Explanation Given information: The diameter of the cable is 0.4 in. The weight of the elevator is 800 lb. The length of the elevator is 150 ft = 1 , 800 in . below the drum. The elevator is descending at the constant rate of 2 ft/sec = 24 in ./sec when the drum suddenly stops. Young’s modulus or modulus of elasticity of the steel cable is E s t = 29 ( 10 3 ) ksi = 29 ( 10 6 ) psi . The permissible stress in the bar is σ Y = 50 ksi = 50,000 psi . The acceleration due to gravity is g = 32.2 ft/sec 2 = ( 32.2 × 12 ) in ./sec 2 . Explanation: Area of the bar: Determine the area of the cable ( A ) using the formula. A = π d 2 4 (1) Here, d is diameter of the cable. Stiffness: Determine the stiffness of the spring when the load is applied on top of the post using the equation. k = A E L (2) Here, E is Young’s modulus of the steel cable and L is the length of the cable. Conservation of energy: Express the equation for conservation of energy as follows: U e = U i 1 2 m v 2 + W Δ max = 1 2 k Δ max 2 (3) Here, m is mass of the elevator, v is constant rate, W is the weight of the elevator, and Δ max is the maximum displacement. Maximum dynamic force developed in the cable ( P max ) : Determine the maximum dynamic force developed in the cable using the equation. P max = k Δ max (4) Maximum stress in the cable ( σ max ) : Determine the maximum stress developed in the cable using the formula. σ max = P max A (5) Mass of the cable: Determine the mass of the cable ( m ) using the relation. m = W g (6) Show the calculation of values as follows: Conclusion: Substitute 0.4 in. for d in Equation (1). A = π × 0.4 2 4 = 0

10th Edition

ISBN: 9780134319650

Author: Russell C. Hibbeler

Publisher: PEARSON

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Chapter 14.4, Problem 14.47P

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The maximum stress developed in the cable.

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Chapter 14.4, Problem 14.46P

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