Structural Dynamics Main CW 423.docx

(1)

Structural Dynamics

(CEGEM071/CEGEG071)

Student: Carmine Russo – 14103106

Main Coursework – Design of a tuned-mass absorber

Introduction

In general, if the beams CD, EF and GH were not rigid the system would have twelve degrees of freedom (two each node). In the present case beams are rigid thus, the system has nine degrees of freedom, but by neglecting the columns' axial deformations (

EA → ∞

) they become three D.O.F. We choose as lagrangian coordinates the horizontal displacements of CD, EF and GH. Collecting these variables in the vector, we have:

´

u=

{

u

₁

u

2

u

₃

}

=

{

u

_CD

u

EF

u

_GH

}

In order to write the equilibrium equations, we have to find the stiffness of each element for each displacement.

In this case, we can use the direct method by solving the differential equation of the elastic beam:



η

IV

=

0



η

' ''

=C

1 

η

' '

=C

1

s +C

2 

η

'

=

1 ₂

C

1

s

2

_+C

2

s+C

3 

η=

1 ₆

C

1

s

3

+

1

2 C

2

s

2

+C

₃

s+C

₄ With

0 ≤ s≤ L

_AC

(2)

With the boundary conditions:

η

(A)

=

0 ;η

(A) '

=0 ;η

(C) '

=

0; η

(C)

=u

We get the constants of integration:

C

1

=

−12

L

3AC

u ;C

2

=

6 L

2AC

u ;C

3

=0 ;C

4

=0 ;

Finally:

T

₍_s₎

=−E I η

'' '

=

12 E I

L

3AC

u M

₍_s₎

=−E I η

' '

=

6 E I

L

3AC

(

2 s−L

_AC

)

u

Since there’s no lateral load applied to the columns, the shear force is constant along their length.

It is convenient to indicate with k the quantity:

k =

12 EI

L

3_AC

=7 ∙ 10

3

[

kN

m

]

Then, the generic expression of the shear force become:

T =k ∙u

Where “k” represent the stiffness of the column subjected to a horizontal displacement.

Part 1-a) Stiffness and mass matrix

We can assemble the stiffness matrix, column by column, simply imposing one deformation at time, while keeping the other one equal to zero, and finding the equilibrium forces.

Displacement

u

1 :

In this case, we have the mass

m

of the beam CD subjected to the displacement

u

1 .

As consequence:

 the beam CD is subjected to the horizontal forces

T

AC

, T

BD ,

T

CE

and

T

DF ;

 the beam EF is subjected to the horizontal forces

F

EF generate by

the deformation of the columns EC and FD.

The total forces for each mass are:

(3)

¿

(

k +k +k +k )∙ u

₁

=

4 k ∙u

₁

F

_EF

=−T

_EC

−T

_DF

=−(

k +k ) ∙u

₁

=−2 k ∙ u

₁

F

GH

=

0

We can summarize these relations as:

{

F

_CD

F

_EF

F

_GH

}

=

(

4 k

−2 k

0 )

∙ u

₁ Displacement

u

2 :

In this case, we consider the mass

m

of the beam EF subjected to the displacement

u

2 .

F

_CD

=−T

_CE

−T

_DF

=

(−k −k )∙ u

₂

=−2 k ∙u

₂

F

_EF

=

T

_CE

+

T

_DF

+

T

_EG

+T

_FH

=

4 k ∙ u

₂

F

_GH

=−T

_EG

−T

_FH

=(−k−k ) ∙u

₂

=−2 k ∙ u

₂

We can summarize these relations as:

{

F

_CD

F

_EF

F

GH

}

=

(

−2 k

4 k

−2 k

)

∙ u

₂ Displacement

u

3 :

In this case, we consider the mass

m

of the beam GH subjected to the displacement

u

3 .

F

_CD

=

0 F

_EF

=−T

_EG

−T

_FH

=−2 k ∙ u

₃

F

_GH

=T

_EG

+

T

_FH

=(

k +k )∙ u

₂

=2 k ∙ u

₃

(4)

{

F

_CD

F

_EF

F

GH

}

=

(

0 −2 k

2 k

)

∙ u

₃

Thus, the static equilibrium is represented by the equations:

{

F

_CD

F

_EF

F

GH

}

=

[

K

]

∙ ´u=

[

4 k

−2 k

0 −2 k

4 k

−2k

0 −2 k

2 k

]

∙

{

u

₁

u

₂

u

3

}

The stiffness matrix (“Lateral Stiffness Matrix”):

[

K

]

=

[

4 k

−2 k

0 −2 k

4 k

−2 k

0 −2 k

2 k

]

=2 k

[

−1

2 −1

0

0 −1

1 ]

=

[

28000

−14000

0 −14000

28000

−14000

0 −14000

14000

]

[

kN

m

]

The mass matrix:

[

M

]

=

[

m 0

0 0 m 0

0 0 m

]

=

[

9000

0

0 9000

0

0 9000

]

[

kg

]

The energetic approach

Instead of following the equilibrium approach, we can find the equation by calculating the total energy of the system:

 Kinetic Energy

T =

1

2 m ´u

1 2

+

1

2 m ´u

2 2

+

1

2 m ´u

3 2

The mass matrix can be found simply by calculating the Jacobian:

[

M

]

=

[

∂

2

_T

∂ ´u

i

∂ ´u

j

]

=

[

∂

2

_T

∂ ´u

₁2

∂

2

_T

∂ ´u

₁

∂ ´u

₂

∂

2

_T

∂ ´u

₁

∂ ´u

₃

∂

2

_T

∂ ´u

₂

∂ ´u

₁

∂

2

_T

∂ ´u

₂2

∂

2

_T

∂ ´u

₂

∂ ´u

₃

∂

2

T

∂ ´u

3

∂ ´u

1

∂

2

T

∂ ´u

3

∂ ´u

2

∂

2

T

∂ ´u

3 2

]

=

[

m 0

0 0 m 0

0 0 m

]

(5)

 Potential Energy

In this exercise the potential elastic energy is given by the horizontal displacements of the beams, therefore by using the stiffnesses already calculated above, we can directly write the expression of this energy without calculate the integrals. Simply remembering that the potential energy of a single spring is:

V

_spring

=

1

2 k ∙ Δ x

2

We just have to sum the potential energies of each deformed element:

V =

_∑

1

2 k ∙ u

i 2

=

¿

1

2 [

k

AC

u

1 2

_+k

BD

u

1 2

+

k

_CE

₍

u

₁

−u

₂

₎

2

+

k

_DF

₍

u

₁

−u

₂

₎

2

+

k

_FH

₍

u

₂

−u

₃

₎

2

+

k

_EG

₍

u

₂

−u

₃

₎

2

]

=

¿

1

2 k

[

u

1 2

+

u

₁2

+

(

u

₁

−

u

₂

)

2

+

(

u

₁

−u

₂

)

2

+

(

u

₂

−u

₃

)

2

+

(

u

₂

−u

₃

)

2

]

=

¿

1

2 k

[

2 u

1 2

+

2 ₍

u

₁

−u

₂

₎

2

₊₂

(

u

₂

−u

₃

₎

2

]

=

¿

1

2 k

[

4 u

1 2

+

4 u

₂2

+2 u

₃2

−

4 u

₁

u

₂

−4 u

₂

u

₃

]

=k

[

2 u

₁2

+2 u

₂2

+

u

₃2

−2 u

₁

u

₂

−2 u

₂

u

₃

]

The stiffness matrix can be found simply by calculating the Jacobian:

[

K

]

=

∂

2

_V

∂u

i

∂ u

j

=

[

∂

2

_V

∂ u

₁2

∂

2

_V

∂u

₁

∂ u

₂

∂

2

_V

∂u

₁

∂ u

₃

∂

2

_V

∂u

₂

∂ u

₁

∂

2

_V

∂u

₂2

∂

2

_V

∂u

₂

∂ u

₃

∂

2

V

∂u

₃

∂ u

₁

∂

2

V

∂u

₃

∂ u

₂

∂

2

V

∂u

3 2

]

=

[

4 k

−2 k

0 −2 k

4 k

−2k

0 −2 k

2 k

]

Exactly the same results obtained with the equilibrium method! The equations of motion:

In case of undamped free vibrations, the equations of motion are:

(6)

[

m 0

0 0 m 0

0 0 m

]

(

´

u

₁₍_t₎

´

u

₂₍_t₎

´

u

₃₍_t₎

)

+

[

_{−2 k}

4 k

−2 k

_{4 k}

_{−2 k}

0

0 −2 k

2 k

]

(

u

₁₍_t₎

u

₂₍_t₎

u

₃₍_t₎

)

=

(

0 ₀

0 )

Part 1-b) Approximate evaluation of eigenvalues:

the Rayleigh quotient

The use of the classic method in the evaluation of eigenvalues and eigenvectors, can lead to a very expensive calculation depending on the dimensions of the matrices [K] and [M]; an alternative approach to this problem consist in approximate numerical methods: one of those makes use of the Rayleigh quotient.

The Generalized Rayleigh Quotient is a scalar field in

R

n defined as the ratio:

R

₍_{K , M ,}_ϕ_r₎

=

{

ϕ

r

}

T

[

K

]

{

ϕ

_r

}

{

ϕ

r

}

T

[

M

]

_{

ϕ

r

}

Where [K] and [M] are real symmetric and positive definite. If

ϕ

i is the

eigenvector of the system, then:

R

₍_{K , M ,}ϕi)

=

ω

i

2

Furthermore,

R

(K , M ,ϕr) is stationary for every eigenvector

ϕ

i of the system:

grad

_[

R

₍_{K , M ,}_ϕ i)

]

=∇ R

(K , M ,ϕi)

=

2 {

ϕ

_i

}

T

[

M

]

{

ϕ

_i

}

(

[

K

]

{

ϕ

i

}

−

R

(K , M ,ϕi)

[

M

]

{

ϕ

i

}

)

⟹

2 {

ϕ

_i

}

T

[

M

]

{

ϕ

_i

}

(

[

K

]

−ω

i 2

_[

_M

_]

)

{

ϕ

i

}

=

{

0 }

Basing on these properties of the Rayleigh Quotient, we can estimate an approximate value of the first natural frequency just trying to guess the first eigenvector (knowing the generic form of the first modal shape of a “shear” frame): Considering that, the stiffness for the first and the second floor is the double of the third one, we can choose as first trial eigenvector:

(7)

{

ϕ

1

}

=

{

1

2 }

Getting the value:

R 1

₍_{K , M ,}_ϕ 1)

=

{

1

2 }

T

[

2.8 ∙ 10

7

_{−1.4 ∙10}

7

₀

−1.4 ∙ 10

7

_{2.8 ∙10}

7

_{−1.4 ∙10}

7

0 −1.4 ∙10

7

1.4 ∙10

7

]

{

1

2 }

{

1

2 }

T

[

9 ∙ 10

3

₀

0 9 ∙ 10

3

₀

0

0 9 ∙ 10

3

]

{

1

2 }

=

¿

518.519 [

ra d

2

s

2

]

As second trial modal shape we can consider the geometry of the system, choosing as shape a linear translation of each floor:

{

ϕ

₂

}

=

{

1

2

3 }

We get:

R 2

₍_{K , M ,}_ϕ 2)

=

{

1

2

3 }

T

[

2.8∙ 10

7

−1.4 ∙ 10

7

0 −1.4 ∙ 10

7

_{2.8 ∙ 10}

7

_{−1.4 ∙ 10}

7

0 −1.4 ∙ 10

7

_{1.4 ∙ 10}

7

]

{

1

2

3 }

{

1

2

3 }

T

[

9∙ 10

3

0

0 9∙ 10

3

0

0 9∙ 10

3

]

{

1

2

3 }

=

¿

333.333 [

ra d

2

s

2

]

Looking at the values of the gradient, we can check “the quality” of the solution: For the first trial

vector:

∇ R 1

(K , M ,ϕ1)

=

2 {

ϕ

1

}

T

[

M

]

{

ϕ

1

}

(

[

K

]

−ω

₁2

[

M

]

)

{

ϕ

₁

}

=

{

345.679 −691.358

172.840 }

(8)

For the second trial vector:

∇ R 2

(K ,M ,ϕ2)

=

2 {

ϕ

2

}

T

[

M

]

{

ϕ

2

}

(

[

K

]

−

ω

₂2

[

M

]

)

{

ϕ

₂

}

=

{

−47.619

−95.238

79.365 }

Clearly, since the second gradient is closer to zero than the first one, if the guess of the mode shape is correct (i.e. we made a choice close to the first effective mode),

{

ϕ

₂

}

_{represent a better approximation of the effective eigenvector.}

The estimate natural frequency is:

ω

_Ray¿

=

√

R 2

(K , M ,ϕ₂)

=

√

333.333 [

ra d

2

s

2

]

=18.257

[

rad

s

]

In general, a good choice for the trial eigenvector is the vector of static displacements, with forces proportional to the weight of each mass [1_{]. In this}

exercise the mass are equal, therefore what plays a key role in the first mode is the geometry of the system (i.e. the position of the masses and the stiffness of columns).

An iterative method to minimize Rayleigh Quotient using the Conjugate Gradient Algorithm

From the property shown above, the Rayleigh quotient is a value between the minimum eigenvalue and the maximum eigenvalue of the system:

λ

₁

≤ R

₍_{K , M ,}_ϕ

r)

≤ λ

n

Therefore it can be used in min-max theorem to get exact values of all eigenvalues analytically (for example with a formulation based on Lagrange multipliers) or, basing on a variational approach, is used in eigenvalue algorithms to obtain an approximation from any trial eigenvector: specifically, this is the basis for the method called “Rayleigh quotient iteration”.

In particular, since the problem asks to find the first eigenvalue, we have to solve a problem of minimum. The basic idea of Rayleigh quotient minimization is to construct a sequence

{

ϕ

r

}

r=1,2,3… such that:

R

₍_{K , M ,}_ϕ

r +1)

≤ R

(K , M ,ϕr)

;

∀ r=0,1,2 …

The hope is that the sequence

{

R

(K , M ,ϕr)

}

converges to

λ

1 and by consequence

the vector sequence

{

ϕ

r

}

towards the corresponding eigenvector. Using a

perturbation approach, defining:

{

ϕ

r +1

}

=

{

ϕ

r

}

+

δ

r

{

p

r

}

Where

p

r represent the “search direction” and the parameter

δ

r is determined

such that the Rayleigh quotient of the new iterate

ϕ

r +1 becomes minimal:

R

₍_{K , M ,}_ϕ

r +1)

=min

δ

[

R

(K , M ,ϕr+δrpr)

]

We can write the Rayleigh quotient of the linear combination

{

ϕ

r

}

+

δ

r

{

p

r

}

of two

(linearly independent) vectors

{

ϕ

r

}

and

{

p

r

}

:

(9)

R

₍_{K , M ,}_ϕ_r_+δ_r_p_r₎

=

{

ϕ

r

}

T

[

K

]

{

ϕ

_r

}

+2 δ

_r

{

ϕ

_r

}

T

[

K

]

{

p

_r

}

+

δ

_r2

{

p

_r

}

T

[

K

]

{

p

_r

}

{

ϕ

r

}

T

[

M

]

_{

ϕ

r

}

+2 δ

r

{

ϕ

r

}

T

[

M

]

_{

p

r

}

+

δ

r 2

{

p

r

}

T

[

M

]

_{

p

r

}

=

¿

{

1 δ

r

}

T

[

{

ϕ

_r

}

T

[

K

]

{

ϕ

_r

} {

ϕ

_r

}

T

[

K

]

{

p

_r

}

{

ϕ

_r

}

T

[

K

]

{

p

_r

} {

p

_r

}

T

[

K

]

{

p

_r

}

]

{

1 δ

r

}

{

1 δ

_r

}

T

[

{

ϕ

r

}

T

[

M

]

{

ϕ

r

} {

ϕ

r

}

T

[

M

]

{

p

_r

}

{

ϕ

r

}

T

[

M

]

_{

p

_r

_{} {}

p

_r

_}

T

[

M

]

_{

p

_r

_}

]

{

1 δ

_r

}

The one above is the Rayleigh Quotient associated with the generalized 2×2 eigenvalue problem:

[

{

ϕ

r

}

T

[

K

]

_{

ϕ

r

} {

ϕ

r

}

T

[

K

]

_{

p

_r

_}

{

ϕ

_r

}

T

[

K

]

{

p

_r

} {

p

_r

}

T

[

K

]

{

p

_r

}

]

{

α

β

}

=

λ

[

{

ϕ

r

}

T

[

M

]

_{

ϕ

r

} {

ϕ

r

}

T

[

M

]

_{

p

_r

_}

{

ϕ

_r

}

T

[

M

]

{

p

_r

} {

p

_r

}

T

[

M

]

{

p

_r

}

]

{

α

β

}

The smaller of the two eigenvalues is the searched value that minimizes the Rayleigh quotient.

There are various ways how to choose the search direction

{

p

r

}

, and every

different method defines a different algorithm. In general, the faster convergence for the generalized eigenvalues problem is obtained with procedures that make use of the gradient: for the purpose of this exercise, we will show an application of the

Conjugate Gradient Algorithm.

We define the search directions as:

{

p

r

}

=−

{

grad

[

R

(K , M ,ϕr)

]

}

+

μ

k

{

p

r−1

}

Where the coefficient

μ

k is determined such that

{

p

r

}

and

{

p

r−1

}

are

conjugate, i.e.:

μ

k

=

{

grad

_[

R

₍_{K , M ,ϕ} r)

]

}

T

[

M

]

_{

grad

_[

R

₍K , M ,ϕr)

]

}

{

grad

_[

R

₍_{K , M ,ϕ} r −1)

]

}

T

[

M

]

_{

grad

_[

R

₍_{K , M ,}_ϕ r−1)

]

}

That is the expression of

μ

that leads to the fastest convergence.

Defined these quantities is possible to build an iterative algorithm that guarantees that

R

₍_{K , M ,}_ϕ

r +1)

≤ R

(K , M ,ϕr)

Unless the quantity (that represent the residual):

o

_r

=

[

K

]

_{

ϕ

_r

_}

−

R

₍_{K , M ,}_ϕ

r)

[

M

]

{

ϕ

r

}

=0

In which case

ϕ

r is the searched eigenvector. In general, if the initial vector

ϕ

0

has a nonvanishing component in the direction of the “smallest” eigenvector

ϕ

1 ,

convergence is toward the smallest eigenvalue

λ

1

=ω

12 [See [1] ]. The Conjugate

(10)

Initialization

Let

ϕ

0 be a unit vector. In case of three D.O.F.

{

ϕ

0

}

=

{

1 1 1

}

T

Calculate the two vectors:

{

v

₀

_}

=

[

K

]

_{

ϕ

0

}

;

{

u

0

}

=

[

M

]

{

ϕ

0

}

;

Calculate the initial Rayleigh Quotient:

R

₍_ϕ₀₎

=

{

v

0

}

T

{

ϕ

0

}

{

u

₀

}

T

{

ϕ

₀

}

Calculate the initial gradient and his norm:

∇

_[

R

₍_ϕ₀₎

_]

=2

₍

{

v

0

}

−R

(ϕ0)

{

u

0

}

)

‖

∇

_[

R

₍_ϕ₀₎

_]

‖

Core operation

While the norm of the gradient is larger than a chosen tolerance value (tol):

‖

∇

_[

R

₍_ϕ_r−1₎

_]

‖

>tol

calculate:

¿

{

p

_r

}

=−∇

_[

R

₍_ϕ r −1)

]

¿

_{

p

_r

_}

=−∇

_[

R

₍_ϕ r −1)

]

+

∇

_[

R

₍_ϕ_{r −1}₎

_]

T

[

M

]

∇

_[

R

₍_ϕ_{r −1}₎

_]

∇

_[

R

₍_ϕ r −2)

]

T

[

M

]

∇

_[

R

₍_ϕ r −2)

]

{

p

_{r −1}

_}

for r =1

¿

for r >1

¿

Define the scalar quantities:

A

_r

=

_{

ϕ

r−1

}

T

[

K

]

_{

ϕ

r−1

}

B

_r

=

{

ϕ

_r−1

}

T

[

K

]

{

p

_r

}

C

r

=

{

p

r

}

T

[

K

]

_{

p

r

}

D

_r

=

{

ϕ

r−1

}

T

[

M

]

{

ϕ

r −1

}

E

_r

=

_{

ϕ

_r−1

}

T

[

M

]

{

p

_r

}

F

_r

=

_{

p

_r

_}

T

[

M

]

_{

p

_r

_}

And the two quantities:

Ψ

1(δ )

=

A

r

+

2 δ

r

B

r

+δ

r 2

C

r

Ψ

2(δ )

=D

r

+

2 δ

r

E

r

+

δ

r2

F

r

Minimize the Rayleigh Quotient to get the value of

δ

r :

d

d δ

_r

[

R

r(δ)

]

=

d

d δ

_r

[

Ψ

_{1(δ )}

Ψ

_{2(δ )}

]

=

d

d δ

_r

[

Ψ

1 (δ )]

Ψ

2(δ)

−Ψ

1(δ)

d

d δ

_r

[

Ψ

2(δ )]

[

Ψ

₂₍_δ₎

_]

2

=0

(11)

⟹

2 [

(

B

r

+

δ

r

C

r

)

(

D

r

+

2δ

r

E

r

+

δ

r 2

F

_r

)

−

(

A

_r

+2 δ

_r

B

_r

+

δ

_r2

C

_r

)

(

E

_r

+

δ

_r

F

_r

)

]

(

D

r

+

2 δ

r

E

r

+

δ

r 2

F

r

)

2

=0

Simplifying:

2 [

δ

r 2

(

C

r

E

r

−B

r

F

r

)

+

δ

r

(

C

r

D

r

−

F

r

A

r

)

+

(

B

r

D

r

−

A

r

E

r

)

]

(

δ

_r2

_F

r

+

2 δ

r

E

r

+

D

r

)

2

=

0

And we get from the numerator:

δ

r 1 2

=

(

F

r

A

r

−C

r

D

r

)

±

√

(

F

r

A

r

−C

r

D

r

)

2

−

4 (

C

_r

E

_r

−

B

_r

F

_r

) (

B

_r

D

_r

−

A

_r

E

_r

)

2 (

C

_r

E

_r

−B

_r

F

_r

)

It is proven that the minimum of

R

r(δ) is reached using the positive sign, with the condition that the roots are not the same of the denominator. Output of each iteration and final results Calculate:

{

ϕ

r +1

}

=

{

ϕ

r

}

+δ

r

{

p

r

}

{

v

_r

_}

=

[

K

]

_{

ϕ

r

}

;

{

u

r

}

=

[

M

]

{

ϕ

r

}

;

R

₍_ϕ_r₎

=

{

v

r

}

T

{

ϕ

r

}

{

u

_r

}

T

{

ϕ

_r

}

∇

_[

R

₍_ϕ r)

]

=2

(

{

v

r

}

−R

(ϕr)

{

u

r

}

)

‖

∇

_[

R

₍_ϕ_r₎

_]

‖

Using the algorithm described above (see the complete Matlab code attached), we get:

The Rayleigh Quotient after 9 iteration is:

R =

308.0969

The first mode shape after 9 iteration is:

fi = 0.5947 1.0716 1.3363

The first natural frequency in [rad/s] is:

Omega_nat_1 = 17.5527

(12)

Compared with the previous value of

ω

natural, the percentage error is

ω

_Ray

¿

−ω

1

ω

₁

∙100=4.012

Part 1-c) Modal Analysis

Starting from the equations of motion, for the undamped free vibrating system:

[

M

]

u

´

₍_t₎

+

[

K

]

u

₍_t₎

=0

[

m 0

0 0 m 0

0 0 m

]

(

´

u

₁₍_t₎

´

u

₂₍_t₎

´

u

₃₍_t₎

)

+

[

4 k

−2 k

0 −2 k

4 k

−2 k

0 −2 k

2 k

]

(

u

₁₍_t₎

u

₂₍_t₎

u

₃₍_t₎

)

=

(

0

0 )

That is a set of linear differential equations with constant coefficients. We may generate a solution by assuming that each generalized coordinate varies exponentially as

e

λt . In principle, the coefficient

λ

could be any constant but, since damping in not taken in account, the system is conservative. If

ℜ

{

λ

}

>0

, the total mechanical energy of the system

(

T +V )

will grow, while

ℜ

{

λ

}

<0

leads to a decaying response: both cases violate conservation of energy.

We construct a solution based on the trial form:

u

_{(t )}

=ℜ

{

B

(

ϕ

)

e

iωt

_}

Where

B

and

ϕ

are constants.

Every term produced by the substitution of the trial solution exhibits the same time dependence, so the equation of motion will be satisfied at all instants only if the coefficients of the exponential terms match. Furthermore, the constant factor

B

is common to every term, so it cancels. Therefore, we have:

−

[

M

]

B (ϕ ) ω

2

_e

iωt

+

[

K

]

B (ϕ ) e

iωt

=

0⟹

{

[

K

]

−

ω

2

_[

_M

_]

_}

₍

_{ϕ )=0}

The nontrivial solution of this system can exist only if the value of

ω

is such that

{

[

K

]

−ω

2

[

M

]

}

is not invertible. We must find the value of

ω

for which the determinant of this matrix is equal to zero (general eigenvalue problem):

det

( [

K

]

−

λ

[

M

])

=0

⟹ det

[

4 k −λm

−2 k

4 k−λm

−2 k

0

0 −2 k

2 k −λm

]

=0

Where we set

ω

2

=

λ

. From this condition, we get the characteristic equation:

(

4 k−λm)

2

(2 k −λm)−4 k

2

(4 k−λm)−4 k

2

(

2 k− λm)=0

⟹ m

3

λ

3

−10 km λ

2

+

24 k

2

mλ−8 k

3

=0

(13)

{

λ

}

=

{

λ

₁

λ

₂

λ

3

}

=

{

308.097 2.419 ∙ 10

3

5.051 ∙ 10

3

}

[

ra d

2

s

2

]

Using the “eig” function in Matlab, the natural frequencies are:

ω

_n

=

√

λ=

(

ω

₁

ω

₂

ω

₃

)

=

(

17.553

49.182

71.069 )

[

rad

s

]

The periods are:

T =

2 π

ω

_n

=

(

T

₁

T

₂

T

₃

)

=

(

0.358

0.128

0.088 )

[

s

]

The frequencies:

f =

1 T

=

(

f

1

f

₂

f

₃

)

=

(

2.794

7.827

11.311 )

[

Hz

]

Eigenvectors:

Now that we have determined the natural frequencies, we proceed to evaluate the mode shapes.

When

ω=ω

1 ,

ω

2 or

ω

3 , at least one of the scalar equations described by

{

[

K

]

−

ω

_i2

_[

_M

_]

_}

₍

_{ϕ )=0}

is not independent of the other. We can retain one of the two conditions (for instance the first one), and add a condition on the norm of the first eigenvector.  Mode shape 1

[

4 k −λ

₁

m

−2 k

0 −2 k

4 k −λ

₁

m

−2 k

0 −2 k

2 k−λ

₁

m

]

(

ϕ

11

ϕ

₂₁

ϕ

₃₁

)

=

(

0

0 )

With the condition:

ϕ

11 2

+ϕ

21 2

+ϕ

31 2

=1

We have:

(14)

{

(

4 k−λ

1

m

)

ϕ

11

−2 k

ϕ

21

=0

−2 k

ϕ

11

+

(

4 k −λ

1

m

)

ϕ

21

−2 k

ϕ

31

=0

ϕ

₁₁2

+

ϕ

₂₁2

+

ϕ

₃₁2

=1

⟹

{

ϕ

₁₁

=

2k

(

4 k− λ

1

m

)

ϕ

₂₁

ϕ

31

=

[

(

4 k− λ

1

m

)

2

−4 k

2

2 k

(

4 k− λ

₁

m

)

]

ϕ

21

ϕ

₂₁

=

{

1 [

2 k

(

4 k −λ

₁

m

)

]

+1+

[

(

4 k−λ

1

m

)

2

−

4 k

2

2 k

(

4 k−λ

₁

m

)

]

2

}

0.5

The first eigenvector is:

(

ϕ

11

ϕ

₂₁

ϕ

₃₁

)

=

(

0.32799

0.59101

0.73698

)

 Mode shape 2

[

4 k −λ

₂

m

−2 k

0 −2 k

4 k −λ

₂

m

−2 k

0 −2 k

2 k− λ

₂

m

]

(

ϕ

₁₂

ϕ

₂₂

ϕ

32

)

=

(

0

0 )

With the condition:

ϕ

12 2

+

ϕ

22 2

+

ϕ

32 2

=1

We have:

{

(

4 k−λ

2

m

)

ϕ

12

−2 k

ϕ

22

=0

−2 k

ϕ

₂₂

+

(

2 k−λ

₂

m

)

ϕ

₃₂

=0

ϕ

₁₂2

_+ϕ

22 2

+

ϕ

₃₂2

=1

⟹

{

ϕ

₁₂

=

2 k

(

4 k −λ

₂

m

₎

ϕ

22

ϕ

₃₂

=

2 k

(

2 k− λ

2

m

)

ϕ

₂₂

ϕ

₂₂

=

{

1 [

2 k

(

4 k −λ

₂

m

₎

]

2

+1+

[

2 k

(

2 k −λ

₂

m

₎

]

2

}

0.5

The second eigenvector is:

(

ϕ

12

ϕ

₂₂

ϕ

₃₂

)

=

(

0.73698

0.32799

−0.59101

)

 Mode shape 3

[

4 k −λ

₃

m

−2 k

0 −2 k

4 k −λ

₃

m

−2 k

0 −2 k

2 k− λ

₃

m

]

(

ϕ

13

ϕ

₂₃

ϕ

₃₃

)

=

(

0

0 )

With the condition:

ϕ

132

+

ϕ

232

+

ϕ

332

=1

(15)

{

(

4 k−λ

3

m

)

ϕ

13

−2 k

ϕ

23

=0

−2 k

ϕ

23

+

(

2 k−λ

3

m

)

ϕ

33

=0

ϕ

13 2

_+ϕ

23 2

+

ϕ

33 2

=1

⟹

{

ϕ

₁₃

=

2k

(

4 k− λ

3

m

)

ϕ

₂₃

ϕ

33

=

2k

(

2 k −λ

3

m

)

ϕ

23

ϕ

23

=

{

1 [

2 k

(

4 k −λ

₃

m

)

]

2

+1+

[

2 k

(

2 k−λ

₃

m

)

]

2

}

0.5

The third eigenvector is:

(

ϕ

13

ϕ

₂₃

ϕ

₃₃

)

=

(

0.59101

−0.73698

0.32799

)

We can normalize both vectors with the respect of the maximum absolute value of each mode shape:

ϕ

₁

=

1 |

ϕ

max (1)

_|

(

ϕ

11

ϕ

₂₁

ϕ

₃₁

)

=

1 ϕ

31

(

ϕ

11

ϕ

₂₁

ϕ

₃₁

)

=

1 0.73698

(

0.32799

0.59101

0.73698

)

=

(

0.445

0.802

1 )

ϕ

₂

=

1 |

ϕ

max(2)

|

(

ϕ

₁₂

ϕ

₂₂

ϕ

₃₂

)

=

1 ϕ

₁₂

(

ϕ

₁₂

ϕ

₂₂

ϕ

₃₂

)

=

1 0.73698

(

0.73698

0.32799

−0.59101

)

=

(

1

0.445 −0.802

)

ϕ

3

=

1 |

ϕ

max (3)

|

(

ϕ

₁₃

ϕ

23

ϕ

33

)

=

1 ϕ

₂₃

(

ϕ

₁₃

ϕ

23

ϕ

33

)

=

1 0.73698

(

0.59101

−0.73698

0.32799

)

=

(

0.802 −1

0.445 )

The modal shape matrix:

[

ϕ

]

=

[

ϕ

1

ϕ

2

ϕ

3

]

=

[

ϕ

₁₁

ϕ

₁₂

ϕ

₁₃

ϕ

21

ϕ

22

ϕ

23

ϕ

₃₁

ϕ

₃₂

ϕ

₃₃

]

=

[

0.445

1

0.802

0.445 −1

1 −0.802 0.445

]

We define the modal masses:

ϕ

iT

[

M

]

ϕ

i

=μ

i

withi=1,2,3

Since each element of an eigenvector is scaled by an arbitrary element, it follows that the modal mass values depend on the choice of that element; moreover, the modal masses occur throughout the evaluation of both responses, thereby compensating this arbitrariness contained in the eigenvector.

μ

₁

=

ϕ

₁T

[

M

]

ϕ

₁

=

(

ϕ

11

ϕ

₂₁

ϕ

₃₁

)

T

[

m 0

0 0 m 0

0 0 m

]

(

ϕ

11

ϕ

₂₁

ϕ

₃₁

)

=

[

₍

ϕ

₁₁

)

2

+

₍

ϕ

₂₁

)

2

+

₍

ϕ

₃₁

)

2

]

m=16570.498

[

kg

]

(16)

μ

₂

=

ϕ

₂T

[

M

]

ϕ

₂

=

(

ϕ

12

ϕ

₂₂

ϕ

32

)

T

[

m 0

0 0 m 0

0 0 m

]

(

ϕ

12

ϕ

₂₂

ϕ

32

)

=

[

₍

ϕ

₁₂

)

2

+

₍

ϕ

₂₂

)

2

+

₍

ϕ

₃₂

)

2

]

m=16570.498

[

kg

]

μ

₃

=

ϕ

3 T

_[

_M

_]

_ϕ

3

=

(

ϕ

₁₃

ϕ

23

ϕ

33

)

T

[

m 0

0 0 m 0

0 0 m

]

(

ϕ

₁₃

ϕ

23

ϕ

33

)

=

[

₍

ϕ

13

)

2

+

₍

ϕ

23

)

2

+

₍

ϕ

33

)

2

_]

_m=16570.498

_[

_kg

_]

Now we can calculate the normal modes (i.e. modes normalized with the respect of the mass matrix)

Φ

1

=

1 √

μ

₁

(

ϕ

11

ϕ

21

ϕ

31

)

=

1

128.726 (

0.445

0.802

1 )

=

(

3.457 ∙10

−3

6.230 ∙10

−3

7.768 ∙10

−3

)

[

k g

−1 2

]

Φ

₂

=

1 √

μ

₂

(

ϕ

12

ϕ

22

ϕ

32

)

=

1

128.726 (

1

0.445 −0.802

)

=

(

7.768 ∙ 10

−3

3.457 ∙ 10

−3

−6.230 ∙10

−3

)

[

k g

−1 2

]

Φ

₃

=

1 √

μ

₃

(

ϕ

13

ϕ

23

ϕ

33

)

=

1

128.726 (

0.802 −1

0.445 )

=

(

6.230 ∙10

−3

−7.768 ∙10

−3

3.457 ∙10

−3

)

[

k g

−1 2

]

The mass normalized shape matrix:

[

Φ

]

=

[

Φ

₁

Φ

₂

Φ

₃

_]

=

[

Φ

₁₁

Φ

₁₂

Φ

₁₃

Φ

₂₁

Φ

₂₂

Φ

₂₃

Φ

₃₁

Φ

₃₂

Φ

₃₃

]

=

[

3.457 ∙ 10

−3

_{7.768 ∙10}

−3

_{6.230 ∙ 10}

−3

6.230 ∙ 10

−3

_{3.457 ∙10}

−3

_{−7.768 ∙10}

−3

7.768 ∙10

−3

_{−6.230∙ 10}

−3

_{3.457 ∙ 10}

−3

]

[

k g

−1 2

]

We can check the orthogonality properties:

Φ

1 T

[

M

]

Φ

1

=1

[

kg

]

Φ

1 T

[

M

]

Φ

2

=0

Φ

1 T

[

M

]

Φ

3

=0

Φ

2 T

[

M

]

Φ

1

=0

Φ

2 T

[

M

]

Φ

2

=1

[

kg

]

Φ

2 T

[

M

]

Φ

3

=0

Φ

3T

[

M

]

Φ

1

=0

Φ

3T

[

M

]

Φ

2

=0

Φ

3T

[

M

]

Φ

3

=1

[

kg

]

Φ

1T

[

K

]

Φ

1

=

ω

12

Φ

1T

[

K

]

Φ

2

=

0 Φ

1T

[

K

]

Φ

3

=

0 Φ

2T

[

K

]

Φ

1

=

0 Φ

2T

[

K

]

Φ

2

=

ω

22

Φ

2T

[

K

]

Φ

3

=

0 Φ

3T

[

K

]

Φ

1

=

0 Φ

3T

[

K

]

Φ

2

=

0 Φ

3T

[

K

]

Φ

3

=

ω

32

(17)

Mode 1 Mode 2 Mode 3

Part 1-d) Rayleigh damping approximation

The expression of total displacement for a structure subjected to ground motion is:

u

_TOT(t)

=u

(t )

+u

g(t)

The displacement of the ground is denoted by

u

g . For a sinusoidal ground motion,

each floor is subjected to a harmonic acceleration:

u

_g₍_t₎

=

X

g

sin (Υ t )⟹ ´u

g(t)

=−X

g

Υ

2

sin(Υ t )

Consequently, the damped equation of motion of the structure subjected to a harmonic lateral ground motion is:

[

M

]

( ´

u

₍_t₎

+ ´

u

_g₍_t₎

)+

[

C

]

u

´

₍_t₎

+

[

K

]

u

₍_t₎

=0

Finally, the equation of motion can be expressed as:

[

M

]

u

´

₍_t₎

+

[

C

]

u

´

₍_t₎

+

[

K

]

u

₍_t₎

=m

(

1

1 )

X

_g

Υ

2

sin (Υ t )

With

Υ =2 πΩ

. We can ignore the sign of the excitation because of it is harmonic and periodic. Assuming a simplified Rayleigh damping matrix

[

C

]

=

α

[

M

]

+

β

[

K

]

, the equation of motion become:

(18)

Where

α

and

β

can be found considering that the system is damped at 3% of the critical damping when oscillating at its first and third natural mode frequency.

Diagonalization by Rayleigh Damping

In order to do that first, we make a modal transformation

u

(t)

=

[

Φ

]

q

(t) by using the

mass normalized shape matrix, and we get:

[

M

][

Φ

]

q

´

(t)

+

(

α

[

M

]

+

β

[

K

]) [

Φ

]

q

´

(t)

+

[

K

] [

Φ

]

q

(t)

=m r F

g(Υ ,t) Then, we pre-multiply by

[

Φ

]

T and we get:

[

Φ

]

T

[

M

][

Φ

]

q

´

(t)

+

[

Φ

]

T

(

α

[

M

]

+

β

[

K

]) [

Φ

]

q

´

(t)

+

[

Φ

]

T

[

K

][

Φ

]

q

(t)

=

[

Φ

]

T

m

(

1

1 )

F

_g(Υ ,t)

Using the orthogonality property of mode shape matrix, the equation become:

´

q

₍_t₎

+

[

α+βω

₁2

₀

0 α+ βω

₂2

₀

0

0 α +βω

₃2

]

´

q

₍_t₎

+

[

ω

₁2

₀

0 ω

₂2

₀

0

0 ω

₃2

]

q

₍_t₎

=

m

(

Φ

₁₁

+Φ

₂₁

+Φ

₃₁

Φ

₁₂

+Φ

₂₂

+Φ

₃₂

Φ

₁₃

+Φ

₂₃

+Φ

₃₃

)

[

Φ

]

T

[

M

][

Φ

]

X

g

Υ

2

sin (Υ t)

Since we are using the modal matrix normalized with the respect to the mass the denominator

[

Φ

]

T

[

M

][

Φ

]

is equal to one for each component of the modal coordinate. Introducing the modal participating factors

Γ

i :

´

q

₍_t₎

+

[

α+βω

1 2

⋯

0 ⋮

⋱

⋮

0 ⋯ α+βω

3 2

]

´

q

₍_t₎

+

[

ω

1 2

⋯ 0

⋮

⋱

⋮

0 ⋯ ω

3 2

]

q

₍_t₎

=

(

Γ

1

Γ

2

Γ

₃

)

X

_g

Υ

2

_{sin (Υ t )}

Usually the form of the equation of a damped motion is:

´

v

₍_t₎

+2 ζ ω

n

´

v

(t)

+

ω

n2

v

(t)

=

0

Simply comparing the system of equations with the common form written above, we get the relations that make us are able to evaluate the damping coefficients

ζ

that comply the Rayleigh approximation:

{

α+βω

₁2

_{=2 ζ}

1

ω

1

α+βω

₂2

_{=2 ζ}

2

ω

2

α+βω

₃2

_{=2 ζ}

3

ω

3

⟹

{

α=2 ζ

1

ω

1

−βω

1 2

ζ

₂

=

α+βω

2 2

2 ω

₂

β=

2 (

ζ

3

ω

3

−ζ

1

ω

1

)

ω

₃2

_−ω

1 2

As specified

ζ

1

=

0.03

and

ζ

3

=

0.03

(damping coefficient at the first natural