Question

The bar has a cross-sectional area $0.25 \text{ in}^2$, length $2 \text{ ft}$, modulus of elasticity $29 \cdot 10^3 \text{ kip/in}^2$, and coefficient thermal expansion $6.6 \cdot 10^{-6} /^\circ\text{F}$. It is constrained to just fit between the two fixed supports when $T_1 = 60 ^\circ\text{F}$. If the temperature is raised to $T_2 = 120 ^\circ\text{F}$, determine the average normal stress developed in the bar.

Name: the bar has a cross sectional area 025 text in2 length 2 text ft modulus of elasticity 29 cdot 103 text kipin2 and coefficient thermal expansion 66 cdot 10 6 circtextf it is constrained to j 49746
Uploaded: 2023-07-22T05:43:32-08:00
Duration: 1 min 2 s
Channel: Adi S
Description: the bar has a cross sectional area 025 text in2 length 2 text ft modulus of elasticity 29 cdot 103 text kipin2 and coefficient thermal expansion 66 cdot 10 6 circtextf it is constrained to j 49746

          The bar has a cross-sectional area $0.25 \text{ in}^2$, length $2 \text{ ft}$, modulus of elasticity $29 \cdot 10^3 \text{ kip/in}^2$, and coefficient thermal expansion $6.6 \cdot 10^{-6} /^\circ\text{F}$. It is constrained to just fit between the two fixed supports when $T_1 = 60 ^\circ\text{F}$. If the temperature is raised to $T_2 = 120 ^\circ\text{F}$, determine the average normal stress developed in the bar.

the bar has a cross sectional area 025 text in2 length 2 text ft modulus of elasticity 29 cdot 103 text kipin2 and coefficient thermal expansion 66 cdot 10 6 circtextf it is constrained to j 49746

Added by Donna R.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Adi S

07/22/2023

Step 1

Cross-sectional area $A = 0.25 \text{ in}^2$ Length $L = 2 \text{ ft} = 2 \times 12 \text{ in} = 24 \text{ in}$ Modulus of elasticity $E = 29 \cdot 10^3 \text{ kip/in}^2 = 29 \cdot 10^6 \text{ psi}$ Coefficient of thermal expansion $\alpha = 6.6 \cdot 10^{-6} Show more…

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The bar has a cross-sectional area $0.25 \text{ in}^2$, length $2 \text{ ft}$, modulus of elasticity $29 \cdot 10^3 \text{ kip/in}^2$, and coefficient thermal expansion $6.6 \cdot 10^{-6} /^\circ\text{F}$. It is constrained to just fit between the two fixed supports when $T_1 = 60 ^\circ\text{F}$. If the temperature is raised to $T_2 = 120 ^\circ\text{F}$, determine the average normal stress developed in the bar.