Finite Element Methods
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Transcript Finite Element Methods
UNIT I
Introduction to Finite Element
Methods
Numerical Methods – Definition
and Advantages
Definition: Methods that seek
quantitative approximations to the solutions
of mathematical problems
Advantages:
What is a Numerical Method – An
Example
Example 1:
du
dt
u
u (0) u 0
What is a Numerical Method – An
Example
Example 1:
du
dt
u
u (0) u 0
What is a Numerical Method – An
Example
Example 2:
du
dt
2u 5e
-t
u (0) u 0
What is a Numerical Method – An
Example
Example 2:
du
dt
2u 5e
-t
u (0) 1
What is a Numerical Method – An
Example
Example 3:
du
dt
u t
2
u (0) 1
What is a Finite Element Method
Discretization
1-D
?-D
2-D
Hybrid
3-D
Approximation
Numerical Interpolation
Non-exact Boundary Conditions
Applications of Finite Element Methods
Structural & Stress Analysis
Thermal Analysis
Dynamic Analysis
Acoustic Analysis
Electro-Magnetic Analysis
Manufacturing Processes
Fluid Dynamics
Lecture 2
Review
Matrix Algebra
• Row and column vectors
• Addition and Subtraction – must have the same dimensions
• Multiplication – with scalar, with vector, with matrix
• Transposition –
• Differentiation and Integration
Matrix Algebra
• Determinant of a Matrix:
• Matrix inversion • Important Matrices
• diagonal matrix
• identity matrix
• zero matrix
• eye matrix
Numerical Integration
b
Calculate:
I
f x dx
a
• Newton – Cotes integration
• Trapezoidal rule – 1st order Newton-Cotes integration
f ( x ) f1 ( x ) f ( a )
b
I
f (b ) f ( a )
ba
b
f ( x ) dx
a
f 1 ( x ) dx ( b a )
(x a)
f ( a ) f (b )
a
2
• Trapezoidal rule – multiple application
xn
b
I
a
f ( x ) dx
x0
x1
f n ( x ) dx
x0
x2
f ( x ) dx f ( x ) dx
x1
n 1
h
I f ( a ) 2 f ( xi ) f (b )
2
i 1
xn
x n 1
f ( x ) dx
Numerical Integration
b
Calculate:
I
f x dx
a
• Newton – Cotes integration
• Simpson 1/3 rule – 2nd order Newton-Cotes integration
f ( x) f2 ( x)
( x x1 )( x x 2 )
( x 0 x1 )( x 0 x 2 )
f ( x0 )
( x x 0 )( x x 2 )
( x1 x 0 )( x1 x 2 )
b
I
a
f ( x1 )
b
f ( x ) dx
a
f 2 ( x ) dx ( x 2 x 0 )
( x x 0 )( x x1 )
( x 2 x 0 )( x 2 x1 )
f ( x2 )
f ( x 0 ) 4 f ( x1 ) f ( x 2 )
6
Numerical Integration
b
Calculate:
I
f x dx
a
• Gaussian Quadrature
Trapezoidal Rule:
I (b a )
Gaussian Quadrature:
f ( a ) f (b )
2
(b a )
2
f (a )
(b a )
I c 0 f ( x 0 ) c1 f ( x1 )
f (b )
2
Choose c 0 , c1 , x 0 , x1
according to certain criteria
Numerical Integration
b
I
Calculate:
f x dx
a
• Gaussian Quadrature
1
I
f x dx
c 0 f x 0 c1 f x 1 c n 1 f x n 1
1
• 2pt Gaussian Quadrature
1
I
1
1
1
f x dx f
f
3
3
•3pt Gaussian Quadrature
1
I
f x dx
0 . 55 f 0 . 77 0 . 89 f 0 0 . 55 f 0 . 77
1
2( x a )
~
x 1
ba
Let:
b
a
f ( x ) dx
1
1
1
(b a ) f ( a b ) (b a ) ~
x d ~
x
2
2
2
1
1
Numerical Integration - Example
1
Calculate:
I
e sin x dx
x
0
• Trapezoidal rule
• Simpson 1/3 rule
• 2pt Gaussian quadrature
• Exact solution
1
I
e sin xdx
x
0
e sin x e cos x
x
x
2
1
0 . 90933
0
Linear System Solver
Solve:
Ax b
• Gaussian Elimination: forward elimination + back substitution
Example:
x1 2 x 2 6 x 3 0
2 x1 2 x 2 3 x 3 3
x1 3 x 2
0
1
2
1
2
2
3
1
0
0
2
1
0
0
2
6
1
6
0
6 x1 0
3 x2 3
0 x 3 2
6 x1 0
9 x2 3
6 x 3 2
6 x1 0
9 x2 3
15 2 x 3 3 2
Linear System Solver
Ax b
Solve:
• Gaussian Elimination: forward elimination + back substitution
Pseudo code:
Forward elimination:
Do k = 1, n-1
Do i = k+1,n
c
a ik
a kk
Do j = k+1, n
a ij a ij ca kj
bi bi cb k
Back substitution:
Do ii = 1, n-1
i = n – ii
sum = 0
Do j = i+1, n
sum = sum + a ij b j
bi
bi sum
a ii
UNIT II
Finite Element Analysis (F.E.A.) of 1-D
Problems
Historical Background
• Hrenikoff, 1941 – “frame work method”
• Courant, 1943 – “piecewise polynomial
interpolation”
• Turner, 1956 – derived stiffness matrice for truss,
beam, etc
• Clough, 1960 – coined the term “finite element”
Key Ideas: - frame work method
piecewise polynomial approximation
Axially Loaded Bar
Review:
Stress:
Stress:
Strain:
Strain:
Deformation:
Deformation:
Axially Loaded Bar
Review:
Stress:
Strain:
Deformation:
Axially Loaded Bar – Governing
Equations and Boundary
Conditions
• Differential Equation
d
du
EA
(
x
)
f ( x) 0
dx
dx
0 x L
• Boundary Condition Types
• prescribed displacement (essential BC)
• prescribed force/derivative of displacement
(natural BC)
Axially Loaded Bar –Boundary
Conditions
• Examples
• fixed end
• simple support
• free end
Potential Energy
• Elastic Potential Energy (PE)
- Spring case
Unstretched spring
PE 0
Stretched bar
PE
1
2
x
- Axially loaded bar
PE 0
undeformed:
PE
deformed:
1
L
Adx
2
0
- Elastic body
PE
1
2
σ ε dv
T
V
kx
2
Potential Energy
• Work Potential (WE)
f
P
B
A
L
WP u fdx P u B
f: distributed force over a line
P: point force
u: displacement
0
• Total Potential Energy
1
L
L
Adx u
2
0
fdx P u B
0
• Principle of Minimum Potential Energy
For conservative systems, of all the kinematically admissible displacement fields,
those corresponding to equilibrium extremize the total potential energy. If the
extremum condition is a minimum, the equilibrium state is stable.
Potential Energy + Rayleigh-Ritz
Approach
Example:
f
P
A
Step 1: assume a displacement field
B
u
a x
i
i
is shape function / basis function
n is the order of approximation
Step 2: calculate total potential energy
i
i 1 to n
Potential Energy + Rayleigh-Ritz
Approach
Example:
f
P
A
B
Step 3:select ai so that the total potential energy is minimum
Galerkin’s Method
Example:
f
P
B
A
d
du
EA ( x )
f (x) 0
dx
dx
u x 0 0
Seek an approximation u~ so
V
EA ( x )
du
dx
P
xL
d u~
d
wi
EA ( x )
f ( x ) dV 0
dx
dx
u~ x 0 0
d u~
EA ( x )
P
dx x L
In the Galerkin’s method, the weight function is chosen to be the same as the shape
function.
Galerkin’s Method
Example:
f
P
B
A
d u~
d
w
EA
(
x
)
f
(
x
)
dV 0
i dx
dx
V
L
d u~ dw i
EA ( x ) dx
0
1
1
2
3
dx
L
dx
w
i
fdx w i EA ( x )
0
2
3
d u~
dx
L
0
0
Finite Element Method – Piecewise
Approximation
u
x
u
x
FEM Formulation of Axially
Loaded Bar – Governing Equations
• Differential Equation
d
du
EA ( x )
f ( x) 0
dx
dx
0 x L
• Weighted-Integral Formulation
L
0
d
w
dx
du
EA
(
x
)
f
(
x
)
dx 0
dx
• Weak Form
L
L
dw
du
du
0
EA ( x )
wf ( x ) dx w EA ( x )
dx
dx
dx 0
0
Approximation Methods – Finite
Element Method
Example:
Step 1: Discretization
Step 2: Weak form of one element
x2
P1
P2
x1
du
du
dw
EA
(
x
)
w
(
x
)
f
(
x
)
dx
w
(
x
)
EA
(
x
)
dx
dx
dx
x1
x2
x2
x2
0
x1
du
dw
EA
(
x
)
w
(
x
)
f
(
x
)
dx w x 2 P2 w x1 P1 0
dx
dx
x1
Approximation Methods – Finite
Element Method
Example (cont):
Step 3: Choosing shape functions
- linear shape functions
u 1 u 1 2 u 2
x
x1
1
x2
l
x2 x
l
; 2
x x1
1
l
x
2
l
x
x1 1; x
x0
x1
x
1 l
2
1x
2
x1
; 2
x1
1x
2
x
Approximation Methods – Finite
Element Method
Example (cont):
Step 4: Forming element equation
E,A are constant
x2
Let
w 1 ,
weak form becomes
1
u 2 u1
l EA l dx
x1
x2
Let
w 2 ,
weak form becomes
EA 1
l 1
1
u 2 u1
l EA l dx
x1
x2
1
f dx 1 P2 1 P1 0
EA
l
u1
EA
l
x2
u2
x1
x2
2
f dx 2 P2 2 P1 0
x1
x2
fdx
1
1 u 1 x1
P1 f 1 P1
1 u 2 x2
P2 f 2 P2
fdx
2
x
1
EA
l
u1
1
f dx P1
x1
EA
l
x2
u2
x1
2
f dx P2
Approximation Methods – Finite
Element Method
Example (cont):
Step 5: Assembling to form system equation
Approach 1:
Element 1:
Element 2:
1
I
I
E A 1
I
0
l
0
II
E A
l
Element 3:
E
II
III
l
II
A
III
III
0
0
0
0
0
0
0
0
1
0
1
0
0
0
0
0
0
0
1
1
1
1
0
0
0
0
0
0
0
1
0
1
0 u1I f 1 I P1 I
0 u 2I f 2I P2I
0 0 0 0
0 0 0 0
0 0
II
0 u1
II
0 u2
0 0
0 0
II II
f 1 P1
II II
f 2 P2
0 0
0 0
0 0
III
1 u1
III
1 u 2
0
0
III
f1
III
f 2
0
0
III
P1
III
P2
Approximation Methods – Finite
Element Method
Example (cont):
Step 5: Assembling to form system equation
Assembled System:
E I AI
I
l
I
I
E A
I
l
0
0
I
E A
l
I
E A
l
I
I
0
I
II
E A
l
II
I
E A
l
0
II
II
II
0
II
E A
II
II
l
II
E
II
0
II
l
E A
II
III
l
E
III
l
A
III
III
A
III
III
E
E
III
A
l
III
III
A
l
III
III
III
I
f1
u 1 f 1 P1
u 2 f 2 P2 f I f II
2
1
II
III
u 3 f 3 P3 f 2 f 1
III
u f P
f2
4 4 4
I
P1
I
II
P2 P1
II
III
P2 P1
III
P2
Approximation Methods – Finite
Element Method
Example (cont):
Step 5: Assembling to form system equation
Approach 2: Element connectivity table
k ij K IJ
e
Element 1 Element 2 Element 3
1
1
2
3
2
2
3
4
local node
(i,j)
global node index
(I,J)
Approximation Methods – Finite
Element Method
Example (cont):
Step 6: Imposing boundary conditions and forming condense system
Condensed system:
II
II
E I AI
E A
I
II
l
l
II
II
E A
II
l
0
II
E A
II
l
II
E
II
0
II
l
E A
II
III
l
E
III
l
A
III
III
A
III
III
E
E
III
A
l
III
III
A
l
III
III
III
u2 f2
u3 f3
u 4 f 4
0
0
P
Approximation Methods – Finite
Element Method
Example (cont):
Step 7: solution
Step 8: post calculation
u u 1 1 u 2 2
du
dx
u1
d 1
dx
u2
d2
dx
E Eu 1
d 1
dx
Eu 2
d2
dx
Summary - Major Steps in FEM
• Discretization
• Derivation of element equation
• weak form
• construct form of approximation solution
over one element
• derive finite element model
• Assembling – putting elements together
• Imposing boundary conditions
• Solving equations
• Postcomputation
Exercises – Linear Element
Example 1:
E = 100 GPa, A = 1 cm2
Linear Formulation for Bar Element
u1
u
x
u2
f(x)
P2
P1
L = x2-x1
x=x1
P1 f 1 K 11
P2 f 2 K 12
K 12 u 1
K 22 u 2
di d j
K ij EA
dx dx
x1
x2
where
1
1
x=x1
x= x2
dx K ji , f i
2
x2
f dx
i
x1
1
x=x2
x
Higher Order Formulation for Bar Element
x
u2
u1
u
u3
3
2
1
u (x) u 1 1 (x) u 2 2 (x) u 3 3 (x)
x
u2
u1
u
u3
3
2
1
u4
4
u (x) u 1 1 ( x ) u 2 2 ( x ) u 3 3 ( x ) u 4 4 ( x )
u1
u
x
1
u2
2
u3
3
u4
4
……………
……………
un
n
u (x) u 1 1 ( x ) u 2 2 ( x ) u 3 3 ( x ) u 4 4 ( x ) u n n ( x )
Natural Coordinates and Interpolation Functions
x
x=-1
x
x=1
x=x1
x= x2
x l
x 0
x x x1
x
Natural (or Normal) Coordinate: x
x
x=-1
x=1
1
2
1
x
x=-1
x=1
1
1
x
2
x x 1
2
3
2
x=-1
1
x 1
3
x=1 1
4
3
2
x1 x 2
2
l/2
, 2
x 1
2
, 2 x 1 x 1 , 3
x
1 x
2
9
1
1
27
1
x 1 x x 1
x x x 1 , 2
16
3
3
16
3
27
16
x
1
9
1
1
x 1 x x
1 x x 1 , 4
3
16
3
3
Quadratic Formulation for Bar Element
P1
P2
P
3
x2
w here K ij
x1
f 1 K 11
f 2 K 12
f K
3 13
K 22
d i d j
EA
dx dx
d i d j 2
EA d x d x l d x K
1
x2
a nd f i
K 23
dx
1
l
i
1
i
K 13 u 1
K 23 u 2
K 33 u 3
1
f dx f 2 d x ,
x1
x=-1
K 12
i , j 1, 2, 3
1
2
x=0
3
x=1
ji
Quadratic Formulation for Bar Element
u1
u2
f(x)
P1
u ( x ) u 1 1 ( x ) u 2 2 ( x ) u 3 3 ( x ) u 1
x x 1
2
dx
2 d 1
l dx
x=1
x x 1
2
u 2 x 1 x 1 u 3
, 2 x 1 x 1 , 3
x1 x 2
x
d 1
x3
x=0
x=-1
x
P3
x2
x1
1
u3
P2
l
2
l/2
2x 1
l
,
d x dx
dx
2 d 2
l dx
4x
l
,
dx
2
2
dx
d 3
1 x
1 x
dx
2
d 2
x
x
2
l
2 d 3
l dx
2x 1
l
Exercises – Quadratic Element
Example 2:
E = 100 GPa, A1 = 1 cm2; A1 = 2 cm2
Some Issues
Non-constant cross section:
Interior load point:
Mixed boundary condition:
k
Finite Element Analysis (F.E.A.) of I-D
Problems – Applications
Plane Truss Problems
Example 1: Find forces inside each member. All members have
the same length.
F
UNIT II
Arbitrarily Oriented 1-D Bar Element on 2-D Plane
Q2 , v2
P2 , u 2
P2 , u2
P1 , u1
P1 , u 1
q
Q1 , v1
Relationship Between Local Coordinates and Global
Coordinates
u 1 cos q
v1 0 sin q
u
2
0
v 0 0
2
sin q
0
cos q
0
0
cos q
0
sin q
u1
v
0
1
sin q u 2
cos q v 2
0
Relationship Between Local Coordinates and Global
Coordinates
P1 cos q
0 sin q
P2 0
0 0
sin q
0
cos q
0
0
cos q
0
sin q
P1
Q
0
1
sin q P2
cos q Q 2
0
Stiffness Matrix of 1-D Bar Element on 2-D Plane
Q2 , v2
P2 , u 2
P2 , u2
P1 , u1
K
q
ij
P1 , u 1
Q1 , v1
P1 cos q
Q 1 sin q
P2 0
Q 0
2
sin q
0
cos q
0
0
cos q
0
sin q
cos 2 q
P1
Q 1 AE sin q cos q
cos 2 q
P
L
2
Q
sin q cos q
2
0
K
ij
sin q
cos q
0
1
AE 0
L 1
0
cos q
sin q
0
0
0
0
0
1
0
0
0
cos q
0
0
cos q
0
sin q
cos q
sin q
sin q cos q
sin q cos q
cos q
sin q
sin q cos q
2
1
sin q
sin q cos q
2
0
2
2
0
0
0
0
u1
v
0
1
sin q u 2
cos q v 2
0
sin q cos q u 1
2
sin q v1
sin q cos q u 2
2
sin q
v 2
Arbitrarily Oriented 1-D Bar Element in 3-D Space
z
2
x-, x-, -x are the Direction
x-
-x
1
Cosines of the bar in the
x-y-z coordinate system
y
x-
x
P1 , u 1
u 1 x
v1 0
y
w 1 0 z
u
2
0
v 0 0
2
w 2 0 0
P2 , u 2
x
x
0
0
y
y
0
0
z
z
0
0
0
0
x
x
0
0
y
y
0
0
z
z
0 u1
0
v1
0 w 1
x u2
y v2
z w 2
P1 x
Q
0
1
y
R1 0 z
P
2
0
Q 0 0
2
R 2 0 0
x
x
0
0
y
y
0
0
z
z
0
0
0
0
x
x
0
0
y
y
0
0
z
z
0 P1
0
Q
1
0 R1
x P2
y Q 2
z R 2
Stiffness Matrix of 1-D Bar Element in 3-D Space
z
2
x-
1
y
-x
x-
x
P1 , u 1
x2
P1
Q1
xx
R1 AE x x
2
P
L
2
x
Q
2
x
x
R 2
x x
P2 , u 2
P1
Q1
R 1
P2
Q
2
R 2
1
0
0
0
AE 0
L
1
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
xx
x x
x
xx
x
x x
xx
x
x x
x
x x
x x
xx
x x
x
xx
x
x x
xx
x
x x
x
x x
x x
2
2
2
2
2
2
2
2
0 u1
0 v1 0
0 w 1 0
0 u2
0 v2 0
w
0
0 2
x x u1
x x v1
2
x w1
x x u2
x x v2
2
x w 2
Matrix Assembly of Multiple Bar Elements
Element I
Element II
Element II I
P1
1
Q
1 AE 0
L 1
P2
Q
0
2
P2
Q 2 AE
4L
P3
Q3
P1
Q 1 AE
4L
P3
Q 3
1
3
1
3
1
3
1
3
0
1
0
0
0
1
0
0
0 u1
0 v1
0 u 2
0 v 2
3
1
3
3
3
1
3
3
3
1
3
3 u 2
3 v2
3 u3
3 v 3
3
3
3
1
3
3 u1
3 v1
3 u 3
3 v 3
Matrix Assembly of Multiple Bar Elements
Element I
Element II
Element II I
P1
Q
1
P2 AE
4L
Q 2
P3
Q 3
4
0
4
0
0
0
P1
Q
1
P2 AE
4L
Q 2
P3
Q 3
0
0
0
0
0
0
P1
Q
1
P2 AE
4L
Q 2
P
3
Q 3
0
4
0
0
0
0
0
0
0
4
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0 u1
0 v1
0 u 2
0 v2
0 u3
0 v 3
0
0
0
0
0
0
0
0
0
1
0
3
3
3
3
0
1
3
0
3
3
1
3
0
0
1
3
3
1
1
3
1
0
0
3
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
3
3
3
3
0 u1
v
0
1
3 u 2
3 v2
3 u3
3 v 3
3 u1
3 v1
0 u 2
0 v2
3 u3
3 v 3
Matrix Assembly of Multiple Bar Elements
R1
S
1
R 2
AE
4L
S2
R
3
S 3
41
0 3
4
0
1
3
0
4
0
03
0
0
0
41
3
0
0
3
3
1
0
3
1
3
3
3
3
3
3
1
3
3
3
03
3
11
3
3
33
3
u1
v1
u 2
v2
3 u3
v 3
Apply known boundary conditions
R1 ?
S 0
1
R 2 F
S2 ?
R ?
3
S 3 ?
AE
4L
5
3
4
0
1
3
4
0
3
0
0
0
5
3
0
3
3
3
1
3
3
3
1
3
1
3
2
3
3
0
3 u1
3 v1
3 u 2
3 v2
0 u3
6 v 3
0
?
?
0
0
0
Solution Procedures
R2 F
S 0
1
R 1 ?
S2 ?
R ?
3
S 3 ?
AE
4L
4
3
5
0
1
3
1
0
5
3
0
0
4
0
1
3
3
3
0
3
3
3
3
1
3
2
3
3
0
3
3 u1
3 v1
3 u 2
3 v2
0 u3
6 v 3
0
?
?
0
0
0
u2= 4FL/5AE, v1= 0
R2 F
S 0
1
R 1 ?
S2 ?
R ?
3
S 3 ?
AE
4L
4
3
5
0
1
3
3
0
0
4
0
1
3
3
0
3
3
3
3
1
5
3
0
3
1
3
2
3
3
0
3 u1
v
1
3
3 u2
3 v2
0 u3
6 v 3
0
0
4 FL
5 AE
0
0
0
Recovery of Axial Forces
Element I
Element II
Element II I
P1
1
Q
0
AE
1
L 1
P2
Q
0
2
1
P2
Q
2 AE 3
4L 1
P3
Q3
3
P1
Q 1 AE
4L
P3
Q 3
1
3
1
3
0
1
0
0
0
1
0
0
0 u1 0
v 0
1
0
4 FL
0 u 2
5 AE
0
v2 0
3
3
3
1
3
3
3
1
3
4 FL
3 u2
5 AE
3
v2 0
3 u 0
3
3
v3 0
1
3
3
3
4
5
0
F
4
5
0
3
1
3
3 u1
3 v1
3 u 3
3
v3
1
5
3
5
F
1
5
3
5
0 0
0 0
0 0
0
0
Stresses inside members
Element I
P1
P3
Element II
1
4F
P2
5
Q3
F
3
5
F
5
3
F
5
P2
Element II I
5A
4F
5
Q2
4F
1
5
F
Lecture 5
FEM of 1-D Problems: Applications
Torsional Shaft
Review
Assumption: Circular cross section
Shear stress:
Shear strain:
Tr
J
Tr
GJ
Deformation:
q
TL
GJ
Finite Element Equation for Torsional Shaft
f 1 T1 GJ 1
f
T
L
2 2
1
1 q 1
1 q 2
Bending Beam
y
Review
M
M
x
Pure bending problems:
Normal strain:
x
Normal stress:
x
y
Ey
Normal stress with bending moment: x ydA
Moment-curvature relationship:
Flexure formula:
x
My
I
I
1
y
M
EI
2
dA
M
M EI
1
2
EI
d y
dx
2
Bending Beam
Review
y
q(x)
x
Relationship between shear force, bending moment and
transverse load:
dV
dM
q
dx
Deflection:
V
dx
4
EI
d y
dx
Sign convention:
4
q
M
+
M
-
V
+
V
-
M
V
Governing Equation and Boundary
Condition
• Governing Equation
2
d v( x)
EI
q ( x ) 0,
2
2
dx
dx
d
2
0<x<L
• Boundary Conditions ----v? &
v? &
dv
2
d
d v
EI
?&
2
dx
dx
?,
at x 0
2
d
d v
EI
?&
2
dx
dx
?,
at x L
2
d v
? & EI
2
dx
dx
dv
d v
2
? & EI
dx
dx
2
Essential BCs – if v or dv is specified at the boundary.
{
Natural BCs – if EI
dx
2
d v
dx
2
2
d
d v
or EI 2
dx
dx
is specified at the boundary.
Weak Formulation for Beam Element
• Governing Equation
2
d v( x)
EI
q ( x ) 0,
2
2
dx
dx
d
2
x1 x x 2
• Weighted-Integral Formulation for one element
d2
0 w( x) 2
x1
dx
x2
2
d v( x)
EI
q ( x ) dx
2
dx
• Weak Form from Integration-by-Parts ----- (1st time)
2
dw d
d v
EI
0
2
dx
dx
dx
x1
x2
2
d
d v
wq dx w
EI
2
dx
dx
x2
x1
Weak Formulation
• Weak Form from Integration-by-Parts ----- (2nd time)
d w
0
2
dx
x1
x2
2
d v
EI
2
dx
2
d
d v
wq dx w
EI
2
dx
dx
2
q(x)
V(x1)
y
x
x2
x1
dw
d v
EI
2
dx
dx
2
M(x2)
x = x1
d 2w
0
2
dx
x1
x1
V(x2)
M(x1)
x2
x2
L = x2-x1
d v
EI
wq dx
2
dx
2
x = x2
dw
wV dx M
x2
x1
Weak Formulation
• Weak Form
d 2w
0
2
dx
x1
x2
Q1
y(v)
x
d 2w
dx 2
x1
dw
wV
M
dx
q(x)
x1
Q4
x = x1
2
d v
EI
2
dx
x2
Q3
Q2
Q1 V x1 ,
x2
d v
EI
wq dx
2
dx
2
L = x2-x1
Q 2 M x1 , Q 3 V x 2 ,
x = x2
Q4 M x2
dw
dw
wq dx w ( x1 ) Q1 w ( x 2 ) Q 3
Q
Q4
2
dx 1
dx 2
Ritz Method for Approximation
q(x)
Q1
y(v)
Q3
Q2
x
Q4
x = x1
x = x2
L = x2-x1
n
Let
v( x)
u
j
j ( x ) and n 4
j 1
where
d 2w
dx 2
x1
x2
EI
u 1 v x1 ; u 2
dv
dx
d 2
i
dx 2
x1
EI
u 3 v x 2 ; u 4
x x1
dx
;
x x2
2
d j
dw
dw
u j dx 2 wq dx w ( x1 ) Q1 w ( x 2 ) Q 3 dx Q 2 dx Q 4
j 1
1
2
4
Let w(x)= i (x),
x2
;
dv
i = 1, 2, 3, 4
2
d j
di
di
u j dx 2 i q dx i ( x1 ) Q1 i ( x 2 ) Q 3 dx Q 2 dx Q 4
j 1
1
2
4
Ritz Method for Approximation
Q1
y(v)
Q4
Q2
x
i
Q3
x = x1
d
i
Q1
dx
x1
K ij
Q
i
2
d
i
Q3
dx
x1
x2
d 2 d 2 j
i
EI
dx 2 dx 2
x1
x2
where
x = x2
L = x2-x1
x2
Q
4
dx and q
i
4
K
j 1
x2
qdx
i
x1
ij
u j q i
Ritz Method for Approximation
Q1
y(v)
x
1
2
3
4
x1
x1
x1
x1
d
1
dx
d
2
dx
d
3
dx
d
4
dx
Q3
Q4
Q2
x = x1
x1
x1
x1
x1
1 x
2
2 x
2
3 x
2
4 x
2
d
1
dx
d
2
dx
d
3
dx
d
4
dx
x = x2
L = x2-x1
x2
Q K
K 12
11
1
K
K 22
x2 Q 2
21
Q 3 K 31 K 32
Q 4 K 41 K 42
x2
where
x2
K 13
K 23
K 33
K 43
K 14 u 1
K 24 u 2
K 34 u 3
K 44 u 4
K ij K
ji
q1
q2
q3
q
4
Selection of Shape Function
The best situation is ----
1
2
3
4
x1
x1
x1
x1
d
1
dx
d
2
dx
d
3
dx
d
4
dx
x1
x1
x1
x1
1 x
2
2 x
2
3 x
2
4 x
2
d
1
dx
d
2
dx
d
3
dx
d
4
dx
Q 1 K 11
Q 2 K 12
K 13
Q3
Q
4 K 14
x2
1
0
x2
0
0
x2
x2
K 12
K 13
K 22
K 23
K 23
K 33
K 24
K 34
0
0
1
0
0
1
0
0
0
0
0
1
K 14 u 1
K 24 u 2
K 34 u 3
K 44 u 4
Interpolation
Properties
q1
q2
q3
q
4
Derivation of Shape Function for Beam
Element – Local Coordinates
How to select i???
v (x ) u11 u 2 2 u 3 3 u 4 4
and
where
dv ( x )
dx
u1 v1
u1
d 1
dx
u2
u2
dv1
dx
d 2
dx
u3
u3 v2
d 3
dx
u4
u4
d 4
dx
dv 2
dx
2
3
Let i a i bix c ix d ix
Find coefficients to satisfy the interpolation properties.
Derivation of Shape Function for Beam
Element
How to select i???
e.g. Let 1 a1 b1x c1x 2 d 1x 3
1
Similarly
2
1
3
1
2
1
4
4
4
1 x 2 1 x
1 x 2 2 x
1 x 2 x
1
1
4
1 x 2 2 x
Derivation of Shape Function for Beam
Element
In the global coordinates:
v ( x ) v11 ( x )
l dv 1
2 dx
2 ( x ) v 2 3 ( x )
l dv 2
2 dx
2
3
x x1
x x1
2
1 3
x 2 x1
x 2 x1
2
2
x x1
1
x x1 1
l
x 2 x1
2
2
3
x
x
x
x
1
1
3 3
2
x x
x x
1
1
2
2
4
2
x x1
x x1
2
x
x
1
x 2 x1
x 2 x1
l
4 ( x )
Element Equations of 4th Order 1-D Model
q(x)
u1
y(v)
u3
u2
x
u4
x = x1
1
L = x2-x1
x = x2
4
1
2
1
3
x=x2
x=x1
Q1
Q 2
Q3
Q
4
d 2 d 2 j
i
EI
dx 2 dx 2
x1
x2
where
K ij
q 1 K 11
q 2 K 12
K 13
q3
q
4 K 14
K 12
K 13
K 22
K 23
K 23
K 33
K 24
K 34
dx K
K 14 u 1
K 24 u 2
K 34 u 3
K 44 u 4
x2
ji
and q i
qdx
i
x1
Element Equations of 4th Order 1-D Model
q(x)
u1
y(v)
x
u3
u2
u4
x = x1
Q 1 q1
Q 2 q 2 2 EI
3
Q
q
L
3
3
Q 4 q 4
x = x2
L = x2-x1
6
3L
6
3L
6
3L
2L
3L
3L
6
2
3L
2
L
x2
where q i
qdx
i
x1
3 L u1
2
u
L
2
3L u3
2
2 L u 4
v1
q1
v2
q 2
Finite Element Analysis of 1-D Problems Applications
Example 1.
F
L
L
L
Governing equation:
2
d v
EI
2
2
dx
dx
d
2
q(x) 0
0 x L
Weak form for one element
x2
2
2
d w d v
dw
EI dx 2 dx 2 wq dx w x1 Q1 dx
x1
where
Q 1 V ( x1 )
Q 2 M ( x1 )
Q 2 w x 2 Q 3
x1
Q 3 V ( x 2 )
dw
dx
Q4 M ( x2 )
Q4 0
x2
Finite Element Analysis of 1-D Problems
Example 1.
Approximation function:
v ( x ) v11 ( x )
l dv 1
2 dx
2 ( x ) v 2 3 ( x )
3
2
x x1
x x1
2
1 3
x 2 x1
x 2 x1
2
x x1
2
1
x x1 1
x 2 x1
l
2
3
2
x
x
x
x
1
1
3 3
2
x 2 x1
x 2 x1
4
2
x x1
x x1
2
x x1
x 2 x1
x 2 x1
l
1
x=x1 2
4
3
x=x2
l dv 2
2 dx
4 ( x )
Finite Element Analysis of 1-D Problems
Example 1.
Finite element model:
Q1
Q 2 2 EI
3
Q
L
3
Q 4
6
3L
6
3L
3L
6
2L
3L
3L
6
2
2
L
3L
3 L v1
2
q1
L
3L v2
2
2 L q 2
Discretization:
P1 , v1
M1 ,
q1
I
P2 , v2
M2 ,
q2
II
P3 , v3
M3 ,
q3
III
P4 , v4
M4 ,
q4
Matrix Assembly of Multiple Beam Elements
Element I
Element II
Q 1I
I
Q 2
Q 3I
I
Q 4
0
0
0
0
2 EI
3
L
6
3L
6
3L
0
0
0
0
0
0
II
Q1
II
Q 2
II
Q 3
Q 4II
0
0
2 EI
3
L
0
0
0
0
0
0
0
0
6
3L
0
0
0
3L
L
2
0
0
0
6
3L
0
0
0
3L
2L
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
3L
2L
2
3L
L
2
2
0
0
0
0
0
0
0
0
0
0
0
0
0
6
3L
6
3L
0
0
3L
2L
3L
L
2
0
0
6
3L
6
3L
0
3L
L
3L
2L
0
0
0
0
0
0
0
0
0
0
0
0
2
2
2
0
0
0 v1
0 q1
0 v2
0 q 2
0 v3
0 q 3
0 v4
0 q 4
0 v1
0 q1
0 v2
0 q 2
0 v3
0 q 3
0 v4
0 q 4
Matrix Assembly of Multiple Beam Elements
0
0
0
0
III
Q1
Q 2III
III
Q 3
Q III
4
Element II I
P1
M1
P2
M 2 2 EI
3
P
L
3
M 3
P
4
M
4
2 EI
3
L
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
6
3L
6
0
0
0
3L
2L
0
0
0
6
3L
0
0
0
3L
L
3L
2
6
3L
2
0 v1
q
0
1
0 v2
0 q 2
3 L v3
2
L q 3
3L v4
2
2 L q 4
6
3L
0
0
0
3L
L
2
0
0
0
66
3L 3L
6
3L
0
3L 3L
2L 2L
3L
L
2
0
0
6
3L
66
3L 3L
6
0
0
3L
L
3L 3L
2L 2L
3L
0
0
0
0
6
3L
0
0
0
0
3L
L
6
3L
3L
2L
6
3L
3L
L
0
2
2
2
2
2
2
2
2
6
3L
0 v1
q
0
1
0 v2
0 q 2
3L v3
2
L q 3
3L v4
2
2 L q 4
Solution Procedures
Apply known boundary conditions
P1 ?
M1 ?
P2 ?
M 2 0 2 EI
3
L
P3 ?
M3 0
P
F
4
M 0
4
6
3L
6
3L
0
0
0
0
M2 0
M3 0
P4 F
M 4 0 2 EI
3
P
?
L
1
M1 ?
P
?
2
P ?
3
3L
0
0
0
6
3L
6
0
6
3L
0
0
0
3L
L
2
0
0
0
12
0
6
3L
0
0
4L
3L
L
2
0
0
6
3L
12
0
0
3L
L
0
4L
0
0
0
6
3L
0
0
0
3L
L
2
3L
2
0
4L
3L
L
2
0
0
3L
L
0
4L
0
0
0
6
3L
0
0
0
3L
L
3L
6
3L
0
0
0
3L
L
2
0
0
0
3L
12
0
6
3L
0
0
6
3L
12
0
6
3L
2L
2
3L
L
L
2
2L
2
2
2
2
2
6
2
2
2
3L
6
3L
6
3L
0 v1
q
0
1
0 v2
0 q 2
3L v3
2
L q 3
3L v4
2
2 L q 4
0
0
0
?
0
?
?
?
0 v1
2
L q1
3L v2
2
2 L q 2
0 v3
0 q 3
0 v4
3 L q 4
0
0
0
?
0
?
?
?
Solution Procedures
M2 0
M3 0
P4 F
M 4 0 2 EI
3
P
?
L
1
M1 ?
P
?
2
P ?
3
M2 0
M 3 0 2 EI
3
P
F
L
4
M 4 0
4 L2
2
L
0
0
L
3L
0
0
0
6
3L
6
0
2
4L
2
3L
4L
0
3L
0
L
0
0
6
0
3L
0
0
3L
0
L
3L
6
0
3L
0
0
3L
0
L
2
0
0
3L
12
6
0
3L
0
0
6
12
3L
0
6
q 2
2
L q 3
3L v4
2
2 L q 4
?
?
?
?
P1 ?
M 1 ? 2 EI
3
L
P2 ?
P3 ?
2
2
2L
0
3L
L
0
L
3L
6
3L
2
0
2
2
L
2
4L
0
2
2
3L
6
3L
0 v1
2
L q1
3L v2
2
2 L v3
0 q 2
0 q 3
0 v4
3 L q 4
3L
2
L
0
3L
0
0
0
0
?
?
?
?
0
0
0
0
3L
0
0
6
0 q 2
0 q 3
0 v4
3 L q 4
Shear Resultant & Bending Moment Diagram
3
F
F
7
P2
9
2
F
7
FL
7
1
FL
7
FL
Plane Flame
Frame: combination of bar and beam
Q1 , v1
E, A, I, L
Q3 , v2
P1 , u1
P 2 , u2
Q4 , q 2
Q2 , q 1
AE
L
P1 0
Q
1
Q 2 0
AE
P2
L
Q 3
0
Q 4
0
0
0
AE
0
0
12 EI
6 EI
L
12 EI
3
L
6 EI
6 EI
2
L
4 EI
2
L
L
0
0
0
0
AE
3
L
2 EI
L
2
L
0
0
L
6 EI
2
L
12 EI
3
L
6 EI
2
L
6 EI
2
L
2 EI
L
0
0
12 EI
3
L
6 EI
2
L
6 EI
2
L
4 EI
L
u1
v1
q1
u
2
v2
q
2
Finite Element Model of an Arbitrarily
Oriented Frame
y
q
x
y
q
x
local
Finite Element Model of an Arbitrarily
Oriented Frame
global
12 EI
L3
P1
0
Q
1 6 EI
Q 2 L2
12 EI
P2
3
L
Q 3
0
Q 4
6 EI
2
L
0
AE
6 EI
2
12 EI
3
L
L
0
0
L
0
0
AE
AE
L
4 EI
6 EI
L
6 EI
L
12 EI
2
2
3
L
L
0
0
L
0
0
0
0
AE
L
2 EI
6 EI
L
L
2
0
6 EI
2
L
0 u1
v1
2 EI
L q1
6 EI u 2
2
L
v2
0 q
2
4 EI
L
Plane Frame Analysis - Example
F
Rigid Joint
Hinge Joint
F F
F
Beam II
Beam I
Bar
Beam
Plane Frame Analysis
Q4 , q2
P2 , u2
P1 , u1
Q2 , q1
12 EI
L3
I
P1
0
Q
1
6 EI
2
Q 2
L
12 EI
P2
3
L
Q 3
0
Q 4
6 EI
2
L
Q3 , v2
Q1 , v1
0
6 EI
L
AE
2
0
12 EI
0
L
0
0
AE
4 EI
L
6 EI
6 EI
2
L
12 EI
2
L
3
0
0
L
0
AE
L
L
0
3
L
0
0
AE
L
2 EI
L
6 EI
L
2
0
6 EI
2
L
0
2 EI
L
6 EI
2
L
0
4 EI
L
I
u1
v
1
q1
u2
v2
q 2
Plane Frame Analysis
Q3 , v3
Q1 , v2
P1 , u2
P2 , u3
Q4 , q3
Q2 , q2
P1
Q
2
Q 2
P2
Q 3
Q 4
II
AE
L
0
0
AE
L
0
0
0
0
AE
0
0
12 EI
6 EI
L
12 EI
3
L
6 EI
6 EI
2
L
4 EI
2
L
L
0
0
0
0
AE
3
L
2 EI
L
2
L
0
0
L
6 EI
2
L
12 EI
3
L
6 EI
2
L
6 EI
2
L
2 EI
L
0
0
12 EI
3
L
6 EI
2
L
6 EI
2
L
4 EI
L
u2
v2
q 2
u
3
v3
q
4
Plane Frame Analysis
12 EI
L3
I
P1
0
Q
1
6 EI
2
Q 2
L
12 EI
P2
3
L
Q 3
0
Q 4
6 EI
2
L
0
6 EI
L
AE
2
0
12 EI
3
L
0
L
0
0
AE
4 EI
6 EI
L
6 EI
L
12 EI
2
2
L
3
0
0
L
0
AE
L
L
0
0
0
AE
L
2 EI
6 EI
L
L
2
0
6 EI
2
L
0
2 EI
L
6 EI
2
L
0
4 EI
L
I
u1
v
1
q1
u2
v2
q 2
P1
Q
2
Q 2
P2
Q 3
Q 4
II
AE
L
0
0
AE
L
0
0
0
0
12 EI
6 EI
AE
0
0
L
3
L
4 EI
2
L
L
6 EI
2
L
0
0
0
AE
0
12 EI
6 EI
3
L
2 EI
L
2
L
0
0
L
6 EI
2
L
12 EI
3
L
6 EI
2
L
6 EI
2
L
2 EI
L
0
0
12 EI
3
L
6 EI
2
L
6 EI
2
L
4 EI
L
u2
v2
q 2
u
3
v3
q
4
Plane Frame Analysis
1
F
2
1
F
2
1
FL
8
3
8
FL
UNIT IV
Finite Element Analysis (F.E.A.) of 1-D
Problems – Heat Conduction
Heat Transfer Mechanisms
Conduction – heat transfer by molecular
agitation within a material without any motion
of the material as a whole.
Convection – heat transfer by motion of a
fluid.
Radiation – the exchange of thermal
radiation between two or more bodies. Thermal
radiation is the energy emitted from hot
surfaces as electromagnetic waves.
Heat Conduction in 1-D
Heat flux q: heat transferred per unit area per unit time (W/m2)
dT
q k
dx
Governing equation:
T
T
A
A
Q
C
A
x
x
t
Q: heat generated per unit volume per unit time
C: mass heat capacity
: thermal conductivity
Steady state equation:
d
dT
A
AQ 0
dx
dx
Thermal Convection
Newton’s Law of Cooling
q h ( T s T )
h : convective heat transfer coefficient ( W m C )
2
o
Thermal Conduction in 1-D
Boundary conditions:
Dirichlet BC:
Natural BC:
Mixed BC:
Weak Formulation of 1-D Heat Conduction
(Steady State Analysis)
• Governing Equation of 1-D Heat Conduction ----
d
dT ( x )
(
x
)
A
(
x
)
AQ ( x) 0
dx
dx
0<x<L
• Weighted Integral Formulation ----L
0
0
dT ( x )
d
w( x)
( x ) A( x )
A Q ( x ) dx
dx
dx
• Weak Form from Integration-by-Parts ----L
0
dw
dx
0
L
dT
dT
A
w
A
Q
dx
w
A
dx
dx
0
Formulation for 1-D Linear Element
x
f1
T1
T2
1
2
x1
x2
f1 ( x ) A
Let
1 ( x )
1T1
x1
f2
T
x
,
f 2 ( x ) A
1
T
x
2
T (x) T11 (x) T2 2 (x)
x2 x
l
,
2 ( x )
x x1
l
2T2
x2
Formulation for 1-D Linear Element
Let w(x)= i (x),
x2
d i d j
0 Tj A
j 1
dx dx
x1
2
2
K
ij
dx
i = 1, 2
x2
A Q dx ( x
i
i
2
) f 2 i ( x1 ) f 1
x1
T j Q i i ( x 2 ) f 2 i ( x1 ) f 1
j 1
f 1 Q 1 K 11
f 2 Q 2 K 12
d i d j
w here K ij A
dx dx
x1
x2
dx , Q i
K 12 T1
K 22 T 2
x2
i A Q dx ,
x1
f1 A
dT
dx
, f2 A
x1
dT
dx
x2
Element Equations of 1-D Linear Element
x
f1
T1
T2
1
2
x1
x2
f 1 Q1 A 1
L 1
f 2 Q 2
x2
w here Q i
f2
i A Q dx ,
x1
f1 A
1 T1
1 T 2
dT
dx
, f2 A
x x1
dT
dx
x x2
1-D Heat Conduction - Example
A composite wall consists of three materials, as shown in the figure below.
The inside wall temperature is 200oC and the outside air temperature is 50oC
with a convection coefficient of h = 10 W(m2.K). Find the temperature along
the composite wall.
1
2
3
t1
t2
t3
T 0 200 C
o
1 70 W
m K , 2
40 W
t1 2 cm ,
t 2 2.5 cm ,
t 3 4 cm
T 50 C
o
x
m K , 3
20 W
m K
Thermal Conduction and
Convection- Fin
Objective: to enhance heat transfer
Governing equation for 1-D heat transfer in thin fin
d
dT
A
c
Ac Q 0
dx
dx
w
t
Q loss
2 h ( T T ) dx w 2 h ( T T ) dx t
x
dx
Ac dx
d
dT
A
c
P h T T Ac Q 0
dx
dx
where
P 2 w t
2 h ( T T ) w t
Ac
Fin - Weak Formulation
(Steady State Analysis)
• Governing Equation of 1-D Heat Conduction ----
d
dT ( x )
(
x
)
A
(
x
)
P h T T A Q 0
dx
dx
0<x<L
• Weighted Integral Formulation ----L
0
0
dT ( x )
d
w( x)
( x ) A( x )
P h ( T T ) A Q ( x ) dx
dx
dx
• Weak Form from Integration-by-Parts ----L
0
dw
dx
0
L
dT
dT
A
w
P
h
(
T
T
)
w
A
Q
dx
w
A
dx
dx
0
Formulation for 1-D Linear Element
Let w(x)= i (x),
i = 1, 2
x2
d i d j
0 Tj A
P h i j dx
dx dx
j 1
x1
2
x2
A Q P hT dx
i
x1
i ( x 2 ) f 2 i ( x 1 ) f 1
2
K
ij
T j Q i i ( x 2 ) f 2 i ( x1 ) f 1
j 1
f 1 Q 1 K 11
f 2 Q 2 K 12
K 12 T1
K 22 T 2
d i d j
w here K ij A
P h i j dx , Q i
dx dx
x1
x2
f1 A
dT
dx
, f2 A
x x1
dT
dx
x x2
x2
A Q P hT dx ,
i
x1
Element Equations of 1-D Linear Element
x
f1
T1
T2
1
2
x=0
x=L
f1 Q1 A 1
f
Q
L
2 2
1
1 P hl
1
6
x2
w here
Qi
i A Q P hT dx ,
x1
f1 A
dT
dx
2
1
f2
1 T1
2 T2
, f2 A
x x1
dT
dx
x x2
Lecture 7
Finite Element Analysis of 2-D Problems
2-D Discretization
Common 2-D elements:
2-D Model Problem with Scalar Function
- Heat Conduction
• Governing Equation
T ( x, y ) T ( x, y )
Q ( x , y ) 0
x
x
y
y
• Boundary Conditions
Dirichlet BC:
Natural BC:
Mixed BC:
in W
Weak Formulation of 2-D Model Problem
• Weighted - Integral of 2-D Problem ----
w x
W
T ( x, y )
T ( x, y )
Q ( x , y ) dA 0
x
y
y
• Weak Form from Integration-by-Parts ---- w
0
x
W
T w
x y
T
w Q ( x , y ) dxdy
y
T
T
w
dxdy
w
dxdy
x
x
y
y
W
W
Weak Formulation of 2-D Model Problem
• Green-Gauss Theorem ----
T
w
dxdy
x
x
T
w
dxdy
y
y
W
W
T
w
n x ds
x
T
w
n y ds
y
G
G
where nx and ny are the components of a unit vector,
which is normal to the boundary G.
n n x i n y j i cos j sin
Weak Formulation of 2-D Model Problem
• Weak Form of 2-D Model Problem ---- w
0
x
W
G
T w
x y
T
w Q ( x , y ) dxdy
y
T
T
w
nx
n y ds
y
x
EBC: Specify T(x,y) on G
NBC: Specify
where
T
T
n
x
ny
y
x
T
T
qn ( s )
i
x
y
on G
j n xi n y j
is the normal
outward flux on the boundary G at the segment ds.
FEM Implementation of 2-D Heat
Conduction – Shape Functions
Step 1: Discretization – linear triangular element
T1
T T11 T 2 2 T3 3
Derivation of linear triangular shape functions:
T3
1 c 0 c1 x c 2 y
Let
T2
c 0 c1 x 1 c 2 y 1 1
Interpolation properties
c 0 c1 x 2 c 2 y 2 0
i 1 at ith node
c 0 c1 x 3 c 2 y 3 0
i 0 at other nodes
1 1
Same
2
1
x
1
y 1
1
x1
x2
x3
y1
y2
y 3
x 3 y1 x1 y 3
y
y 3 y1
2 Ae
x x
1
3
x
c 0 1
c 1
1
c 1
2
1
x1
x2
x3
1
x 2 y 3 x3 y 2
1
x
y
0
y 2 y3
2 Ae
0
x3 x 2
3
1
x1 y 2 x 2 y1
y
y1 y 2
2 Ae
x x
2
1
x
y1
y2
y 3
1
1
0
0
FEM Implementation of 2-D Heat
Conduction – Shape Functions
linear triangular element – area coordinates
T1
1
A2
T3
1
A3
x 2 y 3 x3 y 2
y
A1
y
y
2
3
2 Ae
x x
Ae
3
2
x
A1
T2
2
3
1
1
x 3 y1 x1 y 3
y
A2
y 3 y1
2 Ae
Ae
x x
1
3
x
x1 y 2 x 2 y1
y
A3
y
y
1
2
2 Ae
x x Ae
2
1
x
3
1
2
Interpolation Function - Requirements
• Interpolation condition
• Take a unit value at node i, and is zero at all other nodes
• Local support condition
• i is zero at an edge that doesn’t contain node i.
• Interelement compatibility condition
• Satisfies continuity condition between adjacent elements
over any element boundary that includes node i
• Completeness condition
• The interpolation is able to represent exactly any
displacement field which is polynomial in x and y with the
order of the interpolation function
Formulation of 2-D 4-Node Rectangular Element –
Bi-linear Element
Let u (x , ) 1u1 2 u 2 3 u 3 4 u 4
x
1 1 1
a
b
3
x
a b
2
x
1
a
b
x
4 1
ab
Note: The local node numbers should be arranged in a counter-clockwise sense. Otherwise, the area
Of the element would be negative and the stiffness matrix can not be formed.
1
2
3
4
FEM Implementation of 2-D Heat
Conduction – Element Equation
• Weak Form of 2-D Model Problem ----0
We
w
x
T w T
w Q ( x , y ) dxdy
x y y
n
u ( x, y )
Assume approximation:
wq
n
ds
Ge
u j j ( x , y )
j 1
and let w(x,y)=i(x,y) as before, then
0
We
n
i
T j j
x j 1
x
i
n
T j j
y j 1
y
i Q dxdy
n
K
j 1
where
ij
Tj
Q dxdy q
i
We
i
n
ds
Ge
i j
i j
K ij
dxdy
x
x
y
y
We
q
i
G
n
ds
FEM Implementation of 2-D Heat
Conduction – Element Equation
n
K
ij
Tj
j 1
l 232
K
l 23 l31
4 Ae
l l
23 12
Q dxdy q
i
We
l 23 l 31
2
l31
l31 l12
l 23 l12
l31 l12
2
l12
i
n
ds
Ge
Q1
F Q2
Q
3
q1
q2
q
3
Qi
Q d xd y
i
We
qi
q
i
Ge
n
ds
Assembly of Stiffness Matrices
ne
Fi
(e)
(e) (e)
Q
dxdy
q
ds
K
ij u j
i
i n
W
U 1 u1
(1)
(e)
,U 2 u 2
(1)
G
u1
(2)
j 1
(e)
,U 3 u 3
(1)
u4
(2)
,U 4 u 2
(2)
,U 5 u 3
(2)
Imposing Boundary Conditions
The meaning of qi:
(1)
3
q1
3
1
1
q n 1 ds
(1)
(1)
G1
1
(1)
(1 )
h12
2
(1)
(1)
(1 )
h12
q n 1 ds
(1)
q n 1 ds
q n 1 ds
(1)
(1)
(1 )
h 23
q n 1 ds
(1)
(1)
(1 )
h31
q n 1 ds
(1)
(1)
(1 )
h31
2
q2
(1)
q n 2 ds
(1)
(1)
G1
3
1
1
2
(1 )
h 23
q n 2 ds
(1)
(1)
(1 )
h 23
q n 2 ds
(1)
(1)
(1 )
h31
q n 2 ds
(1)
(1)
(1 )
q n 3 ds
(1)
(1)
G1
2
(1)
h 23
q3
(1)
1
(1)
(1 )
h12
3
(1)
(1 )
h12
q n 2 ds
(1)
q n 2 ds
(1)
(1 )
h12
q n 3 ds
(1)
q n 3 ds
(1)
(1 )
h31
(1 )
h 23
q n 3 ds
(1)
(1)
(1)
q n 3 ds
(1)
(1)
(1 )
h31
q n 3 ds
(1)
(1)
Imposing Boundary Conditions
q2 q
Consider
q
(1)
2
q
(1)
n
(1)
2
ds
(1 )
q
q
(1)
n
q
q3 q3 q 4
(1)
(2)
1
(1)
2
(2)
q1
ds
h2 3
q
(1)
n
(1)
3
ds
(1 )
q
(1)
n
(1)
3
(2)
q4
ds
(1)
qn
Equilibrium of flux:
(2)
(2)
qn
q n 1 ds
(2)
(2)
(2)
h4 1
q n 4 ds
(2)
(2)
(2)
h3 4
h3 1
q n 1 ds
(1 )
h2 3
(2)
(2)
h1 2
(1 )
h1 2
(1)
3
(1)
2
q n 4 ds
(2)
(2)
(2)
h4 1
(2)
(1 )
h 23
(2)
h41
FEM implementation:
q n 2 ds
(1)
(1)
(1 )
q n 1 ds ;
(2)
(2)
(2)
h 23
(1 )
h1 2
q n 2 ds
(1)
(1)
(1)
(1)
(1 )
h 41
q2
q n 3 ds
(2)
h1 2
(2)
(2)
(2)
(2)
(2)
h23
q n 1 ds
q n 4 ds
h41
q3
(1 )
h3 1
q n 3 ds
(1)
(1)
(2)
h3 4
q n 4 ds
(2)
(2)
Calculating the q Vector
Example:
qn 0
T 293K
qn 1
2-D Steady-State Heat Conduction - Example
A
D
qn 0
AB and BC:
CD: convection
DA:
0.6 m
C
B
0.4 m
y
x
T 180 C
o
h 50 W
T 25 C
o
m C
2 o
Finite Element Analysis of Plane Elasticity
Review of Linear Elasticity
Linear Elasticity: A theory to predict mechanical response
of an elastic body under a general loading condition.
Stress: measurement of force intensity
xx
yx
zx
2-D
xy
yy
zy
xz
yz
zz
xx
yx
with
xy
yx
zy
yz
xz zx
xy
yy
Review of Linear Elasticity
Traction (surface force) :
t x xx n x xy n y
t
t y xy n x yy n y
Equilibrium – Newton’s Law
F 0
xx
x
yx
x
xy
y
yy
y
S tatic
fx 0
fy 0
xx
x
yx
x
xy
y
yy
y
fx ux
fy uy
D ynam ic
Review of Linear Elasticity
Strain: measurement of intensity of deformation
xx
u x
xy
x
1
2
xy
1 u x u y
2 y
x
yy
u y
y
Generalized Hooke’s Law
xx
xx
yy
E
yy
zz
xy G xy
zz
xx e 2 G xx
E
yy e 2 G yy
E
xx
yy
E
xx
E
yy
E
E
yz G yz
G
E
zz
E
zz e 2 G zz
e xx yy zz
zx G zx
E
2 1
zz
E
1 1 2
Plane Stress and Plane Strain
Plane Stress - Thin Plate:
x C 11
y C 12
0
xy
C 12
C 22
0
0 x
0 y
C 33 xy
E
1
x
E
y
1
xy
0
E
2
1
E
2
2
1
2
0
x
y
0
E
xy
2 1
0
Plane Stress and Plane Strain
Plane Strain - Thick Plate:
x C 11
y C 12
0
xy
C 12
C 22
0
1 E
x 1 1 2
E
y
1 1 2
xy
0
0 x
0 y
C 33 xy
Plane Strain:
Plane Stress:
Replace E by
E
1
2
and by
1
E
1 1 2
1 E
1 1 2
0
x
0
y
xy
E
2 1
0
Equations of Plane Elasticity
Governing Equations
(Static Equilibrium)
Constitutive Relation
(Linear Elasticity)
x C 11
y C 12
0
xy
C 12
C 22
0
0 x
0 y
C 33
xy
x
x
x
x
xy
x
xy
y
y
y
0
0
Strain-Deformation
(Small Deformation)
x
u
x
xy
y
v
x
u
v
u
v
C 11
C 33
0
C 12
C 33
x
y y
y
x
u
v
u
v
C 33
C 12
0
C 33
C 22
y
x y
x
y
v
y
u
y
Specification of Boundary Conditions
EBC: Specify u(x,y) and/or v(x,y) on G
NBC: Specify tx and/or ty on G
where
T ( s ) t x i t y j ; t x xx n x xy n y ; t y yx n x
yy
ny
is the traction on the boundary G at the segment ds.
UNIT V
Weak Formulation for Plane Elasticity
0
0
v
u
v
u
dxdy
C 33
C 11
C 33
C 12
w 1
x
y
y y
x
x
v
u
v
u
dxdy
C 12
C 33
C 22
C 33
w 2
y
x
x y
y
x
W
W
w 1
0
x
W
w 2
0
x
W
where
t x C 11
t C
y
33
u v
u
v w 1
C 11
dxdy
C 12
C 33
x
y
y
y x
u v w 2
u
v
C 12
dxdy
C 33
C 22
y
x
y
y x
u
v
C 12
n x C 33
x
y
y
u v
u
C 22
n x C 12
y x
x
u
v
n y
x
v
n y
y
w
t ds
1 x
G
w
2
t y ds
G
are components of
traction on the
boundary G
Finite Element Formulation for Plane Elasticity
Let
u( x , y )
v( x , y )
where
and
n
j
( x , y )u j
j1
n
j ( x , y )v j
j1
1
Fi
F 2
i
11
i j
i
C 33
K ij C 11
x x
y
W
12
i j
i
K
C
C
ij
33
12 x y
y
W
22
i j
i
C 22
K ij C 33
x x
y
W
1
F i i t x ds i f x dxdy
G
W
F 2 t ds f dxdy
i y i y
i
G
W
n
K
n
11
ij
uj
j1
12
ij
vj
j1
n
K
n
K
21
ij
uj
j1
j
dxdy
y
j
dxdy K
x
j
dxdy
y
j1
21
ji
22
K ij v j
Constant-Strain Triangular (CST) Element for Plane
Stress Analysis
v 2 , F2 y
u 2 , F2 x
v 3 , F3 y
v1 , F1 y
u 3 , F3 x
u1 , F1 x
Let
u ( x , y ) c 1 c 2 x c 3 y 1 u 1 2 u 2 3 u 3
v ( x , y ) c 5 c 6 x c 7 y 1 v 1 2 v 2 3 v 3
x2 y3 x3 y2
1 x y
1
y2 y3
2 Ae
x x
3
2
x 3 y1 x1 y 3
1 x y
2
y 3 y1
2 Ae
x x
1
3
3
1
x1 y 2 x 2 y1
y
y1 y 2
2 Ae
x x
2
1
x
Constant-Strain Triangular (CST) Element for Plane
Stress Analysis
k 11
k
21
1 k 31
4 Ae k 41
k 51
k 61
k 12
k 13
k 14
k 15
k 22
k 23
k 24
k 25
k 32
k 33
k 34
k 35
k 42
k 43
k 44
k 45
k 52
k 53
k 54
k 55
k 62
k 63
k 64
k 65
k 16 u 1 F1 x
F
k 26
v
1 1y
k 36 u 2 F2 x
k 46 v 2 F2 y
k 56 u 3 F3 x
k 66 v 3 F3 y
k 11 c11 y 2 y 3 c 33 x 3 x 2 ; k 21 c12 y 2 y 3 x 3 x 2 c 33 y 2 y 3 ; k 22 c 22 x 3 x 2 c 33 y 2 y 3
2
2
2
2
2
k 31 c11 y 3 y 1 y 2 y 3 c 33 x 1 x 3 x 3 x 2 ; k 32 c12 y 3 y 1 x 3 x 2 c 33 x 1 x 3 x 3 x 2 ; k 33 c11 y 3 y 1 c 33 x 1 x 3
2
2
k 41 c12 y 2 y 3 c 33 x 1 x 3 x 3 x 2 ; k 42 c 22 x1 x 3 x 3 x 2 c 33 y 2 y 3 y 3 y 1 ; k 43 c12 x 1 x 3 y 3 y 1 c 33 x 1 x 3
2
k 44 c 22 x1 x 3 c 33 y 3 y 1 ; k 51 c11 y 1 y 2 y 2 y 3 c 33 x 2 x 1 x 3 x 2 ; k 52 c12 y 1 y 2 c 33 x 2 x 1 x 3 x 2
2
2
k 53 c11 y 1 y 2 y 3 y 1 c 33 x 2 x 1 x 1 x 3 ; k 54 c12 y 1 y 2 x 1 x 3 c 33 x 2 x 1 x 1 x 3 ; k 55 c11 y 1 y 2 c 33 x 2 x 1
2
2
k 61 c12 y 2 y 3 c 33 x 2 x 1 x 3 x 2 k 62 c 22 x 2 x 1 x 3 x 2 c 33 y 1 y 2 y 2 y 3 k 63 c12 y 3 y 1 c 33 x 2 x 1 x 1 x 3
k 64 c 22 x1 x 3 x 2 x 1 c 33 y 1 y 2 y 3 y 1 k 65 c12 y 1 y 2 x 2 x 1 c 33 x 2 x 1
2
k 66 c 22 x 2 x 1 c 33 y 1 y 2
2
2
4-Node Rectangular Element for Plane Stress Analysis
Let
u ( x , y ) c 1 c 2 x c 3 y c 4 xy 1 u 1 2 u 2 3 u 3 4 u 4
v ( x , y ) c 5 c 6 x c 7 y c 8 xy 1 v 1 2 v 2 3 v 3 4 v 4
x
y
1 1 1
a
b
x y
3
a b
x
y
2 1
a
b
x y
4 1
ab
4-Node Rectangular Element for Plane Stress Analysis
For Plane Strain Analysis:
E
E
1
2
and
1
Loading Conditions for Plane Stress Analysis
1
Fi
F 2
i
n
K
n
11
ij
uj
j1
j1
vj
j1
n
K
12
ij
n
K ij u j
21
j1
22
K ij v j
1
Fi
F 2
i
i t x ds i f x dxdy
G
W
i t y ds i f y dxdy
G
W
Evaluation of Applied Nodal Forces
Fi
1
i
t x ds
G
( A)
F x2
F
1 ( A)
2
t ds
2 x
G
F
( A)
F x3
( A)
x2
8
0
1 ( A)
3
G
F
( A)
x3
8
0
0
2
x
y
y
1 o 1
tdy
a
b
16
2
8
y
y
1 1000 1
0 . 1 dy 100
8
8
16
F
b
t ds
3 x
b
0
0
2
3
y
y
y
1
2
8 16
8 16
2
y
o 1
tdy
a b
16
x y
2
y
1000 1
0 . 1 dy 100
8 8
16
8 y
8
8
0
3
y
y
8 8 16
2
dy 350
2
dy 383 . 3
Evaluation of Applied Nodal Forces
(B )
F x2
F
1 (B )
2
t ds
2 x
G
F
F
(B )
x2
(B )
x3
8
0
1 (B )
3
G
F
(B )
x3
8
0
0
2
x
y
y8
1 o 1
tdy
a
b
16
2
8
y
y8
1 1000 1
0 . 1 dy 100
8
8
16
F
b
t ds
3 x
b
0
0
2
3
3 5y
y
y
4 32 16 2 8 16
2
y8
o 1
tdy
a b
16
x y
2
y8
1000 1
0 . 1 dy 100
8 8
16
8 y
8
8
0
3
3 y 2 y2
y
32 16 2 8 16
2
dy 116 . 7
2
dy 216 . 7
Element Assembly for Plane Elasticity
5
6
B
3
4
4
3
A
1
2
F x1
F y1
Fx
2
F
y2
F x4
Fy
4
F
x3
F
y3
(B )
F x1
F y1
Fx
2
F y2
F x4
Fy
4
F
x3
F
y3
( A)
(B )
u 3
v 3
u 4
v 4
u 5
v 5
u
6
v
6
( A)
u 1
v 1
u 2
v 2
u 3
v 3
u
4
v
4
Element Assembly for Plane Elasticity
5
6
B
4
3
A
2
1
( A )
F x1
( A )
F y1
( A )
F x2
( A )
F
y2
F ( A ) F ( B )
x1
x(4A )
(B )
F y4 F y1
(A)
(B )
F
F
x2
x3
F y( A ) F y( B )
3 (B) 2
F x4
0
(B )
0
F y4
(B )
F x3
0
(B )
F y3
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 u 1
0 v 1
0 u 2
0 v 2
u 3
v 3
u 4
v 4
u 5
v 5
u
6
v 6
Comparison of Applied Nodal Forces
Discussion on Boundary Conditions
•Must have sufficient EBCs to suppress rigid body
translation and rotation
• For higher order elements, the mid side nodes cannot be
skipped while applying EBCs/NBCs
Plane Stress – Example 2
Plane Stress – Example 3
Evaluation of Strains
1 1
3
x y
x
y
1
a
b
a b
x
y
1
a
b
x y
4 1
ab
2
u( x , y ) 1u1 2 u2 3 u3 4 u4
v ( x , y ) 1v 1 2 v 2 3v 3 4 v 4
4
j
u
x u j
j1
x x
4
j
v
v
y
y j
y
j1
xy u v 4
j
j
uj
v j
y x
j1 y
x
Evaluation of Stresses
x
y
xy
1
y
1
a
b
1
y
1
a
b
0
1
1
b
1
1
a
0
1
x
1
b
a
x
a
y
b
x
0
ab
x
ab
0
ab
0
y
0
1
y
1
a
b
y
ab
x
x
ab
y
ab
ab
0
1
x
1
b
a
u1
v
1
0
u2
1
x v2
1
b
a u3
y
v3
ab u
4
v
4
Plane Stress Analysis
E
1
x
E
y
1
xy
0
E
2
1
E
2
2
1
2
0
x
y
0
E
xy
2 1
0
Plane Strain Analysis
1 E
x 1 1 2
E
y
1 1 2
xy
0
E
1 1 2
1 E
1 1 2
0
x
0
y
xy
E
2 1
0
Finite Element Analysis of 2-D Problems – Axisymmetric Problems
Axi-symmetric Problems
Definition:
A problem in which geometry, loadings, boundary
conditions and materials are symmetric about one axis.
Examples:
Axi-symmetric Analysis
Cylindrical coordinates:
r ,
q , z
x r cos q ; y r sin q ; z z
• quantities depend on r and z only
• 3-D problem
2-D problem
Axi-symmetric Analysis
Axi-symmetric Analysis – Single-Variable Problem
1
u (r , z )
u (r , z )
ra
a
11
22
a 00 u f ( r , z ) 0
r r
r
z
z
Weak form:
w
u w
u
0
a 11
a 22
a 00 wu wf ( r , z ) rdrdz
r
r z
z
We
wq
n
ds
Ge
where
q n a11
u (r , z )
r
n r a 22
u (r , z )
z
nz
Finite Element Model – Single-Variable Problem
u
u
j
j (r , z ) j ( x, y )
where
j
j
Ritz method:
w i
n
Weak form
K ij u j f i Q i
e
e
e
e
j 1
where
i j
i j
K a11
a 22
a 00 i j
r r
z z
We
e
ij
fi
e
i
frdrdz
We
Qi
e
q
i
Ge
n
ds
rdrdz
Single-Variable Problem – Heat Transfer
Heat Transfer:
1
T (r , z ) T (r , z )
rk
k
f (r , z ) 0
r r
r
z
z
Weak form
0
We
w
r
wq
n
T w T
k
k
w f ( r , z ) rdrdz
r z z
ds
Ge
where
qn k
T (r , z )
r
nr k
T (r , z )
z
nz
3-Node Axi-symmetric Element
T ( r , z ) T11 T2 2 T3 3
3
1
1
2
1
r2 z 3 r3 z 2
z
z2 z3
2 Ae
r r
2
3
r
r3 z1 r1 z 3
1 r z
2
z 3 z1
2 Ae
r r
3
1
3
1
r1 z 2 r2 z1
z
z1 z 2
2 Ae
r r
1
2
r
4-Node Axi-symmetric Element
4
3
1
2
T ( r , z ) T11 T2 2 T3 3 T4 4
b
a
z
r
x
x
1 1 1
a
b
3
x
a b
2
x
1
a
b
4 1
x
ab
Single-Variable Problem – Example
Step 1: Discretization
Step 2: Element equation
K ij
e
fi
e
We
i j
i j
rdrdz
r r
z
z
We
i frdrdz
Qi
e
q
i
Ge
n
ds
Time-Dependent Problems
Time-Dependent Problems
In general,
u x, t
Key question: How to choose approximate functions?
Two approaches:
u x, t
u
u x, t
u t x
j
j
j x, t
j
Model Problem I – Transient Heat Conduction
u
u
c
a
f x, t
t
x x
Weak form:
x2
u
w u
0 a
cw
wf dx Q1 w ( x1 ) Q 2 w ( x 2 )
x x
t
x1
du
Q1 a
;
dx x1
du
Q2 a
dx x 2
Transient Heat Conduction
u x, t
let:
n
u t x
j
j
and
w i x
j 1
x2
u
w u
0 a
cw
wf dx Q1 w ( x1 ) Q 2 w ( x 2 )
x x
t
x1
K u M u F
x2
K ij
a
x1
i j
x x
ODE!
x2
M ij
dx
x1
i
x1
x2
Fi
c
i
fdx Q i
j
dx
Time Approximation – First Order ODE
a
du
dt
bu f t
0tT
u 0 u 0
Forward difference approximation - explicit
u k 1 u k
t
a
fk
bu k
Backward difference approximation - implicit
u k 1 u k
t
a bt
fk
bu k
Time Approximation – First Order ODE
a
du
dt
bu f t
0tT
u 0 u 0
- family formula:
u k 1 u k t 1 u k u k 1
Equation
u k 1
a 1 tbu k t f k 1 1 f k
a tb
Time Approximation – First Order ODE
a
du
dt
bu f t
0tT
u 0 u 0
Finite Element Approximation
2 tb
a
3
tb
u k 1 a
3
2 f k 1
fk
uk t
3
3
Stability of – Family Approximation
Example
Stability
A 1
a 1 tb
a tb
1
FEA of Transient Heat Conduction
K u M u F
- family formula for vector:
u k 1 u k t 1 u k u k 1
u k 1 M K t
1
M 1 K t u
k
t 1 f k t f k 1
Stability Requirment
t t cri
where
2
1 2 max
K M u Q
Note: One must use the same discretization for solving
the eigenvalue problem.
Transient Heat Conduction - Example
u
t
u
2
u 0 , t 0
u x ,0 1 .0
x
2
0
u
t
0 x 1
1, t 0
t 0
Transient Heat Conduction - Example
Transient Heat Conduction - Example
Transient Heat Conduction - Example
Transient Heat Conduction - Example
Transient Heat Conduction - Example
Transient Heat Conduction - Example
Model Problem II – Transverse Motion of EulerBernoulli Beam
2
2
u
u
EI
A 2 f x, t
2
2
x
x
t
Weak form:
2
x2
2
2
2
w u
u
0 EI
Aw
wf dx
2
2
2
x x
t
x1
w
w
Q 1 w ( x1 ) Q 2
Q3w( x2 ) Q4
x x1
x x2
Where:
Q1
x
2
2
u
u
EI
Q 2 EI
2
2
x
x x
x1
1
Q3
x
2
2
u
u
EI
Q 4 EI
2
2
x
x
x2
x2
Transverse Motion of Euler-Bernoulli Beam
let:
u x, t
n
u t x
j
j
and
w i x
j 1
x2
2
2
2
w u
u
0 EI
Aw
wf dx
2
2
2
x x
t
x1
w
w
Q 1 w ( x1 ) Q 2
Q3w( x2 ) Q4
x x1
x x2
K u M u F
Transverse Motion of Euler-Bernoulli Beam
K u M u F
x2
K ij
EI
x1
i j
x
x2
2
2
2
x
2
M ij
dx
i
x1
x2
Fi
A
x1
i
fdx Q i
j
dx
ODE Solver – Newmark’s Scheme
u s 1 u s
t u s
u s 1 u s us t
where
us q
1
2
t
2
us
1 q us q us 1
Stability requirement:
t t cri
where
1 2
max
2
K M u F
2
1
2
ODE Solver – Newmark’s Scheme
1
, 2
1
2
2
1
1
, 2
2
3
1
, 2 0
2
3
, 2
8
2
5
3
, 2 2
2
Constant-average acceleration method (stable)
Linear acceleration method (conditional stable)
Central difference method (conditional stable)
Galerkin method (stable)
Backward difference method (stable)
Fully Discretized Finite Element Equations
Transverse Motion of Euler-Bernoulli Beam
w
2
t
w 0 , t 0
2
w
t
w
4
x
4
0 x 1
0
0 , t 0
w 1, t 0
w x , 0 sin x x 1 x
w
t
w
t
1, t 0
x ,0 0
Transverse Motion of Euler-Bernoulli Beam
Transverse Motion of Euler-Bernoulli Beam
Transverse Motion of Euler-Bernoulli Beam