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Chapter10 clutches and_brakes

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Chapter10 clutches and_brakes

  1. 1. Mechanical Design PRN Childs, University of Sussex Clutches and brakes Mechanical Design
  2. 2. Mechanical Design PRN Childs, University of Sussex Aims • A clutch is a device that permits the smooth, gradual connection of two shafts rotating at different speeds. • A brake enables the controlled dissipation of energy to slow down, stop or control the speed of a system. • This section describes the basic principles of frictional clutches and brakes and outlines design and selection procedures for disc clutches, disc and drum brakes.
  3. 3. Mechanical Design PRN Childs, University of Sussex Learning objectives At the end of this section you should be able to determine: • the primary dimensions for a single disc clutch, • the principal dimensions and number of discs for a multiple disc clutch, • torque capacity for short or long, internal or external brakes, • the configuration of a brake to be self- energising.
  4. 4. Mechanical Design PRN Childs, University of Sussex Introduction • When a rotating machine is started it must be accelerated from rest to the desired speed. • A clutch is a device used to connect or disconnect a driven component from a prime mover such as an engine or motor.
  5. 5. Mechanical Design PRN Childs, University of Sussex Automotive clutches • A familiar application is the use of a clutch between a car engine’s crankshaft and the gearbox. • The need for the clutch arises from the relatively high torque requirement to get a vehicle moving and the low torque output from an internal combustion engine at low levels of rotational speed. • The disconnection of the engine from the drive enables to engine to speed up unloaded to about 1000 rpm where it is generating sufficient torque to drive the transmission. • The clutch can then be engaged, allowing power to be transmitted to the gearbox, transmission shafts and wheels.
  6. 6. Mechanical Design PRN Childs, University of Sussex Brakes • A brake is a device used to reduce or control the speed of a system or bring it to rest.
  7. 7. Mechanical Design PRN Childs, University of Sussex Typical applications for a clutch and a brake MOTOR OR ENGINE COUPLING CLUTCH GEAR BOX COUPLING DRIVEN MACHINE MOTOR OR ENGINE BRAKE CONNECTION TO DRIVEN MACHINE
  8. 8. Mechanical Design PRN Childs, University of Sussex Classification • Clutches and brakes are similar devices providing frictional, magnetic or mechanical connection between two components. • If one component rotates and the other is fixed to a non-rotating plane of reference the device will function as a brake and if both rotate then as a clutch.
  9. 9. Mechanical Design PRN Childs, University of Sussex Force, torque and energy • Whenever the speed or direction of motion of a body is changed there is force exerted on the body. • If the body is rotating, a torque must be applied to the system to speed it up or slow it down. • If the speed changes, so does the energy, either by addition or absorption.
  10. 10. Mechanical Design PRN Childs, University of Sussex Acceleration • The acceleration, α, of a rotating machine is given by • where • T is the torque (N m) and • I is the mass moment of inertia (kg m2). I T =α
  11. 11. Mechanical Design PRN Childs, University of Sussex Mass moment of inertia • The mass moment of inertia can often be approximated by considering an assembly to be made up of a series of cylinders and discs and summing the individual values for the disc and cylinder mass moments of inertia.
  12. 12. Mechanical Design PRN Childs, University of Sussex Mass moments of inertia • The mass moments of inertia for a cylinder and a disc are given by ( )4 i 4 ocylinder rrL 2 1 I −ρπ= 4 odisc Lr 2 1 I ρπ=
  13. 13. Mechanical Design PRN Childs, University of Sussex Clutch or brake location • Torque is equal to the ratio of power and angular velocity. • In other words torque is inversely proportional to angular velocity. • This implies that it is usually advisable to locate the clutch or brake on the highest speed shaft in the system so that the required torque is a minimum. • Size, cost and response time are lower when the torque is lower.
  14. 14. Mechanical Design PRN Childs, University of Sussex Friction clutches • Friction type clutches and brakes are the most common. • Two or more surfaces are pushed together with a normal force to generate a friction torque. • Normally, at least one of the surfaces is metal and the other a high friction material referred to as the lining. • The frictional contact can occur radially, as for a cylindrical arrangement, or axially as in a disc arrangement. ACTUATOR PUSHES DISCS TOGETHER INPUT OUTPUT DRIVING DISC DRIVEN DISC FRICTION MATERIAL
  15. 15. Mechanical Design PRN Childs, University of Sussex Function • The function of a frictional clutch or brake surface material is to develop a substantial friction force when a normal force is applied. • Ideally a material with a high coefficient of friction, constant properties, good resistance to wear and chemical compatibility is required. • Clutches and brakes transfer or dissipate significant quantities of energy and their design must enable the absorption and transfer of this heat without damage to the component parts of the surroundings.
  16. 16. Mechanical Design PRN Childs, University of Sussex Supply • With the exception of high volume automotive clutches and brakes, engineers rarely need to design a clutch or a brake from scratch. • Clutch and brake assemblies can be purchased from specialist suppliers and the engineer’s task is to specify the torque and speed requirements, the loading characteristics and the system inertias and to select an appropriately sized clutch or brake and the lining materials.
  17. 17. Mechanical Design PRN Childs, University of Sussex Clutches • The function of a clutch is to permit the connection and disconnection of two shafts, either when both are stationary or when there is a difference in the relative rotational speeds of the shafts. • Clutch connection can be achieved by a number of techniques from direct mechanical friction, electromagnetic coupling, hydraulic or pneumatic means or by some combination.
  18. 18. Mechanical Design PRN Childs, University of Sussex Types of clutch METHOD OF ENGAGEMENT MECHANICAL PNEUMATIC AND HYDRAULIC ELECTRICAL MAGNETIC POSITIVE CONTACT FRICTION OVERRUNNING MAGNETIC FLUID COUPLING SQUARE JAW SPIRAL JAW TOOTHED DISC DRUM CONE ROLLER SPRAG SPRING WOUND MAGNETIC PARTICLE HYSTERESIS EDDY CURRENT DRY FLUID HYDRAULIC ACTUATION METHOD OF
  19. 19. Mechanical Design PRN Childs, University of Sussex Requirements Clutches must be designed principally to satisfy four requirements: • the necessary actuation force should not be excessive, • the coefficient of friction should be constant, • the energy converted to heat must be dissipated, • wear must be limited to provide reasonable clutch life.
  20. 20. Mechanical Design PRN Childs, University of Sussex Positive contact clutches • Positive contact clutches have teeth or serrations, which provide mechanical interference between mating components.
  21. 21. Mechanical Design PRN Childs, University of Sussex Over-running clutches • Over-running clutches operate automatically based on the relative velocity of the mating components. • They allow relative motion in one direction only. • If the rotation attempts to reverse the constituent components of the clutch grab the shaft and lock up. • Applications include backstops, indexing and freewheeling. • The range of overrunning clutches the simple ratchet and pawl, roller, sprag and spring wound clutches.
  22. 22. Mechanical Design PRN Childs, University of Sussex Ratchet and pawl clutch
  23. 23. Mechanical Design PRN Childs, University of Sussex Roller clutch
  24. 24. Mechanical Design PRN Childs, University of Sussex Sprag clutch
  25. 25. Mechanical Design PRN Childs, University of Sussex Centrifugal clutches • Centrifugal clutches engage automatically when the shaft speed exceeds some critical value. • Friction elements are forced radially outwards and engage against the inner radius of a mating cylindrical drum. • Common applications of centrifugal clutches include chainsaws, overload-releases and go-karts. RETAINING SPRING DRIVING ELEMENTDRIVEN ELEMENT SHOE
  26. 26. Mechanical Design PRN Childs, University of Sussex Clutch selection criteria (Neale (1994)) Machine tool gearboxes. Numerical control machine tools. Compact. Low wear.Magnetic Electric motor drives. Industrial diesel drives. Automatic engagement at a critical speed.Centrifugal clutch Machine tool head stocks. Motorcycles. The power transmitted can be increased by using more plates allowing a reduction in diameter. Multiple disc clutch Automobile drives.Used when diameter is not restricted. Simple construction Single disc clutch Contractor’s plant. Feed drives for machine tools. Embodies the mechanical principle of the wedge which reduces the axial force required to transmit a given torque. Cone clutch One way operation.One way clutch. Rollers ride up ramps and drive by wedging into place. Roller One way operation. e.g. backstop for hoists. One way clutch. Profiled elements jam against the outer edge to provide drive. High torque capacity. Sprag TYPICAL APPLICATIONSCHARACTERISTICSTYPE OF CLUTCH
  27. 27. Mechanical Design PRN Childs, University of Sussex Clutch design • Clutches are rarely designed from scratch. • Either an existing design is available and is being modified for a new application or a clutch can be bought in from a specialist manufacturer. • In the latter case the type, size and the materials for the clutch lining must be specified. • This requires determination of the system characteristics such as speed, torque, loading characteristic and operating temperatures. • Many of these factors have been lumped into a multiplier called a service factor.
  28. 28. Mechanical Design PRN Childs, University of Sussex Service factor • A lining material is typically tested under steady conditions using an electric motor drive. • The torque capacity obtained from this test is then de-rated by the service factor according to the particular application to take account of vibrations and loading conditions.
  29. 29. Mechanical Design PRN Childs, University of Sussex Disc clutch PRESSURE PLATE HELICAL SPRINGS OR CASTING PRESSING THRUST BEARING MOVE AXIALLY TO DISENGAGE SPLINE OR KEYED DRIVE CLUTCH PLATE DRIVE SPIGOT BEARING FLYWHEEL CASTING FRICTION PLATE CRANKSHAFT OR MOTOR SHAFT DRIVEN OR GEARBOX END COVER
  30. 30. Mechanical Design PRN Childs, University of Sussex Disc clutch Engaged Disengaged PRESSURE PLATE HELICAL SPRINGS OR CASTING PRESSING THRUST BEARING MOVE AXIALLY TO DISENGAGE SPLINE OR KEYED DRIVE CLUTCH PLATE DRIVE SPIGOT BEARING FLYWHEEL CASTING FRICTION PLATE CRANKSHAFT OR MOTOR SHAFT DRIVEN OR GEARBOX END COVER PLATE DRIVE MOTOR SHAFT CRANKSHAFT OR MOVE AXIALLY GEARBOX END DRIVEN OR KEYED DRIVE SPLINE OR TO ENGAGE CLUTCH PLATE PRESSURE
  31. 31. Mechanical Design PRN Childs, University of Sussex Multiple disc clutch BUSH OIL SUPPLY FOR HYDRAULIC CYLINDER BEARING DRIVEN END DRIVING END DRIVEN DISCS DRIVING DISCS HYDRAULIC CYLINDER. PRESSURISE TO ENGAGE CLUTCH SEALS SHAFT SPLINED
  32. 32. Mechanical Design PRN Childs, University of Sussex Disc clutches • Disc clutches can consist of single or multiple discs. • Generally multiple disc clutches enable greater torque capacity but are harder to cool. • Frictional clutches can be run dry or wet using oil. • Typical coefficients of friction are 0.07 for a wet clutch and 0.45 for a dry clutch.
  33. 33. Mechanical Design PRN Childs, University of Sussex Running a clutch wet • While running a clutch wet in oil reduces the coefficient of friction it enhances heat transfer and the potential for cooling of the components. • The expedient solution to the reduction of the friction coefficient is to use more discs and hence the use of multiple disc clutches.
  34. 34. Mechanical Design PRN Childs, University of Sussex Assumptions • Two basic assumptions are used in the development of procedures for disc clutch design based upon a uniform rate of wear at the mating surfaces or a uniform pressure distribution between the mating surfaces. • The equations for both of these methods are outlined in this section.
  35. 35. Mechanical Design PRN Childs, University of Sussex Elemental annular ring • The area of an elemental annular ring on a disc clutch is δA=2πrδr. • Now F=pA, where p is the assumed uniform interface pressure, so δF=2πrpδr. r rδ
  36. 36. Mechanical Design PRN Childs, University of Sussex Normal force • For the disc the normal force acting on the entire face is • Note that F is also the necessary force required to clamp the clutch discs together. o i o i r r 2 r r 2 r p2rpdr2F π=π= ( )2 i 2 o rrpF −π=
  37. 37. Mechanical Design PRN Childs, University of Sussex Friction torque • The friction torque δT that can be developed on an elemental ring is the product of the elemental normal force, given by µδF and the radius: δT=rµδF=2µπr2pδr • where µ is the coefficient of friction which models the less than ideal frictional contact which occurs between two surfaces.
  38. 38. Mechanical Design PRN Childs, University of Sussex Total torque • The total torque is given by integration between the limits of the annular ring, ri and ro: • This equation represents the torque capacity of a clutch with a single frictional interface. In practice clutches use an even number of frictional surfaces. ( )3 i 3 o r r r r 3 2 rrp 3 2 3 r p2pdrr2T o i o i −µπ=µπ=µπ=
  39. 39. Mechanical Design PRN Childs, University of Sussex Torque capacity with N faces • For a clutch with N faces the torque capacity is given by: ( )3 i 3 o rrNp 3 2 T −µπ=
  40. 40. Mechanical Design PRN Childs, University of Sussex Torque capacity • Substituting for the pressure, p, gives an equation for the torque capacity as a function of the axial clamping force. − − µ= 2 i 2 o 3 i 3 o rr rr FN 3 2 T
  41. 41. Mechanical Design PRN Childs, University of Sussex Uniform wear • The equations assuming uniform wear are developed below. • The wear rate is assumed to be proportional to the product of the pressure and velocity. So prω = constant
  42. 42. Mechanical Design PRN Childs, University of Sussex Maximum pressure • For a constant angular velocity the maximum pressure will occur at the smallest radius. pmaxriω = constant • Eliminating the angular velocity and constant gives a relationship for the pressure as a function of the radius: r r pp i max=
  43. 43. Mechanical Design PRN Childs, University of Sussex Dynamic friction coefficients, permissible contact pressures and temperature limits 0.10 - 0.14-Graphite/resin 0.10 - 0.17-Paper based 2600.69- 1.7250.03 - 0.060.15 - 0.25Cast iron 900.345 - 0.620.12 - 0.160.20 - 0.45Wood 800.055 - 0.10.15 - 0.250.30 - 0.50Cork 230 - 6801.035 -2.070.05 - 0.080.15 - 0.45Sintered metal 200 - 2600.345 - 0.690.08 - 0.100.25 - 0.45Woven materials 200 - 2601.035 - 2.070.06 - 0.100.25 - 0.45Moulded compounds T (oC)pmax (MN/m)µoilµdryMATERIAL
  44. 44. Mechanical Design PRN Childs, University of Sussex Axial force • The elemental axial force on an elemental annular ring is given by • Integrating to give the total axial force: rpr2F δπ=δ ( )ioimax r r i max r r rrrp2rdr r r p2prdr2F o i o i −π=π=π=
  45. 45. Mechanical Design PRN Childs, University of Sussex Torque • The elemental torque is given by • Rearranging gives FrT δµ=δ ( )2 i 2 oimax r r imax rrrprdrrp2T o i −µπ=µπ= ( )ioi max rrr2 F p −π =
  46. 46. Mechanical Design PRN Childs, University of Sussex Torque • Substituting gives • For N frictional surfaces ( )io io 2 i 2 o rr 2 F rr rr 2 F T + µ = − −µ = ( )2 i 2 oimax r r imax rrNrprdrrNp2T o i −µπ=µπ= ( )io rr 2 NF T + µ =
  47. 47. Mechanical Design PRN Childs, University of Sussex maximum torque for any outer radius • By differentiating with respect to ri and equating the result to zero, the maximum torque for any outer radius ro is found to occur when • This useful formula can be used to set the inner radius if the outer radius is constrained to a particular value. oi r1/3r =
  48. 48. Mechanical Design PRN Childs, University of Sussex Design • Clutches are usually designed based on uniform wear. • The uniform wear assumption gives a lower torque capacity clutch than the uniform pressure assumption. • The preliminary design procedure for disc clutch design requires the determination of the torque and speed, specification of space limitations, selection of materials, and the selection of principal radii, ro and ri. • Common practice is to set the value of ri between 0.45ro and 0.8ro.
  49. 49. Mechanical Design PRN Childs, University of Sussex Procedure The procedure for determining the initial geometry is itemised below. 1) Determine the service factor. 2) Determine the required torque capacity, T=power/ω. 3) Determine the coefficient of friction µ. 4) Determine the outer radius ro. 5) Find the inner radius ri. 6) Find the axial actuation force required.
  50. 50. Mechanical Design PRN Childs, University of Sussex Materials • The material used for clutch plates is typically grey cast iron or steel. • The friction surface will consist of a lined material which may be moulded, woven, sintered or solid. • Moulded linings consist of a polymeric resin used to bind powdered fibrous material and brass and zinc chips.
  51. 51. Mechanical Design PRN Childs, University of Sussex Example • A clutch is required for transmission of power between a four cylinder internal combustion engine and a small machine. • Determine the radial dimensions for a single face dry disc clutch with a moulded lining which should transmit 5 kW at 1800 rpm. • Base the design on the uniform wear assumption.
  52. 52. Mechanical Design PRN Childs, University of Sussex Example
  53. 53. Mechanical Design PRN Childs, University of Sussex Solution • A service factor of two should be used. • The design will therefore be undertaken using a power of 2×5 kW=10 kW. 3.73.43.23.0Stone crushers, roll mills, heavy mixers, single cylinder compressors overloads, cycling, high inertia starts, high power, pulsating power source 3.22.92.72.5Presses, Punches, piston pumps, Cranes, hoists Frequent start- stops, 2.72.42.22.0Larger conveyor belts, larger machines, reciprocating pumps with some irregularity of load up to 1.5 times nominal power 2.72.42.01.8Light machinery for wood, metal and textiles, conveyor belts Steady power source 2.21.91.71.5Belt drive, small generators, centrifugal pumps, fans, machine tools Steady power source, steady load, no shock or overload SINGLE CYLINDER ENGINE IC ENGINES (2 OR 3 CYLINDER S) IC ENGINES (4 TO 6 CYLINDER S). MEDIUM TO LARGE ELECTRIC MOTORS SMALL ELECTRIC MOTORS, TURBINE TYPICAL DRIVEN SYSTEM DESCRIPTION OF GENERAL SYSTEM TYPE OF DRIVER
  54. 54. Mechanical Design PRN Childs, University of Sussex Solution cont. • The torque is given by • From Table 10.3 taking mid range values for the coefficient friction and the maximum permissible pressure for moulded linings gives µ=0.35 and pmax=1.55 MN/m2. ( ) mN53 60/21800 10000Power T = π× = ω =
  55. 55. Mechanical Design PRN Childs, University of Sussex Solution cont. 0.10 - 0.14-Graphite/resin 0.10 - 0.17-Paper based 2600.69- 1.7250.03 - 0.060.15 - 0.25Cast iron 900.345 - 0.620.12 - 0.160.20 - 0.45Wood 800.055 - 0.10.15 - 0.250.30 - 0.50Cork 230 - 6801.035 -2.070.05 - 0.080.15 - 0.45Sintered metal 200 - 2600.345 - 0.690.08 - 0.100.25 - 0.45Woven materials 200 - 2601.035 - 2.070.06 - 0.100.25 - 0.45Moulded compounds T (oC)pmax (MN/m)µoilµdryMATERIAL
  56. 56. Mechanical Design PRN Childs, University of Sussex Solution cont. • Taking oi r3/1r = ( ) 3 omax 3 o 3 omax 2 o 2 omaxo rp4/27r 3 1 rp1/3 r1/3rpr1/3T =− =−=
  57. 57. Mechanical Design PRN Childs, University of Sussex Solution cont. m0.04324 4/27101.550.35 53.05 4/27p T r 1/3 6 1/3 max o = ×× == m02497.0r3/1r oi ==
  58. 58. Mechanical Design PRN Childs, University of Sussex Solution cont. • So the clutch consists of a disc of inner and outer radius 25 mm and 43 mm respectively, with a moulded lining having a coefficient of friction value of 0.35 and a maximum permissible contact pressure of 1.55 MPa and an actuating force of 4.4 kN. ( ) ( ) N4443 0.024970.04324101.550.024972 rrpr2F 6 iomaxi =−××× =−=
  59. 59. Mechanical Design PRN Childs, University of Sussex Example • A disc clutch, running in oil, is required for a vehicle with a four-cylinder engine. The design power for initial estimation of the clutch specification is 90 kW at 4500 rpm. Determine the radial dimensions and actuating force required. Base the design on the uniform wear assumption.
  60. 60. Mechanical Design PRN Childs, University of Sussex Solution • From Table 10.2 a service factor of 2.7 should be used to account for starts and stops and the four cylinder engine. The design will therefore be undertaken using a power of 2.7×90 = 243 kW. 3.73.43.23.0Stone crushers, roll mills, heavy mixers, single cylinder compressors overloads, cycling, high inertia starts, high power, pulsating power source 3.22.92.72.5Presses, Punches, piston pumps, Cranes, hoists Frequent start- stops, 2.72.42.22.0Larger conveyor belts, larger machines, reciprocating pumps with some irregularity of load up to 1.5 times nominal power 2.72.42.01.8Light machinery for wood, metal and textiles, conveyor belts Steady power source 2.21.91.71.5Belt drive, small generators, centrifugal pumps, fans, machine tools Steady power source, steady load, no shock or overload SINGLE CYLINDER ENGINE IC ENGINES (2 OR 3 CYLINDER S) IC ENGINES (4 TO 6 CYLINDER S). MEDIUM TO LARGE ELECTRIC MOTORS SMALL ELECTRIC MOTORS, TURBINE TYPICAL DRIVEN SYSTEM DESCRIPTION OF GENERAL SYSTEM TYPE OF DRIVER
  61. 61. Mechanical Design PRN Childs, University of Sussex Solution cont. • The torque is given by • From Table 10.3 taking mid range values for the coefficient friction and the maximum permissible pressure for moulded linings gives µ=0.35 and pmax=1.55 MN/m2. ( ) mN7.515 60/24500 243000Power T = π× = ω =
  62. 62. Mechanical Design PRN Childs, University of Sussex Solution cont. • Taking oi r3/1r = m0.07325 4/27101.5520.35 515.7 4/27Np T r 1/3 6 1/3 max o = ××× == m04229.0r3/1r oi ==
  63. 63. Mechanical Design PRN Childs, University of Sussex Solution cont. • So the clutch consists of a disc of inner and outer radius 42.3 mm and 73.3 mm respectively, with a moulded lining having a coefficient of friction value of 0.35 and a maximum permissible contact pressure of 1.55 MPa and an actuating force of 25.5 kN. ( ) ( ) N255000.042290.07325101.55 0.0422922rrprN2F 6 iomaxi =−× ×××=−=
  64. 64. Mechanical Design PRN Childs, University of Sussex Example • A multiple disc clutch, running in oil, is required for a motorcycle with a three- cylinder engine. • The power demand is 75 kW at 8500 rpm. • The preliminary design layout indicates that the maximum diameter of the clutch discs should not exceed 100 mm.
  65. 65. Mechanical Design PRN Childs, University of Sussex Example cont. • In addition previous designs have indicated that a moulded lining with a coefficient of friction of 0.068 in oil and a maximum permissible pressure of 1.2 MPa is reliable. • Within these specifications determine the radii for the discs, the number of discs required and the clamping force.
  66. 66. Mechanical Design PRN Childs, University of Sussex Solution • The torque is given by ( ) mN286.5 /6028500 750003.4 PowerfactorService T = × × = × =
  67. 67. Mechanical Design PRN Childs, University of Sussex Solution cont. • Select the outer radius to be the largest possible, i.e. ro=50 mm. • Using • ri=28.87 mm. oi r3/1r =
  68. 68. Mechanical Design PRN Childs, University of Sussex Solution cont. • The number of frictional surfaces, N • This must be an even number, so the number of frictional surfaces is taken as N=24. This requires thirteen driving discs and twelve driven discs to implement. ( ) ( ) 23.23 0.028870.050.0680.02887101.2 286.5 rrrp T N 226 2 i 2 oimax = −××× = − =
  69. 69. Mechanical Design PRN Childs, University of Sussex Solution cont. • The clamping force can be calculated: ( ) ( ) N4452 02887.005.024068.0 5.2862 rrN T2 F io = +× × = +µ =
  70. 70. Mechanical Design PRN Childs, University of Sussex Brakes • The basic function of a brake is to absorb kinetic energy and dissipate it in the form of heat.
  71. 71. Mechanical Design PRN Childs, University of Sussex Example • An idea of the magnitude of energy that must be dissipated can be obtained from considering the familiar example of a car undergoing an emergency stop in seven seconds from 60 mph (96 km/h).
  72. 72. Mechanical Design PRN Childs, University of Sussex Solution • If the car’s mass is 1400 kg and assuming that 65% of the car’s weight is loaded onto the front axles during rapid braking then the load on the front axle is N892765.081.91400 =××
  73. 73. Mechanical Design PRN Childs, University of Sussex Solution cont. • This will be shared between two brakes so the energy that must be absorbed by one brake is ( )2 f 2 i VVm 2 1 E −= kJ161.80 3600 1096 0.5 9.81 8927 2 1 E 2 23 =− × ×××=
  74. 74. Mechanical Design PRN Childs, University of Sussex Solution cont. • If the car brakes uniformly in seven seconds then the heat that must be dissipated is 161.8×103/7=23.1 kW. • From your experience of heat transfer from say 1 kW domestic heaters you will recognise that this is a significant quantity of heat to transfer away from the relatively compact components that make up brake assemblies.
  75. 75. Mechanical Design PRN Childs, University of Sussex Heat transfer • Convective heat transfer can be modelled by Fourier’s equation: • This equation indicates that the ability of a brake to dissipate the heat generated increases as the surface area increases or as the heat transfer coefficient rises. • For air, the heat transfer coefficient is usually dependent on the local flow velocity and on the geometry. • A method often used for disc brakes to increase both the surface area and the local flow is to machine multiple axial or radials holes in the disc. ( )fs TThAThAQ −=∆=
  76. 76. Mechanical Design PRN Childs, University of Sussex Types of brake METHOD OF AUTOMATIC MAGNETIC AND HYDRAULIC MECHANICAL ELECTRICAL PNEUMATIC ELECTRICALLY ON FULL DISC DRUM DISC ELECTRICALLY OFF CALIPER DISC LONG SHOE SHORT SHOE BAND FRICTION METHOD OF ACTUATION ENGAGEMENT
  77. 77. Mechanical Design PRN Childs, University of Sussex Comparative table of brake performance Vehicles (rear axles on passenger cars) Good if sealedUnstable if humid, ineffective if wet LowHighHigher than external brake Internal drum brake (two leading shoes) Vehicles (rear axles on passenger cars) Good if sealedUnstable if humid, ineffective if wet MediumMediumHigher than external brake Internal drum brake (leading trailing edge) Mills, elevators, winders GoodUnstable if humid, poor if wet MediumMediumLowExternal drum brake (leading trailing edge) Winches, hoist, excavators, tractors GoodUnstable but still effective LowHighLowDifferential band brake TYPICAL APPLICATIO NS DUST AND DIRT DRYNESSSTABILITYBRAKE FACTOR MAXIMUM OPERAT ING TEMPER ATURE TYPE OF BRAKE
  78. 78. Mechanical Design PRN Childs, University of Sussex Self-energising brakes • Brakes can be designed so that, once engaged the actuating force applied is assisted by the braking torque. • This kind of brake is called a self- energising brake and is useful for braking large loads.
  79. 79. Mechanical Design PRN Childs, University of Sussex Self-locking • Great care must be exercised in brake design. • It is possible and sometimes desirable to design a brake, which once engaged, will grab and lock up (called self-locking action).
  80. 80. Mechanical Design PRN Childs, University of Sussex Disc brakes • Disc brakes are familiar from automotive applications where they are used extensively for vehicle wheels. • These typically consist of a cast iron disc, bolted to the wheel hub. • This is sandwiched between two pads actuated by pistons supported in a calliper mounted on the stub shaft. DISC WHEEL HUB STUB-AXLE CALIPER PADS HYDRAULIC CYLINDER SEAL SEAL
  81. 81. Mechanical Design PRN Childs, University of Sussex Torque capacity • With reference to the figure, the torque capacity per pad is given by • where re is an effective radius. ri ro er r ANNULAR PAD CIRCULAR PAD θ R FF eFrT µ=
  82. 82. Mechanical Design PRN Childs, University of Sussex Actuating force • The actuating force assuming constant pressure is given by • or assuming uniform wear by 2 rr pF 2 i 2 o av − θ= ( )ioimax rrrpF −θ=
  83. 83. Mechanical Design PRN Childs, University of Sussex • The relationship between the average and the maximum pressure for the uniform wear assumption is given by oi oi max av r/r1 r/r2 p p + =
  84. 84. Mechanical Design PRN Childs, University of Sussex Effective radius • For an annular disc brake the effective radius is given, assuming constant pressure, by • and assuming uniform wear by ( ) ( )2 i 2 o 3 i 3 o e rr3 rr2 r − − = 2 rr r oi e + =
  85. 85. Mechanical Design PRN Childs, University of Sussex Circular pad disk brake design values (Fazekas, (1972)) • For circular pads the effective radius is given by re=rδ, where values for δ are given in the table as a function of the ratio of the pad radius and the radial location, R/r 1.8750.9380.5 1.5780.9470.4 1.3670.9570.3 1.2120.9690.2 1.0930.9830.1 1.0001.0000 pmax/pavδ=re/rR/r
  86. 86. Mechanical Design PRN Childs, University of Sussex Actuating force • The actuating force for circular pads can be calculated using: av 2 pRF π=
  87. 87. Mechanical Design PRN Childs, University of Sussex Example • A calliper brake is required for the front wheels of a sport’s car with a braking capacity of 820 N m for each brake. • Preliminary design estimates have set the brake geometry as ri=100 mm, ro=160 mm and θ=45o. • A pad with a coefficient of friction of 0.35 has been selected. • Determine the required actuating force and the average and maximum contact pressures.
  88. 88. Mechanical Design PRN Childs, University of Sussex Solution • The torque capacity per pad = 820/2 = 410 N m. • The effective radius is m13.0 2 16.01.0 re = + =
  89. 89. Mechanical Design PRN Childs, University of Sussex Solution cont. • The actuating force is given by • The maximum contact pressure is kN011.9 13.035.0 410 r T F e = × = µ = ( ) ( ) ( ) 2 3 ioi max MN/m1.912 0.10.160.12 /3645 109.011 rrr F p = −××× × = − =
  90. 90. Mechanical Design PRN Childs, University of Sussex Solution cont. • The average pressure is given by 2 oi oi maxav m/MN471.1 r/r1 r/r2 pp = + =
  91. 91. Mechanical Design PRN Childs, University of Sussex Drum brakes • Drum brakes apply friction to the external or internal circumference of a cylinder. • A drum brake consists of the brake shoe, which has the friction material bonded to it, and the brake drum. • For braking, the shoe is forced against the drum developing the friction torque. • Drum brakes can be divided into two groups depending on whether the brake shoe is external or internal to the drum. • A further classification can be made in terms of the length of the brake shoe: short, long or complete band.
  92. 92. Mechanical Design PRN Childs, University of Sussex Short and long shoe internal drum brakes • Short shoe internal brakes are used for centrifugal brakes that engage at a particular critical speed. • Long shoe internal drum brakes are used principally in automotive applications. • Drum brakes can be designed to be self- energising. • Once engaged the friction force increases the normal force non-linearly, increasing the friction torque as in a positive feedback loop.
  93. 93. Mechanical Design PRN Childs, University of Sussex Stability • One problem associated some drum brakes is stability. • If the brake has been designed so that the braking torque is not sensitive to small changes in the coefficient of friction, which would occur if the brake is worn or wet, then the brake is said to be stable. • If a small change in the coefficient of friction causes a significant change to the braking torque the brake is unstable and will tend to grab if the friction coefficient rises or the braking torque will drop noticeably if the friction coefficient reduces.
  94. 94. Mechanical Design PRN Childs, University of Sussex Short shoe external drum brakes a b c r SHOE DRUM PIVOT BRAKE LEVER Fa nF fFy x θ ω
  95. 95. Mechanical Design PRN Childs, University of Sussex Short shoe external drum brakes • If the included angle of contact between the brake shoe and the brake drum is less than 45o, the force between the shoe and the drum is relatively uniform and can be modelled by a single concentrated load Fn at the centre of the contact area. • If the maximum permissible pressure is pmax the force Fn can be estimated by Fn = pmaxrθw
  96. 96. Mechanical Design PRN Childs, University of Sussex Short shoe external drum brakes • The frictional force, Ff, is given by Ff = µFn • where µ is the coefficient of friction. • The torque on the brake drum is T = Ffr = µFnr
  97. 97. Mechanical Design PRN Childs, University of Sussex Short shoe external drum brakes • Summing moments, for the shoe arm, about the pivot gives: =+−= 0cFbFaFM fnapivot a cb F a cFbF F n fn a µ− = − =
  98. 98. Mechanical Design PRN Childs, University of Sussex Short shoe external drum brakes • Resolving forces gives the reactions at the pivot: Rx=-Ff Ry=Fa-Fn a b c r SHOE DRUM PIVOT BRAKE LEVER Fa nF fFy x θ ω
  99. 99. Mechanical Design PRN Childs, University of Sussex Short shoe external drum brakes • Note that for the configuration and direction of rotation shown, the friction moment µFnc adds or combines with the actuating moment aFa. • Once the actuating force is applied the friction generated at the shoe acts to increase the braking torque. • This kind of braking action is called self- energising.
  100. 100. Mechanical Design PRN Childs, University of Sussex Short shoe external drum brakes • If the brake direction is reversed the friction moment term µFnc becomes negative and the applied load Fa must be maintained to generate braking torque. • This combination is called self de- energising.
  101. 101. Mechanical Design PRN Childs, University of Sussex Short shoe external drum brakes • From • note that if the brake is self-energising and if µc>b then the force required to actuate the brake is zero or negative and the brake action is called self-locking. • If the shoe touches the drum it will grab and lock. • This is usually undesirable with exceptions being hoist stops or over-running clutch type applications. a cb F a cFbF F n fn a µ− = − =
  102. 102. Mechanical Design PRN Childs, University of Sussex Long shoe external drum brakes • If the included angle of contact between the brake shoe and the drum is greater than 45o then the pressure between the shoe and the brake lining cannot be regarded as uniform and the approximations made for the short shoe brake analysis are inadequate. • Most drum brakes use contact angles greater than 90o.
  103. 103. Mechanical Design PRN Childs, University of Sussex Long shoe external drum brakes • For a single block brake the force exerted on the drum by the brake shoe must be supported by the bearings. • To balance this load and provide a compact braking arrangement two opposing brake shoes are usually used in a calliper arrangement. PIVOT DRUM r LEVER SHOE Fa 1 2 a b fF nF θθ θ ω SHOE DRUM 1 2 r Fa aF ω θ θθ
  104. 104. Mechanical Design PRN Childs, University of Sussex Long shoe external drum brakes • The braking torque T is given by • This is based on the assumption that the local pressure p at an angular location θ is related to the maximum pressure, pmax, by ( ) ( )21 max max2 coscos sin p wrT θ−θ θ µ= ( )max max sin sinp p θ θ =
  105. 105. Mechanical Design PRN Childs, University of Sussex Long shoe external drum brakes • With the direction of rotation shown (i.e. the brake is self- energising), the magnitude of the actuation force is given by a MM F fn a − = PIVOT DRUM r LEVER SHOE Fa 1 2 a b fF nF θθ θ ω
  106. 106. Mechanical Design PRN Childs, University of Sussex Long shoe external drum brakes • The normal and frictional moments can be determined using ( ) ( ) ( )θ−θ−θ−θ θ = 1212 max max n 2sin2sin 4 1 2 1 sin wrbp M ( ) ( ) ( )θ−θ+θ−θ θ µ = 1221 max max f 2cos2cos 4 b coscosr sin wrp M
  107. 107. Mechanical Design PRN Childs, University of Sussex Long shoe external drum brakes • If the direction of rotation for the drum shown is reversed, the brake becomes self de-energising and the actuation force is given by PIVOT DRUM r LEVER SHOE Fa 1 2 a b fF nF θθ θ ω a MM F fn a + =
  108. 108. Mechanical Design PRN Childs, University of Sussex Example • Design a long shoe drum brake to produce a friction torque of 75 N m to stop a drum rotating at 140 rpm. • Initial design calculations have indicated that a shoe lining with µ=0.25 and using a value of pmax=0.5×106 N/m2 in the design will give suitable life.
  109. 109. Mechanical Design PRN Childs, University of Sussex Solution • First propose trial values for the brake geometry, say r=0.1 m, b=0.2 m, a=0.3 m, θ1=30o, θ2=150o.
  110. 110. Mechanical Design PRN Childs, University of Sussex Solution cont. • Solving for the width of the shoe, • Select the width to be 35 mm as this is a standard size. ( ) ( ) ( ) m0.0346 cos150cos30100.50.10.25 75sin90 coscospr sinT w 62 21max 2 max = −××× = − =
  111. 111. Mechanical Design PRN Childs, University of Sussex Solution cont. • The actual maximum pressure experienced, will be: 26 max m/N494900 035.0 0346.0 105.0p =×=
  112. 112. Mechanical Design PRN Childs, University of Sussex Solution cont. • The moment of the normal force with respect to the shoe pivot is: ( ) mN512.8sin60sin300 4 1 360 2 120 2 1 sin90 100.49490.20.10.035 M 6 n =−−× × ×××× =
  113. 113. Mechanical Design PRN Childs, University of Sussex Solution cont. • The moment of the frictional forces with respect to the shoe pivot is: =75 N m ( ) ( )−+− × ×××× = cos60cos300 4 0.2 cos150cos300.1 sin90 100.49490.10.0350.25 M 6 f
  114. 114. Mechanical Design PRN Childs, University of Sussex Solution cont. • The actuation force is N1459 3.0 758.512 a MM F fn a = − = − =
  115. 115. Mechanical Design PRN Childs, University of Sussex Double long shoe external drum brake • For the double long shoe external drum brake illustrated, the left hand shoe is self- energising and the frictional moment reduces the actuation load. • The right hand shoe, however, is self de-energising and its frictional moment acts to reduce the maximum pressure which occurs on the right hand brake shoe. SHOE DRUM 1 2 r Fa aF ω θ θθ
  116. 116. Mechanical Design PRN Childs, University of Sussex Normal and frictional moments • The normal and frictional moments for a self-energising and self de-energising brake are related by max ' maxn n p pM 'M = max ' maxf f p pM 'M =
  117. 117. Mechanical Design PRN Childs, University of Sussex Example • For the double long shoe external drum brake illustrated in following figure determine the limiting force on the lever such that the maximum pressure on the brake lining does not exceed 1.4 MPa and determine the torque capacity of the brake. 130 o 20 o F a R100 115120 79.37 200 50 20
  118. 118. Mechanical Design PRN Childs, University of Sussex Example cont. • The face width of the shoes is 30 mm and the coefficient of friction between the shoes and the drum can be taken as 0.28.
  119. 119. Mechanical Design PRN Childs, University of Sussex Solution • First it is necessary to calculate values for θ1 and θ2 as these are not indicated directly on the diagram. o1o 1 54.10 120 20 tan20 =−=θ − o1oo 2 5.140 120 20 tan13020 =−+=θ −
  120. 120. Mechanical Design PRN Childs, University of Sussex Solution cont. • The maximum value of sinθ would be sin90=1. • The distance between the pivot and the drum centre, m1217.012.002.0b 22 =+=
  121. 121. Mechanical Design PRN Childs, University of Sussex Solution cont. • The normal moment is given by ( ) ( ) ( )θ−θ−θ−θ θ = 1212 max max n 2sin2sin 4 1 2 1 sin wrbp M ( ) ( ) mN751.1 sin21.08sin281 4 1 360 2 10.54140.5 2 1 sin90 101.40.12170.10.03 6 =−−×− × ×××× =
  122. 122. Mechanical Design PRN Childs, University of Sussex Solution cont. ( ) ( ) ( )θ−θ+θ−θ θ µ = 1221 max max f 2cos2cos 4 b coscosr sin wrp M ( ) ( ) mN179.8 cos21.08cos281 4 0.1217 cos140.5cos10.540.1 sin90 101.40.10.030.28 6 =−+− × ×××× =
  123. 123. Mechanical Design PRN Childs, University of Sussex Solution cont. • The orthogonal distance between the actuation force and the pivot, a=0.12+0.115+0.05=0.285 m. • The actuation load on the left hand shoe is given by N2004 285.0 8.1791.751 a MM F fn shoelefta = − = − =
  124. 124. Mechanical Design PRN Childs, University of Sussex Solution cont. • The torque contribution from the left hand shoe is given by =206.4 N m. ( ) ( )21 max max2 shoeleft coscos sin p wrT θ−θ θ µ= ( )5.140cos54.10cos104.11.003.028.0 62 −××××=
  125. 125. Mechanical Design PRN Childs, University of Sussex Solution cont. • The actuation force on the right hand shoe can be determined by considering each member of the lever mechanism as a free body. C V V C H C H C V B H B V BH B H A V A F 79.37 50 200 F=501 N F a left shoe F a right shoe 14.04 o 200 50 =2004 N =2065 N
  126. 126. Mechanical Design PRN Childs, University of Sussex Solution cont. F-AV+BV=0. AH=BH. BH=CH AH=CH. 0.2F=0.05BH, F=BH/4. BH=2004 N. F=2004/4=501 N. C V V C H C H C V B H B V BH B H A V A F 79.37 50 200 F=501 N F a left shoe F a right shoe 14.04 o 200 50 =2004 N =2065 N
  127. 127. Mechanical Design PRN Childs, University of Sussex Solution cont. • So the limiting lever force is F=501 N. CV=0, BV=0. • The actuating force for the right hand lever is the resultant of F and BH. • The resultant angle is given by tan-1(0.05/0.2)=14.04o. N2065 04.14cos 2004 F shoerighta ==
  128. 128. Mechanical Design PRN Childs, University of Sussex Solution cont. • The orthogonal distance between the actuation force vector and the pivot, is given by • a=(0.235- 0.01969tan14.04)cos 14.04 = 0.2232 m 14.04 o 14.04 o 223.2 235 19.69
  129. 129. Mechanical Design PRN Childs, University of Sussex Solution cont. • The normal and frictional moments for the right hand shoe can be determined using 6 ' max max ' maxn n 104.1 p1.751 p pM 'M × == 6 ' max max ' maxf f 104.1 p8.179 p pM 'M × ==
  130. 130. Mechanical Design PRN Childs, University of Sussex Solution cont. • For the right hand shoe the maximum pressure can be determined from 0.2232101.4 179.8p751.1p 2065 a 'M'M F 6 ' max ' max fn shoerighta ×× − == + = 26' max m/N10130.1p ×=
  131. 131. Mechanical Design PRN Childs, University of Sussex Solution cont. • The torque contribution from the right hand shoe is =166.6 N m ( ) ( )21 max ' max2 shoeright coscos sin p wrT θ−θ θ µ= ( )= × × × × −028 003 01 113 10 1054 14052 6 . . . . cos . cos .
  132. 132. Mechanical Design PRN Childs, University of Sussex Solution cont. • The total torque is given by Ttotal=Tleft shoe+Tright shoe= 206.4+166.6=373 N m
  133. 133. Mechanical Design PRN Childs, University of Sussex Long Shoe Internal Drum Brakes • Most drum brakes use internal shoes that expand against the inner radius of the drum. Long shoe internal drum brakes are principally used in automotive applications. • An automotive drum brake typically comprises two brake shoes and linings supported on a back plate bolted to the axle casing. • The shoes are pivoted at one end on anchor pins or abutments fixed onto the back plate.
  134. 134. Mechanical Design PRN Childs, University of Sussex Long Shoe Internal Drum Brakes c r 1 r 2 DRUM ROTATION RETRACTION SPRING ANCHOR PINS OR BRAKE LINING HYDRAULIC CYLINDER p SHOE LEADING TRAILING SHOE ABUTMENTS HEEL SHOE TOE SHOE θ θ
  135. 135. Mechanical Design PRN Childs, University of Sussex Long Shoe Internal Drum Brakes • The brake can be actuated by a double hydraulic piston expander, which forces the free ends of the brake apart so that the non-rotating shoes come into frictional contact with the rotating brake drum. • A leading and trailing shoe layout consists of a pair of shoes pivoted at a common anchor point.
  136. 136. Mechanical Design PRN Childs, University of Sussex Leading shoe • The leading shoe is identified as the shoe whose expander piston moves in the direction of rotation of the drum. • The frictional drag between the shoe and the drum will tend to assist the expander piston in forcing the shoe against the drum and this action is referred to as self- energising or the self- servo action of the shoe. c r 1 r 2 DRUM ROTATION RETRACTION SPRING ANCHOR PINS OR BRAKE LINING HYDRAULIC CYLINDER p SHOE LEADING TRAILING SHOE ABUTMENTS HEEL SHOE TOE SHOE θ θ
  137. 137. Mechanical Design PRN Childs, University of Sussex Trailing shoe • The trailing shoe is the one whose expander piston moves in the direction opposed the rotation of the drum. • The frictional force opposes the expander and hence a trailing brake shoe provides less braking torque than an equivalent leading shoe actuated by the same force. c r 1 r 2 DRUM ROTATION RETRACTION SPRING ANCHOR PINS OR BRAKE LINING HYDRAULIC CYLINDER p SHOE LEADING TRAILING SHOE ABUTMENTS HEEL SHOE TOE SHOE θ θ
  138. 138. Mechanical Design PRN Childs, University of Sussex Long Shoe Internal Drum Brakes • The equations developed for external long shoe drum brakes are also valid for internal long shoe drum brakes.
  139. 139. Mechanical Design PRN Childs, University of Sussex Example • Determine the actuating force and the braking capacity for the double internal long shoe brake illustrated. • The lining is sintered metal with a coefficient of friction of 0.32 and the maximum lining pressure is 1.2 MPa. • The drum radius is 68 mm and the shoe width is 25 mm. 15 55 48 16 FaaF 120 o o DRUM SHOE LINING R68 20
  140. 140. Mechanical Design PRN Childs, University of Sussex Solution • As the brake lining angles relative to the pivot, brake axis line, are not explicitly shown on the diagram, they must be calculated. • As θ2>90o, the maximum value of sinθ is sin90=1=(sinθ)max. m05701.0055.0015.0b 22 =+= o 1 745.4=θ o 2 7.124=θ
  141. 141. Mechanical Design PRN Childs, University of Sussex Solution cont. • For this brake with the direction of rotation as shown the right hand shoe is self- energising. 15 55 48 16 FaaF 120 o o DRUM SHOE LINING R68 20
  142. 142. Mechanical Design PRN Childs, University of Sussex Solution cont. • For the right hand shoe: ( ) ( )−−×− × ×××× = sin9.49sin249.4 4 1 360 2 4.745124.7 2 1 1 101.20.057010.0680.025 M 6 n mN8.153=
  143. 143. Mechanical Design PRN Childs, University of Sussex Solution cont. ( ) ( )−+− × ×××× = cos9.49cos249.4 4 0.05701 cos124.7cos4.7450.068 1 101.20.0680.0250.32 M 6 f mN57.1=
  144. 144. Mechanical Design PRN Childs, University of Sussex Solution cont. • The actuating force is 938.9 N. m103.0048.0055.0a =+= N9.938 103.0 1.578.153 a MM F fn a = − = − =
  145. 145. Mechanical Design PRN Childs, University of Sussex Solution cont. • The torque applied by the right hand shoe is given by ( ) ( )21 max max 2 shoeright coscos sin pwr T θ−θ θ µ = ( ) mN69.54cos124.7cos4.745 1 101.20.0680.0250.32 62 =− × ×××× =
  146. 146. Mechanical Design PRN Childs, University of Sussex Solution cont. • The torque applied by the left hand shoe cannot be determined until the maximum operating pressure pmax’ for the left hand shoe has been calculated.
  147. 147. Mechanical Design PRN Childs, University of Sussex Solution cont. • As the left hand shoe is self de-energising the normal and frictional moments can be determined. 6 ' max max ' maxn' n 102.1 p8.153 p pM M × == 6 ' max max ' maxf' f 102.1 p1.57 p pM M × ==
  148. 148. Mechanical Design PRN Childs, University of Sussex Solution cont. • The left hand shoe is self de-energising, so, • Fa=938.9 N as calculated earlier. a MM F fn a + = 103.0102.1 p1.57p8.153 9.938 6 ' max ' max ×× + = 26' max m/N105502.0p ×=
  149. 149. Mechanical Design PRN Childs, University of Sussex Solution cont. • The torque applied by the left hand shoe is given by ( ) ( )21 max ' max 2 shoeleft coscos sin pwr T θ−θ θ µ = ( ) mN31.89cos124.7cos4.745 1 100.55020.0680.0250.32 62 =− × ×××× =
  150. 150. Mechanical Design PRN Childs, University of Sussex Solution cont. • The total torque applied by both shoes is: mN101.431.8969.54 TTT shoeleftshoerighttotal =+ =+=
  151. 151. Mechanical Design PRN Childs, University of Sussex Comment • From this example the advantage in torque capacity of using self- energising brakes is apparent. • Both the left hand and the right hand shoes could be made self-energising by inverting the left hand shoe, having the pivot at the top. 15 55 48 16 FaaF 120 o o DRUM SHOE LINING R68 20
  152. 152. Mechanical Design PRN Childs, University of Sussex Comment cont. • This would be advantageous if rotation occurred in just one direction. • If, however, drum rotation is possible in either direction, it may be more suitable to have one brake self-energising for forward motion and one self-energising for reverse motion.
  153. 153. Mechanical Design PRN Childs, University of Sussex Band brakes • One of the simplest types of braking device is the band brake. • This consists of a flexible metal band lined with a frictional material wrapped partly around a drum. • The brake is actuated by pulling the band against the drum FF1 2 F r a a c θ
  154. 154. Mechanical Design PRN Childs, University of Sussex Band brakes • For the clockwise rotation shown the friction forces increase F1 relative to F2. FF1 2 F r a a c θ
  155. 155. Mechanical Design PRN Childs, University of Sussex Band brakes • The relationship between the tight and slack sides of the band is given by • F1 = tension in the tight side of the band (N), • F2 = tension in the slack side of the band (N), • µ = coefficient of friction, • θ = angle of wrap (rad). µθ = e F F 2 1
  156. 156. Mechanical Design PRN Childs, University of Sussex Point of maximum contact pressure • The point of maximum contact pressure for the friction material occurs at the tight end and is given by: • where w is the width of the band (m). rw F p 1 max =
  157. 157. Mechanical Design PRN Childs, University of Sussex Torque braking capacity • The torque braking capacity is given by ( )rFFT 21 −=
  158. 158. Mechanical Design PRN Childs, University of Sussex Moments • The relationship, for the band brake shown between the applied lever force Fa and F2 can be found by taking moments. 0aFcF 2a =− c a FF 2a = FF1 2 F r a a c θ
  159. 159. Mechanical Design PRN Childs, University of Sussex Self-energising band brakes • The brake configuration shown in the top figure opposite is self- energising for clockwise rotation. • The level of self- energisation can be enhanced by using the differential band brake configuration shown in the bottom figure opposite. FF1 2 F r a a c θ ab c F1 2F Fa θ
  160. 160. Mechanical Design PRN Childs, University of Sussex Summation of the moments • Summation of the moments about the pivot gives ab c F1 2F Fa θ 0bFaFcF 12a =+−
  161. 161. Mechanical Design PRN Childs, University of Sussex Force relationship • So the relationship between the applied load Fa and the band brake tensions is given by: c bFaF F 12 a − =
  162. 162. Mechanical Design PRN Childs, University of Sussex Value of b • Note that the value of b must be less than a so that applying the lever tightens F2 more than it loosens F1. • Substituting: ( ) c beaF F 2 a µθ − =
  163. 163. Mechanical Design PRN Childs, University of Sussex Self-locking • The brake can be made self-locking if a<beµθ and the slightest touch on the lever would cause the brake to grab or lock abruptly. • This principle can be used to permit rotation in one direction only as in hoist and conveyor applications.
  164. 164. Mechanical Design PRN Childs, University of Sussex Example • Design a band brake to exert a braking torque of 85 N m. • Assume the coefficient of friction for the lining material is 0.25 and the maximum permissible pressure is 0.345 MPa.
  165. 165. Mechanical Design PRN Childs, University of Sussex Solution • Propose a trial geometry, say r=150 mm, θ=225o and w=50 mm. N258705.015.010345.0rwpF 6 max1 =×××== ( ) N969 e 5.2587 e F F 360/222525.0 1 2 === π×µθ ( ) ( ) mN7.24215.09695.2587rFFT 21 =−=−=
  166. 166. Mechanical Design PRN Childs, University of Sussex Solution cont. • This torque is much greater than the 80 N m desired, so try a different combination of r, θ and w until a satisfactory design is achieved.
  167. 167. Mechanical Design PRN Childs, University of Sussex Solution cont. • Try r=0.1 m, θ=225o and w=50 mm. N172505.01.010345.0rwpF 6 max1 =×××== ( ) N3.646 e 1725 e F F 360/222525.0 1 2 === π×µθ ( ) ( ) mN9.1071.03.6461725rFFT 21 =−=−=
  168. 168. Mechanical Design PRN Childs, University of Sussex Solution cont. • Try r=0.09 m, θ=225o and w=50 mm. N5.155205.009.010345.0rwpF 6 max1 =×××== ( ) N7.581 e 5.1552 e F F 360/222525.0 1 2 === π×µθ ( ) ( ) mN4.8709.07.5815.1552rFFT 21 =−=−=
  169. 169. Mechanical Design PRN Childs, University of Sussex Solution cont. • The actuating force is given by Fa=F2a/c. • If a=0.08 m and c=0.15 m then, N2.310 15.0 08.0 7.581Fa =×=
  170. 170. Mechanical Design PRN Childs, University of Sussex Conclusions • Clutches are designed to permit the smooth, gradual engagement or disengagement of a prime mover from a driven load. • Brakes are designed to decelerate a system. • Clutches and brakes are similar devices providing frictional, magnetic or direct positive connection between two components. • This section has concentrated on rotating clutches and brakes and specifically on the design of friction based devices.
  171. 171. Mechanical Design PRN Childs, University of Sussex Conclusions cont. • The detailed design of a clutch or braking system involves integration of a wide range of skills such as bearings, shafts, splines, teeth, flywheels, casings, frictional surfaces, hydraulics, sensors and control algorithms. • Both brakes and clutches can be purchased from specialist suppliers or alternatively key components such as brake pads or clutch discs can be specified and bought in from specialist suppliers and integrated into a fit for purpose machine design.

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