CHAP 3 WEIGHTED RESIDUAL AND
ENERGY METHOD FOR 1D PROBLEMS
FINITE ELEMENT ANALYSIS AND DESIGN
Nam-Ho Kim
1
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
2
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)
d 2u
+ p( x ) = 0, 0 £ x £ 1
2
dx
u(0) = 0 ü
ïï
ïý Boundary conditions
du
(1) = 1ïï
ïþ
dx
– 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
3
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: d 2u%
+ p( x ) = R( x )
2
dx
– Want to minimize the residual by multiplying with a weight W and
integrate over the domain
1
ò0 R( x )W ( x )dx =
0
Weight function
– 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
4
GALERKIN METHOD
• Approximate solution is a linear combination of trial functions
N
u%( x ) =
å
ci f i ( x )
Trial function
i= 1
– 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
ò0 R( x )f i ( x )dx =
0, i = 1,K , N
1æd 2u
%
ö÷
çç
ò0 çè dx 2 + p( x ) ÷÷øf i ( x )dx = 0, i = 1,K , N
1 d 2u
%
1
ò0 dx 2 f i ( x )dx = - ò0 p( x )f i ( x )dx,
i = 1,K , N
5
GALERKIN METHOD cont.
• Galerkin method cont.
– Integration-by-parts: reduce the order of differentiation in u(x)
du%
f
dx i
1
0
1 du
%d f
1
ò0 dx dx dx = - ò0 p( x )f i ( x )dx,
i
i = 1,K , N
– Apply natural BC and rearrange
du%
dx =
dx dx
1 df
ò0
i
1
ò0
p( x )f i ( x )dx +
du
du
(1)f i (1) (0)f i (0), i = 1,K , N
dx
dx
– Same order of differentiation for both trial function and approx. solution
– Substitute the approximate solution
ò0
N
df j
c
dx =
j
å
dx j = 1 dx
1 df
i
1
ò0
p( x )f i ( x )dx +
du
du
(1)f i (1) (0)f i (0), i = 1,K , N
dx
dx
6
GALERKIN METHOD cont.
• Galerkin method cont.
– Write in matrix form
N
å
K ij c j = Fi , i = 1,K , N
j= 1
K ij =
Fi =
1
ò0
1
ò0
[K] {c} = {F}
(N´ N )(N´ 1)
(N´ 1)
df i df j
dx
dx dx
p( x )f i ( x )dx +
du
du
(1)f i (1) (0)f i (0)
dx
dx
– Coefficient matrix is symmetric; Kij = Kji
– N equations with N unknown coefficients
7
EXAMPLE1
• Differential equation
Trial functions
d 2u
+ 1 = 0,0 £ x £ 1
2
dx
u(0) = 0 ü
ïï
ïý Boundary conditions
du
(1) = 1ïï
ïþ
dx
f 1( x ) = x
f 1¢( x ) = 1
f 2(x) = x2
f 2¢( x ) = 2 x
• Approximate solution (satisfies the essential BC)
2
u%( x ) =
å
ci f i ( x ) = c1x + c2 x 2
i= 1
• Coefficient matrix and RHS vector
K11 =
1
ò0
2
(f 1¢) dx = 1
K12 = K 21 =
K 22 =
1
ò0
1
ò0 (f 1¢f 2¢)dx = 1
2
(f 2¢) dx =
4
3
du
3
(0)
f
(0)
=
1
ò0
dx
2
1
du
4
F2 = ò f 2 ( x )dx + f 2 (1) (0)f 2 (0) =
0
dx
3
F1 =
1
f 1( x )dx + f 1(1) -
8
EXAMPLE1 cont.
• Matrix equation
1 é3 3 ù
ú
[K] = ê
3 êë3 4 ú
û
1 ìï 9 ü
ï
{F} = í ý
6 ïïî 8 ïïþ
ìïï 2 ü
ï
{c} = [K] {F} = í 1 ïý
ïîï - 2 ïþ
ï
- 1
• Approximate solution
x2
u%( x ) = 2x 2
– Approximate solution is also the exact solution because the linear
combination of the trial functions can represent the exact solution
9
EXAMPLE2
• Differential equation
d 2u
+ x = 0,
0£ x£ 1
2
dx
u(0) = 0 ü
ïï
ïý Boundary conditions
du
(1) = 1ïï
ïþ
dx
Trial functions
f 1( x ) = x
f 1¢( x ) = 1
f 2(x) = x2
f 2¢( x ) = 2 x
• Coefficient matrix is same, force vector:
ìï 19 ü
ï
{c} = [K] {F} = ïí 121 ïý
ïîï - 4 ïþ
ï
- 1
1 ìï 16 ü
ï
{F} =
í ý
12 ïïî 15 ïïþ
19
x2
u%( x ) =
x12
4
• Exact solution
3
x3
u( x ) = x 2
6
– The trial functions cannot express the exact solution; thus,
approximate solution is different from the exact one
10
EXAMPLE2 cont.
• Approximation is good for u(x), but not good for du/dx
1.6
1.4
u(x), du/dx
1.2
1.0
0.8
0.6
0.4
0.2
0.0
0
u-exact
0.2
u-approx.
0.4
x
0.6
du/dx (exact)
0.8
1
du/dx (approx.)
11
HIGHER-ORDER DIFFERENTIAL EQUATIONS
• Fourth-order differential equation
d 4w
- p( x ) = 0,
4
dx
0£ x£ L
– Beam bending under pressure load
• Approximate solution
N
w%( x ) =
å
w (0) =
dw
(0) =
dx
d 2w
(L ) =
2
dx
d 3w
(L ) =
3
dx
0 ü
ïï
ïý Essential BC
0 ïï
ïþ
ü
ï
M ïï
ïï
ý Natural BC
ïï
- V ïï
ïþ
ci f i ( x )
i= 1
• Weighted residual equation (Galerkin method)
L æd 4w
%
ö
÷
çç
ò0 çè dx 4 - p( x ) ÷÷øf i ( x )dx = 0, i = 1,K , N
– In order to make the order of differentiation same, integration-by-parts
must be done twice
12
HIGHER-ORDER DE cont.
• After integration-by-parts twice
3
%
d w
f
3 i
dx
L
2%
d w df i
dx 2 dx
0
L d 2w
%d 2f
ò0
dx
2
dx
i
2
dx =
L
+
0
L
ò0
L d 2w
%d 2f
ò0
dx
2
dx
i
2
dx =
3
L
ò0
d w%
p( x )f i ( x )dx f
3 i
dx
p( x )f i ( x )dx, i = 1,K , N
L
+
0
2%
d w df i
dx 2 dx
L
, i = 1,K , N
0
• Substitute approximate solution
L
ò0
N
d 2f j d 2f i
å c j dx 2 dx 2 dx =
j= 1
L
ò0
3
d w%
p( x )f i ( x )dx f
3 i
dx
L
2%
d w df i
dx 2 dx
+
0
L
, i = 1,K , N
0
– Do not substitute the approx. solution in the boundary terms
• Matrix form
K ij =
[K] {c} = {F}
N´ N N´ 1
L d 2f
d 2f
dx
dx
ò0
i
2
j
2
dx
N´ 1
Fi =
L
ò0
3
p( x )f i ( x )dx -
d w
f
3 i
dx
L
2
+
0
d w df i
dx 2 dx
L
0
13
EXMAPLE
dw
(0) = 0
dx
d 2w
d 3w
(1) = 2
(1) = - 1
2
3
dx
dx
• Fourth-order DE
d 4w
- 1 = 0,
4
dx
w (0) = 0
0£ x£ L
• Two trial functions
f 1 = x 2, f 2 = x 3
f 1¢¢= 2, f 2¢¢= 6 x
• Coefficient matrix
K11 =
1
ò0
2
(f 1¢¢) dx = 4
K12 = K 21 =
K 22 =
1
ò0
1
ò0
(f 1¢f¢¢
2¢)dx = 6
é4 6 ù
ú
[K] = ê
êë6 12 ú
û
2
(f 2¢¢) dx = 12
14
EXAMPLE cont.
• RHS
F1 =
F2 =
1
ò0
3
2
d
w
(0)
d
w (0)
16
2
¢
¢
x dx + V f 1(1) +
f 1(0) + Mf 1(1) f 1(0) =
3
2
3
dx
dx
1
ò0
d 3w (0)
d 2w (0)
29
¢
¢
x dx + V f 2 (1) +
f
(0)
+
M
f
(1)
f
(0)
=
2
2
2
4
dx 3
dx 2
3
• Approximate solution
41 ü
ì
ï
- 1
ï
24 ïï
{c} = [K] {F} = í 1 ý
ïîï - 4 ïþ
ï
%( x ) =
w
41 2 1 3
x - x
24
4
• Exact solution
w(x) =
1 4 1 3 7 2
x - x + x
24
3
4
15
EXAMPLE cont.
4
3
w'', w'''
2
1
0
-1
-2
-3
0
w'' (exact)
0.2
0.4
w'' (approx.)
x
0.6
w''' (exact)
0.8
1
w''' (approx.)
16
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
u(x)
Approximate
solution
x
Finite
elements
Analytical
solution
– When more number of finite elements are used, the approximated
piecewise linear solution may converge to the analytical solution
17
FINITE ELEMENT METHOD cont.
• Types of finite elements
1D
2D
3D
• Variational equation is imposed on each element.
1
ò0
dx =
0.1
ò0
dx +
0.2
ò0.1
dx + L +
1
ò0.9
dx
One element
18
TRIAL SOLUTION
– Solution within an element is approximated using simple polynomials.
1
1
2
2
n
n1
3
n1
xi
n
n+1
xi+1
li
– i-th element is composed of two nodes: xi and xi+1. Since two
unknowns are involved, linear polynomial can be used:
u%( x ) = a0 + a1x,
xi £ x £ xi + 1
– The unknown coefficients, a0 and a1, will be expressed in terms of
nodal solutions u(xi) and u(xi+1).
19
TRIAL SOLUTION cont.
– Substitute two nodal values
ìïï u%( xi ) = ui = a0 + a1xi
í
ïïî u%( xi + 1) = ui + 1 = a0 + a1xi + 1
– Express a0 and a1 in terms of ui and ui+1. Then, the solution is
approximated by
u%( x ) =
xi + 1 - x
x - xi
u
+
ui + 1
i
(i )
(i )
L 4443
L 443
14442
1442
Ni ( x )
Ni + 1( x )
– Solution for i-th element:
u%( x ) = Ni ( x )ui + Ni + 1( x )ui + 1,
xi £ x £ xi + 1
– Ni(x) and Ni+1(x): Shape Function or Interpolation Function
20
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.
Ni(x)
Ni+1(x)
xi
xi+1
– The solution is approximated by piecewise linear polynomial and its
gradient is constant within an element.
u
ui+2
ui
xi
ui+1
xi+1
xi+2
du
dx
xi
xi+1
xi+2
– Stress and strain (derivative) are often averaged at the node.
21
GALERKIN METHOD
• Relation between interpolation functions and trial functions
– 1D problem with linear interpolation
ND
u%( x ) =
å
i= 1
ui f i ( x )
0 £ x £ xi - 1
ïìï 0,
ïï
ïï Ni( i - 1) ( x ) = x - xi - 1 , xi - 1 < x £ xi
ïï
L( i - 1)
f i (x) = í
ïï ( i )
xi + 1 - x
N
(
x
)
=
, xi < x £ xi + 1
ïï i
(i )
L
ïï
ïïî 0,
xi + 1 < x £ xND
– Difference: the interpolation function does not exist in the entire
domain, but it exists only in elements connected to the node
• Derivative
ìï 0,
ïï
ïï 1
,
ï
d f i ( x ) ïï L( i - 1)
= í
dx
ïï
1
,
ïï
(i )
ïï L
ïïî 0,
0 £ x £ xi - 1
xi - 1 < x £ xi
1
f i (x )
1/ L(i - 1)
xi -
xi < x £ xi + 1 - 1/ L(i )
xi + 1 < x £ xND
2
xi-
1
xi
xi + 1
d f i (x )
dx
22
EXAMPLE
• Solve using two equal-length elements
d 2u
+ 1 = 0,0 £ x £ 1
2
dx
u(0) = 0 ü
ïï
ïý Boundary conditions
du
(1) = 1ïï
ïþ
dx
• Three nodes at x = 0, 0.5, 1.0; displ at nodes = u1, u2, u3
• Approximate solution u%( x ) = u1f 1( x ) + u2f 2 ( x ) + u3f 3 ( x )
ìï 1 - 2 x, 0 £ x £ 0.5
f 1( x ) = í
ïîï 0,
0.5 < x £ 1
ìï 0,
0 £ x £ 0.5
f 3(x) = í
ïïî - 1 + 2 x,
0.5 < x £ 1
ìï 2 x, 0 £ x £ 0.5
f 2(x) = í
ïîï 2 - 2 x, 0.5 < x £ 1
1
f_1
f_2
f 0.5
f_3
0
0
0.5
x
1
23
EXAMPLE cont.
• Derivatives of interpolation functions
d f 1( x )
=
dx
ìï - 2,
0 £ x £ 0.5
í
ïîï 0, 0.5 < x £ 1
df 3 ( x )
=
dx
ìï 0, 0 £ x £ 0.5
í
ïïî 2, 0.5 < x £ 1
df 2 ( x )
=
dx
ìï 2, 0 £ x £ 0.5
í
ïîï - 2, 0.5 < x £ 1
• Coefficient matrix
0.5
1
df 1 df 2
K12 = ò
dx = ò (- 2)(2)dx + ò (0)(- 2)dx = - 2
0 dx dx
0
0.5
1 df df
0.5
1
2
2
K 22 = ò
dx = ò 4dx + ò 4dx = 4
0 dx dx
0
0.5
1
• RHS
F1 =
F2 =
0.5
ò0
1´ (1 - 2 x )dx +
0.5
ò0
2 xdx +
1
ò0.5
1
ò0.5
1´ (0)dx +
(2 - 2 x )dx +
du
du
du
(1)f 1(1) (0)f 1(0) = 0.25 (0)
dx
dx
dx
du
du
(1)f 2 (1) (0)f 2 (0) = 0.5
dx
dx
24
EXAMPLE cont.
• Matrix equation
é 2 - 2 0 ùìï u1 ü
ï
ê
úïï ïï
ê- 2 4 - 2 úí u2 ý =
ê
úïï ïï
êë 0 - 2 2 û
úîï u3 þ
ï
ìï F1 ü
ïï
ïï
ï
í 0.5 ý
ïï
ïï
ï
îï 1.25 þ
Consider it as unknown
• Striking the 1st row and striking the 1st column (BC)
é 4 - 2 ùìï u2 ü
ï
ê
úí ý =
êë- 2 2 û
úïïî u3 ïþ
ï
ìï 0.5 ü
ï
í
ý
ïîï 1.25 ïþ
ï
• Solve for u2 = 0.875, u3 = 1.5
• Approximate solution
ìï 1.75 x,
%
u( x ) = í
ïïî 0.25 + 1.25 x,
0 £ x £ 0.5
0.5 £ x £ 1
– Piecewise linear solution
25
EXAMPLE cont.
1.6
u(x)
1.2
0.8
u-exact
0.4
u-approx.
0
0
0.2
0.4
x
0.6
0.8
1
2
1.5
du/dx
• Solution comparison
• Approx. solution has about
8% error
• Derivative shows a large
discrepancy
• Approx. derivative is
constant as the solution is
piecewise linear
1
du/dx (exact)
0.5
du/dx (approx.)
0
0
0.2
0.4
x
0.6
0.8
1
26
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
x=
x = xi (1 - x ) + x j x
x - xi
x - xi
=
x j - xi
L(e )
• Approximate solution within the element
N 1( x) N 2 ( x)
u%( x ) = ui N1( x ) + u j N2 ( x )
N1(x ) = (1 - x )
Element e
N2 (x ) = x
æ x - xi
N1( x ) = çç1 è
L( e )
x - xi
N2 ( x ) =
L( e )
xi
ö
÷
÷
÷
ø
xj
x
dN1
=
dx
dN2
=
dx
dN1 d x
1
= - (e )
d x dx
L
dN2 d x
1
= + (e )
d x dx
L
L(e )
27
FORMAL PROCEDURE cont.
• Interpolation property
N1( xi ) = 1, N1( x j ) = 0
N2 ( xi ) = 0, N2 ( x j ) = 1
u%( xi ) = ui
u%( x j ) = u j
• Derivative of approx. solution
du%
=
dx
du%
=
dx
dN1
dN2
+ uj
dx
dx
êdN1 dN2 úìï u1 ü
1 êdN1
ï
ê
úí ý = (e ) ê
dx
dx
úïîï u2 ïþ
L ëê d x
ëê
û
ï
ui
dN2 úìï u1 ü
ï
úí ý
dx û
úïîï u2 ïþ
ï
• Apply Galerkin method in the element level
xj
òx
i
dNi du%
dx =
dx dx
xj
òx
i
du
du
p( x )Ni ( x )dx +
( x )N ( x ) ( x )N ( x ), i = 1,2
dx j i j
dx i i i
28
FORMAL PROCEDURE cont.
• Change variable from x to x
1 1 dNi
L( e ) ò0 d x
êdN1
ê
êë d x
1
dN2 ú ìï u1 ü
ï
(e )
úd x gí ý = L ò p( x )Ni (x )d x
0
dx ú
û ïîï u2 ïþ
ï
du
du
+
( x )N (1) ( x )N (0), i = 1,2
dx j i
dx i i
– Do not use approximate solution for boundary terms
• Element-level matrix equation
ìï
ïï [k( e ) ]{u( e ) } = {f ( e ) }+ ïí
ïï
ïï +
î
ü
du
( xi ) ïïï
dx
ïý
ï
du
( x j ) ïï
ïþ
dx
{f
(e )
(e )
}= L
é ædN ö2
êç 1 ÷
÷
÷
1ê ç
è
ø
d
x
1
(
e
)
ék ù=
ê
ë
û L( e ) ò0 ê
2´ 2
êdN2 dN1
ê dx dx
ë
1
ò0
ìï N1(x ) üï
p( x ) í
ýd x
ïïî N2 (x ) ïïþ
dN1 dN2 ùú
é 1 - 1ù
dx dx ú
úd x = 1 ê
ú
(e ) ê
2 ú
L ë- 1 1 úû
ædN2 ö÷ ú
ç
÷ úû
çè d x ø÷
29
FORMAL PROCEDURE cont.
• Need to derive the element-level equation for all elements
• Consider Elements 1 and 2 (connected at Node 2)
ék11
ê
êëk21
ék11
ê
êëk21
ìï du
üï
ï
(
x
)
ïï dx 1 ïïï
k12 ù ìï u1 üï ìï f1 üï
ú í ý= í ý + í
ý
ïï du
ïï
k22 úû ïîï u2 ïþï ïîï f2 ïþï
+
(
x
)
ïï
ï
î dx 2 ïþ
ìï du
ü
ïï
(2)
(2)
ï
(
x
)
2
ï dx
ï
ìï f2 ü
k12 ù ìï u2 ü
ï
ï
ï
ï
ú í ý= í ý + í
ý
ï
ï
ï
ï
ï
ï
u
f
du
k22 ú
ï îï 3 þ
ï
û îï 3 þ
ïï +
( x3 ) ïï
ïî dx
ïþ
(1)
• Assembly
ék (1)
ê 11
êk (1)
ê 21
ê
ê0
ë
(1)
k12
(1)
(2)
k22
+ k11
(2)
k21
(1)
Vanished
ì
ü
du
ï
ï
0 ùúìï u1 üï ìïï
f1(1) üïï ïï - dx ( x1 ) ïï unknown term
ïï
ï ïï ïï (1)
ï ïïï
(2) úï
(2) ï
k12 úí u2 ý = í f2 + f2 ý + í
0
ýï
ï ï ïï
ïï ïï
ïï
(2) úïï u ïï
(2)
ïï ïï du
ï
k22 ûúî 3 þ ïîï
f3
( x3 ) ïï
þ ï
îï dx
þï
30
FORMAL PROCEDURE cont.
• Assembly of NE elements (ND = NE + 1)
é (1)
(1)
êk11
k12
ê
êk (1) k (1) + k (2)
22
11
ê 21
ê
(2)
k221
ê0
ê
M
ê M
ê
ê0
0
êë
0
K
(2)
k12
L
(2)
(2)
k22
+ k11
M
L
O
0
(NE )
k21
(ND´ ND )
ù
0 úìï u ü
úï 1 ïï
ïï
ïï
0 ú
u
úï 2 ï
ïï
úïï
u
í
ý
0 ú 3 =
ïï
úïï
M úïï Mïï
úï
ï
( NE ) úïï u ïï
k22 úî N þ
û(ND´ 1)
(1)
ìï
ïïü
f
1
ïï
ïï f (1) + f (2) ïïï
2 ï
ïï 2
ïí (2)
(3) ïï +
ý
ïï f3 + f3 ïï
ïï
M ïï
ïï
ïï
( NE )
ïîï fN
ïþ
ï
(ND´ 1)
ïìï ïï
ïï
ï
ïíï
ïï
ïï
ïï
ïï +
îï
ü
du
( x1) ïïï
dx
ïï
ïï
0
ïýï
0
ï
M ïïï
ïï
du
( xN ) ïï
ï
dx
þ
(ND´ 1)
[K]{q} = {F}
• Coefficient matrix [K] is singular; it will become non-singular
after applying boundary conditions
31
EXAMPLE
• Use three equal-length elements
d 2u
+ x = 0,
2
dx
0£ x£ 1
u(0) = 0, u(1) = 0
• All elements have the same coefficient matrix
é 1 - 1ù
ék( e ) ù = 1 ê
=
ë
û2´ 2 L( e ) ê- 1 1 ú
ú
ë
û
é 3 - 3ù
ê
ú, (e = 1,2,3)
êë- 3 3 ú
û
• Change variable of p(x) = x to p(x):
• RHS
{f
(e )
(e )
}= L
1
ò0
ìï
ïï
= L( e ) ïí
ïï
ïï
î
p(x ) = xi (1 - x ) + x j x
1
ìï N1(x ) üï
ìï 1 - x üï
(e )
p( x ) í
ý d x = L ò [xi (1 - x ) + x j x ]í
ýdx
0
ïîï N2 (x ) ïþï
ïîï x ïþï
x j üï
xi
ïï
+
3
6ï
ý,
x j ïï
xi
+
ï
6
3 ïþ
(e = 1,2,3)
32
EXAMPLE cont.
• RHS cont.
• Assembly
é 3 - 3
ê
ê- 3 3 + 3
ê
ê0
- 3
ê
êë 0
0
ìï f (1) üï
ïí 1 ïý = 1 íìï 1 ýüï ,
54 ïîï 2 ïþï
ïï f (1) ïï
î 2 þ
0
- 3
3+ 3
- 3
ìï f (2) üï
ïí 2 ïý = 1 íìï 4 ýüï ,
54 ïîï 5 ïþï
ïï f (2) ïï
î 3 þ
ìï
ïï
ïï
ì
ü
0 ùï u1 ï ï
úïï ïï ïï
0 úïï u2 ïï ïï
úí ý = í
- 3 úïï u3 ïï ïï
úï ï ï
3ú
ûïïî u4 ïïþ ïïï
ïï
ïîï
• Apply boundary conditions
1 du ü
(0) ïïï
54 dx
ïï
ïï
2
4
+
ï
54 54 ïï
ý
ïï
7
5
+
ï
54 54 ïï
8
du ïïï
+
(1)
ï
54 dx ïþ
ìï f (3) üï
ïí 3 ïý = 1 íìï 7 ýüï
54 ïîï 8 ïþï
ïï f (3) ïï
î 4 þ
Element 1
Element 2
Element 3
– Deleting 1st and 4th rows and columns
é 6 - 3 ùìï u2 ü
1 ìï 1 ü
ï
ï
ê
úí ý = í ý
9 îïï 2 þ
ïï
êë- 3 6 û
úïïî u3 ïþ
ï
u2 =
4
81
u3 =
5
81
33
EXAMPLE cont.
• Approximate solution
4
x,
27
4
1æ
çx +
81 27 çè
5
5æ
çx 81 27 çè
1
3
ö 1
1÷
2
,
£
x
£
÷
÷ 3
3ø
3
ö 2
2÷
÷
÷, 3 £ x £ 1
3ø
0£ x£
• Exact solution
0.08
u-approx.
u-exact
0.06
u(x)
ìï
ïï
ïï
ïï
u%( x ) = ïí
ïï
ïï
ïï
ïïî
0.04
0.02
0
0
0.2
0.4
x
0.6
0.8
1
1
u( x ) = x (1 - x 2 )
6
– Three element solutions are poor
– Need more elements
34
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
å
Fx = 0,
å
Fy = 0,
å
Fz = 0
• Virtual displacements: du, dv, and dw
• Virtual work
dW = du å Fx + dv å Fy + dw å Fz = 0
• If the virtual work is zero for arbitrary virtual displacements,
then the particle is in equilibrium under the applied forces
35
PRINCIPLE OF VIRTUAL WORK
• Deformable body (uniaxial bar under body force and tip force)
x
E, A(x)
F
Bx
L
ds x
+ Bx = 0
• Equilibrium equation:
dx
• PVW
This is force equilibrium
L
æd s x
ö
ç
ò ò çè dx + Bx ø÷÷÷du( x )dAdx = 0
0 A
• Integrate over the area, axial force P(x) = As(x)
L
ædP
ö
÷
ç
ò çè dx + bx ø÷÷du( x )dx = 0
0
36
PVW cont.
• Integration by parts
L
L
L
d (du )
dx + ò bx du( x )dx = 0
dx
0
0
0
– 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 de( x ) = d (du )
dx
Pdu -
òP
• PVW:
L
F du(L ) +
L
ò bx du( x )dx = ò Pde( x )dx
0
dWe
0
- dWi
dWe + dWi = 0
37
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
dWe =
ò (t x du + t y dv + tzdw )dS + åi (Fxi dui + Fyi dv i + Fzi dw i )
S
• Internal virtual work
dWi = -
ò (s x dex + s y dey + .... + t xy dg xy )dV
V
38
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)
ut ( x ) = u( x ) + t du( x )
• Variation of displacement
dut ( x )
= du( x )
dt t = 0
Displacement variation
• Variation of a function f(u)
df =
df (ut )
df
=
du
d t t = 0 du
• The order of variation & differentiation can be interchangeable
ædu ö d (du )
dex = d çç ÷
÷ = dx
è dx ø÷
39
PRINCIPLE OF MINIMUM POTENTIAL ENERGY
• Strain energy density of 1D body
U0 =
1
1
s x ex = E ex2
2
2
• Variation in the strain energy density by du(x)
dU0 = E ex dex = s x dex
• Variation of strain energy
L
dU =
L
L
ò ò dU0dAdx = ò ò s x dex dAdx = ò Pdex dx
0 A
0 A
0
dU = - dWi
40
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
dV = - F du(L )
dV = - d(Fu(L) )
F is constant
virtual displacement
– With distributed forces and concentrated force
V = - Fu(L ) -
L
ò0 bxu( x )dx
dV = - dWe
• PVW dU + dV = 0 or d(U + V ) = 0
– Define total potential energy P = U + V
dP = 0
41
EXAMPLE: PMPE TO DISCRETE SYSTEMS
• Express U and V in terms of
displacements, and then
differential P w.r.t displacements
• k(1) = 100 N/mm, k(2) = 200 N/mm 1
k(3) = 150 N/mm, F2 = 1,000 N
F3 = 500 N
• Strain energy of elements (springs)
U
(1)
U
(2)
U
(3)
1
2
= k (1) (u2 - u1 )
2
U
(1)
u2
1
2
3
F3
F3
3
2
u3
u1
é k (1) - k (1) ùìï u1 üï
1
úí ý
= êëu1 u2 úûê (1)
(1)
2 (1´ 2 ) êë- k
k úûïïî u2 ïïþ
é k (2) - k (2) ùìï u1 üï
1
úí ý
= êëu1 u3 úûê (2)
(2)
ê
úûïîï u3 ïþï
2
k
ë- k
( 2´ 2 )
( 2´ 1)
é k (3) - k (3) ùìï u2 üï
1
úí ý
= êëu2 u3 úûê (3)
(3)
ê
úûïîï u3 ïþï
2
k
ë- k
42
EXAMPLE cont.
• Strain energy of the system U =
3
å
U (e )
e= 1
U=
1
êu u
2ë 1 2
ék (1) + k (2)
- k (1)
ê
(1)
ê
u3 ú
k (1) + k (3)
ûê - k
ê - k (2)
- k (3)
êë
1
U = {Q}T [K]{Q}
2
• Potential energy of applied forces
V = - (F1u1 + F2u2 + F3u3 ) = - êëu1 u2
• Total potential energy
P = U+V =
- k (2) ùìï u1 ü
úï ïï
ïu ï
- k (3) ú
úíï 2 ý
ïï
(2)
(3) úï
k + k ú
ûïî u3 ïþ
{Q } = {u1, u2 , u3 }T
ìï F1 üï
ïï ïï
u3 úûí F2 ý = - {Q}T {F}
ïï ïï
ïî F3 ïþ
1
{Q}T [K]{Q} - {Q}T {F}
2
43
EXAMPLE cont.
• Total potential energy is minimized with respect to the DOFs
¶P
= 0,
¶ u1
ìï u1 ü
ïï ïïï
[K ]í u2 ý =
ïï ïï
ïî u3 ïþ
¶P
= 0,
¶ u2
ìï F1 ü
ïï ïïï
í F2 ý
ïï ïï
ïî F3 ïþ
¶P
= 0
¶ u3
or,
¶P
= 0
¶ {Q }
[K]{Q} = {F}
Finite element equations
• Global FE equations
é 300 - 100 - 200 ùìï 0 ü
ï
ê
úïï ïï
ê- 100 250 - 150 úí u2 ý =
ê
úïï ïï
êë- 200 - 150 350 û
úîï u3 þ
ï
ìï F1 ü
ïï
ïï
ï
í 1,000 ý
ïï
ïï
ï
îï 500 þ
u2 = 6.538mm
u3 = 4.231mm
F1 = - 1,500N
• Forces in the springs P (e ) = k (e ) (u j - ui )
P (1) = k (1) (u2 - u1 ) = 654N
P (3) = k (3) (u3 - u2 ) = - 346N
P (2) = k (2) (u3 - u1 ) = 846N
44
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)
u( x ) = c1f1( x ) + L + cn fn ( x )
• Total potential energy in terms of unknown coefficients
P (c1, c2 ,...cn ) = U + V
• PMPE
¶P
= 0, i = 1,K n
¶ ci
45
EXAMPLE
bx
F
• L = 1m, A = 100mm2, E = 100 GPa, F = 10kN, bx = 10kN/m
2
• Approximate solution u( x ) = c1x + c2 x
2
L
L1
L1
æ
ö
du
2
• Strain energy U = ò UL ( x )dx = ò AE ex dx = ò AE çç ÷÷ dx
0
0
2
0
2
çè dx ÷
ø
L
æ
ö
1
1
4
2
U (c1, c2 ) = AE ò (c1 + 2c2 x ) dx = AE ççLc12 + 2L2c1c2 + L3c22 ÷
÷
÷
0
è
ø
2
2
3
• Potential energy of forces
L
V (c1, c2 ) = -
ò bx ( x )u( x )dx 0
L
(- F )u(L ) = -
ò bx (c1x + c2 x 2 )dx + F (c1L + c2L2 )
0
æ
æ 2
L2 ö
L3 ö÷
÷
ç
ç
= c1 ç FL - bx ÷+ c2 ç FL - bx ÷
÷
çè
2ø
3 ø÷
èç
46
EXAMPLE cont.
• PMPE P (c1, c2 ) = U + V
¶P
L2
- b±
2
= AELc1 + AEL c2 + FL - bx
= 0
¶ c1
2
b2 - 4ac
2a
¶P
4
L3
2
3
2
= AEL c1 + AEL c2 + FL - bx
= 0
¶ c2
3
3
107 c1 + 107 c2 = - 5,000
c1 = 0
4 ´ 107
10 c1 +
c2 = - 6,667
3
c2 = - 0.5 ´ 10- 3
7
• Approximate solution u( x ) = - 0.5 ´ 10- 3 x 2
• Axial force P( x ) = AEdu / dx = - 10,000 x
• Reaction force R = - P (0) = 0
47