Mdof

Prof. A. Meher PrasadProf. A. Meher PrasadDepartment of Civil EngineeringDepartment of Civil Engineering

Indian Institute of Technology MadrasIndian Institute of Technology Madras

email: [email protected]

Static Force-Displacement Relationship

Let x1, x2, x3...... xn be the coordinates used to specify the deflection configuration of the system in a dynamic analysis.

Assume that at points of definition of these coordinates the system is acted upon by a set of concentrated forces, F1, F2, F3.......Fn producing the displacements x1,x2, x3...... xn in the direction of the forces.

The relationship between the forces F and the displacements x is defined as follows,

1 11 1 12 2 1

2 21 1 22 2 2

1 1 2 2

........

........

........

n n

n n

n n n nn n

x d F d F d F

x d F d F d F

x d F d F d F

1 11 1 12 2 1

2 21 1 22 2 2

1 1 2 2

........

........

........

n n

n n

n n n nn n

F k x k x k x

F k x k x k x

F k x k x k x

.

...

.

.

.

.

.

...

.

...

(136)

(137)

or

The quantities ‘d’ are known as flexibility coefficients and the ‘k’ as stiffness coefficients.

To establish the meaning of a particular coefficient,consider the equation that contains that coefficient by making the displacement of the term containing the coefficient equal to unity and all the other displacement equal to zero.

Then interpret the meaning of the term on left. For example,to determine the meaning of d21, we consider the second of the equations in (136) and take F1=1 and F2 = F3 = ….. = Fn = 0. We then conclude that ,

21d = displacement at 2 produced by a force of unit magnitude at 1 and no force anywhere else.

Similarly, to determine the meaning of K21 we consider the second equation in (137) and take x1=1 and

2 3 0nx x x

We conclude that 21k = force at 2 produced by a displacement of unit magnitude at 1 and no displacement anywhere else.

In matrix notation,equations 136 and 137 may be written as

{ } [ ] { }{ } [ ] { }x d F

F k x

(138)

(139)

Where [d] and [k] are square matrices of the stiffness coefficients and the flexibility coefficients, respectively, and {F} and {x} are column matrices of the forces at the nodes and of the associated displacements.

The matrix [d] is known as the flexibility matrix of the system,and [k] is known as the stiffness matrix of the system.

Substituting (138) into (139) we obtain

{F} = [k] [d] {F}

Hence we conclude that

[k] [d] =[I] = identity matrix

It follows that [k] and [d] are the inverse of one another. From Maxwell’s law reciprocity it follows ,

dij=dji

And hence the flexibility matrix is symmetric similarly, from Betti’s work principle it follows that

kij=kji

indicating that [k] is also symmetric

(140)

(141)

(142)

(143)

Evaluation of [d] and [k]

1

d41

d31

d21

d11

2

d42

d32

d22

d12 Second column of flexibility matrix

3

4

1

12

3

4

1

1

k41

k31

k21

k11

First column of stiffness matrix

1

k42

k32

k22

k12

Second column of stiffness matrix

. . . . . . . .

. . . . . . . .

Example #1

For the frame shown,

k1

k2

k3

1

1

k

1

1

k

1

1

k

1

1

k

1 2

1 1

k k

1 2

1 1

k k

1 2 3

1 1 1

k k k

1 2

1 1

k k

1

1

k

the flexibility matrix is

[ ]1 1 1

1 1 2 1 2

1 1 2 1 2 3

1 1 1

1 1 1 1 1

1 1 1 1 1 1

k k k

dk k k k k

k k k k k k

￩￪￪￪

￪￪￪

￪￫

x1

x2

x3

The elements of stiffness matrix are obtained as follows

0

-k2

k1+ k2

-k3

k2 +k3

-k2

-k3

0

k3

[ ]1 2 2

2 2 3 3

3 3

0

0

k k k

k k k k k

k k

-￩￪ - -￪￪ -￫

Accordingly the stiffness matrix is

For systems of this type the stiffness matrix is a banded trapezoidal matrix

Example #2

Consider a prismatic cantilever beam with three nodes spaced at intervals h=L/3

3h=L

Flexibility Matrix:

35

6

h

EI38

6

h

EI

31

3

h

EI

38

3

h

EI314

3

h

EI

35

6

h

EI

314

3

h

EI

31 27

3

h

EIﾴ

38

6

h

EI

[ ]3

2 5 8

5 16 286

8 28 54

hd

EI

￩￪￪￪￫

Stiffness matrix:

Formulation is more involved in this case, because a displacement of a mode induces shears not only in the panels connecting to that mode but also in the more distant panels. Such systems are far coupled (have a strong coupling).

Show that

---

-

71612

164446

124680

13

813L

EI]k[

and verify that3

3

80 46 12 2 5 881 1

[ ][ ] 46 44 16 5 16 2813 162

12 16 7 8 28 54

[ ]

EI Lk d

L EI

I

- - -

- Identity matrix

Strain Energy of system

For a system deflected in a configuration defined by x1, x2,…. xm the strain energy ,V, is equal to the work done by the forces F1, F2….. Fn in going through the displacement ,x. that is

[ ]1 1 2 2

1.....

2 n nV F x F x F x

In matrix notation, { } { } { } { }1 1

2 2T T

V F x x F

(144)

(145)

Substituting Equation (138) into the first of equations (145) we obtain,

{ }1[ ]{ }

2

TV F d F

And substituting Equation (139) into the second of equations (145) we obtain,

{ }1[ ]{ }

2

TV x k x

(146)

(147)

Equation (146)expresses V in terms of the flexibility matrix,whereas equations (147) expresses it in terms of the stiffness matrix.

If Equation (138) were substituted into Equation (145) we would obtain,

{ }

{ } [ ]

1 1[ ]{ } ([ ]{ }) { }

2 21

{ }2

T T

TT

F d F d F F

F d F

hence it follows that [ ] [ ]Td d

or that [d] is symmetric. This is a restatement of Maxwell’s reciprocal relation.

(148)

Proceeding in a similar manner and substitute equation 139 into the two equations 145 we obtain,

[ ] { } { } { } [ ] { }

{ } [ ] { } { } [ ] { }

[ ] [ ]

1 1( )

2 21 1

2 2

TT

T TT

T

k x x x k x

x k x x k x

hence k k

or [k] is symmetric. This result was obtained earlier by the application of Betti’s work principle.

Quadratic Form : Equation147 represents an expression of the form.

2 211 1 12 2 12 1 2 13 1 3 23 2 3....... ....V c x c x c x x c x x c x x

Which is the sum of terms involving the squares of the displacements and the products of pairs of displacements, such an expression is known as a quadratic form and represents a scalar quantity.

(149)

If the quadratic form is positive and non-zero for all possible combinations of the unknowns, excluding the trivial case of x1=x2…xm =0 , the form and associated square matrix are said to be definitely positive or positive definite.

The strain energy of the system being a positive quantity, it and the associated matrix are positive. If the system is fixed–base so that the strain energy can be zero only in the trivial case of x1=x2…=xn =0, the quadratic form and the associated stiffness matrix are positive definite.

A quadratic form is said to be positive if its value is positive or zero for all possible combinations of the unknowns. similarly, a square matrix is said to be positive if the associated quadratic form is positive.

A system which can undergo rigid body motion can have zero strain energy for finite values of X. Hence V and [k] in this case are positive but not positive definite.

{ }x& { } 0x &

The mass matrix is positive definite because the associated quadratic form

represents the kinetic energy of the system, a quantity which is non-zero and positive combinations of , except the trivial case of

{ } [ ] { }T1T= x m x

2& &

Mass matrix

Dynamic Equations of motions for MDOF

Mass matrix

( ){ } ( ){ } ( ){ } ( ){ }I d sf t f t f t P t

Inertia force vector

Damping force vector

Elastic force vector

Externally applied force vector

[ ] { } [ ] { } [ ] { } ( ){ }nx1 nx1nxn nxn nxnnx1

M x + C x + K x = P t&& &

Damping matrix

Stiffnessmatrix

A second order non-homogeneous linear ordinary differential equation

x1x2 x3 xi

fIi

Dynamic Equations of motions for MDOF

{ } nx1x

{ } nx1x&

{ } nx1x&&

Displacement vector

Velocity vector

Acceleration vector

[ ] [ ] [ ]M , C , K Are symmetric square matrices

[ ]M

Lumped Mass matrix (Diagonal matrix)

(Concentrated point masses at each of its corners)

Consistent Mass Matrix

(Got from assembly of element mass matrices which is based on shape function used in stiffness formulation- more computational work)

Undamped free vibration of MDOF system

[ ] { } [ ] { } { }M x + K x = 0&&

{ } { }0f ﾹ

[ ] [ ]2Det K -p M =0

{ } if

Equation of motion :

can this system undergo purely harmonic motion ?

(ie) ( ){ } { } ( )sin pt+x t f an admissible solution ?

[ ] { } [ ] { }2K =p Mf fAnswer : Yes (i.e) A generalised eigen value problem

For non-trivial solution: i e

Results in nth order polynomial equation in p2=0

From fundamental theorem of algebra, this polynomial has roots for p2 corresponding to each such root of polynomial in p2,there is a mode shape vector defining the vibration configuration required such that purely harmonic sinusoidal motion is possible at circular natural frequency, pi

Undamped free vibration of MDOF system contd....

0ip { } [ ] { }TM 0 if ( )i j i jp p i jf f

Properties :

(i) For all i = 1,2,....n (all natural frequencies are +ve real numbers)

(ii) Orthogonality conditions

{ } [ ] { }TK 0i jf f

{ } [ ] [ ] [ ] { }b1T

M M K 0 fori j bf f- -

{ } , 1, 2,.....i i nf

{ }if

More generally,

(iii) All n mode shape vectors are linearly independent

Vectors and spans complete n-dimensional vector space – i.e. any vector can be expressed as linear combination of modal shape vectors

2. Orthogonality of Modes

The modes corresponding to any two distinct characteristic values (natural frequencies) are orthogonal both with respect to the inertia matrix and the stiffness matrix.

Properties of Natural Modes and Frequencies

Let and be two distinct natural frequencies and and be the associated modes.

The first orthogonality condition is,

rp sp { }r

f { }s

f

{ } [ ] { }

{ } [ ] { }

T

r s

T

r s

M 0

K 0

f f

f f

(157)

(158)and the second is,

n

i ip isi=1

m =0f fIf [M] is diagonal, as is usually the case, Eq.157 can be written as

In which mi is the concentrated mass at the ith coordinate and and are respectively the amplitudes of the ith coordinate when the system vibrates in the pth and sth natural mode.

(159)

irfisf

Proof 1: The modes and must satisfy{ }r

f { } sf

[ ] { } [ ] { }[ ] { } [ ] { }

2rr r

2ss s

K =p M

K =p M

f f

f f

(160)

(161)

Next we take the transpose of the two members of Eq.160, obtaining

{ } [ ] { } { } [ ] { }T T2sr s r s

K =p Mf f f f

(162){ } [ ] { } [ ]T T2rr r

K =p Mf f

and premultiply the latter equation by . This yields s{ }f

T 2 Tr s r r s{ } [K]{ } p { } [M]{ }f f f f (163)

Finally we subtract Eq.163 from 162 and obtain

2 2 Ts r r s(p p ){ } [M]{ } 0- f f (164)

Since ps has been considered to be different from pr, Eq.164 requires the validity of Eq.157.

Natural modes corresponding to repeated natural frequencies are orthogonal to all other modes, but are not necessarily orthogonal to each other.

premultiply the equation 161 by Tr{ }f

However, by appropriate combination of the non-orthogonal modes it is always possible to construct orthogonal points. Therefore, all modes, including those corresponding to repeated characteristic roots, may be considered to be orthogonal in the sense of Eq.157. Give example.

Having established that Eq.157 is valid, Eq.158 may be proved by reference to Eq.162 and Eq.163, Since the right-hand member of either of these equalities is zero, the left-hand member must also be zero. It is assumed that that the frequency pr or ps is not equal to zero.

Proof 2:

Based on structural principles, inertia forces acting on the system while the system is vibrating in the mode:

r{ }f

2r rp [M]{ }f (a)

The corresponding forces for motion in the mode are, (b)

Now from Betti’s work principle, the work done by the forces (a) acting through (the displacements produced by (b)) is equal to the work done by the forces (b) acting through , i.e. the displacements produced by (a). The first work quantity is

and the second work quantity is

hence,

which is the same as of Eq 164.

[ ] { }2s sp M f

[ ] { }( ) { } { } [ ] { } { } [ ] { }2 2 2T TT T

r r sr s r s r sp M p M p Mf f f f f f

{ } [ ] { }2 T

s r sp Mf f

( ) { } [ ] { }2 2 0T

s r r sp p Mf f-

2. Independence of modes

The natural modes are linearly independent, in the sense that some mode cannot be expressed as a linear combination of all the other modes. Expressed differently, the equation

is satisfied if and only if all the coefficients are zero.

Proof

Pre-multiplying Eq. 165 by we obtain

hence

{ } { } { } { }1 21 2............. .............. 0r nr n

C C C Cf f f f

{ } [ ] { } { } [ ] { } { } [ ] { } { } [ ] { }1 21 1 2 2............. .............. 0

0

T T T T

r nr r n rC M C M C M C Mf f f f f f f f

[ [ [

0 0

{ } [ ] { } 0T

r r rC Mf f

(165)

(164 )

But since [m] is positive definite, the triple product cannot be zero and hence Cr must be zero. By a similar reasoning it is concluded that all the other coefficients must also vanish.

An important consequence of this property of independence is that any n-dimensional vector can be expressed as a linear combination of n characteristics vectors.

3. Reality and Positiveness of Natural Frequency

The natural frequency are real and positive quantities.

Proof of Reality

We start by assuming that the frequencies are not real and show that this assumption leads to a contradictions

Assume that

{ } [ ] { }T

r rMf f

2pp i b (165 )

Since both [k] and [m] are real, it follows that the associated natural modes must also be complex.

i.e.

Further, the complex conjugates of pr and {f}r must also represent a solution.

Letting,

{ } { } { }r r rX if Y

{ } { } { } { }

( ) { } [ ] { }

{ } { }( ) [ ] [ ] { }( )

2 2

2 2

0

2 0

s r

s p rr

T

s r r s

T

p p rp

p p i

i

p p

i X i X

M

M i

b

f f

f f

b

-

C - Y

-

Y - Y

and

and making use of Eq. 164 we obtain

(166 )

(167)

(168 )

(169)

Now it can be verified that the triple product of the matrices in this equation is a real quantity, and since [m] is positive definite, the product is a non-zero positive number. It follows that b must be zero. Hence pr

cannot be complex as assumed in Eqn. 165 but must be real.

Proof of Positiveness:

Pre-multiplying [ ] { } [ ] { }{ }

{ } [ ] { } { } [ ] { }

2

2

,

rr r

T

r

T T

rr r r rK

MK

M

P

P

f f

f

f f f f

by we obtain

Therefore, Pr is either zero or positive. However, if system cannot undergo rigid body motion, the left-handed member of the equation is non-zero and positive. Hence Pr is non-zero and positive is used such cases.

To summarize, provided the system is stable and cannot undergo rigid body motion, the natural frequencies are non-zero positive quantities

(170)

Solution for General Forced Vibration of MDOF systems in Time Domain

a)Direct Integration method b)Modal Superposition method

Modal superposition method:

Both mass matrix,[m] and Stiffness method, [k] can be simultaneously made into diagonal matrices by proper choice of coordinates(known as modal coordinates)

• Transformation of coordinates results in a set of uncoupled SDOF

equations in terms of modal coordinates

• Useful for many practical problems where the response can be

approximated very well by using few eigen modes of the problem

x x x

x x x

x x x

￦

￨

x x x

x x x

x x x

￦

￨

0 0

0 0

0 0

x

x

x

￦

￨

0 0

0 0

0 0

x

x

x

￦

￨

[ ]M [ ]K [ ]K[ ]M

a) Equations of motion:

Modal Superposition Method…

b) Find mode shape and natural frequencies by solving Eigen Value Problem,

[ ] { } [ ] { }2K =p Mf f

{ }i ip , fobtain for i=1,2,…..n

where, = ith mode shape vector { }if

c) Express { } { }n

i ii=1

x(t) = q (t)f￥ (2)

[ ] { } [ ] { } [ ] { } ( ){ }M x + C x + K x = P t&& & (1)

Substitute eqn (2) in (1) and then premultiply by and using Orthogonality Properties, one obtains

{ }jT

f

for j=1,2,……..n(3)

{ } [ ] { } { } [ ] { } { } [ ] { } { } { }nT T T

j i j j i j j i j ji=1

M q + C q + q = P(t)TKf f f f f f f && &

Let {fi}T [C] {fj} = 0 for i j then [C] is said to be classically damped matrix

Special case satisfying orthogonality ,

[C] = [M] + b [K] = Rayleigh Damping

“FOSS” condition [C] [M]-1 [K] = [K] [M]-1[C]

Then division by {fi}T [M] {fj} of Equation (3) leads to ,

2 2

22

( ) 2 ( ) ( ) ( ) ( )

{ } { ( )}

{ } [ ]{ }

j j j j j j j j j

Tj

j Tj j j

q t p q t p q t p c t f t

P tp

p M

z

ff f

￩￪￪￫

&& &

For j=1,2,..n

Where zj = jth modal damping ratio, cj = max2

{ } { ( )}

{ } [ ]{ }

Tj

Tj j j

P t

p M

ff f

Participation factor

(4)

(d) Express initial conditions

1 1

{ } { } { } { }n n

o i i o i ii i

andx a x bf f

&

By pre multiplying above expression with {fj}T [m] and using orthogonality of modes,

{ } { }{ }

{ } [ ]{ }

Tj o

j Tj j

M xa

M

ff f

{ } { }{ }

{ } [ ]{ }

Tj o

j Tj j

M xb

M

ff f

&

aj ,bj - participation factors

(e) Solve qi(t) of the uncoupled SDOF systems

( )

20

( ) cos sin ( )

( ) ( ) sin ( )1

j j

j j

p t j j j jj j dj dj j j

dj

tp tj

j j dj

j

b p aq t e a p t p t c IAF

p

pIAF f ewher p te d

z

z

z

z

-

- -

--

(f) Express { }1

( ) { } ( )n

j jj

x t q tf

In general take first ‘r’ modes only

Elastic forces:

Had the load been static, {fs(t) = {P(t)} and (IAF)j = 1

i.e

( ){ } { } ( )1

r

j jj

x t q tf

( ){ } [ ] ( ){ } [ ]{ } ( )tqKtxKtf jj

r

js f

1

[ ] { } ( )2

1

r

j j jj

M p q tf

￩￫￥

( ){ } ( ){ } [ ] { } { } ( ){ }{ } [ ] { }1

Tn

j j

s Tj

j j

M P tf t P t

M

f f

f f

￥

Implementation issues

Considering static force correction for the neglected modes r+1, ….n

( ){ } { } [ ] { } ( )[ ] { } { } { }

{ } [ ] { }

( ) { } ( ) [ ] { } { }{ } [ ] { }

( ){ }

2

1 1

1

21 1

Tr r

j j

s j j j Tj j

j j

Tr r

j j

j j Tj j

j j j

M Pf t P M p q t

M

x t q t K P tp M

f ff

f f

f ff

f f

-

-

￩￪ -￩￫￪￫

￥￥

￥￥

( ){ } ( ){ } [ ] { }1

2r

s j j j j jj

f t P t M q p qf z

￩ - ￫￥ && &

From Equation (4)

( ) { } ( ){ }{ } [ ] { }

( ) { } ( ){ }{ } [ ] { }

( ){ } [ ] ( ){ } { }

2

2 2

1

21

2

21 1

21

T

j

j j j j j j T

j j

T

j j jj jT

j j jj j

rj j

j jj j j

P tq p q p q t

M

P t qq t q

p p pM

qx t K P t q

p p

fz

f f

f z

f f

zf-

- -

￩ - ￪

￪￫￥

&& &

&&&

&&&

Mode Acceleration Method

m1

m2

m3

k1

k2

k3

x1

x2

x3P3(t)

P2(t)

P1(t) 1 1m x&&

2 2m x&&

3 3m x&&

k3(x3-x2)

k2(x2-x1)

k1x1

• Equations of motion are obtained by application of procedure described earlier which may be summarized as follows:

1. Consider system in a deflected configuration and identify all forces acting on in, including the d’Alembert inertia forces

2. Draw force body diagram for the individual masses and write the equations expressing the equilibrium of the forces

Response of a MDOF System - An Example

• Application of this procedure leads to the following,

For Mass 1

For Mass 2

For Mass 3

( ) ( )1 1 1 1 2 2 1 1m x +k x -k x -x =P t&&

( ) ( ) ( )2 2 2 2 1 3 3 2 2m x +k x -x -k x =P tx-&&

( ) ( )3 3 3 3 2 3m x +k x -x =P t&&

( )( )( )

1 1 1 2 2 1 1

2 2 2 2 3 3 2 2

3 3 3 3 3 3

0

0

m x k k k x P t

m x k k k k x P t

m x k k x P t

- ￬￩￬￩￬￪￪ - - ￭￭￭� � �￪￪￪￪ -￫￮￫￮￮

&&

&&

&&

Or

Response of a MDOF System

(a)

(b)

(c)

(d)

Consider special case in which, k1=k2=k3 and m1=m2=m and m3= ½ m

1 1

2 2

3 3

1 x 2 -1 0 x

m 1 x +k -1 2 -1 x =0

0.5 x 0 -1 1 x

￩￬￩￬￪￪

￭￭� �￪￪￪￪￫￮￫￮

&&

&&

&&

( )1 1

2 2

3 3

x X

x = X sin pt+α

x X

￬￬￭￭� �￮￮

1 12

2 2

3 3

2 -1 0 X 1 X

k -1 2 -1 X = mp 1 X

0 -1 1 X 0.5 X

￩￬￩￬￪￪

￭￭� �￪￪￪￪￫￮￫￮

(132)

(133)

Natural Frequencies and Modes

2mpλ=

kLet

we get

[ ] [ ] [ ] { } { }K -λ M X = 0

( )( )

( )

1

2

3

2-λ -1 0 X

-1 2-λ -1 X 0

0 -1 1-0.5λ X

￩￬￪￭�￪￪￮￫

For a non-trivial solution

( )( )

( )

2-λ -1 0

-1 2-λ -1 =0

0 -1 1-0.5λ

Expanding, we get the frequency equation

( ) ( )32-λ -3 2-λ =0

(134)

( ) ( )22-λ λ -4λ+1 =0

which, upon factoring the term , may be rewritten as( )2-λ

The roots are

21

22

23

k=0.26795

mk

=2m

k=3.73205

m

p

p

p

2

1

3

1

=2.732

=3.732

p

p

p

p

Hence natural frequencies are

λ=2, and

λ=2 3,

. . 0.267 & 3.730i e

ﾱ

(135)

2 1(2 )X Xl -

( )3 1 22X X Xl - -

From the first of these equations,

With the roots (natural frequencies) evaluated, the natural modes are determined from above as follows:

Normalize by taking X1=1, then

1For, λ =2- 3

( ) ( )21

31

X 2 2 3 3

X 1 2 2 3 3 1 3 2

-

- - -

2For, λ =2

22

32

X 2 2 0

X 1 0 1

- - - -

3For, λ =2+ 3

( ) ( )23

33

X 2 2 3 3

X 1 3 3 1 3 2

- - -

- - - -

21 0.2680

kp

m 2

2 2.0k

pm

23 3.7320

kp

m

{ } 1

1

3

2

f

￬￭�

￮

{ } 2

1

0

1

f￬

￭�-￮

{ } 3

1

3

2

f

￬ -￭�

￮

Summary:

Free Vibration

Write expression for the motion of the system due to an initial displacement

{ } ( )0 1 0

1

1

1

x x

￬￭�￮ ( )1 0

x

( )( )

( ) ( ) ( )

( )( )

( ) ( ) ( )

( )( )

( ) ( ) ( )

1 01 1 10 0

1 02 1 0 1 0

1 03 1 0 1 0

1 3 0.5 2 2 30.6220

1 3 0.5 4 6

1 0 0.5 1 0.50.3333

1 0 0.5 1 1.5

1 3 0.5 2 2 30.0447

61 3 0.5 4

m xa x x

m

m xa x x

m

m xa x x

m

-

- -

{ } { }3

1

cosj jjj

x a p tf

￥

Without any initial velocity

where,

Free Vibration

Therefore,

{ } ( ) { } { } { }( )

( ) ( ) ( )

1 1 2 30 1 2 3

1

2 1 1 1 2 1 30 0 0

3

1

2

3

0.62220 cos 0.3333 cos 0.0447 cos

0.6220 0.3333 0.0447

1.0774 cos 0 cos 0.0774 cos

1.2440 0.3333 0.0893

0

x x p t p t p t

x

x x p t

o

x

r

and

p t x p t

x

u

u

u

f f f

- -

( ) ( ) ( )1 1 1 2 1 30 0 0

.6220 0.3333 0.0447

0.4553 cos 0.3333 cos 0.1220 cos

0.1667 0.3333 0.1667

x p t x p t x p t

- - -

Free Vibration

In this case { } { }3

1

sinj jjj

x b p tf

￥

Consider next, the displacement induced by an excitation of the second mass, that is an initial velocity without an initial displacement

{ } ( )0 2 0

0

1

0

v v

￬￭�￮

( ) ( ) ( )

( ) ( ) ( )

2 20 01

1 1

22

2 20 03

3 3

0 3 1 01 3. .

6 6

1 0 0

1.5

0 3 1 01 3. .

6 6

m v vb

P m P

bP

m v vb

P m P

￩￫

￩ - ￫ -

Where,

1

2 0 2 02 1 3

1 33

0.2887 0.2887( ) ( )

0.5 sin 0.5 sin

0.5774 0.5774

xv v

x p t p tP P

x

-￬￬￬￭￭￭� � �

-￮￮￮

Forced Vibration

Consider first the effect of a force P2(t)=P2f(t) applied on the second mass

{ } { }3

1

( . . )j jjj

x c I A Ff

￥

Noting that:

2. P is given as:

4. Making use of the values of obtained earlier we obtain,

{ } 2

0

1

0

P P

￬￭�￮

2 2 21 2 3, andp p p

( )

21 2

2

2 23

1 0 3 01.07735

60.26795

0

310.07735

63.73205

Pc P

k m km

c

P Pc

k m km

- -

Forced Vibration

1

22 1 3

3

1

22 1 3

3

1 1

1.07735 3 ( . . ) 0.07735 3 ( . . )

2 2

1.077 0.077

1.866 ( . . ) 0.134 ( . . )

2.155 0.155

xP

x I A F I A Fk

x

xP

x I A F I A Fk

x

- -

- -

Therefore

or

an

1

22 1 3

3

1.077 0.077

0.789 ( . . ) 0.211 ( . . )

0.289 0.289

uP

u I A F I A Fk

u

- -

d

Maximum values of (I.A.F)j are obtained from the response spectrum applicable to those particular forcing functions considered

• Discuss relative importance of higher models. Note that it depends on response gravity under consideration and on the relative values of (A.F)3 and (A.F.)1. The latter depend, in turn on the value of f1t1 and on the ratio of f3/f1. Finally, note that the participation factors are functions of the spatial distribution of the forcing functions.

• If the force were applied at the first rather than the second floor level, the expression for the response would be as follows:

1 11

1 12

1 13

10.62220

60.26795

10.3333

1.52

10.4447

63.73205

P Pc

k m kmP P

ck km

P Pc

k m km

1

12 1 2 3

3

1

12 1

3

0.62220 0.3333 0.0447

1.0774 ( . . ) 0 ( . . ) 0.0774 ( . . )

1.2440 0.333 0.0893

0.62220 0.

0.4553 ( . . )

0.1667

xP

x I A F I A F I A Fk

x

uP

u I A Fk

u

and

- -

2 3

3333 0.0447

0.333 ( . . ) 0.122 ( . . )

0.333 0.1667

I A F I A F

- - -

Finally if the system were subjected to a set of forces

( ){ } ( )1

1

0.5

P t P f t

Hence,

( )

( )

( )

1

2

3

11 1 1 3 2

1 22.3213

60.26795

11 1 0 1

1 20.1667

1.52

11 1 1 3 2

1 20.0120

63.73205

PP

ck m km

PP

ck m km

PP

ck m km

￦ﾴﾴ￨

￦ -ﾴﾴ￨

￦ - ﾴﾴ￨

The participation factors would be,

1

2 1 2 3

3

1

2

3

2.3213 0.1667 0.0120

4.0208 ( . . ) 0 ( . . ) 0.0208 ( . . )

4.6426 0.1667 0.0240

2.3213

1.6995

0.6

xP

x I A F I A F I A Fk

x

uP

uk

u

- -

with the response given by,

and

1 2 3

0.1667 0.0120

( . . ) 0.1667 ( . . ) 0.0328 ( . . )

218 0.1167 0.0448

I A F I A F I A F

- - -

Note: The contribution of the higher modes is significantly less in this phase than when the system is excited by a single force on the first floor

Analysis of n-DF Damped Systems

In the following analysis of the response of damped systems, damping will be assumed to be viscous. The damping forces, {Fd}, are then a linear function of the velocities of the coordinates, and can be expressed in the form,

d{F } [C]{x} & (204)

In which [c] is the damping matrix of the system. The element cij

corresponding to the ith row and the jth column represents the force at node i induced by a unit velocity of mode j.

The equations of motion of the system can then be written as,

[M]{x} [C]{x} [K]{x} P(t)

[D][M]{x} [D][C]{x} {x} [D]{P(t)}

&& &

&& &

or, in terms of the flexibility of the matrix, [d], as

(206)

(205)

The form of the matrix [C] is obviously a function of the distribution of the damping resistance throughout the structure.

However, unlike the elements of [M], [K] and [D] which can be determined from the basic properties of the structural elements, those of [C] can not be so evaluated, as the relative magnitudes and distributions of the various factors which contribute to the damping of the structure are generally not known.

Factors contributing to the overall damping of the system include,

• the ‘resistance of the fluid’ (air or water) in which the structure is oscillating,

• energy dissipation due to the straining of the elements composing the structure,

• losses at the faying surfaces of the cracked sections,

• friction at joints and supports, and energy absorption in components or elements that may be superimposed on the main structure.

The difficulties involved may be illustrated by reference to the 3DF of the shear-beam type considered earlier.

It is clear that the damping resistance of the system in this, as in any other case can be modeled by a series of dashpots, but how these should be attached to the various floors is not clear.

m2

m1

m3

c1

c2

c3

k1

k2

k3

c1

c2

c3

c1

c2

c3

(a)

External or absolute damping

(b)

Internal or relative damping

(c)

Combination of External and Internal damping

For example, if we consider a set of three dashpots and attach one end of each to a separate floor and the other to the ground, we obtain the arrangement shown in Fig.(a).

Alternatively, if we attach the ends to consecutive floors, we obtain the arrangement shown in Fig.(b)

For case(a), the damping forces would be of the form

1 11

2 2 2

33 3

F xc

F c x

cF x

￬￬￩￪￭￭� �￪￪￫￮￮

&

&

&

(207)

and the damping matrix would be of the same form as the mass matrix. Furthermore, if the ratio (cj / mj) for all the elements were the same, the matrix [c] would be proportional to the mass matrix [m].

For case(b), the damping forces would be of the form,

1 11 2 2

2 2 2 3 3 2

3 33 3

F xc c c

F c c c c x

c cF x

-￬￬￩￪ - -￭￭� �￪￪ -￫￮￮

&

&

&

(208)

and the damping matrix would be of the same form as the stiffness matrix, [k]. Furthermore, if the ratio cj/mj for all the dashpots and springs were the same, the matrix [c] would be proportional to the mass matrix [k].

In general, the form of [c] may be different from either [m] or [k]. An arrangement of dashpots leading to such a matrix is shown in Fig.(c).

If the damping in a structure is due exclusively to the resistance of the fluid in which it is oscillating, it would be reasonable to consider [c] to be of the same form as [m].

Similarly, if damping is exclusively due to energy dissipation due to the straining of the members, [c] can be considered to be of the same form as [k]. A mass-proportional or stiffness-proportional [c] are special cases of the class referred to above.

Proof:

As before, the solution is expressed in the form of Eq.189, i.e., as a linear combination of the natural modes of vibration of the associated undamped systems. Furthermore, we limit ourselves to forces of the form of Eq.179.

Substituting Eqs.189 and 179 into Eq.205, and following the steps used in the analysis of elastic systems, we obtain the following equations, which is the counterpart of Eq.192.

nT T T Tr r r r j j r r r r

j 1

{ } [M]{ } q (t) { } [C] { } q (t) { } [K]{ } q (t) { } {P}f (t)

￦f f f f f f f

￨￥&& & (210)

We shall now investigate if the modal superposition method which we have used for the analysis of undamped systems can also be used for damped systems. It will be shown that this is indeed possible provided the damping matrix, [c], is of a form that satisfies the orthogonality relationship ,

Tr s{ } [C]{ } 0 (for r s)f f ﾹ (209)

Now, if it is assumed that Eq.209 is valid, namely, that pairs of natural modes are also orthogonal with respect to the damping matrix, then all but the rth term of the series in the second term of Eq.210 vanish, and Eq.210 simplifies to,

T T T Tr r r r r r r r r r{ } [M]{ } q (t) { } [C]{ } q (t) { } [K]{ } q (t) { } {P}f (t)f f f f f f f&& & (211)

or r r r r r r rM q (t) C q (t) K q (t) P f (t) %% % %&& & (212)

rC%In which , and are defined by Eqs.196 to 199, respectively, and

the effective damping for the rth mode, defined as ,

rM% rk% rP%

Tr r rC { } [C]{ } f f% (213)

Eq.211 , or its equivalent Eq.212, permits the values of qr(t) to be evaluated independently of each other. In other words, the use of qr(t) uncouples the governing equations of motion, as for undamped systems.

If Eq.211 is now divided by the coefficient of the acceleration term, the resulting coefficient of the displacement term is equal to the square of the rth circular natural frequency of the system, pp

2 (see Eq.170), and the coefficient of the time function on the right hand member of the equation becomes pp

2 cr, in which cr is the participation factor defined by Eq.191.

If we also let,Tr r r

r r Tr r r

{ } [C]{ } C2 p

{ } [M]{ } M

f fz

f f

%

%(214)

we obtain 2 2r r r r r r r rq (t) 2 p q (t) p q (t) c p f (t) z && & (215)

In which zr represents the percent of the critical coefficient of damping for motion in the rth natural mode. Finally, if we compare this equation with the equation of motion of a SDF damped system,

2 2st 0x(t) 2 px p x (x ) p f (t) z && & (216)

And recall that the solution of this equation can be expressed symbolically as

st 0x(t) (x ) (I.A.F) (217)

in which I.A.F. is the Instantaneous Amplification Factor for the particular damped system and excitation considered. We conclude that qr(t) can also be stated as

r r rq (t) c (I.A.F) (218)

The time function (I.A.F.)r is the I.A.F. for a SDF system which has a natural frequency, pr, a damping factor, zr, and is subjected to a forcing function, the time variation of which is defined by f(t).

Substituting Eq.218 into Eq.189, we obtain again Eq.188. It should be emphasized, however that the (I.A.F.)r in this equation refer to damped rather than undamped systems. The other quantities are the same as for the associated undamped systems.

Rayleigh Damping

If [c] is proportional to either [M] or [K], or is a linear combination of [m] and [k] as follows

[c] = [M] + b[K]

In which and b are constants, it is clear that [c] will satisfy the orthogonality relation defined by Eq.209. Indeed, substituting Eq.219 into Eq.209 we obtain,

(219)

T Tr s r s{ } [M]{ } { } [K]{ } f f b f f which is zero because

each of the triple products is zero.

Damping of the form of Equation 21a is known as Rayleigh damping ,after Lord Rayleigh who was the first to demonstrate the applicability of the modal superposition method of analysis in this case.

Substitution of Equation 21a we find that,

22

1 1

2 2

p p p

p pp

wher

p p

pp

e

z b

z b

(220)

Note:

If the damping is mass proportional ,b=0,and the damping factor for a mode is inversely proportional to the frequency of the mode,

If the damping matrix is stiffness proportional ,the damping factor ,zp is proportional to the natural frequency of the mode under consideration.

Implications of Rayleigh Damping

for [c] = [M]

for [c] = b [K]

zp

pp

For a given excitation ,it follows that the higher modes will be damped significant more in the second case than the first (see the sketch).

For example ,if z1 for a system is ,say,0.03 and the first three natural frequencies are in the ratio of 1:3:5 ,the assumption of a mass proportional damping will lead to effective damping factors for the second and third modes of 0.01 and 0.03 (1/5) = 0.006, respectively.

By contrast, the assumption of stiffness proportional damping will lead to effective damping factors of 0.09 and 0.15 for the second and third modes, respectively.

Unless the contribution of these two modes to the overall response of the system is significant, the response values computed in the two cases will be substantially different. It is important, therefore, that great care be exercised in the selection of the form of the damping matrix.

For a two degree of freedom system with arbitrary modal damping factors,z1

and z2 ,it is always possible to determine the damping matrix of the Rayleigh type. Indeed, satisfying Eqn. 220 for p=1 and p=2,we obtain,

1 11

2 22

1

2 2

1

2 2

pp

pp

z b

z b

(221)

which, upon solving simultaneously, yields

11 2

21 2

1

2

12 1

22

2 1

2

2

1

2

1

pp

ppp

pp

p pp

z z

z zb

-

-

-

-

(222)

In the special case in which z2 = z1 (p1 / p2) , b=0 ,indicating that

[c] = [M] , and = 2p1 z1

similarly ,when z2 = z1 (p2 / p1) , = 0 (i.e [c] = b [K]) , and b=2z1/p1.

Fore systems with more than two degree of freedom for which the modal damping factors are arbitrary, it is generally not possible to obtain a matrix [c] of the Rayleigh type.

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