Question 6.9: A timber beam 15 cm wide and 20 cm deep carries a uniformly ...

For the uniformly distributed loading of intensity $w_{ o }$ on a simply supported beam of length L, we have maximum shear force as

$V_{\max }=\frac{w_{ o } L}{2}$

and maximum bending moment as

$M_{\max }=\frac{w_{ o } L^2}{8}$

Now,

$\begin{aligned} \sigma_{\max } & =\frac{M_{\max }}{z}=\frac{6 M_{\max }}{b h^2} \\ \Rightarrow \quad M_{\max } & =\frac{w_{ o } L^2}{8}=\frac{b h^2}{6} \sigma_{\max } \\ \Rightarrow \quad w_{ o } & =\frac{4}{3}\left\lgroup\frac{b h^2}{L^2} \right\rgroup \sigma_{\max } \end{aligned}$

Therefore,

$w_{ o }=\frac{4}{3}\left\lgroup \frac{150 \times 200^2}{400^2} \right\rgroup 30=15  N / mm$

Again for rectangular section the maximum shear stress is:

$\begin{aligned} \tau_{\max } & =\frac{3}{2} \frac{V_{\max }}{A} \\ & =\frac{3}{2} \frac{V_{\max }}{b h} \end{aligned}$

and        $=\frac{w_{ o } L}{2}=\frac{2}{3} \tau_{\max } b h$

Therefore,

$w_{ o }=\frac{4}{3} \frac{b h}{L} \tau_{\max }=\frac{4}{3} \frac{150 \times 200}{4000} \times 3=30  N / mm$

Clearly, the maximum permissible loading intensity is $w_{ o }=15$ N/mm.