Irodov 1.290: Pressure That Cancels a Heated Stretch

Problem Irodov 1.290

A steel cylinder is heated so that its temperature rises by $100^\circ\text{C}$ . Find the pressure that must be applied to its two end faces, directed inward, to keep the cylinder’s length exactly unchanged despite the heating.

The cylinder's two end faces, each carrying an applied pressure p directed inward. The compressive strain this pressure produces exactly cancels the free thermal strain the temperature rise would otherwise produce, so the overall length stays fixed. — The cylinder’s two end faces, each carrying an applied pressure $p$ directed inward. The compressive strain this pressure produces exactly cancels the free thermal strain the temperature rise would otherwise produce, so the overall length stays fixed.

Given

$\Delta T = 100^\circ\text{C}$
steel

Find

p

, the pressure needed to hold the length fixed

Solution

Step 1Balancing the free thermal strain against a mechanical one

Left unconstrained, heating by $\Delta T$ would lengthen the cylinder by the thermal strain $\alpha\Delta T$ , where $\alpha$ is steel’s linear expansion coefficient. Holding the length fixed instead means an applied compressive stress $p$ must produce, by Hooke’s law, a mechanical strain that exactly cancels this: the net fractional length change is the sum of the two contributions, set to zero.

\frac{\Delta l}{l} = \alpha\Delta T - \frac{p}{E} = 0 \tag{1}

Step 2The pressure in atmospheres

Steel: $E = 2.0\times10^{11}$ Pa, $\alpha = 1.1\times10^{-5}$ 1/K. Solving (1) for $p$ and substituting $\Delta T = 100$ K gives the pressure in pascals.

p = \alpha E\Delta T = 1.1\times10^{-5}\times2.0\times10^{11}\times100 = 2.2\times10^{8}\ \text{Pa}

Dividing by one standard atmosphere, $1.013\times10^{5}$ Pa, puts it in the units the answer is usually quoted in.

\frac{2.2\times10^{8}}{1.013\times10^{5}} = 2.2\times10^{3}\ \text{atm}

p = \alpha E\Delta T \approx 2.2\times10^{3}\ \text{atm}

(2)

Check

Dimensionally $[\alpha][E][\Delta T] = (1/\text{K})(\text{Pa})(\text{K}) = \text{Pa}$ , correct for a pressure. And as $\Delta T\to0$ , $p\to0$ : an unheated cylinder needs no clamping force to hold its length, as it must.

Remark

The formula $p=\alpha E\Delta T$ carries no length or cross-sectional area at all: the same roughly 2200 atm would be needed to hold a short stub or a long girder of the same steel at the same $\Delta T$ , because both the thermal strain and the mechanical strain that cancels it are already fractional (per unit length) quantities. It is also why railway rails laid without expansion gaps have to be restrained so carefully: a swing of only 25 K already builds 55 MPa of axial compression in the steel, and that stress is the same in a short rail as in a long one. Whether such a track actually buckles is a separate question, settled by the lateral restraint the ballast and sleepers supply and by how long a run of rail is disturbed, not by the axial stress on its own.