Axial loads are applied with rigid bearing plates to the cylindrical pipes shown. The diameter of aluminum pipe (1) is 1.70 in. with a wall thickness of 0.25 in., the diameter of brass pipe (2) is 1.50 in. with a wall thickness of 0.25 in., and the diameter of steel pipe (3) is 3.00 in. with a wall thickness of 0.25 in. Pipe (2) has a length of 14 in. and deforms 0.22 in. due to the applied loads. (a) Determine the axial stress in rod (2) and state whether it is in compression or tension. (b) Calculate the normal strain in rod (2). -en; d₂ = 1.70in t₁ =₁25in d₂ = 1.501n t₂ =125 in L₂= 14in ad Deforms (10) = .22, h = 3.00in t₂ =25/n compression dz FA 1.501h-2 (₁25in) = lin A = π (1.50² - 1₁n²)=, 98 m² 14iK-22iK 13.78 2.12 LOkips .98in² = /₁98in = E €61.22 ks/ 15 kips D 20 kips 30 kips 35 kips A 15 kips D 8 35 kips

Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
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Axial loads are applied with rigid bearing plates to the cylindrical pipes shown. The diameter of aluminum pipe
(1) is 1.70 in. with a wall thickness of 0.25 in., the diameter of brass pipe (2) is 1.50 in. with a wall thickness of
0.25 in., and the diameter of steel pipe (3) is 3.00 in. with a wall thickness of 0.25 in. Pipe (2) has a length of
14 in. and deforms 0.22 in. due to the applied loads.
(a) Determine the axial stress in rod (2) and state whether it is in compression or tension.
(b) Calculate the normal strain in rod (2).
4₂ = 141n ad Deforms (P) = . 22, h
compression
Given; d = 1.70in t₁ = .25in
d₂ = 1.50in t₂ =125in
= 3.00in t₂ =.25in
dz
1.501h-2 (₁25in) = lin
A = π (1,60² - 1₁m²) = 98 m²
14in-,22iK
13.78
212
60kips
9814²
= 1.981h = E
€61.22 ks/
A
15 kips 15 kips
20 kips
30 kips
35 kips
B
20 kips
35 kips
D
Transcribed Image Text:Axial loads are applied with rigid bearing plates to the cylindrical pipes shown. The diameter of aluminum pipe (1) is 1.70 in. with a wall thickness of 0.25 in., the diameter of brass pipe (2) is 1.50 in. with a wall thickness of 0.25 in., and the diameter of steel pipe (3) is 3.00 in. with a wall thickness of 0.25 in. Pipe (2) has a length of 14 in. and deforms 0.22 in. due to the applied loads. (a) Determine the axial stress in rod (2) and state whether it is in compression or tension. (b) Calculate the normal strain in rod (2). 4₂ = 141n ad Deforms (P) = . 22, h compression Given; d = 1.70in t₁ = .25in d₂ = 1.50in t₂ =125in = 3.00in t₂ =.25in dz 1.501h-2 (₁25in) = lin A = π (1,60² - 1₁m²) = 98 m² 14in-,22iK 13.78 212 60kips 9814² = 1.981h = E €61.22 ks/ A 15 kips 15 kips 20 kips 30 kips 35 kips B 20 kips 35 kips D
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