Section 3 – 6, Mohr’s Circle for Plane Stress
Consider plane stress. The following sketch shows the stresses acting on a small element
of material. Because the element of material is small, the stresses are essentially constant at all
locations within the element.
Y
 XY
X
All stresses act on planes perpendicular to the drawing. All stresses are shown positive.
All other stresses are zero.
Shigley uses “cw” for  XY  0 and “ccw” for  XY  0 . This is an absurd complication.
Learn what I am going to teach you.
The stresses on planes perpendicular to the X - and Y - axes are known. There are no
stresses acting in the Z - direction, and no stresses acting on planes perpendicular to the Z - axis.
Determine the stresses acting on an arbitrary plane perpendicular to the drawing. Make a section
at a counterclockwise angle  (Mr. Hibbeler used  ) with respect to a vertical plane, the plane
normal to the X - axis, and consider a FBD of everything to one side of that section.

X


 XY
Y
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In general, when a section is made through a material, there will be some normal stress
acting on the plane of the section, perpendicular to the section, and some shear stress acting in an
unknown direction in the plane of the section. But in this case there are no forces acting in the
Z - direction, so the shear stress acting on the plane of the section cannot have a component in
the Z - direction because there is no force to balance that component. Therefore, the shear stress
acting on the plane of the section must act in the direction shown, parallel to the plane of the
drawing. Both the shear and normal stresses acting on the plane of the section have been drawn
positive, because these are the directions of positive stresses when   0 .
A denotes the area of the plane section. The area of the left, vertical, plane surface is
equal to A cos  . The area of the bottom, horizontal, plane surface is equal to A sin  .
 FX  0   A cos    A sin   X A cos    XY A sin
(3.1)
 FY  0   A sin   A cos    Y A sin   XY A cos 
(3.2)
Solve equations (3.1) and (3.2) simultaneously for  and  .
   X cos 2    Y sin2   2  XY sin  cos 
(3.3)
  (  Y   X ) sin  cos    XY ( cos 2   sin2  )
(3.4)
Equations (3.3) and (3.4) are two fairly complicated relationships to determine  and  .
These are referred to as the stress-transformation equations for plane stress. The derivation
omitted the details of the algebraic and trigonometric manipulations.
List the double- and half-angle relationships.
cos 2  cos 2   sin2 
(3.5)
1  cos 2
 cos 2 
2
(3.6)
1  cos 2
 sin2 
2
(3.7)
sin 2  2 sin cos 
(3.8)
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All modern authors, “derive” Mohr’s circle in the following manner. Substitute
equations (3.5) through (3.8), into the stress-transformation equations, (3.3) and (3.4), and obtain
relationships for  and  in terms of trigonometric functions of 2 . Do lots of clever rearranging
that no one in his right mind would ever do, unless he knew in advance where he was heading (or
had drunk large amounts of adult beverages). Finally, end up with complicated formulas that
lead to Mohr’s circle.
That is absurd. Mr. Mohr plotted  vs.  for lots of values of  for specific numerical
values of  X ,  Y , and  XY . He discovered that the points ended up on a circle, and that planes
90  apart in the material are represented by points 180  apart on the circle.
Proceed in the following manner. Plot two points on a graph of  vs.  . In this
example,  X ,  Y , and  XY are positive, and  X   Y .  is plotted positive down. Draw a line
connecting the two points. Draw a circle with this line as a diameter.
( Y ,   XY )



(  X ,  XY )

The two points that were plotted, 180  apart on Mohr’s circle, represent the stresses on
two planes that are 90  apart in the material. Angles on Mohr’s circle represent 2 , and are
measured positive counter-clockwise from the (  X ,  XY ) point on Mohr’s circle, corresponding
to the plane perpendicular to the X - axis in the material,   0 . In this sketch,  MAX occurs
for 2  45  . Therefore, in the material, the maximum tensile stress would occur on a plane
about 22.5  counterclockwise from the plane perpendicular to the X - axis.
 MAX
22.5 
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There are different sign conventions for Mohr’s circle. Old-school engineers learned to
plot  positive upward. The angles on Mohr’s circle then go in the opposite direction from the
angles in the material. This sign convention is superior. Be alert if Mohr’s circle is used in other
classes, and be sure to determine the correct sign convention.
The largest and smallest normal stresses are denoted by  1 and  2 ,  1 being the
algebraically larger, at the right edge of the circle. These two stresses are referred to as the
principal stresses. The planes on which the principal stresses act are referred to as the
principal planes. Because the principal stresses occur 180  apart on Mohr’s circle, the
principal planes are always perpendicular to each other in the material. The total shear stress on
any principal plane, including the three principal planes in a three-dimensional state of stress, is
equal to zero.
The maximum and minimum shear stresses on Mohr’s circle, equal to each other in
absolute value, also occur on planes perpendicular to each other. These planes are oriented at
45  to the principal planes.
Any diameter on the circle connects the stresses acting on two mutually perpendicular
planes in the material. The sum of these normal stresses is a constant.
 (  )   (   90  )   X   Y  2  AVG
(3.9)
R
(3.10)
(  X   AVG ) 2   XY 2
 1 ,  2   AVG  R
(3.11)
For the points plotted in this Mohr’s circle, but not in general.

 XY
1  0.5 tan 1 
  X   AVG



(3.12)
The relationship for  1 depends on whether  X or  Y is algebraically larger, and on
whether  XY is positive or negative. The correct approach is to neatly sketch Mohr’s circle and
use the geometry of the circle to determine the various stresses and angles.
 2  1  90 
(3.13)
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Example
5 ksi
2 ksi
3 ksi
 X   3 ksi
(E.1)
 Y  5 ksi
(E.2)
 XY   2 ksi
(E.3)
Draw Mohr’s circle, roughly to scale.
( 3,  2)

( 5, 2 )

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It is now possible to determine stresses and angles from the geometry.
 AVG 
X  Y  3  5

 1.0
2
2
(E.4)
R  ( Y   AVG ) 2   XY 2  (5  1) 2  (  2) 2  4.47
(E.5)
1   AVG  R  1.0  4.47  5.47 ksi 
(E.6)
 2   AVG  R  1.0  4.47   3.47 ksi 
(E.7)
Determining  1 and  2 causes many students difficulty. The following approach will
always work. Begin at the point on Mohr’s circle corresponding to (  X ,  XY ), the point
( 3,  2) in this example. Move counter-clockwise until a principal stress is reached,  2 in this
example. The angle moved on Mohr’s circle will always be less than 180  , so the angle moved
in the material will always be less than 90  .
 2
 2  0.5 tan 1    13.3  
 4
(E.8)
1   2  90  13.3   90   103.3 ,  76.7  
(E.9)
Sketch the planes on which the principal stresses act.
3.47 ksi
2
5.47 ksi
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There are two important special cases.
1.
 X  Y  0
 XY  0
(  0 , 0)



Mohr’s circle is a point at ( 0 , 0 ) . The normal stress is equal to  0 , and the shear stress
is equal to zero, on any plane perpendicular to the drawing of plane stress. The maximum shear
stress in the material is not equal to zero, the radius of Mohr’s circle.
2.
X  Y  0
 XY  0
 XY


Mohr’s circle is centered on the origin. Each principal normal stress has absolute value
equal to the absolute value of  XY . The maximum shear stress in the material is equal to the
radius of Mohr’s circle.
The maximum shear stress in the material may or may not be numerically equal to the
radius of Mohr’s circle. At present, we have not discussed how to decide.
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Three-Dimensional State of Stress
Re-examine the chunk of material used to determine the stress-transformation equations.
Consider a general three-dimensional state of stress, and include only the stresses that were not
considered during the previous analysis.


 XZ 

 ZY

Z
 ZX
 YZ 
The shear stresses on the left vertical surface and on the bottom horizontal surface have
no X - or Y - components. The three stresses on the front face of the material are balanced by
three equal and opposite stresses on the back face of the material that are not shown in the
drawing. Therefore, if all these stresses had been included in the original analysis that required
equilibrium of forces in the X - and Y - directions, the same results would have been obtained.
However, there is an additional stress that is not shown, the component of shear stress on the
inclined face of the material in the Z - direction, perpendicular to the drawing. There was no
such component of shear stress in the plane-stress problem, because there were no applied forces
in the Z - direction, and therefore that component of the shear stress was required to equal zero.
In this problem, because of the other stresses shown in the drawing, it is necessary to include a
component of shear stress on the inclined face of the material in the Z - direction. This shear
stress will also not affect the equilibrium equations in the X - and Y - directions. However, the
maximum and minimum normal stresses computed from Mohr’s circle, will not be acting on
planes with zero total shear stress. Therefore, these normal stresses are not principal stresses.
In general, it is incorrect to use a two-dimensional Mohr’s circle to analyze a threedimensional stress distribution.
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Three-Dimensional State of Stress, Special Case
Consider a small element of material subjected to a plane-stress distribution plus a
normal stress perpendicular to the plane of the drawing.
Y
 XY
Z
X
 Z may be either tensile or compressive.
It is possible to use a plane-stress Mohr’s circle to determine the two principal stresses
acting on planes perpendicular to the drawing. Because equal and opposite normal stresses in the
Z - direction act on the front and back of the sectioned element of material, there will be no shear
stresses in any direction on the inclined planes on which those two principal stresses act.
The third principal stress is then equal to  Z , because that normal stress also acts on a
plane on which no shear stress acts.
This special case is an exception to the general rule that it is incorrect to apply Mohr’s
circle to a three-dimensional state of stress.
After the three principal stresses have been determined, it is possible to determine the
maximum shear stress in the material. This will be explained later in this lecture.
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Re-work the plane-stress problem in a different manner.


X
û
 XY
Y
Set the shear stress on the inclined plane equal to zero. This is not a guess. For positive
 XY , the shear stress at   0  is positive, upward, while for   90  the shear stress is
negative, to the right. Because stresses are continuous, the shear stress must be zero for some
value of  between 0  and 90  . Create a unit vector perpendicular to the inclined plane.
û  u X î  u Y ĵ
(3.14)
Note the following two equalities.
u X  cos 
(3.15)
u Y  sin 
(3.16)
A denotes the area of the inclined plane. The area of the bottom, horizontal surface is
equal to A u Y and the area of the left, vertical surface is equal to A u X .
Sum forces in the X - and Y - directions.
 FX  0    X A u X   XY A u Y   A u X
(3.17)
 FY  0    Y A u Y   XY A u X   A u Y
(3.18)
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Rewrite equations (3.17) and (3.18), canceling a factor of A.
0  (   X ) u X
  XY u Y
(3.19)
0    XY u X  (   Y ) u Y
(3.20)
Equations (3.19) and (3.20) are simultaneous, linear equations in two unknown
variables, u X and u Y . The equations are homogeneous. There are no terms that do not contain
one of the two unknown variables. Therefore, a graph of the two equations results in two
straight lines through the origin, as shown in the following sketch.
uY
uX
The only simultaneous solution of two such homogeneous equations, a point on both
lines, is the origin. This is referred to as the trivial solution. Not only is this an uninteresting
solution, it is impossible, because the values of the two variables must be the components of a
unit vector. The only solution to this dilemma is to choose a value for  that makes the slopes of
the two lines equal to each other. Then both lines will be the same line, and there will be an
infinity of solutions. Every solution will have the same ratio of u Y to u X . The correct solution
will be the values of u Y and u X that create a unit vector.
 X
  XY

  XY
 Y
0  (   X ) (   Y )   XY 2
This is the same equation that results from setting the determinant of the coefficient
matrix equal to zero.
   X   XY
0
  XY    Y
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Example
Use the same values used with Mohr’s circle in the example on page 5.
X   3
(E.1)
Y  5
(E.2)
 XY   2
(E.3)
0  (   3 )(   5 )  4
(E.4)
0   2  2   19
(E.5)
Use the quadratic formula.

 (  2) 
2 2  4 (1) (  19)
 1  4.47  5.47,  3.47
2 (1)
(E.6)
These are the same results previously obtained for the principal stresses. Substitute the
value of  1 , equal to 5.47, back into either equation (3.19) or (3.20). It doesn’t matter which
equation, because they will be the same equation with this particular choice of . Choose
equation (3.19).
0  8.47 u X  2 u Y
(E.7)
u Y   4.24 u X
(E.8)
u Y sin 

 tan    4.24
u X cos 
(E.9)
  103.3 
(E.10)
This is the same result previously obtained for the plane on which  1 , the algebraically
larger principal stress, acts. Substitute the value of  2 , equal to  3.47, back into either
equation(3.19) or (3.20), and obtain  equal to 13.3  , the same result previously obtained for the
plane on which  2 , the algebraically smaller principal stress acts.
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The two values of  are referred to as eigenvalues. The values of the two u X - u Y pairs
are referred to as eigenvectors. Problems of this type show up in many different areas of science.
An important example is vibration analysis, in which the eigenvalues are the squares of the
natural angular frequencies of the system being studied. It is a theory of mathematics that real,
symmetric matrices have real eigenvalues and mutually perpendicular eigenvectors.
Therefore, the principal stresses, the eigenvalues, occur on planes that have mutually
perpendicular normal vectors.
Okay, that was fun, but Mohr’s circle is an excellent way to solve plane-stress problems,
so why bother? The answer is that this analysis makes it possible to simply write the result for a
general three-dimensional stress distribution.
  X
0    XY
  XZ
  XY
  Y
  YZ
  XZ
  YZ
  Z
To determine the principal stresses for a general, three-dimensional state of stress,
substitute values for the six stresses. Expand the determinant, yielding a cubic equation for .
Solve that cubic equation, yielding three real values for the principal stresses. Because the
matrix is real and symmetric, those principal stresses will act on three mutually perpendicular
planes. Plot three principal stresses and draw three Mohr’s circles.
2
3
1


All possible  -  pairs lie in the region outside the smaller circles and inside the largest
circle. The radius of the largest circle is the maximum shear stress in the material.
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It has been stated that, in a plane-stress problem, R, the radius of Mohr’s circle, may or
may not be the largest shear stress in the material. It is now possible to understand why. In a
plane-stress problem, the third principal stress is zero, acting on the plane of the drawing.
Therefore, if  1 and  2 , determined from Mohr’s circle, are both positive, tensile, the following
situation results.
2

1

The maximum shear stress is clearly not the radius of the Mohr’s Circle representing the
plane stress. The maximum shear stress in the material is the radius of the largest circle. The
same situation exists if  1 and  2 , determined from Mohr’s circle, are both negative,
compressive.
2
1

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However, if , after analyzing Mohr’s circle,  1 is positive, tension, and  2 is negative,
compression, the following situation arises.
2
1


The radius of the plane-stress Mohr’s circle is the maximum shear stress in the material.
When analyzing plane-stress problems, always include the third principal stress, equal to
zero, and always draw the other two Mohr’s circles.
In the three drawings on this page and the previous page,  1 and  2 denote the values of
the two principal stresses determined from Mohr’s circle. In general, the three principal stresses
are renumbered.  1 is the algebraically largest principal stress, the farthest to the right.  3 is
the algebraically smallest principal stress, the farthest to the left.  2 is the principal stress
between  1 and  2 .
 MAX 
1   3
2
(3.21)
Homework
Problems 3 – 18 (a), (b) Determine principal normal stresses and maximum shear stress.
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