Now we are going further to start our discussion to understand the derivation of relationship between young’s modulus of elasticity (E), modulus of rigidity (G) and Poisson ratio (ν) with the help of this post.

### Derivation of relationship between young’s modulus of elasticity (E), modulus of rigidity (G) and Poisson ratio (ν)

Let us consider one cube with side length L and let us assume that we have one face of cube which is displayed here by ABCD as mentioned in following figure.

Let us think that cube is subjected with the action of shear stresses and complementary shear stresses as shown in following figure. We can easily say that the diagonal BD will feel a tensile stress and therefore diagonal BD will be elongated. Whereas if we consider the diagonal AC, it will feel compressive stress and therefore the diagonal AC will be shorted.
We will consider the joint effect on the diagonal BD. We can easily conclude that there will be tensile strain in diagonal BD due to tensile stress along the diagonal BD. There will be one more tensile strain in the diagonal BD due to lateral strain produced due to compressive stress along the diagonal AC.

Total Strain in the diagonal BD = (Tensile strain in the diagonal BD due to the tensile stress along the diagonal BD) + (Tensile strain in the diagonal BD due to lateral strain produced due to compressive stress along the diagonal AC)

#### Let us consider

Length of side of the cube = L
Shear stress applied = τ
Young’s modulus of elasticity = E
Modulus of rigidity = G
Poisson ratio = ν
Angle of shear = θ
Longitudinal strain per unit stress = α
Lateral strain per unit stress = β

As we have already discussed the Poisson ratio as the lateral strain to longitudinal strain and therefore we can say that
Poisson ratio, (ν) = β / α

Let us recall the young’s modulus of elasticity, E = Longitudinal stress/Longitudinal strain
E = 1/ [Longitudinal strain/ Longitudinal stress]
E = 1/ α

We can say from above figure that top surface of the cube is moved to C1D1 from CD due to the action of shear stresses and complementary shear stresses. Let us assume that CC1 is l and hence we can also say that DD1=CC1=l

Now consider the triangle BCC1
Tan θ = l/L
Where θ is angle of shear and it will be very small and hence we can say that
Tan θ = θ, (Angle of shear will be measured in radian here)
θ = l/L

Let us consider that we have drawn one perpendicular line DE over the diagonal BD1 from point D as displayed in above figure. Earlier we have seen that BDC = 450 and once shear stresses will be applied and diagonal BD turned to BD1. Angle of shear θ will be very small and therefore we can say that BD1C1 will be very close to BDC and therefore BD1C1 = 450.

Let us consider the small triangle DD1E and from here we will determine D1E. As we can see here that BD and BE will be approximately equal to each other and therefore we can easily say that D1E will be termed as the elongation in the diagonal BD.

D1E = Sin 450 x D1D = 1/ 2
l= D1E x 2

Let us determine the D1E
Total elongation in the diagonal BD i.e. D1E = (Elongation due to tensile stress along the diagonal BD) + (Elongation due to lateral strain produced due to compressive stress along the diagonal AC)
D1E = (α τ x BD) + (β τ x BD)
D1E = τ x BD (α + β)
l= D1E x 2
l= τ x BD (α + β) x 2
l= τ x L2 (α + β) x 2, (Because BD = L2)
l= 2 x τ x L (α + β)
l/L= 2 x τ (α + β)
θ = 2τ (α + β)
1/ [2(α + β)] = τ/θ
τ/θ =1/[2(α + β)]
G= 1/ [2(α + β)],
Because modulus of rigidity = shear stress (τ)/ Shear strain (θ)
G= 1/ [2α (1 + β/ α)]
G= 1/ [2α (1 + ν)]

### G= E/ [2 (1 + ν)]

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Find this post for complete information about centrifugal pump i.e. Centrifugal pump working principle

### Reference:

Strength of material, By R. K. Bansal