The axial assembly consists of a solid 1.01-in-diameter aluminum alloy [E- 8900 ksi, v - 0.33, a = 12.4x 10-6/°F] rod (1) and a solid 1.63-in-diameter bronze [E - 15400 ksi, v- 0.17,a = 7.3 x 10-6/°F] rod (2). Assume L;-15 in., L2-29 in. If the supports at A and Carerigid and the assembly is stress free at 6°F, determine: (a) the normal stresses o1,02 in both rods at 167°F (positive if tensile, negative if compressive). (b) the displacement ug of flange B (positive if to the right, negative to the left). (c) the change in diameter Adı of the aluminum rod. B L1 L2 Answers: (a) o1 26.591 ksi, o2 = i 10.209 ksi (b) ug - i 0.0418 in. (c) Ad, = i in. 0.0003288

The axial assembly consists of a solid 1.01-in-diameter aluminum alloy [E- 8900 ksi, v - 0.33, a = 12.4x 10-6/°F] rod (1) and a solid 1.63-in-diameter bronze [E - 15400 ksi, v- 0.17,a = 7.3 x 10-6/°F] rod (2). Assume L;-15 in., L2-29 in. If the supports at A and Carerigid and the assembly is stress free at 6°F, determine: (a) the normal stresses o1,02 in both rods at 167°F (positive if tensile, negative if compressive). (b) the displacement ug of flange B (positive if to the right, negative to the left). (c) the change in diameter Adı of the aluminum rod. B L1 L2 Answers: (a) o1 26.591 ksi, o2 = i 10.209 ksi (b) ug - i 0.0418 in. (c) Ad, = i in. 0.0003288

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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The axial assembly consists of a solid 1-in.-diameter aluminum alloy [E = 11000 ksi, v = 0.26, α= 10.5 × 10−6/°F] rod (1) and a solid 1.49-in.-diameter bronze [E = 15300 ksi, v = 0.15, α= 7.6 × 10–6/°F] rod (2). Assume L1=11 in., L2=22 in. If the supports at A and C are rigid and the assembly is stress free at 16°F, determine:(a) the normal stresses σ1, σ2 in both rods at 213°F (positive if tensile, negative if compressive).(b) the displacement uB of flange B (positive if to the right, negative to the left).(c) the change in diameter Δd1 of the aluminum rod.
240cos10=236.35 N M, =? 240 N A 15° 15° 65° 240sin 10=41.68 N 350 mm M=236.35(0.35) +41.68(0.03)=83.97 N - m 65° (clockwise direction) 30 mm
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F=800N, d=1.00m, e=2.00m, f=1.90m
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