Belts, Ropes and Chains, Study notes of Mechanical Engineering

Download Belts, Ropes and Chains and more Study notes Mechanical Engineering in PDF only on Docsity! BELTS, ROPES AND CHAINS E Introduction Usually, power is transmitted from one shaft to another by means of belts, ropes, chains and gears, the salient features of which are as follows: 2. Belts, ropes and chains are used where the distance between the shafts is large. For small distances, gears are preferred. . Belts, ropes and chains are flexible type of connectors, i.e., they are bent easily. 3. The flexibility of belts and ropes is due to the property of their materials whereas chains have a number of small rigid elements having relative motion between the two elements. 4. Belts and ropes transmit power dus to friction between them and the pulleys. If the power transmitted exceeds the force of friction, the belt or rope slips over the pulley. 5. Belts and ropes are strained during motion as tensions are developed in them. 6. Owing to slipping and straining action, belts and ropes are not positive type of drives, i.e., their velocity ratios are not constant. On the other hand, chains and gears have constant velocity ratios. This chapter deals with the power transmission by belts, ropes and chains. Power transmission by gears will be dealt in the next chapter. The belts used may be flat or of V shape. The selection of belt drive depends on the speeds of the velocity ratio, power to be transmitted, space available and the service conditions. A flat belt is used for light and moderate power transmission whereas for moderate to huge power transmission, more than one V belt or rope on pulleys with a number of grooves is used. Eg 9.1 BELT AND ROPE DRIVES To transmit power from one shaft to another, pulleys are mounted on the two shafts. The pulleys are then connected by an endless belt or rope passing over the pulleys. The connecting belt or rope is kept in tension so that motion of one pulley is transferred to the other without slip. The speed of the driven shaft can be varied by varying the diameters of the two pulleys. For an unstreched belt mounted on the pulleys, the outer and the inner faces become engaged Effective in tension and compression respectively (Fig. radius Neutral 9.1). In between there is a neutral section which section has no tension or compression. Usually, this is considered at half the thickness of the belt. The | Fis. 91 j Thickness Driving pulley Belts, Ropes and Chains 297 effective radius of rotation of a pulley is obtained by adding half the belt thickness to the radius of the pulley, Belt Drive A belt may be of rectangular section, known as a flat belt [Fig. 9.2(a)] or of trapezoidal section, known as a 7 belt [Fig. 9.2(b)]. In case of a flat belt, the tim of the pulley is slightly crowned which helps to keep the belt running centrally on the pulley rim. The groove on the rim of the pulley of a V-belt drive is made deeper to take the advantage of the wedge action. @ The belt does not touch the bottom of the groove. Owing to wedging action, V-belts Fig. 9.2 need little adjustment and transmit more power, without slip, as compared to flat belts. Also, a multiple V-belt system, using more than one belt in the two pulleys, can be used to increase the power transmitting capacity. Generally, these are more suitable for shorter centre distances. Some advantages of V-belts are * Positive drive as slip between belt and pulley is negligible No joint troubles as V-belts are made endless Operation is smooth and quite High velocity ratio up to 10 can be obtained Due to wedging action in the grooves, limiting ratio of tensions is higher and thus, more power transmission * Multiple V-belt drive increases the power transmission manifold * May be operated in either direction with tight side at the top or bottom « Can be easily installed and removed. Disadvantages of V-belts are « Cannot be used for large centre distances « Construction of pulleys is not simple * Notas durable as flat belts « Costlier as compared to flat belts. Rope Drive For power transmission by ropes, grooved pulleys are used [Fig, 9.2(c)]. The rope is gripped on its sides as it bends down in the groove reducing the chances of slipping, Pulleys with several grooves can also be employed to increase the capacity of power transmission [Fig, 9.2(d)]. These may be connected in either of the two ways: 1, Using a continuous rope passing from one pulley to the other and back again to the same pulley in the next groove, and so on, 2. Using one rope for each pair of grooves, The advantage of using continuous rope is that the tension in it is uniformly distributed. However, in case of belt failure, the whole drive is put out of action. Using one rope for each groove poses difficulty in tightening the ropes to the same extent but with the advantage that the system can continue its operation even if'a rope fails, The repair can be undertaken when it is convenient. Rope drives are, usually, preferred for long centre distances between the shafts, ropes being cheaper as compared to belts, These days, however, long distances are avoided and thus, the use of ropes has been limited. 300 Theory of Machines This is also the speed of the belt on the driven pulley, Peripheral speed of driven pulley -[o(7 2) (te) As Sis the total percentage slip, D,+t)|/f100— Peripheral speed of driven shaft = [«( i ie 0-8 ) 2 100 (ea Nae) Te (100- $,)00-S,) 100-5 or 100 x 100 100 or (100-5, (100—§,) = 100 (100-5) or 10 000 — 100 S,— 100 5; + S, Sy = 10 000-100 S or 100 S = 100 S,+ 100 S)- 5, Sy or S=5,+5,-0.015, 5 (9.2) Effect of slip is to reduce the velocity ratio, vee. -( PHS) (93) N, Dy+t 100 Also, it is to be remembered that slip will first occur on the pulley with smaller angle of lap, i.e., on the smaller pulley, Example9.1 A shaft runs at 80 rpm and Dy= 317.7 mm drives another shaft at 150 v. rpm through belt drive. The Gi) = D +1 (10-5) diameter of the driving pulley N, Dy, +#\ 100 is 600 mm. Determine the 150 (600+5\/100—4 diameter of the driven pulley in the following 30 "Ua es | 100 ca ()) Neglecting belt thickness D,= 304.8 mm (ii) Taking belt thickness as 5 mm (iii) Assuming for case (ii) a total slip of 4% (wy “2-2 { 100- 5) (iv) Assuming for case (ii) a slip of 2% on each N, Dy +e\ 100 pulley where 5 = $, +S, -0.015,S, Solution N,=80rpm —D, = 600 mm =2+2-001%2%2 Ny = 150 rpm = 3.96 ( Nz Pr g, 150 _ 600 150 ( 600+5)/100-3.96 N, Dy 80D, == —_— 80 | D,+5 100 or Dy = 320mm No tt 150 _ 600+5 Dy = 304.9 mm ay ‘Ny Da +t or 80 D,+5 Belts, Ropes and Chains 301] 9.6 MATERIAL FOR BELTS AND ROPES Choice of materials for the belts and ropes is influenced by climate or environmental conditions along with the service requirements, The common materials are as given below: 1, Flat Belts Usual materials for flat belts are leather, canvas, cotton and rubber. These belts are used to connect shafts up to 8-10 m apart with speeds as high as 22 m/s. Leather belts are made from 1.2 to 1.5 m long strips. The thickness of a belt may be increased by cementing the strips together. The belts are specified by the number of layers, i.e., single, double or triple ply. The leather belts are cleaned and dressed periodically with suitable oils to keep them soft and flexible. Fabric belts are made by folding cotton or canvas layers to three or more layers and stitching together. The belts are made waterproof by impregnating with Linseed oil. These are mostly used in belt conveyors and farm machinery. Rubber belts are very flexible and are destroyed quickly on coming in contact with heat, grease or oil. Usually, these are made endless. Rubber belts are used in paper and saw mills as these can withstand moisture. 2. V-Belts These are made of rubber impregnated fabric with the angle of V between 30 to 40 degrees. These are used to connect shafts up to 4 m apart, Speed ratios can be up to 7 to | and belt speeds up to 24 m/s, 3. Ropes The materials for ropes are cotton, hemp, manila or wire, Ropes may be used to connect shafts up to 30. m apart with operating speed less than 3 m/s, Hemp and manila fibres are rough and thus, the ropes made from such materials are not very flexible, Manila ropes are stronger as compared to hemp ropes. Generally, the rope fibres are lubricated with tar, tallow or graphite to prevent sliding of fibres when the ropes are bent over the pulleys. The cotton ropes are soft and smooth and do not require lubrication. These are not as strong and durable as manila ropes. Wire ropes are used when the power transmitted is large over long distances, may be up to 150 m such as cranes, conveyors, elevators, etc, Wire ropes are lighter in weight, have silent operation, do not fail suddenly, more reliable and durable, less costly and can withstand shock loads, 9.7, CROWING OF PULLEYS As mentioned is Section 9.2, the rim of the pulley of a flat- > belt drive is slightly crowned to prevent the slipping off the belt from the pulley. The crowing can be in the form of 1 conical surface or a convex surface. — — Assume that somehow a belt comes over the conical portion of the pulley and takes the position as shown | in Fig. 9.5(a), i.e,, its centre line remains in a plane, the belt will touch the rim surface at its one edge only. This is impractical, Owing to the pull, the belt always tends to stick to the rim surface. The belt also has a lateral stiffness, Thus, @ (b) a belt has to bend in the way shown in Fig, 9.5(b) to be on [ Fis. 95 | the conical surface of the pulley. 4 302 Theory of Machines Let the belt travel in the direction of the arrow. As the belt touches the cone, the point a on it tends to adhere to the cone surface due to pull on the belt, This means as the pulley will take a quarter turn, the point a on the belt will be carried to & which is towards the mid-plane of the pulley than that previously occupied by the edge of the belt. But again, the belt cannot be stable on the pulley in the upright position and has to bend to stick to the cone surface, i.e,, it will occupy the position shown by dotted lines. Thus, if a pulley is made up of two equal cones or of convex surface, the belt will tend to climb on the slopes and will thus, run with its centre line on the mid-plane of the pulley. The amount of crowing is usually 1/96 of the pulley face width, 9.8 TYPES OF PULLEYS Driver 1. Idler Pulleys With constant use, the belt is permanently stretched a little in length. This reduces the initial tension in the belt leading to lower power transmission capacity. However, the tension in the belt can be restored to the original value by using an arrangement shown weights in Fig. 9.6(a). {a) A bell-crank lever, hinged on the axis of the smaller pulley, supports adjustable weights on its one arm and the axis of a pulley on the other. The pulley is free to rotate on its axis and is known as idler pulley. Owing to weights on one arm of the lever, the pulley Driven Driver exerts pressure on the belt increasing the (b} Driver tension and the angle of contact. Thus, life pulley of the belt is increased and power capacity is [ Fig. 9.6 | restored to the previous value. The pressure force on the belt can be varied by changing the weights on the arm of the lever. Motion of one shaft can be transmitted to two or more than two shafis by using a number of idler pulleys. This has been illustrated in Fig. 9.6(b). 2. Intermediate Pulleys When it is required Delving Intermediate or to have — large pulley countershaft pulley velocity _ratios, ordinarily, the size of the larger pulley will be quite big, However, by using an intermediate (or countershafi) [FB 9.7 yf A compound belt drive Driven Belts, Ropes and Chains 305 As CD is tangent to two circles, AC and BD both are perpendicular to CD or AN. Now, 48 L BK and AN 1 BD. ZDBK = ZNAB = B Similarly, as 84 L AJ, NA LAC ZCAI = ZNAB = B L, =2 [Are GC + CD + are Dif] = (Eaoo(S =-[(2-1)r-cooae{Eoa}a| = a(R +r) +2B(R-r)+2C cos B (9.4) This relation gives the exact length of belt required for an open belt drive, In this relation, . (Ror = sin B ( c } (9.5) An approximate relation for the length of belt can also be found in terms of R, rand C eliminating B, if B is small, ie., if the difference in radii of the two pulleys is small and the centre distance is large. For small angle of 8, sin B= B R- “ B= and cos B = J1—sin? B = (sin? py? L, = {1 - sin’ Brion } [By binomial theorem] af{y—-1g2\2)-2/85 cop-{r te ar-t( 2 L, mnie ea[ BP )en-nere|i- 28") (R-ry Cc 2C (R=ry 2 ¢ =H(R+rj+2 +2¢0— (R=rY _(R-1Y =aA(R+ry+2——— +2C Cc Cc =aA(Rtry+ +2C (9.6) (R=ry C 306 Theory of Machines 2. Crossed-Belt As before, let 4 and B be the pulley centres and CD and EF, the common tangents (crossed) to the two pulley circles (Fig. 9.12). ye Draw AW parallel to CD meeting BD produced @ at N so that ZBAN = B We have, ZCAJ = ZDBK = B 5 Let £, = length of belt for crossed-belt drive Sy | K Then N L, = 2[Are GC + CD + Are Dif] Fig, 9.12 =2 (E+e)rvave(Zo a} 2 2 =2 (E+p}r+ceosp-(Z+8)r 2 2 = (n+ 28) (R +7) +2C cos B (9.7) This is the exact length of a crossed-belt drive where p= sin (22) c (9.8) For small angle of 8, sin B= B poor “ SG 2 and cos =(1-3 6°} 1 (22) 2 L,= | (R+r)42€ 1-3(*) Cc 24 C0 2 2 =n(Re rt 2 Fe yoo Re C 2 =n(Rery+ FED 490 (9.9) This is an approximate relation for the length in terms of R, r and C. It can be noted that the length of belt depends only on the sum of the pulley radii and the centre distance in case of crossed-belt drive whereas it depends on the sum as well as the difference of the pulley radii apart from the centre distance in case of open-belt drive. Example 9.2 Two parallel shafis, connected desired to alter the direction of rotation of the by a crossed belt, are provided driven shaft. with pulleys 480 mm and 640 mm in diameters. The distance between the centre C=3m lines of the shafis is 3 m. Find by how much r= 240 mm the length of the helt should be changed if it is Solution R= 320 mm For cross belt psi [ = 10°45’ or 0.1878 1ad cos B = 0.9825 EL, = (1+ 2B) (R +r) + 2€ cos B = (w+ 2X 0.1878) (0.32 + 0.24) + 2 x 3 x 0.9825 =") : (ea = sin” | Belts, Ropes and Chains 307 | ) = sin” 0.1867 = 7.865 m For open belt R- 0.32 —0.24 B=sin™ (42) = sin™ (°22592*) = sin” 0.0267 Cc 3 = 1932’ As the angle is very small, the approximate relation can be used. R-ry 1, =m(R+r+ f= 42¢ Cc (0.32-0,24y° =m (0.324 0.24)+ — 2 x3 = 7.761 m The length of the belt should be reduced by L, - b= 7.865 — 7.761 = 0.104 m or 104 mm 9.11 CONE (STEPPED) PULLEYS Many times, it is required to run the driven shaft at different speeds whereas the driving shaft runs at constant speed which is the speed of the motor. This is facilitated by using a pair of cone or stepped pulleys (Fig. 9.13), A cone pulley has different sets of pulley radii to give varying speeds of the driven shaft, The radii of different steps are so chosen that the same belt can be used at different sets of the cone pulleys, Let # = speed of the driving shaft (constant) speed of the driven shaft when the belt is on #th step r, = radius of the nth step of the driving pulley R,, = radius of the nth step of the driven pulley The subscript # denotes 1, 2, 3, ... #. The ratio of speeds of driving to driven shaft is inversely proportional to the ratio of their pulley radii, ie., Ny ow : 2 OR @ Thus, to get speed N, of the driven shaft from the first pair of steps of the cone pulleys, dimensions of r, and R, can be chosen convenient to the design. For the second pair of steps, 1} 2/3 ae 5 Driving [| pulley Lt (speed n) Driving pulley {speed Ag) | Fig. 9.13 | 310 Theory of Machines 9.12 RATIO OF FRICTION TENSIONS 1. Flat Belt Let T, = tension on tight side T, = tension on slack side 6 = angle of lap or contact of the belt over the pulley H = coefficient of friction between the belt and the pulley Consider a short length of belt subtending an angle 66 at the centre of the pulley (Fig. 9.14), Let R = normal (radial) reaction between the element length of belt and the pulley T = tension on slack side of the element 81 = increase in tension on tight side than that on slack 72 qT side | Fig. 9.14] T +81 = tension on tight side of the element Tensions T and (T + ST) act in directions perpendicular to the radii drawn at the ends of the elements, The friction force R will act tangentially to the pulley rim resisting the slipping of the elementary belt on the pulley. Resolving the forces in the tangential direction, ur +Tcos er + arycos? =0 2 2 As 86s small, cos—— = 1 2 MR+T-T-6F=0 or ST=pR @ Resolving the forces in the radial direction, R-Tsin (r+ 6rysin B= . 60 60 A i W Sa s 60 is small, sin 60 50 STdO R-T 7“ =0 Thus, > > > Neglecting product of two small quantities, R=7 80 (ii) Inserting this value of R in (i), ér OT = 2.760 or 7 Hee Belts, Ropes and Chains 311 Integrating between proper limits, Rar Q Jude [role T, log, + = ne or Be % u i) or Tr =e (9.13) It is to be noted that the above relation is valid only when the belt is on the point of slipping on the pulleys. 2. V-Belt or Rope In case of a V-belt or rope, there are two normal reactions as shown in Fig, 9,15 so that the radial reaction is equal to 2R sin a. aR sin a 2R sin & Thus, total frictional force = 2(uR) = 2uR. Resolving the forces tangentially, 2uRt+ eos (+ 8r)e05 =0 For small angle of 58, 87 = 2uR (iii) Resolving the forces radially, 2Wsina—Tsin (F487 sin = 0 312 Theory of Machines As &@is small, sin 28, 98 2 2 2Rsina—7 2 _ 7 88 _ 2 2 760 R= . or 2sin a iv) From (iii) and (iv), oF =2p 750 2sina@ oT pée or a. Tr sina: Integrating between proper limits, jeje b T sina 8 log, += = or tT sing qT or FLL ueisina (9.14) h The expression is similar to that for a flat-belt drive except that is replaced by u1/sin@, i.e., the coefficient of friction is increased by I/sind, Thus, the ratio 7;/T, is far greater in case of V-belts and ropes for the same angle of lap 0 and coefficient of friction js. Again, it is to be noted that the above expression is derived on the assumption that the belt is on the point of slipping. 9.13 POWER TRANSMITTED | Let 7; = tension on the tight side Ty = tension on the slack side v = linear velocity of the belt P = power transmitted Then, P = Net force x Distance moved/second =(T,-T,) xv (9.15) This relation gives the power transmitted irrespective of the fact whether the belt is on the point of slipping or not. If itis, the relationship between 7, and T, for a flat belt is given by T,/T, = e*®. If it is not, no particular relation is available to calculate T, and 7. Belts, Ropes and Chains 315 | 9.14 CENTRIFUGAL EFFECT ON BELTS While in motion, as a belt passes over a pulley, the centrifugal Fe effect due to its own weight tends to lift the belt from the pulley. Owing to symmetry, the centrifugal force produces equal tensions on the two sides of the belt, i,e., on the tight side as well as on the $al2 ar 3 slack side. Consider a short element of belt (Fig, 9,16). a V7 + Let m = mass per unit length of belt, = . T, = centrifugal tension on tight and slack sides of element F,, = centrifugal force on the element r = radius of the pulley v = velocity of the belt | Fig. 9.16] 650 = angle of lap of the element over the pulley F,, = mass of element x acceleration = (length of element x mass per unit length) x acceleration 2 = (r60 x m)x 2 ; = nv50 @ Also, 7, = 2, sin — As d@is small, . 60 50 sin = = 2 2 p= 2 2 = 7,60 Gi) From (i) and (ii), £,50= nm?50 or T=? (9.16) Thus, centrifugal tension is independent of the tight and slack side tensions and depends only on the velocity of the belt over the pulley. Also, Centrifugal tension Te a Centrifugal stress in the belt = ——=———___ = area of cross section of belt Total tension on tight side = friction tension + centrifugal tension 316 Theory of Machines THT +k, (17) Total tension of slack side = T, + T, It can be shown that the power transmitted is reduced if centrifugal effect is considered for a given value of the total tight side tension T. {a) Centrifugal Tension Considered Friction tension on tight side = 7-7, = 7, Let 7; be the friction tension on the slack side. 7, Then — = e# =k, a constant {b) Centrifugal Tension Neglected Friction tension on tight side = T Let T,’ be the friction tension on slack side. qh ty Power, P = (Z| — Ty = (7-Z}y = r(1-t)y k k As T, is lesser than 7, power transmitted is less when centrifugal force is taken into account, =e =k, or 9.15 MAXIMUM POWER TRANSMITTED BY A BELT ¥§ If it is desired that a belt transmits maximum possible power, two conditions must be fulfilled simultaneously, 1, Larger tension must reach the maximum permissible value for the belt. 2. The belt should be on the point of slipping, i.e., maximum frictional force is developed in the belt. Now, hy P=(-hw=Th- > q 1 where & =1—— = constant ¢ or P=(T-Tjkv = kTv — ky? y = kT — kr? The maximum tension 7 in the belt should not exceed the permissible limit. Hence, treating T as constant and differentiating the power with respect to v and equating the same equal to zero, we get Belts, Ropes and Chains 317 or T=3mv? = 37, T= © or (9.18) Therefore, for maximum power transmission, centrifugal tension in the belt must be equal to one-third of the maximum allowable belt tension and the belt should be on the point of slipping. Also, T 2 q=T-T.=T-—=<7T 303 and An open-belt drive is required to transmit 10 kW of power from a motor running at 600 rpm. Diameter of the driving pulley is 250 mm. The speed of the driven pulley is 220 rpm. The belt is 12 mm thick and has a mass density of 0.001 g/mm?. Safe stress in the beit is not to exceed 2.5 /mm’. The two shafts are 1.25 m apart. The coefficient of friction is 0.25. Determine the width of the belt. Example 9.8 Solution Speed of the driving pulley, N= 600 rpm. Speed of the driven pulley, N,= 220 rpm Thus, smaller pulley is the driver and d@= 250 mm P=10kW; ¢=12mm; p= 0.001 g/mm?= 1000 kg/m?; r =125 mm, C= 1.25 m; N= 220 rpm ; “= 0.25; 6,= 2.5 Némm? = 2.5 x 10° N/m? To calculate the width of the belt, we need to know the maximum tension in the belt which is the sum of the tight side tension and the centrifugal tension, ie, T=T,+T, Calculation of T, P=(0,-Ty hy o +) 2aN +4) where v= @| r++ ]= r+ where 3)= 60 3 (9.19) _ 2x 600 60 10 000 = (F, - 1) x 8.23 or 7, -T,= 1215 (125+ 2) = 8230 mm/s or 8.23 m/s i, Also 2 =e? 2 R- where 9= 1 = 28=n-2sin-[ ra 1) . ~{ 125 x 600/220—125 or =@=#-2sin (2exenetzzo—us 1250 or = #-19.9°= 4-0,347 = 2.79 fF Ty From (i) and (ii), 2.01 T,-T, = 1215 = PT 272.01 or =201 Gi T, = 1203 N T,=2418N Calculation of T, T.=m? = mass per unit length x v” = volume per unit length x density x ¥? x-sectional area x length x density) x v? = (width x thickness x length x density) x v = bx 0.012 x 1 x 1000 x (8.23)? = (812.85) N T=T,+T,=0,x(bxh 2418 + 812.85 =2.5 x 10°x bx 0.012 29 1876 = 2418 6=0.0828m or 82.8mm (bin m) 320 Theory of Machines 9.16 INITIAL TENSION When a belt is first fitted to a pair of pulleys, an initial tension T, is given to the belt when the system is stationary, When transmitting power, the tension on the tight side increases to T; and that on slack side decreases to T;, If it is assumed that the material of the belt is perfectly elastic, i.e., the strain in the belt is proportional to stress in it and the total length of the belt remains unchanged, the tension on the tight side will increase by the same amount as the tension on the slack side decreases, If this change in the tension is ST then tension on tight side, 7; =7,+ 6r tension on slack side, 7; = _ or 7,-5th 2 = mean of the tight and the slack side tensions. (9.20) Initial Tension with Centrifugal Tension Total tension on tight side = T, + T, Total tension on slack side = T, + F, 7 ttt (+h) ‘a 3 +B yp or T)+%-2%-T) Tone al ak Let h e Therefore, AT, +T, = 2(7,- T) 7, = eT) k+l Q(T, -T. and TNA ae) (i) 2 -T) -T) MoE k+l _ 2K = 7) k+l Power transmitted, P=(T,-T)v _ UR=WT =F) k= 1NT, — mv?) k+l a k+l _ k= 1(Fv— mv?) - k+l Belts, Ropes and Chains 321 To find the condition for maximum power transmission, differentiating this expression with respect to v and equating the same to zero, i.e., When the belt drive is started, v = Q and Thus, 7, = 0, 2kT, . na oo From (i) and (ii), it is evident that the maximum tension in the belt is more while starting the drive. Example 9.13 The following data relate to a ; T+T 1392.8+148.8 rope drive: Initial tension = nn =770.8N Power transmitted = 20kW Diameter of pulley = 480 mm Speed 80 rpm Angle of lap on smaller pulley = 160° Number of ropes Mass of rope/m length = 48 G’ kg Limiting working tension = 132G? kN Coefficient of friction =03 Angle of groove = 44 If G is the girth of rope in m, determine the initial tension and the diameter of each rope. Solution Power transmitted /rope = 20 000/8 = 2500 W Velocity of RDN #x048x80 i rope = ———- = ‘elocity of rope 0 0 =2.01 m/s Now, P=(T,-T,)v or 2500 = (T, - 72) x 2.01 or (7, — Tz) = 1244 N a 10x21 03x Also, 2. = ghPisina _ f, From (i) and (ii), 9.359 Ty — Ty = 1244 T, =148.8N T, = 148.8 x 9.359 = 1392.8 N 180. sin 22°= 9.359 (ii) Now, Working tension = Tension on tight side + Centrifugal tension =T,+m* 132 000 G? = 1392.8 + 48 G?. (2.01 131 806 G? = 1392.8 @ = 0.01056 or G = 0.1028 Now girth (circumference) of rope = nd = 0.1028 or 3=d=0.327m Example 9.14 2.5 kW of power is transmitted by an open-belt drive. The linear velocity of the helt is 2.5 m/s. The angle of lap on the smaller pulley is 165°. The coefficient of friction is 0.3. Determine the effect on power transmission in the following cases: (D) Initial tension in the belt is increased by 8% (ii) Initial tension in the belt is decreased by 8% (iii) Angle of lap is increased by 8% by the use of an idler pulley, for the same speed and the tension on the tight side (iv) Coefficient of friction is increased hy 8% by suitable dressing to the friction surface of the belt 322 Theory of Machines Solution P=2.5kW p=03 @= 165° v=2,5 m/s P=(T,-Toy 2500 = (T;-T>) x 2.5 T,-T= 1000 N A gO a Qh 16S _ 9 397 q, or T, =2.37T, 2.37T)—T, = 1000 or T) = 729.9N T, = 729.9 X 2.37 = 1729.9N Initial tension, _T+% _17293+729.9 2 2 (i) When initial tension is increased by 8% Tg = 1229.9 x 1.08 = 1328.3 N T+T; or Ts 1328.3 or T+ Ty = 2656.6 =1229.9N qh As gt and @ remain unchanged, e“? or qh. Lis same. 2 237 Tyt Ty = 2556.6 > = 788.3 N T, = 1868,3N P=(T, Tov = (1868.3 -788,3) x 2.5 = 2700 W or 2.7 kW +. increase in power = a = 0,08 or 8% Gi) When initial tension is decreased by 8% Py = 1229.9 x (1 — 0.08) = 1131.5 T+T; or TST Ls or T+ Ty = 2263 3.37T, = 2263 T,=6715N T,=1591.5N P= (1591.5 — 671.5) x 2.5 = 2300 W or 2.3 kW 2.5-2.3 . Decrease in power = = 0.08 or 8% re) ay A (ii) y T, is the same as before whereas @ increases by 8% 165x108 x. 03 1799 wo 22.54 qr, T= 680.5N P= (1729.9 — 680.5) x 2.5 = 2624 W or 2.624 kW *. Increase in power _ 2.624-2.5 — 2S I65x@ O.ax1.08% of =e 180” = 2,54 = 0,0496 or 4.96% (iv) or 7, =2.54 Ty 229.9 x 2 = 2459.8 ) = 694,9 N T, = 694,9 x 2.54 = 1764.9 N P= (1764.9 — 694.9) x 2.5 = 2675 W or 2.675 kW Increase in power _ 2.675 -2.5 _ 0.07 or 7% 25 In a belt drive, the mass of the belt is 1 kg/m length and its speed is 6 m/s. The drive transmits 9.6 kW of power. Determine the initial tension in the belt and the strength of the belt, The coefficient of friction is 0.25 and the angle of lap is 220°. Solution P=(T,-T,v or 9600 = (T,-7,) x6 Example 9.15 or F.-T,= 1600 (i) 2a 2 or T,=2.61T, Gi) From (i) and (ii), 2.61 T;- T, = 1600 T,= 994N T, = 2594 N Centrifugal tension = mv* = 1 x 127 = 144N A+B ap Initial tension 7, = = aa 144 = 1938. N Solution Larger pulley is the driving pulley, N, = 320 rpm <. D= 800 rpm; d= 250 rpm Ny D If creep is neglected, ™ = a D 300 or NQ=™ X77 = 320x255 = 1024 rpm 9.18 CHAINS A chain is regarded in between the gear drive and the belt drive. Like gears, chains are made of metal and, therefore, occupy lesser space and give constant velocity ratios. Like belts, they are used for longer centre distances, Advantages « Constant velocity ratio due to no slip and thus, it is a positive drive. * No effect of overloads on the velocity ratio. * Oil or grease on surfaces does not affect the velocity ratio. « Chains occupy less space as these are made of metals. * Lesser loads are put on the shafts. * High transmission efficiency due to no slip. « Through one chain only, motion can be transmitted to several shafts. Disadvantages * Itis heavier as compared to the belt, * There is a gradual stretching and increase in length of chains, From time to time some of its links have to be removed, * Lubrication of its parts is required. * Chains are costlier as compared to belts, The wheels over which chains are run, corresponding to the pulleys of a belt drive, Belts, Ropes and Chains 325 N,_ b| E+ fo, N, do Ma Bee, Np = 320 8%, 110+ ¥0.32 250 | 110+ V0.8 = 320x 3.2 x 0,997 = 1021 rpm Speed lost = 1024 - 1021 = 3 rpm A chain drive of a machine 326 Theory of Machines are known as sprockets. The surfaces of sprockets conform to the type of chain used. Usually, a sprocket has projected teeth that fit into the recesses in the chain, Thus, the chain passes round the sprockets asa series of chordal links (Fig. 9.17), The distance between roller centres of two adjacent links is known as the pitch (p) of the chain, A circle through the roller centres of a wrapped chain round a sprocket is called the pitch circle and its diameter as pitch circle diameter, Observe that a chain is wrapped round the sprocket in the form of a pitch polygon and not in the form of, a pitch circle, Let 7= number of teeth on a sprocket, @= angle subtended by chord of a link at the centre of sprocket r= radius of the pitch circle Then p= 2rsin® = 2rsin (20 )=2rsin!? Pp Pp 180° r=——_T5 =F cosec or asin 8° 2 Tr (9.21) 919 CHAIN LENGTH 4a For a given pair of sprockets at a fixed distance apart, the length of the chain may be calculated in the same way as for an open belt. Since the pitch line of a sprocket is a polygon, Eq. 9.6 will give a length slightly more than the actual length. Let R and r be the radii of the pitch circles of the two sprockets having T and ¢ teeth respectively. Also, let Z = length of the chain C = centre distance between sprockets = kp p = pitch of chain From Eq. (9.6), 2 L =n(R+r)+ Fa 5 2¢ The first term in the equation is half the sum of the circumference of the pitch circles, In case of a chain it will be (pT + pa/2. Replacing & and r in the second term by 180° 180° R == cosec—— and r = 2 cosec 2 Tr 2 P 130° p sory = cosec —— — —cosec T 2 prem 3 2 L +2kp kp 130° 180°” = T+t cosec F —cosec ? pests FS ig (9.22) 2 4k Belts, Ropes and Chains 327 | Note that the terms in the square bracket must be an integral number of pitch lengths, In case it is a fraction, it must be rounded off to the next integral number, 9.20 ANGULAR SPEED RATIO The chain is wrapped round the sprocket in the form of a pitch polygon and not as a pitch circle, From Fig, 9.18, it may be observed that the axial line of the chain vibrates between two positions shown by full and dotted lines, Even if the sprocket rotates at an uniform angular velocity «o, the linear velocity of the chain will be varying from a maximum @4C [ Fig. 948] toa minimum @.4D. Thus, the magnitude of the speed variation is the ratio of the distances AC to AD, The variation in the chain speed also causes a variation in the angular speed of the driven sprocket, However, by increasing the number of teeth on the sprocket, the magnitude of the variation in speed may be minimized. It can be shown that at any instant, if the line of transmission cuts the line of centres at O, the angular velocities of the two sprockets will be in the inverse ratio of the distances of their centres from O, i.e., ; _ 04 @, OB The variation of @, will be between OA OA @, — and @ —— OB OB Thus, at any instant, the angular velocity of the driven shaft would be changing. Example 9.19 The center-to-centre distance k= Crp = 0.600/0.0418 = 14.342 between the two sprockets of a chain drive is 600 mm. The ° 180°)? chain drive is used to reduce La plrte, (cose p eosee ) the speed from 180 rpm to 90 5 7 +2k rpm on the driving sprocket has 18 teeth and a pitch circle diameter of 480 mm. Determine the (i) number of teeth on the driven sprocket 36418 | (ii) pitch and the length of the chain 2 NT = 0.0418 180° 180°)" Solution (i) S22 cose — cosee = wi Tao Fag 1 or Ty =7, 2b =18x—~ =36 x143 Ny 90 180° 180° = 0.0418 x (27 + 0.569 + 28.684) (i) p= 2rsin —— =2x0.24xsin =2351m = 0.0418 m or 41.8 mm 330 x 10. an. 42. 23. 14. a5. 16. 17. 18. ag. Theory of Machines Exercises . What are different modes of transmitting power from one shaft to another? Compare them. . Discuss the effect of slip of belt on the pulleys on the velocity ratio of a belt drive. . Name the materials of the flat belts, V-belts and ropes. . What do you mean by crowning of pulleys in flat- belt drives? What is its use? . What are different types of pulleys? Explain briefly. . Explain the following: (i) Idler pulleys (ii) Intermediate pulleys ii) Loose and fast pulleys (iv) Guide pulleys . Define and elaborate the law of belting. . Deduce expressions for the exact and approximate lengths of belt in an open-belt drive. . What is meant by cross-belt drive? Find the length of belt in a cross-belt drive. Where do we use cone (stepped) pulleys? Explain the procedure to design them. Derive the relation for ratio of belt tensions in a flat-belt drive. r Derive the relation, e* for a flat-belt drive with usual notations. * Deduce an expression for the ratio of tight and slack side tensions in case of a V-belt drive. What is the effect of centrifugal tension on the tight and slack sides of a belt drive? Show that it is independent of the tight and slack-side tensions and depends only on the velocity of the belt over the pulley. What is the effect of centrifugal tension on the power transmitted? Derive the condition for maximum power transmission by a belt drive considering the effect of centrifugal tension. What is meant by initial tension in a belt drive? What is creep ina belt drive? Amotor shaft drives a main shaft of a workshop by means of a flat belt, the diameters of the pulleys being s500-mm and 800-mm respectively. Another pulley of 600mm diameter onthe main shaft drives acounter-shaft having a750-mm diameter pulley. If the speed of the motor is 1600 rpm, find the speed of the countershaft neglecting the thickness of the 20. 2a, 22. 23. 24. 25. belt and considering a slip of 4% on each drive. (737.3 rpm) Two pulleys on two shafts are connected by a flat belt. The driving pulley is 250 mm in diameter and runs at 150 rpm. The speed of the driven pulley is to be go rpm. The belt is 120 mm wide, 5-mm thick and weighs 1000 kg/m. Assuming a slip of 2% between the belt and each pulley, determine the diameter of the driven pulley. Also, find the total effective slip. (403 mm; 3.96%) The pulleys of two parallel shafts that 8 m apart are 600 mm and 800 mm in diameters and are connected by a crossed belt. It is needed to change the direction of rotation of the driven shaft by adopting the open-belt drive. Calculate the change in length of the belt. {Shorten the belt by 60 mm) Determine the diameters of the cone pulley joined by a crossed belt. The driven shaft is desired to be run at speeds of 60, 90 and 120 rpm while the driving shaft rotates at 160 rpm. The centre distance between the axes of the two shafts is 2.5 m. The smallest pulley diameter can be taken as 450 mm. (150 mm and 400 mm; 198 mm and 352. mm; 236 mm and 314 mm) Design a set of stepped pulleys to drive a machine from a countershaft running at 300 rpm. It is needed to have the following speeds of the driven shaft: 140 rpm, 280 rpm and 220 rpm. The centre distance between the axes of the two shafts is 5 m. The diameter of the smallest pulley is 300 mm. The two shafts rotate in the same direction. (goo mm and 642 mm; 354mm and 590 mm; 400 mm and 545 mm) A countershaft is to be driven at 240 rpm from a driving shaft rotating at 100 rpm by an open-belt drive. The diameter of the driving pulley is 480 mm. The distance between the centre line of shafts is 2m. Find the width of the belt to transmit 3 kW of power if the safe permissible stress in tension is ag N/mm width of the belt. Take pt = 0.3. (434. mm) A casting having a mass of 100 kg is suspended freely from a rope. The rope makes 2 turns round a drum of 300 mm diameter rotating at 24 rpm. The other end of the rope is pulled by a man. Calculate 4d 26. 27. 28. 29. 30. the force required by the man, power to raise the casting and the power supplied by drum run by a prime-mover. Take {f= 0.3. (226 N; 3.698 kW; 3.623 kW) Aleather belttransmits 10 kW froma motor running at 600 rpm by an open-belt drive. The diameter of the driving pulley of the motor is 350 mm, centre distance between the pulleys is 4 m and speed of the driven pulley is 180 rpm. The belt weighs 1100 kg/m3 and the maximum allowable tension in the belt is 2.5 N/mm?. jt = 0.25. Find the width of the belt assuming the thickness to he 10 mm. Neglect the belt thickness to calculate the velocities. (73.8 mm) Two pulleys mounted on two parallel shafts that are 2 m apart are connected by a crossed belt drive. The diameters of the two pulleys are 500 mm and 240 mm. Find the length of the belt and the angle of contact between the belt and each pulley. Also, find the power transmitted if the larger pulley rotates at 180 rpm and the maximum permissible tension in the belt is 900 N. The coefficient of friction between the belt and pulley is 0.28. {5.23 M, 201.49, 2.658 kW) Determine the maximum power that can be transmitted through a flat belt having the following data: X-section of the belt = 300 mm x12 mm Ratio of friction tensions = 2.2 Maximum permissible tension in belt = 2 N/mm? Mass density of the belt material = 0.0011 g/mm? (64.46 kW) A V-belt weighting 2.6 kg/m run has an area of cross-section of 750 mm?. The angle of lap is 165° on the smaller pulley which has a groove angle of 40°. ft = 0.12. The maximum safe stress in the belt is 9.5 Nimm?, What is the power that can be transmitted by the belt at a speed of 20 m/s? (82.485 kW) Aleatherbelttransmits 8 kW of power froma pulley that is 2.1 m in diameter running at 200 rpm. The angle of lap is 160° and the coefficient of friction between belt and pulley is 0.25. The maximum safe working stress in the belt is 2.2 Nfmm*. The thickness of the belt is 8 mm and the density of 31. 32. 33. 34. 35- Belts, Ropes and Chains 331 leather is 0.002 g/mmi?. Find the width of the belt taking centrifugal tension into account. (78.6 mm) A rope drive transmits 40 kW at 220 rpm by using 15 ropes. The angle of lap on the smaller pulley which is 300 mm in diameter is 165°. Coefficient of friction is 0.25 and the angle of groove is 40°. The rope weighs (50 x 10°) G? kg per metre length of rope and the working tension is limited to 0.14 G?N where G is the girth (circumference) of rope in mm. Determine the initial tension and the diameter of each rope. {903.8 N; 34.2 mm) The smaller pulley of a flat belt drive has a radius of 220 mm and rotates at 480 rpm. The angle of lap is 155°. The initial tension in the belt is 1.8 kN and the coefficient of friction between the belt and the pulley is 0.3. Determine the power transmitted by the belt. (a5.3 kW) A rope drive uses ropes weighing 1.6 kg/m length. The diameter of the pulley is 3.2 m and has 12 grooves of 40° angle. The coefficient of friction between the ropes and the groove sides is 0.3 and the angle of contact is 165°. The permissible tension in the ropes is 870 N. Determine the speed of the pulley and the power transmitted. (80.3 rpm, 86.18 kW) Aman wants to lower an engine weighting 380 kg from atrolley to the ground by using a rope which he passes over a fixed horizontal pipe overhead. The man is capable of controlling the motion with a force of 200 N or less on the free end of the rope. Find the minimum number of times the rope must be passed round the pipe if 4 = 0.22. (2.1 turns) A chain drive is used for speed reduction from 240 rpm to a20 rpm. The number of teeth on the driving sprocket is 22. The centre to centre distance between two sprockets is 540 mm and the pitch circle diameter of the driven sprocket is 480 mm. Determine the number of teeth on the driven sprocket, pitch and the length of the chain. (48, 32.4 mm, 2.2m)