SECTION-15.pdf

LEATHER BELTS DESIGN PROBLEMS

841. A belt drive is to be designed for F₁ F₂ =3, while transmitting 60 hp at 2700 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

psi

s=300 . (b) How does this width compare with that obtained by the ALBA procedure? (c) Compute the maximum stress in the straight port of the ALBA belt. (d) If the belt in (a) stretches until the tight tension F₁=525lb., what is

2 1 F

F ?

Solution:

(a) Table 17.1, Medium Double Ply, SelectD₁=7in. min. in t 64 20 =

( )(

)

fpm n D v_m 4948 12 2700 7 12 1 1 _{= =} =π π fpm fpm fpm 4948 6000 4000 < <

(

)

000 , 33 2 1 F vm F hp= −

(

)(

)

000 , 33 4948 60₌ F1−F2 lb F F₁− ₂ =400 2 1 3F F = lb F F 400 3 ₂− ₂ = lb F₂ =200

(

)

lb F F₁=3 ₂ =3200 =600 sbt F₁= η 300 = d s

For cemented joint, η=1.0

psi s_d =300

(

)( )

      = = 64 20 300 600 1 b F in b=6.4 say b=6.5in

(b) ALBA Procedure

(

hp in.,Table17.1

)

(

bC_mC_p

)

(

C_f1C_f2L

)

hp=

Table 17.1, vm =4948 fpm

Medium Double Ply 448 . 12 = in hp Table 17.2

Squirrel cage, compensator, starting 67 . 0 = m C Pulley Size, D₁=7in 6 . 0 = p C Jerky loads, C_f =0.83

(

12.448

)( )(

0.67

)( )(

0.6 0.83

)

60 b hp= = in b=14.5 say b=15in (c)

( )( )

psi bt F s 128 64 20 15 1 600 1 ₌       = = η (d)

(

)

(

)

2 1 2 1 2 1 2 2 1 1 2 1 200 600 2F_o =F +F = + lb F_o =373.2 lb F₁=525

(

)

(

)

2 1 2 2 1 2 1 525 2 . 373 2 = +F lb F₂ =247 1255 . 2 247 525 2 1 _{= =} F F

842. A 20-hp, 1750 rpm, slip-ring motor is to drive a ventilating fan at 330 rpm. The horizontal center distance must be about 8 to 9 ft. for clearance, and operation is continuous, 24 hr./day. (a) What driving-pulley size is needed for a 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 =70lb in. (§17.10), compute F₁ F₂ and the stress on the tight side if the approximate relationship of the operating tensions and the initial tensions is 2

1 2 1 2 2 1 1 F 2Fo F + = .

Solution: fpm to v_m=4000 4500 assume v_m =4250 fpm 12 1 1n D vm π =

(

)

12 1750 4250₌πD1 in D₁=9.26 say D₁=10in

(b) Using Heavy Double Ply Belt, t in

64 23 =

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

( )(

)

fpm n D vm 4581 12 1750 10 12 1 1 _{= =} =π π ALBA Tables

(

hp in.,Table17.1

)

(

bCmCp

)

(

Cf1Cf2L

)

hp= 8 . 13 = in hp

Slip ring motor, C_m=0.4 Pulley Size, D₁=10in 7 . 0 = p C

Table 17.7, 24 hr/day, continuous 8 . 1 = sf N Assume Cf =0.74

( )( ) (

1.8 20 13.8

)( )( )( )(

b 0.4 0.7 0.74

)

hp= = in b=12.59 use b=13in

(c) General belt equation

      ₋       − = − _θ θ ρ f f s e e v s bt F F 1 2 . 32 12 2 2 1 fps vs 76.35 60 4581 = = . . 035 . 0 lb cu in = ρ for leather in t 64 23 = in b=13

( )( )

lb F F 260 4581 20 8 . 1 000 , 33 2 1− = = 3 . 0 = f on iron or steel C D D₂− ₁ ± ≈π θ ft C=8~9 use 8.5 ft

( )

in D 10 53 330 1750 2  =      =

( )

12 2.72rad 5 . 8 10 53 = − − =π θ

( )(

0.3 2.72

)

=0.816 = θ f 5578 . 0 1 1 816 . 0 816 . 0 = − = − e e e e f f θ θ

( )

(

)(

) (

0.5578

)

2 . 32 35 . 76 035 . 0 12 64 23 13 260 2 2 1       −       = = −F s F psi s=176 (d) 2 1 2 1 2 2 1 1 F 2Fo F + =

(

lb in

)(

in

)

lb F_o = 70 13 =910 lb F F₁− ₂ =260 lb F F₂ = ₁−260

(

260

)

2

(

910

)

2 60.33 1 2 1 1 2 1 1 + F − = = F lb F₁=1045 lb F₂ =1045−260=785

( )

psi bt F s 224 64 23 13 1045 1 ₌       = = 331 . 1 785 1045 2 1 _{= =} F F

843. 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,

in

D₁ =16 ; D₂ =53in, 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 F₁ F₂ =2. Compute the maximum stress in a straight part. (d) The approximate relation of the operating tensions and the

initial tension F_o is 2 1 2 1 2 2 1 1 F 2Fo

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

Solution: (a) Table 17.1

( )(

)

fpm n D v_m 4775 12 1140 16 12 1 1 _{= =} =π π

Use heavy double-ply belt

in t 64 23 = 1 . 14 = in hp

(

hp in.,Table17.1

)

(

bC_mC_p

)

(

C_f1C_f2L

)

hp=

line starting electric motor , Cm =0.5

Table 17.7, squirrel-cage, electric motor, line starting, reciprocating compressor 4 . 1 = sf N in D₁=16 , C_p =0.8 assume, C_f =0.74

( )(

)

hp hp= 1.4 100 =140

(

14.1

)( )( )( )(

0.5 0.8 0.74

)

140 b hp= = in b=33.5 use b=34in (b) §17.6, sd =400η 00 . 1 =

η for cemented joint. psi s_d =400       ₋       − = − _θ θ ρ f f s e e v s bt F F 1 2 . 32 12 2 2 1 fps vs 79.6 60 4775 = = . . 035 . 0 lb cu in = ρ for leather in t 64 23 = in b=34

( )(

)

lb F F 968 4775 100 4 . 1 000 , 33 2 1− = =

( )

(

)(

)

_      ₋       −       = = − _θ θ f f e e F F 1 2 . 32 6 . 79 035 . 0 12 400 64 23 34 968 2 2 1

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) F₁−F₂ =968lb 2 1 2F F = lb F F 968 2 ₂− ₂ =

(

)

lb F F₁=2 ₂ =2968 =1936

( )

psi bt F s 159 64 23 34 1936 1 ₌       = = (d) F₁ =1936lb, F₂ =968lb 2 1 2 2 1 1 2 1 2F_o =F +F

(

)

(

)

2 1 2 1 2 1 968 1936 2Fo = + lb Fo =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 585 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 F₁ and F₁ F₂. (d)

Approximately, 2 1 2 1 2 2 1 1 F 2Fo

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?

Solution: (a) vm 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 =

(

hp in.,Table17.1

)

(

bCmCp

)

(

Cf1Cf2L

)

hp= 86 . 9 = in hp Table 17.2 67 . 0 = m C 8 . 0 = p C

(

0.74

)(

0.80

)

=0.592 = f C

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

( )( )

hp hp= 1.4 50 =70

(

9.86

)( )(

0.67

)( )(

0.8 0.592

)

70 b hp= = in b=22.4 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 2.35rad 6 13 70 = − − =π θ

( )(

0.3 2.35

)

=0.705 = θ f

5059 . 0 1 1 705 . 0 705 . 0 = − = − e e e e f f θ θ

( )( )

lb F F 785 2944 50 4 . 1 000 , 33 2 1− = =

( )

(

)(

) (

0.5059

)

2 . 32 1 . 49 035 . 0 12 64 23 25 785 2 2 1       −       = = −F s F psi s=204 Cemented joint, η=1.0 psi s=204 (c) F sbt

(

)( )

1833lb 64 23 25 204 1 =      = = lb F₂ =1833−785=1048 749 . 1 1048 1833 2 1 _{= =} F F (d) 2 1 2 2 1 1 2 1 2F_o =F +F

(

)

(

)

2 1 2 1 2 1 1048 1833 2F_o = + lb F_o =1413 in lb F_o 56.5 25 1413 = =

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 − + + + ≈

( )( )

(

) (

)

( )( )

in L 286 12 6 4 13 70 13 70 57 . 1 12 6 2 2 = − + + + =

(f) More economical basis 12 1 1n D vm π =

(

)

12 865 4500₌πD1 in D₁=19.87 use D₁ =20in

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

)( )

bCmCpCf 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

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 F₂ =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 F₂ =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 F₁−F₂ =40lb in of width and F₁ F₂ =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) vm Dn

( )(

)

3665 fpm 12 1750 8 12 1 1 _{= =} =π π in b=6

( )( )

lb F F₁− ₂ = 40 6 =240

(

)

(

)(

)

hp v F F hp m 65 . 26 000 , 33 3665 240 000 , 33 2 1− _{= =} = (b) F₁ =3F₂ lb F F 240 3 ₂− ₂ = lb F₂ =120 lb F₁=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 vs 61.1 60 3665 = = lb F F₁− 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

)( )

bCmCpCf hp= fpm v_m=3665 965 . 6 = in hp

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

)( )

bC_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

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 F₁ F₂ v_s 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

( )

( )

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 v2s 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 vs =90.431 fpm vm=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

(

)

000 , 33 60 F₁ F₂ vs 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 v2_s

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 vs =151.6575 fpm vm=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 v2_s 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

( )(

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) F₁ =sbt in b=12 in t 64 20 =

(

)( )

      = 64 20 12 375 1 F (b) m v hp F F₁− ₂ =33,000

( )(

)

fpm n D vm 3770 12 1200 12 12 1 1 _{= =} =π π Table 17.7, Nsf =1.2

( )(

)

lb F F 1050 3770 100 2 . 1 000 , 33 2 1− = = lb F F₂ = 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 vs 62.83 60 3770 = =

. . 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

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 sinα₁+sinα₂ =1406sin35.5+sin37.5 =1672 of force exerted Size of pulley; For medium double leather belt,

fpm

vm=3770 , width = 12in>8in

in D=6+2=8

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

) ( )( ) ( )( )

= F₁ 9 + F₂ 26 150 , 19 26 9F₁+ F₂ =

For compensator started

(

rated hp

)

( )

hp hp=1.4 =1.440 =56 m v hp F F₁− ₂ =33,000

(

)(

)

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 9F₁+ F₁− = lb F₁=1213 lb F₂ =1213−896=317

For medium leather belt, t in

64 20 = sbt F₁=

(

)( )

      = 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

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=1.450 =70 m v hp F F₁− 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

)

=F₁

( )

5 +F₂

( )

25 lb F F₁+5 ₂ =3325 lb F F 5 3325 645+ ₂+ ₂ = lb F₂ =447 lb F F₁=645+ ₂ =645+447=1092 (b) For heavy double leather belt

in t 64 23 = in b=11

( )

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 vs 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)

( )

12 3.040 174o 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

(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 mN 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

(

)(

)

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 = KθKL

(

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

(

)

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

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

( )(

)

fpm n D vm 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 =         − −       =

Adjusted rated hp = K_θK_L

(

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.

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 D_1max: 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

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 =

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 _{− =} = − eX aX 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 vm 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= + − −

(

D D

)

(

)

(

)

in L B=4 −6.28 ₂+ ₁ =4330.8 −6.2879+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 =         − −       =

Adjusted rated hp = K_θK_L

(

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=

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 D₁=10in, D₁ =41in

(

) (

)

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 −6.2841+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, D-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 =         − −       =

Adjusted rated hp = KθKL

(

rated hp

)

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

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 2D D = in D D C 24 2 1 2+ = = 24 2 2 1 1+ = D D in D₁=9.6

( )

in D D₂ =2 ₁=29.6 =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

Minimum number of teeth Assume N₁=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 (b)

(

)

C N N N N C L 40 2 2 2 1 2 2 1+ ₊ − + ≈ pitches 64 8 3 24 =       = C 21 1= N 42 2 = N

( )

(

)

( )

pitches pitches L 159.67 160 64 40 21 42 2 42 21 64 2 2 ≈ = − + + + = Amount of slack

(

)

2 1 2 2 433 . 0 S L h= − in C L= =24

(

)

in in in S 24.062 2 8 3 67 . 159 160 24 =       − + =

(

) ( )

[

]

in in h 4 3 75 . 0 24 062 . 24 433 . 0 2 1 2 2 = = − = (c) p_b = bearing pressure Table 17.8, Chain No. 25

in C=0.141 in E 16 3 = in J =0.05

(

)

(

)

2 04054 . 0 05 . 0 2 16 3 141 . 0 2J in E C A= + = _ + _= hp FV 60 000 , 33 =

(

)

hp F 60 000 , 33 2513 = lb F =787.9 strand lb F 37.5 21 9 . 787 = = psi pb 925 04054 . 0 5 . 37 = =

862. A conveyor is driven by a 2-hp high-starting-torque electric motor through a flexible coupling to a worm-gear speed reducer, whose 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

Use N1=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) N1=12

( )

teeth N n n N 12 50 12 50 1 2 1 2  =      =       = 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) pb = 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 2J in E C A= + = _ + _= hp FV 60 000 , 33 =

( )

lb F 1320 50 2 000 , 33 = =

(

E J

)

₍

₎

psi C F pb 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 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 2 . 1 = sf N

( )

hp hp design =1.25 =6 2 250 500 1 2 _{= =} D D 2 1 2 D D C= + 2 2 16 1 1 D D + = in D1=6.4

( )

in D D₂ =2 ₁=26.4 =12.8

( )(

)

fpm n D v_m 838 12 500 4 . 6 12 1 1 _{= =} =π π

(a) Link Plate Fatigue

P ts n P N hp=0.004 1.08 0.9 3−0.07

( )

P P P D N Nts 11 . 20 4 . 6 1 1≈ = = = π π

(

)

P P P hp 0.9 3 0.07 08 . 1 500 11 . 20 004 . 0  −      = P P1.92 0.07 47 . 27 6= − in P=0.45 use P in 2 1 = , Chain No. 40

( )

40 2 1 4 . 6 1 1 =       = ≈π π P D N 80 2 ₁ 2 = N = N Size of sprocket, N₁=40,N₂ =80, P in 2 1 = . (b) C=16in pitches in in C 32 2 1 16 = =

(

)

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

( )

(

)

( )

32 125.25 40 40 80 2 80 40 32 2 2 = − + + + ≈ L pitches use L=126 pitches (c) Method: v_m =838 fpm.

Use Type II Lubrication (v_max =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 2 . 0 4 . 1 + = sf N (24 hr/day)

( )

hp hp design =1.6 20 =32 (a) 2.85 200 570 2 1 _{= =} n n

85 . 2 2 1 1 2 _{≈ =} n n D D

Considering single strand

P ts n P N hp=0.004 1.08 0.9 3−0.07 min N_ts =17

( ) (

)

P P hp=32=0.004171.08 570 0.9 3−0.07 24 . 1 07 . 0 3 = − P P in P=1.07 use P=1.0in

( ) (

1.08

) ( )

0.9 3 0.07( )1 1 570 1 004 . 0 32= − = N hp 21 1= N

( )

21 60 200 570 2  =      = N Roller width = in 8 5 2 1 2 D D C= +

( )( )

in PN D 1 1 21 6.685 1≈ = = π π

( )( )

in PN D 2 1 60 19.10 2 ≈ = = π π in C 22.44 2 685 . 6 10 . 19 + = = Use C=23in pitches C 1 23 =

Considering multiple strands Assume, P in 2 1 = P ts n P N hp=0.004 1.08 0.9 3−0.07

( ) (

) ( )

( ) hp hp=0.004211.08 570 0.9 0.53−0.070.5 =4.148 No. of strands = 7.7 148 . 4 32 = hp hp Use 8 strands (b) Chain Length

(

)

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

( )

(

)

( )

23 88.15 40 21 60 2 60 21 23 2 2 = − + + + ≈ L pitches use L=88 pitches (c) pb = bearing pressure Table 17.8, P=1in in E 8 5 = in J =0.125 in C=0.312 m v hp F =33,000

(

)(

)

fpm n D v_m 998 12 570 685 . 6 12 1 1 _{= =} =π π

( )

lb F 1058 998 32 000 , 33 = =

(

E J

)

₍

₎

psi C F p_b 3876 125 . 0 2 8 5 312 . 0 1058 2 =     + = + = (d) Factor of Safety = F Fu 4 , based on fatigue lb F_u =14,500 , Table 17.8 Factor of Safety =

(

1058

)

3.43 4 500 , 14 4F = = F_u

865. 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 capacity (N_s =20teeth) in hp? Compare with the power needed at the constant speed. Does it look as though the drive will have a “long” life?

Solution: Table 17.8 in P 8 5 = lb F_u =6100

(a) Factor of Safety = F Fu

(

)( )

lb F = 500 2 =1000 Factor of Safety = 6.1 1000 6100 = (b) Factor of Safety = F F_u 4 (fatigue) Factor of Safety =

(

1000

)

1.5 4 6100 = (c) v_m ft 35 fpm min 1 sec 60 sec 24 14 =       = 20 = s N in P 8 5 =

Rated hp=0.004N_ts1.08n0.9P3−0.07P [Link Plate Fatigue]

( )

fpm n n PN v s m 35 12 20 8 5 12 =       = = rpm n=33.6 Rated hp

( ) (

)

0.6hp 8 5 6 . 33 20 004 . 0 8 5 07 . 0 3 9 . 0 08 . 1 =       =       −

Hp needed at constant speed

(

)( )

hp hp Fv hp m 6 . 0 53 . 0 000 , 33 35 500 000 , 33 = = < =

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 0.2F_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)

(

)

2 1 2 445 . 4 sec 6 sec 60 min 1 1600 fps fpm t v v a =       = − = kips W_h =20 For 6 x 19 IPS, ft lb D w _r2 6 . 1 ≈ kips D kips D wL _r2 0.64 _r2 1000 400 6 . 1  =      = ma W wL F_t− − _h = 2 . 32 64 . 0 20 D_r2 m= +

(

4.445

)

2 . 32 64 . 0 20 20 64 . 0 2 2       ₊ = − − r r t D D F 2 73 . 0 76 . 22 _r t D F = + in D_r 4 3 1 = kips F_t 25 4 3 1 73 . 0 76 . 22 2 =       + = t b u F F F N = − Table AT 28, IPS tons D F_u ≈42 _r2

(

)

tons kips F_u =421.752 =129 =258 with bending load

m b b s A F = s w b D ED s =