Mechanical Vibration (ME 315)
Spring 2024
LECTURE NO: 15
Dr. Taimoor Hassan
Faculty of Mechanical Engineering
1
Chapter 4: Multiple Degree of Freedom Systems
Extending the first 3 chapters to more then one degree of freedom
•2
Two-Degree-of-Freedom System
• DOF: Minimum number of coordinates to specify the position of a system
• Many systems have more than 1 DOF
• Systems that require two independent coordinates to describe their motion are
called two degree-of-freedom systems
• Examples of 2 DOF systems
(a)
(c)
Fig 4.1
(b)
•3
Two-Degree-of-Freedom System
• linear coordinate 𝑥(𝑡), indicating the vertical displacement of the 𝐶. 𝐺. of the mass
• angular coordinate 𝜃(𝑡), denoting the rotation of the mass m about its C.G.
• Instead of 𝑥(𝑡), and 𝜃(𝑡), we can also use 𝑥1 and 𝑥1 , the displacements of points 𝐴 and 𝐵
•4
DoF
• The general rule for the computation of the
number of degrees of freedom can be stated as
follows:
5
Two-Degree-of-Freedom Model
A two degree of freedom system used to base much of the analysis
and conceptual development of MDOF systems on.
•6
Free-Body Diagram of each mass
Fig 4.2
•7
Summing forces yields the equations of motion:
•8
Note
• A two Degree-of-Freedom system has
• Two equations of motion!
• Two natural frequencies!
•9
Two-Degree-of-Freedom System
• Free vibrations, so homogeneous eqs.
• Equations are coupled:
• Both have 𝑥1 and 𝑥2 .
• If only one mass moves, the other follows
•10
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
•11
Solution by Matrix Methods
• The two equations can be written in the form of a single matrix equation:
m1 x1 (t )  ( k1  k 2 ) x1 (t )  k 2 x2 (t )  0
m2 x2 (t )  k 2 x1 (t )  k 2 x2 (t )  0
(4.4), (4.5)
(4.6), (4.9)
•12
Solution by Matrix Methods
•13
Counter check Equations
•14
Initial Conditions
IC’s can also be written in vector form
 x10 
 x10 
x(0)    , and x(0)   
 x20 
 x20 
•15
How to find the Solution:
• For 1DOF we assumed the scalar solution 𝑎𝑒 𝜆𝑡 Similarly, now we
assume the vector form:
(4.15)
(4.16)
(4.17)
•16
This changes the differential equation of motion into algebraic vector equation:
−𝜔2 𝑀 + 𝐾 𝐮 = 𝟎
(4.17)
This is two algebraic equation in 3 uknowns
( 1 vector of two elements and 1 scalar):
𝑢1
𝐮 = 𝑢 , and 𝜔
2
•17
The condition for solution of this matrix equation requires that the matrix inverse does not
exist:
If the inv  - 2 M  K  exists  u  0 : which is the
static equilibrium position. For motion to occur
u  0   - M  K 
2
or
1
det  - 2 M  K   0
does not exist
(4.19)
18
Back to our specific system: the characteristic equation is defined as
det - 2 M  K   0 
 2 m1  k1  k 2
det 
k2


 0 
2
 m2  k 2 
k 2
(4.20)
m1m 2 4  (m1k2  m2 k1  m2 k 2 ) 2  k1 k2  0
(4.21)
•19
Once  is known, use equation (4.17) again to calculate the corresponding vectors
𝑢1 and 𝑢2
This yields vector equation for each squared frequency:
( 12 M  K )u1  0
(4.22)
( 22 M  K )u 2  0
(4.23)
and
•20
Example1.5
Calculating  for physical parameters:
m1=9 kg, m2=1kg, k1=24 N/m and k2=3 N/m
21
Each value of 𝜔2 yields an expression or 𝒖
22
Computing the vectors 𝑢
For 𝜔12 =2,
𝐮1
𝑢11
= 𝑢
12
(−𝜔12 𝑀 + 𝐾)𝐮1 = 𝟎 ⇒
𝑢11
27 − 9(2)
−3
0
=
⇒
−3
3 − (2) 𝑢12
0
9𝑢11 − 3𝑢12 = 0 and − 3𝑢11 + 𝑢12 = 0
23
Only the direction of vectors u can be determined, not the magnitude as it remains
arbitrary
𝑢11 1
1
= ⇒ 𝑢11 = 𝑢12 results from both equations:
𝑢12 3
3
only the direction, not the magnitude can be determined!
This is because:
det(−𝜔12 𝑀 + 𝐾) = 0.
The magnitude of the vector is arbitrary. To see this suppose that
(−𝜔12 𝑀 + 𝐾)𝐮1 = 𝟎, so does 𝑎𝐮1 , 𝑎 arbitrary. So
(−𝜔12 𝑀 + 𝐾)𝑎𝐮1 = 𝟎 ⇔ (−𝜔12 𝑀 + 𝐾)𝐮1 = 𝟎
24
Likewise for the second value of 𝜔2
Note that the other equation is the same
25
What to do about the magnitude!
Several possibilities, here we just fix one element:
 13
u12  1  u1   
1
 1 3 
u 22  1  u 2   
1
26
Return now to the time response:
We have computed four solutions:
x (t )  u1e  j1t , u1e j1t , u 2 e  j2t , u 2 e j2t 
(4.24)
Since linear, we can combine as:
x (t )  au1e  j1t  bu1e j1t  cu 2 e  j2t  du 2 e j2t
 x (t )   ae  j1t  be j1t  u1   ce  j2t  de j2t  u 2
Note that to go from the exponential
form to sine requires Euler’s
Formula for trig functions and uses
up the +/- sign on omega
 A1 sin(1t  1 )u1  A2 sin( 2t  2 )u 2
(4.26)
where A1 , A2 , 1 , and 2 are const ants of integration
determined by initial conditions.
27
Physical interpretation of all that math!
• Each of the TWO masses is oscillating at TWO natural frequencies 𝝎𝟏 and 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 𝑨𝟏 𝐮𝟏 and 𝑨𝟐 𝐮𝟐
• The vectors 𝐮𝟏 and 𝐮𝟐 are called mode shapes because the describe the relative
magnitude of oscillation between the two masses
28
What is a mode shape?
• First note that A1, A2, 1 and 2 are determined by
the initial conditions
• Choose them so that A2 = 1 = 2 =0
• Then:
 x1 (t ) 
 u11 
x(t )  
 A1   sin  1t  A1u1 sin  1t

 x2 (t ) 
 u12 
• Thus each mass oscillates at (one) frequency 1
with magnitudes proportional to u1 the 1st mode
shape
29
A graphic look at mode shapes:
x1
k1
k2
m1
Mode 1:
x1=A/3
x2=A
x1
x2
k1
Mode 2:
x2
m2
 13
u1   
1
m2
 1 3 
u2   
1
k2
m1
x1=-A/3
x2=A
•30
Example 4.1.7
Eq. 4.26:
1 
0 
consider x (0) =   mm, x (0)   
0 
0 
A1
A2


x
(
t
)
 1 




sin
2
t



sin
2
t


1
2 


3
3
 x (t ) 
 2   A sin  2t     A sin 2t    
1
1
2
2
A1
A2


x
(
t
)

 1 




2
cos
2
t



2
cos
2
t


1
2 


3
3
 x (t ) 
 2   A 2 cos  2t     A 2 cos 2t    
1
1
2
2
31
At t=0 we have
32
4 equations in 4 unknowns:
3  A1 sin 1   A2 sin 2 
0  A1 sin 1   A2 sin 2 
0  A1 2 cos 1   A2 2 cos 2 
0  A1 2 cos 1   A2 2 cos 2 
Yields:

A1  1.5 mm , A2  1.5 mm , 1  2  rad
2
33
34
The final solution is:
x1 (t)  0.5 cos 2t  0.5 cos 2t
x2 (t)  1.5 cos 2t  1.5 cos 2t
Figure 4.3
35

Solution as a sum of modes
x(t)  a1u1 cos 1t  a2u2 cos 2 t
Determines how the first
frequency contributes to the
response
Determines how the second
frequency contributes to the
response
36
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
1 
2
k1
 1.63, 2  2 
m1
k2
 1.732
m2
37
Assignment 03:
38