Given    $P=300 N \quad S_{y t}=400 N / mm ^{2} \quad\left(f_{s}\right)=2.5$.

(depth/width) = 3.

Step I Calculation of permissible tensile stress

$\sigma_{t}=\frac{S_{y t}}{(f s)}=\frac{400}{2.5}=160 N / mm ^{2}$.

Step II Calculation of direct and bending stresses
The stresses at section XX consist of direct compressive stress and bending stresses. The tensile stress is maximum at the lower fibre. At the lower fibre,

$\sigma_{c}=\frac{P}{A}=\frac{300}{(t)(3 t)}=\frac{100}{t^{2}} N / mm ^{2}$            (i).

$\sigma_{b}=\frac{M_{b} y}{I}$  .

$=\frac{(300 \times 200)(1.5 t)}{\left[\frac{1}{12}(t)(3 t)^{3}\right]}=\frac{40000}{t^{3}} N / mm ^{2}$              (ii).

Step III Calculation of dimensions of cross-section
Superimposing the two stresses and equating it to permissible stress

$\frac{40000}{t^{3}}-\frac{100}{t^{2}}=160$.

$\text { or } 160 t^{3}+100 t-40000=0$.

Solving the above equation by trail and error method,

$t \cong 6.3 mm$.