# Poisson’s ratio, Strain energy, Thermal Stress

Today we will briefly discuss couple of very important terms and concepts of the same stream. Those are **Poisson’s ratio**, **Strain energy**, **Thermal Stress** etc. We will also derive expressions for these.

This post is the 3rd post of the** Elasticity** series. Already in our last 2 posts we have discussed the meaning of Elasticity and related concepts like Stress, Strain, **Hooke’s Law** and **modulus of elasticity**.

## Poisson’s Ratio formula

When a wire is stretched, its length increases but diameter is reduced.

In other words, both shape and volume change under Longitudinal Stress.

Here comes the **Poisson’s ratio** to measure 2 resulting strains because of this longitudinal stress.

These 2 strains are known as **Lateral Strain** and **Longitudinal Strain**.

**Lateral Strain = (change in diameter) / (Original diameter) = ΔD/D**

**Longitudinal Strain = (change in length) / (original length) = ΔL/L**

Now, the ** Poisson’s ratio (σ)** is derived from the

**ratio of lateral strain and longitudinal strain.**

**σ =**** Lateral Strain / Longitudinal Strain**** ****= (ΔD/D)/(ΔL/L) ****= (ΔD . L ) / ( D. ΔL) ……………. (1)**

**This ratio is dimensionless and unitless as well.**

## Strain energy

### Strain energy definition

We can define **Strain Energy as the energy stored in a strained wire because of longitudinal stress.**

say F is the force applied on the cross sectional surface of area A. This causes a length change of **Δ**L for a wire of original length L.

So work done on the wire = Energy stored in the wire = *Average force . displacement = (F/2) . **Δ**L*

Now, let’s find out the expression of the energy stored in per unit volume of the wire.

Volume of the wire = A L

**Energy stored per unit volume ****is (1/2) (F . ΔL )/(A . L) **

= (1/2) (F/A) (ΔL/L) **= (1/2).(Stress).(Strain) ……….(2)**

Related posts: elasticity **hooke’s law **

## Thermal Stress

As temperature changes, it affects the length of a solid wire, and eventually causes Thermal stress.

Let’s find out the * expression of the Thermal Stress*.

F is the applied force on a surface with area A.

If initial length is L and α be the coefficient of linear expansion, then the change in length ΔL due to Δt is as follows.

ΔL = α L Δt …………….(3)

ΔL = α L Δt …………….(3)

Young Modulus =Y = (FL)/(ΔL.A) …… (4)

**From (3) and (4)**

So F =( Y ΔL.A )/L **= (Y α L Δt .A )/L = Y α Δt .AF = Y α Δt .A …………… (5)**

So, Thermal Stress = F/A

= **Y α Δt .A/A= Y α Δt …….(6)**