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Structural Analysis Lab.
Prof. Hae Sung Lee, http://strana.snu.ac.kr
1
Chapter 1
Fundamentals in Elasticity
E,
tt
uu Su
St
V
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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
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Structural Analysis Lab.
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- 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
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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
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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
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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
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1.2. Domain and Boundary
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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.
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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.
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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.
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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
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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
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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
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Chapter 2
Traction and Stress
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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
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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
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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
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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
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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
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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
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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
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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
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– 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
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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
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Chapter 3
Displacement and Strain
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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
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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
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab.
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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
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab.
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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
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Prof. Hae Sung Lee, http://strana.snu.ac.kr
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
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Prof. Hae Sung Lee, http://strana.snu.ac.kr
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
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab.
Prof. Hae Sung Lee, http://strana.snu.ac.kr
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 ,
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab.
Prof. Hae Sung Lee, http://strana.snu.ac.kr
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
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab.
Prof. Hae Sung Lee, http://strana.snu.ac.kr
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 !!!
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab.
Prof. Hae Sung Lee, http://strana.snu.ac.kr
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
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab.
Prof. Hae Sung Lee, http://strana.snu.ac.kr
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?
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab.
Prof. Hae Sung Lee, http://strana.snu.ac.kr
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.
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab.
Prof. Hae Sung Lee, http://strana.snu.ac.kr
38
Chapter 4
Constitutive Relations
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab.
Prof. Hae Sung Lee, http://strana.snu.ac.kr
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)
Dept. of Civil and Environmental Eng., SNU
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Prof. Hae Sung Lee, http://strana.snu.ac.kr
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
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab.
Prof. Hae Sung Lee, http://strana.snu.ac.kr
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)
Dept. of Civil and Environmental Eng., SNU
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Prof. Hae Sung Lee, http://strana.snu.ac.kr
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
Dept. of Civil and Environmental Eng., SNU
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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
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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
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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
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Chapter 5
Plane Problems in
Cartesian CoordinateSystem
x y
z
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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
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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
–
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– 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
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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
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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
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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
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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
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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(
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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
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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
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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
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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.
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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
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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
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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
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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
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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
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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
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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.
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Chapter 6
Plane Problems in
Polar Coordinate System
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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
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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
Equilibrium Equation
??
)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
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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
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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
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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
)/(
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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
Dept. of Civil and Environmental Eng., SNU
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Prof. Hae Sung Lee, http://strana.snu.ac.kr
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
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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.
Dept. of Civil and Environmental Eng., SNU
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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
Dept. of Civil and Environmental Eng., SNU
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76
– Rigid body rotation
ruur , 0 F
r
xu
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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
Dept. of Civil and Environmental Eng., SNU
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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
Dept. of Civil and Environmental Eng., SNU
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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(
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Prof. Hae Sung Lee, http://strana.snu.ac.kr
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
Dept. of Civil and Environmental Eng., SNU
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Prof. Hae Sung Lee, http://strana.snu.ac.kr
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
Dept. of Civil and Environmental Eng., SNU
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Prof. Hae Sung Lee, http://strana.snu.ac.kr
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
Dept. of Civil and Environmental Eng., SNU
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Prof. Hae Sung Lee, http://strana.snu.ac.kr
83
6.7. Rotating Disk
Centrifugal force : rbr
2
Equilibrium Equation
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
Dept. of Civil and Environmental Eng., SNU
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Prof. Hae Sung Lee, http://strana.snu.ac.kr
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)
Dept. of Civil and Environmental Eng., SNU
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Prof. Hae Sung Lee, http://strana.snu.ac.kr
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
Dept. of Civil and Environmental Eng., SNU
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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
Dept. of Civil and Environmental Eng., SNU
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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
Dept. of Civil and Environmental Eng., SNU
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Prof. Hae Sung Lee, http://strana.snu.ac.kr
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
Dept. of Civil and Environmental Eng., SNU
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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)
Dept. of Civil and Environmental Eng., SNU
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90
Chapter 7
Uniform Torsion (St. Venant Torsion)
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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
Dept. of Civil and Environmental Eng., SNU
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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.
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab.
Prof. Hae Sung Lee, http://strana.snu.ac.kr
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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
Dept. of Civil and Environmental Eng., SNU
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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
Dept. of Civil and Environmental Eng., SNU
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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
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab.
Prof. Hae Sung Lee, http://strana.snu.ac.kr
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
Dept. of Civil and Environmental Eng., SNU
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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
Dept. of Civil and Environmental Eng., SNU
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Prof. Hae Sung Lee, http://strana.snu.ac.kr
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
Dept. of Civil and Environmental Eng., SNU
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99
Chapter 8
Beam Approximation
x1
x2
x3
),,0( 32ttt nnn
)0,0,1(ln
liT
Dept. of Civil and Environmental Eng., SNU
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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
Dept. of Civil and Environmental Eng., SNU
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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
Dept. of Civil and Environmental Eng., SNU
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102
3131232133~~~
MdSTxdSTxedSTxeMS
l
S
l
S
l
kjjk
Dept. of Civil and Environmental Eng., SNU
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Prof. Hae Sung Lee, http://strana.snu.ac.kr
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 ??
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab.
Prof. Hae Sung Lee, http://strana.snu.ac.kr
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
Dept. of Civil and Environmental Eng., SNU
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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
Equilibrium Equation
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
Dept. of Civil and Environmental Eng., SNU
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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
Dept. of Civil and Environmental Eng., SNU
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Prof. Hae Sung Lee, http://strana.snu.ac.kr
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)()(
Dept. of Civil and Environmental Eng., SNU
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Prof. Hae Sung Lee, http://strana.snu.ac.kr
108
Chapter 9
Energy Methods
Dept. of Civil and Environmental Eng., SNU
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Prof. Hae Sung Lee, http://strana.snu.ac.kr
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
Dept. of Civil and Environmental Eng., SNU
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Prof. Hae Sung Lee, http://strana.snu.ac.kr
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
Dept. of Civil and Environmental Eng., SNU
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Prof. Hae Sung Lee, http://strana.snu.ac.kr
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
Dept. of Civil and Environmental Eng., SNU
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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
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab.
Prof. Hae Sung Lee, http://strana.snu.ac.kr
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
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab.
Prof. Hae Sung Lee, http://strana.snu.ac.kr
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
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab.
Prof. Hae Sung Lee, http://strana.snu.ac.kr
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
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab.
Prof. Hae Sung Lee, http://strana.snu.ac.kr
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
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab.
Prof. Hae Sung Lee, http://strana.snu.ac.kr
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