- Email: [email protected]
Structural analysis by partial decomposition
Structural Ana&is by MOSHE
F.
by Partial
Decomposition
RUBINSTEIN
School of Engineering and Applied Science University of California, Los Angeles, California and
VLADIMIR
SIMON&
Docent na Arhitektonsko GradeLen Fakultet-Skopje,
AESTRACT
:A
Yugoslavia
partial decomposition
of
thejlexibility
to yield directly
all the required information
the coordinates,
internal
decomposition all required
of the stiffness information
when zero inertial forces is simple,
forces
effkient
and deformations
matrix,
when no forces
in the analysis. are associated
matrix of a structural
in the analysis from
system is shoun
which displacements
can be obtained.
Similarly,
are applied at some coordinates,
The approach
is also useful in dynamic
with some coordinates.
and most suitable for hand calculation
The partial
and computer
at
a partial yields
analysis
decomposition
applications.
Notation
[Al
[A]a’
[Alij ,$ IDI P’> W’ PI WI* [J$ W [RI ISI
square matrix matrix [A] in step i of pivotal condensat,ion submatrices of [A] flexibility matrix reduced flexibility matrix diagonal matrix in decomposition force vector redundant force vector stiffness matrix reduced stiffness matrix mass matrix reduced mass matrix element force vector submatrix in partial decomposition upper triangular matrix with ones on principal displacement vector auxiliary vector defined by Eqs. (22) or (33) element displacement vector
diagonal
Introduction
The analysis of structures by the flexibility and stiffness methods requires the solution of linear equations. Often, however, the order of the stiffness or flexibility matrix that is formulated exceeds the number of unknowns that must be solved. In such cases it is convenient to reduce these matrices and extract the relevant information required for the solution. The partial decomposition approach proposed here is suitable and efficient for such
429
Mode
F. Rubinstein and Vladimir SimonEe
analyses. All the relevant information is obtained in a single direct operation which is convenient for both hand calculations and computer applications. The method is also useful in dynamic analysis of structures when inertial forces are zero at some coordinates, or when rigid body degrees of freedom must be removed in the analysis. This paper develops the partial decomposition and applies it to the flexibility and stiffness methods. Examples are included to illustrate the ease and efficiency of the method.
Decomposition
A symmetrical matrix [A] can be decomposed of three matrices (1, 6) :
into the following
product
(1)
[Al = [SITLB1LfJl.
[S] is an upper triangular matrix with ones on the principal diagonal and [II] is diagonal. The elements of [S] and [II] are computed by recursion using the following algorithm : 4,
= 4,
Xi, = 1,
i = 1,2 ,..., n,
Ali
i2%
slj=4;9.
i-l
Dii = Aii-
Slj = &
ii
I: S”piDpp,
p=l
Aij - ‘&,,, p=l
i z 2,
S,,*D,,
i>2,
>
jai+l.
Partial Decomposition
Partition matrix [A] into four submatrices r r
[Al =
S
.
s
[A],, and [A],, are square matrices of orders r and s, respectively. A partial decomposition applied to the first r columns of [A] can be written in the followmg partitioned form :
1
Structural Analysis
by Partial Decomposition
in which
[Al,, = [aIT [Dl [Sl, [All, = ISIT [Dl [RI, [Al,, = [AK = [W’ [Dl [fil, - Ml,, = [W’[Dl [RI + [Al”.
(5)
Using these identities, the following useful expressions are generated which relate submatrices of [A] to the submatrices on the right-hand side of Eq. (4) :
(6)
[AId IIAl,.z= [W’ [RI>
(7)
[Al,, [AL-?[Al,, = [W’ [Dl [fil,
(8)
VI,, - CAlal[Al;;1[Al,, = [Al*. Decomposition zeros)
by Pivotal
Condensation:
(Decomposition
by working
for
(2)
The decomposition described in Eqs. (l), (2) and (4) can be achieved by the method of pivotal condensation as follows. Consider again the square matrix [A] with elements Aij. The first pivotal element A,, is set equal to Dll; the first row is divided through by A,, and used to eliminate the remaining elements of the first column. This produces a matrix of the form
(9)
in which A$ = A,/A,, and Ai;) = Aij - A:$) A, for i 2 2. Next the second pivotal element Ai;) is set equal to D22, the elements of the second row of [A](l) are divided through by A$ and then used to reduce to zero the elements of the second column below the main diagonal, yielding
[A]t2) =
1
Ai;)
A$
...
Ail,,
0
1
A$
...
Ai;
0
0
Ag’
...
A;:
.
.
0
0
A$
...
A$
.
(10)
The third pivotal element is A,,(2) = D,, and so on to A&-” = D . Matrix [AI with the number one replacing element AL”,--1)is matriF[S], and elements A!i-1) i= I,2 n, are the elements of matrix [D], i.e. A$ = A,, =?I,,: A;;) = D;,;:: ., Ap;l’ = D,,.
Vol.290,Xo. 5, Rowmber
1970
431
Moshe F. Rubinstein and Vladimir SimoniTe For example, applying the pivotal condensation
,A] _[
J4i4
p
_14
procedure
to the matrix
]
yields
Partial
Decomposition
by Pivotal
Condensation
When the pivotal condensation is applied to the first r columns of matrix [A], then the resulting matrix [A]cn) after r steps has the form 1
Ail,’
A$
. ..
A$
...
...
Ail,,
1
A$
...
A$
...
...
Agj
r
. [A](‘)
zzz
s
.
A(r) 1 ... 0 ... ... ... ... rw .___......__.__. .___.......................... A(T) 0 ... ... ... 0 r+l,r+l ... 4$l,?l . .
0
...
r
...
...
0
A(r) n,r+1
...
A(r) R?l
s
(11) The submatrices of [A]@) are identified as submatrices [Xl, [R] and [A]* in the partial decomposition described by Eq. (4). Matrix [D] consists of the r pivotal elements in the pivotal condensation. The pivotal condensation computations which lead to these results are direct, simple and much easier than the operations called for in Eq. (8) to generate [A]* from submatrices of [A] (including the inversion of [A],,), and simpler than the algorithm of Eq. (2) for generating [AS’]and [II]. For example, applying partial decomposition through pivotal condensation to the first two columns of the matrix
Vln
_:
[Al,,
432
j [Al12
IA],,
Journal of The Franklin Institute
Xtructural Analysis
by Partial Decomposition
yields 1
*;_Q
+-
0
1; : 3_ 11
&
_._._._._ 0 0; () 0;
+* +g
Is 11 g
.
Comparing, for instance, the effort to compute [A]* from Eq. (8) with the computations here (which take 3 min of hand calculations) demonstrates the ease and efficiency of the partial decomposition. Partial
Decomposition
in the Flexibility
Method:
Key Equations
(2)
The basic equation in the flexibility method relates force vector displacement vector (u} through flexibility matrix [a] (4 = Equation
[al-P'>.
(12) can be written in partitioned
(F} to
(12)
form (13)
The superscript zero and asterisk designate respectively coordinates at redundants and external system coordinates. To satisfy the conditions of compatibility at the boundaries, displacements at the redundants are set equal to zero, and Eq. (13) is written as two separate matrix equations, (01 =
bill PI0 + [all2P’I*7 @I* = bL {FI”+ bL2 PI*. The redundant forces are computed V’>” = -
(14) (15)
from Eq. (15)
[ali;’ [all2PI*.
(16)
A reduced flexibility matrix relating forces and displacements at the external system coordinates is obtained by substituting Eq. (16) into (15), {u}*
=
[a]*(F)*
in which
[al* = W2 - [alzl [al;;”[alIz.
(l’i)
(18)
From {F)O the internal forces Pi and internal displacements ai can be computed. The solution for {F(“>‘Jand [a]* can be obtained directly from partial decomposition. Partial Decomposition
of Flexibility
Matrix
The required information in Eqs. (16) and (17) can be generated directly from submatrices [Xl, [R] and [D] which result from a partial decomposition of matrix [a] [see Eqs. (4)-(S)].
[al,-,’[al,, = [W1 [RI, [al* = [aIs - [W[Dl [RI.
(19)
(20) 433
Moshe F. Rub&stein and Vladimir Simonc”e Applying the pivotal condensation to the first r columns of [a] with r equal to the number of redundants yields directly [a]*, [S] and [R] as shown in Eq. (11). {F}O can be computed from (S} and {R} as follows. Substitute Eq. (19) into Eq. (16) and premultiply by [S],
[al @‘I0= P’h
where
(21)
(Y> = -[R](F)*.
(22)
From Eq. (22) compute {Y> for given (Ii’)*, and using {Y} solve Eq. (21) for {F}O by a backward substitution ([S] is triangular). Example I 1
-I
I, uP,FP (cl
(b)
(a)
FIG. 1. (a) Structure choice of redundant6
and applied forces. (b) Reduced {P}O and all system coordinates. coordinates.
primary structure showing (c) Elements and element
The partial decomposition is illustrated in the analysis of the frame shown in Fig. 1. The 5 x 5 flexibility matrix [a] for the coordinates in Fig. l(b) is synthesized from the element flexibility matrices, (2) * 0 _ 0
16
12
-18
; -$
12
20
-24;
-a
36 ... . 4
/ # I . ;*
24
;
24
-18
-24 . . . .. . -3 -2
[al = *
-18
24
- 18
ij
-18
-
- 18 .
Q
Applying partial decomposition to the first three columns (three redundants) gives : D,, = 16, D,, = 11, D,, = s and -1
Q
0
434
1
_Q
-&
_$
1
-N
00
1
00
0
00
0
Journal of The Franklin Institute
Structural Analysis by Partial Decomposition Using given values Pz = 2 and FEJ= 1 in Eq. (22)
{Y}=-[R](F)*=
Use {Y} in Eq. (2 1) to solve for (F}O by backward substitution,
{P}O=;
The element forces Pi [Fig. l(c)] can be computed
from {F}* and {F>O,
1 21
Partial
Decomposition
in the Stifness
Method:
Key
Equations
(2)
The basic equation in the stiffness method relates forces and displacements through a stiffness matrix [k], V’> = Equation
(23)
WIb4
can be written in partitioned
(23)
form
I( w* I
PI11 ! WI,, PI0 .@I0 ..______ (..i@-.] [__..._.......:....__.......
*
=
VI,, ; PI22
in which superscript zero designates coordinates with zero forces, {F}O = (0). Equation (24) can be written as two separate matrix equations :
Vol.290,No. 5, November 1970
(24)
i.e.
(01= WI11{u>”+ PI12 +4*,
(25)
{F}” = WI,,b1° + [kl22 @4*.
(26) 435
Moshe F. Rubinstein and Vladimir SimonZe The displacements
{u}s can be computed in terms of given displacements (u}” = -
mlY [kl,, tu)*.
{u)*
(27)
A reduced stiffness matrix [k]* relating forces and displacements at the coordinates where forces are applied, (*), is obtained by substituting Eq. (27) into (26), {Fj* =
WI*(u)*,
(28)
Where
(29)
WI* = PI22- WI21I312 [kl,,.
From (u}O, the internal forces Pi and displacements ai can be computed. The solution for {u}‘J and [ICI* can be obtained directly from partial decomposition in a way which is analogous to that used in the flexibility method. Partial Decomposition
of Stiffness Matrix
Applying partial decomposition in Eq. (24) gives
to the stiffness matrix [k] partitioned
as
(30)
Flrt CRIB= WY [RI,
(31)
WI* = PI,, - CW’PI [RI.
Using pivotal condensation, [I%]*, [S] and [R] are obtained directly. {u>‘Jis computed from [S] and [R] as follows. Substituting Eq. (30) into Eq. (27) and premultiply by [S] (32)
[XlHo = ma where {Y>=
-
(33)
[RI &I*.
From Eq. (33) compute {Y} for given {u}* [if not given solve from Eq. (ZS)]. Using {Y> solve Eq. (32) for {u}” by backward substitution. Example II The application of the partial decomposition to the stiffness method is illustrated for the structure of Fig. 2. The 5 x 5 stiffness matrix [k] for the coordinates of Fig. 2(b) is synthesized from the element stiffness matrices (2), 0
(
VI = >
436
0
12
-24
4
24
2
-24
108
-24
-96
-6
4
-24
16
0
0
24
-96
0
192
0
2
-6
0
0
8
Journal of The Franklin Institute
Structural Analysis by Partial Decomposition Applying partial decomposition to the first three yields : D,, = 12, Dz2 = 60, D,, = y and - 1
-2
0
1
Q -3%;
;2
-A
0 0 1 ; -2 ................................. 0 164 0 0
-& . -8
0
0
(0) of [k]
$-
-6
0
columns
1-8
(a)
%
_
(cl
(b)
FIG. 2. (a) Structure and applied forces. (b) System coordinates. (c) Element coordinates.
Using given values u z = 149/(8x 336), us* = 65/336, then from Eq. (33) obtain (Y} and from it compute {u>Oby backward substitution from Eq. (32),
[Ij+J. The element displacements from {u}* and (u}O,
2327 52
61 \ 63
f
I$2 65 -ii%
63 64
2327 58 377 -ET
65 hi (
S,
1 >=336
149 8 -T!
S,
65
S9
0
610
0
611 \S
Vol.290,No.5.
6, and forces Pi [Fig. 2(c)] can now be computed
12 )
November 1970
65 377 13
=-
1 21
Moshe F. Rub&stein and Vladimir SimonCe Application
of Partial Decomposition
to Dynamic
Analysis
The method of decomposition is also useful in dynamic analysis of structures (3). Partial decomposition can be used when the inertial forces at some of the coordinates are zero because a reduced stiffness matrix, [k]*, expressed by Eq. (29), is required in the dynamic analysis (4). [k]* can be determined directly from the over-all stiffness matrix [k] by partial decomposition through pivotal condensation. Partial decomposition is also useful in cases where a reduced mass matrix [ml* must be extracted from an original mass matrix [m] which includes rigid body degrees of freedom (5). Namely, by starting with
we obtain The order of [m]i, represents the number of rigid body degrees of freedom. [ml* can be obtained directly from [m] by partial decomposition through pivotal condensation the same way as [k]* is obtained from [k]. Summary
and Conclusions
The method of partial decomposition is developed and applied to structural analysis by the stiffness and flexibility methods. Two examples illustrate the applications. Partial decomposition is also useful in dynamic analysis when inertial forces are zero at some coordinates or when rigid body degrees of freedom must be eliminated in the analysis. It is also useful in stability problems when a reduced stiffness matrix is required. The method is direct and efficient even for hand calculations.
Methods in Engineering”, G. Salvadori and M. L. Baron, “Numerical Englewood Cliffs, N.J., Prentice-Hall, 1952. M. F. Rubinstein, “Matrix Computer Analysis of Structures”, Englewood Cliffs, N.J., Prentice-Hall, 1966. R. Rosen and M. F. Rubinstein, “Dynamic analysis by matrix decomposition”, ASCE Journ. of Eng. Mech. Div., EM2, Vol. 94, pp. 385395, April 1968. W. C. Hurty and M. F. Rubinstein, “Dynamics of Structures”, Englewood Cliffs, N.J., Prentice-Hall, 1964. M. F. Rubinstein and W. C. Hurty, “Dynamic analysis of a rocket model by the component modes method”, Society of Automotive Engineers, Paper No. 9253, Los Angeles, Cal., Oct. 59, 1964. M. F. Rubinstein and R. Rosen, “Structural analysis by matrix decomposition”, J. Franklin Inst., Vol. 286, No. 4, pp. 331-345, Oct. 1968.
(1) M. (2) (3) (4) (5)
(6)
438
Journal
of The Franklin
Institute
F.
by Partial
Decomposition
RUBINSTEIN
School of Engineering and Applied Science University of California, Los Angeles, California and
VLADIMIR
SIMON&
Docent na Arhitektonsko GradeLen Fakultet-Skopje,
AESTRACT
:A
Yugoslavia
partial decomposition
of
thejlexibility
to yield directly
all the required information
the coordinates,
internal
decomposition all required
of the stiffness information
when zero inertial forces is simple,
forces
effkient
and deformations
matrix,
when no forces
in the analysis. are associated
matrix of a structural
in the analysis from
system is shoun
which displacements
can be obtained.
Similarly,
are applied at some coordinates,
The approach
is also useful in dynamic
with some coordinates.
and most suitable for hand calculation
The partial
and computer
at
a partial yields
analysis
decomposition
applications.
Notation
[Al
[A]a’
[Alij ,$ IDI P’> W’ PI WI* [J$ W [RI ISI
square matrix matrix [A] in step i of pivotal condensat,ion submatrices of [A] flexibility matrix reduced flexibility matrix diagonal matrix in decomposition force vector redundant force vector stiffness matrix reduced stiffness matrix mass matrix reduced mass matrix element force vector submatrix in partial decomposition upper triangular matrix with ones on principal displacement vector auxiliary vector defined by Eqs. (22) or (33) element displacement vector
diagonal
Introduction
The analysis of structures by the flexibility and stiffness methods requires the solution of linear equations. Often, however, the order of the stiffness or flexibility matrix that is formulated exceeds the number of unknowns that must be solved. In such cases it is convenient to reduce these matrices and extract the relevant information required for the solution. The partial decomposition approach proposed here is suitable and efficient for such
429
Mode
F. Rubinstein and Vladimir SimonEe
analyses. All the relevant information is obtained in a single direct operation which is convenient for both hand calculations and computer applications. The method is also useful in dynamic analysis of structures when inertial forces are zero at some coordinates, or when rigid body degrees of freedom must be removed in the analysis. This paper develops the partial decomposition and applies it to the flexibility and stiffness methods. Examples are included to illustrate the ease and efficiency of the method.
Decomposition
A symmetrical matrix [A] can be decomposed of three matrices (1, 6) :
into the following
product
(1)
[Al = [SITLB1LfJl.
[S] is an upper triangular matrix with ones on the principal diagonal and [II] is diagonal. The elements of [S] and [II] are computed by recursion using the following algorithm : 4,
= 4,
Xi, = 1,
i = 1,2 ,..., n,
Ali
i2%
slj=4;9.
i-l
Dii = Aii-
Slj = &
ii
I: S”piDpp,
p=l
Aij - ‘&,,, p=l
i z 2,
S,,*D,,
i>2,
>
jai+l.
Partial Decomposition
Partition matrix [A] into four submatrices r r
[Al =
S
.
s
[A],, and [A],, are square matrices of orders r and s, respectively. A partial decomposition applied to the first r columns of [A] can be written in the followmg partitioned form :
1
Structural Analysis
by Partial Decomposition
in which
[Al,, = [aIT [Dl [Sl, [All, = ISIT [Dl [RI, [Al,, = [AK = [W’ [Dl [fil, - Ml,, = [W’[Dl [RI + [Al”.
(5)
Using these identities, the following useful expressions are generated which relate submatrices of [A] to the submatrices on the right-hand side of Eq. (4) :
(6)
[AId IIAl,.z= [W’ [RI>
(7)
[Al,, [AL-?[Al,, = [W’ [Dl [fil,
(8)
VI,, - CAlal[Al;;1[Al,, = [Al*. Decomposition zeros)
by Pivotal
Condensation:
(Decomposition
by working
for
(2)
The decomposition described in Eqs. (l), (2) and (4) can be achieved by the method of pivotal condensation as follows. Consider again the square matrix [A] with elements Aij. The first pivotal element A,, is set equal to Dll; the first row is divided through by A,, and used to eliminate the remaining elements of the first column. This produces a matrix of the form
(9)
in which A$ = A,/A,, and Ai;) = Aij - A:$) A, for i 2 2. Next the second pivotal element Ai;) is set equal to D22, the elements of the second row of [A](l) are divided through by A$ and then used to reduce to zero the elements of the second column below the main diagonal, yielding
[A]t2) =
1
Ai;)
A$
...
Ail,,
0
1
A$
...
Ai;
0
0
Ag’
...
A;:
.
.
0
0
A$
...
A$
.
(10)
The third pivotal element is A,,(2) = D,, and so on to A&-” = D . Matrix [AI with the number one replacing element AL”,--1)is matriF[S], and elements A!i-1) i= I,2 n, are the elements of matrix [D], i.e. A$ = A,, =?I,,: A;;) = D;,;:: ., Ap;l’ = D,,.
Vol.290,Xo. 5, Rowmber
1970
431
Moshe F. Rubinstein and Vladimir SimoniTe For example, applying the pivotal condensation
,A] _[
J4i4
p
_14
procedure
to the matrix
]
yields
Partial
Decomposition
by Pivotal
Condensation
When the pivotal condensation is applied to the first r columns of matrix [A], then the resulting matrix [A]cn) after r steps has the form 1
Ail,’
A$
. ..
A$
...
...
Ail,,
1
A$
...
A$
...
...
Agj
r
. [A](‘)
zzz
s
.
A(r) 1 ... 0 ... ... ... ... rw .___......__.__. .___.......................... A(T) 0 ... ... ... 0 r+l,r+l ... 4$l,?l . .
0
...
r
...
...
0
A(r) n,r+1
...
A(r) R?l
s
(11) The submatrices of [A]@) are identified as submatrices [Xl, [R] and [A]* in the partial decomposition described by Eq. (4). Matrix [D] consists of the r pivotal elements in the pivotal condensation. The pivotal condensation computations which lead to these results are direct, simple and much easier than the operations called for in Eq. (8) to generate [A]* from submatrices of [A] (including the inversion of [A],,), and simpler than the algorithm of Eq. (2) for generating [AS’]and [II]. For example, applying partial decomposition through pivotal condensation to the first two columns of the matrix
Vln
_:
[Al,,
432
j [Al12
IA],,
Journal of The Franklin Institute
Xtructural Analysis
by Partial Decomposition
yields 1
*;_Q
+-
0
1; : 3_ 11
&
_._._._._ 0 0; () 0;
+* +g
Is 11 g
.
Comparing, for instance, the effort to compute [A]* from Eq. (8) with the computations here (which take 3 min of hand calculations) demonstrates the ease and efficiency of the partial decomposition. Partial
Decomposition
in the Flexibility
Method:
Key Equations
(2)
The basic equation in the flexibility method relates force vector displacement vector (u} through flexibility matrix [a] (4 = Equation
[al-P'>.
(12) can be written in partitioned
(F} to
(12)
form (13)
The superscript zero and asterisk designate respectively coordinates at redundants and external system coordinates. To satisfy the conditions of compatibility at the boundaries, displacements at the redundants are set equal to zero, and Eq. (13) is written as two separate matrix equations, (01 =
bill PI0 + [all2P’I*7 @I* = bL {FI”+ bL2 PI*. The redundant forces are computed V’>” = -
(14) (15)
from Eq. (15)
[ali;’ [all2PI*.
(16)
A reduced flexibility matrix relating forces and displacements at the external system coordinates is obtained by substituting Eq. (16) into (15), {u}*
=
[a]*(F)*
in which
[al* = W2 - [alzl [al;;”[alIz.
(l’i)
(18)
From {F)O the internal forces Pi and internal displacements ai can be computed. The solution for {F(“>‘Jand [a]* can be obtained directly from partial decomposition. Partial Decomposition
of Flexibility
Matrix
The required information in Eqs. (16) and (17) can be generated directly from submatrices [Xl, [R] and [D] which result from a partial decomposition of matrix [a] [see Eqs. (4)-(S)].
[al,-,’[al,, = [W1 [RI, [al* = [aIs - [W[Dl [RI.
(19)
(20) 433
Moshe F. Rub&stein and Vladimir Simonc”e Applying the pivotal condensation to the first r columns of [a] with r equal to the number of redundants yields directly [a]*, [S] and [R] as shown in Eq. (11). {F}O can be computed from (S} and {R} as follows. Substitute Eq. (19) into Eq. (16) and premultiply by [S],
[al @‘I0= P’h
where
(21)
(Y> = -[R](F)*.
(22)
From Eq. (22) compute {Y> for given (Ii’)*, and using {Y} solve Eq. (21) for {F}O by a backward substitution ([S] is triangular). Example I 1
-I
I, uP,FP (cl
(b)
(a)
FIG. 1. (a) Structure choice of redundant6
and applied forces. (b) Reduced {P}O and all system coordinates. coordinates.
primary structure showing (c) Elements and element
The partial decomposition is illustrated in the analysis of the frame shown in Fig. 1. The 5 x 5 flexibility matrix [a] for the coordinates in Fig. l(b) is synthesized from the element flexibility matrices, (2) * 0 _ 0
16
12
-18
; -$
12
20
-24;
-a
36 ... . 4
/ # I . ;*
24
;
24
-18
-24 . . . .. . -3 -2
[al = *
-18
24
- 18
ij
-18
-
- 18 .
Q
Applying partial decomposition to the first three columns (three redundants) gives : D,, = 16, D,, = 11, D,, = s and -1
Q
0
434
1
_Q
-&
_$
1
-N
00
1
00
0
00
0
Journal of The Franklin Institute
Structural Analysis by Partial Decomposition Using given values Pz = 2 and FEJ= 1 in Eq. (22)
{Y}=-[R](F)*=
Use {Y} in Eq. (2 1) to solve for (F}O by backward substitution,
{P}O=;
The element forces Pi [Fig. l(c)] can be computed
from {F}* and {F>O,
1 21
Partial
Decomposition
in the Stifness
Method:
Key
Equations
(2)
The basic equation in the stiffness method relates forces and displacements through a stiffness matrix [k], V’> = Equation
(23)
WIb4
can be written in partitioned
(23)
form
I( w* I
PI11 ! WI,, PI0 .@I0 ..______ (..i@-.] [__..._.......:....__.......
*
=
VI,, ; PI22
in which superscript zero designates coordinates with zero forces, {F}O = (0). Equation (24) can be written as two separate matrix equations :
Vol.290,No. 5, November 1970
(24)
i.e.
(01= WI11{u>”+ PI12 +4*,
(25)
{F}” = WI,,b1° + [kl22 @4*.
(26) 435
Moshe F. Rubinstein and Vladimir SimonZe The displacements
{u}s can be computed in terms of given displacements (u}” = -
mlY [kl,, tu)*.
{u)*
(27)
A reduced stiffness matrix [k]* relating forces and displacements at the coordinates where forces are applied, (*), is obtained by substituting Eq. (27) into (26), {Fj* =
WI*(u)*,
(28)
Where
(29)
WI* = PI22- WI21I312 [kl,,.
From (u}O, the internal forces Pi and displacements ai can be computed. The solution for {u}‘J and [ICI* can be obtained directly from partial decomposition in a way which is analogous to that used in the flexibility method. Partial Decomposition
of Stiffness Matrix
Applying partial decomposition in Eq. (24) gives
to the stiffness matrix [k] partitioned
as
(30)
Flrt CRIB= WY [RI,
(31)
WI* = PI,, - CW’PI [RI.
Using pivotal condensation, [I%]*, [S] and [R] are obtained directly. {u>‘Jis computed from [S] and [R] as follows. Substituting Eq. (30) into Eq. (27) and premultiply by [S] (32)
[XlHo = ma where {Y>=
-
(33)
[RI &I*.
From Eq. (33) compute {Y} for given {u}* [if not given solve from Eq. (ZS)]. Using {Y> solve Eq. (32) for {u}” by backward substitution. Example II The application of the partial decomposition to the stiffness method is illustrated for the structure of Fig. 2. The 5 x 5 stiffness matrix [k] for the coordinates of Fig. 2(b) is synthesized from the element stiffness matrices (2), 0
(
VI = >
436
0
12
-24
4
24
2
-24
108
-24
-96
-6
4
-24
16
0
0
24
-96
0
192
0
2
-6
0
0
8
Journal of The Franklin Institute
Structural Analysis by Partial Decomposition Applying partial decomposition to the first three yields : D,, = 12, Dz2 = 60, D,, = y and - 1
-2
0
1
Q -3%;
;2
-A
0 0 1 ; -2 ................................. 0 164 0 0
-& . -8
0
0
(0) of [k]
$-
-6
0
columns
1-8
(a)
%
_
(cl
(b)
FIG. 2. (a) Structure and applied forces. (b) System coordinates. (c) Element coordinates.
Using given values u z = 149/(8x 336), us* = 65/336, then from Eq. (33) obtain (Y} and from it compute {u>Oby backward substitution from Eq. (32),
[Ij+J. The element displacements from {u}* and (u}O,
2327 52
61 \ 63
f
I$2 65 -ii%
63 64
2327 58 377 -ET
65 hi (
S,
1 >=336
149 8 -T!
S,
65
S9
0
610
0
611 \S
Vol.290,No.5.
6, and forces Pi [Fig. 2(c)] can now be computed
12 )
November 1970
65 377 13
=-
1 21
Moshe F. Rub&stein and Vladimir SimonCe Application
of Partial Decomposition
to Dynamic
Analysis
The method of decomposition is also useful in dynamic analysis of structures (3). Partial decomposition can be used when the inertial forces at some of the coordinates are zero because a reduced stiffness matrix, [k]*, expressed by Eq. (29), is required in the dynamic analysis (4). [k]* can be determined directly from the over-all stiffness matrix [k] by partial decomposition through pivotal condensation. Partial decomposition is also useful in cases where a reduced mass matrix [ml* must be extracted from an original mass matrix [m] which includes rigid body degrees of freedom (5). Namely, by starting with
we obtain The order of [m]i, represents the number of rigid body degrees of freedom. [ml* can be obtained directly from [m] by partial decomposition through pivotal condensation the same way as [k]* is obtained from [k]. Summary
and Conclusions
The method of partial decomposition is developed and applied to structural analysis by the stiffness and flexibility methods. Two examples illustrate the applications. Partial decomposition is also useful in dynamic analysis when inertial forces are zero at some coordinates or when rigid body degrees of freedom must be eliminated in the analysis. It is also useful in stability problems when a reduced stiffness matrix is required. The method is direct and efficient even for hand calculations.
Methods in Engineering”, G. Salvadori and M. L. Baron, “Numerical Englewood Cliffs, N.J., Prentice-Hall, 1952. M. F. Rubinstein, “Matrix Computer Analysis of Structures”, Englewood Cliffs, N.J., Prentice-Hall, 1966. R. Rosen and M. F. Rubinstein, “Dynamic analysis by matrix decomposition”, ASCE Journ. of Eng. Mech. Div., EM2, Vol. 94, pp. 385395, April 1968. W. C. Hurty and M. F. Rubinstein, “Dynamics of Structures”, Englewood Cliffs, N.J., Prentice-Hall, 1964. M. F. Rubinstein and W. C. Hurty, “Dynamic analysis of a rocket model by the component modes method”, Society of Automotive Engineers, Paper No. 9253, Los Angeles, Cal., Oct. 59, 1964. M. F. Rubinstein and R. Rosen, “Structural analysis by matrix decomposition”, J. Franklin Inst., Vol. 286, No. 4, pp. 331-345, Oct. 1968.
(1) M. (2) (3) (4) (5)
(6)
438
Journal
of The Franklin
Institute