Question

In limit state design method, the moment of resistance for a balanced section using M20 grade concrete and HYSD steel of grade Fe415 is given by $M_{u,lim}=Kb{d}^2$, what is the value of K?

• A. 2.98
• B. 2.76
• C. 1.19
• D. 0.89

Answer 1

The correct option is B 2.76

Moment of resistance for a balanced section is given by

$M_{u,lim}=0.36f_{ck}\times \frac{x_{u,lim}}{d}\times (1-0.42\times \frac{x_{u,lim}}{d})\times b{d}^2$

But for Fe $415,x_{u,lim}=0.48d$

$\therefore$ $M_{u,lim}=0.36f_{ck}\times 0.48\times (1-0.42\times 0.48)\times b{d}^2$

$\Rightarrow$ $M_{u,lim}=0.13796f_{ck}\times b{d}^2$

On comparison, we get

$K=0.13796f_{ck}$

$=0.13796\times 20=2.76$