APPLICATION OF PARTICLE SWARM OPTIMIZATION TO MICROWAVE TAPERED MICROSTRIP LINES
Three Mass Spring System Amplitude Reduction
1. ME 7820 Group Project
Group No- 6
Submitted by-
Praveen Kumar Kumaresan (FX1628)
Harshada Patil (FX4157)
Sathish Kumar Manjulaushankar (FX6865)
Sushma Mallula (FT5826)
Viswa Sai Manoj Mungara (FZ4883)
Chayan Mishra (FX 5634)
Guided By-
Dr. Emmanuel O. Ayorinde
2. THREE MASS, FOUR SPRING VIBRATORY SYSTEM, WITH AN ATTACHED
VIBRATION ABSORBER WITH A SMALL SPRING AND MASS SYSTEM
M1 = m = 1; K1 = k= 1 (appropriate units)
F_2016 7820 PROJECT GUIDANCE
1.Analyze the horizontal, longitudinal vibration motion of your system (without the appended “absorber”, the
auxiliary system Ma-Ka), assuming frictionless translation, and computing all resonance frequencies.
2.Calculate, Sketch and Anotate (label with values) all the mode shapes.
3. Start the design to obtain values of Ma and Ka that would yield an x2 value (i.e. the motion amplitude of mass
M2) equal to or nearly equal to zero, by trying three sizes (low, medium, large) of values for the auxiliary system,
and reiterating your analyses until you satisfy the project goal (mass and stiffness values that would give us as
near-zero as possible relative amplitude value for M2) for the whole system. In each case of iteration, record the
magnitude of the relative amplitude of mass M2, to be shown as part of your report.
4.Do you think a proper sizing of the auxiliary system can change the amplitude of vibration of the mass to which
the “absorber” system is attached, i.e. M2 here? Give logical, technical reasons.
5.What sizes of Ka and Ma should we use to reduce the amplitude of M2 to practically zero?
6.Comment on the whole exercise and what you think may be learnt from it.
Parameters for Group 6- M2 =1.5m; M3 = 2m; K2 = k; K3 = 2k ; K4 = 3k (Given)
Make a column for these entities.
3. PART-1
The free body diagrams shown below of each of the masses in the mass-spring system shown above.
(Mass-I)
(Mass-II)
(Mass-III)
(Ma-Ka)
Now the system with considering the auxiliary mass using free body diagram analysis, we could set
up the following equation to solve for the spring coefficient matrix below.
a)𝑚1 𝑥̈1= −𝑘1 𝑥1+ 𝑘2 𝑥2− 𝑥1 b) 𝑚2 𝑥̈2= −𝑘2 𝑥2− 𝑥1 + 𝑘3 𝑥3− 𝑥2 + 𝑘 𝑎 𝑥 𝑎− 𝑥2
c)𝑚3 𝑥̈3= −𝑘4 𝑥3− 𝑘3 𝑥3− 𝑥2 d) 𝑚 𝑎 𝑥̈ 𝑎= − 𝑘 𝑎 𝑥 𝑎− 𝑥2
Shifting the variables around, we get the following,
𝑚1 𝑥1+ 𝑘1+ 𝑘2 𝑥1− 𝑘2 𝑥2 = 0 𝑚2 𝑥2− 𝑘2 𝑥1+ 𝑘2+ 𝑘3+ 𝑘 𝑎 𝑥2− 𝑘3 𝑥3 − 𝑘 𝑎 𝑥 𝑎 = 0
𝑚3 𝑥3− 𝑘3 𝑥2+ 𝑘3+ 𝑘4 𝑥3 = 0 𝑚 𝑎 𝑥 𝑎− 𝑘 𝑎 𝑥2+ 𝑘 𝑎 𝑥 𝑎= 0
Now the above equations into matrix form, we obtain the mass and spring coefficient matrices below.
4. [ 𝑚]=[
1 0 0 0
0 1 0 0
0
0
0
0
2 0
0 𝑚4
]
[ 𝑘]=[
𝑘1 + 𝑘2 𝑘2 0 0
−𝑘2 𝑘2 + 𝑘3 −𝑘3 −𝑘𝑎
0
0
−𝑘3
−𝑘𝑎
𝑘3 + 𝑘4 0
0 𝑘𝑎
]
The assigned values of mass and spring stiffness are unity. Therefore, 𝑚=𝑘 =1.
The auxiliary mass and auxiliary spring stiffness equal to zero. The results are shown below. Please
note that lambda is the eigenvalues and omega is the resonant frequencies of the three degree of freedom
system.
Now the auxiliary mass from the system is removed, we have considered the spring stiffness and mass
matrix and set the auxiliary mass and auxiliary spring stiffness equal to zero. Doing this, we have obtain
the following spring stiffness and mass matrices.
[𝒌]=[
𝑘1 + 𝑘2 𝑘2 0
−𝑘2 𝑘2 + 𝑘3 −𝑘3
0 −𝑘3 𝑘3 + 𝑘4
]
[𝑘]=[
2 1 0
−1 3 −2
0 −2 5
]
The mass matrix are as follows
[𝑚]=[
1 0 0
0 1.5 0
0 0 2
]
Using the mass matrix and the spring coefficient matrix, the determinant function is used to solve
for lambda utilizing the equation below.
|[𝑲]− 𝜆[𝑴]|=𝟎
5. By substituting the mass and spring coefficient matrices, the following equation is obtained.
[{
2 1 0
−1 3 −2
0 −2 5
} − λ {
1 0 0
0 1.5 0
0 0 2
}]=[
0 0 0
0 0 0
0 0 0
]
We could utilize MATLAB to solve for the eigenvalues which represent the resonant frequencies of
the system. Below are eigenvalues of the three degree of freedom system.
𝜆=[
λ1
λ2
λ3
]=[
0.7255
2.1622
3.6122
]
The resonant frequencies are found by applying the equation below.
𝜆= 𝜔2
The resonant frequencies of the three degree of freedom system are shown below.
𝝎=[
ω1
ω2
ω3
]=[
0.8517
1.4704
1.9005
]
Therefore, the eigenvectors show in the below matrix for the three degree of freedom system.
[
−0.4731 −0.8163 0.3115
−0.6029 0.1324 −0.5344
−0.3398 0.3920 0.4805
]
6. PART-2
The mode shapes of the three degree of freedom mass-spring system shown below.
I. First Mode Shape-
II. Second Mode Shape-
7. III. Third Mode Shape-
PART-3
For low, middle, and high values we assigned to the auxiliary system are as follows,
Low Middle High
m 7 8 9
k 1.0 2.0 3.0
The results are as follows:
1) For low values,
V=
0.0766 0.5583 0.7655 0.3106
0.1466 0.5082 -0.1394 -0.6062
0.0607 0.3604 -0.4383 0.4176
0.3693 -0.0767 0.0098 0.0227
8. 2) For Middle values,
V=
0.1029 0.6516 0.6983 0.2777
0.1950 0.4068 -0.1438 -0.6652
0.0814 0.3618 -0.4892 0.3509
0.3390 -0.0903 0.0184 0.0401
3) For Maximum values,
V=
0.1125 0.7386 0.6192 0.2416
0.2127 0.3193 -0.1440 -0.7062
0.0890 0.3425 -0.5384 0.2914
0.3169 -0.0862 0.0253 0.0513
PART-4
Many a times, a vibratory system under forced vibration is required to run near resonance i.e. the
excitation frequency is close to the natural frequency of the system. Under such a situation, the response
of the system can be large and we must try to reduce it by taking some measure. By attaching a separate
smaller spring-mass system, an auxiliary system, to the main system the vibration of the main system
can be reduced, drastically, if the mass and the stiffness of the auxiliary system are properly calculated,
i.e. if the auxiliary system is tuned to the natural frequency of the main system and the excitation
frequency.
Adding stiffness and mass to the system, results in change of resonant frequencies and mode shapes of
the system. The reason that we add stiffness to the auxiliary system is to dampen the amplitude of the
second mass. Normally, we are concerned with the first mode, so the objective here is to find an
optimized combination of mass and stiffness of the auxiliary system to reduce the amplitude of mass 2
in the first mode.
10. OBSERVATION:
Value is never zero because there is no damper as we increase mass the value approaches zero but will
get to zero.
In all the three graphs it can be clearly observed that as the stiffness constant increases the amplitude
also increases but here we are more inclined towards finding the right combination which decreases the
amplitude to zero. Even though the amplitude cannot be zero totally we would rather choose the
combination where it approaches to closer to zero.
If we see the trend in the graph, the amplitude of M2 will never be zero but it can get as close to zero
which is 0.02 in this case. So the best combination for lowest amplitude possible is Ma=9.5 & Ka=0.1
and for which the amplitude is 0.25.
PART-6
Amplitude is equation given by
Hypothetically, the reaction of an un-damped framework relies on upon the initial displacement, speed
and natural frequency of the framework. As the initial two terms are not considered in this work, the
main thing influencing the reaction is the regular frequencies of the framework. Expanding mass of
assistant framework does not really bring about decreasing the amplitude of mass 2. Moreover,
expanding the solidness does not roll out any improvement to framework from a specific esteem. The
best mix to diminish the adequacy of mass 2 happens when the mass is 9.5 and stiffness 0.25. It is sure
that expanding mass and firmness all the while doesn’t influence the normal frequency. Evidently in
this issue, impacts of mass are far more than the impact of firmness. That is to state, this framework is
likely to change of mass of helper framework as opposed to the solidness. Since it was watched that
11. after a specific esteem for stiffness of helper framework, there is no adjustment in the sufficiency of
the second mass.
Appendix (MATLAB Code)
clear all
m1=1; m2=1; m3=2; k1=1; k2=1; k3=2; k4=3;
for ka=1:10
ma1=0.75;
K= [k1+k2 -k2 0 0; -k2 k2+k3+ka -k3 -ka; 0 -k3 k3+k4 0; 0
-ka 0 ka];
M= [m1 0 0 0; 0 m2 0 0; 0 0 m3 0; 0 0 0 ma1];
[V1,D1]=eig(K,M);
X(ka)=V1(2,1);
disp(V1);
disp(D1);
end
subplot(2,2,1);
scatter(1:10,X);
grid on;
grid minor;
xlabel('Spring Stiffness');
ylabel('Eigenvector X2');
title('Mass= 0.75');
for ka=1:10
ma1=0.25;
K= [k1+k2 -k2 0 0; -k2 k2+k3+ka -k3 -ka; 0 -k3 k3+k4 0; 0
-ka 0 ka];
M= [m1 0 0 0; 0 m2 0 0; 0 0 m3 0; 0 0 0 ma1];
[V2,D2]=eig(K,M);
X(ka)=V2(2,1);
disp(V2);
disp(D2);
end
subplot(2,2,2);
scatter(1:10,X);
grid on;
grid minor;
xlabel('Spring Stiffness');
ylabel('Eigenvector X2');
title('Mass= 0.25');
for ka=1:10
