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Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Introduction to
Finite Element Method
Fall Semester, 2015
Hae Sung Lee
Dept. of Civil and Environmental Engineering Seoul National University
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Contents
1. Introduction
2. Approximation of Functions & Variational Calculus
3. Differential Equations in One Dimension
4. Multidimensional Problems-Elasticity
5. Discretization
6. Two Dimensional Elasticity Problems
7. Various Types of Elements
8. Numerical Integration
9. Convergence Criteria in Isoparameteric Element
10. Miscellaneous Topics
11. Problems with Higher Continuity Requirement - Beams
12. Mixed Formulation
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Chapter 2
Approximation of Functions and
Variational Calculus
≅ δ
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Fundamental Considerations
What is the best solution for a given problem ?
Needless to say, it is the exact solution…
What is the exact solution?
The solution that satisfies the governing equations as well as boundary conditions if any.
How many do the exact solutions exist?
Of course, one… In case several or infinite numbers of the exact solutions exist for a giv-
en equation, we call the problem as an ill-posed problem, and have great difficulties in
determining a solution of the given problem…
What if it is impossible to determine the exact solution for various reasons?
We need approximate solution.
What is an approximate solution?
Fundamental Questions
- What is the best approximation?
- How can we calculate ai that represents the best approximation?
A Good (or Robust or Well Formulated) Approximation Should
- Yield the best approximation to the exact solution for a given degree of approxima-
tion.
- Converge to the exact solution as higher degree of approximation is employed.
What is the Definition of the Best Approximation?
- May be defined as the closest solution to the exact solution.
- But, how close is “the closest”?
- “Close or Far” implies the distance between two spatial points.
- We should define some sort of ruler to measure the distance between two elements in
a function set…
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Discretization
- Representation of a continuously distributed quantities with some numbers.
)()(1
XX ∑=
=n
iii gaf υ∈∀ )(Xf where gi are the basis functions of a function set, υ
- Set : Collection of some objectives with the same characteristics
- The basis functions should be linearly independent to each other.
0)(1
=∑=
Xn
iii ga if and only if all 0=ia
- Taylor series, Fourier series, etc…
Approximations
υυ ⊂∈=≈= ∑∑==
hm
iii
hn
iii gafgaf )()()()(
11XXXX where nm ≤
Summation Notation: Repeated indices denote summation ii
m
iii baba =∑
=1.
Normes of Functions: A measure of a function set
A function set υ is said to be a normed space if to every υ∈f there associated a
nonnegative real number f , called the norm of f, in a such way that
- 0=f if and only if 0≡f
- ff || α=α for any real number α.
- gfgf +≤+
Every normed space may be regarded as a metric space, in which the distance between
any two elements in the space is measured by the defined norm. Various types of norm
can be defined for a function space. Among them the following norms are important.
- L1 norm: ∫=V
LdVff
1
- L2 norm: 2/12 )(2 ∫=
VL
dVff
- H1 norm: 2/12 ))((1 ∫ ∇⋅∇+=V
HdVffff
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
General Ideas for the Best Approximation
Let’s find out a approximate function that is closest to the given function by use of a
norm defined in the function space. If this is the case, the characteristics of an approxima-
tion method depend on those of the norm used in the approximation.
Least Square Error(LSE) Minimization
Error: hffe −=
Minimize ∫ −=−==ΠV
h
L
hL
dVffffe 222 )(21
21
21
22
mkFaKfdVgadVgg
fdVgdVagg
dVgffdVafff
a
ki
m
iki
Vki
m
i Vik
Vk
V
m
iiik
Vk
h
V k
hh
k
,,1for 0
)()(
11
1
==−=−=
−=
−=∂∂
−=∂
Π∂
∑∫∑∫
∫∫ ∑
∫∫
==
=
Final System Equation : FKa =
If the basis functions are orthogonal, K becomes diagonal.
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Variation of a function
- The variation of a function means a possible change in the function for the fixed x.
Variational Calculus
- if ii gaf = then
the variation of f is defined as ii gaf δ=δ or by definition ii
aaff δ
∂∂
=δ .
- ffFa
af
fFa
aFFfF i
ii
i
δδδδ∂∂
=∂∂
∂∂
=∂∂
= : )(
- hfhf δ+δ=+δ )(
- hffhfh δ+δ=δ )(
- , )()(dx
fdaaf
dxda
dxdf
adxdf
ii
ii
δ=δ
∂∂
=δ∂∂
=δ
- ∫∫∫∫ δ=δ∂∂
=δ∂∂
=δ fdxdxaafafdx
afdx i
ii
i
f
δf
Variation of a function
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Minimization by Variational Calculus
Min ∫ −=Πl
hh dxfff0
2)(21)(
kk
k
l
k
hh
lhh
lh
lhh
aa
adxafff
dxfffdxffdxfff
δ∂
Π∂=δ
∂∂
−=
δ−=−δ=−δ=Πδ
∫
∫∫∫
0
00
2
0
2
)(
)()(21))(
21()(
Min kV
hh adVfff δ=Πδ⇔−=Π ∫ possible allfor 0)(21)( 2
Euler Equation
Min ka
dxxffFfk
l
allfor 0),,()(0
=∂
Π∂⇒′=Π ∫
∫
∫∫∫∫
′∂∂
−∂∂
+′∂
∂=
′′∂
∂+
∂∂
=∂
′∂′∂
∂+
∂∂
∂∂
=∂∂
=′∂∂
=∂
Π∂
l
kk
l
k
l
kk
l
kk
l
k
l
kk
dxgfF
dxdg
fFg
fF
dxgfFg
fFdx
af
fF
af
fFdx
aFdxxffF
aa
00
0000
)(
)()(),,(
In case the basis functions vanish at the boundary, then
0 allfor 0)(0
=′∂
∂−
∂∂
⇔=′∂
∂−
∂∂
=∂
Π∂∫ f
Fdxd
fFkdxg
fF
dxd
fF
a
l
kk
∫∫∫ δ′∂
∂−δ
∂∂
+δ′∂
∂=′δ
′∂∂
+δ∂∂
=′δ=Πδllll
dxffF
dxdf
fFf
fFdxf
fFf
fFdxxffFf
0000
)()(),,()(
In case the variation vanishes at the boundaries, then
kk
k
l
k
l
aa
adxgfF
dxd
fFfdx
fF
dxd
fFf δ
∂Π∂
=δ′∂
∂−
∂∂
=δ′∂
∂−
∂∂
=Πδ ∫∫00
)()()( .
Therefore,
Min 0=Πδ⇔Π
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Example 1
Min ∫ ′+=Π2
1
2)(1)(x
x
dxyy subject to 11 )( yxy = , 22 )( yxy =
0))(1())(1(0 2/122 =′+′−=′+′∂
∂−=
′∂∂
−∂∂ −yy
dxdy
ydxd
fF
dxd
fF
0))(1())(1
)(1())(1(
))(1)(21())(1())(1(
2/322
22/12
2/322/122/12
=′+′′=′+
′−′+′′=
′′′′+−′+′+′′=′+′
−−
−−−
yyy
yyy
yyyyyyyydxd
baxyy +=→=′′ 0 . By applying BC, 12
2112
12
120xx
yxyxxxxyyyy
−−
+−−
=→=′′
Example 2
Min ∫ −′=Πl
dxufuu0
2 ))(21()( subject to 0)0( =u , 0)( =lu
00))(21( 2 =+′′→=′′−−=′−−=′
′∂∂
−−=′∂
∂−
∂∂ fuufu
dxdfu
udxdf
uF
dxd
uF
Homework 1
1. Approximate a cosine function xl
y π=
2cos by polynomials based the minimization of
the least square errors. Use polynomials up to the 20th order. You may use a numerical in-tegration algorithm such as Simpson’s rule, the trapezoidal rule or the rectangular rule. You also need a numerical solver to solve linear simultaneous equations in the “Linpack”. For the accuracy of your calculation, please use “double precision” in your program. You should present proper discussions on your results together with some graphs that show your approximate functions and the given function.
2. Derive Euler equation for the following minimization function, and proper boundary condi-
tions.
Min ∫ ′′′=Πl
dxxfffFf0
),,,()(
3. Derive the governing equation and the boundary conditions for the following problem.
Min ∫ −=Πl
dxwqdx
wdw0
22
2
))(21()(
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Chapter 3
Elliptic Differential Equations in One
Dimension
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
3.1. Problems with homogenous displacement boundary conditions Problem Definition
0)()0( ,0 02
2
==<<=+ luulxfdx
ud
Approximation – Discretization
∑=
=m
iii
h gau1
where 0)()0( == luu hh
Residuals
Verbal Definition : Something left over, or resulting from subtraction...
Equation Residual : lxfdx
udRh
E <<≠+= 0 02
2
Function Residual : lxuuR hF <<≠−= 0 0
Error Estimator :
∫∫ +−==Πl h
hl
EFR dxf
dxuduudxRR
02
2
0
))((21
21
Least Square Error
Error SquareLeast ))((21
))((21))((
21
))((21
))((21
0
00
02
2
2
2
02
2
←Π=−−=
−−−−−=
−−=
+−=Π
∫
∫
∫
∫
LSl hh
l hhlhh
l hh
l hhR
dxdx
dudxdu
dxdu
dxdu
dxdxdu
dxdu
dxdu
dxdu
dxdu
dxduuu
dxdx
uddx
uduu
dxfdx
uduu
Energy Functional – Total Potential Energy
RRl
hl hhl
l hhlhh
lh
lh
lhlh
lh
hh
hl hhR
Cfdxudxdx
dudx
duufdx
dxdx
dudx
dudx
duudxudx
ududxdu
dxduudxfuuf
dxfudx
uduufdx
ududxfdx
uduu
Π+=−+=
+−+−+−=
−−+=+−=Π
∫∫∫
∫∫∫
∫∫
)21(
21
})({21
)(21))((
21
000
0002
2
000
02
2
2
2
02
2
−f
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Minimization Problems RRLSR Π↔Π⇔Π Min Min Min w.r.t. hhu υ∈
RRΠMin : Rayleigh-Ritz Method or Principle of Minimum Potential Energy
- 1st Order Necessary Condition for Minimization Problem
FKa =→==−=
−=
−∂∂
=
−=∂Π∂
∑
∫∑∫
∫ ∑∫ ∑∑
∫∫
=
=
===
mkFaK
fdxgadxdxdg
dxdg
fdxgadaddx
dxga
dxga
dad
fdxdadudx
dxdu
dxdu
dad
a
m
ikiki
l
k
m
ii
lik
l m
iii
k
l m
i
ii
m
i
ii
k
l
k
hl hh
kk
RR
,1for 0
)())((
)(
1
01 0
0 10 11
00
0=Πδ RR : Variational principle or Principle of Virtual Work
0)()( 0)(
)(
)21(
1 1
1 01 0
0000
00
=−δ→=−δ=
−δ=
δ−δ
=δ−δ=
−δ=Πδ
∑ ∑
∑ ∫∑∫
∫∫∫∫
∫∫
= =
= =
FKaa Tm
k
m
ikikik
m
k
l
k
m
ii
lik
k
lh
l hhlh
l hh
lh
l hhRR
FaKa
fdxgadxdxdg
dxdga
fdxudxdx
dudxudfdxudx
dxdu
dxdu
fdxudxdx
dudx
du
Solution Space
- })( , 0)()0(|{0
2 ∞<==≡υ∈ ∫l
dxdxduluuuu
- υ≡υh : The minimization problems yield the exact solution. - υ⊂υh : The minimization problems yield an approximate solution.
Properties of K
- Symmetry : ji
lij
lji
ij Kdxdxdg
dxdg
dxdx
dgdxdg
K === ∫∫00
- Positive Definiteness :
∑∑∑∫∑
∑∫∑∫
= ===
==
≥===
=
m
i
Tj
m
jijij
m
j
jl
im
ii
j
m
j
jl
i
m
i
il h
aKaadxdx
dgdxdg
a
dxadx
dga
dxdg
dxdx
du
1 11 01
10 1
2
0
0
)(
Kaa
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Absolute Minimum Property of Total Potential Energy eh uuu −=
)for only holdssign equality (The )(21
21
21
)(21
21
21
21
21
21
)()()(21
21
0
2
00 0
02
2
0 00
002
2
00 00
000 00
0000
Cudxdx
du
dxdx
dudx
duufdxdxdxdu
dxdu
dxfdx
ududxdx
dudx
duufdxdxdxdu
dxdu
fdxudxdx
ududxduudx
dxdu
dxduufdxdx
dxdu
dxdu
fdxudxdxdu
dxdudx
dxdu
dxduufdxdx
dxdu
dxdu
fdxuudxdx
uuddx
uudfdxudxdx
dudx
du
eEl e
E
l eel l
le
l l eel
le
le
le
l l eel
le
l el l eel
le
l eelh
l hhh
=Π≥+Π=
+−=
+++−=
++−+−=
+−+−=
−−−−
=−=Π
∫
∫∫ ∫
∫∫ ∫∫
∫∫∫ ∫∫
∫∫∫ ∫∫
∫∫∫∫
Weighted Residual Method
mkdxfdx
uddxRl h
k
l
Ekk ,,1for 0)(0
2
2
0
==+φ=φ=π ∫∫
- if kk g=φ : Galerkin Method (elliptic System or self-adjoint system)
FKa =→==+−=
+−=+−=
+−=+φ=π
∫∑∫
∫∫ ∑∫∫
∫∫∫
′
=
′
=
,,1for 0
)(
01 0
00 100
00002
2
mkfdxgadxdxdg
dxdg
fdxgdxadxdg
dxdg
fdxgdxdx
dudx
dg
fdxgdxdx
dudx
dgdx
dugdxfdx
ud
l
ki
m
i
lik
l
k
l m
ii
ikl
k
l hk
l
k
l hk
lh
k
l h
kk
- Identical result to the Rayleigh-Ritz Method or Variational Principle
- if kk g≠φ : Petrov-Galerkin Method (hyperbolic system or non-self adjoint system)
Weighted Residual Method vs. Variational Principle
ii
m
iik aamk δ∀=δπ⇔==π ∑
=
0 ,,1for 01
hl l
hhhl h
h
l h
i
m
ii
m
i
l h
iii
m
ii
ufdxudxdx
dudxuddxf
dxudu
dxfdx
udgadxfdx
udgaa
δ=δ−δ
−=+δ
=+δ=+δ=δπ
∫ ∫∫
∫∑∑ ∫∑===
possible allfor 0)()(
)()(
0 002
2
02
2
11 02
2
1
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Example 1 : 12
2
−=dx
ud , 0)1()0( == uu
- Exact solution : xxu21
21 2 +−=
- First trial : 2321 xaxaau h ++=
Applying BCs : 01 =a , 32 aa −= → )( 23 xxau h +−= , )21(3 xa
dxdu h
+−=
31)21(
1
0
211 =+−= ∫ dxxK ,
61)(1
1
0
21 −=+−⋅= ∫ dxxxF
21
61
31
33 −=→−= aa . Therefore, uu h ≡
- Second Trial : 34
2321 xaxaxaau h +++=
Applying BCs : 01 =a , 432 aaa −−= →
21
)()( 34
23
gg
h xxaxxau +−++−=
)31(),21( 221 xdx
dgxdxdg
+−=+−=
31)21(
1
0
211 =+−= ∫ dxxK ,
54)31(
1
0
2222 =+−= ∫ dxxK
21)31)(21(
1
0
22112 =+−+−== ∫ dxxxKK
61)(1
1
0
21 −=+−⋅= ∫ dxxxF ,
41)(1
1
0
32 −=+−⋅= ∫ dxxxF
System Equation : 0,21
4161
54
21
21
31
43
4
3
=−=→
−
−=
aaa
a
Example 2 : xdx
udππ−= sin2
2
2
, 0)1()0( == uu
- Exact Solution : xu π= sin - First trial : 2
321 xaxaau h ++=
Applying BCs : )( 22 xxau h +−= , )21(2 xa
dxdu h
+−=
31)21(
1
0
211 =+−= ∫ dxxK ,
π−=+−⋅ππ= ∫
4)(sin1
0
221 dxxxxF
π−=→
π−=
12431
22 aa . Therefore, )(12 2xxu h +−π
−=
955.03)5.0( ==π
hu Error = 4.5 %
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
- Second trial : 33
2210 xaxaxaau h +++=
Applying BCs : 01 =a , 432 aaa −−= →
21
)()( 34
23
gg
h xxaxxau +−++−=
)31(),21( 221 xdx
dgxdxdg
+−=+−=
31)21(
1
0
211 =+−= ∫ dxxK ,
54)31(
1
0
2222 =+−= ∫ dxxK
21)31)(21(
1
0
22112 =+−+−== ∫ dxxxKK
π−=+−⋅ππ= ∫
4)(sin1
0
221 dxxxxF ,
π−=+−⋅ππ= ∫
6)(sin1
0
322 dxxxxF
System Equation : 0,126
4
54
21
21
31
32
3
2
=π
−=→
π−
π−
=
aaa
a ????
- In general )(2
∑′
=
+−=m
i
ii
h xxau
Function Error=
2/1
1
0
2
1
0
2
sin
)(sin
π
−π
∫
∫
xdx
dxux h
Derivative Error=
2/1
1
0
22
1
0
2
cos
))(cos(
ππ
′−ππ
∫
∫
xdx
dxux h
To evaluate numerator in the error expressions, the midpoint rule with 100 subinter-
vals is employed.
- Raw Output
***** 2th-order Polynomial ***** a 2 = -0.3819719E+01 Errors for 2th-order polynomial Function error = 0.2009211E+01 % Derivative error = 0.6010036E+01 %
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
***** 3th-order Polynomial ***** a 2 = -0.3819719E+01 a 3 = 0.0000000E+00 Errors for 3th-order polynomial Function error = 0.2009211E+01 % Derivative error = 0.6010036E+01 % ***** 4th-order Polynomial ***** a 2 = 0.4193832E+00 a 3 = -0.7065170E+01 a 4 = 0.3532585E+01 Errors for 4th-order polynomial Function error = 0.4048880E-01 % Derivative error = 0.1956108E+00 % ***** 5th-order Polynomial ***** a 2 = 0.4193832E+00 a 3 = -0.7065170E+01 a 4 = 0.3532585E+01 a 5 = 0.7833734E-12 Errors for 5th-order polynomial Function error = 0.4048880E-01 % Derivative error = 0.1956108E+00 % ***** 6th-order Polynomial ***** a 2 = -0.1405727E-01 a 3 = -0.5042448E+01 a 4 = -0.5128593E+00 a 5 = 0.3640900E+01 a 6 = -0.1213633E+01 Errors for 6th-order polynomial Function error = 0.4444114E-03 % Derivative error = 0.2944068E-02 % ***** 7th-order Polynomial ***** a 2 = -0.1405727E-01 a 3 = -0.5042448E+01 a 4 = -0.5128593E+00 a 5 = 0.3640900E+01 a 6 = -0.1213633E+01 a 7 = 0.5029577E-09 Errors for 7th-order polynomial Function error = 0.4444114E-03 % Derivative error = 0.2944068E-02 % ***** 8th-order Polynomial ***** a 2 = 0.2231430E-03 a 3 = -0.5170971E+01 a 4 = 0.2265606E-01 a 5 = 0.2462766E+01 a 6 = 0.2001274E+00 a 7 = -0.8751851E+00 a 8 = 0.2187963E+00
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Errors for 8th-order polynomial Function error = 0.3045376E-05 % Derivative error = 0.2548945E-04 % ***** 9th-order Polynomial ***** a 2 = 0.2231387E-03 a 3 = -0.5170971E+01 a 4 = 0.2265578E-01 a 5 = 0.2462767E+01 a 6 = 0.2001259E+00 a 7 = -0.8751837E+00 a 8 = 0.2187955E+00 a 9 = 0.1698185E-06 Errors for 9th-order polynomial Function error = 0.3045376E-05 % Derivative error = 0.2548945E-04 % ***** 10th-order Polynomial ***** a 2 = -0.2313690E-05 a 3 = -0.5167665E+01 a 4 = -0.4905381E-03 a 5 = 0.2553037E+01 a 6 = -0.1050486E-01 a 7 = -0.5742830E+00 a 8 = -0.3911909E-01 a 9 = 0.1217931E+00 a10 = -0.2435856E-01 Errors for 10th-order polynomial Function error = 0.1289526E-07 % Derivative error = 0.1450501E-06 % ***** 11th-order Polynomial ***** a 2 = -0.4281311E-05 a 3 = -0.5167629E+01 a 4 = -0.7974500E-03 a 5 = 0.2554541E+01 a 6 = -0.1501611E-01 a 7 = -0.5656904E+00 a 8 = -0.4955267E-01 a 9 = 0.1296181E+00 a10 = -0.2766239E-01 a11 = 0.6006860E-03 Errors for 11th-order polynomial Function error = 0.2263193E-07 % Derivative error = 0.2253144E-06 %
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
- Result plots
Variation of errors
Plot of approximate function for different orders of polynomial
10-8
10-6
10-4
10-2
100
2 3 4 5 6 7 8 9 10 11
Function
Derivative
Erro
r (%
)
Order of polynomial
0
0.2
0.4
0.6
0.8
1
0 0.2 0.4 0.6 0.8 1
Exact
2nd order
4th order
F(X
)
X-Coordinate
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Plot of derivative approximate function for different orders of polynomial
Homework 2
1. Derive RRΠ for the following ODE for a beam subject to axial force P (positive for com-
pression) as well as a traverse load q. Assume homogeneous displacement BCs.
02
2
4
4
=−+ qdx
wdPdx
wdEI
2. On what condition does the principle of minimum potential energy hold for Prob. 1 ?
Discuss the physical and the mathematical meanings of the condition.
3. Approximate solutions for a fixed-fixed end beam with a uniform load by polynomials
with one unknown and two unknowns. No axial force is applied. Compare your solutions including displacement, rotation, moment and shear force to exact solution and discuss.
-4
-3
-2
-1
0
1
2
3
4
0 0.2 0.4 0.6 0.8 1
Exact
2nd order
4th order
F'(X
)
X-Coordinate
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
3.2. Problems with Traction Boundary Conditions Problem Definition
- Differential Equation
lxfdx
ud<<=+ 0 02
2
- Boundary Conditions
at or 0 0 Tdxduux === and at or 0 T
dxduulx ===
Error Minimization: Error Estimator
))((21
))((21))((
21))((
21
))((21))((
21
0
000
002
2
LSl hh
lhh
l hhlhh
lhh
l hhR
dxdx
dudxdu
dxdu
dxdu
dxdu
dxduuudx
dxdu
dxdu
dxdu
dxdu
dxdu
dxduuu
dxdu
dxduuudxf
dxuduu
Π=−−=
−−+−−−−−=
−−++−=Π
∫
∫
∫
Energy Functional – Total Potential Energy
RR
lhl
hl hhl
l
lhhh
h
l
o
hhlhh
lh
lh
lhlh
lhh
lh
hh
h
lhh
l hhR
C
Tufdxudxdx
dudx
duTuufdx
dxduu
dxduu
dxduu
dxduu
dxdx
dudx
dudx
duudxudx
ududxdu
dxduudxfuuf
dxdu
dxduuudxfu
dxuduuf
dxudu
dxdu
dxduuudxf
dxuduu
Π+=
−−++=
+−−+
+−+−+−=
−−+−−+=
−−++−=Π
∫∫∫
∫∫∫
∫
∫
000
00
0
002
2
000
002
2
2
2
002
2
21)(
21
)(21
}{21)(
21
))((21)(
21
))((21))((
21
Minimization Problems RRLSR Π↔Π⇔Π Min Min Min w.r.t. hhu υ∈
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
RRΠMin : Rayleigh-Ritz Method or Principle of Minimum Potential Energy
- 1st Order Necessary Condition of Minimization Problem
miFaKTgfdxgadxdxdg
dxdg
Tgadadfdxga
daddx
dxdg
adxdg
adad
Tdadufdx
dadudx
dxdu
dxdu
dad
a
m
ikiki
lk
l
k
m
ii
lik
lm
iii
k
l m
iii
k
l m
i
ii
m
i
ii
k
l
k
hl
k
hl hh
kk
RR
,1for 0
)())((
)(
10
01 0
010 10 11
000
==−=−−=
−−=
−−=∂Π∂
∑∫∑∫
∑∫ ∑∫ ∑∑
∫∫
==
====
FKa =
0=Πδ RR : Variational Principle or Principle of Virtual Work
0)()( 0)(
)(
)21(
1 1
1 00
1 0
00
00
00
=−δ→=−δ=
−−δ=
δ−δ−δ
=−−δ=Πδ
∑ ∑
∑ ∫∑∫
∫∫∫∫
= =
= =
FKaa Tm
k
m
ikikik
m
k
ll
kk
m
ii
lik
k
llhh
l hhlhl
hl hh
RR
FaKa
Tgfdxgadxdxdg
dxdg
a
Tufdxudxdx
dudxudTufdxudx
dxdu
dxdu
Absolute Minimum Property of Total Potential Energy by the Exact Solution eh uuu −=
)for only holdssign equality (The )(21
21
21
)()(21
21
21
21
21
21
)()()()(21
21
0
2
00
0 0
002
2
0 00
0
000
2
2
00 00
0
0000 0
00
000
000
Cudxdx
du
dxdx
dudx
duTuufdxdxdxdu
dxdu
dxduTudxf
dxududx
dxdu
dxduTuufdxdx
dxdu
dxdu
Tufdxudxdx
ududxduudx
dxdu
dxduTuufdxdx
dxdu
dxdu
Tufdxudxdxdu
dxdudx
dxdu
dxduTuufdxdx
dxdu
dxdu
Tuufdxuudxdx
uuddx
uud
Tufdxudxdx
dudx
du
eEl e
E
l eell l
le
le
l l eell
lel
el
el
el l eel
l
lel
el el l eel
l
lel
el ee
lhl
hl hh
h
=Π≥+Π=
+−−=
−++++−−=
+++−+−−=
++−+−−=
−−−−−−
=
−−=Π
∫
∫∫ ∫
∫∫ ∫∫
∫∫∫ ∫∫
∫∫∫ ∫∫
∫∫
∫∫
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Weighted Residual Method
FKa =→==++−=
++−=
−++−=
−++=−+=π
∫∑∫
∫∫
∫∫
∫∫
=
,,1for 0
)(
)()()(
00
1 0
000
0000
002
2
00
mkTgfdxgadxdxdg
dxdg
Tgfdxgdxdx
dudx
dg
dxdu
dxdugfdxgdx
dxdu
dxdg
dxdug
dxdu
dxdugdxf
dxudg
dxdu
dxdugdxRg
ll
kki
m
i
lik
lk
l
k
l hk
lh
k
l
k
l hk
lh
k
lh
k
l h
k
lh
k
l
Ekk
Weighted Residual vs. Variational Principle
0
)(
)(
possible allfor 0 ,,1for 0
000
10
00
10
00
1
=Πδ=δ+δ+δ
−=
δ+δ+δ
−=
++−δ
δ=δπ⇔==π
∫∫
∑ ∫∫
∑ ∫∫
∑
=
=
=
RRlhl
hl hh
m
k
lkk
l
kk
l hkk
m
k
lk
l
k
l hk
k
ii
m
iik
Tufdxudxdx
dudxud
Tgafdxgadxdx
dudx
gad
Tgfdxgdxdx
dudx
dga
aamk
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
3.3. Integrability Condition - Regularity (continuity) Requirement
- Integration of functions with discontinuity dxxufl
l∫−
))(( ??
Original function u Function with transition zone u
dxxufdxxufl
l
l
l∫∫−
→ε−
= ))((lim))((0
where ε≤≤ε−
≤≤ε
++
ε−
ε−≤≤−
=ε−ε+ε−ε+
x
lxu
uuxuuxlu
u for
for 22
for
, )(ε=ε+ uu , )( ε−=ε− uu
- Can we integrate dxul
l∫−
on what condition ?
∫∫∫∫ε
ε
ε−
ε−
−−
++=l
l
l
l
dxudxudxudxu
ε+=+
+ε
−= ε−ε+
ε
ε−
ε−ε+ε−ε+ε
ε−∫∫ )()
22( uudxuuxuudxu
))((lim)(lim00
ε+++=++= ε−ε+
ε
ε−
−→ε
ε
ε
ε−
ε−
−→ε
−∫∫∫∫∫∫ uudxudxudxudxudxuudxl
l
l
l
l
l
The last integral vanishes as far as ε−u and ε−u are finite, and the integral becomes
∫∫∫ +=−−
l
l
l
l
udxudxudx0
0
u-
u+
u-ε
u+ε
2ε
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
- Can we integrate dxul
l∫−
2 on what condition ?
∫∫∫∫ε
ε
ε−
ε−
−−
++=l
l
l
l
dxudxudxudxu 2222
2)(
6)()
22( 2222 ε
++ε
−=+
+ε
−= ε−ε+ε−ε+
ε
ε−
ε−ε+ε−ε+ε
ε−∫∫ uuuudxuuxuudxu
)2
)(6
)((lim
)(lim
2222
0
222
0
2
ε++
ε−++=
++=
ε−ε+ε−ε+
ε
ε−
−→ε
ε
ε
ε−
ε−
−→ε
−
∫∫
∫∫∫∫
uuuudxudxu
dxudxudxudxu
l
l
l
l
l
l
The last integral vanishes as far as ε−u and ε−u are finite, and the integral becomes
∫∫∫ +=−−
l
l
l
l
dxudxudxu0
20
22
- Can we integrate dxdxdul
l∫−
2)( ??
dxdxuduudx
dxud
dxdxuddxuudx
dxud
dxdxuddx
dxuddx
dxuddx
dxud
l
l
l
l
l
l
l
l
22
2
222
2222
)(2
)()(
)()2
()(
)()()()(
∫∫
∫∫∫
∫∫∫∫
ε
ε−ε+ε−
ε
ε
ε−
ε−ε+ε−
−
ε
ε
ε−
ε−
−−
+ε
−+=
+ε
−+=
++=
ε−
++==ε−ε+
→ε−−
→ε−
∫∫∫∫ 2)(
lim)()()(lim)(2
0
2
0
20
2
0
2 uudxdxdudx
dxdudx
dxuddx
dxdu l
l
l
l
l
l
Therefore, the given definite integral has a finite value if and only if u is continuous.
)(lim)(lim00
ε−=ε+→ε→ε
uu
From the physical point of view, the aforementioned continuity condition represents the compatibility condition, which states that the displacement field in a continuum should be uniquely determined, ie, defined by single valued functions.
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
- Can we integrate dxdxduu
l
l∫−
on what condition ?
∫∫∫∫ε
ε
ε−
ε−
−−
++=l
l
l
l
dxdxududx
dxududx
dxududx
dxudu
2)()()
2)(
22(
ε−ε+ε−ε+
ε
ε−
ε−ε+ε−ε+ε−ε+ε
ε−
+−=
ε−+
+ε
−= ∫∫
uuuudxuuuuxuudxdxudu or
2)()())()((
21 222
ε−ε+ε−ε+
ε
ε−
ε
ε−
ε
ε−
ε
ε−
+−=ε−−ε=→−= ∫∫∫
uuuuuudxudxuddxu
dxududx
dxudu
±±
−
−+−+
−
ε−ε+ε−ε+
ε
ε−
−→ε
ε
ε
ε−
ε−
−→ε
−
++=
+−++=
+−++=
++=
∫∫
∫∫
∫∫
∫∫∫∫
)(][
2)()(
)2
)()((lim
)(lim
0
00
0
0
0
uudxdxduudx
dxduu
uuuudxdxduudx
dxduu
uuuudxdxududx
dxudu
dxdxududx
dxududx
dxududx
dxduu
l
l
l
l
l
l
l
l
l
l
Therefore, the given definite integral has a unique & finite value even if u is discontinu-ous.
Homework 3
1. Derive RΠ , LSΠ , RRΠ for the following ODE with the homogeneous displacement BCs.
04
4
=− qdx
wd
2. Prove the absolute minimum property of RRΠ derived in Prob. 1 by the exact solution.
3. Approximate solutions for a cantilever beam subject to a uniform load (q) over the span and a concentrate load applied at the free end by polynomials with one unknown, two un-knowns and three unknowns. Compare your solutions including displacement, rotation, moment and shear force to exact solution and discuss.
4. Identify the integrability condition for the following integrals, and evaluate the integrals
for the identified condition.
dxdx
wdl2
02
2
)(∫ and dxdx
wddxdwl
∫0
2
2
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
3.4. The Other Side of the Principle of Virtual Work Physical Viewpoint
If a deformable body is in equilibrium under a Q-force system and remains in equilibrium while it is subjected a small virtual deformation, the external virtual work done by exter-nal Q forces acting on the body is equal to the internal virtual work of deformation done by the internal Q-stresses.
∫∫ σδε=δV
ijijS
ii dVdSQu where )(21
i
j
j
iij x
uxu
∂
δ∂+
∂δ∂
=δε
Mathematical Viewpoint - Continuous Problem
If 0)( =uA should hold for a given system, then the following statement should hold. Here, υ∈u , and the order of A should be the same as u.
υ∈δ∀=⋅δ∫ uuAu 0)( dvv
( υ : A proper Function space)
- Example : Beam problem
The governing equation for a beam problem : qdx
wdEI =4
4
The expression for virtual work :
∫∫∫ δ=δ
→=−δvvv
wqdxdxdx
wdEIdx
wddxqdx
wdEIw 2
2
2
2
4
4
0)(
If wδ is the displacement induced by the unit load applied load at xj , the expression for the principle of virtual work becomes as follows.
)()( jv
jvv
Q
xwdxxxwwqdxdxEIMM
=−δ=δ= ∫∫∫µ
where Mµ is the moment induced by the unit load applied at xj and )( jxx −δ is a
delac delta function applied at xj.
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Mathematical Viewpoint - Discrete Problem
υ∈δ∀=⋅δ=⋅δ uuuAu 0 )( )( ii Au ( υ : A proper Function space)
- Example : Truss problem
Equilibrium Equations at joints
0 , 0)(
1
)(
1=+=+ ∑∑
==
iim
j
ij
iim
j
ij YVXH for ni ,,1=
Virtual Work Expression
0))( )((1
)(
1
)(
1=δ++δ+∑ ∑∑
= ==
n
i
iiim
j
ij
iiim
j
ij vYVuXH
0))sin(- )cos((1
)(
1
)(
1=δ+θ+δ+θ−∑ ∑∑
= ==
n
i
iiim
j
ij
ij
iiim
j
ij
ij vYFuXF
∑∑ ∑∑== ==
δ+δ=θδ+θδn
i
iiiin
i
im
j
ij
ij
iim
j
ij
ij
i vYuXFvFu11
)(
1
)(
1) ()sin cos(
iY
iX
iF1
ijF
iimF )(
iY
iX
ivv θδ−δ sin)( 12
iuu θδ−δ cos)( 12
iθ iθ
)( 12 uu δ−δ
)( 12 vv δ−δ
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
∑∑∑∑∑
∑
∑
=µµ
====µ
=
=
+=δ+δ=µ
=µ
=∆
=θδ−δ+θδ−δ
=δ−δθ+δ−δθ
n
i
iiiin
i
iiiinmb
ii
iii
i
iinmb
i
inmb
i
ii
nmb
ii
iii
iii
nmb
i
iii
iiii
i
vYuXvYuXEA
lFEA
lFlF
vvuuF
vvFuuF
11111
11212
11212
) () ()()(
)sin)( cos)((
))(sin )(cos(
If µ force system consists of an unit load applied at k-th joint in arbitrary direction, then
∑=
µµ
µ=θ=θ=+
nmb
ii
iiikkkk
EAlF
vYuX1 )(
coscos uuX
3.5. Finite Element Discretization Domain Discretization
hlh
e l
h
e l
hh
lhn
e l
hn
e l
hh
lhl
hl hh
uTudxufdxdx
dudxud
Tudxufdxdx
dudxud
Tudxufdxdx
dudxud
ee
ee
δ=δ−δ−δ
=
δ−δ−δ
=
δ−δ−δ
=Πδ
∑∫∑∫
∑∫∑∫
∫∫
==
admissible allfor 00
011
000
Interpolation of Displacement Field in an Element
= +
×Leu ×R
eu 1 1 Leu
Reu
el LeN R
eN ex 1+ex
1U 1+nU 1l 2l nl 3l . . .
node 01 =x 2x 3x lxn =+1
2U 3U 4U
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
e
e
ee
eRe
e
e
ee
eLe
eRe
LeR
eLe
Re
Re
Le
Le
he
eRe
LeR
eLe
Re
Re
Le
Le
he
lxx
xxxx
Nl
xxxxxx
N
uu
NNuNuNxu
uu
NNuNuNxu
−−
−=
−−−
=
δ⋅=
δδ
=δ+δ=δ
⋅=
=+=
+
+
+
+ = , =
),()(
),()(
1
1
1
1
uN
uN
Discretized Form of Variational Statement
eRe
Le
Re
LeR
e
ReL
e
Le
he
eRe
Le
Re
LeR
e
ReL
e
Le
he
uu
dxdN
dxdN
udx
dNu
dxdN
dxud
uu
dxdN
dxdN
udx
dNu
dxdN
dxdu
uB
uB
δ⋅=
δδ
=δ+δ=δ
⋅=
=+=
),(
),(
0
))0()0()()(()()( 1
0
=δ−δ−δ=
δ−δ−δ=
δ−δ−δ−δ
=
δ−δ−δ
=Πδ
∑∑
∑ ∫∑ ∫
∑∫∑∫
∑∫∑∫
TuFuuKu
TuNuuBBu
b
ee
Tee
ee
Te
b
e l
TTee
e l
TTe
LRn
e l
The
e l
heT
he
lh
e l
h
e l
hh
ee
ee
ee
fdxdx
TulTlufdxudxdx
dudxud
Tudxufdxdx
dudxud
Compatibility Condition – Continuity Requirement 1
11 , ++− ==== eL
eRe
eRe
Le UuuUuu
UCuUCu δ=δ=
=
=
+
+ eee
n
e
e
Re
Le
e
U
UU
U
uu
, 01000010
1
1
1
Global System Equation – Global Stiffness Equation
.
UFKUU
TCFCUUCKCU
TCUFCUUCKCU
δ=−δ=
−δ−δ=
δ−δ−δ=Πδ
∑∑
∑∑
admissible allfor 0)(
)()(
T
Tb
ee
Te
Te
ee
Te
T
Tb
T
ee
Te
Te
ee
Te
T
→ FKU =
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Rayleigh-Ritz Type Interpretation of FEM
iih Ugu =
hl
hl hh
ufdxudxdx
dudxud
δδ=δ
∫∫ admissible allfor 00
ee
x
x
TTe
n
i
x
xi
ii
iii
ii
x
xn
nn
nnn
x
xn
nn
nnn
x
xn
nn
nnn
x
xi
ii
iii
x
xi
ii
iii
x
xi
ii
iii
x
xi
ii
iii
x
xi
ii
iii
x
xi
ii
iii
x
x
x
x
x
x
n
i
n
j
l
jji
i
l hh
i
i
i
i
n
n
n
n
n
n
i
i
i
i
i
i
i
i
i
i
i
i
dx
dxUdx
dgU
dxdg
dxdg
Udxdg
U
dxUdx
dgU
dxdg
dxdg
U
dxUdx
dgU
dxdg
dxdg
UdxUdx
dgU
dxdg
dxdg
U
dxUdx
dgU
dxdg
dxdg
UdxUdx
dgU
dxdg
dxdg
U
dxUdx
dgU
dxdg
dxdg
UdxUdxdg
Udx
dgdxdg
U
dxUdxdg
Udx
dgdx
dgUdxU
dxdg
Udx
dgdx
dgU
dxUdx
dgU
dxdg
dxdgUdxU
dxdgU
dxdg
dxdgU
dxUdx
dgUdxdg
dxdgU
dxUdx
dgdxdg
Udxdx
dudxud
UBBU∑ ∫
∑ ∫
∫
∫∫
∫∫
∫∫
∫∫
∫∫
∫
∑ ∑∫∫
+
+
+
+
−
+
+
+
+
−
−
−
−
δ=
+δ+δ=
+δ
++δ++δ
++δ++δ
++δ++δ
++δ++δ
++δ++δ
++δ
=δ=δ
=+
+++
+++
+
++
−−
++
+++
++++
+
++
−−
−−−
−−−
−−−
−
+
=
+
=
1
1
1
1
1
2
1
1
1
1
1
1
2
3
2
2
1
2
1
11
111
111
1
11
11
22
111
1111
1
11
11
111
111
221
1
33
222
222
112
2
22
111
1
1
1
1
1 00
))((
)(
)( )(
)( )(
)( )(
)()(
)( )(
)(
gi gi+1
iU 1+iU
gi-1
1−iU
g1 gn+1
1U 1+nU
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Finite Element Procedure 1. Governing equations in the domain, boundary conditions on the boundary.
2. Derive weak form of the G.E. and B.C. by the variational principle or equivalent.
3. Descretize the given domain and boundary with finite elements.
S , 2121 mn SSSVVVV ∪∪∪=∪∪∪=
4. Assume the displacement field by shape functions and nodal values within an element. eee UNu =
5. Calculate the element stiffness matrix and assemble it according to the compatibility.
dVel
e ∫= DBBK T , ∑=e
eKK
6. Calculate the equivalent nodal force and assemble it according to the compatibility.
lT
l
Te dVfe
0TNNF += ∫ , ∑=e
eFF
7. Apply the displacement boundary conditions and solve the stiffness equation.
8. Calculate strain, stress and reaction force. Example with Three Elements and Four Nodes
- Shape Function Matrix
e
e
ee
eRe
e
e
ee
eLe
eeRe
LeR
eLe
Re
Re
Le
Le
he
lxx
xxxx
Nl
xxxxxx
N
uu
NNuNuNxu
−−
−=
−−−
=
⋅=
=+=
+
+
+
+ = , =
),()(
1
1
1
1
uN
]1 , 1[1),( −==e
Re
Le
e ldxdN
dxdN
B
- Element Stiffness Matrix
−
−=−
−
= ∫ 11111]1 , 1[1
111][
eeV ee l
dxll
e
K
1U 1l 2l 3l
01 =x 2x 3x lx =4
2U 3U 4U
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
- Compatibility Matrix
00100001
1
4
3
2
1
1
11 UCu =
=
=
UUUU
uu
R
L
, 01000010
2
4
3
2
1
2
22 UCu =
=
=
UUUU
uu
R
L
10000100
3
4
3
2
1
3
33 UCu =
=
=
UUUU
uu
R
L
- Global Stiffness Matrix
−
−+−
−+−
−
=
−−
+
−−
+
−
−
=
−
−
+
−
−
+
−
−
=
++== ∑
33
3322
2211
11
321
3
21
333222111
1100
11110
01111
0011
1100110000000000
1
0000011001100000
1
0000000000110011
1
10000100
1111
10010000
1
01000010
1111
00100100
1 00100001
1111
00001001
1
ll
llll
llll
ll
lll
l
ll
TTT
eee
Te CKCCKCCKCCKCK
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
- Force Term
=⋅
−−
= ∫+
+
11
211 1
1 ex
x e
e
ee
ldx
xxxx
l
e
e
F
+
+=
+
+
=
++== ∑
2
22
22
2
11
10010000
211
00100100
2
11
00001001
2
3
32
21
1
321
332211
l
ll
ll
l
lll
TTT
ee
Te FCFCFCFCF
- System Equation
+
+=
−
−+−
−+−
−
2
22
22
2
1100
11110
01111
0011
3
32
21
1
4
3
2
1
33
3322
2211
11
l
ll
ll
l
U
U
U
U
ll
llll
llll
ll
- Case 1 : 0)1( , 0)0( 31
321 =====dx
duulll
=
−−−
−→
+
+=
−
−+−
−+−
−
5.011
91
110121012
2
22
22
2
1100
11110
01111
0011
4
3
2
3
32
21
1
4
3
2
1
33
3322
2211
11
UUU
l
ll
ll
l
U
U
U
U
ll
llll
llll
ll
189 ,
188 ,
185
432 === UUU
Displace-
Strain (or Stress)
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
- Case 2 : 0)1( , 0)0( 31
321 ===== uulll
=
−
−→
+
+=
−
−+−
−+−
−
11
91
2112
2
22
22
2
1100
11110
01111
0011
3
2
3
32
21
1
4
3
2
1
33
3322
2211
11
UU
l
ll
ll
l
U
U
U
U
ll
llll
llll
ll
91 ,
91
32 == UU
3.6. Finite Difference Discretization Differential Equation
0 0 22
2
=+→=+ ii fuDf
dxud where D2 is a 2nd-order finite difference operator.
Displace-
Strain (or Stress)
1+ix ix 1−ix
iU
1+iU
1−iU
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Finite Difference Operator (Central Difference)
- Suppose u is approximated by a 2nd-order parabola ,ie, cbxaxu ++≈ 2
)1)11(1(11
11
1121
12
11
+−
−−−
+
−−−
++−+
=→
++=
=+−=
ii
iii
iiii
iii
i
iii
Ul
Ull
Ulll
acblalU
cUcblalU
)1)11(1(22 11
111
22
2
+−
−−−=
++−+
==≈ ii
iii
iiii
ixx
Ul
Ull
Ulll
auDdx
ud
i
Finite Difference Equations for interior nodes
iii
ii
iii
ii
fll
Ul
Ull
Ul 2
1)11(1 11
11
1
+=−++− −
+−
−−
Finite Difference Equations for Boundary Nodes with Displacement BCs
In case that a displacement BC is specified at a boundary nodes, the finite difference
equations need to be set up for only interior nodes. The BC can be applied to the finite
difference equation for the node adjacent to the boundary nodes.
- Example – Case 2
332
43
332
22
221
32
221
11
21)11(1
21)11(1
fll
Ul
Ull
Ul
fllUl
Ull
Ul
+=−++−
+=−++−
Finite Difference Equations for Boundary Nodes with Traction BCs
In case that a traction BC is specified at a boundary nodes, a special treatment for bound-ary condition such as ghost node is introduced.
il 1−− il
iU
1+iU
1−iU
0
1+nx nx 1−nx 2+nx
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
- The finite difference equation at the node n+1 :
11
21
11 2
1)11(1+
++
++
+
+=−++− n
nnn
nn
nnn
n
fll
Ul
Ull
Ul
- Approximation of the traction BC by the finite difference operator.
nnnn
nn
lx
UUllUU
dxdu
=→=+−
≈ ++
+
=2
1
2 0
- Substitution of the FD traction BC into FD equations for the boundary node.
11
111 2)11()11( +
++
++
+=+++− n
nnn
nnn
nn
fll
Ull
Ull
Since the location of the ghost node is arbitrary, nn ll =+1 can be assumed without
loss of generality. The final equation for the boundary node becomes
11 211
++ =+− nn
nn
nn
fl
Ul
Ul
- Example – Case 1
332
43
332
22
221
32
221
11
21)11(1
21)11(1
fll
Ul
Ull
Ul
fllUl
Ull
Ul
+=−++−
+=−++−
43
43
33 2
11 fl
Ul
Ul
=+−
Homework 4
1. Using the continuity requirements for beam problems (4th-order ODE) considered in
homework 2, propose suitable interpolation (shape) functions for a beam element, and de-
rive the element stiffness matrix of the beam element.
2. Derive a FDM equation for a cantilever beam subject to a concentrated load at the free end. Propose FDM equations for all boundary conditions for beam problems using proper ghost nodes. Discretize the beam with 5 nodes including two boundary nodes. Do not solve the system equations.
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Chapter 4
Multidimensional Problems
– Elasticity Problems–
b
a
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
4.1. 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
Strain Energy (Refer to pp. 244 - 246 of Theory of Elasticity by Timoshenko)
dVxu
dVxu
xu
xu
xu
dV
Vij
j
i
jiijV
iji
j
j
i
i
j
j
iij
Vijij
∫
∫
∫
σ∂∂
=
σ=σ←σ∂
∂+
∂∂
=
∂
∂+
∂∂
=ε←σε=Π
21
)(21
21
)(21
21
int
Total Potential Energy
dSTudVbudVS
iiV
iiV
ijij ∫∫∫ −−σε=Π21
V, E , ν
TT =
0== uu Su
St
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
4.2. Error Minimization
Error Estimator : ∫∫ −−++σ−=ΠS
hi
hiii
hjij
V
hii
R dSTTuudVbuu ))((21))((
21
,
Least Square Error
LShlklkijkl
V
hjiji
hklklijkl
V
hjiji
hijij
V
hjiji
S
hi
hii
hi
S
hii
hijij
V
hjiji
S
hi
hiijij
hij
S
hiiij
hij
V
hjiji
S
hi
hiijij
hjij
V
hii
R
dVuuDuu
dVDuu
dVuu
dSTTuudSTTuudVuu
dSTTuudSnuudVuu
dSTTuudVuu
Π=−−=
ε−ε−=
σ−σ−=
−−+−−−σ−σ−=
−−+σ−σ−+σ−σ−−=
−−+σ−σ−=Π
∫
∫
∫
∫∫∫
∫∫∫
∫∫
)()(21
)()(21
))((21
))((21))((
21))((
21
))((21))((
21))((
21
))((21))((
21
,,,,
,,
,,
,,
,,
,,
Energy Functional – Total Potential Energy
RR
Si
hi
Vi
hi
hij
V
hij
Vij
j
i
jS
ijhi
V j
ijhi
hij
V
hij
Vij
j
i
Vij
j
hih
ijV j
hi
Vij
j
i
hij
Vij
j
hi
Vij
j
hi
Vij
j
i
hij
Vij
j
hi
V l
kijkl
j
hi
Vij
j
i
hij
Vij
j
hi
l
hk
ijklV j
i
Vij
j
i
hij
Vij
j
hih
ijV j
i
Vij
j
i
hij
Vij
j
hi
j
iLS
C
dSTudVbudVdVxu
dSnudVx
udVdVxu
dVxu
dVxu
dVxu
dVxu
dVxu
dVxu
dVxu
dVxu
Dxu
dVxu
dVxu
dVxu
Dxu
dVxu
dVxu
dVxu
dVxu
dVxu
xu
t
Π+=
−−σε+σ∂∂
=
σ−∂
∂σ+σε+σ
∂∂
=
σ∂∂
−σ∂∂
+σ∂∂
=
σ−σ∂∂
−σ∂∂
−σ∂∂
=
σ−σ∂∂
−∂∂
∂∂
−σ∂∂
=
σ−σ∂∂
−∂∂
∂∂
−σ∂∂
=
σ−σ∂∂
−σ∂∂
−σ∂∂
=
σ−σ∂∂
−∂∂
=Π
∫∫∫∫
∫∫∫∫
∫∫∫
∫∫∫
∫∫∫
∫∫∫
∫∫∫
∫
21
21
21
21
21
21
)(21
21
21
)(21
21
21
)(21
21
21
)(21
21
21
))((21
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Minimization Problems RRLSR Π↔Π⇔Π Min Min Min w.r.t. h
ihiu υ∈
RRΠMin : Rayleigh-Ritz Method or Principle of Minimum Potential Energy
Find hh υ∈u such that minimize RRΠ
where } , on 0|{ ∞<=≡∈ ∫V l
kijkl
j
iu dV
dxdu
Ddxdu
Suuu υ
- υ≡υh : The exact solution. - υ⊂υh : An approximate solution.
0=Πδ RR : Variational Principle or Principle of Virtual Work
0
)(21
)(21
)21(
=δ−δ−σδε=
δ−δ−εδε=
δ−δ−δεε+εδε=
δ−δ−δσε+σδε=
−−σεδ=Πδ
∫∫∫
∫∫∫
∫∫∫
∫∫∫
∫∫∫
dSTudVbudV
dSTudVbudVD
dSTudVbudVDD
dSTudVbudV
dSTudVbudV
t
t
t
t
t
Si
hi
Vi
hi
V
hij
hij
Si
hi
Vi
hi
V
hklijkl
hij
Si
hi
Vi
hi
V
hklijkl
hij
hklijkl
hij
Si
hi
Vi
hi
V
hij
hij
hij
hij
Si
hi
Vi
hi
hij
V
hij
RR
Absolute Minimum Property of the Total Potential Energy
eii
hi uuu −=
)(
21
21
)()(21
21
21
21
)()()()(21
21
dSTudVbudVxuD
xu
dVxuD
xudSTudVbudV
xuD
xu
dSTuudVbuudVxuD
xu
dVxuD
xudV
xuD
xudV
xuD
xu
dSTuudVbuudVx
uuDx
uu
dSTudVbudV
t
t
t
t
t
Si
ei
Vi
ei
l
k
Vijkl
j
ei
l
ek
Vijkl
j
ei
Sii
Vii
l
k
Vijkl
j
i
Si
eii
Vi
eii
l
ek
Vijkl
j
ei
l
ek
Vijkl
j
i
l
k
Vijkl
j
ei
l
k
Vijkl
j
i
Si
eii
Vi
eii
l
ekk
Vijkl
j
eii
Si
hi
Vi
hi
hij
V
hij
h
∫∫∫
∫∫∫∫
∫∫∫
∫∫∫
∫∫∫
∫∫∫
−−∂∂
∂∂
−
∂∂
∂∂
+−−∂∂
∂∂
=
−−−−∂∂
∂∂
+
∂∂
∂∂
−∂∂
∂∂
−∂∂
∂∂
=
−−−−∂
−∂∂
−∂=
−−σε=Π
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
dVxuD
xu
dVbudVxuD
xu
dVbudVudVxuD
xu
dSTudVbudSTudVudVxuD
xu
dSTudVbudSnudVudVxuD
xu
dSTudVbudVxudV
xuD
xu
l
ek
Vijkl
j
eiE
Vijij
ei
l
ek
Vijkl
j
eiE
Vi
ei
Vjij
ei
l
ek
Vijkl
j
eiE
Si
ei
Vi
ei
Si
ei
Vjij
ei
l
ek
Vijkl
j
eiE
Si
ei
Vi
eijij
ei
Vjij
ei
l
ek
Vijkl
j
eiE
Si
ei
Vi
ei
Vij
j
ei
l
ek
Vijkl
j
eiEh
tt
t
t
∂∂
∂∂
+Π=
+σ+∂∂
∂∂
+Π=
+σ+∂∂
∂∂
+Π=
−−+σ−−∂∂
∂∂
+Π=
−−σ+σ−−∂∂
∂∂
+Π=
−−σ∂∂
−∂∂
∂∂
+Π=Π
∫
∫∫
∫∫∫
∫∫∫∫∫
∫∫∫∫∫
∫∫∫∫
Γ
21
)(21
)(21
)(21
)(21
)(21
,
,
,
,
Since Dijkl is positive definite, ∂∂
∂∂
ux
D ux
ie
jijkl
ke
l
> 0 0 ≡/∂∂
∀j
ei
xu
and
∂∂
∂∂
ux
D ux
dVie
jijkl
V
ke
l∫ = 0 if and only if 0≡
∂∂
j
ei
xu
. Therefore
.)?? =for only holdssign equality The ( +Π≥Π ihi
Eh uu
4.3. 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
iiS
iiV
iil
k
Vijkl
j
i
l
i
Vijkl
j
i
Sii
Vii
l
k
Vijkl
j
i
Siii
Viii
l
i
Vijkl
j
i
l
k
Vijkl
j
i
l
k
Vijkl
j
i
Siii
Viii
l
kk
Vijkl
j
ii
iiRR
vvdSTvdVbvdVxuD
xvg
dVxvD
xvdSTvdVbvdV
xuD
xvg
dSTvudVbvu
dVxvD
xvdV
xuD
xvdV
xuD
xu
dSTvudVbvudVx
vuDx
vuvug
t
t
t
t
υ∈∀=−−∂∂
∂∂
=′
∂∂
∂∂
α+−−∂∂
∂∂
=α′
α+−α+−
∂∂
∂∂
α+∂∂
∂∂
α+∂∂
∂∂
=
α+−α+−∂
α+∂∂
α+∂=
α+Π≡α
∫∫∫
∫∫∫∫
∫∫
∫∫∫
∫∫∫
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
If the principle of virtual work holds, then the principle of minimum potential energy holds
because the boxed equation of the total potential energy vanishes identically. The approximate
version of the principle of virtual work is
hi
Si
hi
Vi
hi
V
hij
j
hi vdSTvdVbvdV
xv
t
admissible allfor 0=−−σ∂∂
∫∫∫
Equivalence to the PDE
- Exact form
υ∈∀=−−+σ ∫∫ iiS
iiiV
jiji vdSTTvdVbvt
0 )()( , → iiijij TTb , 0 , ==+σ
- Approximate form hh
iS
ih
ihi
Vi
hjij
hi vdSTTvdVbv
t
υ∈∀=−−+σ ∫∫ 0 )()( . → iiijij TTb , 0 , ≠≠+σ
Since υ∈∀≠−−+σ ∫∫ iS
ih
iiV
ih
jiji vdSTTvdVbvt
0)()( . .
Equivalence to the Weighted Residual Method
- Discretization : kikhikik
hi gaugav =δ= ,
kdSTTgdVbg
adSTTgadVbga
t
t
Si
hik
Vi
hjijk
ikS
ih
ikikV
ih
jijkik
allfor )()(
possiblefor )()(
.
.
∫∫
∫∫
−=+σ
→δ−δ=+σδ
Uniqueness of solution
If two solutions satisfy the principle of virtual work, then
iS
iiV
iiV
ijj
i
iS
iiV
iiV
ijj
i
vdSTvdVbvdVxv
vdSTvdVbvdVxv
t
t
admissible allfor 0
admissible allfor 0
2
1
=−−σ∂∂
=−−σ∂∂
∫∫∫
∫∫∫
By subtracting two equations,
0 admissible allfor 0)( 2121 =σ−σ→=σ−σ∂∂
∫ ijijiV
ijijj
i vdVxv
00)(2121
=∂∂
−∂∂
→=∂∂
−∂∂
l
k
l
k
l
k
l
kijkl x
uxu
xu
xu
D
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Chapter 5
Discretization
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
5.1. Rayleigh-Ritz Type Discretization
5.2. Finite Element Discretization
5.3. Finite Element Programming (Linear static case)
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
5.1. Rayleigh-Ritz Type Discretization Approximation
∑=
=+++=n
ppipninii
hi gcgcgcgcu
12211
j
pip
n
p j
pip
j
hi
xg
cxg
cxu
∂∂
=∂∂
=∂∂ ∑
=1
Principle of Minimum Potential Energy
dSTgcdVbgcdVxg
cDxg
ctS
ipipV
ipipl
qkq
Vijkl
j
pip
h ∫∫∫ −−∂∂
∂∂
=Π21Min or 0=
∂Π∂
mr
h
c for all m, r
mrfcK
dSTgdVbgdVcxg
Dxg
dSTgdVbgdVxg
cDxg
dSTgdVbgdVxgD
xg
cdVxg
cDxg
dSTgdVbg
dVxg
Dxg
cdVxg
cDxg
c
rmkprmkp
Smr
Vmrkp
V l
pmjkl
j
r
Smr
Vmr
l
pkp
Vmjkl
j
r
Smr
Vmr
l
r
Vijml
j
pip
l
qkq
Vmjkl
j
r
Sirmi
Virmi
l
q
Vrqmkijkl
j
pip
l
qkq
Vijkl
j
prpmi
mr
h
t
t
t
t
and allfor 0
21
21
21
21
=−=
−−∂∂
∂∂
=
−−∂∂
∂∂
=
−−∂∂
∂∂
+∂∂
∂∂
=
δ−δ−
∂∂
δδ∂∂
+∂∂
∂∂
δδ=∂Π∂
∫∫∫
∫∫∫
∫∫∫∫
∫∫
∫∫
Principle of Virtual Work
0=Τδ−δ−σδε ∫∫∫V
ihi
V
hi
hi
V
hij
hij dVudVbudV
ipS
ipV
ipkqV l
qijkl
j
pip
Sip
Vipkq
V l
qijkl
j
pip
Sipip
Vipip
V l
qkqijkl
j
pip
h
cdSTgdVbgdVcxg
Dxg
c
dSTgdVbgdVcxg
Dxg
c
dSTgcdVbgcdVxg
cDxg
c
t
t
t
δ=−−∂∂
∂∂
δ=
−−∂∂
∂∂
δ=
δ−δ−∂∂
∂∂
δ=Πδ
∫∫∫
∫∫∫
∫∫∫
allfor 0)(
)(
and allfor 0 ipfcK pikqpikq =−
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
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
dSdVdVtS
h
V
hh
V
h ∫∫∫ ⋅δ+⋅δ=⋅δ 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
)(
xgc
xgc
xgc
xgc
xgc
xgc
xgc
xgc
xgc
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
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Virtual Strain - Matrix Form
cBδ=
δδδ
δδδδδδ
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
=
δγ
δγ
δγ
δε
δε
δε
=δε
n
n
n
nn
nn
nn
n
n
n
h
h
h
h
h
h
h
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
23
13
12
33
22
11
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
23
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
dSdSdSTu
dVdVdVbu
ttt S
TT
S
Th
Si
hi
V
TT
V
Th
Vi
hi
∫∫∫
∫∫∫
δ=δ=δ
δ=δ=δ
TNcTu
bNcbu
fKccTNbNcDBBc T =δ=−−δ ∫∫∫ or admissible allfor 0)( dSdVdV
tS
T
V
T
V
T
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Example Displacement Field
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
(22223
2
E
lxxlxx
ylx
lxx
ylxx
DBN
Stiffness Matrix
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
x, u
y, v
l
2h
hPTy 2
=
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Load Term
−=
−
−−= ∫ ∫
− Pldxdy
hl
ylh
h
0
32P0
30
02
3
3
2
1
0
f
System Equation
−=
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
bPEI
PEh
a2
22
3
2 1))(1
1351( , 11
23 ν−
−+−=
ν−=
ν−=
Displacement and Stress
)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
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Home Work 5 1. The displacement field of a thin plate is expressed as follows.
),( , ),,( , ),,( yxwwzywzyxvz
xwzyxu =
∂∂
−=∂∂
−=
1) Drive the strain component using the given displacement field.
2) Assume 033 =σ , and the plate is under plane stress condition, drive stress components.
3) Drive expressions for the total potential energy and the virtual work in case the plate is
subject to a traverse load a on the upper surface. (hint: perform analytic integration in the
direction of the thickness)
4) Drive the governing equation and all possible boundary conditions on the four edges.
2. Analyze the structure shown in the following figure under the plane stress condition by
Rayleigh-Ritz method
x
z
y
t
l
2h
q
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
5.2. Finite Element Discretization
Domain Discretization
, 2121 mnVVVV Γ∪∪Γ∪Γ=Γ∪∪∪=
∑ ∫∑ ∫∑ ∫
∫∫∫
δ+δ=δ
Γδ+δ=δΓ
e S
Th
e V
Th
e V
hΤh
Th
V
Th
V
hΤh
dSdVdV
ddVdV
et
ee
t
bubu
Tubu
σε
σε
The displacement field in an element
ee
e
n
n
n
n
n
n
e
e
e
e
uuu
uuuuuu
NN
N
NN
N
NN
N
uuu
UNu =
=
=
3
2
1
32
22
12
31
21
11
2
2
2
1
1
1
3
2
1
000000
00
0000
00
0000
where Ni j ij( )X = δ .
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
The virtual strain field in an element
e
e
n
n
n
nn
nn
nn
n
n
n
ii
ii
ii
ii
ii
ii
ii
ii
ii
ee
ee
ee
e
e
e
e
e
e
e
e
e
e
uuu
uuuuuu
xN
xN
xN
xN
xN
xN
xN
xN
xN
xN
xN
xN
xN
xN
xN
xN
xN
xN
xN
xN
xN
xN
xN
xN
xN
xN
xN
xNu
xNu
xNu
xNu
xNu
xNu
xNu
xNu
xNu
xu
xu
xu
xu
xu
xu
xuxuxu
UBδ=
δδδ
δδδδδδ
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
=
∂∂
δ+∂∂
δ
∂∂
δ+∂∂
δ
∂∂
δ+∂∂
δ
∂∂
δ
∂∂
δ
∂∂
δ
=
∂∂δ
+∂
∂δ∂
∂δ+
∂∂δ
∂∂δ
+∂
∂δ∂
∂δ∂
∂δ∂
∂δ
=
δγδγδγδεδεδε
=δ
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
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
0
0
0
00
00
00
0
0
0
00
00
00
0
0
0
00
00
00
ε
The stress field in an element
ee
e
e
e
e
e
e
e
e
e
e
e
e
e E
DBUD ==
γγγεεε
ν−ν−
ν−ν−
ν−ν−
ν−ν
ν−ν
ν−ν
ν−ν
ν−ν
ν−ν
ν−ν+ν−
=
σσσσσσ
=
ε
σ
23
13
12
33
22
11
23
13
12
33
22
11
)1(22100000
0)1(2
210000
00)1(2
21000
000111
0001
11
00011
1
)21)(1()1(
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Stiffness Equation
0
0
=δ−δ
=δ−δ−δ
∑∑
∑ ∫∑ ∫∑ ∫
e
eTee
e
eTe
e S
TTe
e V
TTe
e
e
V
TTe dSdVdVet
ee
fUUKU
TNUbNUUDBBU
Compatibility Conditions
0fKU
UfTUUTKTU
UTUUTU
=−
δ=δ−δ
δ=δ=
∑∑ admissible allfor 0
,
e
eTeT
e
eeTeT
eeee
Finite Element Procedure
1. Governing equations in the domain, boundary conditions on the boundary.
2. Derive weak form of the G.E. and B.C. by the variational principle or equivalent.
3. Descretize the given domain and boundary with finite elements.
S , 2121 nn SSSVVVV ∪∪∪=∪∪∪=
4. Assume the displacement field by shape functions and nodal values within an element. eee UNu =
5. Calculate the element stiffness matrix and assemble it according to the computability.
dVeV
Te ∫= DBBK , ∑=e
eKK
6. Calculate the equivalent nodal force and assemble it according to the computability.
dSdVet
e S
T
V
Te ∫∫ += TNbNf , ∑=e
eff
7. Apply the displacement boundary conditions and solve the stiffness equation.
8. Calculate strain, stress and reaction force.
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
5.3. Finite Element Programming (Linear static case) Program Structure Data Structure
- Control Data : # of nodes, # of elements, # of support, # of forces applied at nodes...
- Geometry Data : Nodal Coordinates Element information (Type, Material Properties, Incidencies)
- Material Properties - Boundary Condition
Traction boundary condition including forces applied at nodes Displacement boundary condition
- Miscellaneous options
INPUT
Preprocessing
Calculation of Element stiffness matrix and Load Vector
Assembling E.S.M and E.L.V.
Solve Global SM
Postprocessing
Loop over all elements
Global Stiffness Matrix
and Load Vector
- Assemble Nodal Load Vector - Cal. of Destination Array - Cal. of Band width
Cal. Strain & Stress Cal of Reaction Force Dsiplay
Gauss Elimination Band solver Decomposition, etc
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
55
Differential Equation & Boundary Conditions
Principle of Minimum Potential Energy
iS
iiV
iiV
ijij udSTudVbudVt
admissible allfor 21
∫∫∫ −−σε=Π
Principle of Virtual Work
iS
iiV
iiV
ijj
i udSTudVbudVxu
t
δ=δ−δ−σ∂
∂δ∫∫∫ admissible allfor 0
Weighted Residual Method (Galerkin Method)
iiS
iiiV
jiji vdSTTvdVbvt
admissible allfor 0)()( , =−++σ− ∫∫ Minimization of Error
iijV
eij
j
i
j
ei
error vdVxv
xu
admissible allfor ))((21
σ−σ∂∂
−∂∂
=Π ∫
Principle of Minimum Potential Energy hi
Si
hi
Vi
hi
V
hij
hij udSTudVbudV
t
admissible allfor 21
∫∫∫ −−σε=Π
Principle of Virtual Work hi
Si
hi
Vi
hi
V
hij
j
hi udSTudVbudV
xu
t
δ=δ−δ−σ∂
∂δ∫∫∫ admissible allfor 0
Principle of Minimum Potential Energy
0- admissible allfor -21
admissible allfor )(21
=⋅→⋅⋅⋅=
−−σε=Π ∑ ∫∫∫
fUKUfUUKU TT
hi
e Si
hi
Vi
hi
V
hij
hij udSTudVbudV
rt
ee
Principle of Virtual Work
0 admissible allfor 0
admissible allfor 0)(
=−⋅→δ=⋅δ−⋅⋅δ
δ=δ−δ−σ∂
∂δ∑ ∫∫∫
fUKUfUUKU TT
hi
e Si
hi
Vi
hi
V
hij
j
hi udSTudVbudV
xu
et
ee
Strong form
Weak form
Exact Solution
Approximate Solution Rayleigh-Ritz type Discretization
Finite Element Discretization
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
56
Chapter6
Two-Dimensional Elasticity Problems
6.1. Plane Stress
6.2. Plane Strain
6.3. Axisymmetry
x y
z
∞
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
57
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
58
6.1. Plane Stress Stress : σ σ σ33 13 23 0= = =
Strain : εν
ε ε γ γ33 11 22 13 2310 0= −
−+ = =
v( ), ,
Modified Stress-strain Relation
1212
2211222
2211233221111
)1(2
)(1
)(1
))(1
()21)(1(
)1(
γν+
=σ
ε+εν−
=σ
ε+εν−
=ε+εν−
ν+ε
ν−ν+ν−
=σ
E
vE
vEE
e
e
e
e
e
e
e
e Eεσ D=
γεε
ν−ν
ν
ν−=
σσσ
=
12
22
11
2
12
22
11
2100
0101
1
Interpolation of Displacement
ee
e
n
n
n
ne
ee
vu
vuvu
NN
NN
NN
vu
UNu =
=
=
2
2
1
1
2
2
1
1
00
0
0
00
Strain-Displacement Relation
e
e
n
n
nn
n
n
ee
e
e
e
e
e
e
vu
vuvu
xN
yN
yN
xN
xN
yN
yN
xN
xN
yN
yN
xN
xv
yu
yvx
u
BU=
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
=
∂
∂+
∂
∂∂
∂∂
∂
=
γεε
=
2
2
1
1
22
2
2
11
1
1
12
22
11
0
0
0
0
0
0
ε
Principle of Virtual work
∑ ∫∑ ∫∑ ∫
∫∫∫
⋅+⋅=⋅⋅
δ+δ+δ+δ=σδγ+σδε+σδε
e S
T
e A
T
e A
T
yh
Sx
hy
h
Ax
hhhhh
A
hh
eee
t
tdStdAtdA
tdSTvTutdAbvbutdA
TNbNUBDB
)()()( 121222221111
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
59
6.2. Plane Strain Strain : ε γ γ33 13 230 0 0= = =, ,
Stress : σ σ σ ν σ σ13 23 33 11 220= = = − + ( )
Stress-strain Relation
e
e
e
e
e
e
e
e
v
v
vvE
εσ D=
γεε
−ν−
−ν
−ν
ν−ν+−
=
σσσ
=
12
22
11
12
22
11
)1(22100
011
01
1
)21)(1()1(
Interpolation of Displacement
ee
e
n
n
n
ne
ee
vu
vuvu
NN
NN
NN
vu
UNu =
=
=
2
2
1
1
2
2
1
1
00
0
0
00
Strain-Displacement Relation
e
e
n
n
nn
n
n
ee
e
e
e
e
e
e
vu
vuvu
xN
yN
yN
xN
xN
yN
yN
xN
xN
yN
yN
xN
xv
yu
yvx
u
BU=
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
=
∂
∂+
∂
∂∂
∂∂
∂
=
γεε
=
2
2
1
1
22
2
2
11
1
1
12
22
11
0
0
0
0
0
0
ε
Principle of Virtual work
∑ ∫∑ ∫∑ ∫
∫∫∫
⋅+⋅=⋅⋅
δ+δ+δ+δ=σδγ+σδε+σδε
e S
T
e A
T
e A
T
yh
Sx
hy
h
Ax
hhhhh
A
hh
eee
t
tdStdAtdA
tdSTvTutdAbvbutdA
TNbNUBDB
)()()( 121222221111
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
60
6.3. Axisymmetry
Strain : 0 , 0 , , , , =γ=γ∂∂
+∂∂
=γ=ε∂∂
=ε∂∂
=ε θθθθ zrrzzzrr rv
zu
ru
zv
ru
Stress : σ σθ θr z= = 0
Stress-strain Relation
e
erz
e
ezz
err
erz
e
ezz
err
e
v
vv
vv
vv
vEε=
γεεε
−ν−
−ν
−ν
−ν
−ν
−ν
−ν
ν−ν+−
=
σσσσ
=θθθθ
D
)1(221000
0111
01
11
011
1
)21)(1()1(
σ
Interpolation of Displacement
ee
e
n
n
n
ne
ee
vu
vuvu
NN
NN
NN
vu
UNu =
=
=
2
2
1
1
2
2
1
1
00
0
0
00
Strain-Displacement Relation
e
e
n
nnn
n
n
n
ee
e
e
e
erz
e
ezz
err
e
vu
vuvu
rN
zNr
Nz
Nr
N
rN
zNr
Nz
Nr
N
rN
zNr
Nz
Nr
N
rv
zu
ru
zvr
u
BU=
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
=
∂
∂+
∂
∂
∂
∂∂
∂
=
γεεε
=θθ
2
2
1
1
22
2
2
2
11
1
1
1
0
0
0
0
0
0
0
0
0
ε
Principle of Virtual work
( ) ( ) ( )δε σ δε σ δε σ δγ σ π δ δ π δ δ π
π π π
θθ θθ θ θrrh
rrh
Azzh
zzh h h
rh
rh h
rA
hz
hr
S
hz
T
Ae
T
Ae
T
Se
rdA u b v b rdA u T v T rdS
rdA rdA rdS
t
e e e
+ + + = + + +
⋅ ⋅ = ⋅ + ⋅
∫ ∫ ∫
∫∑ ∫∑ ∫∑
2 2 2
2 2 2B D B U N b N T
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
61
Chapter 7
Various Types of Elements
7.1. Constant Strain Triangle (CST) Element
7.2. Isoparametric Formulation
7.3. Bilinear Isoparametric Element
7.4. Higher Order Rectangular Element - Lagrange Family
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
62
7.5. Higher Order Rectangular Element - Serendipity Family
7.6. Triangular Isoparametric Element
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
63
7.1. Constant Strain Triangle (CST) Element Displacement Interpolation
yxyxvyxyxu ee654321 ),( , ),( α+α+α=α+α+α=
33321333
23221222
13121111
),(
),(
),(
yxuyxuyxuyxu
yxuyxu
e
e
e
α+α+α==
α+α+α==
α+α+α==
→
ααα
=
3
2
1
33
22
11
3
2
1
111
yxyxyx
uuu
∆=
=
ααα
−
3
2
1
321
321
321
3
2
1
33
22
11
3
2
1
21
111
1
uuu
cccbbbaaa
uuu
yxyxyx
where a x y x y b y y c x xi j m m j i j m i m j= − = − = − , ,
333322221111
333322221111
)(21)(
21 +)(
21),(
)(21)(
21 +)(
21),(
vycxbavycxbavycxbayxv
uycxbauycxbauycxbayxu
e
e
++∆
+++∆
++∆
=
++∆
+++∆
++∆
=
Strain Components
e
e
ee
e
e
e
e
e
e
vuvuvu
bcc
b
bcc
b
bcc
b
xv
yu
yvx
u
BU=
∆=
∂
∂+
∂
∂∂
∂∂
∂
=
γεε
=
3
3
2
2
1
1
33
3
3
22
2
2
11
1
1
12
22
11
00
00
00
21
ε
Stiffness Matrix
K B D B U B D Be T T e
A
tdA te
= ⋅ ⋅ = ⋅ ⋅∫ ∆
1u
1v
2u
2v
3u
3v
( , )x y1 1
( , )x y2 2
( , )x y3 3
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
64
Homework 6
- Cantilever Beam with three load cases - Implement your own finite element program for 2-D elasticity problems using CST element. When you build your program, consider expandability so that you can easily add other types of elements to your program for next homeworks.
a) Discuss how to simulate two boundary conditions given in the Timoshenkos’s book
(equation(k) and (l) in page 44)
b) For end shear load case, solve the problem for both boundary conditions with 40
CST elements.
c) For other load cases, use one boundary condition of your choice with 40 CST
elements
d) Perform the convergence test with at least 5 different mesh layouts for the end shear
load case. Use the boundary condition of your choice.
e) Discuss local effects, St-Venant effect, Poisson effect stress concentration, etc.
Present suitable plots and tables of displacement and stress to justfy or clearfy your
discussions.
f) Comparision of your results with other solutions such as analytic solutions, one-
dimensional solutions is strongly recommended for your discussion . Assume E = 1.0,
ν = 0.3.
10
2
1 1
1
1
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
65
7.2. Isoparametric Formulation
Interpolation of Geometry
∑
∑
=
=
ηξ=ηξ++ηξ+ηξ=
ηξ=ηξ++ηξ+ηξ=
m
iiimm
m
iiimm
yNyNyNyNy
xNxNxNxNx
12211
12211
),(),(),(),(
),(),(),(),(
ee
e
m
m
m
me
ee
yx
yxyx
NN
NN
NN
yx
XNx =
=
=
2
2
1
1
2
2
1
1
00
0
0
00
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
66
Interpolation of Displacement in a Parent Element
eee
ee
vu
UNu ),( ηξ=
=
Derivatives of the Displacement Shape Functions
→
∂∂∂
∂
∂η∂
∂η∂
∂ξ∂
∂ξ∂
=
∂η∂∂ξ
∂
→
∂η∂
∂∂
+∂η∂
∂∂
=∂η∂
∂ξ∂
∂∂
+∂ξ∂
∂∂
=∂ξ
∂
yNx
N
yx
yx
N
N
yy
Nxx
NN
yy
Nxx
NN
i
i
i
i
iii
iii
NNN
N
N
N
yx
yx
yNx
N
ixi
i
i
i
i
i
η−−
−
∇=∇
∂η∂∂ξ
∂
=
∂η∂∂ξ
∂
∂η∂
∂η∂
∂ξ∂
∂ξ∂
=
∂∂∂
∂11
1
or JJ
XNJ ⋅∇=
∂η∂∂ξ
∂
∂η∂∂ξ
∂
∂η∂∂ξ
∂
=
∂η∂
∂η∂
∂ξ∂
∂ξ∂
=
∂η∂
∂η∂
∂ξ∂
∂ξ∂
= ξ
==
==
∑∑
∑∑
mm
m
m
m
ii
im
ii
i
m
ii
im
ii
i
yx
yxyx
N
N
N
N
N
N
yN
xN
yN
xN
yx
yx
22
11
2
2
1
1
11
11
m > n N N≠ : Superparametric element
m = n N N= : Isoparametric element
m < n N N≠ : Subparametric element
∫ ∫∫− −
ηξηξ⋅⋅ηξ=⋅⋅1
1
1
1
||),(),( ddJttdA T
A
T
e
BDBBDB
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
67
7.3. Bilinear Isoparametric Element
Shape functions in the parent coordinate system.
ξηα+ηα+ξα+α=ηξηξ 4321)),(),,(( yxu
)),(),,(( 8765 ξηα+ηα+ξα+α=ηξηξ yxv
4321444434214444444
4321334333213333333
4321224232212222222
4321114131211111111
)),(),,((),()),(),,((),()),(),,((),(
)),(),,((),(
α−α+α−α=ηξα+ηα+ξα+α=ηξηξ==α+α+α+α=ηξα+ηα+ξα+α=ηξηξ==α−α−α+α=ηξα+ηα+ξα+α=ηξηξ==
α+α−α−α=ηξα+ηα+ξα+α=ηξηξ==
yxuyxuuyxuyxuuyxuyxuu
yxuyxuu
αααα
−−
−−−−
=
4
3
2
1
4
3
2
1
1111111111111111
uuuu
),( , ),( 44
33
22
11
44
33
22
11 vNvNvNvNyxvuNuNuNuNyxu ee +++=+++=
N N N N1 2 3 414
1 1 14
1 1 14
1 1 14
1 1= − − = + − = + + = − +( )( ) ( )( ) ( )( ) ( )( )ξ η ξ η ξ η ξ η , , ,
ξ=−0.
ξ=0.5
η=0.5
η=−0.5
1 2
3 4
ξ
η
η=0.5
η=−0.5
ξ=0.5 ξ=−0.5
1 2
3 4
N1 N2
N4 N3
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
68
7.4. Higher Order Rectangular Element - Lagrange Family
Shape Function of m-th Order for k-th Node in One Dimension
lkm k k m
k k k k k k k m
( ) ( )( ) ( )( ) ( )( )( ) ( )( ) ( )
ξξ ξ ξ ξ ξ ξ ξ ξ ξ ξ
ξ ξ ξ ξ ξ ξ ξ ξ ξ ξ=
− − − − −− − − − −
− + +
− + +
1 2 1 1 1
1 2 1 1 1
lkm k k m
k k k k k k k m
( ) ( )( ) ( )( ) ( )( )( ) ( )( ) ( )
ηη η η η η η η η η η
η η η η η η η η η η=
− − − − −− − − − −
− + +
− + +
1 2 1 1 1
1 2 1 1 1
N N l li IJ Im
Jm= = ×( ) ( )ξ η
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
69
Q9 Element Total number of nodes in an element : (m+1)(m+1)
Total number of terms in m-th order polynomials : ( )( )m m+ +2 12
Total number of the parasitic terms : ( )m m+ 12
The Pascal Polynomials 1
2 2
3 2 2 3
1 1
1 1
ξ η
ξ ξη η
ξ ξ η ξη η
ξ ξ η ξη η
ξ ξ η ξη η
ξ η
m m m m
m m m m
m m
− −
+ +
Complete Polynomial
Parasitic Term
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
70
7.5. Higher Order Rectangular Element - Serendipity Family Q8 Element
)1)(1(21 2
5 η−ξ−=N
)1)(1(21 2
8 η−ξ−=N
)1)(1(41ˆ
1 η−ξ−=N
51 21ˆ NN −
8511 21
21ˆ NNNN −−=
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
71
7.6. Triangular Isoparametric Element
Total number of nodes on sides of a triangle element for m-th order S.F.: 3m
Total number of terms in m-th order polynomials : ( )( )m m+ +2 12
Total number of internal nodes : ( )( )m m+ +2 12
-3m = ( )( )m m− −2 12
The Pascal Polynomials
mmmm ηξηηξξ
ηξηηξξ
ηξηξ
ηξ
−− 11
3223
22
1
Area Coordinate System
α α α α α α11
22
33
1 2 3 1= = = + + =AA
AA
AA
, , ,
Shape functions
- CST Element
N N N1 1 2 2 3 3= = =α α α , ,
- LST Element
316325214
333222111
444 )1(2)1(2)1(2
αα=αα=αα=−αα=−αα=−αα=
NNNNNN
A3
A1 A2
1 2
3
α1 constant line
α2 constant line
α3 constant line
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
72
Interpolation of Geometry
∑∑==
αα=ααα=n
iii
n
iii xNxNx
121
1321 ),(~),,(
∑∑==
αα=ααα=n
iii
n
iii yNyNy
121
1321 ),(~),,(
)(]~[~00~
~00~
~00~
)( 2
2
1
1
2
2
1
1 ee
e
m
m
n
ne
ee XN
yx
yxyx
NN
NN
NN
yx
=
=
=
X
Interpolation of Displacement in a Parent Element
( ) [ ( , , )] ( ) [ ~( , )] ( )uuv
N U N Uee
ee e e e=
= =α α α α α1 2 3 1 2
Derivatives of the Displacement Shape Functions
~
~
~
~
~
~
~
~
~
~
~~~
~~~
2
11
2
1
1
22
11
22
11
2
1
222
111
∂α∂∂α∂
=
∂α∂∂α∂
∂α∂
∂α∂
∂α∂
∂α∂
=
∂∂∂
∂
→
∂∂∂
∂
∂α∂
∂α∂
∂α∂
∂α∂
=
∂α∂∂α∂
→
∂α∂
∂∂
+∂α∂
∂∂
=∂α∂
∂α∂
∂∂
+∂α∂
∂∂
=∂α∂
−
−
i
i
i
i
i
i
i
i
i
i
iii
iii
N
N
N
N
yx
yx
yNx
N
yNx
N
yx
yx
N
N
yy
Nxx
NN
yy
Nxx
NN
J
iix NN ~~
or 1
α− ∇=∇ J
XNJ ⋅∇=
∂α∂∂α∂
∂α∂∂α∂
∂α∂∂α∂
=
∂α∂
∂α∂
∂α∂
∂α∂
=
∂α∂
∂α∂
∂α∂
∂α∂
= α
==
==
∑∑
∑∑ ~
~
~
~
~
~
~
~~
~~
22
11
2
1
2
2
1
2
2
1
1
1
1 21 2
1 11 1
22
11
mm
n
n
m
ii
im
ii
i
m
ii
im
ii
i
yx
yxyx
N
N
N
N
N
N
yN
xN
yN
xN
yx
yx
Homework 7
Drive shape functions qubic serendipity element and triangle element in the parent coordinate.
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
73
Chapter8
Numerical Integration
∫∫∫∫∫∫∫∫∫∫∫∫∫∫∫∫∫∫∫∫∫∫∫∫∫
8.1 Gauss Quadrature Rule
8.2. Reduced Integration
8.3. Spurious Zero Energy mode
8.4. Selective Integration
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
74
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
75
8.1. Gauss Quadrature Rule One Dimension
f d W fi ii
n
( ) ( )ξ ξ ξ≈=−∑∫
11
1
If the given function f ( )ξ is a polynomial, it is possible to construct the quadrature rule
that yields the exact integration.
- f ( )ξ is constant: 0)( af =ξ
2 1 2)( 1
1
1 100 =→==ξξ∫ ∑
− =
W n=Waadfn
ii
- f ( )ξ is first order: ξ+=ξ 10)( aaf One point rule is good enough.
- f ( )ξ is second order: 2210)( ξ+ξ+=ξ aaaf
2 2 , 0 , 32
232)(
111
2
1
1 10
11
1
2202
=→==ξ=ξ
→+ξ+ξ=+=ξξ
∑∑∑
∫ ∑∑∑
===
− ===
nWWW
WaWaWaaadf
n
ii
n
iii
n
iii
n
ii
n
iii
n
iii
21211
1 , 0 ξ−=ξ=→=ξ∑=
WWWn
iii
W W W W Wi ii
ξ ξ ξ ξ ξ2
1
2
1 12
2 22
2 22
2 222 2
313=
∑ = + = = → =
W W W W Wii=∑ = + = = → = → =
1
2
1 2 2 2 22 2 1 0 57735 = 1 / 3 02691 89626ξ .
- f ( )ξ is third order: 33
2210)( ξ+ξ+ξ+=ξ aaaaf Two point rule is enough.
- f ( )ξ is fourth order: 44
33
2210)( ξ+ξ+ξ+ξ+=ξ aaaaaf
3 2 , 0 , 32 , 0 ,
52
232
52)(
111
2
1
3
1
4
10
11
1
22
1
33
1
44
1
1024
=→==ξ=ξ=ξ=ξ
→+ξ+ξ+ξ+ξ
=++=ξξ
∑∑∑∑∑
∑∑∑∑∑
∫
=====
=====
−
nWWWWW
WaWaWaWaWa
aaadf
n
ii
n
iii
n
iii
n
iii
n
iii
n
ii
n
iii
n
iii
n
iii
n
iii
W W W Wi ii
n
i ii
n
ξ ξ ξ ξ ξ3
1 11 3 1 3 20 0 0
= =∑ ∑= = → = = − = , , ,
W W W W Wi ii
ξ ξ ξ ξ ξ4
1
3
1 14
3 34
3 34
3 342 2
515=
∑ = + = = → =
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
76
W W W W W W
W
i ii
ξ ξ ξ ξ ξ ξ
ξ
2
1
2
1 12
3 32
3 32
3 32
3 3
3 3
2 23
13
3 5 59
0 77459 055555 55555 55555=∑ = + = = → = → =
= =
, =
66692 41483 ,
/
. .
W W W W W W Wii=∑ = + + = + = → =
1
2
1 2 3 3 2 22 2 89
= 0.88888 88888 88888
- Because of the symmetry condition, we need to decide only n unknowns for n-points
G.Q..
- We can integrate 2n-1-th polynomials exactly with n-points G.Q. Since for 2m-th or-
der polynomials we have 2m conditions for G.Q. -which means we can determine
(m+1)-point G.Q..
- Stiffness Equation
B D B B D B B D B B D BT
V
T
x
Ti
Ti i i i i
i
n
dV Adx A J d W A Je e
⋅ ⋅ = ⋅ ⋅ = ⋅ ⋅ = ⋅ ⋅∫ ∫ ∫ ∑− =
( ) ( ) ( ) ( )ξ ξ ξ ξ ξ1
1
1
N b N b N b N bT
V
T
x
Ti
Ti i i i
i
n
dV Adx A J d W A Je e
⋅ = ⋅ = ⋅ = ⋅∫ ∫ ∫ ∑− =
( ) ( ) | | ( ) ( ) | |ξ ξ ξ ξ ξ1
1
1
Two-dimensional Case – Rectangular Elements
- Quadrature rule
f d d W f d W W f WW fi ii
n
jj
m
i i ji
n
i j i ji
n
j
m
( , ) ( ) ( ) ( )ξ η ξ η ξ η η ξ η ξ η−− =− = = ==∫∫ ∑∫ ∑ ∑ ∑∑≈ = =1
1
1
1
11
1
1 1 11
- Stiffness equation
∑∑
∫ ∫∫
= =
− −
ηξ⋅⋅ηξ=
ηξηξ⋅⋅ηξ=⋅⋅
n
i
m
jijijjiijji
Tji
T
A
T
JtWW
ddJttdAe
1 1
1
1
1
1
||),(),(
||),(),(
BDB
BDBBDB
∑∫∫
∑∑∫ ∫∫
=−
= =− −
⋅ηξ=ξ⋅ηξ=⋅
⋅ηξ=ηξ⋅ηξ=⋅
n
iijijijpi
Tip
T
S
T
n
i
m
jijijijji
Tji
T
A
T
KtWdKttdS
JtWWddJttdA
e
e
1
1
1
1 1
1
1
1
1
||),(||),(
||),(||),(
TNTNTN
bNbNbN
Two-dimensional Case – Triangular Elements
∑
∫ ∫∫
=
α−
αα⋅⋅αα=
αααα⋅⋅αα=⋅⋅
n
iii
iiij
iiTi
T
A
T
JtW
ddJttdAe
12121
1
0
1
0122121
||),(),(21
||),(),(1
BDB
BDBBDB
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8.2. Reduced Integration Q8 element
27
26
254
23210
27
26
254
23210
ξη+ηξ+η+ξη+ξ+η+ξ+=
ξη+ηξ+η+ξη+ξ+η+ξ+=
bbbbbbbbvaaaaaaaau
276431
276431 22 , 22 η+ξη+η+ξ+=
ξ∂∂
η+ξη+η+ξ+=ξ∂
∂ bbbbbvaaaaau
ξη+ξ+η+ξ+=η∂
∂ξη+ξ+η+ξ+=
η∂∂
72
654272
6542 22 , 22 bbbbbvaaaaau
Terms in stiffness matrix
- From complete polynomials: 22 , , , , , 1 ηξηξηξ
- From parasitic terms: 4433223322 , , , , , , , , ξηξηηξηξξηξηηξ
Reduced Integration
Reduce the integration order by one to eliminate the effect of parasitic terms in the stiffness
matrix
Exact average shear stress
Nodal values extrapolated from Gauss point
Gauss point value by the reduced integration
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8.3. Spurious Zero Energy mode Independent Displacement Modes of a Bilinear Element
Spurious Zero Energy Mode
Hour glass mode
Rigid Body motion – zero energy mode
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Zero Energy mode of Q8 Element Zero Energy Modes of Q9 Element Near Zero Energy Modes
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80
8.4. Selective Integration
SN
EE
E
DD
D
+=
ν++
ν−ν
ν−ν
ν−ν
ν−ν
ν−ν
ν−ν
ν−ν+ν−
=
ν−ν−
ν−ν−
ν−ν−
ν−ν
ν−ν
ν−ν
ν−ν
ν−ν
ν−ν
ν−ν+ν−
=
100000010000001000000000000000000000
)1(2
000000000000000000
000111
0001
11
00011
1
)21)(1()1(
)1(22100000
0)1(2
210000
00)1(2
21000
000111
0001
11
00011
1
)21)(1()1(
nintegratio Reducednintegratio Full
)( dVdVdVdVeeee V
ST
VN
T
VSN
T
V
T ∫∫∫∫ +=+= BDBBDBBDDBDBB
Homework 8
Solve the cantilever beam problem in homework 6 for the end shear load case. Use 40 LST
elements and 20 Q8 elements. For Q8 element, try the full and the reduced integration
scheme. Compare your results including the displacement field and the stress field with those
by CST elements and analytical solution.
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Chapter 9
Convergence Criteria in the
Isoparametric Element
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The interelement Continuity condition
This condition is satisfied because the geometry and the displacement field are defined uniquely on the interelement boundary with the coordinates and displacement field on the element boundary, respectively.
The Completeness Condition φ φ φ φ ξ φ η ξ η φ
ξ η ξ η
ξ η ξ η
ξ
η
( , ) ( ) ( ) ( ) ( , , )
( ) ( ) ( ) ( , , )
( ) ( ) ( ) ( , , )
x y N a a a H
x N x a x a x a x H x
y N y a y a y a y H y
ii
i k k k k
ii
i k k k k
ii
i k k k k
= = + + +
= = + + +
= = + + +
∑
∑
∑
0 1 2
0 1 2
0 1 2
=−
−
− −
− −
= −
1 2 2
1 1
0
0
1 2 1 2
| |( ) ( )( ) ( )
( ) ( , , )
( ) ( , , )
| | ( ) ( ) ( ) ( )
Aa y a xa y a x
x a x H x
y a y H y
A a x a y a y a x
k k
k k
k k
k k
k k k k
ξ η
ξ η
where
φ φ
φ φ
φ φ φ
( , ) ( )
( )| |
( ( ) ( ) ( ) ( )) ( )| |
( ( ) ( ) ( ) ( ))
( ( ) ( ) ( ) ( ))| |
( ( ) ( )
x y a
aA
a x a y a y a x aA
a x a y a y a x
a a y a y a xA
a a x
k
kk k k k
kk k k k
k k k k k k
= −
− + − +
− −
0
10 2 0 2
20 1 0 1
1 2 1 2 1 2
− +a x a yAk k1 2( ) ( ))
| |φ H.O.T.
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83
Chapter 10
Miscellaneous Topics
10.1 Stress Evaluation, Smoothing and Loubignac Iteration
10.2. Incompatible Element - Q6 and QM 6
10.3. Static Condensation & Substructuring
10.4. Symmetry of Structure
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10.5. Constraints in Discrete Problems
10.6. Constraints in Continuous Problems
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10.1. Stress Evaluation, Smoothing and Loubignac Iteration Stress evaluation
- Stress components should be evaluated at the GP’s in each element, not at nodes.
- The stress field is not uniquely determined on inter-element boundaries.
Stress smoothing at nodes
- Continuous stress field can be obtained by extrapolating stresses at the GP’s to nodes, and averaging them.
- The bilinear shape function or the Q9 shape function may be utilized for extrapola-
tion of stress to nodes depending on the integration schemes.
- Mid-side nodes may be treated as either independent nodes or dependent nodes for the stress field
Loubignac iteration
FUKBFUDBB
UDBBBBBUDBB
FTNbNUDBB
∆=∆→−=∆
∆+=∆+==
=Γ+=
∑ ∫∑ ∫
∑ ∫∑ ∫∑ ∫∑ ∫∑ ∫
∑ ∫∑ ∫∑ ∫Γ
e
e V
T
e
e
V
T
e
e
V
T
e V
T
e V
T
e V
T
e
e
V
T
e
T
e V
T
e
e
V
T
ee
eeeee
et
ee
dVdV
dVdVdVdVdV
ddVdV
σ
σσσσ
~
~)~(
where σ~ denotes the extrapolated and averaged stress field.
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10.2. Incompatible Element - Q6
Deformed shape of Bilinear Element for pure bending
Correct deformed shape in pure bending
Behaviors of Bilinear Element for Pure Bending
- Displacement field
u u u u u u
v
= − − − + − + + + − − + =
=
14
1 1 14
1 1 14
1 1 14
1 1
0
( )( ) ( )( ) ( )( ) ( )( )ξ η ξ η ξ η ξ η ξη
- Strain & Stress field
abxuE
abyuE
abyuE
abxu
xv
yu
yu
abyu
au
xu
xyyx
xyyx
)1(2 ,
1 ,
1
, 0 , =
22 ν+=σ
ν−ν
=σν−
=σ
=∂∂
+∂∂
=γ=∂∂
=εη=∂∂
=ε
-Strain Energy - Full Integration
22
222
))1(2(13
1
))()1(2
)(1
(21
)(21
uba
abE
dydxab
xuEab
yuE
dydx
a
a
b
b
xyxyyy
a
a
b
bxxb
ν−+ν−
=
ν++
ν−=
σγ+σε+σε=Π
∫ ∫
∫ ∫
− −
− −
-Strain Energy - Selective Integration
uabE
dydxab
xuEab
yuEa
a
b
b
sb
2
222
132
))()1(2
)(1
(21
ν−=
ν++
ν−=Π ∫ ∫
− −
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Real Behaviors for Pure Bending
- Strain & Stress field
σ σ σ
ε ε γ
x y xy
x y xy
ky
kE
y kvE
y
= = =
= = − =
, ,
, ,
0 0
0
- Displacement field
u kE
xy f y v kvE
y +g x kE
x f y g x
g x kE
x f y C g x kE
x Cx C f y Cy C
xy= + = − = + ′ + ′ =
′ + = − ′ = = − + + = − +
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
, ,
, ,
12
0
12
2
21 2
γ
aub
by
bua
axCx
abuy
abvuvxy
abuu
Cab
uEkuubyax
CCCCyux
CCxxEky
EkvvCCyxy
Eku
2))(1(
2))(1(
21
21 ,
Arbitrary
,
00 021
21 ,
221
22
1
22
122
2
−+−=+−−==
=
=→=→==
==→=+−=→=
++−−=+−=
-Strain Energy – Exact Solution
Π R x xb
b
a
a
y y xy xyb
b
a
a
dydx E uyab
dydx E ba
u= + + = =−− −−∫∫ ∫∫
12
12
23
2 2( ) ( )ε σ ε σ γ σ
Ratio of strain Energy
ΠΠ
ΠΠ
b
R
bs
R
ab
v=+ −
+ =−
≈ =1
11
112
11
11 0 322ν ν ν
( ( ) ) . . , for
The effect of parasitic shear becomes disastrous as the aspect ratio of bilinear element is large.
Q6 Incompatible Element
- Shape function u N u a a
v N v a a
iB
ii
iB
ii
= + − + −
= + − + −
∑
∑
12
22
32
42
1 1
1 1
( ) ( )
( ) ( )
ξ η
ξ η
a1, a2, a3, a4 are called as the “nodeless degrees of freedom”.
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- Element Stiffnes Equation
K K
K K
u
a
f
fbb bi
ib ii
b
i
=
( )
( )
( )
( )
- Static Condensation
)()]([)(][)()])([]][[]([
))]([)((][)()()]([)]([6611
1
QQiii
bibiibibb
ibi
iii
iiib
fuKfKfuKKKKuKfKafaKuK
=→−=−
−=→=+−−
−
10.3. Static Condensation & Substructuring Static Condensation: Eliminate some DOFs prior to a main analysis.
=
r
e
r
e
rrre
eree
PP
uu
KKKK
)()( 1rereeeeeeeerer uKPKuPuKuK −=→=+ −
eeererrereererr
rrereeererrr
rererrr
PKKPuKKKKPuKPKKuK
PuKuK
11
1
)())((
)()(−−
−
−=−
=−+
=+
- From Gauss elimination point of view
−
=
− −−
eeerer
e
r
e
ereererr
eree
PKKPP
uu
KKKKKK
11 )()(0
ue, Pe ur, Pr
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Substructuring
- Substructure 1
=
111
11
111
iii
i
ii PP
uu
KKKK
11
1111
11
1111 )())(( PKKPuKKKK −− −=− iiiiiii
- Substructure 2
=
222
22
222
iii
i
ii PP
UU
KKKK
21
2222
21
2222 )())(( PKKPuKKKK −− −=− iiiiiii
- Assembling
21
22211
11121
21
22211
11121 )()())()(( PKKPKKPPuKKKKKKKK −−−− −−+=−−+ iiiiiiiiiiiii
or
21
22211
11121
22211
11121 )()())()(( PKKPKKPuKKKKKKKK −−−− −−=−−+ iiiiiiiiiiii
u1, P1 u2, P2
ui, Pi
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Partial Substructuring
- Substructure 1
=
111
11
111
iiiii
i
PP
uu
KKKK
11
1111
11
1111 )())(( PKKPuKKKK −− −=− iiiiiii
- Substructure 2
=
222
22
222
iiiii
i
PP
uu
KKKK
- Assembling
−+
=
−+ −−
11
11121
22
11
11112
2
222
)()( PKKPPP
uu
KKKKKKKK
iiiiiiiiiii
i
Pi
u1, P1 u2, P2
ui, Pi
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10.4. Symmetry of Structure Symmetry
In plane displacement 0≠
Out of plane displacement 0=
P P
C L
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Anti-Symmetry In plane displacement 0=
Out of plane displacement 0≠
P P
C L
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Symmetric structures with Non-Symmetric loading
- General loading
- Symmetric loading
- Antisymmetric loading
−=
+=
→
−
+
=
2
221
21
2
1
PPP
PPP
PP
PP
PP
A
S
A
A
S
S
P
P/2 P/2
P/2 P/2
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94
Cyclic Symmetry
1u
iu
2u
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- Structural Resistance force in a segment
=
2
1
22221
21
12111
2
1
uuu
KKKKKKKKK
FFF
i
i
iiii
i
i
- Compatibility
21 uu Γ=
Γ=
22
1
uu
I00I
0
uuu
ii
- Equilibrium
122 FFPFP
Tii
Γ+=
=
Γ
=
→
2
1
2 FFF
I00I0
PP
iTi
Γ
Γ
=
222221
21
12111
2 uu
I00I
0
KKKKKKKKK
I00I0
PP i
i
iiii
i
Ti
- Final Equation
+Γ+Γ+ΓΓ+Γ
+Γ=
+Γ+Γ+Γ
Γ
=
22221121121
21
222212
21
12111
2
uu
KKKKKKKKK
uu
KKKKKKKKK
I00I0
PP
iTT
iiT
iiii
i
i
iiii
i
Ti
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10.5. Constraints in Discrete Problems Minimization Problem
PUKUUu
TT −=Π21Min subject to 0UA =)(
=
r
f
UU
U ,
=
rrrf
frff
KKKK
K
=
r
f
PP
P
1st order Optimality condition
0=∂Π∂U
→ 0=− PKU
Usual homogenous support conditions
0=rU
ffffTfff
Tf
f
PUKPUUKUU
=→−=Π21Min
Non-homogenous support conditions (support Settlement)
rr UU =
rTrf
Tfrrr
Trrfr
Tffff
Tf
rTrf
Tfrrr
Trrfr
Tffrf
Trfff
Tf
PUPUUKUUKUUKU
PUPUUKUUKUUKUUKU
−−++=
−−+++=Π
)2(21
)(21
0MinMin =−+→Π=Π frfrffff
PUKUKUU
rfrffff UKPUK −=
or
00 =∂
Π∂→=δ
∂Π∂
=δ∂
Π∂+δ
∂Π∂
=δ∂Π∂
=Πδf
ff
rr
ff U
UU
UU
UU
UU
General Non-homogenous Linear Constraints
01
0 0)( i
ndof
jjij rur =→=−= ∑
=
rRUUA , i =1 ,…, ncon
0011
0 CCUrRURRUrURURRU +=+−=→=+= −−frffrrffrr
+
=
=
0
0C
UCI
UU
U fr
f
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97
)(21)(
][2121
0000 rT
fTfrr
Trr
Tfr
Tf
frrT
rfT
frffTf
TT
PCPUCKCCKCCKU
UCKCKCCKKU
PUKUU
+−+++
+++=
−=Π
)()(][ 00 CKCCKPCPUCKCKCCKK rrT
frrT
ffrrT
rfT
frff +−+=+++
Lagrange Multiplier
)(21Min 0,
rRUPUKUUU
−+−=Πλ
TTT λ → 0=∂Π∂U
, 0=∂Π∂λ
=
→
=−==∂Π∂
=+−=∂
∂+−=
∂Π∂
00 0)(
0)(
rPU
0RRK
rRUUA
RPKUUUAPKU
Uλ
λ
λλ TTT
)()(0)()(0
0111
01
0
1
rPRKRRKrRPRKrRURPKURPKU
−=→=−−=−
−=→=+−−−−−
−
TT
TT
λλ
λλ
011111111
01111
)))((
))()((
rRRKRKPRKRRKRKKrPRKRRKRPKU
−−−−−−−−
−−−−
+−=
−−=TTTT
TT
(
10.6. Constraints in Continuous Problems Lagrange Multiplier
Πu
Min subject to 0uA =)(
where Π is the original functional derived from the minimization principle or equivalent,
and 0uA =)( denotes an additional set of constraints, which may be defined in some
volume, on some surface, or at some points. For the simplicity of derivation, only con-
straints specified along a surface is considered in this note.
0)( 0 =−= rLuuA on S
dSdSTudVbudV
dSuu
SSi
hi
Vi
hi
hij
V
hij
Siu
t
i
∫∫∫∫
∫
−+−−=
−+Π=Π
)(21
)()(),(Min
0
0,
rLu
rLu
λ
λ
σε
λλ
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By discretizing eΛNλ = in an element and the displacement field in usual way,
qΛRUΛfUKUU
rΛULNΛfUKUU
rΛULNΛfUKUU
TTTT
S
TT
S
TTTT
e S
TTe
e
e
S
TTeTTi
ee
ee
dSdS
dSdSu
−+−=
−+−=
−+−≈Π
∫∫
∑ ∫∑ ∫
21
21
)()(21),(
0
0
ΝΝ
ΝΝλ
Therefore, the stationary condition for modified energy functional becomes
=
→
=−=∂Π∂
=+−=∂Π∂
qf
ΛU
0RRK
qRUΛ
ΛRfKUU T
T
0
0
The solution procedure from this point is exactly the same as that of discrete problems.
The last and the most important question is what kind of interpolation function has to be
employed for the Lagrange multiplier.
Penalty Function
∫ ⋅+Π=ΠS
T
udS)()(
2)()(Min uAuAuu α
)2(22
1
}()(2)(){(22
1
)()(22
1)(
000
00
CqUUKUfUKUU
rrrLUULNLUfUKUU
rLurLuu
+−+−=
+−+−=
−−+−−≈Π
∫∫∑ ∫
∫∫∫∫
TR
TTT
S
T
S
TTe
e
e
S
TTeTT
S
T
Si
hi
Vi
hi
hij
V
hij
eee
t
dSdSdS
dSdSTudVbudV
α
α
ασε
Ν)Ν
qfUKKU
u α+=α+→=∂Π∂
→Π )(0)(Min Ru
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Chapter 11
Problems with Higher
Continuity Requirement
– Beams –
∇4 = ??
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100
11.1. C1 Formulation
11.2. C0 Formuation
11.3. Timoshenko Beam
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101
11.1. C1-Formulation Governing Equation
EI d wdx
q4
4 =
Weak Form of the governing equation
( ) w EI d wdx
q dx wl 4
40
0− =∫ for all admissible
wEI d wdx
dwdx
EI d wdx
d wdx
EI d wdx
dx wqdxl l ll3
30
2
20
2
2
2
200
0− + − =∫∫
d wdx
EI d wdx
dx wqdx wV M d wdx
EI d wdx
dx wqdxl
ll l ll2
2
2
20
00
0
2
2
2
200
= + − → =∫∫ ∫∫θ
Continuity Requirement
The second derivative of the displacement field has to be piecewise-continuous for the valid
finite element formulation. Therefore, w has to be continuous up to the first derivatives.
d wdx
EI d wdx
dx d wdx
EI d wdx
dxle
l
e
2
2
2
2
2
2
2
20
= ∫∑∫
Hermitian Shape Functions
elxrer
xll
rrll
dxdwlww
dxdwww
NwNNwNw
==
=θ==θ=
θ++θ+=
, )( , , )0( where
0
4321
2
32
43
3
2
2
3
2
32
23
3
2
2
1
, 23
2 , 231
eeee
eeee
lx
lxN
lx
lxN
lx
lxxN
lx
lxN
+−=−=
+−=+−=
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102
11.2. C0-Formulation
Energy Consideration
w: total deflection, θ: rotation , dwdx
− θ : shear deformation
Π = + − − −∫ ∫ ∫12 0 0
00
( ( ) ( ) )ddx
EI ddx
dx dwdx
GA dwdx
dx wqdxl l lθ θ
θ θ
ii
iii
i NwNw θ=θ= ∑∑ ,
Tnn
e www ][ 2211 θθθ=→ ∆
eM
eni
i
i
dxdN
dxdN
dxdN
dxdN
dxd
∆∆ B==θ=θ ∑ )](0 0 0[ 21
eS
en
ni
iii
i
i Ndx
dNNdx
dNNdx
dNNwdx
dNdxdw
∆∆ B=−−−=θ−=θ− ∑∑ ] [ 22
11
enNNNw ∆]0 0 0[ 21 =
eTTeTSM
T
e
e
TeeeS
eM
e
Te
l
e
TeeS
lT
SM
lT
Me
Teeee
qdxdxGAdxEI
fKfKK
fKK
NBBBB
∆∆∆∆∆∆
∆∆∆
∆∆∆
−=−+=
−+=
−+=Π
∑∑
∫∑∫∫∑
21)(
21
)()()(21
][)()()(21
00
00
0=−=∂Π∂ efK∆∆
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Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
103
Problems –Functions Space Interlocking
ΠEI
ddx
ddx
dx GA lEI l
dwdx
dwdx
dx wqEI
dxl l l
= + − − −∫ ∫ ∫12
1
0
02
20 0
( ( )( ) )θ θθ θ
beam retangularfor )1(
62
220
tl
vEIlGA
+κ
==λ .
∞→λ As , 0)()(1
02 →θ−θ−∫ dx
dxdw
dxdw
l
l
or θ≡dxdw to keep the total potential energy
finite. This condition may be satisfied in case the entire, exact function spaces for w
and θ are used. For the finite element method, the energy expression is expressed as
∑∫∫∑ ∫ ξ−ξθ−ξ
θ−ξ
+ξξθ
ξθ
=Π
eeee
e
e
eh
dEIwqld
ddw
lddw
lll
EIlGA
ddd
dd
EIl 1
0
21
02
220
1
0
))1()1((21
To keep the energy expression finite as 0→t , there are two alternatives:
1) tl e ≈ , 2) θ≡dxdw . The first condition requires a fine mesh layout because the element
size should be almost the same as the thickness of beam. The second condition requires
the derivative of displacement field have to be exactly the same as the rotation flied,
which is very difficult to satisfy. For example, both the displacement and the rotation
field are interpolated by linear shape functions, ie, baw +ξ= , dc +ξ=θ . The second
condition becomes dca +ξ= for all 10 ≤ξ≤ , which is equivalent to 0=c and
da = . Therefore, the rotation field should be constant within an element. This condi-
tion can be satisfied only when the rotation field is constant for the entire beam because
of the continuity requirement of 0C∈θ , which allows only rigid body rotation of the
beam, and yields a meaningless solution. In case you try to analyze beams or plates,
do not use linear shape functions, which merely results in meaningless solutions.
The aforementioned difficulty may be avoided by using higher order shape functions.
In case shape functions of the same order for both fields are employed, the highest order
term of the rotation field should always vanish, and the function space for the rotation
fields is always a subspace of the displacement field. Since, however, the rotation and
the displacement field are totally independent fields as shown in the next section, the so-
lution space of the rotation field has to be defined independently, and thus the aforemen-
tioned function space constraint yields sub-optimal convergence and unfavorable results.
Be always very careful when you use C1-formulation for the analysis of beams or
plates.
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
104
11.3. Timoshenko Beam
Governing Equation
EI ddx
GA dwdx
2
2 0 0θθ+ − =( ) , GA d w
dxddx
p0
2
2( )− = −θ
Week Form
w
ll
wdxpdxd
dxwdGAwdx
dxdwGA
dxdEI υ∈δ∀υ∈δθ∀=+
θ−δ+θ−+
θδθ θ∫∫ & 0))(())(( 2
2
00
02
2
0
0)())(( 0000
02
2
0
=θ−δθ+θδθ
−θ
δθ=θ−+θ
δθ ∫∫∫ dxdxdwGAdx
dxdEI
dxd
dxdEIdx
dxdwGA
dxdEI
llll
0)()())((0
000
02
2
00
=δ+θ−δ
−θ−δ=+θ
−δ ∫∫∫ pdxwdxdxdwGA
dxwd
dxdwwGAdxp
dxd
dxwdGAw
llll
w
lllll
w
pdxwdxdwwGA
dxdEIdx
dxdwGA
dxwddx
dxdEI
dxd
υ∈δ∀υ∈δθ∀
δ+θ−δ+θ
δθ=θ−δθ−δ
+θδθ
θ
∫∫∫ &
)()()(00
00
000
By assuming homogeneous boundary conditions,
w
lll
wpdxwdxdxdwGA
dxwddx
dxdEI
dxd
υ∈δ∀υ∈δθ∀δ=θ−δθ−δ
+θδθ
θ∫∫∫ & )()(0
000
Boundary Conditions
θθ
θ= = = − =0 0 0 00 or or or EI ddx
w GA dwdx
( )
Elimination of the displacement – Beginning of Nightmare
Differentiation of the first equation and substitution of the second equation into the first
equation
03
3
=+θ p
dxdEI (Oh, My God !!!)
Unfortunately, we have an odd order differential equation, which does not have the min-imum characteristics, and thus is very difficult to solve. At this point, we have to con-sider the Petrov-Galerkin method seriously !!!
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
105
Chapter 12
Mixed Formulation
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Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
106
What is the mixed formulation??? - Definition
Stress or strain fields are treated and interpolated independently!!!
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
Weak statement.
ευ∈ευ∈∀=−−+−εε++σ ∫∫∫ iuiiS
iiijjiijV
ijiV
jiji udSTTudVuudVbut
ˆ ˆ 0 )(ˆ))(21(ˆ)(ˆ ,,,
ευ∈ευ∈∀=−−+−εε++σ ∫∫∫ iuiiS
iikllkklijklV
ijiV
jiji udSTTudVuuDdVbut
ˆ ˆ 0 )(ˆ))(21(ˆ)(ˆ ,,,
ευ∈ευ∈∀=+−εε+−−σ∂∂
∫∫∫∫ iuikllkklijklV
ijS
iiV
iiV
ijj
i udVuuDdSTudVbudVxu
t
ˆ ˆ 0))(21(ˆˆˆ
ˆ,,
ευ∈ευ∈∀=+−εε+−−ε∂∂
∫∫∫∫ iuikllkklijklV
ijS
iiV
iiV
klijklj
i udVuuDdSTudVbudVDxu
t
ˆ ˆ 0))(21(ˆˆˆ
ˆ,,
Interpolations
eee UNu = , eee EN=ε
Finite element discretization
eee
V
T
e
e
V
TTe
T
V
T
e
e
V
TTe
dVdV
ddVdV
e
te
EUUDBNEDNNE
TNbNEDNBU
ˆ and ˆ admissible allfor 0 )()ˆ(
)()ˆ(
=−
+Γ−−
∫∑ ∫
∫∫∑ ∫Γ
FEQ =T , 0=− QUME
−=
−−
→0F
EU
MQQ0 T
FQUMQ =→ −1T
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
107
Important Question
What are the admissible function spaces for the displacement field and the strain field??
Can we choose the interpolation shape functions for the displacement and the strain inde-
pendently ?? Unfortunately, the answer is “No”. In case we choose the function spac-
es arbitrarily, the solutions of the mixed formulation become very unstable, which is
caused by so called “function space interlocking”. The Babuzuka-Brezzi condition (BB
condition) states the required relationship between the individual function spaces. This
issue is out of scope for this class.
Please be very careful when you use the FEM based on the mixed formulation!!!
Possible choices of function spaces
1. 11 ˆ , ˆ HHu ii ∈ε∈
2. 21 ˆ , ˆ LHu ii ∈ε∈
3. ˆ , ˆ 22 LLu ii ∈ε∈
Which one do you like? Can you give a proper explanation for your choice?