# How to derive the coefficient of thermal expansion?

• tony_engin
In summary, Young's Modulus and the coefficient of linear thermal expansion are inversely proportional to each other. This can be derived from basic principles by finding the strain caused by thermal expansion and using the formula for Young's Modulus. This relationship is easier to understand by considering a specific material, where Young's Modulus is given by Y=FL/AL(dl) and the change in length due to thermal expansion is given by dl=L(a)(dT). Substituting these values, we can see that Y is inversely proportional to a.
tony_engin
Hi!
How to relate the Young's Modulus with the coefficient of linear thermal expansion? It is known that they are inversely proportional to each other, but how to prove this?

Hello,

It can be derived from basic principles. You can start by finding the strain caused by
the thermal expansion. The change in length for this part can be found by subbing in the thermal expansion formuale. Hence we have found the strain.

Y=stress/strain
=Fl/A(dl) where dl= change in length
I cannot proceed further than this due to lack of information on which kind of shape it have. ( to calculate A) However, the coefficient of thermal expansion could be seen to be inversely proportional to Y after subbing in the expression for dl.

Ok...For this you need to have a good grip on basic formulae

Since you are dealing with linear Young's modulus.First of all Young's Modulus for a wire of a specific material is given by:

$Y= \frac{FL}{A (dL)}$

where dL=Change in Length

Now change is length due to linear expansion is given by:

$dL=L (a) (dT)$

a=coefficient of linear expansion
dT=Change in Temperature

Now putting the values in Young's modulus equation:

$Y= \frac{F}{A (a) (dT)}$

Therefore Y is inversely proportional to a...easy isn't it?

## 1. What is the coefficient of thermal expansion?

The coefficient of thermal expansion (CTE) is a measure of how much a material expands or contracts when it is heated or cooled. It is typically expressed in units of length per unit temperature (e.g. mm/m°C or in/in°F).

## 2. How do you calculate the coefficient of thermal expansion?

The coefficient of thermal expansion can be calculated by measuring the change in length or volume of a material over a specific temperature range and dividing it by the original length or volume, multiplied by the change in temperature. This can be expressed as: CTE = (ΔL/L₀) / (ΔT), where ΔL is the change in length, L₀ is the original length, and ΔT is the change in temperature.

## 3. What factors can affect the coefficient of thermal expansion?

The coefficient of thermal expansion can be affected by various factors, such as the type of material, its composition, temperature range, and external factors such as pressure and stress. It can also vary depending on the direction of measurement (e.g. longitudinal or transverse).

## 4. How is the coefficient of thermal expansion used in engineering?

The coefficient of thermal expansion is an important factor in engineering as it helps determine the amount of thermal stress a material can withstand and how it will behave under different temperature conditions. It is used in designing structures and components that can withstand thermal expansion and contraction without causing damage or failure.

## 5. Can the coefficient of thermal expansion change over time?

Yes, the coefficient of thermal expansion can change over time due to various factors such as aging, exposure to different temperatures, and changes in the material's structure. This is why it is important to consider the long-term effects of thermal expansion when designing structures and components.

• Introductory Physics Homework Help
Replies
28
Views
1K
• Introductory Physics Homework Help
Replies
4
Views
663
• Introductory Physics Homework Help
Replies
6
Views
2K
• Mechanical Engineering
Replies
16
Views
421
• Mechanical Engineering
Replies
5
Views
2K
• Introductory Physics Homework Help
Replies
6
Views
2K
• Introductory Physics Homework Help
Replies
2
Views
1K
• Introductory Physics Homework Help
Replies
14
Views
469
• Materials and Chemical Engineering
Replies
4
Views
2K
• Introductory Physics Homework Help
Replies
2
Views
2K