Draw shear stress distribution along cross-section of circular beam for 300 mm diameter carrying 400 kN shear force. Also determine the ratio of maximum shear stress to average stress.

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written 5.5 years ago by

teamques10 ★ 69k

$\begin{array}{l}{\text { Data- } \mathrm{d}=300 \mathrm{mm}, \mathrm{S}=400 \mathrm{kN}} \ {\text { Find }-\frac{\mathrm{q}_{\max }}{\mathrm{q}_{\mathrm{avg}}}=?} \ {\mathrm{R}=\frac{\mathrm{d}}{2}=\frac{300}{2}=150 \mathrm{mm}}\end{array}$

$\begin{array}{l}{\mathrm{R}=\frac{\mathrm{d}}{2}=\frac{300}{2}=150 \mathrm{mm}} \ {\mathrm{A}=\frac{\pi}{8} \times \mathrm{d}^{2}=\frac{\pi}{8} \times(300)^{2}=35342.92 \mathrm{mm}^{2}} \ {\overline{Y}=\frac{4 \mathrm{R}}{3 \pi}=\frac{4 \times 150}{3 \times \pi}=63.662 \mathrm{mm}} \ {\mathrm{b}=\mathrm{d}=300 \mathrm{mm}} \ {\mathrm{I}=\frac{\pi \mathrm{d}^{4}}{64}=\frac{\pi \times(300)^{4}}{64}=397607820.2 \mathrm{mm}^{4}} \ {\mathrm{q}_{\max }=\frac{\mathrm{SA} \overline{\mathrm{Y}}}{\mathrm{bI}}}\end{array}$

$\begin{array}{l}{\mathrm{q}_{\max }=\frac{400 \times 10^{3} \times 35342.92 .92 \times 63.662}{300 \times 397607820.2}=7.545 \mathrm{N} / \mathrm{mm}^{2}} \ {\mathrm{q}_{\mathrm{avg}}=\frac{\mathrm{S}}{\mathrm{A}}=\frac{400 \times 10^{3}}{\frac{\pi}{4} \times(300)^{2}}=5.66 \mathrm{N} / \mathrm{mm}^{2}}\end{array}$

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