Shear and Diagonal Tension

Acknowledgement This Powerpoint presentation was prepared by Dr. Terry Weigel, University of Louisville. This work and other contributions to the text by Dr. Weigel are gratefully acknowledged.

Shear Stresses in Concrete Beams Flexural stress Shear stress Principal angle Diagonal tension – Mohr’s circle Principal stresses

Shear Strength of Concrete ACI Code Equation 11-3 – conservative but easy to use ACI Code Equation 11-5 – less conservative but “difficult” to use

Shear Strength of Concrete Equation 11-3 Equation 11-5

Shear Cracking of Reinforced Beams Flexural shear crack – initiate from top of flexural crack For flexural shear cracks to occur, moment must be larger than the cracking moment and shear must be relatively large Flexural shear cracks oriented at angle of approximately 45 degrees to the longitudinal beam axis

Flexural-shear Cracks

Shear Cracking of Reinforced Beams Web shear crack – form independently Typically occur at points of small moment and large shear Occur at ends of beams at simple supports and at inflection points at continuous beam

Web-shear Cracks

Web Reinforcement Stirrups Hangers At a location, the width of a diagonal crack is related to the strain in the stirrup – larger strain = wider crack To reduce crack width, stirrup yield stress is limited to 60 ksi

Web Reinforcement Small crack widths promote aggregate interlock Limiting stirrup yield stress also reduces anchorage problems

Types of Stirrups

Types of Stirrups

Types of Stirrups

Types of Stirrups

Types of Stirrups

Types of Stirrups

Types of Stirrups

Types of Stirrups

Types of Stirrups

Behavior of Beams with Web Reinforcement Truss analogy Concrete in compression is top chord Longitudinal tension steel is bottom chord Stirrups form truss verticals Concrete between diagonal cracks form the truss diagonals

Truss Analogy

Required Web Reinforcement – ACI Code Required for all flexural members except: Footings and solid slabs Certain hollow core units Concrete floor joists Shallow beams with h not larger than 10”

Required Web Reinforcement – ACI Code Required for all flexural members except: Beams built integrally with slabs and h less than 24 in. and h not greater than the larger of 2.5 times the flange thickness or one-half the web width Beams constructed with steel fiber- reinforced concrete with strength not exceeding 6,000 psi and

Stirrups Diagonally inclined stirrups more efficient than vertical stirrups Not practical Bent-up flexural bars can be used instead

Bent-up Bar Web Reinforcement

Shear Cracking The presence of stirrups does not materially effect the onset of shear cracking Stirrups resist shear only after cracks have occurred After cracks occurs, the beam must have sufficient shear reinforcement to resist the load not resisted by the concrete in shear

Benefits of Stirrups Stirrups carry shear across the crack directly Promote aggregate interlock Confine the core of the concrete in the beam thereby increasing strength and ductility Confine the longitudinal bars and prevent cover from prying off the beam Hold the pieces on concrete on either side of the crack together and prevent the crack from propagating into the compression region

Design for Shear Stirrups crossing a crack are assumed to have yielded Shear crack forms at a 45 degree angle

Design for Shear ACI Code Equation 11-15 ACI Code Equation 11-16

Design for Shear

ACI Code Requirements for Shear ACI Code Section 11.4.6.1 – if Vu exceeds one-half fVc, stirrups are required When shear reinforcement is required, ACI Code Section 11.4.6.3 specifies a minimum amount:

ACI Code Requirements for Shear To insure that every diagonal crack is crossed by at least one stirrup, the maximum spacing of stirrups is the smaller of d/2 or 24 in. If maximum spacings are halved. See ACI Code Section 11.4.5.3

ACI Code Requirements for Shear See ACI Code Section 11.4.7.9 ACI Code Section 11.1.2 -> Stirrups should extend as close as cover requirements permit to the tension and compression faces of the member - anchorage

ACI Code Requirements for Shear Stirrup hook requirements shown on the next slide. See ACI Code Section 8.1 and 12.13 In deep beams ( l /d < 4), large shear may affect flexural capacity In most cases, beam can be designed for shear at a distance d from the face of the support . See next three slides for exceptions.

End Shear Reduction Not Permitted

End Shear Reduction Not Permitted

Corbels

ACI Code Requirements Stirrup Hooks

ACI Code Requirements Stirrup Hooks

ACI Code Requirements Stirrup Hooks

Shear Design Examples

Example 8.1 Determine the minimum cross section required for a rectangular beam so that no shear reinforcement is required. Follow ACI Code requirements and use a concrete strength of 4,000 psi. Vu = 38 k.

Example 8.1 Shear strength provided by concrete

Example 8.1 ACI Code Section 11.4.6.1 requires: Use a 24 in x 36 in beam (d = 33.5 in)

Example 8.2 The beam shown in the figure was designed using a concrete strength of 3,000 psi and Grade 60 steel. Determine the theoretical spacing for No 3 U-shaped stirrups for the following values for shear: (a) Vu = 12,000 lb (b) Vu = 40,000 lb (c) Vu = 60,000 lb (d) Vu = 150,000 lb

Example 8.2

Example 8.2 (a) Vu = 12,000 lb (use l = 1)

Example 8.2 (b) Vu = 40,000 lb Stirrups are needed since

Example 8.2 (b) (con’t) Maximum spacing to provide minimum Av

Example 8.2 (b) (con’t) ACI Code maximum spacing

Example 8.2 (c) Vu = 60,000 lb

Example 8.2 (c) (con’t) ACI Code maximum spacing

Example 8.2 (c) (con’t) ACI Code maximum spacing Use s = 7.33 in

Example 8.2 (d) Vu = 150,000 lb

Example 8.3 Select No 3 U-shaped stirrups for the beam shown. The beam carries loads of D = 4 k/ft and L= 6 k/ft. Use normal weight concrete with a strength of 4,000 psi and Grade 60 steel.

Example 8.3

Example 8.3

Example 8.3

Example 8.3 Vu = 100.8 – 14.4x

Example 8.3

Example 8.3

Example 8.3

Example 8.3 At what location along the beam is s = 9 in OK?

Example 8.3 Terminate stirrups when Vu < fVc/2

Distance from face of support (ft) Example 8.3 Distance from face of support (ft) Vu (lb) Vs (lb) Theoretical s (in) 1.875’=22.5” 73,800 55,709 5.33 2’ = 24” 72,000 53,309 5.57 3’ = 36” 57,600 34,109 8.71 3.058’= 36.69” 56,768 33,000 9 4’= 48” 43,200 14,909 > Maximum d/2 = 11.25 5.89’ = 70.66” 16,008 - terminate

Example 8.3 Corrected spacing: Cumulative (in) 1 @ 2 in = 2 in 2

Example 8.4 Determine the value of Vc at a distance 3 ft from the left end of the beam of Example 8.3, using ACI Equation 11-5

Example 8.4

Example 8.4 Using the simplified formula (ACI Eq. 11-3) for Vc results in a value of 42,691 lb1-5

Example 8.5 Select No 3 U-shaped stirrups for the beam of Example 8.3, assuming that the live load is placed to produce maximum shear at the beam centerline, and that the live load is placed to produce the maximum shear at the beam ends.

Example 8.5 Positioning of live load to produce maximum shear at the beam ends is the same as used in Example 8.3; that is, the full factored dead and live load is applied over the full length of the beam. Positioning of the live load to produce maximum positive shear at the beam centerline is shown in the figure on the next slide.

Example 8.5

Example 8.5 Vu at face of left support = 100,800 lb (from Ex. 8.3) From the loads shown on the previous slide, the factored load on the left half of the beam is wu = 1.2 (4 k/ft) = 4.8 k/ft. The left reaction is obtained by summing moments about the right support, and the shear at midspan is:

Example 8.5 Approximating the shear envelope with a straight line between 100.8 k at the face of the support and 16.8 k at midspan 100.8 k 78.3 k Slope = (100.8 -16.8)/7 ft = 12 k/ft fVc = 32.018 k 16.8 k Stirrups carry shear fVc /2 = 16.009 k concrete carries shear

Example 8.5

Example 8.5 Design the stirrups using the shear envelop Terminate stirrups when Vu < fVc/2 This location is past midspan, so stirrups cannot be terminated

Example 8.5 At x = d = 22.5 in Results of similar calculations at other assumed values of x are shown in the table that follows

Example 8.5 At what location along the beam is s = 9 in OK? Results of similar calculations for s = 6 in and s = 11 in are shown in the table that follows

Distance from face of support (ft) Example 8.5 Distance from face of support (ft) Vu (lb) Vs (lb) Theoretical s (in) 0 to d = 1.875 78,300 61,709 4.81 2 76,800 59,709 4.97 2.638’=31.66” 69,143 49,500 6.00 3 64,800 43,709 6.79 3.67’=44” 56,768 33,000 9.00 4 52,800 27,709 10.71 4.04 = 49” 52,268 27,000 11.00 5 40,800 11,709 > Maximum d/2 = 11.25 7.066’= 84.8” 16,009 Terminate

Example 8.5 Selected spacing: Cumulative (in) 1 @ 2 in = 2 in 2

Example 8.6 Select No 3 U-shaped stirrups for a T-beam with bw = 10 in and d = 20 in, if the Vu diagram is shown in the figure. Use normal weight concrete a with a strength of 3,000 psi and Grade 60 steel.

Example 8.6

Example 8.6

Example 8.6

Example 8.6

Example 8.6

Example 8.6 The maximum stirrup spacing is 5 in whenever Vs exceeds 43,820. From the figure, this value is at about 56 in from the left end of the beam.

Example 8.6 The maximum permissible stirrup spacing is the smaller of the two following values:

Distance from face of support (ft) Example 8.6 Distance from face of support (ft) Vu (lb) Vs (lb) Theoretical s (in) Maximum Spacing (in) 0 to d = 1.875 61,330 59,870 4.41 3 56,000 57,760 5.00 6- 44,000 36,760 7.18 6+ 24,000 10,090 26.16 10.00

Example 8.6

Example 8.6 Selected spacing: 1 @ 3 in = 3 in 17 @ 4 in = 68 in