The Invertibility of the p-q Diagram System

We know that we can transform the traditional Mohr-Coulomb $\sigma-\tau$ system to the p-q system by using the equations

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

and

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

Stated formally, this means that, for every set of principal stresses, there is a unique pair of p and q values.

But did you know you can go the other way, if you need to? Let’s start by putting these equations into matrix format, which yields

$\left[\begin{array}{cc} 1/2 & 1/2\\ {\medskip}1/2 & -1/2 \end{array}\right]\left[\begin{array}{c} \sigma_{{1}}\\ {\medskip}\sigma_{{3}} \end{array}\right]=\left[\begin{array}{c} p\\ {\medskip}q \end{array}\right]$

Inverting the matrix and premultiplying the right hand side yields

$\left[\begin{array}{c} \sigma_{{1}}\\ {\medskip}\sigma_{{3}} \end{array}\right]=\left[\begin{array}{cc} 1 & 1\\ {\medskip}1 & -1 \end{array}\right]\left[\begin{array}{c} p\\ {\medskip}q \end{array}\right]$

The inversion is the key step. The fact that the matrix is invertible, square and of the same rank as the vectors means that the transformation is linear, one-to-one and onto. We can also say that, for every set of p and q values, there is a unique set of principal stresses.

Those principal stresses are

$\left[\begin{array}{c} \sigma_{{1}}\\ {\medskip}\sigma_{{3}} \end{array}\right]=\left[\begin{array}{c} p+q\\ {\medskip}p-q \end{array}\right]$