Mohr’s Circle is a graphical method for calculating stresses on different planes of a stressed object. If you are wondering about Mohr’s Circle and you don’t understand the in-depth concept, then this blog is for you. After reading this blog, you will master Mohr’s Circle because this blog will include all the aspects of Mohr’s Circle in stress transformation. 

Stress Transformation

To understand Mohr’s Circle, it is essential to have a basic understanding of stress transformation. So, let’s dive into the depths of stress transformation. Imagine you have a rectangular beam and you are applying a tensile force on it, as shown in Figure 1.

Figure 1 shows a beam on which the “P” tensile force is acting.

Due to this tensile force, tensile stress will be generated, which we can express as 6x, where x represents stress on the x-axis. We can calculate this stress using the simple stress formula and represent it in a stress block, as shown in Figure 2.

Figure 2 shows the stress block of tensile stress due to the force “P”

The concept of a stress block is that we take a small element from the original object to represent the stress acting on this object; these stresses can be normal stresses or shear stresses. Normal stress means those stresses which act perpendicular to a specific plane, such as 6x those that act perpendicular to the vertical plane. And shear stresses are those stresses which act parallel to the plane, but in our example of a beam, there is no shear stress.

At a certain point, due to this stress, this object will fail, as shown in Figure 3.

Figure 3 shows the failure due to tensile stress

In Figure 2, the stress block illustrates simple tensile stress on the x-axis. Now, let me ask you a question. You found the stress on the x-axis by the simple force/area formula, but if you need to find the stress on another plane, such as on the failure plane, how would you calculate it?I have a simple answer to this. We know the stress on the x-axis, which is σx, and you can see it in the stress block in Figure 2. Using this stress block, we can find the stresses on the other planes as well, and this is basically the stress transformation. Let’s understand it with an example;

           In Figure 2, we show a stress block with all the stresses acting on it. We know their magnitude. In our case, we only have 6x. Now, using this 6x, we can find the stresses on the other plane, such as on the failure plane, and this is basically stress transformation. The stress block on the failure plane is shown in Figure 4.

Figure 4 shows the stress block at the failure plane.

Here, 6x’, 6y’, and Tx’y are the transformed stresses because these stresses are generated by a single stress, which is tensile stress on the object on the x-axis. We called them transformed stresses because they have been transformed or changed on the failure plane. As we know, on the x-axis, we have only one stress, but that stress transforms into 6x’, 6y’, and Tx’y on the failure plane. Now you might be thinking how to find the values of these transformed stresses. We can find them using stress transformation equations, which is a mathematical approach. These equations are shown here,

How to apply stress transformation equations

For applying stress transformation equations, the sign convention is essential. If normal stress is tensile, then it is positive; if it is compressive, then it is negative. For shear stress, if it rotates the block anticlockwise, it is positive; if it rotates the block clockwise, then it is negative. In our example, in the failure plane stress block in Figure 4, 6x’ and 6y’ are tensile, which means they are positive. Shear stress, which is acting on the 6x’ plane, is rotating the block in an anticlockwise direction, which is why this is positive, while shear stress acting on the 6y’ plane is negative as it rotates the block in a clockwise direction. For the angle θ, if the block rotates in an anticlockwise direction, then it is positive; for a clockwise direction, it is negative. In Figure 4, the block is rotated anticlockwise at an angle θ, which is why it is positive. This was the generic introduction of the stress transformation.

Mohr’s Circle

Mohr’s circle is a graphical representation of the stresses on the stress block, allowing us to analyse stress conditions on any object. After understanding stress transformation, it becomes easier for us to comprehend Mohr’s circle. Previously, I informed you that you can find transformed stresses using the stress transformation equations, but now I am telling you that you can find these transformed stresses using Mohr’s circle as well. Stress transformation is a mathematical approach, while Mohr’s circle is a graphical approach.

Before understanding how to find these stresses using Mohr’s circle, first let’s know how to draw Mohr’s circle.

How to draw Mohr’s Circle

For your simplicity, I have divided the procedure into different steps using another stress block example. Consider you have a stress block shown in Figure 5, which has 6x, 6y, and Txy.

Figure 5 shows a stress block with 6x, 6y, and Txy stresses.

And you want to make Mohr’s circle of this stress block, then this is the procedure;

1. First, identify all the stresses on the stress block, along with their signs. In our example, four stresses are acting.
   1. 6x and 6y are both normal stresses and are positive. (Positive due to tensile)
   1. Txy and -Txy. Positive Txy is acting on the vertical plane or on that plane on which 6x is acting. While -Txy is acting on the horizontal plane or the plane on which 6y is acting. (Positive means anticlockwise rotation, and negative means clockwise rotation)
2. At the second step, draw the axes for the normal stress and shear stress. This is shown in Figure 6.

Figure 6 shows the first step of creating Mohr’s Circle, creating the normal stress and shear stress axes.

• Now you have four stresses, take 6x from them and mark on the Normal Stress Axis, like in Figure 7.

Figure 7 shows the 3rd step of Mohr’s circle.

• Now, mark the shear stress acting on the plane on which 6x is acting. In our case, that is positive Txy. This is shown in Figure 8.

Figure 8 shows the 4th step of Mohr’s Circle.

• Now trace the point using 6x and Txy, as shown in Figure 9. You will get (6x, +Txy) point.

Figure 9 shows the 5th step of Mohr’s Circle.

• Now, similarly, take 6y and shear stress on the same plane, which is -Txy, and trace the point; this is shown in Figure 10.

Figure 10 shows the 6th step of drawing Mohr’s Circle

• Now join the two points (6x, +Txy) and (6y, -Txy) with a straight line as shown in Figure 11.

Figure 11 shows the 7th step of Mohr’s Circle.

• Now you have got the centre point which is “C”, now put the compass at “C” and (6x, +Txy) point and draw a circle. This will be the Mohr’s Circle of the given stress block, whose radius is from “C” to “(6x, +Txy)” or “C” to “(6y, -Txy)”. This is shown in Figure 12.

Figure 12 shows the complete Mohr’s Circle of the given stress block.

Essential Points of Mohr’s Circle

Specific points in Mohr’s circle are essential, and everyone should know these points.

1. Mohr’s circle will always be constructed in such a manner that half of the circle will be the upper side of the x-axis or the normal stress axis, and the other half will be below the normal stress axis. This is always true in any stress block and stress conditions.
2. You can see in Figure 12 that Mohr’s circle is cutting the x-axis, or normal stress axis, at two locations. These points are mentioned in Figure 13, and these are the principal stresses. Also, remember that in Mohr’s circle, the normal stress axis is also called the principal plane, because on this plane, the shear stress is zero. And stresses on this point, which are 6₁ and 6₂, as shown in Figure 13, are principal stresses because with these stresses, shear stresses are zero. In simple words, if we make the stress block on the principal plane, then on that stress block, the shear stress will be zero. For further clarity, see Figure 14. There is an object whose stress block is given at the bottom of that object, on which normal and shear stresses are acting. Let us assume that at an angle θ in the anticlockwise direction, there is the principal plane, then the stress block on that principal plane will have no shear stress. All these are shown in Figure 14.

Figure 13 shows principal stresses 61 and 62 on Mohr’s circle.

Figure 14 shows the stress block and principal stress block of an object.

• On the Mohr’s circle in Figure 12, we have two peak points, one at the top and the other at the bottom, as shown in Figure 15, which are maximum shear stress points, and this plane is called in-plane, and stresses on this plane are called in-plane stresses. For clarity, see Figure 15.

Figure 15 shows the In-plane in Mohr’s circle.

• One thing is essential: on any stress block, only two normal stresses will act, such as 6x and 6y. If we talk about principal stresses, then 61 and 62, or if we talk about any other plane, then 6x’ and 6y’ will act. Remember one thing: the sum of normal stresses on one plane will be equal to the sum of normal stresses on the other plane. For clarity, see Figure 14. We have two stress blocks: one is for the normal and shear stress plane, and the other is for the principal plane. Now, if you add 6x and 6y in the normal case, that will be equal to 6₁ plus 6₂.

To help you understand this point, I have created a free Mohr’s circle calculator. I have drawn a Mohr’s circle using this calculator, which can be seen in Figure 16

Figure 16 shows the Mohr’s circle drawn with a free Mohr’s circle calculator

Here you can see four types of coloured dots. A black line with X and Y points shows the input values of 6x = 100 and 6y = 200, and the shear value Txy = 100. The calculator has generated a circle which is in front of you. You can see that the orange dots show principal stresses 61 = 261.8 and 62 = 38.2, and stresses on other planes at 31 degrees anticlockwise direction, whose values are 6x’ = 214.8 and 6y’ = 85.2. Now, you can add them each time, and you will get the same answer.

Hope you got my point.

• Last thing in Mohr’s circle, if we need to find the stresses on a new plane, then we have a simple procedure. First, check in which direction the stress block is rotating, clockwise or anticlockwise, and then check the angle θ. Now, in Mohr’s circle, you need to move 2θ from the original reference line, which is black in Figure 16, and this is the (6x, Txy) and (6y, -Txy) line in Figure 15. In Figure 16, the stress block has rotated in an anticlockwise direction with an angle of 31 degrees, but in Mohr’s circle, you can see this angle is 62 degrees because in Mohr’s circle, we need to move 2θ from the original stress line. For further understanding, an actual numerical problem has been solved in the next heading. Please go through it for further concept clearing.

Numerical solution using Mohr’s Circle

Problem Statement:

A structural element is subjected to the following plane stress condition at a critical point as shown in Figure 17.

Figure 17 shows a stress block of numerical.

Required: Using Mohr’s Circle analysis, determine the following:

Part A: Principal Stresses and Orientations

1. Maximum principal stress (σ₁) and its magnitude
2. Minimum principal stress (σ₂) and its magnitude
3. Principal plane orientation.

Part B: Maximum Shear Stress Analysis

1. Maximum in-plane shear stress (τₘₐₓ)
2. Angle of maximum shear stress planes (θₛ)
3. Normal stresses acting on the planes of maximum shear stress

Part C: Stress Transformation

1. Normal and shear stresses on a plane oriented at θ = -20° from the x-axis

Solution:

For the solution of any Mohr’s circle problem, the solver must draw Mohr’s circle first. So, here we will follow the same steps that we learned in the drawing procedure of Mohr’s circle.

1. First of all, identify all the stresses acting on the stress block. Here we have 6x, 6y, and Txy.
   1. 6x = 80 MPa
   1. 6y = -20 MPa
   1. Txy, which is acting on the vertical face on which 6x is acting = 50 MPa
   1. Txy, which is acting on the horizontal face on which 6y is acting = -50 MPa
2. Now draw both axes, the normal stress axis, which is the x-axis and the shear stress axis, which is the y-axis. This is shown in Figure 18.

Figure 18

• Now, take the 6x value and place a dot on that point on the normal stress axis, as shown in Figure 19.

Figure 19

• Now take that shear stress which is acting on 6x’s plane. The value is +50 MPa; now, put the dot on the shear stress axis as shown in Figure 20.

Figure 20

• Now, trace the point using the 6x and Txy points, as shown in Figure 21.

Figure 21

• Now repeat this procedure for 6y and calculate the shear stress on that plane, and trace the point.
  6y = -20 MPa and Txy = -50 MPa as shown in Figure 22.

Figure 22

• Now join these two points with a straight line as shown in Figure 23.

Figure 23

• After joining, you will get the centre point of the circle, take the compass and draw the circle. This will be your Mohr’s circle as shown in Figure 24

Figure 24

Part A Solution.

We need to find principal stresses 61, 62, and their orientation, “Angle on which principal plane is occurring”. See Figure 25, where you can see the location of 61 and 62, and their values can be calculated from the following equations, or you can find them graphically as well.

Figure 25

The angle we found from the Mohr’s circle is 45 degrees, but I already explained that in Mohr’s circle, the angle is always taken twice. Hence, for the stress block, we need to rotate it counterclockwise by 22.5 degrees for the principal plane. From Figure 25, 61 = 100.71 MPa and 62 = -40.71 MPa. I also want to explain another vital point of the Mohr’s circle’s angle, for instance, we have found the θp angle, which is 45 degrees in Mohr’s circle in a counterclockwise direction, but if we see, then the principal plane can also be achieved in a clockwise direction, as shown in Figure 26.

Figure 26

At an angle of -135 degrees, the negative sign is due to the clockwise direction, and in the stress block, the rotation will be half of that angle, which is -67.5 degrees. Hence, we always mention two principal plane angles, and in many books you will find that one is θp1 and the other is θp2, which are 22.5 and -67.5 degrees. If we know θp1, then we can find θp2 as well because there is only a difference of 90 degrees between them, subtract the given angle from 90, and reverse its sign. In our case, 90 – 22.5 = 67.5, after sign reversal -67.5 degrees.  

Part B Solution.

To find the maximum in-plane shear stress, find the value from the circle using the scale on its peaks as shown in Figure 27.

Figure 27

The maximum in-plane shear stress is 70.71 and -70.71 MPa. And remember, maximum in-plane shear stresses are always equal in magnitude and opposite in sign. You can find these values using this simple equation.

Further, we can also have two angles of in-plane, which are θs1 and θs2, which are 67.5 degrees and -22.5 degrees. From them, one angle can be found from the following equation, and the other by subtracting that from 90 and reversing the sign.

We also need to find normal stresses on this plane, which can be seen in Figure 27 in orange colour. These values can be determined by putting θs1 or θs2 in the stress transformation equations, which will give you 6x’ and 6y’.  

Part C Solution.

In this section, we need to find normal and shear stresses at -20 degrees. This can be easily determined by plotting a line at 2θ = 2x-20 = -40 in Mohr’s circle and then determining their coordinates as shown in Figure 28.

Figure 28

Here, 6x’ and 6y’ can be found from the main stress transformation equations, which are given above; we need to put θ = -20 degrees.

Solution with Free Mohr’s Circle Calculator.

This problem can be solved with our free Mohr’s Circle calculator. The solution is shown in Figure 29.

Figure 29 shows a problem solution with a free Mohr’s circle calculator.