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 )
ba
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 
ba
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
xL
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 
x0
x1
x
 1 l
2
1x
2
 x1
; 2 
x1
1x
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
d2
dx
  E   Eu 1
d 1
dx
 Eu 2
d2
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 
 di 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
  xx
 
 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
 xx
 x x
x
 xx
x
 x x
 xx
 x
 x x
x
  x x
  x x
 xx
  x x
x
 xx
 x
  x x
 xx
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 
 41

0  3
 4

 0
 1

  3
0
4
0
03
0
0
0
41
3
0
0

3
3
1

0
3
1
3
3
3
3
3
3
1
3
3
3
03
3

11
3 
3 
33
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 
di
di

 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 )  u11  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 )  v11 ( 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 )  v11 ( 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
66
 3L  3L
6
3L
0
 3L  3L
2L  2L
 3L
L
2
0
0
6
 3L
66
 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)  T11 (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  T11  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  
ab

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
j1
n
  j ( x , y )v j
j1
 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 
j1
12
ij
vj
j1
n

K
n
K
21
ij
uj 
j1
 j 
dxdy
 y 
 j 
dxdy  K
 x 
 j 
dxdy
 y 

j1
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  
ab

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 
j1
j1
vj
j1
n

K
12
ij
n
K ij u j 
21

j1
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 
 y8 
 1   o  1  
  tdy
a 
b 
16

 

2

8
y
 y8 
 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

 y8 
 o 1  
  tdy
a b
 16  

x y
2

 y8 
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  
ab

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

 
j1

  x   x
 
4

 j
v

 
 



v
 y  
 

 y j

y
j1
  
 

 xy    u  v   4   
 j

j

uj 
v j  
  y  x    

  j1  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 )  T11  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 )  T11  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 

ab
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 
0tT
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  bt
 fk
 bu k 
Time Approximation – First Order ODE
a
du
dt
 bu  f t 
0tT
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 
0tT
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  us    t
where
us q
1
2
t
2
us  
 1  q us  q us 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