Theory of Machines

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THEORY OF MACHINES

For MECHANICAL ENGINEERING

THEORY OF MACHINES &

VIBRATIONS

SYLLABUS

Theory of Machines: Displacement, velocity and acceleration analysis of plane

mechanisms; dynamic analysis of linkages; cams; gears and gear trains; flywheels and

governors; balancing of reciprocating and rotating masses; gyroscope.

Vibrations: Free and forced vibration of single degree of freedom systems, effect of

damping; vibration isolation; resonance; critical speeds of shafts.

ANALYSIS OF GATE PAPERS

Exam Year 1 Mark Ques. 2 Mark Ques. Total

2003 6 - 15

2004 8 - 18

2005 6 - 14 2006 9 - 21 2007 1 6 13 2008 1 3 7 2009 2 4 10 2010 5 3 11 2011 1 3 7 2012 2 1 4 2013 3 2 7 2014 Set-1 2 3 8 2014 Set-2 2 3 8 2014 Set-3 2 4 10 2014 Set-4 2 3 8 2015 Set-1 1 2 5 2015 Set-2 2 2 6

2015 Set-3 3 3 9

2016 Set-1 2 3 8

2016 Set-2 1 2 5 2016 Set-3 3 3 9 2017 Set-1 1 3 7 2017 Set-2 2 4 10 2018 Set-1 2 3 8 2018 Set-2 2 1 4

CONTENTS

Topics Page No

1. MECHANICS

1.1 Introduction 01 1.2 Kinematic chain 05 1.3 3-D Space Mechanism 07 1.4 Bull Engine / Pendulum Pump 12 1.5 Basic Instantaneous centers in the mechanism 15 1.6 Theorem of Angular Velocities 16 1.7 Mechanical Advantage of the mechanism 22

2. GEARS

2.1 Introduction 23 2.2 Classification of Gears 23 2.3 Analysis of Involutes Gears 30 2.4 Methods to prevent Interference 32 2.5 Different Types of Involutes Systems 34

3. GEAR TRAIN

3.1 Introduction 37 3.2 Components 37 3.3 Simple Gear Train 37 3.4 Compound Gear Train 37 3.5 Reverted Gear Train 38 3.6 Planetary Gear Trains 42 3.7 Kinetic analysis of single slider crank mechanism 43

4. FLYWHEEL

4.1 Introduction 48 4.2 Turning moment diagram of single cylinder double-acting stream engine 48 4.3 Role of Flywheel 48 4.4 Natural vibrations 54

4.5 Vibrations 55 4.6 Methods of initial disturbances 55 4.7 Springs in combination 56 4.8 Energy Method 56 4.9 Transverse vibrations of the beam 62

5. VIBRATIONS

5.1 Introduction 65 5.2 Logarithmic Decrement(S) 65 5.3 Death of the Shaft 69

6. CAMS & FOLLOWERS

6.1 Introduction 71 6.2 Classification of Cams 71 6.3 Classification of Followers 72 6.4 Application of Cams 74 6.5 Terminology of Cam & Follower 74 6.6 Derivatives of Follower Motion 75 6.7 Motions of Followers 77

7. GOVERNORS

7.1 Introduction 80 7.2 Classification of Governors 80 7.3 Gravity Controlled Centrifugal Governors 81 7.4 Proell Governor 83 7.5 Spring Controlled Centrifugal Governors 83 7.6 Governor Effort and Power 85 7.7 Characteristics of Governors 86 7.8 Controlling Force and Stability of Spring Controlled Governors 86 7.9 Insensitiveness in the Governors 87

8. GATE QUESTIONS 89

9. ASSIGNMENT 160

1.1. INTRODUCTION : 1.1.1 KINEMATIC LINK OR ELEMENT

The subject Theory of Machines may be Every part of a machine which is having the defined as that branch of Engineering- relative motion w.r.t some other part is science, which deals with the study of known as kinematic link or element and for relative motion between the various parts a body to be a link it should be a resistant of a machine, and forces which act on them. body. The knowledge of this subject is very essential for an engineer in designing the various parts of a machine.

Sub-divisions of Theory of Machines

The Theory of Machines may be sub- divided into the following four branches : 1. Kinematics. It is that branch of Theory of Machines which deals with the relative motion between the various parts of the machines. 2. Dynamics. It is that branch of Theory of Machines which deals with the forces and their effects, while acting upon the machine parts in motion. Note: For a link the body should be a 3. Kinetics. It is that branch of Theory of resistant body so that it is capable of Machines which deals with the inertia transmitting Motion from one element to forces which arise from the combined another. effect of the mass and motion of the machine parts. 1.1.2 TYPES OF LINK 4. Statics. It is that branch of Theory of Machines which deals with the forces 1. Rigid Link A rigid link is one which and their effects while the machine does not undergo any deformation parts are at rest. The mass of the parts while transmitting motion. Strictly is assumed to be negligible. speaking, rigid links do not exist. However, as the deformation of a   ds  m connecting rod, crank etc. of a V   s reciprocating steam engine is not dt   appreciable, they can be considered as  dv  d rigid links a2 m F(ext) mv dt s dt    Links in which the deformations are    da  negligible (Microscpoic). j   m 3  Eg :- Connection Link. dt  s V=Velocity 2. Flexible Link : A flexible link is one A=Acceleration which is partly deformed in a manner J=Impact not to affect the transmission of motion.

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission For example, belts, ropes, chains and When the motion between a pair is limited wires are flexible links and transmit to a definite direction irrespective of the tensile forces only direction of force applied, then the motion  Links in which deformations are is said to be a completely constrained there but they are in the permissible motion. For example, the piston & cylinder limit. (in a steam engine) form a pair and the  Eg :- Belt Drives, Rope Drives. motion of the piston is limited to a definite direction Fluid Link A fluid link is one which is Completely Constrained Motion formed by having a fluid in a receptacle For One Input → One output (Without the and the motion is transmitted through help of surrounding) the fluid by pressure or compression only, as in the case of hydraulic presses, jacks and brakes.  Sometimes fluid behaves like a link because the power is transmitted due to the Pressure of the fluid.

 Eg. Hydraulic brake, hydraulic ram, hydraulic lift.

Bernoulli Principle

P1A1=P2A2 P1=Pressure on Left side of tank P2= Presuure on right side of tank A1=Area of left side of tank The motion of a square bar in a square A2=Area on right side of tank hole, as shown in, and the motion of a shaft with collars at each end in a circular hole, as shown in, are also examples of completely constrained motion

2. Successfully Constrained Motion NormalForce Thrust Pressure = = When the motion between the elements, Area Area forming a pair, is such that the constrained Kinematic pair: The two links or elements motion is not completed by itself, but by of a machine, when in contact with each some other means, then the motion is said other, are said to form a pair. If the relative to be successfully constrained motion. motion between them is completely or Consider a shaft in a foot-step bearing as successfully constrained (i.e. in a definite shown in Fig. direction), the pair is known as kinematic The shaft may rotate in a bearing or it may pair. move upwards. This is a case of

incompletely constrained motion. But if the 1.1.3 TYPES OF CONSTRAINED MOTIONS load is placed on the shaft to prevent axial 1. Completely Constrained Motion.

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission upward movement of the shaft, then the another link, the pair is known as turning motion of the pair is said to be successfully pair A shaft withcollars at both ends fitted constrained motion. The motion of an I.C. into a circular hole, the crankshaft in a engine valve (these are kept on their seat journal bearing in an engine, lathe spindle by a spring) and the piston reciprocating supported in head stock, cycle wheels inside an engine cylinder are also the turning over their axles etc. are the examples of successfully constrained motion. examples of a turning pair. A turning pair Motion With the help of Surrounding. also has a completely constrained motion.

Sliding Pair (Prismatic Pair) When the two elements of a pair are connected in such a way that one can only slide relative to the other, the pair is known as a sliding pair. The piston and cylinder, cross-head and guides of a reciprocating steam engine, Footstep Bearing ram and its guides in shaper, tail stock on 3. Incompletely Constrained Motion the lathe bed etc. are the examples of a When the motion between a pair can take sliding pair. place in more than one direction, then the motion is called an incompletely constrained motion. The change in the direction of impressed force may alter the direction of relative motion between the pair. e.g. A

circular bar or shaft in a circular hole Rolling Pair When the two elements of a For one input → two or more output. pair are connected in such a way that one rolls over another fixed link, the pair is known as rolling pair. Ball and roller bearings are examples of rolling pair. When the relative Motion is [Pure Rolling] -> Rolling without slipping & skidding. Important Points  If there is more than one independent motion then it is not a kinematic pair.  Connection should be a non-rigid Connection.  For the kinematic pair there should only one Independent motion. Where, 1.1.4 CLASSIFICATION OF KINEMATIC Vcm = Velocity Due to translation Only PAIR R = Radius of drum Due to rotation Only

1. According to the type of relative motion:-

Turning Pair → Revolute Pair → Pin joint: When the two elements of a pair are connected in such a way that one can only

turn or revolve about a fixed axis of

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Screw Pair :- When the two elements of a H.P=Higher pair pair are connected in such a way that one L.P=Lower pai element can turn about the other by screw threads, the pair is known as screw pair. Wrapping Pair : When one pair wraps on The lead screw of a lathe with nut, and bolt other I is wrapping pair with a nut are examples of a screw pair Belt – Pulley, Rope-Pulley (Propagation is dependent upon the retation)  Nut & Bolt Turning as well as spherical motion

Spherical Pair When the two elements of a pair are connected in such a way that one element (with spherical shape) turns or swivels about the other fixed element, the pair formed is called a spherical pair. The ball and socket joint, attachment of a car mirror, pen stand etc., are the examples of a spherical pair

3D Rotation – Spherical Motion

2. According to the type of Contact: 3. According to the type of Closure :-

Lower Pair When the two elements of a Self Closed Pair (Closed Pair) When the pair have a surface contact when relative two elements of a pair are connected motion takes place and the surface of one together mechanically in such a way that element slides over the surface of the other, only required kind of relative motion the pair formed is known as lower pair. It occurs, it is then known as self closed pair. will be seen that sliding pairs, turning pairs The lower pairs are self closed pair. and screw pairs form lower pairs There is a permanent contact. Surface Contact (Turning, Sliding, Screw Eg . Piston & Cylinder (Sliding Pair),Pin & Spherical Pair) Joint (Turning Pair)  Force Closed Pair (Unclosed Pair) Higher Pair When the two elements of a When the two elements of a pair are not pair have a line or point contact when connected mechanically but are kept in relative motion takes place and the motion contact by the action of external forces, the between the two elements is partly turning pair is said to be a force-closed pair. The and partly sliding, then the pair is known as cam and follower is an example of force higher pair. A pair of friction discs, toothed closed pair, as it is kept in contact by the gearing, belt and rope drives, ball and roller forces exerted by spring and gravity. bearings and cam and follower are the examples of higher pairs. There is forceful contact Point/Line Contact (Rolling Pair Only)  Scooty without Gear (Actually with 1H.P 2L.P Automatic gearing).

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission  Gearing is Controlled by Acceleration & indirectly & the relative motion gear is shifted because of Spring Force between any two links is a constraint (Forced Closed Pair). motion then such a closed chain is known as kinematic chain. 1.1.5 DIFFERENT TYPES OF JOINT  When one link of the kinematic chain is fixed it becomes a mechanism which Binary Joint When two links are joined at can give the desired output w.r.t some the same connection, the joint is known as given input. binary joint.

Ternary Joint When three links are joined  A mechanism or a group of a at the same connection, the joint is known mechanisms when utilised & the as ternary joint desired output is obtained it becomes machine.  Condition for the Kinematic Chain b3 j  l  2 22 (1,2) Where, (2,3) IT 2B j = Number of Binary joints (1,3) h= Number of higher pairs Quaternary Joint: When four links are L= Number of links joined at the same connection, the joint is called a quaternary joint. Case 1 h3 j  L  2 Kinematic Chain 22 L = 4 J = 4 h = 0

(1,2) & (1,3)→ Independent (2,3)→ Dependent

3 4 + 0 = *4 -2 2 44 Kinematic Chain

Result: 1 Quaternary joint = 3 Binary joint Case 2 1.2 KINEMATIC CHAIN h3    j    L  2  Frame / Structure → F 22     When all the links are connected in such j = 3 a way such that the first link is h = 0 connected to the last link directly or L = 3

3 3 + 0 > *3 -2 2 3 2.5

Case 3 h3    j    L  2  →Unconstrained Chain 22    j = 5 J = 13 L = 11 L = 12 h = 0 H = 0 H = 0 H = 0 L = 5 L = 10 J = 15 J = 17 15 > 14.5 17 > 16  Indeterminate J=Joints H = Higher pair L=Lower pair

1.2.2 DEGREES OF FREEDOM

3 5+0 < *5 2  The minimum number of independent 2 variables or parameters required to 5 5.5 define the position or motionof the system is known as D.O.F of the s/m. 1.2.1 FRAME / STRUCTURE

 Frame / Structure are those in which power & Motion cannot be transmitted but force can be transmitted.

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Pair Restraints DOF F=[3(L-1)-2j-h]-Fr → The motion which are 3T + 2R →5 6 – 5 =1 not the part of the mechanism. F=degree of freedom L=Lower pair h=Higher pair

IT = 1 6 – 1 =5 Practice Questions Find the degree of freedom of the following figures L=Lower Pair

IT + IR=2 6 – 2 =4 H-Higher pair F=Degree of freedom

(i)

1 LP = 1 DOF

1HP = 2DOF

1.3 3-D SPACE MECHANISM (ONE LINK FIXED) L =4, j=4, h=0 F = [3(4-1)-2*4-0]-1

No of links → F=6(L-1)-5P1 -4P 2 -3P 3 -2P 4 -1P 5 F0

P1→No. Of those pairs which have D.O.F = 1 P2→No. Of those pairs which have D.O.F = 2 (ii) P3→No. Of those pairs which have D.O.F = 3 P4→No. Of those pairs which have D.O.F = 4 P5→No. Of those pairs which have D.O.F = 5

1.3.1 D-PLANER MECHANISM

(iii)

L=4

(iv)

F = 3(5 - 1) – 2 * 5 – 0 = 12 – 10 F2 To Make it constrained

L = 3 J = 2 h = 1 F = [3(3-1)-2*2-1] – 0

For most of the cases F = [3(L-1)-2j-h]

1.3.2 GRUBLER’S EQUATION

The Grubler’s criterion applies to mechanisms with only single degree of

freedom joints where the overall Four Bar Mechanism (Quadric cycle movability of the mechanism is unity. Mechanism): Substituting n = 1 and h = 0 in Kutzbach equation, we have  Only those mechanism for [F=1] and no higher pair [h=0] F = 3(L-1)-2j-h 1 = 3(L-1)-2j-0 3L 2j  4  0 Grubler’s Equation 3L  Even L Even A little consideration will show that a plane mechanism with a movability of 1 and only single degree of freedom joints can not have odd number of links. The simplest Advantages: possible mechanisms of this type are a four  It governs equally the Input & Output bar mechanism and a slider-crank mechanism in which l = 4 and j = 4. Disadvantages Lmin 4 SimpleMechanisms  It cannot have its own enjoyment. If F=0 Frame / Structure  It cannot transmit any power to the If F < 0 (-1,-2,-3,...) Superstructure output.

1.3.3 INPUT/OUTPUT

Complete Rotation (3600)→Crank

Partial Rotation (Oscillation) Double Rocker →Rocker/lever 1. Double Crank Mechanism 2. S + L = p + q 2. Crank Rocker Mechanism Law Satisfies 3. Double Rocker Mechanism Eg. 2,5,3,4 1.3.4 GRASHOFF’S LAW (BASED ON 7 = 7 GRAPHICAL TECHNIQUE)  Not having pair of equal links.

For the continuous relative motion 3. S + L = p + q between the links & the mechanism , the Law satisfies summation of shortest & longest link Having pair of equal links 2,2,3,3 should not be greater than the summation i) Parallelogram Linkage of other two links in the 4-bar mechanism. (S+L)  (p+q) Best Position  Fixed Max Power of rotation is with the Shortest link Where S=shortest link L=Longest link, p,q =opposite links Double Crank ii) Deltoid Linkage 1. S + L < p + q Law satisfies

i) S →fixed Crank – rocker Double Crank L is fixed - S is fixed. ∵ S is in the best position.  It equally distributes its Power to 4. If (S+L) > (p+q), its adjacent links. Law is not satisfied. ii)S is adjacent to fixed Where, (p+q) = Double –Rocker.

Coupling Rod of Locomotive The mechanism of a coupling rod of a locomotive (also known as double crank mechanism) which consists of four links, In this mechanism, the links having equal length act as cranks and are connected to Crank Rocker

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission the respective wheels. The other link acts as a coupling rod and the link is fixed in order to maintain a constant centre to centre distance between them. This mechanism is meant for transmitting rotary motion from one wheel to the other wheel.

Rotation-rotation Single Slider Crank Mechanism A single  4 bar mechanism slider crank chain is a modification of the  Double Crank basic four bar chain. It consist of one sliding  Parallelogram linkage pair and three turning pairs. It is, usually, found in reciprocating steam engine Beam Engine Mechanism A part of the mechanism. mechanism of a beam engine (also known  This type of mechanism converts rotary as crank and lever mechanism) which motion into reciprocating motion and consists of four links, is shown. In this vice versa. mechanism, when the crank rotates about 4 Link + 3 T.P +1 S.P the fixed centre the lever oscillates about a Cross – Head + Piston Rod + Piston/Slider fixed centre. The end of the lever is  One Link connected to a piston rod which reciprocates due to the rotation of the crank. In other words, the purpose of this mechanism is to convert rotary motion into reciprocating motion.

Watt’s indicator Diagram A *Watt’s Inversions: indicator mechanism (also known as Pendulum pump or Bull engine. In this Watt's straight line mechanism or double mechanism, the inversion is obtained by lever mechanism) which consists of four fixing the cylinder or link 4 (i.e. sliding links. On any small displacement of the pair), as shown in. In this case, when the mechanism, the tracing point Q at the end crank (link 2) rotates, the connecting rod of the linktraces out approximately a (link 3) oscillates about a pin pivoted to the straight line fixed link 4 at A and the piston attached to

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission the piston rod (link 1) reciprocates. The α = 2(B ) cos duplex pump which is used to supply feed 2 water to boilers have two pistons attached OA = 2(BC ) * 1 to link 1, as shown in Fig. 5.23. 1 OB

1) Cylinder Fixed 2 BC  OA   Reciprocating Compressors → Input = OB Crank and output Piston 2 Length of Slotted bar Length of crank  Reciprocating Engines → Input = Piston and output Crank Length of Connecting rod 2) Crank Fixed Examples are

3) Connecting Rod Fixed Examples are 1.3.5 WHITWORTH QUICK RETURN MECHANISM

This mechanism is mostly used in shaping and slotting machines. In this mechanism,

4) Slider Fixed the link forming the turning pair is fixed, as shown The link corresponds to a crank in a reciprocating steam engine. The driving crank rotates at a uniform angular speed.  Crank is Fixed (Shortest link)  Crank is Fixed gives two rotation (Full Rotation)  Link 2 – Driving crank Rotation Rotation     QRMM

 Mechanism is still to the point C. Rotation Oscillation α Return Stroke Angle Cutting. β Cutting Stroke Angle. <  + = 360 Stroke : RR 1 2 Stroke : R1 R 2 = CC12 C C 1 2 = 20C = 2 M

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission  In Crank & Slotted lever QRMM difference, valves opens & fuel is  Stroke cutting length depends on three injected. parameters so it is better than Whitworth in which stroke cutting 1.4 BULL ENGINE/PENDULUM PUMP length depends on one parameter.  Oscillating Cylinder Engine Mechanism In this mechanism, the inversion is  Connecting rod is fixed. obtained by fixing the cylinder or link 4 (i.e.  Rotation  Crank sliding pair), as shown in Fig Oscillation Cylinder  Here crank is not rotating 360 , it is oscillating.  If we need a rotation of crank i.e continuous supply of motor, we need a motor.

1.3.6 RESTRAINT

 To accommodate this system crank size should be very-very small.  If the crank size will be small its strength will be less.  The crank will not sustain the engine force. 1.4.1 DOUBLE SLIDER CRANK CHAIN Rotary in I.C. Engine → Crank is fixed. A kinematic chain which consists of two turning pairs and two sliding pairs is known as double slider crank chain,

 Differential fuel injection system.  Fuel injected through a pressure

difference, when there is a pressure

Circle is drawn (But only one) Having B  2 Inversions of Double Slider mechanism

1.4.3 ELLIPTICAL TRAMMELS

It is an instrument used for drawing ellipses. This inversion is obtained  Why it is not used as an IC Engine? by fixing the slotted plate  When the purpose is fulfilled by single slide crank.  When the link comes in the groove then the motion of the slotted plate is back.

1.4.4 ANOTHER FORM OF ELLPPTICAL TRAMMEL

 In this case only one circle is generated at the midpoint but here  number of ellipses are generated. Therefore it is known as elliptical trammels. It cannot be used in IC engine because of, x cos θ =  More wear & tear of two sliding pair. AQ  Maintenance of an engine is due to wear y & tear in between piston & cylinder. sin θ = QB 22 1.4.5 LINK CONNECTING SLIDERS IS sin θ+cos θ=1 FIXED (OLDHAM’S COUPLING) xy22  =1 AQ22 BQ An oldham's coupling is used for At mid-point (AQ=BQ) connecting two parallel shafts whose axes x2 y 2  AQ 2  BQ 2  (radius)2 are at a small distance apart. The shafts are coupled in such a way that if one shaft rotates, the other shaft also rotates at the same speed.

 Any of the slider is fixed used in  This coupling is basically used to heavy transportation connect the two shafts which have  When the link comes in the groove then lateral mis alignment. the motion of the slotted plate is back.

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission VVVV  AB  C  D . AB AI BI CI DI  In general in the mechanism the motion of any link at any moment is neither pure translationary nor purely rotator, it is a combination of translation & rotation which we say the general motion (Link is in general motion instant).  But any link at any moment can be assumed to be in perfect rotation w.r.t a  Slider works like a shaft because the point in the space known as link connecting the sliders through a instantaneous centre of rotation. pin-point is fixed. Therefore it only  This centre is also known as a virtual allows the slider to have rotation only. centre. Hence the slider behaves like a shaft.  As the link changes its position in general its I-centre keeps on changing. 1.4.6 VELOCITY ANALYSIS & ACCELERATION Therefore the locus of this I-Centre for a ANALYSIS particular link during its whole motion

is known as centrode of link.  Instantaneous Centre Method App.  The locus of instantaneous axis of  Relative Velocity Method App. rotation or a particular link during its whole motion is known as axode of the link.  Axis which posses through the I-Centre or the motion of mechanism.

CENTRODE AXODE In General Curve Curved Surface If the link is in Straight Line Plane Surface pure translation Pure Rotation Point Line

1. Instantaneous Centre Method Approach Number of Instantaneous Centres (IC) in (I.C Approch) the mechanism:  Instantaneous Centre of rotation Total Number of Instantaneous Centres = L(L 1)

2 Where, L= Number of links

Example 1: L=4 IC : 6 12 13 14 23 24 34 Example 2:

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission L = 6 can only slide relative to the other, the pair IC :15 is known as a sliding pair. The piston and 12 13 14 15 16 cylinder, cross-head and guides of a 23 24 25 26 reciprocating steam engine, ram and its 34 35 36 guides in shaper, tail stock on the lathe bed 45 46 etc. are the examples of a sliding pair. A 56 little consideration will show, that a sliding

II12 21 pair has a completely constrained motion.

1.5 BASIC INSTANTANEOUS CENTERS IN THE MECHANISM

Turning Pair : When the two elements of a pair are connected in such a way that one can only turn or revolve about a fixed axis of another link, the pair is known as turning pair. A shaft with collars at both ends fitted into a circular hole, the crankshaft in a journal bearing in an engine, lathe spindle supported in head stock, cycle wheels turning over their axles etc. are the examples of a turning pair. A turning pair also has a completely constrained motion Problem

Rolling Pair : When the two elements of a

pair are connected in such a way that one OE = 1.35m rolls over another fixed link, the pair is OA = 200mm known as rolling pair. Ball and roller AB = 1.5m bearings are examples of rolling pair. BC = 600mm CD = 500mm BE = 400mm Link OA  120 rpm (Clock) Find

VB =? (3.2 m/s) V =? (1.6 m/s) C  The point in a circle where the angular VD =? (1.08 m/s)

velocity is zero is known as centre. ωAB =? (2.99 m/s)  Its I-centre lies on the contact point ω =? (8 rad/s) since at this point contact point BC ωCD =? (2.16 rad/s) RV  CM Sliding Pair When the two elements of a Scale pair are connected in such a way that one 1cm = 200mm

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Links: 6 Link 5: (C , D)I15  IC : 15 VC VD 12 13 14 15 16 5   CD   ICID15 15 23 24 25 26

34 35 36 45 46 56

Take any three random IC’s

Take another Case  Since it is a frame . i.e they do not have relative motion between them.  Therefore their IC’s will not lie on a straight line.

 If three bodies move relative to earth other they have three instantaneous centres lies in a straight line.

1.6 THEOREM OF ANGULAR VELOCITIES

Given

VA   OA  (2 *120)   (0.2)* 2.513m/ s OA 60

Link 3: (A,B) Iβ  →Absolute Approach (w.r.t Fixed).

VVAB 3   AB   V  (I.I)   (II)V  Imn m mn 1m n mn 1n Imn IBB A I .B  The three Concept is absolute. Link 4: (B ,C) I14 

VB VC 4   BC   IBIC14 14

ω2 (I 25 I 12 )=ω 5 (I 25 I 15 ) = +5 – (-14) 24 = +19 (Clockwise)

ω2 (I 24 I 12 )=ω 4 (I 24 I 14 ) 45

ω4 (I 45 I 14 )=ω 5 (I 45 I 15 )

If lmn ,l lies on the same side of mn  Direction is same

Problem 1.

ω23 =(+5)-(+14) = -9 = 9(A.C) When , θ=180o

ω2 =2rad/s(Clockwise)

ω3 =??

ω2 I 23 I 12  ω 3  I 23 I 13 

2a  ω3  2a 

ω3  1rad / sec clockwise

Problem 2.

Problem 3.

What will be the angular velocity of link 2 w.r.t link 3. L = 5 L = 5 ω =5rad/s(Clockwise) J = 6 j = 6 2 H = 0 h = 0 ω =14rad/s 3 F = 3(5-1)-2*6-0 F = 1 F = 0

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 1.6.1 RELATIVE VELOCITY APPROACH In the previous article, we have discused the relative velocity method for the velocity Consider two bodies A and B moving along of any point on a link, whose direction of parallel lines in the same direction with motion and velocity of some other point on absolute velocities vA and vB such that vA> the same link is known. vB , as shown in Fig. The relative velocity of The same method may also be applied for A with respect to B, the velocities in a slider crank mechanism. vAB=Vector difference of vA and vB=vA-vB A slider crank mechanism is shown in Fig. The slider A is attached to the connecting rod AB. Let the radius of crank OB be r and let it rotates in a clockwise direction, about the point O with uniform angular velocity rad/s. Therefore, the velocity of B i.e. vB is known in magnitude and direction. The slider reciprocates along the line of stroke AO. The velocity of the slider A (i.e. vA) may be etermined by relative velocity method as discussed below : 1. From any point o, draw vector ob parallel to the direction of vB (or “Velocity of point B w.r.t point A will be in perpendicular to OB) the direction of  ar to the link AB”. such that ob = vB = .r, to some suitable scale, as shown in Fig Problem 1. a) Slider crank mechanism Same data as take in the IC Approach

b) Velocity diagram

Point W.r.t Procedure A O Line  r to OA B A Line r to AB B E Line r to BE C BC bc

BE be D C Line r to CD 2. Since AB is a rigid link, therefore the D Fixed Line ll to the velocity of A relative to B is motion slider perpendicular to AB. Now draw vector ba perpendicular to 1.6.2 VELOCITY DIAGRAM AB to represent the velocity of A with respect to B i.e. vAB.

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 3. From point o, draw vector oa parallel to Problem 1. the path of motion of the slider A (which is along AO only). The vectors ba and oa intersect at a. Now oa represents the velocity of the slider A i.e. vA, to the scale. V ω=AB AB AB

VBC ω=BC BC VCD ω=CD CD Angular velocity is the velocity of a link OA = 400mm ω =ω Vector quantity. Crank(AB) = 120rpm (Anti-Clockwise) AB BA The direction of vector ab (or ba) AB = 150mm OD = 700mm determines the sense of AB which DE = 200mm shows that it is anticlockwise. Get V =? (2.15 m/s) E Note : The absolute velocity of any other point E on the connecting rod AB Point B Point C may also be found out by dividing Slider Coincident point of vector ba such that be/ba = BE/BA . point B. But not on This is done by drawing any line bA1 Slider is on slotted equal in length of BA. Mark bE1 =BE. bar. Join a A1. From E1 draw a line E1e parallel to aA1. The vector oe now represents the velocity of E and vector ae represents the velocity of E with respect to A.

1.6.3 DIRECTION

1. First see the position of link in the V = Total velocity of slider configuration diag. B 2. Make the rough & make the clock & Point W.r.t Procedure anticlockwise dir. C O Line  r to OC 3. Now see the showing that link. C B Line ll to OC

VC VD ωslottedbar = = VVOC OD

(B A) This centripetal Acceleration is due to change in direction. Find 2 BC ?(34.09rad/ s ) 2 CD ?(79.11rad/ s ) Point w.r.t Procedure B A V2 12 BA BA   (Given)(B A) BA BA BA 2 BA BA  BA (Given)(  ar toRad) C B V2 1 CB  Known  (C B) CB CB 2  CB  (Unkown)( ar toRad) CB   CB 1.6.4 ACCELERATION ANALYSIS C D V2 1 CD  Known  (C D) CD CD 2 CD CD  CD (Unkown)( ar toRad)

2 r2V BA BA  (BA)  BA OAP  Velocity  BA t Velocity Diagram(Rough) BA (BA) BA

1. Take the sense of 

 2. Rotate the V in that sense by 90 .

VVV A= B = BA =ω crank -> given OA OP AP

1.6.5 ACCELERATION DIAGRAM 2)

OARS = Acceleration Acceleration Diag(rough)

 trq a BA  B  BA   2 OA RS OS RA crank

1.6.6 ACCELERATION ( C )

This acceleration is associated with the 4) slider when the slider is sliding on the rotating object. Magnitude C 2v  V=Sliding velocity of slider ω = Angular velocity of body on which slider is sliding.

1.7 MECHANICAL ADVANTAGE OF THE MECHANISM

F M.A= Output FInput T M.A= Output TInput

If ηmechanism =100% Ideal

FOutput FInput .V Input F Output .V Output FInput V  Input VOutput

TOutput T.TInput Input  Output  Output  TInput   Input Output

GEARS 2.1.1 GEARS The effect of slipping is to reduce the velocity ratio of the system. In precision machines, in which a definite velocity ratio is of importance (as in watch mechanism), the only positive drive is by means of gears or toothed wheels. A gear drive is also provided, when the distance between the driver and the follower is very small. As the rollers starts moving dynamic 2.1 INTRODUCTION friction comes into play When there is no relative motion b/w Vinput  M.A = mechanism the rollers (by applying angular speed Voutput to one roller, other does not move) at

winput the moment static friction there M.A = ×ηmechanism woutput

wR12 M.A  R1 w 1 =R 2 w 2  wR21 Advantages and Disadvantages of Gear  Drive Negative drive Positive drive The following are the advantages and disadvantages of the gear drive as Belt Drive Gear drive compared to belt, rope and chain drives: Rope drives (Highway Advantages accurate System) 1. It transmits exact velocity ratio. 2. It may be used to transmit large power. 2.2 CLASSIFICATION OF GEARS SLIP IS SLIP IS (i) according to the axis the shaft’s POSSIBLE IMPOSSIBLE connected. (ii) Both axis are parallel 3. It has high efficiency.  kinematically pure rolling motion b/w 4. It has reliable service. two cylinders in contact. 5. It has compact layout. Spur gear: The two parallel and co-planar shafts connected by the gears is shown in. Disadvantages These gears are called spur gears and the 1. The manufacture of gears requires arrangement is known as spur gearing. special tools and equipment. These gears have teeth parallel to the axis 2. The error in cutting teeth may cause of the wheel vibrations and noise during operation

Teeth are straight & II to the axis of rotation.

Helical Gear Gradual Engagement IMPACT (99%) use

Impact Stress  Instantaneous Engagement and disengagement Bearings can’t be affected This axial trust is due to the component are used for the prevention of axial of the force acting normal to the point movement of shafts. of contact.

2.2.1 HELICAL GEAR: Axial trust mm Another name given to the spur gearing is  helical gearing, in which the teeth are Failed only In 1% due to inclined to the axis. The single and double helical gears connecting parallel shafts Double Helical

 Point contact of engagement & Herringbone disengagement  No impact stresses  2, 3, 4 are helical speed in automobiles  At very-very high speed

Axial Trust is too more Bearing Can’t withstand

90% of the cost of gear is due to its mfg cost or profile cost.  create impact stresses. Prodeing.noise (Eg:-Slapper profile distor)  possible when the speed of operations are very-very less wrist watch, wall clock. Automobile ii) Both axes are intersecting → non – i) First gear of bike, parallel and co-planer

Then the end-section of the hyperboloids were used to Form the spiral gear

Bevel Gears Hypoid Gears Eg: Differentials used in automobiles. Worm & Worm Wheel angle

Teeth are str & II to The axis of rotation Teeth are straight   but inclined to the  Helical Bevel Gear  axis of rotation  99% failed

Worm Special

Driver Very high speed reduction ratio upto 1000 : 1  It is used as speed reduction gear which reduces speed up to 1000: 1

Axes are iii)Neither II for Intersecting  Pure rolling is not possible  The rolling which is possible will be having some partial sliding Classification: According to the type of gearing

1) External Gearing: In external gearing, the gears of the two shafts mesh externally with each other as shown. The larger of these two wheels is called spur wheel and the smaller wheel is called pinion. In an external gearing, the motion of the two wheels is always unlike,

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 1) Pitch Circle  It is an imaginary circle in the gear on which pure rolling motion is observed when mooting gears were transmitting power.  Being an imaginary circle it cannot Bigger : Spur (GEAR) be a physical characteristics of the Smaller : PINION gear but being the most imp circle it →Direction of rotation is opposite. is the biggest specification of the gear. The size of the gear is defined 2) Internal Gearing: by the diameter of the pitch circle. In internal gearing, the gears of the two shafts mesh internally with each other as shown. The larger of these two wheels is called annular wheel and the smaller wheel is called pinion. In an internal gearing, the motion of the two wheels is always like.  Common Tangent at the pitch point. (Fixed Line)  Tangential force acting along this time is responsible for the pure rolling of gear i.e. in the power transmission Bigger: Annular Smaller: Pinion 2) Circular Pitch (Pc): It is the distance measured on the circumference of the 3) According to the tangential velocity pitch circle from a point of one tooth to of gear the corresponding point on the next V < 3 m/s Low velocity gear tooth. It is usually denoted by pc 3  V 15 Medium velocity gear P.C.D V > 15 High velocity gear  Pitch circle dio 4) According to the type of teeth: D P  The teeth on the gear surface may be C T a) straight, Here D can be in any unit. (mm, cm, m) b)inclined For the mating gears c) curved. P = P C1 C2 DD 2.2.2 GEAR TERMINOLOGY 12 TT12    const. DD 12 TT12

3) Module (m): It is the ratio of the pitch circle diameter in millimeters to the number of teeth. It is usually denoted by m.

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission D Sliding in mating teeth m= =mm It is expressed in mm T only. If in the question if No unit is Friction given, take module in mm only. Diameter pitch (Pd): It is the ratio of Heat Generation number of teeth to the pitch circle diameter in millimetres. It is denoted by Thermal Expansion pd. F cos Ø T Sliding ?? Pd  D (inches) Component of the force responsible For the  rotation. *** If Ø F cos Ø P .P   **(in PSU) cd *Line of action must pass through the pitch point.

At every moment. LOR will change

 2.2.4 LAW OF GEARING

Common normal at the point of contact In order to have a constant angular velocity

ratio for all positions of the wheels, the (Line of Action) point P must be the fixed point (called pitch  point) for the two wheels. In other words, Force transmission line the common normal at the point of contact (Tooth Space – Tooth Thickness) between a pair of teeth must always pass through the pitch point. 2.2.3 BACKLASH This is the fundamental condition which must be satisfied while designing the  Backlash is provided for the thermal profiles for the teeth of gear wheels. It is expansion of the gears. As the thermal also known as law of gearing. expansion of the gear  'sec the tooth thickness, sufficient space is provided Notes : b/w the gear teeth. 1. The above condition is fulfilled by teeth  Thermal Expansion occurs due to the of involute form, provided that the root neat generated through the friction. circles from which the profiles are  In mating little bit of sliding occurs b/w generated are tangential to the common the gear tooth thus producing the normal. Friction. 2. If the shape of one tooth profile is arbitrarily chosen and another tooth is designed to satisfy the above condition, then the second tooth is said to be conjugate to the first. The conjugate teeth are not in common use because of difficulty in manufacture, and cost of production. O1O2 (fixed) Line of action

w12 (QP  PM)  w (NP  QP)

w1 QP  WPM 1  w2 NP  W2 QP

Velocity of Sliding Point P fixed. Q Changes at every moment when the contact point Q is at P, the VSliding = 0 , therefore at point P, Therefore at point P, the gears are in  To be in proper contact (Without pure rolling motion. spacing) Or the bodies to be the gear V1 cos  = V2 cos  ↓ OM (O N) (FCommon normal to the profiles.Of (O Q)w 12   (O Q)w 1 1 ON 2 2 OQ both the gears.) 12 Line of action (common normal at point w O N 12 ------(2) of contact Q) w21 O N ↓  O PM O PN Always pass through Pitch Point 12  w O NPN O P 1 2   2 (A) Mating profiles should be designed in w2 O 1 M PM O1 P such a way such that at each & every point law of gearing is satisfied. V (w w )QP Sliding 1 2 If these two bodies are Forms of Teeth GEARS w 1  const. w2 OP 2  const. OP1 P Fixed Point “Line of action b/w two mating gears at any moment should always pass Through the fixed point on the line joining the centres of rotation of the Gears. This fixed point is known as pitch point is known as pitch point. This is known as LAW OF GEARING. If Involute Profiles: this law is satisfied at each & every (By nature conjagate): An involute of a point, then the bodies are known as circle is a plane curve generated by a “GEARS” point on a tangent, which rolls on the

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission circle without slipping or by a point on a taut string which in unwrapped from a reel as shown . In connection with toothed wheels, the circle is known as base circle It is a locus of a point on the line which rolls without Slliping on the fixed Circle In reality

Arc(AP1 ) 0  it’s differential will be zero. Therefore it can be treated as straight line.  “Normal drawn at any point on the involute curve will become tangent to its base circle.”

2.2.6 A BEAUTIFUL REALITY

2.2.5 BASE CIRCLE

 It is a physical characteristic of an involutes profile

 “Base Circle in an external gearing cannot be dedendum circle”  This is because pressure angle 3 sec , due to this FcosØ decreases.  This will reduce the speed of rotation of the gear  But in internal gearing, base circle can be dedendum circle.

 Take o pencil & cover it by using a thread. Stoutly & slowly uncover the Pencil maintaining the tension in the wire.

Arc(AP1 ) A 1 P 1

Arc(AP2 ) A 2 P 2

Arc(AP ) A P 4 4 4

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 2.3 ANALYSIS OF INVOLUTE GEARS KL ↓ Path of Contact ↓ KP PL  Path of appoach Path of recess

O1 M R cos   PM R cos 

O KM 1 2 2 2 2 RA  R cos  (KP  Rsin  ) KP R2  R 2 cos 2   R sin  A   Path of contact PL  r2 r 2 co 2 R sins   A 

2.3.2 ARC OF CONTACT

When a point of contact is travelling from  K Start of engagement Top land of start of engagement to the End of the gear touches the flank of the pinion. engagement the distance travelled by the  L End of engagement Top land of pinion or gear along their pitch circles is the pinion touches the flank of the gear. known as arc of contact  Line of action should be tangent to the Path of contact Arcof contact  base circle. Because it is the normal On cos invdute or point of contact Path of app Arcof approch  2.3.1 LINE OF ACTION cos

Path of recess i) pass through P ]→ Gears Arcof recess  cos ii) Tangent to both the circle ] →Involute if the arc of contact is high, more pairs iii) Line of action Fixed Line of engagement is there    const.  Load will be distributed uniformly iv) Point of contact is gravelling along the line of action. But upto a certain limit because if arc of

contact ' sec v) Motion of pt. of contact Straight line

Total distance travelled by the pt. of contact cos  ,   'es (Q) from start to the end of engagement.

r2 =96+7.5=103.5mm Path of contact (KL): KL  KP  PL (Ra)2  (R 2 cos 2  )  R sin   (ra2  r 2 cos 2  )  rsin   =18.8805mm+17.9037mm =36.7843mm 36.7843  if arc of contact = 1 p.c Arc of contact =  39.1450mm cos200  only one pair is engaged

If arc of contact = 2 p.c

two pairs are engaged.

Arcof contact Contact Ratio  P.C  No.of pairs engaged in one engagement period (1.2 – 1.8) Average value of no. of pairs engaged in one engagement Period. Let contact ratio (1.34) 36.1450 180 one pair is engaged in full engagement i) QPinion =  period, but 34% of time of this period is 96 π like that along. With this pair one more QPinion = 23.36290 pair i.e total Two pairs are engaged. Therefore average value of no. of pairs ii) NP = 450 rpm N T Engaged in one engagement period S × comes out to be 1.34 NRP 450×96 N = =300rpm Problem 4 144 T 36  0 VSliding = (WP +WS )QP  m 8mm,   20 t 24  Gear (102)2  (96cos200 ) 2  (96sin 200  QP) 2 Addendum :- 7.5 mm QP=14.7692mm Gear: VSliding =1.1599m/s mT 8 36 R   144mm 22 DEATH OF INVOLUTE GEAR:

Interference

Pinion: Interference: If the radius of the addendum circle of pinion is increased, the point of

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission contact will move. When this radius is 2.4 METHODS TO PREVENT INTERFERENCE further increased, the point of contact will be on the inside of base circle of wheel and 1) Under Cut Gear not on the involutes profile of tooth on  wheel. The tip of tooth on the pinion will Low power transmission then undercut the tooth on the wheel at the root and remove part of the involutes profile of tooth on the wheel. This effect is known as interference, and occurs when the teeth are being cut. In brief, the In under-cut portion, stress phenomenon when the tip of tooth undercuts the root on its mating gear is concentration will be more due to known as interference. in bending moment. At the roots. 

 This will be the strength of the gear.

2) Ø.  (Increase Pressure Angle)

 Base  . Increase to 200 -220. increase only to a certain limit. Interference decreases but does not

eliminates completely.  O2L rA If rA  point L will shift towards M.

Till M there is no problem  If rA > O2 M

Involute Tip of  Involute profile of gear 1 3) Increase the no. of teeth pinion will touch  in mating with gear 2   non-involute  of non-involute  flank Portion of the gea pr orfile.

Law of gearing not satisfied Tip of pinion will remove some material from non-involute flank portion of the gear.

This removal of the material is known as UNDER CUTTING Similar will be there If No of teeth RA > P1N M, N CRITICAL Points Addendum value  Interference points g the addendum radel

 completely similarly the interference Minimum No. of Teeth n Pinion to avoid interference

If module of pinion = m Addendum =(mAP) ------(ii) mA r 1  G(G  2)sin2   1 P  mt mAmin  1  G(G  2)sin2   1 P 2 

2AP tmin  1 G(G  2)sin2   1 

2AG Tmin  11 2 1 (  2)sin   1 GG

 Velocity ratio can be 1:2 or 2:1  But gear ratio will be always equal to or 2.4.1 BEAUTIFUL REALITY greater than 1. RT = =G.R rt

COS Rule 2 2 2 2 rA  r  R sin  2(r)(Rsin  )cos(90   ) 2 2 2 2 rA  r  (R  2rR)sin  2 99% 22R 2rR r 1  2 sin  r Addendum of pinion/gear same tm = ? 2 2AG = rA  r 1  G(G  2)sin  T  min 11 Addendum of pinion = rA – r 1 (  2)sin2   1 2 GG rA  r 1  G(G  2)sin    1  Tmin AP Fractional addendum of pinion for tmin  1% one module in order to avoid G Interference Addendum of pinion/gear different.

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 2AP Full Depth Involute 14.5 Degree tmin  1 G(G  2)sin2   1  Addendum standard Addendum 2A T  G 1m min 11 1 (  2)sin2   1 GG mAP  T  G  min mAG  tmin  mAR   If it comes AP  20 tmin  A1 35 Tmin G   therefore Tmin = 40 AR  G.R 2 Bending moments are less Min. No. of Teeth to avoid interference in stability can be done within a limit. involute rack-pinion arrangement. because t contact ratio. Stub Involute (200, 250) Addendum < 1m

AP   A1G   for every tooth different base circle of A  same dia R  Best 200 stub Involute i) Lesser interference ii) Gears with less no. teeth (min. no. of teeth requirement will be less to avoid interference) Since pressure angle is 200. some amount of interference is being Reduced after stubbing. Therefore in no. of teeth require iii) Less cost iv) Stranger Tooth

In ΔPMS. Sr. No. System of gear teeth Addendum of rack 1. 1 Sin 14 Composite rsin  2 2. Addendum =rsin2 Ø Full depth involute 2 =mtmin sin Ø, Rack 3. 20 Full depth involute Addendum of Rack =rsinØ 4. 20 Stub involute mt min sin2 --(1) Minimum number of teeth on the pinion 2 12 mt 32 mA min sin2  r 2 18 14 tmin 2A r Problem: 2.5 DIFFERENT TYPES OF INVOLUTES Determine the minimum number of teeth SYSTEMS required on a pinion, in order to avoid interference which is to gear with,

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 1. a wheel to give a gear ratio of 3 to 1 ; and Interference  2. an equal wheel. The variation of centre distance is because The pressure angle is 20° and a standard of vibration addendum of 1 module for the wheel may Base circle radius remains same be assumed. w1 Solution: =const w2 Given :G T / t  3;   20  ; A  1 module w 2 Cycloidal profile :- A cycloid is the curve 1. Minimum number of teeth for a gear traced by a point on the circumference of a ratio of 3:1 circle which rolls without slipping on a We know that minimum number of fixed straight line. When a circle rolls teeth gear required on a pinion without slipping on the outside of a fixed 2A t  w circle, the curve traced by a point on the  112 circumference of a circle is known as epi- G 1  2 sin   1 GG cycloid. On the other hand, if a circle rolls without slipping on the inside of a fixed 21 t  circle, then the curve traced by a point on  112 the circumference of a circle is called hypo- 3 1  2 sin 20   1 33 cycloid. “Locus of a point on the circumference of 2 t 15.04 or 16 the circle which rolls without Slipping on a 0.133 fixed straight line” 2. Minimum number of teeth for equal wheel We know that minimum number of teeth for equal wheel 2A t  w 1 3sin2   1 21 t  1 3sin2 20   1 2 t 12.34 or 13 VR ≠ consts changes 0.162

Method to reduce interference Increase Centre distance

1) Per tooth cost is more as compared to involute But overall cost same. Interference absent

Centre distance  , pitch circle shifts, to a new position. This change in Pitch circle changes the whole specification of the gear.  

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 22 2) Less wear R  3532   85.5   7310 ..(Squaring convex- concave (Life is more) A both side) 2 3) Flank is wide RA   7310  35320  42630 Stronger Tooth Or Ø changing RA  206.5mm Ømax start of engagement ∴ Addendum height for larger gear wheel O pitch point = RA-R =206.5-200 =6.5mm Ømax End of engagement (in the 1 opposite direction) Now path of recess, PL PN 2

Or Problem: 2 22 R.sin  Two mating gears have 20 and 40 involute rA   r cos   rsin   teeth of module 10 mm and 20° pressure 2 angle. The addendum on each wheel is to 2220 rA   100  cos 20 100sin 20 be made of such a length that the line of 200sin 200 contact on each side of the pitch point has 100sin 200 half the maximum possible length. 2 Determine the addendum height for each r 22  100  cos2 20 0  100sin 200  100sin 200 gear wheel, length of the path of contact, A 200  0.342  68.4 arc of contact and contact ratio. 22 Addendum height for each gear wheel rA  8830   68.4   4680 ..(Squaring We know that the pitch circle radius of the both side) smaller gear wheel, r = m.t / 2 = 10 × 20 / 2 r   4680  8830  13510 Or r 116.2mm = 100 mm and pitch circle radius of the A A ∴ Addendum height for smaller gear wheel larger gear wheel, r r  116.2  100  6.2mm R = m.T / 2 = 10 × 40 / 2 = 200 mm A Let RA = Radius of addendum circle for the larger gear wheel, and Length of the path of contact rA = Radius of addendum circle for the We know that length of the path of contact smaller gear wheel. Since the addendum on 11r R  sin KP  PL  MP  PN  each wheel is to be made of such a length 2 2 2 that the line of contact on each side of the 100 200  sin 200 pitch point (i.e. the path of approach and 51.3mm 2 the path of recess) has half the maximum Length of the arc of contact possible length, therefore We know that length of the arc of contact

length of the path of contact Or  2 r.sin  cos R R22 cos   R sin   A  51.3 2 54.6mm Or cos200 22 Contact ratio R 200 cos2 20 0 200sin 200 A    We know that circular pitch 100 sin 200 P  m   10  31.42mm 50sin 200 c 2 length of the path of contact Contact ratio 2 R   35320  50sin 2000  200sin 20 P A 54.6 250  0.342  85.5 1.74say 31.42

3.1 INTRODUCTION circles. When the distance between the two shafts is small, the two gears 1 and 2 are Sometimes, two or more gears are made to made to mesh with each other to transmit mesh with each other to transmit power motion from one shaft to the other, from one shaft to another. Such a combination is called gear train or train of toothed wheels. The nature of the train used depends upon the velocity ratio required and the relative position of the axes of shafts. A gear train may consist of spur, bevel or spiral gears.

3.1.1 TYPES OF GEAR TRAINS

Following are the different types of gear trains, depending upon the arrangement of All the gears are in same module. wheels : → They give contribution only in the 1. Simple gear train, direction 2. Compound gear train, → All the gears should have same module. 3. Reverted gear train, and 1, 2:- wT 4. Epicyclic gear train. 12= ------(1) In the first three types of gear trains, the wT21 axes of the shafts over which the gears are 2, 3:- mounted are fixed relative to each other. w T But in case of epicyclic gear trains, the axes 2 = 3 ------(2) wT of the shafts on which the gears are 32 mounted may move relative to a fixed axis. 3, 4:- w T 3 = 4 ------(3) 3.2 COMPONENTS wT43 (i) X (ii) X (iii) 1) Main DVR wT 2) Main DVN S.R=14 = wT 3) Intermediate Gears 41 →No. of idlers →odd → (same direction) W Main DVR Speed ratio(velocity ratio) →even → (for diff. direction DVR & DVN) WMain DVN →Gear train 3.4 COMPOUND GEAR TRAIN 1 → =Train.Value S.R When there are more than one gear on a shaft, as shown, it is called a compound 3.3 SIMPLE GEAR TRAIN train of gear. But whenever the distance between the When there is only one gear on each shaft, driver and the driven or follower has to be as shown, it is known as simple gear train. bridged over by intermediate gears and at The gears are represented by their pitch the same time a great ( or much less )

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission speed ratio is required, then the advantage Also, the circular pitch or module of all the of intermediate gears is intensified by gears is assumed to be same, therefore providing compound gears on intermediate number of teeth on each gear is directly shafts. In this case, each intermediate shaft proportional to its circumference or radius. has two gears rigidly fixed to it so that they *T1 + T2 = T3 + T4 may have the same speed. One of these two  Compound gear train in which input gears meshes with the driver and the other and output shafts are co-axial with the driven or follower attached to the next shaft.

used in power trak mission power should be conserved. In this type, drives pinion (of smaller radius)  M1 = M2 M3 = M4 M5 = M6 Less torque requirement. 1, 2:- m1 = m2 = m wT m3 = m4 = m’ 12= ------(1) w T ×T wT 1 =S.R= 2 4 21 w T ×T 3, 4:- 4 1 3 w T DVR : (1,3) 3 = 4 ------(2) DVN : (2,4) wT 43 5, 6:- Problem: The speed ratio of the reverted wT 56= ------(3) gear train, as shown in Fig. is to be 12. The wT65 module pitch of gears A and B is 3.125 mm w T ×T ×T and of gears C and D is 2.5 mm. Calculate 1 =S.R= 2 4 6 w T ×T ×T the suitable numbers of teeth for the gears. 6 1 3 5 No gear is to have less than 24 teeth. w S.R= 1 w6 Product of No. of teeth on DVN = Product of No. of teeth on DVR DVR : (1, 3, 5) DVN : (2, 4, 6)

3.5 REVERTED GEAR TRAIN

When the axes of the first gear (i.e. first Since the speed ratio between the gears A driver) and the last gear (i.e. last driven or and B and between the gears C and D are to follower) are co-axial, then the gear train is be same, therefore N N known as reverted gear train as shown A c 12  3.464 r1 + r2 = r3 + r4 ...(i) NNBD

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Also the speed ratio of any pair of gears in gear B meshes with gear A and has its axis mesh is the inverse of their number of on the arm at O2, about which the gear B teeth, therefore can rotate. If the arm is fixed, the gear train TT is simple and gear A can drive gear B or BD3.464 ...(i) vice-versa, but if gear A is fixed and the arm TTAC We know that the distance between the is rotated about the axis of gear A (i.e. O1), then the gear B is forced to rotate upon and shafts x r  r  r  r  200mm ABCD around gear A. Such a motion is called m T . m .Tm .T m .T Or AABB CC DD  200 epicycle and the gear trains arranged in 2 2 2 2 such a manner that one or more of their m.T members move upon and around another r  2 member are known as epicyclic gear trains (epi. means upon and cyclic means 3.125 T T   2.5  T  T   400 AB CD around). The epicyclic gear trains may be mABCD  m andm  m  simple or compound. NTT 100 124 The epicyclic gear trains are useful for ABD  12.3 N T T 28 36 transmitting high velocity ratios with gears DAB of moderate size in a comparatively lesser 400 space. The epicyclic gear trains are used in TAB  T   128 ...(ii) 3.125 the back gear of lathe, differential gears of 400 T  T   160 ....(iii) the automobiles, hoists, pulley blocks, wrist CD2.5 watches etc.

From equation (i), TBA 3.464T . Substituting

this value of TB in equation (ii),

TAA 3.464T 128 or 128 T 28.67say28 Ans A 4.464

TB  128  28  100 Ans Again from equation (i), TD = 3.464TC. Step Conditions of Arm Gear Gear B Substituting this value of TD in equation (iii), No. motion C A TCC 3.464T 160 1. Arm fixed-gear 0 +1 T A or A rotates  TB 160 through +1 TC  35.84say36 Ans. revolution i.e., 4.464 1 rev. Anticlock wise TD  160  36  124 Ans 2. Arm fixed-gear 0 +x T NOTE : The speed ratio of the reverted gear x A A rotates T train with the calculated values of number through +x B of teeth on each gear is revolution Epi-cyclic Gear Train: 3. Add +y +y +y +y We have already discussed that in an revolution to epicyclic gear train, the axes of the shafts, all elements 4. Total motion +y x + y T over which the gears are mounted, may yxA T move relative to a fixed axis. A simple B epicyclic gear train is shown , where a gear A and the arm C have a common axis at O1 All the little consideration will show that about which they can rotate. The when two conditions about the motion of

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission rotation of any two elements are known, the arm, to which the pinions B and C are then the unknown speed of the third attached, make : element may be obtained by substituting 1. when A makes one revolution clockwise the given data in the third column of the and D makes half a revolution forth row. anticlockwise, and 2. when A makes one revolution clockwise Algebric Method : In this method, the and D is stationary ? motion of each element of the epicyclic The number of teeth on the gears A and train relative to the arm is set down in the D are 40 and 90 respectively. form of equations. The number of equations depends upon the number of elements in the gear train. But the two conditions are, usually, supplied in any epicyclic train viz. some element is fixed and the other has specified motion. These two conditions are sufficient to solve all the equations; and hence to determine the Solution: motion of any element in the epicyclic gear First of all, let us find the number of teeth train. let the arm C be fixed in an epicyclic on gears B and C (i.e. TB and TC). Let dA, gear train as shown in the figure above. dB, dC and dD be the pitch circle diameter Therefore speed of the gear A relative to of gears A, B, C and D respectively. the arm C Therefore from the geometry of the figure, NN and speed of the gear B relative dA + dB + dC = dD or dA + 2 dB = dD ...(_ dB AC = dC) to the arm C Since the number of teeth are proportional NN BC to their pitch circle diameters, therefore, Since the gears A and B are meshing TA + 2 TB= TD or 40 + 2 TB = 90 directly, therefore they will revolve in ∴TB = 25, and TC = 25 ...(TB = TC) opposite directions. Step Revolutions of elements NN T No. Conditions Arm Gear A Compound Gear D  BC A of motion gear B – C NNTACB 1. Arm fixed, 0 -1 T TT  A AB gear A    Since the arm C is fixed, therefore its speed, TB TT rotates BD N = 0. T C through-1  A revolution NTBA TD  (i.e., 1 rev. NTAB clockwise) 2. Arm fixed, 0 -x T If the gear A is fixed, then NA = 0 A T gear A x x A NN T TB BC A or rotates TD through-x 0 NCB T revolutions NT BA1 3. Add-y -y -y -y -y NT revolutions AB to all elements 4. Total motion - y -x-y T T Note: The tabular method is easier and xy A  xy A  hence mostly used in solving problems on TB TD epicyclic gear train. 1. Speed of arm when A makes 1 Problem: revolution clockwise and D makes An epicyclic train of gears is arranged as half revolution anticlockwise shown in Fig. How many revolutions does

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Since the gear A makes 1 revolution clockwise, therefore from the fourth row of the table, -x-y=-1 or x+y=1 ...(i) Also, the gear D makes half revolution anticlockwise, therefore T 1 40 1 xy A   or xy   T2D 90 2

 40x × 90y=45 or x-2.25y=1.125 …(ii) From equations (i) and (ii), x = 1.04 1. Number of teeth on wheels A and B and y = - 0.04 Speed of arm= -y=-(-0.04)=+0.04 Let TA = Number of teeth on wheel A, = 0.04 revolution anticlockwise Ans and T = Number of teeth on wheel B B 2. Speed of arm when A makes 1 If dA, dB, dC, dD, dE and dF are the pitch revolution clockwise and D is diameters of wheels A, B, C, D, E and F stationary respectively, then from the geometry of Since the gear A makes 1 revolution the figure drawn above clockwise, therefore from the fourth dACE d 2d and dBDF d 2d row of the table, Since the number of teeth are -x-y=-1 or x+y=1 ….(iii) proportional to their pitch circle Also the gear D id stationary, therefore diameters, for the same module, T 40 therefore x A  y  0 or x  y  0 TD 90 TACE T  2T  28  2  18  64 Ans

40x×90y=0 or x-2.25y=0 …(iv) And TBDF T  2T  26  2  18  62 Ans From equations (iii) and (iv), x = 0.692 and y = 0.308 2. Speed of wheel B when arm G makes ∴ Speed of arm = -y =-(0.308) 100 r.p.m. clockwise and wheel A is =0.308 revolution clockwise Ans fixed First of all, the table of motions is Problem: drawn as given below : In an epicyclic gear train, the internal Step Revolutions of elements No. Conditions of Arm Wheel Wheel E Compound Wheel F Wheel B wheels A and B and compound wheels C motion G A gear C – D 1. Arm fixed, 0 + 1 TA TTTT TTT and D rotate independently about axis O. + AE ADADF   wheel A T      E TTTT TTTC F b The wheels E and F rotate on pins fixed to rotates ECCF T TTAD through + 1  A    the arm G. E gears with A and C and F gears TTCE revolution TC (i.e., 1 rev. with B and D. All the wheels have the same anticlockwise module and the number of teeth are : TC = ) TT 2. Arm fixed, 0 + x T T AD TTAD x A x A x   x  28; TD = 26; wheel A TTCF T TTCE rotates E TC TE = TF = 18. through + x 1. Sketch the arrangement ; revolutions 3. Add + y + y + y + y + y + y + y 2. Find the number of teeth on A and B ; revolutions to all elements 3. If the arm G makes 100 r.p.m. clockwise 4. Total motion + y x + y TA TT TT yx TA yx AD  yx AD  TB yx TT TT and A is fixed, find the speed of B ; and T CF CE 4. If the arm G makes 100 r.p.m. clockwise C Since the arm G makes 100 r.p.m. and wheel A makes 10 r.p.m. counter clockwise, therefore from the fourth clockwise ; find the speed of wheel B. row of the table, y = - 100 ..... (i)

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Also, the wheel A is fixed, therefore Note : The gear at the centre is called the from the fourth row of the table, sun gear and the gears whose axes move x + y = 0 or x = - y = 100 ...... (ii) are called planet gears. TT (By nature they are simple epicyclic) ∴ Speed of wheel B y  x AD  TTCE 64 26  100  100    100  95.8r.p.m. 28 62 = - 4.2 r.p.m. = 4.2 r.p.m. clockwise Ans

3. Speed of wheel B when arm G makes 100 r.p.m. clockwise and wheel A

makes 10 r.p.m. counter clockwise T5 +2TP =TD Since the arm G makes 100 r.p.m. CM Unbalanced clockwise, therefore from the fourth  row of the table Configured force gear y = - 100 ...... (iii)

Also, the wheel A makes 10 r.p.m. Direction counter clockwise, therefore from the fourth row of the table, x + y = 10 or ------(vibration) x = 10 - y = 10 + 100 = 110 ...... (iv) In planetary gear trains ∴ Speed of wheel No. of planets is more than one 64 26  100  110    100  105.4r.p.m. 28 62 = + 5.4 r.p.m. = 5.4 r.p.m. counter clockwise Ans

3.6 PLANETARY GEAR TRAINS

A compound epicyclic gear train is shown It consists of two co-axial shaftsS1 and S2, an annulus gear A which is fixed, the compound gear (or planet gear) B-C, the sun gear D and the arm H. The annulus gear has internal teeth and the compound gear is carried by the arm and revolves freely on a pin of the arm H. The sun gear is co-axial Why more than one planet gear is with the annulus gear and the arm but required ? independent of them. (i) Balancing The annulus gear A meshes with the gear B (ii) High power Transmission and the sun gear D meshes with the gear C. For this we required 2 inputs It may be noted that when the annulus gear is fixed, the sun gear provides the drive and when the sun gear is fixed, the annulus gear provides the drive. In both cases, the arm acts as a follower.

T.D.C

Arm 0 1+x T T 2+x. 1 3-x. . 2 Piston T2 T3 x=B1 B Y Y+x T1 y±x y-x =B1 o-Bo T 3 (r  L)  (BM  Mo)

(r  L)  (Lcos  rcos  ) 3.6.2 FIXING OR HOLDING TORQUE IN l AN EPICYCLIC GEAR TRAIN n  r L nr Text (T input  T output  T fixing )  0 AM Lsin   rsin  T (T T ) ---(i) fixing input output sin  sin n 3.6.3 POWER CONVERSION 2 sin  cos  1 2 n Tinput .w input T output .w output 0 --- (ii) nr n22 sin x r  nr  r cos  Question: n Sol n22 sin Ninput = +100 cos n Noutput = +250 T = +50 r(1  cos  )  r n  n22  sin  input  (50)(100) + Toutput (250) = 0 22 Toutput = -20 KN-m x r (1  cos   n  n  sin    TFixing = - (+50 + (-20))= -(+30) dx dx d V  .  w 3.7 KINEMATIC ANALYSIS OF SINGLE dt d dt SLIDER CRANK MECHANISM sin 2 v rw sin   2 n22 sin Inertia of the connecting rod is not 

considered in this analysis. n  Let, large m⇒mass of the reciprocating parts sin 2 v rw sin   w⇒angular velocity of crank. 2n L⇒Length of connecting rod. dv dv d a  .  w r⇒crank radius dt d dt L n  2 cos 2 (large) Obliquity ratio a rw cos   r n

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 2.) Connecting rod 2 3.33  3.33   4  2  1 d ∴ cosθ 0.26 wCR  dt 22 ……….(taking positive sign) sin  sin Or θ 75 Ans n 2. Maximum velocity of the piston d cos  d  cos  .  w Substituting the value of θ = 75 in dt n dt equation (i), maximum velocity of the w cos piston, wCR  n n22 sin sin150 VP(max) ω.r sin 75 n 2n w cos 0.5  20.95  0.3 0.966  m / s n22 sin 3.33 n  large = 6.54 m/s Ans

dwCR dw CR d CR   . dt d dt PROBLEM: w2 sin  The crank and connecting rod of a steam CR engine are 0.3 m and 1.5 m in length. The n crank rotates at 180 r.p.m. clockwise.

Determine the velocity and acceleration of PROBLEM: the piston when the crank is at 40 degrees If the crank and the connecting rod are 300 from the inner dead centre position. Also mm and 1 m long respectively and the determine the position of the crank for zero crank rotates at a constant speed of 200 acceleration of the piston. r.p.m., determine:1. The crank angle at Solution : which the maximum velocity occurs, and 2. Given : r = 0.3; l = 1.5 m; N = 180 r.p.m. or Maximum velocity of the piston ω π  180 / 60  18.85rad / s;θ 45 Solution : 1. Crank angle at which the maximum Velocity of the piston velocity occurs We know that ratio of lengths of the connecting rod and crank, Let  = Crank angle from the inner dead n = l/r = 1.5/0.3 = 5 center at which the maximum velocity ∴ Velocity of the piston, occurs. We know that ratio of length of connecting rod crank radius, n = l/0.3 = 3.33 sin80 18.85  0.3 sin 40  m / s and velocity of the piston, 25 sin 2θ = 4.19 m/s Ans VP ω.r sinθ ……….(i) 2n Acceleration of the piston For maximum velocity of the piston, We know that acceleration of the piston,

dVP 2 cos 2θ  0 i.e., aP ω .r cosθ dθ n 2cos 2θ  2 cos80 2 ω.r cosθ 0 18.85   0.3 cos40  m / s 2n 5 2 2 Or ncosθ 2cos θ  1  0………. = 85.35 m/s Ans cos2θ 2cos2 θ 1 Position of the crank for zero 2 2cos θ 3.33cosθ  1  0 acceleration of the piston

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Let θ1 = Position of the crank from the A1, A2 Areas of piston at cover & crack inner dead centre for zero acceleration of end.  the piston 2 2 2 A12 D ,A  D  d  We know that acceleration of piston, 44

2 cos2θ1 P1, P2 Gas pressures at cover & crack end. aP1ω .r n cosθ n D Dia of piston d dia of Rod. ω2 .r Or 0 n cosθ11 cos2θ  …a0P   F Fgas  F I  f  mg n 0 ncosθ  cos2θ  0  180   f 11  2 18000 360   f 5cosθ11 2cos 2θ  1  0  or 2cos2 2θ 5cosθ  1  0 FI Inertia force (since it is a prop. Of 11 resistance ) always oppose motion 5  52  4  1  2 ∴ cosθ 0.1862 ………. f kinetic friction 1 22 where

(taking positive sign) Fgas Or θ1  79.27or280.73 Ans

3.7.1 DYNAMIC ANALYSIS OF SINGLE = P A – P A SLIDER CRANK MECHANISM 1 1 2 2 FI =m.a  Effective Driving Force on the Piston cos 2 m.rw2 cos   (Piston Effort ):- (F) n Always calculated from covered to crack end.

It may be noted that in a horizontal engine, the reciprocating parts are accelerated from rest, during the latter half of the stroke (i.e. when the piston moves from inner dead centre to outer dead centre). It is, then, retarded during the latter half of the stroke (i.e. when the piston moves from outer dead centre to inner dead centre). The inertia force due to the acceleration of the reciprocating parts, opposes the force on the piston due to the difference of pressures in the cylinder on the two sides of the piston. On the other hand, the inertia force due to retardation of the reciprocating parts, helps the force on the piston.

1) Force along C.R (FC)

FC cos F F F  C cos

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 2) Normal Trust on Cylinder walls π 2 0.35   500   68730N …(∴ Force = FnC F sin 4 Pressure × Area) Fn  Ftan  Ratio of length of connecting rod and crank, n = l/r = 1.2/0.3 = 4 3) Crank Effort (Ft) F F sin(  ) and accelerating or inertia force on tC reciprocating parts, F Ft  sin(   ) 2 cos2θ cos FIR m .ω r cosθ n

2 cos120 4) Radial Trust on Crank Shaft Bearings :- 250 26.2  0.3 cos60 4 FrC F cos(  ) F = 19306 N Ft  cos(   ) ∴ Piston Effort, cos F F  F  68730  19306 PLI = 49424 N = 49.424 KN 5) Turning moment on Crank shaft :- (Imp)

T F .r t 1. Pressure on slide bars F T  sin(   ).r Let ϕ = Angle of indication of the cos connecting rod to the line of strike. T = f(θ) (θ = Fn(time)) We know that, sinθ sin 60 sin     0.2165 n4 T = f(time) ∴ ϕ = 12.5 I f(Time) We know that pressure on the slide bars, PROBLEM: The crank-pin circle radius of a FNP F tan   49.424  tan12.5 horizontal engine is 300 mm. The mass of = 10.96 KN Ans the reciprocating parts is 250 kg. When the 2. thrust in the connecting rod, crank has travelled 60° from I.D.C., the FP 49.424 difference between the driving and the FQ    50.62K Ans back pressures is 0.35 N/mm2. The cos cos12.5 connecting rod length between centres is 1.2 m and the cylinder bore is 0.5 m. If the PROBLEM : A vertical double acting steam engine runs at 250 r.p.m. and if the effect of engine has a cylinder 300 mm diameter piston rod diameter is neglected, calculate : and 450 mm stroke and runs at 200 r.p.m. 1. pressure on slide bars, 2. thrust in the The reciprocating parts has a mass of 225 connecting rod, kg and the piston rod is 50 mm diameter. The connecting rod is 1.2 m long. When the Solution: crank has turned through 125° from the top dead centre, the steam pressure above Given : r = 300 mm = 0.3 m; mR= 250 Kg; θ 2 the piston is 30 kN/m2 and below the = 60; P1 − P2= 0.35 N/mm ; l = 1.2 m; D = 0.5 m = 500 mm; N = 250 r.p.m. or piston is 1.5 kN/m2. Calculate the effective ω = 2π × 250/60 = 26.2 rad/s. turning moment on the crank shaft. First of all, let us find out the piston effort Solution : Given : D = 300 mm = 0.3 m; L = 450 mm or r = L/2 = 225 mm = 0.225 m; N (FP). We know that net load on the piston, = 200 r.p.m. or ω = 2π × 200/60=20.95 π rad/s; mR= 225 KG; d = 50 mm = 0.05 m; l = FPPD   2 2 L 1 2  1.2 m; θ = 125; P1 = 30 KN/m = 30 × 4 3 2 2 3 2 10 N/m ; P2 = 1.5 KN/m = 1.5 × 10 N/m

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission We know that area of the piston, ππ2 A  D22   0.3   0.0707m I 44 and area of the piston, ππ 2 a  d22   0.05   0.00196m I 44 ∴ Force on the piston due to steam pressure,

FL p 1 A 1  p 2 A 1  a  30  1033  0.0707  1.5  10 0.0707  0.00196 N 2121 103 = 2018 N Ratio of length of connecting rod and crank, n = l/r = 1.2 / 0.225 = 5.33 and inertia force on the reciprocating parts,

2 cos2θ FIR m .ω r cosθ n 2 cos250 FI  225. 20.95   0.225 cos125  n = - 14172 N We know know that for vertical engine, net force on the piston or piston efforts,

FPLIR F  F  m .g = 2018 – (– 14172) + 225 × 9.81 = 18397 N Let ϕ= Angle of inclination of the connecting rod to the line of stroke. We know that, sinθ sin125 sin     0.1537 n 5.33 ∴ ϕ = 8.84 We know that effective turning moment on the crank shaft, F sin θ    TrP cos 183978 sin 125 8.84   0.225N  m cos8.84

= 3021.6 N-m Ans

4.1 INTRODUCTION 4.2 TURNING MOMENT DIAGRAM OF SINGLE CYLINDER DOUBLE- ACTING The turning moment diagram (also known STREAM ENGINE as crankeffort diagram) is the graphical representation of the turning moment or A turning moment diagram for a single crank-effort for various positions of the cylinder double acting steam engine The crank. It is plotted on cartesian co- vertical coordinate represents the turning ordinates, in which the turning moment is moment and the horizontal ordinate taken as the ordinate and crank angle as represents the crank angle. abscissa. A flywheel used in machines serves as a Steam Engine:- reservoir, which stores energy during the period when the supply of energy is more than the requirement, and releases it during the period when the requirement of energy is more than the supply. In case of steam engines, internal combustion engines, reciprocating compressors and pumps, the energy is developed during one stroke and the engine is to run for the whole cycle on the energy produced during this one stroke. For example, in internal combustion engines, the energy is developed only during expansion or power stroke which is much more than the engine load and no energy is being developed during suction, compression and exhaust strokes in case of four stroke engines and during Since the pressure inside the engine compression in case of two stroke engines. cylinder is less than the atmospheric The excess energy developed during power pressure during the suction stroke, stroke is absorbed by the flywheel and therefore a negative loop is formed as releases it to the crankshaft during other shown in Fig. 16.2. During the compression strokes in which no energy is developed, stroke, the work is done on the gases, thus rotating the crankshaft at a uniform therefore a higher negative loop is speed. A little consideration will show that obtained. During the expansion or working when the flywheel absorbs energy, its stroke, the fuel burns and the gases expand, speed increases and when it releases therefore a large positive loop is obtained. energy, the speed decreases. In this stroke, the work is done by the Hence a flywheel does not maintain a gases. During exhaust stroke, the work is constant speed, it simply reduces the done on the gases, therefore a negative fluctuation of speed. In other words, a loop is formed. It may be noted that the flywheel controls the speed variations effect of the inertia forces on the piston is caused by the fluctuation of the engine taken into account turning moment during each cycle of operation 4.3 ROLE OF FLYWHEEL

 Making the turning moment to be const. Area under 1 –θ diag. ↓ WCycle α is const w Cycle  T ↓ 4 mean Fluctuations with jerks will be converted into fluctuation without jerk. 4.3.1 COEFFICIENT OF FLUCTUATION OF ↓ SPEED OF FLYWHEEL device which is directly mounted on the The difference between the maximum and crank-shaft. minimum speeds during a cycle is called

 the maximum fluctuation of speed. The it delivers & stores the energy in the ratio of the maximum fluctuation of speed form of rotational energy. to the mean speed is called the coefficient of fluctuation of speed. Flywheel speed , crankshaft speed . NNmax min CS  since it is one link. N Arm under T-θ diag. NNmax min where, N  2 In a cycle

4%CS = 0.04 WCycle = Tmean × 2π** 5% CS = 0.05 wCycle T  14% CS = 0.08 mean 2 15% CS = 0.1

min Nmin1 , N min 2 , N min3 ......  4.3.2 COEFFICIENT OF STEADINESS FOR  THE FLYWHEEL Nmin wcycle NNmax min   Tmean 4  Coefficient of fluctuation of Speed of max Nmax1 , N max 2 ,......   Flywheel: N max N min N max Cs  N (closed 3-4%) These are the Fluctuation which removed Where, N max N min after installing flywheel. N  Turning Moment Diagram of a Single 2 Cylinder Four Shroke I.C Engine. 4%Cs 0.04

5%Cs 0.05

4% Cs  0.08

5% Cs  0.1

4.3.3 COEFFICIENT OF STEADINESS FOR

THE FLYWHEEL 1  CS 4.3.4 COEFFICIENT OF ENERGY FOR THE FLYWHEEL

It may be defined as the ratio of the maximum fluctuation of energy to the work done per cycle. Mathematically, coefficient of fluctuation of energy EE C  max min E W Cycle In this case we have to integrate The work done per cycle (in N-m or joules) may be obtained by using the following two PROBLEM relations : A horizontal cross compound steam engine Work done per cycle = Tmean× θ develops 300 kW at 90 r.p.m. The where Tmean = Mean torque, and coefficient of fluctuation of energy as found θ= Angle turned (in radians), in one from the turning moment diagram is to be revolution. 0.1 and the fluctuation of speed is to be = 2π, in case of steam engine and two kept within ± 0.5% of the mean speed. Find stroke internal combustion engines the weight of the flywheel required, if the = 4π, in case of four stroke internal radius of gyration is 2 metres combustion engines. Solution Given: 4.3.5 FUNDAMENTAL EQUATION OF THE P=300kW×103; N=90 r.p.m; CE=0.1; k =2m FLYWHEEL We know that the mean angular speed, ω =2π N/60 = 2 π×90/60=9.426 rad/s Let mmass of flywheel ω1 and ω2 = Maximum and minimum K radius of gyration speed respectively. I = mk2 Since the fluctuation of speed is ± 0.5% of Max function of energy mean speed, therefore total fluctuation of  speed , Variation ω1 -ω2 = 1% ω=0.01ω 11 and coefficient of fluctuation of speed E  Iw22  Iw max min 12   22 CS 0.01 2  E Iw C S We know that work done per cycle =P×60/N=300×103×60/90=200×103N.m ∴ Maximum fluctuation of energy, ∆E = Work done per cycle×CE =200×103×0.1=20×103N.m Let m = Mass of the flywheel We know that maximum fluctuation of energy (∆E) 20×103=m.k2.ω2.CS=m×22×(9.426)2×0.01 = 3.554 m ∴ m = 20×103/3.554 = 5630 kg Ans

PROBLEM: The turning moment equation of a particular engine is given by T (20000  9500sin 2  5700cos2  )N  m Where θ Angle turned by the crack from

I.D.C

If the fluctuations in the flywheel are not 105 48180 E(TT)   more than 2% at a mean speed 300 rpm  mean find 15.4818 105.4818 i) power output from the engine.  9500sin 2  5700cos2  ii) Mass of the flywheel required having  15.4818 the radius of gyration 1m. E 11078.8265Joules iii) Angular acceleration of the flywheel at Iw2 C 11078.8626 the position when the crank has Rotated S 450 from I.D.C I=561.2599kg-m2 Assume the resisting torque is constant. Since K =1m. T (20,000  9500sin 2  5700cos2  )N  m I = mk2 m = 561.2599 kg CS  0.02 0 (iii) (T T )  I.  2 300 mean 45 w  10  9500 60  I   sin 2 9500 1  L.C.M of Nr    r  561.2599  H.C.F of D 1 cos 2  α 16.9262 r/s2  1  corrections  Cycle [0, π] Solution sin  2  2  T 5000  1500sin 3   2 3  2  sin 2   2 2 2 1 T 5000  600sin     mean 1   W T.d Cycle → [0, 2, λ] Cycle  0 Paints where T curve cuts Tmean Curve :-  T = T =  (20,000 9500sin 2   5700cos2  )d  mean 0 5000 + 1500sin3θ = 5000 + 600sin θ 5[3sin θ – 4sin θ] = 2sin θ WCycle  (20,000) 13sin θ – 20sin3 θ = 0 20,000 Tmean  sin θ (13~20sin’ θ) = 0  sin  0,180 ,360 13 Sin θ = + → 53.72880, 126.72880 20

Sin θ = - → 233.72880, 306.27110

Points where T curve cuts Tmean line T = Tmean 20000 9500sin 2  5700cos2   20,000 5700 tan 2   tan 2   0.6 9500 306.2711 0 0 0 E  T  T    1656.5094Joules 2θ 30.9637 ,210.9637 ,390.9637 ,   233.7288 mean

Problem I = 44.66 kg – m2 A punching press is required to punch 720 2 110  holes/hour. Each holes has clia of 10 mm 60 and the thickness of the sheet is 5 mm. It 120 100 2 requires 7N-m of energy/mm of the Cs   110 sheared area If the each punching  operation takes 1/10 seconds and the m = 178.645 kg variation in the above problem speed of the flywheel reduces from 120 rpm to 100 rpm during punching. Find the variation 1 mass of the flywheel required having a Cycle Time : 5 sec radius of gyration of 0.5 m in order to have Exact punching time?? clesired punching operation. Stroke : 200 mm 400 mm → 5 sec 5 5 mm → 5 sec 400 variation 2 Cycle Time : 5 sec Exact punching time?? ω ??

720 holes/hour 720 holes/3600sec. Cs → 0.05 sec  1 hole/5sec. 360 → 5 sec Cycle Time : 5 sec.  5 20 → 20 sec 1 360 1 hole / sec. 10 Solution → Hole

A sheared =  × 10 × 5 Exact punching is obtained in 200 of crank = (50 ) mm2 rotation E hole = 50 × 7 = (350 ) Joules

Pmotor = Energy required/ sec

= E hole × No. Of holes/sec 720 = 350  3600 = 70  watt  J/S  → Motor installed 1 → Punching sec 10 2x → 5 sec 1 2 Evaluable = 70 7 Joules  rad 10 5 E hole = (350 ) Joules Solution Cycle Time → 2 sec [350 - 7] =Iω2Cs

1 E min = E - 5 P 1500W  Ehole  motor 2 ∆E = (E + 20) – (E - 5) E hole = 3000 j = 35 mm2 = 35 mm × mm  rad / s,Cs  0.2  1 35   5 Joules Extra punching time = sec  6 3 = Iω2Cs 1 2 3000 - 1500  I  Cs 6  Designing of the flywheel rim:- 2 I = 1393.1662 kg-m I → distribution of mass

Concept of Multi – Cylinder Engines

A → Area of cross Section of rim

The concept of multi – cylinder was introduced not to ↑ the power. It is there to increase the uniformity of the power so that the flywheel requirement is going to be ↓ se up to certain extent.

d 2Tsin  dm  R 2 2 d T  A.R.d   .P.R 2 2 T 22    P. R    P  V  A  VR      Bearing limit of shress q A2 + Ay = A1 + A3 + A5 mm2 A

Let E  E unknown 

Eb = e – A1 Ec = E – A1 + A2 E = E - A + A – A d 1 2 3 EE = E – H1 + H2 – H3 + H4 Highest Energy Storing Capacity EA = E ↓ 1 For Example 2 E max = E + 30 2

↓ Concept of Rotating Unbalancing

4.4 NATURAL VIBRATIONS

Vibrations in which there is no friction at Comparatively the high speed engines are all & there is no external force after the having lighter flywheel & the slow speed initial release of the s/m are known as engines are having heavier flywheels  b natural vibrations. Vibrations were seen at the atomic level. Firstly by the Galileo

1 EI2 2 If M.I of the flywheel is very – very large. It stores the 90 J of energy at 10 rpm only making the engine to be run at 10 rpm only If the M.I of the flywheel is less, it stores the energy (90 J) at 300 rpm making the engine to be run at higher speeds. At mean position Mg = SXo ↓ Initial displacement Inertia Resistive force spliced

F0  L M0 0 sx = ma ma + sx = 0 Lambert Static Balancing Dynamic Balancing

Is always constant in magnitude but always changes its direction Therefore it causes

vibrations. Hence we have to balance it

4.5 VIBRATIONS Natural Angular frequency,

 Those periodic motions which are to & s wn  r / s fro. Motion. m sine wave, cos wave Frequency of Natural vibrations ↓ Periodic & Harmonic sin2 wave ↓ Periodic but not harmonic.

Any Vibrating System 2 Time period T 1. K.E storing device (mass) n wn due to motion of object. 1 Linear frequency, n  wn / 2  (in hertz) Body stores energy in the form of Tn motion due to its mass. R, ∅ will be found by Initial conditions (t=0) 2. P.E storing device (stiffness S) ↓ due to position/stiffness of the body. Const. 3. Friction ≠ 0. 4. Unbalanced Forces 4.6 METHODS OF INITIAL DISTURBANCES (Balancing ≠ 0) Longitudinal vibrations. When the particles Can never to be zero of the shaft or disc moves parallel to the Even relational motion cannot be axis of the shaft, as shown, then the balanced in reality. vibrations are known as longitudinal DAP vibrations.  In this case, the shaft is elongated and Sx + ma = 0 shortened alternately and thus the tensile ma sx 0 and compressive stresses are induced  m.x  sx  0 alternately in the shaft. s Transverse vibrations. When the particles x  x  0 m of the shaft or disc move approximately perpendicular to the axis of the shaft, as The solution of this eqn will be shown then the vibrations are known as Amplitude Max displacement of the s/m transverse vibrations. In this case, the shaft from its position of Equilibrium. is straight and bent alternately and bending for mean stresses are induced in the shaft ↓ To rsional vibrations*. When the particles of the shaft or disc move in a circle about  s the axis of the shaft, as shown in Fig. 23.1 x  R sin t   m (c), then the vibrations are known as ↓ torsional vibrations. Amplitude In this case, the shaft is twisted and ↓ untwisted alternately and the torsional Harmonic Fn . shear stresses are induced in the shaft.  Vibrations are there Note: If the limit of proportionality (i.e. stress proportional to strain) is not Constant & max.E0  exceeded in the three types of vibrations, Where R, ∅ then the restoring force in longitudinal and ↓ transverse vibrations or the restoring Constants.

couple in torsional vibrations which is Parallel exerted on the disc by the shaft (due to the stiffness of the shaft) is directly proportional to the displacement of the disc from its equilibrium or mean position. Hence it follows that the acceleration towards the equilibrium position is directly proportional to the displacement from that position and the vibration is, therefore, simple harmonic.  xx  1) t 0 i R, x  0 velocity     SSS12 x0  2) t 0 R,   xv i  

xx i  Series of Springs  3) t 0  R,  xv  i  s Equation  x x 0 m 5x  2x 0 2 x  wn  x  0 1 2 S  x x 0 L 5 F sx 22 X 1, ww2    nn55 ↓ (Force required to produce unit deflection) 4.7 SPRINGS IN COMBINATION F=s

Series

4.8 ENERGY METHOD

We know that the kinetic energy is due to 1 1 1  the motion of the body and the potential SSS12 energy is with respect to a certain datum S? position which is equal to the amount of S work required to move the body from the w  n m datum position.

In the case of vibrations, the datum 22 122 1 1 mR  v  E Sx  mv     position is the mean or equilibrium 2 2 2 2 R 2 position at which the potential energy of    the body or the system is zero. 122 1 3m E E  Sx   v In the free vibrations, no energy is 2 2 2 transferred to the system or from the Comparing it system. Therefore the summation of kinetic 11 E mv22 Sx energy and potential energy must be a 22 constant quantity which is same at all the times 2s Where wn  , →Considers the whole s/m 3m

11 E Sx22 Iw 22 22 122 1 mR v Sx   m(R) 2 2 2 2 R Mg= SX0 122 1 3m ↓ E Sx v 2 2 2 11 E mV22 Sx it is balance du 22 4.8.1 MOMENT OF INERTIA ↓ ↓ Ring mR2 K.E P.E ↓ In virtual vibrations Hollow Cylinder E = const mR 2 dE Disc  0 2 dt ↓ 1 dv 1 dx  m.2v s.2x. 0 Solid Cylinder 2 dE 2 dt 2 2 mx  sx 0 Hollow Sphere mR 3 s x x 0 m Solid Sphere

1 1 1 E Sx2  mv2  Iw 2 2 2 2

Translational + Rotational motion

11 E sx22  mv  K.E 22 spring The velocity of the spring varies from 0 to v 2 L 1 ms  v  K.E  .dy .y spring     0 2 L  L  3 1 ms2 L 2  .v3  = 1 (mvv ) 2 L 3 11 E sx22  mv  (K.E) 22 spring  (anti-clockwise) 1 1 1 sx2  mv2  m v2 L 2 2 6 s Mg. . 2 1122mS E sx  m  v I. 2 2 3  About the cerotic axis Where , ↑ s 2 w  mL n m I mL m  S 12 3 mL2 I  3 4.8.2 QUE METHOD t  I.  Mg .   0 (Angular oscillations ) 2 mL2 L  Mg. .   0 32  3g     0 2L Where, 3g w  n 2L

I.  I.   Distribution of mass Is taken of the body which is oscillated & around the axis from which it is rotated. If θ is small , sin 0 I. mg.L.   0 L (m.L2 ) mg.L.   0 Mg. . 4  g     0 L mL2  1 Where , Im 12 4 g ↓ wn  L About centurial

Axis GJ GJ TS   L  I.  Mg. .   0 L L. 4 ↓ 2 2 Torque required to give unit defection mL L Im S 12 4 w   n I If this shaft is having M.I→ Ishaft S w   similarly n I I  shaft 3 S  m m  S 3 It is also a two-rotor system. One end is fixed to .earth (ground) w0n  Earth is having  M.I with respect to I. T  n I  ,L  0. 11 Node point will shift to the ground 4.8.3 TENSIONAL VIBRATIONS

T.L  G.J T.M→ Relative movement of fibers with respect to other fibers. SS 12 II12

SS12    I. S .   0 II12 GJ GJ   S     0 LILI1 1 2 2 I LI G.J. (ii) 11 T  LI L 22 Position of node poin

(i) L1 + L2 = L mg  SS S 12 II F 12 F K.x  x  LI x  12 Problem. LI 21 A cantilever shaft 50 mm diameter and 300 Anti-node → are these where amplitude is mm long has a disc of mass 100 kg at its maximum free end. The Young's modulus for the shaft → If the no. of rotors are No. of node. material is 200 GN/m2. Determine the point frequency of longitudinal and transverse → (n-1) vibrations of the shaft → Node point should be shifted towards Given : that rotar. Whose moment of Inertia is d = 50 mm = 0.05 m; I = 300mm = 0.03m; m too large. 2 2 From (i) & (ii) position of node point should be = 100kg; E = 200 TGN/m = 200×m We know that cross-sectional area of the 4.8.4 RAYLEIGH’S METHOD shaft,

2 A= d2  0.05   1.96  103 m 2 In this method, the maximum kinetic 44 energy at the mean position is equal to the And moment of inertia of the shaft maximum potential energy (or strain 2 energy) at the extreme position. Assuming I= d2  0.05   0.03  106 m 4 the motion executed by the vibration to be 64 64 simple harmonic Frequency of longitudinal vibration ↓ We know that static deflection of the shaft, Also known as (Static Deflection of mass W.I 100 9.81 0.3 6 method )    0.751  10 ∆ A.E 1.96 1039  200  10 .... W m.g  ∴ Frequency of longitudinal vibration, 0.4985 0.4985 fn    575Hz  0.751 106

Frequency of transverse vibration, We know that static deflection of the shaft, 3 3 W.I 100 9.81 0.3  4 Natural position (Neither it is in comp. nor    96 0.147  10 m it is in exp) 3E.I 3 200  10  0.3  10 ∴ Frequency of transverse vibration, 0.4985 0.4985 fn    41Hz  0.147 103

Problem A shaft of length 0.75 m, supported freely at the ends, is carrying a body of mass 90 kg at 0.25 m from one end. Find the natural frequency of transverse vibration. Assume

E= 200 GN/m2 and shaft diameter = 50 mm

Solution : load W and if the beam is deflected beyond Given: the static equilibrium position then the d=50mm = 0.05m;m=500kg ;E=200GN/ load will vibrate with simple harmonic motion (as by a helical spring). IfTM is the m2 = 200 × 109N/m2 static deflection due to load W, then the We know that cross –sectional area of shaft, natural frequency of the free transverse ππ A  d2  (0.05)2  1.96  103 m 2 vibration is Vibrations along the length of 44 the beams and moment of inertia of shaft , ππ I  d4  (0.05)4  0.307  106 m 4 64 64 Natural frequency of longitudinal vibration Let m1 = Mass of flywheel carried by the length l1. PL L ∴ m − m1 = Mass of flywheel carried by AE length l2 If L = 1 We known that extension of length l1 AE W .I m .g.l PS I I 1 1 … (i) L A.E A.E ↓ Similarly, compression of length l2 stiffness (WW)I m m g.l I2 12  s  …. (ii) w  A.E A.E n m Since extension of length 푙 must be equal 1   L to compression of length 푙 , therefore 2 PL equating equations (i) and (ii)  m .l m m  l AE 1 1 1 2 mgL  m1  0.9  500  m1  0.6  300  0.6m1 orm 1  200kg  AE ∴ Extension of length 푙1, m .g.l 200 9.81 0.9 g δ 11 wn  A.E 1.96 1039  200  10  4.5 106 m We know that natural frequency of longitudinal vibration, 0.4985 0.4985 fn    235Hz δ A.E 106

4.8.5 LOGITUDINAL VIBRATIONS OF THE AE BEAM S  11 1 L 1 Consider a shaft AB of length l, carrying a point load W at C which is at a distance of AE22 S2  l1 from A and l2 from B,. A little L2 consideration will show that when the SSS shaft is deflected and suddenly released, it 12 will make transverse vibrations. The deflection of the shaft is proportional to the

 ↓  Friction force P12 P  mg    (i)  ↓ LLP,P   1 2  1 2 Denotes the increase in displacement PLPL  but ↓ in velocity 1 1 2 2  (ii) AEAE1 1 2 2 

PLPL1 1 2 2 L12   L   or AEAE1 1 2 2 g w  n 

Friction dissipates the energy from the 4.9TRANSVERSE VIBRATIONS OF THE BEAM system DAP Across the length sx cx  mx   0

c 2  x x   wn  x  0 m The soln of the equation will be ↓ 3 PL 12tt  12  x  Ae  Be 3EI where A & B → constants gs or wn  t  m 12    x  A  Bt  e Conventional  ,  Roots of Auxiliary Equation 1,2,3 (E from 2=200Gpa) 1 2 ↓ Ction ≠ 0 22c     w0n   Ped Vibrations m ∝ displacement Solution ↓ 2 cc 2 Is due to stiffness of string    4wn mm ↓ 1,2 Represent 2 2 P.E of the spring cc 2 1,2    n ↓ 2m 2m Position of the object Friction/ Energy Conservation Damping Term Term 2 c  2m  2 Degree of Dampness n  2 c  2m  Damping factor or Displacement gives  2 C→ Coefficient of damping n Mean/Equilibrium Damping ratio

c2 2 c   4m  s 2 sm m cs2 c 22    = n 2 sm m m Vibration equation will be ↓ 2 x 2 nn  x     x  0

 Solution is

,t 2 ,t X  Ae Be 12  OR t X   A Bt  e 12       2 1  1,2   n  1  Over damped s/ms → No vibrations  1 Critically damped s/ms → No 2) Critically Damped Systems ( 1) vibrations → (At the verge of vibrations)  1 Under damped s/ms→ vibrations α1 = α2 = α = -ωn Solution will be 1) Over damped Systems 1  No periodicity Non-vibratory motion of the system X   A Bt  ent α1, α2 → Different (Real) The whole value → The motion obtained is again non- x Ag 22 1 t Bg  1 t    nn     of x willgoes on ?g vibratory  Non perodic&  Ve term most Ve term → There systems regains their equilibrium Non harmonic position in the shortest possible time without oscillate

→ Critically damped s/ms has a faster are as compared to the over damped system Over damped Solns  2 1  nt    x Ae Be x A Bt  ent The sort of over damped sent has or term –ve & close to tve. But the Son’s

2 of critically damped soils have the term cs –ve therefore Soln curve has a much If> , then the roots k1 and k2 are 2m m faster response as compare to over  damped sol real but negative .This is a case of 3) Under – damped systems overdamping or large damping and the 1,2 2 mass moves slowly to the equilibrium   n i 1  .  n position. This motion is known as

1,2  n id   . aperiodic. When the roots are real, the These are critically damped derice most general solution of the differential The solution will be equation is k12 t k t ,t 2t x C e C e x Ae Be 12  22   nn i  d  t  i  d  t c c  s c c  s = Ae Be     tt       2m 2m  m   2m 2m  m  nt C e  C e   = e ABcos  dd tiABsin      t 12   t Note : In actual practice , the overdamped X xen sin  t   d vibrations are avoided . Where x,∅ → constants 2. When the roots are complex Initial conditions conjugate (underdamping) 2 2 Td = sc d If >.then the radical (i.e. the term 1d m 2m nd  Hz  under the square root ) becomes negative. T2d  The two roots k1 and k2 are then known as complex conjugate. This is a most practical case of damping and it is known as under damping or small damping .The two roots are 2 c s c ki1      2m m 2m And 2 c s c  t ki    x  Xen s ( t   )in 2  d 2m m 2m Decrement ratio → Amplitude of Where i is a Greek letter known as iota and vibrations reduces in a certain ratio its value is −1.For the sake of At t = 0 mathematical calculations, let x0= x sin ∅ cs 2 At t = Td a; ω;  and 2m m n x Xe nTd sin 1 2 At t = 2Td sc 2 2  ωdn  ω   a –  n m 2m x2  Xe  2Td  sin xxxx 0312   ....  een Td  const x1 x 2 x 3 x 4

1. When the roots are real (over damping)

5.1 INTRODUCTION: 3. Damped vibrations. When there is a reduction in amplitude over every cycle of vibration, the motion is said to be When elastic bodies such as a spring, a damped vibration. This is due to the fact beam and a shaft are displaced from the that a certain amount of energy equilibrium position by the application of possessed by the vibrating system is external forces, and then released, they always dissipated in overcoming execute a vibratory motion. This is due to frictional resistances to the motion. the reason that, when a body is displaced, the internal forces in the form of elastic or 5.2 LOGARITHMIC DECREMENT (S) strain energy are present in the body. At release, these forces bring the body to its It is defined as the natural logarithm of the original position. When the body reaches amplitude reduction factor. The amplitude the equilibrium position, the whole of the reduction factor is the ratio of any two elastic or strain energy is converted into successive amplitudes on the same side of kinetic energy due to which the body the mean position. continues to move in the opposite direction. To make the decrement ratio easy to calculate. The whole of the kinetic energy is again Inen TD  converted into strain energy due to which n.2 the body again returns to the equilibrium   nTd  position. In this way, the vibratory motion 12 is repeated indefinitely. 2λΣ δ= 5.1.1 Types of Vibratory Motion 1-Σ2 Of displacement of begin of 3rd cycle to end The following types of vibratory motion are of 7th cycle is 2.5 important from the subject point of view : x 2 =2.5 1. Free or natural vibrations. When no xι external force acts on the body, after x 4 giving it an initial displacement, then To 7 → (from end of end) x the body is said to be under free or 7 natural vibrations. The frequency of the From first to eight → (corves all the free vibrations is called free or natural x complete) 0 frequency. x8 2. Forced vibrations. When the body Cercal damping coefficient (C ) vibrates under the influence of external c force, then the body is said to be under c 2Σ ωn m forced vibrations. The external force = applied to the body is a periodic 2×1× ωn Cc disturbing force created by unbalance. m The vibrations have the same frequency C as the applied force. =ξ Note: When the frequency of the Cc Actual damping coeff external force is same as that of the ξ= natural vibrations, resonance takes place. Critical damping coeff

C, Cc 5.2.1 UNBALANCE Units → N/(m/s)

Question The following data are given for a vibratory system with viscous damping: Mass =2.5kg; spring constant = 3 N/mm & the amplitude decreases to 0.25 of the initial value after five consecutive cycles. Determine the 5.2.2 GAP damping coefficient of the damper in the mx cx   5x   F sin  t  0 system. 0

Solution 2 F0 x 2  n  x    n  x  sin  t Given : m=2.5kg :s=3 N/mm =3000N/m ; m 푥6 = 0.25푥퐼 We know that natural circular frequency of Final Solution vibration, X = CF + PI s 3000 ωn    34.64rad / s m 2.5 Let c = Damping coefficient of the damper in N/m/s, 5.2.3 PARTICULAR INTEGRAL

xI = Initial amplitude , and F x6 = Final amplitude after five consecutive 0 sin t cycles = 0.25푥 …(Given) m 퐼 PI   We know that 22 D 2 Nn  0    x xxx x I 2 354 Or x x x x x o Oprator 2 3 4 5 6  5 D22  x x xxx x x  I I  2 35 4   I F0 x6 x 2 x 3 x 4 x 5 x 6 x 2 sin t 22  m nn    2D    1/5 1/5 PI  x x x 22 22 II I 1/5 nn   2D      2D      (4)  1.32   nn   x x 0.25x 2 6 I F 0 n22   sin  t  2  n  cos  t We know that m     2 x 2π 22 2 I n    2  n  loge  a   2 2 x2 ωa n  F 0 sin t   2π m loge  1.32   a or = 2 2 34.64 a2 2 2   n    2n   a×2π 0.2776  F0 / s 2 PI  1200-a 222   2   Squaring both sides , 1    39.5a2 nn    0.077  or 1200  a2 → vibration in a running system will never 92.4 0.077aa22 39.5 stop Solution of the Forced – Damped a2  2.335or a  1.53 vibrations will be We know that a = c/2m or X = CF + PI c a  2m  1.53  2  2.5  7.65N / m/ s

F / s max at the condition of resonance CF 0 222 ω 1   2   =1 M.F =  1    ωn 0 nn     ↑ damping →  ↑ → If A is high →Internal cracks will be ↓ damping → ↓ high/more→ Deformation is more ↑ Under damping   ↓ ↑ Under damping ↑ → It → ↓↓ (0.8 to 0.6) M.F ↑↑ & the curve

shift to → In case of critical damp & over damped, the second term will dominate, which will bring the curve downward not → These vibrations brings fatigue (cyclic making the value of M.F max loading) to the Product → Vibration in running system will never Running life will be max at critical & stop By virtue of this over damping. → every product is defined by the Running Life of the system → Nature of cud of MF will same as the A A  comparison of strength of MF  F0 / s  vibrations → Strength of Amplitude of steady – state vibrations. 1 MF  22   2   1    n    → If MF ↑↑ , Running Life ↓↓ = 1 → In case of Under damping  1 →   (ω = ωn) → Resonance Firstly the first term dominates the Amplitude of steady state vibration (A) ω curve &after sometime. ↑ I values. ↑ will be max at ωn < 1 → under damping ses the value of second term which will bring the curve down. ↑ under damping    → In case if there is a ↑ in under damping. MF ↑ A ↑ Running life ↓ (i.e. the value   'ses ), the curve shifts Amax (No damping) to the right has side with ↑ in MF. The max of M.F in case of under damping At = 1 ω After some time will lie under the lim of <1 ωn CF → 0 → In case of no damping   0  , the X = PI X = Asin(ωt -f) second will become zero, the first dominates the during the whole process x  A  cos  t   which will dash ↑ the M.F & becomes

 2 A  sin   t    2n  2 F SA 1  T n2 x  A 2 sin  t    2 2 2 F0 sin  t  m  A  sin  t   FT  SA 1  n   c  A  sin   t    2 222    2   F0  SA 1      nn    F  T F0

2 2 1  n  222   2   1    nn      If  0   1 n   For any value of  5.2.4 ACTION ISOLATION If 2   1  n  Safety of Surroundings

5.2.5 FLORENCE OF V.I SYSTEM

Performance F Parameter T F o 0 < 훜 < 1 훜 → 0(Best) 1) ↑ under damping  F SA22 CWA  T       2 n 2 CWA  SA 1   cosnt  2 No change SA   n 2 c     2 SA 1 m  n  s  2) Vibration isolation is effective m

 Question  2 1 n A single cylinder vertical petrol engine of  total mass 300 kg is mounted upon a steel n (Advance level of Spring) chassis frame and causes a vertical static 2 deflection of 2 mm. The reciprocating parts of the engine has a mass of 20 kg and move 3) If through a vertical stroke of 150 mm with simple harmonic motion. A dashpot is Damping becomes detrimental [Spring provided whose damping resistance is Favor] (harmful) directly proportional to the velocity and  amounts to 1.5 kN per metre per second. 4) If  2 n Considering that the steady state of Very high level of damping is required vibration is reached ; determine : 1. the amplitude of forced vibrations, when the 5.3 DEATH OF THE SHAFT driving shaft of the engine rotates at 480 r.p.m., and 2. the speed of the driving shaft at which resonance will occur. → Ultrasonic causes the vibration they

have very sir energy Question → Shaft fails due to t vibrations caused by A single cylinder vertical petrol engine of centrifugal f1 & not by the deflection total mass 300 kg is mounted upon a steel C.M will It does not causes the failure. chassis frame and causes a vertical static m y+e ω2 =S.y deflection of 2 mm. The reciprocating parts of the engine has a mass of 20 kg and move 22 me   y s  m   through a vertical stroke of 150 mm with simple harmonic motion. A dashpot is s m 22 .e  y.m   1 2 provided whose damping resistance is m directly proportional to the velocity and e amounts to 1.5 kN per metre per second. y 2 n Considering that the steady state of 1 vibration is reached ; determine :  1. the amplitude of forced vibrations, when the driving shaft of the engine rotates at 480 r.p.m., and 2. the speed of the driving shaft at which resonance will occur. Solution : 1. Amplitude of the forced vibrations We know that stiffness of the frame, 3 s m.g / δ  300  9.81/ 2  10 1.47 106 N / m Since the length of stroke (l) =150mm = 0.15 m, therefore radius of crank, r = l/2 = 0.15/2 = 0.075 m We know that centrifugal force due to the reciprocating parts or the static force, 22 F mI .ω .r 20(50.3) 0.075

 3795N ∴ Amplitude of the forced vibration (maximum), F x  max 2 c.2ω 2  s mω2  3795  2 2 6 2 2 1500  (50.3) 1.47  10  300(50.3) 3795  5.7 1099  500  10 3795  5.3  103 m 710 103  5.3mm

2. Speed of the driving shaft at which the resonance occurs Let N=Speed of the driving shaft at which the resonance occurs in r.p.m. We know that the angular speed at which the resonance occurs , s 1.47 106 ωω   n m 300  70rad / s

6.1 Introduction: Cams are mechanical devices which are used to generate curvilinear or irregular motion of mechanical elements. They are used to convert rotary motion into oscillatory motion or oscillatory motion into rotary motion. There are two links namely the cam itself which acts as an input member. The other link that acts as 6.2 Classification of cams an output member is called the follower. 1. Wedge and Flat Cams The cam transmits the motion to the follower by direct contact. In a cam- A wedge cam has a wedge of follower pair, the cam usually rotates specified contour and has while the follower translates or translational motion. The oscillates. follower can either translate or oscillate. A spring is used to Cams are widely used in internal maintain the contact between the combustion engines, machine tools, cam and the follower. printing control mechanisms, textile weaving industries, automated machines etc. Necessary elements of a cam mechanism are: • A driver member known as the cam • A driven member called the follower • A frame which supports the cam and guides the follower

2. Plate cam

In this type of cams, the follower moves in a radial direction from the centre of rotation of the cam. They are also known as radial or disc cam. The follower

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission reciprocates or oscillates in a  line of motion of followers plane normal to the cam axis. (I). Classification based on type of surface contact between cam and follower

Figure: 5 Types of follower based on

Figure: 3 Plate cam the surface in contact 1. Knife edge follower 3. Cylindrical cam The contacting end of the follower has a Here a cylinder has a circumferential sharp knife edge. A sliding motion exists contour cut in the surface and the cam between the contacting cam and follower rotates about its axis. The follower surfaces. It is rarely used in practice motion is either oscillating or because the small area of contacting reciprocating type. These cams are also surface results in excessive wear. called drum or barrel cams. 2. Roller follower It consists of a cylindrical roller which rolls on cam surface. Because of the rolling motion between the contacting surfaces, the rate of wear is reduced in comparison with Knife edge follower. The roller followers are extensively used where more space is available such as gas and oil engines.

3. Flat face follower Figure: 4 cylindrical cam The follower face is perfectly flat. It experiences a side thrust due to the 6.3 Classification of followers: friction between contact surfaces of follower and cam. Followers can be classified based on 4. Spherical face follower The contacting end of the follower is of  type of surface contact spherical shape which overcomes the between cam and follower drawback of side thrust as experiences  type of follower motion by flat face follower.

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission (II). Classification based on followers' 1. Radial follower motion The line of movement of the follower passes through the center of the camshaft (Figure: 7 a). 2. Offset follower The line of movement of the follower is offset from the center of the cam shaft (Figure: 7 b).

3. Force closed cam follower system In this type of cam-follower system, an external force is needed to maintain the Figure: 6 Classification of follower contact between cam and follower. based on motion Generally a spring maintains the contact 1. Oscillating follower between the two elements. The follower can be a oscillating type (Figure: 8 a) or In this configuration, the rotary motion of translational type (Figure: 8 b). of the cam is converted into predetermined oscillatory motion of the follower as shown in Figure: 6 a 2. Translating follower These are also called as reciprocating follower. The follower reciprocates in the ‘guide' as the cam rotates uniformly as shown in Figure: 6 b.

(III). Classification based on line of Figure: 8 Force closed cam followers motion 4. Form closed cam follower system

In this system a slot or a groove profile is cut in the cam. The roller fits in the slot and follows the groove profile. These kind of systems do not require a spring. These are extensively used in machine tools and machinery. The follower can be an oscillating type or a translating type.

Figure: 7 Classification of follower based on line of motion

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission motion of tappet against the spring. This translatory motion is used to open or close the valve.

Figure: 9 Form closed cam followers Note: Three-dimensional cam or Camoid Camoid is a combination of radial and axial cams. It has three dimensional surface and two degrees-of-freedom. Two inputs are rotation of the cam about its axis and translation of the cam along its axis. Follower motion is based on both the inputs. Figure :10 shows a typical Camoid. 6.5 Terminology of Cam and Follower

6.4 Applications of cams Figure: 12 Terminology of Cam and Cams are widely used in automation of Follower machinery, gear cutting machines, screw machines, printing press, textile The Cam Profile: industries, automobile engine valves, The working contour of a cam which tool changers of machine centers, comes into contact with the follower to conveyors, pallet changers, sliding fork operate it, is known as the cam profile. in wearhouses etc. Cams are also used in I.C engines to The Base Circle: operate the inlet valves and exhaust The smallest circle, drawn from the valves. The cam shaft rotates by using centre of rotation of a cam, which forms prime moveres. It causes the rotation of part of the cam profile, is known as the cam. This rotation produces translatory base circle and its radius is called the

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission least radius of the cam. A circle with to the highest position is called the angle centre of ascent denoted as θ1 . O and of radius OA forms the base circle. The angle covered by the cam during Size of a cam depends upon the size of which the follower remains at rest at its the base circle. highest position is called the angle of

The Tracing Point: dwell, denoted by θ2 . The point of the follower from which the The angle covered by the cam, for the profile of a cam is determined is called follower to fall from its highest position the tracing point. In case of a knife - edge to the lowest position is called the angle follower, the knife edge itself is the of descent denoted as θ3 . tracing point. In roller follower, the centre of roller is the tracing point. The total angle moved by the cam for the follower to return to its lowest position The Pitch Curve: after the period of ascent, dwell and The locus or path of the tracing point is descent is called the angle of action. It is

known as the pitch curve. In knife-edge the sum of θθ12, and θ 3. follower, the pitch curve itself will be the cam profile. In roller follower, the cam The Pressure Angle: profile will be determined by subtracting The angle included between the normal the radius of the roller radially to the pitch curve at any point and the throughout the pitch curve. line of motion of the follower at the point, The Prime Circle: is known as the pressure angle. This angle represents the steepness of the The smallest circle drawn to the pitch cam profile and as such it is very curve from the centre of rotation of the important in cam design. cam is called as the prime circle. In knife - edge follower, the base circle and the The Pitch Point: prime circle are the same. In roller The point on the pitch curve having the follower, the radius of the prime circle is maximum pressure angle is known as the the base circle radius plus the radius of pitch point. the roller. The Cam Angle: The Lift or Stroke: It is the angle of rotation of the cam for a It is the maximum displacement of the certain displacement of the follower. follower from the base circle of the cam. It is also called as the throw of the cam. 6.6 DERIVATIVES OF FOLLOWER In Figure: 12, distance B’B and C’C is the MOTION lift, for the roller follower. The derivatives of follower motion can The Angles of Ascent, Dwell, Descent be kinematic (with respect to θ ) which and Action relate to the geometry of the cam system or physical (with respect to time) which The angle covered by a cam for the relate to the motion of the follower of the follower to rise from its lowest position cam system.

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Kinematic Derivatives The next derivative can also be taken if desired. A displacement diagram of the follower motion is plotted with the cam angle θ as ds3 f '''()θ= the abscissa and the follower linear or dθ3 angular motion as the ordinate. Thus, it is a graph that relates the input and the It is not easy to describe it geometrically. output of the cam system. However, it should also be committed as Mathematically, if 's' is the displacement far as possible while choosing the shape of the follower then of the displacement diagram for smooth working of the cam. sf=() θ Physical Derivatives where, We have, s= f()() θ and θ= g t θ= Cam angle rotation in radians. However, since the cam rotates at a Taking the first derivative with respect constant angular velocity, to time, • ds ds dθ ds Differentiating it with respect to 9 s ==×=ω provides the first derivative. dt dθθ dt d

ds Which represents the velocity of the f'()θ= dθ follower It represents the slope or the steepness The second derivative is

of the displacement curve at each 22 •• ds ds position of the cam angle. s = = ω2 dt22 dθ A higher value of this means a steep rise It represents the acceleration of the or fall which hampers the smooth follower. A higher value of acceleration running of the cam. means a inertia force. The second derivatives is represented by A third derivatives is known as the jerk 2 ds ••• 33 f ''()θ= ds3 ds dθ2 s = = ω dt33 dθ This derivative is related to the radius of For smooth movement of the follower, curvature of the cam at different points even the high values of the jerk are along its profile and is in inverse undesirable in case of higher speed cams. proportion. Thus, with an increasing in its value, the radius of curvature Note: decreases. If its value becomes infinity, The follower curves can be studied in the cam profile becomes pointed at the terms of the simple diagram shown in position which is undesirable from the Figure: point of view of stresses between the cam and follower surfaces. A general displacement diagram will be made up of three or more parts:

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 1. Rises (1 or more), 2. Returns (1 or more), and 3. Dwells (0 or more). Both the rise and return parts will contain one or more inflection points. These are points where a maximum slope is reached, and they correspond to points on the cam surface with maximum steepness. These points are identified by the locations where the curvature of the diagram changes sign. At the inflection points, the radius of curvature of the curve is infinite. FIGURE:14 Displacement, velocity, and acceleration relations for uniform motion. For a constant velocity curve

let s= C01 +θ C

atθ= 0,s = 0 ⇒ C0 = 0 h atθ=β ,s = h ⇒ C = 1 β FIGURE: 13 Terminology used when h discussing follower- displacement ∴=θDisplacement ()s β programs. • hω Velocity s = = a constant   β •• 6.7 Motions of followers: Acceleration s= 0  Symbols: We see that the displacement is uniform, 1. Uniform motion (constant velocity) the velocity is constant, and the The constant velocity or uniform acceleration is zero during the rise. At displacement is the simplest of all. It has the ends where the dwell meets the a straight-line displacement at a constant curve, however, we have an impractical slope (Fig. 2.2) giving the smallest length condition. That is, as we go from the for a given rise of all the curves. When the dwell (zero velocity) to a finite velocity, straight-line curve is developed for a we have an instantaneous change in radial cam, it becomes the Archimedes velocity, giving a theoretically infinite spiral. acceleration. This acceleration transmits high shock throughout the follower linkage, the magnitude of which depends on its flexibility. 2. Constant Acceleration and Deceleration (Parabolic)

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission In such a follower programme, there is primarily used for low-speed systems if acceleration in the first half of the at all. follower motion whereas it is Simple harmonic Motion (SHM): deceleration during the latter half. The displacement curve is found to be This motion (having a cosine parabolic in this case. The magnitude of acceleration curve) is a very popular the acceleration and the deceleration is choice in combination with other motion. the same and constant in the two halves. In Fig. 2.8, the projection of a radius point P starting at point O moves vertically at point Q along the diameter h of the circle with simple harmonic motion

FIGURE:15Displacement, velocity, acceleration, and jerk relations for parabolic motion during rise. FIGURE:16 Simple harmonic motion The displacement of the first half of curve construction. motion of its symmetrical motion, the Let positive acceleration period is h φ=angle of rotations with radius For a constant acceleration curve 2 β let s =C θ2 for 0 ≤θ≤ The basic harmonic motion 2 displacement function is β h 2h atθ= ,s = ⇒ C = 22 β2 h s=() 1 −φ cos 2 2h ∴=θDisplacement() s 2 β2 Displacements are taken at angular ••4hω 2h ωβ increments moving through angle φ the = θ⇒  = θ= Velocitys2  s at ββ max  2  same increments along the displacement •• 4hω2 curve. The relationship between angle φ , Acceleration s= = a constant   β2 the generating circle, and the cam angle φπ ••• θ is = Jerk s0= , except at the changes in the θβ acceleration where it equals infinity. Since the parabolic curve has discontinuities in the acceleration at the dwell ends and the transition point, it is

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission FIGURE:18 Displacement, velocity, FIGURE:17 Displacement, velocity, and acceleration relations for and acceleration relations for SHM cycloidal motion.

motion. For a Cycloidal motion curve For a Simple harmonic motion curve θ12 πθ h πθ Displacement() s= h − sin Displacement() s = 1− cos  2 β βπ2 β •• hhπω πθ  πω β • Velocitys= sin ⇒  s = at θ=  hω 2πθ 2βββ  22 max Velocity s = 1− cos ••hhπω22πθ  •• πω 22  ββ Acceleration s= cos⇒ s =() at θ= 0  ββ22β  22max • 2hωβ ••• hπω33 πθ ⇒s = at θ= Jerk s = sin   2β3 β max β 2 •• 22hπω2 πθ = The simple harmonic motion has a Acceleration s 2 sin smooth acceleration curve except at the  ββ •• 2hπω2  β dwell ends, where it has discontinuities ⇒ = θ= s 2 at (i.e., infinite jerk). max β 4 ••• 42hπ23ω πθ CYCLOIDAL MOTION : Jerk s= cos  β3 β The cycloidal motion curve, as the name implies, is generated from a cycloid and has a sine acceleration curve. A cycloid is the locus of a point on a circle that is rolled on a straight line. The circumference of the circle is equal to the follower rise h. The cycloid is the first dwell-rise-dwell curve we discuss that does not have any discontinuities in the acceleration curve. Therefore, it can be applied to higher speeds than the foregoing curves even though its maximum acceleration is higher in some cases.

7.1 INTRODUCTION: movement of the sleeve and it reduces the supply. Thus, the energy input (fuel Flywheel minimizes fluctuations of supply in IC engines, steam in steam speed within the cycle but it cannot turbines and water in hydraulic minimize fluctuations due to load turbines) is adjusted to the new load on variation. This means flywheel does not the engine. Thus the governor senses the exercise any control over mean speed of change in speed and then regulates the the engine. To minimize fluctuations in supply. the mean speed which may occur due to load variation, governor is used. The governor has no influence over cyclic speed fluctuations but it controls the mean speed over a long period during which load on the engine may vary. When there is change in load, variation in speed also takes place then governor operates a regulatory control and adjusts the fuel supply to maintain the mean Figure: 1 Governor and Linkages speed nearly constant. Therefore, the governor automatically regulates 7.2 CLASSIFICATION OF GOVERNORS through linkages, the energy supply to The broad classification of governor can the engine as demanded by variation of be made depending on their operation. load so that the engine speed is maintained nearly constant. (a) Centrifugal governors The governor shaft is rotated by the (b) Inertia and flywheel governors engine. If load on the engine increases (c) Pickering governors. the engine speed tends to reduce, as a result of which governor balls move Centrifugal Governors: inwards. This causes sleeve to move In these governors, the change in downwards and this movement is centrifugal forces of the rotating masses transmitted to the valve through due to change in the speed of the engine linkages to increase the opening and, is utilized for movement of the governor thereby, to increase the supply. sleeve. One of this type of governors is On the other hand, reduction in the load shown in Figure: 1.These governors are increases engine speed. As a result of commonly used because of simplicity in which the governor balls try to fly operation. outwards. This causes an upward

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Inertia and Flywheel Governors: governors are comparatively smaller in size. In these governors, the inertia forces caused by the angular acceleration of the 7.3 GRAVITY CONTROLLED engine shaft or flywheel by change in CENTRIFUGAL GOVERNORS: speed are utilized for the movement of There are three commonly used gravity the balls. The movement of the balls is controlled centrifugal governors: due to the rate of change of speed instead of change in speed itself as in case of (a) Watt governor centrifugal governors. Thus, these (b) Porter governor governors are more sensitive than centrifugal governors. (c) Proell governor Pickering Governors: Watt governor does not carry dead weight at the sleeve. Porter governor and This type of governor is used for driving proell governor have heavy dead weight a gramophone. As compared to the at the sleeve. In porter governor balls are centrifugal governors, the sleeve placed at the junction of upper and lower movement is very small. It controls the arms. In case of proell governor the balls speed by dissipating the excess kinetic are placed at the extension of lower energy. It is very simple in construction arms. The sensitiveness of watt governor and can be used for a small machine. is poor at high speed and this limits its 7.2.1 Types of Centrifugal Governors field of application. Porter governor is more sensitive than watt governor. The Depending on the construction these proell governor is most sensitive out of governors are of two types: these three. (a) Gravity controlled centrifugal 7.3.1 Watt Governor governors, and This governor was used by James Watt in (b) Spring controlled centrifugal his steam engine. The spindle is driven governors. by the output shaft of the prime mover. Gravity Controlled Centrifugal The balls are mounted at the junction of Governors: the two arms. The upper arms are connected to the spindle and lower arms In this type of governors there is gravity are connected to the sleeve as shown in force due to weight on the sleeve or Figure: 2(a). weight of sleeve itself which controls movement of the sleeve. These governors are comparatively larger in size. Spring Controlled Centrifugal Governors: In these governors, a helical spring or several springs are utilized to control the movement of sleeve or balls. These Figure: 2 Watt Governor

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission We ignore mass of the sleeve, upper and Let m be the mass of the ball, Governors T and T be tension in upper and lower arms, respectively, lower arms for simplicity of analysis. We 12 F be the centrifugal force, can ignore the friction also. The ball is C subjected to the three forces which are r be the radius of rotation of the ball from axis, and centrifugal force (Fc), weight (mg) and I is the instantaneous centre of the lower arm. tension by upper arm (T). Taking M be the mass of the sleeve moment about point O (intersection of arm and spindle axis), we get

FC h−= mg r 0 g ⇒mr ω22 h − mgr = 0 ⇒ω = h 895 2πN ⇒h = ω= N62 0 where, N is in rpm.

Figure: 3 shows a graph between height ‘h’ and speed ‘N’ in rpm. At high speed the Figure: 4 Porter Governor change in height h is very small which Taking moment of all forces acting on the indicates that the sensitiveness of the ball about I and neglecting friction at the governor is very poor at high speeds sleeve, we get because of flatness of the curve at higher speeds. Mg F× AD − mg ×− ID ×= IC 0 C 2

2 ID Mg  ID+ DC  ⇒mr ω= mg × + ×   AD 2  AD  ID r DC tanα= = & tan β= AD h AD tan β let K = tan α g 2mg++ Mg() 1 K ∴ω2 =  Figure: 3 Graph between Height and h 2mg Speed 895 2mg++ Mg() 1 K ⇒=h 7.3.2 Porter Governor 2  N 2mg A schematic diagram of the porter 2Nπ ω= governor is shown in Figure: 4(a). There 60 are two sets of arms. The top arms OA and OB connect balls to the hinge O. The Note: hinge may be on the spindle or slightly If friction at the sleeve is f, the force at the away. The lower arms support dead sleeve should be replaced by ()Mg+ f weight and connect balls also. All of them rotate with the spindle. We can consider for rising and by ()Mg− f for falling one-half of governor for equilibrium. speed as friction opposes the motion of sleeve. Therefore, if the friction at the sleeve is to be considered, W should be

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission replaced by ()Mg± f . The expression For equilibrium of governor, taking becomes moments about point I, ()Mg± f 895 2mg+()() Mg ± f 1+ K mrω×2 d − mg a + r − r − a + b = 0 ⇒= 11() () h2  ...... () 1 2 N 2mg where, dimensions a, b, c and d are 7.4 PROELL GOVERNOR shown in Figure : 6. In Proell governor, the rotating balls are If the position of extended portion BF is mounted on an extension of the lower vertical (i.e. rr= ), then arm of the 1

2 c a()Mg± f  ab  Porter governor as shown in Figure: 5. mrω= mg  +  +  d c 2  cc  The lower arm CBF is a one piece arm  ar b bent at B. tanα= = & tan β= ch c tanβ Under the normal conditions, the let K = α extended portion of arm BF remains tan 2 gc2mg++ Mg() 1 K vertical: however it changes position ∴ω =  h d 2mg with variation in speed. The forces acting  895 c 2mg+()() Mg ±+ f 1 K on the proell governor are shown in ⇒=h ...... (2) N2  d 2mg figure: 6. Assuming point I as  2Nπ instantaneous centre of rotation of the ω= 60 arm CBF. Comparing Eqn. (1) and (2), we find that for given values of ball mass m, sleeve mass M and height of governor h, the effect of placing balls at point F in Proell governor instead of at point A in Porter governor is to reduce the equilibrium speed of Proell governor. In other words, for the same height, Proell governor requires mass of smaller size compared to Porter governor. Figure: 5 Proell governor

7.5 SPRING CONTROLLED CENTRIFUGAL GOVERNORS

In these governors springs are used to counteract the centrifugal force. They can be designed to operate at high speeds. They are comparatively smaller in size. Their speed range can be changed FIGURE: 6 Forces acting on Proell by changing the initial setting of the Governor spring. They can work with inclined axis

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission of rotation also. These governors may be r2 = Maximum radius of rotation of ball very suitable for IC engines, etc. centre from spindle axis, in m,

The most commonly used spring F = Spring force exerted on sleeve at controlled centrifugal governors are : S1 minimum radius, in N, (a) Hartnell governor FS2 = Spring force exerted on sleeve at (b) Wilson-Hartnell governor maximum radius, in N, (c) Hartung governor m = Mass of each ball, in kg, Hartnell governor: M = Mass of sleeve, in kg,

The Hartnell governor is shown in N = Minimum speed of governor at Figure: 7. The two bell crank levers have 1 minimum radius, in rpm, been provided which can have rotating motion about fulcrums O and O'. The N2 = Maximum speed of governor at frame is connected to the spindle. A maximum radius, in rpm, helical spring is mounted around the

spindle between frame and sleeve. With ω1 and ω2 = Corresponding minimum the rotation of the spindle, all these parts and maximum angular velocities, in rotate. rad/s, With the increase of speed, the radius of ()FC = Centrifugal force corresponding rotation of the balls increases and the 1 to minimum speed = mr ω 2 rollers lift the sleeve against the spring 11 force. With the decrease in speed, the ()F = Centrifugal force corresponding sleeve moves downwards. The C 2 to maximum speed = 2 movement of the sleeve are transferred mr22ω to the throttle of the engine through linkages. S = Stiffness of spring or the force required to compress the spring by one m, r = Distance of fulcrum O from the governor axis or radius of rotation, a = Length of ball arm of bell-crank lever, i.e. distance OA, and b = Length of sleeve arm of bell-crank lever, i.e. distance OC. Figure: 7 Hartnell Governor

Let r1 = Minimum radius of rotation of ball centre from spindle axis, in m,

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Total lift=()() x12 + x = b θ+ 1 b θ 2 ()rr−−() r r =bb() θ +θ =12 + 12 aa b =()rr − a 21 b F− F = Total lift ×= S() r − r S S2 S1a 2 1 2 a {}()()FFCC− stiffness of spring() S= 2 21 Figure: 8 Forces acting on Hartnell b() rr21− Governor 7.6 GOVERNOR EFFORT AND POWER Considering the position of the ball at Governor effort and power can be used radius ‘ r ’, as shown in Figure: 8(a) and 1 to compare the effectiveness of different taking moments of all the forces about O type of governors. ()Mg+ F = θ− θ−S1 θ= MoC() F1 a cos 1 mg a sin 1 b cos 1 0 Governor Effort 2 ()Mg+ F b ⇒ = θ+ S1 It is defined as the mean force exerted on ()FC11 mg tan .....(1) 2a the sleeve during a given change in Considering the position of the ball at speed.

radius ‘ r2 ’ as shown in Figure: 8(b) and When governor speed is constant the net taking the moments of all the forces force at the sleeve is zero. When about O' governor speed increases, there will be a net force on the sleeve to move it ()Mg+ F M '= F a cos θ+ mg a sin θ−S2 b cos θ= 0 oC()2 2 2 2 upwards and sleeve starts moving to the 2 new equilibrium position where net ()Mg+ FS2 b ⇒=()FC2 − mg tan θ ...... () 2 2 2a force becomes zero.

If θ1 and θ2 are very small and mass of Governor Power the ball is negligible as compared to the It is defined as the work done at the

spring force, the terms mg tan θ1 and sleeve for a given change in speed. Therefore, mg tan θ2 may be ignored.

+ Power of  Displacement  ()Mg FS1 b =Governor effort ×   ⇒=()FC  .....(3) governor of sleeve 1 2a    and Note: ()Mg+ F b = S2 C = A factor which when multiplied to ()FC 2  ...... () 4 2a equilibrium speed, gives the increase in speed. ()FFS2− S1 b ∴−=()()FFCC  21 2a Porter Governor Effort C ={}2mg ++ Mg(1 K) ()1K+

Hartnell Governor Effort =C{} Mg ± Fs

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 7.7 CHARACTERISTICS OF Isochronism GOVERNORS A governor is said to be isochronous if Different governors can be compared on equilibrium speed is constant for all the the basis of following characteristics: radii of rotation in the working range. Therefore, for an isochronous governor Stability the speed range is zero and this type of A governor is said to be stable when governor shall maintain constant speed. there is one radius of rotation of the balls Hunting for each speed which is within the speed range of the governor. Whenever there is change in speed due to the change in load on the engine, the Sensitiveness Governors sleeve moves towards the new position The sensitiveness can be defined under but because of inertia if overshoots the the two situations: desired position. Sleeve then moves back but again overshoots the desired (a) When the governor is considered as a position due to inertia. This results in single entity. setting up of oscillations in engine speed. (b) When the governor is fitted in the If the frequency of fluctuations in engine prime mover and it is treated as part of speed coincides with the natural prime mover. frequency of oscillations of the governor, this results in increase of amplitude of (a) A governor is said to be sensitive oscillations due to resonance. The when there is larger displacement of the governor, then, tends to intensity the sleeve due to a fractional change in speed variation instead of controlling it. speed. Smaller the change in speed of the This phenomenon is known as hunting of governor for a given displacement of the the governor. Higher the sensitiveness of sleeve, the governor will be more the governor, the problem of hunting sensitive. becomes more acute.  N1NN12+ 7.8 CONTROLLING FORCE AND ∴==Sensitiveness  NN12−− 2NN 12 STABILITY OF SPRING CONTROLLED GOVERNORS (b) The smaller the change in speed from no load to the full load, the more The resultant external force which sensitive the governor will be. According controls the movement of the ball and to this definition, the sensitiveness of the acts along the radial line towards the axis governor shall be determined by the is called controlling force. This force acts ratio of speed range to the mean speed. at the centre of the ball. It is equal and The smaller the ratio more sensitive the acts opposite to the direction of governor will be centrifugal force. The controlling force () F= mr ω2 NN12−− NN 12 ∴==Sensitiveness 2 2 N NN12+ F 2Nπ ⇒=m r 60 where N21− N = Speed range from no load to full load.

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission For controlling force diagram in which ‘F’ a stable governor must intersect the F controlling force axis (i.e. y-axis) below is plotted against radius ‘r’, represents r the origin, when produced. Then the slope of the curve. equation of the curve will be of the form

F Fb i.e. =tan φα N2 F=−=− ar b or a r rr F Therefore, for a stable governor slope in As r increases increase and thereby controlling force diagram should r increase with the increase in speed. tan φ increases. Therefore, this equation represents stable governor. Stability of Spring-controlled Governors (b) If b in the above equation is zero then The controlling force curve is the controlling force curve OC will pass F approximately straight line for spring through the origin. The ratio will be controlled governors. As controlling r force curve represents the variation of constant for all radius of rotation and controlling force ‘F’ with radius of hence the governor will become rotation ‘r’, hence, straight line equation isochronous. Hence for isochronous, the can be, equation will be F F= ar +b;F = ar or F = ar − b F= ar or = a r Where a and b are constants. In the above equation b may be +ve, or –ve or (c) If b is positive, then controlling force zero. curve AB will intersect the controlling force axis (i.e. y-axis) above the origin. The equation of the curve will be

F Fb =ar +b or = a + r rr F As r increases, speed increases, or r tan φ reduces. Hence this equation cannot represent stable governor but unstable governor. 7.9 INSENSITIVENESS IN THE Figure: 9 Stability of GOVERNORS Spring Controlled Governors The friction force at the sleeve gives rise These three cases are as follows: to the insensitiveness in the governor. At (a) We know that for a stable governor, any given radius there will be two different speeds one being when sleeve F the ratio must increase as r increases. moves up and other when sleeve moves r down. Figure: 10 shows the controlling Hence the controlling force curve DE for force diagram for such a governor.

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Figure: 10 Insensitiveness in the Governors The corresponding three values of speeds for the same radius OA are: (a) The speed N when there is no friction.

(b) The speed N'when speed is increasing or sleeve is on the verge of moving up, and

(c) The speed N"when speed is decreasing or sleeve on the verge of moving down. This means that, when radius is OA, the speed of rotation may vary between the limits N' and N"without causing any displacement of the governor sleeve. The governor is said to be insensitive over this range of speed. Therefore,

N'− N" ∴=Coefficient of insensitiveness  N

1. DISPLACEMENT, VELOCITY AND ACCELERATION

2. PLANER MECHANISMS 90

3. CAMS 100

4. DYNAMIC ANALYSIS OF SLIDER CRANK 113

5. GEAR AND GEAR TRAIN 117

6. FLYWHEEL 121

7. VIBRATION 131

8. BALANCING 135

9. GYROSCOPE 157

159

velocity of link 1 with respect to Common Data for Q.1 and Q.2: The circular disc shown in its plan view in link 2 is 12 m/sec. Link 2 rotates at the figure rotates in a plane parallel to the a constant speed of 120 rpm. The horizontal plane about the point O at a magnitude of Coriolis component of uniform angular velocity ω . Two other acceleration of link 1 is points A and B are located on the line OZ

at distances rA and r B from O respectively.

a) 320 m/s2 b) 604 m/s2 Q.1 The velocity of Point B with respect c) 906 m/s2 d) 1208 m/s2 to point A is a vector of magnitude [GATE-2004] a) 0 Q.4 A link OB is rotating with a b) ω (rB- rA) and direction constant angular velocity of 2 rad/s opposite to the direction of in counter clockwise direction and a motion of point B block is sliding radically outward on c) ω (rB- rA) and direction same as it with an uniform velocity of 0.75 the direction of motion of point m/s with respect to the rod as B shown in the figure. If OA=1 m, the

d) ω(rB- rA) and direction being magnitude of the absolute from O to Z acceleration of the block at location [GATE-2003] a in m/s2 is

Q.2 The acceleration of point B with respect to point A is a vector of magnitude a) 0

b) ω (rB- rA) & direction same as the direction of motion of point a) 3 b) 4 B c) 5 d) 6 2 [GATE-2013] c) ω (rB-rA) & direction opposite to be direction of motion of Q.5 A rigid link PQ of length 2 m rotates point B about the pinned end Q with a 2 d) ω (rB-rA) and direction being constant angular acceleration of from Z to O 12 rad / s2 .When the angular velocity [GATE-2003] of the link is 4 rad/s, the magnitude of the resultant acceleration Q.3 In the figure shown, the relative ()in m / s2 of the end P is ______

constant angular velocity ω2 A slider link 3 moves outwards with a

constant relative velocity VQ/P , where Q is a point on slider 3 and P is a point on link 2. The magnitude and direction of Coriolis’s component of acceleration is given by [GATE-2016(2)] Q.8 A slider crank mechanism with crank radius 200 mm and connecting rod length 800 mm is shown. The crank is rotating at 600 rpm in the counter clockwise direction. In the configuration shown, the crank makes an angle of 90o with the sliding direction of the

a) 2Vω2 Q/P ; direction of VQ/P rotated slider, and a force of 5 kN is acting by 90° in the directionω on the slider. Neglecting the inertia 2 forces, the turning moment on the b) ω V ; direction of V rotated 2 Q/P Q/P crank (in kN-m) is _____

by 90° in the directionω2 .

c) 2Vω2 Q/P ; direction of VQ/P rotated by 90° opposite to the direction of ω 2 . [GATE-2016(1)]

d) ω2 VQ/P ; direction of VQ/P rotated by 90° opposite to the direction Q.9 The number of degrees of freedom in a planar mechanism having ω . n 2 links and j simple hinge joints is [GATE-2015(3)] a) 3(n -3) 2j b) 3(n -1) 2j c) 3n- 2j d) 2j-3n +4 The rod AB, of length 1 m, shown in Q.7 [GATE-2016(3)] the figure is connected to two sliders at each end through pins. Q.10 A rigid link PQ is undergoing plane The sliders can slide along QP and motion as shown in the figure (VP QR. If the velocity VA of the slider at and VQ are non-zero). VQP is the A is 2 m/s, the velocity of the relative velocity of point Q with midpoint of the rod at this instant is respect to point P. m/s. Which one of the following is TRUE?

a) VQP has components along and perpendicular to PQ

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission b) VQP has only one component directed from P to Q c) VQP has only one component directed from Q to P d) VQP has only one component perpendicular to PQ [GATE-2016(3)]

Q.11 For an inline slider- crank mechanism, the length of the crank [GATE-2017(2)] and connecting rod are 3m and 4m, respectively. At the instant when the connecting rod is perpendicular to the crank, if the velocity of the slider Q.14 For an inline slider-crank mechanism, is 1m/s, the magnitude of angular the length of the crank and the connecting velocity (up to 3 decimal point rod are 3m and 4m, respectively. At the accuracy) of the crank is ___radian/s instant when the connecting rod is [GATE-2017(1)] perpendicular to the crank, if the velocity of the slider is 1m/s, the magnitude of the Q.12 In a slider – crank mechanism, the angular velocity (up to 3 decimal points lengths of the crank and the accuracy) of the crank connecting rod are 100mm and is______radian/sec 160mm, respectively. The crank is rotating with an angular velocity of 10 radian/s counter clockwise. The magnitude of linear velocity (in m/s) of the piston at the instant corresponding to the configuration shown in the figure is _____ [GATE-2017(2)] Q.15 The rod PQ of length L= 2m and uniformly distribute mass of M=10 kg, is released from rest at the position √as shown in the figure. The end slide along the frictionless faces OP and PQ. Assume acceleration due to gravity g=10m/s2 .The mass moment inertia of the rod about its [GATE-2017(2)] centre of mass and an axis perpendicular to the plane of the figure is (ML2/12).At Q.13 Block 2 slides outward on link 1 at a this instant, the magnitude of angular uniform velocity of 6m/shown in acceleration (in rad/sec2)of the rod the figure. Link 1 rotating at a is______. constant angular velocity of 20 radian/s counter clockwise. The magnitude of the total acceleration (in m/s2) of point P of the block with respect to fixed point O is ___

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission (a) 8 233 (b) 10 225 [GATE-2017(2)] (c) 10 2170 (d) 8 2170 ∠ 0 [GATE∠-20180 (2)] Q.16 A rigid rod of length 1 m is resting at ∠ ∠ an angle 450 as shown in the figure. The end P is dragged with a velocity of U = 5m/s to the right. At the instant shown, the magnitude of the velocity V (in m/s) of point Q as it moves along the wall without losing contact is (a) 5 (b) 6 (c) 8 (d) 10

[GATE-2018(2)]

Q.17 In a rigid body in plane motion, the point R is accelerating with respect to point P at 10 180 m/ . If the instaneous acceleration0 of point2 Q is zero, the acceleration∠ (in m/𝑠𝑠 ) of point R is 2 𝑠𝑠

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 (c) (d) (a) (c) 40 (a) 1 1 (b) (d) 0.28 1 243.31 0.266 7.5

16 17 (a) (d)

Q.1 (c) Q.4 (c) Given the circular disc rotates about Absolute acceleration of given the point O at a uniform angular linked at location A is velocity . 22 ∂=aaradial + tangential ....(i) 22 ω Where ∂==×radial ω r ω OA = 222×= 1 4m / s

∂tangential =2vω =×× 2 0.75 2 = 3m / s2 So that from equation (i), e get ∂=4322 + = 5 m/s Let VA is the linear velocity of point A &υ s is the linear velocity of point B. Q.5 (40) υ A = A and υ A = A Velocity of point B with respect to pointωτ A is given by. ωτ

υBA =υ B −υ A =ωr B −ω r A =ω (r B − r A ) ....(i) From the given figure,

rrB > A α= × = So, ωrBA -ωr a1 R 2 12 24 2 υBA -υ aR =ω= R 16 ×= 2 32 Therefore, relative velocity ω() r -r 22 BA a= aa1R + in the direction of point B. =2422 += 32 40m / s 2 Q.2 (d) Acceleration of point B with respect Q.6 (a) to point a is given by, Direction is obtained by rotating velocity vector through in the α =ωυ =×−=ω ω(r r )ω 2 (r − r ) …(i) BA BABB A A direction of rotation of the link. 90 This equation (i) gives the value of centripetal acceleration which acts Q.7 (1) always BA act towards to O i.e its direction from Z to O So, α Q.3 (a) Given N2 120 rpm , v1 = 12 m /sec. So, coriolis component of the acceleration of link 1 is. 2π× 120 ∂c =2 ων = 2 × × 120 12 1 1 60 = 301.44 m/s2 320 m/s2

Given AB= 2 Q.8 (1) Since Rod AB is rigid, so Axial velocity of A & B should be same

VAB cos60= V cos60

VAB= V = 2m / sec C is mid point of AB T cos = 5 Moment about crank shaft (M) =T × rϕ sin(90 −φ ) =(T cos φ )r =5 × 0.2 =1kN / m Alternate method: Velocity corresponding to pure translation part λ M=×+ 5 0.2 sin(90) sin(180) + 2 =1 kN / m

Q.9 (b) DOF(F)3(n-1)-2j Where, n =total number of links j =Effective number of binary points

Velocity corresponding to Rotational part Q.10 (d)

Vc =1 ⟹ Alternate Method:

(2)2+− (2) 22 x cos120 = 222×× 2 1x −=−1  2 22

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Let V &V make an angle with axis π PQ Here θ= of link PQ respectively. P Q V& V and 2 Since link PQ is rigid, so the distance between P & Q will never change. v= r ω= .1 × 10 = 1m / s Hence relative velocity between P & OR Q along axial direction should be ∴ zero.

VPQ cosα= V sin β r =VP sin α× | PQ | ⊥ to PQ r =VQ sin β× | PQ | ⊥ to PQ r ()VPQ sinα− V sin β | PQ | ⊥ to PQ Relative velocity between P & Q V After plotting I-centers

Here, I and I will come at the same point 23 24

Q.11 (0.28)

Applying angular velocity theorem at sin αθ  v = rω sin θ+ I 2n sin106.26 24 1=ω+ 3 sin 53.13 (I I ) = V = V 2× 1.333 2 24 12 4 B ω=0.287 rad/sec V = ∴(I ωI ) = 10 × 0.1(I I = 100mm = 0.1m = AB) B 2 24 12 24 12 Q.12 (1m/s) ω V = 1m/s

B Q.13 (243.31)

sin 2θ ∵v velocity= r ω sin of θ+ piston is given by 2n

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Where ( ) = = = 1 c =corial is acceleration ( ) = 1 a 2 24 12 4 𝐵𝐵 = ω= × × ∴ 𝜔𝜔 𝐼𝐼 𝐼𝐼 4 𝑉𝑉 𝑉𝑉 2v 2 6 20 = tan2 24 12= 53.13010 c2 𝜔𝜔 𝐼𝐼 𝐼𝐼 a= 240m / sec 3 = 90−1 = 36.869 a = vertical acceleration 𝜃𝜃 � � r 0 =rw22 = .1 × (20) 𝛼𝛼 − 𝜃𝜃 2 ar = 40m / sec

2C2 ∴ α=Traa +() =4022 + (240)

aT = 243.31rad / sec

3 Q.14 (0.266) cos = 3 = 𝛼𝛼 =12 3.247499 cos 36.869𝐼𝐼 𝐼𝐼

𝐼𝐼12𝐼𝐼24 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 Now , (3.7499) = 1 1 = 2 = 0.266 / 𝜔𝜔(3.7499) Here = 1 / 𝜔𝜔2 𝑟𝑟𝑟𝑟𝑟𝑟 𝑠𝑠𝑠𝑠𝑠𝑠 Let us draw I-centres 𝐵𝐵 =? Q.15 (7.5) 𝑣𝑣 𝑚𝑚 𝑠𝑠 L= 2 𝜔𝜔𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 M=10kg g =√ 10 m/s2 Both the surface are smooth = = sin 3g = × sin45 θ̈ α× 2l θ ( = 34510) 0 α 2 √2 0 θ = × × = = 7.5 rad/sec2 × 3 10 1 30 Q.16 (a) α 2 √2 √2 4

Applying angular velocity theorem at Treating like elliptical trammel

𝐼𝐼24

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Rod motion (P,Q) (By sitting on So, 180 + 36.8698 = 216.8698 I ) So, answer is 8 217 0 0 V V 0 13 = ∠ I PP I QQ 135 V13 = 1 1 Q 2 2

V√ = 5 m/s√

Q.17 (d) Q

As acceleration of point Q is zero, so this rigid body PQR is hinged at Q

a = a a is given 10 m/s2 at angle of 1800,that �⃗RPmeans�⃗R − �only⃗P radial acceleration is there hence at that instant a = (RP) = 10 (20) =210 RP ω1 =2 ω 2 as of whole ωbody remain same so point R has only radial√ acceleration at thatα instant a = (QR) 1 2 = (16R ) = 8 m/s 2 ω and will be in horizontal 2backward direction ,but our reference is PR So the angle of it from reference is (180 + ). From PQR tan = θ = 36.869812 ∆ θ 16 0 θ

Q1. The mechanism used in a shaping Q.4 Match the following with respect to machine is spatial mechanisms. Types of Joint a) A closed 4-bar chain having 4 P. Revolute revolute pairs Q. Cylindrical b) A closed 6-bar chain having 6 R. Spherical revolute pairs c) A closed 4-bar chain having 2 Degree of constraints revolute and 2 sliding pairs 1. Three d) An inversion of the single slider- 2. Five crank chain 3. Four [GATE-2003] 4. Two 5. Zero Q2. The lengths of the links of a 4-bar a) P-1 Q-3 R-3 b) P-5 Q-4 R-3 linkage with revolute pairs are p, q, c) P-2 Q-3 R-1 d) P-4 Q-5 R-3 r, and s units. given that p < q < r < [GATE-2004] s . Which of these links should be the fixed one, for? obtaining a Q.5 The figure below shows a planar "double crank" mechanism? mechanism with single degree of freedom. The instant centre 24 for a) link of length p the given configuration is located at b) link of length q a position c) link of length r d) link of length s [GATE-2003]

Q.3 Match the following Type of Mechanism P. Scott-Russel Mechanism Q. Geneva Mechanism R. Off-set slider-crank Mechanism S. Scotch Yoke Mechanism a) L b) M c) N d) ∞ Motion achieved [GATE-2004] 1. Intermittent Motion 2. Quick return Motion Common Data for Q.6, Q.7, and Q.8 3. Simple Harmonic Motion 4. Straight Line Motion An instantaneous configuration of a four- bar mechanism, whose plane is horizontal a) P-2 Q-3 R-1 S-4 is shown in the figure below. At this b) P-3 Q-2 R-4 S-1 instant, the angular velocity and angular c) P-4 Q-1 R-2 S-3 acceleration of link O2 A are ω = 8 rad/s & d) P-4 Q-3 R-1 S-2 =0, respectively, and the driving torque ( [GATE-2004] τ ) is zero. The link O2 A is balanced so that α its centre of mass falls at O2

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission the three moving links will rotate by 360 if a) S + L P + Q b) S + L> P + Q c) S + P L + Q d) S + P>L + Q ≤ [GATE-2006] ≤ Q.11 Match the items in columns I and II Column I P. Higher Kinematic pair Q. lower kinemation pair Q.6 Which kind of 4-bar mechanism is R. Quick return mechanism O2ABO4? S. Mobility of a linkage a) Double-crank mechanism Column II b) Crank-rocker mechanism 1. Grubler's Equation c) Double-rocker mechanism 2. Line contact d) Parallelogram mechanism 3. Euler's Equation [GATE-2005] 4.Planar 5. Shaper Q.7 At the instant considered, what is 6. surface contact the magnitude of the angular a) P-2, Q-6, R-4, S-3 velocity of O4B? b) P-6, Q-2, R-4, S-1 a) 1 rad/s b) 3 rad/s c) P-6, Q-2, R-5, S-3 64 d) P-2, Q-6, R-5, S-1 c) 8 rad/s d) rad/s 3 [GATE-2006] [GATE-2005] Q.12 The number of inversion for a slider Q.8 At the same instant, if the crank mechanism is component of the force at joint A a) 6 b) 5 along AB is 30 N, then the c) 4 d) 3 magnitude of the joint reaction at [GATE-2006] O2 a) is zero Q.13 Match the item in columns I and II b) is 30 N Column I c) is 78 N P. Addendum d) Cannot be determined from the Q. Instantaneous centre of velocity given data R. Section modulus [GATE-2005] S. Prime circle Column II Q.9 The number of degrees of freedom 1. Cam of a planar linkage with 8 links and 2. Beam 9 simple revolute joints is 3. Linkage S. Prime circle a) 1 b) 2 4. Gear c) 3 d) 4 a) P-4, Q-2, R-3, S-1 [GATE-2005] b) P-4, Q-3, R-2, S-1 c) P-3, Q-2, R-1, S-4 Q.10 In a four-bar linkage, S denotes the d) P-3, Q-4, R-1, S-2 shortest link length, L is the longest [GATE-2006] link length, P and Q are the lengths of other two links. At least one of Common Data for Q.14 and Q.15

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission A quick return mechanism is shown below. [GATE-2007] The crank OS is driven at 2 rev/s in Q.17 A planar mechanism has 8 links and counter-clockwise direction. 10 rotary joints. The number of degrees of freedom of the mechanism, using Gruebler's criterion, is a) 0 b) 1 c) 2 d) 3 [GATE-2008]

Q.18 Match the approaches given below to perform stated kinematics/dynamics analysis of machine. Analysis If the quick return ratio is 1 : 2, Q.14 P. Continuous relative rotation then the length of the crank in mm Q. Velocity and acceleration is R. Mobility a) 500 b) 250 3 S. Dynamic-static analysis c) 250 d) 500 3 Approach [GATE√-2007] 1. D' Alembert's principle √ 2. Grubler's criterion Q.15 The angular speed of PQ in rev/s 3. Grashoff's law when the block R attains maximum 4. Kennedy's theorem speed during forward stroke a) P-1, Q-2, R-3, S-4 (stroke with slower speed) is b) P-3, Q-4, R-2, S-1 1 2 a) b) c) P-2, Q-3, R-4, S-1 3 3 d) P-4, Q-2, R-1, S-3 c) 2 d) 3 [GATE-2009] [GATE-2007] Q.19 A simple quick return mechanism is Q.16 The input link O2P of a four bar shown in the figure. The forward to linkage is rotated at 2 rad/s in return ratio of the quick return counter clockwise direction as mechanism is 2:1. If the radius of

shown below. The angular velocity crank O1P is 125 mm, then the of the coupler PQ in rad/s, at an distance 'd ' (in mm) between the  instant when ∠ O4O 2P= 180 , is crank centre to lever pivot centre point should be

a) 4 b) 22 1 c)1 d) a) 144.3 b) 216.5 2 c) 240.0 d) 250.0 [GATE-2009]

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission d) Gruebler's criterion assumes Q.20 For the configuration shown, the mobility of a planar mechanism angular velocity of link AB is 10 to be one rad/s counter clockwise. The [GATE-2010] magnitude of the relative sliding velocity (in ms-1) of slider B with Q.24 A double-parallelogram mechanism respect to rigid link CD is is shown in the figure. Note that PQ is a Single link. The mobility of the mechanism is

a) 0 b) 0.86 a) -1 b) 0 c) 1.25 d) 2.50 c) 1 d) 2 [GATE-2010] [GATE-2011]

Q.21 Mobility of a statically Q.25 For the four-bar linkage shown in indeterminate structure is the figure, the angular velocity of a) - 1 b) 0 link AB is 1 rad/s. The length of c) 1 d) 2 link CD is 1.5 times the length of ≤ [GATE-2010] link AB. In the configuration shown, ≥ the angular velocity of link CD in Q.22 There are two points P and Q on a rad/s is planar rigid body. The relative velocity between the two points a) Should always be along PQ b) Can be oriented along any direction c) Should always be perpendicular to PQ d) Should be along QP when the 3 body undergoes pure translation a)3 b) [GATE-2010] 2 2 c) 1 d) Q.23 Which of the following statements is 3 INCORRECT? [GATE-2013] a) Grashoff's rule states that for a planar crank-rocker four bar Q.26 A planar closed kinematic chain is mechanism, the sum of the formed with rigid links PQ = 2.0m, shortest and longest link lengths QR= 3.0m cannot be less than the sum of RS = 2.5m and SP = 2.7m with all the remaining two link lengths revolute joints. The link to be fixed b) Inversions of a mechanism are to obtain a double rocker (rocker- created by fixing different links rocker) mechanism is one at a time a) PQ b) QR c) Geneva mechanism is an c) RS d) SP intermittent motion device [GATE-2013]

If= 20 mm,l in = 40mm,l co = 50 mm and l out = 60mm. The suffixes 'f', 'in', 'co' and 'out'

denote the fixed link, the input link, a) the coupler and output link respectively. Which one of the following statements is true about the input and output links? a) Both links can execute full circular motion b) Both links cannot execute full circular motion c) Only the output link cannot execute full circular motion b) d) Only the input link cannot execute full circular motion [GATE -2014(2)] Q.28 A rigid link PQ is 2 m long and oriented at 20o to the horizontal in the figure. The magnitude and

direction of velocity VQ , and the c) direction of velocity Vp are given.

The magnitude of Vp () in m / s at this instant is

a) 2.14 b) 1.89 c) 1.21 d) 0.96 [GATE -2014(1)] Q.29 Consider a slider crank mechanism with nonzero masses and inertia. A [GATE-2015(1)] crank as shown in the figure. Which ofconstant the following torque plots τ is best applied resembles on the The number of degrees of freedom Q.30 of the linage shown in the figure is variation of crank angle, θ versus time

Q.31 A four bar mechanism is made up of links of length 100, 200, 300 and 350 mm. If the 350 m link is fixed, the number of links that can rotate fully is ______. [GATE-2018(1)]

1 2 3 4 5 6 7 8 9 10 11 12 13 14 (d) (a) (c) (c) (d) (b) (b) (d) (c) (a) (d) (c) (b) (a) 15 16 17 18 19 20 21 22 23 24 25 26 27 28 (b) (c) (b) (b) (d) (d) (a) (c) (a) (c) (d) (c) (a) (d) 29 30 31 (d) (c) (1)

EXPLANATIONS

Q.1 (d) Types of Joint A single slider crank chain is a P. Revolute modification of the basic four bar Q. Cylindrical chain. It is find, that four inversions R. Spherical of a single slider crank chain are Degree of constraints possible. From these four 2. Five inversions, crank and slotted lever 3. Four quick return motion mechanism is 1. Three used in shaping machines, slotting So, correct pairs are P-2, Q-3, R-1 machines and in rotary internal combustion engines. Q.5 (d) Given planar mechanism has degree Q.2 (a) of freedom, N = 1and two infinite Given p < q < r < s. parallel lines meet at infinity. So, the “Double crank” mechanism occurs, instantaneous centre I24 will be at N, when the shortest link is fixed. From but for single degree of freedom, the given pairs p is the shortest link. system moves only in one direction. So, link of length p should be fixed. Hence, I24 is located at infinity ( ).

Q.3 (c) Q.6 (b) Types of Mechanisms P. Scott-Russel Mechanism Q. Geneva Mechanism R. Off-set slider-crank Mechanism S. Scotch Yoke Mechanism Motion Achieved 4. Straight Line Motion 1. Intermittent Motion 2. Quick Return Mechanism 3. Simple Harmonic Motion So, correct pairs are, P-4,Q-1,R-2, S-3

Q.4 (c) From Triangle ABC

=67600 = 260mm υ24 = ω 4 ×ll 24 14 = ω× ll 24 14

Length Of shortest link l2 = 60mm ll 144 ω =24 12 ×ω= ×8 Length of longest link 4 l2 = 260mm l24 l 14 (240+ 144) From the Grashof’s law. 144 = ×=rad /sec. 83 ll13+>+/ ll 24 384 60+280 >/ 160 + 240 320 >/ 400 Q.8 (d) So, ll+<+ ll From the given data the component 13 24 of force at joint A along AO is Also, when the shortest link O A will 2 2 necessary to find the joint reaction make a complete revolution relative at O . So, it is not possible to find the to other three links, if it satisfies the 2 magnitude of the joint reaction at Grashof’s law. Such a link is known O2. as crank. The link O4B which makes a partial rotation or oscillates is Q.9 (c) known as rocker. So, crank rocker Given = 8 , j = 9 mechanism is obtained. We know that. Degree of freedom. Here, O2A = l1= 60 mm is crank (fixed link) n= 2(l −− 1) 2 j = 3(8 −−×= 1) 2 9 3 Adjacent link, O2O4= 240 mm is fixed Q.10 (a) So, crank rocker mechanism will be Here P, Q, R, &S are the lengths of obtained. the links. According to Grashof’s law : “For a four bar mechanism, the Q.7 (b) sum of the shortest and longest link Let, 4 is the angular velocity of link lengths should not be greater than O4B the sum of remaining two link Fromω the triangle ABC , lengths, if there is to be continuous 100 5 relative motion between the two tan θ= = 240 12 links SL+≤+ PQ −105 θ=tan = 22.60 12 Also from the triangle O1O2A. OA tan θ= 2 OO12 OA 60 O O=2 = = 144mm 12 tan θ 60 12 Q.11 (d) In this question pair or mechanism is related to contact & machine related to it. Column I P. Higher Kinematic Pair Q. Lower Kinematic Pair

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission R. Quick Return Mechanism Given Quick return ratio = 1:2, S. Mobility of a Linkage OP= 500 mm Column II Here OT= Length of the crank. We 2. Line Contact see that the angle made by the 6. Surface Contact forward stroke is greater than the 5. Shaper angle a described βby the return 1. Grubler’s Equation stroke. Since the crank has uniform So correct pairs are, P-2,Q-6,R-5,S-1 angular speed, therefore. Since the crank has uniform angular Q.12 (c) speed, therefore. For a 4 bar slider crank mechanism, Timeof return stroke Quick Return ratio = there are the number of links or Timeof cuttingstroke inversions are 4. These different 1 αα inversions are obtained by fixing = = different links once at a time for one 2β 360 −α inversion. Hence, the number of 360 – = 2 inversions for a slider crank 3 =360 α α α 120 mechanism is 4. and Angle TOP= = = 600 α 22 Q.13 (b) From the TOP Column I α OT r cos = = = P. Addendum 2∆ OP 500 Q. Instantaneous centre of velocity OT = r R. Section modulus r S. Prime circle cos 600 = 500 Column II 1 4. Gear r= 500 ×= 500mm 3. Linkage 5 2. Beam 1. Cam Q.15 (b) So correct pairs are, P-4,Q-3, R-2,S-1 We know that maximum speed during forward stroke occur when Q.14 (a) QR & QR are perpendicular. . SO, V= OS ×ωas = PQ × w PQ Vr= ω

250× 2 = 750 ×ωPQ 500 2 ω= =rad / sec PQ 750 3

Q.16 (c) 0 Given OOP = 180 , ω 42 OP1 = 2 rad /sec The instantaneous centre diagram is given below. Let, velocity of point P on like O2P is Vp.

Vp=ω×=ω× op 2 OP 2 op 2 (II 12 23 ) = 2a ...(i)

Vp=ω× PQ O 2 P =ω× PQ (I 13 I 23 )...(ii)

Both the links OP2 and QP are runs at the same speed From equation (i) and (ii) we get

2a=ω×PQ 2a

or, ωpq = 1 rad/sec

Given O1P = r = 125mm Forward to return ratio = 2.1 We know that Time of cutting (forward) stroke = time of return stroke β360 −α = = αα Substitute the value of Forward to return ratio we have 2 360 −α = Q.17 (b) 1 α From Gruebler’s criterion, the 2α= 360 −α ⇒α=1200 equation for degree of freedom is And angle given by, 0 α 120 0 n= 3( j −− 1) 2 j − h ...(i) RO12 O= = = 60 Given and j = 10 , h = 0 22 Now we are to find the distance d2 n= 3(8 − 1) −× 2 10 = 1 between the crank centre to lever from equation (i) pivot centre point (O1O2) from the RO2O2

0 α OR1 r Q.18 (b) ∆sin 90 −= = 2 OO OO Analysis 12 12 r P. Continuous relative rotation sin() 9000−= 60 Q. Velocity and Acceleration OO12 R. Mobility r 125 O O= = = 250mm S. Dynamic-static Analysis 12 sin 300 1/ 2 Approach 3. Grashoff law Q.20 (d) 4. Kennedy’s Theorem Let, vB is the velocity of slider B 2. Grubler’s Criterion relative to link CD. 1. D’Alembert’s Principle The crank length AB = 250 mm and So, correct pairs are P-3,Q-4,R-2,S-1 velocity of slider B with respect to rigid link CD is simply velocity of B (because C is a fixed point). Q.19 (d) Hence, −3 υB =(AB) ×ω AB = 250 × 10 × 10 = 2.5m / sec

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Alternate Method : Velocity of any point on a link with From the given figure, direction of respect to another point (relative velocity of CD is perpendicular to velocity) on the same link is always link AB & direction of velocity of AB perpendicular to the line joining is parallel to link CD. these points on the configuration So, direction of relative velocity of (or space) diagram. slider B with respect to C is in line VQP = Relative velocity between P & with link BC. Q

Hence υ=C 0 = vp - vQ =- always perpendicular to Or PQ. ν =ν −ν =AB ×ω − 0 BC B C AB Q.23 (a) = ×= 0.025 10 2.5 m / sec According to Grashof’s law “For a

four bar mechanism, the sum of the

Q.21 (a) shortest and longest link lengths Give figure shows the six bar should not be greater than the sum mechanism. of remaining two link lengths if there is to be continuous relative motion between the two links.

ll4+>+ 2/ ll 13

We know movability or degree of freedom is n=3( - 1) – 2j – h. The mechanism shown in figure has six links and eight binary joints (because there are four ternary joints A, B, C & D, i.e. l = 6, j = 8 h = 0 So, n=3 (6 – 1) – 2× 8 = -1 Therefore Q.24 (c) when n = -1 or less, then there are Given that PQ is a single link. redundant constraints in the chain, Hence : l =5 , j = 5 , h = 1 & it forms a statically indeterminate It has been assumed that slipping is structure. So, From the Given (A) possible between the link l5 & l1. satisfy the statically indeterminate From the kutzbach criterion for a structure n -1 plane mechanism, Numbers of degree of freedom or movability. Q.22 (c) ≤ n = 3 (l-1)-2j-h=3(5-1)2×5-1=1

Q.25 (d)

Given AB=1rad/sec, llcd=1.5 AB

lcd ⇒=ω 1.5 lAB

Let angular velocity of link CD is ωCD From angular velocity ratio theorem. ω l AB = cd ωCDl AB

Q.26 (c)

As given in problem, a planer closed Q.29 (d) kinematic chain is shown in above figure. In given figure, if the link opposite to the shortest link, i.e. link RS = 2.5 m is fixed and the shortest link PQ is made a coupler, the other two links QR and SP would oscillate. The mechanism is known as rocker- rocker or double-rocker or double Net Torque m crank lever or oscillating-oscillating  τ=τ+(CW) F R(CW) converter. ()ext t Newton’s second law   Q.27 (a) τ=θext I S1+< P + Q  τ+ft R = I θ (CW) 20+<+⇒< 60 40 50 80 90 dθ If smaller link is fixed both input =( τ− FR)t and Output link execute full circular dt θ motion. tt θ= τ − ∫∫∫d dt Ft Rdt 00 0 Q.28 (d) t By Instantaneous center method; θ=τ − t∫ Ft Rdt Say = angular velocity of rod 0 t vqq= r. ω ... (1) dθ ω =τ−t F Rdt = ω ... (2) ∫ t vqq r. dt 0 ()()12÷ t t tt θ= τ − ∫d ∫ tdt ∫∫ Ft Rdt dt vrqqvrqq r q 0 0 00 = ; =; vpq = .v 2 tt vrppvrpp r q τt  θ= − F Rdt dt Now by sine law, ∫∫t 2 00 rr pq= sin() 45+ 20 sin(70o ) Q.30 (c) Number of links, N = 6 r sin() 65o Total number as binary joints, j = 7 p = r sin(70o ) F = 3 (N-1) – 2j q = 15-14 = 1. Q.31 (1)

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission The schematic of 4-bar mechanism sum of longest and shortest (450 mm) is less than rest two (500 mm). s = 100, p = 200, l = 350, q = 300 (s + l) = 350 + 100 = 450 < (p + q) 450 < 200 + 300 450 < 500 Hence, Grashof ’s Law is satisfied 350 mm link is fixed Then shortest link 100mm which is “s” adjacent to fixed, will give crank only.

Q.1 In a cam-follower mechanism, the follower needs to rise through 20 mm during 60 of cam rotation, the first 30 with a constant acceleration and then with a deceleration of the same magnitude. The initial and final speeds of the a) 0.05 b) 0.1 follower are zero. The cam rotates c) 5.0 d) 10.0 at a uniform speed of 300 rpm. The [GATE-2012] maximum speed of the follower is a) 0.60 m/s b) 1.20 m/s Q.4 Consider a rotating disk cam and a c) 1.68 m/s d) 2.40 M/s translating roller follower with zero [GATE-2005] off set. Which one of the following pitch curves, parameterized by t, Q.2 In a cam design, the rise motion is lying in the interval 0 to 2 , is given by a simple harmonic associated with the maximum motion (SHM) s=h/2(1-cos( / ))h translation of the follower duringπ where he is total rise, is camshaft one full rotation of the cam rotating angle, is the total angle of πθthe βrise about the center at (x, y) = (0, 0) ? interval. The jerk is given𝜃𝜃 by a) x(t)= cost,y(t) = sin t h 𝛽𝛽 πθ π h πθ b) x(t)= cost,y(t) = 2sin t a) b) 1- cos sin  1 2 β β2 β c) x(t)=+= cost,y(t) 2sin t 2 3 2 π h πθ π h πθ c) cos d) - sin  1 β22  β β23  β d) x(t)=+= cost,y(t) sin t 2 [GATE-2008] [GATE -2014(3)] Q.5 In a cam-follower rises by h as the Q.3 In the mechanism given below, if the can rotates by (radians) at constant angular velocity of the eccentric circular disc is 1 rad/s, the angular angular velocity (radians/s). The δ velocity (rad/s) of the follower link follower is uniformly accelerating during for the instant shown in the figure is the first half of the riseω period and it is (Note. All dimensions are in mm). uniformly decelerating in the latter half of the rise period. Assuming that the magnitude of the acceleration and deceleration are same, the maximum velocity of the follower is

(a) 4h / (b) h

ANSWER KEY: 1 2 3 4 5 (b) (d) (b) (c) (C)

Q.1 (b) dh2πω 22πω()t = cos Given N=300 r.p.m. 22 dt 2 ββ Angular velocity of cam. Again Differentiating w. r. t. t, 2Nπ ω= = 10 π rad / sec ds3 hπ 33 ω πθ = − 60 33sin Time taken to move 300 is dt 2 ββ π 1 Let = 1 rad /sec ×30 33 180 6 1 ds hπ πθ t= = = sec =ω − sin  10π 10 60 dt33 2 ββ Now, Cam moves 30°with a constant acceleration & then with a Q.3 (b) deceleration, so maximum speed of From similar PQO and PQ PO the follower is at the end of first = 30°rotation of the cam and during SR SO ∆ ∆SRO this 30° rotation the distance covered 22 PQ = (50)− (25) = 43.3 mm is 10mm, with initial velocity u = 0. From Newton’s second law of motion, 1 S = ut + at 2 2 2 11 0.01 = 0 + ××a  2 60 a= 0.01 ×× 2 (60)22 = 72m / sec 43.3 50 Maximum velocity, From Eq. (i) = 1 SR 5 V=+=× u at 72 = 1.2m / sec 43.5× 5 max 60 SR= = 4.33mm 50 Q.2 (d) Velocity of Q = Velocity of R Jerk is given by triple differentiation (situated at the same link) of s w r. t. t. VQ= VR = SR =4.33× 1 = 4.33 dS3 m/s. Jerk = 3 ×ω dt Angular velocity of PQ. Give VQ 4.33 ω=PQ = =0.1rad / s. hhπθ πω()t PQ 43.3 s=−=− 1 cos 1 cos ββ 22Q.4 (c) θ=ωt From all the four options the Differentiating above equation w.r.t. maximum amplitudes is in point ’C’ t we get as t=0

ds h πω πω()t (X)t0= (Y) t0 = ==−−sin  a) = = b) = = dt 2 ββ x 1 y 0 x 1 y 0 3 3 Again Differentiating w.r.t. t. c) x= y0 = d) x= y0 = 2 2

and θ0 strokeδ length = h

angular velocity =

V = u + at

t (V ) = 0 + a × = a × 20 0 max δ 2ω (i)

h 1 t a(t ) a( ) = 0 + a × = = 2 2 2 2 4 2 4 2 0 0 δ � � 2 4 h ω a = ( )2 ω 2 δ (ii)

From (i) and (ii) 4 h h (V ) = a × = × = ( )2 δ ω δ 2ω 0 max 2 2ω δ 2ω δ

Q.1 For a mechanism shown below, the figure. The crank is rotating with a mechanical advantage for the given uniform angular speed of 10 rad/s, configuration is counter clockwise. For the given a) 0 b) 0.5 configuration, the speed (in m/s) of c) 1.0 d) the slider is ______[GATE-2004] ∞ Q.2 For a four-bar linkage in toggle position, the value of mechanical advantage is a) 0.0 b) 0.5 c) 1.0 d)∞ [GATE -2014(3)] [GATE-2006] Q.6 In a certain slider-crank mechanism, Q.3 A slider crank mechanism has slider lengths of crank and connecting rod of mass 10 kg, stroke of 0.2 m and are equal. If the crank rotates with a rotates with a uniform angular uniform angular seed of 14 rad/s velocity of 10 rad/s. The primary and the crank length is 300 mm, the inertia forces of the slider are maximum acceleration of the slider partially balanced by a revolving (in m/s2) is ______mass of 6 kg at the crank, placed at a [GATE-2015(2)] distance equal to crank radius, Neglect the mass of connecting rod Q.7 A slider crank mechanism is shown and crank; When the crank angle in the figure. At some instant, the (with respect to slider axis) is 30o , crank angle is 45° and a force of 40 The unbalanced force (in Newton) N is acting towards the left on the normal to the slider axis is ______slider. The length of the crank is 30 [GATE-2014(1)] mm and the connecting rod is 70 mm. Ignoring the effect of Q.4 An offset slider- crank mechanism is gravity, friction and inertial forces, shown in the figure at an instant. the magnitude of the crankshaft Conventionally, the Quick Return torque (in Nm) needed to keep the ratio (QRR) is considered to be mechanism in equilibrium is ____ greater than one. The value of QRR (correct to two decimal places). is _____

[GATE -2014(1)] [GATE-2018(1)] Q.5 A slider-crank mechanism with crank radius 60 mm and connecting rod length 240 mm is shown in

1 2 3 4 5 6 7 (d) (d) 31 1.3 0.68 117.6 1.12

Q.1 (d) 0.2 Mechanical advantage in the form of r= = 0.1 m = 6kg 2 torque is given by, F= mr(102 ) sin θ Toutputω input M.A = = =××6 0.1 100 × sin 30o T ω input output = 30 N Here output link is a slider. So. output = 0 Therefore M. A. = ω Q.2 (d) ∞

Q.4 (1.25) AB= stroke length AO=−= 40 20 20 BO=+= 40 20 60 ∠AOB =∠ BOC −∠ AOC =80.41 − 60 180 +φ = 20.41 QRR= = 1.255 180 −φ

TRω2 MA =4 = = PD Q.5 (0.54 to 0.68) T3ω 4R PA from angular velocity ration Q.6 (117.6) theorem Construct B A and C’ D a= 2r ω2 (when Inner perpendicular to the line PBC. Also max dead centre assign labels and y to the acute 2 θ=0) i.e at angles made by the coupler. =××2 0.3 14 = 2 R R Rβ sin γ amax 117.6m / s PD =CD = CD RPA R SA R BA sinβ Q.7 The torque required to hold the TAω 2 R sin γ So M. A. = = = CD mechanism in equilibrium is equal and T2ωβ 4 R sin BA opposite of turning moment When the mechanism is toggle, then , = 00 and 1800 So M.A. = β Q.3 (20) ∞

1.12 N m

≈ −

30sin45 = 70sin o 30sin45 β sin = = 0.303045 70 o β = 17.6406 0 βF cos = 40

40c β F = = 41.9737N cos c F = Fβsin ( + )

F = 41.9737sint c (45θ + β17.6406 ) o 0 t F = 37.2785N

T = F ×t r = 37.2785 × 0.030

t = 1.1183 N m T 1.12 N m −

≅ OR −

Turning moment on the crankshaft

sin2 T = F × r sin + 2 n sin θ P � θ 2 2 � √ − θ sin90 = 40 × 0.03 sin45 + ⎧ 70 ⎫ 2 sin 45 30 2 ⎨ 2 ⎬ � ⎩ � � − ⎭ 1 1 = 1.2 + ⎧ 2 49 1⎫ 2 9 2 ⎨√ ⎬ � = 1.2⎩{0.707 + 0.225− }⎭

Common Data For Q.1 and Q.2 Q.3 What is the relation between the A compacting machine shown in the figure angular velocities of Gear 1 and below is used to create a desired thrust Gear 4 ? force by using a rack and pinion ω −ω ω −ω a) 15= b) 45= arrangement. The input gear is mounted 6 6 ω45 −ω ω15 −ω on the motor shaft. The gears have in ω −ω 2 ω −ω 8 volute teeth of 2 mm module. c) 12= − d) 25= − ω45 −ω 3 ω45 −ω 9 [GATE-2006]

Q.4 For ω1 = 60 rpm clockwise (CW) when looked from the left, what is the angular velocity of the carrier and its direction so that Gear 4 rotates in counter clockwise (CCW) direction at twice the angular Q.1 If the drive efficiency is 80%, the velocity of Gear 1 when looked torque required on the input shaft from the left? to create 1000 N output thrust is a) 130 rpm, CW b) 223 rpm, CCW a) 20 Nm b) 25 Nm c) 256 rpm, CW d) 156 rpm, CCW c) 32 Nm d) 50 Nm [GATE-2006] [GATE-2004] Q.5 An epicyclical gear train in shown Q.2 If the pressure angle of the rack is schematically in the given figure.  20 , then force acting along the line The run gear 2 on the input shaft is of action between the rack and the a 20 teeth external gear. The planet gear teeth is gear 3 is a 40 teeth external gear. a) 250 N b) 342 N The ring gear 5 is a 100 teeth c) 532 N d) 600 N internal gear. The ring gear 5 is [GATE-2004] fixed and the gear 2 is rotating at Common Data for Q.3 and Q.4 60rpm CCW (CCW=counter- clockwise and CW=clockwise). A planetary gear train has four gears and one carrier. Angular velocities of the gears

are ω123 ,ω ,ω & ω4 respectively. The carrier

rotates with angular velocity ω5 .

The arm 4 attached to the output shaft will rotate at a) 10 rpm CCW b) 10 rpm CW c) 12 rpm CW d) 12 rpm CCW [GATE-2009]

shown in the figure ω2 =100 rad/s

clockwise (CW) and ωarm =80 rad/s counter clockwise (CCW). The a) 40 b) 80 angular velocity (in rad/s) is ω5 c) 120 d) 160 [GATE-2013] Q.10 Gear 2 rotates at 1200 rpm in counter clockwise direction and engages with Gear 3. Gear 3 and Gear 4 are mounted on the same shaft. Gear 5 engages with Gear 4. The numbers of teeth on Gears 2, 3, a) 0 b) 70 CW 4 & 5 are 20,40,15 & 30 respectively. c) 140 CCW d) 140 CW The angular speed of Gear 5 is [GATE-2010] Q.7 Tooth interference in an external involute spur gear pair can be reduced by a) decreasing center distance between gear pair b) decreasing module a) 300 rpm counter clockwise c) decreasing pressure angle b) 300 rpm clockwise d) increasing number of gear teeth c) 4800 rpm counter clockwise [GATE-2010] d)4800 rpm clockwise [GATE -2014(3)] Q.8 The following are the data for two crossed helical gears used for speed Q.11 It is desired to avoid interference in reduction: Gear I: Pitch circle a pair of spur gears having a 20° diameter in the plane of rotation 80 pressure angle. With increase in mm and helix angle 30 .Gear II: Pitch pinion to gear speed ratio, the circle diameter in the plane of minimum number of teeth on the rotation 120mm and helix angle pinion 22.5 . If the input speed is 1440 a) increases rpm, the output speed in rpm is b) decreases a) 200 b) 900 c) first increases and then decreases c) 875 d) 720 d) remains unchanged [GATE-2012] [GATE -2014(4)] Q.9 A compound gear train with gears P, Q.12 The number of degrees of freedom Q, R and S has number of teeth 20, 40, of the planetary gear train shown in 15 and 20, respectively. Gears Q and R the figure is are mounted on the same shaft as shown in the figure below. The diameter of the gear Q is twice that of the gear R. If the module of the gear R is 2mm, The center distance in mm between gears P & S is.

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission a) 0 b) 1 a) 166.7 counter clockwise c) 2 d) 3 b) 166.7 clockwise [GATE-2015(2)] c) 62.5 counter clockwise d) 62.5 clockwise Q.13 A gear train is made up of five spur [GATE-2016(1)] gears as shown in the figure. Gear 2 is driver and gear 6 is driven Q.15 In an epicyclic gear, shown in the member. N2, N3, N4, N5 and N6 figure, the outer ring gear is fixed. represent number of teeth on gears While the sun gear rotates counter 2, 3, 5 and 6 respectively. The clockwise at 100 rpm. Let the gear(s) which act(s) as idler(s) number if teeth on the sum planet is/are and outer gears to 50, 25 and 100, respectively. The ratio of magnitude of angular velocity of the planet gear to the angular velocity of the carrier arm is ____

a) only 3 b) only 4 c) only 5 d) Both 3 and 5 [GATE-2015(3)]

Q.14 In the gear train shown, gear 3 is carried on arm 5. Gear 3 meshes with gear 2 and gear 4. The number of teeth on gear 2, 3, and 4 are 60, 20, and 100, respectively. If gear 2 is fixed and gear 4 rotates with an angular velocity of 100 rpm in the [GATE-2017(1)] counter clockwise direction, the angular speed of arm 5 (in rpm) is Q.16 A gear train shown in the figure consists of gears P,Q, R and S, Gear Q and gear R are mounted on the same shaft. All the gears are mounted on parallel shafts and the number of teeth of P, Q, R and S are 24,45,30 and 80, respectively. Gear P is rotating at 400 rpm. The speed in (rpm) of the gear S is ____

[GATE-2017(2)]

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Q.17An epicycle gear train is shown in the figure below. The number of teeth on the gears A, B and D are 20, 30 and 20, respectively. Gear C has 80 teeth on the inner surface and 100 teeth on the outer surface. If the carrier arm AB is fixed and the sum gear A rotates at 300 rpm in the clockwise direction, then the rpm of D in the clockwise direction is

(a) 240 (b) –240 (c) 375 (d) –375 [GATE-2018(1)]

Q.18 A frictionless gear train is shown in fig. The left most 12-teeth gear is given a torque of 100 N-m. The output Torque from 60-teeth gear on right in N-m is

(a) 50 (b) 20 (c) 500 (d) 2000

[GATE-2018(2)]

1 2 3 4 5 6 7 8 9 10 11 12 13 14 (b) (c) (a) (d) (a) (c) (d) (a) (b) (a) (a) (c) (c) (c) 15 16 17 18 3 120 (C) (d)

Q.1 (b) D 2F×× ω2 = 2 ...(i) T22ω We know velocity ratio is given by N1ω Z 2π N =12 = ω= N2ω21 Z 60 From equation (i) D 2F×× Z η=2 × 1 TZ22

F×× DZ1 500 0.160 20 T1 = × = ×= 25 Nm η Z2 0.08 80

Q.2 (c)

Let, Z is the number of teeth and motor rotates with an angular velocity in clockwise direction &

develops a torque T1. Due to the rotation of motor, the gear 2 rotates Given pressure angle = 200 , in anti-clockwise direction & gear 3 Fr = 500 N from previous question. rotates in clock wise direction with From the given figure∅ we easily see the same angular speed. that force action along the line of Let, T2 is the torque developed by action is gear. Now, for two equal size big F .From the triangle ABC, gears, Module FF D (Pitch circlediameter) cos∅= rr ⇒F = m = = F cos∅ Z (Noofteeths) 500 = = D = mZ = 2 × 80 = 160 mm 0 532N cos 20 (Due to rotation of gear 2 & gear 3 equation force (F) is generated in Q.3 (a) the downward direction because teeth are same for both the gears). For equilibrium condition, we have Downward force = upward force. F+F=1000 Power Output 2T×ω And = = 22 Power Input T11ω Output power is generated by the two gears

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission The table of motions is given below ω −ω60 −ω 13=⇒= 5 Take 66 ω45 −ω −120 −ω 5 CW = + ve CCW = -ve 60 - ω5 = - 720 - 6 ω5 780 ⟹ω=− =−156rpm 5 5 Negative sign show the counter clock wise direction

SO, ω5 =156 rpm. CCW.

Q.5 (a) Given Z2 = 20 Teeth, T3 = 40 Teeth Z5 = 100 Teeth N5 =0, N2 = 60 rpm (CCW)

Note : Speed of drive (i) Speed ration = Speed of drivn Noof teethondriven = Noof teethondirver NZ If gear 2 rotates in the CCW i.e. 11= NZ direction, then gear 3 rotates in the 22 clockwise direction. Let, Arm 4 will CCW =Counter clock wish direction(-ve) rotate at N rpm. The table of CW = Clock wise direction (+ve) 4 motions is given below. Take CCW =+ve, CW = -ve. (ii) Gear 2 & Gear 3 mounted on the same shaft (Compound Gears) So N2 = N2 2Nπ We know, ω= ⇒ωa N 60 Hence= NN− ω −ω ()xy+− y 1 2= 12 = NN− ω −ω ZZ13− 4 5 45yx++ − Y ZZ24− ω −ω x ZZ− 12= = 13 ω −ωZZ13− ZZ − Note. Speed ratio = 45x × 24 − Speed of drive Noof teeth on driven ZZ24 = ω −ω 45× 40 Speed of drivn Noof teeth on dirver 12= =×=32 6 Ring gear 5 is fixed so. ω45 −ω15 × 20 N05 = Q.4 (d) From the table Given ω = 60 rpm (CW), Z 1 yx−=2 0 Z ω=4 2 × 60(CCW) =− 120rpm 3 From the previous part.

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Z 20 x yx−==2 x x ...(i) Speed of drive Noof teeth on driven Z3 100 5 = Give N2 = 60 rpm (CCW) Speed of drivn Noof teeth on dirver y + x = 60 ω N i.e. 12= From table ω21N x +=x 60 Gear 3 & 4 mounted on same shaft 5 So ω −ω x = 10 × 5= 50rpm. 34 And ω=y And from equation (i) arm 50 From the table y= = 10rpm(CCW) y = -80 rad /sec (CCW) 5 x+y = ω =100 From the table the arm will rotate at 2 N= y = 10rpm(CCM) From the table 4 x = 100 – (-80) = 180 rad/sec(CW)

NN24 Q.6 (c) And ω5 = y – x ×× NN35 From the table 20 32 =−−×80 180 × = rad / sec 24 80 Negative sign shows the counter clockwise direction.

Q.7 (d) When gear teeth are produced by a Given N1 = No. of teeth for gear i. generating process, interference is N= 20, N = 24, N = 32, N = 80 2345 automatically eliminated because ω= rad /sec. (CW) 2 100 the cutting tool removes the

ω=arm 80 rad / sec (CCW) interfering portion of the flank. This = -80 rad /sec effect is called undercutting. By The table of the motion given below undercutting the undercut tooth can Take CCW = - ve and CW = +ve be considerably weakened. So, interference can be reduced by using more teeth on the gear. However, if the gears are to transmit a given amount of power, more teeth can be used only by increasing the pitch diameter.

Q.8 (a) Since two nodes are observed at frequency of 1800 rpm, therefore third critical speed of shaft f3 =1800 rpm Because two nodes can be observed only in 3rd mode. The whirling frequency of shaft Note. Speed ratio =

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Q.13 (c) π 2 g fn= × ωωωω N N N NN 2 δ 22=3 5 = 3 5 6 = 36 Where n = Numbers of mode ω655624524 ωωω N N N NN π g only Gear(5) is Idle For n = 1. f = 1 2 δ 2 Q14 (c) Therefore fn1= nf × 2 or f11= 3f × f 1800 or f=3 = = 200rpm 1 (3)2 9

Q.9 (b) Given: x + y = 0 …..(1) We have 3 ZP = 20, ZQ= 40, ZR= 15, ZS=20, y−= x 100 … (2) DQ =2DR, mR=2mm 5 D 3 5× 100 Module m = R y+ y100 ⇒= y = 62.5 R Z 58 R counter clockwise or DR = 2 × 15 = 30 mm and = 2 ×30 = 60 mm DQ Q.15 (3) DZ Also PP= DZQQ 20 Or D= ×= 60 30mm P 40

ZS Again DDSR= × ZR 20 ⇒ ×=30 40mm Given ND=0 of Ns=100 15 x y0−= Thus the entre distance between P α and S is x D y = DDPRQ DS α dps =+++ 2222 & y+=− x 100 = 15 + 30 + 15 +20 = 80 mm x +=−x 100 α Q.10 (a) 3x− 200 N5 TT24× 20× 15 =−⇒=100 x = = α N T×× T 40 30 3 2 35 −100 ⇒=N 1200 × 0.25 = 300rpm ccw ∴=y 5 3 −100 400 Q.11 (a) Now N=−= y 2x + = 100 P 33 − Q.12 (c) 100 Nyarm = = A planetary gear train has 2 DOF 3 and hence requires too input to get desired output.

Q.16 (120) = 24, = 45, = 30, = 80, = 400 , =? 𝑇𝑇𝑃𝑃 𝑇𝑇𝑄𝑄 𝑇𝑇𝑅𝑅 𝑇𝑇𝑆𝑆 𝑁𝑁𝑃𝑃 Here gear R is not meshing at all The speed of compound gear, 45 𝑟𝑟𝑟𝑟𝑟𝑟 𝑁𝑁𝑆𝑆 = = …………………………..( ) = 𝑃𝑃 𝑄𝑄 24 𝑁𝑁 𝑇𝑇 𝑍𝑍1 80 𝑖𝑖 Since the intermediate2 1gear is compound, 𝑁𝑁𝑄𝑄 =𝑇𝑇𝑃𝑃 = ………………………..( ) 𝑁𝑁 −𝑁𝑁 2 𝑄𝑄 𝑆𝑆 45 so 𝑍𝑍 By𝑁𝑁 (i) and𝑇𝑇 (ii) = 𝑆𝑆 𝑄𝑄 𝑖𝑖𝑖𝑖 𝑁𝑁 𝑇𝑇 45 80 The speed of last gear, = × 3 2 24 45 𝑁𝑁 𝑁𝑁 𝑁𝑁𝑃𝑃 = = × 24 24 3 1 3 = 𝑆𝑆 × = 400 × 𝑁𝑁 4 312𝑍𝑍 12 1 𝑍𝑍 𝑍𝑍 80 80 𝑁𝑁 =−𝑁𝑁 4× 𝑁𝑁= 2 4 = 120 . . 48𝑍𝑍 60 𝑍𝑍20 𝑍𝑍 𝑁𝑁𝑆𝑆 𝑁𝑁𝑃𝑃 1 Since gear train in1 friction less.𝑁𝑁 𝑆𝑆 𝑁𝑁 Q.17(c) 𝑁𝑁 𝑟𝑟 𝑝𝑝 𝑚𝑚 Input power = Output power =

= =𝑇𝑇2100𝑁𝑁2 ×𝑇𝑇201𝑁𝑁=1 2000 . 𝑇𝑇1𝑁𝑁1 𝑇𝑇2 𝑁𝑁 𝑚𝑚 𝑁𝑁2

The arm is fixed, y = 0 The gear A rotates at 100 rpm CW. ∴ x + y = 300 x = 300 Rotation of B, 2 2 × 300 = = 0 3 3 𝑥𝑥 = 200 ( ) 𝑁𝑁𝐵𝐵 𝑦𝑦 − − Rotation of C, 300 𝑟𝑟𝑟𝑟𝑟𝑟 𝐶𝐶𝐶𝐶𝐶𝐶 = = 0 = 75 ( ) 4 4 The Gear D meshes𝑥𝑥 with C externally 𝑁𝑁𝐶𝐶 𝑦𝑦 − − 𝑟𝑟𝑟𝑟𝑟𝑟 𝐶𝐶𝐶𝐶𝐶𝐶 = 𝐶𝐶 100 𝐷𝐷 𝐶𝐶 𝑍𝑍 = 75 × =𝑁𝑁 75 ×−𝑁𝑁 5 = 375𝐷𝐷 ( ) 20 𝑍𝑍 𝑟𝑟𝑟𝑟𝑟𝑟 𝐶𝐶𝐶𝐶

Q.1 Which of the following statements is correct? [GATE-2006] a) Flywheel reduces speed fluctuations during a cycle for a Q.4 The speed of an engine varies from constant load, but flywheel does 210 rad/s to 190 rad/s. During the not control the mean speed of cycle the change in kinetic energy is the engine, if the load changes. found to be 400 Nm. The inertia of 2 b) Flywheel does not reduce speed the flywheel in kg/m is fluctuation during a cycle for a a) 0.10 b) 0.20 constant load, but flywheel does c) 0.30 d) 0.40 not control the mean speed of [GATE-2007] the engine, if the load changes. Q5 A flywheel connected to a punching c) Governor controls speed machine has to supply energy of 400 fluctuations during a cycle for a Nm while running at a mean angular constant load, but governor does speed of 20 rad/s. If the total not control the mean speed of fluctuation of speed is not to exceed the engine, if the load changes. ±2%, the mass moment of inertia of d) Governor controls speed the flywheel in kgm2 is fluctuations during a cycle for a a) 25 b) 50 constant load, and governor also c) 100 d) 125 controls the mean speed of the [GATE-2013] engine, if the load changes. [GATE-2001] Q.6 Maximum fluctuation of kinetic energy in an engine has been Q.2 For a certain engine having an average speed of 1200 rpm, a calculated to be 2600 J. flywheel approximated as a solid Assuming that the engine runs at an disc, is required for keeping the average speed of 200 rpm, the polar 2 fluctuation of speed within 2% mass moment of inertia ()in kg.m about the average speed. The of a flywheel to keep the speed fluctuation of kinetic energy per fluctuation within ±0.5% of the cycle is found to be 2 k J. What is the average speed is ______least possible mass of the flywheel if [GATE -2014(2)] its diameter is not to exceed 1m? a) 40 kg b) 51 kg Q.7 Torque and angular speed data over c) 62 kg d) 73 kg one cycle for a shaft carrying a [GATE-2003] flywheel are shown in the figures. The moment of inertia 2 of Q.3 If Cf is the coefficient of speed ()in kg.m Fluctuation of a flywheel then the the flywheel is ______

ratio of ωωmax/ min will be 1− 2C 2C− a) f b) f 1+ 2Cf 2C+ f 1+ 2C 2C+ c) f d) f 1− 2Cf 2C− f

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission inner dead center position. Assuming the resisting torque to be constant, the power (in kW) developed by the engine at 100 rpm is ______. [GATE-2015(3)]

[GATE -2014(4)]

Q.8 The torque (in Nm) exerted on the crank shaft of a two stroke engine can be described as T = 10000 + – is the crank angle as measured from 1000 sin 1θ 1200 cos 2θ, where θ

ANSWER KEY:

1 2 3 4 5 6 7 8 (a) (b) (d) (a) (a) 595 32 104

Q.1 (a) NN+ N= Mean speed in r.p.m. = = 12 The mean speed of the engine is 2 controlled by the governor. If load ... (i) increases then fluid supply Therefore, increases by the governor and vice- NN12−− 2(NN) 12 versa. Flywheel stores the extra Cf = = from N NN+ energy and delivers it when needed. 12 So, Flywheel reduces speed equation (i) . ω −ω fluctuations. = 12 Flywheel reduce speed fluctuations ω during a cycle for a constant load, 2(ω −ω ) C =max min ω −ω but Flywheel does not control the f 1 max ωmax +ω min + NN12 Cω+ω=ω−ω C 22 mean speed N = of the f max f min max max 2 ω()()C − 2 =ω −− 2C engine. max i min i ω ()2C+ 2C+ Hence max =−=i i Q.2 (b) ωminC i −− 2 2C f Given N = 1200 rmp, E=2kJ = 2000 J , D = 1 m, C5 = 0.02. Q.4 (a) Mean angular speed of∆ engine. Given ω1 = 210 rad /sec, ω2 = 190 2π N 2 ×× 3.14 1200 ω= = =125.66 rad /sec , E = 400 Nm 60 60 As the speed of flywheel changes rad sec from ω ∆ to ω the maximum Fluctuation of energy of the flywheel 1 2 fluctuation of energy. is given by 1 22 1 ∆E =l  ω −ω 2 22 ()()22 ∆=E lω Cs2 = mR ω C 2 2 2∆× E 2 400 2 = = mR l 22 2 2 For solid disc l = ()()ω −ω ()()210 − 190  2 22   2∆ E 22000 800 = = 2 m 22 2 = = 0.10 kgm RCω S 1 2 400× 20 ××()125.66 0.02 2 Q.5 (a) 4×× 2 2000 =2 = 50.66kg 51kg We have E = 400 N-m, ω=20 rad/ ()125.66× 0.02 sec, Cs= 0.04 The energy of flywheel is given by 2 Q.3 (d) EIC= ω S The ratio of the maximum E fluctuation of speed to the mean Or I = ω2C speed is called the coefficient of S fluctuation of speed (Cf) Let N1 & N2 400 2 2 =25kg − m = Maximum & Minimum speeds in ()20× 0.04 r.p.m. during the cycle ⇒

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Q.6 (590 to 595) π ∆=ωEIC2 s 4 2 2π× 200  0.5  − 0.5 ⇒=×2600 I   ×−  60  100  100 ⇒=1 595

Q.7 (30 to 32 ) ω +ω 20+ 0 ω=max min = =10 22 ω −ω 20− 0 C2=max min = = s ω 10 E= Area of T-diagram π ∆= ×3000 + 1500 ×π= 3000 π 2 2 ∆=ωEICS ⇒3000 π= I(10)2 × 2 ⇒=I 47.12

Q.8 104

Tmean = 100000N − m 2π ω=×100 mean 60 200π P= T ×ω = 104 × = 104kW mean mean 60

Q.1 The assembly shown in the figure is equivalent mass of the shaft is composed of two mass less rods of included in the rotor mass) length L with two particles, each of mounted at the ends. The bearings mass m. The natural frequency of are assumed to simulate simply this assembly for small oscillations supported boundary conditions. is The shaft is made of steel for which the value of E 2.1 X 1011 Pa. What is the critical speed of rotation of the shaft? a) 60 Hz b) 90 Hz c) 135 Hz d) 180 Hz [GATE-2003] g 2g a) b) Q.4 A mass M , of 20 kg is attached to L (Lcosα ) the free end of a steel cantilever g g(cosα) beam of length 1000 mm having a c) d) (Lcosα ) L cross-section of 25 × 25 mm. Assume the mass of the cantilever [GATE-2001] to be negligible and Esteel = 200 GPa. Q.2 In the figure shown, the spring If the lateral vibration of this A (the system is critically damped using a equilibrium position) when a mass viscous damper, then damping mdeflects is kept by δon to it. position During free constant of the damper is vibration, the mass is at position B at some instant. The charge in potential energy of the spring mass system from position A to position a)1250 Ns/m b)625 Ns/m B is c)312.50 Ns/m d)156.25 Ns/m [GATE-2004] Q.5 A uniform stiff rod of length 300 mm and having a weight of 300 N is pivoted at one end and connected to a spring at the other end. For keeping the rod vertical in a stable a) ½ k 2 b) ½ k 2- mgx position the minimum value of c) ½ k (x+ )2 d) ½ k 2+mgx spring constant k needed is 𝑥𝑥 [GATE𝑥𝑥 -2001] 𝛿𝛿 𝑥𝑥 Q.3 A flexible rotor-shaft system comprises of a 10 kg rotor disc placed in the middle of a mass-less shaft of diameter 30 mm and length a) 300 N/m b) 400 N/m 500 mm between bearings (shaft is c) 500 N/m d) 1000 N/m being taken mass-less as the [GATE-2004]

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Q.6 A vibrating machine is isolated pot with damping coefficient of 15 Ns/m. from the floor using springs. If the ratio of excitation frequency of Q.10 The value of critical damping of the vibration of machine to the natural system is frequency of the isolation system is a) 0.223 Ns/m b) 17.88 Ns/m equal to 0.5, then transmissibility c) 71.4 Ns/m d) 223.6Ns/m ratio of isolation is [GATE-2006] a) 1/2 b) 3/4 c) 4/3 d) 2 Q.11 The value of logarithmic decrement [GATE-2004] is a) 1.35 b) 1.32 Q.7 A rotating disc of 1 m diameter has c) 0.68 d)None of these two eccentric masses of 0.5 kg each [GATE-2006] at radii of 50 mm and 60 mm at   angular positions of 0 and 150 , Q.12 A machine of 250 kg mass is respectively. A balancing mass of supported on springs of total 0.1 kg is to be used to balance the stiffness 100 kN/m. Machine has an rotor. What is the radial position of unbalanced rotating force of 350 N the balancing mass? at speed of 3600 rpm. Assuming a a) 50 mm b) 120 mm damping factor of 0.15, the value of c) 150 mm d) 280 mm transmissibility ratio is [GATE-2005] a) 0.0531 b) 0.9922 Q.8 In a spring-mass system, the mass c) 0.0162 d) 0.0028 is 0.1 kg and the stiffness of the [GATE-2006] spring is 1 kN/m. By introducing a damper, the frequency of oscillation Q.13 The differential equation governing is found to be 90% of the original the vibrating system is value. What is the damping coefficient of the damper? a) 1.2 Ns/m b) 3.4 Ns/m c) 8.7 Ns/m d) 12.0Ns/m [GATE-2005] a) m + c + k (x- y)=0 b) m ( - ) + c ( - )+kx=0 Q.9 There are four samples P, Q, R and c) m𝑥𝑥̈+c(𝑥𝑥̇ - )+kx=0 S, with natura0l frequencies 64, 96, d) m (𝑥𝑥̈ -𝑦𝑦̈ ) + c (𝑥𝑥̇ -𝑦𝑦̇ )+k(x-y)=0 128 and 256 Hz, respectively. They 𝑥𝑥̈ 𝑥𝑥̇ 𝑦𝑦̇ [GATE-2006] are mounted on test setups for 𝑥𝑥̈ 𝑦𝑦̈ 𝑥𝑥̇ 𝑦𝑦̇ conducting vibration experiments. Q.14 The natural frequency of the system If a loud pure note of frequency 144 shown below is Hz is produced by some instrument, which of the samples will show the most perceptible induced vibration? a) P b) Q c) R d) S k k a) b) [GATE-2005] 2m m Common Data For Q.10 and Q.11: 2k 3k c) d) m m A vibratory system consists of a mass 12.5 [GATE-2007] kg, a spring of stiffness 1000 N/m, & a dash

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Q.15 The equation of motion of a Q.19 An automotive engine weighing 240 harmonic oscillator is given by kg is supported on four springs with 2 linear characteristics. Each of the d x dx 2 2 +2ξwnn +w x=0 front two springs have a stiffness of dt dt And the initial conditions at t=0 are 16 MN/m while the stiffness of each dx rear spring is 32 MN/m. The engine x(0)=X , (0) = 0. The amplitude speed (in rpm), at which resonance dt is likely to occur, is of x(t) after n complete cycles is a) 6040 b) 3020 ξ ξ c) 1424 d) 955 a) Xe-2nπ b) Xe2nπ  2 2 [GATE-2009] 1-ξ 1-ξ Q.20 A vehicle suspension system 1-ξ2 c) Xe-2nπ d) X consists of a spring and a damper.  ξ The stiffness of the spring is 3.6 [GATE-2007] kN/m and the damping constant of the damper is 400 Ns/m. If the Q.16 For an under damped harmonic mass is 50 kg, then the damping oscillator, resonance factor (d) and damped natural a) occurs when excitation frequency (fn), respectively, are frequency is greater than a)0.471 and 1.19 Hz undamped natural frequency b) 0.471 and 7.48 Hz b) occurs when excitation c) 0.666 and 1.35 Hz frequency is less than undamped d) 0.666 and 8.50 Hz natural frequency [GATE-2009] c) occurs when excitation frequency is equal to undamped Q.21 The rotor shaft of a large electric natural frequency motor supported between short d) never occurs bearings at both the ends shows a [GATE-2007] deflection of 1.8 mm in the middle of the rotor. Assuming the rotor to Q.17 A uniform rigid rod of mass m= 1 kg be perfectly balanced & supported and length L= 1 m is hinged at its at knife edges at both the ends, the centre and laterally supported at likely critical speed (in rpm) of the one end by a spring of spring shaft is constant k= 300N/m. The natural a) 350 b) 705 frequency in rad/s is ωn c) 2810 d) 4430 a) 10 b) 20 [GATE-2009] c) 30 d) 40 [GATE-2008] Q.22 A mass m attached to a spring is subjected to a harmonic force as Q.18 The natural frequency of the spring shown in figure. The amplitude of mass system shown in the figure is the forced motion is observed to be closest to 50 mm. The value of m (in kg) is

a) 8 Hz b) 10 Hz c) 12 Hz d) 14 Hz [GATE-2008] a) 0.1 b) 1.0

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission c) 0.3 d) 0.5 by a spring of stiffness k. For very [GATE-2010] small amplitude of vibration, neglecting the weights of the rod Q.23 The natural frequency of a spring- and spring, the undammed natural

mass system on earth is ω.n The frequency of the system is natural frequency of this system on k 2k a) b) the moon (gmoon = gearth/6) is m m a) b) ωn 0.408 ωn k 4k c) d) c) d) 0.204 ωn 0.167 ωn 2m m [GATE-2010] [GATE-2012]

Q.24 A mass of 1 kg is attached to two Q.27 If two nodes are observed at a identical springs each with stiffness frequency of 1800 rpm during k= 20kN/m as shown in the figure. whirling of a simply supported long Under the frictionless conditions, slender rotating shaft, the first the natural frequency of the system critical speed of the shaft in rpm is in Hz is close to a) 200 b) 450 c) 600 d) 900 [GATE-2013]

Q.28 Critical damping is the a) Largest amount of damping for which no oscillation occurs in free vibration a) 32 b) 23 b) Smallest amount of damping for c) 16 d) 11 which no oscillation occurs in [GATE-2011] free vibration c) Largest amount of damping for Q.25 A disc of mass m is attached to a which the motion is simple spring of stiffness k as shown in the harmonic in free vibration figure. The disc rolls without d) Smallest amount of damping for slipping on a horizontal surface. The which the motion is simple natural frequency of vibration of the harmonic in free vibration system is [GATE -2014(1)]

Q.29 In vibration isolation, which one of the following statements is NOT correct regarding Transmissibility (T)? 1k 1 2k a) T is nearly unity at small a) b) excitation frequencies 2πm 2πm b) T can be always reduced by 1 2k 1 3k c) d) using higher damping at any 2π 3m 2π 2m excitation frequency [GATE-2011] c) T is unity at the frequency ratio

Q.26 A concentrated mass m is attached of 2 at the centre of a rod of length 2L as d) T is infinity at resonance for shown in the figure. The rod is kept undamped systems in a horizontal equilibrium position [GATE -2014(2)]

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Q.30 Consider a single degree-of-freedom system with viscous damping excited by a harmonic force. At resonance, the phase angle (in degree) of the displacement with respect to the exciting force is a) 0 b) 45 c) 90 d) 135 [GATE -2014(3)] kk+ kk+ a) 12 b) 12 Q.31 A point mass is executing simple 2m 4m harmonic motion with an amplitude kk− kk+ of 10 mm and frequency of 4 Hz. The c) 12 d) 12 maximum acceleration ()m/s2 of m m [GATE -2014(2)] the mass is ______[GATE -2014(4)] Q.34 The damping ratio of a single degree of freedom spring-mass-damper Q.32 A rigid uniform rod AB of length L system with mass of 1 kg, stiffness and mass m is hinged at C such that 100 N/m and viscous damping AC= L/3,CB = 2L/3 .Ends A and B coefficient of 5 Ns/m is ______are supported by springs of spring [GATE -2014(3)] constant k. The natural frequency of the system is given by Q.35 A mass-spring-dashpot system with mass m = 10 kg, spring constant k = 6250 N/m is excited by a harmonic excitation of 10 cos(25t) N. At the steady state, the vibration amplitude of the mass is 40 mm. The damping coefficient (c, in N.s/m) of the dashpot is ______.

k k a) b) 2m m 2k 5k c) d) m m [GATE -2014(1)]

Q.33 What is the natural frequency of the spring mass system shown below? [GATE -2014(3)] The contact between the block and Q.36 A single degree of freedom system the inclined plane is frictionless. The has a mass of 2 kg, stiffness 8 N/m mass of the block is denoted by m and viscous damping ratio 0.02. The and the spring constants are dynamic magnification factor at an denoted by as shown below. k1 & k2 excitation frequency of 1.5 rad/s is______[GATE -2014(4)]

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Q.37 In a spring-mass system, the mass is Q.40 A precision instrument package (m m and the spring constant is k. The = 1 kg) needs to be mounted on a critical damping coefficient of the surface vibrating at 60 Hz. It is system is 0.1kg/s. In another spring- desired that only 5% of the base mass system, the mass is 2m and the surface vibration amplitude be spring constant is 8K. The critical transmitted to the instrument. damping coefficient (in kg/s) of this Assuming that the isolation is system is ______designed with its natural frequency [GATE-2015(2)] significantly lesser than 60 Hz, so that the effect of damping may be Q.38 Which of the following statements ignored. The stiffness (in N/m) of are TRUE for damped vibrations? the required mounting pad is P. For a system having critical ______. damping, the value of damping [GATE-2015(1)] ratio is unity and system does

not undergo a vibratory motion. Q.41 A single-degree freedom spring- Q. Logarithmic decrement method mass system is subjected to a is used to determine the amount sinusoidal force of 10 N amplitude do damping in a physical system. R. In case of damping due to dry the spring. The stiffness of the friction between moving springand frequency is 150 N/m, ω alongdamping the factor axis ofis surfaces resisting force of 0.2 and the undamped natural constant magnitude acts frequency is at steady state, the opposite to the relative motion. amplitude of vibration (in m) is S. For the case of viscous damping, drag force is directly [GATE-2015(2)] proportional to the square of approximately 10 ω. relative velocity. Q.42 Figure shows a single degree of a) P and Q only b) P and S only freedom system. The system c) P, Q and R only d) Q and S only consists of a massless rigid bar OP [GATE-2015(3)] hinged at O and a mass m at end P. The natural frequency of vibration Q.39 Considering massless rigid rod and of the system i small oscillations, the natural frequency (in rad/s) of vibration of the system shown in the figure is

1k 1k a) fn = b) fn = 2π 4m 2π 2m 1k 1 2k c) fn = d) fn = 2mπ 2mπ 400 400 [GATE-2015(3)] a) b) 1 2 Q.43 A single degree of freedom spring- 400 400 c) d) mass system is subjected to a 3 4 harmonic force of constant [GATE-2015(1)] amplitude. For an excitation

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission frequency of the ratio of the [GATE-2016(1)] amplitude of steady state response to the static deflection of the spring Q.47 A solid disc with radius a is is . 3k , m connected to a spring at a point d above the center of the disc. The other end of the spring is fixed to the vertical wall. The disc is free to [GATE-2016(3)] roll without slipping on the ground. The mass of the disc is M and the Q.44 A single degree of freedom mass- spring constant is K. The polar spring-viscous damper system with moment of inertia for the disc about mass m, spring constant k and its centre is 2 J Ma2 / 2 viscous damping coefficient q is critically damped. The correct relation among m, k, and q is a) q= 2km b) q= 2 km 2k k c) q = d) q = m m [GATE-2016(2)] The natural frequency of this system in rad/s is given by 2 Q.45 The system shown in the figure 2K() a+ d 2K a) b) consists of block A of mass 5 kg 2 3Ma 3M connected to a spring through a 2 2 massless rope passing over pulley B 2K() a+ d Ka()+ d c) d) of radius r and mass 20 kg. The Ma 2 Ma 2 spring constant k is 1500 N/m. If [GATE-2016(1)] there is no slipping of the rope over the pulley, the natural frequency of the system is rad/s. Q.48 The static deflection of a spring under gravity, when a mass of 1 kg is suspended from it, is 1 mm. Assume the acceleration due to gravity g =10 m/s2. The natural frequency of this spring-mass system (in rad/s) is____. [GATE-2016(3)]

Q.49 The damping ratio for a viscously [GATE-2016(2)] damped spring mass system, governed by the relationship d2 x dx Q.46 A single degree of freedom spring m+ c += kx F(t), is given by mass system with viscous damping dt2 dt has a spring constant of 10 kN/m. c c The system is excited by a a) b) mk sinusoidal force of amplitude 100 N. 2 km c c If the damping factor (ratio) is 0.25, c) d) the amplitude of steady state km 2mk oscillation at resonance is mm. [GATE-2017(1)]

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission suspension and the centre of mass is Q.50 A thin uniform rigid bar L and M is 250mm. considering the acceleration hinged at point O, located at a due to gravity is 9.81m/s2, the natural distance of L/3 from one of its ends. frequency (in radian/s) of the The bar is further supported using compound pendulum is springs, each of stiffness, k located [GATE-2017(2)] at the two ends. A particle of mass m =M/4 is fixed at one end of the bar, as shown in the figure. For small rotations of the bar about O, the Q.53 A machine of mass m = 200 kg is natural frequency of the system is supported on two mounts, each of stiffness k = 10 kN/m. The machine is subjected to an external force (in N) F(t) = 50 cos 5t. Assuming only vertical translatory motion, the magnitude of the dynamic force (in N) transmitted from each mount to the ground is ______(correct to two decimal places).

5k 5k a) b) M 2M 3k 3k c) d) 2M M [GATE-2017(1)]

Q.51 A mass m is attached to two identical springs having constant k as shown in the figure. The natural

freedom system is frequency ω of this single degree of [GATE-2018(1)]

Q.54 In a single degree of freedom underdamped spring mass-damper system as shown in the figure, and additional damper is added in parallel such that the 2k k a) b) system still remains underdamped. Which m m one of the following statements is ALWAYS k 4k true? c) d) 2m m [GATE-2017(2)]

Q.52 The radius of gyration of a compound pendulum about the point of suspension is 100 mm. The distance between the point of

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission (a) Transmissibility will increase. (b) Transmissibility will decrease. (c) Time period of free oscillations will increase. (d) Time period of free oscillation will decrease. [GATE-2018(2)] Q.55 The equation of motion for a spring mass system excited by a harmonic force is + = cos( ) Where M is the mass, K is the spring stiffness, F 𝑀𝑀𝑥𝑥̈is the force𝑘𝑘𝑘𝑘 amplitude𝐹𝐹 𝜔𝜔𝜔𝜔 e and is the angular frequency of excitation. Resonance occurs when is equal to 𝜔𝜔

(a) (b) 𝜔𝜔 𝑀𝑀 1 𝑘𝑘 (c ) 2� 𝑘𝑘 (d)2𝜋𝜋 �𝑀𝑀 𝑘𝑘 𝑘𝑘 𝜋𝜋�𝑀𝑀 �𝑀𝑀 [GATE-2018(2)]

ANSWER KEY:

1 2 3 4 5 6 7 8 9 10 11 12 13 14

(d) (b) (b) (a) (c) (c) (c) (c) (c) (d) (d) (c) (c) (a) 15 16 17 18 19 20 21 22 23 24 25 26 27 28 (a) (c) (c) (b) (a) (a) (b) (a) (a) (a) (c) (d) (b) (b) 29 30 31 32 33 34 35 36 37 38 39 40 41 42

(a) (c) 6.32 (d) (d) 1.25 10 0.799 0.4 c d 6767.6 b a

43 44 45 46 47 48 49 50 51 52 53 54 55

0.5 b 10 20 a 100 b a 15.66 (33.33) (c) (d)

Q.1 (d) and at B Give a small displacement to the 1 PE= mg L – ( δ+ x) + kx2 assembly. So assembly oscillates B 2 about its mean position. So change in potential energy from position A to position B is AB B - A 1 =mgL −−+ mg6 mgx kx2 −+ mgL mg6 ∆PE = PE PE 2

1 2 = AB = kx - mgx 2 ∆PE Q.3 (b) Given m = 10 kg. d = 30 mm = 0.03m From this a restoring torque is acts l = 500 mm = 0.5 m. along the line of oscillation. Eshift = 2.1× 1011 Pa Net restoring torque, T = mg sin (a + θ× ) L - mg sin (a - θ× ) L T = mgL [sin cos + cos sin - sin cos + cos sin ] T = 2mg L sinα sin θ. α θ Fromα veryθ small deflectionα θ . We know that, stick deflation due to sin α θ 10 kg of Mass oat the centre is given T = 2mgL cos θ by Nowθ from≅ θ newton’s law Wl33 mgl  θ α δ= = ...(i) Iθ + T = 0 48El 48El Iθ cos =0 The moment of inertia of the shaft. 2 d θ ππx 4− 34 2m+ L 2mgLθ2 + () 2mgLα θ cos α θ= 0 l= d = (0.03) = 3.974 × 10 m dt2 64 64 l= m L22 + m L ...(ii) d2θ g cos a Substitute value sin equation (i), we + θ= 2 0 get dt L 2 2 10×× 981 (0.5) On comparing with θ+ω θ=0 δ= n ××11 × ×− 4 we get 48 2.1 10 3.974 10 12.2625 = = 3.06 ×10-5m 2 g cos a 3 ω=n 400.58× 10 L If is the critical or white line g cos a c ω= speed in r.p.s. then n L ω gg ω=cc ⇒2f π = Q.2 (b) δδ Let initial length of the spring L = 1 g 1 9.81 f = − Potential energy at A, c 2πδ 2 × 3.14 3.06 × 10−3 PEA = mg(L −δ )

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 1 9.81 From the similar triangle OCD & =− =90.16Hz ≅ 90 Hz OAB, 6.28 30.6× 10 6 yx tan θ= = Q.4 (a) 0.15 0.30 Given M = 20 kg , l =1000 = 1 m , If is very, very small, then 2 yx A = 25×25 mm tan θ= 9 θ Esteel =200 GPa = 200 10 Pa 0.15 0.30 Mass moment of inertia of a square x = 0.30 and y = 0.15 section is given by, On taking moment about the hinged − point o θ θ b43 (25× 10 ) I= = = 325 × 10−84 m kx × 300+ W × y = 0 12 12 Wy 300 y 1 Deflection of a cantilever. Loaded k = =-×− = =N/mn with a point load placed at the free 300x 300 x 2 end is, From equation (i) Wl33 mgl 20×× 9.81 (1) 3 y 0.15θ N δ= = = = = −500 3EI 3EI 3× 200 ×× 1094 3.25 × 10− x 0.30θ m 196.2 Negative sign shows that the spring = = 0.01m tends to move to the point B. 19500 In magnitude. k = 500 N/m. g 9.81 ωn = = = 31.32rad / sec δ 0.01 Q.6 (c) Therefore, critical damping constant. ω Given = r = 0.5 CC = 3M n =2 × 20 × 31.32 ω = 1252.8 Ns/m 1250 Ns/m n ω And due to isolation damping ration C Q.5 (c) ε= =0 For isolation c = 0 CC We know the transmissibility radio of isolation is given by 2 ω 1+  2s ωn T.R. = 222 ωω   +1 +  2s  ωω nn   10+ 1 4 = = = 2 2 0.75 3 1++ (0.5) 0

Q.7 (c)

Given l = 300 mm = 0.3 m, W = 300N Let, rod is twisted to the left, through an angle

θ.

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 2 ()1−ε ωdn = 0.9 ω On spearing both the side, we get 2 I−ε2 =() 0.9 = 0.81 ε=−2 1 0.81 = 0.19 ε=0.19 = 0.436 And Damping ratio is given by CC ε= = CC 2 km Give m1 = m2 =0.5 kg, r1 =0.5 m, r2 = 0.6 m. Balancing mass m = 0.1 kg C= 2 km ×ε= 2 1000 × 0.1 × 0.436 Let disc rotates with uniform = 8.72 Ns/m  8.7 Ns/m angular velocity and x &y is the position of balancing mass along X Q.9 (c) &Y axis. The speed of sound in air = 332 m/s Resolving the forces in the x - For frequency of instrument of 144 direction, we get Hz Length of sound wave. 332 Fx = 0 L= = 2.30m 0.5[-6-0×06 cos 300+ 0.05 cos 00 ] l 144 Σ 2 = 0.1 × x × 3 For sample P of 64 Hz. 0.5 × (- 0.00196) = 0.1 x 332 2 L= = 5.1875m Fωc= wr ω p 64 x = -0.0098 m = -9.8 mm 332 Similarlyω in y -direction, F = 0 Q of 96 Hz LQ = = 3.458m y 96 0.5 (0.06 × sin 300 + 0.05×sin0) 332 2 3 Σ = 0.1 × y × R of 128 Hz LR = = 2.593m 0.5 × 0.03 = 0.1×y 128 yω = 0.15m = 150ω mm 332 S of 250 Hz LS = = 1.2968m Position of balancing mass is given 256 by, Here, the length of sound wave of

22 22 sample R(LR= 2.593m) is most close r= x +=− y()() 9.8 + 150 to the length of sound wave of = 150.31mm 150 mm Instrument (LI= 2.30 m). Hence, sample R produce most perceptible Q.8 (c) ≃ induced vibration. Given m = 0.1 kg , k = 1kN/m

Let ωd be the frequency of damped Q.10 (d) nbe the natural Give m = 12.5 kg, k = 1000 N/m, frequency of spring mass system. C = 15 Ns/m. Hencevibration &ω Critical Damping.

ω=d90%of ω= nn 0.9 ω (Given) …(i) k CC = 2m = 2 km Fernery of damped vibration m 2 On substitution the values, we get ωd = (1− εω ) n From equation (i) and equation (ii), Cc = 2 ()1000 × 12.5 = 223.6 Ns/m we get Q.11 (d)

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission We know logarithmic decrement. 2πε δ= ...(i) 1−ε2 C 15 And ε= = =0.0671 CC 223.6 Ns C= 223.6 , C m Differential equation governing the Now, from equation (i) we get above system is, 2×× 3.14 0.0671 d2 x dx dy δ= =0.422 =− − − −= 2 ∑ F m2 c kx 0 1− () 0.0671 dt dt dt mx+ c() x −+ y kx = 0 Q.12 (c) Give = = kN /m, m 250kg,k 100 Q.14 (a) C kk N = 3600 rmp ,  = = 0.15 The springs, with stiffness & are CC 22 2π n 2 ×× 3.14 3600 in parallel combination. So their ω= = =20rad/sec 60 60 resultant stiffness will be. ω 376.8 kk So = =18.84 kk1 =+= wn 20 22 T.R. = As & k are in series, so the 2 resultant stiffness will be ω 12+  kk× k2 k F ωn keq = += r = k+ k 2k 2 2222 F ωω  The general equation of motion for + ++ 12 12  undammed free vibration is given ωωnn  as.

mx+= keq x 0 1+× (2 0.15 × 18.84)2 k 2 2 mx+= x 0 1−() 354.945 +×[] 2 0.15 × 18.84 2 k 1+ 31.945 32.945 += = x x0 2 2m −+ 125309 []1 354.945 31.945 Compare above equation with =0.0162 general equation x+ω2 x0 = . we get n Natural frequency of the system is. Q.13 (c) kk Assume any arbitrary relationship ω=2 ⇒ω=0 between the coordinates and their nn2m 2m first derivatives, say x > y and x > y. Alternative Also assume x > 0 and x > 0 . k k = A small displacement gives to the eq 2 system towards the left direction. We know for a spring mass system. Mass m is fixed, so only damper keq k/2 k moves for both the variable x and y . ω=n = = Note that these forces are acting in m m 2m the negative direction.

Give the equation of motion of a Theσ amplitude of x(t) after n harmonic oscillator is compile cycles is.

d2x dx 2 nd x1 +2 ξωnn + ω x0 = e = dt2 dt x(t) + ξω + ω2 = −nb x 22nn x x 0 x(t)= e x Compare equation (i) with the r2πξ general equation. x= xe 1−ξ mx++= cx kx 0 From equation (iv) ck xxx0++= mm Q.16 (c) c We get =2 ξω ...(ii) For an under damped harmonic m n oscillator resonance occurs when excitation frequency is equal to the kk2 =ωnn ⇒ω = ....(iii) undammed natural frequency mm ωω= From equation (ii) & (iii) dn cc ξ= = k 2 km Q.17 (c) 2m× m Logarithmic decrement. x 2cπ ξ=In 1 = x 22 2 ccc − x 2π× 2 ξ km ξ=In 1 = x 22 2 ccc − x Given m = 1kg, L = 1m, k = 300 N/m = In 1 We have to turn the rigid rod at an x2 2π× 2 ξ km 4 π×ξ km rod moves upward at a distance x = = 22 2 andangle also θ about deflect its hingedin the point, opposite then 2 km−ξ 2 km 4km− 4 π×ξ km () () direction with the same amount. Let 2πξ = is very small and take tan 1−ξ2 From AOB θ xL θ≃θ x1 2πξ θ= ⇒x = θ ...(i) = = L/2∆ 2 x 2 2 e1−ξ and ...(ii) If system executes n cycles, the By using the principal of energy logarithmic decrement can be conservation.θ= ωt ⟹ θω

written as 1122 1 x δ lω+ kx =Constant ξ= log, 1 22 2 nxn1+ 12 1L x lkθ+ θ = c end = 1 2 22 xn1+ From equation (i) and (ii) Where x_1 = amplitude at the 11 lθ+2 Lk 22 θ= c starting position. 22

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission On differentiating w.r.t. t we get Given k1 = k2 = 16 MN/m , 1 kL2 k3 = k4 = 32 MN/m, m = 240 kg l×+×= 200 200 0 Here k & k are the front two 24 1 2 For a rigid rod of length L & mass m. springs or k1 & k2 are the rear two hinged at its centre, the moment of springs. inertia. These 4 springs are parallel. So equivalent stiffness. mL2 l = keq=++ k 1 k 2 + k 3 k 4 12 2 Substitute l in equation (iii), we get = 16 + 16 + 32 + 32 = 96 MN /m We know at resonance 1 mL22 kL =××+200 = 0 k 2 12 4 ω=ωn = 3k m θ+ θ=0 ...(iv) 2Nπ k m = eq Compare equation (iv) with the 60 m general equation. N = Engine speed in rpm 2 θ+ω θ= 8 n 0 60k 60 96× 10 N =eq = 3k So, we have = 2ππ m 2 240 m 60 ××102 40 3k 3× 300 2π ω= = =30rad/sec n m1 = 6042.03 6040 rpm

Q.18 (b) Q.20 (a) ≃ Given m= 1.4 kg, k1 = 4000 N/m , Given k = 3.6 kN/m, c = 400 Ns /m, k2 = 1600 N/m m =50 kg In the given system k1 & k2 are in We know that, Natural Frequency parallel combination k 3.6× 1000 ω= = =8.485rad / sec So, Keq = k1 + k2 = 4000 + 1600 = n m 50 5600 N/m And damping factor is given by Natural frequency of spring mass D or system is given by, c c 400 ∈= = = 1keq 1 5600 cC 2 km 2××× 3.6 1000 50 fn = = 2ππ m 2 1.4 400 1 = = 0.471 =×=63.245 10.07 10Hz 2× 424.26 2π Damping Natural frequency. ω =1 −ε2 ω Q19 (a) dn 2 2πfdn = 1-ε ω ω f = d d 2π ω f1=d ×−ε2 d 2π 8.485 = ×−1 (0.471)2 = 1.19 Hz 2× 3.14

Q.21 (b)

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Given = 1.8 mm = 0.0018 m F0 A = 22 The critical or whirling speed is ()()km−ω+ω a given by. Here F(t) =100 cos (100t), Fo=100 g N, ω=2 δ A = 50 mm = 50×10-3 m ω= 2Nπ c g 100rad / sec = -1 60 δ k =3000Nm , c =0 So from equation (i), we get 60 g 60 9.81 = = F Nc = 0 2πδ 2 × 3.14 0.0018 A 2 km−ω = 9.55 5450 = 704.981 705 rpm F km− ω=2 0 A ≃ Q.22 (a) 100 300 – m (100)2 = 50× 10−3 1000 m = 1000 m = 0.1 kg

Q.23 (a) ⟹ We know natural frequency of a Given k = 3000 N/ m , c = 0, A = 50 spring mass system is. mm, F(t) = 100 cos(100t)N ω=t 100t k ω=n ω=100 , It is a forced vibratory m system. This equation (i) does not depend From the Newton’s law, on the g and weight (W = mg) mx+kx=F ….(i) So, the natural frequency of a spring And its general solution will be. mass system is unchanged on the x= A cos ω t moon. dx Hence, it will remain n, =x =−ω A sin ω t i.e. = dt moon n ω k where ω= Q.24 (a) ω ω m Give k = 20 kN/ m , m =1kg dx2 From the Given spring mass system =x =−ω A2 cos ω t dt2 springs are in parallel combination. Substitute these values in equation So, (ii), we get Keq =+= k k 2k ω2 cos ω+ t kAcos ω= t 100cos( ω t) - Natural Frequency of spring mass MAω+2 kA = 100 system is, Now substitute k = 3000 N/m, A = K ω= eq 0.5m in above equation, we get p m m× 0.05×(100)2 + 3000× 0.05 = Keq 100-5m+1.5=1 2fπ=n m = 0.1 kg m Alternative Method f0=Natural Frequency is file. We now that, in forced vibration 1K 1 2k f =eq = amplitude is given by: n 2ππ m2m

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 1 2×× 20 1000 Therefore, natural frequency of = vibration of the system is, 2× 3.14 1 ω 200 n 1 2k = = 31.84 Hz 32 Hz fn = = 6.28 2ππ 2 3m

≃ Q.25 (c) Q.26 (d) For a very small amplitude of vibration

Form above figure change in length of spring x = 2Lsinθ = 2Lθ

Mass moment of inertia of mass (m) (isabout very O smallis so sin θ θ) l = mL2 As no internal force acting on the x system So governing equation of ⇒xr ⇒θ r motion from Newton’s law of motion is. θTotal = energy of the system remains constant. Iθ + kx× 2L = 0 So T.E. = T.E. due to translator or mL2θ ×2L = 0 motion + K.E. due to rotary motion 4kL2θ + P.E of spring θ + = 0+ k2Lθ mL2 1 11 T.E. = 22+ θ+ 2 4kθ mx I kx Or θ + = 0 2 22 m 1 11 = mr22θ+ I θ+ 2 kx 22 θ From Comparing general equation 2 22 θ + ω2 n=0 we have equation (i) xr = θ 4k 4k 1111 ω=2 ⇒ω= = mr22θ+ × mr 22 θ+ kx 22 θ For n m n m 2222

mr 2 a disc I= Q.27 (b) 2 For helical gears, speed ratio is 3 given by = mr22θ+ kx 22 θ= Constant 4 N D COSβ 12=× 2----(i) On differentiating above equation N21 D COSβ 1 w.r.t. t. we get 1= 1440 rpm, D1 = 80 mm, 31 0 mr22×() 2 θθ  + kx() 2 θθ  = 0 D2 = 120 mm, 1 = 30 . 2 = 22.5° 42 Hence𝑁𝑁 from Eq. (i)

3 22 2k D COS𝛽𝛽β 𝛽𝛽 θ+ θ= θ+ θ= 11 mr kx 0 0 N21 = × ×N 4 3m D12 COSβ ⇒ ωn 1 2k 80 COS30° fn = = = × ×1440 2ππ 2 3m 120 COS22.5°

Q.30 (c) Q.36 (2.0 to 2.4) Damping Ratio Q.31 (6.3 to 6.4) CC =0.02 = = Given A=10mm=0.01m 2mω S F=4Hz n 28×× m x= A cos ω t dx 8 v= =−ω A sin ω t ⇒C = 0.02 ××× 2 8 = 0.32 dt 8 dx2 Dynamic magnification factor = =−ω2 ω 2 a A cos t 1 dt = 2 For maximum, t=0 c22ωω 2 22 +−1 ∴a = A ω= 0.01(2 π f ) 22 s ωn ω= π = ×π×22 =  2 f 0.01(2 4) 6.32m / s 1 = = 0.799 22 2 Q.32 (d) (0.32)× (1.5) (1.5) 22+−1 (8) (1) Q.33 (d) It is parallel Q.37 0.4 k= kk + C eq 1 2 ω= C1 n1 2 keq m1 ω=n m CC1= 2Km 1 1 = 2Km kk+ = = × = =×= ω= 12 CC2 2 K 1 m 1 2 8K 2m 4C C1 4 0.1 0.4 n m Q.38 (c) Q.34 (1.24 to 1.26) Damping ratio Q.39 (d) CC Form ()ξ= = 2 Cc 2 km dx ⇒θ⇒2r.dx "= 2rθ C 25 dt = = =1.25 2 1× 100 20 Taking moments 2rθ .m.2r. θ+ " 400 × (r θ×θ= ) r 0 Q.35 (9 to 11) ⇒ 4m.θ " + 400 ×θ F 400 X = ⇒ ω= 2 2 ()km−ω2 +() C ω 4 =40mm = 0.04 Q.40 6767.6 F= 10N ω=2 π N = 2 ×π× 60 = 376.99 ω=25

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 22 1 11 1 ωω  M.F=± =±== 0.5 .05= ⇒− 20 = 1 − ⇒  = 21 2 2 ωω 13− 2 ω nn 1− ω 1−  ωn ωn ω= ω2 2m21 Q.44 (b)

ω=n 82.266rad / s We know that q k = ωn ω=n =82.266 2m m k 1= 6767.6005N / m q=×= 2m 2 km m Q.41 (b) Amplitude of vibration Q.45 10 f /s Displace the block “A” & Release A = 0 22 1 22 ω 2 ξω  (krθ )r + Mr + mr θ= 0 1−+   2 ωωnn    10− 150 = k 2 θ+ θ=0 22 1 11   Mm+ 1− +×  2 0.2 ×  2 10  10   = 0.06605 k 1500  0.07 ω=n  = =10rad / sec 1 + Mm+ 10 5  Q.42 (a) 2 force in the spring F= 2mg [from equilibrium] Alternate method: Deflection as mass at P, Energy of system remain conserved, 11 1 E= J ω22 + mv − mgy + k(y +δ ) 2 ...(1) 22 2

x x=×= 2a 2x Where, 1 a 2mg 4mg =×=2 equilibrium which is calculated as kk follows:δ=Static elongation of spring at gg k mg w;n = = = mg= k δ⇒δ= s x1 4m k Differentiating Eqn.(1)w.r.t time, 1 1k which will be zero because E fnn= ω= 2ππ 2 4m constant dE Q.43 0.5 = 0 dt …(2)

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission dω dv Jω + mv − mgv + k(y +δ )v = 0 dt dt Since there is no slipping between rope & pulley

12  v  1 dv  dv Mr  +mv − mgv + k(y +δ )v = 0 2v=rω r  r dt  dt

M dy2 +m += ky 0 2 dt2 k 1500 Q.48 100 ω=n = =10rad / sec M 10+ 5 −3 + m ∆=st 10 m 2 g 10 ω= = =100rad / sec n ∆ −3 Q.46 20 st 10 Given: Spring constant (k) = 10 kN/m Q.49 (b) = 10,000 N/m

Magnitude of force (F0 )= 100N Damping factor (ξ= ) 0.25

frequency ()ωn Static deflection of Forcing frequency (ω) = Natural F 100 spring =0 = =10−2 m = 10mm k 104 Dynamic deflection Static deflection = 22 ωω   12− +ξ   ωω dw= p.2 π rdr nn   dFr=µ=µπ dw p.2 rdr 10 = dT= dFr.r 2 ()1− (1)22 +× (2 0.25 × 1) dT=µπ p.2 rdr.r = π× 2 10 2 p 0.01r.r dr = = 20mm 3 2× 0.25 dT= 0.0628pr dr R Total torque (T) = Q.47 (a) ∫ dT Apply D'Alembert Principle: 0 Ro [k(a+ d) θ ](a + d) + (I + Ma2 ) θ= 0 3 cm T= ∫ 0.0628pr dr 2 k(a+ d) Ri θ+ θ= R 2 0 4 o Ma 2 r + Ma = 0.0628p  2 4 2 =44 − 2k(a+ d) T 0.0157P Roi R ω=n 3Ma 2 44 T= 0.0157P 40 − 20

Q.51 (a)

By equilibrium method 2L 1 Iθ+ kx × + kx × = 0 0133 2 22 2L   L Ik0θ+  + k  θ= 0 33   By equilibrium method 5kl2 Iθ+ θ= 0 (1) ∑=F0 0 9 y mx ++= kx kx 0 Now for Io mx += 2kx 0 Io= (I o ) bar + (I o ) point mass

2k ω=n 2L L 4L− 3L L m x = −= = 32 6 6 ∴ 2 Q.52 15.66 (I0 ) bar= I C.G + Mx Ml22 Ml = + 12 36 3Ml22+ Ml Ml 2 (I ) = = 0 bar 36 9

2 2l 4ml2 (I0 ) point load mass = M = 39

4M2 (I0 ) point mass =  l 94 Ml2 By equilibrium method (I0 ) point mass = θ+ θ = 9 I mg sin l 0 22 2 Ml Ml 2Ml ∴=I + = & I =mk2 0 99 9 Ifmk θ 2sθ+ very mgl less θ= sin 0 θ ≈ θ Put I0 value in equation (1)

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Hence, increasing damping in underdamped system increases time period. OR Transmissibility gl 9.81× .25 ω= = 1 + 2 n k22 (.1) 2 = 𝜔𝜔 � ω=15.66rad / sec � 𝜉𝜉 𝜔𝜔𝑛𝑛� n 𝜖𝜖 1 + 2 2 2 2 𝜔𝜔 𝜔𝜔 Q.53 (33.33) Additional �Damper� is𝑛𝑛 added� into parallel𝑛𝑛 − �𝜔𝜔 � � 𝜉𝜉 𝜔𝜔 � The Force Transmissibility, Then Damping will increase. 1 + 2 But it is still underdamped. 2 But , will increase. = 𝜔𝜔 = � 𝑠𝑠 Here no unbalance force is there. � 𝜉𝜉 𝜔𝜔𝑛𝑛� 𝐹𝐹 𝐹𝐹𝑡𝑡𝑡𝑡 1 + 2 𝜉𝜉 2 2 2 𝐹𝐹𝑜𝑜 But, = 1 = 𝜔𝜔 𝜔𝜔 2 𝐾𝐾 Since damping�� − is � zero𝑛𝑛� �in the� system,𝜉𝜉 𝑛𝑛� 𝑑𝑑 𝑛𝑛 𝑛𝑛 𝜔𝜔 𝜔𝜔 But as𝜔𝜔 increases𝜔𝜔 � − 𝜉𝜉 �𝜔𝜔 �𝑚𝑚� ∴ ξ = 0 Then will decreases 1 𝜉𝜉 = = Then 𝑑𝑑 = will increases 𝑠𝑠 𝜔𝜔 2𝜋𝜋 1 𝐹𝐹 Q.55 (d)𝑑𝑑 𝐹𝐹𝑡𝑡𝑡𝑡 2 𝑇𝑇 𝜔𝜔𝑑𝑑 Natural frequency of system,𝜔𝜔 𝐹𝐹𝑜𝑜 The given equation is � 𝑛𝑛 � − �𝜔𝜔 � + = cos( ) (10,000) × 2 = = It is the equation of frequency under 2000 damped forced𝑀𝑀𝑥𝑥̈ vibration.𝑘𝑘𝑘𝑘 𝐹𝐹 𝜔𝜔𝜔𝜔 𝑛𝑛 𝐾𝐾 𝜔𝜔 � �= 100 / Also we know that the natural frequency of 1 𝑚𝑚 1 free vibration is given by = = = 1.33 √ 𝑟𝑟𝑟𝑟𝑟𝑟25 𝑠𝑠𝑠𝑠𝑠𝑠 𝑠𝑠 5 ( ) = 𝐹𝐹 1 1 100 100 2 2 𝑘𝑘 𝐹𝐹𝑜𝑜 𝜔𝜔𝑛𝑛 Force� transmitted− � � to� foundation,� − � �� = 𝑀𝑀 = 1.33√ = 1.33 × 50 = 66.66 𝑘𝑘 Force transmitted to ground from each 𝑛𝑛 � 𝑠𝑠 𝑜𝑜 At resonance, 𝜔𝜔 = mounting𝐹𝐹 = 𝐹𝐹= 33.33 𝑁𝑁 Therefore the angular speed𝑀𝑀 at which 𝑛𝑛 𝐹𝐹𝑠𝑠 resonance occurs is 𝜔𝜔 𝜔𝜔 2 Q.54(c) 𝑁𝑁 Adding damper in parallel increases overall = = damping in system such as, 𝑘𝑘 𝜔𝜔 𝜔𝜔𝑛𝑛 � Ceq = C1 + C2 𝑀𝑀 But overall damping in still under damped as given in question. Since the system in free damping system as shown in figure. So the only option is damped time period as, 2 2 = = 1 𝑑𝑑 𝜋𝜋 𝜋𝜋 𝑇𝑇 𝑑𝑑 2 𝜔𝜔 𝜔𝜔𝑛𝑛� − 𝜉𝜉

Q.1 Two masses m are attached to Q.2 Three masses are connected to a opposite sides of a rigid rotating rotating shaft supported on bearing shaft in the vertical plane. Another A and B as shown in the figure, The pair of equal masses m1 is attached system is in a space where the to the opposite sides of the shaft in gravitational effect is absent. the vertical plane as shown in figure. Neglect the mass of shaft and rods Consider m = 1 kg, e= 50 mm, e1 = connecting the masses. For

20 mm, b = 0.3 m, a = 2 m and a1 = m1 = 10kg , m2 = 5kg and m3 = 2.5kg 2.5 m. For the system to be and for a shaft angular speed of dynamically balanced, m1 should be 1000 radian/s the magnitude of the kg. bearing reaction (in N ) at location B is ____

[GATE-2016(3)] [GATE-2017(2)]

ANSWER KEY:

1 2 2 0

EXPLANATIONS

Q.1 2

Couple due to m = couple due to m1 It means all three masses are in same plane Let us calculate the net force mea= m111 e a 50  2  = m1 =××= 1   2kg 2 20  2.5  𝑥𝑥 = (10 × 0.1 × ) � 𝐹𝐹 � 𝑚𝑚𝑟𝑟 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 (5 × 0.2 × 2 60 ) Q.2 (0) (2.5 × 0.4 ×𝜔𝜔2 600 ) . − 1 𝜔𝜔 𝑐𝑐𝑐𝑐𝑐𝑐2 1 0 = 1 5 × 0.2 × 2.5 × 0.4 × − 2 𝜔𝜔 𝑐𝑐𝑐𝑐𝑐𝑐2 2 � − �= � − � �� 𝜔𝜔 =2(5 × 0.2 × 60 ) � 𝐹𝐹𝑦𝑦 � 𝑚𝑚𝑟𝑟 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 (2.5 × 0.4 × 2 600 ) 𝜔𝜔 𝑠𝑠2𝑠𝑠𝑠𝑠 0 − 𝜔𝜔 𝑠𝑠𝑠𝑠𝑠𝑠

= + = 0 2 2 Therefore𝐹𝐹 reaction�� 𝐹𝐹at𝑥𝑥 B is� zero 𝐹𝐹𝑦𝑦 = 0

𝑅𝑅𝐵𝐵

Q.1 A car is moving on a curved horizontal road of radius 100 m with a speed of 20 m/s. The rotating masses of the engine have an angular speed of 100 rad/s in clockwise direction when viewed from the front of the car. The combined moment of inertia of the rotating masses is 10 kg-m2. The magnitude of the gyroscopic moment (in N-m) is . [GATE-2016(1)]

ANSWER KEY: 1 200

EXPLANATIONS

Q.1 200

Given: Spin velocity

(ω=s ) 100 rad/sec Moment of Inertia (MOI) =10kg − m2 Precision Angular Velocity linear speed ()ω= p R 20 = = 0.2rad / sec 100

Gyroscopic moment = MOIωsp ×ω =××10 100 0.2 =200Nm

Q.1 If N is the number of link in a a) Parallel to link mechanism then number of possible b) Perpendicular to link inversion is equal to c) Radically outward along link a) N b) N- 1 d) Coincident with the axis of link c) N + 1 d) N + 2 Q.6 A constrained kinematic chain Q.2 The Kinematic planar chain shown (mechanism) has n number of links. in the given figure is a What is the number of instantaneous centres n(n 1) n(n 1) a) b) 2 2 n(2n 1) n(2n 1) c) d) 2 2

Q.7 The product of diametral pitch and a) Structure circular pitch is b) Mechanism with 2 degree of a)π b) Zero freedom 1 c) One d) c) Mechanism with 1 degree of 2 freedom d) Mechanism with more than 2 Q.8 The velocity of sliding of meshing degree of freedom gear teeth is  a)   x b) 1 x Q.3 The magnitude of the Coriolis 12  component of acceleration of a 2 12   slider .moving at a velocity Von a c) 12   x d) link rotating at angular speed is x a) Vω b) 2Vω Q.9 A rack is a gear of infinite V 2V c) d) a) Pitch b) Module 2  c) Diameter d) Number of teeth

Q.4 Klein's construction can be used Q.10 When two spur gears having when involute profiles on their teeth a) Crank has a uniform angular engaged, the line of action is velocity tangential to the b) Crank has a non-uniform a) Pitch circles velocity b) Dedendum circles c) Crank has uniform angular c) Addendum circles acceleration d) Base circle d) Crank has a uniform angular velocity and angular acceleration Q.11 Common contact ratio of a pair of spur pinion Q.5 Coriolis component of acceleration a) less than 1 b) equal to 1 is always c) greater than 1 d)greater than3

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Q.12 Recommended gear to be used for a a) less sensitive b) highly sensitive speed reduction of 40 : 1 c) high stable d) less stable a) Worm and worm wheel b) Hypoid gear Q.19 In a governor, the controlling force c) Hering bone gear curve shows a relationship between d) Bevel gear a) controlling force and speed of rotation Q.13 The arrangement is called bevel b) controlling force and radius of gearing, when two shafts are rotation a) intersecting and co-planer c) controlling force and height of b) non-intersecting & non-co-planer sleeve lift c) parallel and co-planer d) controlling force and mass of d) parallel and non-co-planer governor ball

Q.14 The type of gears used to connect Q.20 In the case of a flywheel of mass two parallel co-planer shaft is moment of inertia 'I' rotating at an a) straight bevel gears angular velocity 'ro' the expression b) cross helical gears 1/2 lro2 represents the c) spiral gears a) centrifugal force d) straight spur gears b) angular momentum c) torque Q.15 A Hartnell governor is a governor of the d) kinetic energy a) inertia type b) pendulum type Q.21 The amount of energy absorbed by a c) centrifugal type flywheel is determined from the d) dead weight type a) torque-crank angle diagram b) acceleration-crankangle diagram Q.16 Sensitiveness of a governor is c) speed-space diagram defined as d) speed-energy diagram a) range of speed/mean speed b) mean speed/range of speed Q.22 A body of mass m and radius of c) mean speed x range 8f speed gyration K is to be replaced by two d) none of these masses m1 and m2 located at distances and from the C.G. of Q.17 Which one of the following original body. These will be statements is correct? dynamically equivalent to original A governor will be stable if the body. If radius of rotation of the balls a) h h k b) h22 h K a) increases as the equilibrium 12 12 speed decreases c) K h12 h d) K h12 h b) decreases as the equilibrium speed increases Q.23 In case of partial balancing of single c) increases as the equilibrium cylinder reciprocating engine what speed increases 1s the primary disturbing force d) remains unaltered with the along the line of stroke change in equilibrium speed. a) cm2 cos

22 Q.18 The problem of hunting of a b) 1 C  mr  cos  centrifugal governor becomes very c) 1 C  mr 2 cos  acute when the governor becomes

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission d) 1 C  m 22 r cos  c) the point on pitch curve having c = fraction of reciprocating mass to the maximum pressure angle be balanced d) none ω = angular velocity of crank shaft θ= crank angle Q.29 The CAM-follower generally used in automobile engines is Q.24 The rotor of every prime mover a) Knife-edge follower must be balanced b) Flat faced follower a) to eliminate or minimize effect c) Spherical faced follower due to unbalanced forces d) Roller follower b) to regulate the effects due to c) to eliminate or minimize inertial Q.30 Which of the following motion of forces cam roller follower leads to the d) to keep speed constant lowest jerk? a) Uniform b) Simple harmonic Q.25 The governing equation of a specific c) Cycloidal d) Parabolic simple vibrating system is Q.31 How many degrees of freedom do 2xx the kinematic chain as shown in 2 B  Kx   t tt figure above have, when it The system is subjected to undergoes planer motion. a) forced vibrations b) free vibrations c) resonance d) no vibrations

Q.26 Whirling speed of a shaft coincides with the natural frequency of its a) -1 b) 0 a) longitudinal vibration c) 2 d) 3 b) transverse vibration c) torsional vibration Q.32 Which one of the following d) coupled bending torsional statements is correct? Transmission vibration angle is the angle between a) the output link and the frame Q.27 The normal speed of rotation of a b) the output link and the coupler shaft is chosen to be c) the input link and the coupler a) equal to natural frequency of the d) the input link and the frame shaft b) twice the natural frequency of Q.33 The given figure shows below a/an the shaft c) much above or below the natural frequency of the shaft d) a random multiple of natural frequency of the shaft

Q.28 Pitch point on a CAM is a) locked chain a) any point on pitch curve b) constrained kinematic chain b) any point on pitch circle c) unconstrained kinematic chain d) mechanism

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Q. 34 Consider a four-bar mechanism C. Wind shield wiper shown in the given figure. The D. Harmonic vibrators driving link DA is rotation uniformly List-11 at a speed of 100 rpm clockwise 1. Scotch yoke 2. Crank rocker 3. Whitworth 4. Geneva wheel Codes: A B C D a) 1 3 2 4 b) 4 2 3 1 The velocity of A will be c) 4 3 2 1 a) 300 cm/s b) 314 cm/s d) 1 2 3 4 c) 325 cm/s d) 400 cm/s

Q.35 The directions of Coriolis Q.38 The number of links in a planer component of acceleration, 2 , of mechanism with revolute joints· having 10 Instantaneous centre is the slider A with respect to the a) 3 b)4 coincident point B is shown in figure c) 5 d) 6 1, 2, 3 and 4. Directions shown by figures Q.39 Match List-1 with List-It and select the correct answer using the codes given below the lists: List-1 A. Pantograph B. Single slider crank chain C. Double slider crank chain D. Straight line motion List-11 1 . Scotch yoke mechanism 2. Double lever mechanism a) 2 and 4 are wrong b) 1 and 2 are wrong 3. Tchebi cheff mechanism c) 1 and 3 are wrong 4. Double crank mechanism d) 2 and 3 are wrong 5. Hand pump Codes: Q.36 A point on a connecting link of a A B C D double slider crank mechanism a) 4 3 5 1 traces a b) 2 5 1 3 a) straight line path c) 2 1 5 3 b) hyperbolic path d) 4 5 2 1 c) parabolic path d) elliptical path Q.40 The four bar mechanism shown in the figure(Given: OA = 3 cm, AB = 5 Q.37 Match List-1 (Application) with List- cm, BC = 6 cm, OC = 7 cm) is a It (Mechanism) and select the correct answer using the codes given below the lists: List-1 A. Mechanical counters B. Shaping machine

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission a) Double crank mechanism a) constant b) Double rocker mechanism b) maximum at the pitch point c) Crank rocker mechanism c) minimum at the pitch point d) Single slider mechanism d) zero at the pitch point

Q.41 A slider sliding at 10 cm/s on a link Q.46 The maximum efficiency for spiral which is rotating at 60 rpm is gears in mesh is given by subjected to Coriolis acceleration of 1 cos(    ) 1 cos(    ) a) b) magnitude 1 cos(    ) 1 cos(    ) 2 2 2 a) 40π cm/s b) 0.4 π cm/s 1 cos(    ) 1 cos(    ) c) 40π cm/s2 d) 4 π2cm/s2 c) d) 1 cos(    ) 1 cos(    )

Q.42 The mechanism shown in the given figure represents Q.47 Match List-1 (Terms) with List-11 (Definitions) and select the correct answer using the codes given below the lists: List-I A. Module B. Addendum C. Circular pitch List-11 1. Radial distance of a tooth from a) Hart's mechanism the pitch circle to the top of the b) Toggle mechanism tooth c) Watt's mechanism 2. Radial distance of a tooth from d) Beam Engine mechanism the pitch circle to the bottom of

the tooth Q.43 Ratio of pitch circle diameter in 3. Distance on the circumference of millimeters to the number of teeth is the pitch circle from a point of known as one tooth to the corresponding a) module b) circular pitch point on the next tooth c) diametral pitch d) clearance 4. Ratio of pitch circle diameter in

mm to the number of Teeth Q.44 Two parallel shafts whose axes are Codes: separated by a distance of 75 mm A B C D are to be connected a spur gear set a) 1 2 3 4 so that the output shaft rotates at b) 1 4 3 2 50% of the speed of the input shaft c) 3 4 1 2 which of the following could be the d) 3 2 1 4 possible pitch circle diameters of the

gears. Q.48 Consider the following specifications a) 25 mm and 50 mm of gears A, B, C and D b) 30 mm and 60 mm c) 50 mm and 100 mm d) 60 mm and 120 mm

Q.45 In case of a gear pair with involute tooth profile, the pressure angle throughout the contact is

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Which of these gears form the pair and a pitch circle diameter of 210 of spur gears to achieve a gear ratio mm. If the pressure angle of the of 3? worm is 20°, what is the axial pitch a) A and B b) A and D of the worm? c) Band C d) C and D a) 7 mm b) 22 mm c) 14 mm d) 63 mm Q.49 Which one of the following is the correct statement? The consequence Q.54 Match List-1 with List-11 and select of a slight increase in the centre the correct answer using the codes distance between two mating given below involute gears is that List-I a) The law of gearing is not A. Compound gear train satisfied perfectly B. Epicyclic spur gear train with b) Interference occurs brake bands c) Pressure angle increases C. Worm and worm-wheel gear d) Pressure angle decreases train D. Epicyclic bevel gear train Q.50 If an imaginary circle is drawn List-II which by pure rolling action gives 1. Automobile gear box the same motion as the actual gear, 2. Automatic transmission of what is the circle called? automobile a) Addendum circle 3. Speed reducers for lifts b) Pitch circle 4. Automobile differential c) Root circle Codes: d) Dedendum circle A B C D a) 1 2 3 4 Q.51 How does the contact ratio of helical b) 1 4 3 2 gears compare with that of spur c) 3 4 1 2 gears? d) 3 2 1 4 a) It is lesser than that of spur gear b) It is greater than that of spur Q.55 100 kW power is supplied to the gear machine through a gear box which c) Infinite uses an Epicyclic gear train. The d) It is equal to that of spur gear power is supplied at 100 rad/s. The speed of the output shaft of the gear Q.52 In an epicyclic gear train with a sun box is 10 rad/s in a sense opposite gear, planet gear and a moving arm. to the input speed. What is the The ratio of number of teeth of the holding toque on the fixed gear of sun gear to that of the planet gear is the train? 3. If the planet gear is fixed and the a) 8 kNm b) 9 kNm arm has an angular velocity of 150 c) 10 kNm d) 11 kNm RPM the sun gear will rotate with a) 50 RPM b) 100 RPM Q.56 Why is the mass of flywheel c) 150 RPM d) 200 RPM concentrated in the rim? a) to store minimum energy Q.53 Speed reduction in a gear box is b) to make it strong achieved. Using a worm and worm c) to store maximum energy wheel. The worm wheel has 30 teeth d) to let it rotate freely

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Q.57 The safe rim velocity of a flywheel is (assuming negative torque demand influenced by the is negligible) a) centrifugal stresses b) fluctuation of energy c) fluctuation of speed d) mass of the flywheel

Q.58 The crank-effort diagram of an engine running a machine is showing the areas above and below a) 80 Nm b) 120 Nm the mean line (in joules). What is the c) 60 Nm d) 40 Nm maximum fluctuation of energy in the below diagram? Q.62 The moment of inertia of a flywheel is 2000 kgm2. Standing from rest, it is moving with a uniform acceleration of 0.5 rad/S2. After 10 seconds from the start its kinetic energy will be

a) 0 J b) 30 J a) 250 Nm b) 500 Nm c) 50 J d) 55 J c) 5000 Nm d) 25000 Nm

Q.59 A flywheel is fitted to the crank shaft Q.63 The angular speed of a watt's of an engine having indicated work governor when its height is 20 cm, per Revolution E. If the permissible will be equal to limits of the coefficients of fluctuation a) 20 rad/sec b) 10 rad/sec of energy and speed are Ke and Ks c) 6 rad/sec d) 7 rad/sec

respectively, then the kinetic energy Q.64 In a spring-controlled governor the of the flywheel is equal to controlling force curve is straight 2K E KE a) e b) e line. The balls are 400 mm apart Ks 2Ks when the controlling force is 1600 N

KEe KEs and they are 240 mm apart when c) d) the force is 800 N. To make the Ks 2Ke governor isochronous the initial Q.60 In a reciprocating engine, the tension must be increased by maximum fluctuation in energy is a) 100 N b) 200 N found to be 2 J and the variation in c) 400 N d) 800 N speed is limited to 2% of the mean Q.65 The stiffness of spring k used in the speed. If the mean speed is 100 Hartnell governor as shown in the rad/s and the radius of gyration of given figure (F & F are centrifugal wheel is 0.1 m, the mass of the 1 2 forces at maximum & minimum radii flywheel is of rotation r & r respectively) is a) 3 kg b) 2 kg 1 2 c) 1 kg d) 0.5 kg

Q.61 A simplified turning moment diagram of a tour-stroke engine is shown in the given figure. If the mean torque ‘Tm’ is 10 Nm, the

estimated peak torque ‘Tp’ will be

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission force 'F' (=A sinωt) acts on the mass 2 FF12 a) 2(b / a)  and the damper is not used, then rr12

2 FF b) 2 (a / b) 12 rr12

2 F +F c) 2(b / a) 12 rr12

2 F12 +F d) 2 (a / b)  rr 1 12 a)  K/M b)  K/M 2 Q.66 The controlling force curve used to c) 2 K / M d) 2 K / 2M determine the stability and

sensitiveness of a governor in a Q.70 A spring-mass suspension has a graph whose ordinate is controlling natural frequency of 40 rad/s. What force. What is abscissa of this graph? is the damping ratio required if it is a) Speed of rotation of the engine desired ro reduce this frequency to shaft 20 rad/s by adding a damper to it? b) Height of the governor 3 1 c) Lift of sleeve a) b) d) Radius of rotation of balls 2 2 1 1 Q.67 Two spring of equal length but c) d) 2 4 having stiffness of 10 N/mm and 15 N/mm support· a mass of 2 tonnes in series. Find the frequency of Q.71 A reciprocating engine, running at vibration 80 rad/s, is supported on springs. 1 1 The static deflection of the spring is a) 3Hz b) 10Hz 1 mm. Take g = 10 m/s2. When are 2 2 engine runs, what will be the 1 1 c) 15Hz d) 2Hz frequency of vibrations of the 2 2 system? a) 80 rad/s b) 90 rad/s Q.68 Three spring of stiffness K1, K2 and c) 100 rad/s d) 160 rad/s K3 are connected in series. The stiffness of a single spring equivalent to these three springs Q.72 The static deflection of a shaft under would be equal to a flywheel is 4 mm. Take g = 10 m/s2. What is the critical speed in a) KKK 1 2 3 rad/s? KKK b) 1 2 3 a) 50 b) 20 3 c) 10 d) 5 KKK c) 1 2 3 Q.73 A flywheel has a mass of 300 kg and KKKKKK1 2 2 3 3 4 a radius of gyration of 1 m. It is KKKKKK given a spin of 100 rpm about its d) 1 2 2 3 3 4 horizontal axis. The whole assembly KKK 1 2 3 rotates about a vertical axis at 6

Q.69 The given figure show vibrations of rad/sec. The gyroscopic couple a mass 'M' isolated by means of experienced will be springs and a damper. If an external a) 3πkNm b) 6πkNm c) 180 kNm d) 360 πkNm

ANSWER KEY:

1 2 3 4 5 6 7 8 9 10 11 12 13 14 (a) (c) (b) (a) (b) (a) (a) (c) (c) (d) (c) (a) (a) (d) 15 16 17 18 19 20 21 22 23 24 25 26 27 28 (c) (b) (c) (b) (b) (d) (a) (c) (c) (a) (a) (b) (c) (c) 29 30 31 32 33 34 35 36 37 38 39 40 41 42 (c) (b) (a) (b) (c) (b) (a) (d) (c) (c) (b) (c) (c) (b)

43 44 45 46 47 48 49 50 51 52 53 54 55 56 (a) (c) (a) (b) (a) (b) (c) (b) (b) (d) (b) (b) (b) (c) 57 58 59 60 61 62 63 64 65 66 67 68 69 70 (a) (c) (b) (c) (a) (c) (d) (c) (b) (d) (a) (c) (a) (a) 71 72 73 (c) (a) (b)

Q.1 (a) Q.6 (a) If there are in number of inks, then the number of possible invasion will Q.7 (a) be equal to the number of links in T Diametric pitch  the mechanism. D U Q.2 (c) Circular pitch  T This device is known as telescopic TU hand (D.P.) (C.P.)    DT

Q.8 (c)

Q.9 (c) Rack is defined as a great of infinite diameter

It has 4 loop and10 link hence Q.10 (d) degree of freedom s N-9 i.e. one Two important properties of an involute Q.3 (b) i) A normal to an involute is a Magnitude f Conrolis component – 2 tangent to the basic circle. ii) The radius of curvature of an involute is equal to the length Q.4 (a) the tangent to the bases circle. Klein’s construction is used to determine the acceleration of Q.11 (c) various part and this method is used Contract ratio of a pair of spur when crank has uniform angular pinion and gear is usually more than velocity. 1 here appropriate answer is (c)

Q.5 (b) Q.12 (a) Worm and worm wheel is used for heavy speed reduction.

Q.13 (a) To have a gear drive between two intersecting shafts bevel gears are used.

Q.14 (d) Corollas components of acceleration Q.15 (c) appears in the direction Classification of Governors perpendicular to the link.

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Conditions for dynamically equal valence. 1. The sum of their messes is equal to the total mass of the body i.e. m – m1 + m2 2. The centre of gravity of two masses coincide with that of the body. 3. The sum of mass moment o inertia of the masses about their centre of gravity is equal to the mass moment of inertia of the body. Q.16 (b) K h h 12 Q.17 (c) For stable governor Q.23 (c) F = Ar - B And F = Mrω2 Q.24 (a)

2 B  A  Q.25 (a) r Since excitation force is present in

the equation hence it is the equation Q.18 (b) of forced vibration. Sensitiveness of a governor is a

desirable quality However if a Q.26 (b) governor is too sensitive. It may Whirling of shaft occurs when the fluctuate continuously above and natural frequency of transverse below the mean position it is called vibration matches with the hunting of governor. frequency of rotating shaft.

Q.19 (b) Q.27 (c) Controlling force curve is plotted

between controlling force at y – axis Q.28 (c) and radius of rotation of x-axis.

Q.29 (c) Q.20 (d)

Kinetic energy stored in the Q.30 (b) 1 flywheel is K.E. = 2 2 Q.31 (a) Q.21 (a) In this N = 4 The amount of energy absorbed by a P1 = 5 flywheel is determined form the DOF = 3(N-1) -2P1 torque-crank angle diagram. The DOF = 3(4-1) – 2 5 area under torque – crank angle DOF = 9-10 = -1 diagram is known as turning moment. Q.32 (b)

The angle between the output link Q.22 (c) and the couple is known as transmission angle.

Q.33 (c)

Q.36 (d) It is known as elliptical trammel hence, it will trace elliptical path

Q.37 (c)

Q.38 (c) Number of links = 5 No of instantaneous centre = Number of lower pair (joint) = 5 n(n 1) for N= 5 Degree of freedom 2 f = 3 (N-1) – 3p1 We get, No of instantaneous centre = f=3 (5-1) – 2× 5 10 f =2 Since DOF > 1 Hence it is unconstrained kinematic Q.39 (b) chain. Q.40 (c) Q.34 (b) Shortest link OA = 3 cm Longest link OC = 7 cm Sum of shortest and longest link = 10 cm Sum of other two = 5 + 6 = 11 cm If sum of shortest and longest link is lesser than sum of other two then rocker mechanism is obtained.

Q.41 (c) V = rω Carioles acceleration 2 100 2 60 30 = 314 cm /sec 2V   2  10  60 60 40 cm / S2 Q.35 (a) Fcor = 2Vω Q.42 (b) 2 and 4 are wrong according to This is toggle mechanism, this above equation. These should be as mechanism is used to overcome a show below. larger resistance of a barber with a small diving force.

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Where D = Diameter of the pitch Q.43 (a) circle and ν T = Number of teeth on the wheel Module = T Where D = Pitch circle diameter Q.48 (b) T = No of teeth Combination of gear B-C and C-D cannot use beaus pressure angle is Q.44 (c) not same. Similarly combination of Centre distance between the shaft gear – A-B cannot use because = 75 mm module is not some hence Hence combination of A-D should be taken vv because they have some module and 12 75 2 pressure angle with gear ratio of 3.

D D 150 12 Q.49 (c) Also D11 2D Interference is involutes gear can be

3D1  150 avoided by increasing the centre D 50mm distance. Interference can also be 2 eliminating by increasing the D 100mm 1 pressure angle.

Q.45 (a) Q.50 (b) In involutes profile of gear teeth Pitch Circle: The pitch circle is the pressure angle is constant curve of intersection of the pitch throughout the engagement of teeth. surface of revolution and the plane This result is smooth running of the of rotation. It is an imaginary circle gear. that rolls without slipping with the pitch circle of a mating gear. The Q.46 (b) pitch circles of a par of mating gears are tangent to each other Q.47 (a) Module: It is the ratio of the pitch Q.51 (b) circle diameter to the number of teeth. It is usually denoted by m Q.52 (d) Mathematically. Module, m = D/T Addendum: The addendum is the radial distance between pitch and the addendum circles. Addendum, indicates the height of tooth above the pitched circle. Circle Pitch: It is the distance measured on the circumference of the pitch circle from a point of one

tooth of the corresponding point on the next tooth It is usually denoted T2 by Pc. Mathematically.  3 T2 Circular pitch Pc =πD/T If N3 = 0 N1 = 150 rpm

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission  N1 N3 N2 Rim velocity v  Fix the arm and give 0 +1 T   3 +1 rev to adjacent T2 Where gear σ = centrifugal stress of Multiply with x 0 +x T x 2 circumferential shreds. T 1 Ρ = density of rim material T Add y y X + y yx 3 T Hence rim velocity is the 1 function of centrifugal stress and

density of rim material. N = 0 x + y =0 3 N = 150 y = 150 1 Q.58 (c) X = -150

T2 N2 = y – x  T2 1 N2 = 150 + 150  3 EA = E = 150 + 50 = 200 rpm EB = E + 50

EC = E + 50 – 20 = E +30 Q.53 (b) Eo = E + 30 + 5 = E + 35 d Axial pitch p = 2 EE = E + 35 - 35 = E T2 Maximum energy =E + 50 210 Minimum energy = E p = = p = 7π⇒p= 22 mm 30 Maximum fluctuation of energy = E + 50 –E = 50 J Q.54 (b) Q.59 (b) Q.55 (b) Coefficient of fluctuation of energy Holding torque (Ke) Maximumfluctuation of energy 1  T31 T 1 work doneper cycle 2 E P = T1×ω  E 100×103=T1×100 Coefficient of fluctuation of speed T1=1000N-m (K ) 100 s T3=1000 1 Rangeof speed 10  meanspeed T3=1000×9⇒ T3=9kN-m 2 ∆E = Iω Ks Q.56 (c) 1 2 E= 2Ks Flywheel stores energy in the form 2 of inertia and its inertia will be more E.Ks=(K.E)2Ks for heavy mass at its periphery. This EK K.E  e is the reason mass of flywheel is 2K concentrated in the rim. s

Q.60 (c) Q.57 (a) E = IW2CS

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 2 when f = 800 N r = 240 mm 2 = m(0.1)2 (100)2× 100 1600= a(0.4) + b (i) M = 1 kg 800= a(O.24) ÷ b By solving (i) and (ii) we get Q.61 (a) a=5000N/m Area of triangle= Area of rectangle b=—400N 1 3   2    T 2 P

T4m  

2 TPm  4  .T

TP  8  10  80N  m

Q.62 (c) To make the governor isochronous, l = 2000 kgm2 the controlling force line must pass Initial velocity = 0 through the origin. This is possible 2 a= 0.5 rad/s only if the initial tension is T = 10 sec increased by 400 N. ω2+ ω1+at Since initial verity is zero Q.65 (b) ω1= 0 Stiffness of the spring And ω2= 5 rad/s 2 a FF12 Kinetic energy of flywheel K2  1 2 2 r12 r  2 2 1 Q.66 (d)  2000  25  250000N  m 2 Controlling force curve is drawn between controlling force & radiuses Q.63 (d) of rotation 895 h  N2 Q.67 (a) Where h = height of governor N = RPM 895 N2  0.2 N 4475 66.86 2 N 2 66.89    60 60 In Series rad / sec 1 1 1  k K K eq 1 2 Q.64 (c) 1 1 1  The controlling force curve of a keq 10 15 spring controlled governor is k 6 N / mm or 6000 N / m straight line and thus can be eq expressed as 1 k 1 6000 3 ff    f = ar + b 2 M 2 2000 2 when f= 1600N r=400mm HZ

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Q.68 (c) 2 2 100 C 300(1)  6 In series combination of springs 60 1 1 1 1    C = 5×2π×100×6 KKKKeq 1 2 3 C = 10 ×100×6 π 1 KKKKKK C = 6 π kNm  2 3 1 3 1 2 KKKKeq 1 2 3 1 KKK  1 2 3 KKKKKKKeq 1 2 2 3 3 4

Q.69 (a) KK KK   eq 22 k  m

Q.70 (a) ωn = 40 rad/s ωd = 20 rad/s 2 nn  1   20 1  2 40 1 1  2 4 1 2 1  4 3 2 2

Q.71 (c) Frequency of vibration of system g 10    100rad / sec 1 103

Q.72 (a) Critical speed in rad/s2 g   10 50rad / sec 1 103

Q.73 (b) Gyroscopic couple C = 1ωp