A. Safe in shear
B. Is safe with stirrups
C. Is safe with stirrups and inclined members
D. Needs revision of the section
Answer: Option A
This Question Belongs to Civil Engineering >> RCC Structures Design
Distribution of shear intensity over a rectangular section of a beam, follows:
A. A circular curve
B. A straight line
C. A parabolic curve
D. An elliptical curve
If the shear stress in a R.C.C. beam is
A. Equal or less than 5 kg/cm2, no shear reinforcement is provided
B. Greater than 4 kg/cm2, but less than 20 kg/cm2, shear reinforcement is provided
C. Greater than 20 kg/cm2, the size of the section is changed
D. All the above
In a pre-stressed member it is advisable to use
A. Low strength concrete only
B. High strength concrete only
C. Low strength concrete but high tensile steel
D. High strength concrete and high tensile steel
In a simply supported slab, alternate bars are curtailed at
A. $${\frac{1}{4}^{{\text{th}}}}$$ of the span
B. $${\frac{1}{5}^{{\text{th}}}}$$ of the span
C. $${\frac{1}{6}^{{\text{th}}}}$$ of the span
D. $${\frac{1}{7}^{{\text{th}}}}$$ of the span
Step-1 Max Shear force Simply Supported Beam =Vmax= wL/2 = (2000x5)/2 = 5000 kg
Step-2 Nominal Shear Force Produced due to External Loading for Simply Supported Beam =Vmax / bd =5000/25x45 = 4.44 kg./cm2.
Step-3 Check for Permissible Shear Stress in Section
Condition i. if < 5 kg/cm2 i.e Safe in Shear no reinforcement required.
Condition ii. if b/w 5kg/cm2 i.e to 20 kg/ cm2 Safe in shear but shear Reinforcement Needed.
Condition iii. if > 20 kg/cm2 i.e Section needed to redesign.
As Nominal Shear is 4.44 kg/cm2 which is < 5 kg/cm2 so Beam is Safe in Shear with no reinforcement needed.
As,
Max value of shear stress without striups is 5kg/cm^2
Calculate the stress due to load=shearforce/area
=(2000*5/2)/45*25=4.44
Which is less than 5
Hence it is safe in shear
Lever arm = d-(n/3) #depth of NA = d/2
Substituting value of lever arm and n ...d=54cm
Assuming overall depth with Clear cover ( +25)
D= 80 cm approx or 0.8m ...
S.F= WL/2 : S.F= 5000kg/m2 = 50KN/m2
Shear Stress = S.F/bd = 250KN/m2 =0.25N/mm2
Hence Safe ( as Shear stress < 0.5 N/mm2)
tariq meo
SF = 2000 ÷ (500 * 100) = 0.04 kg/sq cm.
Safe shearing stress = L/A
Load = 20 * 500 = 10000 kg.
Area = 25 * 45 = 1125 sq cm.
Therefore L/A = 8.88 kg/sqcm.
Hence Safe shearing stress > shear force.
Result : Beam is safe in shear.
How to it calculate