SECTION-15.pdf

7/30/2019 SECTION-15.pdf

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SECTION 15 FLEXIBLE POWER-TRANSMITTING ELEMENTS

Page 1 of 56

LEATHER BELTS

DESIGN PROBLEMS

841. A belt drive is to be designed for 321 =FF , while transmitting 60 hp at 2700rpm of the driver 1D ; 85.1wm ; use a medium double belt, cemented joint, a

squirrel-cage, compensator-motor drive with mildly jerking loads; center distance

is expected to be about twice the diameter of larger pulley. (a) Choose suitable

iron-pulley sizes and determine the belt width for a maximum permissible

psis 300= . (b) How does this width compare with that obtained by the ALBA

procedure? (c) Compute the maximum stress in the straight port of the ALBA

belt. (d) If the belt in (a) stretches until the tight tension lbF 5251 = ., what is

21 FF ?

Solution:

(a) Table 17.1, Medium Double Ply,

Select inD 71 = . min.

int64

20=

( )( )fpm

nDvm 4948

12

27007

12

11===

fpmfpmfpm 600049484000


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(b) ALBA Procedure

( )( ) L21

1.17., ffpm CCCbCTableinhphp =

Table 17.1, fpmvm 4948=

Medium Double Ply

448.12=inhp

Table 17.2Squirrel cage, compensator, starting

67.0=mC

Pulley Size, inD 71 =

6.0=pC

Jerky loads, 83.0=fC

( )( )( )( )( )83.06.067.0448.1260 bhp == inb 5.14=

say inb 15=

(c)

( )( )psi

bt

Fs 128

64

20151

6001=

==

(d) ( ) ( )21

2

12

1

22

1

12

1

2006002 +=+= FFFo

lbFo 2.373=

lbF 5251 =

( ) ( ) 21

22

1

2

1

5252.3732 F+=

lbF 2472 =

1255.2247

525

2

1==

F

F

842. A 20-hp, 1750 rpm, slip-ring motor is to drive a ventilating fan at 330 rpm. The

horizontal center distance must be about 8 to 9 ft. for clearance, and operation is

continuous, 24 hr./day. (a) What driving-pulley size is needed for a speedrecommended as about optimum in the Text? (b) Decide upon a pulley size (iron

or steel) and belt thickness, and determine the belt width by the ALBA tables. (c)Compute the stress from the general belt equation assuming that the applicablecoefficient of friction is that suggested by the Text. (d) Suppose the belt is

installed with an initial tension inlbFo 70= . (17.10), compute 21 FF and the

stress on the tight side if the approximate relationship of the operating tensions

and the initial tensions is 21

2

1

22

1

1 2 oFFF =+ .


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Solution:

fpmtovm 45004000=

assume fpmvm 4250=

12

11nDvm

=

( )12

17504250 1

D=

inD 26.91 =

say inD 101 =

(b) Using Heavy Double Ply Belt, int64

23=

Minimum pulley diameter for fpmvm 4250 , inD 101 =

Use inD 101 =

( )( ) fpmnDvm 458112

17501012

11===

ALBA Tables

( ) L21


8.13=inhp

Slip ring motor, 4.0=mC

Pulley Size, inD 101 =

7.0=pC

Table 17.7, 24 hr/day, continuous

8.1=

sfN Assume 74.0=fC

( )( ) ( )( )( )( )( )74.07.04.08.13208.1 bhp == inb 59.12=

use inb 13=

(c) General belt equation

=

f

f

s

e

evsbtFF

1

2.32

12 2

21

fpsvs 35.7660

4581 ==

..035.0 inculb= for leather

int64

23=

inb 13=


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( )( )lbFF 260

4581

208.1000,3321 ==

3.0=f on iron or steel

C

DD 12

ftC 9~8= use 8.5 ft

( ) inD 5310330

17502 =

=

( )rad72.2

125.8

1053=

=

( )( ) 816.072.23.0 ==f

5578.011

816.0

816.0

=

=

e

e

e

ef

f

( )

( )( )

( )5578.02.3235.76035.012

64

23

13260

2

21

== sFF

psis 176=

(d) 21

2

1

22

1

1 2 oFFF =+

( )( ) lbininlbFo 9101370 ==

lbFF 26021 =

lbFF 26012 =

( ) ( ) 33.609102260 21

2

1

12

1

1 ==+ FF

lbF 10451 =

lbF 78526010452 ==

( )psi

bt

Fs 224

64

2313

10451=

==

331.1785

1045

2

1==

F

F

843. A 100-hp squirrel-cage, line-starting electric motor is used to drive a Freonreciprocating compressor and turns at 1140 rpm; for the cast-iron motor pulley,

inD 161 = ; inD 532 = , a flywheel; cemented joints;l ftC 8= . (a) Choose an

appropriate belt thickness and determine the belt width by the ALBA tables. (b)

Using the design stress of 17.6, compute the coefficient of friction that would beneeded. Is this value satisfactory? (c) Suppose that in the beginning, the initial

tension was set so that the operating 221 =FF . Compute the maximum stress in

a straight part. (d) The approximate relation of the operating tensions and the


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initial tension oF is2

1

2

1

22

1

1 2 oFFF =+ . For the condition in (c), compute oF . Is it

reasonable compared to Taylors recommendation?

Solution:

(a) Table 17.1 ( )( )fpm

nDvm 4775

12

114016

12

11===

Use heavy double-ply belt

int64

23=

1.14=inhp

( )( ) L21


line starting electric motor , 5.0=mC

Table 17.7, squirrel-cage, electric motor, line starting, reciprocating compressor

4.1=sfN

inD 161 = , 8.0=pC

assume, 74.0=fC

( )( ) hphp 1401004.1 ==

( )( )( )( )( )74.08.05.01.14140 bhp == inb 5.33=

use inb 34=

(b) 17.6, 400=ds

00.1= for cemented joint.psis

d 400=

=

f

f

s

e

evsbtFF

1

2.32

12 2

21

fpsvs 6.7960

4775==


int64

23=

inb 34= ( )( )

lbFF 9684775

1004.1000,3321 ==

( )( )( )

==

f

f

e

eFF

1

2.32

6.79035.012400

64

2334968

2

21


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2496.01=

f

f

e

e

28715.0=f

C

DD 12

ftC 8=

( )rad7562.2

128

1653=

=

( ) 28715.07562.2 =f 3.01042.0


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Solution:

(a)( )( )

fpmnD

vm 294412

86513

12

11===

Table 17.1, use Heavy Double Ply,

inD 9min = for fpmvm 2944= belts less than 8 in wide

int64

23=

( )( ) L21


86.9=inhp

Table 17.2

67.0=mC

8.0=pC

( )( ) 592.080.074.0 ==fC

Table 17.7, electric motor, compensator-started (squirrel cage) and reciprocating

compressor

4.1=sfN

( )( ) hphp 70504.1 ==

( )( )( )( )( )592.08.067.086.970 bhp == inb 4.22=

use inb 25=

(b) General Belt Equation

=

f

f

s

e

evsbtFF

1

2.32

12 2

21

inb 25=

int64

23=


fpsvs 1.4960

2944==

Leather on iron, 3.0=f

C

DD 12 =

( )rad35.2

126

1370=

=

( )( ) 705.035.23.0 ==f


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5059.011

705.0

705.0

=

=

e

e

e

ef

f

( )( )lbFF 785

2944

504.1000,3321 ==

( ) ( )( ) ( )5059.02.321.49035.012

642325785

2

21

== sFF

psis 204=

Cemented joint, 0.1=

psis 204=

(c) ( )( ) lbsbtF 183364

23252041 =

==

lbF 104878518332 ==

749.11048

1833

2

1

==F

F

(d) 21

22

1

12

1

2 FFFo +=

( ) ( )21

2

12

1

104818332 +=oF

lbFo 1413=

inlbFo 5.56

25

1413==

Approximately less than Taylors recommendation ( = 70 lb/in.)

(e) ( )( )

C

DDDDCL

457.12

2

1212

+++

( )( ) ( )( )

( )( )inL 286

1264

1370137057.11262

2

=

+++=

(f) More economical basis

12

11nDvm

=

( )12

8654500 1

D=

inD 87.191 =

use inD 201 =


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CHECK PROBLEMS

846. An exhaust fan in a wood shop is driven by a belt from a squirrel-cage motor that

runs at 880 rpm, compensator started. A medium double leather belt, 10 in. wide

is used; inC 54= .; inD 141 = . (motor), inD 542 = ., both iron. (a) What

horsepower, by ALBA tables, may this belt transmit? (b) For this power,compute the stress from the general belt equation. (c) For this stress, what is

21 FF ? (d) If the belt has stretched until psis 200= on the tight side, what is

21 FF ? (e) Compute the belt length.

Solution:

(a) For medium double leather belt

int64

20=

( )( ) fpm CCCbinhphp=

Table 17.1 and 17.2

67.0=mC

8.0=pC

74.0=fC

inb 10=

( )( )fpm

nDv

m 322512

88014

12

11===

6625.6=inhp

( )( )( )( )( ) hphp 43.2674.08.067.0106625.6==

(b)

=

f

f

s

e

evsbtFF

1

2.32

12 2

21

inb 10=

int64

20=

..035.0 inculb=

fpsvs 75.5360

3225==

C

DD 12 =

rad4.254

1454=

=

Leather on iron 3.0=f

( )( ) 72.04.23.0 ==f


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51325.011

72.0

72.0

=

=

e

e

e

ef

f

( )lbFF 270

3225

43.26000,3321 ==

( ) ( )( ) ( )51325.02.3275.53035.012

642010270

2

21

== sFF

psis 206=

(c) ( )( ) lbsbtF 64464

20102061 =

==

lbF 3742706442 ==

72.1374

644

2

1==

F

F

(d) psis 200=

( )( ) lbsbtF 62564

20102001 =

==

lbF 3552706252 ==

76.1355

625

2

1==

F

F

(e) ( )( )

C

DDDDCL

457.12

2

1212

+++

( ) ( ) ( )( )

inL 222544

1454145457.1542

2

=

+++=

847. A motor is driving a centrifugal compressor through a 6-in. heavy, single-ply

leather belt in a dusty location. The 8-in motor pulley turns 1750 rpm;

inD 122 = . (compressor shaft); ftC 5= . The belt has been designed for a net

belt pull of inlbFF 4021 = of width and 321 =FF . Compute (a) the

horsepower, (b) the stress in tight side. (c) For this stress, what needed value of

f is indicated by the general belt equation? (d) Considering the original

data,what horsepower is obtained from the ALBA tables? Any remarks?

Solution:

(a)( )( )

fpmnD

vm 3665

12

17508

12

11===

inb 6=

( )( ) lbFF 24064021 ==


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Page 11 of 56

( ) ( )( )hp

vFFhp m 65.26

000,33

3665240

000,33

21==

=

(b) 21 3FF =

lbFF 2403 22 =

lbF 1202 =

lbF 3601 =

bt

Fs 1=

For heavy single-ply leather belt

int64

13=

( )psis 295

64

136

360=

=

(c)

=

f

f

s

e

evsbtFF

1

2.32

12 2

21

..035.0 inculb=

fpsvs 1.6160

3665==

lbFF 24021 =

( )( )( )

==

f

f

e

eFF

1

2.32

1.61035.012295

64

136240

2

21

7995.01=

f

f

e

e

C

DD 12 =

( )rad075.3

125

812=

=

9875.4=fe

607.1=f

( ) 607.1075.3=

f 5226.0=f

(d) ALBA Tables (Table 17.1 and 17.2)

( )( )fpm

CCCbinhphp =

fpmvm 3665=

965.6=inhp


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Page 12 of 56

inb 10=

0.1=mC (assumed)

6.0=pC

74.0=fC

( )( )( )( )( ) hphphp 65.266.1874.06.00.16965.6


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Page 13 of 56

rad9987.242

915=

=

Leather on paper pulleys, 5.0=f

( )( ) 5.19987.25.0 ==f

77687.01=

f

f

e

e

fpsvs 72.6860

4123==

( )( )( )

( ) lbFF 82277687.02.32

72.68035.012400

64

2010

2

21 =

=

( ) ( )( )hp

vFFhp m 7.102

000,33

4123822

000,33

21==

=

(c) Table 17.7

6.1=sfN

hphphp 7.1022.646.1

7.102 .

MISCELLANEOUS

849. Let the coefficient of friction be constant. Find the speed at which a leather belt

may transmit maximum power if the stress in the belt is (a) 400 psi, (b) 320 psi.(c) How do these speeds compare with those mentioned in 17.9, Text? (d)

Would the corresponding speeds for a rubber belt be larger or smaller? (HINT:

Try the first derivative of the power with respect to velocity.)

Solution:

=

f

f

s

e

evsbtFF

1

2.32

12 2

21

( )000,33

21 mvFFhp

=

( )000,33

60 21 svFFhp =

=

f

f

ss

e

evs

btvhp

1

2.32

12

000,33

60 2

ss

f

f

vv

se

ebthp

=

2.32

121

000,33

60 2


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( )

( )0

2.32

24

2.32

121

000,33

60 22=

=

ss

f

f

s

vvs

e

ebt

vd

hpd

2.32

36 2svs

=

..035.0 inculb=

(a) psis 400=

( )

2.32

035.036400

2

sv=

fpsvs 105.101=

fpmvm 6066=

(b) psis 320=

( )2.32035.036320

2

sv=

fpsvs 431.90=

fpmvm 5426=

(c) Larger than those mentioned in 17.9 (4000 4500 fpm)

(d) Rubber belt, ..045.0 inculb=

(a) psis 400=

( )2.32

045.0364002

sv=

fpsvs 166.89=

fpmfpmvm 60665350


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( )000,33

60 21 svFFhp

=

ss

f

f

vv

se

ebthp

=

2.32

121

000,33

60 2

( )( )

02.32

24

2.32

121

000,33

60 22=

= ss

f

f

s

vvse

ebt

vd

hpd

2.32

36 2svs

= for maximum power

(a) At zero power:

2.32

12 2svs

=

psis 300=

..035.0 inculb=

( )

2.32

035.012300

2

sv=

fpsvs 6575.151=

fpmvm 9100=

Speed, 40 in pulley,( )

( )rpm

D

vn m 869

40

91001212

2

2 ===

(b) Maximum power

2.32

36 2svs

=

( )

2.32

035.036300

2

sv=

fpsvs 5595.87=

fpmvm 5254=

ss

f

f

vv

se

ebthp

=

2.32

121

000,33

60 2

int64

20=

inb 20=

C

DD 12 =

( )rad0225.3

1214

2040=

=

3.0=f


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( )( ) 90675.00225.33.0 ==f

5962.01=

f

f

e

e

( )

( ) ( )( ) ( ) 64.1185595.872.32 5595.87035.0123005962.0000,33 64

202060 2

=

=hp

fpmvm 5254=

AUTOMATIC TENSION DEVICES

851. An ammonia compressor is driven by a 100-hp synchronous motor that turns

1200 rpm; 12-in. paper motor pulley; 78-in. compressor pulley, cast-iron;

inC 84= . A tension pulley is placed so that the angle of contact on the motor

pulley is 193o

and on the compressor pulley, 240o. A 12-in. medium double

leather belt with a cemented joint is used. (a) What will be the tension in thetight side of the belt if the stress is 375 psi? (b) What will be the tension in the

slack side? (c) What coefficient of friction is required on each pulley as indicated

by the general equation? (d) What force must be exerted on the tension pulley to

hold the belt tight, and what size do you recommend?

Solution:

(a) sbtF =1

inb 12=

int64

20=

( )( )

=

64

20123751F

(b)m

v

hpFF

000,3321 =

( )( )fpm

nDvm 3770

12

120012

12

11===

Table 17.7, 2.1=sfN

( )( )lbFF 1050

3770

1002.1000,3321 ==

lbFF 35610501406105012 ===

(c)

=

f

f

s

e

evsbtFF

1

2.32

12 2

21

fpsvs 83.6260

3770==


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Page 17 of 56

..035.0 inculb=

( )( )( )

=

f

f

e

e 1

2.32

83.62035.012375

64

20121050

8655.01=

f

f

e

e

006.2=f

Motor pulley

rad3685.3180

193193 =

==

o

( ) 006.23685.3 =f 5955.0=f

Compressor Pulley

rad1888.41802402403=

==

o

( ) 006.21888.4 =f 4789.0=f

(d) Force:

Without tension pulley

radC

DD356.2

84

1278121 =

=

=

radC

DD

9273.384

1278122 =

+=

+=


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o5.356197.02

356.2356.23685.3

2

1111 ==

=

= rad

o5.376544.09273.31888.42

9273.3

222

22 ==+

=+

= rad

( ) ( ) lbFQ 16725.37sin5.35sin1406sinsin 211 =+=+= of force exertedSize of pulley; For medium double leather belt,

fpmvm 3770= , width = inin 812 >

inD 826 =+=


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Page 19 of 56

852. A 40-hp motor, weighing 1915 lb., runs at 685 rpm and is mounted on a pivoted

base. In Fig. 17.11, Text, ine 10= ., inh16

319= . The center of the 11 -in.

motor pulley is 11 in. lower than the center of the 60-in. driven pulley;

inC 48= . (a) With the aid of a graphical layout, find the tensions in the belt for

maximum output of the motor if it is compensator started. What should be thewidth of the medium double leather belt if psis 300= ? (c) What coefficient of

friction is indicated by the general belt equation? (Data courtesy of Rockwood

Mfg. Co.)

Solution:

(a)

lbR 1915=

Graphically

inb 26

ina 9

[ ] = 0BM bFaFeR 21 +=

( )( ) ( )( ) ( )( )269191510 21 FF += 150,19269 21 =+ FF

For compensator started

( ) ( ) hphpratedhp 56404.14.1 ===

mv

hpFF

000,3321 =


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( )( )fpm

nDv

m 206212

6855.11

12

11===

( )lbFF 896

2062

56000,3321 ==

89612 = FF

Substituting

( ) 150,19896269 11 =+ FF

lbF 12131 =

lbF 31789612132 ==

For medium leather belt, int64

20=

sbtF =1

( )( )

=

64

203001213 b

inb 13=

(c)

=

f

f

s

e

evsbtFF

1

2.32

12 2

21

fpsvs 37.3460

2062==

..035.0 inculb=

( )( )( )

=

f

f

e

e 1

2.32

37.34035.012300

64

2013896

775.01=

f

f

e

e

492.1=f

radC

DD1312.2

48

5.116012=

=

=

( ) 492.11312.2 =f 70.0=f

853. A 50-hp motor, weighing 1900 lb., is mounted on a pivoted base, turns 1140 rpm,

and drives a reciprocating compressor; in Fig. 17.11, Text, ine4

38= .,

inh16

517= . The center of the 12-in. motor pulley is on the same level as the

center of the 54-in. compressor pulley; inC 40= . (a) With the aid of a graphical

layout, find the tensions in the belt for maximum output of the motor if it is

compensator started. (b) What will be the stress in the belt if it is a heavy double


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leather belt, 11 in. wide? (c) What coefficient of friction is indicated by thegeneral belt equation? (Data courtesy of Rockwood Mfg. Co.)

Solution:

(a) For compensator-started

( ) hphp 70504.1 ==

mv

hpFF

000,3321 =

( )( )fpm

nDvm 3581

12

114012

12

11===

( )lbFF 645

2062

70000,3321 ==

inb 25

ina 5

lbR 1900=

bFaFeR 21 +=

( )( ) ( ) ( )255190075.8 21 FF +=

lbFF 33255 21 =+

lbFF 33255645 22 =++

lbF 4472 =

lbFF1092447645645 21

=+=+=

(b) For heavy double leather belt

int64

23=

inb 11=


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Page 22 of 56

( )psi

bt

Fs 276

64

2011

10921=

==

(c)

=

f

f

s

e

evsbtFF

1

2.32

12 2

21

fpsvs 68.5960

3581==

..035.0 inculb=

( )( )( )

=

f

f

e

e 1

2.32

68.59035.012276

64

2311645

241.1=f

radC

DD092.2

40

125412=

=

=

( ) 492.1092.2=

f 60.0=f

RUBBER BELTS

854. A 5-ply rubber belt transmits 20 horsepower to drive a mine fan. An 8-in., motor

pulley turns 1150 rpm; inD 362 = ., fan pulley; ftC 23= . (a) Design a rubber

belt to suit these conditions, using a net belt pull as recommended in 17.15,

Text. (b) Actually, a 9-in., 5-ply Goodrich high-flex rubber belt was used. Whatare the indications for a good life?

Solution:

(a)( )

o174040.31223

83612==

=

= rad

C

DD

976.0=K

2400

KNbvhp

pm=

976.0=K

( )( )fpm

nDvm 2409

12

11508

12

11===

5=pN

( )( )( )2400

976.05240920

bhp ==

inb 1.4=

min. inb 5=


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Page 23 of 56

(b) With inb 9= is safe for good life.

855. A 20-in., 10-ply rubber belt transmits power from a 300-hp motor, running at 650rpm, to an ore crusher. The center distance between the 33-in. motor pulley and

the 108-in. driven pulley is 18 ft. The motor and crusher are so located that the

belt must operate at an angle 75o

with the horizontal. What is the overloadcapacity of this belt if the rated capacity is as defined in 17.15, Text?

Solution:

2400

pmNbvhp =

inb 20=

( )( )fpm

nDvm 5616

12

65033

12

11===

10=pN

( )( )( ) hphp 4682400

10561620 ==

Overlaod Capacity = ( ) %56%100300

300468=

V-BELTS

NOTE:If manufacturers catalogs are available, solve these problems from catalogs as

well as from data in the Text.

856. A centrifugal pump, running at 340 rpm, consuming 105 hp in 24-hr service, is to

be driven by a 125-hp, 1180-rpm, compensator-started motor; intoC 4943= .

Determine the details of a multiple V-belt drive for this installation. The B.F.

Goodrich Company recommended six C195 V-belts with 14.4-in. and 50-in.

sheaves; inC 2.45 .

Solution:

Table 17.7

4.12.02.1 =+=sfN (24 hr/day)

Design hp = sfN (transmitted hp) = ( )( ) hp1751254.1 =

Fig. 17.4, 175 hp, 1180 rpm

inD 13min = , D-section

4.14

50

340

1180

1

2==

D

D

use ininD 134.141 >=

inD 502 =


24/56


Page 24 of 56

( )( )fpm

nDv

m 444912

11804.14

12

11===

36

2

1

09.03

1010

10 mm

dm

vve

DK

c

vahpRated

=

Table 17.3, D-section788.18=a , 7.137=c , 0848.0=e

Table 17.4, 47.31

2=

D

D

14.1=dK

( )( )( )

( )hphpRated 294.28

10

4449

10

44490848.0

4.1414.1

7.137

4449

10788.18

36

209.03

=

=

Back to Fig. 17.14, C-section must be used.

792.8=a , 819.38=c , 0416.0=e

36

2

1

09.03

1010

10 mm

dm

vve

DK

c

vahpRated

=

( )( )( )

( )hphpRated 0.20

10

4449

10

44490416.0

4.1414.1

819.38

4449

10792.8

36

209.03

=

=

Adjusted rated hp = ( )hpratedKK L Table 17.5,

77.046

4.145012 =

=

CDD

88.0=K

Table 17.6

( )( )

C

DDDDCL

457.12

2

1212

+++

( ) ( )( )

( )inL 200

464

4.14504.145057.1462

2

=

+++=

use C195, inL 9.197=

07.1=L

K

Adjusted rated hp = ( )( )( ) hp83.182007.188.0 =

beltshpratedAdjusted

hpDesignbeltsofNo 3.9

83.18

175. === use 9 belts

Use 9 , C195 V-belts with 14.4 in and 50 in sheaves


25/56


Page 25 of 56

( )

16

322

12

2 DDBBC

+=

( ) ( ) ( ) inDDLB 2.3874.145028.69.197428.64 12 =+=+=

( ) ( )

inC 9.4416

4.1450322.3872.38722

=+

=

857. A 50-hp, 1160-rpm, AC split-phase motor is to be used to drive a reciprocatingpump at a speed of 330 rpm. The pump is for 12-hr. service and normally

requires 44 hp, but it is subjected to peak loads of 175 % of full load; inC 50 .

Determine the details of a multiple V-belt drive for this application. The DodgeManufacturing Corporation recommended a Dyna-V Drive consisting of six

5V1800 belts with 10.9-in. and 37.5-in. sheaves; inC 2.50 .

Solution:

Table 17.7, (12 hr/day)

2.12.04.1 ==sfN

Design hp = ( )( )( ) hp1055075.12.1 = Fig. 17.4, 105 hp, 1160 rpm


2.13

4.46

330

1160

1

2=

D

D

use ininD 132.131 >=

inD 4.462 =

( )( )fpm

nD

vm 400912

11602.13

12

11

===

36

2

1

09.03

1010

10 mm

dm

vve

DK

c

vahpRated

=

Table 17.3, D-section

788.18=a , 7.137=c , 0848.0=e

Table 17.4, 5.32.13

4.46

1

2==

D

D

14.1=dK

( )( )( ) ( ) hphpRated 32.24

104009

1040090848.0

2.1314.17.137

400910788.18

36

209.03

=

=

Back to Fig. 17.14, C-section must be used.

792.8=a , 819.38=c , 0416.0=e

inD 9min =


26/56


Page 26 of 56

1.9

32

330

1160

1

2=

D

D

use inD 1.91 =

( )( )fpm

nDvm 2764

12

11601.9

12

11===

( )( )( )

( )hphpRated 96.10

10

2764

10

27640416.0

1.914.1

819.38

2764

10792.8

36

209.03

=

=


458.050

1.93212=

=

C

DD

935.0=K

Table 17.6

( )( )

C

DDDDCL

457.12

2

1212

+++

( ) ( )( )

( )inL 167

504

1.9321.93257.1502

2

=

+++=

use C158, inL 9.160=

02.1=LK

Adjusted rated hp = ( )( )( ) hp45.1096.1002.1935.0 =


hpDesignbeltsofNo 10

43.10

105. ===

( )

16

322

12

2 DDBBC

+=

( ) ( ) ( ) inDDLB 5.3851.93228.69.160428.64 12 =+=+=

( ) ( )inC 8.46

16

1.932325.3855.38522

=+

=

Use 10-C158 belts, inD 1.91 =

inD 322 = , inC 8.46=

858. A 200-hp, 600-rpm induction motor is to drive a jaw crusher at 125 rpm; starting

load is heavy; operating with shock; intermittent service; intoC 123113= .

Recommend a multiple V-flat drive for this installation. The B.F. Goodrich

Company recommended eight D480 V-belts with a 26-in. sheave and a 120.175-

in. pulley; inC 3.116 .

Solution:


27/56


Page 27 of 56

Table 17.7

4.12.06.1 ==sfN

( )( ) hphp 2802004.1 == Fig. 17.14, 280 hp, 600 rpmUse Section E

But in Table 17.3, section E is not available, use section D

13min =D

8.4125

600

1

2==

D

D

For max1D :

121

2min D

DDC +

+=

111

2

8.4113 D

DD+

+=

inD 281 = 2min DC=

inD 1132 =

inD 5.238.4

1131 ==

use ( ) inD 185.23132

11 =+

( )( ) inD 4.86188.42 ==

( )( )

C

DDDDCL

457.12

2

1212

+++

( ) ( )( )

( )inL 410

1184

184.86184.8657.11182

2

=

+++=

using inD 191 = , inD 2.912 = , inC 118=

( ) ( )( )

( )inL 420

1184

192.91192.9157.11182

2

=

+++=

Therefore use D420 sections

inD 191 = , inD 2.912 =

( )( )fpm

nDvm 2985

12

60019

12

11===

36

2

1

09.03

1010

10 mm

dm

vve

DK

c

vahpRated

=


788.18=a , 7.137=c , 0848.0=e

Table 17.4, 8.41

2=

D

D


28/56


Page 28 of 56

14.1=dK

( )( )( )

( )hphpRated 6.29

10

2985

10

29850848.0

1914.1

7.137

2985

10788.18

36

209.03

=

=

Therefore, Fig. 17.14, section D is used.


612.0118

192.9112=

=

C

DD

83.0=K (V-flat)

Table 17.6, D420

inL 8.420=

12.1=LK




280. ===

Use10 , D420, inD 191 = , inD 2.912 = , inC 118=

859. A 150-hp, 700-rpm, slip-ring induction motor is to drive a ball mill at 195 rpm;

heavy starting load; intermittent seasonal service; outdoors. Determine all details

for a V-flat drive. The B.F. Goodrich Company recommended eight D270 V-

belts, 17.24-in sheave, 61-in. pully, inC 7.69 .

Solution:Table 17.7,

4.12.06.1 ==sfN

Design hp = ( )( ) hp2101504.1 = Fig. 17.4, 210 hp, 700 rpm


36

2

1

09.03

1010

10 mm

dm

vve

DK

c

vahpRated

=

For Max. Rated hp,( )

0

103

=

m

vd

hpd

3

33

1

91.0

3 101010

=

mm

d

m vev

DK

cvahpRated

Let310

mv

X=


29/56


Page 29 of 56

3

1

91.0 eXXDK

caXhp

d

=

( )3

1

3

11

3 1012

700

101210 =

==

DnDvX m

7001012

3

1 XD

=

3

3

91.0

1012

700eX

K

caXhp

d

=

( )

( )0391.0 209.0 == eXaX

Xd

hpd

e

aX

3

91.009.2=


788.18=a , 7.137=c , 0848.0=e

( )( )0848.03

788.1891.0

10

09.2

3

09.2=

= mvX

fpmvm 7488=

748812

11==

nDvm

( )7488

12

7001==

Dvm

inD 86.401 =

max inD 86.401 =

ave. ( ) inD 93.2686.4013211 =+=

use inD 221 =

22

79

195

700

1

2=

D

D

inD 221 = , inD 792 =

Min. inDDD

C 5.72222

7922

21

21=+

+=+

+=

Or Min. inDC 792 ==

( )( )

C

DDDDCL

457.12

2

1212

+++

( ) ( )( )

( )inL 327

794

2279227957.1792

2

=

+++=

use D330, inL 8.330=

( )

16

322

12

2 DDBBC

+=


30/56


Page 30 of 56

( ) ( ) ( ) inDDLB 689227928.68.330428.64 12 =+=+=

( ) ( )inC 12.81

16

22793268968922

=+

=

( )( )fpm

nDv

m 4032

12

70022

12

11===

14.1=dK

( )( )( )

( )hphpRated 124.39

10

4032

10

40320848.0

2214.1

7.137

4032

10788.18

36

209.03

=

=


70.012.81

227912=

=

C

DD

84.0=K (V-flat)

Table 17.6D330

07.1=LK




165.35

210. === use 6 belts

Use 6 , D330 V-belts , inD 221 = , inD 792 = , inC 1.81

860. A 30-hp, 1160-rpm, squirrel-cage motor is to be used to drive a fan. During the

summer, the load is 29.3 hp at a fan speed of 280 rpm; during the winter, it is 24hp at 238 rpm; inC 5044


31/56


Page 31 of 56

or 121

2D

DDC +

+=

056.4286

1160

1

2==

D

D

use1

056.4 DC=

inCin 5044


32/56


Page 32 of 56

Table 17.6

9.175=L , C173

04.1=LK



hpDesign

beltsofNo 5.402.12

54

. === use 5 belts

Use 5 , C173 V-belts , inD 1.101 = , inD 412 =

POWER CHAINS

NOTE:If manufacturers catalogs are available, solve these problems from catalogs as

well as from data in the Text.

861. A roller chain is to be used on a paving machine to transmit 30 hp from the 4-

cylinder Diesel engine to a counter-shaft; engine speed 1000 rpm, counter-shaftspeed 500 rpm. The center distance is fixed at 24 in. The cain will be subjected to

intermittent overloads of 100 %. (a) Determine the pitch and the number ofchains required to transmit this power. (b) What is the length of the chain

required? How much slack must be allowed in order to have a whole number of

pitches? A chain drive with significant slack and subjected to impulsive loading

should have an idler sprocket against the slack strand. If it were possible tochange the speed ratio slightly, it might be possible to have a chain with no

appreciable slack. (c) How much is the bearing pressure between the roller and

pin?

Solution:

(a) ( ) hphpdesign 60302 == intermittent

2500

1000

2

1

1

2==

n

n

D

D

12 2DD =

inD

DC 242

12 =+=

242

2 11 =+D

D

inD 6.91 =

( ) inDD 2.196.922 12 ===

( )( )fpm

nDvm 2513

12

10006.9

12

11===

Table 17.8, use Chain No. 35,

Limiting Speed = 2800 fpm


33/56


Page 33 of 56

Minimum number of teeth

Assume 211 =N

422 12 == NN

[Roller-Bushing Impact]

8.0

5.1100

Pn

NKhp tsr

=

Chain No. 35

inP8

3=

21=tsN

rpmn 1000=

29=rK

( )hphp 3.40

8

3

1000

2110029

8.05.1

=

=

[Link Plate Fatigue]P

ts PnNhp07.039.008.1004.0 =

( ) ( ) hphp 91.28

3100021004.0

8

307.03

9.008.1=

=

No. of strands = 2191.2

60==

hprated

hpdesign

Use Chain No. 35, inP8

3= , 21 strands

(b)( )

C

NNNNCL

4022

2

1221 +

++ pitches

64

8

3

24=

=C

211 =N

422 =N

( )

( )

( ) pitchespitchesL 16067.1596440

2142

2

4221

642

2

=

++

+=

Amount of slack

( )21

22433.0 LSh =

inCL 24==


34/56


Page 34 of 56

( )in

in

inS 062.242

8

367.159160

24 =

+=

( ) ( )[ ] ininh4

375.024062.24433.0 2

122

===

(c) bp = bearing pressure

Table 17.8, Chain No. 25

inC 141.0=

inE16

3=

inJ 05.0=

( ) ( ) 204054.005.0216

3141.02 inJECA =

+=+=

hpFV 60000,33

=

( )hp

F60

000,33

2513=

lbF 9.787=

strandlbF 5.3721

9.787==

psipb 925

04054.0

5.37==

862. A conveyor is driven by a 2-hp high-starting-torque electric motor through a

flexible coupling to a worm-gear speed reducer, whose 35wm , and then via a

roller chain to the conveyor shaft that is to turn about 12 rpm; motor rpm is 1750.Operation is smooth, 8 hr./day. (a) Decide upon suitable sprocket sizes, center

distance, and chain pitch. Compute (b) the length of chain, (c) the bearing

pressure between the roller and pin. The Morse Chain Company recommended

15- and 60-tooth sprockets, 1-in. pitch, inC 24= ., pitchesL 88= .

Solution:

Table 17.70.12.02.1 ==sfN (8 hr/day)

( ) hphpdesign 0.220.1 ==

rpmn 5035

17501 ==

rpmn 122 =

Minimum number of teeth = 12


35/56


Page 35 of 56

Use 121 =N

[Link Plate Fatigue]P

ts PnNhp07.039.008.1004.0

=

( ) ( )

0.1

5012004.0

0.2

004.09.008.19.008.1

307.03===

nN

hpPP

ts

P

Use Chain No. 80, inP 0.1=

To check for roller-bushing fatigue

8.0

5.1100

Pn

NKhp tsr

=

29=rK

( )( ) hphphp 227471

1000

1210017

8.0

5.1

>=

=

(a) 121=N

( ) teethNn

nN 5012

12

501

2

12 =

=

=

2

12

DDC +=

( )( )in

PND 82.3

120.111 ==

( )( )in

PND 92.15

500.112 ==

inC 83.172

82.3

92.15=+

use inC 18=

pitchesC 18=

chain pitch = 1.0 in, Chain No. 80

(b)( )

C

NNNNCL

4022

2

1221 +

++

( )( )

( )69

1840

1250

2

5912182

2

=

++

+L pitches

use pitchesL 70=

(c) bp = bearing pressure

Table 17.8, Chain No. 80

inC 312.0=

inE8

5=


36/56


Page 36 of 56

inJ 125.0=

( )( )( )fpm

nPNv tsm 50

12

50121

12

1===

( ) ( ) 204054.005.0216

3141.02 inJECA =

+=+=

hpFV

60000,33

=

( )lbF 1320

50

2000,33==

( ) ( )psi

JEC

Fpb 4835

125.028

5312.0

1320

2=

+

=+

=

863. A roller chain is to transmit 5 hp from a gearmotor to a wood-working machine,

with moderate shock. The 1-in output shaft of the gearmotor turns rpmn 500=

.The 1 -in. driven shaft turns 250 rpm; inC 16 . (a) Determine the size of

sprockets and pitch of chain that may be used. If a catalog is available, be sure

maximum bore of sprocket is sufficient to fit the shafts. (b) Compute the centerdistance and length of chain. (c) What method should be used to supply oil to the

chain? (d) If a catalog is available, design also for an inverted tooth chain.

Solution:

Table 17.7

2.1=sfN

( ) hphpdesign 652.1 ==

2250

500

1

2==

D

D

2

12

DDC +=

2216 11

DD +=

inD 4.61 =

( ) inDD 8.124.622 12 === ( )( )

fpmnD

vm

83812

5004.6

12

11===

(a) Link Plate FatigueP

ts PnNhp07.039.008.1004.0

=

( )PPP

DNNts

11.204.611 ===


37/56


Page 37 of 56

( ) PPP

hp 07.039.0

08.1

50011.20

004.0

=

PP 07.092.147.276

=

inP 45.0=

use inP21= , Chain No. 40

( )40

2

1

4.611 =

=

P

DN

802 12 == NN

Size of sprocket, 401 =N , 802 =N , inP2

1= .

(b) inC 16=

pitchesin

in

C 32

2

1

16==

( )C

NNNNCL

4022

2

1221 +

++

( )( )

( )25.125

3240

4080

2

8040322

2

=

++

+L pitches

use pitchesL 126=

(c) Method: fpmvm 838= .

Use Type II Lubrication ( fpmv 1300max = ) oil is supplied from a drip lubricator to linkplate edges.

864. A roller chain is to transmit 20 hp from a split-phase motor, turning 570 rpm, to a

reciprocating pump, turning at 200 rpm; 24 hr./day service. (a) Decide upon the

tooth numbers for the sprockets, the pitch and width of chain, and centerdistance. Consider both single and multiple strands. Compute (b) the chain

length, (c) the bearing pressure between the roller and pin, (d) the factor of safety

against fatigue failure (Table 17.8), with the chain pull as the force on the chain.(e) If a catalog is available, design also an inverted-tooth chain drive.

Solution:Table 17.7

2.04.1 +=sfN (24 hr/day)

( ) hphpdesign 32206.1 ==

(a) 85.2200

570

2

1==

n

n


38/56


Page 38 of 56

85.22

1

1

2=

n

n

D

D

Considering single strandP

ts PnNhp07.039.008.1004.0 =

min 17=

tsN ( ) ( ) PPhp 07.039.008.1 57017004.032 ==

24.107.03 = PP

inP 07.1=

use inP 0.1=

( ) ( ) ( ) ( )107.039.008.11 1570004.032

== Nhp

211 =N

( ) 6021200

5702 =

=N

Roller width = in85

2

12

DDC +=

( )( )in

PND 685.6

21111 ==

( )( )in

PND 10.19

60122 ==

inC 44.222

685.610.19 =+=

Use inC 23=

pitchesC1

23=

Considering multiple strands

Assume, inP2

1=

P

ts PnNhp07.039.008.1004.0 =

( ) ( ) ( ) ( ) hphp 148.45.057021004.0 5.007.039.008.1 ==

No. of strands = 7.7148.4

32=

hp

hp

Use 8 strands

(b) Chain Length

( )C

NNNNCL

4022

2

1221 +

++


39/56


40/56


Page 40 of 56

(a) Factor of Safety =F

Fu

( )( ) lbF 10002500 ==

Factor of Safety = 1.61000

6100=

(b) Factor of Safety =F

Fu

4(fatigue)

Factor of Safety =( )

5.110004

6100=

(c) fpmft

vm 35

min1

sec60

sec24

14=

=

20=sN

inP 8

5=

Rated Pts PnNhp07.039.008.1004.0 = [Link Plate Fatigue]

( )fpm

nnPN

v sm 3512

208

5

12=

==

rpmn 6.33=

Rated ( ) ( ) hphp 6.08

56.3320004.0

8

507.03

9.008.1=

=

Hp needed at constant speed

( )( ) hphpFv

hp m 6.053.0000,33

35500

000,33


41/56


Page 41 of 56

Solution:

(a)

( )212 445.4

sec6

sec60

min11600

fps

fpm

t

vva =

=

=

kipsWh 20=

For 6 x 19 IPS,

ftlbDw r26.1

kipsDkipsDwL rr22 64.0

1000

4006.1 =

=

maWwLF ht =

2.32

64.020 2rDm+

=

( )445.42.32

64.0202064.0

22

+=

rrt

DDF

273.076.22 rt DF +=

inDr4

31=

kipsFt 254

3173.076.22

2

=

+=

t

bu

F

FFN

=

Table AT 28, IPS

tonsDF ru242

( ) kipstonsFu 25812975.1422

===

with bending load

mbb AsF =

s

wb

D

EDs =


42/56


Page 42 of 56

s

wmb

D

DEAF =

Table At 28, 6 x 19 Wire Rope

( ) inDD rw 11725.075.1067.0067.0 ===

inftDs

968 ==

ksiE 000,30= 24.0 rm DA

( ) insqAm 225.175.14.02==

( )( )( )

( )kipsFb 45

96

11725.0225.1000,30==

52.825

45258=

=

=

t

bu

F

FFN

without bending load

32.1025258 ===

t

u

F

FN

(b) 35.1=N on fatigue

IPS, ksisu 260

( ) uu

tsr

ssp

NFDD

2=

( )( )( )( )

( )( )2602535.12

9675.1usp

=

0015.0=usp

Fig. 17.30, 6 x 19 IPS

Number of bends to failure = 7 x 105

(c)rmEA

FL=

insqAm 225.1=

ksiEr 000,12 (6 x 19 IPS)

kipsF 14=

inftL 4800400 ==

( )( )( )( )in57.4

000,12225.1480014 ==

( )( ) kipsinFU === 3257.4142

1

2

1

(d) ( ) kipsFF u 6.512582.02.0 ===


43/56


Page 43 of 56

rmEA

FL=

( )( )

( )( )in85.16

000,12225.1

48006.51==

( )( ) kipsinFU === 43485.166.512121

For 1 in, as-rolled 1045 steel rod

ksisu 96=

( ) ( ) kipsAsF uu 9.23075.14

962=

==

( ) kipsFF u 2.469.2302.02.0 ===

AE

FL=

( )( )

( ) ( ) in073.3000,3075.14

48002.46

2

=

=

( )( ) UkipsinFU


44/56


Page 44 of 56

For 6 x 19 IPS,

ftlbDw r26.1


1000

20006.1 =

=

kipsWh

10=

aWwL

WwLF hht

+=

2.32

( ) ( ) ( )102.317267.1102.312.32

56.51

2.32

22+=+

+=+

+= rrht DDWwL

aF

(a)( ) uu

tsr

ssp

NFDD

2=

Fig. 17.30, 200,000 cycles, 6 x 19

0028.0=usp

PS: ksisu

225

inftDs 726 ==

3.1=N

( )( )( )( )

( )( )2250028.0

102.317267.13.1272

2+

=r

r

DD

49.307566.936.45 2 += rr DD

01251.364916.42 =+ rr DD

inDr 815.0=

say inDr

8

7=

(b) by 5=N , Equation (v)

t

bu

F

FFN

=

s

wb

D

EDs =

rw DD 067.0=

( )( )r

rb D

Ds 92.27

72

067.0000,30==

mbb AsF = 24.0 rm DA =

( )( ) 32 17.114.092.27 rrrb DDDF == tonsDF ru

236= for PS

kipsDFru

272=

tbu NFFF =


45/56


Page 45 of 56

( )( )( )102.317267.1517.1172 232 += rrr DDD ( )( )102.38634.517.1172 232 += rrr DDD

inDr 216.1=

use inDr

4

11=

(c)( ) uu

tsr

ssp

NFDD

2=

( )( )( )( ) ( )[ ]

( )( )2251025.12.317267.13.12

7225.12

usp

+=

00226.0=usp

Fig. 17.20

Expected Life = 3 x 105

cycles

(d) kipsF 7=

ksiEr 000,12=

inftL 000,242000 ==

For (a) inDr8

7=

rmEA

FL=

insqDA rm 30625.08

74.04.0

2

3=

=

( )( )

( )( ) in7.45000,1230625.0

000,247==

For (b) inDr4

11=

rmEA

FL=

insqDA rm 625.04

114.04.0

2

3=

=

( )( )

( )( )in4.22

000,12625.0

000,247==

(e) For (a) ( )( ) kipsinFU === 1607.4572

1

2

1

For (b) ( )( ) kipsinFU === 4.784.2272

1

2

1

(f) Limiting pressure, cast-iron sheaves, 6 x19, psip 500= .


46/56


Page 46 of 56

For (a) 0028.0=usp

( ) psipsikipsp 500630630.02250028.0 >=== , not reasonable.

For (b) 00226.0=usp

( ) psipsikipsp 5005.5085085.022500226.0 === , reasonable.

869. For a mine hoist, the cage weighs 5900 lb., the cars 2100 lb., and the load of coal

in the car 2800 lb.; one car loaded loaded at a time on the hoist. The drumdiameter is 5 ft., the maximum depth is 1500 ft. It takes 6 sec. to accelerate the

loaded cage to 3285 fpm. Decide on a grade of wire and the kind and size of rope

on the basis of (a) a life of 5102 cycles and 3.1=N against fatigue failure, (b)

static consideration (but not omitting inertia effect) and 5=N . (c) Make a final

recommendation. (d) If the loaded car can be moved gradually onto the freelyhanging cage, how much would the rope stretch? (e) What total energy has the

rope absorbed, fully loaded at the bottom of the shaft? Neglect the ropes weight

for this calculation. (f) Compute the pressure of the rope on the cast-iron drum. Is

it all right?

Solution:

kipslbWh 8.10800,10280021005900 ==++=

( )212 125.9

sec6

sec60

min13285

fps

fpm

t

vva =

=

=

aWwL

WwLF hht

+=

2.32

Assume 6 x 19 IPS,

ftlbDwr

26.1


1000

15006.1 =

=

( ) ( ) 86.1308.3104.212.32

125.91

2.32

22+=+

+=+

+= rrht DDWwL

aF

(a) Fig. 17.30, 2 x 105

cycles

0028.0=usp


47/56


Page 47 of 56

( ) uu

tsr

ssp

NFDD

2=

inftDs 605 ==

rs DD 45

inDr 33.14560max ==

use inDr4

11=

kipsFt 67.1886.13

4

1108.3

2

=+

=

( )( )

( ) ( )ksisu 231

604

110028.0

67.183.12=

=

Use Plow Steel, 6 x 19 Wire Rope, inDr 4

11= .

(b)t

bu

F

FFN

=

s

wb

D

EDs =

inDD rw 08375.04

11067.0067.0 =

==

inDs 60=

ksiE 000,30=

( )( )ksisb 875.41

60

08375.0000,30==

2

2

2 625.04

114.04.0 inDA rm =

==

( )( ) kipsAsF mbb 17.26625.0875.41 === 5=N

( )( ) tonskipsFNFF btu 76.5952.11917.2667.185 ==+=+=

25.38

4

11

76.59

22

=

=

r

u

D

F

Table AT 28,

Use IPS, 6 x 19, 25.38422

>=

r

u

D

F


48/56


Page 48 of 56

(c) Recommendation:

6 x 19, improved plow steel, inDr4

11=

(d)rmEA

FL=

lbF 490028002100 =+=

psiEr61012

inftL 000,181500 ==

( )( )

( )( )in76.11

1012625.0

000,1849006=

=

(e) ( )( ) lbinFU === 800,2876.1149002

1

2

1

(f) 0028.0=usp

ksisu 231= ( ) psip 8.646000,2310028.0 ==

For cast-iron sheave, limiting pressure is 500 psi

psipsip 5008.646 >= , not al right.

870. The wire rope of a hoist with a short lift handles a total maximum load of 14 kips

each trip. It is estimated that the maximum number of trips per week will be

1000. The rope is 6 x 37, IPS, 1 3/8 in. in diameter, with steel core. (a) On the

basis of 1=N for fatigue, what size drum should be used for a 6-yr. life? (n)

Because of space limitations, the actual size used was a 2.5-ft. drum. What is the

factor of safety on a static basis? What life can be expected ( 1=

N )?

Solution:

(a)

No. of cycles = ( ) cyclescycleswk

trips

days

wk

yr

daysyr 5103857,312

1

1000

7

1

1

3656 =

Figure 17.30, 6 x 37, IPS

00225.0=usp

( ) uu

tsr

ssp

NFDD

2=

For IPS, ksisu 260

kipsFt 14=

0.1=N

inDr 375.1=

( ) uu

tsr

ssp

NFDD

2=


49/56


Page 49 of 56

( )( )( )

( )( )26000225.0

140.12375.1 =sD

inDs 8.34=

(b) inftDs

302 ==

Static Basis

t

bu

F

FFN

=

Table AT 28, 6 x 37

( ) inDD rw 066.0375.1048.0048.0 ==

( ) 222 75625.0375.14.04.0 inDA rm ==

( )( ) kipsAsF muu 6.19675625.0260 ===

( )( )( )kipsD

AED

AsFs

mw

mbb 9.4930

75625.0066.0000,30====

5.105.10

9.496.196=

=

=

t

bu

F

FFN

Life: 0.1=N (fatigue)

( ) uu

tsr

ssp

NFDD

2=

( )( )( )( )

( )( )260

140.1230375.1

usp=

0026.0=usp

Figure 17.30, Life cycles5105.2 , 6 x 37.

871. A wire rope passes about a driving sheave making an angle of contact of 540o, as

shown. A counterweight of 3220 lb. is suspended from one side and the

acceleration is 4 fps2. (a) If 1.0=f , what load may be noised without slipping on

the rope? (b) If the sheave is rubber lined and the rope is dry, what load may be

raised without slipping? (c) Neglecting the stress caused by bending about the

sheave, find the size of 6 x 19 MPS rope required for 6=

N and for the loadfound in (a). (d) Compute the diameter of the sheave for indefinite life with say

1.1=N on fatigue. What changes could be made in the solution to allow the use

of a smaller sheave?


50/56


Page 50 of 56

Problems 871 874.

Solution:

( ) lbfps

fpslbF 2820

2.32

413220

2

2

2 =

=

(a) feFF 21 =

3540 == o 10.0=f

( ) ( )( ) lbeF 72372820 310.01 ==

(b) For rubber lined, dry rope495.0=f

( ) ( )( ) lbeF 466,2492820 3495.01 ==

(c) lbFFt 72371 ==

( )

t

u

t

bu

F

F

F

FFN =

=

0

tonsDFru

232 for MPS

kipsDF ru264

lbDF ru 2000,64=

tuNFF =

( )( )72376000,64 2 =rD

inDr 824.0=

use inDr 875.0=


51/56


Page 51 of 56

(d)( ) uu

tsr

ssp

NFDD

2=

Indefinite life, 0015.0=usp

MPS: psiksisu 000,195195 =

( ) ( )( )( )( )000,1950015.072371.12875.0 =sD

inDs 2.62=

To reduce the size of sheave, increase the size of rope.

872. A traction elevator with a total weight of 8 kips has an acceleration of 3 fps2; the

6 cables pass over the upper sheave twice, the lower one once, as shown..

Compute the minimum weight of counterweight to prevent slipping on the

driving sheave if it is (a) iron with a greasy rope, (b) iron with a dry rope, (c)rubber lined with a greasy rope. (d) Using MPS and the combination in (a),

decide upon a rope and sheave size that will have indefinite life ( 1=N will do).

(e) Compute the factor of safety defined in the Text. (f) If it were decided that5105 bending cycles would be enough life, would there be a significant

difference in the results?

Solution:

( ) kipsfps

fpskipsF 745.8

2.32

318

2

2

1 =

+=

( ) 31803 == o

f

e

FF 12 =

cW = weight of counterweight

22 10274.1

2.32

31

FF

Wc =

=

fce

FW 1

10274.1=

(a) Iron sheave, greasy rope, 07.0=f

( )( )( )

kipse

Wc 986.4745.810274.1

307.0==

(b) Iron sheave, dry rope, 12.0=f ( )

( )( )kips

eWc 112.3

745.810274.1312.0

==

(c) Rubber lined with a greasy rope, 205.0=f

( )( )( )

kipse

Wc 397.1

745.810274.13205.0

==


52/56


Page 52 of 56

(d)( ) uu

tsr

ssp

NFDD

2=


kipsFFt 745.81 == total

kipsFt 458.16745.8 == each rope

lbsFt 1458=

1=N

Table AT 28, 6 x 19

rs DD 45

( )( )( )

( )( )000,1950015.0

14581245 =rr DD

inDr 47.0=

Use ininDr 5.02

1

==

(e)t

bu

F

FFN

=

Table AT 28, MPS

( ) lblblbDtonsDF rru 000,165.0000,64000,6432222

====

s

mwb

D

AEDF =

psiE6

1030= 6 x 19, rw DD 067.0=

rs DD 45

( ) ..1.05.04.04.0 22 insqDA rm ===

( )( )( )lbFb 4467

45

1.0067.01030 6=

=

91.71458

4467000,16=

=N

(f) 5 x 105 cycles

Fig. 17.30, 6 x 19.

0017.0=usp

( ) uut

srssp

NFDD

2=


53/56


Page 53 of 56

( )( )( )

( )( )000,1950017.0

14581245 =rr DD

inDr 44.0=

since ininDr 47.044.0 = as in (d), therefore, no significant difference will result.

873. A 5000-lb. elevator with a traction drive is supported by a 6 wire ropes, each

passing over the driving sheave twice, the idler once, as shown. Maximum valuesare 4500-lb load, 4 fps2 acceleration during stopping. The brake is applied to a

drum on the motor shaft, so that the entire decelerating force comes on the

cables, whose maximum length will be 120 ft. (a) Using the desirable sD in

terms of rD , decide on the diameter and type of wire rope. (b) For this rope and

05.1=N , compute the sheave diameter that would be needed for indefinite life.(c) Compute the factor of safety defined in the Text for the result in (b). (d)

Determine the minimum counterweight to prevent slipping with a dry rope on an

iron sheave. (e) Compute the probable life of the rope on the sheave found in (a)

and recommend a final choice.

Solution:(a)

lbFt 4500=

lbWh 5000=

awLW

FwLW hth

+=+

2.32

assume 6 x 19

ftlbDw r26.1=

( )( ) 22 1921206.1 rr DDwL == per rope( ) 22 11521926 rr DDwL ==

( )42.32

115250004500115250002

2

+=+

rr

DD

22 11.14312.6215001152 rr DD +=+

inDr 3465.0=

say ininDr8

3375.0 ==


54/56


Page 54 of 56

inDD rs8

716

8

34545 =

=

Six 6 x 19 rope, inDr8

3=

(a) ininDr8

3375.0 ==

lbFt 7506

4500==

05.1=N

( ) uut

srssp

NFDD

2=

assume IPS, psiksisu 000,260260 ==


( ) ( )( )( )( )000,2600015.0

75005.12375.0 =sD

inDs 77.10=

(c)t

bu

F

FFN

=

lbFt 750=

IPS

lblblbDtonsDF rru 813,11

8

3000,84000,8442

2

22=

==

s

mwb

D

AEDF =

6 x 19,

inDs 77.10= as in (b)

inDD rw 025.08

3067.0067.0 =

==

..05625.08

34.04.0

2

2

insqDA rm =

==

psiE 61030=

( )( )( )lbFb 3917

77.10

05625.0025.01030 6=

=


55/56


Page 55 of 56

53.10750

3917813,11=

=N

(c) lbFF t 45001 == feFF 21 =

For iron sheave, dry rope, 12.0=f

3540 == o

( )( )lb

ee

FF

f1452

4500312.0

12 ===

22.32

1 Fa

CW =

+

1452

2.32

41 =

+CW

lbCW 1291=

874. A traction elevator has a maximum deceleration of 5 fps2 when being braked on

the downward motion with a total load of 10 kips. There are 5 cables that passtwice over the driving sheave. The counterweight weighs 8 kips. (a) Compute the

minimum coefficient of friction needed between ropes and sheaves for no

slipping. Is a special sheave surface needed? (b) What size 6 x 19 mild-plow-

steel rope should be used for 4=N , including the bending effect? (Static

approach.) (c) What is the estimated life of these ropes ( 1=N )?

Solution:

205.8 fpsa =

(a) kipsF 101 =

( ) kipskipsF 62.32

05.8182 =

=

3=

feF

F=

2

1


56/56


( )3

6

10 fe=

0542.0=f

Special sheave surface is needed for this coefficient of friction, 17.21.

(b)t

bu

F

FFN =

kipsFt 25

10==

s

mwb

D

AEDF =

Table AT 28, 6 x 19, MPS

rw DD 067.0=

rs DD 45

24.0 rm DA

psiE 61030=

( )( )( )kipsD

D

DDF r

r

rrb

226

87.1745

4.0067.01030=

=

kipsDtonsDF rru22 6432 =

2

87.17644

22

rr DDN

==

inDr 4164.0=

use inDr16

7=

(c) inDD rs 2016

74545 =

=

( ) uu

tsr

ssp

NFDD

2=

kipsFt 2= each rope

MPS, ksisu 195=

0.1=N

( )( )( )

( )( )19520.12

2016

7

usp=

00230sp

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