Modeling of square wave

Hello everyone,

I am modeling the position varying gear mesh stiffness function with the following code inside ODE solver.

z(k) = mod(theta_p,2*pi/t_np);

if z(k)>=(2-c)*2*pi/t_np

k_m(k) = k_max;

else

k_m(k) = k_min;

end

where c, k_max, k_min, t_np are the constants:

k_min = 5e7/2; %minimum stiffness;

k_max = 5e7; %maximum stiffness;

c = 1.63; %engagement rate;

t_np = 25; %number of teeth on pinion;

The goal of this function is to make a square wave with t_np number of maximums over 2*pi interval.

This is how the time/varying gear mesh stiffness should look like:

When the gear have a fault, the stiffness of the faulty gear tooth is compromized, so when the broken tooth is in contact, the stiffness will decrease. I am struggling to figure out how to make the value of one of the minima less than others on each 2*pi intervals. Any suggestions highly appreciated.

Thank you!

[Steve Miller]

If you are interested in modeling gears with faulty behavior, you may be interested in the Simple Gear block in Simscape Driveline.

--Steve

Answer 1

[Abhimenyu]

Hi,

I understand that you want to modify the gear mesh stiffness function to include a fault that decreases the stiffness of one tooth in each 2π interval. For this, you can introduce a condition to identify the faulty tooth and adjust its stiffness accordingly. Please follow this example MATLAB code :

% Constants
k_min = 5e7 / 2; % minimum stiffness
k_max = 5e7;     % maximum stiffness
c     = 1.63;    % engagement rate
t_np  = 25;      % number of teeth on pinion

% Faulty tooth index
faulty_tooth_index = 5; % example: 5th tooth is faulty

faulty_tooth_index: This variable specifies which tooth is faulty. In this example, the 5th tooth is considered faulty.

faulty_tooth_stiffness = k_min / 2; % stiffness of faulty tooth (you can adjust this value)

faulty_tooth_stiffness: This specifies the stiffness of the faulty tooth. You can adjust this value to reflect how much the stiffness decreases.

% ODE solver function
%theta_p as given in the diagram shared
for k = 1:length(theta_p)
    z(k) = mod(theta_p(k), 2 * pi / t_np);
    tooth_index = floor(theta_p(k) / (2 * pi / t_np)) + 1; % find current tooth index

tooth_index: This is calculated to determine which tooth is currently engaged based on the angle θp\theta_p​.

    if tooth_index == faulty_tooth_index
        k_m(k) = faulty_tooth_stiffness;
    elseif z(k) >= (2 - c) * 2 * pi / t_np
        k_m(k) = k_max;
    else
        k_m(k) = k_min;
    end
end

I hope this answers your query!