A 50ft long solid steel rod is subjected to a load of 12,000Ib. This load causes the rod to stretch 0.412 in. The modulus of elasticity of the steel is 30,000,000 psi. Determine the diameter of the rod.

0.0003758 m²

Explanation:

The formula to apply is;

ΔL=1/Y *F/A*L₀

Where ΔL=change in length = 0.412 in

Y=modulus of elasticity=30000000 psi

change psi to Pa

1 psi= 6.894757 kPa

30000000=?

6.894757 *30,000,000 =206842710000 Pa

F=applied force = 12000 lb---change to Newtons

1 lb=4.4482189159 N

12000 lb=? 12000*4.4482189159 =53378.6 N

A=cross-sectional area=?

L₀=original length of the rod= 50 ft change to meters

1ft =0.3048

50 ft=50*0.3048=15.24 m

ΔL =0.412 in change to meters

1 in =0.0254

0.412 in =? 0.412*0.0254 =0.0104648 m

Applying the formula

ΔL=1/Y *F/A*L₀

0.0104648 =1/206842710000 *(53378.6/A)*15.24

142031994.2=53378.6/A

A=53378.6/142031994.2

A=0.0003758 m²

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