Question

A doubly reinforced rectangular concrete beam has a width of 300 mm and an effective depth of 500 mm. The beam is reinforced with 2200 mm ${}^2$ of steel in tension and 628 mm${}^2$ of steel in compression. The effective cover for compressive steel is 50 mm. Assume that both tensioned and compressive steel yield. The grades of concrete and steel used are M-20 and Fe-250 respectively. The stress block parameters (rounded off to first two decimal places) for concrete shall be as per IS : 456-2000.

The moment of resistance of the section is

• A. 206.00 kN-m
• B. 209.20 kN-m
• C. 273.80 kN-m
• D. 251.90 kN-m

Answer 1

The correct option is B 209.20 kN-m


Grade of concrete = M20
Grade of steel = Fe-250

Effective depth, d = 500 mm
Width, b = 300 mm
Both tension and compressive steel yield .
So stress in both the bar
$=0.87f_y$
$=0.87\times 250$
Depth of Neutral Axis $\left(x_u\right)$,
C = T
$0.36f_{ck}b.x_u+A_{SC}\times (0.87f_y-0.45f_{ck})$
$=0.87f_y.A_{st}$
$0.36\times 20\times 300\times x_u+628(0.87\times 250-0.45\times 20)$
$=0.87\times 250\times 2200$
$x_u=160.908$ mm
= 160.91 mm
Maximum depth of Neutral axis
= 0.53 d
=0.53$\times$500
$x_{umax}$= 265 mm
$\Rightarrow x_u<x_{umax}$
Hence O.K.
Moment of resistance of the section
$M_u=0.36f_{ck}.b{x}_u(d-0.42x_u)+(0.87f_y-0.45f_{ck})A_{sc}\times (d-50)$
$=0.36\times 20\times 300\times 160.91(500-0.42\times 160.91)+(0.87\times 250-0.45\times 20)\times (500-50)$
= $209.21\times {10}^6$ N.mm
= 209.21 kNm

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Note : If we calculate $M_u$ from tension side, then we have to calculate the point of application of compressive force to calculate the lever arm. So it is better to calculate $M_u$ from compression side.
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