Thermal Expansion: Predict Which Way a Bimetal Strip Bends

Thermal expansion: predict which way a bimetal strip bends

A joined strip does not bend because one side “gets hotter” in every example. Even when both layers reach the same temperature, different materials can have different preferred changes in length. Bonding prevents them from expanding independently. To predict the bend, compare their free length changes, then decide which layer must lie on the longer outside of the curve.

All temperatures, coefficients and diagrams below describe fictional classroom materials. Use them as paper models. They are not directions for heating metal, opening a thermostat or wiring a switch.

Begin with equal lengths at a reference temperature

At 20 °C, strips A and B are straight and each has reference length L₀ = 0.250 m. They are bonded along their lengths, with A above B and the left end fixed. Over the stated interval, use the constant positive coefficients α_A = 24 × 10⁻⁶ per °C and α_B = 12 × 10⁻⁶ per °C. Thus A would change its free length twice as much as B for the same temperature change.

The approximate free-expansion equation is ΔL = αL₀ΔT, where ΔT = final temperature − reference temperature. Heating gives a positive ΔT; cooling gives a negative value. A difference of 60 °C has the same numerical size as a difference of 60 K. That does not mean the temperature 60 °C equals 60 K. The coefficient must describe the material and temperature interval being modelled. OpenStax explains linear expansion and joined-metal bending.

Original schematic. Initially at 20 degrees Celsius, material A lies above B and both layers have equal reference length. A has the larger positive expansion coefficient. At 80 degrees, A occupies the longer outer arc and the free right end bends down toward B. At minus 10 degrees, A contracts more, occupies the shorter inner arc and the free end bends up toward A. All panels fix the left end; dimensions and curvature are schematic.

Worked case 1: compare the free length changes

Both layers warm from 20 °C to 80 °C, so ΔT = 60 °C. For A, ΔL_A = 24 × 10⁻⁶ × 0.250 × 60 = 0.000360 m = 0.360 mm. For B, ΔL_B = 12 × 10⁻⁶ × 0.250 × 60 = 0.000180 m = 0.180 mm. Their preferred free-length increases differ by 0.180 mm.

If they were separate and unconstrained, their final lengths would be 250.360 mm and 250.180 mm. Because they are bonded, those numbers are a comparison of their expansion tendencies, not a claim that every line within the bent strip has those exact lengths. Bonding introduces stresses, and actual curvature also depends on thickness, stiffness and the connection between layers. The simple calculation predicts which material needs the longer path; it does not calculate a real device’s bend angle.

Worked case 2: use outside and inside, then name a direction

For equal angles, an outer circular arc is longer than an inner arc. A expands more on heating, so A belongs on the outside of the bend. In our side-view drawing A starts above B, the left end remains fixed, and the free end extends right. The strip bends downward toward B: A is the upper, longer outer layer and B is the shorter inner layer.

The reliable conclusion is “toward the lower-expansion layer on heating” for this stated pair. Saying only “it bends down” is incomplete because down depends on how the strip is drawn. Rotate the page and the physical result stays the same even though the page direction changes. Label materials, fixed end and reference temperature before using an arrow.

As a paper switching idea, imagine an unpowered contact mark below the free tip. The predicted downward movement is toward that mark. This establishes only a possible direction of approach. It does not show that contact will be reached, specify a switching temperature, or justify any electrical connection. Those would need additional dimensions and material behaviour.

Worked case 3: cool the strip and reverse the layers

Starting from its straight 20 °C reference shape, cool the original strip to −10 °C. Now ΔT = −30 °C. A has ΔL = −0.180 mm and B has ΔL = −0.090 mm. Both prefer to contract, but A prefers the shorter final length. A therefore occupies the inner arc. With A still above B and the left end fixed, the free right end bends upward toward A.

Now consider a different strip with the same materials exchanged: B is above A. Heating it from the same straight reference state makes the lower A layer the longer outer layer. Its free end bends upward toward B. Cooling that reversed arrangement makes A the inner layer and sends the free end downward. These comparisons show why identifying only the larger coefficient is not enough; the arrangement and sign of ΔT also matter.

What happens to the particles?

In an ordinary positive-expansion solid, a temperature rise increases thermal motion and can increase the average separation of neighbouring particles. The particles do not simply swell into larger particles. A useful sketch keeps the particle symbols the same size while increasing their average spacing. This is a model of the size change, not a scale drawing of atoms.

Do not extend the rule to every substance at every temperature. Some materials or intervals behave differently, and a tabulated coefficient may vary with temperature. Our prediction is conditional on the supplied positive, approximately constant coefficients and on the strip staying bonded within its elastic behaviour.

Three solved practice tasks

Task: A separate strip is 0.400 m long at its reference temperature. Its supplied coefficient is 15 × 10⁻⁶ per °C. Find the free expansion for a 40 °C rise.

Solution: ΔL = 15 × 10⁻⁶ × 0.400 × 40 = 0.000240 m = 0.240 mm. The final length is 400.240 mm. Multiplication by 1,000 converts metres to millimetres.
Task: Equal-length joined layers have upper coefficient 8 × 10⁻⁶ per °C and lower coefficient 20 × 10⁻⁶ per °C. The left end is fixed. Predict the free-end direction when heated.

Solution: The lower layer expands more and needs the outer arc. The upper layer is inside the curve, so the free end bends upward toward the upper, smaller-coefficient layer.
Task: Two initially equal-length bonded layers have the same positive coefficient and experience the same temperature change. Does coefficient difference predict a bend?

Solution: No. Their free length changes are equal, so this simple model predicts no differential-expansion bend. Uneven heating, pre-existing stress or other effects would require information not supplied in the task.

Frequently asked questions

Which layer is outside when the strip is heated?: For this positive-coefficient model, the layer with the greater free expansion lies on the longer outside arc.
Which layer is inside when it is cooled?: The layer that contracts more needs the shorter path and lies inside the bend, relative to the straight reference state.
Does a higher coefficient mean a higher temperature?: No. A coefficient describes a material’s expansion response. Both layers can be at the same temperature.
Why must the original length be supplied?: Expansion depends on original length as well as coefficient and temperature change. Equal coefficients do not imply equal changes for unequal lengths.
Can I use kelvin for the temperature change?: Yes. A temperature difference has the same numerical value in kelvin and degrees Celsius when the coefficient uses the corresponding reciprocal unit.
Do these calculations give the tip displacement?: No. They compare free length changes. Predicting the actual bend needs further mechanical and geometric information.
What is the quickest diagram check?: Label the larger-expansion material, decide whether it should be longer or shorter, and check that it is on the correct outer or inner arc.