1. Shear and Diagonal Tension

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.

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

4. 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

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

6. 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

7. Flexural-shear Cracks

8. 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

9. Web-shear Cracks

10. 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

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

12. Types of Stirrups

13. Types of Stirrups

14. Types of Stirrups

15. Types of Stirrups

16. Types of Stirrups

17. Types of Stirrups

18. Types of Stirrups

19. Types of Stirrups

20. Types of Stirrups

21. 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

22. Truss Analogy

23. 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”

24. 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

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

26. Bent-up Bar Web Reinforcement

27. 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

28. 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

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

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

31. Design for Shear

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

33. 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

34. ACI Code Requirements for Shear
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
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.

36. End Shear Reduction Not Permitted

37. End Shear Reduction Not Permitted

38. Corbels

39. ACI Code Requirements Stirrup Hooks

40. ACI Code Requirements Stirrup Hooks

41. ACI Code Requirements Stirrup Hooks

42. Shear Design Examples

43. 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.

44. Example 8.1
Shear strength provided by concrete

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

46. 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

47. Example 8.2

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

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

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

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

52. Example 8.2
(c) Vu = 60,000 lb

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

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

55. Example 8.2
(d) Vu = 150,000 lb

56. 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.

57. Example 8.3

58. Example 8.3

59. Example 8.3

60. Example 8.3
Vu = – 14.4x

61. Example 8.3

62. Example 8.3

63. Example 8.3

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

65. Example 8.3
Terminate stirrups when Vu < fVc/2

66. 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

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

68. 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

69. Example 8.4

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

71. 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.

72. 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.

73. Example 8.5

74. 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:

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

76. Example 8.5

77. 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

78. 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

79. 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

80. 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

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

82. 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.

83. Example 8.6

84. Example 8.6

85. Example 8.6

86. Example 8.6

87. Example 8.6

88. 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.

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

90. 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

91. Example 8.6

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