ANALYSIS OF STRESS AND STRAIN II

MOHR’S CIRCLE FOR PLANE STRESS

If we were only interested in obtaining the normal stress and the shear stress (s, t) on a single

surface, say the x' surface than s and t would be equal to

tssttsss 2222 sincoscossin,sincos2sincos xyyxxyyx

These equations can be rewritten using the trigonometric identities

2cossincos2sin21

cossin

22cos1

sin2

2cos1cos

22

22

So s and t can be written as

tss

ttssss

s 2cos2sin2

,2sin2cos22 xy

yx

xy

yxyx

These are parametric equations which represents a circle in s, t space where is the

parameter. Rearranging the terms give

2

2

2

2

22 xy

yxyxt

sst

sss

x

yR

2

2

2 xy

yxR t

ss

This equation represents the equation of a circle in s, t space where

the radius is

and the center of the circle is displaced in the positive s direction by the average of the

normal stress

2

yx

ave

sss

and 222

Rave tss

From the circle it can be observed that the

maximum and minimum values of s and t are

2

2

minmax 22 xy

yxyx

ave R tssss

ss

and 2

2

minmax 2 xy

yxt

sst

It is important to note that the maximum and

minimum values of s and t determined here may

not be the absolute maximums or minimums. Figure 1 Mohr’s circle for plane stress

If one were to plot the circle graphically with the coordinate axes s and t positive to the right

and up, respectively, the circle would graph 2 in the clockwise direction as increased in the

counterclockwise direction, Figure 1. This opposing direction can lead to errors, therefore,

plotting positive t downwards can prevent errors.

THREE DIMENSIONAL STRESS TRANSFORMATION SIMPLIFIED

The total shear stress acting on a surface depends only on the

particular surface as defined by the surface normal (Figure 2).

If we redefine the surface normal by the n axis, where the

direction cosines of the normal are

zz

yy

xx

n

n

n

cos

cos

cos

The normal stress will be equal to

xzzxzyyzyxxyzzyyxx nnnnnnnnn tttssss 222222

The total shear stress can be found from a summation of forces. The total force in the x

direction due to the stresses on the orthogonal surfaces is

zzxyxyxxx AAAF tts zzyyxx nAAnAAnAA 000 ,, and

zzxyxyxxx nnnAF tts 0

Likewise the forces in the y and z directions due to stresses on the orthogonal surfaces are

zzyyzxzxz

zyzyyxxyy

nnnAF

nnnAF

stt

tst

0

0

Figure 2 Mohr’s circle for plane stress

n

x

y

z

x

y

z

xs

zst

ys

xytyztzxt

ts

The magnitude of t can be found readily since for equilibrium the magnitude of the forces on the

oblique surface must balance the magnitude of forces on the orthogonal surfaces

. t is obtained as

2/122

0 ts A

2/1222

zyx FFF

2/12222

sstttstttst zzyyzxzxzyzyyxxyzzxyxyxx nnnnnnnnn

The direction cosines establishing the direction of t can be defined as tx, ty and tz. For equilibrium in x

direction

000 xxx FnAtA st

Solving for tx

x

xx n

A

Ft s

t 0

1Substituting for Fx gives

zzxyxyxxx nnnt ttsst

1

Likewise for ty and tz

zzyzyxzxz

zyzyyxxyy

nnnt

nnnt

ssttt

tsstt

1

1

n

x

y

z

x

y

z

xs

zst

ys

xytyztzxt

ts

Example

The state of stress at a particular point relative to the xyz coordinate system is given by the

stress matrix

Determine the normal stress and the magnitude and direction of the shear stress on a surface

intersecting the point and parallel to the plane given by the equation

MPa

3507

0107

7714

s

932 zyx

Example 2.1

A point in a state of plane stress is isolated by three surfaces as shown. Determine the values

of s and t using transformation equations and the Mohr’s circle.

The case cannot happen since

PRINCIPLE STRESSES

Principle stresses exist on mutually orthogonal surfaces that contain no shear stress. The

three principle stresses are denoted as s1, s2 and s3, as s1 > s2 > s3. If two of the principle

stresses are equal, there will exist an infinite set of surfaces containing these principle

stresses, where the normals of these surfaces are perpendicular to the direction of the third

principle stress. If all three principle stresses are equal, a hydrostatic state of stress exists and

regardless of orientation, all surfaces contain the same principle stress with no shear stress. If

we assume that on the oblique surface the shear stress

t is zero, than s will be a principle stress denoted by sP.

The components of the force on the oblique surface in

x,y and z directions are

For equilibrium etc. This results in

zPz

yPy

xPx

nAF

nAF

nAF

0

0

0

s

s

s

,0 xx FF

0

0

0

zPzyyzxzx

zyzyPyxxy

zzxyxyxPx

nnn

nnn

nnn

sstt

tsst

ttss

0 zyx nnn 1222 zyx nnn

n

x

y

z

x

y

z

xs

zst

ys

xytyztzxt

ts

This is a cubic equation in the unknown sP, where three solutions result in the three

principle stresses s1, s2 and s3. It can be shown that the principle stresses exist on surfaces

for which the normal stress is stationary. This means that at least one of the principle

stresses is a maximum and at least one is a minimum.

The direction cosines nx, ny and nz associated with a specific principle stress will not be

independent of each other.

In order to avoid the zero solution of the direction cosines the determinant of the coefficients

nx, ny and nz is equated to zero.

Resulting in

02 222

22223

xyzzxyyzxxyzxyzzyx

PxyzxyzxzzyyxPzyxP

tstststttsss

stttsssssssssss

0

Pzyzzx

yzPyxy

zxxyPx

sstt

tsst

ttss

Example

For the stress matrix given below, determine the principle stresses and the direction cosines

associated with the normals to the surfaces of each principle stress.

MPa

021

201

113

s

Stress Invariants

The solutions of sP are independent of the coordinate system used to define the coefficients of

the cubic equation of sP. Therefore the coefficients of sP are constant and are referred to as

the stress invaraints.

02 222

22223

xyzzxyyzxxyzxyzzyx

PxyzxyzxzzyyxPzyxP

tstststttsss

stttsssssssssss

zyzxz

yzyxy

xzxyx

xyzzxyyzxxyzxyzzyx

xyzxyzxzzyyx

zyx

PPP

I

or

I

I

I

III

stt

tst

tts

tstststttsss

tttssssss

sss

sss

3

222

3

222

2

1

12

2

1

3

2

0

MOHR’S CIRCLE IN THREE DIMENSIONS

An element undergoing a general state of three dimensional stress can be

transformed to an element containing only principle stresses s1, s2 and s3

acting along axes 1, 2 and 3, Figure 3. A transformation of stresses in the

12 plane depends only on s1 and s2 ; in the 23 plane depends only on s2

and s3; in the 31 plane depends only on s3 and s1. This means that for

each case, a plane stress analysis describes the state of stress in each of

the three planes and three Mohr’s circles can be constructed to portray

each case shown in Figure 4 (a). Furthermore, it will be shown that all

possible states of stress (s, t) exist either on the circles or within the

shaded are shown in Figure 4 (b).

Figure 3 Principle stress state

For the element in Figure 3 let n1, n2 and n3 be the

direction cosines of an arbitrary surface relative to the 1,

2 and 3 axes, respectively. We can write,

12

3

2

2

2

1

222

3

22

2

22

1

2

2

33

2

22

2

11

321

nnn

nnn

nnn

stsss

ssss

Figure 4 3D Mohr’s circle

These equations can be arranged to form three equations where each equation is

written exclusively in terms of each direction cosine. This can be done by solving the

above simultaneous equations for and . Using Cramer’s rule

Expanding each determinant with respect to the third row

2

3n

Figure 4 3D Mohr’s circle

2

2

2

1 , nn

111

111

2

3

2

2

2

1

321

2

3

2

2

2232

2

1

sss

sss

ssst

sss

n

2

12

2

21

2

13

2

31

2

23

2

32

22

2

2

2

22

3

2

3

2

23

2

322

1ssssssssssss

stsssstsssssss

n

Reducing to 323121

2

23

22

3232232

1

1sssssssss

stsssssss

n

and

3121

2

322

1 ssss

tssss

n

12

3

2

2

2

1

222

3

22

2

22

1

2

2

33

2

22

2

11

321

nnn

nnn

nnn

stsss

ssss

Cross multiplication results in

In a similar manner solving for n2 and n3, the following equations are obtained

Figure 4 3D Mohr’s circle

The equation for the Mohr’s circle connecting s2 and s3 can be obtained as

and

3121

2

1

2

32 sssstssss n

The first term on the left side can be written as

2

32

2

3232 22

sssssssss

2

323121

2

1

2

2

32

22

sssssst

sss n

2

213231

2

3

2

2

21

2

132123

2

2

2

2

13

22

22

sssssst

sss

sssssst

sss

n

n

2

23

2

322

2

32

22R

sst

sss

R23 is the radius of the s2 , s3 circle. Other radii can be determined similarly.

Maximum Shear Stress

Based on the observations on Mohr’s circles in three dimensions the

maximum and minimum shear stresses are found from the radius of the

largest circle (s1, s3) as

231

minmax

sst

STRAIN TRANSFORMATIONS

The strain transformation equations are similar to those for stresses.

Normal strains are obtained as

xzzzzxzzyzyzyzxzxyzzzyzyxzxz

xyzyzxzyyyyzyyxyxyzyzyyyxyxy

xxzxzxzxyxyzyxxxxyzxzyxyxxxx

nnnnnnnnn

nnnnnnnnn

nnnnnnnnn

222

222

222

Shear strains are obtained as

xzzxzzxxzx

yzzxzzyx

yz

xzyxyzxx

xy

zzzxzyzyxyxzxxxxz

xzzyzzxyzx

yzzyzzyy

yz

xzyyyzxy

xy

zzzyzyzyyyxzxyxzy

xyzxzyxxzx

yyzxzyyx

yz

xyyxyyxx

xy

zyzxzyyyxyxyxxxyx

nnnnnnnn

nnnnnnnnnn

nnnnnnnn

nnnnnnnnnn

nnnnnnnn

nnnnnnnnnn

22

2

22

2

22

2

x

y

z

xAzA

yA),0,0( cC

)0,0,(aA

)0,,0( bB

yx

xx

zx

0A

PRINCIPAL STRAINS

Based on the observation of the similarity of the stress and strain transformation

equations, the equations for principle strains will be analogous to the principle stress

equations. In these equations it is necessary to replace s and t with and /2, respectively.

The principle strains will occur on the same axes on which the principle stresses exist

provided the material is isotropic.

EQUILIBRIUM EQUATIONS

Consider a small infinitesimal element of a body of dimensions ∆x, ∆y and thickness t = 1

subjected to stresses varying over distances ∆x and ∆y as shown in Figure 5. X and Y are

body forces per unit volume in x- and y-directions.

Figure 5 A small infinitesimal element in equilibrium

Volume of the body = ∆x × ∆y × 1

Taking the summation of forces in x-direction

or

011

11

yxXxyxy

x

yyxx

y

yx

yx

yx

xx

x

tt

t

ss

s

0

yxXyx

yyx

x

yxxts

Simplifying further

Taking the summation of forces in y-directions

Simplifying further

0

0111

1111

yxYyxx

yxy

or

yxYyyxx

yxyxy

x

xyyy

xyxy

xyyyyy

yy

t

s

tt

tss

s

0

X

yx

yxxts

0

Y

xy

xyy ts

In three-dimensional case equilibrium equations can be written as

where X, Y, Z are body forces per unit volume.

In these equations, mechanical properties have not been used. So, these equations are

applicable whether a material is elastic, plastic or viscoelastic.

0

0

0

Zzyx

Yzyx

Xzyx

zyzxz

zyyxy

zxyxx

stt

tst

tts

In a two-dimensional case, equilibrium equations are

where body forces are zero.

0

0

yx

yx

yxy

yxx

st

ts

Airy’s Stress Function

We may permanently satisfy these equations by expressing stresses in terms of a function

ϕ, called Airy’s stress function, as follows:

In a plane stress case a body subjected to stresses σx, σy, τxy, the strains are

yxxyxyyx

t

s

s

2

2

2

2

2

,,

2

2

2

2

2

2

2

2

2

2

2

21

,1

yxEE

yxEEExyEEE

yxz

xy

y

yxx

ss

ss

ss

Shear strain,

yxGG

xy

xy

t

21

Compatibility Equations

When one is seeking a solution to the stress distribution in a body, the dynamic state of the body

must be satisfied. For example, if the body is in equilibrium, any segment of the body together with

its internal-force distribution must maintain the segment in static equilibrium. At any given section it

is possible to find many stress distributions which will ensure equilibrium. An acceptable stress

distribution is one which ensures a piecewise-continuous-deformation distribution of the body. This

is the essential characteristic of compatibility, i.e., the stress distribution and the resulting deflection

distribution must be compatible with boundary conditions and a continuous distribution of

deformations so that no “holes” or overlapping of specific points in the body occur.

Strain compatibility equation,

yxxy

xyyx

2

2

2

2

2

Substituting the values of εx, εy, xy we get

Putting the value of in the above equation, we get

Simplifying further the equation becomes

a Biharmonic equation

or ∇4 ϕ = 0

Airy’s stress function ϕ chosen for any problem must satisfy the above Biharmonic equation.

22

4

22

4

4

4

22

4

4

411

yxGyxxyxyE

12GE

22

4

22

4

4

4

4

4

122yxyxyx

02

22

4

4

4

4

4

yxyx

εx = 6 + x2 + y2 + x4 + y4

εy = 4 + 3x2 + 3y2 + x4 + y4

xy = 5 + 4xy(x2 + y2 + 2)

= 5 + 4x3y + 4xy3 + 8xy

Example The following strains are given

Determine whether the above strain field is possible. If it is possible determine

displacement components u and v, assuming u = v = 0 at origin.

SECOND DEGREE POLYNOMIAL

Let us consider an Airy’s stress function

∇4 ϕ = 0 satisfies the compatibility condition

stress in x direction

stress in y direction

Shear stress,

(negative, tends to rotate the body in counterclockwise

direction). This represents a plane stress condition as

shown in Figure 6.

Figure 6 Plane stress condition

Ax

y

2

2s

22

22y

CBxyx

A

Cy

x

2

2s

Byxxy

t

2

02

22

4

4

4

4

4

yxyx

compatibility condition

Example

Airy’s stress function ϕ = 40 x2−30 xy + 60 y2 satisfies the compatibility

condition ∇4 ϕ = 0. Determine stresses σx, σy and txy, show graphically the stress

distribution. Stresses are in MPa.

A BEAM SUBJECTED TO PURE BENDING

A beam of depth d is subjected to pure bending moment M and no shear force, as shown in Figure

7. Airy’s stress function can be a third degree polynomial:

Figure 7 Beam subjected to pure bending

∇4 ϕ = 0, for this function

Taking B = 0 as stress σx is independent of x, σx = Ay

y varies from − to + as shown (Figure 8).

Figure 8 Stress distributions

Maxm. σx in tension

where

x

y 3223

6226x

Dyx

Cxy

By

A

002

2

BxAy

yx

s

2d

At s

2d

IM

IM

A

Maxm.σxx in compression

where M = bending moment M, and

I = second moment of area (of cross section of beam) about the neutral axis.

Now

but σy = 0, so constant C = D = 0

Shear stress

but

also because it is a case of pure bending.

So, constants B = C = 0

Finally Airy’s stress function is

.

22d

IMd

Ac s

DxCyx

y

2

2s

CxByyxxy

t

2

0xyt

3

6y

A

Example A bar of circular section of diameter 30 mm is subjected to a pure bending moment of 3 × 105 Nmm. What is the maximum bending stress developed in the beam? What is Airy’s stress function for this case?

Example 2.2 For a beam of rectangular section B = 20 mm, D = 30 mm, Airy’s stress function ϕ = 1.6 y3.

Determine the magnitude of constant bending moment acting on beam section.

Example A beam of rectangular section is subjected to shear force F = 1 kN and bending moment = −1 × 106 Nmm. Section of beam is B = 20 mm, D = 60 mm. Write down the stress tensor for an element located at 15 mm below the top surface .

M = −1 × 106 Nmm (producing convexity)

I = second moment of area about neutral plane

y = 15 mm from neutral axis. Bending stress,

4433

103612

602012

mmBD

I

2

4

6

/66.41151036

101mmNy

IM

x

s

To determine shear stress at y = 15 mm from neutral layer

where area a = 15 × 20 = 300 mm2 = area of section above the layer under consideration,

ӯ = 22.5 mm distance of CG of area a from neutral layer,

b = breadth = 20 mm,

I = 36 × 104 mm4,

F = 1 kN = 1000 N,

txy

=

= 0.9375 N/mm2.

Stress tensor for this state of stress, at a layer 15 mm below top surface

Ib

yFaxy t

201036

5.2230010004

MPa09375.0

9375.066.41

Example A cantilever beam of rectangular section B × D and of length L carries a

concentrated load W at free end. Consider a section at a distance x from fixed end and a layer at a distance y from neutral layer zz; derive expression for Airy’s stress functions, if B=40 mm, D=60 mm, W=1 kN,

x=2 m, L=5 m. Write the stress tensor for the layer bc, if y = 20 mm.

Stresses

Let us assume Airy’s stress function ϕ = C1y3 + C2 xy3 + C3 xy.

Boundary conditions are

For τxy = 0, σyy = 0

, shear force is constant

For x = L, σxx = 0 Everywhere

0

y

xx I

yxLW

s

s

yD

I

W

yD

yD

I

W

yD

yyD

BBI

W

z

z

zxy

42

22

22

2

2

2t

2/

2/

D

D

xy WdyBt

xx I

yMs

Applying these boundary conditions

everywhere ϕ satisfies this condition

So, or

02

2

xy

s

2

3

2

2

2

2

30 DyD

yxy CyCyx

or

t

04

3 3

2

2 CD

C2

23 43

DCC

LCC

xyCyCy

Lx

Lx

Lxx

21

212

2

066

s

ϕ = C1y3 + C2 xy3 + C3 xy

Moreover

or

Putting the value of into above equation 2

234

3DCC

DCD

CB

DDC

DDCBW

yCy

CBW

WBdyCyC

WBdyyx

dyB

D

D

D

D

D

D

D

D

xy

3

3

2

3

33

2

2/

2/

3

3

2

2/

2/

3

2

2

2/

2/

22/

2/

4

2288

33

2

t

WBDCD

BC 3

4

2 4

32

33

2

2

2

3

2

2

41

43

43

4

BD

WC

WBDBDC

WBDDCD

BC

ϕ = C1y3 + C2 xy3 + C3 xy

Then

Finally, Airy’s stress function is

At the layer where x = 2000 mm, L = 5000 mm

and

Stress tensor

BDW

BD

WDC

432

43

3

2

3 321

2

BD

WLLCC

xyBDW

xyBD

Wy

BD

WL4322 3

3

3

3

MPa

I

yxLW

mmBD

I

mmy

xx

z

33.831072

2030001000

107212

604012

20

4

4433

s

MPa

yD

I

W

zxy

347.010144

5001000

204

60

10722

1000

42

4

22

4

22

t

MPa0347.0

347.033.83

Example 2.3 A solid circular shaft of steel is transmitting 100 hp at 200 rpm. Determine shaft diameter if the maximum shear stress in shaft is not to exceed 80 MPa. Write down stress tensor for the surface of the shaft. Show that surface of the shaft is under both plane stress and plane strain conditions E = 200 ×

103 N/mm2, = 0.3.

Hint: HP transmitted = 100

Revolutions per minute, N = 200

angular speed,

Torque transmitted,

sradN

/94.2060

2002

60

2

mm 106.3562T

m 6.356294.20746100

3

N

NT

Example 2.4 A cylindrical bar of length L, area of cross section A is fixed at top end. Write down Airy’s stress function for stress due to self weight in bar, if w

is the weight density of the bar.