ME351

Mechanical Vibrations

Lecture 22

Chap 5.4

Chapter 5 Design for Vibration Suppression

5.1 Acceptable levels of vibration

5.2 Vibration isolation
5.3 Vibration absorbers
5.4 (Effects of )Damping in vibration absorption
5.5 Optimization (Informative)
5.6 Viscoelastic damping treatments
5.7 Critical speeds of rotating disks (skipped)

1
 5.4 Damped vibration absorber
x   ca ca   x   k  ka
 m 0    ka   x (t )   F0 
F (t )=F0sint  0 m   

 
a   xa 
  
 ca ca   xa   ka
     sin t
ka   xa (t )   0 
x  x (t )   X 
      sin  t   
 xa (t )   X a 
X2 ( ka  ma 2 ) 2   2 ca2
m  (5.35)
F02 (k  m 2 )(k  m  2 )  m k  2  2   k  ( m  m ) 2  2 c 2 2
 a a a a   a  a
ca ka xa
X a2 ka2   2 ca2

ma F02 (k  m 2 )(k  m  2 )  m k  2  2   k  ( m  m ) 2  2 c 2 2
k /2 k /2  a a a a   a  a
 k ka m     
 p  , a  ,  a ,  a ,r  r  in Sec.5.3 
 m ma m p p  a 

  k maa2 ca ca a ca
 
a
  ,  
2
     a
k m p2 2ma p 2maa  p 2 ka ma
 
 a
Four Design Variables
: mass ratio Xk
r  
 r      2 r 
2 2 2 2

(5.37) (A)
 : frequency ratio   r  1 r       2 r   r 1 r 2 
2 2
F0 r 
2 2 2 2 2 2 2

 : damping ratio 4 2
r : excitation frequency ratio 1 r2
X ak 4
r  
  r 2  2   r 2  1 r 2   2     2 r 2  r 2  1   r 2 
2 2
F0
 

Figure 5.19, 20, 21 Nondimensional amplitudes of vibration of the

primary mass as a function of the frequency ratio for several values of
the frequency ratio = 𝜔 /𝜔 with the damping ratio as a parameter, and a
mass ratio  = 0.25
(Note that three curves intersect with each other at two points)

=1.0 =0.9 =0.8

Fig. 5.19 Fig. 5.20 Fig. 5.21

Xk 4 2 r 2  ( r 2   2 ) 2
 f (r ,  ,  ,  )  2
(5.37)
F0 4 2 r 2 (r 2   r 2  1) 2   (r 2  1)(r 2   2 )   r 2  2 

2
 Matlab commands for
Dimensionless amplitude of primary mass
Xk 4 2 r 2  (r 2   2 ) 2
 f (r ,  ,  ,  )  2
F0 4 2 r 2 (r 2   r 2  1) 2   (r 2  1)(r 2   2 )   r 2  2 

% Matlab commands to plot nondimensional amplitude of primary mass

% for vibration absorber
r=linspace(0.5,1.5,500);
ze=[0.01;0.05;0.1;0.20;0.50];be=1.0;mu=0.25
num=(2*ze*r).^2+ones(size(ze))*(r.*r‐be*be).^2
den1=ones(size(ze))*(‐1+(1+mu)*r.*r).^2.*(2*ze*r).^2
den2=ones(size(ze))*(mu*be*be*(r.*r)‐(r.*r‐1).*(r.*r‐be*be)).^2
X=sqrt( num ./ ( den1+den2 ));
figure(1)
plot(r,X)% in linear scale
plot(r,20*log10(X))% in dB scale

Magnitude of primary mass vibration in dB scale

(for 0.01, 0.05, 0.1, 0.2, 0.5, =0.25 & =1.0)
50
plot(r,20*log10(X)) Xk
40 r 
F0

 2 r   r2   2 
2 2
30

dB(X)=20*log10(X)

r     r 2  2   r 2  1 r 2   2  
2
2 2
 2 r 
2 2
1  r
20

10

-10

-20

-30
0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5
Region of absorption
As damping increases, the frequency range of absorption decreases, but the resonance peaks come down.
‐ Note that curves for different damping ratios cross with each other at two invariant points.
‐ Can you derive the location of the two points ?
6

3
 120

100

80

60

40

20

0
0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5
Plot in linear scale:
Do you see differences among curves better in comparison with the plot in dB scale?

How to get two invariant points?

50

 2 r    r    2 2
2 2 2
 Xk  40

  
 F0   2 r   r 2  1   r 2     r 2  2   r 2  1 r 2   2  
2 2 2 30

  20

  r 2   2   =1
2
 10

4r 2  2  2 
 4r 
0

 -10
   r 2  2   r 2  1 r 2   2   
2

4r 2  r 2  1   r 2   2   
2 -20
 𝑟 𝑟
 4r  r  1   r 
2 2 2 2
 -30
  0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5

r2   2   r 2  2   r 2  1 r 2   2  
 When   1 𝛽 𝜇𝛽 2𝛽
2r 2r  r 2  1   r 2  𝑟 2
2 𝜇
𝑟
2 𝜇
0

r2 Only plus sign (+) is valid for 2 positive values of rP and rQ

 : Independent of 
r 2  r 2 1  r 2 
2

 
2 Xk
 r  rP , rQ   1  1 r 2  1

1
 F     r rP ,rQ
  1    r 2  1 
0

 
8

4
 Den Hartog’s Optimal Tuning
Prof. Den Hartog suggested that the most favorable case should be such that two fixed
points are adjusted to equal values of the dimensionless amplitude by a proper choice of 𝜷.
𝑿𝒌 𝑿𝒌
Optimum Tuning: 𝒓𝑷 𝒓𝑸
𝑭𝟎 𝑭𝟎
Xk
 r  rP , rQ   1  1 r 2  1
F0     r rP ,rQ
1 𝜇 𝑟 1= 1 𝜇 𝑟 1

for minus sign 𝑟 +𝑟 =

1 𝛽 𝜇𝛽 2𝛽
from 𝑟 2 𝑟 0
2 𝜇 2 𝜇
𝑟 +𝑟 2 Optimal tuning
condition

2 =

Optimum Tuning 𝜔 1
∴𝛽
𝜔 1 𝜇

𝑟 𝑟 = 1
9

Den Hartog’s Optimal Tuning

Find the original article written by Prof. Den Hartog (uploaded in KLMS)

10

5
 With damping in the absorber

• The absorber is less sensitive to variation of the

excitation frequency.
• The peaks at resonances are reduced.
• But, complete absorption (i.e. X=0) of the excitation at
one given frequency is not possible.
• It becomes a design problem to pick the most
favorable values of , , and .

11

Matlab commands for

Nondimensional Amplitude of Absorber Mass, ma
4 2
1 r2
Xak  4
r   (B)

r     r 2  2   r 2  1 r 2   2  
2
 2 r  2 2
2
F0 2
1  r
% Matlab commands to plot normalized amplitude of absorber mass
r=linspace(0.5,1.5,500);
ze=[0.01;0.05;0.1;0.20;0.50];be=1;mu=0.25
num=1+(2*ze*r).^2/be.^4
den1=ones(size(ze))*(‐1+(1+mu)*r.*r).^2.*(2*ze*r).^2
den2=ones(size(ze))*(mu*be*be*(r.*r)‐(r.*r‐1).*(r.*r‐be*be)).^2
X=sqrt( num ./ ( den1+den2 ));
figure(1)
plot(r,X)% in linear scale
plot(r,20*log10(X))% in dB scale

12

6
 Magnitude of absorber mass vibration
(for =0.01, 0.05, 0.1, 0.2, 0.5, =0.25 & =1.0)
• In the operational range, the absorber mass has relatively large motion.
• Deflection limits of the absorber mass must be taken care of!!
50 300

40 250

30 200

20 150

10 100

0 50

-10 0
0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5

dB scale
Range of operation
linear scale
13

Viscous Vibration Absorber

• Applied often to rotating machines
x(t) xa
k • Consisting of fluid damper and
m ma inertia only without absorber stiffness)
ca • Called Houdaille damper
ka  (or Lanchester damper)
a   0,   a  0
ma p

14

7
 Houdaille (or Lanchester) Damper

• Equation of motion:

 J1 0  1   ca ca  1   k 0  1   M 0  jt

      (C)
0 J 2  2   ca ca  2   0 0   2   0 
e


15

Figure 5.24 The amplitude curves for a system with a viscous

absorber for the case μ = 0.25 and three different values of ζ.

Xk 4 2 r 2  (r 2   2 ) 2 
Recall :  f (r ,  ;  ,  )  2
(5.37) a 
ka
 0,   a  0 (D)
F0 4 r (r   r  1) 2  (r 2  1)(r 2   2 )   r 2  2 
2 2 2 2 ma p

1k 4 2  r 2
f (r ,  ;  )   (5.39) (E)
M0 4 2 ( r 2   r 2  1) 2  (r 2  1) 2 r 2
where   ca /2J 2 p  ca /2ma p , r   /  p

16

8
 How to get the invariant point?
2
 Xk  4 2  r 2
  
4 2  r 2  1   r 2   r 2  r 2  1
2 2
 F0 
 r2 
4 2  
 4

 r 2  r 2  1 
2
2  
4  r 2  1   r 2   2  
 4  r 2
 1   r  
2 2

r 2 1 2
 When 2  1: r 2 
r 1 r 2 2   For only minus sign
2
 2
 1  
  
 Xk   2 2
 rP ,    ,1  
 F0   2    

17

Nondimensional Amplitude of Primary Mass

in Lanchester Damper
(for =0.01, 0.05, 0.1, 0.2, 0.5, =0.25 & =0)
50

40

30
Xk 4 2  r 2
r  
4  r  1   r 2   r 2  r 2  1
2 2 2 2
F0 20

10

0  Xk   2 2
 rP ,    ,1  
 F0   2   
-10
0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5

Note that curves for different damping ratios cross with each other at one
invariant point. Which curve is most preferable? Please find the optimal
damping ratio and email me including the
18
solution. (In‐Class Quiz by May 23)