kinematics of machinery unit i basics of...

KINEMATICS OF MACHINERY

UNIT I BASICS OF MECHANISMS 9

Classification of mechanisms – Basic kinematic concepts and definitions – Degree of freedom,

Mobility – Kutzbach criterion, Gruebler‟s criterion – Grashof‟s Law – Kinematic inversions of four-

bar chain and slider crank chains – Limit positions – Mechanical advantage – Transmission

Angle – Description of some common mechanisms – Quick return mechanisms, Straight line

generators, Universal Joint – rocker mechanisms.

UNIT II KINEMATICS OF LINKAGE MECHANISMS 9

Displacement, velocity and acceleration analysis of simple mechanisms – Graphical method– Velocity

and acceleration polygons – Velocity analysis using instantaneous centres – kinematic analysis of

simple mechanisms – Coincident points – Coriolis component of Acceleration – Introduction to linkage

synthesis problem.

UNIT III KINEMATICS OF CAM MECHANISMS 9

Classification of cams and followers – Terminology and definitions – Displacement diagrams –

Uniform velocity, parabolic, simple harmonic and cycloidal motions – Derivatives of follower

motions – Layout of plate cam profiles – Specified contour cams – Circular arc and tangent cams – Pressure angle and undercutting – sizing of cams.

UNIT IV GEARS AND GEAR TRAINS 9

Law of toothed gearing – Involutes and cycloidal tooth profiles –Spur Gear terminology and

definitions –Gear tooth action – contact ratio – Interference and undercutting. Helical, Bevel, Worm, Rack and Pinion gears [Basics only]. Gear trains – Speed ratio, train value – Parallel axis gear trains – Epicyclic Gear Trains.

UNIT V FRICTION IN MACHINE ELEMENTS 9

Surface contacts – Sliding and Rolling friction – Friction drives – Friction in screw threads –

Bearings and lubrication – Friction clutches – Belt and rope drives – Friction in brakes- Band

and Block brakes.


SCE 1 Department of Mechanical Engineering

UNIT I BASICS OF MECHANISMS

Introduction:

Definitions : Link or Element, Pairing of Elements with degrees of freedom, Grubler’s criterion

(without derivation), Kinematic chain, Mechanism, Mobility of Mechanism, Inversions,

Machine.

Kinematic Chains and Inversions:

Kinematic chain with three lower pairs, Four bar chain, Single slider crank chain and Double

slider crank chain and

their inversions.

Mechanisms:

i) Quick return motion mechanisms – Drag link mechanism, Whitworth mechanism and Crank

and slotted lever mechanism

ii) Straight line motion mechanisms – Peacelier’s mechanism and Robert’s mechanism.

iii) Intermittent motion mechanisms – Geneva mechanism and Ratchet & Pawl mechanism.

iv) Toggle mechanism, Pantograph, Hooke’s joint and Ackerman Steering gear mechanism.

Terminology and Definitions-Degree of Freedom, Mobility

• Kinematics: The study of motion (position, velocity, acceleration). A major goal of

understanding kinematics is to develop the ability to design a system that will satisfy specified motion requirements. This will be the emphasis of this class.

• Kinetics: The effect of forces on moving bodies. Good kinematic design should produce

good kinetics.

• Mechanism: A system design to transmit motion. (low forces)

• Machine: A system designed to transmit motion and energy. (forces involved)

• Basic Mechanisms: Includes geared systems, cam-follower systems and linkages (rigid links connected by sliding or rotating joints). A mechanism has multiple moving parts (for

example, a simple hinged door does not qualify as a mechanism).



• Examples of mechanisms: Tin snips, vise grips, car suspension, backhoe, piston engine,

folding chair, windshield wiper drive system, etc.

Key concepts:

• Degrees of freedom: The number of inputs required to completely control a system.

Examples: A simple rotating link. A two link system. A four-bar linkage. A five-bar

linkage.

• Types of motion: Mechanisms may produce motions that are pure rotation, pure translation,

or a combination of the two. We reduce the degrees of freedom of a mechanism by

restraining the ability of the mechanism to move in translation (x-y directions for a 2D

mechanism) or in rotation (about the z-axis for a 2-D mechanism).

• Link: A rigid body with two or more nodes (joints) that are used to connect to other rigid bodies. (WM examples: binary link, ternary link (3 joints), quaternary link (4 joints)

• Joint: A connection between two links that allows motion between the links. The motion

allowed may be rotational (revolute joint), translational (sliding or prismatic joint), or a combination of the two (roll-slide joint).

• Kinematic chain: An assembly of links and joints used to coordinate an output motion with

an input motion.

• Link or element:

A mechanism is made of a number of resistant bodies outof which some may have motions relative

to the others. Aresistant body or a group of resistant bodies with rigidconnections preventing their

relative movement is known as alink.

A link may also be defined as a member or a combination ofmembers of a mechanism, connecting

other members and havingmotion relative to them, thus a link may consist of one or moreresistant

bodies. A link is also known as Kinematic link or anelement.

Links can be classified into 1) Binary, 2) Ternary, 3) Quarternary, etc.

• Kinematic Pair:

A Kinematic Pair or simply a pair is a joint of two links having relative motion between them.



Example:

In the above given Slider crank mechanism, link 2 rotates relative to link 1 and constitutes a revolute

or turning pair. Similarly, links 2, 3 and 3, 4 constitute turning pairs. Link 4 (Slider) reciprocates

relative to link 1 and its a sliding pair.

Types of Kinematic Pairs:

Kinematic pairs can be classified according to

i) Nature of contact.

ii) Nature of mechanical constraint.

iii) Nature of relative motion.

i) Kinematic pairs according to nature of contact:

a) Lower Pair: A pair of links having surface or area contact between the members is known as a

lower pair. The contact surfaces of the two links are similar.

Examples: Nut turning on a screw, shaft rotating in a bearing, all pairs of a slider-crank mechanism,

universal joint.

b) Higher Pair: When a pair has a point or line contact between the links, it is known as a higher pair.

The contact surfaces of the two links are dissimilar.

Examples: Wheel rolling on a surface cam and follower pair, tooth gears, ball and roller bearings,

etc.

ii) Kinematic pairs according to nature of mechanical constraint.



a) Closed pair: When the elements of a pair are held together mechanically, it is known as a closed

pair. The contact between the two can only be broken only by the destruction of at least one of the

members. All the lower pairs and some of the higher pairs are closed pairs.

b) Unclosed pair: When two links of a pair are in contact either due to force of gravity or some

spring action, they constitute an unclosed pair. In this the links are not held together mechanically.

Ex.: Cam and follower pair.

iii) Kinematic pairs according to nature of relative motion.

a) Sliding pair: If two links have a sliding motion relative to each other, they form a sliding pair. A

rectangular rod in a rectangular hole in a prism is an example of a sliding pair.

b) Turning Pair: When on link has a turning or revolving motion relative to the other, they constitute

a turning pair or revolving pair.

c) Rolling pair: When the links of a pair have a rolling motion relative to each other, they form a

rolling pair. A rolling wheel on a flat surface, ball ad roller bearings, etc. are some of the examples

for a Rolling pair.

d) Screw pair (Helical Pair): if two mating links have a turning as well as sliding motion between

them, they form a screw pair. This is achieved by cutting matching threads on the two links.

The lead screw and the nut of a lathe is a screw Pair

e) Spherical pair: When one link in the form of a sphere turns inside a fixed link, it is a spherical

pair. The ball and socket joint is a spherical pair.

Degrees of Freedom:

An unconstrained rigid body moving in space can describe the following independent

motions.

1. Translational Motions along any three mutually perpendicular axes x, y and z,

2. Rotational motions along these axes.

Thus a rigid body possesses six degrees of freedom. The connection of a link with another imposes

certain constraints on their relative motion. The number of restraints can never be zero (joint is

disconnected) or six (joint becomes solid).

Degrees of freedom of a pair is defined as the number of independent relative motions, both

translational and rotational, a pair can have.

Degrees of freedom = 6 – no. of restraints.

To find the number of degrees of freedom for a plane mechanism we have an equation known as

Grubler’s equation and is given by F = 3 ( n – 1 ) – 2 j1 – j2



F = Mobility or number of degrees of freedom

n = Number of links including frame.

j1 = Joints with single (one) degree of freedom.

J2 = Joints with two degrees of freedom.

If F > 0, results in a mechanism with ‘F’ degrees of freedom.

F = 0, results in a statically determinate structure.

F < 0, results in a statically indeterminate structure.

Kinematic Chain:

A Kinematic chain is an assembly of links in which the relative motions of the links is

possible and the motion of each relative to the others is definite (fig. a, b, and c.)

In case, the motion of a link results in indefinite motions of other links, it is a non-kinematic chain.

However, some authors prefer to call all chains having relative motions of the links as kinematic

chains.

Linkage, Mechanism and structure:

A linkage is obtained if one of the links of kinematic chain is fixed to the ground. If motion

of each link results in definite motion of the others, the linkage is known as mechanism. If one of the

links of a redundant chain is fixed, it is known as a structure.

To obtain constrained or definite motions of some of the links of a linkage, it is necessary to know

how many inputs are needed. In some mechanisms, only one input is necessary that determines the

motion of other links and are said to have one degree of freedom. In other mechanisms, two inputs



may be necessary to get a constrained motion of the other links and are said to have two degrees of

freedom and so on.

The degree of freedom of a structure is zero or less. A structure with negative degrees of freedom is

known as a Superstructure.

• Motionand itstypes:

• The threemain typesof constrainedmotion in kinematic pairare,

1. Completely constrained motion : If the motion between a pair of links is limited to a definite

direction, then it is completely constrained motion. E.g.: Motion of a shaft or rod with collars at each

end in a hole as shown in fig.



2. Incompletely Constrained motion : If the motion between a pair of links is not confined to a

definite direction, then it is incompletely constrained motion. E.g.: A spherical ball or circular shaft

in a circular hole may either rotate or slide in the hole as shown in fig.

3. Successfully constrained motion or Partially constrained motion:If the motion in a definite direction is not brought about by itself but by some other means, then it is known as successfully constrained motion. E.g.: Foot step Bearing

• Machine:

It is a combination of resistant bodies with successfully constrained motion which is used to transmit

or transform motion to do some useful work. E.g.: Lathe, Shaper, Steam Engine, etc.

• Kinematic chain with three lower pairs

It is impossible to have a kinematic chain consisting of three turning pairs only. But it is possible to

have a chain which consists of three sliding pairs or which consists of a turning, sliding and a screw

pair.

The figure shows a kinematic chain with three sliding pairs. It consists of a frame B, wedge C and a

sliding rod A. So the three sliding pairs are, one between the wedge C and the frame B, second

between wedge C and sliding rod A and the frame B.



This figure shows the mechanism of a fly press. The element B forms a sliding with A and turning pair with screw rod C which in turn forms a screw pair with A. When link A is fixed, the required fly

press mechanism is obtained.

Kutzbach criterion,

Grashoff's law Kutzbach

criterion:

• Fundamental Equation for 2-D Mechanisms: M = 3(L – 1) – 2J1 – J2

Can we intuitively derive Kutzbach’s modification of Grubler’s equation? Consider a

rigid link constrained to move in a plane. How many degrees of freedom does the link have?

(3: translation in x and y directions, rotation about z-axis)

• If you pin one end of the link to the plane, how many degrees of freedom does it now have?

• Add a second link to the picture so that you have one link pinned to the plane and one free to

move in the plane. How many degrees of freedom exist between the two links? (4 is the

correct answer)

• Pin the second link to the free end of the first link. How many degrees of freedom do you now have?



• How many degrees of freedom do you have each time you introduce a moving link? How

many degrees of freedom do you take away when you add a simple joint? How many

degrees of freedom would you take away by adding a half joint? Do the different terms in

equation make sense in light of this knowledge?

Grashoff's law:

• Grashoff 4-bar linkage: A linkage that contains one or more links capable of undergoing a

full rotation. A linkage is Grashoff if: S + L < P + Q (where: S = shortest link length, L =

longest, P, Q = intermediate length links). Both joints of the shortest link are capable of 360

degrees of rotation in a Grashoff linkages. This gives us 4 possible linkages: crank-rocker

(input rotates 360), rocker-crank-rocker (coupler rotates 360), rocker-crank (follower);

double crank (all links rotate 360). Note that these mechanisms are simply the possible

inversions (section 2.11, Figure 2-16) of a Grashoff mechanism.

• Non Grashoff 4 bar: No link can rotate 360 if: S + L > P + Q

Let’s examine why the Grashoff condition works:

• Consider a linkage with the shortest and longest sides joined together. Examine the linkage

when the shortest side is parallel to the longest side (2 positions possible, folded over on the

long side and extended away from the long side). How long do P and Q have to be to allow

the linkage to achieve these positions?

• Consider a linkage where the long and short sides are not joined. Can you figure out the

required lengths for P and Q in this type of mechanism

Kinematic Inversions of 4-bar chain and slider crank chains:

• Types of Kinematic Chain: 1) Four bar chain 2) Single slider chain 3) Double Slider chain

• Four bar Chain:

The chain has four links and it looks like a cycle frame and hence it is also called quadric cycle

chain. It is shown in the figure. In this type of chain all four pairs will be turning pairs.



• Inversions:

By fixing each link at a time we get as many mechanisms as the number of links, then each

mechanism is called ‘Inversion’ of the original Kinematic Chain.

Inversions of four bar chain mechanism:

There are three inversions: 1) Beam Engine or Crank and lever mechanism. 2) Coupling rod of

locomotive or double crank mechanism. 3) Watt’s straight line mechanism or double lever

mechanism.

• Beam Engine:

When the crank AB rotates about A, the link CE pivoted at D makes vertical reciprocating motion at

end E. This is used to convert rotary motion to reciprocating motion and vice versa. It is also known

as Crank and lever mechanism. This mechanism is shown in the figure below.

2. Coupling rod of locomotive: In this mechanism the length of link AD = length of link C. Also

length of link AB = length of link CD. When AB rotates about A, the crank DC rotates about D. this

mechanism is used for coupling locomotive wheels. Since links AB and CD work as cranks, this

mechanism is also known as double crank mechanism. This is shown in the figure below.

3. Watt’s straight line mechanism or Double lever mechanism: In this mechanism, the links AB

& DE act as levers at the ends A & E of these levers are fixed. The AB & DE are parallel in the

mean position of the mechanism and coupling rod BD is perpendicular to the levers AB & DE. On

any small displacement of the mechanism the tracing point ‘C’ traces the shape of number ‘8’, a



portion of which will be approximately straight. Hence this is also an example for the approximate

straight line mechanism. This mechanism is shown below.

• 2. Slider crank Chain:

It is a four bar chain having one sliding pair and three turning pairs. It is shown in the figure below

the purpose of this mechanism is to convert rotary motion to reciprocating motion and vice versa.

Inversions of a Slider crank chain:

There are four inversions in a single slider chain mechanism. They are:

st

1) Reciprocating engine mechanism (1 inversion)

nd

2) Oscillating cylinder engine mechanism (2 nd

inversion)

3) Crank and slotted lever mechanism (2 inversion) rd

4) Whitworth quick return motion mechanism (3 rd

inversion)

5) Rotary engine mechanism (3 th

inversion)

6) Bull engine mechanism (4 th

inversion)

7) Hand Pump (4 inversion)

• 1. Reciprocating engine mechanism :

In the first inversion, the link 1 i.e., the cylinder and the frame is kept fixed. The fig below shows a

reciprocating engine.



A slotted link 1 is fixed. When the crank 2 rotates about O, the sliding piston 4 reciprocates in the

slotted link 1. This mechanism is used in steam engine, pumps, compressors, I.C. engines, etc.

• 2. Crank and slotted lever mechanism:

It is an application of second inversion. The crank and slotted lever mechanism is shown in figure

below.

In this mechanism link 3 is fixed. The slider (link 1) reciprocates in oscillating slotted lever (link 4)

and crank (link 2) rotates. Link 5 connects link 4 to the ram (link 6). The ram with the cutting tool

reciprocates perpendicular to the fixed link 3. The ram with the tool reverses its direction of motion

when link 2 is perpendicular to link 4. Thus the cutting stroke is executed during the rotation of the

crank through angle α and the return stroke is executed when the crank rotates through angle β or

360 – α. Therefore, when the crank rotates uniformly, we get

Time to cutting = α = α

Time of return β 360 – α

This mechanism is used in shaping machines, slotting machines and in rotary engines.

1.4 . 1 Whitworth quick return motion mechanism:



Third inversion is obtained by fixing the crank i.e. link 2. Whitworth quick return mechanism is an

application of third inversion. This mechanism is shown in the figure below. The crank OC is fixed

and OQ rotates about O. The slider slides in the slotted link and generates a circle of radius CP. Link

5 connects the extension OQ provided on the opposite side of the link 1 to the ram (link 6). The

rotary motion of P is taken to the ram R which reciprocates. The quick return motion mechanism is

used in shapers and slotting machines. The angle covered during cutting stroke from P1 to P2 in

counter clockwise direction is α or 360 -2θ. During the return stroke, the angle covered is 2θ or β.

Therefore, Time to cutting = 360 -2θ = 180 – θ

Time of return 2θθ = α = α . β 360 – α

1. Rotary engine mechanism or Gnome Engine:

Rotary engine mechanism or gnome engine is another application of third inversion. It is a rotary

cylinder V – type internal combustion engine used as an aero – engine. But now Gnome engine has

been replaced by Gas turbines. The Gnome engine has generally seven cylinders in one plane. The

crank OA is fixed and all the connecting rods from the pistons are connected to A. In this mechanism

when the pistons reciprocate in the cylinders, the whole assembly of cylinders, pistons and

connecting rods rotate about the axis O, where the entire mechanical power developed, is obtained in

the form of rotation of the crank shaft. This mechanism is shown in the figure below.



2 Double Slider Crank Chain:

A four bar chain having two turning and two sliding pairs such that two pairs of the same kind are

adjacent is known as double slider crank chain.

3 Inversions of Double slider Crank chain:

It consists of two sliding pairs and two turning pairs. They are three important inversions of double

slider crank chain. 1) Elliptical trammel. 2) Scotch yoke mechanism. 3) Oldham’s Coupling.

4 1. Elliptical Trammel:

This is an instrument for drawing ellipses. Here the slotted link is fixed. The sliding block P and Q in

vertical and horizontal slots respectively. The end R generates an ellipse with the displacement of

sliders P and Q.

The co-ordinates of the point R are x and y. From the fig. cos θ = x.PR

and Sin θ = y. QR Squaring and adding (i) and (ii) we get x2 + y2= cos2 θ + sin2 θ

x2 + y2= 1

2

(PR) 2

(QR)

2 2

(PR) (QR) The equation is that of an ellipse, Hence the instrument traces an ellipse. Path traced by mid-point of

2 2 2 2

PQ is a circle. In this case, PR = PQ and so x +y =1 (PR) (QR) It is an equation of circle with PR = QR = radius of a circle.

5. Scotch yoke mechanism: This mechanism, the slider P is fixed. When PQ rotates above P, the

slider Q reciprocates in the vertical slot. The mechanism is used to convert rotary to reciprocating

mechanism.



5.Oldham’s coupling: The third inversion of obtained by fixing the link connecting the 2 blocks

P & Q. If one block is turning through an angle, the frame and the other block will also turn

through the same angle. It is shown in the figure below.

An application of the third inversion of the double slider crank mechanism is Oldham’s coupling

shown in the figure. This coupling is used for connecting two parallel shafts when the distance

between the shafts is small. The two shafts to be connected have flanges at their ends, secured by

forging. Slots are cut in the flanges. These flanges form 1 and 3. An intermediate disc having

tongues at right angles and opposite sides is fitted in between the flanges. The intermediate piece

forms the link 4 which slides or reciprocates in flanges 1 & 3. The link two is fixed as shown. When

flange 1 turns, the intermediate disc 4 must turn through the same angle and whatever angle 4 turns,

the flange 3 must turn through the same angle. Hence 1, 4 & 3 must have the same angular velocity

at every instant. If the distance between the axis of the shaft is x, it will be the diameter if the circle

traced by the centre of the intermediate piece. The maximum sliding speed of each tongue along its

slot is given by

v=xω where, ω = angular velocity of each shaft in rad/sec v = linear velocity in m/sec

Mechanical Advantage, Transmission angle:

1The mechanical advantage (MA) is defined as the ratio of output torque to the input torque. (or)

ratio

of load to output.

2 Transmission angle.

3 The extreme values of the transmission angle occur when the crank lies along the line of frame.



Description of common mechanisms-Single, Double and offset slider mechanisms - Quick

return mechanisms:

1. Quick Return Motion Mechanisms:

Many a times mechanisms are designed to perform repetitive operations. During these operations

for a certain period the mechanisms will be under load known as working stroke and the remaining

period is known as the return stroke, the mechanism returns to repeat the operation without load. The

ratio of time of working stroke to that of the return stroke is known a time ratio. Quick return

mechanisms are used in machine tools to give a slow cutting stroke and a quick return stroke. The

various quick return mechanisms commonly used are i) Whitworth ii) Drag link. iii) Crank and

slotted lever mechanism

2. Whitworth quick return mechanism:

Whitworth quick return mechanism is an application of third inversion of the single slider

crank chain. This mechanism is shown in the figure below. The crank OC is fixed and OQ rotates

about O. The slider slides in the slotted link and generates a circle of radius CP. Link 5 connects the

extension OQ provided on the opposite side of the link 1 to the ram (link 6). The rotary motion of P

is taken to the ram R which reciprocates. The quick return motion mechanism is used in shapers and

slotting machines.

The angle covered during cutting stroke from P1 to P2 in counter clockwise direction is α or 360 -2θ.

During the return stroke, the angle covered is 2θ or β.

3. Drag link mechanism :

This is four bar mechanism with double crank in which the shortest link is fixed. If the crank AB

rotates at a uniform speed, the crank CD rotate at a non-uniform speed. This rotation of link CD is

transformed to quick return reciprocatory motion of the ram E by the link CE as shown in figure.



When the crank AB rotates through an angle α in Counter clockwise direction during working stroke,

the link CD rotates through 180. We can observe that / α >/ β. Hence time of working stroke is α /β

times more or the return stroke is α /β times quicker. Shortest link is always stationary link. Sum of

the shortest and the longest links of the four links 1, 2, 3 and 4 are less than the sum of the other two.

It is the necessary condition for the drag link quick return mechanism.

4. Crank and slotted lever mechanism:

It is an application of second inversion. The crank and slotted lever mechanism is shown in figure

below.

In this mechanism link 3 is fixed. The slider (link 1) reciprocates in oscillating slotted lever (link 4)

and crank (link 2) rotates. Link 5 connects link 4 to the ram (link 6). The ram with the cutting tool

reciprocates perpendicular to the fixed link 3. The ram with the tool reverses its direction of motion

when link 2 is perpendicular to link 4. Thus the cutting stroke is executed during the rotation of the

crank through angle α and the return stroke is executed when the crank rotates through angle β or

360 – α. Therefore, when the crank rotates uniformly, we get,

Time to cutting = α = α

Time of return β 360 – α

This mechanism is used in shaping machines, slotting machines and in rotary engines.

5. Ratchets and escapements - Indexing Mechanisms - Rocking Mechanisms:

Intermittent motion mechanism:

Ratchet and Pawl mechanism: This mechanism is used in producing intermittent rotary motion

member. A ratchet and Pawl mechanism consists of a ratchet wheel 2 and a pawl 3 as shown in

the figure. When the lever 4 carrying pawl is raised, the ratchet wheel rotates in the counter clock

wise direction (driven by pawl). As the pawl lever is lowered the pawl slides over the ratchet

teeth. One more pawl 5 is used to prevent the ratchet from reversing. Ratchets are used in feed

mechanisms, lifting jacks, clocks, watches and counting devices.



6. Geneva mechanism: Geneva mechanism is an intermittent motion mechanism. It consists of a

driving wheel D carrying a pin P which engages in a slot of follower F as shown in figure.

During one quarter revolution of the driving plate, the Pin and follower remain in contact and

hence the follower is turned by one quarter of a turn. During the remaining time of one

revolution of the driver, the follower remains in rest locked in position by the circular arc.

7. Pantograph: Pantograph is used to copy the curves in reduced or enlarged scales. Hence this

mechanism finds its use in copying devices such as engraving or profiling machines.



This is a simple figure of a Pantograph. The links are pin jointed at A, B, C and D. AB is parallel to

DC and AD is parallel to BC. Link BA is extended to fixed pin O. Q is a point on the link AD. If the

motion of Q is to be enlarged then the link BC is extended to P such that O, Q and P are in a straight

line. Then it can be shown that the points P and Q always move parallel and similar to each other

over any path straight or curved. Their motions will be proportional to their distance from the fixed

point. Let ABCD be the initial position. Suppose if point Q moves to Q1 , then all the links and the

joints will move to the new positions (such as A moves to A1 , B moves to Q1, C moves to Q1 , D

moves to D1 and P to P1 ) and the new configuration of the mechanism is shown by dotted lines. The

movement of Q (Q Q1) will be enlarged to PP1 in a definite ratio.

8. Toggle Mechanism:

In slider crank mechanism as the crank approaches one of its dead centre position, the slider

approaches zero. The ratio of the crank movement to the slider movement approaching infinity is

proportional to the mechanical advantage. This is the principle used in toggle mechanism. A toggle

mechanism is used when large forces act through a short distance is required. The figure below

shows a toggle mechanism. Links CD and CE are of same length. Resolving the forces at C

vertically F Sin α =P Cos α 2

Therefore, F = P . (because Sin α/Cos α = Tan α) 2 tan α Thus for the given value of P, as the links

CD and CE approaches collinear position (αO), the force F rises rapidly.

9. Hooke’s joint:



Hooke’s joint used to connect two parallel intersecting shafts as shown in figure. This can also be

used for shaft with angular misalignment where flexible coupling does not serve the purpose. Hence

Hooke’s joint is a means of connecting two rotating shafts whose axes lie in the same plane and their

directions making a small angle with each other. It is commonly known as Universal joint. In Europe

it is called as Cardan joint.

10. Ackermann steering gear mechanism:

This mechanism is made of only turning pairs and is made of only turning pairs wear and tear of the

parts is less and cheaper in manufacturing. The cross link KL connects two short axles AC and BD

of the front wheels through the short links AK and BL which forms bell crank levers CAK and DBL

respectively as shown in fig, the longer links AB and KL are parallel and the shorter links AK and

BL are inclined at an angle α. When the vehicles steer to the right as shown in the figure, the short

link BL is turned so as to increase α, where as the link LK causes the other short link AK to turn so

as to reduce α. The fundamental equation for correct steering is, CotΦ–Cosθ = b / l

In the above arrangement it is clear that the angle Φ through which AK turns is less than the angle θ

through which the BL turns and therefore the left front axle turns through a smaller angle than the

right front axle. For different angle of turn θ, the corresponding value of Φ and (Cot Φ – Cos θ) are

noted. This is done by actually drawing the mechanism to a scale or by calculations. Therefore for

different value of the corresponding value of and are tabulated. Approximate value of b/l for correct

steering should be between 0.4 and 0.5. In an Ackermann steering gear mechanism, the

instantaneous centre I does not lie on the axis of the rear axle but on a line parallel to the rear axle

axis at an approximate distance of 0.3l above it.

Three correct steering positions will be:

1) When moving straight. 2) When moving one correct angle to the right corresponding to the

link ratio AK/AB and angle α. 3) Similar position when moving to the left. In all other

positions pure rolling is not obtainable.



Some Of The Mechanisms Which Are Used In Day To Day Life.

BELL CRANK: GENEVA STOP:

BELL CRANK: The bell crank was originally used in large house to operate the servant’s bell, hence the

name. The bell crank is used to convert the direction of reciprocating movement. By varying the angle of the

crank piece it can be used to change the angle of movement from 1 degree to 180 degrees.

GENEVA STOP: The Geneva stop is named after the Geneva cross, a similar shape to the main part

of the mechanism. The Geneva stop is used to provide intermittent motion, the orange wheel turns

continuously, the dark blue pin then turns the blue cross quarter of a turn for each revolution of the

drive wheel. The crescent shaped cut out in dark orange section lets the points of the cross past, then

locks the wheel in place when it is stationary. The Geneva stop mechanism is used commonly in film

cameras.

ELLIPTICAL TRAMMEL PISTON ARRANGEMENT

ELLIPTICAL TRAMMEL: This fascinating mechanism converts rotary motion to reciprocating



motion in two axis. Notice that the handle traces out an ellipse rather than a circle. A similar

mechanism is used in ellipse drawing tools.

PISTON ARRANGEMENT: This mechanism is used to convert between rotary motion and

reciprocating motion, it works either way. Notice how the speed of the piston changes. The piston

starts from one end, and increases its speed. It reaches maximum speed in the middle of its travel

then gradually slows down until it reaches the end of its travel.

RACK AND PINION RATCHET

RACK AND PINION: The rack and pinion is used to convert between rotary and linear motion.

The rack is the flat, toothed part, the pinion is the gear. Rack and pinion can convert from rotary to

linear of from linear to rotary. The diameter of the gear determines the speed that the rack moves as

the pinion turns. Rack and pinions are commonly used in the steering system of cars to convert the

rotary motion of the steering wheel to the side to side motion in the wheels. Rack and pinion gears

give a positive motion especially compared to the friction drive of a wheel in tarmac. In the rack and

pinion railway a central rack between the two rails engages with a pinion on the engine allowing the

train to be pulled up very steep slopes.

RATCHET: The ratchet can be used to move a toothed wheel one tooth at a time. The part used to

move the ratchet is known as the pawl. The ratchet can be used as a way of gearing down motion. By

its nature motion created by a ratchet is intermittent. By using two pawls simultaneously this

intermittent effect can be almost, but not quite, removed. Ratchets are also used to ensure that motion

only occurs in only one direction, useful for winding gear which must not be allowed to drop.

Ratchets are also used in the freewheel mechanism of a bicycle.

WORM GEAR WATCH ESCAPEMENT.



WORM GEAR: A worm is used to reduce speed. For each complete turn of the worm shaft the gear

shaft advances only one tooth of the gear. In this case, with a twelve tooth gear, the speed is reduced

by a factor of twelve. Also, the axis of rotation is turned by 90 degrees. Unlike ordinary gears, the

motion is not reversible, a worm can drive a gear to reduce speed but a gear cannot drive a worm to

increase it. As the speed is reduced the power to the drive increases correspondingly. Worm gears

are a compact, efficient means of substantially decreasing speed and increasing power. Ideal for use

with small electric motors.

WATCH ESCAPEMENT: The watch escapement is the centre of the time piece. It is the

escapement which divides the time into equal segments.The balance wheel, the gold wheel, oscillates

backwards and forwards on a hairspring (not shown) as the balance wheel moves the lever is moved

allowing the escape wheel (green) to rotate by one tooth. The power comes through the escape wheel

which gives a small 'kick' to the palettes (purple) at each tick.

GEARS CAM FOLLOWER.

GEARS: Gears are used to change speed in rotational movement. In the example above the blue

gear has eleven teeth and the orange gear has twenty five. To turn the orange gear one full turn the

blue gear must turn 25/11 or 2.2727r turns. Notice that as the blue gear turns clockwise the orange

gear turns anti-clockwise. In the above example the number of teeth on the orange gear is not

divisible by the number of teeth on the blue gear. This is deliberate. If the orange gear had thirty

three teeth then every three turns of the blue gear the same teeth would mesh together which could

cause excessive wear. By using none divisible numbers the same teeth mesh only every seventeen

turns of the blue gear.

CAMS: Cams are used to convert rotary motion into reciprocating motion. The motion created can

be simple and regular or complex and irregular. As the cam turns, driven by the circular motion, the

cam follower traces the surface of the cam transmitting its motion to the required mechanism. Cam

follower design is important in the way the profile of the cam is followed. A fine pointed follower



will more accurately trace the outline of the cam. This more accurate movement is at the expense of

the strength of the cam follower.

STEAM ENGINE.

Steam engines were the backbone of the industrial revolution. In this common design high pressure

steam is pumped alternately into one side of the piston, then the other forcing it back and forth. The

reciprocating motion of the piston is converted to useful rotary motion using a crank.

As the large wheel (the fly wheel) turns a small crank or cam is used to move the small red control

valve back and forth controlling where the steam flows. In this animation the oval crank has been

made transparent so that you can see how the control valve crank is attached.

Straight line generators, Design of Crank-rocker Mechanisms:

• Straight Line Motion Mechanisms:

The easiest way to generate a straight line motion is by using a sliding pair but in precision machines

sliding pairs are not preferred because of wear and tear. Hence in such cases different methods are

used to generate straight line motion mechanisms:

1. Exact straight line motion mechanism.

a. Peaucellier mechanism, b. Hart mechanism, c. Scott Russell mechanism

2. Approximate straight line motion mechanisms

a. Watt mechanism, b. Grasshopper’s mechanism, c. Robert’s mechanism,

d. Tchebicheff’s mechanism

• a. Peaucillier mechanism :

The pin Q is constrained to move long the circumference of a circle by means of the link OQ. The

link OQ and the fixed link are equal in length. The pins P and Q are on opposite corners of a four bar

chain

which has all four links QC, CP, PB and BQ of equal length to the fixed pin A. i.e., link AB = link



AC. The product AQ x AP remain constant as the link OQ rotates may be proved as follows: Join

BC to bisect PQ at F; then, from the right angled triangles AFB, BFP, we have AB=AF+FB and

BP=BF+FP. Subtracting, AB-BP= AF-FP=(AF–FP)(AF+FP) = AQ x AP .

Since AB and BP are links of a constant length, the product AQ x AP is constant. Therefore the point

P traces out a straight path normal to AR.

• b. Robert’s mechanism:

This is also a four bar chain. The link PQ and RS are of equal length and the tracing pint ‘O’ is

rigidly attached to the link QR on a line which bisects QR at right angles. The best position for O

may be found by making use of the instantaneous centre of QR. The path of O is clearly

approximately horizontal in the Robert’s mechanism.

a. Peaucillier mechanism b. Hart mechanism



UNIT II KINEMATICS OF LINKAGE MECHANISMS

Displacement, velocity and acceleration analysis in simple

mechanisms: Important Concepts in Velocity Analysis

1. The absolute velocity of any point on a mechanism is the velocity of that point with reference

to ground.

2. Relative velocity describes how one point on a mechanism moves relative to another point on

the mechanism.

3. The velocity of a point on a moving link relative to the pivot of the link is given by the

equation: V = r, where = angular velocity of the link and r = distance from pivot.

Acceleration Components

• Normal Acceleration:An = 2r. Points toward the center of rotation

• Tangential Acceleration:At = r. In a direction perpendicular to the link

• Coriolis Acceleration:Ac = 2 (dr/dt). In a direction perpendicular to the link

• Sliding Acceleration:As = d2r/dt2. In the direction of sliding.

A rotating link will produce normal and tangential acceleration components at any point a

distance, r, from the rotational pivot of the link. The total acceleration of that point is the vector

sum of the components.

A slider attached to ground experiences only sliding acceleration.

A slider attached to a rotating link (such that the slider is moving in or out along the link as the

link rotates) experiences all 4 components of acceleration. Perhaps the most confusing of these

is the coriolis acceleration, though the concept of coriolis acceleration is fairly simple. Imagine

yourself standing at the center of a merry-go-round as it spins at a constant speed ( ). You begin

to walk toward the outer edge of the merry-go-round at a constant speed (dr/dt). Even though

you are walking at a constant speed and the merry-go-round is spinning at a constant speed, your

total velocity is increasing because you are moving away from the center of rotation (i.e. the edge

of the merry-go-round is moving faster than the center). This is the coriolis acceleration. In

what direction did your speed increase? This is the direction of the coriolis acceleration.

The total acceleration of a point is the vector sum of all applicable acceleration components:

A = An + At + Ac + As

These vectors and the above equation can be broken into x and y components by applying sines



and cosines to the vector diagrams to determine the x and y components of each vector. In this

way, the x and y components of the total acceleration can be found.

Graphical Method, Velocity and Acceleration polygons :

* Graphical velocity analysis:

It is a very short step (using basic trigonometry with sines and cosines) to convert the graphical

results into numerical results. The basic steps are these:

1. Set up a velocity reference plane with a point of zero velocity designated.

2. Use the equation, V = r, to calculate any known linkage velocities.

3. Plot your known linkage velocities on the velocity plot. A linkage that is rotating about

ground gives an absolute velocity. This is a vector that originates at the zero velocity point and

runs perpendicular to the link to show the direction of motion. The vector, VA, gives the velocity

of point A.

4. Plot all other velocity vector directions. A point on a grounded link (such as point B) will

produce an absolute velocity vector passing through the zero velocity point and perpendicular to

the link. A point on a floating link (such as B relative to point A) will produce a relative velocity

vector. This vector will be perpendicular to the link AB and pass through the reference point (A)

on the velocity diagram.

5. One should be able to form a closed triangle (for a 4-bar) that shows the vector equation: VB =

VA + VB/A where VB = absolute velocity of point B, VA = absolute velocity of point A, and VB/A

is the velocity of point B relative to point A.

Velocity and Acceleration analysis of mechanisms (Graphical Methods):

Velocity and acceleration analysis by vector polygons: Relative velocity and accelerations of

particles in a common link, relative velocity and accelerations of coincident particles on separate

link, Coriolis component of acceleration.

Velocity and acceleration analysis by complex numbers: Analysis of single slider crank

mechanism and four bar mechanism by loop closure equations and complex numbers.

✓ Velocity Analysis of Four Bar Mechanisms:

• Problems solving in Four Bar Mechanisms and additional links.

✓ Velocity Analysis of Slider Crank Mechanisms:

• Problems solving in Slider Crank Mechanisms and additional links.

✓ Acceleration Analysis of Four Bar Mechanisms:

• Problems solving in Four Bar Mechanisms and additional links.



✓ Acceleration Analysis of Slider Crank Mechanisms:

• Problems solving in Slider Crank Mechanisms and additional links.

✓ Kinematic analysis by Complex Algebra methods:

• Analysis of single slider crank mechanism and four bar mechanism by loop closure equations

and complex numbers.

✓ Vector Approach:

• Relative velocity and accelerations of particles in a common link, relative velocity and

accelerations of coincident particles on separate link

✓ Computer applications in the kinematic analysis of simple mechanisms:

• Computer programming for simple mechanisms

Coincident points, Coriolis Acceleration:

• Coriolis Acceleration:Ac = 2 (dr/dt). In a direction perpendicular to the link.

A slider attached to ground experiences only sliding acceleration.

A slider attached to a rotating link (such that the slider is moving in or out along the link as the

link rotates) experiences all 4 components of acceleration. Perhaps the most confusing of these

is the coriolis acceleration, though the concept of coriolis acceleration is fairly simple. Imagine

yourself standing at the center of a merry-go-round as it spins at a constant speed ( ). You begin

to walk toward the outer edge of the merry-go-round at a constant speed (dr/dt). Even though

you are walking at a constant speed and the merry-go-round is spinning at a constant speed, your

total velocity is increasing because you are moving away from the center of rotation (i.e. the edge

of the merry-go-round is moving faster than the center). This is the coriolis acceleration. In

what direction did your speed increase? This is the direction of the coriolis acceleration.

Linkage Synthesis

Problem Example:1



UNIT III KINEMATICS OF CAMMECHANISMS

INTRODUCTION

Acam is a mechanical device used to transmitmotion to a follower by direct contact. The

driveris called the cam and the driven member is called the follower.In a cam follower pair,the

cam normally rotates while the follower may translate or oscillate.Afamiliar exampleis the cam

shaft of an automobile engine, where the cams drive the push rods (thefollowers) to open

And close the valves in synchronization withth emotion of the pistons.

• Cams:

Type of cams, Type of followers, Displacement, Velocity and acceleration time curves for cam

profiles, Disc cam with reciprocating follower having knife edge, roller follower, Follower

motions including SHM, Uniform velocity, Uniform acceleration and retardation and Cycloidal

motion.

Cams are used to convert rotary motion into reciprocating motion. The motion created can be simple

and regular or complex and irregular. As the cam turns, driven by the circular motion, the cam

follower traces the surface of the cam transmitting its motion to the required mechanism. Cam

follower design is important in the way the profile of the cam is followed. A fine pointed follower

will more accurately trace the outline of the cam. This more accurate movement is at the expense of

the strength of the cam follower.

Types of cams

Cams can be classified based on their physical shape.

a) DiskorplatecamThedisk(orplate)camhas anirregularcontourtoimpartaspecificmotion

tothefollower.Thefollowermovesinaplaneperpendiculartotheaxis ofrotationofthe camshaftandis

heldincontactwiththecambysprings orgravity.



Fig.3.1Plateordiskcam.

b) Cylindricalcam:Thecylindricalcamhas agroovecutalongits cylindrical surface.The

rollerfollows thegroove,andthefollowermovesinaplaneparalleltotheaxis ofrotationofthe cylinder.

Fig.3.2Cylindricalcam.

c) Translatingcam.Thetranslatingcamis acontouredorgroovedplateslidingonaguiding

surface(s).Thefollowermayoscillate (Fig.3.3a)orreciprocate(Fig.3.3b).Thecontourorthe

shapeofthegrooveis determinedbythespecifiedmotionofthefollower.



Fig.3.3translatingcam

Types offollowers(Fig3.4):

(i)Basedonsurfaceincontact.

(a)Knifeedgefollower

(b)Rollerfollower

(c) Flatfacedfollower

(d) Sphericalfollower



Fig3.4

(ii)Basedontypeofmotion(Fig3.5):

(a)Oscillatingfollower

(b)Translatingfollower

Fig:3.5

Basedonlineofmotion(Fig3.6):

(a) Radialfollower:Thelines ofmovementofin-linecamfollowers pass throughthecenters of

thecamshafts

(b) Off-setfollower:Forthis type,thelines ofmovementareoffsetfromthecenters ofthe

camshafts



nomenclature (Fig.3.7):

CamProfileThecontouroftheworkingsurfaceofthecam.

Tracer PointThepointattheknifeedgeofafollower, orthecenterofa roller, orthecenterofa

sphericalface.

PitchCurveThepathofthetracerpoint.

BaseCircleThesmallestcircledrawn,tangentialtothecamprofile, withits centerontheaxis of

thecamshaft.Thesizeofthebasecircledetermines thesizeof

thecam.

PrimeCircleThesmallestcircledrawn,tangentialtothepitchcurve, withits centeron theaxis

ofthecamshaft.

PressureAngleTheanglebetweenthenormaltothepitchcurveandthedirectionof

motionofthefolloweratthepointofcontact

Fig

3.6

Cam



Fig3.7

Types offollowermotion:

Cam follower systems are designed to achieve a desired oscillatory motion.Appropriate

displacement patterns are to beselected forthis purpose, before designing the cam surface.The

cam is assumed to rotate at a constant speed and the follower raises,dwells, returns to it soriginal

position and dwells again through specified angles of rotation of the cam,during each revolution

of the cam.Some of the standard follower motions are as follows:

Theyare,follower motion with,

(a)Uniformvelocity



(b)Modifieduniformvelocity

(c)Uniformaccelerationanddeceleration

(d)Simpleharmonicmotion

(e)Cycloidalmotion

Displacementdiagrams:

In a cam follower system,the motion of the follower is very important. Its displacement can

be plotted against the angular displacement θ of the cam and it is called as the displacement

diagram.The displacement of the follower is plotted along they-axis and angular displacement θ of

the cam is plotted along x-axis.From the displacement diagram, velocity y and acceleration of the

follower can also be plotted for different angular displacements θ of the cam.The

displacement,velocity and acceleration diagrams are plotted forone cycle of operationi.e., one

rotation of the cam. Displacement diagrams are basic requirements for the construction of cam

profiles.Construction of displacement diagrams and calculation of velocities and accelerations of

followers with different types of motions are discussed in the following sections.

(a) FollowermotionwithUniformvelocity:

Fig.3.8shows the displacement,velocity and acceleration patterns of a follower having uniform

velocity type of motion.Since the follower moves with constant velocity,during riseand fall,the

displacement varies linearly with θ.Also,since the velocity changes from zero to a finite value,

with in no time, theoretically,the acceleration be comes in finite at the beginning and end of rise

and fall.


Follower motion with modified uniform velocity:

Fig

3.8

(b)

41

SCE Department of Mechanical Engineering

ME6401 KINEMATICS OF MACHINERY

42

Itis observedin thedisplacement diagrams ofthefollower with uniform velocity that the

acceleration nof the follower becomes in finit eat the beginning and ending of rise and return

strokes. Inorder to prevent this,the displacement diagrams are slightly modified.In the modified

form,thevelocity of the follower changes uniformly during the beginning and end of each

stroke. Accordingly,the displacemen to the follower varies parabolically during the seperiods.

With this modification, the acceleration becomes constant during the seperiods, instead of being

infinite a sin the uniform velocity type of motion.The displacement, velocity and acceleration

patterns shownin fig

(c)

Follower

motion

with

uniform

SCE Department of Mechanical Engineering



(b) SimpleHarmonicMotion:In fig3.11, the motion executed by point Pl, whichis the

projection of point P on the vertical diameter is called simple harmonicmotion. Here, P moves

With uniform angular velocity ωp, along a circle of radius r (r=s/2).



Fig3.11

(c) Cycloidalmotion:

Cycloid is the path generated by a point on the circumference of a circle, as the circle rolls

without slipping, on a straight/flat surface.The motion executed by the follower here, is similar

to that of the projection of a point moving along a cyloidal curve on a vertical lineas shown in

figure6.12.



Drawthecamprofileforfollowingconditions:

Followertype=Knifeedged,in-line;lift=50mm;basecircle radius =50mm;outstrokewith SHM,for600

camrotation;dwellfor450camrotation;returnstrokewithSHM,for90ocam rotation;dwellforthe

remainingperiod.

(2)Draw the cam profile for the same operating condition so f with the follower offset by 10mm

to the left of cam center.

Camprofile:



Camprofilewith10mmoffset:

(1)Drawthecamprofileforfollowingconditions:

Follower type=roller follower, in-line; lift=25mm; base circle radius=20mm;roller radius=

5mm;out stroke with Uniform acceleration and retardation,for1200cam rotation;dwellfor600

camrotation;return stroke withUniform acceleration and retardation,for900 camrotation; dwell for

the remaining period.

(4)Draw the cam profile for conditions same with follower off set to right

of cam center by 5mm and cam rotating counter clockwise.



DisplacementDiagram:

Cam profile;



Camprofilewith5mmoffset

(2) Drawthecamprofileforfollowingconditions:

Follower type=knife edge d follower, in line; lift=30mm;base circle radius =20mm;outstroke with

uniform velocity in1200of cam rotation;dwell for 600; return stroke with uniform velocity,

during900ofcam rotation;dwell for the remaining period.



DisplacementDiagram

Cam profile



(3) Drawthecamprofileforfollowingconditions:

Followertype=flatfacedfollower,inline;followerrisesby20mmwithSHMin1200of cam

rotation,dwellsfor300 ofcamrotation;returnswithSHMin1200 ofcamrotationanddwells duringthe

remainingperiod.Basecircle radius =25mm.

DisplacementDiagram:

Cam profile



Layout of plate cam profiles:

• Drawing the displacement diagrams for the different kinds of the motions and the plate cam

profiles for these different motions and different followers.

• SHM, Uniform velocity, Uniform acceleration and retardation and Cycloidal motions

• Knife-edge, Roller, Flat-faced and Mushroom followers.

Derivatives of Follower motion:

• Velocity and acceleration of the followers for various types of motions.

• Calculation of Velocity and acceleration of the followers for various types of motions.

Circular arc and Tangent cams:

• Circular arc

• Tangent cam

Standard cam motion:

• Simple Harmonic Motion

• Uniform velocity motion

• Uniform acceleration and retardation motion

• Cycloidal motion

Pressure angle and undercutting:

• Pressure angle

• Undercutting



UNIT IV GEARS AND TRAINS

Introduction

Gears are used to change speed in rotational movement.

In the example above the blue gear has eleven teeth and the orange gear has twenty five. To turn the

orange gear one full turn the blue gear must turn 25/11 or 2.2727r turns. Notice that as the blue gear

turns clockwise the orange gear turns anti-clockwise. In the above example the number of teeth on

the orange gear is not divisible by the number of teeth on the blue gear. This is deliberate. If the

orange gear had thirty three teeth then every three turns of the blue gear the same teeth would mesh

together which could cause excessive wear. By using none divisible numbers the same teeth mesh

only every seventeen turns of the blue gear.

A gear is a rotating machine part having cut teeth,or cogs,which mesh with

another toothed part in order to transmit torque.Two or more gears working intandemare



called a transmission and can produce a mechanical advantage through agearratio and

thus may be considered a simple machine.Geared devices can change the speed,

magnitude,and direction of a powersource.The most common situation is for a gear to

mesh with another gear,however a gear can also mesh a non-rotating toothed part, called

a rack, there by producing translation instead of rotation.

The gears in a transmission are analogous to the wheels in a pulley.Anadvantage of

gears is that theteeth of a gear prevent slipping.

When two gears of unequal number of teeth are combined a mechanical advantage is

produced, with both the rotational speeds and the torques of the two gears differing in

asimple relationship.

In transmissions which offer multiple gear ratios, such as bicycles and cars, the term

gear, as infirstgear, refers to agear ratio rather than an actual physical gear. The

term is used to describe similar devices even when gear ratio is continuous rather than discrete, or

when the device does not actually on tain any gears, as in a continuously variable transmission.

Fundamental LawofGear-Tooth

Pitch point divides the line between the line of

centers and its positiond ecides the velocity ratio of

the two teeth.The above expression is the

fundamental law of gear-tooth action.

Formation of teeth:

Involute teeth

Cycloidal teeth

Involute curve:

The curve most commonly use d for gear-tooth

profiles is the involute of a circle. This involute curve

is the path traced by a pointonalineas the line

Rolls without slipping on the circumference of a circle. It may also be defined asapathtraced by

the end of a string, which is originally wrapped on a circle when the string is un wrapped from the

circle.The circle from which the involute is derived is called the base circle

.



CycloidalCurve



Pathofcontact:

Consider a pinion driving wheel as shown in figure. When the pinion rotates in

clockwise, the contact between a pair of involute teeth begin sat K(on the near the base circle of

pinionortheouterendofthetoothfaceonthewheel)andendsatL(outerendofthetoothface on the pinion

or on the flank near the base circle of wheel).

MN is the common normal at the point of contacts and the common tangent to the base

circles.The point K is the intersection of the addendum circle of wheel and the common tangent.

The point L is the intersection of the addendum circle of pinion and common tangent.

The lengthof path of contact is the length of common normal cut-offbytheaddendum circlesof

thewheelandthepinion.Thusthelengthof partof contactisKLwhichisthesumof thepartsof pathof

contactsKPandPL.ContactlengthKPiscalledaspathofapproachand contactlength PLis

calledaspathofrecess.



Arcofcontact:Arc of contact is the path traced by a point on the pitch circle from the beginning

to the end of engagement of a given pair of teeth. In Figure, the arc of contact is EPF or GPH.

The arc GP is known a sarcofapproachand the arc PH is called arc of recess.The angles

subtended by the searcs at O1arecalled angle of approach and angle of recessrespectively.



Contact Ratio ( or Number of Pairs of Teeth in Contact)

The contactratioor the number of pairs ofteethincontactis defi

ned as the ratio of the length of the arc of contact to the circular pitch.

Continuous motion transfer requires two pairs of teeth in contact at the ends of the

path of contact, though the reisonly one pairincontact in the middle of the path,as in Figure. The

average number of teeth in contact is an important parameter-If itis tool owdue the use of

inappropriate profile shift sortoan excessive centredistance.Them anufacturinginaccuracies

mayleadtolossof kinematiccontinuity-thatistoimpact,vibrationandnoiseTheaverage numberof

teethincontactisalsoaguidetoloadsharingbetweenteeth;itistermedthecontact ratio



The tooth tip of the pinion will then undercut the toot h on the wheel at the

rootanddamages partoftheinvoluteprofile.Thiseffectis knownas

interference,andoccurs when theteetharebeingcutandweakens thetoothatits root.

Ingeneral,thephenomenon,whenthetipoftoothundercuts the rootonitsmatinggearis

knownasinterference.Similarly,if theradiusof theaddendum circlesof thewheelincreases

beyondO2M,thenthetipoftoothonwheelwillcauseinterferencewiththetoothonpinion.The points M

andNarecalledinterferencepoints.

Interference may be avoided if the path of the contact does not extend beyond

interferencepoints.Thelimitingvalueoftheradius oftheaddendumcircleofthepinionisO1N

andofthewheelis O2M.

Theinterferencemayonlybeprevented,ifthepointof contactbetweenthetwoteethis alwayson

theinvoluteprofilesandiftheaddendum circlesofthetwomatinggearscutthe

commontangenttothebasecircles atthepoints oftangency.

MethodstoavoidInterference

1.Heightoftheteethmaybe reduced.

2.undercutofthe radialflankofthepinion.

3. Centredistancemaybeincreased.Itleads toincreaseinpressureangle.

4. Bytoothcorrection,thepressureangle,centredistanceandbasecircles remainunchanged,but

tooththickness ofgear willbegreaterthanthepiniontooththickness.

Minimumnumbers ofteethonthepinionavoidInterference

Thepinionturns clockwiseanddrives thegearas showninFigure.

PointsMandNarecalledinterferencepoints.i.e.,ifthecontacttakesplacebeyondMandN,

interferencewilloccur.

Thelimitingvalueofaddendumcircleradiusof pinionisO1Nandthelimitingvalueof addendum

circleradiusofgearisO2M.Consideringthecriticaladdendum circleradiusofgear,

thelimitingnumberofteethongearcanbecalculated.



equation gives minimum number of teeth required on the pinion to avoid interference.

If the number of teethonpinionandgearissame:G=1

Theequationgivesminimum number of teethrequiredonthewheel toavoidinterference.



Spur Gear Terminology



Addendum:The radialdistancebetweenthePitchCircleandthetopoftheteeth.

Arc of Action: Is the arc of the Pitch Circle between the beginning and the end ofthe

engagementofagivenpairofteeth.

ArcofApproach:Is the arc of the Pitch Circle between thefirstpointofcontactofthegear

teethandthePitchPoint.

ArcofRecession:ThatarcofthePitchCirclebetweenthePitchPointandthelastpointof

contactofthegearteeth.

Backlash:Playbetweenmatingteeth.

BaseCircle:Thecirclefromwhichisgeneratedtheinvolutecurveuponwhichthetoothprofile isbased.

CenterDistance:Thedistancebetweencenters oftwogears.

ChordalAddendum:Thedistancebetweenachord,passingthroughthepointswherethe Pitch

Circlecrosses thetoothprofile, andthetoothtop.

ChordalThickness:Thethicknessofthetoothmeasuredalongachordpassingthroughthe points

wherethePitchCirclecrosses thetoothprofile.

CircularPitch:MillimeterofPitchCirclecircumferencepertooth

CircularThickness:Thethickness ofthetoothmeasuredalonganarcfollowingthePitchCircle

Clearance:Thedistancebetweenthetopofatooth andthebottomofthespaceintowhichitfits

onthemeshinggear.

ContactRatio:The ratioofthelengthoftheArcofActiontotheCircularPitch.

Dedendum:The radialdistancebetweenthebottomofthetoothtopitchcircle.

DiametralPitch:Teethpermmofdiameter.

Face:Theworkingsurfaceofageartooth,locatedbetweenthepitchdiameterandthetopofthe tooth.

FaceWidth:Thewidthofthetoothmeasuredparalleltothegearaxis.

Flank:Theworkingsurfaceofageartooth,locatedbetweenthepitchdiameterandthebottom oftheteeth

Gear: Thelargeroftwomeshedgears.Ifbothgearsarethesamesize,theyarebothcalled

"gears".

Land:Thetopsurfaceofthetooth.

LineofAction:Thatlinealongwhichthepointofcontactbetweengearteethtravels,between



thefirstpointofcontactandthelast.

Module:MillimeterofPitchDiametertoTeeth.

Pinion:Thesmalleroftwomeshedgears.

PitchCircle:Thecircle,theradiusof whichisequaltothedistancefromthecenterofthegear

tothepitchpoint.

Diametralpitch:Teethpermillimeterofpitchdiameter.

PitchPoint:Thepointoftangencyofthepitchcirclesoftwomeshinggears,wheretheLineof

Centers crosses thepitchcircles.

PressureAngle:AnglebetweentheLineofActionandalineperpendiculartotheLineof

Centers.

Profile Shift:AnincreaseintheOuterDiameterandRootDiameterofagear,introduced

tolowerthepracticaltoothnumberor acheiveanon-standardCenterDistance.

Ratio:Ratioofthenumbers ofteethonmatinggears.

RootCircle:Thecirclethatpasses throughthebottomofthetoothspaces.

RootDiameter:ThediameteroftheRootCircle.

WorkingDepth:Thedepthtowhichatoothextendsintothespacebetweenteethonthemating gear.

Worm, Rack and Pinion Gears

RACK AND PINION WORM GEAR



RACK AND PINION: The rack and pinion is used to convert between rotary and linear motion.

The rack is the flat, toothed part, the pinion is the gear. Rack and pinion can convert from rotary to

linear of from linear to rotary. The diameter of the gear determines the speed that the rack moves as

the pinion turns. Rack and pinions are commonly used in the steering system of cars to convert the

rotary motion of the steering wheel to the side to side motion in the wheels. Rack and pinion gears

give a positive motion especially compared to the friction drive of a wheel in tarmac. In the rack and

pinion railway a central rack between the two rails engages with a pinion on the engine allowing the

train to be pulled up very steep slopes.

WORM GEAR: A worm is used to reduce speed. For each complete turn of the worm shaft the gear

shaft advances only one tooth of the gear. In this case, with a twelve tooth gear, the speed is reduced

by a factor of twelve. Also, the axis of rotation is turned by 90 degrees. Unlike ordinary gears, the

motion is not reversible, a worm can drive a gear to reduce speed but a gear cannot drive a worm to

increase it. As the speed is reduced the power to the drive increases correspondingly. Worm gears

are a compact, efficient means of substantially decreasing speed and increasing power. Ideal for use

with small electric motors.

Parallel axis gear trains:

• Simple Gear Trains – A simple gear train is a collection of meshing gears where each gear is

on its own axis. The train ratio for a simple gear train is the ratio of the number of teeth on

the input gear to the number of teeth on the output gear. A simple gear train will typically

have 2 or 3 gears and a gear ratio of 10:1 or less. If the train has 3 gears, the intermediate

gear has no numerical effect on the train ratio except to change the direction of the output

gear.

• Compound Gear Trains – A compound gear train is a train where at least one shaft carries

more than one gear. The train ratio is given by the ratio mV = (product of number of teeth on

driver gears)/(product of number of teeth on driven gears). A common approach to the design

of compound gear trains is to first determine the number of gear reduction steps needed (each



step is typically smaller than 10:1 for size purposes). Once this is done, determine the

desired ratio for each step, select a pinion size, and then calculate the gear size.

• Reverted Gear Trains – A reverted gear train is a special case of a compound gear train. A

reverted gear train has the input and output shafts in –line with one another. Assuming no

idler gears are used, a reverted gear train can be realized only if the number of teeth on the

input side of the train adds up to the same as the number of teeth on the output side of the

train.

Epicyclic gear trains:

• If the axis of the shafts over which the gears are mounted are moving relative to a fixed axis ,

the gear train is called the epicyclic gear train.

• Problems in epicyclic gear trains.

Differentials:

• Used in the rear axle of an automobile.

• To enable the rear wheels to revolve at different speeds when negotiating a curve.

• To enable the rear wheels to revolve at the same speeds when going straight.



Problem:1

Problem:2



UNIT VFRICTIONIN THE MACHINE ELEMENTS

Surface contacts:

• Basic laws of friction

• Pivot and collar, introduction and types.

• Problem on flat pivot, Problems on conical pivot.

Friction

Frictionisa measureofhow harditistoslide one objectoveranother. Take alook at thefigure

below.Bothof theblocksaremadefromthesame material,butone is heavier.Ithink we

allknow which onewillbeharder forthe bulldozer to push.

Friction forceversus weight

To understand whythis is, let's takea close look at one ofthe blocks and thetable:

Because frictionexistsatthemicroscopic level, the amount of

forceittakestomoveagivenblockis proportionaltothat block'sweight.

Eventhoughtheblockslooksmoothtothenakedeye,they areactuallyquiteroughat

themicroscopiclevel.Whenyousettheblockdownonthetable,thelittle peaksand

valleysgetsquishedtogether,andsomeofthemmayactuallyweldtogether.Theweightof the



heavier block causes ittosquishtogether more,soitisevenharder to slide.

Differentmaterialshavedifferentmicroscopic structures;forinstance,itisharder to slide

rubber againstrubber than itistoslidesteelagainststeel.The typeof material

determinesthecoefficient offriction,the ratioof the force requiredtoslide the block

totheblock'sweight.Ifthecoefficientwere1.0inourexample,thenitwouldtake

100poundsof force toslide the 100-pound(45 kg) block, or 400pounds(180 kg) of

forcetoslidethe400-poundblock.Ifthecoefficientwere0.1,thenitwouldtake 10 pounds of

forceto slideto the100-pound blockor40 pounds offorceto slide the400- pound block.

Sothe amountof force ittakestomove agivenblock isproportionaltothatblock's weight.The

more weight,themore force required.Thisconceptappliesfor devices

likebrakesandclutches,whereapadispressedagainstaspinning disc.Themore forcethat

presses on thepad, thegreaterthe stoppingforce

Types orfriction:

1. Staticfriction:Itis experiencedbyabody,whenatrest.

2. Dynamicfriction:It is friction experienced by a body when in motion.

a. Sliding friction:It is friction experienced by a body when it slide so veran other

body.

b. Rolling friction:It is frictionexperiencedbetweenthesurfaceswhichtheballsor

rollersinterposedbetweenthem.

c. Pivot friction:Itis thefrictionexperiencedbyabodyduetomotionofrotation.

Further classified

1.Frictionbetweenunlubricatedsurfaces

2.Frictionbetweenlubricatedsurfaces.

Laws ofdryorsolidfriction:

Theforceoffrictiondirectlyproportionaltothenormalloadbetweenthesurfaces.

Theforceoffrictionisindependentoftheareaofthecontactsurfaceforagivennormal load.

Theforceoffrictiondepends uponmaterialwhichthecontactsurfaces ormade.

Theforceoffrictionisindependentofthevelocityofslidingofonebodyrelativetoother body.



Coefficientoffriction(µ):

Itis as the ratiothelimitingfriction(F)tothenormalreaction(RN)betweenthetwobodies.

µ=F/RN

Angleoffriction:

It may be defined as the angle which the resultant reaction R makes with

normal reactions

tanϕ=F/RN

Friction drives:



Friction in screw threads:

• Friction in screw and nut

• Friction in screw jack

• Problems in screw jack

Screwjack:

The screw jack is adevice lift ingloads.Forlift in gheavy loads by applying a

comparatively smaller effortatits handle.The principle on which as crewjack works in a

smallert ot hatofaninclinedplane.



1. Torquerequiredliftingtheloadbyascrewjack

Let p=pitchofthescrew

d=meandiameterofthescrew

α=helixangle

P=effortappliedatthecircumferenceofthescrewtolifttheload

W=loadtobelifted

µ=coefficientofthefriction

tanα=p/Πd

Torquerequiredovercomingfrictionbetweenthescrewandnut



Thetorquerequiredtoovercomefrictionatthecollar

TotalTorque requiredovercomingfriction

IfanP1isappliedattheendofaleverofarml,thenthetotaltorquerequiredtoovercome

frictionmustbeequaltothetorqueappliedattheendofthelever

.Torquerequiredlowertheloadbyascrewjack

Let p=pitchofthescrew

d=meandiameterofthescrew

α=helixangle

P=effortappliedatthecircumferenceofthescrewtolifttheload

W=loadtobelifted

µ=coefficientofthefriction



Torque requiredovercomingfrictionbetweenthescrewandnut

Efficiencyofthescrewjack:

Thee fficiency of the screwjack may b defined as the ratio be tweentheideal efforts to

actualeffort.

Selflockingandoverhaulingofscrews

Torque requiredtolowertheload



In the above expressions,if ϕ<α,then the torque required to lower the loadwill

be negative,the load will start movingdown ward without application offorce,such

acondit ion is known as overhaulin gofscrew.

If ϕ>α,then the torquerequiredtolowertheloadwillbepositive,indicatingthatan

efforts appliedlowertheload,suchaconditionis knownas self lockingofscrew

Clutch

For other uses, seeClutch

(disambiguation).

Clutchforadriveshaft:The clutchdisc (center) spins with the flywheel (left). To

disengage, the lever is pulled (black arrow),causingawhite pressureplate(right) to

disengagethegreenclutch disc from turningthedriveshaft, which turnswithin the

thrust-bearingringof thelever. Never will all 3 rings connect, with nogaps.

A clutchis a mechanicaldevice, byconvention understood to be rotating,which

provides driving force to another mechanismwhenrequired, typicallybyconnecting

the driven mechanismto thedrivingmechanism.Clutches andbrakesaresimilar; if the

driven memberof aclutch is fixed to themechanismframe, it serves asabrake.

Clutches areuseful in devices that havetworotatingshafts.In thesedevices, oneshaft

is typicallyattached to amotoror other power unit (thedrivingmember),and the

othershaft (thedriven member) provides output power for work to be

done.Inadrill,forinstance, oneshaft isdriven byamotor, and theotherdrivesadrill



chuck. The clutch connects thetwo shafts so that theycaneither belocked

togetherandspin at the same speed(engaged), orbedecoupled and spin at different

speeds (disengaged).

Multiple plate clutch

This typeofclutch has several drivingmembers interleavedwith several driven

members.It is used in racecars includingF1,Indycar, World rallyandeven most

club racing, motorcycles,automatictransmissionsand in some diesel locomotives

with mechanical transmissions.It is also used in some electronicallycontrolled

all- wheel drivesystems.

Vehicular

There aredifferent designs of vehicle clutch, butmostarebased on oneormore

frictiondiscs, pressed tightlytogether or againstaflywheelusingsprings.The friction

material varies incomposition dependingon whether the clutch is dryor wet, and on

other considerations. Friction discs once containedasbestos, but this has been largely

eliminated. Clutches found in heavydutyapplications such as trucks andcompetition

cars useceramic clutchesthat haveagreatlyincreased friction coefficient.However,

thesehavea"grabby" action and areunsuitableforroad cars. Thespringpressureis

released when theclutchpedal is depressed thus either pushingor pullingthe

diaphragm of the pressureplate, dependingon type. However, raisingtheengine speed

too highwhile engagingtheclutch willcause excessive clutch platewear.

Engagingtheclutch abruptlywhen theengine is turningat high speed causes aharsh,

jerkystart. This kind ofstart is necessaryand desirable indrag racingandother

competitions, wherespeed is moreimportant than comfort.

5.4.1 Friction clutches:

CLUTCHFRICTION



Single plate clutches andMulti-plate clutche

Uniform wear theory andUniform

pressuretheory Problems in clutches

BELT

Thebeltorropesareusedtotransmitpowerfromoneshafttoanothershaftby means ofpulleys

whichrotateatthesamespeedoratdifferentspeed.

Types ofbeltdrives.

1. Light drives:belt speed upto10m/s

2. Medium drives: speed 10m/s to22m/s

3. Heavy drives :Speed ove r22m/s

Types ofbelt

1. Flat belt

2.V- belt

3.Circular belt or rope

Typeofflatbeltdrives

Openbeltdriv

e Cross

beltdrive

Quarterturnb

eltdrive

Beltdrivewithidlerpulley



Belt andropedrives:

V Belt Drives

BELT

DRIV

ES



ROPE DRIVES



Friction aspects inBrakes:

BRAKES



QUESTION BANK

UNIT -I

TWO MARKS QUESTION AND ANSWER

1. Define Kinematic link.

It is a resistive body which go to make a part of a machine having relative motion

between them.

2. Define Kinematic pair.

When two links are in contact with each other it is known as a pair. If the pair makes

constrain motion it is known as kinematic pair.

3. Define Kinematic chain.

When a number of links connected in space make relative motion of any point on a link

with respect to any other point on the other link follow a definite law it is

known as kinematic chain.

4. Write the Grubler’s criterion for determining the degrees of freedom of a

mechanism having plane motion.

n=3(l-1)-2j

h-Higher pair

joint l-

Number of

links

j-Lower pair joint

5. Define degree of freedom, what is meant by mobility. (Ap/May-2008)

The mobility of a mechanism is defined as the number of input parameters which must be

independently controlled in order to bring the device into a particular position.

6. Write the Kutzbach’s relation. (Ap/May-2008)

Kutzbach’s criterion for determining the number of degrees of freedom or movability (n)

of a plane mechanism is n=3(l-

1)-2j-h n-Degree of freedom.

l-Number of

links. h-

Higher pair

joint j-Lower

pair joint.



7. Define Grashoff’s law and state its significance. (Ap/May-2008)

It states that in a planar four bar mechanism, the sum of shortest link length and

longest link length is not greater than the sum of remaining two links length, if there is

to be continuous relative motion between two members.

Significance:

Grashoff’s law specifies the order in which the links are connected in a kinematic

chain. Grashoff’s law specifies which link of the four-bar chain is

fixed.(s+1)=(p+q) should be

satisfied, if not, no link will make a complete revolution relative to

another. s= length of the shorter length

l= length of the longest link

p & q are the lengths of the other two links.

8. Define Inversion of mechanism.

The method of obtaining different mechanism by fixing different links in a kinematic chain

is known as inversion of mechanism.

9. What is meant by Mechanical advantage of mechanism?

It is defined as the ratio of output torque to the input torque also defined as the ratio of

load to effort.

M.A ideal = TB / TA

TB =driven (resisting

torque) TA =driving

torque

10. Define Transmission angle.

The acute angle between follower and coupler is known as transmission angle.

11. Define Toggle position.

If the driver and coupler lie in the same straight line at this point mechanical advantage is

maximum. Under this condition the mechanism is known as toggle position.

12. List out few types of rocking mechanism.

Pendulum motion is called rocking mechanism.

1. Quick return motion mechanism.

2. Crank and rocker mechanism.

3. Cam and follower mechanism.

13. Define pantograph.

It is device which is used to reproduce a displacement exactly in a enlarged scale. It

is

used in drawing offices, for duplicating the drawing maps, plans, etc. It works

on the principle of 4 bar chain mechanism.



Eg. Oscillating-Oscillating converter mechanism

14. Name the application of crank and slotted lever quick return motion mechanism.

1. Shaping machines.

2. Slotting mechanism.

3. Rotary internal combustion engine.

15. Define structure.

It is an assemblage of a number of resistant bodies having no relative motion between

them and meant for carrying loads having straining action.

16. What is simple mechanism?

A mechanism with four links is known as simple mechanism.

17. Define mechanism.

When one of the links of a kinematic chain is fixed, the chain is known as a mechanism.

18. Define equivalent mechanism; and spatial mechanism.

Equivalent mechanism: The mechanism, that obtained has the same number of the

degree of freedom, as the original mechanism called equivalent mechanism.

Spatial mechanism: Spatial mechanism have special geometric characteristics in that

all revolute axes are parallel and perpendicular to the plane of motion and all prism

axes lie in the plane of motion.

19. Define double slider crank chain mechanism.

A kinematic chain which consists of two turning pair and two sliding pair is known as

double slider crank mechanism.

20. Define single slider crank chain mechanism.

A single slider crank chain is a modification of the basic four bar chain. It consists of one

sliding pair and three turning pair.

Applications:

Rotary or Gnome engines Crank and slotted lever mechanism

Oscillating cylinder engine Ball engine

Hand pump

21. Define Sliding pair.

In a sliding pair minimum number of degree of freedom is only one.

22. Define Turning pair.

In a turning pair also degree of freedom is one. When two links are connected such that

one link revolves around another link it forms a turning pair.



23. Define cylindrical pair.

In a cylindrical pair degree of freedom is two. If one link turns and slides along another

link it forms a cylindrical pair.

24. Define Rolling pair.

In a rolling pair degree of freedom is two. The object moves both linearly and angularly.

25. Define Spherical pair.

In a spherical pair degree of freedom is three. It can both move left and right,up and

down, and rotate along the same point.

26. Define Lower pair.

If contact between two links is surface contact also having degree of freedom one, then

the pair is known as lower pair.

Example: Sliding pair.

27. Define higher pair.

If contact between two links is either point contact or line contact then the pair is known

as higher pair.

Example: Point contact-Rolling pair.

Line contact-Cylindrical

pair.

PART-B (16 Marks)

1. a) Explain different types ofLink.(8)

b)Classifyand explain the Kinematicpair.(8)

2. a) Explain anytwo inversion of four bar chain.(8)

b)Explain the first inversion ofSingle SliderCrank Chain.(8)

3. a) Explain first inversion ofDoubleSlider crank chain.(8)

b)Explain third inversion of double slider crank chain.(8)

4. a) Explain the offset slider crank mechanism.(8)

b)Explain Straight line mechanismwith neat sketch(8)

5. a)With the help of aneat sketch explain the workingof Oldham‘s coupling.(8)

b)Explain steering gearmechanismwith neat sketch (8)



UNIT -II

6. With the help of aneatsketch explain the workingof Whitworth

quick return mechanism. (16)

7. With the help of aneatsketch explain the workingof Single

slider and double slider crankchain mechanism.(16)


1. What are the components of acceleration?

Radial component of acceleration Tangential component of acceleration

2. Write an expression for find number of instantaneous centers in a mechanism.

N=n(n-1)/2, n-no of links

3. What is expression for Coriolis componenet of acceleration?

a BC

=2r

Where =Angular velocity of

‘OA’ V=Linear

velocity of ‘B’

4. How can we represent the direction of linear velocity of any point on a link with

respect to another point on the same link?

The direction of linear velocity of any point on a link with respect to another point on the

same link the direction is perpendicular to the line joining the points.

5. What is the expression for radial and tangential component of acceleration?

Radial component

Arc

OB=OB*OB

Tangential

component

Arc OB=OB*OB

Where OB=Angular velocity of link OB

OB=Angular acceleration of

link OB OB=Length of link OB. (Radial component of acceleration is perpendicular to the velocity of the component and tangential component is perpendicular to link position)

6. Define instantaneous center and instantaneous axis?



Instantaneous center of a moving body may be defined as that center which

goes on changing from one instant to another.

Instantaneous axis is a line drawn through an instantaneous center and

perpendicular to the plane of motion.

7. What are the names of instantaneous center?

Virtual center Centro Rotopole.

8. How can we apply instantaneous center method to determine velocity?

Consider three points A, B, C on a rigid link. I am being instantaneous Center. Let VA,

VB, VC be the points A, B

& C. Then we have VA/IA=VB/IB=VC/IC

9. What is the objective of Kinematic analysis?

The objective of Kinematic analysis is to determine the Kinematic quantities such as

displacement, velocity and acceleration of the element in a mechanism.

10. Write any two rules to locate Instantaneous center.

When two links are connected by a pin joint the instantaneous center lies on the center of

the pin.

When two links have a sliding contact, the instantaneous center lies at infinity

in a direction perpendicular to the path of motion of slide.

11. Define Kennedy’s theorem.

The Kennedy’s theorem states that if three bodies move relatively to each other, they

have three instantaneous centers and lie on a straight line.

12. What is meant by efficiency of a mechanism?

Efficiency of a mechanism is defined as the ratio of the product of force and velocity

in



driven link to the product of force and velocity in driving link.

13. A pin joins two links A & B. A rotates with A angular velocity and B rotates with

B

angular velocity in opposite direction. What is the rubbing velocity of that point?

Rubbing velocity of pin = ( A + B ) * r

Where ‘r’ is the radius of pin.

14. What is the magnitude of linear velocity of a point B on a link AB relative to A?

The magnitude of linear velocity of a point B on a link AB, which rotates with

‘’ angular velocity with respect to A is:

VB/A = A/B *AB

PART-B (16 Marks)

1. TheCrank ofaslider crank mechanisms rotatesclockwise at a Constant speed of

600 r.p.m. The crank is 125 mmand connectingrodis 500 mmlong.

Determine1. Linear velocityand acceleration ofthe mid Pointof the

connecting rod, and 2. Angular velocityandangular acceleration ofthe

connecting rod, at acrankangle

of45° from innerdead centreposition.

2. Ina four link mechanism, thedimensions of thelinks areAB=200 mm,

BC=400mm, CD=450 mm and AD=600mm. At the instant when DAB=90°,

the link ABhasangularvelocityof36 rad/s in the clockwisedirection.

Determine (i) The velocityof pointC, (ii) Thevelocityof pointEonthe link BC

When BE =200 mm (iii) the angularvelocities oflinks BC and CD,

iv)acceleration oflink oflink BC

3. Thedimensions of thevarious links of amechanism, as shown in fig. are

as follows: OA=300 mm; AB=1200;BC=450 mmand CD=450 mm. ifthe

crank OA rotates at 20 r.p.m. in theanticlockwise direction and gives

motion to the sliding blocks Band D,find, forgivenconfiguration:

(1)Velocityof slidingatBand D, (2) Angular velocityof CD

(3)Linearacceleration ofD and (4)angularacceleration of CD.

4 a)Derivethe expressions forVelocityandacceleration ofpiston in



reciprocating steam engine mechanismwith neat sketch

b).Derivethe expression for Coriolis component of acceleration with neat sketch

5.Inaslider crankmechanism, the length of thecrank and theconnectingrod

are100 mmand 400 mmrespectively./ Thecrank [position is 45° fromIDC,

the crank shaft speed is 600 r.p.m. clockwise.Using analytical method

Determine

(1)Velocityandacceleration of theslider,and (2)Angular

velocityandangular acceleration oftheconnectingrod.

6.Locate allinstantaneous centers of theslider crank mechanism; thelength of crank

OBand Connecting rod ABare125 mmand 500 mm respectively. Thecrank speed is 600

rpm clockwise. When thecrank has turned 45° from theIDC. Determine(i)velocityof

. slider‘A‘ (ii)Angular Velocityofconnecting rod‗AB‘.

UNIT -III


1. What is a cam? Give some examples?

A Cam is a rotating machine element, which gives reciprocating or oscillating motion to

another element known as

follower. Examples are:

Radial or Disc, Cylindrical or Barrel, End or Face cams and Wedge cams.

2. What is the Classification of cams?

• According to cam shape

• According to follower movement

• According to manner of constraint of the follower

3. Classify cams based on their shape.

• Wedge cam • Radial cams • Spiral cams

• Drum cams • Spherical cams

4. What is the classification of followers?

(i) According to follower shape (ii) According to motion of follower

5. What is a roller follower?

In place of a knife edge roller is provided at the contacting end of the follower

6. What is a Spherical follower?

In the contacting end of the follower is of spherical shape.



7. What is meant by angle of ascend?

The angle of rotation of the cam from the position when the follower begins to rise till it

reaches its highest points. It is denoted by θ

8. What is meant by angle of descend?

The angle through which the cam rotates during the time the follower returns to the initial

position. It is denoted by θr.

9. What is angle of dwell?

It is the angle through which the cam rotates while the follower remains stationary at the

highest or the lowest.

10. What is meant by angle of action?

The total angle moved by the cam during its rotation between the beginning of

rise and the end of return of the follower.

11. What is radial or disc cams?

In radial cams the follower reciprocates or oscillates in a direction perpendicular to the

cam axis. The cams are all radial rams. In actual practice, radial cams are widely

used due to their simplicity and compactness.

12. What is dwell?

The zero displacement or the absence of motion of the follower during the motion of the

cam is called dwell.

13. What are the classifications of follower according to follower shape?

• Knife edge follower

• Roller follower

• Mushroom or flat faced follower and

• Spherical faced or curved shoe follower

14. What are the classifications of follower according to the motion of the follower?

• Reciprocating or translating follower

• Oscillating or rotating follower

15. What are the classifications of followers according to the path of motion?

• Radial follower • Offset follower

16. What is the motion of the follower?

The follower can have any of the following four types of motions

• Uniform velocity • Simple harmonic motion



• Uniform acceleration and retardation • Cycloidal motion.

17. What is the application of cam?

Closing and opening of inlet and exit valve operating in IC engine .

18. What are the necessary elements of a cam mechanism?

Cam-The driving member is known as the cam Follower-The driven member is known as the

follower. Frame-It supports the cam and guider the follower.

19. Where are the roller follower extensively used? (Ap/May-2008)

Roller followers are extensively used where more space is available such as in stationary

gas oil engines and aircraft

engines.

20. Write the formula for maximum velocity.

Vo (max) = 2ωs

θo

Vr (max) =

2ωs

PART-B (16 Marks)

1.A cam is to give the followingmotion to aknifeedgedfollower:

(a) Outstrokeduring60°of cam rotation

(b)Dwellfor thenext 45° of cam rotation

(c) Return strokeduring next 90° of cam rotation and

(d)Dwellfor theremainingofcam rotation

Thestrokeof thefollower is 40 mmand the minimumradius of the cam is50 mm.

The followermoves with uniform velocityduringboth theoutstrokeand return

strokes.Draw the profileof thecam when (a) theaxis of the follower passes

through theaxis of the cam shaft, and (b)the axis of the follower is offset

by20mmfrom the axis of the cam shaft.

2. Draw theprofileofacam operatingaKnife-edgedfollower from the following data:

(a)Follower to moveoutward through 40 mm during60° ofa cam rotation;(b)

Follower to dwellforthenext 45° (c) Followertoreturn its original position during

next 90° (d)Follower to dwellforthe rest of camrotation. Thedisplacement of the


SCE

Engineering

100 Department of Mechanical

follower is to takeplacewith simple harmonic motion duringboth theoutward and

return strokes. Theleast radius of the cam is 50mm.If the cam rotates at 300 r.p.m.,

determinethe maximum velocityand accelerationof the follower duringtheoutward

stroke and return stroke.

3. A cam, with aminimum radius of 50 mm, rotatingclockwise at a uniform speed,

is required togiveraknife-edgedfollower themotion as described below: (a) To

moveoutwards through 40 mmduring100° rotation of the cam; (b)to dwellfornext

80° (c) To return to its startingposition duringnext 90 ° and (d)Todwellforthe rest

periodof revolution. Draw theprofileof the cam (i) When thelineof strokeof the

follower passes through the centreof the cam shaft and (ii)When the line ofstrokeof

the follower is to takeplacewith Uniform acceleration and uniform retardation.

Determinethe maximum velocity and acceleration of the followerwhen the cam

shaft rotates at 900 r.p.m.

4. Draw theprofileofacam operatingaroller reciprocatingfollowerand with the

followingdata:

Minimum radius of cam =25 mm;lift=30mm; Rollerdiameter=15mm. Thecam lifts

the follower for 120° with SHM, followed byadwellperiod of30°. Thenthe follower

lowers down during150°of cam rotation with uniform acceleration and retardation

followed byadwellperiod.If thecam rotates atauniform speed of 150 RPM.

Calculate the maximumvelocityand accelerationof follower duringthe descent

period.

5. It is required to set outthe profileof a cam togivethe followingmotion to the

reciprocating follower with a flat mushroom contact surface: (i)Follower to

haveastrokeof 20mm during120° ofcam rotation, (ii) Follower to dwellfor30°

ofcam rotation,(iii) Follower to return to its initial position during120° of cam

rotation, (iv)Follower to dwellfor remaining90°of cam rotation. Theminimum

radius of the cam is 25 mm. Theout strokeof the follower is performed with

SHM and return strokewith equal uniform acceleration andretardation.

6. A tangent cam to drivea roller follower throughatotal lift of 12.5 mm fora cam

rotation of75°.Thecam speed is 600rpm . Thedistancebetweencam centre and


SCE

Engineering


follower centre atfull lift is 45 mm and the roller is 20 mmin diameter. Find the

cam proportions and plot displacement, velocityandacceleration for one full cycle.

UNIT -IV


1. Define spur gear.

A spur gear is a cylindrical gear whose tooth traces are straight line generation of the

reference cylinder. They are used to transmit rotary motion between parallel shafts.

2. Define addendum and dedendum.

Addendum is the radial distance of a tooth from the pitch circle to the top of the tooth.

Dedendum is the radial distance of a tooth from the pitch circle to the bottom of the

tooth.

3. Define circular pitch.

It is the distance measured on the circumference of the pitch circle from a point of one

tooth to the corresponding point on the next tooth. It is denoted by Pc

Circular pitch Pc= /DT

Where D = Diameter of pitch

circle. T = Number of teeth on

the wheel. 4. Define I) path of contact. II) Length of path of contact.

Path of contact: It is the path traced by the point of contact of two teeth from the

beginning to the end of engagement.

Length of path of contact: It is the length of common normal cut- off by the

addendum circles of the wheel and pinion.

5. State the law of gearing.

Law of gearing states that, the common normal at the point of contact between a pair of

teeth must always pass through the pitch point.

6. Define conjugate action.

When the tooth profiles are so shaped so as to produce a constant angular velocity ratio

during Meshing, then the surface are said to de conjugate.

7. Define angle of approach.

The angle of approach is defined as the angle through which a gear rotates from the

instant a pair of teeth comes into contact until the teeth are in contact at the pitch point.


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8. List out the characteristics of in volute action.

a) Arc of contact. b) Length of path of contact. c) Contact ratio.

9. Define contact ratio.

Contact ratio is defined as the ratio of the length of arc of contact to the circular pitch. Mathematically.

Contact ratio = length of arc of conta

[Pc] Where Pc = circular path.

10. What is the advantage of involute gear?

The most important advantage of involutes gear is that the center distance for a pair of

involute gears can be varied within limits without changing the velocity ratio.

11. What are the conditions to be satisfied for interchangeability of all gears?

For interchangeability of all gears, the set must have the same circular pitch, module,

diameter pitch, pressure, angle, addendum and dedendum and tooth thickness must

be one half of the circular pitch.

12. Define gear tooth system.

A tooth system is a standard which specifies the relationship between addendum,

dedendum, working depth, tooth thickness and pressure angle to attain

interchangeability of gears of tooth numbers but of the same pressure angle and pitch

13. Define cycloid.

A cycloid is the curve traced by a point on the circumference of a circle which rolls

without slipping on a fixed straight line.

14. Define clearance.

The amount by which the dedendum of a gear exceeds the addendum of the mating gear

is called clearance.

15. When in volute interference occurs.

If the teeth are of such proportion that the beginning of contact occurs before the

interference point is met then the involute proportion of the driven gear will mate a

non in volute portion of the driving gear and involute interference is said to occur.

16. What is the principle reason for employing non standard gears?

a) To eliminate the undercutting. b) To prevent interference.

c) To maintain reasonable contact ratio.


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17. What are the advantages and disadvantages of gear drive?

Advantages:

a) It transmits exact velocity

ratio. b) It has high

efficiency.

Disadvantages:

The manufacture of gears require special tool and equipment.

The error in cutting teeth may cause vibrations and noise during operation.

18. Define helix angle ().

It is the angle between the line drawn through one of the teeth and the center line of the

shaft on which the gear is maintained.

19. Define gear ratio.

The quotient of the number of teeth on the wheel divided by the number of threads on the

worm.

20. Define gear train.

A combination of gears that is used for transmitting motion from one shaft to another

shaft is known as gear

train. eg. spur gear,

spiral gear.

21. Define velocity ratio.

Velocity ratio of a simple gear train is defined as the ratio of the angular velocity of the

first gear in the train to the angular velocity of the last gear.

PART-B (16 Marks)

1. a) Two matingspurgear with modulepitch of 6.5 mmhave19 ad 47 teeth of

20° pressureangle and 6.5 mm addendum. Determinethe number ofpair of

teethand angleturned through bythelarger wheel foronepair ofteeth in contact.

Determinealso theslidingvelocityat theinstant (i)engagement commences (ii)

engagement terminates. When the pitch line velocityis 1.2 m/s.(8)

b)Thenumberof teeth on each of thetwo spurgears in mesh is 40. Theteeth have

20° involuteprofile and the module is6mm.If thearcof contact is 1.75 times the circularpitch. Find

theaddendum. (8)


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2. a) Two 20° involutespurgears haveamoduleof 10 mm. The addendumis one

module. Thelargergear has 50 teeth and pinion 13 teeth. Does the

interferenceoccur?Ifitoccurs, to whatvalue should thepressureanglebe changed

to eliminate interference?(8)

b)Twomatinginvolute spur gears 20° pressure angle haveagearratio of2.the

numberof teeth on thepinion is 20 and its speed is 250 rpm. Themodulepitch of the

teeth is 12 mm. if theaddendum on each wheelwheel recess on each side arehalf

themaximum possible lengtheach, find(1)the addendum forpinion and

gearwheel(2)the length of arcof contact(30 themaximum velocityof

slidingduringapproachand recess.Assume pinion to bedriver. (8)

3.a) Apair of spurgearwith involuteteeth is to give agear ratio of 4:1. The arcof

approach is not be less than the circular pitch and thesmaller wheel is the driver. The

angleof pressureis 14.5What is the least number ofteeth can beusedoneach wheel?

What is the addendum of thewheel in terms ofcircularpitch?(8)

b). A pair 20° fulldepth involutespur gear having 30 and 50 teeth

respectively module4 mmarcin mesh, thesmaller gearrotates at 1000 rpm.

Determine (a) Slidingvelocities at engagementand disengagement of apair

ofteethand (b) Contact ratio.(8)

3. Inan epicyclicgear train theinternal wheels Aand Band compound wheels Cand

D rotateindependentlyabout axis O. Thewheels E andFrotateon pins fixed to the arm G. E

gears withA and C . Wheel Fgearwith B and D. Allthe wheels havethe same

module and the numberof teeth are: TC=28 TD=26;TE=TF=18.(1) Sketch the

arrangement, (2) Find thenumber ofteeth on A and B,(3)Ifthe armG makes 100

rpm clockwise and A is fixed, find thespeed ofB, and (4)If thearm Gmakes 100

rpm clockwise and wheel A makes 10rpm counter clockwise; Find

the speed ofwheelB. (16)

4. Twogearwheels meshexternallyandaretogiveavelocityratio of 3to 1. The

teeth areof involute form;module=6mm, addendum=onemodule,

pressureangle=20°. The pinion rotates at 90 rpm. Determine (1) thenumber

ofteeth on thepinion toavoid interferenceon itand the correspondingnumber


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ofteeth on thewheel, (2)Thelength of path andarc of contact, (3)the numberof

pairs of teeth in contact.(16)

5. The arm ofan epicyclicgear train rotatesat 100rpm in the anticlock wisedirection.

The arm carries two wheels AandBhaving36 and 45 teeth respectively. Thewheel A

is fixed and the arm rotates about thecentreof wheel A. Find thespeed of wheelB.

What will be the speed ofB, if thewheelA instead of beingfixed, makes 200 rpm

(clockwise).(16)

6. Ina revertedepicyclictrain, the armF carries two wheels Aand D anda compound

wheel B-C. Wheel A meshes with wheelBand Wheel D meshes with wheel C. Ther

numberof teeth on wheel A, Dand C are80,48, and 72. Find thespeed and direction

ofwheel D , when wheel A is fixed and arm Fmakes 200 rpm clockwise.(16)

7. An epicyclictrain is composed of afixed annularwheel A having150 teeth. Meshingwith A is

awheel Bwhich drives wheel D throughan idlewheel C, D being concentricwith A. Wheels B and

C are carried onarm which revolves clockwise at 100 rpm about the axis of A orD.Ifthe wheels

Band D arehaving25 teeth and 40 teeth respectively,find thenumberof teeth of C and the speed

and sense ofrotation of C. (16)

8. Thesun planetgear ofan epicyclicgear train, the annularD has 100 internal teeth,

the sun gear Ahas 50 external teethand planetgear Bhas 25external teeth.Thegear

Bmeshes withgear Dand gear A. ThegearBis carried onarm E, which rotatesabout

the centreof annulargearD.If thegear Dis fixedand arm rotates at 20 rpm, then find

the speeds ofgear AandB.(16)

UNIT -V


1. Define clutch.

Clutch is a transmission device of an automobile which is used to engage and disengage

the power from the engine to the rest of the system.

2. What are the types of friction

clutches?

Types of friction clutches are:

*Disc or plate clutches.

*Cone clutches.

*Centrifugal clutches.


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3. Define centrifugal clutch.

Centrifugal clutch is being increasingly used in automobile and machines obviously it

works on the principle of centrifugal force.

4. What are the types of flat drives?

The types of flat drives are:

*Compound belt drive.

*Stepped or cone pulley drive.

*Fast and loose pulley.

5. Define slip.

Slip is defined as the relative motion between the belt and pulley.

6. Define law of belting.

Law of belting states that the centre line of the belt, as it approaches the pulley lie in a

plane perpendicular to the axis of that pulley or must lie in the plane of the pulley,

otherwise the belt will run off the pulley.

7. What is the use of rope drive?

The rope drives are widely used when large power is to be transmitted continuously from

one pulley to another over a considerable distance. One advantage of rope drives is

that a number of separate driver may be from the driving pulley.

8. What is the use of belt drive?

Belt drive is commonly used for transmission of power when exact velocity ratio is not

required. Generally, belt drives are used to transmit power from one pulley to another,

when the two pulleys are not more than 10 meters apart.

9. What are the types of ropes?

The types of ropes are: *Fiber ropes. *Wire ropes.

10. What is the use of quarter turn left drive?

The quarter turn left drive is used with shafts arranged at right angles and rotating in one

definite direction.

11. Define the velocity ratio of the belt drive.

The velocity ratio of the belt drive is defined as the ratio between the velocities of the

driver and the follower or the driven.


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12. What are the advantages of V-belt?

*Power transmitted is more due to wedging action in the grooved pulleys.

*V-belt is more compact, quite and shock absorbing.

*The V-belt drive is positive because of negligible slip between the belt and the groove.

*High velocity ratio may be obtained.

13. What are the disadvantages of V-belt?

*It cannot be used with large center distances.

*It is not as durable as flat belt.

*It is a costlier system.

14. What is the use of circular belts or

ropes?

*Ropes are circular in cross section. *It is used to transmit more power.

*Distance between two pulleys is more than 8metres.

15. Specify the various belt materials.

BELT TYPES BELT MATERIALS

Flat belts Leather, canvas, cotton & rubber.

V-belts Rubberized fabric & rubber.

Ropes Cotton, hemp & manila.

16. Name the types of friction.

*Static friction. *Dynamic friction.

17. Define the angle of repose.

If the angle of inclination ‘α’ of the plane to horizontal is such that the body begin to

move down the plane. Then the angle α is called the angle of repose.

18. What is meant by frictional force?

Force of friction is always acting in the direction opposite to the direction of motion.

19. Why self locking screws have lesser efficiency?

Self locking needs some friction on the thread surface of the screw and hence it needs

higher effort to lift a body and hence automatically the efficiency decreases.

20. What is static friction?

It is the friction experienced by a body, when at rest.

21. What is dynamic friction?

It is the friction experienced by the body, when in motion. The dynamic friction is also

called as kinematic friction.


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22. What is co-efficient of friction?

It is defined as the ratio of the limiting friction (F) to the normal reaction (RN) between

the two bodies. It is generally denoted by μ. μ=F/RN.

23. Define screw jack.

The screw jack is the device used to lift the heavy loads by applying a comparatively

small effort at its handle. The working principle of screw jack is similar to that of an

inclined plane.

Part B

1. a) For a flat belt, prove that T1/T2=eμ_ Where T1 and T2= Tension in the tight

and slack sides of the belt, _= Angle of contact between the belt and the pulley,

and μ= Coefficient of friction between the belt and the pulley. (8)

b) An open belt running over two pulley of 1.5 m and 1.0 m diameters connects

two parallel shafts 4.8 m apart. The initial ten in the belt is 3000 N. The smaller

pulley is rotating at 600 rpm. The mass of belt is 0.6703 kg/m length. The

coefficient of friction between the belt and pulleys is 0.3. Find (1) the exact length

of the belt required (2) the power transmitted taking c.f tension into account.(8)

2.a) A multiplate disc clutch transmits 55 KW of power at 1800 rpm. Coefficient of

friction for the friction surfaces is 0.1. Axial intensity at pressure is not to exceed

160

KN/m2. The internal radius is 80 mm and is 0.7 times the external radius. Find

the number of plates needed to transmit the required torque (8)

b) A rope drive is required to transmit 230 KW from a pulley of 1m diameter

running at 450 rpm. The safe pull in each rope is 800 N and the mass of the rope is

0.4 kg per meter length. The angle of lap and groove angle 1600 and 450

respectively. If coefficient of friction is 0.3, find the number of ropes required. (8)

3.The mean diameter of the screw jack having pitch of 10 mm is 50 mm. A load of 20

KN is lifted through a distance of 170 mm. Find the work done in lifting the load

and efficiency of the screw jack when (i) the load rotates with the screw, and (ii) the

load rests on the loose head which does not rotate with screw. The external and

internal diameter of the bearing surface of the loose head is 60 mm and 10mm

respectively. The coefficient of friction for the screw as well as the bearing surface

may be taken as


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0.08. (16)

4.a).A leather belt is required to transmit 7.5 kw from a pulley 1.2 m in

diameter, running at 250 rpm. The angle entranced is 1650 and the coefficient

of friction

between the belt6 and the pulley is 0.3. If safe working stress for the leather belt is 1.5

MPa, density of leather is 1 kg/ m3 end thickness of belt is 10 mm. Determine

the width of the belt taking C.F tension into account. (8)

b).Two pulley one 450 mm diameter and other 200mm dia are on parallel shaft 2.1

m apart and are connected by a cross belt. The larger pulley rotates at 225 rpm. The

maximum permissible tension in the belt is 1 KN and the coefficient of friction

between the belt and the pulley is 0.25. Find the length of the belt required and the

power can be transmitted. (8)

5.Two shaft whose centers are 1m apart are connected by a V belt drive. The

driving pulley is supplied with 100 KW and has an effective diameter of 300 mm.

It runs at 375 rpm. The angle of groove on the pulley is 400 The permissible

tension in 400 mm2 cross sectional area of the belt is 2.1 MPa. The density of the

belt is 1100 kg/ mm3 coefficient of friction is 0.28. Estimate number of belts

required. (16)


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UNIVERSITY QUESTIONS

B.E.IB.Tech.DEGREE EXAMINATION,NOVEMBER 2006.

Fourth Semester

(Regulation 2004) Mechanical Engineering

ME 1252- KINEMATICS OFMACHINERY

(Common toB.E. (Part-Time) Third Semester

Regulation 2005)

1. (i)DescribeWhitworth's quick return echanism. (6)

(ii)Distancebetweentwoparallelshaftsconnected by Boldham‘scouplingis25

mm.Determinemaximumspeedofsliding oftongueofintermediatepieceintheslot in the

flangeif drivingshaft is run at 250 rpm. (4)

(iii) DiscusstheapplicationofGrashoff‘slawinidentifyingtheinputandoutput

motions of four-bar echanism.

2. (i)Defineand explain inversion of mechanisms. (4) (ii)

Sketch and explainoldham‘s coupling. (6)

(iii)Designafour-barcrankrockerquickreturnmechanismtogiveatimeratioof

1.25withrockerswingangleas75°clockwise.Assumetheoutputlink(rocker) length as 50 mm and in

theleft extreme position it is vertical.

3. (i)A four bar mechanism DABC has the followingdimensions

:

DA=300mm; CB =AB =360mm; DC=600 mm.LinkDCisfixedand angleADC

isDriving linkDAturnsclockwiseat100rpm.Constantdriving torqueis50N-m.

Determinethefollowing:(1)Velocity ofpointB(2)Angularvelocity ofdrivenlink

CB(3)Mechanical advantageof mechanismin this position (4)Resistingtorque.(12)

(ii) Sketcha four-barcrankrocker mechanismin (1)Maximum transmission angle

position and (2) toggle position wheremechanicaladvantageis infinity. (4)


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4.(i)Findthenumber ofinstantaneouscentresforasixlinkmechanism.Statetheuse

ofinstantaneous centre method in kinematic analysis.(4)

(ii)Crankofaslidercrankmechanismrotatesclockwiseataconstantspeedof

300rpm,crankandconnectingrod are of lengths150mmand600respectively. Determine

thefollowing;ata crankangle of45°frominnerdeadcentreposition(1) Linear velocity

and acceleration of the midpoint of connecting rod (2) Angular velocityand

angularacceleration ofthe connectingrod.(12)

5.(i)Explainthefollowingterms:(1)Primecircle(2)PressureAnglerelatedto cams. (4)

(ii)Drawthe profileof a disc camtogive uniformmotionanduniformvelocity

duringoutstrokeof25mmtoaknifeedgefollowerduringfirsthalfrevolution.

Returnofcamisalsoofsimilaruniformmotionwithuniformvelocity during

remaininghalfrevolution.Minimumradiusofthecamis25mm.Assumethatthe axis of

knife edge follower passes throughcam axis.(12)

6. (i)Brieflyexplain the features of mushroom followers

4)

(ii)Drawtheprofileofacamtogivefollowingmotiontoaflatfacedreciprocating

follower.(1)Followertohaveastroke of20 mmduring 120°ofcamrotation.(2)

Followertodwellfor 30°ofcamrotation(3) Followertoreturntoinitialposition during

120°ofcamrotation.(4)Followertodwellforremaining 90°ofcamrotation.

Thebasecircleradiusis40mmandthefolloweraxiscoincideswith camaxis of rotation.

(12) (i)Discuss the advantages of involutegear toothprofile. 4)

(ii) Describethe advantages and applications of helical, bevel and wormgears.

(6)

(iii) Inanepicyclicgeartrain,thesumgearAandtheplanetgearB are having36 and 45 teeth

respectively.If the arm rotatesat150rpm counterclockwiseabout center ofAwhich

isfixed, determinespeedofgearB.Ifthe arm is locked andgearArotates at 300 rpm what

is the speed ofgearB?(6)


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8.(i)Apinionhaving 25teethdrivesagearof60teeth.Toothprofileisinvolutewith

pressureangle of 20°,module of8mmandaddendumof 1module.Determine:(1) Length

of path of contact(2)Length ofarcof contact and(3)contact ratio.(6)

(ii) Designacompoundgeartrainforanexacttrainratioof180:1.Minimum teeth

onanygearshall be12 toavoid interferenceand Maximum gearratioin anyone

stageis10 : 1. Also sketch the arrangement.(6)

(iii) Discuss thenecessityof differentials in automobiles.

(4)

9. (i)Discuss the advantages of

Vbelts.

(ii) Aropedrivetransmits600kWfromapulley ofeffectivediameter4m,which

runsat90rpm.Angleoflapis160°;angleofgrooveis45°;co-efficientoffrictionis

0.28;massofropeis1.5kg/mandallowabletensionperropeis2400N.Determine the

number ofropes required

(iii) Describethe features of internal expanding brakes used inautomobiles.

(4)

10.(i)Following dataareforascrewjack.Screwpitchis12.5mm;meandiameterof

screwis50mm;co-efficientoffrictionis0.13.Determinetorquestoraiseandlower 20 kN

load and efficiencyofjack. (6)

(ii) Discuss the functions of clutches in automobiles.

(4)

(iii) Derivetheexpression to determinethe ratioof tensions in a flat beltdrive.


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B.E.IB.Tech.DEGREE EXAMINATION, , NOVEMBER 2006.

Fourth Semester


T8248- KINEMATICS OFMACHINERY


Regulation 2005)

11. What is Kutzbach criterion forplanar mechanism?

12. Sketch an exact straight line mechanism, with link proportions.

13. Explain normal component of acceleration.

14. StateCorioli's law.

15. What arethe classifications of cams based oncontact surfaces?

16. Statethe basic requirements forhigh speedcams.

17. Definemoduleofgears and its relation to circularpitch.

18. Explain brieflythe useof differential in an automobile.

19. What aretimingbelts?

20. Explain brieflysignificanceof friction in braking.

11.(i)Whatismeantby interferenceingears?Whatarethemeasurestoeliminatethe same?

(4)

(ii)AnepicyclictrainhasapinionAhaving15teeth,centrally locatedandrigidly

connectedtoshaftofdriving motor.AnothergearBhaving20teethisgearing withA

andalsowithannularfixedwheelD.GearCisintegralwithB andmesheswith another

annular wheelEwhichiskeyedtothe shaftof driven unit.The armrotates aboutdriving

shaftandcarriescompoundgearB,C.Sketchthearrangementand

determinespeedofmachineforamotorspeedof 1000rpm.Alsodeterminetorqueon

machine shaft for amotor torqueof 100 N-m.(12)

12. Twomating gearshave20and40involuteteethofmodule10mmand20° pressure

angle.The addendumoneachwheelistobemade ofsuchalengththatthe line of

contactoneachsideof thepitchpointhashalfthemaximumpossiblelength. Determine the

addendumheightforeachgearwheel,lengthofthepathofcontact,arc of contact


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andcontactratio.

13. (i)Explain the inversions of four barchain, with neat

sketches. (8)

(ii)Explainwithneatsketchesthefollowing:(1) Offsetslider mechanism,(2)An indexingmechanism.

14. (i)Explain theinversionsof single slider crank chains, with neat sketches.

(10)

(ii)Explainmechanicaladvantageandtransmissionanglerelatedtofour-bar mechanisms.

(6)

15. (i) Compare instantaneous center method and velocity polygon method for

velocityanalysisof mechanisms.

(4)

(ii)Afourbarlinkagehasfollowing dimensions:CrankAO2=150mm;LinkAB

= 450mm;LinkBO4= 300mm;LinkO2O4= 200mm.Link O2O4isfixed.Findthe

angular accelerationoflinksAB andBO4whenthecrankisrotating withaconstant

angular velocity of200rad/scounterclockwiseandalsopositionedof45°to horizontal.

(12)

16. (i)Statethe properties of relative velocities ofpoints on kinematic links.

(4)

(ii)Inasmallsteamenginerunning at600rad/mincounterclockwise,lengthof

crankis80mmandratioofconnectingrodlengthtocrankradiusis3.Forthe

positionwhencrankmakes45°tohorizontal,determine accelerationofpiston.Find also

total acceleration ofapointX on connectingrod 80 from crank pin.(12)

17. Drawtheprofileofdisccamtogiveuniformaccelerationandretardationout

strokeof25toaknifeedgefollowerduringfirsthalfofrevolution.Returnofcam also takes

place with uniform motion during remaining half of cam revolution. Assume

minimumradius of cam as 25 mm.


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18. Drawtheprofileofacamtogivefollowingmotiontoareciprocatingfollower

withaflatface:(i)Followertohaveastrokeof20mmduring120°ofcamrotation. (ii) Follower todwell

for 30°ofcamrotation.(iii)Followertoreturntoitsinitial positionduring

120°ofcamrotation.(iv)Followertodwellforremaining90°ofcam

rotation.Minimumradiusofcam=25mm.Outstrokeandreturnstrokeofthefollower aresimple

harmonic.

19. (i) Pitch of 50 mm diameter threaded screw of a screw jack is 12.5 mm.

Coefficientof frictionbetweenscrew andnutis0.10.Determine the torque toraisea load

of 25kNrotatingwiththe screw. Alsofindthe torquerequiredtolower the load and

efficiencyof screw jack.(4)

(ii)100kWistobetransmittedby aropedrivethrougha160cmdiameter45° groovedpulley

runningat200rpm.Angleofoverlap140°andcoefficientoffriction betweenpulley

andropeis0.25.Massofropeis0.7kg/manditcanwithstanda tensionof 800N.

Considering centrifugaltension,findthefollowing:(i)Numberof ropes required,

(ii)Initialtension in the rope.(12)

20. (i)Explain the following:

(1) Crowningof pulleys(3)

(2) Self-lockingofbrakes(3)

(3)Uses of brakesin

automobiles. (2)

(ii) A10kWenginedevelopsamaximumtorque of100N-mandisdrivingacar

havingasingleplateclutchoftwoactivesurfaces.Axialpressureisnottoexceed

0.85bar.Externaldiameter offrictionplateis1.25timesinternaldiameter.Assume

uniformwearandco-efficientoffriction= 0.3.Determine dimensionoffrictionplate and

axial forceexerted bythe springs.

B.E.IB.Tech.DEGREE EXAMINATION, MAY2007

.

Fourth Semester




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Regulation 2005)

21. Enumerate thedifferencebetween aMachineand aStructure.

22.Listout theinversions of adouble slider crankchain.

23. Define rubbingvelocity.

24. DefineCorioli'scomponent of acceleration.

25.Statetheexpressionsformaximumvelocityandaccelerationofafollowermoves with

Cycloidal motion.

26.Whatisprimecircleofacam?Whatistheradialdistancebetweentheprime circle and

basecircle fora cam with knifeedge follower?

27. What is axial pitch ofahelicalgear?

28.Listout the applications of epicyclicgear train.

29. What is the conditionof maximum efficiencyofaScrew jack?

30. What arethe advantages of wireropes over fabric ropes?

R3456, NOVEMBER 2007

31. DefineDegreeofFreedom andgive theDOFforashaft in acircular hole.

32. StateGrashof‘s law fora four bar linkage.

33. What is Corioli's componentof acceleration?

34. Statethe Freudenstein's equation for a four-bar mechanism.

35. What is a circulararc cam?

36.Statetheexpressionsformaximumvelocityandaccelerationofafollowermoves with

cycloidal motion.

37. Differentiate diametral pitch and circular pitchof afriction wheel.

38. What is revertedgeartrain?

39. Diagrammaticallyrepresenttheforcesactingonabodywhenitslideseitherupor down

on an inclined planewith outconsideringthe effect offriction.

40. Listout anyfour desirable characteristics ofbrakeliningmaterial.


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B.E.IB.Tech.DEGREE EXAMINATION, MAY2008

.

Fourth Semester



41. Definedegreeof freedom.

42. DefineGrashoffs law.


Regulation 2005)

43. Explain the coriolis component of acceleration.

44. What is higher pair?

45. Definepitch curveofthe cam.

46. Defineundercuttingin Gears.

47. Defineinterferencein Gears.

48. Definepressureangleand explain the effect ofdifferent pressureangle.

49. What is creep in thecaseof belt?

50. Which typeof screwthread is preferable in power transmission?

B.E.IB.Tech.DEGREE EXAMINATION,MAY/JUNE 2007.

Fourth Semester





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Regulation 2005)

Time :Three hours Maximum: 100marks

- A3sizedrawing sheet willbeissued, ifrequired.

Answer ALLquestions. PARTA- (10x2=20marks)

1. Enumerate the difference between aMachine and aStructure.

2. List outthe inversions ofadouble slider crank chain.

3. Define rubbing velocity.

4. Define Corioli's component ofacceleration.

5. State the expressions for maximum velocity and acceleration of a

follower moves with Cycloidal motion

6. What isprime circle ofacam? What isthe radial distance between the

prime circle and base circle foracamwith knife edgefollower?=

7. What isaxial pitch ofahelical gear?

8. List out the applications ofepicyclic gear train.

9. What isthe condition ofmaximum efficiency ofaScrewjack?

10. What are the advantages ofwire ropes overfabric ropes?

PART B- (5x16=80marks)

11. (a) (i) Define transmission angle ofafour bar linkage.What isthe effect

oftransmission angle onmechanicaladvantage? (4)

(ii) Brieflyexplainvarioustypes ofconstrainedmotions. (4)

(iii) Illustrate a crank and slotted lever mechanism as an inversion of single

slider crank chain. Deducean expressionforlength ofstroke interms oflink

lengths. (8)

Or

(b) Analyticallyperformthe displacement analysisofafour bar mechanism.

12. (a) The driving crank AD of the quick-return mechanism, as shown in

Fig. 12.(a). revolvesat auniformspeed of200 r.p.m. Find the velocity and

accelerationofthe tool-boxR,inthe positionshown,when the crank makes

an angle of60°with the vertical line ofcenters PA.What isthe

accelerationofsliding ofthe blockatBalong the slotted lever PQ?

(b) For the toggle mechanism as shown in Fig. 12(b), the slider D IS

-


SCE

Engineering


constrained to move along horizontal direction. The crank rotates at

180rpm. The dimension ofvarious links are asfollows. OA= 180mm ;

CB=240mm; AB=360mm; BD=540mm. For the given configuration

determine the velocity ofthe slider and angular velocity oflinks AB,BC

and BD.Alsodetermine the linear acceleration ofthe slider D. 360 mm

13. (a) Acam with aminimum radius of25mm, rotating in clockwise direction with

auniform speed of100rpm istobedesigned togivethe motion fora roller follower

asfollows.

(i) Toraise through 50mm during 120°rotation ofcamwith SHM.

(ii) Fully raised through next 30°

(iii) Tolower during next 60°with UAUR (iv)

Dwell forthe remaining period.

Draw the profile ofthe camwhen the line ofstroke ofthe follower is offset

by15mm from the axis ofthe camshaft.

Or

(b) Constructatangent cam and mentionthe important terminologiesonit.

Also derive the expression for displacement, velocity,acceleration ofa

reciprocatingroller followerwhen the roller has contactwith the nose.

14. (a) (i) An epicyclicgear train is shownIn the Fig. 14(a). How many

revolutionsdoesthe arm makes,

(1) when Amakesone revolutionin clockwiseand Dmakesta revolutioninthe

oppositesense

(2) when A makes one revolutionIn clockwiseand D remains stationary.

The number of teeth I gears A and Dare 40 and 90


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Engineering


respectively. (10)

(ii) What is reverted gear train? Explain the arrangement ofvarious gears in a reverted

gear train and express the characteristic equationsused todefine their operation.

(6)

(b) (i) State and prove lawofgearing.

(ii) Apair ofinvolutespur gears with 16°pressure angle and pitch of

module6mm isin mesh. The number ofteeth in pinionis 16and its rotational

speed is 240 rpm. The gear ratio is 1.75. In order to avoid the

interference,determine

(1) addendaonpinionand wheel

(2) length ofpath ofcontact

(3) maximumvelocityofslidingoneither side ofpitch point. (12)

15. (a) (i) Asquarethreaded bolt ofroot diameter22.5mm and pitch 5mm is

tightened by screwing a nut whose mean diameter of bearing

surfaceis50mm. Ifthe coefficientoffrictionbetweennut and bolt is 0.1 and

nut and bearing surface is 0.16, determine the force required at the

end ofspanner 500mm long when the load onthe boltis10kN.

(8)

(ii) A leather faced conical clutch has a cone angle of 30°. If the intensity

of pressure between the contact surfaces is limited to

0.35 Nzmm?and the breadth ofthe conicalsurfaceisnot toexceed

1I3rdofthe mean radius. Determine the dimensionsofthe contact

surfacesto transmit 22.5 KWat 2000 rpm. Assumeuniformwear rate and

Jl=0.15. (8)

Or

(b) (i) Derivethe expressionforFrictional torqu oncone clutch based on


Engineering

uniformpressuretheory. (6)

(ii) The brake whose dimensions are shown in figure 15 (b) has a co-

efficientoffriction of0.3and istohave amaximum pressure of 1000kPa against the friction material.

(1) Using an actuating force of1750N,determine the face width ofthe

Shoes (both shoes have same width) and

(2) What torque willthe brake absorb?

SCE 121 Department of Mechanical


Engineering


kinematics of machinery unit i basics of...

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