Answer:
E = 8.83 kips
Explanation:
First, we determine the stress on the rod:
[tex]\sigma = \frac{F}{A}\\\\[/tex]
where,
σ = stress = ?
F = Force Applied = 1300 lb
A = Cross-sectional Area of rod = 0.5[tex]\pi \frac{d^2}{4} = \pi \frac{(0.5\ in)^2}{4} = 0.1963\ in^2[/tex]
Therefore,
[tex]\sigma = \frac{1300\ lb}{0.1963\ in^2} \\\\\sigma = 6.62\ kips[/tex]
Now, we determine the strain:
[tex]strain = \epsilon = \frac{elongation}{original\ length} \\\\\epsilon = \frac{0.009\ in}{12\ in}\\\\\epsilon = 7.5\ x\ 10^{-4}[/tex]
Now, the modulus of elasticity (E) is given as:
[tex]E = \frac{\sigma}{\epsilon}\\\\E = \frac{6.62\ kips}{7.5\ x\ 10^{-4}}[/tex]
E = 8.83 kips
