- The shear stress develops in beam due to longitudinal- transverse displacement.
Shear stress distributions
(a) Ultimate stage -
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| doubly reinforced concrete |
 |
| singly reinforced beam |
(b) Elastic stage -
shear in RC beam resisted by following sections -
1. Dowel action of main steel
2. Aggregate interlocking
3. Uncracked sections
Critical section for shear -
(a) When supports are in Compression -
- In this case , critical section occurs at 'd' distance from face of support.
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| Supports in compression |
(b) When supports are in tension then critical section occurs at face of support.
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| supports are in tension |
τ = Vu/bd
- If depth is varrying -
τ = (Vu +,- M/d • tanβ) / bd
τ = Nominal shear stress , Vu = Shear force,
b= width , d = depth , M= Bending moment
β = angle between upper and lower fibre in beam.
+ve = when moment decrease with depth
-ve = when moment increase with depth
Maximum shear stress -
- τcmax = 0.62√fck
- when τ > τcmax then " Diagonal Compression failure" occurs.
Design shear stress -
- Design shear stress (τc) is determined based on grade of concrete and percentage of steel. with help of IS 456-2000 code.
(a) If τ < τc/2 - No shear reinforcement provides.
(b) If τc/2 < τ < τc = Minimum shear reinforcement provides -
Sv = (0.87fy• Asv) / 0.4b
(c) If τ-τc < 0.4 , then minimum reinforcement provides.
(d) If τ>τc , then design of stirrups -
- For vertical stirrups -
Sv = 0.87fy•Asv •d / Vs
,
Sv max. = 0.75d or, 300mm whichever is minimum
- For Inclind stirrups -
Sv = 0.87fy•Asv •d( cosα + sinα )/ Vs
- Maximum spacing is " d "
Sv = spacing
d = effective depth
fy = yield stress of steel
α = inclination of stirrups , should not be less than 45°
Asv = Area of stirrups
Vs = shear in steel , Vs = (τ-τc)•bd
Note -
in case of RC beam , when-
Shear stress < 5kg/cm2 = No shear reinforcement provides
Shear stress = 5 - 20Kg/cm2 = minimum shear reinforcement provides.
Shear stress >5kg/cm2 , then Dimension of beam need to be changed.
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