WSM design of 300x450 section

Assumptions: the 30 kN/m includes self weight; effective cover 50 mm so $d=450-50=400$ mm. M20: $\sigma_{cbc}=7$, $m=13.33$; mild steel: $\sigma_{st}=140$ N/mm$^2$.

Step 1: Bending moment

$$M=\frac{wL^2}{8}=\frac{30\times3.6^2}{8}=48.60\ \text{kN·m}$$

Step 2: Balanced section capacity

$$k=\frac{13.33\times7}{13.33\times7+140}=0.400,\quad j=0.867,\quad Q=1.213$$ $$M_{bal}=Qbd^2=1.213\times300\times400^2=58.23\ \text{kN·m}>M$$

So the section is under-reinforced and singly reinforced steel is enough.

Step 3: Area of steel

$$A_{st}=\frac{M}{\sigma_{st}\,j\,d}=\frac{48.60\times10^6}{140\times0.867\times400}=1001\ \text{mm}^2$$

(Using $j$ of the balanced section is slightly conservative.) Minimum steel $0.85bd/f_y=408$ mm$^2$ is satisfied.

Provide 5-16 mm dia bars, $A_{st}=1005$ mm$^2$.

Check: neutral axis $x=149.57$ mm $<x_c=kd=159.98$ mm; $M_r=49.28$ kN·m $>$ 48.60 kN·m, safe.

Answer: Section 300 x 450 mm with 5-16 mm bars at effective depth 400 mm (plus nominal 2-10 mm hanger bars and stirrups).