4Shear Strength of Concrete ACI Code Equation 11-3 – conservative but easy to useACI Code Equation 11-5 – less conservative but “difficult” to use
5Shear Strength of Concrete Equation 11-3Equation 11-5
6Shear Cracking of Reinforced Beams Flexural shear crack – initiate from top of flexural crackFor flexural shear cracks to occur, moment must be larger than the cracking moment and shear must be relatively largeFlexural shear cracks oriented at angle of approximately 45 degrees to the longitudinal beam axis
8Shear Cracking of Reinforced Beams Web shear crack – form independentlyTypically occur at points of small moment and large shearOccur at ends of beams at simple supports and at inflection points at continuous beam
10Web Reinforcement Stirrups Hangers At a location, the width of a diagonal crack is related to the strain in the stirrup – larger strain = wider crackTo reduce crack width, stirrup yield stress is limited to 60 ksi
11Web Reinforcement Small crack widths promote aggregate interlock Limiting stirrup yield stress also reduces anchorage problems
21Behavior of Beams with Web Reinforcement Truss analogyConcrete in compression is top chordLongitudinal tension steel is bottom chordStirrups form truss verticalsConcrete between diagonal cracks form the truss diagonals
23Required Web Reinforcement – ACI Code Required for all flexural members except:Footings and solid slabsCertain hollow core unitsConcrete floor joistsShallow beams with h not larger than 10”
24Required 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 widthBeams constructed with steel fiber- reinforced concrete with strength not exceeding 6,000 psi and
25StirrupsDiagonally inclined stirrups more efficient than vertical stirrupsNot practicalBent-up flexural bars can be used instead
27Shear CrackingThe presence of stirrups does not materially effect the onset of shear crackingStirrups resist shear only after cracks have occurredAfter cracks occurs, the beam must have sufficient shear reinforcement to resist the load not resisted by the concrete in shear
28Benefits of Stirrups Stirrups carry shear across the crack directly Promote aggregate interlockConfine the core of the concrete in the beam thereby increasing strength and ductilityConfine the longitudinal bars and prevent cover from prying off the beamHold the pieces on concrete on either side of the crack together and prevent the crack from propagating into the compression region
29Design for Shear Stirrups crossing a crack are assumed to have yielded Shear crack forms at a 45 degree angle
30Design for ShearACI Code Equation 11-15ACI CodeEquation 11-16
32ACI Code Requirements for Shear ACI Code Section – if Vu exceeds one-half fVc, stirrups are requiredWhen shear reinforcement is required, ACI Code Section specifies a minimum amount:
33ACI 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.Ifmaximumspacingsare halved. See ACI Code Section
34ACI Code Requirements for Shear See ACI Code SectionACI Code Section >Stirrups should extend as close as cover requirements permit to the tension and compression faces of the member - anchorage
35ACI Code Requirements for Shear Stirrup hook requirements shown on the next slide. See ACI Code Section 8.1 and 12.13In deep beams ( l /d < 4), large shear may affect flexural capacityIn most cases, beam can be designed for shear at a distance d from the face of the support . See next three slides for exceptions.
43Example 8.1Determine 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.
45Example 8.1 ACI Code Section 184.108.40.206 requires: Use a 24 in x 36 in beam (d = 33.5 in)
46Example 8.2The 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
66Distance from face of support (ft) Example 8.3Distance from face of support (ft)Vu (lb)Vs (lb)Theoretical s (in)1.875’=22.5”73,80055,7095.332’ = 24”72,00053,3095.573’ = 36”57,60034,1098.713.058’= 36.69”56,76833,00094’= 48”43,20014,909> Maximum d/2 = 11.255.89’ = 70.66”16,008-terminate
67Example 8.3 Corrected spacing: Cumulative (in) 1 @ 2 in = 2 in 2
68Example 8.4Determine 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
70Example 8.4Using the simplified formula (ACI Eq. 11-3) for Vc results in a value of 42,691 lb1-5
71Example 8.5Select 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.
72Example 8.5Positioning 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.
74Example 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:
75Example 8.5Approximating the shear envelope with a straight line between k at the face of the support and 16.8 k at midspan100.8 k78.3 kSlope = ( )/7 ft = 12 k/ftfVc = k16.8 kStirrups carry shearfVc /2 = kconcrete carries shear
77Example 8.5 Design the stirrups using the shear envelop Terminate stirrups when Vu < fVc/2This location is past midspan, so stirrups cannot be terminated
78Example 8.5At x = d = 22.5 inResults of similar calculations at other assumed values of x are shown in the table that follows
79Example 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
80Distance from face of support (ft) Example 8.5Distance from face of support (ft)Vu (lb)Vs (lb)Theoretical s (in)0 to d = 1.87578,30061,7094.81276,80059,7094.972.638’=31.66”69,14349,5006.00364,80043,7096.793.67’=44”56,76833,0009.00452,80027,70910.714.04 = 49”52,26827,00011.00540,80011,709> Maximum d/2 = 11.257.066’= 84.8”16,009Terminate
81Example 8.5 Selected spacing: Cumulative (in) 1 @ 2 in = 2 in 2
82Example 8.6Select 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.
88Example 8.6The 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.
89Example 8.6The maximum permissible stirrup spacing is the smaller of the two following values:
90Distance from face of support (ft) Example 8.6Distance from face of support (ft)Vu (lb)Vs (lb)Theoretical s (in)Maximum Spacing (in)0 to d = 1.87561,33059,8704.41356,00057,7605.006-44,00036,7607.186+24,00010,09026.1610.00