$Data:b = 40 mm, t = 5mm, R = 2.5mm, E = 2 x 10^{5} MPa.$ Find: $\sigma_{b} and BM$ $\mathrm{I}_{\mathrm{xx}}=\frac{\mathrm{bd}^{3}}{12}=\frac{40 \times 5^{3}}{12}=416.67 \mathrm{mm}^{4}$ $\mathrm{Y}=\frac{\mathrm{t}}{2}=\frac{5}{2}=2.5 \mathrm{mm}$ **Bending stress equation:** $\frac{\sigma_{\mathrm{b}}}{\mathrm{Y}}=\frac{\mathrm{M}}{\mathrm{I}}=\frac{\mathrm{E}}{\mathrm{R}}$ $\frac{\sigma_{\mathrm{b}}}{\mathrm{Y}}=\frac{\mathrm{E}}{\mathrm{R}}$ $\sigma_{\mathrm{b}}=\frac{\mathrm{E}}{\mathrm{R}} \times \mathrm{Y}=\frac{2 \times 10^{5}}{2500} \times 2.5=200 \mathrm{N} / \mathrm{mm}^{2}$ $\frac{\mathrm{M}}{\mathrm{I}}=\frac{\mathrm{E}}{\mathrm{R}}$ $\mathrm{M}=\frac{\mathrm{E}}{\mathrm{R}} \times \mathrm{I}=\frac{2 \times 10^{5}}{2500} \times 416.67=3.33 \times 10^{4} \mathrm{N}-\mathrm{mm}$