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

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