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PROBLEM 19.116 (Continued)
PROBLEM 19.117
A 180-kg motor is bolted to a light horizontal beam. The unbalance of
its rotor is equivalent to a 28-g mass located 150 mm from the axis of
rotation, and the static deflection of the beam due to the weight of the
motor is 12 mm. The amplitude of the vibration due to the unbalance
can be decreased by adding a plate to the base of the motor. If the
amplitude of vibration is to be less than 60
μ
m for motor speeds above
300 rpm, determine the required mass of the plate.
SOLUTION
−
PROBLEM 19.118
The unbalance of the rotor of a 400-lb motor is equivalent to a 3-oz weight
located 6 in. from the axis of rotation. In order to limit to 0.2 lb the
amplitude of the fluctuating force exerted on the foundation when the motor
is run at speeds of 100 rpm and above, a pad is to be placed between the
motor and the foundation. Determine (a) the maximum allowable spring
constant k of the pad, (b) the corresponding amplitude of the fluctuating
force exerted on the foundation when the motor is run at 200 rpm.
SOLUTION
400 12.422 lb s /ft
2
(0.005823)(0.5)(10.472) 0.2
+ 525 lb/ftk
PROBLEM 19.118 (Continued)
m
PROBLEM 19.119
A counter-rotating eccentric mass exciter consisting of two rotating 100-g masses
describing circles of radius r at the same speed but in opposite senses is placed on a
machine element to induce a steady-state vibration of the element. The total mass of
the system is 300 kg, the constant of each spring is k = 600 kN/m, and the rotational
speed of the exciter is 1200 rpm. Knowing that the amplitude of the total fluctuating
force exerted on the foundation is 160 N, determine the radius r.
SOLUTION
2
2
22
2
2
2
2, ,
1
f
f
mr
k
mfm n
k
Pmr x
M
ω
ω
ωω
== =
−
PROBLEM 19.120
A 360-lb motor is supported by springs of total constant 12.5 kips/ft. The unbalance of the rotor is equivalent
to a 0.9-oz weight located 7.5 in. from the axis of rotation. Determine the range of speeds of the motor for
which the amplitude of the fluctuating force exerted on the foundation is less than 5 lb.
SOLUTION
2
f
m
ω
f
PROBLEM 19.121
Figures (1) and (2) show how springs can be used to support a
block in two different situations. In Figure (1), they help
decrease the amplitude of the fluctuating force transmitted by
the block to the foundation. In Figure (2), they help decrease
the amplitude of the fluctuating displacement transmitted by the
foundation to the block. The ratio of the transmitted force to
the impressed force or the ratio of the transmitted displacement
to the impressed displacement is called the transmissibility.
Derive an equation for the transmissibility for each situation.
Give your answer in terms of the ratio /
f
n
ω
ω
of the frequency
ω
f of the impressed force or impressed displacement to the
natural frequency n
ω
of the spring-mass system. Show that in
order to cause any reduction in transmissibility, the ratio /
f
n
ω
ω
must be greater than 2.
SOLUTION
m
P
k
n
ω
⎝⎠ 2 Q.E.D.
n
ω
S
O
O
LUTION
P
R
A
es
s
na
t
w
h
a
m
a
m
de
t
is
6
m
δ
⎛
R
OBLEM
1
vibrometer u
s
s
entially of a
t
ural frequenc
y
h
ich is movi
n
m
plitude
m
z
o
f
m
easure of t
h
t
ermine (a) t
h
6
00 Hz, (b) t
h
m
⎞
1
9.122
s
ed to measu
r
box containi
n
y
of 120 Hz.
T
n
g according
t
f
the motion
o
h
e amplitude
h
e percent err
o
h
e frequency a
r
e the amplit
u
n
g a mass-spr
i
T
he box is ri
g
t
o the equati
o
o
f the mass rel
m
δ
of the
v
o
r when the
fr
t which the e
r
u
de of vibrati
o
i
ng system w
i
g
idly attached
o
n
sin
m
y
δ
=
ative to the b
o
v
ibration of
fr
equency of t
r
ror is zero.
o
ns consists
i
th a known
to a surface,
.
ft
ω
If the
o
x is used as
the surface,
he vibration
P
R
A
m
a
T
h
to
th
e
m
e
su
r
vi
b
R
OBLEM
1
certain accel
e
a
ss-spring sy
s
h
e box is rigi
d
the equation
e
mass relati
v
e
asure of the
m
r
face, deter
m
b
ration is 600
1
9.123
e
rometer con
s
s
tem with a
k
d
ly attached t
o
sin
mf
y
t
δ
ω
=
v
e to the box
m
aximum ac
c
m
ine the perc
e
Hz.
s
ists essential
l
k
nown natura
l
o
a surface, w
h
.
t
If the ampli
times a scal
e
c
eleration m
α
e
nt error wh
e
l
y of a box
c
l
frequency
o
h
ich is movin
g
tude
m
z
of t
h
e
factor
2
n
ω
i
2
mf
δω
= of t
h
e
n the frequ
e
c
ontaining a
o
f 2200 Hz.
g
according
h
e motion of
s used as a
h
e vibrating
e
ncy of the
S
O
O
LUTION
PROBLE
M
Block A ca
n
vertical peri
o
20 N.
m
P=
A
22-kg block
steady-state
vibration of
M
19.124
n
move witho
u
o
dic force of
A
spring of c
o
B. Determin
e
vibration of
block B.
u
t friction in t
h
magnitude
P
o
nstant k is at
t
e
(a) the valu
e
block A, (b
)
h
e slot as sho
w
sin ,
mf
Pt
ω
=
t
ached to the
b
e
of the const
a
)
the corresp
w
n and is act
e
where
f
ω
=
b
ottom of blo
c
a
nt k which
w
onding ampl
i
e
d upon by a
2
rad/s and
c
k A and to a
w
ill prevent a
i
tude of the
PR
A
6
vert
i
spri
n
(a)
t
the
s
OBLEM 1
9
6
0-lb disk is
a
i
cal shaft AB,
w
n
g constant
k
t
he angular v
e
s
haft when
f
ω
9
.125
a
ttached with
w
hich revolv
e
k
for horizont
a
e
locity
f
ω
at
1200 rpm.
f
=
an eccentric
i
e
s at a constan
t
a
l movement
which reson
a
i
ty
0.006
i
e=
t
angular velo
c
of the disk i
s
a
nce will occ
u
i
n.
to the mi
c
ity
.
f
ω
Kno
w
s
40,000 lb/f
t
u
r, (b) the de
f
dpoint of a
w
ing that the
t
, determine
f
lection r of
PROB
LEM 19.12
5
5
(Continu
e
e
d)
SO
LUTION
PRO
B
A sma
l
suppor
t
a road
,
an am
p
b
etwe
e
trough
occur,
50 km
/
B
LEM 19.1
l
l trailer and
t
ed by two s
p
,
the surface
o
p
litude of 40
e
n successive
is 80 mm).
(b) the ampl
/
h.
26
its load have
p
rings, each o
f
o
f which can
b
mm and a
w
crests is 5 m
Determine (
a
itude of the
a total mass
f
constant 10
k
b
e approxim
a
w
avelength o
f
and the verti
c
a
) the speed
a
vibration of
t
of 250-kg. T
h
k
N/m, and is
a
ted by a sine
f
5 m (i.e., t
h
c
al distance f
r
a
t which res
o
t
he trailer at
h
e trailer is
pulled over
curve with
h
e distance
r
om crest to
o
nance will
a speed of
P
R
Sh
o
(a)
ar
b
S
O
R
OBLEM 1
o
w that in th
e
if it is relea
s
b
itrary initial
v
O
LUTION
9.127
e
case of hea
v
s
ed with no i
n
v
elocity.
v
y damping
(
c
n
itial velocit
y
),
c
c
c>
a bod
y
from an arb
i
y never pass
e
i
trary positio
n
e
s through its
n
or (b) if it
i
position of e
q
i
s started fro
m
q
uilibrium O
m
O with an
PROB
L
L
EM 19.127
(Continue
d)
P
R
Sh
o
ini
t
S
O
R
OBLEM 1
o
w that in the
t
ial velocity c
a
O
LUTION
9.128
case of heav
y
a
nnot pass m
o
y
damping
(c
o
re than once
t
),
c
c>
a bod
y
t
hrough its eq
u
y
released fro
m
u
ilibrium pos
i
m
an arbitrary
i
tion.
position wit
h
h
an arbitrary
PROBLEM 19.129
In the case of light damping, the displacements 1,
x
2,
x
3,
x
shown in Figure 19.11 may be assumed equal to
the maximum displacements. Show that the ratio of any two successive maximum displacements n
x
and 1n
x
+
is a constant and that the natural logarithm of this ratio, called the logarithmic decrement, is
2
1
2(/ )
ln
1(/)
nc
nc
xcc
xcc
π
+
=−
1
1
nc
c
xmc c
c
+
⎛⎞
−⎜⎟
⎝⎠
()
2
1
1
c
nc
c
x
+
−
PROBLEM 19.130
In practice, it is often difficult to determine the logarithmic decrement of a system with light damping defined
in Problem 19.129 by measuring two successive maximum displacements. Show that the logarithmic
decrement can also be expressed as (1/ ) ln ( / ),
nnk
kxx
+ where k is the number of cycles between readings of
the maximum displacement.
nk
kx
+
P
R
In
a
d
τ
=
lim
i
dis
p
dis
p
gre
a
SO
R
OBLEM 1
9
a
n underdam
p
2/
d
πω
=
corr
i
ting curves
s
p
lacement an
d
p
lacements is
a
ter than
1
4.
d
τ
LUTION
9
.131
p
ed system
(
e
sponding to
s
hown in Fig
u
d
the followi
n
1
2,
d
τ
(c)
b
et
w
(
),
c
cc<
the
p
two successi
v
u
re 19.11. S
h
n
g maximum
w
een a maxi
m
p
eriod of vi
b
v
e points wh
e
h
ow that the
negative dis
p
m
um positive
d
b
ration is co
m
e
re the displa
c
interval of ti
m
p
lacement is
1
2
d
isplacement
a
m
monly defi
n
c
ement-time
c
m
e (a)
b
etw
e
2
,
d
τ
(b)
b
et
w
a
nd the follo
w
n
ed as the ti
m
c
urve touche
s
e
en a maxim
u
w
een two suc
c
w
ing zero dis
p
m
e interval
s
one of the
u
m positive
c
essive zero
p
lacement is
PROB
LEM 19.13
1
1
(Continu
e
e
d)
