CLUTCHES AND BRAKES. Square-jaw clutch

Transcription

1 Clutches: CLUTCHES AND BRAKES A Clutch is a mechanical device which is used to connect or disconnect the source of power from the remaining parts so the power transmission system at the will of the operator. The flow of mechanical power is controlled by the clutch. Types of Clutches (i Positive Clutches (ii Friction clutches Positive Clutches: In this type of clutch, the engaging clutch surfaces interlock to produce rigid joint they are suitable for situations requiring simple and rapid disconnection, although they must be connected while shafts are stationery and unloaded, the engaging surfaces are usually of jaw type. The jaws may be square jaw type or spiral jaw type. They are designed empirically by considering compressive strength of the material used. The merits of the positive clutches are. (i Simple (ii No slip (iii No heat generated compact and low cost. Square-jaw clutch Spiral-jaw clutch Friction Clutches: Friction Clutches work on the basis of the frictional forces developed between the two or more surfaces in contact. Friction clutches are usually over the jaw clutches due to their better performance. There is a slip in friction clutch. The merits are (i (ii (iii They friction surfaces can slip during engagement which enables the driver to pickup and accelerate the load with minimum shock. They can be used at high engagement speeds since they do not have jaw or teeth Smooth engagement due to the gradual increase in normal force. The major types of friction clutches are (i Plate clutch (Single plate (multiple plate (ii Cone clutch (iii Centrifugal clutch (iv Dry (v Magnetic current clutches (vi Eddy current clutches

2 We will be studying about single plate multi-plate and cone clutches. Single plate clutch: A single plate friction clutch consisting of two flanges shown in fig. One flange is rigidly keyed in to the driving shaft, while the other is free to move along the driven shaft due to spliced connection. The actuating force is provided by a spring, which forces the driven flange to move towards the driving flange. The face of the drive flange is linked with friction material such as cork, leather or ferodo Torque transmitted by plate or disc clutch A friction disk of a single plate clutch is shown in above fig The following notations are used in the derivation D o Outer diameter of friction disc (mm D i Inna diameter of friction disc (mm P pressure of intensity N/mm F Total operating force (N (Axial force T torque transmitted by friction (N-mm Consider an elemental ring of radius r and radial thickness dr Area of elemental length πr. dr Axial force length πr r. P

3 (µ or f friction force πr. drp µ Friction torque πr dr Pµ * r Total axial force Fa D / F r drp ( a D i/ Torque Transmitted by friction T D / D/ πr dr µ p ( There are two criteria to obtain the torque capacity uniform pressure and uniform wear. Uniform pressure Theory: In case of new clutches, un playing assumed to be uniformly distributed over the entire surface area of the friction disc. With this assumption, P is regarded as constant. Equation becomes F a D/ D i/ r drp D / Fa π p r dr D i/ r π p D / D / F F a a π D D p π p 4 ( D D 4 Fa or P π D D From Equation - [ ] T D / D/ π µ pr dr T π µ p D / D/ r dr

4 4 r T π µ p D / D / T Do πµ p Di T π µ p( Do D Substituting the value of p from equation πµ ( D D T π 4Fa ( D D o ( D Di T µ Fa ( D D i Where µ Fa Dm T D D i Dm D Di mean diameter Torque transmitted by n- friction surfaces n µfadm T π Axial force π P( R R P( D 4 Uniform Wear Theory: According to this theory, it is assumed that the wear is uniformly distributed over the entire surface --- of the friction disc. This assumption is used for workout clutches. The axial wear of the friction disc is import ional to the frictional work. The work done by the frictional force (µ P and subbing velocity (πrn where N is speed in rpm. Assuming speed N and coefficient of friction µ is constant for given configuration Wear Pr Pr constant C When clutch plate is new and rigid. The wear at the outer radius will be more, which will release the pressure at the outer edge due to the rigid pressure plate this will change the pressure distribution. During running condition, the pressure distribution is adjusted in such a manner that the product pressure is constant, C. D i

5 5 From equation - ( Axial force Fa π D / D / Fa π c Pr dr D / D / dr D / Fa π c[ r] Fa D / D D π c C Fa π [ D D ] - (7 From equation - ( T D / π µ Pr dr π µc D / r π µ c D / D / D / r dr D / D / / D π µc 8 8 π µ c T [ D D ] Substitute the value of C from equation (7 8 T π µ [ D D ] Fa 4 π D µ Fa [ D D ] T [ D ] µ Fa Dm T - (8

6 Where Dm D + D - (9 Torque transmitted by n friction plates T n µ Fa Dm Axial force Fa Average pressure occurs at mean radius r D / ( m m π b Dm ( D Fa D - ( Maximum pressure occurs at inner radius π p Di Fa ( D Di - ( Note: The major portion of the life of friction lining comes under the uniform wear friction lining comes under the uniform wear criterion in design of clutches uniform wear theory is justified. Problems:. A single plate friction clutch of both sides effective has mm outer diameter and mm inner diameter. The coefficient of friction o. and it runs at rpm. Find the power transmitted for uniform wear and uniform pressure distributions cases if allowable maximum pressure is.8 Mpa. Given: N I, D mm D mm µ. N rpm p.8 Mpa.8 N /mm Solution: i. Uniform wear theory Mean Diameter D D + D m mm Axial Force Fa π b D ( D D From DDH. or Equation - (

7 7 Fa π.8 ( N Torque transmitted T µ n Fa Dm T T 9484 N mm Power transmitted π M T P π 9484 P P.5 kw ii. Uniform wear theory Mean Diameter D D D D D m D m D m 7. mm π p ( D Axial Force Fa D 4 π.8 Fa ( 4 Fa 44.4 N Torque transmitted T n µ Fa Dm

8 8 T T 988. N mm Power transmitted π nt P π 988. P P. kw. A car engine develops maximum power of 5 kw at rpm. The clutch used is single plate clutch both side effective having external diameter.5 times internal diameter µ.. Mean axial pressure is not to exceed.85 N/ mm. Determine the dimension of the friction surface and the force necessary to engage the plates. Assume uniform pressure condition. Given p 5 kw, n rpm, I both sides are effective D.5 D, µ., p.85 N/mm Torque transmitted π nt P T P π n 5 T π 4. 9 N mm

9 9 Mean Diameter D m D D m D Di D (.5D i Di (.5D i Di i. D i π Axial Force Fa p ( D Di 4 π Fa.85 (.5D i D i 4 Fa 755 D i Torque transmitted T i µ Fa Dm D i. Di D i 4 mm D 8 mm D m 5 mm Fa 884. N Thickness of disc h mm. Design a single plate clutch consist of two pairs of contacting surfaces for a torque capacity of N-m. Due to space limitation the outside diameter of the clutch is to be 5mm Given: Single plate clutch, Torque x 5 N-mm, D 5mm I (since two pairs of contacting surfaces Solution: Assume suitable friction material leather µ. to.5 P varies from.7 to.9 Mpa select µ.4, P.5 Mpa N /mm. Torque transmitted x 5 N-mm. Mean diameter Assuming uniform wear theory

10 D D + D D i i + m 5. Axial force : For uniform, wear condition Torque transmitted Fa π p Di ( Di Di T π.5 Di (5 D i. D i (5 D i µ Fa Dm i 5 i.e.4. D i (5 Di (5 + Di 5 D - D By trial and error method Inner dia D i 85.4 mm is 8 mm Outer dia D 5 mm given, mean m dia of friction surface Dm 8 mm Fa. x 8 ( N Multiple plate clutch Fig. shows a multiple plate clutch. The driving discs are splined to the driving shaft so that they are free to slip along the shaft but must rotate with it. The driven discs drive the housing by means of bolts along which they are free to slide. The housing is keyed to the driven shaft by a sunk key. In the clutch shown there are five pairs of friction surfaces. The driving discs may be pressed against the driven discs by a suitable mechanism so that the torque may be transmitted by friction between the discs.

11 Multi disc Clutch: Equations derived for torque transmitting velocity of single plate are modified to account for the number of pairs of contacting surfaces in the following way. For uniform pressure T i µ Fa Dm π p ( D Di 4 Uniform wear T ( D D i Where I number of pairs of contacting surfaces. For uniform pressure theory T i µ Fa Dm D m ( D ( D Di D i π Fa p ( D Di 4 Where I number of friction surfaces Uniform wear theory T µ i Fa Dm D m ( D + Di Fa π b Dm ( D Di Maximum pressure occurs at inner radius Fa π p D ( D Di Problem: A multi plate clutch having effective diameter 5mm and 5mm has to transmit kw at rpm. The end thrust is 4.5 kn and coefficient of friction is.8 calculate the number of plates easuming (i Uniform wear and (ii uniform pressure distribution on the plates Given: D 5 mm D i 5mm, P kw, N rpm, Fa 4.5 kn 45N, µ.8

12 π NT P Torque T P π 4775 N mm (i Uniform wear theory Mean diameter Dm D + Di mm T µ i Fa Dm 4775 i.8 45 Number of friction plates, I..4 (even numbers Total number of plates i Uniform pressure D m ( D ( D Di 4.7 mm D i Problem 4: T i Fa µ Dm 4775 i i.99 4 (even number Total number of plates A multi plate clutch of alternate bronze and steel plates is to transmit kw power at 8 rpm. The inner radius is 8 mm and outer radius is 7 mm. The coefficient of friction is. and maximum allowable pressure is 5 kn /m determine (i (ii (iii (iv Axial force required Total number of discs Average pressure and Actual maximum pressure

13 Given: P kw, N 8 rpm, R 8mm, D i 7 mm R 7, D 4mm, µ., P 5 kn / m.5 N / mm. Axial force Fa π p D ( D D - (. DDH Fa π.5 7 ( N Torque to be transmitted π NT P T P π N π 8 75 N mm Assuming uniform wear theory ( D + D Mean Diameter D m mm Torque transmitted T µ i Fa Dm 75 n n 4.9 (even number Number of driving (steel discs n n Number of driven (bronze discs n n Average pressure occurs at mean diameter Axial force Fa π p Dm ( D Di 74. π p 8 (4 7 - Average pressure p.4 N/ mm

14 4 4. For friction surface, torque transmitted T µ i Fa Dm 75. Fa 8 - Fa. N Maximum pressure occur at inner radius Axial force π p D ( D Di. π p 7 (4 7 - Actual maximum pressure P.89 N/ mm Problem 5: In a maultilate clutch radial width of the friction material is to be. of maximum radius. The coefficient of friction is.5. The clutch is KW at rpm. Its maximum diameter is 5mm and the axial force is limited is to N. Determine (i Number of driving and driven plates (ii mean unit pressure on each contact surface assume uniform wear Given: Radial width. Ro, µ.5, P KW, N rpm, D 5mm, Ro 5mm, Fa N uniform wear condition. Solution b Ro- Ri. Ro Ri Ri.8 Ro.8x 5 mm Inner diameter x mm i Number of disc Torque transmitted T P π N π 9N.mm For uniform wear condition Dm Do + Di Torque Transmitted Mean diameter ( ( mm T µ n FaDm

15 5 9.5 n, 5 i.e n. Number of active surfaces n ( n must be even number Number of disc on the driver shaft n n Number disc on the driven shaft n n Total number of plates : n + n 7 ii Mean unit pressure Fa π p Dm ( Do Di π p 5 5 P.4 N/ mm ( iii for actual mean unit pressure Actual axial force T Fa µ n D m n T µ fa D m 9 π 55.9N / \ π PDm 5 Actual mean unit pressure P. N / mm ( 5 A Multiple plate clutch has steel on bronze is to transmit 8 KW at 44 rpm. The inner diameter of the contact is 8mm and outer diameter of contact is4 mm. The clutch operates in oil with coefficient of friction of.. The overage allowable pressure is.5mpa. Assume uniform wear theory and determine the following. a Number of steel and bronze plates b Axial force required c Actual maximum pressure Given P 8KW, N 44 rpm, D 8mm, Do 4mm, µ., P.5 N / mm Uniform Wear Theory.

16 Solution: Number of steel and bronze plates For uniform wear theory Axial force Fa π p ( D o D i π.5 8( 4 8 Fa 8.9 µ N Dm D + Di Mean diameter ( ( mm Torque transmitted T Also T P π N 8 T N mm π 44 µ FaDmn n n.55 No. of Active surface is 4 Number discs on the driver shaft (Bronze 4 n n Number disc on the driven shaft (Steel 4 n n + + Total No. of disc n n + 5 b Axial force required T Fa 4. µ n D m. 4 c Actual maximum pressure since maximum presser occur at inner diameter i o i 4. P.8 Fa π P D ( D D P. N / mm ( 4 8

17 7 Cone clutch A simple form of a cone clutch is shown in fig. it consists of ;i driver or cup and the follower or cone. The cup is keyed to the driving shaft by a sunk key and has an inside conical surface or face which exactly fits the outside conical surface of the cone. The slope of the cone face is made small enough to give a high normal force. The cone is fitted to the driven shaft by a feather key. The follower may be shifted along the shaft by a forked shifting lever in order to engage the clutch by bringing the two conical surfaces in contact. Advantages and disadvantages of cone clutch: Advantages:. This clutch is simple in design.. Less axial force is required to engage the clutch. Disadvantages :. There is a tendency to grab.. There is some reluctance in disengagement. Strict requirements made to the co-axiality of the shafts being connected Torque transmitted by the cone clutch:- Let Di Inner diameter of cone, Ri inner radius of cone, Do Outer diameter of cone, Ro Outer radius of cone, Rm mean radius of cone, Dm mean diameter of cone, Semi cone angle or pitch cone angle or face angle, P Intensity of normal pressure at contact surface, µ Coefficient of friction, Fa Axial force Fa FnNormal force sin α

18 8 Consider an elemental ring of radius r and thickness dr as shown in the figure The sloping length dr sin α dr Area of elementary ring π r sin α dr Normal force on the ring P π r sin α pπ r. dr Axial component of the above force Sinα Sinα Total axial force Fa Ro π pr π pr dr ( Ri dr Frictional force outer ring µ p πr dr Sin α Moment of friction force about the axial dr µ P π r r Sin α π p µ r dr Sin α Ro dr Total torque T π µ Pr ( Ri Sin α Uniform pressure theory: p constant Equation ( becomes Fa Ro π pr dr π p Ri Fa π P ( R o R i Ro Ri dr p Fa π R o R i ( ( Equation becomes Ro T π p µ r ( Ro ri Ri dr Sinα π P µ Sinα Ro Ri r dr

19 9 π µ p T Sinα R a R i Substitutes the value of P from equation ( T π µ P Sin α π a ( R R o Fa R R i π Fa Fa T Sin α π R R a i ( ( R R o R o R π Fa Do D i µ Fa T. Dm (4 Sinα Do D io Sin α Where D m Do Di Do D io Axial force Fa P ( i π D o D i (5 Uniform wear: For uniform wear condition Pr C Constant Equation ( become Fa Ro π P r d r π P Ri Fa π c ( R o R i or Ro Ri dr Fa C π R o R i ( Equation ( become Ro dr π µ C T π P µ r Sinα Sinα Ri π µ c ( R T a R i Substitute for C Sinα π µ ( Ra Ri T Sinα π Fa R Ro Ri ( R o i r d r

20 ( D D µ Fa o + Sinα i µ Fa D m. Sin α Where Dm Mean diameter Dm D o + D i If the clutch is engaged when one member is stationary and other rotating, then the cone faces will tend to slide on each other in the direction of an element of the cone. This will resist the engagement and then force i Axial load Fa Fa ( Sinα + cosα Force width b Do Di Sin α Outer diameter Inner diameter D o Dm + b Sinα D i Dm b Sinα Problem: A cone clutch is to transmit 7.5 KW at rpm. The face width is 5mm, mean diameter is mm and the face angle 5. Assuming co efficient of friction as., determine the axial force necessary to hold the clutch parts together and the normal pressure on the cone surface. Given P 7.5 KW, N rpm, b 5mm, Dm mm, 5 µ. Solution: P 7.5 T 975 N mm π N π Torque transmitted µ Fa Dm T Sin α. Fa 975 Sin 5 Also Fa N Fa π Dm Pb Sin α Equation.7 DDH 9.88 π P 5 Sin 5 P.844N / mm

21 A friction cone clutch has to transmit a torque of N-m at 44 rpm. The longer diameter of the cone is 5mm. The cone pitch angle is.5 the force width is 5mm. the coefficient of friction is.. Determine i the axial force required to transmit the torque. ii The average normal pressure on the contact surface when maximum torque is transmitted. Data T N-m, 5 N-mm N 44 rpm Do 5,.5 b 5mm, µ. Solution I Axial force Outer diameter Do Dm + b Sin α 5 D m + 5 Sin.5 D m 4. 9mm Torque transmitted T µ Fa D m Sin α 5. Fa Sin.5 D m Axial force required Fa 4.94 N ii Average normal pressure Fa π D m Pb Sin α 4.94 π Sin.5 P Average Normal pressure P.8 N / mm An engine developing KW at 5 rpm is filted with a cone clutch. The cone face angle of.5. The mean diameter is 4 rpm µ. and the normal pressure is not to exceed.8 N / mm. Design the clutch Date: P KW, N 5 rpm,.5, Dm 4mm, µ., P.8 N/ mm Solution i Torque transmitted p T π N π 5

22 ii Axial force Fa T 9 N mm T µ Fa Dm. Fa 4 9, Sin α Sin.5 Fa 8.8 N Dimensions Fa π D m Pb Sin α 8.8 π. 4.8 b Sin.5 b 8mm Inner diameter Dm Dm b Sinα 4 8 Sin.5 9mm Outer diameter Dm Dm + b Sinα Sin.5 48mm

23 BRAKES A brake is defined as a machine element used to control the motion by absorbing kinetic energy of a moving body or by absorbing potential energy of the objects being lowered by hoists, elevators, etc. The obsorbed energy appears as heat energy which should be transferred to cooling fluid such as water or surrounding air. The difference between a clutch and a brake is that whereas in the former both the members to be engaged are in motion, the brake connects a moving member to a stationary member. Block or shoe brake A single-block brake is shown in fig. It consists of a short shoe which may be rigidly mounted or pivoted to a lever. The block is pressed against the rotating wheel by an effort Fat one end of the lever. The other end of the lever is pivoted on a fixed fulcrum O. The friclional force produced by the block on the wheel will retard the rotation of the wheel. This type of brake is commonly used in railway trains. When the brake is applied, the lever with the block can be considered as a free body in equilibrium under the action of the following forces.. Applied force F at the end of the lever.. Normal reaction F n between the shoe and the wheel.. Frictional or tangential braking force F θ between the shoe and the wheel. 4. Pin reaction. Let F Operating force M Torque on the wheel r Radius of the wheel θ Angle of contact surface of the block µ Coefficient of friction M F θ Tangential braking force t r Fθ F n Normal force µ Consider the following three cases; a Distance between the fulcrum pin and the center of the shoe b Distance between the center of the shoe to the end of lever where the effort is applied c Distance between the fulcrum pin and the line of action of F g (i Line of action of tangential force F passes through fulcrum

24 4 Taking moments about O, Fθ F ( a + b Fn a a µ F n µ Fθ a Actuating force F µ ( a + b In this case the actuating force is the same whether the direction of tangential force is towards or away from the fulcrum. (ii Line of action of tangential force F θ is in between the center of the drum and the fulcrum (a Direction of F θ is towards the fulcurm : F θ Taking moments about O, F ( a + b F c F n n a Fθ a ( a + b F a Fθ c F c µ F n µ F n θ Fθ a c Actuating force F ( a + b µ a F θ (b Direction of F θ is away from the fulcrum :

25 5 Taking moments about O, Fθ F F ( a + b Fn a Fθ c a + Fθ c µ F θ n µ Fθ a c Actuating force F ( a + b µ a (iii Line of action of tangential force F θ is above the center of the drum and the fulcrum: (a Direction of F θ is towards the fulcrum : Taking moments about O, F a F ( a + b Fn a θ Fθ c Fθ c µ + F F θ n µ Fθ a c Actuating force F ( a + b µ a (b Direction of F θ is away from the fulcrum : Taking moments about O, Fθ a F F ( a + b + Fθ c Fθ a a µ F θ n µ F a F ( a + b F c θ µ Fθ a c Actuating force F ( a + b µ a θ

26 Note: If the direction of F is towards the fulcrum, use the clockwise rotation formula and if the direction of F & is away from the fulcrum, use counter clockwise formula from the data handbook. When the angle of contact between the block and the wheel is less than, we assume that the normal pressure is uniform between them. But when the angle of contact θ is more than, we assume that the unit pressure normal to the surface of contact is less at the ends than at the center and the wear in the direction of applied force is uniform. In such case we employ the equivalent coefficient of friction µ', which is given by. 4sinθ Equivalent coefficient of friction µ µ θ + sin θ Where µ Actual coefficient of friction θ Semi block angle For the given value of θ. The value of 4sinθ θ + sin θ can be found by using the chart given in The brake is self -energizing when the friction force helps to apply the brake. If this effect is great enough to apply the brake with zero external force, the brake is called self-locking i.e., the brake is self locking when the applied force F is zero or negative. Normal load Normal pressure on the shoe p P Pr ojected area of Fn shoe wr sinθ Where w Width of the shoe Example: The block type hand brake shown in fig.. la has a face width of 45 mm. The friction material permits a maximum pressure of. MPa and a coefficient of friction of.4. Determine;. Effort F,. Maximum torque,. Heat generated if the speed of the drum is rpm and the brake is applied for 5 sec. at full capacity to bring the shaft to stop. Data: w 45 mm, p. MPa. N/mm, µ.4, θ 9, θ 45 e, d mm,

27 7 r 5 mm, n rpm Solution: 4sinθ Since θ >, equivalent coefficient of friction µ µ θ + sin θ 4sin π + sin 9 8 Allowable pressure p Fn wr sinθ i.e.,. 45 5sin 45 Normal force F n N Tangential force F θ µ' F n.4 x N The various forces acting on the shoe are shown in fig..b. From the figure, a mm, b mm, c The tangential force F θ, passes through the fulcurm. F n Effort Fθ a F µ ' ( a + b 5..4 ( + 9. N Torque on the drum M F θ r 5. x 5 85 N-mm.85 N-m Power absorbed M n.85 N.75kW.75 kj / sec Heat generated during 5 sec 5 x kj Example: A 4 mm radius brake drum contacts a single shoe as shown in fjg..a, and sustains N-m torque at 5 rpm. For a coefficient of friction.5, determine:. Normal force on the shoe.. Required force F to apply the brake for clockwise rotation.. Required force F to apply the brake for counter clockwise rotation. 4. The dimension c required to make the brake self-locking, assuming the other dimensions remains the same. 5. Heat generated.

28 8 Data; r 4 mm, M - N-m x N-mm, n 5 rpm, µ.5, a 5 mm, i mm, b l - a mm, c 4 mm Solution: Tangential friction force M Fθ 5 N r 4 Normal force on the F 5 F θ θ N µ.5 From the figure, the tangential force F θ lies between the fulcrum and the center of drum.. When the force F θ acts towards the fulcrum (clockwise rotation, Actuating force F θ F a a + b µ c a N For anticlockwise rotation of the drum, the tangential force F s acts away from the fulcrum as shown in figure. Actuating force F θ F a a + b + µ c a N When the drum rotates in clockwise direction, self locking will occur. For self locking effort F.

29 9 i. e., F a a + b µ c a or c a, µ c a µ 5 c mm.5 Heat generated H g µ p Ac v F n v π W.47 kw.47 kj / s Example: The layout of a brake to be rated at 5 N-m at rpm is shown in figure. The drum diameter is mm and the angle of contact of each shoe is. The coefficient of friction may be assumed as. Determine.. Spring force F required to set the brake.. Width of the shoe if the value of pv is N-m/mm -sec Data: M t 5 N-m 5 x N-mm, n rpm, d mm, r mm, θ,, µ., pv N-m/mm -sec

30 Solution: 4Sinθ Since >, equivalent coefficient of friction µ ' µ θ + Sinθ Sin..5 + π + sin 8 The various forces acting on the shoes are shown in figure. Left hand shoe: For left hand shoe, a 5 mm, b 5 mm, c 5 mm The tangential force F θ lies above the center of the drum and fulcrum, and acting away from the fulcrum (counter clockwise. F a c Spring force F a + b µ a F F Tangential force F gi.795 F Right hand shoe: The tangential force F θ lies below the center of the drum and the fulcrum, and acting towards the fulcrum (clockwise. Spring force F F a c a + b µ a F F Tangential force F.85 F Torque M F F r ( + i.e., 5 (.795 F +.85 F Spring force F 75. N The maximum load occurs on left hand shoe.

31 Tangential force F.795 x N Normal force on left hand shoe F 9. F µ.5 N π dn π Surface velocity of drum v.8 m / sec By data pv N-m/mm sec Normal pressure p v.8.8 N / mm Also pressure p Fn wr sin θ.e.,.8 wr sin Width of the shoe w 7.5mm

32 Band brakes A band brake consists of a band, generally made of metal, and embracing a part of the circumference of the drum. The braking action is obtained by tightenting the band. The difference in the tensions at each end of the band determines the torque capacity. Simple band brakes: When one end of the band is connected to the fixed fulcrum, then the band brake is called simple band brake as shown in figure Let T, Tight side tension in N T Slack side tension in N θ Angle of lap in radians µ Coefficient of friction D Diameter of brake drum in mm M Torque on the drum in N-mm T Ratio of tensions T e Braking force F θ T T M D t Clockwise rotation of the drum: Taking moments about the fulcrum O, F a T b T b Force at the end of lever F a T Braking force F θ T T T e T T ( e e T :. Slack side tension T F e θ

33 Substitute the value of T in equation (, we get T b F a e Counter clockwise rotation of the drum: For counter clockwise rotation of the drum, the tensions T and T will exchange their places and for the same braking torque, a larger force F is required to operate the brake. Taking moments about the fulcrum O, F a T b T b Force at the end of lever F a T T Braking force Fθ T T T T ( e e e T Tight side tension T F e e θ Substitute the value of T, in equation (i, we get F Fθ b e a e This type of brake does not have any self-energizing and self-locking properties. Thickness of the band h.5 D T Width of band w h σ d where σ is the allowable tensile stress in the band. d

34 4 Differential band brakes: In differential band brake the two ends of the band are attached to pins on the lever at a distance of b l and b from the pivot pin as shown in figure. It is to be noted that when b > b, the force F must act upwards in order to apply the brake. When b < b, the force F must act downwards to apply the brake. Clockwise rotation of the drum: Taking moments about the fulcrum O, T b T b + F a T e b T b Fa ( T Te Braking force F T T T e T T ( e T T e θ ( Slack side tension T θ e F Substitute the value of T in equation (, we get F θ b e b F a e For the brake to be self locking, the force F at the end of the lever must be equal to zero or negative. F θ b e b a e The condition for self-locking is b e b

35 5 Counter clockwise rotation of the drum: Taking moments about the fulcrum O, T b T b + F a b T e Fa ( T Te T Or T F ( b b e a Braking force F T T T e T T ( e T T e θ ( Slack side tension T θ e F Substitute the value of T in equation (, we get F θ b b e F a e For self-brake locking, the force F must be zero or negative. F θ b b e i.e., a e The condition for self-locking is b b e If b b b, the brake is called two way brake and is shown in figure. This type of brake can be used in either direction of rotation with same effort. Fθ b e Force at the end of lever F a e +

36 Example: A simple band brake operates on a drum. m in diameter rotating at rpm. The coefficient of friction is.5 and the angle of contact of the band is 7. One end of the band is fastened to a fixed pin and the other end to 5 mm from the fixed pin. The brake arm is 75 mm long. (i What is the minimum pull necessary at the end of the brake arm to.stop the wheel if 5 kw is being absorbed? What is the direction of rotation for minimum pull? (ii Find the width of.4 mm thick steel band if the maximum tensile stress is not to exceed 55 N/mm. Data: D. m mm, n rpm, p.5, 7, a 75 mm, b 5 mm, N 5kW, h.4 mm, σ d 55 N/mm Solution: Frictional torque M 955 N n N m 7.5 N mm Braking for M M 7.5 Fθ N R D T.5 7 π8 Ratio of tensions e e. 48 T Clockwise rotation: Force Fθ b F a e

37 7 Counter clockwise rotation: N Force Fθ b e F a e Therefore the minimum pull F 4. N The direction of rotation is clockwise N T Braking force Fθ T T T T ( e e i.e., T. 48 Tight side tension T N T Width of band w h σ d mm mm.4 55 Problem: In a simple band brake, the length of lever is 44 mm. The tight end of the band is attached to the fulcrum of the lever and the slack end to a pin 5 mm from the fulcrum. The diameter of the brake drum is m and the arc of contact is. The coefficient of friction between the band and the drum is.5. The brake drum is attached to a hoisting drum of diameter.5 m that sustains a load of kn. Determine;. Force required at the end of lever to just support the load.. Required force when the direction of rotation is reversed.. Width of stee! band if the tensile stress is limited to 5 N/mnr.

38 8 Data: a 44 mm, b 5 mm, D m mm, D b.5 m 5 mm, θ, µ.5, W kn x N, σ d 5N/mm Solution: WDb Torque on hoisting drum M W Rb 5.5 N mm The hoisting drum and the brake drum are mounted on same shaft. Torque on brake drum M.5 x N-mm M M Braking for Fθ R D.5 N T.5 π8 Ratio of tensions e e. 5 T Clockwise rotation: Fθ b Force at the end of lever F a e N 44.5 Counter clockwise rotation:

39 9 Force Fθ b e F a e N 44.5 Thickness of band h.5 D.5 x 5 mm T Braking force Fθ T T T T ( e e i.e., T. 5 Tight side tension T 547. N T Width of band w h σ d mm mm 5 5 Problem: A band brake shown in figure uses a V-belt. The pitch diameter of the V-grooved pulley is 4 mm. The groove angle is 45 and the coefficient of friction is.. Determine the power rating. Data: D - 4 mm, a 45, α.5, µ -., F N, b 4 mm, a 75 mm, θ 8 π rad, n 4 rpm

40 4 Solution: Ratio of tensions for V-belt, For clockwise rotation, T Sin α e / T e. π / Sin.5.78 Force i.e., Fθ b F a e F θ Braking force F θ.4 N Fθ D Torque on the drum M Fθ R N mm 4. 8 N m Power rating M N n kw 955 Problem Figure shows a two way band brake. It is so designed that it can operate equally well in both clockwise and counter clockwise rotation of the brake drum. The diameter of the drum is 4 mm and the coefficient of friction between the band and the drum is.. The angle of contact of band brake is 7 and the torque absorbed in the band brake is 4 N-m. Calculate;. Force F required at the end of the lever.. Width of the band if the allowable stress in the band is 7 MPa.

41 4 Data: D 4 mm, µ., θ 7, M t 4 N-m 4 x N-mm, σ d 7 MPa 7 N/mm, b b l b 5mm, a mm Solution: Braking force F θ M t 4 T T D 4 N T. 7 π8 Ratio of tensions e e 4. T Fθ b e Force at the end of lever F a e N 4. Band thickness h.5 D.5 x 4 mm T Braking force Fθ T T T ( e e T i.e., T 4. Tight side tension T 4.84 N T Width of band w h σ d mm mm 7