MOHR'S CIRCLE
The formulas developed in the preceding article may be used for any case of plane
stress. A visual interpretation of them, devised by the German engineer Otto Mohr in
1882, eliminates the necessity for remembering them.* In this interpretation a circle is
used; accordingly, the construction is called Mohr's, circle. If this construction is plotted
to scale, the results can be obtained graphically; usually, however, only a rough sketch
is drawn, analytical results being obtained from it by following the rules given later.
We can easily show that Eqs. (1) and (2) define a circle by first rewriting them as
follows:
n 
 x  y
2

 x  y
2
cos 2   xy sin 2
(1)

 x  y
2
sin 2   xy cos 2
(2)
Rewriting the equation (1)
n 
 x  y
2

 x  y
2
cos 2   xy sin 2
(3)
Taking squares of equations (2) & (3)
[ ]  [
2
[ n 
 x  y
 x  y
2
2
] [
2
sin 2   xy cos 2 ]
2
 x  y
2
cos2   xy sin 2 ]2
(4)
(5)
By adding equ.(4) & (5), and simplifying, we obtain
 x  y

  n 
2


  x  y 
2
    
   xy 2

 2 
2
2
(6)
Recall that σx, σy, and τxy are known constants defining the specified state of stress,
whereas σn and τ are variables. Consequently, (σx + σy)/2 is a constant, say, h, and the
right-hand member of Eq. (6) is another constant, say, r. Using these substitutions, we
transform Eq. (6) into
 n  h 
2
2
r
2
(7)
The equation (7) is similar to equation of Circle i.e.,
( x  h)2  ( y  k )2  r 2
Center of circle is
Ch
 x  y
2
From the origin.
Figure 9-14 represents Mohr's circle for the state of plane stress that was analyzed in the
preceding article. The center C is the average of the normal stresses, and the radius
  x  y 
   xy 2
R  r  
 2 
2
From figure
a
 x  y
2
is the hypotenuse of the right triangle CDA. How do the coordinates of points E, F, and
G compare with the expressions derived for σ1,σ2 ,τmax ?We shall see that Mohr's circle
is a graphic visualization of the stress variation given by Eqs. (1) and (2). The following
rules summarize the construction of Mohr's circle.
Figure 9-14 Mohr's circle for general state of plane stress.
Rules for Applying Mohr's Circle to Combined Stresses
1. On rectangular σ-τ axes, plot points having the coordinates (σx, τxy) and (σy, τyx).
These points represent the normal and shearing stresses acting on the x and y faces of
an element for which the stresses are known. In plotting these points, assume tension as
plus, compression as minus, and shearing stress as plus when its moment about the
center of the element is clockwise.*
2. Join the points just plotted by a straight line. This line is the diameter of a circle
whose center is on the a axis.
3. As different planes are passed through the selected point in a stressed body, the
normal and shearing stress components on these planes are represented by the
coordinates of points whose position shifts around the circumference of Mohr's circle.
4. The radius of the circle to any point on its circumference represents the axis directed
normal to the plane whose stress components are given by the coordinates of that point.
5. The angle between the radii to selected points on Mohr's circle is twice the angle
between the normal to the actual planes represented by these points, or to twice the
space angularity between the planes so represented. The rotational sense of this angle
corresponds to the rotational sense of the actual angle between the normal to the planes;
that is, if the n axis is actually at a counterclockwise angle θ from the x axis, then on
Mohr's circle the n radius is laid off at a counterclockwise angle 2θ from the x radius.
x
x-axis
v, v1 plane
2θs1
σx, xy
σ2
2θP2
max
2θP1
2θs2
y-axis
H, H1 plane
σy, -xy
min
y
σy
 x  y


2

 x  y


2





σx
σ1




Example Problem 1
It has been determined that a point in a load-carrying member is subjected to the
following stress condition:
σx=400MPa
σy=-300MPa
τxy=200MPa(CW)
Perform the following
(a) Draw the initial stress element.
(b) Draw the complete Mohr’s circle, labeling critical points.
(c) Draw the complete principal stress element.
(d) Draw the maximum shear stress element.
Solution
The 15-step Procedure for drawing Mohr's circle is used here to complete the problem.
The numerical results from steps 1-12 are summarized here and shown in Figure 11-12.
Step 1. The initial stress element is shown at the upper left of Figure 11-12.
Step 2. Point 1 is plotted at ax = 400 MPa and τxy = 200 MPa in quadrant 1.
Step 3. Point 2 is plotted at ay = -300 MPa and τyx = -200 MPa in quadrant 3.
Step 4. The line from point 1 to point 2 has been drawn.
Step 5. The line from step 4 crosses the σ-axis at the average applied normal stress,
called O in Fig 11-12, is computed from any,
 avg  12  x   y   12 400 (300)  50MPa
Step 6. Point 0 is the center of the circle. The line from point O through point 1 is
labeled as the x-axis to correspond with the x-axis on the initial stress element.
Step 7. The values of G, b, and R are found using the triangle formed by the lines
from point 0 to point 1 to σx = 400 MPa and back to point O.
The lower side of the triangle,
a  12  x   y   12 400 (300)  350MPa
FIG 11-12 Complete Mohr’s circle
The vertical side of the triangle, b, is completed from
b   xy  200MPa
The radius of the circle, R, is completed from:
R  a 2  b 2  (350 ) 2  (200 ) 2  403 MPa
Step 8. This is the drawing of the circle with point 0 as the center at σavg = 50 MPa
and a radius of R = 403 MPa.
Step 9. The vertical diameter of the circle has been drawn through point O. The
intersection of this line with the circle at the top indicates the value of τmax = 403 MPa,
the same as the value of R.
Step 10. The maximum principal stress, σ1, is at the right end of the horizontal
diameter of the circle and the minimum principal stress, σ2, is at the left.
Step 11. The values for al and a2 are
 1  O  R  50  403 453MPa
 2  O  R  50  403 353MPa
Step 12. The angle 2Φ is shown on the circle as the angle from the x-axis to the σ1-axis,
a clockwise rotation. The value is computed from
200
2  tan
 29.74 o
350
1
Note that 2Φ is CW from the x-axis to σ1 on the circle.
29.74o

 14.87o
2
Step 13. Using the results from Steps 11 and 12, the principal stress element is drawn as
shown in Figure 11-13(b). The element is rotated 14.870 CW from the original x-axis to
FIG 11-13 Results for Example Problem 11-2
the face on which the tensile stress σ1 = 453 MPa acts. The compressive stress σ2 = -353
MPa acts on the faces perpendicular to the al faces.
Step 14. The angle 2Φ’ is shown in Figure 11-12 drawn from the x -axis CCW to the
vertical diameter that locates τmax at the top of the circle. Its value can be found in either
of two ways. First using Equation 11-8 and observing that the numerator is the same as
the value of a and the denominator is the same as the value of b from the construction of
the circle. Then
2 '  tan ( b )  tan (
1 a
1 350
)  60.26 CCW
o
200
Or, using the geometry of the circle. we can compute
2 '  90  2  90  29.74  60.26 CCW
o
o
o
o
Then the angle Φ’ is one-half of 2Φ’.
60.26o
'
 30.13o
2
Step 15. The maximum shear stress element is drawn in Figure 11-13(c), rotated 30.13°
CCW from the original x-axis to the face on which the positive τmax acts. The maximum
shear stress of 403 MPa is shown on all four faces with vectors that create the two pairs
of opposing couples characteristic of shear stresses on a stress element. Also shown is
the tensile stress σmax = 50 MPa acting on all four faces of the element.
Summary of Results for Example Problem 1 Mohr's Circle
Given σx=440MPa
σy= -300MPa
τxy=200MPa CW
Results Figures 11-12 and 11-13.
σ1=453MPa
τmax=403MPa
σ2= -353MPa
σavg=50MPa
Φ=14.87o CW from x-axis
Φ’=30.13o CCW fron x-axis
Example Problem 2
Given σx=-120MPa
σy= 180MPa
τxy=80MPa CCW
(a) Draw the initial stress element.
(b) Draw the complete Mohr’s circle, labeling critical points.
(c) Draw the complete principal stress element.
(d) Draw the maximum shear stress element.
Solution:
Results Figures 11-15.
σ1=200MPa
σ2= -140MPa
Φ=75.96o CCW
τmax=170MPa
σavg=30MPa
Φ’=59.04o CW
Figure 11-15 Result for Example Problem 11-4, X-axis in the third quadrant.
Example Problem 3
Given σx=-30ksi
σy=20 ksi
τxy=40 ksi CW
(a) Draw the initial stress element.
(b) Draw the complete Mohr’s circle, labeling critical points.
(c) Draw the complete principal stress element.
(d) Draw the maximum shear stress element.
Solution:
Results Figures 11-5.
σ1=42.17 ksi
σ2= -52.17 ksi
τmax=47.17 ksi
σavg=-5.0 ksi
Comments
The x-axis is in the fourth quadrant.
Φ=61.0o CW
Φ’=16.0o CW
Figure 11-16 Result for Example Problem 11-5, X-axis in the fourth quadrant.
Example Problem4
Given σx=220MPa
σy=-120MPa
τxy=0MPa
Solution:
Results Figures 11-17.
σ1=220MPa
τmax=170MPa
σ2= -120MPa
σavg=50MPa
Φ=0o
Φ’=45.0o CCW
Fig 11-17 Result for Example Problem 11-5,Special case of biaxial stress with no
shear
Example Problem 5:
Given σx=40 ksi
σy=0 ksi
τxy=0ksi
(a) Draw the initial stress element.
(b) Draw the complete Mohr’s circle, labeling critical points.
(c) Draw the complete principal stress element.
(d) Draw the maximum shear stress element.
Solution:
Results Figures 11-18.
σ1=40 ksi
τmax=20 ksi
σ2=0 ksi
σavg=20 ksi
Φ=0o
Φ’=45.0o CCW
Fig 11-18 Results of Example Problem 11-7. Special case of uniaxial tension
Example Problem 6
Given σx=0 ksi
σy=0 ksi
τxy=40ksi CW
Solution:
Results Figures 11-19.
σ1=40 ksi
σ2=-40 ksi
Φ=45o CW
τmax=40 ksi
σavg=0 ksi
Φ’=0o
Fig 11-19 Results of Example Problem 11-8, Special case of Pure shear.
Example Problem 7:
At a certain point in a stressed body, the principal stresses are σx = 80 MPa and σy = -40
MPa. Determine σ and τ on the planes whose normal are at +30° and + 1 20° with the x
axis. Show your results on a sketch of a differential element.
Solution: The given state of stress is shown in Fig. 9- 1 5a. Following the rules given
previously, draw a set of rectangular axes and label them a and r as shown in Fig. 915b. (Note that, for convenience, the stresses are plotted in units of MPa.) Since the
normal stress component on the x face is 80 MPa and the shear stress on that face is
zero, these components are represented by point A which has the coordinates (80, 0).
Similarly, the stress components on the y face are represented by point B (-40, 0).
According to rule 2, the diameter of Mohr's circle is AB. Its center C, lying midway
between A and B, is 20 MPa from the origin O. The radius of the circle is the distance
CA = 80 - 20 = 60 MPa. From rule 4, the radius CA represents the x axis. In accordance
with rules 4 and 5, point D represents the stress components on the face whose normal
is inclined at +30° to the x axis, and point E represents the stress components on the
perpendicular face. Observe that positive angles on the circle are plotted in a
counterclockwise direction from the x axis and are double the angles between actual
planes.
* This special rule of sign for shearing stress makes τx= —τyx in Mohr's circle. From
here on, we use this rule to designate positive shearing stress. However, the
mathematical theory of elasticity uses the convention that shearing stress is positive
when directed in the positive coordinate direction on a positive face of an element, that
is, when acting upward on the right face or rightward on the upper face. This other rule
makes τxy = τyx, which is convenient for mathematical work but confusing when applied
to Mohr's circle.
Figure 9-15
From rule 3, the coordinates of point D represent the required stress components on the
30° face. From the geometry of Mohr's circle, these values are
  OF  OC  CF  20  60cos60o  50MPa
  DF  60o sin 60o  52.0MPa
On the perpendicular 120° face we have
 '  OG  OC  CG  20  60cos60o  10MPa
 '  GE  60sin 60o  52.0MPa
Both sets of these stress components are shown on the differential element in Fig. 9-16.
Observe the clockwise and counterclockwise moments of τ and τ', respectively, relative
to the center of the element (see rule 1). Finally, note that a complete sketch of a
differential element shows the stress components acting on all four faces of the element
and the angle at which the element is inclined.
Figure 9-16