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Truncatjon Errors Numerical Integratjon Multjple Support Excitatjon

Giacomo Boffj

htup://intranet.dica.polimi.it/people/boffj‐giacomo Dipartjmento di Ingegneria Civile Ambientale e Territoriale Politecnico di Milano

April 2, 2020

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Part I How many eigenvectors?

Introductjon Modal partecipatjon factor Dynamic magnifjcatjon factor Statjc Correctjon

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Sectjon 1 Introductjon

Introductjon Modal partecipatjon factor Dynamic magnifjcatjon factor Statjc Correctjon

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How many eigenvectors?

To understand how many eigenvectors we have to use in a modal analysis, we must consider two factors, the loading shape and the excitatjon frequency.

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Introductjon

In the following, we’ll consider only external loadings whose dependance

• n tjme and space can be separated, as in

𝐪(𝐲, 𝑢) = 𝐬 𝑔(𝑢), so that we can regard separately the two aspects of the problem.

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Introductjon

It is worth notjng that earthquake loadings are precisely of this type: ⃗ 𝑞( ⃗ 𝑦, 𝑢) = 𝐍 ̃ ⃗ 𝑠 ̈ 𝑣g where the vector ̃ ⃗ 𝑠 is used to choose the structural dof’s that are excited by the ground motjon component under consideratjon.

̃ ⃗ 𝑠 is an incidence vector, ofuen simply a vector of ones and zeroes where the

• nes stay for the inertjal forces that are excited by a specifjc component of the

earthquake ground acceleratjon.

Multjplicatjon of 𝐍 and division of ̈ 𝑣𝑕 by 𝑕, acceleratjon of gravity, serves to show a dimensional load vector multjplied by an adimensional functjon. ⃗ 𝑞( ⃗ 𝑦, 𝑢) = 𝑕 𝐍 ̃ ⃗ 𝑠

̈ 𝑣g(𝑢) 𝑕

= 𝐬g𝑔

g(𝑢)

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Introductjon

It is worth notjng that earthquake loadings are precisely of this type: ⃗ 𝑞( ⃗ 𝑦, 𝑢) = 𝐍 ̃ ⃗ 𝑠 ̈ 𝑣g where the vector ̃ ⃗ 𝑠 is used to choose the structural dof’s that are excited by the ground motjon component under consideratjon.

̃ ⃗ 𝑠 is an incidence vector, ofuen simply a vector of ones and zeroes where the

• nes stay for the inertjal forces that are excited by a specifjc component of the

earthquake ground acceleratjon.

Multjplicatjon of 𝐍 and division of ̈ 𝑣𝑕 by 𝑕, acceleratjon of gravity, serves to show a dimensional load vector multjplied by an adimensional functjon. ⃗ 𝑞( ⃗ 𝑦, 𝑢) = 𝑕 𝐍 ̃ ⃗ 𝑠

̈ 𝑣g(𝑢) 𝑕

= 𝐬g𝑔

g(𝑢)

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Sectjon 2 Modal partecipatjon factor

Introductjon Modal partecipatjon factor Dynamic magnifjcatjon factor Statjc Correctjon

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Modal partecipatjon factor

Under the assumptjon of separability, we can write the 𝑗‐th modal equatjon of motjon as ̈ 𝑟𝑗 + 2𝜂𝑗𝜕𝑗 ̇ 𝑟𝑗 + 𝜕2

𝑗 𝑟𝑗 = 𝝎𝑈

𝑗 𝐬

𝑁𝑗 𝑔(𝑢) 𝑕 𝝎𝑈

𝑗 𝐍 ̂

𝐬 𝑁𝑗

𝑔

g(𝑢)

= Γ𝑗𝑔(𝑢) with the modal mass 𝑁𝑗 = 𝝎𝑈

𝑗 𝐍𝝎𝑗.

It is apparent that the modal response amplitude depends

• n the characteristjcs of the tjme dependency of loading, 𝑔(𝑢),
• n the so called modal partecipatjon factor Γ𝑗,

Γ𝑗 = 𝝎𝑈

𝑗 𝐬/𝑁𝑗

• r

Γ𝑗 = 𝑕 𝝎𝑈

𝑗 𝐍 ̂

𝐬/𝑁𝑗 = 𝝎𝑈

𝑗 𝐬g/𝑁𝑗

Note that both the defjnitjons of modal partecipatjon give it the dimensions of an acceleratjon.

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Modal partecipatjon factor

Under the assumptjon of separability, we can write the 𝑗‐th modal equatjon of motjon as ̈ 𝑟𝑗 + 2𝜂𝑗𝜕𝑗 ̇ 𝑟𝑗 + 𝜕2

𝑗 𝑟𝑗 = 𝝎𝑈

𝑗 𝐬

𝑁𝑗 𝑔(𝑢) 𝑕 𝝎𝑈

𝑗 𝐍 ̂

𝐬 𝑁𝑗

𝑔

g(𝑢)

= Γ𝑗𝑔(𝑢) with the modal mass 𝑁𝑗 = 𝝎𝑈

𝑗 𝐍𝝎𝑗.

It is apparent that the modal response amplitude depends

• n the characteristjcs of the tjme dependency of loading, 𝑔(𝑢),
• n the so called modal partecipatjon factor Γ𝑗,

Γ𝑗 = 𝝎𝑈

𝑗 𝐬/𝑁𝑗

• r

Γ𝑗 = 𝑕 𝝎𝑈

𝑗 𝐍 ̂

𝐬/𝑁𝑗 = 𝝎𝑈

𝑗 𝐬g/𝑁𝑗

Note that both the defjnitjons of modal partecipatjon give it the dimensions of an acceleratjon.

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Modal partecipatjon factor

Under the assumptjon of separability, we can write the 𝑗‐th modal equatjon of motjon as ̈ 𝑟𝑗 + 2𝜂𝑗𝜕𝑗 ̇ 𝑟𝑗 + 𝜕2

𝑗 𝑟𝑗 = 𝝎𝑈

𝑗 𝐬

𝑁𝑗 𝑔(𝑢) 𝑕 𝝎𝑈

𝑗 𝐍 ̂

𝐬 𝑁𝑗

𝑔

g(𝑢)

= Γ𝑗𝑔(𝑢) with the modal mass 𝑁𝑗 = 𝝎𝑈

𝑗 𝐍𝝎𝑗.

It is apparent that the modal response amplitude depends

• n the characteristjcs of the tjme dependency of loading, 𝑔(𝑢),
• n the so called modal partecipatjon factor Γ𝑗,

Γ𝑗 = 𝝎𝑈

𝑗 𝐬/𝑁𝑗

• r

Γ𝑗 = 𝑕 𝝎𝑈

𝑗 𝐍 ̂

𝐬/𝑁𝑗 = 𝝎𝑈

𝑗 𝐬g/𝑁𝑗

Note that both the defjnitjons of modal partecipatjon give it the dimensions of an acceleratjon.

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Partecipatjon Factor Amplitudes

For a given loading 𝐬 the modal partecipatjon factor Γ𝑗 is proportjonal to the work done by the modal displacement 𝑟𝑗𝝎𝑈

𝑗 for the given loading 𝐬:

if the mode shape and the loading shape are approximately equal (equal signs, component by component), the work (dot product) is maximized, if the mode shape is signifjcantly difgerent from the loading (difgerent signs), there is some amount of cancellatjon and the value

• f the Γ’s will be reduced.

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Example

𝑕𝐍 ̂ 𝐬 𝐬 𝝎1 𝝎2 𝝎3 Consider a shear type building, its fjrst 3 eigenvectors as sketched above, with mass distributjon approximately constant over its height and its earthquake load shape vector ̂ 𝐬 = {1, 1, … , 1}𝑈 → 𝑕 𝐍 ̂ 𝐬 ≈ 𝑛𝑕{1, 1, … , 1}𝑈. Consider also the external, assigned load shape vector 𝐬...

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Example

𝑕𝐍 ̂ 𝐬 𝐬 𝝎1 𝝎2 𝝎3 Consider a shear type building, its fjrst 3 eigenvectors as sketched above, with mass distributjon approximately constant over its height and its earthquake load shape vector ̂ 𝐬 = {1, 1, … , 1}𝑈 → 𝑕 𝐍 ̂ 𝐬 ≈ 𝑛𝑕{1, 1, … , 1}𝑈. Consider also the external, assigned load shape vector 𝐬...

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Example, cont.

𝑕𝐍 ̂ 𝐬 𝐬 𝝎1 𝝎2 𝝎3 For EQ loading, Γ1 is relatjvely large for the fjrst mode, as loading components and displacements have the same sign, with respect to

• ther Γ𝑗’s, where the
• scillatjng nature of the

higher eigenvectors will lead to increasing cancellatjon. On the other hand, consider the external loading, whose peculiar shape is similar to the 3rd mode. Γ3 will be more relevant than Γ𝑗’s for lower or higher modes.

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Modal Loads Expansion

We defjne the modal load contributjon as 𝐬𝑗 = 𝐍 𝝎𝑗𝑏𝑗 and express the load vector as a linear combinatjon of the modal contributjons 𝐬 =

𝑗

𝐍 𝝎𝑗𝑏𝑗 =

𝑗

𝐬𝑗. Premultjplying by 𝝎𝑈

𝑘 the above equatjon we have a relatjon that enables

the computatjon of the coeffjcients 𝑏𝑗: 𝝎𝑈

𝑗 𝐬 = 𝝎𝑈 𝑗 𝑘

𝐍 𝝎𝑘𝑏𝑘 =

𝑘

𝜀𝑗𝑘𝑁𝑘𝑏𝑘 = 𝑏𝑗𝑁𝑗 → 𝑏𝑗 = 𝝎𝑈

𝑗 𝐬

𝑁𝑗

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Modal Loads Expansion

We defjne the modal load contributjon as 𝐬𝑗 = 𝐍 𝝎𝑗𝑏𝑗 and express the load vector as a linear combinatjon of the modal contributjons 𝐬 =

𝑗

𝐍 𝝎𝑗𝑏𝑗 =

𝑗

𝐬𝑗. Premultjplying by 𝝎𝑈

𝑘 the above equatjon we have a relatjon that enables

the computatjon of the coeffjcients 𝑏𝑗: 𝝎𝑈

𝑗 𝐬 = 𝝎𝑈 𝑗 𝑘

𝐍 𝝎𝑘𝑏𝑘 =

𝑘

𝜀𝑗𝑘𝑁𝑘𝑏𝑘 = 𝑏𝑗𝑁𝑗 → 𝑏𝑗 = 𝝎𝑈

𝑗 𝐬

𝑁𝑗

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Modal Loads Expansion

1 A modal load component works only for the displacements

associated with the corresponding eigenvector, 𝝎𝑈

𝑘 𝐬𝑗 = 𝑏𝑗 𝝎𝑈 𝑘 𝐍𝝎𝑗 = 𝜀𝑗𝑘𝑏𝑗𝑁𝑗. 2 Comparing 𝝎𝑈 𝑘 𝐬 = 𝝎𝑈 𝑘 ∑𝑗 𝐍 𝝎𝑗𝑏𝑗 = 𝜀𝑗𝑘𝑁𝑗𝑏𝑗 with the defjnitjon of

Γ𝑗 = 𝝎𝑈

𝑗 𝐬/𝑁𝑗, we conclude that 𝑏𝑗 ≡ Γ𝑗 and fjnally write

𝐬𝑗 = Γ𝑗𝐍 𝝎𝑗.

3 The modal load contributjons can be collected in a matrix: with

𝚫 = diag Γ𝑗 we have 𝐒 = 𝐍 𝛀 𝚫.

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Equivalent Statjc Forces

For mode 𝑗, the equatjon of motjon is ̈ 𝑟𝑗 + 2𝜂𝑗𝜕𝑗 ̇ 𝑟𝑗 + 𝜕2

𝑗 𝑟𝑗 = Γ 𝑗𝑔(𝑢)

with 𝑟𝑗 = Γ

𝑗𝐸𝑗, we can write, to single out the dependency on the modulatjng

functjon, ̈ 𝐸𝑗 + 2𝜂𝑗𝜕𝑗 ̇ 𝐸𝑗 + 𝜕2

𝑗 𝐸𝑗 = 𝑔(𝑢)

The modal contributjon to displacement is 𝐲𝑗 = Γ

𝑗𝝎𝑗𝐸𝑗(𝑢)

and the modal contributjon to elastjc forces 𝐠𝑗 = 𝐋 𝐲𝑗 can be writuen (being 𝐋𝝎𝑗 = 𝜕2

𝑗 𝐍𝝎𝑗) as

𝐠𝑗 = 𝐋 𝐲𝑗 = Γ

𝑗𝐋 𝝎𝑗𝐸𝑗 = 𝜕2 𝑗 (Γ 𝑗𝐍 𝝎𝑗)𝐸𝑗 = 𝐬𝑗𝜕2 𝑗 𝐸𝑗

𝐸 is usually named pseudo‐displacement.

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Equivalent Statjc Forces

For mode 𝑗, the equatjon of motjon is ̈ 𝑟𝑗 + 2𝜂𝑗𝜕𝑗 ̇ 𝑟𝑗 + 𝜕2

𝑗 𝑟𝑗 = Γ 𝑗𝑔(𝑢)

with 𝑟𝑗 = Γ

𝑗𝐸𝑗, we can write, to single out the dependency on the modulatjng

functjon, ̈ 𝐸𝑗 + 2𝜂𝑗𝜕𝑗 ̇ 𝐸𝑗 + 𝜕2

𝑗 𝐸𝑗 = 𝑔(𝑢)

The modal contributjon to displacement is 𝐲𝑗 = Γ

𝑗𝝎𝑗𝐸𝑗(𝑢)

and the modal contributjon to elastjc forces 𝐠𝑗 = 𝐋 𝐲𝑗 can be writuen (being 𝐋𝝎𝑗 = 𝜕2

𝑗 𝐍𝝎𝑗) as

𝐠𝑗 = 𝐋 𝐲𝑗 = Γ

𝑗𝐋 𝝎𝑗𝐸𝑗 = 𝜕2 𝑗 (Γ 𝑗𝐍 𝝎𝑗)𝐸𝑗 = 𝐬𝑗𝜕2 𝑗 𝐸𝑗

𝐸 is usually named pseudo‐displacement.

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Equivalent Statjc Response

The response can be determined by superpositjon of the efgects of these pseudo‐statjc forces 𝐠𝑗 = 𝐬𝑗𝜕2

𝑗 𝐸𝑗(𝑢).

If a required response quantjty (be it a nodal displacement, a bending moment in a beam, the total shear force in a building storey, etc etc) is indicated by 𝑡(𝑢), we can compute with a statjc calculatjon (usually using the FEM model underlying the dynamic analysis) the modal statjc contributjon 𝑡st

𝑗 and write

𝑡(𝑢) = 𝑡st

𝑗 (𝜕2 𝑗 𝐸𝑗(𝑢)) = 𝑡𝑗(𝑢),

where the modal contributjon to response 𝑡𝑗(𝑢) is given by

1

statjc analysis using 𝐬𝑗 as the statjc load vector,

2

dynamic amplifjcatjon using the factor 𝜕2

𝑗 𝐸𝑗(𝑢).

This formulatjon is partjcularly apt to our discussion of difgerent contributjons to response components.

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Equivalent Statjc Response

The response can be determined by superpositjon of the efgects of these pseudo‐statjc forces 𝐠𝑗 = 𝐬𝑗𝜕2

𝑗 𝐸𝑗(𝑢).

If a required response quantjty (be it a nodal displacement, a bending moment in a beam, the total shear force in a building storey, etc etc) is indicated by 𝑡(𝑢), we can compute with a statjc calculatjon (usually using the FEM model underlying the dynamic analysis) the modal statjc contributjon 𝑡st

𝑗 and write

𝑡(𝑢) = 𝑡st

𝑗 (𝜕2 𝑗 𝐸𝑗(𝑢)) = 𝑡𝑗(𝑢),

where the modal contributjon to response 𝑡𝑗(𝑢) is given by

1

statjc analysis using 𝐬𝑗 as the statjc load vector,

2

dynamic amplifjcatjon using the factor 𝜕2

𝑗 𝐸𝑗(𝑢).

This formulatjon is partjcularly apt to our discussion of difgerent contributjons to response components.

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Modal Contributjon Factors

Say that the statjc response due to 𝐬 is denoted by 𝑡st, then 𝑡𝑗(𝑢), the modal contributjon to response 𝑡(𝑢), can be writuen 𝑡𝑗(𝑢) = 𝑡st

𝑗 𝜕2 𝑗 𝐸𝑗(𝑢) = 𝑡st 𝑡st 𝑗

𝑡st 𝜕2

𝑗 𝐸𝑗(𝑢) = ̄

𝑡𝑗𝑡st 𝜕2

𝑗 𝐸𝑗(𝑢).

We have introduced ̄ 𝑡𝑗 =

𝑡st

𝑗

𝑡st , the modal contributjon factor, the ratjo of

the modal statjc contributjon to the total statjc response. The ̄ 𝑡𝑗 are dimensionless, are indipendent from the eigenvector scaling procedure and their sum is unity, ∑ ̄ 𝑡𝑗 = 1.

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Maximum Response

Denote by 𝐸𝑗0 the maximum absolute value (or peak) of the pseudo displacement tjme history, 𝐸𝑗0 = max

𝑢 {|𝐸𝑗(𝑢)|}.

It will be 𝑡𝑗0 = ̄ 𝑡𝑗𝑡st 𝜕2

𝑗 𝐸𝑗0.

The dynamic response factor for mode 𝑗, ℜ𝑒𝑗 is defjned by ℜ𝑒𝑗 = 𝐸𝑗0 𝐸st

𝑗0

where 𝐸st

𝑗0 is the peak value of the statjc pseudo displacement

𝐸st

𝑗 = 𝑔(𝑢)

𝜕2

𝑗

, → 𝐸st

𝑗0 = 𝑔

𝜕2

𝑗

.

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Maximum Response

With 𝑔

0 = max{|𝑔(𝑢)|} the peak pseudo displacement is

𝐸𝑗0 = ℜ𝑒𝑗𝑔

0/𝜕2 𝑗

and the peak of the modal contributjon is 𝑡𝑗0(𝑢) = ̄ 𝑡𝑗𝑡st 𝜕2

𝑗 𝐸𝑗0(𝑢) = 𝑔 0𝑡st

̄ 𝑡𝑗ℜ𝑒𝑗 The fjrst two terms are independent of the mode, the last are independent from each other and their product is the factor that infmuences the modal contributjons. Note that this product has the sign of ̄ 𝑡𝑗, as the dynamic response factor is always positjve.

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Maximum Response

With 𝑔

0 = max{|𝑔(𝑢)|} the peak pseudo displacement is

𝐸𝑗0 = ℜ𝑒𝑗𝑔

0/𝜕2 𝑗

and the peak of the modal contributjon is 𝑡𝑗0(𝑢) = ̄ 𝑡𝑗𝑡st 𝜕2

𝑗 𝐸𝑗0(𝑢) = 𝑔 0𝑡st

̄ 𝑡𝑗ℜ𝑒𝑗 The fjrst two terms are independent of the mode, the last are independent from each other and their product is the factor that infmuences the modal contributjons. Note that this product has the sign of ̄ 𝑡𝑗, as the dynamic response factor is always positjve.

MCF’s example

The following table (from Chopra, 2nd ed.) displays the ̄ 𝑡𝑗 and their partjal sums for a shear‐type, 5 fmoors building where all the storey masses are equal and all the storey stjfgnesses are equal too. The response quantjtjes chosen are ̄ 𝑦5𝑜, the MCF’s to the top displacement and ̄ 𝑊

𝑜, the MCF’s to

the base shear, for two difgerent load shapes. 𝐬 = {0, 0, 0, 0, 1}𝑈 𝐬 = {0, 0, 0, −1, 2}𝑈 Top Displacement Base Shear Top Displacement Base Shear 𝑜 or 𝐾 ̄ 𝑦5𝑜 ∑𝐾 ̄ 𝑦5𝑗 ̄ 𝑊

𝑜

∑𝐾 ̄ 𝑊

𝑗

̄ 𝑦5𝑜 ∑𝐾 ̄ 𝑦5𝑗 ̄ 𝑊

𝑜

∑𝐾 ̄ 𝑊

𝑗

1 0.880 0.880 1.252 1.252 0.792 0.792 1.353 1.353 2 0.087 0.967 ‐0.362 0.890 0.123 0.915 ‐0.612 0.741 3 0.024 0.991 0.159 1.048 0.055 0.970 0.043 1.172 4 0.008 0.998 ‐0.063 0.985 0.024 0.994 ‐0.242 0.930 5 0.002 1.000 0.015 1.000 0.006 1.000 0.070 1.000 Note that (1) for any given 𝐬, the base shear is more infmuenced by higher modes and (2) for any given reponse quantjty, the second, skewed 𝐬 gives greater modal contributjons for higher modes.

MCF’s example

The following table (from Chopra, 2nd ed.) displays the ̄ 𝑡𝑗 and their partjal sums for a shear‐type, 5 fmoors building where all the storey masses are equal and all the storey stjfgnesses are equal too. The response quantjtjes chosen are ̄ 𝑦5𝑜, the MCF’s to the top displacement and ̄ 𝑊

𝑜, the MCF’s to

the base shear, for two difgerent load shapes. 𝐬 = {0, 0, 0, 0, 1}𝑈 𝐬 = {0, 0, 0, −1, 2}𝑈 Top Displacement Base Shear Top Displacement Base Shear 𝑜 or 𝐾 ̄ 𝑦5𝑜 ∑𝐾 ̄ 𝑦5𝑗 ̄ 𝑊

𝑜

∑𝐾 ̄ 𝑊

𝑗

̄ 𝑦5𝑜 ∑𝐾 ̄ 𝑦5𝑗 ̄ 𝑊

𝑜

∑𝐾 ̄ 𝑊

𝑗

1 0.880 0.880 1.252 1.252 0.792 0.792 1.353 1.353 2 0.087 0.967 ‐0.362 0.890 0.123 0.915 ‐0.612 0.741 3 0.024 0.991 0.159 1.048 0.055 0.970 0.043 1.172 4 0.008 0.998 ‐0.063 0.985 0.024 0.994 ‐0.242 0.930 5 0.002 1.000 0.015 1.000 0.006 1.000 0.070 1.000 Note that (1) for any given 𝐬, the base shear is more infmuenced by higher modes and (2) for any given reponse quantjty, the second, skewed 𝐬 gives greater modal contributjons for higher modes.

MCF’s example

The following table (from Chopra, 2nd ed.) displays the ̄ 𝑡𝑗 and their partjal sums for a shear‐type, 5 fmoors building where all the storey masses are equal and all the storey stjfgnesses are equal too. The response quantjtjes chosen are ̄ 𝑦5𝑜, the MCF’s to the top displacement and ̄ 𝑊

𝑜, the MCF’s to

the base shear, for two difgerent load shapes. 𝐬 = {0, 0, 0, 0, 1}𝑈 𝐬 = {0, 0, 0, −1, 2}𝑈 Top Displacement Base Shear Top Displacement Base Shear 𝑜 or 𝐾 ̄ 𝑦5𝑜 ∑𝐾 ̄ 𝑦5𝑗 ̄ 𝑊

𝑜

∑𝐾 ̄ 𝑊

𝑗

̄ 𝑦5𝑜 ∑𝐾 ̄ 𝑦5𝑗 ̄ 𝑊

𝑜

∑𝐾 ̄ 𝑊

𝑗

1 0.880 0.880 1.252 1.252 0.792 0.792 1.353 1.353 2 0.087 0.967 ‐0.362 0.890 0.123 0.915 ‐0.612 0.741 3 0.024 0.991 0.159 1.048 0.055 0.970 0.043 1.172 4 0.008 0.998 ‐0.063 0.985 0.024 0.994 ‐0.242 0.930 5 0.002 1.000 0.015 1.000 0.006 1.000 0.070 1.000 Note that (1) for any given 𝐬, the base shear is more infmuenced by higher modes and (2) for any given reponse quantjty, the second, skewed 𝐬 gives greater modal contributjons for higher modes.

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Sectjon 3 Dynamic magnifjcatjon factor

Introductjon Modal partecipatjon factor Dynamic magnifjcatjon factor Statjc Correctjon

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Dynamic Response Ratjos

Dynamic Response Ratjos are the same that we have seen for SDOF systems. Next page, for an undamped system, harmonically excited, solid line, the ratjo of the modal elastjc force 𝐺

𝑇,𝑗 = 𝐿𝑗𝑟𝑗 sin 𝜕𝑢 to the harmonic

applied modal force, 𝑄

𝑗 sin 𝜕𝑢, plotued against the frequency ratjo 𝛾 = 𝜕/𝜕𝑗.

For 𝛾 = 0 the ratjo is 1, the applied load is fully balanced by the elastjc resistance. For fjxed excitatjon frequency, 𝛾 → 0 for high modal frequencies. dashed line,the ratjo of the modal inertjal force, 𝐺

𝐽,𝑗 = −𝛾2𝐺 𝑇,𝑗 to the load.

Note that for steady‐state motjon the sum of the elastjc and inertjal force ratjos is constant and equal to 1, as in (𝐺

𝑇,𝑗 + 𝐺 𝐽,𝑗) sin 𝜕𝑢 = 𝑄 𝑗 sin 𝜕𝑢.

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Dynamic Response Ratjos

Dynamic Response Ratjos are the same that we have seen for SDOF systems. Next page, for an undamped system, harmonically excited, solid line, the ratjo of the modal elastjc force 𝐺

𝑇,𝑗 = 𝐿𝑗𝑟𝑗 sin 𝜕𝑢 to the harmonic

applied modal force, 𝑄

𝑗 sin 𝜕𝑢, plotued against the frequency ratjo 𝛾 = 𝜕/𝜕𝑗.

For 𝛾 = 0 the ratjo is 1, the applied load is fully balanced by the elastjc resistance. For fjxed excitatjon frequency, 𝛾 → 0 for high modal frequencies. dashed line,the ratjo of the modal inertjal force, 𝐺

𝐽,𝑗 = −𝛾2𝐺 𝑇,𝑗 to the load.

Note that for steady‐state motjon the sum of the elastjc and inertjal force ratjos is constant and equal to 1, as in (𝐺

𝑇,𝑗 + 𝐺 𝐽,𝑗) sin 𝜕𝑢 = 𝑄 𝑗 sin 𝜕𝑢.

• 3
• 2
• 1

1 2 3 4 0.5 1 1.5 2 2.5 3 Modal resistance ratios Frequency ratio, β=ω/ωi FS/Pi FI/Pi

For a fjxed excitatjon frequency and high modal frequencies the frequency ratjo 𝛾 → 0. For 𝛾 → 0 the response is quasi‐statjc. Hence, for higher modes the response is pseudo‐statjc. On the contrary, for excitatjon frequencies high enough the lower modes respond with purely inertjal forces.

• 3
• 2
• 1

1 2 3 4 0.5 1 1.5 2 2.5 3 Modal resistance ratios Frequency ratio, β=ω/ωi FS/Pi FI/Pi

For a fjxed excitatjon frequency and high modal frequencies the frequency ratjo 𝛾 → 0. For 𝛾 → 0 the response is quasi‐statjc. Hence, for higher modes the response is pseudo‐statjc. On the contrary, for excitatjon frequencies high enough the lower modes respond with purely inertjal forces.

• 3
• 2
• 1

1 2 3 4 0.5 1 1.5 2 2.5 3 Modal resistance ratios Frequency ratio, β=ω/ωi FS/Pi FI/Pi

For a fjxed excitatjon frequency and high modal frequencies the frequency ratjo 𝛾 → 0. For 𝛾 → 0 the response is quasi‐statjc. Hence, for higher modes the response is pseudo‐statjc. On the contrary, for excitatjon frequencies high enough the lower modes respond with purely inertjal forces.

• 3
• 2
• 1

1 2 3 4 0.5 1 1.5 2 2.5 3 Modal resistance ratios Frequency ratio, β=ω/ωi FS/Pi FI/Pi

For a fjxed excitatjon frequency and high modal frequencies the frequency ratjo 𝛾 → 0. For 𝛾 → 0 the response is quasi‐statjc. Hence, for higher modes the response is pseudo‐statjc. On the contrary, for excitatjon frequencies high enough the lower modes respond with purely inertjal forces.

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Sectjon 4 Statjc Correctjon

Introductjon Modal partecipatjon factor Dynamic magnifjcatjon factor Statjc Correctjon

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Statjc Correctjon

The preceding discussion indicates that higher modes contributjons to the response could be approximated with the statjc response, leading to a Statjc Correctjon of the dynamic response. For a system where 𝑟𝑗(𝑢) ≈ 𝑞𝑗(𝑢) 𝐿𝑗 for 𝑗 > 𝑜dy, 𝑜dy being the number of dynamically responding modes, we can write 𝐲(𝑢) ≈ 𝐲dy(𝑢) + 𝐲st(𝑢) =

𝑜dy

• 1

𝝎𝑗𝑟𝑗(𝑢) +

𝑂

• 𝑜dy+1

𝝎𝑗 𝑞𝑗(𝑢) 𝐿𝑗 where the response for each of the fjrst 𝑜dy modes can be computed as usual.

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tjon Support Exc. Giacomo Boffj Introductjon Modal partecipatjon factor Dynamic magnifjcatjon factor Statjc Correctjon

Statjc Correctjon

The preceding discussion indicates that higher modes contributjons to the response could be approximated with the statjc response, leading to a Statjc Correctjon of the dynamic response. For a system where 𝑟𝑗(𝑢) ≈ 𝑞𝑗(𝑢) 𝐿𝑗 for 𝑗 > 𝑜dy, 𝑜dy being the number of dynamically responding modes, we can write 𝐲(𝑢) ≈ 𝐲dy(𝑢) + 𝐲st(𝑢) =

𝑜dy

• 1

𝝎𝑗𝑟𝑗(𝑢) +

𝑂

• 𝑜dy+1

𝝎𝑗 𝑞𝑗(𝑢) 𝐿𝑗 where the response for each of the fjrst 𝑜dy modes can be computed as usual.

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Statjc Modal Components

The statjc modal displacement component 𝐲𝑘, 𝑘 > 𝑜dy can be writuen 𝑦𝑘(𝑢) = 𝝎𝑘𝑟𝑘(𝑢) ≈ 𝝎𝑘𝝎𝑈

𝑘

𝐿𝑘 𝐪(𝑢) = 𝐆𝑘𝐪(𝑢) The modal fmexibility matrix is defjned by 𝐆𝑘 = 𝝎𝑘𝝎𝑈

𝑘

𝐿𝑘 and is used to compute the 𝑘‐th mode statjc defmectjons due to the applied load vector. The total displacements, the dynamic contributjons and the statjc correctjon, for 𝐪(𝑢) = 𝐬 𝑔(𝑢), are then 𝐲 ≈

𝑜dy

• 1

𝝎𝑘𝑟𝑘(𝑢) + 𝑔(𝑢)

𝑂

• 𝑜dy+1

𝐆𝑘𝐬.

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Alternatjve Formulatjon

Our last formula for statjc correctjon is 𝐲 ≈

𝑜dy

• 1

𝝎𝑘𝑟𝑘(𝑢) + 𝑔(𝑢)

𝑂

• 𝑜dy+1

𝐆𝑘𝐬. To use the above formula all mode shapes, all modal stjfgnesses and all modal fmexibility matrices must be computed, undermining the effjciency

• f the procedure.

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Alternatjve Formulatjon

This problem can be obviated computjng the total statjc displacements, 𝐲total

st

= 𝐋−1𝐪(𝑢), and subtractjng the statjc displacements due to the fjrst 𝑜dy modes...

𝑂

• 𝑜dy

𝐆𝑘𝐬𝑔(𝑢) = 𝐋−1𝐬𝑔(𝑢) −

𝑜dy

• 1

𝐆𝑘𝐬𝑔(𝑢) = 𝑔(𝑢) 𝐋−1 −

𝑜dy

• 1

𝐆𝑘 𝐬, so that the corrected total displacements have the expression 𝐲 ≈

𝑜dy

• 1

𝝎𝑗𝑟𝑗(𝑢) + 𝑔(𝑢) 𝐋−1 −

𝑜dy

• 1

𝐆𝑗 𝐬, The constant term (a generalized displacement vector) following 𝑔(𝑢) can be computed with the informatjon in our posses at the moment we begin the integratjon of the modal equatjons of motjon.

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Efgectjveness of Statjc Correctjon

In these circumstances, few modes with statjc correctjon give results comparable to the results obtained using much more modes in a straightgorward modal displacement superpositjon analysis. An high number of modes is required to account for the spatjal distributjon of the loading but only a few lower modes are subjected to signifjcant dynamic amplifjcatjon. Refjned stress analysis is required even if the dynamic response involves only a few lower modes.

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tjon Support Exc. Giacomo Boffj Introductjon Modal partecipatjon factor Dynamic magnifjcatjon factor Statjc Correctjon

Efgectjveness of Statjc Correctjon

In these circumstances, few modes with statjc correctjon give results comparable to the results obtained using much more modes in a straightgorward modal displacement superpositjon analysis. An high number of modes is required to account for the spatjal distributjon of the loading but only a few lower modes are subjected to signifjcant dynamic amplifjcatjon. Refjned stress analysis is required even if the dynamic response involves only a few lower modes.

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tjon Support Exc. Giacomo Boffj Introductjon Modal partecipatjon factor Dynamic magnifjcatjon factor Statjc Correctjon

Efgectjveness of Statjc Correctjon

In these circumstances, few modes with statjc correctjon give results comparable to the results obtained using much more modes in a straightgorward modal displacement superpositjon analysis. An high number of modes is required to account for the spatjal distributjon of the loading but only a few lower modes are subjected to signifjcant dynamic amplifjcatjon. Refjned stress analysis is required even if the dynamic response involves only a few lower modes.

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tjon Support Exc. Giacomo Boffj Introductjon Constant Acceleratjon Wilson’s Theta Method

Part II Numerical Integratjon

Introductjon Constant Acceleratjon Wilson’s Theta Method

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Sectjon 5 Introductjon

Introductjon Constant Acceleratjon Wilson’s Theta Method

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Introductjon to Numerical Integratjon

When we reviewed the numerical integratjon methods, we said that some methods are unconditjonally stable and others are conditjonally stable, that is the response blows‐out if the tjme step ℎ is great with respect to the natural preriod of vibratjon, ℎ >

𝑈

𝑜

𝑏 , where 𝑏 is a constant

that depends on the numerical algorithm. For MDOF systems, the relevant 𝑈 is the one associated with the highest mode present in the structural model, so for moderately complex structures it becomes impossibile to use a conditjonally stable algorithm. In the following, two unconditjonally stable algorithms will be analysed, i.e., the constant acceleratjon method, that we already know, and the new Wilson’s 𝜄 method.

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Sectjon 6 Constant Acceleratjon

Introductjon Constant Acceleratjon Wilson’s Theta Method

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Constant Acceleratjon, preliminaries

The initjal conditjons are known: 𝐲0, ̇ 𝐲0, 𝐪0, → ̈ 𝐲0 = 𝐍−1(𝐪0 − 𝐃 ̇ 𝐲0 − 𝐋 𝐲0). With a fjxed tjme step ℎ, compute the constant matrices 𝐁 = 2𝐃 + 4 ℎ𝐍, 𝐂 = 2𝐍, 𝐋+ = 2 ℎ𝐃 + 4 ℎ2 𝐍.

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Constant Acceleratjon, stepping

Startjng with 𝑗 = 0, compute the efgectjve force increment, Δ ̂ 𝐪𝑗 = 𝐪𝑗+1 − 𝐪𝑗 + 𝐁 ̇ 𝐲𝑗 + 𝐂 ̈ 𝐲𝑗, the tangent stjfgness 𝐋𝑗 and the current incremental stjfgness, ̂ 𝐋𝑗 = 𝐋𝑗 + 𝐋+. For linear systems, it is Δ𝐲𝑗 = ̂ 𝐋−1

𝑗 Δ ̂

𝐪𝑗, for a non linear system Δ𝐲𝑗 is produced by the modifjed Newton‐Raphson iteratjon procedure. The state vectors at the end of the step are 𝐲𝑗+1 = 𝐲𝑗 + Δ𝐲𝑗, ̇ 𝐲𝑗+1 = 2Δ𝐲𝑗 ℎ − ̇ 𝐲𝑗

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Constant Acceleratjon, new step

Increment the step index, 𝑗 = 𝑗 + 1. Compute the acceleratjons using the equatjon of equilibrium, ̈ 𝐲𝑗 = 𝐍−1(𝐪𝑗 − 𝐃 ̇ 𝐲𝑗 − 𝐋 𝐲𝑗). Repeat the substeps detailed in the previous slide.

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Modifjed Newton‐Raphson

Initjalizatjon 𝐳0 = 𝐲𝑗 𝐠S,0 = 𝐠S(system state) Δ𝐒1 = Δ ̂ 𝐪𝑗 𝐋T = ̂ 𝐋𝑗 For 𝑘 = 1, 2, … 𝐋TΔ𝐳

𝑘 = Δ𝐒𝑘

→Δ𝐳

𝑘 (test for convergence)

Δ ̇ 𝐳

𝑘 = ⋯

𝐳

𝑘 = 𝐳𝑘−1 + Δ𝐳 𝑘,

̇ 𝐳

𝑘 = ̇

𝐳𝑘−1 + Δ ̇ 𝐳

𝑘

𝐠S,𝑘 = 𝐠S(updated system state) Δ𝐠S,𝑘 = 𝐠S,𝑘 − 𝐠S,𝑘−1 − (𝐋T − 𝐋𝑗)Δ𝐳

𝑘

Δ𝐒𝑘+1 = Δ𝐒𝑘 − Δ𝐠S,𝑘 Return the value Δ𝐲𝑗 = 𝐳

𝑘 − 𝐲𝑗

A suitable convergence test is Δ𝐒𝑈

𝑘 Δ𝐳 𝑘

Δ ̂ 𝐪𝑈

𝑗 Δ𝐲𝑗,𝑘

≤ tol

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Sectjon 7 Wilson’s Theta Method

Introductjon Constant Acceleratjon Wilson’s Theta Method

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Wilson’s Theta Method

The linear acceleratjon method is signifjcantly more accurate than the constant acceleratjon method, meaning that it is possible to use a longer tjme step to compute the response of a SDOF system within a required accuracy. On the other hand, the method is not safely applicable to MDOF systems due to its numerical instability. Professor Ed Wilson demonstrated that simple variatjons of the linear acceleratjon method can be made unconditjonally stable and found the most accurate in this family of algorithms, collectjvely known as Wilson’s 𝜄 methods.

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Wilson’s Theta Method

The linear acceleratjon method is signifjcantly more accurate than the constant acceleratjon method, meaning that it is possible to use a longer tjme step to compute the response of a SDOF system within a required accuracy. On the other hand, the method is not safely applicable to MDOF systems due to its numerical instability. Professor Ed Wilson demonstrated that simple variatjons of the linear acceleratjon method can be made unconditjonally stable and found the most accurate in this family of algorithms, collectjvely known as Wilson’s 𝜄 methods.

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Wilson’s 𝜄 method

Wilson’s idea is very simple: the results of the linear acceleratjon algorithm are good enough only in a fractjon of the tjme step. Wilson demonstrated that his idea was correct, too... The procedure is really simple,

1 solve the incremental equatjon of equilibrium using the linear acceleratjon

algorithm, with an extended tjme step ̂ ℎ = 𝜄 ℎ, 𝜄 ≥ 1,

2 compute the extended acceleratjon increment

̂ Δ ̈ 𝐲 at ̂ 𝑢 = 𝑢𝑗 + ̂ ℎ,

3 scale the extended acceleratjon increment under the assumptjon of linear

acceleratjon, Δ ̈ 𝐲 =

1 𝜄 ̂

Δ ̈ 𝐲,

4 compute the velocity and displacements increment using the reduced

value of the increment of acceleratjon.

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tjon Support Exc. Giacomo Boffj Introductjon Constant Acceleratjon Wilson’s Theta Method

Wilson’s 𝜄 method

Wilson’s idea is very simple: the results of the linear acceleratjon algorithm are good enough only in a fractjon of the tjme step. Wilson demonstrated that his idea was correct, too... The procedure is really simple,

1 solve the incremental equatjon of equilibrium using the linear acceleratjon

algorithm, with an extended tjme step ̂ ℎ = 𝜄 ℎ, 𝜄 ≥ 1,

2 compute the extended acceleratjon increment

̂ Δ ̈ 𝐲 at ̂ 𝑢 = 𝑢𝑗 + ̂ ℎ,

3 scale the extended acceleratjon increment under the assumptjon of linear

acceleratjon, Δ ̈ 𝐲 =

1 𝜄 ̂

Δ ̈ 𝐲,

4 compute the velocity and displacements increment using the reduced

value of the increment of acceleratjon.

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tjon Support Exc. Giacomo Boffj Introductjon Constant Acceleratjon Wilson’s Theta Method

Wilson’s 𝜄 method

Wilson’s idea is very simple: the results of the linear acceleratjon algorithm are good enough only in a fractjon of the tjme step. Wilson demonstrated that his idea was correct, too... The procedure is really simple,

1 solve the incremental equatjon of equilibrium using the linear acceleratjon

algorithm, with an extended tjme step ̂ ℎ = 𝜄 ℎ, 𝜄 ≥ 1,

2 compute the extended acceleratjon increment

̂ Δ ̈ 𝐲 at ̂ 𝑢 = 𝑢𝑗 + ̂ ℎ,

3 scale the extended acceleratjon increment under the assumptjon of linear

acceleratjon, Δ ̈ 𝐲 =

1 𝜄 ̂

Δ ̈ 𝐲,

4 compute the velocity and displacements increment using the reduced

value of the increment of acceleratjon.

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tjon Support Exc. Giacomo Boffj Introductjon Constant Acceleratjon Wilson’s Theta Method

Wilson’s 𝜄 method

Wilson’s idea is very simple: the results of the linear acceleratjon algorithm are good enough only in a fractjon of the tjme step. Wilson demonstrated that his idea was correct, too... The procedure is really simple,

1 solve the incremental equatjon of equilibrium using the linear acceleratjon

algorithm, with an extended tjme step ̂ ℎ = 𝜄 ℎ, 𝜄 ≥ 1,

2 compute the extended acceleratjon increment

̂ Δ ̈ 𝐲 at ̂ 𝑢 = 𝑢𝑗 + ̂ ℎ,

3 scale the extended acceleratjon increment under the assumptjon of linear

acceleratjon, Δ ̈ 𝐲 =

1 𝜄 ̂

Δ ̈ 𝐲,

4 compute the velocity and displacements increment using the reduced

value of the increment of acceleratjon.

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tjon Support Exc. Giacomo Boffj Introductjon Constant Acceleratjon Wilson’s Theta Method

Wilson’s 𝜄 method

Wilson’s idea is very simple: the results of the linear acceleratjon algorithm are good enough only in a fractjon of the tjme step. Wilson demonstrated that his idea was correct, too... The procedure is really simple,

1 solve the incremental equatjon of equilibrium using the linear acceleratjon

algorithm, with an extended tjme step ̂ ℎ = 𝜄 ℎ, 𝜄 ≥ 1,

2 compute the extended acceleratjon increment

̂ Δ ̈ 𝐲 at ̂ 𝑢 = 𝑢𝑗 + ̂ ℎ,

3 scale the extended acceleratjon increment under the assumptjon of linear

acceleratjon, Δ ̈ 𝐲 =

1 𝜄 ̂

Δ ̈ 𝐲,

4 compute the velocity and displacements increment using the reduced

value of the increment of acceleratjon.

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tjon Support Exc. Giacomo Boffj Introductjon Constant Acceleratjon Wilson’s Theta Method

Wilson’s 𝜄 method descriptjon

Using the same symbols used for constant acceleratjon. First of all, for given initjal conditjons 𝐲0 and ̇ 𝐲0, initjalise the procedure computjng the constants (matrices) used in the following procedure and the initjal acceleratjon, ̈ 𝐲0 = 𝐍−1(𝐪0 − 𝐃 ̇ 𝐲0 − 𝐋 𝐲0), 𝐁 = 6𝐍/ ̂ ℎ + 3𝐃, 𝐂 = 3𝐍 + ̂ ℎ𝐃/2, 𝐋+ = 3𝐃/ ̂ ℎ + 6𝐍/ ̂ ℎ2.

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tjon Support Exc. Giacomo Boffj Introductjon Constant Acceleratjon Wilson’s Theta Method

Wilson’s 𝜄 method descriptjon

Startjng with 𝑗 = 0,

1 update the tangent stjfgness, 𝐋𝑗 = 𝐋(𝐲, ̇

𝐲𝑗) and the efgectjve stjfgness, ̂ 𝐋𝑗 = 𝐋𝑗 + 𝐋+, compute ̂ Δ ̂ 𝐪𝑗 = 𝜄Δ𝐪𝑗 + 𝐁 ̇ 𝐲𝑗 + 𝐂 ̈ 𝐲𝑗, with Δ𝐪𝑗 = 𝐪(𝑢𝑗 + ℎ) − 𝐪(𝑢𝑗)

2 solve ̂

𝐋𝑗 ̂ Δ𝐲 = ̂ Δ ̂ 𝐪𝑗, compute ̂ Δ ̈ 𝐲 = 6 ̂ Δ𝐲 ̂ ℎ2 − 6 ̇ 𝐲𝑗 ̂ ℎ − 3 ̈ 𝐲𝑗 → Δ ̈ 𝐲 = 1 𝜄 ̂ Δ ̈ 𝐲

3 compute

Δ ̇ 𝐲 = ( ̈ 𝐲𝑗 + 1 2Δ ̈ 𝐲)ℎ Δ𝐲 = ̇ 𝐲𝑗ℎ + (1 2 ̈ 𝐲𝑗 + 1 6Δ ̈ 𝐲)ℎ2

4 update state, 𝐲𝑗+1 = 𝐲𝑗 + Δ𝐲, ̇

𝐲𝑗+1 = ̇ 𝐲𝑗 + Δ ̇ 𝐲, 𝑗 = 𝑗 + 1, iterate restartjng from 1.

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tjon Support Exc. Giacomo Boffj Introductjon Constant Acceleratjon Wilson’s Theta Method

Wilson’s 𝜄 method descriptjon

Startjng with 𝑗 = 0,

1 update the tangent stjfgness, 𝐋𝑗 = 𝐋(𝐲, ̇

𝐲𝑗) and the efgectjve stjfgness, ̂ 𝐋𝑗 = 𝐋𝑗 + 𝐋+, compute ̂ Δ ̂ 𝐪𝑗 = 𝜄Δ𝐪𝑗 + 𝐁 ̇ 𝐲𝑗 + 𝐂 ̈ 𝐲𝑗, with Δ𝐪𝑗 = 𝐪(𝑢𝑗 + ℎ) − 𝐪(𝑢𝑗)

2 solve ̂

𝐋𝑗 ̂ Δ𝐲 = ̂ Δ ̂ 𝐪𝑗, compute ̂ Δ ̈ 𝐲 = 6 ̂ Δ𝐲 ̂ ℎ2 − 6 ̇ 𝐲𝑗 ̂ ℎ − 3 ̈ 𝐲𝑗 → Δ ̈ 𝐲 = 1 𝜄 ̂ Δ ̈ 𝐲

3 compute

Δ ̇ 𝐲 = ( ̈ 𝐲𝑗 + 1 2Δ ̈ 𝐲)ℎ Δ𝐲 = ̇ 𝐲𝑗ℎ + (1 2 ̈ 𝐲𝑗 + 1 6Δ ̈ 𝐲)ℎ2

4 update state, 𝐲𝑗+1 = 𝐲𝑗 + Δ𝐲, ̇

𝐲𝑗+1 = ̇ 𝐲𝑗 + Δ ̇ 𝐲, 𝑗 = 𝑗 + 1, iterate restartjng from 1.

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tjon Support Exc. Giacomo Boffj Introductjon Constant Acceleratjon Wilson’s Theta Method

Wilson’s 𝜄 method descriptjon

Startjng with 𝑗 = 0,

1 update the tangent stjfgness, 𝐋𝑗 = 𝐋(𝐲, ̇

𝐲𝑗) and the efgectjve stjfgness, ̂ 𝐋𝑗 = 𝐋𝑗 + 𝐋+, compute ̂ Δ ̂ 𝐪𝑗 = 𝜄Δ𝐪𝑗 + 𝐁 ̇ 𝐲𝑗 + 𝐂 ̈ 𝐲𝑗, with Δ𝐪𝑗 = 𝐪(𝑢𝑗 + ℎ) − 𝐪(𝑢𝑗)

2 solve ̂

𝐋𝑗 ̂ Δ𝐲 = ̂ Δ ̂ 𝐪𝑗, compute ̂ Δ ̈ 𝐲 = 6 ̂ Δ𝐲 ̂ ℎ2 − 6 ̇ 𝐲𝑗 ̂ ℎ − 3 ̈ 𝐲𝑗 → Δ ̈ 𝐲 = 1 𝜄 ̂ Δ ̈ 𝐲

3 compute

Δ ̇ 𝐲 = ( ̈ 𝐲𝑗 + 1 2Δ ̈ 𝐲)ℎ Δ𝐲 = ̇ 𝐲𝑗ℎ + (1 2 ̈ 𝐲𝑗 + 1 6Δ ̈ 𝐲)ℎ2

4 update state, 𝐲𝑗+1 = 𝐲𝑗 + Δ𝐲, ̇

𝐲𝑗+1 = ̇ 𝐲𝑗 + Δ ̇ 𝐲, 𝑗 = 𝑗 + 1, iterate restartjng from 1.

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tjon Support Exc. Giacomo Boffj Introductjon Constant Acceleratjon Wilson’s Theta Method

Wilson’s 𝜄 method descriptjon

Startjng with 𝑗 = 0,

1 update the tangent stjfgness, 𝐋𝑗 = 𝐋(𝐲, ̇

𝐲𝑗) and the efgectjve stjfgness, ̂ 𝐋𝑗 = 𝐋𝑗 + 𝐋+, compute ̂ Δ ̂ 𝐪𝑗 = 𝜄Δ𝐪𝑗 + 𝐁 ̇ 𝐲𝑗 + 𝐂 ̈ 𝐲𝑗, with Δ𝐪𝑗 = 𝐪(𝑢𝑗 + ℎ) − 𝐪(𝑢𝑗)

2 solve ̂

𝐋𝑗 ̂ Δ𝐲 = ̂ Δ ̂ 𝐪𝑗, compute ̂ Δ ̈ 𝐲 = 6 ̂ Δ𝐲 ̂ ℎ2 − 6 ̇ 𝐲𝑗 ̂ ℎ − 3 ̈ 𝐲𝑗 → Δ ̈ 𝐲 = 1 𝜄 ̂ Δ ̈ 𝐲

3 compute

Δ ̇ 𝐲 = ( ̈ 𝐲𝑗 + 1 2Δ ̈ 𝐲)ℎ Δ𝐲 = ̇ 𝐲𝑗ℎ + (1 2 ̈ 𝐲𝑗 + 1 6Δ ̈ 𝐲)ℎ2

4 update state, 𝐲𝑗+1 = 𝐲𝑗 + Δ𝐲, ̇

𝐲𝑗+1 = ̇ 𝐲𝑗 + Δ ̇ 𝐲, 𝑗 = 𝑗 + 1, iterate restartjng from 1.

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tjon Support Exc. Giacomo Boffj Introductjon Constant Acceleratjon Wilson’s Theta Method

A fjnal remark

The Theta Method is unconditjonally stable for 𝜄 > 1.37 and it achieves the maximum accuracy for 𝜄 = 1.42.

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tjon Support Exc. Giacomo Boffj Introductjon The Equatjon of Motjon An Example Response Analysis Response Analysis Example

Part III Multjple Support Excitatjon

Introductjon The Equatjon of Motjon An Example Response Analysis Response Analysis Example

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Sectjon 8 Introductjon

Introductjon The Equatjon of Motjon An Example Response Analysis Response Analysis Example

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tjon Support Exc. Giacomo Boffj Introductjon The Equatjon of Motjon An Example Response Analysis Response Analysis Example

Defjnitjons

Consider the case of a structure where the supports are subjected to assigned displacements histories, 𝑣𝑗 = 𝑣𝑗(𝑢). To solve this problem, we start with augmentjng the degrees of freedom with the support displacements. We denote the superstructure DOF with 𝐲𝑈, the support DOF with 𝐲𝑕 and we have a global displacement vector 𝐲, 𝐲 = 𝐲𝑈 𝐲𝑕 .

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Sectjon 9 The Equatjon of Motjon

Introductjon The Equatjon of Motjon An Example Response Analysis Response Analysis Example

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tjon Support Exc. Giacomo Boffj Introductjon The Equatjon of Motjon An Example Response Analysis Response Analysis Example

The Equatjon of Motjon

Damping efgects will be introduced at the end of our manipulatjons. The equatjon of motjon is 𝐍 𝐍𝑕 𝐍𝑈

𝑕

𝐍𝑕𝑕 ̈ 𝐲𝑈 ̈ 𝐲𝑕 + 𝐋 𝐋𝑕 𝐋𝑈

𝑕

𝐋𝑕𝑕 𝐲𝑈 𝐲𝑕 = 𝟏 𝐪𝑕 where 𝐍 and 𝐋 are the usual structural matrices, while 𝐍𝑕 and 𝐍𝑕𝑕 are, in the common case of a lumped mass model, zero matrices.

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Statjc Components

We decompose the vector of displacements into two contributjons, a statjc contributjon and a dynamic contributjon, atuributjng the given support displacements to the statjc contributjon. 𝐲𝑈 𝐲𝑕 = 𝐲𝑡 𝐲𝑕 + 𝐲 𝟏 where 𝐲 is the usual relatjve displacements vector.

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Determinatjon of statjc components

Because the 𝐲𝑕 are given, we can write two matricial equatjons that give us the statjc superstructure displacements and the forces we must apply to the supports, 𝐋𝐲𝑡 + 𝐋𝑕𝐲𝑕 = 𝟏 𝐋𝑈

𝑕𝐲𝑡 + 𝐋𝑕𝑕𝐲𝑕 = 𝐪𝑕

From the fjrst equatjon we have 𝐲𝑡 = −𝐋−1𝐋𝑕𝐲𝑕 and from the second we have 𝐪𝑕 = (𝐋𝑕𝑕 − 𝐋𝑈

𝑕𝐋−1𝐋𝑕)𝐲𝑕

Make a note that the support forces are zero when the structure is isostatjc or the structure is subjected to a rigid motjon.

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Determinatjon of statjc components

Because the 𝐲𝑕 are given, we can write two matricial equatjons that give us the statjc superstructure displacements and the forces we must apply to the supports, 𝐋𝐲𝑡 + 𝐋𝑕𝐲𝑕 = 𝟏 𝐋𝑈

𝑕𝐲𝑡 + 𝐋𝑕𝑕𝐲𝑕 = 𝐪𝑕

From the fjrst equatjon we have 𝐲𝑡 = −𝐋−1𝐋𝑕𝐲𝑕 and from the second we have 𝐪𝑕 = (𝐋𝑕𝑕 − 𝐋𝑈

𝑕𝐋−1𝐋𝑕)𝐲𝑕

Make a note that the support forces are zero when the structure is isostatjc or the structure is subjected to a rigid motjon.

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Going back to the EOM

We need the fjrst row of the two matrix equatjon of equilibrium, 𝐍 𝐍𝑕 𝐍𝑈

𝑕

𝐍𝑕𝑕 ̈ 𝐲𝑈 ̈ 𝐲𝑕 + 𝐋 𝐋𝑕 𝐋𝑈

𝑕

𝐋𝑕𝑕 𝐲𝑈 𝐲𝑕 = 𝟏 𝐪𝑕 substjtutjng 𝐲𝑈 = 𝐲𝑡 + 𝐲 in the fjrst row 𝐍 ̈ 𝐲 + 𝐍 ̈ 𝐲𝑡 + 𝐍𝑕 ̈ 𝐲𝑕 + 𝐋𝐲 + 𝐋𝐲𝑡 + 𝐋𝑕𝐲𝑕 = 𝟏 by the equatjon of statjc equilibrium, 𝐋𝐲𝑡 + 𝐋𝑕𝐲𝑕 = 𝟏 we can simplify 𝐍 ̈ 𝐲 + 𝐍 ̈ 𝐲𝑡 + 𝐍𝑕 ̈ 𝐲𝑕 + 𝐋𝐲 = 𝐍 ̈ 𝐲 + (𝐍𝑕 − 𝐍𝐋−1𝐋𝑕) ̈ 𝐲𝑕 + 𝐋𝐲 = 𝟏.

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Infmuence matrix

The equatjon of motjon is 𝐍 ̈ 𝐲 + (𝐍𝑕 − 𝐍𝐋−1𝐋𝑕) ̈ 𝐲𝑕 + 𝐋𝐲 = 𝟏. We defjne the infmuence matrix 𝐅 by 𝐅 = −𝐋−1𝐋𝑕, and write, reintroducing the damping efgects, 𝐍 ̈ 𝐲 + 𝐃 ̇ 𝐲 + 𝐋𝐲 = −(𝐍𝐅 + 𝐍𝑕) ̈ 𝐲𝑕 − (𝐃𝐅 + 𝐃𝑕) ̇ 𝐲𝑕

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Simplifjcatjon of the EOM

For a lumped mass model, 𝐍𝑕 = 𝟏 and also the effjcace forces due to damping are really small with respect to the inertjal ones, and with this understanding we write 𝐍 ̈ 𝐲 + 𝐃 ̇ 𝐲 + 𝐋𝐲 = −𝐍𝐅 ̈ 𝐲𝑕.

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Signifjcance of 𝐅

𝐅 can be understood as a collectjon of vectors 𝐟𝑗, 𝑗 = 1, … , 𝑂𝑕 (𝑂𝑕 being the number of DOF associated with the support motjon), 𝐅 = 𝐟1 𝐟2 ⋯ 𝐟𝑂𝑕 where the individual 𝐟𝑗 collects the displacements in all the DOF of the superstructure due to imposing a unit displacement to the support DOF number 𝑗.

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Signifjcance of 𝐅

This understanding means that the infmuence matrix can be computed column by column, in the general case by releasing one support DOF, applying a unit force to the released DOF, computjng all the displacements and scaling the displacements so that the support displacement component is made equal to 1,

• r in the case of an isostatjc component by examining the

instantaneous motjon of the 1 DOF rigid system that we obtain by releasing one constraint.

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Sectjon 10 An Example

Introductjon The Equatjon of Motjon An Example Response Analysis Response Analysis Example

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A First Example

𝑛 𝑛 B 𝑤𝐶 = 𝑤𝐶(𝑢) 𝑦3 𝑦1 𝑦2 A 𝑦1 𝑦2 A We want to determine the infmuence matrix 𝐅 for the structure in the fjgure above, subjected to an assigned motjon in B. First step, put in evidence another degree of freedom 𝑦3 corresponding to the assigned vertjcal motjon of the support in B and compute, using e.g. the PVD, the fmexibility matrix: 𝐆 = 𝑀3 3𝐹𝐾 54.0000 8.0000 28.0000 8.0000 2.0000 5.0000 28.0000 5.0000 16.0000 .

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Example, cont.

The stjfgness matrix is found by inversion, 𝐋 = 3𝐹𝐾 13𝑀3 +7.0000 +12.0000 −16.0000 +12.0000 +80.0000 −46.0000 −16.0000 −46.0000 +44.0000 . We are interested in the partjtjons 𝐋𝑦𝑦 and 𝐋𝑦𝑕: 𝐋𝑦𝑦 = 3𝐹𝐾 13𝑀3 +7.0000 +12.0000.0000 +12.0000 +80.0000.0000 , 𝐋𝑦𝑕 = 3𝐹𝐾 13𝑀3 −16 −46 . The infmuence matrix is 𝐅 = −𝐋−1

𝑦𝑦𝐋𝑦𝑕 = 1

16 28.0000 5.0000 , please compare 𝐅 with the last column of the fmexibility matrix, 𝐆.

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Sectjon 11 Response Analysis

Introductjon The Equatjon of Motjon An Example Response Analysis Response Analysis Example

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Response Analysis

Consider the vector of support acceleratjons, ̈ 𝐲𝑕 = ̈ 𝑦𝑕𝑚, 𝑚 = 1, … , 𝑂𝑕 and the efgectjve load vector 𝐪𝑓𝑔𝑔 = −𝐍𝐅 ̈ 𝐲𝑕 = −

𝑂𝑕

• 𝑚=1

𝐍𝐟𝑚 ̈ 𝑦𝑕𝑚(𝑢). We can write the modal equatjon of motjon for mode number 𝑜 ̈ 𝑟𝑜 + 2𝜂𝑜𝜕𝑜 ̇ 𝑟𝑜 + 𝜕2

𝑜𝑟𝑜 = − 𝑂𝑕

• 𝑚=1

Γ𝑜𝑚 ̈ 𝑦𝑕𝑚(𝑢) where Γ𝑜𝑚 = 𝝎𝑈

𝑜𝐍𝐟𝑚

𝑁∗

𝑜

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Response Analysis, cont.

The solutjon 𝑟𝑜(𝑢), with the notatjon we used previously, is hence 𝑟𝑜(𝑢) =

𝑂𝑕

• 𝑚=1

Γ𝑜𝑚𝐸𝑜𝑚(𝑢), 𝐸𝑜𝑚 being the response functjon for 𝜂𝑜 and 𝜕𝑜 due to the ground excitatjon ̈ 𝑦𝑕𝑚.

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Response Analysis, cont.

The total displacements 𝐲𝑈 are given by two contributjons, 𝐲𝑈 = 𝐲𝑡 + 𝐲, the expression of the contributjons are 𝐲𝑡 = 𝐅𝐲𝑕(𝑢) =

𝑂𝑕

• 𝑚=1

𝐟𝑚𝑦𝑕𝑚(𝑢), 𝐲 =

𝑂

• 𝑜=1

𝑂𝑕

• 𝑚=1

𝝎𝑜Γ𝑜𝑚𝐸𝑜𝑚(𝑢), and fjnally we have 𝐲𝑈 =

𝑂𝑕

• 𝑚=1

𝐟𝑚𝑦𝑕𝑚(𝑢) +

𝑂

• 𝑜=1

𝑂𝑕

• 𝑚=1

𝝎𝑜Γ𝑜𝑚𝐸𝑜𝑚(𝑢).

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Response in terms of Forces

For a computer program, the easiest way to compute the nodal forces is 𝑏) compute, element by element, the nodal displacements by 𝐲𝑈 and 𝐲𝑕, 𝑐) use the element stjfgness matrix to compute nodal forces, 𝑑) assemble element nodal loads into global nodal loads. That said, let’s see the analytjcal development...

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Forces, cont.

The forces on superstructure nodes due to deformatjons are 𝐠𝑡 =

𝑂

• 𝑜=1

𝑂𝑕

• 𝑚=1

Γ𝑜𝑚𝐋𝝎𝑜𝐸𝑜𝑚(𝑢) 𝐠𝑡 =

𝑂

• 𝑜=1

𝑂𝑕

• 𝑚=1

(Γ𝑜𝑚𝐍𝝎𝑜)(𝜕2

𝑜𝐸𝑜𝑚(𝑢)) = 𝑠 𝑜𝑚𝐵𝑜𝑚(𝑢)

the forces on support 𝐠𝑕𝑡 = 𝐋𝑈

𝑕𝐲𝑈 + 𝐋𝑕𝑕𝐲𝑕 = 𝐋𝑈 𝑕𝐲 + 𝐪𝑕

• r, using 𝐲𝑡 = 𝐅𝐲𝑕

𝐠𝑕𝑡 = (

𝑂𝑕

• 𝑚=1

𝐋𝑈

𝑕𝐟𝑚 + 𝐋𝑕𝑕,𝑚)𝑦𝑕𝑚 + 𝑂

• 𝑜=1

𝑂𝑕

• 𝑚=1

Γ𝑜𝑚𝐋𝑈

𝑕𝝎𝑜𝐸𝑜𝑚(𝑢)

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Forces

The structure response components must be computed considering the structure loaded by all the nodal forces, 𝐠 = 𝐠𝑡 𝐠𝑕𝑡 .

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Sectjon 12 Response Analysis Example

Introductjon The Equatjon of Motjon An Example Response Analysis Response Analysis Example

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Example

1,7 2,9 3,6 4,8 5,10 m m L L L L The dynamic DOF are 𝑦1 and 𝑦2, vertjcal displacements of the two equal masses, 𝑦3, 𝑦4, 𝑦5 are the imposed vertjcal displacements of the supports, 𝑦6, … , 𝑦10 are the rotatjonal degrees of freedom (removed by statjc condensatjon).

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Example

The stjfgness matrix for the 10x10 model is 𝐋10×10 = 𝐹𝐾 𝑀3 ⎡ ⎢ ⎢ ⎢ ⎢ ⎣

12 −12 6𝑀 6𝑀 −12 24 −12 −6𝑀 6𝑀 −12 24 −12 −6𝑀 6𝑀 −12 24 −12 −6𝑀 6𝑀 −12 12 −6𝑀 −6𝑀 6𝑀 −6𝑀 4𝑀2 2𝑀2 6𝑀 −6𝑀 2𝑀2 8𝑀2 2𝑀2 6𝑀 −6𝑀 2𝑀2 8𝑀2 2𝑀2 6𝑀 −6𝑀 2𝑀2 8𝑀2 2𝑀2 6𝑀 −6𝑀 2𝑀2 4𝑀2

⎤ ⎥ ⎥ ⎥ ⎥ ⎦

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Example, cont.

The fjrst product of the statjc condensatjon procedure is the linear mapping between translatjonal and rotatjonal degrees of freedom, given by ⃗ 𝜚 = 1 56𝑀

71 −90 24 −6 1 26 12 −48 12 −2 −7 42 −42 7 2 −12 48 −12 −26 −1 6 −24 90 −71

⃗ 𝑦.

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Example, cont.

Following statjc condensatjon and reordering rows and columns, the partjtjoned stjfgness matrices are 𝐋 = 𝐹𝐾 28𝑀3 276 108

108 276 ,

𝐋g = 𝐹𝐾 28𝑀3 −102 −264 −18

−18 −264 −102 ,

𝐋gg = 𝐹𝐾 28𝑀3

45 72 3 72 384 72 3 72 45 .

The infmuence matrix is 𝐅 = 𝐋−1𝐋g = 1 32 13 22 −3

−3 22 13 .

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Example, cont.

The eigenvector matrix is 𝛀 = −1 1

1 1

the matrix of modal masses is 𝐍⋆ = 𝛀𝑈𝐍𝛀 = 𝑛 2 0

0 2

the matrix of the non normalized modal partecipatjon coeffjcients is 𝐌 = 𝛀𝑈𝐍𝐅 = 𝑛

− 1

2 1 2 5 16 11 8 5 16

• and, fjnally, the matrix of modal partecipatjon factors,

𝚫 = (𝐍⋆)−1𝐌 =

− 1

4 1 4 5 32 11 16 5 32

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Example, cont.

Denotjng with 𝐸𝑗𝑘 = 𝐸𝑗𝑘(𝑢) the response functjon for mode 𝑗 due to ground excitatjon ̈ 𝑦g𝑘, the response can be writuen 𝐲 =

𝜔11− 1

4𝐸11+ 1 4𝐸13+𝜔12 5 32𝐸21+ 5 32𝐸23+ 11 16𝐸22

𝜔21− 1

4𝐸11+ 1 4𝐸13+𝜔22 5 32𝐸21+ 5 32𝐸23+ 11 16𝐸22

=

− 1

4𝐸13+ 1 4𝐸11+ 5 32𝐸21+ 5 32𝐸23+ 11 16𝐸22

− 1

4𝐸11+ 1 4𝐸13+ 5 32𝐸21+ 5 32𝐸23+ 11 16𝐸22 .