Mohr’s circle is a two-dimensional graphical representation of the transformation law for the Cauchy stress tensor.

That’s why I never understood what it was when I was getting my undergraduate degree. For any real geotechnical engineers, you don’t need to read this article. See you next week.

Mohr’s circle is very simple. In 1882, Christian Otto Mohr developed the graphical method for analyzing stress known as Mohr’s circle and used it to propose an early theory of strength based on shear stress.

Okay, let’s visualize an element of soil in the ground. A piece of dirt if you wish. This soil element is below level ground. The soil element is under a vertical load (i.e., a vertical stress), symbolized by σ_{v}. There is also a horizontal stress. In a normally consolidated soil, the vertical stress is greater than the horizontal stress, symbolized by σ_{h}. In this case, the ratio of horizontal to vertical stress is called coefficient of Earth pressure at rest, and is symbolized as K_{0}, and K_{0} = σ_{h} / σ_{v}. At any depth Z, the vertical pressure, σ_{v} = γZ. Where γ is the unit weight of soil.

The value of K_{0} depends upon the relative density of soil, and the process by which the soil deposition has taken place. If the process does not involve artificial tamping, the value of K_{0} ranges from about 0.4 for loose sand to 0.6 for dense sand. Artificial tamping increases the value of K_{0}.

Commonly, K_{0} is defined by K_{0} = (1 – sin f’), so for a soil with a 30^{o} friction angle, K_{0} = 0.5

Where were we, oh yes, Mohr’s circles.

Let’s work out an example, using U.S. Customary Units. Say the soil density is 115 pounds per cubic foot (pcf). The friction angle is 30^{o}. The depth of our soil element is 10 feet.

So, σ_{v }= γZ = 115 pcf * 10 feet = 1150 pounds per square foot (psf).

If K_{0} = σ_{h} / σ_{v} then σ_{h} = K_{0} * σ_{v}

Given K_{0} = (1 – sin f’), then K_{0} = (1 – sin 30^{o}) = 0.5 (as already stated), so,

σ_{h} = K_{0} * σ_{v} = 0.5 * 1150 psf = 575 psf

And now we can plot this.

Notice the two points on the Mohr circle. They are the vertical and horizontal stresses.

The angle of friction (i.e., the failure envelope) is also plotted. The circle is not touching the failure envelope. It isn’t at failure. It is just in a condition of K_{0} consolidation. If our little soil element becomes buried deeper (without tamping), the circle will grow, shift to the right, and still touch the K_{0} consolidation line. Okay, it will be tangent to the K_{0} consolidation line.

It’s really very simple! And now you know Mohr!