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

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