Question 12.P.2: A simply supported beam of span 3.5 m carries a uniformly di...

The maximum bending moment is

M_{\max }=\frac{46.5 \times 3.5^{2}}{8}=71.2 \mathrm{\ kN} \mathrm{m} (see Ex: 3.8)

Also, for the timber

I_{\mathrm{t}}=2\left(\frac{100 \times 75^{3}}{12}+100 \times 75 \times 112.5^{2}\right)=196.9 \times 10^{6} \mathrm{~mm}^{4}

and for the steel

I_{\mathrm{s}}=\frac{2 t \times 300^{3}}{12}=4.5 \times 10^{6} t \mathrm{~mm}^{4}

Then, from Eq. (12.7) \sigma_{\mathrm{t}}=-\frac{M y}{I_{\mathrm{t}}+\frac{E_{\mathrm{s}}}{E_{\mathrm{t}}} I_{\mathrm{s}}},

8=\frac{71.2 \times 10^{6} \times 150}{(196.9+20 \times 4.5 t) \times 10^{6}}

so that

t = 12.6 mm

From Eq. (12.8) \sigma_{\mathrm{s}}=-\frac{M y}{I_{\mathrm{s}}+\frac{E_{\mathrm{t}}}{E_{\mathrm{s}}} I_{\mathrm{t}}}.

124=\frac{71.2 \times 10^{6} \times 150}{\left(4.5 t+\frac{196.9}{20}\right) \times 10^{6}}

which gives t = 17.0 mm.

Therefore the required thickness of the steel plates is 17 mm.