Question

A cylindrical specimen of some metal alloy 11.2 mm (0.4409 in.) in diameter is stressed elastically in tension. A force of 14100 N (3170 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

Answer:

μ = 0.436

Explanation:

Given:

Change in diameter, ΔD = 7 × 10⁻³ mm

Original diameter, D = 11.2 mm = 11.2 × 10⁻³ m

Applied force, P = 14100 N

Cross-section area of the specimen, A = $\frac{\pi}{4}D^2$ =  $\frac{\pi}{4}(11.2\times 10^{-3})^2$

Now,

elongation due to tensile force is given as:

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

or

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

on substituting the values, we get

$\frac{\delta}{L} = \frac{14100}{\frac{\pi}{4}(11.2\times 10^{-3})^2\times100\times 10^9}$

or

$\frac{\delta}{L} = 0.00143=\epsilon_x$

where,

$\epsilon_x$ is the strain in the direction of force

Now,

$\epsilon_z=\frac{\Delta D}{D}=\frac{7\times 10^{-3}}{11.2}=0.000625$

now, the poisson ratio, μ is given as:

$\mu=\frac{\epsilon_z}{\epsilon_x}$

on substituting the values we get,

$\mu=\frac{0.000625}{0.00143}=0.436$

Answer 2

Answer:

The Poisson's ratio for this material is 0.4370.

Explanation:

Given that,

Diameter of metal = 11.2 mm

Force = 14100 N

Reduction diameter $d=7\times10^{-3}\ mm$

Elastic modulus = 100 GPa

We need to calculate the change in length

Using formula of modulus elasticity

$E=\dfrac{FL}{A\Delta L}$

The change in length is

$\Delta L=\dfrac{FL}{AE}$

$\dfrac{\Delta L}{L}=\dfrac{14100}{\pi\times\dfrac{(11.2\times10^{-3})^2}{4}100\times10^{9}}$

$\dfrac{\Delta L}{L}=0.00143$

We need to calculate the Poisson's ratio

Using formula of Poisson's ratio

$\nu=\dfrac{longitudinal\ strain}{Transverse strain}$

$\nu=\dfrac{-\dfrac{\Delta d}{d}}{-\dfrac{\Delta L}{L}}$

Put the value into the formula

$\nu=\dfrac{\dfrac{7\times10^{-6}}{11.2\times10^{-3}}}{0.00143}$

$\nu=0.4370$

Hence, The Poisson's ratio for this material is 0.4370.