Question

Find the relation between modulus of elasticity(E) and modulus of rigidity (G) from first principle.​

Answer 1

ANSWER:-

(a) Relationship between E and G

Modulus of Rigidity, G = ShearStressShearstrain

Shear Strain = ShearstressG

From the diagram, Shear Strain φ = PR′QR

Since Shear Stress = τ ,

RR′QR=τG.......(i)

From R, drop a perpendicular onto distorted diagonal PR'

The strain experienced by the diagonal = TR′PR(Considering that PT ≈ PR)

=RR′cos45(QR/cos45)=RR′2QR

Strain of the Diagonal PR = RR′2QR=τ2G(FromI)........(ii)

Let f be the Direct Stress induced in the diagonal PR due to the Shear Stress τ

Strain of the diagonal = τ2G=f2G..........(iii)

The diagonal PR is subjected to Direct Tensile Stress while the diagonal RS is subjected to Direct Compressive Stress.

The total strain on Diagonal PR would be = fE+1m(fE)

=fE(1+1m)...........(iv)

Comparing Equations (III) and (IV), we have

f2G=fE(1+1m)

Re – arranging the terms, we have,

E=2G(1+1m)...........(A)

(b) Relationship between E and K

Instead of Shear Stress , let the cube be subjected to direct stress f on all faces of the cube.

We know,

ev=fx+fy+fzE[1−2m]

Since f=fx=fy=fz

ev=3fE[1−2m].............(v)

Also, by the definition of Bulk Modulus,

ev=fK...........(vi)

Equating (V) and (VI), we have:

fK=3fE[1−2m]

E=3K[1−2m]..............(B)

(c) Relationship between E, G and K

From the equation (A),

1m=E−2G2G

From the equation (B)

1m=3K−E6K

Equating both, we get,

E−2G2G=3K−E6K

Simplifying the equation, we get,

E=9KG3K+G

This is the relationship between E, G and K.

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