Section 15

SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS

LEATHER BELTS DESIGN PROBLEMS A belt drive is to be designed for F 1 F 2 = 3 , while transmitting 60 hp at 2700

841.

rpm of the driver D1 ; mw ≈ 1.85 ; 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 s = 300 psi . (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 F 1 = 525 lb ., what is F 1 F 2 ?

Solution: (a) Table 17.1, Medium Double Ply, Select D1 = 7 in . min. t =

20

in 64 π D1n1

π (7 )(2700 ) = 4948 fpm 12 12 4000 fpm < 4948 fpm < 6000 fpm

vm =

hp =

60 =

=

(F 1 − F 2 )vm 33,000 (F 1 − F 2 )(4948 ) 33,000

F 1 − F 2 = 400 lb F 1 = 3F 2

3F 2 − F 2 = 400 lb F 2 = 200 lb F 1 = 3F 2 = 3(200) = 600 lb F 1 = sbt sd = 300η

For cemented joint, η = 1.0 sd = 300 psi

 20    64 

F 1 = 600 = (300)(b ) b = 6.4 in say b = 6.5 in

757


(b) ALBA Procedure hp = (hp in., Table 17.1)(bC mC p ) C f 1 C f 2

L

Table 17.1, vm = 4948 fpm Medium Double Ply hp in = 12.448 Table 17.2 Squirrel cage, compensator, starting C m = 0.67 Pulley Size, D1 = 7 in C p = 0.6

Jerky loads, C f = 0.83

)(0.83) hp = 60 = (12.448)(b )(0.67 )(0.6 )( b = 14.5 in say b = 15 in (c) s =

F 1

η bt

1

=

1

600 = 128 psi  20  (1)(15)   64  1

1

1

(d) 2 F = F 1 + F 22 = (600) 2 + (200)2 2 o

2

F o = 373.2 lb F 1 = 525 lb 1

1

1

2(373.2 )2 = (525)2 + F 22 F 2 = 247 lb F 1 F 2

=

842.

525 247

= 2.1255

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 speed recommended 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 applicable coefficient of friction is that suggested by the Text . (d) Suppose the belt is installed with an initial tension F o = 70 lb in . (§17.10), compute F 1 F 2 and the stress on the tight side if the approximate relationship of the operating tensions 1

1

1

and the initial tensions is F 1 + F = 2F o2 . 2

2 2

758


(b) ALBA Procedure hp = (hp in., Table 17.1)(bC mC p ) C f 1 C f 2

L

Table 17.1, vm = 4948 fpm Medium Double Ply hp in = 12.448 Table 17.2 Squirrel cage, compensator, starting C m = 0.67 Pulley Size, D1 = 7 in C p = 0.6

Jerky loads, C f = 0.83

)(0.83) hp = 60 = (12.448)(b )(0.67 )(0.6 )( b = 14.5 in say b = 15 in (c) s =

F 1

η bt

1

=

1

600 = 128 psi  20  (1)(15)   64  1

1

1

(d) 2 F = F 1 + F 22 = (600) 2 + (200)2 2 o

2

F o = 373.2 lb F 1 = 525 lb 1

1

1

2(373.2 )2 = (525)2 + F 22 F 2 = 247 lb F 1 F 2

=

842.

525 247

= 2.1255

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 speed recommended 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 applicable coefficient of friction is that suggested by the Text . (d) Suppose the belt is installed with an initial tension F o = 70 lb in . (§17.10), compute F 1 F 2 and the stress on the tight side if the approximate relationship of the operating tensions 1

1

1

and the initial tensions is F 1 + F = 2F o2 . 2

2 2

758

SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS Solution: vm = 4000 to 4500 fpm assume vm = 4250 fpm vm =

π D1n1 12 π D1 (1750 )

4250 =

12 D1 = 9.26 in

say D1 = 10 in (b) Using Heavy Double Ply Belt, t =

23

in 64 Minimum pulley diameter for vm ≈ 4250 fpm , D1 = 10 in

Use D1 = 10 in π D1n1 π (10 )(1750 ) vm = = = 4581 fpm 12 12 ALBA Tables hp = (hp in., Table 17.1)(bC mC p ) C f 1 C f 2

L

hp in = 13.8

Slip ring motor, C m = 0.4 Pulley Size, D1 = 10 in C p = 0.7

Table 17.7, 24 hr/day, continuous N sf = 1.8 Assume C f = 0.74 hp = (1.8)(20) = (13.8)(b )(0.4)(0.7 )(0.74 ) b = 12.59 in use b = 13 in

(c) General belt equation

 12 ρ vs2  e f θ − 1   f θ  F 1 − F 2 = bt  s − 32 . 2   e  vs =

4581

= 76.35 fps 60 ρ = 0.035 lb cu. in. for leather

t =

23

in 64 b = 13 in

759


F 1 − F 2 =

33,000(1.8)(20 )

4581 f = 0.3 on iron or steel

θ ≈ π ±

=

260 lb

D2 − D1

C C = 8 ~ 9 ft use 8.5 ft

 1750  (10) = 53 in  330 

D2 = 

θ = π −

53 − 10

= 2.72 rad 8.5(12 ) )(2.72) = 0.816 f θ = (0.3 )( θ

e f − 1 e f θ

=

e0.816 − 1 e0.816

= 0.5578

)(76.35)2   23   12(0.035 )( F 1 − F 2 = 260 = (13)   s −  (0.5578) 32.2  64    s = 176 psi 1

1

1

(d) F 1 + F = 2 F o2 2

2 2

F o = (70 lb in)(13 in ) = 910 lb F 1 − F 2 = 260 lb F 2 = F 1 − 260 lb 1

1

1

F 12 + ( F 1 − 260 )2 = 2(910)2 = 60.33

F 1 = 1045 lb F 2 = 1045 − 260 = 785 lb s=

F 1 F 2

F 1 bt

=

843.

=

1045 = 224 psi  23  (13)   64 

1045 785

= 1.331

A 100-hp squirrel-cage, line-starting electric motor is used to drive a Freon reciprocating compressor and turns at 1140 rpm; for the cast-iron motor pulley, D1 = 16 in ; D2 = 53 in , a flywheel; cemented joints;l C = 8 ft . (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 be needed. Is this value satisfactory? (c) Suppose that in the beginning, the initial tension was set so that the operating F 1 F 2 = 2 . Compute the maximum stress in a straight part. (d) The approximate relation of the operating tensions and the 760

SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS 1

1

1

initial tension F o is F 1 + F = 2 F o2 . For the condition in (c), compute F o . Is it 2

2 2

reasonable compared to Taylor’s recommendation? Solution: (a) Table 17.1 π D1n1 π (16 )(1140 ) vm = = = 4775 fpm 12 12 Use heavy double-ply belt 23 t = in 64 hp in = 14.1 hp = (hp in., Table 17.1)(bC mC p ) C f 1 C f 2

L

line starting electric electr ic motor , C m = 0.5 Table 17.7, squirrel-cage, electric motor, line starting, reciprocating compressor N sf = 1.4 D1 = 16 in , C p = 0.8

assume, C f = 0.74 hp = (1.4)(100) = 140 hp hp = 140 = (14.1)(b )(0.5)(0.8)(0.74) b = 33.5 in use b = 34 in

(b) §17.6, sd = 400η

η = 1.00 for cemented joint. sd = 400 psi

 12 ρ vs2  e f θ − 1   f θ  F 1 − F 2 = bt  s − 32 . 2   e  vs =

4775

= 79.6 fps 60 ρ = 0.035 lb cu. in. for leather

t =

23

in 64 b = 34 in F 1 − F 2 =

33,000(1.4 )(100 ) 4775

= 968 lb

12(0.035)(79.6 )  23   F 1 − F 2 = 968 = (34)  400 − 32.2  64  

2

761

 e f θ − 1   f θ   e 

SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS θ

e f − 1

= 0.2496 θ e f f θ = 0.28715

θ ≈ π ±

D2 − D1 C

C = 8 ft

θ = π −

53 − 16

= 2.7562 rad 8(12) f (2.7562) = 0.28715 f = 0.1042 < 0.3 Therefore satisfactory.

(c) F 1 − F 2 = 968 lb F 1 = 2 F 2

2 F 2 − F 2 = 968 lb F 1 = 2 F 2 = 2(968) = 1936 lb s=

F 1 bt

=

1936 = 159 psi  23  (34)   64 

(d) F 1 = 1936 lb , F 2 = 968 lb 1

1

1

2 F = F 1 + F 22 2 o

2

1

1

1

2 F o2 = (1936 ) 2 + (968)2 F o = 1411 lb F o =

844.

1411 34

=

41.5 lb in of width is less than Taylor’s recommendation and is reasonable.

A 50-hp compensator-started motor running at 865 rpm drives a reciprocating compressor for a 40-ton refrigerating plant, flat leather belt, cemented joints. The diameter of the fiber driving pulley is 13 in., D2 = 70 in ., a cast-iron flywheel; C = 6 ft .11 in. Because of space limitations, the belt is nearly vertical; the surroundings are quite moist. (a) Choose a belt thickness and determine the width by the ALBA tables. (b) Using recommendations in the Text , compute s from the general belt equation. (c) With this value of s , compute F 1 and F 1 F 2 . (d) 1

Approximately,

1

1

F 1 + F = 2 F o2 , where F o is the initial tension. For the 2

2 2

condition in (c), what should be the initial tension? Compare with Taylor, §17.10. (e) Compute the belt length. (f) The data are from an actual drive. Do you have any recommendations for redesign on a more economical basis? 762


Solution: π D1n1 π (13)(865 ) = = 2944 fpm (a) vm = 12 12 Table 17.1, use Heavy Double Ply, Dmin = 9 in for vm = 2944 fpm belts less than 8 in wide 23 t = in 64 hp = (hp in., Table 17.1)(bC mC p ) C f 1 C f 2

L

hp in = 9.86 Table 17.2 C m = 0.67 C p = 0.8 � � = (0�74 )(0�83) = 0�6142

Table 17.7, electric motor, compensator-started (squirrel cage) and reciprocating compressor N sf = 1.4 hp = (1.4)(50) = 70 hp �� = 70 = (9�86)(� )(0�67 )(0�8 )(0�6142) � = 21�6 ��

use b = 25 in (b) General Belt Equation

 12 ρ vs2  e f θ − 1   f θ  F 1 − F 2 = bt  s − 32 . 2   e  b = 25 in t =

23

in 64 ρ = 0.035 lb cu. in. for leather vs =

2944

= 49.1 fps 60 Leather on iron, f = 0.3

θ = π − θ = π −

D2 − D1 C 70 − 13

6(12) + 11

= 2�455 ��

� θ = (0�3)(2�455) = 0�7365

763


� � − 1 �

� θ

=

�

F 1 − F 2 =

0�7365

�

−1

0�7365

= 0�5212

33,000(1.4 )(50) 2944

=

785 lb

2  23   12(0�035)(49�1)   � − (0�5212) 32�2  64   

� 1 − � 2 = 785 = (25)

� = 199 ��

Cemented joint, η = 1.0 � = 199 ��

 23  (c) � 1 = �� = (199)(25)  = 1788 ��  64  � 2 = 1788 − 785 = 1003 �� 1 � 2

=

1788 1003

= 1�783

1

1

1

(d) 2 F = F 1 + F 22 2 o

2

1

1

1

2� �2 = (1788) 2 + (1003) 2 � � = 1367 ��

� � =

1367 25

= 54�7 ��

Approximately less than Taylor’s recommendation ( = 70 lb/in.) (e) L ≈ 2C + 1.57( D2 + D1 ) +

( D2 − D1 )2

� = 2[(6)(12 ) + 11] + 1�57(70 + 13) +

4C

(70 −13)2 = 306 �� 4[(6)(12 ) + 11]

(f) More economical basis π D1n1 vm = 12 π D1 (865 ) 4500 = 12 D1 = 19.87 in use D1 = 20 in

764

SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS 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; C = 54 in .; D1 = 14 in . (motor), D2 = 54 in ., 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 F 1 F 2 ? (d) If the belt has stretched until s = 200 psi on the tight side, what is F 1 F 2 ? (e) Compute the belt length.

Solution: (a) For medium double leather belt 20 t = in 64 hp = (hp in )(b )C mC p C f Table 17.1 and 17.2 C m = 0.67 C p = 0.8 C f = 0.74

b = 10 in vm =

π D1n1

=

π (14 )(880 )

12 hp in = 6.6625

12

=

3225 fpm

hp = (6.6625)(10 )(0.67 )(0.8)(0.74) = 26.43 hp

 12 ρ vs2  e f θ − 1   f θ  (b) F 1 − F 2 = bt  s − 32 . 2   e  b = 10 in t =

20

in 64 ρ = 0.035 lb cu. in. vs =

3225

= 53.75 fps 60 D2 − D1

θ = π −

θ = π −

C 54 − 14

= 2.4 rad 54 Leather on iron f = 0.3

f θ = (0.3)(2.4 ) = 0.72

765


e f − 1

= 0.51325 e0.72 33,000(26.43) = 270 lb F 1 − F 2 = 3225 θ

e f

=

e0.72 − 1

2  20   12(0.035)(53.75 )  F 1 − F 2 = 270 = (10 ) s −  (0.51325 ) 32.2  64   

s = 206 psi

 20   = 644 lb  64 

(c) F 1 = sbt = (206)(10) F 2 = 644 − 270 = 374 lb F 1 F 2

=

644 374

= 1.72

(d) s = 200 psi

 20   = 625 lb 64  

F 1 = sbt = (200)(10)

F 2 = 625 − 270 = 355 lb F 1 F 2

=

625 355

= 1.76

(e) L ≈ 2C + 1.57( D2 + D1 ) +

( D2 − D1 )2 4C

(54 − 14)2 L = 2(54) + 1.57(54 + 14) + 4(54 ) 847.

= 222 in

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; D2 = 12 in . (compressor shaft); C = 5 ft . The belt has been designed for a net belt pull of F 1 − F 2 = 40 lb in of width and F 1 F 2 = 3 . 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: π D1n1 π (8)(1750) = = 3665 fpm (a) vm = 12 12 b = 6 in F 1 − F 2 = (40)(6) = 240 lb

766


hp =

(F 1 − F 2 )vm 33,000

=

(240)(3665 ) 33,000

= 26.65 hp

(b) F 1 = 3F 2 3F 2 − F 2 = 240 lb F 2 = 120 lb F 1 = 360 lb s=

F 1

bt For heavy single-ply leather belt 13 t = in 64 360 = 295 psi s=  13  (6)   64 

 12 ρ vs2  e f θ − 1   f θ  (c) F 1 − F 2 = bt  s − 32 . 2   e  ρ = 0.035 lb cu. in. vs =

3665

= 61.1 fps 60 F 1 − F 2 = 240 lb 2 12(0.035 )(61.1)  e f θ − 1   13   F 1 − F 2 = 240 = (6)   295 −  f θ  64 32 . 2     e 

θ

e f − 1 θ

e f

= 0.7995

θ = π −

θ = π −

D2 − D1 C 12 − 8

5(12)

= 3.075 rad

θ e f = 4.9875 f θ = 1.607

f (3.075) = 1.607 f = 0.5226

(d) ALBA Tables (Table 17.1 and 17.2) hp = (hp in )(b )C mC p C f vm = 3665 fpm hp in = 6.965

767

SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS b = 10 in C m = 1.0 (assumed) C p = 0.6 C f = 0.74

hp = (6.965)(6 )(1.0 )(0.6 )(0.74) = 18.6 hp < 26.65 hp

848.

A 10-in. medium double leather belt, cemented joints, transmits 60 hp from a 9in. paper pulley to a 15-in. pulley on a mine fab; dusty conditions. The compensator-started motor turns 1750 rpm; C = 42 in . This is an actual installation. (a) Determine the horsepower from the ALBA tables. (b) Using the general equation, determine the horsepower for this belt. (c) Estimate the service factor from Table 17.7 and apply it to the answer in (b). Does this result in better or worse agreement of (a) and (b)? What is your opinion as to the life of the belt?

Solution: π D1n1 π (9)(1750 ) vm = = = 4123 fpm 12 12 (a) hp = (hp in )(b )C mC pC f Table 17.1 and 17.2 Medium double leather belt 20 t = in 64 vm = 4123 fpm hp in = 11.15 C m = 0.67 C p = 0.7 C f = 0.74

b = 10 in hp = (11.15)(10)(0.67 )(0.7)(0.74) = 38.7 hp

 12 ρ vs2  e f θ − 1   f θ  (b) F 1 − F 2 = bt  s − 32 . 2   e  b = 10 in

ρ = 0.035 lb cu. in. s = 400η η = 1.0 cemented joint s = 400 psi

θ = π −

D2 − D1 C

768


θ = π −

15 − 9

= 2.9987 rad 42 Leather on paper pulleys, f = 0.5

f θ = (0.5)(2.9987 ) = 1.5 θ

e f − 1 e f θ vs =

= 0.77687

4123 60

=

68.72 fps

12(0.035)(68.72)  20   F 1 − F 2 = (10 )  400 − 32.2  64  

2

hp =

(F 1 − F 2 )vm 33,000

=

(822 )(4123) 33,000

  (0.77687 ) = 822 lb 

= 102.7 hp

(c) Table 17.7 N sf = 1.6 hp =

102.7

= 64.2 hp < 102.7 hp 1.6 Therefore, better agreement

Life of belt, not continuous, 60 hp > 38.7 hp . 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:

 12 ρ vs2  e f θ − 1   f θ  F 1 − F 2 = bt  s − 32 . 2   e  hp = hp =

(F 1 − F 2 )vm 33,000 60(F 1 − F 2 )vs 33,000

60vs bt  12 ρ vs2  e f θ − 1  s −   hp = 33,000  32.2  e f θ  60bt  e f θ − 1  12 ρ vs2    s − vs hp = 33,000  e f θ  32.2 

769

SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS d (hp ) d (vs ) s=

60bt  e f θ − 1   12 ρ vs2  24 ρ vs2     s − − = =0 33,000  e f θ   32.2  32.2 

36 ρ vs2

32.2 ρ = 0.035 lb cu. in. (a) s = 400 psi 400 =

36(0.035)vs2

32.2 vs = 101.105 fps vm = 6066 fpm

(b) s = 320 psi 320 =

36(0.035)vs2

32.2 vs = 90.431 fps vm = 5426 fpm

(c) Larger than those mentioned in §17.9 (4000 – 4500 fpm) (d) Rubber belt, ρ = 0.045 lb cu. in. (a) s = 400 psi 400 =

36(0.045)vs2

32.2 vs = 89.166 fps vm = 5350 fpm < 6066 fpm

Therefore, speeds for a rubber belt is smaller. 850.

A 40-in. pulley transmits power to a 20-in. pulley by means of a medium double leather belt, 20 in. wide; C = 14 ft , let f = 0.3 . (a) What is the speed of the 40-in pulley in order to stress the belt to 300 psi at zero power? (b) What maximum horsepower can be transmitted if the indicated stress in the belt is 300 psi? What is the speed of the belt when this power is transmitted? (See HINT in 849).

Solution:

 12 ρ vs2  e f θ − 1   f θ  F 1 − F 2 = bt  s − 32 . 2   e 

770


hp =

60(F 1 − F 2 )vs 33,000

60bt  e f θ − 1  12 ρ vs2    s − vs hp = 33,000  e f θ  32.2  d (hp ) d (vs ) s=

60bt  e f θ − 1   12 ρ vs2  24 ρ vs2     s − − = =0 33,000  e f θ   32.2  32.2 

36 ρ vs2

for maximum power 32.2 (a) At zero power: s=

12 ρ vs2

32.2 s = 300 psi

ρ = 0.035 lb cu. in. 300 =

12(0.035)vs2

32.2 vs = 151.6575 fps vm = 9100 fpm

Speed, 40 in pulley, n2 =

12vm

π D2

=

12(9100 )

π (40 )

= 869 rpm

(b) Maximum power 36 ρ vs2 s= 32.2 36(0.035)vs2 300 = 32.2 vs = 87.5595 fps vm = 5254 fpm 60bt  e f θ − 1  12 ρ vs2    s − vs hp = 33,000  e f θ  32.2  t =

20

in 64 b = 20 in

θ = π −

D2 − D1

θ = π −

C 40 − 20

14(12)

= 3.0225 rad

f = 0.3

771

SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS f θ = (0.3)(3.0225 ) = 0.90675 e f θ − 1 e f θ

= 0.5962

 20   2  64  (0.5962 )300 − 12(0.035)(87.5595)  (87.5595) = 118.64   33,000 32.2  

60(20 ) hp =

vm = 5254 fpm

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; C = 84 in . A tension pulley is placed so that the angle of contact on the motor o o pulley is 193 and on the compressor pulley, 240 . A 12-in. medium double leather belt with a cemented joint is used. (a) What will be the tension in the tight 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) F 1 = sbt b = 12 in t =

20 64

in

 20    64 

F 1 = (375)(12 )

33,000hp

(b) F 1 − F 2 = vm =

π D1n1

vm

=

π (12 )(1200 )

12 12 Table 17.7, N sf = 1.2 F 1 − F 2 =

= 3770 fpm

33,000(1.2 )(100 )

= 1050 lb 3770 F 2 = F 1 − 1050 = 1406 − 1050 = 356 lb

 12 ρ vs2  e f θ − 1   f θ  (c) F 1 − F 2 = bt  s − 32 . 2   e  vs =

3770 60

=

62.83 fps

772


ρ = 0.035 lb cu. in. 12(0.035)(62.83)  e f θ − 1   20   1050 = (12 )  375 −  e f θ  32.2  64     θ

e f − 1

= 0.8655 θ e f f θ = 2.006 Motor pulley

θ = 193

o

 π   = 3.3685 rad 180  

= 193

f (3.3685 ) = 2.006 f = 0.5955

Compressor Pulley

θ = 2403

o

=

 π   = 4.1888 rad 180  

240

f (4.1888) = 2.006 f = 0.4789

(d) Force:

Without tension pulley 78 − 12 D − D1 θ 1 = π − 2 = π − = 2.356 rad 84 C 78 − 12 D − D1 θ 2 = π + 2 = π + = 3.9273 rad 84 C

773


α 1

= θ 1′ − θ 1 −

α 2

=

θ 2 − π

π − θ 1 2

= 3.3685 − 2.356 −

+ θ 2′ − θ 2 =

3.9273 − π

π − 2.356 2

=

0.6197 rad = 35.5

+ 4.1888 − 3.9273 =

o

0.6544 rad = 37.5

o

2 2 Q = F 1 (sin α 1 + sin α 2 ) = 1406(sin 35.5 + sin 37.5) = 1672 lb of force exerted Size of pulley; For medium double leather belt, vm = 3770 fpm , width = 12 in > 8 in D = 6 + 2 = 8 in

774

SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS 852.

A 40-hp motor, weighing 1915 lb., runs at 685 rpm and is mounted on a pivoted 3 in . The center of the 11 ½-in. base. In Fig. 17.11, Text , e = 10 in ., h = 19 16 motor pulley is 11 ½ in. lower than the center of the 60-in. driven pulley; C = 48 in . (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 the width of the medium double leather belt if s = 300 psi ? (c) What coefficient of friction is indicated by the general belt equation? ( Data courtesy of Rockwood Mfg. Co.)

Solution: (a)

R = 1915 lb Graphically b ≈ 26 in a ≈ 9 in

[∑ M

B

=0

]

eR = F 1a + F 2b

(10)(1915) = ( F 1 )(9) + (F 2 )(26 ) 9 F 1 + 26 F 2 = 19,150 For compensator started hp = 1.4(rated hp ) = 1.4(40 ) = 56 hp 33,000hp F 1 − F 2 = vm 775


vm =

π D1n1 12

F 1 − F 2 =

=

π (11.5)(685)

12 33,000(56)

=

2062 fpm

= 896 lb 2062 F 2 = F 1 − 896 Substituting 9 F 1 + 26(F 1 − 896) = 19,150

F 1 = 1213 lb F 2 = 1213 − 896 = 317 lb

For medium leather belt, t =

20 64

in

F 1 = sbt

 20    64 

1213 = (300)(b ) b = 13 in


2062

= 34.37 fps 60 ρ = 0.035 lb cu. in.

12(0.035)(34.37 )  e f θ − 1   20    f θ  896 = (13)  300 −  32.2  64     e  e f θ − 1

= 0.775 f θ e f θ = 1.492

θ = π −

D2 − D1

= π −

C f (2.1312) = 1.492 f = 0.70

853.

60 − 11.5 48

=

2.1312 rad

A 50-hp motor, weighing 1900 lb., is mounted on a pivoted base, turns 1140 rpm, 3 and drives a reciprocating compressor; in Fig. 17.11, Text, e = 8 in ., 4 5 h = 17 in . The center of the 12-in. motor pulley is on the same level as the 16 center of the 54-in. compressor pulley; C = 40 in . (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 776

SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS leather belt, 11 in. wide? (c) What coefficient of friction is indicated by the general belt equation? ( Data courtesy of Rockwood Mfg. Co.) Solution: (a) For compensator-started hp = 1.4(50 ) = 70 hp F 1 − F 2 = vm =

33,000hp vm

π D1n1 12

F 1 − F 2 =

=

π (12 )(1140 )

12 33,000(70 ) 2062

=

= 3581 fpm

645 lb

b ≈ 25 in a ≈ 5 in R = 1900 lb eR = F 1a + F 2b

(8.75)(1900 ) = F 1 (5) + F 2 (25) F 1+5F 2 = 3325 lb

645 + F 2+5 F 2 = 3325 lb F 2 = 447 lb F 1 = 645 + F 2 = 645 + 447 = 1092 lb

(b) For heavy double leather belt 23 t = in 64 b = 11 in

777


s=

F 1 bt

=

1092 = 276 psi  20  (11)   64 


3581

= 59.68 fps 60 ρ = 0.035 lb cu. in.

12(0.035)(59.68)  e − 1   23   645 = (11)  276 −  e f θ  32.2  64     f θ

f θ = 1.241

θ = π −

D2 − D1

= π −

54 − 12

C f (2.092) = 1.492 f = 0.60

40

=

2.092 rad

RUBBER BELTS 854.

A 5-ply rubber belt transmits 20 horsepower to drive a mine fan. An 8-in., motor pulley turns 1150 rpm; D2 = 36 in ., fan pulley; C = 23 ft . (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. What are the indications for a good life?

Solution: (a) θ = π −

D2 − D1 C

= π −

36 − 8 23(12 )

= 3.040 rad = 174

K θ = 0.976 hp =

bvm N p K θ

2400 K θ = 0.976

vm =

π D1n1

12 N p = 5 hp = 20 =

=

π (8)(1150 ) 12

=

2409 fpm

b(2409 )(5)(0.976 )

2400

b = 4.1 in

min. b = 5 in

778

o

SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS (b) With b = 9 in is safe for good life. 855.

A 20-in., 10-ply rubber belt transmits power from a 300-hp motor, running at 650 rpm, 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 o belt must operate at an angle 75 with the horizontal. What is the overload capacity of this belt if the rated capacity is as defined in §17.15, Text ?

Solution: bvm N p hp = 2400 b = 20 in π D1n1 π (33)(650 ) vm = = = 5616 fpm 12 12 N p = 10 hp =

(20 )(5616)(10 ) 2400

=

Overlaod Capacity =

468 hp 468 − 300 300

(100%) = 56%

V-BELTS NOTE: If manufacturer’s 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; C = 43 to 49 in . 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; C ≈ 45.2 in .

Solution: Table 17.7 N sf = 1.2 + 0.2 = 1.4 (24 hr/day) Design hp = N sf (transmitted hp) = (1.4 )(125) = 175 hp Fig. 17.4, 175 hp, 1180 rpm Dmin = 13 in , D-section D2 D1

=

1180 340

=

50 14.4

use D1 = 14.4 in > 13 in D2 = 50 in

779


vm =

π D1n1 12

=

π (14.4 )(1180) 12

=

4449 fpm

  103  0.09 vm2  vm c  −  3 −e Rated hp =  a 6 10 v K D   m   10 d 1 Table 17.3, D-section a = 18.788 , c = 137.7 , e = 0.0848 D Table 17.4, 2 = 3.47 D1 K d = 1.14 0.09   103  (4449 )2  4449 137.7  −  3 = 28.294 hp − (0.0848) Rated hp = 18.788 6 ( )( ) 4449 1 . 14 14 . 4 10   10  

Back to Fig. 17.14, C-section must be used. a = 8.792 , c = 38.819 , e = 0.0416

  103  0.09 vm2  vm c  −  3 −e Rated hp =  a 6 10 v K D   m   10 d 1 0.09   103  (4449 )2  4449 38.819  −  3 Rated hp = 8.792 − (0.0416 ) 6 ( )( ) 4449 1 . 14 14 . 4 10   10   Adjusted rated hp = K θ K L (rated hp ) Table 17.5, D2 − D1 50 − 14.4 C

=

46

=

0.77

K θ = 0.88

Table 17.6 L ≈ 2C + 1.57( D2 + D1 ) +

( D2 − D1 )2 4C

(50 − 14.4)2 L = 2(46) + 1.57(50 + 14.4 ) + 4(46 )

=

200 in

use C195, L = 197.9 in K L = 1.07

Adjusted rated hp = (0.88)(1.07 )(20 ) = 18.83 hp Design hp 175 No. of belts = = = 9.3 belts use 9 belts Adjusted rated hp 18.83 Use 9 , C195 V-belts with 14.4 in and 50 in sheaves

780

=

20.0 hp


B + B − 32( D2 − D1 )

2

2

C =

16 B = 4 L − 6.28( D2 + D1 ) = 4(197.9) − 6.28(50 + 14.4 ) = 387.2 in 387.2 +

C =

(387.2 )2 − 32(50 − 14.4)2 16

857.

=

44.9 in

A 50-hp, 1160-rpm, AC split-phase motor is to be used to drive a reciprocating pump 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; C ≈ 50 in . Determine the details of a multiple V-belt drive for this application. The Dodge Manufacturing Corporation recommended a Dyna-V Drive consisting of six 5V1800 belts with 10.9-in. and 37.5-in. sheaves; C ≈ 50.2 in .

Solution: Table 17.7, (12 hr/day) N sf = 1.4 − 0.2 = 1.2 Design hp = (1.2 )(1.75)(50 ) = 105 hp Fig. 17.4, 105 hp, 1160 rpm Dmin = 13 in , D-section D2 D1

=

1160 330

≈

46.4 13.2

use D1 = 13.2 in > 13 in D2 = 46.4 in vm =

π D1n1 12

=

π (13.2 )(1160) = 4009 fpm 12

  103  0.09 vm2  vm c  −  3 Rated hp =  a −e 6 v K D 10   m   10 d 1 Table 17.3, D-section a = 18.788 , c = 137.7 , e = 0.0848 D 46.4 Table 17.4, 2 = = 3.5 D1 13.2 K d = 1.14 0.09   103  (4009 )2  4009 137.7  −  3 − (0.0848) Rated hp = 18.788 6 ( )( ) 4009 1 . 14 13 . 2 10   10  

Back to Fig. 17.14, C-section must be used. a = 8.792 , c = 38.819 , e = 0.0416 Dmin = 9 in

781

=

24.32 hp

SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS D2

=

D1

1160 330

≈

32 9. 1

use D1 = 9.1 in π D1n1 π (9.1)(1160 ) vm = = = 2764 fpm 12 12 0.09   103  (2764 )2  2764 38.819  −  3 = 10.96 hp − (0.0416 ) Rated hp = 8.792 6 ( )( ) 2764 1 . 14 9 . 1 10   10  

Adjusted rated hp = K θ K L (rated hp ) Table 17.5, D2 − D1 32 − 9.1 C

=

50

= 0.458

K θ = 0.935

Table 17.6 L ≈ 2C + 1.57( D2 + D1 ) +

( D2 − D1 )2 4C

(32 − 9.1)2 L = 2(50 ) + 1.57(32 + 9.1) + = 167 in 4(50 ) use C158, L = 160.9 in K L = 1.02

Adjusted rated hp = (0.935)(1.02)(10.96 ) = 10.45 hp No. of belts =

Design hp Adjusted rated hp

105 10.43

= 10 belts

B + B − 32( D2 − D1 )

2

2

C =

=

16 B = 4 L − 6.28( D2 + D1 ) = 4(160.9) − 6.28(32 + 9.1) = 385.5 in C =

385.5 +

(385.5)2 − 32(32 − 9.1)2

16 Use 10-C158 belts, D1 = 9.1 in

=

46.8 in

D2 = 32 in , C = 46.8 in

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; C = 113 to 123 in . 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 120.175in. pulley; C ≈ 116.3 in .

Solution: 782

SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS Table 17.7 N sf = 1.6 − 0.2 = 1.4 hp = (1.4)(200) = 280 hp Fig. 17.14, 280 hp, 600 rpm Use Section E But in Table 17.3, section E is not available, use section D Dmin = 13 D2 D1

=

600 125

= 4.8

For D1max : min C = 113 =

D1 + D2

2 D1 + 4.8 D1 2

+ D1 +

D1

D1 = 28 in

min C = D2 D2 = 113 in D1 =

113 4.8

=

23.5 in

1

(13 + 23.5) = 18 in 2 D2 = (4.8)(18) = 86.4 in use D1 ≈

L ≈ 2C + 1.57( D2 + D1 ) +

( D2 − D1 )2 4C

(86.4 − 18)2 L = 2(118) + 1.57(86.4 + 18) + 4(118)

= 410 in

using D1 = 19 in , D2 = 91.2 in , C = 118 in

(91.2 − 19)2 L = 2(118) + 1.57(91.2 + 19) + 4(118)

= 420 in

Therefore use D420 sections D1 = 19 in , D2 = 91.2 in π D1n1 π (19)(600) = = 2985 fpm vm = 12 12 2   103  0.09 vm  vm c  −  3 Rated hp =  a −e 6 v K D 10   m   10 d 1

Table 17.3, D-section a = 18.788 , c = 137.7 , e = 0.0848 D Table 17.4, 2 = 4.8 D1 783

SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS K d = 1.14 0.09   103  (2985)2  2985 137.7  −  3 = 29.6 hp Rated hp = 18.788 − (0.0848) 6 ( )( ) 2985 1 . 14 19 10   10  

Therefore, Fig. 17.14, section D is used. Adjusted rated hp = K θ K L (rated hp ) Table 17.5, D2 − D1 91.2 − 19 =

118 K θ = 0.83 (V-flat) C

=

0.612

Table 17.6, D420 L = 420.8 in K L = 1.12

Adjusted rated hp = (0.83)(1.12 )(29.6 ) = 27.52 hp Design hp 280 No. of belts = = = 10 belts Adjusted rated hp 27.52 Use10 , D420, D1 = 19 in , D2 = 91.2 in , C = 118 in 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 Vbelts, 17.24-in sheave, 61-in. pully, C ≈ 69.7 in .

Solution: Table 17.7, N sf = 1.6 − 0.2 = 1.4 Design hp = (1.4 )(150) = 210 hp Fig. 17.4, 210 hp, 700 rpm Dmin = 13 in , D-section

  103  0.09 vm2  vm c  −  3 −e Rated hp =  a 6 v K D 10   m   10 d 1 For Max. Rated hp,

 v  Rated hp = a m3   10  Let X =

d (hp )

 v  d  m3   10  0.91

=

0

 vm   vm  −  3  − e 3  K d D1  10   10  c

3

vm

103

784


hp = aX 0.91 −

c K d D1

X − eX 3

π D1n1 π D1 (700) = 103 12 × 103 12 × 103 12 ×103 X D1 = 700π 700π c 3 hp = aX 0.91 − − eX 3 12 × 10 K d X =

vm

d (hp ) d ( X )

=

− 0.09

= 0.91aX

X 2.09 =

2

− 3eX = 0

0.91a

3e Table 17.3, D-section a = 18.788 , c = 137.7 , e = 0.0848 2.09

X

 vm  = 3  10 

2.09

=

0.91(18.788) 3(0.0848 )

vm = 7488 fpm

π D1n1 = 7488 12 π D1 (700) = 7488 vm = 12 D1 = 40.86 in max D1 = 40.86 in 1 ave. D1 = (13 + 40.86 ) = 26.93 in 2 use D1 = 22 in D2 700 79 = ≈ D1 195 22 vm =

D1 = 22 in , D2 = 79 in

Min. C =

D1 + D2

+ D1 =

22 + 79

2 Or Min. C = D2 = 79 in

L ≈ 2C + 1.57( D2 + D1 ) +

2

+ 22 = 72.5 in

( D2 − D1 )2 4C

(79 − 22)2 L = 2(79) + 1.57(79 + 22 ) + 4(79 )

= 327 in

use D330, L = 330.8 in B + B − 32( D2 − D1 )

2

2

C =

16 785

SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS B = 4 L − 6.28( D2 + D1 ) = 4(330.8) − 6.28(79 + 22 ) = 689 in C =

689 +

π D1n1 12 K d = 1.14

vm =

(689)2 − 32(79 − 22)2 =

16 π (22 )(700) 12

=

= 81.12 in

4032 fpm

0.09   103  (4032 )2  4032 137.7  −  3 Rated hp = 18.788 − (0.0848) 6 ( )( ) 4032 1 . 14 22 10   10  

= 39.124 hp

Adjusted rated hp = K θ K L (rated hp ) Table 17.5, D2 − D1 79 − 22 C

=

81.12 K θ = 0.84 (V-flat)

= 0.70

Table 17.6 D330 K L = 1.07 Adjusted rated hp = (0.84 )(1.07)(39.124) = 35.165 hp Design hp 210 No. of belts = = = 5.97 belts use 6 belts Adjusted rated hp 35.165 Use 6 , D330 V-belts , D1 = 22 in , D2 = 79 in , C ≈ 81.1 in 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 24 hp at 238 rpm; 44 < C < 50 in .; 20 hr./day operation with no overload. Decide upon the size and number of V-belts, sheave sizes, and belt length. ( Data courtesy of The Worthington Corporation.)

Solution: Table 17.7 N sf = 1.6 + 0.2 = 1.8 Design hp = (1.8)(30) = 54 hp Speed of fan at 30 hp 30 − 24 (280 − 238) + 238 = 286 rpm n2 = 29.3 − 24 at 54 hp, 1160 rpm. Fig. 17.4 use either section C or section D Minimum center distance: C = D2

786


or C = D2

=

D1

D1 + D2

2 1160 286

+ D1

= 4.056

use C = 4.056 D1 44 in < C < 50 in , use C = 47 in D1max =

47

= 11.6 in 4.056 use C-section, Dmin = 9 in Let �1 = 10�1 �� , �2 = 41 ��

L ≈ 2C + 1.57( D2 + D1 ) +

( D2 − D1 )2 4C

(41 − 10.1)2 L = 2(47 ) + 1.57(41 + 10.1) + = 179.3 in 4(47 ) use C137, L = 175.9 in B + B − 32( D2 − D1 )

2

2

C =

16 B = 4 L − 6.28( D2 + D1 ) = 4(175.9) − 6.28(41 + 10.1) = 328.7 in 382.7 +

C =

(382.7 )2 − 32(41 − 10.1)2

16 C173, satisfies 44 in < C < 50 in

 v  Rated hp = a m3   10  π D1n1 12 Table 17.4 vm =

D2 D1

=

0.91

=

45.2 in ≈ 44 in

 vm   vm  −  3  − e 3  K d D1  10   10  c

3

π (10.1)(1160 ) = 3067 fpm 12

= 4.056 , K d = 1.14

Table 17.3, C-section a = 8.792 , c = 38.819 , e = 0.0416 0.09   103  (3067 )2  3067 38.819  −  3 = 12.838 hp − (0.0416 ) Rated hp = 8.792 6 ( )( ) 3067 1 . 14 10 . 1 10   10  

Adjusted rated hp = K θ K L (rated hp ) Table 17.5, D2 − D1 41 − 10.1 C

=

45.2

= 0.68

K θ = 0.90

787

SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS Table 17.6 L = 175.9 , C173 K L = 1.04 Adjusted rated hp = (0.90 )(1.04)(12.838) = 12.02 hp No. of belts =

Design hp Adjusted rated hp

=

54 12.02

= 4.5 belts use

5 belts

Use 5 , C173 V-belts , D1 = 10.1 in , D2 = 41 in POWER CHAINS NOTE: If manufacturer’s 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 4cylinder Diesel engine to a counter-shaft; engine speed 1000 rpm, counter-shaft speed 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 of chains 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 to change 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) design hp = 2(30 ) = 60 hp intermittent D2 D1

≈

n1 n2

=

1000 500

=

2

D2 = 2 D1 C = D2 +

2 D1 +

D1

2

D1

2 =

=

24 in

24

�1�� = 9�6 �� 2�� = 2�1�� = 2(9�6) = 19�2 ��

π D1n1 π (9.6 )(1000 ) = = 2513 fpm 12 12 Table 17.8, use Chain No. 35, Limiting Speed = 2800 fpm vm =

788

SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS Minimum number of teeth Assume N 1 = 21 N 2 = 2 N 1 = 42 [Roller-Bushing Impact] 1.5

 100 N ts  0.8 hp = K r   P  n  Chain No. 35 3 P = in 8 N ts = 21 n = 1000 rpm K r = 29 1.5

100(21)   3  hp = 29    1000   8 

0.8

=

40.3 hp

[Link Plate Fatigue] P − hp = 0.004 N ts1.08 n 0.9 P 3 0.07 hp = 0.004(21)

1.08

0.9  3 

 3    8 

3− 0.07 

(1000 )  

 8 

No. of strands =

design hp rated hp

Use Chain No. 35, P =

=

60 2.91

= 2.91 hp

= 21

3

in , 21 strands 8 Check for diameter and velocity �

�1 =

=

 180    �� 

��

� � =

π �1�1 12

=

0�375 = 2�516 ��  180 

��

  21 

π (2�516)(1000) 12

= 659 ��

Therefore, we can use higher size, Max. pitch  180  180   = 9�6 �� � = �1 ��   = 1�43 ��  21   ��  say Chain no. 80 � = 1 �� 1�08

�� = 0�004(21)

(1000)0�9 (1)3−0�07 (1) = 53�7 ��

A single-strand is underdesign, two strands will give almost twice over design, Try Chain no. 60

789


�=

3 4

��

1�08

�� = 0�004(21) ��

=

23

 180   �  � 

π �1�1

3 strands

0�75 = 5�0 ��  180 

��

  21 

π (5�0 )(1000)

=

12

= 2�61 or

=

��

� � =

= 23 ��

 4 

60

�

�1 =

3  (1000)0�9   

 3    4 

3− 0�07 

12

= 1309 ��

The answer is Chain No. 60, with P = ¾ in and 3 chains, limiting velocity is 1800 fpm (b) L ≈ 2C + � =

N 1 + N 2

2

+

( N 2 − N 1 )2 40C

pitches

24 = 32  3 

   4  N 1 = 21 N 2 = 42

� = 2(32) +

21 + 42 2

(42 − 21)2 + = 95�845 �� ≈ 96 �� 40(32 )

Amount of slack 1 2 2

h = 0.433(S − L 2

)

L = C = 24 in 3  (96 − 95�845)  ��  4  = 24�058�� = 24 �� + 2

[

2

� = 0�433 (24�058) − (24)

2

1

]

2

= 0�7229 ��

(c) pb = bearing pressure Table 17.8, Chain No. 60 � = 0�234 �� =

1

�� 2 � = 0�094 ��

1  2 � = � (� + 2 � ) = 0�234 + 2(0�094) = 0�160992 �� 2  FV

33,000

= 60 hp

790

SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS � (1309)

= 60 �� 33�000 � = 1512�6 ��

� =

1512�6

�� =

= 504�2 ��

3 504�2

0�160992

862.

= 3131 ��

A conveyor is driven by a 2-hp high-starting-torque electric motor through a flexible coupling to a worm-gear speed reducer, whose mw ≈ 35 , 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, C = 24 in ., L = 88 pitches .

Solution: Table 17.7 N sf = 1.2 − 0.2 = 1.0 (8 hr/day) design hp = 1.0(2 ) = 2.0 hp n1 =

1750

= 50 rpm 35 n2 = 12 rpm Minimum number of teeth = 12 Use N 1 = 12 [Link Plate Fatigue] P hp = 0.004 N ts1.08 n 0.9 P 3− 0.07

P 3−0.07 ≈ P 3 =

hp

P

1.08

0.004 N ts n

0.9

=

2.0 1.08

0.004(12)

(50)0.9

Use Chain No. 80, P = 1.0 in To check for roller-bushing fatigue 1.5

 100 N ts  0.8 hp = K r   P  n  K r = 29 1.5

100(12 ) 0.8 ( ) 1 hp = 17    1000 

=

2747 hp > 2 hp

(a) N 1 = 12

 n   50  N 2 =  1  N 1 =  (12 ) = 50 teeth  12   n2  791

= 1.0


C = D2 + D1 ≈ D2 ≈

D1

PN 1

π PN 1

π

2 =

(1.0 )(12) π

(1.0 )(50)

=

C ≈ 15.92 +

π 3.82 2

= 3.82 in = 15.92 in

= 17.83 in

use C = 18 in C = 18 pitches chain pitch = 1.0 in, Chain No. 80 (b) L ≈ 2C + L ≈ 2(18) +

N 1 + N 2

2

12 + 59

2 use L = 70 pitches

+

( N 2 − N 1 )2 40C

(50 − 12)2 + 40(18)

= 69 pitches

(c) pb = bearing pressure Table 17.8, Chain No. 80 C = 0.312 in 5 E = in 8 J = 0.125 in PN ts n1 (1)(12 )(50 ) vm = = = 50 fpm 12 12 3  2 A = C ( E + 2 J ) = 0.141 + 2(0.05) = 0.04054 in 16  FV = 60 hp 33,000 33,000(2 ) F = = 1320 lb 50 1320 F pb = = = 4835 psi C ( E + 2 J ) 5  0.312  + 2(0.125 ) 8  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 n = 500 rpm . The 1 ¼-in. driven shaft turns 250 rpm; C ≈ 16 in . (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 center 792

SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS distance 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 N sf = 1.2 design hp = 1.2(5) = 6 hp D2

=

D1

500

=

250

C = D2 +

2

D1 2 D1

16 = 2 D1 +

2

�1�� = 6�4 �� 2�� = 2�1�� = 2(6�4 ) = 12�8 ��

vm =

π D1n1

π (6.4 )(500 )

=

12 12 (a) Link Plate Fatigue P hp = 0.004 N ts1.08 n 0.9 P 3− 0.07

= 838 fpm

�� = �1 = 21 1�08

�� = 0�004(21)

(500)0�9 � 3−0�07 �

Max. pitch  180  180   = 6�4 �� � = �1 ��   = 0�95 ��  21   ��  Try chain no. 60 �=

5 8

��

1�08

�� = 0�004(21)

0�9  5 

(500)  

 5  3− 0�07    8 

 8 

= 7�17 ��

5

use � = �� , Chain No. 60 �1 =

8 �

=

0�625 = 4�194 ��  180 

=

0�625 = 8�364 ��  180 

 180   ��   21  �  1   �   500  �2 = �1  1  = 21  = 42 250 �    2  ��

�2 =

�

 180   �  2 

��

��

  42 

793


Size of sprocket, �1 = 21 , �2 = 42 , � = �� , �1 = 4�194 �� , �2 = 8�364 �� 8

(b) C = 16 in � =

16 �� = 25�6 �� 5 8

��

L ≈ 2C +

N 1 + N 2

2

� ≈ 2(25�6) +

+

21 + 42 2

( N 2 − N 1 )2 40C

(42 − 21)2 + = 83�13 �� 84 pitches 40(25�6)

use � = 84 �� (c) Method: π �1�1 π (4�194)(500) � � =

12

=

12

= 549 �� .

Use Type II Lubrication ( vmax = 1300 fpm ) – oil is supplied from a drip lubricator to link plate 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 center distance. 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 N sf = 1.4 + 0.2 (24 hr/day) design hp = 1.6(20) = 32 hp

(a) D2 D1

n1 n2 ≈

=

n1 n2

570 200

= 2.85

= 2.85

Considering single strand P hp = 0.004 N ts1.08 n 0.9 P 3− 0.07 min N ts = 17 1.08

hp = 32 = 0.004(17 )

(570 )0.9 P 3

−0.07 P

P 3−0.07 = 1.24 P = 1.07 in P

use P = 1.0 in 794

SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS 1. 08

hp = 32 = 0.004( N 1 )

−0.07 (1)

(570 )0.9 (1)3

N 1 = 21

 570  (21) = 60 200  

N 2 = 

5

Roller width = C = D2 + D1 ≈ D2 ≈

D1 2

PN 1 PN 2

22�44 1

π

=

C = 19.10 + � =

(1)(21)

=

π π

in

8

= 6.685 in

(1)(60 ) π

6.685 2

= 19.10 in

=

22.44 in

= 22�44 ��

Considering multiple strands 5

Assume, � = �� 8

hp = 0.004 N ts1.08 n 0.9 P 3− 0.07 1�08

�� = 0�004(21)

P

(570)0�9 (0�625)3−0�07(0�625 ) = 8�07 ��

No. of strands =

32 �� 8�07 ��

= 4�0

Use 4 strands Roller width = 0�4 ��

 5 ( )   21 ��1  8  �1 ≈ = = 4� 178 �� π

π

 5 ( )   60 ��2  8  �2 ≈ = = 11�937 �� π

� = 11�937 + � =

π

4�178 2

= 14�026 ��

14�026 = 22�44 �� 5 8

(b) Chain Length For single strands

795


L ≈ 2C +

N 1 + N 2

2

� ≈ 2(22�44) +

+

21 + 60 2

( N 2 − N 1 )2 40C

(60 − 21)2 + = 87�08 pitches 40(22�44 )

use � = 88 �� For multiple strands L ≈ 2C +

N 1 + N 2

2

� ≈ 2(22�44) +

+

21 + 60 2

( N 2 − N 1 )2 40C

(60 − 21)2 + = 87�08 pitches 40(22�44 )

use � = 88 ��

(c) pb = bearing pressure Table 17.8, P = 1 in E =

5

in 8 J = 0.125 in C = 0.312 in F = vm = F =

33,000hp vm

π D1n1

=

π (6.685 )(570 )

12 33,000(32 )

12 = 1058 lb

998 F

pb =

= 998 fpm

C ( E + 2 J )

=

1058

5  0.312  + 2(0.125 ) 8 

= 3876 psi

5

Table 17.8, � = �� 8

� =

3

��

8 � = 0�080 �� = 0�200 ��

F = vm =

33,000hp vm

π D1n1 12

=

π (6.685 )(570 ) 12

= 998 fpm

796


F = � =

33,000(32 ) 998

1058

�� =

4

= 1058 lb

= 264�5 ��

� � (� + 2 � )

=

264�5 = 2472 �� 3  0�200  + 2(0�080) 8 

F u

(d) Factor of Safety =

4 F

, based on fatigue

F u = 14,500 lb , Table 17.8

Factor of Safety =

F u

4 F

=

14,500 4(1058 )

= 3.43

4-strand, chain no.50 � � = 6100 �� , Table 17.8 Factor of Safety =

865.

� �

=

4�

6100 4(264�5)

= 5�77

A 5/8-in. roller chain is used on a hoist to lift a 500-lb. load through 14 ft. in 24 sec. at constant velocity. If the load on the chain is doubled during the speed-up period, compute the factor of safety (a) based on the chain’s ultimate strength, (b) based on its fatigue strength. (c) At the given speed, what is the chain’s rated 20 teeth ) in hp? Compare with the power needed at the constant capacity ( N s = speed. Does it look as though the drive will have a “long” life?

Solution: Table 17.8 5 P = in 8 F u = 6100 lb (a) Factor of Safety =

F u F

F = (500)(2 ) = 1000 lb 6100

= 6. 1 1000 F (b) Factor of Safety = u (fatigue) 4 F

Factor of Safety =

797


Factor of Safety =

6100 4(1000 )

= 1.5

14 ft  60 sec 



(c) vm =

 = 35 fpm

24 sec  1 min 

N s = 20 P=

5

in 8 Rated hp = 0.004 N ts1.08 n 0.9 P 3− 0.07 P [Link Plate Fatigue]

vm =

 5   (20)n PN s n  8  = = 35 fpm

12 n = 33.6 rpm

12

Rated hp = 0.004(20 )

1.08

5  (33.6 )0.9   

 5  3−0.07    8 

 8 

= 0.6 hp

Hp needed at constant speed (500)(35) Fvm hp = = = 0.53 hp < 0.6 hp 33,000 33,000 Therefore safe for “long” life. WIRE ROPES 866.

In a coal-mine hoist, the weight of the cage and load is 20 kips; the shaft is 400 ft. deep. The cage is accelerated from rest to 1600 fpm in 6 sec. A single 6 x 19, IPS, 1 ¾-in. rope is used, wound on an 8-ft. drum. (a) Include the inertia force but take the static view and compute the factor of safety with and without allowances for the bending load. (b) If N = 1.35 , based on fatigue, what is the expected life? (c) Let the cage be at the bottom of the shaft and ignore the effect of the rope’s weight. A load of 14 kips is gradually applied on the 6-kip cage. How much is the deflection of the cable due to the load and the additional energy absorbed? (d) For educational purposes and for a load of 0.2 F u , compute the energy that this 400-ft rope can absorb and compare it with that for a 400-ft., 1 ¾-in., as-rolled-1045 steel rod. Omit the weights of the rope and rod. What is the energy per pound of material in each case?

Solution: (a)

798


a=

v2 − v1 t

 1 min   60 sec  

(1600 fpm ) =

6 sec

= 4.445 fps

2

W h = 20 kips

For 6 x 19 IPS, w ≈ 1.6 Dr 2 lb ft 2

 400  2  kips = 0.64 Dr kips  1000 

wL = 1.6 Dr 

F t − wL − W h = ma

20 + 0.64 Dr 2

m=

32.2

 20 + 0.64 Dr 2  (4.445 ) 32 . 2  

F t − 0.64 Dr − 20 =  2

F t = 22.76 + 0.73Dr 2 Dr = 1

3 4

in

 3  F t = 22.76 + 0.731   4 

2

=

25 kips

F u − F b

N =

F t

Table AT 28, IPS F u ≈ 42 Dr 2 tons 2

F u = 42(1.75) = 129 tons = 258 kips

with bending load F b = sb Am sb =

EDw

F b =

Ds EAm Dw Ds

Table At 28, 6 x 19 Wire Rope

799

SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS Dw = 0.067 Dr = 0.067(1.75) = 0.11725 in Ds = 8 ft = 96 in E = 30,000 ksi Am ≈ 0.4 Dr 2 2

Am = 0.4(1.75 ) = 1.225 sq in F b = N =

(30,000)(1.225 )(0.11725) = 45 kips (96) F u − F b

258 − 45

=

25 without bending load F 258 N = u = = 10.32 F t 25 F t

= 8.52

(b) N = 1.35 on fatigue IPS, su ≈ 260 ksi Dr Ds =

2 NF t

( p

su )su

(1.75)(96) =

2(1.35)(25)

( p

su )(260)

p su = 0.0015

Fig. 17.30, 6 x 19 IPS 5 Number of bends to failure = 7 x 10 (c) δ =

FL Am E r

Am = 1.225 sq in E r ≈ 12,000 ksi (6 x 19 IPS) F = 14 kips L = 400 ft = 4800 in

(14)(4800 ) = 4.57 in (1.225)(12,000)

δ = U =

1 2

F δ =

1 2

(14 )(4.57 ) = 32 in − kips

.6 kips (d) F = 0.2 F u = 0.2(258) = 51

δ = δ =

FL Am E r

(51.6)(4800 ) = 16.85 in (1.225)(12,000) 800


1

(51.6)(16.85) = 434 in − kips 2 2 For 1 ¾ in, as-rolled 1045 steel rod su = 96 ksi U =

F δ =

 π  2 (1.75)  4 

F u = su A = (96)

.9 kips = 230

F = 0.2 F u = 0.2(230.9) = 46 .2 kips

δ =

FL AE

δ =

U =

868.

(46.2 )(4800 )  π  2  (1.75 ) (30,000 )  4  1 2

1

F δ =

2

= 3.073 in

(46.2 )(3.073) = 71 in − kips < U of wire rope.

A hoist in a copper mine lifts ore a maximum of 2000 ft. The weight of car, cage, and ore per trip is 10 kips, accelerated in 6 sec. to 2000 fpm; drum diameter is 6 ft. Use a 6 x 19 plow-steel rope. Determine the size (a) for a life of 200,000 cycles and N = 1.3 on the basis of fatigue, (b) for N = 5 by equation ( v), §17.25, Text . (c) What is the expected life of the rope found in (b) for N = 1.3 on the basis of fatigue? (d) If a loaded car weighing 7 kips can be moved gradually onto the freely hanging cage, how much would the rope stretch? (e) What total energy is stored in the rope with full load at the bottom of te shaft? Neglect the rope’s weight for this calculation. (f) Compute the pressure of the rope on the cast-iron drum. Is it reasonable?

Solution:

a=

v2 − v1 t

 1 min   60 sec  

(2000 fpm ) =

6 sec

= 5.56 fps

2

For 6 x 19 IPS, w ≈ 1.6 Dr 2 lb ft 2

 2000  2  kips = 3.2 Dr kips  1000 

wL = 1.6 Dr 

801

SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS W h = 10 kips

 wL + W h  a  32.2   a   5.56  2 2 F t =  + 1(wL + W h ) =  + 1(3.2 Dr + 10 ) = 1.17267 (3.2 Dr + 10 )  32.2   32.2  F t − wL − W h = 

(a) Dr Ds =

2 NF t

( p

su )su

Fig. 17.30, 200,000 cycles, 6 x 19 p su = 0.0028 PS: su ≈ 225 ksi Ds = 6 ft = 72 in N = 1.3

2(1.3)(1.17267 )(3.2 Dr 2 + 10)

Dr (72) =

(0.0028)(225)

45.36 Dr = 9.7566 Dr 2 + 30.49 Dr 2 − 4.64916 Dr + 3.1251 = 0 Dr = 0.815 in

say Dr =

7 8

in

(b) by N = 5 , Equation (v) F − F b N = u F t sb =

EDw Ds

Dw = 0.067 Dr sb =

(30,000 )(0.067 Dr ) 72

=

27.92 Dr

F b = sb Am Am = 0.4 Dr 2 F b = (27.92 Dr )(0.4 Dr 2 ) = 11.17 Dr 3 F u = 36 Dr 2 tons for PS F u = 72 Dr 2 kips F u − F b = NF t

72 Dr 2 − 11.17 Dr 3 = (5)(1.17267 )(3.2Dr 2 + 10 ) 72 Dr 2 − 11.17 Dr 3 = (5.8634 )(3.2 Dr 2 + 10) Dr = 1.216 in

802

SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS 1 use Dr = 1 in 4 2 NF t (c) Dr Ds = ( p su )su

(1.25)(72) =

[

2

2(1.3)(1.17267 ) 3.2(1.25) + 10

( p

]

su )(225 )

p su = 0.00226

Fig. 17.20 5 Expected Life = 3 x 10 cycles (d) F = 7 kips E r = 12,000 ksi L = 2000 ft = 24,000 in

For (a) Dr =

δ =

7 8

in

FL Am E r

 7  Am ≈ 0.4 D = 0.4   8 

2

3 r

δ =

=

0.30625 sq in

(7 )(24,000) = 45.7 in (0.30625)(12,000)

1 For (b) Dr = 1 in 4 FL

δ =

Am E r

 1  Am ≈ 0.4 D = 0.41   4 

2

3 r

δ =

=

0.625 sq in

(7 )(24,000) = 22.4 in (0.625)(12,000)

(e) For (a) U = 1

1 2

F δ =

1 2

(7 )(45.7 ) = 160 in − kips

1

(7 )(22.4 ) = 78. 4 in − kips 2 2 (f) Limiting pressure, cast-iron sheaves, 6 x19, p = 500 psi . For (b) U =

F δ =

For (a) p su = 0.0028 p = 0.0028(225) = 0.630 kips = 630 psi > 500 psi , not reasonable.

For (b) p su = 0.00226 803

SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS p = 0.00226(225) = 0.5085 kips = 508.5 psi ≈ 500 psi , 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 drum diameter 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 2 × 105 cycles and N = 1.3 against fatigue failure, (b) static consideration (but not omitting inertia effect) and N = 5 . (c) Make a final recommendation. (d) If the loaded car can be moved gradually onto the freely hanging 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 rope’s weight for this calculation. (f) Compute the pressure of the rope on the cast-iron drum. Is it all right?

Solution:

W h = 5900 + 2100 + 2800 = 10,800 lb = 10.8 kips

a=

v2 − v1 t

 1 min   60 sec  

(3285 fpm ) =

6 sec

= 9.125 fps

2

 wL + W h  a  32.2 

F t − wL − W h = 

Assume 6 x 19 IPS, w ≈ 1.6 Dr 2 lb ft

 1500  2  kips = 2.4 Dr kips  1000   a   9.125  2 2 F t =  + 1(wL + W h ) =  + 1(2.4 Dr + 10) = 3.08 Dr + 13.86  32.2   32.2  2

wL = 1.6 Dr 

5

(a) Fig. 17.30, 2 x 10 cycles p su = 0.0028 Dr Ds =

2 NF t

( p

su )su

Ds = 5 ft = 60 in

804

SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS Ds ≈ 45Dr Dr max =

60

= 1 .33 in 45 1 use Dr = 1 in 4

 1  F t = 3.081   4  su =

2

+ 13.86 = 18.67

2(1.3)(18.67 )

=

1  (0.0028 ) 1 (60 )  4 

kips

231 ksi

1 Use Plow Steel, 6 x 19 Wire Rope, Dr = 1 in . 4

(b) N = sb =

F u − F b F t

EDw Ds

 1  Dw = 0.067 Dr = 0.0671  = 0.08375 in  4  Ds = 60 in E = 30,000 ksi sb =

(30,000 )(0.08375 ) 60

 1  Am = 0.4 D = 0.41   4  2 r

=

41.875 ksi

2

= 0.625 in

2

F b = sb Am = (41.875)(0.625) = 26 .17 kips N = 5 F u = NF t + F b = (5)(18.67 ) + 26.17 = 119.52 kips = 59.76 tons F u 2 r

D

=

59.76

 1  1   4 

2

= 38.25

Table AT 28, Use IPS, 6 x 19,

F u 2

Dr

= 42 > 38.25

(c) Recommendation: 1 6 x 19, improved plow steel, Dr = 1 in 4

805


(d) δ =

FL Am E r

F = 2100 + 2800 = 4900 lb E r ≈ 12 ×10 6 psi L = 1500 ft = 18,000 in

δ =

(4900 )(18,000) = 11.76 in (0.625)(12 ×106 )

(e) U =

1 2

F δ =

1

2 (f) p su = 0.0028

(4900 )(11.76) = 28,800 in − lb

su = 231 ksi p = 0.0028(231,000) = 646.8 psi For cast-iron sheave, limiting pressure is 500 psi p = 646.8 psi > 500 psi , 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 N = 1 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 ( N = 1 )?

Solution: (a)

 365 days  1 wk  1000 trips  5    = 312,857 cycles ≈ 3 × 10 cycles  1 yr  7 days  1 wk 

No. of cycles = (6 yr )

Figure 17.30, 6 x 37, IPS p su = 0.00225 Dr Ds =

2 NF t

( p

su )su

For IPS, su ≈ 260 ksi F t = 14 kips N = 1.0 Dr = 1.375 in Dr Ds =

2 NF t

( p

(1.375) Ds

=

su )su

2(1.0 )(14 )

(0.00225 )(260) 806

SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS Ds = 34.8 in

(b) Ds = 2 ft = 30 in Static Basis F − F b N = u F t Table AT 28, 6 x 37 Dw ≈ 0.048 Dr = 0.048(1.375) = 0.066 in 2

Am ≈ 0.4 Dr 2 = 0.4(1.375 ) = 0.75625 in 2 F u = su Am = (260)(0.75625 ) = 196 .6 kips F b = sb Am = N =

F u − F b F t

EDw Am Ds =

=

(30,000 )(0.066)(0.75625)

196.6 − 49.9 10.5

30

= 49.9 kips

= 10.5

Life: N = 1.0 (fatigue) 2 NF t Dr Ds = ( p su )su

(1.375)(30) =

2(1.0 )(14 )

( p

su )(260)

p su = 0.0026

Figure 17.30, Life ≈ 2.5 ×105 cycles , 6 x 37.

871.

o

A wire rope passes about a driving sheave making an angle of contact of 540 , as shown. A counterweight of 3220 lb. is suspended from one side and the acceleration is 4 fps 2. (a) If f = 0.1 , 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 N = 6 and for the load found in (a). (d) Compute the diameter of the sheave for indefinite life with say N = 1.1 on fatigue. What changes could be made in the solution to allow the use of a smaller sheave?

807


Problems 871 – 874. Solution:

 4 fps 2   = 2820 lb F 2 = (3220 lb )1 − 2  32 . 2 fps   (a) F 1 = F 2 e f θ

θ = 540

o

= 3π

f = 0.10 F 1 = (2820 )e (0.10)(3π ) = 7237 lb

(b) For rubber lined, dry rope f = 0.495 F 1 = (2820 )e (0.495 )(3π ) = 249,466 lb

(c) F t = F 1 = 7237 lb N =

F u − (F b ≈ 0 ) F t

=

F u F t

F u ≈ 32 Dr 2 tons for MPS F u ≈ 64 Dr 2 kips F u = 64,000 Dr 2 lb F u = NF t

64,000 Dr 2 = (6)(7237 ) Dr = 0.824 in

use Dr = 0.875 in

808


(d) Dr Ds =

2 NF t

( p

su )su

Indefinite life, p su = 0.0015 MPS: su ≈ 195 ksi = 195,000 psi

(0.875) Ds

=

2(1.1)(7237 )

(0.0015 )(195,000)

Ds = 62.2 in

To reduce the size of sheave, increase the size of rope. 872.

2

A traction elevator with a total weight of 8 kips has an acceleration of 3 fps ; 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 ( N = 1 will do). (e) Compute the factor of safety defined in the Text . (f) If it were decided that 5 ×105 bending cycles would be enough life, would there be a significant difference in the results?

Solution:

 3 fps 2   F 1 = (8 kips )1 + 2  = 8.745 kips 32 . 2 fps   θ = 3(180o ) = 3π F 2 =

F 1

e f θ W c = weight of counterweight

W c =

1− W c =

F 2 3

= 1.10274 F 2

32.2 1.10274 F 1

e f θ (a) Iron sheave, greasy rope, f = 0.07 W c =

1.10274 (8.745 )

= 4.986 kips e ( 0.07 )(3π ) (b) Iron sheave, dry rope, f = 0.12

W c =

1.10274 (8.745 )

= 3.112 kips ( )( π ) e 0.12 3 (c) Rubber lined with a greasy rope, f = 0.205

W c =

1.10274 (8.745 ) e (0.205 )(3

π )

= 1.397

kips

809


(d) Dr Ds =

2 NF t

( p su )su

Indefinite life, p su = 0.0015 F t = F 1 = 8.745 kips total F t =

8.745

= 1.458 kips each rope 6 F t = 1458 lbs

N = 1 Table AT 28, 6 x 19 Ds ≈ 45Dr

2(1)(1458 )

Dr (45Dr ) =

(0.0015)(195,000)

Dr = 0.47 in 1

Use Dr =

(e) N =

2

in = 0.5 in

F u − F b F t

Table AT 28, MPS 2

F u = 32 Dr 2 tons = 64,000 Dr 2 lb = 64,000(0.5) lb = 16,000 lb

F b =

EDw Am Ds

E = 30 ×106 psi

6 x 19, Dw = 0.067 Dr Ds ≈ 45Dr 2

Am = 0.4 Dr 2 = 0.4(0.5 ) = 0.1 sq. in.

(30 ×10 )(0.067)(0.1) = 4467 lb F = 6

b

N =

45 16,000 − 4467 1458

= 7.91

(f) 5 x 105 cycles Fig. 17.30, 6 x 19. p su = 0.0017 Dr Ds =

2 NF t

( p su )su

810


Dr (45Dr ) =

2(1)(1458 )

(0.0017 )(195,000 )

Dr = 0.44 in

since Dr = 0.44 in ≈ 0.47 in 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 values 2 are 4500-lb load, 4 fps 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 Ds in terms of Dr , decide on the diameter and type of wire rope. (b) For this rope and N = 1.05 , 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)

F t = 4500 lb W h = 5000 lb

 W h + wL  a  32.2 

W h + wL − F t = 

assume 6 x 19 w = 1.6 Dr 2 lb ft wL = (1.6 Dr 2 )(120) = 192 Dr 2 per rope wL = 6(192 Dr 2 ) = 1152 Dr 2

 5000 + 1152 Dr 2  (4 ) 5000 + 1152 D − 4500 =  32 . 2   2 r

1152 Dr 2 + 500 = 621.12 + 143.11Dr 2 Dr = 0.3465 in

say Dr = 0.375 in =

3 8

in

811

SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS 7  3  Ds ≈ 45 Dr = 45  = 16 in 8  8  3 Six – 6 x 19 rope, Dr = in 8 (a) Dr = 0.375 in = F t =

4500

6 N = 1.05

Dr Ds =

3 8

in

= 750 lb

2 NF t

( p su )su

assume IPS, su = 260 ksi = 260,000 psi Indefinite life, p su = 0.0015

(0.375) Ds

=

2(1.05)(750)

(0.0015 )(260,000)

Ds = 10.77 in

(c) N =

F u − F b F t

F t = 750 lb

IPS 2

 3  F u ≈ 42 D tons = 84,000 D lb = 84,000  lb = 11,813 lb  8  2 r

F b =

2 r

EDw Am Ds

6 x 19, Ds = 10.77 in as in (b)

 3  Dw = 0.067 Dr = 0.067  = 0.025 in  8   3  Am = 0.4 D = 0.4   8  2 r

2

=

0.05625 sq. in.

E = 30 ×106 psi

(30 ×10 )(0.025)(0.05625) = 3917 lb F = 6

b

10.77

812


N =

11,813 − 3917 750

= 10.53

(c) F 1 = F t = 4500 lb F 1 = F 2 e f θ For iron sheave, dry rope, f = 0.12

θ = 540 F 2 =

o

F 1 e f θ

= 3π

=

4500 e (0.12 )(3π )

= 1452

lb

a    = F 2  32.2  4   CW 1 +  = 1452  32.2  CW 1 +

CW = 1291 lb 2

874.

A traction elevator has a maximum deceleration of 8.05 fps when being braked on the downward motion with a total load of 10 kips. There are 5 cables that pass twice 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-plowsteel rope should be used for N = 4 , including the bending effect? (Static approach.) (c) What is the estimated life of these ropes ( N = 1 )?

Solution:

a = 8.05 fps 2

(a) F 1 = 10 kips

 

F 2 = (8 kips )1 −

8.05   = 6 kips 32.2 

θ = 3π F 1 F 2

=e

f θ

813

Section 15

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