It corresponds to the longest natural period, which is called fundamental natural period (T1) of vibration

CHAPTER 10

FREE VIBRATION OF MDOF SYSTEMS System without Damping

The equation of motion of a two-DOF system in free vibration (no external force) is

mu + ku = 0

The displacements of masses are the solution with an initial condition

( )

⁰

=

u u and ^{u u =}

( )

⁰

If a two-DOF system is let to vibrate with an arbitrary initial displacement, the displacement of each mass will be as shown. They are not a simple harmonic as free vibration of SDF system and deflected shape changes with time as the ratio u u₁/ ₂ varies with time.

12 - 2

An undamped structure would undergo simple harmonic motion without change of deflected shape if free vibration is initiated by appropriate displacement shapes. It will maintain that initial shape and both floors will reach their maximum displacement at the same time.

For the first shape above, both floors move in the same direction.

For the second shape, the two floors move in the opposite directions.

The location where displacement remains zero is called “node.” These shapes are called “natural modes of vibration.” The number of modes increases as number of degrees of freedom increases.

For each mode, all floors move like a simple harmonic motion with the same frequency called “natural circular frequency of vibration” (ω_n), or “natural cyclic frequency of vibration” ( f_n) and the time to complete one cycle is called “natural period of vibration.”

2

n

n

T π

= ω 1

n n

f =T

The lowest natural frequency is known as the fundamental natural frequency, denoted by ω₁. It corresponds to the longest natural period, which is called fundamental natural period (T₁) of vibration.

12 - 4

Natural Vibration Frequencies and Modes

Displacement vector ^u

( )

^t can be written as a product of mode shape vector φ_n and modal coordinate q tn

( )

, which is a scalar quantity.

( )

t =q tn

( )

φn

u

Modal coordinate varies with time whereas the mode shape vector does not. In free vibration, the modal coordinate varies like a simple harmonic with frequency equal to the natural frequency.

( )

^{cos sin}

n n n n n

q t = A ω t B+ ω t Therefore, u

( )

t =φn

[

An^cosωnt B+ n^sinωnt

]

Substitute this in the equation of motion, we get

( )

2

n n n q tn

ω φ φ

− +  =

 m k  0

This equation can be satisfied by two ways:

(1) q tn

( )

=⁰ but this is not very useful so it is called trivial solution (2) −ω φ_n²m _n +kφ_n =0 or kφ_n =ω φ_n²m _n

This is called a matrix eigenvalue problem. The matrices k and m are known and we want to determine ω_n² and φ_n

2

n n n

φ ω φ− =

k m 0 k−ω_n²mφ_n =0

Again, this equation can be satisfied by either φ_n =0 (trivial) or detk−ω_n2m =0

This determinant is a polynomial of order N in terms of ω_n². This equation is called “characteristic equation or frequency equation.” It has N real and positive roots for ω_n² because matrices k and mare symmetric and positive definite.

The N roots for ω_n² are known as eigenvalues, or characteristic values, and they correspond to the natural frequencies of N modes of vibration.

For each value of ω_n², φ_n can be solved from the equation

2

n n

ω φ

 −  =

k m 0, but they are not unique because many vectors obtained from scaling a solution of φ_n will also satisfy the equation. The shape of this vector φ_n is called natural mode of vibration, or mode shape. It is also known as eigenvector corresponding to the eigenvalue

2

ωn.

12 - 6

Mode and Spectral Matrices

We can put many mode shapes together in a matrix to have them in compact form.

11 12 1

21 22 2

1 2

...

...

...

N N jn

N N NN

φ φ φ

φ φ φ

φ

φ φ φ

 

 

 

=    = 

 

 

Φ # # % #

The matrix Φ is called modal matrix for the eigenvalue problem.

The N eigenvalues ω_n² can be assembled into a diagonal matrix Ω², which is known as the spectral matrix of the eigenvalue problem.

2 1

2

2 2

2 N

ω ω

ω

 

 

 

= 

 

 

 

Ω %

From the eigenvalue problem in vector form

2

n n n

φ =ω φ

k m

It is possible to write them in matrix form

= 2

kΦ mΦΩ

Orthogonality of Modes

The natural modes corresponding to different natural frequencies can be shown to satisfy the following orthogonality conditions.

When ω_n ≠ω_r,

T 0

n r

φ φk = φ_n^Tmφ_r =0 It can be proved as follows: From

2

n n n

φ =ω φ

k m

Premultiply by φ_r^T

2

T T

r n n r n

φ φk =ω φ mφ

Above is scalar so we can transpose the right hand side

2

T T

n r n n r

φ φk =ω φ mφ Suppose we start from the r^th mode,

2

r r r

φ =ω φ

k m

premultiply by φ_n^T, we get

2

T T

n r r n r

φ φk =ω φ mφ Subtract the two equations, we get

(

^{ωn2 −ω φr2}

)

^{nTmφr =0}

When ω_n² ≠ω_r², φ_n^Tmφ_r =0 and also φ φ_n^Tk _r =0.

12 - 8

The orthogonality condition implies that the matrices

≡ T

K Φ kΦ M ≡Φ^TmΦ are diagonal where the diagonal elements are

T

n n n

K =φ φk M_n =φ_n^Tmφ_n

Because k and m are positive definite so K_n and M_n are positive are related by

2

n n n

K =ω M

by ^{Kn =φ φnTk n =φ ω φnT}

(

^{n2m n}

)

^{=ω φn2}

(

^nTmφn

)

^=ωn2Mn

Interpretation of Modal Orthogonality

Consider a structure vibrating in nth node with displacements

( ) ( )

n t =q tn φn

u

The corresponding accelerations are un

( )

t =q tn

( )

φn and inertia forces are

( )

fI n = −mun

( )

t = −mφn nq t

( )

Next consider displacements of the structures in its r^th mode

( ) ( )

r t =q tr φr

u The work done by inertia force is

( )

fI ^Tn ur = −

(

φn^Tmφr

)

^{q t q tn}

( ) ( )

^r

which is zero because of modal orthogonality.

Another implication is that the work done by equivalent static forces associated with displacement in n^th mode doing through the r^th mode displacement is zero.

( )

fS n =kun

( )

t =kφn nq t

( )

The work done is

( )

fS ^Tn ur =

(

φ φn^Tk r

)

^{q t q tn}

( ) ( )

^r

which is zero because of modal orthogonality.

12 - 10

Normalization of Modes

The eigenvector multiplied by a scalar scaling factor is also the eigenvector corresponding to the same eigenvalue.

Therefore, only the shape of mode is essential. A mode can be scaled without changing the shape. This process is called normalization.

Sometimes the mode shape vector is scaled such that the largest element equal to unity.

For a building, mode is often scaled such that the DOF at the roof is unity.

Computer program usually normalizes the modes such that

T 1

n n n

M =φ mφ = or M =Φ^TmΦ=I

where I is an identity matrix (diagonal with 1 along the main diagonal).

These modes are called a mass orthonormal set. This results in

2 2

T

n n n n n n

K =φ φk =ω M =ω or K =Φ^TkΦ Ω= ²

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12 - 22

Modal Expansion of Displacements

Any set of N independent vectors can be used as a basis for representing any other vectors of order N.

1 N

r r r

φ q

=

=

∑

=

u Φq

where q_r are scalar multipliers called modal coordinates or normal coordinates.

( )

1

T N T

n n r r

r

φ φ φ q

=

=

∑

mu m

Because of modal orthogonality, only the term r n= is nonzero.

( )

T T

n n n qn

φ mu= φ mφ Therefore,

T T

n n

n T

n n n

q M

φ φ

φ φ

= mu = mu

m

This modal expansion is employed to determine the free vibration response and response to excitation of MDF systems.

12 - 24

Free Vibration of Undamped MDF Systems

Given ^u

( )

^{0 and u}

( )

⁰ determine ^u

( )

^t

Use modal expansion of initial conditions

( ) ( )

1

0 ^N _{n n} 0

n

φ q

=

=

∑

u

( ) ( )

1

0 ^N _{n n} 0

n

φ q

=

=

∑

u

Therefore,

( ) ( )

⁰

0

T n n

n

q M

= φ mu

( ) ( )

⁰

0

T n n

n

q M

=φ mu

Then, the free vibration response is

( ) ( )

1 N

n n n

t φ q t

=

=

∑

u

( ) ( ) ( )

1

0 cos 0 sin

N n

n n n n

n n

t φ q ω t q ω t

= ω

 

=  + 

 

∑

u

12 - 26

Free Vibration of Damped MDF Systems

If the damping matrix c of the system is such that C=Φ^TcΦ is diagonal, we said that the system has classical damping.

The equations of motion can be uncoupled and solved by modal analysis.

If C=Φ^TcΦ is not a diagonal matrix, the system is nonclassically damped and must be solved by numerical method and its eigenvalue will be complex numbers.

For classically damped system, the uncoupled equation is

n n n n n n 0

M q +C q +K q = where

T

n n n

C =φ φc The modal damping ratio is

2

n n

n n

C ζ M

= ω

The solution of q tn

( )

will be free vibration of a damped SDF system.

2 2 0

n n n n n n

q + ζ ω q +ω q =

( ) ( ) ( )

⁰

( )

⁰

0 cos sin

n nt n n n n

n n nD nD

nD

q q

q t e ^{ζ ω} q t ζ ω t

ω ω

ω

−  + 

=  + 

 

where

1 2

nD n n

ω =ω −ζ

12 - 28