An R.C.C beam of 25 cm width has a clear span of 5 metres and carries a U.D.L. of 2000 kg/m inclusive of its self weight. If the lever arm of the section is 45 cm., the beam is
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
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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
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