978-0073398242 Chapter 19 Solution Manual Part 19

subject Type Homework Help
subject Pages 9
subject Words 1436
subject Authors Brian Self, David Mazurek, E. Johnston, Ferdinand Beer, Phillip Cornwell

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page-pf1
PROBLEM 19.132
A loaded railroad car weighing 30,000 lb is rolling at a constant velocity v0 when it couples with a spring and
dashpot bumper system (Figure 1). The recorded displacement-time curve of the loaded railroad car after
coupling is as shown (Figure 2). Determine (a) the damping constant, (b) the spring constant. (Hint: Use the
definition of logarithmic decrement given in Problem 19.129.)
SOLUTION
Mass of railroad car:
2
30,000
32.2
931.67 lb s /ft
W
mg
==
=⋅
The differential equation of motion for the system is
1,
x
() ()
1
2
2
1
1
2
c
m
cd
m
t
xe
xe
x
τ
=
page-pf2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 19.132 (Continued)
(a) Damping constant.
From Eq. (1): 1
2
ln
2
d
c
x
mx
τ
⎛⎞
=⎜⎟
⎝⎠
1
2
2ln
(2)(931.67) 0.5
ln
0.41 0.12
6485.9 lb s/ft
d
x
m
cx
τ
=
=
=⋅
6.49 kip s/ft
c=⋅
(b) Spring constant.
Equation for
:
d
ω
2
2
2
d
kc
mm
ω
⎛⎞
=−
⎜⎟
⎝⎠
2
2
2
2
3
4
(6485.9)
(931.67)(15.325) (4)(931.67)
230 10 lb/ft
d
c
km m
ω
=+
=+
230 kips/ftk=
page-pf3
PROBLEM 19.133
A torsional pendulum has a centroidal mass moment of inertia of 0.3 kg-m2 and when given an initial twist
and released is found to have a frequency of oscillation of 200 rpm. Knowing that when this pendulum is
immersed in oil and given the same initial condition it is found to have a frequency of oscillation of 180 rpm,
determine the damping constant for the oil.
SOLUTION
Let the mass be rotated through the small angle
θ
from the equilibrium position.
Couples acting on the mass: Shaft: K
θ
Oil:
C
θ
Equation of motion: :
M
IKCI
θ
θθθ
Σ= =
 

page-pf4
P
R
Th
e
co
n
b
a
r
b
a
r
S
O
(a)
(b)
R
OBLEM 1
e
barrel of a
f
n
stant
110
0
c=
r
rel into firin
g
r
rel to move b
O
LUTION
A critical
l
Then
For a criti
c
9.134
f
ield gun wei
g
0
lb s/ft.
Det
e
g
position in
t
ack two-third
s
l
y damped sys
c
ally damped
g
hs 1500 lb
a
e
rmine (a) the
t
he shortest
p
s
of the way f
r
tem regains i
t
system, Equa
t
a
nd is returne
d
constant k w
h
p
ossible time
w
r
om its maxi
m
t
s equilibrium
1100
2
c
cc
m
=
=
=
()
2
2
c
c
km
=
t
ion (19.43):
n
ω
d
into firing
p
h
ich should
b
w
ithout any
o
m
um-recoil po
position in th
e
k
m
()
2
2
2
1100
2
1500 lb
32.2 ft/s
6
==
n
t
p
osition after
r
b
e used for th
e
o
scillation, (b
)
sition to its fi
r
e
shortest tim
e
6
494 lb/ft
r
ecoil by a re
e
recuperator
t
)
the time ne
e
r
ing position.
e
.
6
k=
cuperato
r
of
t
o return the
e
ded for the
6
490 lb/ft
page-pf5
S
O
n
ω
O
LUTION
8ra
d
k
m
==
c
d
/s, c=
2
=
PROB
L
A 2-kg
b
dashpot
w
in equilib
the bloc
k
decreme
n
equilibri
u
232
N
n
m
ω
=
L
EM 19.13
5
b
lock is supp
o
w
ith a coeffici
e
rium when it
k
an upward
v
n
t, (b) the m
a
u
m after two c
y
N
s/m,
c
c
c
5
o
rted by a s
p
e
nt of viscous
is struck fro
m
v
elocity of 0.
a
ximum up
w
y
cles. (Hint:
S
0.01875=
p
ring of cons
t
damping
c
=
m
below by a
h
4 m/s. Deter
m
w
ard displace
m
S
ee Probs. 19.
t
ant k = 128
0.6 N s/m.
=
T
h
ammer whic
h
m
ine (a) the
m
ent of the
b
128 and 19.1
2
N/m and a
T
he block is
h
imparts to
logarithmic
b
lock from
2
9.)
page-pf6
S
O
Ve
Ve
Co
O
LUTION
locity of Blo
c
locity of Blo
c
nservation of
c
k A just befo
r
c
ks A and B i
m
momentum.
(
4
m
PROBL
E
A 4-kg bl
o
which is a
t
and is at
t
Knowing
t
b
locks wil
l
r
e impact.
A
v
m
mediately af
t
4
)(3.962) 0
A
ABB
m
vmv
+
=
+
=
E
M 19.136
o
ck A is drop
p
t
rest. Block
B
t
ached to a
t
hat there is
l
move after t
h
2
2(9.81)(0.
3.962 m/s
gh=
=
=
t
er impact.
()
(4 9)
AB
mmv
v
=
+
=
+
p
ed from a
h
B
is supported
dashpot of
no rebound,
h
e impact.
8)
h
eight of 800
by a spring
o
damping co
e
determine th
e
mm onto a 9
o
f constant
k
=
e
fficient
c=
e
maximum
d
-kg block B
1500 N/m
=
230 N s/m.
d
istance the
page-pf7
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 19.136 (Continued)
Expression for :
d
ω
2
2
2
8.846
12
2
1500 230
13 (2)(13)
6.094 rad/s
( sin 6.094 cos 6.094 )
d
d
t
kc
mm
x
eC tC t
ω
ω
⎛⎞
=−
⎜⎟
⎝⎠
⎛⎞
=−
⎜⎟
⎝⎠
=
=+
Initial conditions: 0
0
0
012
2
( 8.846)0
1
0.02619 m
( 0) 1.219 m/s
0.02619 [ (0) (1)]
0.02619
(0) 8.846 [ (0) ( 0.02619)(1)]
x
tx
xeCC
C
xeC
=−
==+
=− = +
=−
=− +
'
(8.846)(0)
12
[6.094 (1) (0)] 1.219eCC
++=
1
1
8.846
1.219 ( 8.846)( 0.02619) 6.094
0.16202
(0.16202sin 6.094 0.02619 cos 6.094 )
t
C
C
x
ett
=− +
=
=−
Maximum deflection occurs when 0x
=
8.846
8.846
0 8.846 (0.16202 sin 6.094 0.02619 cos 6.094 )
[6.094][0.1620cos 6.094 0.02619sin 6.094 ]
0 [( 8.846)(0.16202) (6.094)(0.02619)]sin 6.094
[( 8.846)( 0.02619) (6.094)(0.1620)]cos 6.
m
m
t
mm
t
mm
m
x
ett
ett
t
==
++
=− +
+− +
094
0 1.274 sin 6.094 1.219 cos 6.094
1.219
tan 6.094 0.957
1.274
m
mm
m
t
tt
t
=− +
==
1
(8.846)(0.1253)
tan 0.957
Time at maximum deflection 0.1253 s
6.094
[0.1620sin(6.094)(0.1253)
0.02619 cos(6.094)(0.1253)]
(0.3301)(0.1120 0.0189) 0.0307 m
m
m
m
t
xe
x
== =
=
=−=
Blocks move, static deflection m
x
+ Total distance 0.02619 0.0307 0.0569 m 56.9 mm
=
+= =
page-pf8
S
O
Fr
o
O
LUTION
o
m the given
d
PRO
B
A 0.9-
k
shown
coeffic
i
connec
t
spring
d
d
at
a
n
ω
=
Stre
t
B
LEM 19.1
3
k
g block B is
from two sp
r
i
ent
7.5
N
c=
t
ing A and B
d
uring the res
u
215
0
k
m=
t
ch at equilibr
i
3
7
connected b
y
r
ings, each o
f
N
s/m.
Kno
w
is cut, deter
m
u
lting motion
.
0
rad/s,
c
c
(
(
2.4 k
g
i
um 2
=
y
a cord to a
f
constant
k
=
w
ing that th
e
m
ine the mi
n
.
2
5
c
n
m
ω
==
)
()
)
2
g
9.81 m/s
180 N/m
2
2.4-kg block
180 N/m
=
, a
n
e
system is
n
imum tensio
n
5
8.7878 N s/
m
0.0654 m =
A which is s
u
n
d a dashpot
at rest wh
e
n
that will o
c
m
u
spended as
of damping
e
n the cord
c
cur in each
page-pf9
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 19.138
A 0.9-kg block B is connected by a cord to a 2.4-kg block A which is suspended as
shown from two springs, each of constant k = 180 N/m, and a dashpot of damping
coefficient 60 N s/m.c
=
Knowing that the system is at rest when the cord connecting
A and B is cut, determine the velocity of block A after 0.1 s.
SOLUTION
From the given data
2150 rad/s
n
k
m
ω
==
2 58.7878 N s/m
cn
cm
ω
=
=⋅
(
)
(
)
()
2
2.4 kg 9.81 m/s
Stretch at equilibrium 0.0654 m
2180N/m
==
(
)
(
)
()
2
3.3 kg 9.81 m/s
Initial stretch 0.089925 m
2180N/m
==
12
12
:tt
c
cc xce ce
λ
λ
>=+
So
(
)
(
)
2
2.4 kg 60 N s/m 360 0
λλ
+
⋅+=
() ()( )( )
()
2
60 60 4 360 2.4
22.4
λ
−±
=
12
10, 15
λλ
=
−=
12
12
11 2 2
, 0.024524 m solve for ,
0
tx cc cc
xcc
λλ
== =+
== +
12
0.073575; 0.04905cc==
12
0.1 0.1
11 2 2
xce ce
λ
λ
λλ
=+
0.1065 m/sx
=
106.5 mm/sv=
page-pfa
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 19.139
A machine element weighing 800 lb is supported by two springs, each having a
constant of 200 lb/in. A periodic force of maximum value 30 lb is applied to the
element with a frequency of 2.5 cycles per second. Knowing that the coefficient
of damping is 8 lb s/in., determine the amplitude of the steady-state vibration of
the element.
SOLUTION
Equivalent spring: 2(200) 400 lb/in. 4800 lb/ftk
=
==
Undamped natural frequency: 4800 13.90 rad/s
800/32.2
13.90 2.212 Hz
22
n
n
n
k
m
f
ω
ω
ππ
== =
== =
Critical damping coefficient: 800
2 2 (13.90) 691 lb s/ft
32.2
cn
cm
ω
⎛⎞
=
==
⎜⎟
⎝⎠
Damping coefficient: 8 lb s/in. 96 lb s/ftc
=
⋅=
Damping ratio: 96 0.1390
691
c
c
c==
Amplitude:
()
22
2
/
12
ff
ncn
m
m
c
c
Pk
x
ωω
ωω
=
⎡⎤
−+
⎢⎥
⎣⎦
(1)
where 2.5 Hz 1.130
2.212 Hz
ff
nn
f
f
ω
ω
== =
Substituting into Eq. (1) with 30 lb,
m
P=we have
22 2
30 lb/400 lb/in.
[1 (1.130) ] [2(0.1390)(1.130)]
m
x=−+ 0.1791 in.
m
x=

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