What are the different types of brakes and Dynamometers

difference between brakes and dynamometers and distinguish between brakes and dynamometers and write difference between brakes and dynamometers pdf free download
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 732 Theory of Machines 19 Features Brakes and 1. Introduction 2. Materials for Brake Lining. 3. Types of Brakes. Dynamometers 4. Single Block or Shoe Brake. 5. Pivoted Block or Shoe Brake. 6. Double Block or Shoe Brake. 19.1. Introduction 7. Simple Band Brake. 8. Differential Band Brake. A brake is a device by means of which artificial frictional resistance is applied to a moving machine member, 9. Band and Block Brake. in order to retard or stop the motion of a machine. In the process 10. Internal Expanding Brake. of performing this function, the brake absorbs either kinetic 11. Braking of a Vehicle. energy of the moving member or potential energy given up by 12. Dynamometer. objects being lowered by hoists, elevators etc. The energy 13. Types of Dynamometers. absorbed by brakes is dissipated in the form of heat. This heat 14. Classification of Absorption is dissipated in the surrounding air (or water which is circulated Dynamometers. through the passages in the brake drum) so that excessive 15. Prony Brake Dynamometer. heating of the brake lining does not take place. The capacity of 16. Rope Brake Dynamometers. a brake depends upon the following factors : 17. Classification of Transmission 1. The unit pressure between the braking surfaces, Dynamometers. 2. The coefficient of friction between the braking 18. Epicyclic-train surfaces, Dynamometers. 3. The peripheral velocity of the brake drum, 19. Belt Transmission 4. The projected area of the friction surfaces, and Dynamometer-Froude or 5. The ability of the brake to dissipate heat equivalent Throneycraft Transmission to the energy being absorbed. Dynamometer. The major functional difference between a clutch and 20. Torsion Dynamometer. a brake is that a clutch is used to keep the driving and driven 21. Bevis Gibson Flash Light member moving together, whereas brakes are used to stop a Torsion Dynamometer. moving member or to control its speed. 19.2. Materials for Brake Lining The material used for the brake lining should have the following characteristics : 732 Chapter 19 : Brakes and Dynamometers 733 1. It should have high coefficient of friction with minimum fading. In other words, the coeffi- cient of friction should remain constant with change in temperature. 2. It should have low wear rate. 3. It should have high heat resistance. 4. It should have high heat dissipation capacity. 5. It should have adequate mechanical strength. 6. It should not be affected by moisture and oil. The materials commonly used for facing or lining of brakes and their properties are shown in the following table. Table 19.1. Properties of materials for brake lining. Table 19.1. Properties of materials for brake lining. Table 19.1. Properties of materials for brake lining. Table 19.1. Properties of materials for brake lining. Table 19.1. Properties of materials for brake lining. Coefficient of friction (µ) Allowable Material for braking lining pressure ( p ) 2 Dry Greasy Lubricated N/mm Cast iron on cast iron 0.15 – 0.2 0.06 – 0.10 0.05 – 0.10 1.0 – 1.75 Bronze on cast iron – 0.05 – 0.10 0.05 – 0.10 0.56 – 0.84 Steel on cast iron 0.20 – 0.30 0.07 – 0.12 0.06 – 0.10 0.84 – 1.40 Wood on cast iron 0.20 – 0.35 0.08 – 0.12 – 0.40 – 0.62 Fibre on metal – 0.10 – 0.20 – 0.07 – 0.28 Cork on metal 0.35 0.25 – 0.30 0.22 – 0.25 0.05 – 0.10 Leather on metal 0.30 – 0.5 0.15 – 0.20 0.12 – 0.15 0.07 – 0.28 Wire asbestos on metal 0.35 – 0.5 0.25 – 0.30 0.20 – 0.25 0.20 – 0.55 Asbestos blocks on metal 0.40 – 0.48 0.25 – 0.30 – 0.28 – 1.1 Asbestos on metal (Short – – 0.20 – 0.25 1.4 – 2.1 action) Metal on cast iron (Short – – 0.05 – 0.10 1.4 – 2.1 action) 19.3. 19.3. 19.3. 19.3. 19.3. Types of Brakes Types of Brakes Types of Brakes Types of Brakes Types of Brakes The brakes, according to the means used for transforming the energy by the braking elements, are classified as : 1. Hydraulic brakes e.g. pumps or hydrodynamic brake and fluid agitator, 2. Electric brakes e.g. generators and eddy current brakes, and 3. Mechanical brakes. The hydraulic and electric brakes cannot bring the member to rest and are mostly used where large amounts of energy are to be transformed while the brake is retarding the load such as in laboratory dynamometers, high way trucks and electric locomotives. These brakes are also used for retarding or controlling the speed of a vehicle for down-hill travel. The mechanical brakes, according to the direction of acting force, may be divided into the following two groups : (a) Radial brakes. In these brakes, the force acting on the brake drum is in radial direction. The radial brakes may be Simple bicycle brakes. 734 Theory of Machines sub-divided into external brakes and internal brakes. According to the shape of the friction ele- ments, these brakes may be block or shoe brakes and band brakes. (b) Axial brakes. In these brakes, the force acting on the brake drum is in axial direction. The axial brakes may be disc brakes and cone brakes. The analysis of these brakes is similar to clutches. Since we are concerned with only mechanical brakes, therefore, these are discussed, in detail, in the following pages. 19.4. Single Block or Shoe Brake A single block or shoe brake is shown in Fig. 19.1. It consists of a block or shoe which is pressed against the rim of a revolving brake wheel drum. The block is made of a softer material than the rim of the wheel. This type of a brake is commonly used on railway trains and tram cars. The friction between the block and the wheel causes a tangential braking force to act on the wheel, which retard the rotation of the wheel. The block is pressed against the wheel by a force applied to one end of a lever to which the block is rigidly fixed as shown in Fig. 19.1. The other end of the lever is pivoted on a fixed fulcrum O. (a) Clockwise rotation of brake wheel (b) Anticlockwise rotation of brake wheel. Fig. 19.1. Single block brake. Line of action of tangential force passes through the fulcrum of the lever. Let P = Force applied at the end of the lever, R = Normal force pressing the brake block on the wheel, N r = Radius of the wheel, 2θ = Angle of contact surface of the block, µ = Coefficient of friction, and F = Tangential braking force or the frictional force acting at the contact t surface of the block and the wheel. If the angle of contact is less than 60°, then it may be assumed that the normal pressure between the block and the wheel is uniform. In such cases, tangential braking force on the wheel, F = µ.R ...(i) t N and the braking torque, T = F .r = µ.R .r ... (ii) B t N Let us now consider the following three cases : Case 1. When the line of action of tangential brak- ing force (F ) passes through the fulcrum O of the lever, t and the brake wheel rotates clockwise as shown in Fig. 19.1 (a), then for equilibrium, taking moments about the fulcrum O, we have Pl × When brakes are on, the pads grip the Rx×=P×l or R = N N x wheel rim from either side, friction Braking torque, ∴ between the pads and the rim converts Pl.. µ Pl..r the cycle's kinetic energy into heat as TR =µ..r=µ× × =r BN they reduce its speed. xx Chapter 19 : Brakes and Dynamometers 735 It may be noted that when the brake wheel rotates anticlockwise as shown in Fig. 19.1 (b), then the braking torque is same, i.e. µ .. Pl.r TR =µ..r= BN x Case 2. When the line of action of the tangential braking force (F ) passes through a distance t ‘a’ below the fulcrum O, and the brake wheel rotates clockwise as shown in Fig. 19.2 (a), then for equilibrium, taking moments about the fulcrum O, Pl . R × x + F × a = P.l or R × x + µ R × a = P.l or R = N t N N N xa +µ . µ .. pl.r TR =µ .r= and braking torque, BN xa +µ . (a) Clockwise rotation of brake wheel. (b) Anticlockwise rotation of brake wheel. Fig. 19.2. Single block brake. Line of action of F passes below the fulcrum. t When the brake wheel rotates anticlockwise, as shown in Fig. 19.2 (b), then for equilibrium, R .x = P.l + F .a = P.l + µ.R .a ...(i) N t N Pl . or R (x – µ.a) = P.l or R = N N xa −µ . µ .. Pl.r and braking torque, TR =µ..r= BN xa −µ . Case 3. When the line of action of the tangential braking force (F ) passes through a distance t ‘a’ above the fulcrum O, and the brake wheel rotates clockwise as shown in Fig. 19.3 (a), then for equilibrium, taking moments about the fulcrum O, we have R .x = P.l + F . a = P.l + µ.R .a . . . (ii) N t N Pl . or R (x – µ.a) = P.l or R = N N xa −µ . (a) Clockwise rotation of brake wheel. (b) Anticlockwise rotation of brake wheel. Fig. 19.3. Single block brake. Line of action of F passes above the fulcrum. t µ .. Pl.r and braking torque, T = µ.R .r = B N xa −µ . 736 Theory of Machines When the brake wheel rotates anticlockwise as shown in Fig. 19.3 (b), then for equilibrium, taking moments about the fulcrum O, we have Pl . R × x + F × a = P.l or R × x + µ.R × a = P.l or R = N t N N N xa +µ . µ .. Pl.r and braking torque, T = µ.R .r = B N xa +µ . Notes : 1. From above we see that when the brake wheel rotates anticlockwise in case 2 Fig. 19.2 (b) and when it rotates clockwise in case 3 Fig. 19.3 (a), the equations (i) and (ii) are same, i.e. R × x = P.l + µ.R .a N N From this we see that the moment of frictional force (µ.R .a) adds to the moment N of force (P.l). In other words, the frictional force helps to apply the brake. Such type of brakes are said to be self ener- gizing brakes. When the fric- tional force is great enough to apply the brake with no exter- nal force, then the brake is said to be self-locking brake. From the above ex- Shoe brakes of a racing car pression, we see that if xa ≤µ . , then P will be negative or equal to zero. This means no external force is needed to apply the brake and hence the brake is self locking. Therefore the condition for the brake to be self locking is xa ≤µ . The self locking brake is used only in back-stop applications. 2. The brake should be self energizing and not the self locking. 3. In order to avoid self locking and to prevent the brake from grabbing, x is kept greater than µ . a. 4. If A is the projected bearing area of the block or shoe, then the bearing pressure on the shoe, b p = R / A b N b We know that A = Width of shoe × Projected length of shoe = wr (2 sinθ) b 5. When a single block or shoe brake is applied to a rolling wheel, an additional load is thrown on the shaft bearings due to heavy normal force (R ) and produces bending of the shaft. N In order to overcome this drawback, a double block or shoe brake is used, as discussed in Art. 19.6. 19.5. Pivoted Block or Shoe Brake We have discussed in the previous article that when the angle of contact is less than 60°, then it may be assumed that the normal pressure between the block and the wheel is uniform. But when the angle of contact is greater than 60°, then the unit pressure normal to the surface of contact is less at the ends than at the centre. In such cases, the block or shoe is pivoted to the lever, as shown in Fig. 19.4, instead of being rigidly attached to the lever. This gives uniform wear of the brake lining in the direction of the applied force. The braking torque for a pivoted block or shoe brake (i.e. when 2 60°) is θ given by Fig. 19.4. Pivoted block or shoe brake. =× =µ ′.. TF r Rr BN t 4sµθin ′ where µ = Equivalent coefficient of friction = , and 2s θ+in2θ µ = Actual coefficient of friction. These brakes have more life and may provide a higher braking torque. Chapter 19 : Brakes and Dynamometers 737 Example 19.1. A single block brake is shown in Fig. 19.5. The diameter of the drum is 250 mm and the angle of contact is 90°. If the operating force of 700 N is applied at the end of a lever and the coefficient of friction between the drum and the lining is 0.35, determine the torque that may be transmitted by the block brake. Solution. Given : d = 250 mm or r = 125 mm ; 2θ = 90° π/2 = rad ; P = 700 N ; µ = 0.35 All dimensions in mm. Fig. 19.5 Since the angle of contact is greater than 60°, therefore equivalent coefficient of friction, 4µθ sin 4× 0.35× sin 45° ′ µ= = = 0.385 2θ+ sin 2θ π / 2+ sin 90° Let R = Normal force pressing the block to the brake drum, and N ′ µ .R F = Tangential braking force = N t Taking moments about the fulcrum O, we have FF tt 700(250+ 200)+FR ×50= ×= 200 ×= 200 ×= 200 520F tt N µ ′ 0.385 or 520 F – 50F = 700 × 450 or F = 700 × 450/470 = 670 N t t t We know that torque transmitted by the block brake, T = F × r = 670 × 125 = 8 3750 N-mm = 83.75N-m Ans. B t Example 19.2. Fig. 19.6 shows a brake shoe applied to a drum by a lever AB which is pivoted at a fixed point A and rigidly fixed to the shoe. The radius of the drum is 160 mm. The coefficient of friction at the brake lining is 0.3. If the drum rotates clockwise, find the braking torque due to the horizon- tal force of 600 N at B. Solution. Given : r = 160 mm = 0.16 m ; µ = 0.3 ; P = 600 N Since the angle subtended by the shoe at the centre of drum is 40°, therefore we need not to calcu- ′ µ . late the equivalent coefficient of friction Let R = Normal force pressing the N Fig. 19.6 block to the brake drum, and F = Tangential braking force = µ.R t N Taking moments about point A, R × 350 + F (200 – 160) = 600 (400 + 350) N t F t 3 F ×+ 350 40 = 600×750 or 1207 F = 450 × 10 t t 0.3 3 F = 450 × 10 /1207 = 372.8 N ∴ t We know that braking torque, T = F × r = 372.8 × 0.16 = 59.6 N-m Ans. B t 738 Theory of Machines Example 19.3. A bicycle and rider of mass 100 kg are travelling at the rate of 16 km/h on a level road. A brake is applied to the rear wheel which is 0.9 m in diameter and this is the only resistance acting. How far will the bicycle travel and how many turns will it make before it comes to rest ? The pressure applied on the brake is 100 N and µ = 0.05. Solution. Given : m = 100 kg, v = 16 km / h = 4.44 m / s ; D = 0.9 m ; R = 100 N ; µ = 0.05 N Distance travelled by the bicycle before it comes to rest Let x = Distance travelled (in metres) by the bicycle before it comes to rest. We know that tangential braking force acting at the point of contact of the brake and wheel, = µ.R = 0.05 × 100 = 5 N F t N and work done = F × x = 5 × x = 5x N-m . . . (i) t We know that kinetic energy of the bicycle 22 m.ν 100(4.44) == 22 = 986 N-m . . . (ii) In order to bring the bicycle to rest, the work done against friction must be equal to kinetic energy of the bi- cycle. Therefore equating equations (i) and (ii), 5x = 986 or x = 986/5 = 197.2 m Ans. Number of revolutions made by the bicycle before it comes to rest Let N = Required number of revolutions. We know that distance travelled by the bicycle (x), Shoe brake. 197.2=π DN =π×0.9N = 2.83N ∴ N = 197.2 / 2.83 = 70 Ans. Example 19.4. A braking system has its braking lever inclined at an angle of 30° to the horizontal plane, as shown in Fig. 19.7. The mass and diameter of the brake drum are 218 kg and 0.54 m respectively. Fig. 19.7 At the instant the lever is pressed on the brake drum with a vertical force of 600 N, the drum is found to rotate at 2400 r.p.m. clockwise. The coefficient of friction between the brake shoe and the brake drum is 0.4. Assume that the lever and brake shoe are perfectly rigid and possess negligible weight. Find : Chapter 19 : Brakes and Dynamometers 739 1. Braking torque, 2. Number of revolutions the drum will make before coming to rest from the instant of pressing the lever, and 3. Time taken for the drum to come to rest from the instant of pressing the lever. Solution. Given : m = 218 kg ; d = 0.54 m or r = 0.27 m ; P = 600 N ; N = 2400 r.p.m.; µ = 0.4 1. Braking torque Let R = Normal force pressing the block to the brake drum, and N F = Tangential braking force. t The various forces acting on the braking system are shown in Fig. 19.8. Fig. 19.8 Taking moments about the fulcrum O, 600 cos 30° × 1.2 = R × 0.4 or 623.5 = 0.4 R N N R = 623.5/0.4 = 1560 N ∴ N and F = µ.R = 0.4 × 1560 = 624 N t N We know that braking torque, T = F × r = 624 × 0.27 = 168.5 N-m Ans. B t 2. Number of revolutions the drum will make before coming to rest Let n = Required number of revolutions. We know that kinetic energy of the brake drum 22 2 mv . 218ππ d.N × 0.54× 2400    == 109 N-m =    2 2 60 60    3 = 502 × 10 N-m . . . (i) and work done by the brake drum due to braking torque = Tn ×π 2 = 168.5× 2πn= 1060n N-m . . . (ii) B Since the kinetic energy of the brake drum is used to overcome the work done due to braking torque, therefore equating equations (i) and (ii), 3 n = 502 × 10 /1060 = 474 Ans. 3. Time taken for the drum to come to rest We know that time taken for the drum to come to rest i.e. time required for 474 revolutions, n 474 t== = 0.2 min = 12 s Ans. N 2400 740 Theory of Machines 19.6. Double Block or Shoe Brake When a single block brake is applied to a rolling wheel, an additional load is thrown on the shaft bearings due to the normal force (R ). This produces N bending of the shaft. In order to overcome this drawback, a double block or shoe brake, as shown in Fig. 19.9, is used. It consists of two brake blocks applied at the opposite ends of a diameter of the wheel which eliminate or reduces the unbalanced force on the shaft. The brake is set by a spring which pulls the upper ends of the brake arms together. When a force P is applied to the bell crank lever, the spring is compressed and the brake is released. This type of brake is often used on electric cranes and the force P is produced by an electromagnet or solenoid. When the current is switched off, there is no force on the bell crank lever and the brake is engaged automatically due to the spring force and thus there will be no downward movement of the load. Fig. 19.9. Double block or shoe In a double block brake, the braking action is doubled brake. by the use of two blocks and these blocks may be operated practically by the same force which will operate one. In case of double block or shoe brake, the braking torque is given by T = (F + F ) r B t1 t2 where F and F are the braking forces on the two blocks. t1 t2 Example 19.5. A double shoe brake, as shown in Fig. 19.10, is capable of absorbing a torque of 1400 N-m. The diameter of the brake drum is 350 mm and the angle of contact for each shoe is 100°. If the coefficient of friction between the brake drum and lining is 0.4 ; find 1. the spring force necessary to set the brake ; and 2. the width of the brake shoes, if the bearing pressure on the lining 2 material is not to exceed 0.3 N/mm . 3 Solution. Given : T = 1400 N-m = 1400 × 10 N-mm ; B d = 350 mm or r = 175 mm ; = 100° = 100 × /180 = 1.75 rad; All dimensions in mm. π 2θ Fig. 19.10 2 µ = 0.4 ; p = 0.3 N/mm b 1. Spring force necessary to set the brake Let S = Spring force necessary to set the brake. R and F = Normal reaction and the N1 t1 braking force on the right hand side shoe, and R and F = Corresponding values on N2 t2 the left hand side shoe. Since the angle of contact is greater than 60°, therefore equivalent coefficient of friction, 4sµθin 4× 0.4× sin5°0 µ′ = = = 0.45 2θ+ sin 2θ 1.75+ sin100° Brakes on a railway coach. Chapter 19 : Brakes and Dynamometers 741 Taking moments about the fulcrum O , we have 1 F t1 SR ×=× 450 200+F (175− 40)= × 200+F×135= 579.4F N1 t1 tt 1 1 0.45  F t1 Substituting R = . . .  N1 ′ µ  ∴ F = S × 450 / 579.4 = 0.776 S t1 Again taking moments about O , we have 2 F t 2 SF ×+ 450 (175− 40)=R × 200= × 200= 444.4F tt 2N2 2 0.45  F t2 Substituting R = . . . N2 µ ′  444.4 F – 135F = S × 450 or 309.4 F = S × 450 t2 t2 t2 ∴ F = S × 450 / 309.4 = 1.454 S t2 We know that torque capacity of the brake (T ), B 3 1400 × 10 = (F + F ) r = (0.776 S + 1.454 S) 175 = 390.25 S t1 t2 3 ∴ S = 1400 × 10 /390.25 = 3587 N Ans. 2. Width of the brake shoes Let b = Width of the brake shoes in mm. We know that projected bearing area for one shoe, 2 Ab=θ (2r sin )=b(2×175sin 50°)= 268b mm b Normal force on the right hand side of the shoe, F 0.776×× S 0.776 3587 t1 R == = = 6186 N N1 ′ µ 0.45 0.45 and normal force on the left hand side of the shoe, F 1.454×× S 1.454 3587 t2 R == = = 11 590 N N2 ′ µ 0.45 0.45 We see that the maximum normal force is on the left hand side of the shoe. Therefore we shall find the width of the shoe for the maximum normal force i.e. R . N2 We know that the bearing pressure on the lining material ( p ), b R 11 590 43.25 N2 0.3== = 268 Ab b b ∴ b = 43.25 / 0.3 = 144.2 mm Ans. 19.7. Simple Band Brake A band brake consists of a flexible band of leather, one or more ropes,or a steel lined with friction material, which embraces a part of the circumference of the drum. A band brake, as shown in Fig. 19.11, is called a simple band brake in which one end of the band is attached to a fixed pin or fulcrum of the lever while the other end is attached to the lever at a distance b from the fulcrum. When a force P is applied to the lever at C, the lever turns about the fulcrum pin O and tightens the band on the drum and hence the brakes are applied. The friction between the band and the drum provides the braking force. The force P on the lever at C may be determined as discussed below : Let T = Tension in the tight side of the band, 1 T = Tension in the slack side of the band, 2 742 Theory of Machines = Angle of lap (or embrace) of the band on the drum, θ µ = Coefficient of friction between the band and the drum, r = Radius of the drum, t = Thickness of the band, and t r + r = Effective radius of the drum = e 2 Band brake Bands of a brake shown separately (a) Clockwise rotation of drum. (b) Anticlockwise rotation of drum. Fig. 19.11. Simple band brake. We know that limiting ratio of the tensions is given by the relation,  T T 1 µθ 1 = e 2.3log =µθ. or  T T 2  2 and braking force on the drum = T – T 1 2 Braking torque on the drum, ∴ T = (T – T ) r . . . (Neglecting thickness of band) B 1 2 = (T – T ) r . . . (Considering thickness of band) 1 2 e Now considering the equilibrium of the lever OBC. It may be noted that when the drum rotates in the clockwise direction, as shown in Fig. 19.11 (a), the end of the band attached to the fulcrum O will be slack with tension T and end of the band attached to B will be tight with tension T . 2 1 On the other hand, when the drum rotates in the anticlockwise direction, as shown in Fig. 19.11 (b), the tensions in the band will reverse, i.e. the end of the band attached to the fulcrum O will be tight with tension T and the end of the band attached to B will be slack with tension T . Now taking 1 2 moments about the fulcrum O, we have P.l = T .b . . . (For clockwise rotation of the drum) 1 and P.l = T .b . . . (For anticlockwise rotation of the drum) 2 Chapter 19 : Brakes and Dynamometers 743 where l = Length of the lever from the fulcrum (OC), and b = Perpendicular distance from O to the line of action of T or T . 1 2 Notes : 1. When the brake band is attached to the lever, as shown in Fig. 19.11 (a) and (b), then the force (P) must act in the upward direction in order to tighten the band on the drum. 2. If the permissible tensile stress ( σ ) for the material of the band is known, then maximum tension in the band is given by T = σ.. wt 1 where w = Width of the band, and t = thickness of the band. Example 19.6. A band brake acts on the 3/4th of circumference of a drum of 450 mm diam- eter which is keyed to the shaft. The band brake provides a braking torque of 225 N-m. One end of the band is attached to a fulcrum pin of the lever and the other end to a pin 100 mm from the fulcrum. If the operating force is applied at 500 mm from the fulcrum and the coefficient of friction is 0.25, find the operating force when the drum rotates in the (a) anticlockwise direction, and (b) clockwise direction. Solution. Given : d = 450 mm or r = 225 mm = 0.225 m ; T = 225 N-m ; b = OB = 100 mm B = 0.1 m ; l = 500 mm = 0.5 m ; µ = 0.25 Let P = Operating force. (a) Operating force when drum rotates in anticlockwise direction The band brake is shown in Fig. 19.11. Since one end of the band is attached to the fulcrum at O, therefore the operating force P will act upward and when the drum ro- tates anticlockwise, as shown in Fig. 19.11 (b), the end of the band attached to O will be tight with tension T and the 1 end of the band attached to B will be slack with tension T . 2 First of all, let us find the tensions T and T . 1 2 Drums for band brakes. We know that angle of wrap, 33 θ= th of circumference = × 360°= 270° 44 =× 270π /180= 4.713 rad  T 1 and 2.3log . 0.25 4.713 1.178 =µθ= × =  T  1 T  T 1.178 1 1 or . . . (i) ∴ = 3.253 log == 0.5123  T T 2.3 2  2 . . . (Taking antilog of 0.5123) We know that braking torque (T ), B 225 = (T – T ) r = (T – T ) 0.225 1 2 1 2 T – T = 225 / 0.225 = 1000 N . . . (ii) ∴ 1 2 From equations (i) and (ii), we have T = 1444 N; and T = 444 N 1 2 Now taking moments about the fulcrum O, we have P × l = T .b or P × 0.5 = 444 × 0.1 = 44.4 2 P = 44.4 / 0.5 = 88.8 N Ans. ∴ 744 Theory of Machines (b) Operating force when drum rotates in clockwise direction When the drum rotates in clockwise direction, as shown in Fig.19.11 (a), then taking mo- ments about the fulcrum O, we have P × l = T . b or P × 0.5 = 1444 × 0.1 = 144.4 1 P = 144.4 / 0.5 = 288.8 N Ans. ∴ Example 19.7. The simple band brake, as shown in Fig. 19.12, is applied to a shaft carrying a flywheel of mass 400 kg. The radius of gyration of the flywheel is 450 mm and runs at 300 r.p.m. If the coefficient of friction is 0.2 and the brake drum diameter is 240 mm, find : 1. the torque applied due to a hand load of 100 N, 2. the number of turns of the wheel before it is brought to rest, and 3. the time required to bring it to rest, from the moment of the application of the brake. Solution. Given : m = 400 kg ; k = 450 mm = 0.45 m ; N = 300 r.p.m. or = 31.42 rad/s ; µ = 0.2 ; ω=23 π×00/60 d = 240 mm = 0.24 m or r = 0.12 m All dimensions in mm. Fig. 19.12 1. Torque applied due to hand load First of all, let us find the tensions in the tight and slack sides of the band i.e. T and T 1 2 respectively. From the geometry of the Fig. 19.12, angle of lap of the band on the drum, π θ= 360°−150°= 210°= 210× = 3.666 rad 180 We know that  T 1 2.3log . 0.2 3.666 0.7332 =µθ = × =  T  2  T T 0.7332 1 1 = 2.08 log == 0.3188 or . . . (i)  T T 2.3  2 2 ... (Taking antilog of 0.3188) Taking moments about the fulcrum O, T × 120 = 100 × 300 = 30 000 or T = 30 000/120 = 250 N 2 2 T = 2.08T = 2.08 × 250 = 520 N . . . From equation (i) ∴ 1 2 We know that torque applied, T = (T – T ) r = (520 – 250) 0.12 = 32.4 N-m Ans. B 1 2 2. Number of turns of the wheel before it is brought to rest Let n = Number of turns of the wheel before it is brought to rest. We know that kinetic energy of rotation of the drum 11 1 222 2 2 =×Im .ω = × .k .ω =× 400(0.45) (31.42) = 40 000 N-m 22 2 This energy is used to overcome the work done due to the braking torque (T ). B ∴ 40 000 = T × 2πn = 32.4 × 2πn = 203.6 n B or n = 40 000 / 203.6 = 196.5 Ans. Chapter 19 : Brakes and Dynamometers 745 3. Time required to bring the wheel to rest We know that the time required to bring the wheel to rest = n / N = 196.5 / 300 = 0.655 min = 39.3 s Ans. Example 19.8. A simple band brake operates on a drum of 600 mm in diameter that is running at 200 r.p.m. The coefficient of friction is 0.25. The brake band has a contact of 270°, one end is fastened to a fixed pin and the other end to the brake arm 125 mm from the fixed pin. The straight brake arm is 750 mm long and placed perpendicular to the diameter that bisects the angle of contact. 1. What is the pull necessary on the end of the brake arm to stop the wheel if 35 kW is being absorbed ? What is the direction for this minimum pull ? 2. What width of steel band of 2.5 mm thick is required for this brake if the maximum tensile stress is not to exceed 2 50 N/mm ? Solution. Given : d = 600 mm or r = 300 mm ; N = 200 r.p.m. ; µ = 0.25 ; θ= 270° = 270×π /180= 4.713 rad ; 3 2 Power = 35 kW = 35 × 10 W ; t = 2.5 mm ; σ = 50 N/mm 1. Pull necessary on the end of the brake arm to stop the wheel All dimensions in mm Let P = Pull necessary on the end of the brake arm to Fig. 19.13 stop the wheel. The simple band brake is shown in Fig. 19.13. Since one end of the band is attached to the fixed pin O, therefore the pull P on the end of the brake arm will act upward and when the wheel rotates anticlockwise, the end of the band attached to O will be tight with tension T and the end of the 1 band attached to B will be slack with tension T . First of all, let us find the tensions T and T . We 2 1 2 know that  T 1 2.3log =µθ.= 0.2× 5 4.713= 1.178  T  2  T T 1.178 1 1 or ... (Taking antilog of 0.5122) ... (i) ∴ 3.25 log == 0.5122 =  T 2.3 T 2  2 Let T = Braking torque. B We know that power absorbed, 2. π×NT 2π×200×T 3 BB 35×= 10 = = 21 T B 60 60 33 T=× 35 10 / 21=1667 N-m=1667×10 N-mm ∴ B We also know that braking torque (T ), B 3 1667 × 10 = (T – T ) r = (T – T ) 300 1 2 1 2 3 T – T = 1167 × 10 /300 = 5556 N ...(ii) ∴ 1 2 From equations (i) and (ii), we find that T = 8025 N; and T = 2469 N 1 2 746 Theory of Machines Now taking moments about O, we have P × 750 = T × OD = T × 62.5 2 = 2469 × 88.4 = 218 260 2 2 P = 218260 / 750 = 291 N Ans. ∴ 2. Width of steel band Let w = Width of steel band in mm. We know that maximum tension in the band (T ), 1 8025 = σ.. wt = 50 × w × 2.5 = 125 w w = 8025 / 125 = 64.2 mm Ans. ∴ 19.8. Differential Band Brake In a differential band brake, as shown in Fig. 19.14, the ends of the band are joined at A and B to a lever AOC pivoted on a fixed pin or fulcrum O. It may be noted that for the band to tighten, the length OA must be greater than the length OB. (a) Clockwise rotation of the drum. (a) Anticlockwise rotation of the drum. Fig. 19.14. Differential band brake. The braking torque on the drum may be obtained in the similar way as discussed in simple band brake. Now considering the equilibrium of the lever AOC. It may be noted that when the drum rotates in the clockwise direc- tion, as shown in Fig. 19.14 (a), the end of the band attached to A will be slack with tension T and end of the 2 band attached to B will be tight with tension T . On the 1 other hand, when the drum rotates in the anticlockwise direction, as shown in Fig. 19.14 (b), the end of the band attached to A will be tight with tension T and end of the 1 band attached to B will be slack with tension T . Now 2 taking moments about the fulcrum O, we have P.l + T .b = T .a 1 2 Tractors are specially made to move on ... (For clockwise rotation of the drum ) rough terrain and exert high power at or P.l = T .a – T .b ... (i) low speeds. 2 1 Note : This picture is given as additional and P.l + T .b = T .a 2 1 information and is not a direct example of the ... (For anticlockwise rotation of the drum ) current chapter. or P.l = T .a – T .b ... (ii) 1 2 OD = Perpendicular distance from O to the line of action of tension T . 2 OE = EB = OB/2 = 125/2 = 62.5 mm, and ∠DOE = 45° ∴ OD = OE sec 45° = 62.5 2 mm Chapter 19 : Brakes and Dynamometers 747 We have discussed in block brakes (Art. 19.4), that when the frictional force helps to apply the brake, it is said to be self energizing brake. In case of differential band brake, we see from equa- tions (i) and (ii) that the moment T .b and T .b helps in applying the brake (because it adds to the 1 2 moment P.l ) for the clockwise and anticlockwise rotation of the drum respectively. We have also discussed that when the force P is negative or zero, then brake is self locking. Thus for differential band brake and for clockwise rotation of the drum, the condition for self locking is Ta.. ≤T b or TT// ≤b a 21 21 and for anticlockwise rotation of the drum, the condition for self locking is Ta.. ≤T b or TT// ≤b a 12 12 Notes : 1. The condition for self locking may also be written as follows : For clockwise rotation of the drum, Tb ≥T a TT ≥a b .. or // 12 12 and for anticlockwise rotation of the drum, Tb ≥T a TT ≥ab .. or // 21 12 2. When in Fig. 19.14 (a) and (b), the length OB is greater than OA, then the force P must act in the upward direction in order to apply the brake. The tensions in the band, i.e. T and T will remain unchanged. 1 2 Example 19.9. In a winch, the rope supports a load W and is wound round a barrel 450 mm diameter. A differential band brake acts on a drum 800 mm diameter which is keyed to the same shaft as the barrel. The two ends of the bands are attached to pins on opposite sides of the fulcrum of the brake lever and at distances of 25 mm and 100 mm from the fulcrum. The angle of lap of the brake band is 250° and the coefficient of friction is 0.25. What is the maximum load W which can be supported by the brake when a force of 750 N is applied to the lever at a distance of 3000 mm from the fulcrum ? Solution. Given : D = 450 mm or R = 225 mm ; d = 800 mm or r = 400 mm ; OB = 25 mm ; OA = 100 mm ; = 250° = 250 × π/180 = 4.364 rad ; θ µ = 0.25 ; P = 750 N ; l = OC = 3000 mm Since OA is greater than OB, therefore the operating force (P = 750 N) will act downwards. First of all, let us consider that the drum rotates in clockwise direction. We know that when the drum rotates in clock- wise direction, the end of band attached to A will be slack with tension T and the end of the band attached 2 to B will be tight with tension T , as shown in Fig. 19.15. 1 All dimensions in mm. Now let us find out the values of tensions T and T . We 1 2 Fig. 19.15 know that  T 1 2.3log =µθ.= 0.2× 5 4.364= 1.091  T  2 T  T 1.091 1 1 or = 2.98 ... (Taking antilog of 0.4743) ∴ log 0.4743 ==  T T 2.3 2  2 and T = 2.98 T ... (i) 1 2 Now taking moments about the fulcrum O, 750 × 3000 + T × 25 = T × 100 1 2 748 Theory of Machines 3 or T × 100 – 2.98 T × 25 = 2250 × 10 ... ( T = 2.98 T )  2 2 1 2 3 3 3 25.5 T = 2250 × 10 or T = 2250 × 10 /25.5 = 88 × 10 N 2 2 3 3 and T = 2.98T = 2.98 × 88 × 10 = 262 × 10 N 1 2 We know that braking torque, T = (T – T ) r B 1 2 3 3 6 = (262 × 10 – 88 × 10 ) 400 = 69.6 × 10 N-mm ...(i) and the torque due to load W newtons, T = W.R = W × 225 = 225 W N-mm ... (ii) W Since the braking torque must be equal to the torque due to load W newtons, therefore from equations (i) and (ii), 6 3 W = 69.6 × 10 /225 = 309 × 10 N = 309 kN Now let us consider that the drum rotates in anticlockwise direction. We know that when the drum rotates in anticlockwise direction, the end of the band attached to A will be tight with tension T and end of the band attached to 1 B will be slack with tension T , as shown in Fig. 19.16. The 2 ratio of tensions T and T will be same as calculated above, 1 2 i.e. T 1 or T = 2.98 T = 2.98 1 2 T 2 All dimensions in mm. Now taking moments about the fulcrum O, Fig. 19.16 750 × 3000 + T × 25 = T × 100 2 1 3 or 2.98 T × 100 – T × 25 = 2250 × 10 ... ( T = 2.98 T )  2 2 1 2 3 3 273 T = 2250 × 10 or T = 2250 × 10 /273 = 8242 N 2 2 and T = 2.98 T = 2.98 × 8242 = 24 561 N 1 2 ∴ Braking torque, T = (T × T ) r B 1 2 6 = (24 561 – 8242)400 = 6.53 × 10 N-mm ...(iii) From equations (ii) and (iii), 6 3 W = 6.53 × 10 /225 = 29 × 10 N = 29 kN From above, we see that the maximum load (W ) that can be supported by the brake is 309 kN, when the drum rotates in clockwise direction. Ans. Example 19.10. A differential band brake, as shown in Fig. 19.17, has an angle of contact of 225°. The band has a compressed woven lining and bears against a cast iron drum of 350 mm diameter. The brake is to sustain a torque of 350 N-m and the coefficient of friction between the band and the drum is 0.3. Find : 1. The necessary force (P) for the clockwise and anticlockwise rotation of the drum; and 2. The value of ‘OA’ for the brake to be self locking, when the drum rotates clockwise. Solution. Given: θ = 225° = 225 × π /180 = 3.93 rad ; d = 350 mm or r = 175 mm ; 3 T = 350 N-m = 350 × 10 N-mm 1. Necessary force (P) for the clockwise and anticlockwise rotation of the drum When the drum rotates in the clockwise direction, the end of the band attached to A will be slack with tension T and the end of the band attached to B will be tight with tension T , as shown in 2 1 Fig. 19.18. First of all, let us find the values of tensions T and T . 1 2 Chapter 19 : Brakes and Dynamometers 749 All dimensions in mm. Fig. 19.17 Fig. 19.18 We know that  T 1 2.3log . 0.3 3.93 1.179 =µθ= × =  T  2 T  T 1.179 1 1 or = 3.255 ... (Taking antilog of 0.5126 ) ... (i) ∴ log == 0.5126  T 2.3 T 2  2 and braking torque (T ), B 3 350 × 10 = (T – T )r = (T – T ) 175 1 2 1 2 3 T – T = 350 × 10 /175 = 2000 N ... (ii) ∴ 1 2 From equations (i) and (ii), we find that T = 2887 N ; and T = 887 N 1 2 Now taking moments about the fulcrum O, we have 3 P × 500 = T × 150 – T × 35 = 887 × 150 – 2887 × 35 = 32 ×10 2 1 3 P = 32 × 10 /500 = 64 N Ans. ∴ When the drum rotates in the anticlockwise direction, the end of the band attached to A will be tight with tension T and end of the band attached to B will 1 be slack with tension T , as shown in Fig. 19.19. Taking 2 moments about the fulcrum O, we have P × 500 = T × 150 – T × 35 1 2 = 2887 × 150 – 887 × 35 3 = 402 × 10 3 P = 402 × 10 /500 = 804 N Ans. Fig. 19.19 2. Value of ‘OA’ for the brake to be self locking, when the drum rotates clockwise The clockwise rotation of the drum is shown in Fig 19.18. For clockwise rotation of the drum, we know that P × 500 = T × OA – T × OB 2 1 For the brake to be self locking, P must be equal to zero. Therefore T × OA = T × OB 2 1 TO×× B 2887 35 1 OA== and = 114 mm Ans. T 887 2 750 Theory of Machines 19.9. Band and Block Brake The band brake may be lined with blocks of wood or other material, as shown in Fig. 19.20 (a). The friction between the blocks and the drum provides braking action. Let there are ‘n’ number of blocks, each subtending an angle at the centre and the drum rotates in anticlockwise direction. 2θ (a)(b) Fig. 19.20. Band and block brake. Let T = Tension in the tight side, 1 T = Tension in the slack side, 2 µ = Coefficient of friction between the blocks and drum, ′ = Tension in the band between the first and second block, T 1 ′′ TT , etc.= Tensions in the band between the second and third block, 23 between the third and fourth block etc. Consider one of the blocks (say first block) as shown in Fig. 19.20 (b). This is in equilibrium under the action of the following forces : 1. Tension in the tight side (T ), 1 ′ 2. Tension in the slack side ( ) or tension in the band between the first and second block, T 1 3. Normal reaction of the drum on the block (R ), and N 4. The force of friction ( µ.R ). N Resolving the forces radially, we have ′ ... (i) ()+θsin= TT R 11 N Resolving the forces tangentially, we have ′ ... (ii) () TT+θcos=µ.R 11 N Dividing equation (ii) by (i), we have ′ () TT−θcos µ .R 11 N = ′ R () TT+θsin N 11 ′′ or () TT−=µtanθ(T+ T) 11 1 1 T 1t +µ anθ 1 ∴ = ′ 1t −µ an θ T 1 Similarly, it can be proved for each of the blocks that ′ ′′ TTTT 1t +µ an θ 1231 n− === ........ = T 1t −µ an θ ′′ ′ TT T 2 234 Chapter 19 : Brakes and Dynamometers 751 n ′′ TT T T T  1t +µ anθ 11 1 2 n−1 = ×××.........× =  ... (iii) ∴ TT 1t −µ an θ ′′ ′  22 TT T 12 3 Braking torque on the drum of effective radius r , e T = (T – T ) r B 1 2 e = (T – T ) r ... Neglecting thickness of band 1 2 ′ Note : For the first block, the tension in the tight side is T and in the slack side is T and for the second block, 1 1 ′ ′ the tension in the tight side is and in the slack side is . Similarly for the third block, the tension in the T T 1 2 T′ ′ tight side is and in the slack side is and so on. For the last block, the tension in the tight side is T 2 3 T and in the slack side is T . n-1 2 Example 19.11. In the band and block brake shown in Fig. 19.21, the band is lined with 12 blocks each of which subtends an angle of 15° at the centre of the rotating drum. The thickness of the blocks is 75 mm and the diameter of the drum is 850 mm. If, when the brake is in action, the greatest and least tensions in the brake strap are T and T , show that 1 2 12 T 1t +µ an7.°5 1 = , where µ is the  T 1t −µ an7.°5 2 coefficient of friction for the blocks. All dimensions in mm. With the lever arrangement as shown in Fig.19.21, find the least force required at C for the Fig. 19.21 blocks to absorb 225 kW at 240 r.p.m. The coefficient of friction between the band and blocks is 0.4. Solution. Given : n = 12 ; = 15° or = 7.5°; t = 75 mm = 0.075 m ; d = 850 mm 2θ θ 3 = 0.85 m ; Power = 225 kW = 225 × 10 W; N = 240 r.p.m.; µ = 0.4 Since OA OB, therefore the force at C must act downward. Also, the drum rotates clock- wise, therefore the end of the band attached to A will be slack with tension T (least tension) and the 2 end of the band attached to B will be tight with tension T (greatest tension). 1 Consider one of the blocks (say first block) as shown in Fig. 19.22. This is in equilibrium under the action of the following four forces : 1. Tension in the tight side (T ), 1 ′ 2. Tension in the slack side ( ) or the tension in the band between the first and second block, T 1 3. Normal reaction of the drum on the block (R ), and N 4. The force of friction ( µ.R ). N Resolving the forces radially, we have ′ ... (i) () TT+° sin7.5=R 11 N Resolving the forces tangentially, we have ′ ... (ii) () TT−° cos7.5=µ.R 11 N Dividing equation (ii) by (i), we have ′ ′ TT − () TT−° cos7.5 11 11 Fig. 19.22 =µ tan 7.° 5 =µ or ′ ′ TT + () TT+°sin7.5 11 11

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