The composite bar shown in Fig. P-273 is firmly attached to unyielding supports. An axial force P = 50 kips is applied at 60°F. Compute the stress in each material at 120°F. Assume a = 6.5 x 10-6 in/(in.°F) for steel and 12.8 x 10-6 in/(in-°F) for aluminum. Figure P-273 and P-274 Aluminum A = 2 in? E = 10 x 10° psi Steel A = 3 in? E = 29 x 10° psi 15 in 10 in

Mechanics of Materials (MindTap Course List)
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Author:Barry J. Goodno, James M. Gere
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Chapter2: Axially Loaded Members
Section: Chapter Questions
Problem 2.6.9P: A prismatic bar with a length L = 3 ft and cross-sectional area A = 8 in2 is compressed by an axial...
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The composite bar shown in Fig. P-273 is firmly attached to unyielding supports. An
axial force P = 50 kips is applied at 60°F. Compute the stress in each material at 120°F.
Assume a = 6.5 × 10-6 in/(in.°F) for steel and 12.8 × 10-6 in/(in-°F) for aluminum.
Figure P-273 and P-274
Aluminum
A = 2 in?
E = 10 x 10° psi
Steel
A = 3 in?
E = 29 x 10° psi
15 in
10 in
Transcribed Image Text:The composite bar shown in Fig. P-273 is firmly attached to unyielding supports. An axial force P = 50 kips is applied at 60°F. Compute the stress in each material at 120°F. Assume a = 6.5 × 10-6 in/(in.°F) for steel and 12.8 × 10-6 in/(in-°F) for aluminum. Figure P-273 and P-274 Aluminum A = 2 in? E = 10 x 10° psi Steel A = 3 in? E = 29 x 10° psi 15 in 10 in
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I just have one question to clarify. The given problem stated that in the part of value of Eal=10×10(race to -6) psi and the given we have in the problem for the value of Est=10×10(race to -6). Why is that on the solving part in yoir solution you are using 10×10(race to 6) instead of 10×10(race to -6)? I'm just curious, how did it happen? Did the value will still be the same? Can you give me the reason? Thank you!

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