Rods (1) and (2) were supposed to fit exactly between the rigid walls at A and C, as indicated in the figure below. However, rod (2) was manufactured a small amount, 8L, too long. A technician decides to lower the temperature of rod (2) by AT so that he can insert it in the space between B and C. When rod (2) returns to its initial temperature, the length of each rod is no longer the same as the original one, and the resulting strain is often called "residual strain"". Rod (1) has Young modulus E₁ = 74 GPa and cross-sectional area A = 1749 mm². Rod (2) has Young modulus E₂ = 204 GPa and same cross-sectional area A. Both rods have length L = 738 mm. L (1) ₁.₁. A E₂. az. A (2) Determine the resulting "residual normal straine in rod (1) when 8-9-10¹ mm.

Mechanics of Materials (MindTap Course List)
9th Edition
ISBN:9781337093347
Author:Barry J. Goodno, James M. Gere
Publisher:Barry J. Goodno, James M. Gere
Chapter2: Axially Loaded Members
Section: Chapter Questions
Problem 2.7.5P: Determine the strain energy per unit volume (units of psi) and the strain energy per unit weight...
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# 2-12. Normal Strain in rods constrained by walls with misfit
Rods (1) and (2) were supposed to fit exactly between the rigid walls at A and C, as indicated in the figure below. However, rod
(2) was manufactured a small amount, 8L, too long. A technician decides to lower the temperature of rod (2) by AT so that
he can insert it in the space between B and C. When rod (2) returns to its initial temperature, the length of each rod is no longer
the same as the original one, and the resulting strain is often called "residual strain"".
Rod (1) has Young modulus E₁ = 74 GPa and cross-sectional area A = 1749 mm². Rod (2) has Young modulus
E₂ = 204 GPa and same cross-sectional area A. Both rods have length L = 738 mm.
L
L
(1)
€ =
E₁,α₁, A
B
Ez az. A (2)
Determine the resulting "residual" normal strain e in rod (1) when 8M = 9.10-¹ mm.
-10-3 mm/mm
Transcribed Image Text:# 2-12. Normal Strain in rods constrained by walls with misfit Rods (1) and (2) were supposed to fit exactly between the rigid walls at A and C, as indicated in the figure below. However, rod (2) was manufactured a small amount, 8L, too long. A technician decides to lower the temperature of rod (2) by AT so that he can insert it in the space between B and C. When rod (2) returns to its initial temperature, the length of each rod is no longer the same as the original one, and the resulting strain is often called "residual strain"". Rod (1) has Young modulus E₁ = 74 GPa and cross-sectional area A = 1749 mm². Rod (2) has Young modulus E₂ = 204 GPa and same cross-sectional area A. Both rods have length L = 738 mm. L L (1) € = E₁,α₁, A B Ez az. A (2) Determine the resulting "residual" normal strain e in rod (1) when 8M = 9.10-¹ mm. -10-3 mm/mm
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