Mdof

Dynamics Course

1. Prof. A. Meher PrasadProf. A. Meher Prasad
Department of Civil EngineeringDepartment of Civil Engineering
Indian Institute of Technology MadrasIndian Institute of Technology Madras
email: [email protected]

2. Static Force-Displacement Relationship
Let x
1
, x
2
, x
3
...... x
n
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, F
1
, F
2
, F
3
.......F
n
producing the
displacements x
1
,x
2
, x
3
...... x
n
in the direction of the forces.
The relationship between the forces F and the displacements x is defined as
follows,
1 111 122 1
2 211 222 2
11 22
........
........
........
nn
nn
n n n nnn
xdFdF dF
xdFdF dF
xdFdF dF
   
   
   
1 111 122 1
2 211 222 2
11 22
........
........
........
nn
nn
n n n nnn
Fkxkx kx
Fkxkx kx
Fkxkx kx
   
   
   
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
(136)
(137)
or

3. 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 d
21, we consider the second of the equations in (136) and take F
1=1
and F
2
= F
3
= ….. = F
n
= 0. We then conclude that ,
21
d= displacement at 2 produced by a force of unit
magnitude at 1 and no force anywhere else.

4. Similarly, to determine the meaning of K
21
we consider the second equation in
(137) and take x
1
=1 and
2 3
0
n
xxx
We conclude that
21
k= 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
{}[]{}
{}[]{}
xdF
Fkx


(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.

5. 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 ,
d
ij=
d
ji
And hence the flexibility matrix is symmetric similarly, from Betti’s work principle
it follows that
k
ij=
k
ji
indicating that [k] is also symmetric
(140)
(141)
(142)
(143)

6. Evaluation of [d] and [k]
1
d
41
d
31
d
21
d
11
2
d
42
d
32
d
22
d
12Second column of
flexibility matrix
3
4
1
12
3
4
1
1
k
41
k
31
k
21
k
11
First column of
stiffness matrix
1
k
42
k
32
k
22
k
12
Second column of stiffness
matrix
. . . . . . . .
. . . . . . . .

7. Example #1
For the frame shown,
k
1
k
2
k
3
1
1
k
1
1
k
1
1
k
1
1
k
1 2
11
kk

1 2
11
kk

1 2 3
111
kkk

1 2
11
kk

1
1
k
the flexibility matrix is
[]
1 1 1
1 1 2 1 2
1 1 2 1 2 3
1 1 1
111 11
111111
k k k
d
kkk kk
kkkkkk
￩
￪
￪
￪
  ￪
￪
￪
 ￪
￫
x
1
x
2
x
3

8. The elements of stiffness matrix are obtained as follows
0
-k
2
k
1
+ k
2
-k
3
k
2
+k
3
-k
2
-k
3
0
k
3
[]
1 2 2
2 2 3 3
3 3
0
0
kk k
k kkkk
k k
 -￩
￪
-  -
￪
￪ -
￫
Accordingly the stiffness matrix is
For systems of this type the stiffness matrix is a banded trapezoidal matrix

9. Example #2
Consider a prismatic cantilever beam with
three nodes spaced at intervals h=L/3
3h=L
Flexibility Matrix:
3
5
6
h
EI
3
8
6
h
EI
3
1
3
h
EI
3
8
3
h
EI
3
14
3
h
EI
3
5
6
h
EI
3
14
3
h
EI
3
127
3
h
EI
ﾴ
3
8
6
h
EI
[]
3
258
51628
6
82854
h
d
EI
￩
￪

￪
￪
￫

10. 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
81
3
L
EI
]k[
and verify that
3
3
804612258
81 1
[][] 46441651628
13162
1216782854
[]
EI L
kd
L EI
I
-  
  
 - -
  
-  
  
 Identitymatrix

11. Strain Energy of system
For a system deflected in a configuration defined by x
1
, x
2
,…. x
m
the strain
energy ,V, is equal to the work done by the forces F
1
, F
2…..
F
n
in going
through the displacement
,x. that is
[ ]
11 22
1
.....
2
nn
V FxFx Fx  
In matrix notation,
{}{} {}{}
1 1
2 2
T T
V Fx xF 
(144)
(145)
Substituting Equation (138) into the first of equations (145) we obtain,
{}
1
[]{}
2
T
V FdF
And substituting Equation (139) into the second of equations (145) we obtain,
{}
1
[]{}
2
T
V xkx
(146)
(147)

12. 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 2
1
{}
2
T
T
TT
FdF dFF
FdF


hence it follows that [][]
T
dd
or that [d] is symmetric. This is a restatement of Maxwell’s reciprocal
relation.
(148)

13. Proceeding in a similar manner and substitute equation 139 into the two
equations 145 we obtain,
[]{}{} {}[]{}
{}[]{} {}[]{}
[] []
1 1
( )
2 2
1 1
2 2
T
T
T TT
T
kxx xkx
xkx xkx
hencek 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 2
111 122 1212 1313 2323
....... ....Vcxcx cxxcxxcxx      
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)

14. If the quadratic form is positive and non-zero for all possible combinations
of the unknowns, excluding the trivial case of x
1
=x
2
…x
m
=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 x
1
=x
2
…=x
n
=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.

15. {}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
{}[]{}
T1
T=xmx
2
& &
Mass matrix

16. Dynamic Equations of motions for MDOF
Mass
matrix
(){ } (){ } (){ } (){ }
I d s
ftftft Pt  
Inertia
force
vector
Damping
force
vector
Elastic
force
vector
Externally
applied
force
vector
[]{} []{} []{} (){ }
nx1 nx1nxn nxn nxn
nx1
M x+C x+K x=Pt&& &
Damping
matrix
Stiffness
matrix
A second order non-
homogeneous linear
ordinary differential
equation
x
1
x
2 x
3 x
i
f
Ii

17. Dynamic Equations of motions for MDOF
{}
nx1
x
{}
nx1
x&
{}
nx1
x&&
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)

18. Undamped free vibration of MDOF system
[]{}[]{}{}Mx+Kx=0&&
{}{}0fﾹ
[] []
2
DetK-pM=0
{}
i
f
Equation of motion :
can this system undergo purely harmonic motion ?
(ie)
(){ }{} ( )sinpt+xtf 
an admissible solution ?
[]{} []{}
2
K =pMf fAnswer : Yes (i.e)
A generalised eigen
value problem
For non-trivial solution: i e
Results in n
th
order polynomial
equation in p
2
=0
From fundamental theorem of algebra, this polynomial has roots for p
2

corresponding to each such root of polynomial in p
2
,there is a mode shape
vector defining the vibration configuration required such that purely
harmonic sinusoidal motion is possible at circular natural frequency, p
i

19. Undamped free vibration of MDOF system contd....
0
i
p{}[]{}
T
M 0if ()
i j i j
ppijf f  
Properties :
(i) For all i = 1,2,....n (all natural frequencies are +ve real numbers)
(ii) Orthogonality conditions
{}[]{}
T
K 0
i j
f f
{}[][][]{}
b
1T
MMK 0for
i j
bf f
-
   -
 
{},1,2,.....
i
i nf
{}
i
f
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

20. 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,
r
p
s
p {}
r
f {}
s
f
{}[]{}
{}[]{}
T
r s
T
r s
M 0
K 0
f f
f f


(157)
(158)
and the second is,

21. n
iipis
i=1
m =0ff
If [M] is diagonal, as is usually the case, Eq.157 can be written as
In which m
i
is the concentrated mass at the i
th
coordinate and
and are respectively the amplitudes of the i
th
coordinate when
the system vibrates in the p
th
and

s
th
natural mode.

(159)
ir
f
is
f
Proof 1: The modes and must satisfy{}
r
f {}
s
f
[]{} []{}
[]{} []{}
2
rr r
2
ss s
K =pM
K =pM
f f
f f
(160)
(161)

22. Next we take the transpose of the two members of Eq.160, obtaining
{}[]{} {}[]{}
T T
2
sr s r s
K =p Mf f f f
(162){}[] {}[]
T T
2
rr r
K=p Mf f
and premultiply the latter equation by . This yields
s
{}f
T 2 T
r s r r s
{}[K]{}p{}[M]{}f ff f (163)
Finally we subtract Eq.163 from 162 and obtain
2 2 T
s r r s
(pp){}[M]{}0-f f (164)
Since p
s
has been considered to be different from p
r
, 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
T
r
{}f

23. 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 p
r or p
s 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
2
r r
p[M]{}f
(a)

24. 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.
[]{}
2
s s
pMf
[]{}( ){} {}[]{} {}[]{}
2 2 2
T TT T
r r sr s r s r s
pM p M p Mff f f f f 
{}[]{}
2
T
s r s
p Mf f
( ){}[]{}
2 2
0
T
s r r s
pp Mf f- 

25. 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
............. .............. 0
r nr n
C C C Cf f f f   
{}[]{} {}[]{} {}[]{} {}[]{}
1 21 1 2 2
............. .............. 0

0
T T T T
r nr r n r
C M C M C M C Mf f f f f f f f   
[ [ [
0 0
{}[]{}0
T
r r r
C Mf f
(165)
(164 )

26. But since [m] is positive definite, the triple product cannot be
zero and hence C
r
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 r
Mf f
2
p
p ib
(165 )

27. 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 p
r
and {f}
r
must also represent a
solution.
Letting,
{} {} {}
r r r
Xif Y
{} {} {} {}
( ){}[]{}
{} {}( )[][] {}( )
2 2
2 2


0
2 0
s r
s p rr
T
s r r s
T
p p rp
pp i
i
pp
iX i X
M
M i
b
ff
f f
b
-
 C-Y
- 
Y -Y
and
and making use of Eq. 164 we obtain

(166 )
(167)
(168 )
(169)

28. 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 p
r
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 r
K
MK
M
P
P
f f
f
f f f f


by we obtain
Therefore, P
r
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 P
r
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)

29. 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
xxx
xxx
xxx
￦
￨
xxx
xxx
xxx
￦
￨
00
0 0
00
x
x
x
￦
￨
00
0 0
00
x
x
x
￦
￨
[]M []K []K[]M

30. a)Equations of motion:
Modal Superposition Method…
b) Find mode shape and natural frequencies by solving Eigen Value
Problem,

[]{} []{}
2
K =pMf f
{}
i i
p, fobtain for i=1,2,…..n
where, = i
th
mode shape vector {}
i
f
c) Express { } {}
n
ii
i=1
x(t)= q(t)f￥
(2)
[]{}[]{}[]{} (){ }Mx+Cx+Kx=Pt&& & (1)
Substitute eqn (2) in (1) and then premultiply by and using
Orthogonality Properties, one obtains
{}
j
T
f
for j=1,2,……..n
(3)
{}[]{} {}[]{} {}[]{} {}{ }
n
T T T
j ij j ij j ij j
i=1
M q+ C q+ q= P(t)
T
Kf f f f f f f  
&& &

31. Let {f
i}
T
[C] {f
j}

= 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 {f
i
}
T
[M] {f
j
}

of Equation (3)

leads to ,

2 2
2
2
()2 () () ()()
{}{()}
{}[]{}
j jjj jj jj j
T
j
j T
jj j
qt pqtpqtpctft
Pt
p
p M
z
f
f f
  
￩
￪
￪￫
&& &
For j=1,2,..n
Where z
j
= j
th
modal damping ratio, c
j
= max
2
{}{()}
{}[]{}
T
j
T
jj j
Pt
p M
f
f f
Participation factor
(4)

32. (d) Express initial conditions
1 1
{} {} {} {}
n n
o ii o ii
i i
andx a x bf f
 
  
&
By pre multiplying above expression with {f
j
}
T
[m] and using
orthogonality of modes,
{}{}{}
{}[]{}
T
j o
j T
j j
Mx
a
M
f
f f

{}{}{}
{}[]{}
T
j o
j T
j j
Mx
b
M
f
f f

&
a
j
,b
j
- participation factors

33. (e) Solve q
i
(t) of the uncoupled SDOF systems
()
2
0
() cos sin ()
() () sin()
1
jj
jj
pt j jjj
j j dj dj j j
dj
t
ptj
j j dj
j
b pa
qte a pt ptcIAF
p
p
IAF fewher pte d
z
z 
z
 
z
-
- -
  
    
 
   
 -
-

(f) Express{ }
1
() {}()
n
jj
j
xt qtf



34. In general take first ‘r’ modes only
Elastic forces:
Had the load been static, {f
s
(t) = {P(t)} and (IAF)
j
= 1
i.e
(){ } {} ()
1
r
jj
j
xt qtf


(){ }[](){ } []{}()tqKtxKtf
jj
r
j
s f


1
[]{} ()
2
1
r
j jj
j
M pqtf

￩
￫￥
(){ } (){ }
[]{}{} (){ }
{}[]{}
1
T
n
j j
s T
j
j j
M Pt
ft Pt
M
ff
f f
 ￥
Implementation issues

35. Considering static force correction for the neglected modes r+1, ….n
(){ }{} []{} ()
[]{}{}{}
{}[]{}
() {} () []
{}{}
{}[]{}
(){ }
2
1 1
1
2
1 1
T
r r
j j
s j jj T
j j
j j
T
r r
j j
jj T
j j
j j j
M P
ft P M pqt
M
xt qtK Pt
p M
ff
f
f f
ff
f
f f
 
-
 
 -
￩
￪  -￩
￫
￪
￫
￥ ￥
￥ ￥

36. (){ } (){ } []{}
1
2
r
s j j jjj
j
ft Pt M q pqf z

￩ - 
￫￥
&& &
From Equation (4)
()
{} (){ }
{}[]{}
()
{} (){ }
{}[]{}
(){ }[] (){ } {}
2
2 2
1
2
1
2
21 1
21
T
j
j jjj jj T
j j
T
j jj
j jT
j j j
j j
r
jj
j j
j j j
Pt
q pqpqt
M
Pt q
qt q
p p p
M
q
xtKPt q
p p
f
z
f f
f z
f f
z
f
-

  
 - -
￩
 - ￪
￪￫
￥
&& &
&
&&
&
&&
Mode Acceleration Method

37. m
1
m
2
m
3
k
1
k
2
k
3
x
1
x
2
x
3
P
3
(t)
P
2
(t)
P
1
(t) 11
mx&&
22
mx&&
33
mx&&
k
3
(x
3
-x
2
)
k
2
(x
2
-x
1
)
k
1
x
1
•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

38. •Application of this procedure leads to the following,
For Mass 1
For Mass 2
For Mass 3
( ) ()
11 112 21 1
mx+kx-kx-x=Pt&&
( ) ( ) ()
22 2 21 3 3 2 2
mx+kx-x-kx =Ptx-&&
( ) ()
33 3 32 3
mx+kx-x=Pt&&
()
()
()
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 kk k x Pt
m x kkkkx Pt
mx k kx Pt
 - ￬￩ ￬ ￩ ￬
￪ ￪
-  - ￭ ￭ ￭� ��
￪ ￪
￪ ￪ -
￫ ￮ ￫ ￮ ￮
&&
&&
&&
Or
Response of a MDOF System
(a)
(b)
(c)
(d)

39. Consider special case in which, k
1
=k
2
=k
3
and m
1
=m
2
=m and m
3
= ½ m
1 1
2 2
3 3
1 x 2-10x
m 1 x+k-12-1x=0
0.5x 0-11x
￩ ￬ ￩ ￬
￪ ￪
￭ ￭� �
￪ ￪
￪ ￪
￫ ￮ ￫ ￮
&&
&&
&&
( )
1 1
2 2
3 3
x X
x=Xsinpt+α
x X
￬ ￬
￭ ￭��
￮ ￮
1 1
2
2 2
3 3
2-10X 1 X
k-12-1X=mp 1 X
0-11X 0.5X
￩ ￬ ￩ ￬ 
￪ ￪
￭ ￭� �
￪ ￪
￪ ￪
￫ ￮ ￫ ￮
(132)
(133)
Natural Frequencies and Modes

40. 2
mp
λ=
k
Let
we get
[][][]{}{}K-λMX=0
( )
( )
( )
1
2
3
2-λ -1 0 X
-12-λ -1 X 0
0 -11-0.5λX
￩ ￬
￪
￭�￪
￪
￮￫
For a non-trivial solution
( )
( )
( )
2-λ -1 0
-12-λ -1=0
0 -11-0.5λ
Expanding, we get the frequency equation
( ) ( )
3
2-λ-32-λ=0
(134)

41. ( )( )
2
2-λλ-4λ+1=0
which, upon factoring the term , may be rewritten as( )2-λ
The roots are
2
1
2
2
2
3
k
=0.26795
m
k
=2
m
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=23,
..0.267&3.730ie
ﾱ
(135)

42. 2 1
(2)X Xl-
( )
3 1 2
2X 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 X
1
=1, then
1
For, λ=2-3
( )( )
21
31
X2233
X 12233132
-
-- -

43. 2
For, λ=2
22
32
X220
X 101
-
---
3
For, λ=2+3
( )( )
23
33
X223 3
X 1 3 3132
---
-- --

44. 2
1
0.2680
k
p
m

2
2
2.0
k
p
m

2
3
3.7320
k
p
m

{}
1
1
3
2
f
￬
￭�
￮
{}
2
1
0
1
f
￬
￭�
-
￮
{}
3
1
3
2
f
￬
-￭�
￮
Summary:

45. Free Vibration
Write expression for the motion of the system due to an initial displacement
{} ()
0 10
1
1
1
x x
￬
￭�
￮
()
10
x
()
()
()
() ()
()
()
()
() ()
()
()
()
() ()
10
1 1 10 0
10
2 10 10
10
3 10 10
130.52
23
0.6220
130.54 6
100.51 0.5
0.3333
100.51 1.5
130.52
23
0.0447
6130.54
mx
a x x
m
mx
a x x
m
mx
a x x
m
 
 
  
 
 
- 
 
  
 
 
 -
- 
  
 
 
{} {}
3
1
cos
j jj
j
x a ptf

￥
Without any initial velocity
where,

46. Free Vibration
Therefore,
{}() {} {} {}( )
() () ()
1 1 2 30 1 2 3
1
2 1 1 1 2 1 30 0 0
3
1
2
3
0.62220cos 0.3333cos 0.0447cos
0.6220 0.3333 0.0447
1.0774 cos 0 cos 0.0774 cos
1.2440 0.3333 0.0893
0
xx pt pt pt
x
x x pt
o
x
r
and
pt x pt
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 pt x pt x pt
     
     
- -     
     
-
     

47. Free Vibration
In this case {} {}
3
1
sin
j jj
j
x b ptf

￥
Consider next, the displacement induced by an excitation of the second
mass, that is an initial velocity without an initial displacement
{} ()
0 20
0
1
0
v v
￬
￭�
￮
() () ()
() () ()
2 20 0
1
1 1
2
2
2 20 0
3
3 3
0310
1 3
. .
6 6
1 0
0
1.5
0310
1 3
. .
6 6
mv v
b
P m P
b
P
mv v
b
P m P
￩ 
￫
 
 
￩- 
￫
 -
Where,
1
20 20
2 1 3
1 3
3
0.2887 0.2887
() ()
0.5sin 0.5sin
0.5774 0.5774
x
v v
x pt pt
P P
x
-￬ ￬ ￬  
 ￭ ￭ ￭� � �
-
￮ ￮ ￮

48. Forced Vibration
Consider first the effect of a force P
2
(t)=P
2
f(t) applied on the second mass
{} {}
3
1
(..)
j jj
j
x c IAFf

￥
Noting that:
2.P is given as:
4.Making use of the values of obtained earlier we obtain,

{}
2
0
1
0
P P
￬
￭�
￮
22 2
1 2 3
,andpp p
( )
2
1 2
2
2
2
3
1 030
1.07735
6
0.26795
0
3
1
0.07735
6
3.73205
P
c P
k m k
m
c
P
P
c
k m k
m

 

-
 -

49. Forced Vibration
1
2
2 1 3
3
1
2
2 1 3
3
1 1
1.077353(..)0.077353(..)
2 2
1.077 0.077
1.866(..)0.134(..)
2.155 0.155
x
P
x IAF IAF
k
x
x
P
x IAF IAF
k
x
  
  
 - -   
   
   
-     
     
      
     
-
     
Therefore
or
an
1
2
2 1 3
3
1.077 0.077
0.789(..)0.211(..)
0.289 0.289
u
P
u IAF IAF
k
u
-     
     
      
     
-
     
d
Maximum values of (I.A.F)
j
are obtained from the response spectrum
applicable to those particular forcing functions considered

50. •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 f
1
t
1
and on
the ratio of f
3
/f
1
. 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 1
1
1 1
2
1 1
3
1
0.62220
6
0.26795
1
0.3333
1.5
2
1
0.4447
6
3.73205
P P
c
km k
m
P P
c
k k
m
P P
c
km k
m
 
 
 

51. 1
1
2 1 2 3
3
1
1
2 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
x
P
x IAF IAF IAF
k
x
u
P
u IAF
k
u
and
       
       
  -       
       
-
       
  
  
   
  
  
2 3
3333 0.0447
0.333(..) 0.122(..)
0.333 0.1667
IAF IAF
    
    
- -    
    
-
    
Finally if the system were subjected to a set of forces
(){ } ()
1
1
0.5
Pt Pft





Hence,

52. ( )
()
( )
1
2
3
1
1113 2
1 2
2.3213
6
0.26795
1
110 1
1 2
0.1667
1.5
2
1
1113 2
1 2
0.0120
6
3.73205
P
P
c
k m k
m
P
P
c
k m k
m
P
P
c
k m k
m
￦
 ﾴ ﾴ
￨
 
￦
 -ﾴ ﾴ
￨
 
￦
-ﾴ ﾴ
￨
 
The participation factors would be,

53. 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
x
P
x IAF IAF IAF
k
x
u
P
u
k
u
       
       
  -       
       
-
       





with the response given by,
and
1 2 3
0.1667 0.0120
(..) 0.1667(..) 0.0328(..)
218 0.1167 0.0448
IAF IAF IAF
      
      
- -      
      
-
      
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

54. 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, {F
d
}, 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 c
ij
corresponding to the i
th
row and the j
th
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)

55. 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.

56. 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.

57. m
2
m
1
m
3
c
1
c
2
c
3
k
1
k
2
k
3
c
1
c
2
c
3
c
1
c
2
c
3
(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)

58. 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 (c
j
/ m
j
) 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 xcc c
F ccccx
c cF x
 -￬ ￬ ￩
￪
-  -￭ ￭� �
￪
￪ -
￫￮ ￮
&
&
&
(208)
and the damping matrix would be of the same form as the stiffness
matrix, [k]. Furthermore, if the ratio c
j
/m
j
for all the dashpots and springs
were the same, the matrix [c] would be proportional to the mass matrix
[k].

59. 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.

60. 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.
n
T T T T
r rr r jj r rr r
j1
{}[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 ,
T
r s
{}[C]{}0(forrs)f f ﾹ
(209)

61. 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 r
th
term of the series in the second term of Eq.210 vanish, and
Eq.210 simplifies to,
T T T T
r rr r rr r rr r
{}[M]{}q(t){}[C]{}q(t){}[K]{}q(t){}{P}f(t)f f f f f f f&& &
(211)
or
rr rr rr r
Mq(t)Cq(t)Kq(t)Pf(t)  
%% % %&& & (212)
r
C
%
In which , and are defined by Eqs.196 to 199, respectively, and
the effective damping for the r
th
mode, defined as ,
r
M
%
r
k
%
r
P
%
T
r r r
C{}[C]{}f f
% (213)
Eq.211 , or its equivalent Eq.212, permits the values of q
r
(t) to be evaluated
independently of each other. In other words, the use of q
r
(t) uncouples the
governing equations of motion, as for undamped systems.

62. 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 r
th
circular natural frequency of the system, p
p
2
(see Eq.170), and
the coefficient of the time function on the right hand member of the
equation becomes p
p
2
c
r, in which c
r is the participation factor defined
by Eq.191.
If we also let,
T
r r r
rr T
r r r
{}[C]{}C
2p
{}[M]{}M
f f
z 
f f
%
%
(214)
we obtain
2 2
r rrr rr rr
q(t)2pq(t)pq(t)cpf(t)z  && & (215)

63. In which z
r
represents the percent of the critical coefficient of damping for
motion in the r
th
natural mode. Finally, if we compare this equation with
the equation of motion of a SDF damped system,
2 2
st0
x(t)2pxpx(x)pf(t)z&& & (216)
And recall that the solution of this equation can be expressed
symbolically as
st0
x(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 q
r
(t) can also be stated as
r r r
q(t)c(I.A.F)
(218)

64. The time function (I.A.F.)
r
is the I.A.F. for a SDF system which has a
natural frequency, p
r
, a damping factor, z
r
, 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.

65. 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 T
r s r s
{}[M]{}{}[K]{}f fbf 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.

66. Substitution of Equation 21a we find that,
2
2
1 1
2 2
pp p
p p
p
wher
p p
p
p
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 ,z
p
is
proportional to the natural frequency of the mode under consideration.
Implications of Rayleigh Damping

67. for [c] =  [M]
for [c] = b [K]
z
p
p
p
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 z
1 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.

68. 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,z
1
and z
2
,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 1
1
2 2
2
1
2 2
1
2 2
p
p
p
p

z b

z b
 
 
(221)

69. which, upon solving simultaneously, yields
1
1 2
2
1 2
1
2
1
2 1
2
2
2
1
2
2
1
2
1
p
p
p
p
p
p
p
p
p
p
z z

z z
b

-




-



-




-


(222)
In the special case in which z
2
= z
1
(p
1
/ p
2
) , b=0 ,indicating that
[c] =  [M] , and  = 2p
1
z
1
similarly ,when z
2
= z
1
(p
2
/ p
1
) ,  = 0 (i.e [c] = b [K]) , and b=2z
1
/p
1.
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.