Question 6.12: A simply supported beam 100 × 200 mm carries a central conce...

For a simply supported beam carrying a central load W, we know that

\text { Maximum bending moment }=\frac{W L}{4}=M_{\max }

and      \text { Maximum shear force }=\frac{W}{2}=V_{\max }

where L is the beam span.
Now, maximum bending stress for rectangular beam cross-section is given by

\sigma=\frac{6 M_{\max }}{b h^2} \text { (for rectangular beam cross-section) }

Putting values, we get

15=\frac{\left(6 \times W \times 3 \times 10^3\right) / 4}{100 \times 200^2}

or      W = 13.33 kN

Maximum shear stress for rectangular beam cross-section is given by

\tau=\frac{3}{2} \times \frac{V_{\max }}{A}=\frac{3}{2} \times \frac{V_{\max }}{b h}=\frac{3}{2} \times \frac{W / 2}{100 \times 200}=1.2

which gives W = 32 kN. Therefore, to keep both conditions satisfied we must have

W_{\text {safe }}=\operatorname{Min}(13.33  kN , 32.0  kN )=13.33  kN