Hook’s Law and Linear Expansion

Task number: 1290

A steel wire with a temperature of 100 °C is fixed between two clamps. The ambient air temperature is 20 °C.

a) Does the wire breaks before it cools down to the temperature of the ambient air?

b) At what temperature do we have to fix the wire between the clamps, so that it would not tear during the cooling to the ambient air temperature?

Although it does not fully correspond to reality, assume that the deformation is elastic right up to the breaking load.

  • Hint

    Realize what is going on with the wire when it is cooling down and what happens when we fix both ends of the wire.

  • Analysis

    When the wire cools down, it shortens. If both ends are firmly fixed, it cannot shorten and its internal tension increases according to the Hook’s Law.

    The equation, that describes how the internal tension depends on the temperature difference, can be obtained by substituting the formula for the linear thermal expansion into the Hook’s Law.

    To determine whether the wire tears, we will compare the internal tension of the wire induced by the temperature decrease, and the breaking load of the wire. If the internal tension is greater than the breaking load, the wire tears, and vice versa.

    In the second task, we need to determine the temperature, for which it applies, that when we cool the wire from this temperature to the ambient air temperature, the internal tension just reaches the value of the breaking load.

  • Numerical values

    t0 = 100 °C the initial temperature of the wire
    ta = 20 °C the ambient air temperature
    σ = ? the internal tension of the wire after cooling
    tmax = ? the maximum temperature

    From The Handbook of Chemistry and Physics:

    σb = 5.0·108 Pa the breaking load of steel
    E = 21.0·1010 Pa the Young’s modulus of steel
    α = 1.2·10−5 K−1 the coefficient of linear expansion of steel
  • Solution

    When the wire cools down, it shortens. This is, however, not possible due to its fixation between the clamps. Hence, with the decreasing temperature the internal tension in the wire increases, which according to Hook’s Law for elastic deformation is proportional to the relative shortening of the wire ε, which would have occurred if the wire had not been fixed.

    We will now determine from Hook’s Law how the internal tension σ depends on temperature t:

    \[\sigma = E \varepsilon = E\frac{\mathrm{\Delta} l}{l_{0}}=E\frac{l_{0} \alpha (t_{0}-t)}{l_{0}} = E \alpha (t_{0}-t)\,,\]

    where E is the Young’s modulus of elasticity of steel, α is the coefficient of the linear expansion of steel and t0 is the initial temperature of the wire.

    a) The wire tears when the internal tension induced by the cooling of the wire from the initial temperature t0 to the ambient air temperature ta is greater than the breaking load of the wire.

    The tension in the wire is determined by the relation:

    \[\sigma = E \alpha (t_{0}-t_{a})\,.\]

    b)It must apply for the unknown temperature tmax that the internal tension reaches the breaking load σb at the very moment when the wire reaches the ambient air temperature ta. Therefore it is true that (according to the formula set in the part a)):

    \[\sigma_{b} = E \alpha (t_{\mathrm{max}}-t_{a})\,,\]

    from which we can express the unknown temperature tmax:

    \[t_{\mathrm{max}}-t_{a} = \frac{\sigma_{b}}{E \alpha},\] \[t_{\mathrm{max}} = t_{a}+\frac{\sigma_{b}}{E \alpha}.\]
  • Numerical insertion

    a)

    \[\sigma = E \alpha (t_{0}-t_{a}) = 21\cdot {10^{10}}\cdot{1.2}\cdot{10^{-5}}\cdot \left( 100-20 \right) \, \mathrm{Pa} \dot{=} 2\cdot{10^{8}}\, \mathrm{Pa}\] \[\sigma\, <\, \sigma_{b}\]

    b)

    \[t_{\mathrm{max}} = t_{a}+\frac{\sigma_{b}}{E \alpha} = \left(20+\frac{5 \cdot{10^{8}}}{21 \cdot{10^{10}} \cdot{1.2} \cdot{10^{-5}}}\right)\, \mathrm{^{\circ}C}\] \[t_{\mathrm{max}} \dot{=} 218.4\,\mathrm{^{\circ}C}\]
  • Answer

    a) The tension in the wire is less than the breaking load. The wire will therefore not tear.

    b) The wire can be fixed between two clamps, when its temperature is 218 °C at maximum.

Difficulty level: Level 3 – Advanced upper secondary level
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