A polymer cylinder (2) is clamped between rigid heads by two steel bolts (1). The steel [E = 29,000 ksi; a = 6.5x10-6/°F] bolts have a diameter of 0.30 in. The polymer [E = 370 ksi; a = 38.3x 10 6/F] cylinder has an outside diameter of 6.6 in. and a wall thickness of 0.75 in. Assume a = 29 in. and b= 37 in. If the temperature of this assembly changes by AT = 135°F, determine: (a) the normal stress in the polymer cylinder. (b) the normal strain in the polymer cylinder. c) the normal strain in the steel bolts. (1) (2) (1) Answer: (a) 02 = psi. (b) E2 = (c) E1 =

A polymer cylinder (2) is clamped between rigid heads by two steel bolts (1). The steel [E = 29,000 ksi; a = 6.5x10-6/°F] bolts have a diameter of 0.30 in. The polymer [E = 370 ksi; a = 38.3x 10 6/F] cylinder has an outside diameter of 6.6 in. and a wall thickness of 0.75 in. Assume a = 29 in. and b= 37 in. If the temperature of this assembly changes by AT = 135°F, determine: (a) the normal stress in the polymer cylinder. (b) the normal strain in the polymer cylinder. c) the normal strain in the steel bolts. (1) (2) (1) Answer: (a) 02 = psi. (b) E2 = (c) E1 =

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Current Attempt in Progress A polymer cylinder (2) is clamped between rigid heads by two steel bolts (1). The steel [E = 29,000 ksi; a = 6.5x10-6/F] bolts have a diameter of 0.35 in. The polymer [E = 370 ksi; a = 37.2x 10-/°F] cylinder has an outside diameter of 6.9 in. and a wall thickness of 0.65 in. Assume a = 28 in. and b = 34 in. If the temperature of this assembly changes by AT = 130°F, determine: (a) the normal stress in the polymer cylinder. (b) the normal strain in the polymer cylinder. (c) the normal strain in the steel bolts. (1) (2) (1) b. Answer: (a) 02 = psi. (b) €2 = (c) E1 = Save for Later Attempts: 0 of 1 used Submit Answer NWP Assessme. NWP Assessme. O ManyCam Desktop 30°C
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Determine the compressive force (in N) in the bars shown after a temperature rise of 165°F if d = 0.011 in. Round off the final answer to five decimal places. d ... L M Use the following data: Bronze Aluminum A = 2.33 sq. in E = 14988461 psi A = 2.6 sq. in E = 10519262 psi L = 17.3 in L = 12.6 in a = 11.71 x 106/ °F. a = 12.73 x 106/°F