Stress,Strain & Elasticity

7/27/2019 Stress,Strain & Elasticity

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Stress, Strain, & Elasticity

Mostly from Dieter


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Nine quantities are required to define the

state of stress at a point.

Moment balance shows; tij = tji

Six independent quantities

zxyzxyzyx ,,,,,

zyzzx

yzyxy

zxxyx


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Plane Stress

2-D state of stress

Approached when one dimension of the

body is relatively small (example: thin

Plane stress when s3 = 0

Only three stresses are required

xyyx and ,,


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2/122

2/122

4

2/2cos

4

2sin

xyyx

yx

xyyx

xy

Orientation

2/1

22

2min

1max

22

xy

yxyx

2

21max


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State of stress in 3-D

If , triaxialstate of stress

0321

,

If , hydrostatic state of stress

321

321


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Invariants

2223

2222

1

2 xyzzxyyzxzxyzxyzyx

zxyzxyxzzyyx

zyx

I

I

I

The sum of normal stresses for any orientation

in the coordinate system is equal to the sum of the

normal stresses for any other orientation


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Stress Tensor

Stress is a second-rank tensor quantity Vector is a first-rank tensor quantity

jiji SaS '

' aaa

A scalar, which remains unchanged with transformation ofaxes, is a zero-rank tensor

No. components =3n n tensor-rank

3

2

333231

232221'3

'2

S

S

aaa

aaa

S

S


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Kronecker delta, dij

A second-rank unit isotropic tensor

001

If we multiply a tensor of nth rank with dij,

the product tensor will have (n-2)th rank

jiij

0

100

010


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Components of a stress tensor, sij

Stress is asymmetric tensor

First invariant of the stress tensor,I1

I1

is a scalar


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Second Invariant,I2, is the sum of principalminors

Third Invariant,I3, is the determinant of the

matrix. The three invariants are given by the roots

of the following equation.


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2

2

2

0

02

yx

xy

xy

yx

yx

yx

yxy

xyx


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Hydrostatic stress is given by

Decomposition of the stress tensor:


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J1, J2, and J3 are the principal values of thedeviatoric stress tensor.

J1 is the sum of the diagonal terms:

J2 is the sum of the principal minors:

0)()()(1 mzmymxJ

222

J3 is the determinant of the deviatoric stress matrix.

)(66 222

2

xzyzxy

yxxzyJ


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Displacement, ui

Q(x,y,z) to Q(x+u, y+v, z+w)

u =f(u, v, w)


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1-D Strain

AB

ABBA

L

Lex

''

x

u

dx

dxdxx

udx

u = exx


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Generalization to 3-D

For 1-D, u = exx

Generalizezeyexeu

jiji

zzzyxz

yzyyyx

xeu

zeyexew

zeyexev

eij= ui/xj


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Displacement Tensor

Produces both shear strain and rigid body rotation


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Is eij a satisfactory measure of the strain?

If yes, eij=0 when there is no distortion

Consider a rigid body rotation

0

0

ije


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Decomposition

eij needs to be decomposed into shear strain andrigid-body rotation

Any second-rank tensor can be decomposed into

a symmetric and an anti-symmetric tensor.

jiijijjiijij

ijijij

eeandee

e

2

1

2

1

Strain Tensor Rotation Tensor

Generalized Displacement Relation: jijjiji xxu


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Principal Strains

Similar in concept to principal stresses Can identify, principal axes along which there

are no shear strains or rotations, only pureextension or contraction.

,coincide with the principal stress axes

Definition of principal strain axes: Three

mutually perpendicular directions in the bodywhich remain mutually perpendicular duringdeformation.

Remain unchanged if and only ifvij =0


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Dilatation, D

Volume change or dilatation

Note D is the first invariant of the strain tensor

1'

1)1)(1)(1(

321

321

sfor

Mean Strain, em = D/3

Strain deviator, eij, is the part of the strain tensor

that represents shape change at constant volume

ijijmijj

3

'


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Engineering Shear Strains

Shear Strain, g = a/h = tanq ~ q


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Simple Shear

+Rotation

=

Pure Shear

xy

yxxy

yxxyxy ee

2

Tensor shear strains


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Elasticity (for isotropic solids) Equations that relate stresses to strains are

known as Constitutive Equations

Hookes law: sx=Eex

Poissons Relation:

E

xxzy


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zxzxyzzyxyxy GGG ;;Need only two elastic constants, E and n

)1(2 GShear Modulus,

ijkkijij EE

1


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Other Elastic Constants

Bulk Modulus,

pK m

E

Compressibility, b = 1/K

Lams Constant,

)21(3

)21(

2

G


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Inversion

ijkkijijEE

1

ijijmijj

3

'

ijij

1

ijij G 2kkii K 3

Distortion:

Dilatation:

)1(2 G

pK m


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Plane Stress (s3=0):

1222

2121

1

1

E

E

1 ane tra n e3= :

213 E

122

2

212

1

)1(11

)1(11

E

E


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Strain Energy

Elastic strain energy, U= energy spent by theexternal forces in deforming an elastic body

dU=0.5P du = 0.5(sxA)(exdx) = 0.5(sxex)Adx

22

Strain Energy/vol.,222

0 xxE

U

ijijzxzxyzyzxyxy

zzyyxxU

2

1

2

10


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zxzx

yzzy

xyxy

G

G

G

222

2220

21

2

zxyzxy

xzzyyxzyx

G

EEU

ijij

U

0


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Atomistic Aspects(from Ashby and Jones)

E is influenced by two factors(a) the interatomic bonds spring constant

(b) the packing of atoms no. of springs

Different types of Bonds Primary: Metallic, ionic, and covalent

Strong

Secondary: van der Waals and HydrogenWeak


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Ionic Bondni

r

B

r

qUrU

0

2

4

)(

Attractive

part

Repulsive

part

q-chargee0-permitivity

of vacuum

n~12Lacks Directionality


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Covalent Bond )()( nmr

B

rrU

nm

Hi hl

directional

Metallic bond is

similar

Interatomic


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Interatomic

Forces

drdUF,Force

0rrF for small displacements


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2

2

dr

Ud

dr

dF

S

Stiffness,

When stretching is small, S is a constant

0

2

2

0

rrdr

UdS

Spring Constantof the Bond

00 rrSF


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00 rrNS

0rr

n

No. of bonds/area,2

0/1 rN

00

r

SE

n


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Generalized Hookes Law

Fourth-rank tensors (81 components)

klijklij

klijklijC

S

Sijkl Compliance TensorCijkl Stiffness Tensor

(from Nye: Physical Properties of Crystals)

Symmetry: sij = sji and eij = eji

ijlkijkllkijlkklijkl

lkijlkijklijklij

SSSS

SS

Reduces the no. independent

components from 81 to 36


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Contracted Notation

345

426

561

332313

232212

131211

11

345

426

561

332313

232212

131211

2

1

2

12

1

2

122


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565554535251

464544434241

363534333231

262524232221

161514131211

CCCCCC

CCCCCC

CCCCCC

CCCCCC

CCCCCC

Energy consideration Cmn=Cnm

Reduces the no. of independent constants to 21

Possible to reduce no. of independent constants

further by considering crystal symmetry


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nmnm S

Tensor notation 11 22 33 23, 32 31, 13 12, 21

Matrix notation 1 2 3 4 5 6

S = S when m and n are 1 2 or 32Sijkl = Smn when eitherm orn is 4, 5, or 6

4Sijkl = Smn when both m and n are 4, 5, or 6

S1111 = S112S1123 = S144S2323 = S44

Possible to reduce no of


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Example

Measures extension in Ox3 direction when the

S34 in orthorhombic crystal

Possible to reduce no. of

independent constants further by

considering crystal symmetry

crysta s s eare a out t e x1 rect on

Operate a diad axis parallel to Ox2 direction

Crystal remains unaltered because of symmetry So does extension parallel to Ox3, now under

reverse forces

Implies that S34 has to be necessarily equal to zero


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Cubic Crystals

Let Ox, Oy, and Ozbe parallel to [100],

[010] and [001], respectively.

Rotate by 90. The crystal will look the

same.

665544

312312

CCC

CCC

Remaining constants

vanish


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12111211

1212

12111211

121111

2

2

SSSS

SC

SSSS

SS

C

4444

1

SC


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Isotropic Solids

Form of the matrix can be obtained from thecubic matrix by requiring that the components

should be unaltered by a 45 rotation

GSSS

11441211

12112 SSX


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CompositesIsostrain AnalysisCompatibility:

EEEm

m

f

f

c

c

mfc

fAAfAA

PPP

EA

P

EAEA

P

cmcf

mfc

mm

m

ff

f

cc

c

1/;/

mfcEffEE )1(

Rule-of-Mixtures

I A l i


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Isostress Ananlysis

mfc

mfc

ff

ff

)1(

)1(

fm

mfc

EffE

EEE

)1(


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Upper and Lower Bounds

Stress,Strain & Elasticity

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