Module 12 Design of brakes. Version 2 ME, IIT Kharagpur

Transcription

1 Module 1 Design of brakes Version ME, IIT Kharagpur

2 Lesson 1 Design of shoe brakes Version ME, IIT Kharagpur

3 Instructional Objectives: After reading the lesson the students should learn: Different types of shoe brakes and their operating principles Design procedure of different shoe brakes 1. Types of brakes Brakes are devices that dissipate kinetic energy of the moving parts of a machine. In mechanical brakes the dissipation is achieved through sliding friction between a stationary object and a rotating part. Depending upon the direction of application of braking force, the mechanical brakes are primarily of three types Shoe or block brakes braking force applied radially Band brakes braking force applied tangentially. Disc brake braking force applied axially.. Shoe or block brake In a shoe brake the rotating drum is brought in contact with the shoe by suitable force. The contacting surface of the shoe is coated with friction material. Different types of shoe brakes are used, viz., single shoe brake, double shoe brake, internal expanding brake, external expanding brake. These are sketched in figure lever shoe drum Figure 1(a) Single shoe brake Version ME, IIT Kharagpur

4 lever shoe Figure 1(b) Double shoe brake drum shoe drum Figure 1(c): Internal expanding shoe brake Version ME, IIT Kharagpur

5 shoe drum Figure 1(d): External expanding shoe brake Figure 1.1.1: Different shoe brakes Single Shoe brake The force needed to secure contact is supplied by a lever. When a force F is applied to the shoe (see figure 1.1.a ) frictional force proportional to the applied force F fr = μ ' F develops, where μ ' depends of friction material and the geometry of the shoe. A simplified analysis is done as discussed below. O F fr F P Figure 1.1.a: Free body diagram of a brake shoe Though the exact nature of the contact pressure distribution is unknown, an approximation (based on wear considerations) is made as Version ME, IIT Kharagpur

6 p( ) = p cos 0 Where the angle is measured from the centerline of the shoe. If Coulomb s law of friction is assumed to hold good, then f fr ( ) = μp0 cos Since the net normal force of the drum is F, one has 0 Rb p( )cos d = F, 0 Where R and b are the radius of the brake drum and width of the shoe respectively. The total frictional torque is T = b 0 0 f fr ( ) R d If the total frictional force is assumed to be a concentrated one, then the T equivalent force becomes F fr =. A simple calculation yields, R 4μ sin0 μ = + sin 0 0 Figure 1.1.(b): Pressure distribution on brake Version ME, IIT Kharagpur

7 0 It may be seen that for very small value of 0, μ = μ'. Even when 0 = 30, 0 μ ' = μ. Usually if the contact angle is below 60, the two values of friction coefficient are taken to be equal. Consider, now single shoe brakes as shown in figures 1.1.3(a) and 3(b). Suppose a force P is applied at the end of a lever arm with length l. The shoe placed at a distance x from the hinge experiences a normal force N and a friction force F, whose direction depends upon the sense of rotation of the drum. Drawing free body diagram of the lever and taking moment about the hinge one gets (a) for clockwise rotation of the brake wheel, Nx + Fa = Pl (b) for anticlockwise rotation of the brake wheel, Nx Fa = Pl. Where a is the distance between the hinge and the line of action of F and is measured positive when F acts below point O as shown in the figure. Using Coulomb s law of friction the following results are obtained, μpl (a) for clockwise rotation F =, x + μa (b) for anticlockwise rotation μpl F =, x μa It may be noted that for anticlockwise rotating brake, if x μ >, then the force P a has negative value implying that a force is to applied in the opposite direction to bring the lever to equilibrium. Without any force the shoe will, in this case, draw the lever closer to the drum by itself. This kind of brake is known as selflocking, brake. Two points deserve attention. (1) If a < 0, the drum brake with clockwise rotation becomes self-energizing and if friction is large, may be self locking. () If the brake is self locking for one direction, it is never self locking for the opposite direction. This makes the self locking brakes useful for back stop s of the rotors. Version ME, IIT Kharagpur

8 P x l a F N Figure 1.1.3(a): FBD of shoe (CW drum rotation) P x l a F N Figure 1.1.3(b): FBD of shoe (CCW drum rotation) Double shoe brake Since in a single shoe brake normal force introduces transverse loading on the shaft on which the brake drum is mounted two shoes are often used to provide braking torque. The opposite forces on two shoes minimize the transverse loading. The analysis of the double shoe brake is very similar to the single shoe brake. External expanding shoe brake An external expanding shoe brake consists of two symmetrically placed shoes having inner surfaces coated with frictional lining. Each shoe can rotate about respective fulcrum (say, and O ). A schematic diagram with only one shoe is O1 Version ME, IIT Kharagpur

9 presented (figure 1.1.4) When the shoes are engaged, non-uniform pressure develops between the friction lining and the drum. The pressure is assumed to be proportional to wear which is in turn proportional to the perpendicular distance from pivoting point (O 1 N in figure 1.1.4). A simple geometrical consideration reveals that this distance is proportional to sine of the angle between the line joining the pivot and the center of the drum and the line joining the center and the chosen point. This means p( ) = p sin, where the angle is measured from line OO 1 and is limited as 1. Drawing the free body diagram of one of the shoes (left shoe, for example) and writing the moment equilibrium equation about resulted for clockwise rotation of the drum : 0 Fl M M O 1 1 = p f, (say) the following equation is Where F 1 is the force applied at the end of the shoe, and 1 1 M p = p0brδ ( 1) ( sin 1 sin ), + 1 δ M f = μ0brδ R(cos 1 ) ( cos 1 cos ), 4 where δ is the distance between the center and the pivot (OO 1 in figure 1.1.4) and is the distance from the pivot to the line of action of the force F (O1C in 1 the figure). In a similar manner the force to be applied at the other shoe can be obtained from the equation The net braking torque in this case is Fl= M + M p f. T = μ p0br 1 (cos cos ). Version ME, IIT Kharagpur

10 C F B N O A O 1 Figure 1.1.4: Force distribution in externally expanding brake. Internal expanding shoe brake Here the brake shoes are engaged with the internal surface of the drum. The analysis runs in the similar fashion as that of an external shoe brake. The forces required are F1 = ( M + M ) l p f and F = ( M p M f), l respectively. One of the important member of the expanding shoe brakes is the anchor pin. The size of the pin is to be properly selected depending upon the face acting on it during brake engagement. Version ME, IIT Kharagpur

11 Module 1 Design of Brakes Version ME, IIT Kharagpur

12 Lesson Design of Band and Disc Brakes Version ME, IIT Kharagpur

13 Instructional Objectives: After reading this lesson the students should learn: Different types of band brakes Design of band brakes Design of disc brakes Properties of friction materials 1. Band brakes: The operating principle of this type of brake is the following. A flexible band of leather or rope or steel with friction lining is wound round a drum. Frictional torque is generated when tension is applied to the band. It is known (see any text book on engineering mechanics) that the tensions in the two ends of the band are unequal because of friction and bear the following relationship: T T 1 = e μ β, where T 1 T = tension in the taut side, = tension in the slack side, μ = coefficient of kinetic friction and β = angle of wrap. If the band is wound around a drum of radius R, then the braking torque is br ( ) ( ) T = T T R = T e μβ R Depending upon the connection of the band to the lever arm, the member responsible for application of the tensions, the band brakes are of two types, (a) Simple band brake: In simple band brake one end of the band is attached to the fulcrum of the lever arm (see figures 1..1(a) and 1(b) ). The required force to be applied to the lever is b P = for clockwise rotation of the brake drum and T1 l Version ME, IIT Kharagpur

14 b P = for anticlockwise rotation of the brake drum, T l where l = length of the lever arm and b = perpendicular distance from the fulcrum to the point of attachment of other end of the band. b l b l T T 1 T 1 T P P 1(a): Band brake with CW rotating drum 1(b): Band brake with CCW rotating drum Figure 1..1: Band brakes (b) Differential band brake: In this type of band brake, two ends of the band are attached to two points on the lever arm other than fulcrum (see figures 1..(a) and 1..(b)). Drawing the free body diagram of the lever arm and taking moment about the fulcrum it is found that a P = T T1 l b l, for clockwise rotation of the brake drum and a P = T T l 1 Hence, P is negative if b, for anticlockwise rotation of the brake drum. l Version ME, IIT Kharagpur

15 e T T μβ 1 = > for clockwise rotation of the brake drum a b T1 a and e μβ = < for counterclockwise rotation of the brake drum. In T b these cases the force is to be applied on the lever arm in opposite direction to maintain equilibrium. The brakes are then self locking. The important design variables of a band brake are the thickness and width of the band. Since the band is likely to fail in tension, the following relationship is to be satisfied for safe operation. where w = width of the band, t = thickness of the band and s T T = wts 1 T = allowable tensile stress of the band material. The steel bands of the following dimensions are normally used w 5-40 mm mm 80 mm 100 mm mm t 3 mm 3-4 mm 4-6 mm 4-7 mm 6-10 mm a b l T T 1 P a b l T 1 T P Fig.1..(a): Differential Band brake with CW rotation Fig 1..(b): Differential Band brake with CCW rotation Version ME, IIT Kharagpur

16 . Band and block brakes: Sometimes instead of applying continuous friction lining along the band, blocks of wood or other frictional materials are inserted between the band and the drum. In this case the tensions within the band at both sides of a block bear the relation T1 1+ μtan =, T 1 μtan 1 where T 1 = tension at the taut side of any block T 1 = tension at the slack side of the same block = angle subtended by each block at center. If n number of blocks are used then the ratio between the tensions at taut side to slack side becomes T T The braking torque is T ( T T ) br 1 = R 1 1+ μtan =. 1 μtan n 3. Disc brake: Version ME, IIT Kharagpur

17 In this type of brake two friction pads are pressed axially against a rotating disc to dissipate kinetic energy. The working principle is very similar to friction clutch. When the pads are new the pressure distribution at pad-disc interface is uniform, i.e. If F is the total axial force applied then The frictional torque is given by T p = constant. braking F p =, where A is the area of the pad. A μ F A = A rda where μ = coefficient of kinetic friction and r is the radial distance of an infinitesimal element of pad. After some time the pad gradually wears away. The wear becomes uniforms after sufficiently long time, when pr = constant = c (say) where F = pda= c da r. The braking torque is μ AF Tbraking ' = μ prda= μac= da r It is clear that the total braking torque depends on the geometry of the pad. If the annular pad is used then 3 3 R1 R Tbr = μ F 3 R 1 R R1 R T + br = μ F where R and 1 Rare the inner and outer radius of the pad. Version ME, IIT Kharagpur

18 4. Friction materials and their properties. The most important member in a mechanical brake is the friction material. A good friction material is required to possess the following properties: High and reproducible coefficient of friction. Imperviousness to environmental conditions. Ability to withstand high temperature (thermal stability) High wear resistance. Flexibility and conformability to any surface. Some common friction materials are woven cotton lining, woven asbestos lining, molded asbestos lining, molded asbestos pad, Sintered metal pads etc. Review questions and answers: Q.1. A double shoe brake has diameter of brake drum 300mm and contact angle of each shoe 90 degrees, as shown in figure below. If the coefficient of friction for the brake lining and the drum is 0.4, find the spring force necessary to transmit a torque of 30 Nm. Also determine the width of the brake shoe if the braking pressure on the lining material is not to exceed 0.8 MPa. S S 50 5 Figure 1..3 Ans. The friction force required to produce the given torque is Version ME, IIT Kharagpur

19 30 F1+ F = = 00 N ( ) F1 The normal forces on the shoes are N = F 1, N, μ ' = μ ' where 4 μ sin μ = = π = Writing the moment equilibrium equations about 0 ' ( 0 ) sin0 4 the pivot points of individual shoes (draw correct FBDs and verify) Sl Sl + N1x + F1a = 0 F1 = = S, and x a + μ ' Sl Sl Nx + Fa = 0 F = = S x a μ ' This yields S = 98.4(N). Width of the friction lining : According to the pressure distribution assumed for a shoe brake, the maximum bearing pressure occurs at the centerline of the shoe. The width of the brake lining must be selected from the higher values of the normal forces, in this case N. Noting that π /4 = max cos, π /4 N Rbp d Where R = 0.150, p = 0.8 X 10, N = / 0.44, 6 max the value of b is calculated to be 5.4 mm or 6 mm (approx.). Q. A differential band brake has brake drum of diameter 500mm and the maximum torque on the drum is 1000 N-m. The brake embraces /3 rd of the circumference. If the band brake is lined with asbestos fabric having a coefficient of friction 0.3, then design the steel band. The permissible stress is 70 MPa in tesnion. The bearing pressure for the brake lining should not exceed 0. MPa. Version ME, IIT Kharagpur

20 Ans. The design of belt is to be carried out when the braking torque is maximum i.e. T br = 1000 N-m. According to the principle of band brake T br 4π ( 1 μβ T e ) R = T 1 1 e 0.5 = Which yield T = 5587 N, T = e μβ T = 1587 N. 1 1 In order to find the pressure on the band, consider an infinitesimal element. The force balance along the radial direction yields N = T Δ F T Since N = pbrδ so p = T br. N T1 The maximum pressure is pmax =. br 5587 Hence b = = 0.11 m (approx.) T+ΔT The thickness t of the band is calculated from the relation Sbt t = T Which yields t = = m or 1 mm (approx.) Δ Version ME, IIT Kharagpur

21 MODULE III Brakes, Clutches and Flywheel, Brakes A brake is a device by means of which artificial resistance is applied on to a moving machine member in order to retard or stop the motion of the member or machine Types of Brakes Different types of brakes are used in different applications Based on the working principle used brakes can be classified as mechanical brakes, hydraulic brakes, electrical (eddy current) magnetic and electro-magnetic types. Mechanical Brakes Mechanical brakes are invariably based on the frictional resistance principles In mechanical brakes artificial resistances created using frictional contact between the moving member and a stationary member, to retard or stop the motion of the moving member. Basic mechanism of braking The illustration below explains the working of mechanical brakes. An element da of the stationary member is shown with the braked body moving past at velocity v. When the brake is actuated contact is established between the stationary and moving member and a normal pressure is developed in the contact region. The elemental normal force dn is equal to the product of contact pressure p and area of contact da. As one member is stationary and the other is in relative motion, a frictional force df is developed between the members. The

22 magnitude of the frictional force is equal to the co-efficient of friction times the normal force dn da dn=p.da v dff =µ.dn=µ.p.da Figue The moment of the frictional force relative to the point of motion contributes to the retardation of motion and braking. The basic mechanism of braking is illustrated above. Design and Analysis To design, select or analyze the performance of these devices knowledge on the following are required. The braking torque The actuating force needed The energy loss and temperature rise At this beginning stage attention will be focused mainly on some preliminary analysis related to these aspects, namely torque, actuating force, energy absorbed and temperature rise. Torque induced is related to the actuating force, the geometry of the member and other contact conditions. Most mechanical brakes that work on the frictional contact basis are classified based on the geometry. There are two major classes of brakes, namely drum brakes and disc brakes. Design and analysis of drum brakes will be considered in detail in following

23 sections, the discussion that follow on disc or plate clutches will from the basis for design of disc type of brakes. Drum brakes basically consists of a rotating body called drum whose motion is braked together with a shoe mounted on a lever which can swing freely about a fixed hinge H. A lining is attached to the shoe and contacts the braked body. The actuation force P applied to the shoe gives rise to a normal contact pressure distributed over the contact area between the lining and the braked body. A corresponding friction force is developed between the stationary shoe and the rotating body which manifest as retarding torque about the axis of the braked body. Brakes Classification Short Shoe Long shoe Lining Shoe Stationary member Rotating body (drum) ω O ω O ω O G Rigid Pivoted Figure 3.1. Various geometric configurations of drum brakes are illustrated above. Drum Brakes are classified based on the shoe geometry. Shoes are classified as being either short or long. A short shoe is one whose lining dimension in the direction of motion is so small that contact pressure variation is negligible, i.e. the pressure is uniform everywhere.

24 When the area of contact becomes larger, the contact may no longer be with a uniform pressure, in which case the shoe is termed as long shoe. The shoes are either rigid or pivoted, pivoted shoes are also some times known as hinged shoes. The shoe is termed rigid because the shoes with attached linings are rigidly connected to the pivoted posts. In a hinged shoe brake - the shoes are not rigidly fixed but hinged or pivoted to the posts. The hinged shoe is connected to the actuating post by the hinge, G, which introduces another degree of freedom Preliminary Analysis The figure shows a brake shoe mounted on a lever, hinged at O, having an actuating force F a, applied at the end of the lever. On the application of an actuating force, a normal force F n is created when the shoe contacts the rotating drum. And a frictional force F f of magnitude f.f n, f being the coefficient of friction, develops between the shoe and the drum. Moment of this frictional force about the drum center constitutes the braking torque.

25 Fa F a a drum shoe ω r o F n F f c b R x R y o (a) Brake assembly (b) Free-body diagram Rx T Ry Figure Short Shoe Analysis For a short shoe we assume that the pressure is uniformly distributed over the contact area. Consequently the equivalent normal force F n = p.a, where = p is the contact pressure and.a is the surface area of the shoe. Consequently the friction force Ff = f.fn where f is the co-efficient of friction between the shoe lining material and the drum material. The torque on the brake drum is then, T = f Fn. r = f.p.a.r A quasi static analysis is used to determine the other parameters of braking. Applying the equilibrium condition by taking moment about the pivot O we can write M = F a F b+ f F c = 0 O a n n Substituting for F n and solving for the actuating force, we get,

26 F a = F n (b+-fc)/a The reaction forces on the hinge pin (pivot) are found from a summation of forces, i.e. F x = 0,R x = fp a A F = 0,R = p A F y y a a Self- energizing The principle of self energizing and leading and trailing shoes With the shown direction of the drum rotation (CCW), the moment of the frictional force f. F n c adds to the moment of the actuating force, F a As a consequence, the required actuation force needed to create a known contact pressure p is much smaller than that if this effect is not present. This phenomenon of frictional force aiding the brake actuation is referred to as selfenergization. Leading and trailing shoe For a given direction of rotation the shoe in which self energization is present is known as the leading shoe When the direction of rotation is changed, the moment of frictional force now will be opposing the actuation force and hence greater magnitude of force is needed to create the same contact pressure. The shoe on which this is prevailing is known as a trailing shoe

27 Self Locking At certain critical value of f.c the term (b-fc) becomes zero. i.e no actuation force need to be applied for braking. This is the condition for self-locking. Self-locking will not occur unless it is specifically desired. Short and Long Shoe Analysis Foregoing analysis is based on a constant contact pressure p. In reality constant or uniform constant pressure may not prevail at all points of contact on the shoe. In such case the following general procedure of analysis can be adopted General Procedure of Analysis Estimate or determine the distribution of pressure on the frictional surfaces Find the relation between the maximum pressure and the pressure at any point For the given geometry, apply the condition of static equilibrium to find the actuating force, torque and reactions on support pins etc.

28 Drum Brakes Among the various types of devices to be studied, based on their practical use, the discussion will be limited to Drum brakes of the following types which are mainly used in automotive vehicles and cranes and elevators. Drum Brake Types: Rim types with internal expanding shoes Rim types with external contracting shoes Internal expanding Shoe The rim type internal expanding shoe is widely used for braking systems in automotive applications and is generally referred as internal shoe drum brake. The basic approach applied for its analysis is known as long-rigid shoe brake analysis. Long rigid Shoe Analysis A schematic sketch of a single shoe located inside a rotating drum with relevant notations, is shown in the figure below. In this analysis, the pressure at any point is assumed to be proportional to the vertical distance from the hinge pin, the vertical distance from the hinge pin, which in this case is proportional to sine of the angle and thus, p dsin sin Since the distance d is constant, the normal pressure at any point is just proportional to sinθ. Call this constant of proportionality as K

29 Y fd Ncos fdn dn dn sin F X F f dnsin d dncos F Y R X > X < d > R Y Figure Thus p = Ksin It the maximum allowable pressure for the lining material is p max then the constant K can be defined as K p p = = sin sin max max max pmax p = sin sin

30 The normal force dn is computed as the product of pressure and area and the frictional force as the product of normal force and frictional coefficient i.e. f dn. By integrating these over the shoe length in terms of its angle the braking torque T, and other brake parameters are computed. To determine the actuating force F, the moment equilibrium about the pivot point is applied. For this we need to determine the moment of the normal force M N and moment of the frictional force about the pivot point. Moment of the normal force is equal to the normal force times its moment arm about the pivot point. From the figure it is clear that the moment arm in this case is equal to d sin Θ where d is the distance between the drum center and pivot center 1 1 M N = p.b.r.d.d sin = b.p.r.d.sin.d 1 pmax = b.d.r. sin d sin p max b.d.r 1 1 M N = ( 1) (sin sin ) sin a 4 1 On similar lines the moment of friction force is computed 1 ( ) M F = f.p.b.r.d r d sin 1 pmax = f.b.r. sin ( r dsin ) d sin max ( ) ( n 1) f.pmax. b.r d M f = r cos cos 1 sin si sin a

31 The actuating force F is determined by the summation of the moments of normal and frictional forces about the hinge pin and equating it to zero. Summing the moment about point O gives F = M N ± M f c where, M N and M f are the moment of the normal and frictional forces respectively, about the shoe pivot point. The sign depends upon the direction of drum rotation, (- sign for self energizing and + sign for non self energizing shoe)where the lower sigh is for a self energizing shoe and the upper one for a self deenergizing shoe. The reaction forces are determined by applying force summation and equilibrium R x = dn.cos+ df.sin = b.r.p cos d+ f b.r.p sin d 1 1 pmax pmax = b.r. sin cos d+ f b.r. sin d sin sin max 1 1 max ( ) ( ) pmax.br max = ( 1 ) sin sin 1 ± f sin sin sin 4 1 The equations can be simplified and put as p a br R x = (A fb) sin a p a br R y = (B± fa) F sin x a Where ( 1) 1 A = sin sin

32 1 1 1 B = ( ) sin sin The braking torque T on the drum by the shoe is of the frictional forces f. dn times the radius of the drum and resulting equation is, 1 1 T = f.b.p.r.d.r 1 1 p = fbr max sin d sin fdp r a (cos 1 cos ) T = sin a T Double Shoe Brakes Twin Shoe Brakes Behavior of a single shoe has been discussed at length. Two such shoes are combined into a complete practical brake unit, two being used to cover maximum area and to minimize the unbalanced forces on the drum, shaft and bearings. max If both the shoes are arranged such that both are leading shoes in which self energizing are prevailing, then all the other parameters will remain same and the total braking torque on the drum will be twice the value obtained in the analysis. However in most practical applications the shoes are arranged such that one will be leading and the other will be trailing for a given direction of drum rotation

33 If the direction of drum rotation changes then the leading shoe will become trailing and vice versa. Thus this type of arrangement will be equally effective for either direction of drum rotation. Further the shoes can be operated upon using a single cam or hydraulic cylinder thus provide for ease of operation One leading shoe & one trailing Two Leading shoe Figure However the total braking torque will not be the twice the value of a single shoe, if the same normal force is applied or created at the point of force application on both the brake shoes which is the normal practice as they are actuated using a common cam or hydraulic cylinder. This is because the effective contact pressure (force) on the trailing shoe will not be the same, as the moment of the friction force opposes the normal force, there by reducing its actual value as in most applications

34 the same normal force is applied or created at the point of force application on the brake shoe as noted above Consequently we may write the actual or effective pressure prevailing on a trailing shoe ' F.a p a = p a. (M n + M f ) Resulting equation for the braking torque p T f.w.r. a B = (cos 1 -cos )(p p sin a + a ') a below Some pictorial illustrations of the automotive drum brakes are presented Figure 3.1.6

35 Figure Oblong Cam Actuator Leading shoe Trailing shoe Rotating Drum Pivot point(fixed axis) The anatomy of the single leading shoe drum Brake Animation

36 Figure Figure

37 External Contracting Shoe The same analysis can be extended to a drum brake with external contracting type of shoes, typically used in elevators and cranes. A schematic sketch of as single shoe located external to the rotating drum is with all relevant notations is shown in the figure below. Figure Corresponding contact geometry is shown in the figure The resulting equations for moment of normal and frictional force as well as the actuating force and braking torque are same as seen earlier. For convenience they are reproduced here again ( ) fbp r cos cos T = a 1 sin a

38 M N ± M F = f c p a bra 1 1 M N = ( 1) (sin sin ) sin a 4 1 ( ) fpabr d M f = r cos 1 cos (sin sin sin a 1 ) TWIN SHOE BRAKES As noted earlier for the internal expanding shoes, for the double shoe brake the braking torque for one leading and one trailing shoe acted upon a common cam or actuating force the torque equation developed earlier can be applied. i.e p T f.w.r. a B = (cos 1 -cos )(p p sin a + a ') a

39 Brake with a pivoted long shoe When the shoe is rigidly fixed to the lever, the tendency of the frictional force (f.f n ) is to unseat the block with respect to the lever. This is eliminated in the case of pivoted or hinged shoe brake and it also provides some additional advantages. Long Hinged Shoe In a hinged shoe brake - the shoes are not rigidly fixed but hinged or pivoted to the posts. The hinged shoe is connected to the actuating post by the hinge, G, which introduces another degree of freedom - so the shoe tends to assume an optimum position in which the pressure distribution over it is less peaked than in a rigid shoe. o B φ G B G OH=a OG=b F y F x H X Figure As wear proceeds the extra degree of freedom allows the linings to conform more closely to the drum than would be the case to

40 rigid shoes. This permits the linings to act more effectively and also reduces the need for wear adjustment. The extra expense of providing another hinge is thus justified on the grounds of more uniform lining wear and consequently a longer life. This is the main advantage of the pivoted shoe brake This is possible only if the shoe is in equilibrium. For equilibrium of the shoe moments of the forces about the hinge pin should balance - i.e ΣM G =T+F x b y -F y b x =0 where b x = b.cos G b y = b.sin G This needs that the resultant moment due to the frictional force (and due to the normal force) about the pivot point should be zero, so that no rotation of the shoe will occur about the pivot point. To facilitate this location of the pivot is to be selected carefully. The actuating force P is applied to the post HG so the shoe itself is subject to actual and ideal contacts only - the (ideal) at pin G and the actual as distributed contact with the drum. The location is in such a way that the moment of frictional force (and the normal force) about the pivot is zero. i.e the actual distributed contact leads to the ideal (concentrated)contact at the hinge or pivot. i.e the actual distributed contact leads to the ideal contact at the hinge or pivot Further it is desirable to minimize the effect of pin reaction for which the shoe pivot and post pivot points are made con current.

41 Let us now look how this can be met, satisfying the conditions set above and consequently the derive the equations relating the location of the pivot from the center of the drum A schematic sketch of a single shoe is shown in the figure fdn sin fdn cos r dn fdn dn cos rcos h R y (hcos r) R x Force acting on shoe An element of friction lining located at an angle Θ and subtending to a small angle d is shown in figure. The area if the element is ( r.d.b), where b is the width of the friction lining parallel to the axis of the brake drum. If the intensity of pressure at the element is p, the normal reaction dn on the element is given by dn = (rd b)p (a)

42 Distribution of pressure If the shoe is long then the pressure will not be uniform We need to determine the distribution of pressure along the lining; the pressure distribution should be conducive for maintaining a uniform wear Since the brake drum is made of a hard material like cast iron or steel, the wear occurs on the friction lining, which is attached to the shoe. As shown in fig the lining need to retain the cylindrical shape of the brake drum when wear occurs. After the radial wear has take place, a point such as X moves to X in order to maintain contact on the lining with the brake drum. In figure δx is the wear in the X direction and δr is the wear in the radial direction. If it is assumed that the shoe is constrained to move towards the brake drum to compensate to wear, δx should be constant because it need to be same for all points. Therefore, δr δ x = cos = constant (b) Y δx x δx x' wear X wear of friction lining

43 The radial wear δr is proportional to the work done by the frictional force. The work done by the frictional force depends upon the frictional force ( fdn ) and the rubbing velocity. Since the rubbing velocity is constant for all points on friction lining, Or δr ( frdbp) δr fdn Therefore δr p (c) From the expression (b) and (c) p = cons tan t or p = C1 cos cos (d) Where C 1 is the constant of proportionality. The pressure is maximum when. =0 Substituting, From Eqs (d) and (e), p = C (e) max 1 p = pmax cos Substituting this value in Eq. (a dn = (rdb)pmax cos (f) The forces acting on the element of the friction lining are shown in figure. The distance h of the pivot is selected in such a manner that the moment of frictional force about it is zero. Therefore, Mf == 0 dm f =f.dn moment arm moment arm in this case = (h cos r)

44 f M = fdn(hcos r) =0 f M = fp rdcos (h r cos ) max Substituting dn from Eq. (f), 0 ( ) h cos rcos d= 0 1+ cos or h d r cos d = φ+ sin φ or h R ( sin φ ) 0 0 = 0 4R sin h = + sin The elemental torque of frictional force fdn about the axis of brake drum is fdnr. Therefore Substituting the value of dn from Eq.(f) B B T = fdnr B T = fr bp cos d max T = fr bp sin max The reaction R X can be determined by considering two components ( dn cos ) and ( fdn sin ).

45 Due to symmetry, the other two vertical components of the force balances i.e fdn sin = 0 dn sin = 0 Therefore, Note that R is also = x F n R x = dncos = rbp cos d max + sin = rbpmax 4 1 or R x = rbp max(+ sin ) The reaction R y can be determined by considering two components ( dnsin ) and ( fdn cos ) Due to symmetry, dnsin =0 Therefore,

46 R y = fdncos = frbp cos d max 1 or R y = frbp max(+ sin ) As noted earlier, Rewriting it, B T = fr bp sin max + sin 4rsin TB = frbp max. + sin = ffn h DOUBLE BRAKE SHOE figure. A double block brake with two symmetrical and pivoted shoes is show in If the same magnitude of actuating forces are acted upon the posts, then TB = f.(fn1 + F n).h = f.f n. h P P Θ Q Pivoted double block brake

47 Pivoted shoe brakes are mainly used in hoists and cranes. Their applications are limited because of the physical problem in locating pivot so close to the drum surface.

48 Energy Consideration It has been noted that the most common brakes employ friction to transform the braked system's mechanical energy, irreversibly into heat which is then transferred to the surrounding environment - Kinetic energy is absorbed during slippage of either a clutch or brake, and this energy appears as heat. If the heat generated is faster than it is dissipated, then the temperature rises. Thorough design of a brake therefore requires a detailed transient thermal analysis of the interplay between heat generated by friction, heat transferred through the lining and the surrounding metalwork to the environment, and the instantaneous temperature of the surface of the drum as well as the lining. For a given size of brake there is a limit to the mechanical power that can be transformed into heat and dissipated without lthe temperatures reaching damaging levels. Temperature of the lining is more critical and the brake size is characterized by lining contact area, A. The capacity of a clutch or brake is therefore limited by two factors: 1. The characteristics of the material and,. The ability of the brake to dissipate heat. Heat Generated In Braking During deceleration, the system is subjected to an essentially constant torque T exerted by the brake, and in the usual situation this constancy implies constant deceleration too. Application of the work or energy principle to the system enables the torque exerted by the brake and the work done by the brake, U, to be calculated from:-

49 U = E = T () Where E is the loss of system total energy which is absorbed by the brake during deceleration, transformed into heat, and eventually dissipated. The elementary equations of constant rotational deceleration apply, thus when the brake drum is brought to rest from an initial speed ω o :- Deceleration = ω o / (1) = ω m. t ; ω m = ω o / = / t where ω m is the mean drum speed over the deceleration period. The mean rate of power transformation by the brake over the braking period is :- P m = U / t = T ω m ( 3 ) which forms a basis for the selection or the design of the necessary brake dimensions. The rise in temperature in the lining material is also important as rate of wear is also a function of the temperature. Further for any lining material, the maximum allowable temperature is also another performance criteria. Temperature Rise The temperature rise of the brake assembly can be approximated by the classic expression,, E T = C.m Where is temperature T is rise in temperature in o C, C is the specific heat of the brake drum material (500J/Kg for steel or Cast Iron) and m is the mass (kg) of the brake parts dissipating the heat into the surroundings. Though the equation appears to be simple, there are so many variables involved that it would be most unlikely that such an analysis would even approximate experimental results.

50 On the other hand the temperature-rise equations can be used to explain what happens when a clutch or brake is operated frequently. For this reason such analysis are most useful, for repetitive cycling, in pin pointing those design parameters that have the greatest effect on performance. An object heated to a temperature T 1 cools to an ambient temperature T a according to the exponential relation Time-temperature relation T T (T T )e (AU / WC)t i a = 1 a Where T1 = instantaneous temperature at time t, C A= heat transfer area, m U= Heat Transfer coefficient, W/(m.s. C) T1 = Initial temperature, C Ta = Ambient temperature, C C - Specific heat t - time of operation, s

51 T T 1 A T T B T a t t B t A Time t C C Figure Figure shows an application of Eq. (a). At time t A a clutching or braking operation causes the temperature to rise to T 1 at A. Though the rise occurs in a finite time interval, it is assumed to occur instantaneously. The temperature then drops along the decay line ABC unless interrupted by another braking operation. If a second operation occurs at time t B, the temperature will rise along the dashed line to T and then begin an exponential drop as before. About 5-10 % of the heat generated at the sliding interface of a friction brake must be transferred through the lining to the surrounding environment without allowing the lining to reach excessive temperatures, since high temperatures lead to hot spots and distortion, to fade (the fall-off in friction coefficient) or, worse, to degradation and charring of the lining which often incorporates organic constituents

52 In order to determine the brake dimensions the energy need to be absorbed during critical braking conditions is to be estimated. Energy to be Absorbed If t is the time of brake application and ω m the mean or average angular velocity then the energy to be absorbed in braking E E = T. ω m.t = Ek+ Ep+ Ei where E k is the kinetic energy of the rotating system E p is the potential energy of the moving system E i is the inertial energy of the system Energy to be absorbed E k 1 ω = g 1 = mv 1 ω 1 = ( v v1) = I ( ω ω1) g E = mgh =ωh E p i 1 = Iω Frictional Material A brake or clutch friction material should have the following characteristics to a degree, which is dependent upon the severity of the service. A high and uniform coefficient of friction. Imperviousness to environmental conditions, such as moisture. The ability to withstand high temperatures together with good thermal conductivity. Good resiliency. High resistance to wear, scoring, and galling.

53 Linings The choice of lining material for a given application is based upon criteria such as the expected coefficient of friction; fade resistance, wear resistance, ease of attachment, rigidity or formability, cost, abrasive tendencies on drum, etc. The lining is sacrificial - it is worn away. The necessary thickness of the lining is therefore dictated by the volume of material lost - this in turn is the product of the total energy dissipated by the lining throughout its life, and the specific wear rate R w (volume sacrificed per unit energy dissipated) which is a material property and strongly temperature dependent. The characteristics of a typical moulded asbestos lining material is illustrated in the figure below. The coefficient of friction, which may be taken as 0.39 for design purposes, is not much affected by pressure or by velocity - which should not exceed 18 m/s. The maximum allowable temperature is 400 C. However at this temperature the wear is very high. From a lower wear or higher life point, the maximum temperature should not exceed about 00 o C o Temperature ( C ) 400 Figure

54 Linings traditionally were made from asbestos fibers bound in an organic matrix, however the health risks posed by asbestos have led to the decline of its use. Non-asbestos linings generally consist of three components - metal fibers for strength, modifiers to improve heat conduction, and a phenolic matrix to bind everything together. Brake Design Section The braked system is first examined to find out the required brake capacity that is the torque and average power developed over the braking period. - The brake is then either selected from a commercially available range or designed from scratch ff a drum brake has to be designed for a particular system (rather than chosen from an available range) then the salient brake dimensions may be estimated from the necessary lining area, A, together with a drum diameter- tolining width ratio somewhere between 3:1 and 10:1, and an angular extent of 100 C say for each of the two shoes. Worked out Example 1 An improved lining material is being tried on an existing passenger car drum brake shown in Figure. Quality tests on the material indicated permissible pressure of 1.0 MPa and friction co-efficient of 0.3. Determine what maximum actuating force can be applied for a lining width of 40 mm and the corresponding braking torque that could be developed. While cruising on level road at 100 kmph, if it is to decelerated at 0.5g and brought to rest, how much energy is absorbed and what is the expected stopping distance?

55 While cruising on level road at 100 kmph, if it is to decelerated at 0.5g and brought to rest, how much energy is absorbed and what is the expected stopping distance? F F R= Pin Pin AUTOMOTIVE DOUBLE SHOE BRAKE Figure Analysis based on leading shoe P = 1 MPa f = 0.3 a = mm a b = 40 mm 90 0 max = d = = mm = = r =15 mm

56 a B 1 B B a n f 3 6 a p brd 1 M (sin sin ) M a 1 n = 1 sin a 4 n *40*10 *15 *10 * π 1 = * ( sin 40 sin10 ) =40*15* ( 1.03) 4 = N.m ( ) f.b.rpm d M r(cos - cos ) - sin - sin f = sin 1 1 a = 0.3*40*10-3 *15*10-3 * (cos5 - cos10 ) (sin 10 - sin 5 ) M f = 4.85 N - m F*a= M N M f Mn Mf F = = = 174.3N a Max. actutating force p F T = fbr ( cos cos )) 1 + sin α M + M T = 0.3 * 40 *10 * (0.15) *10 (Cos5 Cos10) T = N-m Running at 100 kmph =100*5/8 = 7.7 m/s

57 U= 7.7 m/s Deceleration =0.5*9.8=4.9 V U = as 0 (7.7) = *( 4.9) *S 7.7 S = = 78.9 m *4.9 E = T. ω av.t = = = 138.KJ Worked out example A spring set, hydraulically released double shoe drum brake, schematically shown at Fig is to be designed to have a torque capacity of 600 N.m under almost continuous duty when the brake drum is rotating at 400 rpm in either direction. Assume that the brake lining is to be molded asbestos having a friction coefficient of 0.3 and permissible pressure of 0.8 MPa. The width of the brake shoe is to be third of drum diameter and the remaining proportion's are as shown in figure. a = 14. D 1 b= D D d = = D 0 cos30

58 1.4 D 90 ο 0.6 D 30 ο Double block brake Figure Determine the required brake drum diameter, width of the lining and the spring force required to be set. a = 1.4D 1 b = D 3 0.6D d = = 0.693D 0 cos30

59 6 1 D π sin(10 ) sin 30 = 0.8*10 * D * *0.608D M = D *1.035 N Pbrf a MJ = ( r acos) sind 1 d ( ) ( 1) = Pabrf r cos 1 cos sin sin 6 D D D = 0.8*10 * * * 0.3 ( 1.5) ( 0.866) 3 = 1500D P 'for the trailing shoe Torque due to trailing and leading shoe= total τ= a ' MN MF Pa = Pa M N + M F = Pa = P fdn.r Pa sin = f.r.rd.b sin a a a fbr = P a sin d+ P a ' sin d sin a 1 1 fbr =.P + P'.cos cos sin ( ) ( ) a a 1 D D 0.3* * = 1 = 4334 D T *0.8*10 6 cos15 cos1.5 3 = 600 Nm Therefore D = 40.14mm ( )

60 Actuating force due to spring MN MF * 0.40 F = = = 343.N 1.4 * D 1.4 Actuating force =343. N D Living width = 3 = mm

61 CLUTCH Clutch Introduction A Clutch is ia machine member used to connect the driving shaft to a driven shaft, so that the driven shaft may be started or stopped at will, without stopping the driving shaft. A clutch thus provides an interruptible connection between two rotating shafts Clutches allow a high inertia load to be stated with a small power. A popularly known application of clutch is in automotive vehicles where it is used to connect the engine and the gear box. Here the clutch enables to crank and start the engine disengaging the transmission Disengage the transmission and change the gear to alter the torque on the wheels. Clutches are also used extensively in production machinery of all types Mechanical Model Two inertia s Iand 1 I and traveling at the respective angular velocities ω1 and ω, and one of which may be zero, are to be brought to the same speed by engaging. Slippage occurs because the two elements are running at different speeds and energy is dissipated during actuation, resulting in temperature rise. Clutch or brake ω1 ω Ι1 Ι1 Dynamic Representation of Clutch or Brake Figure 3..1

62 Animated Figure 3.. To design analyze the performance of these devices, a knowledge on the following are required. 1. The torque transmitted. The actuating force. 3. The energy loss 4. The temperature rise FRICTION CLUTCHES As in brakes a wide range of clutches are in use wherein they vary in their are in use their working principle as well the method of actuation and application of normal forces. The discussion here will be limited to mechanical type friction

63 clutches or more specifically to the plate or disc clutches also known as axial clutches Frictional Contact axial or Disc Clutches An axial clutch is one in which the mating frictional members are moved in a direction parallel to the shaft. A typical clutch is illustrated in the figure below. It consist of a driving disc connected to the drive shaft and a driven disc co9nnected to the driven shaft. A friction plate is attached to one of the members. Actuating spring keeps both the members in contact and power/motion is transmitted from one member to the other. When the power of motion is to be interrupted the driven disc is moved axially creating a gap between the members as shown in the figure. Figure 3..3

64 Flywheel Clutch cover Clutch plate Diaphragm spring to transmission Throw out Bearing Pressure plate Animated Figure 3..4 METHOD OF ANALYSIS The torque that can be transmitted by a clutch is a function of its geometry and the magnitude of the actuating force applied as well the condition of contact prevailing between the members. The applied force can keep the members together with a uniform pressure all over its contact area and the consequent analysis is based on uniform pressure condition

65 Uniform Pressure and wear However as the time progresses some wear takes place between the contacting members and this may alter or vary the contact pressure appropriately and uniform pressure condition may no longer prevail. Hence the analysis here is based on uniform wear condition Elementary Analysis Assuming uniform pressure and considering an elemental area da da = Π.r dr The normal force on this elemental area is dn = π.r.dr.p The frictional force df on this area is therefore df = f. π.r.dr.p

66 lining dr < F r di < do > A single-surface Axial Disk Clutch Figure 3..5 Now the torque that can be transmitted by this elemental are is equal to the frictional force times the moment arm about the axis that is the radius r i.e. T = df. r = f.dn. r = f.p.a.r = f.p..π.r. dr.r The total torque that could be transmitted is obtained by integrating this equation between the limits of inner radius ri to the outer radius ro r o T = π pfr dr = pf (r 3 o r i ) r 3 π 3 i Integrating the normal force between the same limits we get the actuating force that need to be applied to transmit this torque.

67 F a r o = πprdr r i ( ) F =π r r.p a o i Equation 1 and can be combined together to give equation for the torque 3 3 o i a o i (r r ) T = ff. 3 (r r ) Uniform Wear Condition According to some established theories the wear in a mechanical system is proportional to the PV factor where P refers the contact pressure and V the sliding velocity. Based on this for the case of a plate clutch we can state The constant-wear rate R w is assumed to be proportional to the product of pressure p and velocity V. R w = pv= constant And the velocity at any point on the face of the clutch is V= r. ω Combining these equation, assuming a constant angular velocity ω pr = constant = K The largest pressure p max must then occur at the smallest radius r i, K = pmaxri Hence pressure at any point in the contact region p = p max r i r

68 In the previous equations substituting this value for the pressure term p and integrating between the limits as done earlier we get the equation for the torque transmitted and the actuating force to be applied. I.e The axial force F a is found by substituting ri p = pmax for p. r and integrating equation dn = π prdr Similarly the Torque r o r o r F prdr p i = π = π max rdr = πp max r i (r o r i ) r r r i i r o T = f p r rdr f p r (r r π maxi = π maxi o i ) r i Substituting the values of actuating force Fa The equation can be given as (r o + r i ) T = ff a. Single plate dry Clutch Automotive application The clutch used in automotive applications is generally a single plate dry clutch. In this type the clutch plate is interposed between the flywheel surface of the engine and pressure plate.

69 Flywheel Friction planes Engine crankshaft Clutch plate (driven disk) Pressure plate Pressure spring Release bearing Housing To transmission To release Figure 3..6 Single Clutch and Multiple Disk Clutch Basically, the clutch needs three parts. These are the engine flywheel, a friction disc called the clutch plate and a pressure plate. When the engine is running and the flywheel is rotating, the pressure plate also rotates as the pressure plate is attached to the flywheel. The friction disc is located between the two. When

70 the driver has pushed down the clutch pedal the clutch is released. This action forces the pressure plate to move away from the friction disc. There are now air gaps between the flywheel and the friction disc, and between the friction disc and the pressure plate. No power can be transmitted through the clutch. Operation Of Clutch When the driver releases the clutch pedal, power can flow through the clutch. Springs in the clutch force the pressure plate against the friction disc. This action clamps the friction disk tightly between the flywheel and the pressure plate. Now, the pressure plate and friction disc rotate with the flywheel. As both side surfaces of the clutch plate is used for transmitting the torque, a term N is added to include the number of surfaces used for transmitting the torque By rearranging the terms the equations can be modified and a more general form of the equation can be written as T = N.f.F a.r m T is the torque (Nm). N is the number of frictional discs in contact. f is the coefficient of friction F a R m is the actuating force (N). is the mean or equivalent radius (m). Note that N = n1 + n -1 Where n1= number of driving discs n = number of driven discs Values of the actuating force F and the mean radius analysis are summarized and shown in the table r m for the two conditions of

71 Clutch Construction Two basic types of clutch are the coil-spring clutch and the diaphragm-spring clutch. The difference between them is in the type of spring used. The coil spring clutch shown in left Fig 3..6 uses coil springs as pressure springs (only two pressure spring is shown). The clutch shown in right figure 3..6 uses a diaphragm spring. Figure 3..6 The coil-spring clutch has a series of coil springs set in a circle. At high rotational speeds, problems can arise with multi coil spring clutches owing to the effects of centrifugal forces both on the spring themselves and the lever of the release mechanism. These problems are obviated when diaphragm type springs are used, and a number of other advantages are also experienced

72 Clutch or Driven Plate More complex arrangements are used on the driven or clutch plate to facilitate smooth function of the clutch The friction disc, more generally known as the clutch plate, is shown partly cut away in Fig. It consists of a hub and a plate, with facings attached to the plate. Figure 3..7 First to ensure that the drive is taken up progressively, the centre plate, on which the friction facings are mounted, consists of a series of cushion springs which is crimped radially so that as the clamping force is applied to the facings the crimping is progressively squeezed flat, enabling gradual transfer of the force On the release of the clamping force, the plate springs back to its original position crimped (wavy) state

73 This plate is also slotted so that the heat generated does not cause distortion that would be liable to occur if it were a plain plate. This plate is of course thin to keep rotational inertia to a minimum. Plate to hub Connection Secondly the plate and its hub are entirely separate components, the drive being transmitted from one to the other through coil springs interposed between them. These springs are carried within rectangular holes or slots in the hub and plate and arranged with their axes aligned appropriately for transmitting the drive. These dampening springs are heavy coil springs set in a circle around the hub. The hub is driven through these springs. They help to smooth out the torsional vibration (the power pulses from the engine) so that the power flow to the transmission is smooth. In a simple design all the springs may be identical, but in more sophisticated designs the are arranged in pairs located diametrically opposite, each pair having a different rate and different end clearances so that their role is progressive providing increasing spring rate to cater to wider torsional damping The clutch plate is assembled on a splined shaft that carries the rotary motion to the transmission. This shaft is called the clutch shaft, or transmission input shaft. This shaft is connected to the gear box or forms a part of the gear box. Friction Facings or Pads It is the friction pads or facings which actually transmit the power from the fly wheel to hub in the clutch plate and from there to the out put shaft. There are

74 grooves in both sides of the friction-disc facings. These grooves prevent the facings from sticking to the flywheel face and pressure plate when the clutch is disengaged. The grooves break any vacuum that might form and cause the facings to stick to the flywheel or pressure plate. The facings on many friction discs are made of cotton and asbestos fibers woven or molded together and impregnated with resins or other binding agents. In many friction discs, copper wires are woven or pressed into the facings to give them added strength. However, asbestos is being replaced with other materials in many clutches. Some friction discs have ceramic-metallic facings. Such discs are widely used in multiple plate clutches The minimize the wear problems, all the plates will be enclosed in a covered chamber and immersed in an oil medium Such clutches are called wet clutches Multiple Plate Clutches Figure 3..8 The properties of the frictional lining are important factors in the design of the clutches