Schaum s outline of mechanical vibrations

-


Covers
vibration
measurement,
finite
element
analysis,
and
eigenvalue
determination
-

learn
how
to
use
Mathcad,
Maple,
Matlab,
and
Mathematica
to
solve
vibrations
problems
-

I
ncludes

313
solved
problems completely
explained
-
~
,-
-
Use
with
your
textbook
or
for
independent
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SCHAUM'S
OUT
LI

NE
OF
THEORY
AND
PROBLEMS
of
MECHANICAL
VIBRATIONS
S.
GRAH
AM KELLY, Ph.
D.
Associate Professor
of
Me
chanical
En
gineering alld Assistant Provost
Th
e University
of
Akr
oll :
SCHAUM'S OUTLINE SERIES
McGRAW-
HI
LL
New York
St.
Louis San Francisco

Auckland
Bogota C
ar
acas
Lisbon Londoll Madrid Mexico City Milan Montreal New Delhi
San
iuan
Singapore Sydney Tokyo Toronto
\
S.
GRAHAM
KELLY is Associate Professor of M
ec
hanical Engineering
and Assistant Provost at
Th
e U
niv
ers
it
y of Akron. He has been on the
facuity at Akron s
in
ce
19
82, serv
in
g before at the University
of
Notre

Dam
e.
He ho
ld
s a B.S.
in
Eng
in
eering Science and Mechanics and an
M.S. a
nd
a Ph.D.
in
Engin
eer
ing Mechanics from Virginia Tech.
He
is
also
th
e author of Fundamenrals
of
Mechanical Vibrations and the
accompany
in
g software
VIBES
, published by McGraw·Hill.
Cred
its

Selected mater
ial
reprinted from
S.
Gra
ham Kelly. Fun
da
me
nt
als
of
Mechanical Vibrations,
© 1993 McGraw-Hili.
Jn
c.
Reprinted
by permission
of
McGr
aw-Hi
li
, In
c.
Maple-based
ma
terial produced by per
mi
ss
ion
of

Water
loo M aple
In
c.
Mapl
e
an
d Maple V
are registered trademarks of Waterl
oo
Maple Inc.:
Mapl
e
sys
tem itself copyrighted by
Wate
rl
oo
Maple Inc.
Mathcad mate
ri
al
produced by permission
of
MathSoft, Inc.
Schaum's Outline
of
T
he
ory

and Problems
of
MEC
H
ANICAL
VI
BRAT
I
ONS
Copyright ©
1996
by
The
McGraw-Hi
li
Compan
ies, Inc. All rights reserved. Prin ted in the
United States of America. Except as p
er
mitted under th e
Co
pyright Act
of
1976. no part
of
this publication may be r
eproduced
or
distributed in any form
or

by any means. or st
ored
in
a data base
or
retrieval syst
em.
with
out
the
prior
written permission
of
the puhlisher.
I
2345
6789
JO
t I
12 13 14
15
16
17
18
1
920
PRS
PRS
9 8 7 6
ISBN

0-07-034041-2
Sponsoring
Edito
r: Jo
hn
A
li
ano
Production Supe
rvi
sor: Suzanne Rapcavage
Project Supervision:
The
T
ota
l Book
Library or Congress Cataloging.in.Publication Data
Ke
ll
y.
S.
Graham.
Schaum's
ou
tline
of
theory
and problems
of
mechanical vibrations I

S.
Graham
Kelly.
p. cm. - (Schaum
's
outl
ine series)
Includes
in
dex.
ISBN 0-07-03
4041
·2
1.
Vibration-Proble
ms. exercises. etc.
2.
Vibration-Out
lines.
syllabi. elc.
I. Title.
OA935.K383 t996
620.3·076-<1c20 95-4
75
54
CIP
McGraw-Hill·
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A
Dili

s
iOJ1
o{T'hcMcG
rawB
iUCompanies
sc
h.
Gl.,p ,
"\35
.
K3.
'23
i"l."\.b
c
~
1
Preface
A student
of
mechanical vibrations must draw upon knowledge of many ar
eas
of engineering science (statics, dynamics, mechanics of materials, and even fluid
mechanics) as well as mathematics (calculus, differential equations, and lin
ear
algebra).
Th
e
student
must then
sy

nthes
iz
e this knowledge to formulate the
solution
of
a mechanical vibrations problem.
Many mechanical systems require modeling before their
vi
brations can be
analyzed. After appropriate assumptions are
made
, including the numb
er
of
degrees
of
free
dom
necessar
y,
bas
ic
conservation laws are applied to derive
governing differential
eq
uations. Appropriate mathematical methods are applied
to sol
ve
the differential equations. Often the modeling results
in

a differential
equation whose solution is well known,
in
which case the existing solution is used.
If
this
is
the case the solution must be studied and written
in
a form which can be
used in anal
ys
is and design applications.
A student
of
mechanical vibrations must learn how to use existing knowledge
to do all of the above.
The
purpose
of
this book is to provide a supplement f
or
a
student studying mechanical vibrations that will guide the stude
nt
through a
ll
aspects
of
vibration analysis. Each chapter has a short introduc

ti
on of the th
eo
ry
used in the chapter, followed
by
a large number
of
so
lv
ed
problems. The so
lv
ed
proble
ms
mostly show how the theory is used
in
design and analysis applications.
A
few
problems
in
each
chapter
examine the th
eo
ry in m
ore
d

eta
il.
Th
e coverage
of
the book is quite
broad
and
includes free and forced
vibrations
of
I-degree-of-freedom, multi-degr
ee
-of-freedom, and continuous sys-
tems.
Undamped
systems and systems with viscous damping are considered.
Systems with Coulomb damping and
hy
steretic damping are considered for
I-degree-of-free
dom
systems. There are several chapters
of
special note.
Chapter
8 focuses on design
of
vibratIon control devices such as vibration isolators and
vibration absorbers.

Chapter
9 introduces the finite element method from an
- analytical viewpoint.
Th
e problems in
Chapter
9
us
e the finite element method
using only a few elements to analyze the vibrations of bars and beams.
Chapter
10
focuses
on
nonlinear vibration
s,
mainly discussing the differences between line
ar
and nonlinear systems including self-excited vibrations and chaotic motion.
Chapter
11
shows how applications software can be used in vibration analysis and
design .

The book can be used to supplement a course using any of the popular
vibrations textbooks,
or
can be used as a textb
oo
k in a course where theoretical

development
is
limited. In any case the book
is
a good source for studying the
solutions
of
vibrations problems.
The author would like to thank the staff
at
McGr
aw-Hill, especially
John
Aliano, for making this book po
ss
ible. He would also like to thank
his
wife and
son, Seala and
Graham
, for patie
nc
e during
preparation
of
the manuscript and
Gara
Alderman and Peggy
Du
ckworth for clerical help.

S.
GRAHAM
K E
LLY
v
Contents
PROBLEMS AND EXAMPLES ALSO FO
U1\"
D IN
TH
E COMPANION
SCHAUM'S
IlVTER4.CTIVE
OuTLI
NE

. .




ix
Chapter 1
Chapter 2
Chapter 3
C
hapter
4
C
hapt

er 5
Chapter 6
MECHANICAL SYSTEM
ANALySiS
. . . .


•
.•

. .
.•
. .

1.1
Degrees
of
Freedom
and
Generalized Coordinates.
1.2
Mecha
ni
cal
System Components. 1.3 Equival
ent
Systems
Ana
lysis.
1.4

Torsional
Sy
stems
.
1.5
Static
Eq
uili
brium
P
os
ition.
FR
EE
VIBRATIO:-IS
OF
I·DEGREE·OF·FREEDOM
SYSTEMS
2.1
Derivation
of
Differential
Equat
ions. 2.2
Standard
Form
of
Diff
erentia
l

Equations. 2.3
Undamped
Respons.!. 2.4
Dam
ped Re sponse. 2.5 Fr
ee
Vibration
Rt:
spon
st!
for Sy
ste
ms Subject to Coulomb Damping.
HAR:VIONIC EXCITATION
OF
l
·DEG
R
EE
·OF·FREEDOM
36
SYSTEMS

.

. .
•


.

.
.••
.
•.

•


64
3. 1 Dcriva:ion
of
Differenti
a!
Equa
tion
s.
3.2 -H
armo
nic Excitation.
3.3 L
:ndamped
SY
S[~
:-:1
Response.
3

Dnmpt:d

System Response.
3.5 Frequt!ncy
Squared
Exc
i~J.[ions.
3.6 Harmonic Support Excitation.
3.
7
Multifrequency
Excit
ations.
3.8
General
Periodic
Excitations: Fouri!!:-
Series.
3.9
Coulomb
Dampi:1g.
3.:0
Hyste
retic
Damping.
GE
NERAL FORCED
RESPONSE
OF
I-DEGREE·OF
-F
REEDOM

SYSTEMS .

. .
•





. .



•

• 109
4.1
Gc!
neral
Differentia
l E
quation.

t2
Convo
l
utio
n Integra
l.
4.3

LaP
b.:
"!
Transform
Solutions. 4A
Cnit
I
m!=,ulse
Function
and
L'nit Step
Funct
ic:1

4.5
;\u
mer
ical
Methods.
4.6
Response
Sp~ctrum.
FREE
VIBRATIONS
OF
MULTI·DEGREE·OF
·F
REEDOM
SY
STEMS

•
.

.
••



.
•.
• •
13
6
5.1
LaGr
ange's EqL:J;ions. 5.2
M:::.trix
Formul
atio
n
of
Di
ffere ntial
Eq
u:l:!ons
for Li ne
ar
Systems. 5.3 Stiffness
Influence
Cod

ficients. 5.4
F1e:~ibili[v
~latrix
.
5.5
Normal
Mode
Solu
ti
on
. 5.6
Mode
S
hap
e Orthogonality. "
5.7
~Iatrix
Iterati
on.
5.8
Damped
Systems.
F
ORCED
VIBRA
TIO
NS
OF
o\IULTI-DEGREE·OF·FREEDOi\l
SYSTEMS

.
.•
. . .
• •
••

. .
.•.
.

. . . .
180
6.1
General System. 6.2
HJrmon
ic Excitation. 6

3 LaPlace
Transform
Solutions.
6.4 M
oda
l
Analysis
for
Systc!ms with
Proportio
nal D
amp

ing

6.5
~Iodal
Analy
s
is
fo
r
Sysr~ms
with
General
Damping.
vi
i
v
iii
Ch
ap
ter 7
Ch
apter
8
Chapter 9
Ch
apte
r 10
, Chapter
11
Append

ix
CO
NTEN
TS
V
IBR
ATIONS OF CONTINUOUS
SySTEMS
. .

. . .


. . .

201
7
.1
Wave Equa
ti
on.
7.2
Wave Solution.
7.3
Normal Mode
So
lution.
7.4 Beam Equat
io
n. 7.5 Modal Superp

os
ition.
7.
6 Ray leigh's Quo
ti
ent.
7
.7
Rayleigh-Ritz Method.
VIBRATION
CONTROL





235
8.1 Vibrat
io
n I
so
la
ti
o
n.
8.2 I
so
la
ti
on f

ro
m Ha
rm
o
ni
c E
xci
tatio
n.
8.3
Sh
oc
k
Is
ol
a
ti
o
n.
8
.4
Impulse Isolation. 8.5 Vibra
ti
on Ab
so
rbers. 8.6 Damped
Absorber
s.
8.7 Houda
ill

e Damper
s.
8.8
Wh
irlin
g.
FINITE ELEMENT
METHOD
. . .

. . .




. 264
9.
1
Gen
eral Method.
9.
2 Forced Vibra
ti
on
s.
9
.3
Bar Element. 9
.4
Beam

Elemen
t.
NONLI
NEA
R SYSTEMS

.

. .


.



. .

285
10.1
Di
ffe
rences from
Lin
ear System
s.
10.2 Qual
it
ati
ve
Analysis.

10
.3 Du
ffi
ng's Equat
io
n. 10
.4
Sel
f-
E
xc
it
ed Vibrat
io
n
s.
COMPUTER
APPLICATIONS .

.



. .

. 301
11.1
Software Speci
fic
to Vibra

ti
ons Appli
ca
ti
ons.
11
.2
Spreadsheet
Programs.
11.3
Electro
ni
c Notepad
s.
11
.4
S
ym
bo
li
c Processors.
SAMPLE
SCREENS FROM
THE
COMPANION
INTERACTIVE
OUTLINE

.

. . .

.

.

.

. . .

. . .

.

. . . .

. . 333
Index
•
.

.

••.
•
• •
• .

. . . • . .

.

.

349
Problems and Examples Also Found
in
the Companion
SCHAUM'S ELECTRONIC TUTOR
Some of the problems and examples
in
this book have software components
in
the companion
Schaum's Electronic' Tutor. The Mathcad Engine, which "drives" t
he
Electronic Tutor, allows every
number, formula, and graph chosen
to
be
completely l
ive
and interactive.
To
identify those
items that are
avai
lable in the Electronic Tutor software, please look for the Mathcad icons,

~,
placed under the problem number or adjacent to a numbered item. A complete list of these Mathcad
entries follows below. For more information about the software, including the sample screens, see
Appendix
on
page 333.
Problem
1.4
Problem 3.10 Problem 4.26 Problem
7.1
Problem 8.24
Problem
1.5
Problem 3.12 Problem 5.19
Problem 7.4
Problem 8.25
Problem
1.7
Problem
3.
15 Problem 5.20
Problem 7.5
Problem 8.26
Problem 1.12 Problem 3.18 Problem 5.25
Problem 7.6 Problem 8.27
Problem 1.14 Problem 3.
19
Problem 5.26
Problem 7. [3 Problem 8.28
Problem 1.19 Problem

3.20 Problem 5.27
Problem
7.[6
Problem 8.32
Problem 2
.8
Problem 3.23
Problem 5.28 Problem 7.22
Prob[em 8.34
Problem 2.9
Problem 3.24 Problem 5.30 Problem 7.23 Problem 8.35
Problem 2.14 Problem 3.25
Problem 5.3 [ Problem 7.25 Problem 8.37
Problem
2.
[5
Problem 3.26
Problem 5.32 Problem 8.3 Problem 9.5
Problem
2.
[6
Problem 3.27
Problem 5.35 Problem 8.4
Problem 9.6
Problem 2. [7
Problem 3.28
Problem 5.38
Problem 8.5 Problem 9.7
Problem
2.[8

Problem 3.34
Problem 5.40
Problem 8.6 Problem 9.13
Problem 2.19 Problem 3.35
Problem 5.41
Problem
8.
[0
Problem
9.[4
Probl
em
2.20
Problem 3.36 Problem 5.42
Problem
8.
[ [
Problem 10.6
Problem 2.2 [ Problem 3.38 Problem 5.44
Problem
8.
12 Problem 10.8
Problem 2.22
Problem 3.40 Problem 5.45
Problem 8.
13
Problem 10.
11
Problem 2.23 Problem 4.3
Problem 6.3 Problem 8.14

Problem 11.4
Proble",! 2.25 Problem 4.5
Problem 6.9 Problem 8.
15
Problem 11.5
Problem 2.29
Prob[em 4.6
Problem 6.10 Problem 8.16
Problem 11.6
Probl
em
3.4 Problem 4.13
Problem 6.1 [ Problem 8.17
Problem 11.7
Problem 3.5
Problem 4.18 Problem 6.12
Problem 8.18 Problem
1l
.8
Problem 3.7
Problem 4.19 Problem 6.15 Problem 8.19
Problem
11.
9
Pro
bl
em
3.8
Problem 4.24 Problem 6.16
ix

Chapter 1
Mechanical System Analysis
1.1
DEGREES OF FREEDOM
AND
GENERALIZED COORDINATES
The number of degrees
of
freedom used
in
the analysis of a mechanical system
is
the number
of
ki
nematica
ll
y independent coordinates necessary to completely describe the motion of every
particle
in
the system. Any such set
of
coordinates is called a set
of
generaliz
ed
coordinat
es
.
The

choice of a set of
ge
nera
li
zed coordinates is not unique. Kinemat
ic
qu
antities such as
di
splacements,
ve
loci
ti
es, and accelera
ti
ons are written as functions of the generalized
coordinates and their
tim
e derivatives. A system with a finite number of degrees
of
fr
eedom
is
ca
ll
ed a discrete system, while a system with an infinite number
of
degrees of freedom
is
called a

continuous system
or
· a distributed parameter system.
1.2
MECHANICAL SYSTEM COMPONENTS
A mecha
ni
cal system comprises
in
ertia components, sti
ff
ness component
s,
and damping
components.
Th
e
in
ertia compon
en
ts have kine
ti
c energy when the system is in motion. The
kinetic energy of a rigid body un
de
rgo
in
g
pl
anar motion is

_
T
=!
mv
2+U
w
2
~
(I.
I)
where v is the velocity
of
the body's mass center, w
is
its angul
ar
velocity about an axis
perpendi
cu
l
ar
to the plane of motion, m
is
the body's mass, and I
is
it
s mass moment of
in
ertia
about an

aXIS
parallel to the axis of
rot
ation through the mass cente
r.
A linear stiffness component
(a
linear spring) has a fo
rc
e displacement rela
ti
on
of
the form
F=kx
(1
.2
)
where
F is
app
lied force and x is the component's change in length from its unstretched length.
The stiffness
k has dimensions of force p
er
length.
A
dashpot is a mechanical device that adds
vi
scous damping to a mechanical system. A

lin
ear
viscous damping
co
mponent has a force-velocity relation of the form

F=cv<ft-
(1.3)
wh
ere
c::,
is the damping coefficient
of
dimensions mass per time.
1.3
EQUIVALENT SYSTEMS ANALYSIS
All lin
ear
I-degree
-o
f-
fre
edom systems with
vi
scous damping can be modeled
by
the simple
mass-spr
in
g-

dashpot
sys
tem of
Fi
g.
1-1. Let x be the chosen generalized coordinate.
The
kinetic
energy
of
a linear system can be written in the form
(1.4)
2
MECHANICAL
SYSTEM ANALYSIS [CHAP. 1
The
potential energy
of
a lin
ear
system can
be
written
in
the form
(1.5)
The
work done by the viscous damping force in a lin
ear
system between two arbitrary locations

X,
and
X,
can be written as


W = - f
co.x
dx

(1.6)
x,
Co.
Fig. 1·1
1.4 TORSIONAL SYSTEMS
When an ang
ul
ar
coordinate is used as a generalized coordinate for a lin
ea
r system,
the
system can be modeled by the equivalent torsional system
of
Fig. 1·2.
The
moment applied to a
linear torsional spring is proportional to its ang
ul
ar rotation while the moment applied to a

linear torsional viscous damper
is
proportiona
l to its ang
ul
ar
velocity.
The
equivalent system
coefficients for a
tor
sional system
are
determined by calculating the total kinetic energy,
potential energy, and work done by viscous damping forces for the original syst
em
in terms
of
the chosen generalized coordinate and sett
in
g them equal to
v = !
k,,,e
'
8,
W = - f c
tJde
'"
8,
'

1'

'Fig. 1·2
(1.7)
(1.8)
(1
.9)
CHAP.1J
MECHANICAL
SYSTEM ANALYSIS
3
1.5
STATIC EQUILIBRIUM POSITION
Systems, such as the one in Fig. 1·3, have
e
l
~J.tic
elements that are subject to force when the
system is
in
equilibrium.
The
resulting deflecllon
in
the elastic
el
ement is ca
ll
ed its static
deflection,

usually
den
oted by
11.
".
The static deflec
ti
on of an elastic elem,ent
in
a lin
ear
sys
tem
has no effect on the system's
eq
ui
va
lent s
ti
ff
ness.
Fig. 1·3
Solved Problems
1.1 Determine the number
of
degrees
of
freedom to be used
in
the vibra

ti
on analysis of the
rigid b
ar
of F
ig
. 1-4, and specify a set of generalized
coor
dinates that can be used
in
its
vibration anal
ys
i
s.
f-
30em
-t
70em
1:
k=2
000N/m
(11U
'
\
m
-2.3
kg
Fig. 1·4
Since

the bar
is
r
igid,
the
sys
tem h
as
only 1 degree of freedom. One
po
ssib
le c
hoice
for
th
e
ge
nera
li
zed
coordinate is 8,
th
e angular displacement of
th
e bar
me
as
ur
ed
positiv

e
clockwi
se
fr
om
the
syste
m's
eq
uil
ibri
um
pos
it
ion.
1.2 I?etermine the number
of
degrees
of
freedom needed f
or
the analysis of the mechanical
system
of
Fi
g.
1·5, and specify a set
of
generalized coordinates that can be used
in

its
vibration analysi
s.
Fig. 1·5
4
MECHANICAL
SYSTEM
ANALYSIS
[CHA
P.
I
Let
x be the displacement of the mass center
of
the
ri
g
id
bar,
mea
s
ur
ed from the syst
em's
equilibrium posi
ti
o
n.
Kn
ow

ledge of x, by itself, is not s
uffi
cient
to
determine the di
sp
la
ceme
nt of
any
other
particle on
th
e bar.
Thu
s
the
system has
more
than I deg
ree
of
fr
ee
dom
.
Let
8 be the clockwise angular rotation
of
the

bar
with rl;Spect to the axis
of
the
bar
in its
equilibrium positio
n.
If
8
is
sma
ll
, then the displacement
of
the
right
end
of the
bar
is x +
(L/2)8.
Thus
the
sys
tem has 2
de
grees of freedom,
and
x

and
8
are
a
po
ss
ible
se
t
of
generalized
coordinates, as illus
trate
d in Fi
g.
1
-6
.
Fig. 1-6
1.3
Detennine
the
numb
er
of
d
eg
rees
of
freedom

used
in
th
e a
nalysis
of
th
e m
ec
ha
nic
al
sys
tem
of
Fi
g. 1-7.
Specify
'a
se
t
of
ge
ne rali
ze
d
coordinates
that
can be u
se

d in
th
e
sys
tem
's
vibr
atio n
analysis.
1.4
rf+
sa.
M~hcad
Fig. 1-7
The
system
of
Fig. 1
-7
ha
s 4 degrees
of
freedom. A
po
ssible set
of
generalized c
oor
dinates
are

8" the
cl
ockwise angul
ar
displacement from equilibrium
of
the disk whose
center
is at
0,
; 8" the
cl
oc
kwise angular displacement from equilibrium
of
the
disk whose
center
is
at
0
,;
x"
the
downward displace
ment
of
block B; and x"
the
downward

displace
ment
of
block
C.
Note
that the
upward
displaceme
nt
of
bl
oc
k A is given by T,8, and hence is not kinematica
ll
y ind
epen
dent
of
the
motion
of
the disk.
A
tightly
wound
heli
ca
l
coi

l
spr
ing is
made
fr
om
an
18-mm-diameter
bar
of
0.2
p
erce
nt
harden
ed
steel
(G
=
80
X
10
9
N/m
2).
The
s
prin
g h
as

80
active
coils
with
a
coil
di
ame
ter
of
16
cm.
What
is
the
chang
e in
length
of
the
spring
when
it
hangs
vertically
with
one
end
fixed
and

a
200-kg
block
attached
to
its
other
end?
CHAP.
1]
MECHANICAL
SYSTEM
ANALYSIS
The
stiffness
of
a helical coil spring
is
where D
is
the
bar
diameter, r
is
the coil radius, and N
is
the number
of
active turns. Substituting
known

va
lu
es leads to
~
(80
X
10"
-m
N
, )(0.018
m)'
k 3.20 X
10
'
~
= 64(80)(0.08
m)
' m
Using Eq. (1.2), the
change
in
length
of
the spring
is
F m (200 kg)(9.81
~)
x = - =
'l
= 0.613 m

k k 3.20 X
10
'
~
m
1.5
Determine
th
e
longitudinal
stiffness
of
the
bar
of
Fig. 1-8.
~f+
sa.
Matl1cad
f-
-

- L

-
1
Fig. 1·8
The
longitudinal motion
of

the block
of
Fig.
I·g
can be modeled by an
undamped
system
of
th
e form
of
Fig.
I-\.
When
a force F
is
applied
to
the
end
of the bar, its change in l
ength
is

0 =
FL
4:-
AE
"""'II:"'"
or

F =
AE
0
L
which
is
in
the form
of
Eq. (1.2).
Thus
k
=AE
<q
L
1.6
Determine
the
torsional
stiffness
of
the
shaft
in
the
system
of
Fig.
1·9
.

1
1.4
m

1
Fig.
1·9
ri =
15mm
ro=25
mm
G=80
X IO'.!:':
m'
6
1.7
~(
+
ila.
Mathcad
M
EC
HANI
CA
L
SYSTEM
ANALYSIS
[C
HAP
. I

If
a moment M is applied to the end of the shaft, the angle of
twi
st at the end
of
the sha
ft
is
determined from mechanics
of
mate
ri
als
as
IJ=ML
JG
where J is the polar moment of
in
ert
ia
of the shaft·s cross section. Thus
M =
JG
IJ
L
and
the
shaft's equivalent torsion
al
stiffness

is
For
the sh
af
t of Fig. 1
-9
k = JG
, L
J =
~
(,:
-,,')
=
~
[(0.025
m)'
- (0.015 m
)'
] = 5.34 X 10-'
m'
(5.34 X W-' m
')(
80 X
10'
~)
k = m
,
1.4
m
Thu

s
3.05 X 10' N-m
rad
A machine whose mass is much larger than the ma
ss
of
the beam shown
in
Fi
g.
J-
lO
is
bolted to the beam. Since the inertia of the beam is sma
ll
compared to the
in
erti a of the
machine, a 1-degree-of-freedom model is used to analyze the vibra
ti
ons
of
the mach
in
e.
The
system is modeled by the
sys
tem of
Fi

g.
1-3. Determine the equivalent spring
stiffness if the machine is bolted to the beam at
(a) Z = 1 m
(b)
z = 1.5 m
L~_""',
_______
2l
E =
210
X
10
'
~
F
1=1.5X
1O
·'
m'
f

3 m
~
Fig. 1-10
Let w(z; a) be the deflection of the beam at a location z due to a un
it
concentrated load
app
li

ed
at z =
a.
From mechanics of materials, the beam deflection
is
lin
ear, and thus the deflection
due to a concentrated load of magnitude
F
is
given by
y(z; a) = Fw(z; a)
If
the machine is bolted to the beam at z =
a,
the deflection at this loca
ti
on is
which is sim
il
ar to Eq. (1.2) with
y ea;
a)
=
Fw(
a;
a)
k
=_l_
w(a;a)

(J.1O)
CHAP. I]
MECHANICAL
SYSTEM
ANALYS
IS
7
From mechanics of material
s,
th
e de Rec
ti
on of a beam
fix
ed at z = 0 and pinned at z = L due
to
a
un
it
co
nc
entra
ted
load at z = a, for z < a is
(a) For a = I m,
aIL
=
\.
Then u
si

ng Eq
s.
(
1.1
0)
a
nd
(1.1
1),
k
=_
1
_=
81
E1
<,
w(a;a)
!l
a'
81(210 X
1O
' ;;')(l.5 X
10
-'
m')
11(1
m)'
(b)
For a =
1.5

m,
aIL
=
I.
Then us
in
g Eqs. (1.10) and (1.11),
k
=_1_=96
£1
<q
w(a;
a)
7a'
2.32 x
10
'
~
m
96(210 x
10
'
;;')(1.5
x
10
-'
m')
7(1.5
m)
'

1.28 x
10
'
~
m
(I.
ll
)
1.8 A m
ac
hine of mass m
is
atta
ched
to the midspan
of
a s
impl
y su
pp
o
rted
beam
of
length L ,
elastic
modulu
s E, and cross-sectional m
ome
nt

of
inerti
a I.
Th
e mass
of
th
e m
ac
hine
is
much
greater
than
the
mass of
the
beam
; thus the syst
em
can be modeled using 1
degree
of
freedo
m.
What
is the e
quivalent
stiffness
of

the
be
am using the midsp
an
deflection as
the
genera
li
zed
coordin
at
e?
The deflection of a s
im
ply suppo
rt
ed beam at its midspan
du
e to a
co
ncentrated load F
a
ppli
ed at
th
e
mid
span is
F
L'

{j
= 48£1
The equivalent s
tiffn
ess is
th
e r
ec
iprocal of
th
e
mid
span deflection
du
e to a
mid
span concentrated
unit load. Thus
k = 48£1
cq
LJ
1.9
Th
e s
pring
s
of
Fi
g.
1·

11
are
sa
id to be in parallel.
Deriv
e an
equation
for the
eq
uivale
nt
stiffness
of
the
par
a
ll
el
combination
of
springs if
the
system
of
Fig. 1-11
is
to be modeled
by
th
e

equivalent
system
of
Fig. 1·1.
Fig
.
I·ll
8
MECHANICAL SYSTEM ANALYSIS [CHAP. 1
If
the block is subject to
an
arbitrary displacement x, the change
in
length
of
each spring
in
the parallel combination
is
x. The free body diagram of Fig. 1-12 shows that the total force acting
on the block
is
F = k ,x + k,x + k,x +

+ knx =
(i
kl
)x
,-,

(l
.
12)
• k"x
Fig.
]·12
The system
of
Fig.
1-1
can
be
us
ed
to
model the system
of
Fig.
1-11
if
the force acting on the block
of Fig.
1-1
is
equal to the force of Eq. (1.12) when the spring has a displacement
x.
If
the spring of
Fi
g.

1-1 has a displacement x, then the force act
in
g on the block of Fig.
1-1
is
F = kc
qX
(1.13)
Then for the forces from Eqs. (1.12) and (1.13) to be equal:
1.10
The
sp
ring
s in
the
system
of
Fig.
1·13
ar
e said
to
be
ill series.
Derive
an
eq
ua
tion
for

the
series
combination
of
springs if
the
s
ystem
of
Fig. 1-13 is
to
be
mod
el
ed
by
the
equivalent
sys
tem
of
Fig. 1-1.
k,
k,
. k,
J\~"I\~
~
-
vvvLJ
Fig. ]-13

Let
x be the displacement of the block of
Fig.
1-13 at an arbitrary
in
stant. Let x, be the
change in length
of
the ith
sp
rin
g from the fixed support. If each spring
is
assumed massles
s,
then
the force developed
at
each end of the spring has the same magnitude but oppos
it
e
in
direction, as
shown
in
Fig. 1-14. Thus the force is the same
in
each spring:
In addition,
Solving for

x,
fr
om Eq. (1.14) and substituting into Eq. (1.15) leads
to
F=_x_
±.!.
i _ I k;
(1.14)
(J.J5)
(1.16)
CHAP
.I
J
ME
C
HANI
C
AL
SYST
EM A N
ALY
SIS
9
Not
in
g that the force acting on the
bl
oc
k of the system
of

Fi
g.
1-1 f
or
an arbitrary x is k
.,
x and
equating this to the force from Eq. (1.16) l
eads
to
k
=_1-
cq " 1
L -
,. , k,
Fig. 1-14
1_11
Model
th
e s
ys
tem
sho
wn
in
Fi
g. 1-15
by
a
bl

o
ck
a
tt
ac
hed to a single s
prin
g o f a n
equival
e
nt
s
ti
ff
n
ess
.
Fig.
1-15
Th
e first st
ep
is to replace the para
ll
el combina
ti
ons by springs of equ
iva
lent s
ti

lfnesses using
the results of
Pr
oblem 1.9.
Th
e result is shown
in
Fi
g.
1-16a.
Th
e springs on the le
ft
of
the block are
in s
er
ies with one anothe
r.
Th
e result of Problem
1.1
0 is used to r
ep
lace these springs by a spring
whose s
tiffn
ess
is
calculated

as
k
1 I 1 I
:2
:ik
+
:ik+:k+:ik
Th
e sprin
gs
att
ached
to
the
ri
g
ht
of
the
bl
ock are in
se
ri
es
and are replaced by a spring
of
sti lfness
1 I
-
+-

k 2k
2k
Th
e result is the sys
tem
of
Fi
g. l -16b.
Wh
en the block has an arbitrary
di
splacement x, the
displacements in each of the springs of Fig.
1-
16
b are the
sa
me, and the total force acting
on
the
block is the s
um
of the forces developed in the springs.
Thu
s these sprin
gs
behave as if they are in
par
a
ll

el and ca n be replaced
by
a spring
of
stilfness
k
2k
7k
- + - =-
2 3 6
as illus
tr
ated in Fig.
1-1
6c.
10
1.12
rf
+
sa.
Mathcad
MECHANICAL SYS
TEM
ANALYS
IS
[CHAP. I
(a)
~
k
~

2 3
m
(b)
(e)
Fig. 1-16
Model the torsional system of
Fi
g.
1-17 by a d
isk
attached to a torsional spring of an
equivalent stiffness.
~6ocm j
-
80cm
==1l
,1 120 lcm r1
~===
= =
~
I
=========tJ==1
~
.
A B
CD
E
AB
: Steel shaft wi
th

aluminum core
BC:
So
li
d
Sleel
shaft
'IA
.
B;;
20
mm
'UB:
40 mm
DE
:
So
li
d a
luminum
sh
af
t
'BC = 18 mm
G
,,
=
80X
10'~
m'

Fig. 1-17
The st
iff
ness of each of the sha
ft
s of F
ig.
1-17 are ca
lcul
ated as
'DE
;;
25
mm
, N
G., =
40X
I
O;;;;-
' G
~
[
(0.04
m
)·
-
(0.02m)·
1
(SOXIO'~)
k _

AS
"
"'Bn
_ m
A8.,
- L
AB
-
0.6
m
= 5.03 X 10'
N-
m
rad
' G
~(
002m
)'(
40
X
I
O'~)
N
k =
ABA
I
A8
A
I
= m =

1.68
X
10"
~
ABA
I
LAB
0.6 m rad
k _
'B
CG BC
Be - L ac
~(O.O
I
S
m)'(sox
10
'
;,)
N
-m
O.S
m
1.
65
X 10'
~
' G
~
(0.025 m

)'(
40 X 10'
~)
kD£=~=
m
Lo. 12 m
N-m
2.05 X
10'
~
CHAP.
11
MECHANICAL
SYSTEM ANALYSIS
11
The angle of
twi
st of the end of aluminum core of shaft
AB
is
the same as the angle of
twi
st of
the end of the steel shell of shaft
AB
. Also, the total torque
on
the end
of
shaft

AB
is
th
e sum
of
the resisting torque
in
the aluminum core and
th
e resisting torque
in
the steel shell. Hence the
aluminum core and st
ee
l shell
of
shaft A B behave as torsional springs
in
parallel with
an
equivalent
stiffness
of
N-m N-m N-m
kAB
= k
AB
• + k
AB
A<

=
5.03
x
10
'
7ad
+ 1.68 x 10'
7ad
= 5.20 x 1
0'
7ad
Th
e torques developed
in
shafts
AB
and
Be
are the same, and the angle of rotation of
th
e disk is
8
A8
+ e
BC
- Thus shafts
AB
and
Be
behave as torsional springs

in
series whose combination acts
in
parallel with shaft DE. Hence the equivalent stiffness
is
1 N-m
koq = -1

1-+
kD£
= 3.65 x 10'
7ad
-+-
k
A8
k8C
1.13
Derive
an
expression
for
the
equivalent
stiffness
of
the
system
of
Fig. 1-18
when

the
deflection
of
the
machine
is
used as
the
generalized
coordinat
e.
~
~

+-

2
Fig. 1-18
L
2
I
E,I
Consider a concentrated downward load
F,
applied
to
the midspan of the simply supported
beam leading to a midspan deflection
x.
A compressive force

kx
is
developed in the spring.
The
total downward force acting on the beam at its midspan
is
F,
- kx. As noted
in
Problem
1.
8,
the
midspan deflection
of
a simply supported beam due to a concentrated load at its midspan
is
Thus for the beam
of
Fig. 1-18,
which leads to
FL'
x=
48£1
L'
x =
(F,
-
kx)
48

EI
F,
x
=k
48EI
+u
The
equivalent stiffness
is
obtained by setting
F,
= I, leading to
12
1.14
MECHANICAL
SYSTEM
ANALYSIS
[CHA
P.
1
Using the resuits
of
Problem
1.
9, it is observed that the beam and the spring act as two sprin
gs
in
paralle
l.
What

is
the
equiyalent
stiffness
of
the
system
of
Fig.
1-19 using
th
e
displacement
of
the
blo
ck
as
the
generalized
coordinate?
E=2tO
x J
O'!::I.
m'
1=
1.5
x
IO
-

s
m"
2
.5
m
Fig. 1·19
The
de
fl
ec
ti
on of a fixed-free beam at its free end
due
to
a unit concentrated l
oad
at
it
s (ree
end is
L'
/(3£1).
Thus
the equivalent stiffness
of
th
e cantilever beam is
_
3£
1 3(2

10
X
10
'
;')(1.5
X
10
-'
m
')
, N
kb - U (2.5 m
)'
6.05 X
10
;;;
The
anal
ys
is
of
Problem
1.1
3 suggests that the b
ea
m and the upper spring act
in
parallel. This
parallel combination is in series with the spring placed between the beam and the block. This se
ri

es
combination is in
par
allel with the spring between
th
e block and the fixed surface. Thus us
in
g the
formulas for parallel and series combinations, the
equ
ivalent stiffness is calculated
as
N N
k
••
= 1 1
+3X
IO' ;;; = 4.69 X
10
' ;;;
- -
"
- -
+
6.05X10'~+5xlO'~
2
X
10
'
~

m m m
1.15
The
viscous
da
mp
er s
hown
in
Fig
. 1-20
contain
s a
reser
vo
ir
of
a
vi
s
cous
fluid
of
vi
scos
ity
J.L
and
depth
h. A

plate
slid
es
ove
r
th
e s
urf
ace
of
th
e
reservoir
with
an
area
of
contact
A.
Wh
at
is
th
e
damping
coefficie
nt
for
this
vi

sco
us
damper?
Fig. 1-20
CHA
P.
1]
M
EC
H
AN
I
CAL
SYSTEM
ANALYSIS
13
Let y be a coordinate
up
into
th
e fluid, m
eas
ur
ed
from
th
e bottom
of
th
e reservoir.

If
h is
sma
ll
and
unsteady effects
are
n
eg
lected,
th
e
ve
loc
it
y profile
u(y)
in
the fluid is lin
ea
r with
u(O)
= 0
and
u(h)
=
u,
as shown in Fig. 1·21 where u is the velocity of the plate.
The
math

ema
tical
fo
rm of
the
ve
locity profile is
u(y)=u~
.
The shear s
tr
ess acting on
th
e s
ur
face
of
the plate is calculated us
in
g Newton's viscosity law,
du
le
ading to
r =
I-'
d;
I-'u
r=-
h
Th

e total viscous force
is
th
e
re
sultant
of
th
e shear stress distribution
F=rA
= I-'A
u
h
The
co
ns
tant
of
pr
oporti
ona
lity b
etween
the force and the
plate
ve
locity is the d
amp
in
g coefficient

I-'A
c=-
h
Plate
or
area A
II
.;_~,
I~
v
p,
J
P"~y)
=
~
h
Fig. 1·21
1.
16
Th
e
tor
sional viscous damper
of
Fig. 1·22 consists
of
a thin disk attached to a rotating
shaft.
The
face of the disk has a radius R and

rotat
es
in
a dish
of
fluid of depth
hand
viscosity
1-'
. Determine the torsional viscous damping coefficient for this damper.
OJ
Fig. 1·22
14
ME
C
HANICAL
SYSTEM
ANALYSIS
[C
HAP
. 1
Let r be the s
he
ar s
tr
ess acting
on
the differential
area
dA

= r
dr
d8
on
the
surfa
ce
of the dis
k,
as illustrated in Fig. 1-
23.
Th
e resullant
moment
about the axis
of
rotation
due to the shear s
tr
ess
di
s
tribution
is
M = r J
rr(rdrd8)
(l.17)
o 0
If
w is the angular velocity of the shaft and disk,

then
the velocity
of
the differential
element
is rw.
Let y be a c
oo
rdinate, measured
upward
into the fluid from
the
bOllom
of
the
dis
h.
Negl
ec
ting
unsteady e
ff
ec
ts
and
assuming the de
pth
of
the fluid is small, the velocity distribution
u(r,

y)
in the
fluid
is
approximately linear in y with u(r, 0) = 0
and
u(r
,
h)
= rw, leading to
Th
e sh
ear
stress acting
on
the fact
of
the
disk is calculated from
Newton
's
vi
scosity law,
au
rJ Lw
t =
JJ.
ay
(r,
h)

=-
h-
which,
wh
en substituted into Eq. (1.17), l
ea
ds to
Th
e torsional
dampin
g coefficient
is
the constant of
prop
ortionality
between
the
mom
ent and the
angular
ve
locity,
JJ."R'
c,
=
z;;-
Fig. 1-23
dM
=
rrdA

R
CHAP.
1]
MECHANICAL
SYSTEM
ANALYSIS
15
1.17
Show
that
the
inertia
effects
of
a
linear
spring
connecting
a fixed
support
and
a
1-degree-of-freedom
system
can
be
approximated
by
placing
a

particle
of
mass
equal
to
one-third
the
mass
of
the
spring
at
the
system
location
where
the
spring
is
attached.
Let m, be the mass
of
a uniform spacing
of
unstretched length e that
is
connected
between
a
fixed

support
and
a particle
in
a I-degree-of-freedom system whose displacement
is
given by x(c).
Let m
o
, be
the
mass
of
a particle placed at
the
end
of
the spring. This particle can
be
used to
approximate the inertia effects
of
the spring
if
the kinetic energy
of
the spring
is
T =
~m"q.f

Let z be a coordinate along the axis
of
the spring
in
its
unstretched position,
0::::;
z
:5
e,
as
illustrated in Fig. 1-24. Assume
the
displacement function u(
z,
c)
is
linear along the length of the
spring at any instant with
u(O) = 0 and
u(t)
=
x:
x
u(z,
t)
=
eZ
The
kinetic energy

of
a differential spring
element
is
dT=H~),
dm
=H?z)'9dZ
from which the total kinetic energy
of
the spring
is
calculated
T = f
dT
=
i;;"
f
z'
dz
=
~
'?
i'
o
Thus
if
a particle
of
ma
ss

m,/3
is
placed
at
the location on the system where the spring
is
attached,
it
s
kin
etic energy
is
the sa
me
as that
of
th
e
lin
ear spring assuming a linear displacement function.
~l ~
f
x
(0)
. x
u(t)=x
u(O)~
(b)
Fig. 1-24
1.18

What
is
the
mass
of
a
particle
that
should
be
added
to
the
block
of
the
system
of
Fig.
1-25
to
approximate
the
inertia
effects
of
the
series
combination
of

springs?
Fig. 1-25
16
1.19
r{.
sa
Mathcad
MECHANICAL
SYSTEM
ANALYSIS
[CHAP
. I
Using the results
of
Pr
ob
lem 1.17,
th
e inertia effects
of
the l
ef
t
sp
ring can
be
ap
pro
ximated by
placing a particle

of
mass m
.l
3 at the junc
ti
on betw
ee
n
th
e two springs. Let x
be
th
e displacement
of
the block at an arbitrary instant. Let
z,
a
nd
z,
be
coordina
tes al
ong
the axes
of
the left and
ri
ght
springs, respectively. Let u
,(z"

I) and u,
(z"
I)
be
the
displacement functions for these spring
s.
It is
known that
u,(O,
I) =
0,
u,
(f"
I) =
x.
Also, u,(f" I) =
u,(O,
I) =
IV.
Assum
ing lin
ea
r displacemer.t
functions for
eac
h spring, this leads
to
IV
u

,(z"
I) =
7.
z,
x -
IV
u,(z
" I) = T
z,
+ w
Since the springs are in se
ri
es, the forces in
ea
ch spring are the sam
e:
klV
=
2k(x
- w) > w = Ix
Thu
s the kinet
ic
ene
rgy
of
the se
ri
es combination is
T

-
I
(",
,
)(2
.)'
1 J'
(1
Z,
2)'.,
m. d

-
-x
+-
+-
x - z
2 3 3
2"
3 f, 3 2(, ,
w
hi
ch leads
to
an
added
particle ma
ss
of m./2.
Use

th
e sta
tic
deflection
functi
on
of
th
e s
impl
y s
upported
b
ea
m
to
d
ete
rmin
e
th
e m
ass
of
a
particl
e
that
s
hould

be
a
ttach
ed
to th
e
block
of
th
e syst
em
of
Fig.
1-
26
to
approxima
te
in
e
rtia
effec
ts
of
the
b
ea
m.
L 2L
+

-

'3

1
Fig. 1-26
£.1
Th
e static deflection
y(z)
of
a simply s
upport
ed
b
ea
m
due
to
a
concentrated
load F applied at
z =
LI3
is
Th
e force required
to
cause a static deflection x at z =
L/3

is calc
ul
ated
as
x =
(~)
=
4FL
' > F = 243
Elx
y 3
243£
1
4L
'
C
HAP.
IJ
MECHAN
I
CA
L
SYSTEM
ANALYSI
S
17
The kinetic ener
gy
of
th

e beam is
T = J
~
j'PA
dz
,
= O.586pAL = O.586m,
Hence the inertia effects of the beam are approximated
by
adding a particle of mass 0.586m,
to
the
m
ac
hin
e.
1.20
An
a
pproximati
on to
the
de
fl
e
ction
of
a
fix
ed-fixed

beam
due
to
a
conce
ntr
at
ed
load at its
mids
pan
is
x (
27r
Z)
y(z)
=2
I-cosT
where x is the mids
pan
deflection. Use this
approximat
ion to
determine
the
ma
ss
of
a
particle to be

pl
aced
at the
midspan
of
a b
ea
m to
app
roxim
ate
the
beam's
inertia
effects.
Th
e k
in
e
ti
c energy
of
th
e beam
is
L
f
l .
T =
2:

y'(z)pA
dz
"
where p
is
the bea
m·
s
ma
ss
dens
ity
and A is
it
s cro
ss
-sectional ar
ea.
Substituting the suggested
approxima
ti
on
,
I
x'
f'
· (
21rZ)
'
T =

2:
pA
"4
1 - cos L dz
"
=!(~PAL)x'=
!(
~
m
)x
'
2 8
28
",,
·m
The inertia e
ffect
s of
th
e beam can .be approximated
by
adding a particle of mass 3/8m
"".m
at its
mid
span.
1_21
Let
x be the displ
acemen

t of the block
of
Fig. 1-27,
measured
positive
downw
ard
from
the sys
tem'
s equi
li
brium
po
sition.
Show
that
the sys
tem
's diffe
rence
in
potential
ene
rgies
betw
ee
n two
arbitrary
pos

itions
is
independent
of
the mass
of
the
block.
cbl
x
Fig. 1-27
18
MECHANICAL
SYSTEM
ANALYSIS
[C
HAP.
1
Let
x be
the
downward displacement
of
the bl
oc
k from the system's equilibrium position.
When the system
is
in equi
li

br
ium,
the
spring has a static deflection
/1
= mg/k.
If
the datum for the
potential energy
due
to gravity
is
taken as the system's equilibrium position,
the
potential energy of
the system
at
an arbitrary
in
stant
is
V = l
k(x
+
/1)
' - mgx
= l
kx
' + (k/1 - mg)x +
lk/1'

=
~kx2
+
~
kA
2
Thus the difference
in
potential energies as the block
moves
between
XI
and X
2
is
v - V
=!kx
'+!
/1
'-
!kx
'-!k/1'
, , 2 ' k
2'
2
=
!k(x
,-x
' )
2 ' ,

which is
independent
of
the mass
of
the
block.
Th
e results
of
this problem
are
used to infer that
the
static de
fl
ection
of
a spring and the gravity force causing the static defl
ec
tion ca
nc
el with
one
another
in
the potential
energy
difference.
1.22

Determine
m
eq
and
k
eq
for
the
system
of
Fig. 1
-2
8
whe
n x,
the
downward
displacement
of
the
block
,
measured
from
the
system's
equi
librium
po
sition, is u

se
d
as
th
e
ge
nera
li
ze
d
coordinate.
2k
J
Fig. 1-28
From
the results
of
Problem
1.21
, it is evident that
the
effects
of
gravity
and
stat
ic de
fl
ections
cancel in potential energy calculations

and
can thus
be
ignored.
Th
e potential energy
of
th
e system
is
v = l
kx'
+ l(2k
)x
' = l(3k )x'
+
k
.q
= 3k
Th
e kinetic energy
of
the
system is
T = !
mi
'
+!
l
(~)

'
=
!(m
+ !")i '
+
+ I
2 2 r 2 ,2
m
~q
-;:::
m
~