Chapter 1 Fundamentals in...

Dept. of Civil and Environmental Eng., SNU

Structural Analysis Lab.

Prof. Hae Sung Lee, http://strana.snu.ac.kr

1

Chapter 1

Fundamentals in Elasticity

E,

tt

uu Su

St

V




2

1.0 Introduction

Classification of Classic Mechanics

Definition of Mechanics by Encyclopedia Britannica, 2004

Science concerned with the motion of bodies under the action of forces, including the

special case in which a body remains at rest. Of first concern in the problem of motion

are the forces that bodies exert on one another. This leads to the study of such topics as

gravitation, electricity, and magnetism, according to the nature of the forces involved.

Given the forces, one can seek the manner inwhich bodies move under the action offorces;

this is the subject matter of mechanics proper.

What is Classic Mechanics?

The motion of bodies follows the Newton’s laws of motion.

- The law of inertia

- The law of acceleration : maF

- The law of action and reaction

Who started?

역 학(Mechanics)

분 체 역 학(Granular Mechanics)

열 역 학(Thermo Mechanics)

고 체 역 학(Solid Mechanics)

유 체 역 학(Fluid Mechanics)

구조해석(Structural Analysis)

Continuum Mechanics




3

- Galilei Galileo (1564 ~ 1642)

He tried to explain motions of bodies based on observation or experimentation. His

formulation of (circular) inertia, the law of falling bodies, and parabolictrajectories

marked the beginning of a fundamental change in the study of motion. He insisted

that the rules ofnature should be written in the language of mathematics, which

changed natural philosophy from a verbal, qualitative account to a mathematical one

He utilized experimentation as a recognized method for discovering thefacts of nature.

Unfortunately, he did not have mathematical tools known as “calculus”that deals with

differentiation and integration, and thus he failed to describe motions of bodies in a

complete mathematical way.

- Issac Newton(1642~ 1727)

He completed the mathematical formulations of the classic mechanics that Galileo

tried by the virtue of differentiation and integration, which also independently

proposed by a German mathematician Gottfried Wilhelm Leibniz (1646~1716)

almost at the same time..

- Who is next?

We can name hundreds of great scientists such as

R. Hooke (1635~1703): Hooke’s Law

Jakob. Bernoulli (1655~1705): beam theory, study on catenary

Daniel Bernoulli (1700~1782): Bernoulli’s principle for the inviscid flow

L. Euler (1707~1783): Euler equation, Euler buckling load…

T. Young (1773~1829): Young’s modulus, interference of light…

A. Cauchy(1789~1857): Cauchy’s strain, functions ofa complex variable…

Fluids and Solids

- Difference in Material Properties

Fluid flows because EG .

Solid does not because EG .

Inviscid fluid: 0G

Viscous fluid: EG 0

- Fundamental principles in the fluid and solid mechanics are identical except material




4

properties.

- The fluid and solid mechanics can be formulated from exactly the same framework.

- The textbook by Malvern (Introduction to the mechanics of a continuous media)deals

with the solid, fluid and thermo mechanics simultaneously.

Elasticity, Plasticity and Linear Problems

- Bodies with Elastic Material: Recovers the original shapes of a body when external

loads are removed.

- Bodies with Plastic Material: Cannot recover the original shapes of a body when

external loads are removed. Permanent deformation remains in bodies.

- Linear Problems: The principle of the superposition is valid.

BA , then BA

- The elasticity problems may be either linear or nonlinear.

Classification of Engineering Problems

Direct Problems : (k u) = f (Analysis)

Inverse Problems : Reconstruction

Inverse Problems : System Identification

SystemInput Response

SystemInput Measured Response

SystemInput Measured Response

ErrorError

Error

Error




5

1.1 Problem Definition

Prescribed Values

- Domain V and Boundary S

- Material properties

- Boundary Conditions : uu on Su , tt on St, where SSS tu and

tu SS

Unknowns in domain:

- Stress ()

- Strain ()

- Displacement (u)

We have 15 unknowns, and thus have to derive 15 equations to solve the elasticity

problems. Since we determine the 15 unknowns in the domain from the prescribed

boundary conditions, the elasticity problems are a type of mixed boundary value

problems.

What we have to study during this class:

- Stress (): force per unit area developed in the body by the action of external forces

- Strain (): Deformation of body

- Displacement (u): Changes in positions of the body caused by the motion of the body

- Relationships among the unknowns such as equilibrium, strain-displacement

relationship, strain-stress relationship

E,

tt

uu Su

St

V




6

Equilibrium equation:

Mainly concerns the equilibrium state of a given body under the actions of external forces,

and expressed in terms of stress.

0 dSS

t

Strain-displacement relation:

Defines deformation of a given body, and relate strain to displacement.

)(uf

Stress-strain relation:

Represent the material properties of a given body such as the Hooke’s Law.

)( g

Types of Engineering Problems

- Boundary Value Problem (BVP): The unknowns are determined from the prescribed

values on the boundary of a given domain. Especially when the boundary

conditions are expressed in terms of the unknowns themselves as well as their

derivative, then the problems are referred to the mixed BVP is given as:

0)(

0)0( and 0for 0)(

2

2

dx

ldlxf

dx

xd

- Initial value problems (IVP): The unknowns of a given problem are determined from

initial values prescribed at a reference time, usually at 0t .

0)0(

0)0( and 0for 0)()(

2

2

dt

dttf

dt

td

- Initial-boundary Value Problems: we have to utilize both the initial and boundary

values to solve a given problem.

0),()0,( and 0 and 0),0(

0),0( and 0for ),(),(),(

2

2

2

2

lttlxdt

xdxtxtf

dx

xtd

dt

xtd




7

1.2. Domain and Boundary




8

1.3. Definition of Continuum

Definition of a continuous set

A set A is called as a continuous set if there always exists Ac which lies between a

and b for all Aba , .

Example: real numbers

Continuous distribution

A physical quantity is said to be continuously

distributed if the following limit is uniquely defined

everywhere in the given domain V.

n

n

Vn V

MP

n

lim0 ,

)(

where Mn is the sum of the quantity in Vn, and

1 nn VV , nVP , VV 0.

Continuous body or continuum in a mathematical sense

A body is called as continuum if material particles are continuously distributed in the

body or there exists the continuous density function.

Continuous body or continuum in a real sense

Instead of 0nV as n , nV approaches a finite number as n . Then,

n

n

Vn V

MP

n

)(lim ,

where is an acceptable variability. You may consider as the smallest volume you

can differentiate with your own naked eyes.A body is referred to as a continuum if

material particles are distributed in a continuous fashion. Therefore, conceptually, you

can pick a material particle between any two material particles in the continuous body.




9

1.4. Fundamental Laws

Newton’s Laws of Motion

- The momentum of an object (mv) is constant unless an external force acts on the

object; this means that any object either remains at rest or continues uniform motion

in a straight line unless acted on by a force.

- The change of the momentum with respect to time of an object is equal to the force

acting on the object.

- For every action (force) there is an equal and opposite reaction (force).

The 1st law of thermodynamics : The conservation of energy

The 2nd law of thermodynamics

Defines the direction of energy flow. That is, energy always flows from the high level to

the low level spontaneously. Work should be applied to reverse the spontaneous energy

flow.




10

1.5. Axioms

1. Newton’s laws of motion and the 1st and 2

nd law of thermodynamics are valid.

2. A material continuum remains a continuum under the action of force.

3. Stress and strain can be defined everywhere in the body.

4. The stress at a point is related to the strain and the rate of change of strain at the same point.

))(),(()( PPP f VP

For a rate-independent material, ))(()( PP f VP where the temporal rate =dt

d ) (

and spatial change =x

) (,

y

) (,

z

) (.

The consequence of the last axiom?

If the stress at a point were influenced by strains at the other points, the stress-strain

relation should include the spatial derivatives of the strain to represent the spatial change

of strain, which results in differential equations. Since, however the last axiom states that

the stress at a point depends only the strain at the same point, the stress at a point is

independent of the spatial derivatives of strain. The stress-strain relation representing

material properties of a given domain is expressed algebraically in the spatial domain,

and 6 equations out of 15 governing equations become algebraic equations rather than

differential equations, which make the elasticity problems much simpler. Since the rate-

dependent material indicates that the stress at a point depends on temporal rate change of

strain at the point, the stress-strain relation becomes differential equations in the time

domain.




11

1.6. Tensors and Operations

Tensorial notation : A

Indicial notation :Ai , Aij , Aijk , Aijkl

Summation notation : the repeated subscripts or superscripts in a term denotes summation.

The repeated index is called as “dummy index”.

kkiinn

n

i

ii BABABABABABA

2211

1

222

2

2

1 inii AAAAAA

ninii

n

j

jijjij BABABABABA

2211

1

n

i

ininiiii

n

i

n

j

ijijijij BABABABABA1

2211

1 1

)(

Dot product

A dot product of any two variables represents the summation on the last index of the first

variables and the first index of the second variable.

nnii BABABABA 2211BA

njnjjiji BABABABA 2211BA

njnjjiji ABABABAB 2211AB

nnjjjiij

T BABABABA 2211BA

njinjijikjik BABABABA 2211BA

Cross product of two vectors

BAC

321

321

321

BBB

AAA

iii

C , sinBAC




12

Kronecker delta

for 1

for 0

ji

jiij

Permutation tensor

equal are indices Any two

npermutatio oddfor

npermutatioeven for

0

1

1

ijke

Determinant of a matrix A

)(Det)(Det 321321

T

kjiijkkjiijk AAAeAAAe AA

kjijkiikjijkkjiijk BAeCBAeBAe

BA

BA

BA

BBB

AAA ii

i

i

iiii

BAC

333

222

111

321

321

321

In general, pnkjipijk AAAAe 321)(Det A

Tensor fields

In case the coordinate rotation is defined by ijij xx

Scalar field (tensor field of rank 0) : ),,(),,( zyxzyx

Vector field (tensor field of rank 1) : ),,(),,( zyxzyx ijij

Tensor field of rank 2 : jnmnimij zyxzyx ),,(),,(

Tensor field of rank 4 : lskrpqrsjqipijkl zyxzyx ),,(),,(

x

y

z

x

y

z

x




13

Gradient operator

),,(321 xxx

ixxxxxxx

),,(),,(

321321

22

2

3

2

2

2

2

2

1

2

3

2

2

2

2

1

2

321321

)(),,(),,(

ii xx

xxxxxxxxxxxx

iiii xxxx

2

3

3

2

2

1

1

xxxxx

jjj

i

ij

ij

i

Divergence (Gauss) Theorem

3

3

2

2

1

1 divx

F

x

F

x

F

x

F

i

i

FF

dSdVSV

nFF

dSgdVgdVgdVgSVVV

nFFFF )(

dVgdSgdVgVSV

FnFF




14

Chapter 2

Traction and Stress




15

2.1. Traction and Stress

Body Force: dV

d

VV

PPb

lim

0

Traction: dS

d

SΔSn

FFT

0lim

Stress : x, y, z components of traction

x

n

Tn

S

F

P




16

2.2. Definition of Stress

z, x3

xx

xz

xy

zx

zz

zy

yx yz

yy

x,x1

y, x2

xx

xz

xy

zx zz

zy

yx

yz

yy

x,x1

z, x3

y, x2




17

2.3. Cauchy’s Tetrahedron and Relation

Equilibrium Condition

33

3

222

111

321

2

2

332211

2

2

3

3

2

2

1

1

),cos( ,),cos( ,),cos(

3

1

3

1

3

1

3

1

0

nxnOC

h

S

Snxn

OB

h

S

Snxn

OA

h

S

S

SOCSOBSOAShV

S

V

dt

ud

S

Vb

S

St

S

St

S

StT

Vdt

udVbStStStST

i

iiii

n

i

i

iiii

n

i

Cauchy’s relation

By the definition of stress ji

j

it and as 0V

jji

n

iiii

n

i nTntntntT 3

3

2

2

1

1

t2

t3 t1

Tn n

O A

B

C h

n

O A

C h

B

x1 x1

x2 x2

x3 x3




18

2.4. Equilibrium Equation - Integral Approach

Force Equilibrium: 0 iF for i=1, 2, 3

02

2

V

i

S V

ii dVt

udVbdST or 0

2

2

VS V

dVt

dVdSu

bT

By the divergence theorem, and the Cauchy’s relationship

V j

ji

S

jji

S

i dVx

dSndST

Therefore, the force equilibrium equation becomes

0)(2

2

2

2

V

ii

j

ji

V

i

V

i

V j

jidV

t

ub

xdV

t

udVbdV

x

Since the above equation should satisfy for all bodies under equilibrium, the integrand

should vanish everywhere in the domain.

02

2

t

ub

x

ii

j

ji 02

2

,

t

ub i

ijji 02

2

t

ub

2

3

2

3

3

33

2

23

1

13

2

2

2

2

3

32

2

22

1

12

2

1

2

1

3

31

2

21

1

11

t

ub

xxx

t

ub

xxx

t

ub

xxx




19

Moment Equilibrium: 0 iM for i=1, 2, 3

02

2

VVVS

dVt

dVdVdSu

xmbxTx or the indicial form becomes

02

2

V

nmimn

V

i

V

nmimn

S

nmimn dVt

uxedVmdVbxedSTxe

V j

jn

mmnimn

V j

jn

mjnmjimn

V j

jn

mjn

j

m

imn

V j

jnmimn

S

jjnmimn

S

nmimn

dVx

xedVx

xe

dVx

xx

xedV

x

xedSnxedSTxe

)()(

)(

0)(2

2

V

n

mimn

V

i

V

nmimn

V j

jn

mmnimn dVt

uxedVmdVbxedV

xxe

0)()(2

2

V

n

j

jn

nmimn

V

imnimn dVt

u

xbxedVme

00)( imnimn

V

imnimn medVme

In case there is no body moment, 0mnimne

21123

31132

32231

0

0

0

0

mnmn

mnmn

mnmn

mnimn

e

e

e

e




20

2.5. Conservation and Potential Problems

Conservation in General

Only the vector component normal to the surface (or boundary) flows into or out the

volume. The tangential component just flows along the boundary without any effect on

the vector field in the volume. Therefore, the conservation of the vector field is

expressed as:

S V

fdVdS 0nv

The minus sign implies that in-flow direction is taken as the positive direction. As the

outward normal vector always points outside of the volume, the inflow direction should

be opposite to the outward normal vector of the surface.

– By divergence theorem,

VS

dVdS vnv where ),,(),,(321 xxxzyx

.

VVS VV

dVffdVdVfdVdS 0)( vvnv

– Since the integral equation should hold for all systems,

0 fv

v

n

dS




21

Potential Problems

– In a potential problem, the vector field of a system is defined by a gradient of a scalar

function referred to as a potential function

kv

– The famous Laplace equation for a conservative system.

0)( ff kv

– the system properties are homogeneous and isotropic, kv

0)( 2 fkfkfv or

0)(2

2

2

2

2

2

2

f

xxkf

zyxk

ii




22

2.6. Equilibrium and Potential Problems

Equilibrium

– Force Equilibrium: 0zyx FFF or 0F

S V

dVdS 0bT or S V

ii dVbdST 0 for i = 1,2,3

– Suppose nT or

3

1j

ijiji nT n

3

2

1

333231

232221

131211

,

3

2

1

n

n

n

n

– Divergence Theorem

0)(

V

ii

V V

ii

S V

ii

S V

ii

dVb

dVbdVdVbdSdVbdST

n

for i = 1,2,3

– The integral equation should hold for all systems in equilibrium.

0 ii b for i = 1, 2, 3

– Moment Equilibrium: 0 iM for i=1, 2, 3 or 0M

0 VVS

dVdVdS mbxTx or 211231133223 , ,

Potential Problems

– In case elasticity problems are potential problems

i

i

i uC or k

ii

jkijx

uC

where repeated index idoes not indicate the summation. However, in general,

k

jj

ikji

k

ii

jkijx

uC

x

uC

because ui and uj are independent potential functions.

Therefore, to maintain symmetry condition of stress, each stress component should

be a function of the gradients of all components of the potential functions, and

furthermore the material properties should be defined as a fourth order tensor rather

than a second order tensor:

uC : or l

kijklij

x

uC

and jiklijkl CC




23

– In case ijlkijkl CC

klijkl

k

l

l

kijkl

k

lijkl

l

kijkl

l

kijlk

l

kijkl

l

kijklij

Cx

u

x

uC

x

uC

x

uC

x

uC

x

uC

x

uC

)(2

1

)(2

1)(

2

1

– Equilibum equation in terms of the potential functions: 0):( buC

2.7. Rotation of Axis and Stress

Suppose a new primed coordinate system is defined, and we want to expresss each

component of a vector in the primed coordinate system.

Rotation of Axis

332211332211 eeeeeex xxxxxx

iijjiijkkjiijkkjkkii xxxxxxxx eeeeeeeeee

formin tensor or ) ,cos( where βxxee ijijjiijij xxxx

xβxeeeeeeeeee T

kkjkjkjkkjijikkjiijkkii xxxxxxxxx

Orthogonality

ijkikjikikjijikikjkkjj xxxx 0)(

ΤΤijkjki 1

Diffential Operator

k

kimi

k

km

i

mkm

ki

k

ki xxx

x

xx

x

xx

()()()()()

x

y

z

x

y

z

x




24

Tractions in the directions of the primed (new) axes in the original CS.

- Stress in each coordinate system by the deficition: i

j

x

xij T , i

j

x

xij T

- Cauchy’s relation: jkki

x

kki

x

x

xx

x

xx

x

xx

x

x

xjj

i

j

i

j

i

j

i

j

inTnTnTnTT

321321

- Normal vector: jkkjkj

x

kk

x

k

xxx

j

xxxnnnnn jjjjjj

),cos(332211 eeeeeeen

The above tractions are given in the original CS, and thus each traction vector should be

transformed to represent the stress in the primed CS.

Stress in the primed coordinate system

T

kpqpmqmqqpkp

x

xkp

x

xmkm

p

m

kTT

ljklkiij

T

Equilibrium Equation in primed coordinate system

- Stress: liklkjji , Body force: llii bb , Diplacement: llii uu

0)(2

2

2

2

2

2

2

2

t

ub

xt

ub

x

t

ub

xt

ub

x

ll

k

klli

llillikmli

m

kl

llillimjlikj

m

klllillimj

m

liklkj

2

2

t

ub

x

ll

k

kl

- The equilibrium equation is independent of a selection of coordinate system.

x2

x1

x3

O A

B

C

jx

jxT

1xT

2xT

3xT




25

Chapter 3

Displacement and Strain




26

3.0. Kinematic Description

Kinematics?

A subdivision of classical mechanics concerned with the geometrically possible motion

of abody or system of bodies without consideration of the forces involved (i.e., causes

and effects of the motions).

Kinematic Description

Description of the spatial position of bodies or systems of material particles, the rate at

which the particles are moving (velocity), and the rate at which their velocity is changing

(acceleration)

Do the wind velocity and the velocity of a car have the same physical meaning?

How to describe the traffic condition of Olympic Highway?

t t+t




27

Method I: Drive your own car on Olympic highway, and measure the velocity of your

car with time. If you differentiate the measured velocity with respect to time, the

acceleration of your car can be obtained.

aVVV

t

ttt

t

t

tc

)()(),car(lim

0fixedar

If the velocity of every car on the highway is measured and reported to you for all

time, then you have a complete description of the traffic condition of the Olympic

highway. No difficulty is encountered in formulating problems with this kinematic

description (solid mechanics).

Method II: Sit down at any location of your choice near the highway and observe and

record the velocities of cars passing in front of you.If the measurement is taken place

at every location on the highway, then you also have a complete description of the

traffic condition of the Olympic highway. If you differentiate the recorded velocity at

a location, the true acceleration defined by Sir Isaac Newton cannot be obtained.

This is because you just calculated change of velocities of two cars passing though

you at a specification location at two different time.

aVVxV

x

t

ttt

t

t

t

)()(),(lim

0fixed

Notice that )( ttV and )(tV are velocoities of two different cars measured at

different times at the same spatial location. The true acceleration becomes (out of

scope of this class)

VVxV

ax

fixed

),(

t

t

This method is very convenient way to TBS, but results in a very, very, very (…)

complicate situation in solving physical problems with this kinematic description

(fluid mechanics).

Total change in the velocity of a fixed material particle

ttxVx

ttxVt

t

txVtxV

),(

),(),(),(

fixed fixed Material

x

),(),(),(),(

lim

fixed fixed Material0

txVx

txV

t

txV

t

txVa

t

x




28

Referential (Lagrangian) Description: Method I

Independent variables are the position x of the material particle in an arbitrarily chosen

reference configuration, and the time t. In elasticity, the reference configuration is

usually chosen to be the natural or unstrained position (known). When the reference

configuration is chosen to be the actual initial configuration at 0t , the referential

description is called the Lagrangian description. The reference position x is used for a

label (or name) for the material particle occupying the position x in the reference

configuration. All variables are considered as functions of x. Notice that x denotes the

material particle occupying the position x in the reference configuration, not just a spatial

point.

x

3

x

1

x

2 x X

Reference configuration

t

),( txXX

Velo

city

Observation Point

V(t)

V(t+ t)

V t

tt

txV

fixed

),(

x

tVx

ttxV

fixed

),(

x




29

Spatial (Eulerian) Description: Method II

The independent variables are the spatial position vectors x. The spatial description

fixes attention on a given region of space (control volume) instead of on a given body of

matter. The material particles occupying a fixed position change for different time. It is

the description most used in fluid mechanics, often called the Eulerian description, which

gives us “Navier-Stokes equation”, the most notorious partial differential equation in the

engineering field.

3.1. Displacement and Strain

Displacement is defined as the change of the spatial position of a fixed material particle.

Displacement

xxaxu )()( or )()( axaau

x1

x2

x3

u(x)

x a

Control Volume: analysis domain

x

3

x

1

x

2 x

t+t

t




30

Strain

iidxdxdxdxdxds 2

3

2

2

2

1

2

0 , iidadadadadads 2

3

2

2

2

1

2

k

k

i

i dxx

ada

,

k

k

i

i daa

xdx

lk

l

j

k

i

ijl

l

j

k

k

i

ijjiijii dadaa

x

a

xda

a

xda

a

xdxdxdxdxds

2

0

lk

l

j

k

i

ijl

l

j

k

k

i

ijjiijii dxdxx

a

x

adx

x

adx

x

adadadadads

2

jiijjiij

ji

jiijji

ji

jiijlk

l

j

k

i

ij

dxdxEdxdxx

a

x

a

dxdxdxdxx

a

x

adxdxdxdx

x

a

x

adsds

2)(

2

0

2

Eij is called as Green’s strain.

jiijji

ji

ij

ji

ji

jiijlk

l

j

k

i

ijjiij

dadaedadaa

x

a

x

dadaa

x

a

xdadadada

a

x

a

xdadadsds

2)(

2

0

2

eij is called as Almansi strain.

x1

x2 u(x) x

a

dx

da

x3




31

Green’s Strain in terms of displacement

)(2

1

)(2

1

2

1))((

2

1

))()(

(2

1)(

2

1

jii

j

j

i

ij

jii

j

j

iji

ij

j

j

i

i

ij

ji

ij

ji

ij

x

u

x

u

x

u

x

u

x

u

x

u

x

u

x

u

x

u

x

u

x

ux

x

ux

x

a

x

aE

)(2

1)(

2

1)(

2

1

)(2

1)(

2

1)(

2

1

)(2

1)(

2

1)(

2

1

])()()[(2

1)(

2

1

])()()[(2

1)(

2

1

])()()[(2

1)(

2

1

2

3

3

3

2

2

3

2

2

1

3

1

2

3

3

2

232

3

3

232

3

3

1

3

3

2

1

2

3

1

1

1

3

1

1

3

313

1

1

3

13

2

3

1

3

2

2

1

2

2

1

1

1

2

1

1

2

212

1

1

212

2

3

32

3

22

3

1

3

3

333

3

33

2

2

32

2

22

2

1

2

2

112

222

2

1

32

1

22

1

1

1

1

111

111

x

u

x

u

x

u

x

u

x

u

x

u

x

u

x

u

x

u

x

u

x

u

x

uE

x

u

x

u

x

u

x

u

x

u

x

u

x

u

x

u

x

u

x

u

x

u

x

uE

x

u

x

u

x

u

x

u

x

u

x

u

x

u

x

u

x

u

x

u

x

u

x

uE

x

u

x

u

x

u

x

u

x

u

x

u

x

uE

x

u

x

u

x

u

x

u

x

u

x

u

x

uE

x

u

x

u

x

u

x

u

x

u

x

u

x

uE

Cauchy’s infinitesimal strain tensor – Small deformation

)(2

1

i

j

j

i

ijx

u

x

u

for small displacement gradient 1

j

i

x

u

In this case, ijijij eE

All kinds of strain are symmetric.

Rigid body motion : 0dsds

jiijjiij dadaedxdxEdsds 222

0

2 for dx and da. 0 ijij eE for whole domain.

For small strain, 0ij.

Engineering strain : ijij 2 for ji




32

Strain Component in primed coordinate system

– Diplacement : kkii uu

– Strain

mjkmkimj

m

l

k

l

k

m

m

k

ki

mj

m

l

k

l

kiki

k

m

mjmj

m

k

ki

nj

n

l

m

l

miki

k

m

mjmj

m

k

ki

nj

n

l

mi

m

k

klki

k

mmj

mj

m

kki

nj

n

llmi

m

kkmi

m

kkj

mj

m

kki

j

ll

i

kk

i

kkj

j

kki

ij

Ex

u

x

u

x

u

x

u

x

u

x

u

x

u

x

u

x

u

x

u

x

u

x

u

x

u

x

u

x

u

x

u

x

u

x

u

x

u

x

u

x

u

x

u

x

u

x

uE

)(2

1

)(2

1

)(2

1

)(2

1

)(2

1

)(2

1

TT EEEE ,




33

3.2. Physical Meanings

Normal Strain

For x1 direction, )0,0,( 1dx

2

0111111

2

0

2 222 dsEdxdxEdxdxEdsds jiij , 01121 dsEds

0

0

0

0

2

0

2

0

2

1122 ds

dsds

ds

dsds

ds

dsdsE

Since 0dsds for small strain, then0

0

1111ds

dsdsE

Shear Strain

),,(),,(),,(

),,(),,(),,(

2

2

3

2

2

22

2

1321321

1

1

3

1

1

21

1

1321321

xdx

axd

x

axd

x

axd

x

axd

x

axd

x

aadadadd

dxx

adx

x

adx

x

adx

x

adx

x

adx

x

adadadad

i

i

i

i

i

i

i

i

i

i

i

i

a

a

211221

21

2

cos

xddxExddxx

a

x

axddx

x

a

x

a

xdx

adx

x

aaddasdsddd

kknm

n

k

m

k

n

n

km

m

kkk

aa

→sdsd

xddxE 21122cos

222022111011 2121 , 2121 xdEsdEsddxEdsEds

2211

1212

2121

2)sin(cos

EE

E

, 2121)sin(

2

1 22111212 EEE

In case 1 , 2211 EE , 1212122

1 )sin(

2

1 E

a

ds

)0,0,( 1dxd x

0ds

)0,,0( 2xdd x

a

aa d

aa d

sd

ds

)0,0,( 1dxd x

aa d




34

3.3. Compatibility in 2-D Problems

In case the strain field is given,does the unique displacement field possibly exist ?

, , 3

333

2

222

1

111

x

u

x

u

x

u

)(2

1 , )(

2

1 , )(

2

1

2

3

3

232

3

1

1

313

2

1

1

212

x

u

x

u

x

u

x

u

x

u

x

u

- In case that displacement components are given, six independent strain components

can be defined without any ambiguity.

- What if six independent strain components are given to define displacement of a body?

Can we always calculate three independent displacement components uniquely?

- There are six strain components, but only three displacement components exist.

- We have to determine three unknowns using six equations, which generally is

impossible to solve. Therefore, all the strain components are not independent.

- Three additional relationships between strain components should be defined.

- We refer to the above conditions as the compatibility condition!

Ex: Two dimensional Case (Plane strain condition)

0231333 , )(2

1 , ,

1

2

2

112

2

222

1

111

x

u

x

u

x

u

x

u

)( , )( 1222221111 xfdxuxgdxu

))()(

(2

1)(

2

1

1

1222

12

2111

21

2

2

112

x

xfdx

xx

xgdx

xx

u

x

u

- Compatibility condition

)(2

1222

2

2

1

2

111

1

2

2

2

21

12

dx

xxdx

xxxx

2

22

2

2

11

2

21

12

2

2

22

2

2

11

2

21

12

2

1212

, )(2

1

xxxxxxxx

The strain field can not be defined independently !!!




35

3.4. General Case

Path Independent condition for displacement

- The displacement field should be single-valued functions, which are independent of

path.

- This condition is related to the second axiom and the finiteness of strain energy.

C

kjk

C

kjkj

C

k

j

k

k

j

C

k

j

k

k

j

j

C

k

k

j

j

C

jj

p

j

dxdxu

dxx

u

x

udx

x

u

x

uudx

x

uuduuu

0

000 )(2

1)(

2

1

C

m

m

jkp

kkjkk

p

k

C

p

kkjk

C

kjk dxx

xxxxxxddx )()()( 00

j

km

k

mj

k

m

m

k

jj

m

m

j

k

kj

m

kj

m

j

k

k

j

mm

jk

xxx

u

x

u

xx

u

x

u

x

xx

u

xx

u

x

u

x

u

xx

)(2

1)(

2

1

2

1

2

1)(

2

122

C

mjmjkk

p

kj

C

m

j

km

k

mjp

kkjmjkk

p

kj

C

kjk

C

kjkj

p

j

dxFxxu

dxxx

xxxxu

dxdxuu

000

000

0

)(

)))((()(

u0

up

x0

xp




36

For the path independent condition

n

jm

m

jn

x

F

x

F

)])(([)])(([j

km

k

mjp

kkjm

nj

kn

k

njp

kkjn

m xxxx

xxxxx

x

0])[(

)()(

2222

jn

km

kn

mj

jm

kn

km

njp

kk

j

km

k

mj

kn

n

jm

j

kn

k

nj

km

m

jn

xxxxxxxxxx

xxxxxx

0])[(

)()(

2222

jn

km

kn

mj

jm

kn

km

njp

kk

j

nm

n

mj

n

jm

j

mn

m

nj

m

jn

xxxxxxxxxx

xxxxxx

02222

jn

km

kn

mj

jm

kn

km

nj

xxxxxxxx(by E. Cesaro)

St.-Venant’s Compatibility Equations

0)(

0)(

0)(

02

02

02

3

12

2

13

1

23

321

33

2

3

3

12

2

13

1

23

231

22

2

2

3

12

2

13

1

23

132

11

2

1

21

12

2

2

1

22

2

2

2

11

2

3

31

13

2

2

3

11

2

2

1

33

2

2

32

23

2

2

2

33

2

2

3

22

2

1

xxxxxxU

xxxxxxU

xxxxxxU

xxxxR

xxxxR

xxxxR

We need only three equations, but six equations are resulted in. What happens?




37

Bianchi formulas

0 , 0 , 03

3

2

1

1

2

3

1

2

2

1

3

3

2

2

3

1

1

x

R

x

U

x

U

x

U

x

R

x

U

x

U

x

U

x

R

The Bianchi formulas show that the St. Veanat’s equations are not independent, but three

compatibility equations out of six are dependent. Unfortunately, we still have one more

issue on the selection of three independent relationships out of six!

Dependence of Compatibility(K. Washizu, 1957)

Suppose one set of the compatibility equations are satisfied in the domain, while the other

set are satisfied on the boundary. Then, the compatibility equations specified on the

boundary are automatically satisfied in the domain. Therefore, only one set of the

compatibility equations is required to be satisfied either in the domain or on the boundary.

- Case 1

0321 UUU in V and 0321 RRR on S. By Bianchi condition

03

3

2

2

1

1

x

R

x

R

x

R in V 0321 RRR in V

- Case 2

0321 RRR in V and 0321 UUU on S. By Bianchi condition

0 ,0 ,02

1

1

2

3

1

1

3

3

2

2

3

x

U

x

U

x

U

x

U

x

U

x

U in V .

For arbitrary ),,( 321 xxxF , ),,( 321 xxxG and ),,( 321 xxxH

0

)()()([

])()()[(

)()()([

21

3

31

2

32

1

321

2

1

1

2

3

1

1

3

3

2

2

3

dVx

F

x

GU

x

F

x

HU

x

G

x

HU

dSUmFlGUlHnFUmHnG

dVx

U

x

UH

x

U

x

UG

x

U

x

UFI

V

S

V

Since ),,( 321 xxxF , ),,( 321 xxxG and ),,( 321 xxxH arearbitrary,

0321 UUU in V.




38

Chapter 4

Constitutive Relations




39

4.1. Constitutive Law

Generalized Hooke’s Law

klijklklijklij CEC

Governing Equations in solid mechanics

– Equilibrium : 2

2

,t

ub i

ijij

– Strain-displacement relation : )(2

1

i

j

j

iij

x

u

x

u

– Constitutive law : klijklklijklij CEC

Equilibrium Equations in terms of displacement

2

22

t

ub

xx

uC i

i

lj

kijkl

Boundary conditions

– Displacement BC :uii uu on

– Traction BC :tii TT on

Traction BC in terms of displacement

ij

l

k

ijkljiji Tnx

uCnT

4.2. Isotropic Material

Isotropy ?

“Iso” means identical or same, and “tropy” means “tendency to change in response to a

stimulus”. “Isotropy” stands for identical tendency to change in response to a stimulus.

Therefore, “an isotropic material” indicates a material exhibiting identical propertieswhen

measured along axes in all directions. In other words, the material properties are

independent of the selection of axes.

Characteristics of Elasticity Tensor ijklC

– Total number of coefficient: 81 ( )3333

– Symmetry on and : jiklijkl CC ,

ijlkijkl CC (36)

– Symmetry w.r.t ij and kl : klijijkl CC (21=(36-6)/2+6)




40

Independent Relations

Rel. Coefficient # of Coeff. # of Ind. Rel.

N-N 333322221111 CCC 3 1

N-O.N. 113322331122 CCC 3 1

S-S 131323231212 CCC 3 1

S-O.S 132323121213 CCC 3 1

N-S

(S-N) 332333133312

222322132212

112311131112

CCC

CCC

CCC

9 1

Total 21 5

Normal-Shear Relation =0

Case I Case II

– Case I ( 0 , 0 231312332211 )11121112 CI

– Case II ( 0 , 0 231312331122 )22122212 CII

By isotropic condition, III

1212 , for 2211 .

0020)( 12111112111112221211 CCCC

11

12

22

12




41

Shear-Other ShearRelation

Case I Case II

– Case I ( 0 , 0 131233221123 )23122312 2 CI

– Case II ( 0 , 0 231333221112 )12231223 2 CII

By isotropic condition, III

2312 , for 2312 .

0040)(2 12231212231223121223 CCCC

31

23

12

33

22

11

3131

23312323

123112231212

3331332333123333

22312223221222332222

113111231112113311221111

31

23

12

33

22

11

.

C

CCsymm

CCC

CCCC

CCCCC

CCCCCC

90 Rotation

N-O.N (A2) N-S (S-N)

N-N (A1)

S-O.S

S-S (A3)




42

expressions

Rel. Coefficient relation Value

N-N 333322221111 CCC lpkpjpip A1

N-O.N. )( 331133222211

113322331122

CCC

CCC

lpkpjpipklij A2

S-S

)(

)(

)(

313132322121

133123321221

311332232112

131323231212

CCC

CCC

CCC

CCC

lpkpjpipjkiljlik 2 A3

The expressions in parentheses represent dependent relations.

Superposition

)2( )( 321 lpkpjpipjkiljliklpkpjpipklijlpkpjpipijkl AAAC

lpkpjpipjkiljlikklijijkl AAAAAC )2()( 32132

Rotation of Stress and Strain

qlpqpkkljnmnimij ,

klijkl

kllpkpjpipkljkiljlikklklij

kllpkpjpipkljkiljlikklklij

pqqrprjrirpqjpiqjqippqpqij

pqqrprjrir

pqqmpnjnimqnpmjnimpqqkpkjmim

pqqlpklrkrnrmrjnim

pqqlpknkmlnlmkjnimpqqlpkklmnjnim

pqqlpkmnkljnimqlpqpkjnmnklimij

C

AAAAA

AAAAA

AAAAA

AAA

AA

AAA

AA

CC

)2(][

)2(][

)2(][

)2(

][

)2(

)(

32132

32132

32132

321

32

321

32

For isotropy, ijklijkl CC

lpkpjpiplpkpjpip AAAAAA )2()2( 321321

321321 202 AAAAAA

Final Relation: two independent coefficient

)()(32 jkiljlikklijjkiljlikklijijkl AAC




43

Lame Elastic Constant, and

)1(2 ,

)21)(1( v

E

vv

vE

Detailed Expressions

ijijkkjiijijkk

kljkiljlikklijij

2

))((

23233223

13133113

12122112

33221133

33221122

33221111

2

2

2

)2(

)2(

)2(

Equilibrium equation in terms of Lame constant

2

22

2

22

2

2

2

2

2

2

,

)()(

)(

)(

t

t

ub

xx

u

x

u

x

t

ub

x

u

xx

u

xx

u

x

t

ub

x

u

x

u

xx

u

x

t

ub

ii

jj

i

k

k

i

ii

j

j

ij

i

jk

k

i

ii

i

j

j

i

j

ij

k

k

j

iijij

ubuu

or

2

2

2

3

2

2

2

2

2

1

2

3

3

2

2

1

1 )()()(t

ub

x

u

x

u

x

u

x

u

x

u

x

u

x

ii

iii

i




44

4.3. Physical Meanings of Material Properties

Young’s Modulus, Poisson’s Ratio& Lame Constant

)(

)(

)(

22113333

33112222

33221111

EEE

EEE

EEE

))()1(()21)(1(

))()1(()21)(1(

))()1(()21)(1(

22113333

33112222

33221111

vvv

E

vvv

E

vvv

E

Gv

EAAA

vv

EA

vv

EvA

)1(22 ,

)21)(1( ,

)21)(1(

)1( 21321

(1+11)L

h (1-11)h

L




45

Physical Derivation of Shear Modulus

)1(max

EEE

1)sin(121)sin(1)1(

)2

cos(2)]1(2[

max

2

max

2222

max hhhh

)1(2

)1(2)1(

2max

EG

GE

E

)1(2 maxh

h




46

Chapter 5

Plane Problems in

Cartesian CoordinateSystem

x y

z




47

5.1. Plane Stress

Assumption

– The thickness of the plate is small compared to the other

dimensions of the plate.

– No z- directional traction is applied on the surface.

– The variation of stress through the thickness is neglected.

Traction boundary condition

jiji nT

– On yx plane: )1,0,0( n

– 0 , 0 , 0 zzzyzyxzx TTT

– By the third assumption

0 zzyzxz in V

Equilibrium Equation

0

0

0

zzzzyzx

y

yzyyyx

xxzxyxx

bzyx

bzyx

bzyx

00

0

0

y

yyxy

x

xyxx

byx

byx

x

y

z




48

Strain-Displacement relations

),( , ),( yxuuyxuu yyxx , zu : no effect on solution

)(2

1 , ,

x

u

y

u

y

u

x

u yxxy

y

yyx

xx

Constitutive law

xyxy

xxyyyy

yyxxxx

xy

xy

yyxx

yyxxzz

xxyy

yy

yyxxxx

G

vv

E

vv

E

G

EE

EE

EE

2

)(1

)(1

2

)(1

)(

2

2

Displacement method

– Stress in terms of displacement

)(

)(1

)(1

2

2

x

u

y

uG

x

uv

y

u

v

E

y

uv

x

u

v

E

yxxy

xy

yy

yxxx

– Equilibrium equation in terms of displacement

0)1(2)()1()(2

0)1(2)()1()(2

2

2

2

2

2

2

2

2

2

2

E

bv

y

u

x

u

xv

y

u

x

u

E

bv

x

u

y

u

yv

y

u

x

u

yxyyy

xyxxx

– Compatibility equations are automatically satisfied.

Force method

– Body Force: Potential Field

, y

bx

b yx

– Airy’s Stress function

, , 2

2

2

2

2

yxxyxyyyxx

–




49

– Equilibrium Equation: satisfied identically

000)()(

000)()(

2

22

2

2

2

yxyyxx

xyxyyx

– Strains in terms of stress function

yxEG

yExE

xEyE

xy

xy

yy

xx

2

2

2

2

2

2

2

2

2

)1(2

)()(1

)()(1

– Compatibility equation

2

2

2

22

xyyx

yyxxxy

0)()()()()1(22

2

22

4

2

2

4

4

2

2

22

4

2

2

4

4

22

4

xyxxxyxyyyyx

24

2

2

2

2

4

4

22

4

4

4

)1(

0))(1(2yxyyxx




50

5.2. Plane Strain

Assumption

– Infinite Structure in z-direction with an identical section.

– Identical traction in x,y-direction

– No z- directional traction is applied on the surface.

Strain-Displacement Relation

),( , ),( yxuuyxuu yyxx , 0zu

)(2

1 , ,

x

u

y

u

y

u

x

u yx

xy

y

yy

x

xx

0)(2

1 , 0)(

2

1 , 0

y

u

z

u

x

u

z

u

x

u zy

yzzx

xzz

zz

Constitutive law

0)(

EEE

yyxxzzzz )( yyxxzz

xyxy

xxyy

yy

yyxxxx

E

EEv

EEv

)1(2

)()1(

)()1(

22

22

Equilibrium & Compatibility equation → Homework #5 - 1

x y

z




51

5.3 Polynomial Stress Functions

Compatability Condition w/o body forces

024

4

22

4

4

44

yyxx

Stress Components

, , 2

2

2

2

2

yxxyxyyyxx

Do we have any robust strategy to solve the compatibility equation? Unfortunately, the

answer is “No”! The basic strategy to find the solutions of the C.E. is the trial and error

method.

Trial and Error Solution Procedures

1. Predefine several Airy’s stress functions satisfying the compatibility equation.

2. Combine the predefined functions to satisfy given traction boundary conditions.

3. Calculate strain and displacement.

4. Check the displacement boundary conditions.

5. If the DBCs are satisfied, you have the solution. If not, try another combination!

Second order

, ,

22

2

2

22

2

22

2

222

222

byx

ax

cy

yc

xybxa

xyyyxx

– Constant stress status




52

Third Order

y cxbyx

y bxax

ydxcy

yd

xyc

yxb

xa

xy

yy

xx

33

2

332

2

332

2

332323333

232223

– Linear varying stress status in x-y direction

– d3 0 or a3 0 : pure bending cases

– b3 0 : constant normal stress and linearly varying shear stress




53

Fourth order

yd

xycxb

yx

ycxybxax

yacxydxcy

ace

ye

xyd

yxc

yxb

xa

xy

yy

xx

244

24

2

2

44

2

42

2

2

444

2

42

2

444

443422434444

22

2

)2(

)2(

342322334

– d4 0 :

yd

xyd xyyyxx

244

2 , 0 ,

Moment by S. stress = lcdclcd

clcd 3

4

2

4

2

4

3

22

23

12

2

Moment by N. stress = lcdclcd 3

4

2

4

3

22

3

2

2




54

Fifth order

yacxydyxcxb

yx

yd

xycyxbxax

ydbxyacyxdxc

y

dbface

yf

xye

yxd

yxc

yxb

xa

xy

yy

xx

3

55

2

5

2

5

35

2

352

5

2

5

3

52

2

3

55

2

55

2

5

3

5

2

2

555555

55453252354555

5

)32(3

1

3

3

)2( 3

1)32(

3

)2(3

1 , )32(

453423233445

– d5 0 :

xydyd

yy xd xyyyxx

2

5

3532

5 , 3

,)3

2(




55

Final Solution

i

i

– Coefficient of each polynomial should be selected so that the traction boundary

conditions are satisifed.

jiji nT

– The tractions on the surface where displacement is specified are determined once the

S.F. is obtained.

– The displacement field is obtained by integrating the strain field.

Saint-Venant’s Principle

In the elsatostatics, if the boundary tractions ( T ) on a part S1 of the boundary S are

replaced by a statically equilvalent traction distribution ( T ), the effects on the stress

distribution in the body are negligible at points whose distance from S1 is large compared

to the maximum distance between points of S1.

11 SS

dSdS TT ,

11 SS

dSdS rTrT




56

5.4. Cantilever Beam

Traction Boundary Conditions

– on cy

0 xyixix nT , 0 yyiyiy nT ,

– on 0x

0 xxixix nT , yxiyiy nT

Assumption

– Stress distributions can be determined by the beam theory.

xydxx 4 , 0yy ,

2

2

42

ydbxy

Boundary condition on cy

2

24

2

42 20

2)(

c

bd

cdbcyxy

Statical equivalence

ct

PbPtdyy

c

bbtdytdyT

c

c

c

c

xy

c

c

xy4

3)()( 2

2

2

220

Stress

)(2

0

2

3

22

3

ycI

P

yI

My

I

Pxxy

tc

P

xy

yy

xx

L

2c

y

x




57

Displacement

– Strain

)(2

22 ycIG

P

Gx

u

y

u

xyEI

P

EEy

u

xyEI

P

EEx

u

xyyxxy

yyxxy

yy

yyxxxxx

– Integration

)(2

, )(2

1

22 xfxyEI

Puyfyx

EI

Pu yx

– By shear strain

)(2

))(2

())(2

( 22

1

22 ycIG

Pxfxy

EI

P

xyfyx

EI

P

y

IG

Pcy

IG

P

dy

dfy

EI

P

dx

dfx

EI

P

ycIG

P

dx

dfy

EI

P

dy

dfx

EI

P

2)

22()

2(

)(2

)2

()2

(

22212

22122

IG

PcyFxF yx

2)()(

2

– Fx and Fy should be constant.

eyIG

P

dy

dfy

EI

PFd

dx

dfx

EI

PF yx

2212

22 ,

2

– Integration of Fx and Fy

geyyIG

Py

EI

Pfhdxx

EI

Pf

333

166

, 6

– Displacement components

hdxxEI

Pxy

EI

Pu

geyyIG

Py

EI

Pyx

EI

Pu

y

x

32

332

62

662




58

Displacement Boundary Conditions

– on lx

0xu , 0yu

0)66

()2

(

662

32

332

gyIG

P

EI

Pyel

EI

P

geyyIG

Py

EI

Pyl

EI

Pux

IG

P

EI

P

EI

Pleg

66,

2 , 0

2

– however, the third condition is impossible.

2)1(266

EEGE

IG

P

EI

P

– y-Displacement on x = l

062

32

hdllEI

Ply

EI

Pu y 0

6 and 0

2

3

hdllEI

Pl

EI

P

0or 0 Pv

We end up with auseless solution and the displacement BC cannot be satisfied!!!

Why ???

Question on DBC

Since we have 4 integration constants and only one condition (IG

Pcde

2

2

) for them,

three conditions at some points (not on a surface) are required to determine the

integration constant. Boundary conditions specified on a surface can be enforced

exactly with a function on the surface, rather than with just a few constants. Generally,

the integration constants for a partial differential equation with boundary conditions

defined on a surface should be a function defined on the surface. Therefore, it is

impossible to enforce boundary conditions defined on a surface with a few constants.

The assumptions on the stress field for this problem based on the beam solution cannot

satisfy the fixed end condition on the whole right boundary, and thus in principle we have

to guess another stress field, which is very difficult. Anyway, let’s try to determine the

integration constant with three displacement boundary conditions given at the center of

the fixed end.




59

DBC on the neutral axis at the support.

– at lx and 0y 0 yx uu

dllEI

Phhdll

EI

Plu

glu

y

x

33

60

6)0,(

0)0,(

– at lx and 0y 0

x

u y

GI

Pcl

EI

Pe

lEI

Phl

EI

Pddl

EI

P

x

uy

22

320

22

2

222

EI

Plx

EI

Plx

EI

Pxy

EI

Pu

yGI

Pc

EI

Ply

IG

Py

EI

Pyx

EI

Pu

y

x

3262

)22

(662

3232

22332

– Verification of solution

%45.02

3

3/

2 2

232

l

c

EI

Pllc

EI

P for 1.0

l

c and =0.3

)3(66

)(2

)22

(662

23322

22332

ycyGI

Py

EI

Pylx

EI

P

yGI

Pc

EI

Ply

IG

Py

EI

Pyx

EI

Pux

Beam solution ylxEI

Puy

EI

Pxxxx )(

2

22

%1.0)(3

1

2/

6

%7.1))(1(3

4

2/

3

223

223

l

cv

EI

cPl

EI

Pc

l

cv

EI

cPl

GI

Pc

for 1.0l

c and =0.3

– at lx and 0y 0

y

ux

GI

lPcl

EI

Ph

GI

Pcl

EI

Pd

EI

Plee

EI

Pl

y

ux

23

2220

2

23

22

22




60

)(22326

66)(

22

232

3

3322

xlGI

Pcxy

EI

P

EI

Plx

EI

Plx

EI

Pu

yIG

Py

EI

Pylx

EI

Pu

y

x




61

5.5. Simple Beam

Traction boundary conditions

i. on y = -c

001 xyxyjxjnT ,

qqnT yyyyjyj 2

ii. on y = c

01 xyjxjnT , 02 yyjyjnT

iii. on x = -l

01 xxjxjnT , 02 yxjyjnT

iv. on x= l

01 xxjxjnT , 02 yxjyj nT

Symmetry conditions

i. 22 should not vary along x on constant y.

ii. 12 should be anti-symmetry with respect to x should be odd function wrt x.

Selection of functions

i. n = 2

212222211 , , bac

From the SM condition ii, 02 b

ii. n = 3

ycxbybxaydxc 331233223311 , ,

From the SM condition i, 03 a

From the SM condition ii, 03 c

2l

x 2c

y




62

iii. n = 4

2

44

2

412

2

44

2

422

2

444

2

411

22

)+2(

ydxycxb

ycxybxa

yacxydxc

From the SM condition 1, 0 , 0 44 ba

From the SM condition 2, 04 d

iv. n = 5

3

55

2

55

3

512

3

52

5

2

5

3

522

3

55

2

55

2

5

3

511

)3+2(3

1

3

3

)2+(3

1)3+2(

3

yacxydxycxb

ydxycyxbxa

ydbxyacyxdxc

From the SM condition 1, 0 , 0 , 0 555 cba

Final Expressions of Stress components

2

54312

3

52

43222

32

5

22

43211

2

3

)3

2()2(

xydxycxb

ydycyba

yyxdyxcydc

Apply the given boundary conditions

– The boundary condition 1

0202 2

543

2

543 cdccbxcdcxcxb

qcd

cccba 3

3

52

432

– The boundary condition 2

0202 2

543

2

543 cdccbxcdcxcxb

03

3

52

432 cd

cccba

– Solve the simultaneous equations

I

q

c

qdc

I

qc

c

qb

qa

24

3 , 0 ,

24

3 ,

2 354

2

32




63

Apply the static equilibrium conditions

– The static equilibrium condition 1: 011

c

c

lxdy

002))3

2(

2( 22

32

32

cccdyyylI

qydc

c

c

– The static equilibrium condition 2: 011

c

c

lxydy

)

5

2(

2

0)15

2

3(

3

2))

3

2(

2(

22

3

5323

34222

3

clI

qd

ccl

I

qcddyyyl

I

qyd

c

c

The final expression for stress

xycI

q

yycc

I

q

ycyI

qyxl

I

q

)(2

)33

2(

2

)5

2

3

2(

2)(

2

22

12

323

22

2322

11

The static equilibrium conditions:

qldy

c

c

lx

12 qlldyycI

qc

c

)(2

22




64

5.6. Series Solution

024

4

22

4

4

44

yyxx

)()cossin( 21 yfl

xmX

l

xmXm

,

0m

m

In case either sine function or cosine function is used,

)(sin)(sin yxfyfl

xmm

,

0m

m

Compatibility equation for one sine function.

0)()(2)( 24 yfyfyf

General solution

yyCyyCyCyCyf sinhcoshsinhcosh)( 4321

Stress components

)]sinhcosh2(

)coshsinh2(sinhcosh[sin

4

32

2

1

2

yyyC

yyyCyCyCxxx

]sinhcosh sinhcosh[sin 4321

2 yyCyyCyCyCxyy

)]cosh(sinh

)sinhcosh(coshsinh[cos

4

321

yyyC

yyyCyCyCxxy

Traction Boundary conditions

– on cy

0xy , xBmyy sin

– on cy

0xy , xAmyy sin

Integration constants

cc

cBAC

cc

cBAC

cc

cccBAC

cc

cccBAC

mm

mm

mm

mm

22sinh

sinh

22sinh

cosh

22sinh

sinhcosh

22sinh

coshsinh

24

23

22

21




65

Fourier coefficients

l

l

um dxl

xmxqA sin)( ,

l

l

bm dxl

xmxqB sin)(

Refer to pp. 53-63 of Theory of Elasticity by Timoshenko.




66

Chapter 6

Plane Problems in

Polar Coordinate System




67

6.1. Polar Coordinate system

Basics

222

1

, sin

tan , cos

yxrry

x

yrx

z = z

rr

y

x

r

x

x

r

sin

cos

2

,

rr

x

y

r

y

y

r

cos

sin

2

()cos()sin

()()()

()sin()cos

()()()

rryy

r

ry

rrxx

r

rx

Transformation matrix

100

0cossin

0sincos




68

Stress

pTc =

100

0cossin

0sincos

100

0cossin

0sincos

zzzzr

zr

rzrrr

cossin

sincos

)sin(coscossin)(

2sincossin

2sinsincos

22

22

22

zzrzy

zzrzx

rrrxy

zz

rrryy

rrrxx


??

)2sinsincos( 22

xx

rrrxx

??

))sin(coscossin)(( 22

yy

rrrxy

??

x

xyxx byx

Physical Derivation of the Equilibrium Equation.

02

cos)(

2sin)())((

rdrdbd

drd

ddrdrdddrrdr

rF

rrr

r

rrrr

rrr

02

sin)(

))((2

cos)((

rdrdbd

drd

rdddrrdrr

ddrdF

r

r

r

r

r

r

By neglecting higher order terms,

021

0)(11

brrr

brrr

rr

rrrrrr




69

Stress Functions in Cartesian Coordinate System

, , 2

2

2

2

2

yxxyxyyyxx

Second Derivatives

??

)()sin()

(cos()

rrxxx

??

)()cos()

(sin()

rryyy

??

)()cos()

(sin()

rrxyx

Laplace Operator

2

2

22

2

2

2

2

2 ()1()1()()()

rrrryx




70

Stress Function

??

2sinsincos

2sinsincos

22

2

22

2

2

22

yxxy

xyyyxxrr

??

2sincossin

2sincossin

22

2

22

2

2

22

yxxy

xyyyxx

??

)sin(coscossin)(

)sin(coscossin)(

222

2

2

2

2

22

yxyx

xyxxyyr

)1

(11

11

2

2

2

2

2

2

2

rrrrr

r

rrr

r

rr




71

Displacement

zz

yx

yxr

uu

uuu

uuu

cossin

sincos

Strain

pTc =

100

0cossin

0sincos

100

0cossin

0sincos

zzzzr

zr

rzrrr

cossin

sincos

)sin(coscossin)(

2sincossin

2sinsincos

22

22

22

zzrzy

zzrzx

rrrxy

zzzz

rrryy

rrrxx

- By chain rule

)1

(2sin2

1)

1(sincos

)sincos)(sin

(cos

22

r

uu

rr

uu

rr

u

r

u

uurrx

u

rrr

rx

xx

)1

(2

1 , )

1( ,

r

uu

rr

uu

rr

u

r

u rr

rrrr

Geometric derivation of strain components

uvuu r ,

AB

ABHA

AB

ABBArr

''

drr

udrurdr

r

uudrrHA

drAB

rr

rr

)(

r

u

dr

drdrrudr rrrr

)/(




72

0 , 0 ruu (Red line)

r

u

rd

rddur

AC

ACDE rr

)(

0 , 0 ruu (Green line)

u

r

rd

rdudu

urd

AC

ACFA

1

)(

0 , 0 ruu

u

rr

u

rd

rddudur rr 1)/()(

0 , 0 ruu

rrrr u

rrd

urduur

FA

CF 1)()/(2

0 , 0 ruu

r

u

dr

urdruu rr

/

314 , r

u3

0 , 0 ruu

r

u

r

uu

r

rr

1)( 34212

)1

(2

1)(

2

1342

r

u

r

uu

r

rr

u

du

urd

1drr

uu

u

3

d

uur r

r

ru

2




73

6.2. Governing Equation and Boundary Conditions

Equilibrium

021

0)(11

brrr

brrr

rr

rrrrrr

Stress Function

)1

(11

11

2

2

2

2

2

2

2

rrrrr

r

rrr

r

rr

Compatibility

2

2

22

2

2

2

2

2 ()1()1()()()

rrrryx

0)11

)(11

())((2

2

22

2

2

2

22

2

2

2

2

2

2

2

2

24

rrrrrrrryxyx

Strain Components

r

urrr

,

u

rr

ur 1 , )

1(

2

1

r

u

r

uu

r

rr

Constitutive Law p

klijkl

p

ij C

Displacement Boundary Conditions

p

i

c

jij

c

jij

p

i

c

i

c

i uuuuuu pp

uu

Cauchy’s Relation

p

m

p

jm

c

kmkmn

c

jnij

c

kmnmk

c

jnij

c

k

c

jkij

c

jij

p

i

c

j

c

ij

c

i nnnnTTnT

pppnT

Traction Boundary Condition

p

i

c

jij

c

jij

p

i

c

i

c

i TTTTTT pp

TT




74

6.3. Solutions in the Polar Coordinate System

Compatibility Equation

0)11

)(11

(2

2

22

2

2

2

22

2

rrrrrrrr

In case is a function of r only

0112

0)1

)(1

(32

2

23

3

4

4

2

2

2

2

dr

d

rdr

d

rdr

d

rdr

d

rrrrrr

Solution of the Compatibility Equation

dtedrer tt 0

)6116(1

)23(1

))(1

(

)(1

)1

(

1

2

2

3

3

4

4

44

4

2

2

3

3

32

2

23

3

2

2

22

2

dt

d

dt

d

dt

d

dt

d

rdr

d

dt

d

dt

d

dt

d

rdr

dt

dt

d

dt

d

rdt

d

dr

d

dt

d

dt

d

rdr

dt

dt

d

rdt

d

dr

d

dt

d

rdr

dt

dt

d

dr

d

0)44(1

11)(

11)23(

12)6116(

1

112

2

2

3

3

4

4

4

32

2

222

2

3

3

32

2

3

3

4

4

4

32

2

23

3

4

4

dt

d

dt

d

dt

d

r

dt

d

rrdt

d

dt

d

rrdt

d

dt

d

dt

d

rrdt

d

dt

d

dt

d

dt

d

r

dr

d

rdr

d

rdr

d

rdr

d

DCrrBrrADCeBteAt tt 2222 lnln

Stress Components

0)1

(

2)ln23(

2)ln21(1

22

2

2

rr

CrBr

A

dr

d

CrBr

A

dr

d

r

r

rr

- If no hole exists at the origin of coordinates, A=B=0

- The term associated with B is not a single-valued function.




75

Displacement (plane stress)

)(])1(2)1(ln)1(2)1(

[1

])1(2)31(ln)1(2)1(

[1

)(1

2

fCrBrrBrr

A

Eu

CBrBr

A

Er

u

E

r

rrrrr

)()(4

)(4

])1(2)3(ln)1(2)1(

[1

)(1

1

2

rfdfE

Bruf

E

Bru

CBrBr

A

Er

u

r

u

E

rrr

0)(1

)(1)()(1

0)1

(2

11

1

rfr

dfrr

rff

rr

u

r

uu

r

rr

pFrrfprfr

rfr

KHfff

pdff

)()()(

cossin)(0)()(

)()(

111

2

2

Substitution of the above solutions into the original Eq., p = 0.

FrKHE

Bru

KHCrBrrBrr

A

Eur

sincos4

cossin])1(2)1(ln)1(2)1(

[1

– Rigid body translation

cossin , sincos yxyxr uuuuuu xuK , yuH

xu

xu

yu

xu




76

– Rigid body rotation

ruur , 0 F

r

xu




77

6.4. Thick Pipe Problem

Traction BC

– On ar : irrjiji pnT

– On br : orrjiji pnT

orr

irr

pCb

Ab

pCa

Aa

2)(

2)(

2

2

22

22

22

22

2

)(

ab

bpapC

ab

ppbaA

oi

io

Stress components

22

22

222

22

22

22

222

22

1)(

1)(

ab

bpap

rab

ppba

ab

bpap

rab

ppba

oiio

oiiorr

Displacement

0 , ])1()1()(

[1

22

22

22

22

ur

ab

bpap

rab

ppba

Eu oiio

r

a

b

po pi




78

End Conditions

– Plane strain

22

22

2)(

0

ab

bpap oirrzz

zz

– Plane stress (open end)

22

22

2)(

0

ab

bpap

EE

oirrzz

zz

– Closed End with a Rigid End Plate

)()2)((

))()(()(

22

22

2222

2222

oioi

zz

rrzzzzz

pbpaab

bpapEab

EababF

22

22

22

22

)(

)21()21(

ab

pbpa

abE

pbpav

oizz

oizz

0op

– Stress : )1( , )1(2

2

22

2

2

2

22

2

r

b

ab

pa

r

b

ab

pa iirr

ii p

ab

abp

22

22

max

)()(

– Displacement : ])1()1[(22

2

b

r

r

b

ab

a

E

bpu i

r




79

6.5. Special Cases

Thin pipe with an internal pressure: btab

t

pr

r

b

ab

pa

r

b

ab

pa

trb

tra

ii

i

rr

0

2

2

22

2

2

2

22

2

00

)1(

0 )1(

2 ,

2

Et

rp

b

r

r

b

ab

a

E

bpu ii

r

2

0

22

2

])1()1[(

Infinite Region pipe with an internal pressure: b

)11

()/(1

)1(

)11

()/(1

)1(

222

2

2

2

22

2

222

2

2

2

22

2

rbba

pa

r

b

ab

pa

rbba

pa

r

b

ab

pa

ii

iirr

])1(1

)1[()/(1

])1()1[(22

2

22

2

b

r

rba

a

E

p

b

r

r

b

ab

a

E

bpu ii

r

as b ,

2

2

2

2

r

pa

r

pa

i

irr

, r

a

E

pau i

r )1(

Infinite Region pipe with a Lining

– Region 1

22

22

222

22

22

22

222

22

1)(

1)(

ab

bpap

rab

ppba

ab

bpap

rab

ppba

oiio

oiiorr

0 , ])1()1()(

[1

122

22

1

22

22

1

1

ur

ab

bpap

rab

ppba

Eu oiio

r

– Region 2

2

0

2

2

2

r

pb

r

pb orr

, r

b

E

pbu o

r

2

2

2 )1(




80

– Compatibility

)()( 12 bubu rr

])1()1()(

[1

)1( 122

22

1

22

22

12

2 bab

bpap

bab

ppba

EE

pb oiioo

)]1()1([1

)]1()1([1

)1( 122

2

122

2

1

122

3

122

2

12

2

ab

bap

ab

bpa

Eab

bp

ab

bpa

EE

pb iiooo

22

2

1

122

3

122

2

12

2

21)]1()1([

1)1(

ab

bpa

Eab

bp

ab

bpa

EE

pb iooo

))(())(1(

222

1

22

2

22

21

2

2babaEabE

paEp i

o

– A thin lining

i

i

o ptr

tr

trEtE

ptrEp

0

0

10221

0

2)()1(

)(

a

b

pi

po

po




81

6.6. Curved Beam

Boundary Condition

– On ar and b : 00 rrrjiji nT

– On 0 : 0 nT rr and nT

Applied Moment

b

a

b

a

rdrrdrTM , 0 b

a

b

a

drdrT

Stress Component

CrBr

A

dr

d

CrBr

A

dr

d

rrr

2)ln23(

2)ln21(1

22

2

2

Application BC on ar and b

02)ln21(

02)ln21(

2

2

CbBb

A

CaBa

A

Moment Condition

)()(2

2

2

abrrdr

d

drdr

dr

dr

drdr

dr

drdrM

b

a

b

arr

b

a

b

a

b

a

b

a

b

a

b

a

DCrrBrrA 22 lnln

)()lnln(ln 2222 abCaabbBa

bAM

a

b

M M




82

Solution

))lnln(2(

)(2

ln4

2222

22

22

aabbabN

MC

abN

MB

a

bba

N

MA

where 222222 )(ln4)(

a

bbaabN .

Stress

)lnlnln(4

)lnlnln(4

2222

2

22

22

2

22

abr

aa

b

rb

a

b

r

ba

N

M

r

aa

b

rb

a

b

r

ba

N

Mrr

Displacement

FrKHE

Bru

KHCrBrrBrr

A

Eur

sincos4

cossin])1(2)1(ln)1(2)1(

[1

Displacement B.C.

0

r

uuur at 0 and 2/)( bar 0 HF

A ring with cut out.

– Cut out angle :

– Compatibility condition

88)2(

EBr

E

Bru

– Moment required to close the ring

8)(

2 22 Eab

N

MB

)(16 22 ab

ENM




83

6.7. Rotating Disk

Centrifugal force : rbr

2


22

021

0)(11

rr

r

brrr

brrr rr

rr

rrrrrr

Stress and strain

r

urrr

,

r

uu

rr

u rr

1

)(1

)(1

)(1

)(1

22

22

r

u

r

uEE

r

u

r

uEE

rrrr

rrrrrr

Equilibrium Equation in terms of displacement

322

2

22 1

rE

vu

dr

dur

dr

udr r

rr

(Euler-Cauchy Equation)

General Solution

]8

11)1()1[(

1 322

1 rr

CvCrvE

ur

b




84

22

21

22

21

222

21

22

2

222

21

222

212

2

8

)31(1

8

)3(1

]8

1)3(

1)1()1[(

1

1

)]8

11)1()1((

8

13

1)1()1[(

1

1

)(1

rv

rCC

rv

rCC

rvr

CC

rr

CvCvvrr

CvCv

r

u

r

uE rrrr

Solid disk

2222

18

)3(0

8

)3()( , 0 b

vCb

vCC brrr

2222222

8

)31(

8

)3( , )(

8

)3(r

vb

vrb

vrr

22

maxmax8

)3()()( b

vrr

at r = 0

Disk with a circular hole at the center

222

1

222

22

2

1

22

2

1

8

)3(

)(8

)3(

08

)3()(

08

)3()(

bav

C

bav

C

bv

b

CC

av

a

CC

brrr

arrr

)3

31(

8

3

)(8

3

2

2

22222

2

2

22222

rv

v

r

baba

v

rr

baba

vrr

arav

vb

v

abrabv

rr

at )3

1(

4

3)(

at )(8

3)(

222

max

22

max

7200 rpm 3.5 inch (8.9 cm) hard disk

222

4222

max

/)(9.1/0.189342sec/0.189342

10)5.03.03

3.015.4(1207850

4

3.3)(

cmfkgmNmkg

Solution by force method and Disk with two different material (homework)




85

6.8. Plate with Circular hole

Stress on remote field

Sxx , 0yy , 0xy

(Cartesian)

2sin2

)sin(coscossin)(

)2cos1(2

2sincossin

)2cos1(2

2sinsincos

22

22

22

S

S

S

xyxxyyr

xyyyxx

xyyyxxrr

Solution for the uniform stress (Thick pipe problem)

)1(2

)1(2

)(2

)(2

2

2

2

2

2

2

22

2

22

2

2

2

22

2

22

2

r

aS

r

aS

r

a

ab

b

ab

bS

r

a

ab

b

ab

bSrr

ab

rr

0 r

Solution for non-uniform stress field

– BC on r = b : 2sin2

, 2cos2

SSrrr

– Stress Function : 2cos)(rf

)1

(11

, 11

2

2

2

2

2

2

2

rrrrr

rrrr

r

rr

b

a




86

– Compatibility Equation

Dr

CBrArD

rCr

BArf

DH

BK

rDrBHKr

rKrKrHKrf

fr

CArf

rDrBf

rdr

df

rdr

fd

rDrBF

Frdr

dF

rdr

Fd

Frdr

dF

rdr

Fd

frdr

df

rdr

fdF

frdr

df

rdr

fd

d

d

rdr

d

rdr

d

p

p

2

42

2

42

2

24

2

224

2

2

2

2

22

2

2

2

22

2

22

24

22

2

22

2

2

2

22

24

1

4

1

12

4 ,

12

1)(

4412

1

1411

041

02cos)41

(

41

02cos)41

)(11

(

– Stress component

2sin)26

62()1

(

2cos)6

12(2A

2cos)46

(2A 11

24

2

4

2

2

2

242

2

2

r

D

r

CBrA

rr

r

CBr

r

r

D

r

C

rrr

r

rr

– Applying BCs

026

62

2

2662

046

2A

2

462A

24

2

24

2

24

24

a

D

a

CBaA

S

b

D

b

CBbA

a

D

a

C

S

b

D

b

C

– As b

4

SA , 0B , S

aC

4

4

, Sa

D2

2




87

Final Solution

2sin)23

1(2

2cos)3

1(2

)1(2

2cos)43

1(2

)1(2

2

2

4

4

4

4

2

2

2

2

4

4

2

2

r

a

r

aS

r

aS

r

aS

r

a

r

aS

r

aS

r

rr

Stress concentration on ar

2cos2

0

SS

rrr

– at 2

3 ,

2

1 : S3

– at , 0 : S (Compression)

– at 2

1 : )

2

3

2

11(

4

4

2

2

r

a

r

aS (rapidly decaying)

Tensile traction in both x- and y- direction

0

)1(

)1(

)2cos()43

1(2

)1(2

2cos)43

1(2

)1(2

2

2

2

2

2

2

4

4

2

2

2

2

4

4

2

2

r

rr

r

aS

r

aS

r

a

r

aS

r

aS

r

a

r

aS

r

aS

Opposite tractions (pure shear case)

2cos)3

1(

)2cos()3

1(2

)1(2

2cos)3

1(2

)1(2

4

2

4

4

2

2

4

4

2

2

r

aS

r

aS

r

aS

r

aS

r

aS

S4)( max




88

r

6.9. Flamant solution (2-D Boussinesq Solution)

Boundary condition

0 r on

2

Equilibrium condition

Prdrr

2/

2/

cos for all r.r

frr

)(

0)(

)11

()11

)(11

(

0 , )(11

2

2

22

2

2

2

22

2

2

2

22

2

2

2

2

2

2

r

f

rrrrrrrrrrrr

rr

f

rrrrr

sincos)(0)()( BAfff

By symmetry of stress, B = 0

PAdAdA

2

)2cos1((2

1cos

2/

2/

2/

2/

2

Stress components (Polar)

r

Prr

cos2, 0 r

Stress components (Cartesian)

3

222

432

cossin2

cossin

sincos2

sin

cos2

cos2

cos

a

P

a

P

a

P

r

P

rrxy

rryy

rrxx




89

Convolution Integral (Influence line)

b

a

dgyxfRdgyxfdR )(),()(),(

– Moment Case

x

fM

yxfyxfM

yxfyxfPyxPfyxPfR

)),(),(

(lim

)),(),(

(lim)),(),((lim

0

00

f(x,y) f(x-y)

g(x)




90

Chapter 7

Uniform Torsion (St. Venant Torsion)




91

7.1. Basics

Displacement

yr

yrr

rrux

sinsin

)cossinsincos(cos)cos)(cos(

xr

xrrrru y cossin)sinsincoscos(sin)sin)(sin(

z , : rotation angle per unit length

Warping function

),( yxuz

Strain components

0),(

0

0

z

yx

z

u

y

zx

y

u

x

zy

x

u

zzz

y

yy

x

xx

)(

)(

0

xyyz

zx

y

u

z

u

yxxz

zy

x

u

z

u

x

zx

y

zy

x

u

y

u

zy

yz

zx

xz

yx

xy

z

x

y

x

y




92

Stress components

)(

)(

0

xy

G

yx

G

yz

xz

xyzzyyxx

Equilibrium equation (z-direction only)

00)(2

2

2

2

2

2

2

2

yxyxG

zyx

zzyzxz

Traction free BC on the longitudinal surface

0 jiji nT for all i , )0,,( yx nnn

0

0

0

yzyxzxz

yyyxyxy

yxyxxxx

nnT

nnT

nnT

The first two BCs are automatically satisfied, and

0)()(

0

ds

dxx

yds

dyy

x

ds

dx

ds

dynn zyzxyzyxzx

Stress function

)(

)(

xy

Gx

yx

Gy

yz

xz

by eliminating ,

G

yx2

2

2

2

2

The traction BC becomes

0

ds

d

ds

dx

xds

dy

y is constant on the boundary.




93

BCs at the end of the bar

0

0

S

x

AA

zy

S

y

AA

zx

dSndAx

dA

dSndAy

dA

AASA

AAA

zx

A

zyt

dAdAdSdA

ydAy

xdAx

ydAxdAM

22nXX

Summary

– Governing equation :

G

yx2

2

2

2

2

– Boundary condition : 0

– Twisting moment : A

t dAM 2




94

7.2. Torsion in an Elliptic Section

Elliptic section

012

2

2

2

b

y

a

x

Stress function

)1(2

2

2

2

b

y

a

xm

Gba

bamG

bam

Gb

y

a

x

yxmG

yx

22

22

22

2

2

2

2

2

2

2

2

2

2

2

2

2)22

(

2)1)((2

Gb

y

a

x

ba

ba)1(

2

2

2

2

22

22

Twisting angle

22

33

2

3

2

3

22

22

2

2

2

2

22

22

)44

(2

)1(2

ba

baG

abb

ab

a

ba

ba

baG

dAb

y

a

x

ba

baGM

A

t

G

M

ba

ba t

33

22

)1(2

2

2

2

b

y

a

x

ab

M t

Stress

xba

M

xy

ab

M

y

tyz

txz

33

2 ,

2

Warping function

)()(2

)(2

)()(2

)(2

33

33

xgxyGxyba

Mx

yGx

ba

M

yfxyGxyab

My

xGy

ab

M

tt

tt

cyfxg

yfxgxyabba

Mabxy

ba

M tt

)()(

)()()(2

)(2 22

33

22

33




95

cabba

Mabxy

ba

M tt 2)(2

)(2 22

33

22

33

22

33

22

22

ab

ba

M

cxy

ab

ab

t

00)0,0( c

22

22

xyab

ab

For a circular section, the warping does not occur




96

7.3. Rectangular Section(Series Solution)

Governing equation

G

yx2

2

2

2

2

BCs: 0 on ax and by

Series expansion

n

n

Ya

xn

,5,3,1 2cos (BCs on ax are satisfied.)

a

xn

nGG n

n 2cos)1(

422 2/)1(

,5,3,1

ODE in y-direction

))1(8

)2

((2

cos 2/)1(2

,5,3,1

n

nn

n nGYY

a

n

a

xn

=0

2/)1(2 )1(8

)2

(

n

nnn

GYa

nY for 5,3,1n

2/)1(

3

2

)1()(

32

2cosh

2sinh

n

nn

aG

a

ynB

a

ynAY

Application BCs on by

0)1()(

32

2cosh

2sinh)( 2/)1(

3

2

n

bynn

aG

a

bnB

a

bnAY

0A , 2/)1(

3

2

)1()2/cosh(

1

)(

32

n

abnn

aGB

))2/cosh(

)2/cosh(1()1(

)(

32 2/)1(

3

2

abn

ayn

n

aGY n

n

2b

2a




97

Final Solution

))2/cosh(

)2/cosh(1(

2cos)1(

132

,5,3,1

2/)1(

33

2

abn

ayn

a

xn

n

aG

n

n

Stress

))2/cosh(

)2/cosh(1(

2sin)1(

116

,5,3,1

2/)1(

22 abn

ayn

a

xn

n

aG

x n

n

yz

)2/cosh(

)2/sinh(

2cos)1(

116

,5,3,1

2/)1(

22 abn

ayn

a

xn

n

aG

y n

n

xz

Maximum Stress

)87

1

5

1

3

11( ,

)2/cosh(

11162

))2/cosh(

11(

116

))2/cosh(

11(

2sin)1(

116)(

2

222,5,3,1

22

,5,3,122

,5,3,1

2/)1(

22)0,(max

n

n

n

n

yaxyz

abnn

aGaG

abnn

aG

abn

n

n

aG

Twisting angle

)967

1

5

1

3

11( , )

)2/tanh(1921(

3

2)2(

)2/tanh()2(6412)2(32

))2/cosh(

)2/sinh(2(2

22

164

))2/cosh(

)2/cosh(1(

2cos)1(

164

2

4

444,5,3,1

55

3

,5,3,155

4

,5,3,144

3

,5,3,133

2

,5,3,1

2/)1(

33

2

n

nn

n

b

bn

a

a

n

A

t

n

abn

b

abaG

n

abnaG

n

baG

abn

abn

n

ab

n

a

n

aG

dyabn

ayndx

a

xn

n

aG

dAM

Narrow section 3a

b

19998.00090.03178.111

0090.03178.111)2/tanh(

ab

)6274.01(3

2)2())

3125

1

243

11(6274.01(

3

2)2( 33

b

abaG

b

abaGM t




98

Square section

9171.02079.08105.4

2079.08105.4)2/tanh(

4

4

4

)2(1415.0

)9171.06274.01(3

)2(

))3125

1

243

19171.0(6274.01(

3

)2(

aG

aG

aGM t

Open section with constant thickness

i

iit twG

M 3

3




99

Chapter 8

Beam Approximation

x1

x2

x3

),,0( 32ttt nnn

)0,0,1(ln

liT




100

8.1. Equilibrium Equation

Force Resultant

A

i

A

l

jji

A

l

ii dAdAndATQ 1

i

A

i

j

ji

S

t

i

A

i

A

i

j

ji

S

t

i

t

i

A

i

A

i

j

ji

A

ii

A

i

A

i

j

ji

A

ii

j

ji

A

ii

qdAbx

dSTdAbdAbx

dSnndAbdAbx

dAxx

dAbdAbx

dAxxx

dAxx

Q

)()(

)()(

)()(

)(

3322

3

3

2

2

3

3

2

2

1

1

1

0)(1

A

i

j

ji

ii dAb

xq

x

Q

x1

x2

x3

),,0( 32ttt nnn

)0,0,1(ln

liT

x2

x3

S

A




101

Moment Resultant

A

kjijk

A

l

mkmjijk

A

l

kjijk

A

l

i dAxedAnxedATxedAM 1~~~~~

Tx where ),,0(~32 xxx

k

l

jijk

S

t

kjijk

A

kjijk

A

k

m

kmjijk

A

kkikki

S

t

mkmjijk

A

kjijk

A

k

m

kmjijk

A

kkikki

A

kjijkkjijk

A

kjijk

A

k

m

kmjijk

A

kkjijk

A m

kmjijk

A

kk

m

kmjijk

A

kjijk

i

QnedSTxedAbxedAbx

xe

dAeedSnxedAbxedAbx

xe

dAee

dAx

xe

x

xedAbxedAb

xxe

dAxx

xedAx

xe

dAxxx

xe

dAx

xex

M

~~)(~

)(~~)(~

)(

)~~

(~)(~

)(~~

)(~

~~

3322

3322

3

3

2

2

3

3

2

2

3

3

2

2

1

1

1

0~~~

1

k

l

jijk

S

t

kjijk

A

kjijki QnedSTxedAbxe

x

M

Sign Convention for Moment

For i =1

For i =2

2131323122~~~

MdSTxdSTxedSTxeMS

l

S

l

S

l

kjjk

For i =3

x1

x2

T1

x1 T1

x3

12332231323212311 )()~~(~~MdSTxTxdSTxeTxedSTxeM

S

ll

S

ll

S

l

kjjk

x2

x3

T3

T2




102

3131232133~~~

MdSTxdSTxedSTxeMS

l

S

l

S

l

kjjk




103

Component Form

01

1

1

q

x

Q S

t

A

dSTdAbq 111

02

1

2

q

x

Q S

t

A

dSTdAbq 222

03

1

3

q

x

Q S

t

A

dSTdAbq 333

01

1

1

m

x

M

S

tt

A

dSTxTxdAbxbxm )~()( 233223321

032

1

2

Qm

x

M

S

t

A

dSTxdAbxm 13132

023

1

3

Qm

x

M

S

t

A

dSTxdAbxm 12123 ( 33

~MM )

Planar Beam 03131 tt TTbb

02

1

2

q

x

Q , 02

1

3

Q

x

M 22

1

3

2

qx

M

01

1

1

m

x

M

S

t

A

dSTxdAbxm 23231

What condition(s) should be satisfied in order that the vertical loads do not cause torsion

of beam ??




104

8.2. Bernoulli Beams

Assumption on displacement field

– A plane section remains plane after deformation.

– Normal remains normal.

– The vertical displacement is constant through the depth of a beam.

Displacement Components by assumption

yx

uu

y

x

, )(xuu yy , ??zu

Strain-displacement relation

??

0

2

2

z

u

yx

u

x

u

zzz

yy

yxxx

??

??

0

x

u

x

u

z

u

y

u

z

u

y

u

x

u

y

u

zzxxz

zyzyz

yxxy

Strain-Stress relation

)(

)(

)(

22113333

33112222

33221111

EEE

EEE

EEE




105

Assumption on stress

– By traction BC, 0zz , 0yy

EEE

zzzz

xxyy

xxxx

, ,

Since, however, 0 yy by the strain-displacement relation, the Poisson ratio should be

zero, which is equivalent to neglecting the Poisson effect.

2

2

2

2

2

2

2

2

2

2

x

uEIdAy

x

uEdAy

x

uEydAM

yx

uE

y

A

y

A

y

A

xxz

y

xx


0

0

y

yyxy

x

xyxx

byx

byx

0

0

y

xy

xyxx

bx

yx

000)(

000

00)(0)(

2

1

qx

VdAbdA

xdAb

x

Vx

MdAdzy

x

MydA

yx

M

ydAy

ydAx

ydAyx

ydAyx

A

y

A

xy

A

y

xy

A

xy

z

h

hxy

A

xy

A A

xy

xx

A

xyxx

A

xyxx

02

2

q

x

M

qx

uEI

xq

x

M y

2

2

2

2

2

2

, yI

Mxx

Ib

VQhybdy

yydA

I

VdA

yI

y

x

M

yxxyxy

h

y

xy

h

y

h

y

xyxyxx

)()(00)(0 2

222

Simply supported beam with a uniformly distributed load

qh

qlyyxx max2

2

max )( , 8

6)(

6

8

)(

)(2

2

max

max

l

h

xx

yy




106

5.3. Timoshenko Beams

Assumption on displacement field

– A plane section remains plane after deformation.

– Normal remains normal.

– The vertical displacement is constant through the depth of a beam.

Displacement Components by assumption

yxux )( , )(xuu yy , ??zu

Strain Components

??

0

z

u

yxx

u

zzz

yy

xxx

??

??

x

u

x

u

z

u

y

u

z

u

y

u

x

u

x

u

y

u

zzxxz

zyzyz

yyxxy

Equilibrium equation

Vx

M

, 0

q

x

V




107

By energy consideration

)( 0

22

2

2

2

2

GA

V

GA

fVdA

yb

Q

GI

VdA

GVdAdA s

AA

xy

xyxy

A

xyxy

A

xyxy

which yield xyGAV 0

)(

0

2

x

uGAV

xEIdAy

xEdAyM

y

AA

xx

qx

EI

qxx

uGAq

x

V

x

uGA

xEI

x

uGA

xEIV

x

M

y

yy

3

3

2

2

0

02

2

02

2

)(0

0)()(




108

Chapter 9

Energy Methods




109

9.1. Problem Definition

Governing Equations and Boundary Conditions

Equilibrium Equation : 0 b in V

Constitutive Law : :D in V

Strain-Displacement Relationship : ))((2

1 Tuu

in V

Displacement Boundary condition : 0 uu on uS

Traction Boundary Condition : 0TT on tS

Cauchy’s Relation on the Boundary : nT on S

Energy Conservation

int2

1

2

1

2

1

2

1

2

1

2

1

2

1

2

1

2

1

2

1

2

1

2

1

V

ijij

V

ij

j

i

S

ii

V

ij

j

i

S

ii

S

jiji

V

ij

j

i

S

ii

V j

ij

i

S

ii

V

ii

S

iiext

dVdVx

u

dSTudVx

udSTu

dSnudVx

udSTu

dVx

udSTu

dVbudSTu

V, E,

TT

0 uu Su

St




110

For physical interpretation of the strain energy, please refer to pp. 244 - 246 of Theory of

Elasticity by Timoshenko

Total Potential Energy

dSTudVbudV

tS

ii

V

iiij

V

ij 2

1




111

9.2 . Principle of Minimum Potential Energy

dVx

uD

x

u

dVbudVx

uD

x

u

dVbudVudVx

uD

x

u

dSTudVbudSTudVudVx

uD

x

u

dSTudVbudSnudVudVx

uD

x

u

dSTudVbudVx

u

dVx

uD

x

udSTudVbudV

dSTudVbudVx

uD

x

u

dVx

uD

x

udSTudVbudV

x

uD

x

u

dSTuudVbuu

dVx

uD

x

udV

x

uD

x

udV

x

uD

x

udV

x

uD

x

u

dSTuudVbuudVx

uuD

x

uu

dSTudVbudV

uuu

l

e

k

V

ijkl

j

e

iE

V

ijij

e

i

l

e

k

V

ijkl

j

e

iE

V

i

e

i

V

jij

e

i

l

e

k

V

ijkl

j

e

iE

S

i

e

i

V

i

e

i

S

i

e

i

V

jij

e

i

l

e

k

V

ijkl

j

e

iE

S

i

e

i

V

i

e

i

S

jij

e

i

V

jij

e

i

l

e

k

V

ijkl

j

e

iE

S

i

e

i

V

i

e

i

V

ij

j

e

i

l

e

k

V

ijkl

j

e

i

S

ii

V

ii

V

ijij

S

i

e

i

V

i

e

i

l

k

V

ijkl

j

e

i

l

e

k

V

ijkl

j

e

i

S

ii

V

ii

l

k

V

ijkl

j

i

S

i

e

ii

V

i

e

ii

l

e

k

V

ijkl

j

e

i

l

e

k

V

ijkl

j

i

l

k

V

ijkl

j

e

i

l

k

V

ijkl

j

i

S

i

e

ii

V

i

e

ii

l

e

kk

V

ijkl

j

e

ii

S

i

h

i

V

i

h

i

h

ij

V

h

ij

h

e

ii

h

i

tt

t

t

t

t

t

t

t

t

2

1

)(2

1

)(2

1

)(2

1

)(2

1

)(

2

1

2

1

)(

2

1

2

1

)()(

2

1

2

1

2

1

2

1

)()()()(

2

1

2

1

,

,

,

,

Since Dijkl is positive definite,

u

xD

u

x

i

e

j

ijklk

e

l

0 for all

u

x

i

e

j

0 . Therefore,

u

xD

u

xdVi

e

j

ijkl

V

k

e

l

0 for all

u

x

i

e

j

0 , and

u

xD

u

xdVi

e

j

ijkl

V

k

e

l

0 iff

u

x

i

e

j

0 .

.)?? =for only holdssign equality The ( i

h

i

Eh uu




112

9.3. Principle of Virtual Work

If the following inequality is valid for all real number , the principle of virtual work

holds.

ii

RR

ii

RR vuvu )()(

) admissible allfor ( 0)0(

)(

)()(

2

1

2

1

)()()()(

2

1

)()(

2

iiii

V

ii

l

k

V

ijkl

j

i

l

i

V

ijkl

j

i

ii

V

ii

l

k

V

ijkl

j

i

iii

V

iii

l

i

V

ijkl

j

i

l

k

V

ijkl

j

i

l

k

V

ijkl

j

i

iii

V

iii

l

ii

V

ijkl

j

ii

ii

RR

vvdTvdVbvdVx

uD

x

vg

dVx

vD

x

vdTvdVbvdV

x

uD

x

vg

dTvudVbvu

dVx

vD

x

vdV

x

uD

x

vdV

x

uD

x

u

dTvudVbvudVx

vuD

x

vu

vug

t

t

t

t

v

xdV v b dV v T d vi

j

ij

V

i i

V

i i i

t

0 for all admissible

If the principle of virtual work holds, then the principle of minimum potential energy holds

because the term in the parenthesis in the boxedequation of the principle of minimum

potential energy vanishes identically. The approximate version of the principle of virtual

work is

v

xdV v b dV v T d vi

h

j

ij

h

V

i

h

i

V

i

h

i i

h

t

0 for all admissible




113

9.4. Rayleigh-Ritz Type Discretization

Approximation

n

p

pipninii

h

i gcgcgcgcu1

2211

Principle of Minimum Potential Energy

dTgcdVbgcdVx

gcD

x

gc

x

gc

x

gc

x

u

t

ipip

V

ipip

l

q

kq

V

ijkl

j

p

ip

h

j

p

ip

n

p j

p

ip

j

h

i

2

1Min

1

or 0

mr

h

c for all m, r

mrfcK

dTgdVbgdVcx

gD

x

g

dTgdVbgdVx

gcD

x

g

dTgdVbgdVx

gD

x

gcdV

x

gcD

x

g

dTgdVbgdVx

gD

x

gcdV

x

gcD

x

g

c

rmkprmkp

mr

V

mrkp

V l

p

mjkl

j

r

mr

V

mr

l

p

kp

V

mjkl

j

r

mr

V

mr

l

r

V

ijml

j

p

ip

l

q

kq

V

mjkl

j

r

irmi

V

irmi

l

rmk

V

ijkl

j

p

ip

l

q

kq

V

ijkl

j

rmi

mr

h

t

t

t

t

and allfor 0

2

1

2

1

2

1

2

1

Principle of Virtual Work

and allfor 0

allfor 0)(

)(

1

2211

ipfcK

cdTgdVbgdVcx

gD

x

gc

dTgdVbgdVcx

gD

x

gc

dTgcdVbgcdVx

gcD

x

gc

gcgcgcgcgcuv

pikqpikq

h

ipip

V

ipkq

V l

q

ijkl

j

p

ip

ip

V

ipkq

V l

q

ijkl

j

p

ip

ipip

V

ipip

V l

q

kqijkl

j

p

ip

h

pip

n

p

pipninii

h

i

h

i

t

t

t




114

Matrix Form – Virtual Work Expression

dVdV

dV

dV

dV

hT

V

hhhhhhhhhhh

V

hh

hhhhhhhhhh

V

hh

hhhhhhhhhhhhhhhh

V

hh

V

h

ij

h

ij

)(

)222(

)(

232313131212333322221111

232313131212333322221111

323223233131131321211212333322221111

ddVdV

t

h

V

hh

V

hTubu

Displacement

Ncu

n

n

n

n

n

n

h

h

h

h

c

c

c

c

c

c

c

c

c

g

g

g

g

g

g

g

g

g

u

u

u

3

2

1

32

22

12

31

21

11

2

2

2

1

1

1

3

2

1

00

00

00

00

00

00

00

00

00

Virtual Strain

2

3

3

2

1

3

3

1

1

2

2

1

3

3

2

2

1

1

2

3

3

2

1

3

3

1

1

2

2

1

3

3

2

2

1

1

23

13

12

33

22

11

)(

x

gc

x

gc

x

gc

x

gc

x

gc

x

gc

x

gc

x

gc

x

gc

x

u

x

u

x

u

x

u

x

u

x

u

x

u

x

u

x

u

i

i

i

i

i

i

i

i

i

i

i

i

ii

i

i

ii

hh

hh

hh

h

h

h

h

h

h

h

h

h

h

ij

h

ij

h

V

i

h

i

V

i

h

idV u b dV u T d

t

0




115

cB

n

n

n

nn

nn

nn

n

n

n

c

c

c

c

c

c

c

c

c

x

g

x

g

x

g

x

g

x

g

x

g

x

g

x

g

x

g

x

g

x

g

x

g

x

g

x

g

x

g

x

g

x

g

x

g

x

g

x

g

x

g

x

g

x

g

x

g

x

g

x

g

x

g

3

2

1

32

22

12

31

21

11

23

13

12

3

2

1

2

2

3

2

1

2

3

2

1

2

2

2

3

2

2

2

1

2

2

1

3

1

1

1

3

1

1

1

2

1

3

1

2

1

1

1

0

0

0

00

00

00

0

0

0

00

00

00

0

0

0

00

00

00

Stress-strain (displacement) Relation

DBcD

h

h

h

h

h

h

h

h

h

h

h

h

h

h

v

v

v

v

v

vv

v

v

vv

v

v

vv

v

v

v

vv

E

23

13

12

33

22

11

13

13

12

33

22

11

)1(2

2100000

0)1(2

210000

00)1(2

21000

000111

0001

11

00011

1

)21)(1(

)1()(

Final System Equation

t

TTTTh

i

h

i

b

T

V

TT

V

Th

V

i

h

i

T

V

TT

V

hTh

V

h

ij

h

ij

dddTu

dVdVdVbu

dVdVdV

ttt

fcTNcTu

fcbNcbu

KcccDBBc

→

fcKccfcfcKcc

Tubu

TT

t

T

b

TT

h

V

hh

V

hh ddVdV

t

2

1

2

1

2

1

Principle of Minimun Potential Energy

fKcfKcc

0

hh




116

9.5 Example

By the elementary beam solution, the displacement field of the structure is assumed as

)2

6

(

)2

(

23

2

lxx

bv

ylxx

au

2

100

01

01

1 ,

22

00

0)(

,

260

0)2

(

222

26

2

E

lxx

lxx

ylx

lxx

ylxx

DBN

)2(15

2

2

1)2(

15

2

2

1

)2(15

2

2

1)2(

15

2

2

1

332

1

)2

(2

1)

2(

2

1

)2

(2

1)

2(

2

1)(

1

22

00

0)(

2

100

01

01

200

20)(

1

55

5533

2

0

1

0 22

22

22

22

22

2

0

1

022

2

2

2

hlhl

hlhlhl

E

dzdydx

lxx

lxx

lxx

lxx

ylxE

dzdydx

lxx

lxx

ylx

lxx

lxx

ylxE

l h

h

l h

h

K

P

ldy

hl

yl

h

h

0

32

P0

30

02

3

3

2

1

0

f

x, u

y, v

l

2h

h

PTy

2




117

P

l

b

a

hlhl

hlhlhl

E 0

3)2(

15

2

2

1)2(

15

2

2

1

)2(15

2

2

1)2(

15

2

2

1

332

1

3

55

5533

2

PEIl

h

vb

PEI

PEh

a

22

2

3

2

1))(

1

1

3

51(

11

2

3

)2

6

(

1))(

1

1

3

51(

)2

(

1

2322

22

lxx

PEIl

h

vv

ylxx

PEI

u

Pxlx

Il

h

yI

M

yI

My

I

lxP

xy

yy

xx

)2

(

1)(

6

5

)(

22

Chapter 1 Fundamentals in...

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