FEM for Heat Transfer Problems (Finite Element Method) Part 3
Introduction to Finite Element Method -...
Transcript of Introduction to Finite Element Method -...
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, 2013
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
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
given 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 solutiona.
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 )(Xf where gi are the basis functions of a function set,
- 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 0ia
- Taylor series, Fourier series, etc…
Approximations
hm
iii
hn
iii gafgaf )()()()(
11
XXXX where nm
Summation Notation: Repeated indices denote summation ii
m
iii baba
1
.
Normes of Functions: A measure of a function space
A function space is said to be a normed space if to every f there associated a
nonnegative real number f , called the norm of f, in a such way that
- 0f if and only if 0f
- ff || for any real number .
- 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 V
h
L
h
LdVffffe 222
)(2
1
2
1
2
122
mkFaKfdVgadVgg
fdVgdVagg
dVgffdVa
fff
a
ki
m
iki
V
ki
m
i V
ik
V
k
V
m
iiik
V
kh
V k
hh
k
,,1for 0
)()(
11
1
Final System Equation : FKa
If the basis functions are orthogonal, K becomes diagonal matrix.
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 or by definition
ii
aa
ff
.
- ff
Fa
a
f
f
Fa
a
FFfF i
ii
i
: )(
- hfhf )(
- hffhfh )(
- , )()(dx
fda
a
f
dx
da
dx
df
adx
dfi
ii
i
-
fdxdxaa
fafdx
afdx i
ii
i
f
f
Variation of a function
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Minimization of LSE by Variational Calculus
Min l
hh dxfff0
2)(2
1)(
k
l
k
hh
lhh
lh
lhh
adxa
fffdxfff
dxffdxfff
00
0
2
0
2
)()(
)(2
1))(
2
1()(
Min kh
V
hh afdVfff possible allfor 0)()(2
1)( 2
Euler Equation
Min ka
dxxffFfk
l
allfor 0),,()(0
l
kk
l
k
l
kk
l
kk
l
k
l
kk
dxgf
F
dx
dg
f
Fg
f
F
dxgf
Fg
f
Fdx
a
f
f
F
a
f
f
Fdx
a
FdxxffF
aa
00
0000
)(
)()(),,(
In case the basis functions vanish at the boundary, then
0 allfor 0)(0
f
F
dx
d
f
Fkdxg
f
F
dx
d
f
F
a
l
kk
llll
dxff
F
dx
df
f
Ff
f
Fdxf
f
Ff
f
FdxxffFf
0000
)()(),,()(
In case the variation vanishes at the boundaries, then
0)()()(00
kk
k
l
k
l
aa
adxgf
F
dx
d
f
Ffdx
f
F
dx
d
f
Ff . Therefore,
Min 0
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Example 1
Min 2
1
2)(1)(x
x
dxyy subject to 11 )( yxy , 22 )( yxy
2/122 ))(1())(1(0
yydx
dy
ydx
d
f
F
dx
d
f
F
0))(1())(1
)(1())(1(
))(1)(2
1())(1())(1(
2/322
22/12
2/322/122/12
yyy
yyy
yyyyyyyydx
d
baxyy 0
By applying BC, 12
2112
12
120xx
yxyxx
xx
yyyy
Example 2
Min l
dxufuu0
2 ))(2
1()( subject to 0)0( u , 0)( lu
00))(2
1( 2
fuufudx
dfu
udx
df
u
F
dx
d
u
F
Homework 1
1. Approximate a cosine function xl
y
2
cos by polynomials based the minimization of
the least square errors. Use polynomials up to the 20th order. You may use a numerical integration 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 l
dxxfffFf0
),,,()(
3. Derive the governing equation and the boundary conditions for the following problem.
Min l
dxwqdx
wdw
0
22
2
))(2
1()(
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
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
udR
h
E 0 02
2
Function Residual : lxuuR hF 0 0
Error Estimator :
l h
hl
EFR dxf
dx
uduudxRR
02
2
0
))((2
1
2
1
Least Square Error
Error SquareLeast ))((2
1
))((2
1))((
2
1
))((2
1
))((2
1
0
00
02
2
2
2
02
2
LSl hh
l hhlhh
l hh
l hhR
dxdx
du
dx
du
dx
du
dx
du
dxdx
du
dx
du
dx
du
dx
du
dx
du
dx
duuu
dxdx
ud
dx
uduu
dxfdx
uduu
Energy Functional – Total Potential Energy
RRl
hl hhl
lhh
hhllhh
lh
lh
lh
hh
hl hhR
Cfdxudxdx
du
dx
duufdx
dxfudx
udu
dx
du
dx
duufdx
dx
duuu
dx
du
dx
duu
dxfudx
uduuf
dx
ududxf
dx
uduu
)2
1(
2
1
)(2
1
2
1
2
1
2
1
2
1
)(2
1))((
2
1
000
02
2
0000
02
2
2
2
02
2
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Minimization Problems
RRLSR Min Min Min w.r.t. hhu
RRMin : Rayleigh-Ritz Method or Principle of Minimum Potential Energy
- First Order Necessary Condition for Minimization Problem
FKa
mkFaK
fdxgadxdx
dg
dx
dg
fdxgdxgadx
d
dx
dg
fdxda
dudx
dx
du
da
du
dx
d
a
m
kikiki
l
k
m
ii
lik
l
k
l
i
m
ii
k
l
k
hl h
k
h
k
RR
,1for 0
)(
1
01 0
00 1
00
0 RR : Variational principle or Principle of Virtual Work
0)()( 0)(
)(
)(2
1
)(2
1)
2
1(
1 1
1 01 000
00
0000
FKaa Tm
k
m
ikikik
m
k
l
k
m
ii
lik
k
lh
l hh
lh
l hhhh
lh
l hhlh
l hhRR
FaKa
fdxgadxdx
dg
dx
dgafdxudx
dx
du
dx
ud
fdxudxdx
ud
dx
du
dx
du
dx
ud
fdxudxdx
du
dx
dufdxudx
dx
du
dx
du
Solution Space
- })( , 0)()0(|{0
2 l
dxdx
duluuuu
- h : The minimization problems yield the exact solution. - h : The minimization problems yield an approximate solution.
Properties of K
- Symmetry : ji
lij
lji
ij Kdxdx
dg
dx
dgdx
dx
dg
dx
dgK
00
- Positive Definiteness :
m
i
Tj
m
jijij
m
j
jl
im
ii
j
m
j
jl
i
m
i
il h
aKaadxdx
dg
dx
dga
dxadx
dga
dx
dgdx
dx
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 by the Exact Solution
eh uuu
)for only holdssign equality (The )(2
1
2
1
2
1
)(2
1
2
1
2
1
2
1
2
1
2
1
)()()(
2
1
2
1
0
2
00 0
02
2
0 00
002
2
00 00
000 00
0000
Cudxdx
du
dxdx
du
dx
duufdxdx
dx
du
dx
du
dxfdx
ududx
dx
du
dx
duufdxdx
dx
du
dx
du
fdxudxdx
udu
dx
duudx
dx
du
dx
duufdxdx
dx
du
dx
du
fdxudxdx
du
dx
dudx
dx
du
dx
duufdxdx
dx
du
dx
du
fdxuudxdx
uud
dx
uudfdxudx
dx
du
dx
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
Weighted Residual Method
mkdxfdx
uddxR
l h
k
l
Ekk ,,1for 0)(0
2
2
0
- if kk g : Galerkin Method (elliptic System or self-adjoint system)
FKa
,,1for 0
)(
01 0
00 100
00002
2
mkfdxgadxdx
dg
dx
dg
fdxgdxadx
dg
dx
dgfdxgdx
dx
du
dx
dg
fdxgdxdx
du
dx
dg
dx
dugdxf
dx
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 : Petrov-Galerkin Method (hyperbolic system or non-self adjoint system)
Weighted Residual Method vs. Variational Principle
ii
m
iik aamk
0 ,,1for 01
hl l
hhhl h
h
l h
i
m
ii
m
i
l h
iii
m
ii
ufdxudxdx
du
dx
uddxf
dx
udu
dxfdx
udgadxf
dx
udgaa
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 : xxu2
1
2
1 2
- First trial : 2321 xaxaau h
Applying BCs : 01 a , 32 aa )( 23 xxau h , )21(3 xa
dx
du h
3
1)21(
1
0
211 dxxK ,
6
1)(1
1
0
21 dxxxF
2
1
6
1
3
133 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
dgx
dx
dg
3
1)21(
1
0
211 dxxK ,
5
4)31(
1
0
2222 dxxK
2
1)31)(21(
1
0
22112 dxxxKK
6
1)(1
1
0
21 dxxxF ,
4
1)(1
1
0
32 dxxxF
System Equation : 0,2
1
4
16
1
5
4
2
12
1
3
1
43
4
3
aaa
a
Example 2 : xdx
ud sin22
2
, 0)1()0( uu
- Exact Solution : xu sin
- First trial : 2321 xaxaau h
Applying BCs : )( 22 xxau h , )21(2 xa
dx
du h
3
1)21(
1
0
211 dxxK ,
4
)(sin1
0
221 dxxxxF
124
3
122 aa . Therefore, )(
12 2xxu h
955.03
)5.0(
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
dgx
dx
dg
3
1)21(
1
0
211 dxxK ,
5
4)31(
1
0
2222 dxxK
2
1)31)(21(
1
0
22112 dxxxKK
4
)(sin1
0
221 dxxxxF ,
6
)(sin1
0
322 dxxxxF
System Equation : 0,12
6
4
5
4
2
12
1
3
1
32
3
2
aaa
a
????
- In general )(2
m
i
ii
h xxau
Function Error=
2/1
1
0
2
1
0
2
sin
)(sin
xdx
dxux h
Derivative Error=
2/1
1
0
22
1
0
2
cos
))(cos(
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
Err
or (
%)
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 RR 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
wdP
dx
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 0
2
2
- Boundary Conditions
at or 0 0 Tdx
duux and at or 0 T
dx
duulx
Error Minimization: Error Estimator
))((2
1
))((2
1))((
2
1))((
2
1
))((2
1))((
2
1
0
000
002
2
LSl hh
lhh
l hhlhh
lhh
l hhR
dxdx
du
dx
du
dx
du
dx
du
dx
du
dx
duuudx
dx
du
dx
du
dx
du
dx
du
dx
du
dx
duuu
dx
du
dx
duuudxf
dx
uduu
Energy Functional – Total Potential Energy
RR
lhl
hl hh
ll
lhhh
h
l
o
hhlhh
lh
lh
lhlh
lhh
lh
hh
h
lhh
l hhR
C
Tufdxudxdx
du
dx
duTuufdx
dx
duu
dx
duu
dx
duu
dx
duu
dxdx
du
dx
du
dx
duudxu
dx
udu
dx
du
dx
duudxfuuf
dx
du
dx
duuudxfu
dx
uduuf
dx
udu
dx
du
dx
duuudxf
dx
uduu
000
00
0
002
2
000
002
2
2
2
002
2
2
1)(
2
1
)(2
1
}{2
1)(
2
1
))((2
1)(
2
1
))((2
1))((
2
1
Minimization Problems
RRLSR Min Min Min w.r.t. hhu
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
RRMin : Rayleigh-Ritz Method or Principle of Minimum Potential Energy
- 1st Order Necessary Condition of Minimization Problem
miFaKTgfdxgadxdx
dg
dx
dg
Tgada
dfdxga
da
ddx
dx
dga
dx
dga
da
d
Tda
dufdx
da
dudx
dx
du
dx
du
da
d
a
m
ikiki
l
k
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
FKa
0 RR : Variational Principle or Principle of Virtual Work
0)()( 0)(
)(
)2
1(
1 1
1 00
1 0
00
00
00
FKaa Tm
k
m
ikikik
m
k
ll
kk
m
ii
lik
k
llhh
l hhlh
lh
l hhRR
FaKa
Tgfdxgadxdx
dg
dx
dga
Tufdxudxdx
du
dx
udTufdxudx
dx
du
dx
du
Absolute Minimum Property of Total Potential Energy by the Exact Solution
eh uuu
)for only holdssign equality (The )(2
1
2
1
2
1
)()(2
1
2
1
2
1
2
1
2
1
2
1
)()()()(
2
1
2
1
0
2
00
0 0
002
2
0 00
0
000
2
2
00 00
0
0000 0
00
000
000
Cudxdx
du
dxdx
du
dx
duTuufdxdx
dx
du
dx
du
dx
duTudxf
dx
ududx
dx
du
dx
duTuufdxdx
dx
du
dx
du
Tufdxudxdx
udu
dx
duudx
dx
du
dx
duTuufdxdx
dx
du
dx
du
Tufdxudxdx
du
dx
dudx
dx
du
dx
duTuufdxdx
dx
du
dx
du
Tuufdxuudxdx
uud
dx
uud
Tufdxudxdx
du
dx
du
eEl e
E
l eel
l l
le
le
l l eel
l
lel
el
el
el l ee
ll
lel
el el l ee
ll
lel
el ee
lhl
hl hh
h
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Weighted Residual Method
FKa
,,1for 0
)(
)()()(
00
1 0
000
0000
002
2
00
mkTgfdxgadxdx
dg
dx
dg
Tgfdxgdxdx
du
dx
dg
dx
du
dx
dugfdxgdx
dx
du
dx
dg
dx
dug
dx
du
dx
dugdxf
dx
udg
dx
du
dx
dugdxRg
ll
kki
m
i
lik
l
k
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
RRlhl
hl hh
m
k
l
kk
l
kk
l hkk
m
k
l
k
l
k
l hk
k
ii
m
iik
Tufdxudxdx
du
dx
ud
Tgafdxgadxdx
du
dx
gad
Tgfdxgdxdx
du
dx
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
))((lim))((0
where
x
lxu
uux
uuxlu
u for
for 22
for
, )( uu , )( uu
- Can we integrate dxul
l
on what condition ?
l
l
l
l
dxudxudxudxu
)()
22( uudx
uux
uudxu
))((lim)(lim00
uudxudxudxudxudxuudxl
l
l
l
l
l
The last integral vanishes as far as u- and u+ are finite, and the integral becomes
l
l
l
l
udxudxudx0
0
u-
u+
u-
u+
2
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 ?
l
l
l
l
dxudxudxudxu 2222
2)(
6)()
22( 2222
uuuudx
uux
uudxu
)2
)(6
)((lim
)(lim
2222
0
222
0
2
uuuudxudxu
dxudxudxudxu
l
l
l
l
l
l
The last integral vanishes as far as u- and u+ are finite, and the integral becomes
l
l
l
l
dxudxudxu0
20
22
- Can we integrate dxdx
dul
l
2)( ??
dxdx
uduudx
dx
ud
dxdx
uddx
uudx
dx
ud
dxdx
uddx
dx
uddx
dx
uddx
dx
ud
l
l
l
l
l
l
l
l
22
2
222
2222
)(2
)()(
)()2
()(
)()()()(
2
)(lim)()()(lim)(
2
0
2
0
20
2
0
2 uudx
dx
dudx
dx
dudx
dx
uddx
dx
du l
l
l
l
l
l
Therefore, the given definite integral has a finite value if and only if u is continuous.
uu limlim
00
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 dxdx
duu
l
l
on what condition ?
l
l
l
l
dxdx
ududx
dx
ududx
dx
ududx
dx
udu
2
)()()
2)(
22(
uu
uudxuuuu
xuu
dxdx
udu or
2
)()())()((
2
1 222
uuuuuudxu
dx
uddxu
dx
ududx
dx
udu
)(][
2
)()(
)2
)()((lim
)(lim
0
0
0
0
0
0
uudxdx
duudx
dx
duu
uuuudx
dx
duudx
dx
duu
uuuudx
dx
ududx
dx
udu
dxdx
ududx
dx
ududx
dx
ududx
dx
duu
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 R , LS , RR for the following ODE with the homogeneous displacement BCs.
04
4
qdx
wd
2. Prove the absolute minimum property of RR 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
wd
dx
dwl
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.
V
ijij
S
ii dVdSQu where )(2
1
i
j
j
iij x
u
x
u
Mathematical Viewpoint - Continuous Problem
If 0)( uA should hold for a given system, then the following statement should hold. Here, u , and the order of A should be the same as u.
uuAu 0)( dvv
( : A proper Function space)
- Example : Beam problem
The governing equation for a beam problem : qdx
wdEI
4
4
The expression for virtual work :
vvv
wqdxdxdx
wdEI
dx
wddxq
dx
wdEIw
2
2
2
2
4
4
0)(
If w is the displacement induced by the unit load applied load at xj , the expression for the principle of virtual work becomes as follows.
)()( j
v
j
vv
Q
xwdxxxwwqdxdxEI
MM
where M is the moment induced by the unit load applied at xj and )( jxx 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
uuuAu 0 )( )( ii Au ( : 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
n
i
iiim
j
ij
iiim
j
ij vYVuXH
0))sin( )cos((1
)(
1
)(
1
n
i
iiim
j
ij
ij
iiim
j
ij
ij vYFuXF
n
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 sin)( 12
iuu cos)( 12
i i
)( 12 uu
)( 12 vv
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
n
i
iiiin
i
iiiinmb
ii
iii
i
iinmb
i
inmb
i
ii
nmb
ii
iii
iii
nmb
i
iii
iiii
i
vYuXvYuXEA
lF
EA
lFlF
vvuuF
vvFuuF
11111
11212
11212
) () ()()(
)sin)( cos)((
))(sin )(cos(
If force system constists of an unit load applied at k-th joint in arbitrary direction, then
nmb
ii
iiikkkk
EA
lFvYuX
1 )(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
du
dx
ud
Tudxufdxdx
du
dx
ud
Tudxufdxdx
du
dx
ud
ee
ee
admissible allfor 00
011
000
Interpolation of Displacement Field in a Element
= +
Leu R
eu 1 1 Leu
Reu
elLeN R
eN
ex 1ex
1U 1nU 1l 2l nl3l . . .
node01 x 2x 3x lxn 1
2U 3U4U
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
eRe
LeR
eLe
Re
Re
Le
Le
he
e
e
ee
eRe
e
e
ee
eLe
eRe
LeR
eLe
Re
Re
Le
Le
he
u
uNNuNuNxu
l
xx
xx
xxN
l
xx
xx
xxN
u
uNNuNuNxu
uN
uN
),()(
= , =
),()(
1
1
1
1
Discretized Form of Variational Statement
eRe
Le
Re
LeR
e
ReL
e
Le
he
eRe
Le
Re
LeR
e
ReL
e
Le
he
u
u
dx
dN
dx
dNu
dx
dNu
dx
dN
dx
ud
u
u
dx
dN
dx
dNu
dx
dNu
dx
dN
dx
du
uB
uB
),(
),(
0
))0()0()(()()( 1
0
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
TulTufdxudxdx
du
dx
ud
Tudxufdxdx
du
dx
ud
Compatibility Condition – Continuity Requirement
111 , eL
eRe
eRe
Le UuuUuu
UCuUCu
eee
n
e
e
Re
Le
e
U
U
U
U
u
u ,
0100
0010
1
1
1
Global System Equation – Global Stiffness Equation
.
UFKUU
TCFCUUCKCU
TCUFCUUCKCU
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
du
dx
ud
admissible allfor 00
ee
x
x
TTe
n
i
x
x
ii
iii
ii
i
x
x
nn
nnn
n
x
x
nn
nnn
n
x
x
nn
nnn
n
x
x
ii
iii
i
x
x
ii
iii
i
x
x
ii
iii
i
x
x
ii
iii
i
x
x
ii
iii
i
x
x
ii
iii
i
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
dx
dg
dx
dgU
dx
dgU
dxUdx
dgU
dx
dg
dx
dgU
dxUdx
dgU
dx
dg
dx
dgUdxU
dx
dgU
dx
dg
dx
dgU
dxUdx
dgU
dx
dg
dx
dgUdxU
dx
dgU
dx
dg
dx
dgU
dxUdx
dgU
dx
dg
dx
dgUdxU
dx
dgU
dx
dg
dx
dgU
dxUdx
dgU
dx
dg
dx
dgUdxU
dx
dgU
dx
dg
dx
dgU
dxUdx
dgU
dx
dg
dx
dgUdxU
dx
dgU
dx
dg
dx
dgU
dxUdx
dgU
dx
dg
dx
dgU
dxUdx
dg
dx
dgUdx
dx
du
dx
ud
UBBU
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 1iU
gi-1
1iU
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
l
xx
xx
xxN
l
xx
xx
xxN
u
uNNuNuNxu
= , =
),()(
1
1
1
1
uN
]1 , 1[1
),( e
Re
Le
e ldx
dN
dx
dNB
- Element Stiffness Matrix
11
111]1 , 1[
1
1
11][
eeV ee l
dxll
e
K
1U 1l 2l 3l
01 x 2x 3x lx 4
2U 3U4U
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
- Compatibility Matrix
0010
00011
4
3
2
1
1
11 UCu
U
U
U
U
u
uR
L
, 0100
00102
4
3
2
1
2
22 UCu
U
U
U
U
u
uR
L
1000
01003
4
3
2
1
3
33 UCu
U
U
U
U
u
uR
L
- Global Stiffness Matrix
33
3322
2211
11
321
3
21
333222111
1100
11110
01111
0011
1100
1100
0000
0000
1
0000
0110
0110
0000
1
0000
0000
0011
0011
1
1000
0100
11
11
10
01
00
00
1
0100
0010
11
11
00
10
01
00
1
0010
0001
11
11
00
00
10
01
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
1
1
21
1 11 e
x
x e
e
ee
ldx
xx
xx
l
e
e
F
2
22
22
2
1
1
10
01
00
00
21
1
00
10
01
00
2
1
1
00
00
10
01
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( 3
1321
dx
duulll
5.0
1
1
9
1
110
121
012
2
22
22
2
1100
11110
01111
0011
4
3
2
3
32
21
1
4
3
2
1
33
3322
2211
11
U
U
U
l
ll
ll
l
U
U
U
U
ll
llll
llll
ll
18
9 ,
18
8 ,
18
5432 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( 3
1321 uulll
1
1
9
1
21
12
2
22
22
2
1100
11110
01111
0011
4
3
3
32
21
1
4
3
2
1
33
3322
2211
11
U
U
l
ll
ll
l
U
U
U
U
ll
llll
llll
ll
9
1 ,
9
132 UU
3.6. Finite Difference Discretization Differential Equation
0 0 22
2
ii fuDf
dx
ud where D2 is a 2nd-order finite difference operator.
Displace- Strain (or Stress)
1ixix1ix
iU
1iU
1iU
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
(1
11
1112
1
12
11
ii
iii
iiii
iii
i
iii
Ul
Ull
Ulll
a
cblalU
cU
cblalU
)1
)11
(1
(2
2 11
111
22
2
ii
iii
iiii
i
xx
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
2
1)
11(
1
2
1)
11(
1
fll
Ul
Ull
Ul
fll
Ul
Ull
Ul
Finite Difference Equations for Boundary Nodes with Traction BCs
In case that a traction BC is specified at a boundary nodes, some special treatment for
boundary condition such as ghost node is needed.
il1 il
iU
1iU
1iU
0
1nx nx1nx 2nx
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
UUll
UU
dx
du
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 2
11 n
nn
nn
n
fl
Ul
Ul
- Example – Case 1
332
43
332
22
221
32
221
11
2
1)
11(
1
2
1)
11(
1
fll
Ul
Ull
Ul
fll
Ul
Ull
Ul
43
43
33 2
11f
lU
lU
l
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 –
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 b in V
Constitutive Law : :D in V
Strain-Displacement Relationship : ))((2
1 Tuu
in V
Displacement Boundary condition : 0 uu on uS
Traction Boundary Condition : 0 TT on tS
Cauchy’s Relation on the Boundary : nT on S
Strain Energy (Refer to pp. 244 - 246 of Theory of Elasticity by Timoshenko)
dVx
u
dVx
u
x
u
x
u
x
udV
V
ijj
i
jiij
V
iji
j
j
i
i
j
j
iij
V
ijij
2
1
)(2
1
2
1
)(2
1
2
1int
Total Potential Energy
dSTudVbudVS
ii
V
ii
V
ijij 2
1
V, E ,
TT
0 uuSu
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 : S
hi
hiii
hjij
V
hii
R dSTTuudVbuu ))((2
1))((
2
1,
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
)()(2
1
)()(2
1
))((2
1
))((2
1))((
2
1))((
2
1
))((2
1))((
2
1))((
2
1
))((2
1))((
2
1
,,,,
,,
,,
,,
,,
,,
Energy Functional – Total Potential Energy
RR
S
ihi
V
ihi
hij
V
hij
V
ijj
i
j
S
ijhi
V j
ijhi
hij
V
hij
V
ijj
i
V
ijj
hih
ij
V j
hi
V
ijj
i
hij
V
ijj
hi
V
ijj
hi
V
ijj
i
hij
V
ijj
hi
V l
kijkl
j
hi
V
ijj
i
hij
V
ijj
hi
l
hk
ijkl
V j
i
V
ijj
i
hij
V
ijj
hih
ij
V j
i
V
ijj
i
hij
V
ijj
hi
j
iLS
C
dSTudVbudVdVx
u
dSnudVx
udVdVx
u
dVx
udV
x
udV
x
u
dVx
udV
x
udV
x
u
dVx
udV
x
uD
x
udV
x
u
dVx
udV
x
uD
x
udV
x
u
dVx
udV
x
udV
x
u
dVx
u
x
u
t
2
1
2
1
2
1
2
1
2
1
2
1
)(2
1
2
1
2
1
)(2
1
2
1
2
1
)(2
1
2
1
2
1
)(2
1
2
1
2
1
))((2
1
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Minimization Problems
RRLSR Min Min Min w.r.t. hi
hiu
RRMin : Rayleigh-Ritz Method or Principle of Minimum Potential Energy
Find hh u such that minimize RR
where } , on 0|{ V l
kijkl
j
iu dV
dx
duD
dx
duSuuu
- h : The exact solution. - h : An approximate solution.
0 RR : Variational Principle or Principle of Virtual Work
0
)(2
1
))(2
1
)2
1(
dSTudVbudV
dSTudVbudVD
dSTudVbudVDD
dSTudVbudV
dSTudVbudV
t
t
t
t
t
S
ihi
V
ihi
V
hij
hij
S
ihi
V
ihi
V
hklijkl
hij
S
ihi
V
ihi
V
hklijkl
hij
hklijkl
hij
S
ihi
V
ihi
V
hij
hij
hij
hij
S
ihi
V
ihi
hij
V
hij
RR
Absolute Minimum Property of the Total Potential Energy
eii
hi uuu
)(
2
1
2
1
)()(2
1
2
1
2
1
2
1
)()()()(
2
1
2
1
dSTudVbudVx
uD
x
u
dVx
uD
x
udSTudVbudV
x
uD
x
u
dSTuudVbuudVx
uD
x
u
dVx
uD
x
udV
x
uD
x
udV
x
uD
x
u
dSTuudVbuudVx
uuD
x
uu
dSTudVbudV
t
t
t
t
t
S
iei
V
iei
l
k
V
ijklj
ei
l
ek
V
ijklj
ei
S
ii
V
iil
k
V
ijklj
i
S
ieii
V
ieii
l
ek
V
ijklj
ei
l
ek
V
ijklj
i
l
k
V
ijklj
ei
l
k
V
ijklj
i
S
ieii
V
ieii
l
ekk
V
ijklj
eii
S
ihi
V
ihi
hij
V
hij
h
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
dVx
uD
x
u
dVbudVx
uD
x
u
dVbudVudVx
uD
x
u
dSTudVbudSTudVudVx
uD
x
u
dSTudVbudSnudVudVx
uD
x
u
dSTudVbudVx
udV
x
uD
x
u
l
ek
V
ijklj
eiE
V
ijijei
l
ek
V
ijklj
eiE
V
iei
V
jijei
l
ek
V
ijklj
eiE
S
iei
V
iei
S
iei
V
jijei
l
ek
V
ijklj
eiE
S
iei
V
ieijij
ei
V
jijei
l
ek
V
ijklj
eiE
S
iei
V
iei
V
ijj
ei
l
ek
V
ijklj
eiEh
tt
t
t
2
1
)(2
1
)(2
1
)(2
1
)(2
1
)(2
1
,
,
,
,
Since Dijkl is positive definite,
u
xD
u
xie
jijkl
ke
l
0 0
j
ei
x
uand
u
xD
u
xdVi
e
jijkl
V
ke
l 0 if and only if 0
j
ei
x
u. Therefore
.)?? =for only holdssign equality The ( ihi
Eh uu
4.3. Principle of Virtual Work If the following inequality is valid for all real number , the principle of virtual work holds.
iiRR
iiRR vuvu )()(
) admissible allfor ( 0)0(
)(
)()(
2
1
2
1
)()()()(
2
1
)()(
2
ii
S
ii
V
iil
k
V
ijklj
i
l
i
V
ijklj
i
S
ii
V
iil
k
V
ijklj
i
S
iii
V
iii
l
i
V
ijklj
i
l
k
V
ijklj
i
l
k
V
ijklj
i
S
iii
V
iiil
kk
V
ijklj
ii
iiRR
vvdSTvdVbvdVx
uD
x
vg
dVx
vD
x
vdSTvdVbvdV
x
uD
x
vg
dSTvudVbvu
dVx
vD
x
vdV
x
uD
x
vdV
x
uD
x
u
dSTvudVbvudVx
vuD
x
vu
vug
t
t
t
t
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
S
ihi
V
ihi
V
hij
j
hi vdSTvdVbvdV
x
v
t
admissible allfor 0
Equivalence to the PDE
- Exact form
ii
S
iii
V
jiji vdSTTvdVbvt
0 )()( , iiijij TTb , 0 ,
- Approximate form
hhi
S
ih
ihi
V
ih
jijhi vdSTTvdVbv
t
0 )()( . iiijij TTb , 0 ,
Since i
S
ih
ii
V
ih
jiji vdSTTvdVbvt
0)()( . .
Equivalence to the Weighted Residual Method
- Discretization : kikhikik
hi gaugav ,
kdSTTgdVbg
adSTTgadVbga
t
t
S
ih
ik
V
ih
jijk
ik
S
ih
ikik
V
ih
jijkik
allfor )()(
possiblefor )()(
.
.
Uniqueness of solution
If two solutions satisfy the principle of virtual work, then
i
S
ii
V
ii
V
ijj
i
i
S
ii
V
ii
V
ijj
i
vdSTvdVbvdVx
v
vdSTvdVbvdVx
v
t
t
admissible allfor 0
admissible allfor 0
2
1
By subtracting two equations,
0 admissible allfor 0)( 2121 ijiji
V
ijijj
i vdVx
v
00)(2121
l
k
l
k
l
k
l
kijkl x
u
x
u
x
u
x
uD
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Chapter 5
Discretization
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
x
gc
x
gc
x
u
1
Principle of Minimum Potential Energy
dSTgcdVbgcdVx
gcD
x
gc
tS
ipip
V
ipipl
qkq
V
ijklj
pip
h
2
1Min or 0
mr
h
c for all m, r
mrfcK
dSTgdVbgdVcx
gD
x
g
dSTgdVbgdVx
gcD
x
g
dSTgdVbgdVx
gD
x
gcdV
x
gcD
x
g
dSTgdVbg
dVx
gD
x
gcdV
x
gcD
x
g
c
rmkprmkp
S
mr
V
mrkp
V l
pmjkl
j
r
S
mr
V
mrl
pkp
V
mjklj
r
S
mr
V
mrl
r
V
ijmlj
pip
l
qkq
V
mjklj
r
S
irmi
V
irmi
l
q
V
rqmkijklj
pip
l
qkq
V
ijklj
prpmi
mr
h
t
t
t
t
and allfor 0
2
1
2
1
2
1
2
1
Principle of Virtual Work
0 V
ihi
V
hi
hi
V
hij
hij dVudVbudV
ip
S
ip
V
ipkq
V l
qijkl
j
pip
S
ip
V
ipkq
V l
qijkl
j
pip
S
ipip
V
ipip
V l
qkqijkl
j
pip
h
cdSTgdVbgdVcx
gD
x
gc
dSTgdVbgdVcx
gD
x
gc
dSTgcdVbgcdVx
gcD
x
gc
t
t
t
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
)(
)222(
)(
232313131212333322221111
232313131212333322221111
323223233131131321211212333322221111
dSdVdVtS
h
V
hh
V
h Tubu
Displacement
Ncu
n
n
n
n
n
n
h
h
h
h
c
c
c
c
c
c
c
c
c
g
g
g
g
g
g
g
g
g
u
u
u
3
2
1
32
22
12
31
21
11
2
2
2
1
1
1
3
2
1
00
00
00
00
00
00
00
00
00
Virtual Strain
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
)(
x
gc
x
gc
x
gc
x
gc
x
gc
x
gc
x
gc
x
gc
x
gc
x
u
x
ux
u
x
ux
u
x
ux
ux
ux
u
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
cB
n
n
n
nn
nn
nn
n
n
n
h
h
h
h
h
h
h
c
c
c
c
c
c
c
c
c
x
g
x
gx
g
x
gx
g
x
gx
gx
gx
g
x
g
x
gx
g
x
gx
g
x
gx
gx
gx
g
x
g
x
gx
g
x
gx
g
x
gx
gx
gx
g
3
2
1
32
22
12
31
21
11
23
13
12
3
2
1
2
2
3
2
1
2
3
2
1
2
2
2
3
2
2
2
1
2
2
1
3
1
1
1
3
1
1
1
2
1
3
1
2
1
1
1
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
h
h
h
h
h
h
h
h
h
h
h
h
h
h
v
vv
vv
vv
v
v
vv
v
v
vv
v
v
v
vv
E
23
13
12
33
22
11
13
13
12
33
22
11
)1(2
2100000
0)1(2
210000
00)1(2
21000
000111
0001
11
00011
1
)21)(1(
)1()(
Final System Equation
cDBBc dVdVdVV
TT
V
hTh
V
hij
hij
dSdSdSTu
dVdVdVbu
ttt S
TT
S
Th
S
ihi
V
TT
V
Th
V
ihi
TNcTu
bNcbu
fKccTNbNcDBBc T or admissible allfor 0)( dSdVdVtS
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
lxx
bv
ylxx
au
2
100
01
01
1 ,
22
00
0)(
,
260
0)2
(
22226
2
E
lxx
lxx
ylx
lxx
ylxx
DBN
Stiffness Matrix
)2(15
2
2
1)2(
15
2
2
1
)2(15
2
2
1)2(
15
2
2
1
332
1
)
2(
2
1)
2(
2
1
)2
(2
1)
2(
2
1)(
1
22
00
0)(
2
100
01
01
200
20)(
1
55
5533
2
0
1
0 22
22
22
22
22
2
0
1
022
2
2
2
hlhl
hlhlhl
E
dzdydxlx
xlx
x
lxx
lxx
ylxE
dzdydx
lxx
lxx
ylx
lxx
lxx
ylxE
l h
h
l h
h
K
x, u
y, v
l
2h
h
PTy 2
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
Load Term
P
ldy
hl
yl
h
h
0
32
P0
30
02
3
3
2
1
0
f
System Equation
P
l
b
a
hlhl
hlhlhl
E 0
3)2(15
2
2
1)2(
15
2
2
1
)2(15
2
2
1)2(
15
2
2
1
332
1
3
55
5533
2
PEIl
h
vbP
EIP
Eha
22
2
3
2 1))(
1
1
3
51( ,
11
2
3
Displacement and Stress
)2
6
(
1))(
1
1
3
51(
)2
(
1
2322
22
lxx
PEIl
h
vv
ylxx
PEI
u
Pxlx
Il
h
yI
M
yI
My
I
lxP
xy
yy
xx
)2
(
1)(
6
5
)(
22
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.
),( , ),,( , ),,( yxwwzy
wzyxvz
x
wzyxu
1) Drive the strain component using the given displacement field.
2) Assume 033 , 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
0
e S
Th
e V
Th
e V
hΤh
Th
V
Th
V
hΤh
dSdVdV
ddVdV
et
ee
t
bubu
Tubu
The displacement field in an element
ee
e
n
n
n
n
n
n
e
e
e
e
u
u
u
u
u
u
u
u
u
N
N
N
N
N
N
N
N
N
u
u
u
UNu
3
2
1
32
22
12
31
21
11
2
2
2
1
1
1
3
2
1
00
00
00
00
00
00
00
00
00
where Ni j ij( )X .
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
u
u
u
u
u
u
u
u
u
x
N
x
Nx
N
x
Nx
N
x
Nx
Nx
Nx
N
x
N
x
Nx
N
x
Nx
N
x
Nx
Nx
Nx
N
x
N
x
Nx
N
x
Nx
N
x
Nx
Nx
Nx
N
x
Nu
x
Nu
x
Nu
x
Nu
x
Nu
x
Nu
x
Nu
x
Nu
x
Nu
x
u
x
ux
u
x
ux
u
x
ux
ux
ux
u
UB
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
The stress field in an element
ee
e
e
e
e
e
e
e
e
e
e
e
e
e E
DBUD
23
13
12
33
22
11
23
13
12
33
22
11
)1(2
2100000
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
e
eTee
e
eTe
e S
TTe
e V
TTe
e
e
V
TTe dSdVdVet
ee
fUUKU
TNUbNUUDBBU
Compatibility Conditions
0fKU
UfTUUTKTU
UTUUTU
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 Detination 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
52
Differential Equation & Boundary Conditions
Principle of Minimum Potential Energy
i
S
ii
V
ii
V
ijij udSTudVbudVt
admissible allfor 2
1
Principle of Virtual Work
i
S
ii
V
ii
V
ijj
i udSTudVbudVx
u
t
admissible allfor 0
Weighted Residual Method (Galerkin Method)
ii
S
iii
V
jiji vdSTTvdVbvt
admissible allfor 0)()( , Minimization of Error
errorie
j
i
jije
V
ij i
u
x
v
xdV v
1
2( )( )
for all admissible
Principle of Minimum Potential Energy hi
S
ihi
V
ihi
V
hij
hij udSTudVbudV
t
admissible allfor 2
1
Principle of Virtual Work
hi
S
ihi
V
ihi
V
hij
j
hi udSTudVbudV
x
u
t
admissible allfor 0
Principle of Minimum Potential Energy
0- admissible allfor -2
1
admissible allfor )(2
1
fUKUfUUKU TT
hi
e S
ihi
V
ihi
V
hij
hij udSTudVbudV
rt
ee
Principle of Virtual Work
0 admissible allfor 0
admissible allfor 0)(
fUKUfUUKU
TT
hi
e S
ihi
V
ihi
V
hij
j
hi udSTudVbudV
x
u
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
53
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
54
6.1. Plane Stress Stress : 33 13 23 0
Strain :
33 11 22 13 2310 0
v( ), ,
Modified Stress-strain Relation
1212
2211222
2211233221111
)1(2
)(1
)(1
))(1
()21)(1(
)1(
E
vE
vEE
e
e
e
e
e
e
e
e E D
12
22
11
2
12
22
11
2
100
01
01
1
Interpolation of Displacement
ee
e
n
n
n
ne
ee
v
u
v
u
v
u
N
N
N
N
N
N
v
uUNu
2
2
1
1
2
2
1
1
0
0
0
0
0
0
Strain-Displacement Relation
e
e
n
n
nn
n
n
ee
e
e
e
e
e
e
v
u
v
u
v
u
x
N
y
Ny
Nx
N
x
N
y
Ny
Nx
N
x
N
y
Ny
Nx
N
x
v
y
uy
vx
u
BU
2
2
1
1
22
2
2
11
1
1
12
22
11
0
0
0
0
0
0
Principle of Virtual work
e S
T
e A
T
e A
T
yh
S
xh
yh
A
xhhhhh
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
55
6.2. Plane Strain Strain : 33 13 230 0 0 , ,
Stress : 13 23 33 11 220 ( )
Stress-strain Relation
e
e
e
e
e
e
e
e
v
v
vvE
D
12
22
11
12
22
11
)1(2
2100
011
01
1
)21)(1(
)1(
Interpolation of Displacement
ee
e
n
n
n
ne
ee
v
u
v
u
v
u
N
N
N
N
N
N
v
uUNu
2
2
1
1
2
2
1
1
0
0
0
0
0
0
Strain-Displacement Relation
e
e
n
n
nn
n
n
ee
e
e
e
e
e
e
v
u
v
u
v
u
x
N
y
Ny
Nx
N
x
N
y
Ny
Nx
N
x
N
y
Ny
Nx
N
x
v
y
uy
vx
u
BU
2
2
1
1
22
2
2
11
1
1
12
22
11
0
0
0
0
0
0
Principle of Virtual work
e S
T
e A
T
e A
T
yh
S
xh
yh
A
xhhhhh
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
56
6.3. Axisymmetry
Strain : 0 , 0 , , , ,
zrrzzzrr r
v
z
u
r
u
z
v
r
u
Stress : r z 0
Stress-strain Relation
e
erz
e
ezz
err
erz
e
ezz
err
e
v
vv
vv
vv
vE
D
)1(2
21000
0111
01
11
011
1
)21)(1(
)1(
Interpolation of Displacement
ee
e
n
n
n
ne
ee
v
u
v
u
v
u
N
N
N
N
N
N
v
uUNu
2
2
1
1
2
2
1
1
0
0
0
0
0
0
Strain-Displacement Relation
e
e
n
nnn
n
n
n
ee
e
e
e
erz
e
ezz
err
e
v
u
v
u
v
u
r
N
z
Nr
Nz
Nr
N
r
N
z
Nr
Nz
Nr
N
r
N
z
Nr
Nz
Nr
N
r
v
z
ur
uz
vr
u
BU
2
2
1
1
22
2
2
2
11
1
1
1
0
0
0
0
0
0
0
0
0
Principle of Virtual work
( ) ( ) ( )
rrh
rrh
A
zzh
zzh h h
rh
rh h
r
A
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
57
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
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
58
7.1. Constant Strain Triangle (CST) Element Displacement Interpolation
yxyxvyxyxu ee654321 ),( , ),(
33321333
23221222
13121111
),(
),(
),(
yxuyxu
yxuyxu
yxuyxu
e
e
e
→
3
2
1
33
22
11
3
2
1
1
1
1
yx
yx
yx
u
u
u
3
2
1
321
321
321
3
2
1
33
22
11
3
2
1
2
1
1
1
11
u
u
u
ccc
bbb
aaa
u
u
u
yx
yx
yx
where a x y x y b y y c x xi j m m j i j m i m j , ,
333322221111
333322221111
)(2
1)(
2
1 +)(
2
1),(
)(2
1)(
2
1 +)(
2
1),(
vycxbavycxbavycxbayxv
uycxbauycxbauycxbayxu
e
e
Strain Components
e
e
ee
e
e
e
e
e
e
v
u
v
u
v
u
bc
c
b
bc
c
b
bc
c
b
x
v
y
uy
vx
u
BU
3
3
2
2
1
1
33
3
3
22
2
2
11
1
1
12
22
11
0
0
0
0
0
0
2
1
Stiffness Matrix
K B D B U B D Be T T e
A
tdA te
1u
1v
2u
v 2
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
59
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,
= 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
60
7.2. Isoparametric Formulation
Interpolation of Geometry
m
iiinm
m
iiinm
yNyNyNyNy
xNxNxNxNx
12211
12211
),(),(),(),(
),(),(),(),(
ee
e
m
m
m
me
ee
y
x
y
x
y
x
N
N
N
N
N
N
y
xXNx
2
2
1
1
2
2
1
1
0
0
0
0
0
0
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
61
Interpolation of Displacement in a Parent Element
eee
ee
v
uUNu ),(
Derivatives of the Displacement Shape Functions
y
Nx
N
yx
yx
N
N
y
y
Nx
x
NN
y
y
Nx
x
NN
i
i
i
i
iii
iii
NNN
N
N
N
yx
yx
y
Nx
N
ixi
i
i
i
i
i
11
1
or JJ
XNJ
mm
m
m
m
ii
im
ii
i
m
ii
im
ii
i
yx
yx
yx
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
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
62
7.3. Bilinear Isoparametric Element
Shape functions in the parent coordinate system.
4321)),(),,(( yxu
)),(),,(( 8765 yxv
4321444434214444444
4321334333213333333
4321224232212222222
4321114131211111111
)),(),,((),(
)),(),,((),(
)),(),,((),(
)),(),,((),(
yxuyxuu
yxuyxuu
yxuyxuu
yxuyxuu
4
3
2
1
4
3
2
1
1111
1111
1111
1111
u
u
u
u
),( , ),( 44
33
22
11
44
33
22
11 vNvNvNvNyxvuNuNuNuNyxu ee
N N N N1 2 3 4
1
41 1
1
41 1
1
41 1
1
41 1 ( )( ) ( )( ) ( )( ) ( )( ) , , ,
1 2
3
4
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
63
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
( )( )( ) ( )( ) ( )
( )( ) ( )( ) ( )
1 2 1 1 1
1 2 1 1 1
lkm k k m
k k k k k k k m
( )( )( ) ( )( ) ( )
( )( ) ( )( ) ( )
1 2 1 1 1
1 2 1 1 1
N N l li IJ Im
Jm ( ) ( )
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
64
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 1
2
Total number of the parasitic terms : ( )m m1
2
The Pascal Polynomials
1
2 2
3 2 2 3
1 1
1 1
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
65
7.5. Higher Order Rectangular Element - Serendipity Family Q8 Element
)1)(1(2
1 25 N
)1)(1(2
1 28 N
)1)(1(4
1ˆ1 N
51 2
1ˆ NN
8511 2
1
2
1ˆ NNNN
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
66
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 1
2
Total number of internal nodes : ( )( )m m 2 1
2-3m =
( )( )m m 2 1
2
The Pascal Polynomials
mmmm
11
3223
22
1
Area Coordinate System
11
22
33
1 2 3 1 A
A
A
A
A
A , , ,
Shape functions
- CST Element
N N N1 1 2 2 3 3 , ,
- LST Element
316325214
333222111
444
)1(2)1(2)1(2
NNN
NNN
A3
A1A2
1
2
3
constant line
constant line
constant line
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
67
Interpolation of Geometry
n
iii
n
iii xNxNx
121
1321 ),(
~),,(
n
iii
n
iii yNyNy
121
1321 ),(
~),,(
)(]~
[~0
0~
~0
0~
~0
0~
)( 2
2
1
1
2
2
1
1 ee
e
m
m
n
ne
ee XN
y
x
y
x
y
x
N
N
N
N
N
N
y
x
X
Interpolation of Displacement in a Parent Element
( ) [ ( , , )] ( ) [~
( , )] ( )uu
vN U N Ue
e
ee e e e
1 2 3 1 2
Derivatives of the Displacement Shape Functions
~
~
~
~
~
~
~
~
~
~
~~~
~~~
2
11
2
1
1
22
11
22
11
2
1
222
111
i
i
i
i
i
i
i
i
i
i
iii
iii
N
N
N
N
yx
yx
y
Nx
N
y
Nx
N
yx
yx
N
N
y
y
Nx
x
NN
y
y
Nx
x
NN
J
iix NN~~
or1
J
XNJ
~
~
~
~
~
~
~
~~
~~
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
yx
yx
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
68
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
69
8.1. Gauss Quadrature Rule One Dimension
f d W fi ii
n
( ) ( )
11
1
If the given function f ( ) is a polynomial, it is possible to construct the quadrature rule
that yields the exact integration.
- f ( ) is constant: 0)( af
2 1 2)( 1
1
1 100
W n=Waadxfn
ii
- f ( ) is first order: 10)( aaf One point rule is good enough.
- f ( ) is second order: 2210)( aaaf
2 2 , 0 , 3
2
23
2)(
111
2
1
1 10
11
1
2202
nWWW
WaWaWaaaf
n
ii
n
iii
n
iii
n
ii
n
iii
n
iii
W W Wi ii
n
1
11 2 1 20
,
W W W W Wi ii
2
1
2
1 12
2 22
2 22
2 222
2
3
1
3
W W W W Wii
1
2
1 2 2 2 22 2 1 057735 = 1 / 3 02691 89626 .
- f ( ) is third order: 33
2210)( aaaaf Two point rule is enough.
- f ( ) is fourth order: 44
33
2210)( aaaaaf
3 2 , 0 , 3
2 , 0 ,
5
2
23
2
5
2)(
111
2
1
3
1
4
10
11
1
22
1
33
1
44
1
1
024
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
3
1 11 3 1 3 20 0 0
, , ,
W W W W Wi ii
4
1
3
1 14
3 34
3 34
3 342
2
5
1
5
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70
W W W W W W
W
i ii
2
1
2
1 12
3 32
3 32
3 32
3 3
3 3
22
3
1
33 5
5
9
0 77459 0 55555 55555 55555
, =
66692 41483 ,
/
. .
W W W W W W Wii
1
2
1 2 3 3 2 22 28
9 = 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
( ) ( ) ( ) ( ) 1
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
( ) ( ) | | ( ) ( ) | | 1
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
( , ) ( ) ( ) ( ) 1
1
1
1
11
1
1 1 11
- Stiffness equation
n
i
m
jijijjiijji
Tji
T
A
T
JtWW
ddJttdAe
1 1
1
1
1
1
||),(),(
||),(),(
BDB
BDBBDB
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
n
iii
iiij
iiTi
T
A
T
JtW
ddJttdAe
12121
1
0
1
0
122121
||),(),(2
1
||),(),(1
BDB
BDBBDB
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8.2. Reduced Integration
Q8 element
27
26
254
23210
27
26
254
23210
bbbbbbbbv
aaaaaaaau
276431
276431 22 , 22
bbbbbv
aaaaau
72
654272
6542 22 , 22 bbbbbv
aaaaau
Terms in stiffness matrix
- From complete polynomials: 22 , , , , , 1
- From parasitic terms: 3322 , , , , , 443322 , , , ,
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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8.4. Selective Integration
SN
EE
E
DD
D
100000
010000
001000
000000
000000
000000
)1(2
000000
000000
000000
000111
0001
11
00011
1
)21)(1(
)1(
)1(2
2100000
0)1(2
210000
00)1(2
21000
000111
0001
11
00011
1
)21)(1(
)1(
nintegratio Reducednintegratio Full
)( dVdVdVdVeeee V
ST
V
NT
V
SNT
V
T BDBBDBBDDBDBB
Homework 8
Solve the cantilever beam problem in homework 7 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
( , ) ( ) ( ) ( ) ( , , )
( ) ( ) ( ) ( , , )
( ) ( ) ( ) ( , , )
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
| |
( ) ( )
( ) ( )
( ) ( , , )
( ) ( , , )
| | ( ) ( ) ( ) ( )
A
a y a x
a 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
where
( , ) ( )
( )
| |( ( ) ( ) ( ) ( ))
( )
| |( ( ) ( ) ( ) ( ))
( ( ) ( ) ( ) ( ))| |
( ( ) ( )
x y a
a
Aa x a y a y a x
a
Aa x a y a y a x
a a y a y ax
Aa 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 ay
Ak k1 2( ) ( ))| |
H.O.T.
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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
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
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 S
T
e V
T
e
e
V
T
ee
eeeee
et
ee
dVdV
dVdVdVdVdV
dSdVdV
~
~)~(
where ~ 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
1
41 1
1
41 1
1
41 1
1
41 1
0
( )( ) ( )( ) ( )( ) ( )( )
- Strain & Stress field
x y xy
x y xy
u
x
u
a
uy
ab
u
y
u
y
v
x
ux
ab
E uy
ab
E uy
ab
E ux
ab
= , ,
, ,
0
1 1 2 12 2 ( )
-Strain Energy - Full Integration
22
222
))1(2(13
1
))()1(2
)(1
(2
1
)(2
1
ub
a
a
bE
dydxab
xuE
ab
yuE
dydx
a
a
b
b
xyxyyy
a
a
b
b
xxb
-Strain Energy - Selective Integration
ua
bE
dydxab
xuE
ab
yuEa
a
b
b
sb
2
222
13
2
))()1(2
)(1
(2
1
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Real Behaviors for Pure Bending
- Strain & Stress field
x y xy
x y xy
ky
k
Ey
kv
Ey
, ,
, ,
0 0
0
- Displacement field
uk
Exy f y v
kv
Ey +g x
k
Ex f y g x
g xk
Ex f y C g x
k
Ex Cx C f y Cy C
xy
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
, ,
, ,
1
20
1
2
2
21 2
a
ub
b
y
b
ua
a
xCx
ab
uy
ab
vuvxy
ab
uu
Cab
uEkuubyax
CCCCyux
CCxxE
ky
E
kvvCCyxy
E
ku
2))(1(
2))(1(
2
1
2
1 ,
Arbitrary
,
00 02
1
2
1 ,
221
22
1
22
122
2
-Strain Energy
R x x
b
b
a
a
y y xy xy
b
b
a
a
dydx Euy
abdydx E
b
au
1
2
1
2
2
32 2( ) ( )
Ratio of strain Energy
b
R
bs
R
a
bv
1
1
1
1
1
2
1
111 0 32
2 ( ( ) ) . . , 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
( ) ( )
( ) ( )
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
f
bb bi
ib ii
b
i
( )
( )
( )
( )
- Static Condensation
[ ]( ) [ ]( ) ( ) ( ) [ ] (( ) [ ]( ))
([ ] [ ][ ] [ ])( ) ( ) [ ] ( ) [ ]( ) ( )
K u K a f a K f K u
K K K K u f K f K u f
ib ii iii
ib
bb bi ii ibb
iii Q Q
1
1 1 6 6
10.3. Static Condensation & Substructuring Static Condensation: Eliminate some DOFs prior to a main analysis.
r
e
r
e
rrre
eree
P
P
u
u
KK
KK
)()( 1rereeeeeeeerer uKPKuPuKuK
eeererrereererr
rrereeererrr
rererrr
PKKPuKKKK
PuKPKKuK
PuKuK
11
1
)())((
)()(
- From Gauss elimination point of view
eeerer
e
r
e
ereererr
eree
PKKP
P
u
u
KKKK
KK11 )()(0
ue, Pe ur, Pr
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Substructuring
- Substructure 1
111
11
111
iii
i
ii P
P
u
u
KK
KK
11
1111
11
1111 )())(( PKKPUKKKK iiiiiii
- Substructure 2
222
22
222
iii
i
ii P
P
U
U
KK
KK
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
P
P
u
u
KK
KK
11
1111
11
1111 )())(( PKKPuKKKK iiiiiii
- Substructure 2
222
22
222
iiiii
i
P
P
u
u
KK
KK
- Assembling
11
11121
22
11
11112
2
222
)()( PKKPP
P
u
u
KKKKKK
KK
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
2
21
21
2
1
PPP
PPP
P
P
P
P
P
P
A
S
A
A
S
S
P
P/2 P/2
P/2 P/2
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Cyclic Symmetry
U1 U2
Ui
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- Structural Resistance force in a segment
2
1
22221
21
12111
2
1
U
U
U
KKK
KKK
KKK
F
F
F
i
i
iiii
i
i
- Compatibility
21 UU
22
1
U
U
I0
0I
0
U
U
Ui
i
- Equilibrium
122 FFP
FPT
ii
2
1
2 F
F
F
I0
0I0
P
PiT
i
222221
21
12111
2 U
U
I0
0I
0
KKK
KKK
KKK
I0
0I0
P
P i
i
iiii
i
T
i
- Final Equation
22221121121
21
222212
21
12111
2
U
U
KKKKKK
KKK
U
U
KKK
KKK
KKK
I0
0I0
P
P
iTT
iiT
iiii
i
i
iiii
i
T
i
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10.5. Constraints in Discrete Problems Minimization Problem
PUKUUu
TT 2
1Min subject to 0UA )(
r
f
U
UU ,
rrrf
frff
KK
KKK
r
f
P
PP
1st order Optimality condition
0U
0 PKU
Usual homogenous support conditions
0rU
ffffTfff
Tf
f
PUKPUUKUU
2
1Min
Non-homogenous support conditions (support Settlement)
rr UU
rTrf
Tfrrr
Trrfr
Tffff
Tf
rTrf
Tfrrr
Trrfr
Tffrf
Trfff
Tf
PUPUUKUUKUUKU
PUPUUKUUKUUKUUKU
)2(2
1
)(2
1
0MinMin frfrffff
PUKUKUU
rfrffff UKPUK
or
00
f
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
0
CU
C
I
U
UU f
r
f
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90
)(2
1)(
][2
12
1
0000 rT
fTfrr
Trr
Tfr
Tf
frrT
rfT
frffTf
TT
PCPUCKCCKCCKU
UCKCKCCKKU
PUKUU
)()(][ 00 CKCCKPCPUCKCKCCKK rrT
frrT
ffrrT
rfT
frff
Lagrange Multiplier
)(2
1Min 0
,rRUPUKUU
U
TTT 0U
, 0
00 0)(
0)(
r
UU
0R
RK
rRUUA
RPKUU
UAPKU
U
TTT
)()(0)(
)(0
0111
01
0
1
rPRKRRKrRPRKrRU
RPKURPKU
TT
TT
011111111
01111
)))((
))()((
rRRKRKPRKRRKRKK
rPRKRRKRPKU
TTTT
TT
10.6. Constraints in Continuous Problems Lagrange Multiplier
u
Min subject to 0uA )(
where 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
SS
ihi
V
ihi
hij
V
hij
S
iu
t
i
)(2
1
)()(),(Min
0
0,
rLu
rLu
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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
2
1
2
1
)()(2
1),(
0
0
Therefore, the stationary condition for modified energy functional becomes
q
f
Λ
U
0R
RK
qRUΛ
ΛRfKUU
TT
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
S
T
udS)()(
2)()(Min uAuAuu
)2(22
1
}()(2)(){(22
1
)()(22
1)(
000
00
CqUUKUfUKUU
rrrLUULNLUfUKUU
rLurLuu
TR
TTT
S
T
S
TTe
e
e
S
TTeTT
S
T
S
ihi
V
ihi
hij
V
hij
eee
t
dSdSdS
dSdSTudVbudV
qfUKKU
u
)(0)(Min Ru
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Chapter 11
Problems with Higher Continuity Requirement
– Beams –
4 = ??
11.1. C1 Formulation
11.2. C0 Formuation
11.3. Timoshenko Beam
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11.1. C1-Formulation Governing Equation
EId w
dxq
4
4
Weak Form of the governing equation
( ) w EId w
dxq dx w
l 4
40
0 for all admissible
wEId w
dx
dw
dxEI
d w
dx
d w
dxEI
d w
dxdx wqdx
l l ll3
30
2
20
2
2
2
200
0
d w
dxEI
d w
dxdx wqdx wV M
d w
dxEI
d w
dxdx wqdx
ll
ll
ll2
2
2
20
00
0
2
2
2
200
Continuity Condition
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 w
dxEI
d w
dxdx
d w
dxEI
d w
dxdx
le
l
e
2
2
2
2
2
2
2
20
Hermitian Shape Functions
elxlel
xll
rrll
dx
dwlww
dx
dwww
NwNNwNw
, )( , , )0( where
0
4321
2
32
43
3
2
2
3
2
32
23
3
2
2
1
, 23
2 , 231
eeee
eeee
l
x
l
xN
l
x
l
xN
l
x
l
xxN
l
x
l
xN
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11.2. C0-Formulation
Energy Consideration
w: total deflection, : rotation , dw
dx : shear deformation
1
2 0 0
0
0
( ( ) ( ) )d
dxEI
d
dxdx
dw
dxGA
dw
dxdx wqdx
l l l
ii
iii
i NwNw ,
Tnn
e www ][ 2211
eM
eni
i
i
dx
dN
dx
dN
dx
dN
dx
dN
dx
d B
)](0 0 0[ 21
eS
en
ni
iii
i
i Ndx
dNN
dx
dNN
dx
dNNw
dx
dN
dx
dw B ] [ 2
21
1
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
2
1)(
2
1
)()()(2
1
][)()()(2
1
0
0
00
0 efK
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Problems –Functions Space Interlocking
EI
d
dx
d
dxdx
GA l
EI l
dw
dx
dw
dxdx
wq
EIdx
l l l
1
2
1
0
02
20 0
( ( )( ) )
beam retangularfor )1(
62
220
t
l
vEI
lGA
.
t As , 1
02
0l
dw
dx
dw
dxdx
l
( )( ) or dx
dw to keep the total potential energy
finite. This condition may be satisfied in case the entire, exact function spaces for w and
are used. For the finite element method, the energy expression is expressed as
eeee
e
e
eh
dEI
wqld
d
dw
ld
dw
lt
l
vd
d
d
d
d
EI
l 1
0
21
02
21
0
))1
()1
()1(
6(
2
1
To keep the energy expression finite as 0t , there are two alternatives:
1) tl e , 2) dx
dw. 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 , dc . The second
condition becomes dca for all 10 , which is equivalent to 0c 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 , 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
96
11.3. Timoshenko Beam
Governing Equation
EId
dxGA
dw
dx
2
2 0 0
( ) , GAd w
dx
d
dxp0
2
2( )
Week Form
w
ll
wdxpdx
d
dx
wdGAwdx
dx
dwGA
dx
dEI
& 0))(())((
2
2
0
0
02
2
0
0)())(( 0
00002
2
0
dxdx
dwGAdx
dx
dEI
dx
d
dx
dEIdx
dx
dwGA
dx
dEI
llll
0)()())((0
0
0002
2
0
0
pdxwdxdx
dwGA
dx
wd
dx
dwwGAdxp
dx
d
dx
wdGAw
llll
w
lllll
w
pdxwdx
dwwGA
dx
dEIdx
dx
dwGA
dx
wddx
dx
dEI
dx
d
&
)()()(00
00
0
00
By assuming homogeneous boundary conditions,
w
lll
wpdxwdxdx
dwGA
dx
wddx
dx
dEI
dx
d
& )()(
0
0
00
Boundary Conditions
0 0 0 00 or or or EId
dxw GA
dw
dx( )
Elimination of the displacement – Beginning of Nightmare
Differentiation of the first equation and substitution of the second equation into the first
equation
03
3
pdx
dEI (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
97
Chapter 12
Mixed Formulation
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
98
What is the mixed formulation??? - Definition
Stress or strain fields are treated and interpolated as independent variables!!!
Governing Equations and Boundary Conditions
Equilibrium Equation : 0 b in V
Constitutive Law : :D in V
Strain-Displacement Relationship : ))((2
1 Tuu in V
Displacement Boundary condition : 0 uu on uS
Traction Boundary Condition : 0 TT on tS
Cauchy’s Relation on the Boundary : nT on S
Weak statement.
iuii
S
iiijjiij
V
iji
V
jiji udSTTudVuudVbut
ˆ ˆ 0 )(ˆ))(2
1(ˆ)(ˆ ,,,
iuii
S
iikllkklijkl
V
iji
V
jiji udSTTudVuuDdVbut
ˆ ˆ 0 )(ˆ))(2
1(ˆ)(ˆ ,,,
iuikllkklijkl
V
ij
S
ii
V
ii
V
ijj
i udVuuDdSTudVbudVx
u
t
ˆ ˆ 0))(2
1(ˆˆˆ
ˆ,,
iuikllkklijkl
V
ij
S
ii
V
ii
V
klijklj
i udVuuDdSTudVbudVDx
u
t
ˆ ˆ 0))(2
1(ˆˆˆ
ˆ,,
Interpolations
eee UNu , eee EN
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(
FEQ T , 0QUME
0
F
E
U
MQ
Q0 T
FQUMQ 1T
Dept. of Civil and Environmental Eng., SNU
Structural Analysis Lab. Prof. Hae Sung Lee, http://strana.snu.ac.kr
99
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
2. 21 ˆ , ˆ LHu ii
3. ˆ , ˆ 22 LLu ii
Which one do you like? Can you give a proper explanation for your choice?