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Shear and Diagonal Tension 1

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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. 2

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Shear Stresses in Concrete Beams 3 Diagonal tension – Mohr’s circle Principal stresses Flexural stress Shear stress Principal angle

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Shear Strength of Concrete 4 ACI Code Equation 11-3 – conservative but easy to use ACI Code Equation 11-5 – less conservative but “difficult” to use

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

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Shear Cracking of Reinforced Beams 6 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

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Flexural-shear Cracks 7

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Shear Cracking of Reinforced Beams 8 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

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Web-shear Cracks 9

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Web Reinforcement 10 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

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Web Reinforcement 11 Small crack widths promote aggregate interlock Limiting stirrup yield stress also reduces anchorage problems

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Types of Stirrups 12

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Types of Stirrups 13

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Types of Stirrups 14

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Types of Stirrups 15

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Types of Stirrups 16

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Types of Stirrups 17

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Types of Stirrups 18

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Types of Stirrups 19

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Types of Stirrups 20

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Behavior of Beams with Web Reinforcement 21 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

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Truss Analogy 22

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Required Web Reinforcement – ACI Code 23 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”

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Required Web Reinforcement – ACI Code 24 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

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Stirrups 25 Diagonally inclined stirrups more efficient than vertical stirrups Not practical Bent-up flexural bars can be used instead

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Bent-up Bar Web Reinforcement 26

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Shear Cracking 27 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

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Benefits of Stirrups 28 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

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Design for Shear 29 Stirrups crossing a crack are assumed to have yielded Shear crack forms at a 45 degree angle

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Design for Shear 30 ACI Code Equation 11-15 ACI Code Equation 11-16

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Design for Shear 31

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ACI Code Requirements for Shear 32 ACI Code Section 11.4.6.1 – if V u exceeds one-half V c, stirrups are required When shear reinforcement is required, ACI Code Section 11.4.6.3 specifies a minimum amount:

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ACI Code Requirements for Shear 33 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

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ACI Code Requirements for Shear 34 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

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ACI Code Requirements for Shear 35 Stirrup hook requirements shown on the next slide. See ACI Code Section 8.1 and 12.13 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. In deep beams ( l / d < 4), large shear may affect flexural capacity

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End Shear Reduction Not Permitted 36

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End Shear Reduction Not Permitted 37

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Corbels 38

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ACI Code Requirements Stirrup Hooks 39

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ACI Code Requirements Stirrup Hooks 40

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ACI Code Requirements Stirrup Hooks 41

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Shear Design Examples 42

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Example 8.1 43 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. V u = 38 k.

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Example 8.1 44 Shear strength provided by concrete

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Example 8.1 45 ACI Code Section 11.4.6.1 requires: Use a 24 in x 36 in beam (d = 33.5 in)

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Example 8.2 46 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) V u = 12,000 lb (b) V u = 40,000 lb (c) V u = 60,000 lb (d) V u = 150,000 lb

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Example 8.2 47

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Example 8.2 48 (a) V u = 12,000 lb (use = 1)

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Example 8.2 49 (b) V u = 40,000 lb Stirrups are needed since

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Example 8.2 50 (b) (con’t) Maximum spacing to provide minimum A v

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Example 8.2 51 (b) (con’t) ACI Code maximum spacing

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Example 8.2 52 (c) V u = 60,000 lb

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Example 8.2 53 (c) (con’t) ACI Code maximum spacing

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Example 8.2 54 (c) (con’t) ACI Code maximum spacing Use s = 7.33 in

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Example 8.2 55 (d) V u = 150,000 lb

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Example 8.3 56 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.

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Example 8.3 57

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Example 8.3 58

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Example 8.3 59

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Example 8.3 60 V u = 100.8 – 14.4x

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Example 8.3 61

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Example 8.3 62

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Example 8.3 63

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Example 8.3 64 At what location along the beam is s = 9 in OK?

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Example 8.3 65 Terminate stirrups when V u < V c /2

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Example 8.3 66 Distance from face of support (ft) V u (lb)V s (lb)Theoretical s (in) 1.875’=22.5”73,80055,7095.33 2’ = 24”72,00053,3095.57 3’ = 36”57,60034,1098.71 3.058’= 36.69”56,76833,0009 4’= 48”43,20014,909> Maximum d/2 = 11.25 5.89’ = 70.66”16,008-terminate

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Example 8.3 67 Corrected spacing:Cumulative (in) 1 @ 2 in = 2 in 2 7 @ 5 in = 35 in37 > 36.69 4 @ 9 in = 36 in73 >70.66

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Example 8.4 68 Determine the value of V c at a distance 3 ft from the left end of the beam of Example 8.3, using ACI Equation 11-5

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Example 8.4 69

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Example 8.4 70 Using the simplified formula (ACI Eq. 11-3) for V c results in a value of 42,691 lb1-5

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Example 8.5 71 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.

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Example 8.5 72 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.

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Example 8.5 73

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Example 8.5 74 V u 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:

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Example 8.5 75 Approximating the shear envelope with a straight line between 100.8 k at the face of the support and 16.8 k at midspan V c = 32.018 k V c /2 = 16.009 k 100.8 k 78.3 k 16.8 k Stirrups carry shear concrete carries shear Slope = (100.8 -16.8)/7 ft = 12 k/ft

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Example 8.5 76

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Example 8.5 77 Design the stirrups using the shear envelop Terminate stirrups when V u < V c /2 This location is past midspan, so stirrups cannot be terminated

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Example 8.5 78 At x = d = 22.5 in Results of similar calculations at other assumed values of x are shown in the table that follows

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Example 8.5 79 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

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Example 8.5 80 Distance from face of support (ft) V u (lb)V s (lb)Theoretical s (in) 0 to d = 1.87578,30061,7094.81 276,80059,7094.97 2.638’=31.66”69,14349,5006.00 364,80043,7096.79 3.67’=44”56,76833,0009.00 452,80027,70910.71 4.04 = 49”52,26827,00011.00 540,80011,709> Maximum d/2 = 11.25 7.066’= 84.8”16,009Terminate

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Example 8.5 81 Selected spacing:Cumulative (in) 1 @ 2 in = 2 in 2 8 @ 4 in = 32 in34 > 32” 2 @ 6 in = 12 in46 > 44” 5 @ 9 in = 45 in89 > 85”

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Example 8.6 82 Select No 3 U-shaped stirrups for a T-beam with b w = 10 in and d = 20 in, if the V u diagram is shown in the figure. Use normal weight concrete a with a strength of 3,000 psi and Grade 60 steel.

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Example 8.6 83

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Example 8.6 84

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Example 8.6 85

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Example 8.6 86

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Example 8.6 87

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Example 8.6 88 The maximum stirrup spacing is 5 in whenever V s exceeds 43,820. From the figure, this value is at about 56 in from the left end of the beam.

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Example 8.6 89 The maximum permissible stirrup spacing is the smaller of the two following values:

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Example 8.6 90 Distance from face of support (ft) V u (lb)V s (lb)Theoretic al s (in) Maximum Spacing (in) 0 to d = 1.875 61,33059,8704.41 356,00057,7605.00 6-44,00036,7607.185.00 6+24,00010,09026.1610.00

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Example 8.6 91

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Example 8.6 92 Selected spacing: 1 @ 3 in = 3 in 17 @ 4 in = 68 in 5@ 10 in = 50in

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