Chap5 Elasticity

8/13/2019 Chap5 Elasticity

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Theory of Elasticity

Introduction

Elasticity of Solids

Field Equations of Elasticity - DifferentialFormulation

Prismatic Rods

Plane Prob lems Theory and Solut ion s

Plane Problems

Applications

Variational Formulation of Elasticity

Three-dimensional Problems

Index

0


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Plane Problems Theory and

Solutions

Plane Strain and Plane Stress

Planar Anisotropic Case

Planar Isotropic Case

Biharmonic

Equation

Solution in Cartesian Coordinates

Solution in Polar Coordinates

Kolosov-Muskhelishvili Method

Chapter 5

0


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Plane Strain

Chapter 5.1

1

The state of plane strain exists as a rigorous and

non-conflicting case.

Assumptions

(1) A prismatic body whose length is much larger

than any in-plane dimension, .

(2) In-plane loads are independent of the out-of-plane coordinatex3.

(3) Absence of normal strain , in a direction

perpendicular to the plane.

ma xRL

033

Tend

R

A

end

X

Y

Z

Tside

V

L

Aside


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Plane Strain

Chapter 5.1

2

Examples


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Plane Stress


Combination of the last two

assumptions dictates that03 i

Non-vanishing field variables:

2133 ,,,, xxu

1X

2X

3X

L

Chapter 5.1

4


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Consistence Check by Compatibility Equations


Compatibility equation 0, klijnilmjkee

In a planeproblem 0, ijnimj ee

For a plane strainproblem 0, nm ee

Non-trivial components3 nm

0,,

Chapter 5.1

5


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For a plane stressproblem 033

Alternatives:

(1) If both mand nare not equal to 3, the indices iandjhas to be

assigned as 3 to avoid the trivial identity of 00;

(2) If both m and nare equal to 3, the indices iandjhas to be

assigned as not 3 to avoid the trivial identity of 00, andconsequently producing ;

(3) If either mor nis equal to 3 and the other is not, can

only produce the trivial identity of 00.

0,,

0, ijnimj ee

Chapter 5.1

6


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Additional strain compatibility condition

0,3333 ee

The most general form of 33

CBxAx 2133

ijklijklij SS

Constitutive law under the plane stresssituation

Restriction CBxAxS 2133

Chapter 5.1

7


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Generalized Plane Stress


L

xL 03d

1

The averaged in-plane components of stress

and strain satisfy all in-plane equations.

Chapter 5.1

8


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References

Chapter 5.2

9

1. Cherkaev A., Lurie K., Milton G.W.,

Proc. Roy. Soc. London.,A438(1992),519-529

2. Zheng Q.S., Hwang K.C., Proc. Roy.

Soc.London.,A452(1996), 2493-2503

3. Yang W., Ma C-C., Proc. Roy. Soc.

London,A454(1998), 1843-1855


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Airy Stress Func t ion

Equilibrium equation

21

2

122

1

2

222

2

2

11 ,,xxxx

0, 21,xx

Chapter 5.2

10


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J-Tensor

Cherkaev, Lurie and Milton (1992 )

I11J

,J

0, ggJ

1is the unit tensor of rank two.

Iis the unit tensor of rank four.

1

I :Chapter 5.2

11


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Planar Anisotropy


Two dimensional linear

elastic constitutive law:

S

4222

11D1DD1JIS v

E

222 Re ZD C 444 Re ZD C

n

zzzn eeeZ 21 eee iz

Chapter 5.2

12


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Planar Anisotropy

Stretch a fiber along

a direction


1X

2X

q

)(qE

qq

q

ii eCeCE

E 44

2

2Re1

2C 4Ctwo-fold symmetry four-fold symmetry

Chapter 5.2

13


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Planar Anisotropy


planar tetragonal 02C 04 C

planar orthotropic0ImIm

42

CC

isotropic 042 CC EE q

qq

q4cos2cos 42 aa

EE

Chapter 5.2

14


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Governing Equations


0,

S

422

21Re1:: D1DD1JIJSJS v

E

can be regarded as a rank four rotation tensor.

the consequence of Srotated by 90 degrees.

J

S

Chapter 5.2

15


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Governing Equations


Governing equation can be expressed in a

much simpler form (Yang and Ma, 1998)

0Re2

2

2

2

4

2

22

2

zzC

zzC

z

Chapter 5.2

16


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CLM Shift

For an isotropic material, the governing

equation will be reduced to :

0Re2

2

2

2

zz

NoMaterial

constants!!!

TJ , TractionBCs

C

L

M

One may change the Youngs modulus

Eand the Poissons ratiov arbitrarily

without any influence on the stress state.

Chapter 5.2

17


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Planar Isotropic CaseBiharmonic EquationBasic Equation for Planar Isotropic Materials

Chapter 5.3

18

,,

,

2

1

2,0

uu

f kk

Plane strain kk

Plane stress 022 333333 kk

233

2

22

kk


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A unified planar constitutive relation

2

1

3

Planar Isotropic CaseBiharmonic EquationBasic Equation for Planar Isotropic Materials

4

3

2

1a

43

1

3

Plane strain

Plane stress

Chapter 5.3

19


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Biharmonic Equation

,2 V

,Vf 04

3

2

1

4

1 2,

2

0, f

StressPlaneStrainPlane

11

21

1

12

2222

VV

V22

4

1

0,,

4

3

2

1a

V22

Chapter 5.3

20


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Solution in CartesianCoordinatesRectangular Regions

Chapter 5.4

21

X

Y

a a

b

b

Astrong boundarycondition has to be prescribedpoint-wise.

Aweak boundarycondition can be relaxed toan integral form.

The errorinduced

by the weak BCsis

characterized by theSaint-Venant

corrective solution.


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Polynomial Solution

Solution in CartesianCoordinates

022

,2

0

,

N

N yxP

N

i

iiN

iN

yxAP0

P0andP1induce no stress.

SUGGEST :

Chapter 5.4

22


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Polynomial Solution


04

0

44

N

i

iiN

iN yxBP 4,,0,0 NiBi

Solution familyNP~

NPP~

,,~

3

32

~,

~PP

4 independent coefficients

3 independent coefficients

Chapter 5.4

23


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Examples of Lower Degree Polynomials


Polynomial of degree 2 Y

X

-A1

2A0

2A2

2

21

2

0 yAxyAxA

22Axx 02Ayy

1Axy

Chapter 5.4

24


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Examples of Lower Degree Polynomials


Polynomial of degree 3

X

Y

Pure bending

3

3

2

2

2

1

3

0 yAxyAyxAxA

yAxAyy 10 26 yAxAxy 21 22 yAxAxx 32 62

0,0 3210 AAAA

Pure bending

yAxx 36

0 yyxy

Chapter 5.4

24


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Lateral Bending of a Slender Rectangle


X

Y

b

b

a

F

BCs :

0xxFdy

b

b

xy

:0x

by

0yy 0xy

Chapter 5.4

26


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2

2

y

xx

main stress, proportional toxy

TRY :3

1xyC xyCxx 16 yCxy 130yy

byxy at0NOTsatisfy the BC :

CORRECT : xyCxyC 23

1

Chapter 5.4

27


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The final stress

function :

3

22

4

3

b

byFyx

Stresses :2

2

3

b

Fxyxx

322

43

bybFxy

0yy

Chapter 5.4

28


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QUESTION:

Does the global equilibrium maintain?

BCs at the right end:

b

b

xx y 0d Fy

b

b

xy

d Fayy

b

b

xy

d

REASON: Local equilibrium plus suitable boundaryconditions along three side naturally leads to the global

balance for the boundary traction over the remaining side.

Chapter 5.4

29


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Higher Degree PolynomialsReduction by

Symmetry


B

C

Symmetricpart to x (or) y

Stress function contains only the

evendegree terms ofx (or)y

Anti-symmetricpart to x(or) yStress function contains only the

odddegree terms ofx (or)y.

Chapter 5.4

30


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Reduction by Symmetry

Bending of a slender

rectangle by uniformly

distributed pressure:


X

Y

a a

b

b

yxxx2

Without more complications, the Airy stress functionwould involve polynomials of degree 5 and below.

Undetermined constantants : 6 + 5 + 4 + 3 = 18

Chapter 5.4

31


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Higher Degree Polynomials

Bending of a slender rectangle by uniformly

distributed pressure:


X

Y

a a

b

b

PaPa

2Pyy

2Pyy 2Pyy

2Pyy

Chapter 5.4

32


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Solution in CartesianCoordinatesPaPa

2Pyy

2Pyy

Stress function :

3

5

2

4

5

3

32

2

4

1 yCyxCyCyxCyxC

Strong BCs : byp

yyxy at2

,0

Weak BCs : axyyyb

b

xx

b

b

xx

at0d,0d

Chapter 5.4

33


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Solution in CartesianCoordinatesX

Y

a a

b

b

Airy stress function :

233232225323

10251554

xbybyayxbyyxabp

Stresses :

ybyayyxbp

xx

2232

3 615101520

2234

3yb

b

pxxy 3233 234

bybyb

pyy

Chapter 5.4

34


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Solution in CartesianCoordinatesSolut io n by Fourier Series

q1(x)

q(x)

Y

X

Symmetric

1

cosk

kk xyf

Anti-symmetric

1

sink

kk xyf

Chapter 5.4

35


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Solution by Fourier Series


1

cos

k

kk xyf

1

sink

kk xyf

022

yyDCyyBAyf

kkk

kkkk

sinh

cosh

Chapter 5.4

36


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Solution by Fourier Series


For an infinite or a semi-infinite beam

0

dcos,, xyfyx

yyDCyyBAyf sinhcosh,

Chapter 5.4

37


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Polar Coordinates

Solution in PolarCoordinates

Chapter 5.5

38

q

r

X

Y

qcosrx qsinry

22 yxr x

y1tanq

q

qq

q

q

rrxrx

r

x

sincos

q

qq

q

q

rryry

r

y

cossin


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Polar Coordinates


Chapter 5.5

39

q

r

X

Y

qqqq

qqq

rrrrrrrx

2

22

2

2

2

2

22

2

2 11cossin2

11sincos

qqqq

qqq

rrrrrrry

2

22

2

2

2

2

22

2

2 11cossin2

11cossin

qqqq

qqq

rrrrrrryx

2

2

22

2

2

22

22 11sincos

11cossin


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Basic Equations in Polar Coordinates


Chapter 5.5

40

Stresses

2

2

2

11

q

rrrrr 2

2

rqq

qq

rrr

1

Laplace operators 2

2

22

2

2

2

2

22 11

q

rrrryx

2

2

2

22

24 11

qrrrr

Biharmonic

operators


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Chapter 5.5

41

Displacementsqq qsincos uuu rx

qq qcossin uuu ry

Strains

qq

qqq

q qqq 22 sin1cossin1cos

u

rruu

rru

ru

ru rrrxx

qq

qqq

q qqq 22 cos1

cossin1

sin

u

rr

uu

rr

u

r

u

r

u rrryy

qqq

qqq

qqq 22 sincos1

2

1cossin

1

rrrxy

u

rr

u

r

uu

rr

u

r

u


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Chapter 5.5

42

Transformation

rur

xyyyxxrr qqqq cossin2sincos 22

r

u

r

uu

r

rr

qqq

q

1

2

1

r

uu

r

r

q qqq

1


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Plane Problems with Axi-symmetry


Chapter 5.5

43

A plane geometry with axi-symmetry

(1) Assumption of plane strain or plane stress holds ;

(2) Geometry does not depend on the polar angle.

bq

r

aX

Y

Single-valued-ness

qq uu ,,2,,


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Chapter 5.5

44

Stress function

00

sincosn

n

n

n nrgnrf qq

0d

d1

d

d2

2

2

2

2

rg

rf

r

n

rrr n

nBiharmonic equation

1,0n The solution has the form of rk.

Characteristic

equation

02 2222 nknk

nnk 2,


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Chapter 5.5

45

Solutions to Eulers function :

1n3,1kSimple roots 1kDuplex root

11413123

111 ln rArArrArArf

0n2,0kDuplex roots

rAArArArf lnln 0403022

010

n

n

n

n

n

n

n

nn

rArArArArf 43

2

2

2

11n


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Chapter 5.5

46

BCs :

ar q1

Frr

qq 3

Fr

br q 2Frr qq 4Fr

1,2,3,4,sincos00

jnbnaFn

nj

n

njj qqq

For each trignomic terms, one can always finds 4

equations to solve for the 4 coefficients 431 ,,, nnn2n AAAA


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Chapter 5.5

47

Axi-symmetry


010

Stresses 2

040201 ln212

r

ArAArr

2040201 ln232 rArAA qq

0qr


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Example : A Hole under Remote Shear


Chapter 5.5

48

a

a

rq

BCs :

arrrr at0q

r

yyxx

xyat

0


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Chapter 5.5

49

q

2sin2

2rxy Uniform

solution

Perturbation

solution qq 2sin2sin 224222* rAArg

q

2sin

2

2

2422

2*

rAAr

q 2sin64

4

24

2

22

r

A

r

Arrqq 2cos

624

24

2

22

r

A

r

Ar qqq 2sin

64

24

r

A


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Chapter 5.5

50

a

Stresses

q 2sin3414

4

2

2

ra

ra

rr

qq 2cos32

14

4

2

2

r

a

r

arqqq 2sin

31

4

4

r

a

Along the rim of the hole qqq 2sin4ar

The maximum hoop stress4

3,

4

q


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Degeneracy (Solution)


Chapter 5.5

51

Terms that do not induce a stress :

qq sincos CrBrACyBxA

1303 and AA can be neglected.

1n2

2

nn rA

n

nrA 3

111111 rrD

rrCBrArrf1n

rDrCrrf ln2,lim0


54/98

Degeneracy


Chapter 5.5

52

0n 2nr 2nr

rrrn

n

nln

d

dlim

22

0

nr nr

rrn

n

nln

d

dlim

0

Other qq sincos BrAr

qlnln BrA

0 qqrr

2r

Br q

1n qqqq cossin CrBr

0 qqq r

r

C

r

Brr

qq

sin2cos2


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Michells Solution


Chapter 5.5

53

J. H. Michell, Proc. London. Math. Soc. Vol.31

(1899), 100-124

2

43

2

2

2

1

243

2

2

2

1

13

1

1412

3

11

13

1

1412

3

11

0403

2

02

2

01

sin

cos

sinsinln

sincosln

lnln

n

n

n

n

n

n

n

n

n

n

n

n

n

n

n

n

n

n

nrBrBrBrB

nrArArArA

rBrBrrBrB

rArArrArA

ArArrArA

q

q

qqq

qqq

q


56/98

A Circular Hole under Tension


Chapter 5.5

54

BCs :

arrrr at0q

r

yyxy

xxat

0

Uniformsolution

q 2cos142

22 ry


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A Circular Hole under Tension(Details)


Chapter 5.5

55

From Michellssolution, the

terms that might match the

uniform solution:

rrrrrrrr

ln,ln,2cos

ln,ln,22

22

q

Terms that lead to decaying stresses :

q2cos,1,ln 2rr

Stress function

qqq 2cos2cosln2cos14

22 CrBrAr
http://localhost/var/www/apps/conversion/tmp/scratch_6/Chap5/A%20Circular%20Hole%20under%20Tension.pdfhttp://localhost/var/www/apps/conversion/tmp/scratch_6/Chap5/A%20Circular%20Hole%20under%20Tension.pdf


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Chapter 5.5

56

143

22cos1

22

2

4

4

2

2

ra

ra

ra

rrq

1

3

2

2cos1

2 4

4

2

2

r

a

r

a qqq

1

23

2

2sin2

2

4

4

r

a

r

ar

qq

Stresses


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Displacements for Circular Hole under Shear


Chapter 5.5

57

Stresses

q 2sin34

14

4

2

2

r

a

r

arr

qq 2cos32

14

4

2

2

r

a

r

ar

qqq 2sin3

14

4

r

a


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Chapter 5.5

58

Strain q

2sin

1341

4

4

2

2

r

a

r

av

Er

urrr

qqq 2sin134

14

4

2

2

r

a

r

a

E

q

q 2cos

!3121

4

4

2

2

r

a

r

a

Er

)(2sin14

14

42

qq

fr

av

r

ar

Eur

r

urrr


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Chapter 5.5

59

ruru

qq

q q

rgFr

a

r

ar

Eu

qq

q 2cos1

121

3

42

qqq dfF

r

u

r

uu

r

rr

qqq

q

1

2

1 0 CrgrrgFF qq

qqq sincos BAF Crrg


62/98



Chapter 5.5

60

Displacement fields

qqq

sincos2sin

14

1 3

42

BAr

a

r

a

rEur

CrBAr

a

r

ar

Eu

qqq

q cossin2cos1

121

3

42

Translation : A=B= 0

Rotation : C= 0


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Chapter 5.5

61

Radial displacement

along the hole :

E

au

arr

q 2sin4

4

5,

4

q

4

7,

4

3 q

MAX :

MIN :


64/98

Axi-symmetric Problems(Details)


Chapter 5.5

62

pi

po

Displacement fields:

ruu rr 0qu

0d

d1

d

d22

2

r

u

r

u

rr

u rrr

r

CrCur

21

Solution in Polar
http://localhost/var/www/apps/conversion/tmp/scratch_6/Chap5/Axi-symmetric%20Problems.pdfhttp://localhost/var/www/apps/conversion/tmp/scratch_6/Chap5/Axi-symmetric%20Problems.pdf


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Displacements of Michells Solution

Coordinates

Chapter 5.5

63

BCs :

ar q 1Frr qq 3Fr

br q 2Frr qq 4Fr

Zero resultant forces and zero resultant torque:

qqqqqqqqqq

2

0

42

2

0

31 dsincosdsincos FFbFFa

qqqqqqqqqq2

0

42

2

0

31 dcossindcossin FFbFFa

qqqq

2

0

4

2

2

0

3

2 dd FbFa

Solution in Polar


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Displacements of Michells Solution

Coordinates

Chapter 5.5

64

Orthogonality of Fourier series dictates that the

above conditions only bears influence on the

terms of 1,0

n

QUESTION: Three independent equationsWHILE four sets of coefficients?

141114110401 ,, BBAAAA

ANSWER: Single-valued-ness requirement of thedisplacement fields, and that imposes restrictions on the

multi-valued terms in uru that are linear in .

Kolosov Muskhelishvilli


67/98

MethodComplex Variables

Chapter 5.6

65

Recall: in-plane elasticity formulation of an

isotropic solid without body forces :

02 2

022

In-plane stress trace is harmonic

The Airy stress function is biharmonic

Analogy to the anti-plane problems

iyxz iyxz



68/98

Complex Variables

Method

Chapter 5.6

66

Derivatives

yi

xyzy

xzx

z 21

yi

xyz

y

xz

x

z 2

1

zzyi

xyi

x

22 4



69/98

Kolosov-Muskhelishivilli Potentials

Method

Chapter 5.6

67

zz

2

zzzzgz

Re42

2

zzzz

24

22

zzzzz

2

1

zzzz dRe



70/98

Fields by Complex Potentials

Method

Chapter 5.6

68

Kolosov and Muskhelishivilli potentials : ,

Stress zyyxxm Re22

zzz

i xyxxyy

d

2

Strain zmyyxx

m

Re

2

1

4

1

2

zzzi dxyxxyy

d

2

1

2

1

2



71/98


Method

Chapter 5.6

69

Displacement zzzzfiuuu yx

2

1c

xy

xxyy

d i

2

x

u

y

ui

x

u

y

u yxxy

2

1

yx iuu

yi

x

2

1 yx iuu

z



72/98


Method

Chapter 5.6

70

yyxxm 2

zuz

iuuy

ixy

u

x

u

c

yx

yx

Re1

2Re

Re

01Re zzzf 0Re zzf

izizzf 2



73/98


Method

Chapter 5.6

71

Rigid translation

Rigid body rotation

Complex displacement

Strain energy density

zzzzuc 2

22Re122

1

2

1

zzW ddmm



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Coordinate Transform

Method

Chapter 5.6

72

q

Y

XY

zeiyxeiyxz ii ***

c

i

c ueu *

mm *

d

i

d e 2*



75/98


Method

Chapter 5.6

73

yx

i

r iuueiuu q

q

yyxxrr qq

xy

xxyyi

r

rr

iei

qq

qq

22

2

q X

Y

XY



76/98


Method

Chapter 5.6

74

q X*

Y*

z0

z

z*

Y

X

*

0 zzz

yyxxyyxx ~~~~

0~Re zz

zz ~

yx

xxyy

xy

xxyyii ~~

~~~~

22

zzzzzz ~~~

zzzz 0~



77/98

Disks

Method

Chapter 5.6

75

Uc=U

0(q)

aBy Taylor series:

0n

n

nzz

0n

n

nzz

Complex displacement:

0

22n

in

n

ni

n

in

n

n

c eenerzzzzu qqq

0

2

0 )(2n

in

n

ni

n

in

n

n eeneau qqq qBCs:



78/98

Holes

Method

Chapter 5.6

76

0d

0 d zzz

0 z 0 z

0 m 0 zz

iAz 2 iz 2ln 0ln zAz

zSzAz ln

0 z

Ais REAL.



79/98

Holes

Method

Chapter 5.6

77

Single-valued analytical function can be expressed

by Laurent series.

nnzzBzAzz lnln

nnzzCz ln

Ais REAL,B, Cand are COMPLEX.nn ,

LINKS TO

the Michells solutions for holes

qirz lnln



80/98

Increments of Complex Displacement by

Circling the Hole

Method

Chapter 5.6

78

Kolosov-Muskhelishvilli functions

zzBzAzz *lnln zzCz *ln

are single-valued analytic functions ofz. zz ** ,

CBAziuc

1

Increment of the

complex displacement



81/98

Inserted Wedge

Method

Chapter 5.6

79

A

qq

iuueAz

iu r

i

c 1

0 ru Ar

Gu

1 q

Be interpreted as inserting

a wedge of an angle :

A

1

zzln

1

0



82/98

Method

Chapter 5.6

80

CB

yx ibb

iCB

xx ub yy ub

yx ibb

iBC

Burgers vector

i

yx beibb

22

yx bbb

x

y

b

b1tan

Dislocat ions

http://uet.edu.pk/dmems/EdgeDislocation.gif


83/98

Solutions of Single-valued Displacements

Method

Chapter 5.6

81

For a problem free of inserted wedge or dislocations,

The Kolosov-Muskhelishvilli solutionis reduced to

zzBz *ln

zzBz *ln



84/98

MethodConcentrated Forces

Chapter 5.6

82

Following the derivation in the textbook

by Lu and Luo, Section 8.1

zzzziRRi yx For the single-valued-ness of , and

the relation

zz ** ,

iz 2ln

)1(2 iBiRRi yx

zziRR

z yx *ln

12

zziRRz yx *ln12



85/98

Solutions without Wedges, Dislocations and

Concentrated Forces

Method

Chapter 5.6

83

Kolosov-Muskhelishvilli functions

1

0

*

k

k

kzazazz

1

0

*

k

k

kzbzbzz

Stresses at infinity :

0Re am 0bd



86/98

Stresses along the Hole (Details)

Method

Chapter 5.6

84

Tractiondistribution along a circle:

zzzezzi irrr qq 2

BCscan be expanded by Fourier series

in complex form:

m

im

mrrr eCi q

q~~

http://localhost/var/www/apps/conversion/tmp/scratch_6/Chap5/Stresses%20along%20hole.pdfhttp://localhost/var/www/apps/conversion/tmp/scratch_6/Chap5/Stresses%20along%20hole.pdf


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Method

Chapter 5.6

85

Remote tension

2

m

2

d

Kolosov-Muskhelishvilli stress functions

1224

zazz

2412222

zazazz

http://localhost/var/www/apps/conversion/tmp/scratch_6/Chap5/A%20Circular%20Hole%20under%20Tension1.pdfhttp://localhost/var/www/apps/conversion/tmp/scratch_6/Chap5/A%20Circular%20Hole%20under%20Tension1.pdf


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Method

Chapter 5.6

86

Stresses

q 2cos21

2 2

2

ra

m

qq

q

qqq

ii

i

rrr

er

a

r

ae

r

a

z

a

z

a

z

azei

2

4

4

2

22

2

2

4

4

2

2

3

22

32

322

Compare with

Michells Solution



89/98


Method

Chapter 5.6

87

Great benefitsof the complex variable approach lie

in the evaluation of displacement field.

zzzzeGu ic q 2

q 2cos112

1

44

4

2

2

2

2

r

a

r

a

r

a

G

rur

qq 2sin114

4

4

2

2

r

a

r

a

G

ru

Solution in Polar


90/98


Coordinates

Chapter 5.5

36

Solutions to Eulers function :

nnn

n

n

n

n

nn rArArArArf 43

2

2

2

1

0n2,0kDuplex roots

1n3,1kSimple roots 1kDuplex root

11413123

111 ln rArArrArArf


010

1n

back

Degen

Michell solution


91/98

qcos3r qcos2r qsin2r qcos6r

qqsinr r/cos2 q 0 0

qcoslnrr r/cos2 q r/sinq r/cosq

r/cosq 3/cos2 rq 3/sin2 rq 3/cos2 rq

qsin3

r qsin2r qcos2r qsin6r

qqcosr rr /sin2 q 0 0

qsinln rr r/sinq r/cosq r/sinq

r/sinq 3/sin2 rq 3/cos2 rq 3/sin2 rq

Michell solution

rr qr qq

2r 2 0 2

rr ln2 1ln2 r 0 3ln2 r

rln 2/1 r 0 2/1 r

q 0 2/1 r 0

STRESS

back

Disp

Circular

Michel l

1

Appendix

Michell solution


92/98

Michell solution

STRESS

qnrn cos2 qnrnn n cos21 qnrnn n sin1 qnrnn n cos21

qnr n cos2 qnrnn n cos12 qnrnn n sin1 qnrnn n cos21

qnrn cos qnrnn n cos1 2 qnrnn n sin1 2 qnrnn n cos1 2

qnr n cos qnrnn n cos1 2 qnrnn n sin1 2 qnrnn n cos1 2

qnrn sin2 qnrnn n sin21 qnrnn n cos1 qnrnn n sin21

qnr n sin2 qnrnn n sin12 qnrnn n cos1 qnrnn n sin21

qnrn sin qnrnn n sin1 2 qnrnn n cos1 2 qnrnn n sin1 2

qnr n sin qnrnn n sin1 2 qnrnn n cos1 2 qnrnn n sin1 2

2

Appendix

back

Disp

Circular

Michel l

Michell solution


93/98

Michell solution

DISPLACEMENT

ru2 qu2

2r r1 0

rr ln2 rrr ln1 q r1

rln r/1 0

q 0 r/1

3

Appendix

back

Disp

Circular

Michel l

Michell solution


94/98

Michell solution

DISPLACEMENT

qcos3r q cos2 2r q sin2 2r

qqsinr

]cosln1

cossin1[2

1

q

qqq

r

]sinln1

sincos1[2

1

q

qqq

r

qcoslnrr

]cosln1

cossin1[2

1

q

qqq

r

]sinln1

sincos1[2

1

q

qqq

r

r/cosq 2/cos rq 2/sin rq

qsin3r q sin2 2r q cos2 2r

qqcosr

]sinln1

sincos1[2

1

q

qqq

r

]cosln1

cossin1[2

1

q

qqq

r

qsinln rr

]sinln1

sincos1[2

1

q

qqq

r

]cosln1

cossin1[2

1

q

qqq

r

r/sinq 2/sin rq 2/cos rq 4

Appendix

back

Disp

Circular

Michel l

Michell solution


95/98

Michell solution

DISPLACEMENT

qnrn cos2 q nrn n cos1 1 q nrn n sin1 1

qnr n cos2 q nrn n cos1 1 q nrn n sin1 1

qnrn cos qnnrn cos1 qnnrn sin1

qnr n cos qnnr n cos1 qnnr n sin1

qnrn sin2 q nrn n sin1 1 q nrn n cos1 1

qnr n sin2 q nrn n sin1 1 q nrn n cos1 1

qnrn sin qnnrn sin1 qnnrn cos1

qnr n sin qnnr n sin1 qnnr n cos1

5

Appendix

back

Disp

Circular

Michel l

Solution in Polar


96/98

Michells Solution

Coordinates

Chapter 5.5

51

J. H. Michell, Proc. London. Math. Soc. Vol.31

(1899), 100-124

2

43

2

2

2

1

2

43

2

2

2

1

13

1

1412

3

11

13

1

1412

3

11

0403

2

02

2

01

sin

cos

sinsinln

sincosln

lnln

n

n

n

n

n

n

n

n

n

n

n

n

n

n

n

n

n

n

nrBrBrBrB

nrArArArA

rBrBrrBrB

rArArrArA

ArArrArA

q

q

qqq

qqq

q

back

HOLES

Solution in Polar


97/98


Coordinates

Chapter 5.5

53

From Michells solution, the

terms that might match the

uniform solution: rrrr

rrrr

ln,ln,2cos

ln,ln,22

22

q

Terms that lead to decaying stresses :

q2cos,1,ln 2rr

Stress function

qqq 2cos2cosln2cos14

22 CrBrAr

back

HOLES

Solution in Polar
http://localhost/var/www/apps/conversion/tmp/scratch_6/Chap5/A%20Circular%20Hole%20under%20Tension.pdfhttp://localhost/var/www/apps/conversion/tmp/scratch_6/Chap5/A%20Circular%20Hole%20under%20Tension.pdf


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Coordinates

1432 2cos12 2

2

4

4

2

2

ra

ra

rarr q

1

3

2

2cos1

2 4

4

2

2

r

a

r

a qqq

1

23

2

2sin2

2

4

4

r

a

r

ar

qq

Stresses

back

HOLES

Documents

Chap5 Elasticity