Copy of Finite Element Formulation

8/14/2019 Copy of Finite Element Formulation

1/38

Dr. B. S. SARMA


2/38

Overview

semi-discretization-Variational Equations of Linear Solid Mechanics

Finite Element Discretization

Analytical Solution Procedures

-Modal Superposition Method-Eigen value economizer Method

-Free Response , Damped Eigen values

Forced Periodic Response

Transient Response

-Frequency Response Method

-Modal Decomposition Method


3/38

With FEM, many dynamic and nonlinear numerical simulations

are now possible within the realm of engineering analysis andconsiderable effort has been directed towards the area of structuraldynamics.

The most common approach to structural dynamics is the one in

which the spatial and temporal solutions are performedindependently. This is called SEMI-DISCRETIZATION.The governing partial differential equations are first discretized in

space yielding a coupled system of ordinary differential equationsin time which are solved using various integration methods.

Finite Element Method is an example ofSEMI-DISCRETIZATION approach for a multi-dimensional linearcontinuum.


4/38

In the FINITE ELEMENT FORMULATION, one startswith the equation of a continuum and their variationalforms and use the finite element method to construct the

discrete equations of motion.

The continuum formulation can be extended to includestructural mechanics theories such as Beams, Plates and

Shells.


5/38

In the analysis of small deformations of a solid continuum, thegoverning equations are obtained from the MOMENTUMCONSERVATION LAW and they are referred to as EQUATIONSOF MOTION. In the absence of body forces they are written as

ij,j = i in (1) ij is the stress tensor

is the density

u is the displacementand represents the continuum domain.


6/38

The INFINITESIMAL STRAIN is related to displacements

by ij = (ui,j + uj,i ) (2)

The constitutive law relating stress to strain for elastic materialis given by

ij = Cijklkl (3)where Cijkl is the elastic tensor.

The continuum has a boundary which can be divided intou where displacements are prescribed and where stresses are prescribed.

These boundary conditions are written asui = ui* on u and ijnj = i* on (4)


7/38

In addition we have initial conditionsui(0) = u0i and i(0)= 0i (5)

Equations (1) (4) and (5) represent the strong form of theinitial/ boundary value problem ofElasto-Dynamics.

To facilitate the finite element discretization, the weakform of the Variational statement corresponding to theseequations is needed.


8/38

Let a test function w be defined such that it is a function of spaceonly and is identically zero on the boundary u. Then thevariational statement is as follows:

Given the prescribed tractions i* , initial displacements uoi and

velocity oi , find ui (t) such that for all wi

M (w,u) + Fint

(w,u) = Fext

(w, ) (6)

Where M(w,u) = wi i d (7)

Fint = wi,j ij d (8)

Fext

= wi i* d (9)


9/38

Equations (7) to (9) correspond to the work done by the inertial,internal and external forces respectively.

It is assumed that ui

is chosen such that the essential boundaryconditions are exactly satisfied.

For a linear elastic case equation (8) can alternatively be written byusing the equation (3) and the symmetry of the stress tensor as

F

int=

w(i,j) Cijkl u(k,l) d (10)


10/38

In Finite element semi-discretization, the domain is subdividedinto elements e The displacement field in each element is approximated by

ui(x,t) = NI (x) diI (t) (11)where

the shape function NI(x) are functions of space alone anddiI (t) include the time dependence.

Introducing eqn.(11) in matrix formu = N de (12)

where de

are the displacements local to element e. The discrete matrix form of the gradient u (i,j) can then be written

as

s u = B de (13)

where B contains the appropriate derivatives of Ni


11/38

Considering that the same shape functions are used for both uiandwi( as in the Galerkin approximation), the semi-discretization of eqn

(6) yields

dT (M + Fint - Fext ) = 0 (14)

where d is a vector of nodal parameters resulting from theinterpolation.

Global mass matrix M; internal force vector Fint ; external force vectorF ext and nodal displacement vector d are assembled from elementcontributions where

M = A

e

M

e

(15)F

int= A

eF

int .e(16)

F

ext= A

eF

ext.e(17)

d = Aede (18)


12/38

The element matrices are defined using Eqns.(11), (12) and (13)to be

Me = [ Me

ijIJ] = e ij Ni Nj d (19)

Fint,e

= e Bt d (20)

Fext,e

= {FiIext,e

} = > e Ni *i d (21)

Since Eqn. (14) must hold for any d, the equation of motion becomes

M + Fint = Fext (22)

For a linear elastic material, the equation of motion can alternativelybe written by using Eqn (10) as

M + Kd = Fext (23)

where K is the global stiffness matrix assembled from element

contribution i .e, K = AeK

e(24)

and K = e Bt DB d (25)


13/38

To include viscous damping in a consistent manner one can considerdamping matrices of the form (Rayleighs Damping )

C = a M + b Kwhere the coefficients a and b control the amount of damping

proportionality with mass and stiffness respectively a and b aredetermined experimentally.

Now the damped equation of motion is given byM + C + Kd = F

ext(27)

To account for mass proportional damping effect, Eqn (1) to be

replaced by ij,j = i + a in (28)and for the stiffness proportional damping, Eqn (3) is to be augmented

as ij = Cijkl (kl + b ck,l) (29)

Since M and C are derived from a variational form,they are referred to


14/38

CONSISTENT AND LUMPED MASS MODELS Consistent Mass Matrices obtained from Eqns.(15) and (19)are one

form of mass matrices used in current FE S/W packages. In the context of Direct Time integration solutions Lumped Mass

matrices are often considered. They are diagonal matrices with off-diagonal terms as zeros.

Another alternative is the so called higher order matrices whichare a combination of both consistent and lumped matrices.

Implicit methods can operate on larger time steps but require thesolution of equations at each time step. Consistent mass matrices have

the same sparcity and bandwidth as that of the stiffness matrix. So thecomputational work required for the solution is the same forconsistent or lumped matrices.

Explicit time integration schemes require a diagonal mass matrix to be

truly explicit. Therefore lumped mass models are used.


15/38

MASS LUMPING SCHEMES

For Low order elements in continuous formulation the simplest procedure is toassign an equal fraction of total mass of each element to each node.

Me = [ MeijIJ ] = ij IJ/ nen e d

where nen is the number of nodes per element. Another simple procedure is the row sum technique

[ MeijIJ ] = MeijIJThe same can be written for continuum elements as

[ MeijIJ ] = ij e NI d The above lumped mass scheme conserves mass in that

MeijIJ = ij e

nen

J=1

J=1

nen


16/38

For higher order elements nodal wuadrature method is normallyemployed. Here the quadrature points coincides with the nodesresulting in a lumped mass.

Labatto Quadrature rules such as Trapezoidal rule and Simpsonsrule are used to diagonalize the mass amtrex.

By definition of shape functions, Numerical quadrature finallybecomes

[ MeijIJ ] = ij (1)J(1)wi A draw back of nodal quadrature is that, in some cases, zero or

negative masses are obtained.

A Scheme that always results in positive lumped masses suggestedis

[ MeijIJ ] = ij Ni2 d

Where is so chosen that the total element mass is conserved.


17/38

ANALYTICAL SOLUTION PROCEDURES As a result of semi-discretization, many time dependent problems

can be reduced to a system of ordinary differential equations of thecharacteristic form given by

M + C + Kd = Fext

(30) Systems of linear ordinary differential equations with constant

coefficients can always be solved in principle analytically without theintroduction of additional approximations.

Here we shall deal with

a. Determination of free vibration response (i.e Fext

(t)=0)

b. Determination of periodic response (i.e (i.e Fext

(t) is periodic)c. Determination of transient response (i.e (i.e F

ext(t) is arbitrary)

In the first two cases, initial conditions of the systems are of no

importance and a general solution is sought.


18/38

MODAL ANALYSIS

Regardless of the type of structure analyzed or semi-discretization

method used, Eqn (30) represents a standard form of the discreteequation of motion for linear structural dynamics

The coupled system of ordinary differential equations possessescertain attributes such as M, C, K are symmetric and M as positive

definite and K are semi-definite matrices.

Two Techniques to solve these equations, for the given initialconditions are

* Direct Time Integration and Modal Superposition

In Direct Time Integration, the solution is marched through timestep by step starting with the initial conditions and progressing to

find the final solution.


19/38

Modal Analysis is a procedure to transform the coupled system ofordinary differential equations described by Eqn (27) into a set ofuncoupled equations representing the natural modes of the system.

Basic understanding of Modal Analysis is essential even if DirectTime Integration is used as a solution scheme.

WithModal Superposition, the solution for all time is obtainedsimultaneously by a superposition of contributions from thenatural modes of the system.


20/38

THE MODAL SUPERPOSITION METHOD

Consider the homogeneous, undamped equation of motion

M + kd = 0 (31)Assume possible solutions to Eqn (31) of the form

d (t) = sin( t + ) (32)where is the natural frequency and is a constant phase angle

Differentiating Eqn (32) approximately and substituting it into Eqn(31), we get

(K - 2M ) sin ( t + ) = 0 (33)

i.e [K 2M] = 0 (34)

This is an Eigen value problem and its non- trivial solution exists ifDet |(K-2M)| = 0 (35)


21/38

When expanded out, Eqn (35) becomes an neqth degree algebraic

equation called the characteristic equation.

The neqroots of this equation are1,22232 . . . n2 and thesefrequencies are generally placed in the order

12 < 2

2


22/38

From the property of modal orthogonality, we have

rT

M s = 0 for r s (38)

rT

K s = 0 for r s (39) To prove the above statements, we can write, from Eqn (34) for

any moder

2 Mr = K r (39a)

s2

Ms = K s (39b)

Pre-multiplying the first by sT

and the second by rT

and thensubtracting, we get (Noting that M is symmetry)

(r2

- s2

)(rT

M s) = 0 (39c) And if r s, the orthogonality condition for matrix M is

proved. Immediately from this, orthogonality of the matrix Kfollows.

Finally the modal matrix defined as = [1, 2, 3, 4.. n eq] (40)


23/38

The coupled system of equations can now be transformed to a setof uncoupled equations of modal components,

Expanding the solution d(t) as the sum

d(t) = 1Z1(t) + 2Z2(t) + n eqZ n eq(t) = Z(t) (41)

where z(t) are the normal coordinates that incorporate the timedependence.

Substituting Eqn (41) into the Eqn (23) and pre-multiplying by

rT

, we have representing each mode of vibration

rTM Z(t) + r

TK Z(t) = r

TF

ext(42)


24/38

In the modal superposition method, the global system of Eqns (23) istransformed into the set of single degree of freedom problems (43) bythe normal coordination transformation.

Each of the single degree of freedom problems is solved using asuitable transformation of initial conditions.

Finally the solution in physical coordinates is found by the normalcoordinate transformation which is a superposition of various modalequations.

Using orthogonality conditions Eqn (38,39) and definitions ofEqn(36,37) on Eqn (42), we obtain the uncoupled set of ordinarydifferential equations in time

Z(t) + r2 Zr(t) = Fr for r = 1,2,3.. n eqns (43)

where Fr = rT Fext (44)


25/38

If Rayleigh damping Eqn (26) is used, the same normal modecoordinate transformation may be applied to Eqn (27) to obtain aset of damped single degree of freedom equations of he form

Zr(t) + 2 r rZr(t) + r2

Zr(t) = fr , for r =1,2,3n eq (45)

The modal fraction of critical damping r is determined by themass and stiffness proportionality coefficients a, b and rby therelation

r = (a/ r+ b r)/2 (46)


26/38

EIGEN VALUE ECONOMIZER METHOD:

Computer effort of determining the Eigen values and Eigenvectors is larger by an order of magnitude than the solution of anequivalent static problem.

Fortunately, reasonably good Eigen values can be determined withfewer degrees of freedom than needed for static solutions.

If a rather fine sub-division is used in the dynamic analysis, a

number of degrees of freedom can be eliminated and the mass anddamping effects can be lumped at a reduced number ofnodalparameter.

[similar to the sub-structure analysis used in static problems]


27/38

Let the nodal vector d be divided into two partsd = { } (47)

Assume that the displacements ds

depend in some uniqueway on the displacement dm. i.e ds is called master and dmis called slave variables.If dm = T ds (48)

We haved = [ ] dm => T* dm (49)

Where T is the matrix specifying the dependence.

Now the dynamic equation of the whole systemM + k d = 0 (50)

can be reduced by applying the constraint on the

deformation freedom implied in Eq (49)

ds

dm

I

T


28/38

Pre-multiplying Eqn (50) by T*T (principles of transformation) aftersubstituting in Eqn (49), we get

M* m + K* dm = 0

Where K = T*T K T and M* = T*T M T (51)Eqn (51) involves smaller number of variables.

How to determine the relationship between the slave and the

master deflections:-- It is reasonable to assume that the general pattern ofdeformation will follow that which would be obtained byimposing displacements dm on an otherwise unloaded structure in

static condition.- Thus partitioning with = = 0, the standard equation can

be written as

Kd = [ ] { } = { } (52)

Kss Ksm

Ksm Kmm

ds

dm

0

fm


29/38

As the slave loads are unloaded, we can get

Kss ds + Ksm dm = 0 (53)

or, ds = - Kss -1 Ksm dm

i.e T = - Kss -1Ksm (54)

The best choice of the master nodes

- The obvious choice is to eliminate first the nodes with little or nomass attached.

- In an automatic procedure, the ratio of diagonal stiffness andmass terms be first calculated (kii / Mii) and the nodes with the

largest values of this ratio be first eliminated .-One drawback of the condensation is that the mass matrix in thereduced problem is never diagonal .


30/38

FREE RESPONSE, DAMPED EIGENVALUES:

For Free vibration conditions, the standard equation (27) iswritten as

M + c + K = 0 (55)

And substituting d= et

The characteristic equation becomes

[(2

M + C + K) { }] = 0 (56)

Where and in general are found to be complex


31/38

FORCED PERIODIC RESPONSE If the forcing term in the standard equation (30) is periodic or

more generally if we can express it asf = f e

t(57)

Then a general solution can be written as

d = et

(58)

Where d is a complex value ( i.e = 1 + i 2 )

Substituting the above equation in Eqn (30), we get[2 M + c + k ] => D = - f (59)

It is no longer an Eigen value problem. But it can be solved byinverting the matrix D i.e = -D

-1f (60)

Eqn(60) is precisely of the same form as that used for staticproblems. But, now it has to be determined in terms of complex

quantities.

-

The computation can be arranged directly noting that


32/38

The computation can be arranged directly, noting that

et

= et

(cos 2t + i sin 2t)f = f1 + if2 (61)

= 1 + i 2in which 1 , 2 ,f1 , f2 ,1, 2 are real quantities

Inserting the above quantities in equation (59)

[ ] { } = -{ } 62)

Eqn(62) form a system in which all quantities are real and fromwhich the response to any periodic input can be obtained by thedirect solution.

The system is no longer positive definite (It is skew symmetric) With periodic Inputs , the solution after an initial transient is not

sensitive to the initial conditions and this guessed solutionrepresents the final solution.

It is valid for problems of dynamic structural response

- - -

((12 2

2 )M + 1 C +K) (-212M -1c)

(212 M + 2C) ((12 22 )M + 1C + K )

1

2

f1

f2


33/38

Steady state general solutions neither took care of any account ofinitial conditions nor of the non-periodic form of the forcing terms

The response of dynamic system taking these features requireseither a full-time discretization or special analytical procedures.

Two broad possibilites include

* FREQUENCY RESPONSE PROCEDURE* THE MODAL ANALYSIS PROCEDURE

TRANSIENT RESPONSE BY ANALYTICAL PROCEDURES


34/38

FREQUENCY RESPONSE PROCEDURE

The response of the system to any forcing terms of the general

periodic type or in particular, to a periodic forcing functionf = f eit was already discussed.

As a completely arbitrary forcing function can be represented bya Fourier series

f(t) = f0 + (am cos mt + bm sinm mt)or in the limit exactly as Fourier Integral, the response to such aninput can be obtained by a synthesis of a curve representing theresponse of any quantity. e.g: Displacement at a particular pointto all frequencies ranging from zero to infinity.

In fact only a limited number of such forcing frequencies need tobe considered and the results can be efficiently synthesized by fastFourier transform techniques

The technique of frequency response is adapted to problems where

damping matrix C is of an arbitrary specified form.

-

i=1

m


35/38

MODAL DECOMPOSTION ANALYSIS

This procedure is the most important and is widely used in

practice. It provides insight into the behavior of the whole system,where strictly numerical procedures are used. Starting with general problem of equation(2 7) i.e

M + C + Kd = F(t) (64)

where f(t) is an arbitrary function of time. The general solution for the free response is of the form

d = et

i.e i eit

(65)Where di are the Eigen values and i are Eigen vectors. For Forced response, assume that the solution can be written in a

linear combination of modes asd = i yi = [ 1 2 3 n] y (66)

Where the scalar mode participation factor Y(t) is a

function of

i=1

m


36/38

Such an assumed solution clearly gives the proportions of eachmode occurring in the response. This assumption does not presentany restriction since all the modes are linearly independent vectors

Substituting Eqn (66) into equation(64) and pre-multiplying theresult by it (i = 1,2,3,4.. n), then the resulting equations aresimply a set of scalar independent equations.

miyi + Ciy + ki yi = fi (67)

mi = it M i => 1 (if modes normalized)

Ci = it C i ; Ki = it C i and fi = it f(t).

As for true Eigen values iit Mj = it C j = it K j = 0(68)

Each scalar equation of eqn (67) can be solved by usual


37/38

Assuming that the modes have been normalized so that mi = 1,the eqn (66) can be written in a standard form as of second degree:

y + 2ci yi + i2 yi = fi (69)

The general solution can be of the formyi = fi e sin i(t-) d (70)

Such integration can be carried out numerically and the modalresponse is obtained.

In principle, superposition will result in the full transient response.

Often a single calculation is carried out for each mode todetermine the maximum responses and a suitable addition of theseresults is used as a standard procedure.

-c i (t- )


38/38

Thank You

Date post:	30-May-2018
Category:	Documents
Upload:	ntrjn
View:	220 times
Download:	1 times

Copy of Finite Element Formulation

Documents