Dynamics of multiple degree of freedom linear systems

© Eng. Vib, 3rd Ed.
1/58 @ProfAdhikari, #EG260
Chapter 4 Multiple Degree of
Freedom Systems
Extending the first 3 chapters to
more then one degree of freedom
The Millennium bridge required
many degrees of freedom to model
and design with.

2/58 College of Engineering
The first step in analyzing multiple
degrees of freedom (DOF) is to look at 2
DOF
• DOF: Minimum number of coordinates to specify the
position of a system
• Many systems have more than 1 DOF
• Examples of 2 DOF systems
– car with sprung and unsprung mass (both heave)
– elastic pendulum (radial and angular)
– motions of a ship (roll and pitch)
Fig 4.1

4.1 Two-Degree-of-Freedom
Model (Undamped)
A 2 degree of freedom system used to base
much of the analysis and conceptual
development of MDOF systems on.

Free-Body Diagram of each mass
x1 x2
m1
m2k1 x1
k2(x2 -x1)
Figure 4.2

Summing forces yields the
equations of motion:
m1
x1(t) = -k1x1(t)+ k2 x2 (t)- x1(t)( )
m2
x2 (t) = -k2 x2 (t)- x1(t)( )
(4.1)
Rearranging terms:
m1
x1(t)+(k1 + k2 )x1(t)- k2 x2 (t) = 0
m2
x2 (t)- k2 x1(t)+ k2 x2 (t) = 0
(4.2)

Note that it is always the case
that
• A 2 Degree-of-Freedom system has
– Two equations of motion!
– Two natural frequencies (as we shall see)!

The dynamics of a 2 DOF system consists
of 2 homogeneous and coupled equations
• Free vibrations, so homogeneous eqs.
• Equations are coupled:
– Both have x1 and x2.
– If only one mass moves, the other follows
– Example: pitch and heave of a car model
• In this case the coupling is due to k2.
– Mathematically and Physically
– If k2 = 0, no coupling occurs and can be solved
as two independent SDOF systems

Initial Conditions
• Two coupled, second -order, ordinary
differential equations with constant
coefficients
• Needs 4 constants of integration to solve
• Thus 4 initial conditions on positions and
velocities
x1(0)= x10, x1(0)= x10, x2(0)= x20, x2(0)= x20

Solution by Matrix Methods
x(t) =
x1(t)
x2 (t)
é
ë
ê
ê
ù
û
ú
ú
, x(t) =
x1(t)
x2 (t)
é
ë
ê
ê
ù
û
ú
ú
, x(t) =
x1(t)
x2 (t)
é
ë
ê
ê
ù
û
ú
ú
M =
m1 0
0 m2
é
ë
ê
ê
ù
û
ú
ú
, K =
k1 + k2 -k2
-k2 k2
é
ë
ê
ê
ù
û
ú
ú
Mx + Kx = 0 m1
x1
(t)+(k1
+ k2
)x1
(t)- k2
x2
(t) = 0
m2
x2
(t)- k2
x1
(t)+ k2
x2
(t) = 0
The two equations can be written in the form of a
single matrix equation (see pages 272-275 if matrices and
vectors are a struggle for you) :
(4.4), (4.5)
(4.6), (4.9)

Initial Conditions
x(0) =
x10
x20
é
ë
ê
ê
ù
û
ú
ú
, and x(0) =
x10
x20
é
ë
ê
ê
ù
û
ú
ú
IC’s can also be written in vector form

The approach to a Solution:
2
2
Let ( )
1, , , unknown
-
-
j t
j t
t e
j
M K e
M K
x u
u 0 u
u 0
u 0
For 1DOF we assumed the scalar solution aeλt
Similarly, now we assume the vector form:
(4.15)
(4.16)
(4.17)

This changes the differential
equation of motion into algebraic
vector equation:
2
1
2
- (4.17)
This is two algebraic equation in 3 uknowns
( 1 vector of two elements and 1 scalar):
= , and
M K
u
u
u 0
u

The condition for solution of this matrix
equation requires that the the matrix
inverse does not exist:
2
1
2
2
If the inv - exists : which is the
static equilibrium position. For motion to occur
- does not exist
or det - (4.19)
M K
M K
M K
u 0
u 0
0
The determinant results in 1 equation
in one unknown ω (called the characteristic equation)

Back to our specific system: the
characteristic equation is defined as
2
2
1 1 2 2
2
2 2 2
4 2
1 2 1 2 2 1 2 2 1 2
det - 0
det 0
( ) 0
M K
m k k k
k m k
m m m k m k m k k k
Eq. (4.21) is quadratic in so four solutions result:
(4.20)
(4.21)
2 2
1 2 1 2and and

Once ω is known, use equation (4.17) again to
calculate the corresponding vectors u1 and u2
2
1 1
2
2 2
( ) (4.22)
and
( ) (4.23)
M K
M K
u 0
u 0
This yields vector equation for each squared frequency:
Each of these matrix equations represents 2
equations in the 2 unknowns components of the
vector, but the coefficient matrix is singular so
each matrix equation results in only 1 independent
equation. The following examples clarify this.

Examples 4.1.5 & 4.1.6:calculating u
and ω
• m1=9 kg,m2=1kg, k1=24 N/m and k2=3 N/m
• The characteristic equation becomes
ω4-6ω2+8=(ω2-2)(ω2-4)=0
ω2 = 2 and ω2 =4 or
1,3 2 rad/s, 2,4 2 rad/s
Each value of ω2 yields an expression for u:

Computing the vectors u
For 1
2
=2, denote u1
u11
u12
then we have
(- 1
2
M K)u1 0
27 9(2) 3
3 3 (2)
u11
u12
0
0
9u11 3u12 0 and 3u11 u12 0
2 equations, 2 unknowns but DEPENDENT!
(the 2nd equation is -3 times the first)

u11
u12
1
3
u11
1
3
u12 results from both equations:
only the direction, not the magnitude can be determined!
This is because: det( 1
2
M K) 0.
The magnitude of the vector is arbitrary. To see this suppose
that u1 satisfies
( 1
2
M K)u1 0, so does au1, a arbitrary. So
( 1
2
M K)au1 0 ( 1
2
M K)u1 0
Only the direction of vectors u can be
determined, not the magnitude as it
remains arbitrary

Likewise for the second value of ω2
For 2
2
= 4, let u2
u21
u22
then we have
(- 1
2
M K)u 0
27 9(4) 3
3 3 (4)
u21
u22
0
0
9u21 3u22 0 or u21
1
3
u22
Note that the other equation is the same

What to do about the magnitude!
1
3
12 1
1
3
22 2
1
1
1
1
u
u
u
u
Several possibilities, here we just fix one element:
Choose:
Choose:

Thus the solution to the algebraic
matrix equation is:
1
3
1,3 1
1
3
2,4 2
2, has mode shape
1
2, has mode shape
1
u
u
Here we have introduce the name
mode shape to describe the vectors
u1 and u2. The origin of this name comes later

Return now to the time response:
1 1 2 2
1 1 2 2
1 1 2 2
1 1 2 2
1 1 2 2
1 2
1 1 1 1 2 2 2 2
1 2 1 2
( ) , , ,
( )
( )
sin( ) sin( )
where , , , and are const
j t j t j t j t
j t j t j t j t
j t j t j t j t
t e e e e
t a e b e c e d e
t ae be ce de
A t A t
A A
x u u u u
x u u u u
x u u
u u
ants of integration
We have computed four solutions:
Since linear, we can combine as:
determined by initial conditions.
(4.24)
(4.26)
Note that to go from the exponential
form to to sine requires Euler’s formula
for trig functions and uses up the
+/- sign on omega

Physical interpretation of all that
math!
• Each of the TWO masses is oscillating at TWO
natural frequencies ω1and ω2
• The relative magnitude of each sine term, and
hence of the magnitude of oscillation of m1 and m2
is determined by the value of A1u1 and A2u2
• The vectors u1 and u2 are called mode shapes
because the describe the relative magnitude of
oscillation between the two masses

What is a mode shape?
• First note that A1, A2, Φ1 and Φ2 are determined by
the initial conditions
• Choose them so that A2 = Φ1 = Φ2 =0
• Then:
• Thus each mass oscillates at (one) frequency 1
with magnitudes proportional to u1 the 1st mode
shape
x(t)
x1(t)
x2 (t)
A1
u11
u12
sin 1t A1u1 sin 1t

A graphic look at mode shapes:
Mode 1:
k1
m1
x1
m2
x2
k2
Mode 2:
k1
m1
x1
m2
x2
k2
x2=A
x2=Ax1=A/3
x1=-A/3
u1
1
3
1
u2
1
3
1
If IC’s correspond to mode 1 or 2, then the response is purely in
mode 1 or mode 2.

Example 4.1.7 given the initial
conditions compute the time response
consider x(0)= 1
0
é
ë
ê
ù
û
ú mm, x(0) = 0
0
é
ë
ê
ù
û
ú
x1
(t)
x2
(t)
é
ë
ê
ê
ù
û
ú
ú
=
A1
3
sin 2t +j1( )-
A2
3
sin 2t +j2( )
A1
sin 2t +j1( )+ A2
sin 2t +j2( )
é
ë
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
x1
(t)
x2
(t)
é
ë
ê
ê
ù
û
ú
ú
=
A1
3
2 cos 2t +j1( )-
A2
3
2cos 2t +j2( )
A1
2 cos 2t +j1( )+ A2
2cos 2t +j2( )
é
ë
ê
ê
ê
ê
ù
û
ú
ú
ú
ú

1 2
1 2
1 1 2 2
1 2
1 2
1 1 2 2
sin sin1 mm
3 3
0
sin sin
2 cos 2 cos0
3 3
0
2 cos 2 cos
A A
A A
A A
A A
At t = 0 we have

1 1 2 2
1 1 2 2
1 1 2 2
1 1 2 2
1 2 1 2
3 sin sin
0 sin sin
0 2 cos 2cos
0 2 cos 2cos
1.5 mm, 1.5 mm, rad
2
A A
A A
A A
A A
A A
4 equations in 4 unknowns:
Yields:

The final solution is:
1
2
( ) 0.5cos 2 0.5cos2
( ) 1.5cos 2 1.5cos2
x t t t
x t t t
These initial conditions gives a response that is a combination
of modes. Both harmonic, but their summation is not.
Figure 4.3
(4.34)

Solution as a sum of modes
1 1 1 2 2 2( ) cos cost a t a tx u u
Determines how the first
frequency contributes to the
response
Determines how the second
frequency contributes to the
response

Things to note
• Two degrees of freedom implies two natural
frequencies
• Each mass oscillates at with these two frequencies
present in the response and beats could result
• Frequencies are not those of two component
systems
• The above is not the most efficient way to
calculate frequencies as the following describes
1 2
k1
m1
1.63, 2 2
k2
m2
1.732

Some matrix and vector reminders
A =
a b
c d
é
ë
ê
ù
û
úÞ A-1
=
1
ad -cb
d -b
-c a
é
ë
ê
ù
û
ú
xT
x = x1
2
+ x2
2
M =
m1 0
0 m2
é
ë
ê
ê
ù
û
ú
ú
Þ xT
Mx = m1x1
2
+ m2 x2
2
M > 0 Þ xT
Mx > 0 for every value of x except 0
Then M is said to be positive definite

4.2 Eigenvalues and Natural
Frequencies
• Can connect the vibration problem with the
algebraic eigenvalue problem developed in
math
• This will give us some powerful
computational skills
• And some powerful theory
• All the codes have eigen-solvers so these
painful calculations can be automated

Some matrix results to help us use
available computational tools:
A matrix M is defined to be symmetric if
M MT
A symmetric matrix M is positive definite if
xT
Mx 0 for all nonzero vectors x
A symmetric positive definite matrix M can
be factored M LLT
Here L is upper triangular, called a Cholesky matrix

If the matrix L is diagonal, it defines
the matrix square root
The matrix square root is the matrix M1/2
such that
M1/2
M1/2
M
If M is diagonal, then the matrix square root is just the root
of the diagonal elements:
L M1/2
m1 0
0 m2
(4.35)

A change of coordinates is introduced to
capitalize on existing mathematics
M =
m1 0
0 m2
é
ë
ê
ê
ù
û
ú
ú
, M-1
=
1
m1
0
0 1
m2
é
ë
ê
ê
ù
û
ú
ú
, M-1/2
=
1
m1
0
0 1
m2
é
ë
ê
ê
ù
û
ú
ú
Let x(t) = M-1/2
q(t) and multiply by M-1/2
:
M-1/2
MM-1/2
I identity
   q(t)+ M-1/2
KM-1/2
K symmetric
   q(t) = 0 (4.38)
or q(t)+ Kq(t) = 0 where K = M-1/2
KM-1/2
K is called the mass normalized stiffness and is similar to the scalar
k
m
used extensively in single degree of freedom analysis. The key here is that
K is a SYMMETRIC matrix allowing the use of many nice properties and
computational tools
For a diagonal, positive definite matrix M:

How the vibration problem relates to the
real symmetric eigenvalue problem
Assume q(t) = vejwt
in q(t)+ Kq(t) = 0
-w2
vejwt
+ Kvejwt
= 0, v ¹ 0 or
Kv =w2
v
vibration problem
(4.40)
   Û Kv = lv
real symmetric
eigenvalue problem
(4.41)
  v ¹ 0
Note that the martrix K contains the same type of information
as does wn
2
in the single degree of freedom case.

Important Properties of the n x n Real
Symmetric Eigenvalue Problem
• There are n eigenvalues and they are all real
valued
• There are n eigenvectors and they are all real
valued
• The set of eigenvectors are orthogonal
• The set of eigenvectors are linearly
independent
• The matrix is similar to a diagonal matrix
Window 4.1 page 285

Square Matrix Review
• Let aik be the ikth element of A then A is symmetric
if aik = aki denoted AT=A
• A is positive definite if xTAx > 0 for all nonzero x
(also implies each λi > 0)
• The stiffness matrix is usually symmetric and
positive semi definite (could have a zero
eigenvalue)
• The mass matrix is positive definite and
symmetric (and so far, its diagonal)

Normal and orthogonal vectors

x
x1
M
xn
, y
y1
M
yn
, inner product is xT
y xi yi
i 1
n
x orthogonal to y if xT
y 0
x is normal if xT
x 1
if a the set of vectores is is both orthogonal and normal it
is called an orthonormal set
The norm of x is x xT
x (4.43)

Normalizing any vector can be
done by dividing it by its norm:
x
xT
x
has norm of 1
To see this compute
(4.44)
x
xT
x
xT
xT
x
x
xT
x
xT
x
xT
x
1

Examples 4.2.2 through 4.2.4
K = M-1/2
KM-1/2
=
1
3 0
0 1
é
ë
ê
ê
ù
û
ú
ú
27 -3
-3 3
é
ë
ê
ù
û
ú
1
3 0
0 1
é
ë
ê
ê
ù
û
ú
ú
so K =
3 -1
-1 3
é
ë
ê
ù
û
ú which is symmetric.
det( K - lI) = det
3-l -1
-1 3-l
é
ë
ê
ù
û
ú= l2
- 6l +8 = 0
which has roots: l1 = 2 =w1
2
and l2 = 4 =w2
2

( K - l1I)v1 = 0 Þ
3- 2 -1
-1 3- 2
é
ë
ê
ù
û
ú
v11
v12
é
ë
ê
ê
ù
û
ú
ú
=
0
0
é
ë
ê
ù
û
úÞ
v11 - v12 = 0 Þ v1 =a 1
1
é
ë
ê
ù
û
ú
v1 = a2
(1+1) =1Þ a = 1
2
v1 =
1
2
1
1
é
ë
ê
ù
û
ú
The first normalized eigenvector

v2
1
2
1
1
, v1
T
v2
1
2
(1 1) 0
v1
T
v1
1
2
(1 1) 1
v2
T
v2
1
2
(1 ( 1)( 1)) 1
vi are orthonormal
Likewise the second normalized eigenvector is
computed and shown to be orthogonal to the first,
so that the set is orthonormal

Modes u and Eigenvectors v are
different but related:
u1 v1 and u2 v2
x M 1/2
q u M 1/2
v
Note
M1/2
u1
3 0
0 1
1
3
1
1
1
v1
(4.37)

This orthonormal set of vectors is
used to form an Orthogonal Matrix
P = v1 v2
é
ë
ù
û
PT
P =
v1
T
v1 v1
T
v2
v2
T
v1 v2
T
v2
é
ë
ê
ê
ù
û
ú
ú
=
1 0
0 1
é
ë
ê
ù
û
ú= I
PT KP = PT Kv1
Kv2
é
ëê
ù
ûú= PT
l1v1 l2v2
é
ë
ù
û
=
l1v1
T
v1 l2v1
T
v2
l1v2
T
v1 l2v2
T
v2
é
ë
ê
ê
ù
û
ú
ú
=
l1 0
0 l2
é
ë
ê
ê
ù
û
ú
ú
= diag(w1
2
,w2
2
) = L
P is called an orthogonal matrix
P is also called a modal matrix
called a matrix of eigenvectors (normalized)
(4.47)

Example 4.2.3 compute P and show
that it is an orthogonal matrix
From the previous example:
1 1
1 11
1 12
1 1 1 11 1
1 1 1 12 2
1 1 1 1 2 01 1
2 1 1 1 1 2 0 2
T
P
P P
I
v v

Example 4.2.4 Compute the square of
the frequencies by matrix manipulation
PT KP =
1
2
1 1
1 -1
é
ë
ê
ù
û
ú
3 -1
-1 3
é
ë
ê
ù
û
ú
1
2
1 1
1 -1
é
ë
ê
ù
û
ú
=
1
2
1 1
1 -1
é
ë
ê
ù
û
ú
2 4
2 -4
é
ë
ê
ù
û
ú
=
1
2
4 0
0 8
é
ë
ê
ù
û
ú=
2 0
0 4
é
ë
ê
ù
û
ú= L =
w1
2
0
0 w2
2
é
ë
ê
ê
ù
û
ú
ú
1 2 rad/s and 2 2 rad/s
In general:
L = PT KP = diag li( )= diag(wi
2
) (4.48)

Example 4.2.5
Figure 4.4The equations of motion:
m1
x1 +(k1 +k2 )x1 -k2x2 = 0
m2
x2 -k2 x1 +(k2 +k3)x2 = 0
(4.49)
In matrix form these become:
m1 0
0 m2
é
ë
ê
ê
ù
û
ú
ú
x+
k1 +k2 -k2
-k2 k2 +k3
é
ë
ê
ê
ù
û
ú
ú
x = 0 (4.50)

Next substitute numerical values
and compute P and Λ
m1 1 kg, m2 4 kg, k1 k3 10 N/m and k2 =2 N/m
Þ M =
1 0
0 4
é
ë
ê
ù
û
ú, K = 12 -2
-2 12
é
ë
ê
ù
û
ú
Þ K = M-1/2
KM-1/2
= 12 -1
-1 12
é
ë
ê
ù
û
ú
Þ det K - lI( )= det
12 - l -1
-1 12 - l
é
ë
ê
ù
û
ú= l2
-15l +35 = 0
Þ l1 = 2.8902 and l2 =12.1098
Þw1 =1.7 rad/s and w2 =12.1098 rad/s

Next compute the eigenvectors
For 1 equation (4.41 ) becomes:
12 - 2.8902 1
1 3- 2.8902
v11
v21
0
9.1089v11 v21
Normalizing v1 yields
1 v1 v11
2
v21
2
v11
2
(9.1089)2
v11
2
v11 0.1091, and v21 0.9940
v1
0.1091
0.9940
, likewise v2
0.9940
0.1091

Next check the value of P to see
if it behaves as its suppose to:
P = v1 v2
é
ë
ù
û=
0.1091 -0.9940
0.9940 0.1091
é
ë
ê
ù
û
ú
PT KP =
0.1091 0.9940
-0.9940 0.1091
é
ë
ê
ù
û
ú
12 -1
-1 3
é
ë
ê
ù
û
ú
0.1091 -0.9940
0.9940 0.1091
é
ë
ê
ù
û
ú=
2.8402 0
0 12.1098
é
ë
ê
ù
û
ú
PT
P =
0.1091 0.9940
-0.9940 0.1091
é
ë
ê
ù
û
ú
0.1091 -0.9940
0.9940 0.1091
é
ë
ê
ù
û
ú=
1 0
0 1
é
ë
ê
ù
û
ú
Yes!

A note on eigenvectors
In the previous section, we could have chosed v2 to be
v2
0.9940
0.1091
instead of v2
-0.9940
0.1091
because one can always multiple an eigenvector by a constant
and if the constant is -1 the result is still a normalized vector.
Does this make any difference?
No! Try it in the previous example

All of the previous examples can and should
be solved by “hand” to learn the methods
However, they can also be solved on
calculators with matrix functions and
with the codes listed in the last section
In fact, for more then two DOF one must
use a code to solve for the natural
frequencies and mode shapes.
Next we examine 3 other formulations for
solving for modal data

Matlab commands
• To compute the inverse of the square matrix
A: inv(A) or use Aeye(n) where n is the
size of the matrix
• [P,D]=eig(A) computes the eigenvalues and
normalized eigenvectors (watch the order).
Stores them in the eigenvector matrix P and
the diagonal matrix D (D= )

More commands
• To compute the matrix square root use sqrtm(A)
• To compute the Cholesky factor: L= chol(M)
• To compute the norm: norm(x)
• To compute the determinant det(A)
• To enter a matrix:
K=[27 -3;-3 3]; M=[9 0;0 1];
• To multiply: K*inv(chol(M))

An alternate approach to normalizing
mode shapes
From equation (4.17) M 2
K u 0, u 0
Now scale the mode shapes by computing such that
iui
T
M iui 1 i
1
ui
T
ui
wi iui is called mass normalized and it satisfies:
i
2
Mwi Kwi 0 i
2
wi
T
Kwi , i 1,2
(4.53)

There are 3 approaches to computing
mode shapes and frequencies
(i) 2
Mu Ku (ii) 2
u M 1
Ku (iii) 2
v M
1
2
KM
1
2
v
(i) Is the Generalized Symmetric Eigenvalue Problem
easy for hand computations, inefficient for computers
(ii) Is the Asymmetric Eigenvalue Problem
very expensive computationally
(iii) Is the Symmetric Eigenvalue Problem
the cheapest computationally

Dynamics of multiple degree of freedom linear systems

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