UnderDamped Free SDOF System

%%%%%%%%%%%%%%%% Equation of Motion: Damped SDOF System %%%%%%%%%%%%%%%%%%%

%%%%%% Example Problem - 2.5 (Dynamics of Structures - Ashok K.Jain) %%%%%%

clc;

clear all;

close all;

%% INPUTs:

m = 5*10^3; % Lumped Mass (kg)

k = 10^5; % Stiffness (N/m)

c = 0.05; % Damping Coefficient

x0 = 0.02; % Initial Displacement (in m)

v0 = 0.05; % Initial Velocity (in m/s)

%% OUTPUTs:

z = c/m;

wn = sqrt(k/m) % Natural Circular Frequency (rad/s)

f = wn/(2*pi()) % Natural Cyclic Frequency (Hertz-Hz)

T = 1/f % Fundamental Time-Period (sec)

Z = c/(2*m*wn) % Damping Ratio

wd = wn*(sqrt(1-Z^2)) % Damped Frequency

A = sqrt((x0^2) + (v0/wn)^2) % Amplitude (m)

vm = A*wn % Maximum Velocity (m/s)

am = vm*wn % Maximum Acceleration (m/s/s)

Phi = atand(x0*wn/v0) % Phase Angle (in degree)

p = [1 z wn^2];

roots(p)

syms X(t)

E = diff(X,t,2) + diff(X,t)*z + (wn^2)*X == 0;

x = dsolve(vpa(E)) % C1 & C2 are constant and can be determined by BCs

dX = diff(X,t);

conds =[X(0)==x0,dX(0)==v0];

x = dsolve(vpa(E),conds)

xD = rad2deg(x)

%% Plots:

fplot(xD,[0 5],'k','LineWidth',1.25);

xlabel('displacment (in m)');

ylabel('time (t)');

title('Displacement Response Curve');

Problem-1: The graph obtained through this code is looks similar to undamped system, while this is a code of undamped system and has to degrad with each cycle?

Answer 1

[Manikanta Aditya]

Hey,

The code you’ve written is for a damped system, but the damping coefficient c you’ve chosen is quite small (0.05). This means the system is lightly damped and the response will look similar to that of an undamped system. In a damped system, the amplitude of oscillation decreases over time, which is known as decay.

The rate of this decay is determined by the damping ratio Z. In your case, the damping ratio is calculated as Z = c/(2*m*wn), which turns out to be a small value due to the small damping coefficient. Therefore, the decay of oscillation is slow and might not be clearly visible within the first few cycles.

If you want to see a noticeable decay in the oscillation, you could increase the damping coefficient c. This will increase the damping ratio Z, leading to a faster decay of the oscillation amplitude. Try adjusting these parameters and observe the changes in the system response.

%%%%%%%%%%%%%%%% Equation of Motion: Damped SDOF System %%%%%%%%%%%%%%%%%%%
%%%%%% Example Problem - 2.5 (Dynamics of Structures - Ashok K.Jain) %%%%%%
clc;
clear all;
close all;
%% INPUTs:
m = 5*10^3;     % Lumped Mass (kg)
k = 10^5;       % Stiffness (N/m)
c = 500;        % Increased Damping Coefficient
x0 = 0.02;      % Initial Displacement (in m)
v0 = 0.05;      % Initial Velocity (in m/s)
%% OUTPUTs:
z = c/m;   
wn = sqrt(k/m);               % Natural Circular Frequency (rad/s)
f = wn/(2*pi());              % Natural Cyclic Frequency (Hertz-Hz)
T = 1/f;                      % Fundamental Time-Period (sec)
Z = c/(2*m*wn);               % Damping Ratio
wd = wn*(sqrt(1-Z^2));        % Damped Frequency
A = sqrt((x0^2) + (v0/wn)^2); % Amplitude (m)
vm = A*wn;                    % Maximum Velocity (m/s)
am = vm*wn;                   % Maximum Acceleration (m/s/s)
Phi = atand(x0*wn/v0);        % Phase Angle (in degree)
p = [1 z wn^2];
roots(p)

ans = 
  -0.0500 + 4.4719i
  -0.0500 - 4.4719i

syms X(t)   
E = diff(X,t,2) + diff(X,t)*z + (wn^2)*X == 0;
x = dsolve(vpa(E));   % C1 & C2 are constant and can be determined by BCs
dX = diff(X,t);
conds =[X(0)==x0,dX(0)==v0];
x = dsolve(vpa(E),conds); 
xD = rad2deg(x);
%% Plots:
fplot(xD,[0 5],'k','LineWidth',1.25);
xlabel('displacment (in m)');
ylabel('time (t)');
title('Displacement Response Curve');

In this code, I’ve increased the damping coefficient c from 0.05 to 500. This should make the damping more noticeable in the system response.

Hope it helps!

[Sam Chak]

@Parvesh, It should be noted that the computed damping ratio (ζ or Z) is extremely small at the scale of . Thus, it probably takes millions of seconds to observe the damping effect. What is the unit for the parameter c?

m   = 5*10^3;           % Lumped Mass (kg)
k   = 10^5;             % Stiffness (N/m)
c   = 0.05;             % Increased Damping Coefficient
wn  = sqrt(k/m);        % Natural Circular Frequency (rad/s)
Z   = c/(2*m*wn)        % Damping Ratio

Z = 1.1180e-06

wd  = wn*(sqrt(1-Z^2)); % Damped Frequency

[Parvesh Deepan]

Sorry my bad. Actually the question has given the data for damping ratio as 0.05 and I've take damping coefficient as 0.05.

Thanks for your input as it suggest me to look into my problem with a different approach.

[Parvesh Deepan]

Can you resolve one more query related to the same problem stated above?

What sort of change is needed in this code to determine velocity vs time graph ?