Shear modulus

Prerequisites

In our consideration of Young's modulus and the Bulk modulus for solids, we examined how to describe how solids responded to normal forces squeezing them. But because solids have some rigidity, they can be deformed in another way by forces that are parallel to a surface, rather than perpendicular to it. We could do this using friction by squeezing another object to one side of our solid and trying to slide it. This is usually how it happens in the everyday world.

But let's think about a simpler way to talk about it. Let's imagine that we have glued something to the top of a block and are trying to drag it along the surface while holding the bottom of block fixed as in the figure. How does the block respond to this kind of twisting force — which we call shear?

This is more complicated since we have multiple directions involved. Let's consider a cube as shown in the figure and take the horizontal plane on the top as the x-y plane. And we'll push our glued object on top in the x direction. We'll let the dimension of the block in the x dimension be $L$ and the area of the top be $A_{xy} = A$ (remember it's an x-y plane). And let's let $F$ be the amount of force we exert.

Let's set up a stress-strain like relation. We're exerting a force $F$ extended over an area $A$, so let's define the shear stress, τ, as the applied force divided by the surface area,

$$\tau = \frac{F}{A}$$

However, it is now the surface area parallel to the force, rather than perpendicular to it. If you think about it, in shear, $A$ represents the area of molecules that has to move past other ones as the object is distorted.

Just as strain is the result of stress, let's define the shear strain, γ (gamma), as the fractional change of shape as a result of the shear stress. We define the shear strain to be the displacement resulting from the applied shear force divided by the initial length of the solid along the direction of the displacement:

$$\gamma = \frac{\Delta x}{L}$$

And just as stress and strain were related by Young's modulus, we expect the shear stress and the shear strain to be related by a constant shear modulus, $G$. (As long as the strain is not too great.)

The shear modulus is then related to the ratio of the shear stress and the shear strain.  In this case:

$$\tau = G \gamma$$

or

$$G = \frac{\tau}{\gamma} = \frac{F/A}{\Delta x/L}$$

For most materials, the shear modulus is two to three times less than Young’s modulus. Therefore, shear motion is larger for the same applied force. This suggests that it is easier to shear most materials than it is to compress or stretch them.

The forces applied to many biological structures are often applied along multiple directions such that there can be both longitudinal and tangential forces. If one end of an object is pushed relative to the other, there may also be shear forces. As a dog runs, much of the force is directed along the length of the leg, where the bone is strongest. However, some of the force is perpendicular to the end of the bone closest to the ground and will result in a shear force along this direction.  Biological materials must be strong enough in this dimension to withstand these forces to enable organisms to move about in their environment.

Julia Gouvea & Joe Redish 8/21/13

Article 407