182_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

182_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

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176 CHAPTER 2 Axially Loaded Members Problem 2.5-25 A polyethylene tube (length L ) has a cap which is held in place by a spring (with undeformed length L 1 ± L ). After installing the cap, the spring is post-tensioned by turning an adjustment screw by amount ² . Ignore deforma- tions of the cap and base. Use the force at the base of the spring as the redundant. Use numerical properties in the boxes below. (a) What is the resulting force in the spring, F k ? (b) What is the resulting force in the tube, F t ? (c) What is the final length of the tube, L f ? (d) What temperature change ³ T inside the tube will result in zero force in the spring? Cap (assume rigid) Tube ( d 0 , t , L , a t , E t ) Spring ( k , L 1 < L ) Adjustment screw d = L L 1 Modulus of elasticity Polyethylene tube ( E t = 100 ksi) Coefficients of thermal expansion a t = 80 ´ 10 –6 / ° F, a k = 6.5 ´ 10 –6 / ° F d 0 = 6 in. in. 1 8 t = k = 1.5
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Unformatted text preview: kip in. ––– L = 12 in. Properties and dimensions L 1 = 11.875 in. Solution 2.5-25 The figure shows a section through the tube, cap and spring Properties & dimensions d o µ 6 in. in. E t µ 100 ksi L µ 12 in. ¶ L 1 µ 11.875 in. · k µ 6.5(10 ¸ 6 ) ± t µ 80 ´ (10 ¸ 6 ) A t µ 2.307 in 2 A t µ p 4 [ d o 2 ¸ 1 d o ¸ 2t 2 2 ] k µ 1.5 kip in t µ 1 8 Pretension & temperature spring is 1/8 in. shorter than tube ² µ L ¸ L 1 µ 0.125 in. ³ T µ note that Q result below is for zero temp. (until part (d)) Flexibilities (a) Force in spring (F k ) µ redundant (Q) follow solution procedure outlined in Prob. 2.5-23 solution Q µ d + ¢ T 1 ¸ a k L 1 + a t L 2 f + f t µ F k f t µ L E t A t f µ 1 k Sec_2.5.qxd 9/25/08 3:00 PM Page 176...
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This note was uploaded on 12/22/2011 for the course MEEG 310 taught by Professor Staff during the Fall '11 term at University of Delaware.

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