isoparametric elements and solution techniques. advanced design for mechanical system - lec...
Post on 15-Jan-2016
218 views
TRANSCRIPT
Isoparametric elements and solution techniques
Advanced Design for Mechanical System - Lec 2008/10/09 2
Advanced Design for Mechanical System - Lec 2008/10/09 3
= ½ d1-2Tk1-2d1-2 +
+ ½ d2-4Tk2-4d2-4 +….=
= ½ DTKD
Advanced Design for Mechanical System - Lec 2008/10/09 4
R=KD
• gauss elimination
• computation time:
(n order of K, b bandwith)
Advanced Design for Mechanical System - Lec 2008/10/09 5
recall: gauss elimination
Advanced Design for Mechanical System - Lec 2008/10/09 6
rotations
Advanced Design for Mechanical System - Lec 2008/10/09 7
Advanced Design for Mechanical System - Lec 2008/10/09 8
isoparametric elements
isoparametric: same shape functions for both displacements and coordinates
Advanced Design for Mechanical System - Lec 2008/10/09 9
computation of B
x = du / dX
• but u=u(, ), v=v (, )
Advanced Design for Mechanical System - Lec 2008/10/09 10
Advanced Design for Mechanical System - Lec 2008/10/09 11
• J11* and J12* are coefficients of the first row of J-1
and
Advanced Design for Mechanical System - Lec 2008/10/09 12
Advanced Design for Mechanical System - Lec 2008/10/09 13
gauss quadrature
Advanced Design for Mechanical System - Lec 2008/10/09 14
Advanced Design for Mechanical System - Lec 2008/10/09 15
Advanced Design for Mechanical System - Lec 2008/10/09 16
Advanced Design for Mechanical System - Lec 2008/10/09 17
Advanced Design for Mechanical System - Lec 2008/10/09 19
Advanced Design for Mechanical System - Lec 2008/10/09 20
no strain at the Gauss points
so no associated strain energy
Advanced Design for Mechanical System - Lec 2008/10/09 21
The FE would have no resistance to loads that would activate these modes
Global K singularUsually such modes superposed to ‘right’ modes
Advanced Design for Mechanical System - Lec 2008/10/09 22
Advanced Design for Mechanical System - Lec 2008/10/09 23
calculated stress =EBd are accurate at Gauss points
Advanced Design for Mechanical System - Lec 2008/10/09 24
• the locations of greatest accuracy are the same Gauss points that were used for integration of the stiffness matrix
Advanced Design for Mechanical System - Lec 2008/10/09 25
Advanced Design for Mechanical System - Lec 2008/10/09 26
Rayleigh-Ritz method
Guess a displacement set that is compatible and satisfies the
boundary conditions
Advanced Design for Mechanical System - Lec 2008/10/09 27
• define the strain energy as function of displacement set
• define the work done by external loads• write the total energy as function of the
displacement set• minimize the total energy as function of
the displacement and find• simulataneous equations that are solved
to find displacements
Advanced Design for Mechanical System - Lec 2008/10/09 28
= (d)
d / d d1 = 0d / d d2 = 0d / d d3 = 0d / d d4 = 0……d / d dn = 0
Advanced Design for Mechanical System - Lec 2008/10/09 29
Advanced Design for Mechanical System - Lec 2008/10/09 30
Advanced Design for Mechanical System - Lec 2008/10/09 31
patch tests
• only for those who develops FE
Advanced Design for Mechanical System - Lec 2008/10/09 32
substructures
Advanced Design for Mechanical System - Lec 2008/10/09 33
• divide the FEmodel in more parts
• create a FE model of each substructure
• Assemble the reduced equations KD=R
• Solve equations
Advanced Design for Mechanical System - Lec 2008/10/09 34
Simmetry
Advanced Design for Mechanical System - Lec 2008/10/09 35
Advanced Design for Mechanical System - Lec 2008/10/09 36
Advanced Design for Mechanical System - Lec 2008/10/09 37
Constraints
CD – Q =0
C is a mxn matrix
m is the number of constraints
n is the number of d.o.f.
How to impose constraints on KD=R
Advanced Design for Mechanical System - Lec 2008/10/09 38
way 1 – Lagrange multipliers
=[1 2 …. m]T
T [CD-Q]=0
= 1/2DTKD – DTR + T [CD-Q]
Advanced Design for Mechanical System - Lec 2008/10/09 39
• remember
dAD / dD = AT
dDTA/ dD = A
Advanced Design for Mechanical System - Lec 2008/10/09 40
example
Advanced Design for Mechanical System - Lec 2008/10/09 41
Advanced Design for Mechanical System - Lec 2008/10/09 42
way 2- penalty method
Advanced Design for Mechanical System - Lec 2008/10/09 43
½tT t = ½ [(CD-Q)T(CD-Q)]== ½ [(CD-Q)T(CD- Q)]== ½ [(CD-Q)TCD- (CD-Q)T Q)]== ½ [(DTCTCD-QTCD-DTCTQ+QTQ)]= ½[·];d(½[·])/dD==½[2(CTC)-(QTC)T- CTQ]==½[2(CTC)-(C)TQ- CTQ]==½[2(CTC)-CT Q- CTQ]=
Advanced Design for Mechanical System - Lec 2008/10/09 44
=½[2(CTC)-CT Q- CTQ]== CTC-CT Q
(= T)
Advanced Design for Mechanical System - Lec 2008/10/09 45