# What is Thermal Stress?

By BYJU'S Exam Prep

Updated on: September 25th, 2023

**Thermal stress** is the stress induced in a body when that body undergoes temperature changes,i.e when a body expands due to a rise in temperature or contracts due to falling in temperature. Thermal stress is highly dependent on the coefficient of thermal expansion (α) of the material of which the body is made. The greater the coefficient of thermal expansion (α) is, the more will the expansion and vice-versa.

In this article, we will try to understand thermal stress, thermal stress in simple bars, thermal stress in varying cross-sections, and thermal stress in a composite bar.

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## What is Thermal Stress?

The stress which is developed in a body due to the change in the temperature is known as thermal stress. Thermal stress is developed in the body when there is a change in temperature ( it might be a temperature rise or temperature drop ) and there is no option for the body to expand (when the temperature is raised) and contract (when the temperature is lowered).

Suppose a bar that is free to expand. Consider the below figure; assume the cross-sectional area A, length of bar L, and coefficient of thermal expansion is ‘α’.

‘t’ is the change in temperature

Δ is the free thermal expansion

Free thermal expansion for the above case is given by

Δ = Lαt

If the temperature is reducing then there will be a contraction given by

Δ = Lαt

In this particular condition, no thermal stress is developed as there is no restriction for expansion or contraction. Stress is developed only when there is restriction or resistance to free motion.

## Thermal Stress and Strain

As we know, thermal stress is the stress induced in a body due to a change in temperature. Due to this development of thermal stress, thermal strain is induced in that body. It can be said that thermal stress and strains are developed in the body when it is heated or cooled. Generally, a thermal strain developed due to thermal stress is very small. As the material will expand or contract in all directions, the thermal strain will be developed in all directions.

## Thermal Stress in Simple Bars

**Case 1: ****When boundaries are fixed**– Consider Figure A, as there is no restriction for expansion or contraction when there is a change in temperature, the thermal stress developed is zero and the free expansion is Lαt. Also Consider Figure B, as both ends are fixed, free movement is not possible when there is a change in temperature, this leads to the development of thermal stress in the body.

Deformation due to force P_{T}= P_{T}L/AE

Deformation due to force P_{T}= ∆

P_{T}L/AE=Lαt

σ_{T}=αtE

where σ_{T} = Thermal Stress

σ_{T} is compressive stress when there is an increase in temperature because when there is a temperature rise, the body tries to expand but boundaries will try to keep it in its original position, so the thermal forces will be compressive.

σ_{T} is tensile when there is a decrease in temperature because when there is a temperature drop, the body tries to contract, but the boundaries will try to keep it in its original position, so the thermal forces will be tensile.

**Case 2: ****Thermal Stress when Support Yield**

Free Expansion =Lαt

Net Deformation = Δ – δ (where δ is the support yield or gap)

Net Deformation = Lαt – δ

Deformation due to force P_{T}= P_{T}L/AE

Deformation due to force P_{T} = Net Deformation

Lαt – δ= P_{T}L/AE

Lαt – δ= σ_{T}L/E

## Thermal Stress in Varying Cross-Section

Consider a series of bars that have different cross-section areas and different lengths, different materials, to find the stress in different bars. Let us assume that the temperature is increasing. When there is an increase in temperature, the bars will try to expand, but the boundaries will restrict the expansion by exerting force (P).

The force in both bars is the same.

(Force)_{1} = (Force)_{2}

σ_{1}A_{1}= σ_{2}A_{2}

σ_{1}= σ_{2}A_{2}A_{1}

Free Expansion = Δ = Δ_{1} + Δ_{2}

= L_{1}α_{1}t + L_{2}α_{2}t

Deformation due to force P= PL_{1}/A_{1}E_{1}+ PL_{2}/A_{2}E_{2}

Deformation in bars due to P = Free Expansion

PL_{1}/A_{1}E_{1}+ PL_{2}/A_{2}E_{2}= L_{1}α_{1}t + L_{2}α_{2}t

σ_{1}L_{1}/E_{1}+ σ_{2}L_{2}/E_{2}=(L_{1}α_{1}+ L_{2}α_{2})t

## Thermal Stress in Composite Bars

Composite bars are the bars in which there will be bars of different materials whose coefficient of thermal expansion is different and are clubbed together so that they act as a single unit and there will be the same explanation in both bars. As both bars act as a single unit, the deformation in both bars will be the same. Thermal stress in composite bars can be given by the following formula:-

## Thermal Stress Problems

**Question 1.** A steel rail is 13m long and is laid at a temperature of 25ºC. The maximum temperature is expected to be 45ºC. Given, E = 2 x 105 N/mm2, α = 12 x 10-6 /ºC. Estimate the minimum gap to be left between the rails so that the temperature stress doesn’t develop.

**Solution:** Given, E = 2 x 105 N/mm^{2}, α = 12 x 10^{-6} /ºC

T = 45º– 25º= 20ºC

We know that, δ = αtL

= 12 x 10^{-6}x 20 x 13000

= 3.12mm

Temperature stress won’t get developed if the minimum gap left on the rails is equal to free expansion.

**Question 2.** A steel rod is firmly held between two rigid supports as shown. Find the stress developed in the two portions of the rod, when it is heated through 20^{o}C. Given Given, E = 2 x 10^{5} N/mm^{2}, α = 12 x 10^{-6} /ºC.

**Solution:** We know that,

σ_{1}A_{1}= σ_{2}A_{2}

σ_{1}×400= σ_{2}×500

σ_{1}= 1.252σ_{2}

σ_{1}L_{1}/E_{1}+σ_{2}L_{2}/E_{2}=αLt

=12 × 10^{-6} ×1100 ×20

σ_{1}= 43.1 MPa

σ_{2}=53.87 MPa

**Question 3.** A bimetallic strip is made of two metals with one metal having twice the area as that of another. Due to temperature change, the stress in metal strips with a lesser area is -30 MPa. What is the stress developed in another component of the composite bar?

**Solution:** Metal 1 has twice the area of Metal 2.

A_{1}= 2A_{2}

Metal 2 has a lesser area, as given in the question. Metal 2 has a stress of -30MPa.

Let us consider A_{2} = x and A_{1} = 2x

σ_{1} x 2x = 30 x x

σ_{1}= 15 MPa

The stress developed in the first bar is equal to σ_{1} = 15MPa with the opposite sign. Stress in composite bars is always the opposite.

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