Thermal Expansion — Question 7

Question 7

A nonuniform rod expands according to its temperature integral

A rod has initial length $L_0=2.00\,\mathrm m$ and constant $\alpha=12.0\times10^{-6}\,\mathrm{K^{-1}}$. Its initial uniform temperature is $20^\circ\mathrm C$. The final field, indexed by initial coordinate $x$, is $C(x)=20+30x$, with $x$ in metres. Each element expands freely to first order: $d(\Delta L)=\alpha[C(x)-20]dx$.

1. Find the total extension and the position shift of the midpoint when the left end is fixed.

2. Would uniformly heating the rod to its final spatial-average temperature produce the same total extension and midpoint shift?

Question 7 - Solution

A nonuniform rod expands according to its temperature integral

(a) Method and calculation

$$\Delta L=\int_0^2\alpha(30x)dx=60\alpha=.000720\,\mathrm m.$$$$u(1)=\int_0^1\alpha(30x)dx=15\alpha=.000180\,\mathrm m.$$Thus extension is $.720$ mm and midpoint shift is $.180$ mm.

(b) Interpretation and further calculation

$$\overline C=20+30(2)/2=50^\circ\mathrm C.$$$$\Delta L_{\mathrm{uniform}}=\alpha(30)(2)=.720\,\mathrm{mm},\quad u_{\mathrm{uniform}}(1)=\alpha(30)(1)=.360\,\mathrm{mm}.$$The total agrees, but the displacement field does not. A total-length measurement cannot recover the full temperature profile.

Physical check

Integrating only the left half explains why its motion is smaller for the ramp than for uniform heating.