1

CHAP 3 WEIGHTED RESIDUAL AND ENERGY METHOD FOR 1D PROBLEMS

FINITE ELEMENT ANALYSIS AND DESIGNNam-Ho Kim

2

INTRODUCTION

• Direct stiffness method is limited for simple 1D problems• FEM can be applied to many engineering problems that are

governed by a differential equation• Need systematic approaches to generate FE equations

– Weighted residual method– Energy method

• Ordinary differential equation (second-order or fourth-order) can be solved using the weighted residual method, in particular using Galerkin method

• Principle of minimum potential energy can be used to derive finite element equations

3

EXACT VS. APPROXIMATE SOLUTION

• Exact solution– Boundary value problem: differential equation + boundary conditions– Displacements in a uniaxial bar subject to a distributed force p(x)

– Essential BC: The solution value at a point is prescribed (displacement or kinematic BC)

– Natural BC: The derivative is given at a point (stress BC)– Exact solution u(x): twice differential function– In general, it is difficult to find the exact solution when the domain

and/or boundary conditions are complicated– Sometimes the solution may not exists even if the problem is well

defined

2

2( ) 0, 0 1

(0) 0Boundary conditions

(1) 1

d up x x

dxu

dudx

+ = £ £

ü= ïïïýï= ïïþ

4

EXACT VS. APPROXIMATE SOLUTION cont.

• Approximate solution– It satisfies the essential BC, but not natural BC– The approximate solution may not satisfy the DE exactly– Residual:

– Want to minimize the residual by multiplying with a weight W and integrate over the domain

– If it satisfies for any W(x), then R(x) will approaches zero, and the approximate solution will approach the exact solution

– Depending on choice of W(x): least square error method, collocation method, Petrov-Galerkin method, and Galerkin method

2

2( ) ( )

d up x R x

dx+ =

%

1

0( ) ( ) 0R x W x dx =ò Weight function

5

GALERKIN METHOD

• Approximate solution is a linear combination of trial functions

– Accuracy depends on the choice of trial functions– The approximate solution must satisfy the essential BC

• Galerkin method– Use N trial functions for weight functions

1

( ) ( )N

i ii

u x c xf=

=å% Trial function

1

0( ) ( ) 0, 1, ,iR x x dx i Nf = =ò K

21

20( ) ( ) 0, 1, ,i

d up x x dx i N

dxf

æ ö÷ç + = =÷ç ÷÷çè øò%

K

21 1

20 0( ) ( ) ( ) , 1, ,i i

d ux dx p x x dx i N

dxff =- =ò ò

%K

6

GALERKIN METHOD cont.

• Galerkin method cont.– Integration-by-parts: reduce the order of differentiation in u(x)

– Apply natural BC and rearrange

– Same order of differentiation for both trial function and approx. solution– Substitute the approximate solution

1 1 1

0 00

( ) ( ) , 1, ,ii i

du du ddx p x x dx i N

dx dx dxf

ff - = - =ò ò% %

K

1 1

0 0( ) ( ) (1) (1) (0) (0), 1, ,i

i i id du du du

dx p x x dx i Ndx dx dx dxf

ff f= + - =ò ò%

K

1 1

0 01

( ) ( ) (1) (1) (0) (0), 1, ,N

jij i i i

j

dd du duc dx p x x dx i N

dx dx dx dx

ffff f

=

= + - =åò ò K

7

GALERKIN METHOD cont.

• Galerkin method cont.– Write in matrix form

– Coefficient matrix is symmetric; Kij = Kji– N equations with N unknown coefficients

1

, 1, ,N

ij j ij

K c F i N=

= =å K( )( ) ( )1 1[ ] { } { }N N N N´ ´ ´

=K c F

1

0

jiij

ddK dx

dx dx

ff=ò

1

0( ) ( ) (1) (1) (0) (0)i i i i

du duF p x x dx

dx dxff f= + -ò

8

EXAMPLE1

• Differential equation Trial functions

• Approximate solution (satisfies the essential BC)

• Coefficient matrix and RHS vector

2

21 0,0 1

(0) 0Boundary conditions

(1) 1

d ux

dxu

dudx

+ = £ £

ü= ïïïýï= ïïþ

1 1

22 2

( ) ( ) 1

( ) ( ) 2

x x x

x x x x

ff

ff

¢= =

¢= =

22

1 21

( ) ( )i ii

u x c x c x c xf=

= = +å%

( )

( )

( )

1 211 1

0

1

12 21 1 20

1 222 2

0

1

1

43

K dx

K K dx

K dx

f

ff

f

¢= =

¢¢= = =

¢= =

ò

ò

ò

1

1 1 10

( ) (1) (0)du

F x dxdx

ff= + -ò 1

1

2 2 20

3(0)

2

( ) (1) (0)du

F x dxdx

f

ff

=

= + -ò 24

(0)3

f =

9

EXAMPLE1 cont.

• Matrix equation

• Approximate solution

– Approximate solution is also the exact solution because the linear combination of the trial functions can represent the exact solution

3 31[ ]

3 3 4

é ùê ú=ê úë û

K91

{ }6 8

ì üï ïï ï= í ýï ïï ïî þF 1

12

2{ } [ ] { }-

ì üï ïï ï= = í ýï ï-ï ïî þc K F

2

( ) 22x

u x x= -%

10

EXAMPLE2

• Differential equation Trial functions

• Coefficient matrix is same, force vector:

• Exact solution

– The trial functions cannot express the exact solution; thus, approximate solution is different from the exact one

2

20, 0 1

(0) 0Boundary conditions

(1) 1

d ux x

dxu

dudx

+ = £ £

ü= ïïïýï= ïïþ

1 1

22 2

( ) ( ) 1

( ) ( ) 2

x x x

x x x x

ff

ff

¢= =

¢= =

161{ }

12 15

ì üï ïï ï= í ýï ïï ïî þF

191 12

14

{ } [ ] { }-ì üï ïï ï= = í ýï ï-ï ïî þ

c K F219

( )12 4

xu x x= -%

33( )

2 6x

u x x= -

11

EXAMPLE2 cont.

• Approximation is good for u(x), but not good for du/dx

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

0 0.2 0.4 0.6 0.8 1x

u(x

), d

u/d

x

u-exact u-approx. du/dx (exact) du/dx (approx.)

12

HIGHER-ORDER DIFFERENTIAL EQUATIONS

• Fourth-order differential equation

– Beam bending under pressure load

• Approximate solution

• Weighted residual equation (Galerkin method)

– In order to make the order of differentiation same, integration-by-parts must be done twice

4

4( ) 0, 0

d wp x x L

dx- = £ £ 2

2

3

3

(0) 0Essential BC

(0) 0

( )Natural BC

( )

w

dwdx

d wL M

dx

d wL V

dx

ü= ïïïýï= ïïþüïï= ïïïýïïï= - ïïþ

1

( ) ( )N

i ii

w x c xf=

=å%

4

40( ) ( ) 0, 1, ,

L

id w

p x x dx i Ndx

fæ ö÷ç - = =÷ç ÷÷çè øò

%K

13

HIGHER-ORDER DE cont.

• After integration-by-parts twice

• Substitute approximate solution

– Do not substitute the approx. solution in the boundary terms

• Matrix form

3 2 2 2

3 2 2 20 00 0

( ) ( ) , 1, ,L L

L Li i

i id w d w d d w d

dx p x x dx i Ndxdx dx dx dx

ffff - + = =ò ò

% % %K

2 2 3 2

2 2 3 20 00 0

( ) ( ) , 1, ,L L

L Li i

i id w d d w d w d

dx p x x dx i Ndxdx dx dx dx

ffff= - + =ò ò

% % %K

2 2 3 2

2 2 3 20 01 0 0

( ) ( ) , 1, ,L LNL Lj i i

j i ij

d d d w d w dc dx p x x dx i N

dxdx dx dx dx

f ffff

=

= - + =åò ò% %

K

1 1[ ] { } { }N N N N´ ´ ´

=K c F

22

2 20

L jiij

ddK dx

dx dx

ff=ò

3 2

3 200 0

( ) ( )L L

Li

i i id w d w d

F p x x dxdxdx dx

fff= - +ò

14

EXMAPLE

• Fourth-order DE

• Two trial functions

• Coefficient matrix

4

41 0, 0

d wx L

dx- = £ £ 2 3

2 3

(0) 0 (0) 0

(1) 2 (1) 1

dww

dx

d w d w

dx dx

= =

= = -

2 31 2 1 2, 2, 6x x xff ff¢¢ ¢¢= = = =

( )

( )

( )

1 211 1

0

1

12 21 1 20

1 222 2

0

4

6

12

K dx

K K dx

K dx

f

ff

f

¢¢= =

¢¢¢¢= = =

¢¢= =

ò

ò

ò

4 6[ ]

6 12

é ùê ú=ê úë û

K

15

EXAMPLE cont.

• RHS

• Approximate solution

• Exact solution

312

1 1 130

(0)(1) (0)

d wF x dx V

dxff= + +ò

2

1 12

(0)(1) (0)

d wM

dxff¢ ¢+ -

313

2 2 230

163

(0)(1) (0)

d wF x dx V

dxff

=

= + +ò2

2 22

(0)(1) (0)

d wM

dxff¢ ¢+ -

294

=

2 341 1( )

24 4w x x x= -%

411 24

14

{ } [ ] { }-ì üï ïï ï= = í ýï ï-ï ïî þ

c K F

4 3 21 1 7( )

24 3 4w x x x x= - +

16

EXAMPLE cont.

-3

-2

-1

0

1

2

3

4

0 0.2 0.4 0.6 0.8 1x

w'',

w'''

w'' (exact) w'' (approx.) w''' (exact) w''' (approx.)

17

FINITE ELEMENT APPROXIMATION

• Domain Discretization– Weighted residual method is still difficult to obtain the trial functions

that satisfy the essential BC– FEM is to divide the entire domain into a set of simple sub-domains

(finite element) and share nodes with adjacent elements– Within a finite element, the solution is approximated in a simple

polynomial form

– When more number of finite elements are used, the approximated piecewise linear solution may converge to the analytical solution

Analytical solution

Approximate solution

x

u(x)

Finite elements

18

FINITE ELEMENT METHOD cont.

• Types of finite elements

1D 2D 3D

• Variational equation is imposed on each element.

One element

1 0.1 0.2 1

0 0 0.1 0.9dx dx dx dx= + + +ò ò ò òL

19

TRIAL SOLUTION

– Solution within an element is approximated using simple polynomials.

– i-th element is composed of two nodes: xi and xi+1. Since two unknowns are involved, linear polynomial can be used:

– The unknown coefficients, a0 and a1, will be expressed in terms of nodal solutions u(xi) and u(xi+1).

1 2 3 n 1 n+1 n

1 2 n 1 n

xi xi+1

il

0 1 1( ) , i iu x a a x x x x += + £ £%

20

TRIAL SOLUTION cont.

– Substitute two nodal values

– Express a0 and a1 in terms of ui and ui+1. Then, the solution is approximated by

– Solution for i-th element:

– Ni(x) and Ni+1(x): Shape Function or Interpolation Function

0 1

1 1 0 1 1

( )

( )i i i

i i i

u x u a a x

u x u a a x+ + +

ì = = +ïïíï = = +ïî

%

%

1

11( ) ( )

( ) ( )

( )

i i

i ii ii i

N x N x

x x x xu x u u

L L+

++

- -= +%144424443 1442443

1 1 1( ) ( ) ( ) ,i i i i i iu x N x u N x u x x x+ + += + £ £%

21

TRIAL SOLUTION cont.

• Observations– Solution u(x) is interpolated using its nodal values ui and ui+1.

– Ni(x) = 1 at node xi, and =0 at node xi+1.

– The solution is approximated by piecewise linear polynomial and its gradient is constant within an element.

– Stress and strain (derivative) are often averaged at the node.

Ni(x) Ni+1(x)

xi xi+1

xi xi+1 xi+2 xi xi+1 xi+2

ui ui+1

ui+2 dudx

u

22

GALERKIN METHOD

• Relation between interpolation functions and trial functions– 1D problem with linear interpolation

– Difference: the interpolation function does not exist in the entire domain, but it exists only in elements connected to the node

• Derivative

1

( ) ( )DN

i ii

u x u xf=

=å%

1

( 1) 11( 1)

( ) 11( )

1

0, 0

( ) ,

( )( ) ,

0,D

i

i ii ii i

ii i

i ii i

i N

x x

x xN x x x x

Lx

x xN x x x x

Lx x x

f

-

- ---

++

+

ì £ £ïïïï -ï = < £ïïï= íï -ï = < £ïïïïï < £ïî

1

1( 1)

1( )

1

0, 0

1,

( )1

,

0,D

i

i iii

i ii

i N

x x

x x xd x Ldx

x x xL

x x x

f

-

--

+

+

ì £ £ïïïïï < £ïïï= íïï - < £ïïïïï < £ïî

ix 1ix +1ix -

( )id xdxf

( )i xf

2ix -

( )1/ iL-

( )11/ iL -

1

23

EXAMPLE

• Solve using two equal-length elements

• Three nodes at x = 0, 0.5, 1.0; displ at nodes = u1, u2, u3

• Approximate solution

2

2

(0) 01 0,0 1 Boundary conditions

(1) 1

ud u

x dudx

dx

ü= ïïï+ = £ £ ýï= ïïþ

1 1 2 2 3 3( ) ( ) ( ) ( )u x u x u x u xff f= + +%

1 2

3

1 2 , 0 0.5 2 , 0 0.5( ) ( )

0, 0.5 1 2 2 , 0.5 1

0, 0 0.5( )

1 2 , 0.5 1

x x x xx x

x x x

xx

x x

ff

f

ì ì- £ £ £ £ï ïï ï= =í íï ï< £ - < £ï ïî îì £ £ïï= íï - + < £ïî

0

0.5

1

0 0.5 1x

f

f_1f_2f_3

24

EXAMPLE cont.

• Derivatives of interpolation functions

• Coefficient matrix

• RHS

1 2

3

2, 0 0.5 2, 0 0.5( ) ( )

0, 0.5 1 2, 0.5 1

0, 0 0.5( )

2, 0.5 1

x xd x d xdx dxx x

xd xdx x

ff

f

ì ì- £ £ £ £ï ïï ï= =í íï ï< £ - < £ï ïî îì £ £ïï= íï < £ïî

1 0.5 11 2

120 0 0.5

( 2)(2) (0)( 2) 2d d

K dx dx dxdx dxff

= = - + - = -ò ò ò1 0.5 1

2 222

0 0 0.54 4 4

d dK dx dx dx

dx dxff

= = + =ò ò ò

0.5 1

10 0.5

1 (1 2 ) 1 (0) (1)du

F x dx dxdx

= ´ - + ´ +ò ò 1 1(1) (0) (0) 0.25 (0)du dudx dx

ff - = -

0.5 1

2 20 0.5

2 (2 2 ) (1) (1)du

F xdx x dxdx

f= + - +ò ò 2(0) (0)dudx

f- 0.5=

25

EXAMPLE cont.

• Matrix equation

• Striking the 1st row and striking the 1st column (BC)

• Solve for u2 = 0.875, u3 = 1.5

• Approximate solution

– Piecewise linear solution

1 1

2

3

2 2 0

2 4 2 0.5

0 2 2 1.25

u F

u

u

é ùì ü ì ü- ï ï ï ïï ï ï ïê úï ï ï ïê ú- - =í ý í ýê úï ï ï ïï ï ï ïê ú- ï ï ï ïë ûî þ î þ

Consider it as unknown

2

3

4 2 0.5

2 2 1.25

u

u

ì üé ù ì ü- ï ï ï ïï ï ï ïê ú =í ý í ýï ï ï ïê ú- ï ïï ïë û î þî þ

1.75 , 0 0.5( )

0.25 1.25 , 0.5 1

x xu x

x x

ì £ £ïï= íï + £ £ïî%

26

EXAMPLE cont.

• Solution comparison• Approx. solution has about

8% error• Derivative shows a large

discrepancy• Approx. derivative is

constant as the solution is piecewise linear

0

0.4

0.8

1.2

1.6

0 0.2 0.4 0.6 0.8 1

u(x

)

x

u-exact

u-approx.

0

0.5

1

1.5

2

0 0.2 0.4 0.6 0.8 1

du

/dx

x

du/dx (exact)

du/dx (approx.)

27

FORMAL PROCEDURE

• Galerkin method is still not general enough for computer code• Apply Galerkin method to one element (e) at a time• Introduce a local coordinate

• Approximate solution within the element

ix jx

1( )N x 2( )N x

x ( )eL

Element e

(1 )i jx x xx x= - + ( )i i

ej i

x x x xx x L

x- -

= =-

1 2( ) ( ) ( )i ju x u N x u N x= +%

( )1

2

( ) 1

( )

N

N

x x

x x

= -

=

1 ( )

2 ( )

( ) 1

( )

ie

ie

x xN x

Lx x

N xL

æ ö- ÷ç= - ÷ç ÷çè ø

-=

1 1( )

2 2( )

1

1

e

e

dN dN ddx d dx LdN dN ddx d dx L

xx

xx

= =-

= =+

28

FORMAL PROCEDURE cont.

• Interpolation property

• Derivative of approx. solution

• Apply Galerkin method in the element level

1 1

2 2

( ) 1, ( ) 0

( ) 0, ( ) 1

i j

i j

N x N x

N x N x

= =

= =

( )

( )i i

j j

u x u

u x u

=

=

%

%

1 2i j

du dN dNu u

dx dx dx= +

%

1 11 2 1 2( )

2 2

1e

u udu dN dN dN dNdx dx dx d du uL x x

ì ü ì üê ú ê úï ï ï ïï ï ï ï= =ê ú ê úí ý í ýï ï ï ïê ú ê úë û ë ûï ï ï ïî þ î þ

%

( ) ( ) ( ) ( ) ( ) ( ), 1,2j j

i i

x xi

i j i j i i ix x

dN du du dudx p x N x dx x N x x N x i

dx dx dx dx= + - =ò ò

%

29

FORMAL PROCEDURE cont.

• Change variable from x to x

– Do not use approximate solution for boundary terms

• Element-level matrix equation

1 111 2 ( )( ) 0 02

1( ) ( )

( ) (1) ( ) (0), 1,2

i eie

j i i i

udN dN dNd L p x N d

d d d uL

du dux N x N i

dx dx

x x xx x x

ì üê ú ï ïï ï =ê ú í ýï ïê úë û ï ïî þ

+ - =

ò òg

{ }( ) ( ) ( )( )

[ ]{ }

( )

ie e e

j

dux

dxdu

xdx

ì üï ïï ï-ï ïï ï= +í ýï ïï ï+ï ïï ïî þ

k u f

21 1 2

1( )

( ) ( )202 2 2 1 2

1 11 1

1 1e

e e

dN dN dNd d d

dL LdN dN dN

d d d

x x xx

x x x´

é ùæ öê ú÷ç ÷ç ÷çê úè ø é ù-é ù ê ú ê ú= =ë û ê ú ê ú-ë ûæ öê ú÷ç ÷ê úç ÷çè øë û

òk

1 1( ) ( )

0 2

( ){ } ( )

( )e e N

L p x dN

xx

x

ì üï ïï ï= í ýï ïï ïî þòf

30

FORMAL PROCEDURE cont.

• Need to derive the element-level equation for all elements• Consider Elements 1 and 2 (connected at Node 2)

• Assembly

(1) (1) 11 111 12

2 221 222

( )

( )

duxu fk k dx

u f duk k xdx

ì üï ïï ï-é ù ï ïì ü ì üï ï ï ïï ï ï ï ï ïê ú = +í ý í ý í ýê ú ï ï ï ï ï ïï ï ï ïî þ î þë û ï ï+ï ïï ïî þ

(2) (2) 22 211 12

3 321 223

( )

( )

duxu fk k dx

u f duk k xdx

ì üï ïï ï-é ù ï ïì ü ì üï ï ï ïï ï ï ï ï ïê ú = +í ý í ý í ýê ú ï ï ï ï ï ïï ï ï ïî þ î þë û ï ï+ï ïï ïî þ

(1) (1) (1) 111 12 11(1) (1) (2) (2) (1) (2)

221 22 11 12 2 2(2) (2) (2)

321 22 3 3

( )0

0

0 ( )

duxk k fu dx

k k k k u f f

u duk k f xdx

ì üï ïï ï-é ù ì üï ï ï ïì üï ï ï ïê ú ï ïï ï ï ï ï ïê úï ï ï ï ï ï+ = + +í ý í ý í ýê úï ï ï ï ï ïê úï ï ï ï ï ïï ï ï ï ï ïê úî þ ï ï ï ïë û î þ ï ïï ïî þ

Vanished unknown term

31

FORMAL PROCEDURE cont.

• Assembly of NE elements (ND = NE + 1)

• Coefficient matrix [K] is singular; it will become non-singular after applying boundary conditions

( )

( )( )

(1) (1) (1)11 12 11(1) (1) (2) (2) (1) (2)21 22 11 12 2 22

(2) (2) (2) (2) (3)3221 22 11 3 3

( )( )21 22 1

0 0

0

0 0

0 0 0 E EE

D

D D

N NN N NN

N N

k k fuk k k k f fu

uk k k f f

u fk k´

´

é ùìê úì üï ïê úï ïï ïê úï ï+ +ê úï ïï ïê úï ï =í ý í+ +ê úï ïê úï ïï ïê úï ïê úï ïï ïê úï ïî þê úë û

K

L

LMM M M O M M

( )( )

1

11

( )

0

0

( )D

D

NN

N

dux

dx

dux

dx´´

ì üï ïï ïü -ï ï ï ïï ï ï ïï ï ï ïï ï ï ïï ï ï ïï ï ï ïï ïï ï ï ï+ý í ýï ï ï ïï ï ï ïï ï ï ïï ï ï ïï ï ï ïï ï ï ïï ï ï ï+ï ïî þ ï ïï ïî þ

M

[ ]{ } { }=K q F

32

EXAMPLE

• Use three equal-length elements

• All elements have the same coefficient matrix

• Change variable of p(x) = x to p(x):• RHS

2

20, 0 1 (0) 0, (1) 0

d ux x u u

dx+ = £ £ = =

( )( )2 2

1 1 3 31, ( 1,2,3)

1 1 3 3e

ee

L´

é ù é ù- -é ù ê ú ê ú= = =ë û ê ú ê ú- -ë û ë ûk

( ) (1 )i jp x xx x x= - +

( )[ ]1 11( ) ( ) ( )

0 02

( )

( ) 1{ } ( ) 1

( )

3 6 , ( 1,2,3)

6 3

e e ei j

ji

e

ji

NL p x d L x x d

N

xx

L exx

x xx x x x

x x

ì ü ì ü-ï ï ï ïï ï ï ï= = - +í ý í ýï ï ï ïï ïï ï î þî þì üï ïï ï+ï ïï ï= =í ýï ïï ï+ï ïï ïî þ

ò òf

33

EXAMPLE cont.

• RHS cont.

• Assembly

• Apply boundary conditions– Deleting 1st and 4th rows and columns

(2)(1) (3)21 3(2)(1) (3)3 42

1 4 71 1 1, ,

54 54 542 5 8

ff f

ff f

ì üì ü ì üï ïï ï ï ïì ü ì ü ì üï ï ï ï ï ïï ï ï ï ï ï ï ï ï ï ï ï= = =í ý í ý í ý í ý í ý í ýï ï ï ï ï ï ï ï ï ï ï ïï ï ï ï ï ïî þ î þ î þï ï ï ï ï ïî þî þ î þ

1

2

3

4

1(0)

543 3 0 0 2 43 3 3 3 0 54 54

0 3 3 3 3 7 554 540 0 3 38

(1)54

dudx

u

u

u

ududx

ì üï ïï ï-ï ïï ïï ïì üé ù- ï ï ï ïï ï ï ïê úï ï +ï ïï ïê ú- + - ï ïï ï ï ïê ú =í ý í ýê úï ï ï ï- + - ï ï ï ïê ú +ï ï ï ïï ï ï ïê ú- ï ï ï ïë ûî þ ï ïï ïï ï+ï ïï ïî þ

Element 1

Element 2

Element 3

2

3

6 3 1193 6 2

u

u

ì üé ù ì ü- ï ï ï ïï ï ï ïê ú =í ý í ýï ï ï ïê ú- ï ïï ïë û î þî þ

42 81

53 81

u

u

=

=

34

EXAMPLE cont.

• Approximate solution

• Exact solution

– Three element solutions are poor– Need more elements

4 1, 0

27 34 1 1 1 2

( ) ,81 27 3 3 3

5 5 2 2, 1

81 27 3 3

x x

u x x x

x x

ìïï £ £ïïïï æ öïï ÷ç= + - £ £í ÷ç ÷çè øïïïï æ öï ÷ç- - £ £ï ÷ç ÷çï è øïî

%

0

0.02

0.04

0.06

0.08

0 0.2 0.4 0.6 0.8 1x

u(x)

u-approx.u-exact

( )21( ) 1

6u x x x= -

35

ENERGY METHOD

• Powerful alternative method to obtain FE equations• Principle of virtual work for a particle

– for a particle in equilibrium the virtual work is identically equal to zero– Virtual work: work done by the (real) external forces through the virtual

displacements– Virtual displacement: small arbitrary (imaginary, not real) displacement

that is consistent with the kinematic constraints of the particle

• Force equilibrium

• Virtual displacements: du, dv, and dw• Virtual work

• If the virtual work is zero for arbitrary virtual displacements, then the particle is in equilibrium under the applied forces

0, 0, 0x y zF F F= = =å å å

0x y zW u F v F w Fd d d d= + + =å å å

36

PRINCIPLE OF VIRTUAL WORK

• Deformable body (uniaxial bar under body force and tip force)

• Equilibrium equation:• PVW

• Integrate over the area, axial force P(x) = As(x)

xE, A(x)

Bx

F

L

0xx

dB

dxs

+ = This is force equilibrium

0

( ) 0L

xx

A

dB u x dAdx

dxs

dæ ö÷ç + =÷ç ÷çè øòò

0

( ) 0L

xdP

b u x dxdx

dæ ö÷ç + =÷ç ÷çè øò

37

PVW cont.

• Integration by parts

– At x = 0, u(0) = 0. Thus, du(0) = 0– the virtual displacement should be consistent with the displacement

constraints of the body– At x = L, P(L) = F

• Virtual strain• PVW:

( )0

0 0

( ) 0L L

L

xd u

P u P dx b u x dxdxd

d d- + =ò ò

( )( )

d ux

dxd

de =

0 0

( ) ( ) ( )L L

xF u L b u x dx P x dxd d de+ =ò ò

iWd-eWd

0e iW Wd d+ =

38

PVW cont.

• in equilibrium, the sum of external and internal virtual work is zero for every virtual displacement field

• 3D PVW has the same form with different expressions• With distributed forces and concentrated forces

• Internal virtual work

( ) ( )e x y z xi i yi i zi iiS

W t u t v t w dS F u F v F wd d d d d d d= + + + + +åò

( )....i x x y y xy xy

V

W dVd s de s de t dg=- + + +ò

39

VARIATION OF A FUNCTION

• Virtual displacements in the previous section can be considered as a variation of real displacements

• Perturbation of displ u(x) by arbitrary virtual displ du(x)

• Variation of displacement

• Variation of a function f(u)

• The order of variation & differentiation can be interchangeable

( ) ( ) ( )u x u x u xt td= +

0

( )( )

du xu x

dt

td

t == Displacement variation

0

( )df u dff u

d dut

td d

t == =

( )x

du d udx dx

dde d

æ ö÷ç= =÷ç ÷çè ø

40

PRINCIPLE OF MINIMUM POTENTIAL ENERGY

• Strain energy density of 1D body

• Variation in the strain energy density by du(x)

• Variation of strain energy

20

1 12 2x x xU Es e e= =

0 x x x xU Ed e de s de= =

0

0 0 0

L L L

x x x

A A

U U dAdx dAdx P dxd d s de de= = =òò òò ò

iU Wd d=-

41

PMPE cont.

• Potential energy of external forces– Force F is applied at x = L with corresponding virtual displ du(L)– Work done by the force = Fdu(L)– The potential is reduced by the amount of work

– With distributed forces and concentrated force

• PVW

– Define total potential energy

( )V F u Ld d=- ( )( )V Fu Ld d=- F is constant virtual displacement

0( ) ( )

L

xV Fu L b u x dx=- - ò eV Wd d=-

( )0 or 0U V U Vd d d+ = + =

U VP = +

0dP =

42

EXAMPLE: PMPE TO DISCRETE SYSTEMS

• Express U and V in terms of displacements, and thendifferential P w.r.t displacements

• k(1) = 100 N/mm, k(2) = 200 N/mmk(3) = 150 N/mm, F2 = 1,000 NF3 = 500 N

• Strain energy of elements (springs)

F3

3

2

1

2

3

1F3

u1

u2

u3

( )2(1) (1)2 1

12

U k u u= -( )

( ) ( )

(1) (1)1(1)

1 2 (1) (1)1 2 2

2 12 2

12

uk kU u u

uk k´´´

é ùì ü- ï ïï ïê úê ú= í ýë ûê úï ï- ï ïî þë û(2) (2)

1(2)1 3 (2) (2)

3

12

uk kU u u

uk k

é ùì ü- ï ïï ïê úê ú= í ýë ûê úï ï- ï ïî þë û

(3) (3)2(3)

2 3 (3) (3)3

12

uk kU u u

uk k

é ùì ü- ï ïï ïê úê ú= í ýë ûê úï ï- ï ïî þë û

43

EXAMPLE cont.

• Strain energy of the system

• Potential energy of applied forces

• Total potential energy

3( )

1

e

e

U U=

=å

(1) (2) (1) (2)1

(1) (1) (3) (3)1 2 3 2

(2) (3) (2) (3)3

12

k k k k u

U u u u k k k k u

k k k k u

é ùì ü+ - - ï ïê úï ïï ïê úê ú= - + - í ýë ûê úï ïï ïê ú- - + ï ïê úî þë û1

{ } [ ]{ }2

TU = Q K Q

( )1

1 1 2 2 3 3 1 2 3 2

3

{ } { }T

F

V Fu F u F u u u u F

F

ì üï ïï ïï ïê ú= - + + = - = -í ýë ûï ïï ïï ïî þ

Q F

1{ } [ ]{ } { } { }

2T TU VP = + = -Q K Q Q F

1 2 3{ } { , , }Tu u u=Q

44

EXAMPLE cont.

• Total potential energy is minimized with respect to the DOFs

• Global FE equations

• Forces in the springs

{ }1 2 30, 0, 0 or, 0

u u u¶P ¶P ¶P ¶P

= = = =¶ ¶ ¶ ¶ Q

[ ]1 1

2 2

3 3

u F

u F

u F

ì ü ì üï ï ï ïï ï ï ïï ï ï ï=í ý í ýï ï ï ïï ï ï ïï ï ï ïî þ î þ

K

1

2

3

300 100 200 0

100 250 150 1,000

200 150 350 500

F

u

u

é ùì ü ì ü- - ï ï ï ïï ï ï ïê úï ï ï ïê ú- - =í ý í ýê úï ï ï ïï ï ï ïê ú- - ï ï ï ïë ûî þ î þ

[ ]{ } { }=K Q F

Finite element equations

2

3

1

6.538mm

4.231mm

1,500N

u

u

F

=

=

=-

( )( ) ( )e ej iP k u u= -

( ) ( )

( )

(1) (1) (2) (2)2 1 3 1

(3) (3)3 2

654N 846N

346N

P k u u P k u u

P k u u

= - = = - =

= - = -

45

RAYLEIGH-RITZ METHOD

• PMPE is good for discrete system (exact solution)• Rayleigh-Ritz method approximates a continuous system as a

discrete system with finite number of DOFs• Approximate the displacements by a function containing finite

number of coefficients • Apply PMPE to determine the coefficients that minimizes the

total potential energy• Assumed displacement (must satisfy the essential BC)

• Total potential energy in terms of unknown coefficients

• PMPE

1 1( ) ( ) ( )n nu x c f x c f x= + +L

1 2( , ,... )nc c c U VP = +

0, 1,i

i nc¶P

= =¶

K

46

EXAMPLE

• L = 1m, A = 100mm2, E = 100 GPa, F = 10kN, bx = 10kN/m• Approximate solution• Strain energy

• Potential energy of forces

F

bx

21 2( )u x c x c x= +

22

0 0 0

1 1( )

2 2

L L L

L xdu

U U x dx AE dx AE dxdx

eæ ö÷ç= = = ÷ç ÷çè øò ò ò

( ) ( )2 2 2 3 21 2 1 2 1 1 2 2

0

1 1 4, 2 2

2 2 3

LU c c AE c c x dx AE Lc L c c L c

æ ö÷ç= + = + + ÷ç ÷çè øò

( ) ( ) ( )2 21 2 1 2 1 2

0 0

2 32

1 2

, ( ) ( ) ( ) ( )

2 3

L L

x x

x x

V c c b x u x dx F u L b c x c x dx F c L c L

L Lc FL b c FL b

=- - - = - + + +

æ ö æ ö÷ ÷ç ç= - + -÷ ÷ç ç÷ ÷÷ ÷ç çè ø è ø

ò ò

47

EXAMPLE cont.

• PMPE

• Approximate solution• Axial force• Reaction force

2 22

1 21

32 3 2

1 22

40

2 2

40

3 3

x

x

L b b acAELc AEL c FL b

c a

LAEL c AEL c FL b

c

¶P - ± -= + + - =

¶

¶P= + + - =

¶

( )1 2,c c U VP = +

7 71 2

77

1 2

10 10 5,000

4 1010 6,667

3

c c

c c

+ =-

´+ = -

1

32

0

0.5 10

c

c -

=

=- ´

3 2( ) 0.5 10u x x-=- ´

( ) / 10,000P x AEdu dx x= =-

(0) 0R P=- =