Question

A 1-ft rod with a diameter of 0.5 in. is subjected to a tensile force of 1,300 lb and has an elongation of 0.009 in. The modulus of elasticity of the material is most nearly:

Answer 1

Answer:

E = 8.83 kips

Explanation:

First, we determine the stress on the rod:

$\sigma = \frac{F}{A}$

where,

σ = stress = ?

F = Force Applied = 1300 lb

A = Cross-sectional Area of rod = 0.5$\pi \frac{d^2}{4} = \pi \frac{(0.5\ in)^2}{4} = 0.1963\ in^2$

Therefore,

$\sigma = \frac{1300\ lb}{0.1963\ in^2} \\\\\sigma = 6.62\ kips$

Now, we determine the strain:

$strain = \epsilon = \frac{elongation}{original\ length} \\\\\epsilon = \frac{0.009\ in}{12\ in}\\\\\epsilon = 7.5\ x\ 10^{-4}$

Now, the modulus of elasticity (E) is given as:

$E = \frac{\sigma}{\epsilon}\\\\E = \frac{6.62\ kips}{7.5\ x\ 10^{-4}}$

E = 8.83 kips