Mohr’s circle

In two dimensional stress analysis, Mohr’s circle is a graphical representation of the stress state of a point in a body under static equilibrium. The two dimensional loading state can be either plane stress or plane strain loading. For three dimensional analysis we may use the extended Mohr circle for three dimensions (Mohr’s circle 3d).

Consider a body in equilibrium under two dimensional loading (cf. Fig. 1). The stress tensor for this case is given by:

\[ \sigma_{ij}=\left[\begin{array}{cc}\sigma_{11} & \sigma_{12}\\ \sigma_{21} & \sigma_{22}\end{array}\right] \]
(1)
Normal and shear stress on a plane in two dimensional analysis
Figure 1: Free body diagram of a point under static equilibrium in two dimensional analysis. Illustration of the normal and shear stresses acting on an arbitrary plane.

In this article we will use the following sign convention: Compression stresses are positive and shear stresses are positive when they have a direction as shown in Fig. 1.

Consider now an arbitrary plane inside this body. The normal vector \( \vec{n} \) of this plane angle \( \theta \) with respect to the horizontal axis \( x_{1} \). The stress vector \( T \) that acts on this plane may be calculated by (see also Calculation of normal and shear stress on a plane):

\[ \begin{array}{l}T_{i}=\sigma_{ij}n_{j}\Leftrightarrow\\ \bigg\{\begin{array}{l}T_{1}=\sigma_{11}n_{1}+\sigma_{12}n_{2}\\ T_{2}=\sigma_{21}n_{1}+\sigma_{22}n_{2}\end{array}\Leftrightarrow\\ \bigg\{\begin{array}{l}T_{1}=\sigma_{11}\cos\theta+\sigma_{12}\sin\theta\\ T_{2}=\sigma_{21}\cos\theta+\sigma_{22}\sin\theta\end{array}\end{array} \]
(2)

where \( (n_{1},n_{2}) \) are the direction cosines of the plane. The normal stress \( \sigma_{n} \) that acts on this plane is given by:

\[ \sigma_{n}=T_{1}n_{1}+T_{2}n_{2}=T_{1}\cos\theta+T_{2}\sin\theta \]
(3)

Expanding the above equation (3) and taking under consideration that:

\[ \begin{array}{l}\sigma_{12}=\sigma_{21}\\ \sin^{2}\theta=\frac{1}{2}(1-\cos2\theta)\\\cos^{2}\theta=\frac{1}{2}(1+\cos2\theta)\\ \sin2\theta=2\cos\theta \sin\theta\end{array} \]
(4)

we derive the following expression:

\[ \sigma_{n}=\frac{1}{2}(\sigma_{11}+\sigma_{22})+\frac{1}{2}(\sigma_{11}-\sigma_{22})\cos2\theta+\sigma_{12}\sin2\theta \]
(5)

The shear stress on the plane may be derived from the Pythagorean theorem:

\[ \tau_{n}^{2}=T^{2}-\sigma_{n}^{2}=T_{1}^{2}+T_{2}^{2}-\sigma_{n}^{2} \]
(6)

Expanding the above equation (6) and also taking under consideration equations (3) leads to:

\[ \tau_{n}=T_{2}\cos\theta-T_{1}\sin\theta \]
(7)

or

\[ \tau_{n}=-\frac{1}{2}(\sigma_{11}-\sigma_{22})\sin2\theta+\sigma_{12}\cos2\theta \]
(8)

Equations (5) and (8) are the parametric equations of a circle. To prove the latter we rewrite equation (5) as follows:

\[ \sigma_{n}-\frac{1}{2}(\sigma_{11}+\sigma_{22})=\frac{1}{2}(\sigma_{11}-\sigma_{22})\cos2\theta+\sigma_{12}\sin2\theta \]
(9)

If equations (8) and (9) are squared and added by parts, then after some calculations lead to:

\[ \left(\sigma_{n}-\frac{1}{2}(\sigma_{11}+\sigma_{22})\right)^{2}+\tau_{n}^{2}=\left(\frac{1}{2}(\sigma_{11}-\sigma_{22})\right)^{2}+\sigma_{12}^{2} \]
(10)

If we set

\[ C=\frac{1}{2}(\sigma_{11}+\sigma_{22}) \]
(11)

and

\[ R^{2}=\left(\frac{1}{2}(\sigma_{11}-\sigma_{22})\right)^{2}+\sigma_{12}^{2} \]
(12)

then equation (10) becomes:

\[ \left(\sigma_{n}-C\right)^{2}+\tau_{n}^{2}=R^{2} \]
(13)

which is the equation of a circle with center \( C \) and radius \( R \) on a coordinate system with the abscissa representing the normal stress \( \sigma \) and the ordinate the shear stress \( \tau \) (cf. Fig. 2).

Mohr's circle
Figure 2: In two dimensional analysis, Mohr’s circle represents the state of stress of a point in a body under static equilibrium.

In order to construct Mohr’s circle for the stress tensor (1) we must firstly define a special rule. The shear stresses that yield a clockwise moment about the center point of the element (cf. Fig. 1) will be taken positive. The shear stresses that tend to rotate the element counterclockwise will be considered negative. Thus, for the case of Fig. 1 we firstly draw the two points \( (\sigma_{11},\sigma_{12}) \) and \( (\sigma_{22},\sigma_{21}) \) as shown in Fig. 2. Next, we connect these to points by a straight line. This line is the diameter of the circle. The interception point between this line and the horizontal axis is the center of the circle. The distance between the center and \( (\sigma_{11},\sigma_{12}) \) or \( (\sigma_{22},\sigma_{21}) \) is equal to the radius. Hence, we may now draw Mohr’s circle.

According to equations (5) and (8) a plane with angle \( \theta \) as shown in Fig. 1, angle \( 2\theta \) on Mohr’s circle as shown in Fig. 2. Hence, for a given plane it is suffice to measure angle \( 2\theta \) on Mohr’s circle and the coordinates of the point on the circle are the normal and shear stress that act on that particular plane. It should be noted that the angle \( 2\theta \) is measured from the straight line connecting¬† the center of the circle and \( (\sigma_{11},\sigma_{12}) \) with the same direction as in the free body diagram (cf. Fig. 2).

Subsequently, we may derive the principal stresses and their directions. If we set:

\[ \tau_{n}=0 \]
(14)

then from equation (8) we derive:

\[ \tan2\theta=\frac{2\sigma_{12}}{\sigma_{11}-\sigma_{22}} \]
(15)

Equation (15) has two solutions:

\[ \begin{array}{l}\theta=\theta_{1}\\ \theta=\theta_{2}=\theta_{1}+\frac{\pi}{2}\end{array} \]
(16)

The two angles \( \theta_{1} \) and \( \theta_{2} \) are called principal directions and represent the direction of the two principal stresses. Substituting angles \( \theta \) from equation (15) into equation (5) and taking under consideration that:

\[ \begin{array}{l}\sin(\arctan(x))=\dfrac{x}{\sqrt{1+x^{2}}}\\ \cos(\arctan(x))=\dfrac{1}{\sqrt{1+x^{2}}}\end{array} \]
(17)

leads to:

\[ \begin{array}{c}\sigma_{1}\\ \sigma_{2}\end{array}=\frac{1}{2}(\sigma_{11}+\sigma_{22})\pm\sqrt{\left(\frac{\sigma_{11}-\sigma_{22}}{2}\right)^{2}+\sigma_{12}^{2}} \]
(18)

where \( \sigma_{1} \) and \( \sigma_{2} \) are the two principal stresses. In Fig. 2 the two principal stresses are the intersection points between Mohr’s circle and the horizontal axis.

Maximum shear stress occurs when the derivative of shear stress with respect to \( \theta \) equals to zero (see equation (8)). The maximum shear stress angle \( \pm 45^{o} \) with respect to the principal directions. Its absolute value is equal to the radius of Mohr’s circle:

\[ \tau_{max}=\frac{\sigma_{1}-\sigma_{2}}{2}=\sqrt{\left(\frac{\sigma_{11}-\sigma_{22}}{2}\right)^{2}+\sigma_{12}^{2}} \]
(19)

It must be noted that the above equation (19) holds true for planes parallel to z-axis. If inclined planes with respect to \( z \)-axis are considered then higher shear stresses may occur.

Pole points

So far we have discussed two methods to calculate the normal and shear stress on an arbitrary plane. The first method is to use equations (5) and (8) and the second method is to measure angle \( 2\theta \) on Mohr’s circle. An alternative graphical method to calculate the normal and shear stress is to use the pole point on Mohr’s circle.

Two pole points can be established on Mohr’s circle. The first pole point is related with the directions of the stresses and the second is related with the planes on which the stresses act. Consider the case of Fig. 1. We construct Mohr’s circle as shown in Fig. 2. Then, the Pole for stresses may be found if we draw a horizontal line from point \( (\sigma_{11},\sigma_{12}) \) parallel to \( \sigma_{11} \) direction, or by drawing a vertical line from point \( (\sigma_{22},\sigma_{21}) \) parallel to \( \sigma_{22} \) direction (cf. Fig. 3). Both lines intersect Mohr’s circle to the same point which is the pole for stresses (\( P_{s} \) ). Pole for planes may be found if we draw a vertical line from point \( (\sigma_{11},\sigma_{12}) \) which is parallel to the plane where \( \sigma_{11} \) acts. Similarly we may draw a horizontal line from point \( (\sigma_{22},\sigma_{21}) \) which is parallel to the plane where \( \sigma_{22} \) acts. Both lines intersect Mohr’s circle to the same point, the pole for planes (\( P_{p} \) ).

Mohr's circle and pole points
Figure 3: There are two special points on each Mohr’s circle, called the pole points. The normal and shear stress on an arbitrary plane can be found by virtue of these points by simply drawing normal or parallel lines to the plane.

Consider now an arbitrary plane with normal vector \( \vec{n} \) that angle \( \theta \) as shown in Fig. 3. We can use any of the two poles to calculate the normal and shear stresses that act on this plane. If we use the pole for stresses,¬† we draw a straight line from the pole point perpendicular to the plane (parallel to normal vector \( \vec{n} \) ). The intersection point between the line and Mohr’s circle gives the normal and shear stresses that act on this plane. Alternatively we may draw a straight line with angle \( \theta \) from \( P_{s} \) and find the intersection point. \( \theta \) is measured from the horizontal axis counterclockwise as shown in Fig. 3. If we use the pole for planes, we draw a straight line from the pole point parallel to the plane of interest. The intersection point between the straight line and Mohr’s circle gives the normal and shear stresses that act on this plane.

Both pole points can be used to calculate the directions of the principal stresses. We will use pole for stresses \( P_{s} \). From point \( P_{s} \) we draw straight lines to the two principal stresses \( \sigma_{1} \) and \( \sigma_{2} \) on Mohr’s circle as shown in Fig. 4. The angle \( \alpha \) between \( \sigma_{1} \) and the horizontal stress \( \sigma_{11} \) is measured counterclockwise from the horizontal axis on Mohr’s diagram as shown in Fig. 4.

Principal stresses on Mohr's circle and their direction.
Figure 4: The principal stresses on Mohr’s circle are the points where the circle crosses the horizontal axis (shear stresses vanish). The pole points may be used to find their direction.

Important note

The sign of shear stress calculated by equation (8) gives the direction of the shear stress for a local right handed coordinate system attached on the plane of interest.¬† The positive direction of the horizontal axis on this coordinate system is the direction of the normal vector \( \vec{n} \). The sign of shear stress calculated by Mohr’s circle gives the direction of the shear stress according to the special rule that we defined previously. Shear stresses that tend to rotate the element clockwise are positive and negative otherwise. Thus, a positive shear stress calculated by equation (8) is represented as negative on Mohr’s circle and the other way around.

Exercise

Consider a solid body under plane stress loading as shown in Fig. 1. The stresses acting on this body are given by the following stress tensor:

\[ \sigma_{ij}=\left[\begin{array}{cc}5&2\\2&3\end{array}\right] \]
(20)

Consider also an inclined plane inside this body. The normal vector of the plane angle \( 45^{o} \) with respect to horizontal axis. Calculate the normal and shear stress on this plane by using equations (5) and (8) and by using the pole on Mohr’s circle. Also, calculate the principal stresses and the principal directions.

Show solution…

By using equation (5) we calculate the normal stress:

\[ \sigma_{n}=\frac{1}{2}(5+3)+\frac{1}{2}(5-3)\cos90^{o}+2\sin90^{o}=6 \]
(21)

And for the shear stress we use equation (8):

\[ \tau_{n}=-\frac{1}{2}(5-3)\sin90^{o}+2\cos90^{o}=-1 \]
(22)

We use equation (18) to calculate the principal stresses:

\[ \begin{array}{c}\sigma_{1}\\ \sigma_{2}\end{array}=\frac{1}{2}(5+3)\pm\sqrt{\left(\frac{5-3}{2}\right)^{2}+2^{2}} \]
(23)
\[ \sigma_{1}\approx 6.25, \sigma_{2}\approx 1.75 \]
(24)

and the directions of the principal stresses are calculated by equation (15):

\[ \tan2\theta=\frac{2\cdot 2}{5-3}\Rightarrow \theta_{1}\approx 32^{o}, \theta_{2}\approx 122^{o} \]
(25)

To construct Mohr’s circle we draw firstly the two points from stress tensor (20). Note that the shear stress on the vertical sides of the element tend to rotate it counterclockwise thus according to the special rule we will consider them negative. Then we connect these two points and find the center of the circle (cf. Fig. 5). With this center and radius equal to the distance between the center and any of these two points, we draw the circle.

Example of stress calculation with the Mohr's circle
Figure 5: Example of calculation of the normal and shear stress on a plane with the help of Mohr’s circle. Calculation, also, of the principal stresses and their direction.

Next we draw a horizontal line from point \( (5,2) \) parallel to \( \sigma_{11} \) direction. The intersection point between the line and the circle is the pole for stresses. The shear and normal stress on the plane of interest can be found by either measuring angle \( 45^{o} \) from the pole point or by measuring angle \( 90^{o} \) from the center. The intersection points of the circle with the horizontal axis are the principal stresses \( \sigma_{1} \) and \( \sigma_{2} \). We connect the pole point with these points and we measure the angle of the principal stresses (principal directions) with respect to \( Oxy \) coordinate system of the stress tensor (20). Note that the shear stress on Mohr’s circle is positive which means that it tends to rotate the element clockwise. Shear stress from equation (22) is negative which means that its direction is opposite to the positive direction of the local right handed coordinate system attached on the plane. Both methods show the same direction for the shear stress. In Fig. 5 we have drawn on the plane the directions of \( \sigma_{n} \) and \( \tau_{n} \).