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Thermal stress and strain, applications of thermal stress and strain, practice problems, FAQs

Thermal stress and strain, applications of thermal stress and strain, practice problems, FAQs

If you heat a glass on a flame and immediately dip it into cold water, what do you think will happen to the glass? It will break due to fracture and this fracture is not the same as it would have happened if a stone is thrown at the glass. To know the reason behind this fracture you might need to know about thermal stress first.

Table of contents

  • Thermal stress and strain
  • Experiment for understanding thermal strain
  • Time Period of Simple Pendulum under Temperature Variation
  • Practice problems
  • FAQs

Thermal Stress and strain

The mechanical stress developed due to the change in temperature in restricted conditions is known as thermal stress.

Unlike the case of external forces being exerted on the rod and producing strain, the thermal stress is caused due to the rod’s tendency of changing length but not having the freedom to do so. Therefore, despite the null elongation/compression, the rod remains under stress and experiences strain. 

In the given figure, a rod of length L0 at temperature T0 is clamped at both ends. The temperature of this rod is increased by ΔT. The corresponding strain developed in the rod as shown in the figure.

Strain (𝜀),in this case, will be,

The minus sign in the expression of strain (𝜀)is due to the stress being compressive.

Here, the observed length (Lobserved) and the natural length (Lnatural) are measured at the same temperature. (Lobserved) is the existing length of the rod at temperature (T0 + ΔT), when (Lnatural) should have been its length at temperature (T0+ΔT) if it was allowed to expand.

Since 𝛼Δ<<1, the denominator can be approximated as


As we know Young’s modulus, 

Therefore, thermal stress,

Force exerted by the rod on the clamps,

We can better understand thermal strain from the following experiment. 

Experiment for understanding thermal strain:

Case I: The rod is free to change its length

Let us take a rod of length L0at temperature T0 and heat it up to temperature (T0 + ΔT). 

Change in observed length,

However, , since,

Case II: The rod is fixed at both ends

Similar to Case I, let us take a rod of length L0at temperature T0 and heat it up to temperature (T0 + ΔT).

Change in observed length, Δ= 0

However, , since the rod is not allowed to expand up to its natural length.

Practical applications of thermal stress and strain

  • Thermostat: Thermostat is a device used to regulate the temperature in various devices like refrigerators, air conditioners etc. The working of a thermostat is based on the concept of thermal expansion.

In some cases it uses a bimetallic strip that is formed by bonding two different metal strips together. Since the two metals will be having different expansion coefficients, they react differently to the same rise in temperature i.e. one of them expands more relative to the other. This difference in expansion produces stress in the strip and the bimetallic strips get bent. The mechanism is designed in such a way that as soon as it bends, it opens the electric circuit. It remains open until the strip gets back to its normal shape.

  • Thermometers: Mercury thermometers use the property of thermal expansion to measure the temperature. With the change in temperature of the mercury, the volume expansion or contraction is used to measure the temperature.

Practice problems

Q 1. A steel rod is clamped at its two ends and rests on a fixed horizontal surface. The rod is at its natural length at 20 °C. Find the longitudinal strain developed in the rod if the temperature rises by 30 °C. (𝛼steel = 1.2 × 105 °C−1)



T0= 20 °C, = 50 °C, 𝛼steel = 1.2 × 10−5 °C−1

Hence, the strain produced in the rod due to an increase in temperature comes out to be 3.6 × 104.

Q 2. A steel wire of cross-sectional area 0.5 mm2 is welded between two fixed supports. If the wire is just taut at 20 °C, determine the tension when the temperature falls to 0 °C. Young’s modulus of steel is 2 ×1011 N m−2.



A = 0.5 mm2, T0 = 20 °C, T1= 0 °C,𝛼steel = 1.2 × 105 °C−1,=2 × 1011 N m−2

Longitudinal strain (𝜀)is as follows:

Using the properties of solids, the tension in the rod can be found as following,

Q 3. Tensile force having magnitude is applied on a steel rod of area of cross section which results in a change in length of the rod. If the steel rod is heated, what will be the change of temperature required to produce the same elongation?

(The modulus of elasticity is and the coefficient of linear expansion of steel is .


We know ,

Modulus of elasticity,



Change in length (thermal strain),

Q 4. Two elastic rods are joined and put together between fixed supports as shown in figure. Find the condition for no change in the lengths of the individual rods with increase in temperature.[Given linear expansion coefficient, area of cross-section of rod, Young's modulus and length of rod]

                               (Recreated the image)

A. As per question,

The compressive forces in both rods due to thermal expansion should be equal since there is no change in the lengths of individual rods. Then, the net force will be zero on each rod.

[ Using formula of linear expansion and Young's modulus ]

Since temperature difference is same,


Q 1. What kind of stress does heating a rod generate if the rod is restricted to any change in length?
A. Since heating a rod leads to expansion, therefore the restriction will apply compressive stress on the rod. Hence the rod will be under compressive stress.

Q 2. A steel rod of length 1 m rests on a smooth horizontal surface. If it is heated from 0 °C to 100 °C, what is the stress developed?
A. Since the rod is resting on the surface without clamps, it is free to expand due to increased temperature. Hence, the stress (𝜎) produced in the rod will be zero. 

Q 3. Does cooling generate any stress in a fully constrained bar?
A. Yes, the bar tries to contract upon cooling and thus the wall while trying to resist the deformation pulls the bar from both sides and thus the bar experiences tensile stress.

Q 4. What are the units and dimensions of thermal strain?
A. Thermal strain is a unitless and dimensionless quantity. 

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