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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

Where support loading conditions ensure inclined cracks form outward

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

Pattern Loading Effect on Shear at

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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