Introduction to Finite Element Method - strana.snu.ac.krstrana.snu.ac.kr/lecture/fem_2017/Note17_FEM.pdf · Dept. of Civil and Environmental Eng., SNU. Structural Analysis Lab. Prof

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



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



Chapter 2

Approximation of Functions and

Variational Calculus

≅ δ





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…



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



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.



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



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



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()(



Chapter 3

Elliptic Differential Equations in One

Dimension





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



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



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



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 %



- 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 %



***** 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



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 %



- 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



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



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

∫

∫

∫


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∈




- 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

∫

∫∫ ∫

∫∫ ∫∫

∫∫∫ ∫∫

∫∫∫ ∫∫

∫∫

∫∫



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



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



- 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.



- 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



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.



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



∑∑∑∑∑

∑

∑

=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



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 =



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



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



- 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



- 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)



- 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



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



- 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.



Chapter 4

Multidimensional Problems

– Elasticity Problems–

b

a



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



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

,,,,

,,

,,

,,

,,

,,


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



Minimization Problems RRLSR Π↔Π⇔Π Min Min Min w.r.t. h

ihiu υ∈


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



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

∫∫∫

∫∫∫∫

∫∫

∫∫∫

∫∫∫



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



Chapter 5

Discretization



5.1. Rayleigh-Ritz Type Discretization

5.2. Finite Element Discretization

5.3. Finite Element Programming (Linear static case)



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 =−



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



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



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

=



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



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



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 .



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(



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.



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



55

Differential Equation & Boundary Conditions


iS

iiV

iiV

ijij udSTudVbudVt


∫∫∫ −−σε=Π


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


∫∫∫ −−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


0- admissible allfor -21

admissible allfor )(21

=⋅→⋅⋅⋅=

−−σε=Π ∑ ∫∫∫

fUKUfUUKU TT

hi

e Si

hi

Vi

hi

V

hij

hij udSTudVbudV

rt

ee


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



56

Chapter6

Two-Dimensional Elasticity Problems

6.1. Plane Stress

6.2. Plane Strain

6.3. Axisymmetry

x y

z

∞



57



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



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(


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


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


∑ ∫∑ ∫∑ ∫

∫∫∫

⋅+⋅=⋅⋅

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



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


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


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


( ) ( ) ( )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



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



62

7.5. Higher Order Rectangular Element - Serendipity Family

7.6. Triangular Isoparametric Element



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



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



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



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



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



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



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



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 −−=



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



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.



73

Chapter8

Numerical Integration

∫∫∫∫∫∫∫∫∫∫∫∫∫∫∫∫∫∫∫∫∫∫∫∫∫

8.1 Gauss Quadrature Rule

8.2. Reduced Integration

8.3. Spurious Zero Energy mode

8.4. Selective Integration



74



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=

∑ = + = = → =



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



77

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



78

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



79

Zero Energy mode of Q8 Element Zero Energy Modes of Q9 Element Near Zero Energy Modes



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.



81

Chapter 9

Convergence Criteria in the

Isoparametric Element



82

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.



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



84

10.5. Constraints in Discrete Problems

10.6. Constraints in Continuous Problems



85

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.



86

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 ∫ ∫

− −



87

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”.



88

- 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



89

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


- Assembling

21

22211

11121

21

22211

11121 )()())()(( PKKPKKPPuKKKKKKKK −−−− −−+=−−+ iiiiiiiiiiiii

or

21

22211

11121

22211

11121 )()())()(( PKKPKKPuKKKKKKKK −−−− −−=−−+ iiiiiiiiiiii

u1, P1 u2, P2

ui, Pi



90

Partial Substructuring

- Substructure 1

=

111

11

111

iiiii

i

PP

uu

KKKK

11

1111

11


- 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



91

10.4. Symmetry of Structure Symmetry

In plane displacement 0≠

Out of plane displacement 0=

P P

C L



92

Anti-Symmetry In plane displacement 0=

Out of plane displacement 0≠

P P

C L



93

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



94

Cyclic Symmetry

1u

iu

2u



95

- 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



96

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



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



98

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



99

Chapter 11

Problems with Higher

Continuity Requirement

– Beams –

∇4 = ??



100

11.1. C1 Formulation

11.2. C0 Formuation

11.3. Timoshenko Beam



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

+−=−=

+−=+−=



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∆∆



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.



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 !!!



105

Chapter 12

Mixed Formulation



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



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?

Documents

Introduction to Finite Element Method - strana.snu.ac.krstrana.snu.ac.kr/lecture/fem_2017/Note17_FEM.pdf · Dept. of Civil and Environmental Eng., SNU. Structural Analysis Lab. Prof