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Calculate longitudinal stress developed in 2 cm diameter bar undergo tensile force of 120 kN.
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written 4.6 years ago by
teamques10
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Data: d = 2 cm, P = 120 kN
Find: $\sigma$
$\sigma=\frac{P}{A}$
$\sigma=\frac{120 \times 10^{3}}{\frac{\pi \times 20^{2}}{4}}$
$\sigma=381.97 \mathrm{N} / \mathrm{mm}^{2}$
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