COMPUTATIONAL METHODS IN ENGINEERING AND SCIENCE
EPMESC X, Aug. 21-23, 2006, Sanya, Hainan,China
©2006 Tsinghua University Press & Springer
A New Iterative Method for Solution of Rectangular Elastic Structure
Fuyong Lin *
Huaqiao University, Quanzhou, Fujian, 362021 China
Email: [email protected]
Abstract Kernel solution (with force acting at zeros point) of elastic problem in periodic zone is obtained by
orthogonal element method, the solution of elastic problem in other area can be gotten by expanding the area to
periodic zone with acting the forces in the boundary of given structures. By using shift operator Cni ,,mj one can get the
solution of displacement of elastic problem with forces acting at any point in periodic zone. By satisfying the boundary
conditions one can get the solution of given problem. A new iterative algorithm to solution rectangular elastic structure
problem is proposed by orthogonalizing the boundary conditions. By the proposed method the calculation of inverse of
large matrix is transformed to calculate a serial inverses of Matrix of 4×4 for plate system, 6×6 three dimension
problem. It is potential to apply proposed method to construct new domain decomposition and iterative algorithms.
Key words: boundary element method, orthogonal Spline function, structure analysis, elastic structure , group
INTRODUCTION
Group theory have widely used in the physics and chemistry. Recently the method also be used in the engineering field
for analysis of symmetry structures. The first application of group representation theory to structural analysis was by
Zlokovic [5, 6], who formulated the so-called G-vector analysis based on group representation theory, and applied it to
problems in static, vibration and stability. Bossavit [7] exploited the symmetry for domains of partial differential
equation and a lot work have been done by using representation theory of Group, all these group-theoretic studies are
bases on a mathematical pricnciple---isotypic decomposition---that the solution space for symmetric structure is to be
partitioned into a series of orthogonal subspaces, see, e.g. Serre and Fujiii and Yamaguti [8]. This principle guarantee
the existence of a transformation matric H that can put elastic stiffness matrix K of symmetric structures into a block
diagonal form.
Lin [9-11] using group theory to construction orthogonal real bases of segmentation function and by expanding the
defined areas to periodic areas to get solution of bending problems of beam, plate. In this paper we will construction
orthogonal complex bases of spline space, based on the based a kernel solution for elastic structure is proposed, and a
new iterative method to solution rectangular elastic structure is presented, some simple example is presented in the
paper. In section one we construct orthogonal spline function, in section two three dimension elastic problem is dealt,
and in section three bending problem is considered. The method is expected being used to develop a new kind domain
decomposition theory.
ORTHOGONAL SPLINE [1-3]
Let U be a set of node points U : 0, 1, 2, 3, ..., k the spline is defined as follow:
⎧1 i ≤ x ≤ i + 1
N i ,1 ( x) = ⎨
⎩ 0 other
N i ,2 ( x) =
x−i
⎧
x−i
i+2− x
N i ,1 ( x) +
N i +1,1 ( x) = ⎨
1
1
⎩i + 2 − x
(1)
i ≤ x ≤ i +1
i +1≤ x ≤ i + 2
⎯ 499 ⎯
N i ,k ( x) =
x−i
i+k−x
N i , k −1 ( x) +
N i +1, k −1 ( x)
k −1
k −1
N k ( x) = N 0, k ( x − k / 2)
(2)
Where k –1 is the order of spline, and N k ( x) be called spline local base.
Now let us use shift operator Cns to shift the spline local base to zone [ s − k / 2, s + k / 2], s = 0, 1, 2, …, n − 1 , that:
Cns N k ( x) = N k ( x − s )
So we get the spline bases in all periodic zone [ 0, n ] N k ( x − s), s = 0, 1,
space:
n −1
⎧
Vnk = ⎨ f ( x) = ∑ ai N k ( x − i ),
i =0
⎩
, n − 1 . All these bases build a spline
⎫
∀ai ∈ R ⎬
⎭
It is clear that all the periodic spline function of order k can be linearly expressed by these bases, that is any periodic
spline function of order k can be written as:
n −1
f ( x ) = ∑ as N k ( x − s )
s =0
We know the set of shift operator of Cn = ( e, Cn1 ,
, Cnn −1 ) in periodic zone [ 0, n ] by calculation relation
Cni Cnj = Cni + j is an Abel Group. Now let us consider Abel Group Cn = ( e, Cn1 ,
, Cnn −1 ) , by using every element of Abel
⎧
⎫
, Cnn −1 ) as a base we can obtain a Group space that is Vg = ⎨∑ ai Cni , ∀ai ∈ R ⎬ each element of
⎩ i
⎭
i
j
i+ j
Abel is one base of the Group spaces and the relation between the two bases is Cn Cn = Cn .
group Cn = ( e, Cn1 ,
Using the operator R = Cn1 to transform the bases ( Cn = ( e, Cn1 ,
, Cnn −1 ) ) of Abel Group space to orthogonal bases by
the same way as to get eigen-vector of a Matrix ( RQ = λ Q , where Q = ∑ ak Cnk ). We can easily get orthogonal bases
k
of Abel Group space [3-7]
Qi , n = ∑ dik, nCnk
k
Where: dil, n = exp(α
2ilπ
2ilπ
2ilπ
) = cos(
) + α sin(
)
n
n
n
α is the sign of complex number.
Acting the orthogonal bases of Abel Group space on the spline base, so we get the orthogonal bases:
n −1
n −1
l =0
l =0
Eni , k ( x) = ∑ dil, n Cnl N k ( x) = ∑ dil, n N k ( x − l )
i = 0,1,
, n −1
Now let us construct the orthogonal complex bases spline in the periodic zone [0, L] as the following:
n −1
Ein , k ( x) = ∑ exp(α
l =0
2ilπ
nx
⎫
) N k ( − l ) ⎬ i = 0, 1,
n
L
⎭
, n −1
(3)
Remark 1: The orthogonal complex spline segments Ein , k ( x) are continuously differential up to order k-2 order, and
they are also orthogonal to each other:
v n, k
d u Ein , k ( x) d E j ( x)
dx = 0, i ≠ j
∫0 dxu
dx v
L
μ,ν ≤ k
k
d u Ein , k ( x) d v Ein , k ( x)
n
2ilπ
dx = 2 L( )v + u ∑ cos(
)F (u , v, l , k )
v
∫0 dxu
dx
L
n
l =0
L
⎯ 500 ⎯
(4)
μ ,ν ≤ k
Where:
∂ u N k ( x) ∂ v N k ( x − l )
dx
∫ ∂xu
∂x v
−k / 2
k/2
F (u , v, l , k ) =
∂ u N k ( x) ∂ v N k ( x)
dx / 2
∫ ∂xu
∂x v
−k / 2
k/2
F (u , v, 0, k ) =
Prove:
s = 0, 1, 2, , n − 1 are continuously differential up to order k-2 order, so
Because all spline bases N k ( x − s ),
those orthogonal spline bases which are linearly function of spline bases N k ( x − s), s = 0, 1, 2, , n − 1 are also
continuously differential up to order k-2.
nx
nx
υ n, k
∂ μ N k ( − l ) n −1
∂υ N k ( − s )
L n −1
js
∂ μ Ein , k ( x) ∂ E j ( x)
2ilπ
2
π
L
L
dx = ∫ ∑ exp(α
dx
)
× ∑ exp(−α
)
μ
υ
∫0 ∂x μ
n
x
n
x
∂xυ
∂
∂
s =0
0 l =0
L
∂ μ N k ( x − s ) ∂υ N k ( x − l )
2 jsπ
2ilπ ⎛ n ⎞
= ∫ ∑∑
exp(−α
) exp(α
)⎜ ⎟
μ
υ
n
n ⎝L⎠
∂x
∂x
0 l =0 s =0
n n −1 n −1
μ +υ −1
dx
2 jsπ
2isπ
2i (l − s )π ⎛ n ⎞
∂ μ N k ( x − s ) ∂υ N k ( x − l )
exp(−α
) exp(α
)exp(α
)⎜ ⎟
μ
υ
∂x
n
n
n
∂x
⎝L⎠
0 l =0 s =0
n n −1 n −1
= ∫ ∑∑
⎛n⎞
=⎜ ⎟
⎝L⎠
μ +υ −1 n −1 n
⎛n⎞
=⎜ ⎟
⎝L⎠
μ +υ −1 n −1 n
⎛n⎞
=⎜ ⎟
⎝L⎠
μ +υ −1
μ +υ −1
dx
2itπ n −1
2 jsπ
2isπ
∂ μ N k ( x − t ) ∂υ N k ( x)
dx
exp(
α
)∑ exp(−α
) exp(α
)
∑
μ
υ
∫
n s =0
n
n
∂x
∂x
s =0 0
2itπ n −1
2(i − j ) sπ
∂ μ N k ( x − t ) ∂υ N k ( x)
dx
exp(
α
)∑ exp(α
)
υ
μ
∂x
n s =0
n
∂x
t =0 0
∑∫
∂ μ N k ( x − t ) ∂υ N k ( x )
2itπ 1 − exp(2α (i − j )π )
dx exp(α
)
∑
μ
υ
∫
n 1 − exp(α 2(i − j )π )
∂x
∂x
t =− k / 2 − k / 2
n
k/2
k/2
⎧⎛ n ⎞ μ +υ k / 2 k / 2
2itπ
nx
nx
)
⎪⎜ ⎟ L ∑ ∫ N k ( − t )N k ( )dx exp(α
= ⎨⎝ L ⎠
L
L
n
t =− k / 2 − k / 2
⎪
0
i≠ j
⎩
⎧ ⎛ n ⎞ μ +υ k / 2
2itπ
L ∑ F (u , v, t , k ) cos(
)
⎪2
= ⎨ ⎜⎝ L ⎟⎠
n
t =0
⎪0
⎩
i= j
(5)
i= j
i≠ j
Let us call E in ,k ( x) as orthogonal complex base spline functions
iL
(n − 1) L ⎤
⎡ L
If we use the value of node points ⎢ 0, , , , ,
to construct periodic spline functions, all these periodic
n
n ⎥⎦
⎣ n
spline function build a spline space Vnk , the bases of the spaces are N k ( x − s ), s = 0, 1, , n − 1 , and also one can
also transform to orthogonal bases Ein , k ( x), i = 0, 1,
, n −1 .
Prove:
For a spline there n points value is given, we write the spline as:
n −1
n −1
i =0
i =0
f ( x) = ∑ ai N k ( x − i ) = ∑ ai ' Ein ,k ( x)
It is clear that above function is a spline function of order k, there are n coefficients to be decided, consider every the
point value, one can obtain n linear equations, one can obtain the coefficient directly from solution these equation
(because these bases is linear independent). And if the value of node points is presented, there is only one spline
function in the periodic zone [0, n], therefore the spline function which obtained by solution the linear equation of
coefficient is exactly the one we wish to obtained. #
⎯ 501 ⎯
Therefore the orthogonal bases build up a space that all periodic spline of order k in zone [ 0, L ] are included in this
space. That is:
n −1
⎧
Vnk = ⎨ f ( x) = ∑ ai Enk,i ( x),
i =0
⎩
⎫
∀ai ∈ R ⎬
⎭
APPLICATION IN THREE DIMENSION ELASTIC STRUCTURE PROBLEMS
1. Three dimension periodic elastic structure In this section we use orthogonal spline to seek a spline solutions
for three dimension elastic problems. First let us consider periodic isotropy elastic structure, we evenly divide the
defined rectangular area, the number of sub-area is [l×m×n], where l is division in x direction, m is division in y
direction, n is division in z direction, and we use spline function to approaches the displacement of elastic structure.
For an isotropy elastic space problem, the displacement (u , v, w) of elastic structure satisfies the following equations:
⎫
∂θ
+ X = 0⎪
∂x
⎪
⎪⎪
∂θ
μ∇ 2v + (λ + μ ) + Y = 0 ⎬
∂y
⎪
⎪
θ
∂
μ∇ 2 w + (λ + μ ) + Z = 0 ⎪
∂y
⎪⎭
μ∇ 2u + (λ + μ )
Ω
in
(6)
Where:
∇2 =
∂2
∂2
∂2
,
+
+
∂x 2 ∂y 2 ∂z 2
μ=
E
Eν
,λ =
2(1 + ν )
(1 + ν )(1 − 2ν )
θ=
∂u ∂v ∂w
,
+
+
∂x ∂y ∂z
E is elastic modal and ν is passion coefficient. Ω is periodic rectangular area.
Eq. (6) can be rewritten as:
⎧u ⎫
⎧X ⎫
⎪ ⎪
⎪ ⎪
H ⎨v ⎬ = − ⎨ Y ⎬
⎪z⎪
⎪Z ⎪
⎩ ⎭
⎩ ⎭
(6a)
Where:
⎡ 2
∂2
(
λ
μ
)
∇
+
+
⎢
∂x 2
⎢
⎢
∂2
H = ⎢ (λ + μ )
∂x∂y
⎢
⎢
∂2
⎢ (λ + μ )
∂x∂z
⎢⎣
(λ + μ )
∂2
∂x∂y
∇ 2 + (λ + μ )
(λ + μ )
∂2
∂y 2
∂2
∂y∂z
∂2 ⎤
⎥
∂x∂z ⎥
∂2 ⎥
(λ + μ )
⎥
∂y∂z ⎥
∂2 ⎥
∇ 2 + (λ + μ ) 2 ⎥
∂z ⎥⎦
(λ + μ )
In the following we use orthogonal spline bases to solution elastic problem in periodic areas, in order to simplify the
expression in late expression we note N k ( x) as N (x) . First let us consider the defined periodic area Ω
(0 ≤ x ≤ L;0 ≤ y ≤ T ;0 ≤ z ≤ S ) , so the displacement of elastic structure can be approached by the following spline
functions:
⎯ 502 ⎯
⎧ u ⎫ l −1 m −1 n −1 ⎧ ui , j ,k ⎫
l −1 m −1 n −1
⎪
⎪
⎪ ⎪
U( x, y , z ) = ⎨ v ⎬ = ∑∑∑ ⎨ vi , j ,k ⎬ Eil ( x)E mj ( y ) Ekn ( z ) = ∑∑∑ U i , j ,k Eil ( x)E mj ( y ) Ekn ( z )
i −1 j = 0 k = 0
⎪ w⎪ i −1 j =0 k =0 ⎪ w ⎪
⎩ ⎭
⎩ i , j ,k ⎭
(7)
Where:
Ui , j ,k
⎧ ui , j ,k ⎫
⎪
⎪
= ⎨ vi , j , k ⎬
⎪w ⎪
⎩ i , j ,k ⎭
l −1
Eil ( x) = ∑ exp(α
s =0
m −1
E mj ( x) = ∑ exp(α
s =0
n −1
Ekn ( x) = ∑ exp(α
s =0
2isπ
lx
) N ( − s)
l
L
2 jsπ
my
)N (
− s)
m
T
2ksπ
nz
) N ( − s)
n
S
i = 0,1,
l −1
j = 0,1,
k = 0,1,
m −1
(8)
n −1
Take Eq. (7) into Eq. (6) and multiply by U i , j ,k ( x, y, z ) = Eil ( x) E mj ( y ) Ekn ( z ) , and integrate in the periodic area,
considering orthogonality of orthogonal spline bases (Eqs. (4, 5)), we can get the coefficient equation:
l −1 m −1 n −1
l
m
n
l
m
n
∫∫∫ Eh ( x)E f ( y ) Eg ( z )∑∑∑ HEi ( x) E j ( y) Ek ( z )dxdydzUi , j ,k =
i −1 j = 0 k = 0
= ∫∫∫ Ehl ( x)E mf ( y ) Egn ( z ) HEhl ( x) E mf ( y ) Egn ( z ) dxdydzU f ,h , g = Fh , f , g U h , f , g
(9)
= − ∫∫∫ XEhl ( x) E mf ( y ) Egn ( z ) dxdydz = X h , f , g
Where:
Fh , f , g = ∫∫∫ Ehl ( x)E mf ( y ) Egn ( z )HEhl ( x) E mf ( y ) Egn ( z )dxdydz
Xi , j , k = − ∫∫∫ XEil ( x) E mj ( y ) Ekn ( z )dxdydz
(10)
From the Eq. (9), one can get ui , j ,k , vi , j ,k , wi , j , k for every (i, j , k ) :
Ui , j , k = Fi , j , k −1Xi , j , k
(11)
And finally we get the solution of displacement for the periodic zones as the following:
l −1 m −1 n −1
l −1
e=0 f =0 g =0
i =0
U ( x, y, z ) = ∑∑∑ U i , j , k (∑ exp(α
m −1
n −1
2eiπ
lx
2 fjπ
my
2 gkπ
nz
) N ( − i )) × (∑ exp(α
)N (
) N ( − k ))
− j )) × (∑ exp(α
l
L
m
S
n
L
j =0
k =0
lx
my
nz
2eiπ
2 fjπ
2 gkπ
− j ) N ( − k ))
) exp(α
) exp(α
)) ×N ( − i ) N (
= ∑ ∑∑ (∑∑∑ U i , j , k exp(α
L
T
S
l
m
n
i −1 j = 0 k = 0 e = 0 f = 0 g = 0
l −1 m −1 n −1
l −1 m −1 n −1
(12)
Remark 2: Eq. (11) is obtained by considering orthogonal relationship (4), and Eq. (10) can use Fast Fourier
Transform (FFT) to reduce the computation. Eq. (12) also can use FFT to reduction the calculation. So by use FFT
method the calculation of proposed method is O(l×n×m×log(lnm)), where l×n×m is number of node points.
2. Kernel solution (displacement) of periodic elastic structure In order to get the solution of elastic structure
problem without periodicity, let us first consider the kernel solution of periodic structure. In simple, in the following
we take 3-order spline as example. We try to get kernel solution of displacement of elastic structure for periodic zone
in orthogonal spline base series. To simplify the solution, first let the forces, acting at the point,
(( x ', y ', z ')
lx ' < L; my ' < T ; nz ' < S ) be:
⎯ 503 ⎯
⎧X ⎫
⎪ ⎪
F = δ ( x ')δ ( y ')δ ( z ') ⎨ Y ⎬ ,
⎪Z ⎪
⎩ ⎭
(13)
Let displacement of periodic three dimension structure expressed as:
⎧ u ⎫ l −1 m −1 n −1
⎪ ⎪
U = ⎨ v ⎬ = ∑∑∑ U i , j ,k Eil ( x)E mj ( y ) Ekn ( z )
⎪ w⎪ i −1 j = 0 k = 0
⎩ ⎭
(14)
Considering Eq. (10) and Eq. (11), we get kernel solution of displacement in periodic area:
Ui , j , k = Fi , j , k −1Xi , j , k
(15)
Where:
Xi , j , k = ∫∫∫ XU i , j , k ( x, y, z )dv
=
1
∫∫∫ δ ( x ')δ ( y ')δ ( z ') Ei ( x) E j ( y) Ek ( z )dxdydz [ X ,Y , Z ] =( ∑ N k (
τ
s =−2
1
( ∑ Nk (
s =−2
lx '
2 siπ
− s )exp(−α
)) ×
L
l
1
my '
2 sjπ
nz '
2 skπ
τ
− s )exp(−α
− s )exp(−α
))( ∑ N k (
)) [ X , Y , Z ]
T
m
S
n
s =−2
= Li , j , k [ X , Y , Z ]
τ
Where:
1
Li , j , k = ( ∑ N k (
s =−2
1
1
lx '
2 siπ
my '
2 sjπ
nz '
2 skπ
)) × ( ∑ N k (
)) × ( ∑ N k (
))
− s ) exp(−α
− s ) exp(−α
− s ) exp(−α
L
l
T
m
S
n
s =−2
s =−2
(16)
Consider Eq. (8), (15) and (16) we finally get the solution in spline serial
Ui , j , k = Fi , j , k −1 Xi , j , k = Fi , j , k −1 Li , j , k [ X , Y , Z ] = Li , j , k Fi , j , k −1 [ X , Y , Z ] = Li , j , k U i0, j , k
τ
τ
(17)
Where:
Ui0, j ,k = Fi , j , k −1 [ X , Y , Z ]
τ
(18)
From the equation (17) we know any forces act at point (( x ' , y ' , z ' )
e
l
f
m
lx ' < L; my ' < T ; nz ' < S ) , can directly
g
n
obtained by using shift operator C C C and basic kernel solution Specially, when the unit force act at zeros point,
the solution can be expressed as follow (use three order spline as an example):
l −1 m −1 n −1
U 0 = ∑∑∑ (2 / 3 + 1/ 3cos(
i =0 j =0 k =0
2iπ
2 jπ
2iπ
2 jπ
)) × (2 / 3 + 1/ 3cos(
))(2 / 3 + 1/ 3cos(
)) × (2 / 3 + 1/ 3cos(
)) ×
l
m
l
m
2 kπ
×(2 / 3 + 1/ 3cos(
)) Eil ( x) E mj ( x) Ekn ( x))Ui0, j , k
n
(19)
Remark 3: In getting the kernel solution, we ignore (i = 0, j = 0, k = 0) by using even distributed forces in all periodic
zone to balance the unit force, so we will find the solution may be adding some constant.
Forces acting on the other span one can use shift operator to shift the solution, in the following we will develop it in
detail. Let us act unit force at any given point ( x '', y '', z '') lx '' = eL + lx '; my '' = fT + my '; my '' = fT + my ';
nz '' = gS + nz ', lx ' < L; my ' < T ; nz ' < S ) in the same way as we develop Eq. (16) we can get:
so we obtain the solution that forces acting in other zone:
⎯ 504 ⎯
Xi , j , k = ∫∫∫ XU i , j , k ( x, y, z )dv = ∫∫∫ δ ( x '')δ ( y '')δ ( z '') Ei ( x) E j ( y ) Ek ( z )dxdydz [ X , Y , Z ]
τ
1
= ( ∑ Nk (
s =−2
1
lx '
2(e + a )iπ
my '
2( f + b) jπ
− e) exp(−α
)) × ( ∑ N k (
− f ) exp(−α
))
L
l
T
m
s =−2
nz '
2k ( g + c)π
τ
× ( ∑ Nk (
− g ) exp(−α
)) [ X , Y , Z ]
S
n
s =−2
1
= exp(−α
(20)
2aiπ
2bjπ
2kcπ
τ
) exp(−α
) exp(−α
) Li , j , k [ X , Y , Z ]
l
m
n
Especially when forces acting at the node point (e, f, g) the solution can be written as:
−1
U i , j ,k = Fi , j ,k X i , j ,k = exp(−α
2aiπ
2bjπ
2ckπ
)) × exp(−α
) × exp(−α
) Li , j ,k U i0, j ,k
l
m
n
(21)
⎧ u ⎫ l −1 m −1 n −1 ⎧ ui , j , k ⎫
⎪
⎪
⎪ ⎪
U( x, y, z ) = ⎨ v ⎬ = ∑∑∑ ⎨ vi , j , k ⎬ Eil ( x)E mj ( y ) Ekn ( z )
⎪ w⎪ i −1 j = 0 k = 0 ⎪ w ⎪
⎩ ⎭
⎩ i, j ,k ⎭
l −1 m −1 n −1
= ∑∑∑ exp(−α
i −1 j = 0 k = 0
l −1 m −1 n −1
= ∑∑∑ exp(−α
i −1 j = 0 k = 0
2aiπ
2bjπ
2ckπ
)) × exp(−α
) × exp(−α
) Li , j , k U i0, j ,k Eil ( x)E mj ( y ) Ekn ( z )
l
m
n
(22)
2bjπ m
2ckπ n
2aiπ l
) E j ( y ) × exp(−α
)Ek ( z )Li , j ,k U i0, j , k
) Ei ( x) × exp(−α
l
m
n
= Cla Cmb Cnc U 0
Where:
l −1 m −1 n −1
U 0 = ∑∑∑ Li , j , k U i0, j , k Eil ( x)E mj ( y ) Ekn ( z )
i −1 j = 0 k = 0
3. Iterative solution for three dimension rectangular elastic structure We know it in above that for three
dimension periodic elastic problem the calculation of solution is of O(l×n×m) multiply, one also wish that for
rectangular elastic problem, the calculation of solution would also be O(l×n×m), in the following we will proposed
iterative method to reduce the calculation to O(l×n×m). First let us consider elastic problem of is rectangular zone, the
displacement satisfy the following equation:
⎫
∂θ
+ X = 0⎪
∂x
⎪
⎪⎪
∂θ
+Y = 0 ⎬
μ∇ 2 v + (λ + μ )
∂y
⎪
⎪
∂θ
+ Z = 0⎪
μ∇ 2 w + (λ + μ )
∂y
⎪⎭
μ∇ 2u + (λ + μ )
in
Ω
And boundary condition is:
(1) The force boundary conditions
Xσ = σ x ⎫
⎪⎪
Yσ = τ yx ⎬
⎪
Yσ = τ zx ⎪⎭
in Γσ
u = us ; v = vs ; w = ws
in Γ s
(23)
especially when the we deal the rectangular structure, we can get:
⎯ 505 ⎯
Xσ = σ x ⎫
⎪⎪
Yσ = τ yx ⎬ x = a1 , a2
⎪
Yσ = τ zx ⎪⎭
X σ = τ yx ⎫
⎪⎪
Yσ = σ y ⎬ y = b1 , b2
⎪
Yσ = τ zy ⎭⎪
X σ = τ zx ⎫
⎪⎪
Yσ = τ zy ⎬ z = c1 , c2
⎪
Yσ = σ z ⎪⎭
(23a)
(2) Displacement boundary conditions:
u = us ; v = vs ; w = ws x = a1 , a2 or
y = b1 , b2 or
z = c1 , c2
in Γ s
(23b)
In order to get solution, we first expand the defined area to periodic one by acting additional forces on the boundary [9,
10, 12], or outside the defined areas. We know that for any elastic problem, if solution satisfies the government
Eq. (6) and boundary condition (23a, 23b), the solution is one and only one. We expect that we could expand the
defined areas to a periodic areas and get solution in expand areas, if the solution also satisfies the boundary conditions,
we can assure that solution is the solution demanded. Fig. 1 is an example that simply supported beam be expanded to
periodic beam.
Figure 1: An expanded beam system
Now let us expand the area Ω ( a1 ≤ x ≤ a2 , b1 ≤ y ≤ b2 , c1 ≤ y ≤ c2 ) to periodic zone Ω *(0 ≤ x ≤ L;0 ≤ y ≤ T ;
0 ≤ z ≤ S ; ) by acting the F( X x '', y '', z '' , Yx '', y '', z '' , Z x '', y '', z '' ) in the boundary, so we transform the problem to the following
problem:
⎧u ⎫
⎧X ⎫
⎪ ⎪
⎪ ⎪
H ⎨ v ⎬ = − ⎨ Y ⎬ + ∑ δ ( xi )δ ( y j )δ ( zk ) Fi , j ,k
⎪z⎪
⎪ Z ⎪ i , j ,k
⎩ ⎭
⎩ ⎭
in
Ω∗
(24)
i
j
k
Where Fi , j ,k is additional force acting at point x = L, y = T , z = S outside the given areas.
l
m
n
Consider Eq. (21) and Eq. (24)
*
⎧u ⎫ ⎧u ⎫
⎪ ⎪ ⎪ *⎪
i
j k
⎨ v ⎬ = ⎨ v ⎬ + ∑ Cl Cm Cn U 0 Fi , j , k
⎪ w⎪ ⎪ w* ⎪ i , j ,k '
⎩ ⎭ ⎩ ⎭
(25)
⎧ u* ⎫
⎪ ⎪
Where ⎨ v* ⎬ is special solution of periodic elastic structure, direct obtained by Eq. (9). One can use Eq. (25) and
⎪ w* ⎪
⎩ ⎭
boundary condition (23) to obtain Fi , j ,k and finally get the solution of given problem. If use the boundary condition
(23) directly one must boundary linear equation which calculation is O(L3), L is the number of additional forces
outside the given areas. If the structure is rectangular, one can use iterative method to get the solution the calculation of
boundary condition is O(m) or O(n). in the following we present the method.
Now let us consider the boundary condition for rectangular boundary condition, we can expand the conditions into
orthogonal serial, so we wish the additional forces could expand orthogonal series. So let us act additional forces on the
boundary and the additional forces be expand to Fourier series:
x = a1 , a2
⎧ X ⎫ m −1 n −1
m −1
n −1
2 jsπ
s
2ktπ
t
⎪ ⎪
F1 = ⎨ Y ⎬ = ∑∑ ( A1j , k (∑ exp(α
)δ ( y − ))(∑ exp(α
)δ ( z − )))
m
M t =0
n
N
s =0
⎪ Z ⎪ j =0 k =0
⎩ ⎭
⎯ 506 ⎯
⎧ X ⎫ m −1 n −1
m −1
n −1
2 jsπ
s
2ktπ
t
⎪ ⎪
)δ ( y − ))(∑ exp(α
)δ ( z − )))
F2 = ⎨ Y ⎬ = ∑∑ ( A 2j ,k (∑ exp(α
m
M t =0
n
N
s =0
⎪ Z ⎪ j =0 k =0
⎩ ⎭
y = b1 , b2
⎧ X ⎫ l −1 n −1
m −1
2isπ
s n −1
2ktπ
t
⎪ ⎪
)δ ( x − ))(∑ exp(α
)δ ( z − )))
F3 = ⎨ Y ⎬ = ∑∑ (B1i ,k (∑ exp(α
l
L t =0
n
N
s =0
⎪ Z ⎪ i =0 k =0
⎩ ⎭
⎧ X ⎫ l −1 n −1
m −1
2isπ
s n −1
2ktπ
t
⎪ ⎪
)δ ( x − ))(∑ exp(α
)δ ( z − )))
F4 = ⎨ Y ⎬ = ∑∑ (Bi2,k (∑ exp(α
l
L t =0
n
N
s =0
⎪ Z ⎪ i =0 k =0
⎩ ⎭
z = c1 , c2
⎧ X ⎫ l −1 m−1
m −1
2isπ
s n −1
2 jtπ
t
⎪ ⎪
F5 = ⎨ Y ⎬ = ∑∑ (C1i ,k (∑ exp(α
)δ ( x − ))(∑ exp(α
)δ ( y − )))
l
L
m
M
s =0
t =0
⎪ Z ⎪ i =0 j =0
⎩ ⎭
(26)
⎧ X ⎫ l −1 m−1
m −1
2isπ
s n −1
2 jtπ
t
⎪ ⎪
F6 = ⎨ Y ⎬ = ∑∑ (Ci2,k (∑ exp(α
)δ ( x − ))(∑ exp(α
)δ ( y − )))
l
L t =0
m
M
s =0
⎪ Z ⎪ i =0 j =0
⎩ ⎭
Take Eq. (26) into Eq. (25) and consider Eq. (21), we get:
*
⎧ u ⎫ ⎧ u ⎫ m −1 n −1
⎪ ⎪ ⎪ *⎪
⎨ v ⎬ = ⎨ v ⎬ + ∑∑
⎪ w⎪ ⎪ w* ⎪ f =0 g =0
⎩ ⎭ ⎩ ⎭
lm −1 n −1
+ ∑∑
+
f =0 g =0
i =0 j =0 k =0
l −1 m −1 n −1
∑∑∑ exp(α
i =0 j =0 k =0
l −1 m −1 n −1
= ∑∑∑ (l exp(−α
i =0 j =0 k =0
+ m exp(−α
l −1 m −1 n −1
∑∑∑ exp(α
2π fj
2π kg
2π ia1
2π fj
2π kg
) exp(α
) exp(−α
) exp(−α
) exp(−α
)U i , j ,k A1j ,k
m
n
l
m
n
2π kc2
2π fj
2π kg
2π fi
2π jg
) exp(−α
) exp(−α
)U i , j ,k Ci2, j
) exp(α
) exp(−α
l
m
n
m
n
(27)
2π ia1
2π ia2
2π jb1
)U i , j ,k A1j ,k + l exp(−α
)U i , j , k A 2j ,k + + m exp(−α
)U i , j ,k B1j ,k
l
l
m
2π kb2
2π jc1
2π kc2
)U i , j ,k B 2j ,k + n exp(−α
)U i , j ,k C1j , k + n exp(−α
) U i , j ,k C2j ,k )
n
m
n
Now let us consider the boundary condition (23a, 23b), in the following only consider the boundary condition in
x = a1 , the other boundary condition can be simplified in the same way.
Take Eq. (27) into Eq. (23a, 23b), and multiplied by E j ( y ) Ek ( z ), Ei ( x ) Ek ( z ), Ei ( z ) E j ( y ) separately and integrate the
boundary condition, we can get:
Boundary condition on boundary x = a1
f ∗,1 j , k exp(α
l −1
l −1
l −1
2ia1 )π
2i (a2 − a1 )π
2ia1 )π
2 jb1π
) + ∑ fi1, j , k A1j , k + ∑ fi1, j , k A 2j , k exp(−α
) + ∑ fi1, j , k B1i , k exp(α
) exp(−α
)
L
L
L
M
i =0
i =0
i =0
l −1
+ ∑ fi1, j , k Bi2, k exp(α
i =0
l −1
+ ∑ fi1, j , k Ci2, j exp(α
i =0
l −1
2kc1π
2ia1 )π
2 jb2π
2ia1 )π
) exp(−α
)
) exp(−α
) + ∑ fi1, j , k C1i , j exp(α
L
N
L
M
i =0
2ia1 )π
2kc2π
) exp(−α
)=0
L
N
j = 0,
, m − 1; k = 0,
, n −1
Where f ∗j ,,k1 and f i1, j , k is related to boundary condition, for example, given displacement condition, we can get:
⎯ 507 ⎯
(28a)
f
*,1
j ,k
=∫
M
0
∫
N
0
⎧u *⎫
⎧ us ⎫
⎪ ⎪
⎪ ⎪
( ⎨ v * ⎬ − ⎨ vs ⎬) E j ( y ) Ek ( z )dydz
⎪ w *⎪
⎪ ⎪
⎩ ⎭ x = a1 ⎩ ws ⎭
fi1, j ,k = exp( −
(28b)
M N
2ia1π
) Ei (a1 )(ai , j , k ) −1 ∫ ∫ ( E j ( y )) 2 ( Ek ( z )) 2 dydz
0
0
l
Eq. (28) can be simplified as follow
1
1,2
2
1
E1,1
j ,k A j ,k + E j ,k A j ,k = G j ,k
1
2,2
2
2
E2,1
j , k A j , k + E j , k A j , k = G j , k j = 0,
,, m − 1; k = 0,
,n −1
E1i ,,1k B1i ,k + E1i ,,k2 B i2,k = G 1i ,k ;
E i2,,k1B1i ,k + E i2,,k2 B i2,k = G i2,k ; i = 0, L, l − 1; k = 0,L, n − 1
(28)
E1i ,,1j C1i , j + E1i ,, 2j C i2, j = G 1i , j ;
E i2,,1j C1i , j + E i2,,j2 C i2, j = G i2, j ; i = 0, L, l − 1; j = 0,L, m − 1
Where:
2i (a 2 − a1 )π
); E 2j ,,1k = E 1j,,2k ;
L
i =0
i =0
l −1
2ia s )π
2 jb1π
) exp(−α
)+
G sj ,k = −(f ∗, s j ,k + ∑ f i1, j ,k B1i ,k exp(α
L
M
i =0
l −1
2ia s )π
2 jb2π
+ ∑ f i1, j ,k B i2,k exp(α
) exp(−α
)+
L
M
i =0
l −1
2ia s )π
2kc1π
+ ∑ f i1, j ,k C1i , j exp(α
) exp(−α
)+
L
N
i =0
l −1
2ia s )π
2kc 2π
+ ∑ f i1, j ,k C i2, j exp(α
) exp(−α
))
L
N
i =0
s = 1,2
l −1
l −1
E1j,,1k = E 2j ,,k2 = ∑ f i1, j ,k ; E1j,,2k = ∑ f i1, j ,k exp(−α
k = 0, L , n − 1; j = 0, L , m − 1
m −1
m −1
E1i ,,1k = E i2,,k2 = ∑ f i2, j ,k ; E1i ,,k2 = ∑ f i2, j ,k exp(−α
j =0
j =0
2 j (b2 − b1 )π
); E i2,,k1 = Ei1,,k2
M
2 jbs π
2i ( a1 )π
) exp(α
)+
L
M
i =0
l −1
2 jbs π
2i (a1 )π
f i2, j ,k A 2j ,k exp(−α
) exp(α
)+
∑
L
M
i =0
n −1
2 jbs π
2kc1π
+ ∑ f i2, j ,k C1i , j exp(α
) exp(−α
)+
M
N
k =0
n −1
2 jbs π
2kc2π
+ ∑ f i2, j ,k C i2, j exp(α
) exp(−α
))
M
N
k =0
s = 1,2
i = 0,L , l − 1; k = 0, L , n − 1
l −1
G is,k = −(f ∗, s i ,k + ∑ f i2, j ,k A 1j ,k exp(−α
n −1
n −1
2,2
3
1,2
3
E1,1
i , j = Ei , j = ∑ fi , j , k ; Ei , j = ∑ fi , j , k exp( −α
i =0
i =0
l −1
G1i , j = −(f ∗s i , j , k + ∑ fi3, j , k A1j , k exp(−α
i =0
l −1
l −1
i =0
i =0
+ ∑ fi3, j , k B1i , k + ∑ fi3, j , k Bi2, k exp(α
2k (c2 − c1 )π
1,2
); Ei2,1
, j = Ei , j
N
l −1
2i ( a1 )π
2kc1π
2k (c1 )π
2 jb1π
) exp(α
) + ∑ fi3, j , k A 2j , k exp(α
) exp(−α
)
L
N
N
M
i =0
2kc1 )π
2 jb2π
) exp( −α
))
L
M
i = 0,
⎯ 508 ⎯
, l − 1; j = 0,
, m −1
Therefore we can get the following step to get solution the structure problem
(1) Expand the structure to rectangular zone and balance the forces in expand areas by add some forces outside given
areas, the boundary condition also be expanded to rectangular area.
(2) First give A 1j,,0k , A 2j ,,k0 , B 1i ,,k0 , B i2,,k0 , C1i ,, 0j , C i2,,j0
(3) Take B 1i ,,k0 , B i2,,k0 , C1i ,, 0j , C i2,,j0 into Eq. (28) to get A 1j,,1k , A 2j ,,1k , Take A 1j,,0k , A 2j ,,0k , C1i ,, 0j , C i2,,j0 into (28) to get B 1i ,,1k , B i2,,k1 ,
and Take A 1j,,0k , A 2j ,,0k , B1i ,,k0 , B i2,,k0 into (28) to get C1i ,,1j , C i2,,1j
(4) Further take B 1i ,,kn −1 , B i2,,kn −1 , C1i ,,nj −1 , C i2,,jn −1 into Eq. (28) to get A 1j,,nk , A 2j ,,nk , Take A 1j,,nk−1 , A 2j ,,nk −1 , C1i ,,nj −1 , C i2,,jn −1 into
(28) to get B 1i ,,kn , B i2,,kn , and Take A 1j,,nk−1 , A 2j ,,nk −1 , B 1i ,,kn −1 , B i2,,kn −1 into (28) to get C1i ,,nj ,C i2,,jn
(5) If the accuracy is satisfied then stop, if is not satisfied replete step four.
The calculation of every step take calculation O(n×m×l), in every equation (i,k) need computation of O(m), (j,k) of
O(l),(i,j) of O(n), so we can obtain the calculation of proposed method is O(l×n×m), we also can expect the iterative
method is convergence, because we know that the local balance forces will take little influence on other place that is on
great distance from the local forces.
4. Example A simply supported rectangular plate (a× a) is acted by given forces.
We expand the plate (1.6a×1.6 a) , and divide the plate into 32×32 zones, Using iterative method to get the solution,
we carry eight circle step the error of boundary force. condition to less than O(10-3).
Fig. 2 is the displacement of the plate acted by distributed force: f = sin(πx / a ) sin(πy / a ) , Fig. 3 is the displacement
of the plate acted by constant distributed
Figure 2:
Figure 3:
CONCLUSION AND DISCUSSION
(1) For convergence problem, let consider ellipse equation boundary problem. We know that for the ellipse differential
equation when the boundary condition is satisfied, the solution is one and only one, so we can expect the solution
gotten by expanding the problem to large zone, and letting the solution satisfy the boundary condition will be the exact
solution itself. And we can also expect that when the boundary condition is totally satisfied, the solution obtained by
proposed method will be the same solution gotten by element method. From the numerical experiment (use Matlab in
personal computer), we find the method is convergence, but it only a numerical experiment, and still need exact
mathematic proving.
(2) the calculation of proposed method equation, Eq. (7, 9) can be using the Fast Fourier Transform to reduce the
calculation, for two dimension problem the calculation to get displacement in periodic system without consider
boundary condition be reduced to O(N×M)log(N×M)) [7]. Where N, M are the division in x dimension and y
dimension, the calculation every step also be O(N×M), compare the calculation of boundary element method
(O(N+M)3) it is a great reduction.
⎯ 509 ⎯
(3) The potential application of this method, for large complex problem we can divided the problem zone into a serial
regular zone and in every zone one can use proposed method to carry out the solution, so we can expect great reduction
in calculation. Another advantage of this method is that it is easily to carry out parallel calculation. A note must be bear
in mind that when one use this method, one must first balance force by adding forces outside problem zone, and when
consider the boundary condition one can add a constant displacement to satisfy the displacement condition [9,12].
The proposed method has potential application in Engineering’s computation, and future investigation is in need.
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⎯ 510 ⎯