Question

A circular copper wire is put in tension under a weight of 7000N. What is the ratio of its diameter after and before the load is applied, if the initial cross section of the wire is 0.01m2 and its Poisson ratio is 0.3

Answer 1

Answer:

$\frac{d_f}{d} =0.9999983$

Explanation:

Given:

• force applied on the copper wire, $F=7000\ N$

• cross sectional area of the wire, $a=0.01\ m^2$

• Poisson's ratio, $\mu=0.3$

• we have, Young's modulus, $E=128\times 10^3\ MPa$

Stress induced due to the applied force:

$\sigma=\frac{F}{a}$

$\sigma=\frac{7000}{0.01}$

$\sigma=700000\ Pa=0.7\ MPa$

Now the longitudinal strain:

$\epsilon=\frac{\sigma}{E}$

$\epsilon=\frac{0.7}{128\times 10^3}$

$\epsilon=5.468\times 10^{-6}$

Now from the relation of Poisson's ratio:

$\mu=\frac{\nu}{\epsilon}$

where:

$\nu=$ lateral strain

$0.3=\frac{\nu}{5.468\times 10^{-6}}$

$\nu=1.6406\times 10^{-6}$ ..................(1)

Now we find the diameter of the wire:

$a=\pi.\frac{d^2}{4}$

$0.01=\pi\times \frac{d^2}{4}$

$\frac{0.04}{\pi} =d^2$

$d=0.1128\ m=112.8\ mm$

When the tensile load is applied its diameter decreases:

The lateral strain is also given as,

$\nu=\frac{\Delta d}{d}$

$1.6406\times 10^{-6}=\frac{\Delta d}{112.8}$

$\Delta d=0.000185\ mm$

Now the final diameter will be:

$d_f=d-\Delta d$

$d_f=112.8-0.000185$

$d_f=112.799815\ mm$

Now the ratio:

$\frac{d_f}{d} =\frac{112.799815}{112.8}$

$\frac{d_f}{d} =0.9999983$