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

E = 8.83 kips

Step-by-step explanation:

First, we determine the stress on the rod:

`\sigma = (F)/(A)\\\\`

where,

σ = stress = ?

F = Force Applied = 1300 lb

A = Cross-sectional Area of rod = 0.5
`\pi (d^2)/(4) = \pi ((0.5\ in)^2)/(4) = 0.1963\ in^2`

Therefore,

`\sigma = (1300\ lb)/(0.1963\ in^2) \\\\\sigma = 6.62\ kips`

Now, we determine the strain:

`strain = \epsilon = (elongation)/(original\ length) \\\\\epsilon = (0.009\ in)/(12\ in)\\\\\epsilon = 7.5\ x\ 10^(-4)`

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

`E = (\sigma)/(\epsilon)\\\\E = (6.62\ kips)/(7.5\ x\ 10^(-4))`

E = 8.83 kips