### Poisson’s ratio, Strain energy & Thermal Stress – expressions

##### December 17, 2017

# Poisson’s ratio, Strain energy, Thermal Stress

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**.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.

## Poisson’s Ratio

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 is 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 X displacement = (F/2) X **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 = (F/2) X ΔL = (1/2) (F X ΔL )/(A X L) = (1/2) (F/A) (ΔL/L) = (1/2) X (Stress) X (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)

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 .A
F = Y α Δt .A …………… (5)**

So, Thermal Stress = F/A = **Y α Δt .A/A= Y α Δt …………………..(6)**

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