Answered: 150 ft | bartleby

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Question

A steel cable having a diameter of 0.4 in. wraps over a drum and is used to lower an elevator having a weight of 80 lb. The elevator is 150 ft below the drum and is descending at the constant rate of 2 ft>s when the drum suddenly stops. Determine the maximum stress developed in the cable when this occurs. Est = 29(103) ksi, sY = 50 ksi.

Expert Solution

Step 1

Given Data

The diameter of the steel cable is: $d = 0.4 in$
Weight of the elevator is: $W = 80 lb$
The distance of the elevator from the drum is: $L = 150 ft$
The descending of the elevator is: $v = 2 ft / s = 24 in / s$
The modulus of elasticity of the steel is: $E_{s t} = 29 \times 10^{3} ksi = 29 \times 10^{6} psi$
yield strength is: $σ_{y} = 50 ksi = 50000 psi$

The stiffness of the cable is

$\begin{array}{rcl} k & = & \frac{A E}{L} \\ = & \frac{\frac{π}{4} {(0.4)}^{2} \times (29 \times 10^{6})}{150 \times 12} \\ = & 2024.58 lb / in \end{array}$

Step 2

The strain energy in the cable is

$U_{c} = \frac{1}{2} k δ^{2}$

Here $δ$ is the maximum deformation

The elevator energy is

$\begin{array}{rcl} U_{c} & = & \frac{1}{2} m v^{2} + W δ \\ \frac{1}{2} (2024.58) δ^{2} & = & \frac{1}{2} \times \frac{80}{386.4} {(24)}^{2} + 80 δ \\ 1012.29 δ^{2} & = & 59.62 + 80 δ \\ 1012.29 δ^{2} - 80 δ - 59.52 & = & 0 \end{array}$

The roots for this quadratic equation are 0.2851 and -0.206 As the deformation can not be negative in this case so the deformation in the cable is $0.2851 in$ .

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