3- Chapter Mechanical.pdf

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THEORY OF MACHINES It is the branch of Engineering Science, which deals with the study of relative motion between the various parts of machine along with the forces acting on the parts is known as the Theory of Machines (TOM). Kinematic Link: Each resistant body in a machine which moves relative to another resistant body is called kinematic link or element. A resistant body is which donot go under deformation while transmitting the force. Kinematic Pair: If the relative motion between the two elements of a machine in contact with each other is completely or successfully constrained then these elements together is known as kinematic pair. CONSTRAINED MOTIONS Constrained motion (or relative motion) can be broadly classified is to three types. 1. Completely Constrained: Constrained motion in which relative motion between the links of a kinematic pair occurs in a definite direction by itself irrespective of the external forces applied. For example a square bar in a square hole undergoes completely constrained motion. 2. Incompletely Constrained: Constrained motion in which the relative motion between the links depend on the direction of external forces acting on them. These motions between a pair can take place in more than one direction. For example a shaft inside a circular hole. 3. Partially (or Successfully) Constrained Motion: If the relative motion between its links occurs in a definite direction, not by itself, but by some other means, then kinematic pair is said to be partially or successfully constrained. For example a piston reciprocating inside a cylinder in an internal combustion engine.

2.

3.

Based on the type of mechanical constraint (or mechanical contact) (a) Self Closed Pair: If the links in the pair have direct mechanical contact, even without the application of external force. (b) Force Closed Pair: If the links in the pair are kept in contact by the application of external forces. Based on the type of relative motion between the elements of the pair (a) Sliding Pair: A kinematic pair in which each element has sliding contact with respect to the other element. (b) Rolling Pair: In a rolling pair, one element undergoes rolling motion with respect to the other. (c) Turning Pair: In a turning pair, one link undergoes turning motion relative to the other link. (d) Screw Pair: It consists of links that have both turning and sliding motion relative to each other. (e) Cylindrical Pair: A kinematic pair in which the links undergo both rotational and translational motion relative to one another. (f) Spherical Pair: In a spherical pair, a spherical link turns inside a fixed link. It has three degrees of freedom.

DEFINITION OF KINEMATIC CHAIN Combination of kinematic pairs joined in such a way that the last link is joined to the first link and the relative motion between them is definite. There are two equations to find out. Whether the chain is kinematic or not. l = 2p – 4 where l = number of links p = number of pairs also

TYPES OF KINEMATIC PAIRS The classified of kinematic pairs is listed as below: 1. Based on the nature of contact between the pairing elements. (a) Lower Pair: Links in the pair have surface or area contact between them. The surface of one element slides over the surface of the other. For example: a piston along with cylinder. (b) Higher Pair: In which the links have point or line contact and motions are partly turing and partly sliding. For example: ball bearings, can and follower.

j=

3 l 2

2

where j = number of joints To determine the nature of chain we use equation h 3 = l–2 2 2 h = No. of higher pairs L.H.S > R.H.S. then it is a locked chain j+

where If

44 L.H.S. = R.H.S. then it is a kinematic chain L.H.S. < R.H.S. then it is an unconstrained chain GRUBLER’S CRITERION In a mechanism total no. of degrees of freedom is given by F = 3(n – 1) – 2j where n is no. of links and j = no. of joints (simple hinges) most of the mechanism are constrained so F = 1 which produces 1 = 3(n – 1) – 2j 2j – 3n + 4 = 0 this is called Grubler’s criterion. If there are higher pairs also no. of degrees of freedom is given by F = 3(n – 1) – 2j – h where h = no. of higher pairs. Also known as Kutz Bach criterion to determine the number of degree of freedom. This statement says that if the higher pairs are present in the mechanism like as slider crank mechanism or a mechanism in which slipping is possible between the wheel and fixed links. Higher pair: When the two element of a pair have a line or point contact when relative motion takes place and the motion between two elements is partly turning and partly sliding. E.g. Cam and follower, bale and bearing, belt and rope drive etc. Number of degree of freedom (movability): The number of independent parameters that define its configuration. The number of input parameters which must be independently controlled in order to bring the mechanism into useful engineering purpose.

velocity. The plane motion of all the particles of the body may be considered as pure rotation about the point. Such a point is called the instantaneous centre of the body. If there are three rigid bodies in relative planar motion and share three instantaneous centre, all lie on the straight line, called Kennedy’s theorem. Instantaneous axis of rotation: The axis passing through the instaneous centre of the body at right angles to the plane of motion is called instantaneous axis of rotation. Axode: The instantaneous centre changes every moment, its locus is called centrods, and the surface generated by the instantaneous axis is called the axode.

Methods to Locate Instantaneous Centre Locating the instantaneous centre of a body depends on the situation given. Following are some examples: (i) The instantaneous centre I lies at a distance

Va

along the

perpendicular to the direction of velocity Va at point A on a rigid body shown in Fig. IA =

Va

I B

I

GRASHOF’S CRITERIA

Vb

Grashof’s criteria is applied to pinned four bar linkages and states that the sum of the shortest and longest link of a planar four-bar linkage cannot be greater than the sum of remaining two links if there is to be continuous relative motion between the links.

B Va

A

A

Va

A

l p

q s

Fig. Linkage shown in Fig. 1 is Grashof type if s+l
Inversion of a Mechanism We can obtain different mechanisms by fixing different links in a kinematic chain, this method is known as inversion of a mechanism. INSTANTANEOUS CENTRE A point located in the plane (of motion of a body) which has zero

Fig. Fig. (ii) The instantaneous centre I is the point of intersection of the lines perpendicular to the direction of velocities at the given points on the body as shown in Fig., we can write as Va = IA Vb = IB where is the angular velocity with which the body shall appear to rotate about the instantaneous centre I. (iii) If the two links have a pure rolling contact, the instantaneous centre lies on their point of contact. (iv) If the slider moves on a fixed link having straight surface, the instantaneous centre lies at infinity and each point on the slider have the same velocity.

Number of Instantaneous Centres in a Constrained Kinematic Chain If n are the number of links in a constrained kinematic chain, then the number of instantaneous centre (N) is given by N=

n (n 1) 2

VELOCITY AND ACCELERATION OF MECHANISMS To analyse velocity and acceleration of a mechanism we proceed

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45 link by link associated in the mechanism. Let us consider two points P and Q on a rigid link PQ, as shown in Fig. Let point Q of the link moves in clockwise direction relative to point P. In this case the relative velocity of point Q with respect to P would be perpendicular to the line PQ. VQP

P

A

v C 2 v

B

Fig. where v is the velocity of the particle C with respect to coincident point C. ACKERMAN STEERING GEAR MECHANISM

Q

Fig. Now if the point Q moves with respect to P with an angular velocity and angular acceleration , thus velocity has two components, perpendicular to each other. (a) Radial or centripetal component (b) Tangential component These components of velocity can be determined by calculating linear accelerations in radial and tangential directions. Figure shows the link representing both the components of acceleration.

P

r

aQP

Q

All the four wheels must turn about the same instantaneous centre to fulfill the condition for correct steering. Equation for the correct steering is cot – cot = c/b where c = Distance between the pivots of the front axles b = Wheel base and are angle through which the axis of the outer wheel and inner whel turns respectively. For approximately correct steering, value of c/b should be in between 0.4 and 0.5. DAVIS STEERING GEAR MECHANISM According to Davis Steering gear the condition for the correct steering is given by tan = c/2b where c = Distance between the pivots of the front axles b = Wheel base = Angle of inclination of the links to the vertical HOOK’S JOINT To connect two shafts intersecting at a small angle (say ) we use Hook’s joint also known as universal joint. Velocity ratio of the shafts connected by hooks joint is given by

VQP t

aQP

N Driving

Fig. Radial component of the linear acceleration of Q with respect to P is given by r a QP

2

=

PQ =

VQP PQ

2

PQ =

2 VQP

PQ

N Driven

=

1 cos 2 sin2 cos

where NDriving = Speed of the driving shaft in r.p.m. NDriven = Speed of the driven shaft in r.p.m. = Angle through which the arms of the cross turn = Angle of intersection of two shafts Hook’s joint can be represented as shown in Fig.

and the tangential component of the linear acceleration of Q with respect to P is given by a tQP =

=

t a QP

PQ

PQ

which is perpendicular to the link PQ

CORIOLIS COMPONENT OF ACCELERATION If a particle C moves with a velocity v on a link AB rotating with angular velocity , as shown in Fig., then the tangential component of the acceleration of the particle C with respect to the coincident point on the link AB is called coriolis component of acceleration which is given by t a CCC = a CC =2 v

Fig. Condition for the unit velocity ratio is given by tan =

cos

FRACTIONAL TORQUE IN PIVOT AND COLLAR BEARING Pivot and Collar bearings are used to take axial thrust of the rotating shaft. While studying the friction in bearing it is assumed that 1. The pressure over the rubbing surfaces is uniformly distributed through out the bearing surface. 2. The wear is uniform throughout the bearing surface. (i) Frictional torque transmitted in a flat bearing is given by


46 2 WR while considering uniform pressure 3 And in case of uniform wear

T2

T=

1 WR 2 where = Coefficient of friction W = Load transmitted to the bearing R = Radius of the shaft Frictional torque transmitted in a Conical Pivot bearing is given by

Slack Side

T2

T=

(ii)

Driving Pulley

T1

Fig.: Belt drive-open system

T2

2 WR cosec 3 while considering uniform pressure And in case of uniform wear

T=

T=

1 2

WR cosec

where = semi angle of the cone (iii) Frictional torque trnsmitted in a trapezoidal or truncated conical pivot bearing is given by T=

2 3

W

r 13

r 32

r 12

r 22

Driven Pulley

Side T1 Tension

T1

T2

T1 Driver Pulley

Fig. : Belt drive-cross system When a number of pulleys are used to transmit the power from one shaft to another, then a compound drive is used as shown in Fig.

cosec

while considering uniform pressure. And in case of uniform wear 1 W (r1 + r2 ) cosec = WR cosec 2 where r1 and r2 are the external and internal radii of the conical bearing respectively

T=

R=

r1

r2

is the mean radius of the bearing. 2 (iv) Frictional torque transmitted in a flat collar bearing is given by T=

2 3

W

r 13

r 32

r 12

r 22

Fig.: Belt drive-compound system Types of Belts There are three types of belts (a) Flat belts: Cross section of a flat belt is shown in Fig. 14. Flat belts are easier to use and are subjected to minimum bending stress. The load carrying capacity of a flat belt depends on its width.

while considering uniform pressure And in case of uniform wear 1 W (r1 r2 ) 2 The frictional torque transmitted by a disc or plate clutch is same as that of flat collar bearing and by a cone clutch is same as that of truncated conical pivot bearing.

T=

BELT DRIVE The transmission of power from one rotating shaft to another lying at a considerable distance, is achieved using belts and ropes. Two parallel shafts may be connected by open belt or by cross belt. In the open belt system, the rotation of both the pulleys is in the same direction. If a crossed belt system is used, the rotation of pulleys will be in the opposite direction. Fig. 11 and Fig. shows open and crossed system respectively.

Fig. : Flat belt The ratio of driving tensions for flat belt drive is given by T1 =e T2

2.3 log where pulley

T1 T2

=

= coefficient of friction between the belt and the

= angle of contact in radians Material used for flat belt is generally leather of various Buy books : http://www.dishapublication.com/new-release/guide-to-drdo-ceptam-mechanical-engineer-exam-with-practice-set.html

47

(b)

types having ultimate tensile strength between 4.5 to 7 N per cm width. For heavy duty, two or three piles of leather are cemented and pressed one above the other such belts are called double or triple ply belts. V-belts: Fig. shows the cross section of the V-belts. V-belts are available in five sections designed A, B, C, D, and E and there are used in order of increasing loads. Section A is used for light loads only and section E is used for heavy duty machines. The angle of V-belt for all sections is about 40°. In order to increase the power output, several V-belts may be operated side by side. In multiple V-belt drive, all the belts should stretch at the same rate so that the load is equally divided between them. If one of the set of belts break, the entire set should be replaced at the same time. The groove angle in the pulley for running the belt is between 40°to 60°. Due to reduced slipping, V-belts offer a more positive drive. V-belt drives run quietly at high speeds and are capable of absorbing high shock.

Fig. V-belt The ratio of driving tension for the V-belt drive is given by T1 = e( T2

2.3 log

T1 T2

=

cosec )

cosec

where (c)

= Semi-angle of the groove = Angle of contact in radians V-belts are usually made of cotton fabric, cards and rubber. Circular belts: The cross section of a circular belt is shown in Fig. The circular belts are also known as round belts. These are employed when low power is to be transmitted, for example in house hold appliances, table top tools and machinery of the clothing. Round belts are made of leather, canvas and rubber.

r2 )2 + 2x x where r1 and r2 are radii of the two pulleys x = distance between the centres of two pulleys In a crossed belt drive the length of the belt is given by L =

(r1

r2 ) 2 + 2x x When the belt passes from the slack side to the tight side a certain portion of the belt extends and when the belt passes from the tight to slack side the belt contracts. Due to these changes in length, there is relative motion (called creep) between the belt and pulley surfaces. Creep reduces the velocity of the belt drive system like slip do. Centrifugal Tension The centrifugal tension (Tc) is given by Tc = mV2 where m = Mass per unit length of the belt V = Linear velocity of the belt The power transmitted can be calculated as below: The total tension on the tight side = T1 + Tc The total tension on the slack side = T2 + Tc Power Transmitted = [(T1 + Tc) – (T2 + Tc)] V = (T1 – T2) V Which is equal to the value of power transmitted given by effective turning force (T1 – T2), that is the centrifugal tension has no effect on the power transmitted. The maximum power transmitted by the belt is given by the maximum total tension in the tight side of the belt when it is three times the centrifugal tension. T = 3Tc T = 3 mV2 So velocity for the maximum power transmitted is given by L =

(r1 + r2) +

(r1

T 3m

v=

Velocity Ratio The velocity ratio of speeds of driver and driven pulleys is given by 2 1

where

Fig. Circular Belt The ratio of driving tensions in round belts and rope drive is same as V-belt drive. Length of Belt In an open belt drive system the length of the belt is given by

(r1 + r2) +

=

N2 d1 = N1 d 2

t S 1 t 100

d1, d2 = diameters of driver and driven pulleys 1, 2 = angular velocities of driver and driven pulleys N1, N2 = rotational speeds of driver and driven pulleys expressed in revoluations per minute (r.p.m.) S = S1 + S2 + 0.01S1S2 is percentage of total effective slip S1 = Percentage slip between driver and the belt S2 = Percentage slip between belt and the follower (driven pulleys)

GEARS AND GEAR DRIVE A wheel with teeth on its periphery is known as gear. The gears are used to transmit power from one shaft to another when the shafts are at a small distance apart.


48

Types of Gears Commonly used gear are as below: (a) Spur gear: A cylindrical gear whose tooth traces are straight lines parallel to the gear axis. These are used for transmitting motion between two shafts whose axis are parallel and coplanar. (b) Helical gear: A cylindrical gear whose tooth traces are straight helices, teeth are inclined at an angle to the gear axis. Double helical gears called herringbone gears. The helical gears are used in automobile gear boxes and in steam and gas turbines for speed reduction. The herringbone gears are used in machinery where large power is transmitted at low speeds. (c) Bevel gear: The bevel gear wheels conform to the frusta of cones having a common vertex,tooth traces are straight line generators of the cone. Bevel gears are used to connect two shafts whose axis are coplanar but intersecting when the shafts are at right angles and the wheels equal in size, the bevel gears are called mitre gears. When the bevel gears have their teeth inclined to the face of the bevel, they are known as helical bevel gears.

Pinion

Rack

(g)

Fig. : Rack and pinion Internal and external gearing: Two gears on parallel shaft may gear either externally or internally as shown in Fig. Annular wheel

Spur gear

Pinion Rack Pinion

Pinion (a)

(b)

(c)

Fig.

(d)

(e)

Fig. : Bevel gear Spiral gear: These are identical to helical gears with the difference that these gears have a point contact rather than a line contact.These gears are used to connect intersecting and coplanar shafts. Worm gear: The system consists of a worm basically part of a screw. The warm meshes with the teeth on a gear wheel called worm wheel. It is used for connecting two non-parallel, non-intersecting shafts which are usually at right angles.

Gear Terminology Terms associated with profile of a gear tooth are illustrated in Fig. Pitch circle: Essentially an imaginary circle which by pure rolling action gives the same motion as the actual gear. Pressure angle or angle of obliquity: Angle between the common normal to two gear teeth at the point of contact and the common tangent at the pitch point (common point of contact between two pitch circles). It is usually denoted by . The standard pressure angles are 14

1 and 20°. 2

Addendum: Radial distance of a tooth from the pitch circle to the top of the tooth.

Worm wheel

Worm

(f)

Fig. : Worm gear Rack and pinion: Rack is a straight line spur gear of infinite diameter. It meshes,both internally and externally, with a circular wheel called pinion. Rack and pinion is used to convert linear motion into rotary motion and vice versa.

Fig. Dedendum: Radial distance of a tooth from the pitch circle to the bottom of the tooth. Addendum circle: Circle drawn through the top of the teeth and is concentric with the pitch circle. Dedendum circle: Circle drawn through the bottom of the teeth. It is also called root circle.


49 Root circle diameter = Pitch circle diameter cos where is the pressure angle. Circular pitch: Distance measured on the circumference of the pitch circle from a point of one tooth to the corresponding point on the next tooth. It is denoted by Pc, mathematically Pc can be calculated as D T where D = Pitch circle diameter T = Number of teeth on the wheel. For two gears to mesh correctly their circular pitch should be same

Pc =

Pc =

D1 D2 = T1 T2

D1 D = 2 T1 T2

Velocity ratio of two meshing gears is given by V1 = D1 N1 V2 = D2 N2 Linear speed of the two meshing gears is equal So

D1 N1 =

D2 N2

N2 D = 1 N1 D2 N2 T = 1 N1 T2

Diametral pitch: It represents the number of teeth on a wheel per unit of its diameter. Diameter pitch Pd =

T = D Pc

Pc Pd = Module: It represents the ratio of pitch circle diameter (in mm) to the number of teeth. D 1 m= = T Pd

m Pd = 1 Recommended series of modules in Indian Standards are 1, 1.25, 1.5, 2, 2.5, 3, 4, 5, 6, 8, 10, 12, 16, and 20. Modules of second choice are 1.125, 1.375, 1.75, 2.25, 2.75, 3.5, 4.5, 5, 5.5, 7, 9, 11, 14 and 18. Total depth: Radial distance between the addendum and the dedendum circles of gear. Tooth depth = Addendum + dedendum Clearance: Radial distance from the top of the tooth to the bottom of the tooth in a meshing gear. Circle passing through the top of the meshing gear is known as clearance circle. Standard value of clearance is 0.157 m, where m is module. Dedendum = Addendum + 0.157 m = m + 0.157 m = 1.157 m Working depth: Radial distance from the addendum circle to the clearance circle.

Working depth = Addendum of first gear + Addendum of second gear Back lash: Difference between the tooth space and tooth thickness, measured along the pitch circle. In actual practice somebacklash must be allowed to prevent jamming of the teeth due to tooth errors and thermal expansion. Path of contact: Path traced by the point of contact of two teeth from the begining to the end of engagement. Length of the path of contact: Length of the common normal cutoff by the addendum circles of the wheel and pinion. Arc of contact: The path traced by a point on the pitch circle from the beginning to the end of engagement of a given pair of teeth. It consists of (a) Arc of approach (b) Arc of recess Arc of approach: Portion of the path of contact from the beginning of the engagement to the pitch point. Arc of recess: Portion of the path of contact from the pitch point to the end of the engagement of a pair of teeth. Contact Ratio =

Length of arc of contact Circular Pitch

Contact ratio is the number pairs of teeth in contact. Length of Arc of contact: Length of the arc of contact can be calculated as Length of Arc of contact =

Length of path of contact cos

where is pressure angle Interference: The phenomenon, when the tip of a tooth under cuts the root on its mating gear. It may only be avoided, if the addendum circles of the two mating gears cut the common tangent to the base circles between the points of tangency. Law of gearing: According to the law of gearing, the common normal at the point of contact between a pair of teeth must always pass through the pitch point.

Gear Trains Any combination of gear wheels by means of which power and motion is transmitted from one shaft to another is known as gear train. Various types of gear train are 1. Simple gear train: A gear train in which each shaft carries one wheel only. Fig. shows the arrangement of a simple gear train. Driver

1

Idle gears 2

4

3

Driven or follower

Fig. : Simple gear train


50 Velocity ratio =

one or the other gears in the train. Fig. shows an arrangement of an epicyclic gear train.

Speed of the driving wheel Speed of the driven wheel

Q Wheel B

no. of teeth on the driven wheel = no. of teeth on the driving wheel N1 N = 1 N4 N2

Train value = 2.

N2 N3

N3 T2 = N 4 T1

T3 T2

T4 T4 = T3 T1

6

Driven or follower

4 L

M

1

2 P 3

4 5

Q 6

Fig. : Compound gear train N1 N1 N3 N5 T2 T4 T6 = = N6 N 2 N 4 N6 T1 T3 T5 Reverted gear train: A reverted gear train manifests when the first driving gear and the last driven gear are on the same axis. Axes are coincidental and coaxial. Fig. shows an arrangement of the reverted gear train. If D1, D2, D3, D4 be the pitch circle diameters of the respective gears and corresponding speeds are N1, N2,N3, N4 then D1 D2 D3 D 4 = 2 2 D1 + D2 = D3 + D4

Velocity ratio =

3.

Velocity ratio =

N1 T2 T4 = N 4 T1 T3 1 4

Wheel A

NB NC

Velocity Ratio

TA TB

1

FLY WHEEL

Compound gears 5

3

1

P

Fig. : Epicyclic gear train

N 4 T1 = N1 T4

Compound gear train: A compound gear train includes two gears mounted on the same shaft as shown in Fig. Driver

Arm C

A wheel used in machines to control the speed variations caused by the fluctuation of the engine turning moment during each cycle of operation. These wheels are known as flywheel. It absorbs energy when crank turning moment is greater than resisting moment and gives energy when turning moment is less than resisting moment. The speed of a flywheel increases during it absorbs energy and decreases when it gives up energy. This way flywheel supplies energy from the power source to the machine at a constant rate throughout the operation. Coefficient of fluctution of energy: Ratio of the maximum fluctuation of energy to the work done per cycle, is called coefficient of fluctuation of energy. Emax = Emax – Emin Cenergy =

E max Wper cycle

where Emax = maximum fluctuation of energy Cenergy = coefficient of fluctuation of energy Coefficient of fluctuation of speed: Ratio of the maximum fluctuation of speed to the mean speed is called the coefficient of fluctuation of speed. max = max – min Cs =

Compound gear 3 2

max mean

Relation between maximum fluctution of energy coefficient of fluctuation of speed. Emax = I mean ( max – min) =I

( mean

min )

max

E max and

mean

mean

Co-axial shafts

= I

2 mean

max mean

Emax = I

2 mean

Cs

…(1)

also energy stored is a flywheel is given by Fig. : Reverted gear train 1 In a clock mechanism a reverted gear train is used to connect E = I 2mean 2 hour hand to minute hand in a clock mechanism. 4. Epicyclic gear train: A special type of gear train in which Buy books http://www.dishapublication.com/new-release/guide-to-drdo-ceptam-mechanical-engineer-exam-with-practice-set.html axis of: rotation of one or more of the wheels is carried on an arm and this arm is free to rotate about the axis of rotation of

51 2 mean

=

2E I

…(2)

from (1) and (2) Emax = 2ECs where Cs = coefficient of fluctuation of speed.

increases and decreases as the sleeve moves upwards or downwards respectively. If hp is the height of porter governor (when length of arms and links are equal). and hw is height of watt’s governor then hp

GOVERNORS The function of a governor is to regulate the mean speed of an engine within mentioned speed limits for varying type of load condition. Terms Used in Governors (a) Height of Governor: Vertical distance from the centre of the ball to a point where arms intersect on the spindle axis. (b) Equilibrium Speed: The speed at which the governor balls, arms etc. are in complete equilibrium and the sleeve does not tend to move upwards or downwards. (c) Sleeve Lift: Vertical distance with the sleeve travels because of change in equilibrium speed. (d) Mean Equilibrium Speed: The speed at the mean position of the balls or sleeve. (e) Maximum and Minimum Equilibrium Speeds: The speeds at the maximum and minimum radius of rotation of the balls, without tending to move either way are known as maximum and minimum equilibrium speeds respectively. If N1 and N2 are maximum and minimum speeds then Sensitiveness =

2 (N1 – N 2 ) (N1 N 2 )

(f)

Sensitiveness: A governor is said to be sensitive, if its change of speed is from no load to full load may be small a fraction of the mean equilibrium speed as possible and the corresponding sleeve lift may be as large as possible. (g) Stability: If for every speed within the working range there is a configuration of governor balls, then it is said that governor is stable. For a stable governor, the radius of governor balls must increase with increase in the equilibrium speed. (h) Hunting: Fluctuation in the speed engine continuously above and below the mean speed is called hunting. (i) Isochronism: A governor is isochronous provided the equilibrium speed is constant for all radii of rotation of the balls upto the working range. (j) Governor Effort: The average force required on the sleeve to make it rise or come down for a given change in speed. (k) Power of Governor: The work done at sleeve for a given percentage change in speed. Mathematically Power = Mean effort Lift of sleeve Types of Governors Different types of Governors are: (1) Simple governor-Watt type: The simplest type a centrifugal governor is known as watt type or watt governor. Height of the governor is given by h=

895

metres N2 where N = speed of the arm and ball about the spindle axis. (2) Porter governor: It is obtained by modifying a Watt governor with a central load attached to the sleeve. The governor speed

hw

=

m+M m

where m = mass of the ball M = mass of the sleeve (3) Hartnell governor: This is a spring controlled governor. If lsleeve is the lift of the sleeve and Xcompression is the compression of the spring then y x where r1 = Minimum radius of rotation r2 = Maximum radius of rotation x = Length of ball arm of lever y = Length of sleeve arm of lever Stiffness of the spring is given by

lsleeve = Xcompression = (r 2 – r 1)

S=

S2

S1

h where S1 = Spring force at minimum radius of rotation S2 = Spring force at maximum radius of rotation

BRAKES A device which applies artificial frictional resistance to a moving machine member, inorder to retard or stop the motion of machine, is known as brake.

Types of Brakes There are four types of brakes commonly used listed below: 1. Shoe brake 2. Band and block brake 3. Internal expanding brake Dynamometers It is a brake having a device to measure the frictional resistance. We may obtain the torque transmitted and have power of the engine by knowing the frictional resistance. There are two types of dynamometers, used for measuring the brake power of the engine. 1. Absorption dynamometers: A dynamometer in which the entire energy or power produced by the engine is absorbed by the friction resistances of the brakes and is transformed into heat, during process of measurement. There are two types of absorption dynamometers. (i) Prony brake dynamometer (ii) Rope brake dynamometer. 2. Transmission dynamometers: A dynamometer in which the energy is not wasted in friction but is used for doing work. The energy or power produced by the engine is transmitted through the dynamometer to some other machines where the power developed is suitably measured. There are three types of transmission dynamometer (i) Epicyclic-train dynamometer (ii) Belt transmission dynamometer (iii) Torsion dynamometer


52 CAMS

9.

A rotating machine element which gives reciprocating or oscillating motion to another element called follower is known as cam. These are mainly used for inlet and exhaust values of I.C. engines, lathes etc. Types of Cams 1. Radial cam: A cam in which follower reciprocates or oscillates in a direction perpendicular to the axis of the cam. Radial cam is further classified as (a) Reciprocating cam (b) Tangent cam (c) Circular cam 2. Cylindrical Cam: A cam in which the follower reciprocates or oscillates in a direction parallel to the cam axis. CAMS TERMINOLOGY A radial cam with reciprocating roller follower is shown in Fig. 1. Base Circle: Smallest circle that can be drawn to the cam profile. 2. Trace Point: The reference point on the follower which is used to generate the pitch curve that varies from case to case. For example, in case of knife edge follower, the knife edge represents the trace point and the pitch curve corresponds to the cam profile while in case of roller follower, the centre of the roller represents the trace point. 3. Pressure Angle: The angle between the direction of the follower motion and a normal to the pitch curve. Keeping the pressure angle too large will lead to joining of reciprocating follower. 4. Pitch Point: A point on the pitch pitch curve having the maximum pressure angle. Reciprocating roller follower

Trace point Maximum pressure angle Pitch point

Total follower travel Pitch circle Cam shaft Base circle Prime circle Pitch point Cam profile

5. 6. 7.

8.

Fig. Pitch Circle: A circle drawn from the centre of the cam through the pitch points. Pitch Curve: The curve generated by the trace point as the follower moves relative to the cam. Prime Circle: Smallest circle that can be drawn from the centre of the cam and tangent to the pitch curve. For a roller follower, the prime circle is larger than the base circle by the radius of the roller while in case of knife edge and a flat face follower it is equal. Lift or Stroke: The maximum travel of follower from its lowest position to the topmost position is called life or stroke.

10. 11. 12.

Angle of Ascent: It is the angle moved by cam from the time the follower begins to rise till it reaches the highest point. Angle of Descent: Angle during which follower returns to its initial position. Angle of Action: It is the total angle moved by cam from the beginning of ascent to finish of descent. Under Cutting: The situation of a Cam Profile which has an inadequate curvature to provide correct follower movement, is known as under cutting.

VELOCITY AND ACCELERATION OF THE FOLLOWER (a)

(b)

Tangent Cam with Reciprocating Roller Follower In the tangent cam flanks of the cam are straight and tangential to the base circle and nose circle. Tangent cams are used for operating the inlet and exhaust values of I.C. engines. Displacement of the follower is given by yf = (r1 + r2) (1 – cos ) sec Velocity of the follower is given by vf = (r1 + r2) sin sec2 and acceleration of the follower is given by af = 2 (r1 + r2) (2 – cos2 ) sec2 where r1 = Minimum value of the radius of the cam r2 = Radius of the roller follower = Angle turned by the cam, from the beginning of the follower displacement = Angular velocity of the cam Circular Arc Cam with Flat-faced Follower In the circular arc cam the flanks of the cam connecting the base circle and nose are of convex circular arcs. Displacement of the flat faced follower is given by yf = (R – r 1) (1 – cos ) Velocity of the follower is given by vf = (R – r1) sin and acceleration of the follower is given by af = 2 (R – r1) cos where R = Radius of circular flank r1 = Minimum radius of the cam = Angle turned through by the cam = Angular velocity of the cam

BALANCING The main aim of balancing is to remove the effects of the inertial force due to acceleration of various parts of the engine. (a) Balancing of rotating masses The process of providing the second mass in order to counter act the effect of the centrifugal force of the first mass is called balancing of rotating masses. m1 2 r1=m2 2 r2 where = angular speed of the shaft m1= mass attached to rotating shaft m2 =balancing mass In order to put the system in complete balance it has to have (i) Static balancing (ii) Dynamic balancing (i) Static balancing: In this balancing the net dynamic force acting on the shaft is equal to zero. (ii) Dynamic balancing: In this balancing the net couple due to dynamic forces acting on the shaft is equal to zero along with the static balancing. If several masses revolve in different planes then in order to


53

(b)

have a complete balance FResultant = 0 and CoupleResultant = 0 Balancing of reciprocating masses To eliminate the shaking force and a shaking couple is called balancing of reciprocating masses. It is usually not practical to eliminate shaking force and shaking couple completely, only we can do if we can reduce them. The reciprocating masses are only partially balanced. Inertia force due to reciprocating parts is given by FI

= FR = m

2

r cos

cos 2 n

where = angle made by the crank. The horizontal component of the force exerted on the crank shaft bearing = Fu (unbalanced force)

F u = FI In magnitude Fu = FI Fu = m

2

r cos

cos 2 n

FTmax =

2 (1 – C) m

2

r

FTmin = 2 (1 – C) m 2 r Swaying Couple: This is the effect caused by the couple that is produced by the unbalanced forces. This effect tends to sway the engine alternately clock-wise and anti-clockwise. Swaying couple is given by a (cos + sin ) 2 where a = distance between the central line of two cylinders Swaying couple is maximum or minimum when = 45° or 225° Value of minimum and maximum swaying couple

Swaying couple = (1 – C) m

2

r

a (1 – C) m 2 r 2 Hammer Blow: It is the maximum unbalanced force in perpendicular direction to the line of stroke. If P is the downward pressure on rails and B is the balancing mass at radius b then the value of downward pressure so that wheels do not lift from the rails is given by P=B 2 b =

Fu = FP + FS whereFP (Primary unbalanced force) = m

2

r cos

cos 2 n Tractive Force: Unbalanced forces produced due to two cylinders along the line of the stroke. The tractive force (FT) in a locomotive with two cylinders is given by

FS (Secondary unbalanced force) = m

FT = (1 – C) m 2 r (cos – sin ) where m = mass of the reciprocating parts = angular velocity of crank r = radius of crank C = fraction of balanced reciprocating mass FTmax or FTmin occurs when = 135° or 315°

2

r

EXERCISE 1.

2.

3.

4.

A rotating disc of 1 m diameter has two eccentric masses of 0.5 kg each at radii of 50 mm and 60 mm at angular positions of 0° and 150°, respectively. A balancing mass of 0.1 kg is to be used to balance the rotor. What is the radial position of the balancing mass? (a) 50mm (b) 120 mm (c) 150 mm (d) 280 mm The number of degrees of freedom of a planar linkage with 8 links and 9 simple revolute joints is (a) 1 (b) 2 (c) 3 (d) 4 Match the items in Column I and Column II. Column I Column II P. Higher kinematic pair 1. crubler’s equation Q. Lower kinematic pair 2. Line contact R. Quick return mechanism 3. Euler’s equation S. Mobility of a linkage 4. Planar 5. Shaper 6. Surface contact (a) P-2, Q-6, R-4, S-3 (b) P-6, Q-2, R-4, S-1 (c) P-6, Q-2, R-5, S-3 (d) P-2, Q-6, R-5, S-1 Match the items in Column I and Column II. Column I Column II P. Addendum 1. Cam Q. Instantaneous centre 2. Beam of velocity R. Section modulus 3. Linkage

5. 6.

7.

8.

9.

S. Prime circle 4. Gear (a) P-4, Q-2, R-3, S-1 (b) P-4, Q-3, R-2, S-1 (c) P-3, Q-2, R-1, S-4 (d) P-3, Q-4, R-1, S-2 The number of inversions for a slider crank mechanism is (a) 6 (b) 5 (c) 4 (d) 3 For a four-bar linkage in toggle-position, the value of mechanical advantage is (a) zero (b) 0.5 (c) 1.0 (d) infinite The speed of an engine varies from 210 rad/s to 190 rad/s. During a cycle, the change in kinetic energy is found to be 400 N-m. The inertia of the flywheel in kg-m2 is (a) 0.10 (b) 0.20 (c) 0.30 (d) 0.40 The rotor shaft of a large electric motor supported between short bearings at both deflection of 1.8 mm in the middle of the rotor. Assuming the rotor to be perfectly balanced and supported at knife edges at both the ends, the likely critical speed (in rpm) of the shaft is (a) 350 (b) 705 (c) 2810 (d) 4430 Which of the following statements is incorrect? (a) Gashoff’s rule states that for a planar crank-rocker four bar mechanism, the sum of the shortest and longest link lengths cannot be less than the sum of remaining two link lengths


54 (b) Inversions of a mechanism are created by fixing different links one at a time (c) Geneva mechanism is an intermittent motion device (d) Grubler’s criterion assumes mobility of a planar mechanism to be one 10. Mobility of a statically indeterminate structure is (a) –1 (b) zero (c) 1 (d) 2 11. A double-parallelogram mechanism is shown in the figure. Note that PQ is a single link. The mobility of the mechanism is P Q

18.

19.

12.

13.

14.

15.

16.

17.

(a) –1 (b) zero (c) 1 (d) 2 A circular object of radius r rolls without slipping on a horizontal level floor with the centre having velocity V. The velocity at the point of contact between the object and the floor is (a) zero (b) V in the direction of motion (c) V opposite to the direction of motion (d) V vertically upward from the floor For the given statements: I. Mating spur gear teeth is an example of higher pair. II. A revolute joint is an example of lower pair. Indicate the correct answer. (a) Both I and II are false (b) I is true and II is false (c) I is false and II is true (d) Both I and II are true In a mechanism, the fixed instantaneous centres are those centres which (a) Remain in the same place for all configuration of mechanism (b) Large with configuration of mechanism (c) Moves as the mechanism moves, but joints are of permanent nature (d) None of the above Maximum fluctuation of energy is the (a) Ratio of maximum and minimum energies (b) sum of maximum and minimum energies (c) Difference of maximum and minimum energies (d) Difference of maximum and minimum energies from mean value In full depth 1/4 degree involute system, the smallest number of teeth in a pinion which meshes with rack without interference is (a) 12 (b) 16 (c) 25 (d) 32 The two-link system, shown in the figure, is constrained to move with planer motion. It possesses

20.

21.

22.

23.

24.

25.

26.

(a) 2 degrees of freedom (b) 3 degrees of freedom (c) 4 degrees of freedom (d) 6 degrees of freedom If the ratio of the length of connecting rod to the crank radius increases, then (a) primary unbalanced forces will increase (b) primary unbalanced forces will decrease (c) secondary unbalanced forces will increase (d) secondary unbalanced forces will decrease In a cam mechanism with reciprocating roller follower, the follower has a constant acceleration in the case of (a) cycloidal motion (b) simple harmonic motion (c) parabolic motion (d) 3 - 4 - 5 polynomial motion A flywheel fitted in a steam engine has a mass of 800 kg. Its radius of gyration is 360 mm. The starting torque of engine is 580 N-m. Find the kinetic energy of flywheel after 12 seconds? (a) 233.3 kJ (b) 349.8 kJ (c) 487.5 kJ (d) None of these In a slider-crank mechanism, the maximum acceleration of slider is obtained when the crank is (a) at the inner dead centre position (b) at the outer dead centre position (c) exactly midway position between the two dead centres (d) none of these If the rotating mass of a rim type flywheel is distributed on another rim type flywheel whose mean radius is half the mean radius of the former, then energy stored in the later at the same speed will be (a) four times the first one (b) same as the first one (c) one fourth of the first one (d) one and a half times the first one What will be the number of pair of teeth in contact if arc of contact is 31.4 mm and module is equal to 5. (a) 3 pairs (b) 4 pairs (c) 2 pairs (d) 5 pairs The distance between the parallel shaft is 18 mm and they are conntected by an Oldham’s couling. The driving shaft revalues at 160 rpm. What will be the maximum speed of sliding the tongue of the intermediate piece along its grow? (a) 0.302 m/s (b) 0.604 m/s (c) 0.906 m/s (d) None of these Two spur gears have a velocity ratio of 1/3. The driven gear has 72 teeth of 8 mm module and rotates at 300 rpm. The pitch line velocity will be (a) 3.08 m/s (b) 6.12 m/s (c) 9.04 m/s (d) 12.13 m/s Instantaneous centre of a body rolling with sliding on a


55 stationary curved surface lies (a) at the point of contact (b) on the common normal at the point of contact (c) at the centre of curvature of the stationary surface (d) Both (b) and (c) 27. If Cf is the coefficient of speed fluctuation of a flywheel then the ratio of max/ min will be (a)

1 2Cf 1 2Cf

(b)

2 Cf 2 Cf

1 Cf 2 Cf (d) 1 Cf 2 Cf 28. A rotor supported at A and B, carries two masses as shown in the given figure. The rotor is

(c)

(a) dynamically balanced (b) statically balanced (c) statically and dynamically balanced (d) not balanced 29. A body of mass m and radius of gyration k is to be replaced by two masses m1 and m2 located at distances h1 and h2 from the CG of the original body. An equivalent dynamic system will result, if (a) h1 + h2 = k

(b)

h12

h 22

k2

(c) h1h2 = k2 (d) h1 h 2 k 2 30. A cord is wrapped around a cylinder of radius ‘r’ and mass ‘m’ as shown in the given figure. If the cylinder is releasd from rest, velocity of the cylinder, after it has moved through a distance ‘h’ will be

(a)

2 gh

(b)

gh

4 gh gh (d) 3 3 31. There are six gears A, B, C, D, E, F, in a compound train. The number of teeths in the gears are 20, 60, 30, 80, 25 and 75 respectively. The ratio of the angular speeds of the driven (F) to the driver (A) of the drive is

(c)

(a)

1 24

(b)

1 8

2 and 4 have equal lengths. The point P on the coupler 3 will generate a/an (a) ellipse (b) parabola (c) approximately straight line (d) circle 33. A system of masses rotating in different parallel planes is in dynamic balance if the resultant (a) force is equal to zero (b) couple is equal to zero (c) force and the resultant couple are both equal to zero (d) force is numerically equal to the resultant couple, but neither of them need necessarily be zero. 34. A bicycle remains stable in running through a bend because of (a) Gyroscopic action (b) Corioliss’ acceleration (c) Centrifugal action (d) Radius of curved path 35. The maximum fluctuation of energy Ef, during a cycle for a flywheel is (a) l( 2max – 2min) (b) 1/2 l av ( 2max – 2min) 1 lK 2 2 es av (d) lKes 2av (where l = Mass moment of inertia of the flywheel av = Average rotational speed Kes = coefficient of fluctuation of speed) The road roller shown in the given figure is being moved over an obstacle by a pull ‘P’. The value of ‘P’ required will be the minimum when it is

(c)

36.

(a) horizontal (b) vertical (c) at 45° to the horizontal (d) perpendicular to the line CO 37. Two gear 20 and 40 teeth respectively are in mesh. Pressure angle is 20°, module is 12 and line of contact on each side of the pitch point is half the maximum length. What will be the height of addendum for the gear wheel (a) 4 mm (b) 6mm (c) 8 mm (d) 10mm 38. In a slider-bar mechanism, when does the connecting rod have zero angular velocity? (a) When crank angle = 0° (b) When crank angle = 90° (c) When crank angle = 45° (d) Never 39. A disc of mass m is attached to a spring of stiffness k as shown in the figure. The disc rolls without slipping on a horizontal surface. The natural frequency of vibration of the system is

4 (d) 12 15 In the four-bar mechanism shown in the given figure, links

(c) 32.


56 (a)

1 2

k m

(b)

1 2

2k m

1 2k 1 3k (d) 2 3m 2 2m For a four bar linkage in toggle position, the value of mechanical advantage is (a) 0.0 (b) 0.5 (c) 1.0 (d) What will the normal circular pitch and axial pitch of helical gear if circular pitch is 15 mm and helix angle is 30° (a) 13 mm and 39 mm (b) 26 mm and 39 mm (c) 26 mm and 13 mm (d) 13 m and 26 mm The speed of an engine varies from 210 rad/s to rad/s. During cycle the change in kinetic energy is found to be 400 Nm. The inertia of the flywheel in kgm2 is (a) 0.10 (b) 0.20 (c) 0.30 (d) 0.40 If first and last gear having teeth 30 and 50 respectively of a simple gear train, what will be the train value and speed ratio gear respectively if first gear is driving gear (a) 3/5 and 5/3 (b) 3/5 and 4/5 (c) 5/3 and 3/5 (d) 4/5 and 3/5 The centre of gravity of the coupler link in a 4-bar mechanism would experience (a) no acceleration (b) only linear acceleration (c) only angular acceleration (d) both linear and angular accelerations In a four-bar linkage, S denotes the shortest link length, L is the longest link length, P and Q are the lengths of other two links. At least one of the three moving links will rotate by 360° if (a) S + L P + Q (b) S + L > P + Q

(c) 40.

41.

42.

43.

44.

45.

46.

(c) S + P L + Q (d) S + P > L + Q An involute pinion and gear are in mesh. If both have the same size of addendum, then there will be an interference between the (a) tip of the gear teeth and flank of pinion

1 2 3 4 5 6 7 8 9 10

(c) (c) (d) (b) (c) (d) (a) (b) (a) (d)

11 12 13 14 15 16 17 18 19 20

(c) (a) (d) (a) (c) (d) (a) (d) (c) (a)

(b) tip of the pinion and flank of gear (c) flanks of both gear and pinion (d) tip of both gear and pinion. 47. ABCD is a four-bar mechanism in which AB = 30 cm and CD = 45 cm. AB and CD are both perpendicular to fixed link AD, as shown in the figure. If velocity of B at this condition is V, then velocity of C is

(a) V

(b)

3 V 2

2 9 V V (d) 3 4 The transmission angle is maximum when the crank angle with the fixed link is (a) 0° (b) 90° (c) 180° (d) 270° In the given figure, ABCD is a four-bar mechanism. At the instant shown, AB and CD are vertical and BC is horizontal. AB is shorter than CD by 30 cm, AB is rotating at 5 rad/s and CD is rotating at 2 rad/s. The length of AB is

(c)

48.

49.

(a) 10 cm (b) 20 cm (c) 30 cm (d) 50 cm 50. A link OP is 0.5 m long and rotate about point O. It has a slider at permit B. Centripetal acceleration of P relative to O is 8 m/sec2. The sliding velocity of slider relative to P is 2 m/ sec. The magnitude of Coriolis component of acceleration is (a) 16 m/sec2 (b) 8 m/sec2 2 (c) 32 m/sec (d) Data insufficient

ANSWER KEY 21 (a) 22 (c) 23 (c) 24 (a) 25 (c) 26 (d) 27 (d) 28 (c) 29 (c) 30 (a)

31 32 33 34 35 36 37 38 39 40

(a) (a) (c) (c) (d) (c) (c) (b) (c) (d)

41 42 43 44 45 46 47 48 49 50

(d) (a) (a) (d) (a) (a) (b) (c) (b) (a)


57

HINTS AND EXPLANATIONS 1.

(c) Since, all the masses lie in the single plane of the disc. So, we have a force polygon. 7.

For a four bar linkage in toggle position Effort = 0 Mechanical advantage = (a) For flywheel which controls the fluctuations in speed during a cycle at constant output load, E

1 I 2

2 2

2 1

1 .I 210 2 1902 2 I = 0.1 kg-m2. (b) The critical or whirling speed of centrally loaded shaft between two bearings 400

Let

be angular velocity of disc F1 = m1r1 2. = 0.5 × 0.05 × 2 = 0.025 2 N F2 = m2r2 2.= 0.5 × 0.06 × 2 = 0.030 2 N If r is the radial position of balancing mass 0.1 kg, so

F2

R

8.

R 150°

9.

F1

r = mr 2 = 0.1 r 2 N From the above force polygon, F11

R

=

F22

0.025

2

2

0.030

2

0.025 0.030

2.

5.

6.

10.

2F1 F2 cos 150 2 4

2 0.866

= 0.015033 2. Now 0.1r 2 = 0.015033 2. r = 0.150 m r = 150 mm (c) According to Grubler’s criterion, the number of degrees of freedom of a mechanism is given by F = 3(n – 1) – 2j – h = 3(8 – 1) –2 × 9 – 0 = 21 – 18 = 3 (c) For a 4-bar chain/mechanism like slider-crank mechanism, there are as number of inversions as the number of links or bars. These different inversions are obtained by fixing different links one at a item for one inversion. Hence, number of inversions for a slider-crank mechanism will be four. Load to be lifted (d) Mechanical advantage = Effort applied

=

Output force Input force

k m

c

n

c

9.81 0.0018

g

73.82 rad / s

2 Nc 73.82 60 Nc = 704.96 705 rpm (a) According to Grsashoff’s rule for a planar crank-rocker four bar mechanism, the sum of lengths of shortest and longest links should be less than the sum of lengths of other two remaining links. So, statement (a) is incorrect and rest are correct. (d) The mobility or degrees of freedom of a plane structure is the number of inputs (i. e., number of independent coordinates required to determine the configuration or position of all the links of the mechanism w.r.t. fixed link. It is determined by Grubler’s equation as F = 3(n – 1) – 2j – h where F = degrees of freedom or movability of mechanism n = number of links j = number of lower pairs h = numbers of higher pairs Now, a 5-bar chain is the simplest statically indeterminate structure in which link 1 is fixed as shown. Hence to specify the position of all links, two coordinates 1 and 2 are required. So two inputs are required to give a unique output. So, F = 2 or the mobility is 2.

3

4

2

5

1

2

1

Similarly, for 6-bar or more chains, F > 2 Hence, for a statically indeterminate structures, Mobility 2


58 11.

(c)

12. (a)

14.

15.

16.

P

F = 3(n – 1) + 2f1 – f2. = 3(5 – 1) – 2 × 5 – 1 = 12 – 10 – 1 = 1 As we know that, velocity at point of contact between object and floor will be R. While, radius ‘R’ will be equal to zero an instantaneous centre is situated at the intersection point of object (radius ‘r’) and floor.

(a) Type of instantaneous centres: (a) Fixed instantaneous centres (b) Permanent instantaneous centres (c) Neither fixed nor permanent instantaneous centres (a) Fixed instantaneous centre : They remains in the same place for all configuration of the mechanism. (b) Permanent instantaneous centres : They move when the mechanism move, but the joints are of permanent mature. (c) Neither fixed nor permanent instantaneous centre:They vary with the configuration of the Mechanism. (c) The fluctuation of energy may be determined by the turning moment diagram for one complete cycle of operation. The difference between the maximum and minimum energies is known as maximum fluctuation of energy. AE = Maximum energy – Minimum energy (d) The minimum number of teeth on a pinion is found on the basis of consideration of avoiding interference. In case of 14½° involute system, the minimum number of teeth in a pinion which meshes with rack

t min 17.

Q

(a)

2 sin 2

18. (d) FP = Primary unbalanced force = mr 2cos Fs = Secondary unbalanced force mr 2 l cos 2 n n r 19. (c) For uniform acceleration and retardation the velocity of the follower must change at a constant rate and hence the velocity diagram of the follower consists of sloping straight lines. The velocity diagram represents everywhere the slope of the displacement diagram, the latter must be curve whose slope changes at a constant rate. Hence the displacement diagram consists of double parabola. T mk 2

20. (a) 2

Y

higher pair

link 2

5.59 rad/s 2

t = 0. + 5.59 × 12 = 67.08 rad/s

1

1 1 800 (0.36) 2 (67.08) 2 mk2 2 2 2 = 233270 N = 233.3 kJ

KE = 21. (a)

fp = r

cos

2

At IDC

1

cos 2 n

=0

fp = r

1 n

1

2

At ODC fp = – r

32

580 800 (0.36)2

= 180° 2

1

1 n

22. (c) Energy stored in flywheel is dependent on moment of inertia given by : I = (w/g)k2 where k = radius of gyration In case of rim type of flywheel, k = radius of flywheel. k 2 23. (c) Arc of contact = 31.4 mm Module (m) = 5 Circular pitch = m = 5

Since, k =

link 1

X O Kutzbach criterion for movability of a mechanism, Number of degree of freedom = 3 (l – 1) – 2j – h = 3(2 – 1) – 2 × 0 – 1 =3–1=2 Hence, it possesses 2 degree of freedom.

So, No. of pair of teeth in contact =

=

Arc of contact Circular pitch 31.4 = 2 pairs. 5


59 From the equations (ii) and (iii); we get

2 N 2 160 = 16.75 rad/s 60 60 maximum velocity of sliding = × d = 16.75 × 0.018 = 0.302 m/s (c) T2 = 72 1 VR = 3 N2 = 300 rpm

24. (a)

25.

N2 N1

T1 T2

T1 T2

1 T1 3

m1h12

min

)

min

2 max + CS min = 2

max min

max – 2 min

2 CS 2 CS

mk 2 h1 (h1 h 2 )

m 2 h 22

mk 2

...(iii)

mh 2 (h1 h 2 )

k2 = h1h2 30. (a) Since cylinder falls freely under effect of gravity, it follows basic law of motion and 2

= 2gh and =

2gh 31. (a) Ratio of angular speeds of F to A TA TC TE TB TD TF

20 30 25 60 80 75

1 24

32. (a) Point P being rigidly connected to point 3, will trace same path as point 3, i.e. ellipse. 33. (c) A system of masses rotating in different parallel planes is in dynamic balance if the resultant force and the resultant couple are both to zero. This is known as dynamic balancing. 34. (c) A bicycle remains stable in running through a bond because of centrifugal action. 35. (d)

28. (c) Static balance is a balance of forces due to the action of gravity. Consider a rigid rotor with the shaft laid on horizontal parallel ways. if it is in static balance, the shaft will not on the ways whatever may be the angular position of the rotor. For this to happen, the centre of gravity of the system of masses must lie at the axis of rotation of the shaft. For the centre of gravity to be at the axis of the shaft, the horizontal and vertical moments of the rotors must be equal to zero Wr sin = 0, Wr cos = 0 The above equations are also true with the dynamic balance of the inertia forces. Thus if the conditions for the dynamic balance are met, the conditions for static balance are also met. 29. (c) For dynamically equivalent m1 + m2 = m ...(i) m1h1 = m2h2 ...(ii) m1h12

...(iv)

From equations (iii) and (iv)

d2 8 72 = 2 300 2 2 = 542867 mm/min = 9.04 m/s 26. (d) The position of the instantaneous centre changes with the motion of the body. Instantaneous centre of a body rolling with sliding on a stationary curved surface lies (i) on the common normal at the point of contact, and also (ii) at the centre of curvature of the stationary surface 27. (d) We know that coefficient of fluctuation of speed (CS) is

or, CS

m

mh 2 m1 = h h 1 2

= 2 N2

max

m1h1 h2

m1

24

max

...(iii)

From the equations (i) and (ii) we get

1 3

(

mk 2 h1 (h1 h 2 )

m1 =

Pitch line velocity = w1r1 or w2r2

CS

mk 2

(m1h1 ) h 2

1 l( 2

e max 1 l( 2 e max

2 max

max

1

2 av

min

2 min

)(

)

max

min

)

l

av

CS

av

CS

36. (c) From the figure, it shows that the value of ‘P’ required will be minimum when it is at 452 to the horizontal. This can be solved by resolution of forces. 37. (c)

R

mT 2

r=

mt 2

r sin 2

40 12 = 240 mm 2

20 12 2 R 2a

R 2 cos 2

R sin

120 sin 20 R 2a (240 cos 20) 2 2 Ra = 248 mm addendum = 248 – 240 = 8 mm

240sin 20


60 38.

cos

(b)

cr

n

2

43. (a) Train values =

2

sin

cos n Angular velocity is maximum at = 0, 180° Angular velocity is zero at = 90°

If n is large

39.

44. (d) 45. (a)

46. (a)

Taking moments about instantaneous centre ‘A’

(kx) r

47. (b)

0

+

+

mr 2

kr 2 3 2 mr 2

+ k ( r2) = 0 =0

2k =0 3m

41.

1 2k 2 3m (d) In toggle position, for a four bar linkage, the mechanical advantage will be infinity. (d) = 30° Normal circular pitch = circular pitch × cos = 15 × cos 30° = 13 mm

Circular pitch 15 15 = = tan 30 tan 30 1 3 = 26 mm 42. (a) We know that Axial pitch =

E=

1 l 2

400

2 1

2 2

400 l

l = 0.1 kg-m 2

VC VB

VC VB

CD AB

VC

VB

W .CD W . AB

CD AB

V

CD AB

45 30

V

3 2

3 V 2

3 V. 2 48. (c) The transmission angle is maximum when crank angle with fixed link is 180°. The transmission angle is minimum when crank angle with fixed link is 0°. The transmission angle is optimum when crank angle with fixed link is 90°. 49. (b) CD = AB + 30 cm Rotation of AB, 1 = 5 rad/s Rotation of CD, 2 = 2 rad/s So, 1AB = 2CD 5 AB = 2 (AB + 30) AB = 20 cm 50. (a) aC = r 2 8 = 0.5 × 2 2 = 16 = 4 rad/sec Coridis component of acceleration = 2 v = 2 × 4 × 2 = 16 m/s2

Velocity of C =

n=

40.

3 5

1 5 Train value 3 The centre of gravity of the coupler link in a 4 bar mechanism would experience both linear and angular accelerations. According to Grashof’s law for a four bar mechanism. The sum of shortest and longest link lengths should not be greater than the sum of the remaining two link length. i.e. S + L P + Q An involve pinion and gear in mesh. If both have the same size of addendum, then there will be interference between the tip of the gear teeth and blank of renion. This is a phenomenon of interference. We know that, VB = V, CD = 45 cm, AB = 30 cm

VCD VBA

(IO + mr2) + kx ( r) r = 0

1 2 mr 2

30 50

Speed ratio =

cr =

(c)

Ia

Tfirst Tlast

400

1 2

210

2

190

2


3- Chapter Mechanical.pdf

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