Elastic moduli describe different kinds of stiffness. Young’s modulus describes axial stretching, shear modulus
describes shape distortion, and bulk modulus describes resistance to volume change.
1. Young’s modulus
Young’s modulus \(E\) relates normal stress to normal strain:
\[
\boxed{
E=\frac{\sigma}{\varepsilon}
}.
\]
A large \(E\) means the material is stiff in tension or compression.
2. Shear modulus
Shear modulus \(G\), also called modulus of rigidity, relates shear stress to shear strain:
\[
\boxed{
G=\frac{\tau}{\gamma}
}.
\]
A large \(G\) means the material strongly resists shape change.
3. Bulk modulus
Bulk modulus \(K\) relates pressure change to volumetric strain:
\[
\boxed{
K=-\frac{\Delta p}{\Delta V/V}
}.
\]
A large \(K\) means the material strongly resists compression of volume.
4. Poisson’s ratio
Poisson’s ratio \(\nu\) compares lateral strain with axial strain:
\[
\boxed{
\nu=-\frac{\varepsilon_{\mathrm{lat}}}{\varepsilon_{\mathrm{long}}}
}.
\]
For many ordinary solids, \(\nu\) is between about \(0.2\) and \(0.35\). Rubber-like materials often have
\(\nu\) close to \(0.5\), meaning they are nearly incompressible.
5. Isotropic elastic relations
For an isotropic linear elastic material, only two independent elastic constants are needed. The others can be
calculated from them.
\[
\boxed{
E=2G(1+\nu)
}
\]
\[
\boxed{
E=3K(1-2\nu)
}
\]
\[
\boxed{
E=\frac{9KG}{3K+G}
}
\]
6. Calculating \(G\) and \(K\) from \(E\) and \(\nu\)
If \(E\) and \(\nu\) are known, the shear modulus is
\[
\boxed{
G=\frac{E}{2(1+\nu)}
}.
\]
The bulk modulus is
\[
\boxed{
K=\frac{E}{3(1-2\nu)}
}.
\]
7. Calculating \(\nu\) from \(E\) and \(G\)
Starting from \(E=2G(1+\nu)\):
\[
\boxed{
\nu=\frac{E}{2G}-1
}.
\]
8. Calculating \(\nu\) from \(E\) and \(K\)
Starting from \(E=3K(1-2\nu)\):
\[
\boxed{
\nu=\frac{3K-E}{6K}
}.
\]
9. Calculating \(E\) and \(\nu\) from \(G\) and \(K\)
If \(G\) and \(K\) are known:
\[
\boxed{
E=\frac{9KG}{3K+G}
}.
\]
\[
\boxed{
\nu=\frac{3K-2G}{2(3K+G)}
}.
\]
10. Stability conditions
For a stable isotropic linear elastic material:
\[
G>0,
\qquad
K>0,
\qquad
-1<\nu<0.5.
\]
If \(\nu\) approaches \(0.5\), the denominator \(1-2\nu\) in the bulk modulus formula becomes very small:
\[
K=\frac{E}{3(1-2\nu)}.
\]
Therefore \(K\) becomes very large. This corresponds to nearly incompressible behavior.
11. Typical values
| Material |
Typical \(E\) |
Typical \(\nu\) |
Typical behavior |
| Steel |
\(\approx200\ \mathrm{GPa}\) |
\(\approx0.30\) |
Very stiff structural metal |
| Aluminum |
\(\approx69\ \mathrm{GPa}\) |
\(\approx0.33\) |
Lighter, less stiff than steel |
| Copper |
\(\approx110\ \mathrm{GPa}\) |
\(\approx0.34\) |
Ductile metal |
| Concrete |
\(\approx20{-}40\ \mathrm{GPa}\) |
\(\approx0.15{-}0.25\) |
Strong in compression, brittle in tension |
| Rubber-like material |
Much smaller than metals |
\(\approx0.49\) |
Nearly incompressible |
| Auxetic foam |
Often small |
\(<0\) |
Expands laterally when stretched |
12. Consistency checks
If more than two constants are supplied, the values should agree with the isotropic relations. For example, if
\(E\), \(G\), and \(\nu\) are supplied, then
\[
E_{\mathrm{calc}}=2G(1+\nu)
\]
should be close to the supplied \(E\). If not, the values may describe an anisotropic material, a measurement
error, or inconsistent source data.
13. Worked example: steel
Suppose steel has
\[
E=200\ \mathrm{GPa},
\qquad
\nu=0.30.
\]
The shear modulus is
\[
G=\frac{E}{2(1+\nu)}.
\]
\[
G=\frac{200}{2(1+0.30)}\ \mathrm{GPa}.
\]
\[
G=\frac{200}{2.6}\ \mathrm{GPa}
\approx76.9\ \mathrm{GPa}.
\]
The bulk modulus is
\[
K=\frac{E}{3(1-2\nu)}.
\]
\[
K=\frac{200}{3(1-2(0.30))}\ \mathrm{GPa}.
\]
\[
K=\frac{200}{1.2}\ \mathrm{GPa}
\approx166.7\ \mathrm{GPa}.
\]
So for steel,
\[
\boxed{
G\approx76.9\ \mathrm{GPa}
},
\qquad
\boxed{
K\approx166.7\ \mathrm{GPa}
}.
\]
Key idea: for isotropic materials, any two independent elastic constants determine the others.
