# 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

Strictly speaking, Young's modulus is not always greater than the shear modulus, but it does tend to work out that way. You can see the reason why if you look at the relation between the two quantities (and Poisson's ratio).

$$G = \frac{E}{2(1+\nu)}$$

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.

For an explanation of why the Poisson's ratio must fall within the above range, I invite you to check out some of my previous answers.

High-level: Range of poissons ratio

Detailed explanation: Limits of Poisson's ratio in isotropic solid