Steel railroad reels 10 m long are laid with a clearance of 3 mm at a temperature of 15°C. At what temperature will the rails just touch? What stress would be induced in the rails at that temperature if there were no initial clearance? Assume α = 11.7 µm/(m·°C) and E = 200 GPa.
Question P.263: Steel railroad reels 10 m long are laid with a clearance of ...
- Length of steel railroad reels: 10 m
- Clearance at a temperature of 15°C: 3 mm
- Coefficient of linear expansion (α): 11.7 µm/(m·°C)
- Young’s modulus (E): 200 GPa
Step 1:
We are given the equation for thermal expansion, which is δT = αL(ΔT), where δT is the change in length, α is the coefficient of linear expansion, L is the original length, and ΔT is the change in temperature.
Step 2:
We substitute the given values into the equation. In this case, δT = 3mm, α = 11.7 x 10^-6, L = 100000, and ΔT = Tf - Ti.
Step 3:
We solve the equation for Tf by rearranging the equation and isolating Tf. In this case, we have 3 = (11.7 x 10^-6)(100000)(Tf - 15).
Step 4:
We calculate Tf by solving the equation. In this case, Tf = 40.64°C.
Step 5:
We move on to calculating the required stress. We are given the equation δ = δT, where δ is the stress, δT is the change in length, and E is the Young's modulus.
Step 6:
We substitute the given values into the equation. In this case, δ = 3mm, α = 11.7 x 10^-6, L = 100000, ΔT = Tf - Ti, and E = 200000.
Step 7:
We solve the equation for σ by rearranging the equation and isolating σ. In this case, we have σ = αE(Tf - Ti).
Step 8:
We calculate σ by solving the equation. In this case, σ = (11.7 x 10^-6)(200000)(40.64 - 15).
Step 9:
We find that the required stress is 60 MPa.
Final Answer
Temperature at which \delta_{T}=3\operatorname{mm} :
\begin{aligned}& \delta_{T}=\alpha\,L\left(\Delta\,T\right)\\& \delta_{T}=\alpha\,L\left(T_{f}-T_{i}\right)\\& 3=(11\,.7\,\times\,10^{-6})(10\,0000)(T_{f}\!-\!15)\\& T_{f}=40.64^{\circ}\rm C\end{aligned}
Required stress:
\begin{aligned}& \delta=\delta_T\\& \frac{\sigma {L}}{E}=\alpha\,{L}\,(\Delta T)\\& \sigma=\alpha E\left(T_{f}-T_{i}\right)\\& \sigma=(11.7\times10^{-6})(200\,000)(40.64-15)\\& \sigma=60\, \rm MPa\end{aligned}
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