Relation between Modulus of elasticity (E), Modulus of Rigidity (G) ,Bulk modulus (K) and Poisson’s Ratio (1/m or µ):

Relation between E, G and Poisson’s Ratio (1/m or µ):

Consider a cubic element ABCD fixed at the bottom face and subjected to a shearing force at the top face. The block experiences the following effects due to this shearing load:

1. Shearing stress τ is induced at the faces DC and AB

2. Complimentary shearing stress of the same magnitude is set up on the faces AD and BC.

3. The block distorts to a new configuration ABC'D'.

4. The diagonal AC elongates (tension) and diagonal BD shortens (compression). Longitudinal strain ‘e’ in diagonal AC = (AC’ – AC)/AC

= (AC’ – AE)/AC

= EC’/AC---------- (1)

Where CE is perpendicular from C onto AC’. Since CC’ is too small, assume

Angle ACB = Angle ACB = 45⁰ Therefore EC’ = CC’cos45⁰ = CC’/√2 Longitudinal strain ‘e’ = CC’/AC√2

= CC’/√2.BC.√2

= tanΦ/2 = Φ/2 = e_S/2--------------- (2)

Where, Φ = CC’/BC represents the shear strain (e_S)

In terms of shear stress τ and modulus of rigidity G, shear strain (e_S) = τ/G----------- (3)

Putting shear strain (e_S) = 2.

Longitudinal strain Longitudinal the strain of diagonal AC = τ/2---(4)

The strain in diagonal AC is also given by

= strain due to tensile stress in AC - strain due to compressive stress in BD

= τ/E – (–τ/mE) = τ/E (1 + 1/m)------------- (5)

From equation (4) and (5), we get

τ/2G = τ/E(1 + 1/m)

or E = 2G(1 + 1/m) or E = 2G(1+µ)--------------- (6)

Relation between E, G and K:

With reference to the relations (1) and (6) derived above,

E = 2G (1 + µ) = 3K (1- 2 µ)

To eliminate 1/m from these two expressions for E, we have

E = 9KG/ (G + 3K)

Finally; E = 2G (1 + µ) = 3K (1 – 2 µ) or E = 9KG/ (G+3K)

Civil Engineering