Finite Element Methods · Approximation Methods – Finite Element Method Example (cont): ......
Transcript of Finite Element Methods · Approximation Methods – Finite Element Method Example (cont): ......
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Prepared by
Prof. V.V.S.H. Prasad, Madhav, Professor, Dept of Mechanical Engineering
Mr . C. Labesh Kumar, Assistant Professor, Dept of Mechanical Engineering
Finite Element Methods III B Tech II Semester
INSTITUTE OF AERONAUTICAL ENGINEERING (Autonomous)
Dundigal, Hyderabad - 500 043
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UNIT - 1
• Introduction to FEM:
– Stiffness equations for a axial bar element in local co-ordinates
bars
using Potential Energy approach and Virtual energy principle
– Finite element analysis of uniform, stepped and tapered
subjected to mechanical and thermal loads
– Assembly of Global stiffness matrix and load vector
– Quadratic shape functions
– properties of stiffness matrix
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Axially Loaded Bar Review:
Stress:
Strain:
Deformation:
Stress:
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Strain:
Deformation:
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Axially Loaded Bar
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Review:
Stress:
Strain:
Deformation:
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Axially Loaded Bar – Governing Equations
and Boundary Conditions
• Differential Equation
• Boundary Condition Types
• prescribed displacement (essential BC)
•prescribed force/derivative of displacement
(natural BC)
dx dx
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d EA(x)
du f (x) 0 0 x L
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Axially Loaded Bar –Boundary Conditions
• Examples
• fixed end
• simple support
• free end
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Potential Energy
- Axially loaded bar
- Elastic body
x
Stretched bar
• Elastic Potential Energy (PE) - Spring case
Unstretched spring
PE 0
2 1
2 kx PE
undeformed:
deformed:
0
7
PE 0
1 L
PE Adx 2
2 V
PE 1
σT εdv
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Potential Energy
• Work Potential (WE)
B
L
WP u fdx P u
0
P
f
f: distributed force over a line
P: point force
u: displacement
A B
• Total Potential Energy L
0 0
1 L
Adx u fdx P uB
2
• 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.
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Potential Energy + Rayleigh-Ritz Approach
P
f
A B
Example:
Step 1: assume a displacement field i
i 1 to n u aii x
is shape function / basis function
n is the order of approximation
Step 2: calculate total potential energy
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Potential Energy + Rayleigh-Ritz
Approach
f Example:
P
A B
Step 3:select ai so that the total potential energy is minimum
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Galerkin’s Method
P
f
A
Example:
EA( x) du
P dx xL
d du
dx EA( x)
dx f ( x) 0
ux 0 0 dx w
V
i
P
d du
dx
dx xL
du~ EA( x)
u~x 0 0
f ( x)dV 0
EA( x)
B
Seek an approximation u~ so
~
In the Galerkin’s method, the weight function is chosen to be the same as the shape
function.
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Galerkin’s Method
P
f
A B
Example:
f ( x)dV 0
dx
du~ EA( x)
dx
d
V
wi 0 0
0
L L
dx 0
du~ L du~ dw EA( x) i dx wi fdx wi EA( x) dx dx
1 2 3
1
2
3
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Finite Element Method – Piecewise
Approximation
x
u
x
u
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FEM Formulation of Axially Loaded
Bar – Governing Equations
• Weak Form
• Differential Equation
d du
dx EA(x)
dx f (x) 0 0 x L
• Weighted-Integral Formulation
0
d du w EA(x) f (x) dx 0 dx dx
L
0
L
du
0
du
L dw 0 dx
EA(x) dx
wf (x)dx w EA(x) dx
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Approximation Methods – Finite
Element Method
Example:
Step 1: Discretization
Step 2: Weak form of one element P2 P 1 x1 x2
1 1
0
x x
x
du 2
du dx
EA( x) dx
w( x) f ( x)dx w( x) EA( x) dx
x2 dw
1 1 2 2
P w x P 0 dx
EA( x) w( x) f ( x) dx w x dx
x2 dw du
x1 15
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Approximation Methods – Finite
Element Method
Example (cont):
Step 3: Choosing shape functions
- linear shape functions u 1u1 2u2
l x1 x2
x
1
l 2
2
l 1
x x x x ; 2
16
2
1 ; 2
1 1
1 1 2 l
x 1l
x 2 x x 1;
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Approximation Methods – Finite
Element Method
Example (cont):
Step 4: Forming element equation
weak form becomes Let w 1, 1 1 2 1 1
1 1
l
1 u u
l
x2
x
x2
x
EA 2 1 dx f dx P P 0
2
1 l
EA EA
l
x
x
u1 u2 1 f dx P1
weak form becomes Let w 2, 2 2 2 2 1
1 1
l
u u l
x2
x
x2 1
x
EA 2 1 dx f dx P P 0 1
l
EA
l
EA x2
x
u1 u2 2 f dx P2
1 1 1
2
P f P
x2
EA 1 1 u1
x1
P
f
P
1 u
x2 l 1 2 2 2 2 fdx
x1
1 fdx
E,A are constant
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Approximation Methods – Finite
Element Method
Example (cont):
Step 5: Assembling to form system equation
Approach 1:
Element 1: 0 I I I I I u f P E A
l I
f I PI
1
1
1
1 2
2
2
0 0 0 0 0 0
0 0 0 0
Element 2: 1
2 2
f
u II f II P II l II
0 0
0 0
E II AII 0 0 uII
II
PII
1
1
2
0 0 0 0
Element 3:
1 1
III III III III
III III III
f P
1 u f P
0 0 0 0
E III AIII 0 0 0 0 0 1
0
1 1 0 0 uI
1 0
0 0
0 0
0 0 0
1 1
0 1 1 0 0 0 0
0 0 0
0 0
l 0 0 1 1 u
0 1 2 2 2
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Approximation Methods – Finite
Element Method
Example (cont):
Step 5: Assembling to form system equation
Assembled System:
4 4
19
0 0
0
0
I I II II
l I
l
0
l l l
E II AII E II AII E III AIII E III AIII
l II
0
l III
E III AIII E III AIII
l III l III
E I AI I I
E A l I
f1 u1 E I AI E I AI
E II AII
II II
E A
3 l II l III
2 1 2 1
2 1 2 1
I I
II
II III II III
f1 P1
f III P III
P1 f I PI u f P f II
P
2
2
2
u
f
P
f f P P 3 3
u f P 4 2 2
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Approximation Methods – Finite
Element Method
Example (cont):
Step 5: Assembling to form system equation
Approach 2: Element connectivity table Element 1 Element 2 Element 3
1 1 2 3
2 2 3 4
global node index
(I,J)
local node
(i,j)
ij
20
IJ k e K
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Approximation Methods – Finite
Element Method
Example (cont):
Step 6: Imposing boundary conditions and forming condense system
Condensed system:
0 l I l II l II
l II
0
l III l III
E III AIII
l III l III
E I AI
E II AII
II II
E A u2 f2 0
E II AII
E III AIII
E A
III III
u3 f3 0 f P u 4
E II AII
E III AIII
4
l II
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Approximation Methods – Finite
Element Method
Example (cont):
Step 7: solution
Step 8: post calculation
2 dx 1 dx
d2
dx
du u
d1 u 1 1 2 2
u u u 2 1 dx
d2
dx E Eu
d1 Eu
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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 23
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Exercises – Linear Element
Example 1:
E = 100 GPa, A = 1 cm2
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Linear Formulation for Bar Element
2 2 12 22 u2 K
K
K12 u1
P
f
P1
f1
K11
1 1
x2
x
i
x2
x
ji i ij dx dx EA i j dx K , f where K f dx d d
x=x1 x=x2
2 1
x
x=x1 x= x2
u1 u2
P1 P2
f(x)
L = x2-x1
u
x
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Higher Order Formulation for Bar Element
u(x) u11 (x) u22 (x) u33 (x)
1 3
u1 u3 u
x
u2
2
u1 u4 u
x
u2 u3
1 2 3 4
u(x) u11 ( x ) u22 ( x ) u33 ( x ) u44 ( x )
u(x) u11 ( x ) u22 ( x ) u33 ( x ) u44 ( x ) unn ( x )
n
u1 un u
x
u2 u3 u4
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……………
1 2 3 4 ……………
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Natural Coordinates and Interpolation Functions
1
2 , 2
1
2 1
Natural (or Normal) Coordinate:
x=x1 x= x2
=-1 =1
x
x 0 x l
l /2
x x x1
x x1 x2
2
1 3 2
1 =-1
2 =1
=-1 =1
1 4 2
=-1 =1
3
2 2 3 2 1
1 1 1,
1 ,
2 1
3
3 16
3
16
1 , 27 1
1 1
9 1
1
27
3
3 16
3 16
4 3 27 1
1 1, 9 1
1 1
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Quadratic Formulation for Bar Element
i i i
l
2 f and f f dx
1
1
1 x2
x 1
d d d d 2
dx dx
1
d d l
d , i, j 1, 2, 3
where Kij EA i
dx EA i
d K j j
ji
2
33 3 13 23
23 22 12
3 3
2 2
u
u
K
K13 u1 K11 K12
P f
P
f
K K K
K K
P1 f1
=-1 =0 =1
x2
x1
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Quadratic Formulation for Bar Element
u1 u3 u2 f(x)
P3 P1
P 2
x1
=-1
x2
=0
x3
=1
2 2 3 2 1 3 3 2 2 1 1
1 u 1 1 u
1 ) u ( ) u ( ) u ( u( ) u
2 2 3 2 1
1 , 1 1,
1
l l dx l d dx l d l dx l d
d1 2 d1
2 1 ,
d2 2 d2
4 ,
d3 2 d3
2 1
l /2
x x1 x2
2 l d dx 2
d 2 dx l
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Exercises – Quadratic Element
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Example 2:
E = 100 GPa, A1 = 1 cm2; A1 = 2 cm2
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Some Issues
Non-constant cross section:
Interior load point:
Mixed boundary condition:
k
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UNIT – 2
Finite Element Analysis of Trusses
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Plane Truss Problems
33
Example 1: Find forces inside each member. All members have
the same length.
F
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Arbitrarily Oriented 1-D Bar Element on 2-D Plane
Q2 , v2
P1 , u1
P2 , u2
P2 , u2
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Q1 , v1
P1 , u1
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Relationship Between Local Coordinates and Global
Coordinates
1
2
1
0
0
0
0 v u1
v2 0
u
u1
cos v2
sin u2
v 0 sin
sin 0
cos 0
0 cos
0 sin
cos
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Relationship Between Local Coordinates and Global
Coordinates
36
2
1
2 0
0
0
0
0
0
Q P1
P
P1
cos Q2
sin
P
sin
sin 0
cos 0
0 cos
0 sin
cos
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Stiffness Matrix of 1-D Bar Element on 2-D Plane
1
2
1
0
0
0
0
0
0
0
u2
v u1
K
Q2
P
Q P1
ij sin
0 cos
0 sin cos v2
0 sin
sin 0
cos 0
cos
sin
cos
0 cos
0 sin
sin
sin 0
cos 0
cos
0
0
0 0 0 0
1 0 1 0
L 1 0 1 0
0 0
AE K ij
Q2 , v2
P1, u1
P2 , u2
P2 , u2
Q1 , v1
P1 , u1
1 2 2
2
2
1
v2
v
L
AE
Q2
P
Q P1
sin 2 sin cos sin cos
sin cos u2 cos sin cos
sin 2
cos
sin 2 sin cos sin cos sin
sin cos u1 cos2 sin cos cos2
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Arbitrarily Oriented 1-D Bar Element in 3-D Space
2
2
1
y y y
x x x
y y y
2
2
1
0
0
0 w1
0 v 0 u1
w2 0 0
0 v
u
w1 0
0 v
u1
z w2
v
u
0 0
0 0
0 0 z z
z z z
x x 0 0
0 0
0 0
x
-x, -x, -x are the Direction
Cosines of the bar in the x-
y-z coordinate system
P1 , u1
2 2 P , u
x
x y
z 2
1
- x-
x-
38
2
1
2
1
1
0
0
0 R1
0 Q 0 P1
R2 0
Q2 0
P
R 0
Q 0
P1
x x x
z
y y y
z R2
y Q2
P
0 0
0 0 y y
0 0 0 z z
z z
x x x 0 0
0 0
0 0
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Stiffness Matrix of 1-D Bar Element in 3-D Space
2
2
1
2
1
w
v 2
u
w1
v1
u
L
AE
R2
Q2
P2
R
Q1
P1
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
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 2
2
x x x 2
1 1 P , u
2 2 P , u
x
x
x y
z 2
1
-
-
x-
39
0
0
0
0 0 0 0 0w 0
0 v 0
0 0 0 1 0
0 0 0 0
1
L 1 0 0 1 0
AE 0 R 0
Q 0
0 0 0 0 0 v2 0
0 0 0 0 0 0w2 0
u2
1
1
u1
R2 0
P2 Q2 0
1
1
P1
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Matrix Assembly of Multiple Bar Elements
1
2
1
0v2
0 u2
0v 0u1
L 1
AE 0
P
Q P1
Element I
Element II
Element III
1 0 1
0 0
0 1
0 0 0
2
3
2
1 3 1
v3
3u3
3 v
3 u2
4L
1
AE 3 3 3
3 1
3 3 3 3 Q3
P
Q
Q2
P2
40
1
3
1
3 1
3 3 3 3
1
v3
3 u3
3 v
3 1 3u1
4L
1
AE
3 3 3
Q3
P
Q P1
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Matrix Assembly of Multiple Bar Elements
1
2
1
4 0 4 0 0 0
0 0 0 0 0 0
4 0 4 0 0 0
0 0 0 0 0
0
0 0 0 0 0 0
0 0 0 0 0 0
u3
v2
u2
v u1
4L
AE
P3
Q2
P
Q P1
Element I
1
2
1
0
0 1 3 1
0
3 3 3
0 1 3 1
0
0 0
0
v3
0
0
0
0
0 0 0
0 0
v3
3 v2
3u3
3 u2
v u1
4L 0
AE
P3
Q2
P
Q
Q3
P1
0 0 3 3 3 3
1
2
1
1
3
3
3
0
0
0
0
1
3
3
3
0
0
0
0
0
0
0
0
0
0
0
0
1 3
3
3
0
0
0
0
1 3
3
3 v3
v2 u
3
u2
v u1
4L
AE
Q3
P3
Q2
P
Q
Q3
P1
Element II
Element III
41
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Matrix Assembly of Multiple Bar Elements
2
1
2
1
v3
v
u2
v u1
3 3 3
3 u3 3 1 1 3
3 3 3 3 3
1 3 1
3
3
0 4 1 0 3 1
0 0 3 0 3 3
3 3 0 3 0 3
3 1 0
0
4 1 0 3 4
0
4 L
S3
R3
S
R2
AE 4
0
S R1
42
0
v
v ? 5 3
3 3
4
0
0
0
1
3
3
3
4 0
0 0
5
3
3
3
1
3
3
3
1 3
3 3
1 3
3
3
2
0
0
6
?
4 L
S
S 0
v3 0
u 0 3
2
u2 ?
1
u1 0
S3 ?
R3 ? 2
R2 F
AE 1
R1 ?
Apply known boundary conditions
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Solution Procedures
0
?
3 3 3
0
6
1 3 1 3 2
3 0
3 v ?
0 3 3
5 3 1
0
4 0
3
?
4 L
S
S 0
v3 0
u 0 3
2
5 3 4 0 1 3 u2
0 0 3 3 3 3
v
1
3 u1 0
S3 ?
R3 ? 2
R1 ?
AE 1
R2 F
u2= 4FL/5AE, v1= 0
43
u 0
5 AE
4FL
v 0
6 3 3
1
0
3 4 0 1
0 3 3 3 3 3
3 2 0
1
3 3 0
5
3
3 5 3 1
0 0 3 3
4 0
3
?
4 L
S
3
v3 0
v2 0
3 u2
1
u1 0
S3 ?
R3 ? 2
R1 ?
AE S1 0
R2 F
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Recovery of Axial Forces
0
4
4 5
0
0
1
2
v2 0 5AE
0
0 4FL F
u2
v 0
u1 0 1 0 1
0 0 0
L 1 0 1
0 0 0 Q2
P
Q1
AE 0
P1
Element I
Element II
Element III
5
5 3
5 3
1 5
3 3 3
3
3 3 3 3
3 1 3 1
3
1
3
3
2
F 1
v 0
2
u3 0
v 0 5AE
4FL
u2
AE 3
Q3
P
4L 1
Q2
P
44
0 0
0
5
0
3 1
3 3 3 3
3 1 1
1
3
1
v3 0
3 u3 0
3 v 0
3u1 0
4L
1
AE
3 3 3
Q3
P
Q P1
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Stresses inside members
Element I
Element II
Element III
5A
4F
4F P1
5
4F P2
5
5 2 P
1 F
3 Q2
5 F
3 F
5 3 Q
5
45
3 P 1
F
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46
Finite Element Analysis of
Beams
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Bending Beam
Normal strain:
Pure bending problems:
Normal stress:
Normal stress with bending moment:
Moment-curvature relationship:
Flexure formula:
y Review
x
y
Ey x
M M x
x ydA M
EI
1 M
I x
My
I y2dA
1 d 2 y M EI
EI
dx2
47
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Bending Beam
Review
Deflection:
Sign convention:
q(x)
x
Relationship between shear force, bending moment and
transverse load:
y
dx
dV q
dx
dM V
q EI dx4
d 4 y
+ - M M M
+ - V
48
V V
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Governing Equation and Boundary Condition
• Governing Equation
• Boundary Conditions -----
2 ? ,
at x 0 dx
d 2v EI
dx d
? & EI ? & dv d 2v
v ? &
Essential BCs – if v or
Natural BCs – if or
is specdifvied at the boundary.
dx { dx
dx
dv dx dx
d at x L ? , ? & EI v ? & ? & EI
2
d 2v
dx2
dx2
d 2v
d 2v EI dx2
2 dx
EI dx
isdsp
ecifi
ded2v
at the boundary.
2
q(x) 0,
dx2
d 2v(x) EI
dx
d 2
0<x<L
49
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Weak Formulation for Beam Element
• Governing Equation
2
q(x) 0, dx2
d 2v(x) EI
dx
d 2
• Weighted-Integral Formulation for one element
2
x2
dx2
d 2v(x) EI q(x)dx
dx
d 2
0 w(x) x1
• Weak Form from Integration-by-Parts ----- (1st time)
2
1 50
2 dx dx dx dx dx x
x
d 2v 2
EI
d
x1
x2
EI wq dx w
dw d d 2v 0
2 1 x x x
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Weak Formulation
• Weak Form from Integration-by-Parts ----- (2nd time)
1 1
2 2 2
x x dx dx dx dx dx
x x
d 2v 2
dw d 2v 2
EI EI
d
x1
wq dx w
x2 d 2 w d 2v 0
dx2 EI
V(x2)
x = x1
M(x2)
q(x) y
x x = x2
V(x1)
M(x1)
L = x2-x1
2
51
x1 dx
dw
dx EI
x
2
wV M
x1 wq dx
d 2v
x2 d 2 w 0
dx2
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Weak Formulation
• Weak Form
Q4 M x2 Q1 V x1 ,
2 1
1 1 2 3 2 dx
dw dw
dx Q2
dx Q4 )Q
x1
EI wq dx w(x )Q w(x
x2 d 2 w d 2v
dx2
2
x1 dx
dw
dx
x
2
M
x1 wq dx wV
d 2v
x2 d 2 w 0
dx2 EI
Q3
Q4
q(x) y(v)
x x = x2
Q1
Q2
x = x1 L = x2-x1
Q2 M x1 , Q3 V x2 ,
52
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Ritz Method for Approximation
n
j 1
Let v(x) u j j (x) and n 4
4
2
2
1
1 1 2 3
4
Q dx
dw
dx
dw
dx2
j
dx2 j Q
x1
wq dx w(x )Q w(x )Q
x2 d 2 w EI u
j 1
d 2
Let w(x)= i (x), i = 1, 2, 3, 4
4
2
2
1
4
Q Q u EI i 1 1 i 2 3 i
j
dx2
i
dx2
d d i i
dx dx (x )Q (x )Q
x1
q dx
d 2
x2 d 2
j
j 1
Q3
Q4
y(v)
x x = x2
Q1 q(x)
Q2
x = x1 L = x2-x1
where 3 2 4
53
1 1 2 dx dx x x2 x x1
u vx ; u dv
; u vx ; u dv
;
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Ritz Method for Approximation
i i q
Q K u
dx i
Q
dx Q
4
j 1
4 ij j 2 1
x2 i x2
3
x1 i x1
d Q
d
x2
x1
i i
x2
x1
j
ij dx2 dx2 EI i where K dx and q qdx
d 2 d 2
Q3
x = x1
y(v)
x x = x2
Q1
Q2
L = x2-x1
54
Q4
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Ritz Method for Approximation
2
4 4 4
3 3
2 2 2
1 1
2 1
2 1
2 2
1 1
q3
q1
u2 q2
K44 u4 q4 K41
K34 u3 3 K31
K14 u1 K12 K13
K21 K22 K23 K24
K32 K33
K42 K43
K11
Q4
Q
Q
Q1
dx dx
dx dx
dx dx
dx dx
x x2
x x1
x x2
x x1
x x x
x2
1 x2
x1 1 x1
d d
d 3
4
d 3
d 2 x
d
d d
where K K ij ji
Q3
x = x1
y(v)
x x = x2
Q1
Q2
L = x2-x1
55
Q4
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Selection of Shape Function
0
0
1 0 0 0
0 1 0 0
0 1
0 0 0 1
2 1
2 1
2 2
1 1
2
1
4 4 4
3 3
2 2 2
1 1 1
x x2
x x1
x x2
x x1
x x x
x x2
x x1
dx dx
dx dx
dx dx
dx dx
d d
d 3
4
d
d 2 x
d
d 1 d
3
3
2
2
34 3 33 23 13
24 23 22 12
3
2
q1
u q
u q
K24 K34 K44 u4 q4
K
K K K
K K K
K14 u1 K12 K13 K11
Q4 K14
Q
Q K
Q1
The best situation is -----
Interpolation
Properties
56
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v( ) u11 u22 u33 u44
Derivation of Shape Function for Beam Element –
Local Coordinates
How to select i???
2 4 2 4
d 1 d d 3 d d
dv( ) u
d1 d u
d3 d u u
and
where 1 2 4
dv dv
d d u1 v1 u2 u3 v2 u
Let
57
3 2
i i ai bi ci d
Find coefficients to satisfy the interpolation properties.
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Derivation of Shape Function for Beam Element
How to select i???
e.g. Let 3
1
2
1 1 1 1 a b c d
2 1 1
4
2
1
Similarly 1
4
1
4
1
4 58
2
2
2
3
2
2
1 1
2 1
1 1
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Derivation of Shape Function for Beam Element
In the global coordinates:
4
59
2 1 1 1
2 dx
l dv
2 dx
l dv ( x) 2 ( x) v23 ( x) v( x) v ( x)
2
2 1 2 1
1 1
3
2
2
1 1
3
2
3
2
1
2
2 1 3 1
2
2 1 1 3 1
x x
1
x x x x
x x x x l
x1 x2
x x
x1 x2
x x
x2 x1
x x x x 1
l
x1 x2
x x
x1 x2
x x
4
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Element Equations of 4th Order 1-D Model
u3
x = x1
u4
y(v)
x x = x2
u1 q(x)
u2
L = x2-x1
2
33 23 13
24 23 22 12
3 3
2
2
34 u3
u
K44 u4 K24 K34
K
K K K
K K K
K14 u1 K12 K13 K11 q1
Q4 q4
K14
Q q
Q q K
Q1
x2
x1
i i
x2
x1
ji ij
j dx K dx2 dx2
EI i where K and q qdx d 2 d 2
x=x2 x=x1
1 1 1
3 2
4
60
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Element Equations of 4th Order 1-D Model
u3
x = x1
u4
y(v)
x x = x2
u1 q(x)
u2
L = x2-x1
61
x2
where qi i qdx x1
4 2
2 3
2 1
2
3
4 4
3
2 2
u
u
3Lu v
L2
3L u1 v1 6 3L 6
2L2 3L
6 3L 6 3L L2 3L 2L
L3
2EI 3L
q
q
q
q1
Q
Q
Q Q1
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Finite Element Analysis of 1-D Problems -
Applications F
L L L
Example 1.
Governing equation:
dx dx
EI
0 x L q(x) 0
2
d 2v 2
d 2
Weak form for one element
1 1
62
1
Q4 0
dx
dw Q2 wx2 Q3
dx
dw wx Q wq dx
dx2 dx2
d 2 w d 2v EI
x2 x1
x2
x
where Q1 V (x1 ) Q2 M (x1 ) Q3 V (x2 ) Q4 M (x2 )
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Finite Element Analysis of 1-D Problems Example 1.
Approximation function:
3
1
x=x1 2
x=x2
2
2 1 2 1
4
1 1
3
2
2
1 1
3
2
3
2
1
2
3 1 2 1
1 2
2 1 1 3 1
x x
1
x x x x
x x x x l
x1 x2
x x
x1 x2
x x
x2 x1
x x x x
l
x1 x2
x x
x1 x2
x x
4
4
63
2 2 3 2
1 1 1
2 dx 2 dx
l dv l dv ( x) ( x) v ( x) v( x) v ( x)
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Finite Element Analysis of 1-D Problems Example 1.
Finite element model:
2
2
1
2
4
3
2
3Lv
L2 3L v1 6 3L 6
2L2 3L
6 3L 6 3L L2 3L 2L
L3
2EI 3L
Q
Q
Q Q1
P1 , v1 P2 , v2 P3 , v3 P4 , v4
M1 , 1 M2 , 2 M3 , 3 M4 , 4
I II III
64
Discretization:
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Matrix Assembly of Multiple Beam Elements
Element I
Element II
2 2
1 0 Q I
L2
L3
Q I 0 v1
1
2 Q I
6 3L 6 3L 0 0 0
3L 2L 3L L 0 0 0
6 3L 6 3L 0 0 0 0 v2
3
Q I 2EI 3L 3L 2L2 0 0 0 0
4
2
0 0 v3
0
0 0 0
3
0 0 0 v4
0 0 0
1
2
3 3
4
0 0 0 0 0 0
0 0 0
0 0 0
0
0
0 0
0 0
II
II
Q
Q L3
Q II 0 0 3L L2
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
4
0 v1 0 0 0
1 Q II 3L 0 0 v
2
2EI 0 0
2
0
0 0 6 3L 6 0 3L 2L2 3L L2 0
0 6 3L 6 3L 0 0 v
0 3L 2L2 0
3
0 0 0 0 4
0 v4
0
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Matrix Assembly of Multiple Beam Elements
Element III 3
2
3
2
0 0 0 0 0 0
III
L2
Q III
Q
0 0 v1 0 0 0 1 0
0 0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0 0 0 v2 0
2EI 0 0
2 III
Q1 Q III
0 3L v
L 0 0 0 0 6 3L 6 3L
v3
0 0 0 0 3L 2L2 3L
0 0 0 6 3L 6
0 0 0 0 3L L2 3L 2L
3
4
4
3
3
2
1
2
2 2
4
3
3
2
1
0
0
0
0
0 0
0 0
0 0
0 0
0 0
0
0
4
0 0
0
0
0
0
0
6
4
6
0 v2
v1
6
3L
3L 6 3Lv4
L2 3L 2L
L2
6
3L
6 3L
v
3L
6 6 3L 3L
3L 3L 2L2 2L2
3L
L2
L2 3L L 2L 2 3L 3L L2 3L
3L 6 6 3L 3L
3L
3L
L2
6 3L
3L 2L2
6 3L
L3
2EI
M
P4
M
P
M P2
M P1
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Solution Procedures
0
0
0
0
6
0
0
0
0
4
3
3
2
1
2
4
3
3
2
1
?
?
0 ?
0
v2 0
0 v1 0
0 0 0 6
0 0 0 3L
0 4L2 3L L2 L2 0 0 3L
12 0 6 3L v 0
0 6 3L
0 4L2 3L L2 L2
0 0 0 0
6 3L 0 0 3L 12 0
L2 3L 2L2 3L
6 3L 6 3L 0 0
L3
2EI 3L
M
P4 F
M
P ?
M
P2 ?
M ? P1 ?
Apply known boundary conditions
67
0
0
0
0 0 0
0 0 0
6
0
0
0
0
4
3
2
1
2
2
1
4
3
3 ?
?
0
v ?
v 0
v1 0
3L 4 ?
6 3L 0
12 0 6
3L 6 3L
3L 2L2 3L L2
6 3L 12 0 0 0 6 3L
L2 0 0 0 3L 3L 2L
3Lv2 0
0
L2
3L 6 3Lv4 ?
L2 3L 2L
3L L2 0 4L2 3L L2 0
0 3L L2 0 4L2 3L
0 0 0 6 3L 6
L3
2EI
0
P3 ?
P ?
M1 ?
P ?
M
P4 F
M
M 2 0
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Solution Procedures
0
4
3
2
4
3
2
4 ?
?
?
3L v ?
L2
4L2 L2 0 0
4L2 3L L2
3L 6
0 L2 3L 2L2
L3
2EI
M4 0
P F
M 0 M 0
4
3
2
2
3
2
1
1
0
4
v
3L
0 0 0
0 0 0
3L 0 0
0 6 3L
L
3L
L3
2EI
P ?
P ?
M ? P ?
68
0
0
0
6
0
0
0
0
4
2
3
1
2
2
1
4
3
3 ?
?
0
v ?
v 0
v1 0
3L 4 ?
3L 0 0
L2 0 0
6 0 3L 0
12 3L 0 6
L2 0 0 3L 0 3L 2L
3L 6 0
3L 2L2 3L 0 6 3L 12 0 0 6
3Lv2 0
0
L2
4L2 L2 0
L2 4L2 3L
0 3L 6
3L L2 0 3L
0 3L 0
0 0 6
L3
2EI
0
P3 ?
P ?
M1 ?
P ?
M
P4 F
M
M 2 0
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Shear Resultant & Bending Moment Diagram
1 FL
7
FL
9 F
2 FL
7
7
3 F
7
P2
F
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Plane Flame
Frame: combination of bar and beam
E, A, I, L Q1 , v1 Q3 , v2
Q2 , 1
P1 , u1
Q4 , 2
P2 , u2
70
2
1
1
2
2
2
1
0 0
0 0 0 0
0
0 0
2
1
v u2
v
L
4EI
L2
6EI
L
2EI
L2
L2
6EI 12EI
L3 L2
12EI 6EI
L3
6EI
L
0
0
AE
L
0
AE L L2 L L
L2
6EI u
L3
6EI 2EI
12EI
L2
4EI L3
6EI
0 0
12EI 6EI L
0
AE
L
0
AE
Q4
Q3
P
Q Q1
P
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x
Finite Element Model of an Arbitrarily Oriented
Frame y
x
y
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Finite Element Model of an Arbitrarily Oriented
Frame
2
1
1
3
2
1
0
2
1
v u2
v
u
L
4EI
L2 L L2
6EI
L
0
0
AE
L
0
0
AE L2
0
6EI
L3
0
6EI
L2
0
2EI
L3
0
12EI L
2EI
L L2
6EI 12EI
6EI L
0
AE
L
0
0
AE L2
0
6EI
L3
0
6EI
12EI
L2
0
4EI
6EI
L3
0
12EI
Q4
Q
P
Q2
L2
Q1
P
local
72
global
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Plane Frame Analysis - Example
Rigid Joint Hinge Joint
Beam II
Beam I Beam
Bar
F F F F
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Plane Frame Analysis
2
1
1
3
2
1
0
0
2
1
v
u2
v
u
L
4EI
L2 L L2
6EI
L
0
0
AE
L
0
AE L2
0
6EI
L3
0
6EI
L2
0
2EI
L3
0
12EI L
2EI
L L2
6EI 12EI
4EI 6EI L
0
AE
L
0
0
AE L2
0
L3
0
6EI
L2
0
6EI 12EI
L3
0
12EI
Q4
Q
P
Q2
L2
Q1
6EI I
P I
P2 , u2
Q , 4 2
P1 , u1
Q2 , 1
Q1 , v1
Q3 , v2
74
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Plane Frame Analysis
P1 , u2
Q3 , v3
Q2 , 2 Q4 , 3
Q1 , v2
P2 , u3
75
3
3
2
2
2
1
0 0
0 0 0 0
0
0 0
4
2
v
u
v
L
4EI
L2
6EI
L
2EI
L2
L2
6EI 12EI
L3 L2
12EI 6EI
L3
6EI
L
0
0
AE
L
0
AE L L2 L L2
6EI L2
6EI u
L3
6EI 2EI
12EI
L2
4EI L3
0 0
12EI 6EI L
0
AE
L
0
AE
Q4
Q3
P
Q2 Q2
P II
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Plane Frame Analysis
2
1
3
2
1
0
2
1
v
u2
v
u
AE
L
0
L2
0
2EI
L
6EI
L3
0
6EI
L2
12EI
12EI 0
6EI
L3 L2
0 AE
0 L
6EI 0
4EI
L2 L
12EI
0 6EI
L3 L2
0 AE
0 L
6EI 0
2EI
L2 L 12EI
0 L3
0 AE
L 6EI
0 L2
6EI
L2
0
4EI
L
Q4
Q
P
Q2
Q1
I
P I
76
2
2
2
3
1
1
0
0 0
2
v u3
v
u
L L
L2 4
6EI 3
12EI
L3 6EI 4EI
AE
L
0
AE
L
0
0
0
12EI
L3
6EI
L2
0
6EI
AE
L
0
0
0
12EI
L3
6EI
L2
0
6EI
L2
2EI
L
L2
4EI
L
AE
L
0
0
0
12EI
L3
6EI
L2
0
6EI
L2
2EI
L
Q4
Q P2
Q2 Q2
P II
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Plane Frame Analysis
1 FL
8
1 F
2
1 F
2
3 FL
8
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UNIT- 3
Finite element analysis constant
strain triangle
78
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Finite element formulation for 2D:
x
y
S u
S T u
v
x
px
Step 1: Divide the body into finite elements connected to each
other through special points (“nodes”)
py
3
2
1
4
x
2
u1
u 2
u3
u 4
v 4
v3
3
v 2 Element ‘e’
4
v1
y
1 v
u
79
4
3
2
2
1
u 4 v
u 3 v
v
v1 u
u
d
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N3(x, y) u3 N4(x, y) u 4
N3(x, y) v3 N4(x, y) v4
u (x, y) N1(x, y) u1 N2(x, y) u 2
v (x, y) N1(x, y) v1 N2(x, y) v2
4
2
1
4 3 3 2 1
4 3 2 1
v
v3
u 4
u 2
v u1
N u
N 0 N 0 N 0
0
0
v
0 N 0 N 0 N
v (x, y)
u (x, y) N u
u N d
80
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TASK 2: APPROXIMATE THE STRAIN and STRESS WITHIN EACH
ELEMENT
1
81
1 xy
4 3 2
4 3 2
...... N1(x, y)
v
N4 (x, y)
v N3(x, y)
v N2 (x, y)
v
N4 (x, y)
u
x
u (x, y)
v (x, y)
N1(x, y) u
y x y
y y y y y 1 y
x x x x 1 x x
v (x, y)
N1(x, y)
v
Approximation of the strain in element ‘e’
u (x, y)
N1(x, y)
u N2 (x, y)
u N3(x, y)
u
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4
4
3
2
1
B
1
v
u
4 v3
u
v u1
, y) N (x, y)
N4 (x, y)v2
0 u N (x, y)
0
N3(x, y)
N (x, y) 0
N2 (x, y)
N (x, y) 0
N1(x, y)
N (x, y)
N (x, y) N (x, y) N (x, y) N (x, y) N (x, y) N (x, y) N (x
x y x y x y x y
y y y y
1 1 2 2 3 3 4
4
x
0
3
x
0
2
x
0
x
0
y
xy
x
ε B d
82
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Strain approximation in terms of strain-displacement matrix
Stress approximation
Summary: For each element
Displacement approximation in terms of shape functions
Element stiffness matrix
Element nodal load vector
u N d
DB d
ε B d
e V k B D B dV
T
S
e
b f
ST S
f
V
T T N T dS f e
N X dV 83
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Constant Strain Triangle (CST) : Simplest 2D finite element
y
v v1
u 1
3
1
(x,y)
84
v u
1 1
v2
(x ,y )
2 (x ,y ) u2
2 2
x
• 3 nodes per element
• 2 dofs per node (each node can move in x- and y- directions)
• Hence 6 dofs per element
3
(x3,y3)
u3
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u 21 N26 d61
1
1
3 2 2 1
v3
u 3
v u
N v
0
0 u 2 0 N 2 0 N3
N 0 N 0
v (x, y) u
u (x, y)
N1
The displacement approximation in terms of shape functions is
85
3 2 1 N 0
0 N 2 0 N3 0
N 0 N 0 N
N1
u (x,y) N1u1 N2u2 N3u3
v(x,y) N1v1 N2 v2 N3v3
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N
2A
3
2
1 1 1 1
N a2 b2 x c2 y
2A
a3 b3 x c3 y
2A
a b x c y N
2
x 3
2
b1 y2
b2 y3
b3 y1
x3 y2
x1 y3
x2 y1
a1 x2 y3
a2 x3 y1
a3 x1 y2
y 2
y3
1
y3 c1 x3 x2
y1 c2 x1 x3
y2 c3 x2 x1
1 x1 y1
A area of triangle 1
det1 x
where
x
y
u 3
v3
Formula for the shape functions are v1
u1
u2
3
1
(x,y)
86
v u
1 1
v2
(x ,y )
2 2 2 (x ,y )
3 3 (x ,y )
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Properties of the shape functions:
1. The shape functions N1, N2 and N3 are linear functions of x and
y
x
y
2
3
1
1
N1
2
3
1
N2
1 2
3
1
1
N3
at node 'i'
0 at other nodes N
1 i
87
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2. At every point in the domain
y
x
3
i1
3
Ni yi
i1
Ni xi
i1
3
1 Ni
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3. Geometric interpretation of the shape functions
At any point P(x,y) that the shape functions are evaluated,
y
2
x
3
1 P (x,y)
A 3
A2
A1
3
2
1
N
N
N A1
A
A2
A
A3
A
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Approximation of the strains
xy
x
u
v B d y
x
y u
v
y x
90
2 1
3
N (x, y) N (x, y) N (x, y) N (x, y) N (x, y) N (x, y)
N (x, y)
N (x, y) N (x, y)
0 N3(x, y)
0 N2(x, y)
0 N1(x, y)
3 b3 c1
c 1 0 c
b1 0
2A
x
3 3 y
1 1 2 2
x y x
b2 0 b3 0
0 c 0
b1 c2 b2 c3
y
y
2
y
1
y
x
0
x
0
x
B 0
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91
Element stresses (constant inside each element)
DB d
Inside each element, all components of strain are constant: hence
the name Constant Strain Triangle
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92
IMPORTANT NOTE:
1.The displacement field is continuous across element boundaries
2.The strains and stresses are NOT continuous across element
boundaries
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Element stiffness matrix
e V k B D B dV
T
e V
T dV BT D B At k B D B
t=thickness of the element
A=surface area of the element
Since B is constant
t
93
A
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S
e
b f
ST S
T
f
V
T dS f e
N X dV N T
Element nodal load vector
94
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Element nodal load vector due to body forces
e Ae V
T T
b f N X dV t N X dA
e
e
e
A
A
A
b
N X dA t
N X dA t
N X dA t
f
f f
f
f
t Ae
N3 X b
t Ae N 2 X b dA
t Ae N3 X a dA
dA
2 a
1 b
1 a
b3 y
fb3x
b2 y
b 2 x
fb1y
b1x
x
y
f b3x
f b3y
fb1y
f b1x
fb2x
fb2y
2
3
1
(x,y)
95
X b X a
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EXAMPLE:
If Xa=1 and Xb=0
96
0 3
3 0
0 3
0
0
3
2
1
tA
tA
tA
N dA t
N dA
0
t
N dA t
t
t
N X dA
N X dA
t
N X dA t
N X dA t
N X dA t
f
f f
e
e
e
e
e
e
e
e
A
A
A
A N3 X b dA
A 3 a
A 2 b
A 2 a
A 1 b
A 1 a
fb3 y
b3x
fb2 y
b 2 x e
fb1x
fb1y
b
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Element nodal load vector due to traction
e ST
S
T
S dS f N T
EXAMPLE:
y
fS3x
fS3y
fS1y
fS1x
2
x
3
1
97
e l
S S T dS f t
13 along 13
NT
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Element nodal load vector due to traction
EXAMPLE:
y
f S3x
2
3
S T S dS f t
l23 along 23
NT
e
fS3y
fS2x
fS2y
(2,0)
(2,2)
1 (0,0)
0
x
1
S T
t 2 1 t 1
23
2 x l fS t N (1) dy
e 2 along 23
fS 0 2 y
fS t 3 x
fS 0 3 y
2 Similarly, compute
1
98
2
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99
Recommendations for use of CST
1.Use in areas where strain gradients are small
2.Use in mesh transition areas (fine mesh to coarse mesh)
3.Avoid CST in critical areas of structures (e.g., stress
concentrations, edges of holes, corners)
4.In general CSTs are not recommended for general analysis
purposes as a very large number of these elements are required
for reasonable accuracy.
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Example
x
y
El 1
El 2
1
2 3
4
300 psi
1000 lb
3 in
100
2 in Thickness (t) = 0.5 in
E= 30×106 psi
=0.25
(a) Compute the unknown nodal displacements.
(b) Compute the stresses in the two elements.
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Realize that this is a plane stress problem and therefore we need to use
E
101
7
2
0 10 psi
1.2
0
0 0.8
0
1
1
0 3.2 0.8 0
1 3.2
0 0
D 1 2
Step 1: Node-element connectivity chart
ELEMENT Node 1 Node 2 Node 3 Area
(sqin)
1 1 2 4 3
2 3 4 2 3
Node x y
1 3 0
2 3 2
3 0 2
4 0 0
Nodal coordinates
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Step 2: Compute strain-displacement matrices for the elements
1 1 2 2 3 3
3 2 1
b b c b c c
c b2 0 b3 0
0 c 0
b1 0
2A B
1 0 c
Recall b1 y2 y3 b2 y3 y1 b3 y1 y2
c1 x3 x2 c2 x1 x3 c3 x2 x1
with
For Element #1:
1(1)
2(2)
4(3)
(local numbers within brackets)
y1 0; y2 2; y3 0
x1 3; x2 3; x3 0
Hence b3 2
c3 0
b2 0
c2 3
b1 2
c1 3
3
0 2 0 0 0 2 0
3 0 3 0 2 3 0 0 2
6 B(1)
1 0
Therefore
For Element #2:
2 3
0 2 0 0 0 2 0
3 0 3 0
2 3 0 0 6
B(2) 1 0
102
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Step 3: Compute element stiffness matrices
0
0
0.3
0.2 0.2 0.5333
1.2 0.2
0 0 0.3 107
1.2 0.2
0.5333
0.9833 0.5 0.45
1.4 0.3
0.45
k (1) AtB(1)T D B(1) (3)(0.5)B(1)T
D B(1)
u1
103
u2 u4
0.2 v4 v2 v1
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0
0
0.3
0.2 0.2 0.5333
1.2 0.2
0 0 0.3 107
1.2 0.2
0.5333
0.9833 0.5 0.45
1.4 0.3
0.45
k (2) AtB(2)T D B(2) (3)(0.5)B(2)T
D B(2)
u3
104
u4 u2
0.2 v2 v4 v3
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105
Step 4: Assemble the global stiffness matrix corresponding to the nonzero degrees of
freedom
Notice that
u3 v3 u4 v4 v1 0
Hence we need to calculate only a small (3x3) stiffness matrix
K
1
0.983 0.45 0.2 u
0.45 0.983 0 107 1u 2
0.2
u
0
u2
1.4 v2
v2
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Step 5: Compute consistent nodal loads
f
f 2 x
f f
2 y 2 y
0
f1x 0
f2 y 1000 fS 2 y
The consistent nodal load due to traction on the edge 3-2
N dx 3
3
3
S2 y
9
0
x2 3
50 2 50 2
225 lb
(300)(0.5)
N (300)tdx f
150 x
dx 3
x0 3
x0 32
x0 3 32
3 2
N
106
x
3 2 32
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Hence
f2 y 1000 fS 2 y
1225 lb
Step 6: Solve the system equations to obtain the unknown nodal loads
K d f
107
0
0
0.2
0
0.983
2
107 0.45 u2
0.45 0.2u1
0.983
0 1.4v 1225
Solve to get
v
u 4
2
2
1
0.9084 104 in
u 0.1069 10 in
0.2337 104 in
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108
0.9084 104 0 0 0.1069 104 0.2337 104 0
1 1 2 2 4 v4 v u v u u d(1)T
76.1
Calculate
114.1
(1) 1391.1 psi
Step 7: Compute the stresses in the elements
In Element #1
(1) D B(1) d(1)
With
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109
3
0.9084 104 3
0 0 0 0 0.1069 104
2 v2 u v u4 v4 u d(2)T
Calculate
114.1
(2) 28.52 psi
In Element #2
(2) D B(2) d(2)
With
363.35
Notice that the stresses are constant in each element
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110
UNIT - 4
Heat transfer analysis
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Thermal Convection
Newton’s Law of Cooling
q h(Ts T )
h: convective heat transfer coefficient (W m2 Co )
111
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112
Thermal Conduction in 1-D
Boundary conditions:
Dirichlet BC:
Natural BC:
Mixed BC:
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Weak Formulation of 1-D Heat Conduction
(Steady State Analysis)
• Governing Equation of 1-D Heat Conduction -----
d ( x) A( x)
dT ( x) AQ( x) 0 dx dx
0<x<L
• Weighted Integral Formulation -----
• Weak Form from Integration-by-Parts -----
0
L
dx
d dT (x) AQ( x) dx
0 w( x)
dx ( x) A( x)
L dw dT
113
L
dT
0 0 0 dx
A dx wAQ dx w A
dx
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Formulation for 1-D Linear Element
Let T (x) T11(x) T22 (x)
f2 1 x1
2
x2
x
T1 T2
f1
2 1 1 2
l l
x x ,
x x ( x) ( x)
x2 x1
1T1
2T2
2
2
1
1 x x
, f (x) A T
f (x) A T
114
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Formulation for 1-D Linear Element
Let w(x)= i (x), i = 1, 2
2 x2
i j
j i i 2 2 i 1 1
j1
x2 d d ( x ) f dx AQ dx ( x ) f dx dx
x 1 1
0 T A x
2
Qi i (x2 ) f2 i (x1 ) f1 KijTj
j 1
11 1 1
K22 T2 K12
K12 T1 K
2 Q2
f
f Q
1
115
1
2
x2 x2
i j
ij i i 1
x1 x2 x
dT dT
dx dx
d d where K A dx, Q AQ dx, f A , f A dx dx
x
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Element Equations of 1-D Linear Element
1 1
1 T2
1 1T1
L 1 2 Q2
Q
f
f A
i i 1
dT dT
dx x x1 xx2
where Q AQ dx, f A , f2 A dx
x2
x1
f2 1 x1
2
x2
T1
x
T2
116
f1
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1-D Heat Conduction - Example
t1 t2 t3
0 T 200o C T 50o C
x
1
2
3
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 70W m K , 2 40W m K , 3 20W m K
t1 2cm, t2 2.5cm, t3 4cm
117
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Thermal Conduction and Convection-
Fin
Objective: to enhance heat transfer
dx
t
x
w
loss Q 2h(T T ) dx w 2h(T T ) dx t
2h(T T ) w t
Ac dx Ac
Governing equation for 1-D heat transfer in thin fin
c c
d A dT A Q 0
dx dx
c
118
c
d A dT Ph T T A Q 0
dx dx
P 2 w t where
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Fin - Weak Formulation
(Steady State Analysis)
• Governing Equation of 1-D Heat Conduction -----
d ( x) A( x) dT ( x) Ph T T AQ 0
dx dx
0<x<L
• Weighted Integral Formulation -----
• Weak Form from Integration-by-Parts -----
0
L
dx
d dT (x) Ph(T T ) AQ( x) dx
0 w( x)
dx ( x) A( x)
dx
119
dx dx
L dw dT L
dT 0 A wPh(T T ) wAQ dx w A
0 0
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Formulation for 1-D Linear Element
Let w(x)= i (x), i = 1, 2
2
dx dx
d d
j1
x2 i AQ PhT dx A i j Ph dx
i j x 1 1
i ( x2 ) f2 i ( x1 ) f1
0 T j
x2
x
2
i (x1 ) f1 Qi i (x2 ) f2 KijTj
j 1
11 1 1
K22 T2 K12
K12 T1 K
2 Q2
f
f Q
1 1
1 2
x2
ij i i i j
dx dx
dx dx
d d A
i
x x1 x x2
x2 where K Ph dx, Q AQ PhT dx,
j x
f A dT
, f A dT
x
120
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Element Equations of 1-D Linear Element
1 1 f Q 1 1 T1
A 1
Phl 2
f
2 Q2 1 6 1 L 1 2 T2
f2 1 x=0
2
x=L
T1
x
T2
f1
x2
121
dT dT
x x1 x x2
f1 A dx
, f2 A dx
where Qi i AQ PhT dx, x1
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122
Time-Dependent Problems
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123
Time-Dependent Problems
In general,
Key question: How to choose approximate functions?
Two approaches:
ux, t
ux, t u j j x, t
ux, t u j t j x
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Model Problem I – Transient Heat Conduction
Weak form:
t x
x c
u
a
u f x, t
1 1
1
a ) Q2 w(x2 )
x2 wf dx Q w(x
u
t cw
w u
x x 0
x
2
124
1
x2 x1 dx
Q
a
du
dx
Q a
du ;
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let:
) 1 1 2 2
1
a
x2 wf dx Q w(x ) Q w(x
u
t cw
w u
x x 0
x
Transient Heat Conduction n
ux, t u j t j x j 1
and w i x
K u M u
F
x2
x1
ij x x
K a i j dx
M ij
x2
ci j dx x1 x2
Fi i fdx Qi
x1
ODE!
125
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Time Approximation – First Order ODE
dt
126
0 t T 0 a du
bu f t u0 u
Forward difference approximation - explicit
Backward difference approximation - implicit
k k k a
k 1 u u
t f bu
k k k k 1 u u
t f bu a bt
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Stability Requirment
max
127
2
1 2 cri t t
K M u Q where
Note: One must use the same discretization for solving
the eigenvalue problem.
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Transient Heat Conduction - Example
0
128
u
2u 0 x 1
t
u 1, t 0
t x2
u0, t 0
ux,0 1.0
t 0
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Transient Heat Conduction - Example
129
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Transient Heat Conduction - Example
130
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Transient Heat Conduction - Example
131
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Transient Heat Conduction - Example
132
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Transient Heat Conduction - Example
133
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UNIT – 5
Dynamic Analysis
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Axi-symmetric Analysis
Cylindrical coordinates:
x r cos ; y r sin ; z z
r, , z
• quantities depend on r and z only
• 3-D problem 2-D problem
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Axi-symmetric Analysis
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Axi-symmetric Analysis – Single-Variable Problem
00 22 11
z
u(r, z) a u f (r, z) 0
z
r
u(r, z)
a
1 ra
r r
Weak form:
e
wqn ds e
z
z
r
u
w a r
w
a u a wu wf (r, z)rdrdz 0 00 22 11
where nz
137
z r
u(r, z) u(r, z) nr a22 qn a11
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Finite Element Model – Single-Variable Problem
u u j j j
Ritz method:
e n
e e K u f Qe ij j i i
j 1
where j (r, z) j (x, y)
w i
Weak form
22
138
e
j j a
i i K e a a rdrdz
00 i j r r z z
ij 11
where
i i
e
f e frdrdz
i i n
e
Qe q ds
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Single-Variable Problem – Heat Transfer
Weak form
Heat Transfer:
1
rk T (r, z)
k T (r, z)
f (r, z) 0 z z
r r r
e
0
w k
T
w k
T wf (r, z) rdrdz
r r z z
wqn ds e
nz
139
T (r, z) T (r, z) qn k nr k
r z where
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3-Node Axi-symmetric Element
T (r, z) T11 T22 T33
1
2
3
140
1 2 3
e
1 r z
2 A
r2 z3 r3 z2
z z r r 3 2
3 1 1 3
2 3 1
e
1 r z
2 A
r z r z
z z
r r 1 3
1 2 2 1
3 1 2
e
1 r z
2 A
r z r z
z z r r 2 1
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4-Node Axi-symmetric Element
T (r, z) T11 T22 T33 T44
1 2
3 4
a
b
r
141
z
2
3 a b
1
1
1
1 a b a b
1
4
a b
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Single-Variable Problem – Example
Step 1: Discretization
Step 2: Element equation
e
142
j i ij
K e
i j
rdrdz r r z z
i i
e
f e frdrdz i i n
e
Qe q ds
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Review of CST Element
• Constant Strain Triangle (CST) - easiest and simplest finite element – Displacement field in terms of generalized coordinates
– Resulting strain field is
– Strains do not vary within the element. Hence, the name constant strain triangle (CST)
• Other elements are not so lucky.
• Can also be called linear triangle because displacement field is linear in x and y - sides remain straight.
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Constant Strain Triangle
• The strain field from the shape functions looks like:
– Where, xi and yi are nodal coordinates (i=1, 2, 3)
– xij = xi - xj and yij=yi - yj
– 2A is twice the area of the triangle, 2A = x21y31-x31y21
• Node numbering is arbitrary except that the sequence 123 must go clockwise around the element if A is to be positive.
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Constant Strain Triangle
• Stiffness matrix for element k =BTEB tA
• The CST gives good results in regions of the FE
model where there is little strain gradient
– Otherwise it does not work well.
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Linear Strain Triangle
• Changes the shape functions and results in
quadratic displacement distributions and
linear strain distributions within the element.
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Linear Strain Triangle
• Will this element work better for the problem?
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Example Problem
• Consider the problem we were looking at:
5 in.
1 in.
0.1 in.
I 0.113 / 12 0.008333 in4
M c
1 0.5
I 0.008333 60 ksi
E
0.00207
2
ML
25
2EI 2 29000 0.008333 0.0517 in.
148
1k
1k
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Bilinear Quadratic
• The Q4 element is a quadrilateral element
that has four nodes. In terms of generalized
coordinates, its displacement field is:
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Bilinear Quadratic • Shape functions and strain-displacement
matrix
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Example Problem
• Consider the problem we were looking at:
5 in.
1 in.
0.1 in.
E
151
I 0.008333
0.0345in. 3EI 3 29000 0.008333
0.2 125
60ksi 1 0.5
I 0.1 13 / 12 0.008333in4
PL3
0.00207
M c
0.1k
0.1k
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Quadratic Quadrilateral Element
• The 8 noded quadratic quadrilateral element
uses quadratic functions for the displacements
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Quadratic Quadrilateral Element
• Shape function examples:
• Strain distribution within the element
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154
Quadratic Quadrilateral Element
• Should we try to use this element to solve our
problem?
• Or try fixing the Q4 element for our purposes.
– Hmm… tough choice.
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155
Isoparametric Elements and Solution
• Biggest breakthrough in the implementation of the finite element method is the development of an isoparametric element with capabilities to model structure (problem) geometries of any shape and size.
• The whole idea works on mapping. – The element in the real structure is mapped to an
‘imaginary’ element in an ideal coordinate system
– The solution to the stress analysis problem is easy and known for the ‘imaginary’ element
– These solutions are mapped back to the element in the real structure.
– All the loads and boundary conditions are also mapped from the real to the ‘imaginary’ element in this approach
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Isoparametric Element
3
1 (x1, y1)
2 (x2, y2)
(x3, y3) 4 (x4, y4)
X, u
Y,v
2 (1, -1)
1 (-1, -1)
4 (-1, 1)
3 (1, 1)
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Isoparametric element
Y
1 2 3 4
0 0 0 N1
X N
0
• The mapping functions areqx1u ite simple:
4
x
2
x3
N N N 0 0 0 0 x
N2 N3 N4 y1 y2
1 4
N 1
(1 )(1 )
2 4
N 1
(1 )(1 )
3 4
N 1
(1 )(1 )
4 4
N 1
(1 )(1 )
y3 y4
Basically, the x and y coordinates of any point in the
element are interpolations of the nodal (corner)
coordinates.
From the Q4 element, the bilinear shape functions
are borrowed to be used as the interpolation
functions. They readily satisfy the boundary values
too. 157
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Isoparametric element
v
1 2 3 4
0 0 0 N1
u N
0
• Nodal shape functions for d ui1splacements
4
u
2
u3
N N N 0 0 0 0 u
N2 N3 N4 v1 v2
v3 v4
1 4
N 1
(1 )(1 )
2 4
N 1
(1 )(1 )
3 4
N 1
(1 )(1 )
4 4
N 1
(1 )(1 ) 158
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• The displacement strain relationships:
x
u
u
u
X X X
y
v
v
v
Y Y Y
x
xy
y
u
X v
Y
Y X
u
v
Y Y
0
0 0
X X
0
Y Y X X
u
v
u
v
But , i t is too diff icult to obtain
and
X X 159
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Isoparametric Element
X Y
u
u
X Y u
X Y u
X
Y
It is easier to obtain X
and Y
X Y
J X Y Jacobian
defines coordinate transformation
Hence we will do it another way
u
u
X
u Y
X Y
u
u
X
u
Y
X
Ni X i
Y N
i Yi
X N
i X
i
Y N
i Y
i
X
u
Y u
1
u
J
u
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Gauss Quadrature
• The mapping approach requires us to be able to evaluate the integrations within the domain (-1…1) of the functions shown.
• Integration can be done analytically by using closed- form formulas from a table of integrals (Nah..)
– Or numerical integration can be performed
• Gauss quadrature is the more common form of numerical integration - better suited for numerical analysis and finite element method.
• It evaluated the integral of a function as a sum of a
becomes n
I Wii i 1
finite numb1er of terms I d
1
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Gauss Quadrature
• Wi is the ‘weight’ and i is the value of f(=i)
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Numerical Integration
Calculate: b
I f xdx a
• Newton – Cotes integration
• Trapezoidal rule – 1st order Newton-Cotes integration
• Trapezoidal rule – multiple application
1 b a
f (x) f (x) f (a) f (b) f (a)
(x a)
2
f (a) f (b) b b
I f (x)dx f1 (x)dx (b a) a a
b
163
xn1
xn x1 x2 xn
a x0 x0 x1
I f (x)dx fn (x)dx f (x)dx f (x)dx f (x)dx
2
n1
i1
i f (x ) f (b) f (a) 2
h I
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Numerical Integration
Calculate: b
I f xdx a
• Newton – Cotes integration
• Simpson 1/3 rule – 2nd order Newton-Cotes integration
2
2 0 1 1
0 2 0 1 2 f (x )
(x2 x0 )(x2 x1 )
(x x0 )(x x1 ) f (x ) f (x ) (x0 x1 )(x0 x2 ) (x x )(x x )
(x x1 )(x x2 ) (x x )(x x ) f (x) f (x)
b b
a a 6
f (x ) 4 f (x ) f (x ) I f (x)dx f2 (x)dx (x2 x0 )
0 1 2
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Numerical Integration
Calculate: b
I f xdx
2
(b a)
f (a) (b a)
f (b) 2 2
165
I (b a) f (a) f (b)
I c0 f (x0 ) c1 f (x1)
a
• Gaussian Quadrature
Trapezoidal Rule: Gaussian Quadrature:
Choose according to certain criteria c0 , c1, x0 , x1
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Numerical Integration
Calculate: b
I f xdx a
• Gaussian Quadrature
3
1 1 1
1
I f xdx f 3
f
1
• 3pt Gaussian Quadrature
I f xdx 0.55 f 0.77 0.89 f 0 0.55 f 0.77 1
1
1
• 2pt Gaussian Quadrature
I f xdx c0 f x0 c1 f x1 cn1 f xn1
b a
~x 1 2(x a) Let:
b 1
1 a
166
~ ~ 1 1 1 f (x)dx
2 (b a) f 2
(a b) 2
(b a)x dx