Question

A circular bar is 800mm in length and 32mm in diameter. The bar is made from a material with a modulus of elasticity E = 150 GPa and Poisson’s ratio of 0.27. When a force is applied to the bar, it extends by 0.7mm. What is the change in diameter and the force applied?

Answer 1

Answer:

For any material if ∈ is the axial strain then the lateral strain is given by -μ∈ is the lateral strain in the object

Where,

μ is the poisson's ratio of the material

The longitudinal strain is calculated as follows

$\varepsilon _{axial}=\frac{\Delta length}{Length_{original}}\\\\\therefore \varepsilon _{axial}=\frac{0.7}{800}=8.75\times 10^{-4}$

Thus the lateral strain becomes

$\varepsilon _{lateral}=-\mu\varepsilon _{axial}\\\\\varepsilon _{lateral}=-0.27\times 8.75\times 10^{-4}=-2.36\times 10^{-4}$

now by definition of lateral strain we have

$\varepsilon _{lateral}=\frac{\Delta diameter}{diameter_{original}}\\\\\Rightarrow \Delta Diameter=-2.36\times 10^{-4}\times 32=-7.56\times 10^{-3}\\\\D_{f}-D_{i}=-7.56\times 10^{-3}\\\\D_{f}=32-7.56\times 10^{-3}=31.992mm$

By hookes law the stress developed due to the given strain is given by

$\sigma =\varepsilon _{axial}E$

Applying values we get

$\sigma =8.75\times 10^{-4}\times 150\times 10^{9}\\\\\sigma =131.25MPa$

Thus the force is calculated as

$Force=\sigma \times Area\\\\Force=131.25\times 10^{6}\times \frac{\pi (32\times 10^{-3})^{4}}{4}\\\\Force=105.55kN$