Explanation Given information: Weight of the block is 100 lb. Height of fall of the block on the cylindrical post and spring assembly is h = 3 ft = 36 in . . Speed of the block falling is v = 20 ft/sec = 240 in ./sec The post is made of 2014-T6 aluminum. Length of the post is L = 1.5 ft = 18 in . . The diameter of the post is d = 2 in . . Young’s modulus or modulus of elasticity of the 2014-T6 aluminum is E a l = 10.6 ( 10 3 ) ksi . The acceleration due to gravity is g = 32.2 ft/sec 2 = ( 32.2 × 12 ) in ./sec 2 . The stiffness in the spring is k s p = 500 kip/in . Assumption: The maximum allowable stress in the 2014-T6 aluminum post is σ Y = 60 ksi . Area of the cylinder: Determine the area of the post ( A ) using the formula. A = π d 2 4 (1) Here, d is diameter of the post. Substitute 2 in. for d in Equation (1). A = π × 2 2 4 = π in . 2 Determine the equivalent spring constant using the equation. k p = A E L (2) Here, E is Young’s modulus of the aluminum and L is the length of the post. Substitute π in . 2 for A , 10.6 ( 10 3 ) ksi for E a l , and 1.5 ft for L in Equation (2). k p = π × 10.6 ( 10 3 ) 1.5 ft × 12 in . 1 ft = 1 , 850.05 kip/in . Determine the mass of the block using the relation. m = W g (3) Here, W is weight of the block and g is acceleration due to gravity. Substitute 100 lb for W and ( 32.2 × 12 ) in ./sec 2 for g in Equation (3). m = 100 32.2 × 12 = 0.258799 slug Conservation of energy: Express the equation for conservation of energy as follows: U e = U i 1 2 m v 2 + W ( h + Δ max ) = 1 2 k p Δ max 2 + 1 2 k s p Δ max 2 (4) Here, v is velocityand h is the height of fall

9th Edition

ISBN: 9780133254426

Author: Russell C. Hibbeler

Publisher: Prentice Hall

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Chapter 14.4, Problem 14.57P

To determine

The maximum normal stress developed in the post.

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