Walk along an older railway line in India and look at the joints. Two rails meet end to end, bolted to a steel plate, with a small gap between their ends. On a winter morning the gap is a few millimetres wide. On a hot afternoon in May it has almost closed. The track was built with the gap on purpose.
Heated solids lengthen
Most solids lengthen when they are heated. The chapters on Thermal Physics study this in detail; here only one result is needed.
A rod of length \(L\) heated through a small rise of temperature \(\Delta T\) lengthens by \(\Delta L = L\alpha\,\Delta T\). The constant \(\alpha\) is the coefficient of linear expansion of the material: the fractional change in length for each kelvin.
For steel, \(\alpha = 1.2 \times 10^{-5}\,\text{K}^{-1}\). A steel rail \(10\,\text{m}\) long that warms by \(40\,\text{K}\) from a winter night to a summer afternoon lengthens by \(1.2 \times 10^{-5} \times 10 \times 40 = 4.8 \times 10^{-3}\,\text{m}\). That is almost five millimetres, far more than the elastic stretches of the last four lessons.
When the rod is not free
Now clamp a rod between two rigid walls, just touching them at the starting temperature, and heat it by \(\Delta T\). The rod tends to lengthen by \(L\alpha\,\Delta T\), but the walls do not let it. They push inward on its ends and hold it at its old length \(L\). The rod has been compressed by exactly the expansion it was denied.
Think of it in two steps. First let the rod expand freely to its new natural length, \(L + L\alpha\,\Delta T\). Then push it back to \(L\). The push-back is a compression of \(L\alpha\,\Delta T\) on a rod of length very nearly \(L\). So the strain is \(\alpha\,\Delta T\), and by Hooke's law the stress is \(Y\alpha\,\Delta T\). This stress is called thermal stress.
\[ \Delta L_{\text{free}} = L\alpha\,\Delta T \]
Why. the expansion the rod would have if nothing held it.
\[ \text{strain} = \dfrac{L\alpha\,\Delta T}{L} = \alpha\,\Delta T \]
Why. the walls push the rod back by the whole free expansion; divide by the length, since \(L\alpha\,\Delta T \ll L\).
\[ \text{stress} = Y\alpha\,\Delta T \]
Why. Hooke's law, stress \(= Y \times\) strain. The length \(L\) has cancelled.
\[ F = YA\alpha\,\Delta T \]
Read it. the force on each wall is the stress times the area of cross-section.
A steel rod of area \(2\,\text{cm}^2\) is clamped between two rigid walls and heated by \(50\,\text{K}\). For steel \(\alpha = 1.2 \times 10^{-5}\,\text{K}^{-1}\) and \(Y = 2 \times 10^{11}\,\text{Pa}\). Find the thermal stress and the force on each wall.
The strain is \(\alpha\,\Delta T = 1.2 \times 10^{-5} \times 50 = 6 \times 10^{-4}\), whatever the length of the rod.
Stress \(= Y\alpha\,\Delta T = 2 \times 10^{11} \times 6 \times 10^{-4} = 1.2 \times 10^8\,\text{Pa}\).
Force \(= 1.2 \times 10^8 \times 2 \times 10^{-4} = 2.4 \times 10^4\,\text{N}\), the weight of about \(2.4\) tonnes, from a mere \(50\,\text{K}\) of warming.
The length has dropped out of the stress. A short rod expands little, but a small push-back is a large strain for it. A long rod expands much, but its push-back is spread over a long length. The strain, and so the stress, is \(\alpha\,\Delta T\) either way.
In the welded state of the figure above, the stress rises from the first kelvin: \(96\,\text{MPa}\) at \(+40\,\text{K}\), where \(1\,\text{MPa}\) is \(10^6\,\text{Pa}\). On each support that is \(720\,\text{kN}\), the weight of some \(72\) tonnes. Now choose the gap. The rail lengthens freely until its expansion equals the gap, at \(\Delta T = 2 \times 10^{-3}/(10 \times 1.2 \times 10^{-5}) \approx 17\,\text{K}\). Only the expansion beyond that is stopped.
A steel rail is \(10\,\text{m}\) long at the temperature of laying, with a gap of \(2\,\text{mm}\) at its joint. On a hot day it is \(40\,\text{K}\) warmer. Take \(\alpha = 1.2 \times 10^{-5}\,\text{K}^{-1}\) and \(Y = 2 \times 10^{11}\,\text{Pa}\). Find the stress in the rail.
The free expansion would be \(L\alpha\,\Delta T = 10 \times 1.2 \times 10^{-5} \times 40 = 4.8\,\text{mm}\).
The first \(2\,\text{mm}\) closes the gap freely. The remaining \(2.8\,\text{mm}\) is stopped, so the strain is \(2.8 \times 10^{-3}/10 = 2.8 \times 10^{-4}\).
Stress \(= Y \times \text{strain} = 2 \times 10^{11} \times 2.8 \times 10^{-4} = 5.6 \times 10^7\,\text{Pa}\).
The gap cuts the stress from \(96\) to \(56\,\text{MPa}\). A gap of \(5\,\text{mm}\) would have removed it entirely on this day.
A steel rod \(2\,\text{m}\) long and \(10\,\text{cm}^2\) in section is fixed rigidly at both ends and heated by \(40\,\text{K}\). Take \(Y = 2 \times 10^{11}\,\text{Pa}\) and \(\alpha = 1.2 \times 10^{-5}\,\text{K}^{-1}\). Find the force on each support, in kN.
Two steel rods of the same area, one \(1\,\text{m}\) long and one \(4\,\text{m}\) long, are clamped between rigid walls. Both are heated by the same amount. How do their thermal stresses compare?
The sign of the stress
A clamped rod that is heated wants to be longer than the space between the walls. The walls push inward, and the stress is compressive. A clamped rod that is cooled wants to be shorter. The walls must pull outward to hold it at its old length, and the stress is tensile. Drag the slider of the figure above below zero to see the forces on the rail reverse.
A wire fixed between two poles on a warm day, just taut, tightens as it cools. A steel wire of \(1\,\text{mm}^2\) cooled by \(20\,\text{K}\) gains a tension of \(YA\alpha\,\Delta T = 2 \times 10^{11} \times 10^{-6} \times 1.2 \times 10^{-5} \times 20 = 48\,\text{N}\). That is why overhead electric wires are strung with a sag in summer. The sag lets them shorten in winter without snapping.
A rod clamped between rigid walls is heated. Its length does not change. Why does it carry a stress, and of what kind?
Two rods between the walls
Join two different rods end to end between rigid walls, and heat both. They behave like two springs in series, from the lesson on wires as springs. One force \(F\) squeezes both. Each rod ends up shorter than its free length by \(F\,L/(YA)\). Together the two push-backs must use up the whole free expansion of the pair.
For Fig 5.1 take \(\alpha = 1.2 \times 10^{-5}\,\text{K}^{-1}\) and \(Y = 2 \times 10^{11}\,\text{Pa}\) for steel, \(\alpha = 1.7 \times 10^{-5}\,\text{K}^{-1}\) and \(Y = 1 \times 10^{11}\,\text{Pa}\) for copper. Find the force in the rods and how far the junction moves.
Free expansions: steel \(1.2 \times 10^{-5} \times 1 \times 30 = 0.36\,\text{mm}\), copper \(1.7 \times 10^{-5} \times 1 \times 30 = 0.51\,\text{mm}\). Together \(0.87\,\text{mm}\) must be pushed back.
One force \(F\) pushes back each rod by \(F L/(YA)\): \(F\left(\dfrac{1}{2 \times 10^7} + \dfrac{1}{10^7}\right) = 0.87 \times 10^{-3}\).
\(F \times 1.5 \times 10^{-7} = 8.7 \times 10^{-4}\), so \(F = 5800\,\text{N}\), a stress of \(5.8 \times 10^7\,\text{Pa}\) in both.
The steel is pushed back \(5800/(2 \times 10^7) = 0.29\,\text{mm}\), less than its free \(0.36\,\text{mm}\). So it ends \(0.07\,\text{mm}\) longer, and the junction moves \(0.07\,\text{mm}\) towards the copper.
Joints that let structures breathe
Engineers give every long structure room to change its length. Old railway lines leave gaps at the joints. Modern tracks are welded into long lengths instead, laid at a chosen temperature and held down firmly enough to carry the stress. Bridges have expansion joints: interlocking steel teeth across a gap in the road, so that the deck can slide as it lengthens and shortens.
When the expansion has nowhere to go, the stress finds its own way out. A rail laid without enough room can bow sideways on a very hot day, a failure called buckling. A thick glass tumbler can crack when boiling water is poured in. Its hot inner surface expands, the cold outer layer holds it back, and the stress between them breaks the glass.
The main span of the Howrah Bridge in Kolkata is a steel structure about \(457\,\text{m}\) long. Estimate how much its length changes between a winter night and a summer afternoon, a difference of about \(30\,\text{K}\), in cm. Take \(\alpha = 1.2 \times 10^{-5}\,\text{K}^{-1}\).
A steel wire of section \(1\,\text{mm}^2\) is fixed between two rigid supports, just taut, at \(35\,°\text{C}\). Take \(Y = 2 \times 10^{11}\,\text{Pa}\) and \(\alpha = 1.2 \times 10^{-5}\,\text{K}^{-1}\). Find its tension at \(15\,°\text{C}\), in N.
Drive across a long bridge such as the Howrah Bridge or the Vidyasagar Setu in Kolkata. You feel a bump at a few places. Look down and you may see a row of interlocking steel teeth across the road. These are expansion joints. On a winter night the teeth stand apart; on a summer afternoon they close up. The deck moves, and no stress builds up in it.
Return to the gaps between the rails. A \(10\,\text{m}\) steel rail warmed by \(40\,\text{K}\) wants to grow \(4.8\,\text{mm}\). Welded to its neighbours without room, it would carry \(96\,\text{MPa}\), hundreds of kilonewtons on each end, and might buckle. A gap of a few millimetres lets the rail take most or all of its new length freely, and the stress stays small.
- Putting the length \(L\) into the thermal stress.The stress is \(Y\alpha\,\Delta T\); the length cancels.
- Calling the stress in a heated clamped rod tensile.The walls push inward on a heated rod: compression. Cooling gives tension.
- Using the whole free expansion when a gap is left.Only the expansion beyond the gap becomes strain: \(L\alpha\,\Delta T - \delta\).
- Taking a fall of \(20\,°\text{C}\) to be a different size in kelvin.A change of \(1\,°\text{C}\) is a change of \(1\,\text{K}\).