Chapter 1 Fundamentals in Elasticitystrana.snu.ac.kr/lecture/elasticity_2017/Note/Note_all... ·...

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 , ν

tt =

uu = Su

St

V

http://strana.snu.ac.kr/



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 in which bodies move under the action of

forces; 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?

- Galilei Galileo (1564 ~ 1642)

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

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

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

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

that the rules of nature 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 the facts 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 of a complex variable…

Fluids and Solids

- Difference in Material Properties

Fluid flows because EG << .

Solid does not because EG ≈ .

Inviscid fluid: 0≈G

Viscous fluid: EG <<<<0

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

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




4

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.

β→α→ BA , then β+α→+ 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

ErrorError




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 (σ) - Strain (ε) - 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 (σ): force per unit area developed in the body by the action of external forces - Strain (ε): 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 , ν

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=ε

Stress-strain relation:

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

)(ε=σ 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

=φ

=φ<<=+φ

dxldlxf

dxxd

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

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

0)0(0)0( and 0for 0)()(2

2

=φ

=φ<=−φ

dtdttf

dttd

- 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

=φ=φ<<=φ

=φ<+φ

=φ lttlx

dtxdxtxtf

dxxtd

dtxtd




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 ρ is said to be continuously

distributed if the following limit is uniquely defined

everywhere in the given domain V.

n

n

Vn VM

Pn

lim0 ,

)(→∞→

=ρ

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 0→nV as ∞→n , nV approaches a finite number ω as ∞→n . Then,

ε<−ρω→∞→ n

n

Vn VM

Pn

)(lim ,

where ε is an acceptable variability. You may consider ω 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.




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1.4. Fundamental Laws

Newton’s Laws of Motion

- The momentum of an object (mv) is constant unless an outside 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 time rate of change of the momentum 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.




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1.5. Axioms 1. Newton’s laws of motion and the 1st and 2nd 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 εεσ f= VP ∈∀

For a rate-independent material, ))(()( PP εσ 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

iii BABABABABABA ==+++=∑

=

22111

2222

21 inii AAAAAA ≠+++=

ninii

n

jjijjij BABABABABA +++== ∑

=

22111

∑∑∑== =

+++==n

iininiiii

n

i

n

jijijijij BABABABABA

12211

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

nnjjjiijT BABABABA +++==⋅ 2211BA

njinjijikjik BABABABA +++==⋅ 2211BA

Cross product of two vectors

BAC ×=

321

321

321

BBBAAAiii

C = , θ= sinBAC

Gradient operator




12

),,(321 xxx ∂

∂∂∂

∂∂

=∇

ixxxxxxx ∂φ∂

=∂

φ∂∂

φ∂∂

φ∂=φ

∂∂

∂∂

∂∂

=φ∇ ),,(),,(321321

φ∇=∂∂φ∂

=

∂φ∂

+∂

φ∂+

∂φ∂

=φ∂∂

+∂∂

+∂∂

=φ∂∂

∂∂

∂∂

⋅∂∂

∂∂

∂∂

=φ∇⋅∇

22

23

2

22

2

21

2

3

2

2

2

21

2

321321

)(),,(),,(

ii xx

xxxxxxxxxxxx

iiii xxxx ∂∂φ∂

=φ∂∂

∂∂

=φ∇⋅∇2

3

3

2

2

1

1

xxxxxjjj

i

ijij

i ∂

σ∂+

∂

σ∂+

∂

σ∂=

∂

σ∂=σ

∂∂

=⋅∇ σ

Kronecker delta

for 1for 0

=≠

=δjiji

ij

Permutation tensor

equal are indices Any two

npermutatio oddfor npermutatioeven for

011

−=ijke

Determinant of a matrix A

)(Det)(Det 321321T

kjiijkkjiijk AAAeAAAe AA ===

kjijkiikjijkkjiijk BAeCBAeBAeBABABA

BBBAAA =→====×= ii

iiiiii

BAC

333

222

111

321

321

321

In general, pnkjipijk AAAAe 321)(Det =A




13

Tensor fields In case the coordinate rotation is defined by ijij xx β=′ Scalar field (tensor field of rank 0) : ),,(),,( zyxzyx φ=′′′φ′

Vector field (tensor field of rank 1) : ),,(),,( zyxzyx ijij φβ=′′′φ′

Tensor field of rank 2 : jnmnimij zyxzyx βφβ=′′′φ′ ),,(),,(

Tensor field of rank 4 : lskrpqrsjqipijkl zyxzyx ββφββ=′′′φ′ ),,(),,(

Divergence (Gauss) Theorem

3

3

2

2

1

1 divxF

xF

xF

xF

i

i

∂∂

+∂∂

+∂∂

=∂∂

=⋅∇= FF

dSdVSV

nFF ⋅=⋅∇ ∫∫

dSgdVgdVgdVgSVVV

nFFFF ⋅=⋅∇+⋅∇=⋅∇ ∫∫∫∫ )( →

dVgdSgdVgVSV∫∫∫ ⋅∇−⋅=⋅∇ FnFF

x

y

z

x′

y′

z′

x




14

Chapter 2

Traction and Stress




15

2.1. Traction and Stress

Body Force: dVd

VV

PPb =∆∆

=→∆

lim0

Traction: dSd

SΔSnFFT =

∆∆

=→0

lim Stress : x, y, z components of traction

x

n Tn

∆S

∆F




16

2.2. Definition of Stress

z, x3

σxx

σxz

σxy

σzx σzz

σzy

σyx σyz

σyy

x,x1

y, x2

σxx σxz

σxy σzx

σzz

σzy

σyx σyz

σyy x,x1

z, x3

y, x2




17

2.3. Cauchy’s Tetrahedron and Relation

Equilibrium Condition

333

222

111

321

2

2332211

2

2

33

22

11

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

31

31

31

31

0

nxnOCh

SS

nxnOBh

SSnxn

OAh

SS

SOCSOBSOAShV

SV

dtud

SVb

SS

tSSt

SStT

Vdt

udVbStStStST

iiiii

ni

iiiii

ni

===∆∆

===∆∆

===∆∆

∆=∆=∆=∆=∆

∆∆

ρ+∆∆

−∆∆

+∆∆

+∆∆

=

=∆ρ−∆+∆−∆−∆−∆

Cauchy’s relation

By the definition of stress jij

it σ= and as 0→∆V

jji

niiii

ni nTntntntT σ=→++= 3

32

21

1

t2

t3 t1 Tn

n

O A

B

C h

n

O A C h

B




18

2.4. Equilibrium Equation - Integral Approach Force Equilibrium: 0=∑ iF for i=1, 2, 3

02

2

=∂∂

ρ−+ ∫∫ ∫V

i

S Vii dV

tu

dVbdST or 02

2

=∂∂

ρ−+ ∫∫ ∫VS V

dVt

dVdS ubT

By the divergence theorem, and the Cauchy’s relationship

∫∫∫ ∂σ∂

=σ=V j

ji

Sjji

Si dV

xdSndST

Therefore, the force equilibrium equation becomes

0)( 2

2

2

2

=∂∂

ρ−+∂σ∂

=∂∂

ρ−+∂σ∂

∫∫∫∫V

ii

j

ji

V

i

Vi

V j

ji dVtub

xdV

tudVbdV

x

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

should vanish everywhere in the domain.

02

2

=∂∂

ρ−+∂σ∂

tub

xi

ij

ji → 02

2

, =∂∂

ρ−+σtub i

ijji → 02

2

=∂∂

ρ−+⋅∇tubσ

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

02

2

=∂∂

×ρ−+×+× ∫∫∫∫VVVS

dVt

dVdVdS uxmbxTx or the indicial form becomes

02

2

=∂

∂ρ−++ ∫∫∫∫

V

nmimn

Vi

Vnmimn

Snmimn dV

tuxedVmdVbxedSTxe




19

∫∫

∫∫∫∫

∂

σ∂+σ=

∂

σ∂+σδ

=∂

σ∂+σ

∂∂

=∂

σ∂=σ=

V j

jnmmnimn

V j

jnmjnmjimn

V j

jnmjn

j

mimn

V j

jnmimn

Sjjnmimn

Snmimn

dVx

xedVx

xe

dVx

xxx

edVxxe

dSnxedSTxe

)()(

)(

0)( 2

2

=∂

∂ρ−++

∂

σ∂+σ ∫∫∫∫

V

nmimn

Vi

Vnmimn

V j

jnmmnimn dV

tu

xedVmdVbxedVx

xe

0)()( 2

2

=∂

∂ρ−

∂σ∂

+++σ ∫∫V

n

j

jnnmimn

Vimnimn dV

tu

xbxedVme

00)( =+σ→=+σ∫ imnimnV

imnimn medVme

In case there is no body moment, 0=σmnimne

σ=σ→=σσ=σ→=σσ=σ→=σ

→=σ

21123

31132

32231

000

0

mnmn

mnmn

mnmn

mnimn

eee

e

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:

v n

dS




20

∫ ∫ =+⋅−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

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

Φ∇⋅−= kv

– The famous Laplace equation for a conservative system.

0)( =+Φ∇⋅⋅∇=+⋅∇− ff kv

– the system properties are homogeneous and isotropic, Φ∇−= kv

0)( 2 =+Φ∇=+Φ∇⋅∇=+⋅∇− fkfkfv or

0)(2

2

2

2

2

2

2

=+∂∂Φ∂

=+∂

Φ∂+

∂Φ∂

+∂

Φ∂ fxx

kfzyx

kii




21

2.6. Equilibrium and Potential Problems

Equilibrium

– Force Equilibrium: ∑∑∑ === 0zyx FFF or 0=∑F

∫ ∫ =+S V

dVdS 0bT or ∫ ∫ =+S V

ii dVbdST 0 for i = 1,2,3

– Suppose nT ⋅= σ or ∑=

⋅=σ=3

1jijiji nT nσ

=

σσσσσσσσσ

=

3

2

1

333231

232221

131211

σσσ

σ ,

=

3

2

1

nnn

n

– Divergence Theorem

0)( =+⋅∇=

+⋅∇=+⋅=+

∫

∫ ∫∫ ∫∫ ∫

Vii

V Vii

S Vii

S Vii

dVb

dVbdVdVbdSdVbdST

σ

σσ n for i = 1,2,3

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

0=+⋅∇ ii bσ for i = 1, 2, 3

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

0=+×+× ∫∫∫VVS

dVdVdS mbxTx or 211231133223 , , σ=σσ=σσ=σ

Potential Problems

– In case elasticity problems are potential problems

ii

i u∇⋅= Cσ or k

iijkij x

uC∂∂

=σ

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

k

jjikji

k

iijkij x

uC

xuC

∂

∂=σ≠

∂∂

=σ 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 ∇= :σ or l

kijklij x

uC∂∂

=σ and jiklijkl CC =




22

– In case ijlkijkl CC =

klijklk

l

l

kijkl

k

lijkl

l

kijkl

l

kijlk

l

kijkl

l

kijklij

Cxu

xuC

xuC

xuC

xuC

xuC

xuC

ε=∂∂

+∂∂

=

∂∂

+∂∂

=∂∂

+∂∂

=∂∂

=σ

)(21

)(21)(

21

– 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 ⋅′=′→⋅′=′δ→⋅′=′′⋅′→′′=

formin tensor or ) ,cos( where βxxee =′′=⋅′=ββ=′ ijijjiijij xxxx

xβxeeeeeeeeee ′=→′β=′⋅′=→′′⋅=δ→′′⋅=⋅→′′= Tkkjkjkjkkjijikkjiijkkii xxxxxxxxx

Orthogonality

ijkikjikikjijikikjkkjj xxxx δ=ββ→=ββ−δ→ββ=′β= 0)(

ΤΤijkjki ββΙββ =→=→δ=ββ −1

Diffential Operator

kkimi

kkm

i

mkm

ki

k

ki xxxx

xxx

xx ′∂∂

β=δ′∂

∂β=

∂β∂

′∂∂

=∂

′∂′∂

∂=

∂∂ ()()()()()

x

y

z

x′

y′

z′

x




23

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

- Stress in each coordinate system by the deficition: i

j

xxij T=σ , i

j

xxij T ′′=σ′

- Cauchy’s relation: jkkixkki

xx

xxx

xxx

xxx

xx

jj

i

j

i

j

i

j

i

j

inTnTnTnTT βσ=σ=→++= ′′′′′′

321321

- Normal vector: jkkjkjxkk

xk

xxxj

x xxnnnnn jjjjjj β=′=⋅′=→=++=′= ′′′′′′ ),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

Tkpqpmqmqqpkp

xxkp

xxmk

m

p

m

kTT βσβσ =′→βσβ=βσβ=β==σ′ ′′

′

ijklkiijT βσ′β=σ→′= βσβσ

Equilibrium Equation in primed coordinate system

- Stress: liklkjji βσ′β=σ , Body force: llii bb ′β= , Diplacement: llii uu ′β=

0)( 2

2

2

2

2

2

2

2

=∂

′∂−′+

′∂σ′∂

β→∂

′∂β=′β+δβ

′∂σ′∂

→∂

′∂β=′β+βββ

′∂σ′∂

→∂

′β∂=′β+β

′∂βσ′β∂

tub

xtub

x

tub

xtub

x

ll

k

klli

llillikmli

m

kl

llillimjlikj

m

klllillimj

m

liklkj

2

2

tub

xl

lk

kl

∂′∂

=′+′∂

σ′∂

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

x2

x1

x3

O A

BB

C

jx′

jx′T 1xT

2xT

3xT




24

Chapter 3

Displacement and Strain




25

3.0. Kinematic Description Kinematics?

A subdivision of classical mechanics concerned with the geometrically possible motion

of a body 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?

- 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.

t t+∆t




26

aVVV=

∆−∆+

=∂

∂

→∆ tttt

tt

tc

)()(),car( lim0fixedar

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

≠∆

−∆+=

∂∂

→∆ tttt

tt

t

)()(),( lim0fixed

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)

VVxVax

⋅∇+∂

∂=

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

ttxVtt

txVtxV ∆∂

∆+∂+∆

∂∂

=∆ ),(),(),(),(fixed

fixed Materialx

),(),(),(),(limfixed fixed Material

0txV

xtxV

ttxV

ttxVa

t ∂∂

+∂

∂=

∆∆

=→∆

x




27

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 0=t , 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 X

Reference configuration t

),( txXX =

Velocity

Observation Point

V(t)

V(t+∆ t) V∆ t

tt

txV∆

∂∂

fixed

),(x

tV

xttxV

∆∂

∆+∂

fixed

),(x




28

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.

Control Volume: analysis domain

x

t

t+∆t

t




29

3.1. Displacement and Strain

Displacement

xxaxu −= )()( or )()( axaau −=

Strain

ii dxdxdxdxdxds =++= 23

22

21

20 , ii dadadadadads =++= 2

322

21

2

kk

ii dx

xa

da∂∂

= , kk

ii da

ax

dx∂∂

=

lkl

j

k

iijl

l

jk

k

iijjiijii dada

ax

ax

daax

daax

dxdxdxdxds∂

∂

∂∂

δ=∂

∂

∂∂

δ=δ==20

lkl

j

k

iijl

l

jk

k

iijjiijii dxdx

xa

xa

dxxa

dxxa

dadadadads∂

∂

∂∂

δ=∂

∂

∂∂

δ=δ==2

jiijjiijji

jiijjiji

jiijlkl

j

k

iij

dxdxEdxdxxa

xa

dxdxdxdxxa

xa

dxdxdxdxxa

xa

dsds

2)(

20

2

=δ−∂

∂

∂∂

δ=

δ−∂

∂

∂∂

δ=δ−∂

∂

∂∂

δ=−

βααβ

βααβ

Eij is called as Green’s strain.

x1

x2

x3

u(x)

x a

x1

x2 u(x) x a

dx

da




30

jiijjiji

ij

jiji

jiijlkl

j

k

iijjiij

dadaedadaax

ax

dadaax

ax

dadadadaax

ax

dadadsds

2)(

20

2

=∂

∂

∂∂

−=

∂

∂

∂∂

−=∂

∂

∂∂

−=−

βααβ

βααβ

δδ

δδδδ

eij is called as Almansi strain. Green’s Strain in terms of displacement

)(21

)(21

21))((

21

))()(

(21)(

21

jii

j

j

i

ijjii

jj

iji

ijj

ji

i

ijji

ijji

ij

xu

xu

xu

xu

xu

xu

xu

xu

xu

xu

xux

xux

xa

xa

E

∂∂

∂∂

+∂

∂+

∂∂

=

δ−∂

∂

∂∂

δ+∂∂

δδ+∂

∂δδ+δδδ=

δ−∂

∂+δ

∂∂

+δδ=

δ−∂

+∂

∂+∂

δ=δ−∂

∂

∂∂

δ=

αα

βααβ

αβαβ

βααββααβ

ββ

αααβ

ββαααβ

βααβ

)(21)(

21)(

21

)(21)(

21)(

21

)(21)(

21)(

21

])()()[(21)(

21

])()()[(21)(

21

])()()[(21)(

21

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

313

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

333

2

2

32

2

22

2

1

2

2

112

222

2

1

32

1

22

1

1

1

1

111

111

xu

xu

xu

xu

xu

xu

xu

xu

xu

xu

xu

xuE

xu

xu

xu

xu

xu

xu

xu

xu

xu

xu

xu

xu

E

xu

xu

xu

xu

xu

xu

xu

xu

xu

xu

xu

xuE

xu

xu

xu

xu

xu

xu

xu

E

xu

xu

xu

xu

xu

xu

xuE

xu

xu

xu

xu

xu

xu

xuE

∂∂

∂∂

+∂∂

∂∂

+∂∂

∂∂

+∂∂

+∂∂

=∂∂

∂∂

+∂∂

+∂∂

=

∂∂

∂∂

+∂∂

∂∂

+∂∂

∂∂

+∂∂

+∂∂

=∂∂

∂∂

+∂∂

+∂∂

=

∂∂

∂∂

+∂∂

∂∂

+∂∂

∂∂

+∂∂

+∂∂

=∂∂

∂∂

+∂∂

+∂∂

=

∂∂

+∂∂

+∂∂

+∂∂

=∂∂

∂∂

+∂∂

=

∂∂

+∂∂

+∂∂

+∂∂

=∂∂

∂∂

+∂∂

=

∂∂

+∂∂

+∂∂

+∂∂

=∂∂

∂∂

+∂∂

=

αα

αα

αα

αα

αα

αα

Cauchy’s infinitesimal strain tensor – Small deformation

)(21

i

j

j

iij x

uxu

∂

∂+

∂∂

=ε for small displacement gradient 1<<∂∂

j

i

xu

In this case, ijijij eE ε==

All kinds of strain are symmetric.




31

Rigid body motion : 0dsds =

jiijjiij dadaedxdxEdsds 2220

2 ==− for dx and da. 0≡=↔ ijij eE for whole domain.

For small strain, 0≡ε ij .

Engineering strain : ijij ε=γ 2 for ji ≠

Strain Component in primed coordinate system

– Diplacement : kkii uu ′β=

– Strain

mjkmkimjm

l

k

l

k

m

m

kki

mjm

l

k

lkiki

k

mmjmj

m

kki

njn

l

m

lmiki

k

mmjmj

m

kki

njn

lmi

m

kklki

k

mmjmj

m

kki

njn

llmi

m

kkmi

m

kkjmj

m

kki

j

ll

i

kk

i

kkj

j

kkiij

Exu

xu

xu

xu

xu

xu

xu

xu

xu

xu

xu

xu

xu

xu

xu

xu

xu

xu

xu

xu

xu

xu

xu

xu

E

β′β=β′∂′∂

′∂′∂

+′∂′∂

+′∂′∂

β=

β′∂′∂

′∂′∂

β+β′∂′∂

β+β′∂′∂

β=

β′∂′∂

′∂′∂

β+β′∂′∂

β+β′∂′∂

β=

β′∂′∂

β′∂′∂

δ+β′∂

′β∂+β

′∂′β∂

=

β′∂

′β∂β

′∂′β∂

+β′∂

′β∂+β

′∂′β∂

=

∂′β∂

∂′β∂

+∂

′β∂+

∂′β∂

=

αα

αα

)(21

)(21

)(21

)(21

)(21

)(21

TT ββββ EEEE =′′= ,




32

3.2. Physical Meanings

Normal Strain

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

20111111

20

2 222 dsEdxdxEdxdxEdsds jiij ===− , 01121 dsEds +=→

0

0

0

020

20

2

11 22 dsdsds

dsdsds

dsdsds

E+−

=−

=

Since 0dsds ≈ for small strain, then0

01111 ds

dsdsE

−=ε=

Shear Strain

21122121

2

cos

xddxExddxxa

xaxddx

xa

xa

xdxadx

xaaddasdsddd

kknm

n

k

m

k

nn

km

m

kkk

=∂∂

∂∂

=∂∂

∂∂

=

∂∂

∂∂

==θ=⋅ aa

sdsdxddxE 21122cos =θ

222022111011 2121 , 2121 xdEsdEsddxEdsEds +=+=+=+=

2211

1212 2121

2)sin(cosEE

E++

=α=θ

2121)sin(21 22111212 EEE ++α=

In case 1 , 2211 <<EE , 121212 21 )sin(

21 α≈α=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+




33

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

uxu

xu

∂∂

=ε∂∂

=ε∂∂

=ε

)(21 , )(

21 , )(

21

2

3

3

232

3

1

1

313

2

1

1

212 x

uxu

xu

xu

xu

xu

∂∂

+∂∂

=ε∂∂

+∂∂

=ε∂∂

+∂∂

=ε

- 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! What conditions is required to do so ?

Two dimensional Case :

0231333 =ε=ε=ε , )(21 , ,

1

2

2

112

2

222

1

111 x

uxu

xu

xu

∂∂

+∂∂

=ε∂∂

=ε∂∂

=ε

∫∫ +ε=+ε= )( , )( 1222221111 xfdxuxgdxu

))()((21)(

21

1

1222

12

2111

21

2

2

112 x

xfdxxx

xgdxxx

uxu

∂∂

+ε∂∂

+∂

∂+ε

∂∂

=∂∂

+∂∂

=ε ∫∫

222

2

211

2

21

122

222

2

211

2

21

122

1212

, )(21

xxxxxxxx ∂ε∂

+∂

ε∂=

∂∂γ∂

∂ε∂

+∂

ε∂=

∂∂ε∂

The strain field can not be defined independently !!!




34

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.

∫∫

∫∫∫∫

Ω+ε+=

∂∂

−∂

∂+

∂∂

+∂

∂+=

∂

∂+=+=

Ckjk

Ckjkj

Ck

j

k

k

j

Ck

j

k

k

jj

Ck

k

jj

Cjj

pj

dxdxu

dxxu

xu

dxxu

xu

udxxu

uduuu

0

000 )(21)(

21

∫∫∫ ∂

Ω∂−−Ω−=−Ω=Ω

Cm

m

jkpkkjkk

pk

C

pkkjk

Ckjk dx

xxxxxxxddx )()()( 00

j

km

k

mj

k

m

m

k

jj

m

m

j

k

kj

m

kj

m

j

k

k

j

mm

jk

xxxu

xu

xxu

xu

x

xxu

xxu

xu

xu

xx

∂ε∂

−∂

ε∂=

∂∂

+∂∂

∂∂

−∂∂

+∂

∂

∂∂

=

∂∂∂

−∂∂

∂+

∂∂

−∂

∂

∂∂

=∂

Ω∂

)(21)(

21

21

21)(

21 22

∫

∫

∫∫

+Ω−+=

∂ε∂

−∂ε∂

−−ε+Ω−+=

Ω+ε+=

Cmjmjkk

pkj

Cm

j

km

k

mjpkkjmjkk

pkj

Ckjk

Ckjkj

pj

dxFxxu

dxxx

xxxxu

dxdxuu

000

000

0

)(

)))((()(

For the path independent condition

n

jm

m

jn

xF

xF

∂∂

=∂∂

u0

up

x0

xp




35

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

km

k

mjpkkjm

nj

kn

k

njpkkjn

m xxxx

xxxxx

x ∂ε∂

−∂

ε∂−−ε

∂∂

=∂ε∂

−∂

ε∂−−ε

∂∂

0])[(

)()(

2222

=∂∂ε∂

+∂∂

ε∂−

∂∂ε∂

−∂∂

ε∂−

−∂ε∂

−∂

ε∂δ+

∂

ε∂−

∂ε∂

−∂

ε∂δ−

∂

ε∂

jn

km

kn

mj

jm

kn

km

njpkk

j

km

k

mjkn

n

jm

j

kn

k

njkm

m

jn

xxxxxxxxxx

xxxxxx

0])[(

)()(

2222

=∂∂ε∂

+∂∂

ε∂−

∂∂ε∂

−∂∂

ε∂−

−∂ε∂

−∂

ε∂+

∂

ε∂−

∂ε∂

−∂

ε∂−

∂

ε∂

jn

km

kn

mj

jm

kn

km

njpkk

j

nm

n

mj

n

jm

j

mn

m

nj

m

jn

xxxxxxxxxx

xxxxxx

02222

=∂∂ε∂

+∂∂ε∂

−∂∂ε∂

−∂∂ε∂

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

332

3

3

12

2

13

1

23

231

222

2

3

12

2

13

1

23

132

112

1

21

122

21

222

22

112

3

31

132

23

112

21

332

2

32

232

22

332

23

222

1

=∂ε∂

−∂ε∂

+∂ε∂

∂∂

+∂∂ε∂

−=

=∂ε∂

+∂ε∂

−∂ε∂

∂∂

+∂∂ε∂

−=

=∂ε∂

+∂ε∂

+∂ε∂

−∂∂

+∂∂ε∂

−=

=∂∂ε∂

−∂

ε∂+

∂ε∂

=

=∂∂ε∂

−∂

ε∂+

∂ε∂

=

=∂∂ε∂

−∂

ε∂+

∂ε∂

=

xxxxxxU

xxxxxxU

xxxxxxU

xxxxR

xxxxR

xxxxR

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

Bianchi formulas

0 , 0 , 03

3

2

1

1

2

3

1

2

2

1

3

3

2

2

3

1

1 =∂∂

+∂∂

+∂∂

=∂∂

+∂∂

+∂∂

=∂∂

+∂∂

+∂∂

xR

xU

xU

xU

xR

xU

xU

xU

xR




36

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

0321 =∂∂

=∂∂

=∂∂

zR

yR

xR in V → 0321 === RRR in V

- Case 2

0=== zyx RRR in V and 0=== zyx UUU on S. By Bianchi condition

0 ,0 ,02

1

1

2

3

1

1

3

3

2

2

3 =∂∂

+∂∂

=∂∂

+∂∂

=∂∂

+∂∂

xU

xU

xU

xU

xU

xU in V .

For arbitrary ),,( zyxF , ),,( zyxG and ),,( zyxH

0

)()()([

])()()[(

)()()([

213

312

321

321

2

1

1

2

3

1

1

3

3

2

2

3

=

∂∂

+∂∂

+∂∂

+∂∂

+∂∂

+∂∂

−

+++++=

∂∂

+∂∂

+∂∂

+∂∂

+∂∂

+∂∂

=

∫

∫

∫

dVxF

xGU

xF

xHU

xG

xHU

dSUmFlGUlHnFUmHnG

dVxU

xUH

xU

xUG

xU

xUFI

V

S

V

Since ),,( zyxF , ),,( zyxG and ),,( zyxH are arbitrary, 0≡== zyx UUU in V.




37

Chapter 4

Constitutive Relations




38

4.1. Constitutive Law Generalized Hooke’s Law

klijklklijklij CEC ε==σ

Governing Equations in solid mechanics

– Equilibrium : 2

2

, tub i

ijij ∂∂

ρ=+σ

– Strain-displacement relation : )(21

i

j

j

iij x

uxu

∂∂

+∂∂

=ε

– Constitutive law : klijklklijklij CEC ε==σ

Equilibrium Equations in terms of displacement

2

22

tub

xxuC i

ilj

kijkl ∂

∂ρ=+

∂∂∂

Boundary conditions

– Displacement BC : uii uu Γ= on

– Traction BC : tii TT Γ= on Traction BC in terms of displacement

ijl

kijkljiji Tn

xu

CnT =∂∂

=σ=

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 properties

when 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 ×××=




39

– Symmetry on σ and ε: jiklijkl CC = , ijlkijkl CC = (36)

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

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

CCCCCCCCC

========

9 1

Total 21 5

Normal-Shear Relation = 0

Case I Case II

– Case I ( 0 , 0 231312332211 =ε=ε=ε=ε=ε≠ε ) → 11121112 ε=σ CI

– Case II ( 0 , 0 231312331122 =ε=ε=ε=ε=ε≠ε ) → 22122212 ε=σ CII

By isotropic condition, III1212 σ−=σ , for 2211 ε=ε .

0020)( 12111112111112221211 =→=ε→=ε+ CCCC

ε11

σ12 ε22

σ12




40

Shear-Other Shear Relation

Case I Case II

– Case I ( 0 , 0 131233221123 =ε=ε=ε=ε=ε≠ε ) → 23122312 2 ε=σ CI

– Case II ( 0 , 0 231333221112 =ε=ε=ε=ε=ε≠ε ) → 12231223 2 ε=σ CII

By isotropic condition, III2312 σ−=σ , for 2312 ε=ε .

0040)(2 12231212231223121223 =→=ε→=ε+ CCCC

εεεεεε

=

σσσσσσ

31

23

12

33

22

11

3131

23312323

123112231212

3331332333123333

22312223221222332222

113111231112113311221111

31

23

12

33

22

11

.CCCsymmCCCCCCCCCCCCCCCCCC

90° Rotation

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

N-N (A1)

S-O.S

S-S (A3)




41

δ expressions

Rel. Coefficient δ relation Value

N-N 333322221111 CCC == lpkpjpip δδδδ A1

N-O.N. )( 331133222211

113322331122

CCCCCC

====

lpkpjpipklij δδδδ−δδ A2

S-S

)()()(

313132322121

133123321221

311332232112

131323231212

CCCCCCCCCCCC

========

lpkpjpipjkiljlik δδδδ−δδ+δδ 2 A3

The expressions in parentheses represent dependent relations.

Superposition

)2( )( 321 lpkpjpipjkiljliklpkpjpipklijlpkpjpipijkl AAAC δδδδ−δδ+δδ+δδδδ−δδ+δδδδ=

lpkpjpipjkiljlikklijijkl AAAAAC δδδδ−−+δδ+δδ+δδ= )2()( 32132 Rotation of Stress and Strain

qlpqpkkljnmnimij βε′β=εβσβ=σ′ ,

klijkl

kllpkpjpipkljkiljlilklklij

kllpkpjpipkljkiljlilklklij

pqqrprjrirpqjpiqjqippqpqij

pqqrprjrir

pqqmpnjnimqnpmjnimpqqkpkjmim

pqqlpklrkrnrmrjnim

pqqlpknkmlnlmkjnimpqqlpkklmnjnim

pqqlpkmnkljnimqlpqpkjnmnklimij

CAAAAAAAAAA

AAAAAAAA

AAAAA

AACC

ε′′=

ε′δδδδ−−+ε′δδ+δδ+ε′δδ=

ε′ββββ−−+ε′δδ+δδ+ε′δδ=

ε′ββββ−−+ε′δδ+δδ+ε′δδ=

ε′ββββ−−

+ε′ββββ+ββββ+ε′ββββ=

ε′ββδδδδββ−−

+ε′ββδδ+δδββ+ε′ββδδββ=

ε′ββββ=βε′βββ=σ′

)2( ][

)2( ][

)2( ][

)2(

][

)2(

)(

32132

32132

32132

321

32

321

32

For isotropy, ijklijkl CC ′=

lpkpjpiplpkpjpip AAAAAA δδδδ−−=ββββ−− )2()2( 321321

321321 202 AAAAAA +=→=−−

Final Relation: two independent coefficient

)()(32 jkiljlikklijjkiljlikklijijkl AAC δδ+δδµ+δλδ=δδ+δδ+δδ=




42

Lame Elastic Constant, λ and µ

)1(2 ,

)21)(1( vE

vvvE

+=µ

−+=λ

Detailed Expressions

ijijkkjiijijkk

kljkiljlikklijij

µε+δλε=µε+µε+δλε=

εδδ+δδµ+δλδ=σ

2

))((

23233223

13133113

12122112

33221133

33221122

33221111

222

)2()2(

)2(

µγ=µε=σ=σµγ=µε=σ=σµγ=µε=σ=σ

εµ+λ+λε+λε=σλε+εµ+λ+λε=σλε+λε+εµ+λ=σ

Equilibrium equation in terms of Lame constant

2

22

2

22

2

2

2

2

2

2

,

)()(

)(

)(

t

tub

xxu

xu

x

tub

xu

xxu

xxu

x

tub

xu

xu

xxu

x

tub

ii

jj

i

k

k

i

ii

j

j

ij

i

jk

k

i

ii

i

j

j

i

jij

k

k

j

iijij

∂∂

ρ=+∇µ+⋅∇∇µ+λ

∂∂

ρ=+∂∂

∂µ+

∂∂

∂∂

µ+λ

∂∂

ρ=+∂∂

∂∂

µ+∂∂

∂∂

µ+∂∂

∂∂

λ

∂∂

ρ=+∂∂

+∂∂

∂∂

µ+δ∂∂

∂∂

λ

∂∂

ρ=+σ

ubuu

or

2

2

23

2

22

2

21

2

3

3

2

2

1

1 )()()(tub

xu

xu

xu

xu

xu

xu

xi

iiii

i ∂∂

ρ=+∂∂

+∂∂

+∂∂

µ+∂∂

+∂∂

+∂∂

∂∂

µ+λ




43

4.3. Physical Meanings of Material Properties

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

)(

)(

)(

22113333

33112222

33221111

EEE

EEE

EEE

σ+

σν−

σ=ε

σ+

σν−

σ=ε

σ+

σν−

σ=ε

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

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

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

22113333

33112222

33221111

ε+εν+ε−−+

=σ

ε+εν+ε−−+

=σ

ε+εν+ε−−+

=σ

vvv

E

vvv

E

vvv

E

Gv

EAAAvv

EAvv

EvA =+

=−

=µ=−+

ν=λ=

−+−

=)1(22

, )21)(1(

, )21)(1(

)1( 21321

(1+ε11)L

h (1-νε11)h

L




44

Physical Derivation of Shear Modulus

)1(max ν+τ

=τ

ν+τ

=εEEE

γ+≈γ+≈ε+→γ+=ε+

γ+π

−+=ε+

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

)2

cos(2)]1(2[

max2

max

2222max hhhh

)1(2

)1(2)1(

2max

ν+=

γ=τ→γν+

=τ→ν+τ

=γ

=ε

EG

GEE

τ

τ

τ

τ

τ τ

τ τ

)1(2 maxε+h

h




45

Chapter 5

Beam Approximation

≅

x1

x2

x3

),,0( 32ttt nn=n

)0,0,1(=ln

liT




46

5.1. Equilibrium Equation

Force Resultant

∫∫∫ σ=σ==A

iA

ljji

A

lii dAdAndATQ 1

iA

ij

ji

S

ti

Ai

Ai

j

ji

S

ti

ti

Ai

Ai

j

ji

A

ii

Ai

Ai

j

ji

A

ii

j

ji

A

ii

qdAbx

dSTdAbdAbx

dSnndAbdAbx

dAxx

dAbdAbx

dAxxx

dAxx

Q

−+∂

σ∂=−−+

∂

σ∂=

σ+σ−−+∂

σ∂=

∂σ∂

+∂σ∂

−−+∂

σ∂=

∂σ∂

−∂σ∂

−∂

σ∂=

∂σ∂

=∂∂

∫∫∫∫

∫∫∫

∫∫∫

∫∫

)()(

)()(

)()(

)(

3322

3

3

2

2

3

3

2

2

1

1

1

0)(1

=+∂σ∂

=+∂∂

∫A

ij

jii

i dAbx

qxQ

x1

x2

x3

),,0( 32ttt nn=n

)0,0,1(=ln

liT

x2

x3

S




47

Moment Resultant

∫∫∫ σ=σ==A

kjijkA

lmkmjijk

A

lkjijki dAxedAnxedATxeM 1

~~~ where ),,0(~32 xx=x

kljijk

S

tkjijk

Akjijk

Ak

m

kmjijk

Akkikki

S

tmkmjijk

Akjijk

Ak

m

kmjijk

Akkikki

A

kjijkkjijk

Akjijk

Ak

m

kmjijk

A

kkjijk

A m

kmjijk

A

kk

m

kmjijk

A

kjijk

i

QnedSTxedAbxedAbx

xe

dAeedSnxedAbxedAbx

xe

dAee

dAxxe

xxe

dAbxedAbx

xe

dAxx

xedAx

xe

dAxxx

xe

dAx

xexM

−−−+∂

∂=

++−−+∂

∂=

+

+∂

∂+

∂

∂−−+

∂∂

=

∂∂

+∂

∂−

∂∂

=

∂∂

−∂

∂−

∂∂

=

∂∂

=∂∂

∫∫∫

∫∫∫∫

∫

∫∫∫

∫∫

∫

∫

~~)(~

)(~~)(~

)(

)~~

(~)(~

)(~~

)(~

~

3322

3322

3

3

2

2

3

3

2

2

3

3

2

2

1

1

1

σ

σσσσ

σσ

σσσ

σσσ

σσσ

σ

0~~1

=+++∂∂

∫∫ kljijk

S

tkjijk

Akjijk

i QnedSTxedAbxexM

Sign Convention for Moment

For i =1

12332231323212311 )()~~(~~ MdATxTxdATxeTxedATxeMA

ll

A

ll

A

lkjjk ∫∫∫ =−=+==

For i =2

2131323122

~~~ MdATxdATxedATxeMA

l

A

l

A

lkjjk ∫∫∫ ====

For i =3

x2

x3

T3

T2

x1

x2

T1

x1 T1

x3




48

3131232133~~~ MdATxdATxedATxeM

A

l

A

l

A

lkjjk −=−=== ∫∫∫




49

Component Form

011

1 =+∂∂ q

xQ ∫∫ +=

S

t

A

dSTdAbq 111

021

2 =+∂∂ q

xQ ∫∫ +=

S

t

A

dSTdAbq 222

031

3 =+∂∂ q

xQ ∫∫ +=

S

t

A

dSTdAbq 333

011

1 =+∂∂ m

xM ∫∫ −+−=

S

tt

A

dSTxTxdAbxbxm )~()( 233223321

0321

2 =−+∂

∂ Qmx

M ∫∫ +=S

t

A

dSTxdAbxm 13132

0231

3 =+−∂

∂ Qmx

M ∫∫ +=S

t

A

dSTxdAbxm 12123

Planar Beam 03131 ==== tt TTbb

021

2 =+∂∂ q

xQ , 02

1

3 =−∂

∂ Qx

M → 221

32

qxM

−=∂

∂

011

1 =+∂∂ m

xM ∫∫ −−=

S

t

A

dSTxdAbxm 23231

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

of beam ??




50

5.2. 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

yxu

u yx ∂

∂−= , )(xuu yy = , ??=zu

Strain-displacement relation

??

0

2

2

=∂∂

=ε

=ε∂∂

−=∂∂

=ε

zu

yxu

xu

zzz

yy

yxxx

??

??

0

=∂∂

=∂∂

+∂∂

=γ

=∂∂

=∂

∂+

∂∂

=γ

=∂∂

+∂∂

=γ

xu

xu

zu

yu

zu

yu

xu

yu

zzxxz

zyzyz

yxxy

Strain-Stress relation

)(

)(

)(

22113333

33112222

33221111

EEE

EEE

EEE

σ+

σν−

σ=ε

σ+

σν−

σ=ε

σ+

σν−

σ=ε




51

Assumption on stress

– By traction BC, 0=σzz , 0=σ yy (??)

EEEzz

zzxx

yyxx

xxσ

ν−=εσ

ν−=εσ

=ε , ,

Since, however, 0=ε yy by the strain-displacement relation, the Poisson ratio should be zero, which is equivalent to neglecting the Poisson effect.

2

22

2

22

2

2

2

2

xu

EIdAyxu

EdAyxu

EydAM

yxu

E

y

A

y

A

y

Axxz

yxx

∂

∂−=

∂

∂−=

∂

∂−=σ=

∂

∂−=σ

∫∫∫

Equilibrium Equation

0

0

=+∂σ∂

+∂σ∂

=+∂σ∂

+∂σ∂

yyyxy

xxyxx

byx

byx

→ 0

0

=+∂σ∂

=∂σ∂

+∂σ∂

yxy

xyxx

bx

yx

000)(

000

00)(0)(

2

1

=+∂∂

→=+σ∂∂

→=+∂σ∂

=−∂

∂→=σ−σ+

∂∂

→=∂σ∂

+∂

∂

→=∂σ∂

+σ∂∂

→=∂σ∂

+∂σ∂

→=∂σ∂

+∂σ∂

∫∫∫

∫∫∫

∫ ∫∫∫

qxVdAbdA

xdAb

x

Vx

MdAdzyx

MydAyx

M

ydAy

ydAx

ydAyx

ydAyx

Ay

Axy

Ay

xy

Axy

z

h

hxyA

xy

A A

xyxx

A

xyxx

A

xyxx

02

2

=+∂

∂ qxM

qxu

EIx

qxM y =

∂∂

∂∂

→−=∂

∂2

2

2

2

2

2

, yI

Mxx =σ

IbVQhybdy

yydA

IVdA

yIy

xM

yx xyxy

h

y

xyh

y

h

y

xyxyxx =σ−σ→=∂σ∂

+→=∂σ∂

+∂

∂→=

∂σ∂

+∂σ∂

∫∫∫ )()(00)(0 2

222

Simply supported beam with a uniformly distributed load

qhql

yyxx ≈σ=σ max2

2

max )( , 86)(

68

)()(

2

2

max

max

lh

xx

yy ≈σσ




52

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 )(θ−= , )(xuu yy = , ??=zu

Strain Components

??

0

=∂∂

=ε

=ε∂θ∂

−=∂∂

=ε

zu

yxx

u

zzz

yy

xxx

??

??

=∂∂

=∂∂

+∂∂

=γ

=∂∂

=∂

∂+

∂∂

=γ

∂∂

+θ−=∂∂

+∂∂

=γ

xu

xu

zu

yu

zu

yu

xu

xu

yu

zzxxz

zyzyz

yyxxy

Equilibrium equation

Vx

M=

∂∂

, 0=+∂∂ q

xV




53

By energy consideration

)(

~0

22

2

2

2

22

xys

AA

xy

Axyxy V

GAV

GAfVdA

ybQ

GIVdA

GdA γ====

σ=γσ ∫∫∫

which yield xyGAV γ= 0

0)()(

)(

02

2

02

2

0

2

=θ−∂∂

+∂

θ∂→θ−

∂∂

=∂

θ∂−

θ−∂∂

=

∂θ∂

−=∂θ∂

−=σ= ∫∫

xu

GAx

EIxu

GAx

EI

xu

GAV

xEIdAy

xEdAyM

yy

y

AAxx

qxx

uGA y −=

∂θ∂

−∂∂

)( 2

2

0




54

Chapter 6

Plane Problems in

Cartesian Coordinate System

x y

z

∞




55

6.1. 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 σ=

– On yx −± plane: )1,0,0( ±=n

– 0 , 0 , 0 =σ±==σ±==σ±= zzzyzyxzx TTT

– By the third assumption

0=σ=σ=σ zzyzxz in V


0

0

0

33

33

2

32

1

31

23

23

2

22

1

21

13

13

2

12

1

11

=+∂σ∂

+∂σ∂

+∂σ∂

=+∂σ∂

+∂σ∂

+∂σ∂

=+∂σ∂

+∂σ∂

+∂σ∂

bxxx

bxxx

bxxx

→

00

0

0

=

=+∂

σ∂+

∂

σ∂

=+∂

σ∂+

∂σ∂

yyyxy

xxyxx

byx

byx

Strain-Displacement relations

x

y

z




56

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

)(21 , ,

xu

yu

yu

xu yx

xyy

yyx

xx ∂∂

+∂∂

=ε∂∂

=ε∂∂

=ε

Constitutive law

ε=σ

ε+ε−

=σ

ε+ε−

=σ

→

σ=ε

ε+εν−

ν−=

σ+

σν−=ε

σν−

σ=ε

σν−

σ=ε

xyxy

xxyyyy

yyxxxx

xyxy

yyxxyyxx

zz

xxyyyy

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

xu

yuG

xuv

yu

vE

yu

vxu

vE

yxxy

xyyy

yxxx

∂∂

+∂∂

=σ

∂∂

+∂∂

−=σ

∂∂

+∂∂

−=σ

– Equilibrium equation in terms of displacement

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

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

22

2

2

2

22

2

2

2

=−+∂∂

−∂

∂∂∂

+−∂∂

+∂∂

=−+∂

∂−

∂∂

∂∂

+−∂∂

+∂∂

Eb

vyu

xu

xv

yu

xu

Ebv

xu

yu

yv

yu

xu

yxyyy

xyxxx

– Compatibility equations are automatically satisfied.

Force method

– Body Force: Potential Field

, y

bx

b yx ∂ψ∂

−=∂ψ∂

−=

– Airy’s Stress function

, , 2

2

2

2

2

yxxy xyyyxx ∂∂φ∂

−=σ∂

φ∂=ψ−σ

∂φ∂

=ψ−σ

– Equilibrium Equation: satisfied identically




57

000)()(

000)()(

2

22

2

2

2

=→=∂ψ∂

−ψ+∂

φ∂∂∂

+∂∂φ∂

∂∂

−

=→=∂ψ∂

−∂∂φ∂

∂∂

−ψ+∂

φ∂∂∂

yxyyxx

xyxyyx

– Strains in terms of stress function

yxEG

yExE

xEyE

xyxy

yy

xx

∂∂φ∂ν+

−=σ

=γ

ψ+∂

φ∂ν−ψ+

∂φ∂

=ε

ψ+∂

φ∂ν−ψ+

∂φ∂

=ε

2

2

2

2

2

2

2

2

2

)1(2

)()(1

)()(1

– Compatibility equation

222

2

211

2

21

122

12xxxx ∂ε∂

+∂

ε∂=

∂∂γ∂

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

2

22

4

2

2

4

4

2

2

22

4

2

2

4

4

22

4

=∂

ψ∂+

∂∂φ∂

ν−∂

ψ∂+

∂φ∂

+∂

ψ∂+

∂∂φ∂

ν−∂

ψ∂+

∂φ∂

+∂∂φ∂

ν+xyxxxyxyyyyx

ψ∇ν−−=φ∇

=∂

ψ∂+

∂ψ∂

ν−+∂

φ∂+

∂∂φ∂

+∂

φ∂

24

2

2

2

2

4

4

22

4

4

4

)1(

0))(1(2yxyyxx




58

6.2. 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 == , 0=zu

)(21 , ,

xu

yu

yu

xu yx

xyy

yyx

xx ∂

∂+

∂∂

=ε∂

∂=ε

∂∂

=ε

0)(21 , 0)(

21 , 0 =

∂∂

+∂

∂=ε=

∂∂

+∂∂

=ε=∂∂

=εyu

zu

xu

zu

xu zy

yzzx

xzz

zz

Constitutive law

0)( =σ

+σ

ν−σ

=εEEE

yyxxzzzz → )( yyxxzz σ+σν=σ

xyxy

xxyyyy

yyxxxx

E

EEv

EEv

σν+

=γ

σν+ν−

σ−=ε

σν+ν−

σ−=ε

)1(2

)()1(

)()1(

22

22

x y

z

∞




59

6.3 Polynomial Stress Functions Compatability Condition w/o body forces

02 4

4

22

4

4

44 =

∂φ∂

+∂∂φ∂

+∂

φ∂=φ∇

yyxx

Stress Components

, , 2

2

2

2

2

yxxy xyyyxx ∂∂φ∂

−=σ∂

φ∂=σ

∂φ∂

=σ

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

ycxybxa

xyyyxx −=∂∂φ∂

−=σ=∂

φ∂=σ=

∂φ∂

=σ

++=φ

– Constant stress status




60

Third Order

y cxbyx

y bxax

ydxcy

ydxycyxbxa

xy

yy

xx

33

2

332

2

332

2

332323333

232223

−−=∂∂φ∂

−=σ

+=∂

φ∂=σ

+=∂

φ∂=σ

⋅+++

⋅=φ

– 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




61

Fourth order

ydxycxbyx

ycxybxax

yacxydxcy

ace

yexydyxcyxbxa

xy

yy

xx

244

242

244

242

2

2444

242

2444

443422434444

22

2

)2(

)2(342322334

−−−=∂∂φ∂

−=σ

++=∂

φ∂=σ

+−+=∂

φ∂=σ

+−=⋅

+⋅

++⋅

+⋅

=φ

– d4 0≠ :

ydxyd xyyyxx24

4 2 , 0 , −=σ=σ=σ

Moment by S. stress = lcdclcdclcd 34

24

24

322

2312

2=−

Moment by N. stress = lcdclcd 34

24

322

32

2=




62

Fifth order

yacxydyxcxb

yx

yd

xycyxbxax

ydbxyacyxdxc

y

dbface

yf

xye

yxd

yxc

yxb

xa

xy

yy

xx

355

25

25

352

3525

25

352

2

355

255

25

35

2

2

555555

554532523545555

)32(31

3

3

)2( 31)32(

3

)2(31 , )32(

453423233445

++−−−=∂∂

∂−=

+++=∂∂

=

+−+−+=∂∂

=

+−=+−=

⋅+

⋅+

⋅+

⋅+

⋅+

⋅=

φσ

φσ

φσ

φ

– d5 0≠ :

xydyd yy xd xyyyxx2

53532

5 , 3

,)32( −=σ=σ−=σ




63

Final Solution ∑φ=φ

ii

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

conditions are satisifed.

jiji nT σ=

– 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




64

6.4. Cantilever Beam

Traction Boundary Conditions

– on cy ±= 0=σ±=σ= xyixix nT , 0=σ±=σ= yyiyiy nT ,

– on 0=x 0=σ−=σ= xxixix nT , yxiyiy nT σ−=σ=

Assumption

– Stress distributions can be determined by the beam theory.

xydxx 4=σ , 0=σ yy , 2

24

2ydbxy −−=σ

Boundary condition on cy ±=

22

4

24

2 202

)(cbdcdbcyxy −=→=−−=σ ±=

Statical equivalence

ctPbPtdyy

cbbtdytdyT

c

c

c

cxy

c

cxy 4

3)()( 22

22

20 =→=−=σ−= ∫∫∫−−−

=

Stress

)(2

023

22

3

ycI

P

yI

MyI

Pxxytc

P

xy

yy

xx

−−=σ

=σ

−=−=−=σ

L

2c

y

x




65

Displacement

– Strain

)(2

22 ycIGP

Gxu

yu

xyEIP

EEyu

xyEIP

EExu

xyyxxy

yyxxyyy

yyxxxxx

−−=σ

=∂∂

+∂∂

=γ

ν=

σ+

σν−=

∂∂

=ε

−=σ

ν−σ

=∂∂

=ε

– Integration

)(2

, )(2 1

22 xfxyEIPuyfyx

EIPu yx +

ν=+−=

– By shear strain

)(2

))(2

())(2

( 221

22 ycIGPxfxy

EIP

xyfyx

EIP

y−−=+

ν∂∂

++−∂∂

IGPcy

IGP

dydfy

EIP

dxdfx

EIP

ycIGP

dxdfy

EIP

dydfx

EIP

2)

22()

2(

)(2

)2

()2

(

22212

22122

−=−+ν

++−

−−=+ν

++−

IGPcyFxF yx 2

)()(2

−=+

– Fx and Fy should be constant.

eyIGP

dydfy

EIPFd

dxdfx

EIPF yx =−+

ν==+−= 2212

22 ,

2

– Integration of Fx and Fy

geyyIGPy

EIPfhdxx

EIPf +++

ν−=++= 333

1 66 ,

6

– Displacement components

hdxxEIPxy

EIPu

geyyIGPy

EIPyx

EIPu

y

x

+++ν

=

+++ν

−−=

32

332

62

662




66

Displacement Boundary Conditions – on lx =

0=xu , 0=yu

0)66

()2

(

66232

332

=++ν

−++−=

+++ν

−−=

gyIGP

EIPyel

EIP

geyyIGPy

EIPyl

EIPux

→ IGP

EIP

EIPleg

66,

2 , 0

2

=ν

==

– however, the third condition is impossible.

2)1(266

−=ν→ν+

ν=→ν=→=

ν EEGEIGP

EIP

– y-Displacement on x = l

062

32 =+++ν

= hdllEIPly

EIPu y → 0

6 and 0

23 =++=

ν hdllEIPl

EIP

0or 0 == Pv We end up wirh a useless solution and the displacement BC cannot be satisfied !!!

Why ???

Question on DBC

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

Pcde2

2

−=+ ) for them,

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

integration constant. However, the additional three displacement boundary conditions are

required to prevent a rigid body motions of the beam, and are not specific boundary

conditions for this problem. If so, what happened to our displacement boundary

conditions ?? Anyway, let’s try the aforementioned displacement boundary conditions.

DBC on the neutral axis at the support.

– at lx = and 0=y 0== yx uu

dllEIPhhdll

EIPlu

glu

y

x

−−=→=++=

==

33

60

6)0,(

0)0,(

– at lx = and 0=y 0=∂

∂x

uy




67

GIPcl

EIPe

lEIPhl

EIPddl

EIP

xuy

22

320

22

2

222

−=

=→−=→=+=∂∂

EIPlx

EIPlx

EIPxy

EIPu

yGI

PcEI

PlyIGPy

EIPyx

EIPu

y

x

3262

)22

(662

3232

22332

+−+ν

=

−++ν

−−=

– Verification of solution

%45.023

3/

2 2

232 ≈= νν

lc

EIPllc

EIP for 1.0=

lc

and ν=0.3

)3(66

)(2

)22

(662

23322

22332

ycyGIPy

EIPylx

EIP

yGI

PcEI

PlyIGPy

EIPyx

EIPux

−+ν

−+−=

−++ν

−−=

Beam solution ylxEIPuy

EIPx

xxx )(2

22 +−−=→−=ε

%1.0)(31

2/

6

%7.1))(1(34

2/

32

23

223

≈=

≈+=

lcv

EIcPl

EIPc

lcv

EIcPl

GIPc

ν for 1.0=

lc

and ν=0.3

– at lx = and 0=y 0=∂∂

yux

GIlPcl

EIPh

GIPcl

EIPd

EIPlee

EIPl

yux

23

2220

22

3

22

22

+=

−−=→=→=+−=∂∂

)(22326

66)(

22

232

3

3322

xlGI

PcxyEIP

EIPlx

EIPlx

EIPu

yIGPy

EIPylx

EIPu

y

x

−+ν

++−=

+ν

−−−=




68

6.5. Simple Beam Traction boundary conditions

i. on y = -c 001 =σ→=σ−=σ= xyxyjxjnT ,

qqnT yyyyjyj −=σ→=σ−=σ=2

ii. on y = c 01 =σ=σ= xyjxjnT , 02 =σ=σ= yyjyjnT

iii. on x = -l 01 =σ−=σ= xxjxjnT , 02 =σ−=σ= yxjyjnT

iv. on x= l 01 =σ=σ= xxjxj nT , 02 =σ=σ= yxjyj nT

Symmetry conditions

i. 22σ should not vary along x on constant y. ii. 12σ should be anti-symmetry with respect to x → should be odd function wrt x.

Selection of functions

i. n = 2

212222211 , , bac −=σ=σ=σ

From the SM condition ii, 02 =b

ii. n = 3

ycxbybxaydxc 331233223311 , , −−=σ+=σ+=σ

From the SM condition i, 03 =a From the SM condition ii, 03 =c

2l

x 2c

y




69

iii. n = 4

244

24

12

244

2422

2444

2411

22

)+2(

ydxycxbycxybxa

yacxydxc

−−−=σ

++=σ

−+=σ

From the SM condition 1, 0 , 0 44 == ba From the SM condition 2, 04 =d

iv. n = 5

355

255

35

12

352

52

53

522

355

255

25

35

11

)3+2(31

3

3

)2+(31)3+2(

3

yacxydxycxb

ydxycyxbxa

ydbxyacyxdxc

+−−−=σ

+++=σ

−−+=σ

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

Final Expressions of Stress components

254312

352

43222

325

2243211

23

)32()2(

xydxycxb

ydycyba

yyxdyxcydc

−−−=σ

+++=σ

−+−++=σ

Apply the given boundary conditions

– The boundary condition 1 0202 2

5432

543 =+−→=−+− cdccbxcdcxcxb

qcdcccba −=−+−3

352

432

– The boundary condition 2 0202 2

5432

543 =++→=−−− cdccbxcdcxcxb

03

352

432 =+++cdcccba

– Solve the simultaneous equations

I

qcqdc

Iqc

cqbqa

243 , 0 ,

243 ,

2 354

2

32 −=−====−=




70

Apply the static equilibrium conditions

– The static equilibrium condition 1: 011 =σ∫−

±=

c

clxdy

002))32(

2( 22

3232 =→==−−+∫

−

cccdyyylI

qydcc

c

– The static equilibrium condition 2: 011 =σ∫−

±=

c

clxydy

)

52(

2

0)152

3(

32))

32(

2(

223

5323

342223

clI

qd

cclIqcddyyyl

Iqyd

c

c

−=

=−−=−−∫−

The final expression for stress

xycI

q

yyccI

q

ycyI

qyxlI

q

)(2

)33

2(2

)52

32(

2)(

2

2212

323

22

232211

−−=σ

+−−=σ

−+−=σ

The static equilibrium conditions:

qldyc

clx

=σ∫−

±=12

qlldyyc

Iqc

c

=−∫−

)(2

22




71

6.6. Series Solution

02 4

4

22

4

4

44 =

∂φ∂

+∂∂φ∂

+∂

φ∂=φ∇

yyxx

)()cossin( 21 yfl

xmXl

xmXmπ

+π

=φ , ∑∞

=

φ=φ0m

m

In case either sine function or cosine function is used,

)(sin)(sin yxfyfl

xmm α=

π=φ , ∑

∞

=

φ=φ0m

m

Compatibility equation for one sine function.

0)()(2)( 24 =′′′′+′′α−α yfyfyf

General solution

yyCyyCyCyCyf α+α+α+α= sinhcoshsinhcosh)( 4321

Stress components

)]sinhcosh2( )coshsinh2(sinhcosh[sin

4

322

12

yyyCyyyCyCyCxxx

αα+αα+αα+αα+αα+ααα=σ

]sinhcosh sinhcosh[sin 43212 yyCyyCyCyCxyy α+α+α+ααα−=σ

)]cosh(sinh

)sinhcosh(coshsinh[cos

4

321

yyyCyyyCyCyCxxy

ααα

ααααααααασ

++

+++−=

Traction Boundary conditions

– on cy = 0=σxy , xBmyy α−=σ sin

– on cy −= 0=σxy , xAmyy α−=σ sin

Integration constants

cccBA

C

cccBA

C

cccccBA

C

cccccBA

C

mm

mm

mm

mm

α+ααα

α+

−=

α−ααα

α−

=

α−ααα+α

α−

−=

α+ααα+α

α+

=

22sinhcosh

22sinhsinh

22sinhsinhcosh

22sinhcoshsinh

24

23

22

21




72

Fourier coefficients

∫−

π=

l

lum dx

lxmxqA sin)( , ∫

−

π=

l

lbm dx

lxmxqB sin)(

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




73

Chapter 7

Plane Problems in

Polar Coordinate System




74

7.1. Polar Coordinate system

Basics

222

1

, sin

tan , cos

yxrryxyrx

+=θ=

=θθ= −

z = z

rry

x

rx

xr

θ−=−=

∂θ∂

θ==∂∂

sin

cos

2

,

rrx

y

ry

yr

θ==

∂θ∂

θ==∂∂

cos

sin

2

θ∂∂θ

+∂∂

θ=∂θ∂

θ∂∂

+∂∂

∂∂

=∂∂

θ∂∂θ

−∂∂

θ=∂θ∂

θ∂∂

+∂∂

∂∂

=∂∂

()cos()sin()()()

()sin()cos()()()

rryyr

ry

rrxxr

rx

Transformation matrix

θθ−θθ

=β1000cossin0sincos

Stress

βσβ=σ pTc =

θθ−θθ

σσσσσσσσσ

θθθ−θ

θ

θθθθ

θ

1000cossin0sincos

1000cossin0sincos

zzzzr

zr

rzrrr




75

θσ+θσ=σθσ−θσ=σ

θ−θσ+θθσ−σ=σ

σ=σ

θσ+θσ+θσ=σ

θσ−θσ+θσ=σ

θ

θ

θθθ

θθθ

θθθ

cossinsincos

)sin(coscossin)(

2sincossin

2sinsincos

22

22

22

zzrzy

zzrzx

rrrxy

zz

rrryy

rrrxx

Physical Derivation of the Equilibrium Equation.

02

cos)(

2sin)())((

=θ+θ

σ−θθ∂

σ∂+σ

+θ

σ+θθ∂

σ∂+σ−θσ−θ+

∂σ∂

+σ=

θθ

θ

θθθθ

θθ∑

rdrdbddrd

ddrdrdddrrdrr

F

rrr

r

rrrr

rrr

02

sin)(

))((2

cos)((

=++∂

∂+

+−+∂

∂++−

∂∂

+=∑

θθσθθ

σσ

θσθσ

σθσθθ

σσ

θθθ

θ

θθ

θθθθθ

θθθ

rdrdbddrd

rdddrrdrr

ddrdF

rr

r

rr

r

By neglecting higher order terms,

021

0)(11

=+σ

+θ∂

σ∂+

∂σ∂

=+σ−σ+θ∂

σ∂+

∂σ∂

θθθθθ

θθθ

brrr

brrr

rr

rrrrrr




76

Second Derivatives

??

)()sin()(cos()

=θ∂

∂θ−

∂∂

θ∂∂

=∂∂

∂∂

rrxxx

??

)()cos()(sin()

=θ∂

∂θ+

∂∂

θ∂∂

=∂∂

∂∂

rryyy

??

)()cos()(sin()

=θ∂

∂θ+

∂∂

θ∂∂

=∂∂

∂∂

rrxyx

Stress Function

??

2sinsincos

2sinsincos2

22

22

2

2

22

=

θ∂∂φ∂

−θ∂

φ∂+θ

∂φ∂

=

θσ+θσ+θσ=σ

yxxy

xyyyxxrr

??

2sincossin

2sincossin2

22

22

2

2

22

=

θ∂∂φ∂

+θ∂

φ∂+θ

∂φ∂

=

θσ−θσ+θσ=σθθ

yxxy

xyyyxx

??

)sin(coscossin)(

)sin(coscossin)(

222

2

2

2

2

22

=

θ−θ∂∂φ∂

−θθ∂

φ∂−

∂φ∂

=

θ−θσ+θθσ−σ=σ θ

yxyx

xyxxyyr

)1(11

11

2

2

2

2

2

2

2

θ∂φ∂

∂∂

−=θ∂∂φ∂

−θ∂φ∂

=σ

∂φ∂

=σ

θ∂φ∂

+∂φ∂

=σ

θ

θθ

rrrrr

r

rrr

r

rr


??

)2sinsincos( 22

=∂

θσ−θσ+θσ∂=

∂σ∂ θθθ

xxrrrxx

??

))sin(coscossin)(( 22

=∂

θ−θσ+θθσ−σ∂=

∂

σ∂θθθ

yyrrrxy

??=+∂

σ∂+

∂σ∂

xxyxx b

yx




77

Displacement

zz

yx

yxr

uuuuu

uuu

=

θ+θ−=

θ+θ=

θ cossin

sincos

Strain

βεβ=ε pTc =

θθ−θθ

εεεεεεεεε

θθθ−θ

θ

θθθθ

θ

1000cossin0sincos

1000cossin0sincos

zzzzr

zr

rzrrr

θε+θε=εθε−θε=ε

θ−θε+θθε−ε=ε

ε=ε

θε+θε+θε=ε

θε−θε+θε=ε

θ

θ

θθθ

θθθ

θθθ

cossinsincos

)sin(coscossin)(

2sincossin

2sinsincos

22

22

22

zzrzy

zzrzx

rrrxy

zz

rrryy

rrrxx

- By chain rule

)1(2sin21)1(sincos

)sincos)(sin(cos

22

ruu

rruu

rru

ru

uurrx

u

rrr

rx

xx

θθθ

θ

−θ∂

∂+

∂∂

θ−θ∂

∂+θ+

∂∂

θ=

θ−θθ∂

∂θ−

∂∂

θ=∂∂

=ε

)1(21 , )1( ,

ruu

rruu

rru

ru r

rrr

rrθθ

θθ

θθ −θ∂

∂+

∂∂

=εθ∂

∂+=ε

∂∂

=ε

Geometric derivation of strain components

θ== uvuu r ,

ABABHA

ABABBA

rr−

≈−′

=ε''

drrudrurdr

ruudrrHA

drAB

rrr ∂

∂+=+−

∂∂

+++=′

=

)(

ru

drdrdrrudr rr

rr ∂∂

=−∂∂+

=ε)/(




78

0 , 0 ≠=θ ruu (Red line)

ru

rdrddur

ACACDE rr =

θθ−θ+

=−

=εθθ

)(

0 , 0 =≠θ ruu (Green line)

θ∂∂

=

θ

θ−−θθ∂

∂++θ

=−′′′′

=ε

θ

θθ

θ

θθ

ur

rd

rduduurd

ACACFA

1

)(

0 , 0 ≠≠θ ruu

θ∂∂

+=θ

θ−θθ∂∂+θ+=ε θθ

θθ

urr

urd

rddudur rr 1)/()(

0 , 0 ≠=θ ruu

θ∂∂

=θ

+−θθ∂∂++=

′′

=θ rrrr urrd

urduurFA

CF 1)()/(2

0 , 0 =≠θ ruu

ru

drurdruu rr

∂∂

=−∂∂+

=θ+θ=θ θθ /314 ,

ruθ=θ3

0 , 0 ≠≠θ ruu

ru

ruu

rr

rθθ

θ −∂∂

+θ∂

∂=θ−θ+θ=θ+θ=γ

1)( 34212

)1(21)(

21

342 ru

ruu

rr

rθθ

θ −∂

∂+

θ∂∂

=θ−θ+θ=ε

θu

θθ∂

∂++θ θ

θ duurd

1θ drruu∂∂

+ θθ

θu

3θ

θθ∂

∂++ duur r

r

ru

2θ




79

7.2. Governing Equation and Boundary Conditions Equilibrium

021

0)(11

=+σ

+θ∂

σ∂+

∂σ∂

=+σ−σ+θ∂

σ∂+

∂σ∂

θθθθθ

θθθ

brrr

brrr

rr

rrrrrr

Stress Function

)1(11

11

2

2

2

2

2

2

2

θφ

θφ

θφσ

φσ

θφφσ

θ

θθ

∂∂

∂∂

−=∂∂

∂−

∂∂

=

∂∂

=

∂∂

+∂∂

=

rrrrr

r

rrr

r

rr

Compatibility

2

2

22

2

2

2

2

2 ()1()1()()()θ∂

∂+

∂∂

+∂∂

=∂∂

+∂∂

rrrryx

0)11)(11())(( 2

2

22

2

2

2

22

2

2

2

2

2

2

2

2

24 =φ

θ∂∂

+∂∂

+∂∂

θ∂∂

+∂∂

+∂∂

=φ∂∂

+∂∂

∂∂

+∂∂

=φ∇rrrrrrrryxyx

Strain Components

rur

rr ∂∂

=ε , θ∂

∂+=ε θ

θθu

rrur 1 , )1(

21

ru

ruu

rr

rθθ

θ −∂∂

+θ∂

∂=ε

Constitutive Law

pklijkl

pij C ε=σ

Displacement Boundary Conditions

pi

cjij

cjij

pi

ci

ci uuuuuu =β=β=→= → pp uu =

Cauchy’s Relation

pm

pjm

ckmkmn

cjnij

ckmnmk

cjnij

ck

cjkij

cjij

pi

cj

cij

ci nnnnTTnT σ=ββσβ=ββσβ=σβ=β=→σ=

ppp nT σ=

Traction Boundary Condition

pi

cjij

cjij

pi

ci

ci TTTTTT =β=β=→= → pp TT =




80

7.3. Solutions in the Polar Coordinate System

Compatibility Equation

0)11)(11( 2

2

22

2

2

2

22

2

=φθ∂∂

+∂∂

+∂∂

θ∂∂

+∂∂

+∂∂

rrrrrrrr

In case φ is a function of r only

01120)1)(1( 32

2

23

3

4

4

2

2

2

2

=φ

+φ

−φ

+φ

→=φ∂∂

+∂∂

∂∂

+∂∂

drd

rdrd

rdrd

rdrd

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

dtd

dtd

dtd

dtd

rdrd

dtd

dtd

dtd

rdrdt

dtd

dtd

rdtd

drd

dtd

dtd

rdrdt

dtd

rdtd

drd

dtd

rdrdt

dtd

drd

φ−

φ+

φ−

φ=

φ

φ+

φ−

φ=

φ−

φ=

φ

φ−

φ=

φ=

φ

φ=

φ=

φ

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

=φ

+φ

−φ

=φ

+φ

−φ

−φ

+φ

−φ

+φ

−φ

+φ

−φ

=φ

+φ

−φ

+φ

dtd

dtd

dtd

r

dtd

rrdtd

dtd

rrdtd

dtd

dtd

rrdtd

dtd

dtd

dtd

r

drd

rdrd

rdrd

rdrd

DCrrBrrADCeBteAt tt +++=+++=φ 2222 lnln

Stress Components

0)1(

2)ln23(

2)ln21(1

22

2

2

=θ∂φ∂

∂∂

−=σ

+++−=φ

=σ

+++=φ

=σ

θ

θθ

rr

CrBrA

drd

CrBrA

drd

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.




81

Displacement (plane stress)

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

])1(2)31(ln)1(2)1([1)(12

θ+ν−+ν+−ν−+ν+

−=

ν−+ν−+ν−+ν+

=∂∂

=νσ−σ=ε θθ

fCrBrrBrr

AE

u

CBrBr

AEr

uE

r

rrrrr

∫ +θθ−θ

=→θ−=θ∂

∂

ν−+ν−+ν−+ν+

−=θ∂

∂+=νσ−σ=ε

θθ

θθθθθ

)()(4)(4

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

1

2

rfdfEBruf

EBru

CBrBr

AEr

ur

uE

rrr

0)(1)(1)()(10)1(21

11 =−θθ+∂

∂+

θ∂θ∂

→=−∂

∂+

θ∂∂

=ε ∫θθθ rf

rdf

rrrff

rru

ruu

rr

r

pFrrfprfrrfr

KHfffpdff

−=→−=−∂

∂

θ+θ=θ→=θ+θ∂

θ∂→=θθ+

θ∂θ∂

∫

)()()(

cossin)(0)()()()(

111

2

2

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

FrKHEBru

KHCrBrrBrr

AE

ur

+θ−θ+θ

=

θ+θ+ν−+ν+−ν−+ν+

−=

θ sincos4

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

– Rigid body translation

θ+θ−=θ+θ= θ cossin , sincos yxyxr uuuuuu → xuK = , yuH =

xu

yu




82

– Rigid body rotation

θ== θ ruur , 0 → θ=F

θr




83

Displacement (plane strain)

0)1(

2)ln23(

2)ln21(11

22

2

22

2

=θ∂φ∂

∂∂

−=σ

+++−=∂

φ∂=σ

+++=θ∂φ∂

+∂φ∂

=σ

θ

θθ

rr

CrBrA

r

CrBrA

rrr

r

rr

)(])21(2)1(ln)21(2)1([1

])21(2)431(ln)21(2)1[(1

)]2)ln23()(()2)ln21()(1[(1

))()1((1))((1)(1

22

2222

22

22

222

θ+ν−ν−+ν+−ν−ν−+ν+

−=

ν−ν−+ν−ν−+ν−ν−+ν+=

+++−ν+ν−+++ν−=

σν+ν−σν−=σ+σν−νσ−σ=νσ−νσ−σ=∂∂

=ε θθθθθθθθ

fCrBrrBrr

AE

u

CBrBrA

E

CrBrACrB

rA

E

EEEru

r

rrrrrrzzrrr

rr

∫ +θθ−θ−

=→θ−−

=θ∂

∂

ν−ν−+ν−ν−+ν−ν−+ν+−=

+++ν+ν−+++−ν−=

σν+ν−σν−=σ+σν−νσ−σ=θ∂

∂+=ε

θθ

θθθθθθθ

θθ

)()()1(4)()1(4

)]21(2)43(ln)21(2)1([1

)]2)ln21()(()2)ln23()(1[(1

))()1((1))((1

1

22

2222

22

22

222

rfdfE

BrvufE

Brvu

CBrBrA

E

CrBrACrB

rA

E

EEru

ru

rrrrrrr

0)(1)(1)()(10)1(21

11 =−θθ+∂

∂+

θ∂θ∂

→=−∂∂

+θ∂

∂=ε ∫θθ

θ rfr

dfrr

rffrr

uruu

rr

r

pFrrfprfrrfr

KHfffpdff

−=→−=−∂

∂

θ+θ=θ→=θ+θ∂

θ∂→=θθ+

θ∂θ∂

∫

)()()(

cossin)(0)()()()(

111

2

2

Substitution of the above solutions into the original equation leads to p = 0.

FrKHEBru

KHCrBrrBrr

AE

ur

+θ−θ+θ

=


−=

θ sincos4

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




84

7.4. Thick Pipe Problem

Traction BC

– On ar = : irrjiji pnT −=σ→σ= – On br = : orrjiji pnT −=σ→σ=

orr

irr

pCbAb

pCaAa

−=+=σ

−=+=σ

2)(

2)(

2

2→

22

22

22

22

2

)(

abbpapC

abppbaA

oi

io

−−

=

−−

=

Stress components

22

22

222

22

22

22

222

22

1)(

1)(

abbpap

rabppba

abbpap

rabppba

oiio

oiiorr

−−

+−

−−=σ

−−

+−

−=σ

θθ

Displacement

0 , ])1()1()([1

22

22

22

22

=ν−−−

+ν+

−−

−= θurab

bpaprab

ppbaE

u oiior

a

b

po pi




85

End Conditions

– Plane strain

22

22

2)(

0

abbpap oi

rrzz

zz

−−

ν=σ+σν=σ

=ε

θθ

– Plane stress (open end)

22

22

2)(

0

abbpap

EEoirr

zz

zz

−−ν

−=σ+σν

−=ε

=σ

θθ

– Closed End with a Rigid End Plate

)()2)((

))()(()(

2222

2222

2222

oioi

zz

rrzzzzz

pbpaab

bpapEab

EababF

−π=−−

ν+ε−π=

σ+σν+ε−π=σ−π= θθ

22

22

22

22

)()21()21(

abpbpa

abEpbpav

oizz

oizz

−−

=σ

−ν−−−

=ε

0=op

– Stress : )1( , )1( 2

2

22

2

2

2

22

2

rb

abpa

rb

abpa ii

rr +−

=σ−−

=σ θθ

ii p

ababp

>−+

=σθθ 22

22

max)()(

– Displacement : ])1()1[(22

2

br

rb

aba

Ebpu i

r ν−+ν+−

=




86

7.5. Special Cases

Thin pipe with an internal pressure: btab <<=−

tpr

rb

abpa

rb

abpa

trbtra

ii

irr

02

2

22

2

2

2

22

2

00

)1(

0 )1(

2 ,

2

=+−

=σ

≈−−

=σ

+=−=

θθ

Etrp

br

rb

aba

Ebpu ii

r

20

22

2

])1()1[( =ν−+ν+−

=

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

rbbapa

rb

abpa

rbbapa

rb

abpa

ii

iirr

+−

=+−

=σ

−−

=−−

=σ

θθ

])1(1)1[()/(1

])1()1[( 22

2

22

2

br

rbaa

Ep

br

rb

aba

Ebp

u iir ν−+ν+

−=ν−+ν+

−=

as ∞→b ,

2

2

2

2

rpar

pa

i

irr

=σ

−=σ

θθ

, ra

Epau i

r )1( ν+=

Infinite Region pipe with a Lining

– Region 1

22

22

222

22

22

22

222

22

1)(

1)(

abbpap

rabppba

abbpap

rabppba

oiio

oiiorr

−−

+−

−−=σ

−−

+−

−=σ

θθ

0 , ])1()1()(

[1122

221

22

22

1

1 =ν−−−

+ν+

−−

−= θurab

bpaprab

ppbaE

u oiior

– Region 2

20

2

2

2

rpbrpb o

rr

=σ

−=σ

θθ

, rb

Epbu o

r2

22 )1( ν+=




87

– Compatibility )()( 12 bubu rr =

])1()1()([1)1( 122

221

22

22

122 b

abbpap

babppba

EEpb oiioo ν−

−−

+ν+

−−

−=ν+

)]1()1([1)]1()1([1)1( 122

2

122

2

1122

3

122

2

122 ν−

−+ν+

−=ν−

−+ν+

−+ν+

abbap

abbpa

Eabbp

abbpa

EEp

b iiooo

22

2

1122

3

122

2

122

21)]1()1([1)1(ab

bpaEab

bpab

bpaEE

pb iooo

−=ν−

−+ν+

−+ν+

))(())(1(2

221

222

2221

2

2 babaEabEpaEp i

o −ν+++−ν+=

– A thin lining

0221

02

10221

02 )1()()1(

)(rEtE

prE

trEtEptr

Ep iio +ν+

≈ν−+ν+

−=

a

b

pi

po

po




88

7.6. Curved Beam

Boundary Condition

– On ar = and b : 00 =σ=σ→=σ= θrrrjiji nT – On 0=θ : 0=σ= θθnT rr and θθθθθθ σ=σ= nT

Applied Moment

∫∫ θθθ σ−==b

a

b

a

rdrrdrTM , 0=σ= ∫∫ θθθ

b

a

b

a

drdrT

Stress Component

CrBrA

drd

CrBrA

drd

rrr

2)ln23(

2)ln21(1

22

2

2

+++−=φ

=σ

+++=φ

=σ

θθ

Application BC on ar = and b

02)ln21(

02)ln21(

2

2

=+++

=+++

CbBbA

CaBaA

Moment Condition

)()(2

2

2

abrrdrd

drdrdr

drdrdr

drdrdrM

b

a

b

arrb

a

b

a

b

a

b

a

b

a

b

a

φ−φ=φ+σ−=φ+φ

−=

φ+

φ−=

φ−=σ−= ∫∫ ∫θθ

DCrrBrrA +++=φ 22 lnln

)()lnln(ln 2222 abCaabbBabAM −+−+=

a

b

M M




89

Solution

))lnln(2(

)(2

ln4

2222

22

22

aabbabNMC

abNMB

abba

NMA

−+−=

−−=

−=

where 222222 )(ln4)(abbaabN −−= .

Stress

)lnlnln(4

)lnlnln(4

22222

22

222

22

abraa

brb

ab

rba

NM

raa

brb

ab

rba

NM

rr

−+++−−=σ

++−=σ

θθ

Displacement

FrKHEBru

KHCrBrrBrr

AE

ur

+θ−θ+θ

=


−=

θ sincos4

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

Displacement B.C.

0=∂∂

== θθ r

uuur at 0=θ and 2/)( bar += 0==→ HF

A ring with cut out.

– Cut out angle : α

– Compatibility condition

πα

=→α=π=πθ 88)2( EBr

EBru

– Moment required to close the ring

πα

=−−=8

)(2 22 EabNMB

)(16 22 abENM

−πα

−=→




90

7.7. Rotating Disk Centrifugal force : rbr

2ρω=


22

021

0)(11

rr

r

brrr

brrr rr

rr

rrrrrr

ρω−=σ−∂σ∂

→

=+σ

+θ∂

σ∂+

∂σ∂

=+σ−σ+θ∂

σ∂+

∂σ∂

θθ

θθθθθ

θθθ

Stress and strain

rur

rr ∂∂

=ε , ruu

rru rr =

θ∂∂

+=ε θθθ

1

)(1

)(1

)(1

)(1

22

22

ru

ruEE

ru

ruEE

rrrr

rrrrrr

+∂∂

νν−

=ε+νεν−

=σ

ν+∂∂

ν−=νε+ε

ν−=σ

θθθθ

θθ

Equilibrium Equation in terms of displacement

322

2

22 1 r

Evu

drdur

drudr r

rr ρω−

−=−+ (Euler-Cauchy Equation)

General Solution

]8

11)1()1[(1 322

1 rr

CvCrvE

ur ρων−

−+−−=

ω

b




91

2221

2221

222

2122

2

222

2122

2

212

2

8)31(1

8)3(1

]8

1)3(1)1()1[(1

1

)]8

11)1()1((8

131)1()1[(1

1

)(1

rvr

CC

rvr

CC

rvr

CC

rr

CvCvvrr

CvCv

ru

ruE rr

rr

ρω+

−−=σ

ρω+

−+=

ρων−

+−ν−+ν−ν−

=

ρων−

−+−−+ρων−

−++−ν−

=

ν+∂

∂ν−

=σ

θθ

Solid disk

22221 8

)3(08

)3()( , 0 bvCbvCC brrr ρω+

=→=ρω+

−=σ= =

2222222

8)31(

8)3( , )(

8)3( rvbvrbv

rr ρω+

−ρω+

=σ−ρω+

=σ θθ

22maxmax 8

)3()()( bvrr ρω

+=σ=σ θθ at r = 0

Disk with a circular hole at the center

ρω+

−=

+ρω+

=→

=ρω+

−+=σ

=ρω+

−+=σ

=

=

2221

222

2221

2221

8)3(

)(8

)3(

08

)3()(

08

)3()(

bavC

bavC

bvbCC

avaCC

brrr

arrr

)3

31(8

3

)(8

3

22

22222

22

22222

rvv

rbabav

rrbabav

rr

++

−++ρω+

=σ

−−+ρω+

=σ

θθ

aravvbv

abrabvrr

=+−

−ρω+

=σ

=−ρω+

=σ

θθ at )31(

43)(

at )(8

3)(

222max

22max

7200 rpm 3.5 inch (8.9 cm) hard disk

222

4222max

/)(9.1/16.188381sec/16.188381

10)5.03.033.015.4(1207850

43.3)(

cmfkgmNmkg ≈==

×+−

−×=σ −θθ

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




92

7.8. Plate with Circular hole

Stress on remote field

Sxx =σ , 0=σ yy , 0=σxy (Cartesian)

θ−=θ−θσ+θθσ−σ=σ

θ−=θσ−θσ+θσ=σ

θ+=θσ+θσ+θσ=σ

θ

θθ

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)

+=σ

−=σ →

−+

−=σ

−−

−=σ

θθ

>>

θθ )1(2

)1(2

)(2

)(2

2

2

2

2

2

2

22

2

22

2

2

2

22

2

22

2

raSraS

ra

abb

abbS

ra

abb

abbS

rr

ab

rr

0=σ θr

Solution for non-uniform stress field

– BC on r = b : θ−=σθ=σ θ 2sin2

, 2cos2

SSrrr

– Stress Function : θ=φ 2cos)(rf

)1(11

, 11

2

2

2

2

2

2

2

θ∂φ∂

∂φ∂

−=θ∂∂φ∂

−θ∂φ∂

=σ

∂φ∂

=σθ∂φ∂

+∂φ∂

=σ

θ

θθ

rrrrr

rrrr

r

rr

b

a




93

– Compatibility Equation

Dr

CBrArDr

CrBArf

DHBK

rDrBHKr

rKrKrHKrf

fr

CArf

rDrBf

rdrdf

rdrfd

rDrBF

Frdr

dFrdr

Fd

Frdr

dFrdr

Fd

frdr

dfrdr

fdF

frdr

dfrdr

fddd

rdrd

rdrd

p

p

+++=′

−+′

+=

′−=

′=

′+′=+−+→+=

++=

′+′=−+→′+′=

=−+

=θ−+=φ∇

−+=

=θ−+θ

++=φ∇

242

242

224

2224

22

22

22

2

22

22

2

22

24

22

2

22

2

2

2

22

24

14

112

4 ,

12

1)(4412

1

1411

041

02cos)41(

41

02cos)41)(11(

– Stress component

θ−−+=θ∂φ∂

∂φ∂

=σ

θ++=∂

φ∂=σ

θ++−=θ∂φ∂

+∂φ∂

=σ

θ

θθ

2sin)2662()1(

2cos)612(2A

2cos)46(2A 11

242

42

2

2

242

2

2

rD

rCBrA

rr

rCBr

r

rD

rC

rrr

r

rr

– Applying BCs

02662

22662

0462A

2462A

242

242

24

24

=−−+

−=−−+

=++

−=++

aD

aCBaA

SbD

bCBbA

aD

aC

SbD

bC

– As ∞→b

4SA −= , 0=B , SaC

4

4

−= , SaD2

2

=




94

Final Solution

θ+−−=σ

θ+−+=σ

θ−++−=σ

θ

θθ

2sin)231(2

2cos)31(2

)1(2

2cos)431(2

)1(2

2

2

4

4

4

4

2

2

2

2

4

4

2

2

ra

raS

raS

raS

ra

raS

raS

r

rr

Stress concentration on ar =

θ−=σ=σ=σ

θθ

θ

2cos20SS

rrr

– at ππ=θ23 ,

21

: S3=σθθ

– at π=θ , 0 : S−=σθθ (Compression)

– at π=θ21

: )23

211( 4

4

2

2

ra

raS ++=σθθ (rapidly decaying)

Tensile traction in both x- and y- direction

0

)1(

)1(

)2cos()431(2

)1(2

2cos)431(2

)1(2

2

2

2

2

2

2

4

4

2

2

2

2

4

4

2

2

=σ

+=σ

−=

π+θ−++−

+θ−++−=σ

θ

θθ

r

rr

raS

raS

ra

raS

raS

ra

raS

raS

Opposite tractions (pure shear case)

θ+−=

π+θ+++

−θ+−+=σθθ

2cos)31(

)2cos()31(2

)1(2

2cos)31(2

)1(2

2

2

2

2

2

2

2

2

2

2

raS

raS

raS

raS

raS

S4)( max ±=σ→ θθ




95

θ r

7.9. Flamant solution (2-D Boussinesq Solution)

Boundary condition

0=σ=σ θθθ r on 2π

±=θ

Equilibrium condition

Prdrr −=θθσ∫π

π−

2/

2/

cos for all r.r

frr

)(θ=σ→

0)()11()11)(11(

0 , )(11

2

2

22

2

2

2

22

2

2

2

22

2

2

2

2

2

2

=θ

θ∂∂

+∂∂

+∂∂

=θ∂φ∂

+∂φ∂

+∂

φ∂θ∂

∂+

∂∂

+∂∂

=∂

φ∂=σ

θ=

θ∂φ∂

+∂φ∂

=σ θθ

rf

rrrrrrrrrrrr

rrf

rrrrr

θ+θ=θ→=θ′′+θ sincos)(0)()( BAfff

By symmetry of stress, B = 0

PAdAdA −=π

=θθ+=θθ ∫∫π

π−

π

π− 2)2cos1((

21cos

2/

2/

2/

2/

2

Stress components (Polar)

rP

rrθ

π−=σ

cos2, 0=σ=σ θθθ r

Stress components (Cartesian)

θθπ

−=θθσ=σ

θθπ

−=θσ=σ

θπ

−=θπ

−=θσ=σ

3

222

432

cossin2cossin

sincos2sin

cos2cos2cos

aP

aP

aP

rP

rrxy

rryy

rrxx




96

Convolution Integral (Influence line)

∫ ξξξ−=→ξξξ−=b

a

dgyxfRdgyxfdR )(),()(),(

– Moment Case

xfMyxfyxfM

yxfyxfPyxPfyxPfR

∂∂

=ε

ε−−=

=ε

ε−−ε=ε−−=

→ε

→ε→ε

)),(),((lim

)),(),((lim)),(),((lim

0

00

ξ

f(x,y) f(x-ξ,y)

g(x)




97

Chapter 8

Uniform Torsion (St. Venant Torsion)




98

8.1. Basics

Displacement

yryrr

rrux

α−=α−≈ϕα−≈

ϕ−ϕα−ϕα=ϕ−ϕ+α=

sinsin

)cossinsincos(cos)cos)(cos(

xrxrrrru y α=α≈ϕα≈ϕ−ϕα+ϕα=ϕ−ϕ+α= cossin)sinsincoscos(sin)sin)(sin(

zθ=α , θ : rotation angle per unit length Warping function

),( yxuz θψ=

Strain components

0),(

0

0

=∂

θψ∂−=

∂∂

=ε

=∂θ∂

=∂

∂=ε

=∂θ∂

−=∂

∂=ε

zyx

zu

yzx

yu

xzy

xu

zzz

yyy

xxx

)(

)(

0

xyyz

zxyu

zu

yxxz

zyx

uz

uxzx

yzy

xu

yu

zyyz

zxxz

yxxy

+∂ψ∂

θ=∂θψ∂

+∂θ∂

=∂∂

+∂

∂=γ

−∂ψ∂

θ=∂θψ∂

+∂θ∂

−=∂

∂+

∂∂

=γ

=∂θ∂

+∂θ∂

−=∂

∂+

∂∂

=γ

z

x

y

x

y




99

Stress components

)(

)(

0

xy

G

yx

G

yz

xz

xyzzyyxx

+∂ψ∂

θ=σ

−∂ψ∂

θ=σ

=σ=σ=σ=σ

Equilibrium equation (z-direction only)

00)( 2

2

2

2

2

2

2

2

=∂

ψ∂+

∂ψ∂

→=∂

ψ∂+

∂ψ∂

θ=∂σ∂

+∂σ∂

+∂σ∂

yxyxG

zyxzzyzxz

Traction free BC on the longitudinal surface

0=σ= jiji nT for all i , )0,,( yx nn=n

0

0

0

=σ+σ=

=σ+σ=

=σ+σ=

yzyxzxz

yyyxyxy

yxyxxxx

nnTnnTnnT

The first two BCs are automatically satisfied, and

0)()(

0

=+∂ψ∂

−−∂ψ∂

=σ−σ=σ+σ

dsdxx

ydsdyy

x

dsdx

dsdynn zyzxyzyxzx

Stress function

)(

)(

xy

Gx

yx

Gy

yz

xz

+∂ψ∂

θ=∂φ∂

−=σ

−∂ψ∂

θ=∂φ∂

=σ

by eliminating ψ ,

θ−=∂

φ∂+

∂φ∂ G

yx22

2

2

2

The traction BC becomes

0=φ

=∂φ∂

+∂φ∂

dsd

dsdx

xdsdy

yφ→ is constant on the boundary.




100

BCs at the end of the bar

0

0

=φ−=∂φ∂

−=σ

=φ=∂φ∂

=σ

∫∫∫

∫∫∫

Sx

AAzy

Sy

AAzx

dSndAx

dA

dSndAy

dA

∫∫∫∫

∫∫∫∫φ=φ+⋅φ−=⋅φ∇−=

∂φ∂

−∂φ∂

−=σ−σ=

AASA

AAAzx

Azyt

dAdAdSdA

ydAy

xdAx

ydAxdAM

22nXX

Summary

– Governing equation : θ−=∂

φ∂+

∂φ∂ G

yx22

2

2

2

– Boundary condition : 0=φ – Twisting moment : ∫φ=

At dAM 2




101

8.2. Torsion in an Elliptic Section

Elliptic section

012

2

2

2

=−+by

ax

Stress function

)1( 2

2

2

2

−+=φby

axm

θ+

−=→θ−=+

θ−=−+∂∂

+∂∂

→θ−=∂

φ∂+

∂φ∂

Gba

bamGba

m

Gby

ax

yxmG

yx

22

22

22

2

2

2

2

2

2

2

2

2

2

2

2

2)22(

2)1)((2

θ−++

−=φ Gby

ax

baba )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

babaG

abb

ababa

babaG

dAby

ax

babaGM

At

+π

θ=

π−π

+π

+θ−=

−++

θ−= ∫

GM

baba t

π+

=θ→ 33

22

)1( 2

2

2

2

−+π

−=φby

ax

abMt

Stress

xbaM

xy

abM

yt

yzt

xz π=

∂φ∂

−=σπ

−=∂φ∂

=σ 332 , 2

Warping function

)()(2 )(2

)()(2)(2

33

33

xgxyGxybaMx

yGx

baM

yfxyGxyab

Myx

Gyab

M

tt

tt

++ψθ=π

→+∂ψ∂

θ=π

+−ψθ=π

−→−∂ψ∂

θ=π

−

cyfxg

yfxgxyabbaMabxy

baM tt

==

−++π

=+π

)()(

)()()(2)(2 2233

2233




102

cabbaMabxy

baM tt 2)(2)(2 22

3322

33 +ψ+π

=−π

22

33

22

22

abba

Mcxy

abab

t +π

−+−

=ψ

00)0,0( =→=ψ c

22

22

xyabab

+−

=ψ

For a circular section, the warping does not occur




103

8.3. Rectangular Section (Series Solution)

Governing equation

θ−=∂

φ∂+

∂φ∂ G

yx22

2

2

2

BCs: 0=φ on ax ±= and by ±=

Series expansion

nn

Yaxn∑

∞

=

π=φ

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

axn

nGG n

n 2cos)1(422 2/)1(

,5,3,1

π−

πθ−=θ− −

∞

=∑

ODE in y-direction

))1(8)2

((2

cos 2/)1(2

,5,3,1

−∞

=

−π

θ+′′+π

−π∑ n

nnn n

GYYa

naxn

=0

2/)1(2 )1(8)2

( −−π

θ−=π

−′′ nnn n

GYa

nY for 5,3,1=n

2/)1(3

2

)1()(

322

cosh2

sinh −−π

θ+π

+π

= nn n

aGaynB

aynAY

Application BCs on by ±=

0)1()(

322

cosh2

sinh)( 2/)1(3

2

=−π

θ+π

+π

±= −±=

nbyn n

aGabnB

abnAY

0=A , 2/)1(3

2

)1()2/cosh(

1)(

32 −−ππ

θ−= n

abnnaGB

))2/cosh()2/cosh(1()1(

)(32 2/)1(

3

2

abnayn

naGY n

n ππ

−−π

θ= −

2b

2a




104

Final Solution

))2/cosh()2/cosh(1(

2cos)1(132

,5,3,1

2/)1(33

2

abnayn

axn

naG

n

n

ππ

−π

−π

θ=φ ∑

∞

=

−

Stress

))2/cosh()2/cosh(1(

2sin)1(116

,5,3,1

2/)1(22 abn

aynaxn

naG

x n

nyz π

π−

π−

πθ

=∂φ∂

−=σ ∑∞

=

−

)2/cosh()2/sinh(

2cos)1(116

,5,3,1

2/)1(22 abn

aynaxn

naG

y n

nxz π

ππ−

πθ

−=∂φ∂

=σ ∑∞

=

−

Maximum Stress

)87

151

311( ,

)2/cosh(11162

))2/cosh(

11(116

))2/cosh(

11(2

sin)1(116)(

2

222,5,3,1

22

,5,3,122

,5,3,1

2/)1(22)0,(max

π=++++

ππθ

−θ=

π−

πθ

=

π−

π−

πθ

=σ=σ

∑

∑

∑

∞

=

∞

=

∞

=

−==

n

n

n

nyaxyz

abnnaGaG

abnnaG

abnn

naG

Twisting angle

)967

151

311( , ))2/tanh(1921(

32)2(

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

))2/cosh()2/sinh(2(222164

))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

π=++++

ππ

−θ

=

ππθ

−π

θ=

ππ

π−

ππθ

=

ππ

−π

−π

θ=

φ=

∑

∑∑

∑

∫∑ ∫

∫

∞

=

∞

=

∞

=

∞

=

−

∞

= −

−

n

nn

n

b

bn

a

a

n

At

nabn

babaG

nabnaG

nbaG

abnabn

nab

na

naG

dyabnayndx

axn

naG

dAM

Narrow section 3≥ab

19998.00090.03178.1110090.03178.111)2/tanh( ≈=

+−

=π ab

)6274.01(3

2)2())3125

124311(6274.01(

32)2( 33

babaG

babaGM t −

θ≈+++−

θ=




105

Square section

9171.02079.08105.42079.08105.4)2/tanh( =

+−

=π

4

4

4

)2(1415.0

)9171.06274.01(3

)2(

))3125

124319171.0(6274.01(

3)2(

aG

aG

aGM t

θ=

×−θ

≈

+++−θ

=

Open section with constant thickness

∑θ≈

iiit twGM 3

3




106

Chapter 9

Energy Methods




107

9.1. Problem Definition

Governing Equations and Boundary Conditions

Equilibrium Equation : 0=+⋅∇ bσ in V

Constitutive Law : εσ :D= in V

Strain-Displacement Relationship : ))((

21 Tuu ∇+∇=ε

in V

Displacement Boundary condition : 0=− uu on uS

Traction Boundary Condition : 0=− TT on tS

Cauchy’s Relation on the Boundary : nT ⋅= σ on S

Energy Conservation

int21

21

21

21

21

21

21

21

21

21

21

21

Π==∂∂

=

−∂∂

+=

−∂∂

+=

∂

∂−=

+=Π

∫∫

∫∫∫

∫∫∫

∫∫

∫∫

Vijij

Vij

j

i

Sii

Vij

j

i

Sii

Sjiji

Vij

j

i

Sii

V j

iji

Sii

Vii

Siiext

dVdVxu

dSTudVxu

dSTu

dSnudVxu

dSTu

dVx

udSTu

dVbudSTu

σεσ

σ

σσ

σ

V, E , ν

TT =

0== uu Su

St




108

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

Total Potential Energy

Γ−−σε=Π ∫∫∫Γ

dTudVbudVt

iiV

iiijV

ij21




109

9.2 . Principle of Minimum Potential Energy

dVxu

Dxu

dVbudVxu

Dxu

dVbudVudVxu

Dxu

dTudVbudTudVudVxu

Dxu

dTudVbudnudVudVxu

Dxu

dTudVbudVxu

dVxu

Dxu

dTudVbudV

dTudVbudVxu

Dxu

dVxu

Dxu

dTudVbudVxu

Dxu

dTuudVbuu

dVxu

Dxu

dVxu

Dxu

dVxu

Dxu

dVxu

Dxu

dTuudVbuudVx

uuD

xuu

dTudVbudV

uuu

l

ek

Vijkl

j

eiE

Vijij

ei

l

ek

Vijkl

j

eiE

Vi

ei

Vjij

ei

l

ek

Vijkl

j

eiE

iei

Vi

eii

ei

Vjij

ei

l

ek

Vijkl

j

eiE

iei

Vi

eijij

ei

Vjij

ei

l

ek

Vijkl

j

eiE

iei

Vi

ei

Vij

j

ei

l

ek

Vijkl

j

ei

iiV

iiV

ijij

iei

Vi

ei

l

k

Vijkl

j

ei

l

ek

Vijkl

j

ei

iiV

iil

k

Vijkl

j

i

ieii

Vi

eii

l

ek

Vijkl

j

ei

l

ek

Vijkl

j

i

l

k

Vijkl

j

ei

l

k

Vijkl

j

i

ieii

Vi

eii

l

ekk

Vijkl

j

eii

ihi

Vi

hi

hij

V

hij

h

eii

hi

tt

t

t

t

t

t

t

t

t

∂∂

∂∂

+Π=

+σ+∂∂

∂∂

+Π=

+σ+∂∂

∂∂

+Π=

Γ−−Γ+σ−−∂∂

∂∂

+Π=

Γ−−Γσ+σ−−∂∂

∂∂

+Π=

Γ−−σ∂∂

−∂∂

∂∂

+Γ−−σε=

Γ−−∂∂

∂∂

−

∂∂

∂∂

+Γ−−∂∂

∂∂

=

Γ−−−−

∂∂

∂∂

+∂∂

∂∂

−∂∂

∂∂

−∂∂

∂∂

=

Γ−−−−∂

−∂∂

−∂=

Γ−−σε=Π

−=

∫

∫∫

∫∫∫

∫∫∫∫∫

∫∫∫∫∫

∫∫∫

∫∫∫∫

∫∫∫

∫∫∫∫

∫∫

∫∫∫∫

∫∫∫

∫∫∫

ΓΓ

ΓΓ

Γ

Γ

Γ

Γ

Γ

Γ

Γ

21

)(21

)(21

)(21

)(21

)(

21

21

)(

21

21

)()(

21

21

21

21

)()()()(

21

21

,

,

,

,

Since Dijkl is positive definite, ∂∂

∂∂

ux

D ux

ie

jijkl

ke

l

> 0 for all ∂∂ux

ie

j

≠ 0 . Therefore,

∂∂

∂∂

ux

D ux

dVie

jijkl

V

ke

l∫ > 0 for all

∂∂ux

ie

j

≠ 0 , and ∂∂

∂∂

ux

D ux

dVie

jijkl

V

ke

l∫ = 0 iff

∂∂ux

ie

j

≡ 0 .

.)?? =for only holdssign equality The ( +Π≥Π ihi

Eh uu




110

9.3. Principle of Virtual Work If the following inequality is valid for all real number α , the principle of virtual work

holds.

υ∈∀Π≥α+Π iiRR

iiRR vuvu )()(

) admissible allfor ( 0)0(

)(

)()(

21

21

)()()()(

21

)()(

2

iiiiV

iil

k

Vijkl

j

i

l

i

Vijkl

j

iii

Vii

l

k

Vijkl

j

i

iiiV

iii

l

i

Vijkl

j

i

l

k

Vijkl

j

i

l

k

Vijkl

j

i

iiiV

iiil

ii

Vijkl

j

ii

iiRR

vvdTvdVbvdVxu

Dxv

g

dVxv

Dxv

dTvdVbvdVxu

Dxv

g

dTvudVbvu

dVxv

Dxv

dVxu

Dxv

dVxu

Dxu

dTvudVbvudVx

vuD

xvu

vug

t

t

t

t

υ∈∀=Γ−−∂∂

∂∂

=′

∂∂

∂∂

α+Γ−−∂∂

∂∂

=α′

Γα+−α+−

∂∂

∂∂

α+∂∂

∂∂

α+∂∂

∂∂

=

Γα+−α+−∂

α+∂∂

α+∂=

α+Π≡α

∫∫∫

∫∫∫∫

∫∫

∫∫∫

∫∫∫

Γ

Γ

Γ

Γ

∂∂

σvx

dV v b dV v T d vi

jij

Vi i

Vi i i

t

∫ ∫ ∫− − =Γ

Γ 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 boxed equation of the principle of minimum

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

work is

∂∂

σvx

dV v b dV v T d vih

jijh

Vih

iV

ih

i ih

t

∫ ∫ ∫− − =Γ

Γ 0 for all admissible




111

9.4. Rayleigh-Ritz Type Discretization Approximation

∑=

=+++=n

ppipninii

hi gcgcgcgcu

12211

Principle of Minimum Potential Energy

Γ−−∂

∂

∂

∂=Π

∂

∂=

∂

∂=

∂∂

∫∫∫

∑

Γ

=

dTgcdVbgcdVxg

cDxg

c

xg

cxg

cxu

t

ipipV

ipipl

qkq

Vijkl

j

pip

h

j

pip

n

p j

pip

j

hi

21Min

1

mrfcK

dTgdVbgdVcxg

Dxg

dTgdVbgdVxg

cDxg

dTgdVbgdVxgD

xg

cdVxg

cDxg

dTgdVbgdVxgD

xg

cdVxg

cDxg

c

rmkprmkp

mrV

mrkpV l

pmjkl

j

r

mrV

mrl

pkp

Vmjkl

j

r

mrV

mrl

r

Vijml

j

pip

l

qkq

Vmjkl

j

r

irmiV

irmil

rmk

Vijkl

j

pip

l

qkq

Vijkl

j

rmi

mr

h

t

t

t

t

and allfor 0

21

21

21

21

=−=

Γ−−∂

∂

∂∂

=

Γ−−∂

∂

∂∂

=

Γ−−∂∂

∂

∂+

∂

∂

∂∂

=

Γδ−δ−∂∂

δ∂

∂+

∂

∂

∂∂

δ=∂Π∂

∫∫∫

∫∫∫

∫∫∫∫

∫∫∫∫

Γ

Γ

Γ

Γ

Principle of Virtual Work

and allfor 0

allfor 0)(

)(

1

2211

ipfcK

cdTgdVbgdVcxg

Dxg

c

dTgdVbgdVcxg

Dxg

c

dTgcdVbgcdVxg

cDxg

c

gcgcgcgcgcuv

pikqpikqh

ipipV

ipkqV l

qijkl

j

pip

ipV

ipkqV l

qijkl

j

pip

ipipV

ipipV l

qkqijkl

j

pip

h

pip

n

ppipninii

hi

hi

t

t

t

=−=Πδ

δ=Γ−−∂

∂

∂

∂δ=

Γ−−∂

∂

∂

∂δ=

Γδ−δ−∂

∂

∂

∂δ=Πδ

δ=δ=δ++δ+δ=δ=

∫∫∫

∫∫∫

∫∫∫

∑

Γ

Γ

Γ

=




112

Matrix Form – Virtual Work Expression

dVdV

dV

dV

dV

hT

V

hhhhhhhhhhh

V

hh

hhhhhhhhhh

V

hh

hhhhhhhhhhhhhhhh

V

hh

V

hij

hij

σε∫∫

∫

∫

∫

δ=σδγ+σδγ+σδγ+σδε+σδε+σδε

=σδε+σδε+σδε+σδε+σεδ+σδε

=σδε+σδε+σδε+σδε+σδε+σδε+σδε+σεδ+σδε

=σδε

)(

)222(

)(

232313131212333322221111

232313131212333322221111

323223233131131321211212333322221111

Γ⋅δ+⋅δ=⋅δ ∫∫∫Γ

ddVdVt

h

V

hh

V

h Tubuσε

Displacement

Ncu =

=

=

n

n

n

n

n

n

h

h

h

h

ccc

cccccc

gg

g

gg

g

gg

g

uuu

3

2

1

32

22

12

31

21

11

2

2

2

1

1

1

3

2

1

000000

00

0000

00

0000

Virtual Strain

∂∂

δ+∂∂

δ

∂∂

δ+∂∂

δ

∂∂

δ+∂∂

δ

∂∂

δ

∂∂

δ

∂∂

δ

=

∂∂δ

+∂

∂δ∂

∂δ+

∂∂δ

∂∂δ

+∂

∂δ∂

∂δ∂

∂δ∂

∂δ

=

δγδγδγδεδεδε

=δε

23

32

13

31

12

21

33

22

11

2

3

3

2

1

3

3

1

1

2

2

1

3

3

2

2

1

1

23

13

12

33

22

11

)(

xg

cxg

c

xg

cxg

c

xg

cxg

c

xg

c

xg

c

xg

c

xu

xu

xu

xu

xu

xu

xuxuxu

ii

ii

ii

ii

ii

ii

ii

ii

ii

hh

hh

hh

h

h

h

h

h

h

h

h

h

h

δε σ δ δijh

ijh

Vih

iV

ih

idV u b dV u T dt

∫ ∫ ∫− − =Γ

Γ 0




113

cBδ=

δδδ

δδδδδδ

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

=

n

n

n

nn

nn

nn

n

n

n

ccc

cccccc

xg

xg

xg

xg

xg

xg

xg

xg

xg

xg

xg

xg

xg

xg

xg

xg

xg

xg

xg

xg

xg

xg

xg

xg

xg

xg

xg

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 ==

γγγεεε

−−

−−

−−

−−

−−

−−

−+ν−

=

σσσσσσ

=

h

h

h

h

h

h

h

h

h

h

h

h

h

h

vv

vv

vv

vv

vv

vv

vv

vv

vv

vvE

ε

σ

23

13

12

33

22

11

13

13

12

33

22

11

)1(22100000

0)1(2

210000

00)1(2

21000

000111

0001

11

00011

1

)21)(1()1()(

Final System Equation

cDBBc dVdVdVV

TT

V

hTh

V

hij

hij ∫∫∫ δ=δ=σδε σε

Γδ=Γδ=Γδ

δ=δ=δ

∫∫∫

∫∫∫

ΓΓΓ

dddTu

dVdVdVbu

ttt

TTThi

hi

V

TT

V

Th

Vi

hi

TNcTu

bNcbu

fKccTNbNcDBBc T =δ=Γ−−δ ∫∫∫Γ

or admissible allfor 0)( ddVdVt

T

V

T

V

T




114

9.5 Example

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

)2

6

(

)2

(

23

2

lxxbv

ylxxau

−=

−=

ν−ν

ν

ν−=

−−

−=

−

−=

2100

0101

1 ,

22

000)(

,

260

0)2

(22226

2

E

lxxlxx

ylx

lxx

ylxx

DBN

ν−ν−

ν−ν−+

ν−=

−ν−

−ν−

−ν−

−ν−

+−

ν−=

−−

−

ν−ν

ν

−

−−

ν−=

∫ ∫ ∫

∫ ∫ ∫

−

−

)2(152

21)2(

152

21

)2(152

21)2(

152

21

332

1

)

2(

21)

2(

21

)2

(2

1)2

(2

1)(

1

22

000)(

2100

0101

200

20)(

1

55

5533

2

0

1

0 22

22

22

22

22

2

0

1

0222

2

2

hlhl

hlhlhlE

dzdydxlxxlxx

lxxlxxylxE

dzdydx

lxxlxx

ylx

lxx

lxxylxE

l h

h

l h

h

K

−=

−

−= ∫ ∫

− Pldy

hl

ylh

h

0

32P0

30

02

3

3

2

1

0

f

x, u

y, v

l

2h

hPTy 2

=




115

−=

ν−ν−

ν−ν−+

ν− Pl

ba

hlhl

hlhlhlE 0

3)2(

152

21)2(

152

21

)2(152

21)2(

152

21

332

1

3

55

5533

2

PEIl

hv

b

PEI

PEh

a

22

2

3

2

1))(1

1351(

1123

ν−−

+−=

ν−=

ν−=

)2

6

(1))(1

1351(

)2

(1

2322

22

lxxPEIl

hv

v

ylxxPEI

u

−ν−

−+−=

−ν−

=

PxlxIl

h

yI

M

yI

MyI

lxP

xy

yy

xx

)2

(1)(65

)(

22 −−=τ

ν=σ

=−

=σ


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