179_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

# 179_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

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SECTION 2.5 Thermal Effects 173 Problem 2.5-23 A polyethylene tube (length L ) has a cap which when installed compresses a spring (with undeformed length L 1 ± L ) by amount ²³ ( L 1 ´ L ). Ignore deformations of the cap and base. Use the force at the base of the spring as the redundant. Use numerical properties in the boxes given. (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 µ T inside the tube will result in zero force in the spring? Cap (assume rigid) Tube ( d 0 , t , L , a t , E t ) d = L 1 L Spring ( k , L 1 > L ) Modulus of elasticity Polyethylene tube ( E t = 100 ksi) Coefficients of thermal expansion
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Unformatted text preview: a t = 80 Â¶ 10 â€“6 / Â° F, a k = 6.5 Â¶ 10 â€“6 / Â° F d = 6 in. 1 8 â€” t = in. k = 1.5 kip in. â€“â€“â€“ L 1 = 12.125 in. > L = 12 in. Properties and dimensions (b) R EQUIRED LENGTH OF SOLDER JOINT â‰ˆ A s Â³ Â· d 2 s s reqd Â³ 25.7 mm (c) F INAL ELONGATION Â² f Â³ net of elongation of bolt ( b ) & shortening of sleeve ( s ) b Â³ 0.413 mm d b Â³ P B a L 1 + L 2 Â´ s E s A b b s reqd Â³ P B p d 2 t aj t Â³ P A s s Â³ Â´ 0.064 mm f Â³ b Â¸ s f Â³ 0.35 mm ; d s Â³ P s c L 1 Â´ s E c A 1 + L 2 Â´ s E c A 2 + s E c (A 1 + A 2 ) d Sec_2.5.qxd 9/25/08 3:00 PM Page 173...
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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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