Axial loading: stress, strain and held-cold rods

Pull on it

• The running case is the bracket's tie rod: $d = 12\ \mathrm{mm}$ steel, $L = 1.5\ \mathrm{m}$, carrying $P = 25\ \mathrm{kN}$ tension. Normal stress 正应力 is $\sigma = P/A$ and strain 应变 is $\varepsilon = \delta/L$; Hooke's law $\sigma = E\varepsilon$ ties them with $E = 200\ \mathrm{GPa}$.
• With $A = 1.131 \times 10^{-4}\ \mathrm{m^2}$: $\sigma = \mathbf{221.0}$ MPa - safely under the $250\ \mathrm{MPa}$ yield - and the rod stretches $\delta = PL/AE = \mathbf{1.658}$ mm.

The squeeze beside the pull

• Poisson's ratio 泊松比 $\nu = 0.30$: pull axially and the sides cinch, $\varepsilon_{lat} = -\nu\varepsilon$. At $\varepsilon = 0.001105$ the lateral strain is $\mathbf{0.000332}$, so the 12 mm diameter thins by $\mathbf{3.98}\ \mu\mathrm{m}$ - small, but it is why press fits loosen under tension.

Held cold

• Fit the same rod between rigid walls and cool it $40\ \mathrm{K}$ ($\alpha = 12 \times 10^{-6}/\mathrm{K}$): it cannot contract, so $\sigma_T = E\alpha\Delta T = \mathbf{96.0}$ MPa tension - a load with no external force anywhere. That is the signature of a statically indeterminate problem: the constraint, not the load, sets the stress.

Three axial slips

Recap