Question

Determine displacement (in) of a 1.37 in diameter steel bar, which is 50 ft long under a force of 27,865 lb if elasticity modulus is 30 x 100 psi. Make sure you do not round up your answer to less than 4 significant (non- zero) figures

Answer 1

Answer:

So displacement in inch will be $\Delta l=3.78249\times 10^{-5}inch$

Explanation:

We have given length = 50 feet

We know that 1 feet = 12 inches

Force F = 27865 LB

Modulus of elasticity $E=30\times 10^{10}psi$

So 50 feet = 50×12 = 600 inches

Diameter d = 1.37 inch

So radius $r=\frac{d}{2}=\frac{1.37}{2}=0.685inch$

So area $A=\pi r^2=3.14\times 0.685^2=1.4733inch^2$

We know that stress $=\frac{force}{area}=\frac{27865}{1.4733}=18912.47lb/inch^2$

Now we know that modulus of elasticity $E=\frac{stress}{strain}$

$30\times 10^{10}=\frac{18912.47}{strain}$

$strain=630.4156\times 10^{-10}inch$

Now we know that $strain=\frac{\Delta l}{l}$

$630.4156\times 10^{-10}=\frac{\Delta l}{600}$

$\Delta l=3.78249\times 10^{-5}inch$