# Mohr's circle

Introduction

How would you break a metal rod if you could only use your hands?

1. Pull it apart or compress it (not usually the easiest way)

2. Twist it

3. Bend it

Each of these methods induce stress into the rod in a different way. Mohr's circle helps analyse this. For now, only the first two ways will be analysed.

1.Tension/compression:

Below is a diagram of a rod with a circular cross-section that is subjected to a tensile force at either end.(When you pull the rod apart you are exerting a tensile force). Let us look at a square element at the surface of the rod:

How do you imagine the square element would react to the tensile force?

The force will 'stretch' the rod and the square element, as shown below:

Therefore the force applied at either end has produced a tensile stress, ${\displaystyle \sigma _{x}}$, on the element, as shown below:

where: ${\displaystyle \sigma _{x}={\frac {F}{A}}}$

Similarly, a Compressive force on the rod would induce a compressive stress on the square element, as shown below:

2. Torsion

In this case, the rod is subjected to a torque (twisting force) at either end:

How do you imagine the element would stretch this time?

Answer: The left side of the square element would stretch upwards, the right side would stretch downwards, as shown below. The stress induced by this shearing motion is called shear stress. It's symbol is, ${\displaystyle \tau }$. It is shown on the square element below:

Although it may be more difficult to visualize, there is also shear stress on the horizontal edges of the square element. The square element is drawn as follows:

If the torque was applied in the opposite direction, the shear stress on the element would look like this:

calculate shear stress using: ${\displaystyle \tau ={\frac {Tr}{J}}}$

where:

T = Torque

r = Radius of the rod

J = polar moment of area. For a rod with a circular cross-section : ${\displaystyle J={\frac {\pi r^{4}}{2}}}$

Tensile Force and Torque

If there is both a tensile force and torque applied at either end, you superimpose the two solutions and the square element would look like this:

The idea behind Mohr's circle - It's not essential that you read this

Imagine you rotated the square element by ${\displaystyle \theta }$ degree as shown below.

Take ${\displaystyle \theta }$ to be 45 degrees. How do you imagine the element would react to the torque in this case? This one is more difficult to imagine. It would stretch as shown below:

Thus, you would draw the stresses on the element like so:

Note that there is no shear stress acting on the element at this orientation. When there is no shear stress acting on the element, the element is called the "principal element", and the 2 stresses on the element ${\displaystyle \sigma _{1}}$ and ${\displaystyle \sigma _{2}}$ are known as the principal stresses.

The objective of the Mohr's circle method is to find the orientation of the principal element (i.e.${\displaystyle \theta }$, which for this simple case was 45 degrees), and find the values of ${\displaystyle \sigma _{1}}$ and ${\displaystyle \sigma _{2}}$.

Finally - The method of Mohr's Cicle

Consider a square element that experiences the following stresses:

(${\displaystyle \sigma _{y}}$ = 0, in the examples previously shown )

At A:

${\displaystyle \sigma }$ = ${\displaystyle \sigma _{y}}$

${\displaystyle \tau }$ = ${\displaystyle -\tau _{1}}$.

We take the shear stress as negative, because the shear stress at surface A tries to rotate the square element in an anticlockwise direction (about the centre of the element). This is the general convention used.

At B:

${\displaystyle \sigma }$ = ${\displaystyle \sigma _{x}}$

${\displaystyle \tau }$ = ${\displaystyle +\tau _{1}}$ (it causes a clockwise rotation - hence, it is positive)

Plot points A and B as shown below and draw a straight line across them:

Now draw a circle with a centre C and radius R, such that circle passes through points A and B as shown below:

As you can see from the diagram, C is the midpoint of A and B, hence its co-ordinate is calculated as: C = ${\displaystyle {\frac {\sigma _{x}+\sigma _{y}}{2}}}$

Observe the right angled triangle BC${\displaystyle \sigma _{x}}$. Using Pythagoras' theorem, R can be calculated: ${\displaystyle R^{2}}$ = ${\displaystyle (\sigma _{x}-C)^{2}}$ + ${\displaystyle \tau _{1}^{2}}$

Also, draw an angle of 2${\displaystyle \theta }$, going from the line AB to the ${\displaystyle \sigma _{x}}$ axis. Notice that in this case, this is a clockwise angle.

${\displaystyle \theta }$ can be calculated through ${\displaystyle \tan(2\theta )={\frac {\tau _{1}}{\sigma _{x}-C}}}$

Finally, the principal stresses ${\displaystyle \sigma _{1}}$ and ${\displaystyle \sigma _{2}}$ occur where the circle meets the ${\displaystyle \sigma }$ -axis. (Notice that these points have zero shear stress.)

From the diagram:

${\displaystyle \sigma _{1}}$ = C + R

${\displaystyle \sigma _{2}}$ = C - R

Finally, to transform into the coordinate system of the principal axes, rotate the original square element by ${\displaystyle \theta }$ degrees clockwise (because you draw 2${\displaystyle \theta }$ as a clockwise angle in the above diagram):

## Example

A rod is subjected to a tensile force and a torque, as shown below. Use Mohr's circle to work out the principal stresses and draw the rotated square element. (I recommend you try this first before seeing the answer)

F = 2000N

T = 10 Nm

r = 0.005m

A (cross sectional area) = ${\displaystyle \pi r^{2}=8\cdot 10^{-5}}$ m2

${\displaystyle \sigma _{x}={\frac {F}{A}}=}$ 26,000,000 Pa = 26 MPa

${\displaystyle J={\frac {\pi r^{4}}{2}}=9.8\cdot 10^{-10}}$

${\displaystyle \tau ={\frac {T\cdot r}{J}}=}$ 51 MPa

At A:

${\displaystyle \sigma =0\,}$

${\displaystyle \tau =51\,}$

At B:

${\displaystyle \sigma =26\,}$

${\displaystyle \tau =-51\,}$

${\displaystyle C={\frac {0+26}{2}}=13}$

${\displaystyle R^{2}=(26-13)^{2}+51^{2}\;\;\Rightarrow \;R=53}$

${\displaystyle \sigma _{1}=13+53=66\;\mathrm {MPa} \,}$

${\displaystyle \sigma _{2}=13-53=-40\;\mathrm {MPa} \,}$

${\displaystyle \tan(2\theta )={\frac {51}{26-13}}\;\;\Rightarrow \;\theta =38^{\circ }\,}$

Notice that 66MPa is drawn as a tensile stress (as ${\displaystyle \sigma _{1}\,}$ is positive), and 40MPa as a compressive stress (as ${\displaystyle \sigma _{2}\,}$ is negative)

That's it! If you have found this article useful, please comment in the discussion section (at the top of the page), as this will help me decide whether to write more articles like this. Also please comment if there are other topics you want covered, or you would like something in this article to be written in more detail.