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A load of 6 kN is to be raised with the help of steel cable. Determine the minimum diameter of steel if stress is not to exceed 110 $N/mm^{2}$

written 5.5 years ago by teamques10 ★ 69k

modified 2.9 years ago by RakeshBhuse • 3.2k

strength of materials

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written 5.5 years ago by teamques10 ★ 69k

$$\begin{array}{l}{\text { Data } : \mathrm{P}=6 \mathrm{kN} \quad \sigma=110 \mathrm{N} / \mathrm{mm}^{2}} \ {\text { Find: } \mathrm{d}_{\min }=?} \ {\sigma_{\max }=\frac{p_{\max }}{A_{\min }}} \ {A_{\min }=\frac{p_{\max }}{\sigma_{\min }}}\end{array}$$ $$\begin{array}{l}{A_{\min }=\frac{6 \times 10^{3}}{110}=54.54 m m^{2}} \ {\frac{\pi}{4} \times d^{2} \min =54.54 m m^{2}} \ {d^{2} \min =54.54 \times \frac{4}{\pi}=69.442}\end{array}$$ $$d_{\min }=8.33 m m$$

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