PROBLEM 19.116 (Continued)
Out of phase motion with | | 0.06 in.
m
x=
(
)
()
2
2
0.00125
0.06
1
f
n
f
n
ω
ω
ω
ω
<
−
f
f
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
Before the plate is added, 3
1180 kg, 28 10 kg
150 mm 0.150 m
Mm
r
−
==×
==
Equivalent spring constant: 11
WMg
k
δδ
==
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
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
Mass of motor. 2
400 12.422 lb s /ft
32.2
M== ⋅
Unbalance mass. 2
31 0.005823 lb s /ft
16 32.2
m⎛⎞⎛ ⎞
==⋅
⎜⎟⎜ ⎟
⎝⎠⎝ ⎠
Eccentricity. 6 in. 0.5 ftr
=
=
Equation of motion: 2
sin sin
mf f f
M
xkxP t mr t
ω
ωω
+= =
22
2
2
()
fm f
f
m
f
Mkxmr
mr
xkM
ωω
ω
ω
−+=
=−
Transmitted force.
2
2
f
mm
f
kmr
Fkx kM
ω
ω
==
−
For out of phase motion,
2
2
||
f
m
f
kmr
F
M
k
ω
ω
=
−
(1)
(a) Required value of k.
Solve Eq. (1) for k. 22
||( )
mf f
F
Mkkmr
ω
ω
−=
22
(||)||
f
mmf
kmr F F M
ω
ω
+=
2
2
||
||
mf
f
m
FM
kmr F
ω
ω
=+
Data: ||0.2lb
m
F= 100 rpm 10.472 rad/s
f
ω
==
2
2
(0.2)(12.422)(10.472) 524.65
(0.005823)(0.5)(10.472) 0.2
k==
+ 525 lb/ft
k
=
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 19.118 (Continued)
(b) Force amplitude at 200 rpm. 20.944 rad/s
f
ω
=
From Eq. (1),
2
2
(524.65)(0.005823)(0.5)(20.944)
|| (12.422)(20.944) 524.65
m
F=−
| | 0.1361 lb
m
F=
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
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
n
mr
k
mfm n
k
Pmr x
M
ω
ω
ω
ωω
== =
−
With 2
2
2
2
2 160 N , 40 rad/s
1f
f
mf
M
k
mr
kx
ω
ωωπ
==± =
−
Solving for r,
()
()
2
1
300 kg 40 s
1200000 N/m
12
160 N 1
0.1493 m
2(0.1 kg)(40 s )
r
π
π
−
−
⎡⎤
⎢⎥
−
⎢⎥
⎣⎦
=± =
149.3 mm
r
=
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
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
From Problem 19.113
()
(
)
()
2
2
1
f
n
f
n
m
M
m
r
x
ω
ω
ω
ω
=
−
And
()
2
2
2
() , , ()
1f
n
f
Tm m n Tm
rm
k
Fkx F
M
ω
ω
ω
ω
== =
⎡
⎤
−
⎢
⎥
⎣
⎦
Then
(
)
0.9
16 2
2
lb
7.5 ft 0.0010918 lb s
12 32.2 ft/s
rm
⎛⎞
⎛⎞
⎜⎟
=
=⋅
⎜⎟
⎜⎟
⎝⎠
⎝⎠
2
22
360 lb
32.2 ft/s
12500 lb/ft 1118.1 s
n
k
M
ω
−
== =
2
2
2
1118.1
( ) (0.0010918 lb s )
1
f
f
Tm
F
ω
ω
=⋅
−
or
2
22
( ) 1 (0.0010918 lb s )
1118.1
f
Tm f
F
ω
ω
⎡⎤
−= ⋅
⎢⎥
⎢⎥
⎣⎦
()
2
()
( ) 0.0010918
1118.1
Tm
Tm f
F
F
ω
⎡
⎤
=+
⎢
⎥
⎣
⎦
Then 21118.1( )
( ) 1.2207
Tm
f
Tm
F
F
ω
=+
(a)
(
)
22
1118.1 5
( ) 5: 898.69 s ,
5 1.2207
Tm f
F
ω
−
=+ = =
+
29.978 rad/s
f
ω
≤
286.26 rpm
≤
286 rpm
f
ω
≤
(b) 22
1118.1 ( 5)
( ) 5: 1479.2 s ,
51.2207
Tm f
F
ω
−
−
=− = =
−+
38.461 rad/s
f
ω
>
367.27 rpm> 367 rpm
f
ω
>
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
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
(1) From Equation (19.33):
()
2
1
m
f
n
P
k
m
x
ω
ω
=
−
Force transmitted:
()
2
()
1
m
f
n
P
k
Tm m
Pkxk
ω
ω
⎡
⎤
⎢
⎥
==
⎢
⎥
−
⎢
⎥
⎣
⎦
Thus, Transmissibility
()
2
() 1
1f
n
Tm
m
P
P
ω
ω
==
−
(2) From Equation
(19.33 ):
′
Displacement transmitted:
()
2
1f
n
m
m
x
ω
ω
δ
=
−
Transmissibility
()
2
1
1f
n
m
m
x
ω
ω
δ
==
−
For () or to be less than 1,
Tm m
mm
Px
P
δ
()
2
11
1f
n
ω
ω
<
−
(
)
2
11 f
n
ω
ω
<−
2
2
f
n
ω
ω
⎛⎞
>
⎜⎟
⎝⎠ 2 Q.E.D.
f
n
ω
ω
>
S
O
O
LUTION
P
R
A
es
s
na
t
w
h
a
m
a
m
de
t
is
6
1
si
n
relat
i
δ
m
m
x
y
z
z
x
y
ω
ω
δ
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
2
2
2
2
sin
n
i
ve motion
1
f
n
f
n
m
f
f
m
t
t
y
ω
ω
ω
ω
ω
ω
δ
δ
⎞
⎟
⎟
⎠
⎡
−
⎢
=−
⎢
⎣
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
sin
mf
t
δ
ω
⎤
⎥
⎥
⎦
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,
.
f
t
ω
If the
o
x is used as
the surface,
he vibration
7
SO
LUTION
P
R
A
m
a
T
h
to
th
e
m
e
su
r
vi
b
m
x
y
z
z
z
=
=
=
=
=
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
2
2
2
2
sin
1
sin
relative moti
o
1
1
1
1
f
n
f
n
m
mf
m
m
t
xy
ω
ω
ω
ω
ω
ω
δ
ω
δω
δ
δ
⎛⎞
⎜⎟
−
⎜⎟
⎝⎠
⎡
⎢
−=
−
⎢
⎣
⎡
−
⎢
−
⎢
⎣
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.
2
2
2
2
2
2
o
n
sin
1
1
f
n
f
n
f
n
f
m
f
m
t
ω
ω
ω
ω
ω
ω
ω
δ
ω
δ
⎤
−⎥
⎥
⎦
⎤
⎥=
⎥−
⎦
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
f
t
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
e
e
t
t
n
n
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
o
o
o
s
i
o
o
o
i
k
n
d
k
d
n
m
e
n
e
s
s
l
p