lecture10 mohr's circle

1

FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES

Slide No. 1

Example 2The stresses shown in Figure 12a act at a point on the free surface of a stressed body. Determine the normal stresses σnand σt and the shearing stress τnt at this point if they act on the rotated stress element shown in Figure 12b.

The Stress Transformation Equations for Plane Stress

2

Slide No. 2


Example 2 (cont’d)

MPa 70

MPa 40

MPa 10

nσtσ ntτ

015

ntFigure 12

(a) (b)

Slide No. 3


Example 2 (cont’d)The given values are as follows:

000t

0 1059015 ,15

MPa 40 MPa, 70 MPa, 10

=+==

+=−=−=

θθ

τσσ

n

xyyx

nσtσ ntτ

015

nt

t

015

090n

3

Slide No. 4


Example 2 (cont’d)Applying Eq. 12 for the given values

( ) ( ) ( ) ( )

(Tension) MPa 981.5MPa 981.5

15cos15sin)40(215sin7015cos10

cossin2sincos22

22

==

+−−=

++=

n

xyyxn

σ

θθτθσθσσ

( ) ( ) ( ) ( )

n)(comprssio MPa 86MPa 98.85

105cos105sin)40(2105sin70105cos10

cossin2sincos22

22

=−=

+−−=

++=

t

xyyxt

σ

θθτθσθσσ

Slide No. 5


Example 2 (cont’d)( ) ( )( ) ( ) ( )( )

MPa 64.1915sin15cos40)15cos()15sin()70(10

sincoscossin22

22

=−+−−−−=

−+−−=

nt

xyyxnt

τ

θθτθθσστ

MPa 70

MPa 40

MPa 10

MPa 98.5=nσ

MPa 86=tσ MPa 64.19=ntτ

015

nt

4

Slide No. 6

Principal Stresses and Maximum Shearing Stress

Principal Stresses– The transformation equations (Eq. 12 or

13) provides a means for determining the normal stress σn and the shearing stress τnton different planes through a point O in stressed body.

– Consider, for example, the state of stress at a point O of the free surface of a structure or machine component (Fig. 13).

Slide No. 7


Principal Stresses

x

y

z

(b)

ksi 7

ksi 12

ksi 25⋅ o

(a)

⋅

Surfaces perpendicular to z-axisare stress-free.

Figure 13

5

Slide No. 8


Principal Stresses– As the element is rotated through an angle θ about an axis perpendicular to the stress-free surface, the normal stress σn and the shearing stressτnt on different planes vary continuously as shown in Figure 14.

– For design purposes, critical stress at the point are usually the maximum tensile (or compressive) and shearing stresses.

Slide No. 9


Principal Stresses– The principal stresses are the maximum

normal stress σmax and minimum normal stress σmin.

– In general, these maximum and minimum or principal stresses can be determined by plotting curves similar to those of Fig. 14.

– But this process is time-consuming, and therefore, general methods are needed.

6

Slide No. 10


Variation of Stresses as Functions of θ

-20

-10

0

10

20

30

40

0 60 120 180 240 300 360

Angle θ (Degrees)

Stre

ss (K

si)

Stress snStress tnt

nt

n

τσ

ksi 12

ksi 7

ksi 25x

y nt

θ

Figure 14

Slide No. 11ENES 220 ©Assakkaf


Principal Stresses– Principal Stresses for Special Loading

Conditions:• Bar under axial load

• Shaft under Pure BendingA

PAP

2 and maxmax == τσ (14)

JcTmax

maxmax ==τσ (15)

7

Slide No. 12


Principal Stresses for Axially Loaded Bar

θ

P

PF

Original Area, AInclined Area, An

Figure 15

Slide No. 13



PN

Vθ

θ

θ

θθ

cos

sincos

AA

PVPN

n =

==

θ

( )θθ

θ

θσ 2cos12

cos

cos

cos 2 +====A

PAP

AP

AN

nn

Figure 16a

8

Slide No. 14



PN

Vθ

θ

θ

θθ

cos

sincos

AA

PVPN

n =

==

θ

θθθ

θ

θτ 2sin2

cossin

cos

sinA

PAP

AP

AV

nn ====

Figure 16b

Slide No. 15


σn is maximum when θ = 00 or 1800

τn is maximum when θ = 450 or 1350

– Also

Therefore2max

maxστ =

AP

AP

2 and maxmax == τσ

(16)

(17)


9

Slide No. 16


Principal Stresses for Shaft under Pure Torsion

x

y A

x

y

xyτxyτ

yxτ

yxτ

(a)

(b)x

yt

α

α

n

σn dA

τn t dA

α

τyx dA sin α

τ x y

dA

cosα

(c)

Fig.17

Slide No. 17


Principal Stresses for Shaft under Pure Torsion yt

α

n

σn dA

τn t dA

α

τyx dA sin α

τ x y

dA

cosα

( ) ατααττ 2cossincos 22xyxynt =−=

(18)αταατσ 2sincossin2 xyxyn ==

(19)

JcTmax

maxmax ==τσ (20)

10

Slide No. 18


Development of Principal Stresses Equations

Recall Eq 13

θτθσσ

τ 2cos2sin2 xy

yxnt +

−−=

θτθσσσσ

σ 2sin2cos22 xy

yxyxn +

−+

+= (13a)

(13b)

Slide No. 19



Differentiating the first equation with respect to θ and equate the result to zero, gives

( ) 0 2cos22sin

2sin2cos22

=+−−=

+

−+

+=

θτθσσ

θτθσσσσ

θθ

xyyx

xyyxyxn

dd

ddσ

2

tan2or 2

2tan

or

1

−=

−= −

yx

xyp

yx

xyp σσ

τθ

σστ

θ (21)

set

11

Slide No. 20



Substituting the expression for 2θp into Eq. 13a, yields

Eq. 21 gives the two principal stresses in the xy-plane, and the third stress σp3 = σz = 0.

22

2 ,1 22 xyyxyx

pp τσσσσ

σ +

−±

+= (22)

Slide No. 21


Principal StressesPrincipal stresses σmax and σmin can be computed from

where subscript p refers to the planes of maximum and minimum values of σn.

22

2 ,1 22 xyyxyx

pp τσσσσ

σ +

−±

+= (22a)

12

Slide No. 22


Location of the Plane of Principal Stresses

2

tan21 1

−= −

yx

xyp σσ

τθ (22b)

Slide No. 23


Notes on Principal Stresses Equation1. Eq. 22 gives the angle θp and θp + 900

between x-plane (or y-plane) and the mutually perpendicular planes on which the principal stresses act.

2. When tan 2θp is positive, θp is positive, and the rotation is counterclockwise from the x- and y-planes to the planes on which the two principal stresses act.

13

Slide No. 24


Notes on Principal Stresses Equation3. When tan 2θp is negative, θp is negative,

and the rotation is clockwise.4. The shearing stress is zero on planes

experiencing maximum and minimum values of normal stresses.

5. If one or both of the principal stresses from Eq.22 is negative, the algebraic maximum stress can have a smaller absolute value than the minimum stress.

Slide No. 25

Development of Maximum Shearing Stress Equation

Recall Eq 13b: θτθσσ

τ 2cos2sin2 xy

yxnt +

−−=

( ) 0 2sin22cos

2cos2sin2

=−−−=

+

−−=

θτθσσ

θτθσσ

θθτ

xyyx

xyyxnt

dd

dd

2

tan2or 2

2tan

or

1

−=

−−= −

xy

yx

xy

yx

τσσ

θτσσ

θ ττ

set

(23)


14

Slide No. 26

Development of Principal Shearing Stress Equation

Substituting the expression for 2θτ into Eq. 13b, yields

Eq. 24 gives the maximum in-plane shearing stress.

22

2 xyyx

p τσσ

τ +

−±= (24)


Slide No. 27

Maximum In-Plane Shearing StressMaximum in-plane shearing stress can be computed from

where the subscript p refers to the plane of maximum in-plane shearing stress τp.

(24a)22

2 xyyx

p τσσ

τ +

−±=


15

Slide No. 28


Location of the Plane of Maximum Shearing Stress

2

tan21 1

−= −

xy

yx

τσσ

θτ (24b)

Slide No. 29


Notes on Principal Stresses and Maximum In-Plane Shearing Stress Equation

1. The two angles 2θp and 2θτ differ by 900, therefore, θp and θτ are 450 apart.

2. This means that the planes in which the maximum in-plane shearing stress occur are 450 from the principal planes.

16

Slide No. 30


Notes on Principal Stresses and Maximum In-Plane Shearing Stress Equation

3. The direction of the maximum shearing stress can be determined by drawing a wedge-shaped block with two sides parallel to the planes having the maximum and minimum principal stresses, and with the third side at an angle of 450. The direction of the maximum shearing stress must oppose the larger of the two principal stresses.

Slide No. 31

pθ

xyτ

xσ

pθ

yσ

1pσ

2pσ

nσ

maxτ

maxτ

nσ

1pσ

2pσ

21 Pp σσ >

045

045

x

y Wedge-shaped BlockFig.18


17

Slide No. 32


Useful Relationships• The maximum value of τnt is equal to one half

the difference between the two in-plane principal stresses, that is

• For plane stress, the sum of the normal stresses on any two orthogonal planes through a point in a body is a constant or in invariant.

221 pp

p

σστ

−=

yxpp σσσσ +=+ 21

(25)

(26)

Slide No. 33


Useful Relationships– When a state of plane exists, one of the

principal stresses is zero.– If the values of σp1 and σp2 from Eq. 25

have the same sign, then the third principal stress σp3 equals zero, will be either the maximum or minimum normal stresses.

– Three possibilities:( ) ( ) ( ) 2/0max ,2/0max ,2/max 2121 pppp σσσσ −=−=−=

18

Slide No. 34


Example 3Normal and shearing stresses on horizontal and vertical planes through a point in a structural member subjected to plane stress are shown in Figure 19. Determine and show on a sketch the principal and maximum shearing stresses.

Slide No. 35



ksi 4

ksi 12

ksi 6

Fig.19The given values for usein Eqs. 22 and 24 are:

σx = +12 ksiσy = - 4 ksiτxy = - 6 ksi

19

Slide No. 36


Example 3 (cont’d)Using Eq. 22a for the given values:

Therefore,

( ) ( ) 10462

4122

)4(12

22

22

22

±=−+

−−

±−+

=

+

−±

+= xy

yxyxp τ

σσσσσ

0

(C) ksi 6 ksi 6104(T) ksi 14ksi 14104

3

2

1

==

=−=−==+=+=

zp

p

P

σσ

σσ

Slide No. 37


Example 3 (cont’d)Since the σp1 and σp2 have opposite sign, the maximum shearing stress is

The location θp of the principal stresses is computed from Eq. 22b

( ) ksi 102

202

6142

21max +==

−−=

−= pp σσ

τ

( )( )

01 18.43 412

62tan21

=

−−−

= −

yxpθ

20

Slide No. 38


Example 3 (cont’d)Sketch of principal and max shearing stressesksi 6

ksi 12

043.18

ksi 4

ksi 10

ksi 4

ksi 14

ksi 6

045

x

y

ksi 6

maxksi 10 ττ ==p

(T) ksi 42

4122

=−

=+

= yxn

σσσ

Slide No. 39


Example 4Normal and shearing stresses on horizontal and vertical planes through a point in a structural member subjected to plane stress are shown in Figure 20. Determine and show on a sketch the principal and maximum shearing stresses

21

Slide No. 40



MPa 36

MPa 72

MPa 24

Fig.20The given values for usein Eqs. 22 and 24 are:

σx = +72 MPaσy = +36 MPaτxy = - 24 MPa

Slide No. 41

Example 4 (cont’d)Using Eq. 22a for the given values:

Therefore,

( ) ( ) 3054242

36722

)36(72

22

22

22

±=−+

+−

±++

=

+

−±

+= xy

yxyxp τ

σσσσσ

0

(T) MPa 24 ksi 243054(T) MPa 84ksi 843054

3

2

1

==

=+=−==+=+=

zp

p

P

σσ

σσ


22

Slide No. 42

Example 4 (cont’d)Since the σp1 and σp2 have the same sign, the maximum shearing stress is

The location θp of the principal stresses is computed from Eq. 22b

MPa 422

842

0842

01max +==

−=

−= pσ

τ

011 57.263672

)24(2tan212

tan21

−=

−−

=

−= −−

yx

xyp σσ

τθ


Slide No. 43

Example 4 (cont’d)Sketch of principal and max shearing stressesksi 6

MPa 72

057.26

MPa 36

MPa 30

MPa 54

MPa 84

MPa 24

045

x

y

MPa 24

MPa 42

MPa 302

24842

max

21

max

=

=−

=−

=

≠

τ

σστ

ττ

ppp

p

(T) MPa 542

36722

=+

=+

= yxn

σσσ

045

MPa 42

MPa 84

03 =pσ

MPa 42


23

Slide No. 44

Mohr’s Circle for Plane Stress

Introduction– Mohr’s circle is a pictorial or graphical

interpretation of the transformation equations for plane stress.

– The process involves the construction of a circle in such a manner that the coordinates of each point on the circle represent the normal and shearing stresses on one plane through the stressed

Slide No. 45ENES 220 ©AssakkafMaximum Shearing Stress

Maximum shearing stress occurs for avex σσ =′

2

45by fromoffset

and 90by separated angles twodefines :Note

22tan

2

o

o

22

max

yxave

p

xy

yxs

xyyxR

σσσσ

θ

τσσ

θ

τσσ

τ

+==′

−−=

+

−==

24

Slide No. 46


IntroductionPoint, and the angular position of the radius to the point gives the orientation of the plane.

– The proof that normal and shearing components of stress on arbitrary plane through a point can be represented as points on a circle follows from Eqs. 13a and 13b.

Slide No. 47


Plane Stresses using Mohr’s Circle• Recall Eqs. 13a and 13b,

• Squaring both equations, adding, and simplifying gives

θτθσσ

τ 2cos2sin2 xy

yxnt +

−−=

θτθσσσσ

σ 2sin2cos22 xy

yxyxn +

−+

+= (13a)

(13a)

22

22

22 xyyxyx

n ntτ

σστ

σσσ +

−=+

+− (27)

25

Slide No. 48


Plane Stresses using Mohr’s Circle– The previous equation is indeed an

equation of a circle in terms of the variable σn and τnt. The circle is centered on the the σ axis at a distance (σx - σy)/2 from the τaxis, and the radius of the circle is given by

22

2 xyyxR τ

σσ+

−= (28)

Slide No. 49


Plane Stresses using Mohr’s Circle– Normal stresses are plotted as horizontal

coordinates, with tensile stresses (positive) plotted to the right of the origin and compressive stresses (negative) plotted to the left.

– Shearing stresses are plotted as vertical coordinates, with those tending to produce a clockwise rotation of the stress element

26

Slide No. 50


Plane Stresses using Mohr’s Circleplotted above the σ-axis, and those tending to produce counterclockwise rotation of the stress element plotted below the σ-axis.

– Sign conventions for interpreting the normal and shearing stresses will be provided, and illustrated through examples.

Slide No. 51


Plane Stresses using Mohr’s Circle• Mohr’s circle for any point subjected to plane

stress can be drawn when stresses on two mutually perpendicular planes through the point are known.

x

y

n

t

θ

yσ

xσ

xyτ

yxτ

xyτ

yxτxσ

yσ

A

A

θ

27

Slide No. 52


Mohr’s Circle

yxτ

yσ

2pσ

2yx σσ +

xσnσ

1pσ

τ

σθ2

pθ2

pτ

ntτ

xyτ2yx σσ −

),( yxyH τσ

),( xyxV τσ

Fig.21

C

Slide No. 53


Mohr’s Circle

Maximum shearing stress occurs for avex σσ =′

2

45by fromoffset and 90by separated angles twodefines :Note

22tan

2

o

o

22

max

yxave

p

xy

yxs

xyyxR

σσσσ

θ

τσσ

θ

τσσ

τ

+==′

−−=

+

−==

28

Slide No. 54


Plane Stresses using Mohr’s CircleDrawing Procedure for Mohr’s Circle

1. Choose a set of x-y coordinate axes.2. Identify the stresses σx, σy and τxy = τyx and list

them with proper sign.3. Draw a set of στ-coordinate axes with σ and τ

positive to the right and upward, respectively.4. Plot the point (σx, -τxy) and label it point V

(vertical plane).

Slide No. 55


Plane Stresses using Mohr’s CircleDrawing Procedure for Mohr’s Circle (cont’d)

5. Plot the point (σy, τyx) and label it point H(horizontal plane).

6. Draw a line between V and H. This establishes the center and the radius R of Mohr’s circle.

7. Draw the circle.8. An extension of the radius between C and V

can be identified as the x-axis or reference line for the angle measurements (I.e., θ =0).

29

Slide No. 56


Sign Conventions– In a given face of the stressed element, the

shearing stresses that tends to rotate the element clockwise will be plotted above the σ-axis in the circle.

– In a given face of the stressed element, the shearing stresses that tends to rotate the element counterclockwise will be plotted below the σ-axis in the circle.

Slide No. 57


Sign Conventions

σ

τ

(a) Clockwise → Above

σ

τ

(b) Counterclockwise → Below

Fig.22

The following jingle maybe helpful in rememberingthis conventions:“In the kitchen, the clockis above, and the counteris below.”

Beer and Johnston (1992)

σ

τ•

στ

•

30

Slide No. 58


Points of Interests on Mohr’s Circle1. Point D that provides the principal stress

σp1.2. Point E that gives the principal stress σp2.3. Point A that provides the maximum in-

plane shearing stress -τp and the accompanied normal stress σavg that acts on the plane.


Mohr’s Circle for Plane Stress• With the physical significance of Mohr’s circle

for plane stress established, it may be applied with simple geometric considerations. Critical values are estimated graphically or calculated.

• For a known state of plane stressplot the points X and Y and construct the circle centered at C.

xyyx τσσ ,,

22

22 xyyxyx

ave R τσσσσ

σ +

−=

+=

• The principal stresses are obtained at A and B.

yx

xyp

ave R

σστ

θ

σσ

−=

±=

22tan

minmax,

The direction of rotation of Ox to Oa is the same as CX to CA.

31


Mohr’s Circle for Plane Stress• With Mohr’s circle uniquely defined, the state

of stress at other axes orientations may be depicted.

• For the state of stress at an angle θ with respect to the xy axes, construct a new diameter X’Y’ at an angle 2θ with respect to XY.

• Normal and shear stresses are obtained from the coordinates X’Y’.


Mohr’s Circle for Plane Stress• Mohr’s circle for centric axial loading:

0, === xyyx AP τσσ

AP

xyyx 2=== τσσ

• Mohr’s circle for torsional loading:

JTc

xyyx === τσσ 0 0=== xyyx JTc τσσ

32

Slide No. 62

Mohr’s Circle for Plane StressExample 5

The stresses shown in Figure 23 act at a point on the free surface of a stressed body. Use Mohr’s circle to determine the normal and shearing stresses at this point on the inclined plane AB shown in the figure.

ksi 14

ksi 20B

A

125Fig.23

Slide No. 63



The given values for use indrawing Mohr’s circle are:

( ) ksi 89.14 )7.0(31724.45 cos

32

1420radius

ksi 172

1420

76.134512tan22 0

ksi 14

ksi 20

013

2

1

=−=−=

=−

==

=+

=

−=

−===

==

==

−

RC

R

C

n

pz

py

px

σ

θσσ

σσ

σσ

076.13424.45

3=R

σ

τ C

3=Rnσ

( ) ( ) ksi 13.2 24.45sin324.45sin === Rntτ

ntτ

33

Slide No. 64


Example 6

For the state of plane stress shown, (a) construct Mohr’s circle, determine (b) the principal planes, (c) the principal stresses, (d) the maximum shearing stress and the corresponding normal stress.

Slide No. 65



SOLUTION:

• Construction of Mohr’s circle( ) ( )

( ) ( ) MPa504030

MPa40MPa302050

MPa202

10502

22 =+==

==−=

=−+

=+

=

CXR

FXCF

yxave

σσσ

34


Example 6 (cont’d)• Principal planes and stresses

5020max +=+== CAOCOAσ

MPa70max =σ

5020max −=−== BCOCOBσ

MPa30max −=σ

°=

==

1.53230402tan

p

p CPFX

θ

θ

°= 6.26pθ



• Maximum shear stress

°+= 45ps θθ

°= 6.71sθ

R=maxτ

MPa 50max =τ

aveσσ =′

MPa20=′σ

lecture10 mohr's circle

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