Relation Between Elastic Constants - A Complete Guide

Relation between young’s modulus, bulk modulus and Poisson's ratio:-

Let us consider a cube whose side is unit length to derive the relation between Y, K and σ

I.e., OA = OB = OC = 1

So, the volume of the cube

V = 1 (without normal stress)

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When a normal stress is applied on a every face of cube so the increasement in unit length is α and decreasement is β due to unit force.

So, the side of the cube

OA = 1 + F.1.α - F.1.? - F.1.?

= 1 + F.α - 2F?

= 1 + F ( α - 2? )

So, OA = OB = OC = 1 + F ( α - 2? )

Then the volume of the cube

V’ = OA.OB.OC

= [1 + F ( α - 2? )]³

V’ = [ 1 + 3F ( α - 2? )]

So, change in the volume of the cube

ΔV = V’ - V

= 1 + 3F(α - 2? ) - 1

ΔV = 3F(α - 2? )

And the volume strain of the cube = change in volume/initial volume

= ΔV/V = 3F(α - 2? )/1

= 3F(α - 2? )

And the bulk modulus of elasticity of the cube (K) = normal stress/normal strain

= Force/area/ΔV/V = F/1/3F(α - 2? )

K = 1/3(α - 2? ) = 1/3α(1 - 2?/α)

= 1/α/3(1 - 2σ) ( ∵ σ = β/α)

K = Y / 3 ( 1 - 2σ)

∵ Y = long. stress/long. Strain

Y = F/A/Δl/l = F/1/F.1.α/1 = 1/α

……(A)

Relation between E and K is

E = 3K( 1 - 2σ)

This is the relation between elastic constants- relation between young modulus and bulk modulus and poisson ratio/ relation between bulk modulus and young’s modulus.

Relation between relation between young's modulus and modulus of rigidity

For derive the relation between elastic constants - Young’s modulus (Y), modulus of rigidity (η) and the poission’s ratio. Considered a face of cube ABCD side of this face is L. Fixed the side CD and applied the tangential stress (T) on the side AD due to to this A and B point is displaced linearly (l) distance and angularly ? angle and obtained a new position A’ and B’.

So, tangential stress on the side AB

T = force/area = F/L²

And shearing strain

tan? = l/L

But ? is very small tan? ∝ ?

? = l/L

So, the modulus of rigidity of face ABCD

η = T/?

η = F/l²/l/L

η = F/lL …..(i)

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Due to tangential stress T diagonal BD of ABCD face changes to B’D. To determine the change of diagonal BD to B’D a perpendicular line is drawn which cut at N point. In this position, NB’ diagonal represent the increment in the B’D diagonal.

In ΔBNB’

Cos45° = NB’/BB’

NB’ = lcos45°

NB’ = l/√2 ….(ii)

This eqn shows that the increment in the diagonal

Due to tangential stress T increments in the BD diagonal and decrement in AC diagonal. Due to unit force increment and decrement in unit length is represented by

α and β.

Due to tangential stress T increasement in BD diagonal = T.BD.α

= F/L².L√2.α

= F/L.√2.α

Due to tangential stress T decreasement in AC diagonal = T.AC.?

= F/L².L√2.?

= F/L.√2.?

So, the total change in the diagonals of ABCD face = F/L.√2.α + F/L.√2.?

= F√2/L (α + ? ) ….(iii)

From equation (ii) and (iii)

l/√2 = F√2/L (α + ? )

F/lL = 1/2(α + ? )

F/lL = 1/2α(α + ?/α) (∵ σ = ?/α)

F/lL = 1/2α(1 + σ)

From eqn (i)

η = 1/α/2(1 + σ ) ( ∵ Y = 1/α)

η = Y/2(1 + σ )

…….(B)

This is the relation between young modulus and shear of modulus ( or modulus of rigidity ).

And the relation between young’s modulus modulus of rigidity and Poisson’s ratio/ relation between young's modulus and poisson's ratio.

Relation between E and G is

E = 2G(1 + σ)

This is the relation between elastic constants- relation between young’s modulus and rigidity modulus

Relation between Bulk modulus, Modulus of rigidity and Poisson’s ratio:-

From eqn (A) and (B)

2η(1 + σ) = 3K(1 - 2σ)

2η + 2ησ = 3K - 6Kσ

6Kσ + 2ησ = 3K - 2η

σ( 6K + 2η) = 3K - 2η

………(C)

Also read :

• NCERT notes Class 11 Physics Chapter 9 Mechanical Properties of Solids
• NCERT solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids
• NCERT Exemplar Class 11 Physics Solutions Chapter 9 Mechanical Properties of Solids

This is the relation between bulk modulus of elasticity, poission’s ratio and modulus of

elasticity/ relation between elastic constants.

Relation between Young’s modulus bulk modulus and modulus of rigidity:-

From eqn (A)

Y = 3K ( 1 - 2σ )

Y/3K = 1 - 2σ

σ = 1/2 - Y/6K …….(a)

From eqn (B)

Y = 2η(1 + σ)

Y/2η = (1 + σ)

σ = Y/2η -1 ………(b)

From eqn (1) and (2)

1/2 - Y/6K = Y/2η -1

Y/2η + Y/6K = 1/2 + 1

Y[ 1/2η + 1/6K] = 3/2

Y/2 [1/η + 1/3K] = 3/2

……(D)

or, Y = 9Kη/3K + η

This is the relation between elastic constants Young’s modulus, bulk modulus and modulus of rigidity.

Relation between E, G and K Young’s modulus, Bulk modulus and Modulus of rigidity is

E = 9KG/3K + G .

These are the elastic constant formula and elastic constants relation.

Also check-

• NCERT Exemplar Class 11th Physics Solutions
• NCERT Exemplar Class 12th Physics Solutions
• NCERT Exemplar Solutions for All Subjects