**What do you mean by Poisson’s Ratio?**

The exact ratio of Lateral Strain to the Longitudinal Strain is known as Poisson’s ratio. It is denoted by μ. So, you can write μ = The lateral strain / The longitudinal strain.

**Uni – axial stress system **

To understand this in a proper way, you just need to go through the following diagram

You can see here a bar is lengthened by an axial tension where a reduction will be taken place in transverse dimensions. The expert Poinssion concluded and showed that in case of the unit deformation ratio or you can say the strain in these sides or directions is constant for the exact stresses as there is proportional limit.

And now you can say that Stress/ Strain = E

Now, if the direction of Z is getting increased with its dimension, then the dimension in the direction x and y will be decreased.

So, clearly the law can easily be written as –

Lateral Strain/longitudinal strain

= – εx / εz = -εy / εz

If you want to know that what is E here, then it is the property of a material. And μ is required to find out properly as it is also a property of the material.

µ = – ε_{x} /ε_{z }

ε_{x} = – µ ε_{z = – }µ x σ_{z} / E

ε_{y}= – µ . σ_{z} / E

ε_{z }= σ_{z} / E.

Here is the strain in x, y and z direction while you have the perfect loading on z direction.

**Biaxial Stress System**

** **

In bi-axial stress is described in two different directions suppose σ_{x }and σ_{z. }In this z direction dimension increases and σ_{x }influences the length in X direction.

ε_{z }= σ_{z} / E. – µ . σ_{x} / E

ε_{x }= σ_{x} / E. – µ . σ_{z} / E

ε_{y }= σ_{x}/ E. – µ . σ_{z} / E

This is explained properly because of the loading in x and z direction. The strain can be seen in x, y and z direction.

**Tri-axial Stress System**

The strain can be determined as the summation Poisson’s effect and direct effect.

+ Direct effect

-Poisson’s effect

ε_{x }= σ_{x} / E. – µ. σ_{y} / E . – µ σ_{z} / E

ε_{y }= σ_{y} / E. – µ . σ_{x} / E. – µ σ_{z} / E

ε_{z }= σ_{z}/ E. – µ . σ_{x} / E – µ σ_{z} / E

ε_{x =} 1/ E [ σ_{x} – µ (σ_{y} + σ_{z})]

_{ }ε_{y =}1/ E [ σ_{y} – µ (σ_{x} + σ_{z})]

ε_{z }=1/ E [ σ_{z} – µ (σ_{z} + σ_{y})]

These are the equations of the Triaxial Stress system. All the stresses present in these equations are tensiles. In case of some particular conditions when the stresses are compressive, then it can be changed adequately in case of signs.

**ax and ay.for biaxial stresses**

**There are different conditions in which the stresses are bi-axials. Thus, to acquire the perfect solution, you can easily get **

σ_{x} = ( ε_{x} + µε_{y}) E / 1-µ^{2}

for biaxial stresses

σ_{y} = (ε_{y} + µε_{x}) E / 1-µ^{2}

**Links of Previous Main Topic:-**

- Rectilinear motion in kinetics of particles
- Work and energy
- Linear momentum
- Force mass acceleration
- Simple stress introduction
- Normal strain
- Stress strain diagram ductile material mild steel

**Links of Next Mechanical Engineering Topics:-**