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CE 334 –
Fall 2018 Week 10 Homework Quiz
Name:________________
1. True or false: Stirrups prevent shear cracks from starting.
2. Why don’t the stirrups start to contribute to the shear strength until after inclined cracks are
formed?
3. List three benefits from using stirrups in a beam.
4. What is the stress in the stirrups assumed to be at failure?
5. What is the shear area Av in in2 of a #4 bar used as a double leg stirrup?
6. Why do we put bends on the tops of U-shaped stirrups?
7. What do we assume is the angle of the diagonal crack when calculating the shear strength of the
stirrups?
8. How do we calculate the number of stirrups n crossing a shear crack in the equation for the
shear strength of the stirrups Vs = Avfytn (in other words, what replaces n in the equation)?
9. What are the most common bar sizes used for stirrups?
CE 334 –
Fall 2018 Week 10 Homework Quiz
Name:________________
10. List four exceptions to the requirement to provide at least a minimum amount of stirrups
wherever Vu > ½ ϕVc.
11. True or false: The requirement for minimum area of shear reinforcement is identical to the
equation used for the minimum area of moment reinforcement.
12. Why does the Code limit the maximum spacing used for stirrups to s = d/2?
13. When is the maximum stirrup spacing requirement cut in half from s = d/2 to s = d/4?
14. Why does the Code limit the amount of shear reinforcement so that Vs < 8√f’c bwd? 15. What must the designer do if Vs must be larger than 8√f’c bwdto satisfy Vu < ϕVc + ϕVs? 16. Where do we find the maximum design shear required by the Code in beams that have uniformly distributed loads and compression (i.e. upward) reactions? CE 334 – Fall 2018 Week 10 Homework Quiz Name:________________ 17. True or false: Full uniformly distributed dead and live load over the entire span of a simply supported beam is the load pattern that causes maximum shear at the supports. 18. How do we apply uniformly distributed pattern live load on a simply supported beam to get maximum shear at mid-span? 19. How is the approximate shear force envelope for pattern loading in a simply supported beam drawn? 20. What is the maximum value for Vu from uniformly distributed pattern loading at the midpoint of a simply supported beam? Shear Reinforcement and ACI Code Design for Shear 1 ACI Code Equations for Shear Strength of Concrete Vn = Vc + Vs Vu  Vn  Vc + Vs Vc = 2 f b d ' c w ϕ = 0.75 for shear ACI Equation 22.5.5.1  Vu d  ' Vc = 1.9 f c + 2500  w  bw d Mu    3.5 f b d ' c w ACI Table 22.5.5.1 As w = bw d Vu d 1 Mu 2 Concrete vs Stirrup Contribution to Shear Resistance The presence of stirrups does not affect the onset of shear cracking Prior to cracking, strain in stirrups ≈ strain in concrete, so stirrups resist shear only after inclined cracks have occurred After cracks occur, the shear reinforcement must be able to pick up the shear that the concrete can’t 3 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 of concrete on either side of the crack together and prevent the crack from propagating into the compression region 4 Shear Strength of Stirrups • Stirrups are used to ensure full flexural capacity can be developed • Stirrups minimize size of diagonal tension cracks and carry tension stress from one side of the crack to the other • Prior to inclined cracking, stirrups carry very little tension due to strain level • Stirrups crossing a crack are assumed to yield at failure Single Leg Types of Stirrups Double Leg Closed 6 Shear Strength of Stirrups Vs = Av f yt n where n = number of stirrups crossing a 45o crack d n= s d Vs = Av f yt s Av f yt d s= Vs Vs = ACI Code Equation 22.5.10.5.3 Av f yt ( sin  + cos  ) d s ACI Code Equation 22.5.10.5.4 for inclined stirrups 7 ACI Code Requires Web Reinforcement for all Flexural Members Except: ACI Code Section 9.6.3.1 – if Vu exceeds one-half Vc, stirrups are required except for: • Footings and solid slabs • Certain hollow core units • Concrete floor joists • Shallow beams with h not larger than 10” • Beams built integrally with slabs with h < 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 Vu  2 f c' bw d 8 ACI Code Minimum Requirements for Shear When shear reinforcement is required, ACI Code Section 9.6.3.3 specifies a minimum amount: Av ,minimum = 0.75 f c' bw s f yt 50bw s  f yt 9 ACI Code Requirements for Stirrup Spacing 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 → Vs  4 f c' bw d maximum spacings are halved (ACI Code Section 9.7.6.2.2) 10 ACI Code Requirements for Shear Vs  8 f c' bw d See ACI Code Section 22.5.1.2 Prevents excessive shear crack widths at service loads, as well as shear failure due to crushing of the web ACI Code Section 22.5.3.2 ->
f c’  100 psi
fyt is limited to 60 ksi
Stirrups should extend as close as cover requirements permit to
the tension and compression faces of the member – anchorage
11
Critical Sections for Maximum Shear in
Beams

Maximum shear is typically found at supports
from the support into the span, loads applied within distance d from the
support are transferred directly to the support by compression concrete
above the crack. Under such conditions, ACI permits sections located less
than a distance d from face of the support to be designed for the same
shear Vu as that computed at distance d.
Reduction must be applied carefully, as not applicable in all cases
Mid-span
•Maximum shear at span ends/
supports occurs with live load over
entire span
•Maximum shear at other points
occur with live load acting over the
portion of the span from that
location to the furthest support
•Maximum shear at mid-span occurs
with live load acting over half span
•Can approximate maximum shear
at other locations by linear shear
force envelope connecting maximum
shear at supports and mid-span

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