Answer:
[tex]\sigma=11,700\ psi[/tex] (Compression)
Compressive stress is 11,700 psi
Explanation:
Lets calculate the change in temperature first:
[tex]\Delta T=120^o-60^o\\\Delta T=60^o F[/tex]
Given:
Coefficient of thermal expansion [tex]\alpha=6.5*10^{-6}/oF[/tex]
Modulus of elasticity [tex]E=30*10^6 psi[/tex]
Find:
[tex]\sigma=?[/tex]
Solution:
[tex]\Delta l=-l\alpha\Delta T[/tex]
Now,[tex]\frac{\Delta l}{l}= \epsilon[/tex]
The above equation will become:
[tex]\epsilon=-\alpha*\Delta T[/tex]
[tex]\epsilon=\frac{\sigma}{E}[/tex]
Putting above two equations equal, we will get:
[tex]-\alpha*\Delta T=\frac{\sigma}{E}[/tex]
Rearranging the equation:
[tex]\sigma=-E*\Delta T*\sigma[/tex]
[tex]\sigma=-30*10^6*60*6.5*10^{-6}\\\sigma=-11,700\ psi[/tex]
-ve Sign shows the compressive stress
Compressive stress is 11,700 psi
The compressive stress that is produced in the rails is; -11700 Psi
We are given;
Coefficient of thermal expansion; α = 6.5 × 10⁻⁶ /°F
Modulus of Elasticity; E = 30 × 10⁶ psi
Initial Temperature; T_i = 60°F
Final Temperature; T_f = 120 °F
We know that;
Modulus of elasticity = Stress/Strain
Thus; E = σ/ε
where ε = ΔL/L
We also know that; ΔL = -L*α * Δt
Thus;
ε = -L*α * Δt/L
ε = -α * Δt
We can rewrite as; σ/E = -α * Δt
σ = -E * α * Δt
σ = -30 × 10⁶ * 6.5 × 10⁻⁶ * (120 - 60)
σ = -11700 Psi
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