# Linear dilation: what is it, formula and coefficients, example

The **linear expansion** occurs when an object undergoes expansion due to a temperature change, predominantly in one dimension. This is due to the characteristics of the material or its geometric shape.

For example, in a wire or bar, when there is a rise in temperature, it is the length that changes the most due to thermal expansion.

Birds perched on wires. Source: Pixabay

The cables on which the birds in the previous figure land stretch when the temperature increases; rather, they contract when they cool. So, for example, do the bars that make up the rails of a railway.

**What is linear dilation?**

In a solid material, atoms keep their relative positions more or less fixed around an equilibrium point. However, due to the thermal agitation, they are always revolving around it.

As the temperature increases, the thermal swing also increases, causing the average swing positions to change. This is because the binding potential is not exactly parabolic and has asymmetry around the minimum.

Below is a figure that describes the energy of chemical bonding as a function of interatomic distance. It also shows the total oscillation energy at two temperatures and how the oscillation center moves.

Graph of chemical bond energy versus interatomic distance. Source: own elaboration.

**Linear expansion formula and its coefficient**

To measure linear expansion, we start with an initial length L and an initial temperature T of the object from which you want to measure its expansion.

Suppose the object is a bar whose length is L and the cross-section dimensions are much smaller than L.

First, said object is subjected to a temperature variation AT, so that the final temperature of the object, once thermal equilibrium with the heat source has been established, is T ‘ = T + AT.

During this process, the object’s length will also change to a new value L ‘= L + ΔL, where ΔL is the change in length.

The linear expansion coefficient α is defined as the ratio of the relative change in length per unit of temperature change. The following formula defines the linear expansion coefficient *α* :

The dimensions of the coefficient of linear expansion are those of the inverse of the temperature.

**Linear expansion coefficient for various materials**

Below we will list the coefficient of linear expansion for some typical materials and elements. The coefficient is calculated at normal atmospheric pressure, based on an ambient temperature of 25 °C; and its value is considered constant over a ΔT range up to 100 °C.

The unit of coefficient of linear expansion will be (°C) ^{-1} .

– Steel: α = 12 ∙ 10 ^{-6} (° C) ^{-1}

– Aluminum: α = 23 ∙ 10 ^{-6} (°C) ^{-1}

– Gold: α = 14 ∙ 10 ^{-6} (°C) ^{-1}

– Copper: α = 17 ∙ 10 ^{-6} (°C) ^{-1}

– Brass: α = 18 ∙ 10 ^{-6} (°C) ^{-1}

– Iron: α = 12 ∙ 10 ^{-6} (°C) ^{-1}

– Glass: α = (7 to 9) ∙ 10 ^{-6} (° C) ^{-1}

– Mercury: α = 60.4 ± 10 ^{-6} (°C) ^{-1}

– Quartz: α = 0.4 ± 10 ^{-6} (°C) ^{-1}

– Diamond: α = 1.2 ∙ 10 ^{-6} (°C) ^{-1}

– Lead: α = 30 ∙ 10 ^{-6} (°C) ^{-1}

– Oak wood: α = 54 ∙ 10 ^{-6} (°C) ^{-1}

– PVC: α = 52 × 10 ^{-6} (°C) ^{-1}

– Carbon fiber: α = -0.8 ± 10 ^{-6} (°C) ^{-1}

– Concrete: α = (8 to 12) ∙ 10 ^{-6} (° C) ^{-1}

Most materials stretch with an increase in temperature. However, some special materials, such as carbon fiber, shrink with increasing temperature.

**Solved examples of linear expansion**

**Example 1**

A copper wire is hung between two poles, and its length on a cold day at 20 °C is 12 m. Calculate the value of your duration on a hot day at 35 °C.

**Solution**

Starting from the definition of the coefficient of linear expansion and knowing that for copper this coefficient is valid: α = 17 ∙ 10 ^{-6} (° C) ^{-1}

Copper wire has an increase in length, but this is only 3 mm. That is, the cable extends from 12,000 m to 12,003 m.

**Example 2**

In the steel industry, an aluminum bar comes out of the furnace at 800 degrees Celsius, measuring a length of 10.00 m. When it cools down to an ambient temperature of 18 degrees Celsius, determine how long the bar will be.

**Solution**

That is, the bar, once cold, will have a total length of:

9.83 m.

**Example 3**

A steel rivet has a diameter of 0.915 cm. A gap of 0.910 cm is made in an aluminum plate. These are the initial diameters when the ambient temperature is 18°C.

At what minimum temperature must the plate be heated for the rivet to pass through the hole? The purpose is that when the iron returns to room temperature, the rivet is fitted to the plate.

**Solution**

Although the plate is a surface, we are interested in expanding the diameter of the hole, which is a one-dimensional quantity.

let’s call D of the original diameter of the aluminum plate and D that it will have heated once.

Clearing the final temperature T, you have:

The result of the above operations is 257 °C, which is the minimum temperature at which the plate must be heated for the rivet to pass through the hole.

**Example 4**

The rivet and plate from the previous exercise are placed together in the oven. Determine what minimum temperature the oven must be for the steel rivet to pass through the hole in the aluminum plate.

**Solution**

In this case, the rivet and hole will be expanded. But the expansion coefficient of steel is α = 12 ∙ 10 ^{-6} (°C) ^{-1} , while that of aluminum is α = 23 ∙ 10 ^{-6} (°C) ^{-1} .

We then look for a final temperature T such that the two diameters match.

If we call 1 rivet and 2 aluminum sheet, we look for an end temperature T such that D _{1} = D _{2} .

If we clear the final temperature T, we have:

Then we put in the corresponding values.

The bottom line is that the oven must be at least 520.5 °C for the rivet to pass through the hole in the aluminum plate.