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Determine the displacement (in inches) of a 1.37-inch diameter steel bar, which is 50 feet long, under a force of 27,865 lb, if the modulus of elasticity is [tex]30 \times 10^6[/tex] psi. Make sure you do not round your answer to less than 4 significant figures.

Answer :

Answer:

So displacement in inch will be [tex]\Delta l=3.78249\times 10^{-5}inch[/tex]

Explanation:

We have given length = 50 feet

We know that 1 feet = 12 inches

Force F = 27865 LB

Modulus of elasticity [tex]E=30\times 10^{10}psi[/tex]

So 50 feet = 50×12 = 600 inches

Diameter d = 1.37 inch

So radius [tex]r=\frac{d}{2}=\frac{1.37}{2}=0.685inch[/tex]

So area [tex]A=\pi r^2=3.14\times 0.685^2=1.4733inch^2[/tex]

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

Now we know that modulus of elasticity [tex]E=\frac{stress}{strain}[/tex]

[tex]30\times 10^{10}=\frac{18912.47}{strain}[/tex]

[tex]strain=630.4156\times 10^{-10}inch[/tex]

Now we know that [tex]strain=\frac{\Delta l}{l}[/tex]

[tex]630.4156\times 10^{-10}=\frac{\Delta l}{600}[/tex]

[tex]\Delta l=3.78249\times 10^{-5}inch[/tex]

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