NetNews Usenet Archive 1993 #3/NN_1993_3.iso/spool/sci/math/numanal/3910

Text File | 1993-01-21 | 5.3 KB | 103 lines

Newsgroups: sci.math.num-analysis
Path: sparky!uunet!cs.utexas.edu!csc.ti.com!tilde.csc.ti.com!pan.mc.ti.com!a722756
From: [e-mail] (W. Donald Rolph)
Subject: Re: HELP!!(Cholesky Decompsition for negative-definite matrix ???)
Message-ID: <[e-mail]>
Originator: [e-mail]
Sender: [e-mail] (USENET News System)
Organization: Texas Instruments / Attleboro Mass / USA
References: <[e-mail]> <[e-mail]> <[e-mail]>
Date: Thu, 21 Jan 1993 13:41:21 GMT
Lines: 90


te pages
experiment in TI quickly.  Certain of the mail handlers (the rnate pages
experiment in TI quickly.  Certain of the mail handlers (the rnaIn article <[e-mail]>, [e-mail] (Chen, Shen Yeh) writes:
|> In article <[e-mail]>, [e-mail] (Makarand A. Kulkarni) writes...
|> >In article <[e-mail]>, [e-mail] (Chen, Shen Yeh) writes:
|> >|>
|> >|>    I need to solve a linear system equation. In fact, that's the stiffness
|> >|>  matrix [K] from finite element method. The system is like :
|> >|>  [K]*{U}={P}.  where [K](nxn)
|> >|>
|> >|>    [K] IS ALWAYS SYMETRIC, but not necessarily positive-definite. If it is
|> >|>  singular, we think this system to be unstable. Because of the symetry, I
|> >|>  always use Cholesky Decompsition to reduce the size of [K]. However, now
|> >|>  I have some situation that [K] is negative-definite but I still have to
|> >|>  solve the equation. I have not try Gauss method yet, however.
|> >|>
|> >|>    Does anyone kow :
|> >|>    1) Is there any method I can use to reduce [K], and still solve the system
|> >|>       when [K] is negative-definite ? (Note that [K] IS ALWAYS SYMETRIC)
|> >|>    2) If such method exists, what's the criteria for positive- or negative-
|> >|>       definite?
|> >|>    3) If no such method exits,( We can not take advantage of symetry to
|> >|>       reduce the size of [k], I mean.), does Gauss method work? What's
|> >|>       the criteria for question (2)?
|> >|>
|> >|>  NOTE : (1) In such system, we do not do pivotting among the rows of [K],
|> >|>             if using Gauss method. If there is a zero on the diagonal,
|> >|>             we consider it as unstable.


|>
|>   This is from Shrn-yeh Chen, for further discription of the problem.
|>
|>   This is hard to explain here. In fact, if a structure is under 'unloading'
|> condition, [k] will be negative-definite. That is, in general condition,
|> an incremental load will cause the total strain energy increase-if the
|> structure is 'intended' to be stable. Other wise, there is condition call
|> 'snapping through ' happening.
|>
|>
|>    For example, a loaded sturcture like the FIG1   will have the behavior
|> described    above:
|>
|>  ###########################################################################
|>  ##   @                   ##   |      *  *                                ##
|>  ## A  \                  ##   |   *   |    *                      *      ##
|>  ##      \                ##   | *     |      *                   *       ##
|>  ##        \              ##  L|*      |        *  <DEFLECTION>  *        ##
|>  ##          \ <----------##  O|-------|------------------------*-----    ##
|>  ##          /    LOAD    ##  A|       |          *            *          ##
|>  ##        /              ##  D|       |            *         *           ##
|>  ##      /                ##   |       |              *     *             ##
|>  ## B  /                  ##   |       |                 *                ##
|>  ##   @                   ##           |C               D|                ##
|>  ########< FIG1 >##########################< FIG2 >#########################
|>
|>    Please see FIG2.    [K] will be sigular on C & D
|>                        [K] SHOULD be negative-definite from C to D
|>
|>   HOWEVER, SOMEONE TELL ME THAT,  IF [K] IS NEGATIVE, THEN  -[K] SHOULD BE
|>   POSITIVE!! SO THERE IS NO PROBLEM!!!!!!!
|>   IS THIS CORRECT ???   IF YES(OR NO), CAN ANYONE PROVE IT???
|>
|>   OR MAYBE [K] IS NOT NEGATIVE FROM C TO D. I HAVE PROVE IT TO BE NEGATIVE
|>   , ALTHOUGH IT MAY BE WRONG.
|>
|>
|>   THANK ALL OF YOU AGAIN !!
|>

You have what is typically considerd a buckling problem.  I analyze these classes
of problems regularly in production.  They are trivial  to analyze if the force
loading is aligned with the deflection (I push sideways on the end of a beam, and
it deflects sideways, mathematically the inner  product of the force vector and
the displacement vector arenot orthogonal).  In  that case simply apply the
loading as a displacement based loading and the problem is unconditionally stable
(so long as the material tensor remains positive definite tough luck with perfect
plasticity).

If the force vector and displacement vector are orthoganal (buckling collapse of
beam under end loading) then one needs to go to the various buckling algorithm
which have been developed to stablize this problem.  ADINA and ABAQUS both
include this algorithm as part of the delivered code, though I am having trouble
at the moment tracking down a journal reference.
--

Regards.

Don Rolph  [e-mail] WD3 MS10-13 (508)-699-1263