Post on 18-Jan-2016
transcript
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=- =