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Thermal Stress Learning Goal: Most materials change in size when subjected to a temperature change. For a temperature change of a homogenous isotropic material, the change in length of a bar of length L due to a temperature change AT can be calculated as braATL, where a is the linear coefficient of thermal expansion-a property of the material the bar is made from For a member that is not constrained, this expansion can occur freely. However, if the member is statically indeterminate, the deflection is constrained. Thus, changes in temperature will induce internal thermal stresses. The compatibility condition is that the changes in length due to the temperature change and to the induced stress must cancel each other out, &+5p=0, where y with N being the induced internal normal force, positive for tension NL AE The round bar shown (Figure 1) has a diameter of 7.6 cm and a length of 10 m. The modulus of elasticity is E=130 GPa and the linear coefficient of thermal expansion is 1.9x10-5 1 K Part A Calculate thermal stress If the bar originally has no internal normal forces and the temperature decreases by AT 40 K, what is the thermal stress developed in the bar? Express your answer with appropriate units to three significant figures. ▸ View Available Hint(s) Figure B <1 of 2 Submit Value Part B - Multiple materials t Units HANDWRITTEN SOLUTION REQUIRED He 10 of 21 > Ferm 1 but the same modulus of elasticity (Figure 2). What is the thermal stress developed for the entire bar when the The right half of the bar from Part A is replaced with a material that has a 3.2×105 temperature decreases by AT 40 K from a temperature where there is no stress in the bar? Express your answer with appropriate units to three significant figures. ▸ View Available Hint(s) ? Value Units Submit Provide Feedback

Thermal Stress Learning Goal: Most materials change in size when subjected to a temperature change. For a temperature change of a homogenous isotropic material, the change in length of a bar of length L due to a temperature change AT can be calculated as braATL, where a is the linear coefficient of thermal expansion-a property of the material the bar is made from For a member that is not constrained, this expansion can occur freely. However, if the member is statically indeterminate, the deflection is constrained. Thus, changes in temperature will induce internal thermal stresses. The compatibility condition is that the changes in length due to the temperature change and to the induced stress must cancel each other out, &+5p=0, where y with N being the induced internal normal force, positive for tension NL AE The round bar shown (Figure 1) has a diameter of 7.6 cm and a length of 10 m. The modulus of elasticity is E=130 GPa and the linear coefficient of thermal expansion is 1.9x10-5 1 K Part A Calculate thermal stress If the bar originally has no internal normal forces and the temperature decreases by AT 40 K, what is the thermal stress developed in the bar? Express your answer with appropriate units to three significant figures. ▸ View Available Hint(s) Figure B <1 of 2 Submit Value Part B - Multiple materials t Units HANDWRITTEN SOLUTION REQUIRED He 10 of 21 > Ferm 1 but the same modulus of elasticity (Figure 2). What is the thermal stress developed for the entire bar when the The right half of the bar from Part A is replaced with a material that has a 3.2×105 temperature decreases by AT 40 K from a temperature where there is no stress in the bar? Express your answer with appropriate units to three significant figures. ▸ View Available Hint(s) ? Value Units Submit Provide Feedback

Principles of Heat Transfer (Activate Learning with these NEW titles from Engineering!)
Principles of Heat Transfer (Activate Learning with these NEW titles from Engineering!)
8th Edition
ISBN:9781305387102
Author:Kreith, Frank; Manglik, Raj M.
Publisher:Kreith, Frank; Manglik, Raj M.
Chapter2: Steady Heat Conduction
Section: Chapter Questions
Problem 2.24P
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Thermal Stress Learning Goal: Most materials change in size when subjected to a temperature change. For a temperature change of a homogenous isotropic material, the change in length of a bar of length L due to a temperature change AT can be calculated as braATL, where a is the linear coefficient of thermal expansion-a property of the material the bar is made from For a member that is not constrained, this expansion can occur freely. However, if the member is statically indeterminate, the deflection is constrained. Thus, changes in temperature will induce internal thermal stresses. The compatibility condition is that the changes in length due to the temperature change and to the induced stress must cancel each other out, &+5p=0, where y with N being the induced internal normal force, positive for tension NL AE The round bar shown (Figure 1) has a diameter of 7.6 cm and a length of 10 m. The modulus of elasticity is E=130 GPa and the linear coefficient of thermal expansion is 1.9x10-5 1 K Part A Calculate thermal stress If the bar originally has no internal normal forces and the temperature decreases by AT 40 K, what is the thermal stress developed in the bar? Express your answer with appropriate units to three significant figures. ▸ View Available Hint(s) Figure B <1 of 2 Submit Value Part B - Multiple materials t Units HANDWRITTEN SOLUTION REQUIRED He 10 of 21 > Ferm 1 but the same modulus of elasticity (Figure 2). What is the thermal stress developed for the entire bar when the The right half of the bar from Part A is replaced with a material that has a 3.2×105 temperature decreases by AT 40 K from a temperature where there is no stress in the bar? Express your answer with appropriate units to three significant figures. ▸ View Available Hint(s) ? Value Units Submit Provide Feedback
Transcribed Image Text:Thermal Stress Learning Goal: Most materials change in size when subjected to a temperature change. For a temperature change of a homogenous isotropic material, the change in length of a bar of length L due to a temperature change AT can be calculated as braATL, where a is the linear coefficient of thermal expansion-a property of the material the bar is made from For a member that is not constrained, this expansion can occur freely. However, if the member is statically indeterminate, the deflection is constrained. Thus, changes in temperature will induce internal thermal stresses. The compatibility condition is that the changes in length due to the temperature change and to the induced stress must cancel each other out, &+5p=0, where y with N being the induced internal normal force, positive for tension NL AE The round bar shown (Figure 1) has a diameter of 7.6 cm and a length of 10 m. The modulus of elasticity is E=130 GPa and the linear coefficient of thermal expansion is 1.9x10-5 1 K Part A Calculate thermal stress If the bar originally has no internal normal forces and the temperature decreases by AT 40 K, what is the thermal stress developed in the bar? Express your answer with appropriate units to three significant figures. ▸ View Available Hint(s) Figure B <1 of 2 Submit Value Part B - Multiple materials t Units HANDWRITTEN SOLUTION REQUIRED He 10 of 21 > Ferm 1 but the same modulus of elasticity (Figure 2). What is the thermal stress developed for the entire bar when the The right half of the bar from Part A is replaced with a material that has a 3.2×105 temperature decreases by AT 40 K from a temperature where there is no stress in the bar? Express your answer with appropriate units to three significant figures. ▸ View Available Hint(s) ? Value Units Submit Provide Feedback
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