Answer:
The minimum shear strain in the tube is [tex]4.162\times10^{-3}[/tex]
The shear strain at the median line of the tube section is [tex]9.15\times10^{-3}[/tex]
Explanation:
Given that,
Maximum shear strain = 0.005
We need to calculate the minimum shear strain
Using formula of maximum shear strain
[tex]\gamma_{max}=\dfrac{d_{2}}{2}\times\theta[/tex]
[tex]\theta=\dfrac{2\times\gamma_{max}}{d_{2}}[/tex]
Where, [tex]\gamma_{max}[/tex]=maximum shear strain
[tex]\theta[/tex]=angle of twist
[tex]d_{2}[/tex]= diameter
Put the value into the formula
[tex]\theta=\dfrac{2\times0.005}{3}[/tex]
[tex]\theta=0.00333\ rad[/tex]
[tex]\theta=3.33\times10^{-3}\ rad[/tex]
Now, Using formula of minimum shear strain
[tex]\gamma_{min}=\dfrac{d_{1}}{2}\times\theta[/tex]
Put the value into the formula
[tex]\gamma_{min}=\dfrac{2.5}{2}\times3.33\times10^{-3}[/tex]
[tex]\gamma_{min}=0.0041625[/tex]
[tex]\gamma_{min}=4.162\times10^{-3}[/tex]
We need to calculate the shear strain at the median line of the tube section
Using formula of shear strain at the median line
[tex]\gamma=\dfrac{d_{1}+d_{2}}{2}\times\theta[/tex]
Put the value into the formula
[tex]\gamma=\dfrac{2.5+3}{2}\times3.33\times10^{-3}[/tex]
[tex]\gamma=0.0091575[/tex]
[tex]\gamma=9.15\times10^{-3}[/tex]
Hence, The minimum shear strain in the tube is [tex]4.162\times10^{-3}[/tex]
The shear strain at the median line of the tube section is [tex]9.15\times10^{-3}[/tex]
