An R.C.C. beam of 6 m span is 30 cm wide and has a lever arm of 55 cm. If it carries a U.D.L. of 12 t per m and allowable shear stress is 5 kg/cm2, the beam
A. Is safe in shear
B. Is safe with stirrups
C. Is safe with stirrups and inclined bars
D. Needs revision of section
Answer: Option D
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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
Step 1 Find Maximum shear Force for Simply Supported Beam =Vmax = wl/2= 12x6/2= 36 Ton= 36000 Kg.
Step 2 Find Maximum Shear Force produced in beam due to Loading = Force/Area = Vmax/ b x lever arm
= 36000/30x55 = 21.8 kg/cm2
Step 3 As According to permissible Stress in section
if < 5 kg/cm2 meansSafe in Shear No shear Reinforcemnt needed
if between 5kg/cm2 and 20 kg/cm2 means not safe in shae and Shear Reinforment needed.
if > 20 kg/cm2 then need revise of Section.
As Shear Stress produced in section is 21.8 which >20 kg/cm2 So need revise of section. Option D is Coreect.
Step 1. Find Maximum Shear Force= Vmax for Simlpy Supported beam which is = wl/2
= 12x6/2 = 36 ton = 36000 kg.
we know that moment M=V*LA where LA is liver arm (for given signs F=verticle shear force, v=horizontal shear force, x=longitudinal distance)
now v=M/LA , also M=F*x
shear stress S=V/(B*x)
S=F/(B*LA)
for question:
S=12000*6/2*(30*55)=21.82 >8 kg/cm^2 hence unsafe
Divide 12000 kg with 55*30 we will get 7.27 kg/sqcm but in question allowable shear stress is 5kg/sqcm so it will fail and section needs to be redesign.
can any one explain this question