Youngs Modulus = Axial Stress/ Axial Strain ==> E = σ/ε ............. {1}
Shear modulus = Shear Stress/ Shear Strain ==> G = τ / γ .......... {2}
Poissons Ratio = Transverse Strain/ Axial Strain ==> ν = t /ε ....... {3}
We can get Shear Strain γ = AC / AX ........................................... {4}
Now, as the strain angle γ is small, ∠AOX = ∠BOX
 OA=OB
Strain along diagonal ε’= BC/OB = BC/OA ................................... {5}
sin 45° = BC/AC ==> BC = AC sin 45° = AC / √2
Again, sin 45° = AX/OA ==> OA = AX /sin 45° = AX √2
Hence, Diagonal Strain ε’ = (AC / √2)/(AX √2) = γ/2 = τ / 2G .........{6}
[Substituting {2}]
Now, Normal Stress due to shear σ = τ
 Diagonal Strain ε’ = σ / 2G ..................................................... {7}
But, Diagonal Strain = Axial Strain + Transverse Strain
 ε’ = ε + t
= ε + ε ν [Substituting {3}]
= ε (1 + ν)
= (σ/E) (1 + ν) [Substituting {1}]
 ε’ = (σ/E) (1 + ν) ................................................................... {8}
σ /2G = (σ/E) (1 + ν)
 G = E / [2*(1 + v)]
Hence Proved
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