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

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)=(\Delta length)/(Length_(original))\\\\\therefore \varepsilon _(axial)=(0.7)/(800)=8.75* 10^(-4)`

Thus the lateral strain becomes

`\varepsilon _(lateral)=-\mu\varepsilon _(axial)\\\\\varepsilon _(lateral)=-0.27* 8.75* 10^(-4)=-2.36* 10^(-4)`

now by definition of lateral strain we have

`\varepsilon _(lateral)=(\Delta diameter)/(diameter_(original))\\\\\Rightarrow \Delta Diameter=-2.36* 10^(-4)* 32=-7.56* 10^(-3)\\\\D_(f)-D_(i)=-7.56* 10^(-3)\\\\D_(f)=32-7.56* 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* 10^(-4)* 150* 10^(9)\\\\\sigma =131.25MPa`

Thus the force is calculated as

`Force=\sigma * Area\\\\Force=131.25* 10^(6)* (\pi (32* 10^(-3))^(4))/(4)\\\\Force=105.55kN`