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For isotropic materials, only 2 of the 4 elastic constants (Y, G, B, nu) are independent. Three key relations connect them:

1. Y = 2G(1 + nu) — connects Young's modulus with shear modulus
2. Y = 3B(1 - 2nu) — connects Young's modulus with bulk modulus
3. Y = 9$\frac{BG}{3B + G}$ — eliminates Poisson's ratio

Useful rearrangements: nu = $\frac{Y}{2G}$ - 1 = [1 - $\frac{Y}{3B}$]/2. G = Y/[2(1+nu)]. B = Y/[3(1-2nu)].

Special cases: For nu = 0: Y = 2G = 3B. For nu = 0.25: Y = 2.5G, B = Y/1.5. For nu = 0.5: Y = 3G, B = infinity.

Limiting behaviors: If B >> G, then nu approaches 0.5 (nearly incompressible, like rubber). If G >> B, then nu approaches -1 (theoretical limit). For most metals, G < Y < 3G and B is comparable to Y.

These relations are frequently tested in JEE — common question types include finding the third constant given two, computing Poisson's ratio from two moduli, and identifying which materials satisfy given relationships.