ME-603 AEP Stress Function

8/10/2019 ME-603 AEP Stress Function

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ME 603: Applied Elasticity and Plasticity

STRESS FUNCTIONSTRESS FUNCTIONCartesian coordinates

Prof. S.K.Sahoo


2/17

Plane

Stress

1,1 xyyyxxEE

0 yzxzz

Stress-strain

relationship

The equilibrium equations

Field equations

),(),,(),,( yxyxyx xyxyyyxx

0

0

Yyx

Xyx

yxy

xyx

Plane

Strain

0),,(),,( wyxvvyxuu

2)( xyxx G

0

0

Yyx

Xyx

yxy

xyx

212

13

32

GKGK

= Lams constantG = shear modulus

0 yzxzz

0,1

1

yzxzxyxy

yxyxz

E

E

)(2

1

0)(2

1

)(2

1

,,

x

w

z

u

y

w

z

xy

u

z

w

y

v

x

u

xz

yz

xy

zyx

strain-displacement

equations 0,2

)()(

yzxzxyxy

yxyxz

yyxy

G

0

)(2

1

,,

yzxzz

xy

yx

xy

u

yx

u

)21)(1(

E

xyxy

xyy

yxx

E

v

v

v

E

v

v

v

E

v

)1(

)1

(1

)1(

1

2

2

xyxy

xyy

yxx

v

E

vv

E

vv

E

)1(

)(1

)(1

2

2


3/17

Simplified Elasticity Formulations

Displacement Formulation

Eliminate the stresses and strains

from the general system of

equations. This generates a system

Stress Formulation

Eliminate the displacements and

strains from the general system of

equations. This generates a system

A practical elasticity problem may requires to solve 15 equations with 15

unknowns. It is very difficult to solve many real problems, so modifiedformulations have been developed.

unknown displacement components.

Equations are on u, v, w terms.

unknown stress components.

Equations are on

termsF(z)

G(x,y)

z

x

y Even using displacement or stress formulations

three-dimensional problems are difficult to solve.

So most solutions are assumed to of two-dimensional problems

],,,,,[ zxyzxyzzyyxx


4/17

Displacement Formulation

0)1(2

0)1(2

2

2

Yy

v

x

u

yv

EvG

Xy

v

x

u

xv

EuG

Plane stress

+

Plane strain

+

0)(

0)(

2

2

Yy

v

x

u

yGvG

Xy

v

x

u

xGuG

),(),,( yxvvyxuu bb

(B.C.)

),(),,( yxvvyxuu bb

(B.C.)

)21)(1(

21

2

13

3

2

E

GKGK

= Lams constant

G = shear modulus


5/17

Stress Formulation

222

0

0

Yyx

Xyx

yxy

xyx

+

Plane stress Plane strain

0

0

Yyx

Xyx

yxy

xyx

222 +

yxxy

xyyx

22

y

YxXvyx )1()(

2

or

+

y

YxX

vyx

)1(1)(2

yyxy

yxxx

mlY

mlX

yyxy

yxxx

mlY

mlX

(B.C.)

or

yxxy

xyyx

22

+


6/17

Stress Function Approach

0

0

Yyx

Xyx

yxy

xyx

yxxy

xyyx

2

2

2

2

2

Solution of plain problems:

y

Y

x

X

vyx

1

1)(2

Plain Strain

y

Y

x

X

vyx )1()(

2

Plain Stress

0)(2

2

2

2

yxyx

3 different equations

with 3 unknowns

Airy (An English Mathematician) for2Dproblems proposes a single stress

function (function of space coordinates)that will satisfy all equations

yyxy

yxxx

mlY

m

When No body force

0)(2 yx or

Philosophy:Stress = f (strain)

Strain = f (displacement)

= f (space coordinate)= f (x, y, z)

=> Stress = f (x, y, z)


7/17

Airy Stress Function Method

Plane Problems with No Body Forces

0

0

yx

yx

yxy

xyx

Stress Formulation

yxxy xyyx

2

2

2

2

2

,,

Airy Representation

Let assume a stress function =(x,y)

such that:

Putting it on equilibrium equations, it

can be checked that both equations are

0)(2 yx

02 44

4

22

4

4

4

yyxx

Biharmonic Governing Equation

(Single Equation with Single function satisfy equilibrium and

compatibility equations. The unknown parameters of stress function canbe find out by putting boundary conditions. )

0)(2

2

2

2

yx

yx

or

.

Putting on compatibility equation

gives


8/17

Steps of the Stress Function Approach

1. Choose a stress function,

2. Confirm is biharmonic (compatibility is satisfied)

3. Derive stresses from second derivatives of4. Use boundary conditions to identify unknown parameters of

5. Derive strains from stress-strain relationships

6. Integrate strain-displacement relations to get displacements


9/17

Example of biharmonic Polynomial Stress Functions

2

0211

2

20011000

0 0

),( yAxyAxAyAxAAyxAyxm n

nm

mn

terms do not contribute to the stresses and are therefore dropped1 nmterms will automatically satisfy the biharmonic equation3 nmterms require constantsAmn to be related in order to satisfybiharmonic equation

3 nm

cybyax ,If

04

stress function have no meaning, as it gives zero stress in

22If, cybxyax

...........,If 2234222322 ykxyjxhxgyfxyyexdxcybxyax

cy

xx 2002

2

0022

2

ax

yy

00

2

b

yxxy

Stress distribution can be obtained but

all to satisfy biharmonic equation,which requires that there exist some

relations among coefficients.

..........2006200200 2kxgyfxcxx

...........26120026002 22 kyjxyhxeydxayy

..........430022000 2 kxyjxfyexbxy

.

Any quadratic function of x and y will automatically satisfy the biharmonic equation

but give constant values of stresses.


10/17

Stress functionswithbody forces(Plane strain)

0

Xzyx

zxxyxx

0

Y

zyx

yzyyxy

0

Z

zyx

zzyzzx

0

Xyx

xyxx

0

Y

yx

yyxy

0

z

zz

We have general equilibrium equation; For plane strain condition;

The com onent of bod force X and Y

The third equation will be

dropped from the

discussion for the time

being, since its solution

only indicates

, which value may later be

determined as a function

of

yxfzz ,

yyxx &

independent Z & Z=0. The body force can be

expressed by a potential function , so that; xX

y

Y

,0

yx

xy

xx

0

yy

xy

yy

So equilibrium equation becomes,

2

2

yxx

yx

xy

2

2

2

xyy

If we now choose a stress functionso that,

We can verify that the equilibrium equations are

identically satisfied. For complete solution, we

must consider the compatibility condition, i.e.

y

Y

x

X

vyyxx

1

1)(2


11/17

)())(( 2

2

2

2

2

2

2

2

2

2

2

2

1

1

yxxyyx

Stress functionswithbody forces(Plane strain)..

xyyyxx &,Now substituting the in the terms of stress function. This will

result a differential equation will be such that any solution to it will satisfy

both equilibrium & compatibility condition.

Simplifying,

It gives,

01

122 )()()( 2

2

2

2

22

4

4

4

4

4

2

2

2

2

yxyxyxyx

021222444

If there is no body force,

yxyxyx

01

21 222

0

1

21 24

04

This gives the condition of stress function equation for the solution of

problem of plane strain, provided the proper boundary conditions are also

satisfied.


12/17

Stress functionswithbody forces(Plane stress)

0

Xzyx

zxxyxx

0

Y

zyx

yzyyxy

0

Z

zyx

zzyzzx

0

Xyx

xyxx

0

Y

yx

yyxy

We have general equilibrium equation; For plane stress condition;

The com onent of bod force X and Y

independent Z & Z=0. The body force can be

expressed by a potential function , so that; xX

y

Y

,0

yx

xy

xx

0

yy

xy

yy

So equilibrium equation becomes,

2

2

yxx

yx

xy

2

2

2

xyy

If we now choose a stress functionso that,

We can verify that the equilibrium equations are

identically satisfied. For complete solution, we

must consider the compatibility condition, i.e.

y

Y

x

Xvyyxx )1()(

2


13/17

)())(( 2

2

2

2

2

2

2

2

2

2

2

2

)1(yxxyyx

Stress functionswithbody forces(Plane stress)..

xyyyxx &,Now substituting the in the terms of stress function. This will

result a differential equation will be such that any solution to it will satisfy

both equilibrium & compatibility condition.

Simplifying,

It gives,

0)1(22 )()()( 2

2

2

2

22

4

4

4

4

4

2

2

2

2

yxyxyxyx

0)1(222444


yxyxyx

0)1( 222 0)1( 24

04

This gives the condition of stress function equation for the solution of

problem of plane stress, provided the proper boundary conditions are also

satisfied.


14/17

Summary: Solutions to Plane Problems in Cartesian Coordinates

Airy Representation for stresses444

Biharmonic Governing Equation

0)1( 24 0

1

21 24


Plane Stress Plane Strain

yxxy xyyyxx

2

2

2

2

2

,,02 4

4224

yyxx

),(,),( yxfYyxfX yx

Boundary Conditions

RS

x

y

yyxy

yxxx

mlY

mlX


15/17


16/17

22 ByAx 3Ax 224 3 yxxA

Problems:Prove that the followings are Airys stress function & examine the stress

distribution represented by them: , ,

22 ByAx ,2Ax

x

Byy

2

Ax 22

2

By

22

2

04

4

x

04

4

y

022

4

yx

00002

4

4

22

4

4

44

yyxx

0,2,22

2

2

2

2

yxA

xB

y xyyyxx

Ans:

Hence it is a Airys stress function.

It is a uniform two-dimensional stress function.422 3 4 4

,

,

,3 2Axx

042

yyy

Axx

62

A

x6

3

04 x04 0,2,0 xyyxx Ax One-dimensional linear stress field.

022

yx

23 64 xyxAx

yAxyxAy

22 66

2

2

2

612 yxAx

222

2

66 AxxAy

Ax

x24

3

3

04

4

y

Ax

244

4

Axyyx

122

Ayyx

122

3

A

yx12

22

4

0024244 AA

26Axxx

22 612 yxAyy Axyxy 12So it is a Airys stress function, and

c)


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ME-603 AEP Stress Function