18/04/2023

CE502: Finite Element Analysis

By

Dr. A. ChakrabortyDepartment of Civil Engineering

Indian Institute of Technology Guwahati, India

1

Variational Principle

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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)

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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:

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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

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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)

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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

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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)

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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Dirac delta function:

0 0

0

0 0

( ) 0

= 1

( ) ( ) ( )

x x x x

x x

f x x x dx f x

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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

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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

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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

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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

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, ,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

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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.

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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

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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

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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

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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.

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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

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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

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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.

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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

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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

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Thank You!!!