This question was previously asked in

ESE Civil 2015 Paper 1: Official Paper

Option 3 : 2.0

CT 1: Ratio and Proportion

2672

10 Questions
16 Marks
30 Mins

__Concept:__

The moment of resistance for balanced section is given as

M = Q Bd^{2}

Where

B is the width of member and d is the effective depth

Q is design coefficient and it depends on the method of design.

**As per Limit State Method of Design:**

Q = 0.148f_{ck} for Fe 250

** Q = 0.138f _{ck} for Fe 415**

Q = 0.133f_{ck} for Fe 500

Where f_{ck} is the characteristic compressive strength.

As per Working Stress Method of Design:

Q = 0.174 σ_{cbc} for Fe 250

**Q = 0.131 σ _{cbc} for Fe 415**

Q = 0.116 σ_{cbc} for Fe 500

Where σ_{cbc} is the permissible bending strength for concrete.

__Calculation:__

For M20: σ_{cbc} = 7 MPa and f_{ck} = 20 MPa

Working moment by LSM = \(\frac{{0.138\;{f_{ck}}\;b\;{d^2}}}{{1.5}}\) = 1.84 bd^{2}

Working moment by WSM = 0.131 σ_{cbc} b d^{2} = 0.131 × 7 bd^{2} = 0.917 bd^{2}

∴ \(\frac{{M\left( {LSM} \right)}}{{M\left( {WSM} \right)}} = A = \frac{{1.84}}{{0.917}} = 2\)

**∴**** A = 2 **