Mechanics of machines_1.pdf

7/27/2019 Mechanics of machines_1.pdf

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C H A P T E R~BELT DRIV tES AN D BAND BRAK ES

10.1 Ratio of belt tensions. Consider a flat belt partly . woundround a pulley so that the angle of lap is fJ, Fig . 10.1, and let T 1 and T 2be the tens ions in the belt when it is about to slip in the directi on shown.

I f the tensions at the en ds of an element subtending an angle dfJ atthe centre are T and T + dT and the reaction between the belt and thepulley is R, then, resolving forc es radially,

(T + dT)dO + TdfJ = R2 2

Th erefore, neglec ting the seco nd order of small quantities,TdO=R

Fla. 10.1

Resolving forces tangentially,(T + dT) - T = p.R

1e. dT =p.R= p.1' dO from equation (10.1)

dTi.e. T = p, dO

J.e.

or

J', dT Jo- = p.dO

'1' T o

log, T1 = p.OT2

Tl = e''oT2

176

(10.1)

(10.2)

.I

BELT DRIVl!lS AND BAND BRAKES 177

I f the belt is used to transmit power between two pulleys, Fig. 10.2,T1 and T 2 are the tight and slack side tensions respectively. I f he pulleysare of unequal diameter s, the belt will slip first on the pulley having thesmaller angle of lap, i .e. on the smaller pulley .

Fla. 10.2

l f v is the speed of the belt in m/ s and T 1 and 1'2 arc in ncwLous, then

power transm it t e d = ( 1 \ - T 2) v W (10.3}

(10 .4 )

10.2 Modification for V-grooved pulley. For a V-grooved

pulley, the normal force betwe en the belt or rope and the pulley isiucreascd since the radial component of this force must equa l R. Thus,if the semi-angle of the groove is {3, Fig. 10.3,

RN = '2 cosec{3

frictional resistance = 2p.N

= f ~ - Rcosec {3

Fro. 10 .3


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178 MECHAN I CS OF MACHINES

Th e friction forc e is ther efore inc reased in th e ratio cosec {3 : 1, so thatthe V-grooved pulley is equiva lent to a flat p ulley having a coefficientof friction of fJ, cosec ,8.

Hence T1 = e"o cosec /IT2

(10.5)

10.3 Effect of centrifug a l tension. Consid er a belt, of ma ssnt per unit leng th , wound round a pulle y of radiu s r, Fig. 10.4. Let th espeed of the belt be v and th e centrifug al tension be T 0

p

FIG. 10.4

I f F is the centrifugal force act ing on an element of th e belt subtendingan an gle dO at th e centre, then, resolving forc es radially,dO

F = 2T0 2' \. v2 \.

1.e. mrd{}. - = T 0 dfl"

or (10 .6)

This is t he te nsion caused by centrifuga l for ce on th e belt and isa ddi tional to the tens ion due to t he transmiss ion of power.

I f allowance is mad e for t hi s additional tension in determining t heratio of the belt tensions, equati on (10.1) be comes

so that

i.e .

. v2TdO = R + F = R + rr;rilO. -, r

dT = !J-R = fJ- ( T - mv 2 ) dO = !J-(T - T,) dOdT- - = ~ J - d O

T - T 0whence, integrating as before,

T1 - Tc = ef'o (or e"'ocosec ll)T 1 - Te

(10.7)

BELT DRIVES AND BAND BRA KES 179

T 1 - Tc an d T 2 - T 0 are the effective dr ivin g tensions and T 1 andT 2 are now th e total t ensions in the belt .

I .e .

I.e .

Allowing for centrif ugal t ension, equat ion (10. 4) beco mes

power = (T 1 - T.) (1 - _!_)v (10.8)e"'oFrom eq ua tion (10.8) th e po wer tra nsmitted is a maxim um wh en

d

dv{(T1 - T 0 )v} = 0

:v(T 1v- mv 3 ) = 0mv2 = lT I

or T. = f l \ (10.9)Th e maxim um power is th en obtained by substit uting this va lue of T 0

an d th e corresponding va lue of v in equa t ion (10. 8) .

10 .4 Initial tension . The belt is assem bl ed with an init ial tens ion ,T 0 Wh en power is being transmitted, the tension in th e tight sideincr eases from T0 to T 1 and on th e slack s ide decr ease s from T 0 to T 2 I f t he belt is assum ed to obey Hook e's Law and its lengt h to remaincon st a nt, then the increase in length of t he t ight side is equal to thedecrease in lengt h of the s lack side,

I.e. . T1 - T0 = T 0 - T2since th e lengths an d cross -sect iona l areas of th e belt are t he sa me oneach side.

H ence . (10.10)

l. T tvo pulleys , one 450 mm diameter and the other 200 mm diameter,are on para llel shafts 1,95 m apart, and me connected by a mossed belt. Findthe length of belt r equi red and the angle of contact between the belt and eachpulley .

What powe1 can be tran smitted by th e belt when the larger pulley r otatesat 200 rev/ mi n if he maximum permissib le tension in the belt is 1 kN, andthe coefficient of f riction between belt and pu l le1.Jis 0, 25? (U. Lond.)

From triangle ABC, Fig. 10.5,

c o s ~=225 + 100 = 0 1 667

2 1950 '

. ~ = 80 24 '"2


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180 MECHANICS OB' MACHINES

) ' Cp- ,

I ' . . ._J ....

....

Fro. 10.5

angle of lap for each pulley,

0 = 360 - 2 X 80 24'= 199 12' = 3,474 ra u

' ' '

L ength of bel t= 0,225 X 3,474 + 0,10 X 3,474 + 2 X 1,95 sin 80 24'= 4,97 5 m

T 1 =1000N

T = 1000_ = 1000 = 419 3 N2 e0,25 x ~ , 4 i 4 2,384 '

power = (T - T 2)v from equation (10.3)

2:n;= (1000- 419,3) X

60X 200 X 0,225

= 2740 W or 2,74 kW

2. A belt drive consists of two V -belts in JJamllel, on grooved pulley s ofthe same size. The angle of the groove is 30 . The cross -sectional area ofeach belt is 750 mm 2 and p, = 0,1 2. The density of the belt material is1,2 Mg j m 3 and the maximum safe stress in the mate ,rial is 7 MN m2

Calculate the power that can be transmitted between pulleys 300 mmdiameter rotating at 1500 rev j min. Find also the shaft speed in rev/ minat which the power tmnsmitted would be a maximum. (U . Lon d .)

8 = :n; rad

2nv = - X 1500 X 0,15 = 23,6 m /s

60

m = 1,2 X 103 X 1 X 750 X 10- 6 = 0,9 kgjm

i.e.

from which

BELT DRIVES AND BAND BRAKES

T 0 = mv 2 from equation (10.6)

= 0,9 X 23,G 2 = f>02 NT1 = 7 X 10 6 X 750 X 10- 6 = 5250 N

ep.Ocose c fl = e0,12 x "x cooecls = 4,291

T - T 0 = ep.Oco.ccfl from equation (10.7)T

2- T

0

5250 - 502 = 4;291T 2 - 502

T 2 = 1610N

power= (T 1 - T 2)v X 2

= (5250- 1610) X ~ 3, 6~= 172 000 W or 172 kW

Fo r maximum power,

1750 = 0,9v 2

:. v = 44,1 m /s

N = 1500 X44

1

= 2800 rev min_23-,6

------

181

3. An open belt drive connects two pulleys 1,2 and 0,5 m diameter, onparallel shafts 3,6 m apart. The belt ha s a ma,ss of 0, 9 kgjm length, and themaximum tension in i t i s not to exceed 2 leN.

The 1, 2 m pulley, which is the driver, rttns at 200 rev jmin. Due to beltslip on one of he pulleys, the velocity of the driven shaft is only 450 rev/min.Calculate the torque on each of the two shafts, the power tmnsmitted, andthe power lost in friction. p, = 0,3.

What is the efficiency of the drive? (U. Lond.)

From Fig. 10.6, 0- 0 , 6 - 0,25 - 0 0972COS -- - ,2 3,6

~ = 84 25' = I 472 rad. . 2

angle of lap on sma ller pulley, /

0 = 2,944 rad


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182 MEOHANIOB OF MACHINES

FIG. 10. 6

The b elt speed is th at correspo ndi ng to the peripheral speed of the larger(drivi ng) pulley.

1.e .

2n:. v = 200 X - X 0,6 = 12,57 m js

60

= 0,9 X 12,57 2 = 142,3 N

Tl - Tc = e ~ ' 6T 2 - To

2 0 0 0 - 142,3 = eO,S X 2,944 = 2,419

T 2 - 142,3fro m which T 2 = 910 N

:. to rque on driver = (2 0 0 0 - 910) X 0,6 = 654 N m

and torque on fo llower = (2000 - 910) X 0,25 = 272,5 N m

f d. (2000- 910) X 12,57

13 7I W .., f .,_-r .. .,

power o n ve r = = , c woo v ~T - . ~r- I . . ~

I f her e were no slip, spee d of follower would be ~ ~X 200= 480 rev min .

450 .power tr an smitted to follower= 13,7 X

480= 12,85 kW

power lost in friction= 1 3 , 7 - 12,85 = 0,85 kW

. f d . 12,85 8Effimency o nve = -- X 100 = 9 3 . . . : . ' - - - ' p ~ e _ r _ c _ e _ n _ t13,7

BELT DRIVES AND BAND BRAKES 183

4. A small air compressor is belt-driven from a lay shaft in a workshop,the pulley on the compressor being 300 mm diamete1, and the a ngle of lap ofthe belt is 165 . Wlien the belt is moved from the loose to the fast pulley,it slips for 8 s until the compressor attains its constant speed of 300 rev/min .Th e l y w l ~ e e lof the compressor has a moment of inertia of 4 kg m 2 and thefriction requires a constant torque of 4 N m . I f the coifficient of riction is0,28 duri ng the accelemting period, find the tensions in both 1eaches of thebelt, and also the distance that the belt slips and the energy lost in that timed1te to the belt slip. (U. Glas.)

i.e.

Whi le slip is taking place, ratio of belt tensions,. /

T1 = eo,zs x (166 x 1~0) = 2 24Ta '

300 x 2nAcceleration of compressor= ~ S )60 = 5n r a d j ~ 2I

I .4---- - "\..,... ... " . __... .. ---

Net torque on compressor= ~ ~ 2) x@--:,.!- ..; f.' 5m

:. (T1 - T 2) X 0,15 - 4 = 4 X - 1 v4;T1 - T 2 = 131,4 N

Therefore, from equations (1) and (2),

T 1 = 237 N and T 2 = 106 N

Belt velocity= ( 300 X ~ ) X 0,15 = 4,71 mjs '

distance moved in 8 s = 8 X 4,71 = 37,68 m

(1)

(2)

Since the pulley accelerates uniformly until its circumferential speedis 4,71 mjs , distance moved by a point on the c ircumference

= ! X 37-t-68 = 18,84 m . /slip of belt r e l a ti

~= 37,6 8 ~ 18,84 = 18,84 m "

Energy lost due to slip = (T - T 2 ) X distance slipped

= 131 ,4 X 18,84 = 2470 J

5. Two parallel horizontal shafts, whose centre lines are ' 4,8 m apart,one being vertically above the other, are connected by an open belt dr ive . Thepulley on the upper shaft is 1,05 m diameter, that on the lower shaft 1,5 mdiameter. T he belt is 150 mm wide and the initial tension in i t whe n stationaryand when no torque is being transmitted is 3 kN. The belt has a mass of1, 5 kg m length ; the gravitational force on it ma y be neglected but centrifugalforce must be taken into account. The material of he belt ma y be assumed to


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184 MECHANICS OF MACHINES

obey H ooke's Law, and the free leng ths of the belt between pulleys may beassumed to be straight . Th e coefficient of riction between the belt and eitherpulley is 0,3. Calculate

(a) the pressure in N j m 2 between the belt and the upper pulley whenthe belt and pulleys are stationary and no torque is being tran smitted ;

(b) the tension in the belt and the pressure between the belt and the upperpulley if he upper shaft rotates at 400 rev / mi n and there is no re sisting torqu eon th e lower shaft, hence no power being transmitted ;

(c) the greatest tension in the belt if he upper shaft rotates at 400 rev j mi nand the maximum possible power is being transmitted to the lower shaft .

(U. Lond.)

(a) L et the pre ssure on an element s ub-~

ending an angle d() a t the centre, Fig. J O o O N T T 300000N10.7, be p N/ m 2 Then, re solving force s pradially,

d()p X 0,525 d() X 0,15 = 2 X 3000 2 d6

:. p = 38 100 N/ m 2 Fw . 10.7

(b) T 0 = mv 2

(2n )21,5 X60

X 400 X 0,525 = 725 N

Th e total length of the belt remains constant, and since t he ma t er ialobeys H ooke' s Law, the ten sion remains constant at 3000 N. Part ofthis force i s now due to centrifu gal tension, how ever, an d the react ionbetw een the belt and th e pulley is reduced.

Eff ect ive te nsion = 3000 - 725 = 2275 N

p = 2275 X 38 100 = 28 900 Nj 1n 2. . 3000

0 0,75 - 0,525 0 0468 . E I 3(c}cos- = = , as m xam p c2 4,8

Also

: . 0 = 174 38' = 3,05 rad

T1 + T 2 = 2T 0 = 6000 N!_1 - 725 = eo 3 x s,o6 = 2,5T 2 - 725

Th erefore, from equations (1) and (2), T 1 = 3970 N

(1)

(2)

BELT DRIVES AND BA ND BRAKES 185

6. F ig. 10.8 shows a belt drive fitted with a gravity idler. The driverrotates anticlockwise at 360 rev /min and the coefficient of ricti on between beltand pulley is 0,3. Determine the initial belt tension and the power t:rans-mit ted (neglect an y sag in the belt). (U. Lond.)

Fla. 10.8

Iq

FIG. 10.9

Let P = norm al reaction at 0 between idler and lever, Fig. 10 .9R = resultant of initial belt tensions, T 0 , at idler

= y' 2 T 0Taking moments abo ut lever pivot,

P X 300 = 180 X 400:. P = 240 N

:. R = 240 sec 15 = 249 N

249T0 = - = 176Ny' 2 --

Angl e of lap on driving pull ey :J ,2l0

= 3,66 radT___! = eO,S x 3,66 = 3,0Tz

Due to the ac tion of the idler, the slack side tension remains constantat 176 N.

UM E - 0

:. T1 = 176 X 3 = 528 Npower= (T 1 - T 2)v

2n= (528- 176) X

60X 360 X 0,15 = 1990 W


6/8


7/8

188 MECHANICS OF MACHINES

Taking moments about D, Fig . 10.14,

mg X 750+ 2T 2 X 3 0 0 = 2T 1 X 300t .e . mg = t (TI - T2)

509 m = . ! x -

6 9,81= 41,5 kg

T2 T 2 T 1 T1

mg

Fro. 10.14

(b) When belt is about to slip,T__..! = e0,2" = 1,875T2

T 1 = 1100 N

1100T 2 = -- = 586,5 N1,875 m = ~ (1100 - 586,5) = 41 9 k. . 5 9,81 _ , _ g

9. For the simple brake shown in Fig. 10.15, find a relationship between

~ 3 7 5 m m ~I

I

~O O mC ~ - - ; -

P N

Fxo. 10.15

the braking tmque an d the applied orce PN, if the coefficient of fr i ction betweenthe brake drum and brake bloclc is 0,35.

What is the braking torque i f P= 500 N '! Also find the magnitudeand direction of the resultant force ateach of the .,tinge s A and B.

(U. Lond.)

The reaction, R, between the blockand the drum passes through the poin tB, Fig. 10.16, and at the point ofintersection with the drum periphery,Q, it is inclin ed at the friction angle,, to the radius at that point, OQ.

4>= tan- 1,u = tan- 1 0,350= 19 18' =: lq. 2,

BELT DRIVES AND BAND BRAKES

From triangle OQB,300 375

sin() - sin (180 - 19 18')

: . ()= 15 20' -::. \.-E>-'32 ()H

"

450mm

p~ - - - - - ' C

Fro. 10.16

Taking moments about A,P X 825= R X x

. R = 825 P 375 COB()

friction torque , T = Rf t

When P = 500 N,

_ 825P375

. ()- - - - - X am

375 COB()= 825P X ta n 15 20'= 226P N mm

T = 226 x 500 = 113 N m1000

. 825 X 500Resultant force at hinge B, R =0

= 1142 N375 cos 15 20'

Inclination to horizontal, () = 15 20'Horizontal component of reaction at A,

H = R c o s O - P= 1142 cos 15 20' - 500 = 600 N

189


8/8

190 MEOHANIOS OF MACHINES

Vert ic a l component of reaction at A,V = RsinO

= 1142 sin 15 20' = 302 N:. resultant reaction at A, F = y'(H 2 + V2)

= y'(600 2 + 302 2 ) = 671,7 NInclination to horizontal, ex = ta n - 1 ~

= ta.n- 1 302 = 26 41'600

10 . A ship i s dr agged throu gh a lo ck by means of a cap sta n and rope . Thecapstan, which has a diam eter of 500 mm , turns at 30 rev min. The rope makes3 complete turns around the capstan, p. being 0,25, and at the free end of therope a pull of 100 N is app lied. Find (a) the pull on th e ship , (b) t he powerrequired to drive the capstan. (U . Lond . ) (Ans.: 11,1 kN ; 8,65 kW)

11 . A leath er b elt , 125 mm wide an d 6 mm thick, transmits power from apulley 750 mm diameter which runs a t 500 rev m in. The angle of lap is 150an d p. = 0,3. I f he ma ss of 1 m 3 of leather is 1 Mg an d the stress in the belt isno t to exceed 2,75 MN / m 2, find the maximum power which can be tr a nsmitted .(1. Meek. E.) (Ana.: 18,94 kW)

12 . In a belt drive, the angle of lap of the belt on the small pulley is 150 .With a belt speed of20 m /s and a tension in t he tig ht s ide ofthe belt of 1,35 kN ,

the greatest pow er whi ch can be transmitted without slip is 10 kW. Whatincrease of power would b e obtained for th e same belt speed an d maximumtension by using an idler pulley so as to increase the a ngle of lap to 210 ? Takeinto acco un t th e centrifugal effect , the mass of t he belt being 0 ,75 kg/m. (U.Lond.) (Ans.: 2,6 kW)

13. A pull ey is dri ven by a flat belt, t he ang le of lap bein g 120. The belt is100 mm wide by 6 mm thick and has a mass of 1 Mg/m 3 I f p. = 0,3 and thema ximum stress in the belt is not to exceed 1,5 MN/m 2 , find th e greatest powerwhich the belt can transmit and th e correspo nding speed of the belt . (U . Lond.)

(A118. : 6,265 kW ; 22,36 m/s)14. P ower is tr ansmitted betw een two shafts , 4,5 m apart , b y an open wire

rope passing round two pulleys, of 3 m an d 2 m diame ter respectively, t he grooveangle being 40. I f he rope h as a mass of 4 kg / m, and the maximum workingtension i s 20 kN, determine th e maximum power th at the rope can transmit,an d the corres pondin g spee d of the sma ller pulley. p. = 0,2. (U . Lond .)

(Ans.: 446 k W; 390 re v fmin)15. Power is transmitted from an electric motor to a ma chine tool by an open

be lt drive. The effective diam eter of the pulley on the motor shaft is 150 mmwhile that on the ma chine tool is 200 mm with a centre distance of 600 mm.I f h e motor speed is 1440 rev/ min an d the maximum permis sib le belt tens ion is900 N, then the maximum power transmis sible is 6 kW .

It is necessary t hat the power transmis sible be in creased to 6,75 kW , using thesa me pulleys, centre distanc e and motor speed . Th e b elt s treated with a specialpreparation that increases it s coeffi cient of fri ction b y 10 per cent of its existingva lue, and in addition a jockey pulley ma y be fitted. Determine,

(a) th e existing coefficient of friction; (b ) th e new angle of lap. (U. Lond.)(A M . : 0,29; 195)

BELT DRIVES AN D BAND BRAKES 191

16 . A vertical open-belt drive connects two pull eys A an d B, the centres ofwhich are 4 m apart. The belt has a mass of 1,15 kg/m. Pulley A is 1 mdi ameter, ha.s radius of gyrati on of 420 mm, and a mass of 25 kg. Pulley B is 0,5 mdiameter, has radius of gyrat ion of 225 mm, and a mass of 18 kg. Wh en at restthe tension in the bel t is 700 N . Assuming that the belt obeys Hooke's Law,determine the tensions in the two portions of the belt between the pulleys when1,5 kW is being trans mitted, th e speed of A being 180 rev/min. Neglect beltstretch over th e pulleys. Find also the kineti c ene rgy of the belt and pulley sunder these conditions. ( U. Lond.)

(A M . : 779,6 N; 620,4 N ; 530 J ; 1430 J)17. A belt drive consists of a V-belt working on a grooved pulley, with an

angle oflap of 160. Th e cross-sectional area of the belt is 650 mm 8 , th e grooveangle is 30 and p. = 0;1. The density of the belt material is 1 Mgjm 3 and itamaximum safe stress is 8 MN/m 8 of cross-section.

Derive an expression for th e ratio of the te nsions on the two sides of th e driv ewhen the belt is about to slip.

Calculate the power that can b e tr ans mitted at a belt speed of 25 m/ s . (U .Lond . ) (.Ans. : 79 kW)

18. Th e following par t icul ars apply to one pulley of a rope drive b etween twoparallel shafts :Effective diamet er of pull eyMinimum angle of lap .Mass of rope per m run

1,5m180

0,45 kg

Total ang le of grooveMaximum permitted load

per rope . . . 650 NCoefficien t o f friction 0,25

(a) Find the power transmitted per rope at a pulley speed of 200 rev/min, i fcentrifuga l tension may be ne glected.

(b) Find t he pulley speed when centrifu ga l tension a cconnts for half th e permitted load in the rope , and t he power which can be transmitted at that speed.(U . Lond.) (A 118.: 8,9 kW ; 342 rev min ; 7,62 kW)

19. Power is tr a nsmitted from a shaft rotating at 250 .rev / min by 10 ropesrunning in grooves in t he p er iph er y of a wheel of effective diamete r 1,65 m(to th e centre line of the rope). The groove angle is 50, an d the arc of contactround the wheel rim is 180 . The maximum pennissible load in each rope is900 N and it s mass is 0,55 kg / m.

I f he coefficient of friction between th e rope and wheel surface is o,.a, whatpower can be transmitted und er the above conditions? (U. Lond.)

(AM.: 124 kW)20. A rope drive is required to transmit 35 kW at 160 rev/min. The grooved

pulley has a mean di ameter to the rope centre of 1,2 m and the groove angles are45 . Taking p. as 0, 25, and the arc of con ta ct of the ropes as 190, d ete rmin e thenumber of ropes re quir ed i f th e greatest pull in eaoh rope is limited to 700 N .(U. Lond.) (A M . : 6)

21. A sma ll generator is dr iven by means of a V-belt which has a to t al angleof 60 between t he faces of the V. The ang le of lap on the pulley is 120 a.nd themean radius of t he be lt as it passes round th e pulley is 50 mm. I f p. = 0,2 an dthe mass of the belt is 0,45 kg/ m, find t he tens ion in eac h aide of th e belt when750 W is being tr a nsmitted at a pulley speed of 1800 rev/min. (U. Lond.)

(A118.: 180 N; 100;5 N)22. A 4 to 1 speed reduction drive b etween two parallel sha f ts at 2 m cent r es

is provided by means of five parallel V- belt s running on suitable pulleys moun tedon the shaft s. The effective diamet er of the driving pulley is 350 mm a.nd th edriving sh af t rotates at 740 rev min. Th e included an gle of each pulle y groove

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