Indeterminate Structures — Question 6

Question 6

Prevented thermal expansion produces a compressive self-stress

A straight rod is fixed between immovable walls and heated by $40\,\mathrm K$. Its modulus is $E=200\,\mathrm{GPa}$, area $A=100\,\mathrm{mm^2}$ and expansion coefficient $\alpha=12\times10^{-6}\,\mathrm{K^{-1}}$. Use $\Delta L=NL/(EA)+\alpha L\Delta T$, with tension positive; ignore buckling.

1. Find the axial stress and force.

2. Explain why the corresponding freely expanding rod has no thermal stress.

Question 6 - Solution

Prevented thermal expansion produces a compressive self-stress

(a) Derivation and calculation

The fixed-end compatibility condition is $$0=\frac{NL}{EA}+\alpha L\Delta T\Rightarrow\sigma=\frac NA=-E\alpha\Delta T.$$ $$\boxed{\sigma=-(200\times10^9)(12\times10^{-6})(40)=-96.0\ \mathrm{MPa}},$$ $$\boxed{N=\sigma A=(-96\times10^6)(100\times10^{-6})=-9.60\ \mathrm{kN}}.$$

(b) Result and interpretation

For a freely expanding rod with unloaded ends, $$N=0,\qquad\sigma=0,\qquad\boxed{\Delta L=\alpha L\Delta T}.$$ In the constrained rod, the elastic strain $N/(EA)=-\alpha\Delta T$ cancels the thermal strain. Thermal expansion produces stress only when the prescribed compatibility prevents that free strain.

Physical check

The negative signs indicate compression, consistent with walls opposing the preferred expansion of the rod.