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ESE Civil 2020: Official Paper

Option 1 : 128 kN

ST 1: Building Material and Concrete Technology

18104

20 Questions
20 Marks
12 Mins

__Concept:__

The equivalent shear force, V_{e} as per Limit state is given by:

\({V_e} = {V_u} + \frac{{1.6{T_u}}}{b}\)

Where,

V_{u} is factored shear force and T_{u} is factored twisting moment.

B is width of beam.

__Calculation:__

Given,

Width of the beam, b = 200 mm

and depth of the beam, d = 300 mm

A beam is subjected to a limit state of shear of 80 kN and torsional moment of 6 kNm means that it is factored shear force of 80 kN and factored twisting moment of 6 kNm i.e. V_{u} = 80 kN and T_{u} = 6 kNm

Now,

\({V_e} = 80 + \frac{{1.6 \times 6}}{{0.2}}\)

**∴ ****V _{e} = 128 kN**