Answer:
0.0735 in
Explanation:
Data provided in the question:
Length of each rod, L = 5 ft = 60 in
Diameter of rod 1, d₁ = [tex]\frac{3}{4}\ in[/tex] = 0.75 in
Diameter of rod 2, d₂ = [tex]\frac{1}{2}\ in[/tex] = 0.5 in
Applied tensile load, P = 5000 lb
Now,
For steel, Modulus of elasticity, E = 30 × 10⁶ psi
Also,
Elongation due to load P = [tex]\frac{PL}{AE}[/tex]
here,
L is the length
A is the are of the cross-section
Therefore,
Total elongation = [tex]\frac{PL}{A_1E}+\frac{PL}{A_2E}[/tex]
or
Total elongation = [tex]\frac{PL}{E}(\frac{1}{\frac{\pi d_1^2}{4}}+\frac{1}{\frac{\pi d_2^2}{4}})[/tex]
or
= [tex]\frac{5000\times60}{30\times10^6}(\frac{1}{\frac{\pi 0.75^2}{4}}+\frac{1}{\frac{\pi 0.5^2}{4}})[/tex]
= 0.01 × ( 2.264 + 5.093 )
= 0.0735 in
