1. No category

ME 429 Fall 2014 HW3 Solution
You are expected to provide
 clear explanation of each step of your solution,
 units,
 well annotated, scaled plots (title, axis labels, units, legend), not random hand sketches,
 source code attached to your solution if you use a software package in your calculations.
Your grades are subject to these items as well as your calculations.
Problem-1: Consider the single-DOF lumped mass-spring system with Coulomb damping
shown below.
g
m = 12 kg
k = 5 kN/m
µ = 0.13
(a) If F(t) = 0 and an initial displacement of +35 mm is given to the mass m without
any initial velocity, determine:
(i) the response x(t) for any nth half-cycle and plot x(t) for the first 6 half-cycles
(ii) the position and the time at which the mass stops
(b) If F(t) = Fosin(40t) [N] and the amplitude of the steady-state oscillations of the
mass is 30 mm, determine the amplitude of the harmonic force Fo applied to the mass.
Solution:
(a) This is the free vibrations case with δ = +35 mm.
(i) The response x(t) for any nth half-cycle is formulated as:
(1)
Substituting the given values;
(2)
Note that since δ > 0,
equals “-1” for odd n values (i.e. n = 1, 3, 5, …) and “+1” for
even n values (i.e. n = 2, 4, 6, …). You can observe this phenomenon by looking at the drawn
x(t) vs. t graph in the lecture notes.
In addition, the time elapsed up to nth half-cycle is:
(3)
1/5
Thus, the time elapsed between two subsequent half-cycles is:
(4)
40
30
x(t) [mm]
20
10
0
-10
-20
-30
0
0.1
0.2
0.3
0.4
0.5
t [sec]
0.6
0.7
0.8
0.9
1
(ii) In order to determine the position and the time at which the mass stops, the maximum
number of half-cycles that can be traveled until stopping should be determined first. The
smallest integer n which satisfies the below inequality gives nmax:
(5)
Thus;
nmax = 6 cycles
Substituting nmax into Eqn. (3);
tnmax = 0.92 sec
The position at the end of the nth half-cycle is:
(6)
where
2/5
Substituting nmax into Eqn. (6);
x(tnmax) = -1.73 mm
%ME 429 Fall 2014 HW-3 Prob-1 Part-a Solution
clc; clear all; close all;
%System parameters
m=12; %kg
k=5000; %N/m
wn=sqrt(k/m) %rad/s
mu=0.13;
g=9.81; %m/s^2
%Initial displacement
delta=35*10^(-3); %m
%The response of the system
for n=1:6
for t=(n-1)*pi/wn:0.001:n*pi/wn
if mod(n,2)==0
x=(delta-(2*n-1)*mu*m*g/k)*cos(wn*t)-mu*m*g/k;
plot(t,x*10^3,'o'); hold on
xlabel('t [sec]')
ylabel('x(t) [mm]')
grid on
else
x=(delta-(2*n-1)*mu*m*g/k)*cos(wn*t)+mu*m*g/k;
plot(t,x*10^3,'o'); hold on
xlabel('t [sec]')
ylabel('x(t) [mm]')
grid on
end
end
end
%Number of half-cycles traveled before the system stops
n_max=ceil((delta-mu*m*g/k)/(2*mu*m*g/k))
%The time at which the mass stops
tn_max=n_max*pi/wn %sec
%The position at which the mass stops
if mod(n_max,2)==0
x_max=(delta-2*n_max*mu*m*g/k)*10^3 %mm
else
x_max=-(delta-2*n_max*mu*m*g/k)*10^3 %mm
end
b) This is the forced vibrations case with X = 0.030 m.
The amplitude of the steady-state x(t) is:
(7)
3/5
where
(8)
since Ff = µmg.
From Eqn’s (7) and (8);
(9)
Substituting the given values into Eqn. (9);
Fo = 426.58 N
Note that µmg/Fo = 0.036 < π/4.
Problem-2: A structure showing hysteresis damping characteristics is subjected to a load of 3
kN which causes a static displacement of 1.5 cm. When the same structure is subjected to a
harmonic force of amplitude 450 N at resonance, the resulting steady-state response amplitude
is 25 cm. Determine:
(a) the loss factor η
(b) the steady-state response amplitude at ω = 0.4ωn as a result of the same harmonic
force amplitude
(c) the steady-state response amplitude at ω = 4ωn as a result of the same harmonic
force amplitude
Solution:
(a) At static equilibrium;
F = kx
(10)
Thus;
k = 3000/0.015 = 200 kN/m
At resonance where ω = ωn;
(11)
Thus;
4/5
(b) The steady-state response amplitude is:
(12)
At ω = 0.4ωn, thus,
X = 2.68 mm
(c) Substituting ω = 4ωn and the necessary numerical values into Eqn. (12);
X = 0.15 mm
5/5