beam portion) alone in resisting the torsion.

3.3.2.2

Longitudinal bars

Provide additional longitudinal bars having at least the following total area:

These bars should be distributed evenly around the inside perimeter of the links and should be spaced at no more than 300 mm. There should be at least one bar in each corner of the link.

3.4

COMBINED SHEAR AND TORSION

Situation Provide minimum shear reinforcement; no torsion reinforcement

Design torsion reinforcement but not less than minimum shear reinforcement

Design shear reinforcement; no torsion reinforcement

Design both shear and torsion reinforcement

3.45


3.4.1

G.K. Parrott

Module 3 - Beams

EXAMPLE 3.6 (combined shear and torsion)

A 450 x 300 beam is 5.5 m long and 'fixed' to the supports at either end. The beam will be cast monolithically with a cantilever slab150 mm thick which will project 1 100 mm from the face of the beam as shown below.

The slab is to be designed to carry a live load of 3.5 kN/m². Grade 25/26 concrete is to be used and exposure conditions are moderate. Two Y16 bars are required top and bottom for bending. Calculate suitable reinforcement for shear and torsion.

SABS Shear Self-weight

4.2.2.1

4.3.4.1.1

3.46


G.K. Parrott

Module 3 - Beams

SABS

4.3.4.1.2

Torsion Slab self-weight

4.3.5.3

We will let this torsion be resisted by the rectangular portion of beam alone: T8

Combined shear and torsion

3.47


G.K. Parrott

Module 3 - Beams

SABS

4.3.5.3.5

4.11.4.5.3

4.3.5.3.6 4.11.4.5.4

4.3.5.3.5

Adopt

Changing the provided 4 Y16 bars to 4 Y20, the increase in area is 453 mm2 . Further area required

3.48

.


3.5

G.K. Parrott

Module 3 - Beams

DEFLECTION

If the deflection of any part of a structure is excessive, it becomes noticeable and may cause damages to finishes such as partitions, windows, etc. Reasonable limits for deflection are set out in SABS 0100 and for a reinforced concrete beam may be summarised as follows: -

To prevent the deflection from becoming noticeable, the total final deflection should not exceed span/250.

-

To prevent damage to finishes, the deflection that takes place after the construction of any finishes should not exceed span/350 or 20 mm .

The deflections calculated must be the long-term deflections, which must include the effects of temperature, creep and shrinkage. Deflections may be calculated and compared with the above requirements, but in normal cases it will be adequate to limit the span to effective depth ratio. To calculate the actual deflections we may either carry out exact mathematical calculations or use the deflection equations given in various handbooks, e.g. South African Steel Construction Handbook (LSD), pg’s 5.71 to 5.83 Remember that deflection is a serviceability limit state check and is therefore calculated using service loads. The value of Young's modulus to be used in the calculation of the actual deflection may be obtained from table 1 for different grades of concrete. It is normal practice to check the basic span to ef fective depth ratio before commencing with the complete design to eliminate any obvious deflection failures, and then to do a more accurate check, or calculate the actual service deflections once t he analysis is complete and all reinforcement has been calculated.

3.5.1

RECTANGULAR BEAMS

[4.3.6.2]

Table 10 in SABS 0100 gives the basic allowable span/d ratios for r ectangular beams with a span of less than 10 m. This table may be used for spans exceeding 10 m but only when it is not necessary to limit the increase in deflection after the construction of partitions and finishes. Otherwise, in order to prevent damage to finishes and partitions, the values obtained in table 10 should be multiplied by 10/span. The deflection of any cantilever over 10 m in length should be justified by calculation.

3.49


G.K. Parrott

Module 3 - Beams

Table 10 - Basic span/effective depth ratios for rectangular beams Support conditions

Ratio

Truly simply supported

16

Single span with nominal end restraint

20

One end continuous

24

Both ends continuous

28

Cantilever

7

Multiply these ratios by a factor of 0.85 if poor quality concrete or workmanship is expected. Deflection will be influenced by the moment applied to a section as well as the actual stress occurring in the tension reinforcement (service stress ). For this reason the value obtained from table 10 should be modified by the factors obtained from table 11, which is derived from the following equation: Modification factor for tension steel =

fs M b d

3.50

= = = =

2.0



service stress in reinforcement (see page 3.25), but not exceeding 300 MPa ultimate bending moment effective width of section effective depth of section


G.K. Parrott

Module 3 - Beams

Sagging moments in a beam seldom require compression reinforcement. However, if it is required then the amount of compression reinforcement provided will also influence the deflection. Table 12 in SABS 0100 gives the modification factor to take the effect of this reinforcement into account. The breadth b to be used in this table is the width of the web. Note: Tables 10 to 12 take into account the effects of normal creep and shrinkage.

3.5.2

FLANGED BEAMS

[4.3.6.5]

Because the moment of inertia of a cross section has a direct influence on the amount of deflection that will take place, the allowable span/d ratio for a rectangular beam must be reduced for flanged beams to take into account the absence of concrete below the flange. For a flanged beam, multiply the ratio obtained from tables 10 to 12 by a further factor, which may be read off the table given below:

Consider the three cross-sections shown below:

(a)

rectangular section, therefore no reduction in span/depth ratio

(b)

flanged section with high web / flange width ratio, therefore small reduction in moment of inertia and small reduction in span / depth ratio

(c)

flanged section with low web / flange width ratio, therefore large reduction in moment of inertia and large reduction in span / depth ratio

3.51


3.5.3

G.K. Parrott

Module 3 - Beams

EXAMPLE 3.7 (deflection)

The dimensions and reinforcement for a 'T'-beam which is truly simply supported over a span of 11 m is given below. A dry-wall partition will be fitted to the soffit of the beam. The cover to all reinforcement is 20 mm and the concrete is grade 25/26 Live load Dead load

Ln = 12 kN/m Dn = 8 kN/m

Check whether the beam satisfies the code requirements for deflection.

In designing the beam for flexure, the following results were obtained:

3.52


G.K. Parrott

Module 3 - Beams

SABS T10

T11

4.3.6.5

We could increase the depth of the beam, but although the beam does not satisfy the requirements of span/d ratio we could still calculate the expected service deflection and check this against the allowable limits:

3.53


G.K. Parrott

Module 3 - Beams

SABS T1

3.2.3.2.1.1

3.2.3.2.1.2 This is very close to the expected working deflection therefore can accept!

3.54


3.6

G.K. Parrott

Module 3 - Beams

DESIGN EXAMPLE

The drawing below shows the plan of a suspended reinforced concrete floor slab to be utilised for an office with data processing equipment. The top of the slab will be finished with a screed 35 mm thick and the soffit will receive plaster 25 mm thick. Exposure conditions may be taken as mild and no fi re rating is required. The slab will be cast monolithically with the 450 X 300 beams supporting it. All columns are 300 X 300. Design beam (2) using grade 25/26 concrete. Include all the necessary checks, and provide a schedule of reinforcement.

3.55


G.K. Parrott

SABS

LOADING Slab Loading (kN/m2)

SABS 0160 Table 4

Beam loading (kN/m)

Span 4-5

Span 5-6 (slab continuity factor = 1.2)

3.56

Module 3 - Beams


G.K. Parrott

Module 3 - Beams

SABS

The combinations of load shown above, in accordance with clause 4.3.2.1.2, are used to determine the ultimate shear f orce and bending moment envelopes below. (Note that redistribution of moments has not been carried out in this example.)

3.57


G.K. Parrott

SABS

BENDING - positive moment on span 5-6

4.3.1.5

LA curve

4.3.3.4

4.11.4

4.11.5.1

3.58

Module 3 - Beams


G.K. Parrott

Module 3 - Beams

SABS

4.11.8.1 T 25

Consider curtailing the centre Y32 .

LA equation

Let ‘x’ be the distance from the RHS of span 5-6 to the theoretical cut-off point. Considering load case 2:

4.11.6.2

3.59


G.K. Parrott

SABS

DEFLECTION - span 5-6 T 10

4.3.6.3.1

T 11

4.3.6.5

BENDING - positive moment on span 4-5

4.3.1.5

3.60

Module 3 - Beams


G.K. Parrott

Module 3 - Beams

SABS

LA curve

4.3.3.4

4.11.4

4.11.8.1

BENDING - negative moment at support 5

3.61


G.K. Parrott

Module 3 - Beams

SABS

4.3.3.4

Utilise the Y32's provided in the bottom of the beam for compression steel.

Use the two Y32 bars (As = 1 608) and ensure adequate compression anchorage

4.11.5.1

3.62


G.K. Parrott

Module 3 - Beams

SABS

Consider curtailing the lower 2 Y25 bars and reducing the central Y32 to a Y20 .

LA equation

Let ‘a’ be the distance from the LHS of span 4-5 to the theoretical cut-off point. Considering load case 2:

Let ‘b’ be the distance from the RHS of span 5-6 to the theoretical cut-off point. Considering load case 1:

3.63


G.K. Parrott

SABS

4.11.6.2

4.3.4

3.64

SHEAR - just to right of support 5

Module 3 - Beams


G.K. Parrott

Module 3 - Beams

SABS

4.11.4.5.3 4.11.4.5.4

Adopt

Consider curtailing the stirrups to Y10 @ 250 c/c

Consider 2Y25 and 1Y20 to determine vc

4.3.4.1.2

4.3.4.1.3

3.65


G.K. Parrott

Module 3 - Beams

SABS

The shaded portion of the shear force envelope shown above represents the length of the beam over which the Y10 stirrups are required to be spaced at 125 centres. In between these shaded portions the stirrups may be spaced at 250 centres .

3.66


G.K. Parrott

Module 3 - Beams

3.67

MODULE 4

SLABS Slabs are the elements that generally receive the load first and transfer it to the columns directly, or through beams to the columns. Slabs are designed for ‘flexure’ (bending), ‘shear’, ‘torsion’ and ‘deflection’ and may span in one or both directions.


G.K. Parrott

Module 4 - Slabs

CONTENTS Page 4.1

INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1

4.2 ANALYSIS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.1 One-way spanning . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.2 Two-way spanning . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.2.1 Simply supported unrestrained slabs . . . . . . . . . . . . . . . . . . . . . 4.2.2.2 Restrained . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.3 Flat slabs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.3.1 General configuration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.3.2 Division of panels . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.3.3 Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

4.2 4.3 4.3 4.4 4.5 4.6 4.6 4.7 4.8

4.3 DESIGN FOR FLEXURE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3.1 Reinforcement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3.2 Nominal reinforcement and spacing of bars . . . . . . . . . . . . . . . . . . . . 4.3.2.1 Minimum area of tension reinforcement . . . . . . . . . . . . . . . . . . 4.3.2.2 Minimum spacing of bars . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3.2.3 Maximum spacing of bars . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3.3 Placement and curtailment of bars . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3.3.1 One-way spanning slabs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3.3.2 Two-way spanning slabs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3.3.3 Flat slabs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

4.11 4.11 4.12 4.12 4.12 4.12 4.12 4.13 4.13 4.14

4.4 DEFLECTION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4.1 One-way spanning slabs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4.2 Two-way spanning slabs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4.3 Flat slabs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

4.15 4.15 4.15 4.16

4.5 DESIGN FOR SHEAR . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.5.1 Beam shear . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.5.2 Punching shear . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.5.2.1 Point load on solid slabs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.5.2.2 Flat slabs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

4.16 4.16 4.17 4.17 4.19

4.6 DESIGN EXAMPLES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.6.1 Simply supported slab . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.6.2 Restrained slab . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.6.3 Flat slab . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

4.19 4.19 4.24 4.27

4.7 OTHER SLAB SYSTEMS . . . . . . . . . . . . . . . . . . . . . . . 4.7.1 Rib and hollow block . . . . . . . . . . . . . . . . . . . . . . 4.7.2 Coffer or waffle . . . . . . . . . . . . . . . . . . . . . . . . . . 4.7.3 Voided construction . . . . . . . . . . . . . . . . . . . . . . . 4.7.4 Permanent formwork . . . . . . . . . . . . . . . . . . . . . . 4.7.5 Ribbed beam . . . . . . . . . . . . . . . . . . . . . . . . . . . .

4.35 4.35 4.36 4.37 4.37 4.38

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


4.1

G.K. Parrott

Module 4 - Slabs

INTRODUCTION

Slabs are those elements forming floors and roofs of buildings, and usually support an uniformly distributed load (kN/m²). Slabs are usually relatively thin and deflection is often a major factor in the design. The end (edge) conditions (supports) to a slab may be f ree, simply supported or continuous. To identify these conditions clearly in our design calculations for any particular slab panel under consideration, the following convention will be adopted.

Most slabs are designed assuming that they consist of a series of rectangular beams one metre (1000 mm) in width.

Slabs may span in one or two directions depending on the location of the supports provided.

One-way spanning

Two-way spanning

4.1


4.2

G.K. Parrott

Module 4 - Slabs

ANALYSIS

The methods that follow for analysing any particular type of slab are based on a uniform load over the entire slab panel area. Non-uniform loads such as a point load (from a column) and linear load (from a wall) may however occur. When non-uniform loads are present, it is best to first calculate the applied moment due to the uniform loading alone, and then to add this to the moment resulting from the nonuniform load using the principle of superposition. Although it is possible to share the non-uniform load between the long and short span of two-way spanning slabs, it is easiest (although a bit conservative) to consider the short span as carrying all of the non-uniform portion of the load. Consider the following two cases that follow where the slab is loaded with a wall producing a linear load. - Wall across short span It may be assumed that only one strip of slab will carry the full load. This single strip would therefore require more reinforcement than the other strips.

- Wall across long span In this case it may be assumed that the wall will act as a point load on the short span, and each strip of slab will carry a point load equal to the weight of a 1 m length of wall.

4.2


4.2.1

G.K. Parrott

Module 4 - Slabs

ONE-WAY SPANNING

A one-way spanning slab is that supported on opposite ends of the panel only, or where the panel is supported on all four edges. If the short span is less than half of the l ength of the long span, the slab is designed across the short span only. Slabs spanning in one direction only (simply supported or continuous) may be analysed as for beams. For continuous one-way spanning slabs the bending moments and shear forces may be obtained from table 13 in SABS 0100:1 provided that: - the area of each bay exceeds 30 m² Table 13 -

Ultimate bending moments and shear forces in one-way spanning slabs Position

Moment

Shear

At outer support

0

0.4 F

Near middle of end span

0.086 F L

-


-0.086 F L

0.6 F


0.063 F L

-

At interior supports

-0.063 F L

0.5 F

NOTE: F is the total ultimate load

4.2.2

TWO-WAY SPANNING

Slabs spanning in two directions are more difficult to analyse, and various options of analysis are available: -

Yield line Strip Finite element Moment coefficients

The use of moment coefficients is by far the simplest method, and these tables are given in SABS 0100. Table 14 Table 15 Table 16

-

simply supported unrestrained slabs restrained slabs flat slabs 4.3


4.2.2.1

G.K. Parrott

Module 4 - Slabs

Simply supported unrestrained slabs

An unrestrained slab is a single rectangular slab panel which is simply supported on all four edges and has no means of restraint to prevent the corners from lifting. Using moment coefficients, the moments per unit width may be obtained from the following equations (note that the coefficients are given such that both the long and the short span moments are calculated using the length of the short span):

where:

=

maximum moment on short span

=

maximum moment on long span

=

total ultimate load

and

are the coefficients given below, which is a slight modification to Table 14 given in the code.

1.0

0.045

0.045

1.1

0.061

0.038

1.2

0.071

0.031

1.3

0.080

0.027

1.4

0.087

0.023

1.5

0.092

0.020

1.6

0.097

0.017

1.7

0.100

0.016

1.8

0.102

0.016

1.9

0.103

0.016

2.0

0.104

0.016

Note that if the ratio of long span to short span exceeds 2.0, then the slab is assumed to span in one direction across the short span only, and

4.4


G.K. Parrott

Module 4 - Slabs

The maximum shear force need only be calculated on the short span by assuming it to be simply supported and spanning in that direction only.

4.2.2.2

Restrained slabs

When the corners of a panel are prevented from lifting and adequate provision to resist torsion at the corners is made, the maximum moments per unit width are given by the following equations:

where:

= and

total ultimate load are the coefficients from table 15

The diagram below shows the plan of a building constructed using two-way spanning slab panels supported on beams, and indicates each slab panel as a particular case referring to Table 15.

Shear forces may be calculated across the short span using the shear coefficients given in Table 13, assuming that the short span is continuous and one-way spanning. By doing this, the shear forces across the long span can be ignored.

4.5


4.2.3 4.2.3.1

G.K. Parrott

Module 4 - Slabs

FLAT SLABS General configuration

A flat slab is a reinforced concrete slab which is generally supported by columns only. Although not common, the perimeter of the slab may be supported by walls or beams.

The column and slab may be of constant dimension.

The column may be increased in dimension towards the soffit of the slab to form a ‘column head ’. The angle of slope of the column head, for the purposes of analysis, must not exceed 45o from the vertical. The diameter of the column head is taken at a distance of 40 mm below the soffit of the slab or drop, and may not be taken as greater than 0.25 the average dimension of the panel. If the column head is rectangular, consider a circle of the same area as the cross-section of the head.

The slab may be of constant thickness or may be increased in thickness locally around the column or column head. This is referred to as a ‘drop’.

A combination of a ‘column head’ and ‘drop’ may also be adopted.

4.6


G.K. Parrott

Module 4 - Slabs

Although the thickness of a flat slab is usually greater than that required for beam and slab construction, the omission of beams gives shorter storey heights and construction is simpler.

The slab is divided into panels contained by the lines joining the column centres.

4.2.3.2

Division of panels

Flat slabs are two-way spanning and each panel is divided into strips in both directions as follows: Where there are no drops provided, the flat slab panel is to be divided into column strips and middle strips as shown below.

Where drops are provided, the width of the column strips is made equal to the size of the drop, except that drops less than one-third the size of the smaller panel dimension are ignored.

4.7


4.2.3.3

G.K. Parrott

Module 4 - Slabs

Analysis

Continuous frame analysis For this method, the structure is divided into a series of frames or sub-frames and analysed in accordance with Module 2.

Simplified analysis As long as the following restrictions are met, a flat slab may be analysed using bending moment and shear force coefficients given in table 16 of SABS 0100:1. -

Stability must be provided by bracing or shear walls.

-

The ratio of the characteristic imposed load to the characteristic self-weight load may not exceed 1.25.

-

The characteristic imposed load may not exceed 5 kN/m².

-

The slab must comprise a series of rectangular panels in at least three rows in each direction.

-

Each panel must be of approximately equal span in the direction under consideration (within say 15%).

-

The ratio of long span to short span may not exceed 2.

Table 16 is used to determine the total panel bending moments and shear forces as well as the total moment to be distributed between the columns above and below. Table 16 - Bending moments and shear force coefficients for flat slabs of three or more equal spans Position

Moment

Shear

Total column moment

Outer support:

4.8

Column

-0.040 F L

0.045 F

0.04 F L

Wall

-0.020 F L

0.40 F

-

Near middle of end span

0.083 F L

-

-


-0.063 F L

0.60 F

0.022 F L


0.071 F L

-

-

At interior supports

-0.055 F L

0.50 F

0.022 F L


G.K. Parrott

Module 4 - Slabs

The moment thus calculated is then divided between the middle and column strips in the following proportion: Moments

Column strip

Middle strip

Negative

75 %

25 %

Positive

55 %

45 %

The bending moment coefficients obtained from table 16 can be presented on a bending moment diagram as shown below:

4.9


G.K. Parrott

Module 4 - Slabs

It must be noted, however, that the maximum moment that can be developed at the outside edge of a column strip cannot be greater than the moment that can be transferred into the column at that point. It will be assumed that the maximum moment that can be transferred into an exterior column by the column strip will be equal in magnitude to the flexural capacity of a strip of slab having an effective width (be) equal to the width of the column parallel to the edge of the slab (Cx) plus the distance measured from the edge of the slab to the internal edge of the column (y). i.e.

(bs cannot be taken greater than the width of the column strip)

The maximum moment that can be developed at the outside edge of a column strip is therefore given as: ML max = 0.156 fcu b d2 If, in the analysis of the slab, a moment exceeding ML max is obtained at this position, then the moment must be reduced to ML max and the positive moment in the exterior span adjusted accordingly. The normal limitations on redistribution of moment may be ignored in this case.

4.10


G.K. Parrott

Module 4 - Slabs

Where the column strip is taken as equal to the width of the drop, and the middle strip is thereby increased in width, increase the moments to be resisted by the middle strip in proportion to its increased width. The moments to be resisted by the column strip may then be decreased by an amount such that there is no reduction in either the total positive or the tot al negative moments resisted by the column strip and middle strip together. These moments are applied to the full strip width; therefore a half-column strip parallel to the edge of the slab would carry half of the column strip moment calculated. Where the edge of the slab is supported by a load-bearing wall or beam, the moments on the half-column strip adjacent to that edge are taken as one-quarter of the moments in the interior column strip parallel to the edge. The wall or beam should be designed to carry the direct load plus a uniformly distributed load equal to one-quarter of the total load on the panel.

4.3

DESIGN FOR FLEXURE

4.3.1

REINFORCEMENT

The calculation of the required area of reinforcement to resist the applied ultimate bending moment is no different to that of beams, except that in thin slabs (thickness less than say 250 mm) it is preferable not to require compression reinforcement. Because it is convenient to use a design width of 1 m, the f ollowing table proves useful in choosing a suitable arrangement of reinforcement. Area for different diameters and spacing of bars

Spacing

BAR DIAMETER 8

10

12

16

20

25

32

50

1005

1571

2262

4021

6283

9817

16085

75

670

1048

1508

2681

4189

6545

10723

10 0

503

785

1131

2011

3142

4909

8042

12 5

402

628

905

1608

2513

3927

6434

15 0

335

524

754

1340

2094

3272

5362

17 5

287

449

646

1149

1795

2805

4596

20 0

251

393

564

1005

1571

2454

4021

22 5

226

353

508

904

1414

2208

3619

25 0

201

314

452

804

1257

1963

3217

27 5

184

288

414

737

1152

1799

2949

30 0

168

262

377

670

1047

1636

2681

32 5

156

243

350

622

972

1519

2489

35 0

144

224

323

574

898

1402

2298

37 5

134

209

302

536

838

1309

2145

40 0

126

196

283

503

785

1227

2011

4.11


4.3.2

Module 4 - Slabs

NOMINAL REINFORCEMENT AND SPACING OF BARS

4.3.2.1

Minimum area of tension reinforcement

[4.11.4]

Ensure that, in both directions, the area of tension reinforcement in a slab is not less than 0.13% b h when using high yield reinforcement, or 0.24% b h when using mild steel, where: b = the breadth of the slab ( 1000 mm ) h = overall depth of concrete

4.3.2.2

Minimum spacing of bars

It is essential for the proper compaction of the concrete that the spacing of bars allows: a) b)

for the stone in the concrete to pass between the bars and that the reinforcement in the top of the slab allows for a vibrator to be inserted.

See clause 4.11.8.1 of S.A.B.S 0100 for a further guide to bar spacing.

4.3.2.3

Maximum spacing of bars

[4.11.8.2.2]

These requirements provide control of cracking. When a slab is not more than 200 mm thick when using high tensile steel, or 250 mm thick when using mild steel, no check is required providing the area of reinforcement p rovided is less than 0.3 % and the bars have a maximum spacing of three times the effective depth. When the slab is thicker than that given above or the amount of tension reinforcement exceeds 1 %, the clear distance between bars should be limited to the values given in table 25. For percentages between 0.3 % and 1 %, the clear distance between bars should be limited to the values given in table 25 divided by the percentage of reinforcement.

4.3.3

PLACEMENT AND CURTAILMENT OF BARS

In all types of slabs, it is convenient to curtail every second bar if the requirements of maximum spacing and minimum area are still satisfied.

Consider the two detailing options (shown adjacent) of curtailing 50% of the reinforcement in a single span. In the option on the left only one bar type is used and alternate bars are reversed. In the second option two bar types are used which alternate in position. 4.12


Module 4 - Slabs

Due to the allowed tolerances in the overall dimensions of the bent bars, the second option is not preferred, since there will be inadequate cover if the bars are bent longer than scheduled. Also, if the bars are bent longer than scheduled there could be inadequate anchorage at the supports.

4.3.3.1

One-way spanning slabs

The reinforcement is curtailed the same as beams. i.e. the bending moment equation can be used to determine the theoretical cut-off points of the remaining bars, or the simplified rules of curtailment may be adopted.

4.3.3.2

Two-way spanning slabs

In each particular direction, the calculated required area of reinforcement must be placed in a central strip taken as being equal to three-quarters of the full width perpendicular to the direction of the span. Only nominal reinforcement is required in the remaining strips (shown shaded above), which are partially supported by the parallel edges. Detailing as mentioned here can however become cumbersome, and in practice the required area of reinforcement is often provided over the full width perpendicular to the direction of the span.

Paragraphs a) to d) of clause 4.4.4.2.2 set out the simplified rules for curtailing the reinforcement in the bottom and top of the slab. Although the reinforcement may be curtailed to less than the minimum area required (0.13%), it is preferable for the reinforcement in the bottom of the slab at an exterior edge of each panel to be at least the minimum required. The diagram below shows how this curtailment may be applied in each particular direction.

4.13


Module 4 - Slabs

Torsion steel In restrained two-way spanning slabs, because the corners of the slab are prevented from lifting, torsional stresses will develop at the external corners of the exterior panels. Torsion reinforcement as detailed below will be required to resist these torsional stresses and to prevent cracking in that region.

Both edges simply supported : Provide torsion reinforcement consisting of top and bottom bars as shown, where the area of reinforcement in each of these four layers must be three-quarters of the area required for the maximum mid-span moment in the slab.

One edge continuous: Provide torsion reinforcement equal to half that described above.

4.3.3.3

Flat slabs

[4.6.5.4] & [4.11.7.3.3]

Tension steel should extend over the full width of each strip and may be curtailed as follows: Top steel :

Two-thirds of the amount of reinforcement required to resist the negative moment in the column strip should be placed in a width equal to half that of the column strip and central to the column strip. 50% of all tension reinforcement in the top of the slab should extend 0.3 beyond the line joining the centre of the columns. The remaining 50% should extend the greater of 0.15  or 45 times the diameter of the bar beyond the line joining the centre of the columns.

Bottom steel:

4.14

Half the bottom reinforcement should extend 20 diameters beyond the line joining the centre of the columns. The other half shall extend to within a distance of 0.2 from the line joining the centre of internal columns and 0.1 from the line joining the centre of external columns.


4.4 4.4.1

Module 4 - Slabs

DEFLECTION ONE-WAY SPANNING SLABS

SABS 0100 states that it will be sufficient to restrict the span to effective depth ratio (table 10, modified by tables 11 and 12). Consider only the reinforcement at the centre of the span in the width of slab under consideration to influence deflection. Base the ratio on the short span and the amount of reinforcement in that direction.

4.4.2

TWO-WAY SPANNING SLABS

For two-way spanning slabs, checking deflection using the span to effective depth ratio is usually too conservative since it does not take into account the stiffening effect of the perpendicular long span. Because of this it often becomes necessary to check the actual working deflection against the normal allowable limits of span/250 for total service deflection and span/300 for working service deflection. The determination of the expected deflections in two-way spanning slabs is very difficult to calculate and is usually done using a finite element computer software package. Although not perfectly accurate, it will generally be acceptable to assume that the deflection of a two-way spanning slab panel compared to a similar simply supported one-way spanning panel would be reduced by an equal proportion to what the bending moment would reduce. In other words:

The deflection can be calculated assuming the slab to be simply supported and spanning one way across the short span only, and then reduced by a factor equal to the bending moment coefficient that was used in calculating the bending moment in the short span divided by 0.125 (where 0.125 is the bending moment coefficient for a simply supported one-way spanning element).

estimated deflection =

(The co-efficient is obtained in table 14 or table 15.)

4.15


4.4.3

Module 4 - Slabs

FLAT SLABS

For slabs with drops of total width in both directions equal to at least one-third of the respective spans: It will be sufficient to restrict the span to eff ective depth ratio. (table 10, modified by tables 11 and 12). Consider only the reinforcement at the centre of the span in the strip under consideration to influence deflection. Base the ratio on the longer span and the amount of reinforcement in that direction. For slabs with no drops, or where the drops are of total width in both directions less than one-third of the respective spans, multiply the allowable span/effective depth ratio by 0.9.

4.5

DESIGN FOR SHEAR

Where the actual ultimate shear stress in the slab stress , shear reinforcement will be required.

exceeds the ultimate allowable shear

It is however preferable not to require shear reinforcement in thin slabs (thickness less than say 250 mm) since the efficiency of the short vertical legs of the links is reduced. Note that nominal shear reinforcement is not required in slabs when beams.

as it is in

Two types of shear stresses need to be considered in slabs, viz. one-way shear (as in beams) and two-way shear (punching shear).

4.5.1

BEAM SHEAR

The ultimate shear stress spacing ratio

, allowable shear stress

, and the required link area to

are calculated as for beams (except that:

= 1000 mm).

One-way shear reinforcement in a slab may be in the form of beam cages in one direction and loose bars in the other direction as shown below.

4.16


Module 4 - Slabs

The width and spacing of the cages will determine the spacing of the shear reinforcement ( ), and the number of legs per metre-wide strip will determine the area of the links ( ). Note that the spacing of the legs

4.5.2

must not exceed 0.75 d.

PUNCHING SHEAR

4.5.2.1

[4.4.5.2]

Point load on solid slabs

When a concentrated load is applied to a slab, a portion of the slab will tend to be 'punched' out as shown.

The shear stress is calculated by dividing the ultimate concentrated load by an effective shear plane. This shear plane is taken as having a depth equal to the average effective depth of the slab considering both layers of tension reinforcement , and a breadth equal to the length of a perimeter .

The maximum shear stress, which is calculated using the actual perimeter of the loaded area The perimeter

, should not exceed the lesser of may be defined as:

or

.

The boundary of the smallest rectangle that can be drawn around the loaded area such that no part of the rectangle comes closer than a specified distance .

Note: Refer to the code for cases where the perimeter may need to be reduced to an effective length of a perimeter to allow for the effects of openings or external edges.

4.17


Module 4 - Slabs

The distance has a minimum value of 1.5 0.75 until, on such a perimeter, required. The calculation of

and is taken to increase in increments of at which point no shear reinforcement is

is as before, in which the term

is given by the average for

the two directions.

When, on a particular perimeter , reinforcement is required to resist the shear stress, this reinforcement should be evenly distributed in a zone of slab between that perimeter and a perimeter inside of it.

This reinforcement should be in the form of vertical links, with each leg not greater than apart and spaced at no more than . This reinforcement may be calculated as follows:

for

for

Avoid requiring shear reinforcement in slabs less than 200 mm thick and for any slab thickness where , in which cases a rational design is required. Note that in these equations, 4.18

may not be taken as less than 0.4.


Module 4 - Slabs

The simplest method, yet slightly conservative, of designing this shear reinforcement is to design a suitable beam cage to satisfy the requirements of t he first failure zone, and then to extend these beam cages to the perimeter just before that where .

4.5.2.2

Flat slabs

[4.6.2]

Punching shear is a critical design consideration in flat slab design. The effective shear force depends on the position of the column under consideration and may be obtained from the table below, which includes an allowance for non-symmetrical loading and moment transfer from the slab to the column. Column position

where:

internal columns

1.15

corner columns

1.25

edge columns

1.40

is the force obtained by considering each column to take one-quarter of the ultimate design load on each adjoining panel.

The maximum shear stress for flat slabs is slightly more that that for point load on a solid slab, and should not exceed the lesser of

or

.

The methods for checking the shear stresses or designing shear reinforcement are as mentioned previously.

4.6 4.6.1

DESIGN EXAMPLES SIMPLY SUPPORTED SLAB

Analyse, design and detail the roof slab for a single garage using grade 25/26 reinforced concrete. The slab is to be 150 mm thick, and will be supported on 220 walls on three sides, and on a 400 X 220 beam over the opening. The internal dimensions of the garage are to be 4 000 X 6 500. The roof is to be used as an open verandah and be capable of supporting a live load of 3 kN/m². A screed with an average thickness of 50 mm will be placed on top of the concrete, and will be finished with 35 mm mastic asphalt waterproofing.

4.19


SABS

4.3.1.2

Table 14

SHORT SPAN 4.3.3.4.1

4.20

Bending

Module 4 - Slabs


Module 4 - Slabs

SABS

LA curve

4.3.3.4

4.11.4

4.11.8.2.2 Try:

Try:

Adopt: Note that the first option could work out to be more econo mical even although curtailmen t is not possible due to the maximum spacing being exceeded, but the second option has been adopted in any case, to dem onstrate the detailing of the curtailed steel.

4.4.5.1

Shear

4.21


Module 4 - Slabs

SABS

Deflection

Table 10

Table 11

Alt ho ug h the sla b do es no t s ati sfy th e require men ts of sp an /de pth ratio , th e actual ex pe cte d long-term service deflection can be calculated and compared with the recommended limits. (This is more often than n ot the case with most slabs.)

4.22


Module 4 - Slabs

SABS

Table 1

3.2.3.2.1.1

LONG SPAN 4.3.3.4.1

Bending

LA curve

4.11.4

4.11.8.2.2 Adopt: 4.23


Module 4 - Slabs

SABS

4.4.4.1

4.6.2

RESTRAINED SLAB

Determine the ultimate bending moments and shear forces in the panel marked ‘A’ in the floor layout shown overleaf. The slab is supported by beams that are 250 mm wide and cast monolithically with the slab. The slab is to be 165 mm thick and finishes to the floor may be taken as 1.0 kN/m². The occupancy of the floor is offices for general use.

4.24


Module 4 - Slabs

SABS

0160 Table 4

4.25


Module 4 - Slabs

SABS

where:

Table 15

Panel is case 4 Position

Table 13

4.26

Coefficient

Moment (kNm)

short span positive

+0.065

14.59

short span negative

-0.087

19.53

long span positive

+0.034

7.63

long span negative

-0.045

10.10


4.6.3

Module 4 - Slabs

FLAT SLAB

Flat slab construction is to be used for the floor of an office block with a plan dimension of 40 m x 30 m. The columns will be spaced at 8 m centres over the length of 40 m and at 6 m centres over the length of 30 m. The perimeter of the slab is supported on 220 mm loadbearing brickwork and stability to the structure will be provided by shear walls. 25 mm cover is required to all reinforcement. The finishes to the slab will amount to approx. 1.0 kN/m² and the slab is to support a live load of 4 kN/m².

The columns are to be 350 x 350 square with 1.6 m square heads as shown.

Use grade 25/26 concrete and high tensile reinforcement to design suitable reinforcement for the interior edge panel on the 40 m length (shown shaded below). Check one of the interior columns for punching shear and check the panel f or deflection.

4.27


SABS

4.6.1.3

Check requirements for simplified analysis: 4.6.5.2

4.28

Module 4 - Slabs


Module 4 - Slabs

SABS BENDING - 8 m span

4.6.1.4

4.29


Module 4 - Slabs

SABS T16 & T17

POSITION

Total strip moment

b

Moment per metre

middle positive

132.3

3 000

44.1

middle negative

57.0

3 000

19.0

column positive

161.8

3 000

53.9

column negative

170.9

3 000

57.0

POSITION

4.11.4

4.30

*

K

Z/d

middle positive

0.0368

0.950

middle negative

0.0158

column positive column negative

As

Reinf.

As prov.

541

Y12@200

564

0.950

233*

Y10@225

353

0.0450

0.947

664

Y12@150

754

0.0475

0.944

704

Y12@150

754

req.


Module 4 - Slabs

SABS BENDING - 6 m span

4.6.1.4

T16 & T17

4.31


Module 4 - Slabs

SABS T16 & T17

POSITION

Total strip moment

b

Moment per metre

middle positive

103.2

5 000

20.6

Ext. middle negative

13.8

5 000

2.76

Int. middle negative

43.5

5 000

8.70

column positive

126.1

3 000

42.0

Ext. column negative

41.4

3 000

13.8

Int. column negative

130.5

3 000

43.5

POSITION

4.11.4 *

4.32

K

Z/d

middle positive

0.0193

0.950

Ext. middle negative

0.0026

Int. middle negative

As

Reinf.

As prov.

268*

Y10@225

353

0.950

36*

Y10@225

353

0.0081

0.950

113*

Y10@225

353

column positive

0.0392

0.950

546

Y12@200

564

Ext. column negative

0.0129

0.950

179*

Y10@225

353

Int. column negative

0.0406

0.950

565

Y12@200

564

req.


Module 4 - Slabs

SABS SHEAR - internal column 4.6.2.2.1

Extend all reinforcement in the top of the slab over the column a minimum distance of an effective depth beyond the exterior design perimeter.

4.3.4.1.2

4.6.2.4

4.4.5.2.2

4.33


Module 4 - Slabs

SABS 4.4.5.2.3

No shear reinforcement is required!

DEFLECTION - long span middle strip T10

4.3.6.3.1

T11

4.6.3

Increase the slab depth or check the actual deflection using computer software

4.34


4.7

Module 4 - Slabs

OTHER SLAB SYSTEMS

There are various methods of constructing in-situ slabs in such a way that maximum use is made of our design theory to reduce the amount of concrete and reinforcement over a given floor area. At times a system may be required where a conventional solid slab is not practical or economical. These slabs are usually deeper than the conventional slab and generally require little or no temporary support. Each system has its advantages and disadvantages which should be considered before adopting any particular system.

4.7.1

RIB AND HOLLOW BLOCK

[4.5.1.1 a) 1)]

This system consists of concrete ribs cast in situ between blocks that remain part of the completed structure. The tops of the ribs are connected by a topping of concrete forming flanges to the ribs. The main reinforcement is provided in reinforced or prestressed concrete planks which are propped to support the blocks during the placing of the in situ concrete. The rib and hollow block slab can span in one direction only, and is usually designed as simply supported single spans. The blocks are available in a variety of sizes, are generally hollow and are shaped to provide a suitable rib above the planks into which the in-situ concrete may be placed.

The thickness of the topping should be a minimum of 40 mm or one-tenth of the clear distance between the ribs. Design and detailing A single layer of mesh reinforcement should be provided to control cracking in the topping. The mesh should have a cross-sectional area in each direction of at least 0.12% of the cross-sectional area of the topping, and the spacing of wires should not exceed half the centre-to-centre distance between the ribs. Where the slab has been designed as simply supported but is in fact continuous, the reinforcement provided in the top of the slab should be at least 25% of that required in the middle of adjoining spans.

4.35


Module 4 - Slabs

It is common to ignore the strength of the hollow blocks in the design of such a system and the slab is designed as a series of flanged beams. The same methods for determining the ultimate moment of resistance of beams may be used. The same methods for determining the ultimate shear resistance of beams may be used except that where < , no shear reinforcement is required.

4.7.2

COFFER OR WAFFLE

[4.5.1.1 a) 2)]

As the name suggests, these slabs have the appearance of a waffle when viewing the finished slab soffit. This system uses moulded fibre-glass elements as temporary formwork. These form moulds are available in a variety of sizes. The space between the moulds will form the rib in which the tension reinforcement is placed, and the topping of concrete above the moulds will form the compression flange. The moulds are square in plan, which makes this system capable of spanning in two directions.

A typical cross-section is shown here:

The thickness of the topping should be a minimum of 50 mm or one-tenth of the clear distance between the ribs. Design and detailing A single layer of mesh reinforcement should be provided to control cracking in the topping. The mesh should have a cross-sectional area in each direction of at least 0.12% of the cross-sectional area of the topping, and the spacing of wires should not exceed half the centre-to-centre distance between the ribs. The slab is designed as a series of flanged beams, and the same methods for determining the ultimate moment of resistance of beams may be used. At least 50% of the tension reinforcement should be continued to the supports and anchored accordingly (clause 4.11.7). The same methods for determining the ultimate shear resistance of beams may also be used except that where < , no shear reinforcement is required.

4.36


4.7.3

Module 4 - Slabs

VOIDED CONSTRUCTION

This system is constructed with a continuous top and bottom containing voids of various shapes. Generally the shapes used are either rectangular or oval. The material used to form these voids may be any relatively strong light-weight material such as cardboard or polystyrene.

Design and detailing A single layer of mesh reinforcement should be provided to control cracking in the topping. The mesh should have a cross-sectional area in each direction of at least 0.12% of the cross-sectional area of the topping, and the spacing of wires should not exceed half the centre-to-centre distance between the ribs. The slab is designed as a series of f langed beams, and the same methods for determining the ultimate moment of resistance of beams may be used. An advantage of this system is that it contains a bottom flange which increases the resistance to negative moments. At least 50% of the tension reinforcement should be continued to the supports and anchored accordingly (clause 4.11.7). The same methods for determining the ultimate shear resistance of beams may also be used except that where < , no shear reinforcement is required.

4.7.4

PERMANENT FORMWORK

This system incorporates a concept of composite co-action between concrete and steel, which is dependent upon unique deformations of the profiled steel unit. The strength of this system is difficult to analyse accurately, but the suppliers of the system provide design charts giving the allowable span/load for the different profil es available.

The profiled steel units are usually of sufficient strength to support the weight of the wet concrete and construction loads such that no propping is required.

4.37


4.7.5

Module 4 - Slabs

RIBBED BEAM

This system consists of precast 'T' beams (usually prestressed) which, when laid side by side, form the basis of a suspended concrete slab. The added advantage of this system is that it requires no propping. This means that, for example, a domestic house may be altered into a double storey without disturbing the existing ceiling.

4.38

MODULE 5

STAIRCASES Stairs are generally designed as solid slabs spanning in one direction. A stair may be designed to: spa span lon long gitu itudin dinally ally betw betwee een n lan landi ding ngs s, wal walls ls or beam beams s span laterally between wall alls stri tringer beams cant cantil ilev ever er later ateral allly from from a stif stifff wal walll or spin spine e beam beam


Module 5 - Staircases

CONTENTS Page 5.1

.................................................... General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Imposed load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Dead load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

LOADING

5.1.1 5.1.2 5.1.3 5. 5.2 2

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Effective sp span . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2

ANALY NALYSI SIS S

5.2.1 5. 3

5.1 5.1 5.1 5.2

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . Longitudinally spanning . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Example 5.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Laterally spanning . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

DESIGN

5.3.1 5.3.1.1 5.3.2

5.3 5.3 5.4 5.7


5.1 5.1 5.1.1


LOA LOADING ING GENERAL

Although a staircase is inclined, the loading in taken on plan area and the span is taken as a horizontal distance.

When the two spans of stair landings are at right angles to each other, the ultimate load on the common areas may be assumed to be divided equally between the two spans. Consider the plan of a staircase surrounding a lift shaft shown:

5.1.2

IMPOSED LOAD

The imposed load on a stair should be taken as at least equal to t he imposed load on any area giving access to the stair. Where staircases are nominally built in to walls by at least 110 mm, a strip of 150 mm adjacent to the wall may be neglected in calculating the loading - i.e. considering the diagram below, the imposed load is only applied to t he shaded portion.

5.1 5. 1


5 .1 . 3


DEAD LOAD

The dead load should include the self-weight self -weight of the stair as well as any applied finishes. f inishes. The self-weight of a staircase is best calculated by considering one 'step' and then converting that into either a lateral or longitudinal horizontal linear load.

In the figure above, the area of a single 'step' is given by:

5. 5.2 2

ANALY NALYSI SIS S

Staircases are usually simply supported and therefore the normal rules of static equilibrium may be used to calculate the bending moments. etc. Special stairs such as spirals will, however, require a more rational analysis. 5 .2 . 1

EFFECTIVE SPAN

[4.9.1.3]

For stairs spanning onto landings which span at right angles to the span of the staircase, the effective support position shall be taken as that to the centre of the landing subject to a maximum distance from the end of the stair of 900 mm. i.e.

5.2 5. 2



In the case of a stair supported by a distinct element, the effective support position shall be taken as that to the centre of the support width subject to a maximum distance from the inside face of the support of half the effective depth of the stair.

5.3 5.3.1

DESIGN LONGITUDINALLY SPANNING

These stairs are the most common and are designed as solid slabs using the waist of the stair as the overall slab depth. Distribution reinforcement should preferably be spaced at one bar per step. The positioning of longitudinal reinforcement at the flight/landing intersection is of the utmost importance. The following diagrams may be used as a guide.

Supported at end of landing (simply supported)

Supported by landing (semi-continuous)

5.3



Supported at end of flight (continuous)

5.3.1.1

Example 5.1

Design and detail the staircase shown below, which spans longitudinally between beams at the end of each landing. The landing, risers and treads of the staircase will be finished with a screed 20 mm thick (the soffit finish will be ‘off-shutter’). Consider an imposed live load of 1.5 kN/m2.

5.4



SABS

UDL 1

UDL 2

ANALYSIS

5.5


S.A.B.S.

DESIGN FOR FLEXURE Longitudinal reinforcement:

Adopt

Transverse reinforcement:

Adopt

DESIGN FOR SHEAR

This is very low (< 0.4) therefore shear is okay!

5.6



5.3.2


LATERALLY SPANNING

In this case each step is designed as an independent beam using a cross-section as shown below.

5.7

MODULE 6

COLUMNS Columns are those elements acting as the vertical supports to beams and slabs. The load is transferred from beams and slabs through the columns down to the foundations. Columns are primarily compression members although they may also have to r esist bending moments applied to one or both axes.


Module 6 - Columns

CONTENTS Page 6.1

INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1

6.2

BRACING . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1

6.3 EFFECTIVE HEIGHT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3.1 General method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3.2 Rigorous method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3.3 Nomograph . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

6.2 6.2 6.3 6.4

6.4 SLENDERNESS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.5 6.4.1 Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.6 6.4.2 Example 6.1 (slenderness classification) . . . . . . . . . . . . . . . . . . . . . . . 6.7 6.5

DESIGN AXIAL LOAD . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.8

6.6 DESIGN MOMENT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.9 6.6.1 Additional moments in slender columns . . . . . . . . . . . . . . . . . . . . . . . . 6.9 6.6.2 Uni-axial bending . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.11 6.6.2.1 Braced slender columns . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.11 6.6.2.2 Unbraced slender columns . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.13 6.6.2.3 Short columns . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.14 6.6.3 Bi-axial bending . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.14 6.6.4 Example 6.2 (design moment ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.15 6.7 LONGITUDINAL REINFORCEMENT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.7.1 Derivation of design equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.7.1.1 Full compression . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.7.1.2 Partial compression . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.7.2 Design charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

6.17 6.17 6.18 6.19 6.20

6.8 OTHER CONSIDERATIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.8.1 Minimum area of main reinforcement . . . . . . . . . . . . . . . . . . . . . . . . . 6.8.2 Maximum area of main reinforcement . . . . . . . . . . . . . . . . . . . . . . . . . 6.8.3 Minimum requirement for links . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

6.28 6.28 6.28 6.28

6.9

EXAMPLE 6.3 (reinforcement ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.29


6.1

Module 6 - Columns

INTRODUCTION

The cross-section of a column can be either rectangular, circular or irregular.

Reinforcement is in the form of longitudinal bars which are prevented from buckling by closely spaced links.

6.2

BRACING

[4.7.1.3]

Braced columns are defined as those where the stability of the whole structure is provided by shear walls or other bracing designed to resist all lateral loads. A structure may be braced in one plane and unbraced in the other. Consider the basic building layouts shown in plan below:

6.1


6.3

Module 6 - Columns

EFFECTIVE HEIGHT

[4.7.1.6]

The effective height ( ) of a column is by far the most critical design consideration and will determine the mode of failure. Three methods for obtaining the effective height are given: general method, rigorous method and nomograph.

6.3.1

GENERAL METHOD

[4.7.1.6.1]

Referring to a particular axis (x-x or y-y), the effective height Le is obtained by multiplying the clear height between lateral restraints Lo by a coefficient  .

Note that where beams are used in one direction only, the clear height need not be the same for the different axes. The diagram below shows an elevation of the columns in the x-x plane. ie. buckling visible in this plane will be about the y-y axis.

The coefficient  is obtained from table 18 for a braced axis and from table 19 for an unbraced axis. These tables are based on the degree of end fixity, which is measured on a scale of 1 to 4.

End condition 1

-

beam or slab depth exceeds column dimension or foundation designed to resist moment.

End condition 2

-

column dimension exceeds beam or slab depth.

End condition 3

-

nominal end restraint only (pinned).

End condition 4

-

totally unrestrained (e.g. tip of cantilever).

6.2


Module 6 - Columns

Table 18 - Values for

End condition at top

End condition at bottom 1

2

3

1

0.75

0.80

0.90

2

0.80

0.85

0.95

3

0.90

0.95

1.00

Table 19 - Values for

End condition at top

6.3.2

for braced columns

for unbraced columns

End condition at bottom 1

2

3

1

1.2

1.3

1.6

2

1.3

1.5

1.8

3

1.6

1.8

-

4

2.2

-

-

RIGOROUS METHOD

[4.7.1.6.2]

This method for calculating the effective height of a column is dependent on the actual ratio of column-to-beam stiffness at either end. For a framed structure, the effective height may be obtained from the following equations: a)

For a braced column, the lesser of and

b)

For an unbraced column, the lesser of and

6.3


Module 6 - Columns

is the ratio of the sum of column stiffness to the sum of in-plane beam stiffness at one end of the column. is the ratio of sum of column stiffness to sum of in-plane beam stiffness at the other end of the column. is the lesser of

and

.

The stiffness of a member should be obtained by dividing the second moment of area of the section by the distance centre-to-centre of restraints. The following simplifying assumptions may be made:

6.3.3

-

For flat slab construction, use the stiffness of the column strip.

-

For a 'pinned' column end,

= 10.

-

For a 'fixed' column end,

= 1.

NOMOGRAPH

[4.7.1.6.3]

As an alternative method to using the above equations, the nomograph given in the code (figure 20) may be used.

The adjacent figure shows the nomograph for a Braced frame  is the ratio of the sum of column stiffness to the sum of in-plane beam stiffness at the end of the column (each end to be calculated separately).

A value for is to be read off from the relevant nomograph, which is multiplied by the clear column height to obtain the effective height.

6.4


6.4

SLENDERNESS

Module 6 - Columns

[4.7.1.4]

Compression failure of the concrete and or steel reinforcement will occur in columns that are short and stocky.

Buckling failure will occur in columns that are long and slender.

It is important, therefore, to be able to define a column as ' short' or 'slender ' since the failure mode for each case is different and hence the design procedure will also be different. To determine how slender a column is we shall examine the slenderness ratio, which is defined as the effective height about a particular axis divided by the depth of the column measured perpendicular to that axis (dimension in the plane of bending).

Slenderness ratio about the x-x axis =

Slenderness ratio about the y-y axis =

6.5


Module 6 - Columns

A column that is unbraced about a particular axis may be considered short if its slenderness ratio about that axis does not exceed .

A column that is braced about about a particular axis a xis may be considered short if its slenderness ratio about that axis does not exceed

where:

M1

=

smaller initial end moment which should be taken as negative for a column bent in double curvature.

M2

=

larger initial end moment, taken as positive.

Double curvature

6.4.1

Single curvature

LIMITS

Generally, the clear height

[4.7.1.5] should satisfy the following:

and For a cantilever column, however, ensure that

.

Ensure also that the average slenderness ratio of unbraced columns at any particular level does not exceed 30.

6.6 6. 6


6.4.2

Module 6 - Columns

classification) EXAMPLE 6.1 (slenderness classification)

Calculate the effective height about each axis for the column shown below and state whether the column is short or slender. All beams are 500 x 220 and the slabs are 150 thick. The storey height measured from floor to floor is 4.5 m. The structure is braced about the y-y axis only. only.

SABS

Major axis (x-x) - unbraced

Condition at top =1 Condition Condition at bottom bottom = 1

T 19 4.7.1.4

Short limit = 10

6.7 6. 7


Module 6 - Columns

SABS

Minor Axis (y-y) - unbraced

Assume for the purpose of this example that the column is bent in double curvature about the minor axis with end moments of 7 kNm and 10 kNm.

Condition at top =2 Condition Condition at bottom bottom = 1 T18

4.7.1.4

6.5

Short limit =

DESIGN AXIAL LOAD

The ultimate ultimate load (kN) applied to a column will be the sum of the load from the column immediately above, plus the reactions from the beams/slab at the top of the column, plus the self-weight of the column. The axial load may be calculated on the assumption that beams and slabs transmitting force into the column are simply supported.

6.8 6. 8


6.6

Module 6 - Columns

DESIGN MOMENT

Moment may be applied to either axis or both axes simultaneously. The primary ultimate moment should be determined from an elastic first-order analysis of the structure. This moment will not necessarily be the moment for which the column will be designed. -

In slen slende derr col colum umns ns a hig highe herr mom momen entt may may be requ requir ired ed to make make allo allow wance ance for for additional moments arising out of slenderness.

-

In colu column mns s tha thatt are are bent bent abou aboutt bot both h axe axes, s, the the col colum umn n wil willl be be des desig igne ned d to to resist an enhanced moment about one axis only.

-

The The des desig ign n mom momen entt abo about ut eith either er axis axis shou should ld not not be be les less s tha than n tha thatt res resul ulti ting ng from the design ultimate axial load acting at a minimum eccentricity equal to 0.05 times the depth of the column about the axis under consideration. This minimum eccentricity should not, however, be taken to exceed 20 mm.

The moment thus obtained will be used to calculate the reinforcement in the column and will will be refer referre red d to as the the desi design gn momen momentt ( or ).

6.6.1

ADDITIONAL MOMENTS IN SLENDER COLUMNS

[4.7.3]

Account has to be taken of the additional moment induced in the column by its deflection. This is often referred to as the effect.

Consider the axial load a deflection .

applied to a slender axis resulting in

An additional moment equal to

will result.

This induced deflection is referred to as

and is calculated as follows:

6.9 6. 9


where:

Module 6 - Columns

= (The slenderness ratio to be used here is generally the maximum, except for bi-axial bending, where the slenderness ratio for the relevant axis is used.) =

the the dim dimen ensi sion on perp perpen endi dicu cula larr to to the the axis axis unde underr con consi side dera rati tion on..

is a reduction factor that corrects the deflection to allow for the influence of axial load and may be calculated as follows: =

where:

or if this gives a value greater than unity, then =

where: = initia initiall desi design gn moment moment before before allowa allowance nce for for additional moments slenderness.

where:

arising

out

of

is the area of longitudinal reinforcement is the net area of concrete, i.e. is the dimension parallel to the axis of bending is the dimension perpendicular to the axis of bending

equations are only true true for symmetrically reinforced rectangular rectangular sections. sections. Note: These equations

You will notice that, to calculate this reduction factor, the area of longitudinal reinforcement is required. The appropriate value of may however be found iteratively, taking an initial value of 1.0. Alternatively it would always always be conservative conservative to assume that = 1.0, but this could result in providing an area of reinforcement far exceeding that required.

The additional moment in a slender column is therefore given by:

6.10


6.6.2

6.6.2.1

Module 6 - Columns

UNI-AXIAL BENDING

Braced slender columns

[4.7.3.2.1]

If a column is slender about one axis only, the additional moment need only be considered in that plane. The full additional moment Half of the additional moment

is assumed to occur near mid-height. is assumed to occur at fixed ends.

The figures below show the shape of the additional moment diagram for different end conditions.

The design moment (used for calculating the required longitudinal reinforcement) is obtained by combining the additional bending moment and the primary bending moment diagrams. The combination of these moment diagrams for a braced column fixed at both ends might look something like that shown below:

6.11


Module 6 - Columns

The curvature of the additional moment diagram is orientated in such a way that the maximum additional moment at mid-height occurs on the same side (i.e. same sign) as the maximum primary moment. Depending on the shape of t he primary and additional moment diagrams, the critical design moment could be either at the top or bottom, or at mid-height. Near mid-height the additional moment is added to an initial primary moment

which may

be given by: but is the smaller end moment due to ultimate loads and should be taken as negative for a column bent in double curvature. is the larger end moment due to ultimate loads and is assumed to be positive.

Note that regardless of the maximum value produced by the combined bending moment diagram, the column must be designed for a moment of at least . The design moment

in a braced column bent uni-axially will therefore be given by the

greatest of (a), (b) or (c) below. a)

or

[ critical at ends ]

b)

[ critical at mid-height ]

c)

[ critical at mid-height ]

6.12


6.6.2.2

Module 6 - Columns

Unbraced slender columns

[4.7.3.2.2]

For slender columns the additional moment is assumed to occur at the end of the column having the larger primary moment (i.e. the end having the larger stiffness). The magnitude of the additional moment occurring at the other end of the column may be reduced in proportion to the end stiffness. The figure below shows the shape of the additional moment diagram for the only two possible combinations of end conditions in an unbraced slender column.

The design moment (used for calculating the required longitudinal reinforcement) is obtained by adding the additional bending moment diagram to the primary bending moment diagram. The curvature of the additional moment diagram is orientated in such a way that the maximum additional end-moment occurs on the same side (i.e. same sign) as the maximum primary moment. The critical design moment will always occur at the end of the column having the greatest stiffness. The design moment

in an unbraced column bent uni-axially will therefore be given by

the greatest of (a) or (b) below. a) b)

NOTE:

The equations given in the code are more complex, and include vertical and horizontal components of the primary moment. This approach may be adopted once further explanation is found.

6.13


6.6.2.3

Module 6 - Columns

Short columns

For short columns, where no additional moment arises out of slenderness, the design moment will be given by the greatest of (a) or (b) below. a) b)

6.6.3

BI-AXIAL BENDING

[4.7.4.4]

For symmetrically reinforced rectangular columns subjected to bending about bot h axes, the section may be designed to withstand an enhanced design moment about one of the axes only.

The design axis will be that having the greater moment/depth ratio.

For x-x design axis:

For y-y design axis:

Table 21 - Values of coefficient

0.000

0.50

0.075

0.60

0.150

0.70

0.250

0.70

0.300

0.65

0.400

0.53

0.500

0.42

0.600

0.30



6.14


6.6.4

Module 6 - Columns

EXAMPLE 6.2 (design moment )

For the column given in the example on page 135, a first order linear analysis of the structure produced the following results (ULS) for the load case considered to be the most critical.

Determine the design axis and give the ultimate moment for which the section should be designed.

SABS

X-AXIS (unbraced) This axis was found to be ‘slender’ with a slenderness ratio of 13.71.

4.7.3.1

4.7.2.3

6.15


Module 6 - Columns

SABS

Y-AXIS (braced) This axis was found to be ‘slender’ with a slenderness ratio of 15.82 .

4.7.3.1

4.7.3.2.1

4.7.2.3

BI-AXIAL BENDING

6.16


6.7

odule 6 - Columns

LONGITUDINAL REINFORCEMENT

6.7.1

DERIVATION OF DESIGN EQUATIONS

To derive suitable design equations we shall consider two cases of stress distribution which may occur in a column cross-section. (Remember that depth of concrete in compression = , where is the depth to the neutral axis.) Firstly, when

the concrete is in compression over the full depth of the crosssection, and all steel is in compression

and when

compression exists over only a portion of the cross-section, and the steel opposite the more highly compressed face will be in tension.

Consider the diagram below, which will be used to derive the necessary design equations.

C T f 

= = = =

compression force tension force stress strain

The subscript

refers to the reinforcement in the more highly compressed face.

The subscript

refers to the reinforcement in the less highly compressed face. 6.17


6.7.1.1

Module 6 - Columns

Full compression

,

and

For force equilibrium, the applied ultimate axial load forces.

Dividing by

must be balanced by the internal

we get:

For moment equilibrium, the applied ultimate moment must be balanced by the internal moment of resistance taken about the centre of the cross-section.

Dividing by

6.18

we get:


6.7.1.2

Module 6 - Columns

Partial compression ,

and

For force equilibrium, the applied ultimate axial load forces.

Dividing by

must be balanced by the internal

we get:

For moment equilibrium, the applied ultimate moment must be balanced by the internal moment of resistance taken about the centre of the cross-section.

Dividing by

we get:

6.19


6.7.2

Module 6 - Columns

DESIGN CHARTS

From the foregoing equations, a set of design charts can be produced, each for different values of

,

and

Using these parameters and a chosen proportion of reinforcement values for

and

can be calculated from the foregoing equations by using the

stress in the reinforcement ( depths expressed as

, a range of

or

) corresponding to a range of depth of compression

.

The stress in the reinforcement (

or

) may be determined by calculating the strain in

the steel and multiplying it by Young's modulus (E = 200 000 N/mm²). Note, however, that the maximum ultimate stress in the steel is given as for high tensile steel in compression and

For full compression:

is compressive and by similar 's,

6.20

in tension,

for mild steel in compression.


Module 6 - Columns


for partial compression:


is tensile and by similar 's,

6.21


Values for

Module 6 - Columns

and

can then be plotted for a variety of

values, the intersection

of these two values representing the chosen of reinforcement. To use the design charts, for known values of

and

, a percentage of reinforcement

may be read

off.

If we consider only high tensile reinforcement ( grade of concrete for columns (

= 450 N/mm²) and the most common

= 30 N/mm²), the following five design charts are

produced.

Note that where the actual

ratio for the section being designed lies between two charts,

both charts may be read and the percentage of reinforcement found by linear interpolation.

6.22


Module 6 - Columns

6.23


6.24

Module 6 - Columns


Module 6 - Columns

6.25


6.26

Module 6 - Columns


Module 6 - Columns

6.27


6.8 6.8.1

Module 6 - Columns

OTHER CONSIDERATIONS MINIMUM AREA OF MAIN REINFORCEMENT

[4.11.4.2.2]

Ensure that the minimum number of longitudinal bars provided in a column is four in rectangular columns and six in circular columns and that the diameter of those bars is at least 12 mm. Ensure that the total cross-sectional area of the longitudinal bars will be at least 0.4% of the cross-sectional area of the column.

6.8.2

MAXIMUM AREA OF MAIN REINFORCEMENT

[4.11.5.2]

The area of longitudinal reinforcement in a vertically cast column may not exceed 6%. The area of longitudinal reinforcement in a horizontally cast column may not exceed 8%. For both vertically and horizontally cast columns, the area of r einforcement at laps may not exceed 10%.

6.8.3

MINIMUM REQUIREMENT FOR LINKS

[4.11.4.5]

Provide links at least one-quarter the size of the largest bar at a maximum spacing of twelve times the size of the smallest bar. Ensure that all other bars or groups are within 150 mm of a restrained bar.

Very often a down pipe will prevent clips (as shown above) from being used to restrain those bars that are further away than 150 mm from a fully restrained bar. In this case an alternate arrangement of links will need to be considered. Refer to the diagram overleaf:

6.28


Module 6 - Columns

Arrange links so that every corner and alternate bar or group in an outer layer is supported by a link passing passing round the bar and having having an inclined inclined angle of not more than 135o. In the case of a circular column, provide a spiral tie (circular link) passing round all the reinforcement.

6.9

EXAMPLE 6.3 (reinforcement (reinforcement )

If the required cover is 40 mm, calculate suitable reinforcement for the column in the previous example and provide a sketch to show the placing of this reinforcement.

SABS

We saw s aw on page 6.16 that the X-axis is the design axis with a design moment M = 119 kNm. The design axial load was given as N = 785 kN.

4.7.4.2

Assume R8 links and Y20 Y20 longitudinal longitudinal bars:

Design charts

6.29


SABS

4.11.4.2.2 4.11.5.2

4.11.4.5

4.11.4.5

6.30

Module 6 - Columns


Module 6 - Columns

6.31

MODULE 7

FOUNDATIONS Foundations are the elements of a structure that transfer all loads into the ground and therefore form a critical component of design. Foundations may be deep (as in the case of piles) or shallow (which is the type that will be covered in this text).


Module 7 - Foundations

CONTENTS Page 7.1 7.1

INTR INTROD ODUC UCTI TION ON . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1

7.2 ISOLA ISOLATE TED D BASE BASES S ............................................. 7.2.1 Base dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2.2 Bending . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2.2.1 Layout of reinforcement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2.3 Shear . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2.3.1 Beam shear . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2.3.2 Punching shear . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2.4 Example 7.1 (centric axial load ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2.5 Example 7.2 (centric axial load and applied moment ) . . . . . . . . . . . . . .

7.1 7.2 7.2 7.3 7.3 7.3 7.4 7.4 7.8

7.3 7.3 COMB COMBIN INED ED BASE BASES S ........................................... 7.3.1 Base dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.3.2 Bending . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.3.3 Shear . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.3.4 Example 7.3 (rectangular combined base) base) . . . . . . . . . . . . . . . . . . . . .

7.13 7.13 7.14 7.14 7.14

7.4 STRA STRAPP PPED ED BASE BASES S ........................................... 7.4.1 Base Dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.4.2 Bending . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.4.3 Shear . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.4.4 Example 7.4 (strapped base) base) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

7.20 7.21 7.21 7.21 7.22


7.1


INTRODUCTION

There are various types of foundations, and generally they may be classified as follows: 1) Isolated 2) Combined 3) Strapped 4) Strip 5) Piled 6) Rafts and basements The first four mentioned in this list are defined as ‘shallow’ foundations. The choice of the type of foundation to be adopted depends on numerous f actors such as: a) soil conditions b) proximity of boundaries c) allowable differential settlement d) types of load e) site conditions In choosing the type of foundation to adopt, we must consider those which are safe, practical and least costly.

7.2

ISOLATED BASES

Isolated bases may also be referred to as 'spot' or 'pad' foundations. They may be required to resist axial load alone or a combination of axial load and moment. The foundation is usually rectangular in plan and supports a single column load which is, wherever possible, in the centre of the base. Due to the proximity of boundaries, however, it may be necessary to position the column eccentrically on the base, which will result in an additional moment applied to the base resulting from the eccentricity of the axial load. Four possible cases arise: -

centric column with axial load only centric column with axial load and moment eccentric column with axial load only eccentric column with axial load and moment

7.1


7.2.1


BASE DIMENSIONS

Choose a suitable base area such that the maximum allowable working ground pressure is not exceeded. The working loads should include all applied dead and live loads and an estimate of the self-weight of the base

.

For bases that support an axial load only the pressure is assumed to be uniform below the base, and the required area may be calculated from the following formula:

These bases are usually square in plan, but where the column has a depth much greater than its width, the base should (whenever possible) extend equal distances from the column face in both directions. For bases that are designed to resist moment in addition to axial load, the pressure below the base will be assumed to vary linearly between a minimum and maximum value in the plane of bending. The base should be proportioned such that the eccentricity of the resultant force Ecc falls within the middle third of the base, and that the maximum working pressure does not exceed the allowable. The working pressure may be calculated from the following formula:

The depth of the base shear.

7.2.2

is dependent on the ultimate limit state design for bending and

BENDING

[4.10.2]

Bending is designed at the ultimate limit state, so the pressure needs to be recalculated using the appropriate load factors.

The ground pressure below the foundation will tend to 'curl' the base up around the column.

The base may be designed as an inverted cantilever slab bending about each face of the column. That is, the reinforcement must be designed to resist the ultimate moment about the X-X and the Y-Y axes. It is convenient to choose a width of 1 m for design purposes.

7.2


7.2.2.1


Layout of reinforcement

[4.10.3.2]

In a square base the reinforcement is spaced equally in each direction.

For a rectangular base, the reinforcement may be spaced equally in the direction of the long span. In the direction of the short span we must concentrate most of the reinforcement over a band with a width equal to the smaller dimension of the base and centred on the column.

= the ratio of the longer side to the shorter side, i.e. L/B The remainder of the reinforcement not placed in the centre band must be spread out over the outer parts, but remember that in no case must the reinforcement anywhere in the base be less than the minimum allowable. A full anchorage length must be provided to all tension reinforcement beyond the point of maximum stress. ie. 40 times the bar diameter beyond the face of the column.

7.2.3

SHEAR

[4.10.3.3]

It is preferable not to have to use shear reinforcement in bases but to rather ensure adequate depth. Design for shear generally governs the depth of the foundation. 7.2.3.1

Beam shear

[4.10.3.3.2 a)]

The vertical shear force is taken as the sum of the vertical loads on the outside of the critical section, which may be taken at 1.5 times the effective depth of the base from the face of the column.

The shear stress is then checked as for slabs (see 4.5.1).

7.3


7.2.3.2


Punching shear

[4.10.3.3.2 b)]

The effective shear force Veff needs to be calculated, which includes allowance for the foundation slab bearing on an elastic medium. The effective shear force is calculated using the ultimate pressure acting on the area outside of the critical perimeter. The punching shear stresses are checked on the actual perimeter of the column and on the first perimeter from the column face. The methods for checking the shear stresses will be as given on page 4.17.

7.2.4

EXAMPLE

7.1 (centric axial load )

This is the most basic of all foundation designs and should clearly demonstrate the design considerations applicable to most isolated foundations. A 850 x 250 column carries a dead load of 900 kN and a live load of 400 kN. The allowable soil pressure is 200 kPa. Design a suitable rectangular base for the column using grade 25/26 concrete and high tensile reinforcement. Use grade 30/26 concrete and high tensile reinforcement and assume that the base is to have a depth of 500 mm. SABS

For equal moments in each direction, the distance from the face of the column to the edge of the base should be approximately equal in each direction.

7.4



SABS

Adopt a base 3 000 x 2 400

Note that the weight of the base has been omitted in calculating the ultimat e pressure since, although it would increase the pressure, it would be acting in the opposite direction to the pressure when we calculate the shear force or moments.

Bending

4.3.3

4.11.4

4.10.3.2

7.5



SABS For central strip of width b = 2.4 m:

Adopt Note that the reinforcement required outside of this central strip is less than that provided within the central strip width, but for practical purposes the Y16 bars at 200 centres will be continued over the full length of the base.

Beam shear 4.10.3.3.2

4.3.4.1.2

Punching shear

7.6



SABS. Load perimeter:

First perimeter

7.7


7.2.5


EXAMPLE 7.2 (centric axial load and applied moment )

A 450 X 450 column is to support the applied nominal loads as tabled below:

Dead Live

Axial load 800 kN 300 kN

Moment 60 kNm 40 kNm

It is proposed that the base be 3.75 m long, 2.75 m wide and 500 mm deep. The safe bearing pressure is 150 kPa. Check that the size of the base is suitable for the applied loads and calculate suitable reinforcement using grade 25/26 concrete. SABS WORKING PRESSURE

ULTIMATE PRESSURE

Note that the weight of the base has been omitted in calculating the ultimate pressure since, although it would increase the pressure, it would be acting in the opposite direction to the pressure when calculating the shear force and moment . 7.8



SABS

BENDING - long direction

(assuming Y16 bars in the first layer)

4.3.3

7.9


SABS LA curve

4.11.4 Adopt

BENDING - short direction

(assuming Y16 bars in the second layer)

4.3.3

LA curve

Adopt

7.10




SABS 4.10.3.3.2

BEAM SHEAR

4.3.4.1.2

PUNCHING SHEAR

4.3.4.1.2

7.11


SABS

4.4.5.2.2

4.4.5.2.3

No shear reinforcement is required!

7.12



7.3 7.3


COMB COMBIN INED ED BASE BASES S

Initially, foundation sizes are based on compact spot bases. W here these overlap as shown below, long and narrow spot bases may be considered.

7.3.1

BASE DIMENSIONS

Long narrow spot foundations are acceptable only if the length of the foundation does not exceed twice the width. Where this ratio of length to width is exceeded, the foundation tends to become uneconomical. Generally, a combined base is more practical and economical. Non-uniform pressure below a combined base may m ay result in uneven settlement and rotation rotat ion of the foundation. To prevent this, we ensure that the pressure is evenly distributed below the base by so proportioning the base that its centroid coincides with the centroid of the column loads. Owing to the proximity of boundary lines and other obstructions, it may not be possible to maintain a rectangular shape to the foundation and still maintain even pressure. Consider the example given below where the ideal dimension of a combined rectangular base would encroach beyond the site boundary. Two possible solutions are offered in the form of either a 'trapezoidal' or a 'stepped' foundation.

7.13


7.3.2


BENDING

These foundations are generally designed as a 'beam-in-slab' spanning parallel to the columns centred on the columns. The remaining portion of the base will then be designed to cantilever from this beam. Note that the beam is required to carry the full pressure, and not only that acting over its width.

Where the bending moments are high, it may sometimes be necessary to provide an inverted beam to offer adequate ultimate bending resistance. 7.3.3

SHEAR

The beam-in-slab should be checked for beam shear as for a rectangular beam. Note that multiple leg shear links may be used. Punching Punching shear shear is also require required d to be checked. checked. The effectiv effective e shear force, force, in this case case Veff , needs to be calculated, which includes allowance for the reduction in shear force resulting from the ultimate soil pressure acting within the perimeter under consideration.

7.3.4

EXAMPLE 7.3 (rectangular (rectangular combined base) base)

Two 350 x 350 columns 'A' and 'B' ' B' are spaced 2 m apart and carry the nominal loads tabled below: Dead load Live load Column 'A' 650 kN 370 kN Column 'B' 800 kN 550 kN If the maximum ground pressure is 135 kPa, design a suitable foundation for the two columns. There are no boundary or other restrictions to the shape of the foundation.

7.14



SABS In an attempt to design two separate spot bases it is found that the required length of the base for column 'B' exceeds twice the width. A rectangular combined base will therefore be designed.

Taking moments about ‘B’

Assume self-weight self-weight of base = 175 kN

Let ‘x’ be the equal overhang at column ’B’, which is the column having the largest load.

7.15


SABS

Adopt a base size size as shown:

ULTIMATE LIMIT STATE

7.16




SABS BENDING - column B Try a 1 m wide beam-in-slab: Choose a depth such that no compression steel will will be required.

Try h = say 650 mm

(assuming Y16 stirrups and Y32 longitudinal bars)

4.3.3

LA curve

Adopt

BENDING - column A

LA curve

7.17


SABS

Adopt

BENDING - cantilever portion

(assuming Y16 bars in first layer)

4.3.3

LA curve

Adopt

7.18




SABS 4.4.5.1

BEAM SHEAR - column B

4.3.4.1.2

4.11.4.5.3

4.11.4.5.4

Adopt

PUNCHING SHEAR - column B Although not calculated here, it can be shown that punching shear stresses are within the allowable limits.

7.19


7.4


STRAPPED BASES

In the CBD it is generally impossible to avoid large eccentricities in the external foundations due to the fact that the maximum use must be made of all available ground, and hence the outside edges of the building will be positioned on the site boundary. Such an eccentricity would result in the soil being grossly over-loaded, and a strap footing is provided between the external foundation and the adjacent internal foundation (or counter-weight if there is no suitable internal foundation) t o remove the eccentricity from the foundation.

To obtain an uniform pressure below the base, a model will be adopted which assumes a stiff strap and an imaginary sub-foundation below the external base with a central stub column. This stub column replaces the uniform load below the sub-foundation with a single resultant force applied centrally below the actual base as shown in the diagram below. In this way the pressure at the actual founding level may be assumed to be uniform.

The foundation system will tend to rotate about the assumed resultant force. Using the total actual load (Dn + Ln) at P1 and the minimum load (0.9 Dn ) at P2 , the factor of safety for overturning must be at least 1.5. The strap is designed as a rectangular beam. Note that the width of the strap should not be less than 500 mm to allow for easy excavation.

7.20


7.4.1


BASE DIMENSIONS

The internal base is designed as an isolated base and sized accordingly. The external base should be rectangular in plan with the larger dimension parallel to the boundary. The dimensions of this base must be calculated to resist a force equal to the reaction from the stub column which will be greater than the column load due to the eccentricity. As a starting point, the reaction may be assumed to be approximately 12% greater than the column load.

7.4.2

BENDING

The internal foundation is designed as an isolated base. The strap will be designed to resist the ultimate bending moment resulting from an elastic analysis. The strap is to extend to the extreme ends of each base. The portion of the external base beyond the width of the strap will be designed as a cantilever slab.

7.4.3

SHEAR

The effective width to be used in calculating the shear stress may be taken as that shown by the shaded area below. Shear forces are high, and fi llets may be required to increase the shear width.

Punching shear is also checked, but be aware of the possible reduced perimeter around the external column.

7.21


7.4.4


EXAMPLE 7.4 (strapped base)

The drawing below shows the plan of the column layout to a building where the outside edge of the columns will lie on the site boundary. The permissible ground pressure is 200 kPa

Draw the shear force and bending moment diagrams applied to the strap of a suitable strapped base for columns 1C and 2C. All columns are 600 x 300. The characteristic loads from the two columns at foundation level are given below: Column 1C

Dn = 475 kN Ln = 375 kN

Column 2C

Dn = 940 kN = 720 kN

SABS BASE 2 C

7.22



SABS BASE 1 C

7.23


SABS

ULTIMATE LIMIT STATE

7.24


MODULE 8

RETAINING WALLS Retaining walls are structures used to retain material which would not be able to stand vertically unsupported.


Module 8 - Retaining Walls

CONTENTS Page 8.1 TYPES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1.1 Gravity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1.2 Cantilever . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1.3 Counter-fort or buttress . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2

8.1 8.1 8.1 8.1

COMPONENTS OF A RETAINING WALL . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2

8.3 MODES OF FAILURE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3.1 Overturning . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3.2 Sliding . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3.3 Soil failure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3.4 Fracture of elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

8.2 8.2 8.3 8.3 8.3

8.4 SOIL PARAMETERS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.4.1 Angle of internal friction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.4.2 Soil density . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.4.3 Sliding friction coefficient . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.4.4 Maximum ground pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

8.4 8.4 8.4 8.4 8.4

8.5 APPLIED PRESSURES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.5.1 Active soil pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.5.2 Passive soil pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.5.3 Surcharge pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.5.4 Water pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.5.5 Resulting pressure distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

8.4 8.4 8.5 8.5 8.5 8.6

8.6

DESIGN APPROACH . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.8

8.7

EXAMPLE 8.1 (cantilever retaining wall ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.10


8.1


TYPES

There are many different forms that a retaining wall may take. A few of the most common are discussed here.

8.1.1

GRAVITY

Stability is provided by the weight of the structure alone and would not require the design of reinforcing bars.

8.1.2

CANTILEVER

Stability is provided by the weight of the wall plus the earth on the base. The concrete needs to be reinforced to resist the forces acting on it.

8.1.3

COUNTER-FORT OR BUTTRESS

Stability is also provided by the weight of the structure plus the earth on the base. The concrete needs to be reinforced to resist the forces acting on it. The wall spans between the counter-forts (or buttresses) as well as off the base. Design moments are lower than for cantilever walls of the same height.

8.1


8.2


COMPONENTS OF A RETAINING WALL

TYPICAL SECTION THROUGH A CANTILEVER RETAINING WALL

8.3 8.3.1

MODES OF FAILURE OVERTURNING

The wall will be assumed to overturn about the lower tip of the toe if there is not sufficient weight on the base to resist the overturning moment applied by the active soil pressure behind the wall. This mode is checked under working loads by ensuring a factor of safety against overturning of 1.5.

8.2


8.3.2


SLIDING

The force tending to slide the wall forward is given by the active force of the soil behind the wall (area of the active pressure diagram). Resistance to sliding is off ered by: the friction between the base of the retaining wall and the founding material the passive force offered by the soil in front of the wall This mode of failure is checked under working loads by ensuring a factor of safety against sliding of 1.5.

8.3.3

SOIL FAILURE

Excessive settlement will occur if the resulting pressure under the base of the retaining wall exceeds the maximum allowable ground pressure. This mode of failure is also checked using working loads.

8.3.4

FRACTURE OF WALL ELEMENTS

The elements that make up the retaining wall (heel, toe, key, wall, etc.) are required to resist all the forces and moments applied to them. These elements are designed at the ultimate limit state.

8.3


8.4


SOIL PARAMETERS

There are certain properties of the soil that we need to know before we can commence with our design. These are as follows:

8.4.1

ANGLE OF INTERNAL FRICTION

(degrees)

This is the angle at which the material will safely support itself. The value of  is normally between 30o and 45o depending on the material.

8.4.2

SOIL DENSITY

(kN/m3)

This is the weight of the soil. The value is normally upwards of 15 kN/m3 .

8.4.3

SLIDING FRICTION COEFFICIENT

(dimensionless)

This is the measure of the frictional resistance between the concrete and the founding material. The value is normally between 0.3 and 0.5. A value of tan ( - 5) is often assumed.

8.4.4

MAXIMUM GROUND PRESSURE

(kPa)

This is a measure of the maximum pressure that the ground can support bef ore excessive settlements will begin to take place. A typical value is 150 kPa for reasonable sands.

8.5

APPLIED PRESSURES

8.5.1

ACTIVE SOIL PRESSURE

The pressure from the soil behind the wall will become 'active' as it begins to move forward. The distribution of pressure will be triangular, with a value of zero at the top of the wall and increasing to a maximum value at the bottom of the wall. According to Rankine's theory, the active soil pressure at any depth is given by:

where :

8.4

=

active soil pressure

=

active pressure coefficient

=

density of the retained material

=

depth to point being considered



Where the backfill is level behind the wall:

where :

=

angle of internal friction of retained material

For a sloping backfill at an angle of

measured from the horizontal:

where:

n

=

8.5.2

PASSIVE SOIL PRESSURE

The passive pressure is that pressure which will develop when the wall 'pushes' against the soil in front of it. The distribution of pressure will be triangular, and is given by:

where:

8.5.3

SURCHARGE PRESSURE

The pressure resulting from a uniform surcharge applied to the top of the backfill behind the wall will have a rectangular pressure distribution which will be given by:

where :

8.5.4

=

pressure due to surcharge

=

surcharge (kN/m2)

WATER PRESSURE

Weep-holes are provided to relieve any build up of water pressure behind the wall, and therefore the effects of water may be ignored as long as the weep-holes are correctly constructed.

8.5


8.5.5


RESULTING PRESSURE DISTRIBUTION

To calculate the pressure distribution below the base, we simply apply the basic combined stress formula:

where:

= = = =

pressure ( ) total vertical weight ( ) width of base x 1 m ( )

=

section modulus

The formula may be rewritten as:

The eccentricity

is the distance at which the resultant force acts from the centre of the

base and may be calculated as follows:

8.6



=

horizontal force acting at a lever arm of from ' '

=

vertical weight acting at a lever arm of from ' '

= overturning moment = = stabilising moment =

By symmetry:

But:

From the above diagram:

It follows that:

Note : If

or

, then the resultant force will not fall within the middle third of the base,

and the full length of the base will not be under compression. However, tensile stress cannot develop between the base of the retaining wall and the soil, the maximum pressure in this case being given by:

where the length of the pressure diagram =

Note that it is advisable for at least 75% of the base length to be under compression.

8.7


8.6


DESIGN APPROACH

The stability checks and calculation of ground pressures are the same for all types of retaining walls, the only difference in design being the analysis of the ultimate forces and moments on the retaining wall elements. A 1 m length of wall is considered for design purposes, and all elements of a retaining wall are generally designed as for slabs. The general design procedure is to assume dimensions for the wall elements (toe, heel, base, etc.) and then check the retaining wall for stability using those dimensions. Once suitable dimensions have been established, the reinforcement i n the retaining wall elements will be designed at the ultimate limit state. The following may be used as guidelines for determining the t rial dimensions for a cantilever wall. It must be remembered, however, that these guidelines apply only to a wall that is not restricted by boundaries or subjected to unusual loading. Thickness of wall at base : Depth of base : Toe : Heel :

0.09 h 0.09 h + 75 mm 0.125 h 0.4 h

The design of retaining walls is very methodical in approach and is therefore suitable for the development of a simple spreadsheet programme. The tables that follow show the general calculation format for checking the stability of a retaining wall based on the symbols indicated in the sketch below:

8.8



OVERTURNING Description

Force F

Lever arm LA

Moment Mo

Weight W

Lever arm LA

Moment Ms

backfill

surcharge

STABILISING Description

backfill

bank

cover

stem rectangle

stem triangle

base

surcharge

where:

= =

surcharge (kN/m2) unit weight of soil

=

unit weight of concrete (typically 24 kN/m2 )

Note that any other form of load (e.g. concentrated point load on top of stem) not listed in the above tables must be appropriately included.

8.9


8.7


EXAMPLE 8.1 (cantilever retaining wall )

Design a suitable reinforced concrete cantilever retaining wall where the difference in ground levels is to be 3.7 m. The top of the foundation is to be 300 mm below the finished ground level in front of the wall. The soil behind the wall will be inclined at 10o and will carry a live load surcharge of 2.0 kN/m². The top of the wall is to extend 200 mm above the level of the backfill immediately behind the wall. Use grade 25/26 concrete and high tensile reinforcement. The soil properties are given as follows: = 35o = 16 kN/m3 = 0.4 The working ground pressure may not exceed 120 kPa. SABS Trial dimensions

Try the following wall layout:

8.10



SABS OVERTURNING

Overturning Description

F (kN)

LA (m)

Mo (kNm)

W (kN)

LA (m)

Ms (kNm)

backfill surcharge

Stabilising Description backfill bank cover stem rectangle stem triangle base surcharge

8.11


SABS SLIDING

SOIL PRESSURE

8.12




SABS DESIGN AT ULTIMATE LIMIT STATE The following partial load factors at the ultimate limit state will be considered: SABS 0160 Table 2

Concrete Surcharge Soil

= = =

1.2 1.6 1.4

(taken as the avarage of 1.2 and 1.4 due to the uncertainty of considering the soil load as either ‘permanent’ or ‘live’)

W actual

i

Wultimate

102.40

1.4

143.36

bank

3.61

1.4

5.05

cover

2.40

1.4

3..36

stem rectangle

20.16

1.2

24.19

stem triangle

7.56

1.2

9.07

base

23.52

1.2

28.22

surcharge

3.20

1.6

5.12

Description backfill

W = 218.37 kN

8.13



SABS

Heel Description

Shear (kN)

LA (m)

Moment (kNm)

Shear (kN)

LA (m)

Moment (kNm)

rectangular pressure

triangular pressure heel backfill bank surcharge

Toe Description rectangular pressure

triangular pressure toe cover

8.14



SABS Stem Description

Shear (kN)

LA (m)

Moment (kNm)

backfill

surcharge

4.3.3.4.1

BENDING - stem

LA curve

4.11.4 4.11.8.2.2 Adopt

Consider curtailing the Y10's:

8.15



SABS

Let ‘h’ be the distance from the top of the backfill to the theoretical cut-off point.



8.16

Y10 bars can stop 850 mm above the base as shown in the elevation below:



S.A.B.S SHEAR - stem

4.4.5.1

4.3.3.4.1

BENDING - heel

LA curve

4.3.3.4

4.11.4 Adopt Note that this reinforcement has been chosen to match the spacing of the bars in the stem. This reinforcemen t could be curtailed to Y10 @ 250, but note this would then be less than the required minimum area of steel.

8.17



SABS

SHEAR - heel

4.4.5.1

TOE Extend the reinforcement provided in the heel through to the t oe.

4.11.4

SECONDARY REINFORCEMENT Wall:

Base:

8.18

MODULE 9

SILOS This section is limited to the very basic analysis and design of a reinforced concrete silo storing granular material and will in no way equip you with all of the required information necessary to analyse or design the entire structure.


Module 9 - Silos

CONTENTS Page 9.1

INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.1

9.2 PARAMETERS REQUIRED FOR DESIGN . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.1 9.2.1 Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.1 9.2.2 Properties of retained material . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.1 9.3 ANALYSIS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3.1 Horizontal pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3.2 Ring tension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3.3 Vertical pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3.3.1 Walls . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3.3.2 Floor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.4

9.2 9.3 9.4 9.4 9.4 9.4

EXAMPLE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.5


9.1

Module 9 - Silos

INTRODUCTION

Material-containing structures are classified as ‘shallow’ or ‘deep’ and the contents of such a structure can consist of any granular material. The pressures acting on a shallow container would be calculated in a similar manner to the active pressure acting on a retaining wall. When a container is classified as deep there is a tendency for the retained material to ‘arch’ between the vertical surfaces, and hence the pressures will vary from that of a shallow container. A silo is a deep container.

9.2 9.2.1

PARAMETERS REQUIRED FOR DESIGN GEOMETRY

The overall height h1 and internal diameter d are chosen to satisfy the required capacity of the silo. The thickness of the concrete wall t is a strength requirement which will be checked at the ultimate limit state. The hydraulic radius r is another parameter that is required for the design of a silo, and this is given as the cross-sectional area divided by the internal perimeter.

therefore

9.2.2

PROPERTIES OF RETAINED MATERIAL

The unit weight D (kN/m3), the angle of internal friction of the retained material  (degrees) and the angle of friction between the retained material and the concrete  (degrees) must be known for the silo shell to be analysed.

9.1


Module 9 - Silos

The following table gives approximate values for  and D for some of the common types of retained material.

Material

 (degrees)

D (kN/m3 )

Wheat

25

8.40

Maize

27

8.40

Barley

27

7.50

Oats

28

6.30

Cement

10

14.2

Coal (crushed)

40

8.80

Ash

35

9.43

The angle of friction between the contained material and the concrete wall  is required for periods of both filling f and emptying e , and can be obtained from the table below, which gives values for granular material (particle size > 0.2 mm) and powdered material (particle size < 0.06 mm). Linear interpolation may be used for intermediate particle sizes.

9.3

Grain size (mm)

f

e

> 0.20

0.75 

0.60 

< 0.06





ANALYSIS

The load effects required are:

Horizontal pressure Hoop tension Vertical wall pressure Vertical floor pressure

The magnitude of the applied stresses will depend on whether the silo is filling, emptying or in a state of rest. Table 21 of the Reinforced concrete designer’s handbook by Reynolds and Steedman will be used here to calculate these stresses.

9.2


9.3.1

Module 9 - Silos

HORIZONTAL PRESSURE

The horizontal pressure qh will vary with depth and obviously depend on whether the silo is filling or emptying. A graph showing the variation of pressure with height is shown below:

Note that the horizontal pressure during discharge is reduced over a height of 1.2 d (but not exceeding 0.75 h1) above the outlet owing to the proximity of the compartment floor. The equation given below applies only to deep containers where

:

(KN/m2) Note that this equation is derived from Janssen’s formula which is:

= ho

=

reference depth

= =

for filling for emptying

9.3


9.3.2

Module 9 - Silos

RING TENSION

The horizontal ring tension T (or hoop tension) is given by the following formula:

(KN per m of wall height)

9.3.3

VERTICAL PRESSURE

9.3.3.1

Walls

The vertical pressure on the walls qv is caused by the self-weight of the walls together with the vertical load of the retained material supported by wall friction. The m aximum value will occur at the level of the floor of the silo. The self-weight of the walls W (kN) is a straightforward calculation and needs no explanation. The vertical load of the retained material supported by wall friction (kN/m2 ) is given by the product of the unit weight of the material and the hydraulic radius. This load is then multiplied by the surface contact area ( d h) to obtain the load in kN.

(N/mm2) where A

9.3.3.2

=

total cross-sectional area of the wall

Floor

The vertical pressure on the floor qv is caused by the vertical load of the retained material and will be equal to the horizontal pressure during emptying and twice the horizontal pressure during filling.

9.4


9.4

Module 9 - Silos

EXAMPLE 9.1 (silo)

Design a cylindrical reinforced concrete silo with an external diameter of 10 m and a wall thickness of 180 mm. The silo is required to store 2 600 tons of crushed coal having an average particle size of 0.125 mm. Use grade 30/26 concrete and high tensile reinforcement.

SABS Unit weight

Unit mass

Required volume

Internal diameter

Internal area

Required height

Design at ultimate limit state

Hydraulic radius Angle of repose

During filling: (by interpolation)

9.5


SABS

At h = 40 m :

During emptying:

At h = 28.43 m :

At h = 10 m :

9.6

Module 9 - Silos


Module 9 - Silos

SABS At h = 5 m :

Horizontal reinforcement

Adopt

Vertical reinforcement

9.7


SABS

4.8.4.1

4.11.4

Adopt

9.8

Module 9 - Silos

MODULE 10

WATER-RETAINING STRUCTURES These structures are no different from any other type of structure except for the fact that they must be impermeable and reasonably waterproof. The structures that can be designed following the guidelines of this module exclude dams, aqueducts, structures retaining liquids that have a severe corrosive effect on the concrete, and structures with special shapes.


Module 10 - Water-retaining structures

CONTENTS Page 10.1

LOADS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.1

10.2 ANALYSIS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2.1 Slabs and floors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2.2 Walls of cylindrical tanks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2.3 Walls of rectangular tanks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

10.1 10.1 10.1 10.1

10.3 LIMIT STATE REQUIREMENTS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.3.1 Ultimate limit state . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.3.2 Serviceability limit state . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.3.2.1 Deflection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.3.2.2 Cracking . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

10.2 10.2 10.2 10.2 10.2

10.4 CRACK CONTROL . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.3 10.4.1 Direct tension in mature concrete . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.3 10.4.2 Direct tension in immature concrete . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.4 10.4.2.1 Minimum reinforcement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.4 10.4.2.2 Calculation of crack spacing . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.5 10.4.2.3 Calculation of crack width . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.5 10.4.3 Flexural tension in mature concrete . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.6 10.4.4 Minimum reinforcement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.10 10.5 DESIGN AND DETAILING OF JOINTS . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.5.1 Construction joints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.5.2 Sliding joints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.5.3 Movement joints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.5.3.1 Contraction joint . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.5.3.2 Induced Contraction joint . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.5.3.3 Expansion joint . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.5.3.4 Spacing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

10.10 10.10 10.10 10.11 10.11 10.12 10.12 10.12

10.6 SPECIFICATION OF MATERIALS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.6.1 Blinding layer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.6.2 Concrete . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.6.2.1 Cement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.6.2.2 Aggregate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.6.3 Formwork . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.6.4 Joining materials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.6.4.1 Joint fillers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.6.4.2 Waterstops . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.6.4.3 Joint sealing compounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

10.14 10.14 10.14 10.14 10.14 10.14 10.14 10.14 10.15 10.15

10.7

DESIGN EXAMPLE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.16

ANNEXURE A

Bending moments in circular tanks . . . . . . . . . . . . . . . . . . . . . 10.20

ANNEXURE B

Ring tension in circular tanks . . . . . . . . . . . . . . . . . . . . . . . . . . 10.24



10.1 LOADS Structures should be designed for both full and empty conditions, and the arrangement of load should be such as to cause the most critical effects. For a structure subjected to lateral soil pressure, the active pressure should be considered when the structure is empty, but no passive soil pressure may be taken into account when considering the structure full. Water and lateral soil loads should be considered as live loads, but the soil covering a roof of a structure may be taken as a dead load.

10.2 ANALYSIS 10.2.1

Slabs and floors

These elements are generally analysed as for solid or flat slabs, and the method of analysis given in SABS 0100-1:2000 may be adopted (see Module 4).

10.2.2

Walls of cylindrical tanks

Tables produced by the Portland Cement Association of America are available which give coefficients for calculating the vertical moment as well as the ring tension developed at any required height in a wall for a variety of end conditions. These tables conveniently allow for rectangular, triangular or trapezoidal loading patterns (see Annexure A).

10.2.3

Walls of rectangular tanks

These walls may be designed using a simple elastic analysis. However care should be exercised at all points of intersection. Tables are available which give bending moment coefficients for both horizontal and vertical moments in wall panels. The walls may be divided into panels and may span vertically, horizontally or in both directions.

10.1



10.3 LIMIT STATE REQUIREMENTS 10.3.1


The recommendations given in SABS 0100-1:2000 should be applied. For combined bending and tension (as in the case of a circular tank), the areas of steel can be calculated separately and then added together.

10.3.2


10.3.2.1

Deflection

The limits for deflection should follow the recommendations given in SABS 0100-1:2000 for non-liquid-retaining structures since only in very exceptional circumstances will deflection be critical with regard to the effects of liquid. 10.3.2.2

Cracking

The allowable crack widths will depend on the class of structure, and for reinforced concrete structures should be controlled as shown in the following table. Classes A and B are designed as water-retaining structures according to BS 5337, and class C structures are designed as normal reinforced concrete structures according to SABS 0100-1:2000.

CLASS

DEFINITION

LIMITING CRACK WIDTH (mm)

A

Exposed to a moist or corrosive atmosphere or subjected to alternate wetting and drying

0.1

B

Exposed to continuous or almost continuous contact with liquid

0.2

C

Generally not exposed to liquid or to moist corrosive conditions

0.3

For a class C structure exposed to an aggressive environment, the surface width of cracks at points nearest to the main reinforcement should not in general exceed 0.0004 times the nominal cover to the main reinforcement. Where a wall or slab has a thickness exceeding 225 mm, the class of each face may be considered separately. For a wall or slab 225 mm or less, the class of both faces should be taken as the most severe of the two exposure conditions.

10.2



10.4 CRACK CONTROL To make a concrete structure impermeable we need to control the spacing and width of cracks which may be caused by early thermal contraction, temperature and moisture effects and flexure. Remember that crack control is a serviceability limit state requirement.

10.4.1

Direct tension in mature concrete

This is a simple check where the crack widths may be deemed to be satisfactory if the steel stress under service conditions does not exceed the appropriate value given in the table below.

Class of exposure

Allowable stress (N/mm2) Plain bars

Deformed bars

A

85

100

B

115

130

The service stress may be estimated as shown below (clause 4.11.8.2.1.4 of SABS 01001:2000), or determined by the method given in 1.4.3 of this module.

where:

= = = =

self-weight factor for SLS imposed load factor for SLS self-weight factor for ULS imposed load factor for ULS

(1.1) (1.0) (1.2) (1.6)

Where all applied loads are of a similar nature (e.g. all UDL), the following equation will more accurately calculate the service stress by allowing for the varying proportions of dead and live loads.

Where the loading is not uniform the service and ultimate loads in this equation can be replaced by the service and ultimate moments.

10.3


10.4.2


Direct tension in immature concrete

Changes in the temperature of the concrete and reinforcement and in the moisture content of the concrete cause dimensional changes which, if resisted internally or externally, may crack the concrete. The heat evolved by the hydration of cement will raise the temperature for a day or more after casting, and then fall towards ambient. Cracking usually occurs at this time in the immature weak concrete. 10.4.2.1

Minimum reinforcement

A minimum area of reinforcement is required to distribute cracking due to the direct tension resulting from early thermal and moisture effects in the immature concrete. This reinforcement is placed in ‘surface zones’ as described below: -

For ground floor slabs less than 200 mm thick, assume that the top surface zone is 100 mm thick and ignore the bottom zone.

-

For ground floor slabs between 200 mm and 400 mm thick, assume that the top surface zone is half of the slab thickness and ignore the bottom zone.

-

For ground floor slabs between 400 mm and 500 mm thick, assume that the top surface zone is half the slab thickness and the bottom surface is half the slab thickness less 150 mm.

-

For ground floor slabs exceeding 500 mm thick, assume that the top surface zone is 250 mm thick and the bottom surface zone is 100 mm thick.

-

For walls exceeding 500 mm thick, assume that each surface zone is 250 mm thick.

The critical steel/concrete ratio

where:

is given by:

=

tensile strength of the immature concrete, which may be taken as 1.15 for grade 25 MPa concrete and as 1.3 N/mm2 for grade 30 MPa concrete

=

width of section (generally 1 000 mm)

=

overall thickness of the slab or thickness of the surface zone under consideration

Note that only 2/3 is required where the cracking is controlled by closely spaced movement joints (see 10.5.3.4 option 3).

10.4


10.4.2.2


Calculation of crack spacing

When sufficient reinforcement is provided to distribute cracking, the likely maximum spacing of cracks is given by:

where:

=

diameter of each reinforcing bar

=

actual steel/concrete ratio

=

average bond strength between the steel and concrete obtained from clause 4.11.6.3 of SABS 0100-1:2000 and reduced by 30% for bars in direct tension

Alternatively, the value of

may be taken as 1.0 for round bars and 2/3 for deformed bars.

This equation may be expressed for design purposes as:

where: 10.4.2.3

=

the number of bars in the width of section

Calculation of crack width

For elements exposed to normal climatic conditions the estimated maximum crack width for temperature effects is given by:

where:

=

the fall in temperature between the hydration peak and ambient (which is dependent on the type of cement, cement content and type of formwork) should be obtained from specialist literature. (A value of 230 for floors and 280 for walls may be assumed.)

=

a further fall in temperature due to seasonal variations. This further fall in temperature should not be taken any less than 150. Note that if partial movement joints are provided at not more than 7.5 m centres, or complete movement joints are provided at not more than 15 m centres, this subsequent temperature fall need not be considered.

=

coefficient of thermal expansion of concrete (typically 10 x10-6 m/mm/0c.

10.5


10.4.3


Flexural tension in mature concrete

The width of flexural cracks on the surface of an element depends primarily on three factors: -

proximity of the reinforcing bars magnitude of the applied moment (and position of neutral axis) average strain at the surface being considered

The following formula for calculating the crack width due to flexural tension in mature concrete may be used, provided that the service stress in the tension reinforcement is limited to 0.8 and the stress in the concrete is limited to 0.45 :

SECTION

where:

10.6

STRAIN DIAGRAM

=

distance from point being considered to nearest bar

=

strain in the tension reinforcement

=

strain at the level being considered

= =

average strain at the level being considered minimum cover to reinforcement



The strain in the tension reinforcement is given by: where:

=

service stress in the tension reinforcement

=

Young’s modulus for steel reinforcement (200 GPa)

The strain at the level being considered is given by:

To calculate the distance to the neutral axis (x), consider the diagram below.

SECTION

STRAIN

ELASTIC STR ESS

By similar triangles: For equilibrium:

Hence:

Dividing by

we get:

. . . . . . . . . . . . . . . . . . . . . . . . eq. 1

is the modular ratio for which we shall use the symbol

Now multiplying eq. 1 by

.

we get: or:

. . . . . . . . . . eq. 2 10.7


Dividing eq. 2 by

If we let


gives:

, we can rewrite the above equation as:

Solving this quadratic equation produces:

Rewriting we get:

where:

= =

(usually in the order of 15)

Note that the value for should be taken as half of that given in table 1 of SABS 01001:2000 to make allowance for the long-term effects of creep and shrinkage.

The average strain at the surface is given by:

(The second term in this equation allows for the stiffening effect of the concrete in the tension zone.) To calculate the service stress

, consider the previous diagram.

The internal moment of resistance is given by:

where:

Therefore:

The service stress in the steel is therefore given by:

10.8



Alternatively, the service stress in the tension reinforcement and the depth to the neutral axis may be obtained from the following table, which is extracted from table 2 of TDH 5158.

0.10

0.159

0.095

0.58

0.339

0.514

1.06

0.427

0.909

0.12

0.173

0.113

0.60

0.344

0.531

1.08

0.430

0.925

0.14

0.185

0.131

0.62

0.348

0.548

1.10

0.433

0.941

0.16

0.196

0.150

0.64

0.353

0.565

1.12

0.436

0.957

0.18

0.207

0.168

0.66

0.357

0.581

1.14

0.438

0.973

0.20

0.217

0.186

0.68

0.361

0.598

1.16

0.441

0.989

0.22

.0226

0.203

0.70

0.365

0.615

1.18

0.444

1.005

0.24

0.235

0.221

0.72

0.369

0.631

1.20

0.446

1.021

0.26

0.243

0.239

0.74

0.373

0.648

1.22

0.449

1.037

0.28

0.251

0.257

0.76

0.377

0.665

1.24

0.452

1.053

0.30

0.258

0.274

0.78

0.381

0.681

1.26

0.454

1.069

0.32

0.266

0.292

0.80

0.384

0.697

1.28

0.457

1.085

0.34

0.272

0.309

0.82

0.388

0.714

1.30

0.459

1.101

0.36

0.279

0.217

0.84

0.392

0.730

1.32

0.462

1.117

0.38

0.286

0.344

0.86

0.395

0.747

1.34

0.464

1.133

0.40

0.292

0.361

0.88

0.398

0.763

1.36

0.467

1.149

0.42

0.298

0.378

0.90

0.402

0.779

1.38

0.469

1.164

0.44

0.303

0.396

0.92

0.405

0.796

1.40

0.471

1.180

0.46

0.309

0.413

0.94

0.408

0.812

1.42

0.474

1.196

0.48

0.314

0.430

0.96

0.412

0.828

1.44

0.476

1.212

0.50

0.319

0.447

0.98

0.415

0.845

1.46

0.478

1.227

0.52

0.325

0.464

1.00

0.418

0.861

1.48

0.480

1.242

0.54

0.330

0.481

1.02

0.421

0.877

1.50

0.483

1.259

0.56

0.334

0.498

1.04

0.424

0.893

is the moment calculated using service loads.

10.9


10.4.4


Minimum reinforcement

The reinforcement in each direction should be at least 0.3% of the gross cross-section for high tensile steel and 0.5% of the gross cross-section for mild steel. Where the thickness of a slab is less than 200 mm, the reinforcement may be placed in a single layer. Where the thickness of a slab (wall or floor) is 200 mm or greater, half of this reinforcement should be provided in each face. All reinforcement should consist of bars of small diameter at relatively close spacing.

10.5 DESIGN AND DETAILING OF JOINTS 10.5.1

Construction joints

A construction joint is a joint in the concrete introduced for convenience in construction at which measures are taken to achieve subsequent continuity.

10.5.2

Sliding joints

This joint has complete discontinuity in both reinforcement and concrete, and special provision is made to facilitate relative movement in the plane of the joint.

10.10


10.5.3


Movement joints

A movement joint is specially formed to accommodate movement between adjoining parts of the structure with special provision being made for maintaining water tightness. Movement joints in adjacent elements should preferably be aligned. 10.5.3.1

Contraction joint

There is no initial gap between the concrete on each side of the joint. The joint may be partial (reinforcement is continued through the joint) or complete (both the concrete and reinforcement are interrupted).

Partial contraction joint in floor

Complete contraction joint in wall

Note that the water-stop is positioned centrally in a wall and on the under-surface of a floor. 10.11


10.5.3.2


Induced contraction joint

The concrete is continuous at the time of casting. It is provided with a reduced cross-section at which a crack should be induced. For this type of joint to be effective, a minimum reduction in section of 33% for floors and 25% for walls should be provided.

10.5.3.3

Expansion joint

This joint has complete discontinuity in both the reinforcement and the concrete, and is intended to accommodate either expansion or contraction.

The above figure shows a typical detail in a floor. For a wall the waterstop would be positioned centrally in the section.

10.5.3.4

Spacing

The provision of movement joints and their spacing is dependent on the design philosophy adopted. Three options of construction method and control are given. At the one extreme (option 1), no joints may be provided in the concrete, but a substantial amount of small diameter reinforcement bars will be required. At the other extreme (option 3), close movement joints are provided and less reinforcement is required. In between the two (option 2), control can be exercised by varying the reinforcement and joint spacing.

10.12



Option 1:

Continuous construction where the control of temperature and moisture effects is done by offering full restraint to movement. To achieve this an area of reinforcement well in excess of the critical area will be required, and will consist of closely spaced bars of small diameter.

Option 2:

Semi-continuous construction where the control of temperature and moisture effects is done by offering partial restraint to movement with joints to control the balance of movement. The reinforcement will be less than that required in option 1, but will still be in excess of . Complete joints are provided at a spacing not exceeding 15 m, or partial joints are provided at a spacing not exceeding 7.5 m. A combination of complete and partial joints may be considered, where the maximum spacing of joints would be interpolated between 7.5 m and 15 m. Note that by using this option, the fall in temperature due to seasonal variations (

) may be neglected in calculating the crack width in immature

concrete. Option 3:

Here the control of temperature and moisture effects is done by allowing for freedom of movement. It is adequate to provide an amount of rei nforcement exceeding only 2/3 . The joint spacing should not exceed the following: For complete joints: For alternate complete and partial joints: For partial joints:

10.13



10.6 SPECIFICATION OF MATERIALS 10.6.1

Blinding layer

A layer of blinding of at least 75 mm thick should be placed over the ground below any floor. When casting a floor over clay, the last 150 mm or so of excavation should be done by hand just before the blinding layer is cast to prevent the clay from drying out and later expanding as it regains moisture.

10.6.2

Concrete

The concrete grade should be at least 25 MPa and should be durable and well compacted. Curing should continue for at least 4 days after casting, and it is desirable to prevent the concrete from drying by draping suitable material over the walls or ponding the floor and roof. Keep the cement content down to the minimum permitted and ensure that it does not exceed 400 kg/m3. Avoid thermal shock or over-rapid cooling of concrete surfaces. 10.6.2.1

Cement

Use cements with low rates of heat evolution. 10.6.2.2

Aggregate

Use aggregates having low coefficients of thermal expansion and avoid the use of shrinkable aggregate.

10.6.3

Formwork

Ties used to secure and align the formwork should not pass completely through any part of a structure that will be retaining water as this will lead to a potential weakness in water tightness. The cover to a tie should be at least that provided to the reinforcement, and the gap left from the end of the tie to the face of the concrete should be effectively sealed.

10.6.4

Joining materials

10.6.4.1

Joint fillers

The material should be compressible and of sufficient thickness to maintain the initial gap between the concrete surfaces under pressure from the newly placed concrete. It is preferable to use non-rotting cork-based joint fillers rather than f ibrous jonit fillers which can absorb moisture.

10.14


10.6.4.2


Waterstops

Waterstops must be durable and impermeable and be capable of allowing for the whole range of joint movements. The waterstop must be sufficiently wide to ensure that the water path through the concrete past the waterstop is not unduly short. The full concrete cover to all reinforcement should be maintained in the vicinity of a waterstop, and it is essential to ensure the proper compaction of concrete. 10.6.4.3

Joint sealing compounds

Joint sealing compounds (sealants) are impermeable ductile materials which are required to provide a watertight seal by adhesion to the concrete throughout the range of joint movements.

10.15



10.7 DESIGN EXAMPLE The ground floor slab of a reservoir is 250 mm thick and subjected to a service moment of 28 kNm and an ultimate moment of 45 kNm. The structure is to be class B and the minimum cover required to all reinforcement is 40 mm. Use grade 25/26 concrete and mild steel bars to calculate suitable reinforcement if complete movement joints are provided at 12 m centres. SABS 4.3.3.4.1

Bending

LA curve

4.3.3.4

4.11.4

Flexural tension in the mature concrete Table 1

10.16



BS 5337

10.17


BS 5337

Direct tension in immature concrete

10.18




BS 5337

10.19


ANNEXURE A

-


Bending moments in circular tanks

The figure below shows the typical distribution of the bending moments induced in the vertical section.

No moment is induced in the vertical section when the top is free and the base is free to slide.

When the top of the wall is free and the base is either ‘pinned’ or ‘fixed’, the moments induced in the vertical section at any depth may be obtained f rom Tables A1, A2 or A3 that follow. The load from retained liquid is triangular. The load from gas pressure or prestress is uniform (constant throughout the depth). Further tables should be consulted for the cases where the top of the wall has shear restraint from a roof or for moments applied to the base of the wall.

10.20


TABLE A1


Fixed base, free top

Triangular load

Bending moment coefficients 0.1 h

0.2 h

0.3 h

0.4 h

0.5 h

0.6 h

0.7 h

0.8 h

0.9 h

1.0 h

0.4

+0.05

+0.14

+0.21

+0.07

-0.42

-1.50

-3.02

-5.29

-8.16

-12.1

0.8

+0.11

+0.37

+0.63

+0.08

+0.70

+0.23

-0.68

-2.24

-4.65

-7.95

1.2

+0.12

+0.42

+0.77

+1.03

+1.12

+0.90

+0.22

-1.08

-3.11

-6.02

1.6

+0.11

+0.41

+0.75

+1.07

+1.21

+1.11

+0.58

-0.51

-2.32

-5.05

2.0

+0.10

+0.35

+0.68

+0.99

+1.20

+1.15

+0.75

-0.21

-1.85

-4.36

3.0

+0.06

+0.24

+0.47

+0.71

+0.90

+0.97

+0.77

+0.21

-1.19

-3.33

4.0

+0.03

+0.15

+0.28

+0.47

+0.66

+0.77

+0.69

-0.23

-0.80

-2.68

5.0

+0.02

+0.08

+0.16

+0.29

+0.46

+0.59

+0.59

-0.28

-0.58

-2.22

6.0

+0.01

+0.03

+0.08

+0.19

+0.32

+0.46

+0.51

-0.29

-0.41

-1.87

8.0

0.00

+0.01

+0.02

+0.08

+0.16

+0.28

+0.38

-0.29

-0.22

-1.46

10.0

0.00

0.00

+0.01

+0.04

+0.07

+0.19

+0.29

-0.28

-0.12

-1.22

12.0

0.00

+0.01

+0.01

+0.02

+0.03

+0.13

+0.23

-0.26

-0.05

-1.04

14.0

0.00

0.00

0.00

0.00

+0.01

+0.08

+0.19

-0.22

-0.01

-0.90

16.0

0.00

0.00

-0.01

-0.02

-0.01

+0.04

+0.13

-0.19

-0.01

-0.79

Note that a positive coefficient indicates tension on the outer face.

where:

= = = = = =

vertical moment at given depth internal tank diameter thickness of tank wall bending moment coefficient unit weight of retained liquid depth measured from top

kNm per m m m kN/m3 m

10.21


TABLE A2


Fixed base, free top

Uniform load

Bending moment coefficients 0.1 h

0.2 h

0.3 h

0.4 h

0.5 h

0.6 h

0.7 h

0.8 h

0.9 h

1.0 h

0.4

-0.23

-0.93

-0.27

-4.39

-7.10

-10.2

-14.6

-20.0

-25.9

-33.1

0.8

0.00

-0.06

-0.25

-0.83

-1.85

-3.62

-5.94

-9.17

-13.3

-18.4

1.2

+0.08

+0.26

+0.37

+0.29

-0.09

-0.89

-2.27

-4.68

-8.15

-11.8

1.6

+0.11

+0.36

+0.62

+0.77

+0.68

+0.11

-0.93

-2.67

-5.29

-8.76

2.0

+0.10

+0.36

+0.66

+0.88

+0.89

+0.59

-0.19

-1.68

-3.89

-7.19

3.0

+0.07

+0.26

+0.51

+0.74

+0.91

+0.83

+0.42

-0.53

-2.23

-4.83

4.0

+0.04

+0.15

+0.33

+0.52

+0.68

+0.75

+0.53

-0.13

-1.45

-3.65

5.0

+0.02

+0.08

+0.19

+0.35

+0.51

+0.61

+0.52

+0.07

-1.01

-2.93

6.0

+0.01

+0.04

+0.11

+0.22

+0.36

+0.49

+0.48

+0.17

-0.73

-2.42

8.0

0.00

+0.01

+0.03

+0.08

+0.18

+0.31

+0.38

+0.24

-0.40

-1.84

10.0

0.00

-0.01

0.00

+0.02

+0.09

+0.21

+0.30

+0.26

-0.22

-1.47

12.0

0.00

0.00

+0.01

0.00

+0.04

+0.14

+0.24

+0.22

-0.12

-1.23

14.0

0.00

0.00

0.00

0.00

+0.02

+0.10

+0.18

+0.21

-0.07

-1.05

16.0

0.00

0.00

0.00

-0.01

-0.01

+0.06

+0.12

+0.20

-0.05

-0.91


where:

10.22

= = = = = =

vertical moment at given depth internal tank diameter thickness of tank wall bending moment coefficient depth measured from top uniform horizontal pressure

kNm per m m m m kN/m3


TABLE A3


Pinned base, free top:

Combined triangular and uniform Load

Bending moment co-efficients 0.1 h

0.2 h

0.3 h

0.4 h

0.5 h

0.6 h

0.7 h

0.8 h

0.9 h

1.0 h

0.4

+0.20

+0.72

+1.51

+2.30

+3.01

+3.48

+3.57

+3.12

+1.97

0.00

0.8

+0.19

+0.64

+1.33

+2.07

+2.71

+3.19

+3.29

+2.92

+1.87

0.00

1.2

+0.16

+0.58

+1.11

+1.77

+2.37

+2.80

+2.96

+2.63

+1.71

0.00

1.6

+0.12

+0.44

+0.91

+1.45

+1.95

+2.36

+2.55

+2.32

+1.55

0.00

2.0

+0.09

+0.33

+0.73

+1.14

+1.58

+1.99

+2.19

+2.05

+1.45

0.00

3.0

+0.04

+0.18

+0.40

+0.63

+0.92

+1.27

+1.52

+1.53

+1.11

0.00

4.0

+0.01

+0.07

+0.16

+0.33

+0.57

+0.83

+1.09

+1.18

+0.92

0.00

5.0

0.00

+0.01

+0.06

+0.16

+0.34

+0.57

+0.80

+0.94

+0.78

0.00

6.0

0.00

0.00

+0.02

+0.08

+0.19

+0.39

+0.62

+0.78

+0.68

0.00

8.0

0.00

0.00

-0.02

0.00

+0.07

+0.20

+0.38

+0.57

+0.54

0.00

10.0

0.00

0.00

-0.02

-0.01

+0.02

+0.11

+0.25

+0.43

+0.45

0.00

12.0

0.00

0.00

-0.01

-0.02

0.00

+0.05

+0.17

+0.32

+0.39

0.00

14.0

0.00

0.00

-0.01

-0.01

-0.01

0.00

+0.12

+0.26

+0.33

0.00

16.0

0.00

0.00

0.00

-0.01

-0.02

-0.04

+0.08

+0.22

+0.29

0.00


where:

= = = = = = =

vertical moment at given depth internal tank diameter thickness of tank wall bending moment coefficient unit weight of retained liquid depth measured from top uniform horizontal pressure

kNm per m m m kN/m3 m kN/m3

10.23


ANNEXURE B

-


Ring tension in circular tanks

The figure below shows the typical distribution of ring tension throughout the height of the wall.

When the top is free and the base is free to slide, t he ring tension at any depth is given by:

where:

= = =

ring tension internal tank diameter unit weight of retained liquid

kN m kN/m3

=

depth measured from top

m

When the top of the wall is free and the base is either ‘pinned’ or ‘fixed’, the ring tension at any depth may be obtained from tables B1 or B2.

10.24


TABLE B1


Pinned base, free top:

Triangular load

Ring tension coefficients 0.0 h

0.1 h

0.2 h

0.3 h

0.4 h

0.5 h

0.6 h

0.7 h

0.8 h

0.9 h

0.4

+47.4

+44.0

+39.5

+35.2

+30.8

+26.4

+21.5

+16.5

+11.1

+5.70

0.8

+42.3

+40.2

+38.1

+35.8

+33.0

+29.7

+24.9

+20.2

+14.5

+7.60

1.2

+35.0

+35.5

+36.1

+36.2

+35.8

+34.3

+30.9

+25.6

+18.6

+9.30

1.6

+27.1

+30.3

+24.1

+36.9

+38.5

+38.5

+36.2

+31.4

+23.3

+12.4

2.0

+20.5

+25.0

+32.1

+37.3

+41.1

+43.4

+41.9

+36.9

+28.0

+15.1

3.0

+7.40

+17.9

+28.1

+37.5

+44.9

+50.6

+51.9

+47.9

+37.5

+21.0

4.0

+1.70

+13.7

+25.2

+36.7

+46.9

+54.5

+57.9

+55.3

+44.7

+25.6

5.0

-0.80

+11.4

+23.5

+35.6

+46.9

+56.2

+61.7

+60.6

50.3

+29.4

6.0

-1.10

+10.3

+22.3

+34.3

+46.3

+56.6

+63.9

+64.3

+54.7

+32.7

8.0

-1.50

+9.60

+20.8

+32.4

+44.3

+56.4

+66.1

+69.7

+62.1

+38.6

10.0

-0.80

+9.50

+20.0

+31.1

+42.8

+55.2

+66.6

+73.0

+67.8

+43.3

12.0

-0.20

+9.70

+19.7

+30.2

+41.7

+54.1

+66.4

+75.0

+72.0

+47.7

14.0

0.00

+9.80

+19.7

+29.9

+40.8

+53.1

+65.9

+76.1

+75.2

+51.3

16.0

+0.20

+10.0

+19.8

+29.9

+40.3

+52.1

+65.0

+76.4

+77.6

+53.6

Note that a positive coefficient indicates tension.

where:

= = =

ring tension internal tank diameter unit weight of retained liquid

kN m kN/m3

=

depth measured from top

m

10.25

UNISA Reinforced Concrete Design Study Guide

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