Section 15

(1)

LEATHER BELTS LEATHER BELTS DESIGN PROBLEMS DESIGN PROBLEMS 841.

841. A belt A belt drive drive is is to to be be designed designed for for F F ₁₁ F F ₂₂ ==33, while transmitting 60 hp at 2700, while transmitting 60 hp at 2700 rpm of the driver

rpm of the driver D D₁₁; ; mm_w_w ≈≈11..8585; use a medium double belt, cemented joint, a; use a medium double belt, cemented joint, a squirrel-cage, compensator-motor drive with mildly jerking loads; center distance 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 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 iron-pulley sizes and determine the belt width for a maximum permissible

psi psi

ss==300300 . (b) How does this width compare with that obtained by the ALBA. (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 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

belt. (d) If the belt in (a) stretches until the tight tension F F ₁₁ ==525525lblb., what is., what is 2 2 1 1 F F F F ?? Solution: Solution:

(a) Table 17.1, Medium Double Ply, (a) Table 17.1, Medium Double Ply, Select

Select D D₁₁ ==77inin. min.. min.

in in t t 64 64 20 20 = =

(

( ))(

(

))

fpm fpm n n D D vv_m_m 49484948 12 12 2700 2700 7 7 12 12 1 1 1 1 ₌₌ ₌₌ = =π π π π fpm fpm fpm fpm fpm fpm 49484948 60006000 4000 4000 << <<

( (

))

000 000 ,, 33 33 2 2 1 1 F F vvmm F F hp hp== −−

(

))( )

( )

000 000 ,, 33 33 4948 4948 60 60₌₌ F F 11−−F F 22 lb lb F F F F ₁₁−− ₂₂ ==400400 2 2 1 1 33F F F F == lb lb F F F F 400400 3 3 ₂₂ −− ₂₂ == lb lb F F ₂₂ ==200200

( (

))

lblb F F F F ₁₁ ==33 ₂₂ ==33 200200 ==600600 sbt sbt F F 11 == η η 300 300 = = d d ss For

For cemented cemented joint, joint, η η ==11..00

psi psi ss_d_d==300300

(

( ))(

( ))













= = = = 64 64 20 20 300 300 600 600 1 1 bb F F in in b b==66..44 say say bb==66..55inin

(2)

(b) ALBA Procedure (b) ALBA Procedure

( (

))

( (

))

LL 2 2 1 1 1 1 .. 17 17 ., .,TableTable bC bC _m_mC C _p_p C C _f_fC C _f_f in in hp hp hp hp = = Table 17.1, Table 17.1, vv_m_m ==49484948 fpm fpm

Medium Double Ply Medium Double Ply

448 448 .. 12 12 = = in in hp hp Table 17.2 Table 17.2

Squirrel cage, compensator, starting Squirrel cage, compensator, starting

67 67 .. 0 0 = = m m C C Pulley Size, Pulley Size, D D₁₁ ==77inin 6 6 .. 0 0 = = p p C C Jerky loads, Jerky loads, C C _f_f ==00..8383

(

1212..448448

))(

( ))(

(

00..6767

))( )

( )(

00..66

(

00..8383

))

60 60 bb hp hp== == in in b b==1414..55 say say bb==1515inin (c) (c)

(

( ))(

( ))

psi psi bt bt F F ss 128128 64 64 20 20 15 15 1 1 600 600 1 1 ₌₌













= = = = η η (d) (d)

( ) ( )

22 1 1 2 2 1 1 2 2 1 1 2 2 2 2 1 1 1 1 2 2 1 1 200 200 600 600 2 2F F _o_o ==F F ++F F == ++ lb lb F F _o_o ==373373..22 lb lb F F ₁₁ ==525525

(

)

) (

( ))

22 1 1 2 2 2 2 1 1 2 2 1 1 525 525 2 2 .. 373 373 2 2 == ++F F lb lb F F ₂₂ ==247247 1255 1255 .. 2 2 247 247 525 525 2 2 1 1 ₌₌ ₌₌ F F F F 842.

842. A 20-hp, 1750 rpm, slip-ring motor is to drive a ventilating fan at 330 rpm. TheA 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 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 continuous, 24 hr./day. (a) What driving-pulley size is needed for a speed recommended as about optimum in the

recommended as about optimum in the Text Text ? (b) Decide upon a pulley size (iron? (b) Decide upon a pulley size (iron or steel) and belt thickness, and determine the belt width by the ALBA tables. (c) 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 Compute the stress from the general belt equation assuming that the applicable coefficient of friction is that suggested by the

coefficient of friction is that suggested by the Text Text . (d) Suppose the belt is. (d) Suppose the belt is installed with an initial tension

installed with an initial tension F F _o_o ==7070lblb inin. (§17.10), compute. (§17.10), compute F F ₁₁ F F ₂₂ and the and the stress on the tight side if the approximate relationship of the operating tensions stress on the tight side if the approximate relationship of the operating tensions and the initial tensions is

and the initial tensions is 22 1 1 2 2 1 1 2 2 2 2 1 1 1 1 F F 22F F oo F F ++ == ..

(3)

(b) ALBA Procedure (b) ALBA Procedure

( (

))

( (

))

LL 2 2 1 1 1 1 .. 17 17 ., .,TableTable bC bC _m_mC C _p_p C C _f_fC C _f_f in in hp hp hp hp = = Table 17.1, Table 17.1, vv_m_m ==49484948 fpm fpm

Medium Double Ply Medium Double Ply

448 448 .. 12 12 = = in in hp hp Table 17.2 Table 17.2

Squirrel cage, compensator, starting Squirrel cage, compensator, starting

67 67 .. 0 0 = = m m C C Pulley Size, Pulley Size, D D₁₁ ==77inin 6 6 .. 0 0 = = p p C C Jerky loads, Jerky loads, C C _f_f ==00..8383

(

1212..448448

))(

( ))(

(

00..6767

))( )

( )(

00..66

(

00..8383

))

60 60 bb hp hp== == in in b b==1414..55 say say bb==1515inin (c) (c)

(

( ))(

( ))

psi psi bt bt F F ss 128128 64 64 20 20 15 15 1 1 600 600 1 1 ₌₌













= = = = η η (d) (d)

( ) ( )

22 1 1 2 2 1 1 2 2 1 1 2 2 2 2 1 1 1 1 2 2 1 1 200 200 600 600 2 2F F _o_o ==F F ++F F == ++ lb lb F F _o_o ==373373..22 lb lb F F ₁₁ ==525525

(

)

) (

( ))

22 1 1 2 2 2 2 1 1 2 2 1 1 525 525 2 2 .. 373 373 2 2 == ++F F lb lb F F ₂₂ ==247247 1255 1255 .. 2 2 247 247 525 525 2 2 1 1 ₌₌ ₌₌ F F F F 842.

842. A 20-hp, 1750 rpm, slip-ring motor is to drive a ventilating fan at 330 rpm. TheA 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 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 continuous, 24 hr./day. (a) What driving-pulley size is needed for a speed recommended as about optimum in the

recommended as about optimum in the Text Text ? (b) Decide upon a pulley size (iron? (b) Decide upon a pulley size (iron or steel) and belt thickness, and determine the belt width by the ALBA tables. (c) 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 Compute the stress from the general belt equation assuming that the applicable coefficient of friction is that suggested by the

coefficient of friction is that suggested by the Text Text . (d) Suppose the belt is. (d) Suppose the belt is installed with an initial tension

installed with an initial tension F F _o_o ==7070lblb inin. (§17.10), compute. (§17.10), compute F F ₁₁ F F ₂₂ and the and the stress on the tight side if the approximate relationship of the operating tensions stress on the tight side if the approximate relationship of the operating tensions and the initial tensions is

and the initial tensions is 22 1 1 2 2 1 1 2 2 2 2 1 1 1 1 F F 22F F oo F F ++ == ..

(4)

Solution: Solution: fpm fpm to to vv_m_m ==40004000 45004500 assume assume vv_m_m == 42504250 fpm fpm 12 12 1 1 1 1nn D D vv_m_m ==π π

( (

))

12 12 1750 1750 4250 4250₌₌π π D D11 in in D D₁₁ ==99..2626 say say D D₁₁ ==1010inin

(b) Using Heavy Double Ply Belt,

(b) Using Heavy Double Ply Belt, t t inin 64 64 23 23 = =

Minimum pulley diameter for

Minimum pulley diameter for vv_m_m ≈≈ 42504250 fpm fpm,, D D₁₁ ==1010inin

Use Use D D₁₁ ==1010inin

(

( ))(

(

))

fpm fpm n n D D vv_m_m 45814581 12 12 1750 1750 10 10 12 12 1 1 1 1 ₌₌ ₌₌ = = π π π π ALBA Tables ALBA Tables

( (

))

( (

))

LL 2 2 1 1 1 1 .. 17 17 ., .,TableTable bC bC _m_mC C _p_p C C _f_fC C _f_f in in hp hp hp hp = = 8 8 .. 13 13 = = in in hp hp Slip

Slip ring ring motor, motor, C C mm ==00..44 Pulley Size, Pulley Size, D D₁₁ ==1010inin 7 7 .. 0 0 = = p p C C

Table 17.7, 24 hr/day, continuous Table 17.7, 24 hr/day, continuous

8 8 .. 1 1 = = sf sf N N Assume Assume C C _f_f ==00..7474

(

( ))( ) (

11..88

( ) ( ))(

2020 1313..88

( ))(

bb

( ))(

00..44

( ))(

00..77

(

00..7474

))

hp hp== == in in b b==1212..5959 use use bb==1313inin

(c) General belt equation (c) General belt equation













₋₋













− − = = − − _θ_θ θ θ ρ ρ f f f f ss ee ee vv ss bt bt F F F F 11 2 2 .. 32 32 12 12 22 2 2 1 1 fps fps vv_ss 7676..3535 60 60 4581 4581 = = = = .. .. 035 035 .. 0 0 lblb cucu inin = = ρ

ρ for leather for leather

in in t t 64 64 23 23 = = in in b b==1313

(5)

(

( ))(

( ))

lb lb F F F F 260260 4581 4581 20 20 8 8 .. 1 1 000 000 ,, 33 33 2 2 1 1−− == == 3 3 .. 0 0 = = f

f on iron or steel on iron or steel

C C D D D D₂₂ − − ₁₁ ± ± ≈ ≈π π θ θ ft ft C C ==88~~99 use 8.5 ft use 8.5 ft

( ( ))

inin D D 1010 5353 330 330 1750 1750 2 2



==











= =

( ( ))

1212 22..7272rad rad 5 5 .. 8 8 10 10 53 53 = = − − − − = =π π θ θ

( )

( )(

00..33

( ))

22..7272 ==00..816816 = = θ θ f f 5578 5578 .. 0 0 1 1 1 1 816 816 .. 0 0 816 816 .. 0 0 = = − − = = − − ee ee ee ee f f f f θ θ θ θ

( ( ))

(

)

)(

(

))

( (

00..55785578

))

2 2 .. 32 32 35 35 .. 76 76 035 035 .. 0 0 12 12 64 64 23 23 13 13 260 260 2 2 2 2 1 1













− −













= = = = − −F F ss F F psi psi ss==176176 (d) (d) 22 1 1 2 2 1 1 2 2 2 2 1 1 1 1 F F 22F F oo F F ++ ==

(

lblb inin

))(

( ))

inin lblb F F _o_o == 7070 1313 ==910910 lb lb F F F F ₁₁−− ₂₂ ==260260 lb lb F F F F ₂₂ == ₁₁−−260260

(

260260

) (

)

22

( ))

910910 22 6060..3333 1 1 2 2 1 1 1 1 2 2 1 1 1 1 ++ F F −− == == F F lb lb F F ₁₁ ==10451045 lb lb F F ₂₂ ==10451045−−260260==785785

( ( ))

psi psi bt bt F F ss 224224 64 64 23 23 13 13 1045 1045 1 1 ₌₌













= = = = 331 331 .. 1 1 785 785 1045 1045 2 2 1 1 ₌₌ ₌₌ F F F F 843.

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

in in D

D₁₁ ==1616 ;; D D₂₂ ==5353inin, a flywheel; cemented joints;l, a flywheel; cemented joints;l C C ==88 ft ft . (a) Choose an. (a) Choose an appropriate belt thickness and determine the belt width by the ALBA tables. (b) 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 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 needed. Is this value satisfactory? (c) Suppose that in the beginning, the initial tension

(6)

initial tension initial tension F F _o_o isis 22 1 1 2 2 1 1 2 2 2 2 1 1 1 1 F F 22F F oo F

F ++ == . For the condition in (c), compute. For the condition in (c), compute F F _o_o. Is it. Is it reasonable compared to Taylor’s recommendation?

reasonable compared to Taylor’s recommendation? Solution: Solution: (a) Table 17.1 (a) Table 17.1

(

( ))(

(

))

fpm fpm n n D D vv_m_m 47754775 12 12 1140 1140 16 16 12 12 1 1 1 1 ₌₌ ₌₌ = =π π π π

Use heavy double-ply belt Use heavy double-ply belt

in in t t 64 64 23 23 = = 1 1 .. 14 14 = = in in hp hp

( (

))

( (

))

LL 2 2 1 1 1 1 .. 17 17 ., .,TableTable bC bC _m_mC C _p_p C C _f_fC C _f_f in in hp hp hp hp = = line

line starting starting electrelectric ic motor motor , , C C mm ==00..55

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

4 4 .. 1 1 = = sf sf N N in in D D₁₁ ==1616 ,, C C p p ==00..88 assume, assume, C C f f ==00..7474

(

( ))(

( ))

hphp hp hp== 11..44 100100 ==140140

(

( ))(

1414..11

( ))(

00..55

( ))(

00..88

(

00..7474

))

140 140 bb hp hp== == in in b b==3333..55 use use bb==3434inin (b) §17.6, (b) §17.6, ssd d ==400400η η 00 00 .. 1 1 = = η

η for cemented joint. for cemented joint.

psi psi ss_d_d==400400













₋₋













− − = = − − _θ_θ θ θ ρ ρ f f f f ss ee ee vv ss bt bt F F F F 11 2 2 .. 32 32 12 12 22 2 2 1 1 fps fps vv_ss 7979..66 60 60 4775 4775 = = = = .. .. 035 035 .. 0 0 lblb cucu inin = = ρ

ρ for leather for leather

in in t t 64 64 23 23 = = in in b b==3434

(

( ))(

( ))

lb lb F F F F 968968 4775 4775 100 100 4 4 .. 1 1 000 000 ,, 33 33 2 2 1 1−− == ==

( ( ))

(

))(

( ))

_











₋₋













− −













= = = = − − _θ_θ θ θ f f f f ee ee F F F F 11 2 2 .. 32 32 6 6 .. 79 79 035 035 .. 0 0 12 12 400 400 64 64 23 23 34 34 968 968 2 2 2 2 1 1

(7)

2496 . 0 1 = − θ θ f f e e 28715 . 0 = θ f C D D₂ − ₁ ± ≈π θ ft C =8

( )

12 2.7562rad 8 16 53 = − − =π θ

(

2.7562

)

=0.28715 f 3 . 0 1042 . 0 < = f Therefore satisfactory. (c) F ₁−F ₂ =968lb 2 1 2F F = lb F F 968 2 ₂ − ₂ =

(

)

lb F F ₁ =2 ₂ =2 968 =1936

( )

psi bt F s 159 64 23 34 1936 1 ₌













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

( ) ( )

2 1 2 1 2 1 968 1936 2F o = + lb F _o =1411 in lb F _o 41.5 34 1411 =

= of width is less than Taylor’s recommendation and is reasonable.

844. 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 =70in., a cast-iron flywheel;

. 11 . 6 ft in

C = 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)

Approximately, 2 1 2 1 2 2 1 1 F 2F o

F + = , where F _o is the initial tension. For the 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?

(8)

Solution: (a) v_m Dn

( )( )

2944 fpm 12 865 13 12 1 1 ₌ ₌ =π π

Table 17.1, use Heavy Double Ply,

in

D_min =9 for v_m =2944 fpm

belts less than 8 in wide

in t 64 23 =

(

)

(

)

L 2 1 1 . 17 .,Table bC _mC _p C _fC _f in hp hp = 86 . 9 = in hp Table 17.2 67 . 0 = m C 8 . 0 = p C ( )( )0�74 0�83 ₌0�6142 = � �

Table 17.7, electric motor, compensator-started (squirrel cage) and reciprocating compressor 4 . 1 = sf N

( )( )

hp hp= 1.4 50 =70

(

986

)( )(

0 67

)( )(

08 06142

)

70 � � � � � ��= = �� =21�6 use b=25in

(b) General Belt Equation













₋













− = − _θ θ ρ f f s e e v s bt F F 1 2 . 32 12 2 2 1 in b= 25 in t 64 23 = . . 035 . 0 lb cu in = ρ for leather fps v_s 49.1 60 2944 = = Leather on iron, f =0.3 C D D₂ − ₁ − =π θ ( )12 11 2455�� 6 13 70 � = + − − =π θ

( )(

0�3 2�455

)

=0�7365 = θ �

(9)

5212 0 1 1 7365 0 7365 0 � � � = − = − � � � � � � θ θ

( )( )

lb F F 785 2944 50 4 . 1 000 , 33 2 1− = = ( ) ( )( ) ( 05212) 2 32 1 49 035 0 12 64 23 25 785 2 2 1 � � � �       −       = = −� � � �� =199 Cemented joint, η =1.0 �� =199 (c) � �� ( )( ) 1788�� 64 23 25 199 1 =      = = �� ₂ =1788−785=1003 783 1 1003 1788 2 1 ₌ ₌ _� � � (d) 2 1 2 2 1 1 2 1 2F _o =F +F ( ) ( )2 1 2 1 2 1 1003 1788 2� � = + �� _� =1367 �� _� 547 25 1367 � = =

Approximately less than Taylor’s recommendation ( = 70 lb/in.)

(e)

(

)

(

)

C D D D D C L 4 57 . 1 2 2 1 2 1 2 − + + + ≈

( )( )

[

]

(

)

(

)

( )( )

[

]

�� 306 11 12 6 4 13 70 13 70 57 1 11 12 6 2 2 = + − + + + + = �

(f) More economical basis

12 1 1n D v_m = π

(

)

12 865 4500₌π D1 in D₁ =19.87 use D₁ =20in

(10)

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 =54in.; D₁ =14in. (motor), D₂ =54in., 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

2 1 F

F ? (d) If the belt has stretched until s=200 psi on the tight side, what is 2

1 F

F ? (e) Compute the belt length. Solution:

(a) For medium double leather belt

in t 64 20 =

(

hp in

)( )

b C _mC _pC _f hp = Table 17.1 and 17.2 67 . 0 = m C 8 . 0 = p C 74 . 0 = f C in b=10

( )( )

fpm n D v_m 3225 12 880 14 12 1 1 ₌ ₌ = π π 6625 . 6 = in hp

(

)( )( )( )( )

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

_











₋













− = − _θ θ ρ f f s e e v s bt F F 1 2 . 32 12 2 2 1 in b=10 in t 64 20 = . . 035 . 0 lb cu in = ρ fps v_s 53.75 60 3225 = = C D D₂ − ₁ − =π θ rad 4 . 2 54 14 54 = − − =π θ Leather on iron f =0.3

( )( )

0.3 2.4 =0.72 = θ f

(11)

51325 . 0 1 1 72 . 0 72 . 0 = − = − e e e e f f θ θ

(

)

lb F F 270 3225 43 . 26 000 , 33 2 1− = =

( )

(

)(

)

(

0.51325

)

2 . 32 75 . 53 035 . 0 12 64 20 10 270 2 2 1













−













= = −F s F psi s= 206 (c) F sbt

( )( )

644lb 64 20 10 206 1



=











= = lb F ₂ =644−270=374 72 . 1 374 644 2 1 ₌ ₌ F F (d) s= 200 psi

( )( )

lb sbt F 625 64 20 10 200 1



=











= = lb F ₂ =625−270=355 76 . 1 355 625 2 1 ₌ ₌ F F (e)

(

)

(

)

C D D D D C L 4 57 . 1 2 2 1 2 1 2 − + + + ≈

( )

(

)

(

)

( )

in L 222 54 4 14 54 14 54 57 . 1 54 2 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;

in

D₂ =12 . (compressor shaft); C =5 ft . The belt has been designed for a net belt pull of F ₁−F ₂ =40lb 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: (a) v_m Dn

( )(

)

3665 fpm 12 1750 8 12 1 1 ₌ ₌ = π π in b=6

( )( )

lb F F ₁− ₂ = 40 6 =240

(12)

(

)

( )(

)

hp v F F hp m ₂₆_.₆₅ 000 , 33 3665 240 000 , 33 2 1− ₌ ₌ = (b) F ₁ =3F ₂ lb F F 240 3 ₂ − ₂ = lb F ₂ =120 lb F ₁ =360 bt F s= 1

For heavy single-ply leather belt

in t 64 13 =

( )

psi s 295 64 13 6 360 =













= (c)

_











₋













− = − _θ θ ρ f f s e e v s bt F F 1 2 . 32 12 2 2 1 . . 035 . 0 lb cu in = ρ fps v_s 61.1 60 3665 = = lb F F ₁− ₂ =240

( )

(

)( )

_











₋













−













= = − _θ θ f f e e F F 1 2 . 32 1 . 61 035 . 0 12 295 64 13 6 240 2 2 1 7995 . 0 1 = − θ θ f f e e C D D₂ − ₁ − =π θ

( )

12 3.075rad 5 8 12 = − − =π θ 9875 . 4 = θ f e 607 . 1 = θ f

(

3.075

)

=1.607 f 5226 . 0 = f

(d) ALBA Tables (Table 17.1 and 17.2)

(

hp in

)( )

b C _mC _pC _f hp = fpm v_m =3665 965 . 6 = in hp

(13)

in b=10 0 . 1 = m C (assumed) 6 . 0 = p C 74 . 0 = f C

(

)( )( )( )(

)

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

848. A 10-in. medium double leather belt, cemented joints, transmits 60 hp from a 9-in. paper pulley to a 15-9-in. pulley on a mine fab; dusty conditions. The compensator-started motor turns 1750 rpm; C =42in. 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:

( )(

)

fpm n D v_m 4123 12 1750 9 12 1 1 = = = π π (a) hp =

(

hp in

)( )

b C mC pC f Table 17.1 and 17.2

Medium double leather belt

in t 64 20 = fpm v_m =4123 15 . 11 = in hp 67 . 0 = m C 7 . 0 = p C 74 . 0 = f C in b=10

(

)( )(

)

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

_











₋













− = − _θ θ ρ f f s e e v s bt F F 1 2 . 32 12 2 2 1 in b=10 . . 035 . 0 lb cu in = ρ η 400 = s 0 . 1 = η cemented joint psi s= 400 C D D₂ − ₁ − =π θ

(14)

rad 9987 . 2 42 9 15 = − − =π θ

Leather on paper pulleys, f =0.5

( )(

0.5 2.9987

)

=1.5 = θ f 77687 . 0 1 = − θ θ f f e e fps v_s 68.72 60 4123 = =

( )

(

)(

)

(

)

lb F F 0.77687 822 2 . 32 72 . 68 035 . 0 12 400 64 20 10 2 2 1



=











−













= −

(

)

( )(

)

hp v F F hp m ₁₀₂_.₇ 000 , 33 4123 822 000 , 33 2 1− ₌ ₌ = (c) Table 17.7 6 . 1 = sf N hp hp hp 64.2 102.7 6 . 1 7 . 102 < = =

Therefore, better agreement

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













₋













− = − _θ θ ρ f f s e e v s bt F F 1 2 . 32 12 2 2 1

(

)

000 , 33 2 1 F vm F hp= −

(

)

000 , 33 60 F 1 F 2 vs hp= −













₋













− = _θ θ ρ f f s s e e v s bt v hp 1 2 . 32 12 000 , 33 60 2 s s f f v v s e e bt hp

_











−













₋ = 2 . 32 12 1 000 , 33 60 ρ 2 θ θ

(15)

( )

32.2 0 24 2 . 32 12 1 000 , 33 60 2 2 =













−













−













₋ = s s f f s v v s e e bt v d hp d ρ ρ θ θ 2 . 32 36 v_s2 s= ρ . . 035 . 0 lb cu in = ρ (a) s=400 psi

(

)

2 . 32 035 . 0 36 400 2 s v = fps v_s =101.105 fpm v_m =6066 (b) s=320 psi

(

)

2 . 32 035 . 0 36 320 2 s v = fps v_s =90.431 fpm v_m =5426

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

(

)

2 . 32 045 . 0 36 400 2 s v = fps v_s =89.166 fpm fpm v_m =5350 <6066

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:













₋













− = − _θ θ ρ f f s e e v s bt F F 1 2 . 32 12 2 2 1

(16)

(

)

000 , 33 60 F ₁ F ₂ v_s hp= − s s f f v v s e e bt hp

_











−













₋ = 2 . 32 12 1 000 , 33 60 ρ 2 θ θ

( )

32.2 0 24 2 . 32 12 1 000 , 33 60 2 2 =













−













−













₋ = s s f f s v v s e e bt v d hp d ρ ρ θ θ 2 . 32 36 v_s2

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

2 . 32 12 v_s2 s= ρ psi s=300 . . 035 . 0 lb cu in = ρ

(

)

2 . 32 035 . 0 12 300 2 s v = fps v_s =151.6575 fpm v_m =9100 Speed, 40 in pulley,

(

)

( )

rpm D v n m ₈₆₉ 40 9100 12 12 2 2 = = = π π (b) Maximum power 2 . 32 36 v_s2 s= ρ

(

)

2 . 32 035 . 0 36 300 2 s v = fps v_s =87.5595 fpm v_m =5254 s s f f v v s e e bt hp

_











−













₋ = 2 . 32 12 1 000 , 33 60 ρ 2 θ θ in t 64 20 = in b= 20 C D D₂ − ₁ − =π θ

( )

12 3.0225rad 14 20 40 = − − =π θ 3 . 0 = f

(17)

( )(

0.3 3.0225

)

=0.90675 = θ f 5962 . 0 1 = − θ θ f f e e

( )

(

)

(

)(

)

(

87.5595

)

118.64 2 . 32 5595 . 87 035 . 0 12 300 5962 . 0 000 , 33 64 20 20 60 2 =













−













= hp fpm v_m =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;

in

C =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 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 in b=12 in t 64 20 =

( )( )













= 64 20 12 375 1 F (b) m v hp F F ₁ − ₂ = 33,000

( )(

)

fpm n D v_m 3770 12 1200 12 12 1 1 ₌ ₌ = π π Table 17.7, N _sf =1.2

( )( )

lb F F 1050 3770 100 2 . 1 000 , 33 2 1− = = lb F F ₂ = ₁−1050=1406−1050=356 (c)

_











₋













− = − _θ θ ρ f f s e e v s bt F F 1 2 . 32 12 2 2 1 fps v_s 62.83 60 3770 = =

(18)

. . 035 . 0 lb cu in = ρ

( )

(

)(

)

_











₋









−













= _θ θ f f e e 1 2 . 32 83 . 62 035 . 0 12 375 64 20 12 1050 8655 . 0 1 = − θ θ f f e e 006 . 2 = θ f Motor pulley rad 3685 . 3 180 193 193



=











= = π θ o

(

3.3685

)

=2.006 f 5955 . 0 = f Compressor Pulley rad 1888 . 4 180 240 2403



=











= = π θ o

(

4.1888

)

= 2.006 f 4789 . 0 = f (d) Force:

Without tension pulley

rad C D D 356 . 2 84 12 78 1 2 1 = − − = − − =π π θ rad C D D 9273 . 3 84 12 78 1 2 2 = − + = − + =π π θ

(19)

o 5 . 35 6197 . 0 2 356 . 2 356 . 2 3685 . 3 2 1 1 1 1 = = − − − = − − − ′ =θ θ π θ π rad α o 5 . 37 6544 . 0 9273 . 3 1888 . 4 2 9273 . 3 2 2 2 2 2 + − = = − = − ′ + − = θ π θ θ π rad α

(

)

(

)

lb F

Q= ₁ sinα ₁ +sinα ₂ =1406 sin35.5+sin37.5 =1672 of force exerted Size of pulley; For medium double leather belt,

fpm

v_m =3770 , width = 12in>8in in

(20)

852. A 40-hp motor, weighing 1915 lb., runs at 685 rpm and is mounted on a pivoted base. In Fig. 17.11, Text , e=10in., h in

16 3 19

= . The center of the 11 ½-in. motor pulley is 11 ½ in. lower than the center of the 60-in. driven pulley;

in

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

[

_∑

M _B =0

]

b F a F eR = ₁ + ₂

( )(

10 1915

) ( )( ) ( )( )

= F ₁ 9 + F ₂ 26 150 , 19 26 9F ₁+ F ₂ =

For compensator started

(

rated hp

) ( )

hp hp=1.4 =1.4 40 =56 m v hp F F ₁− ₂ = 33,000

(21)

( )( )

fpm n D v_m 2062 12 685 5 . 11 12 1 1 ₌ ₌ = π π

( )

lb F F 896 2062 56 000 , 33 2 1− = = 896 1 2 = F − F Substituting

(

896

)

19,150 26 9F ₁+ F ₁ − = lb F ₁ =1213 lb F ₂ =1213−896=317

For medium leather belt, t in 64 20 = sbt F 1 =

( )( )













= 64 20 300 1213 b in b=13 (c)

_











₋













− = − _θ θ ρ f f s e e v s bt F F 1 2 . 32 12 2 2 1 fps v_s 34.37 60 2062 = = . . 035 . 0 lb cu in = ρ

( )

(

)(

)

_











₋









−













= _θ θ f f e e 1 2 . 32 37 . 34 035 . 0 12 300 64 20 13 896 775 . 0 1 = − θ θ f f e e 492 . 1 = θ f rad C D D 1312 . 2 48 5 . 11 60 1 2 − ₌ ₋ − ₌ − =π π θ

(

2.1312

)

=1.492 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, e in

4 3 8 = ., in h 16 5 17

= . The center of the 12-in. motor pulley is on the same level as the center of the 54-in. compressor pulley; C =40in. (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

(22)

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 hp=1.4 50 =70 m v hp F F ₁− ₂ = 33,000

( )(

)

fpm n D v_m 3581 12 1140 12 12 1 1 ₌ ₌ = π π

( )

lb F F 645 2062 70 000 , 33 2 1− = = in b≈ 25 in a≈5 lb R=1900 b F a F eR = ₁ + ₂

( )(

8.75 1900

) ( ) ( )

=F ₁ 5 +F ₂ 25 lb F F ₁+5 ₂ =3325 lb F F 5 3325 645+ ₂+ ₂ = lb F ₂ = 447 lb F F ₁ =645+ ₂ = 645+447=1092

(b) For heavy double leather belt

in t 64 23 = in b=11

(23)

( )

psi bt F s 276 64 20 11 1092 1 ₌













= = (c)

_











₋













− = − _θ θ ρ f f s e e v s bt F F 1 2 . 32 12 2 2 1 fps v_s 59.68 60 3581 = = . . 035 . 0 lb cu in = ρ

( )

(

)(

)

_











₋









−













= _θ θ f f e e 1 2 . 32 68 . 59 035 . 0 12 276 64 23 11 645 241 . 1 = θ f rad C D D 092 . 2 40 12 54 1 2 − ₌ ₋ − ₌ − =π π θ

(

2.092

)

=1.492 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; D₂ =36in., 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)

( )

o 174 040 . 3 12 23 8 36 1 2 − ₌ ₋ − ₌ ₌ − = rad C D D π π θ 976 . 0 = θ K 2400 θ K N bv hp= m p 976 . 0 = θ K

( )(

)

fpm n D v_m 2409 12 1150 8 12 1 1 ₌ ₌ = π π 5 = p N

(

)( )(

)

2400 976 . 0 5 2409 20 b hp= = in b= 4.1 min. b=5in

(24)

(b) With b=9in 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 belt must operate at an angle 75o with the horizontal. What is the overload capacity of this belt if the rated capacity is as defined in §17.15, Text ?

Solution: 2400 p m N bv hp = in b= 20

( )( )

fpm n D v_m 5616 12 650 33 12 1 1 = = = π π 10 = p N

( )(

)( )

hp hp 468 2400 10 5616 20 = = Overlaod Capacity =

(

100%

)

56% 300 300 468 = − 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 =43to49in. 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 ≈45.2in. Solution: Table 17.7 4 . 1 2 . 0 2 . 1 + = = sf N (24 hr/day) Design hp = N _sf(transmitted hp) =

( )( )

1.4 125 =175hp Fig. 17.4, 175 hp, 1180 rpm in D_min =13 , D-section 4 . 14 50 340 1180 1 2 ₌ ₌ D D use D₁ =14.4in>13in in D₂ =50

(25)

( )(

)

fpm n D v_m 4449 12 1180 4 . 14 12 1 1 ₌ ₌ = π π 3 6 2 1 09 . 0 3 10 10 10 _m _m d m v v e D K c v a hp Rated













− −













= Table 17.3, D-section 788 . 18 = a ,c=137.7, e=0.0848 Table 17.4, 3.47 1 2 ₌ D D 14 . 1 = d K

( )( )

(

)

(

)

hp hp Rated 28.294 10 4449 10 4449 0848 . 0 4 . 14 14 . 1 7 . 137 4449 10 788 . 18 ₆ ₃ 2 09 . 0 3 =













− −













=

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

792 . 8 = a ,c=38.819, e=0.0416 3 6 2 1 09 . 0 3 10 10 10 _m _m d m v v e D K c v a hp Rated













− −













=

( )( )

(

)

(

)

hp hp Rated 20.0 10 4449 10 4449 0416 . 0 4 . 14 14 . 1 819 . 38 4449 10 792 . 8 ₆ ₃ 2 09 . 0 3 =













− −













=

Adjusted rated hp = K _θK _L

(

rated hp

)

Table 17.5, 77 . 0 46 4 . 14 50 1 2 − ₌ − ₌ C D D 88 . 0 = θ K Table 17.6

(

)

(

)

C D D D D C L 4 57 . 1 2 2 1 2 1 2 − + + + ≈

( )

(

)

(

)

( )

in L 200 46 4 4 . 14 50 4 . 14 50 57 . 1 46 2 2 = − + + + = use C195, L=197.9in 07 . 1 = L K Adjusted rated hp =

(

0.88

)( )( )

1.07 20 =18.83hp belts hp rated Adjusted hp Design belts of No 9.3 83 . 18 175 . = = = use 9 belts

(26)

(

)

16 32 ₂ ₁ 2 2 D D B B C = + − −

(

D D

)

(

)

(

)

in L B= 4 −6.28 ₂ + ₁ =4197.9 −6.2850+14.4 =387.2

(

)

(

)

in C 44.9 16 4 . 14 50 32 2 . 387 2 . 387 2 2 = − − + =

857. 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 ≈50in. 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 ≈50.2in.

Solution: Table 17.7, (12 hr/day) 2 . 1 2 . 0 4 . 1 − = = sf N Design hp =

( )( )( )

1.2 1.75 50 =105hp Fig. 17.4, 105 hp, 1160 rpm in D_min =13 , D-section 2 . 13 4 . 46 330 1160 1 2 ₌ _≈ D D use D₁ =13.2in>13in in D₂ = 46.4

( )(

)

fpm n D v_m 4009 12 1160 2 . 13 12 1 1 ₌ ₌ = π π 3 6 2 1 09 . 0 3 10 10 10 m m d m v v e D K c v a hp Rated













− −













= Table 17.3, D-section 788 . 18 = a ,c=137.7, e=0.0848 Table 17.4, 3.5 2 . 13 4 . 46 1 2 ₌ ₌ D D 14 . 1 = d K

( )( )

(

)

(

)

hp hp Rated 24.32 10 4009 10 4009 0848 . 0 2 . 13 14 . 1 7 . 137 4009 10 788 . 18 ₆ ₃ 2 09 . 0 3 =













− −













=

Back to Fig. 17.14, C-section must be used. 792 . 8 = a ,c=38.819, e=0.0416 in D_min =9

(27)

1 . 9 32 330 1160 1 2 ₌ _≈ D D use D₁ =9.1in

( )(

)

fpm n D v_m 2764 12 1160 1 . 9 12 1 1 = = = π π

( )( )

(

)

(

)

hp hp Rated 10.96 10 2764 10 2764 0416 . 0 1 . 9 14 . 1 819 . 38 2764 10 792 . 8 ₆ ₃ 2 09 . 0 3 =













− −













=

(

rated hp

)

Table 17.5, 458 . 0 50 1 . 9 32 1 2 − ₌ − ₌ C D D 935 . 0 = θ K Table 17.6

(

)

(

)

C D D D D C L 4 57 . 1 2 2 1 2 1 2 − + + + ≈

( )

(

)

(

)

( )

in L 167 50 4 1 . 9 32 1 . 9 32 57 . 1 50 2 2 = − + + + = use C158, L=160.9in 02 . 1 = L K Adjusted rated hp =

(

0.935

)( )(

1.02 10.96

)

=10.45hp belts hp rated Adjusted hp Design belts of No 10 43 . 10 105 . = = =

(

)

16 32 ₂ ₁ 2 2 D D B B C = + − −

(

D D

) ( )

(

)

in L B= 4 −6.28 ₂ + ₁ =4160.9 −6.2832+9.1 =385.5

( ) (

)

in C 46.8 16 1 . 9 32 32 5 . 385 5 . 385 2 2 = − − + = Use 10-C158 belts, D₁ =9.1in in D₂ =32 ,C =46.8in

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 =113to123in. 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; C ≈116.3in.

(28)

Table 17.7 4 . 1 2 . 0 6 . 1 − = = sf N

( )( )

hp hp= 1.4 200 =280 Fig. 17.14, 280 hp, 600 rpm Use Section E

But in Table 17.3, section E is not available, use section D 13 min = D 8 . 4 125 600 1 2 ₌ ₌ D D For D₁_max: 1 2 1 2 minC = D + D +D 1 1 1 2 8 . 4 113= D + D +D in D₁ = 28 2 minC = D in D₂ =113 in D 23.5 8 . 4 113 1 = = use D

(

13 23.5

)

18in 2 1 1 ≈ + =

( )( )

in D₂ = 4.8 18 =86.4

(

)

(

)

C D D D D C L 4 57 . 1 2 2 1 2 1 2 − + + + ≈

( ) (

)

(

)

(

)

in L 410 118 4 18 4 . 86 18 4 . 86 57 . 1 118 2 2 = − + + + = using D₁ =19in, D₂ =91.2in, C =118in

( ) (

)

(

)

(

)

in L 420 118 4 19 2 . 91 19 2 . 91 57 . 1 118 2 2 = − + + + =

Therefore use D420 sections

in D₁ =19 , D₂ =91.2in

( )( )

fpm n D v_m 2985 12 600 19 12 1 1 ₌ ₌ = π π 3 6 2 1 09 . 0 3 10 10 10 _m _m d m v v e D K c v a hp Rated













− −













= Table 17.3, D-section 788 . 18 = a ,c=137.7, e=0.0848 Table 17.4, 4.8 1 2 ₌ D D

(29)

14 . 1 = d K

( )( )

(

)

(

)

hp hp Rated 29.6 10 2985 10 2985 0848 . 0 19 14 . 1 7 . 137 2985 10 788 . 18 ₆ ₃ 2 09 . 0 3 =













− −













=

Therefore, Fig. 17.14, section D is used. Adjusted rated hp = K _θK _L

(

rated hp

)

Table 17.5, 612 . 0 118 19 2 . 91 1 2 − ₌ − ₌ C D D 83 . 0 = θ K (V-flat) Table 17.6, D420 in L=420.8 12 . 1 = L K Adjusted rated hp =

( )( )( )

0.83 1.12 29.6 = 27.52hp belts hp rated Adjusted hp Design belts of No 10 52 . 27 280 . = = = Use10 , D420, D₁ =19in, D₂ =91.2in, C =118in

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 flat drive. The B.F. Goodrich Company recommended eight D270 V-belts, 17.24-in sheave, 61-in. pully, C ≈69.7in.

Solution: Table 17.7, 4 . 1 2 . 0 6 . 1 − = = sf N Design hp =

( )( )

1.4 150 =210hp Fig. 17.4, 210 hp, 700 rpm in D_min =13 , D-section 3 6 2 1 09 . 0 3 10 10 10 _m _m d m v v e D K c v a hp Rated













− −













=

For Max. Rated hp,

( )

0 103 =













v_m d hp d 3 3 3 1 91 . 0 3 10 10 10













−













−













= m m d m v e v D K c v a hp Rated Let ₃ 10 m v X =

(30)

3 1 91 . 0 eX X D K c aX hp d − − =

(

)

3 1 3 1 1 3 10 12 700 10 12 10 = × = × = v Dn D X m π π π 700 10 12 3 1 X D = × 3 3 91 . 0 10 12 700 eX K c aX hp d − × − = π

( )

0.91 3 0 2 09 . 0 = − = _aX− _eX X d hp d e a X 3 91 . 0 09 . 2 = Table 17.3, D-section 788 . 18 = a ,c=137.7, e=0.0848

(

)

(

0.0848

)

3 788 . 18 91 . 0 10 09 . 2 3 09 . 2 =













= vm X fpm v_m =7488 7488 12 1 1 ₌ = Dn v_m π

(

)

7488 12 700 1 ₌ = D v_m π in D₁ = 40.86 max D₁ =40.86in ave. D

(

13 40.86

)

26.93in 2 1 1 = + = use D₁ =22in 22 79 195 700 1 2 ₌ _≈ D D in D₁ = 22 , D₂ =79in Min. C D D D 22 72.5in 2 79 22 2 1 2 1 + ₊ ₌ + ₊ ₌ = Or Min. C = D₂ =79in

(

)

(

)

C D D D D C L 4 57 . 1 2 2 1 2 1 2 − + + + ≈

( )

(

)

(

)

( )

in L 327 79 4 22 79 22 79 57 . 1 79 2 2 = − + + + = use D330, L=330.8in

(

)

16 32 ₂ ₁ 2 2 D D B B C = + − −

(31)

(

D D

) ( )

(

)

in L B= 4 −6.28 ₂ + ₁ =4 330.8 −6.28 79+22 =689

( ) (

)

in C 81.12 16 22 79 32 689 689 2 2 = − − + =

( )( )

fpm n D v_m 4032 12 700 22 12 1 1 ₌ ₌ = π π 14 . 1 = d K

( )( )

(

)

(

)

hp hp Rated 39.124 10 4032 10 4032 0848 . 0 22 14 . 1 7 . 137 4032 10 788 . 18 ₆ ₃ 2 09 . 0 3 =













− −













=

(

rated hp

)

Table 17.5, 70 . 0 12 . 81 22 79 1 2 − ₌ − ₌ C D D 84 . 0 = θ K (V-flat) Table 17.6 D330 07 . 1 = L K Adjusted rated hp =

( )( )(

0.84 1.07 39.124

)

=35.165hp belts hp rated Adjusted hp Design belts of No 5.97 165 . 35 210 . = = = use 6 belts Use 6 , D330 V-belts , D₁ =22in, D₂ =79in, C ≈81.1in

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 <50in.; 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 8 . 1 2 . 0 6 . 1 + = = sf N Design hp =

( )( )

1.8 30 =54hp Speed of fan at 30 hp

(

)

rpm n 280 238 238 286 24 3 . 29 24 30 2 − + = − − = at 54 hp, 1160 rpm. Fig. 17.4 use either section C or section D Minimum center distance:

2

D C =

(32)

or 1 2 ₁ 2 D D D C = + + 056 . 4 286 1160 1 2 ₌ ₌ D D use C =4.056D₁ in C in 50 44 < < , use C =47in in D 11.6 056 . 4 47 max 1 = =

use C-section, D_min =9in

Let �₁ =10�1��, �₂ =41��

(

)

(

)

C D D D D C L 4 57 . 1 2 2 1 2 1 2 − + + + ≈

( )

(

)

(

)

( )

in L 179.3 47 4 1 . 10 41 1 . 10 41 57 . 1 47 2 2 = − + + + = use C137, L=175.9in

(

)

16 32 ₂ ₁ 2 2 D D B B C = + − −

(

D D

) ( ) (

)

in L B= 4 −6.28 ₂ + ₁ =4175.9 −6.28 41+10.1 =328.7

(

)

(

)

in in C 45.2 44 16 1 . 10 41 32 7 . 382 7 . 382 2 2 ≈ = − − + = C173, satisfies 44in<C <50in 3 3 3 1 91 . 0 3 10 10 10













−













−













= m m d m v _e v D K c v a hp Rated

( )(

)

fpm n D v_m 3067 12 1160 1 . 10 12 1 1 ₌ ₌ = π π Table 17.4 056 . 4 1 2 ₌ D D , K d =1.14 Table 17.3, C-section 792 . 8 = a ,c=38.819, e=0.0416

( )( )

(

)

(

)

hp hp Rated 12.838 10 3067 10 3067 0416 . 0 1 . 10 14 . 1 819 . 38 3067 10 792 . 8 ₆ ₃ 2 09 . 0 3 =













− −













=

(

rated hp

)

Table 17.5, 68 . 0 2 . 45 1 . 10 41 1 2 − ₌ − ₌ C D D 90 . 0 = θ K

(33)

Table 17.6 9 . 175 = L , C173 04 . 1 = L K Adjusted rated hp =

( )( )(

0.90 1.04 12.838

)

=12.02hp belts hp rated Adjusted hp Design belts of No 4.5 02 . 12 54 . = = = use 5 belts Use 5 , C173 V-belts , D₁ =10.1in, D₂ = 41in 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 4-cylinder 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) designhp =2

( )

30 =60hp intermittent 2 500 1000 2 1 1 2 _≈ ₌ ₌ n n D D 1 2 2 D D = in D D C 24 2 1 2 + = = 24 2 2 1 1+ = D D �� ₁_�� ₌9�6

( )

�� ₂_�� =2 ₁_�� =2 9�6 =19�2

( )(

)

fpm n D v_m 2513 12 1000 6 . 9 12 1 1 ₌ ₌ = π π

Table 17.8, use Chain No. 35, Limiting Speed = 2800 fpm

(34)

Minimum number of teeth Assume N ₁ =21 42 2 ₁ 2 = N = N [Roller-Bushing Impact] 8 . 0 5 . 1 100 P n N K hp ts r













= Chain No. 35 in P 8 3 = 21 = ts N rpm n=1000 29 = r K

( )

hp hp 40.3 8 3 1000 21 100 29 8 . 0 5 . 1 =





















=

[Link Plate Fatigue]

P ts n P N hp=0.004 1.08 0.9 3−0.07

( ) ( )

hp hp 2.91 8 3 1000 21 004 . 0 8 3 07 . 0 3 9 . 0 08 . 1 =













=       − No. of strands = 21 91 . 2 60 = = hp rated hp design

Use Chain No. 35, P in

8 3

= , 21 strands Check for diameter and velocity

�� 516 2 21 180 375 0 180 1 � �� =       =       = ( )( ) �� _� 659 12 1000 516 2 12 1 1 ₌ ₌ =π π �

Therefore, we can use higher size, Max. pitch �� 43 1 21 180 6 9 180 1�� _= �     =       =

say Chain no. 80

�� =1

( ) ( ) ( ) ( ) ��

��₌0�004211�08 1000 0�9 13−0�071 ₌53�7

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

(35)

�� 4 3 =

( ) ( )

�� 23 4 3 1000 21 004 0 4 3 07 0 3 9 0 08 1 =       =       − � � � � 61 2 23 60 � = = �� or 3 strands �� 0 5 21 180 75 0 180 1 � �� =       =       =

( )(

)

�� _� 1309 12 1000 0 5 12 1 1 ₌ ₌ =π π �

The answer is Chain No. 60, with P = ¾ in and 3 chains, limiting velocity is 1800 fpm

(b)

(

)

C N N N N C L 40 2 2 2 1 2 2 1 + ₊ − + ≈ pitches 32 4 3 24 =       = � 21 1 = N 42 2 = N

( )

(

)

( )

�� 95845 96 32 40 21 42 2 42 21 32 2 2 ≈ = − + + + = � Amount of slack

(

)

2 1 2 2 433 . 0 S L h= − in C L= =24

(

)

�� 24058 2 4 3 845 95 96 24 � � =













− + = ( ) ( )

[

]

�� 0433 24 058 24 2 07229 1 2 2 � � � ₋ ₌ = (c) p_b = bearing pressure Table 17.8, Chain No. 60

�� =0�234 �� 2 1 = �� =0�094 ( ) ( ) 2 160992 0 094 0 2 2 1 234 0 2 � �� = + = � _ + � _= � hp FV 60 000 , 33 =

(36)

( ) �� 60 000 33 1309 = � �� =1512�6 �� 5042 3 6 1512 � � = = �� _� 3131 160992 0 2 504 = = � �

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 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 =24in., L=88 pitches.

Solution: Table 17.7 0 . 1 2 . 0 2 . 1 − = = sf N (8 hr/day)

( )

hp hp design =1.0 2 = 2.0 rpm n 50 35 1750 1 = = rpm n₂ =12

Minimum number of teeth = 12 Use N ₁ =12

[Link Plate Fatigue]

P ts n P N hp=0.004 1.08 0.9 3−0.07

( ) ( )

12 50 1.0 004 . 0 0 . 2 004 . 0 1.08 0.9 1.08 0.9 3 07 . 0 3 = = = ≈ − n N hp P P ts P

Use Chain No. 80, P =1.0in

To check for roller-bushing fatigue 8 . 0 5 . 1 100 P n N K hp ts r













= 29 = r K

( )

hp hp hp 1 2747 2 1000 12 100 17 0.8 5 . 1 > =









= (a) N ₁ =12

( )

teeth N n n N 12 50 12 50 1 1 2



=











=













=

(37)

2 1 2 D D C = +

( )( )

in PN D 1 1.0 12 3.82 1 ≈ = = π π

( )( )

in PN D 1 1.0 50 15.92 2 ≈ = = π π in C 17.83 2 82 . 3 92 . 15 + = ≈ use C =18in pitches C =18

chain pitch = 1.0 in, Chain No. 80

(b)

(

)

C N N N N C L 40 2 2 2 1 2 2 1 + ₊ − + ≈

( )

(

)

( )

18 69 40 12 50 2 59 12 18 2 2 = − + + + ≈ L pitches use L=70 pitches (c) p_b = bearing pressure Table 17.8, Chain No. 80

in C =0.312 in E 8 5 = in J =0.125

( )( )( )

fpm n PN v ts m 50 12 50 12 1 12 1 ₌ ₌ =

(

)

(

)

2 04054 . 0 05 . 0 2 16 3 141 . 0 2 J in E C A= + =

_



+

_



= hp FV 60 000 , 33 =

( )

lb F 1320 50 2 000 , 33 = =

(

E J

)

₍

₎

psi C F p_b 4835 125 . 0 2 8 5 312 . 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 n=500rpm. The 1 ¼-in. driven shaft turns 250 rpm; C ≈16in. (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