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Dept. of Civil and Environmental Eng., SNU 1
Chapter 1
Fundamentals in Elasticity
E , ν
tt =
uu =
Su
St
V
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 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
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 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
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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.
- The elasticity problems may be either linear or nonlinear.
-
- 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
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Dept. of Civil and Environmental Eng., SNU 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) - Strain (ε) - Displacement (u) - Relationships among the unknowns such as equilibrium, strain-displacement
relationship, strain-stress relationship
E , ν
tt =
uu =
Su
St
V
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Dept. of Civil and Environmental Eng., SNU 6
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 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. 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.
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1.5. 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
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Dept. of Civil and Environmental Eng., SNU 10
),,(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
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Dept. of Civil and Environmental Eng., SNU 11
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
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Chapter 2
Traction and Stress
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2.1. Traction and Stress
Traction: dSd
SΔSnFFT =
∆∆
=→0
lim Stress : x, y, z components of traction
x
n Tn
∆S
∆F
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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
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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
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Dept. of Civil and Environmental Eng., SNU 16
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
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
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Dept. of Civil and Environmental Eng., SNU 17
∫∫
∫∫∫∫
∂
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:
∫ ∫ =+⋅−S V
fdVdS 0nv
v n
dS
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Dept. of Civil and Environmental Eng., SNU 18
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
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Dept. of Civil and Environmental Eng., SNU 19
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)( =+⋅∇=
+⋅∇=+⋅=+
∫
∫ ∫∫ ∫∫ ∫
Sii
S 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 =
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Dept. of Civil and Environmental Eng., SNU 20
– 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→⋅′=′′⋅′→′′=
ijjiijij xx ee ⋅′=ββ=′ where
kkjkjkjkkjijikkjiijkkii xxxxxxxxx ′β=′⋅′=→′′⋅=δ→′′⋅=⋅→′′= eeeeeeeeee
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′
P
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Dept. of Civil and Environmental Eng., SNU 21
Tractions in primed coordinate system
jjijjixjji
xi
jjijjixjji
xi
jjijjixjji
xi
xxnT
xxnT
xxnT
33
22
11
),cos(
),cos(
),cos(
33
22
11
βσ=′σ=σ=
βσ=′σ=σ=
βσ=′σ=σ=
′′
′′
′′
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 primed coordinate
system
jjikix
ikik
jjikix
ikik
jjikix
ikik
T
T
T
33
22
11
3
2
1
βσβ=β=σ′
βσβ=β=σ′
βσβ=β=σ′
′
′
′
βσβσ
βσβσ
′=→βσ′β=σ
=′→βσβ=σ′T
ljklkiij
Tkijimjmk
1xiT ′
3xiT ′
2xiT ′
1xiT ′
3xiT ′
2xiT ′
1x′n
2x′n
3x′n
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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.
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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 a body or system of bodies without consideration of the forces involved (i.e., causes
and effects of the motions).
Kinematics 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
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 25
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).
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
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 26
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.
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.
x X
Reference configuration t
Control Volume: analysis domain
x
t
t+∆t
t
),( txXX =
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 27
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
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 28
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.
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 29
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 =′′= ,
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 30
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
aa d+
a
ds
)0,0,( 1dxd =x
0ds
)0,,0( 2xdd =x
a
aa d+
aa d+
sd
ds
)0,0,( 1dxd =x
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 31
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 :
0231323 =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 !!!
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 32
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
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 33
)])(([)])(([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
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 34
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
xMU
xF
xHU
xG
xHU
dSUmFlGUlHnFUmHnF
dVxU
xUH
xU
xUG
xU
xUFI
V
S
V
Since ),,( zyxF , ),,( zyxG and ),,( zyxH are arbitrary, 0≡== zyx UUU in V.
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 35
Chapter 4
Constitutive Relations
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 36
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 ×××=
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 37
– 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
ε22 σ12 σ12
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 38
Shear-Other Shear Relation
Case I Case II
– Case I ( 0 , 0 131233221123 =ε=ε=ε=ε=ε≠ε ) → 23122312 ε=σ CI
– Case II ( 0 , 0 231333221112 =ε=ε=ε=ε=ε≠ε ) → 12231223 ε=σ CII
By isotropic condition, III2312 σ−=σ , for 2312 ε=ε .
0020)( 12231212231223121223 =→=ε→=ε+ CCCC
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.
90° Rotation
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 39
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=
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
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 40
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
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
(1+ε11)L
h (1-νε11)h
L
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 41
Gv
EAAAvv
EAvv
EvA =+
=−
=µ=−+
ν=λ=
−+−
=)1(22
, )21)(1(
, )21)(1(
)1( 21321
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
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 42
Chapter 5
Beam Approximation
≅
x1
x2
x3
),,0( 32ttt nn=n
)0,0,1(=ln
liT
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 43
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
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 44
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
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 =+∂
∂ mx
M ∫∫ −+−=S
tt
A
dSTxTxdAbxbxm )~()( 233223321
0321
2 =++∂
∂ Qmx
M ∫∫ +=S
t
A
dSTxdAbxm 13132
0231
3 =−−∂
∂ Qmx
M ∫∫ +=S
t
A
dSTxdAbxm 12123
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 45
Planar beam 03131 ==== tt TTbb
021
2 =+∂∂ q
xQ , 02
1
3 =−∂
∂ Qx
M → 221
32
qxM
−=∂
∂
011
1 =+∂
∂ mx
M ∫∫ −−=S
t
A
dSTxdAbxm 23231
What condition(s) should be satisfied in order that the vertical loads do not cause torsion
of beam ??
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 46
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
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 47
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
∫∫∫
qxu
EIx
qxM y =
∂∂
∂∂
→−=∂
∂2
2
2
2
2
2
, yI
Mxx =σ
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
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 48
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
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 49
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
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 50
Chapter 6
Plane Problems in
Cartesian Coordinate System
x y
z
∞
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 51
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
Equilibrium Equation
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
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 52
),( , ),( 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
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 53
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
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 54
6.2. Plane Strain
Assumption
– Infinite Structure in z-direction with an identical section.
– Identical traction in z-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
∞
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 55
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
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 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
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 56
– 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 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
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 57
Moment by S. stress = lcdclcdclcd 34
24
24
322
2312
2=−
Moment by N. stress = lcdclcd 34
24
322
32
2=
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
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 58
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
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 59
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
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 60
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
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 61
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
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 62
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
−−−=
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 63
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
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 64
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 −=−====−=
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 65
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
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 66
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
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 67
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.
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 68
Chapter 7
Plane Problems in
Polar Coordinate System
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 69
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
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 70
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
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 71
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
Equilibrium Equation
??
)2sinsincos( 22
=∂
θσ−θσ+θσ∂=
∂σ∂ θθθ
xxrrrxx
??
))sin(coscossin)(( 22
=∂
θ−θσ+θθσ−σ∂=
∂
σ∂θθθ
yyrrrxy
??=+∂
σ∂+
∂σ∂
xxyxx b
yx
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 72
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 ,
drruHB
duCF
drrudrHA
r
r
∂∂
=′
θθ∂
∂=′
∂∂
+=′
θ
)()( θθ∂
∂++−θ+=
+′−=′
θθθ duuudur
EFADDEFA
r
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 73
ru
drdrurdrruudrr
ABABBA rrrr
rr ∂∂
=−+−∂∂+++
=−′
=ε)()/('
θ∂∂
+=ε→θθ∂
∂+θ+=′≈θε+=′′
θε+=′′→−′′
=ε
θθθ
θθθ
θθθθ
urr
ududurFArdCA
rdCAAC
ACCA
rr
1)()1(
)1(
ru
ruu
rdudurdu
FACF
ru
ruu
rr
r
r
rr
θθ
θ
θθθ
=θ∂∂
=θθ∂
∂≈
θθ∂∂+θ+θθ∂∂
=′
′=θ
−∂∂
+θ∂
∂=θ−θ+θ=θ+θ=ε
342
34212
,,1)/()(
)/(
)1(21)(
21)(
21
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
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 74
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
0)11)(11( 2
2
22
2
2
2
22
24 =φ
θ∂∂
+∂∂
+∂∂
θ∂∂
+∂∂
+∂∂
=φ∇rrrrrrrr
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 =
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 75
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.
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 76
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
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 77
– Rigid body rotation
θ== θ ruur , 0 → θ=F
θr
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 78
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
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 79
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+−
=
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 80
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+=
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 81
– 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
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 82
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
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 83
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+θ+ν−+ν+−ν−+ν+
−=
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
−=→
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 84
7.7. Rotating Disk Centrifugal force : rbr
2ρω=
Equilibrium Equation
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
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 85
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)
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 86
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
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 87
– 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
=
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 88
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
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 89
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
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 90
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)
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 91
Chapter 8
Uniform Torsion (St. Venant Torsion)
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 92
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
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 93
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.
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 94
BCs at the end of the bar
0
0
=φ−=∂φ∂
−=σ
=φ=∂φ∂
=σ
∫∫∫
∫∫∫
Ay
AAzy
Sx
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
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 95
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
cabbaMabxy
baM tt 2)(2)(2 22
3322
33 +ψ+π
=−π
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 96
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
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 97
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
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 98
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=
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 99
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
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 100
Chapter 9
Energy Methods
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 101
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
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 102
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
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 103
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
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 104
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
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 105
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
=
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 106
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
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 107
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
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 108
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
=
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU 109
−=
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
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr