WSM steel area and resistance, M15

Assumption: the section is given as $b\times d$, so $b=250$ mm and effective depth $d=450$ mm. For M15 concrete $\sigma_{cbc}=5$ N/mm$^2$, $m=\dfrac{280}{3\times5}=18.67$; for mild steel $\sigma_{st}=140$ N/mm$^2$.

Step 1: Balanced section constants

$$\begin{aligned} k&=\frac{m\sigma_{cbc}}{m\sigma_{cbc}+\sigma_{st}}=\frac{18.67\times5}{18.67\times5+140}=0.400\\ j&=1-\frac{k}{3}=0.867\\ Q&=\frac12\sigma_{cbc}\,k\,j=\frac12\times5\times0.400\times0.867=0.867\ \text{N/mm}^2 \end{aligned}$$

Step 2: Moment of resistance (balanced section, maximum for singly reinforced)

$$M_r=Q\,b\,d^2=0.867\times250\times450^2=43.88\ \text{kN·m}$$

Step 3: Area of steel

$$A_{st}=\frac{M_r}{\sigma_{st}\,j\,d}=\frac{43.88\times10^6}{140\times0.867\times450}=803.7\ \text{mm}^2$$

Minimum steel check (IS 456, cl. 26.5.1.1): $0.85bd/f_y=0.85\times250\times450/250=382$ mm$^2$ $<A_{st}$, OK.

Provide 4-16 mm dia bars ($804$ mm$^2$).

Answer: $A_{st}=803.7$ mm$^2$ (say 4-16 mm bars) and $M_r=43.88$ kN·m.