The elastic limit of brass is 379 MPa. What should be the minimum diameter of a brass rod if it is to support a 400 N load without exceeding its elastic limit ?

Engineering

Physics

Stress Strain and Hookes Law New

Question

1.00 mm

1.16 mm

1.36 mm

0.90 mm

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$\frac{400}{\frac{π}{4} d^{2}} = 379 \times 10^{6}$

$d^{2} = \frac{4 \times 400 \times 10^{- 6}}{π \times 379} = 0 .336 \times 10^{- 6} \times 4$

$d = 2 \sqrt{0 .336} \times 10^{- 3} M ≃ 1 .16 mm$