Matrix iteration repear, modal constirbutions and static correction

Truncation Errors,CorrectionProcedures

Giacomo Boffi

Rayleigh-RitzExample

Subspace iteration

How manyeigenvectors?

Truncation Errors, Correction Procedures

Giacomo Boffi

Dipartimento di Ingegneria Civile e Ambientale, Politecnico di Milano

May 20, 2015


Giacomo Boffi


Subspace iteration


Outline

Rayleigh-Ritz Example

Subspace iteration

How many eigenvectors?Modal Partecipation FactorDynamic magnification factorStatic Correction

RR Example

m

m

m

m

k

k

k

k

m

k

x5

x4

x3

x2

x1

The structural matrices

K = k

+2 1 0 0 01 +2 1 0 00 1 +2 1 00 0 1 +2 10 0 0 1 +1

M = m1 0 0 0 00 1 0 0 00 0 1 0 00 0 0 1 00 0 0 0 1

The Ritz base vectors and the reduced matrices,

=

0.2 0.50.4 1.00.6 0.50.8 +0.01.0 1.0

K = k

[0.2 0.20.2 2.0

]M = m

[2.2 0.20.2 2.5

]

Red. eigenproblem ( = 2m/k):[2 22 2 22 2 20 25

]{z1z2

}=

{00

}The roots are 1 = 0.0824, 2 = 0.800, the frequencies are1 = 0.287

k/m [ = 0.285], 2 = 0.850

k/m [ = 0.831], while the

k/m normalized exact eigenvalues are [0.08101405, 0.69027853].The first eigenvalue is estimated with good approximation.

RR Example

m

m

m

m

k

k

k

k

m

k

x5

x4

x3

x2

x1


K = k

+2 1 0 0 01 +2 1 0 00 1 +2 1 00 0 1 +2 10 0 0 1 +1

M = m1 0 0 0 00 1 0 0 00 0 1 0 00 0 0 1 00 0 0 0 1


=

0.2 0.50.4 1.00.6 0.50.8 +0.01.0 1.0

K = k

[0.2 0.20.2 2.0

]M = m

[2.2 0.20.2 2.5

]


]{z1z2

}=

{00

}

The roots are 1 = 0.0824, 2 = 0.800, the frequencies are1 = 0.287

k/m [ = 0.285], 2 = 0.850



RR Example

m

m

m

m

k

k

k

k

m

k

x5

x4

x3

x2

x1


K = k

+2 1 0 0 01 +2 1 0 00 1 +2 1 00 0 1 +2 10 0 0 1 +1

M = m1 0 0 0 00 1 0 0 00 0 1 0 00 0 0 1 00 0 0 0 1


=

0.2 0.50.4 1.00.6 0.50.8 +0.01.0 1.0

K = k

[0.2 0.20.2 2.0

]M = m

[2.2 0.20.2 2.5

]


]{z1z2

}=

{00

}The roots are 1 = 0.0824, 2 = 0.800, the frequencies are1 = 0.287

k/m [ = 0.285], 2 = 0.850




Giacomo Boffi


Subspace iteration


Rayleigh-Ritz Example

The Ritz coordinates eigenvector matrix is Z =[

1.329 0.031700.1360 1.240

].

The RR eigenvector matrix, and the exact one, :

=

+0.3338 0.6135+0.6676 1.2270+0.8654 0.6008+1.0632 +0.0254+1.1932 +1.2713

, =

+0.3338 0.8398+0.6405 1.0999+0.8954 0.6008+1.0779 +0.3131+1.1932 +1.0108

.

The accuracy of the estimates for the 1st mode is very good, on thecontrary the 2nd mode estimates are in error starting from the seconddigit.

It may be interesting to use = K1M as a new Ritz base to get anew estimate of the Ritz and of the structural eigenpairs.


Giacomo Boffi


Subspace iteration


Introduction to Subspace Iteration

We have seen that the Rayleigh-Ritz procedure can offer agood estimate for p M/2 modes, mostly because of thearbitrariness in the choice of the Ritz reduced base .Solving a M = 2p order eigenvalue problem to get peigenvalues is very onerous as the operation count isO(M3).

If we could reduce the arbitrariness in the choice of the Ritzbase vectors, we could use less vectors and solve a muchsmaller (in terms of operations count) eigenvalue problem.If one thinks of it, with a M = 1 base we could neverthelesscompute within arbitrary accuracy one eigenvector usingMatrix Iteration, isnt it? the trick is changing the base atevery iteration...It happens that Matrix Iteration can be applied to a set oftrial vectors at once, under the name of Subspace Iteration.


Giacomo Boffi


Subspace iteration



We have seen that the Rayleigh-Ritz procedure can offer agood estimate for p M/2 modes, mostly because of thearbitrariness in the choice of the Ritz reduced base .Solving a M = 2p order eigenvalue problem to get peigenvalues is very onerous as the operation count isO(M3).If we could reduce the arbitrariness in the choice of the Ritzbase vectors, we could use less vectors and solve a muchsmaller (in terms of operations count) eigenvalue problem.

If one thinks of it, with a M = 1 base we could neverthelesscompute within arbitrary accuracy one eigenvector usingMatrix Iteration, isnt it? the trick is changing the base atevery iteration...It happens that Matrix Iteration can be applied to a set oftrial vectors at once, under the name of Subspace Iteration.


Giacomo Boffi


Subspace iteration



We have seen that the Rayleigh-Ritz procedure can offer agood estimate for p M/2 modes, mostly because of thearbitrariness in the choice of the Ritz reduced base .Solving a M = 2p order eigenvalue problem to get peigenvalues is very onerous as the operation count isO(M3).If we could reduce the arbitrariness in the choice of the Ritzbase vectors, we could use less vectors and solve a muchsmaller (in terms of operations count) eigenvalue problem.If one thinks of it, with a M = 1 base we could neverthelesscompute within arbitrary accuracy one eigenvector usingMatrix Iteration, isnt it? the trick is changing the base atevery iteration...It happens that Matrix Iteration can be applied to a set oftrial vectors at once, under the name of Subspace Iteration.


Giacomo Boffi


Subspace iteration


Statement of the procedure

The first M eigenvalue equations can be written in matrixalgebra, in terms of an N M matrix of eigenvectors andan M M diagonal matrix that collects the eigenvalues

KNN

NM

= MNN

NM

MM

Using again the hat notation for the unnormalized iterate,from the previous equation we can write

K1 = M0

where 0 is the matrix, N M, of the zero order trialvectors, and 1 is the matrix of the non-normalized firstorder trial vectors.


Giacomo Boffi


Subspace iteration


Orthonormalization

To proceed with iterations,

1. the trial vectors in n+1 must be orthogonalized, sothat each trial vector converges to a differenteigenvector instead of collapsing to the firsteigenvector,

2. all the trial vectors must be normalized, so that theratio between the normalized vectors and theunnormalized iterated vectors converges to thecorresponding eigenvalue.

These operations can be performed in different ways (e.g.,ortho-normalization by Gram-Schmidt1 procedure).Another possibility to do both at once is solving aRayleigh-Ritz eigenvalue problem, defined in the Ritz baseconstituted by the vectors in n+1.

1Next week, more on Gram-Schmidt procedure


Giacomo Boffi


Subspace iteration


Associated Eigenvalue Problem

Developing the procedure for n = 0, with the generalized matrices

K?1 = 1TK1

andM?1 = 1

TM1the Rayleigh-Ritz eigenvalue problem associated with theorthonormalisation of 1 is

K?1Z1 = M?1Z1

21.

After solving for the Ritz coordinates mode shapes, Z1 and thefrequencies 21, using any suitable procedure, it is usually convenient tonormalize the shapes, so that Z1TM?1Z1 = I. The ortho-normalized setof trial vectors at the end of the iteration is then written as

1 = 1Z1.

The entire process can be repeated for n = 1, then n = 2, n = . . .until the eigenvalues converge within a prescribed tolerance.


Giacomo Boffi


Subspace iteration


Convergence

In principle, the procedure will converge to all the M lowereigenvalues and eigenvectors of the structural problem, butit was found that the subspace iteration method convergesfaster to the lower p eigenpairs, those required for dynamicanalysis, if there is some additional trial vector; on theother hand, too many additional trial vectors slow down thecomputation without ulterior benefits.

Experience has shown that the optimal total number M oftrial vectors is the minimum of 2p and p + 8.The subspace iteration method makes it possible tocompute simultaneosly a set of eigenpairs within anyrequired level of approximation, and is the preferred methodto compute the eigenpairs of a complex dynamic system.


Giacomo Boffi


Subspace iteration


Convergence

In principle, the procedure will converge to all the M lowereigenvalues and eigenvectors of the structural problem, butit was found that the subspace iteration method convergesfaster to the lower p eigenpairs, those required for dynamicanalysis, if there is some additional trial vector; on theother hand, too many additional trial vectors slow down thecomputation without ulterior benefits.Experience has shown that the optimal total number M oftrial vectors is the minimum of 2p and p + 8.

The subspace iteration method makes it possible tocompute simultaneosly a set of eigenpairs within anyrequired level of approximation, and is the preferred methodto compute the eigenpairs of a complex dynamic system.


Giacomo Boffi


Subspace iteration


Convergence

In principle, the procedure will converge to all the M lowereigenvalues and eigenvectors of the structural problem, butit was found that the subspace iteration method convergesfaster to the lower p eigenpairs, those required for dynamicanalysis, if there is some additional trial vector; on theother hand, too many additional trial vectors slow down thecomputation without ulterior benefits.Experience has shown that the optimal total number M oftrial vectors is the minimum of 2p and p + 8.The subspace iteration method makes it possible tocompute simultaneosly a set of eigenpairs within anyrequired level of approximation, and is the preferred methodto compute the eigenpairs of a complex dynamic system.


Giacomo Boffi


Subspace iteration


Standard Form

In algebra textbooks, the eigenproblem is usually stated as

Ay = y

and all the relevant algorithms to actually compute the eigenthings(Jacobi method, QR method, etc) are referred to the abovestatement of the problem.Our problem is, instead, formulated as

Kx = Mx.

Any symmetric, definite positive matrix B can be subjected to a uniqueCholeski Decomposition (CD), B = LLT where L is a lower triangularmatrix. Applying CD to M, the eigenvector equation is,

Kx = K (LT )1LT I

x = LLTM

x.

Premultiplying by L1, with y = LTx

L1K(LT )1 A

LTxy

= L1L I

LTxy

Ay = y.


Giacomo Boffi


Subspace iteration

How manyeigenvectors?Modal Partecipation Factor

Dynamic magnificationfactor

Static Correction

How Many Eigenvectors

are needed to correctly represent the response of a MDOFsystem to a time-varying load?


Giacomo Boffi


Subspace iteration



Static Correction

Introduction

To understand how many eigenvectors we have to use in adynamic analysis, we must consider two aspects, theloading shape and the excitation frequency.In the following, well consider only external loadings whosedependance on time and space can be separated, as in

p(x, t) = r f (t),

so that we can discuss separately the two aspects of theproblem.


Giacomo Boffi


Subspace iteration



Static Correction

Introduction

It is worth noting that earthquake loadings are precisely ofthis type:

p(x, t) = Mr ug(t)

where the vector r is used to choose the structural dofsthat are excited by the ground motion component underconsideration.

Usually r is simply a vector of zeros and ones that represents thedegrees of freedom that are excited by a component of the groundacceleration.

Multiplication of M by g (the acceleration of gravity) anddivision of ug by g, serves to show a dimensional loadvector multiplied by an adimensional function.

p(x, t) = gMr ug(t)g= rgfg(t)


Giacomo Boffi


Subspace iteration



Static Correction

Modal Partecipation Factor

Under the assumption of separability, we can write the i-th modalequation of motion as

qi + 2ii qi + 2i qi =

Ti rMi

f (t)g

Ti MrMi

fg(t)= i f (t)

with the modal mass Mi = Ti Mi .It is apparent that the modal response amplitude depends

I on the characteristics of the time dependency of loading,f (t),

I on the so called modal partecipation factor i ,

i = Ti r/Mi

= gTi Mr/Mi = Ti r

g/Mi

= M1T r

Note that both the definitions of modal partecipation give it thedimensions of an acceleration.


Giacomo Boffi


Subspace iteration



Static Correction

Partecipation Factor Amplitudes

For a given loading r the modal partecipation factor i is proportionalto the work done by the modal displacement qiTi for the givenloading r:I if the mode shape and the loading shape are approximately equal

(equal signs, component by component), the work (dot product)is maximized,

I if the mode shape is significantly different from the loading(different signs), there is some amount of cancellation and thevalue of the s will be reduced.


Giacomo Boffi


Subspace iteration



Static Correction

Example

Consider a shear type building, with mass distributionapproximately constant over its height:

r = {1, 1, . . . , 1}T and gMr mg{1, 1, . . . , 1}T .

an external loading and the first 3 eigenvectors as sketchedbelow:

gMr r 1 2 3


Giacomo Boffi


Subspace iteration



Static Correction

Example, cont.

gMr r 1 2 3

For EQ loading, 1 is relatively large for the first mode, asloading components and displacements have the same sign,with respect to other i s, where the oscillating nature ofthe higher eigenvectors will lead to increasing cancellation.On the other hand, consider the external loading, whosepeculiar shape is similar to the 3rd mode. 3 will be morerelevant than i s for lower or higher modes.


Giacomo Boffi


Subspace iteration



Static Correction

Modal Loads Expansion

We define the modal load contribution as

ri = Miai

and express the load vector as a linear combination of the modalcontributions

r =i

Miai =i

ri .

If we premultiply by Tj the above equation,

Tj r = Tj

i

Miai =i

ijMiai = Mjaj

1. a modal load component works only for the displacementsassociated with the corresponding eigenvector,

2. comparing with the definition of i = Ti r/Mi , we conclude that

ri = Mii

R = M diag()


Giacomo Boffi


Subspace iteration



Static Correction

Equivalent Static Forces

For mode i , the equation of motion is

qi + 2ii qi + 2i qi = i f (t)

with qi = iDi , we can write, to single out thedependency on the modulating function,

Di + 2ii Di + 2i Di = f (t)

The modal contribution to displacement is

xi = iiDi(t)

and the modal contribution to elastic forces f i = Kxi canbe written (being Ki = 2i Mi) as

f i = Kxi = iKiDi = 2i (iMi)Di = ri2i Di

1 Di (dimensionally the square of a time), is calledpseudo-displacement.


Giacomo Boffi


Subspace iteration



Static Correction

Equivalent Static Response

The response can be determined by superposition of the effects ofthese pseudo-static forces f i = ri2i Di(t).If a required response quantity (be it a nodal displacement, a bendingmoment in a beam, the total shear force in a building storey, etc etc)is indicated by s(t), we can compute with a static calculation (usuallyusing the FEM model underlying the dynamic analysis) the modalstatic contribution ssti and write

s(t) =

ssti (2i Di(t)) =

si(t),

where the modal contribution to response si(t) is given by

1. static analysis using ri as the static load vector,

2. dynamic amplification using the factor 2i Di(t).

This formulation is particularly apt to our discussion of differentcontributions to response components.


Giacomo Boffi


Subspace iteration



Static Correction

Modal Contribution Factors (MCF)

Say that the static response due to r is denoted by sst, thensi(t), the modal contribution to response s(t), can bewritten

si(t) = ssti 2i Di(t) = s

st sstisst

2i Di(t) = sisst 2i Di(t).

We have introduced si =sstisst , the modal contribution factor,

the ratio of the modal static contribution to the total staticresponse.The si are dimensionless, are indipendent on the eigenvectorscaling procedure and their sum is unity,

si = 1.


Giacomo Boffi


Subspace iteration



Static Correction

Maximum Response

Denote by Di0 the maximum absolute value (or peak) ofthe pseudo displacement time history,

Di0 = maxt{|Di(t)|}.

It will besi0 = sisst 2i Di0

Eventually, the dynamic response factor for mode i , Rdi isdefined by

Rdi =Di0Dsti0

where Dsti0 is the peak value of the static pseudo-displacement

Dsti =f (t)2i

, Dsti0 =f02i


Giacomo Boffi


Subspace iteration



Static Correction

Maximum Response

With f0 = max{|f (t)|} the peak p-displacement is

Di0 = Rdi f0/2i

and the peak of the modal contribution is

si0 = sisst 2i Di0 = f0sst si Rdi

The first two terms are independent of the mode, the lastare independent from each other and their product is thefactor that influences the modal contributions.Note that this product has the sign of si , as the dynamicresponse factor is always positive.

MCFs exampleThe following table (from Chopra, 2nd ed.) displays the si and theirpartial sums for a shear-type, 5 floors building where all the storeymasses are equal and all the storey stiffnesses are equal too.The response quantities chosen are x5n, the MCFs to the topdisplacement and Vn, the MCF s to the base shear, for two differentload shapes.

r = {0, 0, 0, 0, 1}T r = {0, 0, 0,1, 2}T

Top Disp. Base Shear Top Disp. Base Shear

n or i x5nix5j Vn

iVj x5n

ix5j Vn

iVj

1 0.880 0.880 1.252 1.252 0.792 0.792 1.353 1.3532 0.087 0.967 -0.362 0.890 0.123 0.915 -0.612 0.7413 0.024 0.991 0.159 1.048 0.055 0.970 0.043 1.1724 0.008 0.998 -0.063 0.985 0.024 0.994 -0.242 0.9305 0.002 1.000 0.015 1.000 0.006 1.000 0.070 1.000

Note that:1. for any given r, the base shear is more influenced by higher modes, and2. for any given response quantity, the second, skewed r gives greater

modal contributions for higher modes.


Giacomo Boffi


Subspace iteration



Static Correction

Dynamic Response Ratios

Dynamic Response Ratios are the same that we have seen for SDOFsystems.Next page, for an undamped system,

I solid line, the ratio of the modal elastic force FS,i = Kiqi sint tothe harmonic applied modal force, Pi sint, plotted against thefrequency ratio = /i .For = 0 the ratio is 1, the applied load is fully balanced by theelastic resistance.For fixed excitation frequency, 0 for high modal frequencies.

I dashed line,the ratio of the modal inertial force, FI ,i = 2FS,i tothe load.

Note that for steady-state motion the sum of the elastic and inertialforce ratios is constant and equal to 1, as in

(FS,i + FI ,i) sint = Pi sint.

-3

-2

-1

0

1

2

3

4

0 0.5 1 1.5 2 2.5 3

Moda

l res

istan

ce ra

tios

Frequency ratio, =/i

FS/PiFI/Pi

I For a fixed excitation frequency and high modal frequencies thefrequency ratio 0.

I For 0 the response is quasi-static.I Hence, for higher modes the response is pseudo-static.I On the contrary, for excitation frequencies high enough the lower

modes respond with purely inertial forces.

-3

-2

-1

0

1

2

3

4

0 0.5 1 1.5 2 2.5 3

Moda

l res

istan

ce ra

tios


FS/PiFI/Pi


I For 0 the response is quasi-static.

I Hence, for higher modes the response is pseudo-static.I On the contrary, for excitation frequencies high enough the lower


-3

-2

-1

0

1

2

3

4

0 0.5 1 1.5 2 2.5 3

Moda

l res

istan

ce ra

tios


FS/PiFI/Pi


I For 0 the response is quasi-static.I Hence, for higher modes the response is pseudo-static.

I On the contrary, for excitation frequencies high enough the lowermodes respond with purely inertial forces.

-3

-2

-1

0

1

2

3

4

0 0.5 1 1.5 2 2.5 3

Moda

l res

istan

ce ra

tios


FS/PiFI/Pi


I For 0 the response is quasi-static.I Hence, for higher modes the response is pseudo-static.I On the contrary, for excitation frequencies high enough the lower



Giacomo Boffi


Subspace iteration



Static Correction

Static Correction

The preceding discussion indicates that higher modescontributions to the response could be approximated withthe static response, leading to the idea of a StaticCorrection of the dynamic response

For a system where qi(t) pi(t)Ki for i > ndy,ndy being the number of dynamically responding modes,we can write

x(t) xdy(t) + xst(t) =ndy1

iqi(t) +N

ndy+1

ipi(t)Ki

where the response for each of the first ndy modes can becomputed as usual.


Giacomo Boffi


Subspace iteration



Static Correction

Static Modal Components

The static modal displacement component xj , j > ndy canbe written

xj(t) = jqj(t) j

Tj

Kjp(t) = Fjp(t)

The modal flexibility matrix is defined by

Fj =j

Tj

Kj

and is used to compute the j-th mode static deflections dueto the applied load vector.The total displacements, the dynamic contributions and thestatic correction, for p(t) = r f (t), are then

x ndy1

jqj(t) + f (t)N

ndy+1

Fj r.


Giacomo Boffi


Subspace iteration



Static Correction

Alternative Formulation

Our last formula for static correction is

x ndy1

jqj(t) + f (t)N

ndy+1

Fj r.

To use the above formula all mode shapes, all modalstiffnesses and all modal flexibility matrices must becomputed, undermining the efficiency of the procedure.


Giacomo Boffi


Subspace iteration



Static Correction

Alternative Formulation

This problem can be obviated computing the total staticdisplacements and expressing it in terms of modalcontributions: xst = K1rf (t) =

N1 Fj rf (t).

Subtracting the static displacements due to the first ndymodes to both members it isN

ndy+1

Fj rf (t) = K1rf (t)ndy1

Fj rf (t) = f (t)

(K1

ndy1

Fj

)r.

The corrected total displacements have hence theexpression

x ndy1

iqi(t) + f (t)

(K1

ndy1

Fi

)r,

Note that the constant term following f (t) can becomputed with information already in our possess at theend of the dynamic analysis.


Giacomo Boffi


Subspace iteration



Static Correction

Effectiveness of Static Correction

In these circumstances, few modes with static correctiongive results comparable to the results obtained using muchmore modes in a straightforward modal displacementsuperposition analysis.

I An high number of modes is required to account forthe spatial distribution of the loading but only a fewlower modes are subjected to significant dynamicamplification.

I Refined stress analysis is required even if the dynamicresponse involves only a few lower modes.


Giacomo Boffi


Subspace iteration



Static Correction

Effectiveness of Static Correction

In these circumstances, few modes with static correctiongive results comparable to the results obtained using muchmore modes in a straightforward modal displacementsuperposition analysis.

I An high number of modes is required to account forthe spatial distribution of the loading but only a fewlower modes are subjected to significant dynamicamplification.

I Refined stress analysis is required even if the dynamicresponse involves only a few lower modes.

Rayleigh-Ritz ExampleSubspace iterationHow many eigenvectors?Modal Partecipation FactorDynamic magnification factorStatic Correction

Documents

Matrix iteration repear, modal constirbutions and static correction