CH.3. COMPATIBILITY EQUATIONS - PRESENTACIOmmc.rmee.upc.edu/documents/Slides/Ch3_v13.pdf · exists a scalar function (named ... xy S zx S yz ... The compatibility equations can be

CH.3. COMPATIBILITY EQUATIONSContinuum Mechanics Course (MMC) - ETSECCPB - UPC

Overview

Compatibility Conditions

Compatibility Equations of a Potential Vector Field

Compatibility Conditions for Infinitesimal Strains

Integration of the Infinitesimal Strain Tensor

Integration of the Deformation Rate Tensor

24/10/2014MMC - ETSECCPB - UPC2

Ch.3. Compatibility Equations

3.1 Compatibility Conditions


Introduction

24/10/2014MMC - ETSECCPB - UPC

Given a displacement field, the corresponding strain field is found:

Is the inverse possible?

, tU X

, tu x

1 , 1,2,32

ji k kij

j i i j

UU U UE i jX X X X

1 , 1,2,32

jiij

j i

uu i jx x

, tx , tu x

4



Given an (arbitrary) symmetric second order tensor field, , a displacement field, , fulfilling cannot always be obtained:

For to match a symmetrical strain tensor: It must be integrable. There must exist a displacement field from which it comes from.

, tx , tu x ( , ) ,s t tu x x

1 , 1,2,32

jiij

j i

uu i jx x

6 PDEs3 unknowns

OVERDETERMINED SYSTEM

, tx

COMPATIBILITY CONDITIONS must be satisfied

REMARKGiven , there will always exist an associated strain tensor, , obtainable through differentiation, which will automatically satisfy the compatibility conditions.

, tx , tu x

5



The compatibility conditions are the conditions a symmetrical 2nd

order tensor must satisfy in order to be a strain tensor and, thus, exist a displacement field which satisfies:

They guarantee the continuity of the continuous medium during the deformation process.

t,XE

Incompatible strain field

1 , 1,2,32

jiij

j i

uu i jx x

6


3.2 Compatibility Equations of a Potential Vector Field


Preliminary example: Potential Vector Field


A vector field will be a potential vector field if there exists a scalar function (named potential function) such that:

Given a continuous scalar function there will always exist a potential vector field .

Is the inverse true?

, tv x , t x

, ,

,v , 1,2,3i

i

t t

tt i

x

v x x

xx

, t x , tv x

, tv x , t x such that , ,t t x v x

8

Potential Field


In component form,

Differentiating once these expressions with respect to :

, tv x , t x such that , ,t t x v x

, ,v , v , 0 1,2,3i i

i i

t tt t i

x x

x xx x 3 eqns.

1 unknown

OVERDETERMINED SYSTEM

ix

2 ,v , 1,2,3i

j i j

ti j

x x x

x9 eqns.

9

Schwartz Theorem


The Schwartz Theorem or equality of mixed partial derivatives guarantees that, given a continuous function with continuous derivatives, the following holds true:

1 2, ,..., nx x x

2 2

,i j j i

i jx x x x

10

Compatibility Equations


Considering the Schwartz Theorem,

In this system of 9 equations, only 6 different 2nd derivatives of the unknown appear:

They can be eliminated and the following identities are obtained:

2 2 2

2

2 2 2

2

2 2 2

2

v v v

v v v

v v v

x x x

y y y

z z z

x x y x y z x z

x y x y y z y z

x z x y z y z z

v vv v v vy yx x z z

y x z x z y

, t x2

2

x

2

2

y

2

2

z

yx 2

zx 2

zy 2

, , , , and

11



A scalar function which satisfies will exist if the vector field verifies:

, t x , tv x

, ,t t x v x

y

z

v v 0

v v 0

vv 0

defx

z

defx z

y

defy

x

Sx y

Sz x

Sy z

1 2 3ˆ ˆ ˆ

v v v

x

y

zx y z

SS

x y zS

e e e

S vwhere

vv 0 , 1,2,3ji

j i

i jx x

v 0INTEGRABILITY(COMPATIBILITY)

EQUATIONS of a potential vector field

REMARKA functional relation can be established between these three equations.

0 v

12


3.3 Compatibility Conditions for Infinitesimal Strains


Infinitesimal Strains


The infinitesimal strain field can be written as:

1 12 2

12

yx x x z

xx xy xzy y z

xy yy yz

xz yz zz

z

uu u u ux y x z x

u u uy z y

usymmetricalz

6 PDEs3 unknowns

14

Infinitesimal Strains


The infinitesimal strain field can be written as:

6 PDEs3 unknowns

10 02

10 02

10 02

yx xxx xy

y x zyy xz

yz zzz yz

uu ux y x

u u uy z x

uu uz z y

The system will have a solution only if certain compatibility conditions are satisfied.

15



The compatibility conditions for the infinitesimal strain field are obtained through double differentiation (single differentiation is not enough).

2

2 2 2

2

2 2 2

, , , , ,

12

, , , , ,

xxx

y zyz

ux

x y z xy xz yz

u uz y

x y z xy xz yz

6 equations

6 equations

6x6=36 equations

16


The compatibility conditions for the infinitesimal strain field are obtained through: 2 32 3 3

2 3 2 2 2

2 32 3 3

2 2 2 2 3

2 32 3 3

2 2 2 3 2

2 32 3

2

12

12

1...2

12

yz yxx x z

yz yxx x z

yz yxx x z

yz yxx x

uu ux x x z x y x

uu uy x y y z y y

uu uz x z z z y z

uux y x y x y

3

2

2 32 3 3

2 2

2 32 3 3

2 2

12

12

z

yz yxx x z

yz yxx x z

uz x y y x

uu ux z x z x z z x y x z

uu uy z x y z y z z y y z

, ,xx yy zz 18 equations for

, ,xy xz yz 18 equations for

17 24/10/2014MMC - ETSECCPB - UPC



All the third derivatives of and appear in the equations:

which constitute 30 of the unknowns in the system of 36 equations:

3

3 2 2 3 2 2 3 2 2

3

3 2 2 3 2 2 3 2 2

3

3 2 2 3 2 2 3 2 2

, , , , , , , , ,

, , , , , , , , ,

, , , , , , , , ,

x

y

z

ux x y x z y y x y z z z x z y xyz



,x y zu u u

10 derivatives

10 derivatives

10 derivatives

23

, 1, 2,...,36ijin

j k l k l

uf nx x x x x

3018



Eliminating the 30 unknowns , , 6 equations are obtained:

2 22

2 2

2 22

2 22 22

2 2

2

2

2 0

2 0

2 0

0

defyy yzzz

xx

defxx xzzz

yy

defyy xyxx

zz

defyz xyxzzz

xy

defyy yz xxz

xz

Sz y y z

Sx z x z

Sy x x y

Sx y z x y z

Sx z y x y

2

0

0

y

defyz xyxx xz

yz

z

Sy z x x y z

3i

j k l

ux x x

COMPATIBILITY EQUATIONS

for the infinitesimal strain tensor

S ε 0

19



The six equations are not functionally independent. They satisfy the equation,

In indicial notation:

S ε 0

0

0

0

xyxx xz

xy yy yz

yzxz zz

SS Sx y z

S S Sx y z

SS Sx y z

20



The compatibility equations can be expressed in terms of the permutation operator, .

Or, alternatively:

, 0ml mjq lir ij qrS e e

ijke 1ijke

1ijke

, , , , 0 , , , 1,2,3ij kl kl ij ik jl jl ik i j k l

REMARKAny linear strain tensor (1st order polynomial) with respect to the spatial variables will be compatible and, thus, integrable.

21


3.4 Integration of the Infinitesimal Strain Tensor


Preliminary Equations


Rotation tensor :

Rotation vector :

,tΩ x

,tθ x

1( ) ( )2

1 , 1, 2,32

jiij

j i

skew

uu i jx x

Ω u u u

23

1 23 3 2

2 31 3 1

3 12 2 1

01 ( ) 02

0

yz

zx

xy

u



Differentiating with respect to :

Adding and subtracting the term :

kx

1 12 2

j ij ji iij

j i k k j i

u uu ux x x x x x

212

k

i j

ux x

2 21 1 12 2 2

1 12 2

ij ji k k

k k j i i j i j

j jki k k ik

j k i i k j j i

uu u ux x x x x x x x

uu u ux x x x x x x x

,tx

jkik

24

1

11

1

yz xyxz

yz yz yy

yz zyzz

x x y z

y y y z

z z y z



Using the previous results, the derivative of is obtained:

2

22

2

zx xx xz

xy yzzx

zx xz zz

x x z x

y y z x

z z z x

3

33

3

xy xy xx

xy yy xy

xy yz xz

x x x y

y y x y

z z x y

,tθ x

25



Considering the displacement gradient tensor ,

Introducing the definition of , the components of are rewritten:

,

1 1 , 1, 2,32 2

j ji i iij ij ij

j j i j i

t

u uu u uJ i jx x x x x

u xJ

x

, tJ x

ij

ij

,tθ x , tJ x

3 2

3 1

2 1

1 2 3

1:

2 :

3:

x x xxx xy xz

y y yxy yy yz

z z zxz yz zz

u u ux y zu u ux

j j j

i

iy z

u u ux z

iy

1

2

3

23

31

12

yz

zx

xy

26

Integration of the Strain Field

The integration of the strain field is performed in two steps: 1. Integration of derivative of using the1st order PDE system

derived for . The solution will be of the type:

The integration constants can be obtained knowing the value of the rotation vector in some points of the medium (boundary conditions).

2. Known and , is integrated using the 1st order PDE system derived for . The solution will be:

The integration constants can be obtained knowing the value of the displacements in some point of space (boundary conditions)

,tε x

,tθ x1 2 3, and

, , , 1,2,3i i ix y z t c t i

ic t

,tε x ,tθ x u

, , , 1,2,3i i iu u x y z t c t i

ic t

REMARKIf the compati-bility equations are satisfied, these equations will be integra-ble.

u


The integration constants that appear imply that an integrable strain tensor will determine the movement in any instant of time except for a rotation and a translation :

A displacement field can be constructed from this uniform rotation and translation:

This corresponds to a rigid solid movement.

ˆˆ ˆ( , ) ( ( )) ( )t t t u x x u

Integration of the Strain Field


,tε xˆ( ) ( )

nott tc ˆ( ) ( )

nott t c u

ˆ, ,,

ˆ, ,

t t tt

t t t

x xx

u x u x u

1 12 2

ˆˆ ˆ( *) ( ( ) ) ( )S T T

uu u u 0

28


3.5 Integration of the Deformation Rate Tensor


Compatibility Equations in a Deformation Rate Field


There is a correspondence between

The concept of compatibility conditions can be extended to deformation rate tensor .

( )

12

12

12

jiij

j i

jiij

j i

uux x

uux x

uu

u

( )

vv12

vv1w2

12

jiij

j i

jiij

j i

dx x

x x

vd v

v

d v

30

Example

Deduce the velocity field corresponding to the deformation rate tensor:

In point the following holds true:


0 0

, 0 00 0

ty

ty

tz

tet te

te

d x

1, 1, 1

1,1,1

2,

t

t

t

et e

e

xv x

1,1,1

01, 02

t

tte

xω x v

31

Example - Solution

Consider the correspondence:

Take the expressions derived for substitute with and with :


( )12

uu

u

( )12

vd v

v

1 2 3, and , tθ x , tω x , tx , td x

1

11

1

0 0

0 0

0 0

xyxz

yz yy

zyzz

ddx y z

d dy y z

ddz y z

0 0

, 0 00 0

ty

ty

tz

tet te

te

d x

1 1t C t

32

Example - Solution


2

22

2

0 0

0 0

0 0

xx xz

xy yz

xz zz

d dx z x

d dy z x

d dz z x

3

233

3

0 0

0

0 0

xy xx

yy xy ty

yz xz

d dx x y

d dt e

y x yd d

z x y

0 0

, 0 00 0

ty

ty

tz

tet te

te

d x

2 2t C t

23 3, ty tyy t t e dy te C t

33

Example - Solution

For point :

So,

Therefore, for any point,


1 1C t

2 2C t

3 3tyte C t

1, 1, 1

0

1, 02

t

tte

ω x v

2 20 C t

1 10 C t

3 31,1,1

t tyte te C t

x

1

2

3

0

0

0

C t

C t

C t

0

, 0ty

tte

ω x

34

Example - Solution

Taking the expressions

The components of the velocities can be obtained:


3 2

3 1

2 1

v v v

v v v

v

1 2 3

1:

2 :

3: v v

x x xxx xy xz

y y yxy yy yz

z z zxz yz zz

j j j

i

i

d d dx y z

d d dx y z

d dy

i dx z

0 0

, 0 00 0

ty

ty

tz

tet te

te

d x 0

, 0 ;ty

tte

ω x

3

2

v 0

v 2

v 0 0

xxx

ty ty tyxxy

xxz

dx

d te te tey

dz

1v , 2 2ty tyx y t te dy e C t

35

Example - Solution

The components of the velocities can be obtained:


0 0

, 0 00 0

ty

ty

tz

tet te

te

d x 0

, 0 ;ty

tte

ω x

3

1

v0

v0

v0 0

y ty tyxy

yyy

yyz

d te tex

dy

dz

2v y t C t

2

1

v 0 0

v 0 0

v

zxz

zyz

tzzzz

dx

dy

d tez

3v , tz tzz z t te dz e C t

36

Example - Solution

For point :

So,

Therefore, for any point,


1, 1, 1

2

,

t

t

t

et e

e

v x

11,1,1

v 2 2t tyx e e C t

x 2v t

y e C t

31,1,1

v t tzz e e C t

x

1

2

3

0

0

t

C t

C t e

C t

2

,

ty

t

tz

et e

e

v x

3v tzz e C t

2v y C t 1v 2 ty

x e C t

37


Summary


Given always exists:

Given will exist only if the compatibility conditions are satisfied.

Compatibility conditions: Conditions that a symmetrical 2nd order tensor must satisfy in order to be an

“infinitesimal strain tensor” and, thus, to exist a displacement field which satisfies:

They guarantee the continuity of the continuous medium during the deformation process.

Summary


u ε 1 , 1,2,32

jiij

j i

uu i jx x

ε u

1 , 1,2,32

jiij

j i

uu i jx x

39

Compatibility equations for the infinitesimal strain tensor:

Summary


2 22

2 2

2 22

2 2

2 22

2 2

2

2

2 0

2 0

2 0

0

defyy yzzz

xx

defxx xzzz

yy

defyy xyxx

zz

defyz xyxzzz

xy

defyy yz xxz

xz

Sz y y z

Sx z x z

Sy x x y

Sx y z x y z

Sx z y x y

2

0

0

y

defyz xyxx xz

yz

z

Sy z x x y z

S ε 0

, 0ml mjq lir ij qrS e e

1ijke

1ijke

, , , , 0

, , , 1,2,3ij kl kl ij ik jl jl ik

i j k l

40

Summary


Rotation tensor:

Rotation vector:

Derivative of :

1 ( )2

Ω u u

1 23

2 31

3 12

12

yz

zx

xy

θ u

,tθ x

1

11

1

yz xyxz

yz yz yy

yz zyzz

x x y z

y y y z

z z y z

2

22

2

zx xx xz

xy yzzx

zx xz zz

x x z x

y y z x

z z z x

3

33

3

xy xy xx

xy yy xy

xy yz xz

x x x y

y y x y

z z x y

41

Summary


The integration of the strain field : 1. Integration of derivative of using the expressions derived for

.

2. Known and , is integrated using:

The solution will be:

,tε x ,tθ x

1 2 3, and

, , , 1,2,3i i ix y z t c t i

,tε x ,tθ x u

, , , 1,2,3i i iu u x y z t c t i

3 2

3 1

2 1

1 2 3

1:

2 :

3 :

x x xxx xy xz

y y yxy yy yz

z z zxz yz zz

u u ux y zu u ux

j j j

i

iy z

u u ux z

iy

42

There is a correspondence between:

The concept of compatibility conditions can be extended to deformation rate tensor .

Summary


( )

12

12

12

jiij

j i

jiij

j i

uux x

uux x

uu

u

( )

vv12

vv1w2

12

jiij

j i

jiij

j i

dx x

x x

vd v

v

d v

43

CH.3. COMPATIBILITY EQUATIONS - PRESENTACIOmmc.rmee.upc.edu/documents/Slides/Ch3_v13.pdf · exists a scalar function (named ... xy S zx S yz ... The compatibility equations can be

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