978-0073398242 Chapter 19 Solution Manual Part 19

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

1

2

c

m

cd

m

t

xe

x

τ

−

=

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

d

kc

mm

ω

⎛⎞

=−

⎜⎟

⎝⎠

2

3

4

(6485.9)

(931.67)(15.325) (4)(931.67)

230 10 lb/ft

d

c

km m

ω

=+

=× 230 kips/ftk= 

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= − − =

  

 

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

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

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



m

n

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

⋅

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.)

s

m

g

S

O

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

a

n

q

PROBLEM 19.136 (Continued)

Expression for :

d

ω

2

8.846

12

2

1500 230

13 (2)(13)

6.094 rad/s

( sin 6.094 cos 6.094 )

d

t

kc

mm

x

eC tC t

ω

−

⎛⎞

=−

⎜⎟

⎝⎠

⎛⎞

=−

⎜⎟

⎝⎠

=

=+

Initial conditions: 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

8.846

1.219 ( 8.846)( 0.02619) 6.094

0.16202

(0.16202sin 6.094 0.02619 cos 6.094 )

t

C

x

ett

−

=− − +

=

=−

Maximum deflection occurs when 0x

=



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

t

mm

t

mm

m

x

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

t

xe

x

−

== =

=

−

=−=

Blocks move, static deflection m

x

+ Total distance 0.02619 0.0307 0.0569 m 56.9 mm

=

+= = 

S

O

Fr

o

O

LUTION

o

m the given

d

PRO

B

A 0.9-

k

shown

coeffic

i

connec

t

spring

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

(

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

i

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

:tt

c

cc xce ce

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

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

λλ

=+



0.1065 m/sx

=

−



106.5 mm/sv= 

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

k

m

f

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==

Amplitude:

()

22

2

/

12

ff

ncn

m

c

Pk

x

ωω

=

⎡⎤

⎡

⎤

−+

⎢⎥

⎢

⎥

⎣

⎦

⎣⎦

(1)

where 2.5 Hz 1.130

2.212 Hz

ff

nn

f

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= 