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:

θτθσσσσ

σ 2sin2cos22

xy

yxyx

n −−

++

= (1)

θτθσσσσ

σ 2sin2cos22

xy

yxyx

n −−

=+

−

θτθσσ

τ 2cos2sin2

xy

yx +−

=

(3)

Rewriting the equation (1)

(2)

Taking squares of equations (2) & (3)

22 ]2cos2sin2

[][ θτθσσ

τxy

yx +−

=

22 ]2sin2cos2

[]2

[ θτθσσσσ

σ xy

yxyx

n −−

=+

−

(4)

(5)

By adding equ.(4) & (5), and simplifying, we obtain

( )22

2

2

22xy

yxyx

n τσσ

τσσ

σ +

−=+

+− (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

( ) 222rhn =+− τσ (7)

The equation (7) is similar to equation of Circle i.e.,

222 )()( rkyhx =−+−

Center of circle is

2

yxhC

σσ +==

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

( )22

2xy

yxrR τ

σσ+

−==

From figure

2

yxa

σσ −=

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.

2θP22θP1

σx, τxy

σy, -τxy

σ2

σ1

x-axis

v, v1 plane

xy-a

xis

H, H

1p

lane y

−

2

yx σσ

σx

+

2

yx σσ

σy

2θs1

ττττmax

ττττmin

2θs2

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,

( ) [ ] MPayxavg 50)300(40021

21 =−+=+= σσσ

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,

( ) [ ] MPaa yx 350)300(40021

21 =−−=−= σσ

FIG 11-12 Complete Mohr’s circle

The vertical side of the triangle, b, is completed from

MPab xy 200==τThe radius of the circle, R, is completed from:

MPabaR 403)200()350( 2222 =+=+=

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

MPaRO

MPaRO

35340350

45340350

2

1

−=−=−=

=+=+=

σ

σ

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

o74.29350

200tan2 1 == −φ

Note that 2Φ is CW from the x-axis to σ1 on the circle.

oo

87.142

74.29==φ

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

CCWob

a 26.60)(tan)(tan'2 20035011 === −−φ

Or, using the geometry of the circle. we can compute

CCWoooo 26.6074.2990290'2 =−=−= φφThen the angle Φ’ is one-half of 2Φ’.

oo

13.302

26.60' ==φ

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 σ2= -353MPa Φ=14.87o CW from x-axis

τmax=403MPa σavg=50MPa Φ’=30.13o CCW fron x-axis

Example Problem 2

Given σx=-120MPa σy= 180MPa τxy=80MPa CCW

Results Figures 11-15.

σ1=200MPa σ2= -140MPa Φ=75.96o CCW

τmax=170MPa σavg=30MPa Φ’=59.04o 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:

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

Results Figures 11-5.

σ1=42.17 ksi σ2= -52.17 ksi Φ=61.0o CW

τmax=47.17 ksi σavg=-5.0 ksi Φ’=16.0o CW

Comments The x-axis is in the fourth quadrant.

(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:

Figure 11-16 Result for Example Problem 11-5, X-axis in the fourth quadrant.

Example Problem4

Given σx=220MPa σy=-120MPa τxy=0MPa

Results Figures 11-17.

σ1=220MPa σ2= -120MPa Φ=0o

τmax=170MPa σavg=50MPa Φ’=45.0o CCW

Solution:

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

Results Figures 11-18.

σ1=40 ksi σ2=0 ksi Φ=0o

τmax=20 ksi σavg=20 ksi Φ’=45.0o CCW

Solution:

(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.

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

Results Figures 11-19.

σ1=40 ksi σ2=-40 ksi Φ=45o CW

Solution:

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. 9-

15b. (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

MPaCFOCOF o 5060cos6020 =+=+==σ

MPaDF oo 0.5260sin60 ===τ

On the perpendicular 120° face we have

MPaCGOCOG o 1060cos6020' −=−=−==σ

MPaGE o 0.5260sin60' −=−==τBoth 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