A timber beam 15 cm wide and 20 cm deep carries a uniformly distributed load over a span of 4 m and is simply supported. If the permissible stresses are 30 N/m² longitudinally and 3 N/mm² in transverse shear, calculate the maximum load which can be carried by the timber beam.

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Accepted Answer

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

[latex] V_{\max }=\frac{w_{ o } L}{2} [/latex]

and maximum bending moment as

[latex] M_{\max }=\frac{w_{ o } L^2}{8} [/latex]

Now,

[latex] \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} ight group \sigma_{\max } \end{aligned} [/latex]

Therefore,

[latex] w_{ o }=\frac{4}{3}\left\lgroup \frac{150 imes 200^2}{400^2} ight group 30=15 N / mm [/latex]

Again for rectangular section the maximum shear stress is:

[latex] \begin{aligned} au_{\max } & =\frac{3}{2} \frac{V_{\max }}{A} \ & =\frac{3}{2} \frac{V_{\max }}{b h} \end{aligned} [/latex]

and [latex] =\frac{w_{ o } L}{2}=\frac{2}{3} au_{\max } b h [/latex]

Therefore,

[latex] w_{ o }=\frac{4}{3} \frac{b h}{L} au_{\max }=\frac{4}{3} \frac{150 imes 200}{4000} imes 3=30 N / mm [/latex]

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

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

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