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STERA_FEM
Technical Manual
Basic Theory of Finite Element Method
Version 2.0
Taiki SAITO
Toyohashi University of Technology, Japan
1
UPDATE HISTORY
2013/11/11 STERA_FEM Technical Manual Ver.1.1 is uploaded.
2013/12/11 STERA_FEM Technical Manual Ver.1.2 is uploaded.
2014/11/16 STERA_FEM Technical Manual Ver.1.3 is uploaded.
2015/06/11 STERA_FEM Technical Manual Ver.2.0 is uploaded.
2
CHAPTER 1
ELASTIC THEORY
3
Index of Chapter 1 1. Introduction
1-1. Section
1-2. Stress and Strain
1-3. Beam Theory
1-4. Properties of Reinforced Concrete Structure
2. Simple Example for FEM Formulation
3. Triangular Element for Plane Analysis
4. Stiffness Matrix for Triangular Element
5. From Element Stiffness Matrix to Global Stiffness Matrix
6. Higher Order Element
7. Interpolation Function
8. Natural Coordinate
9. Isoparametric Element
10. Systematic Formulation of Interpolation Function
11. Stiffness Matrix for Isoparametric Element
12. Stress and Strain at Gaussian Points
13. Independent Freedom
14. Skyline Method
15. Incompatible Element
16. Hexahedron Element
17. Plain Stress and Plain Strain
4
1. INTRODUCTION
1-1. Section
1-2. Stress and Strain
1) One-Dimensional Problem
bDA D
b b
Area Moment of Inertia
12
3bDI x
12
3DbI y
L δ
bD N N
A
N
L
E
E
Stress Strain Hooke’s Law
E: Young’s Modulus
L
EAN
Force – Deformation Relationship
5
2) Two-Dimensional Problem
xy
y
x
xy
y
x
G
EE
EE
100
011
011
or
xy
y
x
xy
y
xE
2
100
01
01
1 2
E
E
xy
xx
: Poisson Ratio
xx x x
y
y
EE
EE
xyy
yxx
xy
xyxy
xy
xy
Normal Stress and Strain
Shear Stress and Strain
xyxy G
EG)1(2
1
: Shear Modulus
6
1-3. Beam Theory
P
P
Q
Q M
M
Q : Shear Force
M : Moment
P
M(x)
x
P
y
)(1
2
2
xMEIdx
yd
)()(
xQdx
xdM
Example )
213
12
2
2
2
2
6
12
1
)(),(1
cxcxEI
Py
cxEI
P
dx
dy
xEI
P
dx
yd
PxxMxMEIdx
yd
EI
PLc
EI
PLc
Therefore
yanddx
dyLxat
3
2
2
1 3
1,
2
,
:00,
L
EI
PLx
EI
PLx
EI
Py
323
3
1
2
1
6
1
7
1.4 Properties of Reinforced Concrete Structure
Unit Weight
Concrete Type Nominal Strength
（N/mm2 = MPa) Unit Weight （kN/m3）
Normal Concrete Fc≦36 24
Material Parameters
Young’s Modulus
（N/mm2 = MPa) Poisson’s Ratio
Thermal
Expansion
Coefficient（1/℃）
Steel Bar 200 000 1/4 1 x 10-5
Concrete
22 000 ( Fc = 18 )
25 000 ( Fc = 24 )
28 000 ( Fc = 30 )
1/6 1 x 10-5
8
2. SIMPLE EXAMPLE FOR FEM FORMULATION
Step.1: Description of the Problem
The problem is to obtain the deformation of a simple supported beam under various load
conditions.
If you change the load condition, you will get the different deformation pattern. Actually,
there are infinite variations for the deformation pattern.
Step.2: Assumption of deformation function
We assume a particular function for the deformation pattern to fix the variation, such
as the following function:
)sin()( xL
axv
(2-1)
Step.3: Relation between nodal displacement and element deformation
From Equation (2-1), The displacement δ at the center node A is calculated as
aLv )5.0( (2-2)
The relation between nodal displacement and element deformation is then expressed as,
)sin()( xL
xv (2-3)
A
δ
x = 0 x = L x
v
etc.
9
Step.4: Constitutive equation at the node
We obtain the relation between the nodal force and the nodal displacement, for example,
by using the “Principle of Virtual Work Method.”
KP (2-4)
The process is summarized as follows:
Translate external forces into
equivalent nodal force, P.
Calculate nodal displacement,δ ,
from the constitutive equation,
PK 1
Obtain the element deformation
from the nodal displacement.
)sin()( xL
xv
The above example tells the essence of the finite element analysis, which is:
“Assume the deformation pattern to reduce the degree of freedom of the element, then,
obtain the deformation from the limited number of nodal displacements.”
v
δ
P
P
A
δ
P
10
3. TRIANGULAR ELEMENT FOR PLANE ANALYSIS
Step.1: Description of the Problem
The problem is to obtain the deformation of a simple triangular element.
There are infinite variations for the deformation patterns.
Step.2: Assumption of deformation function
To fix the variation for the deformation patterns, we assume a linear function for the
deformation pattern.
yxyxv
yxyxu
654
321
),(
),(
(3-1)
In a matrix form,
6
5
4
3
2
1
1000
0001
yx
yx
v
u (3-2)
Step.3: Relation between nodal displacement and element deformation
The displacements of the element nodes are expressed as,
etc.
x
y
11
Node 1:
6
5
4
3
2
1
11
11
1
1
1000
0001
yx
yx
v
u (3-3)
Node 2:
6
5
4
3
2
1
22
22
2
2
1000
0001
yx
yx
v
u (3-4)
Node 3:
6
5
4
3
2
1
33
33
3
3
1000
0001
yx
yx
v
u (3-5)
It is summarized as,
6
5
4
3
2
1
33
22
11
33
22
11
3
2
1
3
2
1
1000
1000
1000
0001
0001
0001
yx
yx
yx
yx
yx
yx
v
v
v
u
u
u
(3-6)
U = A α
xx1 x3 x2
y3
u1
v1 u2
v2 u3
v3
1
2
3
y
y2
y1
12
We can obtain the coefficients 61 , from the nodal displacements as,
α = A-1 U (3-7)
Substituting Equation (3-7) into Equation (3-2), the relation between nodal
displacement and element deformation is,
3
2
1
3
2
1
1
1000
0001
v
v
v
u
u
u
Ayx
yx
v
u (3-8)
u(x,y) = H(x,y) U
Step.4: Constitutive equation at the node
We obtain the relation between the nodal force and the nodal displacement, for example,
by using the “Principle of Virtual Work Method.”
3
2
1
3
2
1
3
2
1
3
2
1
v
v
v
u
u
u
K
Q
Q
Q
P
P
P
(3-9)
F = K U
The process is summarized as follows:
(1) Translate external forces into equivalent nodal force,
F = {P1, P2, P3, Q1, Q2, Q3}T
(2) Calculate the nodal displacements from the constitutive equation,
U = K-1 F
(3) Obtain the element deformation from the nodal displacement.
u(x,y) = H(x,y)U
P1
Q1 P2
Q2 P3
Q3
1
2
3
13
4. STIFFNESS MATRIX FOR TRIANGULAR ELEMENT
Stiffness matrix in Equation (3-9) can be obtained from the “Principle of Virtual Work
Method,” which is expressed in the following form:
V
TT FUdv (4-1)
where, is a virtual strain vector, is a stress vector, U is a virtual displacement
vector and F is a load vector, respectively.
In case of the plane problem, the strain vector is defined as,
x
v
y
uy
vx
u
xy
y
x
(4-2)
Substituting Equation (3-8) into Equation (4-2), the strain vector is calculated from the
nodal displacement vector as,
3
2
1
3
2
1
1
010100
100000
000010
v
v
v
u
u
u
A
x
v
y
uy
vx
u
xy
y
x
(4-3)
ε = B U
In the plane stress problem, the stress-strain relationship is expressed as,
xy
y
x
xy
y
xE
2
100
01
01
1 2 (4-4)
σ = D ε
14
Substituting Equation (4-3) into Equation (4-4),
σ= D B U (4-5)
From the Principle of Virtual Work Method,
FUUDBdvBUdvDBUUB T
V
TTT
V
(4-6)
Therefore, the constitutive equation is obtained as,
V
T DBdvBKKUF , (4-7)
15
5. FROM ELEMENT STIFFNESS MATRIX TO GLOBAL STIFFNESS MATRIX
1) Force control
Element Stiffness Matrix:
Element (1) …
4
2
1
4
2
1
)1(66
)1(65
)1(64
)1(63
)1(62
)1(61
)1(56
)1(55
)1(54
)1(53
)1(52
)1(51
)1(46
)1(45
)1(44
)1(43
)1(42
)1(41
)1(36
)1(35
)1(34
)1(33
)1(32
)1(31
)1(26
)1(25
)1(24
)1(23
)1(22
)1(21
)1(16
)1(15
)1(14
)1(13
)1(12
)1(11
4
2
1
4
2
1
v
v
v
u
u
u
kkkkkk
kkkkkk
kkkkkk
kkkkkk
kkkkkk
kkkkkk
Q
Q
Q
P
P
P
(5-1)
Element (2) …
4
3
1
4
3
1
)2(66
)2(65
)2(64
)2(63
)2(62
)2(61
)2(56
)2(55
)2(54
)2(53
)2(52
)2(51
)2(46
)2(45
)2(44
)2(43
)2(42
)2(41
)2(36
)2(35
)2(34
)2(33
)2(32
)2(31
)2(26
)2(25
)2(24
)2(23
)2(22
)2(21
)2(16
)2(15
)2(14
)2(13
)2(12
)2(11
4
3
1
4
3
1
v
v
v
u
u
u
kkkkkk
kkkkkk
kkkkkk
kkkkkk
kkkkkk
kkkkkk
Q
Q
Q
P
P
P
(5-2)
Global Stiffness Matrix:
4
3
4
3
)2(66
)1(66
)2(65
)2(63
)1(63
)2(62
)2(56
)2(55
)2(53
)2(52
)2(36
)2(36
)2(35
)2(33
)1(33
)2(32
)2(26
)2(25
)2(23
)2(22
4
3
4
3
v
v
u
u
kkkkkk
kkkk
kkkkkk
kkkk
Q
Q
P
P
(5-3)
F = K U
① ③
② ④
P
(2)
(1)
4
3
4
3
4
3
2
1
4
3
2
1
v
v
u
u
ConditionBoundary
v
v
v
v
u
u
u
u
① fixed … u1=v1=0
② fixed … u2=v2=0
16
In case of force control, set the load condition,
0
0
0
4
3
4
3
P
Q
Q
P
P
(5-4)
The displacement vector is then obtained by solving the constitutive equation,
0
0
0
1
4
3
4
3
PK
v
v
u
u
(5-5)
17
2) Displacement Control
Imposing the control displacement Dv 3 to the Global Stiffness Matrix:
4
4
3
)2(66
)1(66
)2(65
)2(63
)1(63
)2(62
)2(56
)2(55
)2(53
)2(52
)2(36
)2(36
)2(35
)2(33
)1(33
)2(32
)2(26
)2(25
)2(23
)2(22
4
3
4
3
v
D
u
u
kkkkkk
kkkk
kkkkkk
kkkk
Q
Q
P
P
(5-6)
Subtracting the force related to the control displacement from force vector, and other
external loads to be zero,
4
4
3
)2(66
)1(66
)2(63
)1(63
)2(62
)2(36
)2(36
)2(33
)1(33
)2(32
)2(26
)2(23
)2(22
)2(65
)2(35
)2(25
v
u
u
kkkkk
kkkkk
kkk
Dk
Dk
Dk
(5-7)
This process can be done before making the Global Stiffness Matrix. Imposing the
control displacement Dv 3 to the element stiffness matrix of Element (2),
4
1
4
3
1
)2(66
)2(65
)2(64
)2(63
)2(62
)2(61
)2(56
)2(55
)2(54
)2(53
)2(52
)2(51
)2(46
)2(45
)2(44
)2(43
)2(42
)2(41
)2(36
)2(35
)2(34
)2(33
)2(32
)2(31
)2(26
)2(25
)2(24
)2(23
)2(22
)2(21
)2(16
)2(15
)2(14
)2(13
)2(12
)2(11
4
3
1
4
3
1
v
D
v
u
u
u
kkkkkk
kkkkkk
kkkkkk
kkkkkk
kkkkkk
kkkkkk
Q
Q
Q
P
P
P
D
k
k
k
k
k
k
f
)2(65
)2(55
)2(45
)2(35
)2(25
)2(15
)2( (5-8)
The force vector )2(f will be summed together in the same manner of the element
stiffness matrix.
① ③
② ④
D
(2)
(1)
4
3
4
3
4
3
2
1
4
3
2
1
v
v
u
u
ConditionBoundary
v
v
v
v
u
u
u
u
③ fixed … u1=v1=0
④ fixed … u2=v2=0
18
6. HIGHER ORDER ELEMENT
The linear triangular element assumes the
deformation pattern to be a linear function
between two nodes.
It requires a large number of elements at the
place where deformation changes largely.
To reduce the number of elements, we
introduce the higher order elements, such as
the following second order elements where
the deformation pattern is assumed to be the
second order function of coordinate.
21211
210987
265
24321
yxyxyxv
yxyxyxu
(6-1)
In a matrix form,
12
2
1
22
22
1000000
0000001
yxyxyx
yxyxyx
v
u (6-2)
In order to define the second order function, we need
an additional node in the middle of each side of the
triangle. At the result, the total number of nodes in
one element is 6.
Before deformation
After deformation
Before
After
Before
After
v3
2
6
3
4
5
1
u3 v2
u2
19
The displacement of the element nodes are then expressed as,
12
8
7
6
2
1
2666
2666
2222
2222
2111
2111
2666
2666
2222
2222
2111
2111
6
2
1
6
2
1
1|
|0
1|
1|
|
|1
|
0|1
|1
yyxxyx
yyxxyx
yyxxyx
yyxxyx
yyxxyx
yyxxyx
v
v
v
u
u
u
(6-3)
u = A α
From Equations (6-1) and (6-2), we obtain
6
2
1
6
2
1
1
22
22
1000000
0000001
v
v
v
u
u
u
Ayxyxyx
yxyxyx
v
u
(6-4)
u(x,y) = H(x,y) U
As the same as the linear triangular element, the constitutive equation is obtained as
6
2
1
6
2
1
6
2
1
6
2
1
v
v
v
u
u
u
K
Q
Q
Q
P
P
P
(6-5)
20
F = K U
The process is summarized as follows:
(1) Translate external forces into equivalent nodal force,
F = {P1, …, P6, Q1, …, Q6}T
(2) Calculate the nodal displacements from the constitutive equation,
U = K-1 F
(3) Obtain the element deformation from the nodal displacement.
u(x,y) = H(x,y)U
21
7. INTERPOLATION FUNCTION
Suppose we have one dimensional element under loading. As discussed before, we
assume a linear function for the deformation pattern after loading,
xaaxu 10)(
or
1
01)(a
axxu (7-1)
The next step is to obtain the coefficients, a0, a1, from the nodal displacements. From
the relations:
1101 xaau
2102 xaau
or
1
0
2
1
2
1
1
1
a
a
x
x
u
u (7-2)
U = A α
The coefficients are obtained as, α = A-1 U. Then, the relation between the deformation
and the nodal displacements is,
2
111)(u
uAxxu (7-3)
Instead of the previous procedure, we introduce the interpolation functions to express
the deformation directly from the nodal displacements:
2211 )()()( uxhuxhxu (7-4)
The interpolation functions, h1 and h2, have the following characteristics:
1
11 ,0
,1)(
ux
uxxh ,
2
22 ,0
,1)(
ux
uxxh (7-5)
x1 x2 x
1 2 u1
u2
l
22
From these characteristics, the functions are easily obtained as,
l
xxxh
2
1 )( , l
xxxh 1
2 )(
(7-6)
One of the advantages of using interpolation functions is to reduce the burden to
calculate the inverse matrix of A in Equation (7-3).
In the same manner, if we assume a second order function for the deformation pattern,
the deformation can be directly expressed using interpolation functions as follows:
332211 )()()()( uxhuxhuxhxu (7-7)
x1 x2 x
1 2 u1
u2
l
x1 x2 x
u1
x1 x2 x
u2
x1 x2 x
1 2 u1
u2
l
x1 x2 x
u1
x1 x2 x
u2
x1 x2 x
u3
h1(x)u1
h2(x)u2
3
h3(x)u3
h1(x)u1
h2(x)u2
Second order interpolation function First order interpolation function
23
8. NATURAL COORDINATE
1) Natural coordinate
When we measure the coordinate of
the pencil, the result is different
depending on the scale we use. In
this example, the coordinate of the
head of the pencil is 5.0 in x-scale
and 9.5 in t-scale.
As long as we have one-to-one
relationship between two scales,
we can translate the value in one
scale to the value in another scale
anytime.
The total weight of the pencil will be calculated in x-axis as,
5
0
)( dxxwW (8-1)
To translate it into t-axis, we use the following relationships:
Global relationship:
)7(2 tx (8-2)
Local relationship:
dtdx 2 (8-3)
Substituting Equations (8-2) and (8-3) into (8-1), the total weight is expressed as,
5.9
5.7
)(2 dttxwW (8-4)
x
t = 7 + 0.5 x
t
x = 2 ( t – 7 )
1 2 3 4 5 60 x
8 9 107 t
6
w(x) : distribution of weight
x
x
w(x
x x+dx
dx
t 1
2
3
4
5
6 7 8 9 10
dt
2
x
0
24
Next we consider a more complicated scale to measure the total weight of the pencil.
The relationships between x-axis and t-axis are:
Global relationship: )(txx (8-5)
Local relationship: dtdt
tdxdx
)( (8-6)
Where dx(t)/dt represents the first derivative of x(x) by the variable t, which correspond
to the slope of x(t) at t. Substituting Equations (8-5) and (8-6) into (8-1), the total weight
will be expressed in t-axis as,
dtdt
tdxtxwW
)()( (8-7)
Setting α=-1, β=1,
dt
tdxtxwtfdttfW
)()()(,)(
1
1
(8-8)
Such coordinate is called “natural coordinate.”
1 2 3 4 5 60 x
βα t
w(x) : distribution of weight
x
x
w(x
x x+dx
dx
t 1
2
3
4
5
dt
0α β t
x = x(t)
x
25
2) Gaussian quadrature rule
If the integration range is [-1, 1], the integration can be evaluated approximately by
n-points Gaussian quadrature rule which is generally expressed in the following form:
)()()()(1
1
2211 nn tfwtfwtfwdttf
(8-9)
where, nwww ,,, 21 are the weighting coefficients. This formula requires a limited
number of function values, )(,),(),( 21 ntftftf , at the sampling points, nttt ,,, 21 , to
evaluate the integration.
For example, the 3 points Gaussian quadrature rule is defined as:
)()()(
)7746.0(5556.0)0(8889.0)7746.0(5556.0)(
332211
1
1
tfwtfwtfw
fffdttf
(8-10)
where, 5556.09/5,8889.09/8,5556.09/5 321 www
7746.05/3,0,7746.05/3 321 ttt
-1 -0.7746 0 +0.7746 +1
f(t)
f(-0.7746) f(0)
f(0.7746)
t
26
9. ISOPARAMETRIC ELEMENT
We now introduce the natural coordinate for the example of one dimensional element.
If we assume the linear transfer function x(t) between x-axis and t-axis, x(t) will be
expressed as
2211 )()()( xthxthtx (9-1)
where
)1(2
1)(),1(
2
1)( 21 tthtth (9-2)
Actually, it satisfies the fact that
21 )1(,)1( xxxx (9-3)
The deformation of the element is also
expressed as,
2211 )()()( uthuthtu (9-4)
Therefore, the functions )(),( 21 thth are the
interpolation functions we introduced before.
The element where both the coordinate
transfer function x(t) and the deformation
function u(t) are expressed using the same
interpolation functions on the natural
coordinate is called “Isoparametric element.”
x1 x2 x
1 2 u1
u2
-1 +1 t -1 +1
t
x
x1
x2 x(t)
t
-1 +1
u1 u2
u1
u2
-1 +1
-1 +1
h1(t)u1
h2(t)u2
t
t
27
Advantages of using isplarametric elements are summarized below:
(1) The relation
n
iii uthtu
1
)()( does not require the calculation of inverse matrix.
(2) The relation
n
iii xthtx
1
)()( enables to use the numerical integration method.
(3) Both functions u(t) and x(t) are expressed using the same interpolation functions.
28
10. SYSTEMATIC FORMULATION OF INTERPOLATION FUNCTION
(1) One dimensional element
2 Node
rh 12
11
1r 1r
1 rh 12
12
1r 1r
1
)1(2
1
)1(2
1
2
1
rh
rh
3 Node
)1(2
1
)1(2
1
2
1
rh
rh
)1(2
1
)1(2
1
2
2
r
r
23 1 rh
As presented here, if you increase a node to define the second order function for the
deformation, the interpolation function changes in the following manners:
- Modify the existing interpolation functions, h1 and h2,
- Define a new interpolation function, h3.
29
(2) Two dimensional element
30
11. STIFFNESS MATRIX FOR ISOPARAMETRIC ELEMENT
A two dimensional isoparametric element is adopted below.
The coordinate transfer function {x, y} is expressed using the interpolation functions as
4321
4
1
4321
4
1
)1)(1(4
1)1)(1(
4
1)1)(1(
4
1)1)(1(
4
1),(),(
)1)(1(4
1)1)(1(
4
1)1)(1(
4
1)1)(1(
4
1),(),(
ysrysrysrysrysrhsry
xsrxsrxsrxsrxsrhsrx
iii
iii
(11-1)
The deformation function {u, v} is also expressed using the same interpolation functions.
4321
4
1
4321
4
1
)1)(1(4
1)1)(1(
4
1)1)(1(
4
1)1)(1(
4
1),(),(
)1)(1(4
1)1)(1(
4
1)1)(1(
4
1)1)(1(
4
1),(),(
vsrvsrvsrvsrvsrhsrv
usrusrusrusrusrhsru
iii
iii
(11-2)
In a matrix form,
4
4
3
3
2
2
1
1
4321
4321
),(0),(0),(0),(0
0),(0),(0),(0),(
),(
),(
v
u
v
u
v
u
v
u
srhsrhsrhsrh
srhsrhsrhsrh
srv
sru
u(r, s) = H(r, s) U (11-3)
x, u
y, v
x4
y4
r
s Node 1
Node 2
Node 3 Node 4
31
Stiffness matrix can be obtained from the “Principle of Virtual Work Method,” which is
expressed in the following form:
V
TT FUdv (11-4)
where, is a virtual strain vector, is a stress vector, U is a virtual displacement
vector and F is a load vector, respectively. In case of the plane problem, the strain
vector is defined as,
x
v
y
uy
vx
u
xy
y
x
(11-5)
Substituting Equation (11-2), the strain vector is calculated as,
4
4
3
3
2
2
1
1
44332211
4321
4321
4
1
4
1
4
1
4
1
0000
0000
v
u
v
u
v
u
v
u
x
h
y
h
x
h
y
h
x
h
y
h
x
h
y
hy
h
y
h
y
h
y
hx
h
x
h
x
h
x
h
vx
hu
y
h
vy
h
ux
h
x
v
y
uy
vx
u
ii
i
ii
i
ii
i
ii
i
xy
y
x
ε = B U (11-6)
In the plane stress problem, the stress-strain relationship is expressed as,
xy
y
x
xy
y
xE
2
100
01
01
1 (11-7)
σ = D ε
32
Substituting Equation (11-6) into Equation (11-7),
σ= D B U (11-8)
From the Principle of Virtual Work Method,
FUUDBdvBUdvDBUUB T
V
TTT
V
(11-9)
Therefore, the constitutive equation is obtained as,
V
T DBdvBKKUF , (11-10)
If we assume the constant thickness of the plate (= t), using the relation tdxdydv ,
),( yxV
T DBdxdyBtK (11-11)
The following relationship is also derived from Equation (11-9)
FUdvBUdvUB T
V
TTT
V
(11-12)
Therefore, the relationship between stress vector and nodal force is
V
T dvBF (11-13)
Since this integration is defined in x-y coordinate, we must transfer the coordinate into
r-s coordinate to use the numerical integration method. Introducing the Jacobian
matrix,
MatrixJacobian
s
y
s
xr
y
r
x
J ;
(11-14)
the above integration is expressed in r-s coordinate as,
1
1
1
1 ),(
),(,,,,,, drds
sr
yxsrysrxDBsrysrxBtK T (11-15)
where
33
s
y
s
xr
y
r
x
Jsr
yx
det),(
),( (11-16)
1) Evaluation of Jacobian Matrix
4
1
4
1
4
1
4
1
ii
i
ii
i
ii
i
ii
i
ys
hx
s
h
yr
hx
r
h
s
y
s
xr
y
r
x
J (11-17)
2) Evaluation of the matrix B
From Equation (11-6),
x
h
y
h
x
h
y
h
x
h
y
h
x
h
y
hy
h
y
h
y
h
y
hx
h
x
h
x
h
x
h
B
44332211
4321
4321
0000
0000
(11-18)
The derivatives y
h
y
h
x
h
x
h
4141 ,,,,, are calculated as,
y
s
s
h
y
r
r
h
y
h
y
s
s
h
y
r
r
h
y
hx
s
s
h
x
r
r
h
x
h
x
s
s
h
x
r
r
h
x
h
444111
444111
,,
,,,
In a matrix form,
s
h
s
h
s
h
s
hr
h
r
h
r
h
r
h
y
s
y
rx
s
x
r
y
h
y
h
y
h
y
hx
h
x
h
x
h
x
h
4321
4321
4321
4321
s
h
s
h
s
h
s
hr
h
r
h
r
h
r
h
J4321
4321
1 (11-19)
34
3) Evaluation of partial derivatives of the interpolation functions
)1(4
1
)1(4
1
)1(4
1
)1(4
1
4
3
2
1
sr
h
sr
h
sr
h
sr
h
,
)1(4
1
)1(4
1
)1(4
1
)1(4
1
4
3
2
1
ss
h
rs
h
rs
h
rs
h
(11-20)
4) Numerical integration
Using the 3 points Gaussian quadrature rule, the stiffness matrix is calculated
numerically as follows:
3
1
3
1
1
1
1
1
1
1
1
1
),(
),(
),(
),(,,,,,,
i jjiji
T
srFwwt
drdssrFt
drdssr
yxsrysrxDBsrysrxBtK
(11-21)
where
),(
),(,,,,,,),(
sr
yxsrysrxDBsrysrxBsrF T
5556.09/5,8889.09/8,5556.09/5 321 www
7746.05/3,0,7746.05/3 332211 srsrsr
5) Assemble of finite element
To total stiffness matrix can be obtained to assemble the element stiffness matrix over
the areas of all finite elements.
m
mKK (11-22)
where m denotes the m-th element.
35
12. STRESS AND STRAIN AT GAUSSIAN POINTS
1) Stress and strain at Gaussian point
If you use the 3-points Gaussian Integration Method, there are nine Gaussian points
3,2,1,3,2,1, jisr ji in an element.
The stress and strain at the Gaussian point, ji sr , , is obtained from Equations (11-5)
and (11-7) as
UDBij
ijxy
y
x
ij
(12-1)
UBij
ijxy
y
x
ij
(12-2)
s = -1
s = +1
s = s2 = 0
r = -1
r = +1
×
×
×
r = r2 = 0
×
×
×
r = r3
×s = s3
s = s1
×
×
× : Gaussian points
x, u
y, v
x4
y4
r
s Node 1
Node 2
Node 3 Node 4 r = r1
36
2) Principal stress at Gaussian point
2
2
1 22 xyyxyx
(12-3)
2
2
2 22 xyyxyx
(12-4)
yx
xyp
2arctan
2
1 (12-5)
x
y
xy 12
x
y
xy
1
2
Note) Range of arctan is [-π/2, π/2], and 2θp is in this range Note) Range of arctan is [-π/2, π/2], and 2θp is in this range.
Note) Range of arctan is [-π/2, π/2], and 2θp is in this range.
37
2sin2
2cos22
2cos22
21
2121
2121
xy
y
x
(12-6)
x
y
xy
1
2
x
y
xy
1
2
Note) Range of arctan is [-π/2, π/2], and 2θp is not in this range.
So, the angle must be 2θp = π - |arctan|.
Note) Range of arctan is [-π/2, π/2], and 2θp is not in this range.
So, the angle must be 2θp = arctan - π.
2 1
y
x 2
221
xy
38
3) Displacement at Gaussian point
After obtaining the nodal displacement, the displacement at the Gaussian point
ji sr , , is obtained from Equation (11-2) as
4
1
4
1
),(),(
),(),(
iijiiji
iijiiji
vsrhsrv
usrhsru
(12-7)
39
13. INDEPENDENT FREEDOM 1) Freedom Vector
Exercise 1) Please obtain the freedom vector of the following structure.
21
3 4
X
Y
200 mm
P = 500 N
①
1000 mm
P
P
50 mm
E = 22000 N/mm/mm, = 0.1666
Initial Restrained freedom = 1 Numbering
1X 0 1 0
1Y 0 1 0
2X 0 0 1
{ F } = 2Y 0 0 2
3X 0 1 0
3Y 0 1 0
4X 0 0 3
4Y 0 0 4
21 3X
Y
① ②
54 6
40
2) Location Matrix
21
3 4
X
Y
200 mm
P = 500 N
①
1000 mm
P
P
50 mm
E = 22000 N/mm/mm, = 0.1666
1’X 1’ Y 2’ X 2’ Y 3’ X 3’ Y 4’ X 4’ Y
P1’ K11 K12 K13 K14 K15 K16 K17 K18 U1’
Q1’ K21 K22 K23 K24 K25 K26 K27 K28 V1’
P2’ K31 K32 K33 K34 K35 K36 K37 K38 U2’
Q2’ = K41 K42 K43 K44 K45 K46 K47 K48 V2’
P3’ K51 K52 K53 K54 K55 K56 K57 K58 U3’
Q3’ K61 K62 K63 K64 K65 K66 K67 K68 V3’
P4’ K71 K72 K73 K74 K75 K76 K77 K78 U4’
Q4’ K81 K82 K83 K84 K85 K86 K87 K88 V4’
①Element stiffness matrix
②Total stiffness matrix
U2 V2 U4 V4
P2 K33 K34 K35 K36 U2
Q2 = K43 K44 K45 K46 V2
P4 K53 K54 K55 K56 U4
Q4 K63 K64 K65 K66 V4
1’ 2’
3’4’
0
0
1
2
3
4
0
0
Location matrix
Element node number should be in anti-clockwise order.
41
Exercise 2) Please obtain the location matrix of the element ② of the following structure.
Freedom vector
1’ 2’
3’ 4’
21
3 4
①
1X 0
1Y 0
2X 1
{ F } = 2Y 2
3X 0
3Y 0
4X 3
4Y 4
Node number
1 > 1’
2 > 2’
3 > 4’
4 > 3’
1’X 0
1’Y 0
2’X 1
2’Y 2
3’X 3
3’Y 4
4’X 0
4’Y 0
Location matrix
21 3X
Y
① ②
54 6
42
Answers 1) and 2)
1) Freedom vector
2) Location matrix
21 3X
Y
②
4 6
①
5
2
1
3
4
Initial Restrained freedom = 1 Numbering
1X 0 1 0
1Y 0 1 0
2X 0 0 1
{ F } = 2Y 0 0 2
3X 0 1 0
3Y 0 1 0
4X 0 1 0
4Y 0 1 0
5X 0 0 3
5Y 0 0 4
6X 0 1 0
6Y 0 1 0
1’ 2’
3’ 4’
①
1’ 2’
3’ 4’
②
1’X 0
1’Y 0
2’X 1
2’Y 2
3’X 3
3’Y 4
4’X 0
4’Y 0
1’X 1
1’Y 2
2’X 0
2’Y 0
3’X 0
3’Y 0
4’X 3
4’Y 4
43
Element①
1 3X
Y
②
4 6
①
2
5
2 1
3
4
1’X 1’ Y 2’ X 2’ Y 3’ X 3’ Y 4’ X 4’ Y
P1’ K11 K12 K13 K14 K15 K16 K17 K18 U1’
Q1’ K21 K22 K23 K24 K25 K26 K27 K28 V1’
P2’ K31 K32 K33 K34 K35 K36 K37 K38 U2’
Q2’ = K41 K42 K43 K44 K45 K46 K47 K48 V2’
P3’ K51 K52 K53 K54 K55 K56 K57 K58 U3’
Q3’ K61 K62 K63 K64 K65 K66 K67 K68 V3’
P4’ K71 K72 K73 K74 K75 K76 K77 K78 U4’
Q4’ K81 K82 K83 K84 K85 K86 K87 K88 V4’
U2 V2 U5 V5
P2 K33 K34 K35 K36 U2
Q2 = K43 K44 K45 K46 V2
P5 K53 K54 K55 K56 U5
Q5 K63 K64 K65 K66 V5
0
0
1
2
3
4
0
0
Location matrix
44
Element②
Total stiffness matrix
1’X 1’ Y 2’ X 2’ Y 3’ X 3’ Y 4’ X 4’ Y
P1’ K11 K12 K13 K14 K15 K16 K17 K18 U1’
Q1’ K21 K22 K23 K24 K25 K26 K27 K28 V1’
P2’ K31 K32 K33 K34 K35 K36 K37 K38 U2’
Q2’ = K41 K42 K43 K44 K45 K46 K47 K48 V2’
P3’ K51 K52 K53 K54 K55 K56 K57 K58 U3’
Q3’ K61 K62 K63 K64 K65 K66 K67 K68 V3’
P4’ K71 K72 K73 K74 K75 K76 K77 K78 U4’
Q4’ K81 K82 K83 K84 K85 K86 K87 K88 V4’
U2 V2 U5 V5
P2 K11 K12 K17 K18 U2
Q2 = K21 K22 K27 K28 V2
P5 K71 K72 K77 K78 U5
Q5 K81 K82 K87 K88 V5
1
2
0
0
0
0
3
4
Location matrix
U2 V2 U5 V5
P2 K331+K112 K341+K122 K351+K172 K361+K182 U2
Q2 = K431+K212 K441+K222 K451+K272 K461+K282 V2
P5 K531+K712 K541+K721 K551+K771 K561+K781 U5
Q5 K631+K812 K641+K822 K651+K871 K661+K882 V5
45
In case of displacement control Treating the control direction 3Y and 6Y as restrained freedom, the freedom vector will
be the same as the previous example.
2
0
0
0
0
0
1
0
0
0
0
0
6
6
5
5
4
4
3
3
2
2
1
1
,
0
0
4
3
0
0
0
0
2
1
0
0
1
1
0
0
1
1
1
1
0
0
1
1
0
0
0
0
0
0
0
0
0
0
0
0
6
6
5
5
4
4
3
3
2
2
1
1
Y
X
Y
X
Y
X
Y
X
Y
X
Y
X
P
Y
X
Y
X
Y
X
Y
X
Y
X
Y
X
FF
Element②
0
0
2
0
1
0
0
0
'4
'4
'3
'3
'2
'2
'1
'1
,
4
3
0
0
0
0
2
1
'4
'4
'3
'3
'2
'2
'1
'1
Y
X
Y
X
Y
X
Y
X
P
Y
X
Y
X
Y
X
Y
X
FF
21 3X
Y
① ②
54 6 D
D
Freedom vector Control vector
Location matrix Location of control
2(1’) 3(2’)
6(3’) 5(4’)
46
Element②
Total equilibrium equation is
1’X 1’ Y 2’ X 2’ Y 3’ X 3’ Y 4’ X 4’ Y
P1’ K11 K12 K13 K14 K15 K16 K17 K18 U1’
Q1’ K21 K22 K23 K24 K25 K26 K27 K28 V1’
P2’ K31 K32 K33 K34 K35 K36 K37 K38 U2’
Q2’ = K41 K42 K43 K44 K45 K46 K47 K48 V2’
P3’ K51 K52 K53 K54 K55 K56 K57 K58 U3’
Q3’ K61 K62 K63 K64 K65 K66 K67 K68 V3’
P4’ K71 K72 K73 K74 K75 K76 K77 K78 U4’
Q4’ K81 K82 K83 K84 K85 K86 K87 K88 V4’
U2 V2 U5 V5
P2 K11 K12 K17 K18 U2
Q2 = K21 K22 K27 K28 V2
P5 K71 K72 K77 K78 U5
Q5 K81 K82 K87 K88 V5
1 0
2 0
0 0
0 1
0 0
0 2
3 0
4 0
Location matrix
Location of control
2(1’) 3(2’)
②
5(4’) 6(3’) D
D
V3 V6
P2 K14 K16
Q2 = K24 K26 -1 D
P5 K74 K76 -1
Q5 K84 K86
U2 V2 U5 V5
-K14-K16 K331+K112 K341+K122 K351+K172 K361+K182 U2
-K24-K26 D = K431+K212 K441+K222 K451+K272 K461+K282 V2
-K74-K76 K531+K712 K541+K721 K551+K771 K561+K781 U5
-K84-K86 K631+K812 K641+K822 K651+K871 K661+K882 V5
47
14. SKYLINE METHOD
Usually, the total stiffness matrix [ K ] is symmetric and sparse as shown below.
Therefore, to save memory size and to reduce calculation time for linear equation solver,
the elements in the upper triangular part of the matrix under the Skyline (thick line)
are stored in an one-dimension vector.
1 K11
2 K22
3 K21
4 K33
5 K23
6 K44
7 K43
8 0
9 K14
10 K55
11 K45
12 K35
13 K25
14 K66
15 K56
16 K77
17 K67
18 K57
19 K47
20 K37
21 K27
22 K17
23 K88
24 K78
25 K68
26 K58
27
0 1 1 3 3 1 6 3
Skyline height
1 2 4 6 10 14 16 23 27
Diagonal element order
Total stiffness matrix [ K ] Stiffness vector Skyline
6
Band width ( = the maximum skyline height)
48
Exercise 3) Using the same structure in Exercise 1), please compare
1) skyline height, 2) diagonal element order and 3) band width
between the following two cases with different node order.
Case 1
Case 2
2 1 4X
Y
① ② ③ ④
3 5
7 6 98 10
3 1 7X
Y
① ② ③ ④
5 9
4 2 86 10
49
15. INCOMPATIBLE ELEMENT
In an ideal situation, a beam under a pure bending moment experiences a curved shape
change. The angle between the curved horizontal dotted line and the straight vertical
line remains at 90 degree after bending. Therefore, no shear strain occurs inside a
material.
To model the ideal shape change, an element should have the ability to assume the
curved shape. The second order element with eight nodes enables to represent the
curved shape. On the contrary, the first order element is not able to bend to curves and
all dotted lines remain straight and the angle is no longer at 90 degree. To cause the
angle to change under pure moment, an incorrect artificial shear strain and stress have
been introduced. Therefore, the strain energy of the element is larger than ideal
situation. As we demonstrated in the principle of virtual work method, overestimate of
strain energy causes overestimate of stiffness matrix. This is the reason that the first
order element with four nodes becomes overly stiff under the bending moment. This
problem is called shear locking.
no shear strain
no shear strain
shear strain
(a) Second order element (b) First order element
50
To solve the problem in the first order element, we introduce the new displacement
shape functions to add curved displacement modes.
42
32
4
1
22
12
4
1
11),(),(
11),(),(
srvsrhsrv
srusrhsru
iii
iii
(15-1)
It can avoid over stiff in bending; however there might be incompatibility of deformation
at the boundary. Therefore, this element is called incompatible element.
Equation (15-1) can be written in a matrix form as,
u(r, s) = H(r, s) U + G(r, s) A (15-2)
where,
4321
4321
0000
0000
hhhh
hhhhH ,
21
21
00
00
gg
ggG ,
22
21 1,1 sgrg
44332211 vuvuvuvuU T , 4321 TA
21 sr
s
21 r
incompatible
51
Stiffness matrix can be obtained from the “Principle of Virtual Work Method,” which is
expressed in the following form:
V
TT FUdv (15-3)
where, is a virtual strain vector, is a stress vector, U is a virtual displacement
vector and F is a load vector, respectively.
The strain vector is calculated from the nodal displacement vector as,
4
3
2
1
2121
21
21
4
4
3
3
2
2
1
1
44332211
4321
4321
4
14
23
12
21
14
1
4
14
23
1
4
12
21
1
00
00
0000
0000
x
g
x
g
y
g
y
gy
g
y
gx
g
x
g
v
u
v
u
v
u
v
u
x
h
y
h
x
h
y
h
x
h
y
h
x
h
y
hy
h
y
h
y
h
y
hx
h
x
h
x
h
x
h
x
g
x
gv
x
h
y
g
y
gu
y
hy
g
y
gv
y
hx
g
x
gu
x
h
x
v
y
uy
vx
u
ii
i
ii
i
ii
i
ii
i
xy
y
x
ε = B U + G A (15-4)
In the plane stress problem, the stress-strain relationship is expressed as,
xy
y
x
xy
y
xE
2
100
01
01
1 (15-5)
σ = D ε
Substituting Equation (15-4) into Equation (15-5),
σ= D ( B U + G A ) (15-6)
52
From the Principle of Virtual Work Method,
FU
ADGdvGAUDBdvGAADGdvBUUDBdvBU
dvGABUDAGUB
T
V
TT
V
TT
V
TT
V
TT
T
V
(15-7)
It can be written in a Matrix form as;
0
FAU
A
U
DGdvGDBdvG
DGdvBDBdvB
AU TT
V
T
V
TV
T
V
T
TT
Therefore, the equilibrium equation is obtained as,
0
F
A
U
KK
KK
AAAU
UAUU (15-8)
where
V
TAA
V
TAU
V
TUA
V
TUU
DGdvGKDBdvGK
DGdvBKDBdvBK
,
,
From the second equation in (15-8), we can eliminate the incompatible displacement
modes as;
UKKA
AKUK
AUAA
AAAU
1
0
(15-9)
Then, the element stiffness matrix is given by:
AUAAUAUU KKKKK
KUF1
(15-10)
53
1) Evaluation of Jacobian Matrix
4
1
4
1
4
1
4
1
ii
i
ii
i
ii
i
ii
i
ys
hx
s
h
yr
hx
r
h
s
y
s
xr
y
r
x
J (15-11)
2) Evaluation of the matrix G
From Equation (15-4),
x
g
x
g
y
g
y
gy
g
y
gx
g
x
g
G
2121
21
21
00
00
(15-12)
The derivatives y
g
y
g
x
g
x
g
2121 ,,, are calculated as,
y
s
s
g
y
r
r
g
y
g
y
s
s
g
y
r
r
g
y
gx
s
s
g
x
r
r
g
x
g
x
s
s
g
x
r
r
g
x
g
222111
222111
,
,,
In a matrix form,
s
g
s
gr
g
r
g
J
s
g
s
gr
g
r
g
y
s
y
rx
s
x
r
y
g
y
gx
g
x
g
21
21
1
21
21
21
21
(15-13)
3) Evaluation of partial derivatives of the interpolation functions
0
2
2
1
r
g
rr
g
,
ss
gs
g
2
0
2
1
(15-14)
54
4) Numerical integration
Using the 3 points Gaussian quadrature rule, the stiffness matrices are calculated
numerically as follows:
3
1
3
1
),(i j
jiUUjiUU srFwwtK
3
1
3
1
),(i j
jiUAjiUA srFwwtK
3
1
3
1
),(i j
jiAUjiAU srFwwtK
3
1
3
1
),(i j
jiUAjiAA srFwwtK (15-15)
where
),(
),(,,,,,,),(
sr
yxsrysrxDBsrysrxBsrF T
UU
),(
),(,,,,,,),(
sr
yxsrysrxDGsrysrxBsrF T
UA
),(
),(,,,,,,),(
sr
yxsrysrxDBsrysrxGsrF T
AU
),(
),(,,,,,,),(
sr
yxsrysrxDGsrysrxGsrF T
AA
5556.09/5,8889.09/8,5556.09/5 321 www
7746.05/3,0,7746.05/3 332211 srsrsr
Then, the element stiffness matrix is calculated as,
AUAAUAUU KKKKK
KUF1
(15-16)
5) Assemble of finite element
To total stiffness matrix can be obtained to assemble the element stiffness matrix over
the areas of all finite elements.
m
mKK (15-17)
where m denotes the m-th element.
55
6) Strain and Stress at Gaussian point
Strain at Gaussian point is obtained from Equations (15-4) and (15-9) as:
UKGKBGABU AUAA )( 1 (15-8)
Stress at Gaussian point is obtained as:
UKDGKDBD AUAA )( 1 (15-9)
56
16. HEXAHEDRON ELEMENT
A hexahedron element is extensively used in modeling three-dimensional solids. It has
eight corners, twelve edges, and six faces. By introducing natural coordinates tsr ,, ,
which vary from -1 to +1, it is possible to use the Gauss integration formulas.
1 (1,1,1)
t
r
18 (-1,1,0)
3 (-1,-1,1)
s
4 (1,-1,1)
6 (-1,1,-1)
5 (1,1,-1)
7 (-1,-1,-1)
8 (1,-1,-1)
t
r
s
9 (0,1,1)
10 (-1,0,1)
11 (0,-1,1)
12 (1,0,1)
13 (0,1,-1)14 (-1,0,-1)
16 (1,0,-1)
15 (0,-1,-1)
t
r
s
(b) Natural coordinates
17 (1,1,0)
2 (-1,1,1)
19 (-1,-1,0)
20 (1,-1,0)
57
The corner numbering rule is described as follows to guarantee a positive Jacobian
determinant.
1) Chose starting corner (number 1) and select one face pertaining to that corner.
2) Number the other corners on the face in counterclockwise order (number 2, 3, 4).
3) Number the corners of the opposite face to be opposite 1,2,3,4 as 5,6,7,8.
4) Number the middle point on the edge following the number of corner.
The interpolation function is given by
22
22
22
222
11114
1
11114
1
11114
1
21118
1,,
tssrrt
srrtts
rttssr
ttssrrttssrrtsrtsrh
iii
iii
iii
iiiiiiiiii
(16-1)
where
i 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
ri 1 -1 -1 1 1 -1 -1 1 0 -1 0 1 0 -1 0 1 1 -1 -1 1
si 1 1 -1 -1 1 1 -1 -1 1 0 -1 0 1 0 -1 0 1 1 -1 -1
ti 1 1 1 1 -1 -1 -1 -1 1 1 1 1 -1 -1 -1 -1 0 0 0 0
(16-2)
The partial derivative of the interpolation function is
tssrrttsrrstrssrt
ssrrttssrrrtsrt
tsrh
trrtssrrttsrttrs
ttrrttssrrstsrs
tsrh
tsstrsttsrrttssr
ttssttssrrrtsrr
tsrh
iiiiiiiiii
iiiiiiiiii
iiiiiiiii
iiiiiiiiii
iiiiiiiii
iiiiiiiiii
1112
1111
4
1111
4
1
11128
1,,
1114
1111
2
1111
4
1
11128
1,,
1114
1111
4
1111
2
1
11128
1,,
22222
222
22222
222
22222
222
(16-3)
58
The coordinate transfer function {x, y, z} is expressed using the interpolation functions
20
1
20
1
20
1
),,(),,(
),,(),,(
),,(),,(
iii
iii
iii
ztsrhtsrz
ytsrhtsry
xtsrhtsrx
(16-4)
The deformation function {u, v, w} is expressed using the same interpolation functions.
20
1
20
1
20
1
),,(),,(
),,(),,(
),,(),,(
iii
iii
iii
wtsrhtsrw
vtsrhtsrv
utsrhtsru
(16-5)
In a matrix form,
20
20
20
1
1
1
201
201
201
0000
0000
0000
w
v
u
w
v
u
hh
hh
hh
w
v
u
u(r, s,t) = H(r, s,t) U (16-6)
In case of the plane problem, the strain vector is defined as,
20
20
20
1
1
1
202011
202011
202011
201
201
201
20
1
20
1
20
1
20
1
20
1
20
1
20
1
20
1
20
1
00
00
00
0000
0000
0000
w
v
u
w
v
u
x
h
z
h
x
h
z
hy
h
z
h
y
h
z
hx
h
y
h
x
h
y
hz
h
z
hy
h
y
hx
h
x
h
uz
hw
x
h
wy
hv
z
h
vx
hu
y
h
wz
h
vy
h
ux
h
z
u
x
wy
w
z
vx
v
y
uz
wy
vx
u
ii
i
ii
i
ii
i
ii
i
ii
i
ii
i
ii
i
ii
i
ii
i
zx
yz
xy
z
y
x
ε = B U (16-7)
59
The stress-strain relationship is expressed as,
EGGGG
EEEEEEEEE
zxzx
yzyz
xyxy
yxzz
zxyy
zyxx
)1(2
1,,,
,,,
(16-8)
In a matrix form,
zx
yz
xy
z
y
x
zx
yz
xy
z
y
x
G
G
G
EEE
EEE
EEE
100000
010000
001000
0001
0001
0001
(16-9)
From the inverse matrix, the constitutive matrix is obtained as,
zx
yz
xy
z
y
x
zx
yz
xy
z
y
x
G
2/2100000
02/210000
002/21000
0001
0001
0001
21
2
σ = D ε (16-10)
Substituting Equation (16-7) into Equation (16-10),
σ= D B U (16-11)
From the Principle of Virtual Work Method, the stiffness matrix is obtained as,
V
T DBdvBKKUF , (16-12)
Since this integration is defined in x-y-z coordinate, we must transfer the coordinate
into r-s-t coordinate to use the numerical integration method. Introducing the Jacobian
matrix,
60
MatrixJacobian
t
z
t
y
t
xs
z
s
y
s
xr
z
r
y
r
x
J ;
(16-13)
The above integration is expressed in r-s coordinate as,
1
1
1
1
1
1 ),,(
),,(,,,, drdsdt
tsr
zyxtsrDBtsrBK T (16-14)
where
t
z
t
y
t
xs
z
s
y
s
xr
z
r
y
r
x
Jtsr
zyx
det),,(
),,( (16-15)
1) Evaluation of Jacobian Matrix
20
1
20
1
20
1
20
1
20
1
20
1
20
1
20
1
20
1
ii
i
ii
i
ii
i
ii
i
ii
i
ii
i
ii
i
ii
i
ii
i
zt
hy
t
hx
t
h
zs
hy
s
hx
s
h
zr
hy
r
hx
r
h
t
z
t
y
t
xs
z
s
y
s
xr
z
r
y
r
x
J (16-16)
2) Evaluation of the matrix B
From Equation (16-7),
x
h
z
h
x
h
z
hy
h
z
h
y
h
z
hx
h
y
h
x
h
y
hz
h
z
hy
h
y
hx
h
x
h
B
202011
2012011
202011
201
201
201
00
00
00
0000
0000
0000
(16-17)
61
The derivatives z
h
z
h
y
h
y
h
x
h
x
h
201201201 ,,,,,,,, are calculated as,
z
t
t
h
z
s
s
h
z
r
r
h
z
h
z
t
t
h
z
s
s
h
z
r
r
h
z
h
y
t
t
h
y
s
s
h
y
r
r
h
y
h
y
t
t
h
y
s
s
h
y
r
r
h
y
hx
t
t
h
x
s
s
h
x
r
r
h
x
h
x
t
t
h
x
s
s
h
x
r
r
h
x
h
202020201111
202020201111
202020201111
,,
,,,
,,,
(16-18)
In a matrix form,
t
h
t
hs
h
s
hr
h
r
h
J
t
h
t
hs
h
s
hr
h
r
h
z
t
z
s
z
ry
t
y
s
y
rx
t
x
s
x
r
z
h
z
hy
h
y
hx
h
x
h
201
201
201
1
201
201
201
201
201
201
(16-19)
3) Stress and strain at Gaussian point
If you use the 3-points Gaussian Integration Method, there are 27 Gaussian points
3,2,1,3,2,1,3,2,1,, kjitsr kji in an element. The stress and strain at the
Gaussian point kji tsr ,, is obtained from
UDBij
ijzx
yz
xy
z
y
x
ij
, UBij
ijzx
yz
xy
z
y
x
ij
(16-20)
4) Numerical integration
To reduce calculation time, the 2 points Gaussian quadrature rule is adopted to
calculate the stiffness matrix as follows:
62
2
1
2
1
2
1
1
1
1
1
1
1
1
1
1
1
1
1
),,(
),,(
),,(
),,(,,,,
i j kkjikji
T
tsrFwww
drdsdttsrF
drdsdttsr
zyxtsrDBtsrBK
(16-21)
where
),,(
),,(,,,,),,(
tsr
zyxtsrDBtsrBtsrF T
0.121 ww
0.577353/1,0.577353/1 222111 tsrtsr
5) Assemble of finite element
To total stiffness matrix can be obtained to assemble the element stiffness matrix over
the areas of all finite elements.
m
mKK (16-22)
where m denotes the m-th element.
63
17. PLAIN STRESS AND PLAIN STRAIN
1) Plain stress assumption
If a thin plate in the x-y coordinate is loaded by forces at the boundary and distributed
uniformly over the thickness, the stress components along z-axis, yzxzz ,, , are zero
on both faces of the plate. It can be assumed that they are also zero within the plate.
Then, from Equation (16-9), the stress-strain relationship is expressed as,
0
0
0
100000
010000
001000
0001
0001
0001
zx
yz
xy
z
y
x
zx
yz
xy
z
y
x
G
G
G
EEE
EEE
EEE
(17-1)
Thus,
xy
y
x
xy
y
x
G
EE
EE
100
011
011
(17-2)
or
xy
y
x
xy
y
xE
2
100
01
01
1 2 (17-3)
x
z
y
Plain stress condition
64
2) Plain strain assumption
When the dimension of the body in z-direction is very large such as a retaining wall
with lateral pressure, it may be assumed that the displacement in the z-direction is
prevented.
Since the longitudinal displacement is zero, from Equation (16-10),
0
0
0
2100000
0210000
0021000
0001
0001
0001
21
2
zx
yz
xy
z
y
x
zx
yz
xy
z
y
x
G
(17-4)
Thus,
xy
y
x
xy
y
xG
2/2100
01
01
21
2 (17-5)
or
xy
y
x
xy
y
x
G
200
01
01
2
1 (17-6)
Note that, from Equation (17-4), the normal stress along z-axis is not zero.
0z
which will be considered in case of nonlinear problem.
Plain strain condition
xz
y
65
CHAPTER 2
DYNAMIC ANALYSIS
66
Index of Chapter 2 1. Mass Matrix for Isoparametric Element
2. Eigen Value Problem
3. Classical Damping
4. Equation of Motion under Earthquake Ground Motion
67
1. MASS MATRIX FOR ISOPARAMETRIC ELEMENT
1) Formulation
Under dynamic loading, the “Principle of Virtual Work Method in dynamic problem =
D’Alembert’s principle” is expressed in the following form:
0 WQ (1-1)
V
T dvQ (1-2)
V
T dvt
u
t
uFuW
2
2
(1-3)
where F: body force, ρ: density, μ: damping coefficient
Substitute following relationships into above equations:
u(r, s) = H(r, s) U
ε= B U
σ= D B U
UDBdvBUdvDBUUBQV
TTT
V
(1-4)
UHdvHUUHdvHUdvFHU
dvt
u
t
uFuW
V
TT
V
TT
V
TT
V
T
2
2
(1-5)
Therefore, from Equation (1-1),
V
TT
V
T
V
T
V
T dvFHUUCBdvBUHdvHUHdvH (1-5)
That is, the equilibrium equation is expressed as,
V
T
V
T
V
T
V
T dvFHRDBdvBKHdvHCHdvHM
RKUUCUM
,,,
(1-6)
68
2) Evaluation of the matrix H and HTH
For a two dimensional isoparametric element,
4321
4321
0000
0000
hhhh
hhhhH (1-7)
24
24
4323
4323
423222
423222
4131212
1
4131212
1
4321
4321
4
4
3
3
2
2
1
1
0.
0
00
00
000
000
0000
0000
0000
0
0
0
0
0
0
0
0
h
hsym
hhh
hhh
hhhhh
hhhhh
hhhhhhh
hhhhhhh
hhhh
hhhh
h
h
h
h
h
h
h
h
HH T
(1-8)
For a three-dimensional hexahedron element,
201
201
201
0000
0000
0000
hh
hh
hh
H
(1-9)
69
220
220
220
20222
20222
20222
201212
1
201212
1
201212
1
201
201
201
20
20
20
1
1
1
0
00.
00
000
0000
0000
00000
000000
0000
0000
0000
00
00
00
00
00
00
h
h
hsym
hhh
hhh
hhh
hhhhh
hhhhh
hhhhh
hh
hh
hh
h
h
h
h
h
h
HH T
(1-10)
3) Numerical integration
The integration for mass matrix can be expressed in r-s coordinate as,
1
1
1
1
),(
det drdsJHHt
HdxdyHt
HdvHM
T
yxV
T
V
T
(1-11)
The integration can be evaluated by the Gaussian Integration Formula as,
ijjijiT
ji
i jjiji
JsrHsrHsrG
srGwwtM
det,,,
,3
1
3
1
(1-12)
70
4) Lumped mass model
The mass matrix obtained from the density of material is called the consistent mass
matrix using the same interpolation functions for stiffness matrix, mass matrix and
load vectors. Instead of performing the integrations, we may evaluate an approximate
mass matrix by lumping equal parts of the total element mass to the nodal points which
is called the lumped mass matrix. An important advantage of using a lumped mass
matrix is that the matrix is diagonal and the numerical operations for the solution of
the dynamic equations are reduced significantly.
Suppose that the consistent mass matrix is expressed as,
CnnCnCn
nCCC
nCCC
C
mmm
mmm
mmm
M
21
22221
11211
(1-13)
The lumped matrix can be evaluated from the consistent mass matrix from the
following formula:
Lnn
L
L
L
m
m
m
M
00
00
00
22
11
,
n
jCijLii mm
1
(1-14)
5) Gravity force
Gravity force is considered as a body force as
g
FdrdsJFHtFdxdyHtdvFHR T
yxV
T
V
T
0
,det1
1
1
1),(
(1-15)
where )/8.9( 2smg is the gravity acceleration.
ijjiT
ji
i jjiji
Jg
srHsrI
srIwwtR
det0
,,
,3
1
3
1
(1-16)
71
2. EIGEN VALUE PROBLEM
The free vibration equilibrium equation without damping is
0 KUUM (2-1)
where K is the stiffness matrix and M is the lumped mass matrix in the form,
nm
m
m
M
00
00
00
2
1
(2-2)
The solution can be postulated to be in the form
tieU (2-3)
where is a vector of order n, is a frequency of vibration of the vector .
Then, the generalized eigenproblem is,
MK 2 (2-4)
This eigenproblem yields the n eigensolutions nn ,,,,,, 22
221
21 where the
eigenvevtors are M-orthonormalized as,
ji
jimM
n
kkjkikj
Ti ;0
;1
1,, (2-5)
222
210 n (2-6)
The vector i is called the i-th mode shape vector, and i is the corresponding
frequency of vibration.
Defining a matrix whose columns are the eigenvectors and a diagonal matrix 2
which stores the eigenvalues on its diagonal as,
72
n 21 ,
2
22
21
2
n
(2-7)
We introduce the following transformation on the displacement vector of the
equilibrium Equation (1-6),
)()( tXtU (2-8)
Then,
RXKXCXM (2-9)
Multiplying T ,
RXKXCXM TTTT (2-10)
Using 2, KIM TT ,
RXXCX TT 2 (2-11)
A damping matrix that is diagonalized by is called a classical damping matrix.
nn
T
h
h
h
CC
2
2
2
22
11
(2-12)
where ih is the modal damping ratio of the i-th mode.
Then, Equation (2-11) reduce to n- equations of the form
)()()(2)( 2 trtxtxhtx iiiii (2-13)
where )()( tRtr Tii
The initial conditions on X(t) are obtained from Equation (2-8) as,
0000 , tT
ttT
t UMXMUX (2-14)
73
3. CLASSICAL DAMPING
Three procedures for constructing a classical damping matrix are described as follow:
1) Proportional damping
Consider first mass-proportional damping and stiffness-proportional damping,
MaC 0 and KaC 1 (3-1)
where the constants 10 , aa have units of sec-1 and sec, respectively.
For a system with mass-proportional damping, the generalized damping for the i-th
mode is,
ii mac 0 , iiii hmc 2/ (3-2)
Therefore,
iiha 20 , i
i
ah
1
20 (3-3)
Similarly, for a system with stiffness-proportional damping, the generalized damping
for the i-th mode is,
iii mac 21 , iiii hmc 2/ (3-4)
Therefore,
i
iha
2
1 , ii
ah
21 (3-5)
ih
KaC 1
ii
ah
21
MaC 0
ii
ah
1
20
74
2) Rayleigh damping
A Rayleigh damping matrix is proportional to the mass and stiffness matrices as,
KaMaC 10 (3-6)
The modal damping ratio for the i-th mode of such a system is,
ii
i
aah
2
1
210 (3-7)
The coefficients 10 , aa can be determined from specified damping ratios 21, hh modes,
respectively. Expressing Equation (3-3) for these two modes in matrix form leads to:
2
1
1
0
22
11
/1
/1
2
1
h
h
a
a
(3-8)
Solving the above system, we obtain the coefficients 10 , aa :
2
22
1
12211
22
21
1221210
2
2
hha
hha
(3-9)
ih
KaMaC 10
ii
i
aah
2
1
210
75
3) Modal damping
It is an alternative procedure to determinate a classical damping matrix from modal
damping ratios. From the definition of a classical damping matrix,
nn
T
h
h
h
CC
2
2
2
22
11
11 CC T (3-10)
Since ,IMT
MT 1, MT1 (3-11)
Therefore,
MCMC T (3-12)
76
4. EQUATION OF MOTION UNDER EARTHQUAKE GROUND MOTION
1) Equation of motion under earthquake ground motion
Earthquake ground motions are defined as two components acceleration; 0X and 0Y , in X and Y
directions. The inertia forces at node i are defined as,
0
0
0
0
0
0
10
01
Y
XMIUM
Y
XM
v
uM
YvM
XuM
i
i
ii
ii
(4-1)
where
10
01,
00
00
00
, 2
1
I
m
m
m
Mv
uU
n
i
i (4-2)
Equilibrium condition of the structure under earthquake ground motion is:
Finally the equation of motion is obtained as:
RY
XMIKUUCUM
0
0
(4-3)
Inertia force Damping force
Restoring force
0
0
Y
XMIUMKUUC
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2) Numerical integration by Newmark-β method
The incremental formulation for the equation of motion of a structural system is,
iiii pdKvCaM (4-4)
where, M , C and K are the mass, damping and stiffness matrices. id , iv ,
ia and ip are the increments of the displacement, velocity, acceleration and external force
vectors, that is,
iii ddd 1 , iii vvv 1 ,
iii aaa 1 , iii ppp 1 (4-5)
Using the Newmark-β method,
tatav iii 2
1 (4-6)
22
2
1tatatvd iiii (4-7)
From Equation (4-7), we obtain
iiii avt
dt
a 2
1112
(4-8)
Substituting Equation (4-7) into Equation (4-6) gives
tavdt
v iiii
4
11
2
1
2
1 (4-9)
Equations (4-8) and (4-9) are substituted into Equation (4-4), and we obtain
tavCavt
Mp
KCt
Mt
d
iiiii
i
14
1
2
1
2
11
2
112
(4-10)
The equation can be rewritten as,
ii pdK ˆˆ (4-11)
where,
Mt
Ct
KK2
1
2
1ˆ
(4-12)
tavCav
tMpp iiiiii 1
4
1
2
1
2
11ˆ
(4-13)
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