This question was previously asked in

UPPSC AE Civil 2019 Official Paper I (Held on 13 Dec 2020)

Option 3 : 268 kN

__Concept__

Equivalent Shear force (Ve) is given by

\({V_e} = V + \dfrac{{1.6T}}{B}\)

Where, T = Torque

B = Width of the section, V = shear force

__Calculation:__

Given,

B = 250 mm = 0.25 m,

V = 140 kN, T = 20 kN-m

Equivalent shear force is given by,

\({V_e} = V + \dfrac{{1.6\:\times \:T}}{B}\)

\({V_e} = 140 + \dfrac{{1.6 \times 20}}{{0.25}}\) = **2****68** kN

__Additional Information__

Equivalent Moment (Me) is given by

\({M_e} = M + \dfrac{T}{{1.7}}\left( {1 + \dfrac{D}{B}} \right)\)

Where, M = Bending moment, D = Overall dept, T is Torque