1 Gyroscope. Course Contents

Transcription

1 1 Gyroscope Course Contents 1.1 Principle of Gyroscope 1. Angular Velocity 1.3 Angular Acceleration 1.4 Gyroscopic Torque (Couple) 1.5 Gyroscopic Effect on Aeroplanes 1.6 Gyroscopic Effect on Naval Ships 1.7 Stability of an Automobile 1.8 Stability of two-wheel vehicle 1.9 Rigid Disc at an Angle Fixed To a Rotating Shaft 1.10 Summary Prepared By: Subhesh Pansuria Darshan Institute of Engineering & Technology, Rajkot Page 1.1

2 1. Gyroscope Theory of Machines (15190) 1.1 Principle of Gyroscope If the axis of spinning or rotating body is given an angular motion about an axis perpendicular to the axis of spin, an angular acceleration acts on the body about the third perpendicular axis. The torque required to produce this acceleration is known as active gyroscopic torque. A reactive gyroscopic torque or couple also acts similar to the concept of centripetal and centrifugal forces on a reacting body. The effect produced by the reactive gyroscopic couple is known as gyroscopic effect. Thus aeroplanes, ships, automobiles, etc., that have rotating parts in the form of wheels or rotors of engines experiences this effect while taking turn, i.e., when the axes of spin is subjected to some angular motion. 1. Angular Velocity The angular velocity of a rotating body is specified by the magnitude of velocity the direction of the axis of rotor the sense of rotation of the rotor, i.e., clockwise or counter-clockwise Angular velocity is represented by a vector in the following manner: (i) Magnitude of the velocity is represented by the length of the vector. (ii) Direction of axis of the rotor is represented by drawing the vector parallel to the axis of the rotor or normal to the plane of the angular velocity. (iii) Sense of rotation of the rotor is denoted by taking the direction of the vector in a set rule. The general rule is that of a right-handed screw, i.e., if a screw is rotated in the clockwise direction, it goes away from the viewer and vice-versa. For example, Fig. (a) shows a rotor which rotates in the clockwise direction when viewed from the end l. Its angular motion has been shown vectorially in Fig. (b). The vector has been taken to Prepared By: Subhesh Pansuria Darshan Institute of Engineering & Technology, Rajkot Page 1.

3 1. Gyroscope Theory of Machines (15190) a scale parallel to the axis of the rotor. The sense of direction of the vector is from a to b according to the screw rule. However, if the direction of rotation of the rotor is reversed, it would be from b to a [Fig.(c)]. 1.3 Angular Acceleration Let a rotor spin (rotate) about the horizontal axis Ox at a speed of ω rad/s in the direction as shown in Fig.(a). Let oa represent its angular velocity Fig. (b). Now, if the magnitude of the angular velocity changes to (ω+δω) and the direction of the axis of spin to Ox (in time δt). The vector ob would represent its angular velocity in the new position. Join ab which represents the change in the angular velocity of the rotor. The vector ab can be resolved into two components: (i) ac representing angular velocity change in a plane normal to ac or x-axis, and (ii) cb representing angular velocity change in a plane normal to cb or y-axis. Change of angular velocity, ac = (ω + δω) cos δθ ω Rate of change of angular velocity (ω + δω) cos δθ ω = δt Angular acceleration lim δt 0 (ω + δω) cos δθ ω δt Prepared By: Subhesh Pansuria Darshan Institute of Engineering & Technology, Rajkot Page 1.3

4 1. Gyroscope Theory of Machines (15190) As δt 0, δθ 0 and cos δθ 1 Angular acceleration ω + δω ω = lim = dω δt 0 δt dt Change of angular velocity, cd = (ω + δω) sin δθ Rate of change of angular velocity (ω + δω) sin δθ = δt Angular acceleration = lim δt 0 (ω + δω) sin δθ δt As δt 0, δθ 0 and cos δθ δθ Angular acceleration (ω + δω)δθ = lim = ω dθ δt 0 δt dt Total angular acceleration, α = dω dt + ω dθ dt This shows that the total angular acceleration of the rotor is the sum of 1. dω/dt, representing change in the magnitude of the angular velocity of the rotor. ω dθ/dt, Representing change in the direction of the axis of spin, the direction of cb is from c to b in the vector diagram (being a component of ab), the acceleration acts clockwise in the vertical plane xy. (When viewed from front along they-axis) Prepared By: Subhesh Pansuria Darshan Institute of Engineering & Technology, Rajkot Page 1.4

5 1. Gyroscope Theory of Machines (15190) 1.4 Gyroscopic Torque (Couple) Let I be the moment of inertia of a rotor and ω its angular velocity about a horizontal axis of spin Ox in the direction as shown in Fig. (a). Let this axis of spin turn through a small angle δθ in the horizontal plane (xy) to the position Ox' in time δt, Figure (b) shows the vector diagram. oa represents the angular velocity vector when the axis is ox and ob when the axis is changed to ox'. Then ab represent the change in the angular velocity due to change in direction of the axis of spin of the rotor. This change in the angular velocity is clockwise when viewed from a towards b and is in the vertical plane xz. This change results in angular acceleration, the sense and direction of which are the same as that of the change in the angular velocity. Change in the angular velocity, ab = ω δθ Angular acceleration, α = ω dθ dt In the limit, whn δt 0, α = ω dθ dt Usually, dθ/dt the angular velocity of the axis of spin is called the angular velocity of precession and is denoted by ω p Prepared By: Subhesh Pansuria Darshan Institute of Engineering & Technology, Rajkot Page 1.5

6 1. Gyroscope Theory of Machines (15190) Angular acceleration α = ω ω p The torque required to produce this acceleration is known as the gyroscopic torque and is a couple which must be applied to the axis of spin to cause it to rotate with angular velocity ω p about the axis of precession Oz. Acceleration torque, T = I α = I ω ω p For the configuration of Fig.(a) Ox is known as the axis of spin Oz is known as the axis of precession Oy is known as the axis of gyroscopic couple 00 yz is plane of spin xy is plane of precession yz is plane of gyroscopic couple The torque obtained above is that which is required to cause the axis of spin to precess in the horizontal plane and is known as the active gyroscopic torque or the applied torque. A reactive gyroscopic torque or reaction torque is also applied to the axis which tends to rotate the axis of spin in the opposite direction i.e., in the counter-clockwise direction in the above case. Just as the centrifugal force on a rotating body tends move the body tends to move the body outwards, while a centripetal acceleration (and thus centripetal force) acts on it inwards, in the same way, the effects of active and reactive gyroscopic torques can be understood. The effect of the gyroscopic couple on a rotating body is known as the gyroscope effect on the body. A gyroscope is a spinning body which is free to move in other directions under the action of external forces. Prepared By: Subhesh Pansuria Darshan Institute of Engineering & Technology, Rajkot Page 1.6

7 1. Gyroscope Theory of Machines (15190) 1.5 Gyroscopic Effect on Aeroplanes Figure (a) shows an aeroplane in space. Let the propeller be rotating in the clockwise direction when viewed from the rear end. The angular momentum vector oa due to the angular velocity is shown in Fig. (b). If the plane takes a left turn, the angular momentum vector is shifted and may be represented by the vector ob. The change is shown by the vector ab and is the active gyroscopic couple. This vector is in the horizontal plane and is perpendicular to the vector oa in the limit. The reactive vector is given by b'a' which is equal and opposite to the vector ab. The interpretation of this vector shows that the couple acts in the vertical plane and is counter-clockwise when viewed from the right-hand side of the plane. This indicates that it tends to raise the nose and depress the tail of the aeroplane. Figure (c) shows the gyroscopic effect, when the aeroplane takes the right turn. The change is shown by the vector cd and is the active gyroscopic couple. It is perpendicular to the vector oc in the limit in the horizontal plane. The reactive couple is given by d'c'. The couple acts in the vertical plane and is clockwise when viewed from the right-hand side of the plane. Thus, it tends to dip the nose and raise the tail of the aeroplane. If the rotation of the engine is reversed, i.e., it rotates counter-clockwise when viewing from the rear end, the angular momentum vector is oe as shown in Fig. (d). On taking a left turn, it changes to of. The active gyroscopic vector is ef and the reactive fe'. Viewing from the right-hand side of the plane, it indicates that the nose is dipped and the tail is raised. Similarly, when the Prepared By: Subhesh Pansuria Darshan Institute of Engineering & Technology, Rajkot Page 1.7

8 1. Gyroscope Theory of Machines (15190) plane takes a right tum, the effect is indicated in Fig. (e). The nose is raised and the tail is depressed. It can be concluded from the above cases that if the direction of either the spin of the rotor or of the precession is changed, the gyroscopic effect is reversed, but if both are changed, the effect remains same. 1.6 Gyroscopic Effect on Naval Ships Some of the terms used in connection with the motion of naval ships or sea vessels are given below. Bow is the fore or the front end. Stern or aft is the rear end. Starboard is the right hand side when looking from stern. Port is the left-hand side when looking form the stern. Steering is turning on the side when viewing from the top. Pitching is limited angular motion of the ship about the transverse axis. Rolling is limited angular motion of the ship about the longitudinal axis. Let the plane of spin the rotor and other rotating masses be horizontal and across the breath of the ship. Assume ω to the angular velocity of the rotor in the clockwise direction when viewed from stern. Prepared By: Subhesh Pansuria Darshan Institute of Engineering & Technology, Rajkot Page 1.8

9 1. Gyroscope Theory of Machines (15190) Gyroscopic Effect during Turning When the ship turns left the angular momentum vector changes from oa to ob. The reaction couple is found to be b a which tends to raise the bow and lower the stern. On the turning right the reaction couple is revered. So that bow is lower and stern is raised. Gyroscopic Effect on Pitching Pitching of the ship is usually considered to take place with simple harmonic motion. A simple harmonic motion is represented by x = X sin ω 0t Such a motion is obtain by the projection of a rotating vector X on a diameter while rotating around a circle with a constant angular velocity ω 0 and where x is a displacement from the time mean position in time t. in such the way angular displacement θ of the axis of the spin from its mean position is given by θ = φ sin ω 0 t Where φ = amplitude (angular) of swing or the maximum angle turned from the mean position in radius Angular velocity of precession, ω 0 = angular velocity of SHM = π Time Period dθ dt = φ ω 0 cos ω 0 t This is maximum when cos ω 0 t = 1 Therefore, maximum angular velocity of precession ω p = φ ω 0 Gyroscopic Couple, π I ω ω p = I ω (φ Time Period ) Prepared By: Subhesh Pansuria Darshan Institute of Engineering & Technology, Rajkot Page 1.9

10 1. Gyroscope Theory of Machines (15190) When the bow is rising, the reaction couple is clockwise on viewing from top and thus the ship would move towards right or starboard side. Similarly, when the bow is lowered, the ship turns toward left or port side. Angular acceleration = φ ω 0 sin ω 0 t Maximum angular acceleration = φ ω 0 Gyroscopic Effect on Rolling As the axes of the rolling of the ship of the rotor are parallel, there is no precession of the axis of spin and thus there is no gyroscopic effect. In the same way, the effect on steering, pitching or rolling can be observed when the plane of the spin of the rotating masses is horizontal but along the longitudinal axis of the vessel or the axis is vertical. 1.7 Stability of an Automobile Prepared By: Subhesh Pansuria Darshan Institute of Engineering & Technology, Rajkot Page 1.10

11 1. Gyroscope Theory of Machines (15190) In case of a four wheeled vehicle, it is essential that that no wheel is lifted off the ground while the vehicle takes a turn. The condition is fulfilled as long as the vertical reaction of the ground on any the wheels is positive Fig. shows a four wheeled vehicle having a mass m. assuming that the weight is equally divided among the four wheels, Weight on each wheel = w 4 = mg 4 (downwards) Reaction of ground on each wheel, R w = w 4 = mg 4 (upwards) Effect of Gyroscopic Couple Gyroscopic couple due to four wheels, C w = 4 I w ω w ω p = 4 I w v rr Where I w = mass moment of inertia of each wheel ω w = angular velocity of wheel = v r ω p = angular velocity of precession = v R v = linear velocity of the vehicle R = radius of curvature Gyroscopic couple due to engine rotating parts, C e = I e ω e ω p = I e G ω w ω p Where G is the gear ratio, G = ω e ω w Prepared By: Subhesh Pansuria Darshan Institute of Engineering & Technology, Rajkot Page 1.11

12 1. Gyroscope Theory of Machines (15190) Total gyroscopic couple, C G = c w ± c e Positive sign is used when the engine parts rotate in same direction as the wheels and the negative sign when they rotate in the opposite. Assuming that C G is positive and the vehicle takes a left turn, the reaction gyroscopic couple on it is clockwise when viewed from the rear of the vehicle. The reaction couple is provided by equal and opposite forces on the outer and the inner wheels of the vehicle. Forces on two outer wheels = c G w (downwards) Forces on the two inner wheels = c G w (upwards) Forces on each of the outer wheels = c G w (downwards) Forces on each of the inner wheels = c G w (upwards) Thus the forces on each of the outer wheels is similar to the weight. On the inner wheels it is in the opposite direction. Thus, Reaction of ground on each outer wheel, R G = c G w (upwards) Reaction of ground on each inner wheel, R G = c G w (downward) Effect of Centrifugal Couple As the vehicle moves on a curved path, a centrifugal force also acts on the vehicle in the outward direction at the centre of mass of the vehicle. Centrifugal force = m R ω p = m R ( v R ) = m v R This force would tend to overturn the vehicle outwards and the overturning couple will be C c = m R ω p = m v R h Prepared By: Subhesh Pansuria Darshan Institute of Engineering & Technology, Rajkot Page 1.1

13 1. Gyroscope Theory of Machines (15190) This is equivalent to a couple to equal and opposite forces on outer and inner wheels. Forces on each outer wheel = c c w (downwards) Forces on each inner wheel = c c w (upwards) Again, the force on each of the outer wheels is similar to the weight on each of the inner wheels, it is opposite. Reaction of ground on each outer wheel, R c = c c w (upwards) Reaction of ground on each inner wheel, R c = c c w (downwards) Vertical reaction on each outer wheel = w 4 + c G w + c c w (upwards) Vertical reaction on each inner wheel = w 4 + c G w + c c w (upwards) It can be observed that there are chances that the reaction of the ground on the inner wheels may not be upwards and thus the wheels are lifted from the ground. For positive reaction, the Conditions will be w 4 c G w c c w 0 or w 4 c G + c c w or R w R G + R c 1.8 Stability of two-wheel vehicle Prepared By: Subhesh Pansuria Darshan Institute of Engineering & Technology, Rajkot Page 1.13

14 1. Gyroscope Theory of Machines (15190) The case of two-wheel vehicle can be taken in the same way as that of an auto mobile. However, it easier to tilt such a vehicle inwards to neutralise the overturning effect and the vehicle can stay in equilibrium while taking a turn. Let a vehicle take a left turn as shown in fig. (a). The vehicle is inclined to the vertical (inwards) for equilibrium. The angle of inclination of the vehicle to the vertical is known as the angle of heel. Let v = linear velocity of vehicle on the track r = radius of the wheel R = radius of the Track I x = moment of inertia of each wheel I e = moment of inertia of rotating parts of the engine m = total mass of the vehicle and the rider ω w = angular velocity of the wheel ω e = angular velocity of rotating parts of engine G = gear ratio h = height of center of mass of the vehicle and the rider θ = inclination of vehicle to the vertical (angle of heel) As the axis of spin is not horizontal but inclined to the vertical at an angle θ and the axis of precession is vertical, it is necessary to take the horizontal component of the spin vector. Spin vector (horizontal) = I w cos θ = ( I w ω w + I e ω e ) cos θ and Gyroscopic Couple = ( I w ω w + G I e ω w ) cos ω p = ( I w + G I e ) v r v R cos θ = ( I w + G I e ) v rr cos θ The reaction couple b a is clockwise when viewed for the rear (back) of the vehicle and tends to overturn it in the outward direction. Prepared By: Subhesh Pansuria Darshan Institute of Engineering & Technology, Rajkot Page 1.14

15 1. Gyroscope Theory of Machines (15190) Overturning couple due to centrifugal force = (m v ) h cos θ R Total overturning = ( I w + G I e ) v v cos θ + m h cos θ rr R = v R ( I w + G I e r + mh) cos θ Rightening (balancing)couple due to the weight = mgh sin θ For Equilibrium, v R ( I w + G I e r + mh) cos θ = mgh sin θ From This Relation, The angle of heel θ can be determined to avoid skidding of the vehicle. 1.9 Rigid Disc at an Angle Fixed To a Rotating Shaft Consider a circular disc fixed rigidly to a rotating shaft in such a way that the polar axis of the shaft makes angel θ with the axis of the shaft (Fig.). Assume that the shaft rotates clockwise with angular velocity ω when viewed along the left end of the shaft. Let OX be the axis of the shaft OP be the polar axis of the disc and OD the horizontal diametral axis of the disc Also, let m, r and t be the mass, radius and the thickness of the disc. Then I p = moment of inertia of disc about polar Axis OP = mr Prepared By: Subhesh Pansuria Darshan Institute of Engineering & Technology, Rajkot Page 1.15

16 1. Gyroscope Theory of Machines (15190) I d = moment of inertia of disc about diametral axis = m ( t 1 + r 4 ) = (mr 4 ) (If the disc is thin and t is neglected) First consider the spinning about the polar axis Angular velocity of spin = Angular velocity of Disc about the polar axis OP = ω cos θ Angular velocity of precession = Angular velocity of disc about the diametral axis OD = ω sin θ Gyroscopic couple = I p ω cos θ ω sin θ = 1 I p ω sin θ Its effect is to rotate the disc counter-clockwise when viewing from the top. Now consider the spinning about the diametral axis Angular velocity of spin = Angular velocity of disc about the diametral axis OD = ω sin θ Angular velocity of precession = = Angular velocity o disc about the polar axis OP = ω cos θ Gyroscopic couple = I d ω sin θ ω cos θ = 1 I d ω sin θ Its effect is to rotate the disc clockwise when viewing from the top. (Angular velocity of precession is counter-clockwise when viewing from the right end along OP.) Resultant gyroscopic couple on the disc, C = 1 (I p I d ) ω sin θ = 1 (mr mr 4 ) ω sin θ = mr 8 ω sin θ 1.10 Summary 1. The angular velocity is represented by a vector by drawing the parallel to the axis of the rotor and representing the magnitude by the length of the vector to some scale. Sense of rotation of the rotor is denoted by the rule of a right- handed screw, i.e., if a screw is rotated in the clockwise direction, it goes away from the viewer and vice-versa.. The axis of spin, the axis of precession and the axis of gyroscopic couple are in three perpendicular planes. Prepared By: Subhesh Pansuria Darshan Institute of Engineering & Technology, Rajkot Page 1.16

17 1. Gyroscope Theory of Machines (15190) 3. The torque required to cause the axis of spin to process in a plane is known as the active gyroscopic torque or the applied torque. 4. A reactive gyroscopic torque or reaction torque tends to rotate the axis of spin in the opposite direction. 5. The effect of the gyroscopic couple on a rotating body is known as the gyroscopic effect on the body. 6. A gyroscopic is a spinning body which is free to move in other directions under the action of external forces. 7. A four-wheel vehicle tends to turn outwards when taking a turn due to the effect of gyroscopic couple and the centrifugal force. 8. A two-wheel vehicle stabilises itself by tilting towards inside while taking a turn to nullify the effects of gyroscopic couple and the centrifugal force. Prepared By: Subhesh Pansuria Darshan Institute of Engineering & Technology, Rajkot Page 1.17

18 1. Gyroscope Theory of Machines (15190) Example 1 An aeroplane flying at 40 km/h turns towards the left and completes a quarter circle of 60 m radius. The mass of the rotary engine and the propeller of the plane is 450 kg with a radius of gyration of 30 mm. The engine speed is 000 rpm clockwise when viewed from rear. Determine the gyroscopic couple on the aircraft and state its effect. In what way is the effect changed when the (1) aeroplane turns towards right () engine rotates clockwise when viewed from the front (nose end) and the aeroplane turns (a) left (b) right? Ans. m = 450 kg k = 0.3 m N = 000 rpm w = 09.4 rad/sec v = 40 km/h = m/s I = m.k = kg.m R = 60 m wp = v / R = 1.11 rad/sec C Turning = I ω ω p = = N. m (1) When engine rotates clockwise from rear and aeroplane takes right turn, the nose is depressed and the tail is raised. () When engine rotates clockwise from front (a) aeroplane takes left turn, the nose is depressed and the tail is raised (b) aeroplane takes right turn, the nose is raised and the tail is depressed. Example The turbine rotor of a sheep has a mass of. tonnes and rotates at 1800 rpm clockwise when viewed from the aft. The radius of gyration of the rotor is 30 mm. Determine the gyroscopic couple and its effect when the (1) ship turns right at a radius of 50 m with a speed of 5 km/h () ship pitches with the bow rising at an angular velocity of 0.8 rad/s (3) ship rolls at an angular velocity of 0.1 rad/s. Ans. m =. tonne = 00 kg N = 1800 rpm R = 50 m k = 0.3 m r = 0.3 m v = 5 km/h = 6.94 m/s I = m.k = 5.3 kg.m w = rad/sec Prepared By: Subhesh Pansuria Darshan Institute of Engineering & Technology, Rajkot Page 1.1

19 1. Gyroscope Theory of Machines (15190) (1)wp = v / R = rad/sec ()wp = 0.8 rad/sec (3)wp = 0.1 rad/sec C Turning = I ω ω p = = 1180 N. m The effect is to lower the bow and raise the stern when ship turns right. C Pitching = I ω ω p = = 3397 N. m The effect is to turn the ship right towards starboard when the bow is rising. C Rolling = I ω ω p = = N. m No gyroscopic effect as the axis of spin is parallel to axis of precession. Example 3 Each wheel of a four wheeled rear engine automobile has a moment of inertia of.4 kg.m and an effective diameter of 660 mm. The rotating parts of the engine have a moment of inertia of 1. kg.m. The gear ratio of engine to the back wheel is 3 to 1. The engine axis is parallel to the rear axle and the crankshaft rotates in the same sense as the road wheels. The mass of the vehicle is 00 kg and the centre of the mass is 550 mm above the road level. The track width of the vehicle is 1.5 m. determine the limiting speed of the vehicle around a curve with 80 m radius so that all the four wheels maintain contact with the road surfaces. Ans. Iw =.4 kg.m m = 00 kg r = 0.33 m h = 0.55 m Ie = 1. kg.m w = 1.5 m G = 3 R = 80 m Reaction due to weight, R w = Gyroscopic Couple due to wheel, m g 4 = = N ( ) v C w = 4 I w ω ω p = 4 I w r R == 4.4 v = 0.364v Prepared By: Subhesh Pansuria Darshan Institute of Engineering & Technology, Rajkot Page 1.1

20 1. Gyroscope Theory of Machines (15190) Gyroscopic Couple due to engine, v C e = Ie G ω ω p = Ie G r R = 1. 3 v = 0.136v Total Gyroscopic Couple, C G = C w + C e = 0.364v v = 0.5v Reaction due to Gyroscopic Couple, R Go = C G w = 0.5v 1.5 = 0.167v ( ) Centrifugal Couple, R Gi = C G w = 0.5v 1.5 = 0.167v ( ) Reaction due to Centrifugal Couple, m v 00 v C c = h = 0.55 = 15.15v R 80 R co = C c w = 15.15v 1.5 = 5.04v ( ) For Maximum safe speed, R ci = C c w = 15.15v 1.5 = 5.04v ( ) R w = R Gi + R ci = ( ) v v = v = 3.18 m/s = km/h Prepared By: Subhesh Pansuria Darshan Institute of Engineering & Technology, Rajkot Page 1.1

21 1. Gyroscope Theory of Machines (15190) Example 4 Each wheel of a motorcycle is of 600 mm diameter and has a moment of inertia of 1. kg.m. The total mass of the motorcycle and the rider is 10 kg and the combined centre of mass is 580 mm above the ground level when the motor cycle is upright. The moment of inertia of rotating parts of the engine is 0. kg.m. The engine speed is 5 times the speed of wheels and is in the same sense. Determine the angle of heel necessary when the motorcycle takes a turn of 35 m radius at a speed of 54 km/h. Ans. m = 180 kg Iw = 1. kg/m r = 0.3 m Ie = 0. kg/m G = 5 h = 0.85 m v = 54 km/h = 15 m/s R = 35 m Gyroscopic Couple, C G = ( Iw + G Ie) v cos θ r R 15 C G = ( ) cos θ = 7.86 cos θ Centrifugal Couple, C e = m v h cos θ R C e = cos θ = cos θ 35 Total Overturning Couple = ( ) cos θ = 744 cos θ Rightening Couple = m g h sin θ = sin θ = 104 sin θ 104 sin θ = 744 cos θ tan θ = = 0.77 θ = 36 Prepared By: Subhesh Pansuria Darshan Institute of Engineering & Technology, Rajkot Page 1.1

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23 FRICTION DEVICES: CLUTCHES, BRAKES AND DYNAMOMETERS Course Contents.1 Introduction to Clutch. Classification of Clutches Positive Contact Clutch Friction Clutch.3 Brakes.4 Classification of Brakes.5 Energy Equations Block or Shoe Brake Band Brake Band and Block Brake Internal Expanding Brake.6 Braking of Vehicle.7 Dynamometer.8 Types of Dynamometer Prony Brake Dynamometer Rope Brake Dynamometer Belt Transmission Dynamometer Epicyclic Train Dynamometer Bevis Gibson Torsion Dynamometer Prepared By: Vimal Limbasiya Darshan Institute of Engineering & Technology, Rajkot Page.1

24 . Friction Devices: Clutches, Brakes and Dynamometers Theory of Machines (15190).1 Introduction to clutch The clutch is a mechanical device, which is used to connect or disconnect the source of power from remaining parts of power transmission system at the will of the operator. An automotive clutch can permit the engine to run without driving the car. This is desirable when the engine is to be started or stopped, or when the gears are to be shifted. Very often, three terms are used together, namely, couplings, clutches and brakes. There is a basic difference between the coupling and the clutch. A coupling, such as a flange coupling, is a permanent connection. The driving and driven shafts are permanently attached by means of coupling and it is not possible to disconnect the shafts, unless the coupling is dismantled. On the other hand, the clutch can connect or disconnect the driving and driven shafts, as and when required by the operator. Similarly, there is a basic difference between initial and final conditions in clutch and brake operations. In the operation of clutch, the conditions are as follows: (i) Initial Condition The driving member is rotating and the driven member is at rest. (ii) Final Condition Both members rotate at the same speed and have no relative motion. In the operation of brake, the conditions are as follows: (i) Initial Condition One member such as the brake drum is rotating and the braking member such as the brake shoe is at rest. (ii) Final Condition Both members are at rest and have no relative motion. Functions of clutches (i) It maintains constant load on shaft (ii) It transmits or starts high inertia load of the machine with a small motor. (iii) It acts like emergency device to disengage the shaft of the machine from the shaft of the motor in case of condition of machine jam.. Classification of clutches Clutches are classified into the following four groups: (i) (ii) Positive contact Clutches: They include square jaw clutches; spiral jaw clutches and toothed clutches. In these clutches, power transmission, is achieved by means of interlocking of jaws or teeth. Their main advantage is positive engagement and once coupled, they can transmit large torque with no slip. Friction Clutches: They include single and multi-plate clutches, cone clutches and centrifugal clutches. In these clutches, power transmission is achieved by means of friction between contacting surfaces. (iii) Fluid Clutches and Couplings: In these clutches, power transmission is achieved by means of hydraulic pressure. A fluid coupling provides extremely smooth starts and absorbs shock. Prepared By: Vimal G. Limbasiya Page. Darshan Institute of Engineering & Technology, Rajkot

25 Theory of Machines (15190). Friction Devices: Clutches, Brakes and Dynamometers (iv) Electromagnetic Clutches: They include magnetic particle clutches, magnetic hysteresis clutches and eddy current clutches. In these clutches, power transmission is achieved by means of the magnetic field. These clutches have many advantages, such as rapid response time, ease of control, and smooth starts and stops. Positive Contact Clutches The simplest form of positive contact clutches is the square jaw clutch as shown in Fig..1. It consists of two halves carrying projections or jaws. One clutch half is fixed and the other can move along the axis of the shaft over either a feather key or splines by means of shift lever. During the engagement, the jaws of the moving half enter into the sockets of the mating half. The engaging surfaces of jaw and socket form a rigid mechanical junction. Jaw clutches can be used to connect shafts, when the driving shaft is stationary or rotating at very low velocity. There are two types of jaws, namely, square and spiral. The spiral jaws can be engaged at slightly higher speed without clashing. Frequent engagement results in wear of jaws. The jaw clutches have the following advantages: (a) They do not slip and engagement is positive. (b) No heat is generated during engagement or disengagement. Fig..1 Square Jaw Clutch The jaw clutches have the following drawbacks: (a) Jaw clutches can be engaged only when both shafts are stationary or rotate with very small speed difference. (b) They cannot be engaged at high speeds because engagement of jaws and sockets results in shock. In general, positive contact clutches are rarely used as compared with friction clutches. However, they have some important applications where synchronous operation is required like power presses and rolling mills. Prepared By: Vimal G. Limbasiya Darshan Institute of Engineering & Technology, Rajkot Page.3

26 . Friction Devices: Clutches, Brakes and Dynamometers Theory of Machines (15190) Friction Clutches Single Plate Clutch A single plate friction clutch consisting of two flanges is shown in Fig... One flange is rigidly keyed to the driving shaft, while the other is connected to the driven shaft by means of splines. The splines permit free axial movement of the driven flange with respect to the driven shaft. This axial movement is essential for engagement and disengagement of clutch. The actuating force is provided by a helical compression spring, which forces the driven flange to move towards the driving flange. Power is transmitted from the driving shaft to the driving flange by means of the key. Power is then transmitted from the driving flange to the driven flange by means of frictional force. Finally, power is transmitted from the driven flange to the driven shaft by means of the splines. Since the power is transmitted by means of frictional force between the driving and driven flanges, the clutch is called 'friction' clutch. In order to disengage the clutch, a fork is inserted in the collar on the driven flange to shift it axially to the right side. This relieves the pressure between the driving and the driven flanges and no torque can be transmitted. In the working condition, the clutch is in an engaged position under the action of spring force. Levers or forks are operated to disengage the clutch. Fig.. Single Plate Clutch Advantages of Friction Clutch The main advantages of friction clutch are as follows: (i) The engagement is smooth. (ii) Slip occurs only during engaging operation and once the clutch is engaged, there is no slip between the contacting surfaces. Therefore, power loss and consequent heat generation do not create problems, unless the operation requires frequent starts and stops. Prepared By: Vimal G. Limbasiya Page.4 Darshan Institute of Engineering & Technology, Rajkot

27 Theory of Machines (15190). Friction Devices: Clutches, Brakes and Dynamometers (iii) In certain cases, the friction clutch serves as a safety device. It slips when the torque transmitted through it exceeds a safe value. This prevents the breakage of parts in the transmission chain. Depending upon the number of friction surfaces, the friction clutches are classified as single-plate or multi-plate clutches. Depending upon the shape of the friction material, the clutches are classified as disk clutches, cone clutches or expanding shoe clutches. The following factors should be considered while designing friction clutches: (i) Selection of a proper type of clutch that is suitable for the given application (ii) Selection of suitable friction material at the contacting surfaces (iii) Designing the clutch for sufficient torque capacity (iv) Engagement and disengagement should be without shock or jerk (v) Provision for holding the contacting surfaces together by the clutch itself and without any external assistance (vi) Low weight for rotating parts to reduce inertia forces, particularly in high-speed applications (vii) Provision for taking or compensating wear of rubbing surfaces (viii) Provision for carrying away the heat generated at the rubbing surfaces Torque Transmitting Capacity A friction disk of a single plate clutch is shown in Fig.3. The following notations are used in the derivation: r O = outer diameter of friction disk (mm) r i = inner diameter of friction disk (mm) p = intensity of pressure at radius r (N/mm ) W = total operating force (N) F t = frictional force applied on plate T = torque transmitted by the clutch (N-mm) Fig..3 Friction Disk Assume elemental ring at radius r whose thickness is dr. dw = axial force on elemental ring dw = p x r dr Prepared By: Vimal G. Limbasiya Darshan Institute of Engineering & Technology, Rajkot Page.5

28 . Friction Devices: Clutches, Brakes and Dynamometers Theory of Machines (15190) Elemental area = r dr Elemental axial force = p (r dr) ro Total axial force, W dw ri ro Total axial force, W p ( r) dr ri ro Total axial force, W p r dr ri Two theories are used to obtain the torque capacity of the clutch. They are called uniform pressure theory and uniform wear theory. (i) Uniform Pressure Theory: In case of new clutches employing a number of springs, the pressure remains constant over the entire surface area of the friction disk. With this assumption, p is assumed to be constant. This constant pressure distribution is illustrated in Fig..4. Fig..4 Pressure Distribution Axial force on elemental ring ro W p r dr W ri r O r p W p ro r i W p ro r i Total torque transmitted, t ro ri ri ro M T dt df r ro ri pr ri t dr ro dw r ri (i) Prepared By: Vimal G. Limbasiya Page.6 Darshan Institute of Engineering & Technology, Rajkot

29 Theory of Machines (15190). Friction Devices: Clutches, Brakes and Dynamometers ro p r dr ri r 3 O r p pro r i 3 Substituting value of p from equation (i) W T r r O i 3 r r O ri 3 3 O i 3 3 ri i ro T 3 W.(ii) r r M t = T = μ.w.r m Where, r r r 3 3 O i m 3 ro ri The above equations have been derived for a single pair of contacting surfaces. When there are a number of friction surfaces in contact, as in the case of the multi-disk clutch, Eq. (ii) should be multiplied by the number of pairs of contacting surfaces to obtain the resultant torque transmitting capacity. (ii) Uniform Wear Theory: According to the second theory it is assumed that the wear is uniformly distributed over the entire surface area of the friction disk. This assumption is used for worn-out clutches. The axial wear of the friction disk is proportional to the frictional work. The work done by the friction force at radius r is proportional to the frictional force (μp) and rubbing velocity (rn) where n is speed in rev/min. Fig..5 Pressure Distribution Wear (μp) (r n) Assuming speed n and the coefficient of friction μ as constant for a given configuration, Wear pr When the wear is uniform, pr = constant Prepared By: Vimal G. Limbasiya Darshan Institute of Engineering & Technology, Rajkot Page.7

30 . Friction Devices: Clutches, Brakes and Dynamometers Theory of Machines (15190) The pressure distribution according to uniform wear theory is illustrated in Fig..5. In this case, p is inversely proportional to r. Therefore, pressure is maximum at the inner radius and minimum at the outer periphery. The maximum pressure intensity at the inner radius r i is denoted by p a. It is also the permissible intensity of pressure. ro Total axial force, W p ( r) dr pr dr Total torque transmitted, ri W c r r O O i i ro ri r O ri c r c W r r..(iii) t ro M T dt ro ri ri dw r ro pr ri ro ri ro ri df r dr (r) p r dr r O r c c r ri r O i c r r O i Substituting value of c from equation (iii) W r r t r r O i O (r r) O i W i Conclusions M t = T = μ.w.r m Where, r m (r O r) i (i) The uniform-pressure theory is applicable only when the friction lining is new. (ii) The uniform-wear theory is applicable when the friction lining gets worn out. (iii) The friction radius for new clutches is slightly greater than that of worn-out clutches. (iv) The torque transmitting capacity of new clutches is slightly more than that of wornout clutches. Prepared By: Vimal G. Limbasiya Page.8 Darshan Institute of Engineering & Technology, Rajkot

31 Theory of Machines (15190). Friction Devices: Clutches, Brakes and Dynamometers (v) A major portion of the life of friction lining comes under the uniform wear criterion. (vi) It is more logical and safer to use uniform-wear theory in the design of clutches. Therefore, the torque transmitting capacity can be increased by three methods: (a) Use friction material with a higher coefficient of friction (μ) (b) Increase the plate pressure (p) (c) Increase the mean radius of the friction disk (r m ) In design of clutches, the following factors should be considered: (a) Service Factor In order to start the machine from rest and accelerate it to the rated speed, the clutch should have torque capacity substantially higher than the nominal torque rating. In most of the cases, the accelerating or starting torque is much more than the running torque. If the clutch is not designed for this increased torque, it will slip under the load and no power can be transmitted. There is another factor to account for additional torque. In many applications, the torque developed by the prime mover fluctuates and also, the torque requirement by driven machinery fluctuates as in the case of presses. These two factors are accounted by means of service factor. (b) Location of clutch Let us consider a mill driven by a diesel engine. The optimum operating speed of the engine is too high for direct connection to the mill shaft. Therefore, a gearbox is provided to reduce the speed. In this set-up, a clutch is also required so that the engine can be started and brought up to the full speed before connecting to the mill shaft. In such applications, the question arises about the location of clutch-whether the clutch should be located between the engine and the gearbox or between the gearbox and the mill? The clutch is required to transmit a given power. The power transmitted by the clutch is the product of torque and speed. Therefore, greater the speed, lower is the torque to be transmitted. It is, therefore, logical to place the clutch at the high-speed side, that is, between the engine and the gearbox. Since the torque capacity is low, the cost of the clutch is also low. On the other hand, the speed is low between the gearbox and the mill and, the clutch will have to transmit high torque, increasing the cost. (c) The coefficient of friction for automotive clutches, which use asbestos lining in contact with a cast iron surface, is taken from 0.3 to 0.4.The allowable pressure on the friction lining varies from 0.1 N/mm for large heavy-duty double-plate clutches to 0.5 N/mm for an average passenger car clutch. The allowable pressure for clutches with metal plates is from 0.7 to 1.05 N/mm. Prepared By: Vimal G. Limbasiya Darshan Institute of Engineering & Technology, Rajkot Page.9

32 . Friction Devices: Clutches, Brakes and Dynamometers Theory of Machines (15190) Multi-Disk Clutches A multi-disk clutch, as shown in Fig..6, consists of two sets of disks A and B. Disks of Set A are usually made of hardened steel, while those of set B are made of bronze. Disks of Set A are connected to the driven shaft by means of splines. Because of splines, they are free to move in an axial direction on the splined sleeve. There are four through bolts, which pass through the holes in the disks of Set B. A clearance fit between the bolt and the holes in the plates allows disks of Set B to move in an axial direction. The bolts are rigidly fixed to a rotating drum, which is keyed to the driving shaft. The axial force P (or W), which is required to hold the disks together, is provided by means of levers or springs. When the driving shaft rotates, the drum, along with bolts and disks of Set B, also rotate. Power is transmitted from the disks of Set B to those of A by means of friction. When the disks of Set A rotate, they transmit the power to the driven shaft through splined sleeve. Fig..6 Multi-Disk Clutch Equations derived for torque transmitting capacity of the single plate clutch are modified to account for the number of pairs of contacting surfaces in the following way: For the uniform pressure criterion, r r T Wn 3 r r 3 3 O i O i For the uniform wear criterion, (r O r) i T Wn Where n is the number of pairs of contacting surfaces. Prepared By: Vimal G. Limbasiya Page.10 Darshan Institute of Engineering & Technology, Rajkot

33 Theory of Machines (15190). Friction Devices: Clutches, Brakes and Dynamometers Fig..7 Number of Disks In the design of multi-disk clutches, very often it is required to determine the number of disks rather than the number of pairs of contacting surfaces. In multi-disk clutch, illustrated in Fig..6, there are two types of disks called disks of Set A and disks of Set B. We can use steel disks for Set A and bronze disks for Set B. Or, we can use plane steel disks for Set A and Set B consisting of steel disks with asbestos lining. Let us consider 5 disks 3 disks of Set A and disks of Set B. As shown in Fig.7, the number of pairs of contacting surfaces is 4. Therefore, Number of disks = Number of pairs of contacting surfaces +1 = n + 1.(a) Suppose, n 1 = number of disks on driving shaft n = number of disks on driven shaft Substituting in (a), n 1 + n = Number of pairs of contacting surfaces + 1 or number of pairs of contacting surfaces = n 1 + n 1 It should be noted that the two outer disks have contacting surface on one side only. Difference between Single and Multi-Plate Clutches The difference between single and multi-plate clutches is as follows: (i) (ii) The number of pairs of contacting surfaces in the single plate clutch is one or at the most two. There are more number of contacting surfaces in the multi-disk clutch. As the number of contacting surfaces is increased, the torque transmitting capacity is also increased, other conditions being equal. In other words, for a given torque capacity, the size of the multi-plate clutch is smaller than that of the single plate clutch, resulting incompact construction. (iii) The work done by friction force during engagement is converted into heat is generated in the multi-plate clutch due to increased number of contacting surfaces. Heat dissipation is a serious problem in the multi-plate clutch. Therefore, multi-plate clutches are wet clutches, while single plate clutches are dry. Prepared By: Vimal G. Limbasiya Darshan Institute of Engineering & Technology, Rajkot Page.11

34 . Friction Devices: Clutches, Brakes and Dynamometers Theory of Machines (15190) (iv) The coefficient of friction decreases due to cooling oil, thereby reducing the torque transmitting capacity of the multi-plate clutch. The coefficient of friction is high in dry single plate clutches. (v) Single plate clutches are used in applications where large radial space is available, such as trucks and cars. Multi-disk clutches are used in applications where compact construction is desirable, e.g., scooter and motorcycle. Difference between Dry and Wet Clutches The difference between dry and wet clutches is as follows: (i) (ii) A dry clutch has higher coefficient of friction. In wet clutches, the coefficient of friction is reduced due to oil. The coefficient of friction for dry operation is 0.3 or more, while it is 0.1 or less for wet operation. The torque capacity of dry clutch is high compared with the torque capacity of wet clutch of the same dimensions. (iii) For dry clutch, it is necessary to prevent contamination due to moisture or near by lubricated machinery, by providing seals. Such a problem is not serious in wet clutches. (iv) Heat dissipation is more difficult in dry clutches. In wet clutches, the lubricating oil carries away the frictional heat. (v) Rate of wear is far less in wet clutches compared to dry clutches. The wear rate in wet clutches is about 1% of the rate expected in dry clutches. (vi) The engagement in wet clutch is smoother than in the case of dry clutch. (vii) In wet clutches, the clutch facings are grooved to provide for passage of lubricant. This reduces the net face area for transmitting torque. Friction Material For light loads and low speeds, wood, cork and leather are used as friction materials. The present trend for high speeds and heavy loads has given a stimulus to the development of new friction materials, which are capable of withstanding severe service conditions. The desirable properties of a good friction material are as follows: (i) It should have high coefficient of friction. (ii) The coefficient of friction should remain constant over the entire range of, temperatures encountered in applications. (iii) It should have good thermal conductivity. (iv) It should remain unaffected by environmental conditions like moisture, or dirt particles. (v) It should have high resistance to abrasive and adhesive wear. (vi) It should have good resilience to provide good distribution of pressure at the contacting surfaces. Prepared By: Vimal G. Limbasiya Page.1 Darshan Institute of Engineering & Technology, Rajkot

35 Theory of Machines (15190). Friction Devices: Clutches, Brakes and Dynamometers The coefficient of friction depends upon a number of factors. They include materials of contacting surfaces, surface finish, surface temperature, rubbing speed, foreign particles on rubbing surfaces and atmospheric conditions like moisture. There are two types of friction materials in common use asbestos-base and sintered metals. There are two types of asbestos friction materials-woven and moulded. A woven asbestos friction disk consists of asbestos fiber woven around brass, copper or zinc wires and impregnated with rubber or asphalt. Asbestos material whether woven or moulded is anorganic material and is subject to destruction by heat at comparatively low temperature. Sintered-metal friction materials solve this difficulty. There are two varieties of friction disks made from sintered metals bronze-base and ironbase, depending upon the major constituents. The advantages of sintered-metal friction disks are as follows: (i) They have higher wear resistance. (ii) They can be used at high temperatures. (iii) The coefficient of friction is constant over a wide range of temperature and pressure. (iv) They are unaffected by environmental conditions, such as dampness, salt water or fungi. Sintered-metal friction materials offer an excellent design with lighter, cheaper and compact construction. The maximum permissible intensity of pressure for woven and moulded asbestos materials is 0.3 N/mm and 1 N/mm respectively, while for sintered metals it can be taken between 1 and N/mm. It has been found that if asbestos dust is inhaled, it may lead to cancer. The body cells which come in contact with asbestos particles are agitated and develop into cancer cells. Lung cancer is common among operators working in atmospheres of asbestos. There are federal regulations in a number of countries, which prohibit the use of asbestos in clutch or brake linings. Nowadays, metallic or semi-metallic fibres or powder is used in place of asbestos fibres. Modern friction lining consists of four basic ingredients, namely, fibres, filler, binder and friction modifiers. Fibres provide rigidity and strength for the friction lining Nowadays, steel wool or aramid is used as fibre material instead of asbestos. A filler fills the space between the fibres and extend the lining life. Filler materials are barytes, clay and calcium carbonate. In case of metallic lining, fine metal power is used as filler material. Binder is a glue that holds the lining ingredients together. Phenolformaldehyde is extensively used as binder material. Friction modifier improves frictional and wear properties. Brass and zinc particles are added as friction modifiers to control the abrasive properties of the lining. Prepared By: Vimal G. Limbasiya Darshan Institute of Engineering & Technology, Rajkot Page.13

36 . Friction Devices: Clutches, Brakes and Dynamometers Theory of Machines (15190) Centrifugal Clutches Whenever it is required to engage the load after the driving member has attained a particular speed, a centrifugal clutch is used. The centrifugal clutches permit the drivemotor or engine to start, warm up and accelerate to the operating speed without load. Then the clutch is automatically engaged and the driven equipment is smoothly brought up to the operating speed. These clutches are particularly useful with internal combustion engines, which can not start under load. The centrifugal clutch works on the principle of centrifugal force. The centrifugal force increases with speed. Construction of centrifugal clutch is shown in Fig..8. It consists of a spider, which is mounted on the input shaft, and which is provided with four equally spaced radial guides. A sliding shoe is retained in each guide by means of a spring. The outer surface of the sliding shoe is provided with a lining of friction material like asbestos. The complete assembly of spider, shoes and springs is enclosed in a coaxial drum, which is mounted on the output shaft. As the angular speed of the input shaft increases, the centrifugal force acting on the sliding shoes increases, causing the shoes to move in a radially outward direction. The shoes continue to move with increasing speed until they contact the inner surface of the drum. Power is transmitted due to frictional force between the shoe lining and the inner surface of the drum. The clutch is disengaged automatically. When the angular velocity of the shoes decreases, the centrifugal force decreases. This reduces the normal farce between the friction lining and the drum. The friction force, which is proportional to normal force, also reduces. The lining slips with respect to the drum and no torque can be transmitted. Fig..8 Centrifugal Clutch Prepared By: Vimal G. Limbasiya Page.14 Darshan Institute of Engineering & Technology, Rajkot

37 Theory of Machines (15190). Friction Devices: Clutches, Brakes and Dynamometers Fig..9 Forces on Shoe The forces acting on the shoe are shown in Fig..9, where following notations are used: m = mass of each shoe in kg r = distance of centre of gravity of shoe from centre of spider at the time of engagement in m R = inner radius of rim in m z = No. of shoes N = driving speed in rpm = angular speed of driving shaft (rad/s) g = angular speed at time of engagement (rad/s) x = g / = angle subtended by each shoe at centre of spider F C = centrifugal force in N F S = spring force in N F = outward force on each shoe in N F F = Friction force in N T = M t = torque transmitted l = contact length b = face width Centrifugal force, F C = mr Spring force, F S = mr g Total or Net outward force on shoe, F = F C F S = mr mr g F mr 1 = mr ( g ) g F mr 1 x Tangential frictional force, F F = μ x F = μ mr (1 x ) Torque transmitted, T = F F x R x z = μ mr z R (1 x ) (If not given, assume r R and = 60) Prepared By: Vimal G. Limbasiya Darshan Institute of Engineering & Technology, Rajkot Page.15

38 . Friction Devices: Clutches, Brakes and Dynamometers Theory of Machines (15190) Example 1:- A single plate clutch with both sides effective is to transmit 75 kw at 900 rpm. The axial pressure is limited to 0.07 MPa. The coefficient of friction may be taken as 0.5. The ratio of face width to mean radius is 0.5. Determine the outer and inner radii of clutch plate. Solution: P = 75 KW N = 900 rpm p = 0.07 MPa μ = 0.5 b/r m = 0.5 NT Power, P T T = N.m Assuming uniform wear theory, T Wrm n T (Ap) rm n T (r bp) r n rm T rm prm n 4 3 T rm pn rm 0.07 r m = mm m ro ri rm ro ri r O + r i = mm m.(i) face width of plate is equal to difference of outer and inner radii b = r O r i b 0.5 r m b = mm r O r i = mm.(ii) From equation (i) and (ii) r i = 13.4 mm r O = mm Prepared By: Vimal G. Limbasiya Page.16 Darshan Institute of Engineering & Technology, Rajkot

39 Theory of Machines (15190). Friction Devices: Clutches, Brakes and Dynamometers Example :- Design a single plate clutch considering uniform wear criterion and effective two pair of contacting surfaces from the following specification: Power to be transmitted = 40 kw Speed = 1560 r.p.m. Service factor = 1.5 Permissible pressure for the lining = 0.4 MPa Coefficient of friction = 0.30 Outer diameter = 300 mm Permissible stress for shaft material = 45 MPa Solution: P = 40 KW N = 1560 rpm p = 0.4 MPa μ = 0.03 d O = 300 mm τ = 45 MPa Assuming uniform wear theory, NT Power, P T T = N.m T Wrm n T (Ap) rm n T (r bp) r n m r r r r T O i r O i Ori pn 3 ro r i ror ipn 4 r i = 14.4 mm m i i r 150 r 0.4 Example 3:- A car engine has a maximum torque of 100 N.m. The clutch is a single plate with two acting surfaces; the axial pressure is not to exceed 0.85 bar. The ratio of outer diameter to inner diameter can be considered as 1.. Find the dimensions of friction plate & axial force required by springs. Assume µ = 0.3. Solution: T = 100 N.m n = p = 0.85 bar = 0.85 x 10 5 N/m d O /d i = 1. μ = 0.3 Prepared By: Vimal G. Limbasiya Darshan Institute of Engineering & Technology, Rajkot Page.17

40 . Friction Devices: Clutches, Brakes and Dynamometers Theory of Machines (15190) Assuming uniform wear theory, O T Wrm n r r O i W i W r r r O Wri ri i Wr Wri W c r r O c = p x r i = x r i W (85000 r)r 1. 1 W r i i i i (i) (ii) From equation (i) and (ii) ri r i = m r O = m W = N Example 4:- A multiple disc clutch is to transmit 4 kw at 750 rpm. Available steel and bronze discs are 40 mm inner radius and 70 mm outer radius are to be assembled alternately in appropriate numbers. The clutch is to operate in oil with an expected coefficient of friction of 0.1 and maximum allowable pressure is not to exceed 350 KPa. Assume uniform wear condition to prevail and specify the number of steel (driving) and bronze (driven) discs required. Also determine what axial force is to be applied to develop the full torque. Solution: P = 4 kw N = 750 rpm r i = 40 mm μ = 0.1 r O = 70 mm p = 350 KPa NT Power, P T T = N.m Assuming uniform wear theory, T Wrm n Prepared By: Vimal G. Limbasiya Page.18 Darshan Institute of Engineering & Technology, Rajkot

41 Theory of Machines (15190). Friction Devices: Clutches, Brakes and Dynamometers W n Wn = 960 W c r r O i W (p r) r r max i O i W W = N (i) Wn = 960 n = 3.5 n 4 No. of plates required = n + 1 = 5 No. of steel (driving) discs = 3 No. of bronze (driven) discs = Example 5:- A multiplate clutch is used to transmit 5 kw power at 1440 rpm. The inner & outer diameters of contacting surfaces are 50 mm and 80 mm respectively. The coefficient of friction and the average allowable pressure intensity for the lining may be assumed as 0.1 and 350 kpa respectively. Determine (i) Number of friction plates & pressure plates (ii) Axial force required to transmit power (iii) The actual average pressure (iv) Actual maximum pressure intensity after wear. Solution: P = 5 kw N = 1440 rpm d i = 50 mm μ = 0.1 d O = 80 mm p = 350 KPa NT Power, P T T = N.m Assuming uniform wear theory, T Wrm n W n Wn = W c r r O i (i) Prepared By: Vimal G. Limbasiya Darshan Institute of Engineering & Technology, Rajkot Page.19

42 . Friction Devices: Clutches, Brakes and Dynamometers Theory of Machines (15190) W (p r ) r r avg O i W W = N Wn = n 10 No. of plates required = n + 1= 11 Plates on drivingshafts = 6 Plates on driven shafts = 5 Actual number of active surfaces = 10 T W r n act Wact m W act = N W c r r act O i c c = c = p avg x r avg = p avg x p avg = N/m c = p max x r i = p max x 0.05 p max = N/m Example 6:- A Centrifugal clutch is to be designed to transmit 15 kw at 900 rpm. The shoes are four in number. The speed at which the engagement begins is 3/4 th of the running speed. The inside radius of the pulley rim is 150 mm. The shoes are lined with Ferrodo for which the coefficient of friction may be taken as 0.5. Determine: (i) Mass of the shoes and (ii) Size of the shoes. Solution: P = 15 kw N = 900 rpm z = 4 g = ¾ R = 150 mm μ = 0.5 Assume, r =10 mm NT Power, P T T = N.m Prepared By: Vimal G. Limbasiya Page.0 Darshan Institute of Engineering & Technology, Rajkot

43 Theory of Machines (15190). Friction Devices: Clutches, Brakes and Dynamometers N rad / sec T mr zr(1 x ) m10 (94.6) m =.7 kg Assuming that the arc of contact of the shoes subtend an angle of = 60 at the centre of the spider, therefore, Contact length, l R l = 157 mm Assuming intensity of pressure (p n ) exerted on shoes is 0.1 N/mm p n F mr (1 x ) lb l b 3 4 l b (94.6) 1 l x b = b = m Example 7:- A centrifugal clutch consists of four shoes, each having a mass of Kg. Inner radius of the drum is 140 mm in the engaged position. The distance of C.G. of the shoe from the axis of rotation of the spider is 115 mm. µ = 0.. The spring force at the beginning of the engagement is 1400 N. Calculate (i) The speed at which the engagement begins (ii)the power transmitted by the clutch at 100 rpm. Solution: z = 4 m = kg R = 140 mm r = 115 mm μ = 0. F S = 1400 N N = 100 rpm Engagement occurs only whenever the spring force is overcomes by centrifugal force. F S = mr g 1400 = x x g g = rad/sec N rad / sec g x Prepared By: Vimal G. Limbasiya Darshan Institute of Engineering & Technology, Rajkot Page.1

44 . Friction Devices: Clutches, Brakes and Dynamometers Theory of Machines (15190) T mr zr(1 x ) T (15.66) = 50.4 N.m NT Power, P P 60 P = 31.46kW Example 8:- Find out mass of a shoe, volume of a shoe and maximum load of a spring of a centrifugal clutch for following data: Power to be transmitted = 15 kw No. of shoes = 4 Engagement begins at 75 % of the Running Speed Inside diameter of pulley rim = 3.5 cm Distance of C.G. of shoe from the centre of the spider = 10 mm Coefficient of friction the shoe and drum = 0.5 Running Speed = 70 r.p.m. Shoe is made up of C.I. for which the density is 760 Kg/m 3 Solution: P = 15 kw z = 4 x = 0.75 D = 3.5 cm r = 10 mm μ = 0.5 N = 70 rpm ρ = 760 Kg/m 3 NT Power, P T T = N.m N rad / sec T mr zr(1 x ) m0.1 (75.398) m = 4.1 kg m 4.1 Volume, V 760 = 5.65 x 10 4 m 3 g = 0.75 = rad/sec Prepared By: Vimal G. Limbasiya Page. Darshan Institute of Engineering & Technology, Rajkot

45 Theory of Machines (15190). Friction Devices: Clutches, Brakes and Dynamometers F Smax = mr g F Smax = 4.1 x 0.1 x = N Example 9:- The following specification refers to a centrifugal clutch: Power to be transmitted = 30 kw No. of shoes = 4 Running Speed = 950 r.p.m. The speed at which the engagement starts is 80 % of the Running Speed Inner radius of drum = 10 mm The radial distance of centre of gravity from the axis of spider is 100 mm The normal intensity of pressure between friction lining and the drum = 0. MPa The arc of contact subtended by friction lining of shoe at centre of spider is 60 0 The material of friction lining is Ferrodo with coefficient of friction = 0.5 Find (i) capacity of the clutch Mass of each shoe and (ii) size of each shoe. Solution: P = 30 kw z = 4 N = 950 rpm x = 0.8 R = 0.1 m r = 0.1 m p = 0. MPa μ = 0.5 NT Power, P T T = N.m T mr zr(1 x ) = 60 N rad / sec m0.1 (99.48) m = 7.05 kg Contact length, l R l = 0.15 m p n F mr (1 x ) lb l b l x b = Face width, b = m Prepared By: Vimal G. Limbasiya Darshan Institute of Engineering & Technology, Rajkot Page (99.48) l b

46 . Friction Devices: Clutches, Brakes and Dynamometers Theory of Machines (15190).3 Brakes A brake is defined as a mechanical device which is used to absorb the energy possessed by a moving system or mechanism by the friction. The primary purpose of the brake is to slow down or completely stop the motion of moving system such as rotating drum, machine or vehicle. It is also used to hold the parts of the system in position at rest. An automobile brake is used either to reduce the speed of the car or bring it to rest. It also used to keep the car stationary on the downhill road. The energy absorbed by the brake can be either kinetic or potential or both. In automobile application, the brake absorbs kinetic energy of moving vehicle. In hoists and elevators, the brake absorbs the potential energy released by the object during the braking period. The energy absorbs by the brake converts into heat energy and dissipated to surrounding. Heat dissipation is a serious problem in brake application..4 Classification of brakes Brakes are classified into the following three groups: (a) Mechanical brakes which is operated by mechanical means such as levers, springs and pedals. Depending upon the shape of the friction material, the mechanical brakes are classified as Block brakes, Band Brakes, Block and Band Brake and internal or external shoes brakes. (b) Hydraulic brakes and pneumatic brakes which are operated by fluid pressure such as oil pressure or air pressure. (c) Electrical brakes which are operated by magnetic force and which include magnetic particle brakes, hysteresis brakes and eddy current brakes. Brake capacity depends upon the following factor. (i) (ii) (iii) (iv) (v) The unit pressure between braking surface. The contacting area of braking surface. The radius of the brake drum The coefficient of friction The ability of the brakes to dissipate heat that is equivalent to the energy being absorbed..5 Energy Equations The first step in the design of a mechanical brake is to determine the braking-torque capacity for the given application. The braking-torque depends upon the amount of energy absorbed by the brake. When a mechanical system of mass m moving with a velocity v 1 is slowed down to the velocity v during the period of braking, the kinetic energy absorbed by the brake is given by Prepared By: Vimal G. Limbasiya Page.4 Darshan Institute of Engineering & Technology, Rajkot

47 Theory of Machines (15190). Friction Devices: Clutches, Brakes and Dynamometers 1 KE mv 1 v..(i) Where, KE = kinetic energy absorbed by the brake (J) m = mass of the system (kg) v 1 and v = initial and final velocities of the system (m/s) Similarly, the kinetic energy of the rotating body is given by 1 KE I1..(ii) 1 KE mk 1.(iii) where, I =mass moment of inertia of the rotating body (kg-m ) k = radius of gyration of the body (m) ω 1, ω = initial and final angular velocities of the body (rad/s) In certain applications like hoists, the brake absorbs the potential energy released by the moving weight during the braking period. When a body of mass m falls through a distance h, the potential energy absorbed by the brake during the braking period is given by PE =mgh.(iv) where, g =gravitational constant (9.81 m/s ) Depending upon the type of application, the total energy absorbed by the brake is determined by adding the respective quantities of Eqs (i) to (iv). This energy is equated to the work done by the brake. Therefore, E = T b θ where, E = total energy absorbed by the brake (J) T b = braking torque (N-m) θ = angle through which the brake drum rotates during the braking period (rad) Block or Shoe Brake: Prepared By: Vimal G. Limbasiya Darshan Institute of Engineering & Technology, Rajkot Page.5

48 . Friction Devices: Clutches, Brakes and Dynamometers Theory of Machines (15190) Fig..10 Block or Shoe Brake A block or shoe brake consists of a block or shoe which is pressed against a rotating drum. The force on the drum is increased by using a lever [Fig..10(a)]. If only one block is used for the purpose, a side thrust on the bearing of the shaft supporting the drum will act. This can be prevented by using two blocks on the two sides of the drum [Fig..10(b)]. This also doubles the braking torque. A material softer than that of the drum or the rim of the wheel is used to make the blocks so that these can be replaced easily on wearing. Wood and rubber are used for light and slow vehicles and cast steel for heavy and fast ones. Let r Radius of drum Co efficient of friction F Radial force applied on the drum r R F f n Normalreaction on the block Force applied at the lever end F Frictional force R n Assuming that normal reaction R and frictional force n F f act at the mid point of the block. Breaking Torque Frictional force Radius R n r The direction of the frictional force on the drum is to be opposite to that of its rotation while on the block it is in the same direction. Taking moments about the pivot O [Fig..10(a)], F a R b R c 0 R n n Fa b c n bc F R n (1) a When b c, F = 0 which implies that the force needed to apply the brake is virtually zero, or that once contact is made between the block and the drum, the brake is applied itself. Such a brake is known as a self locking brake. Prepared By: Vimal G. Limbasiya Page.6 Darshan Institute of Engineering & Technology, Rajkot

49 Theory of Machines (15190). Friction Devices: Clutches, Brakes and Dynamometers As the moment of the force F f about O is in the same direction as that of the applied force F, F f aids in applying the brake. Such a brake is known as a self enegised brake. If the rotation of the drum is reversed, i.e., it is made clockwise, F Rn b c / a which shows that the required force F will be far greater than what it would be when the drum rotates counter clockwise. a If the pivot lies on the line of action of E f, i.e., at O, c = 0 and F R n, b which is valid for clockwise as well as for counter clockwise rotation. If c is made negative, i.e., if the pivot is at O, and bc F Rn for counter clockwise rotation a bc F Rn for clockwise rotation a In case the pivot is provided on the same side of the applied force and the block as shown in Fig..10 (c), the equilibrium condition can be considered accordingly. In the above treatment, it is assumed that the normal reaction and the frictional force act at the mid point of the block. However, this is true only for small angles of contact. When the angles of contact is more than 40, the normal pressure is less at the ends than at the centre. In that case, µ has to be replaced by an equivalent coefficient of friction µ given by Band Brake: 4 sin ' sin Fig..11 Band Brake Prepared By: Vimal G. Limbasiya Darshan Institute of Engineering & Technology, Rajkot Page.7

50 . Friction Devices: Clutches, Brakes and Dynamometers Theory of Machines (15190) It consists of a rope, belt or flexible steel band (lined with friction material) which is pressed against the external surface of a cylindrical drum when the brake is applied. The force is applied at the free end of a lever (Fig..11). Brake torque on the drum = ( T 1 T ) r where r is the effective radius of the drum. The ratio of the tight and slack side tensions is given by T 1 / T that the band is on the point of slipping on the drum. The effectiveness of the force F depends upon the direction of rotation of the drum ratio of length a and b direction of the applied force F. e on the assumption To apply the brke to the rotating drum, the band has to be tightened on the drum. This is possible if 1. F is applied in the downward direction when a > b. F is applied in the upward direction when a < b If the force applied is not as above, the band is further loosened on the drum which means no braking effect is possible. 1. a > b, F Downwards (a) Rotation Counter Clockwise For counter clockwise rotation of the drum, the tight and the slack sides of the band will be as shown in Fig..11. Considering the forces acting on the lever and taking moments about the pivot, F l T a T b 0 or 1 T1a T b F l (1) As T 1 > T and a > b under all conditions, the effectiveness of the brake will depend upon the force F. (b) Rotation Clockwise Fig..1 Prepared By: Vimal G. Limbasiya Page.8 Darshan Institute of Engineering & Technology, Rajkot

51 Theory of Machines (15190). Friction Devices: Clutches, Brakes and Dynamometers In this case, the tight and the slack sides are reversed as shown in Fig..1. Now, F l T a T b 0 or 1 Ta T1b F l () As T < T 1 and a > b, the brake will be effective as long as T. a is greater than T 1. b Or T a T b or 1 T T 1 b a i.e., as long as the ratio of T to T 1 is greater than the ratio b/a. T a When T1 b, F is zero or negative, i.e., the brake becomes self locking as no force is needed to apply the brake. Once the brake has been engaged, no further force is required to stop the rotation of the drum.. a < b, F upwards Fig..13 (a) Rotation Counter Clockwise The tight and the slack sides will be as shown in Fig..13(a). Therefore, F l T a T b 0 or 1 Tb T1 a F l As T < T 1 and b > a, the brake is operative only as long as T a Tb T1 a or T b 1 Once T / T 1 becomes equal to a/b, F required is zero and the brake becomes self locking. Prepared By: Vimal G. Limbasiya Darshan Institute of Engineering & Technology, Rajkot Page.9

52 . Friction Devices: Clutches, Brakes and Dynamometers Theory of Machines (15190) (b) Rotation Clockwise The tight and the slack sides are shown in Fig..13(a). T1b Ta From Fig..13(b), Fl T1b T a 0 or F l As T 1 > T and b > a, under all conditions, effectiveness of the brake will depend upon the force F. When a = b, the band cannot be tightened and thus, the brake cannot be applied. Fig..14 The band brake just discussed is known as a differential band brake. However, if either a or b is made zero, a simple band brake is obtained. If b = 0 (Fig..14) and F downwards, F l T a 0 or 1 a F T1 l Similarly, the force can be found for the other cases. Note that such a brake can neither have self energising properties nor it can be self locked. The brake is said to be more effecctive when maximum braking force is applied with the least effort F. For case (i), when a > b and F is downwards, the force (effort) F required is less when the rotation is clockwise assuming that the brake is effective. For case (ii), when a < b and F is upwards, F required is less when the rotation is counter clockwise assuming that the brake is effective. Thus, for the given arrangement of the differential brake, it is more effective when (a) a > b, F downwards, rotation clockwise (b) a > b, F upwards, rotation counter clockwise. The advantages of self locking is taken in hoists and conveyers where motion is permissible in only one direction. If the motion gets reversed somehow, the self locking is engaged which can be released only by reversing the applied force. Prepared By: Vimal G. Limbasiya Page.30 Darshan Institute of Engineering & Technology, Rajkot

53 Theory of Machines (15190). Friction Devices: Clutches, Brakes and Dynamometers It is seen that a differential band brake is more effective only in one direction of rotation of the drum. Fig..15 However, a two-way band brake can also be deigned which is equally effective for both the directions of rotation of the drum (Fig..15). In such a brake, the two lever arms are made equal. For both directions of rotation of the drum, F l T a T a 0 or 1 F T T 1 a l Band brake offers the following advantages: (i) Band brake has simple construction. It has small number of parts. These features reduce the cost of band brake. (ii) Most equipment manufacturers can easily produce band brake without requiring specialized facilities like foundry or forging shop. The friction lining is the only part which must be purchased from outside agencies. (iii) Band brake is more reliable due to small number of parts. (iv) Band brake requires little maintenance. The disadvantages of band brake are as follows: (a) The heat dissipation capacity of a band brake is poor. (b) The wear of friction lining is uneven from one end to the other. Band brakes are used in applications like bucket conveyors, hoists and chain saws. They are more popular as back-stop devices. Prepared By: Vimal G. Limbasiya Darshan Institute of Engineering & Technology, Rajkot Page.31

54 . Friction Devices: Clutches, Brakes and Dynamometers Theory of Machines (15190) Band and Block Brake: Fig..16 Band and Block brake A band and block brake consists of a number of wooden blocks secured inside a flexible steel band. When the brake is applied, the blocks are pressed against the drum. Two sides of the band becomes tight and slack as usual. Wooden blocks have a higher coefficient of friction. Thus, increasing the effectiveness of the brake. Also, such blocks can be easily replaced on being worn out [ Fig..16(a)]. Each block subtends a small angle of θ at the centre of the drum. The frictional force on the blocks acts in the direction of rotation of the drum. For n blocks on the brake, Let T T T T 0 1 n n Tension on the slack side. Tension on the tight side after one block. Tension on the tight side after two blocks. Tension on the tight side after nblocks. Coefficient of friction. R Normalreaction on the block. Resolving the forces horizontally and vertically. T T cos R (1) 1 0 n T T sin R () 1 0 n T T cos R T T sin R 1 0 n 1 0 n 1 0 T1 T0 tan T T T1 T0 T1 T0 T T T T tan 1 tan 1 Prepared By: Vimal G. Limbasiya Page.3 Darshan Institute of Engineering & Technology, Rajkot

55 Theory of Machines (15190). Friction Devices: Clutches, Brakes and Dynamometers Similarly T1 1 tan T 1 tan 0 0 T1 1 tan T 1 tan T 1 tan and so on T1 1 tan Tn 1 tan T 1 tan n1 Internal Expanding Brake T T T T T T T T T T n n n1 1 0 n1 n tan 1 tan n The construction of an internal expanding brake is shown in Fig..17. It consists of a shoe, which is pivoted at one end and subjected to an actuating force P at the other end. A friction lining is fixed on the shoe and the complete assembly of shoe lining and pivot is placed inside the brake drum. Fig..17 Internal Expanding Brake Internal shoe brakes, with two symmetrical shoes, are used on all automobile vehicles. The actuating force is usually provided by means of a hydraulic cylinder or a cam mechanism. The analysis of the internal shoe brake is based on the following assumptions: (a) The intensity of normal pressure between the friction lining and the brake drum at any point is proportional to its vertical distance from the pivot. (b) The brake drum and the shoe are rigid. (c) The centrifugal force acting on the shoe is negligible. (d) The coefficient of friction is constant. Prepared By: Vimal G. Limbasiya Darshan Institute of Engineering & Technology, Rajkot Page.33

56 . Friction Devices: Clutches, Brakes and Dynamometers Theory of Machines (15190).6 Braking of Vehicle: In a four wheeled moving vehicle, the brakes may be applied to 1. The rear wheels only,. The front wheels only, and 3. All the four wheels. In all the above mentioned three types ofbraking, it is required to determine theretardationof the vehicle when brakes are applied. Since thevehicle retards, therefore it is a problem of dynamics. But it may be reduced to an equivalentproblem of statics byincluding the inertia force inthe system of forces actually applied to the vehicle.the inertia force is equal and opposite to thebraking force causing retardation. Fig..18 Consider a vehicle moving up an inclined plane shown in Fig..18. Let Angle of inclination of the plane to the horizontal. m mass of vehicle (Such that its weight mg innewton). h Height of C.G. of the vehicle above the road surface (mtr). x Perpendicular dis tance of C.G. from the rear axle (mtr). L Dis tance between centre of rear and front wheel(wheel Base)(mtr). R,R A B Reactions of the ground on the front & rear wheel(n). Coefficient of frictionbetween tyre androad surface. a Retardation of the vehicle (m / s ). a. Brakes applied to Rear wheel only: It is a common way of braking a vehicle in which the braking froce acts at the rear wheel. Resolving the forces F m g sin m a (1) B R R m g cos () A B Prepared By: Vimal G. Limbasiya Page.34 Darshan Institute of Engineering & Technology, Rajkot

57 Theory of Machines (15190). Friction Devices: Clutches, Brakes and Dynamometers Taking moment about G.. and From equation (1) B B A F F h R x R L x R h R x (m g cos R ) L x B B B B B m g cosl x R h x mg cos L x R L x R h x L x m g cos L x B R B Lh RA m g cos RB m g cosl x m g cos Lh m g cos x h RA L h FB m g sin m a FB m g sin m F m R m B g sin B Totalbreaking force acting at the rear wheel due to application of brakes R B B a g sin Putting value of RB in equation m g cos (L x) Lh a g sin m g cos (L x) a g sin Lh Note : (1) On Road level α = 0 so.. g (L x) a Lh () When vehicle moves down a plane equation (1) becomes F B m g sin ma FB RB RA m g cos R B Prepared By: Vimal G. Limbasiya Darshan Institute of Engineering & Technology, Rajkot Page.35

58 . Friction Devices: Clutches, Brakes and Dynamometers Theory of Machines (15190) F m RB g sin m g cos (L x) a g sin Lh B a g sin b. Brake applied to Front wheels only: It is a very rare way of braking the vehicle, in which the braking froce acts at the front wheels only. Let F A Braking force acting at the front wheel. R Resolving the forces horizontally and vertically.. A A B A F m g sin m a (1) R R m g cos () Taking moments about G From equation () From equation (1) FA h RB x R A (L x ) R h mg cos R x R L x A A A R h mg cos x R L A A R L h m g cos x R A A m g cosx L h R R mg cos B B R m g cos R m g cosx m g cos Lh x m g cos 1 L h L h x RB m g cos Lh F m g sin ma A A A FA RA RB m g cos RA Putting values... Prepared By: Vimal G. Limbasiya Page.36 Darshan Institute of Engineering & Technology, Rajkot

59 Theory of Machines (15190). Friction Devices: Clutches, Brakes and Dynamometers FA m g sin a m RA m g sin Putting Value of R m m g cosx m g sin Lh m g cos x a g sin Lh Note : (1) On a Road level α = 0 m g cosx m g x RA cos 0 1 L h L h L h x m g L h x RB m g cos L h L h g cos x cos 0 1 a g sin L h sin 0 0 gx L h () When vehicle moves down the plane.. FA m g sin ma FA m g sin a m m RA g sin m g cos x g sin Lh c. Brakes applied to all four wheels This is the most common way of braking the vehicle, in which the braking force acts on both the rear and front wheels. F Braking force for front wheels R A F Braking force for rear wheels R B Note : A little consideration will show that whenthe brakes are applied to all the four wheels, thebraking distance (i.e. the distance in which thevehicle is brought to rest after applying thebrakes) will be the least. It is due to this reasonthat the brakes are applied to all the four wheels. Resolving the forces. F F m g sin m a (1) A B R R m g cos () A B Prepared By: Vimal G. Limbasiya Darshan Institute of Engineering & Technology, Rajkot Page.37 A B A

60 . Friction Devices: Clutches, Brakes and Dynamometers Theory of Machines (15190) Taking moment about G.. F F h R x R (L x ) (3) A B B A R R h mg cos R x R (L x ) A B A A A A A A FA RA FB R B RB m g cos RA Putting values in eq.(3)... R mg cos R h mg cos R x R (L x ) R A m g cos h x L R m g cos R B A (4) hx m g cos m g cos L L h x m g cos (5) L From equation (1) F F mg sin ma A A B B R R m g sin m a m g cos m g sin m a R R m g cos a g cos sin (6) A B Note : (1) On a Road level α = 0 R R A B m g cos h x mg h x L L L h x m g L ag () If vehicle moves down the plane, then equation (1) may be written as.. F F mg sin ma A B R R m g sin m a A B m g cos m g sin m a R R m g cos a g cos sin A B Prepared By: Vimal G. Limbasiya Page.38 Darshan Institute of Engineering & Technology, Rajkot

61 Theory of Machines (15190). Friction Devices: Clutches, Brakes and Dynamometers.7 Dynamometer A Dynamometer is a brake but in addition it has a device to measure the frictional resistance. Knowing the frictional resistance, may obtain the torque transmitted and hence the power of the engine..8 Types of Dynamometers: There are mainly two types of dynamometers: 1. Absorption Dynamometers: In this type, the work done is converted into heat by friction while being measured. They can be used for the measurement of moderate powers only. Examples are prony brake dynamometer and rope brake dynamometer.. Transmission Dynamometers: In this type, the work is not absorbed in the process, but is utilised after the measurement. Examples are the belt transmission dynamometer and the trosion dynamometer. Prony Brake Dynamometer: Fig..19 Prony Brake Dynamometer A prony brake dynamometer consists of two wooden blocks clamped together on a revolving pulley carrying a lever (Fig..19). The friction between the blocks and the pulley tends to rotate the blocks in the direction of rotation of the shaft. However, the weight due to suspended mass at the end of the lever prevents this tendency. The grip of the blocks over the pulley is adjusted using the bolts of the clamp until the engine runs at the required speed. The mass added to the scale pan is such that the arm remains horizontal in the equilibrium position; the power of the engine is thus absorbed by the friction. Frictional torque Wl Mg l N Power of themachine under test T Mg l 60 MNk where k is a constant for a particular brake. Note that the expression fro power is independent of the size of the pulley and the coefficient of friction. Prepared By: Vimal G. Limbasiya Darshan Institute of Engineering & Technology, Rajkot Page.39

62 . Friction Devices: Clutches, Brakes and Dynamometers Theory of Machines (15190) Rope Brake Dynamometer: In a rope brake dynamometer (Fig..0), a rope is wrapped over the rim of a pulley keyed to the shaft of the engine. The diameter of the rope depends upon the power of the machine. The spacing of the ropes on the pulley is done by 3 to 4 U-shaped wooden blocks which also prevent the rope from slipping off the pulley. The upper end of the rope is attached to a spring balance where as the lower end supports the weight of suspended mass. Power of the machine T Ft r N M g s r 60 Fig..0 Rope Brake Dynamometer If the power produced is high, so will be the heat produced due to friction between the rope and the wheel, and a cooling arrangement is necessary. For this, the channel of the flywheeel usually has flanges turned inside in which water from a rope is supplied. An outlet pipe with a flattenend end takes the water out. A rope brake dynamometer is frequently used to test the power of engines. It is easy to manufacture, inexpensive, and requires no lubrication. If the rope is wrapped several times over the wheel, the tension on the slack side of the rope, i.e., the spring balance reading can be reduced to a negligible value as compared to the tension of the tight side (as T 1 / T = e µθ and θ is increaased). Thus, one can even do away with the spring balance. Prepared By: Vimal G. Limbasiya Page.40 Darshan Institute of Engineering & Technology, Rajkot

63 Theory of Machines (15190). Friction Devices: Clutches, Brakes and Dynamometers Belt Transmission Dynamometer: Fig..1 Belt Transmission Dynamometer The belt transmission dynamometer occupies a prominent position among transmission dynamometers. When a belt transmits power from one pulley to another, there exists a difference in tensions between the tight and slack sides. A dynamometer measures directly the difference in tensions (T 1 T ) while the belt is running. Fig..1 shows a Tatham dynamometer. A continuous belt runs over the driving and the driven pulleys through two intermediate pulleys. The intermediate pulleys have their pins fixed to a lever with its fulcrum at the midpoint of the two pulley centres. As the lever is not pivoted at its midpoint, a mass at the left end is used for its initial equilibrium. When the belt transmits power, the lever tends to rotate in the counter clockwise direction due to the difference of tensions on the tight and slack sides. To maintain its horizontal position, a weight of the required amount is provided at the right end of the lever. Two stops, one on each side of the lever arm, are used to limit the motion of the lever. Taking moments about the fulcrum, M g l T a T a M g l a T T 0 T 1 M g l T a Power,P T T v 1 where v =belt speed in metres per second. Prepared By: Vimal G. Limbasiya Darshan Institute of Engineering & Technology, Rajkot Page.41

64 . Friction Devices: Clutches, Brakes and Dynamometers Theory of Machines (15190) Epicyclic - Train Dynamometer: An epicyclic train dynamometer is another transmission type of dynamometer. As shown in Fig.., it consists of a simple epicyclic train of gears. A spur gear A is the driving wheel which drives an annular driven wheel B through an intermediate pinion C. Fig.. Epicyclic - Train Dynamometer The intermediate gear C is mounted on a horizontal lever, the weight of which is balanced by a counterweight at the left end when the system is at rest. When the wheel A rotates counter clockwise, the wheel B as well as the wheel C rotates clockwise. Two tangential forces, each equal to F, act at the ends of the pinion C, one due to the driving force by the wheel A and the other due to reactive force of the driven wheel B. Both forces are equal if friction is ignored. This tends to rotate the lever in the counter clockwise direction and it no longer remains horizontal. To maintain it in the same position as earlier, a balancing weight W is provided at the right end of the lever. Two stops, one on each side of the lever arm, are used to limit the motion of the lever. For the equilibrium of the lever, Wl F a W l or F a and torque transmitted = F. r where r is the radius of the driving wheel Thus power, N P T Fr 60 Prepared By: Vimal G. Limbasiya Page.4 Darshan Institute of Engineering & Technology, Rajkot

65 Theory of Machines (15190). Friction Devices: Clutches, Brakes and Dynamometers Bevis Gibson Torsion Dynamometer: Fig..3 Bevis Gibson Torsion Dynamometer A Bevis Gibson torsion dynamometer consists of two discs A and B, a lamp and a movable torque finder arranged as shown in Fig..3(a). The two discs are similar and are fixed to the shaft at a fixed distance from each other. Thus, the two discs revolve with the shaft. The lamp is masked and fixed on the bearing of the shaft. The torque finder has an eyepiece capable of moving circumferentially. Each disc has a small radial slot near its periphery. Similar slots are also made at the same radius on the mask of the lamp and on the torque finder. When the shaft rotates freely and does not transmit any torque, all the four slots are in a line and a ray of light from the lamp can be seen through the eyepiece after every revolution. When a torque is transmitted, the shaft twists and the slot in the disc B shifts its position. The ray of light can no longer pass through the four slots. However, if the eyepiece is moved circumferentially by an amount equal to the displacement, the flash will again be visible once in each revolution of the shaft. The eyepeice is moved by a micrometer spindle. The angle of twist may be measured up to one hundredth of the degree. In case the torque is varied during each revolution of the shaft as in reciprocating engines and it is required to measure the angle of twist at different angular position, then each disc can be perforated with several slots arranged in the form of a spiral at varying radii [Fig..3(b)]. The lamp and the torque finder have to be moved radiallly to and from the shaft so that they come opposite each pair of slots in the discs. Prepared By: Vimal G. Limbasiya Darshan Institute of Engineering & Technology, Rajkot Page.43

66 . Friction Devices: Clutches, Brakes and Dynamometers Theory of Machines (15190) Example 10:- A brake as shown in Fig..4 is fitted with a C.I. brake shoe. The braking torsional moment = 360N.m The drum diameter = 300mm The coefficient of friction = 0.3 Find: (i) force P for counter clockwise rotation (ii) force P for clockwise rotation (iii) where must pivot be placed to make brake self locking with clockwise rotation? Solution: T b = 360 N.m d = 300 mm μ = 0.3 Fig..4 (a) Fig..5 for the clockwise rotation of the brake drum, the frictional force or the tangential force (F t ) acting at the contact surfaces is shown in Fig..5 (a) Braking torque, T b = F t x r = μ R N x r 360 = 0.3 x R N x 0.15 Normal force, R N = 8000 N Now taking moments about the fulcrum O P ( ) + F t x 50 = R N x 00 P x (0.3 x 8000) 50 = 8000 x 00 P x 800 = 1480 x 10 3 P = 1850 N (b) Prepared By: Vimal G. Limbasiya Page.44 Darshan Institute of Engineering & Technology, Rajkot

67 Theory of Machines (15190). Friction Devices: Clutches, Brakes and Dynamometers for the counter clockwise rotation of the brake drum, the frictional force or the tangential force (F t ) acting at the contact surfaces is shown in Fig..5 (b) Now taking moments about the fulcrum O, P ( ) = F t x 50 + R N x 00 P x 800 = (0.3 x 8000) x 00 P x 800 = 170 x 10 3 P = 150 N Location of the pivot or fulcrum to make the brake self-locking The clockwise rotation of the brake drum is shown in Fig..5 (a). Let x be the distance of the pivot or fulcrum O from the line of action of the tangential force (F t ). Taking moments about the fulcrum O, P ( ) + F t x x R N x 00 = 0 In order to make the brake self-locking, F t x x must be equal to R N x 00, so that the force P is zero. F t x x = R N x x x = 8000 x 00 x = 667 mm Example 11:- A simple band brake is shown in Fig..6 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 rpm. The co-efficient of friction is 0. and the brake drum diameter is 40 mm. Take b = 10 mm, l = 40 mm, = 10, then find out the followings: (i) The torque applied due to hand load of 100 N (ii) The number of turns of the flywheel before it is brought to rest (iii) The time required to bring it to rest from the moment of the application of the brake. Fig..6 Solution: m = 400 kg k = 450 mm N = 300 rpm μ = 0. d = 40 mm = 10 = rad Prepared By: Vimal G. Limbasiya Darshan Institute of Engineering & Technology, Rajkot Page.45

68 . Friction Devices: Clutches, Brakes and Dynamometers Theory of Machines (15190) Taking moments about the fulcrum O, T 1 x b = P x l T 1 x 0.1 = P x 0.4 T 1 x 0.1 = 100 x 0.4 T 1 = 350 N T Now, e e T T1.081 T T1 350 T T = N Braking torque, T b = F t x r = (T 1 T ) x r = ( ) x 0.1 = N.m Work done against friction due to absorption of K.E. 1 Ft S mv 1 Ft S m(r ) m dn S F 60 S = D x n S No. of turns, n D = 0.73 t S = m T b = I α = (mk ) α = (400 x ) α α = 0.69 = o + αt N 0 t 60 Prepared By: Vimal G. Limbasiya Page.46 Darshan Institute of Engineering & Technology, Rajkot

69 Theory of Machines (15190). Friction Devices: Clutches, Brakes and Dynamometers t 60 t = sec Example 1:- A band brake shown in Fig..7 is used to balance a torque of 980 N-m at the drum shaft. The drum diameter is 400 mm (rotating in clockwise direction) and the allowable pressure between lining and drum is 0.5 MPa. The coefficient of friction is 0.5. Design the steel band, shaft, brake lever and fulcrum pin, if all these elements are made from steel having permissible tensile stress 70 MPa and shear stress 50 MPa. Fig..7 Solution: T b = 980 N.m d = 400 mm p = 0.5 MPa μ = 0.5 σ t = 70 MPa τ = 50 MPa = 10 =.094 rad Braking torque, T b = F t x r = (T 1 T ) x r 980 x 10 3 = (T 1 T ) x 00 T 1 T = 4.9 x 10 3 Now, T1 e e T T T T T = 4.9 x T = N T 1 = N (i) (ii) Taking moments about the fulcrum O, T x 80 = P x x 80 = P x 600 P = N Prepared By: Vimal G. Limbasiya Darshan Institute of Engineering & Technology, Rajkot Page.47

70 . Friction Devices: Clutches, Brakes and Dynamometers Theory of Machines (15190) Design of Shaft Since the shaft has to transmit torque equal to braking torque 3 Tb ds ds d S = mm Shaft diameter, d S 50 mm Design of Lever t 1 = thickness of the lever B = width of the lever Maximum bending moment at fulcrum O due to force P M = P x l = x 600 = N.mm Section modulus, 1 1 Z t1 B t 1 (t 1) 6 6 = 0.67 t 3 1 mm 3 (Assuming B = t 1 ) M Bending stress Z t t 1 =.988 mm 3 mm B = t 1 = 46 mm Design of Pins d 1 = diameter of pins l 1 = length of pins = 1.5 d 1 The pins are designed for maximum tension in the band (i.e. T 1 = N) Considering bearing of the pins, maximum tension (T 1 ), = d 1.l 1.p = d 1 x 1.5 d 1 x 0.5 d 1 = mm d mm l 1 = 1.5 d 1 = 1.5 x 140 = 175 mm 1 Check the pin for induced shearing stress. Since the pin is in double shear, therefore maximum tension (T 1 ), (d 1 ) (140) Prepared By: Vimal G. Limbasiya Page.48 Darshan Institute of Engineering & Technology, Rajkot

71 Theory of Machines (15190). Friction Devices: Clutches, Brakes and Dynamometers = 0.39 MPa This induced stress is quite within permissible limits. The pin may be checked for induced bending stress. 5 5 Maximum bending moment, M W.l = N-mm (W= T 1 ) section modulus, Z (d 1 ) (140) mm 3 3 M Bending stress Z = 1.67 MPa The induced bending stress is within safe limits of 70 MPa. The lever has an eye hole for the pin and connectors at band have forked end. Thickness of each eye l 175 1, t 87.5mm Outer diameter of the eye, D= d 1 = x 140 = 80 mm A clearance of 1.5 mm is provided on either side of the lever in the fork. A brass bush of 3 mm thickness may be provided in the eye of the lever. Diameter of hole in the lever = d 1 + x 3 = x 3 = 146 mm The boss is made at pin joints whose outer diameter is taken equal to twice the diameter of the pin and length equal to length of the pin. The inner diameter of the boss is equal to diameter of the hole in the lever. Outer diameter of the boss = d 1 = (140) = 80 mm Length of the boss = l 1 = 175 mm Check the bending stress induced in the lever at the fulcrum. Maximum bending moment at the fulcrum, M = P x l = x 600 = N-mm Section modulus, Z 1 80 / = mm 3 M Bending stress 0.9 MPa Z The induced bending stress is within safe limits of 70 MPa. Fig. 6.8 Prepared By: Vimal G. Limbasiya Darshan Institute of Engineering & Technology, Rajkot Page.49

72 . Friction Devices: Clutches, Brakes and Dynamometers Theory of Machines (15190) Example 13:- A differential band brake has a drum with a diameter of 800 mm. The two ends of the band are fixed to the pins on the opposite sides of the fulcrum of the lever at distances of 40 mm and 00 mm from the fulcrum. The angle of contact is 70 and the coefficient of friction is 0.. Determine the brake torque when a force of 600 N is applied to the lever at a distance of 800 mm from the fulcrum. Solution: d = 800 mm θ = 70 μ = 0. F = 600 N l = 800 mm Fig..9 Assuming a = 00 mm, b = 40 mm, i.e., a > b, F must act downward direction to apply the brake (Fig..9). T e e T T1.57 T (a) Anticlockwise Rotation Taking moment about fulcrum O. F l T a T b 0 (Fig..9) (.57 T ) 00 T T 40 T T T N and T 60.53N Braking Torque, TB T1 T r N m (b) Clockwise Rotation Taking moment about fulcrum O. (Fig..30) Prepared By: Vimal G. Limbasiya Page.50 Darshan Institute of Engineering & Technology, Rajkot

73 Theory of Machines (15190). Friction Devices: Clutches, Brakes and Dynamometers F l T b T a T 40 T 00 0 T 4938 N and T.57 T 1691N Braking Torque, TB T1 T r N m Fig..30 Assuming a = 40 mm, b = 00 mm, i.e., a < b, F must act upward to apply the brake. (a) Anticlockwise Rotation Fig..31 Taking moment about fulcrum O. (Fig..31) F l T a T b T (40) T 00 0 T 4938 N and T 1691N Braking Torque, TB T1 T r N m Prepared By: Vimal G. Limbasiya Darshan Institute of Engineering & Technology, Rajkot Page.51

74 . Friction Devices: Clutches, Brakes and Dynamometers Theory of Machines (15190) (b) Clockwise Rotation Taking moment about fulcrum O. (Fig..3) F l T a T b Fig T T 00 0 T N and T 60.5N Braking Torque, TB T1 T r Nm Note: The above results show that the effectiveness of the brake in one direction of rotation is equal to the effectiveness in the other direction if the distance of the pins on the opposite sides of the fulcrum are changed and the force is applied in the proper direction so that the band is tightened. Example 14:- A simple band brake (Fig..33) is applied to a shaft carrying a flywheel of 50 kg mass and of radius of gyration of 300 mm. the shaft speed is 00 rpm. The drum diameter is 00 mm and the coefficient of friction is 0.5. Determine the 1. brake torque when a force of 10 N is applied at the lever end. number of turns of the flywheel before it comes to rest 3. time taken by the flywheel to come to rest. Solution: Fig..33 Prepared By: Vimal G. Limbasiya Page.5 Darshan Institute of Engineering & Technology, Rajkot

75 Theory of Machines (15190). Friction Devices: Clutches, Brakes and Dynamometers m 50 kg k 300 mm N 00 rpm d 00 mm rad Taking moment about O. 1 F l T a T 100 T 336 N T1 T 15.8 N.67 Braking Torque, TB T1 T r Nm. K.E of flywheel 1 1 N I mk Nm T e e T T1.67 T This K.E is used to overcome the work done by the braking torque in n revolutions. K. E of flywheel T Angular displacement B n n 37.4 Revolution 3. For uniform retardation, 00 Average speed 100 rpm n 37.4 Time taken min. N = 100 / sec Example 15:- A simple band brake is applied to a drum of 560 mm diameter which rotates at 40 rpm. The angle of contact of the band is 70. One end of the band is fastened to a fixed pin and the other end to the brake lever, 140 mm from the fixed pin. The brake lever is 800 mm long and is placed perpendicular to the diameter that bisects the angle of contact. Assuming the coefficient of friction as 0.3, determine the necessary pull at the end of the lever to stop the drum if 40 kw of power is being absorbed. Also, find the width of the band if its thickness is 3 mm and the maximum tensile stress is limited to 40 N/mm. Prepared By: Vimal G. Limbasiya Darshan Institute of Engineering & Technology, Rajkot Page.53

76 . Friction Devices: Clutches, Brakes and Dynamometers Theory of Machines (15190) Solution: d 560 mm N 40 rpm 70 a 140 mm, l 800 mm P 40 kw 0.3 thickness t 3mm 40N/mm Fig..34 Note: It can be observed from the figure that to tighten the band, the force is to be applied upwards. If the drum rotates counter clockwise, the tight and slack sides will be as shown. and From equation (1) & () 1 P T B N T1 T r T1 T T T 5684 (1) T T T T 1 e e 4.11 () 188 N 7514 N Taking the moment about O Let maximum tension T F l T 140 cos 45 F cos 45 1 F b t 6. N b 3 b 6.6 mm Prepared By: Vimal G. Limbasiya Page.54 Darshan Institute of Engineering & Technology, Rajkot

77 Theory of Machines (15190). Friction Devices: Clutches, Brakes and Dynamometers Note: If drum rotates clockwise, the brake is less effective as in that case tight and slack sides are interchange and the force requied to apply the same braking torque is more. F l T 140 cos 40 1 F cos 40 F 930 N Example 16:- A crane is required to support a load of 1. tonnes on the rope round its barrel of 400 mm diameter (Fig..35). The brake drum which is keyed to the same shaft as the barrel has a diameter of 600 mm. The angle of contact of the band brake is 75 and the coefficient of friction is 0.. Determine the force required at the end of the lever to support the load. Take a = 150 mm and l = 750 mm. Solution: W N R 300 mm r 00 mm Fig..35 For equilibrium position.. T1 T R W r.87 T T T 4197N m and T Nm Taking moment about O.. F l T a 1 F F 409 N T T e e.87 Prepared By: Vimal G. Limbasiya Darshan Institute of Engineering & Technology, Rajkot Page.55

78 . Friction Devices: Clutches, Brakes and Dynamometers Theory of Machines (15190) Example 17:- A band and block brake has 14 blocks. Each block subtends an angle of 14 at the centre of the rotating drum. The diameter of the drum is 750 mm and the thickness of the blocks is 65 mm. The two ends of the band are fixed to the pins on the lever at distances of 50 mm and 10 mm from the fulcrum on the opposite sides. Determine the least force required to be applied at the lever at a distance of 600 mm from the fulcrum if the power absorbed by the blocks is 180 kw at 175 rpm. Coefficient of friction between the blocks and drum is Solution: n 14 blocks 14 d 750 mm t 65 mm a 10 mm, b 50 mm l 600 mm P 180 kw N 175 rpm 0.35 F? Let Let T14 T P T T v T T DN T T 33N (1) T14 1 tan T0 1 tan T tan7 T tan7 n () From equation (1) and () T T N N Assume a > b, F must be downward and clockwise rotation for maximum braking torque. Moment about O. (Fig. in theory) F l T a T b F F 690 N Prepared By: Vimal G. Limbasiya Page.56 Darshan Institute of Engineering & Technology, Rajkot

79 Theory of Machines (15190). Friction Devices: Clutches, Brakes and Dynamometers Example 18:- A band and block brake having 1 blocks, each of which subtends an angle of 16 at the centre, is applied to a rotating drum with a diameter of 600 mm. The blocks are 75 mm thick. The drum and the flywheel mounted on the same shaft have a mass of 1800 kg and have a combined radius of gyration of 600 mm. The two ends of the band are attached to pins on the opposite sides of the brake fulcrum at distances of 40 mm and 150 mm from it. If a force of 50 N is applied on the lever at a distance of 900 mm from the fulcrum, find the 1. Maximum braking torque. Angular retardation of the drum 3. Time taken by the system to be stationary from the rated speed of 300 rpm. Solution: 16 n 1 blocks d 600 mm t 75mm m 1800 kg k 600 mm a 150 mm, b 40 mm F 50 N l 900 mm N 300 rpm 0.3 Refer the Fig T1 1 tan T0 1 tan T tan8 T tan8.75 n 1 Assume a = 150 mm, b = 40 mm as a > b, F must be downwardsand rotation is clockwise. Taking moment about fulcrum O. F l T a T b T (150) T T T T 5636 N 15511N Maximum Braking torque. B 1 0 T T T r N m d d t r Prepared By: Vimal G. Limbasiya Darshan Institute of Engineering & Technology, Rajkot Page.57

80 . Friction Devices: Clutches, Brakes and Dynamometers Theory of Machines (15190). Let Braking torque T I mk B rad / sec 4. Initial Angular Speed 0 t t t 5.5 sec Final Angular speed 0 N rad / sec Example 19:- A car moving on a level road at a speed 50 km/h has a wheel base.8metres, distance of C.G. from ground level 600 mm, and the distance of C.G. from rear wheels 1.metres. Find the distance travelled by the car before coming to rest when brakes are applied, 1. to the rear wheels,. to the front wheels, and 3. to all the four wheels. The coefficient of friction between the tyres and the road may be taken as 0.6. Solution: u 50 km / hr m / sec L.8 m h 600 mm x 1. m Rear Wheels Here vehicle moves on a level road, so retardation of car is.. If retardation is uniform. Front Wheels g (L x) a.98 m / sec L h v u s a 0 u s a u s 3.4 m a.98 Here vehicle moves on a level road, so retardation of car is. a g x m / sec L h Prepared By: Vimal G. Limbasiya Page.58 Darshan Institute of Engineering & Technology, Rajkot

81 Theory of Machines (15190). Friction Devices: Clutches, Brakes and Dynamometers For uniform retardation. u s a s m 3. All the four wheels Here vehicle moves on a level road, so retardation of car is.. For uniform retardation. a = gμ = 9.81 x 0.6 = m/sec u s a m Example 0:- A vehicle moving on a rough plane inclined at 10 with the horizontal at a speed of 36 km/h has a wheel base 1.8 metres. The centre of gravity of the vehicle is 0.8 metre from the rear wheels and 0.9 metre above the inclined plane. Find the distance travelled by the vehicle before coming to rest and the time taken to do so when 1. The vehicle moves up the plane, and. The vehicle moves down the plane. The brakes are applied to all the four wheels and the coefficient of friction is 0.5. Solution: 10 u 36 km / h 10 m / sec L 1.8 m x 0.8 m h 0.9 m The vehicle moves up (All Four wheels) a g cos sin cos 10 sin m / sec For uniform retardation. u 10 s a m Final velocity of vehicle. v u a t ve sign due to retardation t t 1.53 sec. The vehicle moves down the plane (All Four wheels) Prepared By: Vimal G. Limbasiya Darshan Institute of Engineering & Technology, Rajkot Page.59

82 . Friction Devices: Clutches, Brakes and Dynamometers Theory of Machines (15190) a g cos sin cos 10 sin m / sec For uniform retardation. Final velocity of vehicle. v u a t u 10 s 16 m a t t 3. sec Example 1:- The following data refer to a laboratory experiment with a rope brake: Diameter of the flywheel = 800 mm Diameter of the rope = 8 mm Dead weight on the brake = 40 kg Speed of the engine = 150 rpm Spring balance reading = 100 N Find the power of the engine. Solution: N P M g s r W Example :- In a belt transmission dynamometer, the driving pulley rotates at 300 rpm. The distance between the centre of the driving pulley and the dead mass is 800 mm. The diameter of each of the driving as well as the intermediate pulleys is equal to 360 mm. Find the value of the dead mass required to maintain the lever in a horizontal position when the power transmitted is 3 kw. Also, find its value when the belt just begins to slip on the driving pulley, µ being 0.5 and the maximum tension in the belt 100 N. Solution: N 300 rpm a 0.36 m l 0.8 m P 3000 W (i) P T1 T v M g l r a M g l N r a 60 M M 48.7 kg Prepared By: Vimal G. Limbasiya Page.60 Darshan Institute of Engineering & Technology, Rajkot

83 Theory of Machines (15190). Friction Devices: Clutches, Brakes and Dynamometers (ii) T 100 N, 0.5, rad 1 T T e e T 548 N.19 M g l T1 T a M M 59.8 kg Example 3:- A torsion dynamometer is fitted to a propeller shaft of a marine engine. It is found that the shaft twists o in a length of 0 meters at 10 r.p.m. If the shaft is hollow with 400 mm external diameter and 300 mm internal diameter, find the power of the engine. Take modulus of rigidity for the shaft material as 80 GPa. Solution: l 0 m N 10 rpm D 400 mm d 300 mm G 80 GPa 0.035rad 180 Polar moment of Inertia of the shaft, 4 4 J (D d ) J [(0.4) (0.3) ] 3 J = m 4 T G J l G T J l T Torque applied to the shaft, T = 38 x 10 3 N.m NT P P 60 Power of the engine, P = 990 kw 3 Prepared By: Vimal G. Limbasiya Darshan Institute of Engineering & Technology, Rajkot Page.61

84

85 3 Flywheels Course Contents 3.1 Turning-Moment Diagram 3. Single-Cylinder Double-Acting Steam Engine 3.3 Single cylinder Four Stroke Engine 3.4 Multi Cylinder Engines 3.5 Fluctuation of energy 3.6 Flywheel 3.7 Dimensions of Flywheel Rim 3.8 Punching Press 3.9 Summary Prepared By: Subhesh Pansuria Darshan Institute of Engineering & Technology, Rajkot Page 3.1

86 3. Flywheels Theory of Machines (15190) 3.1 Turning-Moment Diagrams During revolution of the crankshaft of a steam engine or IC engine, the torque on it varies and is given by, sin θ T = F t r = F r (sin θ + n sin θ ) Where F is the net piston effort. A plot of T vs θ is known as the turning-moment diagram. The inertia effect of the connecting rod is, usually ignored while drawing these diagrams, but can be taken into account if desired. As T=Ft x r, a plot of Ft vs θ (known as crank effort diagram) is identical to a turning-moment diagram. The turning-moment diagram for different types of engines are being given below: 3. Single-Cylinder Double-Acting Steam Engine Figure shows a turning-moment diagram for a single-cylinder double-acting steam engine. The crank angle θ is represented along x-axis & the turning-moment along y-axis. It can be observed that during the outstroke (ogp) the turning moment is maximum when the crank angle is a little less than 90. & zero with the crank angle is zero & 180. A somewhat similar turning moment diagram is obtained and the angle turned. Note that the area of the turning-moment diagram is proportional to the work done per revolution as the work is the product of the turning-moment diagram and the angle turned. The mean torque against which the engine works is given by oe = Area ogpkp π Prepared By: Subhesh Pansuria Darshan Institute of Engineering & Technology, Rajkot Page 3.

87 3. Flywheels Theory of Machines (15190) Where oe is the mean torque & is the mean height of the turning-moment diagram. When the crank turns from the angle oa to ob, the work done bt the engine is represented by the area afghb. But the work done against the resisting torque is represented by afhb. Thus, the engine has work done more work than what has been taken from it. The excess work is represented by the area fgh. This excess work increases the speed of the engine & is stored in the flywheel. During the crank travel from the ob or oc, the work needed for the external resistance is proportional to bhjc whereas the work produced by the engine is represented by the area under hpj. Thus, during this period, more work has been taken from the engine that is produced. The loss is made up by the flywheel which gives up some of its energy& the speed decreases during this period. Similarly, during the period of crank travel from oc to od, excess work is again developed and is stored in the flywheel & the speed of the engine increases. During the crank travel from od to oa, the loss of work is made up by the flywheel and the speed again decreases. The areas fgh, hpj, jkl & lqf represent fluctuations of energy of the flywheel. When the crank is at b, the flywheel has absorbed energy while the crank has moved from a to b & thereby, the speed of the engine is maximum. At c, the flywheel has given out energy while the crank has moved from b to c & thus the engine has a maximum speed. Similarly, the engine speed is again maximum at d & minimum at a. Thus, there are two maximum & two minimum speeds for the turning-moment diagram. The greatest speed is the greater of the two maximum speeds & the least speed is the lesser of the two minimum speeds. The difference between the greatest & the least speeds of the engine over one revolution is known as fluctuation of speed. 3.3 Single cylinder Four Stroke Engine Prepared By: Subhesh Pansuria Darshan Institute of Engineering & Technology, Rajkot Page 3.3

88 3. Flywheels Theory of Machines (15190) 3.4 Multi Cylinder Engines 3.5 Fluctuation of energy Let a1, a, and a5 be the areas in work units of the portions above the mean torque ab of the turning moment diagram (fig.). These areas represent quantities of energies added to flywheel. Similarly, areas a, a4 and a6 below ab represent quantities of energies taken from the flywheel. The energies of the flywheel corresponding to positions of the crank are as follows: Crank position c d E E + a 1 Flywheel energy e f g h c E + a 1 a E + a 1 a + a 3 E + a 1 a + a 3 a 4 E + a 1 a + a 3 a 4 + a 5 E + a 1 a + a 3 a 4 + a 5 a 6 From the two values of the energies of the flywheel corresponding to the position c, it is concluded that E + a 1 a + a 3 a 4 + a 5 a 6 = 0 The greatest of these energies is the maximum kinetic energy of the flywheel and for the corresponding crank position, the speed is maximum. The least of these energies is the least kinetic energy of the flywheel and for the corresponding crank position, the speed is minimum. Prepared By: Subhesh Pansuria Darshan Institute of Engineering & Technology, Rajkot Page 3.4

89 3. Flywheels Theory of Machines (15190) The different between the maximum and minimum kinetic energies of the flywheel is known as the maximum fluctuation of energy whereas the ratio of this maximum fluctuation of energy to the work done per cycle is define as the coefficient fluctuation of energy. The different between the greatest speed and the least speed is known as the maximum fluctuation of speed and the ratio of maximum fluctuation of speed to the mean speed is the coefficient of fluctuation of speed. 3.6 Flywheel A flywheel is used to control the variations in speed during each cycle of an engine. A flywheel of suitable dimensions attached to the crankshaft, makes the moment of the rotating part quite large and thus, acts as a reservoir of energy. During the periods when the supply of energy is more than required, it stores energy and during the periods the requirements is more than the supply, it release energy. I = moment of inertia of the flywheel ω 1 = maximum speed ω = minimum speed ω = mean speed E = kinetic energy of the flywheel at mean speed e = maximum fluctuation of energy K = coefficient of fluctuation of speed = ω 1 ω ω Maximum fluctuation of energy, e = 1 Iω 1 1 Iω = 1 I(ω 1 ω ) = I ( ω 1 + ω ) (ω 1 ω ) = I ω (ω 1 ω ) = I ω ( ω 1 ω ω ) = I ω K K = e I w = e 1 = e E Iω Prepared By: Subhesh Pansuria Darshan Institute of Engineering & Technology, Rajkot Page 3.5

90 3. Flywheels Theory of Machines (15190) 3.7 Dimensions of Flywheel Rims The inertia of a flywheel is provided by the hub, spokes and the Rim. However, as the inertia due to the hub and the spokes is very small, usually it is ignored. In case it is known, it can be taken into account. Consider a rim of flywheel as shown in Fig Let ω = angular velocity r = mean radius t = thickness of the rim ρ = density of the material of the rim Consider an element of the rim, Centrifugal force on the element/unit length = = [ρ (r dθ) t] r ω Total vertical force /unit length π = ρ r 0 π dθ t ω sin θ = ρ. r. t. ω sin θ dθ 0 = ρ r t ω ( cos θ) π = ρ 0 r t ω Prepared By: Subhesh Pansuria Darshan Institute of Engineering & Technology, Rajkot Page 3.6

91 3. Flywheels Theory of Machines (15190) Let σ = circumferential stress induces in the rim (Circumferential stress is also known as hoop stress.) Then for equilibrium, σ(t) l = ρ r t ω σ = ρ r ω = ρ v The above relation provides the limiting tangential velocity at the mean radius of rim of the flywheel. Then the diameter can be calculated from the relation, v = πdn/60 Also, mass = density volume = density circumference cross-sectional area m = ρ πd b t The relation can be used to find the width and the thickness of the rim. 3.8 Punching Press From the previous discussion, it can be observed that when the load on the crankshaft is constant or varies and the input torque varies continuously during a cycle, a flywheel is used to reduce the fluctuation of speed. A flywheel can perform the same purpose in a punching press or a riveting machine in which the torque available is constant but the load varies during the cycle. Prepared By: Subhesh Pansuria Darshan Institute of Engineering & Technology, Rajkot Page 3.7

92 3. Flywheels Theory of Machines (15190) Figure shows the sketch of a punching press. It is a slider crank mechanism in which a punch replaces the slider. A motor provides a constant torque to the crankshaft through a flywheel. It may be observed that the actual punching process is performed only during the downward stroke of the punch and that also for a limiting period when the punch travels through the thickness of the plate. Thus, the load is applied during the actual punching process only and during the rest of the downward stroke and the return stroke, there is no load on the crankshaft. In the absence of a flywheel, the decrease in the speed of the crankshaft will be very large during the actual punching period whereas it will increase to a much higher value during the no-load period as the motor will continue to supply the energy all the time. 3.9 Summary 1. Dynamic forces are associated with accelerating masses. As all machines have some accelerating parts, dynamic forces are always present when the machines operate.. D' Alembert's principle states that the inertia forces and couples, and the external forces and torques on a body together give statically equilibrium. 3. In graphical solutions, it is possible to replace inertia force and inertia couple by an equivalent offset inertia force which can account for both. This is done by displacing the line of action of the inertia force from the centre of mass. 4. The sense of angular acceleration of the connecting rod is such that it tends to reduce the angle of the connecting rod with the line of stroke. 5. The piston effort is the net or effective force applied on the piston. 6. Inertia force on the piston, cos θ F b = m f = m r ω (cos θ + n ) 7. Crank effort is the net effort (force) applied at the crankpin perpendicular to the crank which gives the required turning moment on the crankshaft. 8. Turning moment due to force F on the piston sin θ T = F r (sin θ + n sin θ ) 9. A dynamically equivalent system means that the rigid link is replaced by a link with two point masses in such a way that it has the same motion as the rigid link when subjected to the same force, i.e., the centre of mass of the equivalent link has the same linear acceleration and the link has the same angular acceleration. 10. The distributed mass of a rod can be replaced by two point masses to have the same Prepared By: Subhesh Pansuria Darshan Institute of Engineering & Technology, Rajkot Page 3.8

93 3. Flywheels Theory of Machines (15190) dynamical properties if the sum of the two masses is equal to the total mass, the combined centre of mass coincides with that of the rod and the moment of inertia of two point masses about the perpendicular axis through their combined centre of mass is equal to that of the rod. 11. In the analysis of the connecting rod, the two point masses are considered to be located at the center of the two end bearings and then a correction is applied for the error involved. 1. A plot of T vs. θ is known as the turning- moment diagram. 13. The difference between the maximum and minimum kinetic energies of the flywheel is known as the maximum fluctuation of energy. 14. The difference between the greatest speed and the least speed is known as the maximum fluctuation of speed. 15. A flywheel is used to control the variations in speed during each cycle of an engine. 16. Coefficient of fluctuation of speed is given by K = e Iω = e E. Prepared By: Subhesh Pansuria Darshan Institute of Engineering & Technology, Rajkot Page 3.9

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95 4 Governors Course Contents 4.1 Introduction. 4. Types of Governors. 4.3 Terms used in Governor. 4.4 Watt Governor. 4.5 Porter Governor. 4.6 Proell Governor. 4.7 Terms related with Governor. 4.8 Controlling Force. 4.9 Problems. Prepared By: Prof. A. J. Makadia Darshan Institute of Engineering & Technology, Rajkot Page 4.1

96 4. Governors Theory of Machines (15190) 4.1 Introduction Function of Governor: The function of a governor is to regulate the meanspeed of an engine, when there are variations in the load e.g.when the load on an engine increases, its speed decreases,therefore it becomes necessary to increase the supply of workingfluid. On the other hand, when the load on the enginedecreases, its speed increases and thus less working fluid isrequired. The governor automatically controls the supply ofworking fluid to the engine with the varying load conditionsand keeps the mean speed within certain limits. A little consideration will show, that when the loadincreases, the configuration of the governor changes and avalve is moved to increase the supply of the working fluid ;conversely, when the load decreases, the engine speed increasesand the governor decreases the supply of workingfluid. 4.Types of Governors 1. Centrifugal Governor.. Inertia Governor. Centrifugal Governor Pendulum type Loaded Type Watt Governor Dead Weight Governor Spring Controlled Governor Porter Governor Proell Governor Hartnell Hartung Wilson Hartnell Pickering Governor Governor Governor Governor Prepared By: Prof. A. J. Makadia Page 4. Darshan Institute of Engineering & Technology, Rajkot

97 Theory of Machines (15190) 4. Governors Centrifugal Governor Fig. 4.1 Centrifugal Governor The centrifugal governors are based on the balancing of centrifugal force on the rotating balls by an equal and opposite radial force, known as the controlling force. It consists of two balls of equal mass, which are attached to the arms as shown in figure. These balls are known as governor balls or fly balls. The balls revolve with a spindle, which is driven by the engine through bevel gears. The upper ends of the arms are pivoted to the spindle, so that the balls may rise up or fall down as they revolve about the vertical axis. The arms are connected by the links to a sleeve, which is keyed to the spindle. This sleeve revolves with the spindle; but can slide up and down. The balls and the sleeve rise when the spindle speed increases, and falls when the speed decreases. In order to limit the travel of the sleeve in upward and downward directions, two stops S, S are provided on the spindle. The sleeve is connected by a bell crank lever to a throttle valve. The supply of the working fluid decreases when the sleeve rises and increases when it falls. When the load on the engine increases, the engine and the governor speed decreases. This results in the decrease of centrifugal force on the balls. Hence the balls move inwards and the sleeve moves downwards. The downward movement of the sleeve operates a throttle valve at the other end of the bell crank lever to increase the supply of working fluid and thus the engine speed is increased. In this case, the extra power output is provided to balance the increased load. When the load on the engine decreases, the engine and the governor speed increases, which results in the increase of centrifugal force on the balls. Thus the balls move outwards and the sleeve rises upwards. This upward movement of the sleeve reduces the supply of the working fluid and hence the speed is decreased. In this case, the power output is reduced. Prepared By: Prof. A. J. Makadia Darshan Institute of Engineering & Technology, Rajkot Page 4.3

98 4. Governors Theory of Machines (15190) 4.3Terms used in Governor 1. Height of governor (h): It is the verticaldistance from the centre of the ball to a pointwherethe axes of the arms (or arms produced) intersecton the spindle axis. It is usually denoted by h.. Equilibrium speed: It is the speed atwhich the governor balls, arms etc., are in complete equilibrium and the sleeve does not tend to moveupwards or downwards. 3. Mean equilibrium speed: It is the speedat the mean position of the balls or the sleeve. 4. Maximum and Minimum equilibrium speed: The speeds at the maximum and minimum radius of rotation of the balls, without tending tomove either way are known as maximum and minimumequilibrium speeds respectively. 5. Sleeve lift: It is the vertical distance which the sleeve travels due to change in equilibrium speed. 4.4Watt Governor Fig. 4. Watt Governor The simplest form of a centrifugal governor is a Watt governor, as shown in Figure. It is basically a conical pendulum with links attached to a sleeve of negligible mass. The arms of thegovernor may be connected to the spindle in the following three ways : 1. The pivot (P), may be on the spindle axis as shown in figure 4. (a).. The pivot (P), may be offset from the spindle axis and the arms when produced intersect at O, as shwon in figure 4. (b). 3. The pivot (P), may be offset, but the arms cross the axis at O, as shown in figure 4. (c). Let m Mass of ballkg, w Weight of ball innewtons mg, T Tension in the arm innewtons, Prepared By: Prof. A. J. Makadia Page 4.4 Darshan Institute of Engineering & Technology, Rajkot

99 Theory of Machines (15190) 4. Governors Angular velocity of the arm & ball about the spindle axis inrad / s, r F C h Radius of the path of rotation of the ball i.e. Horizontal dist. from centre of the ball to the spindle axis inmetres, Centrifugal force acting on the ball innewtons m r, and Height of governor inmetres. It is assumed that the weight of arms, link & sleeve are negligible as comparedto the weight of the balls. Governor ball is equilibrium under the following 1. Centrifugal force F C. Tension (T) 3. Weight of ball (w) Taking moment about O. FC h w r (m r )h m g r g where g gravitational force 9.81m / s h N h metres N N Porter Governor Fig. 4.3 Porter Governor The Porter governor is a modification of a Watt s governor, with central load attached to the sleeve as shown in Fig. 4.3 (a). The load moves up and down the central spindle. This Prepared By: Prof. A. J. Makadia Darshan Institute of Engineering & Technology, Rajkot Page 4.5

100 4. Governors Theory of Machines (15190) additional downward force increases the speed of revolution required to enable the balls to rise to any predetermined level. Consider the forces acting on one half of the governor as shown in Fig. 4.3 (b). Let m Mass of eachballinkg, w Weight of eachball in newtons mg, W Weight of Centralload in newtons mg, r h N radius of rotation in metres, Height of governor in metres, Speed of the balls inr.p.m., Angular speed of the balls inrad / s N/ 60, FC Centrifugal force acting on the ball m r, T T 1 Force in the arm innewtons, Force in the link innewtons, Angle of inclination of the arm (or upper link) to the vertical, and Angle of inclination of the link (or lower link) to the vertical. Though there are several ways of determining the relation between the height of the governor (h) and the angular speed of the balls (ω), yet the following two methods are important from the subject point of view: 1. Method of resolution of forces; and. Instantaneous centre method. 1. Method of resolution of forces Considering the equilibrium of the forces acting at D, W M g T cos M g or T...(i) cos Again, considering the equilibrium of the forces acting on B. the point B is in equilibrium under the action of the following forces, as shown in figure 4.3(b). i. The weight of ball (w = m. g), ii. The centrifugal force (F C ), iii. The tension in the arm (T 1 ), and iv. The tension in the link (T ). Resolving the forces vertically, Mg T1 cos T cos w m g...(ii) Prepared By: Prof. A. J. Makadia Page 4.6 Darshan Institute of Engineering & Technology, Rajkot

101 Theory of Machines (15190) 4. Governors Resolving the forces horizontally, T1 sin T sin FC Mg T cos w m g Mg Mg T1 sin sin FC T cos cos T sin Mg tan F 1 C Mg T1 sin FC tan...(iii) Dividing equation (3) by equation () M g tan FC T1 sin T Mg 1 cos mg Mg Mg tan m g tan F C Mg FC Mg tan tan mg Take q,tan tan tan tan FC m r r h Mg m r Mg m g q r h Mg m h m g 1 q Mg m g 1 q h...(iv) m M g 1 q mg hm g 1 q m M hm Prepared By: Prof. A. J. Makadia Darshan Institute of Engineering & Technology, Rajkot Page 4.7

102 4. Governors Theory of Machines (15190) N g M m 1 q 60 hm 60 g M N m 1 q hm Take g 9.81m / s 895 M N m 1 q...(v) hm Note: 1. When the length of arms are equal to the length of links and the points P & D lie on the same vertical line, then tan tan tan then q 1 tan Therefore, the equation (5) becomes 895 N (m M)...(vi) hm.when the loaded sleeve moves up & down the spindle, the frictional force acts on it in a direction opposite to that of the motion of sleeve. If F = Frictional force acting on the sleeve in newtons, then the equations (5) and (6) may be written as 895 Mg F N mg 1 q hm g The + sign is used when the sleeve moves upwards or the governor speed increases and negative sign is used when the sleeve moves downwards or the governor speed decreases.. Instantaneous centre method In this method, equilibrium of the forces acting on the link BD are considered. The instantaneous centre I lies at the point of intersection of PB produced and a line through D perpendicular to the spindle axis, as shown in Fig Taking moments about the point I, W FC BM wim ID Mg FC BM mg IM IM MD Fig. 4.4 Instantaneous centre method Prepared By: Prof. A. J. Makadia Page 4.8 Darshan Institute of Engineering & Technology, Rajkot

103 Theory of Machines (15190) 4. Governors IM Mg IM MD FC mg BM BM BM Mg FC mg tan tan tan IM MD tan, tan BM BM Dividing the equation by tan FC Mg tan mg 1 tan tan m r Mg mg 1 q r h tan r tan h q, tan,fc m r Mg m h mg 1 q Mg mg 1 q h m g M h m 1 q m m M h g m When tan tan tan then q 1 tan 4.6Proell Governor The Proell governor has the balls fixed at B and C to the extension of the links DF and EG, as shown in Fig. 4.5 (a). The arms FP and GQ are pivoted at P and Q respectively. Consider the equilibrium of the forces on one-half of the governor as shown in Fig. 4.5 (b). The instantaneous centre (I) lies on the intersection of the line PF produced and the line from D drawn perpendicular to the spindle axis. The perpendicular BM is drawn on ID. Prepared By: Prof. A. J. Makadia Darshan Institute of Engineering & Technology, Rajkot Page 4.9

104 4. Governors Theory of Machines (15190) Fig. 4.5 Proell Governor Taking moment about I W FC BM wim ID Mg FC BM mg IM IM MD...(i) IM Mg IM MD FC mg BM BM BM FM IM Mg IM MD mg BM FM FM FM Multiplying and dividing by FM FM mg tan Mg tan tan IM tan, MD tan BM FM FM FM Mg tan tan mg 1 BM tan tan tan r h Putting q, tan, FC m r FM r Mg m r mg 1 q BM h FM g M m 1 q BM hm...(ii) N, g 9.81m / s 60 FM 895 M N m 1 q BM mh...(iii) Prepared By: Prof. A. J. Makadia Page 4.10 Darshan Institute of Engineering & Technology, Rajkot

105 Theory of Machines (15190) 4. Governors 4.7 Terms related with Governor 1. Sensitiveness of Governors Consider two governors A and B running at the same speed. When this speed increases or decreases by a certain amount, the lift of the sleeve of governor A is greater than the lift of the sleeveof governor B. It is then said that the governor A is more sensitive than the governor B. In general, the greater the lift of the sleeve corresponding to a given fractional change in speed, the greater is the sensitiveness of the governor.it may also be stated in another way that for agiven lift of the sleeve, the sensitiveness of the governor increases as the speed range decreases. Thisdefinition of sensitiveness may be quite satisfactory when the governor is considered as an independentmechanism. But when the governor is fitted to an engine, the practical requirement is simply that thechange of equilibrium speed from the full load to the no load position of the sleeve should be as smalla fraction as possible of the mean equilibrium speed. The actual displacement of the sleeve is immaterial,provided that it is sufficient to change the energy supplied to the engine by the required amount. Forthis reason, the sensitiveness is defined as the ratio of the difference between the maximum andminimum equilibrium speeds to the mean equilibrium speed. Let N N N 1 Minimum equilibrium speed, Maximum equilibrium speed, Mean equilibrium speed N1 N. N 1 Sensitiveness N N N1 N N N (In terms of angular speeds). Stability of Governors A governor is said to be stable when for every speed within the working range there is a definite configuration i.e. there is only one radius of rotation of the governor balls at which the governor is in equilibrium. For a stable governor, if the equilibrium speed increases, the radius ofgovernor balls must also increase. A governor is said to be unstable, if the radius of rotation decreases as the speed increases. Prepared By: Prof. A. J. Makadia Darshan Institute of Engineering & Technology, Rajkot Page 4.11

106 4. Governors Theory of Machines (15190) 3. Isochronous Governors A governor is said to be isochronous when the equilibrium speed is constant (i.e. range of speed is zero) for all radii of rotation of the balls within the working range, neglecting friction. The isochronism is the stage of infinite sensitivity. Let us consider the case of a Porter governor running at speeds N 1 and N r.p.m. 895 M N1 m 1 q...(i) hm 895 M N m 1 q...(ii) hm For isochronism, range of speed should be zero i.e. N N 1 = 0 or N = N 1. Therefore from equations (i) and (ii), h 1 = h, which is impossible in case of a Porter governor. Hence a Porter governor cannot be isochronous. Now consider the case of a Hartnell governor running at speeds N 1 and N r.p.m. M g S y F x x N 1 x M g S1 FC1 m r1 y 60 y 1 C1...(iii) M g S y F x x N x M g S FC m r y 60 y C...(iv) For isochronism, N = N.Therefore from equation (iii) and (iv), Mg S r Mg S r 1 1 (condition of Isochronous governor) 4. Hunting A governor is said to be hunt if the speedof the engine fluctuates continuously above andbelow the mean speed. This is caused by a toosensitive governor which changes the fuel supplyby a large amount when a small change in thespeed of rotation takes place. For example, whenthe load on the engine increases, the engine speeddecreases and, if the governor is very sensitive,the governor sleeve immediately falls to its lowestposition. This will result in the opening of thecontrol valve wide which will supply the fuel to the engine in excess of its requirement so that theengine speed rapidly increases again and the governor Prepared By: Prof. A. J. Makadia Page 4.1 Darshan Institute of Engineering & Technology, Rajkot

107 Theory of Machines (15190) 4. Governors sleeve rises to its highest position. Due to thismovement of the sleeve, the control valve will cut off the fuel supply to the engine and thus the enginespeed begins to fall once again. This cycle is repeated indefinitely. Such a governor may admit either the maximum or the minimum amount of fuel. The effectof this will be to cause wide fluctuations in the engine speed or in other words, the engine will hunt. 4.8 Controlling Force We have seen earlier that when a body rotates in a circularpath, there is an inward radial force or centripetal force acting on it.in case of a governor running at a steady speed, the inward forceacting on the rotating balls is known as controlling force. It is equaland opposite to the centrifugal reaction. Controlling force, F C = m.ω.r Fig. 4.6 Controlling force diagram The controlling force is provided by the weight of the sleeve and balls as in Porter governor and by the spring and weight as in Hartnell governor (or spring controlled governor). When the graph between the controlling force (F C ) as ordinate and radius of rotation of the balls (r) as abscissa is drawn, then the graph obtained is known as controlling force diagram. This diagram enables the stability and sensitiveness of the governor to be examined and also shows clearly the effect of friction. 1. Controlling Force diagram for Porter Governor The controlling force diagram for a porter governor is a curve as shown in Fig N FC m r m r 60 Or 1 60 FC 1 60 FC N tan tan m r m r 60 tan N m 1 Where φ is the angle between the axis of radius of rotation and a line joining a given point(say A) on the curve to the origin O.. Controlling Force diagram for Spring - controlled Governors The controlling force diagram for the spring controlled governors is a straight line, as shownin Fig We know that controlling force, Prepared By: Prof. A. J. Makadia Darshan Institute of Engineering & Technology, Rajkot Page 4.13

108 4. Governors Theory of Machines (15190) F m r or F lr m C C The following points, for the stability of spring controlled governors, may be noted: 1. For the governor to be stable, the controlling force (F C ) must increase as the radius of rotation(r) increases, i.e. F C / r must increase as r increases. Hence the controlling force line AB when produced must intersect the controlling force axis below the origin, as shown in Fig The relation between the controlling force (FC) and the radius of rotation (r) for the stability of spring controlled governors is given by the following equation FC ar b...(i) Where a and b are constants. Fig The value of b in equation (i) may be made either zero or positive by increasing the initial tension of the spring. If b is zero, the controlling force line CD passes through the origin and the governor becomes isochronous because F C /r will remain constant for all radii of rotation. The relation between the controlling force and the radius of rotation, for an isochronous governor is, therefore, FC ar...(ii) 3.If b is greater than zero or positive, then FC /r decreases as r increases, so that the equilibrium speed of the governor decreases with an increase of the radius of rotation of balls, which is impracticable. Such a governor is said to be unstable and the relation between the controlling force and the radius of rotation is, therefore FC ar b...(iii) Prepared By: Prof. A. J. Makadia Page 4.14 Darshan Institute of Engineering & Technology, Rajkot

109 Theory of Machines (15190) 4. Governors 4.9 Problems Problem 1:-Calculate the vertical height of a Watt governor when it rotates at 60 r.p.m. Also find the change in vertical height when its speed increases to 61 r.p.m. Solution: N N 1 Let 60 rpm 61rpm h 0.48 m h 1 N m N 61 Change inheight m Problem :-The arms of a Porter governor are each 50 mm long and pivoted on the governor axis. The mass of each ball is 5 kg and the mass of the central sleeve is 30 kg. The radius of rotation of the balls is 150 mm when the sleeve begins to rise and reaches a value of 00 mm for maximum speed. Determine the speed range of the governor. If the friction at the sleeve is equivalent of 0 N of load at the sleeve, determine how the speed range is modified. Solution: BP BD 50 mm m 5kg M 30 kg r 150 mm,r 00 mm 1 F 0 N(Friction) For Minimum Speed Prepared By: Prof. A. J. Makadia Darshan Institute of Engineering & Technology, Rajkot Page 4.15

110 4. Governors Theory of Machines (15190) 895 M N1 m 1 q h m N rpm Arms are same length andmounted on the same axis tan q 1 tan h mm 0. m For Maximum speed 895 M N m 1 q h m N rpm tan Here q 1 tan h mm 0.15m Speed range of governor N N rpm (a) Considering friction F(N) for minimum speed. 895 Mg F tan N1 mg 1 q q 1 h1m g tan 895 mg Mg F h m g N 17 rpm Prepared By: Prof. A. J. Makadia Page 4.16 Darshan Institute of Engineering & Technology, Rajkot

111 Theory of Machines (15190) 4. Governors (b) Considering friction F = 0N for maximum speed. 895 N mg Mg F h m g N 10 rpm Speed range of governor N N rpm Problem 3:-In an engine governor of the Porter type, the upper and lower arms are 00mmand 50 mm respectively and pivoted on the axis of rotation. The mass of the central load is 15 kg, the mass of each ball is kg and friction of the sleeve together with the resistance of the operating gear is equal to a load of 4 N at the sleeve. If the limiting inclinations of the upper arms to the vertical are 30 and 40, find, taking friction into account, range of speed of the governor. Solution: PB 00 BD 50 M 15kg m kg F 4 N For Minimum Condition Figure (a) Prepared By: Prof. A. J. Makadia Darshan Institute of Engineering & Technology, Rajkot Page 4.17

112 4. Governors Theory of Machines (15190) For Maximum Condition Figure (b) 895 Mg F N1 mg 1 q1 h 1m g N rpm 895 Mg F N mg 1 q m g h N From fig. h1 cos30 h m 00 r mm 0.1m tan1 tan sin tan 1 tan tan1 q tan rpm From fig. h cos40 h m 00 tan q tan Here tan tan r sin40 r m sin tan tan tan 0.59 q tan Range o f speed N N rpm 1 Prepared By: Prof. A. J. Makadia Page 4.18 Darshan Institute of Engineering & Technology, Rajkot

113 Theory of Machines (15190) 4. Governors Problem 4:-The arms of a Porter governor are 300 mm long. The upper arms are pivoted on the axis of rotation. The lower arms are attached to a sleeve at a distance of 40 mm from the axis of rotation. The mass of the load on the sleeve is 70 kg and the mass of each ball is 10 kg. Determine the equilibrium speed when the radius of rotation of the balls is 00 mm. If the friction is equivalent to a load of 0 N at the sleeve, what will be the range of speed for this position? Solution: BP BD 300 mm DH 40 mm M 70 kg m 10 kg r 00 mm N? F 4 N N N? 1 (1) Equilibrium speed at r = 00 mm 895 M N m 1 q hm N 167 rpm From figure... h m tan 00 q,tan tan 4 DF mm 0.54 m tan 0.630, q () Range of speed when F = 0N. Prepared By: Prof. A. J. Makadia Darshan Institute of Engineering & Technology, Rajkot Page 4.19

114 4. Governors Theory of Machines (15190) (a) For minimum condition 895 Mg F N1 mg 1 q1 h1m g N (b) For maximum condition rpm Here h1 0.4 h tan q tan F 0 N 895 Mg F N mg 1 q hm g N 169 rpm Range of speed N N rpm Here h 0.4 h q q F 0 N Problem 5:-A loaded Porter governor has four links each 50 mm long, two revolving masses each of 3 kg and a central dead weight of mass 0 kg. All the links are attached to respective sleeves at radial distances of 40 mm from the axis of rotation. The masses revolve at a radius of 150mm at minimum speed and at a radius of 00 mm at maximum speed. Determine the range of speed. Prepared By: Prof. A. J. Makadia Page 4.0 Darshan Institute of Engineering & Technology, Rajkot

115 Theory of Machines (15190) 4. Governors Solution: PB BD50 m 3kg M 0 kg DH 40 mm r 150 mm 1 r 00 mm N N? Let M N1 m 1 q h m N rpm tan Here q 1 tan 110 and sin tan6.10 h m h1 Then 895 M N m 1 q h m N rpm Prepared By: Prof. A. J. Makadia Darshan Institute of Engineering & Technology, Rajkot Page 4.1

116 4. Governors Theory of Machines (15190) Range of speed N N rpm tan Here q 1 tan 160 and sin tan39.8 h 0.40 m h Problem 6:-In a spring controlled governor, the curve of controlling force is a straight line. When balls are 400 mm apart, the controlling force is 100 N and when 00 mm apart, the controlling force is 450 N. At what speed will the governor run when the balls are 50 mm apart? What initial tension on the spring would be required for isochronism and what would then be the speed? The mass of each ball is 9 kg. Solution: When r 00 mm F 100 N C r 100 mm F 450 N 1 C1 (1) For the stability of spring controlled governor F ar b C When r r 100 mm 1 0.1m 450 a 0.1 b...(1) and r r 00 mm By solving equation (1) & () a 7500 Now we have 0.m 100 a 0. b...() b 300 FC r F C increase as r increase so far stability r conditionfc ar b Prepared By: Prof. A. J. Makadia Page 4. Darshan Institute of Engineering & Technology, Rajkot

117 Theory of Machines (15190) 4. Governors 50 FC ar b r 15 given 637.5N F m r C rad / sec N N 7.3rpm 60 () Initial tension on spring for Isochronism An Isochronism governor, the controlling force line passed through the origin (i.e. b=0). The value of b is made zero by increasing the initial tension of spring to 300 N. Initial tension on the spring for Isochronism = 300 N (3) Isochronism speed For Isochronism...F ' ar C m ' r ar N' N' 75 rpm Prepared By: Prof. A. J. Makadia Darshan Institute of Engineering & Technology, Rajkot Page 4.3

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119 5 Introduction to Dynamics Course Contents 5.1 Force Convention 5. Free-Body Diagrams 5.3 Equilibrium of Two-And-Three Force Members 5.4 Member with Two Forces and a Torque 5.5 Equilibrium of Four-Force Members 5.6 Superposition 5.7 Mass Moments and Products of Inertia 5.8 Inertia Forces and D alembert s Principle 5.9 The Principle of Superposition 5.10 Measuring Mass Moment of Inertia Prepared By: Subhesh Pansuria Darshan Institute of Engineering & Technology, Rajkot Page 5.1

120 5. Introduction to Dynamics Theory of Machines (15190) 5.1 Force Convention The force exerted by the member i on the member i is represented by Fij. (a) Points of application of forces F 1 and F to a rigid body may or may not be important. (b) The rectangular components of a force vector. 5. Free-Body Diagrams A free-body diagram is a sketch or diagram of a part isolated from the mechanism in order to determine the nature of forces acting on it. Prepared By: Subhesh Pansuria Darshan Institute of Engineering & Technology, Rajkot Page 5.

121 5. Introduction to Dynamics Theory of Machines (15190) Fig. (a) Shows a four-link mechanism. The free-body diagrams of its members, 3 and 4 are shown in Figs (b) (c) and (d) respectively. Various forces acting on each member are also shown. As the mechanism is in static equilibrium, each of its members must be in equilibrium individually. Member 4 is acted upon by three forces F, F34 and F14. Member 3 is acted upon by two forces F3 and F43. Member is acted upon by two forces F3 and F1 and a torque T. Initially, the direction and the sense of some of the forces may not be known. Assume that the force F on the member 4 is known completely. To know the other two forces acting on this member completely, the direction of one more force must be known. Link 3 is a two-force member and for its equilibrium, F3 and F43 must act along BC. Thus. F34, being equal and opposite to F43, also acts along BC. For the member 4 to be in equilibrium, F14 passes through the intersection of F and F34, by drawing a force triangle (F is completely known), magnitudes of F14 and F34 can be known [(e)]. Now, F34 = F43 = F3 = F3 Member will be in equilibrium if F1 is equal, parallel and opposite to F3 and T = F1 h = F3 h 5.3 Equilibrium of Two-And-Three Force Members A member under the action of two forces will be in equilibrium if The forces are of the same magnitude, The force acting along the same line, and The forces are in opposite directions. Figure shows such a member. A member under the action of three forces will be in equilibrium if The resultant of the forces is zero, and The lines of action of the forces intersect at a point (known as point of concurrency). Prepared By: Subhesh Pansuria Darshan Institute of Engineering & Technology, Rajkot Page 1.3

122 5. Introduction to Dynamics Theory of Machines (15190) (a) Two-force member not in equilibrium; (b) two-force member in equilibrium if FA and FB are equal, opposite, and share the same line of action; (c) three-force member not in equilibrium; and (d) three-force member in equilibrium if F A, F B, and F C are coplanar, if their lines of action intersect at a common point O, and if their vector sum is zero. Figure (a) shows a member acted upon by three forces F1, F, and F3 and is in equilibrium as the lines of action of forces intersect at one point O and the resultant is zero. This is verified by adding the forces vector ally [(b)]. As the head of the last vector F3 meets the tail of the first vector F1, the resultant is zero. It is not necessary to add the three vectors in order to obtain the resultant as is shown in Fig. (c) in which F is added to F3 and then F1 is taken. Figure shows a case where the magnitudes and directions of the forces are the same as before, but the lines of action of the forces do not intersect at one point. Thus, the member is not in equilibrium. Consider a member in equilibrium in which the force F1 is completely known, F is known in direction only and F3 is completely unknown. The point of application of F1, F, and F3 are A, B and C respectively. To solve such a problem, first find the point of concurrency O from the two forces with known directions, i.e., from F1 and F. Joining O with C gives the line of action of the third force F3. To know the magnitudes of the forces F and F3, take a vector of proper magnitude and direction to represent the force F1. From its two ends, draw lines parallel to the lines of action of the forces F and F3 forming of force triangle. Mark arrowheads on F and F3 so that F1, F Prepared By: Subhesh Pansuria Darshan Institute of Engineering & Technology, Rajkot Page 5.4

123 5. Introduction to Dynamics Theory of Machines (15190) and F3 are in the same order. If the lines of the action of the two forces are parallel then the point of concurrency lies at infinity and, therefore, the third force is also parallel to the first two. 5.4 Member with Two Forces and a Torque A member under the action of two forces and an applied torque will be in equilibrium if The forces are equal in magnitude, parallel in direction and opposite in sense, and The forces form a couple which is equal and opposite to the applied torque. Figure shows a member acted upon by two equal forces F1 and F and an applied torque T. For equilibrium, T = F1 * h = F * h Where T, F1 and F are the magnitudes of T, F1 and F respectively is clockwise whereas coupled formed by F1 and F is counter clockwise. 5.5 Equilibrium of Four-Force Members Normally, in most of the cases the above conditions for equilibrium of a member are found to be sufficient. However, in some problems, it may be found that the number of forces on a member is four or even more than that. In such cases, first look for the forces completely known and combine them into a single force repenting the sum of the known forces. This may reduce the number of forces acting on a body to two or three. However, in planer mechanism, a four-force system is also solvable if one force is known completely along with lines of action of the others. The following example illustrates the procedure. 5.6 Superposition In linear systems, if a number of loads act on a system of forces, the net effect is equal to the superposition of the effects of the individual loads taken one at a time. A linear system is one in which the output force is directly proportional to the input force, i.e., in mechanisms where coulomb or dry friction is neglected. Prepared By: Subhesh Pansuria Darshan Institute of Engineering & Technology, Rajkot Page 1.5

124 5. Introduction to Dynamics Theory of Machines (15190) 5.7 Mass Moments and Products of Inertia Another problem that often arises when forces are distributed over an area or volume is that of calculating their moment about a specified point or axis of rotation. Sometimes the force intensity varies according to its distance from the point or axis of rotation. Although we will save a more through derivation of this equation until previous Section and later, we will point out here that such problems always give rise to integrals of the form ʃ(distance) dm In three-dimensional problems, three such integrals are defined as follows: (I) xx = (Î R)(Î R)dm = ʃ [(R y ) + (R z ) ]dm (I) yy = (j R)(j R)dm = ʃ [(R z ) + (R x ) ]dm (I) zz = (k R)(k R)dm = ʃ [(R x ) + (R y ) ]dm These three integrals are called the mass moment for inertia of the body. Another three similar integrals are (I) xy = (I) yx = (Î R)(j R)dm = ʃ (R x R y )dm (I) yz = (I) zy = (j R)(k R)dm = ʃ (R y R z )dm (I) zx = (I) xz = (k R)(Î R)dm = ʃ (R z R x )dm And these three integrals are called the mass products of inertia of body. Sometimes it is convenient to arrange these mass moments of inertia and the mass product of inertia into a symmetric square array or matrix format called the inertia tensor of the body: I xx I xy I xz I = [ I yx I yy I yz ] I zx I zy I zz A careful look at the above integrals will indicate that they represents the mass distribution of the body with respect to the coordinate system about which they are determine, but that they change if evaluated in a different coordinate system.to keep their meaning direct and simple,we assume that the coordinate system chosen for each body is attached to that body in a convenient location and orientation.therefore, for rigid bodies,the mass moments and products of inertia are Prepared By: Subhesh Pansuria Darshan Institute of Engineering & Technology, Rajkot Page 5.6

125 5. Introduction to Dynamics Theory of Machines (15190) constant properties of body and it s mass distribution and they do not changed when the body is moves; they do,however,depend on the coordinate system chosen. An interesting property of these integrals is that it is always possible to choose the coordinate system so that it is origin it located at the center of mass of the body and oriented such that all of the mass product of inertia become zero such choice of coordinate axes of the body is called is its principal axes, and the corresponding values of eqs. (14.7) are then called the principal mass moment of inertia. A variety of simple geometric solid, the orientation of their principal axes, and formulas for their principal mass moments of inertia are included in Appendix A. If we note that mass moments of inertia have units of mass times distance squared, it seems natural to define a radius value of the body as I G = k m or k = I m This distance k is called the radius of gyration of the bodies, and it is always calculated or means med from the center of mass of the part about one of the principal axes. For three-dimensional motions of parts there are three radii of gyration k x, k y and k z associated with the three principal axes, I xx, I yy and I zz. It is often necessary to determine the moments and products of inertia of bodies, which are composed of several simpler sub shapes for which formula are known, such as those given in table 5 in Appendix A. The easiest method of finding these is to compute the mass moment about the principal axes of each sub shape, then to shift the origins of each to the mass center of the composite body, and then to sum the results.this require that we develop methods of redefining mass moments and products of inertia when the axes are translated to a new position. The form of the transfer, or parallel-axis theorem for mass moment of inertia, is written I = I G + md Where I G is one of the principal mass moments of inertia about some known principal axis and I is mass moment of inertia about a parallel axis at distance d from that principal axis. Equation must be used for translation of inertia axes starting from a principal axis. Also, the rotation of these axes results in the introduction of product of inertia terms. More will be said in general transformation of inertia in section. Only one mass moment of inertia I zz was requested or terminated. This does not mean that I xx and Prepared By: Subhesh Pansuria Darshan Institute of Engineering & Technology, Rajkot Page 1.7

126 5. Introduction to Dynamics Theory of Machines (15190) I yy are zero, but rather that will likely not be needed for further analysis.in problems with only planer motion, only I zz is needed because I xx and I yy are used only rotation out of the xy plan. These other mass moment and products of inertia are used in a section and later, where we treat problems with spatial motion, and they are determine in identical fashion. 5.8 Inertia Forces and D alembert s Principle Next, let us consider a moving rigid body of mass m acted upon by any system of forces, say F1, F, and F3, as illustrated in fig. We designate the center of mass of the body as point G, and we find the resultant of the system of forces from the equation F = F 1 + F + F 3 (a) An unbalanced set of forces on a rigid body. (b) The accelerations that result from the unbalanced forces. In the general case, the line of action of this resultant will not be through the mass center but will be displaced by some distance, illustrated in fig. (a) as distance h. We demonstrated in Eq. that the effect of this unbalanced force system is to produce an acceleration of the center of mass of the body: F ij = m i A G j In a very similar way, it has been proven that the unbalanced moment effect of this resultant force about the center of mass causes angular acceleration of the body that obeys the equation: M G = I G j α j Prepared By: Subhesh Pansuria Darshan Institute of Engineering & Technology, Rajkot Page 5.8

127 5. Introduction to Dynamics Theory of Machines (15190) However, this equation is restricted to use in taking moments about the center of mass G. It cannot be used for taking moments about an arbitrary point. The quantity F is the resultant of all external forces acting upon the body, and MG is the sum of all applied external moments and the moments of all externally applied forces about point G. The mass moment of inertia is designated as IG, Signifying that it must be taken with respect to the mass center G. Equations demonstrate that when an unbalanced system of forces acts upon a rigid body, the body experiences a rectilinear acceleration AG of its mass center in the same direction as the resultant force F. The body also experiences an angular acceleration α in the same direction as the resultant moment MG, caused by the moments of the forces and the torques about the mass center. This situation is illustrated in fig. (b). If the forces and moments are known, Eqs. may be used to determine the resulting acceleration pattern that is, the resulting motion of the body. During engineering design, however, the motions of the machine members are often specified in advance by other machine requirements. The problem then is this: given the motions of the machine elements, what forces are required to produce these motions? The problem requires (1) a kinematic analysis to determine the translational and rotational accelerations of the various members and () definitions of the actual shapes, dimensions, and material specifications to determine the centroids and mass moments of inertia of the members. In the examples presented here, only the results of the kinematic analysis are included because methods of finding these have been presented in previous Chapter. The selection of the materials, shapes and many of the dimensions of machine members form the subject of machine design and are not further discussed here. In the dynamic analysis of machines, the acceleration vectors are usually known; therefore, an alternative form of Equations. Are often convenient in determining the forces required to produce these known accelerations. Thus, we can write F + ( ma G ) = 0 And M G + ( I G α) = 0 Both of these are vector equations applying to the planar motion of a rigid body. Equation states that vector sum of all external forces acting upon the body plus the fictitious force -mag sum to Prepared By: Subhesh Pansuria Darshan Institute of Engineering & Technology, Rajkot Page 1.9

128 5. Introduction to Dynamics Theory of Machines (15190) zero. This new fictitious force -mag is called an inertia force. It has the same line of action as the absolute acceleration AG, but is opposite in sense. Equation states that the sum of all external moments and the moments of all external forces acting upon the body about an axis through G and perpendicular to the plane of motion plus the fictitious torque -IGα sum to zero. This new fictitious torque -IGα is called an inertia torque. The inertia torque is opposite in sense to the angular acceleration vector α. We recall that Newton s first law states that a body perseveres in its state of uniform motion except when compelled to change by impressed; in other words, bodies resist any change in motion. In a sense, we can picture the fictitious inertia force and inertia torque vectors as resistances of the body to the change of motion required by the net unbalanced forces. (a) Unbalanced forces and resulting accelerations. (b) Inertia force and inertia couple. (c) Inertia force offset from center of mass. Equations are known as d Alembert s principle, because d Alembert was the first to call attention to the fact that addition of the inertia force and inertia torque to the real system of forces and torques enables a solution from the equations of static equilibrium. We note that the equations can also be written F = 0 and M = 0 Where it is understood that both the external and the inertia forces and torques are to be included in the summations. Equations are useful because they permit us to take the summation of moments about any axis perpendicular to the plane of motion. D Alembert s principle is summarized as follows: The vector sum of all external forces and inertia forces acting upon a system of rigid bodies is zero. The vector sum of all external moments and inertia torques acting upon a system of rigid bodies is also separately zero. When a graphical solution by a force polygon is desired, Eqs. can be combined. In Fig. (a), link 3 is acted upon by the external forces F3 and F43. The resultant F3 + F43 produces an acceleration Prepared By: Subhesh Pansuria Darshan Institute of Engineering & Technology, Rajkot Page 5.10

129 5. Introduction to Dynamics Theory of Machines (15190) of the center of mass AG and an angular acceleration of link α3 because the line of action of the resultant does not pass through the center of mass. Representing the inertia torque -IGα3 as a couple, as illustrated in Fig. (b), we intentionally choose the two forces of this couple to +be m3ag. For the moment of the couple to be of magnitude -IGα, the distance between the forces of the couple must be h = I Gα 3 m 3 A G Because of this particular choice for the couple, one force of the couple exactly balances the inertia forces itself and leaves only a single forces, as illustrated in Fig. (c). This force includes the combined effects of the inertia force and the inertia torque, yet appears as only a single inertia force offset by the distance h to give the effect of the inertia torque. 5.9 The Principle of Superposition Linear systems are those in which effect is proportional to cause. This means that the response or output of a linear system is directly proportional to the drive or input to the system. An example of a linear system is a spring, where the deflection (output) is directly proportional to the force (input) exerted on the spring. The principle of superposition may be used to solve problems involving linear systems by considering each of the inputs to the systems separately. If the system is linear, the responses to each of these inputs can be summed or superposed on each other to determine the total response of the system. Thus, the principle of superposition states that for a linear system the individual responses to several disturbances, or driving functions, can be superposed on each other to obtain the total response of the system. The principle of superposition does not apply to non-linear systems. Some examples of non-linear systems, where superposition may not be used, are systems with static or Coulomb friction, systems with clearances or backlash, or systems with springs that change stiffness as they are deflected. We have now reviewed all of the principles necessary for making a complete dynamic-force analysis of a planar motion mechanism. The steps in using the principle of superposition for Prepared By: Subhesh Pansuria Darshan Institute of Engineering & Technology, Rajkot Page 1.11

130 5. Introduction to Dynamics Theory of Machines (15190) making such an analysis are summarized as follows: 1. Perform a kinematic analysis of the mechanism. Locate the centre of mass of each link and find its acceleration; also find the angular acceleration of each link.. If the inertia forces are attached to all links simultaneously, along with other applied forces and moments, then there are often no two-or-three-force members and it may become difficult to find the lines of action of unknown constraint forces. Instead of doing this, it is sometimes more convenient to ignore the masses and applied forces and moments on all but one or two links and to leave other links as two-or-three-force members. By choosing in this manner, a solution may become possible for the constraint forces caused by the masses or applied forces and moments being considered, but without those caused by the masses and applied forces and moments being ignored. 3. Those masses and applied forces and moments considered in Step can now be ignored while a solution is obtained for additional constraint force components caused by some of the previously ignored masses or applied forces and moments. This process can be continued until constraint force components caused by all masses and all applied forces and moments are found. 4. The results of Steps and 3 can now be vectorially added to obtain the resultant forces and torques on each link caused by the combined effects of all masses and all applied forces and moments Measuring Mass Moment of Inertia (a) Simple pendulum. (b) Torsional pendulum. Sometimes the shapes of machine parts are so complicated that it is extremely tedious and time Prepared By: Subhesh Pansuria Darshan Institute of Engineering & Technology, Rajkot Page 5.1