Posted in Geotechnical Engineering, Soil Mechanics

# p-q Diagrams and Mohr-Coulomb Failure

Students and practicioners of soil mechanics alike are used to seeing triaxial test results that look like this (from DM 7.01):

Ideally, the Mohr-Coulomb failure line should be straight, but with real soils it doesn’t have to be that way.  With the advent of finite element analysis we also have the failure function to consider, thus (from Warrington (2016)):

All of these involve constructing (or using) a line which is tangent to a circle at failure.  This can be confusing to understand completely.  The biggest problem from a “newbie” standpoint is that the maximum shear defined by the circle of stress (its radius) and the failure shear stress defined by the intersection of the circle with the Mohr-Coulomb failure envelope are not the same.

Is there a better graphical way to represent the interaction of stresses with the Mohr-Coulomb failure criterion?  The answer is “yes” and it involves the use of p-q diagrams.  These have been around for a long time and are used in such things as critical state soil mechanics and stress paths.  A broad explanation of these is found in our new publication, Geotechnical Site Characterization.  The purpose of this article is to present these as a purely mathematical transformation of the classic Mohr-Coulomb diagram.  This is especially important since their explanation is frequently lacking in textbooks.

## The Basics

Consider the failure function, which is valid throughout the Mohr-Coulomb plot.  It can be stated as follows:

$f=\sigma_{{1}}-\sigma_{{3}}-2\,c\cos(\phi)-\left (\sigma_{{1}}+\sigma_ {{3}}\right )\sin(\phi)$

(The main difference between the two formulations is multiplication by 2; the failure function can either be diametral or radial relative to Mohr’s Circle.  With a purely elasto-plastic model, the results are the same.)

Now let us define the following terms:

$p=1/2\,\sigma_{{1}}+1/2\,\sigma_{{3}}$

$q=1/2\,\sigma_{{1}}-1/2\,\sigma_{{3}}$

We should also define the following:

$\sin(\phi)=\tan(\delta)$

The physical significance of the last one is discussed in this post.  In any case we can start with $\phi$ and solve for $\delta$ or vice versa.  Solving for $\phi$ and substituting this and the equations for p and q into the failure functions yields

$f=2\,q-2\,c\sqrt {1-\left (\tan(\delta)\right )^{2}}-2\,p\tan(\delta)$

For the failure line, $f = 0$.  Let us set the p axis as the abscissa (x-axis) and the q axis as the ordinate (y-axis.)  For the failure line, if we substitute for $f$ and solve for q, we have

$q = p\tan(\delta) + c\sqrt {1-\left (\tan(\delta)\right )^{2}}$

This is a classic “slope-intercept” form like $y = mx + b$, where in this case $q = mp + b$, $m = \tan(\delta)$ and $b = c\sqrt {1-\left (\tan(\delta)\right )^{2}}$.  A sample plot of this kind is shown below.

### Some Observations

1. For the case of a purely cohesive soil, where $\phi = \delta = 0$, the failure envelope is horizontal, just like with a conventional Mohr-Coulomb diagram.
2. For the case of a purely cohesionless soil, where $c = 0$, the y-intercept is in both cases through the origin.
3. The two diagrams are thus very similar visually, it’s just that the p-q diagram eliminates the circles and tangents, reducing each case to a single point.

## Examples of Use

### Drained Triaxial Test in Clay

Consider the example of a drained triaxial test in clay with the following two data points:

1. Confining Pressure = 70 kPa; Failure Pressure = 200 kPa.
2. Confining Pressure = 160 kPa; Failure Pressure = 383.5 kPa.

Determine the friction angle and cohesion using the p-q diagram.

We first start by computing p and q for each case as follows:

$p_1 = 200/2+70/2 = 135\,kPa$

$p_2 = 383.5/2 + 160/2 = 271.75\,kPa$

$q_1 = 200/2-70/2 = 65\,kPa$

$q_2 = 383.5/2 - 160/2 = 111.75\,kPa$

The slope is simply

$m = \frac {q_2 - q_1}{p_2 - p_1} = \frac {111.75 - 65}{271.5 - 135} = 0.342 = \tan(\delta)$

from which

$\delta = 18.9^o$

$\phi = sin^{-1}(tan(\delta)) = sin^{-1}(0.342) = 20.03^o$

$b = q - mp = 65 - 0.342 \times 135 = 18.83$ (using values from the first point, just as easy to use the second one.)

$b = c\sqrt {1-\left (\tan(\delta)\right )^{2}} = c \sqrt {1-0.342^{2}} = 0.94 c$

$b = 18.83 = 0.94 c$

$c = 20.03\,kPa$

Use of this method eliminates the need to solve two equations in two unknowns, and the repetition of the quantity $tan(\delta)$ makes the calculations a little simpler.  When $c = 0$, the calculations are even simpler, as $p_1 = q_1 = 0$.

### Stress Paths

As mentioned earlier, p-q diagrams are commonly used with stress paths.  An example of this from DM 7.01 is shown below.

We note that p and q are defined here exactly as we have them above.  (That isn’t always the case; examples of other formulations of the p-q diagram are here.  We should note, however, that for this diagram $\phi" = \delta$)  With this we can track the stress state of a sample from the start (where the deviator stress is zero, at the start of the triaxial test) around to its various points of stress.

As an example, consider the stress path example from Verruijt, A., and van Bars, S. (2007). Soil Mechanics. VSSD, Delft, the Netherlands.  The basic data from Test 1 are below:

 $\sigma_3$ Deviator Stress Pore Water Pressure 40 0 0 40 10 4 40 20 9 40 30 13 40 40 17 40 50 21 40 60 25

Using the p-q diagram and performing some calculations (which are shown in the spreadsheet Stress Paths Verruijt Example)  the stress paths can be plotted as follows:

It’s worth noting that the q axis is unaffected by the drainage condition because the pore water pressures cancel each other out.  Only the p-axis changes.

## Conclusion

The p-q diagram is a method of simplifying the analysis of triaxial and other stress data which are commonly used in soil mechanics.  It can be used in a variety of applications and solve a range of problems.