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Shear and Diagonal Tension 1. Acknowledgement This Powerpoint presentation was prepared by Dr. Terry Weigel, University of Louisville. This work and other.

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Presentation on theme: "Shear and Diagonal Tension 1. Acknowledgement This Powerpoint presentation was prepared by Dr. Terry Weigel, University of Louisville. This work and other."— Presentation transcript:

1 Shear and Diagonal Tension 1

2 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

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

4 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

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

6 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

7 Flexural-shear Cracks 7

8 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

9 Web-shear Cracks 9

10 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

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

12 Types of Stirrups 12

13 Types of Stirrups 13

14 Types of Stirrups 14

15 Types of Stirrups 15

16 Types of Stirrups 16

17 Types of Stirrups 17

18 Types of Stirrups 18

19 Types of Stirrups 19

20 Types of Stirrups 20

21 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

22 Truss Analogy 22

23 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”

24 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

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

26 Bent-up Bar Web Reinforcement 26

27 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

28 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

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

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

31 Design for Shear 31

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

33 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

34 ACI Code Requirements for Shear 34 See ACI Code Section ACI Code Section > Stirrups should extend as close as cover requirements permit to the tension and compression faces of the member - anchorage

35 ACI Code Requirements for Shear 35 Stirrup hook requirements shown on the next slide. See ACI Code Section 8.1 and 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

36 End Shear Reduction Not Permitted 36

37 End Shear Reduction Not Permitted 37

38 Corbels 38

39 ACI Code Requirements Stirrup Hooks 39

40 ACI Code Requirements Stirrup Hooks 40

41 ACI Code Requirements Stirrup Hooks 41

42 Shear Design Examples 42

43 Example 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.

44 Example Shear strength provided by concrete

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

46 Example 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

47 Example

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

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

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

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

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

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

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

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

56 Example 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.

57 Example

58 Example

59 Example

60 Example V u = – 14.4x

61 Example

62 Example

63 Example

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

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

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

67 Example Corrected spacing:Cumulative (in) 2 in = 2 in 2 5 in = 35 in37 > in = 36 in73 >70.66

68 Example 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

69 Example

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

71 Example 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.

72 Example 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.

73 Example

74 Example 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:

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

76 Example

77 Example 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

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

79 Example 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

80 Example Distance from face of support (ft) V u (lb)V s (lb)Theoretical s (in) 0 to d = ,30061, ,80059, ’=31.66”69,14349, ,80043, ’=44”56,76833, ,80027, = 49”52,26827, ,80011,709> Maximum d/2 = ’= 84.8”16,009Terminate

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

82 Example 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.

83 Example

84 Example

85 Example

86 Example

87 Example

88 Example 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.

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

90 Example Distance from face of support (ft) V u (lb)V s (lb)Theoretic al s (in) Maximum Spacing (in) 0 to d = ,33059, ,00057, ,00036, ,00010,

91 Example

92 Example Selected spacing: 3 in = 3 in 4 in = 68 in 10 in = 50in


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