Question

A pipe is subjected to a tension force of P = 90 kN. The pipe outside diameter is 45 mm, the wall thickness is 5 mm, and the elastic modulus is E = 150 GPa. Determine the normal strain in the pipe. Express the strain in mm/mm.

Answer 1

Answer:

The strain is $1.79\times10^{-3}$

Explanation:

Given that,

Force = 90 kN

Outside diameter = 45 mm

Thickness = 5 mm

Elastic modulus = 150 GPa

We need to calculate the inner diameter

Using formula of inner diameter

$D_{in}=D_{o}-t$

Where, t = thickness

Put the value into the formula

$D_{in}=45-5$

$D_{in}=40\ mm$

We need to calculate the area of the pipe

Using formula of area

$A=\dfrac{\pi}{4}(D_{o}^2-D_{in}^2)$

Put the value into the formula

$A=\dfrac{\pi}{4}((45)^2-(40)^2)$

$A=333.79\ mm^2$

We need to calculate the stress

Using formula of stress

$\sigma=\dfrac{P}{A}$

Put the value into the formula

$\sigma=\dfrac{90\times10^{3}}{333.79}$

$\sigma=269.63\ N/mm^2$

We need to calculate the strain in the pipe

Using formula of elastic modulus

$E=\dfrac{stress}{strain}$

Put the value into the formula

$150\times10^{3}=\dfrac{269.63}{strain}$

$strain=\dfrac{269.63}{150\times10^{3}}$

$strain=0.00179$

$strain=1.79\times10^{-3}$

Hence, The strain is $1.79\times10^{-3}$