Respuesta :
Answer:
The strain is [tex]1.79\times10^{-3}[/tex]
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
[tex]D_{in}=D_{o}-t[/tex]
Where, t = thickness
Put the value into the formula
[tex]D_{in}=45-5[/tex]
[tex]D_{in}=40\ mm[/tex]
We need to calculate the area of the pipe
Using formula of area
[tex]A=\dfrac{\pi}{4}(D_{o}^2-D_{in}^2)[/tex]
Put the value into the formula
[tex]A=\dfrac{\pi}{4}((45)^2-(40)^2)[/tex]
[tex]A=333.79\ mm^2[/tex]
We need to calculate the stress
Using formula of stress
[tex]\sigma=\dfrac{P}{A}[/tex]
Put the value into the formula
[tex]\sigma=\dfrac{90\times10^{3}}{333.79}[/tex]
[tex]\sigma=269.63\ N/mm^2[/tex]
We need to calculate the strain in the pipe
Using formula of elastic modulus
[tex]E=\dfrac{stress}{strain}[/tex]
Put the value into the formula
[tex]150\times10^{3}=\dfrac{269.63}{strain}[/tex]
[tex]strain=\dfrac{269.63}{150\times10^{3}}[/tex]
[tex]strain=0.00179[/tex]
[tex]strain=1.79\times10^{-3}[/tex]
Hence, The strain is [tex]1.79\times10^{-3}[/tex]
