Dimensionless variables of the differential equation
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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
19 Comments
Torsten
on 6 Oct 2024
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.
Accepted Answer
Sam Chak
on 7 Oct 2024
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:
3 Comments
Sam Chak
on 7 Oct 2024
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.
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