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 800 lb. The elevator is 150 ft below the drum and is descending at the constant rate of 3 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

$Load developed (P) in the cable is given by : \to P = \frac{∆ L \times AE}{L} where ∆ L = extension / deformation in cable L = Length of cable = 150 ft = (150 \times 12) in A = area of crosssection of cable = \frac{π}{4} \times 0 . 4^{2} E = 29 \times 10^{3} ksi = 29 \times 10^{6} psi Now, \to stiffness of cable (K) = \frac{AE}{L} \Rightarrow K = \frac{\frac{π}{4} \times 0 . 4^{2} \times 29 \times 10^{6}}{150 \times 12} \Rightarrow K = 2024.58 \frac{lb}{in} Now, strain energy stored in cable (U_{C}) is given by : \to U_{C} = \frac{1}{2} \times K \times {(∆ L_{MAX})}^{2} = \frac{1}{2} \times 2024.58 \times {(∆ L_{MAX})}^{2} where, ∆ L_{MAX} = maximum deformation in cable due to sudden loading$

Step 2

$Strain energy stored in elevator (U_{E}) is given by : \to U_{E} = \frac{1}{2} \times m_{E} \times V^{2} + (W_{E} \times ∆ L_{MAX}) here, m_{E} = mass of elevator = \frac{W_{E}}{g} W_{E} = weight of elevator = 800 lb g = acceleration due to gravity = 386.04 \frac{in}{s^{2}} \Rightarrow U_{E} = \frac{1}{2} \times \frac{W_{E}}{g} \times V^{2} + (W_{E} \times ∆ L_{MAX}) here, V = velocity of elevator = 3 \frac{ft}{s} = 36 \frac{in}{s} \Rightarrow U_{E} = \frac{1}{2} \times \frac{800}{386.04} \times 36^{2} + (800 \times ∆ L_{MAX})$

Step 3

$Applying conservation of energy : \to U_{C} = U_{E} \Rightarrow \frac{1}{2} \times 2024.58 \times (∆ L$ _MAX)2=12×800386.04×362+(800×∆LMAX)⇒1012.29 (∆L_MAX)2 -800∆LMAX-1342.86=0⇒∆LMAX=1.6128 in

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