2
$\begingroup$

I am very puzzled by the motivation for Mohr's circle in Wikipedia here. Please, explain why we need something called "Mohr's circle". Use as little words as possible and be precise.

Helper questions

  • To illustrate normal stresses and shear stresses?
  • Why is it used? When is it really used?
  • Some example to show why it may have some usage scenarios?
  • Is it just a tool to show two-dimensional stresses on a two-dimensional plane?
$\endgroup$

3 Answers 3

5
+25
$\begingroup$

It provides a convenient graphical means of finding the maximum and minimum shear stress, which are important for determining material failure. You don't absolutely need it, but the graphical interpretation of the circular relationship between normal and shear stress is somewhat convenient. I've read good solid mechanics books that give little if any attention directly to Mohr's circle. Well, at least they don't mention it by name, but determining the maximum shear stress is always important. As an undergraduate I was introduced to Mohr's circle in my first solid mechanics course. Undergraduate courses, at least in the USA, generally avoid introducing tensors. Since finding the maximum shear stress is important for predicting failure, Mohr's circle was presented as a way to find it for 2D problems without understanding tensors.

$\endgroup$
1
  • $\begingroup$ Exactly! What is missing from the Mohr's circle is the curve where the Von-Mises stress is constant. This creates an ellipse centered around the origin, which when the Mohr's circle touches (or crosses) the part might break. I call this the Leela plot, because of Leela Turanga from Futurama has one large elliptical eye which resembles this plot. $\endgroup$ Commented Nov 17, 2013 at 15:23
3
$\begingroup$

The Cauchy stress matrix $\Sigma$ is a $3 \times 3$ real symmetric matrix. It is interesting that we may without problems generalize $\Sigma$ to a $3 \times 3$ Hermitian matrix. It has three mutually orthogonal principal stress directions with principal stresses (eigenvalues) $\lambda_1\geq\lambda_2\geq \lambda_3$.

Consider an arbitrary unit vector

$$\tag{1} |{\bf n}\rangle~=~\begin{bmatrix} n_1\\ n_2\\ n_3\end{bmatrix}~\in~\mathbb{C}^3,$$

where $\langle {\bf n}|{\bf n}\rangle=1$. Defined the directional stress as the "expectation value"

$$\tag{2} \sigma_{\bf n}~:=~E_{\bf n}(\Sigma)~:=~ \langle {\bf n}| \Sigma |{\bf n}\rangle. $$

Let us define the unsigned shear-stress as the "standard deviation"

$$\tag{3} \tau_{\bf n}~:=~\sqrt{{\rm Var}_{\bf n}(\Sigma)} ~:=~\sqrt{\langle {\bf n}| \Sigma^2 |{\bf n}\rangle -\langle {\bf n}| \Sigma |{\bf n}\rangle^2} ~=~\sqrt{\langle {\bf n}| \Sigma^2 |{\bf n}\rangle -\sigma_{\bf n}^2}~\geq~0.$$

Let us now choose a coordinate system such that the $x$-axis is parallel to the unit vector $n$. In other words, the "expectation value" $\sigma_{\bf n}$ is then the first diagonal element in the $\Sigma$ matrix.

According to the Schur-Horn Theorem, a diagonal element $\sigma_{\bf n}$ of $\Sigma$ must lie between the eigenvalues

$$\tag{4} \lambda_3~\leq~ \sigma_{\bf n}~\leq~ \lambda_1. $$

The fact that diagonal elements of a Hermitian matrix always are constrained by its eigenvalues has profound consequences in many areas of physics, see e.g. this Phys.SE post.

Equation (4) shows that the point $(\sigma_{\bf n},\tau_{\bf n})$ must belong to a half-strip in the $(\sigma_{\bf n},\tau_{\bf n})$ plane. A half-strip is not quite Mohr's three semicircles with centers $(\frac{\lambda_i+\lambda_j}{2},0)$ and radii $\frac{|\lambda_i-\lambda_j|}{2}$ for $i\neq j$; but we can do better, cf. below.

$\uparrow$ Figure 1: Mohr's three circles in the $(\sigma_{\bf n},\tau_{\bf n})$ plane. Possible $(\sigma_{\bf n},\tau_{\bf n})$ points are within the green region. The principal stresses $\lambda_i$ are called $\sigma_i$.

Similarly according to the Schur-Horn Theorem, a diagonal element $\sigma_{\bf n}^2+\tau_{\bf n}^2$ of the Hermitian matrix $\Sigma^2$ must lie between the eigenvalues

$$\tag{5} \min_{i\in\{1,2,3\}}\lambda_i^2~\leq~ \sigma_{\bf n}^2+\tau_n^2~\leq ~ \max_{i\in\{1,2,3\}}\lambda_i^2. $$

More generally, a diagonal element $(\sigma_{\bf n}-\lambda)^2+\tau_{\bf n}^2$ of the Hermitian matrix

$$\tag{6} (\Sigma-\lambda{\bf 1}_{3\times 3})^2 ~=~\Sigma^2-2\lambda\Sigma+\lambda^2{\bf 1}_{3\times 3} $$

must lie between the eigenvalues

$$\tag{7} \min_{i\in\{1,2,3\}}(\lambda_i-\lambda)^2~\leq~ (\sigma_{\bf n}-\lambda)^2+\tau_{\bf n}^2~\leq~\max_{i\in\{1,2,3\}}(\lambda_i-\lambda)^2 , $$

for a real parameter $\lambda\in \mathbb{R}$.

Equation (7) leads to the three Mohr's semicircles if we choose $\lambda = \frac{\lambda_i+\lambda_j}{2}$ with $i\neq j$.

The three prongs that touch the horizontal $\sigma_{\bf n}$-axis can in a quantum mechanical analogy be understood as the uncertainty $\tau_{\bf n}$ can only become small if the unit vector $\bf n$ is close to one of the three principal stress directions.

$\endgroup$
2
  • $\begingroup$ Great! +1 some references to read further? $\endgroup$
    – hhh
    Commented Nov 16, 2013 at 22:40
  • $\begingroup$ What a great answer and what an elegant little theorem (Schur-Horn)- I'd never heard of that one, thanks! $\endgroup$ Commented Feb 10, 2014 at 2:43
1
$\begingroup$

Mohr's circle is a graphical technique for getting principal stresses. It was useful in the age when most design engineers sat at a drawing board. In an era when most engineers have cellphones and tablets, Mohr's circle is an anachronism. I show my students Mohr's circle because some companies think it's 'essential knowledge', then I give them an Excel spreadsheet that does the tensor transformation properly.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.