KIRCHHOFF PLATE MODELLING USING FINITE ELEMENT METHOD
NAZIHAH BINTI ISMAIL
UNIVERSITI TEKNOLOGI MALAYSIA
KIRCHHOFF PLATE MODELLING USING FINITE ELEMENT METHOD
NAZIHAH BINTI ISMAIL
A dissertation submitted in partial fulfillment of the
requirements for the award of the degree of
Master of Science (Mathematics)
Faculty of Science
Universiti Teknologi Malaysia
DECEMBER 2010
iii
Dedicated to my beloved,
Abang,
dearest Mak and Ayah,
my brothers, Abe Yie, Abe We, Abe Pan,
Ran, Acah and Aqil,
&
my supervisor,
Prof. Dr. Shaharuddin Salleh
iv
ACKNOWLEDGEMENT
In the name of Allah S.W.T, The Most Merciful and Beneficent, Syukur and
Alhamdulillah that I have finally succeeded to complete this dissertation. In preparing
this dissertation, I was in contact with many individuals, who have contributed to the
accomplishment of this dissertation. Without their helps and guidance, I would have
never achieved this level. Specifically, I wish to recognize the very helpful insights
provided by my supervisor, Prof. Dr. Shaharuddin Salleh, who has generously provided
ideas, valuable advice, motivation, patient guidance, and great encouragement
throughout the duration of the attachment. Without his continued support and interest,
this thesis would not have attained its scope.
I am also indebted to Universiti Teknologi Mara (UiTM) for funding my M.Sc
study. My special thank also dedicated to my dear husband, Rosnaidi Deris for his
unconditional loving support, understanding and encouragement. Not forgotten to my
parents, Ismail Harun and Aripah Hassan. My thesis would not have proceeded smoothly
without their blessing and support.
Last but not least, I would like to express my sincere appreciation to all my
fellow friends for all the understanding and assistance they have given to me. A special
note of gratitude goes to Yana, Jue, James, N, and Yong. Once again, thank you to all of
you. Thank you.
v
ABSTRACT
The Kirchhoff plate theory works well for thin plates where the real shear strains
are small. In this study, the development of Kirchhoff plate theory using FEM is
presented.
2 M xx
x 2
2
The
2 M xy
xy
equilibrium
2 M yy
y 2
condition
of
the
problem
defined
as
q 0 is investigated in providing the appropriate boundary
conditions, hence to the establishment of the FE formulation of the problem. The plate
elements developed are the two-dimensional triangular element.
To meet the
convergence criteria, the quadratic interpolation function is adopted and the six nodes
triangular
element
is
developed.
The
deflection
w
takes
the
form
of
wx, y 1 2 x 3 y 4 x 2 5 xy 6 y 2 . The numerical results of two neighbouring
six nodes triangular elements are studied. These elements are considered to be
interconnected at specified nodes which lie on the element boundaries where adjacent
elements are considered to be connected. In each piece or element, the element shape
function N i , the stiffness matrix K , and the load vector f l are derived. The assemblage
of these matrices together with the derivation of boundary vector f b will yield to an
approximate solution for the displacement of the problem. The computational scheme is
developed by using Matlab programming language on the Windows environment for
computing the problem studied.
vi
ABSTRAK
Teori plat Kirchhoff berjalan dengan baiknya untuk plat-plat nipis di mana
ricihan ketegangan sebenar adalah kecil. Dalam kajian ini, pembangunan teori plat
Kirchhoff menggunakan FEM dibentangkan. Keadaan keseimbangan masalah yang
didefinisikan sebagai
2 M xx
x 2
2
2 M xy
xy
2 M yy
y 2
q 0 diselidiki dalam menyediakan
syarat-syarat sempadan yang bersesuaian , seterusnya untuk penubuhan perumusan FE
masalah tersebut. Elemen-elemen plat yang dikaji adalah berunsurkan segitiga dua
dimensi. Dalam menepati kriteria penumpuan, fungsi interpolasi kuadratik dipilih dan
unsur
segitiga
enam
nodus
dibangunkan.
Pesongan
sebagai wx, y 1 2 x 3 y 4 x 2 5 xy 6 y 2 .
w
mengambil
Penyelesaian
berangka
bentuk
dua
segitiga enam nodus yang berjiran adalah dikaji. Unsur-unsur ini dianggap saling
berkait pada nodus-nodus yang ditetapkan yang mana nodus-nodus ini berada di garisan
sempadan unsur-unsur berjiran tersebut. Dalam setiap bahagian atau unsur tersebut,
fungsi bentuk unsur N i , matriks kekukuhan K , dan vektor beban f l diterbitkan.
Himpunan matrik-matrik ini bersama dengan penerbitan vektor sempadan f b akan
menghasilkan satu penyelesaian hampiran kepada masalah yang dikaji. Perisian
berangka untuk menyelesaikan masalah yang dikaji dibangunkan dengan menggunakan
bahasa pengaturcaraan Matlab dan diaplikasikan pada persekitaran Windows.
vii
TABLE OF CONTENTS
CHAPTER
1
TITLE
PAGE
DECLARATION
ii
DEDICATION
iii
ACKNOWLEDGEMENT
iv
ABSTRACT
v
ABSTRAK
vi
TABLE OF CONTENTS
vii
LIST OF TABLES
x
LIST OF FIGURES
xi
LIST OF SYMBOLS
xiii
LIST OF APPENDICES
xv
RESEARCH FRAMEWORK
1.1
Introduction
1
1.2
Problem Statement
4
1.3
Objectives of the Study
4
1.4
Scope of the Study
5
1.5
Significance of the Study
5
1.6
Research Methodology
6
1.7
Thesis Organization
6
viii
2
LITERATURE REVIEW
2.1
Introduction
8
2.2
Review on the Development of Plate Theory
9
2.3
Plate Theory
11
2.3.1
Equilibrium Conditions
12
2.3.2
Kinematic Relations
16
2.3.3
Constitutive Relations
18
2.4
Further Derivations
19
2.5
Differential Equations for Plate Theory
22
2.6
Moments and Shear Forces Acting on an Arbitrary
24
Plane
2.7
3
Boundary Conditions
29
FINITE ELEMENT FORMULATION OF
KIRCHHOFF PLATE
3.1
Introduction
32
3.2
Strong Form and Weak Form
32
3.3
Green-Gauss Theorem
33
3.4
Weak Formulation – Proper Static Boundary
36
Conditions
3.5
Advantages of the Weak Form Compared to the
39
Strong Form
3.6
4
FE Formulation
41
NUMERICAL RESULTS
4.1
Introduction
45
4.2
General Procedure of Finite Element Method
45
4.2.1
46
Discretization of the Domain
ix
4.2.2
Interpolation Polynomials
47
4.2.3
Derivation of Element Stiffness Matrices and
56
Vectors
4.2.4
Assemblage
of
Element
Matrices
and
57
Vectors and Derivation of System Equations
4.2.5
Solution of the Finite Element System
58
Equations
4.3
5
6
59
COMPUTATIONAL RESULTS
5.1
Introduction
70
5.2
Features of an FE Computer Program
70
5.3
The Program Code
72
CONCLUSIONS AND SUGGESTIONS
6.1
Introduction
75
6.2
Summary of the Work
75
6.3
Suggestions for Future Study
77
REFERENCES
Appendix A
Test Example of the Problem
78
80-82
x
LIST OF TABLES
TABLE NO.
TITLE
PAGE
5.1
The declaration of the variables
72
5.2
The derivation of stiffness matrix and load vector for
73
element 1
5.3
The expanded matrix of stiffness matrix and load
74
vector for element 1
5.4
The assemblage vectors and matrices of all elements
74
xi
LIST OF FIGURES
FIGURE NO.
TITLE
PAGE
1.1
Steps in engineering mechanic analysis
1
1.2
Illustration of modelling steps
2
1.3
Finite element mesh of the structural part of a car
3
1.4
A thin plate subdivided into finite elements
3
2.1
Configuration and loading of plate
11
2.2
Illustration of M xx , M xy and V xz
13
2.3
Illustration of M yy , M yx , and V yz
13
2.4
Illustration of horizontal forces N xx , N yy , and N xy
14
2.5
Vertical shear forces and moments acting on an
15
infinitesimally small part of the plate
2.6
Illustration of stresses xx , yy and xy
18
2.7
Illustration of stress components nn , nm , and nz in a
24
plane defined by the unit normal vector n
2.8
Two-dimensional problem formulation of plate theory
29
2.9
Kinematics quantities w , n and m along the boundary
30
4.1
Two neighbouring triangular elements
47
4.2
Quadratic triangular element with six nodes wi , at
48
corners i 1,2,3 and midpoints i 4,5,6
xii
4.3
Area coordinates for a triangular element
49
4.4
Simply supported square plate
60
4.5
Finite element mesh of the problem
60
4.6
Global and local nodal points for element 1 and 2
60
5.1
The program flow of the FE numerical simulation of
71
the problem
xiii
LIST OF SYMBOLS
D
-
flexural rigidity of a plate
t
-
plate thickness
q
-
transverse loading of the plate
w
-
deflection of the plate
ij
-
stress component
M ij
-
moment component
V ij
-
vertical force component
N ij
-
horizontal force component
u0
-
displacement of the mid-plane in the x -directions
0
-
displacement of the mid-plane in the y -directions
ij
-
strain component
ij
-
shear strain component
-
stress components matrix
-
strain components matrix
D
-
plane stress constitutive matrix
E
-
Young’s modulus coefficient
v
-
Poisson’s ratio coefficient
-
curvature matrix
M
-
moments matrix
-
matrix differential operator
xiv
n
-
a unit normal vector located in the xy -plane
m
-
a unit vector that is orthogonal to n
t
-
traction vector
R
-
square matrix
r
-
unit vector defined in the xy -plane
-
Parameter
-
Parameter
-
the gradient of -
two-dimensional quantity [ x, y ]
A
-
region on mid-plane at two-dimensional problem
L
-
boundary of A
n
-
slope of a straight line normal to L
m
-
slope of a straight line tangential to L
x, y -
weight function
Ni
-
the element shape function
ui
-
nodal values
c
-
parameter
K
-
the stiffness matrix
fb
-
the boundary vector
fl
-
the load vector
f
-
the force vector
Li
-
triangular coordinate system i 1,2,3
A
-
area of the triangle
N
-
element shape function matrix
xv
LIST OF APPENDICES
APPENDIX
TITLE
PAGE
A
APPENDIX A
80
CHAPTER I
RESEARCH FRAMEWORK
1.1
Introduction
A variety of specializations under the umbrella of the mechanical engineering
discipline such as aeronautical, biomechanical, and automotive industries are modelled
by differential equations. Usually, the problem addressed is too complicated to be
solved by classical analytical methods. The finite element method (FEM) is a numerical
approach by which general differential equations can be solved in an approximate
manner. In other words, FEM is an approximate numerical procedure for analyzing
large structures and continua (Cook et al., 1989). Figure 1.1 illustrates generally how
the physical phenomenon encountered in engineering mechanics is modelled.
Figure 1.1 Steps in engineering mechanic analysis
2
As the FEM is a numerical, means of solving general differential equations, it
can be applied to various physical phenomena. Furthermore, FEM became popular with
the advancements in digital computers since they allow engineers to solve large systems
of equations quickly and efficiently. The method becomes a very useful tool for the
solution of many types of engineering problems such as the analysis of the plate and
beam structures, heat transfer and fluid flow. The method is also widely used in the
design of air frames, ships, electric motors, heat engines and spacecraft.
Although the finite element model does not behave exactly like the actual
physical structure, to obtain sufficiently accurate results for most practical applications
become possible. In FEM, the finite element model is created by dividing the structure
into smaller parts, called finite elements. Each element is interconnected by nodes and
the selection of elements for modelling the structure depends upon the behavior and
geometry of the structure being analyzed. The modelling pattern, which is generally
called mesh, is a very important part of the modelling process. This is because; the
results obtained depend upon the selection of the finite elements and the mesh size.
After having determined the behavior of all elements, these elements are then patched
together to form the entire region, which enable to obtain an approximate solution for
the behavior of the entire body. The situation discussed is shown in Figure 1.2 while
Figure 1.3 shows the finite element mesh of the structural part of a car.
Figure 1.2 Illustration of modelling steps
3
Figure 1.3 Finite element mesh of the structural part of a car
As mentioned in the second paragraph before, one of the applications of FEM is
the formulation of plate elements. Plate elements can be formulated and modelled
mathematically based on the Kirchhoff plate theory. The focus of this dissertation is to
develop the triangular elements for the finite element analysis of Kirchhoff plate
problem. An important aspect of the work is to implement the problem on the computer
using Matlab programming language. Figure 1.4 shows a region of thin plate is divided
into finite elements.
Figure 1.4 A thin plate subdivided into finite elements
4
1.2
Problem Statement
The subject of plates was one of the first to which the finite element method was
applied in the early of 1960’s. At that time the various difficulties that were to be
encountered were not fully appreciated and the topic remains one in which research is
active to the present day. The first convincing plate theory was established by Kirchhoff
which therefore also termed Kirchhoff plate theory as described by Boresi et al. (1978)
and Timoshenko and Woinowsky-Krieger (1959). In this study, it will be concentrated
in deriving a numerical solution for plate problem (for triangular elements), given their
boundary conditions by using finite element method. Great effort also will be
concentrated in developing the computational scheme of the problem by using Matlab
programming language.
1.3
Objectives of the Study
The objectives of this study are:
1. To study the various aspects of plate theory and its finite element (FE)
formulation.
2. To set up a numerical scheme by using FEM in solving the Kirchhoff plate
problem.
3. To develop a computational scheme of Kirchhoff plate problem by using
Matlab programming language.
5
1.4
Scope of the Study
The dissertation will be focusing on the formulation by using FEM for 2dimensional problem. The simplest plate element for the analysis of plates of arbitrary
shape which is the six nodes triangular element mesh with quadratic interpolation
functions is considered in this study. Both numerical and the computational scheme of
the problem then will be carried out. Effort will be concentrated on developing the
computational scheme/simulation of the problem by using Matlab programming
language.
1.5
Significance of the Study
The significance of the study is stated as follows:
1. The derivation of numerical codes and efficient algorithms of the Kirchhoff
plate problem help to solve the related problems in the future. The results
hence will contribute towards an enhanced understanding of the problem.
2. The simulation of the problem gives a significant results and solutions for
validation purposes in related problems.
6
1.6
Research Methodology
In this study, there are five steps that will be concentrated in order to get the
computational scheme for six nodes triangular Kirchhoff plate problem. The steps are:
1. Literature review on the Kirchhoff plate theory.
2. Comprehend the various aspects of plate theory. This covers equilibrium
conditions, kinematic relations, constitutive relation, differential equations,
and boundary conditions.
3. Next, the derivation of the differential equations of plate theory will be
conducted. Hence to the establishment of the FE formulation of the
Kirchhoff plate theory.
4. Then, the numerical solution of the problem is carried out.
5. Lastly, the simulation of the problem is presented. The computational code
is developed by using Matlab programming language and running on the
windows environment.
1.7
Thesis Organization
The dissertation is organized into six chapters. Chapter I is the research
framework. This chapter describes in detail some discussion with the introduction of the
7
study, a description of the problem, the objectives of the study, scope of the study,
significance of the study, research methodology, and chapter organization.
Chapter II starts with a brief literature review on the development of plate
theory. This chapter also contains a review and discussion the various aspects of plate
theory. The establishment of the differential equations of the problem is also presented
in this chapter.
Chapter III discusses in detail the FE formulation of the Kirchhoff plate theory.
It starts with the derivation of the weak form of the problem, and followed with the
establishment of stiffness matrix, the boundary vector, and the load vector.
Chapter IV presents the numerical results of the Kirchhoff plate problem. The
process of the FEM in obtaining the numerical scheme is outlined in detail in this
chapter.
In Chapter V, the computational scheme of the problem is presented. This
chapter also highlights on the analysis and discussion of the simulation model
developed with Matlab programming language.
Lastly, we will make some conclusions of this study in Chapter VI. This chapter
presents a summary of the important results and a discussion of the results. Suggestions
for future research are also given in this chapter. All the references quoted are listed in
the reference section after this chapter.
CHAPTER II
LITERATURE REVIEW
2.1
Introduction
Since Clough (1960) started the evolution of the term finite element, there has
been significant developments in finite element method. A large number of different
finite elements have been developed and extensively has been used for solving
problems in different fields of engineering. The finite element method became even
more popular with the advancement of microcomputers and development of various
efficient programming languages.
It would be a formidable task to identify and classify all published documents on
the subject that is certainly a few thousands. Nevertheless, the subject remains an active
area of research because of the importance of plate structures and the difficulty to
develop accurate and robust plate finite elements. This is because, in real world
scenario, we can roughly estimate that around 70 percent of finite element modelling of
industrial solid mechanics problems is dealing with plates and shells structures.
9
A very large number of plate elements have been developed and the most
commonly used plate theories are Kirchhoff plate theory (Timoshenko and WoinowskyKrieger, 1959) and Mindlin-Reissner plate theory (Hughes, 1987). For the present time,
we will limit our discussion to Kirchhoff plate theory. In this chapter, the review of the
development of plate theory and the establishment of the differential equations of the
problem will be discussed.
2.2
Review on the Development of Plate Theory
Tracing the family tree of plate theory to its roots, one travels back to the year
1776 where the earliest performing free vibration of plate problems was discovered.
This first mathematical statement of plate problems was probably done by Euler
(Ventsel and Krauthammer, 2001). According to Venstel and Krauthammer (2001),
Chladni a German physicist, then performed experiments on horizontal plates to
quantify their vibratory modes. In the experiment, he sprinkled sand on the plates, and
struck them with a hammer, and noted the regular patterns that formed along the nodal
lines where no vertical displacements occurred. Bernoulli then attempted to justify
theoretically the experimental results of Chaldni. The solution was based on the
previous work developed in the Euler-Bernoulli’s bending beam theory, but his results
did not capture the full dynamics.
As written in a book by Venstel and Krauthammer (2001), the French
mathematician named Germain, developed a plate differential equation that lacked a
warping term. By the way, she was awarded a prize by the Parisian Academy in 1816
for her work. But this result has been corrected by one of the reviewers of the
10
Germain’s work, Lagrange. Thus, Lagrange was the first person to present the general
plate equation properly.
Cauchy together with Poisson developed the problem of plate bending using
general theory of elasticity. Then, in year 1829, Poisson successfully expanded the
Germain-Lagrange plate equation to the solution of a plate under static loading. In this
solution however, the plate flexural rigidity D was set equal to a constant term (Ventsel
and Krauthammer, 2001).
Some of the greatest contributions toward thin plate theory came from
Kirchhoff’s thesis in 1850. This theory which presented by Kirchhoff is considered as
the first convincing plate theory. This plate theory is therefore also termed Kirchhoff
plate theory as described by Boresi et al. (1978) and Timoshenko and WoinowskyKrieger (1959). In analogy with the Bernoulli beam theory, the Kirchhoff plate theory
works well for thin plates where the real shear strains xz and yz are small. For thicker
plates, the Kirchhoff theory becomes questionable, and more refined plate theories must
be invoked which allow for the existence of non-zero shear strains. Among such refined
plate theories, the one proposed by Mindlin and Reissner is the most prominent
(Ottosen and Petersson, 1992). Of the numerous plate theories that have been developed
since the late 1800s, these two plate theories, Kirchhoff plate theory and Mindlin –
Reissner plate theory are widely accepted and used in engineering. In this study, we
only concentrated on Kirchhoff plate theory.
11
2.3
Plate Theory
Plates are initially flat structural members bounded by two parallel planes, called
faces, and a cylindrical surface, called an edge or boundary. The generators of the
cylindrical surface are perpendicular to the plane faces. The distance between the plane
faces is called the thickness t of the plate. It will be assumed that the plate thickness is
small compared with all other dimensions of the plate (faces, length, width, diameter,
etc.).
In addition, a plate is loaded by forces normal to the plane of the plate.
Specifically, a coordinate system is introduced and we assume the configuration of the
plate is symmetric about the xy -plane, as shown in Figure 2.1. This means that the xy plane is located in the mid-plane of the plate. Even though the plate thickness t in the
z -direction may vary; it will vary symmetrically about the xy -plane. The plate is
loaded by a transverse loading q , measured as positive in the z -direction. The
deflection w of the plate is also measured as positive in the z -direction.
Figure 2.1 Configuration and loading of plate
12
2.3.1
Equilibrium Conditions
For sections normal to the x and y -axes, the stress components xx , xy , xz
and yx , yy , yz exist. These stress components give rise to the following forces and
moments:
V xz V yz t
t
2
t2
2
t
xz dz
(2.1)
yz dz
2
and
M xx M yy t
t
t
t
2
z xx dz
2
2
z yy dz
(2.2)
2
M xy M yx t
t
2
z xy dz
2
By referring to Figure 2.2, it shows a plane cut in the plate and normal to the x axis. This illustrates the quantities defined by (2.1) and (2.2).
The stress
components xx , xy , and xz acting on this plane create the quantities M xx , M xy and
V xz which were defined by (2.1) and (2.2), and shown in Figure 2.2. It appears that
M xx is the bending moment per unit length, M xy is the twisting moment per unit length
and V xz is the vertical shear force per unit length.
13
Figure 2.2 Illustration of M xx , M xy and V xz
Next, we consider a plane cut in the plate and normal to the y -axis, the stress
components yx , yy , and yz acting on this plane create the twisting moment M yx per
unit length, the bending moment M yy per unit length and the vertical shear force V yz per
unit length. The quantities described are shown in Figure 2.3.
Figure 2.3 Illustration of M yy , M yx , and V yz
Besides these moments and vertical shear forces, the stress components xx , yy ,
and xy also result in the following horizontal forces:
N xx N yy t
t
t
t
2
xx dz
2
2
yy dz
2
N xy N yx t
t
2
2
xy dz
(2.3)
14
As shown in Figure 2.4, these forces act in the xy -plane. It appears that N xx is
the normal force per unit length in the x -direction, N yy is the normal force per unit
length in the y -direction and N xy is the horizontal shear force per unit length.
Figure 2.4 Illustration of horizontal forces N xx , N yy , and N xy
To establish the equilibrium conditions for the plate, plate is assumed to be
loaded by transverse forces only. Therefore, as no resulting forces (or restraints) act in
the xy -plane, horizontal equilibrium requires
N xx N yy N xy 0
(2.4)
Consider an infinitesimally small part of the plate as shown in Figure 2.5. As no
resulting forces act in the xy -plane (2.4), all forces and moments acting on this small
part of the plate appear from Figure 2.5. Vertical equilibrium requires
V yz
V
dy dx V xz dy 0
q dxdy V yz dx V xz xz dx dy V yz y
x
which can be written as
Vxz V yz
q 0
y
x
(2.5)
15
Figure 2.5 Vertical shear forces and moments acting on an infinitesimally small part of
the plate
Next let considering the moment equilibrium about the right side of the small
part of the plate, which is parallel to the x -axis
V
1
1
1
dy V yz dxdy V xz xz dx dy dy V xz dy dy M yy dx
2
x
2
2
M xy
M yy
M xy
dx dy M yy
dy dx M xy dy 0
x
y
q dxdy
that is,
q
M xy M yy
V 1
1
dy V yz xz dy 0
2
x 2
x
y
16
As dy is an infinitely small quantity, hence we have
M xy
x
M yy
y
V yz
(2.6)
Considering the moment equilibrium about one of the sides parallel to the y -axis, in a
similar way we have
M xx M xy
V xz
x
y
2.3.2
(2.7)
Kinematic Relations
The plate is assumed to deform in accordance with Bernoulli’s assumption
which explains that plane sections normal to the mid-plane remain plane and normal to
the mid-plane during deformation. Hence these reach the following displacement in
the x , y , and z -plane directions:
ux u0 z
uy 0 z
w
x
w
y
(2.8)
uz w
where u 0 and 0 are the displacements of the mid-plane in the x and y -directions
respectively. It follows that
17
u 0 u 0 x, y ;
0 0 x, y and
w wx, y (2.9)
where the assumption that the deflection w is independent of z have been introduced
by (2.9), that is w wx, y . Given the normal strains are defined:
xx yy zz u x
x
u y
y
(2.10)
u z
z
and the derivation of shear strains are:
xy u x u y
y
x
xz u x u z
z
x
yz u y
z
(2.11)
u z
y
Hence, from (2.8) and (2.9), the strains as given by (2.10) and (2.11) become
xx u 0
2w
z 2
x
x
yy 0
2w
z 2
y
y
xy u 0 0
2w
2z
y
x
xy
(2.12)
18
and
zz xz yz 0
(2.13)
Noted that the kinematic assumptions imply that the shear strains xz and yz are zero.
2.3.3
Constitutive Relations
Assuming that Hookes’s law is applicable, it is not possible to obtain a
correspondence between the non-zero shear stresses xz and yz necessary to maintain
equilibrium in (2.5) and the zero shear strains xz and yz . However, as discussed
before, the plate is assumed to be thin, and the largest stresses will be xx , yy and xy
that are illustrated in Figure 2.6.
Figure 2.6 Illustration of stresses xx , yy and xy
This observation suggests that the assumption of plane stress is most applicable, that is
D
(2.14)
19
where
xx yy xy and
xx yy xy (2.15)
For isotropic elasticity, the plane stress constitutive matrix D is given by,
1 v
E v 1
D
1 v2 0 0
0 1
1 v 2
0
(2.16)
However, the D -matrix may take any form. It is even possible to consider initial
strains, 0 for instance, in terms of thermal strains. Due to this, (2.14) is replaced by
D D 0
2.4
(2.17)
Further Derivations
Previously we have the strains as defined by (2.12). This may be written in the
form
0 z
(2.18)
20
where
u 0
x
0
0
y
0
0 u y
x and
2w 2
x2 w 2 y2 w
2 xy (2.19)
and here is denoted the curvature matrix, while 0 expresses the straining of the midplane. Using (2.18) and (2.14), we have
D 0 zD
(2.20)
The matrix M is defined by
M xx M M yy M xy (2.21)
Previously we have the expression for the moments given by (2.2). These
expressions can be written in the compact form
M
t
2
(2.22)
z dz
t2
By inserting (2.20) into (2.22) and recalling that neither 0 nor depends on the
coordinate z , produce
M D 0
t
2
t2
z dz - D
t
2
t2
z 2 dz
(2.23)
21
where it was assumed that D is independent of z , which is
D D( x, y )
(2.24)
We have
t
2
t2
z dz 0
and
t
2
t2
z 2 dz t3
12
that is, irrespective of the value 0 -strains, (2.23) reduces to
~
M - D
where
~ t3
D D
12
(2.25)
with defined by (2.15), the horizontal forces given by (2.3) can be expressed as
N xx N yy N xy t
2
dz
t2
Use of (2.20) yields
N xx 0
N yy D t
N xy (2.26)
As discussed in (2.4) before, in the present case, there is no resulting horizontal
forces act in the mid-plane. Hence, it can be concluded that
0 0
(2.27)
22
In accordance with expectations, this shows that there is no straining of the midplane. But even in the situation where the horizontal forces are different from zero, from
(2.25) and (2.26) explain that although the moments M are controlled by the curvature
matrix , the horizontal forces are determined by the in-plane strains 0 . This implies
that the phenomena of bending and straining of the mid-plane are uncoupled
phenomena, which can be treated separately if (2.24) holds. Moreover, if the mid-plane
is deformed, means 0
2.5
0 , the phenomenon is called the membrane-action.
Differential Equations for Plate Theory
Previously, we have discussed there is no resulting horizontal forces act in the
mid-plane. In this present situation, (2.27) holds, hence (2.18) and (2.20) reduce to
z
zD
and
(2.28)
Recall that the equilibrium conditions were given by (2.5) - (2.7). Since the
moments can only be expressed in terms of kinematic quantities, the shear forces
V xz and V yz are eliminated from (2.5) - (2.7). Differentiate (2.6) with respect to y and
(2.7) with respect to x; adding the results and use (2.5) to obtain
2 M xx
x 2
2
2 M xy
xy
2 M yy
y 2
q0
(2.29)
23
This equilibrium condition holds irrespective of the constitutive assumption.
Introduce the matrix differential operator , which is defined by
2 2
x2 2 y 2 2 xy (2.30)
It happens that the equilibrium condition (2.29) can be written as
T M q 0
(2.31)
Moreover, rewrite (2.19) as
w
(2.32)
Hence (2.25) takes the form
~ M D w
(2.33)
Inserting this expression into (2.31), yields the following differential equation for plate
theory:
~ T D w q
(2.34)
When the deflection w has been determined from this differential equation, all
quantities of interest can be derived. Assume that the thickness t is constant and that
24
isotropic elasticity is considered, which D is given by (2.16). If, in addition, D does
not depend on x and y , (2.34) takes the follows differential equation which is also
known as the biharmonic equation. This fourth- order differential equation was derived
in year 1811 by Langrange (Reddy, 2007).
4w
4w
4 w 12 1 v 2
q
2
Et 3
x 4
x 2 y 2 y 4
2.6
(2.35)
Moments and Shear Forces Acting on an Arbitrary Plane
Consider a section in the plate normal to the mid-plane and defined by the unit
normal vector n , which is located in the xy -plane as shown in Figure 2.7. A unit vector
m located in the xy -plane is also defined and taken to be tangential to the section, that
is orthogonal to n. Moreover, the vector m is chosen so that n and m , after a suitable
rigid-body rotation, follow the same directions as the x and y -axes, respectively.
Figure 2.7 Illustration of stress components nn , nm , and nz in a plane defined by the
unit normal vector n
25
We have
n x n x n n y n y n z 0 n m 1;
and
mx mx m m y m y m z 0 (2.36)
and
n !m 0
(2.37)
and
The traction vector t acting on the section defined by n is given by
t x xx n x xy n y xz n z
t y yx n x yy n y yz n z
(2.38)
t z zx n x zy n y zz n z
and as n z 0 , we obtain
t x xx n x xy n y t t y yx n x yy n y t z zx n x zy n y (2.39)
In defining stress components, we consider a plane normal to the x -axis, and the
stress components xx , yx and xz were defined as the components of the traction
vector t in the x , y and z -directions respectively. In an analogy to this, for the plane
identified by the unit normal vector n , we define the stress components nn , nm , nz as
the components of the traction vector in the n, m , and z -directions respectively. These
stress components are illustrated in Figure 2.7, and we obtain
26
nn n ! t
nm m ! t
nz "0 0 1# t
With n and m defined by (2.36) and the traction vector t by (2.39), the result becomes
nn n x2 xx n y2 yy 2n x n y xy
nm n x m x xx n y m y yy n y m x n x m y xy
(2.40)
nz n x xz n y yz
As in the definitions (2.1) and (2.2), the bending moment M nn per unit length,
twisting moment M nm per unit length and vertical shear force V nz per unit length acting
on the section defined by the unit vector n are defined according to
2
t
M nm V nz t
M nn t
t
t
t
2
z nn dz
2
2
z nm dz
(2.41)
2
nz dz
2
Inserting (2.40) into (2.41) and making use of the definitions (2.1) and (2.2), produce
M nn n x2 M xx n y2 M yy 2n x n y M xy
M nm n x mx M xx n y m y M yy n y mx nx m y M xy
Vnz n xVxz n yV yz
(2.42)
27
For future purposes, some useful relations will be derived from (2.42). The
relations are used for FE formulation which will be discussed in details in the next
chapter. Previously we have the three-dimensional vectors n and m defined by (2.36)
and (2.37), now we define the two-dimensional orthogonal vectors n and m by
n x n ,
n y m x m m y where
n m 1,
n !m 0
(2.43)
These two vectors are located in the xy plane. By carrying out the matrix
multiplications it follows that
M nn n ! An,
M nm m ! An
and
(2.44)
where
M xx
A
M xy
M xy M yy (2.45)
Using (2.43) – (2.45), the following expression is derived
nM nn mM nm nn ! mm T An
Introduce the square matrix R defined by
R nn ! mm !
As n and m are orthogonal, any vector r in the xy plane can be expressed as
r n m
(2.46)
28
where and are some parameters. Since n and m are orthogonal unit vectors, it
appears that
R I r 0
Rr r;
As this relation holds for arbitrary vectors r , we conclude that R I . Hence we have
nn ! mm ! I
(2.47)
where I is the unit square matrix. (2.46) reduces to
nM nn mM nm An
Multiplication by the quantity ! , where x, y is an arbitrary function and denotes the gradient of , leads
! nM nn ! mM nm ! An
(2.48)
We have the definition of
d
T
m
dm
(2.49)
Use of (2.49) and (2.45), (2.48) leads to the following relation
d
d
M nn M nm dm
dn
x
M xx
y M xy
M xy n x M yy n y 29
In a simplified form we have
d
d
M nn M nm dn
dm
x
M xx n x M xy n y y M xy n x M yy n y (2.50)
The relation derived by (2.50) carries the significant aspect since it will later be
used in the weak formulation. The weak formulation is in fact a reformulation of the
strong form and it is from the weak form that the FE approach is established.
2.7
Boundary Conditions
The identification of proper boundary conditions had already been established
by Kirchhoff himself in 1850 (Vitor, 2001). From previous discussion, we have arrived
at a two-dimensional problem, where everything depends on the coordinates x and y in
the mid-plane of the plate. By referring Figure 2.8, it illustrates that the region spanned
by this mid-plane is denoted by A and its boundary by L.
Figure 2.8 Two-dimensional problem formulation of plate theory
30
The assumption for the kinematics of the plate is such that a straight line normal
to the mid-plane remains straight during deformation. Along the boundary L , the
movement of such a straight line is given by its slope n , normal to the boundary and its
slope m tangential to the boundary. The situation is shown in Figure 2.9. Therefore, we
can identify three kinematic quantities along the boundary which are; the deflection w
and the slopes n and m .
Figure 2.9 Kinematics quantities w , n and m along the boundary
In Kirchhoff plate theory, it is assumed that a straight line normal to the midplane remains straight, and normal to the mid-plane during deformation; we have
n dw dn and m dw dm . That is, the three kinematic quantities identified
are w, dw dn , and dw dm . Suppose that the deflection w is known along part of the
boundary, hence we may have w w(m) . This is shown in Figure 2.8. As w(m) is known
along the boundary, then the slope dw dm is also known. This means there are only two
independent kinematic quantities namely w and dw dn . These quantities are defined as
the independent kinematic boundary conditions;
w
and
dw
dn
(2.51)
31
Consider next the boundary conditions given in terms of moments and shear
forces, which denoted as the static boundary conditions. Referring to (2.41), along the
boundary it is able to prescribe the bending moment M nn , the twisting moment M nm and
the shear force vector V nz are the static boundary conditions;
M nn ,
M nm
and
Vnz
(2.52)
Whereas there are two independent kinematic boundary conditions given by
(2.51), apparently three independent static boundary conditions are defined (2.52). To
investigate this problem, the differential equation in w given by (2.34) is considered.
Considering the biharmonic differential equation defined by (2.35), it appears that this
differential equation is of fourth order. Assuming that the boundary L consists of two
regions, one part where the kinematic boundary conditions are prescribed, and another
part where the static boundary conditions are prescribed. This condition led to the
conclusion that along each of these parts, the solution of the fourth-order differential
equation requires the specification of two independent boundary conditions. This
criteria is fulfilled for the kinematic boundary conditions (2.51), and it implies that the
three static quantities (2.52) must in such a way that they appear in the form of two
independent terms.
Hence, this is the significant of the weak form. The weak form by itself gives
the natural boundary conditions. Thus, by deriving the weak form of the plate problem,
the proper static boundary conditions will be identified. The derivation of the weak
form will be discussed in detail in the next chapter.
CHAPTER III
FINITE ELEMENT FORMULATION OF KIRCHHOFF PLATE
3.1
Introduction
In Chapter 2, an overview of plate theory and the establishment of Kirchhoff
plate theory have been discussed in detail. The main purpose of this chapter is to present
the derivation of the FE formulation of Kirchhoff plate. In this chapter, we shall
therefore discuss the weak formulation, hence to the establishment of the FE
formulation of the problem.
3.2
Strong Form and Weak Form
The differential equation developed in Chapter 2, given the equilibrium
condition defined by (2.29), is known as strong form for the Kirchhoff plate problem.
33
The strong form, in contrast to a weak form, requires strong continuity on the dependent
field variables (in this case we have M xx , M xy , and M yy ). These field variables have to
be differentiable up to the order of the differential equation that defined in the strong
form. Computing the exact solution for a strong form is usually very difficult for
practical engineering problems. The finite difference method can be used to solve the
strong form to obtain an approximated solution. However, the method usually works
well for problems with simple and regular geometry and boundary conditions.
The weak form is often an integral form and requires a weaker continuity on the
field variables. Due to this criterion and with the integral operation, a formulation
based on a weak form usually produces a set of discretized system equations that give
much more accurate results, especially for problems of complex geometry. Hence the
weak form is preferred for obtaining an approximated solution. The FEM is a typical
example of successfully using weak form formulation. The weak form usually leads to a
set of well-behaved algebraic system equations. As the problem domain can be
discretized into different types of elements, FEM can be applied for many practical
engineering problems with most kinds of complex geometry and boundary conditions.
3.3
Green-Gauss Theorem
In obtaining the weak form, the integration by parts is necessary. Addition to
this, in two or three dimensions, a similar type of integration needs to be performed.
Since the problem considered is two dimensional problems, for this purpose we shall
make use the Green-Gauss theorem, which is based on Gauss’ divergence theorem.
These theorems are of fundamental importance applied for many applications in
engineering.
34
Given the functions x, y and $ $ x, y , and the region A with boundary
L , the arbitrary vector q is defined by
q x q q y $ where qx and q y are the components in the x and y -directions respectively. Given the
Gauss’ divergence theorem is defined as
div q dA L
A
q T n dL
(3.1)
where div q is the divergence of q defined by
div q q x q y
x
y
(3.2)
Assume now that instead of q , consider the vector q where is a scalar. Using the
definition (3.2) we obtain
div q q
q
q x q y x y q x q y
x
y
x
y
x
y
Introduce the definition of the gradient given by
x y (3.3)
35
Using the definition of the gradient , equation (3.3) can be written as
div q div q q
T
(3.4)
Inserting q instead of q in (3.1) and making use of (3.4), the Green-Gauss theorem is
obtained according to
div q dA L
A
q T n dL -
T
A
q dA
(3.5)
The Green-Gauss theorem can be seen as the two dimensional counterpart for
integration by parts. Most of the cases, it is important to write the Green-Gauss theorem
in a slightly different way. If the arbitrary vector q is chosen so that q x $ and q y 0 ,
equation (3.5) reduces to
A
$
dA x
L
$n x dL -
A
$ dA
x
(3.6)
Alternatively, if q is chosen such that q x 0 and q y $ , equation (3.5) takes the form
A
$
dA y
L
$n y dL -
$ dA
A y
(3.7)
In the next section, the weak formulation of the problem will be carried out in
detail. The equations defined by (3.6) and (3.7) hold the important aspect which the
definitions are repeatedly used in deriving the weak form of the problem.
36
Weak Formulation – Proper Static Boundary Conditions
3.4
As already touched upon, we lost one static boundary condition due to the
reformulation of the equilibrium equations. Originally, we had the three first-order
differential equations (2.5) – (2.7) corresponding to three static boundary conditions.
However, when the shear forces were eliminated we arrived at the second-order
differential equation (2.29) considered as the strong form, which only requires two
static boundary conditions.
Now we shall establish the weak form of equation (2.29). The equation (2.29)
can be written as
M xx M xy
x x x y
M xy
y x
M yy
y y
q0
(3.8)
To obtain the weak form, we multiply (3.1) by an arbitrary weight function x, y and
integrate over the region A ;
M xx dA x x A
A
M yy
y y
A
dA M xy
x y
q dA 0
A
dA A
M xy
y x
dA
37
Applying the integration by parts and the use of Green-Gauss theorem given in the form
of (3.6) and (3.7), hence the results give
L
M xy
M xx
M xx
M xy
n x dL dA n x dL dA
L
A x
A x
x
x
y
y
M xy
M yy
M xy
M yy
n y dL dA n y dL dA
A y
A y
L
x
x
y
y
q dA 0
L
A
Using (2.6) and (2.7) in the boundary integral gives
L
M xx
dA A x
x
V xz n x dL V yz n y dL L
M xy
dA A y
x
M yy
dA A y
y
M xy
dA
A x
y
q dA 0
A
In the boundary integrals, applying of (2.42), and the result becomes
M xx
dA A x
x
L
M xy
dA A x
y
M xy
dA A y
x
M yy
dA
A y
y
Vnz dL q dA 0
A
Repeated use the Green-Gauss theorem ((3.6) and (3.7)), yields
M xx n x dL x
M xy n x dL y
L
L
2
M yy dA A y 2
L
2
M xx dA A x 2
2
M xy dA A xy
L
L
M xy n y dL x
M yy n y dL
y
Vnz dL q dA 0
A
2
M xy dA
A yx
38
rearrange the differential equation, hence it can be written as
2
2
2
2
M
M
M xy dA
xx
yy
2
2
A x
xy
y
M xx n x M xy n y M xy n x M yy n y
y
x
Vnz dL q dA 0
L
L
dL
(3.9)
A
With the definitions of (2.21) and (2.30), we have
!
2
2
2
M xy
M 2 M xx 2 M yy 2
xy
y
x
(3.10)
Moreover, the relation given by (2.50) is also can be written as
d
d
M nn M nm M xx n x M xy n y M xy n x M yy n y
x
y
dn
dm
(3.11)
Where the unit vector m is orthogonal to n , that is m is tangential to the boundary.
With (3.10) and (3.11), (3.9) reduces to
!
M dA A
d
d
M nn M nm dL d
n
d
m
L
L
Vnz dL q dA
A
Now, as m is tangential to the boundary, we may take dL dm, thus we have
L
d
M nm dL dm
dM nm
d
M nm dL dL
L dm
L
dm
dM nm
d
M nm dL
L
L
dm
dM nm
dL
L
dm
(3.12)
39
Hence, (3.12) takes the form
!
M dA A
d
M nn dL dn
d
M nn dL dn
Vnz L
L
dM nm
dL dm
L
Vnz dL q dA
A
simplified to
!
M dA A
L
L
dM nm dL dm q dA
A
(3.13)
The term defined in (3.13) is the weak form of the equilibrium equation considered
(2.29). Hence, the natural boundary conditions are given by;
M nn
and
Vnz dM nm
dm
(3.14)
Therefore, it is not the quantities M nn , M nm , and V nz given by (2.52) determine
the static boundary conditions. The proper static boundary conditions defined by (3.14)
are called the Kirchhoff’s boundary conditions and the term Vnz dM nm
is known the
dm
effective shear force.
3.5
Advantages of the Weak Form Compared to the Strong Form
Generally, FE method is a numerical method to solve arbitrary differential
equations. To achieve this objective, it is a characteristic feature of the FE approach that
the differential equations in question are first reformulated into an equivalent form,
40
called weak formulation. This is because; the fundamental for the establishment of the
FE method is on the weak form that the FE formulation is based.
In order to understand this point, we consider the differential equation in the
strong form defined in (2.29). It appears that the unknown functions M xx , M xy and
M yy are differentiated twice. In other hand, the FE approach is an approximate method,
so in one way or another we have to replace the functions by an approximate one. If this
approximation is made directly in (2.29), we need to deal with an approximating
function, which is at least twice differentiable. However, in the obtained weak form
(3.13), only the first derivative enters. That is, if we choose the weak form as the basis
for the approximation, we may deal with approximating functions which only need to
be differentiable once.
This aspect clearly favours the weak form compared with the strong form, and it
also suggests the terminology of weak and strong form. Another point, closely related to
the matter discussed before, is that the weak form provides in fact a more general
formulation than the strong form. We have previously seen that before deriving the
weak form, we have three independent static boundary conditions (2.52). So it is
tempting to obtain the weak form of the plate problem discussed since the equation
itself gives the natural boundary conditions. After the weak formulation, from the static
boundary conditions defined in (2.52), we able to decide on the proper static boundary
conditions defined by (3.14).
41
3.6
FE Formulation
Following this, we have a detailed discussion of various aspects of plate theory.
Now, we arrive to the establishment of FE formulation of the problem. Previously, as
the deflection w is the unknown function in the plate theory (2.34) and (2.35), at some
stage the approximation for w quite generally as
w Na
(3.15)
where,
N "N1 N 2 N n #
u1 u
a 2
u n and
(3.16)
and n is the number of unknowns for the entire plate. From (3.15) and (2.30), produce
w Ba
where
B N
(3.17)
which is
2 N 2 N1
2
2
2x 2x
N N1
B
2 2
y
2 y2
N N1
2 xy 2 xy
2N2
x 2
2N2
y 2
2N2
2
xy
2Nn x 2 2Nn y 2 2Nn 2
xy (3.18)
42
With the Galerkin method, the following expression is used for the arbitrary weight
function :
Nc
(3.19)
As is arbitrary, the parameters c are also arbitrary. It appears that
Bc
(3.20)
Use of (2.49) provides
d
!
!
n c ! N n
dn
(3.21)
where,
N1
N Nx
1
y
N 2
x
N 2
y
N n x N n y (3.22)
Since T , (3.19) is also can be written as
c! N!
(3.23)
Previously we have the weak formulation of the equilibrium condition given by
(3.13), which holds for arbitrary weight functions . Inserting (3.20), (3.21) and (3.23),
and noting that c ! is independent, thus the weak form of the equilibrium equation gives
43
c ! B ! M dA A
N !
L
nM nn dL N
L
!
dM nm Vnz dL dm N
!
A
q dA 0
As c ! is arbitrary, therefore
B
!
A
M dA N
!
L
nM nn dL N
L
!
dM nm Vnz dL dm N
!
A
q dA 0
(3.24)
which is an expression for the equilibrium condition for the plate. It is recalled that the
kinematic boundary conditions are w and
boundary conditions are M nn and Vnz dw
(defined by (2.51)), and the natural
dn
dM nm
(defined by (3.14)).
dm
Moreover, it is not always possible to divide the boundary into parts where the
kinematic boundary conditions are prescribed and parts where the static boundary
conditions are prescribed. Indeed, it is entirely possible to have kinematic and static
boundary conditions prescribed along the same part. For instance, along a simply
supported we have w 0 and M nm 0 . At this point, introduce the constitutive and
kinematic assumptions adopted. From (2.33) and (3.17) we have
~
M DBa
(3.25)
Insertion into (3.24) gives the FE-formulation,
B D~ B dAa N V
!
A
!
L
nz
dM nm dL dm !
N nM
L
nm
dL N
A
!
q dA
(3.26)
44
Written in a more compact fashion we have the stiffness matrix K, the boundary
vector f b , and the load vector f l as follows,
~
K B ! DB d A
A
fb N
fl N
!
dM nm
Vnz dm
L
A
!
dL L
!
N
nM nn dL
(3.27)
q dA
The formulation given by (3.26) can be written as
Ka f b f l
(3.28)
The force vector f is defined by
f fb fl
(3.29)
Thus, we have
Ka f
(3.30)
Previously, the strong form and the weak form, hence to the establishment of FE
formulation of the Kirchhoff plate problem have been discussed in detail. We have seen
that once the FE formulation is comprehended, it is a straightforward task to derive the
numerical result of the problem. In the next chapter, we shall discuss some general
guidelines for the FEM of the problem. In addition to this, the numerical process will be
investigated in detail.
CHAPTER IV
NUMERICAL RESULTS
4.1
Introduction
In the previous chapter, a very detailed of FE formulation of the problem has
been dealt. This Chapter 4 presents the numerical solution of the problem discussed.
But before we arrive to the numerical result of the problem, there are some procedures
of FEM must be adopted in obtaining the results.
4.2
General Procedure of Finite Element Method
In FEM, the actual continuum or body of matter is represented as an assemblage
of subdivisions called finite elements. These elements are considered to be
interconnected at specified joints which are called nodes or nodal points. The nodes
46
usually lie on the element boundaries where adjacent elements are considered to be
connected. Since the actual variation of the field variable (in this study is deflection w )
inside the continuum is not known, we assume that the variation of the field variable
inside a finite element can be approximated by a simple function. These approximating
functions are defined in terms of the values of the field variables at the nodes.
When the field equations (equilibrium equation defined by (2.29)) for the whole
continuum are written, the new unknowns will be the nodal values of the field variable.
By solving the field equations, which are generally in the form of matrix equations, the
nodal values of the field variable will be known. Once these are known, the
approximating functions define the field variable throughout the assemblage of
elements. The solution of a general continuum problem by the FEM always follows an
orderly step-by-step process. Without exception, the derivation of the numerical result
of the Kirchhoff plate problem also follows the same procedures. In this study, steps
involved generally can be outlined as discretization of the domain, interpolation
polynomials, derivation of element stiffness matrices and vectors, assemblage of
element matrices and vectors and derivation of system equations, finally the solution of
finite element system equations.
4.2.1
Discretization of the Domain
The discretization of the domain or solution region into subregions (finite
elements) is the first step in FEM. This is equivalent to replace the domain having an
infinite number of degrees of freedom, by a system having finite number of degrees of
freedom. The process of discretization is essentially an exercise of engineering
judgment. The shape, size, number and configuration of the element have to be chosen
47
carefully such that the original body or domain is simulated as closely as possible
without increasing the computational effort needed for the solution.
In this dissertation, the triangular element for two dimensional problem is
considered. The domain of the body is defined by assembling two triangular elements
with equal subdivisions. This is shown in Figure 4.1.
Figure 4.1 Two neighbouring triangular elements
4.2.2
Interpolation Polynomials
The basic idea of FEM is piecewise approximation, that is, the solution of a
complicated problem is obtained by dividing the region of interest into small regions
and approximating the solution over each subregion by a simple function. Thus it is a
necessary and important step to choose a simple function for the solution in each
element. The function used to represent the behavior of the solution within an element
is usually called the interpolation function or interpolation model. The chosen of
polynomial type of interpolation function depends on some reasons.
48
Firstly, it is easier to formulate the finite element equations with polynomial
type of interpolation functions. Specifically, it is easier to perform differentiation or
integration with polynomials. The second reason, it is possible to improve the accuracy
of the results by increasing the order of the polynomial. Theoretically a polynomial of
infinite order yields to obtain an exact solution. But in practice, we take polynomials of
finite order only as an approximation. In this study, the quadratic function is selected as
the interpolation function. The domain is idealized to discretize into triangular element
with six nodes. This is shown in Figure 4.2.
Figure 4.2 Quadratic triangular element with six nodes wi , at corners i 1,2,3 and
midpoints i 4,5,6
As two dimensional problem with quadratic model is chosen, hence the
deflection w take the form
wx, y 1 2 x 3 y 4 x 2 5 xy 6 y 2
(4.1)
Where 1 , 2 ,..., 6 , are parameters. In a moment these parameters are expressed in
terms of the deflection at the nodal points of the element. As we have six parameters,
each element contains six nodal points.
49
A natural coordinate system for a triangular element is shown in Figure 4.3. The
natural coordinates are defined as
L1 A1
,
A
L2 A2
,
A
and
L3 A3
A
(4.2)
Where A1 is the area of the triangle formed by the points P, 2 and 3; A2 is the
area of the triangle formed by the points P, 1 and 3; A3 is the area of the triangle formed
by the points P, 1 and 2; and A is the area of the triangle 1, 2 and 3 . As Li are defined in
terms of areas, they are also known as area coordinates.
Figure 4.3 Area coordinates for a triangular element
Since A1 A2 A3 A, hence we have
A
A1
A
2 3 L1 L2 L3 1
A
A
A
(4.3)
The properties of L1 , L2 , and L3 shows that they are also the shape functions for the
two dimensional simplex (linear) triangular element. Thus we can define
50
N i L1 ,
N j L2 , and
N k L3
(4.4)
The relation between the natural and Cartesian coordinates is given by
x x1 L1 x2 L2 x3 L3
and
y y1 L1 y 2 L2 y3 L3
(4.5)
To every set of natural coordinates L1 , L2 , and L3 , there corresponds a unique
set of Cartesian coordinates ( x, y) . At node 1, L1 1 and L2 L3 0 ; at node 2,
L2 1 and L1 L3 0 ; and at node 3, L3 1 and L1 L2 0 . The linear relationship
between Li i 1,2,3 and ( x, y) implies that the contours of L1 are equally placed
straight lines parallel to the side 2, 3 of the triangle on which L1 0 ; the contours of L2
are equally placed straight lines parallel to the side 1, 3 of the triangle on which L2 0 ;
and the contours of L3 are equally placed straight lines parallel to the side 1, 2 of the
triangle on which L3 0 . Equations (4.3) and (4.5) can be expressed in matrix form as
1 1
x x
1
y y1
1
x2
y2
1
x3 y 3 L1 L 2
L3 (4.6)
Inverted the equation (4.6) to obtain
L1 x 2 y 3 x 3 y 2 L 1 x y x y 3 1
1 3
2
2A L3 x1 y 2 x 2 y1 y 2 y3 x3 x 2 y3 y1 x1 x3 y1 y 2 x 2 x1 1 x
y (4.7)
Where A is the area of the triangle 1, 2 and 3 given by
1 x1
1
A 1 x2
2
1 x3
y1
y2
y3
(4.8)
51
Equation (4.8) can be written as
A
1
x2 y1 x1 y 2 x3 y 2 x2 y3 x1 y3 x3 y1 2
(4.9)
Previously we have the equations defined by (3.15) and (3.16). Hence (3.16) can be
defined
N "N1 N 2 N 3 N 4 N 5 N 6 #
and
u1 u 2
u a 3
u 4 u 4 u6 (4.10)
Where N1 , N 2 , ... , N 6 are called the element shape functions or element
interpolation functions, while u1 , u 2 ,..., u6 are the displacements placed at every nodal
point in each element defined. Since we choose quadratic function as the interpolation
model, the values of the field variable at three corner nodes and three midpoints nodes
(shown in Figure 4.2) are taken as the nodal unknowns and the interpolation model can
be expressed as ( x, y) N "N1 N 2 N 3 N 4 N 5 N 6 # where N i can be derived from the
general quadratic relationship
N i a1i L1 a 2i L2 a3i L3 a 4i L1 L2 a 5i L2 L3 a 6i L1 L3
where a1i , a 2i ,..., a 6i are constants.
(4.11)
52
By requiring that N i be equal to one at node i and zero at each of the other nodes, for
N1 we impose the requirements;
( 1 at node 1 L1 1, L2 L3 0
% 0 at node 2 L 1, L L 0
2
1
3
%
%% 0 at node 3 L3 1, L1 L2 0
N1 '
1
%0 at node 4 L1 L2 2 , L3 0
%0 at node 5 L2 L3 12 , L1 0 %
1
&%0 at node 6 L1 L3 2 , L2 0
Substitute the values into equation (4.11) and solve the system of equations, we have
the value of constants
a1 1, a 2 0, a 3 0, a 4 2, a 5 0, and a 6 2
So that equation (4.11) becomes
N1 L1 2L1 L2 2L1 L3
By using the condition L1 L2 L3 1 , we obtain
N1 L1 2L1 1
Similarly for N 2 ;
( 0 at node 1 L1 1, L2 L3 0
% 1 at node 2 L 1, L L 0
2
1
3
%
%% 0 at node 3 L3 1, L1 L2 0 N2 '
1
%0 at node 4 L1 L2 2 , L3 0 %0 at node 5 L2 L3 12 , L1 0 %
%&0 at node 6 L1 L3 12 , L2 0 53
the value of constants derived are
a1 0, a 2 1, a 3 0, a 4 2, a 5 2, and a 6 0
So that equation (4.11) becomes
N 2 L2 2L1 L2 2L2 L3
By using the condition L1 L2 L3 1 , we obtain
N 2 L2 2L2 1
For N 3 we have;
( 0 at node 1 L1 1, L2 L3 0
% 0 at node 2 L 1, L L 0
2
1
3
%
%% 1 at node 3 L3 1, L1 L2 0
N3 '
1
%0 at node 4 L1 L2 2 , L3 0
%0 at node 5 L2 L3 12 , L1 0
%
%&0 at node 6 L1 L3 12 , L2 0
the value of constants derived are
a1 0, a 2 0, a 3 1, a 4 0, a 5 2, and a 6 2
So that equation (4.11) becomes
N 3 L3 2L2 L3 2L1 L3
54
By using the condition L1 L2 L3 1 , we obtain
N 3 L3 2L3 1
Hence, the element shape functions at nodes 1, 2, and 3 generally can be defined as
Ni Li 2Li 1,
i 1, 2, 3
Previously touched upon, the element shape functions defined at midpoints we
have N 4 , N 5 , and N 6 . Derivation follows the same process and for N 4 we impose the
requirements;
( 0 at node 1 L1 1, L2 L3 0
% 0 at node 2 L 1, L L 0
2
1
3
%
%% 0 at node 3 L3 1, L1 L2 0
N4 '
1
%1 at node 4 L1 L2 2 , L3 0
%0 at node 5 L2 L3 12 , L1 0
%
1
&%0 at node 6 L1 L3 2 , L2 0
Substitute the values into equation (4.11) and solve the system of equations, we have
the values of constants
a1 0, a 2 0, a 3 0, a 4 4, a 5 0, and a 6 0
So that equation (4.11) becomes
N 4 4L1 L2
55
The same procedures go for N 5 ;
( 0 at node 1 L1 1, L2 L3 0
% 0 at node 2 L 1, L L 0
2
1
3
%
%% 0 at node 3 L3 1, L1 L2 0
N5 '
1
%0 at node 4 L1 L2 2 , L3 0
%1 at node 5 L2 L3 12 , L1 0
%
%&0 at node 6 L1 L3 12 , L2 0
The values of constants obtained are as follows
a1 0, a 2 0, a 3 0, a 4 0, a 5 4, and a 6 0
So that equation (4.11) becomes
N 5 4L2 L3
Similarly for N 6 ;
( 0 at node 1 L1 1, L2 L3 0 % 0 at node 2 L 1, L L 0 2
1
3
%
%% 0 at node 3 L3 1, L1 L2 0 N6 '
1
%0 at node 4 L1 L2 2 , L3 0
%0 at node 5 L2 L3 12 , L1 0 %
%&1 at node 6 L1 L3 12 , L2 0 We obtained the values of constants
a1 0, a 2 0, a 3 0, a 4 0, a 5 0, and a 6 4
56
Hence, equation (4.11) becomes
N 6 4L1 L3
Generally we have the element shape functions defined are
N i Li 2 Li 1,
i 1, 2, 3
N 4 4 L1 L2
(4.12)
N 5 4 L2 L3
N 6 4 L1 L3
4.2.3
Derivation of Element Stiffness Matrices and Vectors
In previous subsection, the six element shape functions have been successfully
derived. Hence we arrive to the computation of the stiffness matrix K, and the load
~
vector f l defined by (3.27). Previously we have matrix B defined by (3.18), D and D
given by (2.16) and (2.25) respectively. With derivation of element shape functions in
(4.12), and the stiffness matrix K given by
~
K B ! DB d A
(4.13)
A
Since all the terms in matrix B and matrix D are constant, equation (4.13) can be
rewritten as,
~
K B ! DB dA
A
57
or
K
t3
AB ! DB
12
(4.14)
The situation goes differ with the computation of the load vector f l , due to the
element shape functions defined are vary quadratically in x and y . Hence the integral of
the load vector f l has to be computed numerically.
fl 4.2.4
N
A
!
q dA
Assemblage of Element Matrices and Vectors and Derivation of System
Equations
Since the structure is composed of several finite elements, the individual
element stiffness matrices and load vectors are to be assembled in a suitable manner.
Thus, we will have the overall system of equations. The procedure for constructing the
system equations from the element characteristics is the same regardless of the type of
problem, number, and type of element used. For this dissertation, we concentrate over
two neighbouring triangular elements. Hence we have the assembled stiffness matrix
K , and load vector f l as
K K 1e K e2
f l f1ee f 2ee
(4.15)
58
Once the assembled stiffness matrix K , and load vector f l have completely
derived, the boundary vector f b of the problem can be computed. Previously we have,
fb 4.2.5
N
L
!
dM nm Vnz dL dm !
N nM
L
nm
dL
Solution of the Finite Element System Equations
The overall equilibrium equations have to be modified to account for the
boundary conditions of the problem. At this point, we have defined all vectors which
are K , f l and f b . Hence we have
Ka f b f l
and
f fb fl
After the incorporation of the boundary conditions, the equilibrium equations can be
expressed as
Ka f
59
The assembled defined above will lead to the final matrix equation where the solution
u1 u 2
u 3 u 4 for the unknown displacements placed at global points, a u 5 will be carried out.
u 6 u 7
u 8 u 9 4.3
Test Example of the Problem
Previously, we have discussed in general the FEM’s procedures of the problem
considered. In followings, the numerical example is provided to investigate the
computation of the solution of the problem. Consider a thin simply supported square
plate of side a 2 m and thickness t 0.1 m with a uniformly distributed load
q 1 ) 10 11 Pa as shown in Figure 4.4. The plate is meshed into two element
discretization of six-nodes triangular elements. The discretization can be referred to
Figure 4.5. The element numbers and the global nodal numbers are also indicated on
this figure. The material has a Young modulus E 30 ) 108 Pa and Poisson ratio v 0.3 .
In addition to this, since we have a thin simply supported square plate hence the
displacements lie on some global points are known. By referring to Figure 4.4 and
Figure 4.5, here we can conclude that the displacements of u1 , u3 , u 4 , u5 , u7 and u8 for
global points 1,3,4,5,7, and 8 respectively are zeros.
60
M nn 0, Vnz dM nm
0
dm
w 0, M nn 0
w 0, M nn 0
Figure 4.4 Simply supported square plate
5
6
7
8
4
9
2
3
Figure 4.5 Finite element mesh of the problem
5
a)
3
2,2
b)
6
0,2 7
3
2
5
2
9
5
6
4
8
6
1
1
0,0
1
4
2
2
3
2,0
1
4
9
1 0,0
Figure 4.6 Global and local nodal points for element 1 and 2
5
2,2
61
Element 1:
Area of the triangle of element 1 is given by :
A1 1
x2 y1 x1 y 2 x3 y 2 x2 y3 x1 y3 x3 y1 2
A1 2
We have,
( L1 ,
x 2 y 3 x3 y 2 1 % %
' L2 + x3 y1 x1 y 3 % L % 2 A1 x y x y 2 1
& 3*
1 2
y 2 y3 x3 x2 (1 ,
y3 y1 x1 x3 %' x %+
y1 y 2 x 2 x1 %& y %*
Hence ,
1
x
2
1
1
L2 x y
2
2
1
L3 y
2
L1 1 Defining the element shape functions:
N 11 L1 2 L1 1
1
3
1
x x2
2
2
N 21 L2 2 L2 1
1 2
1
1
1
x xy x y 2 y
2
2
2
2
N 31 L3 2 L3 1
1 2 1
y y
2
2
N 41 4 L1 L2
2 x 2 y x 2 xy
62
N 51 4 L2 L3
xy y 2
N 61 4 L1 L3
2 y xy
Matrix B previously is defined as,
2 N 2 N1
2
2
2x 2x
N N1
B
2 2
y2 y2
N N1
2 xy 2 xy
2N2
x 2
2N2
y 2
2N2
2
xy
2 N6 x 2 2 N6 y 2 2 N6 2
xy Hence we have, matrix B for element 1 is,
0 1 1 0 2 0
B1 0 1 1 0 2 0 0 2 0 2
2 2
The element stiffness matrix K 1 ,
K1 t3
~
B1T DB1 dA1 A1B1T DB1
A1
12
1 v
E Where D v 1
1 v2 0 0
0 1
1 v 2
0
Substituting the values of t , E , and v into the equation; thus the element
stiffness matrix K 1 ,
63
10.9890 3.2967
7.1429
1.6484
0
5.4945
7.1429
7.1429 21.9780 21.9780 7.6923 21.9780
1.6484
7.1429
5.4945
0
3.2967 10.9890
K 1 1 ) 10 5 29.6703
14.2857 7.6923
10.9890 21.9780 3.2967
3.2967 21.9780 10.9890 14.2857
29.6703 7.6923
0
7.6923
0
7.6923 7.6923 7.6923 Differ compared to the derivation of the stiffness matrix K 1 , load vector f1 has
to be computed numerically.
!
f1 N q dA1 A1
A1
A1
A1
A1
A1
A1
dA1 q
N 12 q dA1 q
1
N 3 q dA1 q
N 14 q dA1 q
N 15 q dA1 q
N 16 q dA1 q
N 11 q
A1
A1
A1
A1
A1
A1
2 2
3
1 2
1 ) 10 11
1 x x dA1
0
y
2
2 dA1
1 ) 10 11 2 2 1 x 2 xy 1 x 1 y 2 1 y dA 1
1
N 2 dA1
0 y 2
2
2
2 2 2 1
1 2
N 13 dA1 1 ) 10 11
y y dA1
0 y 2
2
N 14 dA1 2 2
11
2
1
10
2
x
2
y
x
xy
d
A
)
1
0 y
N 15 dA1 2 2
11
2
1
10
xy
y
d
A
)
1
0 y
N 16 dA1 2 2
11
2 y xy dA1
1 ) 10
0 y
N 11
We obtained,
f1 1 ) 10 11
0 0 0 0.6667
0.6667
0.6667
Similar to element 1, the computation for element 2 is derived with the same
procedures.
64
Element 2:
Area of the triangle of element 2 is given by :
A2 1
x2 y1 x1 y 2 x3 y 2 x2 y3 x1 y3 x3 y1 2
A2 2
x 2 y 3 x 3 y 2 ( L1 ,
1 % %
' L2 + x3 y1 x1 y 3 % L % 2 A2 x y x y 2 1
1 2
& 3*
y 2 y 3 x3 x 2 (1 ,
y3 y1 x1 x3 %' x %+
y1 y 2 x 2 x1 %& y %*
Hence ,
L1 1 1
y
2
1
x
2
1
1
L3 x y
2
2
L2 The element shape functions:
3
1
y y2
2
2
1
1
N 22 x 2 x
2
2
1
1
1
1
N 32 x 2 xy x y 2 y
2
2
2
2
N 42 2 x xy
N 12 1 N 52 x 2 xy
N 62 2 x 2 y xy y 2
2 N 2 N1
2
2 2x 2x
N N1
Matrix B 2
2 y2 y2
2 N N1
xy 2 xy
2 N2
x 2
2 N2
y 2
2 N2
2
xy
2 N6 x 2 2 N6 y 2 2 N6 2
xy 65
Thus matrix B for element 2,
0 2 0 0 1 1
0
0 2
B 2 1 0 1
0 0 2 2 2
2 The element stiffness matrix K 2 ,
3.2967 10.9890
1.6484
7.1429
0
5.4945
1.6484
10.9890 3.2967 5.4945
7.1429
0
7.1429
7.1429
21.9780
7.6923 21.9780 21.9780
K 2 1 ) 10 5 0
0
7.6923
7.6923
7.6923 7.6923 3.2967 10.9890 21.9780 7.6923 29.6703
14.2857 29.6703 10.9890 3.2967 21.9780 7.6923 14.2857
Computing the load vector f 2 , we have
!
f 2 N q dA1 A1
f 2 1 ) 10 11
A1
A1
A1
A1
A1
A1
2 y
3
1 2
1 ) 10 11
1 y y dA1
0 0 dA1
2
2
2 y1
1
2
1 ) 10 11
d
x
x
A
N 12 q dA1 1
0 0 2
2
1
y
2
1
1 2 1 N 3 q dA1 1 2
11
)
1
10
x
xy
x
y
y
A
d
1
0 0 2
2
2
2 1
N 4 q dA1 2 y
1 ) 10 11
2 x xy dA1
1
0 0
N 5 q dA1 2 y
2
11
)
x
xy
A
1
10
d
1
0 0
N 16 q dA1 2 y
2 x 2 y xy y 2 dA1
1 ) 10 11
0 0
N 11 q
0 0 0 0.6667
0.6667
0.6667
66
Hence generally for the global coordinate systems we have,
( N 1 element 1
N4 ' 5
& 0 elsewhere
( N 2 element 2
N7 ' 3
& 0 elsewhere
( N 1 element 1
N2 ' 4
& 0 elsewhere
( N 31 element 1
%
N 5 ' N 22 element 2
% 0 elsewhere
&
( N 2 element 2
N8 ' 6
& 0 elsewhere
( N 1 element 1
N3 ' 2
& 0 elsewhere
( N 2 element 2
N6 ' 5
& 0 elsewhere
( N 11 element 1
%
N 1 ' N 12 element 2
% 0 elsewhere
&
( N 61 element 1
%
N 9 ' N 42 element 2
% 0 elsewhere
&
This is explained by comparing Figures 4.6(a) and Figure 4.5, local points
1, 2, 3, 4, 5, 6 correspond to the global points 1, 3, 5, 2, 4, 9 respectively. That is, after the
contributions from the first element, the global stiffness matrix K 1 and the global load
vector f1 are given by
10.9890
3.2967
7.1429
1.6484
5.4945
10
.
9890
29
.
6703
21
.
9780
14
.
2857
3.2967
7.1429
21.9780 21.9780 21.9780 7.1429
3.2967 14.2857 21.9780 29.6703 10.9890
5 K1 1 ) 10
3.2967
10.9890
7.1429
5.4945
1.6484
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
7.6923
7.6923
0
7.6923
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0 0 7.6923 0 7.6923 0 7.6923 0
0 0
0 0
0 0
0 0 7.6923 f 1 1 ) 10 11
0 0.6667 0 0.6667 0 0 0 0 0.6667 Similarly, by comparing Figure 4.6(b) and Figure 4.5, the topology data for this
element is that the local nodal points 1, 2, 3, 4, 5, 6 correspond to the global nodal points
1, 5, 7, 9, 6, 8 respectively. Using this topology of data, the expanded stiffness matrix
K 2 and the expanded element load vector f 2 are
67
5.4945
0
0
0
5 K 2 1 ) 10
1.6484
3.2967
7.1429
10.9890
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
3.2967
10.9890
0 1.6484
7.1429
0 0
0
0
0
0
0 0
0
0
0
0
0 0
0
0
0
0
0 10.9890
3.2967
0 5.4945
7.1429
0 0 10.9890 29.6703 21.9780 14.2857 7.6923 21.9780 21.9780 21.9780 7.6923 0 7.1429
0 3.2967 14.2857 21.9780 29.6703 7.6923 7.6923
7.6923
0
0
7.6923
7.6923 f 2 1 ) 10 11
0 0 0 0 0 0.6667 0 0.6667 0.6667 Adding the results and assemblage the vectors and matrices of element 1 and 2,
we then obtain
K K1 K 2
7.1429
3.2968
7.1429
3.2967
3.2967
10.9890
10.9890 10.9890
0
0
0
3.2967
10.9890 29.6703 21.9780 14.2857
7.1429 21.9780 21.9780 21.9780 7.1429
0
0
0
0
0
0
3.2967 14.2857 21.9780 29.6703 10.9890
5 K 1 ) 10
3.2968
7.1429
7.1429
3.2967
10.9890 10.9890 10.9890
3.2967
0
0
0
10.9890 29.6703 21.9780 14.2857
3.2967
7.1429
0
0
0
7.1429 21.9780 21.9780 21.9780
10.9890
0
0
0
3.2967 14.2857 21.9780 29.6703
0
7.6923
7.6923
7.6923
0
7.6923
7.6923
7.6923
f l f1 f 2
f l 1 ) 10 10
0 0.0667
0 0.0667
0 0.0667
0 0.0667
0.1333
0 7.6923 7.6923 7.6923 0 7.6923 7.6923 7.6923 15.3846 68
Now we evaluate the boundary vector f b given by (3.27). As discussed earlier,
the boundary shown in Figure 4.4 and Figure 4.5 can simply be known. Since we have a
thin simply supported square plate, along the boundaries 123 , 765 , 345 , and 187 give the
values of zeros. Hence we obtained
0 0 0 0 f b 0 0 0 0 0 Thus, the FE equations for the body are obtained as
Ka f b f l
3.2967
3.2967
10.9890
7.1429
3.2968
7.1429
10.9890 10.9890
3.2967
0
0
0
10.9890 29.6703 21.9780 14.2857
7.1429
21.9780 21.9780 21.9780
7.1429
0
0
0
0
0
0
3.2967 14.2857 21.9780 29.6703 10.9890
1 ) 105 3.2968
3.2967
10.9890 10.9890 10.9890
3.2967
7.1429
7.1429
3.2967
10.9890 29.6703 21.9780 14.2857
0
0
0
21.9780 21.9780 21.9780
0
0
0
7.1429
7.1429
10.9890
3.2967 14.2857 21.9780 29.6703
0
0
0
7.6923
7.6923
7.6923
7.6923
0
7.6923
0
7.6923
0
0 0 u1 7.6923 u 2 0.0667 0
0 0 7.6923 u3 7.6923 u 4 0.0667 0
10
0
0 0 u5 1 ) 10
0.0667 0
7.6923 u6
7.6923 u7 0 0 0.0667 0
7.6923 u8
15.3846 u9 0.1333 0
As mentioned in the first paragraph, the displacements placed at particular
global points are already known. By referring to Figure 4.4 and Figure 4.5, here we can
conclude that the displacements of u1 , u3 , u 4 , u5 , u7 , and u8 for global points 1,3,4,5,7,
and 8 respectively are zeros. Due to this, the system of linear equations yields to
69
3.2967
3.2967
10.9890
7.1429
3.2968
7.1429
10.9890 10.9890
10
.
9890
29
.
6703
21
.
9780
14
.
2857
3
.
2967
0
0
0
7.1429
7.1429
0
0
0
21.9780 21.9780 21.9780
0
0
0
3.2967 14.2857 21.9780 29.6703 10.9890
1 ) 105 3.2968
7.1429
7.1429
3.2967
10.9890 10.9890 10.9890
3.2967
0
0
0
10.9890 29.6703 21.9780 14.2857
3.2967
0
0
0
7.1429
21.9780 21.9780 21.9780
7.1429
10.9890
0
0
0
3.2967 14.2857 21.9780 29.6703
0
7.6923
0
7.6923
7.6923
7.6923
7.6923
7.6923
0
0 7.6923 u 2 7.6923 0 7.6923 0 0 1 ) 10 10
0
7.6923 u6 7.6923 0 7.6923 0 15.3846 u9 0 0.0667
0 0.0667
0 0.0667
0 0.0667
0.1333
Hence, the solution of linear equations produces
u 2 0.0463
u 0.0921 6 u 9 0.0123 Where the displacements u2 , u6 , and u9 lie on the global nodal points 2, 6 and
9 respectively.
Generally, we have described several procedures of FEM to be considered for
the Kirchhoff plate problem and its numerical result example. The computation
illustrated that solution of the entire body is established as a sum of the contributions of
all elements. In the next chapter, we will over look the computational solution of the
problem. The simulation is developed by using Matlab programming language.
CHAPTER V
COMPUTATIONAL RESULTS
5.1
Introduction
Nowadays, a usage of computer software is important in order to make the
information and results distributed faster and efficiently. In a previous chapter, the
numerical scheme developed by using FEM in solving the Kirchhoff plate problem has
been discussed in detail. In this chapter, the computational code of the problem
developed by using Matlab Programming language will be described.
5.2
Features of an FE Computer Program
In order how to see the program works, in follows, Figure 5.1 illustrates the
program flow of the FE numerical simulation of the problem.
71
Input data:
Number of elements
Type of elements
Position of nodal points
Boundary Conditions
Derivation:
Ai , L1, L2 , L3 , Ni , B , D
Calculate K e and fl e
Expanded matrices:
Local points correspond to
global points.
K e and fl e
Calculate fb e
Assemblage the vectors and matrices
K e K e1 K e 2
fl e f1e f2e
Establish
Ka fb fl
Modify system of equations according to the boundary conditions and solve
the modified system of equations
Calculate and write results (displacement)
Figure 5.1 The program flow of the FE numerical simulation of the problem
72
The program starts when a sequence of input data is read. Then the area of
triangle and the element shape functions for each nodal point in the particular element
mesh is derived. Once the definitions completed, matrix B is computed. For each
element, the element stiffness matrix K and the element load vector f l are calculated
and added to the proper components K and f l using the topology data. This is also
known as the assembling process. Then the boundary conditions are read and the
boundary vector f b is calculated. Hence, the entire system of equations is established,
and after the introduction of the boundary conditions, the modified system of equations
is solved. Finally, the results in terms of the displacement are calculated and written out.
In the next subsections, the flow of the program code is described.
5.3
The Program Code
In this very first phase, the user will initialize all the data or variables needed in
the computation of the problem. For each element, the nodal point of each corner sides
will be generated automatically. Table 5.1 shows the declaration of the variables of the
problem discussed.
Table 5.1 The declaration of the variables
clear
%-----------------------%INPUT DATA
%-----------------------a=2;
% length of the plate
b=2;
% width of the plate
E=30*(10^8);
% Young modulus
v=0.3;
% Poisson ratio
t=0.1;
% thickness of the plate
q=1*(10^-11);
% transverse load
73
Once the data and variables are initialized, the element shape functions is about
to be computed. But the process must be followed with the derivation of the area of the
triangle, and the triangular coordinates L1 , L2 , and L3 defined for each element. The
computation of element 1 is shown in Table 5.2 below.
Table 5.2 The derivation of stiffness matrix and load vector for element 1
A1=0.5*abs((x2*y1-x1*y2)+(x3*y2-x2*y3)+(x1*y3-x3*y1)); % area of the
triangle
% triangular coordinates
L1=(1/(2*A1))*((x2*y3-x3*y2)+(y2-y3)*x+(x3-x2)*y);
L2=(1/(2*A1))*((x3*y1-x1*y3)+(y3-y1)*x+(x1-x3)*y);
L3=(1/(2*A1))*((x1*y2-x2*y1)+(y1-y2)*x+(x2-x1)*y);
% the element shape functions
N1=L1*(2*L1-1);
N2=L2*(2*L2-1);
N3=L3*(2*L3-1);
N4=4*L1*L2;
N5=4*L2*L3;
N6=4*L1*L3;
% stiffness matrix and load vector
D=(E/(1-v^2))*[1 v 0;v 1 0;0 0 0.5*(1-v)];
B1=[diff(N1,x,2) diff(N2,x,2) diff(N3,x,2) diff(N4,x,2) diff(N5,x,2)
diff(N6,x,2);diff(N1,y,2) diff(N2,y,2) diff(N3,y,2) diff(N4,y,2)
diff(N5,y,2) diff(N6,y,2); 2*diff(diff(N1,y),x) 2*diff(diff(N2,y),x)
2*diff(diff(N3,y),x) 2*diff(diff(N4,y),x) 2*diff(diff(N5,y),x)
2*diff(diff(N6,y),x)]
K1=double(((t^3)/12)*A1*B1'*D*B1)
f1=double(q*[int(int(N1,x,((y3-y1)/(x3x1))*y,x2),y,0,y3);int(int(N2,x,((y3-y1)/(x3x1))*y,x2),y,0,y3);int(int(N3,x,((y3-y1)/(x3x1))*y,x2),y,0,y3);int(int(N4,x,((y3-y1)/(x3x1))*y,x2),y,0,y3);int(int(N5,x,((y3-y1)/(x3x1))*y,x2),y,0,y3);int(int(N6,x,((y3-y1)/(x3-x1))*y,x2),y,0,y3)])
Both element meshes considered in the problem will follows the same
derivation. Then the expanded matrix for both stiffness matrix and load vector of
element 1 and 2 will be computed. This is due as the local nodal points lie in each
element corresponds to the global nodal points. In follows, Table 5.3 shows the
computation code of the expanded matrix of stiffness matrix and load vector for
element 1.
74
Table 5.3 The expanded matrix of stiffness matrix and load vector for element 1
fprintf('EXPANDED MATRIX FOR STIFFNESS MATIX AND LOAD VECTOR FOR ELEMENT
1,\n\n')
DD=(E/(1-v^2))*[1 v 0 0 0 0 0 0 0;v 1 0 0 0 0 0 0 0;0 0 0.5*(1-v) 0 0 0 0
0 0;0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0
0;0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0];
B1=[diff(N1,x,2) diff(N4,x,2) diff(N2,x,2) diff(N5,x,2) diff(N3,x,2) 0 0 0
diff(N6,x,2);diff(N1,y,2) diff(N4,y,2) diff(N2,y,2) diff(N5,y,2)
diff(N3,y,2) 0 0 0 diff(N6,y,2); 2*diff(diff(N1,y),x) 2*diff(diff(N4,y),x)
2*diff(diff(N2,y),x) 2*diff(diff(N5,y),x) 2*diff(diff(N3,y),x) 0 0 0
2*diff(diff(N6,y),x);0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0
0;0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0];
K1=double(((t^3)/12)*A1*B1'*DD*B1)
f1=double(q*[int(int(N1,x,((y3-y1)/(x3x1))*y,x2),y,0,y3);int(int(N4,x,((y3-y1)/(x3x1))*y,x2),y,0,y3);int(int(N2,x,((y3-y1)/(x3x1))*y,x2),y,0,y3);int(int(N5,x,((y3-y1)/(x3x1))*y,x2),y,0,y3);int(int(N3,x,((y3-y1)/(x3x1))*y,x2),y,0,y3);0;0;0;int(int(N6,x,((y3-y1)/(x3-x1))*y,x2),y,0,y3)])
Next is the assemblage process for all elements. The completed process will
produce the FE system of linear equations of the problem. The process is derived with
the code shown in Table 5.4. The displacement vector is first declared before the result
is calculated and written out.
Table 5.4 The assemblage vectors and matrices of all elements
fprintf('ASSEMBLAGE THE VECTORS AND MATRICES:\n\n')
K=K1+K2
FL=f1+f2
fprintf('SOLUTION OF THE FINITE ELEMENT SYSTEM EQUATIONS:\n\n')
syms u1 u2 u3 u4 u5 u6 u7 u8 u9
a=[u1;u2;u3;u4;u5;u6;u7;u8;u9]
a=K\FL
CHAPTER VI
CONCLUSIONS AND SUGGESTIONS
6.1
Introduction
In this chapter, some conclusions according to Kirchhoff plate modelling by
using FEM are discussed. Since this study quite generally discussed the problem,
therefore, in this chapter, we conclude some suggestions for further research in the
future.
6.2
Summary of the Work
The basic idea in FEM is to find the solution of a complicated problem by
replacing it by a simpler one. In FEM, the solution region is considered as built up of
76
many small, interconnected subregions called finite elements. A large number of
different finite elements have been developed and extensively used in the field of
structural mechanics and other types of engineering problems. The widely discussed of
the finite elements developed is plate elements. The plate element is one of the more
structural elements and is used to model and analyze structures like automobile parts.
The purpose of this study was to present the Kirchhoff plate element using FEM.
This study discussed in detail the FE formulation of the problem which comprises the
plate theory, differential equations of the plate theory, boundary conditions, and weak
formulation.
In this study, the problem considered was two dimensional cases, with quadratic
interpolation functions. The plate elements developed adopted the basic element or the
simplest plate element for the analysis of plates of arbitrary shape which was the
triangular element.
An example of numerical result was also presented to review the derivation of
stiffness matrix K , the load vector f l , and the boundary vector f b , which yielded to the
approximate solution for the displacements of the problem. Previously touched upon,
the finite element model was two dimensional with quadratic functions; hence it
produced six nodes triangular elements.
The major contribution comes from the computational scheme of the problem
developed by using Matlab programming language. The program computes the
displacements at each node of the finite element model studied. Several test examples
were analyzed using the program and the results obtained were studied.
77
6.3
Suggestions for Future Study
This study illustrates the use of Matlab programming language to compute the
two dimensional problem of six nodes triangular plate elements. There are some
suggestions for future work which include:
1. Discretizating the domain with a small size of finite elements. The size of
elements influences the convergence of the solution directly. If the size of
the elements is small, more number of elements is derived; the final solution
is expected to be more accurate.
2. Improvising the element shapes such as the quadrilateral element. Although
a quadrilateral or its special forms, rectangle and parallelogram element can
be obtained by assembling two or four triangular elements, in some cases the
use of quadrilateral elements proves to be advantageous.
3. Increasing the order of polynomial type of interpolation function.
Theoretically, a polynomial of infinite order corresponds to the exact
solution. A cubic model for an example produces the ten nodes triangular
element for two dimensional cases.
4. Modifying the computational program to include a graphical user interface.
The program can be developed using Visual C++ programming language
which supports parser feature.
78
REFERENCES
Boresi, A. P., Sidebottom, O. M., Seely, F. B. and Smith, J. O. (1978). Advanced
Mechanics of Materials. (3rd ed.). John Wiley & Sons: New York.
Clough R. W. (1960). The Finite Element Method in Plane Stress Analysis.
Proceedings American Society of Civil Engineers. 2nd Conference on Electronic
Computations. Pittsbutgh, Pennsylvania: pp. 345-378.
Cook R. D., Malkus D. S., and Plesha, M. E. (1989). Concepts and Applications of
Finite Element Analysis. (3rd ed.). John Wiley & Sons: New York.
Hughes, T. J. R. (1987). The Finite Element Method. Linear Static and Dynamic Finite
Element Analysis. Prentice Hall: New Jersey.
Ottosen, Niels and Petersson, Hans. (1992). Introduction to the Finite Element Method.
Prentice Hall Inc: UK.
Reddy, J.N. (2007). Theory and Analysis of Elastic Plates and Shells. (2nd ed.). CRC
Press : US.
Timoshenko, S. P. and Woinowsky-Krieger, S. (1959). Theory of Plates and Shells. (2nd
ed.). McGraw-Hill: New York.
79
Ventsel, Eduard and Krauthammer, Theodor. (2001). Thin Plates and Shells: Theory,
Analysis, and Applications. Marcel Dekker, Inc: New York.
Vitor, M.A Leitao. (2001). A Meshless Method for Kirchhoff-plate bending Problems.
International Journal for Numerical Methods in Engineering. 52:1107-1130.
80
APPENDIX A
clear
%-----------------------%INPUT DATA
%-----------------------a=2;
% length of the plate
b=2;
% width of the plate
E=30*(10^8);
% Young modulus
v=0.3;
% Poisson ratio
t=0.1;
% thickness of the plate
q=1*(10^-11);
% transverse load
syms x y
fprintf('ELEMENT 1:\n\n')
x1=0;y1=0;
x2=x1+a;y2=y1;
x3=x2;y3=y2+b;
A1=0.5*abs((x2*y1-x1*y2)+(x3*y2-x2*y3)+(x1*y3-x3*y1)); % area of the triangle
% triangular coordinates
L1=(1/(2*A1))*((x2*y3-x3*y2)+(y2-y3)*x+(x3-x2)*y);
L2=(1/(2*A1))*((x3*y1-x1*y3)+(y3-y1)*x+(x1-x3)*y);
L3=(1/(2*A1))*((x1*y2-x2*y1)+(y1-y2)*x+(x2-x1)*y);
% the element shape functions
N1=L1*(2*L1-1);
N2=L2*(2*L2-1);
N3=L3*(2*L3-1);
N4=4*L1*L2;
N5=4*L2*L3;
N6=4*L1*L3;
% stiffness matrix and load vector
D=(E/(1-v^2))*[1 v 0;v 1 0;0 0 0.5*(1-v)];
B1=[diff(N1,x,2) diff(N2,x,2) diff(N3,x,2) diff(N4,x,2) diff(N5,x,2)
diff(N6,x,2);diff(N1,y,2) diff(N2,y,2) diff(N3,y,2) diff(N4,y,2) diff(N5,y,2)
diff(N6,y,2); 2*diff(diff(N1,y),x) 2*diff(diff(N2,y),x) 2*diff(diff(N3,y),x)
2*diff(diff(N4,y),x) 2*diff(diff(N5,y),x) 2*diff(diff(N6,y),x)]
81
K1=double(((t^3)/12)*A1*B1'*D*B1)
f1=double(q*[int(int(N1,x,((y3-y1)/(x3-x1))*y,x2),y,0,y3);int(int(N2,x,((y3y1)/(x3-x1))*y,x2),y,0,y3);int(int(N3,x,((y3-y1)/(x3x1))*y,x2),y,0,y3);int(int(N4,x,((y3-y1)/(x3x1))*y,x2),y,0,y3);int(int(N5,x,((y3-y1)/(x3x1))*y,x2),y,0,y3);int(int(N6,x,((y3-y1)/(x3-x1))*y,x2),y,0,y3)])
fprintf('EXPANDED MATRIX FOR STIFFNESS MATIX AND LOAD VECTOR FOR ELEMENT
1,\n\n')
DD=(E/(1-v^2))*[1 v 0 0 0 0 0 0 0;v 1 0 0 0 0 0 0 0;0 0 0.5*(1-v) 0 0 0 0 0
0;0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0;0 0
0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0];
B1=[diff(N1,x,2) diff(N4,x,2) diff(N2,x,2) diff(N5,x,2) diff(N3,x,2) 0 0 0
diff(N6,x,2);diff(N1,y,2) diff(N4,y,2) diff(N2,y,2) diff(N5,y,2) diff(N3,y,2)
0 0 0 diff(N6,y,2); 2*diff(diff(N1,y),x) 2*diff(diff(N4,y),x)
2*diff(diff(N2,y),x) 2*diff(diff(N5,y),x) 2*diff(diff(N3,y),x) 0 0 0
2*diff(diff(N6,y),x);0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0;0
0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0];
K1=double(((t^3)/12)*A1*B1'*DD*B1)
f1=double(q*[int(int(N1,x,((y3-y1)/(x3-x1))*y,x2),y,0,y3);int(int(N4,x,((y3y1)/(x3-x1))*y,x2),y,0,y3);int(int(N2,x,((y3-y1)/(x3x1))*y,x2),y,0,y3);int(int(N5,x,((y3-y1)/(x3x1))*y,x2),y,0,y3);int(int(N3,x,((y3-y1)/(x3x1))*y,x2),y,0,y3);0;0;0;int(int(N6,x,((y3-y1)/(x3-x1))*y,x2),y,0,y3)])
fprintf('ELEMENT 2:\n\n')
x11=0;y11=0;
x22=x3;y22=y3;
x33=x11;y33=y11+b;
A2=0.5*abs((x2*y1-x1*y2)+(x3*y2-x2*y3)+(x1*y3-x3*y1));
L11=(1/(2*A2))*((x22*y33-x33*y22)+(y22-y33)*x+(x33-x22)*y);
L22=(1/(2*A2))*((x33*y11-x11*y33)+(y33-y11)*x+(x11-x33)*y);
L33=(1/(2*A2))*((x11*y22-x22*y11)+(y11-y22)*x+(x22-x11)*y);
N11=L11*(2*L11-1);
N22=L22*(2*L22-1);
N33=L33*(2*L33-1);
N44=4*L11*L22;
N55=4*L22*L33;
N66=4*L11*L33;
B2=[diff(N11,x,2) diff(N22,x,2) diff(N33,x,2) diff(N44,x,2) diff(N55,x,2)
diff(N66,x,2);diff(N11,y,2) diff(N22,y,2) diff(N33,y,2) diff(N44,y,2)
diff(N55,y,2) diff(N66,y,2); 2*diff(diff(N11,y),x) 2*diff(diff(N22,y),x)
2*diff(diff(N33,y),x) 2*diff(diff(N44,y),x) 2*diff(diff(N55,y),x)
2*diff(diff(N66,y),x)]
K2=double(((t^3)/12)*A2*B2'*D*B2)
f2=double(q*[int(int(N11,x,0,((y3-y1)/(x3x1))*y),y,0,y3);int(int(N22,x,0,((y3-y1)/(x3x1))*y),y,0,y3);int(int(N33,x,0,((y3-y1)/(x3x1))*y),y,0,y3);int(int(N44,x,0,((y3-y1)/(x3x1))*y),y,0,y3);int(int(N55,x,0,((y3-y1)/(x3x1))*y),y,0,y3);int(int(N66,x,0,((y3-y1)/(x3-x1))*y),y,0,y3)])
82
fprintf('EXPANDED MATRIX FOR STIFFNESS MATIX AND LOAD VECTOR FOR ELEMENT
2:\n\n')
B2=[diff(N11,x,2) 0 0 0 diff(N22,x,2) diff(N55,x,2) diff(N33,x,2)
diff(N66,x,2) diff(N44,x,2);diff(N11,y,2) 0 0 0 diff(N22,y,2) diff(N55,y,2)
diff(N33,y,2) diff(N66,y,2) diff(N44,y,2); 2*diff(diff(N11,y),x) 0 0 0
2*diff(diff(N22,y),x) 2*diff(diff(N55,y),x) 2*diff(diff(N33,y),x)
2*diff(diff(N66,y),x) 2*diff(diff(N44,y),x);0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0
0;0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0];
K2=double(((t^3)/12)*A2*B2'*DD*B2)
f2=double(q*[int(int(N11,x,0,((y3-y1)/(x3x1))*y),y,0,y3);0;0;0;int(int(N22,x,0,((y3-y1)/(x3x1))*y),y,0,y3);int(int(N55,x,0,((y3-y1)/(x3x1))*y),y,0,y3);int(int(N33,x,0,((y3-y1)/(x3x1))*y),y,0,y3);int(int(N66,x,0,((y3-y1)/(x3x1))*y),y,0,y3);int(int(N44,x,0,((y3-y1)/(x3-x1))*y),y,0,y3)])
fprintf('ASSEMBLAGE THE VECTORS AND MATRICES:\n\n')
K=K1+K2
FL=f1+f2
fprintf('SOLUTION OF THE FINITE ELEMENT SYSTEM EQUATIONS:\n\n')
syms u1 u2 u3 u4 u5 u6 u7 u8 u9
a=[u1;u2;u3;u4;u5;u6;u7;u8;u9]
a=K\FL