# What is Thermal Stress? - Know about Thermal Stress and Strain

By Aina Parasher|Updated : May 16th, 2022

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 fall 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 be 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.

## 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 PT= PTL/AE

Deformation due to force PT= ∆

PTL/AE=Lαt

σT=αtE

where σT = Thermal Stress

σT 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.

σT 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 PT= PTL/AE

Deformation due to force PT = Net Deformation

Lαt – δ= PTL/AE

Lαt – δ= σTL/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

σ1A1 = σ2A2
σ1= σ2A2A1

Free Expansion = Δ = Δ1 + Δ2

= L1α1t + L2α2t

Deformation due to force P= PL1/A1E1+ PL2/A2E2

Deformation in bars due to P = Free Expansion

PL1/A1E1+ PL2/A2E2= L1α1t + L2α2t

σ1L1/E+ σ2L2/E2= (L1α1 + L2α2)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:-

P= [A1A2E1E2/(A1E1+ A2E2)] (α1 - α2)t

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## FAQs on Thermal Stress

• Thermal stress is the stress that develops in a body as a result of temperature changes. Composite bars are made up of bars made of various materials with varying coefficients of thermal expansion that have been joined together to behave as a single unit, with some explanation in both bars. Because both bars work as a single unit, their deformation will be identical. The bars with reduced α will suffer tension if the temperature rises.

• In this 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.

Deformation due to force PT= ∆

PTL/AE=Lαt

σT=αtE

where σT = Thermal Stress.

• Given, E = 2 x 105 N/mm2, 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.

• We know that,

σ1A1= σ2A2

σ1×400= σ2×500

σ1= 1.252σ2

σ1L1/E+ σ2L2/E2αLt

=12 × 10-6 ×1100 ×20

σ1= 43.1 MPa

σ253.87 MPa

• Metal 1 has twice the area of Metal 2.

A1= 2A2

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

Let us consider A2 = x and A1 = 2x

σ1 x 2x = 30 x x

σ1= 15 MPa

The stress developed in the first bar is equal to σ1 = 15MPa with the opposite sign. Stress in composite bars is always the opposite.

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