Given:

For rectangular S.S. beam, b = 100 mm. L = 4m, $\sigma_{b} = 40 N /mm^{2} , q_{max} = 80N/mm^{2}$

Solution:

Max B.M = M = $\frac{wl^{2}}{8} \times 10^{6} Nm$

Max S.F = S = $R_{A} = \frac{wL}{2} \times 10^{3}N$

$For beam section, I = \frac{bd^{3}}{12} = \frac{100 \times 150^{3}}{12} = 28.125 \times 10^{6 }mm^{4}$

Value of 'w' for bending stress criteria

$M = \frac{\sigma}{y} \times I$

$10^{6} \times \frac{wl^{2}}{8} = \frac{28 \times 28.125 \times 10^{6}}{75}$

$10^{6} \times \frac{w \times 2^{2}}{8} = \frac{28 \times 28.125 \times 10^{6}}{75}$

$\therefore w = 21 kN/m$

Value of 'w' for shear stress criteria

$q_{max} = 1.5 \frac{S}{A}$

$2 = \frac{1.5 \times w \times 2 \times 10^{3}}{2 \times 100 \times 150}$

$w = 20 kN/m$

$\therefore UDL = minimum of A$ & $ B$

$W = 20 kN/m$