Question

A cylindrical specimen of some alloy 8 mm (0.31 in.) in diameter is stressed elastically in tension. A force of 15,700 N (3530 lbf) produces a reduction in specimen diameter of 5 ´ 10-3 mm (2 ´ 10-4 in.). Compute Poisson's ratio for this material if its modulus of elasticity is 140 GPa (20.3 ´ 106 psi).

Answer 1

Answer:

Poisson's ratio of the given wire is 0.28

Explanation:

Longitudinal extension in the length of the wire is given as

$\frac{\Delta L}{L} = \frac{F}{AY}$

now we have

$\frac{\Delta L}{L} = \frac{15700}{\pi/4 (0.008)^2 (140\times 10^9)}$

so we will have

$\frac{\Delta L}{L} = 2.23 \times 10^{-3}$

now we know that lateral extension of the wire is given as

$\frac{\Delta r}{r} = \frac{5 \times 10^{-3}}{8}$

$\frac{\Delta r}{r} = 6.25 \times 10^[-4}$

Now we know that poisson's ratio is given as

$\eta = \frac{\Delta r/r}{\Delta L/L}$

$\eta = \frac{6.25 \times 10^{-4}}{2.23 \times 10^{-3}}$

$\eta = 0.28$