Elasticity Edited

7/29/2019 Elasticity Edited

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Stress at a Point

P1Fn F Pn

S2

Fs2P2

An

Fs1F

S1

P1 A

P2F is the internal force acting on the elemental area A at the point p(n)

We assume LIM F/ AA 0

Exists and unique.

S1 and S2 arbitrary directions but orthogonal

n: Normal stress component at the point P(n) In the direction n

n = LIM Fn/ AA 0

Similarly s1 = LIM Fs1/ A

A 0 s1 = LIM Fs2/ A

A 0Implied in the notation is

a) Stress varies with location of Pb) Stress varies with orientation of n

Stress on any three independent planes at apoint define the state of stress at pointuniquely.

In Matrix notation : At

t =

T

zyx


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Tkji ,,

zzyzx

yzyyx

xzxyx

A

The matrix A obeys the mathematical properties which are typical of the entities called tensors.Hence A is called stress tensor.

Stress vector on an inclined plane

n : Normal to the inclined plane = l i +m j +n k

Y j h : Altitude of tetrahedron

dAn : Area of inclined face

P : Body force vector /unit massn

P )( z dAz dv : Volume of tetrahedron

n )( x : Stress vector on negative x-plane )( x

dAx x i

)( y dAy

z k

Consider equilibrium of the tetrahedron

amF )( x dAx+ )( y dAy + )( z dAz+ n dAn + dvp = adv ---------------(1)

a = Acceleration vectorWe can show thatdAx = ldAn, dAy = mdAn, dAz = ndAnEquation (1) becomes

)( x ldAn + )( y mdAn + )( z ndAn + n dAn+ 1/3h ndAp =1/3h adAn

As h 0 Stress vector over inclined plane becomes stress vector at the origin in the directionof n .

As h 0 0)()()( zyxn nml -----------------------(2)

Let : n = xn (Along X-Axis)Then : l =1, m = n =0

xn


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From Equation (2) : 0)( xx

)( xx

Similarly yy )( ,

zz )(

Equation (2) becomes zyxn nml ---------------(3)This equation is the basis for analysis of stress at a point.

Analysis of Stress at a Point

Let kTjTiT zyxn --------------------------------------(4)

Tx, Ty, Tz :Components of n along (x,y,z)

Substituting for x , y & z in Eq.(3) in terms of xzxyx ,, etc and equating components we

obtain

nmlT xzxyxx

nmlT yzyxyy

nmlTzyzxzz

or in Matrix Notation

uAt -------------------------------------------------------------(5) Tzyx TTTt ,, A :Stress TensorSymmetric A = TA

T

nmlu ,,

1222 nmluu T

n can also be resolved into components in its own plane : Normal and Shear Components

Normal Component of n

utn Tn . uAu T

ln222222xzyzxyzyx

mnlmnml -------------(6)

is a quadratic in (l,m,n) Since l2+m

2+n

2=1 n n

Only two of (l,m,n) are independent n

To obtain shear component we use triangle law. n

n n

nnn


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222

n

22 n

)(1

nn n

To find the extreme values of the normal stress on a plane

ln222),,( 222xzyzxyzyx

mnlmnmlnmlF

Where 01222 nml

i.e f (l, m, n) = 0

Problem : Find the extrememum value of F(l, m, n) subject to constraint f(l, m, n) = 0

This is the problem of Constrained Extremisation

Lagrange Multiplier Method.

The solution of problem of extremizing 0),,( 321 F

Subject to: 0),( 3,21 f

Coincides with the problem of extremizing fFG ),,( ,321

With respect to and321 ,,

Present case ),,(),,( nmlfnmlFG to be extremised :

G

n

G

m

G

l

G;0;0;0

nmlll

Gxzxyx

20 --------------------------(a)

nmlm

Gyzyxy

)(20 -------------------------(b)

nmln

Gzyzxz

)(20

-------------------------(c)

0

G 1222 nml

Rearranging equations (a) to (c)

0)( uIA ----------------------------------------------------------------------(7)

Equation (7) uAu

Recall uATTTt Tzyx ,, i.e lTx , mTy , nTz

or n

T

m

T

l

T zyx


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Also kTjTiT zyxn and nknjmiln )( --------(8)

For direction which maximizes n

Equation (8) shows that the extremum value of n has component in the normal directions only.That is the shear component is zero.

The directions in which the stress vector reaches extreme value are called principal directions.The corresponding stress vector has normal component only. These normal stresses are referredto as principal stresses.

The principal stress values and the corresponding principal directions are contained in thesolution of equation (9)

0)( uIA

Solution exists only if

Det 0 IA

Expanding 0322

1

3 III

zyxI 1

zyz

yzy

zxz

xzx

yxy

xyxI

2

zyzxz

yzyxy

xzxyx

AI

3

I1, I2 and I3 are stress invariants 3,2,1 are principal stresses

From theory of equations

1321 I

2133221 I

3321 I

Corresponding to 1

0)( 11 uIA

11111 ,, nnmlu TT

1n , The Eigen vectors, is the principal stress direction corresponding to 1 .Similarly

0)( 22 uIA

22222 ,, nnmlu T

0)( 33 uIA


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33333 ,, nnmlu T Properties of Eigen Vectors

1 Since Det 0)( IA , Eigen Vectors are determined as direction ratos. i.e.

magnitude is indeterminate.

2 Eigen Vectors are orthogonal

031

23

1

12

1

1 uuuuuu

Example Problem

529

2162

925

zyzxz

yzyxy

xzxyx

A

I1= 5+ 16+ 5 = 26

52

216

59

95

162

25

2

I

= 76+ (-56)+ (76) = 96

864

529

2162

925

3

I

Characteristic Equation

08649626 23

0)4)(18)(12(

121 ; 182 ; 43 Check:

1321 26 I

2133221 96 I

3321 864 I

And 0

729

242

927

)(

1

1

1

11

n

m

l

uIA

111 nml and 12

1

2

1

2

1 nml

111 nml =3

1 take +ve sign

T

u

3

1

3

1

3

1


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kjin3

1

3

1

3

11

Similarly 0

1329

222

9213

)(

2

2

2

22

n

m

l

uIA

2222 2, lmml

12

2

2

2

2

2 nml

6

1,

6

2,

6

12u take +ve sign

kjin3

1

6

2

6

12

0

929

2202929

)(

3

3

3

33

n

ml

uIA

0, 333 mnl and

12

3

2

3

2

3 nml

kjin2

10

2

13

6

1

6

2

6

13

1

3

1

3

1* 21

kji

nn = kji18

30

18

3

= 32

10

2

1nkji

We verified 321 * nnn


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Stress Transformation

Y

(x,y,z) : Original Coordinate System

(x1,y

1,z

1) : Rotated System

yI

( ),,

111kji

are the corresponding

x1

Unit vectors

1

j 1

i x

z1

1

k

z

We have

T

uknjmili 11111

Tuknjmilj 22221

Tuknjmilk 33331

Where )3,2,1,,,( inimili are the direction cosines defining the rotation of the ),,( 111 zyxsystem from (x,y,z) system.

Using the Eqn (5) t=Au, The stress vector in the directions ),,(111

zyx can be expressed as

zyxx nml 1111

zyxy nml 2221

zyxz nml 3331

Now consider1

.1 ixx

tunml T

z

y

x

x 11111

AuT

1 -------------(a)

Tkji

Also knjmili 1111

= 11

1

1

u

n

m

l

kjiT

-------------------------(b)

Using (a) and (b)


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11

1

.11 uAuiTT

xx

111 AuuT

x

Similarly

12

1

21

1

.

.

11

111

Auui

Auuj

yxy

T

xyx

etc

Let 1A Stress matrix referred ),,( 111 zyx - Axes

11111

11111

11111

1

zyx

yyx

zxyxx

zz

zyA

332313

32212

3121111

AuuAuuAuu

AuuAuuAuu

AuuAuuAuu

ATTT

TTT

TTT

Let

321

31

321

321 ;;

nnn

mmm

lll

uuuU

Then UAUA T1 --------------------------------------------------(10)The transformation represented in equation (10) is so-called second-order Tensor Transformation

Some Properties

1) 1UUT TUU 1

Transpose of U is its inverse i.e U is Orthogonal Matrix2) Det 1 IUUT

i.e 1UUT

12U

1U For ),,( 111 zyx Right Handed System veU

If we transform (x,y,z) to the principal directions with

3

1

2

1

1

1

;; nknjni

Then stress tensor becomes diagonal


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3

2

1

1

00

00

00

A

To show that I1, I2, I3 are Invariants

IUUAUUDetIADet TT 1

LHS = - )(1

3

1

2

21

1

3III -----------------------------(a)

RHS= UIAUT )(

IA

UIAUT

)(

)( 322

1

3III ---------------------------------------(b)

LHS and RHS should be equal for all values of

3

1

3

2

1

2

1

1

1

II

II

II

Invariant under coordinate transformation

Invariant quantities are characteristics of second-order tensor quantities

Extreme Values of Shear Stress

Normal stress vector : nn n n

Total stress vector Tzyxn TTTt ,, n

Shear vector nn n n

From vector triangle nnn

nnn

Problem : To find direction n for which n is an extremum

We have

222

n

22

2 n ---------------------------------------(1)

tt Tn 2

Where t = Au

)()(2

AuAu Tn


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And AuuutTT

We transform the problem in terms of principal stresses and principal directions

Let A* Principal Stress Tensor

n * Orientation of the plane on which is sought with respect to principal directions.

Then

3

2

1

*

00

00

00

A

A*U* = Tnml 3.21 ,,

And )()(2

UAUA Tn

22

3

2

2

22

1

2 nmln -----------------------------------(2)

2

3

2

2

2

1

nml ------------------------------------(3)Using equation (2) and (3) in (1)

)(),,( 232

2

2

1

22

3

22

2

22

1

2 nmlnmlnmlF -------------(4)

Problem : Extremise F(l*,m*,n*)

Subject to 01222 nml

i.e 0),,( nmlf

Using Lagrange multiplier. The above problem is identical to:

Extremise : 0),,,( fFnmlG

W r t ,,, nml

02)2(22 12322212

1

llnmlll

G --------------------------------(5)

02)2(22 22322212

2

mmnmlmm

G --------------------------(6)

02)2(22 32322212

3

nnnmlnn

G -------------------------------(7)

10 222

nmlf

G

Equations (5)(7) are simultaneous equations for ),,( nml leading to directions of maximum

Shear Stress.

These are Non-Linear equations. We can solve by considering particular cases of

nml ,, -

method of exhaustion.

1.

nml ,, cannot be zero simultaneously


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2. Two of

nml ,, are zero

Then n* = 1, This corresponds to principal plane 0

3. one of

nml ,, zero

Let n*=0 122 ml

Eliminate from (5) and (6)

0)(2)( 22212

21 l

Two cases :

(3a) :2

121

l

2

1m

This defines four planes and2

21

Normal stress on this plane 2/)( 21 2

Max Shear occurs on planes 45 deg. To 1 and 2 directions

Similarly m* = 02

1 nl

2

1313

and 1

l* = 02

1 nm

232

23

For the case n* = 0. If 21 all planes such that

122* ml are planes of maximum shear

then 0)()(22

1

2

1

222

1

mlml

Lastly we note that if 321 equations (5) to (7) are satisfied for all values of

i.e All planes are planes of extreme shear. ( )0(

Such a state of stress is referred to as hydrostatic state of stress. Here all planes are principal

planes.Decomposition of stress tensor into Hydrostatic and Deviatoric components

Hydrostatic or Spherical component: All the principal stresses are equal and all planes areprincipal planes

Pure Shear Component or Deviatoric Component: A state of pure shear exists. If there exists

directions such that 321 = 0 in those directions. i.e only shear stress components exist.


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The necessary and sufficient condition for a stress tensor to represent pure shear is :

0zzyyxx

Octahedral shear Stresses

Planes equally inclined to the principle planes are called

3n

Octahedral planes )3

1,3

1,3

1(0 n )3

1,3

1,3

1(0 n

The direction cosines are measured wrt principal directions

2n

),,( 321

nnn

1n

These are eight such planes representing different sign combination. They together form faces of

A Octahedral.

We have the following relations for octahedral planes

2) Normal stresses on octahedral planes

)(3

1332211 oct SPR

ii .

2) Shear stress component on OHP :

2

13

2

32

2

21 )()()(3

1 oct

)(3

1 23

2

2

2

1 SSSoct

(S1, S2, S3) : Principal Deviatoric Streesses

Any stress tensor can be decomposed into hydrostatic and pure shear stress tensors

Spherical part :

S

A

)(3

1zzyyxx

S

zz

S

yy

S

xx

S

zz

S

yy

S

xx

SA

00

00

00

Deviatoric part :DA

S

ijij

D

ij AAA

)(

)(

)(

S

zzzzyzxz

yz

S

yyyyxy

xzxy

S

xxxx

DA

DS AAA SA Hydrostatic state


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DA Pure shear state

oct is also called as vonmises stress used in Mises- Hencky failure criterion

oct Plays important role in plasticity theory.

Equilibrium Equations

Derived from equations of motion applied to an infinitesimal element of the body under (Dynamic)equilibrium

Let ZYX ,, : Body forces /unit mass

dydxdAdzdxdAdzdydA zyx .;.;.

Law of Motion:

1. xx maF

2. zzpzp IM

xF = Total force along X-direction

zpM = Moment of all the forces about z-axis

m = Mass of the body

ax= Acceleration in x-direction

Izp = Mass M.I about z-axis

z = Angular acceleration about z-axis


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Applying second equation

dvdkrtermsHigherordedy

dAdy

dAyddx

dAdx

dAd zpyyxyxyxxxyxxyxy 2)(

22)(

22)(

Dividing by Dv =DxDydz and taking limit as dv 0

Lim dv 0 dvdkddzpyxxyyxxy

2)(

yxxy

yxxy

0

Stress tensor is symmetric

Applying xx maF

xzzxzzxzxyyxyyxyxxxxxx advdAdAddAdAddAdAddvX )()()()(

Using Ax =dydz etc and dividing by dv

xzxyxx a

dz

d

dy

d

dx

dx

As the elemental volume shrinks to a point in the limit we have

x

xzxyx axzyx

y

yzyyxay

zyx

z

zyzxz azzyx

Where we have used yxxy etc

RECAP2

22 n

22

3

2

2

2

1

22

3

22

2

22

1 )( nmlnml

nml ,, : Direction cosines wrt principal directions

We showed that maximum shear stress as on

n* = 02

)( 21

m* = 02

)( 13

l* = 02

)( 32

If ,321


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Then, greatest shear stress =2

)( 31

Normal stress on maximum shear stress plane

0n

2

)( 21 0l 2

)( 32 and 0m

2

)( 13

Extreme values of shear stress occur on planes each of which is perpendicular to one of theprincipal planes (i.e one DC = 0) and lies at 45 deg to other two.

Planes of maximum

stress

ABCD, EBFD and AFCEThese planes intersect at G which is onthe cross diagonal OP

Maximum shear stress planes arediagonal planes of the cube.

Indeed G is the body center of the cube, the lines CG, EG and DG being equal in length and allperpendicular to OG

Octahedral Shear Stress

With respect to principal directions

Octahedral planes equally inclined to all the principal planes nml


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1222 nml or 13 2 l ,3

1l

knjmilnOH

We have :

3321

oct (Mean Stress)

2

13

2

32

2

21 )()()(3

1 oct

We can also show [See Ford and Alexander,pp.44]

}666)()(){(3

1 222222zxyzxyyyzzzzyyyyxxoct

2

3

2

2

2

1 333

oct

i Principal Deviatoric Stress

11,

22 ,

33

3

)( 321

Mean stress

)(6 133221

oct

Deviatoric StressIt is some times convenient to measure the stress wrt some False Zero and not their absolutevalues.The reference stress is taken as hydrostatic stress or means stress which produces only volumechange:

3)(

3

1)(

3

1 1321

Izzyyxx

Principal stress 1 :


18/30

1 = Mean stress +Deviation from mean

= Hydrostatic stress + Deviatoric stress

1 Prime represents D.5

)2(3

1)(

3

132132111

Similarly

)2(3

1)(

3

131232122

)2(3

1)(

3

121332133

For a general stress state:

)2(3

1zzyyxxxx

etc

zxzxyzyzxyxy ;;

Observe that :zzyyxx

0321

i are called principal Deviatoric components and represents a state of pure shear on

octahedral planes:

octoctoct ;0

Also observe that : is the normal stress on the octahedral plane

Mean Stress =3

)(3

1 1321

Ioct

Deviatoric stress = oct


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2212132322212 39

2)()()(

9

1IIoct

The method of representing stresses in terms of and the deviation from amounts to

considering the normal and shear stress on the octahedral planes.

The Deviatoric stress satisfies the requirements of plasticity theory:

(a) The volume of material must remain constant under plastic deformation

(b) The hydrostatic stress component, , does not cause yielding of the material.

(c) The hydrostatic stress component does not influence the point at which yielding occurs.

From the above points it is clear that Deviatoric stress or octahedral shear stress must govern theyield behavior of material under plastic conditions.

The invariants of Deviatoric stress components are denoted as

321 ,, JJJ

Note

zzyyxxJ 01

We can show that :

2

2

2 3 IJ

3

233

JIJ Also2

13

2

32

2

21 )()()(

2

3

2

2

2

1 333

2133221 6(6 J

Hence )3

2( 2

Joct

Failure Theories Based on Distortion Energy and OHS Stress

tU Total S.E. )(22

1133221

2

3

2

2

2

1 E

per unit volume

Volumetric S.E. = S.E. due to volume change

Governed by

)3/(2

3321

E

UV per unit volume

Now : SVt UUU

SU : S.E due to distortion

213232221 )()()(6

)1(

EUS per unit volume

In uniaxial state of stress


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ySU 1

1(Yield stress)

032

)2(6

)1( 2y

E

D.E theory or Von-mises theory of failure postulates that failure is deemed to have occurredwhen:

D.E. in general stress state = D.E. in uniaxial tension

From equations above

213232221 )()()(6

)1(

E= )2(

6

)1( 2y

E

213232221 )()()(2

1 y

= von : Von Mises Stress

D.E. theory is, therefore, also referred to as vonmises theory

In a two-dimensional stress state

;03 2

21

2

2

2

1

2)( von

Octahedral Theory:

1

octoct

i.e OHS in general stress = OHS in uniaxial tension

213232221 )()()(3

1 oct

yoct 3

2

3

21

1

213232221 )()()(2

1 y

Same as von-mises theory

vonoct 3

2)1(

Deformation: Strain

We seek to express the deformation of a body under loads

Rigid body under motion No change in relative position between particles or nochange of lengths of segments. This suggests that change of distance between any two pointscan serve as a measure of deformation- change of shape and size of the body.


21/30

Strain Deformation

1ds

dsdsE

)(

: Engineering Strain

2 2

22 )(

2

1

ds

dsds

: Lagrangian Strain

32

22 )(

2

1

ds

dsdse : Eulerean Strain

When 1E and 1e then eE

Also Ee

e

)21(2

1 2

Shearing Strain

Consider two directions initially at right angles to eachother.

Shearing strain between two perpendicular directionsis the change (Decrease) in the right angle.

12122

Shearing strain is measured in radians for smallstrains:

1212 sin

)2

sin( 12

1212 cos

Shearing strain between directions (-1 & 2) or (1,-2) is 12

Also 2,112


22/30

Strain Displacement Relations

Let

kzyxwjzyxvizyxuzyxu ),,(),,(),,(),,(

Be the deformation vector at any point (x, y, z). Then we can show that:

ET

n where

DDDDETT

2

1)(

2

1

where: n Lagrangian strain measure on an element in the direction ),,( nmln

Tnmln ,, E Finite Lagrangian strain tensor (Matrix) D Displacement gradient matrix

z

w

y

w

x

w

z

v

y

v

x

v

zu

yu

xu

D

zyzxz

yzyxy

xzxyx

E

Note E is symmetric and non-linear in displacement gradientsIf strains are small nn and we can use E

T

n

For small strain and small rotations displacement gradients are small

Under these assumptions ETn where )(2

1DDE

T

We have set 02

1DDT (Non-Linear part) Or

z

w

y

w

z

v

x

w

z

u

y

w

z

v

y

v

x

v

y

ux

w

z

u

x

v

y

u

x

u

E

)(2

1)(

2

1

)(2

1)(

2

1

)(

2

1)(

2

1

E Trans forms as a second order tensor


23/30

Shearing strain between directions nn &

0. nn

2

0 T

cos

For small strains we can show that

ET

2

1

thus ET

n

ET

2

1

Interpretation of terms in E Let Tin 0,0,1

x

uE

E

xx

T

n

Similarly Tjn 0,1,0

y

vyn

Tkn 1,0,0

z

wEzzn

Diagonal terms of E are the direct strains of line elements along x, y and z directionsConsider

Tin 0,0,1; Tjn 0,1,0;


24/30

Then xyxyxy EE 2

1

)(2

1

x

v

y

u

Similarly

)(2

1

2

1

)(21

21

x

w

z

u

yw

zv

xzxz

yzyz

xzyzxy ,, are tensorial shearing strain components

xzyzxy ,, are engineering shearing strain components

zyzxz

yzyxy

xzxyx

E

2

1

2

12

1

2

12

1

2

1

Strain-Displacement Relations

z

w

y

v

x

uzyx

,,

)(2

1),(),(

x

w

z

u

z

v

y

w

x

v

y

uxzyzxy

Strains are derived from pure geometricconsiderations.

Compatability Equations

Deriving strains from displacement components is simple and straightforward. Whereas,

Obtaining displacements from integrating strains (The reverse problem) is not simple

We have six strains and three displacement components

We need to put restrictions on xy .

i.e we cannot arbitrarily specify xy and expect to derive them from a single-valued continuous

displacement function.

Physically deformations are single-valued and continuousNo gaps in the deformed body.


25/30

The restrictions that make the strain field derivable from a single-valued continuous displacementfield are called compatability conditions.

There are six independent compatability equations:

yxxy

xyyyxx

2

2

2

2

2

----------------------------------------------(a)

zyyz

yzzzyy

2

2

2

2

2

----------------------------------------------(b)

zxzx

xzxxzz

2

2

2

2

2

----------------------------------------------(c)

)(22

zyxxzy

xyxzyzxx

---------------------------(d)

)(2

2

xzyyxz

yzyxzxyy

----------------------------(e)

)(22

yxzzyx

zxzyxyzz

----------------------------(f)

Compatability equations are both necessary and sufficient conditions for existence of single-valued continuous displacement field from which the subject strain field can be derived.

Constitutive Relations or Stress-Strain Relations

Isotropic-Elastic material

Hooks law- Stress StrainFor an elastic material

Youngs Modulus,

E

Shear Stress:G

xy

xy

Rigidity Modulus, G

ExtensionalLongitudin

nContractioLateral =Poissons ratio

)1(2

EG


26/30

In three-dimensions

)(EEE

zzyyxx

xx

G

xy

xy

)(EEE

zzxxyy

yy

G

zy

yz

)(EEE

yyxxzzzz

G

zxzx

xz

yz

xy

zz

yy

xx

zx

yz

xy

zz

yy

xx

E

E

E

EEE

EEE

EEE

)1(200000

)1(20000

00)1(

2000

0001

0001

0001

C C : Compliance MatrixThree elastic constants two are independent


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Boundary value problem in linear elasticity.

We have:

1. 6 Stress Variables/Components

2. 6 Strain Variables/Components

3. 6 Displacement Components

4. 6 Equilibrium Equations

5. 6 Strain-Stress Equations

6. 6 Strain- Displacement Equations

15 Variables

15 Equations

Well-posed problem

Auxiliary Equations

7. 6 Compatability Equations

The solutions must satisfy boundary conditions:

Stresses: Stress or Force boundary conditions

Displacements: Displacement or Kinematic B.C

It is also under stood that of explicit use of displacements are not made, the strain distributionsspecified must also satisfy the compatability equations.

The body forces and surface tractions must be statically compatible.

There are three classes of Boundary Value Problems: (BVP)

1 (Determine the distribution of stresses and displacements in the interior of the body)under a surface traction over the boundary: BVP- First kind

2 (Determine the distribution of stresses and displacements in the interior of the body)under a prescribed displacement distribution over the entire boundary: BVP- Second kind

3 (Determine the distribution of stresses and displacements in the interior of the body)under the action of a given body force distribution with a given traction distribution overthe part of the boundary S1 and a prescribed displacement distribution over the remainingpart of the boundary S2. Mixed BVP.

It should be noted that where tractions)(r

T are specified, it must satisfy the equilibrium equation

Or Cauchys formula on the boundary.

332211

)(vvvT iii

v

i

kvjvivv 321

yx TTT

jTiT yx

xyxxmlT

xyyy lmT


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Solution Approaches

BVP First kind

Consider 2-D case (Plane stress 0 yzxzz )

Equilibrium equations.

0

Xyxxyx -------------------------(1a)

0

Y

yx

yxy ---------------------------(1b)

Strain-Stress

xy

xy

xy

xy

y

yx

x

EG

EE

EE

)1(2

---------------------(2)

Compatability Equations

yxxy

xyyx

2

2

2

2

2

--------------------------(3)

Using (2) in (3)

yxxy

xy

xyyx

2

2

2

2

2 )1(2)()( ---------------------(4)

Compatability equations in terms of stress Components:

Let

yx

x

y

xy

y

x

2

2

2

2

2

--------------------------------------------------------------------------(5)

),( yx

is called Airys Stress Function

The stress components assumed in this form satisfy the equilibrium equations permanently.

We must choose to satisfy the compatability equation (4) above

Using equations (5) in (4)


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22

4

22

4

22

4

2

2

2

2

2

2

2

2 )1(2)()()(

yxyxyxxxyy

or

0222

4

4

4

4

4

yxyx

--------------------------------------------------------------(6)

0;0 224

)(2

2

2

22

yx

Laplacian or Harmonic operator

04 Bi-Harmonic Equation

We have reduced number of differential equations. But, the order of differential equations hasincreased.

BVP of Second kind

Express equilibrium in terms of displacements.

C Strain in terms of stress ------------------------------------------(7)Or

D Stress in terms of strains------------------------------------------(8) 1 CD C = Compliance Matrix

D = Stiffness Matrix

2)1(00

01

01

)1(2

ED ---------------------------------------------------(9)

xyxyxy

yxy

yxx

EE

E

E

2

)1(

)1()1(2

)1(

)1(

2

2

2

-------------------------(10)

Substitute Strain- Displacement Relations

)()1(2

)()1(

)()1(

2

2

x

v

y

uE

y

v

x

uE

y

v

x

uE

xy

y

x

---------------------------------------------------------(11)


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Substituting in equilibrium equations

0

)1(22

)1(

)1(

0

2

2

2

2

2

2

X

yx

vE

y

u

x

uE

Xyx

xyx

-------------------(12)

Similarly:

0)1(22

)1(

)1(

0

2

2

2

2

2

2

Yyx

uE

y

v

x

vE

Yxy

xyy

--------------------(13)

If single valued displacements field satisfying kinematics boundary conditions can be foundsatisfying equations (12) and (13), then exact solution would have been found. The stress fieldderived from such a field should also satisfy the stress B.C.

Uniqueness Theorem

If, an addition to the body forces, either the surface displacements or surface tractions arespecified, there exists only one from of stress and strain distributions in the body.

Applies to:

*Linear Elasticity Equations.

*Small displacements and small displacement gradients.

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Elasticity Edited