WSM tensile steel for 230x425 beam

Assumptions: effective cover $d'=40$ mm (top and bottom), so $d=425-40=385$ mm. M20: $\sigma_{cbc}=7$ N/mm$^2$, $m=13.33$. Fe500: $\sigma_{st}=275$ N/mm$^2$ (IS 456 Table 22 / Annex B). $M=64$ kN·m.

Step 1: Balanced moment of the singly reinforced section

$$\begin{aligned} k&=\frac{13.33\times7}{13.33\times7+275}=0.2533,\quad j=1-\frac{k}{3}=0.9156\\ Q&=\tfrac12\sigma_{cbc}kj=0.8118\ \text{N/mm}^2\\ M_{bal}&=Qbd^2=0.8118\times230\times385^2=27.68\ \text{kN·m} \end{aligned}$$

Since $M=64>M_{bal}=27.68$ kN·m, a singly reinforced section is not enough. The beam must be doubly reinforced.

Step 2: Moment shared by compression steel

$$M_2=M-M_{bal}=64-27.68=36.32\ \text{kN·m}$$

Step 3: Compression steel

Depth of neutral axis (balanced) $x_c=kd=97.54$ mm. Stress in concrete at the compression steel level:

$$\sigma_c'=\sigma_{cbc}\frac{x_c-d'}{x_c}=7\times\frac{97.54-40}{97.54}=4.13\ \text{N/mm}^2$$ $$A_{sc}=\frac{M_2}{(1.5m-1)\,\sigma_c'\,(d-d')}=\frac{36.32\times10^6}{(1.5\times13.33-1)\times4.13\times(385-40)}=1342\ \text{mm}^2$$

Stress in compression steel $\sigma_{sc}=1.5m\sigma_c'=82.57$ N/mm$^2$, below the permissible value, OK.

Step 4: Tension steel

$$A_{st}=\frac{M_{bal}}{\sigma_{st}jd}+\frac{M_2}{\sigma_{st}(d-d')}=\frac{27.68\times10^6}{275\times0.9156\times385}+\frac{36.32\times10^6}{275\times(385-40)}$$ $$A_{st}=286+383=668\ \text{mm}^2$$

Answer: tensile reinforcement $A_{st}=668$ mm$^2$ (e.g. 3-20 mm bars); compression reinforcement $A_{sc}=1342$ mm$^2$ (e.g. 7-16 mm bars).