Dimensionless variables of the differential equation

Question

Yu 2024-10-5

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编辑： Yu 2024-10-7

Hello all,

I hope you are all doing well. I have established a 2DOF differetial equation. Basically, the equations are the 2 spring-mass-damper system. But I am wondering how to define the variable X to be a dimentionless variable, such as X/L. In my equations, the variable X should be dimensionless but when I exclude the force (F_o the dimensionless force), the results of the equation are not dimensionless which I have already checked. My codes are attached. I think you can easily run it.

I appreciate your support.

Best wishes,

Yu

clc;
clear all;
tspan = 0:0.25:200;
% Units:
% stiffness: N/m, mass:kg, damping(c_1):Ns/m
% displacement(X):m
% alpha (stiffness ratio):k_1/k_2
% gamma (dimension ratio):a/L (structure length)
% f_opt (frequency ratio): f_1/f_2
X0 = [0 0 0 0];
%parameters-------
mu = 0.02; % mass ratio
f_opt = 1/(1+mu); %frequency ratio
xi_2 = sqrt(3*mu/8*(1+mu));
Omega_1 = 0.188; % 0.03 Hz k_1^2/m_1^2 (stiffness^2/mass^2)
Omega_2 = Omega_1*f_opt; % frequency of the second DOF k_2^2/m_2^2 (stiffness^2/mass^2)
xi_1 = 0.01; % damping ratio of first DOF xi_1=c_1/2*m_1*Omega_1
%matrix--------
M = [1+mu mu;
     1     1];
C = [2*xi_1*Omega_1 0;
     0         2*xi_2*Omega_2];
K = [Omega_1^2 0;
     0         Omega_2^2];
% state space model-------------------
% M: mass matrix, K:stiffness matrix, C:damping matrix
O = zeros(2,2);
I = eye(2);
A = [O I; -inv(M)*K -inv(M)*C];
B = [O; inv(M)];
E = [I O];
D = zeros(2,2);
% solve the equations-------------------
options = odeset('RelTol',1e-10,'AbsTol',1e-10);
[t,X] = ode45(@(t,X) QZSdamper(t,A,B,X),tspan,X0,options);
x = X(:,1:4);
% F_o = 2*Omega_2.^2.*alpha.*(sqrt(1-gamma.^2)+X(:,2)).*(-1+sqrt(1/(X(:,2).^2+2.*sqrt(1-gamma^2).*X(:,2)+1)));
% 
% f_s = Omega_2.^2*(X(:,2)-2*alpha*(sqrt(1-gamma^2)+X(:,2))*(-1+sqrt(1/(X(:,2).^2+2*sqrt(1-gamma^2)*X(:,2)+1))));
% 
% f_1 = f_s(:,1);
% 
% figure,
% plot(t,f_1);
figure,
plot(t,x(:,1)); xlabel('Time/s'),ylabel('Dimensionless displacement of primary structure')
figure,
plot(t,(x(:,2)))
sys = ss(A,B,E,D);
figure,
bodeplot(sys(1,1)) % from input 1 to output 3
bp.FrequencyScale = "linear";
[mag,phase,wout] = bode(sys(1,1));
mag = squeeze(mag);
phase = squeeze(phase);
fout = wout/(2*pi);
BodeTable = table(fout,mag,phase);
function dXdt = QZSdamper(t,A,B,X)
mu = 0.025;
alpha = 1;
gamma = 2*alpha/(2*alpha+1);
f_opt = 1/(1+mu);
Omega_1 = 0.188; % 0.03 Hz
Omega_2 = Omega_1*f_opt;
w = 0.180;
F = 0.001*sin(w*t);
F_o = 2*Omega_2^2*alpha*(sqrt(1-gamma^2)+X(2))*(sqrt(1/(X(2)^2+2*sqrt(1-gamma^2)*X(2)+1))-1);
% F_o = Omega_2^2*(sqrt(1-gamma^2)+2*alpha*(X(2))*(1/((X(2)-sqrt(1-gamma^2))^2+2*sqrt(1-gamma^2)*(X(2)-sqrt(1-gamma^2))+1)^(1/2)-1))
F = [F;F_o];
dXdt = A*X+B*(F);
end

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Sam Chak 2024-10-6

在 MATLAB Online 中打开

Hi @Yu

I ran the code, and the dimensionless displacement becomes unstable due to disturbance from the oscillating force (in QZSdamper). For your information, F_o can stabilize the system but is insufficient to handle the oscillating force.

What is the expected outcome?

tspan = 0:0.25:200;

% Units:

% stiffness: N/m, mass:kg, damping(c_1):Ns/m

% displacement(X):m

% alpha (stiffness ratio):k_1/k_2

% gamma (dimension ratio):a/L (structure length)

% f_opt (frequency ratio): f_1/f_2

X0 = [0; 0; 0; 0];

% X0 = [1; 0; 0; 0]; % test if F_o can stabilize when sinusoidal force, F = 0.

% Parameters-------

mu = 0.02; % mass ratio

f_opt = 1/(1 + mu); % frequency ratio

xi_2 = sqrt(3*mu/8*(1 + mu));

Omega_1 = 0.188; % 0.03 Hz k_1^2/m_1^2 (stiffness^2/mass^2)

Omega_2 = Omega_1*f_opt; % frequency of the second DOF k_2^2/m_2^2 (stiffness^2/mass^2)

xi_1 = 0.01; % damping ratio of first DOF xi_1=c_1/2*m_1*Omega_1

% Matrices

% M: mass matrix,

M = [1+mu, mu;

1 , 1];

% C: damping matrix

C = [2*xi_1*Omega_1, 0;

0 , 2*xi_2*Omega_2];

% K: stiffness matrix,

K = [Omega_1^2, 0;

0 , Omega_2^2];

% state space model-------------------

O = zeros(2,2);

I = eye(2);

A = [O, I;

-inv(M)*K, -inv(M)*C]

A = 4×4

0 0 1.0000 0 0 0 0 1.0000 -0.0353 0.0007 -0.0038 0.0006 0.0353 -0.0347 0.0038 -0.0329

ev = eig(A) % eigenvalues, stable but easily becomes unstable when disturbed

ev =

-0.0099 + 0.1972i -0.0099 - 0.1972i -0.0085 + 0.1753i -0.0085 - 0.1753i

B = [O;

inv(M)];

E = [I O];

D = zeros(2,2);

% solve the equations-------------------

options = odeset('RelTol', 1e-10, 'AbsTol', 1e-10);

[t, x] = ode45(@(t, X) QZSdamper(t, A, B, X), tspan, X0, options);

% x = X(:,1:4);

% F_o = 2*Omega_2.^2.*alpha.*(sqrt(1-gamma.^2)+X(:,2)).*(-1+sqrt(1/(X(:,2).^2+2.*sqrt(1-gamma^2).*X(:,2)+1)));

%

% f_s = Omega_2.^2*(X(:,2)-2*alpha*(sqrt(1-gamma^2)+X(:,2))*(-1+sqrt(1/(X(:,2).^2+2*sqrt(1-gamma^2)*X(:,2)+1))));

%

% f_1 = f_s(:,1);

%

% figure,

% plot(t,f_1);

figure,

plot(t, x(:,1)); grid on

xlabel('Time/s'),

ylabel('Dimensionless displacement of primary structure')

% figure,

% plot(t,(x(:,2)))

% sys = ss(A,B,E,D);

% figure,

% bodeplot(sys(1,1)) % from input 1 to output 3

% bp.FrequencyScale = "linear";

%

% [mag,phase,wout] = bode(sys(1,1));

%

% mag = squeeze(mag);

% phase = squeeze(phase);

% fout = wout/(2*pi);

% BodeTable = table(fout,mag,phase);

function dXdt = QZSdamper(t,A,B,X)

mu = 0.025;

alpha = 1;

gamma = 2*alpha/(2*alpha+1);

f_opt = 1/(1+mu);

Omega_1 = 0.188; % 0.03 Hz

Omega_2 = Omega_1*f_opt;

w = 0.180;

% oscillating force

F = 0.001*sin(w*t);

% Unknown force (comes from somewhere)

F_o = 2*Omega_2^2*alpha*(sqrt(1 - gamma^2) + X(2))*(sqrt(1/(X(2)^2 + 2*sqrt(1 - gamma^2)*X(2) + 1)) - 1);

% F_o = Omega_2^2*(sqrt(1-gamma^2)+2*alpha*(X(2))*(1/((X(2)-sqrt(1-gamma^2))^2+2*sqrt(1-gamma^2)*(X(2)-sqrt(1-gamma^2))+1)^(1/2)-1))

F = [F; % same F = F? bad coding practice from Engineering perspective

F_o]; % ?

% F = [0; % no force to test stability

% F_o]; % no force

dXdt = A*X + B*F;

end

Yu 2024-10-6

编辑：Yu 2024-10-6

@Sam Chak @Torsten Hi, thank you for your comments. I think I would explain my model in more details. I am not sure whether you have time to review it. But I suppose they could be helpful.

My model is the 2-DOF system, including the primary structure and the damper (seen below)

Therefore, the equations:

f_o is the nonlinear stiffness force from oblique spring the strcuture below. k_tmd i from the vertical spring. So if f_o is removed, it will become the tuned mass damper not QZS.

Fig.1

The above structure (Fig.1) can be expressed by (k_v vertical spring stiffness, k_o oblique spring)

Then, the first eqn is divided by m_1, and second is divided by m_tmd. The eqns are now:

Actually, the above equations are checked in symbolic tools. As the force generate by the QZS (Fig.1) is related to the dimensional parameters, therefore, I just want to use the ratios as I mentioned in the questions. I do not want to use the specific numbers at this moment.

Regarding the unstable problem, I removed the f_o and the tspan is 1000. The system are stable? Therefore, I guess my expression may be correct? Also if you see the bodeplot, there is a peak reduction I believe and the structural damping ratio is also low.

For employing F_o (pic below), it seems it not that stable. I guess the stiffness has a 'zero' point that there is a position the total stiffness is 0. For the equations of the F_o, I also verified. It seems it is correct.

The final question is that I am not sure the results of the displacment is dimensionless or not. Especially when removing F_o.

Torsten 2024-10-6

@John D'Errico

Again, force would have units of mass*distance/time^2.

Yes. Dividing each equation by m and the constant L/T^2 (which results from the non-dimensionalization of displacement and time by x~(t~) = 1/L * x(t) ) makes force dimensionless.

Yu 2024-10-6

@Torsten I am really sorry that I have presented a mess description of my questions. I made some screenshots from word.doc last night when I got the comments from you.

I am aware that it may waste your time to undestand the mess variables and apologize for this.

I have editted my previous reply. I also clarify here

F(t) is the harmonic excitation on the m_1. It should be [f(t) 0].
Yes, F(t) is asin(omega*t) --- the harmonic excitation
f_o = . This also is modified in my previous reply.
Yes, x_tmd = x_2
I just nondimensionalize the nonlinear stiffness force f_o in the second equation. But I have no clue about how to make the first equation nondimensionalized. Additionally, I am not sure I nondimensionalize the second equation successfully.
Yes I need to present the clear description of my question to the professor. It was my mistake.

Again, I am sorry for the inconvenience and thank you for your patient help.

Best wishes,

Yu

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Answer 1

Sam Chak 2024-10-7

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2157595-dimensionless-variables-of-the-differential-equation#answer_1527900

Hi @Yu

The aim is to render the entire dynamical system dimensionless (by finding the dimensionless time derivative), rather than focusing solely on a single state. I didn't study your tuned mass damper with with quasi-zero-stiffness (QZS), so I provide a relatively simple example below:

Example:

Consider the following example, which presents a simplified set of coupled dynamic equations for a tethered space system (TSS):

where

l is the tether length (unit: m),
θ is the libration angle (unit: radians, but considered dimensionless),
T is the tether tension (unit: kg·m/s²)
ω is the orbital angular velocity (a constant) of TSS (unit: rad/s²), and
is the mass equivalent (a constant), analogous to adding two "parallel resistors"; the mothership and the tethered probe (unit: kg).

To non-dimensionalize the dynamics of tether length, consider making the tether tension dimensionless by dividing the entire equation by a constant,

. Note that

is a nominal tether length chosen to satisfy a physical constraint.

If the following dimensionless quantities are defined:

dimensionless tension,
dimensionless length, , such that
dimensionless time,
dimensionless time derivative,

then the dimensionless form of the TSS dynamic equations can written as:

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Sam Chak 2024-10-7

Hi @Yu

Please do not confuse the TSS equations with your TMD equations; they are two distinct systems, though both share Newton's

. I reviewed the "scattered equations" (which was very time-consuming), but I did not find the nominal tether length or the orbital angular velocity in your work. You should determine your own nominal displacement and the angular velocity of the TMD system.

I suggest consolidating all equations into a single post to enhance readability. This approach will also facilitate your explanation of the modifications made to the equations and make it easier for others to review them.

Yu 2024-10-7

编辑：Yu 2024-10-7

@Sam Chak Hi Sam,

Many thanks for your suggestions. And I am sorry that the unclear desciption have taken your much time. I have posted another questions with the description of my system. I guess it is clearer.

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