My EMA 545 Mechanical Vibrations, Spring 2013 material

Transcription

My EMA 545 Mechanical Vibrations, Spring 2013 material
Nasser M. Abbasi
March 20, 2015
Contents
1 introduction
2 HW’s
2.1 HW1
2.1.1
2.1.2
2.1.3
2.1.4
2.1.5
2.2
2.1.6
2.1.7
2.1.8
2.1.9
HW2
2.2.1
2.2.2
2.2.3
2.2.4
2.2.5
2.2.6
2.2.7
2.2.8
. . . . . . . . . . . . . .
Problem 1 (1.1 book) .
Problem 2 . . . . . . .
Problem 3 . . . . . . .
2.1.3.1 part(a) . . .
2.1.3.2 part(b) . . .
2.1.3.3 part(c) . . . .
2.1.3.4 part(d) . . .
2.1.4.1 part(a) . . .
2.1.4.2 part(b) . . .
2.1.4.3 part(c) . . . .
2.1.5.1 part(a) . . .
part(b) . . . . . . . . .
part(c) . . . . . . . . .
part(d) . . . . . . . . .
Problem 6 (2.10 book)
. . . . . . . . . . . . . .
problem 1 . . . . . . .
2.2.1.1 Part(a) . . .
2.2.1.2 Part(b) . . .
2.2.1.3 Part(c) . . .
Problem 2 . . . . . . .
2.2.2.1 Part(a) . . .
2.2.2.2 Part(b) . . .
Problem 3 . . . . . . .
Problem 4 . . . . . . .
2.2.4.1 case 1 . . . .
case 2 . . . . . . . . .
Problem 5 . . . . . . .
Part(a) . . . . . . . . .
2.2.7.1 Part(b) . . .
2.2.7.2 Part c . . . .
2.2.7.3 Part d . . . .
Problem 6 . . . . . . .
2.2.8.1 Part a . . . .
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CONTENTS
2.3
2.4
HW3
HW4
2.4.1
2.4.2
2.4.3
2.4.4
2.4.5
2.4.6
2.5
HW5
2.5.1
2.5.2
2.5.3
2.5.4
2.5.5
2.6
HW6
2.6.1
2.6.2
2.6.3
2.7
HW7
2.7.1
2.7.2
iii
2.2.8.2 Part b . . . . . . . . . . . . . . . . . .
2.2.8.3 Part(c) . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . .
problem 1 . . . . . . . . . . . . . . . . . . . . .
Problem 2 . . . . . . . . . . . . . . . . . . . . .
2.4.2.1 part(a) . . . . . . . . . . . . . . . . .
2.4.2.2 Part(b) . . . . . . . . . . . . . . . . .
2.4.2.3 Part(c) . . . . . . . . . . . . . . . . .
Problem 3 . . . . . . . . . . . . . . . . . . . . .
Problem 4 . . . . . . . . . . . . . . . . . . . . .
2.4.4.1 First part . . . . . . . . . . . . . . . .
2.4.4.2 Extra part . . . . . . . . . . . . . . . .
Problem 5 . . . . . . . . . . . . . . . . . . . . .
Problem 6 . . . . . . . . . . . . . . . . . . . . .
2.4.6.1 Part (1) . . . . . . . . . . . . . . . . .
2.4.6.2 Part (2) . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . .
problem 1 . . . . . . . . . . . . . . . . . . . . .
subproblem 2 . . . . . . . . . . . . . . . . . . .
2.5.2.1 Part(a) . . . . . . . . . . . . . . . . .
2.5.2.2 part(b) . . . . . . . . . . . . . . . . .
2.5.2.3 Part(c) . . . . . . . . . . . . . . . . .
Problem 3 . . . . . . . . . . . . . . . . . . . . .
2.5.3.1 part(a) . . . . . . . . . . . . . . . . .
2.5.3.2 Part(b) . . . . . . . . . . . . . . . . .
2.5.3.3 Part(c) . . . . . . . . . . . . . . . . .
problem 4 . . . . . . . . . . . . . . . . . . . . .
2.5.4.1 Part(a) . . . . . . . . . . . . . . . . .
Part(b) . . . . . . . . . . . . . . . . . . . . . .
2.5.5.1 Part(c) . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . .
problem 1 . . . . . . . . . . . . . . . . . . . . .
2.6.1.1 Verification using Matlab ffteasy.m . .
problem 2 . . . . . . . . . . . . . . . . . . . . .
2.6.2.1 Part(a) . . . . . . . . . . . . . . . . .
problem 3 . . . . . . . . . . . . . . . . . . . . .
2.6.3.1 Finding Yn for τ = 3ωπnat . . . . . . . .
2.6.3.2 Plot for the steady state . . . . . . . .
2.6.3.3 Repeating the calculations for τ = ω3π
nat
2.6.3.4 Plot for the steady state . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . .
problem 1 . . . . . . . . . . . . . . . . . . . . .
2.7.1.1 Part(a) . . . . . . . . . . . . . . . . .
2.7.1.2 Part(b) . . . . . . . . . . . . . . . . .
Problem 2 . . . . . . . . . . . . . . . . . . . . .
2.7.2.1 part(a) . . . . . . . . . . . . . . . . .
2.7.2.2 Part(b) . . . . . . . . . . . . . . . . .
2.7.2.3 Part(c) . . . . . . . . . . . . . . . . .
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29
30
32
33
33
34
35
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39
41
46
46
48
49
50
51
52
54
54
55
56
57
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61
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62
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65
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93
94
iv
CONTENTS
2.7.3
2.7.4
2.7.5
Conclusions . . . . . . . . . . . . .
Problem 3 . . . . . . . . . . . . . .
Part(a) . . . . . . . . . . . . . . . .
2.7.5.1 Part(b) . . . . . . . . . .
2.8 HW8 . . . . . . . . . . . . . . . . . . . . .
2.8.1 problem 1 . . . . . . . . . . . . . .
2.8.1.1 part(a) . . . . . . . . . .
2.8.1.2 part(b) . . . . . . . . . .
2.8.1.3 part(c) . . . . . . . . . . .
2.8.1.4 part(d) . . . . . . . . . .
2.8.2 problem 2 . . . . . . . . . . . . . .
2.8.3 problem 3 . . . . . . . . . . . . . .
2.8.4 problem 4 . . . . . . . . . . . . . .
2.8.5 problem 5 . . . . . . . . . . . . . .
2.8.6 problem 6 . . . . . . . . . . . . . .
2.9 HW9 . . . . . . . . . . . . . . . . . . . . .
2.9.1 problem 1 . . . . . . . . . . . . . .
2.9.1.1 verification using Matlab .
2.9.2 Sketch of each mode . . . . . . . .
2.9.3 problem 2 . . . . . . . . . . . . . .
2.9.4 Problem 3 . . . . . . . . . . . . . .
2.9.5 Problem 4 . . . . . . . . . . . . . .
2.9.6 side question . . . . . . . . . . . .
2.9.7 power balance method . . . . . . .
2.9.7.1 Lagrangian method . . . .
2.10 HW10 . . . . . . . . . . . . . . . . . . . .
2.10.1 problem 1 . . . . . . . . . . . . . .
2.10.2 Problem 2 . . . . . . . . . . . . . .
2.10.3 Part (a) k = 0.05 mg
. . . . . . . .
L
2.10.3.1 Part (b) k = 2 mg
. . . . .
L
2.10.4 Problem 3 . . . . . . . . . . . . . .
2.10.5 Problem 4 . . . . . . . . . . . . . .
2.10.6 Problem 5 . . . . . . . . . . . . . .
2.10.6.1 Part(a) . . . . . . . . . .
2.10.6.2 Part(b) . . . . . . . . . .
2.10.6.3 Part(c) . . . . . . . . . .
2.10.7 Problem 6 . . . . . . . . . . . . . .
2.11 HWA1 . . . . . . . . . . . . . . . . . . . .
2.11.1 problem 1 . . . . . . . . . . . . . .
2.11.2 Problem 2 . . . . . . . . . . . . . .
2.11.2.1 Part(b) . . . . . . . . . .
2.11.3 Problem 3 . . . . . . . . . . . . . .
2.11.4 Problem 4 . . . . . . . . . . . . . .
2.11.5 Problem 5 . . . . . . . . . . . . . .
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95
96
96
97
100
100
100
102
104
106
108
111
113
115
117
120
120
124
124
125
131
135
141
142
144
147
147
151
152
155
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165
168
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170
174
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176
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178
179
180
181
185
3 Design project
189
3.1 Design project, team 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189
3.2 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189
3.3 Discussion and results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189
CONTENTS
3.4
3.5
v
3.3.0.1 Notations used in the report . .
3.3.0.2 Mathematical model . . . . . .
3.3.0.3 Design results . . . . . . . . . .
3.3.1 vibration isolation system . . . . . . . .
3.3.2 vibration isolation system Cost estimate
Appendix . . . . . . . . . . . . . . . . . . . . .
3.4.1 Design values . . . . . . . . . . . . . . .
3.4.2 Cost values . . . . . . . . . . . . . . . .
3.4.3 Simulation program description . . . . .
3.4.4 Derivation of the transfer function . . . .
3.4.5 References . . . . . . . . . . . . . . . . .
3.4.6 software . . . . . . . . . . . . . . . . . .
team 3 peer review . . . . . . . . . . . . . . . .
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189
190
190
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196
196
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199
200
4 Exams
4.1 Practice exams . . . . .
4.2 practice exam 2 . . . . .
4.2.1 Problem 1 . . . .
4.2.2 Problem 2 . . . .
4.2.2.1 Part a .
4.2.2.2 Part b .
4.2.3 Problem 3 . . . .
4.3 finals 2nd practice exam
4.3.1 Problem 1 . . . .
4.3.2 Problem 2 . . . .
4.3.2.1 part(a)
4.3.2.2 part(b)
4.3.2.3 part(c) .
4.3.3 Problem 3 . . . .
4.3.3.1 part(a)
4.3.3.2 part(b)
4.3.3.3 part(c) .
4.3.4 Problem 4 . . . .
4.3.5 Problem 5 . . . .
4.3.6 Problem 6 . . . .
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201
202
203
203
204
204
207
208
213
213
215
215
216
217
218
219
219
220
221
222
223
5 Quizes
5.1 my solution to Quiz 2 .
5.1.1 problem . . . .
5.1.2 Answer part (1)
5.1.3 Answer part (2)
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225
226
226
226
231
6 appendix
6.1 cheat sheet . . . . . .
6.2 study notes . . . . .
6.2.1 trig identities
6.3 my lecture notes . . .
6.4 course related links .
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235
236
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240
241
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vi
CONTENTS
Chapter 1
introduction
I took this course in Spring 2013 part of masters degree in engineering mechanics. University of Wisconsin,
Madison.
Instructor is professor Matt Allen
External class web page http://courses.engr.wisc.edu/ema/ema545.html
Text book: Mechanical and Structural Vibrations: Theory and Applications, Jerry H. Ginsberg, 1st
Edition, Wiley, 2001.
Syllubus is PDF
1
2
CHAPTER 1. INTRODUCTION
Chapter 2
HW’s
HW
grade
about
1
95%
series/parallel stiffness, How to use x = Re {Xeiωt } to analyze systems earliest time
to reach maximum value/speed, complex exponential
2
95.70% eq. of motion cube in water, more use of complex exponential analyzing in complex
plane. Logarithmic decrement from graph. Impulse problem
3
100%
4
98.75% force applied for small period, find response. Impulse force on system. analyse in
complex plane. resonance problem, students on bridge.
5
88%
6
97.50% Find complex Fourier series.Verify using fft. Simple model of car moving on ground,
find EQM. Fourier series. rectangle force, Fourier series.
7
95%
Using FFT to find response. Transfer functions Compare to analytical. Using Lagrange
to find EQM, 2 DOF.
8
99%
non-linear EQM, spring stiff approximation. Lagrangian. Model of wing. cart on
spring with sliding mass on it with spring. Lagrangian. Finding ωn for 2DOF
A1
100%
more spring stiff approximation. manipulation of complex form of solution. half-power
point, finding phase lag, solving step response using appendix B method
HW9 95%
small lab
2 DOF system, shock observer on spring.Force transmission to base. off center motion,
find EQM. 2 counter-rotating masses
Full solution in modal coordinates. Mass normalized. Initial conditions in modal
coordinates. All problems done in power balance method. Double physical pendulum
3
4
CHAPTER 2. HW’S
HW10 92.5% Full solution in modal coordinates. 3 DOF problem
HW11 93.3% modal analysis, with damping using specific modal damping. Structual damping
Compare transfer functions for each damping method used. Ritz method, shape
functions. plot mode shapes.
Table 2.1: Homeworks summary table
2.1. HW1
2.1
5
HW1
2.1.1
k3 and k2 are in parallel, hence the effective stiffness is
k23 = k2 + k3
k23 and k1 are now in series, hence the effective stiffness is
1
k123
=
1
1
k23 + k1
k2 + k3 + k1
k2 + k3 + k1
+
=
=
=
k1 k23
k1 k23
k1 (k2 + k3 )
k1 k2 + k1 k3
Therefore
k1 k2 + k1 k3
k2 + k3 + k1
and k4 are now in parallel, hence the effective stiffness is
k123 =
k123
k1234 = k4 + k123
k1 k2 + k1 k3
= k4 +
k2 + k3 + k1
Hence the final effective stiffness is
keq =
2.1.2
k4 (k2 + k3 + k1 ) + k1 k2 + k1 k3
k2 + k3 + k1
Problem 2
We start by drawing a free body diagram and taking displacement of mass from the static equilibrium
position. Let the displacement of the mass be x and positive pointing upwards.
Let 41 be the downward deflection at right end of the bottom beam. Let 42 be the downward
deflection at right end of top beam. The free body diagram is
6
CHAPTER 2. HW’S
P
Applying equilibrium of vertical forces
Fv = 0 for mass m and noting that inertial forces opposes
motion, results in the equation of motion
mx00 + k (x − 41 ) = F
(2.1)
To find an expression for 41 in terms of x, we apply equilibrium of vertical forces at the right end of the
lower beam1
k (x − 41 ) = kb 41 + k (41 − 42 )
(2.2)
Similarly, applying equilibrium of vertical forces at the right end of the top beam
k (41 − 42 ) = kb 42
(2.3)
Solving for 41 , 42 from Eqs 2.111,2.96 (2 equations, 2 unknowns) gives
41 =
k2
k (k + kb )
x
+ 3kkb + kb2
Substituting the above value into Eq 2.110 results in the equation of motion
k (k + kb )
00
mx + kx 1 − 2
=F
k + 3kkb + kb2
2.1.3
Problem 3
Assuming periodic motion, the period is T = 6 ms, or 6 × 10−3 sec. Hence ω =
this as a cosine signal with phase gives
x(t) = A cos(ωt + θ)
1
kb is beam stiffness against vertical displacement at the end and is given as kb =
3EI
L3
π
3
rad/ms Representing
2.1. HW1
7
Then
x(t) = Re[A + cos(ωt + θ)]
= Re[Aeiθ eiωt ]
¯ iωt ]
= Re[Ae
(2.4)
Where now Aˆ = Aeiθ .Using phasor diagram
Hence from the diagram we see that for x(t0 ) to be zero when t0 = 1 ms, we need to have
π
ωt0 + θ =
2
π
But ω = 3 rad/ms, hence
π π
π
θ= − =
2
3
6
To find A we see that the maximum absolute value of x(t) is 20 mm hence A = 20 mm or 20 × 10−3
meter. The equation of x(t) when substituting all numerical values becomes
π
π
x(t) = 20 cos
t+
(2.5)
3
6
Where units used are radians, milliseconds and mm. This is a plot of the above function
parms = f -> 1/(6 10^-3);
Plot[0.02 Cos[2 Pi f t + (Pi/6)] /. parms, {t,0,0.005},
AxesLabel -> {t,x[t]}, ImageSize -> 300]
2.1.3.1
part(a)
ˆ = A cos(θ) = 20 cos( π ) hence
At t = 0, from 2.112 x(0) = Re[A]
6
x(0) = 17.321 mm
ˆ iωt ] hence x0 (0) = Re[ω A]
ˆ = ωA cos(θ) = 20 π cos( π ) giving
From 2.112 x0 (t) = Re[ω Ae
3
6
x0 (0) = 18.138 m/sec
8
CHAPTER 2. HW’S
2.1.3.2
part(b)
This can be solved using calculus2
x0 (t) = −2πf A sin (2πf t + θ)
π
0 = −2πf A sin 2πf t +
6
2π
2π
π
−3
sin
20 × 10
=−
t+
6 × 10−3
6 × 10−3
6
2π
π
0 = sin
t
+
6 × 10−3
6
We solve for t and find t=2.5 ms. But this can be solved more easily by looking at the phasor diagram
The minimum x(t) (in negative sense and not in absolute value sense) occurs when ωtmin + θ = π, hence
tmin = π−θ
, therefore
ω
tmin = 2.5
2.1.3.3
part(c)
This is solved
in a similar
way by treating the speed as the rotating vector in complex plan. Since
i(ωt+θ+ π2 )
0
then in complex plan as follows
x (t) = Re Aωe
The difference is that the velocity vector has phase of θ + π2 instead of θ as was the case with the
position vector, and the amplitude is Aω instead of A. Hence the first time the speed vector will have
the maximum value is when
π
θ + + ωt = 2π
2
2
Taking derivative of x(t) and setting the result to zero and solving for t
2.1. HW1
9
Hence
t=
=
2π − π2 − θ
ω
2π − π2 − π6
π
3
Hence t = 4 ms and the amplitude is given by Aω = 20 π3 hence Aω = 20.944 meter/sec
2.1.3.4
part(d)
Now treating the acceleration as the rotating vector in complex plan
x(t) = Re Aei(θ+ωt)
x0 (t) = Re iAωei(θ+ωt)
x00 (t) = Re −Aω 2 ei(θ+ωt)
But −1 = eiπ This adds a π to the phase resulting in
x00 (t) = Re Aω 2 ei(θ+ωt+π)
Representing x00 (t) in complex plan gives
The first time the x00 (t) vector will have the maximum value is when
θ + π + ωt = 2π
Hence
t=
=
2π − π − θ
ω
π − π6
π
6
Hence t = 2.5 ms and the amplitude is
Aω 2 = 20 mm
π
2
rad/msec
3
= 21.933 × 103 meter/sec2
10
CHAPTER 2. HW’S
2.1.4
2.1.4.1
part(a)
The function of the signal is converted to complex exponential. A sin or cos can be used to represent the
signal as long as we are consistent. Assuming the signal is x(t) = A cos(ωt + θ), plotting the general
representation of the position vector in complex plan gives
The complex representation of the position vector is
x(t) = Re Aei(ωt+θ)
We are given that ω = 2π
=
T
the above diagram that
2π
,
16
and since x(t0 ) has first zero at t0 = 5.5 ms this means from looking at
θ + ωt0 =
Hence θ =
π
2
− (ωt0 ) =
π
2
π
2
− ( π8 55
) which gives
10
θ=
−3π
radians
16
Hence the signal is
i
h
3π
π
= Re 1.2ei( 8 t− 16 )
h
i
−i 3π
i π8 t
16
= Re 1.2e
e
h
i
ˆ i π8 t
= Re Ae
2.1. HW1
11
3π
Where Aˆ = 1.2e−i 16 is the complex amplitude in polar coordinates. In rectangular coordinates it becomes
3π
Aˆ = 1.2e−i 16
3π
3π
− i sin
= 1.2 cos
16
16
= 1.2 (0.831 − i0.5556)
= 0.9977 − i0.6667
Hence
h
π
π i
x(t) = Re (0.998 − i0.668) cos t + i sin t
8
8
h
π
π = Re 0.998 cos t + 0.668 sin t +
8
8
π
π i
i 0.998 sin t − 0.668 cos t
8
8
Here is a plot of the signal for 20 ms
w = Pi/8;
f = 1.2 Cos[w t - 3 Pi/16];
Plot[f, {t, 0, 20}, AxesLabel -> {t, x[t]},
ImageSize -> 300,
GridLines -> Automatic,
GridLinesStyle->{{Dashed,Gray},{Dashed,Gray}},
PlotStyle -> Red]
2.1.4.2
part(b)
From above it was found that
Hence
x0 (t) = Re iωAei(ωt+θ)
π
= Re ei 2 ωAeiθ eiωt
π
= Re ωAei( 2 +θ) eiωt
h
i
ˆ iωt
= Re Ae
12
CHAPTER 2. HW’S
π
π
3
Where Aˆ = ωAei( 2 +θ) Replacing numerical values gives Aˆ = π8 (1.2) ei( 2 − 16 π) = 0.471ei0.983 and
x0 (t) = Re 0.471ei0.983 eiωt
π = Re 0.471ei0.983 ei 8 t
= Re 0.471ei0.983 ei0.3923t
In rectangular coordinates, the above becomes
h
0
x (t) = Re 0.471 (cos 0.983 + i sin 0.983)
i
(cos 0.3923t + i sin 0.3923t)
= Re [(0.261 + 0.392i) (cos 0.392t + i sin 0.392t)]
h
= Re (0.261 cos 0.392t − 0.392 sin 0.392t)
i
+i (0.261 sin 0.392t + 0.392 cos 0.392t)
2.1.4.3
part(c)
To find the maximum rate of the signal
h
i
ˆ iωt
x0 (t) = Re Ae
Then the maximum x0 (t) is Aˆ which is
ˆ
A = |0.261 + 0.392i|
√
= 0.2612 + 0.3922
= 0.471
Hence maximum x0 (t) is 0.471 v/ms or 471 volt/sec.
Maximum velocity in simple harmonic motion occurs when x (t) = 0. This occurs at t = 5.5 ms and
at 8 ms henceforth. Hence maximum speed occurs at
t = 5.5 + n (8)
for n = 0, 1, 2, · · · this results in
t = 5.5, 13, 5, 21.5, · · · ms
0
Here is a plot of x (t) in units of volt/ms
f = 0.261 Cos[0.392 t] - 0.392 Sin[0.392 t];
Plot[f, {t, 0, 30},
AxesLabel -> {Row[{t, "(ms)"}], x'[t]},
ImageSize -> 300, GridLines -> Automatic,
GridLinesStyle -> {{Dashed, Gray}, {Dashed, Gray}},
PlotStyle -> Red]
2.1. HW1
13
2.1.5
2.1.5.1
part(a)
This is a plot of the signal
f = 0.01 Sin[50 t] - 0.02 Cos[50 t - 0.3 Pi];
Plot[f, {t, 0, 0.2},
AxesLabel -> {Row[{t , " (sec)"}], x[t]},
ImageSize -> 300,
GridLines -> Automatic,
GridLinesStyle->{{Dashed,Gray},{Dashed,Gray}},
PlotStyle -> Red]
q = 0.01 sin (50t) − 0.02 cos (50t − 0.3π)
0.01 i50t
i(50t−0.3π)
= Re
e − 0.02e
i
π
= Re 0.01e−i 2 ei50t − 0.02ei50t e−i0.3π
π
= Re 0.01e−i 2 − 0.02e−i0.3π ei50t
h
i
ˆ i50t
= Re Ae
Hence the complex amplitude is
π
Aˆ = 0.01e−i 2 − 0.02e−i0.3π
14
2.1.6
CHAPTER 2. HW’S
part(b)
From above, we see that
ω = 50 rad/sec
50
Hence f = 2π
Hz, or the period T =
0.126
= 0.063 sec or 63 ms
2
2.1.7
2π
50
= 0.126 sec, therefore the time period separating the zeros is
part(c)
π
3π
The complex phase Aˆ can be found by adding the vector 0.01e−i 2 and −0.02e−i 10 by completing the
parallelogram as shown in this diagram. Aˆ = −0.02 cos 0.7π + i (−0.01 + 0.02 sin 0.7π), hence the angle
α that Aˆ makes with the horizontal is
−0.01 + 0.02 sin 0.7π
−1
= arctan (0.526)
tan
−0.02 cos 0.7π
= 0.484 radian
= 27.73 degree
and the amplitude is
q
(−0.01 + 0.02 sin 0.7π)2 + (0.02 cos 0.7π)2 = 0.0133V
To find the earliest time q will be zero, we need to find the time the complex position vector will take to
rotate and reach the imaginary axis.
2.1. HW1
15
Hence we need to solve
3
π − α + ωt0 = π
2
3
π − π + 0.48402
t0 = 2
50
= 0.0411 s
Therefore
t = 41.1 ms
2.1.8
part(d)
The largest value of q is the absolute value of its complex amplitude. We found this above as
ˆ
A = 0.0133 Volt
To find when this occur first time, the time the position vector will align with the real axis in the positive
direction is found. Hence solving for t0 from
π − α + ωt0 = 2π
2π − π + 0.484
t0 =
50
Gives t = 72.5 ms. Another way would be to take derivative of qt) and set that to zero and solve for first
t which satisfy the equation.
2.1.9
5
x1 = 8 sin 10t − π
6
x2 = 12 cos (10t + φ)
Let ω = 10, hence
x = x1 + x2
8 i(ωt− 56 π)
= Re e
+ Re 12ei(ωt+φ)
i
8 i(ωt− 56 π)
i(ωt+φ)
= Re e
+ 12e
i
i
h
−i π2 i(ωt− 56 π )
i(ωt+φ)
= Re 8e e
+ 12e
h
i
π
5
= Re 8e−i 2 eiωt e−i 6 π + 12eiωt eiφ
h
i
−i( 43 π )
iφ
iωt
+ 12e e
= Re 8e
h
i
ˆ iωt
= Re Ae
(2.6)
16
CHAPTER 2. HW’S
Where
4
Aˆ = 8e−i( 3 π) + 12eiφ
= (−4 + 6.928i) + 12 (cos φ + i sin φ)
= (−4 + 12 cos φ) + i (6.928 + sin φ)
Hence Eq 2.6 becomes
x = Re {(−4 + 12 cos φ) + i (6.928 + sin φ)} eiωt
To convert to sin we multiply and divide by i hence
eiωt
x = Re {(−4 + 12 cos φ) + i (6.928 + sin φ)} i
i
eiωt
= Re {− (6.928 + sin φ) + i (−4 + 12 cos φ)}
i
(2.7)
The complex
number − (6.928 + sin φ) + i (−4 + 12 cos φ) can be written in polar form as keiβ where
q
−4+12 cos φ
K = (6.928 + sin φ)2 + (−4 + 12 cos φ)2 and β = tan−1 −(6.928+sin
, hence Eq 2.7 becomes
φ)
iωt
iβ e
x = Re ke
i
i(ωt+β) e
= Re k
i
= k sin (ωt + β)
or in full form
q
x = (6.928 + sin φ)2 + (−4 + 12 cos φ)2
−4 + 12 cos φ
−1
sin ωt + tan
− (6.928 + sin φ)
For pure sine function we need
−4+12 cos φ
−(6.928+sin φ)
= 0 or 12 cos φ = 4 or cos φ = 31 , hence
φ = 1.23096 radian
= 70.529◦
The amplitude can also be found from the complex amplitude above when φ = 1.23096 as follows
−i( 34 π)
+ 12ei1.23096 = −6.592 × 10−6 + 18.242i
8e
q
= (−6.592 × 10−6 )2 + (18.242)2
= 18.242
2.2. HW2
2.2
17
HW2
2.2.1
problem 1
2.2.1.1
Part(a)
We assume the cube is displaced downwards from its static equilibrium position and it is currently at
distance x below the static position.
The buoyant force Fb will push the cube upwards. This force will equal the weight of water displaced
which is xa2 ρw g where ρw is density of water and g is the gravitational constant. The free body diagram
is
Applying F = mx00 we obtain equation of motion
M x00 = −Fb
M x00 + Fb = 0
(2.8)
(2.9)
18
CHAPTER 2. HW’S
M = a3 ρ where ρ is density of pine. The above can be simplified to
a3 ρx00 + xa2 ρw g = 0
ρw g
x00 +
x=0
aρ
x00 + ωn2 x = 0
2.2.1.2
Part(b)
Hence from the above equation
r
ωn =
2.2.1.3
ρw g
aρ
Part(c)
Given ρ = 400 kg/m3 and ρw = 1000 kg/m3 and a = 0.1m then
r
r
rad
ρw g
1000 × 9.81
=
= 15.66
ωn =
aρ
0.1 × 400
sec2
Hence frequency in Hz is
f=
2.2.2
Problem 2
2.2.2.1
Part(a)
ωn
15.66
=
= 2.492 hz
2π
2π
(2.10)
(2.11)
(2.12)
2.2. HW2
19
applying F = mq 00 , we obtain equation of motion
M q 00
M q 00 + kq
k
q 00 + q
M
q 00 + ωn2 q
= −kq
=0
(2.13)
(2.14)
=0
(2.15)
=0
(2.16)
ˆ iωn t where Aˆ is the complex amplitude, which is a complex number that
Let solution be q (t) = Re Ae
ˆ At t = 0, let q (0) = q0
can be written as Aˆ = a + ib. We use initial conditions to determine A.
ˆ iωn t
q0 = Re Ae
(2.17)
= Re Aˆ
(2.18)
=a
ˆ iωn t , then t = 0 we have
Hence a = q0 And since q 0 (t) = Re iωn Ae
0
ˆ
q0 = Re iωn A
(2.19)
= Re (iωn (a + ib))
= Re (iωn a − ωn b)
= −ωn b
(2.20)
(2.21)
(2.22)
(2.23)
q0
Hence b = − ω0n therefore the general solution is
ˆ iωn t
q (t) = Re Ae
= Re (a + ib) eiωn t


ˆ
A
z }| {


q00

iωn t 
= Re  q0 − i
e 
ωn


(2.24)
(2.25)
(2.26)
q0 r
0 2
0
ˆ
q0
ωn
−1
2
Hence A = q0 + ωn and arg Aˆ = θ = tan
. We have 2 complex quantities above being
q0
multiplied. The first is Aˆ and the second is eiωn t , therefore the result is obtained by adding the angles
and by multiplied the magnitudes. The magnitude of eiωn t is one. Hence on the complex plan, the above
expression for q (t) is represented as vector of length Aˆ and phase φ = θ + ωn t
20
CHAPTER 2. HW’S
From the above diagram we see that the maximum value of
qmax (t) = Aˆ
which occurs when
φ = θ + ωn t = 0
solving for t gives
t=
−θ
ωn
Notice that θ is negative, hence we will get positive value for t. Substituting the numerical values given
we find thatAnd the earliest time this occurs is
t=
1.3724
= 2.7303 × 10−3 = 2.73 ms
2π (80)
We confirm this by noticing that the initial position vector was at about 41 cycle away from the positive
x-axis (we found the phase of Aˆ above to be about −80 degrees), and the rotational speed is given as 80
cycles per second. Hence it takes 12.5 ms to make one cycle and 41 of this is about 3ms.
2.2.2.2
Part(b)
Since we found q (t) = Re
q0
q0 − i ω0n eiωn t , then
q00
iωn t
e
q (t) = Re iωn q0 − i
ωn
π
π
= Re ωn q0 − ei 2 q00 ei 2 eiωn t
π
= Re ωn q0 ei 2 − eiπ q00 eiωn t


ˆ
B
z
}|
{
π


= Re  ωn q0 ei 2 + q00 eiωn t 
0
(2.27)
(2.28)
(2.29)
(2.30)
q
ˆ
0
ˆ
Where now B is the complex amplitude of q (t). Hence B = (ωn q0 )2 + (q00 )2 and its phase is
ˆ = tan−1 ωn0q0 . The complex plane representation of q 0 (t) is
arg B
q
0
2.2. HW2
21
ˆ
From the above diagram we see that maximum magnitude of q (t) is B
given by
0
q
ˆ
2
2
B = (ωn q0 ) + (q00 )
q
= (2π (80) (20 × 10−3 ))2 + (−50)2
= 51.001m/s
(2.31)
(2.32)
(2.33)
The earliest time it occurs is found by solving for t in
ωn t + θ = 2π
2π − θ
t=
ωn
2π − tan−1
=
=
2.2.3
(2.34)
(2.35)
ωn q0
q00
2π (80)
2π(80)(20×10−3 )
2π − tan−1
−50
=
2π (80)
2π − tan−1 (−0.20106)
= 1.2895 × 10−2
2π (80)
= 0.129 ms
(2.36)
(2.37)
(2.38)
Problem 3
Adding 2 kg caused deflection of 50 mm, hence from F = k∆ we can find k as follows
k=
F
2g
2 (9.81)
=
=
= 392 N/m
∆
0.05
0.05
(2.39)
where g is the gravitational constant. We also told that f2 = f1 − 5 where f2 is the natural frequency
22
CHAPTER 2. HW’S
after adding the second mass and where f1 =
1
ω
2π 1
and f2 =
1
ω,
2π 2
hence
f2 = f1 − 5
1
1
ω2 =
ω1 − 5
2π
2π
ω2 = ω1 − 10π
But ω1 =
q
k
m
and ω2 =
q
k
,
m+2
(2.40)
(2.41)
(2.42)
hence
r
k
=
m+2
r
k
− 10π
m
From Eq 2.112 the above becomes
r
392
=
m+2
r
392
− 10π
m
Solving numerically gives m = 0.1955kg
2.2.4
Problem 4
To show that x (t) = Beλt is solution to the differential equation, we substitute this solution into the
LHS of the differential equation and see if we obtain zero.
x0 (t) = λBeλt = λx (t)
00
2
λt
2
(2.43)
x (t) = λ Be = λ x (t)
(2.44)
x00 + 2ζωn x0 + ωn2 x = 0
λ2 x (t) + 2ζωn λx (t) + ωn2 x (t) = 0
λ2 + 2ζωn λ + ωn2 x (t) = 0
(2.45)
(2.46)
Then
(2.47)
Hence x (t) = Beλt is a non-trivial solution to the differential equation provided λ2 + 2ζωn λ + ωn2 = 0
since then we obtain 0 = 0.
Now we find λ for the different cases.
2.2.4.1
case 1
The roots of λ2 + 2ζωn λ + ωn2 = 0 are
λ1,2 = −ζωn ± ωn
For underdamped ζ < 1, hence
p
ζ2 − 1
p
ζ 2 − 1 < 0 and we write the above as
p
λ1,2 = −ζωn ± iωn 1 − ζ 2
= −ζωn ± iωd
(2.48)
(2.49)
2.2. HW2
23
where
ωd = ωn
p
1 − ζ2
Let λ = −ζωn + iωd and its complex conjugate λ∗ = −ζωn − iωd , hence the solution is
x (t) = B1 eλt + B2 eλ
∗t
ˆ and B2 = B
ˆ ∗ . Hence the above can be
To obtain a real solution we must have B1 be complex say B
written as
ˆ λt + B
ˆ ∗ e λ∗ t
x (t) = Be
λt
ˆ
= 2 Re Be
ˆ (−ζωn +iωd )t
= Re 2Be
(2.50)
(2.51)
(2.52)
Therefore
ˆ −ζωn t eiωd t
x (t) = Re Ae
(2.53)
ˆ = a + ib. Hence
Where Aˆ = 2B
x (t) = Re (a + ib) e−ζωn t eiωd t
To find a, b we need to use initial conditions. Assuming x (0) = x0 and x0 (0) = x00 then from Eq 2.53 we
obtain
x0 = Re (a + ib) = a
Hence
a = x0
and taking derivative of 2.53
x(t) = Re (a + ib) e−ζωn t eiωd t
0
−ζωn t iωd t
(2.54)
−ζωn t iωd t
x (t) = Re −ζωn (a + ib) e
e
+ iωd (a + ib) e
0
x (0) = Re (−ζωn (a + ib) + iωd (a + ib))
= −ζωn a − ωd b
Hence
b=
x00 + ζωn a
ωd
b=
x00 + ζωn x0
ωd
e
(2.55)
(2.56)
(2.57)
But a = x0 , hence
Hence 2.53 becomes
x (t) = Re (a + ib) e−ζωn t eiωd t
x00 + ζωn x0 −ζωn t iωd t
= Re
x0 + i
e
e
ωd
and this is the general solution. In complex plan it is
(2.58)
(2.59)
24
CHAPTER 2. HW’S
Hence the rotating vector will have its length become smaller with time since Aˆ is multiplied by e−ζωn t .
The real part, which is the solution will eventually damp down to zero. Hence it is a damped sinusoid
oscillation as follows
2.2.5
case 2
From
λ1,2 = −ζωn ± ωn
For overdamped ζ > 1, hence
p
ζ2 − 1
p
ζ 2 − 1 > 0 and we write the above as
λ1,2 = −ζωn ± ωn
p
1 − ζ2
Hence the solution is
x (t) = B1 eλ1 t + B2 eλ2 t
p
p
where λ1 = −ζωn + ωn 1 − ζ 2 and λ2 = −ζωn − ωn 1 − ζ 2 . We see that both roots are negative always,
hence we have 2 exponentially damped solution being added with no oscillation. A sketch of the solution
is
2.2. HW2
2.2.6
Problem 5
2.2.7
Part(a)
25
From looking at the plot above, here are the values estimated for displacement positive peaks and time
they occur
t
y (t)
0.07
16
0.17
12
0.27
9
0.37
6
From the above we estimate the natural period T ≈ 0.1 sec hence f = 10 hz hence ωn = 2πf = 60.3 rad/sec
The log decrement is
yi
δ = ln
yi+N
Select i = 1 and N = 3 gives
16
6
= 0.981
δ = ln
To find ζ we use the log decrement method
δ = 2πN ζ
(2.60)
(2.61)
26
CHAPTER 2. HW’S
Hence
δ
0.98083
=
2πN
2π (3)
ζ = 0.052
ζ=
(2.62)
(2.63)
Hence
ζ = 5.2%
2.2.7.1
Part(b)
ln
y1
y1+N
= 2πN ζ
Where now we write y1 = 16 and yN +1 = 0.01, and hence we need to find N the only unknown in the
equation above
16
ln
= 2πN (0.052)
0.01
Hence
16
ln 0.01
= 22.581
N=
2π (0.052)
We take N = 23. What this says is that after 23 periods beyond the first peak, we will satisfy the
requirement. But T = 0.1 sec, and the first peak was at t = 0.05 sec, therefore
t = 0.07 + 23(0.1)
= 2.37sec
2.2.7.2
(2.64)
(2.65)
Part c
Since δN = 2πN ζ and ζ = ccr = 2√ckm , then if we double k and half the mass m, then ζ would remain the
same since c is held constant. Therefore the answer in part b would not change.
2.2.7.3
Part d
Since this is an underdamped system, the solution is
(−ωn ζ+iωd )t
ˆ
q (t) = Re Ae
Where Aˆ is the complex amplitude, say (a + ib), hence
q (t) = Re (a + ib) e(−ωn ζ+iωd )t
At t = 0 we find that
a = q (0) = q0
Hence
a = −0.01
and the general solution is
q (t) = Re (q0 + ib) e(−ωn ζ+iωd )t
2.2. HW2
27
Now taking derivatives of the above gives
q 0 (t) = Re (−ωn ζ + iωd ) (q0 + ib) e(−ωn ζ+iωd )t
At t = 0 then, assuming q00 is the initial velocity
q00 = Re ((−ωn ζ + iωd ) (q0 + ib))
= −ωn ζq0 − ωd b
Hence
(2.66)
(2.67)
q00 + ωn ζq0
b=−
ωd
Therefore the general solution is
q00 + ωn ζq0 (−ωn ζ+iωd )t
e
q (t) = Re
q0 − i
ωd
and
q00 + ωn ζq0 (−ωn ζ+iωd )t
q (t) = Re (−ωn ζ + iωd ) q0 − i
e
ωd
0
Now at t0 = 0.07 sec the velocity is zero, since this is where the displacement is maximum (first peak).
Hence now we have one equation with one unknown q00 that we can solve for from the above
q00 + ωn ζq0 (−ωn ζ+iωd )t0
0 = Re (−ωn ζ + iωd ) q0 − i
e
(2.68)
ωd
q00 + ωn ζq0
q00 + ωn ζq0
iωd t0
−ωn ζt0
ωn ζ + iq0 ωd +
ωd e
(2.69)
=e
Re
−ωn ζq0 + i
ωd
ωd
0
q0 + ωn ζq0
0
iωd t0
−ωn ζt0
ωn ζ + q0 ωd + q0 e
=e
(2.70)
Re
i
ωd
0
q0 + ωn ζq0
iωd t0
−ωn ζt0
0 iωd t0
ωn ζ + q0 ωd e
=e
+ q0 e
Re i
(2.71)
ωd
−1 q00 + ωn ζq0
i(ωd t0 )
0 iωd t0
−ωn ζt0
ωn ζ + q0 ωd e
+ q0 e
=e
(2.72)
Re
i
ωd
−1 q00 + ωn ζq0
−ωn ζt0
i(ωd t0 )
0 iωd t0
ωn ζ + q0 ωd e
+ Re q0 e
(2.73)
=e
Re
i
ωd
0
q0 + ωn ζq0
0
−ωn ζt0
ωn ζ + q0 ωd sin (ωd t0 ) + q0 cos (ωd t0 )
(2.74)
=e
−
ωd
p
√
But q0 = −0.01 m/sec, ζ = 0.052, ωn = 60.3 rad/sec and ωd = ωn 1 − ζ 2 = 60.3 1 − 0.0522 = 60.
218, therefore ωd t0 = 60.218 × 0.07 = 4.2153 and q0 ωd = (60.218) (−0.01) = −0.60218 hence the above
equation becomes
0
q0 + (60.3) (0.052) (−0.01)
−(60.3)(0.052)(0.07)
0
60.3 × 0.052 − 0.602 sin (4.215) + q0 cos (4.215)
0=e
−
60.218
(2.75)
0
−2
q − 3.136 × 10
= 0.80293 − 0
(3.1356) − 0.602 (−0.879) + q00 (−0.477)
(2.76)
60.218
Solving for q00 gives
q00 = −1.231 m/sec
28
CHAPTER 2. HW’S
Now that we q00 ,
Now that we q00 , we can find the numerical value for b and write the general solution again.
q00 + ωn ζq0
ωd
−1.231 + 60.3 (0.052) (−0.01)
=−
60.218
= 2.0963 × 10−2
b=−
(2.77)
(2.78)
(2.79)
Hence from
q (t) = Re (q0 + ib) e(−ωn ζ+iωd )t
= Re (−0.01 + i0.0209) e(−ωn ζ+iωd
(2.80)
)t
(2.81)
√
givingAˆ = 0.012 + 0.02092 = 0.023
2.2.8
Problem 6
2.2.8.1
Part a
Assume the system is underdamped.
When the package hits the ground, its speed becomes zero. Therefore the impulse generated on it is
the change of linear momentum. Since it speed was v just before impact, then impulse= mv.
Hence the EQM is
mz 00 = −cz 0 − kz − Fimpulse + mg
mz 00 + cz 0 + kz = mg − Fimpulse
(2.82)
(2.83)
2.2. HW2
29
With the initial conditions now being z = 0 and z 0 = 0.
The response due to the force mg can be found from the response to a unit step of amplitude mg
Hence the response due to the force mg is
ζωn
mg
−ζωn t
cos ωd t +
u(t) =
1−e
sin ωd t
k
ωd
The response due to the impulse is the response of a free system with zero initial position but with initial
in the upward (negative) direction. Hence the response due to the impulse only is
velocity impulse
m
mv −ζωn t
e
sin ωd t
mωd
v
= e−ζωn t sin ωd t
ωd
g (t) =
Hence the downward displacement is given by
mg
ζωn
ve−ζωn t
−ζωn t
z(t) =
cos ωd t +
1−e
sin ωd t −
sin ωd t
k
ωd
ωd
2.2.8.2
(2.84)
(2.85)
(2.86)
Part b
Now that the impulse have taken place and we have accounted for it in the z (t) solution, then we can
use this expression to find the spring force since Fspring = kz (t) and the damping force on the mass
Fdamper = cz 0 (t). When resultant net force F is negative then the mass will rebound from the ground.
F = mg − kz(t) − cz 0 (t)
But
mg
z (t) =
k
ζωn
ve−ζωn t
−ζωn t
1−e
cos ωd t +
sin ωd t −
sin ωd t
ωd
ωd
(2.87)
Hence
mg
z (t) =
k
0
ζωn
ζωn ve−ζωn t
−ζωn t
−ζωn t
ζωn e
cos ωd t +
sin ωd t − e
[−ωd sin ωd t + ζωn cos ωd t] −
sin ωd t − ve−ζω
ωd
ωd
(2.88)
or
z 0 (t) =
e−ζωn t
kωd
kvωn ζ + gm ωd2 + ωn2 ζ 2
sin ωd t − kvωd cos (ωd t)
(2.89)
30
CHAPTER 2. HW’S
Hence
F = mg − kz (t) − cz 0 (t)
ζωn
mg
ve−ζωn t
e−ζωn t
−ζωn t
cos ωd t +
= mg − k
1−e
sin ωd t −
sin ωd t − c
k
ωd
ωd
kωd
(2.90)
kvωn ζ + gm ωd2 + ωn2 ζ 2
(2.91)
To find when this force will turn negative first time, we can take the derivative with respect to time and
set it to zero and solve for first t = t0 that will make it zero. Since the force was positive first, then it has
to become zero before turning negative.
2.2.8.3
Part(c)
p
k
Let m = 1 kg, ωn = 5 rad/sec3 , v = 4 m/s. Hence ωd = 5 1 − ζ 2 . Since ωn2 = m
, hence k = 25 N/m.
Also c = ζccr = ζ2mωn = 10ζ
Using these values, the force in part(b) is plotted for different values of ζ. For example, setting
ζ = 5% gives this plot of F (t) for t = 0 to 20 seconds.
The maximum force is seen as little over 20N.Therefore, to find which ζ gives the smallest value of
maximum force, we can try different values of ζ and see how the maximum force changes as a function of
ζ. Using software the following values of maximum for for different ζ are generated along with t = tmax
when this maximum occurs and with the time t = t0 when the mass rebounds first time from z = 0
3
maximum force (N)
ζ%
tmax (sec)
t0 (sec)
22.1
1
0.21
0.84
21.85
3
0.206
0.834
21.45
7
0.184
0.81
21.27
10
0.16
0.779
21.4
20
0.11
0.75
25.8
40
0.01
0.68
30
50
0.001
0.64
typo in book. hz is assumed to mean rad/sec
2.2. HW2
Most protection when damping ratio is below 10%
31
32
2.3
CHAPTER 2. HW’S
HW3
Resonance at 9 Hz
Node location
Resonance at 39 Hz
Lab observation 2/14/13
EME 545 Spring 2013
Nasser M. Abbasi
Resonance at 180 Hz
3 nodes counted in each ring
when it was in resonance
2.4. HW4
2.4
2.4.1
33
HW4
problem 1
Assuming zero initial conditions. The input to the system is made up of two inputs. We find the
response to the first input, then add this response to the response due to the second input. The first
input is
u1 (t) = F0 h (t) − F0 h (t − T )
= F0 (h (t) − h (t − T ))
Which is a rectangular pulse of width T starting at t = 0. For example for T = 10 sec. and F0 = 1
Assuming the response to unit step is gs (t) then the response to u1 (t) is
z }| { z
}|
{
g1 (t) = F0 gs (t) h (t) − gs (t − T ) h (t − T )
From appending B, gs (t) =
1
2
mωn
(1 − cos (ωn t)), hence the above becomes
z
}|
{ z
}|
{
1
1
g1 (t) = F0 
(1 − cos (ωn t)) h (t) −
(1 − cos (ωn (t − T ))) h (t − T )
2
mωn
mωn2
Looking at the second input given by u2 (t) = F0 e−β(t−T ) h (t − T )
(2.92)
34
CHAPTER 2. HW’S
From appendix B, the response to an exponential F0 e−βt h (t) is
F0
m (ωn2 + β 2 )
−βt
e
β
− cos (ωn t) −
sin (ωn t)
h (t)
ωn
Therefore the response to u2 (t) is
F0
g2 (t) =
m (ωn2 + β 2 )
β
−β(t−T )
sin (ωn (t − T ))
h (t − T )
e
− cos (ωn (t − T )) −
ωn
(2.93)
Adding Eqs 2.102 and 2.93 results in the final response
g (t) = g1 (t) + g1 (t)
1
1
(1 − cos (ωn t)) h (t) −
(1 − cos (ωn (t − T ))) h (t − T ) +
= F0
mωn2
mωn2
F0
β
−β(t−T )
e
− cos (ωn (t − T )) −
sin (ωn (t − T ))
h (t − T )
m (ωn2 + β 2 )
ωn
For illustration, the following plot shows the response using some values. Using m = 1 kg, ωn = 1
rad/sec, T = 10 sec, β = 1, F0 = 1 Volt.
2.4.2
Problem 2
2.4. HW4
2.4.2.1
35
part(a)
The differential equation is
1
M L2 θ00 (t) + cθ0 (t) + kθ (t) = F0 (h (t) − h (t − T ))
3
(2.94)
The initial conditions are not given, and assumed to be zero, therefore θ (0) = 0◦ and θ0 (0) = 0
rad/sec. The system is underdamped, hence
ωd = ωn
p
1 − ζ2
36
CHAPTER 2. HW’S
Let Td , be the damped period of oscillation given by
Td =
2π
2π
= p
ωd
ωn 1 − ζ 2
To obtain an expression for ωn , Eq 2.110 is changed to a standard form θ00 (t) + 2ζωn θ0 (t) + ωn2 θ (t) =
F0 (h(t)−h(t−T ))
1
M L2
3
ω2
2ζω
z }|n {
z }|n{
3c 0
3k
F0 (h (t) − h (t − T ))
θ00 (t) +
θ (t) +
θ (t) =
1
2
2
ML
ML
M L2
3
(2.95)
Therefore
ωn2 =
3k
M L2
Using k = 20 N/rad, L = 0.02 meter, M = 0.003 kg
ωn2 =
or
ωn =
3 (20)
= 5.0 × 107
(0.003) (0.02)2
√
5.0 × 107 = 7071 rad/sec
and
Td =
2π
√
= 0.8888 ms
7071.1 1 − 0.022
Therefore
T = 2.5Td = 2.5 × 0.88857 = 2.221 ms
To find F0 it is assumed the head was initially at rest. Therefore
F0 = kθ0
π = 10.472 N-meter
= 20
6
Eq 2.111 becomes
θ00 (t) + 2ζωn θ0 (t) + ωn2 θ (t) =
θ00 (t) + 2 (0.02) (7071) θ0 (t) + 5 × 107 θ (t) =
F0 (h (t) − h (t − 2.5Td ))
1
M L2
3
3 × 20 π6 (h (t) − h (t − 2.5Td ))
(0.003) (0.02)2
θ00 (t) + 283θ0 (t) + 5 × 107 θ (t) = 2.618 × 107 (h (t) − h (t − 0.0022219))
This is solved numerically for 0 < t < 10T with the initial conditions θ (0) = 0◦ and θ0 (0) = 0 rad/sec.
Here is a plot of the solution and the input on a second plot.
2.4. HW4
37
A computational software was used to numerically solve the above differential equation for the solution
θ (t).
2.4.2.2
Part(b)
From appendix B the response to underdamped second order system to a unit step u (t) is
1
qs (t) =
M ωn2
ζωn
−ζωn t
1−e
cos (ωd t) +
sin (ωd t)
h (t)
ωd
38
CHAPTER 2. HW’S
Hence the response U (t) due to
F0
( 13 mL2 )
F0
U (t) = 1 2 L
3
Notice the factor
00
0
F0
1 2
L
3
(h (t) − h (t − T )) is given by
z }| { z
}|
{
qs (t) h (t) − qs (t − T ) h (t − T )
. This was used since appendix B solution on based on equation of motion
θ (t) + 2ζωn θ (t) + ωn2 θ (t) = m1 while in this case, the equation of motion is θ00 (t) + 2ζωn θ0 (t) + ωn2 θ (t)
F0
, hence a factor of 1FL02 is needed to scale the solution. Therefore the analytical solution is
1
mL2
3
3
=
ζωn
3F0 /L2
−ζωn t
cos (ωd t) +
1−e
sin (ωd t)
h (t)
U (t) =
M ωn2
ωd
3F0 /L2
ζωn
−ζωn (t−2T )
−
1−e
sin (ωd (t − T ))
h (t − T )
cos (ωd (t − T )) +
M ωn2
ωd
To compare this solution with the numerical solution found in part(a), the two solutions are plotted
side-by-side for the case T = 2.5Td
We see that solutions are in good approximate. Here is a plot of the difference. The error is in the
order of 10−7
2.4. HW4
2.4.2.3
39
Part(c)
The analytical solutions for T = 2.5Td and T = 3.0Td are
We see when the step load duration is T = 2.5Td , the disk head will vibrate with larger amplitudes
than when the step duration was T = 3Td .
To understand the reason for this, analysis was done on the undamped version of the solution for
part b
From appendix B the response to undamped second order system to a unit step u (t) is
qs (t) =
1
(1 − cos (ωn t)) h (t)
M ωn2
40
CHAPTER 2. HW’S
Therefore the solution for 0 < t < T is
/L2
3F0 /L2
2
M ωn
(1 − cos (ωn t)). This means at t = T which is when the
2
step load is removed, θ (T ) = 3FM0ω2 (1 − cos (ωn T )) and θ0 (T ) = − 3FM0ω/L2 (ωn sin (ωn T )).
n
n
For t > T , the load is not present any more and we have free vibration response but with the above
initial conditions obtained at the end of the T . The solution to free vibration of an undamped system for
t˜ = t − T ≥ 0 is given by
θ0 (T )
sin ωn t˜ + θ (T ) cos ωn t˜
θ t˜ =
ωn
2
− 3FM0ω/L2 (ωn sin (ωn T ))
3F0 /L2
sin ωn t˜ +
(1 − cos (ωn T )) cos ωn t˜
ωn
M ωn2
2
2
3F0 /L2
3F
3F
0 /L
0 /L
=−
sin (ωn T ) sin ωn t˜ +
−
cos (ωn T ) cos ωn t˜
M ωn2
M ωn2
M ωn2
2
2
3F0 /L2
˜ − 3F0 /L cos (ωn T ) cos ωn t˜ + 3F0 /L cos ωn t˜
=−
sin
(ω
T
)
sin
ω
t
n
n
M ωn2
M ωn2
M ωn2
2
3F0 /L2
˜ + cos (ωn T ) cos ωn t˜ + 3F0 /L cos ωn t˜
sin
(ω
T
)
sin
ω
t
=−
n
n
M ωn2
M ωn2
=
n
(2.96)
We have obtained a solution for the time after the step load was removed. We now investigate the
result observed. We see that when T is close to an integer multiple of the period of the system, where we
call the period of the system T˜ to differentiate it from T , then
2π ˜
˜
sin ωn nT = sin
nT = sin (n2π) = 0
T˜
Also
cos ωn nT˜ = cos
2π ˜
nT
T˜
= cos (n2π) = 1
Hence the response given by equation 2.96 becomes
2
3F0 /L2
˜ + 3F0 /L cos ωn t˜
θ t˜ = −
cos
ω
t
n
M ωn2
M ωn2
=0
(2.97)
But if T occurs at multiple of halves of the period of the system (for example, T = 0.5T˜, 1.5T˜, 2.5T˜,
etc...) then now
!!
!!
T˜
T˜
2π
n
sin ωn n
→ sin
→ sin (nπ) → 0
2
2
T˜
However
T˜
cos ωn n
2
!
→ cos
2π T˜
n
T˜ 2
!
→ cos (nπ) → −1
We notice that the sign is now negative. This means equation 2.96 becomes
2
3F0 /L2
˜ − 3F0 /L cos ωn t˜
θ t˜ = −
cos
ω
t
n
M ωn2
M ωn2
6F0 /L2
cos ωn t˜
=−
M ωn2
(2.98)
2.4. HW4
41
Comparing Eqs 2.97 and 2.98 we see that when T is an integer multiple of the period of the system ,
then the response after T is minimal (zero for the case on undamped)
While when T occurs at multiple of halves of the period of the system ,the response is large beyond
the time T .
The above analysis was done for undamped system, but the same idea carries to the underdamped
case. This explains why the response dies out quickly when T = 3Td while it was large when T = 2.5Td
2.4.3
Problem 3
First lets look at the free vibration response (zero input response,
q
q called uzi ). The damping ratio
c
c
660
k
−2
ζ = cr = 2√km = 2√11000×1000 = 9.9499×10 ≈ 0.1 and ωn = m = 10000
hence ωn = 3.162 rad/sec
1000
p
√
and ωd = ωn 1 − ζ 2 = 3.1623 1 − 0.12 hence ωd = 3.146 rad/sec . The damped period of the system is
2π
Td = ω2πd = 3.146
= 1.997 seconds and the natural period is Tn =
Hence the system is underdamped and the solution is
(iωd −ζωn )t
ˆ
uzi = Re Ae
2π
ωn
=
2π
3.162
= 1.987 seconds.
Where Aˆ = a + ib is the complex amplitude. At t = 0 we have
a = uzi (0) = −0.1
and u0zi (0) ≡ u00 = Re ((iωd − ζωn ) (a + ib)) = −bωd − aζωn therefore b =
−u00 −aζωn
.
ωd
Since car was
dropped from rest, then we take u00 = 0 which leads to b = − (−0.1)(0.1)3.162
= 0.1
3.146
0
Hence, since a = u0 (0) ≡ u0 and
b=
−u00 − aζωn
= 0.1
ωd
then
uzi (t) = Re (a − ib) e(iωd −ζωn )t
0
u0 + aζωn
−ζωn t
iωd t
= Re e
u0 − i
e
ωd
0
u0 + aζωn
−ζωn t
=e
u0 (0) cos ωd t +
sin ωd t
ωd
(2.99)
42
CHAPTER 2. HW’S
For the numerical values gives, we now can plot this solution
uzi (t) = e−0.1(3.162)t (−0.1 cos 3.146t + 0.1 sin 3.146t)
The phase is given by tan−1
b
a
= tan−1
0.1
−0.1
= 2.356 rad = 1350 , In complex plane, uh (t) is
Now we add the zero initial conditions response, also called zero state response uzs for an input which
is an impulse using appendix B.
F0
−ζωn t
uzs (t) = e
sin ωd t h (t)
mωd
Hence uzs for an impulse that occurs at time T is
F0
−ζωn (t−T )
uzs (t) = e
sin ωd (t − T ) h (t − T )
mωd
(2.100)
2.4. HW4
43
Hence the solution is found by combining Eq. 2.112 and Eq 2.113
u (t) = uzi + uzs
0
u0 + aζωn
F0
−ζωn t
−ζωn (t−T )
u0 (0) cos ωd t +
=e
sin ωd t h (t) + e
sin ωd (t − T ) h (t − T )
ωd
mωd
We need now to solve for T and F0 in order to meet the requirements that u (t) should become zero
between for 2 < t < 5. To do this in the complex plane, we draw the zero state response as a vector
F0
−ζωn (t−T )
uzs = e
sin ωd (t − T ) h (t − T )
mωd
F0 e−ζωn (t−T ) 1 iωd (t−T )
h (t − T )
= Re
e
mωd
i
F0 e−ζωn (t−T ) i(ωd (t−T )− π2 )
= Re
e
h (t − T )
mωd
F0 e−ζωn (t−T )
π
and phase ωd t − ωd T −
Now to solve the problem
mωd
2
of finding T and F0 : To make the response become zero we need the magnitude of uzs to be equal but
opposite in sign to the magnitude of uzi so that the projection on the x-axis cancel out (the projection
on the x-axis of the vector is the real part which is the solution). Therefore, for the projection of uzs to
be the same as the projection of uzi but of different sign, the following diagram shows all the possible T
values that allows this. We will pick the first T value which is larger than 2 seconds to use.
Hence uzs vector has magnitude
44
CHAPTER 2. HW’S
2π− π − π4
4
From the above diagram, we need ωd T + π2 = 2π − π4 , hence T =
ωd
this value of T is not acceptable. We now look for the next possible T .
=
3
π
2
π
= 1.5 seconds. Hence
2π+ π − π
4
2
From the above diagram we see it will be ωd T + π2 = 2π + π4 hence T =
= 1.75 seconds.
π
Hence this is still too early to apply the impulse. We look at the next possible case. We see that now we
must rotate the vector all the way it was in the first diagram above to get the projection on the x-axis
canceling the projection of the free vibration vector. Hence now the relation to solve for is
π
π
= 4π −
2
4
ωd T +
Where in the above we added full 2π to the first case we considered above. This gives
T =
4π −
π
4
π
−
π
2
= 3.25 sec
.We have found T which brings the system to halt after at least 2 seconds has elapsed. Now we find F0
This is done by equating the amplitudes of the vectors as follows
F0 e−ζωn (t−T )
−ζωn t ˆ
=e
A mωd
2.4. HW4
45
Now fort = T = 3.25 second, plug-in numerical values
√
F0
= e−(0.1)(3.162)(3.25) 0.12 + 0.12
1000 (3.146)
F0
= 5.0607 × 10−2
3146.0
F0 = 159.21
To verify, here is a plot of the response when the impulse hit with
F0 = 159.21 N at t = 3.25 seconds
46
CHAPTER 2. HW’S
2.4.4
Problem 4
2.4.4.1
First part
Let A be the area of the cross section and ρ the mass density and L the length, then actual mass is
mactual = LAρ
= 50 (18 × 0.0254) (4 × 0.3048) (7800)
= 217393 kg
Hence we will use
m=
217393
= 72464 kg
3
The actual stiffness for a simply supported mean with loading at the center is
area moment of inertia. Hence
I=
wh3
(4 × 0.3048) (18 × 0.0254)3
=
= 0.00971 m4
12
12
48EI
L3
where I is the
2.4. HW4
47
Therefore the stiffness of the beam is
48EI
L3
48 (210 × 109 ) (0.00971)
=
= 783014 N/m
503
k=
The natural frequency is
r
r
k
783014
=
= 3.287rad/sec
ωn =
m
72464
fn = 0.523 Hz
ω t) where ω
Therefore, assuming the loading is given by F0 cos (¯
¯ is the forcing frequency. The dynamic
response at any time is given by
F0 /k
ˆ
X = q
(1 − r2 )2 + (2ζr)2
ˆ
ω
¯
Where r = ωn . We start by drawing X
¯ for the load of 1000 N by changing ω
¯ from 0 to 8π,
vs. ω
Hence for a single student the displacement vs. forcing frequency is
48
CHAPTER 2. HW’S
Hence we see that for one student, the maximum displacement is around 6 cm when the student is
jumping at resonance frequency.
To answer the question of how many students are needed to cause |X| to be 50 cm then that will
depend on what forcing frequency is used. Now we will find the minimum number of students needed.
The minimum number will be when they all jump at the resonance frequency which is found from
solving for ω
¯ resonance in
ω
¯ resonance p
= 1 − 2ζ 2
ωn
p
ω
¯ resonance = ωn 1 − 2ζ 2
q
= 3.287 1 − 2 (0.01)2
= 3.28667 rad/sec
Therefore, at this forcing frequency, we now solve for F0 to determine the number of students
ˆ
X = s
1−
ω
¯ resonance
ωn
F0 /k
2 2
2
+ 2ζ ω¯ resonance
ωn
v
2 !2 2
u
ω
¯
ω
¯ resonance
ˆ u
resonance
t
F 0 = k X 1−
+ 2ζ
ωn
ωn
v
u
2 !2 2
u
3.28667
3.28667
t
1−
= (783014) (0.5)
+ 2 (0.01)
3.287
3.287
= 7829.75 N
Therefore we need at least 8 students all jumping at 3.287 rad/sec to cause a displacement of at
least 50 cm.
2.4.4.2
Extra part
To make the structure avoid resonance, we need to make sure the ratio ω$n stays away from one. This is
the ratio of the forcing frequency to the natural frequency. One way is to make ωn much larger than any
expected $ that can occur is typical use of this structure.
q
k
But to make ωn = m
large, means either making m small or making k large. It is hard to reduce
the mass of the structure. Therefore, making the structure more stiff will be a better solution.
The bridge can be made more stiff in many ways, such as by adding additional truss structure to
it (assuming
this will add minimal weight). For this example, suppose we double the stiffness. Hence
q
q
ωn =
2k
m
=
2(783014)
72464
= 4.649rad/sec.
q
p
2
Therefore now ω
¯ resonance = ωn 1 − 2ζ = 4.649 1 − 2 (0.01)2 = 4.65 rad/sec. Now the same number
2.4. HW4
49
of students (8) as before, jumping at same frequency of 3.28667 will cause displacement of
8F0 /k
ˆ
X = s 2 2 2
1 − ωω¯n
+ 2ζ ωω¯n
= r
1−
8000/783014
2
2
3.28667 2
+ 2 (0.01) 3.28667
4.649
4.649
= 0.02 meter
Therefore by making the bridge twice as stiff, now the same 8 students at ω
¯ = 3.287 will cause only 2 cm
displacement instead of 50 cm.
2.4.5
Problem 5
A radar display is to be tested by mounting it on spring-dashpot suspension and subjecting it to harmonic
force Q = F cos (¯
ω t). The mounted mass is 8 kg and ζ = 0.25. A free vibration shows that damped
natural frequency fd = 5hz.It is observed that when the force is applied at very low frequency the
displacement amplitude is 2 mm. The test is to be performed at 5.2 Hz.What will be the steady state
response?
We are given are the following
m = 8 kg
ζ = 0.25
p
ωd = ωn 1 − ζ 2 = 2π (5) rad/sec
F0 /k = 0.002 meter
ω
¯ = 2π (5.2) rad/sec
Hence ωn = √ωd
1−ζ 2
=
√
2π(5)
1−0.252
= 32.446 rad/sec. The steady state response is given by
ˆ i¯ωt
uss = Re Xe
ˆ = X
ˆ eiθ . Hence
where X
ˆ
X = s
1−
= s
F0 /k
2 2
ω
¯
ωn
2
+ 2ζ ωω¯n
0.002
2 2 2
2π(5.2)
1 − 32.446
+ 2 (0.25) 2π(5.2)
32.446
= 0.00397
50
CHAPTER 2. HW’S
and
2ζr
θ = tan
1 − r2
2 (0.25)
−1
= tan
0
−1
tan (∞)
−1
Since 0 ≤ θ ≤ π then the phase is
θ=
π
2
Hence
ˆ i¯ωt
u = Re Xe
π
= Re 0.00397ei 2 ei¯ωt
π
= 0.00397 cos ω
¯t +
2
= −0.00397 sin (¯
ω t)
2.4.6
Problem 6
A one degree of freedom system whose mass is 10 kg and whose natural frequency is 1 khz is subjected
to a harmonic excitation 1.2 sin ω
¯ t kN. The steady state amplitude when ω
¯ = 1 khz is observed to be 2.4
mm. Determine the steady state response at ω
¯ = 0.95 khz and ω
¯ = 1.05 khz.
We are given
m = 10 kg
ωn = 2π (1000) rad/sec
F0 = 1200 N
|X| = 2.4 × 10−3 meter when ω
¯ = ωn
Since ωn2 =
k
,
m
hence k = ωn2 m = (2π (1000))2 (10), therefore
k = 3.949 × 108 N/m
2.4. HW4
51
Now when ω
¯ = ωn we have
F0 /k
ˆ
X = s
2 2 2
1 − ωω¯n
+ 2ζ ωω¯n
2.4 × 10−3 =
1200/ (3.949 × 108 )
q
(2ζ)2
3.039 × 10−6
=
2ζ
Hence
3.039 × 10−6
ζ=
2 × 2.4 × 10−3
= 0.000633
2.4.6.1
Part (1)
when ω
¯ = 2π (950) now r = ωω¯n < 1 hence dynamic magnification factor is positive. Therefore loading
and displacement will be in phase with each others. (i.e. displacement is in same direction as force).
Since the force is sin then the response will be sin with same frequency but different phase and amplitude.
Hence let
uss = X sin (¯
ω t − θ)
Where
X = s
1−
= s
F0 /k
2 2
ω
¯
ωn
2
+ 2ζ ωω¯n
1200/ (3.949 × 108 )
2 2 2
2π(950)
2π(950)
1 − 2π(1000)
+ 2 (0.000633) 2π(1000)
= 3.116 × 10−5 meter
and
2ζr
θ = tan
1 − r2


2π(950)
 2 (0.000633) 2π(1000) 
= tan−1 
2 
2π(950)
1 − 2π(1000)
= tan−1 1.234 × 10−2
= 0.01235 radians
= 0.71◦
−1
Hence steady state response is
uss = 3.116 × 10−5 sin (¯
ω t − 0.71◦ )
Hence we see that the displacement is lagging the load by 0.71◦ . On complex plane it looks as follows
52
CHAPTER 2. HW’S
2.4.6.2
Part (2)
When ω
¯ = 2π (1050) now r = ωω¯n > 1 hence dynamic magnification factor is negative. Therefore loading
and displacement will be out of phase with loading. (i.e .displacement is in opposite direction to force).
Doing the same calculations are done as above
uss = X sin (¯
ω t − θ)
where X
X = s
1−
= s
F0 /k
2 2
ω
¯
ωn
+
2ζ ωω¯n
2
1200/ (3.949 × 108 )
2 2 2
2π(1050)
1 − 2π(1000)
+ 2 (0.000633) 2π(1050)
2π(1000)
= 2.964 × 10−5 meter
2.4. HW4
53
and
2ζr
θ = tan
1 − r2


2π(1050)
 2 (0.000633) 2π(1000) 
= tan−1 
2 
1 − 2π(1050)
2π(1000)
0.0013293
−1
= tan
−0.1025
= 3.12862 radians
= 179.257◦
−1
Hence steady state response is
uss = 2.964 × 10−5 sin (¯
ω t − 179.257◦ )
On complex plane it looks as follows
Here is a plot by hand for the above 2 cases. First, the period that the loading is using T =
1.0526 × 10−3 sec
T = 1.053 ms
2π
$
=
1
950
=
54
CHAPTER 2. HW’S
2.5
2.5.1
HW5
problem 1
Assuming the 2 masses move together (else we will have 2 systems and 2 equations of motions. Hence
I assumed that they move together as one body).
(m1 + m2 ) y 00 + y 0 µ + ky = f (t)
Since y (t) = A sin (ωt) hence
y (t) = Re
A iωt
e
i
Let
f (t) = Re
Fˆ i(ωt)
e
i
!
Where Fˆ is the complex amplitude of the force. Now we substitute all these in the differential equation
above.
y 0 = Re ωAeiωt
y 00 = Re iω 2 Aeiωt
2.5. HW5
55
(m1 + m2 ) y 00 + y 0 µ + ky = Re
Re iω 2 Ae
iωt
A iωt
e
= Re
i
1
iωt
2
= Re
Re iω (m1 + m2 ) + ωµ + k Ae
i
Fˆ
1
2
iω (m1 + m2 ) + ωµ + k A =
i
i
(m1 + m2 ) + Re ωAeiωt µ + k Re
Fˆ i(ωt)
e
i
!
Fˆ i(ωt)
e
i
!
Fˆ i(ωt)
e
i
!
Hence
Fˆ = −ω 2 (m1 + m2 ) + iωµ + k A
k = 3.2 × 103 Nm, µ = 40 Ns/m,A = 0.02 meter. When ω = 75rad/sec the above becomes
Fˆ = −752 (1.5) + i75 × 40 + 3.2 × 103 0.02
= −104.75 + 60.0i
60.0
Hence Re Fˆ = −104.75N and the phase is tan−1 − 104.75
= 2.62 rad/sec
When ω = 85
40
2
ˆ
F = 1.5 −85 + i85
+ 2133.3 0.02
1.5
= −152.75 + 68.0i
68
Re Fˆ = −152.75N and the phase is tan−1 − 152.75
= 2.722 rad/sec
2.5.2
subproblem 2
56
CHAPTER 2. HW’S
2.5.2.1
Part(a)
Force transmitted to floor is given by
Ftr = cz 0 + kz
Let f (t) = F cos (ωt) = Re (F eiωt ) = Re (F eiωt ) where we are given that F = 20 × 103 N. ω =
= 60π = 188.50 rad/sec or 30 Hz .
2π 1800
60
2ζr
1
Let zss = Re Fk |D| ei(ωt−φ) where φ = tan−1 1−r
and |D| = √
. and r = ωωn Hence
2
(1−r2 )2 +(2ζr)2
i(ωt−φ+ π2 )
F
F
0
i(ωt−φ)
. Therefore
z = Re iω k |D| e
= Re ω k |D| e
F
F
i(ωt−φ+ π2 )
i(ωt−φ)
Ftr = c Re ω |D| e
+ k Re
|D| e
k
k
Where c = 2ζωn m and When Ftr = 2 × 103 N . We now solve for k from
F
F
i(ωt−φ+ π2 )
i(ωt−φ)
3
|D| e
2 × 10 ≥ 2ζωn m Re ω |D| e
+ k Re
k
k
Taking the maximum case for RHS where exponential are unity magnitude, hence
F
2 × 103 = 2ζωn mω |D| + F |D|
k F
= 2ζωn mω
+ F |D|
k
F 1 + 2ζωn m
ω
k
=q
2
2
(1 − r ) + (2ζr)2
2.5. HW5
57
Where r =
ω
ωn
= √ω k . Hence the above becomes
m
F 1 + 2ζωn m
ω
k
2 × 10 = s
2
2 ω2
ω
1− k
+ 2ζ √ k
3
m
m
In the above everything
is known except for k which we solve for. Plugging the numerical values
given. ω = 2π 1800
,m
=
450,
F = 20 × 103 , ζ = 0.03 hence
60
q
k 450
3
20 × 10 1 + 2 (0.03) 450
(60π)
k
2 × 103 = s
2
2
450(60π)2
60π
1−
+ 2 (0.03) √ k
k
450
Hence k = 1.2135 × 106 N/m Hence ωn =
2.5.2.2
q
k
m
=
q
1.2135×106
450
= 51.929 rad/sec or 8.265 Hz
part(b)
The total displacement is given by
z (t) = ztransient (t) + zss (t)
=e
−ζωn t
(A cos ωd t + B sin ωd t) + Re
F
|D| ei(ωt−φ)
k
Where
ztransient (t) = e−ζωn t (A cos ωd t + B sin ωd t)
Assuming at t = 0 the system is relaxed hence z (0) = 0 and z 0 (0) = 0 we can determine A, B from
Eq ??.
At t = 0,
z (0) = 0
= A + Re
Hence
A = − Re
F
|D| e−iφ
k
F
|D| e−iφ
k
and
z 0 (t) = −ζωn e−ζωn t (A cos ωd t + B sin ωd t) + e−ζωn t (−ωd A sin ωd t + ωd B cos ωd t)
F
i(−φ+ π2 )
+ Re ω |D| e
k
Hence at t = 0
z 0 (0) = 0
F
i(−φ+ π2 )
= −ζωn A + ωd B + Re ω |D| e
k
58
CHAPTER 2. HW’S
Hence
ζωn
1
F
i(−φ+ π2 )
B=
A−
Re ω |D| e
ωd
ωd
k
1
ζωn
F
F
i(−φ+ π2 )
−iφ
−
=−
Re
|D| e
Re ω |D| e
ωd
k
ωd
k
Therefore the displacement is
F
ζωn
1
F
F
i(−φ+ π2 )
−iφ
−iφ
− Re
cos ωd t + −
−
z (t) = e
|D| e
Re
|D| e
Re ω |D| e
sin ωd t
k
ωd
k
ωd
k
F
i(ωt− π2 −φ)
+ Re
|D| e
k
−ζωn t
Hence expressed in sin and cos
z (t) = −
ζωn F
F
1
F
−ζωn t
−ζωn t
|D| cos φ e
cos ωd t + e
|D| cos φ −
sin ωd t
−
ω |D| sin φ
k
ωd
k
ωd
k
F
|D| cos (ωt − φ)
k
F
F
F |D|
=−
|D| cos φ e−ζωn t cos ωd t + e−ζωn t
(−ζωn cos φ − ω sin φ) sin ωd t + |D| cos (ωt − φ)
k
ωd k
k
F
1
F
= |D| e−ζωn t − cos φ cos ωd t +
(−ζωn cos φ − ω sin φ) sin ωd t + |D| cos (ωt − φ)
k
ωd
k
+
p
√
Since ζ = 0.03 then ωd = ωn 1 − ζ 2 = ωn 1 − 0.032 = 0.99955 (ωn ). Therefore in the above we can
just replace ωd by ωn with very good approximation, hence we now obtain
F
F
1
−ζωn t
z (t) = |D| e
(−ζωn cos φ − ω sin φ) sin ωn t + |D| cos (ωt − φ)
− cos φ cos ωn t +
k
ω
k
n
F
ω
F
= |D| e−ζωn t − cos φ cos ωn t − ζ cos φ +
sin φ sin ωn t + |D| cos (ωt − φ)
k
ωn
k
2ζ
ω
ωn
, and φ = tan−1
2 . The
2
1−
( ωωn )
1−( ωω )
+(2ζ ( ωω ))
n
n
transient solution usually goes away after 5 or 6 cycles. Hence let us assume that the start up time takes
2π
6 × ω2πn = 6 × √2πk = 6 × q 1.2135×10
= 0.72597 seconds. Or 1 second at worst.
6
This is the amplitude. In the above |D| =
m
r
1
2 2
450
Therefore we can now plot the amplitude for t = 0 to t = 1 second in increments of 0.1 second, and
each time advance, we can increment ω from 0 to 60π in linear fashion, hence each 0.1 second we update
ω by an amount 6π. After 1 second has passed, the system is assumed to be in steady state, and then we
keep ω fixed at 60π rad/sec. This is a plot showing z (t) for t = 0 to 2 seconds given the above method
of changing ω
To avoid going over 10mm, this means we have to avoid the case of r = 1 or ω = ωn . When I first
just incremented ωn such that r = 1 was not avoided, resonance caused the amplitude to go over 10mm
as given in this plot. The transient solution itself stayed just below 10mm but the steady state solution
went over 10mm due to resonance
2.5. HW5
2.5.2.3
59
Part(c)
To insure that the amplitude does not go over 10mm, we need to add mass to the generator. Maximum
20×103
1
amplitude is given by Fk 2ζ1 = 1.2135×10
6 2(0.03) = 0.27469 meter or 274mm
3
1
So to insure maximum does not exceed 10mm , solve for new k from 0.01 = 20×10
, hence kn = 3.
kn
2(0.03)
q
7
3333 × 107 . Since ωn = 51.929 = mknn then new mass is mn = 3.3333×10
= 12361 kg using these values,
51.9292
the above plot now are redone. This is the result
60
CHAPTER 2. HW’S
We see that now the maximum displacement remained below 10 mm.
2.5.3
Problem 3
2.5. HW5
61
Let ε = 50mm = 0.05m be the distance of the unbalance mass m. Let M = 80kg be the mass of the
motor. The equation of motion is given by
(M + m) y 00 + cy 0 + ky = mεΩ2 sin (Ωt)
m
1 iΩt
00
0
2
2
y + 2ζωn y + ωn y =
εΩ Re
e
m+M
i
Where ωn =
q
k
m+M
and ζ =
c
.
2(M +m)ωn
Y =
Let y = Re
Y
i
eiΩt . This leads to
m
εΩ2
m + M ωn2 − Ω2 + 2iζωn Ω
Since static deflection is 40mm, then
But ωn2 =
2.5.3.1
k
m+M
=
(M +m)g
0.04(M +m)
(M + m) g
= 0.04
k
(M + m) g
k=
0.04
q
g
= 0.04
, hence ωn = 9.81
= 15.66 rad/sec or 2.492 Hz
0.04
part(a)
Since at steady state the displacement is 10 mm, then Ω = 2π 145
= 15.184 or 2.4167 Hz hence
60
Y iΩt
εm
r2
i(Ωt− π2 )
y = Re
e
e
= Re
i
m + M (1 − r2 + 2iζr)
εmr2
1
−iφ i(Ωt− π2 )
q
=
Re
e
e
m+M
(1 − r2 )2 + (2ζr)2
Where φ = tan−1
2ζr
1−r2
0.01 =
=
. r=
Ω
ωn
=
15.184
15.66
= 0.9696 hence the above becomes, at steady state
(0.05) m
0.96962
q
Re ei(15.184
m + 80
(1 − 0.96962 )2 + (2ζ0.9696)2
t− π2 −φ)
(0.05) m
0.96962
q
sin (15.184 t − φ)
m + 80
2
2
2
(1 − 0.9696 ) + (2ζ0.9696)
(2.101)
We are now told that at Ω = 15.184 and when Ωt = 750 then the displacement is zero, hence
0=
(0.05) m
q
m + 80
0.96962
(1 − 0.96962 )2 + (2ζ0.9696)2
or
sin 750 − φ = 0
750 − φ = 0
φ = 750
sin 750 − φ
62
CHAPTER 2. HW’S
Since φ = tan−1
2ζr
1−r2
then
π 2ζ0.9696
−1
75
= tan
180
1 − 0.96962
Hence
−1
tan
2ζ0.9696
= 1.3090
1 − 0.96962
2ζ0.9696
= tan (1.3090)
1 − 0.96962
2ζ0.9696
= 3.7321
1 − 0.96962
Hence ζ = 0.11523
2.5.3.2
Part(b)
From Eq 2.101
0.01 =
(0.05) m
0.96962
q
sin (15.184 t − φ)
m + 80
2
2
2
(1 − 0.9696 ) + (2ζ0.9696)
The maximum amplitude is when
0.01 =
(0.05) m
q
m + 80
0.96962
(1 − 0.96962 )2 + (2ζ0.9696)2
But ζ = 0.11523, hence we now solve for m
0.01 =
(0.05) m
0.96962
q
m + 80
(1 − 0.96962 )2 + (2 (0.11523) 0.9696)2
Hence
m = 4.1 kg
Hence εm = (0.05) (4.1) = 0.20 kg meter
2.5.3.3
Part(c)
since
y=
εmr2
1
i(Ωt− π2 −φ)
q
Re
e
m+M
(1 − r2 )2 + (2ζr)2
2
As Ω becomes much larger than ωn then (1 − r2 ) → r4 . Now dividing numerator and denominator
by r2 gives
y=
εm
1
q
sin (Ωt − φ)
4
m + M (r +(2ζr)2 )
r4
=
εm
1
r
m+M
1+
4ζ 2
r2
sin (Ωt − φ)
2.5. HW5
63
as r becomes large then
4ζ 2
r2
→ 0 hence
y'
εm
sin (Ωt − φ)
m+M
The smallest possible amplitude is
0.20
4.1 + 80
= 2.3781 × 10−3 meter
|y| =
or
|y| = 2.38 mm
2.5.4
problem 4
2.5.4.1
Part(a)
(note: total mass of system includes the small unbalanced masses) Since static deflection is 8.5mm, then
But ωn2 =
2.5.5
k
M
=
Mg
0.0085M
Mg
= 0.0085
k
Mg
k=
0.0085
q
g
9.81
= 0.0085 , hence ωn = 0.0085
= 33.972 rad/sec or 5.4068 Hz
Part(b)
The equation of motion is (angle Ω is now measured from horizontal, anti-clock wise positive)
00
0
2
M y + cy = 2mεΩ sin (Ωt) = Re
1
2mεΩ2 ei(Ωt)
i
64
CHAPTER 2. HW’S
Let y (t) = Re
1
Y
i
eiΩt hence y 0 (t) = Re Y ΩeiΩt ,y 00 (t) = Re iY Ω2 eiΩt , hence the above becomes
1 2mεΩ2 iΩt
c
iΩt
= Re
Re iY Ω e
+
Re Y Ωe
e
M
i M
1 2mεΩ2 iΩt
cΩ
2
iΩt
= Re
Re
iΩ +
Ye
e
M
i M
cΩ
1 2mεΩ2
2
iΩ +
Y =
M
i M
2 iΩt
2
1 2mεΩ
M
Y =
2
i iΩ + cΩ
M
=
2mεΩ2
icΩ − M Ω2
Hence
1 i(Ωt)
yss (t) = Re
Ye
i
1 2mεΩ2
i(Ωt)
= Re
e
i icΩ − M Ω2
Now we are told when Ωt = π2 (upright position) then y = 0(since it passes static equilibrium). At
this moment Ω = 2π 900
= 94.248 rad/sec , At this moment the centripetal forces equal the damping
60
force downwards (since the mass was moving upwards). Hence
mεΩ2 = cy 0 (t)
But from above we found that
2mεΩ2
iΩt
Ωe
y (t) = Re
icΩ − M Ω2
(0.5) 94.2482
i π2
= Re
94.248e
ic (94.248) − 200 (94.248)2
8.3718 × 105
i π2
= Re
e
94.248ic − 1.7765 × 106
0
Hence
mεΩ2 = c |y 0 (t)|
(0.5) 94.2482 = c q
8.3718 × 105
(94.248c)2 + (1.7765 × 106 )2
c
4441.3 = 8.3718 × 105 √
8882.7c2 + 3.1560 × 1012
Solving numerically for c gives
c = 1.0882 × 104 N second per meter
2.5. HW5
2.5.5.1
65
Part(c)
When Ω = 2π 1000
= 104.72 rad/sec or 16.667 Hz. From
60
1 2mεΩ2
i(Ωt)
y = Re
e
i icΩ − M Ω2
2 (0.5) (104.72)2
i(104.72t− π2 )
= Re
2e
4
i (1.0882 × 10 ) 104.72 − 200 (104.72)
10966.
i(104.72t− π2 )
= Re
e
i1.1396 × 106 − 2.1933 × 106
|y| = 4.4 mm
!
66
CHAPTER 2. HW’S
2.6
2.6.1
HW6
problem 1
3.41 in text: A periodic disturbance consists of a sequence of exponentially pulse repeated at intervals
−λt
T , such that Q (t) = F e T for 0 < t < T , and Q (t ± T ) = Q (t). The parameter λ is nondimensional.
Determine the complex Fourier series representing the force. Evaluate the first 5 coefficients when
λ = 0.1, 1, 10. What does this reveal regarding the influence of λ on the frequency spectrum?
˜ (t) be the Fourier series approximation to Q (t) given by
Let Q
∞
2π
1 X
˜
Q (t) =
Fn ein T t
2 n=−∞
(2.102)
Where
2
Fn =
T
ZT
2π
Q (t) e−in T t dt
0
2
=
T
ZT
Fe
−λt
T
e
t
−in 2π
T
2F
dt =
T
0
ZT
e
− Tλ )
−t(in 2π
T
2F
dt =
T
0
2π
λ
e−t(in T − T )
in 2π
− Tλ
T
!T
0
2π
λ
2F
e−T (in T − T ) − 1
=
in2π − λ
2F
=
e−in2π e−λ − 1
in2π − λ
But e−in2π = 1, hence
Fn =
2F
e−λ − 1
in2π − λ
∞
2π
1 X
2F
˜
Q (t) =
e−λ − 1 ein T t
2 n=−∞ in2π − λ
∞
X
e−λ − 1 in 2π t
=F
e T
in2π
−
λ
n=−∞
=F
∞
X
1 − e−λ in 2π t
e T
λ
+
in2π
n=−∞
For n = −2, −1, 0, 1, 2 we obtain
2
X
1 − e−λ in 2π t
e T
λ
+
in2π
n=−2
1 − e−λ −i 4π t 1 − e−λ −i 2π t 1 − e−λ 1 − e−λ i 2π t 1 − e−λ i 4π t
=F
e T +
e T +
+
e T +
e T
λ − i4π
λ − i2π
λ
λ + i2π
λ + i4π
˜ (t) = F
Q
2.6. HW6
67
For λ = 0.1
−0.1
4π
1 − e−0.1 −i 2π t 1 − e−0.1
1 − e−0.1 i 2π t 1 − e−0.1 i 4π t
˜ (t) = F 1 − e
Q
e−i T t +
e T +
+
e T +
e T
0.1 − i4π
0.1 − i2π
0.1
0.1 + i2π
0.1 + i4π
4π
= F { 6.026 × 10−5 + 7.572 × 10−3 i e−i T t
2π
+ 2.41 × 10−4 + 1.514 × 10−2 i e−i T t
+ 0.952
2π
+ 2.4099 × 10−4 − 1.5142 × 10−2 i ei T t
4π
+ 6.026 × 10−5 − 7.572 × 10−3 i ei T t }
For λ = 1
−1
−1
−1
−1
−1
4π
2π
2π
4π
1
−
e
1
−
e
1
−
e
1
−
e
1
−
e
−i
−i
i
i
t
t
t
t
˜ (t) = F
Q
e T +
e T +
+
e T +
e T
1 − i4π
1 − i2π
1
1 + i2π
1 + i4π
4π
2π
2π
= F {(0.00398 + 0.05i) e−i T t + (0.016 + 0.098i) e−i T t + 0.632 + (0.016 + 0.098i) ei T t + (0.00398 + 0.05i) ei
For λ = 10
1 − e−10 i 2π t 1 − e−10 i 4π t
1 − e−10 −i 4π t 1 − e−10 −i 2π t 1 − e−10
e T +
e T +
+
e T +
e T
10 − i4π
10 − i2π
10
10 + i2π
10 + i4π
4π
= F { 3.877 × 10−2 + 4.872 × 10−2 i e−i T t
2π
+ 7.169 × 10−2 + 4.505 × 10−2 i e−i T t
+ 0.1
2π
+ 7.169 × 10−2 − 4.505 × 10−2 i ei T t
4π
+ 3.877 × 10−2 − 4.872 × 10−2 i ei T t }
˜ (t) = F
Q
We notice that as λ became larger, the DC term became smaller. Since the DC term represents
average value of the whole signal, then we can say that as λ gets larger, then the average becomes smaller.
This means the energy of the signal becomes smaller as λ becomes larger.
2.6.1.1
Verification using Matlab ffteasy.m
From above, we found for λ = 1
2F
e−λ − 1
in2π − λ
2F
=
e−1 − 1
in2π − 1
Fn =
and the first 5 found to be
n
Fn
−2
0.00398 + 0.05i
−1
0.016 + 0.098i
0
0.632
1
0.016 − 0.098i
2
0.00398 − 0.05i
68
CHAPTER 2. HW’S
To verify the result with ffteasy.m using λ = 1, Using F = 1, and using T = 1. This below shows the
result for F0 , F1 , F2 and we see that the DC term F0 agrees, and that complex component of F1 , F2 also
agrees. The real parts are little larger than what I obtained using the above. This might be a scaling
issue, and I was not able to determine the reason for it at this time.
EDU>> T=1; del=0.01; t=0:del:T; lambda=1; xt=exp(-lambda*t/T);
EDU>> (1/length(t))*fft_easy(xt,t)
ans =
0.6326 + 0.0000i
0.0190 - 0.0986i
0.0072 - 0.0502i
2.6.2
problem 2
2.6. HW6
69
We are given that m = 1200 kg, f = 5 Hz, ζ = 0.4 and
(
z (x) =
x − 5x2
0 < x < 0.2
0
0.2 < x < 4
A plot of z (x) for first 20 meters is
z[x_] := Piecewise[{{x - 5 x^2, 0 <= x < 0.2}, {0, 0.2 <= x <= 4}}]
z[x_] /; x > 4 := z[Mod[x, 4]];
Table[{x, z[x]}, {x, 0, 21, .1}];
ListLinePlot[%, PlotRange -> {All, {0, .07}}, Frame -> True,
FrameLabel -> {{"z(x) hight or road (mm)", None}, {"meter",
"bumps on road"}}]
n
o
2π
We need to be able to express z (t) as Re Zei T t where T is the period of the function z (t). Hence
we need to represent z (x) as Fourier series approximation then replace x = vt and use the result.
The period T = 4 meter. Let z˜ (x) be the Fourier series approximation to z (x), hence
N
X
2π
1
z˜ (x) = F0 + Re
Fn ein T x
2
n=1
!
Where
2
Fn =
T
ZT
−in 2π
x
T
z (x) e
1
dx =
2
0
Using integration by parts
2/10
Z
x − 5x
0
R
udv = uv −
2
−in π2 x
e
1
dx =
2
2/10
2/10
Z
Z
π
π
5
xe−in 2 x dx −
x2 e−in 2 x dx
2
0
R
0
π
vdu, letting u = x and dv = e−in 2 x then v =
R
π
e−in 2 x dx =
70
CHAPTER 2. HW’S
π
ie−in 2 x
hence
n π2
2/10
2/10
2
Z
Z
π
−in π2 x 10
π
ie−in 2 x
ie
−in 2 x
xe
dx = x
−
dx
n π2 0
n π2
0
0
π 2
2 ie−in 2 10
2
=
−
π
10 n 2
nπ
π
−in 10
4 ie
=
10 nπ
i2
−
nπ
2/10
Z
π
ie−in 2 x dx
0
π
e−in 2 x
−in π2
π
−in 10
102
0
4 ie
4
+ 2 2 e 2 0
10 nπ
nπ
π
π 2
4 ie−in 10
4 =
+ 2 2 e−in 2 10 − 1
10 nπ
nπ
π
4
4
4i −in π
e 10 + 2 2 e−in 10 − 2 2
=
10nπ nπ nπ
π
2i
4
4
+
= e−in 10
− 2 2
2
2
nπ
5nπ
nπ
=
2
−in π x 10
2/10
Z
π
Now we do the second integral
x2 e−in 2 x dx.
0
Integration by parts,
R
udv = uv −
R
π
vdu, letting u = x2 and dv = e−in 2 x then v =
2/10
2/10
2
Z
Z
π
−in π2 x 10
ie−in 2 x
2 −in π2 x
2 ie
xe
dx = x
−
2x
dx
n π2
n π2
0
0
0
π
8 ie−in 10
4i
−
=
100 nπ
nπ
2/10
Z
π
xe−in 2 x dx
0
2/10
Z
π
But
xe−in 2 x dx was solved before and its results is Eq 2.1, hence
0
2/10
Z
π
π
4 ie−in 10
4i
4
2i
4
2 −in π2 x
−in 10
xe
dx =
−
e
+
− 2 2
100 n π2
nπ
n2 π 2 5nπ
nπ
0
π
8i −in π
16i
8
16i
−in 10
10
−e
=
e
− 2 2 + 3 3
3
3
100nπ
nπ
5n π
nπ
π
8i
16i
8
16i
= e−in 10
− 3 3+ 2 2 + 3 3
100nπ n π
5n π
nπ
π
ie−in 2 x
n π2
hence
2.6. HW6
71
Putting all the above together, we obtain Fn as
2/10
2/10
Z
Z
π
π
5
xe−in 2 x dx −
x2 e−in 2 x dx
2
0
0
4
4
5 −in π
8i
16i
16i
2i
8
1 −in π
10
10
e
− 2 2 −
e
−
+ 3 3
+
+
=
2
n2 π 2 5nπ
nπ
2
100nπ n3 π 3 5n2 π 2
nπ
π
π
2
2
20i
40i
i
20
40i
− 2 2 − e−in 10
− 3 3+ 2 2 − 3 3
+
= e−in 10
2
2
nπ
5nπ
nπ
100nπ n π
5n π
nπ
π
2
20i
40i
2
i
4
40i
= e−in 10
−
+ 3 3− 2 2 − 2 2− 3 3
+
2
2
nπ
5nπ 100nπ n π
nπ
nπ
nπ
π
40i
2
2
40i
−
= e−in 10
−
−
n3 π 3 n2 π 2
n2 π 2 n3 π 3
1
Fn =
2
Now
2
F0 =
T
ZT
0
1
z (x) dx =
2
2/10
Z
1
x − 5x2 dx =
300
0
Hence
!
N
X
1
in 2π
x
Fn e T
z˜ (x) = F0 + Re
2
n=1
!
N X
π
π
1
40i
2
40i
2
=
+ Re
e−in 10
− 2 2 − 2 2 − 3 3 ein 2 x
3π3
600
n
nπ
nπ
nπ
n=1
!
N
X
nπ
nπ
π
1
40i
2
40i
2
=
+ Re
ei( 2 x− 10 )
− 2 2 − ein 2 x
+
3π3
600
n
nπ
n2 π 2 n3 π 3
n=1
N
X
nπ
nπ
nπ
π
−40 1 i( nπ
2
2
1
40 1 π
=
+ Re
e 2 x− 10 ) − 2 2 ei( 2 x− 10 ) − 2 2 ein 2 x + 3 3 ein 2 x
3
3
600
nπ i
nπ
nπ
nπ i
n=1
!
But x = vt, hence
N
X
nπ
nπv
nπ
πv
2
2
1
−40 1 i( nπv
40 1 πv
z˜ (t) =
+ Re
e 2 t− 10 ) − 2 2 ei( 2 t− 10 ) − 2 2 ein 2 t + 3 3 ein 2 t
3
3
600
nπ i
nπ
nπ
nπ i
n=1
Therefore the forcing frequency is n$1 = n πv
or from 2πf1 =
2
πv
,
2
!
hence f1 = v4 Hz.The above can be
72
CHAPTER 2. HW’S
written as
X
N
N
X
nπ
nπv
nπ
1
−40 1 i( nπv
2
i
t−
t−
)
(
)
z˜ (t) =
+
Re
e 2 10
−
Re
e 2 10
2π2
600 n=1
n3 π 3 i
n
n=1
N
N
X
X
2 in πv t
40 1 in πv t
−
Re
+
Re
e 2
e 2
2π2
3π3 i
n
n
n=1
n=1
=
N
N
X
1
−40
nπ X 2
nπ +
−
sin
n$
t
−
cos
n$
t
−
1
1
600 n=1 n3 π 3
10
n2 π 2
10
n=1
N
N
X
X
2
40
−
cos
(n$
t)
+
sin (n$1 t)
1
n2 π 2
n3 π 3
n=1
n=1
Where $1 =
=
N
N
1
40 X 1
nπ 2X1
nπ − 3
sin
n$
t
−
−
cos
n$
t
−
1
1
600 π n=1 n3
10
π 2 n=1 n2
10
−
N
N
2X1
40 X 1
cos
(n$
t)
+
sin (n$1 t)
1
π 2 n=1 n2
π 3 n=1 n3
πv
2
To verify the above, here is a plot for different number of fourier series terms showing that approximation improves as N increases. This was done for v = 5m/s and for 5 seconds.
2.6. HW6
2.6.2.1
73
Part(a)
The equation of motion is
my 00 + c (y 0 − z 0 ) + k (y − z) = 0
my 00 + cy 0 + ky = cz 0 + kz
(2.1)
74
CHAPTER 2. HW’S
From earlier, we found that fourier series approximation to z (t) is
!
∞
X
nπ
nπ
1
−40 1 i(n$t− 10 )
2
2
40 1
z (t) =
+ Re
e
− 2 2 ei(n$t− 10 ) − 2 2 ein$t + 3 3 ein$t
3
3
600
nπ i
nπ
nπ
nπ i
n=1
!
∞
X
1
−40 −i nπ 1 in$t
2 −i nπ in$t
2 in$t
40 1 in$t
=
+ Re
− 2 2 e 10 e
e 10 e
− 2 2e
+ 3 3 e
3π3
600
n
i
nπ
nπ
nπ i
n=1
!
∞
X
nπ
π
nπ
π
−40
1
2
2
40
=
+ Re
ein$t 3 3 e−i( 10 + 2 ) − 2 2 e−i 10 − 2 2 + 3 3 e−i 2
600
nπ
nπ
nπ
nπ
n=1
Let
π
nπ
π
−40 −i( nπ
2
2
40
e 10 + 2 ) − 2 2 e−i 10 − 2 2 + 3 3 e−i 2
3
3
nπ
nπ
nπ
nπ
Then above can be simplified to
!
∞
X
1
z (t) =
+ Re
ein$t Zn
600
n=1
Zn =
Where $ =
πv
,
2
hence
∞
X
0
in$ein$t Zn
z (t) = Re
!
n=1
Hence, let
yss (t) = Re
∞
X
Yn ein$t
n=1
∞
X
2
2
in$t
− mn $ Yn e
+
∞
X
in$t
icn$Yn e
n=1
n=1
∞
X
2
2
∞
∞
∞
X
X
X
k
in$t
in$t
icn$e
Zn +
kYn e
=
+
kein$t Zn
+
600
n=1
n=1
n=1
in$t
−mn $ + icn$ + k Yn e
(icn$ + k) Zn ein$t +
n=1
n=1
∞
X
=
∞
X
k
600
∞
2
2
in$t
−mn $ + icn$ + k Yn e
n=1
Hence
Yn =
X
k
=
+
(icn$ + k) Zn ein$t
600 n=1
(icn$ + k)
Zn
−m (n$)2 + icn$ + k
Let
icn$ + k
−m (n$)2 + icn$ + k
2
i2ζmωnat n$ + ωnat
m
=
2
2
−m (n$) + i2ζmωnat n$ + ωnat
m
$
i2ζn ωnat + 1
= 2
− n ω$
+ i2ζn ω$
+1
nat
nat
D (rn , ζ) =
=
1 + i2ζrn
(1 − rn2 ) + i2ζrn
(2.103)
2.6. HW6
75
Where in the above rn = ωn$
where $ is 2π
which means it is the fundamental frequency of the
T
nat
forcing function and ωnat is the natural frequency.
Then Eq 2.111 becomes
Yn = D (rn , ζ) Zn
And the steady state solution yss (t) becomes
∞
X
k
yss (t) =
+ Re
D (rn , ζ) Zn ein$t
600
n=1
!
Now we can answer the question. When c = 0 then D (rn , ζ) reduces to
k
−m(n$)2 +k
=
1
2
1− n ω $
=
1
2 ,
1−rn
nat
hence
yss (t) =
k
+ Re
600
∞
X
1
Zn ein$t
2
1 − rn
n=1
So the displacement yss (t) will be resonant when rn = 1 or
nπv
2ωnat
!
= 1 or v =
2ωnat
nπ
Hence
v=
2 (2π5)
20
=
nπ
n
Hence v = 20, 10, 5, 2.5, 1.25, · · · meter/sec will each cause resonance. To verify, here is a plot of yss (t)
with no damper for speed near resonance v = 19.99 and comparing this for speeds away from resonance
speed. This plot shows that when speed v is close to any of the above speeds, then the displacement
yss (t) becomes very large. Once the speed is away from those values, then yss (t) quickly comes down to
steady state F/k value.
76
2.6.3
CHAPTER 2. HW’S
problem 3
2.6. HW6
77
The function is periodic with period T = 2τ
f (t) =
P
t
τ
0<t<τ
0
τ < t < 2τ
and f (t ± T ) = f (t). Let f˜ (t) be the Fourier series approximation to f (t), hence
N
X
2π
1
˜
f (t) = F0 + Re
Fn ein T x
2
n=1
!
(2.104)
78
CHAPTER 2. HW’S
Where
2
Fn =
T
ZT
2π
f (t) e−in T t dt
0
2
=
2τ
P
τ2
=
Zτ
0
Zτ
P −in π t
te τ dt
τ
π
te−in τ t dt
0
Using integration by parts
π
ie−in τ t
hence
nπ
R
udv = uv −
R
π
vdu, letting u = t and dv = e−in τ t then v =
τ
Fn =
P
τ2
=
P
τ2
P
τ2
P
= 2
τ
=


−in π t τ
Zτ
τ
π
i
 t ie
− π e−in τ t dt
π
nτ
nτ
0
0
−in π t τ −in π τ i
e τ
ie τ
− π
τ
π
nτ
n τ −in πτ 0
−inπ
τ
τ2
−in π
t
2 ie
τ
+ 2 2 e τ 0
nπ
nπ
−inπ
τ2
2 ie
−inπ
τ
+ 2 2 e
−1
nπ
nπ
e−inπ = cos (nπ) = (−1)n , hence
P
Fn = 2
τ
n
τ2
n
2 i (−1)
τ
+ 2 2 ((−1) − 1)
nπ
nπ
Hence for even n
2 i
τ
nπ
i
=P
nπ
P
Fn = 2
τ
and for odd n
τ2
2 i
−τ
−2 2 2
nπ
nπ
P
2
=−
+i
nπ nπ
P
Fn = 2
τ
P
F0 = 2
τ
Zτ
tdt
0
P
= 2
τ
P
=
2
2 τ
t
P τ2
=
2 0 τ2 2
R
π
e−in τ t dt =
2.6. HW6
79
Now Eq 2.112 becomes
N
X
2π
1
˜
Fn ein T x
f (t) = F0 + Re
2
n=1
1
= F0 + Re
2
!
!
X
in 2π
t
T
Fn e
even n
+
X
Fn e
t
in 2π
T
odd n
!
2π
i in 2π t X
2
P
P
e T +
+ i ein T t
−
nπ
nπ
nπ
even n
odd n
!
2π
P X i in 2π t P X 1 2
p
e T −
+ i ein T t
= + Re
4
π even n n
π odd n n nπ
!
2π
P
P X i in 2π t P X
2
i
= + Re
e T −
+
ein T t
4
π even n n
π odd n n2 π n
p
= + Re
4
X
To verify, here is a plot of the above, using P = 1 and τ = 0.5 sec for t = 0 · · · 2 seconds. This
shows as more terms are added, the approximation becomes very close to the function. At N = 40 the
approximation appears very good.
80
CHAPTER 2. HW’S
Now we need to write f (t) as sum of exponential to answer the question.
N
X
2π
1
˜
f (t) = F0 + Re
Fn ein T x
2
n=1
where $ is the fundamental frequency of the force given by
∞
X
Hence, let yss =
Yn ein$t , then
2π
T
!
=
2π
2τ
=
π
τ
n=−∞
Re m
∞
X
n=−∞
− (n$)2 Yn ein$t + c
∞
X
in$Yn ein$t + k
n=−∞
∞
X
n=−∞
∞
X
!
Yn ein$t
n=−∞
2
−m (n$) + icn$ + k Yn ein$t
1
= F0 + Re
2
N
X
!
2π
Fn ein T
x
n=1
N
X
2π
1
= F0 + Re
Fn ein T x
2
n=1
!
2.6. HW6
81
Hence
Yn =
=
Hence
2.6.3.1
where r =
Fn
k
1
2 1 − n ω$
+ i2ζn ω$
nat
nat
1
Fn
k 1 − (nr)2 + i2ζnr
∞
X
1
yss = F0 + Re
Yn ein$t
2
n=1
Finding Yn for τ =
$
.
ωnat
!
π
3ωnat
When ζ = 0.04 and τ =
π
,
3ωnat
hence now r =
2π
(2τ )ωnat
=
2π 2 3ωπ
ωnat
= 3, therefore
nat
1
Fn
2
k 1 − (3n) + i6 (0.04) n
Fn
1
=
2
k (1 − 9n ) + i0.24n
Yn =
The largest
qYn will occur when the denominator of the above is smallest. Plotting the modulus of the
denominator (1 − 9n2 )2 + (0.24n)2 for different n values shows that n = 1 is the values which makes it
minimum.
This happens since for any n > 1 the denominator will become larger due to n2 and hence Yn will
become smaller. So n = 1 will be used.
For n = 1, we obtain
Y1 =
F1
1
k (1 − 9) + i6 (0.04)
82
CHAPTER 2. HW’S
But F1 = − Pπ
2
π
+ i , hence
2
+
i
1
−P
π
=
Y1 =
k
(1 − 9) + i6 (0.04)
πk −8 + i0.24
2
+ i (8 + i0.24)
P π2 + i
P
π
=
=
πk 8 − i0.24
πk (8 − i0.24)(8 + i0.24)
P
=
(0.075759 + 0.12727i)
πk
− Pπ
2
π
+i
Therefore
Y1 =
P
(0.024115 + 0.0405i)
k
Here is a list of Yn for n = 1 · · · 10 with the phase and magnitude of each (this was done for
p
k
= 1)
From the above we see that most of the energy in the response will be contained in Y1 and adding
more terms will not have large effect on the response shape. This is confirmed by the plot that follows.
2.6.3.2
Plot for the steady state
Since
∞
X
1
yss = F0 + Re
Yn ein$t
2
n=1
Where now r =
$
.
ωnat
π
,
3ωnat
!
hence now r =
2π
(2τ )ωnat
=
2π 2 3ωπ
ωnat
nat
therefore
2.6. HW6
83
r=3
p
yss = + Re
4
∞
X
n=1,3,5···
p
4
=
p
4
!
Yn ein$t
n=2,4,6···
!
∞
X
Fnodd
1
1
F
neven
ein$t +
ein$t
2
2
k
k
1 − (nr) + i2ζnr
1 − (nr) + i2ζnr
n=1,3,5···
n=2,4,6···
!
∞
∞
2
P
i
X
X
− nπ
+
i
P
1
1
nπ
nπ
+ Re
ein$t +
ein$t
2
2
k
k 1 − (nr) + i2ζnr
1 − (nr) + i2ζnr
n=2,4,6···
n=1,3,5···
!
∞
∞
1
2
i
X
X
+
i
p
nπ nπ
nπ
+ Re
−
ein$t +
ein$t
2
2
k
1
−
(nr)
1
−
(nr)
+
i2ζnr
+
i2ζnr
n=1,3,5···
n=2,4,6···
p
= + Re
4
=
Yn ein$t +
∞
X
∞
X
Now let r = 3, ζ = 0.04. Normalizing the equation for $ = 1 which implies τ = π and k = 1 and
p = 1, then the above becomes
1
yss = + Re
4
∞
X
n=1,3,5···
−
2
nπ
1
nπ
2
1 − (3n)
+i
+ i2 (0.04) 3n
eint +
∞
X
n=2,4,6···
i
nπ
1 − (3n)
Here is a plot of the above for t = 0 · · · 20 seconds for different values of n
2
+ i2 (0.04) 3n
!
eint
84
CHAPTER 2. HW’S
We see from the above plot, that yss (t) does not change too much as more terms are added, since
when r = 3, then Yn for n = 1 contains most of the energy, hence adding more terms did not have an
effect.
2.6. HW6
2.6.3.3
r=
$
.
ωnat
85
Repeating the calculations for τ =
3π
,
ωnat
3π
ωnat
hence now r =
2π
(2τ )ωnat
=
2π
2 ω3π ωnat
= 13 , therefore
nat
1
Fn
k 1 − (nr)2 + i2ζnr
1
Fn
=
2
k 1 − 1n
+ i 2 (0.04) n
Yn =
3
=
Fn
k 1−
n2
9
3
1
+ i0.0267n
The largest Yn will occur when the denominator of the above is smallest. Similar to above, we can
either find n which minimizes the denominator (by taking derivative and setting it to zero and solve for
n) or we can make a plot and see how the function behaves. Making a plot shows this
From the above we see that the smallest value of the denominator happens when n = 3
so using n = 3 we find
F3
1
k 1 − (3r)2 + i2ζ3r
F3
1
=
2
k 1 − 31
+ i2 (0.04) 3 1
Y3 =
3
=
P
But Fn = − nπ
2
nπ
3
F3 1
k i0.08
+ i , hence
P
F3 = −
3π
2
+i
3π
86
CHAPTER 2. HW’S
Therefore
Y3 =
P
− 3π
2
3π
k
+i
1
i0.08
Hence
p
(−1.3263 + 0.28145i)
k
Here is a list of Yn for n = 1 · · · 10 with the phase and magnitude of each (this was done for
Y3 =
p
k
= 1)
We see from the above that |Y3 | is the largest harmonic.
2.6.3.4
Plot for the steady state
Since
∞
X
1
yss = F0 + Re
Yn ein$t
2
n=1
Where now r =
$
.
ωnat
3π
,
ωnat
!
hence now r =
2π
(2τ )ωnat
=
2π
2 ω3π ωnat
= 13 ,
nat
therefore from above
p p
yss = + Re
4 k
∞
X
1
−
nπ
n=1,3,5···
2
+i
nπ
∞
X
1
i
1
in$t
e
+
ein$t
2
2
nπ
1 − (nr) + i2ζnr
1
−
(nr)
+
i2ζnr
n=2,4,6···
π
Now let r = 13 , ζ = 0.04, and assuming τ = 0.5 then $ = 2π
= 0.5
, and assuming k = 1, then the
2τ
above becomes


∞
X
π
1 1
1
2
1
in 0.5
t
−
yss = + Re 
+i e
2
1
1
4 k
nπ nπ
1 − n3
+ i2 (0.04) 3 n
n=1,3,5···


∞
X
π
1
i
1
+ Re 
ein 0.5 t 
2
k
nπ 1 − n 1
+ i2 (0.04) 1 n
n=2,4,6···
3
3
Here is a plot of the above for t = 0 · · · 20 seconds for different values of n
!
2.6. HW6
87
We see now that after n = 3 that the response did not change much by adding more terms, this is
because more of the energy are contained in the first 3 harmonics with Yn being the the largest.
88
CHAPTER 2. HW’S
2.7
HW7
2.7.1
problem 1
2.7.1.1
Part(a)
Vibration isolation was based on reducing absolute acceleration of passenger under turbulent external
forces. This was done by isolating the passenger from the base motion subjected to external absolute
acceleration. Hence the model is based on the following diagram
Hence EQM of motion is
my 00 + c (y 0 − z 0 ) + k (y − z) = 0
my 00 + cy 0 + ky = cz 0 + kz
(2.105)
We are given the time history of the turbulent acceleration. Hence in frequency domain we can write
z 00 = Re Znacc ei(ω1 n)t
2.7. HW7
89
Where Znacc is the complex amplitude of the nth harmonic component in the acceleration data. Let
ω1 n ≡ $n then using the above, In frequency domain Eq 3.1 becomes
acc
Zn
Znacc
2
i$n t
i$n t
Re −m$n + i$n c + k Yn e
= Re
c
+k
e
i$n
−$n2
!
c
k
−
2
i$n
$n
Yn =
Znacc
−m$n2 + i$n c + k
The above gives the transfer function between the displacement of the passenger and the external
acceleration. In otherwords
!
)
(
c
k
−
2
i$n
$n
Znacc ei(ω1 n)t
y (t) = Re
−m$n2 + i$n c + k
Let
c
− $k2
i$n
n
−m$n2 + i$n c
Yn =
!
Znacc
+k
then the transfer function is
−ic
− $k2
Yn
$n
n
=
acc
2
Zn
−m$n + i$n c + k
1
(k + ic$n )
=− 2
$n (k − m$n2 ) + i$n c
H ($n ) =
Hence phase is
arg (H ($n )) = tan
−1
c$ n
k
−1
− tan
$n c
k − m$n2
and magnitude is
p
Yn k 2 + c2 $n2
1
q
|H ($n )| = acc =
Zn
$n
(k − m$n2 )2 + ($n c)2
(k+ic$n )
These can be written in terms of ζ and ωnat as follows. From H ($n ) = − $12 (k−m$
, dividing
2
n
n )+i$n c
2
numerator and denominator by k = mωnat and using c = 2ζmωnat then
i2ζmωnat $n
i2ζ$n
1
+
1
+
2
ωnat
mω
1
1
nat
H ($n ) = − 2 =− 2
2
2
nat
$n 1 − m$2 n + i$n 2ζmω
$n 1 − $2 n + i$n 2ζ
2
mωnat
Let rn =
$n
ωnat
mωnat
ωnat
then the above becomes
H ($n ) = −
(1 + i2ζrn )
1
2
$n (1 − rn2 ) + i2rn ζ
Hence
|H ($n )| =
q
1 + (2ζrn )2
1
q
$n
(1 − rn2 )2 + (2rn ζ)2
arg (H ($n )) = tan−1 (2ζrn ) − tan−1
2rn ζ
1 − rn2
ωnat
90
CHAPTER 2. HW’S
The following is a plot showing the passenger absolute acceleration y 00 (t) over the period of 80 seconds
against the turbulent acceleration z 00 (t). We now see that passenger absolute acceleration is close to the
nominal acceleration. This was done using the following values for the vibration isolation
M
100000 kg
ζ
0.72
k
38924 N/m
c
57746 Ns/m
The plot on the right side is the absolute acceleration of the passenger during flight in the turbulent
case.
2.7.1.2
Part(b)
The length of first class cabinet was estimated to be L = 15 meters from looking at Boeing web page.
Using Steel, Structural ASTM-A36 I beam as a cantilever beam for the implementation, then using
k = 3EI
results in
L3
3 (200 × 109 ) I
153
I = 2.1895 × 10−4 m4
38924 =
Using rectangle cross section I =
32 cm.
bh3
.
12
Letting h = 20 cm, then b =
(2.1895×10−4 )12
0.23
= 0.32843 meter or
2.7. HW7
2.7.2
Problem 2
2.7.2.1
part(a)
91
Q = 2000t
(T − t)
[h (t) − h (t − T )]
T2
m = 0.5 kg
ωn = 2πfn
fn = 100 Hz
Hence pulse duration is
1
f
= 0.01 sec.
my 00 + cy 0 + ky = Q (t)
In the frequency domain assuming that the force Q (t) can be represented in its Fourier series as
!
X
Q (t) = Re
Qn eiω1 nt
n
where ω1 is the fundamental frequency for Q (t) which depends on the period we choose to select to
sample over. In this example, I selected 3T as the overall period to sample over so that it covers the
pulse duration and an additional time to show the free vibration part as well and to compare to the
analytical solution. Hence the EQM becomes
Yn =
Qn
−m (nω1 ) + ic (nω1 ) + k
2
k = ωn2 m hence dividing the numerator and denominator by k we obtain
Yn = =
1−
Qn
K
m(nω1 )2
2m
ωn
+
ic(nω1 )
2m
ωn
1
1
Qn
2
k (1 − rn ) + i2ζrn
92
CHAPTER 2. HW’S
where rn =
nω1
.Hence
ωn
response is
!
y (t) = Re
X
Yn eiω1 nt
n
= Re
X1
n
1
Qn eiω1 nt
2
k (1 − rn ) + i2ζrn
!
P
1
y (t) is found by taking the IFFT of n k1 (1−r2 )+i2ζr
Qn .
n
n
Qn values are found by taking the FFT of Q (t). We start by sampling Q (t). To obtain the solution
for say t = 0 · · · 3T , then we have to assume that the period of the signal is actually 3T and sample over
this whole time from 0 · · · 3T − delt. Then we use FFT on the result. Then find the response by doing
IFFT. Using N = 128 over t = 0 · · · 0.03 seconds, the following solution was obtained
%by Nasser M. Abbasi, HW 7, EMA 545
close all;
T
= 0.01; %sec
duration = 3*T; %duration to find solution over
N
= 128;
delT = duration/(N-1);
w1
= 2*pi/duration; %fundamental freq rad/sec
t
= linspace(0,(duration-delT),N);
Qt
= @(t) (2000*t.*(T-t))/T^2.*(t<=T)+0*(t>T)
subplot(2,1,1)
plot(t,Qt(t),'r-o');
2.7. HW7
93
hold on;
plot(0:delT:duration,Qt(0:delT:duration),'r');
title(sprintf('force Q(t) and its reponse. 16 samples, delT=%f',delT));
xlabel('time sec');
grid;
m
wn
k
[Q,ws]
=
=
=
=
0.5; %mass kg
2*pi*100; %natural freq
wn^2*m; %stiffness N/meter
fft_easy(Qt(t),delT);
zeta
I
y
= 0.002;
= sqrt(-1);
= ifft_easy( (Q/k)./( (1-(ws/wn).^2) + 2*I*zeta*ws/wn),ws);
subplot(2,1,2);
plot(t,y,'r');
title(sprintf('reponse at zeta=%f',zeta));
xlabel('time sec');
grid;
2.7.2.2
Part(b)
For ζ = 0.002 the above Matlab script was modified and the following solution resulted.
Now we compare the above with the analytical solution.
94
CHAPTER 2. HW’S
2.7.2.3
Part(c)
The pulse can be written as
F = Q (t) [h (t) − h (t − T )]
= Q (t) h (t) − Q (t) h (t − T )
Let t0 = t − T , hence t = t0 + T , therefore the above becomes
F = Q (t) h (t) − Q (t0 + T ) h (t0 )
But Q (t) =
the above F as
2000t(T −t)
.
T2
Let
2000
T2
= β since it is a constant. Hence Q (t) = βt (T − t). Now we write
F = βt (T − t) h (t) − β (t0 + T ) (T − (t0 + T )) h (t0 )
= βT t − βt2 h (t) − β (t0 + T ) (−t0 ) h (t0 )
2
= βT t − βt2 h (t) + β (t0 ) + T t0 h (t0 )
2
= βT th (t) − βt2 h (t) + βT (t0 ) + βT t0 h (t0 )
(2.106)
So we see that the response to F will be the response to a unit impulse h (t) with forcing basis
functions that are 1, t, t2 . Now we can use the solution from back of the book appendix B to sum the
responses in order to find the final response and compare to the FFT method.
From appendix B, the response to unit ramp th (t)is
1
−ζωn t
2 ωn
r (th (t)) =
ωn t − 2ζ + e
sin ωd t h (t)
2ζ cos ωd t − 1 − 2ζ
mωn3
ωd
and the response to quadratic t2 h (t) is
1
2
2
2
−ζωn t
2
3 ωn
sin ωd t h (t)
s t h (t) =
(ωn t) − 4ζωn t − 2 1 − 4ζ + e
2 1 − 4ζ cos ωd t + 6ζ − 8ζ
mωn4
ωd
Now that we have the basis solutions, we can apply them to EQ 2.110
F = βT (r (t) + r (t0 )) − βT (s (t) − s (t0 ))
= β (r (t) + r (t − T )) − βT (s (t) − s (t − T ))
1
−ζωn t
2 ωn
= (βT )
ωn t − 2ζ + e
sin ωd t h (t)
mωn3
ωd
1
0
−ζωn t0
0
2 ωn
0
+ (βT )
ωn t − 2ζ + e
sin ωd t
h (t0 )
mωn3
ωd
1
2
2
−ζωn t
2
3 ωn
− (β)
(ωn t) − 4ζωn t − 2 1 − 4ζ + e
2 1 − 4ζ cos ωd t + 6ζ − 8ζ
sin ωd t h (t)
mωn4
ωd
1
0 2
0
2
−ζωn t0
2
0
3 ωn
0
+ (β)
(ωn t ) − 4ζωn t − 2 1 − 4ζ + e
2 1 − 4ζ cos ωd t + 6ζ − 8ζ
sin ωd t
h (t0 )
4
mωn
ωd
p
In the above, ωd = ωn 1 − ζ 2 . To plot this solution, the following small script was used and was run
for both ζ = 0.2 and ζ = 0.002
For ζ = 0.2
2.7. HW7
95
For ζ = 0.002
2.7.3
Conclusions
The analytical solution, using superposition agreed with the FFT solution for ζ = 0.2. However, for some
reason which I am not able to determine why yet, the FFT solution when ζ = 0.002 did not agree with
the analytical solution. The analytical solution was verified to be correct using another numerical ODE
solver. So the FFT method for some reason is not giving accurate result for ζ = 0.002. The same Matlab
script was used for both cases. I tried increasing the sampling rate but that did not change the result.
Please see Appendix for verification and the code used to plot the analytical solutions.
96
CHAPTER 2. HW’S
2.7.4
Problem 3
2.7.5
Part(a)
Let T be the kinetic energy and V be the potential energy. Then equation of motion for a generalized
coordinate qi is given by
∂L
d ∂L
−
= Qi
dt ∂ q˙i
∂qi
Where L is the Lagrangian L = T − V and Qi is the generalized force in the qi direction.
Assuming x2 > x1 and masses are moving to the right. For x1 we obtain
1
1
T = m1 x˙ 21 + m1 x˙ 22
2
2
1
1
1
1
V = k1 x21 + k2 (x2 − x1 )2 + k4 x21 + k3 x22
2
2
2
2
Q1 = F
Q2 = 0
Hence
L=T −V
1
1
1
1
1
1
2
2
2
2
2
2
k1 x1 + k2 (x2 − x1 ) + k4 x1 + k3 x2
= m1 x˙ 1 + m1 x˙ 2 −
2
2
2
2
2
2
∂L
= m1 x˙ 1
∂ x˙ 1
d ∂L
= m1 x¨1
dt ∂ x˙ 1
∂L
= −k1 x1 − k2 (x2 − x1 ) (−1) − k4 x1
∂x1
and
2.7. HW7
97
∂L
= m1 x˙ 2
∂ x˙ 2
d ∂L
= m1 x¨2
dt ∂ x˙ 2
∂L
= −k2 (x2 − x1 ) (1) − k3 x2
∂x2
Hence the 2 EOM are for x1
d ∂L
∂L
−
=F
dt ∂ x˙ 1
∂x1
m1 x¨1 − (−k1 x1 + k2 (x2 − x1 ) − k4 x1 ) = F
m1 x¨1 + k1 x1 − k2 (x2 − x1 ) + k4 x1 = F
Therefore EOM 1
m1 x¨1 + (k1 + k2 + k4 ) x1 − k2 x2 = F
and for x2
∂L
d ∂L
−
=0
dt ∂ x˙ 2
∂x2
m1 x¨2 − (−k2 (x2 − x1 ) − k3 x2 ) = 0
m1 x¨2 + k2 (x2 − x1 ) + k3 x2 = 0
Hence EOM 2
m1 x¨2 + (k2 + k3 ) x2 − k2 x1 = 0
Hence in Matrix form EOM are
2.7.5.1
m1
0
0
m2
!
x001
x002
!
+
M X 00 + KX = Q
!
!
!
(k1 + k2 + k4 )
−k2
x1
F
=
−k2
(k2 + k3 )
x2
0
Part(b)
If m2 do not exist, then this means the springs k2 and k3 do not have a mass between them and so these
need to be replaced by single spring, say k5 found by finding equivalent spring in series
98
CHAPTER 2. HW’S
1
1
1
=
+
k5
k2 k3
k3 + k2
k5 =
k2 k3
From above, EQM for m1 becomes


m1 x¨1 + 
k1 +
z

k
}|5 {

k3 + k2
+ k4 
 x1 = F
k2 k3
So now k4 and k4 are in parallel, hence we replace k5 + k4 by k6 found from
k6 = k5 + k4
k3 + k2
+ k4
=
k2 k3
k3 + k2 + k2 k3 k4
k6 =
k2 k3
Hence EQM for m1 now becomes

k
z
}|6
{

k3 + k2 + k2 k3 k4 
 x1 = F
m1 x¨1 + 
k
+
1


k2 k3

2.7. HW7
99
and finally
m1 x¨1 +
k3 + k2 + k2 k3 k4 + k1 k2 k3
x1 = F
k2 k3
100
CHAPTER 2. HW’S
2.8
HW8
2.8.1
problem 1
2.8.1.1
part(a)
Let initial length of the spring (un stretched length) be L0 and when the mass m has moved to the right
by an amount x then let the current length be Lcur .
Therefore the stretch in the spring is
∆ = Lcur − L0
Let the height of the bar by H, where tan θ0 =
H
L
or H = L tan θ0
2.8. HW8
101
Hence from the above diagram we see that L0 =
√
H2
+
L2
and Lcur
q
= H 2 + (L + x)2 , therefore
√
H 2 + (L + x)2 − H 2 + L2
q
2
√
2
2
∆ =
H 2 + (L + x) − H 2 + L2
∆=
q
Now we can derive the equation of motion using energy methods.
Let T be the current kinetic energy in the system, and let V be the current potential energy. This
system is one degree of freedom, since we only need one generalized coordinate to determine the position
of the mass m. This coordinate is x.
1
T = mx˙ 2
2
1
V = k∆2
2
q
2
√
1
2
H 2 + (L + x) − H 2 + L2
= k
2
Hence the Lagrangian Φ is
Φ=T −V =T −V
q
2
√
1
1
2
2
2
2
2
= mx˙ − k
H + (L + x) − H + L
2
2
Now the equation of motion for coordinate x is (using the standard Lagrangian form)
d ∂Φ
∂Φ
−
= Qx
dt ∂ x˙
∂x
But Qx , then generalized force, is zero since there is no external force and no damping. Now we just
need to evaluate each part of the above expression to obtain the EOM.
∂Φ
= mx˙
x˙
∂
d ∂Φ
= m¨
x
dt ∂ x˙
102
CHAPTER 2. HW’S
and
q
2 !
√
1
1
2
mx˙ 2 − k
H 2 + (L + x) − H 2 + L2
2
2
q
√
−1
1
2
2
2
2
= −k
H + (L + x) − H + L
H 2 + (L + x)2 2 2 (L + x)
2

q
√
H 2 + (L + x)2 − H 2 + L2
 (L + x)
q
= −k 
2
2
H + (L + x)
∂Φ
∂
=
∂x
∂x
Hence EOM becomes
∂Φ
∂Φ
−
=0
∂ x˙
∂x
q

√
H 2 + (L + x)2 − H 2 + L2
 (L + x) = 0
q
m¨
x+k
2
2
H + (L + x)
d
dt
2.8.1.2
part(b)
√
For small x we need to expand f (x) = k
√
H 2 +(L+x)2 − H 2 +L2
√
H 2 +(L+x)2
(L + x) around x = 0 in Taylor series
and let higher powers of x go to zero.
f (x) = f (0) + xf 0 (0) +
x2 f 00 (0)
+ HOT.
2!

q
√
H 2 + (L + 0)2 − H 2 + L2
 (L + 0)
q
f (0) = k 
2
H 2 + (L + 0)
!
√
√
H 2 + L2 − H 2 + L2
√
=k
L
H 2 + L2
=0
and now for f 0 (0)
d
f (x)x=0
dx
 q


√
2 + (L + x)2 −
2 + L2
H
H
d
 (L + x)
q
= k 
dx
2
2
H + (L + x)
x=0
q


 q


√
√
2 + (L + x)2 −
2 + L2
2 + (L + x)2 −
2 + L2
H
H
H
H
d 
+
 d (L + x)
q
q
= k (L + x)
dx
dx
H 2 + (L + x)2
H 2 + (L + x)2
x=0


√
! q
2
H 2 + (L + x) − H 2 + L2
√
L+x
2
2



q
= k (L + x)
H +L
+
3
2
2
2
2
2
2
(H + L + 2Lx + x )
H + (L + x)
f 0 (0) =
x=0
2.8. HW8
103
Now we evaluate it at x = 0

√
f 0 (0) = k (L + 0)
H 2 + L2
q

√
2 + (L + 0)2 −
2 + L2
H
H
L+0

q
+
3
2
2
2
2
(H + L + 2L0 + 0)
H 2 + (L + 0)
!
!!
√
√
2 + L2 −
√
H
H 2 + L2
L
√
H 2 + L2
+
=k L
3
H 2 + L2
(H 2 + L2 ) 2
!!
1
(H 2 + L2 ) 2
2
=k L
3
(H 2 + L2 ) 2
L2
=k
(H 2 + L2 )
!
Therefore, EOM of motion becomes (notice we ignored higher order terms, which contains x2 in them)
m¨
x + (f (0) + xf 0 (0)) = 0
Hence the linearized EOM is
m¨
x+k
L2
x=0
(H 2 + L2 )
Or in terms of θ0 the EOM can be written as
L2
x = 0
(L tan θ0 )2 + L2
1
x=0
m¨
x+k
1 + tan2 θ0
m¨
x+k
This is the linearized EOM around x = 0. Using numerical values given in the problem L = 1, m =
1, k = 1000N/m, θ0 = π4 , it becomes
x¨ + 1000
1
1 + tan π4
2 x = 0
x¨ + 500x = 0
Therefore the linearized stiffness is 500x while the nonlinearized stiffness is
 q


√
H 2 + (L + x)2 − H 2 + L2
k 
 (L + x)
q
2
2
H + (L + x)
L=1,θ=450
q
q


2
2
tan π4 + (1 + x)2 −
tan π4 + 1
 (1 + x)
q
= 1000 
2
π 2
tan 4 + (1 + x)
q

2
(x + 1.0) + 1.0 − 1.4142
 (1 + x)
q
= 1000 
2
(x + 1.0) + 1.0
Here is a plot of linearized vs. non-linearized stiffness for x = −1 · · · 1
104
2.8.1.3
CHAPTER 2. HW’S
part(c)
The spring extension ∆ is first found by assuming there is a point A at x = 0 and point B where the
spring is attached to the ceiling. Hence
2.8. HW8
105
˙ = (u˙ A − u˙ B ) eA/B
∆
= (xˆ
˙ ı − 0ˆ
) · (cos θ0ˆı − sin θ0 ˆ)
= x˙ cos θ0
Therefore
∆ = x cos θ0
Now we repeat the same calculations but using ∆ = x cos θ0 for the spring extension.
1
T = mx˙ 2
2
1 2
V = k∆
2
1
= k (x cos θ0 )2
2
Hence the Lagrangian Φ is
Φ=T −V
1
1
= mx˙ 2 − k (x cos θ0 )2
2
2
Now the equation of motion for coordinate x is (using the standard Lagrangian form)
d ∂Φ
∂Φ
−
=0
dt ∂ x˙
∂x
It is equal to zero above, since there is no generalized force associated with coordinate x. Now we just
need to evaluate each part of the above expression to obtain the EOM.
∂Φ
= mx˙
x˙
∂
d ∂Φ
= m¨
x
dt ∂ x˙
106
CHAPTER 2. HW’S
and
∂Φ
∂ 1
1
2
2
=
mx˙ − k (x cos θ0 )
∂x
∂x 2
2
= −k (x cos θ0 ) cos θ0
= −kx cos2 θ0
Hence EOM becomes
d ∂Φ
∂Φ
−
=0
dt ∂ x˙
∂x
m¨
x + kx cos2 θ0 = 0
But cos θ0 =
√ L
H 2 +L2
hence
m¨
x + kx
L2
=0
H 2 + L2
This is the same as the EOM for the linearized case found in part(c)
2.8.1.4
part(d)
Now we need to solve numerically the nonlinear EOM found in part(a) which is
q

√
H 2 + (L + x)2 − H 2 + L2
 (L + x) = 0
q
m¨
x+k
2
H 2 + (L + x)
using m = 1, k = 1000, L = 1, θ0 = 450 . For IC we use x (0) = 0.1, x0 (0) = 0 for first case, and for
second case using x (0) = 0.5, x0 (0) = 0. This is a plot showing both responses on same diagram
2.8. HW8
107
The period for the response for case of IC given by x(0) = 0.5 is seen to be about 0.375 seconds and
for the case x (0) = 0.1 it is 0.275 sec
√
The linearized EOM is x¨ + 500x = 0 and hence ωn2 = 500 or ωn = 500 = 22.361 rad/sec, hence
2π
T = ω2πn = 22.361
= 0.281 sec .
We notice this agrees well with the period of the response of the nonlinear equation for only the case
x = 0.1.This is because x = 0.1 is very close to x = 0 the point at which the linearization happened.
Therefore, the linearized EOM gave an answer of 0.281 sec that is very close the more exact value of
0.275 seconds. But when the initial conditions changed to x (0) = 0.5, then T found from linearized
EOM does not agree with the exact value of 0.375 seconds.
This is because x = 0.5 is far away from the point x = 0 where the linearized was done. Hence the
linearized EOM can be used for only initial conditions that are close to the point where the linearization
was done.
Additionally, the nonlinearity manifests itself in the response of the system by noticing that the
frequency of the free vibration response has actually changed depending on initial conditions. In a linear
system, only the phase and amplitude of the free vibration response will change as initial conditions is
108
CHAPTER 2. HW’S
changed, while the natural frequency of vibrations does not change.
2.8.2
problem 2
Use y and θ as generalized coordinates as shown in this diagram in the positive direction
2.8. HW8
109
Using Lagrangian method, we start by finding the kinetic energy of the system, then the potential
energy.
2
1
1
2
T = my˙ 2 +
θ˙
mrG
2
2
For the potential energy, there will be potential energy due to ky spring extension and due to kT
spring angle of rotation in system. From the diagram above, we see that, for small angle θ
1
1
V = ky ∆2 + kT θ2
2
2
To find ∆ we use the stiff spring approximation. Let the point the spring is attached at the top be B,
then
˙ = (u˙ E − u˙ B ) eE/B
∆
˙
= lθ − y˙ ˆ − 0 · (−ˆ
)
= lθ˙ − y˙ ˆ · (−ˆ
)
= y˙ − lθ˙
Hence
∆ = y − lθ
Therefore, the potential energy now can be found to be
1
1
V = ky (y − lθ)2 + kT θ2
2
2
Therefore, the Lagrangian Φ is
Φ=T −V
2 1
1
1
1
2
= my˙ 2 +
mrG
θ˙ − ky (y − lθ)2 − kT θ2
2
2
2
2
We now find the equations for each coordinate. For y
∂Φ
= my˙
∂ y˙
d ∂Φ
= m¨
y
dt ∂ y˙
∂Φ
= −ky (y − lθ)
∂y
110
CHAPTER 2. HW’S
Hence EOM is
d ∂Φ ∂Φ
−
= Qy
dt ∂ y˙
∂y
m¨
y + ky (y − lθ) = Qy
We just need to find Qy the generalized force in the y direction. Using virtual work, we make small
virtual displacement δy in positive y direction while fixing all other generalized coordinates from moving
(in this case θ) and then find out the work done by external forces. In this case, there is only one external
force which is L. Hence
δW = Lδy
Therefore Qy = L since that is the force that is multiplied by δy. Hence EOM for y is now found
m¨
y + ky (y − lθ) = L
verification: As L increases, then we see that y 00 gets larger. This makes sense since y is upwards
acceleration, so wing accelerates in the same direction.
Now we find EOM for θ
∂Φ
2 ˙
= mrG
θ
∂ θ˙
d ∂Φ
2 ¨
= mrG
θ
dt ∂ θ˙
∂Φ
= ky l (y − lθ) − kT θ
∂θ
Therefore the EOM is
d ∂Φ ∂Φ
−
dt ∂ θ˙
∂θ
2 ¨
mrG θ − ky l (y − lθ) + kT θ
2 ¨
mrG
θ − ky ly + ky l2 θ + kT θ
2 ¨
mrG
θ − ky ly + kT + ky l2 θ
= Qθ
= Qθ
= Qθ
= Qθ
We just need to find Qθ the generalized force in the θ direction. Using virtual work, we make small
virtual displacement δθ in positive θ direction (i.e. anticlock wise) while fixing all other generalized
coordinates from moving (in this case y) and then find out the work done by external forces. In this
case, there is only one external force which is L. When we make δθ rotation in the positive θ direction,
the displacement where the force L acts is (l + s) δθ for small angle. But this displacement is in the
downward direction, hence it is negative, since we are using y as positive upwards. Hence
δW = −L (l + s) δθ
Therefore Qθ = −L (l + s) since that is the force that is multiplied by δθ. Hence EOM for θ is now
found
2 ¨
mrG
θ − ky ly + kT + ky l2 θ = −L (l + s)
Verification: As L gets larger, then θ¨ gets negative (since L has negative sign). This makes sense,
since as L gets larger, the rotation as shown in the positive direction will change sign and the wing will
now swing the opposite direction (i.e. anticlockwise).
2.8. HW8
111
Now we can make the matrix of EOM
m
0
0
2
mrG
!
!
y¨
+
θ¨
ky
−lky
M X 00 + kX = Q
! !
!
−lky
y
L
=
kT + ky l2
θ
−L (l + s)
Notice that for [k] the matrix is symmetric as expected, and also positive on the diagonal as expected.
The mass matrix [m] is symmetric and positive definite as well.
2.8.3
problem 3
Let θ be the small angle of rotation that the rod rotates by in the anti clockwise direction. Let the
point the spring is fixed be B and the moving point where the spring is attached to the rod be A.To find
spring extension ∆ we use the stiff spring approximation. Let the angle α = 53.130 , hence
˙ = (u˙ A − u˙ B ) · eA/B
∆
L˙
=
θ ˆ − 0 · (cos αˆı + sin αˆ
)
3
L
= θ˙ sin α
3
Hence
L
θ sin α
3
Using Lagrangian method, we start by finding the kinetic energy of the system, then the potential
2
energy. θ is the only generalized coordinate. Assume bar has mass m and hence I = mL
3
∆=
1
T = I θ˙2
2
112
CHAPTER 2. HW’S
For the potential energy, there will be potential energy due to k spring extension. From the diagram
above, we see that
2
1
L
V = k
θ sin α
2
3
Therefore, the Lagrangian Φ is
Φ=T −V
1 L2
1
= I θ˙2 − k θ2 sin2 α
2
2 9
Now we find EOM for θ
∂Φ
= I θ˙
∂ θ˙
d ∂Φ
= I θ¨
dt ∂ θ˙
kL2 sin2 α
∂Φ
=−
θ
∂θ
9
Therefore the EOM is
d ∂Φ ∂Φ
= Qθ
−
dt ∂ θ˙
∂θ
kL2 sin2 α
¨
Iθ +
θ = Qθ
9
We now need to find the generalized force due to virtual δθ rotation using the virtual work method.
There are 2 external forces, the damping force which will have negative sign since it takes energy away
from the system, and the external force F which will add energy hence will have positive sign.
We start by making δθ and then find the work done by these 2 forces.
Work done
F is F Lδθ since the displacement is Lδθ for small angle. Now the work done by
by
L ˙ L
damping is c 3 θ 3 δθ hence total work is
L˙ L
δW = F Lδθ − c θ
δθ
3
3
L2 ˙
= F L − c θ δθ
9
Notice that work due to damping was added with negative sign since damping removes energy from
the system.
2
Hence Qθ = F L − c L9 θ˙ therefore the EOM is
kL2 sin2 α
L2 ˙
¨
Iθ +
θ = FL − c θ
9
9
2
2
2
L
kL
sin
α
I θ¨ + c θ˙ +
θ = FL
9
9
2
Hence the damping coefficient is c L9 .
2.8. HW8
2.8.4
113
problem 4
Let x1 and x2 be the generalized coordinates as shown in this diagram
Let mass of cart be m1 and mass of small sliding block be m2 (at the end, they will be replaced by
values given). Let k for spring attached to wall be k1 and k for spring for small block be k2 .We start by
finding the kinetic energy of the system
1
1
T = m1 x˙ 21 + m2 v 2
2
2
where v is the velocity of the block. To find this v it is easier to resolve components on the x and y
direction. Therefore we find that
~v = x˙ 2 sin θj + (x˙ 2 cos θ + x˙ 1 ) i
114
CHAPTER 2. HW’S
Hence
|~v |2 = (x˙ 2 sin θ)2 + (x˙ 2 cos θ + x˙ 1 )2
= x˙ 22 sin2 θ + x˙ 22 cos2 θ + x˙ 21 + 2x˙ 2 x˙ 1 cos θ
= x˙ 22 sin2 θ + cos2 θ + x˙ 21 + 2x˙ 2 x˙ 1 cos θ
= x˙ 22 + x˙ 21 + 2x˙ 2 x˙ 1 cos θ
Therefore
1
1
T = m1 x˙ 21 + m2 x˙ 22 + x˙ 21 + 2x˙ 2 x˙ 1 cos θ
2
2
Now we find the potential energy.
1
1
V = k1 x21 + k2 x22 − m2 g (x2 sin θ)
2
2
There are no external forces, hence generalized forces Qx1 , Qx2 are zero. The Lagrangian Φ is
Φ=T −V
1
1
1
1
= m1 x˙ 21 + m2 x˙ 22 + x˙ 21 + 2x˙ 2 x˙ 1 cos θ − k1 x21 − k2 x22 + m2 g (x2 sin θ)
2
2
2
2
Now we find EOM for x1 is
∂Φ
= m1 x˙ 1 + m2 x˙ 1 + m2 x˙ 2 cos θ
∂ x˙ 1
d ∂Φ
= m1 x¨1 + m2 x¨1 + m2 x¨2 cos θ
dt ∂ x˙ 1
= (m1 + m2 ) x¨1 + m2 x¨2 cos θ
∂Φ
= −k1 x1
∂x1
Therefore the EOM for x1 is
d ∂Φ
∂Φ
−
=0
dt ∂ x˙ 1 ∂x1
(m1 + m2 ) x¨1 + m2 x¨2 cos θ + k1 x1 = 0
Now we replace the actual values for m1 = 2m, m2 = m, k1 = 3k hence
3m¨
x1 + m¨
x2 cos θ + 3kx1 = 0
Now we find EOM for x2 is
∂Φ
= m2 (x˙ 2 + x˙ 1 cos θ)
∂ x˙ 2
d ∂Φ
= m2 (¨
x2 + x¨1 cos θ)
dt ∂ x˙ 1
= m2 cos θ¨
x1 + m2 x¨2
∂Φ
= −k2 x2 + m2 g sin θ
∂x2
2.8. HW8
115
Therefore the EOM for x2 is
d ∂Φ
∂Φ
−
=0
dt ∂ x˙ 2 ∂x2
m2 cos θ¨
x1 + m2 x¨2 + k2 x2 − m2 g sin θ = 0
Now we replace the actual values for m1 = 2m, m2 = m, k2 = k hence
m cos θ¨
x1 + m¨
x2 + kx2 = m2 g sin θ
3m
m cos θ
m cos θ
m
!
x¨1
!
x¨2
+
M X 00 + kX = Q
!
!
!
3k 0
x1
0
=
0 k
x2
m2 g sin θ
Hence
m
3
cos θ
cos θ
1
!
x¨1
x¨2
!
+k
3 0
0 1
!
x1
x2
!
=
0
!
m2 g sin θ
Notice that there zeros now off diagonal in the [K] matrix, which means the springs are not coupled.
(which is expected, as motion of one is not affected by the other). But mass matrix [m] has non-zeros off
the diagonal. So the masses are coupled. i.e. EOM is coupled. This means we can’t solve on EOM on its
own and both have to be solved simultaneously.
2.8.5
problem 5
There are 2 degrees of freedom, θ1 and θ2 as shown in this diagram, using anticlock wise rotation as
positive
116
CHAPTER 2. HW’S
The Lagrangian Φ = T − V where
1
1
T = I1 θ˙12 + I2 θ˙22
2
2
Where I1 =
m1 L2
and I2 =
3
m2 L2
12
+m2
L 2
4
(using parallel axis theorem). Hence I2 =
m2 L2
12
2
+m2 L16 =
7 2
L m2
48
Now we find the potential energy, assuming springs remain straight (stiff spring assumption) and
assuming small angles

2

2
∆1
∆
z
z }|2 {
}|
{


3L 
1  3L
 + 1 k2 Lθ1 + L θ2  + m1 g L θ1 − m2 g L θ2
θ
+
θ
V = k1 
1
2
2  4
4 
2 
2 
2
4
Hence
Φ=T −V
1 ˙2 1 ˙2
=
I1 θ + I2 θ −
2 1 2 2
1
k1
2
3L
3L
θ1 +
θ2
4
4
2
!
2
1
L
L
L
+ k2 Lθ1 + θ2 + m1 g θ1 − m2 g θ2
2
2
2
4
Now we find EOM for θ1
∂Φ
= I θ˙1
˙
∂ θ1
d ∂Φ
= I θ¨1
dt ∂ θ˙1
∂Φ
3L
3L
3L
L
L
= −k1
θ2 +
θ1
− k2
θ2 + Lθ1 (L) − m1 g
∂θ1
4
4
4
2
2
3L
3L
3L
L
L
= − k1
θ2 +
θ1 − k2 L
θ2 + Lθ1 − m1 g
4
4
4
2
2
Therefore the EOM for θ1 is
d ∂Φ
∂Φ
−
= Qθ1
˙
dt ∂ θ1 ∂θ1
3L
3L
3L
L
L
¨
I1 θ1 +
k1
θ2 +
θ1 + k2 L
θ2 + Lθ1 + m1 g = 0
4
4
4
2
2
2.8. HW8
117
The generalized force is zero, since there is no direct external force acting on top rod.
Hence EOM for θ1 is from above
m1 L2 ¨
θ1 + θ1 k1
3
3L
4
2
!
+ k2 L2
+ θ2 k1
3L
4
2
L2
+ k2
2
!
= −m1 g
L
2
∂Φ
= I θ˙2
˙
∂ θ2
d ∂Φ
= I θ¨2
dt ∂ θ˙2
∂Φ
3L
3L
3L
L
L
L
= −k1
θ2 +
θ1
− k2
θ2 + Lθ1
+ m2 g
∂θ2
4
4
4
2
2
4
d ∂Φ
∂Φ
−
= Qθ2
dt ∂ θ˙2 ∂θ2
3L
L
3L
L
L
3L
¨
θ2 +
θ1
+ k2
θ2 + Lθ1
− m2 g = Qθ1
I2 θ2 + k1
4
4
4
2
2
4
Now Qθ2 is found by virtual work. Making a virtual displacement δθ2 while fixing θ1 and finding the
work done by all external forces.
L
δW = F δθ2
2
Hence Qθ2 = F L2 with positive sign since it add energy to the system. Hence EOM for θ2 is
7 2 ¨
L m2 θ2 + θ1 k1
48
3L
4
2
L2
+ k2
2
!
+ θ2 k1
3L
4
2
2 !
L
L
L
+ k2
= m2 g + F
2
4
2
m1 L2
3
0
0
7 2
L m2
48
!
θ¨1
θ¨2
!
2
+
k1 9L
+ k2 L2
16
2
2
k1 3L
+ k2 L2
4
M X 00 + kX = Q
!
!
!
2
L2
L
+
k
k1 3L
θ
−m
g
2 2
1
1 2
4
=
3L 2
L 2
L
k1 4 + k2 2
θ2
m2 g 4 + F L2
The matrix [k] is coupled but the mass matrix [m] is not.
2.8.6
problem 6
"
The inertia and stiffness matrices for a system are [M ] =
4 0
#
"
200 200
#
kg, [K] =
N/m. determine
0 2
200 800
the corresponding natural frequencies and modes of free vibration.
[k] − ω 2 [M ] {Φ} = {0}
118
CHAPTER 2. HW’S
Solving for eigenvalues
"
#
200 200
"
#!
4
0
2
−ω
=0
200 800
0 2
"
#
200 − 4ω 2
200
det
=0
200
800 − 2ω 2
200 − 4ω 2 800 − 2ω 2 − 2002 = 0
8ω 4 − 3600ω 2 + 120000 = 0
det
Hence, taking the positive square root only we find
ω1 = 20.341 rad/sec
ω2 = 6.0211 rad/sec
When ω = ω1
"
200 − 4ω12
200
200
800 − 2ω12
#( )
Φ11
Φ21
=
( )
0
0
Let Φ11 be the arbitrary value 1 hence
"
#( ) ( )
200 − 4ω12 200
1
0
=
×
×
Φ21
×
200 − 4ω12 + 200Φ21 = 0
Φ21 =
−200 + 4 (20.341)2
−200 + 4ω12
=
= 7.2751
200
200
Hence the first mode associated with ω = 20.341 rad/sec is
(
)
1
7.2751
When ω = ω2
"
200 − 4ω22
200
200
800 − 2ω22
#( )
Φ12
Φ22
=
( )
0
0
"
#( ) ( )
200 − 4ω22 200
1
0
=
×
×
Φ22
×
200 − 4ω22 + 200Φ22 = 0
Φ22 =
−200 + 4ω22
−200 + 4 (6.0211)2
=
= −0.27493
200
200
Hence the first mode associated with ω = 6.0211 rad/sec is
(
)
1
−0.27493
2.8. HW8
119
Summary
ωn (rad/sec)
6.0211
20.341
mode shape
(
)
1
−0.27493
(
)
1
7.2751
Verification using Matlab:
EDU>> M=[4 0;0 2]; K=[200 200;200 800];
EDU>> [phi,omega]=eig(K,M);
EDU>> sqrt(omega)
6.0211
0
0
20.3407
EDU>> phi(:,1)/abs(phi(1,1))
-1.0000
0.2749
1.0000
7.2749
Which matches the result derived. One mode shape has both displacement in phase, and the other
mode shape shows the displacements to be out of phase.
120
CHAPTER 2. HW’S
2.9
2.9.1
HW9
problem 1
After solving problem 4.3 in text, Sketch the deformation of the system when it moves in each of the
modes.
positive
We solved this problem in HW8, using classical Lagrangian method. This problem will now be solved
using power balance method. The static equilibrium position must be chosen so that all generalized
coordinates have value zero. Hence, using the above diagram as the static equilibrium, we take θ1 = θ2 = 0
in this position.
Now, as in Lagrangian method, we always start by finding kinetic energy T
1
1
T = I1 θ˙12 + I2 θ˙22
2
2
Where I1 =
m1 L2
and I2 =
3
m2 L2
12
+m2
L 2
4
7 2
L m2
48
Now we compare the above expression to the quadratic form
T =
1
M11 θ˙12 + M22 θ˙22 + 2M12 θ˙1 θ˙2
2
m2 L2
12
2
+m2 L16 =
2.9. HW9
121
Hence we see that M11 = I1 , M22 = I2 ,M12 = M21 = 0, therefore the mass matrix is
[M ] =
I1
0
0
I2
!
We now find the potential energy due to springs. For this, we need to write down the relative
displacement between end points of each spring. Let ∆1 be the relative displacement in the first spring
k1 and let ∆2 be the relative displacement in the second spring k2 . Hence (and assuming springs remain
straight, since we are assuming very stiff springs and small angles) then
3L
3L
θ1 +
θ2
4
4
L
∆2 = Lθ1 + θ2
2
∆1 =
Then
1
1
Vspring = k1 ∆21 + k1 ∆22
2
2
2
2
L
3L
3L
1
1
θ1 +
θ2 + k2 Lθ1 + θ2
= k1
2
4
4
2
2
1
9 2 2
9 2 2 9 2
1
1 2 2
2 2
2
= k1
L θ1 + L θ1 θ2 + L θ2 + k2 L θ1 + L θ1 θ2 + L θ2
2
16
8
16
2
4
9
1
1
9
1
9
= L2 θ12 k1 + L2 θ12 k2 + L2 θ22 k1 + L2 θ22 k2 + L2 θ1 θ2 k1 + L2 θ1 θ2 k2
32
2
32
8
2
16 9 2
1 2
1
1 2
9
9
2
2
2
2
2
=
L k1 + L k2 θ1 +
L k1 + L k2 θ2 +
L k1 + L k2 θ1 θ2
32
2
32
8
16
2
Now we compare the above to quadratic form
Vspring =
1
K11 θ12 + K22 θ22 + 2K12 θ1 θ2
2
We see that
9 2
L k1 + L2 k2
16
9
1
= L2 k1 + L2 k2
16
4
L2
9 2
= L k1 + k2
16
2
K11 =
K22
K12
Now we need to find Vgravity .Taking the static equilibrium position as the datum, then upward
displacement of center of gravity will be positive and downward displacement is negative. This means
the left bar will add positive potential energy due to gravity and the right bar will add negative potential
energy, hence
L
L
Vg = m1 g sin θ1 − m2 g sin θ2
2
4
122
CHAPTER 2. HW’S
Now we need to find the components of the gravity potential energy stiffness matrix. Notice that
each term is evaluated at static equilibrium
2
∂Vg
L
Vg11 =
= −m1 g sin θ1
=0
∂θ12 θ1 =0
2
θ1 =0
θ =0
2 2
∂Vg
L
= −m1 g sin θ2
Vg22 =
=0
∂θ22 θ1 =0
2
θ2 =0
θ2 =0
2 ∂Vg
Vg12 =
=0
∂θ1 ∂θ2 θ1 =0
θ2 =0
Hence, no contribution from gravity is added to the stiffness matrix. All contribution comes from the
springs potential energy. Therefore, the stiffness matrix is
!
9 2
L2
9 2
2
L
k
+
L
k
L
k
+
k
1
2
1
16
2 2
[K] = 16
9 2
L2
1 2
9 2
L k1 + 2 k2 16 L k1 + 4 L k2
16
Now since there is no damping, then Pdisp = 0. To find Pin we need to find
Pin = Q1 θ1 + Q2 θ2
The only external force is F which generates a torque F L2 θ2 , hence by comparing to the above
L
θ2
2
L
Q2 = F
2
Pin = F
M X 00 + kX = Q
!
!
!
!
!
9 2
9 2
L2
2
θ¨1
θ
0
I1 0
L
k
+
L
k
L
k
+
k
1
1
2
1
2
16
2
+ 16
=
9 2
L2
9 2
1 2
¨
0 I2
θ2
L
k
+
k
L
k
+
L
k
θ
F L2
2
1
2
1
2
16
2
16
4
!
!
!
!
!
m1
9
9
1
¨1
0
θ
k
+
k
k
+
k
θ
0
1
2
1
16 1
2 2
L2 3
+ L2 916
=
7
1
9
1
¨
0 48 m2
θ2
k + 2 k2 16 k1 + 4 k2
θ2
F L2
16 1
Now we can solve the problem given.
When m1 = m, m2 = 2m, k1 = k, k2 =
mL2
1
3
0
0
7
24
!
k
2
θ¨1
θ¨2
we obtain
!
+ kL2
17
16
13
16
13
16
11
16
!
θ1
!
θ2
To find modes of free vibration, let the RHS {0} then we write
h
i
2m
[K] − ω
[M ] {Φ} = {0}
k
Let λ = ω 2 m
, hence
k
[[K] − λ [M ]] {Φ} = {0}
=
0
F L2
!
2.9. HW9
123
"
det
17
16
13
16
"
det
13
16
11
16
!
17
16
− 13 λ
−λ
13
16
1
3
0
0
7
24
!#
#
13
16
11
16
−
=0
7
λ
24
−2
9.7222 × 10−2 λ2 − 0.53906λ + 7.0313 × 10
=0
=0
Hence, taking the positive square root only we find
λ1 = 0.13366
λ2 = 5.4110
When λ1 = 0.13366
"
17
16
[[k] − λ1 [M ]] {Φ}1 = {0}
#( ) ( )
13
− 31 λ1
Φ11
0
16
=
13
11
7
Φ21
0
− 24
λ1
16
16
"
#( ) ( )
17
1
13
−
λ
1
0
16
3 1
16
=
13
7
11
− 24
λ1
Φ21
×
16
16
17 1
13
− λ1 + Φ21 = 0
16 3
16
17
16 1
17
16 1
λ1 −
=
(0.137) −
= −1.253
Φ21 =
13 3
16
13 3
16
Hence the first mode associated with λ1 = 0.13366 is
(
)
1
−1.253
When λ2 = 5.4110
"
17
16
[[k] − λ1 [M ]] {Φ}1 = {0}
#( ) ( )
13
Φ12
0
− 31 λ2
16
=
13
11
7
− 24
λ2
Φ22
0
16
16
"
#( ) ( )
17
1
13
−
λ
1
0
2
16
3
16
=
13
11
7
− 24 λ2
Φ22
×
16
16
17 1
13
− λ2 + Φ22 = 0
16 3
16
16 1
17
16 1
17
Φ22 =
λ2 −
=
(5.411) −
= 0.912
13 3
16
13 3
16
124
CHAPTER 2. HW’S
Hence the second mode associated with λ2 = 5.411 is
(
1
)
0.912
Summary
ω (rad/sec)
r
λ = ω2 m
⇒ω=
k
q √
k
0.137 = 0.366
m
r
⇒ω=
λ = ω2 m
k
2.9.1.1
q √
k
5.411 = 2.326
m
k
m
k
m
mode shape
(
)
1
−1.253
(
)
1
0.912
verification using Matlab
EDU>> M=[1/3 0;0 7/24]; K=[17/16 13/16;13/16 11/16];
EDU>> [phi,omega]=eig(K,M);
EDU>> sqrt(omega)
0.3656
0
0
2.3262
1.0000
-1.2529
-1.0000
-0.9122
2.9.2
(
1
Sketch of each mode
)
0.912
means that θ1 and θ2 are in phase, and for each 1 unit rotation of θ1 there will be 0.912 units
(
of rotation of θ2 , while
1
)
means that θ1 and θ2 are out of phase, and for each 1 unit rotation of
−1.253
θ1 there will be 1.253 units of rotation of θ2 but in the opposite direction. This is a sketch of both modes
2.9. HW9
2.9.3
problem 2
125
126
CHAPTER 2. HW’S
Using power balance method, we start by finding the kinetic energy.
Since the top bar does not have a point that is fixed in inertial space as the lower bar does, then we
take its moment of inertia around its center of mass, and add a translational kinetic energy due to the
motion of its center of mass in space. For the lower bar, since it has a point that is fixed in space, then
we take the moment of inertia around that point, and we do not need to account for translational kinetic
energy for the lower bar. To find the speed of the center of mass of the top bar, we can either use its
coordinates system x, y differentiate these w.r.t time, or we can use the angular motion of the base of the
second bar and add it to the speed of the center of mass of the second bar relative to the base. This is
what will be done next:
2.9. HW9
127
Therefore, the speed components of the center of mass of the top bar is
L˙
θ2 cos θ2 + Lθ˙1 cos θ1
2
L
vy = − θ˙2 sin θ2 − Lθ˙1 sin θ1
2
So the velocity of the center of mass is
q
vc.g. = vx2 + vy2
vx =
Now that we have the translation velocity of the top bar, and we know its moment of inertia around
its c.g. then we have all the terms needed to obtain the kinetic energy.
1
1
1 2
T = I1 θ˙12 + I2 θ˙22 + mvc.g.
2
2
2
Again, the important thing is to note that I1 is taken around the base of lower rod while I2 is taken
around the center of mass of the top rod. Hence
1 mL2 ˙2
T =
θ +
2 3 1
1 mL2 ˙2
θ +
=
2 3 1
1 mL2 ˙2
θ +
2 12 2
1 mL2 ˙2
θ +
2 12 2
1
m vx2 + vy2
2
2 2 !
1
L˙
L
m
θ2 cos θ2 + Lθ˙1 cos θ1 + − θ˙2 sin θ2 − Lθ˙1 sin θ1
2
2
2
1 mL2 ˙2 1 mL2 ˙2 1
θ1 +
θ + m
=
2 12 2 2
2
2 2 3
L ˙2 2
L ˙2
2
2 ˙2
2
2˙ ˙
2 ˙2
2
2˙ ˙
θ cos θ2 + L θ1 cos θ1 + L θ2 θ1 cos θ2 cos θ1 +
θ sin θ2 + L θ1 sin θ1 + L θ2 θ1 sin θ2 sin θ1
4 2
4 2
Simplifying the last term, and using cos2 θ1 + sin2 sin2 θ1 = 1 we obtain
2
1 mL2 ˙2 1 mL2 ˙2 1
L ˙2
2 ˙2
2˙ ˙
T =
θ +
θ + m
θ + L θ1 + L θ2 θ1 (cos θ2 cos θ1 + sin θ2 sin θ1 )
2 3 1 2 12 2 2
4 2
To compare with the quadratic form, we collect all terms as follows
1 mL2 1 2
1 mL2 1
1
2
2
2
2
˙
˙
˙
˙
T = θ1
+ L m + θ2
+ mL + θ2 θ1
mL (cos θ2 cos θ1 + sin θ2 sin θ1 )
2 3
2
2 12
8
2
Using cos θ2 cos θ1 + sin θ2 sin θ1 = cos (θ2 − θ1 ) the above becomes
4
1
1
2
2
2
2
2
˙
˙
˙
˙
T = θ1
mL + θ2
mL + θ2 θ1
mL cos (θ2 − θ1 )
6
6
2
We now compare the above to
1
1
T = M11 θ˙12 + M22 θ˙22 + M12 θ˙2 θ˙1
2
2
Therefore
4
M11 = mL2
3
1
M22 = mL2
3
1
M12 = M21 = mL2 cos (θ2 − θ1 )
2
128
CHAPTER 2. HW’S
I am not sure how to get the same answer given for the mass matrix. Even if I assume that θ2 − θ1 is
very small, hence M12 = 12 mL2 then the mass matrix is
!
1
2
1
3
4
3
1
2
[M ] = mL2
Now we find the Vspring the potential energy due to springs.
1
Vspring = βmgLθ12 +
2
1
= βmgLθ12 +
2
1
βmgL (θ2 − θ1 )2
2
1
βmgL θ22 + θ12 − 2θ2 θ1
2 1
2
2
= θ1 (βmgL) + θ2
βmgL + θ2 θ1 (−βmgL)
2
Comparing to quadratic form
1
1
Vspring = K11 θ12 + K11 θ12 + K12 θ1 θ2
2
2
Then
K11 = 2βmgL
K22 = βmgL
K12 = K21 = −βmgL
Hence the stiffness matrix due to springs only is
[K] = mgL
2β
−β
−β
β
!
We know need to find the gravity contribution to stiffness. We start by finding the Vgravity . We take
the datum as the horizontal line at the bottom the lower bar.
L
L
Vgravity = mg cos θ1 + mg L cos θ1 + cos θ2
2
2
Hence
V11 =
∂ 2 Vg
L
= −mg cos θ1 − mg (L cos θ1 )
2
∂ θ1
2
evaluate at θ1 = 0 gives
V11 = −mg
L
− mgL
2
3
= − mgL
2
and
V22
∂ 2 Vg
= 2 = −mg
∂ θ2
evaluate at θ2 = 0 gives
V22 = −mg
L
cos θ2
2
L
2
2.9. HW9
129
and
V12 =
∂ 2 Vg
=0
∂θ1 ∂θ2
Hence the stiffness matrix due to gravity is
[K] = mgL
− 23
0
0
− 12
!
Combine the above with the stiffness matrix due to springs we obtain
!
!
2β −β
− 23 0
+ mgL
[K] = mgL
0 − 12
−β β
!
2β − 23 −β
= mgL
−β
β − 12
There is no Pdisp and no Pin hence the equations of motion are
!
!
!
!
3
4
1
¨1
θ
2β
−
−β
θ
1
2
2
+ mgL
mL 31 21
=
¨2
θ
−β
β − 12
θ2
2
3
For β = 4
mL2
4
3
1
2
1
2
1
3
!
θ¨1
θ¨2
!
+ mgL
!
!
−4
θ1
13
2
7
2
−4
θ2
=
To find modes of free vibration, we write
[k] − ω 2 [M ] {Φ} = {0}
"
det mgL
!
−4
13
2
−4
"
det
7
2
− ω 2 mL2
!
−4
13
2
7
2
−4
L
− ω2
g
4
3
1
2
1
2
1
3
!#
4
3
1
2
1
2
1
3
!#
=0
Let ω 2 Lg = η, hence
"
det
13
2
−4
"
det
Hence η = 55.084, η = 0.63023
!
−4
7
2
13
2
−η
− 43 η
4
3
1
2
1
2
1
3
!#
−4 − 12 η
=0
#
=0
−4 − 12 η 72 − 13 η
7 2 65
27
η − η+
=0
36
6
4
=0
!
0
0
!
0
0
130
CHAPTER 2. HW’S
When η = 55.084
[[k] − η [M ]] {Φ}1 = {0}
#( ) ( )
"
4
1
13
−
η
−4
−
η
Φ11
0
2
3
2
=
1
7
1
−4 − 2 η 2 − 3 η
Φ21
0
"
#( ) ( )
−66.945 −31.542
Φ11
0
=
−31.542 −14.861
Φ21
0
"
#( )
−66.945 −31.542
1
−31.542 −14.861
Φ21
=
( )
0
×
−66.945 − 31.542Φ21 = 0
Φ21 = −
66.945
= −2.1224
31.542
Hence the first mode associated with η = 55.084 is
(
)
1
−2.1224
When η = 0.63023
[k] − ω22 [M ] {Φ}2 = {0}
"
#( ) ( )
13
4
1
−
η
−4
−
η
Φ12
0
2
3
2
=
−4 − 12 η 72 − 13 η
Φ22
0
"
#( ) ( )
5.6597 −4.3151
Φ12
0
=
−4.3151 3.2899
Φ22
0
"
#( )
5.6597 −4.3151
1
−4.3151
3.2899
Φ22
=
( )
0
×
5.6597 − 4.3151Φ22 = 0
−5.6597
= 1.3116
Φ22 =
−4.3151
Hence the second mode associated with η = 0.630 is
(
)
1
1.3116
Summary, ω 2 Lg
√ p
= η hence ω = η Lg
ωn (rad/sec)
p
7.422 Lg
p
0.794 Lg
mode shape
(
)
1
−2.1224
(
)
1
1.3116
2.9. HW9
131
For β = 2
mL2
4
3
1
2
1
2
1
3
!
θ¨1
θ¨2
!
3
2
4−
+ mgL
!
−2
−2
θ1
1
2
2−
!
θ2
=
!
0
0
[k] − ω 2 [M ] {Φ} = {0}
"
!
−2
5
2
det mgL
−2
"
det
3
2
− ω 2 mL2
!
−2
5
2
3
2
−2
L
− ω2
g
4
3
1
2
1
2
1
3
!#
4
3
1
2
1
2
1
3
!#
=0
=0
Let ω 2 Lg = η, hence
"
det
5
2
−2
"
det
!
−2
3
2
5
2
4
3
1
2
−η
− 34 η
1
2
1
3
!#
=0
−2 − 12 η
#
=0
−2 − 12 η 32 − 13 η
1
7 2 29
η − η− =0
36
6
4
Hence η = 24.909, η = −5.162 × 10−2
√ p
Since ω 2 Lg = η hence when η = −5.162 × 10−2 then ω = η Lg which means there will a
complex number for ω which is not possible as the frequency must be positive. This means such a
system is not stable .It is not possible to obtain the shape functions when ω is complex.
2.9.4
Problem 3
"
problem 4.11 in text: the mass and stiffness matrices of a system are [M ] =
"
300
1
4 1
1 3
#
kg, [K] =
#
kN/m, determine the system natural frequencies and normal vibration modes. (hint, normal
1 200
modes means mass normalized modes).
Answer:
[k] − ω 2 [M ] {v} = {0}
""
det
1000
1000
200 × 103
"
det
#
300 × 103
"
− ω2
4 1
##
=0
1 3
300 × 103 − 4ω 2
1000 − ω 2
1000 − ω 2
200 × 103 − 3ω 2
#
=0
11ω 4 − 1698000ω 2 + 59999000000 = 0
132
CHAPTER 2. HW’S
Hence the positive roots are ω = 234.02, ω = 315.59
When ω1 = 234.02 rad/sec then
[k] − ω12 [M ] {v}1 = {0}
"
"
300 × 103 − 4ω12
1000 − ω12
#( )
v11
1000 − ω12
200 × 103 − 3ω12
v21
#( )
v11
300 × 103 − 4 (234.02)2
1000 − (234.02)2
1000 − (234.02)2
200 × 103 − 3 (234.02)2
v21
"
#( )
80939. −53765.
v11
−53765.
Let v11 be the arbitrary value 1 hence
"
#( )
80939. −53765.
1
−53765.
35704.
=
v21
35704.
v21
=
=
=
( )
0
0
( )
0
0
( )
0
0
( )
0
×
80939. − 53765v21 = 0
v21 = −
80939
= 1.5054
53765
Hence
(
{v}1 =
1
)
1.5054
When ω2 = 315.59 rad/sec then
[k] − ω22 [M ] {v}2 = {0}
"
"
300 × 103 − 4ω22
1000 − ω22
#( )
v12
1000 − ω22
200 × 103 − 3ω22
v22
#( )
v12
300 × 103 − 4 (315.59)2
1000 − (315.59)2
1000 − (315.59)2
200 × 103 − 3 (315.59)2
v22
"
#( )
−98388. −98597.
v12
−98597. −98791.
"
#( )
−98388. −98597.
1
−98597. −98791.
v22
=
v22
( )
0
×
−98388 − 98597v22 = 0
v22 = −
98388
= −0.998
98597
=
=
=
( )
0
0
( )
0
0
( )
0
0
2.9. HW9
133
Hence
(
{v}2 =
)
1
−0.998
To obtain the mass normalized shape functions:
µ1 = {v}T1 [M ] {v}1
(
)T "
#(
)
1
4 1
1
=
1.5054
1 3
1.5054
(
)
h
i
1
= 5.5054 5.5162
1.5054
= 13.809
And
µ2 = {v}T2 [M ] {v}2
(
)T "
#(
)
1
4 1
1
=
−0.999
1 3
−0.998
)
(
h
i
1
= 3.002 −1.994
−0.998
= 4.992
Hence
(
1
)
1.505
{v}
{Φ}1 = √ 1 = √
=
µ1
13.809
(
)
0.269
0.405
and
(
{v}
{Φ}2 = √ 2 =
µ2
1
)
(
−0.999
√
4.992
=
0.446
)
−0.447
Summary
ωn (rad/sec)
234.02
315.59
original mode shape
(
)
1
normal mode shapes
(
)
0.269
1.5054
(
)
1
0.405
(
)
0.44759
0.999
−0.44665
Hence
"
[Φ] =
0.2691
0.448
#
0.40511 −0.447
134
CHAPTER 2. HW’S
To verify
"
[Φ]T [M ] [Φ] =
"
=
=
Which is approximately
1.0
0
0.448
#T "
4 1
#"
0.269
0.446
#
0.405 −0.447
1 3 0.405 −0.447
#"
#"
#
0.269 0.405
4 1 0.269 0.448
0.448 −0.447
"
"
0.269
1.0
8.840 × 10−5
0.405 −0.447
#
8.840 × 10−5
1 3
1.0
#
as expected. calculations were not done with high enough accuracy,
0 1.0
so that is why the off-diagonal numerical values were not an exact zeros.
To verify with the [K] matrix
"
#T "
#"
#
0.269 0.448
300 × 103
1000
0.269 0.448
T
[Φ] [K] [Φ] =
0.405 −0.447
1000
200 × 103 0.405 −0.447
"
#
54765. 6.594
=
6.594 99600.0
Note ω12 = 234.022 = 54765. and ω22 = 315.592 = 99597 and these are the values on the diagonal as
expected. The values off the diagonal should be an exact zero, since the [K] matrix should be decoupled.
Due to low precision in the above calculations, the values did not come out to be zero.
Verify using Matlab. Note that Matlab eig() returns the shape function that are mass normalized
EDU>> M=[4 1;1 3];
EDU>> K=[300*10^3 1000;1000 200*10^3];
EDU>> [eig,lam]=eig(K,M)
eig =
-0.2691
-0.4051
-0.4476
0.4467
lam =
1.0e+04 *
5.4764
0
0
9.9600
EDU>> eig'*M*eig
1.0000
0
0
1.0000
EDU>> eig'*K*eig
2.9. HW9
135
1.0e+04 *
5.4764
0
2.9.5
0
9.9600
Problem 4
problem 4.29 in text.
The mapping between the generalized coordinates y1 , y2 and yg , θ is given by
yg (A + B) = y1 A + y2 B
θ (A + B) = y2 − y1
136
CHAPTER 2. HW’S
In our care, A = B = L2 , hence the above becomes
y1 L2 + y2 L2
y1 + y2
=
L
2
y2 − y1
θ=
L
yg =
Hence taking derivative
y˙ 1 + y˙ 2
2
y
˙
−
y˙ 1
2
θ˙ =
L
y˙ g =
Using the power balance method, we start by finding the kinetic energy T
1
1
T = my˙ g2 + Icg θ˙2
2
2
2
y˙ 2 − y˙ 1 2
1
y˙ 1 + y˙ 2
1
2
= m
+
mrG
2
2
2
L
where rG is the radius of gyration 0.4L, hence
2 !
4
1
1
1
m
L
y˙ 22 + y˙ 12 − 2y˙ 1 y˙ 2
T = m y˙ 12 + y˙ 22 + 2y˙ 1 y˙ 2 +
2
8
2
10
L
1
8
= m y˙ 12 + y˙ 22 + 2y˙ 1 y˙ 2 +
m y˙ 22 + y˙ 12 − 2y˙ 1 y˙ 2
8 100
8
1
8
16
2
1
2
2
m+
m + y˙ 2
m+
m + y˙ 1 y˙ 2
m−
m
= y˙ 1
8
100
8
100
8
100
41
41
9
=
my˙ 12 +
my˙ 22 +
my˙ 1 y˙ 2
200
200
100
Comparing the above to quadratic form T = 12 M11 y˙ 12 + 12 M22 y˙ 22 + M12 y˙ 1 y˙ 2 then
41
m = 0.41m
100
41
=
m = 0.41m
100
= 0.09m
M11 =
M22
M12
Hence the mass matrix is
"
[M ] = m
#
0.41 0.09
0.09 0.41
1
1
Vspring = klef t y12 + kright y12
2 2
1 3
1
=
k y12 + ky12
2 2
2
Comparing to quadratic form 12 K11 y12 + 21 K22 y22 + K12 y1 y2 then
"
#
1.5 0
[kspring ] = k
0 1
2.9. HW9
137
y1 + y2
2
Since this will be evaluated at y1 = y2 = 0 then we see right away that there is no contribution to
potential energy to the stiffness matrix. Hence the EOM are
"
#"
#
"
#"
# " #
0.41 0.09 y¨1 (t)
1.5 0 y1 (t)
0
m
+k
=
0.09 0.41 y¨2 (t)
0 1 y2 (t)
0
Vgravity = mgyg = mg
To convert to t0 space, given by t0 =
d2 y
dt2
q
k
t
m
as required, then we see that
dy
dt
d2 y
k
= dt02 m
Hence the ODE becomes
"
#"
#
"
0.41 0.09 y¨1 (t0 ) k
1.5
m
+
k
0.09 0.41 y¨2 (t0 ) m
0
#
"
#"
"
1.5
0.41 0.09 y¨1 (t0 )
+k
k
0
0
0.09 0.41 y¨2 (t )
Since k 6= 0 we can divide by it, hence
"
#"
#
0.41 0.09 y¨1 (t0 )
y¨2 (t0 )
0.09 0.41
"
+
1.5 0
0
1
0
#"
#
y1 (t0 )
y2 (t0 )
#
#"
0 y1 (t0 )
1
y2 (t0 )
1
#"
#
y1 (t0 )
y2 (t0 )
=
=
=
" #
0
0
" #
0
0
" #
0
0
[K] − ω 2 [M ] {v} = {0}
""
det
det
1.5 0
#
"
− ω2
##
0.41 0.09
0 1
0.09 0.41
"
#
1.5 − 0.41ω 2 −ω 2 0.09
−ω 2 0.09
1 − ω 2 0.41
=0
=0
0.16ω 4 − 1.025ω 2 + 1.5 = 0
Hence the positive roots are ω1 = 2.0357, ω2 = 1.5041
When ω1 = 2.0357 then
[k] − ω12 [M ] {v}1 = {0}
"
1.5 − 0.41 (2.0357)2
− (2.0357)2 0.09
− (2.0357)2 0.09
#( )
v11
1 − (2.0357)2 0.41
v21
"
#( )
−0.199 −0.373
v11
−0.373 −0.699
v21
=
=
( )
0
0
( )
0
0
=
dy dt0
dt0 dt
=
dy
dt0
q
k
,
m
and
138
CHAPTER 2. HW’S
"
#( )
−0.199 −0.373
1
−0.373 −0.699
v21
=
( )
0
×
−0.199 − 0.373v21 = 0
v21 = −
Hence
(
{v}1 =
0.199
= −0.534
0.373
)
1
−0.534
When ω2 = 1.5041 then
[k] − ω22 [M ] {v}2 = {0}
"
1.5 − 0.41 (1.504)2
− (1.504)2 0.09
− (1.504)2 0.09
1 − (1.504)2 0.41
v22
"
#( )
0.572 −0.204
v12
−0.204
"
0.572
−0.204
#( )
v12
0.072
#( )
−0.204
1
0.072
v22
=
v22
=
=
( )
0
×
0.572 − 0.204v22 = 0
0.572
= 2.812
v22 =
0.204
Hence
(
{v}2 =
1
)
2.812
To obtain the mass normalized shape functions:
µ1 = {v}T1 [M ] {v}1
(
)T "
#(
)
1
0.41 0.09
1
=
−0.534
0.09 0.41
−0.534
= 0.43073
And
µ2 = {v}T2 [M ] {v}2
(
)T "
#(
)
1
0.41 0.09
1
=
2.812
0.09 0.41
2.812
= 4.1569
( )
0
0
( )
0
0
2.9. HW9
139
Hence
(
)
1
(
−0.53374
√
0.43073
{v}1
{Φ}1 = √
=
0.43073
and
(
=
1.5237
)
−0.81326
)
1
2.812
{v}
=
{Φ}2 = √ 2 = √
µ2
4.1569
(
)
0.491
1.379
Summary
ω (rad/sec)
2.0357
1.5041
original mode shape
(
)
1
normal mode shapes
(
)
1.524
−0.534
(
)
1
−0.813
(
)
0.491
2.812
1.379
Hence
"
[Φ] =
#
0.491
1.524
−0.813 1.379
To verify
"
[Φ]T [M ] [Φ] =
−0.813 1.379
"
=
"
Which is approximately
1.0
#T "
#"
0.491
0.41 0.09
1.524
1.524
0
#
0.491
−0.813 1.379
#
0.09 0.41
1.0
−1.9688 × 10−4
−1.9688 × 10−4
1.0
#
as expected. calculations were not done with high enough accuracy,
0 1.0
so that is why the off-diagonal numerical values were not an exact zeros.
To verify with the [K] matrix
"
[Φ]T [K] [Φ] =
1.5237
−0.81326
"
=
#T "
#"
0.49047
1.5 0
1.5237
1.3790
0
−4.9183 × 10−4
−4.9183 × 10−4
2.2625
EDU>> K=[1.5 0;0 1]; M=[0.41
EDU>> [eig,lam]=eig(K,M)
-1.5238
−0.81326
#
4.1439
Verify using Matlab
eig =
-0.4905
1
0.09;0.09 0.41];
#
0.49047
1.3790
140
CHAPTER 2. HW’S
-1.3789
0.8130
lam =
2.2624
0
0
4.1439
Now we can solve the problem. Using {x} = [Φ] {η}, where
{η} = [Φ]−1 {x}
= [Φ]T [M ] {x}
Hence, initial conditions in the {η} space is
{η}0 = [Φ]T [M ] {x}0
(
)
y
(0)
1
= [Φ]T [M ]
y2 (0)
"
#T "
#( )
mg
1.524 0.491
0.41 0.09
k
=
0
−0.813 1.379
0.09 0.41
(
)
0.552 kg m
=
0.325 kg m
and
{η 0 }0 = [Φ]T [M ] {x0 }0
)
(
0
y
(0)
= [Φ]T [M ] 10
y2 (0)
( )
0
= [Φ]T [M ]
0
( )
0
=
0
So, we need to solve
"
1 0
0 1
#( )
η¨1
η¨2
"
+
#( )
η1
4.1439
0
0
2.2625
η2
=
( )
0
with the initial conditions
(
)
η1 (0)
η2 (0)
(
)
η10 (0)
η20 (0)
The solution is given by
=
=
(
)
0.552 kg m
0.325 kg m
( )
0
0
√
√
η1 (t) = A cos 4.144t + B sin 4.144t
0
2.9. HW9
141
When t = 0, η1 (0) = 0.55152 kg m = A. Taking derivative gives η10 (t) = −A sin
hence when t = 0 we have 0 = B, therefore
√
√
4.144t+B cos 4.144t,
√
g
η1 (t) = 0.552 m cos 4.144t
k
Now we solve for η2 (t) ,The solution is given by
√
√
η2 (t) = A cos 2.263t + B sin 2.263t
When t = 0, η2 (0) = 0.3252 kg m = A. and 0 = B, therefore
√
g
η2 (t) = 0.325 m cos 2.263t
k
Now we obtain the solution in the y space
{y} = [Φ] {η}
)
(
) "
#(
√
y1 (t0 )
1.524 0.491
0.552 kg m cos 4.1439t
=
√
0.325 kg m cos 2.2625t
y2 (t0 )
−0.813 1.379
(
)
0.840 kg m cos (2.036t0 ) + 0.1595 kg m cos (1.504t0 )
=
0.448 kg m cos (1.504t0 ) − 0.449 kg m cos (2.036t0 )
We are supposed to obtain the answer
(
)
y1 (t)
y2 (t)
(
=
0.16 cos (1.5t0 ) + 0.84 cos (2t0 )
)
0.45 cos (1.5t0 ) − 0.45 cos (2t0 )
The answers agree. The scalar kg m for some reason is not shown in the key solution.
2.9.6
side question
After solving problem 4.3 in text, Sketch the deformation of the system when it moves in each of the
modes.
142
CHAPTER 2. HW’S
2.9.7
power balance method
positive
The static equilibrium position must be chosen so that all generalized coordinates have value zero.
Hence, using the above diagram as the static equilibrium, we take θ1 = θ2 = 0 in this position.
Now, as in Lagrangian method, we always start by finding kinetic energy T
1
1
T = I1 θ˙12 + I2 θ˙22
2
2
Where I1 =
m1 L2
and I2 =
3
m2 L2
12
+m2
L 2
4
m2 L2
12
2
+m2 L16 =
7 2
L m2
48
Now we compare the above expression to the quadratic form
T =
1
M11 θ˙12 + M22 θ˙22 + 2M12 θ˙1 θ˙2
2
Hence we see that M11 = I1 , M22 = I2 ,M12 = M21 = 0, therefore the mass matrix is
[M ] =
I1
0
0
I2
!
We now find the potential energy due to springs. For this, we need to write down the relative
displacement between end points of each spring. Let ∆1 be the relative displacement in the first spring
k1 and let ∆2 be the relative displacement in the second spring k2 . Hence (and assuming springs remain
straight, since we are assuming very stiff springs and small angles) then
3L
3L
θ1 +
θ2
4
4
L
∆2 = Lθ1 + θ2
2
∆1 =
Then
2.9. HW9
143
1
1
Vspring = k1 ∆21 + k1 ∆22
2
2
2
2
1
3L
3L
L
1
= k1
θ1 +
θ2 + k2 Lθ1 + θ2
2
4
4
2
2
1
9 2 2 9 2
9 2 2
1
1 2 2
2 2
2
= k1
L θ1 + L θ1 θ2 + L θ2 + k2 L θ1 + L θ1 θ2 + L θ2
2
16
8
16
2
4
9
1
9
1
9
1
= L2 θ12 k1 + L2 θ12 k2 + L2 θ22 k1 + L2 θ22 k2 + L2 θ1 θ2 k1 + L2 θ1 θ2 k2
32
2
32
8
2
16 9
9
9 2
1 2
1
1 2
2
2
2
2
2
=
L k1 + L k2 θ1 +
L k1 + L k2 θ2 +
L k1 + L k2 θ1 θ2
32
2
32
8
16
2
Now we compare the above to quadratic form
Vspring =
1
K11 θ12 + K22 θ22 + 2K12 θ1 θ2
2
We see that
9 2
L k1 + L2 k2
16
9
1
= L2 k1 + L2 k2
16
4
9
L2
= L2 k1 + k2
16
2
K11 =
K22
K12
Now we need to find Vgravity .Taking the static equilibrium position as the datum, then upward
displacement of center of gravity will be positive and downward displacement is negative. This means
the left bar will add positive potential energy due to gravity and the right bar will add negative potential
energy, hence
L
L
sin θ1 − m2 g sin θ2
2
4
Now we need to find the components of the gravity potential energy stiffness matrix. Notice that
each term is evaluated at static equilibrium
Vg = m1 g
Vg11 =
Vg22 =
Vg12 =
∂Vg2
∂θ12
θ1 =0
θ2 =0
2
∂Vg
∂θ22
θ1 =0
θ2 =0
∂Vg2
∂θ1 ∂θ2
L
= −m1 g sin θ1
=0
2
θ1 =0
L
= −m1 g sin θ2
=0
2
θ2 =0
=0
θ1 =0
θ2 =0
Hence, no contribution from gravity is added to the stiffness matrix. All contribution comes from the
springs potential energy. Therefore, the stiffness matrix is
!
9 2
9 2
L2
2
L
k
+
L
k
L
k
+
k
1
2
1
16
2 2
[K] = 16
9 2
L2
9 2
1 2
L k1 + 2 k2 16 L k1 + 4 L k2
16
144
CHAPTER 2. HW’S
Now since there is no damping, then Pdisp = 0. To find Pin we need to find
Pin = Q1 θ1 + Q2 θ2
The only external force is F which generates a torque F L2 θ2 , hence by comparing to the above
L
θ2
2
L
Q2 = F
2
Pin = F
I1
0
m1 L2
3
0
2.9.7.1
M X 00 + kX = Q
!
!
!
9 2
L2
9 2
2
L
k
+
L
k
L
k
+
k
θ
0
θ¨1
0
1
2
1
1
16
2 2
=
+ 16
9 2
L2
9 2
1 2
¨
L
k
+
k
L
k
+
L
k
θ
F L2
θ2
I2
1
2
1
2
2
16
2
16
4
!
!
!
!
!
9
9
1
0
θ¨1
k
+
k
k
+
k
θ
0
1
2
1
2
1
16
2
+ L2 916
=
7 2
1
1
9
¨
θ
L
m
k
+
k
k
+
k
θ2
F L2
2
2
48
16 1
2 2 16 1
4 2
!
!
Lagrangian method
The Lagrangian Φ = T − V where
1
1
T = I1 θ˙12 + I2 θ˙22
2
2
Where I1 =
m1 L2
and I2 =
3
m2 L2
12
+m2
L 2
4
m2 L2
12
2
+m2 L16 =
7 2
L m2
48
Now we find the potential energy, assuming springs remain straight (stiff spring assumption) and
assuming small angles and assuming zero PE datum as above

2

2
∆1
∆
}|
{
z
z }|2 {
1  3L
3L 
L 
L
1 
L
V = k1 
Lθ1 + θ2 
θ1 +
θ2 
+ k2 
+ m1 g sin θ1 − m2 g sin θ2




2
4
4
2
2
2
4
Hence
Φ=T −V
1 ˙2 1 ˙2
I1 θ + I2 θ −
=
2 1 2 2
1
k1
2
3L
3L
θ1 +
θ2
4
4
2
!
2
1
L
L
L
+ k2 Lθ1 + θ2 + m1 g sin θ1 − m2 g sin θ2
2
2
2
4
2.9. HW9
145
∂Φ
= I θ˙1
∂ θ˙1
d ∂Φ
= I θ¨1
dt ∂ θ˙1
∂Φ
3L
3L
3L
L
L
= −k1
θ2 +
θ1
− k2
θ2 + Lθ1 (L) − m1 g cos θ1
∂θ1
4
4
4
2
2
3L
3L
3L
L
L
= − k1
θ2 +
θ1 − k2 L
θ2 + Lθ1 − m1 g cos θ1
4
4
4
2
2
d ∂Φ
∂Φ
= Qθ1
−
dt ∂ θ˙1 ∂θ1
3L
3L
3L
L
L
¨
I1 θ1 +
k1
θ2 +
θ1 + k2 L
θ2 + Lθ1 + m1 g cos θ1 = 0
4
4
4
2
2
The generalized force is zero, since there is no direct external force acting on top rod.
Hence EOM for θ1 is from above
m1 L2 ¨
θ1 + θ1
3
k1
3L
4
2
!
+ k2 L2
+ θ2
k1
3L
4
2
L2
+ k2
2
!
= −m1 g
L
cos θ1
2
∂Φ
= I θ˙2
∂ θ˙2
d ∂Φ
= I θ¨2
dt ∂ θ˙2
3L
L
3L
3L
L
L
∂Φ
θ2 +
θ1
θ2 + Lθ1
= −k1
− k2
+ m2 g cos θ2
∂θ2
4
4
4
2
2
4
d ∂Φ
∂Φ
−
= Qθ2
˙
dt ∂ θ2 ∂θ2
3L
3L
3L
L
L
L
¨
I2 θ2 + k1
+ k2
− m2 g cos θ2 = Qθ1
θ2 +
θ1
θ2 + Lθ1
4
4
4
2
2
4
Now Qθ2 is found by virtual work. Making a virtual displacement δθ2 while fixing θ1 and finding the
work done by all external forces.
L
δW = F δθ2
2
Hence Qθ2 = F L2 with positive sign since it add energy to the system. Hence EOM for θ2 is
7 2 ¨
L m2 θ2 + θ1 k1
48
3L
4
2
L2
+ k2
2
!
+ θ2 k1
3L
4
2
2 !
L
L
L
+ k2
= m2 g cos θ2 + F
2
4
2
146
CHAPTER 2. HW’S
m1 L2
3
0
0
7 2
L m2
48
!
θ¨1
θ¨2
!
2
+
k1 9L
+ k2 L2
16
2
2
+ k2 L2
k1 3L
4
M X 00 + kX = Q
!
!
!
2
L2
L
k1 3L
+
k
cos
θ
θ
−m
g
2 2
1
1
1 2
4
=
3L 2
L 2
L
k1 4 + k2 2
m2 g 4 cos θ2 + F L2
θ2
2.10. HW10
147
2.10
HW10
2.10.1
problem 1
Generalized coordinates are y3 , y2 , y1 . Kinetic energy is T = 12 m3 (y30 )2 + 12 m2 (y20 )2 + 12 m1 (y10 )2 .
Potential energy due to springs is Vspring = 21 k3 y32 + 21 k2 (y2 − y3 )2 + 21 k1 (y1 − y2 )2 . Therefore
1
1
1
Vspring = k3 y32 + k2 y22 + y32 − 2y2 y3 + k1 y12 + y22 − 2y1 y2
2 2
2
1
1
1
1
1
2
2
2
k3 + k2 + y2
k2 + k1 + y1
k1 + y1 y2 (−k1 ) + y1 y3 (0) + y2 y3 (−k2 )
= y3
2
2
2
2
2
The EOM is
   
  

00
k1
−k1
0
y1
0
y
m1 0
0
   
  1 





 0 m2 0  y 00  + −k1 k2 + k1
−k2  y2  = 0

  2 

00
0
y3
0
−k2
k3 + k2
y3
0
0 m3
Following values are for mass (units in kg) m1 = 100, m2 = 200, m3 = 300. Following values are for
spring constants (units in N/m) k1 = 402 (100) , k2 = 502 (200) , k3 = 602 (300) . EOM becomes

  
   
100 0
0
y100
160000 −160000
0
y1
0

  
   
 0 200 0  y 00  + −160000 660000 −500000 y2  = 0

  2 
   
00
0
0 300
y3
0
−500000 1580000
y3
0
Characteristic equation is
det [K] − ω 2 [M ] = 0




160000 −160000
0
100 0
0




−160000 660000 −500000 − ω 2  0 200 0  = 0
det 




0
0 300
0
−500000 1580000


160000 − 100ω 2
−160000
0


2
=0
det 
−160000
660000
−
200ω
−500000


2
0
−500000
1580000 − 300ω
−6 × 106 ω 6 + 6.1 × 1010 ω 4 − 1.54 × 1014 ω 2 + 8.64 × 1016 = 0
148
CHAPTER 2. HW’S
Positive roots of the above polynomial are the natural frequencies (units in rad/sec).
ω1 = 28.1
ω2 = 52.6
ω3 = 81.3
To obtain mode shapes, the eigenvector associated with each eigenvalue is found. Starting with
ω1 = 28.1



    
160000 −160000
0
100 0
0
1
0









−160000 660000 −500000 − 28.12  0 200 0  ϕ21  = 0



    
0
−500000 1580000
0
0 300
ϕ31
0
Hence

   
8.1 × 104 −160000
0
1
0

   
5
 −160000 5.02 × 10
   
−500000 

 ϕ21  = 0
0
−500000 1.34 × 106
ϕ31
0

  
8.1 × 104 − 1.6 × 105 ϕ21
0

  
5.02 × 105 ϕ21 − 5.0 × 105 ϕ31 − 1.6 × 105  = 0

  
6
5
1.34 × 10 ϕ31 − 5 × 10 ϕ21
0
Solving gives ϕ21 = 0.506 and ϕ31 = 0.188. First eigenvector is


1



ϕ1 = 
0.506


0.188
For ω2 = 52.6,
    



0
1
100 0
0
160000 −160000
0









−160000 660000 −500000 − 52.62  0 200 0  ϕ22  = 0
    



0
ϕ32
0
−500000 1580000
0
0 300
Hence

   
−1.17 × 105 −1.6 × 105
0
1
0
   

 −1.6 × 105 1.07 × 105 −5.0 × 105  ϕ22  = 0

   
0
−5.0 × 105 7.50 × 105
ϕ32
0

  
−1.6 × 105 ϕ22 − 1.17 × 105
0
  

1.07 × 105 ϕ22 − 5.0 × 105 ϕ32 − 1.6 × 105  = 0

  
5
5
7.5 × 10 ϕ32 − 5.0 × 10 ϕ22
0
Solving gives ϕ22 = −0.731 and ϕ32 = −0.476 .Second eigenvector is


1



ϕ2 = 
−0.731


−0.476
2.10. HW10
149
For ω3 = 81.3


    
100 0
0
1
0



    
2
−160000 660000 −500000 − 81.3  0 200 0  ϕ23  = 0



    
0
−500000 1580000
0
0 300
ϕ33
0
160000
−160000
0

Hence

   
−5. 01 × 105 −1.6 × 105
0
1
0

   
 −1.6 × 105 −6.62 × 105 −5.0 × 105  ϕ23  = 0

   
0
−5.0 × 105 −4.03 × 105
ϕ33
0

  
−1.6 × 105 ϕ23 − 5.01 × 105
0

  
−6.62 × 105 ϕ23 − 5.0 × 105 ϕ33 − 1.6 × 105  = 0

  
5
5
0
−5.0 × 10 ϕ23 − 4.03 × 10 ϕ33
Solving gives ϕ23 = −3.13 and ϕ32 = ϕ33 = 3.82 . Third eigenvector is

1




ϕ3 = 
−3.13


3.82
Eigenvectors are mass normalized. Mass normalization factors µi are found for each eigenvector
µ1 = ϕT1 [M ] ϕ1


T 

1
100 0
0
1


 






= 0.506  0 200 0  0.506
 = 162.
0.188
0
0 300
0.188
and
µ2 = ϕT2 [M ] ϕ2

T
1



=
−0.731


−0.476



100 0
0
1



 0 200 0  −0.731 = 275.



0
0 300
−0.476
and
µ3 = ϕT3 [M ] ϕ3

T
1



=
−3.13


3.82



100 0
0
1



 0 200 0  −3.13 = 6.44 × 103



0
0 300
3.82
150
CHAPTER 2. HW’S
Normalized eigenvectors are

1


7.86 × 10−2

 

ϕ1
1 
0.506 = 3.98 × 10−2 
Φ1 = √ = √
 

µ1
162 
0.188
1.48 × 10−2

 

1
6.03 × 10−2
 

ϕ2
1 
−0.731 = −4.41 × 10−2 
Φ2 = √ = √
 

µ2
275. 
−2
−0.476
−2.87 × 10

 

−2
1
1.25 × 10

 

1
ϕ3



=  −0.039 
Φ3 = √ = √
−3.13



µ3
6.44 × 103
−2
3.82
4.76 × 10
Verification of the above result follows
EDU>> k=[160000 -160000 0;-160000 660000 -500000;0 -500000 1580000];
EDU>> M=[100 0 0;0 200 0;0 0 300];
EDU>> [eigV,lam]=eig(k,M)
eigV =
0.0786
0.0398
0.0148
0.0606
-0.0437
-0.0289
0.0124
-0.0389
0.0477
lam =
1.0e+03 *
0.7897
0
0
2.7528
0
0
EDU>> sqrt(diag(lam))
ans =
28.1013
52.4674
81.3889
0
0
6.6242
2.10. HW10
2.10.2
151
Problem 2
Initial conditions are θi (0) = 0 for i = 1, 2, 3 and θ10 (0) = θ30 (0) = 0 but θ20 (0) = 2 rad/sec.
The generalized coordinates are shown above. kinetic energy is
1
1
1
2
2
2
T = I (θ10 ) + I (θ20 ) + I (θ30 )
2
2
2
where I = 13 mL2 . Mas matrix becomes
 

1 0 0 θ100
 

1
 00 
[M ] = mL2 
0 1 0

 θ2 
3
0 0 1 θ300
Spring potential energy is
1
1
Vspring = k (Lθ2 − Lθ1 )2 + k (Lθ3 − Lθ2 )2
2
2
1
1 2 2
2
= kL θ2 + θ1 − 2θ1 θ2 + kL2 θ32 + θ22 − 2θ2 θ3
2 2
1
1
1 2
1 2
2
2
2
2
2
= θ1
kL + θ2
kL + kL + θ3
kL + θ1 θ2 −kL2 + θ1 θ3 (0) + θ2 θ3 −kL2
2
2
2
2
Hence stiffness matrix due to spring is

1

[K]spring = kL2 
−1
0
−1
0


−1

−1 1
2
152
CHAPTER 2. HW’S
Assume the zero PE for gravity is taken as the top of the bar. Stiffness due to gravity is
Vg = −mg
V11 =
∂ 2 Vg
∂θ12
L
(cos θ1 + cos θ2 + cos θ3 )
2
= mg L2 (cos θ1 ) . Evaluate this at static position θ1 = 0,hence V11 = m L2 . Similarly,
V22 = V33 = m L2 . All other terms are zero.
Hence stiffness matrix due to gravity is


1 0 0

L

[K]g = mg 
0
1
0

2
0 0 1
Therefore, complete stiffness matrix is




1 0 0
1 −1 0




 + mg L 0 1 0
kL2 
−1
2
−1



2
0 0 1
0 −1 1
There are no generalized forces. Hence EOM is


  
    


1 −1 0
0
θ1
1 0 0 θ100
1 0 0














L
1
 θ00  + kL2 −1 2 −1 + mg 0 1 0 θ2  = 0
mL2 
0
1
0


  2 
    

3
2
00
0 −1 1
0
θ3
0 0 1 θ3
0 0 1
2.10.3
Part (a) k = 0.05 mg
L
mg
, Hence for σ = 0.05 then k = σ mg
. EOM becomes
L
L


  
    


1 −1 0
θ1
0
1 0 0 θ100
1 0 0














L
1
 θ00  + σmgL −1 2 −1 + mg 0 1 0 θ2  = 0
mL2 
0
1
0


  2 

    
3
2
00
0 −1 1
0 0 1 θ3
0 0 1
θ3
0

 

   
1
1 0 0 θ100
+σ
−σ
0
θ1
0

 
   
2
1
2
1
00








mL 0 1 0 θ2  + mgL  −σ 2 + 2σ −σ  θ2  = 0

3
1
00
0 0 1 θ3
0
−σ
+σ
θ3
0
2
 


   
1
1 0 0 θ100
+σ
−σ
0
θ1
0

   3g  2
   
1
0 1 0 θ00  +

   
  2  L  −σ 2 + 2σ −σ  θ2  = 0

1
0 0 1 θ300
0
−σ
θ3
0
+σ
2
For case k = 0.05
Let L = 1, g = 10.

1

0

0
The above becomes
  
   
0 0 θ100
15 + 30σ
−30σ
0
θ1
0
  
   
 00  
   
1 0
15 + 60σ
−30σ 
 θ2  +  −30σ
 θ2  = 0
0 1 θ300
0
−30σ
15 + 30σ
θ3
0
2.10. HW10
153
Natural frequencies of the system are found by solving the eigenvalue problem.




1 0 0
15 + 30σ
−30σ
0




 −30σ
 − ω 2 0 1 0 = 0
det 
15
+
60σ
−30σ




0 0 1
0
−30σ
15 + 30σ
Substituting σ = 0.05 gives



1 0 0
16.5 −1.5
0




2




det −1.5 18.0 −1.5 − ω 0 1 0
 = 0
0 0 1
0
−1.5 16.5


16.5 − ω 2 −1.5
0


2
=0
det 
−1.5
18
−
ω
−1.5


0
−1.5 16.5 − ω 2

−ω 6 + 51ω 4 − 861.75ω 2 + 4826.3 = 0
Positive roots of this polynomial are ω = 3.87, ω = 4.062, ω = 4.416.
Associated eigenvectors are found by solving for ϕi in ([K] − ω 2 [M ]) ϕi = 0 for each eigenvalue ωi .
For ω1 = 3.87
   

2
1
0
16.5 − ω1
−1.5
0
   

2




 −1.5
−1.5  ϕ21  = 0
18 − ω1


2
0
0
−1.5 16.5 − ω1
ϕ31
   

0
1
16.5 − 3.872
−1.5
0
   

 ϕ21  = 0

−1.5
18 − 3.872
−1.5
   

0
ϕ31
0
−1.5
16.5 − 3.872
   

1
0
1.523 −1.5
0

   
 −1.5 3.023 −1.5  ϕ21  = 0
   

0
0
−1.5 1.523 ϕ31
  

0
1.523 − 1.5ϕ21

  
3.023 ϕ21 − 1.5ϕ31 − 1.5 = 0
  

1.523 ϕ31 − 1.5ϕ21
0
Solving gives ϕ21 = 1.015 3 and ϕ31 = 1.046 2. First eigenvector is
  

1
1
  

 = 1.0153
ϕ1 = 
ϕ
21
  

ϕ31
1.0462
Similarly, second and the third eigenvectors are found.
154
CHAPTER 2. HW’S
Eigenvectors are mass normalized. First the mass normalization factors µi are found for each
eigenvector
µ1 = ϕT1 [M ] ϕ1

T
1



=
1.015
3


1.046 2



1 0 0
1



0 1 0 1.0153 = 3.125 4



0 0 1 1.0462
Normalized eigenvector is



0.51353

 

ϕ1
1
1.015 3 = 0.52139
Φ1 = √
=√
 

3.125 4
3.792 
1.046 2
0.53726
1

Verification of the above result (Matlab result is more accurate due to more accurate method used)
EDU>> k=[0.55 -0.05 0;-0.05 0.6 -0.05;0 -0.05 0.55];
EDU>> M=eye(3);
EDU>> [eigV,lam]=eig(k,M)
eigV =
-0.5774
-0.5774
-0.5774
-0.7071
-0.0000
0.7071
0.4082
-0.8165
0.4082
ans =
0.7071
0.7416
0.8062
Transformation matrix (based on Matlab more accurate result) is


−0.577 −0.7073 0.4082



Φ = [Φ1 Φ2 Φ3 ] = 
−0.577
0
−0.8165


−0.577 0.706 9
0.4082
Mapping from physical coordinates θ to modal coordinates η is
θ = [Φ] η
2.10. HW10
155
Bold face is used to indicate a column vector. EOM’s are written in modal coordinates resulting in
   

  
2
00
η1
0
ω
0 0
1 0 0 η1
   

   1
2
00
   
0 1 0 η  +  0 ω
0
2
 η2  = 0

  2 
0 0 ω22
η3
0
0 0 1 η300

  
   
1 0 0 η100
0.70712
0
0
η1
0

  
   
0 1 0 η 00  +  0
   
0.74162
0 

  2 
 η2  = 0
0 0 1 η300
0
0
0.80622
η3
0
Initial conditions are transformed to modal coordinates using η (0) = [Φ]T [M ] θ (0) and η 0 (0) =
[Φ]T [M ] θ0 (0), since θ (0) = 0 then η (0) = 0, however θ0 (0) is not all zero, hence
 
T 
 

0
η (0)
−0.577 −0.7073 0.4082
1 0 0 0
 1  
 
 
η 0 (0) = −0.577


 
0
−0.8165 0 1 0
 2  
 2
η30 (0)
−0.577 0.7069
0.4082
0 0 1 0


−1.154



=
0


−1.633
Initial conditions in modal coordinates are found. The solution can be
found. The solution to
0
η 00 + ω 2 η = 0 with initial conditions η (0) and η 0 (0) is η (t) = η (0) cos ωt + η ω(0) sin ωt. Therefore modal
solutions are
−1.154
sin (0.7071t) = −1.632 sin (0.707 t)
0.7071
η2 (t) = 0
−1.633
sin (0.8062t) = −2.026 sin (0.8062t)
η3 (t) =
0.8062
η1 (t) =
Solution in the normal coordinates is
 



−0.577 −0.7073 0.4082
−1.632 sin (0.707 t)
θ1 (t)

 


θ2 (t) = −0.577


0
−0.8165
0
 



θ3 (t)
−0.577 0.706 9
0.4082
−2.026 sin (0.8062t)


0.94166 sin (0.707 t) − 0.82701 sin (0.8062t)



=
0.94166
sin
(0.707
t)
+
1.6542
sin
(0.8062t)


0.94166 sin (0.707 t) − 0.82701 sin (0.8062t)
2.10.3.1
Part (b) k = 2 mg
L
Using part (a), but with σ = 2 results in




15 + 30σ
−30σ
0
1 0 0




 −30σ
 − ω 2 0 1 0 = 0
det 
15
+
60σ
−30σ




0
−30σ
15 + 30σ
0 0 1
156
CHAPTER 2. HW’S




1 0 0
15 + 30 (2)
−30 (2)
0




 −30 (2)
 − ω 2 0 1 0 = 0
det 
15
+
60
(2)
−30
(2)




0 0 1
0
−30 (2)
15 + 30 (2)


2
75.0 − ω
−60.0
0


2

det  −60.0
135.0 − ω
−60.0 
=0
0
−60.0
75.0 − ω 2
Similar steps as repeated as part (a) above. The final result are shown below using Matlab
EDU>> k=[75 -60 0;-60 135 -60;0 -60 75]
EDU>> M=eye(3);
[eigV,lam]=eig(k,M)
eigV =
-0.5774
-0.7071
-0.5774
0.0000
-0.5774
0.7071
0.4082
-0.8165
0.4082
3.8730
8.6603
13.9642


−0.577 −0.7071 0.4082



Transformation matrix is Φ = [Φ1 Φ2 Φ3 ] = −0.577
0
−0.816 5
 . Mapping from θ to modal
−0.577 0.7071
0.4082
coordinates η is
θ = [Φ] η
Bold face is used to indicate a column vector. EOM’s are written in modal coordinates resulting in

  
   
00
2
1 0 0 η1
ω
0 0
η1
0

   1
   
00
2
0 1 0 η  +  0 ω
   
0
2

  2 
 η2  = 0
0 0 1 η300
0 0 ω22
η3
0

  
   
1 0 0 η100
3.87302
0
0
η1
0

  
   
0 1 0 η 00  +  0
 η2  = 0
8.66032
0

  2 
   
0 0 1 η300
0
0
13.96422
η3
0
Initial conditions are transformed to modal coordinates using η (0) = [Φ]T [M ] x (0) and η 0 (0) =
[Φ]T [M ] θ0 (0), since θ (0) = 0 then η (0) = 0, however θ0 (0) is not all zero. Similar to part (a), initial
conditions are found

 

η10 (0)
−1.154

 

η 0 (0) =  0 
 2  

0
η3 (0)
−1.633
2.10. HW10
157
The solution to η 00 + λ2 η = 0 with initial conditions η (0) and η 0 (0) is given by η (t) = η (0) cos λt +
sin λt. The solutions are
η 0 (0)
λ
−1.154
sin (3.873t) = −0.297 96 sin (3.873 t)
3.8730
η2 (t) = 0
−1.633
η3 (t) =
sin (13.9642) = −0.116 94 sin (13.9642)
13.9642
η1 (t) =
Solution in the physical coordinates is


 

−0.297 96 sin (3.8730 t)
−0.577 −0.7071 0.4082
θ1 (t)


 



θ2 (t) = −0.577
0
0
−0.816
5


 

−0.116 94 sin (13.9642t)
−0.577 0.7071
0.4082
θ3 (t)


−2
0.171 92 sin (3.873t) − 4.773 5 × 10 sin (13.964t)


−2

=
9.548 2 × 10 sin (13.964t) + 0.171 92 sin (3.873t)
0.171 92 sin (3.873t) − 4.773 5 × 10−2 sin (13.964t)
Summary table
k
0.05 mg
L
2 mg
L
frequencies





0.7071

0.7416



0.8062





3.8730




8.6603



13.9642

[Φ]


−0.5774 −0.7071 0.4082


−0.5774

0
−0.8165


−0.5774 0.7071
0.4082


−0.5774 −0.7071 0.4082


−0.5774

0
−0.8165


−0.5774 0.7071
0.4082
solutions in θ


0.94166 sin (0.707 t) − 0.82701 sin (0.8062t)


 0.94166 sin (0.707 t) + 1.6542 sin (0.8062t) 


0.94166 sin (0.707 t) − 0.82701 sin (0.8062t)


0.17192 sin (3.873t) − 4.773 5 × 10−2 sin (13.964t)


9.5482 × 10−2 sin (13.964t) + 0.171 92 sin (3.873t)


−2
0.17192 sin (3.873t) − 4.773 5 × 10 sin (13.964t)
Even though the normalized natural frequencies are different, the shape functions are the same.
Plots of the solutions of θi (t) for both cases are made. For the case of k = 0.05 mg
L
158
CHAPTER 2. HW’S
For k = 2 mg
L
2.10. HW10
159
In addition, a small program is written to animate both the full solution and the modal solutions.
The program to animate the full solution is at http://12000.org/my_courses/univ_wisconson_madison/
spring_2013/EMA_545_Mechanical_Vibrations/HWs/HW10/HW10p2.m.txt while the program that animate
the modal solution is number 112 at bottom of this page my_matlab_functions
160
CHAPTER 2. HW’S
2.10.4
Problem 3
EOM is

  

 
  

0
00 





x
200
cos
(16t)
300
0
−200
x
500
300
−400
x
600 400 200 








1

  
  1



  1 
3

400 1200 0  x00 + 300 900 600  x0 +10  0
0
500 300  x2 =
  2

  2 



















0
00 
−200 300 700
x3
0
x3
−400 600 1300
x3
200
0
800





x1 (0)

 



0



0



x1 (0)

 



0

x2 (0) = 0 and x02 (0) = 0 .

 


 



x (0)
 
0

x0 (0)
 
0

3
3
Solve the eigenvalue problem to determine the natural frequencies of the system
Initial conditions are

300 × 103
0


det 
0
500 × 103

−200 × 103 300 × 103
det [K] − ω 2 [M ] = 0



−200 × 103
600 400 200



2
 = 0
−
ω
300 × 103 
400
1200
0



3
700 × 10
200
0
800
−4.0 × 108 ω 6 + 1.044 × 1012 ω 4 − 4.72 × 1014 ω 2 + 5.8 × 1016 = 0
2.10. HW10
161
Positive roots are {ω = 15.052, ω = 17.562, ω = 45.552} . For each natural frequency the corresponding
eigenvector is found. A program is now used to compute these values.
EDU>> k = [300 0 -200;0 500 300;-200 300 700]*10^3;
M = [600 400 200;400 1200 0;200 0 800];
C = [500 300 -400;300 900 600;-400 600 1300];
[PHI,lam] = eig(k,M);
PHI
lam
= sqrt(diag(lam))
CC
= PHI'*C*PHI;
zeta1 = CC(1,1)/(2*lam(1))
zeta2 = CC(2,2)/(2*lam(2))
zeta3 = CC(3,3)/(2*lam(3))
PHI =
-0.0216
0.0203
-0.0220
0.0232
0.0168
0.0023
-0.0373
0.0201
0.0302
lam =
15.0519
17.5624
45.5522
zeta1 =
0.0018
zeta2 =
0.0219
zeta3 =
0.0376


−0.0216 0.0232 −0.0373


 . In modal coordinates EOM is
[Φ] = 
0.0203
0.0168
0.0201


−0.0220 0.0023 0.0302
162
CHAPTER 2. HW’S


 
00 
500
1 0 0 
η




  1
T
0 1 0 η 00 + [Φ]  300


  2
 
−400
0 0 1 η300 

  
1 0 0 
5.419 × 10−2 5.331 × 10−2

η 00 

  1 
0 1 0 η 00 + 5.331 × 10−2
0.768

  2 


0 0 1 η300 
−0.4156
−3.52 × 10−4

  
1 0 0 
2ζ1 ω1

η 00 

  1 
0 1 0 η 00 +  0

  2 
 
0 0 1 η 00 
0
3

  
 

0
2




η
η
ω
0
0
300 −400





1
200 co
 1  1
 

T
0
2


900 600 
 [Φ] η2  +  0 ω2 0  η2  = [Φ] 

 



η 0 
0 0 ω32 η3 
600 1300
3

 
  
0
2




η
−0.416
η
ω
0
0

 1
 
200 co
 
  1  1
T
0
2


−3.52 × 10−4 
 η2  +  0 ω2 0  η2  = [Φ] 

 



η 0 
3.428
0 0 ω32 η3 
3

  
 
0
2




0
0
0 0 
η1 
ω
η 

200 co
  1  1
 
T
0
2



2ζ2 ω2
0  η2 +  0 ω2 0  η2 = [Φ]




 
 



0
2ζ3 ω3 η30 
0 0 ω32 η3 
In the above 2ζ1 ω1 = 0.0542, 2ζ2 ω2 = 0.7676 and 2ζ3 ω3 = 3.424 7. Hence ζ1 =
0.76755
2(17.5624)
5.419 3×10−2
2(15.0519)
= 0.0018
3.424 7
2(45.5522)
and ζ2 =
= 0.0219 and ζ3 =
= 0.0376
Final EOM in modal coordinates is

  
  
  

0
00 





η
−4.32
cos
(16.0t)
226.56
0
0
η
0.0542
0
0
η
1 0 0 








1

  
  1 
  1 

0



0 1 0 η 00 + 0
4.64 cos (16.0t)
308.44
0  η2 =
0.768
0  η2 + 0
  2 




















0
00 
−7.46 cos (16.0t)
0
0
2075
η3
η3
0
0
3.425
η3
0 0 1
EOM’s to solve are
η100 + 2ζ1 ω1 η10 + ω12 η1 = −4.32 cos (16.0t)
η200 + 2ζ2 ω2 η20 + ω22 η2 = 4.64 cos (16.0t)
η300 + 2ζ3 ω3 η30 + ω32 η3 = −7.46 cos (16.0t)
Initial conditions are zero. The solution in modal coordinates is given in appendix B for underdamped
case. Complete solution for the case of underdamped is given in appendix B as
F0
ζωβ
−ζωt
η (t) = 2
β cos ($t) + 2ζω$ sin ($t) − e
sin (ωd t) h (t)
β cos (ωd t) +
β + 4ζ 2 ω 2 $2
ωd
p
β = (ω 2 − $2 ), ωd = ω 1 − ζ 2 .
The solutions in modal coordinates are now found. Recall that ω1 = 15.0519, ω2 = 17.5624, ω3 =
45.5522 and ζ1 = 0.0018,ζ2 = 0.0219 and ζ3 = 0.0376
3
X
Next step is to transform the solution to the physical coordinates using qj =
Φ (j, m) η (m), or
m=1
q = [Φ] η
In component form
q1 (t) = Φ (1, 1) η1 (t) + Φ (1, 2) η2 (t) + Φ (1, 3) η3 (t)
q2 (t) = Φ (2, 1) η1 (t) + Φ (2, 2) η2 (t) + Φ (2, 3) η3 (t)
q3 (t) = Φ (3, 1) η1 (t) + Φ (3, 2) η2 (t) + Φ (3, 3) η3 (t)
Program was written to complete the computation and make plots. Here is the result showing plots
of each of the above qi (t) vs. time
2.10. HW10
163
function nma_HW10_problem_3_EMA_545()
%solve for q(t) using modal analysis, by Nasser M. Abbasi
close all;
syms t;
N = 3;
k = [300 0 -200;0 500 300;-200 300 700]*10^3;
M = [600 400 200;400 1200 0;200 0 800];
C = [500 300 -400;300 900 600;-400 600 1300];
wF = 16;
F = [200*cos(wF*t); 0; 0];
[PHI,lam] = eig(k,M);
lam
= sqrt(diag(lam));
CC
= PHI'*C*PHI
F = PHI.'*F;
eta = sym(zeros(N, 1));
time_constant = zeros(3,1);
for i=1:N
w
= lam(i);
b
= w^2-wF^2;
zeta = CC(i,i)/(2*w);
wd
= w*sqrt(1-zeta^2);
eta(i) = F(i)/(b^2+4*zeta^2*w^2*wF^2) * ...
( b*cos(wF*t)+2*zeta*w*wF*sin(wF*t)- ...
exp(-zeta*w*t)* ( b*cos(wd*t)+ zeta*w*b/wd * sin(wd*t)
);
time_constant(i) = 1/(zeta*w);
end
) ...
q=PHI*eta;
time_constant
time_constant = sum(time_constant);
% plot the generalized solutions
lims= [-0.004 0.003;
-0.002 0.007;
-0.006 0.002
];
for i=1:N
subplot(3,1,i);
ezplot(q(i),[0,100]);
ylim(lims(i,:));
title(sprintf('q(%d) solution, time constant = %f',i,time_constant));
xlabel('time (sec)');
ylabel('q(t) Newton');
164
CHAPTER 2. HW’S
end
end
From above, the time to reach steady state is about 90 seconds based on looking at q1 (t) since that
takes the longest time to each steady state out of the three coordinates.
The time constant for each ηi (t) solution was calculated giving τ1 = ζ11ω1 = 37.4471 and τ2 = 2.602
and τ3 = 0.58. The first time constant τ1 = 37.4 seconds dominated the result in the response in the
physical coordinates.
This means the dominant time constant found in modal analysis is one to use to estimate how long it
will take for the response in physical coordinates to reach steady state. Each modal solution contributes
to each physical solution. The one with the longest time constant affects more than any other mode how
long the physical solution takes to reach steady state.
2.10. HW10
2.10.5
Problem 4
"
[M ] =
165
5
#
−3
−3
4
(
kg,ω1 = 15.68 rad/sec,ω2 = 40.78 rad/sec. ϕ1 =
(
µ1 =
µ2 =
1
)T "
#(
)
−3
1
5
1.366
−3 4
1.366
(
)T "
#(
1
5 −3
1
−0.366
−3
1.366
= 4.267 8
)
= 7.731 8
−0.366
4
)
1
Normalized eigenvectors are
1
1
Φ1 = √ ϕ1 = √
µ1
4.267 8
1
1
Φ2 = √ ϕ2 = √
µ2
7.731 8
Hence
"
[Φ] = [Φ1 Φ2 ] =
(
1
)
−0.366
(
)
1
−0.366
0.484 06
(
=
=
0.484 06
)
−0.177 17
(
)
0.359 63
−0.131 63
0.359 63
#
−0.177 17 −0.131 63
(
, ϕ2 =
1
)
−0.366
166
"
CHAPTER 2. HW’S
EOM in modal coordinates is
#( ) "
#( ) "
#( )
(
)
1 0
η100
2 (0.08) (15.68)
0
η10
15.682
0
η1
50
sin
(20t)
+
+
= [Φ]T
00
0
2
0 1
η2
0
2 (0.08) (40.78)
η2
0
40.78
η2
100 cos (20t)
#( ) "
#( ) "
"
#( ) "
245.86
0
η1
24.203 sin (20.0t) − 17.71
2.509
0
η10
1 0
η100
+
=
+
0
00
η2
0
1663
η2
17.982 sin (20.0t) − 13.16
η2
0
6.524 8
0 1
The two EOMs to solve are
24.203 i20t
(t) +
(t) + 245.86η1 (t) = 24.203 sin (20t) − 17.717 cos (20t) = Re
+ Re −17.717ei20t
e
i
17.982 i20t
0
00
η2 (t) + 6.525η2 (t) + 1663η2 (t) = 17.982 sin (20t) − 13.163 cos (20t) = Re
+ Re −13.163ei20t
e
i
η100
2.509η10
Hence
η100 (t) + 2.509η10 (t) + 245.86η1 (t) = 24.203 sin (20t) − 17.717 cos (20t) = Re (−24.203i − 17.717) ei20t
η200 (t) + 6.525η20 (t) + 1663η2 (t) = 17.982 sin (20t) − 13.163 cos (20t) = Re (−17.982i − 13.163) ei20t
In matrix form
[I] η 00 + [C] η 0 + [K] η = Re Fei$t
Where $ = 20 rad/sec. F is the complex amplitude of the input
(
)
−24.203i − 17.717
F=
−17.982i − 13.163
Using method of transfer functions (since steady state response is needed), response is
η = Re Xei20t
Where
Xj =
−$2
Fj
+ 2iζj ωj $ + ωj2
Steady state solutions in modal coordinates is
−24.203i − 17.717
i$t
η1 (t) = Re
e
−$2 + 2.5088i$ + 245.86
−24.203i − 17.717
i$t
= Re
e
−400 + 50.176i + 245.86
= Re 5.77 × 10−2 + 0.176i ei$t
−17.982i − 13.163
i$t
η2 (t) = Re
e
−$2 + 6.525i$ + 1663
−17.982i − 13.163 i$t
= Re
e
−400 + 130.5i + 1663
= Re −1.178 × 10−2 − 1.302 × 10−2 i ei$t
Solutions are transformed back to normal coordinates
q = [Φ] η
2.10. HW10
167
Hence
qj (t) =
=
2
X
n
2
X
Φ (j, n) η (n)
Φ (j, n) Re X (n) ei$t
n
= Re
2
X
Φ (j, n) X (n) ei$t
n
"
Since[Φ] =
0.484 06
0.359 63
#
then
−0.177 17 −0.131 63
0.484 06 5.77 × 10−2 + 0.176i + 0.359 63 −1.178 × 10−2 − 1.302 × 10−2 i ei20t
q2 (t) = Re −0.177 17 5.77 × 10−2 + 0.176i − 0.131 63 −1.178 × 10−2 − 1.302 × 10−2 i ei20t
q1 (t) = Re
or
2.369 × 10−2 + 8.051 × 10−2 i ei20t
q2 (t) = Re −8.672 × 10−3 − 2.947 × 10−2 i ei20t
q1 (t) = Re
Therefore
Y1 = 2.369 × 10−2 + 8.051 × 10−2 i
Y2 = −8.672 × 10−3 − 2.947 × 10−2 i
168
CHAPTER 2. HW’S
2.10.6
Problem 5
2.10.6.1
Part(a)
First step is to determine EOM. The kinetic energy T is
1
1
2
2
T = I (θ0 ) + m2 (x0 )
2
2
I = 13 m1 L2 .Assuming small angle, stiff spring approximation and zero gravity datum at the level
where pendulum is hinged, spring potential energy V is
2
1
L
V = k x− θ
2
2
1
L2 2
2
= k x + θ − xLθ
2
4
2 L
1
kL
2
2
=θ
k +x
k + xθ −
8
2
2
2.10. HW10
169
Stiffness matrix due to spring is
"
Kspring =
L2
k
4
− kL
2
− kL
2
#
k
Potential energy due to gravity is Vg = −mg L2 cos θ. Hence Vg11 =
other terms are zero. The stiffness matrix due to gravity is
#
"
mg L2 0
Kspring =
0
0
Combined stiffness matrix is
"
K=
EOM is
"
I
#( )
θ00
0
x00
0 m2
+ mg L2 − kL
2
− kL
2
"
+
L2
k
4
L2
k
4
∂ 2 Vg
∂2θ
= mg L2 cos θ
θ=0
= mg L2 . All
#
k
+ mg L2 − kL
2
− kL
2
k
#( )
θ
x
(
=
Qθ
)
Qx
Generalized forces are now found. Qθ = F L since F is only external forces acting on the first d.o.f. θ
and the work done by this force is F Lδθ for small virtual angle. For Qx work is done only by damper
and acts to remove energy, hence negative in sign. Qx = −cx0 . The above becomes
"
#( ) " 2
#( ) (
)
L
L
kL
I 0
θ00
k
+
mg
−
θ
F
L
2
2
+ 4
=
− kL
k
x
−cx0
0 m2
x00
2
Rearranging
"
I
0
0 m2
#( )
θ00
x00
"
+
0 0
#( )
θ0
0 c
x0
"
+
L2
k
4
+ m2 g L2 − kL
2
− kL
2
#( )
θ
k
x
=
( )
FL
0
Each EOM is
00
Iθ +
L
L2
k + m1 g
4
2
m2 x00 + cx0 −
θ−
kL
x = FL
2
kL
θ + kx = 0
2
2
Units checking: First EOM. each term must have units of torque. L4 kθ have units of torque OK.
L
mg 2 θ have units of torque OK. kLx have units of torque OK.
second EOM Each term must have units of force. cx0 have units of force OK. kLθ have units of force,
OK. kx have units of force, OK.
n
o
Transfer function is now found Let x = Re {Xei$t } , θ = Re {Y ei$t } , F = Re Fˆ ei$t .Substitute in
the above EOM
2
n
o
L
L
kL
2
i$t
−I$ Y +
Y −
= Re Fˆ Lei$t
Re
k + m2 g
X e
4
2
2
kL
2
i$t
Re −m2 $ X + ic$X −
Y + kX e
=0
2
170
CHAPTER 2. HW’S
Simplify
L2
L
kL
2
−I$ + k + m2 g
Y −
X = Fˆ L
4
2
2
kL
−m2 $2 + ic$ + k X =
Y
2
(2.107)
(2.108)
The above two equations are solved to obtain the required transfer functions X/F and Y /F . To
obtain Y /F, the second equation solved for X in terms of Y
X=
kL
2
−m2 $2 + ic$ + k
Y
X in first equation is replaced by the giving
kL
L2
L
kL
2
2
−I$ + k + m2 g
Y −
Y = Fˆ L
4
2
2 −m2 $2 + ic$ + k
!
k2 L2
2
L
L
4
−I$2 + k + m2 g −
Y = Fˆ L
4
2
−m2 $2 + ic$ + k
Hence
Y =
1
− 13 m1 L$2
+
L
k
4
+
m2 g
2
−
k2 L/4
−m2 $2 +ic$+k
Fˆ
To obtain the transfer function X/F , the second equation is solved for Y in terms of X
Y =
(−m2 $2 + ic$ + k)
X
kL/2
This is substituted in the first equation giving
L (−m2 $2 + ic$ + k)
kL
L2
2
X−
X = Fˆ L
−I$ + k + m2 g
4
2
kL/2
2
"
#
− 31 m1 L$2 + L4 k + m22 g (−m2 $2 + ic$ + k) kL
−
X = Fˆ L
k/2
2
Hence
X=
− 13 m1 L$2
+
L
k
4
+
m2 g
2
kL
Fˆ
(−m2 $2 + ic$ + k) − k 2 L
This complete part(a). These are the analytical expressions for the transfer functions.
2.10.6.2
Part(b)
Let m1 = m2 = 1 kg, k = 3 N/m, L = 1 m, g = 9.81 m/s2 , c = 0.1 N-s/m.
A program was written to plot the magnitude and phase spectrums of x (t) and θ (t) using the above
numerical values. This was done for a range of forcing frequencies to cover both natural frequencies and
beyond. Natural frequencies are found by solving the eigenvalue problem det ([K] − ω 2 [M ]) = 0
ω1 = 1.1308 rad/sec
ω2 = 4.3228 rad/sec
2.10. HW10
171
The magnitude and phase of each transfer function are evaluated when $ = ω1 and when $ = ω2 .F = 1
was assumed since its numerical value was not given. Result is shown below. From these plots, magnitude
and phase values are determined at the natural frequencies.
x (t) = Re Xei$t
θ (t) = Re Y ei$t
Table of results
response
magnitude at ω1
phase at ω1
magnitude at ω2
phase at ω2
0
x (t)
4.25
−83
2.62
131.70
θ (t)
2.55
−800
11.5
−500
ratio
4.25/2.55 = 1.666 7
Plots used to obtain these results
11.5/2.62 = 4. 389 3
172
CHAPTER 2. HW’S
The function used to generate the plots
function nma_HW10_problem_5_EMA_545_spectrum()
%plots the spectrums of problem 5, HW10, by Nasser M. Abbasi
close all;
c
g
L
k
m1
m2
F
=
=
=
=
=
=
=
0.1;
9.81;
1;
3;
1;
1;
1;
M
K
C
= [1/3*m1*L^2 0;0 m2];
= [L^2/4*k+m2*g*L/2 -k*L/2;-k*L/2 k];
= [0 0;0 c];
[PHI,w] = eig(K,M);
lam
= sqrt(diag(w))
I = sqrt(-1);
X = @(wf) ((k*L)./((-1/3*m1*L*wf.^2+L/4*k+m2*g/2).*(-m2*wf.^2+I*c*wf+k)- (k^2*L)))*F;
Y = @(wf) (1./((-1/3*m1*L*wf.^2+L/4*k+m2*g/2-( k^2*L./(-m2*wf.^2+I*c*wf+k)))))*F;
N = 2;
2.10. HW10
173
for i=1:N
figure(i);
wf = 0:0.1:6.5;
if i==1
name_='X';
tf_ = X(wf);
else
name_='Y';
tf_ = Y(wf);
end
subplot(2,1,1);
plot(wf,abs(tf_));
hold on;
line([lam(1) lam(1)],[0 5],'LineStyle','-.');
line([lam(2) lam(2)],[0 5],'LineStyle','-.');
title(sprintf('magnitude spectrum of %c, $\\omega_1=%f$, $\\omega_2=%f$',name_,lam(1),lam(
xlabel('forcing frequency (rad/sec)');
ylabel(sprintf('$|%c|$',name_),'interpreter','latex','FontSize',12);
grid;
subplot(2,1,2);
plot(wf,angle(tf_));
line([lam(1) lam(1)],[-5 5],'LineStyle','-.');
line([lam(2) lam(2)],[-5 5],'LineStyle','-.');
title(sprintf('phase spectrum of X, $\\omega_1=%f$, $\\omega_2=%f$',lam(1),lam(2)),'interpr
xlabel('forcing frequency (rad/sec)');
ylabel(sprintf('$arg(%c)$',name_),'interpreter','latex','FontSize',12);
grid;
end
end
Eigenvectors Φ1 and Φ1 are now found, using modal analysis, which de-couples the EOM. The ratio
of one component of the same eigenvector to its other component is found and compared with the result
found above. The eigenvectors found are
Φ1 =
Φ2 =
(
)
−0.5446
−0.9493
)
(
−1.6442
0.3145
The ratios are 0.9493/0.5446 = 1.743 1 and 1.6442/0.3145 = 5.228 0. Compare these to the ratios
found
174
CHAPTER 2. HW’S
response
magnitude at ω1
phase at ω1
magnitude at ω2
phase at ω2
0
x (t)
4.25
−83
2.62
131.70
θ (t)
2.55
−800
11.5
−500
ratio
4.25/2.55 = 1.666 7
11.5/2.62 = 4.389 3
These ratios are close to each others. Ratio Φ1j /Φ2j shows how much one dof (1) will change relative
to dof (2) in mode j
2.10.6.3
Part(c)
Transfer functions are plotted in part(a). From magnitude spectrum q
of Y it is
qseen that |Y | = 0 when $
k
between 1.5 and 2.0 rad/sec and also when $ > 6 rad/sec. ωcart = m2 = 31 = 1.732 1 rad/sec. This
q
agrees with range found in plots. When mk2 = 1.73, top mass acts as vibration absorber, and rod will
not oscillate when F (t) is at this specific frequency.
2.10.7
Problem 6
From HW6, problem 3
f (t) =
P
t
τ
0<t<τ
0
τ < t < 2τ
2.10. HW10
175
Let yss (t) be the solution from problem 3 found using FFT technique. Let the full solution for
deflection of the above pillar be
χ (y, t) = y (t) ψ (y)
y (t) is the time dependent (dynamic) part of the solution. This solution is yss (t) found in problem 3.
ψ (y) is solution due to static loading. Also called the shape function. For cantilever beam with static
force P at its end, deflection curve due to static loading P at end is
ψ (x) =
P
3Lx2 − x3
6EI
2
Internal bending moment M (x, t) = EI d χ(x,t)
and direct stress σ =
dx2
h
modulus. Assume c = 2 . For yield, let σ = 40M P a, then
M (x,t)c
I
where c is the section
σI
c
d2 χ (x, t)
σI
EI
=
dx2
c
M (x, t) =
I=
1
bh3 .
12
d2 χ (x, t)
d2 P
2
3
=
y
(t)
3Lx
−
x
ss
dx2
dx2 6EI
PL
= yss (t)
EI
Solve for P at yield
yss (t)
σyield I
PL
=
EI
c
σyield I
EI
P =
yss (t) h2 L
yss (t) from problem 3 has maximum value of 1.8 at t = 10 sec. Given numerical values in the problem
and using this maximum value of yss (t) then P can be found from above.
I am not sure this is the correct approach to solve this.We did not have any practice or examples on
solving this type of vibration problem before. Need more time to study this subject.
176
CHAPTER 2. HW’S
2.11
HWA1
2.11.1
problem 1
A system has mass M = 20kg and ωn = 100 rad/sec. It is observed that steady state response is
q = 20 cos (110t − 1.5) mm, where t is in seconds. Determine harmonic excitation causing this response
for ζ = 0 and ζ = 0.4
Let the harmonic excitation be
n
o
F (t) = Re Fˆ eiωt
where Fˆ is its complex amplitude. Also let
n
o
iωt
ˆ
q = Re Qe
be the steady state response. We are given that q = 20 × 10−3 cos (110t − 1.5), therefore
q = Re 20 × 10−3 ei(100t−1.5)
= Re 20 × 10−3 e−1.5i ei100t
Therefore
ˆ = 20 × 10−3 e−1.5i
Q
But the transfer function for second order system is
ˆ
1
ˆ=F
Q
k (1 − r2 ) + 2iζr
where r =
ω
,
ωn
hence we can now solve for Fˆ from the above.
ˆ k
Fˆ = Q
1 − r2 + 2iζr
But k = M ωn2 hence
ˆ M ω2
Fˆ = Q
n
1 − r2 + 2iζr
When ζ = 0 we find
Fˆ = 20 × 10−3 e−1.5i
20 × 1002 1 −
110
100
= 20 × 10−3 e−1.5i (−42000.0)
= −42000.0 × 20 × 10−3 e−1.5i
= −840.0e−1.5i
Hence
n
o
= Re −840.0e−1.5i ei110t
= Re −840.0ei(100t−1.5)
2 !!
2.11. HWA1
177
Therefore
F (t) = −840 cos (100t − 1.5)
When ζ = 0.4 we find
ˆ M ω2
Fˆ = Q
n
1 − r2 + 2iζr
= 20 × 10−3 e−1.5i
20 × 1002
1−
110
100
2 !
= 20 × 10−3 e−1.5i 20 × 1002 (−0.21 + 0.88i)
+ i2 (0.4)
= 4000e−1.5i (−0.21 + 0.88i)
q
.88
2
2 i tan−1 ( −0.21
−1.5i
)
= 4000e
(0.21) + (0.88) e
In[4]:= ArcTan[-0.21, 0.88]
Out[4]= 1.80505
Hence
Fˆ = 4000e−1.5i 0.90471ei1.80505
= 3618.8e−1.5i+1.80505i
= 3618.8e0.30505i
Therefore
n
o
= Re 3618.8e0.30505i ei110t
= Re 3618.8ei(100t+0.30505)
Hence
F (t) = 3618.8 cos (100t + 0.30505)
110
100
!!
178
CHAPTER 2. HW’S
2.11.2
Problem 2
Let
n
o
P (t) = Re Fˆ eiωt
where Fˆ is the complex amplitude of the excitation. Hence by comparing this to P (t) = F cos ωt =
Re {F eiωt } we see that Fˆ = F .

 ˆ
Q

z }| π{ 

n
o
iωt
ˆ
When ω = 2π100 then the response was q = Re Qe
= 4 sin (2π100t) hence q = Re 4e−i 2 eiωt =




o
n
π
q = Re 4ei(ωt− 2 ) therefore
ˆ = 4e−i π2
Q
But, from the transfer function of second order system we know that
ˆ
1
ˆ=F
Q
2
k (1 − r ) + 2iζr
Hence
π
Fˆ
1
4e−i 2 = 2
k
2π100
1 − 2π100
+
2iζ
ωn
ωn
Fˆ
1
−i tan−1 2ζr2
1−r
s
=
e
k 2 2 2
1 − 2π100
+ 2ζ 2π100
ωn
ωn
(2.109)
By comparing sides we see that
π
2ζr
=
2
1 − r2
2ζ 2π100
ω
=
n 2
1 − 2π100
ωn
(2.110)
2.11. HWA1
179
When ω = 105Hz we are told it is half power point, which means the amplitude there is 0.707 of the
maximum amplitude which occurs when r = 1. Hence
0.707
Fˆ
q
k
Fˆ
1
1
s
=
2
k 2 2 2
1 − (1)2 + (2ζ (1))2
2π105
2π105
1 − ωn
+ 2ζ ωn
0.707
1
= s
2ζ
1−
2π105
ωn
1
2 2
(2.111)
+ 2ζ
2π105
ωn
2
We now have 2 equations 2.110 and 2.111 to solve numerically for ζ and ωn . Solving and keeping the
positive solutions results in
ζ = 0.0309
ωn = 640.8 rad/sec
= 101.987 Hz
Hence at ω = 105 hz the phase is


tan−1
2ζr
1 − r2
2π(105)
 2 (0.0309) 640.8 
0
= tan−1 
2  = 133.305
2π(105)
1 − 640.8
In[35]:= ArcTan[1 - ((2 Pi 105)/640.8)^2, 2 (0.0309) ((2 Pi 105)/640.8)]*180/Pi
Out[35]= 133.305
2.11.2.1
Part(b)
When ω = 100 Hz we found from Eq 2.109 that
4=
Fˆ
s
k 1−
2π100
ωn
1
2 2
+ 2ζ
2π100
ωn
2
But we found that ωn = 640.8 rad/sec and ζ = 0.0309, hence
v
u
2 !2 2
u
ˆ
F
2π100
2π100
t
=4
1−
+ 2 (0.0309)
k
640.8
640.8
= 0.28733
At ω = 110 Hz
Fˆ
1
ˆ
Q = s
k
2 2 2
2π110
2π110
1 − ωn
+ 2ζ ωn
= 0.28733 r
= 1.6288
1
1−
2
2π110 2
640.8
+ 2 (0.0309)
2π110
640.8
2
180
CHAPTER 2. HW’S
The phase is
tan−1
2ζr
1 − r2
= tan−1
2π110
640.8
2π110 2
640.8
2 (0.0309)
1−
!
= 157.7980
In[37]:= ArcTan[1 - ((2 Pi 110)/640.8)^2, 2 (0.0309) ((2 Pi 110)/640.8)]*180/Pi
Out[37]= 157.798
2.11.3
Problem 3
The two equations are
mx001 + 2kx1 − kx2 = 0
mx002 − kx1 + kx2 = f (t)
Since the responses are harmonic and the input is harmonic, then we can write
n
o
iωt
ˆ
x1 (t) = Re X1 e
n
o
ˆ 2 eiωt
x1 (t) = Re X
Therefore the two equations can be written in terms of the complex amplitudes as
ˆ 1 + 2k X
ˆ1 − kX
ˆ2 = 0
−mω 2 X
ˆ2 − kX
ˆ1 + kX
ˆ2 = F
−mω 2 X
(2.112)
(2.113)
From Eq 2.112
(−mω 2 + 2k) ˆ
ˆ2
X1 = X
k
Substitute the above into Eq 2.113 gives
−mω 2
2
(−mω 2 + 2k) ˆ
ˆ 1 + k (−mω + 2k) X
ˆ1 = F
X1 − k X
k
k
(−m2 ω 4 − mω 2 2k)
2
ˆ1 = F
+ k − mω X
k
ˆ 1 = kF
X
1
(−m2 ω 4
−
mω 2 2k
+ k 2 − kmω 2 )
2.11. HWA1
181
Dividing the numerator and denominator of the RHS by k 2 , and using k 2 = ωn4 m2 and using r =
ˆ1 = F X
k −m2 ω4 −
ω 4 m2
n
ω
ωn
1
mω 2 2
2m
ωn
+1−
mω 2
2m
ωn
1
ˆ1 = F
X
4
k (−r − 2r2 + 1 − r2 )
Hence the transfer function is
1
ˆ1 = F
X
4
k (1 − r − 3r2 )
2.11.4
Problem 4
Summary of method of solution: There are 2 ways to solve these problem. We will solve it using
both methods. The first method is using known standard solution for step input, the solution y (t) is
found for the period of 0 < t < ω3πn using zero initial conditions. Next, the solution y (t) and y 0 (t) is
evaluated again at t = ω3πn . These values are now used as the initial conditions for the solution for t > ω3πn .
The solution for t > ω3πn will have the same form, but the step input now is 200N instead of 100N.
The second method as follows: Let F (t) = 100h (t) + 100h t − ω3πn or F (t) = 100h (t) + 100h t˜
where t˜ = t− ω3πn , then assuming the transient solution to h (t) is s (t) then the solution to F (t) is
100s (t) + 100s t˜ . The second method is simplet than the first method.
Solution using first method:
The system is
my 00 (t) + ky (t) = F (t)
When F (t) is a fixed input, such as a step input of magnitude F then the response is given by
F
y00
F
y (t) = y0 −
cos ωn t +
sin ωn t +
k
ωn
k
Where in the above, y0 and y00 are the initial position and initial velocity. For 0 < t < 1.5Tn the
solution is
y (t) = −
=
F
F
cos ωn t +
k
k
F
(1 − cos ωn t)
k
182
CHAPTER 2. HW’S
Let F = Q1 = 100N, and since k = mωn2 then the above becomes
y (t) =
Q1
(1 − cos ωn t)
mωn2
Now we need first to evaluate y t =
3π
ωn
y 0 (t) =
and y 0 t =
Q1
sin ωn t
mωn
0<t<
3π
ωn
3π
ωn
. From the above
0<t<
3π
ωn
Hence
3π
Q1
Q1
Q1
2Q1
3π
y t=
=
=
(1 − cos 3π) =
(1 − (−1)) =
1 − cos ωn
2
2
2
ωn
mωn
ωn
mωn
mωn
mωn2
and
Q1
3π
Q1
3π
sin ωn
sin (3π) = 0
=
=
y t=
ωn
mωn
ωn
mωn
0
Now let t˜ = t −
3π
Hence
ωn
the solution for t˜ > 0 is
y 0 t˜ = 0
Q
2
˜ + Q2
˜+
y t˜ = 0 −
sin
ω
t
cos
ω
t
n
n
mωn2
ωn
k
2Q1
Q2
Q2
=
−
cos ωn t˜ +
2
2
mωn mωn
k
y t˜ =
Therefore, we have obtain the complete solution, which is
time
0<t<
t˜ = t −
solution
3π
ωn
3π
ωn
Q1
100
2
2 (1 − cos ωn t) =
mω
5(50)
n
2Q1
Q2
cos ωn t˜ + Qk2
2 − k
mωn
(1 − cos 50t) = 0.008 (1 − cos 50t)
200
200
= 2(100)
cos ωn t˜ + 5(50)
2 −
2
2 = 0.016
5(50)
5(50)
This is a plot of the solution. Then a numerical ODE solver is used to verify the result
2.11. HWA1
Now a numerical ODE solver was used to verify. Here is the result
183
184
CHAPTER 2. HW’S
We can see the solutions match very well.
Solution using second method:
Let F (t) = 100h (t) + 100h t − ω3πn then assuming the transient solution to h (t) is s (t) then the
solution to F (t) is 100s (t) + 100s t − ω3πn h t − ω3πn . From appendix B, the solution to h (t) is given
by
1
s (t) =
(1 − cos ωn t)
mωn2
3π
hence the solution to F (t) = 100h (t) + 100h t − ωn is
3π
3π
y (t) = 100s (t) + 100s t −
h t−
ωn
ωn
100
100
3π
3π
=
(1 − cos ωn t) +
1 − cos ωn t −
h t−
mωn2
mωn2
ωn
ωn
To verify, this is a plot of the above solution. We see it is the same as the first analytical solution,
and it is the same solution as the one using numerical ODE solver as well.
2.11. HWA1
2.11.5
185
Problem 5
The input can be written as F0 h (t) − F0 h (t − T ) + F0 e−β(t−T ) h (t − T ) or, by letting t˜ = t − T , the
input becomes
˜
F0 h (t) − F0 h t˜ + F0 e−β t h t˜
186
CHAPTER 2. HW’S
If the response to h (t) is s (t) and the response to e−β t˜ is s1 t˜ then the response to the above
becomes
F0 s (t) − F0 s t˜ h t˜ + F0 s1 t˜ h t˜
From appendix B, we see that
s (t) =
1
(1 − cos ωn t)
mωn2
and
s1
t˜ =
1
2
m (ωn + β 2 )
−β
−β t˜
˜
˜
e
− cos ωn t +
sin ωn t
h t˜
ωn
Therefore the the final response is
˜
y (t) = F0 h (t) − F0 h t˜ + F0 e−β t h t˜
1
1
= F0
(1 − cos ωn t) h (t) − F0
1 − cos ωn t˜ h
2
2
mωn
mωn
−β
1
−β t˜
F0
e
− cos ωn t˜ +
sin ωn t˜ h
m (ωn2 + β 2 )
ωn
t˜ +
t˜
1
1
(1 − cos ωn t) − F0
(1 − cos (ωn (t − T ))) h (t − T ) +
2
mωn
mωn2
1
−β
−β(t−T )
F0
e
− cos (ωn (t − T )) +
sin ωn (t − T )
h (t − T )
m (ωn2 + β 2 )
ωn
= F0
To plot this, we need to choose values for parameters. Let F0 = 100, ωn = 50rad/ sec, m = 5kg, β =
1, T = 1, then a plot of the above is below, followed by solution from numerical ODE solver.
Plot of the analytical solution
2.11. HWA1
To verify, this is the result from numerical ODE solver
187
188
CHAPTER 2. HW’S
We can see that the solutions agree.
Chapter 3
Design project
3.1
Design project, team 4
Donny Kuettel III Nasser M. Abbasi Paul Frisch
March 20, 2015
3.2
Introduction
This report outlines a simple passive vibration isolation system design for use in the first class cabin
of a Boeing 757-200 airplane with the goal of reducing the vibrations felt by the passengers in the first
class cabin. This was done by simulation in order to select suitable design parameters that produced an
acceptable absolute acceleration time history compared the rest of the airplane during a turbulent flight.
3.3
3.3.0.1
Discussion and results
Notations used in the report
M
mass of first class cabin
k
spring constant
ζ
critical damping constant
r
ratio of external load frequency to the natural frequency of first class cabin
rn
ratio of external load nth harmonic frequency to the natural frequency of first class cabin
Tr
Transmissibility. The ratio of cabin absolute displacement to base absolute displacement
ωnat
Natural frequency of first class cabin
ω1
Fundamental frequency of the external load frequency.
EOM
Equation Of Motion
c
damping constant for damper under first class cabin
Znacc
the complex amplitude of the term associated with the nth harmonic of the frequency z 00 (t)
Zndisp
the complex amplitude of the term associated with the nth harmonic of the displacement z (t)
Yn
the complex amplitude of the term associated with the nth harmonic of the displacement of y (t)
189
$
ωnatural
$n
ωnatural
190
CHAPTER 3. DESIGN PROJECT
Table 1. Description of mathematical notations used in report
3.3.0.2
Mathematical model
Reducing the vibration effect felt by the passengers in the first class cabin was based on reducing the
transmissibility ratio (Tr ) of the absolute acceleration of the airplane to that of the first class cabin. A
passive vibration isolation system was used for its ease of implementaion and its low cost. The model is
based on figure 1 below
Figure 1. Mechanical model view of vibration isolation system in place.
The absolute acceleration of the first class cabin, y 00 (t), was calculated with the vibration isolation
system in place and then compared to the absolute acceleration, z 00 (t), of the rest of the airplane. The
goal was to produce a smooth absolute acceleration time history when compared to the rest of the
airplane. This was done by adjusting M, ζ and K and running a simulation of the motion of the plane
with our vibration isolation system in place. A plot of Tr vs. r was also used to insure that the maximum
Tr remained small as the frequency ratio r was increased.
Assuming the mass of cabin is M, which includes the live load (passengers), then applying Newton’s
laws the the first class cabin results in the equation of motion
my 00 + c (y 0 − z 0 ) + k (y − z) = 0
my 00 + cy 0 + ky = cz 0 + kz
The transfer function between y (t) and z (t) in the frequency domain can now be derived (Appendix
contains complete derivation) resulting in
q
1 + (2ζrn )2
Yn T (r) = disp = q
Zn
(1 − rn2 )2 + (2ζrn )2
To compare the absolute acceleration of the first class cabin with the rest of the airplane, the absolute
acceleration, y 00 (t), is now found from Yn . Since y (t) = Re {Yn ei$n t } then y 00 (t) = Re {−$n2 Yn ei$n t }.
3.3.0.3
Design results
z 00 (t) (given) and y 00 (t) (computed) are now plotted on the same plot in order to compare the effect of
our vibration isolation system to the comfort of the first class passengers. The final design parameters
used are (Appendix 5.1)
3.3. DISCUSSION AND RESULTS
191
M Mass of first class cabin (dead+live)
3050 kg
ζ
0.7
k
9700 N/M
Table 3.1: Final values of design parameters
Figure 2 below shows the result using the above parameters
Figure 2. First class cabin absolute acceleration compared to rest of airplane.
We see from figure 2 that the absolute acceleration of the first class cabin has much less variation and
is much smoother than the absolute acceleration of the rest of the airplane. From this we can see that
the first class passengers experience a much more comfortable flight than the rest of the airplane. In
addition, the transmissibility plot was found to be acceptable since Tr decreases with increasing r
Figure 3. Transmissibility plot of first class cabin.
In addition to producing a smooth absolute acceleration time history, the goal was also to insure
that Tr decreased as r increased. This implies that at higher external acceleration relative to the
natural frequency, our vibration isolation system remained effective. The simulation program generated
a mechanical view showing the absolute position of the first class cabin, with an offset, and the absolute
position of the airplane during the flight as shown in figure 4 below.
192
Figure 4. Animation of vibration isolation during flight.
The force shown in Figure 4. below the airplane is the numerical value of −M z 00 where z 00 is the
absolute acceleration of the airplane and M is the total mass of the first class cabin.
3.3.1
vibration isolation system
The vibration dampening system proposed for the first class cabin is a simple spring dashpot system
that utilizes the additive properties of springs and dashpots to dampen the vibration of the first class
cabin in the Boeing 757-200.
Figure 5. Schematic diagram of vibration isolation system in place
The design of our passive vibration isolation system is simple and effective with a minimal costs. It
starts by defining the area that represents the first class cabin, which is at the front of the plane right
behind the cockpit.
3.3. DISCUSSION AND RESULTS
193
The cabin spans the entire inside width of the airplane body, which is 3.53 m (11.58 ft), and then
extends down the body of the plane roughly 3.35 m (11 ft) giving the first class cabin a total area of
3.53 × 3.35 m2 (11.58 × 11 ft2 ).
The next step in our design is to define the area that will actually be part of the vibration isolation
system. We cannot use the whole floor of the first class cabin because the rounded body of the airplane
would not allow the floor to travel up and down rendering our whole system ineffective. To solve this
problem we started at the center of the plane’s cross section and went out 1.524 m (5 ft) in either direction
giving a total area of the platform used in our vibration isolation system 3.048 × 3.35 m2 (10 × 11 ft2 ) as
seen above in figure 5.
To begin the actual design, additional support must be given to the aluminum floor of the cabin. The
use of 6061 T6 Aluminum I-beams (specifications are given in appendix 5) spanning the width of the
platform provides the needed support. In addition the I-beams provide a sturdy surface for the spring
and dashpot system to contact the cabin floor.
The key component of the vibration dampening system is the use of carbon fiber leaf springs. We
chose carbon fiber leaf springs in place of steel for several reasons. They provide a softer ride at a lower
noise level and excellent stability due to better damping characteristics than steel. Placed in series, the
use of 5 carbon fiber leaf springs provides the spring constant required (9700 N/m) and a low increase in
weight.
The dashpots needed for our design, 2K325 Dashpots, can be purchased from many manufactures.
When added in parallel they provided the necessary damping coefficient of 7425 N*s/m needed when the
first class cabin is full and 5800 N/s*m when the cabin is empty.
Our design for this passive vibration isolation system works whether the first class cabin is full, empty,
or half way in-between. The system works best when the cabin is fully loaded with passengers, and has
almost identical results with no passengers on board. Even though the results are slightly diminished
with fewer passengers, the system still creates a noticeably smoother flight.
3.3.2
vibration isolation system Cost estimate
The total cost of our vibration isolation system is around $16500 (appendix 5). The cost of the aluminum
support beams, dashpots and carbon fiber leaf springs make up the majority of the material cost totaling
only about $3000. The majority of the total cost comes from the additional weight of the system and the
resulting price of fuel used during the planes lifetime. The additional weight results in an expected cost
of about $13500 over the lifetime of the plane.
The damping effects of the system could be improved if weight were added to the cabin. However the
additional cost of the added weight over the lifetime of the plane would outweigh the benefits for the
passengers. If however some heavy components of the plane were to be attached to the first class cabin,
the system could be redesigned for an even better ride. This would require further investigation into the
balance of the plane, flight dynamics and a deeper knowledge of the various components of the plane so
it falls out of the scope of this project.
194
3.4
3.4.1
Appendix
Design values
weight:
This table shows the design values based on weight
Description of item
Mass (kg)
15 Chairs @ 100 kg/chair
1500
10 ft by 11 ft aluminum flooring
200
5 Aluminum I-beams @ 20 kg/beam
100
5 Carbon Fiber leaf springs @ 5 kg/spring
25
Miscellaneous weight
25
15 Passengers @ 80 kg
1200
Weight of First Class Cabin before vibration isolation system
1700
Weight of First Class Cabin after vibration isolation system
1850
Weight of First Class Cabin with maximum passengers
3050
Table 3.2: Mass of items used in design calculations
The total mass M has the value of 3050 kg. For our ζ value we choose the value 0.7 as it worked well
in simulations to provide a smooth ride for the passengers while still keeping Tr small.
Leaf springs and spring K value:
The most important aspect of picking a k value is the total allowed clearance the first class cabin
floor has to move. The first class cabin’s floor has a displacement relative to the body of the aircraft and
if that gets too large the floor will make contact with the body of the airplane. The lower the k value
we choose, the larger the displacement of the first class cabin relative to the body of the airplane will
become. The maximum travel distance of the first class cabin is around 20 cm (7.87 in) and we can use
this value to pick an appropriate k value. A k value around 10000 N/m keeps the first class cabin floor
within this tolerance. The following plot shows the absolute acceleration of the cabin vs. the rest of the
airplane during the turbulent flight 1 .
1
absolute position of the first class cabin was computed from the absolute acceleration of the cabin in the frequency
domain. Hence the average value was not used due to the division by zero problem with this method. We do not have
another method to find absolute position from absolute acceleration (unless we use more advanced numerical integration
method in time domain, which is beyond the scope of this course)
3.4. APPENDIX
195
Figure 6. absolute acceleration of first class cabin compared to rest of airplane
To keep the weight of our vibration isolation system as small as possible we opted to use carbon fiber
leaf springs. The k value of any leaf spring system can be calculated by the equation
k=
8Enbt3
3L3
Figure 7. Leaf spring design used in vibration isolation system
Since our springs are in parallel, the k values add to give a total equivalent k. We took the k value
selected (10000 N/m) and divided it by 5 giving us an individual k value of 2000 N/m. Using the
following dimensions for the leaf spring resulted in a k value of 1940 N/m for a total k value of 9700 N/m.
196
• E = 17 Gpa
• n=3
• L = 3.048 meter (10 ft)
• b = 0.1016 meter (4 in)
• t = 0.015875 meter (5/8 in)
3.4.2
Cost values
After finding the materials we needed, the following describes how we calculated the total cost of our
vibration isolation system.
• 5 @ 10 ft 6061 T6 Aluminum I-beams @ $180/beam results in $900.
Width 6 in, Flange 4 in, Web 0.19 in, Thickness 0.28 in.
• 5 Carbon Fiber leaf Springs @ $300/spring results in $1500.
• 150 kg of extra weight, total weight of the airplane is 59350 kg.
Fuel costs for this aircraft was estimated to be $3, 500/hour and a typical aircraft operates 3000
hours per year. An increase of 1% in the weight of the aircraft is expected to increase fuel costs by
0.5%
150 kg
× $3500 × 3000 = $13270
2∗59350 kg
• 5 2k325 Dashpots @ $100/dashpot = $500
Needed c is around 2000 N*s/m. These dashpots have an adjustable c from 0 to 7000 N*s/m
• Total cost estimate $16500
3.4.3
Simulation program description
The simulation program was a GUI program written in Matlab version 2013a, which made it easier to
determine the parameters to use for the design. The following is a screen shot of the program. The
program can be downloaded from the project web site
3.4. APPENDIX
197
Figure 6. Simulation Matlab program used for obtaining the design parameters.
The first step is to load the Matlab .mat file which contains the acceleration time history. Then one
can use the sliders to adjust the system parameters and see the effect on the absolute acceleration of
the first class cabin. Computation was done in the FFT domain using the functions f f t easy() and
if f t easy() in the class web site. The absolute displacement was found from the absolute acceleration in
the frequency domain. Due to the problem of division by zero for the first component in the frequency
vector, this was set to zero before using if f t easy().
3.4.4
Derivation of the transfer function
Assuming the mass of cabinet is M which includes passengers weight, by applying Newton’s laws the
EOM for the first class cabin is
my 00 + c (y 0 − z 0 ) + k (y − z) = 0
my 00 + cy 0 + ky = cz 0 + kz
(3.1)
The time history of the turbulent acceleration z 00 (t) was given to us in the matlab mat file. Therefore
in the frequency domain, and assuming the time history represents one period we can write
z 00 = Re Znacc ei(ω1 n)t
Substituting back into Eq 3.1 and simplifying, the magnitude of the absolute displacement of the first
class cabin relative to absolute displacement of airplane is found to be
q
1 + (2ζrn )2
Yn = q
Z disp n
(1 − rn2 )2 + (2ζrn )2
198
Where Znacc is the complex amplitude of the nth harmonic component in the acceleration data. Letting
ω1 n ≡ $n then in the frequency domain Eq 3.1 becomes
Re
−m$n2
Znacc
Znacc
i$n t
= Re
c
+k
e
i$n
−$n2
!
c
− $k2
i$n
n
Znacc
Yn =
−m$n2 + i$n c + k
i$n t
+ i$n c + k Yn e
=−
Where
rn =
Znacc
1 + i2ζrn
2
$n (1 − rn2 ) + 2iζrn
$n
ωnatural
acc
But − Z$n2 is the absolute displacement of the airplane, say Zndisp , hence the transfer function between
n
the absolute displacement of first class cabin and the absolute displacement of the airplane is
Yn =
1 + i2ζrn
Z disp
(1 − rn2 ) + 2iζrn n
The magnitude of the absolute displacement of first class cabinet relative to absolute displacement of
the airplane is
q
1 + (2ζrn )2
Yn = q
Z disp n
(1 − rn2 )2 + (2ζrn )2
3.4.5
References
1. Aluminum data www.onlinemetals.com
2. Airpot Dashpot Performance Specifications. N.p., n.d. Web. 15 Apr. 2013. http://www.airpot.
com/html/dashpot.html
3. Boeing Commercial Airplanes. 757 Program. n.d. Web. 5 Apr. 2013. http://www.boeing.com/
boeing/commercial/757family/index.page
4. Engineering ToolBox. Young’s Modulus. Fabrication Extrusion Company, n.d. Web. 9 Apr. 2013.
5. Ginsberg, Jerry H. Mechanical and structural vibrations: theory and applications. New York:
Wiley, 2001.
6. Online Metal Store Metal Product Guides at OnlineMetals.com. Metal Product Guides at OnlineMetals.com. N.p., n.d. Web. 15 Apr. 2013. http://www.onlinemetals.com/merchant.cfm?
id=980
7. 7575-200 Airliner flugzeuginfo.net-the aircraft encyclopedia. N.p., n.d. Web. 10 Apr. 2013
http://www.flugzeuginfo.net/acdata_php/acdata_7572_en.php
3.4. APPENDIX
3.4.6
199
software
The following zip file contains the current version of Matlab software to use to design the vibration
isolation system.
This program allows one to easily select k and ζ for a given mass so that the absolute acceleration of
the passengers (mass) when subjected to turbulent acceleration is as close as possible to the nominal
acceleration.
To use, download the above zip file, unzip it and run the m file.
Note that fft_easy.m and ifft_easy.m are needed but they are included inside the m file.
200
3.5
team 3 peer review
by Nasser M. Abbasi, from team 4
Team 3 report was well written and included
all relevant parts of the design, from the equation
of motion to final Matlab plots showing how the
absolute acceleration felt by passengers are reduced
compared to the rest of the airplane.
Report included a table that shows comfort levels felt for different values of absolute acceleration.
There was also a detailed section on spring design
specification which was useful to read.
The cost analysis was detailed and through.
Team 3 seems to have covered and discussed all
important aspects of the design.
1. How would you rate the professionalism of the
presentation? A
2. How would you rate the technical content? A
• Is it correct and clearly explained?
Yes. The report included the correct
equation of motion for vibration isolation and showed the numerical values for
the parameters used for the design.
• Was an additional figure or schematic
needed to understand their design?
Report included needed Matlab plots and
spring/mass/damper diagram. It would
be useful to include the value of critical
damping ζ used and not just the damping coefficient c.
• Were the results well summarized within
the body of the report? When additional
detail was needed, was it easy to find in
the Appendix?
Yes. Report included table that summarized the comfort level for different
acceleration values. (figure 3). Appendix
included the Matlab code and additional
useful plots.
3. How would you rate the team’s discussion of
how their design will perform? A
• Is it clear that the design team understands what they have done and has made
sure that their results are correct and
make sense?
The team talked about level of comfort
felt by passengers. They explained that
the FFT method was used for obtaining
the solution shown.
• Are there any situations that they have
not anticipated that might compromise
their design?
The design was based on equation of motion for a single seat and not for the
whole first class cabin as the object. If using the whole first class cabin as the mass,
then the same method will work, but the
mass will increase depending on number
of seats. It might be useful to consider
applying the design method given in the
report to the whole first class cabin and
not for individual seats.
• Is their cost/weight analysis reasonable
and clearly presented?
The report included detailed section of
cost of material and types of springs
needed in addition to the mention of the
material to be used.
Chapter 4
Exams
Exam
link
grade
first mid
term
TBD
86%
second
midterm
TBD
88%
finals
Hard and long
80%
about
Table 4.1: Exams table
201
202
4.1
CHAPTER 4. EXAMS
Practice exams
1. practice exam 1 PDF
2. first practice exam for finals (has key) questions: PDF key solution PDF
3. extra one problem practice question: PDF
4.2. PRACTICE EXAM 2
4.2
203
practice exam 2
practice exam 2 . checking validity of K matrix, derive EOM for 2 DOF system, bar, spring, damper full
modal analysis, find solution due to impulse.
problems PDF
4.2.1
Problem 1
Taking x as positive as shown, and y as positive as shown, then the middle spring is in compression
with change of length ∆ = (x + Lθ) and the right most spring is in tension with change of length ∆ = x,
hence
1
1
1
Vspring = ky 2 + k (x + Lθ)2 + kx2
2
2
2
1
1 2 1
= ky + k x2 + L2 θ2 + 2xLθ + kx2
2 2
2
1
1
= θ2
kL2 + x2 (k) + y 2
k + xθ (kL)
2
2
Compare to quadratic form
1
1
1
Vspring = K11 θ2 + K22 x2 + K33 y 2 + K12 xθ + K13 θy + K23 xy
2
2
2
Then
K11
K22
K33
K12
K13
K23
= kL2
=k
=k
= kL
=0
=0
204
CHAPTER 4. EXAMS
Hence the K matrix due to stiffness is

 
kL2 kL 0
θ

 
 kL k 0 x

 
0
0 k
y
Therefore, K12 had the wrong units.
This reason is as follows: result of multiplying the first row of
 
θ
 

the Kspring matrix with the column x
 should have units of torque. Therefore the units should be
y
f orce × meter and hence K12 x should come out as N m units. But as given in the problem, it has units
N only, ie. units of force. But now, the units will come out to be N
m.
θ
 

Similarly, the second row of the K matrix when multiplied by 
x should have units of force only
y
(not torque). We can see this this is the case with this correction. So the sign was correct, but the units
did not match before.
4.2.2
Problem 2
4.2.2.1
Part a
This is a 2 degrees of freedom system. The first generalized coordinate is taken as α which the angle of
rotation of the top bar around joint A. The second degree of freedom is taken as x which is the sliding
distance that mass m2 moves as it slides over the lower bar
205
Static equilibrium is at α = 0 and x = 0.
We start by finding the kinetic energy. Since bar m1 is fixed at one point to inertial space, then only
its rotational kinetic energy is added to the system kinetic energy
1 1
1
2
2
2
m1 L (α0 ) + m2 (x0 )
T =
2 12
2
Now we find the potential energy, assuming springs remain straight. Spring k1 will extend by amount
∆1 =
L
α
2
and spring k2 will extend by amount
∆2 = Lα − x sin θ2
Hence potential energy of the system is
1
L
1
V = k1 (∆1 )2 + k2 (∆2 )2 + m1 g sin α + m2 gx sin θ2
2
2
2
Therefore the Lagrangian Φ is
Φ=T −V
1 1
2
2
=
m1 L (α0 ) +
2 12
1 1
2
2
=
m1 L (α0 ) +
2 12
=
1
2
m2 (x0 ) −
2
1
2
m2 (x0 ) −
2
1
1
L
2
2
k1 (∆1 ) + k2 (∆2 ) + m1 g sin α + m2 gx sin θ2
2
2
2
!
2
1
L
1
L
2
k1
α + k2 (Lα − x sin θ2 ) + m1 g sin α + m2 gx sin θ2
2
2
2
2
1
L
1
L2
1
2
2
m1 L2 (α0 ) + m2 (x0 ) − k1 α2 − k2 L2 α2 + x2 sin2 θ2 − 2Lαx sin θ2 − m1 g sin α − m2 gx sin θ2
24
2
8
2
2
EOM for x is
d
dt
∂Φ
∂x0
−
∂Φ
= Qx
∂x
206
CHAPTER 4. EXAMS
where Qx is the generalized for for the x coordinate. To find Qx we make virtual displacement
δx while fixing all other coordinates and obtain virtual work done by non-conservative forces. Only
non-conservative force acting on m2 is the friction force f = c2 v where v is the speed of the mass m2 . The
speed of the mass m2 is the vertical direction is v = x0 sin θ2 , hence the non-conservative force acting on
m2 is c2 (x˙ sin θ2 ) and is acting in negative direction. Hence taking projection of this force along x gives
δW = −c2 (x0 sin θ2 ) sin θ2 δx
Therefore
Qx = −c2 x0 sin2 θ2
Hence
d
dt
∂Φ
= −c2 x0 sin2 θ2
∂x
d
(m2 x0 ) − −k2 x sin2 θ2 + 2k2 Lα sin θ2 − m2 g sin θ2 = −c2 x0 sin2 θ2
dt
m2 x00 + c2 x0 sin2 θ2 + k2 x sin2 θ2 − 2k2 Lα sin θ2 = −m2 g sin θ2
EOM for α is
d
dt
∂Φ
∂α0
−
∂Φ
∂x0
−
∂Φ
= Qα
∂α
where Qα is the generalized for for the α coordinate. To find Qα we make virtual displacement δα
while fixing all other coordinates and obtain virtual work done by non-conservative forces. We see that
the work is
L
δW = −c (Lα0 ) δα + (F sin θ1 ) Lδα
2
cL2 0
= F L sin θ1 −
α δα
2
Hence
Qα = F L sin θ1 −
cL2 0
α
2
Therefore
∂Φ
∂Φ
cL2 0
−
=
F
L
sin
θ
−
α
1
∂α0
∂α
2
d
1
L2
L
cL2 0
2 0
2
m1 L α − −k1 α − k2 L α + 2Lx sin θ2 − m1 g cos α = F L sin θ1 −
α
dt 12
4
2
2
1
L2
L
cL2 0
m1 L2 α00 + k1 α + k2 L2 α − 2k2 Lx sin θ2 + m1 g cos α = F L sin θ1 −
α
12
4
2
2
1
cL2 0
L2
L
2 00
m1 L α +
α + k1
+ k2 L2 α − 2k2 Lx sin θ2 = F L sin θ1 − m1 g cos α
12
2
4
2
d
dt
Hence the 2 EOM are
cL2 0
L2
L
1
2 00
2
m1 L α +
α + k1
+ k2 L α − 2k2 Lx sin θ2 = F L sin θ1 − m1 g cos α
12
2
4
2
207
Linearize around static equilibrium, α = 0, x = 0 then we obtain
1
L
cL2 0
L2
2 00
2
m1 L α +
α + k1
+ k2 L α − 2k2 Lx sin θ2 = F L sin θ1 − m1 g
12
2
4
2
In Matrix form
!
!
1
2
00
m
L
0
α
12 1
+
0
m2
x00
cL2
2
0
0
c2 sin2 θ2
!
α0
x0
!
+
2
k1 L4 + k2 L2 −2k2 L sin θ2
−2k2 L sin θ2
!
k2
α
x
!
=
F L sin θ1 − m1 g L2
−m2 g sin θ2
I think the weight contributions should be zero. So I need to look more into this, but I think the
OEM should be as follows
!
!
!
!
!
!
!
1
cL2
L2
2
00
0
2
m
L
0
α
0
α
+
k
L
−2k
L
sin
θ
α
F
L
sin
θ
k
2
2
2
1
1 4
12 1
2
=
+
+
2
00
0
0
m2
x
0 c2 sin θ2
x
−2k2 L sin θ2
k2
x
0
4.2.2.2
Part b
Checking the Damping matrix units. First row of C
α0
x0
!
should give units of torque. looking at
cL2 0
α . viscous damping coefficient c has
2
T
N L (L)2 T1 = N L, in other words, a torque.
2
units of N TL , hence the units of the expression cL2 α0 are
(in here, L stands for length units, T stands for time units
and N stands for force units). Now to verify the second row of C. We see it is c2 sin2 θ2 x0 which has
units of force (given in the problem). Since the second must have !
units of force, this is verified.
α
Now checking the stiffness matrix units. First row of K
should have units of torque. But
x
2
L
2
k1 4 + k2 L α has units of torque since k has units of force per unit length. and 2k2 L sin θ2 x has units
of torque also (note α has no units as it is an angle).
For the second row of K, it should have units of force, which it does, since k2 x has units of force and
−2k2 L sin θ2 α has units of force. Hence verified.
Check signs on the x EOM:
m2 x00 + c2 x0 sin2 θ2 + k2 x sin2 θ2 − 2k2 Lα sin θ2 = 0
m2 x00 + c2 x0 sin2 θ2 + k2 x sin2 θ2 = 2k2 Lα sin θ2
x00 > 0, x0 > 0, x > 0 then α > 0,checks OK, since when x > 0 then the top bar will be rotating in
the positive direction and α > 0, i.e. the top bar will be above the horizontal.
Check signs on the α EOM:
1
cL2 0
L2
2 00
2
m1 L α +
α + k1
+ k2 L α − 2k2 Lx sin θ2 = F L sin θ1
12
2
4
1
cL2 0
L2
2 00
2
m1 L α +
α + k1
+ k2 L α = F L sin θ1 + 2k2 Lx sin θ2
12
2
4
α00 > 0, α0 > 0, α > 0 then x > 0,checks OK, since when α > 0 then the top bar will be rotating in
the positive direction and x > 0, means the lower mass m2 is moving upwards.
!
208
CHAPTER 4. EXAMS
4.2.3
Problem 3
We solve this in modal coordinates so to de-couple the EOM’s. First find the 2 natural frequencies
!
!
1
−1
1
0
− ω2m
=0
k
−1 1
0 2 !
!
1 −1
1 0 2m
−ω
=0
−1 1
k 0 2 Let ω 2 m
= η 2 then
k
!
!
1 −1
1
0
− η2
=0
−1 1
0 2 !
1 − η2
−1
=0
−1
1 − 2η 1 − η 2 1 − 2η 2 − 1 = 0
Hence taking positive roots η = 1.2247, η = 0 . When η = 0
!( )
ϕ11
1 − η2
−1
−1
1 − 2η 2
1
ϕ12
!( )
−1
1
−1
Hence 1 − ϕ12 = 0 or ϕ12 = 1, therefore ϕ1 =
1
ϕ12
( )
1
1
=
=
( )
0
0
( )
0
0
209
When η = 1.2247
1 − η2
−1
−1
1 − 2η 2
!( )
ϕ12
ϕ22
!( )
−0.5 −1
1
−1
−2
ϕ22
(
Hence −0.5 − ϕ22 = 0 or ϕ22 = −0.5, therefore ϕ2 =
=
=
1
( )
0
0
( )
0
0
)
. Now do mass normalization
−0.5
µ1 = {ϕ}T1 [M ] {ϕ}1
( )T
!( )
1
1 0
1
=
1
0 2
1
=3
and
µ2 = {ϕ}T2 [M ] {ϕ}2
(
)T
!(
)
1
1 0
1
=
−0.5
0 2
−0.5
= 1.5
Hence
( )
1
(
)
1
0.57735
{ϕ}1
{Φ}1 = √ = √ =
µ1
3
0.57735
(
)
1
(
)
−0.5
0.81650
{ϕ}2
=
{Φ}2 = √ = √
µ2
1.5
−0.40825
Hence
[Φ] =
0.57735
0.81650
!
0.57735 −0.40825
Then the modal EOM are
1
0
1
0
[Φ]T [M ] [Φ] + [Φ]T [K] [Φ] = [Φ]T {F }
!( )
!( )
!(
)
0
η¨1
η12 0
η1
0.57735 0.57735
0
+
=
1
η¨2
0 η22
η2
0.81650 −0.40825
F0 δ (t)
!( )
!( ) (
)
0
η¨1
0 0
η1
0.57735F0 δ (t)
+
=
1
η¨2
0 1.5
η2
−0.40825F0 δ (t)
210
CHAPTER 4. EXAMS
For the first mass, EOM is
η¨1 = 0.57735F0 δ (t)
Z t
η˙ 1 =
0.57735F0 δ (t) dt + C1
0
1
= 0.57735F0 h (t) −
+ C1
2
Z t
1
+ C1 dt + C2
η1 (t) =
0.57735F0 h (t) −
2
0
1
= 0.57735F0 t h (t) −
+ tC1 + C2
2
(
) ( )
(
) ( )
(
)
x1 (0)
0
x01 (0)
0
η1 (0)
Now initial conditions are zero since
=
and also
=
then
=
x2 (0)
0
x02 (0)
0
η2 (0)
( )
(
) ( )
η˙ 1 (0)
0
0
and also
=
0
η˙ 2 (0)
0
Initial conditions η1 (0) = 0 implies
C2 = 0
while and η˙ 1 (0) = 0 implies
C1 = −0.57735F0
1
h (t) −
2
Hence the solution is
1
η1 (t) = 0.57735F0 t h (t) −
+ tC1 + C2
2
1
1
− 0.57735F0 h (t) −
= 0.57735F0 t h (t) −
2
2
1
= 0.57735F0 h (t) −
(t − 1)
2
Now the second EOM is solved.
η¨2 + 1.5η2 = −0.40825F0 δ (t)
Which has solution (using appendix B) and using M = 1 and ωD = ωn =
hence
−0.40825F0
η2 (t) =
sin (1.2247t)
1.2247
Now to obtain the solution in normal coordinates
(
)
(
)
x1 (t)
η1 (t)
= [Φ]
x2 (t)
η2 (t)
√
1.5 = 1.2247 since ζ = 0,
Then
(
)
x1 (t)
x2 (t)
=
0.57735
0.81650
!(
)
0.57735F0 h (t) − 12 (t − 1)
0.57735 −0.40825
−0.40825F0
1.2247
sin (1.2247t)
211
So
x1 (t) = 0.57735 0.57735F0 h (t) −
x2 (t) = 0.57735 0.57735F0 h (t) −
1
0.40825F0
(t − 1) − 0.81650
sin (1.225t)
2
1.2247
1
0.40825F0
(t − 1) + 0.40825
sin (1.225t)
2
1.2247
For example, if F0 = 1 then
x1 (t) = 0.57735 0.57735 h (t) −
x2 (t) = 0.57735 0.57735 h (t) −
0.40825
1
sin (1.2247t)
(t − 1) − 0.81650
2
1.2247
1
0.40825
sin (1.2247t)
(t − 1) + 0.40825
2
1.2247
Here is a plot of the solution x1 (t) and x2 (t) . The 2 masses move to the right after the impulse,
while in sinusoidal motion at the same frequency, but different amplitudes.
212
CHAPTER 4. EXAMS
4.3. FINALS 2ND PRACTICE EXAM
4.3
213
finals 2nd practice exam
no key. Question PDF
4.3.1
Problem 1
This is a 2 D.O.F. system. The degrees of freedom are θ1 and θ2 shown above in the positive sense.
The method of power balance is used to obtain the EOM.
2
2
The system kinetic energy is T = 12 m1 L3 (θ10 )2 + 12 m1 L3 (θ20 )2 , hence by comparing term to the
quadratic form, the mass matrix part of the EOM is obtained
"
#( )
θ100
L2 m1 0
3
0 m2
θ200
To find spring stiffness, the spring deformation is found using stiff spring approximation.
∆0 = (VB − VA ) · eB/A
= (Lθ20 i−Lθ10 j) · (cos βi − sin βj)
Where eB/A is unit vector oriented to B from A and tan β = Lh . The above becomes
∆0 = Lθ20 cos β+Lθ10 sin β
Hence, integrating, squaring and collecting terms gives
∆ = Lθ2 cos β+Lθ1 sin β
∆2 = L2 θ22 cos2 β + L2 θ12 sin2 β + 2L2 θ1 θ2 sin β cos β
= θ12 L2 sin2 β + θ22 L2 cos2 β + θ1 θ2 2L2 sin β cos β
214
CHAPTER 4. EXAMS
Using the quadratic form of the power balance method, the spring stiffness matrix part of the EOM
is found from Vspring = 12 k (∆2 ) and by comparing quadratic terms, which leads to
"
Vspring = kL
2
sin2 β
2 sin β cos β
2 sin β cos β
cos2 β
#( )
θ1
θ2
But sin β cos β = 12 (sin 2β) hence
"
Vspring = kL2
sin2 β sin 2β
#( )
θ1
sin 2β cos2 β
θ2
Stiffness due to gravity Vg is now found. Let datum for zero potential energy be at the horizontal
level of the top bar, hence Vg = m1 g L2 sin θ1 − m2 g L2 cos θ2 . Since the derivatives are evaluated at static
equilibrium θ1 = 0 and θ2 = 0, the only term that remains is m2 g L2 which is now added to the k22 term
of the stiffness matrix. F L is the generalized force for θ2 since work done by F in making virtual δθ2 is
F Lδθ2 . Therefore, the EOM becomes
"
#( )
#( ) ( )
"
2
θ100
θ1
sin
β
sin
2β
0
L2 m1 0
+ kL2
=
L
00
2
3
0 m2
θ2
sin 2β cos β + m2 g
θ2
FL
2
To check units of the above EOM, looking at the first EOM from above
L2
m1 θ100 + kL2 sin2 β θ1 + kL2 (sin 2β) θ2 = 0
3
2
Let θ1 = 0. Hence L3 m1 θ100 = −kL2 (sin 2β) θ2 . Assume θ2 ≥ 0 and the system is now released to move.
We should expect the top bar to accelerate down (negative), since the spring is stretched. Looking at the
above, we see that θ100 ≤ 0. hence this is correct. 2
Now let θ2 = 0. Hence L3 m1 θ100 = −kL2 sin2 β θ1 . Assume θ1 ≥ 0 and the system is now released
to move. We should expect the top bar to accelerate down (negative) since the spring was stretched.
Looking at the above, we see that θ100 ≤ 0. This is correct.
Checking the second EOM
L2
L
m2 θ200 + kL2 (sin 2β) θ1 + kL2 cos2 β θ2 = F L − m2 g θ2
3
2
Let θ1 = 0 and F = 0 then
L2
L
m2 θ200 = −m2 g θ2 − L2 cos2 β θ2
3
2
Assume θ2 ≥ 0 and the system is now released to move. We would expect the right bar to accelerate
back (negative) when released to move. From the equation we see that θ200 ≤ 0. This is correct.
Now let θ2 = 0 and F = 0 then
L2
m2 θ200 = −kL2 (sin 2β) θ1
3
Assume θ1 ≤ 0 and the system is now released to move. We would expect the bar to accelerate to the
right (positive) since the spring was compressed. From the equation we see that θ200 > 0. This is correct.
4.3.2
Problem 2
4.3.2.1
part(a)
215
det [k] − ω 2 [m] = 0
"
#
"
#!
k −k
2m
0
det
− ω2
=0
0 m
−k k
"
#
"
#!
1 −1
2
0
m
det
− ω2
=0
k 0 1
−1 1
For normalization, let t0 = ωt then
dt0
dt
= ω and using t0 instead of t as the independent variable the
216
CHAPTER 4. EXAMS
above becomes
"
det
1
#
−1
−1
1
det
q
The roots are ω = 0 and ω =
( )
q
1
ϕ1 =
. When ω = 32 then
1
"
3
.
2
"
− ω2
"
1 − 2ω 2
2 0
#!
=0
0 1
#!
−1
=0
−1
1 − ω2
1 − 2ω 2 1 − ω 2 − 1 = 0
When ω = 0 it is a rigid body motion, So any ϕ will do. Let
1
#
−1
−1
1
"
#! ( ) ( )
ϕ12
0
3 2 0
−
=
2 0 1
ϕ22
0
"
#( ) ( )
−2 −1
ϕ12
0
=
1
−1 − 2
ϕ22
0
(
let ϕ12 = 1 then −2 − ϕ22 = 0 or ϕ22 = −2 hence ϕ2 =
µ1 = ϕT1 [M ] ϕ1 =
µ2 = ϕT2 [M ] ϕ2 =
1
)
−2
.
( )T "
#( )
1
2 0
1
=3
1
0 1
1
#( )
( )T "
1
2 0
1
−2
−2
0 1
( )
1
(
)
0.57735
=6
Hence
ϕ1
1
Φ1 = √ = √
µ1
3
=
1
0.57735
( ) (
)
1
0.40825
ϕ2
1
Φ2 = √ = √
=
µ2
6 −2
−0.81650
4.3.2.2
part(b)
"
Φ=
√1
3
√1
3
√1
6
√
− 26
#
The EOM is
"
1 0
0 1
#( )
η100
η200
"
+
0 0
0
3
2
#( )
η1
η2
(
= ΦT
0
)
−f (t)
"
=
√1
3
√1
3
√1
6
√
− 26
#T (
0
)
−f (t)
( √
)
− 13 3f (t)
=
√
1
6f (t)
3
initial conditions are
(
)
η1 (0)
η2 (0)
)
(
η10 (0)
η20
(0)
=
( )
0
= ΦT [M ]
0
217
and
( )
v0
v0
"
=
√1
3
1
√
3
√1
6
− √26
#T "
#( )
2 0
v0
0 1
v0
(√
=
3v0
)
0
Therefore, the first ODE is
η100 = −
with IC η1 (0) = 0 and η10 (0) =
1√
3F0 (h (t) − h (t − T ))
3
√
3v0 . The second ODE is
3
1√
η200 + η2 =
6F0 (h (t) − h (t − T ))
2
3
with IC η1 (0) = 0 and η10 (0) = 0
4.3.2.3
part(c)
x1 (t) = Φ11 η1 (t) + Φ12 η2 (t)
1
1
= √ η1 (t) + √ η2 (t)
3
6
Therefore, x1 (t) solution has contribution from η1 (t) and η2 (t). But η1 (t) is linear with positive
slope of ν0 and η2 (t) is a sinusoidal, with no damping. So adding both together, here is a sketch of
possible solution
218
4.3.3
CHAPTER 4. EXAMS
Problem 3
4.3.3.1
219
part(a)
"
ΦT [K] ΦT =
"
0.85
1.1
#T "
0.65 −0.5
#"
−100 0.85
100
−100
200
1.1
#
0.65 −0.5
"
#
46.25 −0.5
−0.5 281.0
"
=
"
=
ω12
0
0
ω22
ω12
0
0
ω22
ω12
0
0
ω22
#
#
#
Hence ω12 = 46.25 or ω1 = 6.8 rad/sec
4.3.3.2
part(b)
Using the first natural frequency, since this has the longest time constant τ =
number of periods using logarithmic decrement method
1
ln
N
y1
yN
1
and
ζ1 ω1
solving for the
= 2πζ1
(1)
ζ1 is not known but can be found by evaluating ΦT [C] ΦT
"
ΦT [K] ΦT =
0.85
1.1
#T "
0.65 −0.5
#"
0.85
0.04
0
0
0.05
1.1
#
0.65 −0.5
"
=
0.05
#
0.021
0.021 0.061
0.05
and assuming small damping approximation, then 2ζ1 ω1 = 0.05. Hence ζ1 = 0.05
= 2(6.8)
= 0.0038.
2ω1
Now that the critical damping ratio for the first mode is found, we can use the method of logarithmic
decrement to find how many periods it takes to attenuate by 99%
1
Let yyN1 = 0.01
= 100 then Eq (1) becomes
1
ln (100) = 2π (0.0038)
N
(4.605)
N=
= 192.87
2π (0.0038)
= 193
Where N is the number or periods needed. But T =
t = N T = 192T = 192
2π
,hence
ω1
the time needed is
2π
2π
= 192
= 177.41 sec
ω1
6.8
So it takes 178 seconds for the first modal (decoupled) solution to attenuate in amplitude by 99%.
Since this is the dominant time constant, we expect the physical solution to attenuate in approximately
the same amount of time as well.
220
CHAPTER 4. EXAMS
4.3.3.3
part(c)
The EOM is, in modal coordinates
(
)
#( )
"
# ( ) "
"
#( )
i$t
2
0
Re
(Ae
)
η
ω
0
0.04
0
η
1 0
η100
1
= ΦT
Φ 10 + 1
+ ΦT
2
00
0
η2
0 ω2
0 0.05
η2
0 1
η2
But
"
ΦT
#
0.04
0
0
0.05
Φ=
"
0.85
1.1
#T "
0.65 −0.5
#"
0.85
0.04
0
0
0.05
1.1
#
0.65 −0.5
"
=
0.05
#
0.021
0.021 0.061
Hence EOM in modal coordinates become
"
#( ) "
#( ) "
#( ) "
#T (
)
6.82
0
1 0
η100
0.05 0.021
η10
η1
0.85 1.1
Re (Aei$t )
+
+
=
0 16.92
η2
0.65 −0.5
0
0 1
0.021 0.061
η20
η200
and using small damping approximation
"
#( ) "
#( ) "
#( ) (
)
1 0
η100
0.05
0
η10
6.82
0
η1
0.85 Re (Aeit$ )
+
+
=
0 1
η200
0 0.061
η20
0 16.92
η2
1.1 Re (Aeit$ )
Hence the 2 EOM’s are
η100 + 0.05η10 + 46.24η1 = Re 0.85Aei$t
η200 + 0.061η20 + 285.61η2 = Re 1.1Aei$t
Let η1 = Re (X1 eit$ ) then X1 =
then
0.85A
and
−$2 +i0.05$+46.24
η2 = Re (X2 eit$ )then X2 =
1.1A
−$2 +i0.0609$+285. 61
x = Φ1 η1 + Φ2 η2
(
) (
)
(
)
x1 (t)
0.85
1.1
=
Re X1 ei$t +
Re X2 ei$t
x2 (t)
0.65
−0.5
Hence
x1 (t) = 0.85 Re X1 ei$t + 1.1 Re X2 ei$t
x2 (t) = 0.65 Re X1 ei$t − 0.5 Re X2 ei$t
hence
1.1A
0.85A
i$t
i$t
e
+ 1.1 Re
e
x1 (t) = 0.85 Re
−$2 + i0.05$ + 46.24
−$2 + i0.061$ + 285.61
0.85A
1.1A
i$t
i$t
x2 (t) = 0.65 Re
e
− 0.5 Re
e
−$2 + i0.05$ + 46.24
−$2 + i0.061$ + 285.61
These can be combined to
0.852 A
1.12 A
i$t
x1 (t) = Re
e
+
−$2 + i0.05$ + 46.24 −$2 + i0.0609$ + 285.61
(0.65) (0.85) A
(0.5) (1.1) A
i$t
x2 (t) = Re
−
e
−$2 + i0.05$ + 46.24 −$2 + i0.061$ + 285.61
4.3.4
221
Problem 4
The transient response is given in appendix B as
x (t) =
F0
(1 − cos ωn t) h (t)
k
Hence maximum amplitude of the response is umax = 2Fk 0 . Compare this to static deflection which is
ustatic = Fk0 then we can say that dynamic load is twice as large as the static load. Therefor using 2F0 in
place of F in the expression for stress gives the result needed
−M c
I
− (2F0 ) L4
=
I/c
3 c
= − L F0
8 I
σy =
3
4
222
CHAPTER 4. EXAMS
Therefore
F0 = −
4.3.5
8I
σy
3Lc
Problem 5
2π At t = 0 then x (t) = Re ei 3 which is − cos (600 ) = − 12 . Using ω = 2π rad/sec then x (t) can be
traced. Here is a plot
I=sqrt(-1);
w=2*pi;
x=@(t) real(exp(I*2*pi/3)*exp(I*w*t))
t=0:.01:1;
plot(t,x(t))
grid
xlabel('time (sec)'); ylabel('x(t)');
4.3.6
223
Problem 6
Damped resonances are seen at ω = 8.5, 14 and 23 rad/sec. This is where r = ω$i is close to unity,
where $ is the forcing frequency and ωi is the natural frequency. Since this is a 3 dof system, it will have
3 natural frequencies.
The response of each dof will take contributions from each mode of vibration. Each mode vibrates
at different natural frequency. From the plot above it is seen that the response of x1 (t) has the largest
response when the forcing frequency is close to the ω2 = 14 rad/sec.
The new force now has the following set of discrete harmonics in it: (n = 0 is not counted, DC).
100 3t 99 6t 98 9t 97 12t 96 15t 95 18t 94 21t 98 24t
e
, 2 e , 3 e , 4 e , 5 e , 6 e , 7 e , 8 e , · · · or
1
f (t) = 100e3t , 49.5e6t , 32.7e9t , 24.3e12t , 19.2e15t , 15.8e18t , 13.4e21t , 12.3e24t
So the input force has only discrete frequencies. Since linear sum, each fi (t) will cause the response
|X| at that specific forcing frequency as shown in the plot. Looking the plot it can be seen that when
forcing frequency is 9 rad/sec, this will cause the largest |X| among all these set of discrete frequencies.
Hence the dominant harmonic is 9 rad/sec and will have amplitude around 2.4 from looking at the plot.
224
CHAPTER 4. EXAMS
Chapter 5
Quizes
1. First quiz key_PQ1_Solution_S2013.pdf
225
226
5.1
CHAPTER 5. QUIZES
my solution to Quiz 2
problems to solve problem_description.txt
5.1.1
problem
Consider this 3 DOF system
Suppose a harmonic force f (t) = A cos($t) is applied to the mass in the center. Use modal analysis
to do the following:
1. Find the uncoupled modal equations of motion. Consider the steady state solution for each of
these equations. Sketch the modal amplitude (Xj in the book on page 275) for each mode versus
frequency. A hand sketch is sufficient.
2. Use that result to sketch the frequency response of each of the masses, in other words the complex
amplitude Yn versus ω
5.1.2
Answer part (1)
A summary of the steps needed for full modal analysis is first given. In these steps, a column vector is
shown as bold letter Y and a matrix is shown as [M ]. In this summary, the system is assumed to have n
degree of freedom.
The steps are
1. Determine the system of equations of motion and set up [M ] Y00 + [C] Y0 + [K] Y = F in matrix
form.
2. Solve the eigenvalue problem det ([K] − ω 2 [M ]) = 0 in order to determine the n natural frequencies.
3. For each natural frequency ωj determine the corresponding j th eigenvector ϕj by solving [K] − ωj2 [M ] ϕj =
0. In this step, the first component of ϕj is set to 1 and the other components are solved relative
to it.
4. Obtain the normalized eigenvectors Φj for each ϕj using Φj =
will be a scalar.
ϕj
√
uj
where uj = ϕTj [M ] ϕj . Each uj
5. Set up the modal transformation matrix [Φ] = [Φ1 Φ2 · · · Φn ]. This will be an n × n matrix.
6. The transformation from normal solution y (t) to modal η (t) will be Y = [Φ] η and η = [Φ]−1 Y =
[Φ]T [M ] Y
5.1. MY SOLUTION TO QUIZ 2
227
7. Apply the above transformation on the original equations of motions in matrix form to obtain the
T
T
T
T
00
0
equations of motion in modal
h i coordinates
h i[Φ] [M ] [Φ] Y +[Φ] [C] [Φ] Y + [Φ] [C] [Φ] Y =
h [Φ]
i F.
˜ η (t) = [Φ]T F where I is the identity matrix, C˜ is a
This becomes Iη 00 (t) + C˜ η 0 (t) + K
h i
˜
diagonal damping matrix obtained using a method such as weak damping approximation and K
is diagonal matrix with diagonal that contains the natural frequencies squared ωj2 in each of entries.
8. For steady state
solution
in modal coordinates, the loading vector [Φ]T F is assumed to be Q =
ˆ i$t where Q
ˆ is the complex amplitude of the loading vector in modal coordinates.
[Φ]T F = Re Qe
ˆ i$t where X
ˆ is the complex amplitude
Therefore, the steady state solution is ηss (t) = Re Xe
T
ˆ j = 2 Φj F
of each modal response is X
. For a system with no damping this simplifies to
−$ +i2ζj ωj $+ω 2
j
T
ˆ j = Φ2j F 2 . In here, ΦT represents the transpose of the j th column of the modal transformation
X
j
−$ +ω
j
matrix [Φ], or the transpose of the j th mass normalized eigenvector, and ωj is the j th natural
frequency.
9. Now the steady state solution in modal coordinate
thesolution
is used
to obtain
in normal
i$t
i$t
ˆ
ˆ
ˆ i$t . In
coordinates since Y = [Φ] η. Therefore Yss = Re Xe
= Re [Φ] Xe
= Re Ye
!
!
n
X
ˆ j ei$t
component form Yss = Re
Φj X
j=1
The EOM are derived in the hand out given. The force f (t) acting on the second mass is now added,
resulting in the following equations of motion for the system


  
  
00 





m 0 0 
q 
k1 + k2
−k2
0
0
q 1  



  1 
  
00
0 m 0 q
 −k2

+
=
k1 + 2k2
−k2  q2
A cos($t)

  2 















00 
0 0 m
q3
0
−k2
k1 + k2
q3
0
The first step is to obtain the natural frequencies of the system. This is done by solving the eigenvalue
problem det ([K] − ω 2 [M ]) = 0. The solutions are also given in handout. They are ω12 = km1 , ω22 =
k1 +k2
2
, ω32 = k1 +3k
. The non mass normalized eigenvectors associated with these eigenvalues are found as
m
m
 
 
 






1
1



 

 

1

ϕ1 = 1 , ϕ2 =
0 , ϕ3 = −2









1

−1

1

228
CHAPTER 5. QUIZES
The next step is to mass normalize the eigenvectors as follows
 T 
 




1
m
0
0



  
1


T


µ1 = ϕ1 [M ] ϕ1 = 1
0 m 0  1 = 3m

 


 

1

0 0 m 1
 T 
 




1
m
0
0



  
1


T


µ2 = ϕ2 [M ] ϕ2 =
= 2m
0
0 m 0 0

 





−1



0 0 m
−1
 T 
 




m
0
0
1




1
  

 0 m 0  −2 = 6m
µ3 = ϕT3 [M ] ϕ3 = −2
 
 

 


1
0 0 m 1
Hence the mass normalized eigenvectors are
 




1
ϕ1
1
Φ1 = √ = √
1

µ1
3m 


1
 


1
ϕ2
1  
Φ2 = √ = √
0

µ2
2m 


−1
 

1

ϕ3
1  
Φ3 = √ = √
−2

µ3
6m 

1

Hence the modal transformation matrix [Φ] is

1
[Φ] = [Φ1 Φ2 Φ3 ] = √
m
√1
 3
 √1
 3
√1
3
√1
2
0
−1
√
2

√1
6
−2 
√
6
√1
6


0.577 0.707
0.408

1 

=√ 
0.577
0
−0.816

m
0.577 −0.707 0.408
229
The modal EOM’s are now found using the modal transformation matrix [Φ]

1

0

0

1

0

0
0
1
0
0
1
0
[Φ]T [M ] [Φ] {η 00 } + [Φ]T [K] [Φ] {η} = [Φ]T Q



 
  
00 
2





η1 
0
1 0 0 
0 0 
η 
ω





 
  1  1
0 1 0 η 00 +  0 ω 2 0  η2 = [Φ]T A cos($t)
2

 
  2 










00 
2 
0
0 0 1
η3
0 0 ω3 η3 
 

T

 
00 


0 
η
0.577
0.707
0.408
k
0
0
η
 
 1
1
  1 1 

 
1 
00





√
=
0 η2 +  0 k1 + k2
0.577
0
−0.816
0
η


  2


m
m







00 
1
η3
0.577 −0.707 0.408
0
0
k1 + 3k2
η3


 

 
00 





0 
η 
k1
0
0

η 1 

 0.577 A cos ($t) 
  1 1 
 
1
00



√
+
=
0 η2
0
0 k1 + k2
0
 η 2 
 m


m










00 
η
1
−0.816 A cos ($t)
0
0
k + 3k
η
3
1
2







0




A cos($t)



0
3
Therefore, the 3 uncoupled modal EOM’s are
k1
0.577 A
η100 (t) + η1 (t) = √
cos ($t)
m
m
k1 + k2
η200 (t) +
η2 (t) = 0
m
k1 + 3k2
0.816 A
η300 (t) +
η3 (t) = − √
cos ($t)
m
m
To complete the solution, the above EOM are written as follows by using complex form for the loading
vector
0.577 A i$t
k1
00
√
e
η1 (t) + η1 (t) = Re
m
m
k1 + k2
η2 (t) = 0
η200 (t) +
m
−0.816 A i$t
k1 + 3k2
00
√
η3 (t) +
η3 (t) = Re
e
m
m
Assuming the steady state solution is
ˆ i$t
η = Re Xe
or in expanded form
ˆ 1 ei$t
η1 (t) = Re X
ˆ 2 ei$t
η2 (t) = Re X
ˆ 3 ei$t
η3 (t) = Re X
Where
ˆ1 =
X
0.577
√ A
m
ω12 + 2iζ1 ω1 $ − $2
ˆ2 = 0
X
ˆ3 =
X
−0.816
√ A
m
ω32 + 2iζ3 ω3 $ − $2
230
CHAPTER 5. QUIZES
Dividing the numerator and the denominator by ωi2 where i = 1, 2, 3 and using ri =
ζ = 0 since no damping exists, results in
√
A m
ˆ
X1 =
k1
0.577
2
1 − m $k1
$
ωi
and letting
!
ˆ2 = 0
X
√
A m
ˆ
X3 =
k1 + 3k2
−0.816
2
1 − m k1$
+3k2
!
To sketch these amplitudes, the equations are normalized. This is in effect the same as setting
m = 1, k1 = k2 = 1, A = 1 resulting in

  
0.577


2
ˆ






X 1 
 
 1−$

0
ˆ
ˆ
X = X2 =








1
−0.816
X



ˆ


2
3
4
1− $4
Here is a plot of each Xi vs $. The x-axis is the nondimensional forcing frequency Ω
231
Since there is no damping, resonance will occur at Ω = 1 in first mode and at Ω = 2 for mode 3.
5.1.3
Answer part (2)
The transformation from modal coordinates to normal coordinates is
q = [Φ] η
In expanded form
 




q 1 


T


Φ
{η}


 1

T
q2 = Φ2 {η}

 


q 
 
ΦT {η}

3
3
232
CHAPTER 5. QUIZES

But [Φ] =
√1
m

0.577 0.707
0.408


0.577
 and η = Re Xe
ˆ i$t hence the above becomes
0
−0.816


0.577 −0.707 0.408
 
 
T  ˆ 1 ei$t 



Re
X






0.577















i$t
ˆ


Re
X
e


0.577
2



















i$t  
ˆ



0.577
Re
X
e


3









  
T

i$t  
ˆ



Re
X
e








1
0.707
q







1
  
  
 
i$t
ˆ
Re X2 e
q2 =
0

 










q 
 




i$t 
ˆ




−0.707
3
Re
X
e


3










T


i$t
ˆ1e




Re
X






0.408
















i$t
ˆ


Re
X
e


−0.816
2



















ˆ 3 ei$t 
 0.408
Re X



it$
it$
it$ 

0.577
Re
(X
e
)
+
0.577
Re
(X
e
)
+
0.577
Re
(X
e
)


1
2
3


it$
it$
=
0.707 Re (X1 e ) − 0.707 Re (X3 e )




0.408 Re (X eit$ ) − 0.816 Re (X eit$ ) + 0.408 Re (X eit$ )
1
2
3




0.577 X1 + 0.577 X2 + 0.577 X3 




= Re 
ei$t 
0.707
X
−
0.707
X
1
3




0.408 X − 0.816 X + 0.408 X 

1
2
3
Comparing the above to qss = Re (Yei$t ) shows that
Y=




0.577
X
+
0.577
X
+
0.577
X

1
2
3


0.707 X1 − 0.707 X3



0.408 X − 0.816 X + 0.408 X 

1
2
3
To plot each Yi , let m = 1, k1 = 1, k2 = 1, A = 1, and letting X2 = 0 as found earlier, results in




0.577
−0.816
0.577


0.577 1−$2 + 4


2


1− $4






0.577
0.707
−0.816
Y = 0.707 1−$2 − 4
$2



1− 4 





0.577
0.408
−0.816 


0.408 1−$2 + 4

2
$
1−
Here is a plot of each Yn for n = 1, 2, 3
4
233
The above shows that when the nondimensional frequency Ω is not close to a one of the nondimensional
natural frequencies, then the Y values have comparable magnitudes. For nondimensional frequency Ω
larger than 3 all amplitude are zero, which means the whole system does not oscillate any more in steady
state.
234
CHAPTER 5. QUIZES
Chapter 6
appendix
235
236
6.1
CHAPTER 6. APPENDIX
cheat sheet
EMA545_final_cheat_sheet.pdf
6.2. STUDY NOTES
6.2
237
study notes
1. analysis.nb Misc calculations Mathematica notebook
2. verify_spring_stiffness_formula.nb Misc calculations Mathematica notebook
6.2.1
trig identities
h
i
1 iωt
i(ωt− π2 )
sin ωt = Re e
= Re e
i
cos ωt = Re eiωt
1 iωt
cos ωt =
e + e−iωt
2
1 iωt
sin ωt =
e − e−iωt
2i
When 2 harmonics have same amplitude, we can write then as envolope of one in another
A cos (ω1 t − φ1 ) + A cos (ω2 t − φ2 ) = 2A cos (∆ω t − ∆φ ) cos (ωav t − φav )
Here is an example of the above. We first draw the two signals on their own, then plot the additions
of them
f1 = a Cos[w1 t - p1];
f2 = a Cos[w2 t - p2];
parms = {a -> 1, w1 -> 1, p1 -> Pi/3, w2 -> 10, p2 -> Pi/4};
Plot[Evaluate[{f1, f2} /. parms], {t, 0, 10},
PlotStyle -> {Red, Blue}]
238
CHAPTER 6. APPENDIX
Now we add them to see the envelope effect
Plot[Evaluate[{f1, f1 + f2} /. parms], {t, 0, 10},
PlotStyle -> {Red, Blue}]
Now we plot the same signal addition, but using the form after converting to use the mean and delta
notation as shown above just to confirm it is the same signal
avW = Mean[{w1, w2}];
avP = Mean[{p1, p2}];
delW = w2 - avW;
delp = p2 - avP;
g = 2 a Cos[delW t - delp] Cos[avW t - avP];
Plot[g /. parms, {t, 0, 10}]
π
The beat period is ∆ω
(this is the time between each beat to the next beat). The whole signal will be
ω1
periodic only when ω2 is rational.
6.2. STUDY NOTES
239
Beat shows up when we have 2 harmonics added, that has same amplitude. The beat signal itself will
be period when the ratio between the frequencies of the two harmonics is rational. In the context of
response of a system, we can think of the steady state response as one signal and the transient response
as another singnal. The response will then show a beating signal when the amplitude of the steady state
and transient singnals is the same. Here is an example of that from one of my demos
240
6.3
CHAPTER 6. APPENDIX
my lecture notes
There are my scanned notes and my scanned graded HW’s and exams
1. notes.pdf
6.4. COURSE RELATED LINKS
6.4
course related links
1. final exam schedule
2. Syllabus
3. public course web page
4. internal course web page
5. Lectures download
241

My EMA 545 Mechanical Vibrations, Spring 2013 material

Transcription

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