## How to Calculate and Solve for Poisson’s Ratio | Rock Mechanics

The image above represents Poisson’s ratio.

To compute for Poisson’s ratio, two essential parameters are needed and these parameters are lateral strain (εland axial strain (εa).

The formula for calculating Poisson’s ratio:

v = εl / εa

Where:

v = Poisson’s Ratio
εl = Lateral Strain
εa = Axial Strain

Let’s solve an example;
Find the Poisson’s ratio when the lateral strain is 80 and the axial strain is 4.

This implies that;

εl = Lateral Strain = 80
εa = Axial Strain = 4

v = εl / εa
v = 80 / 4
v = 20

Therefore, the Poisson’s ratio is 20.

Calculating the Lateral Strain when the Poisson’s Ratio and Axial Strain is Given.

εl = v x εa

Where:

εl = Lateral Strain
v = Poisson’s Ratio
εa = Axial Strain

Let’s solve an example;
Find the lateral strain when the Poisson’s ratio is 40 and the axial strain is 9.

This implies that;

v = Poisson’s Ratio = 40
εa = Axial Strain = 9

εl = v x εa
εl = 40 x 9
εl = 360

Therefore, the lateral strain is 360.

## How to Calculate and Solve for Young’s Modulus | Rock Mechanics

The image above represents young’s modulus.

To compute for young’s modulus, two essential parameters are needed and these parameters are axial stress (σa) and axial strain (εa).

The formula for calculating the young’s modulus:

E = σa / εa

Where:

E = Young’s Modulus
σa = Axial Stress
εa = Axial Strain

Let’s solve an example;
Given that the axial stress is 65 and the axial strain is 45. Calculate for the young’s modulus?

This implies that;

σa = Axial Stress = 65
εa = Axial Strain = 45

E = σa / εa
E = 65 / 45
E = 1.44

Therefore, the young’s modulus is 1.44.

Calculating the Axial Stress when the Young’s Modulus and the Axial Strain is Given.

σa = E x εa

Where;

σa = Axial Stress
E = Young’s Modulus
εa = Axial Strain

Let’s solve an example;
Given that the young’s modulus is 25 and the axial strain is 10. Calculate for the axial stress?

This implies that;

E = Young’s Modulus = 25
εa = Axial Strain = 10

σa = E x εa
σa = 25 x 10
σa = 250

Therefore, the axial stress is 250.