Relation between Modulus of Elasticity and Modulus of Rigidity

Modulus of Elasticity (E) 

It is the ratio between Normal stress to Normal strain within the elastic limit.

Elastic Modulus E = Normal stress/Normal strain

E = s/e

Modulus of Rigidity (G) 

It is the ratio between Shear stress to Shear strain within the elastic limit.

Rigidity Modulus G = Shear stress/ Shear strain

G =  Ƭ/ø 

Relation between Modulus of Elasticity and Modulus of Rigidity:

Consider a solid cube PQRS is subjected to a shearing force F. 

Let Ƭ    be the shear stress produced in the faces PQ and RS due to this shear force. The complementary shear stress consequently produced in the vertical faces PS and RQ is also equal to same and shown in figure asƬ 

Due to the pure shearing force, the cube is deformed PQRS to PQR'S'. The point S moved to S' and point R moved to R' as shown in fig. 

The shear strain = The angle of distortion ø

                    ø = RR'/ RQ ---(1)

Shear strain = Shear stress /Rigidity modulus

                   ø = RR'/ RQ =  Ƭ/G

From R, drop a perpendicular RT onto distorted diagonal PR'

The normal strain experienced by the diagonal  PR is   Ԑ = TR′/PT

 (Considering that PT ≈ PR)

          Ԑ = TR′/PR

             = (RR′/cos 45)/(RQ/cos 45) = RR′/(2RQ) 

Substitute (RR'/RQ) form equation (1) in the above expression,

The normal Strain of the Diagonal PR =

                       Ԑ = [RR′/RQ] /2 = Ƭ/2G  ----(2)

Due to the action of pure shear stress in the block PQRS,the diagonal PR is subjected to Direct Tensile Stress σ1 and the diagonal RS is subjected to Direct Compressive Stress σ2. Both these stresses are equal to shear stress Ƭ

If the Poisson's ratio is n, 

the net Strain in the diagonal PQ =

Ԑ  =  (σ1 /E) - n.(σ2 /E)  

But the normal stress σ1 = + Ƭ and σ2 = - Ƭ

Ԑ  =  ( Ƭ /E) - n.(-  Ƭ /E)     

Ԑ =  (Ƭ /E) (1+ n)    ---(3)

The equation (2) and equation (3) are same, then we equate the both 

        Ԑ =  Ƭ/2G = (Ƭ /E) (1+ n) 

E = 2G(1+ n) 

Relations between Elastic constants E,G,K and n

The total number of elastic constants are four. i.e E, G, K and v. It may be seen that not all of these constants are independent of the others. Infact given any two of them, the other two can be determined. Further, it may be noted that the value of the elastic constants E, G and K are always be positive values.

1.The relation between modulus of elasticity (E) and modulus of rigidity(G)  is given by 

                                       E = 2G (1 + n ) or

G = E/[2(1 +n)] 

2.The relation between modulus of elasticity (E) and Bulk modulus (K) is given by

E = 3K (1 - 2 n )

Using the above two relations we may derive antheor relation without poisson's ratio.

3.The relation between modulus of elasticity (E), modulus of rigidity(G) and Bulk modulus (K) is given by

E = 9KG/ (3K+G)  or

1/E =1/3G  + 1/9K

4.The relation between modulus of rigidity(G) and Bulk modulus (K) is given by

K= 2G (1 + n )/ 3(1 - 2 n )

Assignment:

1.For a certain material E= 2.8K. Calculate the Poisson’s ratio. Also calculate the ratio of modulus of elasticity to modulus of rigidity.

(Ans: 0.033, E/G=2.066)

2.If Modulus of rigidity is 75.018 MPa and Poisson's ratio ⅓. Find E and K.

(Ans: E=200 GPa and K=199.9E6 N/mm2 )

3. A metal rod 20 mm diameter and 2 m long is subjected to a tensile force of 60 kN, it showed and elongation of 2 mm and reduction of diameter by 0.006 mm. Calculate the Poisson's ratio and three moduli of elasticity.

(Ans: Poisson’s ratio = 0.3, E=190.99e3 N/mm2 ,G=73.45e3 N/mm2 ,K=159.15e3 N/mm2 )