# Shear Stress, Shear Strain and Shear Modulus

#### Shear Stress

In previous articles we have discussed what a shear force is, but as with normal force and normal stress there is clearly a question of scale we need to examine. If you were to pick up a small amount of soil, it would be easy to shear it between your thumb and forefinger, yet the same soil is usually strong enough for you to run and walk on.

As with forces and normal stresses, the key is that although the similar shear forces have been applied, the shear force has been applied over much different areas. We define a shear force applied over an area as a shear stress.

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

Where, $$\tau$$ is the shear stress that would arise due to a shear force, $$V$$, applied in a direction parallel to the plane of the area, $$A$$. For information on units of stress, click

In the metric system, the typical units for shear force is the Newton, N, and the typical units for area is m², this gives the units of shear stress equal to N/m², this is sometimes called the Pascal, or 1Pa. However, as a Newton is around 100g a stress equal to 1 N/m² is a very low stress. In Engineering we often instead deal with units of shear stress of N/mm² (1N/mm² = 1MPa), this is around 100g of mass applied to an area 1mm x 1mm and is much easier to handle in Engineering applications.

The key word in the above description is "parallel", shear stress is the stress that arises due to a force acting parallel to a plane. Forces that are applied at any angle other than perpendicular to a plane will give rise to normal forces as well as shear force, normal force and normal stress is discussed in a separate article

#### Shear Strain

When an object undergoes a shear stress it will deform by deflecting sideways in the same direction as the shear stress. Some object will deflect more than others, to quantify the amount of deformation, we use the variable “shear strain”:

Shear strain is defined mathmatically as:

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

Where $$\gamma$$ is the value of shear strain, $$\Delta x$$ is the amount of displacement of the area under load and $$l$$ is the height of the element under load. The ratio of $$\Delta x$$ to $$l$$ can be seen as an indication of the angle of deformation, a higher angle of rotation indicating a higher deformation and thus a higher shear strain. For information on units of strain, click

In the metric system, the typical units for both the height of the element, l and the amount of displacement, δx are metres. As a displacement is divided by length the units cancel out, so a value of shear strain is unit less. For typical working loads of Structural Engineering we will use “microstrain”, μγ, 1micro strain = 1x10⁻⁶ strain.

#### Shear Modulus

When a shear force is applied to an object it will deform, as the load increases, the deformation will also increase, so there must be some relationship between shear stress and shear strain. You would also expect that less stiff materials will deflect more than stiffer objects under the same load. The parameter that links shear stress and shear strain is the “Shear Modulus”, which is a fundamental property of a given material and is defined as:

$$G = \frac{\tau}{\gamma} = \frac {V/A}{\Delta x / l}$$

The higher the value of shear modulus stiffer a material is against shear loads, with a higher load required to produce a given shear strain. For information on the units of shear modulus click The above relationship takes a similar form to the Young’s modulus or elastic modulus of a material and the two values are indeed related, this will be discussed in a separate article.

As shear stress has units of Pascals (or N/m²) and shear strain is unitless then the units of shear modulus (Young’s modulus) is also the Pascal (or N/m²)

By plotting values of shear stress against shear strain for different materials it is possible to see immediately the difference between a stiff or flexible material.