A steel cable 1.25 in. in diameter and 50 ft long is to lift a 20-ton load without permanently deforming. What is the length of the cable during lifting? The modulus of elasticity of the steel is 30 3 106 psi.

Respuesta :

Answer:

50.0543248872 ft

Explanation:

F = Load = 20 ton = [tex]20\times 2000\ lb[/tex]

d = Diameter = 1.25 in

[tex]L_1[/tex] = Initial length = 50 ft

[tex]L_2[/tex] = Final length

A = Area = [tex]\dfrac{\pi}{4}d^2[/tex]

Y = Young's modulus = [tex]30\times 10^6\ psi[/tex]

Young's modulus is given by

[tex]Y=\dfrac{FL}{A\Delta L}\\\Rightarrow Y=\dfrac{FL_1}{\dfrac{\pi}{4}d^2(L_2-L_1)}\\\Rightarrow L_2=\dfrac{4FL_1}{Y\pi d^2}+L_1\\\Rightarrow L_2=\dfrac{4\times 40000\times 50}{30\times 10^6\times \pi\times 1.25^2}+50\\\Rightarrow L_2=50.0543248872\ ft[/tex]

The length during the lift is 50.0543248872 ft

ACCESS MORE
EDU ACCESS