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a_sh-v [17]
3 years ago
5

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 modulu

s is 30 x 100 psi. Make sure you do not round up your answer to less than 4 significant (non- zero) figures
Engineering
1 answer:
madreJ [45]3 years ago
3 0

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

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