Structural Analysis and Design Lectures

Lecture 10 – Column Base Plates Columns must transmit vertical loads to the concrete footing. An intermediary steel base plate is used to distribute this column load without crushing the concrete. Applied load “P”

Steel column

Anchor rods (4 min. per OSHA) See AISC p. 14-9 Steel base plate Base Plate thickness

Concrete resistance to crushing

The design of steel base plates is based on the following: • •

AISC Spec. Chapter J8 (p. 16.1-70) AISC Part 14

Lecture 10 - Page 1 of 6

The design of a base plate involves the following steps: Pp = Nominal bearing strength of concrete = 0.85f’cA1 Design Bearing strength of concrete: φcPp where φc = 0.60 LRFD Pp

where Ωc = 2.50 ASD Ωc where: f’c = specified compressive strength of concrete, KSI A1 = area of steel base plate concentrically loaded on conc, in2 = BN (where B and N use whole inches if possible) B bf

See AISC p. 14-5 m 0.95d

d

N

m n

m=

n=

0.80bf

n

N − 0.95d 2 B − 0.80b f 2


tmin = minimum base plate thickness per AISC p. 14-6: t min = L

2 f pu

t min = L

0 .9 F y

where: f pu =

Pu BN

3.33 f pa

Fy

where: f pa =

Pu = factored axial load, kips

Fy = base plate steel yield stress

Pa BN

Pa = service axial load, kips

Fy = base plate steel yield stress

m

m

L = larger of

L = larger of n

=

n

LRFD

ASD


Example (LRFD) GIVEN: A W14x82 A992 column has a factored axial load Pu = 700 KIPS. It bears on a steel base plate using A36 steel. The footing has concrete f’c = 3000 PSI. REQUIRED: Design the column base plate. Step 1 – Determine required base plate area, A1 to avoid conc. crushing:

φcPp = Design bearing strength of concrete = 0.6Pp = 0.6(0.85f’cA1) Re-arranging to solve for A1: A1 =

=

Pu 0.6(0.85 f ' c )

700 KIPS 0.6(0.85(3KSI ))

A1 = 457.5 in2 Step 2 – Determine “Optimized” base plate dimensions:

∆=

0.95d − 0.8b f

=

d and bf → from properties p. 1-22

2 0.95(14.3" ) − 0.8(10.1" ) 2

= 2.75” N≈

≈

A1 + ∆ 457.5in 2 + 2.75"

≈ 24.14” TRY N = 24” and B = 20” (Area = 480 in2 > 457.5 in2)


Step 3 – Determine “m” and “n”: m=

=

N − 0.95d 2

24"−0.95(14.3" ) 2

= 5.2”

B − 0.80b f

n=

2 20"−0.80(10.1" ) 2

=

= 5.96” Step 4 – Determine minimum base plate thickness, tmin: t min = L

2 f pu 0 .9 F y

where: Pu = factored axial load, kips = 700 Kips f pu =

=

Pu BN

700 Kips = 1.46 KSI (20" )(24" ) m = 5.2”

L = larger of n = 5.96”

t min = 5.96"

2(1.46 KSI ) 0.9(36 KSI )

use

Base plate yield stress

= 1.79” → use 1⅞” thick plate


Step 5 – Draw “Summary Sketch”:

20”

W14x82 A992 col. centered on plate

24”

7 1 " thick A36 steel base plate 8


Lecture 12 – Bolted Connections Below is a typical bolt and the terms given to the parts of a bolt:

Bolts used in structural steel fasteners fall within 2 categories – see AISC Table 2-5 p. 2-41: 1) Carbon steel bolts – These bolts achieve their total strength from shear (or tension) strength across the diameter of the bolt. They are relatively low-strength and are used primarily for low-load applications such as for anchor rods. The typical carbon steel bolt used in structural steel buildings is ASTM A307 and F1554 for use in anchor rods. 2) High-strength bolts – These bolts are used for high-load connections and obtain their total strength from the shear strength across the diameter of the bolt PLUS the friction developed between the nut and joined steel surfaces. In order to achieve the friction capacity, these bolts are tensioned to 70% of the ultimate tensile strength of the material according to the table below. ASTM A325 and A490 bolts are typically used. The LRFD references the design of bolted connections in the following: • • • •

AISC Spec. Chapter J3 (p. 16.1-102) AISC Part 7 AISC Part 9 AISC Part 10


Possible Bolted Shear Failure Mechanisms: There are 4 basic types of failure mechanisms for bolted connections under shear: 1) Bolt Shear: This is probably the most obvious failure mode. It occurs when the applied load exceeds the shear capacity through the bolt. The design shear strength is dictated in AISC Table J3.2 p. 16.1-104 and AISC Table 7-1 p. 7-22. Possible remedies include using a larger diameter bolt, higher grade of bolt or more bolts.

Result

Bolt shear failure


2) Edge Tear-Out: This occurs when the bolt is located too close to the edge of the plate in the direction of load. A minimum required edge distance, Le, is dictated in AISC Table J3.4 p. 16.1-107. Possible remedies include increasing the edge distance or reducing the bolt diameter.

Le

Edge Distance failure


3) Bearing Failure: This type of failure occurs when one of the plates is too thin or not strong enough for the applied loads. The design bearing strength at bolt holes is dictated in AISC p. 16.1-111 and AISC Table 7-5 p. 7-28 and AISC Table 7-6 p. 7-30. Possible remedies increasing the plate thickness, use a higher grade of steel or using larger diameter bolts.

Thin plate

Bearing failure


4) Net Section Failure: A net section failure occurs when there are too many bolt holes perpendicular to the line of action – resulting in too little material to carry the load. Think of Swiss cheese. The minimum spacing of bolts is dictated in AISC J3.2 p. 16.1-106 as not less than 2⅔ times the nominal bolt diameter, preferably 3 times the bolt diameter. Usually 3” is used as the nominal bolt spacing for bolts < 1” in diameter.

Net section failure


Types of Bolted Connections 1) Bearing-Type Connections: A bearing-type connection is the most common type of bolted connection. It is used in most simple-shear connections and in situations when loosening or fatigue due to vibration or load fluctuations are NOT design considerations. In these connections, bolts are tightened to the “snug-tight” condition, as defined as the tightness attained by a few impacts of an impact wrench or the full effort of an iron worker using an ordinary spud wrench. The design strength of bearing-type fasteners is per AISC Eq. J3-1 p. 16.1-108. 2) Slip-Critical Connections: A slip-critical connection is one in which loosening due to vibration or load reversals are to be considered. Also, holes that are oversize or slotted shall be designed as slip-critical connections. Bolts that are used in slip-critical connections must be pre-tensioned per AISC Table J3.1 p. 16.1-103. In addition, the design strength of the connection must be checked in accordance with AISC J3.8, J3.9 and J3.10 p. 16.1-109 thru 111. As an alternative, AISC Table 7-3 and 7-4 p. 7-24 thru 27 can be used.

See AISC Table J3.3 p. 16.1-105 for hole dimensions


Design Strength of Bearing-Type Fasteners From AISC J3.6 p. 16.1-108, the design tension or shear strength of a high-strength bolt or threaded part is: Design strength of bolt = φRn LRFD Allowable strength of bolt =

Rn ASD Ω

where: Rn = FnAb φ = 0.75 LRFD Ω = 2.00 ASD Fn = nominal tensile or shear stress of fastener, KSI = from Table J3.2 p. 16.1-104 Ab = x-sect. nominal area of unthreaded body of bolt, in2 Shear Plane: The shear plane is the plane in which the various connected parts are in contact.

Threads Not excluded from shear plane “N” Threads eXcluded from shear plane “X”

Single-shear

Double-shear

Load

Load


Example 1 (LRFD) GIVEN: A ¾” diameter ASTM A325-N bolt in single-shear is subjected to a factored load of 14 KIPS. REQUIRED: Determine the design shear strength of the bolt considering bolt shear ONLY, and comment if the bolt is acceptable.

¾” dia. A325-N bolt

Pu = 14 KIPS Step 1 – Determine design shear strength of bolt: Design shear strength = φRn where: φ = 0.75 Rn = nominal shear strength of fastener = FnAb Fn = from Table J3.2 p. 16.1-104 = 48 KSI (threads Not excluded) Ab = nominal area of unthreaded body of bolt, in2 =

=

π

4

π 4

D2

(0.75" ) 2

= 0.44 in2 Design shear strength = (0.75)(48 KSI)(0.44 in2) Design shear strength = 15.8 KIPS > 14 KIPS → Acceptable


Example 2 (LRFD) GIVEN: Same as Example 1 REQUIRED: Determine bolt design shear using AISC Table 7-1 p. 7-22. Step 1 – Refer to Table 7-1: ASTM A325 Thread condition = “N” Design shear strength = 15.9 KIPS Loading = “S” (Single shear) Bolt Diameter, db = ¾”


Lecture 13 – Bolted Connections (cont.) In the previous lecture, we looked at general strength considerations of bolted connections. In this lecture we will look at a typical all-bolted beam-to-girder shear connection to see practical bolted connection considerations. K

Beam Cope Lev S S

Girder Connection angles Angle gage “g” from AISC p. 1-46 = Leh where: Cope = cut distance of beam flange necessary to clear girder flange and “K” distance, usually 1½”, 2” or 3” K = distance between top of flange to edge of start of flat web = from beam properties AISC Part 1 Lev = required minimum vertical edge distance in direction of load = from AISC Table J3.4 p. 16.1-107 S = bolt center-to-center spacing from AISC J3.3 p. 16.1-106 = 2⅔ times nominal bolt diameter (minimum) = 3 times bolt diameter (preferred) = 3” (typical for bolts up to 1” diameter)


Example (LRFD) GIVEN: A W16x40 A992 steel beam “A” frames into a W18x55 A992 steel girder “B”. The applied floor Service DL = 80 PSF and the applied floor Service LL = 100 PSF. Use ¾” diameter A325-X bolts with standard bolt holes and doubleangle A36 L3x3x¼ connection angles. The beam is coped at top flange only. REQUIRED: Design the all-bolted beam-to-girder connection and provide a summary sketch.

W16x40 Beam “A”

4 @ 6’-0” = 24’-0”

W18x55 Girder “B”

30’-0”

Step 1 – Determine factored beam end reaction: wu = 1.2[6’(80 PSF) + 40 PLF] + 1.6[6’(100 PSF)] = 1584 PLF = 1.6 KLF Beam weight Beam end reaction =

=

wu L 2

1.6 KLF (30'−0" ) 2

= 24 KIPS


Step 2 – Use AISC Table 10-1 “All-Bolted Double-Angle Connections”, p. 10-22:

These tables incorporate all design considerations for typical allbolted double-angle connections. ¾” Bolts

See Step 3 See Step 9 W16x40 Beam

See Step 5

See Step 7 See Step 5

See Step 8


Step 3 – Check Bolt and Angle Design Strength: From Table above, ASTM A325 Bolt and angle design strength = 76.4 KIPS > 24 KIPS

Thread Cond. = X Angle thickness = ¼”

Step 4 – Determine minimum required cope: The minimum required vertical edge distance must be greater than the “K” distance for either the girder or the beam. W18x55 girder “Kdet” = 1

5 " from AISC p. 1-18 16 Use cope = 1½”

W16x40 beam “Kdet” = 1

3 " from AISC p. 1-20 16

Step 5 – Determine vertical edge distance, Lev: For compactness, use Lev = 1¼” (See Table J3.4 p. 16.1-107) Step 6 – Determine angle gage for L3x3x¼ = Leh: From AISC p. 1-46 → g1 = Leh = 1¾” Step 7 – Check Beam Web Design Strength: From Table above, Hole Type = STD Leh = 1¾”

Beam web design strength = 200 KIPS per inch thickness

Lev = 1¼” The web thickness, tw of a W16x40 = 0.305” from AISC p. 1-20 W16x40 web design strength = 0.305”(200 KIPS/inch) = 61 KIPS > 24 KIPS → OK


Step 8 – Check girder Support Design Strength: From Table above, Support Design Strength per Inch Thickness = 526 KIPS The web thickness, tw of a W18x55 = 0.390” from AISC p. 1-18 W18x55 web design strength = 0.390”(526 KIPS/inch) = 205 KIPS > 24 KIPS → OK Step 9 – Determine bolt spacing S: Preferred bolt spacing S = 3 x bolt diameter = 3(¾”) = 2¼” Use S = 3” from Table above > 2¼” → OK Step 10 – Draw summary sketch of connection design:

W16x40 Beam

W18x55 Girder

Cope = 1½” Lev = 1¼” S = 3” S = 3” 1¼”

2 - L 3x3x¼ x 8½” long A36 connection angles with 9 - ¾” A325-X bolts in STD holes Angle gage = 1¾”


Lecture 14 – Welded Connections Welding is a procedure that involves fusing two pieces of steel together by melting a sacrificial “flux” electrode to two pieces, thereby joining the pieces permanently together. They have some distinct advantages over bolted connections including: • • • •

Welded joints are more rigid than bolted joints Can directly connect pieces without the need for connection plates Welds do not create holes in member (i.e., no need to check fracture on net area) Can join odd-shaped pieces together

Welds also have some disadvantages which may preclude their use, including: • • • • •

Welds are brittle, not ductile like bolted connections Very labor intensive Skilled labor required Quality control is difficult to inspect Potential fire hazard in areas of welding

Fillet Welds: The most common type of weld for structural steel connections is the “fillet” weld. This type of weld joins 2 pieces with flat faces at 900 angles. Some examples of fillet welds and their weld symbols are shown below:



A closer inspection through the fillet weld itself is shown below to indicate some of the dimensions of a weld:

The most common type of fillet welding process is “Arc” welding, or sometime called “stick” welding. This process involves running an electric current through a sacrificial electrode creating an arc of extremely high temperature that fuses the steel pieces together. The electrode (stick) has a coating called a “flux” that, when subject to heat, produces a cloud acting as a barrier to impurities in the air entering the molten metal. A diagram of arc welding process is below:


Some examples of other welds are shown below:


Minimum & Maximum Size of Fillet Welds: Below is a table relating the minimum size of fillet weld to the thickness of material to be welded per AISC Table J2.4 p. 16.1-96: Material thickness of the thicker part joined: Up to ¼” inclusive Over ¼” to ½” Over ½” to ¾” Over ¾”

Minimum size of fillet weld: 1/8” 3/16” ¼” 5/16”

Maximum size of a fillet weld = See AISC p. 16.1-96 paragraph 2b = Thickness of thinner part up to ¼” thick = Thickness – 1/16” over ¼” thick

Design SHEAR Strength of Fillet Welds: From AISC p. 16.1-98 Weld available strength = Rn = FwAw

LRFD Design Strength = φRn R ASD Allowable strength = n Ω

where: φ = 0.75 for shear from AISC Table J2.5 p. 16.1-100 Ω = 2.00 for shear from AISC Table J2.5 p. 16.1-100 Fw = nominal strength of weld electrode, Table J2.5 = 0.60FEXX FEXX = weld electrode strength = 70 KSI for E70XX electrodes Aw = effective cross-sectional area of weld, in2 = cos(450) x (Weld Size) x (Weld Length)


Example (LRFD) GIVEN: Two ¼” thick A36 steel plates fillet welded as shown below. Use E70XX weld electrodes. REQUIRED: Determine the maximum factored load, Pu, that can be applied based on shear strength of the welds.

Pu

3/16

4”

Step 1 – Determine total length of fillet welds: Total length = 2(4”) = 8” Step 2 – Determine design shear strength of welds: Weld design strength = φRn where: φ = 0.75 for shear from AISC Table J2.5 Rn = FwAw Fw = nominal strength of weld electrode = 0.60FEXX FEXX = weld electrode strength, Table 8-3 p. 8-65 = 70 KSI for E70XX electrodes Aw = effective cross-sectional area of weld, in2 = cos(450) x (Weld Size) x (Weld Length) Weld design strength = 0.75(0.60(70 KSI))(cos(450)(3/16”)(8”)) Weld design strength = Pu = 33.4 KIPS


Lecture 15 – Welded Connections (cont.) The design of a typical all-welded double-angle simple shear connection will be investigated. Similar to an all-bolted connection, the AISC manual makes use of a one-stop-shopping design aid Table 10-3 p. 10-48 for all design considerations. An example design of a typical all-welded double-angle simple shear connection is as follows: Example 1 (LRFD) GIVEN: A W21x44 A992 girder with a factored end reaction = 87 KIPS has 2 – L3x3x3/8 connection angles shop-welded to the girder web and field-welded to the W12x58 A992 column flange as shown. REQUIRED: Design the connection and provide a summary sketch.

W21x44 Girder

Minimum Web thickness

L=?

W12x58 Column Weld A 3/16

2X weld size

L2x2x¼ erection angle shop-welded to col. flange

L=?

2 – L3x3x3/8 connection angles

Weld B

¼


Step 1 – Refer to AISC Table 10-3 p. 10-48 for design info:


See Step 4

See Step 3


Step 2 – Determine minimum length “L” by checking Weld B strength: From Table above, using a ¼” weld, choose a length “L” such that φRn > 87 KIPS Use Lmin = 12” → φRn = 99 KIPS > 87 KIPS OK Step 3 - Determine minimum length “L” by checking Weld A strength: From Table above, using a 3/16” weld, choose a length “L” such that φRn > 87 KIPS Use Lmin = 7” → φRn = 92.9 KIPS > 87 KIPS OK Step 4 – Check minimum required column flange thickness for Weld B: The flange thickness, tf for a W12x58 column = 0.640” From Table above, using a ¼” weld, the required minimum support thickness = 0.190” < 0.640” OK Step 5 – Check minimum girder web thickness for Weld A: The web thickness, tw for a W21x44 beam = 0.350” From Table above, using a 3/16” weld, the required minimum web thickness = 0.286” < 0.350” OK Step 6 – Determine angle length: Since the minimum length of Weld B = 12” which is greater than the minimum length of Weld A → use 12” long angles.


Step 7 – Draw summary sketch: W21x44 Girder 1¼” 12”

W12x58 Column Weld A 3/16

2(¼”) = ½”

L2x2x¼ erection angle shop-welded to col. flange

12”

2 – L3x3x3/8 connection angles

Weld B

¼


Eccentric Loading on Weld: Table 10-3 assumes that the loading is approximately concentric. In other words, there is no moment acting on the weld. In industrial or other situations, connections are applied eccentrically to a column which creates moment on the column as well as the connection. AISC Tables 84 thru 8-11 can be used for design of eccentrically-loaded welded connections.

Pu Pu

Concentric Load on Weld Group

Eccentric Load on Weld Group

Example 2 (LRFD) GIVEN: A ½” plate is welded to the flange of a column as shown below. This plate carries a cantilevered factored load of 25 KIPS. REQUIRED: Determine if the weld is adequate to carry the eccentric loading. Pu = 25 KIPS 16” ¼ ¼

4”

Column flange

L = 8” ½” Plate


Step 1 – Refer to AISC Table 8-5 p. 8-72:


See Step 4

See Step 3


Step 2 – Determine coefficient “k”: Since “L” = 8” kL = 4” k(8”) = 4” k = 0.5 Step 3 – Determine coefficient “a”: Since “L” = 8” aL = 16” k(8”) = 16” a = 2.0 Step 4 – Determine coefficient “C”: k = 0.5 Use “C” = 0.821 a = 2.0

Step 5 – Determine maximum permissible Pu: From Table above Lmin =

Pu φCC1 D

ERROR in AISC Manual!!!

Rearranging and solving for Pu:

Pu = LφCC1D

where: φ = 0.75 C = 0.821 C1 = 1.0 (AISC p. 8-65) D = 4 - 1/16ths

= (8”)(0.75)(0.821)(1.0)(4) Pu = 19.7 KIPS < 25 KIPS → UNACCEPTABLE


Lecture 16 – AISC Code of Standard Practice Design, fabrication, and erection of steel-framed buildings should incorporate provisions of the AISC “Code of Standard Practice for Steel Buildings and Bridges” and is found in AISC Spec. Section 16.3 and also online at www.AISC.org. It was first published in 1924 and is now in its 6th edition, dated March 7, 2005. It provides a useful framework for the understanding of the acceptable standards for the construction of structural steel structures. It is useful for owners, architects, engineers, contractors, fabricators, construction managers and anyone else involved with construction using structural steel. The Code also serves as a basis for technical project specifications, typically CSI Specification Section 05100 – Structural Steel (see Lecture 17). A summary of the Code of Standard Practice is given below. GLOSSARY Definitions and abbreviations of relevant terms used throughout the Code. Some important, (but often vague) definitions include: • • • • • • • • • • • • • • • • • • •

AESS – Architecturally Exposed Structural Steel Contract Documents Design Drawings EOR – Engineer-of-Record Erection Drawings Erector Fabricator Inspector Owner Owner’s Designated Representative for Construction Owner’s Designated Representative for Design RCSC – Research Council on Structural Connections RFI – written Request for Information SER – Structural Engineer-of-Record Shop Drawings Specifications SSPC – Society for Protective Coatings (formerly Steel Structures Painting Council) Steel Detailer Structural Steel


Section 1 – GENERAL PROVISIONS 1.1 – Scope The Code shall govern the fabrication and erection of structural steel (unless otherwise noted in Contract Documents). 1.2 – Referenced Specifications, Codes and Standards • • • • • • •

AISC Manual of Steel Construction AISC Seismic Provisions AISC Specification ASTM (lots of referenced standards) AWS D1.1 – Structural Welding Code RCSC Specification – Specification for Structural Joints using ASTM A325 or A490 Bolts SSPC – Steel Structures Painting Council

1.3 – Units Either U.S. customary or metric units will be used. Each system shall be independent of the other. 1.4 – Design Criteria The AISC Specification shall be used in the absence of other design criteria. 1.5 – Responsibility for Design •

If the Owner’s Designated Representative for Design provides the design, the Fabricator and Erector are NOT responsible for building code conformance of the design.

•

If the Owner enters a contract with the Fabricator for design/build, then the Fabricator IS responsible for building code conformance of the design.

1.6 – Patents and Copyrights The EOR is responsible for obtaining patents and copyrights of design.


1.7 – Existing Structures Demolition, protection, field dimensions and/or abatement or removal of hazardous are NOT the responsibility of the Fabricator or Erector. 1.8 – Means, Methods and Safety of Erection • •

Erector is responsible for erection of frame. SER is responsible for structural adequacy of completed project.

Section 2 – CLASSIFICATION OF MATERIALS Structural Steel shall consist of the following typical elements: • • • • • • • • • • • •

Anchor Rods that will receive structural steel Base Plates & bearing plates Beams Bracing (permanent) Columns Connections Fasteners for connecting structural steel Girders Hangers Lintels Shear stud connectors Trusses

Section 3 – DESIGN DRAWINGS AND SPECIFICATIONS 3.1 – Structural Design Drawings and Specifications Structural design drawings shall consider design loads and forces in the completed project. Drawings must show: • • • • • • • • •

Size, section, location and material grade of all members Geometry and working points necessary for layout Floor elevations Column centers and offsets Camber requirements for beams (if required) Permanent bracing, stiffeners, reinforcement Connection details or data that can be used by fabricator for design including ASD or LRFD methodology Data relating to non-Structural steel elements that interact with frame Painting requirements of Structural steel Lecture 16 - Page 3 of 13

3.2 – Architectural, Electrical and Mechanical Design Drawings Other trades’ design drawings may be used to show structural steel elements for purposes of defining detail configurations and other construction information, however, ALL STRUCTURAL INFO. MUST BE SHOWN ON STRUCTURAL DRAWINGS. 3.3 – Discrepancies •

Discrepancies discovered in the Contract Documents shall be resolved by the EOR in a timely manner so as not to delay the Fabricator’s work.

Discrepancies between: Design Drawings Specifications Scaled graphic drawings Written info in drawing Arch., Elect., Mech. Structural Drawings Drawings

Which Governs: Design Drawings Written info in drawings Structural Drawings

3.4 – Legibility of Design Drawings Design Drawings must be legible and drawn to a scale of not smaller than 1/8” = 1’-0” (unless clarity of the drawing is carefully considered), larger as necessary to convey detailed information. 3.5 – Revisions to Design Drawings and Specifications All revisions must be communicated either by issuing new Design Drawings and Specifications or by re-issuing existing Design Drawings and Specifications. Revisions must be clearly and individually indicated, dated and identified by a revision number. These revised sketches become “amendments” to the Contract Drawings. 3.6 – Fast-Track Project Delivery Release of structural Design Drawings and Specifications shall constitute a release for construction, regardless of the status of the architectural, electrical, mechanical, or any other trades’ documents.


Section 4 – SHOP AND ERECTION DRAWINGS 4.1 – Owner Responsibility The Owner shall furnish the complete structural Design Drawings and Specifications to the Fabricator in a TIMELY MANNER. 4.2 – Fabricator Responsibility Fabricator shall produce Shop Drawings and Erection Drawings. Fabricators are permitted to use the services of independent detailers. 4.3 – Use of CAD and/or Copies of Design Drawings Fabricator shall NOT reproduce any part of the Design Drawings as part of the Shop or Erection Drawings without the express written permission of Owner’s Designated Representative for Design. 4.4 – Approval Shop and Erection Drawings must be submitted to Owner’s Representative for Design for review and approval and returned to Fabricator within 14 calendar days.


Section 5 – MATERIALS 5.1 – Mill Materials Fabricator is permitted to order materials upon receipt of Contract Documents that have been issued for construction. If mill materials do not meet ASTM A6 tolerances, Fabricator is permitted to make corrective procedures. 5.2 – Stock Materials Fabricator may use stock materials if they meet with required ASTM specifications. Certified mill test reports are used as evidence of record of quality of material. Section 6 – SHOP FABRICATION AND DELIVERY 6.1 – Identification of Material Materials used for special requirements shall be marked by the supplier as specified by ASTM A6 prior to delivery to Fabricator’s shop or point of use. 6.2 – Preparation of Material Thermal cutting of material is permitted. Surfaced specified as “finished” shall have a roughness in accordance with ANSI/ASME B46.1 that is less than or equal to 500. 6.3 – Fitting and Fastening Projecting elements of connection materials need not be straightened in the connecting plane. Backing bars and runoff tabs shall be used to produce as required to produce sound welds, and do not need be removed unless specifically designated in the Contract Documents.


6.4 – Fabrication Tolerances The following tolerances are to be used: Member Type: Both ends finished for contact bearing Members < 30’-0” Members > 30’-0” All members Beam length < 50’-0” Beam length > 50’-0”

Tolerance Variation: Length = ± 1/32” Length = ± 1/16” Length = ± 1/8” Straightness < 1/1000 axial length Camber variation = 0 → ½” Camber variation = 0 → ½” + 1/8” per 10’-0” additional length beyond 50’-0”

6.5 – Shop Cleaning and Painting Structural steel that does not require shop paint shall be cleaned of oil, grease, dirt and any foreign material. Structural steel requiring shop painting shall be free of oil, grease, dirt and any foreign material; as well as meeting the requirements of SSPC-SP2. 6.6 – Marking and Shipping of Materials Erection marks shall be painted to all structural steel members. Connection members shall be shipped in separate closed containers according to grade, length and diameter. 6.7 – Delivery of Materials Steel shall be delivered in a sequence that will permit efficient and economical fabrication and erection. Section 7 – ERECTION 7.1 – Method of Erection Structural steel shall be erected using methods and a sequence that will permit efficiency and economy.


7.2 – Job-Site Conditions Owner’s Representative for Construction shall provide the following: • • •

Access road for deliveries and movement of materials Adequate obstruction-free space for operation of Erector’s equipment Adequate storage space

7.3 – Foundations, Piers and Abutments Owner’s Representative for Construction shall be responsible for accurate location, suitability and access to all foundations, piers and abutments. 7.4 – Building Lines and Bench Marks Owner’s Representative for Construction shall be responsible for accurate location of building lines and benchmarks and shall furnish the Fabricator with a plan containing such information. 7.5 – Installation of Anchor Rods and Other Embedded Items Owner’s Representative for Construction shall be responsible for setting in accordance with Embedment Drawings. The variation in location shall be as follows: Item: Centers of any 2 anchor rods within an anchor rod group Centers of adjacent anchor rod groups Elevation of tops of anchor rods Accumulated variation between centers of anchor rod groups

Variation in Dimension: < 1/8” < ¼” ± ½” < ¼” per 100’-0” not to exceed a total of 1”

7.6 – Installation of Bearing Devices All leveling plates, nuts, washers and bearing plates that can be handled without crane are set to line and grade by the Owner’s Representative for Construction (otherwise set by Erector). 7.7 – Grouting Grouting shall be the responsibility of the Owner’s Representative for Construction. The usual method for supporting columns during erection is by use of leveling nuts and washers or shims.


7.8 – Field Connection Material Fabricator shall provide field connection details consistent with Contract Documents. 7.9 – Loose Material Unless otherwise noted, loose structural steel items that are not connected to the steel frame by the Owner’s Representative for Construction without assistance from Fabricator. 7.10 – Temporary Support of Structural Steel Frames The Owner’s Designated Representative for Design shall identify the following: • •

Lateral load resisting system and connecting diaphragm elements providing stability in the completed structure Any special erection conditions that are required by the design concept, such as use of jacks, shores, etc.

7.11 – Safety Protection The Erector shall provide floor coverings, handrails, walkways and other protection for the Erector’s personnel in accordance with all applicable safety regulations. Unless otherwise specified, the Erector is permitted to remove such safety protection form areas where the erection operations are completed. Safety protection for other trades that are not under the direct employment of the Erector shall be the responsibility of the Owner’s Representative for Construction. 7.12 – Structural Steel Frame Tolerances The accumulation of the mill tolerances (Section 6.4) and fabrication tolerances shall not cause the erection tolerances (Section 7-13) to be exceeded.


7.13 – Erection Tolerances Erection tolerances are referenced from Work Points and Work Lines defined as: • • •

Members other than horizontal members, the member work point is the actual center of the member at each end of the shipping piece. Horizontal members work point shall be the actual centerline of the top flange or top surface at each end. Work line is defined a s a straight line that connects the member work points.

Member: Column tolerance deviation from plumb Individual straight piece (other than column) connecting to column Adjustable members

Erection Tolerance: ± 1/500 distance between work points not to exceed 1” total for first 20 stories + 3/16” - 5/16” ± 3/8”

7.14 – Correction of Errors Correction of minor misfits by means of reaming, grinding, drawing of elements into line by drift pins, welding or cutting shall be considered normal erection operations. Errors that cannot be corrected by these means must be promptly reported to the Owner’s Designated Representative for Design and Construction. 7.15 – Cuts, Alterations and Holes for Other Trades The Fabricator or the Erector may NOT cut, drill or otherwise alter their work to accommodate other trades unless work is specified in the Contract Documents. 7.16 – Handling and Storage The Erector shall take reasonable care in the proper handling and storage of structural steel during erection to avoid excess dirt and foreign matter. However, it is not the Erector’s responsibility to remove dirt or other foreign material that may accumulate during normal erection procedures.


7.17 – Field Painting The Fabricator or the Erector is NOT responsible to paint field bolts, or to touch-up abrasions of the shop coat, or to perform any field painting. 7.18 – Final Cleaning Up Upon completion and acceptance, the Erector shall remove any of the Erector’s falsework, scaffolding, rubbish and temporary structures. Section 8 – QUALITY ASSURANCE 8.1 – General The Fabricator shall maintain a quality assurance program to assure that the work is performed in accordance with this Code. 8.2 – Inspection of Mill Material Certified mill test reports shall constitute sufficient evidence that the mill product satisfies material order requirements. 8.3 – Non-Destructive Testing As per Contract Documents. 8.4 – Surface Preparation and Shop Painting Inspection As per Contract Documents. 8.5 – Independent Inspection •

• • • •

Fabricator and Erector shall provide the Inspector with access to all places where work is being performed, and a minimum of 24 hours notice must be given prior to commencement of work. Inspector shall inspect work at shop as much as possible. Field inspections should be performed as promptly as possible. Deficiencies discovered by Inspector shall be reported to Fabricator and Erector as soon as possible. The Inspector shall NOT approve of any deviations from the Contract Documents without written approval from the Owner’s Designated Representative for Design and Construction.


Section 9 – CONTRACTS 9.1 – Types of Contracts • • • •

Lump sum price Price per pound Price per item Unit price

9.2 – Calculation of Weights Weight is determined by calculation of gross weight of materials as shown on the Shop Drawings. This does NOT include shop and field weld metal or protective coatings. Deductions shall not be made for holes, copes, drilling or other removals for connections. Steel unit weight = 490 lb/ft3. 9.3 – Revisions to Contract Documents Revisions to Contract Drawings shall be confirmed by change order or extra work order, and shall constitute authorization by the Owner that the revision is released for construction. 9.4 – Contract Price Adjustment When the scope of work and responsibilities of the Fabricator and the Erector are changed, an appropriate modification to the contract price shall be made. Requests for contract price adjustments shall be presented to the Owner and approved/disapproved in a timely manner. 9.5 – Scheduling The contract schedule shall state when the Design Drawings will be released for construction so that erection can start at the designated time and continue without interference or delay. 9.6 – Terms of Payment Terms of payment shall be outlined in the Contract Documents.


Section 10 – ARCHITECTURALLY EXPOSED STRUCTURAL STEEL The rapidly increasing use of exposed structural steel as an aesthetic design medium has prompted the use of additional requirements that apply to these members. Typically they call for closer dimensional tolerances and smoother finished surfaces than for ordinary structural steel. 10.1 – General Requirements When members are specifically designated in the Design Drawings as “Architecturally Exposed Structural Steel”, the requirements in Sections 1 through 9 shall apply as modified in Section 10. 10.2 – Fabrication Permissible tolerances shall conform to ASTM A6. •

All copes, miters and cuts in surfaces exposed to view shall be made with uniform gaps of 1/8”.

•

All welds exposed to view shall not project more than 1/16” above the surface.

•

Seams of hollow structural sections (HSS) shall be oriented away from view.

10.3 – Delivery of Materials Fabricator shall take extra care and precautions to avoid bending, twisting or otherwise damaging the structural steel. 10.4 – Erection Erector shall take extra care and precautions to minimize damage during handling and erection procedures. Unless otherwise noted, AESS members shall be plumbed, leveled and aligned to a tolerance that is ½ that of non-AESS members.


Lecture 17 – Structural Steel Specifications Project-specific construction documents generally consist of two items: • •

Design Drawings Specifications

The Design Drawings graphically present the specific design of the structure. However, they do not indicate the specific requirements relating to: • • • • •

Materials Submittals Job conditions Testing & inspection Execution of work

CSI – Construction Specifications Institute The CSI was founded in 1948 in an effort to organize trade-specific specifications into a uniform, industry accepted format. It developed the “MasterFormat”, a breakdown of all construction-related activities into 16 divisions as follows: Division 1 – General Requirements Division 2 – Site Construction Division 3 – Concrete Division 4 – Masonry Division 5 – Metals Division 6 – Wood and Plastics Division 7 – Thermal and Moisture Protection Division 8 – Doors and Windows Division 9 – Finishes Division 10 – Specialties Division 11 – Equipment Division 12 – Furnishings Division 13 – Special Construction Division 14 – Conveying Systems Division 15 – Mechanical Division 16 – Electrical

05050 – Basic Metal Materials 05100 – Structural Steel 05200 – Metal Joists 05300 – Metal Deck 05400 – Cold-Formed Metal Framing 05500 – Metal Fabrications 05600 – Hydraulic Fabrications 05650 – Railroad Track & Accessories 05700 – Ornamental Metal 05800 – Expansion Control 05900 – Metal Restoration & Cleaning

Each division has been further refined into multiple sub-divisions (as shown for Division 5 above). To obtain samples of specifications, go to http://www.ogs.state.ny.us/dnc/generalInfo/masterspecdefault.htm In addition to technical specifications, the CSI MasterFormat is used by most of the construction industry for purposes of cost estimating, contractor qualifications, product research and supply ordering.


Section 05100 – Structural Steel PART 1 - GENERAL 1.1

WORK INCLUDED A. Labor, materials, equipment, services and transportation required to complete structural steel work on the Drawings, as specified herein or both. Structural steel work is that work defined in AISC “Code of Standard Practice for Steel Buildings and Bridges”, dated March 7, 2005, plus work listed below and shown on structural drawings. 1. 2.

3. 4. 5. 6.

7.

1.2

Structural steel beams, columns, girders, trusses, and other main structural components and systems. Furnishing and installation of bracing (temporary and permanent), struts, brackets, stiffeners, anchors, support angles for metal deck, hangers, shear studs, and all other miscellaneous steel support members necessary to complete this Section. Design, fabrication and installation of bolted and welded connections and splices. Furnishing and installation of column base plates and bearing plates. Furnishing and installation of anchor rods and loose leveling plates. Furnishing and installation of openings (unreinforced and reinforced) in structural steel required to accommodate mechanical, plumbing, and electrical work. Furnishing and application of shop primer, paint, including finish coat(s) when required, and field touch-up paint for designated structural steel items.

QUALITY ASSURANCE A. Comply with latest editions of: 1. American Institute of Steel Construction (AISC) Publications: a. Manual of Steel Construction: Includes "Specification for Structural Steel Buildings – Load and Resistance Factor Design (LRFD)", "Code of Standard Practice for Steel Buildings and Bridges", "Specification for Structural Joints Using ASTM A325 or A490 Bolts". b. “Building Code of New York State” by New York State Department of State Division of Code Enforcement and Administration. 2. American Welding Society, Inc. (AWS): AWS D1.1 "Structural Welding Code - Steel". 3. American Hot Dip Galvanizers Association, Inc.; Zinc Institute Inc.: "Inspection Manual for Hot Dip Galvanized Products". 4. Society for Protective Coatings (SSPC): "Surface Preparation Specifications". Lecture 17 - Page 2 of 8

5.

Exposed Structural Steel: All exposed structural steel is classified as Architecturally Exposed Structural Steel (AESS) as defined by AISC. Comply with AESS quality requirements for all exposed structural steel.

B. Qualifications for Welding Work 1. Qualify welding processes and welding operators in accordance with AWS Standards. 2. Provide certification that welders to be employed in the Work have satisfactorily passed AWS qualification tests to perform the type of welding within previous 12 months. C. Qualifications for Fabricator and Erector 1. Fabricator and erector of structural steel shall have not less than 3 years experience in fabrication and erection of structural steel. 2. Submit written description of ability. 1.3

TESTING SERVICES A. The Contractor shall employ a testing laboratory acceptable to Architect to perform the following tests: 1. Visual inspection of all welds according to AWS. 2. Magnetic particle inspection according to ASTM E709 for 10 percent of all shop and field welds. 3. Ultrasonic inspection according to ASTM E587 for all shop or field full penetration welds. 4. Inspection of field-assembled high-strength bolted connections. 5. Inspection of erected columns for plumbness within tolerances specified. 6. Inspection of headed studs. 7. Visual inspection of all erected steel for damage. B. Weld Inspector shall be certified in accordance with AWS. 1. Submit resumes of technicians who will perform work showing evidence of one year minimum experience on similar work.

1.4

SUBMITTALS A. General: Review of submittals will be for general consideration only. Compliance with requirements for materials, fabrication, erection and dimensioning of structural steel shall be Contractor's responsibility. B. Connections: Submit proposed connection types for review before preparing detailed shop drawings. C. Shop Drawings - Submit detailed drawings showing: (NOTE: Design drawings shall NOT be used as shop drawings) 1. Column layout plans. 2. Floor and roof framing plans. Lecture 17 - Page 3 of 8

3. 4. 5. 6. 7.

Shop erection details including cuts, copes, connections, holes, bolts and other pertinent information. Welds with size, length and type. Anchor bolt locations. Location of shop welded masonry anchors. Coordinate with Division 4. Shop finishing information.

D. Material Data: Submit laboratory test reports and other data as required to show compliance with Specifications. Submit producer's or manufacturer's specifications and installation instructions for the following products. 1. Structural steel, including certified copies of mill reports covering chemical and physical properties. 2. High-strength bolts including nuts and washers. 3. Unfinished bolts and nuts. 4. Structural steel primer paint. 5. Welding electrodes. 1.5

JOB CONDITIONS A. B. C. D.

Store material in horizontal position on supports above ground. Protect from elements and keep free of dirt and debris. Handle material carefully so as not to bend or mar. Repair or replace damaged materials.

PART 2 - PRODUCTS 2.1

MATERIALS A. Rolled Steel Plates, Angles, Channels, M shapes, HP shapes and Bars: ASTM A36. B. W Shapes: ASTM A992. C. HSS Steel Rectangular, Square and Round: ASTM A500, Grade B. D. Steel Pipe: ASTM A53 Grade B. E. Unfinished Bolts, Nuts and Washers: ASTM A307, Grade A. F. High-strength Bolts, Nuts and Washers: ASTM A325 or A490. G. Direct Tension Indicating Washers: ASTM F959-85. H. Headed Studs: ASTM A108, Grades 1015 – 1020, minimum field = 50 KSI. I. Anchor Rods, Nuts and Washers: ASTM F1554. J. Non-Shrink Bedding Mortar for Bearing and Base Plates: CRD-C 621, Type D “Masterflow 713” from Master Builders (or equivalent). K. Neoprene Bearing Pads: ASTM D412; 70 Durometer Hardness, 2500 PSI Tensile. L. Weld Electrodes: E70XX and in accordance with AWS. M. Expansion Bolts: ¾” Diameter stainless steel with ultimate capacities in 4000 PSI concrete of 16,000 lbs. in shear and 16,000 lbs. in tension; minimum embedment of 6” “Kwik Bolt II” from Hilti Corp. (or equivalent). Lecture 17 - Page 4 of 8

N. Steel Primer Paint: Fabricator's standard rust-inhibitive primer. or None. Bare steel only except where exposed items to be primed are identified on Drawings. or Series 10-1009 grey primer by Tnemec or accepted equal. O. Hot Dipped Galvanizing: Hot-dip galvanize after fabrication in accordance with ASTM A123. Restraighten members after galvanizing, if necessary, to be square and true. P. Weld-on Masonry Anchors: No. 317 continuous weld-on anchor rod by Heckmann Building Products for columns and No. 315 anchor rod for beams, plain steel or accepted equal. Q. Below Grade Coating: #46H-413 coal tar epoxy by Tnemec or accepted equal. R. Cold Galvanizing: Galvilite Cold Galvanizing Compound by Z.R.C. Products Company or accepted equal. 2.2

FABRICATION A. Fabricate structural steel in strict accordance with reviewed shop drawings and referenced standards. B. Fabricate and assemble structural material in shop to greatest extent possible. C. Provide camber as indicated on Drawings. Where no camber is indicated, fabricate steel with mill camber up. D. Provide holes for securing other work to structural steel framing. Cut, drill or punch holes perpendicular to metal surfaces. Do not flame cut holes or enlarge holes by burning. Drill holes in base and bearing plates. E. Finish and weld column bases to column base plates. F. Anchor Rods: Furnish anchor rods, leveling plate and/or other devices necessary for setting anchoring rods required for securing structural steel to foundation, concrete or masonry. G. Hot dip galvanize all lintels in exterior masonry work or as noted on drawings.

2.3

SHOP PAINTING A. Shop paint only structural steel work which will be exposed to view and finish painted. Do not paint steel embedded in concrete or mortar or receive a spray on fireproofing. Do not paint surfaces which are to be Lecture 17 - Page 5 of 8

welded, including metal deck. Do not paint contact surfaces of highstrength bolted connections or finished bearing surfaces such as bearing plates and column base plates. B. For steel to be shop primed and not exposed to view, remove loose rust and mill scale by mechanical means in accordance with SSPC-SP3 "Power Tool Cleaning". For steel to be galvanized or primed and finish painted, remove all dirt, grease, rust and loose mill scale in accordance with SSPC-SP6 “Commercial Blast Cleaning”, unless recommended otherwise by paint manufacturer. C. Immediately after surface preparation, apply structural steel primer paint in accordance with manufacturer's instructions but not less than a uniform dry-film thickness of 2 mils. Use painting methods which will result in full coverage of joints, corners, edges and exposed surfaces. D. Apply below grade coating to column bases and columns to be placed below top of finished floor. E. Apply two coats of cold galvanizing compound to achieve a minimum dry-film thickness of 3 mils. in accordance with manufacturer’s recommendations. 2.4

CONNECTIONS A. Weld or bolt shop connections. B. Bolt field connections as shown on drawings. C. No one-sided or other eccentric connections will be permitted, unless shown on Drawings. D. Minimum Capacity of Beam Connections: For connections not detailed, provide connection capacity of the nominal full section shear capacity Vn for the given steel member as dictated in AISC Steel Construction Manual. A minimum factored shear capacity of 10 kips shall be provided for all secondary beams. For beam and girders with shear studs, provide a connection capacity of at least 125 percent of uniform load values unless indicated otherwise on drawings. E. Provide snug-tight unfinished threaded fasteners for bolted bearing connections of secondary framing members to primary members; including, but not limited to, girts, door framing systems, roof opening and other framing systems taking only nominal stresses and in no way reacting in stress on primary members. F. Provide high-strength fasteners for all principal bolted connections, unless otherwise indicated. Lecture 17 - Page 6 of 8

G. Provide bearing bolt (X) fastener for all structural connections. H. Use only connections which are published by AISC. Do not modify published connection details unless accepted by Engineer. I.

Use AISC “Single-Plate Shear Connections” for beam connections to face of tubes and column flanges which have a width of 6 inches or less.

J.

Use AISC “Framed Beam Connections” for beam connections to face of tubes and column flanges which have a width greater than 6 inches, and for beam-to-beam connections.

PART 3 - EXECUTION 3.1

INSPECTION A. Examine conditions under which work shall be erected. Do not proceed until all unsatisfactory conditions are corrected.

3.2

ERECTION A. Set structural frames accurately to lines and elevations indicated. Align and adjust various members forming part of a complete frame or structure before permanently fastening. B. Clean bearing surfaces and other surfaces before assembly that will be in permanent contact after assembly. C. Perform necessary adjustments to compensate for discrepancies in elevations and alignment. Level and plumb individual members of structure within specified tolerances. D. Splice members only where shown or specified. E. Maintain work in a stable condition during erection. F. The use of gas cutting torches in field to correct fabricating errors is prohibited. G. Tighten bearing bolt (X) connections to snug-tight condition.

3.3

TOLERANCES A. Tolerances shall be within limits in AISC "Code of Standard Practice". B. Fabrication and mill tolerance shall be within limits in AISC Standard Mill Practice. Lecture 17 - Page 7 of 8

3.4

TOUCH-UP PAINTING A. After erection is complete, touch-up paint damaged shop priming coats and welded areas. Remove weld slag before applying touch-up paint. B. Touch-up below grade coatings to all portions of structural steel embedded within concrete slabs on grades.

3.5

TEMPORARY SHORING AND BRACING A. Provide temporary shoring and bracing members as required, with connections of sufficient strength, to bear imposed loads. B. Remove temporary members and connections when permanent members are in place and final connections are made. C. Provide temporary guy lines to achieve proper alignment of structures as erection proceeds.

3.6

PROTECTION A. Do not use members for storage or working platforms until permanently secured. B. Do not exceed load capacity of members with construction loads. END OF SECTION


Lecture 18 – Open Web Steel Joists Open web steel joists, or “Bar Joists” are very efficient structural members commonly used to support roofs, and to a lesser degree, floors.

Roof construction of Clark Field House facility at SUNY Delhi Steel joists are NOT considered structural steel. As such, they are manufactured as proprietary structural members by various manufacturers. The Steel Joist Institute, SJI, is an organization founded in 1928 that was established to set standards for manufacture, design and construction of joists. It recognizes manufacturers who comply with their standards. Some of the larger SJI recognized manufacturers include Vulcraft, Canam Steel Corp. and SMI Joist Company.


K-Series Joists The most commonly-used joist style is the so-called “K” series. It has a depth ranging from 8” up to 30” and is used economically to span up to 60’-0”. A typical K series joist is as shown below:

A typical designation is 18K3 Actual depth in inches

Section Number = Relative size of members K series


Steel joists are fastened to its supporting members usually by fieldwelding as shown below:

Unlike structural steel beams, steel joists must use bridging placed perpendicular to the span to obtain its stability. This bridging can be one of 2 types: • •

Horizontal Bridging Diagonal Bridging

Bridging requirements are shown in the Vulcraft Joist Catalog p. 9 and 35 and is a function of the Section Number and span. Joists using horizontal bridging is shown below:


LH and DLH Series Joists The LH series joists have depths ranging between 18” and 48” and are suitable for spans up to 96’-0”. The DLH series joists have depths ranging between 52” and 72” and are suitable for spans up to 144’-0”. They are not as commonly used as K series joists, but provide an inexpensive alternative to spanning longer distances than the K series joists. One difference between K series joists is the required end bearing width and height are 6” and 5” respectively for the LH and DLH (vs. 4” and 2½” for the K series).

A typical designation is 32LH10 Actual depth in inches

Section Number = Relative size of members LH series


Joist Girders Joist girders are designed to carry the end reactions from equally-spaced joists applied to the panel points. Typical depths of joist girders range from 20” up to 96” with spans of 100’-0” or more.

A typical joist girder connection to steel column is shown below:


Example 1 GIVEN: A roof framing bay is as shown below. The service loads are as follows: • • • •

Service Dead Load = 16 PSF Service Roof Live Load = 25 PSF Service Snow Load = 35 PSF Service Wind Uplift = -12 PSF

REQUIRED: Design the K series joists assuming the maximum joist spacing = 6’-0” (based on metal roof deck). Assume the joist + accessories weighs 10 PLF.

36’-0”

60’-0”

Step 1 – Determine joist orientation and spacing: It is best to orient the joists in the short direction for strength. For economy, use a joist spacing = 6’-0” giving 10 even spaces.


Step 2 – Determine maximum uniformly distributed total service load: Utilizing the 6 allowable stress design load combinations from the IBC Section 1605.3.1: 1) 2) 3) 4) 5) 6)

D D+L D + L + (Lr or S or R) D + (W or 0.7E) + L + (Lr or S or R) 0.6D + W 0.6D + 0.7E where: D = Dead Load = 6’(16 PSF) + 10 PLF = 106 PLF Lr = Roof Live Load = 6’(25 PSF) = 150 PLF S = Snow Load = 6’(35 PSF) = 210 PLF W = Wind Load = 6’(-12 PSF) = -72 PLF

Check all 6 load combinations and select “worst” case total load: 1) 2) 3) 4) 5) 6)

D = 106 PLF D + L = 106 PLF D + L + (Lr or S or R) = 106 + 150 = 256 PLF D + (W or 0.7E) + L + (Lr or S or R) = 106 + 210 = 316 PLF

0.6D + W = 0.6(106) + (-72) = -8.4 PLF 0.6D + 0.7E = 0.6(106) = 64 PLF Use


Step 3 – Select lightest joist from Vulcraft K series Load Table p. 12: Total Load = 316 PLF Live Load = Total Load – Dead Load = 316 PLF – 106 PLF = 210 PLF Span = 36’-0”

Possibilities: Joist Size: 24K8 26K7 28K6 30K7

Total Load: 346 340 330 395

Live Load: 222 240 252 323

Wt/ft: 11.5 10.9 11.4 12.3

Select 26K7 joist as the lightest from the list above Step 4 – Select lightest joist from “Economical Joist Guide” p. 109: For length = 36’, start at the top of the list and read down until Total load > 316 PLF and Live load > 210 PLF Select 28K6 → Total load = 330, Live load = 252

Use 28K6 Joist for Final Design Step 5 – Determine Bridging requirements, assume horizontal bridging: From Vulcraft p. 9 → Section Number = 6 Joist spacing = 6’-0”

From Vulcraft p. 35 → Section Number = 6 Span = Over 29’ thru 39’


Use 1¼ x 7/64 equal leg angle bridging, good for up to joist spacing = 6’-3”

Use 3 Rows of Bridging

Step 6 – Draw Summary Sketch of Roof Framing Plan:

28K6 @ 6’-0”

3 rows of L1¼x7/64 horizontal bridging equally spaced

36’-0”

10 spaces @ 6’-0” = 60’-0”


Example 2 GIVEN: The roof framing bay from Example 1. REQUIRED: Design the lightest weight 60’-0” span joist girder.

28K6 @ 6’-0”

Joist Girder

36’-0”

10 spaces @ 6’-0” = 60’-0”

Step 1 – Determine 28K6 joist end reactions: Joist end reaction =

=

wL 2

(316 PLF )(36'−0" ) 2

= 5688 LBS. = 5.7 KIPS → USE 6 KIPS Step 2 – Select joist girder depth from Vulcraft p. 87: Girder span = 60’-0” Joist Spaces = 10N @ 6.00’ Load on Each Panel Point = 6 KIPS

Select 72” girder depth → wt. = 35 PLF

Use 72G 10N 6.0K Joist Girder


Lecture 19 – Steel Deck Steel deck, or sometimes called “metal deck” is used in steel framed construction as an intermediate structural system to distribute floor and roof loads to supporting beams. Decking is typically fastened to the steel supporting members by either puddle welds or powder-actuated fasteners. Although made of steel, it is NOT considered to be structural steel. Decking is corrugated having a typical cross-section resembling:

Top flute

Deck height

Bottom flute Panel width = 24” → 36”

The Steel Deck Institute, SDI, was established in 1939 in an effort to regulate the design, manufacture and installation of steel deck. Manufacturers complying with SDI specifications include Vulcraft, Canam Steel Corp. and United Steel Deck, Inc. Types of Steel Deck There are 3 general types of steel deck → roof deck, non-composite floor deck and composite deck. 1. Roof Deck Roof deck is used primarily to carry lightweight roof construction. It is characterized by having relatively narrow bottom flutes so that there is a wider top flute to maximize the surface contact with rigid insulation. It comes in heights ranging from 1” up to 3” and in thicknesses ranging from 24 gage (thinnest) up to 16 gage (thickest). Depending on the section, roof decking can span as much as 15’-0”. Acoustical deck is available to control sound transmission through the decking. It is used for auditoriums, schools, etc., and is obtained by adding fiber sound-absorbing batts between the vertical webs of the decking. In addition, roof deck is available as “cellular” deck for use in placing electrical services or exposed underside. Data relating to roof deck may be found in the Vulcraft catalog p. 3 – 18.


Built-up roof membrane Rigid Insulation Roof Deck screwed or puddle-welded to top chord of steel joist

Roof Deck Fastened to Steel Bar Joist


2. Non-Composite Floor Deck This type of deck essentially acts as a form to carry the concrete slab. It offers no additional strength to the structural steel beam as composite construction would. It ranges in height from 5/8” up to 3” and thicknesses of 26 gage up to 16 gage with spans up to 15’-0”. It is also available as acoustical deck or as “cellular” deck. Data relating to roof deck may be found in the Vulcraft catalog p. 19 - 40.

Welded wire mesh in concrete slab


3. Composite Floor Deck Similar to non-composite deck, except composite deck is used for composite steel construction. Typically, the decking has built-in perforations that aids in the bonding to concrete.

Composite Floor Deck with headed shear studs welded to beams


Roof Deck Example GIVEN: A 1½” Type “F” (intermediate rib) roof deck is to be used in a 3-span condition with a 7’-0” span. The SERVIVE roof loads are as follows: • • • •

SERVIVE roof Dead Load = 15 PSF SERVICE roof Live Load = 20 PSF SERVICE roof Snow Load = 40 PSF SERVICE roof Wind Load = -8 PSF (uplift)

REQUIRED: Design the lightest-weight 1½” Type “F” roof deck using the Vulcraft catalog. Steel roof deck Steel support beams

7’-0”

7’-0”

7’-0”

3 spans (min.) Step 1 – Determine maximum unif. load on deck: Utilizing the 6 allowable stress design load combinations from the IBC Section 1605.3.1: 1) 2) 3) 4) 5) 6)

D D+L D + L + (Lr or S or R) D + (W or 0.7E) + L + (Lr or S or R) 0.6D + W 0.6D + 0.7E where: D = Dead Load = 15 PSF Lr = Roof Live Load = 20 PSF S = Snow Load = 40 PSF W = Wind Load = -8 PSF


Check all 6 load combinations and select “worst” case total load: 1) 2) 3) 4) 5) 6)

D = 15 PSF D + L = 15 PSF D + L + (Lr or S or R) = 15 + 20 = 35 PSF D + (W or 0.7E) + L + (Lr or S or R) = 15 + 40 = 55 PSF

0.6D + W = 0.6(15) + (-8) = 1 PSF 0.6D + 0.7E = 0.6(15) = 9 PSF Use

Step 2 – Refer to the Vulcraft Catalog page 4 for 1½” Type F deck:

7’-0” span

3 span

From Table above, use Vulcraft 1½” Type F 19 Gage Roof Deck → Allow. Load = 59 PSF > 55 PSF


Non-Composite Floor Deck Example GIVEN: A floor framing plan for an office building is as shown below. The slab is 5” normal-weight concrete over “2.0 C Conform” non-composite 2” deck as manufactured by Vulcraft. The superimposed SERVICE live load = 50 PSF and a total superimposed SERVICE dead load (excluding slab weight) = 38 PSF. REQUIRED: Design the lightest weight 2.0 C Conform non-composite deck assuming 3-span condition.

4 @ 6’-0” = 24’-0” Step 1 – Determine the uniform load on the decking: Utilizing the 6 allowable stress design load combinations from the IBC Section 1605.3.1: 1) 2) 3) 4) 5) 6)

See Vulcraft catalog p. 28 for slab wt.

D D+L D + L + (Lr or S or R) D + (W or 0.7E) + L + (Lr or S or R) 0.6D + W 0.6D + 0.7E where: D = Dead Load = Slab wt. + Superimposed Dead Load = 51 PSF + 38 PSF = 89 PSF L= FloorLive Load = 50 PSF

Using Load Combination 2 from above: Total Uniform Load = D + L = 89 PSF + 50 PSF = 139 PSF Lecture 19 - Page 7 of 8

Step 2 – Refer to “Allowable Uniform Load” table from Vulcraft p. 29: No. of Spans = 3 Clear Span = 6’-0” Total Uniform Load = 139 PSF

Use 2C20 → Allowable unif. load = 173 PSF > 139 PSF

Step 3 – Refer to “Reinf. Conc. Slab Allow. Loads” table Vulcraft p. 28: Total Slab Depth = 5” Clear Span = 6’-0” Superimposed Unif. Load = Total Load – Slab Wt. = 139 PSF – 51 PSF = 88 PSF

Use 6x6-W2.1xW2.1 W.W.F. → Allow. load = 107 PSF > 88 PSF

5” conc. slab over 2” - 20 Gage non-composite metal deck reinf. with 6x6-W2.1xW2.1 W.W.F.

5”

Steel support beam


Lecture 2 – Steel properties and ASD & LRFD principles General Steel Properties: Structural steel used in buildings uses carbon steel. It is a mixture of mostly iron (98%+) and carbon (0.15% - 0.59%), as well as a percentage of other alloys used to enhance certain properties. Some of the alloys used include silicon, manganese, nickel, sulfur and phosphorus. Different steels exhibit stress-strain relationships as shown below:

As designers of structural steel, we are most interested in the following 3 properties: a) Modulus of elasticity = 29,000 KSI b) Yield stress = Fy c) Ultimate stress = Fu


Typical structural shapes: (ref. AISC p. 1-3 thru 1-8) 1) 2) 3) 4) 5) 6)

I shapes are categorized into 4 groups – “W”, “M”, “S”, and “HP” Channels – “C” and “MC” Angles – “L” Structural Tees – cut from I shapes, “WT”, “MT” and “ST” Hollow Structural Section – “HSS” rectangular and round Steel pipe

Refer to AISC p. 1-10 thru 1-115 for dimensions and structural properties of the above-noted structural shapes and combination of shapes. Refer to AISC Table 2-3 (page 2-39) for a list of appropriate structural steel ASTM designations for various structural shapes. In general, the following ASTM designations are commonly used: ASTM Designation A36 A572 A992 A53 A500, Gr. B A500, Gr. B A588

Fy (KSI)

Fu (KSI)

36 50 50 35 42 46 50

58 65 65 60 58 58 70

Applicable structural shapes M, S, C, MC, L, plates HP W Pipe HSS round HSS rectangular Corrosion-resistant for all rolled shapes

Advantages of Steel-Framed Structures: 1) 2) 3) 4) 5) 6)

High-strength Excellent quality control, predictability Ductility Speed of erection Lightweight Can be easily modified

Disadvantages of Steel-Framed Structures: 1) 2) 3) 4) 5) 6)

Fireproofing Corrosion Need for bracing Semi-skilled labor (ironworkers, welders) Subject to vibration Temperature effects - brittle below -600F - rapid reduction of “E” above 7000F


Load and Resistance Factor Design (LRFD) see AISC p. 2-6 and 16.1213 thru 217

The LRFD was developed in the 1980s as an alternative design method to the tried-and-true Allowable Stress Design (ASD) method. It is based on a “limit state” philosophy. A limit state is a term used to describe a condition in which the structure ceases to perform as intended. A “strength limit state” defines the safety against failure due to loading and a “serviceability limit state” is a functional requirement such as deflection, drift or vibration. In general, the LRFD method uses a statistical approach in determining factored loads that are compared against ultimate member strengths. In other words:

ΣλiQi < φRn where: λI =load factor Qi = working or service load (see IBC ch. 16) φ = reduction factor, see AISC p. 2-10 = 0.90 for limit-states involving yielding = 0.75 for limit-states involving rupture Rn = nominal resistance strength of member If plotted on a probability graph, 2 bell curves would emerge, one being the probability of loads and the other being the probability of a member’s strength being realized.

Probability of loads, ΣQi

Probability of member strength, Rn

Overlap = failure


LRFD Load Factors: see AISC p. 2-8 The following 6 load factors are used to obtain the most severe “factored” loads: 1) 2) 3) 4) 5) 6)

1.4D 1.2D + 1.6L + 0.5(Lr or S or R) 1.2D + 1.6(Lr or S or R) + (0.5L or 0.8W) 1.2D + 1.6W + 0.5L + 0.5(Lr or S or R) 1.2D + 1.0E + 0.5L + 0.2S 0.9D + (1.6W or 1.0E)

where: D = service dead loads L = service floor live load Lr = service roof live load S = snow load R = rainwater load W = wind load E = earthquake load Example 1 GIVEN: A flat roof is framed with 24’-0” long W18x40 beams spaced 8’-0” o.c. The service applied roof dead load is 25 PSF and the applied service roof live load = 20 PSF (per IBC ch. 16). The service wind load on the flat roof is -8 PSF (uplift).

W18x40

W18x40

W18x40

wu

24’-0” 24’-0”

Beam Loading F.B.D Roof Framing Plan

8’-0” Typ.

REQUIRED: 1) Determine the maximum LRFD factored uniform load on the beam, wu.

2) Determine the maximum LRFD factored moment on the beam, Mu.


Step 1 – Determine D, Lr and W in terms of PLF: D = DL(Trib. Width) + Beam wt. = 25 PSF(8 ft) + 40 PLF = 240 PLF Lr = LL(Trib. Width) = 20 PSF(8 ft) = 160 PLF W = -8 PSF(8 ft) = -64 PLF Step 2 – Determine maximum FACTORED uniform load, wu: 1) 1.4D 1.4(240 PLF) = 336 PLF 2) 1.2D + 1.6L + 0.5(Lr or S or R) 1.2(240 PLF) + 0.5(160 PLF) = 368 PLF 3) 1.2D + 1.6(Lr or S or R) + (0.5L or 0.8W) 1.2(240 PLF) + 1.6(160 PLF) = 544 PLF ← USE 4) 1.2D + 1.6W + 0.5L + 0.5(Lr or S or R) 1.2(240 PLF) + 1.6(-64 PLF) + 0.5(160 PLF) = 266 PLF 5) 1.2D + 1.0E + 0.5L + 0.2S 1.2(240 PLF) = 288 PLF 6) 0.9D + (1.6W or 1.0E) 0.9(240 PLF) + 1.6(-64) = 114 PLF OR 318 PLF Step 3 – Determine maximum FACTORED moment on beam, Mu: wu L2 Mu = 8 =

(544 PLF )(24'−0" ) 2 8

= 39,168 ft-lb Mu = 39.2 kip-ft


Allowable Stress Design (ASD) See AISC p. 2-7 and 16.1-216 thru 217 The Allowable Stress Design (ASD) method is based on the concept that the stress levels in a component do not exceed established specific allowable stresses under service loads. For any single component, there may be several different allowable stress limits that must be checked. The basic design equation for ASD is as follows:

ΣQi ≤

Rn Ω

where: λI =load factor Qi = working or service load (see IBC ch. 16) Rn = nominal resistance strength of member Ω = safety factor, see AISC p. 2-10 = 1.67 for limit-states involving yielding = 2.00 for limit-states involving rupture 1.5 =

φ

ASD Load Factors: see AISC p. 2-9 The following 7 load factors are used to obtain the most severe loads: 1) 2) 3) 4) 5) 6) 7)

D D+L D+(Lr or S or R) D+0.75L+0.75(Lr or S or R) D+(W or 0.7E) D+0.75(W or 0.7E)+0.75L+0.75(Lr or S or R) 0.6D+(W or 0.7E)

where: D = service dead loads L = service floor live load Lr = service roof live load S = snow load R = rainwater load W = wind load E = earthquake load


W18x40

W18x40

W18x40

Example 2 GIVEN: Similar to Example 1, a flat roof is framed with 24’-0” long W18x40 beams spaced 8’-0” o.c. The service applied roof dead load is 25 PSF and the applied service roof live load = 20 PSF (per IBC ch. 16). The service wind load on the flat roof is -8 PSF (uplift).

8’-0”

24’-0”

Roof Framing Plan

Typ. REQUIRED: 1) Determine the maximum ASD service uniform load on the beam, w. 2) Determine the maximum ASD service moment on the beam, Mmax. Step 1 – Determine D, Lr and W in terms of PLF: D = DL(Trib. Width) + Beam wt. = 25 PSF(8 ft) + 40 PLF = 240 PLF Lr = LL(Trib. Width) = 20 PSF(8 ft) = 160 PLF W = -8 PSF(8 ft) = -64 PLF Step 2 – Determine maximum SERVICE uniform load, w: 1) D 240 PLF 2) D+L 240 PLF + 0 = 240 PLF 3) D+(Lr or S or R) 240 PLF + 160 PLF = 400 PLF ← USE


4) D+0.75L+0.75(Lr or S or R) 240 PLF + 0 + 0.75(160 PLF) = 360 PLF 5) D+(W or 0.7E) 240 PLF + (-64 PLF) = 176 PLF 240 PLF – (-64 PLF) = 304 PLF 6) D+0.75(W or 0.7E)+0.75L+0.75(Lr or S or R) 240 PLF + 0.75(-64 PLF) + 0 + 0.75(160 PLF) = 312 PLF 7) 0.6D+(W or 0.7E) 0.6(240 PLF) + (-64 PLF) = 80 PLF 0.6(240 PLF) – (-64 PLF) = 208 PLF Step 3 – Determine maximum SERVICE moment on beam, Mmax: Mmax =

wL2 8

(400 PLF )(24'−0" ) 2 = 8 = 28,800 ft-lb Mmax = 28.8 kip-ft


Lecture 21 – Reinforced Concrete Properties Reinforced concrete structures are typified by their strength, beauty, bulk and longevity. It is the material of choice for many structures where these characteristics are required. Concrete-framed structures have many desirable advantages over other construction materials including: • • • • • • • •

Concrete can be “molded” to form almost any imaginable shape The entire building can be made of concrete – walls, floors, structure Concrete frames are inherently stable (vs. steel & wood) Concrete structures are heavy – excellent for wind-prone areas Concrete is a readily-available material Concrete is very fire-resistant Weather-resistant (if built properly) Relatively inexpensive material

However, reinforced concrete structures have several shortcomings which may preclude it as a building material, including: • • • • • •

Very labor-intensive Quality control Formwork Longer construction schedule due to curing time Much larger, heavier member sizes (vs. steel-framed) Poor insulation values


Concrete Materials: Concrete is a mixture of the following materials: 1. Portland Cement – The active ingredient that “glues” the other materials together, conforming to ASTM C 150-99a. The raw materials used in portland cement consist mainly of limestone, and clays & shales. Different types of Portland cement include: a) Type I – General purpose b) Type II – Moderate sulfate protection and lower heat of hydration c) Type III – High-early strength d) Type IV – Low heat of hydration used for massive concrete structures such as dams e) Type V – High sulfate resistance 2. Water – Water is necessary to create the chemical reaction of hardening the cement called “hydration.” It should be clean and free from any impurities (i.e., potable). 3. Aggregates – Fine (sand) and coarse (gravel). Conforming to ASTM C 33. 4. Admixtures – Other ingredients added to enhance properties: a) Air Entrainment – Tiny bubbles used to reduce cracking in concrete subject to freeze-thaw cycles. Conforming to ASTM C 260 with an air content of 4% - 8% by volume. b) Superplasticizers – Also called “High Range Water Reducers”, used to increase concrete’s flow (workability) instead of adding water. Conforming to ASTM C 494 Type F. c) Retarders – Used to slow the hydration process. Conforming to ASTM C494 Type D. d) Accelerators – Used to speed-up the curing process, conforming to ASTM C494 Type C or E. e) Insulating beads – Increases the “R” value, but diminishes strength. f) Fly Ash – The byproduct of coal-burning electric generating plants. Used to decrease the amount of portland cement required. Conforming to ASTM C 618 Class F. g) Colors – Can be mixed to produce any desirable color.


Reinforced Concrete Properties: 1) Compressive Strength The specified concrete compressive strength, f’c, is actually a stress. It is the most important structural property of concrete and is VERY DEPENDENT upon the water-to-cement ratio. This is the ratio of the weight of water divided by the weight of cement. A low w/c ratio = high f’c and high w/c ratio = low f’c. A low w/c ratio is very stiff and difficult to work with, therefore necessitating the need for superplasticizers. Normal concrete has w/c ratios ranging from about 0.23 (very strong) up to a maximum of about 0.50 but preferably should not exceed 0.45. Values of f’c are based on 28 days of curing. Typical ranges of f’c are: f’c = 3000 PSI (slab-on-grade, footings, foundation walls) = 3500 – 5000 PSI (beams, framed slabs) = 4000 – 14000 PSI (columns) The condition in which concrete cures affects the ultimate strength of the hardened concrete’s f’c. Allowing the freshly-placed concrete to have continuous moisture applied will significantly increase the strength, f’c. Conversely, subjecting the freshly-placed concrete to constant air will decrease the f’c. See the graph below:

Affect of moist curing on concrete strength


2) Tensile Strength Concrete is a brittle material and has very small tensile strength (about 10% of f’c). It is usually assumed that concrete has zero tensile strength. 3) Modulus of Elasticity – Determined by formula below: Econc = 57000

f 'c

where f’c = concrete specified compressive stress in PSI Example: GIVEN: Concrete with f’c = 4000 PSI. REQUIRED: Determine Econc Econc = 57000

f 'c

= 57000 4000 PSI = 3,605,000 PSI Econc = 3605 KSI


4) Reinforcing Bars – Used to carry ALL of the tension in a concrete member, as well as helping to carry shear and compression. The steel uses for bars is typically new “billet” steel having the usual modulus of elasticity “E” = 29,000 KSI. The size of a bar refers to its diameter in 1/8ths. For example a #5 bar is ⅝” in diameter (see table below). Rebar should conform to ASTM A615 for deformed (ribbed) bars. Typical grades include: a) Grade 60 – Has a yield stress Fy = 60 KSI, used for all bars b) Grade 40 – Has a yield stress Fy = 40 KSI, used for low-strength applications only Bar Size: #3 #4 #5 #6 #7 #8 #9 #10 #11

Diameter: ⅜” ½” ⅝” ¾” ⅞” 1” 1⅛” 1¼” 1⅜”

Area (in2): 0.11 0.20 0.31 0.44 0.60 0.79 1.00 1.27 1.56

The following diagram shows the typical markings on a deformed reinforcing bar:


Epoxy-coated reinforcing bars are regular bars with a shopapplied coating of epoxy. These bars have exceptional resistance to corrosion and are used in situations where there is high water/salt exposure (such as road bridge decks, marine structures, etc.). They are smooth to the touch and usually green in color. Codes allow the placement of epoxy-coated bars to be closer to the surface than regular bars because of the increased resistance to corrosion. This usually results in a thinner, lighter concrete beam or slab. Unfortunately, the epoxy coating is often scratched or damaged during construction, leaving exposed bare steel. This exposed bare steel is ripe for allowing moisture and salt deposits to enter and actually INCREASING the propagation of corrosion. For this reason, many state Departments of Transportation do not allow epoxy-coated reinforcing bars for use in bridge decks.

Bridge deck constructed with epoxy-coated reinforcing bars


Reinforcing bars are placed a certain minimum distance away from the edge of the member to ensure that it will not be susceptible to water/salt infusion. This is referred to as cover distance. The cover distance requirements shown below are obtained from ACI 318-02 “Building Code Requirements for Structural Concrete.”

Required minimum cover distance

Concrete member Reinforcing bars

Required minimum cover distance

Minimum Concrete Cover Over Reinforcing Bars Condition: Concrete cast against and permanently exposed to earth Concrete exposed to earth or No. 6 through No. 18 bars weather No. 5 and smaller bars Concrete NOT exposed to Slabs, walls No. 14 & earth or weather & joists No. 18 No. 11 and smaller Beams, Main reinf., columns stirrups, ties, spirals No. 6 and Shells, larger folded plates No. 5 and smaller


Minimum cover: 3” 2” 1½” 1½” ¾” 1½” ¾” ½”

5) Slump – Fresh concrete uses a slump test to determine the workability of the concrete as per ASTM C 143. It is, however, not a very useful measure of the concrete’s strength. It is possible to get very workable concrete with high slump (i.e., very fluid) with the use of superplasticizers. The test involves taking a cone-shaped mold and pouring a sample of concrete into it. Next, the cone is removed upward and the vertical displacement of the concrete is measured.

Technicians performing a slump test on fresh concrete

Recommended Slumps for Various Types of Construction Type of Concrete Member: Foundation walls & footings Beams and walls Columns Pavements and slabs Mass concrete

Slump: Maximum Minimum 3” 1” 4” 1” 4” 1” 3” 1” 2” 1”


Lecture 22 – Introduction to ACI 318-02 The American Concrete Institute (ACI) is the governing agency for all concrete construction in the U.S. It was established in 1904 to serve and represent user interests in the field of concrete. The ACI publishes many different standards, but the most commonly referenced standard used by architects and engineers is the ACI 318 “Building Code Requirements for Structural Concrete.” It is updated every 7 years and the latest version is ACI 318-02 updated in 2002. Almost all Building Codes, including the IBC, refer to ACI 318 as the basis for structural design of concrete members.

Contents: PART 1—GENERAL CHAPTER 1—GENERAL REQUIREMENTS .................................................318-9 CHAPTER 2—DEFINITIONS........................................................................ 318-19 PART 2—STANDARDS FOR TESTS AND MATERIALS CHAPTER 3—MATERIALS.......................................................................... 318-27


PART 3—CONSTRUCTION REQUIREMENTS CHAPTER 4—DURABILITY REQUIREMENTS............................................ 318-41 CHAPTER 5—CONCRETE QUALITY, MIXING, AND PLACING .................318-47 CHAPTER 6—FORMWORK, EMBEDDED PIPES, AND CONSTRUCTION JOINTS .............. 318-63 CHAPTER 7—DETAILS OF REINFORCEMENT........................................ 318-69 PART 4—GENERAL REQUIREMENTS CHAPTER 8—ANALYSIS AND DESIGN - GENERAL CONSIDERATIONS ............................318-85 CHAPTER 9—STRENGTH AND SERVICEABILITY REQUIREMENTS ...................................318-95 CHAPTER 10—FLEXURE AND AXIAL LOADS............................................318-109 CHAPTER 11—SHEAR AND TORSION....................................................... 318-139 CHAPTER 12—DEVELOPMENT AND SPLICES OF REINFORCEMENT ..............................318-187 PART 5—STRUCTURAL SYSTEMS OR ELEMENTS CHAPTER 13—TWO-WAY SLAB SYSTEMS............................................ 318-213 CHAPTER 14—WALLS.............................................................................. 318-233 CHAPTER 15—FOOTINGS........................................................................ 318-241 CHAPTER 16—PRECAST CONCRETE .................................................... 318-249 CHAPTER 17—COMPOSITE CONCRETE FLEXURAL MEMBERS ........ 318-257 CHAPTER 18—PRESTRESSED CONCRETE........................................... 318-261 CHAPTER 19—SHELLS AND FOLDED PLATE MEMBERS...................... 318-289 PART 6—SPECIAL CONSIDERATIONS CHAPTER 20—STRENGTH EVALUATION OF EXISTING STRUCTURES.......................... 318-297 CHAPTER 21—SPECIAL PROVISIONS FOR SEISMIC DESIGN................318-303 PART 7—STRUCTURAL PLAIN CONCRETE CHAPTER 22—STRUCTURAL PLAIN CONCRETE ....................................318-343 COMMENTARY REFERENCES......................................................318-353 APPENDIXES APPENDIX A—STRUT-AND-TIE MODELS ..................................................318-369 APPENDIX B—ALTERNATIVE PROVISIONS FOR REINFORCED AND PRESTRESSED CONCRETE FLEXURAL AND COMPRESSION MEMBERS ..............................................318-385 APPENDIX C—ALTERNATIVE LOAD AND STRENGTH REDUCTION FACTORS................318-393 APPENDIX D—ANCHORING TO CONCRETE.................................................318-399 APPENDIX E—NOTATION................................................................................318-427 APPENDIX F—STEEL REINFORCEMENT INFORMATION ............................318-437 INDEX.................................................................................................................318-439


Analysis and Design – General Considerations Design Basis: Similar to the LRFD method in steel, concrete is designed on the basis of “Ultimate” loading. This is often referred to as “Strength” design. Factors are applied to service loads in accordance with ACI 318 Section 9.2. These factored loads are used to determine maximum factored moments, shears and other effects which are then compared to the strength of the member. Strength of member is reduced by a strength reduction factor. Factored Load Effects < (φ)Member Strength Load Factors: 1) 2) 3) 4) 5) 6) 7)

1.4(D + F) 1.2(D + F + T) + 1.6(L + H) + 0.5(Lr or S or R) 1.2D + 1.6(Lr or S or R) + (1.0L or 0.8W) 1.2D + 1.6W + 1.0L + 0.5(Lr or S or R) 1.2D + 1.0E + 1.0L + 0.2S 0.9D + 1.6W + 1.6H 0.9D + 1.0E + 1.6H where: D = service dead loads L = service live load Lr = service roof live load S = snow loads W = wind loads R = rainwater loads E = earthquake loads F = fluid loads H = soil loads T = Temperature, creep, settlement, shrinkage loads

Strength Reduction Factors, φ Member Type: Tension member Compression member Spiral reinforced Tied reinforced Flexural members (beams) Shear and torsion Bearing


φ 0.90 0.70 0.65 0.85 0.75 0.65

Example 1 GIVEN: The interior column of a 2-story concrete-framed building has the following applied service loads to the 1200 ft2 tributary area as shown: Roof live load = 20 PSF Snow load = 45 PSF Roof superimposed dead load (not including 8” thick slab) = 16 PSF Roof wind uplift = -8 PSF Floor live load = 100 PSF Floor superimposed dead load (not including 10” thick slab) = 42 PSF REQUIRED: Determine the maximum factored load, Pu, at the bottom of the 20” x 20” square column.

Trib. area = 1200 ft2


Step 1 – Determine the total service loads on the roof: a) Service roof live load, Lr = Trib. area(Roof PSF) = 1200 ft2(20 PSF) = 24,000 lbs. = 24.0 KIPS b) Service snow load, S = Trib. area(Floor PSF) = 1200 ft2(45 PSF) = 54,000 lbs. = 54.0 KIPS c) Service wind uplift load, W = 1200 ft2(-8 PSF) = -9,600 lbs. = -9.6 KIPS d) Service roof dead load, Droof = (Superimposed loads) + (slab wt.) = 1200 ft2(16 PSF) + 1200 ft2 ⎛⎜ 8" (150 PCF ) ⎞⎟ ⎝ 12

⎠

= 19,200 lbs. + 120,000 lbs. = 139,200 lbs. = 139.2 KIPS Step 2 – Determine the total service loads on the 2nd floor: a) Service floor live load, L = 1200 ft2(100 PSF) = 120,000 lbs. = 120.0 KIPS b) Service floor dead load, Dfloor = (Superimposed loads) + (slab wt.) = 1200 ft2(42 PSF) + 1200 ft2 ⎛⎜ 10" (150 PCF ) ⎞⎟ ⎝ 12

= 50,400 lbs. + 150,000 lbs. = 200,400 lbs. = 200.4 KIPS Step 3 – Determine the total service dead load of the concrete column:

(

⎛ 20" ⎞⎛ 20" ⎞ 3 Column dead load, Dcolumn = ⎜ ⎟⎜ ⎟(28 ft ) 150lb / ft 12 12 ⎝ ⎠⎝ ⎠ = 11,667 lbs. = 11.7 KIPS


)

⎠

Step 4 – Sum all service dead loads together: Total service dead load, D = Droof + Dfloor + Dcolumn = 139.2 KIPS + 200.4 KIPS + 11.7 KIPS = 351.3 KIPS Step 5 – Check all 7 load factors, select “worst” case: 1) 1.4(D + F) 1.4(351.3) = 491.8 KIPS 2) 1.2(D + F + T) + 1.6(L + H) + 0.5(Lr or S or R) 1.2(351.3) + 1.6(120.0) + 0.5(54.0) = 640.6 KIPS ← USE 3) 1.2D + 1.6(Lr or S or R) + (1.0L or 0.8W) 1.2(351.3) + 1.6(54.0) + (1.0(120.0)) = 628.0 KIPS 4) 1.2D + 1.6W + 1.0L + 0.5(Lr or S or R) 1.2(351.3) + 1.6(-9.6) + 1.0(120.0) + 0.5(54) = 553.2 KIPS 5) 1.2D + 1.0E + 1.0L + 0.2S 1.2(351.3) + 1.0(120) + 0.2(54.0) = 552.4 KIPS 6) 0.9D + 1.6W + 1.6H 0.9(351.3) + 1.6(-9.6) = 300.8 KIPS 7) 0.9D + 1.0E + 1.6H 0.9(351.3) = 316.2 KIPS

From above, Pu = 640.6 KIPS


Example 2 GIVEN: The cantilevered floor balcony beam/slab as shown below. The service superimposed dead load (not including concrete) = 14 PSF and the superimposed service live load = 75 PSF. REQUIRED: Determine the maximum factored moment, Mu on the cantilevered beam.

14’-0”

18”

5” slab 10”

Beam 16’-0”

wu

14’-0”

Side view of cantilevered beam


Step 1 – Determine service dead load, D acting on beam: Since there are 2 beams, each supports ½ of the balcony: 5” slab

18”

10” 8’-0”

18” – 5” slab

16’-0” 10" 13" ⎞ 150lb ⎛ 5" (8' )( ) + ( )( ) ⎟ 3 ⎜ 12 12 ⎠ 12 ft ⎝ = 635.4 PLF

Weight of concrete (shaded area) =

Superimposed dead load on beam = 8’(14 PSF) = 112 PLF Total dead load acting on beam, D = 635.4 PLF + 112 PLF = 747.4 PLF

Step 2 – Determine service live load, L acting on beam: Live load acting on beam = 8’(75 PSF) = 600 PLF Step 3 – Determine factored uniform load on beam, wu: By inspection, use load factor 1.2D + 1.6L wu = 1.2(747.4 PLF) + 1.6(600 PLF) = 1857 PLF = 1.9 KLF Step 4 – Determine maximum factored moment on beam, Mu: wu L2 For a cantilevered beam, Mmax = Mu = 2 =

(1.9 KLF )(14'−0" ) 2 2

Mu = 186.2 KIP-FT


Lecture 23 – Flexural Members Flexural members are those that experience primarily bending stresses, such as beams. A typical reinforced concrete beam is shown below:

Depth to steel “d”

Hanger bars (#4 or #5 bars) Stirrup bars (used to prevent diag. tension cracks) spaced at d/2 apart

Concrete cover = ¾” → 2” as per ACI reqmts.

Height “h”

Width “b”

Tension bars “As” Section A-A


Sometimes, 2 (or more) rows of main tension bars are necessary. It is important to provide minimum adequate cover around all reinforcing bars so that these bars can properly bond with the concrete. ACI 318 dictates that the minimum spacing between bars is 1.5 times the maximum concrete aggregate size. Typical concrete batches use a maximum aggregate size of ¾” diameter, so then the minimum bar spacing = 1.5(¾”) = 1⅛”.

Height “h”

Depth to centroid of steel “d”

Below is a sketch of a typical concrete beam with 2 rows of tension bars:

Min. bar spacing Tension bars “As” Min. bar spacing


As = Total cross-sectional area of all tension bars, in2 d = depth to center of tension bars, inches = h – (concrete cover) – (stirrup bar dia.) – ½(tension bar dia.) fy = yield stress of reinforcing bars = 60 KSI for ASTM A615 Grade 60 bars = 40 KSI for ASTM A615 Grade 40 bars ρactual = actual ratio of tension steel to effective concrete area A = s bd ρmin = minimum allowable ratio of tension steel per ACI 318 200 = where fy = PSI fy


Example 1 GIVEN: A rectangular concrete beam is similar to the one shown above. Use the following: • Height h = 20” • Width b = 12” • Concrete f’c = 4000 PSI • Concrete cover = ¾” • All bars are A615 – Grade 60 (fy = 60 KSI) • Stirrup bar = #3 • 4 - #7 Tension bars REQUIRED: 1) Determine total area of tension bars, As. 2) Determine depth to center of tension bars, d. A 200 3) Determine ρactual = s where ρmin = and state if it is acceptable. bd fy Step 1 – Determine area of tension bars, As: As = 4 bars(0.60 in2 per bar)

See Lect. 21 notes

As = 2.40 in2 Step 2 – Determine depth to tension bars, d: d = depth to center of tension bars, inches = h – (concrete cover) – (stirrup bar dia.) – ½(tension bar dia.) = 20” – ¾” – ⅜” – ½(⅞”) d = 18.44” Step 3 – Determine ρactual and ρmin : ρactual =

=

As bd

ρmin =

2.40in 2 (12" )(18.44" )

=

200 fy

200 60000 PSI

ρmin = 0.0033

ρactual = 0.0108

Since ρactual > ρmin → beam is acceptable


A basic understanding of beam mechanics is necessary to study concrete beam behavior. Consider a simply-supported homogeneous rectangular beam loaded by a uniformly-distributed load as shown below:

Applied loads

Span L

Taking a section through the beam at any place along the length reveals the following stress distribution about the cross-section of the beam:

Compression Neutral Axis

The stress distribution varies linearly from zero stresses at the neutral axis, to a maximum tensile or compressive stress at the extreme edges.

Homogeneous Beam

Tension


In a reinforced concrete beam, the stress distribution is different. Above the neutral axis, the concrete carries all the compression, similar to the homogeneous beam. Below the neutral axis however, the concrete is incapable of resisting tension and must rely on the reinforcing bars to carry all the tension loads.

b


The actual stress distribution in the compression side varies non-linearly from zero stresses at the neutral axis, to a maximum compressive stress at the extreme edge.

Reinforced Concrete Beam Tension = T Reinforcing bars Looking at a side view of the stress distribution of the reinforced concrete beam:

C Neutral Axis

½ (a)

0.85f’cb

a = β 1C

“Whitney” stress block

d Moment arm = Z

T = Asfy Actual Stress Distribution

T = Asfy Idealized Stress Distribution


Assuming an idealized beam, tension equals compression: Tension = Compression Asfy = Area of Whitney stress block Asfy = 0.85f’cab Solve for a: a=

As f y 0.85 f ' c b

= β 1C

β1 = 0.85 for f’c < 4000 PSI = 0.80 for f’c = 5000 PSI = 0.75 for f’c > 6000 PSI C = depth to neutral axis from extreme compression edge Mn = Nominal moment capacity of concrete beam = Asfy(Moment arm) = AsfyZ a = Asfy(d - ) 2 Mu = Usable moment capacity of concrete beam = φMn = 0.9Mn a Mu = 0.9(Asfy(d - ) ) 2

⎡ ⎛ ρ act f y Mu = 0.9Asfyd(1 - ⎢0.59⎜⎜ ⎢⎣ ⎝ f 'c

⎞⎤ ⎟⎟⎥ ) ⎠⎥⎦

ρbal = balanced ratio of tension steel reinforcement ⎛ 0.85β1 f ' c =⎜ ⎜ fy ⎝

⎞⎛ 87,000 ⎟⎜ ⎟⎜ 87,000 + f y ⎠⎝

⎞ ⎟ where fy = PSI ⎟ ⎠

ρmax = maximum allowable ratio of tension steel reinforcement per ACI 318 = 0.75ρbal


Example 2 GIVEN: The concrete beam from Example 1 is used to support the loading as shown below. REQUIRED: 1. Determine the maximum factored applied moment, Mmax. 2. Determine the usable moment capacity of the beam, Mu, and determine if it is acceptable based on Mmax. 3. Determine if the beam is acceptable based on ρmax.

Factored uniform load wu = 3000 PLF (incl. beam wt.)

20’-0”

Step 1 – Determine maximum factored applied moment, Mmax: Mmax =

=

wu L2 8 (3KLF )(20'−0" ) 2 8

Mmax = 150 KIP-FT Step 2 - Determine the usable moment capacity of the beam, Mu:


⎞⎤ ⎟⎟⎥ ) ⎠⎥⎦

where ρact = 0.0108 (see Ex. 1)

⎡ ⎛ (0.0108)(60 KSI ) ⎞⎤ = 0.9(2.40 in2)(60 KSI)(18.44”)(1 - ⎢0.59⎜ ⎟⎥ ) 4 KSI ⎝ ⎠⎦ ⎣ = 2161.4 KIP-IN Mu = 180.1 KIP-FT Since Mu = 180.1 KIP-FT > Mmax = 150 KIP-FT → beam is acceptable


Step 3 – Determine if the beam is acceptable based on ρmax:

ρmax = maximum allowable ratio of tension steel reinforcement per ACI 318 = 0.75ρbal ρbal = balanced ratio of tension steel reinforcement ⎛ 0.85β1 f ' c =⎜ ⎜ fy ⎝

⎞⎛ 87,000 ⎟⎜ ⎟⎜ 87,000 + f y ⎠⎝

⎞ ⎟ where fy = PSI ⎟ ⎠

where β1 = 0.85 since f’c = 4000 PSI

87,000 ⎞ ⎛ 0.85(0.85)(4 KSI ) ⎞⎛ =⎜ ⎟ ⎟⎜ 60 KSI ⎝ ⎠⎝ 87,000 + 60000 PSI ⎠ = 0.0285

ρmax = 0.75(0.0285) ρmax = 0.0214 > ρact = 0.0108 → beam is acceptable


Lecture 24 – Flexural Members (cont.) Determining the usable moment capacity, Mu, of a rectangular reinforced concrete beam is accomplished by using the formula below: (see Lect. 23)


⎞⎤ ⎟⎟⎥ ) ⎠⎥⎦

Designing a beam using the equation above is much more difficult. Assuming the material properties and dimensions are known, the equation above still has 2 unknown variables – As and ρact. Therefore, design of steel reinforcement for a given beam is largely one of trial-and-error.

Beam Design Design of concrete beam members is often one of trial-and-error. It’s difficult to directly solve for all the variables in a reinforced concrete beam. Usually, material properties are known as well as maximum applied factored moment, Mmax. The following Table is useful to get a “trial” beam size: Minimum Suggested Thickness “h” of Concrete Beams & One-Way Slabs Member: End Conditions Simply One end Both ends Cantilever supported continuous continuous Solid one-way slab L/20 L/24 L/28 L/10 Beam L/16 L/18.5 L/21 L/8 Span length L = inches Beams are usually rectangular having the width typically narrower than the height. The diagram below shows typical beam aspect ratios: b

h ≈ 1.5b → 2.5b


Beam Design Aid It is still difficult to directly design a reinforced concrete beam even if dimensions and material properties are known. The use of design aids are commonly used to streamline the design process instead of laboriously using a trial-and-error approach. Mu φbd 2 are in units of PSI. It can be used to directly solve for ρact knowing factored actual moment Mu, f’c, fy, b and d.

The design aid shown below is used for design or analysis. Values of

Table 1 - Concrete f’c = 3000 PSI, Grade 60 Bars


Table 2 – Concrete f’c = 4000 PSI, Grade 60 Bars


Example 1 GIVEN: A rectangular concrete beam with dimensions is shown below (stirrup bars not shown). Use concrete f’c = 4000 PSI and grade 60 bars. REQUIRED: 1) Determine the usable moment capacity Mu of the beam using formula. 2) Determine the usable moment capacity Mu of the beam using Table 2. b =12”

d = 18”

3 - # 7 bars

Step 1 – Determine usable moment capacity Mu of the beam using formula: ρact =

=

As bd

3(0.60in 2 _ per _ bar ) (12" )(18" )

ρact = 0.0083 ⎡ ⎛ ρ act f y Mu = 0.9Asfyd(1 - ⎢0.59⎜⎜ ⎢⎣ ⎝ f 'c

⎞⎤ ⎟⎟⎥ ) ⎠⎥⎦

⎡ ⎛ (0.0083)(60 KSI ) ⎞⎤ = 0.9(1.80 in2)(60 KSI)(18”)(1 - ⎢0.59⎜ ⎟⎥ ) 4 KSI ⎝ ⎠⎦ ⎣ = 1621 KIP-IN Mu = 135 KIP-FT


Step 2 - Determine the usable moment capacity Mu of the beam using Table 2: From Table 2: At ρ = 0.0083 →

Mu = 461.4 PSI φbd 2

Solving for Mu: Mu = 461.4 PSI(φbd2) = 461.4 PSI[(0.9)(12”)(18”)2] = 1,614,531 LB-IN = 1615 KIP-IN Mu = 134.6 KIP-FT NOTE: This answer is the same as in Step 1.


Example 2 GIVEN: The concrete beam below. Use the following: • • • • •

Concrete f’c = 4000 PSI Steel grade 60 Concrete cover = ¾” #8 bars are to be used for main tension bars #3 stirrups

REQUIRED: Design the rectangular beam such that h ≈1.5b and ρact ≈ ½ (ρmax). wu = 2 KLF

28’-0”

Step 1 – Determine maximum factored moment, Mmax: Mmax =

=

wu L2 8 (2 KLF )(28'−0" ) 2 8

Mmax = 196 KIP-FT = 2352 KIP-IN = 2,352,000 LB-IN


Step 2 – Select values from Table 2: a) Select ρact = ½(ρmax) = ½(0.0214) TRY ρact = 0.0107 b) At ρ = 0.0107 →

Mu = 581.2 PSI φbd 2

Step 3 – Solve for “b” and “d” by substituting Mmax for Mu in above equation: Mu = 581.2 PSI φbd 2

where: Mu = Mmax = 2,352,000 LB-IN φ = 0.9 d = 1.5b

2352000 = 581.2 PSI (0.9)(b)(1.5b) 2 Solve for “b”:

2352000 = 581.2 PSI (0.9)(2.25b 3 ) b=3

2352000 (0.9)(2.25)(581.2)

b = 12.6” → Use b = 12” d = 1.5b = 1.5(12”) d ≈ 18”


Step 4 – Select beam dimensions: From above, use b = 12” and d ≈ 18” #8 main bar dia. = 1”

#3 stirrup bar dia. = 3/8”

h = d + conc. cover + stirrup bar dia. + ½(main bar dia.) = 18” + ¾” + 3/8” + ½(1”) = 19.625” → Use h = 20” Revised d = 20” – ¾” – ⅜” – ½(1”) = 18.375” Step 5 – Determine required area of main tension bars: From above, ρact = 0.0107 =

As bd

Solve for As: As = 0.0107(b)(d) = 0.0107(12”)(18.375”) As = 2.36 in2 Step 6 – Determine number of #8 main tension bars: No. of bars =

As Area _ of _ one _ bar

2.36in 2 = 0.79in 2 _ per _#8 _ bar = 2.99 bars → USE 3 - #8 bars


Step 7 – Check beam height with “Minimum Thickness of Beams” Table: From Table: Member type = Beam

h≈

End Condition = Simply-supported

h≈

≈

L 16

L 16

(28'−0" )(12" / ft ) 16

h ≈ 21” which is approximately = 20” as designed Step 8 – Draw “Summary Sketch” labeling all information necessary to build it:

12”

2 - #4 hanger bars 20”

#3 stirrup bars @ 9” o.c.

¾” concrete cover

3 - #8 main bars Section A-A Notes: 1) Concrete f’c = 4000 PSI normal-weight 2) All bars ASTM A615 Grade 60


Lecture 25 – T- Beams Concrete beams are often poured integrally with the slab, forming a much stronger “T” – shaped beam. These beams are very efficient because the slab portion carries the compressive loads and the reinforcing bars placed at the bottom of the stem carry the tension. A T-beam typically has a narrower stem than an ordinary rectangular beam. These stems are typically spaced from 4’-0” apart to more than 12’-0”. The slab portion above the stem is designed as a one-way slab spanning between stems (see Lecture 26). Slab

Main tension bars at bottom of stem

A typical T-beam has the following dimensions and notations:

b = Effective flange width

Overhang width

bw

Clear distance

hf = Slab thickness

d

bw


NOTE: Stirrups in T-beam are required (not shown in this sketch)

Assuming T-beams are symmetrical, the following design dimensions are used: 8hf or

Overhang width = smaller

½(Clear distance) ¼(Beam span) b = smaller

or (2 x overhang width) + bw

T-Beam Analysis T-beams are analyzed similarly to rectangular beams, except the compression area is a narrow “strip” usually located in the slab. a = Effective conc. compressive thickness

b = Effective flange width

hf Z = (d Ac = Shaded area = Effective concrete compression area = (a)(b)

bw


a ) 2

d

As = Total area of main tension bars

Mu = Usable moment capacity of T-beam = φTZ where: φ = 0.9 T = Tension force developed in main bars = Asfy Ac = Effective concrete compression area T = 0.85 f ' c a = Effective concrete compressive thickness A = c b Z = Moment arm distance between center of compression to center of tension =d-

a 2


Example 1 GIVEN: A commercial building has T-beams spaced 6’-6” (center-to-center) with a 4” concrete slab as shown in the framing plan and cross-section views below. Use the following information: • • • •

Superimposed service floor dead load (NOT including conc. wt.) = 40 PSF Superimposed service floor live load = 100 PSF Concrete f’c = 3000 PSI ASTM A615 Grade 60 bars

REQUIRED: 1) Determine the maximum factored moment, Mmax, on the T-beam. 2) Determine the usable moment capacity, Mu, for the T-beam.

A

A

T-beam span = 20’-0”

T-beam

6’-6” Perimeter girder

Typ. Framing Plan


Column

6’-6” 4”

16”

8”

2 - #9 bars Section “A-A” Thru T-Beams

Step 1 – Determine maximum factored moment, Mmax, on T-beam: Determine area of T-beam = Slab area + Stem area = (6.5’)(0.333’) + (1’)(0.666’) = 2.83 ft2 Determine service weight of T-beam = Area of T-beam x Conc. unit wt. = 2.83 ft2(150 lb/ft3) = 425 PLF Det. factored uniform load on T-beam wu = 1.2D + 1.6L Service Dead Load

Service Live Load

= 1.2[(6.5’)(40 PSF) + 425 PLF] + 1.6[(6.5’)(100 PSF)] = 822 PLF + 1040 PLF = 1862 PLF → Use wu = 1.9 KLF Det. Maximum factored moment, Mmax =

=

wu L2 8 (1.9 KLF )(20'−0" ) 2 8

Mmax = 95 KIP-FT


Step 2 – Determine effective concrete slab width “b”: 8hf = 8(4”) = 32” ← USE Overhang width = smaller

or ½(Clear distance) = ½(78” – 8”) = 35”

¼(Beam span) = ¼(20’-0” x 12”/ft) = 60” ← USE b = smaller

or (2 x overhang width) + bw = (2 x 32” + 8”) = 72”

Step 3 – Determine effective conc. compression area Ac: T = Tension force developed in main bars = Asfy = 2 bars(1.00 in2 per #9 bar)(60 KSI) = 120 KIPS Ac = Effective concrete compression area =

T 0.85 f ' c

=

120 KIPS 0.85(3KSI )

= 47.1 in2


Step 4 – Determine usable moment capacity, Mu for the T-beam: a = Effective concrete compressive thickness A = c b =

47.1in 2 60"

a = 0.79” Z = Moment arm distance between center of compression to center of tension =d-

a 2

= 16” -

0.79" 2

Z = 15.6” Mu = φTZ = 0.9(120 KIPS)(15.6”) = 1685 KIP-IN Mu = 140.4 KIP-FT NOTE: Since Mu = 140.4 KIP-FT > Mmax = 95 KIP-FT, Tbeam is ACCEPTABLE.


Heavily-Reinforced T-Beams T-beams with a lot of tension reinforcement may have a portion of the effective concrete area located within the stem as shown below: b

hf

Z

d

Ac = Shaded area = Effective concrete compression area bw

As

The location of the centroid of the effective concrete compression area is found by methods discussed in AECT 210 – Structural Theory (see Lecture 5). After the location is found, analysis is exactly the same as ordinary T-beams. Similar to ordinary rectangular reinforced concrete beams, the ACI 318 limits the amount of tension steel in T-beams so that the steel will yield prior to concrete compression failure. The maximum area of steel, As is shown in the table below.

Maximum Tensile Steel Permitted in T-Beams Concrete and Steel Properties: Concrete f’c = 3000 PSI Steel fy = 40 KSI Concrete f’c = 3000 PSI Steel fy = 60 KSI Concrete f’c = 4000 PSI Steel fy = 40 KSI Concrete f’c = 4000 PSI Steel fy = 60 KSI

Formula (As = in2) As max = 0.0478[bhf + bw(0.582d – hf)] As max = 0.0319[bhf + bw(0.503d – hf)] As max = 0.0638[bhf + bw(0.582d – hf)] As max = 0.0425[bhf + bw(0.503d – hf)]


Example 2 GIVEN: The T-beam from Example 1. REQUIRED: Determine the maximum area of tension steel permitted, As max: Step 1 – Determine As max: From Example 1: Concrete f’c = 3000 PSI Steel fy = 60 KSI b = 60” hf = 4” bw = 8” As max = 0.0319[bhf + bw(0.503d – hf)] = 0.0319[(60”)(4”) + 8”(0.503(16”) – 4”)] As max = 8.7 in2 NOTE: This area of tension steel As = 8.7 in2 is a LOT!! In order to supply this much steel the beam would require 9 - #9 bars, 15 - #7 bars or 20 - #6 bars! It would be far better to change the beam dimensions than to try to squeeze this many bars into the beam.


Lecture 26 – One-Way Slabs A one-way slab is supported by parallel walls or beams, and the main tension reinforcing bars run parallel to the span. It looks like the following: Tension rebar (parallel to span) Slab thickness Shrinkage bar spacing

Tension bar spacing

Wall (or beam) Shrinkage rebar (perpendicular to span)

Slab span

The slab is designed as a series of 1’-0” wide beam “strips”. The analysis is similar to rectangular beams, except the width b = 12” and the height is usually on the order of 4” → 10”. The main tension bars are usually #4, #5 or #6 bars. There are no stirrups in slabs, however, additional bars are placed perpendicular to the main tension bars to prevent cracking during the curing process. These bars are referred to as “shrinkage” or “temperature” bars and are also usually #4 or #5 bars. Spacing reqd. for moment

Maximum spacing between main tension bars = smaller of

or 3 x slab thickness or 12” Spacing reqd. by analysis

Maximum spacing between shrinkage bars = smaller of


or 5 x slab thickness or 18”

Design of Main Tension Bars: As previously mentioned, slabs are designed as a series of 1’-0” wide rectangular beam “strips” as shown below:

Slab strip Support beam

h

d Span b = 12”

Tension bars

Assuming the slab strip is a rectangular beam, then:


⎞⎤ ⎟⎟⎥ ) ⎠⎥⎦

where: Mu = Usable moment capacity of slab strip As = Area of tension bars per 1’-0” width of slab fy = yield stress of rebar f’c = specified compressive strength of concrete A ρact = s bd Alternatively, the “Design Aid” Tables 1 and 2 from Lecture 24 may be used for analysis OR design.


Example 1 GIVEN: A one-way slab has a simple span = 8’-0” and the following materials and loads:

• • • • •

Concrete f’c = 4000 PSI #4 Grade 60 main tension bars and shrinkage bars Concrete cover = ¾” Superimposed floor SERVICE dead load = 38 PSF (not incl. slab wt.) Superimposed floor SERVIVE live load = 125 PSF

REQUIRED: Design the slab, including thickness, main tension bars & shrinkage bars. Slab strip Support beam

h=?

d Span = 8’-0” b = 12”

Tension bars

Step 1 – Determine slab thickness “h” based on Table below: Minimum Suggested Thickness “h” of Concrete Beams & One-Way Slabs Member: End Conditions Simply One end Both ends Cantilever supported continuous continuous Solid one-way slab L/20 L/24 L/28 L/10 Beam L/16 L/18.5 L/21 L/8 Span length L = inches h=

=

L 20

8'−0" (12" / ft ) 20

= 4.8” → USE 5” Thick Slab


Step 2 – Determine maximum factored moment, Mmax, on slab: Factored uniform load wu = 1.2D + 1.6L = 1.2(superimposed dead load + slab wt.) + 1.6(live load) = 1.2(1’(38 PSF) + (5/12)(150 PCF)) + 1.6(1’(125 PSF)) = 120.6 PLF + 200 PLF = 320.6 PLF = 0.32 KLF wu L2 Maximum factored moment Mmax = 8 (0.32 KLF )(8'−0" ) 2 = 8 Mmax = 2.56 KIP-FT = 30.72 KIP-IN = 30,720 LB-IN Step 3 – Determine depth to tension bars “d”: d

5”

b = 12”

d = h” – conc. cover – ½(Tension bar dia.) = 5” – ¾” – ½(4/8”) = 4” NOTE: No stirrups are required in slabs


Step 4 – Using Design Aid Table 2 (see Lecture 24), determine As: Determine the corresponding ρ from

Mu : φbd 2

Mu 30,720 LB − IN = 2 φbd (0.9)(12" )(4" ) 2

= 177.8 PSI

Use ρact = ρmin = 0.0033

ρact =

As bd

Solve for As: As = ρact(b)(d) = 0.0033(12”)(4”) = 0.16 in2 per 1’-0” width of slab Step 5 – Determine #4 bar tension bar spacing reqd. for moment: ⎛ A _ per _ one _ bar ⎞ ⎟⎟ Bar spacing reqd. = 12” ⎜⎜ s As ⎠ ⎝ ⎛ 0.20in 2 _ per _#4 _ bar ⎞ ⎟⎟ = 12” ⎜⎜ 0.16in 2 ⎠ ⎝

= 15” apart


Step 6 – Determine MAXIMUM tension bar spacing requirements per ACI-318:

Spacing reqd. for moment

Maximum spacing between main tension bars = smaller of

= smaller of

Use #4 Tension bars @ 12” o.c. Step 7 – Determine “Shrinkage bar requirements by analysis”: Shrinkage bar As = 0.0020bh = 0.0020(12”)(5”) = 0.12 in2 per 1’-0” width ⎛ A _ per _ one _ bar ⎞ ⎟⎟ Bar spacing reqd. = 12” ⎜⎜ s A s ⎠ ⎝ ⎛ 0.20in 2 _ per _#4 _ bar ⎞ ⎟⎟ = 12” ⎜⎜ 0.12in 2 ⎠ ⎝

= 20” apart


or 3 x slab thickness or 12” 15” (see prev. page) or 3 x 5” = 15” or 12” ← USE

Step 8 – Determine MAXIMUM shrink. bar spacing requirements per ACI-318: Spacing reqd. by analysis

Maximum spacing between shrinkage bars = smaller of

= smaller of

or 5 x slab thickness or 18”

20” (see prev. pages) or 5 x 5” = 25” or 18” ← USE

Use #4 Shrinkage bars @ 18” o.c. Step 9 – Draw “Summary Sketch”:

#4 Tension rebar placed at bottom of slab 5” #4 Shrinkage bar spacing = 18” o.c.

Tension bar spacing = 12” o.c.

Wall (or beam) #4 Shrinkage rebar placed in center of slab

Slab span = 8’-0”


Continuous (Mult-span) Slabs Concrete structural members are typically poured integrally together. Beams and slabs often span multiple supports and are not “simplysupported” as steel and wood framed beams are. As discussed in AECT 210 Lecture 15, these concrete beams and slabs are continuous and have both positive moments and negative moments. The location of tension bars in the members is related to the location of moment: Tension bars are located in the BOTTOM for Mpos Tension bars are located in the TOP for Mneg

2- Equal Span Condition:

Bars at top w

R1 L

M pos =

R3

R2

9 ( wL2 ) 128

L

M pos =

9 ( wL2 ) 128 0.375L

0.375L

1 M neg = − ( wL2 ) 8


Moment Diagram

3- Equal Span Condition: Bars at top w

R2

R1

R4

R3

L

L

Mpos = 0.08(wL2)

L

2

Mpos = 0.08(wL2)

Mpos = 0.025(wL )

0.4L

0.4L

0.5L Mneg = -0.1(wL2)

0.5L Mneg = -0.1(wL2)

Rebar Placement: At the transition between the Mpos and Mneg zones, a minimum overlap of bars is required per ACI 318. These overlaps are required for developing the full bar strength in tension. The friction developed between the concrete and the ribs of the rebar must equal the tensile strength of the bar. The necessary length of the bar embedment to achieve this friction force is called the “Development Length”, Ld, and is specified as a multiple of bar diameters. For example, the Ld for a Grade 60 rebar and concrete f’c = 4000 PSI = 38 x bar diameter. Tensile Strength

Rebar Tensile strength = Friction force Ld

Friction force


Below are schematic cross-sections of required overlap dimensions for bar placement in continuous slabs (beams are similar):


Example 2 GIVEN: A 1’-0” wide “strip” concrete slab that is 6” thick and a 3-span condition is shown below. All loads shown are already factored and includes slab weight. Use concrete f’c = 4000 PSI and Grade 60 bars. Use “d” = 5”. REQUIRED: Determine if the slab reinforcing steel is adequate for both the positive moments and negative moments. #4 @ 8” o.c. bars at bottom of slab

#5 @ 10” o.c. bars at top of slab wu = 1.0 KLF 6”

R2

R1

L = 9’-0”

L = 9’-0”

Mpos = 0.08(wL2)

0.4L

R3

L = 9’-0”

2

Mpos = 0.08(wL2)

Mpos = 0.025(wL )

0.4L 0.5L Mneg = -0.1(wL2)

0.5L Mneg = -0.1(wL2)

Step 1 – Determine maximum factored POSITIVE moment, Mpos: From above, Mpos = 0.08(wL2) = 0.08(1.0 KLF)(9’-0”)2 = 6.48 KIP-FT Step 2 - Determine maximum factored NEGATIVE moment, Mneg: From above, Mpos = -0.1(wL2) = -0.1(1.0 KLF)(9’-0”)2 = -8.1 KIP-FT


Step 3 – Determine usable moment capacity of slab in POSITIVE moment:

The slab is reinforced with #4 @ 8” o.c.: d = 5”

6” 8” b = 12”

#4 @ 8” o.c. at BOTTOM of slab

We must get the reinforcement As in terms of 12” width of slab:

⎛ A _ per _ bar ⎞ ⎟⎟ As per 1’-0” width = 12” ⎜⎜ s ⎝ Spacing ⎠ ⎛ 0.20in 2 _ per _#4 _ bar ⎞ ⎟⎟ = 12” ⎜⎜ 8" ⎠ ⎝

As = 0.30 in2 per 1’-0” width of slab Determine ρact =

=

As bd

0.30in 2 (12" )(5" )

= 0.005 Determine Mu by formula:


⎞⎤ ⎟⎟⎥ ) ⎠⎥⎦

⎡ ⎛ (0.005)(60) ⎞⎤ = 0.9(0.30 in2)(60 KSI)(5”)(1 - ⎢0.59⎜ ⎟⎥ ) ⎝ 4 KSI ⎠⎦ ⎣ = 77.4 KIP-IN Mu = 6.45 KIP-FT ≈ Mpos = 6.48 KIP-FT → ACCEPTABLE


Step 4 – Determine usable moment capacity of slab in NEGATIVE moment:

The slab is reinforced with #5 @ 10” o.c.: 6”

d = 5” 10” b = 12”

#5 @ 10” o.c. at TOP of slab

We must get the reinforcement As in terms of 12” width of slab:

⎛ A _ per _ bar ⎞ ⎟⎟ As per 1’-0” width = 12” ⎜⎜ s ⎝ Spacing ⎠ ⎛ 0.31in 2 _ per _#5 _ bar ⎞ ⎟⎟ = 12” ⎜⎜ 10" ⎠ ⎝

As = 0.37 in2 per 1’-0” width of slab Determine ρact =

=

As bd

0.37in 2 (12" )(5" )

= 0.0061 Determine Mu by formula:


⎞⎤ ⎟⎟⎥ ) ⎠⎥⎦

⎡ ⎛ (0.0061)(60) ⎞⎤ = 0.9(0.37 in2)(60 KSI)(5”)(1 - ⎢0.59⎜ ⎟⎥ ) 4 KSI ⎝ ⎠⎦ ⎣ = 94.5 KIP-IN Mu = 7.88 KIP-FT < Mneg = 8.1 KIP-FT → NOT ACCEPTABLE


Lecture 27 – Two-Way Slabs Two-way slabs have tension reinforcing spanning in BOTH directions, and may take the general form of one of the following:

Types of Two-Way Slab Systems


The following Table may be used to determine minimum thickness of various twoway slabs based on deflection:

Minimum Suggested Thickness “h” for Two-Way Slabs Two-Way Slab System: Minimum Thickness h: Flat plate Ln/30 Flat plate with spandrel beams Ln/33 Flat slab Ln/33 Flat slab with spandrel beams Ln/36 Two-way beam-supported slab Ln/33 Ln = clear distance in long direction Flat Plates Flat plates are the most common type of two-way slab system. It is commonly used in multi-story construction such as hotels, hospitals, offices and apartment buildings. It has several advantages: • • •

Easy formwork Simple bar placement Low floor-to-floor heights

Direct Design Method of Flat Plates per ACI 318-02 Two-way slabs are inherently difficult to analyze by conventional methods of statics because of the two-way bending occurring. Accurately determining the moments on a two-way slab is typically accomplished by finite element computer analysis.

Computer analysis of two-way slab


The ACI 318 code allows a direct design method that can be used in most typical situations. However, the following limitations apply: 1. Must have 3 or more continuous spans in each direction. 2. Slab panels must be rectangular with a ratio of the longer span to shorter span(measured as centerline-to-centerline of support) not greater than 2.0. 3. Successive span lengths in each direction must not differ by more than 1/3 of the longer span. 4. Columns must not be offset by more than 10% of the span (in direction of offset) from either axis between centerlines of successive columns. 5. Loads must be uniformly distributed, with the unfactored live load not more than 2 times the unfactored dead load (L/D < 2.0). Design Strips a) L1 > L2:

L2/4 L2/4

Column (typ.)

Exterior Column Strip

Middle Strip

L2

Interior Column Strip

Middle Strip


L2

L2/4


L1

b) L2 > L1:

Exterior Column Strip

Middle Strip


L2

Middle Strip


L2

L1

L1/4

L1/4 L1/4

Design Moment Coefficients for Flat Plate Supported Directly by Columns Slab End Span Interior Span Moments 1 2 3 4 5 Exterior Positive First Positive Interior Negative Interior Negative Negative Total 0.26Mo 0.52Mo 0.70Mo 0.35Mo 0.65Mo Moment Column 0.26Mo 0.31Mo 0.53Mo 0.21Mo 0.49Mo Strip Middle 0 0.21Mo 0.17Mo 0.14Mo 0.16Mo Strip Mo = Total factored moment per span

End Span 1

Mo =

wu L2 Ln 8

2

Interior Span 3

4

5

2

where Ln = clear span (face-to-face of cols.)in the direction of analysis Lecture 27 - Page 4 of 13

Bar Placement per ACI 318-02 The actual quantity of bars required is determined by analysis (see Example below). However, usage of the Direct Design Method prescribes bar placement as shown below:


Example 1 GIVEN: A two-way flat plate for an office building is shown below. Use the following: • • • • • •

Column dimensions = 20” x 20” Superimposed service floor Dead load = 32 PSF (not including slab weight) Superimposed service floor Live load = 75 PSF Concrete f’c = 4000 PSI #4 Grade 60 main tension bars Concrete cover = ¾”

REQUIRED: Use the “Direct Design Method” to design the two-way slab for the design strip in the direction shown.

L2 = 16’-0”

L2 = 16’-0”

L2 = 16’-0”

20’-0”

Ln

20’-0”

20’-0”

L2/4 L2/4 ½ Middle strip = ½(16’ – Col. strip)

Col. strip

½ Middle strip = ½(16’ – Col. strip)

Design Strip = 16’


Step 1 – Determine slab thickness h: Since it is a flat plate, from Table above, use h =

Ln 30

where Ln = clear span in direction of analysis = (20’-0” x 12”/ft) – 20”

Column size

= 220” = 18.33’ h=

220" 30

= 7.333” Use 8” thick slab Step 2 – Determine factored uniform load, wu on the slab: wu = 1.2D + 1.6L

Slab weight

= 1.2[(32 PSF) + (8/12)(150 PCF)] + 1.6[(75 PSF)] = 278.4 PSF = 0.28 KSF Step 3 – Check applicability of “Direct Design Method”: 1) Must have 3 or more continuous spans in each direction. YES 2) Slab panels must be rectangular with a ratio of the longer span to shorter span(measured as centerline-to-centerline of support) not greater than 2.0. YES 3) Successive span lengths in each direction must not differ by more than 1/3 of the longer span. YES 4) Columns must not be offset by more than 10% of the span (in direction of offset) from either axis between centerlines of successive columns. YES 5) Loads must be uniformly distributed, with the unfactored live load not more than 2 times the unfactored dead load (L/D < 2.0). YES


Step 4 – Determine total factored moment per span, Mo: Mo =

=

wu L2 Ln 8

2

(0.28KSF )(16' )(18.33' ) 2 8

Mo = 188 KIP-FT Step 5 – Determine distribution of total factored moment into col. & middle strips:

Design Moment Coefficients for Flat Plate Supported Directly by Columns Slab End Span Interior Span Moments 1 2 3 4 5 Exterior Positive First Positive Interior Negative Interior Negative Negative 0.70M 0.26M = 48.9 0.52M = 97.8 0.35Mo = 65.8 0.65Mo = 122.2 o = 131.6 o o Total Moment 0.26Mo = 48.9 0.31Mo = 58.3 0.53Mo = 99.6 0.21Mo = 39.5 0.49Mo = 92.1 Column Strip 0 0.21Mo = 39.5 0.17Mo = 32.0 0.14Mo = 26.3 0.16Mo = 30.1 Middle Strip Mo = Total factored moment per span = 188 KIP-FT Step 6 – Determine tension steel bars for col. & middle strips: a) Column strip for region 1 : Factored NEGATIVE moment = 48.9 KIP-FT (see Table above) = 586.8 KIP-IN = 586,800 LB-IN

b = 96” 8”

d

d = 8” – conc. cover – ½(bar dia.) = 8” – ¾” – ½(4/8”) = 7”


Mu 586,800 LB − IN = 2 φbd (0.9)(96" )(7" ) 2

= 138.6 PSI From Lecture 24 Table 2:

Use ρmin = 0.0033 ρ=

As bd

Solve for As: As = ρbd = (0.0033)(96”)(7”) = 2.22 in2 Number of bars required =

=

As As _ per _ bar

2.22in 2 0.20in 2 _ per _#4 _ bar

= 11.1 → Use 12 - #4 TOP bars


b) Column strip for region 2 : Factored POSITIVE moment = 58.3 KIP-FT (see Table above) = 699,600 LB-IN b = 96” 8”

d

d = 8” – conc. cover – ½(bar dia.) = 8” – ¾” – ½(4/8”) = 7” Mu 699,600 LB − IN = 2 φbd (0.9)(96" )(7" ) 2

= 165.2 PSI From Lecture 24 Table 2: Use ρ = 0.0033 As = 2.22 in2 (see calcs. above) Use 12 - #4 BOTTOM bars


c) Middle strip for region

2 :

Factored POSITIVE moment = 39.5 KIP-FT (see Table above) = 474,000 LB-IN b = 96” 8”

d

d = 8” – conc. cover – ½(bar dia.) = 8” – ¾” – ½(4/8”) = 7” Mu 474,000 LB − IN = 2 φbd (0.9)(96" )(7" ) 2

= 112.0 PSI From Lecture 24 Table 2: Use ρ = 0.0033 As = 2.22 in2 (see calcs. above) Use 12 - #4 BOTTOM bars Use 6 - #4 Bottom bars at each ½ Middle Strip


Step 7 – Draw “Summary Sketch” plan view of bars:

16’-0”

16’-0”

16’-0”

Col. strip for region 1 12 - #4 TOP bars

½ Middle strip for region 2 6 - #4 BOTTOM bars

20’-0”

Col. strip for region 2 12 - #4 BOTTOM bars

20’-0”

20’-0”

4’-0” ½ Middle strip = 4’-0”

4’-0”

Col. strip 16’ – 0”


½ Middle strip = 4’-0”

Example 2 GIVEN: The two-way slab system from Example 1. REQUIRED: Design the steel tension bars for design strip shown (perpendicular to those in Example 1).

16’-0”

16’-0”

20’-0”


20’-0”

16’-0”

Col. strip = 8’-0”


20’-0”

20’-0”


Lecture 28 – Shear in Beams Heavy loads on concrete beams produce diagonal shear cracks as shown below:

Shear cracks in areas of high shear

Column (typ.)

Cracking in beams is normal and indicates the tension bars are actually working. Excessive cracking needs to be controlled by additional bars called stirrups placed perpendicular to the cracks as shown below:

Stirrup bars spanning crack Stirrup bar spacing, s


Stirrups may take the shape of the following typical configurations:

Design for Shear in Concrete Beams Vertical shear is carried by the concrete shear capacity and the shear capacity provided by the stirrups. In other words: Vu (from shear diagram) < φVc + φVs where: Vu = factored vertical shear determined at a distance “d” away from the face of support φ = 0.75 Vc = shear strength of concrete = 2 f ' c bw d f’c = concrete strength, PSI bw = width of beam Vs = shear strength provided by stirrups bars =

Vu − φVc

φ


d 2

If Vs < 4 f ' c bw d then Max. stirrup spacing smax = smaller of

or 24”

If 4 f ' c bw d < Vs < 8 f ' c bw d then Max. stirrup spacing smax = smaller of

d 4 or

Stirrups are required when Vu > ½(φVc)

12”

If stirrups are required, s=

Av f y d Vs

< Smax

where: Av = area of stirrup bars crossing crack fy = yield strength of stirrup bar (i.e., 60 KSI) d = depth to center of tension bars


Example GIVEN: A simply-supported concrete beam having the following: (Assume the beam is adequate based on flexure)

• •

Concrete f’c = 4000 PSI #3 Grade 60 U-shaped stirrup bars

REQUIRED: Determine stirrup requirements along the length of the beam.

wu = 5.6 KLF (includes beam wt.) Conc. column

12” x 20” concrete beam 20’-0”

12”

#3 U-shaped stirrup bars

d = 18”

Beam Cross-Section


h = 20”

Step 1 – Determine maximum factored shear Vu at “d” away from face of support:

Simply-supported beam end reaction =

=

wu L 2

(5.6 KLF )(20'−0" ) 2

= 56 KIPS

d = 18” = 1.5’

Vu at distance “d” from face of support = 56 KIPS – 1.5’(5.6 KLF) = 47.6 KIPS

wu = 5.6 KLF

Vu = 47.6 KIPS 56 KIPS

0

0 d = 1.5’

Step 2 – Determine shear strength of concrete φVc:

φVc = 0.75( 2 f ' c bw d ) = 0.75( 2 4000 PSI (12" )(18" ) = 20,492 Lbs. = 20.5 KIPS


Shear Diagram

Step 3 – Determine range where stirrups are required: Stirrups are required when Vu > ½(φVc) ½(φVc) = ½(20.5 KIPS) = 10.25 KIPS wu = 5.6 KLF


½(φVc) = 10.25 KIPS 0

0

-47.6 KIPS -10.25 KIPS Stirrups required

d = 1.5’

Stirrups not required

Stirrups required

Step 4 – Determine shear strength provided by stirrups, Vs: Vs = shear strength provided by stirrups bars =

=

Vu − φVc

φ 47.6 KIPS − 20.5KIPS 0.75

Vs = 36.1 KIPS


d = 1.5’

Step 5 – Determine stirrup spacing, s: a) Check if Vs < 4 f ' c bw d 36.1 KIPS < 4 4000 PSI (12" )(18" ) 36.1 KIPS < 54,644 Lbs. → YES

d 18" = = 9” 2 2

Then: Max. stirrup spacing smax = smaller of

or 24” b) Check spacing requirement:

Av f y d

s=

Vs

Av = 2(0.11 in2 per #3 bar) = 0.22 in2

=

Assumed shear crack

< Smax


2 - #3 bars spanning crack

(0.22in 2 )(60 KSI )(18" ) 36.1KIPS

= 6.6” → round down to 6” o.c. 6” < 9” Use #3 Stirrup bars spaced 6” o.c.


Step 6 – Determine number of stirrups – assuming only ONE spacing: wu = 5.6 KLF


½(φVc) = 10.25 KIPS 0

0

-47.6 KIPS -10.25 KIPS d = 1.5’

Stirrups required

Stirrups not required

Stirrups required

d = 1.5’

⎛ Dis tan ce _ where _ stirrups _ required ⎞ ⎟⎟ + 1 Number of stirrups required = ⎜⎜ Stirrup _ spacing ⎠ ⎝ ⎞ ⎛ ⎛ 47.6 KIPS − 10.25KIPS ⎞ ⎜⎜ ⎟12" / ft ⎟ 5.6 KLF ⎠ ⎟ +1 = ⎜⎝ ⎟ ⎜ 6" o.c. ⎟ ⎜ ⎠ ⎝ = 14.3 stirrups → round up to 15 stirrups Use 15 - #3 U-shaped stirrups @ 6” o.c. ea. end of beam



20’-0”

15 - #3 U-shaped stirrups at ea. end of beam Stirrup bar spacing = 6” o.c. 12”

15 - #3 Grade 60 Ushaped stirrup bars @ 6” o.c. at each end of beam

20”

Beam Cross-Section


Lecture 29 – Shear in Beams (cont.) As a continuation of our discussion of shear in concrete beams, let’s look at an example of a concrete girder with point loads. Example GIVEN: The 20” x 34” concrete girder as shown below. Use the following: • •

Concrete f’c = 4000 PSI #3 – Grade 60 U-shaped stirrups

REQUIRED: Determine the stirrup bar requirements. Assume only one spacing for the beam. 25 KIPS

25 KIPS

25 KIPS

25 KIPS

wu = 1.2 KLF

Conc. column

8’-0”

8’-0”

8’-0”

8’-0”

8’-0”

40’-0”

20”

d = 32”


Beam Cross-Section


h = 34”

Step 1 – Draw shear diagram: 25 KIPS

25 KIPS

25 KIPS

25 KIPS

wu = 1.2 KLF

Conc. column

8’-0”

8’-0”

8’-0”

8’-0”

8’-0”

40’-0”

74

64.4 29.8

39.4 4.8

NOTE: All loads are in Kips

-4.8 -29.8

-29.8 -39.4

Step 2 – Determine Vu at distance “d” from face of support: Vu = 74 KIPS -

32" (1.2 KLF ) 12" / ft

= 70.8 KIPS Step 3 – Determine shear capacity of concrete φVc: φVc = 0.75( 2 f ' c bw d ) = 0.75( 2 4000 PSI (20" )(32" )) = 60,716 Lbs. φVc = 60.7 KIPS


-74

Step 4 – Determine range where stirrups are required: Stirrups are required when Vu > ½(φVc) ½(φVc) = ½(60.7 KIPS) = 30.4 KIPS

25 KIPS

25 KIPS

25 KIPS

25 KIPS

wu = 1.2 KLF

8’-0”

8’-0”

8’-0”

8’-0”

8’-0”

Vu = 70.8 74

64.4 39.4

½(φVc) = 30.4 29.8 4.8

Shear Diagram

-4.8 -29.8

X

32”

-29.8 -39.4

-74

Stirrups reqd.

Because of symmetry → Same condition on right side of beam as left side of beam Distance “X” =

39.4kips − 30.4kips 1.2kips _ per _ foot

= 7.5 feet = 90”


Step 5 – Determine shear strength provided by stirrups, Vs: Vs = shear strength provided by stirrups bars =

=

Vu − φVc

φ 70.8KIPS − 60.7 KIPS 0.75

Vs = 13.5 KIPS Step 6 – Determine stirrup spacing, s: a) Check if Vs < 4 f ' c bw d 13.5 KIPS < 4 4000 PSI (20" )(32" ) 13.5 KIPS < 161,900 Lbs. → YES

d 32" = = 16” 2 2

Then: Max. stirrup spacing smax = smaller of

or 24” Use


b) Check spacing requirement:

Av f y d

s=

Vs

Assumed shear crack

< Smax

2 - #3 bars spanning crack

Av = 2(0.11 in2 per #3 bar) = 0.22 in2

=


(0.22in 2 )(60 KSI )(32" ) 13.5KIPS

= 31.3” > 16” Use #3 Stirrup bars spaced 16” o.c. Step 7 – Determine number of stirrups required:

⎛ Dis tan ce _ where _ stirrups _ required ⎞ ⎟⎟ + 1 Number of stirrups required = ⎜⎜ Stirrup _ spacing ⎠ ⎝ ⎛ (96"−32" ) + ( Dis tan ce _" X " ) ⎞ =⎜ ⎟ +1 16" o.c. ⎠ ⎝

⎛ (96"−32" ) + 90" ⎞ =⎜ ⎟ +1 16" o.c. ⎝ ⎠

See shear diagram

= 10.625 stirrups → round up to 11 stirrups Use 11 - #3 U-shaped stirrups @ 16” o.c. ea. end of beam



40’-0”

11 - #3 U-shaped stirrups at ea. end of beam Stirrup bar spacing = 16” o.c. 20”

11 - #3 Grade 60 Ushaped stirrup bars @ 16” o.c. at each end of beam

34”

Beam Cross-Section


Lecture 3 – Tension members Steel tension members are perhaps the simplest members to design. There is no compressive, bending, shear or other stresses involved. Typical structural members that are under tension loads are: • • •

Trusses Bracing Hangers

The design of steel tension members is found in the following locations in the LRFD Manual: • • •

AISC Part 5 AISC SPEC Chapter D p. 16.1-26 AISC SPEC Chapter D p. 16.1-249

There are 2 types of failure mechanisms for tension members. The first is yielding on the gross area and the second is fracture on the net section. Yielding on Gross Area Yielding on the gross area refers to “stretching” of the gross crosssectional area of the member:

The LRFD design strength for yielding on gross section in tension = φtPn P The ASD allowable strength for yielding on gross section in tension = n Ω where: φt = 0.90 (LRFD) Ω = 1.67 (ASD) Pn = nominal strength of member = FyAg Fy = yield stress of steel (see AISC p. 2-39) Ag = gross cross-sect. area (see AISC Part 1)


Fracture on Net Section Fracture on the net section refers to “breaking” the section perpendicular from the direction of force through the reduced cross-sectional area of a member, typically across the bolt holes.

Bolt holes Material thickness

Hole dia. + 1/16” Ag = Gross area of angle

An = Net area of angle

The LRFD design strength for fracture on net section in tension = φtPn P The ASD allowable strength for fracture on net section in tension = n Ω where: φt = 0.75 (LRFD) Ω = 2.00 (ASD) Pn = nominal strength of member = FuAe Fu = ultimate stress of steel (see AISC p. 2-39) Ae = effective net area = AnU An = net area (see AISC p. 16.1-14) = Ag – [No. of holes{(hole dia. + 1/16”)(matl. thk.)}] U = reduction factor considering “shear lag” = See AISC Table D3.1 p. 16.1-29 = 1.0 if tension load is transmitted directly to each element by means of fasteners or welds ⎛ x⎞ = 1 − ⎜ ⎟ if tension load is transmitted to some but ⎝l⎠ not all of the elements by use of fasteners or welds

where: x = connection eccentricity, inch l = length of connection in the direction of loading, inch


Example 1 (LRFD) GIVEN: The double-angle L4x4x¼ A36 bracing member as shown below is under a FACTORED tensile load of 54 kips. The angles are bolted to the steel gusset plate using 2 – ¾” diameter bolts. REQUIRED: 1) Determine if the angles are acceptable based on yielding on gross area. 2) Determine if the angles are acceptable based on fracture on net area.

l = 3”

FACTORED Step 1 – Determine LRFD design strength of member considering yielding:

Design strength = φtPn = 0.90(FyAg) = 0.90(36 KSI)[2 angles(1.94 in2 per angle)] = 125.7 KIPS Since 125.7 KIPS > 54 KIPS → member is acceptable


Step 2 – Determine LRFD design strength considering fracture on net area:

Design strength = φtPn = 0.75(FuAe) Ae = AnU Taking a cross-section perpendicular through the double-angle:

N.A.

N.A. x

An = Ag – [No. of holes{(hole dia. + 1/16”)(mat. th)}] = (2)(1.94 in2) – [2 holes{(¾” + 1/16”)(¼”)}] = 3.47 in2

⎛ x⎞ U = 1− ⎜ ⎟ ⎝l⎠ where x = 1.08” see properties AISC p. 1-43 l = Connection length = 3” (see sketch above) ⎛ 1.08" ⎞ = 1− ⎜ ⎟ ⎝ 3" ⎠ = 0.64 Ae = AnU = 3.47 in2(0.64) = 2.22 in2 Design strength = 0.75(58 KSI)(2.22 in2) = 96.6 KIPS Since 96.6 KIPS > 54 KIPS → member is acceptable


In the design of steel tension members the AISC recommends a maximum L < 300 for members other than rods, straps or permissible slenderness ratio, rmin HSS (see AISC p. 16.1-26). This is not so much for structural strength – rather to provide some stiffness to reduce the undesirable effects of lateral movement. Also, since truss members and other tension members may see some loadreversal and experience some compressive loads, it is to reduce the likelihood of buckling. Example 2 (ASD) GIVEN: The bottom chord of the truss is to be designed using a single A992 W8 member. The truss has welded connections, so there is no need to check for fracture on the net area. REQUIRED: Design the lightest weight W8 member checking: 1) Slenderness ratio is not exceeded 2) Yielding on gross area

W8 Bottom chord 15’-0”

22’-0”

22’-0”

SERVICE load = 100 Kips Step 1 – Determine rmin such that slenderness ratio is not exceeded:

L rmin

≤ 300

L = 22’-0”(12” per ft) = 264” Rearranging and solve for rmin:

264" ≤ rmin 300 Required rmin > 0.88 in. → looking at AISC p. 1-27, the smallest W8 that will work is W8x18 with ry = 1.23 in. > rmin of 0.88 in.


Step 2 – Determine SERVICE tensile load on member: By simple truss analysis, the force in the member is:

VertForce HorzForce = VertLength HorzLength 100 KIPS HorzForce = 15'−0" 22'−0" Horz Force = 146.7 KIPS Step 3 – Design lightest W8 member considering yielding on gross area: ASD Allowable strength for yielding =

Fy Ag Ω

> 146.7 KIPS

where: Ω = 1.67 Fy = 50 KSI (see AISC p. 2-39) Solve for Ag: Ag >

>

146.7 kips(Ω) Fy

146.7kips(1.67) 50 KSI

Ag > 4.9 in2

AISC p. 1-26

Use W8x18 → Area = 5.26 in2 > 4.9 in2 Step 4 – Determine the maximum allowable tensile yielding load on W8x18:

ASD Allowable strength for yielding =

=

Fy Ag Ω (50 KSI )(5.26in 2 ) 1.67

ASD Allowable strength for yielding = 157.5 Kips


Step 5 – Determine the allowable tensile yielding load on W8x18 using AISC Table 5-1:

From AISC p. 5-13:

ASD

W8x18

Allowable yielding SERVICE tensile load = 157 Kips


Lecture 30 – Development of Reinforcement, Splices, Hooks Reinforcing bars must be embedded a minimum distance into the concrete in order to achieve the full tensile capacity, T of the bar. This length is referred to as “Development Length”, Ld.

Concrete Rebar T = Asfy

Ld

The development length is based upon the BOND between the rebar and the concrete. Factors affecting this bond include the following: • • • • •

Type of ribbing on the bar Presence of epoxy (or other ) coating Concrete quality Distance between bar and edge of concrete Type of end anchorage into the concrete


Determining Ld for Tension Bars: a) #6 and smaller bars: Ld =

d b f y αβλ 25 f ' c

b) #7 and larger bars: Ld =


where: db = diameter of bar fy = yield strength of bar, PSI f’c = specified concrete compressive strength, PSI α = alpha = Bar location factor = 1.3 for top reinforcement = 1.0 for all other locations β = beta = Coating factor = 1.5 for epoxy coated bars = 1.0 for uncoated bars λ = lambda = Lightweight aggregate factor = 1.3 for lightweight aggregate = 1.0 for normal weight aggregate


Example 1 GIVEN: A #6 rebar under tension force. Assume the following conditions: • • • • • •

Concrete f’c = 4000 PSI Normal weight concrete (λ = 1.0) ASTM A615 Grade 60 rebar #6 rebar (γ = 0.8) Uncoated bar (β = 1.0) Bar location is bottom of beam (α = 1.0)

REQUIRED: Determine the development length, Ld to achieve full tensile strength of the bar. Step 1 – Use the formula above to determine Ld: Ld =


⎛6⎞ ⎜ ⎟(60,000 PSI )(1.0)(1.0)(1.0) 8 Ld = ⎝ ⎠ 25 4000 PSI Ld = 28.5”

Concrete #6 Rebar

Ld = 28.5”


Assuming “normal” conditions, the following table may be used to determine development lengths of bars in tension: Development Length Ld of Grade 60 bottom bars in normal weight concrete

Condition Clear spacing of bars > db, clear cover > db

All other cases

Concrete f’c 3000 PSI 4000 PSI 5000 PSI 3000 PSI 4000 PSI 5000 PSI

No. 6 and smaller bars 44db 38db 34db 66db 57db 51db

No. 7 and larger bars 55db 47db 42db 82db 71db 64db

Example 2 GIVEN: The same information as Example 1. REQUIRED: Using the table above, determine the Ld for a #6 bar. Step 1 – Use table above to determine Ld: Concrete f’c = 4000 PSI Clear spacing of bars > db, clear cover > db


Ld = 38db = 38(6/8”) Ld = 28.5”

Determining Ld for Compression Bars: The development length of bars in compression is not as large as the development length in tension because of the absence of tension cracking in the concrete. Ldc = Development length in compression Ldc = 0.02 = larger of

db f y f 'c

or Ldc = 0.0003dbfy

Example 3 GIVEN: A #6 bar in compression. Use f’c = 4000 PSI and Grade 60 bars. REQUIRED: Determine the Ldc for the bar. Ldc = Development length in compression

Ldc = 0.02 = larger of

db f y f 'c

6 (60,000 PSI ) 8 = 0.02 = 14.2” ← Use 4000 PSI

or Ldc = 0.0003dbfy = 0.0003(6/8”)(60,000 PSI) = 13.5”

Ldc = 14.2”


Lap Splices of Bars Bars are generally fabricated to lengths of about 60’-0”, but transportation, workability and other concerns often require bars to be less than about 40’-0” long. For long walls, beams, slabs and other situations requiring long lengths of bars, lap splicing is commonly used. It is good practice to place laps at regions of small tension, i.e., low moment.

Concrete

Ls

Ls = Length of lap splice = 1.0Ld for “Class A” splice if the area of reinforcement provided through the splice > twice that required by analysis and not more than 50% of the total reinforcement is spliced within the lap length = 1.3Ld for “Class B” splice if reinforcement does not meet Class A requirements


Hooked and Bent Bars Hooks are used in concrete members where there is not sufficient straight length to achieve the full development length Ld. The following is a diagram showing the required lengths of bends and hooks:

Ldh = Lhbλ


Where: Lhb = Basic development length of hook in tension = 1200

db f 'c

λ = 1.0 unless otherwise specified below: fy if using other than Grade 60 bars = 60,000 = 0.7 if side concrete cover > 2½” or end cover > 2” = 0.8 if ties or stirrups spacing < 3db = 1.3 if lightweight concrete =

Re quired _ As Pr ovided _ As

Example 4 GIVEN: A #5 Grade 40 bar is in tension as shown below. Use LIGHTWEIGHT concrete with f’c = 4000 PSI. REQUIRED: Determine the min. required hook dimensions “X”, “Y” and “Z”. Z = Ldh

Side cover = 1½”

X

End cover = 1½”

Y

Critical section

Step 1 – Determine dimension “X”: X = 12db = 12(5/8”) X = 7½”


Step 2 – Determine dimension “Y”: Y = 4db since it is a #6 bar = 4(5/8”) Y = 2½” Step 3 – Determine length of hooked bar, Lhb: Lhb = 1200

db f 'c

5 8" = 1200 4000 PSI = 11.9” Step 4 – Determine total development length, Z = Ldh: Ldh = Lhbλ Where: λ = 1.0 since side cover = 1½” < 2½” = 1.3 since lightweight concrete =

fy 60000 PSI

=

40000 PSI = 0.67 60000 PSI

Ldh = Lhbλ = 11.9”(1.0)(1.3)(0.67) Ldh = 10.4”


Lecture 31 – Serviceability Serviceability refers to the structural “performance” of the finished building under service loads. • • •

Beam deflection Lateral drift Vibration

We will be focusing our discussion on beam deflection. The ACI 318-02 Code dictates that the deflections be checked on the basis of effective moment of inertia, Ie, under service loads. Before we can determine the value of the effective moment of inertia, we must first have an understanding of the gross moment of inertia, Ig, and the cracked moment of inertia, Icr. Gross Moment of Inertia Ig: The gross moment of inertia is not appropriate for reinforced concrete beams because the concrete under the neutral axis is in tension and is ineffective. Since tension is carried by the steel rebar, the beam becomes composite and therefore must be analyzed as such (See AECT 210 – Lecture 6). The calculated value of gross moment of inertia is higher than what is actually present. b

h

Ig = Gross moment of Inertia =

bh 3 12


Cracked Moment of Inertia Icr: The cracked moment of inertia takes into consideration the composite action between the concrete and steel rebar. This assumes that the concrete in the tension zone is totally ineffective, which is overly conservative. However, the cracked moment of inertia is far closer to predicting the actual moment of inertia of a reinforced concrete beam than the gross moment of inertia. b

y N.A.

d

nAs

Icr =

by 3 + nAs (d − y ) 2 3

Where: n = Modular ratio E 29,000,000 PSI = steel = E conc 57,000 f ' c As = Area of steel rebar in tension, in2

⎡ ⎤ bd nAs ⎢ 1 + 2 − 1⎥ nAs ⎣ ⎦ y= b


Effective Moment of Inertia, Ie: The effective moment of inertia is typically used to determine the section property of the member at a specific point along the moment diagram. In most cases, the effective moment of inertia is used to determine the actual deflection of the member when comparing to Code-dictated maximums.

⎧⎪⎛ M Ie = ⎨⎜⎜ cr ⎪⎩⎝ M a

3 ⎡ ⎛M ⎞ ⎟⎟ I g + ⎢1 − ⎜⎜ cr ⎢⎣ ⎝ M a ⎠

⎞ ⎟⎟ ⎠

3

⎤ ⎫⎪ ⎥ I cr ⎬ ≤ I g ⎥⎦ ⎪⎭

where: Mcr = moment that would initially crack the section fr I g = yt fr = modulus of rupture for the concrete = 7.5 f ' c yt = dist. from N.A. of uncracked cross-section to extreme tension fiber =

h 2

Ma = maximum unfactored moment at specific location along the moment diagram

Ig = gross moment of inertia Icr = cracked moment of inertia


Example 1 GIVEN: A simply-supported rectangular beam is shown below. The loads indicated are SERVICE loads. Use concrete f’c = 4000 PSI and grade 60 bars. REQUIRED: 1) Determine the gross moment of inertia Ig of the beam. 2) Determine the cracked moment of inertia Icr of the beam. 3) Determine the maximum allowable mid-span deflection of the beam assuming ∆allow = L/360. 4) Determine the actual mid-span deflection of the beam using Ie.

Wservice = 1500 PLF

∆

25’-0”

12”



¾” concrete cover

3 - #8 main bars Section A-A

d = h – conc. cover – stirrup bar dia. – ½(main tension bar dia.) = 20” – ¾” – ⅜” – ½(8/8”) = 18.375”


Step 1 – Determine gross moment of inertia Ig:

bh 3 Ig = 12 =

(12" )(20) 3 12

Ig =8000 in4 Step 2 – Determine cracked moment of inertia Icr:

Icr =

by 3 + nAs (d − y ) 2 3 Where: n = Modular ratio E 29,000,000 PSI 29,000,000 PSI = = steel = E conc 57,000 f ' c 57,000 4000 PSI = 8.04 As = Area of steel rebar in tension, in2 = 3 bars(0.79 in2 per #8 bar) = 2.37 in2

⎡ ⎤ bd nAs ⎢ 1 + 2 − 1⎥ nAs ⎣ ⎦ y= b ⎡ ⎤ (12" )(18.375" ) − 1⎥ (8.04)(2.37in 2 ) ⎢ 1 + 2 2 (8.04)(2.37in ) ⎦ ⎣ = 12" = 6.2”


Icr =

=

by 3 + nAs (d − y ) 2 3 (12" )(6.2" ) 3 + 8.04(2.37in 2 )(18.375"−6.2" ) 2 3

Icr = 2904 in4 Step 3 - Determine the maximum allowable mid-span deflection of the beam assuming ∆allow = L/360:

∆allow =

=

L 360 (25'−0" )12" / ft 360

∆allow = 0.83” Step 4 - Determine the effective moment of inertia Ie: fr = modulus of rupture for the concrete = 7.5 f ' c = 7.5 4000 PSI = 474.3 PSI Mcr = moment that would initially crack the section fr I g = yt =

(474.3PSI )(8000 PSI ) 20" 2

= 379,473 Lb-In = 379.4 KIP-In = 31.6 KIP-FT


Ma = maximum unfactored moment at specific location along the moment diagram =

wL2 8

=

(1.5KLF )(25'−0" ) 2 8

Ma = 117.2 KIP-FT

⎧⎪⎛ M Ie = ⎨⎜⎜ cr ⎪⎩⎝ M a

3 ⎡ ⎛M ⎞ ⎟⎟ I g + ⎢1 − ⎜⎜ cr ⎢⎣ ⎝ M a ⎠

⎞ ⎟⎟ ⎠

3

⎤ ⎫⎪ ⎥ I cr ⎬ ⎥⎦ ⎪⎭

⎧⎪⎛ 31.6 ⎞ 3 ⎫ ⎡ ⎛ 31.6 ⎞ 3 ⎤ 4 4 ⎪ = ⎨⎜ ( 8000 in ) + ⎟ ⎥ (2904in )⎬ ≤ I g ⎟ ⎢1 − ⎜ ⎪⎩⎝ 117.2 ⎠ ⎪⎭ ⎢⎣ ⎝ 117.2 ⎠ ⎥⎦

Ie = 3004 in4 Step 5 – Determine actual mid-span deflection using Ie:

5wL4 ∆act = 384 E conc I e ⎛ 1500 PLF ⎞ 4 5⎜ ⎟(25'−0" x12" / ft ) 12 ⎠ = ⎝ 384(57000 4000 PSI )(3004in 4 )

∆act = 1.22” Since ∆act = 1.22” > ∆allow = 0.83” → member is NOT acceptable


Lecture 33 – Columns Concrete Columns: Concrete is good at resisting compression but poor in resisting tension. So, it might make sense that concrete would be the material of choice for columns. It is true that concrete IS used for compression members such as columns, piers, bearing walls and pedestals. Members under pure compression could then (theoretically) be unreinforced. These members are often subject to additional forces such as moment that would put some tensile forces into the member and would thus necessitate the addition of tension reinforcement. Most columns have combined compression and bending. They are essentially a “beam-column”. The compression capacity of a reinforced column is reduced by the bending stresses on the column and vice-versa. A graph of the axial load capacity of a column vs. the moment capacity of a typical column is shown below:

Axial compression capacity

Pure compression

Pure bending Bending moment capacity


Types of Concrete Columns: There are two types of reinforced concrete columns – tied and spiral – and refer to the type of confining bars used to contain the interior core of concrete. It has been shown that unconfined concrete core will carry MUCH LESS load than a confined core as shown below: Load Load

Concrete & vert. bars “explode” outward

Unconfined core

Insufficient ties


A confined concrete core will carry substantially more load and will NOT explode outward like the unconfined concrete section will. Load

Load

Minor concrete spalling, core remains intact

Confined core Tied Column: A “Tied” concrete column is one in which individual rebar ties are used to wrap completely around the vertical bars to confine the interior core. These ties are usually #3 or #4 bars spaced per ACI requirements: Least column dimension Tie Spacing = smaller of

16 x Vert. Bar Dia. 48 x Tie Bar Dia.

Additionally, the ACI dictates that there must be a minimum of 4 vertical bars having a minimum area of 1% of the column cross-sectional area and a maximum of 8% of the column cross-sectional area. From a constructability standpoint, 4% is the upper maximum that can be readily achieved because of rebar congestion.


A typical tied concrete column looks like the following:

Vertical bars (4 minimum)

Tie bars Tie spacing

Concrete


Spiral Column: A spiral column has a single rebar wrapped around the vertical bars in a spiral and is stronger than a comparable tied column. It is more laborintensive to build than a tied column. The ACI requires a minimum of 6 vertical bars, with the same minimum and maximum areas of steel as a tied column.

Concrete Pitch = 1” → 3” Spiral rebar

Spiral columns have greater ductility than a tied column. This means that a spiral column is capable of sustaining larger strain before failure than a tied column. For this reason, spiral columns are used in areas with high seismicity to prevent brittle failures. The stress-strain curve below shows the relative yield strength of a tied column and a spiral column are similar, however the strain capacity of the spiral column is much greater than that of a tied column.

Spiral column Yield

Stress

Fracture

Fracture

Tied column

Strain


Column Load Capacity – Small eccentricity: The ratio of applied moment to the applied axial load is its “eccentricity.” A small eccentricity “e” means small moment, where a large eccentricity means large moment.

Pu = applied factored axial load

Pu

e

e

Side view of column

h

h

Concentric load (e = small) Mu = Pue

Eccentric load (e = large) Mu = Pue

= Small

= Large


1) Tied columns with small eccentricity: Small eccentricity if

e ≤ 0.10 h

φPn = Usable axial strength of tied column, KIPS = 0.80φc[0.85f’c(Ag-As)+fyAs] where: φc = strength reduction factor for column = 0.65 for tied columns Ag = gross cross-sectional area of column, in2 As = total area of vertical steel bars, in2 = minimum of 4 vertical bars = 0.01Ag → 0.08Ag f’c = specified concrete compressive strength, KSI fy = yield strength of vertical steel bars, KSI 2) Spiral columns with small eccentricity: Small eccentricity if

e ≤ 0.10 h

φPn = Usable axial strength of tied column, KIPS = 0.85φc[0.85f’c(Ag-As)+fyAs] where: φc = strength reduction factor for column = 0.70 for spiral columns Ag = gross cross-sectional area of column, in2 As = total area of vertical steel bars, in2 = minimum of 6 vertical bars = 0.01Ag → 0.08Ag f’c = specified concrete compressive strength, KSI fy = yield strength of vertical steel bars, KSI


Spiral requirements:

⎛ Ag ⎞ f' − 1⎟⎟ c ⎝ Ac ⎠ fy

ρ s = 0.45⎜⎜ s=

4 As (d c − d b )

ρsdc2 where: ρs = ratio of spiral bar to gross area s = spacing of spiral bends (pitch), in Ag = gross area of column, in2 Ac = area of core, in2 = area measured out-to-out of spiral As = area of spiral bar, in2 dc = diameter of core measured out-to-out of spiral db = diameter of spiral bar


Example 1 GIVEN: A 16” x 16” square tied concrete column has an applied factored axial load Pu = 300 KIPS and an applied factored moment Mu = 35 KIP-FT. In addition, use the following:

• • • •

Concrete f’c = 4000 PSI 8 - #7 grade 60 vertical bars #3 individual tie bars Concrete cover = 1½”

REQUIRED: 1) Determine the eccentricity “e” and state whether or not it is considered to be “small eccentricity.” 2) Determine the required tie spacing. 3) Determine if the vertical bars are acceptable based on ACI requirements. 4) Determine the usable axial strength of the tied column, φPn, and state whether or not it is acceptable. 16”

16”

8 - #7 vertical bars Step 1 – Determine eccentricity and whether it is “small.” Eccentricity e =

=

M u (12" / ft ) Pu

35 KIP − FT (12" / ft ) 300 KIPS

e = 1.4” Small eccentricity if e/h < 0.10 e/h = 1.4”/(16”) e/h = 0.0875 → it is small eccentricity


Step 2 – Determine the required tie spacing:

Least column dimension = 16” Tie Spacing = smaller of

16 x Vert. Bar Dia. = 16(⅞”) = 14” ← USE 48 x Tie Bar Dia. = 48(⅜”) = 18”

Step 3 – Determine if vertical bars are acceptable based on ACI: 8 vertical bars are used > 4 bars OK As = 8 bars(0.60 in2 per #7 bar) = 4.8 in2 Min. As = 0.01Ag = 0.01(16” x 16”) = 2.56 in2 < 4.8 in2 OK Max. As = 0.08 Ag = 0.08(16” x 16”) = 20.48 in2 > 4.8 in2 OK Step 4 - Determine the usable axial strength of the tied column, φPn: Since e/h < 0.10 then the formula can be used:

φPn = Usable axial strength of tied column, KIPS = 0.80φc[0.85f’c(Ag-As)+fyAs] where: φc = 0.65 since it is a tied column Ag = 16” x 16” = 256 in2 As = 8 bars(0.60 in2 per #7 bar) = 4.8 in2 = 0.80(0.65)[0.85(4 KSI)(256 in2 – 4.8 in2) + (60 KSI)(4.8 in2)]

φPn = 594 KIPS Since φPn = 594 KIPS > 300 KIPS → column is acceptable


Example 2 GIVEN: A 14” diameter spiral column has a factored axial load Pu = 400 KIPS and a factored moment = 45 KIP-FT. In addition, use the following:

• • • •

Concrete f’c = 4000 PSI 6 - #9 grade 60 vertical bars #3 spiral bar Concrete cover = 1½”

REQUIRED: 1) Determine the eccentricity “e” and state whether or not it is considered to be “small eccentricity.” 2) Determine the required spiral pitch. 3) Determine the usable axial strength of the spiral column, φPn, and state whether or not it is acceptable.


Step 1 – Determine eccentricity and whether it is “small.” Eccentricity e =

=

M u (12" / ft ) Pu

45 KIP − FT (12" / ft ) 400 KIPS

e = 1.35” e < 0.10 h

Small eccentricity if

e 1.35" = h 14" e/h = 0.096 → it is small eccentricity Step 2 – Determine the required spiral pitch “s”: ⎞ f' ⎛ Ag − 1⎟⎟ c ⎠ fy ⎝ Ac

ρ s = 0.45⎜⎜

where: Ag =

=

π

(Outside _ Dia ) 2

4

π 4

Pitch

(14" ) 2

Ag = 153.9 in2 Ac =

=

π 4

π 4

(Core _ Dia ) 2

(11" ) 2

Ac = 95.0 in2


⎞ f' ⎛ Ag − 1⎟⎟ c ⎠ fy ⎝ Ac

ρ s = 0.45⎜⎜

⎞ 4 KSI ⎛ 153.9in 2 − 1⎟⎟ 2 ⎠ 60 KSI ⎝ 95.0in

ρ s = 0.45⎜⎜

ρs = 0.0186 Pitch = s s=

4 As (d c − d b )

ρsdc2

⎛ 3⎞ 4(0.11in 2 )(11"−⎜ ⎟" ) ⎝8⎠ s= (0.0186)(11" ) 2

Pitch s = 2.08” Use pitch = 2”


Step 3 - Determine the usable axial strength of the spiral column, φPn: Since e/h < 0.10 then the formula can be used:

φPn = Usable axial strength of spiral column, KIPS = 0.85φc[0.85f’c(Ag-As)+fyAs] where: φc = 0.70 since it is a spiral column Ag = 153.9 in2 As = 6 bars(1.00 in2 per #9 bar) = 6.0 in2 = 0.85(0.70)[0.85(4 KSI)(153.9 in2 – 6.0 in2) + (60 KSI)(6.0 in2)]

φPn = 513 KIPS Since φPn = 513 KIPS > 400 KIPS → column is acceptable


Lecture 34 – Columns (cont.) In the previous lecture, we talked about columns having small eccentricity (i.e., small applied moment). While this may be the case for interior columns with offsetting moments, the majority of concrete columns do experience applied moments. Concrete framed buildings typically have columns that are poured monolithically with beams and slabs, thus creating a statically-indeterminate frame such as the one shown below:

The analysis of such a frame is usually quite complex and requires computer software such as STAAD or approximate analysis methods such as the Portal Method and others discussed in AECT 210 – Structural Theory Lecture 17.


Axial compression capacity

The compression capacity of a reinforced column is reduced by the bending stresses on the column and vice-versa. A graph of the axial load capacity of a column vs. the moment capacity of a typical column is shown below (from Lecture 33):

Pure compression (no applied moment)

Pure bending (no applied axial load) Bending moment capacity

Determining points along the curve is quite laborious and typically not done using hand calculations. Instead, computer programs or design guides are used to perform column analysis and design. Below are some “Column Interaction Diagrams” that are used for column analysis and design.


Rectangular TIED Column Interaction Diagrams


Circular SPIRAL Column Interaction Diagrams


“Short” Column Design: Short columns are not considered to be susceptible to the effects of buckling as are long columns. The ACI dictates that short columns satisfy the slenderness ratio as shown below: Short column IF

KL ≤ 22 rmin

Where: K = end fixity factor

L = unbraced length in inches rmin = least radius of gyration, inches = 0.3h for rectangular or square columns = 0.25dout for circular columns where dout = outer dia.


Example GIVEN: A 20” x 20” square interior tied column is shown below. Use concrete f’c = 4000 PSI and 8 - #8 grade 60 vertical bars and #3 ties. Concrete cover = 1½”. All loads are factored and includes beam weight. Assume the moment from the 2 w L beams acting on the column is u n where Ln = clear span of attached beam, 12 and wu = factored unif. load. REQUIRED: 1) Determine the applied factored axial load, Pu on the middle column. Be sure to add column weight. Assume the loads from the roof above are 75% of the floor load. KL 2) Determine the slenderness ratio using K = 0.7 and determine if rmin column qualifies as “short.” 3) Determine total factored moments Mtotal applied to the column. 4) Determine if the column is acceptable based on “Column Interaction Diagram.” 22’-0” 46 K

46 K wu = 2.5 KLF

wu = 1.7 KLF

14’-0”

30’-0”

Beam 1

Beam 2

20” x 20” Middle column


Step 1 – Determine the total factored axial load, Pu on the middle column:

a) Beam 1: End reaction = ½(46 K + 46 K + 1.7 KLF(22’-0”)) = 64.7 KIPS b) Beam 2: End reaction = ½(2.5 KLF(30’-0”)) = 37.5 KIPS c) Column weight: Weight =

20" x 20" (14'−0" )(150 PCF ) 144in 2 / ft 2

= 5833 Lbs. = 5.8 KIPS Total Pu = 1.75(64.7 KIPS + 37.5 KIPS + 5.8 KIPS) Total Pu = 189 KIPS Step 2 - Determine the slenderness ratio

KL using K = 0.7: rmin

(0.7)(14'−0" x12" / ft ) KL = rmin 0.3(20" ) KL = 19.6 rmin Since KL/rmin < 22 → it is a “short” column


Step 3 – Determine total factored moment Mtotal applied to the column: a) Beam 1: 22’-0” 46 K

46 K

wu = 1.7 KLF

Ln = 20.33’

2

w L To determine Mmax = u n we must convert point loads into 12 equivalent uniform load wu. From AISC LRFD Manual p. 5160:


wu = wuniform + wpoint = 1.7 KLF +

fP L

= 1.7 KLF +

2.667(46 KIPS ) 22'−0"

= 7.3 KLF


Mmax =

=

wu Ln 12

2

7.3KLF (20.33' ) 2 12

MBeam 1 = 251 KIP-FT b) Beam 2:

30’-0” wu = 2.5 KLF

Ln = 28.33’

w L Mmax = u n 12 =

2

2.5KLF (28.33' ) 2 12

MBeam 2 = 167 KIP-FT c) Determine Mtotal: Since these moments are offsetting each other, Mtotal = MBeam 1 – MBeam 2 = 251 – 167 Mtotal = 84 KIP-FT


Step 4 - Determine if the column is acceptable based on “Column Interaction Diagram.” a) Determine e/h ratio: e = eccentricity =

M total (12" / ft ) Pu

=

84 KIP − FT (12" / ft ) 189 KIPS

e = 5.33”

e 5.33" = h 20" e = 0.27 > 0.10 → CANNOT use small eccentricity formula h b) Determine γ: h = 20” X

λh

X

20”

8 - #8 vertical bars X = concrete cover + stirrup dia. + ½(vert. bar dia.) = 1½” + ⅜” + ½(8/8”) = 2.375”

γh = 20” – (X + X) = 20” – (2.375” + 2.375”) = 15.25” γ(20”) = 15.25” γ = 0.76 → USE γ = 0.75 Lecture 34 - Page 11 of 12

c) Use Interaction Diagram R4-60.75:

φPn Ag

φPn Ag

φPn Ag

=

Pu Ag

=

189 KIPS 20" x 20"

= 0.47 KSI

x

e Pu e x = h Ag h = (0.47 KSI)(0.27)

φPn Ag

x

e = 0.13 KSI h ρg =

As Ag

2 = 8 _ bars (0.79in _ per _#8 _ bar )

20" x 20"

ρg = 0.016 > 0.01 → col. is acceptable Lecture 34 - Page 12 of 12

Lecture 35 – Wall Footings All structural load-bearing walls must bear on footings. These footings in turn distribute the loads to the soil. If the loading on the soil exceeds the soil bearing capacity, undesirable settlement will occur, resulting in cracked walls, uneven floors, or potentially structural failure in extreme overloading. Wall Footing Requirements: 1) Must distribute the loads evenly to the soil. 2) Bottom of footing must be deeper than the frost penetration or else frost heave will occur. 3) Must be well-drained to prevent wash-out. 4) Must bear on undisturbed (or engineered) soil.


Typical Allowable Soil Bearing Values per IBC Table 1804.2 Allowable Soil Bearing (qa)*:

Soil Type: Bedrock Sedimentary and foliated rock Sandy gravel and/or gravel

12,000 PSF 4,000 PSF 3,000 PSF Sand, silty sand, clayey sand, silty gravel, clayey gravel 2,000 PSF Clay, sandy clay, silty clay, clayey silt, silt, sandy silt 1,500 PSF Organics, peat, top soil NOT Recommended for use as footing bearing * = An increase of one-third is permitted when using alternate load combinations in IBC Section 1605.3.2 that include wind or earthquake loads. Example GIVEN: An 8” wide load-bearing poured-in-place concrete wall is to bear on a wall footing with the following: • • • • • • • • •

Service dead load on wall (incl. wall weight) = 8000 PLF Service live load on wall = 6000 PLF Allowable soil bearing = qa = 3000 PSF Depth to frost = 3’-0” below finished grade Soil unit weight = 110 PCF Concrete f’c = 3000 PSI Main steel tension reinforcing = #5 grade 60 bars Longitudinal shrinkage/temperature bars = #4 grade 60 bars Concrete cover = 3” per ACI

REQUIRED: Design the wall footing, including dimensions, and bar requirements. Step 1 – Determine preliminary thickness of footing: Preliminary footing thickness ≈ 1½(Wall width) ≈ 1½(8”) ≈ 12” TRY Preliminary footing thickness = 12”


Step 2 – Determine effective soil bearing, qe: qe = effective soil bearing = reduced soil bearing capacity resulting from weight of footing and weight of soil above footing (overburden) = (Allowable soil bearing) – (weight of footing) – (weight of soil) = qa – ftg. wt. – soil wt. Soil overburden

2’-0”

1’-0”

Depth to frost = 3’-0”

Finished grade

qe qe = qa – (ftg. wt.) – (soil wt.) = 3000 PSF – (1’-0” thick(150 lb/ft3)) + ((2’-0” thick(110 lb/ft3)) = 3000 PSF – 150 PSF – 220 PSF qe = 2630 PSF


Step 3 – Determine required width of footing, wf:

Ptotal = PDL + PLL = 8000 PLF + 6000 PLF = 14,000 PLF

qe = 2630 PSF Footing width = wf

Footing width = wf =

wf =

Ptotal qe

14000 PLF 2630 PSF

wf = 5.32 feet USE footing width = 5’-6”


Step 4 – Determine soil bearing pressure for strength design, qu: qu = Factored soil bearing pressure acting on footing =

1.2( PDL ) + 1.6( PLL ) wf

=

1.2(8000 PLF ) + 1.6(6000 PLF ) 5.50 ft

qu = 3490 PSF Step 5 – Determine minimum depth of footing, dmin based on shear:

Footing thickness

d

#5 bar d = footing thickness – concrete cover – ½(Bar dia.) = 12” – 3” – ½(5/8”) = 8.69” ⎛ w f Wall _ Width d ⎞ Vu = qu ⎜⎜ − − ⎟⎟ 2 2 2⎠ ⎝

8.69" 8" ⎛ ⎜ 5.5' 12" / ft 12" / ft = 3490 PSF ⎜ − − ⎜ 2 2 2 ⎜ ⎝

⎞ ⎟ ⎟ ⎟ ⎟ ⎠

Vu = 7170 Lbs.


dmin = Minimum depth to reinforcement based on shear =

=

Vu (0.85)2 f ' c (12" ) 7170 Lbs. (0.85)2 3000 PSI (12" )

dmin = 6.4” < 8.69” OK Step 6 – Determine factored soil pressure moment, Mu at face of wall:

Footing overhang = 2.42’ 0.67’

Point “A”

qu = 3490 PSF Isolate footing overhang and treat as if it were a cantilevered beam

Footing width = 5.5 ft.

MA = Moment at point “A”

⎛ Footing _ overhang ⎞ = qu(Footing overhang) ⎜ ⎟ 2 ⎠ ⎝

⎛ 2.42' ⎞ = 3490 PSF(2.42’) ⎜ ⎟ ⎝ 2 ⎠ MA = 10,219 FT-LB


Step 7 – Determine required tension steel, As in footing: Recalling from Lecture 24 – Design Aid Table 1:

Concrete f’c = 3000 PSI, Grade 60 bars Mu (10219 Ft − Lb)(12" / ft ) = 2 φbd 0.9(12" )(8.69" ) 2

= 150.4 PSI < 190.3 PSI Use ρmin = 0.0033 As = ρbd = (0.0033)(12”)(8.69”) As = 0.34 in2 per 1’-0” length of footing


Step 8 – Determine spacing of #5 tension bars: ⎛ A _ per _ bar ⎞ ⎟⎟ Spacing = 12" ⎜⎜ s As ⎠ ⎝ ⎛ 0.31in 2 _ per _#5 _ bar ⎞ ⎟⎟ = 12" ⎜⎜ 0.34in 2 ⎠ ⎝

= 10.9” Use #5 bars spacing = 10” o.c. Step 9 – Determine longitudinal shrinkage/temperature bars:

h = 12”

Footing width “b” = 5.5 ft.

As temp = 0.0018bh = 0.0018(5.5’ x 12”/ft)(12”) = 1.43 in2 Number of bars =

=

As As _ per _ bar

1.43in 2 0.20in 2 _ per _#4 _ bar

= 7.15 bars Use 8 - #4 longitudinal temperature/shrinkage bars



3’-0” minimum

8” poured-in-place conc. wall centered over footing

8 - #4 longitudinal bars

3” conc. cover

12” #5 bars @ 10” o.c.

NOTES: 1) 2) 3) 4)

Finished grade

Conc. f’c = 3000 PSI. All bars grade 60 Wall reinf. & dowels not shown Allow. Soil bearing = 3000 PSF

5’-6”


Lecture 36 – Column Footings All columns must bear on footings. Usually these footings are square or circular for ease of design and formwork. The same general requirements and design fundamentals pertain to column footings as for wall footings, i.e., Column Footing Requirements: 1) Must distribute the loads evenly to the soil. 2) Bottom of footing must be deeper than the frost penetration or else frost heave will occur. 3) Must be well-drained to prevent wash-out. 4) Must bear on undisturbed (or engineered) soil. A typical column footing is shown below:


Example GIVEN: A steel column rests on a 12” x 12” base plate and has the following: • • • • • • • •

PDEAD = 120 KIPS (Service Load) PLIVE = 100 KIPS (Service Load) Concrete f’c = 3000 PSI Depth to frost = 4’-0” Allowable soil bearing = qa = 4000 PSF Soil unit weight = 100 PCF Use #7 grade 60 bars for all bars Concrete cover = 3” per ACI

REQUIRED: Design the square concrete footing, including all dimensions and bars. Provide a “Summary Sketch” showing all information necessary to build it. Step 1 – Determine “Trial” footing thickness: Assume footing thickness ≈ 1¼(Column Width OR Base Plate Width) ≈ 1¼(12”) TRY Footing Thickness = 15” Step 2 – Determine effective soil bearing qe: qe = effective soil bearing pressure = reduced soil bearing due to wt. of footing & soil above footing = qa – (wt. of footing) – (wt. of soil) Soil overburden

2’-9”

1’-3”

Depth to frost = 4’-0”

Finished grade

qe qe = 4000 PSF – (1.25’(150 PCF)) – (2.75’(100 PCF)) = 3538 PSF = 3.5 KSF


Step 3 – Determine length & width of footing: Area of footing =

=

=

Ptotal qe PDEAD + PLIVE qe

(120 KIPS ) + (100 KIPS ) 3.5KSF

= 62.9 ft2 Length x Width = 62.9 ft2 Since Length = Width (Length)2 = 62.9 ft2 Length = 7.93 feet Use 8’-0” x 8’-0” square footing → Area = 64 ft2 > 62.9 ft2 Step 4 – Determine bearing pressure for strength design, qu: qu =

=

1.2( PDEAD ) + 1.6( PLIVE ) Footing _ Area 1.2(120 KIPS ) + 1.6(100 KIPS ) 64 ft 2

qu = 4.75 kips/ft2


Step 5 – Determine depth to tension bars “d”:

Footing thickness = 15”

d

#7 bar d = Footing thickness – (concrete cover) – ½(bar diameter) = 15” – 3” – ½(7/8”) = 11.56” Step 6 – Determine “d” required to resist 2-way punching shear:

Footing thickness = 15”

d

d 2

Col. width

bo 4 bo d = col. width + 2( ) 4 2

= 12” + 2(

11.56" ) 2

= 23.56” = 1.96 feet bo = 94.24”


d 2

Vu2 = 2-way punching shear 2

⎛b ⎞ = qu(Footing area - ⎜ o ⎟ ) ⎝4⎠

= 4.75 KSF(64 ft2 - (1.96') ) 2

= 285.8 KIPS = 285,800 Lbs. dreqd = required depth to tension steel for shear =

=

Vu 2 (0.85)4 f ' c bo 285,800 Lbs. (0.85)4 3000 PSI (94.24" )

dreqd = 16.3” Since dreqd = 16.3” > 11.56” → must increase footing thickness Revised footing thickness = h = dreqd + 3” + ½(7/8”) = 16.3” + 3” + ½(7/8”) = 19.73” Revised footing thickness = 20”


Step 7 – Determine moment on footing overhang, Mu: Footing overhang

Column baseplate

8’-0”

12”

3’-6”

3’-6”

8’-0”

Mu = (Footing width)(qu)

= (8’-0”)(4.75KSF)

( Footing _ Overhang ) 2 2

(3.5' ) 2 2

= 232.8 Kip-Ft


Step 8 – Determine area of tension steel As:

Mu = φbd 2

232.8Kip − Ft (12000

Lb − Inch ) Kip − Ft

1 ⎡ ⎤ 0.9(8'−0" x12" / ft ) ⎢20"−3"− (7 / 8" )⎥ 2 ⎣ ⎦

2

d = 117.9 PSI From Lecture 24 Table at f’c = 3000 PSI → use ρmin = 0.0033 since 117.9 PSI < 190.3 PSI. As = ρbd = (0.0033)(8’-0” x 12”/ft)(16.56”) As = 5.25 in2 Step 9 – Determine tension bar layout: Number of bars =

As As _ per _ bar

5.25in 2 = 0.60in 2 _ per _#7bar = 8.75 bars Use 9 - #7 bars both ways at bottom of footing



4’-0” minimum

Steel column on 12” x 12” base plate centered over footing

9 - #7 bars both ways

3” conc. cover

20”

NOTES: 1) 2) 3) 4) 5)

Finished grade

Conc. f’c = 3000 PSI. All bars grade 60 Allow. Soil bearing = 4000 PSF Anchor rods not shown for clarity Column protection not shown for clarity

8’-0”


Lecture 37 – Load-Bearing Walls The ACI recognizes the “Empirical Design Method” for the design of loadbearing concrete walls having the following limitations:

Load-bearing Wall

1. Wall must be solid, with a rectangular cross-section Lu 25

2. Minimum wall thickness “tw” = larger of

4” ← Interior walls only 7½” ← Exterior walls

3. Applied load eccentricity “e” (including lateral load moments) < 4. Wall must be adequately anchored at top and bottom


tw 6

5. Must have horizontal & vertical reinforcement, with 2 rows of reinforcing for walls with tw > 10” Vert. bars As = 0.0012(tw)(12”) #3, #4, #5 grade 60 bars →

Horz. bars As = 0.0020(tw)(12”) Vert. bars As = 0.0015(tw)(12”)

#6 and bigger grade 60 bars →

Horz. bars As = 0.0025(tw)(12”) 3tw

Maximum bar spacing = smaller of

or 18”

6. Effective length of wall for beam reaction = Leff Center-to-center spacing of beams Leff = smaller of Width of bearing + 4tw 7. Design factored axial strength of wall = Pn ⎡ ⎛ KL ⎞ 2 ⎤ u ⎟⎟ ⎥ Pn = 0.55φf ' c Ag ⎢1 − ⎜⎜ ⎢⎣ ⎝ 32t w ⎠ ⎥⎦

where: φ = 0.70 f’c = specified concrete strength, PSI K = end fixity of walls = 1.0 for pinned-pinned = 0.70 for fixed-pinned Ag = Gross effective area of wall section, in2 = Leff x tw Lu = unbraced height of wall, inches tw = thickness of wall, inches


Example GIVEN: A poured-in-place concrete wall supports W18x35 steel beams spaced 6’-0” apart and rests on 10” wide steel bearing plates. Use the following: •

Beam end reaction = 22 KIPS Service Dead Load (Not incl. wall wt.) = 15 KIPS Service Live Load

• • • •

Wind pressure on wall = 25 PSF (service load) Concrete f’c = 4000 PSI Use #5 grade 60 vertical and horizontal bars Assume “K” = 1.0 for wall end fixity

REQUIRED: Design the wall using the Empirical Design Method. PDEAD = 22 KIPS PLIVE = 15 KIPS

Wall

Lu = 13’-0”

10” bearing plate

Wind = 25 PSF

Steel beam

Fin. Grade Slab on grade


Step 1 – Determine “Trial” thickness of wall, tw: Lu 13'−0" (12" / ft ) = = 6.24” 25 25

Minimum wall thickness “tw” = larger of

4” 7½” ← Largest

TRY tw = 8” which is > 7½” Step 2 – Determine maximum SERVICE wind moment Mwind on wall: Assume a 1’-0” wide “strip” of wall: Mwind =

wL2 8

(25PSFx1'−0" )(13' ) 2 = 8 Mwind = 528 Ft-Lb per 1’-0” length of wall = 528 Ft-Lb(6’-0”) for beam spacing = 3168 Ft-Lb per 6’-0” length of wall Step 3 – Determine the maximum SERVICE vertical load on wall Ptotal: Ptotal = PDEAD + PLIVE = (PDEAD + wall weight) + PLIVE ⎛ 8" ⎞ = (22 KIPS + (6’-0” ⎜⎜ 0.150 KCF ) ⎟⎟13'−0" )) + 15 KIPS ⎝ 12" / ft ⎠ = 29.8 KIPS + 15 KIPS Ptotal = 44.8 KIPS


Step 4 – Determine if e/h ratio is acceptable for Empirical Design: e=

=

M P

Service wind load moment

3.168Kip − Ft (12" / ft ) 44.8KIPS

Service axial load (see above)

= 0.85 inches The empirical design method dictates that e < tw 8" = 6 6

tw 6

Wall thickness

tw = 1.33” > 0.85” → OK to use Empirical Design Method 6

Step 5 – Determine applied factored axial load, Pu: Pu = 1.2(PDEAD) + 1.6(PLIVE) = 1.2(29.8 KIPS) + 1.6(15 KIPS) = 59.8 KIPS Step 6 – Determine factored bearing strength of concrete: A1 = bearing contact area from beam, in2 = (bearing plate width)(tw) = (10”)(8”) = 80 in2 Concrete bearing capacity = 0.9(0.85f’cA1) = 0.9(0.85(4000 PSI)(80 in2)) Concrete bearing capacity = 244,000 Lbs > 59.8 KIPS → OK


Step 7 – Determine design axial strength of wall Pn: Leff = smaller of

Center-to-center spacing of beams = 6’(12”/ft) = 72” Width of bearing + 4tw = 10” + 4(8”) = 42” ← USE

Ag = Gross effective area of wall section, in2 = Leff x tw = (42”)(8”) = 336 in2 ⎡ ⎛ KL ⎞ 2 ⎤ u ⎟⎟ ⎥ Pn = 0.55φf ' c Ag ⎢1 − ⎜⎜ ⎢⎣ ⎝ 32t w ⎠ ⎥⎦

⎡ ⎛ (1.0)(13' x12" / ft ) ⎞ 2 ⎤ ⎟⎟ ⎥ = 0.55(0.70)(4000 PSI )(336in ) ⎢1 − ⎜⎜ 32(8" ) ⎢⎣ ⎝ ⎠ ⎥⎦ 2

Pn = 325,300 Lbs. > 59.8 KIPS → wall is acceptable Step 8 – Determine vertical and horizontal bars: #3, #4, #5 grade 60 bars →

Vert. bars As = 0.0012(tw)(12”) Horz. bars As = 0.0020(tw)(12”)

a) Vert. bars → As = 0.0012(8”)(12”) = 0.115 in2 per 1’-0” length of wall ⎛ 0.31in 2 _ per _#5 _ bar ⎞ ⎟⎟ spacing = 12” ⎜⎜ 0.115in 2 ⎠ ⎝

= 32.3” 3tw = 3(8”) = 24” Maximum bar spacing = smaller of

or 18” ← USE

USE #5 vertical bars @ 18” o.c.


b) Horizontal bars → As = 0.0020(tw)(12”) = 0.0020(8”)(12”) = 0.192 in2 ⎛ 0.31in 2 _ per _#5 _ bar ⎞ ⎟⎟ spacing = 12” ⎜⎜ 0.192in 2 ⎠ ⎝

= 19.4” 3tw = 3(8”) = 24” or

Maximum bar spacing = smaller of

18” ← USE USE #5 horizontal bars @ 18” o.c. Step 9 – Draw “Summary Sketch”:

18” Lu = 13’-0”

W18x35 Steel beam 10” bearing plate

8” thick concrete Wall

#5 @ 18” ea. way

Slab on grade

NOTES:

Fin. Grade

1) Conc. f’c = 4000 PSI. 2) All bars grade 60 3) Footing dowels not shown


Lecture 39 – Prestressed Concrete Prestressed concrete refers to concrete that has applied stresses induced into the member. Typically, wires or “tendons” are stretched and then blocked at the ends creating compressive stresses throughout the member’s entire crosssection. Most Prestressed concrete is precast in a plant. Advantages of Prestressed concrete vs. non-Prestressed concrete: • • • •

More efficient members (i.e., smaller members to carry same loads) Much less cracking since member is almost entirely in compression Precast members have very good quality control Precast members offer rapid field erection

Disadvantages of Prestressed concrete vs. non-Prestressed concrete: • • • • •

More expensive in materials, fabrication, delivery Heavy precast members require large cranes Somewhat limited design flexibility Small margin for error More complicated design

Typical Precast Prestressed concrete members


Pre-Tensioned Prestressed Concrete: Pre-tensioned concrete is almost always done in a precast plant. A pretensioned Prestressed concrete member is cast in a preformed casting bed. The BONDED wires (tendons) are tensioned prior to the concrete hardening. After the concrete hardens to approximately 75% of the specified compressive strength f’c, the tendons are released and axial compressive load is then transmitted to the cross-section of the member. Tendons tensioned between bulkheads

“Dead” end

Casting bed

“Live” end

Prestress force Ps

Step 1

Fresh concrete placed in bed

Tendons anchored at “Live” end and “Dead” end

Step 2 Hardened concrete Prestress force Ps Tendons released at “Live” end and “Dead” end creating an axial force along length of precast member


Step 3

Post-Tensioned Prestressed Concrete: A post-tensioned member has UNCOATED tendons cast into concrete in draped patterns. After the concrete hardens to about 75% f’c, the tendons are tensioned and try to straighten out. This creates an upward camber of the member which offsets anticipated downward deflection due to gravity loads. Post-tensioning can be accomplished on-site as necessary.

Tendons tensioned between bulkheads Prestress force Ps “Dead” end

“Live” end

Step 1

Tendons creating an upward force along length of member

Tendons anchored at “Live” end

“Dead” end

Step 2


Analysis of Rectangular Prestressed Members: The analysis of a member is typically done for various stages of loading under SERVICE LOADS. Stresses “f” are obtained as follows: f=

Ps Ps ey ± Ag Ig

where: Ps = prestress force Ag = gross cross-sectional area of member e = eccentric distance between prestressing tendons and member centroid y = distance from centroid to extreme edge of member Ig = gross moment of inertia of member about N.A.

Mu = 0.9Apsfps(dp -

a ) 2

where: Mu = usable moment capacity of prestressed beam Aps = area of prestressed tendons

⎛ ⎡γ p f pu ⎤ ⎞ fps = f pu ⎜⎜1 − ⎢ ρ p ⎥ ⎟⎟ ' β f c ⎦⎠ ⎝ ⎣ 1 fpu = ultimate tensile strength of prestressing tendon

γp = factor based on the type of prestressing steel = 0.40 for ordinary wire strand = 0.28 for low-relaxation wire strand β1 = 0.85 for concrete f’c = 4000 PSI = 0.80 for concrete f’c = 5000 PSI ρp =

A ps bd p


a=

A ps f ps

0.85 f ' c b b

yt h

dp N.A. e

yb

Aps Rectangular Prestressed Beam


Example GIVEN: The rectangular prestressed concrete beam as shown below. Use the following:

• • • • • • • •

Concrete f’c = 5000 PSI Concrete strength = 75%(f’c) at time of prestressing Aps = 3 – ½” dia. 7-wire strands @ 0.153 in2 per strand = 0.459 in2 fpu = 270 KSI (using an ordinary 7-wire strand) Initial prestress force, Ps = 70%(fpu)(Aps) Service dead load, (NOT including beam weight) = 400 PLF Service beam weight = 188 PLF Service live load = 1500 PLF

REQUIRED: 1) Determine the location of the neutral axis and prestress eccentricity “e”. 2) Determine the moment of inertia about the neutral axis, Ig. 3) Determine the stresses during prestressing. 4) Determine the stresses during initial applied service beam weight. 5) Determine the stresses due to service applied dead load + live load. 6) Determine the final stresses due to all service loads and prestressing. 7) Determine the maximum actual factored moment on the beam Mmax. 8) Determine the factored usable moment capacity Mu of the beam.

22’-0” 10”

yt 18”

16” N.A. e

yb

Aps = 0.459 in2 Rectangular Prestressed Beam Lecture 39 - Page 6 of 12

Step 1 - Determine the location of the neutral axis and prestress eccentricity “e”: Using a datum as measured from the top of the beam: 10”

Datum 18”/2 = 9”

yt

dp =16”

h = 18” e

Yb

nAps = 3.30 in2

n=

=

E steel E conc

29,000,000 PSI 57,000 f ' c

= 7.2 nAps = 7.2(0.459 in2) = 3.30 in2

9" (10" x18" ) + 16" (3.30in 2 ) yt = (10" x18" ) + (3.30in 2 ) yt = 9.13” yb = 18” – 9.13” yb = 8.87” e = dp – yt = 16” – 9.13” e = 6.87”


Step 2 - Determine the moment of inertia about the neutral axis, Ig: 2

bh 3 h⎞ ⎛ + bh⎜ y t − ⎟ + nA ps (e) 2 Ig = 12 2⎠ ⎝ 2

=

(10" )(18" ) 3 18" ⎞ ⎛ 2 2 + (10" )(18" )⎜ 9.13"− ⎟ + (3.30in )(6.87" ) 12 2 ⎠ ⎝

= 4860 in4 + 3.0 in4 + 155.7 in4 Ig = 5018.7 in4 Step 3 – Determine the stresses during prestressing: f= −

Ps Ps ey ± Ag Ig where: Ps = prestress force = 70%(fpu)(Aps) = 0.70(270 KSI)(0.459 in2) = 86.8 KIPS y = yt for tensile stresses at top of beam = yb for compressive stresses at bottom of beam

a) Check stresses at TOP of beam: ftop = stress at top of beam =-

Ps Ps ey t + Ag Ig

= −

86.8KIPS (86.8KIPS )(6.87" )(9.13" ) + (10" x18" ) 5018.7in 4

= -0.48 KSI + 1.08 KSI ftop = 0.60 KSI Tension


b) Check stresses at BOTTOM of beam: fbottom = stress at bottom of beam = −

Ps Ps eyb − Ag Ig

= −

86.8KIPS (86.8 KIPS )(6.87" )(8.87" ) − (10" x18" ) 5018.7in 4

= -0.48 KSI - 1.05 KSI fbottom = -1.53 KSI Compression Step 4 - Determine the stresses during initial applied service beam weight: f= ±

M beam ( y ) Ig where: Mbeam = maximum unfactored moment due to beam wt. =

wbeam ( L) 2 8

=

(188PLF )(22'−0" ) 2 8

= 11,374 Lb-Ft = 11.4 KIP-FT y = yt for compression in top = yb for tension in bottom a) Check stresses at TOP: ftop = −

= −

M beam ( y t ) Ig (11.4 KIP − FT (12" / ft ))(9.13" ) 5018.7in 4 ftop = -0.25 KSI Compression


b)

Check stresses at BOTTOM: Fbottom = +

= +

M beam ( y b ) Ig

(11.4 KIP − FT (12" / ft ))(8.87" ) 5018.7in 4 fbottom = 0.24 KSI Tension

Step 5 - Determine the stresses due to service applied dead load + live load: f= ±

M DL + LL ( y ) Ig where: MDL+LL = maximum unfactored moment due to DL+LL =

wDL + LL ( L) 2 8

(400 PLF + 1500 PLF )(22'−0" ) 2 = 8 = 114,950 Lb-Ft = 115.0 KIP-FT y = yt for compression in top = yb for tension in bottom a) Check stresses at TOP: ftop = −

= −

M DL + LL ( y t ) Ig

(115.0 KIP − FT (12" / ft ))(9.13" ) 5018.7in 4 ftop = -2.51 KSI Compression


b) Check stresses at BOTTOM: fbottom = +

= +

M DL + LL ( y b ) Ig

(115.0 KIP − FT (12" / ft ))(8.87" ) 5018.7in 4 fbottom = 2.44 KSI Tension

Step 6 - Determine the final stresses due to all service loads and prestressing:

-0.48C

18”

+1.08T

+

-0.48C

− Ps Ag

-0.25C

+

-1.05C

+ Ps ey Ig

+0.35T

+

=

+0.24T Beam Wt.

-2.51C

-2.16C

=

-1.29C

+2.44T

+1.15T

Transfer

DL + LL

Final

Initial Prestress

Step 7 – Determine the maximum actual factored moment on the beam Mmax:

wu L2 Mmax = 8 wu = 1.2D + 1.6L = 1.2(400 PLF + 188 PLF) + 1.6(1500 PLF) = 3106 PLF = 3.1 KLF Mmax =

3.1(22'−0) 2 8

Mmax = 188 KIP-FT Lecture 39 - Page 11 of 12

Step 8 - Determine the factored usable moment capacity Mu of the beam: a Mu = 0.9Apsfps(dp - ) 2 where: ⎛ ⎡γ p f pu ⎤ ⎞ fps = f pu ⎜1 − ⎢ ρ p ⎥ ⎟⎟ ⎜ β f ' c ⎦⎠ ⎝ ⎣ 1 fpu = ultimate tensile strength of prestressing tendon = 270 KSI

γp = factor based on the type of prestressing steel = 0.40 for ordinary wire strand β1 = 0.80 for concrete f’c = 5000 PSI ρp =

=

A ps bd p 0.453in 2 (10" )(16" )

= 0.00283

⎛ ⎡ 0.40 270 KSI ⎤ ⎞ fps = 270 KSI ⎜⎜1 − ⎢ (0.00283) ⎟ 5KSI ⎥⎦ ⎟⎠ ⎝ ⎣ 0.80 = 249.4 KSI a=

=

A ps f ps 0.85 f ' c b (0.453in 2 )(249.4 KSI ) 0.85(5 KSI )(10" )

= 2.66” Mu = 0.9Apsfps(dp -

a ) 2

= 0.9(0.453 in2)(249.4 KSI)(16” -

2.66" ) 2

= 1492 Kip-In Mu = 124.3 KIP-FT < Mmax = 188 KIP-FT → NOT ACCEPTABLE Lecture 39 - Page 12 of 12

Lecture 4 – Beam Design for Moment Beams, or sometimes referred to as “flexure” members, are designed on the basis of moment. In ASD and LRFD, the design strength of a beam in flexure is called the “nominal flexural moment”, Mn. This Mn must be greater than the maximum applied factored (LRFD) or service (ASD) moment. Beams are designed on the basis of the following LRFD references: • •

AISC Part 3 AISC Spec Chapter F p. 16.1-44

1. Beam Design Considering Yielding: Assuming a beam is adequately laterally braced, it will fail by yielding on the compression flange. Most beams are laterally braced by the metal decking that is attached to the compression flange as shown below: Metal decking puddle-welded to top flange of beam

Concrete slab over metal decking

1) LRFD Beam Design: Design Flexural Strength = φbMn Where: φb = 0.90 Mn = nominal flexural moment = Mp = Plastic moment = FyZx Zx = plastic section modulus = from properties


2) ASD Beam Design: Allowable Flexural Strength =

Mn Ωb

Where: Ωb = 1.67 Mn = nominal flexural moment = Mp = Plastic moment = FyZx Zx = plastic section modulus = from properties Example 1 (LRFD) GIVEN: A W16x26 steel beam using A992 steel is continuously laterally braced, and experiences a FACTORED moment = 104 KIP-FT. REQUIRED: 1) Determine the design flexural moment, φbMn for the beam. 2) Determine if the beam is adequate. Step 1 – Determine φbMn for the beam: φbMn = 0.90(FyZx) since it is continuously laterally braced = 0.90(50 KSI)(44.2 in3) = 1989 KIP-IN From properties, φbMn = 165.8 KIP-FT AISC p. 1-21 Step 2 – Determine if the beam is adequate: Since φbMn = 165.8 KIP-FT > 104 KIP-FT → beam is adequate

Example 2 (LRFD) GIVEN: The W16x26 beam from Example 1. REQUIRED: Determine the design flexural moment, φbMn for the beam using the LRFD “Zx” Table 3-2 (see AISC p. 3-11 thru 3-19) Step 1 – Refer to AISC p. 3-18 for W16x26: Look in the LRFD column φbMpx = 166 KIP-FT


Example 3 (ASD) GIVEN: An A992 steel beam “A” is continuously laterally braced and carries a superimposed SERVICE (i.e., not factored) floor live load = 100 PSF and a superimposed SERVICE dead load = 85 PSF. Assume initially the beam weighs 30 PLF and check your results. REQUIRED: 1) Design the lightest weight steel beam using the “Zx” Table 3-2. 2) Design the lightest weight W14 steel beam using the “Maximum Total Uniform Load Table 3-6” (see AISC p. 3-33 thru 3-95).

Beam “A”

3@7’-0” = 21’-0”

Girder “A”

32’-0”

Step 1 – Determine SERVICE uniform load, w on beam: By inspecting the ASD load factors AISC p. 2-9, the maximum applied uniform load w = D + L Where: D = uniform dead load, PLF = 7’(85 PSF) + 30 PLF = 625 PLF L = uniform live load, PLF = 7’(100 PSF) = 700 PLF w = (625 PLF) + (700 PLF) = 1325 PLF = 1.325 KLF


Assumed beam wt.

Step 2 – Determine maximum SERVICE moment, Ma: w = 1.325 KLF R1 = 21.2 KIPS

R2 = 21.2 KIPS 32’-0”

wL2 Ma = 8 =

1.325 KLF (32' ) 2 8

Ma = 169.6 KIP-FT Step 3 – Select lightest weight beam from Table 3-2: From AISC p. 3-17 look under ASD column

M px

to find a Ωb BOLD moment that is equal or larger than the calculated Ma: Use W16x40 →

M px Ωb

= 182 KIP-FT > 169.6 KIP-FT

→ (NOTE: If the assumed beam weight of 30 PLF were increased to 40 PLF, the revised Ma = 170.9 KIP-FT which is still less than 182 KIP-FT)


Step 4 – Determine total SERVICE uniform load on beam: W = Total service load on beam, KIPS = w x span = 1.325 KLF(32’-0”) = 42.4 KIPS Step 5 – Select lightest W14 beam using Table 3-6 “Maximum Total Uniform Load” tables: From AISC p. 3-67, look find span = 32’ then read across to find the lightest weight beam having maximum ASD total uniform load > 42.4 KIPS. Use W14x43 → max. total unif. load = 43.4 KIPS > 42.4 KIPS

→ (NOTE: Using this method, the lightest possible beam is W16x40 → max. total unif. load = 45.5 KIPS > 42.4 KIPS)


2. Beam Design Considering “Lateral-Torsional Buckling” The compression flange of a beam behaves like a column – it is susceptible to buckling if not adequately laterally braced. This phenomenon of flange buckling of a beam is referred to as “lateraltorsional buckling.” Consider a skinny yard stick under loading. If it is not properly braced, it will twist and fail under a much smaller load than if it were adequately braced. Therefore, wider-flanged beams are better at resisting lateraltorsional buckling than narrow-flanged beams.

Moment capacity

It has been determined that the relationship between distance between lateral bracing and moment capacity looks like the following:

Distance between lateral supports


The AISC has developed graphs of LRFD and ASD beam design Mn moments, φbMn, and respectively, for beam shapes that has been Ωb mathematically altered from the graph above (for simplicity), and looks like the following: (See AISC p. 5-7 and AISC p. 16.1-33)

Moment capacity

φbMp and

φbMr and

Real curve

Mp Ωb

AISC curve

Mr Ωb

Lp

Lr

Distance between lateral supports Where: φb = 0.90 (LRFD) Ωb = 1.67 (ASD) E Fy

AISC p. 16.1-48

E 0.7 Fy

AISC p. 16.1-48

Lp = 1.76ry

Lr = πrts

E = 29000 KSI Fy = yield stress, KSI Mp = (FyZx)

AISC p. 16.1-47

Mr = 0.7(FySx)

AISC p. 16.1-269

Sx = Section modulus about “x” axis from properties


Example 4 (LRFD) GIVEN: A W14x43 steel girder using A992 steel. It is laterally braced at Lb = 8’-0” increments by beams framing into the side. It experiences a maximum FACTORED moment, Mu = 250 KIP-FT. REQUIRED: 1) Draw the graph of design moments vs. unbraced length. 2) Determine the design moment, φbMn for an actual unbraced length Lb = 10’-0” using formula F2-2 from AISC p. 16.1-47. 3) Determine the design moment using the “Beam Available Moments vs. Unbraced Length” graphs on AISC p. 3-125. Step 1 – Draw graph of design moments vs. unbraced length: From AISC p. 3-17, pick values of Lp, Lr, φbMp, and φbMr

Real curve

φbMpx = 261 KIP-FT

AISC curve

φbMn = ???? φbMrx = 164 KIP-FT

Lp = 6.68 ft.

Lr = 20.0 ft. Lb = 10’-0”

Step 2 - Determine the design moment, φbMn for an unbraced length Lb = 8’-0” using formula F2-2 from AISC p. 16.1-32 ⎡ ⎛ Lb − L p ⎞⎤ ⎟ Mn = C b ⎢ M p − ( M p − 0.7 Fy S x )⎜ ⎜ L − L ⎟⎥⎥ ⎢⎣ r p ⎝ ⎠⎦ Where: Cb = 1.0 (conservative) Mp = FyZx = (50KSI)(69.6 in3) = 3480 Kip-In


⎡ 10'−6.68' ⎞ ⎤ Mn = 1.0 ⎢3480 Kip − In − ((3480 − 0.7(50 KSI )(62.6in 3 )⎛⎜ ⎟) ⎥ ⎝ 20'−6.68' ⎠ ⎦ ⎣

= 3159 Kip-In = 263.3 Kip-Ft Therefore, since φ = 0.9: φMn = 0.9(263.3 Kip-Ft) φMn = 237 Kip-Ft Since φMn = 237 Kip-Ft < Mu = 250 Kip-Ft, beam is

UNACCEPTABLE


Step 3 – Determine FACTORED moment strength using Table 3-10:

From AISC p. 3-125: For a W14x43 with Fy = 50 KSI Unbraced length Lb = 10’-0” LRFD

φMn ≅ 237 Kip-Ft

Lb = 10’-0”


Example 5 (ASD) and (LRFD) GIVEN: A steel girder is laterally braced at Lb = 10’-0, and experiences the service loads as shown below. REQUIRED: Design the lightest weight A992 wide-flange beam using the “Available Moment vs. Unbraced Length” graphs. Service DL = 900 PLF (incl. beam wt.) Service LL = 1700 PLF

30’-0”

Step 1 – Determine ASD maximum SERVICE moment: w=D+L = (900 PLF) + (1700 PLF) = 2600 PLF = 2.6 KLF Ma = Maximum applied SERVICE moment wL2 = 8 (2.6 KLF )(30'−0" ) 2 = 8 Ma = 292.5 Kip-Ft.


Step 2 – Determine LRFD maximum FACTORED moment: wu = 1.2D +1.6 L = 1.2(900 PLF) + 1.6(1700 PLF) = 3800 PLF = 3.8 KLF Mu = Maximum applied FACTORED moment w L2 = u 8 (3.8 KLF )(30'−0" ) 2 = 8 Mu = 427.5 Kip-Ft. Step 3 – Design lightest beam using “Available Moment vs. Unbraced Length” graph: From AISC p. 3-121 and 3-119, read up from the bottom at Lb = 10’-0” and look up until you hit a solid line for the lightest weight beam exceeding 292.5 Kip-Ft (ASD) and 427.5 Kip-Ft (LRFD). ASD: Use W21x62 →

LRFD:

Mn ≈ 316 Kip − ft Ω

Use W21x62 → φM n ≈ 474 Kip − ft


Lecture 40 – Concrete Specifications Project-specific construction documents generally consist of two items: • •

Design Drawings Specifications

The Design Drawings graphically present the specific design of the structure. However, they do not indicate the specific requirements relating to: • • • • •

Materials Submittals Job conditions Testing & inspection Execution of work

CSI – Construction Specifications Institute The CSI was founded in 1948 in an effort to organize trade-specific specifications into a uniform, industry accepted format. It developed the “MasterFormat”, a breakdown of all constructionrelated activities into 16 divisions as follows:

Division 1 – General Requirements Division 2 – Site Construction Division 3 – Concrete Division 4 – Masonry Division 5 – Metals Division 6 – Wood and Plastics Division 7 – Thermal and Moisture Protection Division 8 – Doors and Windows Division 9 – Finishes Division 10 – Specialties Division 11 – Equipment Division 12 – Furnishings Division 13 – Special Construction Division 14 – Conveying Systems Division 15 – Mechanical Division 16 – Electrical

03050 – Basic Concrete Materials and Methods 03100 – Concrete Forms and Accessories 03200 – Concrete Reinforcement 03300 – Cast-in-Place Concrete 03400 – Precast Concrete 03500 – Cementitious Decks and Underlayment 03600 – Grouts 03700 – Mass Concrete 03900 – Concrete Restoration and Cleaning

Each division has been further refined into multiple sub-divisions (as shown for Division 3 above). To obtain samples of specifications, go to http://www.ogs.state.ny.us/dnc/masterspec/default.htm In addition to technical specifications, the CSI MasterFormat is used by most of the construction industry for purposes of cost estimating, contractor qualifications, product research and supply ordering.


Section 03300 – Cast-inPlace Concrete PART 1 - GENERAL 1.1

SUMMARY A. Work Included: Cast-in-place concrete.

1.2

QUALITY ASSURANCE A. Comply with the latest edition of the following: 1. ACI 301 "Specifications for Structural Concrete for Buildings" 2. ACI 302 "Guide for Concrete Floor and Slab Construction" 3. ACI 304 "Recommended Practice for Measuring, Mixing, Transporting, Placing Concrete" 4. ACI 305 "Hot Weather Concreting" 5. ACI 306 "Cold Weather Concreting" 6. ACI 311 "Recommended Practice for Concrete Inspection" 7. ACI 315 "Details and Detailing of Concrete Reinforcement" 8. ACI 318 "Building Code Requirements for Reinforced Concrete" 9. ACI 347 "Recommended Practice for Concrete Formwork" 10. ACI SP-15 Field Reference Manual 11. CRSI "Manual of Standard Practice"

1.3

SUBMITTALS A. Shop Drawings: Submit Shop Drawings for fabrication, bending and placement of concrete reinforcement. Show bar schedules, stirrup spacing, diagrams of bent bars, arrangements of concrete reinforcement. Include special reinforcement required for openings through concrete. Show elevations of reinforcement for all members at minimum 1/4 inch = 1'-0" scale. Show locations all construction and control joints. B. Mix Design: Submit proposed mix designs for concrete at least 15 days before start of concreting. Submittal shall include: cement content and type, admixture content and type, aggregate source and gradation, water content, air content, slump, and documentation of average strength by field experience or laboratory prepared trial mixtures in accordance with ACI 318 Article 4.3. C. Product Data: Submit data and installation instructions for proprietary material. D. Material Certificates: Submit materials certificates certifying that each material complies with Specifications.

1.4

TESTING SERVICES A. Employ, at Contractor's expense, a testing laboratory acceptable to Architect to perform material evaluation tests and for quality control during placement. Lecture 40 - Page 2 of 14

B. Sample and test concrete for quality control during placement as follows: 1. Sampling Fresh Concrete: ASTM C172 except modified for slump to comply with ASTM C94. 2. Slump: ASTM C143 - one for each concrete truck, measured at point of discharge. 3. Air Content: ASTM C231 pressure method - one for each truck load of readymixed air-entrained concrete. 4. Temperature: Test concrete temperature hourly when ambient temperature is 40oF and below, and when 80oF and above. 5. Compressive Test Specimen: ASTM C31, one set of six standard cylinders for each compressive strength test. Store undisturbed and in an insulated box during cold weather. Deliver cylinders to lab between 24 and 48 hours after making. 6. Compressive Strength Tests: ASTM C39, one set of six cylinders for each 50 cubic yards or fraction thereof, of each concrete class placed in any one day, two lab specimens tested at 7 days, two lab specimens tested at 28 days and two specimens retained in reserve for later testing if required. C. Test Reports 1. Forward results to Architect, Engineer and Contractor on same day that tests are made. 2. Reports of compressive strength tests shall contain the general information of project identification name and number, date of concrete placement, name of Contractor, name of concrete supplier, truck number and delivery ticket number, name of concrete testing service, concrete type and class, name of individual making specimen, location of concrete batch in structure, design compressive strength at 28 days, concrete mix proportions and materials; and the specific information of slump, air content, temperature, compressive strength and type of break for both 7-day and 28-day tests. 3. Field reports of concrete inspection shall contain general information noted above, plus ambient temperature, concrete temperature, weather, slump, air content and cylinder numbers. D. Additional Testing 1. Testing service shall make additional tests of in-place concrete when test results indicate specified concrete strengths and other characteristics have not been attained in structure. 2. Testing service shall conduct tests to determine adequacy of concrete by cored cylinders complying with ASTM C42 or by other methods acceptable to Architect. 3. Contractor shall pay for such tests conducted, and any other additional testing required, if concrete testing confirms specified strengths have not been met. 1.5

JOB CONDITIONS


A. Store materials so as to ensure preservation of their quality and fitness for the Work. Store reinforcement and formwork in a manner to prevent damage and accumulation of dirt. B. Contractor shall be responsible for correction of concrete work which does not conform to specified requirements, including strength, tolerances and finishes. Correct deficient concrete as directed by Architect. PART 2 - PRODUCTS 2.1

MATERIALS A. Formwork 1. Exposed Concrete: Unless otherwise shown or specified, construct formwork for concrete surfaces, which will be exposed to view in the completed project, with form plywood, metal or other acceptable panel-type material, to provide continuous, straight, smooth exposed surfaces. Furnish in largest practicable sizes to minimize number of joints and to conform to joint system shown on Drawings. Provide form material with sufficient thickness to withstand pressure of newly-placed concrete without bow or deflection. 2. Unexposed Concrete: Form concrete surfaces which will be unexposed to view in the completed Project with plywood, lumber, metal or other acceptable material. Provide lumber dressed on at least 2 edges and 1 side for tight fit. B. Form Ties 1. Exposed Concrete: Plastic cone snaptie, Type 3M by Superior or accepted equal. 2. Unexposed Concrete: Snap-off metal ties, designed to prevent form deflection and prevent spalling surfaces upon removal. Portion remaining after removal shall be at least 1" from concrete surface. C. Form Coatings: Commercial formulation form-coating compounds shall not bond with, stain, nor adversely affect concrete surfaces, and shall not impair subsequent treatments of concrete surfaces requiring bond or adhesion, nor impede wetting of surfaces to be cured with water or curing compound. D. Reinforcement 1. Deformed bars: ASTM A615, Grade 60. 2. Welded Wire Fabric: ASTM A185. Flat sheets only. 3. All chairs, spacers, clips, wire anchors and related items necessary to accurately space and secure reinforcement. 4. Additional bars, if required, to anchor or space reinforcement. 5. Chairs shall be plastic booted at points of bearing on forms for exposed concrete. 6. Minimum 16-gage annealed tie wire, ASTM A82. E. Cement: ASTM C150, Type I or Type II. F. Aggregates: ASTM C33 and as herein specified. Lecture 40 - Page 4 of 14

1.

2.

Fine Aggregate: Clean, sharp, natural sand free from loam, clay, lumps or other deleterious substances with less than 10% passing the #100 sieve and less than 3% passing the #200 sieve. Coarse Aggregate: Clean, uncoated, processed aggregate containing no clay, mud, loam or foreign matter, as follows: a. Crushed stone: Processed from natural rock or stone for concrete slabs meeting NYSDOT 703-0201. b. Clean, sharp, natural or processed gravel, or crushed stone, free from loam, clay, lumps, or other deleterious substances for footings and miscellaneous concrete. c. Maximum Aggregate Size: Footings and walls size 467. Slabs size 57 with minimum of 1800 pounds per cubic yard.

G. Water: Clean, fresh, potable. H. Air Entraining: ASTM C260. I.

Water Reducing Admixture: ASTM C494, Type A.

J.

Non Corrosive, Non Chloride Accelerator: ASTM C494, Type C or E.

K. Prohibited Admixtures: Calcium chloride, thiocyanates. Admixtures containing more than 0.05% chloride ions are not permitted. L. Evaporation Retarder: Confilm by Master Builders, or accepted equal. M. Curing Sheet Materials: ASTM C171, including waterproof paper, polyethylene film or polyethylene coated burlap. N. Liquid Membrane Curing/Sealing Compound: Masterkure by Master Builders or accepted equal. O. Exterior Anti-Spalling Sealer: Penetrating Sealer 40 by Sonneborn or accepted equal. P. Hardener: Lapidolith by Sonneborn or accepted equal. Q. Mineral Aggregate Floor Surface Hardener: Colorcron by Master Builders or accepted equal. French gray color. Apply at rate of 1.00 pounds per square foot. R. Joint Filler: 1/2" thick ASTM D994 premolded expansion joint filler strips; full slab depth. Where joint is exposed and scheduled to receive sealant, provide 1/2" vinyl removable filler cap strip, 940 Series by Green Streak or accepted equal. S. Sleeves: ASTM A120, hot-dipped galvanized. T. Anchor Bolts: Furnished in Section 05100 and installed under this Section.


U. Dowel Bars: 1-inch square steel bars with ¼-inch compressible foam on vertical faces; or 1-inch-diameter steel bars, greased. V. Non-shrink Grout: Sonogrout 14 by Sonneborn, or accepted equal. W. Water Stop: Volclay Waterstop RX, 1" x 3/4" by American Colloid Co. or accepted equal. X. Dovetail Slot: Standard Dovetail Slot #180, 26 gauge galvanized steel with foam filler by Heckmann Building Products or accepted equal. 2.2

PROPORTIONING AND MIX DESIGN A. Prepare design mixes for concrete. Use independent testing facility acceptable to Architect for preparing and reporting proposed mix designs. B. Where the concrete production facility can establish the uniformity of its production for concrete of similar strength and materials based on recent test data, the average strength used as a basis for determining mix design proportions shall exceed the specified design strength by the requirements of ACI-318, Section 4.3.2 or ACI-301, Section 3.9. When a concrete production facility does not have field test records for calculation of standard deviation, the required average strength shall be at least 1,200 psi greater than the specified design strength. C. Concrete Quality Required 28 Location

Maximum

day

Water

Compressive

Cement Ratio

Air Content

Weight

3,000

0.55

4% - 6%

147 - 153 pcf

3,500

0.50

0%

147 - 153 pcf

4,000

0.45

5% - 7%

147 - 153 pcf

4,000

0.45

4% - 6%

114 - 120 pcf

Strength Footings, foundation walls and all other below grade concrete, miscellaneous concrete. Interior slabs on grade and suspended slabs. Exterior concrete subjected to freezing and thawing, exterior slab on grade. Interior light weight suspended slab.

D. Slump 1. Footings and Foundation Walls: 3" to 5". 2. Slabs: 4" maximum. E. Ready Mix Concrete: ASTM C94. Lecture 40 - Page 6 of 14

F. The quantity of coarse aggregate in pounds must be in the range of 1.25 to 1.5 times the quantity of fine aggregate in pounds. G. Fly ash may be substituted for cement for interior slabs only, at a maximum rate of 15 percent by weight. Submittals shall include actual mix design, including percentage of fly ash and test results showing that mix meets specified compressive strength, and air content. Fly ash is not permitted in cold weather concreting unless extended protection is provided. Protection and heat shall be maintained until 70 percent of specified design strength is achieved. H. Pumping of concrete is permitted only if mix designs specifically prepared and used previously for pumping are submitted. Pumpline shall have a 5-inch minimum inside diameter and shall be used with 5-inch pumps. 2.3

REINFORCING FABRICATION A. Fabricate bars to required lengths, shapes and bends. Do not rebend or straighten reinforcement in a manner that shall weaken the material.

2.4

FORMWORK A. Design formwork to support vertical and lateral loads that might be applied until such loads can be supported by concrete structure.

PART 3 - EXECUTION 3.1

INSPECTION A. Examine conditions under which concrete shall be placed. Do not proceed with work until all unsatisfactory conditions are corrected.

3.2

NOTIFICATION A. Notify Architect 24 hours before anticipated time of completion of reinforcement in any section. B. Do not place concrete until reinforcement has been observed and corrections, if any, made.

3.3

FORMWORK INSTALLATION A. Erect, brace, and maintain formwork to support vertical and lateral loads. B. Construct forms to sizes, lines and dimensions shown to obtain accurate alignment, location, grades, level and plumb work in finished structure.


C. Provide for openings, offsets, keys and other features required in work. Accurately position and support items. D. Solidly butt joints and provide backup at joints to prevent leakage of cement paste. E. Provide crush plates or wrecking plates where stripping may damage cast concrete surfaces. F. Kerf wood inserts for forming keys and the like to prevent swelling and for easy removal. G. Provide openings in concrete form to accommodate work of other trades. Determine size and location of openings, recesses and chases from trades providing such. H. Cleaning and Tightening: Thoroughly clean forms and adjacent surfaces to receive concrete. Remove chips, wood, sawdust, dirt or other debris just before concrete is placed. Retighten forms after concrete placement if required to eliminate concrete leaks. I.

3.4

Reuse of Forms: Clean and repair surfaces of forms to be reused in the work. Split, frayed, delaminated, or otherwise damaged form facing material is not acceptable. Apply new form coating compound material. When forms are reused for successive concrete placement, thoroughly clean surfaces, remove fins and laitance, and tighten forms to close all joints. Align and secure joints to avoid offsets.

REINFORCEMENT PLACING A. Clean reinforcement of loose rust, mill scale, earth, ice and other materials which reduce or destroy bond with concrete. B. Accurately position, support and secure reinforcement against displacement by formwork, construction or concrete placement operations. Locate and support reinforcement by metal chairs, runners, bolsters, spacers and hangers as required. Do not use brick. C. Place reinforcement to obtain at least the minimum coverages for concrete protection. D. Arrange, space and securely tie bars and bar supports to hold reinforcement in position during concrete placement. Set wire ties so ends are directed into concrete, not toward exposed concrete surfaces. E. Lap bar splices as indicated. Stagger splices in adjacent bars. Wire tie all splices.

3.5

WELDED WIRE FABRIC REINFORCEMENT PLACEMENT A. Place welded wire fabric one-third of slab thickness below top surface of slab.


B. Place flat sheets in as long lengths as practical. Lap adjoining sheets at least one full mesh. Offset laps to prevent continuous laps in either direction. C. Do not continue welded wire fabric through any control joints or construction joints for slabs on grade. 3.6

CONCRETE PLACEMENT A. Before placing concrete, inspect and complete formwork installation, reinforcing steel and items to be embedded or cast in. B. Notify other trades to permit installation of their work. Cooperate with other trades in setting such work as required. C. Install anchor bolts and sleeves. D. Deposit concrete continuously or in layers of such thickness that no concrete shall be placed on concrete which has hardened sufficiently to cause formation of seams or planes of weakness within section. Provide construction joints if section cannot be placed continuously. E. Deposit concrete as nearly as practicable to its final location to avoid segregation caused by rehandling or flowing. F. Keep excavations free of water. Do not deposit concrete in water, mud, snow or on frozen ground. G. Maximum drop of concrete shall not exceed 5 feet. Use hopper and trunk for greater drops. H. Contractor shall be responsible for controlling the proper placing of all embedded pipe, conduit and other embedded items. I.

3.7

Contractor shall be responsible for finishing of all concrete slabs to proper elevations to insure that all surface moisture will drain freely to floor drains, and that no puddle areas exist. During finishing operation, Contractor shall pay particular attention to this criterion, and shall make all efforts to obtain this. Any cost of corrections to provide for this positive drainage will be the responsibility of Contractor.

CONSOLIDATION A. Consolidate placed concrete by mechanical vibrating equipment supplemented by hand spading, rodding or tamping. B. Do not use vibrators to transport concrete inside formwork.


C. At each insertion, limit duration of vibration to time necessary to consolidate concrete and complete embedment of reinforcement and other embedded items without causing segregation of mix. D. Do not allow vibrator to come in contact with form. 3.8

SURFACE FINISHES A. Finish of Formed Surfaces 1. Rough Form Finish: For formed concrete surfaces not exposed-to-view in the finish work or concealed by other construction unless otherwise indicated. This is the concrete surface having texture imparted by form facing material used, with tie holes and defective areas repaired and patched and fins and other projections exceeding 1/4" in height removed. 2. Smooth Form Finish: For formed concrete surfaces exposed-to-view. This is ascast concrete surface obtained with selected form facing material, arranged orderly and symmetrically with a minimum of seams. Repair and patch defective areas with fins or other projections completely removed and smoothed. Lightly rub all exposed surfaces to achieve a uniform appearance. or Lightly sandblast to expose fine aggregate with occasional exposure of coarse aggregate and to make the color uniform. B. Monolithic Slab Finishes 1. Scratch Finish: Apply scratch finish to monolithic slab surfaces to receive concrete floor topping or mortar setting beds for tile, and other bonded applied cementitious finish flooring material. After placing slabs, roughen surface before final set with stiff brushes, brooms or rakes. 2. Trowel Finish: Apply trowel finish to monolithic slab surfaces to be exposed-toview, and slab surfaces to be covered with carpet, resilient flooring, paint or other thin film finish coating system. After floating, begin first trowel finish operation using a power-driven trowel. Begin final troweling when surface produces a ringing sound as trowel is moved over surface. Consolidate concrete surface by final hand-trowel operation, free of trowel marks. 3. Non-Slip Broom Finish: Apply non-slip broom finish to exterior concrete platforms, steps and ramps, and elsewhere as indicated. Immediately after trowel finishing, slightly roughen concrete surface by brooming with fiber bristle broom perpendicular to main traffic route.

3.9

APPLICATION OF FLOOR SURFACE HARDENER A. Bleed water shall not be present before or during the application of this shake. B. Apply first shake to hand floated concrete adjacent to forms, entryways, columns and walls where moisture will be lost first. Apply two-thirds of the specified total shake


immediately following floating of total area. Distribute evenly by hand broadcasting in all areas. C. Finishing machines with float shoes shall be used as soon as shake has absorbed moisture (indicated by darkening of surface and when surface is firm enough to support a float machine and operator). Float just sufficiently to bring moisture from base slab through the shake. Immediately following floating, apply remaining onethird of total specified shake in the same manner, allow the hardener to darken and machine float as specified. D. At no time shall water be added to the surface. E. As surface further stiffens, indicated by loss of sheen, it shall be hand or mechanically trowled with blades relatively flat. All marks and pinholes shall be removed during the final trowel operation. Finish troweling to produce a light swirl finish to provide skid resistance. 3.10 CURING AND PROTECTION A. Concrete shall be protected from premature drying, excessively hot or cold temperature, and mechanical injury according to provisions of ACI 301, Chapter 12. During placing, all concrete flatwork exposed to or subject to rapid evaporation of moisture under drying conditions (including hot weather, low humidity, wind and/or sunlight) shall be protected immediately following screeding with evaporation retarder applied in accordance with recommendations of manufacturer. Application shall precede and shall be in addition to curing specified below. B. Concrete shall be maintained in a continuously moist condition for at least 7 days after placement. Curing shall begin as soon as possible after concrete has been placed and finished. Materials and methods of curing shall be submitted to Architect for acceptance. C. Curing and Protection: Surfaces not in contact with forms and surfaces in contact with forms for less than seven days. Curing shall be by water curings, application of liquid membrane curing/sealing compound or by application of curing sheet materials. Curing compounds shall be applied in accordance with manufacturer's recommendations. Liquid membrane curing compound used on floor slabs receiving applied finish flooring shall be guaranteed by the manufacturer, in writing, not to impair bonding of adhesive. For slabs use a curing treatment of water curing, curing sheet materials, or by applying and removing curing/sealing compound. The curing compounds must be applied immediately after final finishing. For curing by water curing or curing sheet materials, the concrete must be continually moist-cured for at least 7 days. Curing shall begin immediately after finishing.


For other surfaces (footing, walls, etc.) curing shall be by one of the accepted curing treatments listed above. Restore curing protection on all freshly cut joint edges and faces when sawing joints or removing forms. D. Concrete placed under cold weather conditions shall be cured by completely covering exposed surface of concrete with curing sheet materials with sheeting completely sealed around edges. All concrete shall be cured for a minimum of 14 days with temperatures at or above 40oF or for a minimum of 7 days with temperatures at or above 70oF. 3.11 COLD WEATHER CONCRETING A. Place concrete during cold weather in accordance with ACI 306. B. For cold weather concreting, (defined as a period when for more than three successive days the mean daily temperature drops below 40oF) concrete temperature shall be maintained in accordance with ACI 306. 3.12 HOT WEATHER CONCRETING A. Place concrete in accordance with ACI 305. B. Cool ingredients before mixing to maintain concrete temperature below 90oF at time of placement. C. Cover reinforcing steel with water-soaked burlap if temperature of reinforcing steel exceeds ambient air temperature. D. Wet forms thoroughly before placing concrete. 3.13 WALL JOINTS A. Construction Joints: Locate and install construction joints as shown on Drawings. Where construction joints are not shown, locate joints at masonry control joints. Install joints maximum of 60 feet on center in locations acceptable to Architect. 3.14 INTERIOR SLAB JOINTS A. Construction Joints: Locate and form construction joints as shown on Drawings. Where construction joints are not shown place in locations acceptable to Architect. B. Contraction Joints: Sawcut joints as soon as possible after finishing, generally within 4 to 16 hours. Make sample cut to determine if concrete surface is firm enough so that it is not torn or damaged by the blade.


C. Isolation Joints: Construction isolation in slabs on grade at all points of contact with vertical surfaces and elsewhere as indicated. 3.15 EXTERIOR SLAB JOINTS A. Expansion Joints: Locate and install expansion joints as shown on Drawings. Where expansion joints are not shown, locate and install joints a maximum of 20 feet on center in either direction. B. Contraction Joints: Tool joints during final finishing with edging tool. C. Isolation Joints: Construct isolation joints in slabs on grade at all points of contact with vertical surfaces and elsewhere as indicated. 3.16 TOLERANCES A. Footings 1. Variation of dimensions in plan: plus 2" or minus 1/2". 2. Variation of center from specified center in plan: 2 percent of footing width in direction of variation, plus or minus 2" maximum variation. 3. Variation of bearing surface from specified elevation: plus or minus 1/2". B. Anchor Bolts and Sleeves 1. Variation from specified location in plan: plus or minus 1/4". 2. Variation from specified elevation: plus or minus 1/2". C. Slab on Grade 1. Surface Flatness: FF20 or greater. 2. Surface Levelness: FL17 or greater. 3. Variation from specified elevation: plus or minus 1/4". D. Stairs 1. Variation in riser: 1/8". 2. Variation in tread: 1/8". 3.17 SLAB SEALERS A. Interior Exposed Slabs: Apply two coats of hardener after slabs have cured a minimum of 28 days at a rate of 100 square feet/gallon; in accordance with manufacturer’s recommendations. B. Exterior Exposed Slabs: Apply two coats of Penetrating Exterior Anti-Spalling Sealer after slabs have cured a minimum of 28 days in accordance with manufacturer’s recommendations. 3.18 REPAIR OF SURFACES


A. Contractor shall be responsible for cost of repairing defects. B. Repair defective wall areas with cement mortar or proprietary patching compound, when acceptable to Architect. Cut out honeycomb, rock pockets and voids over 1/2inch diameter back to solid concrete but in no case to a depth of less than 1 inch. Make edges of cuts perpendicular to concrete surface. C. Repair defective interior slab areas as follows: 1. Correct flatness and levelness defects by grinding or removal and replacement of slab. Patching of low spots will not be permitted. 2. For cracks less than 1/32 inch, no repairs are required. For cracks greater than 1/32 inch, use crack repair material. For cracks over 1/8 inch, fill crack with oven-dried sand prior to application of crack repair material, as recommended by manufacturer. Contractor also has option to remove and rebuild areas of cracking. Mask cracks to limit crack repair material to crack only. 3. Curling at slab edges which exceeds 1/4 inch when measured with a 10-foot straightedge shall be made level by grinding or planing. Straightedge shall be located with its end at the slab edge, and the space between the straightedge and the slab shall be measured. If curling exceeds 1/4 inch, core drill slab at 3-foot intervals and inject non-shrink grout to fill void beneath slab. 4. Repair edge spalls which occur from shrinkage cracking or from contractor’s operations. D. Remove and replace all exterior slabs which are cracked or do not properly drain.

END OF SECTION


Lecture 41 – Metric Metric units are used everywhere in the world EXCEPT the USA. It is based on scientific measurements, not the length of a king’s foot or the weight of a chicken.

Metric Conversion Factors Length

Area

Volume

Weight Force

Multiply: inches inches feet feet yards in2 ft2 ft2 yd2 in3 ft3 ft3 gallons gallons pounds pounds kips pounds per foot (PLF)

Moment

Unit weight Stress (pressure)

Lb-In Lb-Ft Kip-Ft Kip-Ft Kip-In Lb per ft3 (PCF) Lb per in2 (PSI) Lb per in2 (PSI) Lb per in2 (PSI) Kips per in2 (KSI) Kips per in2 (KSI) Lb per ft2 (PSF) Lb per ft2 (PSF) Kips per ft2 (KSF) Kips per ft2 (KSF)

By: 25.4 0.0254 304.8 304.8 0.9144 645.2 92,909 0.0929 0.836 16,387 28,317,000 0.0283 0.003785 3.785 0.454 4.448 4448 14.594 0.11299 1.3559 1355.9 1.3559 112.99 0.1571 6895 6.895 0.0006895 6895 6.895 47.88 0.04788 47.88 0.04788


To Obtain: millimeters (mm) meters (m) millimeters (mm) meters (m) meters (m) mm2 mm2 m2 m2 mm3 mm3 m3 m3 = 1000 liters liters kilograms (kg) Newtons (N) Newtons (N) Newtons per meter (N/m) N-m N-m N-m kN-m N-m kN/m3 Pascals = Pa = N/m2 kPa = kilopascals = 1000 Pa mPa = megapascals = N/mm2

kPa mPa Pa kPa kPa mPa

Example 1 GIVEN: All road construction projects in New York State have drawings with metric measurements. All metric drawings always use units of mm for everything. (Reminder: 1 m = 1000 mm) REQUIRED: Convert 1500 mm as shown on the drawings into feet/inch/sixteenths. Use the conversion 1 inch = 25.4 mm Number of inches =

1500mm 25.4mm _ per _ inch

= 59.055118 inches = 48” + 11.055118” = 4’ + 11.055118” = 4’ + 11” + (0.055118 x 16th) = 4’ + 11” +

1500 mm = 4’-11

0.88 16ths

1 " 16

Example 2 GIVEN: The x-sectional area of a W30x211 steel beam = 62.0 in2. REQUIRED: Determine the area of the beam in units of mm2. Use the conversion 1 inch = 25.4 mm 1 inch x 1 inch = 25.4 mm x 25.4 mm 1 in2 = 645.2 mm2 Number of mm2 = 62.0 in2 x 645.2 mm2 per in2 Area = 40,002 mm2


Example 3 GIVEN: The tank below. REQUIRED: How many liters are in the tank?

7’-0”

Dia. = 18’-0” Vol. =

π 4

(18' ) 2 (7' )

= 1781 ft3 = 1781 ft3(7.48 gals per ft3) Vol. = 13324 gal. Use the conversion 1 ft3 = 0.0283 m3 Volume = 1781 ft3(0.0283 m3 per ft3) = 50.4 m3 Volume = 50.4 m3(1000 liters per m3) Volume = 50,400 liters


Example 4 GIVEN: An ASTM A325 steel bolt is capable of resisting an allowable shear stress of 30 kips per square inch (KSI) REQUIRED: What is the allowable shear stress for the bolt in units of KPa? Use the conversions of 1 inch = 25.4 mm 1 Pa = 1 N/m2 1 kPa = 1000 Pa 1 lb = 4.448 N 30 KSI =

30000lb 1in 2

30 KSI =

(30000lb)(4.448 N _ per _ lb) (0.0254m)(0.0254m)

30 KSI =

(133440 N ) (0.00064516mm 2 )

30 KSI = 206,832,414 N/m2 30 KSI = 206,832,414 Pa 30 KSI = 206,832 kPa


Metric Loads All metric loads, forces and derivatives of those are in units of Newtons. Below is a sampling of typical minimum design live loads for buildings:

Minimum Uniformly Distributed Live Loads per ASCE 7-02 Occupancy: Assembly area – fixed seats Assembly area – movable seats Balconies - residential Bowling alleys, poolrooms, rec. areas Dining rooms and restaurants Gymnasium Hospital – operating rooms, labs Hospital – private rooms Library – reading room Library – stack rooms Office Residential Schools - classrooms Stadium - bleachers Stores – 1st floor retail

Pounds per ft2 60 100 60 75 100 100 60 40 60 150 50 40 40 100 100

kN/m2 (soft metric)

2.87 4.79 2.87 3.59 4.79 4.79 2.87 1.92 2.87 7.18 2.40 1.92 1.92 4.79 4.79

Material Properties Material: Steel – A36 Steel – A992 Aluminum Concrete (4000 psi)

Wood - SYP Earth – sandy

Water

Unit Weight: Lb/ft3 kN/m3 490 77.2 490 77.2 170 26.7 150 23.6 37 5.8 100 15.7 62.4 9.8

Modulus Of Elasticity (E):

Kips/in2 29,000 29,000 10,000 3,600 1,600 -

kN/mm2 205 205 70 25.4 11.3 -


Yield Stress (fy): Kips/in2 N/mm2 36 248 50 345 28 190 6 42 -

Metric Dimensions All metric dimensions are in units of millimeters. Below are a few examples of architectural drawings in metric dimensions:



Example 5 – LRFD Steel Beam Analysis GIVEN: A simply-supported ASTM A992 steel W18x35 beam is loaded as shown (all loads are factored and includes beam weight). Assume the beam is continuously laterally braced. REQUIRED: 1) Determine the maximum factored moment in units of kN-m. 2) Determine the plastic section modulus, Zx of the beam in units of mm3. 3) Determine the LRFD design moment of the beam φMn in units of kN-m. 4) Determine if the beam is adequate in moment. w = 3 kN/m

11 m

Step 1 - Determine the maximum factored moment in units of kN-m:

wu L2 Mmax = 8 =

(3kN / m)(11m) 2 8

Mmax = 45.4 kN-m Step 2 - Determine the plastic section modulus of the beam in units of mm3:

From LRFD properties W18x35 → Zx = 66.5 in3

⎛ 25.4mm 25.4mm 25.4mm ⎞ Zx = 66.5 in3 x ⎜ x x ⎟ in in ⎠ ⎝ in Zx = 1,089,742 mm3


Step 3 - Determine the LRFD design moment of the beam φMn in units of kN-m:

From LRFD: See table above

φMn = 0.9FyZx where: Fy = 345 N/mm2 Zx = 1,089,742 mm3

φMn = 0.9(345 N/mm2)(1,089,742 mm3) = 338,360,000 N-mm

m ⎞ ⎛ kN x = 338,360,000 N-mm x ⎜ ⎟ ⎝ 1000 N 1000mm ⎠ φMn = 338.4 kN-m Step 4 - Determine if the beam is adequate in moment: Since φMn = 338.4 kN-m > Mmax = 45.4 kN-m → beam is OK


Example 6 – ACI Concrete Beam Analysis GIVEN: The rectangular concrete beam shown below. All loads are factored and includes beam weight. Use f’c = 25 mPa and 3 – 20 mm diameter bars having fy = 400 mPa. REQUIRED: 1) Determine the ACI 318 factored moment capacity, Mu of the beam in units of kN-m. 2) Determine the ACI 318 factored moment capacity of the beam in units of Kip-Feet. 250

350

Step 1 – Det. moment capacity Mu in units of kN-m:


⎞⎤ ⎟⎟⎥ ) ⎠⎥⎦

⎛ π (20mm) 2 where: As = 3 bars ⎜⎜ 4 ⎝ 2 = 942.5 mm

ρact =

=

⎞ ⎟⎟ ⎠

As bd

942.5mm 2 (250mm)(350mm)

= 0.0108


⎡ ⎛ ρ act f y Mu = 0.9Asfyd(1 - ⎢0.59⎜⎜ ⎝ f 'c ⎣⎢

⎞⎤ ⎟⎟⎥ ) ⎠⎦⎥

⎡ ⎛ (0.0108)(400 N / mm 2 = 0.9(942.5 mm2)(400 N/mm2)(350 mm)(1 - ⎢0.59⎜⎜ 25 N / mm 2 ⎝ ⎣

⎞⎤ ⎟⎟⎥ ) ⎠⎦

= 106,640,000 N-mm

m ⎞ ⎛ kN x = 106,640,000 N-mm x ⎜ ⎟ ⎝ 1000 N 1000mm ⎠ Mu = 106.6 kN-m Step 2 – Det. moment capacity Mu in units of Kip-Feet:

⎛ Kip − Feet ⎞ Mu = 106.6 kN-m x ⎜ ⎟ ⎝ 1.3559kN − m ⎠ Mu = 78.6 Kip-Feet


See conversion table above

Lecture 5 – Beams – Design for Shear & Deflection Steel beams are usually designed solely on the basis of moment. This means that bending stresses are the critical design factor. However, under certain circumstances, shear and deflection must also be checked. 1. Design for Shear Shear in steel beams generally does not control the design EXCEPT in the following two situations: •

Reduced beam cross-sectional area, as with “coped” beams

Shear plane

Reduced Shear plane

Normal beam •

Coped beam

Very heavy loads on short-span beam Very heavy loads

Short span


Shear in steel beams is assumed to be carried entirely by the area of the web, Aw:

d

tw

Beam X-Sect

Aw = Shear area in normal beam (shaded)

Aw = Shear area in coped beam (shaded)

Design for shear is dictated in AISC Spec. G p. 16.1-64 as follows: LRFD Factored Design shear strength = φvVn ASD Service Allowable shear strength =

Vn Ωv

where: φv = 1.00 (LRFD) Ωv = 1.50 (ASD) Vn = nominal shear strength = 0.6FyAwCv Aw = area of web (see sketch above) = twd Cv = Web shear coefficient = 1.0 for webs of rolled “I” – shaped shapes (Conservative) = see AISC Eq. G2-3, G2-4 and G2-5 p. 16.1-65 for other conditions


Example 1 (LRFD) GIVEN: A W14x26 A992 steel beam. REQUIRED: Determine the FACTORED design shear strength, φvVn for the beam. Step 1 – Determine area of web, Aw: Aw = twd

tw

d

= 0.255 in(13.9 in) = 3.54 in2 Step 2 – Determine FACTORED design shear strength: φvVn = 1.00(0.6FyAw)Cv = 1.00(0.6(50 KSI)(3.54 in2)1.0) φvVn = 106.2 KIPS (NOTE: The values of φvVn can be found directly in the “Maximum Total Factored Uniform Load” Table 3-6 AISC p. 3-67)


Example 2 (ASD) GIVEN: The W12x30 A992 steel beam has a 4” cope and has the SERVICE load as shown below. Disregard beam weight. REQUIRED: 1) Determine if the beam is acceptable on the basis of shear at the coped end. 2) Determine if the beam is acceptable on the basis of shear at the location of the point load.

W12x30 R2

R1


Step 1 – Determine the reaction at the left support R1: R1 =

120KIPS (7' ) 10'

= 84 KIPS Step 2 – Determine ALLOWABLE shear strength in coped web at R1: 0.6 Fy Aw C v Vn = Ωv 1.50 Aw = Area of web = twh

h = d - cope

tw

d

From properties, tw = web thickness = 0.260 in. d = 12.3 in. h = d – cope = 12.3” – 4” = 8.3” Aw = twh = 0.260”(8.3”) = 2.16 in2 Vn 0.6(50 KSI (2.16in 2 )(1.0) = Ωv 1.50 Vn = 43.2 KIPS < 84 KIPS → UNACCEPTABLE Ωv


Step 3 – Determine ALLOWABLE shear strength at point of load: Since the beam is not coped at this location, the design shear V strength n can be found from AISC Table 3-6 p. 3-71. Ωv Vn = 64.2 KIPS < 120 KIPS → UNACCEPTABLE Ωv

(Note: This example looks ONLY at shear. Very high loads also require a detailed look at connections, which will be investigated later)

Possible Fixes for High Shear: By modifying any of the variables in the design shear equation, the capacity may be increased. These include: • • •

Using a higher grade of steel (increase Fy) Use a bigger beam (increase Aw) Weld additional plates to the web (increase Aw)

New plate welded to web of beam


2. Design for Deflection Building codes (such as the IBC) require that deflections in beams be held to a minimum for occupancy comfort as well as to reduce likelihood of cracking ceiling finishes such as plaster. It is considered to be a “serviceability” check. Allowable deflection limits are dictated by the codes, such as L/360 of the span. Actual deflection is calculated using SERVICE LOADS in the formulas given in the AISC p. 3-208 thru 3-226. These actual deflections are then compared against the allowable deflection.

The following allowable deflection limits for steel construction are used by the IBC: Construction Roof member supporting plaster ceiling Roof member supporting nonplaster ceiling Roof member supporting no ceiling Floor members

Live Load L/360 L/240 L/180 L/360

Snow or Wind L/360 L/240 L/180 -

It should be noted that the above-noted allowable deflections are minimums. Architects and engineers often reduce the deflection limit to L/480 or even L/540 to ensure that floors are not “bouncy.”


Example 3 GIVEN: The floor framing plan below. The total superimposed service dead load = 86 PSF (not including beam weight) and the service live load = 125 PSF. REQUIRED: Determine the maximum actual mid-span deflection and compare with a Live load limit = L/480 and a Dead load + Live load limit = L/240 on the W18x35 steel beam.

do

do

W18x35 beam

W24x62 girder

25’-0”

4@8’-0” = 32’-0”

Step 1 – Determine the actual max. deflection considering Live load only:

From AISC p. 3-211, the maximum deflection formula is: 5wL4 ∆max = 384 EI w = uniform live load on beam = 8’(125 PSF) = 1000 PLF

⎛ 1000 PLF ⎞ 4 5⎜ ⎟(25' x12" / ft ) 12 ⎠ ∆max = ⎝ 384(29000000 PSI )(510in 4 ) = 0.59 in.


Step 2 – Determine allowable deflection of L/480 for live load only:

L 480

∆allow =

=

25'−0" (12" / ft ) 480

= 0.625 in. Since ∆allow = 0.625” > 0.59” → beam is acceptable Step 3 – Determine max. defl. considering Deal load + Live load:

5wL4 384 EI

∆max =

w = uniform dead load + live load on beam = 8’(125 PSF + 86 PSF) + 35 PLF = 1723 PLF

⎛ 1723PLF ⎞ 4 5⎜ ⎟(25' x12" / ft ) 12 ⎠ ∆max = ⎝ 384(29000000 PSI )(510in 4 ) = 1.02 in. Step 4 – Determine allowable deflection of L/240 for LL + DL:

∆allow =

=

L 240 25'−0" (12" / ft ) 240

= 1.25 in. Since ∆allow = 1.25” > 1.02” → beam is acceptable


Lecture 6 – Plate Girders A plate girder is a built-up beam comprised of plates welded together in the shape of an “I.” Usually they are used in situations when standard W shapes are not adequate. They are fabricated in a shop and are typically much more expensive than rolled shapes, but often cheaper than trusses. Typically, the depth of plate girders range from about 1/10 to 1/12 of the span. Plate girders are designed in AISC Spec. chapter G p. 16.1-37 and 16.1-107.

The design of a plate girder will be explained in the example below. Example GIVEN: The continuously laterally-braced welded unstiffened plate girder below. All loads shown are FACTORED and includes anticipated girder weight. Use F y = 50 KSI for web and flanges. REQUIRED: Design the plate girder. wu = 10 KLF

R2

R1 75’-0”

Step 1 – Determine maximum factored moment, Mu and shear, V u: wu L2 Mu = 8 =

(10KLF )(75' ) 2 8

Mu = 7031 KIP -FT Vu = R1 = R2 =

=

wu L 2

(10KLF )(75' ) 2

Vu = 375 KIPS Step 2 – Determine preliminary depth of girder: Assume depth =

=

Span 10 75'−0" (12" / ft ) 10

depth

= 90”

Step 3 – Determine height of web: Assume the flanges are approximately 1½” thick, so: h = depth – 2(Flange thickness) = 90” – 2(1½”) h = 87”

h = web ht.

Step 4 – Determine web thickness assuming girder is unstiffened (i.e., no web stiffeners): From AISC Spec F2 p. 16.1-35 - 36: If 3.07

E h < ≤ 260 then design shear = φ v Vn Fy t w     4.52 E   Vn = Aw  2   h    t    w  

Determine absolute smallest tw assuming

tw =

h ≤ 260 : tw

87" 260

= 0.33” Try tw = 3/8”:

Aw = Area of web

    4.52 E   Vn = Aw  2   h    t    w  

    4.52(29000 KSI )  = (87” x 3/8”)  2    87"       3 / 8"    = 79.5 KIPS < 375 KIPS → NO GOOD Try tw = 5/8”:

Vn = 368 KIPS < 375 KIPS → NO GOOD

Try tw = ¾”:

Vn = 636 KIPS > 375 KIPS → OK

Use 87” x ¾” web

Step 5 – Preliminary flange design: Mu Fy h

Area of flange = A f ≈

≈

7031KIP − FT (12" / ft ) (50 KSI )(87")

≈ 19.4 in2 TRY 1½” x 14” flange (Area of flange = 21 in2 > 19.4 in2) Step 6 – Check tension (bottom) flange for yield strength (AISC 16.1-107: Design strength for tension flange = φ bMn where: φ b = 0.90 Mn = S xtReFy

I total yb

Sxt = Yt

t wh3 + 2( A f )( y b ) 2 12

Itotal =

87” yb

= Af = Area of flange

0.75" (87) 3 + 2(21in 2 )( 44.25") 2 12

= 123,395 in4 Sxt =

123395in 4 44.25"

= 2788.6 in3 Re = Hybrid girder factor = 1.0 if web & flange steel F y are the same φ bMn = 0.9SxtReFy = 0.9(2788.6 in3)(1.0)(50 KSI) = 125,487 KIP -IN = 10,457 KIP -FT > Mu = 7031 KIP -FT → OK

Step 7 – Check compression (top) flange for buckling (AISC p. 16.1-108): The design strength for compression flange buckling = φ bMn where: φ b = 0.90 Mn = Sxc RPGReFy Sxc = S xt = 2788.6 in3 RPG = 1 -

ar 1200 + 300a r

 hc   − 5.70 E  ≤ 1.0  tw Fy  

ar = ratio of web area to comp. flange area (< 10) =

Aw Af

=

(87" )(0.75" ) 21in 2

= 3.1 hc =

=

h 2 87" 2

= 43.5”  43.5" 3 .1 29000    ≤ 1 .0 RPG = 1 − 5 . 70 1200 + 300(3.1)  0.75" 50  = 1.11 → use 1.0 Re = 1.0 φ bMn = 0.9Sxc RPGReFy = 0.9(2788.6 in3)(1.0)(1.0)(50 KSI) = 125,487 KIP -IN = 10,457 KIP -FT > Mu = 7031 KIP -FT → OK

Step 8 – Summary sketch:

1½” Plate

90”

¾” Plate

1½” Plate 14”

Lecture 7 – Composite Steel Beams Steel beams are usually used to carry a metal deck-supported concrete slab. In non-composite construction, the beam does NOT interact structurally with the slab – the slab is simply dead weight. This is because the slab is not adequately bonded to the beam.

Non-Composite Construction The word “composite” means 2 or more different materials. In composite construction, the slab is adequately bonded to the steel beam by means of headed “shear studs” resulting in a composite beam. The concrete acts like a large “flange” in compression, while a much greater portion of the steel beam acts in tension. The result is a VERY efficient beam – as much as 40% to 60% lighter weight steel than non-composite. Headed shear studs welded thru metal deck to beam flange

Composite Construction


Comp.

Comp.

N.A

N.A

Ten.

Ten. Composite Section

Non-Composite Section

Notice that the neutral axis (N.A.) in the non-composite beam is located in the middle of the section. This indicates that half of the beam section is in tension and half is in compression. In the composite section, the compression is carried ENTIRELY by the concrete, while the tension is carried by the beam. Composite Design per LRFD • • •

AISC Spec Ch. I3 (p. 16.1-83) AISC Part 3 Table 3-19 (p. 3-156 thru 189) AISC Part 3 Table 3-21 (p. 3-207)

From a side view, a composite beam looks like the following: Concrete

Metal Deck

Shear stud Ycon

d

Beam


For LRFD composite design, the following terms are used: b = Effective slab width

a Y2

where: b = Effective slab width (from AISC Spec. p. 16.1-83)

BeamSpan 8 = smaller of

BeamSpacing 2 Distance to edge of slab

a = Effective concrete thickness ≈ 2” for initial trial size =

As Fy

0.85 f ' c b As = Cross-sectional area of beam = from properties f’c = Specified concrete compressive strength = 4 KSI (usually)

Y2 = Ycon -

a 2


Ycon

The best way to see how the design of a composite beam is accomplished is thru an example problem: Example 1 (LRFD) GIVEN: A floor structure using A992 steel beams and the following superimposed SERVICE loads: • •

Service live load = 75 PSF Service dead load (not incl. beam weight) = 63 PSF

In addition, use ¾” diameter shear studs, 4” concrete over 1½” metal deck and normal weight concrete f’c = 4000 PSI. REQUIRED: Design the lightest weight W14 beam “A” using composite construction.

3@8’-0” = 24’-0”

30’-0”

Beam “A”

Step 1 – Determine factored maximum moment Mu: Assumed beam weight

wu = 1.2D + 1.6L = 1.2[8’(63 PSF + 30 PLF)] + 1.6[8’(75 PSF)] = 1601 PLF = 1.6 KLF Mu =

=

wu L2 8

1.6 KLF (30' ) 2 8

= 180 KIP-FT


Step 2 – Determine “Trial” approx. beam weight:

⎤ ⎡ ⎥ ⎢ M u (12" / ft ) ⎥ Approx. Beam weight = 3.4⎢ a ⎞⎥ ⎛d ⎢ ⎢ 0.85Fy ⎜⎝ 2 + Ycon − 2 ⎟⎠ ⎥ ⎦ ⎣ Assume d = 14” (since using W14) a = 2” Ycon = 4”

⎡ ⎤ ⎢ 180 KIP − FT (12" / ft ) ⎥ ⎥ Approx. Beam weight = 3.4⎢ 2" ⎞ ⎥ ⎛ 14" ⎢ ⎢ 0.85(50 KSI )⎜⎝ 2 + 4"− 2 ⎟⎠ ⎥ ⎣ ⎦ = 17.28 PLF From AISC Table 3-19 p. 3-184 → TRY W14x22 Step 3 – Determine concrete flange force, ΣQn:

ΣQn = AsFy where: As = Area of beam = 6.49 in2 (from properties for W14x22)

ΣQn = (6.49 in2)(50 KSI) = 324.5 KIPS (NOTE: From AISC Table 3-19 p. 3-184 the value of ΣQn can be found looking at Y1 = 0.000” → ΣQn = 325 KIPS)


Step 4 – Determine effective concrete slab width, b:

BeamSpan 30' (12" / ft ) = = 45” ← USE 8 8 BeamSpacing 8' (12" / ft ) = = 48” 2 2

b = smaller of

Distance to edge of slab = N/A

Step 5 – Determine Y2 for usage in AISC Table 3-19 p. 3-184: Y2 = Ycon -

a 2

where a =

=

As Fy 0.85 f ' c b 325KIPS 0.85(4 KSI )(45" )

= 2.12” Y2 = 4” -

2.12" 2

= 2.94” Use Y2 = 3” Step 6 – Determine required beam size from AISC Table 3-19: W14x22 Y2 = 3” ΣQn = 325 KIPS

Design strength in flexure = 240 KIP-FT > 180 KIP-FT

See Table 3-19 Below


Page 3-184 Lecture 7 - Page 7 of 10

Step 7 – Determine number of ¾” shear studs required: Number of studs required =

2ΣQn Qn ( stud )

where: ΣQn = AsFy = 325 KIPS Qn(stud) = Nominal horz. shear strength of stud = From AISC Table 3-21 p. 3-207

= 17.2 KIPS

Number of studs required =

Normal wt. conc. f’c = 4 KSI Deck perpendicular 1 weak stud per rib

2(325KIPS ) 17.2 KIPS _ per _ stud

= 37.8 studs Use 38 – ¾” dia. studs Step 8 – Check beam shear at coped end: Assume beam is coped 1½” 1½”

13.7”

W14x22 beam Girder Factored beam end reaction =

1.6 KLF (30' ) 2

= 24 KIPS


Set up a ratio of φvVn for the full W14x22 beam section to the reduced beam section:

φ vV n Full _ depth

=

φ vV n Re duced _ depth

From AISC p. 3-67 → φvVn = 94.8 KIPS for full depth

φ vV n 94.8KIPS = 13.7" (13.7"−1.5" ) φvVn = 84.4 KIPS > 24 KIPS for reduced section OK Step 9 – Draw summary sketch: 38 – ¾” Dia. shear studs welded thru metal deck along center of beam spaced evenly along length of beam

W14x22 A992 beam 30’-0” span


Example 2 GIVEN: The beam from Example 1. All loads and other conditions are the same. REQUIRED: Design lightest weight W14 Beam “A” using NONCOMPOSITE construction. Step 1 – Determine Mu: Mu = 180 KIP-FT Step 2 – Design lightest weight W14 beam: From AISC Zx Table p. 3-18: Use W14x34 → φbMpx = 205 KIP-FT > 180 KIP-FT (Note: The W14x34 non-composite beam is 55% heavier than a W14x22 beam that is used as a composite beam.)


Lecture 8 – Compression members – Axially loaded Compression members are not limited to just columns. Trusses and bracing members may also be designed as compression members. We will look at compression members subject to axial compression ONLY (i.e., no other stresses). Essentially, this means that loads act thru the centroid of the member along the longitudinal length of the member as shown below:

Axial Load

Compression members are designed in accordance with: • •

AISC Spec. Chapter E p. 16.1-32 AISC Part 4


A compression member’s strength is GREATLY determined by its unbraced length, referred to its “Effective Length” KL where: K = Effective length factor = see AISC p. 16.1-240 = usually 1.0 assuming ends are pinned L = Distance between lateral bracing, inches. The design strength of an axially-loaded compression member = φcPn LRFD =

Pn ASD Ωc

where: φc = 0.90 Ωc = 1.67 Pn = nominal compressive strength = AgFcr Ag = Gross cross-sectional area, in2 = From properties Fcr = Elastic critical buckling stress Fy ⎛ ⎞ KL E ⎜ ≤ 4.71 = 0.658 Fe ⎟ Fy IF ⎜ ⎟ rmin Fy ⎝ ⎠ = 0.877Fe

IF

KL E > 4.71 rmin Fy

K = From AISC p. 16.1-240 L = Unbraced length, inches rmin = Minimum radius of gyration (from properties) E = Modulus of elasticity = 29,000 KSI Fe =

π 2E ⎛ KL ⎞ ⎜⎜ ⎟⎟ ⎝ rmin ⎠

2

The limiting slenderness ratio


KL ≤ 200 rmin

Example 1 (LRFD) GIVEN: A W12x50 using A992 is used as a column in a building. It is laterally braced at 14’-0” about both the X axis and Y axis. Use “K” = 1.0. REQUIRED: Determine the design axial compressive strength, φcPn for the column. KL E Step 1 – Determine if ≤ 4.71 : rmin Fy

KL (1.0)(14'−0" X 12" / ft ) = rmin 1.96in = 85.7

4.71

29000 KSI E = 4.71 Fy 50 KSI

= 113.4

KL E ≤ 4.71 rmin Fy

Therefore,

Step 2 – Determine Fe: Fe =

=

π 2E ⎛ KL ⎞ ⎜⎜ ⎟⎟ ⎝ rmin ⎠

2

π 2 (29000 KSI )

(85.7 )2

= 38.97 KSI


Step 3 – Determine Fcr: Fy ⎛ ⎞ ⎜ Fcr = 0.658 Fe ⎟ Fy ⎜ ⎟ ⎝ ⎠ 50 ksi ⎛ ⎞ = ⎜⎜ 0.658 38.97 ksi ⎟⎟50 KSI ⎝ ⎠ = 29.2 KSI

Step 4 – Determine design axial compressive strength φcPn:

φcPn = 0.90(AgFcr) = 0.90(14.6 in2)(29.2 KSI)

φcPn = 383.7 KIPS


Example 2 (LRFD) GIVEN: The truss shown below uses single A36 steel L6x4x½ angle for all members. All loads shown are factored. REQUIRED: Determine if the truss member “A” is adequate based on axial compression.

L6x4x½ Member “A” 6’-0”

10’-0”

10’-0”

30 KIPS Step 1 – Determine force in Member “A”: Length of member “A” = (10' ) 2 + (6' ) 2 = 11.66 feet = 139.9 inches

⎛ 11.66' ⎞ Force in member “A” = 15 KIPS ⎜ ⎟ ⎝ 6' ⎠ = 29.2 KIPS Step 2 – Determine slenderness factor (KL/rmin):

KL 1.0(139.9" ) = rmin 0.864" = 161.9 < 200 → OK


Step 3 – Determine Fcr from AISC Table 4-22 p. 4-322: At Fy = 36 KSI and

KL = 161 → rmin

φcFcr = 8.72 KSI Interpolating at

At Fy = 36 KSI and

KL = 162 → rmin

φcFcr = 8.61 KSI

Page 4-322


KL = 161.9 → φcFcr = 8.62 KSI rmin

Step 4 – Determine design compressive strength φcPn:

φcPn = φc (AgFcr) = Ag(φcFcr) = 4.75 in2(8.62 KSI)

φcPn = 40.9 KIPS > 29.2 KIPS → It is acceptable


Example 3 (LRFD) GIVEN: A column has an unbraced length = 12’-0” and is subject to a factored axial load = 450 KIPS. Use “K” = 1.0. REQUIRED: 1) Design the lightest weight A992 W shape using AISC Table 4-1. 2) Design the lightest weight Square HSS shape using AISC Table 4-4. 3) Design the lightest weight Round HSS shape using AISC Table 4-5. Step 1 – Design lightest W shape: From AISC Table 4-1, the following possibilities may be used: Shape W8x58 W10x49 W12x53 W14x53

Weight (PLF) 58 49 53 53

Design strength φcPn (KIPS) 545 513 547 465

KL 12’ 12’ 12’ 12’

Use W10x49 → φcPn = 513 KIPS > 450 KIPS Step 2 – Design lightest square HSS shape: From AISC Table 4-4, the following possibilities may be used: Shape HSS8x8x½ HSS10x10x3/8 HSS12x12x5/16

Weight (PLF) 48.7 47.8 48.8

KL 12’ 12’ 12’

Design strength φcPn (KIPS) 479 497 505

Use HSS10x10x3/8 → φcPn = 497 KIPS > 450 KIPS Step 3 – Design lightest Round HSS shape: From AISC Table 4-5, the following possibilities may be used: Shape HSS10.000x0.500 HSS10.750x0.500 HSS12.750x0.375

Weight (PLF) 50.8 54.8 49.6

KL 12’ 12’ 12’

Design strength φcPn (KIPS) 471 516 481

Use HSS12.750x0.375 → φcPn = 481 KIPS > 450 KIPS


Example 4 (LRFD) GIVEN: Same as Example 1. A W12x50 using A992 is used as a column in a building. It is laterally braced at 14’-0” about both the X axis and Y axis. Use “K” = 1.0. REQUIRED: Determine the design axial compressive strength, φcPn for the column using AISC Table 4-1.

From Table above, φcPn = 384 Kips (same as Example 1)


Lecture 9 – Compression Members – Combined compression & Bending In Lecture 8, we discussed axially-loaded columns and compression members loaded through the centroid. This is referred to as “concentric” loading. In reality, columns and compression members are usually loaded off the centroid referred to as “eccentric” loading, i.e., moments applied. P

P

Concentric Loading (no moment)

e

Eccentric Loading (applied moment)

M=0

M = Pe

Eccentrically loaded columns are most common in corners or in asymmetrical loadings. Equally-loaded beam end reactions about the same axis offset each other while un-equal loads create moments as shown below:

32 K

Beam end reaction

20 K 32 K

20 K 32 K

20 K

Concentrically Loaded Column

Eccentrically Loaded Column


The LRFD addresses columns and compression members under combined compression and bending in the following: • •

AISC Spec Chapter H (p. 16.1-38) AISC Part 6

In particular, AISC Spec H1 (p. 16.1-70) dictates the following interaction formulas: a) For

Pr ≥ 0 .2 Pc

Pr 8 ⎛⎜ M rx M ry ⎞⎟ + + ≤ 1.0 Pc 9 ⎜⎝ M cx M cy ⎟⎠ b) For

Pr < 0 .2 Pc

=

Pr ⎛⎜ M rx M ry ⎞⎟ + + ≤ 1.0 2 Pc ⎜⎝ M cx M cy ⎟⎠

where: Pr = Applied axial compressive load, KIPS Pc = Available axial compressive strength of column, KIPS = φcPn where φc = 0.90 LRFD P = n where Ωc = 1.67 ASD Ωc

Pn = FcrAg (see Lect. 8)

Mrx = Applied Moment about strong axis, KIP-FT = Pe Mcx = Available moment strength about strong axis, KIP-FT = from AISC Beam Design Moments Chart Mry = Applied Moment about weak axis, KIP-FT = Pe Mcy = Available moment strength about weak axis, KIP-FT = from AISC Spec Chapter F


As an alternative to the above equations, AISC Table 6-1 (p. 6-5 thru 6-95) can be used. In particular the interaction formulas from above may be re-written as: a) For

Pr ≥ 0 .2 Pc

pPr + bxMrx + byMry < 1.0 b) For

See AISC p. 6-4

Pr < 0 .2 Pc

1 9 pPr + (bxMrx + byMry) < 1.0 2 8 where: bx by p

= Formulas from AISC Table 6-1 page 6-3 = From AISC Table 6-1 p. 6-5 thru 6-95

Example 1 (LRFD) GIVEN: A W14x68 A992 steel corner column for a two-story building has an unbraced height = 12’-0” and is subjected to the FACTORED loads applied from beam end reactions as shown in plan view below. REQUIRED: Determine if the column section is adequate under combined compression and bending.

146 K

180 K


Step 1 – Determine total applied factored axial load Pr: Pr = Σ(Beam end reactions) = 180 KIPS + 146 KIPS = 326 KIPS Step 2 – Determine W14x68 column axial design strength, φcPn: From AISC Column Load Table 4-1 on p. 4-14: At KL = 12’-0” → Pc = φcPn = 700 KIPS Step 3 – Determine applied factored moment about strong axis, Mrx: P = 146 KIPS e

Beam

146 K

is equivalent to →

W14x68 col. L4x4x¼ connection angle

d = 14+”

Workable Gages in Angle Legs, in. Leg: g g1 g2

8 4½ 3 3

7 4 2½ 3

6 3½ 2¼ 2½

5 3 2 1¾

4 2½

3½ 2

3 1¾

2½ 1⅜

g g1 g2


2 1⅛

1¾ 1

1½ ⅞

1⅜ ⅞

1¼ ¾

1 ⅝

Mrx = Pe where: P = Beam end reaction From AISC p. 1-46 and above

e = eccentricity = ½(d) + angle gage = ½(14”) + 2½” = 9½”

Since L4x4 angle

⎛ 9.5" ⎞ ⎟⎟ Mrx = (146 KIPS) ⎜⎜ ⎝ 12" / ft ⎠ = 115.6 KIP-FT Step 4 – Determine factored moment about weak axis, Muy: P = 180 KIPS e 180 K

Beam

is equivalent to →

Column web tw L4x4x¼ connection angle Mry = Pe where: P = Beam end reaction e = eccentricity = ½(tw) + angle gage = ½(½”)+ 2½” = 2¾”

⎛ 2.75" ⎞ ⎟⎟ Mry = (180 KIPS) ⎜⎜ ⎝ 12" / ft ⎠ = 41.2 KIP-FT


From AISC p. 1-46

Step 5 – Determine column adequacy using interaction formula: Check

Pr ≥ 0 .2 Pc Pr 326 KIPS = Pc 700 KIPS

= 0.47 > 0.20 Then: pPr + bxMrx + byMry < 1.0 From AISC p. 6-71 for W14x68 @ KL = 12’: bx = 2.19 x 10-3 (KIP-FT)-1 by = 6.42 x 10-3 (KIP-FT)-1 p = 1.43 x 10-3 (KIPS)-1


See Below


Check: pPr + bxMrx + byMry < 1.0 1.43 x 10-3(326 KIPS) + 2.19 x 10-3(115.6 KIP-FT) + 6.42 x 10-3(41.2 KIP-FT) < 1.0

= 0.47 + 0.25 + 0.26 = 0.98 < 1.0 → ACCEPTABLE Use W14x68


18 K

Example 2 (LRFD) GIVEN: A 5-story interior column with non-symmetric loads as shown typically below. Use KL = 14’-0”. Connection angles are L3x3x¼. REQUIRED: Design the lightest weight W12 column for the lowest level.

28 K 34 K

14 K

Typical column loading (4 framed levels) Step 1 – Determine TOTAL applied factored axial load Pr: Pr = 4 floors(34 KIPS + 18 KIPS + 14 KIPS + 28 KIPS) = 376 KIPS Step 2 – Determine applied factored moment about strong axis, Mrx: Mrx = Peqe Peq = equivalent difference in load along axis = 34 KIPS – 18 KIPS = 16 KIPS e = eccentricity = ½(d) + angle gage = ½(12”) + 1¾” = 7¾”

⎛ 7.75" ⎞ ⎟⎟ Mrx = 16 KIPS ⎜⎜ ⎝ 12" / ft ⎠ = 10.3 KIP-FT


Step 3 – Determine applied factored moment about weak axis, Mry: Mry = Peqe Peq = equivalent difference in load along axis = 28 KIPS – 14 KIPS = 14 KIPS e = eccentricity = ½(tw) + angle gage = ½(½”) + 1¾” = 2”

⎛ 2" ⎞ ⎟⎟ Mry = 14 KIPS ⎜⎜ ⎝ 12" / ft ⎠ = 2.3 KIP-FT Step 4 – Select “Trial” W12 column size: Referring to AISC Table 4-1 p. 4-18: @ KL = 14’ → Try W12x50 φcPn = Pc = 443 KIPS > 376 KIPS Step 5 – Determine column adequacy using interaction formula: Check

Pr ≥ 0 .2 Pc Pr 376 KIPS = Pc 443KIPS

= 0.85 > 0.20 Then: pPr + bxMrx + byMry < 1.0


From AISC p. 6-81 for W12x50 @ KL = 14’: bx = 3.91 x 10-3 (KIP-FT)-1 by = 11.1 x 10-3 (KIP-FT)-1 p = 2.61 x 10-3 (KIPS)-1 Check: pPr + bxMrx + byMry < 1.0 2.61 x 10-3(376 KIPS) + 3.91 x 10-3(10.3 KIP-FT) + 11.1 x 10-3(2.3 KIP-FT) < 1.0

= 0.98 + 0.04 + 0.03 = 1.05 > 1.0 → Unacceptable cannot use W12x50

→ (NOTE: A W12x53 column is acceptable since 0.80 < 1.0)


Lecture 3 – Flexural Members Flexural members are those that experience primarily bending stresses, such as beams. A typical rectangular reinforced concrete beam is shown below:


Hanger bars (usually #4 or #5 bars) Stirrup bars (used to prevent diag. tension cracks) spaced at d/2+ apart

Concrete cover = ¾”  2” as per ACI reqmts.

Height “h”

Width “b”

Tension bars “As” Section A-A

Lecture 3 – Page 1 of 10

Sometimes, 2 (or more) rows of main tension bars are necessary. It is important to provide minimum adequate cover around all reinforcing bars so that these bars can properly bond with the concrete. ACI 318 dictates that the minimum spacing between bars is 1.5 times the maximum concrete aggregate size. Typical concrete batches use a maximum aggregate size of ¾” diameter, so then the minimum bar spacing = 1.5(¾”) = 1⅛”.

Height “h”

Depth to centroid of steel “d”

Below is a sketch of a typical concrete beam with 2 rows of tension bars:

Min. bar spacing Tension bars “As” Min. bar spacing


As = Total cross-sectional area of all tension bars, in2 d = depth to center of tension bars, inches = h – (concrete cover) – (stirrup bar dia.) – ½(tension bar dia.) fy = yield stress of reinforcing bars = 60 KSI for ASTM A615 Grade 60 bars = 40 KSI for ASTM A615 Grade 40 bars actual = Rho actual = actual ratio of tension steel to effective concrete area A = s bd min = Rho minimum = minimum allowable ratio of tension steel per ACI 318 200 where fy = PSI = fy


Example 1 GIVEN: A rectangular concrete beam is similar to the one shown above. Use the following:  Height h = 20”  Width b = 12”  Concrete f’c = 4000 PSI  Concrete cover = ¾”  All bars are A615 – Grade 60 (fy = 60 KSI)  Stirrup bar = #3  4 - #7 Tension bars

#3 Stirrup bars

Height “h” = 20”


Width “b” = 12”

4 - #7 tension bars REQUIRED: 1) Determine total area of tension bars, As. 2) Determine depth to center of tension bars, d. A 200 and state if it is acceptable. 3) Determine actual = s where min = bd fy Step 1 – Determine area of tension bars, As: As = 4 bars(0.60 in2 per #7 bar) As = 2.40 in2


See Lect. 1 notes

Step 2 – Determine depth to tension bars, “d”: d = depth to center of tension bars, inches = h – (concrete cover) – (stirrup bar dia.) – ½(tension bar dia.) = 20” – ¾” – ⅜” – ½(⅞”) d = 18.44” Step 3 – Determine actual and min : actual =

=

As bd

min =

2.40in 2 (12" )(18.44" )

=

200 fy

200 60000 PSI

min = 0.0033

actual = 0.0108

Since actual > min  beam is acceptable


A basic understanding of beam mechanics is necessary to study concrete beam behavior. Consider a simply-supported homogeneous rectangular beam loaded by a uniformly-distributed load as shown below:

Applied loads

Span L

Taking a section through the beam at any place along the length reveals the following stress distribution about the cross-section of the beam:


The stress distribution varies linearly from zero stresses at the neutral axis, to a maximum tensile or compressive stress at the extreme edges.

Homogeneous Beam

Tension


In a reinforced concrete beam, the stress distribution is different. Above the neutral axis, the concrete carries all the compression, similar to the homogeneous beam. Below the neutral axis however, the concrete is incapable of resisting tension and must rely on the reinforcing bars to carry all the tension loads.

b


The actual stress distribution in the compression side varies non-linearly from zero stresses at the neutral axis, to a maximum compressive stress at the extreme edge.

Reinforced Concrete Beam Tension = T Reinforcing bars Looking at a side view of the stress distribution of the reinforced concrete beam:

C Neutral Axis

½ (a)

0.85f’cb

a =  1C

“Whitney” stress block

d Moment arm = Z

T = Asfy Actual Stress Distribution

T = Asfy Idealized Stress Distribution


Assuming an idealized beam, tension equals compression: Tension = Compression Asfy = Area of Whitney stress block Asfy = 0.85f’cab Solve for a: a=

As f y 0.85 f ' c b

Beta

=  1C

1 = 0.85 for f’c < 4000 PSI = 0.80 for f’c = 5000 PSI = 0.75 for f’c > 6000 PSI C = depth to neutral axis from extreme compression edge Mn = Nominal moment capacity of concrete beam = Asfy(Moment arm) = AsfyZ a = Asfy(d - ) 2 Mu = Usable moment capacity of concrete beam = Mn = 0.9Mn a Mu = 0.9(Asfy(d - ) ) 2

   act f y Mu = 0.9Asfyd(1 - 0.59   f 'c

  ) 

bal = balanced ratio of tension steel reinforcement  0.851 f ' c =  fy 

 87,000   87,000  f y 

  where fy = PSI  

max = maximum allowable ratio of tension steel reinforcement per ACI 318 = 0.75bal


Example 2 GIVEN: The concrete beam from Example 1 is used to support the loading as shown below. REQUIRED: 1. Determine the maximum factored applied moment, Mmax. 2. Determine the usable moment capacity of the beam, Mu, and determine if it is acceptable based on Mmax. 3. Determine if the beam is acceptable based on max.

Factored uniform load wu = 3000 PLF (incl. beam wt.)

20’-0”

Step 1 – Determine maximum factored applied moment, Mmax: Mmax =

=

wu L2 8 (3KLF )(20'0" ) 2 8

Mmax = 150 KIP-FT Step 2 - Determine the usable moment capacity of the beam, Mu:

   act f y Mu = 0.9Asfyd(1 - 0.59   f 'c

  ) 

where act = 0.0108 (see Ex. 1)

  (0.0108)(60 KSI )  = 0.9(2.40 in2)(60 KSI)(18.44”)(1 - 0.59  ) 4 KSI    = 2161.4 KIP-IN Mu = 180.1 KIP-FT Since Mu = 180.1 KIP-FT > Mmax = 150 KIP-FT  beam is acceptable


Step 3 – Determine if the beam is acceptable based on max:

max = maximum allowable ratio of tension steel reinforcement per ACI 318 = 0.75bal bal = balanced ratio of tension steel reinforcement  0.851 f ' c =  fy 

 87,000   87,000  f y 

  where fy = PSI  

where 1 = 0.85 since f’c = 4000 PSI

87,000   0.85(0.85)(4 KSI )  =   60 KSI   87,000  60000 PSI  = 0.0285

max = 0.75(0.0285) max = 0.0214 > act = 0.0108  beam is acceptable


Lecture 4 – Flexural Members (cont.) Determining the usable moment capacity, Mu, of a rectangular reinforced concrete beam is accomplished by using the formula below: (see Lect. 3)

   act f y Mu = 0.9Asfyd(1 - 0.59   f 'c

  ) 

Designing a beam using the equation above is much more difficult. Assuming the material properties and dimensions are known, the equation above still has 2 unknown variables – As and act. Therefore, design of steel reinforcement for a given beam is largely one of trial-and-error.

Beam Design Design of concrete beam members is often one of trial-and-error. It’s impossible to directly solve for all the variables in a reinforced concrete beam. Usually, material properties are known as well as maximum applied factored moment, Mmax. L The following Table is useful to get a “trial” beam size:

Minimum Suggested Thickness “h” of Concrete Beams & One-Way Slabs Member: End Conditions Simply One end Both ends Cantilever supported continuous continuous Solid one-way slab L/20 L/24 L/28 L/10 Beam L/16 L/18.5 L/21 L/8 Span length L = inches Beams are usually rectangular having the width typically narrower than the height. The diagram below shows typical beam aspect ratios: b NOTE: Beam cross-section dimensions “b” and “h” are USUALLY in multiples of 2” or 4” for ease of formwork.

h  1.5b  2.5b


h

Beam Design Aid It is still difficult to directly design a reinforced concrete beam even if dimensions and material properties are known. The use of design aids are commonly used to streamline the design process instead of laboriously using a trial-and-error approach. Mu bd 2 are in units of PSI. It can be used to directly solve for act knowing factored actual moment Mu, f’c, fy, b and d.

The design aid shown below is used for design or analysis. Values of

Table 1 - Concrete f’c = 3000 PSI, Grade 60 Bars


Table 2 – Concrete f’c = 4000 PSI, Grade 60 Bars


Example 1 GIVEN: A rectangular concrete beam with dimensions is shown below (stirrup bars not shown). Use concrete f’c = 4000 PSI and grade 60 bars. REQUIRED: 1) Determine the usable moment capacity Mu of the beam using formula. 2) Determine the usable moment capacity Mu of the beam using Table 2. b =12”

d = 18”

3 - # 7 bars

Step 1 – Determine usable moment capacity Mu of the beam using formula:

act =

As bd

3(0.60in 2 _ per _ bar ) = (12" )(18" )

act = 0.0083    act f y Mu = 0.9Asfyd(1 - 0.59  f 'c 

  ) 

  (0.0083)(60 KSI )  = 0.9(1.80 in2)(60 KSI)(18”)(1 - 0.59  ) 4 KSI    = 1621 KIP-IN Mu = 135 KIP-FT


Step 2 - Determine the usable moment capacity Mu of the beam using Table 2:

From Table 2 above: At  = 0.0083 

Mu  461.4 PSI bd 2

Solving for Mu: Mu = 461.4 PSI(bd2) = 461.4 PSI[(0.9)(12”)(18”)2] = 1,614,531 LB-IN = 1615 KIP-IN Mu = 134.6 KIP-FT NOTE: This answer is the same as in Step 1.


Example 2 GIVEN: The concrete beam below. Use the following:     

Concrete f’c = 4000 PSI Steel grade 60 Concrete cover = ¾” #8 bars are to be used for main tension bars #3 stirrups

REQUIRED: Design the rectangular beam such that h 1.5b and act  ½ (max). wu = 2 KLF (includes anticipated beam weight = 300 PLF)

28’-0”

Step 1 – Determine maximum factored moment, Mmax: wu L2 Mmax = 8 =

(2 KLF )(28'0" ) 2 8

Mmax = 196 KIP-FT = 2352 KIP-IN = 2,352,000 LB-IN


Step 2 – Select values from Table 2: a) Select act = ½(max) = ½(0.0214) TRY act = 0.0107 b) At  = 0.0107 

Mu = 581.2 PSI bd 2

Step 3 – Solve for “b” and “d” by substituting Mmax for Mu in above equation:

Mu = 581.2 PSI bd 2

where: Mu = Mmax = 2,352,000 LB-IN  = 0.9 d = 1.5b

2352000 Lb  in  581.2 PSI (0.9)(b)(1.5b) 2 Solve for “b”:

2352000  581.2 PSI (0.9)(2.25b 3 ) b3

2352000 (0.9)(2.25)(581.2)

b = 12.6”  Use b = 12” d = 1.5b = 1.5(12”) d  18”


Step 4 – Select beam dimensions: From above, use b = 12” and d  18” #8 main bar dia. = 1”


h = d + conc. cover + stirrup bar dia. + ½(main bar dia.) = 18” + ¾” + 3/8” + ½(1”) = 19.625”  Use h = 20” Revised d = 20” – ¾” – ⅜” – ½(1”) = 18.375” Step 5 – Determine required area of main tension bars: From above, act = 0.0107 =

As bd

Solve for As: As = 0.0107(b)(d) = 0.0107(12”)(18.375”) As = 2.36 in2 Step 6 – Determine number of #8 main tension bars: No. of bars =

=

As Area _ of _ one _ bar 2.36in 2 0.79in 2 _ per _#8 _ bar

= 2.99 bars  USE 3 - #8 bars


Step 7 – Check beam height with “Minimum Thickness of Beams” Table: From Table: Member type = Beam

h

End Condition = Simply-supported

h



L 16

L 16

(28'0" )(12" / ft ) 16

h  21” which is approximately = 20” as designed Step 8 – Check originally assumed beam weight: It was assumed that the beam weight = 300 PLF (See Step 1). Actual beam dimensions:

20”

12”

Recall: UnitWt 

Weight → Rearranging: Volume

BeamWeight  (UnitWt )(Volume)  12" x 20" = (150 PCF)  2 2  144in / ft

 (1'0" ) 

Beam Weight = 250 PLF < 300 PLF (assumed) OK


Step 9 – Draw “Summary Sketch” labeling all information necessary to build it:

12”



¾” concrete cover all around

3 - #8 main bars Section A-A Notes: 1) Concrete f’c = 4000 PSI normal-weight 2) All bars ASTM A615 Grade 60


Example 3 GIVEN: A continuous 10” x 16” concrete beam is shown below. Assume the following:

    

Concrete f’c = 4000 PSI Steel grade 60 Concrete cover = ¾” 2 - #9 bars are to be used for main tension bars #3 stirrups

REQUIRED: 1) Determine the magnitude of the maximum positive and negative moments (Refer to Structural Theory notes) 2) Determine if the beam is adequate based upon flexure only. Check positive & negative moments. wu = 2400 PLF (incl. beam wt.) R1

R2

L = 20’-0”

Mpos = 0.08(wL2)

0.4L

R4

R3

L = 20’-0”

L = 20’-0”

2

Mpos = 0.025(wL )

0.5L

Mpos = 0.08(wL2)

0.5L

Mneg = -0.1(wL2)

Mneg = -0.1(wL2)

R1 = 0.4wL R2 = 1.1wL R3 = 1.1wL R4 = 0.4wL


0.4L

Step 1 – Determine maximum positive and maximum negative moments: a) Maximum positive moment, “Mpos”: From above, the largest Mpos = 0.08(wL2) = 0.08(2.4 KLF)(20’-0”)2 = 76.8 Kip-Ft b) Maximum negative moment, “Mneg”: From above, the largest Mneg = -0.1(wL2) = -0.1(2.4 KLF)(20’-0”)2 = 96.0 Kip-Ft Step 2 – Determine beam’s “d” dimension: Depth “d” to center of tension steel is shown below: 10”

d


¾” concrete cover all around

16”

2 - #4 hanger bars

2 - #9 main bars Beam cross-section in Mpos region (upside-down in Mneg region)


#9 main bar dia. =  9 "

d = h - conc. cover - stirrup bar dia. - ½(main bar dia.)

9  = 16” - ¾” - 3/8” - ½  "  8  d = 14.3125”


8 

Step 3 – Determine usable moment capacity Mu of the beam using formula:

act =

=

As bd 2(1.00in 2 _ per _ bar ) (10" )(14.3125" )

act = 0.01397 which is > min = 0.0033 and < max = 0.0214 OK

   act f y Mu = 0.9Asfyd(1 - 0.59  f 'c 

  ) 

  (0.01397)(60 KSI )  = 0.9(2.00 in2)(60 KSI)(14.3125”)(1 - 0.59  ) 4 KSI    = 1354.6 KIP-IN Mu = 112.9 KIP-FT Step 4 – Check if beam is adequate based upon flexure only: a) Maximum positive moment, “Mpos”: As

From above, the largest Mpos = 76.8 Kip-Ft < 112.9 Kip-Ft ACCEPTABLE

As

b) Maximum negative moment, “Mneg”: From above, the largest Mneg = -96.0 Kip-Ft < 112.9 Kip-Ft ACCEPTABLE


Structural Analysis and Design Lectures

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