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Engineering: The challenge of temperature
Engineering: The challenge of temperature

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3.4 Summary of Section 3

  • The temperature of an object is intimately linked to the average kinetic energy of the atoms from which it is made. As a result, some materials properties such as electrical resistance and mean atom spacings change gradually with temperature. These properties can be modelled with a simple linear equation like the following one that describes thermal expansion:

    where α is the temperature coefficient of expansion.

  • If the thermal expansion of solids is constrained, stresses are induced within them. When not managed properly, thermal stresses can lead to distortions and cracks. However, useful engineering components and devices can be constructed that take advantage of 'gradual' thermal sensitivity.

Box 4 Homogeneous body, uniform ΔT, external constraints

The classic example of the buckled railway line (Figure 6a) is an easy geometry to calculate.

Figure 6
Figure 6 (a) buckled rails, (b) calculating the strain

Suppose the length of a rail is L0 when its ends just touch adjacent rails and then the temperature goes up by ΔT (Figure 6b). The length ought to become L0(1 + αΔT) but is constrained to remain L0. It is therefore as if the rail had expanded and then suffered a compressive strain αΔT which would have needed an (elastic) stress EαΔT. This could cause the rail to buckle; some of the wiggle in Figure 6(a) was probably caused by the passage of the train – but it needn't be as bad as this picture suggests before it's serious enough to cause a derailment. One way out of the problem is to make provision for expansion and so remove the constraint: for example, rail gaps and bridge roller bearings. However, modern high-speed trains require smooth tracks with continuous welded rails, so another solution is called for here. One answer is to clamp and weld up the rails either in hot weather or else when mechanically stretched, so that stresses in the rails in service are always tensile – buckling is associated with compressive stresses.

Box 5 Composite body, uniform ΔT, internal constraints

Pretty pots

A relatively simple example to model is the glaze on a pot. In Figure 7(a) a thin glassy sheet (of thickness x1), the glaze, is stuck onto a thicker slab (x2).A look back at data in Table 4 suggests that a glass will have a higher expansion coefficient than porcelain, so α1 > α2. So, the glaze will be inclined to expand on heating (and contract on cooling) more than the pot to which it is bonded. There will have to be some kind of compromise.

Figure 7
Figure 7 (a) thin glaze on a thick ceramic pot, (b) Chinese pot

In fact, the problem begins at the stage where the glaze is fired onto the pot and initially both glaze and pot contract on cooling. As the pot cools from its firing, the glaze begins to harden and it can no longer accommodate to the body. A thermal stress develops. The glaze should contract more than the pot, but both must move together. So the glaze is in tension because it is prevented from contracting as much as it would if free of the pot, as if the pot is 'pulling' it somewhat – that's tension. The pot is in compression because the glaze is tending to induce more contraction of the surface, 'pushing' atoms closer, than would occur if the pot were unglazed – that's compression. To calculate the size of the thermal stresses you would have to convert the resulting tensile and compressive strains into stresses using the Young's modulus for each material. We don't need to do that to predict what might happen. Because there's much more pot than glaze, the thermal stress is likely to pull the glaze so strongly that it cracks, which might produce a pretty pot (Figure 7b) but it is not the way to make a strong one.

Bimetal strips

We can use the same ideas to anticipate what happens when two thin strips of metals with markedly different expansivities are bonded together to form a flat composite body. When such a bimetal strip is heated one layer will be trying to expand more than the other – it will be held back a little, effectively compressed slightly by the second layer. In turn, the second layer will be in tension as it is expanded beyond the thermal strain owing to the presence of the first layer. If neither metal yields, the result is that the strip must bend.

Bimetal strips are used as a cheap, robust way of indicating a temperature. Central heating pumps are often triggered to switch on and off as relays are tripped by the bending of a strip in response to changes in room temperature.

Simple, non-digital instruments in motor vehicles are based on the deflection of an indicator needle in response to the heating of a bimetal strip. Such instruments sense the strength of an electric current. A higher current causes more heating of the strip and that in turn causes the bimetal strip to bend more; the needle movement is driven by this bending (Figure 8).

Figure 8
Figure 8 A bimetallic instrument gauge


Propose two materials from which to construct the most responsive bimetallic strip, based only on the materials and data in Table 4, giving reasons for the choice made.


The most responsive strip will have the greatest mismatch between the thermal expansion coefficients of the two materials, provided neither metal yields. You should therefore choose from the extremes, though the yield criterion probably means that lead should be excluded. So, Invar and aluminium ought to work well, with Invar and brass a close second.

Box 6 Homogeneous body, non-uniform ΔT

Thermal shock arises because heat flow is not instantaneous. Suppose the surfaces of a plate are suddenly cooled, energy being extracted at some rate (measured in watts per square metre). The temperature inside cannot be uniform and will vary with time, somewhat in the manner shown in Figure 9.

Figure 9
Figure 9 Variation of temperature with time within body subject to rapid cooling from two opposite surfaces

At time t = 0, when cooling begins, and after an infinite time, there are no internal stresses. At all other times every slice of the object should be a slightly different size from its adjacent slice, but is constrained by its neighbour. Complex non-uniform stresses are the result.

We can, however, see what factors influence the magnitude of these stresses. Clearly, very steep temperature gradients make for high stress, and the gradient depends on how much energy has to flow from the inside to the surface and how quickly it can do so. The cooling of a thick, hot slab made of a material which has poor thermal conductivity, high heat capacity per unit volume, and high expansion coefficient, will induce intense stresses. If, additionally, the material efficiently converts the 'effective strain to high stress via a high Young's modulus, and if it is crack-sensitive, you've got a problem on hand. Consider the chances of a ceramic turbine component when the engine is shut down; or a more homely example, how a ceramic hob survives having cold water spilt on it. The trick in that case is rather neat, as we shall see.

Ceramic hob materials

The photograph of a domestic ceramic hob (Figure 10) suggests a puzzling suite of properties.

The hot spots for cooking must be localised, so the material should not be a good thermal conductor or that will spread the heat sideways. If the hob is a bad conductor it must be thin to allow sufficient power transfer for its function. So how can it be strong enough to survive the following?

  • thermal shock if cold water is poured onto a hot ring

  • thermal stress due to steep temperature gradients at the edges of the hot spots

  • having a full saucepan dropped on it from one metre height – this is used as a standard acceptance test.

Figure 10
Figure 10 Ceramic hob

What is your recommendation for thermal expansion coefficient?

It must be as small as possible; thermal shock resistance can be achieved by keeping thermal strain low and by reducing temperature gradients (by having good conduction). The latter characteristic is denied us, so low expansion is imperative.

Fused silica glass has a very low expansion coefficient and a low enough thermal conductivity. What is unacceptable about a glass for this purpose?

Glasses are very brittle. As glaziers well know, once a piece of glass is scratched, cracks easily spread through it and it will break. Glass objects are bound to be scratched in service.

One solution is a ceramic based on lithium aluminium silicate, LAS, which crystallises in hexagonal planes of atoms leading to a curious response to heat. Its expansion coefficient depends on direction within the crystal: +6.5 × 10−6 K−1 in the plane of the hexagon and −2.0 × 10−6 K−1 perpendicular to the hexagon. Yes – a negative coefficient of expansion. As the atoms spread sideways the layers can pack down into their hollows a bit more. This unusual behaviour confers near-zero coefficient upon specimens of LAS that are made up of a large number of randomly oriented crystals.

To make a ceramic hob in the required dimensions would be impossible by ordinary ceramic techniques, but plate glass is easily made. So the material is first prepared as a glass and sheets are made by hot rolling (it is very viscous). Then a heat treatment encourages crystallisation with the hexagon axes randomly oriented. If the individual crystals are very small they can accommodate anisotropic shape changes without inducing large thermal stresses between differently oriented crystals. Also, small crystal size confers strength on the ceramic body by limiting the size of intrinsic flaws which could act as cracks.


In cold climates, the temperature on the outward-facing side of a window pane is often many degrees cooler than that on the side facing inwards. Explain the nature of thermal stresses arising in the glass of a pane that is tightly constrained.


The warmer side will be trying to expand more than the cooler side. Both sides are assumed to be constrained and cannot change their dimensions (in fact, a metal or plastic frame may add additional stress to the glass). The result may be tensile stresses in the cooler side – it is stretched back from its would-be contracted state – and compressive stresses in the warmer side – it is squashed back from its would-be expanded state. Either or both would apply, depending on the temperature at which the installation was done, establishing the constraint.

Box 7 What about welding?

Welding is a classic example of where internal stresses generated by temperature effects can lead to eventual failure. A welding process may involve several passes of molten metal being laid down, with each pass having its own heating and cooling cycle. There is likely to be constraint associated with the parts being welded. There will be a temperature gradient from the weld into the cooler base material.

So whenever welding is carried out, it leaves the joint with a legacy of built-in stresses. As the fusion zone – the weld metal and melted metal of the component – cools after solidifying, it contracts more than the adjacent metal which has not been subjected to the same high temperature. This may in itself cause cracks to form, and if it doesn't there still must be a matching of strains between the weld metal and its surroundings. The extra thermal contraction of the weld leads to stresses because of constraint effects of material away from the weld. The stresses cannot normally exceed the yield stress, but below about 500 K the yield stress of most steels is high enough for appreciable stresses to build up. We call these residual stresses.

A simple way to visualise the formation of residual stresses is the 'three-bar model of the welded joint. The fusion zone corresponds to a hot central bar which is flanked by two cold bars (see Figure 11a). The outer bars correspond to the material surrounding the weld. The bars (all initially of the same length, L0) are regarded as being fixed together by rigid end pieces – in other words, the bars must all remain the same length (because the fusion zone remains the same length as the adjoining material to which it is attached).

Figure 11
Figure 11 Three-bar model for analysing stresses in a weld

In this model of a weld the central bar is initially stress-free and hot; but stresses build up as the weld solidifies, cools and contracts. If the central bar has a coefficient of thermal expansion α and cross-sectional area ΔT, and cools through a temperature range ΔT, if free it would undergo a thermal contraction of L0αΔT. However, it must be constrained to be the same length as the outer bars, and there will be an elastic strain of ε2 needed to ensure this (Figure 11b). I'll assume that the two outer bars have identical cross-sectional area A2 and undergo an elastic strain ε2. Now we can invoke two conditions. First, the total length change in the inner bar must equal that in the outer bars (the condition of compatibility):

Using Hooke's Law, σ = Eε, this expression can be written in terms of the Young's modulus E of the bars and the residual stresses (σ1, σ2) in the bars:

The second condition is that the force σ1A1 (remember: force = stress × area) exerted by the inner bar on the outer bars must be equal and opposite to (2σ2A2 the force exerted by the outer bars on the central one (the condition for equilibrium):

where σ1 and σ2 are the stresses in the inner and outer bars respectively. These equations can be solved for the residual stresses (σ1, σ2) in the bars.

Exercise 5

Using the three-bar model just described, calculate the values of σ1 and σ2 for a steel weld where ΔT= −200 K (i.e. a drop in temperature), α= 11 × 10−6 K−1; and E = 210 GN m−2. You will have to solve the two simultaneous equations containing σ1 and σ 2. Assume that A1 = A2


Starting off with

we can rearrange this to give:

Plugging in the numbers gives us:

Then using σ1A 1 = −2σ2A 2 with A 1 = A 2, we have that σ1 = −2σ2, which gives:

σ2 = −154 MPa (negative sign implies compression)

and σ1 = +308 MPa (positive sign implies tension)

Box 7 contd

The residual tensile stress of 308 MN m−2 in the central bar may be of the order of the room temperature yield strength for many steels – and it requires the weld to cool by only some 200 K more than the adjacent metal. (In practice the weld cools by much more than this, but the residual stress cannot rise too much above the yield strength because at such stresses the weld metal can strain extensively by plastic deformation.) In practice, of course, instead of the uniform residual stresses in the rods (Figure 11c) the residual stresses parallel to a welded joint will vary smoothly across the weld (Figure 11d).

Residual stresses can be reduced by heating the welded structure, because the yield strength falls with increasing temperature. The thermal stresses should not recur during cooling, because of the more uniform temperature distribution. This procedure is called post-weld heat treatment or stress relieving.

Figure 11(e) shows an example of where residual stresses led to cracking near a weld. In this example, a weld for a power plant pipe was found, after initial fabrication, to contain a crack – probably just from shrinkage effects, as described earlier. The cracked area was machined out, and the weld was repaired with a new layer of weld material, as shown. However, the repair process complicates the residual stress generation even further, and after a period of operation at high temperature a so-called 'reheat crack' formed in the material near to the repair.