Upload
allan-marbaniang
View
241
Download
2
Embed Size (px)
DESCRIPTION
Variational Principle
Citation preview
18/04/2023
CE502: Finite Element Analysis
By
Dr. A. ChakrabortyDepartment of Civil Engineering
Indian Institute of Technology Guwahati, India
1
Variational Principle
18/04/2023
2
Optimization of a function
22
2
( ) 1 ( )( ) ( ) ( ) ( )
2
df a d f af x f a x a x a
dx dx
Let, has extremum (min or max) at . Expand in Taylor series at around i.e. in the neighborhood of a
( )f x x a ( )f xx a
For minimum value( ) ( ) 0 f x f a x a x
Here, is either ‘+’ve or ‘-’ve i.e. nonzerox
𝑎 𝑏 𝒙
𝒇 (𝒙)
18/04/2023
3
Let us examine equal to zero first: 1st term will be dominant and here for minimum, therefore ( ) ( ) 0f x f a
( )0
x a
df a
dx
This would be same for both max and min. Next, considering second term
22
2
2
2
2
2
( ) ( ) 0
1 ( )( ) 0
2
( )0 min
( )0 max
x a
x a
x a
f x f a
d f ax a
dx
d f a
dx
d f a
dx
( )0
x a
df a
dx
Note:
18/04/2023
4
Observation: • Max or min of depends on the values of • Only differentiation is enough to identify extremum• Extremum are points (like ) on the space that defines
independent variable
( )f x
,a b
18/04/2023
5
2
11 1 2 2( ( ), ) ( ) & ( )
x
xT F f x x dx given f x f f x f
Optimum of a functionalT
i iiV
dv Pu is a function of & is a function of . is a functional ( , , )u x y z
Note: We are looking for optimal , so that is minimum. Individually speaking, one can find out maximum or minimum of and get . However, our objective is to optimize the following functional
u
1 2 3 4, , ,f f f f x
𝒙
𝑓 4 (𝑥)
𝑓 2(𝑥)
𝑓 3(𝑥 )
𝑥1 𝑥2
𝒇 (𝒙) 𝑓 1(𝑥)
18/04/2023
6
Delta operator : Let us consider a function in ( ) ( )f x 1 2x x x
[ ( )] ( ) ( )f x f x f x
𝑥∗𝒙
𝒇 (𝒙)
𝑥1 𝑥2
~𝑓 (𝑥)
𝑓 (𝑥)𝛿
𝑑𝑑𝑥
Here, is an admissible function that satisfies boundary condition
( )f x
1 1 2 2
1 1 1
2 2 2
( ) ( ) and ( ) ( )
[ ( )] ( ) ( ) 0
[ ( )] ( ) ( ) 0
f x f x f x f x
f x f x f x
f x f x f x
is any admissible candidate function( )f x
18/04/2023
7
Strong & Weak VariationStrong Variation: if we impose no restriction of the derivative of and then is a strong variation.Weak Variation: if we impose restriction that not much difference in derivative of and then is a weak variation
( )f x ( )f x ( )f x
( )f x ( )f x( )f x
𝒙
𝒇 (𝒙)
𝒙
𝒇 (𝒙)
18/04/2023
8
Properties of Delta Operator: • Differentiation of variation
• Variation of integration
•
•
•
•
( )( )
d f d d d dff f f f
dx dx dx dx dx
( )fdx fdx fdx f f dx fdx
1 2
1 2 1 2 1 2
( ) ( ) ( )
( ) [ ] [ ]
g x f x f x
g x g g f f f f f f
1 2 1 2 1 2[ ] [ ]f f f f f f
1 2 1 212
2 2
[ ] [ ]f f f fff f
f f
18/04/2023
9
Functional: Taylor series expansion
( , , ) ( , , ) . . .F F
F u v x F u v x u v h o tu v
Where, and . Neglecting the effect of the higher order terms
u u u v v v
( , , ) ( , , )F F
F u v x F u v x u vu vF F
F u vu v
Note: 1st variation has similar structure like differentiation but there is significant difference
18/04/2023
10
Let, ( , ) ( , , )
( , ) ( , , )
( , , )
I u u F u u x dx
I u u F u u x dx
I F u u x dx
F Fu u dx
u u
Example: . Find 2 ( )( , ) sin ( ) ( ) ev xI u v xu x u x I
2 ( )
( ) ( )
( ) ( )
sin ( ) ( ) e
sin 2 e e
2 sin e e
v x
v x v x
v x v x
I xu x u x
x u u u u v
u x u u v
18/04/2023
11
Theorem of Functional: Let,2
1
2
1
2
1
( , , , ) ( , , , )
( , , , ) ( , , , ) . . .
x
x
x
x
x
x
I u v u v F u v u v dx
F F F FI u v u v I u v u v u v u v dx h o t
u v u v
F F F FI u v u v dx
u v u v
du u
dx Note: by property of Delta operator
2 2
1 1
x x
x x
F F F d F F d FI u u u v dx
u v u dx u v dx v
18/04/2023
12
Therefore, for minimize equate it to zero. Also admissible & means &
( , , , ) ( , , , ) 0I u v u v I u v u v u v 2
10
x
xu 2
10
x
xv
F d F F d FI u v dx
u dx u v dx v
Apply localization theorem, the governing equations are
0
0
F d F
u dx uF d F
v dx v
18/04/2023
13
Example: . Given 2
0
1( )
2
L
L
duu EA qu dx Pu
dx
0(0) 0u u
0
0
0 0
0 0
0
12 0
2
0
0
0
L
L
L
L
L L
L
L
L
du duEA q u dx P u
dx dx
d uduEA q u dx P u
dx dx
du d duEA u EA q udx P u
dx dx dx
du du d duEA P u EA u EA q udx
dx dx dx dx
0
18/04/2023
14
Governing equation:
0 0d du
EA q x Ldx dx
Boundary condition:0
duEA P x L
dx
Essential Boundary Condition Natural Boundary Condition
Dirichlet BVP Neuman BVP
Displacement BVP Force BVP
(0)ux L
duEA P
dx
18/04/2023
15
Example: The total potential energy of a linear elastic body in 3D subjected to body and traction is given by
2
1( )
2 ij ij i i i i
V S
u f u dv t u ds
Given, on is . Using variational principle find governing equation and force boundary condition. Material is isotropic.
iu 1S ˆiu
: Differential equation of equilibrium & Cauchey’s stress-strain relation
Note: Variational principle provides – a) Homogenous form of the governing equation using
essential boundary conditionb) Natural boundary condition needs to be satisfied
18/04/2023
16
Integral Formulation: Let us consider the problem -
( ) ( ) 0d dv
a x f x x Ldx dx
Given, 0(0) & La L
duu u a Q
dx
Note: A large class of problem may be classified in this format
01
( ) ( ) ( ) ( )N
N j jj
u x u x C x x
Where, are unknown coefficients, satisfies the differential equation
jC ( )Nu x
( ) ( ) 0
( , ) ( ) ( ) 0 0
N
Nj
duda x f x x L
dx dx
dudR x C a x f x x L
dx dx
18/04/2023
17
The above equation is not exactly zero as is approximate. is residue. But we need to be zero at points to solve
( )Nu x( , )jR x C R N
jC( , ) 0 , & 1,2, ,j iR x C for x x i N
𝒙
𝒚
𝒛
𝒏𝒙
𝒏𝒚
𝒏𝒛
𝒔𝟐𝒔𝟏
18/04/2023
18
To achieve this we apply Weighted Residual MethodLet, are the weights at ith points-
0
( ) ( , ) 0 1, 2, ,L
i jw x R x C dx i N
iw
Note: For different we have different methodsiw
NamePetrov-Galerkin
Galerkin
Least square
Collocation
iw
i i iw
i iw
( ) ii
ddw a x
dx dx
*( )i iw x x *( ) is Dirac delta function; not operator
18/04/2023
19
Dirac delta function:
0 0
0
0 0
( ) 0
= 1
( ) ( ) ( )
x x x x
x x
f x x x dx f x
18/04/2023
20
Development of Weak Form
0
00
00
0
0
0
L
LL
L
L
d dvw a f
dx dx
dw du dva wf dx wadx dx dx
dw dua wf dx wQ wQdx dx
The above eq. for the weak form of the differential eq. in the given weighted-integral statement. ‘Weak’ refers to reduced order differential of (i.e. 1st order and not in 2nd order). Hence, must satisfies the homogeneous form of the essential boundary condition:
u
w(0) 0w
18/04/2023
21
Weak form:
0
0L
L
dw dua wf dx wQdx dx
Let,
0
0
( , ) bilinear
( ) linear
L
L
L
dw duB w u a dx
dx dx
l w wfdx wQ
Hence, weak form equation cane rewritten
( , ) ( )B w u l w
18/04/2023
22
Note: if w u
2
0 0 0
1( , )
2 2
1( , ) ( , )
2
L L Ld u du a du du duB u u a dx dx a dx
dx dx dx dx dx
B u u B u u
and
0 0
( ) ( )Q
( ) ( )
L L
t Ll u ufdx u t ufdx uQ
l u l u
1( ) ( , ) ( )
2( , ) ( )
I u B u u l u
I B u u l u
18/04/2023
23
For minima for 2 ( , ) 0I B u u 0u 2
0
0
1( )
2
0
L
L
L
L
duu EA qu dx Pu
dx
d du duEA q udx EA P u
dx dx dx
In above eq. u w
Check the Standard Form
0d du
a fdx dx
Subjected to and 0(0)u u Lx L
dua Qdx
18/04/2023
24
, ,a EA q f w u and
0
0
0
( )
L
L
L
L
dw duB a dx
dx dx
dw duEA dx
dx dx
dul EA P u w q dx
dx
using relation of in , one can getB
( , ) ( )
1 ( , ) ( )
21
( , ) ( )2
B u u l u
B u u l u
B u u l u
18/04/2023
25
Then it can be represented by
2
0
1( , ) ( )
2
1( )
2
L
L
I B u u l u
duu EA qu dx Pu
dx
here, is homogenous essential condition. Applying, (0) 0u ( ) 0u
0
d duEA q
dx dx
duEA P at x L
dx
The above two equations are represents governing equation and natural boundary condition respectively.
18/04/2023
26
Example: Solve the above governing equation and fixed weak form. Also solve using Ritz’s method with N=2
Weighted Residual Method:
0
0 0
0 0 0
00 0
0
0
0
0
L
L L
L L L
LL L
d duw EA q dx
dx dx
d duw EA dx wqdxdx dx
du dw duwEA EA dx wqdx
dx dx dx
dw du duEA dx wEA wqdx
dx dx dx
18/04/2023
27
here
0
0 0
( , )
( ) ( ) (0) Weak Form
L
L
L
dw duB w u EA dx
dx dx
du dul u w L EA w EA wqdx
dx dx
hence
( , ) ( )B w u l w
Ritz Solution:Let,
01
( ) ( )N
j jj
u x u x C
Note: already zero. So we can set (0) 0u 0 0
1 1 2 2( ) 2u x C C N Now satisfying the above conditionLet,
1 2(0) 0; (0) (0) 0u 2
1 2( ) and ( )x x x x
18/04/2023
28
Note: You may assume any other polynomial that satisfy the essential boundary condition. Here and are unknown
2C1C
21 2
21 2
1 2
1 2
( )
,
2
2
i i
u x C x C x
w w x w x
duC x C x
dxdw dw
xdx dx
0
1 2 1 2
0 0 0
21 2 1 2
0 0 0
( , )
( 2 ) 2
2 ( 2 ) 2 4
L
L L L
L L L
dw duB w u EA dx Ky
dx dx
EA C C x dx EAC dx EAC xdx
EA x C C x dx EAC xdx EAC x dx
18/04/2023
29
here1 2
0 0
22 2
0 0
0 0 1
22
0 0
2
( , )
2 4
2
2 4
L L
L L
L L
L L
EAC dx EAC xdx
B w u
EAC xdx EAC x dx
dx xdxC
EAC
xdx x dx
Note: The above matrix is symmetric.
18/04/2023
30
00
0
0
( ) ( ) (0)
( )
( )
L
L
L
L
L
du dul u wqdx w L EA w EA
dx dx
duwqdx w L EA
dx
wqdx w L P
In above equation is replaced by . From here, one assume L
duEA
dxP
1
0
2 22
0
L
L
w xqdx PL
w Lx qdx PL
18/04/2023
31
Hence, 0
2 2
0
( )
L
L
xqdx PL
l u
Lx qdx PL
In above equation, the integral part is component due to body force and second part due to point load. Matrix Form:
0 0 01
22 2 2
0 0 0
22
1
2 3 32 2
2
2 4
24
33
L L L
L L L
dx xdx xqdx PLC
EAC
xdx x dx Lx qdx PL
qLPLL L C
EACL L qL
PL
18/04/2023
32
Now solve and 1C 2C
29.81 /g m sLet, . If only is applied and not body force . Use these values in above equations
2 2100 , 1 , 12.7 126.68 , 2.1 08 /P kN L m d mm A mm E E kN m P
5 2 3/ 7.89 10 / , 3.76 10P A kN m
1
72
7 2
0.0038
1.8219 10
( ) 0.0038 1.8219 10
( ) 3.8
C
C
u x L L
u L mm
Note: Variational formulation tell that impact of gravitational pull is negligible.
18/04/2023
33
Note:
( ) ( ) 0
( ) 0
b
a
G x x dx a x b
G x
for arbitrary ( )x
Lemma:
2
( ) ( )
( ) ( ) 0 ( ) 0
( ) 0 ( , )
b b
a a
G x x
G x x dx G x dx
G x a b
18/04/2023
34
General Statement:
is arbitrary in is arbitrary in
( )a a x b ( )G a a x b
2 2
( ) ( ) ( ) ( ) 0
( ) ( ) 0
b
a
b
a
G x x dx G a a
G x dx G a
Sum of two ‘+’ve quantity
( ) 0 & ( ) 0G a G x
18/04/2023
35
Thank You!!!