Fundamentals of Elasticity Theory

Professor M. H. Sadd

Reference: Elasticity Theory Applications and Numerics, M.H. Sadd, Elsevier/Academic Press, 2009

Theory of ElasticityBased Upon Principles of Continuum Mechanics, Elasticity Theory Formulates Stress Analysis Problem As Mathematical Boundary-Value Problem for Solution of Stress, Strain and Displacement Distribution in an Elastic Body.

Governing Field Equations Model Physics Inside Region (Same For All Problems)

Boundary Conditions Describe Physics on Boundary(Different For Each Problem)

R

Su

St

Value of Elasticity Theory

- Develops “Exact” Analytical Solutions For Problems of Limited Complexity

- Provides Framework for Understanding Limitations of Strength of Materials Models

- Establishes Framework for Developing Linear Finite Element Modeling

- Generates Solutions for Benchmark Comparisons with FEA Solutions

Deformation and Strain

u(x,y)

u(x+dx,y)

v(x,y)

v(x,y+dy)

dx

dy

A B

C D

A'

B'

C'

D'dy

y

u

dxx

v

x

y

xyxy

y

x

x

v

y

ue

y

ve

x

ue

2

1

2

1

Strain Displacement Relations

zzyzx

yzyyx

xzxyx

eee

eee

eee

][eeThree-Dimensional Theory

Two-Dimensional Theory

Deformation and Strain Example

AxCyz

u

x

we

CxCxy

w

z

ve

BxBxx

v

y

ue

z

we

Byy

ve

Azx

ue

CBACxywyxBvAxzu

zx

yz

xy

z

y

x

2

1

2

1

2

10

2

1

2

1

202

1

2

1

0

2

_________________________________________

constants are ,, where, ,)(,

fieldnt displaceme following for thestrain of components theDetermine22

Rigid Body MotionTwo-Dimensional Example

uo

vo

dx

dy

A B

C D

y

uz

x

vz

x

y

xvv

yuu

zo

zo

*

*

Fieldnt Displaceme of Form General

Zero Strains!

Strain Compatibility

yx

e

x

e

y

e xyyx

2

2

2

2

2

2

2

3

1

4

Undeformed Configuration

2

3

1

4

Deformed ConfigurationContinuous Displacements

2

3

1

4

Deformed ConfigurationDiscontinuous Displacements

Discretized Elastic Solidx

yCompatibility Equation

Strain Compatibility Example

BA

BABA

yxBAxAyyx

e

x

e

y

e

yxBxyeAxeAye

xyyx

xyyx

3

2ith equation w satisfiesonly

3

246

)22(2662

__________________________________________

equation ity compatibil ldimensiona- two thesatisfies

)(,,

fieldstrain following theif see Check to

2

2

2

2

2

33

Body and Surface Forces

Sectioned Axially Loaded Beam

Surface Forces: T(x)

SCantilever Beam Under Self-Weight Loading

Body Forces: F(x)

Traction and Stress

F

n

A

(Sectioned Body)P1

P2

P3

p

(Externally Loaded Body)

AA

FnxT n

0lim),(

Traction Vector

Note that ordinary elasticity theory does not include nor allow concentrated moments to exist at a continuum point

Stress Components

3213

3212

3211

eeeenxT

eeeenxT

eeeenxT

zzyzx

yzyyx

xzxyx

),(

),(

),(

n

n

n

zzyzx

yzyyx

xzxyx

][

3

2

1

e

e

eT

)σττ(

)τστ(

)ττσ(n

zzyyzxxz

zzyyyxxy

zzxyyxxx

nnn

nnn

nnn

x

z

y

y

x

yx

z

xy

xz

zy

yz

zx

Stress Transformation

333

222

111

),cos(

nml

nml

nml

xx ji

)()()(

)()()(

)()()(

)(2

)(2

)(2

131313131313131313

323232323232323232

212121212121212121

33333323

23

23

22222222

22

22

11111121

21

21

nllnmnnmlmmlnnmmll

nllnmnnmlmmlnnmmll

nllnmnnmlmmlnnmmll

lnnmmlnml

lnnmmlnml

lnnmmlnml

zxyzxyzyxzx

zxyzxyzyxyz

zxyzxyzyxxy

zxyzxyzyxz

zxyzxyzyxy

zxyzxyzyxx

e1

e3 e2

e3

e2e1

x3

x1

x2

x1

x2

x3

)sin(coscossincossin

cossin2cossin

cossin2sincos

22

22

22

xyyxxy

xyyxy

xyyxx

Two-Dimensional Transformation

Three-Dimensional Transformation

x

y

x'

y'

0 10 20 30 40 50 60 70 80 90-0.5

0

0.5

1

(degrees)

Dim

ensi

on

less

Str

ess

Stress Transformation Example

xx

cossin

cos2

x

x

2cos/ x

cossin/ x

Principal Stresses and Directions

00

)(

)(

)(

Solution Trival-Non Equations, Algebraic of System sHomogeneou

0

)(

)(

)(

0)(

0)(

0)(

322

13

3

2

1

321

321

221

III

n

n

n

nnn

nnn

nnn

zyzxz

yzyxy

xzxyx

zyzxz

yzyxy

xzxyx

zyzxz

yzyxy

xzxyx

Roots of the characteristic equation are the principal stresses Corresponding to each principal stress is a principal direction nnn

that can be used to construct a principal coordinate systemy

(General Coordinate System)

1

3

2

(Principal Coordinate System)

n

n

n

x

z

y x

yx

z

xy

xz

zy

yz

zx

1

3

2

Ii = Fundamental Invariants

Equilibrium Equations

yxxy

yyxy

y

xyxx

x

M

Fyx

F

Fyx

F

0

00

00

x

y

dxx

xx

xy

yx

dyy

yy

dxxxy

xy

dyyyx

yx

xF

yF

Body Forces

Equilibrium Equation Example

0000

02

3

2

30

_______________________________________

)1(4

3,0,

22

3

equations mequilibriu esatisfy th stresses

following that theshow forces,body no Assuming

33

2

2

3

yx

c

Py

c

Py

yx

c

y

c

P

c

N

c

Pxy

yxy

yxx

xyyx

Hooke’s Law

zxyzxyzyxzx

zxyzxyzyxyz

zxyzxyzyxxy

zxyzxyzyxz

zxyzxyzyxy

zxyzxyzyxx

eCeCeCeCeCeC

eCeCeCeCeCeC

eCeCeCeCeCeC

eCeCeCeCeCeC

eCeCeCeCeCeC

eCeCeCeCeCeC

666564636261

565554535251

464544434241

363534333231

262524232221

161514131211

222

222

222

222

222

222

Isotropic Homogeneous Materials

zxzx

yzyz

xyxy

zzyxz

yzyxy

xzyxx

e

e

e

eeee

eeee

eeee

2

2

2

2)(

2)(

2)(

zxzxzx

yzyzyz

xyxyxy

yxzz

xzyy

zyxx

Ee

Ee

Ee

Ee

Ee

Ee

2

11

2

11

2

11

)(1

)(1

)(1

= Lamé’s constant = shear modulus or modulus of rigidityE = modulus of elasticity or Young’s modulusv = Poisson’s ratio

Orthotropic Materials(Three Planes of Material Symmetry)

xy

zx

yz

z

y

x

xy

zx

yz

z

y

x

e

e

e

e

e

e

EEE

EEE

EEE

2

2

2

1

01

001

0001

0001

0001

12

31

23

32

23

1

13

3

32

21

12

3

31

2

21

1

Nine Independent Elastic Constants for 3-DFour Independent Elastic Constants for 2-D

Physical Meaning of Elastic Constants

(Simple Tension)

(Pure Shear)

p

p

p

(Hydrostatic Compression)

000

000

00

ij

E

E

E

eij

00

00

00

000

00

00

ij

000

002/

02/0

ije

xyxye /2/xeE /

ijij p

p

p

p

00

00

00

pE

pE

pE

eij

2100

021

0

0021

pE

ekk

)21(3 kp

ModulusBulk )21(3

pE

k

Relations Among Elastic Constants

Typical Values of Elastic Constants

Basic FormulationFundamental Equations (15) - Strain-Displacement (6) - Compatibility (3) - Equilibrium (3) - Hooke’s Law (6)

Fundamental Unknowns (15) - Displacements (3) - Strains (6) - Stresses (6)

Displacement Conditions Mixed ConditionsTraction Conditions

R

S

R

Su

St

T(n)

R

S

u

Typical Boundary Condtions

Basic Problem Formulations

Problem 1 (Traction Problem) Determine the distribution of displacements, strains and stresses in the interior of an elastic body in equilibrium when body forces are given and the distribution of the tractions are prescribed over the surface of the body.

Problem 2 (Displacement Problem) Determine the distribution of displacements, strains and stresses in the interior of an elastic body in equilibrium when body forces are given and the distribution of the displacements are prescribed over the surface of the body.

Problem 3 (Mixed Problem) Determine the distribution of displacements, strains and stresses in the interior of an elastic body in equilibrium when body forces are given and the distribution of the tractions are prescribed over the surface St and the distribution of the displacements are prescribed

over the surface Su of the body.

Displacement Conditions Mixed ConditionsTraction Conditions

R

S

R

Su

St

T(n)

R

S

u

Basic Boundary Conditions

r

r

r

r

rx

xy=Tx

y=Ty

x

y x=Tx

xy=Ty

y

(Cartesian Coordinate Boundaries) (Polar Coordinate Boundaries)

Coordinate Boundary Examples

Non-Coordinate Boundary Example

x

y

),()( yxFnnT xyxyxxn

x

),()( yxFnnT yyyxxyn

y n = unit normal vector

Boundary Condition Examples

Fixed Conditionu = v = 0

Traction Free Condition

x

y

a

b S

Traction Condition

0)( nyT

x

y

l

0)( nxT

Fixed Conditionu = v = 0

Traction Condition

(Coordinate Surface Boundaries) (Non-Coordinate Surface Boundary)

Traction Free Condition

S

0, )()( xyn

yxn

x TSTSTT y

nyxy

nx )()( ,0

0,0 )()( yn

yxyn

x TT

Symmetry Boundary Conditions

Symmetry Line

0

0)(

n

yT

u

x

y

Rigid-SmoothBoundary Condition

Example Solution – Beam Problem

x - Contours

x

Saint-Venant’s PrincipleThe Stress, Strain and Displacement Fields Due to Two Different Statically Equivalent Force Distributions on Parts of the Body Far Away From the Loading Points Are Approximately the Same.

x

y

P

x

y

P/2 P/2

xy

y

x

xy

y

x

Stresses Approximately Equal

Strain Energy

xxxx

x

eEe

Edxdydz

dUU

dxdydzE

dxdydzE

ddxdydz

x

ud

dydzdudydzdxx

uuddU

xx

xx

2

1

22

Volume

EnergyStrain

2

)(

)(

22

2

00

00

dxu

dz

dxx

uu

dy

x

y

z

Strain Energy = Energy Stored Inside an Elastic Solid Due to the Applied Loadings

One-Dimensional Case

Three-Dimensional Case

0)2

1

2

1

2

1()(

2

12

1)(

2

1

2222222

zxyzxyzyxzyx

ijijzxzxyzyzxyxyzzyyxx

eeeeee

eeeeU

Principle of Virtual Work

0

WUdVuFdSuTdVU TiV iS i

niV t

Change in Potential Energy (UT-W) During a Virtual Displacement from Equilibrium is Zero.

V zxzxyzyzxyxyzzyyxx

V ijijT

dVeee

dVeU

)(

Energy Strain Virtual

The virtual displacement ui = {u, v, w} of a material point is a fictitious

displacement such that the forces acting on the point remain unchanged. The work done by these forces during the virtual displacement is called the virtual work.

dVuFdSuT

W

iV iS in

it

sBody Force andSurface by Done Work Virtual

dVuFdSuTdVe iV iS in

iijV ijt

Virtual Strain Energy = Virtual Work Done by Surface and Body Forces