Isoparametric elements and solution techniques

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= ½ d1-2Tk1-2d1-2 +

+ ½ d2-4Tk2-4d2-4 +….=

= ½ DTKD

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R=KD

• gauss elimination

• computation time:

(n order of K, b bandwith)

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recall: gauss elimination

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rotations

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isoparametric elements

isoparametric: same shape functions for both displacements and coordinates

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computation of B

x = du / dX

• but u=u(, ), v=v (, )

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• J11* and J12* are coefficients of the first row of J-1

and

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gauss quadrature

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no strain at the Gauss points

so no associated strain energy

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The FE would have no resistance to loads that would activate these modes

Global K singularUsually such modes superposed to ‘right’ modes

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calculated stress =EBd are accurate at Gauss points

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• the locations of greatest accuracy are the same Gauss points that were used for integration of the stiffness matrix

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Rayleigh-Ritz method

Guess a displacement set that is compatible and satisfies the

boundary conditions

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• 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

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= (d)

d / d d1 = 0d / d d2 = 0d / d d3 = 0d / d d4 = 0……d / d dn = 0

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patch tests

• only for those who develops FE

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substructures

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• divide the FEmodel in more parts

• create a FE model of each substructure

• Assemble the reduced equations KD=R

• Solve equations

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Simmetry

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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

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way 1 – Lagrange multipliers

=[1 2 …. m]T

T [CD-Q]=0

= 1/2DTKD – DTR + T [CD-Q]

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• remember

dAD / dD = AT

dDTA/ dD = A

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example

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way 2- penalty method

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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]=

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=½[2(CTC)-CT Q- CTQ]== CTC-CT Q

(= T)

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