8 Mohr’s Circles For Stress

FIGURE 3.13

Typical Mohr’s circles for stress.



2

n2 =



(σ N − σ III ) (σ N − σ I ) + σ S2

(σ II − σ III ) (σ II − σ I )



(3.8-2b)



2

n3 =



(σ N − σ I ) (σ N − σ II ) + σ S2

(σ III − σ I ) (σ III − σ II )



(3.8-2c)



In these equations, σI , σII , and σIII are known; σN and σS are functions of the

direction cosines ni. Our intention here is to interpret these equations graphically by representing conjugate pairs of σN, σS values, which satisfy Eq 3.8-2,

as a point in the stress plane having σN as absicca and σS as ordinate (see

Figure 3.13).

To develop this graphical interpretation of the three-dimensional stress

state in terms of σN and σS, we note that the denominator of Eq 3.8-2a is

2

positive since both σ I − σ II > 0 and σ I − σ III > 0 , and also that n1 > 0 , all of

which tells us that

2

(σN – σII)(σN – σIII) + σ S ≥ 0



(3.8-3)



For the case where the equality sign holds, this equation may be rewritten,

after some simple algebraic manipulations, to read



[σ



N



−



1

2



]



2



2

σ II + σ III  + σ S =





[ (σ

1

2



II



]



− σ III )



2



(3.8-4)



which is the equation of a circle in the σN, σS plane, with its center at the

1

1

point (σ II + σ III ) on the σN axis, and having a radius (σ II − σ III ) . We label

2

2

this circle C1 and display it in Figure 3.13. For the case in which the inequality

sign holds for Eq 3.8-3, we observe that conjugate pairs of values of σN and

σS which satisfy this relationship result in stress points having coordinates

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exterior to circle C1. Thus, combinations of σN and σS which satisfy Eq 3.8-2a

lie on, or exterior to, circle C1 in Figure 3.13.

Examining Eq 3.8-2b, we note that the denominator is negative since

σ II − σ III > 0 , and σ II − σ I < 0 . The direction cosines are real numbers, so

2

that n2 ≥ 0 and we have



(σ N − σ III )(σ N − σ I ) + σ S2 ≤ 0



(3.8-5)



which for the case of the equality sign defines the circle

2



1





1



2

σ N − 2 (σ I + σ III ) + σ S =  2 (σ I − σ III )











2



(3.8-6)



in the σN, σS plane. This circle is labeled C2 in Figure 3.13, and the stress

points which satisfy the inequality of Eq 3.8-5 lie interior to it. Following the

same general procedure, we rearrange Eq 3.8-2c into an expression from

which we extract the equation of the third circle, C3 in Figure 3.13, namely,

2



1

 1



2





σ N − 2 σ I + σ II   + σ S =  2 σ I − σ II  











2



(3.8-7)



Admissible stress points in the σN, σS plane lie on or exterior to this circle.

The three circles defined above, and shown in Figure 3.13, are called Mohr’s

circles for stress. All possible pairs of values of σN and σS at P which satisfy

Eq 3.8-2 lie on these circles or within the shaded areas enclosed by them.

Actually, in conformance with Figure 3.11 (which is the physical basis for

Figure 3.13), we see that the sign of the shear component is arbitrary so that

only the top half of the circle diagram need be drawn, a practice we will

occasionally follow hereafter. In addition, it is clear from the Mohr’s circles

diagram that the maximum shear stress value at P is the radius of circle C2,

which confirms the result presented in Eq 3.7-11.

In order to relate a typical stress point having coordinates σN and σS in the

stress plane of Figure 3-13 to the orientation of the area element ∆S (denoted

by ni in Figure 3.12) upon which the stress components σN and σS act, we

consider a small spherical portion of the continuum body centered at P. As

the unit normal ni assumes all possible directions at P, the point of intersection of its line of action with the sphere will move over the surface of the

sphere. However, as seen from Eqs 3.7-5 and 3.7-6 the values of σN and σS

are functions of the squares of the direction cosines, and hence do not change

for ni’s reflected in the principal planes. Accordingly, we may restrict our

attention to the first octant of the spherical body, as shown in Figure 3.14A.

Let Q be the point of intersection of the line of action of ni with the spherical

surface ABC in Figure 3.14A and note that

ˆ

ˆ*

ˆ2

ˆ3

n = cos φ e1 + cos β e * + cos θ e *

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(3.8-8)



FIGURE 3.14A

Octant of small sphercal portion of body together with plane at P with normal ni referred to

principal axes Ox1 *x *.

*x 2 3



FIGURE 3.14B

Mohr’s stress semicircle for octant of Figure 3.14A.



ˆ ˆ*

If n = e1 so that its intersection point Q coincides with A, σN = σI. Likewise,

when Q coincides with B, σN = σII, and with C, σN = σIII. In all three cases, σS

will be zero. In the Mohr’s circle diagram (Figure 3.14B), these stress values are

π

located at points a, b and c, respectively. If now θ is set equal to

and φ

2

π

π

allowed to vary from zero to

(β will concurrently go from

to zero), Q

2

2

will move along the quarter-circle arc AB from A to B. In the stress space of

Figure 3.14B, the stress point q (the image point of Q) having coordinates σN

and σS will simultaneously move along the semicircle of C3 from a to b. (Note

that as Q moves 90° along AB in physical space, q moves 180° along the

semicircle, joining a to b in stress space.) Similarly, when Q is located on the

quarter circle BC, or CA of Figure 3.14A, point q will occupy a corresponding

position on the semicircles of bc and ca, respectively, in Figure 3.14B.

π

Now let the angle φ be given some fixed value less than

, say φ = φ1, and

2

imagine that β and θ take on all values compatible with the movement of Q

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FIGURE 3.15A

Reference angles φ and β for intersection point Q on surface of body octant.



FIGURE 3.15B

Mohr’s semicircle for stress state displayed in Figure 3.15A.



along the circle arc KD of Figure 3.15A. For this case, Eq 3.8-2a becomes

2

(σN – σII)(σN – σIII) + σ S = (σI – σII)(σI – σIII) cos2φ1, which may be cast into

the standard form of a circle as

2



2



1

 1



2

2

2





σ N − 2 σ II + σ III   + σ S = (σ I − σ II )(σ I − σ III ) cos φ1 +  2 σ II − σ III   = R1











(3.8-9)



This circle is seen to have its center coincident with that of circle C1 in stress

space and to have a radius R1 indicated by Eq 3.8-9. Therefore, as Q moves

on circle arc KD in Figure 3.15A, the stress point q traces the circle arc kd

π

shown in Figure 3.15B. (Notice that if φ1 =

so that cos φ1 = 0, R1 reduces

2

1

π

to (σ II − σ III ) , the radius of circle C1.) Next, let β = β1 <

and then, as φ

2

2

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and θ range through all admissible values, point Q moves along the circle arc

EG of Figure 3.15A. For this case Eq 3.8-2b may be restructured into the form

2



2



1

 1



2

2

2





σ N − 2 σ I + σ III   + σ S = (σ II − σ III ) (σ II − σ I ) cos β1 +  2 σ I − σ III   = R2 (3.8-10)











which defines a circle whose center is coincident with that of circle C2, and

π

1

, the radius R2 reduces to (σ I − σ III ) ,

having a radius R2. Here, when β1 =

2

2

which is the radius of circle C2. As Q moves on the circle arc EG of

Figure 3.15A, the stress point q traces out the circle arc eg, in Figure 3.15B.

ˆ

In summary, for a specific n at point P in the body, point Q, where the

ˆ

line of action of n intersects the spherical octant of the body (Figure 3.15A),

is located at the common point of circle arcs KD and EG, and at the same

time, the corresponding stress point q (having coordinates σN and σS) is

located at the intersection of circle arcs kd and eg, in the stress plane of

Figure 3.15B. The following example provides details of the procedure.



Example 3.8-1

The state of stress at point P is given in MPa with respect to axes Px1x2x3 by

the matrix



[σ ]

ij



25



= 0

 0





0

−30

−60



0



−60

5





(a) Determine the stress vector on the plane whose unit normal is

ˆ 1 ˆ ˆ

ˆ 

n = 3  2e 1 + e 2 + 2e 3  .

(b) Determine the normal stress component σN and shear component σS on

the same plane.

(c) Verify the results of part (b) by the Mohr’s circle construction of

Figure 3.15B.



Solution



ˆ

(a) Using Eq 3.4-8 in matrix form gives the stress vector ti( n )

ˆ

t ( n )  25

 1n )  

(ˆ

t 2  =  0

ˆ

t ( n )   0

3  



0

−30

−60



0 2 3

 50



 1  1 

−60  3 = −150

3

 −50

5 2 



  3



 



or

t ( n) =

ˆ



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1

ˆ

ˆ

ˆ 

 50e 1 − 150e 2 − 50e 3 

3



FIGURE E3.8-1

Three-dimensional Mohr’s circle diagram.



(b) Making use of Eq 3.7-1, we can calculate σN conveniently from the matrix

product



[2 3



1



3



25

2  0

3 

 0





]



σN =



0

−30

−60



0 2 3

 

−60  1 3 = σ N

5 2 

  3

 



100 150 100

−

−

9

9

9



150

= –16.67 MPa. Note that the same result could have been

9

obtained by the dot product

so that σN = −



ˆ

σ N = t(n ) ⋅ n =

ˆ



1

1

ˆ

ˆ

ˆ   2e1 + e2 + 2e 3 

ˆ ˆ

ˆ 

 50e1 − 150e 2 − 50e 3  ⋅



3

3



The shear component σS is given by Eq 3.7-2, which for the values of σN

ˆ

n

and ti( ) calculated above, results in the equation,

2

σS =



2, 500 + 22, 500 + 2, 500 22, 500

−

= 2, 777

9

81



or, finally,



σS = 52.7 MPa

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(c) Using the procedure of Example 3.6-1, the student should verify that for

the stress tensor σij given here the principal stress values are σI = 50, σII = 25,

*

*

and σIII = –75. Also, the transformation matrix from axes Px1x2x3 to Px1 x* x3 is

2



0

aij =  1



0







[ ]



−



3

5

0

4

5



4

5

0

3



5





ˆ

so that the components of n are given relative to the principal axes by

 *  

 n1   0

 



 *  

n2  =  1



 



 n*   0

 3  

 





−



3

5

0

4

5



4  2   1 



 



5  3   3 





 



0   13  =  2 3 



 



3  2   2 

5  3   3 





 



Therefore, with respect to Figure 3.14A, φ = cos–1(1/3) = 70.53°; β = θ =

cos–1(2/3) = 48.19°, so that — following the procedure outlined for construction

of Figure 3.15B — we obtain Figure E3.8-1, from which we may measure the

coordinates of the stress point q and confirm the values σN = –16.7 and σS =

52.7, both in MPa.



3.9



Plane Stress



When one — and only one — principal stress is zero, we have a state of plane

stress for which the plane of the two nonzero principal stresses is the designated

plane. This is an important state of stress because it represents the physical

situation occurring at an unloaded point on the bounding surface of a body

under stress. The zero principal stress may be any one of the three principal

stresses as indicated by the corresponding Mohr’s circles of Figure 3.16.

If the principal stresses are not ordered and the direction of the zero

principal stress is arbitrarily chosen as x3, we have plane stress parallel to

the x1x2 plane and the stress matrix takes the form



[σ ]

ij



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σ 11



= σ 12

 0





σ 12

σ 22

0



0



0

0





(3.9-1a)



FIGURE 3.16A

Mohr’s circle for plane stress (a) σI = 0.



FIGURE 3.16B

Mohr’s circle for plane stress (b) σII = 0.



FIGURE 3.16C

Mohr’s circle for plane stress (c) σIII = 0.

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FIGURE 3.17A

Plane stress element having nonzero x1 and x2 components.



FIGURE 3.17B

Mohr’s circle for the in-plane stress components.



FIGURE 3.17C

General Mohr’s circles for the plane stress element. Dashed lines represent out-of-plane Mohr’s

circles. Note the maximum shear can occur out-of-plane.



or, with respect to principal axes, the form



[σ ]

*

ij



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σ (1)



= 0

 0





0

σ (2)

0



0



0

0





(3.9-1b)



FIGURE 3.18A

Representative rotation of axes for plane stress.



FIGURE 3.18B

Transformation table for general plane stress.



The pictorial description of this plane stress situation is portrayed by the

block element of a continuum body shown in Figure 3.17A, and is sometimes

represented by a single Mohr’s circle (Figure 3.17B), the locus of which

identifies stress points (having coordinates σN and σS) for unit normals lying

in the x1x2 plane only. The equation of the circle in Figure 3.17B is

2



2



2

2

σ + σ 22 



 σ − σ 22 

 + (σ S ) =  11

 + (σ 12 )

 σ N − 11









2

2



(3.9-2)



1

(σ + σ 22 ) , σS =

2 11

0, and the maximum shear stress in the x1x2 plane to be the radius of the

circle, that is, the square root of the right-hand side of Eq 3.9-2. Points A and

B on the circle represent the stress states for area elements having unit

ˆ

ˆ

normals e1 and e2 , respectively. For an element of area having a unit normal

in an arbitrary direction at point P, we must include the two dashed circles

shown in Figure 3.17C to completely specify the stress state.

′ ′ ′

With respect to axes Ox1x2 x3 rotated by the angle θ about the x3 axis

relative to Ox1x2x3 as shown in Figure 3.18A, the transformation equations

for plane stress in the x1x2 plane are given by the general tensor transformation formula, Eq 2.5-13. Using the table of direction cosines for this situation

as listed in Figure 3.18B, we may express the primed stress components in

terms of the rotation angle θ and the unprimed components by

from which the center of the circle is noted to be at σ N =



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FIGURE E3.9-1

Mohr’s circle for principal stresses, σI = 2σO, σII = σIII = 0.



σ 11 =

′



σ 11 + σ 22 σ 11 − σ 22

+

cos 2θ + σ 12 sin 2θ

2

2



(3.9-3a)



σ 22 =

′



σ 11 + σ 22 σ 11 − σ 22

−

cos 2θ − σ 12 sin 2θ

2

2



(3.9-3b)



σ 12 = −

′



σ 11 − σ 22

sin 2θ + σ 12 cos 2θ

2



(3.9-3c)



In addition, if the principal axes of stress are chosen for the primed directions,

it is easily shown that the two nonzero principal stress values are given by

2

σ (1)  σ + σ

2



 σ − σ 22 

22

= 11

±  11

 + (σ 12 )

σ (2) 





2

2







(3.9-4)



Example 3.9-1

A specimen is loaded with equal tensile and shear stresses. This case of plane

stress may be represented by the matrix



[σ ]

ij



σ o



= σ o

0





σo

σo

0



0



0

0





where σo is a constant stress. Determine the principal stress values and plot

the Mohr’s circles.

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