Stress, Strain, & Elasticity

Mostly from Dieter

Stress

• 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

Plane Stress

• 2-D state of stress

• Approached when one dimension of the body is relatively small (example: thin plates loaded in the plane of the plate)plates loaded in the plane of the plate)

• Plane stress when s3 = 0

• Only three stresses are required

xyyx and ,,

Principal Stresses• For any state of stress, we can find a set of

planes on which only normal stresses act and the shearing stresses are zero.

• Called Principal Planes and the normal stresses acting on these planes are Principal stresses acting on these planes are Principal Stresses denoted as s1, s2 and s3

• Convention, s1> s2 > s3

• The principal directions are orthogonal to each other

2/122

2/122

4

2/2cos

42sin

xyyx

yx

xyyx

xy

Orientation

Magnitude2/1

22

2min

1max

22

xyyxyx

Magnitude

221

max

State of stress in 3-D

• If , triaxial state of stress

• If , cylindrical state of stress

0321

321 • If , cylindrical state of stress

• If , hydrostatic state of stress

321

321

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 thenormal stresses for any other orientation

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

jiji SaS '

1131211

'1 SaaaS

• A scalar, which remains unchanged with transformation of axes, is a zero-rank tensor

• No. components =3n n –tensor-rank

3

2

1

333231

232221

131211

'3

'2

1

S

S

S

aaa

aaa

aaa

S

S

S

Kronecker delta, dij

• A second-rank unit isotropic tensor

ji

1

010

001

• If we multiply a tensor of nth rank with dij, the product tensor will have (n-2)th rank

ji

jiij

0

1

100

010

• Components of a stress tensor, sij

• Stress is a symmetric tensor

• First invariant of the stress tensor, I1

I1 is a scalar

• Second Invariant, I2, is the sum of principal minors

• Third Invariant, I3, is the determinant of the matrix.

• The three invariants are given by the roots of the following equation.

2

2

20

02

yxxy

xyyx

yx

yx

yxy

xyx

• Hydrostatic stress is given by

• Decomposition of the stress tensor:

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

• J1 is the sum of the diagonal terms:

• J2 is the sum of the principal minors:

0)()()(1 mzmymxJ

)()()(1222

yxxzzy

• J3 is the determinant of the deviatoric stress matrix.

)(6

)()()(

61

2222

xzyzxy

yxxzzyJ

Strain

• A point can be displaced by translation, rotation, and deformation.

• Deformation can be made up of • Deformation can be made up of – dilatation---change in volume

– distortion-change in shape

Displacement, ui

Q(x,y,z) to Q’(x+u, y+v, z+w)

u =f(u, v, w)

1-D Strain

uAB

ABBA

L

Lex

''

x

udx

dxdxxu

dx

u = exx

Generalization to 3-D

• For 1-D, u = exx

• Generalizexzxyxx zeyexeu

jiji

zzzyxz

yzyyyx

xzxyxx

xeu

zeyexew

zeyexev

zeyexeu

eij= ui/ xj

Displacement Tensor

Produces both shear strain and rigid body rotation

• 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

Decomposition• eij needs to be decomposed into shear strain and

rigid-body rotation

• Any second-rank tensor can be decomposed into a symmetric and an anti-symmetric tensor.

e

jiijijjiijij

ijijij

eeandee

e

2

1

2

1

Strain Tensor Rotation Tensor

Generalized Displacement Relation: jijjiji xxu

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

are no shear strains or rotations, only pure extension or contraction.

• For isotropic solids, principal strain axes • For isotropic solids, principal strain axes coincide with the principal stress axes

• Definition of principal strain axes: Three mutually perpendicular directions in the body which remain mutually perpendicular during deformation.

• Remain unchanged if and only if vij =0

Dilatation, D• Volume change or dilatation

• Note D is the first invariant of the strain tensor

• Mean Strain, e = D/3

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

ijijmijíj 3

'

Engineering Shear Strains

h

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

h

Simple Shear

+Rotation

=

Pure Shear

xy

yxxy

yxxyxy ee

2

Tensor shear strains

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

known as “Constitutive Equations”

• Hooke’s law: sx=Eex

• Poisson’s Relation:E

xxzy

zxzxyzzyxyxy GGG ;;

Need only two elastic constants, E and n

)1(2

EGShear Modulus,

ijkkijij EE

1

Other Elastic Constants

• Bulk Modulus,

p

K m

E

K

• Compressibility, b = 1/K

• Lamé’s Constant,

)21(3

EK

)21(

2

G

Inversion

ijijE

ijkkijij EE

1

E

ijijmijíj 3

'

ijij 1

ijij G 2

kkii K 3

Distortion:

Dilatation:

)1(2

EG

p

K m

• Plane Stress (s3=0):

• Plane Strain (e =0):

1222

2121

1

1

E

E

2133 01 E• Plane Strain (e3=0):

213

2133 0

E

12

22

212

1

)1(11

)1(11

E

E

Strain Energy• Elastic strain energy, U = energy spent by the

external forces in deforming an elastic body

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

• Strain Energy/vol., 1 22

xx EU

• Strain Energy/vol., 222

10

xxxx

E

EU

ijijzxzxyzyzxyxy

zzyyxxU

2

1

2

10

1

zxzx

yzzy

xyxy

G

G

G

222

2220

2

12

1

zxyzxy

xzzyyxzyx

G

EEU

ijij

U

0

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– Primary: Metallic, ionic, and covalent

Strong– Secondary: van der Waals and Hydrogen

Weak

Ionic Bondni

r

B

r

qUrU

0

2

4)(

Attractive

part

Repulsivepart

q-chargee -permitivity e0-permitivity of vacuumn~12

Lacks Directionality

Covalent Bond )()( nmr

B

r

ArU

nm

Highly Highly directional

Metallic bond issimilar

InteratomicForces

dr

dUF ,Force

0rrF for small displacements

2

2

dr

Ud

dr

dFS Stiffness,

When stretching is small, S is a constant

2

0

2

2

0rrdr

UdS

Spring Constant

of the Bond

00 rrSF

00 rrNS

0

0

r

rrn

No. of bonds/area, 20/1 rN

0r

0

0

r

SE

n

Generalized Hooke’s Law

• Fourth-rank tensors (81 components)

• Symmetry: s = s and e = e

klijklij

klijklij

C

S

Sijkl – Compliance Tensor

Cijkl – 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

Contracted Notation

345

426

561

332313

232212

131211

56111

345

426

561

332313

232212

131211

2

1

2

12

1

2

122

666564636261

565554535251

464544434241

363534333231

262524232221

161514131211

CCCCCC

CCCCCC

CCCCCC

CCCCCC

CCCCCC

CCCCCC

666564636261 CCCCCC

Energy consideration Cmn=Cnm

Reduces the no. of independent constants to 21Possible to reduce no. of independent constants further by considering crystal symmetry

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

Matrix notation 1 2 3 4 5 6

Sijkl = Smn when m and n are 1, 2, or 3 Sijkl = Smn when m and n are 1, 2, or 3 2Sijkl = Smn when either m or n is 4, 5, or 64Sijkl = Smn when both m and n are 4, 5, or 6

S1111 = S11

2S1123 = S14

4S2323 = S44

Example

• Measures extension in Ox3 direction when the crystal is sheared about the Ox direction

S34 in orthorhombic crystal

Possible to reduce no. of independent constants further by considering crystal symmetry

crystal is sheared about the Ox1 direction

• 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

33

Cubic Crystals• Let Ox, Oy, and Oz be parallel to [100],

[010] and [001], respectively.

• Rotate by 90º. The crystal will look the same.

332211 CCC

665544

312312

332211

CCC

CCC

CCC

Remaining constants vanish

12111211

1212

12111211

121111

2

2

SSSS

SC

SSSS

SSC

4444

12111211

1

SC

Isotropic Solids

Form of the matrix can be obtained from thecubic matrix by requiring that the componentsshould be unaltered by a 45º rotation

GS

ES

ES

11441211

12112 SSX

Composites—Isostrain AnalysisCompatibility:

P

EEE m

m

f

f

c

c

mfc

fAAfAA

PPP

EA

P

EA

P

EA

P

cmcf

mfc

mm

m

ff

f

cc

c

1/;/

mfc EffEE )1( Rule-of-Mixtures

Isostress Ananlysis

mfc

mfc

mfc

E

f

E

f

E

ff

)1(

)1(

mfc EEE

fm

mfc EffE

EEE

)1(

Upper and Lower Bounds