Answer:
0.00018784
Explanation:
[tex]L_0[/tex] = Initial length of cylinder = 0.26 m
[tex]\Delta L[/tex] = Change in length
[tex]r_o[/tex] = Outer radius = [tex]1.07\times 10^{-2}\ m[/tex]
[tex]r_i[/tex] = Inner radius = [tex]3.7\times 10^{-3}\ m[/tex]
m = Mass of arm = 57 kg
g = Acceleration due to gravity = 9.81 m/s²
A = Area
Y = Young's modulus = [tex]9.4\times 10^9\ N/m^2[/tex]
When we divide stress by young's modulus we get compressional strain
[tex]\frac{\Delta L}{L_0}=\frac{F}{A}\times \frac{1}{Y}\\\Rightarrow \frac{\Delta L}{L_0}=\frac{mg}{\pi(r_o^2-r_i^2)}\times \frac{1}{Y}\\\Rightarrow \frac{\Delta L}{L_0}=\frac{57\times 9.81}{\pi((1.07\times 10^{-2})^2-(3.7\times 10^{-3})^2)}\times \frac{1}{9.4\times 10^9}\\\Rightarrow \frac{\Delta L}{L_0}=0.00018784[/tex]
Compressional strain of the humerus is 0.00018784
