Question

A cylindrical specimen of some metal alloy 11.1 mm (0.4370 in.) in diameter is stressed elastically in tension. A force of 14000 N (3147 lbf) produces a reduction in specimen diameter of 7 × 10-3 mm (2.756 × 10-4 in.). Compute Poisson's ratio for this material if its elastic modulus is 100 GPa (14.5 × 106 psi).

Answer 1

To develop this problem it is necessary to apply the concepts related to the uni-axial deflection of bodies.

From the expression of Hooke's law we have to

$\sigma = E\epsilon$

Where,

E= Young's modulus

$\epsilon =$ The strain

And substituting P/A for stress and $\delta/L$ for strain gives that

$\frac{P}{A} = E\frac{\delta}{L}$

Where,

P = Force

A = Area

L = Length

Therefore this can be re-arranged to give

$\delta = \frac{PL}{AE}$

If we want to calculate the deformation per unit area then we can also rewrite the equation as

$\frac{delta}{L} = \frac{P}{AE}$

Replacing with our values we have to

$\frac{delta}{L} = \frac{14400}{(\pi/4 (11.1*10^{-3})^2)(100*10^9)}$

$\frac{delta}{L} = 0.001488$

Therefore the posion ratio would be

$\upsilon = \frac{\frac{\delta_{decreased}}{d}}{\frac{\delta_l}{L}}$

$\upsilon = \frac{\frac{7*10^{-3}}{11.1}}{0.001488}$

$\upsilon = 0.4238$

Therefore the Poisson's ratio for this material is 0.4238