Question

Write a relations between elastic constants

Answer 1

Answer:

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Consider a solid cube, subjected to a Shear Stress on the faces PQ and RS and complimentary Shear Stress on faces QR and PS. The distortion of the cube, is represented by the dotted lines. The diagonal PR distorts to PR’.

(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.