Axial deformation of bars

By the end of this topic you will be able to

Use δ = PL/AE to find the change in length of a bar.
Find the change in length caused by a temperature change.
Explain why a bar that is held between rigid walls develops thermal stress.

Combining the definitions

Start from the two definitions in the previous topic, $\sigma = P/A$ and $\varepsilon = \delta/L$ , and Hooke's law $\sigma = E\varepsilon$ . Substituting gives the axial deformation of a uniform bar:

\delta = \frac{PL}{AE}

The product $AE$ is the axial stiffness: a long, thin, flexible bar has a large $\delta$ ; a short, thick, stiff bar has a small one. This formula assumes the load, area and modulus are constant along the length. For a stepped bar, or one with several loads, add the contribution of each segment:

\delta = \sum \frac{P_i L_i}{A_i E_i}

Temperature change

A free bar that is heated expands. The change in length is

\delta_T = \alpha L\,\Delta T

where $\alpha$ is the coefficient of thermal expansion (steel about $12\times10^{-6}$ per °C; aluminium about $23\times10^{-6}$ ). Free expansion produces no stress.

Thermal stress

If the bar is prevented from expanding, for example between two rigid walls, it cannot change length, so the walls must push back. The compressive stress that results is

\sigma = -E\alpha\,\Delta T

The magnitude is independent of the length. This is why railway tracks have expansion gaps and bridges have expansion joints.

Interactive modelAdd load or heat. Switch to rigid walls to see thermal stress.

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Worked example

A 2 m long, 25 mm diameter steel bar ( $E = 200$ GPa, $\alpha = 12\times10^{-6}$ /°C) carries a tensile load of 100 kN. (a) Find its elongation. (b) Find the extra elongation if its temperature rises by 40 °C while free to expand. (c) Find the stress if the same temperature rise acts on the bar held between rigid walls.

(a) $A = \pi\times25^2/4 = 490.9$ mm².

\delta = \frac{100\,000\times2000}{490.9\times200\,000} = 2.04 \text{ mm}

(b) $\delta_T = 12\times10^{-6}\times2000\times40 = 0.96$ mm. Total elongation $= 2.04 + 0.96 = 3.00$ mm.

(c) $\sigma = -200\,000\times12\times10^{-6}\times40 = -96$ MPa, a compressive stress of 96 MPa, with no change in length.

Practice problems

A steel wire 5 m long and 4 mm in diameter carries 2 kN. Find the elongation ( $E = 200$ GPa).

Show answer

$A = 12.57$ mm². $\delta = 2000\times5000/(12.57\times200\,000) = 3.98$ mm.
A 10 m steel girder warms by 30 °C. How much does it lengthen if free? ( $\alpha = 12\times10^{-6}$ /°C)

Show answer

$\delta_T = 12\times10^{-6}\times10\,000\times30 = 3.6$ mm.
The same girder is fixed between rigid abutments. What stress results? ( $E = 200$ GPa)

Show answer

$\sigma = 200\,000\times12\times10^{-6}\times30 = 72$ MPa compressive.
Using the model, which of steel, aluminium or copper gives the largest elongation for the same load and size?

Show answer

Aluminium would stretch more than steel (lower $E$ = 70 GPa). Copper ( $E$ = 117 GPa) is between them. Elongation is inversely proportional to $E$ .

Key formulae

Axial deformation: $\delta = \frac{PL}{AE}$
Free thermal expansion: $\delta_T = \alpha L \,\Delta T$
Thermal stress (fully restrained bar): $\sigma = -E\alpha\,\Delta T$

Lecturer notes: suggested plan, activity and assessment

Suggested plan (40 min). 10 min: derive $\delta = PL/AE$ ; 10 min: stepped bar example on the board; 10 min: thermal expansion and restraint; 10 min: model (switch between free and fixed ends).

Activity. Ask: "At what temperature rise does a restrained steel bar yield at 250 MPa?" ( $\Delta T = 250/(200\,000\times12\times10^{-6}) = 104$ °C.)

Discussion. Where have you seen expansion joints?

Assessment idea. A stepped bar with two materials loaded at three points.