Mechanics of Materials

118/3337–11FE Reference Handbook 10.4 · Mechanics of Materials · Elastic constants

Handbook formula

Shear Modulus from E and ν

G = \frac{E}{2 ( 1 + ν )}

Isotropic linear-elastic identity among Young’s modulus, shear modulus, and Poisson’s ratio. G is always less than E for ν > 0. Typical steel: E = 200 GPa, ν ≈ 0.3, G ≈ 77 GPa. Incompressible limit ν = 0.5 gives G = E/3.

$G$ (Pa): Shear modulus
$E$ (Pa): Modulus of elasticity
$ν$ (—): Poisson’s ratio

Step-by-step solved example

E = 200 GPa, ν = 0.30. Compute G.

G from torsion of a circular shaft: τ = Gγ.

1. Denominator
2(1+ν) = 2(1.30) = 2.60.
$2 (1 + ν) = 2.60$
2. G
G = 200/2.60 = 76.9 GPa.
$G = 76.9 GPa$

Answer: 76.9 GPa

10 practice questions

0/10 correct

1.E = 200 GPa, ν = 0.30. G ≈

2.For ν > 0, G compared with E is

3.Typical Poisson’s ratio for steel is about

4.ν = 0.5 (incompressible) gives G =

5.E = 70 GPa, ν = 0.33. G ≈

6.The relation G = E/[2(1+ν)] assumes

7.If ν = 0, then G =

8.Units of G are

9.E = 200 GPa, G = 80 GPa. ν =

10.Rubber has ν near 0.5 and