Calculate longitudinal stress developed in 2 cm diameter bar undergo tensile force of 120 kN.

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written 4.7 years ago by

teamques10 ★ 65k

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written 4.7 years ago by

teamques10 ★ 65k

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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