# Why is elastic modulus greater than shear modulus?

I was looking at data for elastic modulus $E$ and shear modulus $G$, and found that $G$ is always lower than $E$. So I'm wondering what are the underlying principles that may be the cause of this.

$$G = \dfrac{T\cdot L}{J \cdot \phi}$$

where $T= \text{torque}, \quad J = \text{polar moment of inertia}, \quad \phi = \text{angle of twist}, \quad L = \text{lever arm}$.

$$E = \dfrac{F \cdot L^3}{4bd^3 \cdot v}$$

Where $F = \text{force}, \quad L = \text{length of beam}, \quad v =\text{deflection}, \quad b = \text{width}, \quad d =\text{depth}$

• You should find this useful. Apr 7 '15 at 18:28

$$G = \frac{E}{2(1+\nu)}$$
Combined with the knowledge that $\nu$ can be anywhere in the range $(-1, \frac{1}{2})$, one can see that G can be greater than E for $\nu < -1/2$. That being said, materials with such a negative Poisson's ratio are extremely uncommon, and it is safe to assume that the shear modulus is less than half of Young's modulus.