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CPS Question and answers

Examination: 2017 SUMMER
Que.No Question/Problem marks Link
Q 2 a )

Question:

Draw a neat sketch and explain working of beam engine.

Answer:

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Q 2 c )

Question:

Draw and explain in short, types of followers used in cam and follower.

Answer:

Types of followers :

The followers may be classified as discussed below:

1. According to the surface in contact.

(a)Knife edge follower. When the contacting end of the follower has a sharp knife edge, it is called a knife edge follower.

(b) Roller follower. When the contacting end of the follower is a roller, it is called a roller follower.

(c) Flat faced or mushroom follower. When the contacting end of the follower is a perfectly flat face, it is called a flat faced follower and when the flat faced follower is circular, it is then called a mushroom follower.

2.According to the motion of the follower.

(a) Reciprocating or translating follower. When the follower reciprocates in guides as the cam rotates uniformly, it is known as reciprocating or translating follower.

(b) Oscillating or rotating follower. When the uniform rotary motion of the cam is converted into predetermined oscillatory motion of the follower, it is called oscillating or rotating follower.

3. According to the path of motion of the follower.

(a) Radial follower. When the motion of the follower is along an axis passing through the centre of the cam, it is known as radial follower

(b) Off-set follower. When the motion of the follower is along an axis away from the axis of the cam centre, it is called off-set follower.

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Q 2 d )

Question:

Explain condition for maximum power transmission.

Answer:

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Q 2 e )

Question:

Explain the compound gear train with neat sketch and write down the velocity ratio’s equation.

Answer:

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Q 3 a )

Question:

Differentiate between mechanism and machine.

Answer:

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Q 3 b )

Question:

Explain the working of Whitworth quick return mechanism.

Answer:

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Q 3 e )

Question:

Explain the working of Watt governor with neat diagram.

Answer:

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Q 3 f )

Question:

Explain the working of centrifugal clutch with neat sketch.

Answer:

centrifugal Clutch:

• A centrifugal clutch is a clutch that uses centrifugal force to connect two concentric shafts, with the driving shaft nested inside the driven shaft.

• It consists of number of shoe on the inside of a rim of pulley. The outer surface of pulley is covered with friction material.

• These shoes move radially in guides.

• As the speed of the shaft increase, the centrifugal force on the shoes increases.

• When the centrifugal force is less than the spring force, the shoes remain in the same position as when the driving shaft was stationary, but when the centrifugal force is equal  to the spring force, the shoes are just floating.

• When the centrifugal force exceeds the spring force, the shoes move outward and come into contact with the driven member presses against it.

• The force with which the shoe presses against the driven member is the difference of the centrifugal force and the spring force.

• The increase of speed causes the shoe to press harder and enable more torque to be transmitted.

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Q 4 a )

Question:

Explain the working of freewheel mechanism of bicycle with sketch.

Answer:

A freewheel mechanism on a bicycle allows the rear wheel to turn faster than the pedals. If there is no freewheel on a bicycle, a simple ride could be exhausting, because one could never stop pumping the pedals. And going downhill would be downright dangerous, because the pedals would turn on their own, faster than one could keep up with them.

Power Train of a bicycle: The power train of a simple bicycle consists of a pair of pedals, two sprockets and a chain. The pedals are affixed to one sprocket — the front sprocket, which is mounted to the bike below the seat. The second sprocket is connected to the hub of the rear wheel. The chain connects the two sprockets. When you turn the pedals, the front sprocket turns. The chain transfers that rotation to the rear sprocket, which turns the rear wheel, and the bicycle moves forward. The faster you turn the pedals, the faster the rear wheel goes, and the faster the bike goes.

Coasting: At some point — when going downhill, for instance — speed is high enough so that the rear wheel is turning faster than the pedals. That's when coasting: we stop working the pedals and let the bike's momentum keep moving forward. It's the freewheel that makes this possible. On a bicycle, instead of being affixed to the wheel, the rear sprocket is mounted on a freewheel mechanism, which is either built into the hub of the wheel — a "freehub" — or attached to the hub, making it a true freewheel.

Now when you have to move forward, the pawl acts like a hook and gets locked with the teeth - called ratchet and transmits the torque. The complete mechanism is called ratchet and pawl mechanism.

But when you reverse pedal, it falls back and becomes "free". A spring prevents it from falling permanently. This is the reason why you hear the distinct "click-click" sound when you reverse pedal. Also, there are multiple "pawls" placed along the circumference too.

 

 

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Q 4 c )

Question:

What are the advantages of ‘V’ belt drive over flat belt drive ?

Answer:

Advantages of V-belt drive over flat belt drive :

1. The V-belt drive gives compactness due to the small distance between the centres of pulleys.

2. The drive is positive, because the slip between the belt and the pulley groove is negligible.

3. Since the V-belts are made endless and there is no joint trouble, therefore the drive is smooth.

4. It provides longer life, 3 to 5 years.

5. It can be easily installed and removed.

6. The operation of the belt and pulley is quiet.

7. The belts have the ability to cushion the shock when machines are started.

8. The high velocity ratio (maximum 10) may be obtained.

9. The wedging action of the belt in the groove gives high value of limiting ratio of tensions. Therefore the power transmitted by V-belts is more than flat belts for the same coefficient of friction, arc of contact and allowable tension in the belts.

10. The V-belt may be operated in either direction with tight side of the belt at the top or bottom. The centre line may be horizontal, vertical or inclined.

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Q 4 d )

Question:

Explain the working of flywheel with the help of turning moment diagram.

Answer:

Working of Flywheel with the help of Turning moment diagram:

A flywheel used in machines serves as a reservoir, which stores energy during the period when the supply of energy is more than the requirement, and releases it during the period when the requirement of energy is more than the supply.

The fluctuation of energy may be determined by the turning moment diagram for one complete cycle of operation. Consider the turning moment diagram for a single cylinder double acting steam engine as shown in Fig. We see that the mean resisting torque line AF cuts the turning moment diagram at points B, C, D and E. When the crank moves from a to p, the work done by the engine is equal to the area aBp, whereas the energy required is represented by the area aABp. In other words, the engine has done less work (equal to the area a AB) than the requirement. This amount of energy is taken from the flywheel and hence the speed of the flywheel decreases. Now the crank moves from p to q, the work done by the engine is equal to the area pBbCq, whereas the requirement of energy is represented by the area pBCq. Therefore, the engine has done more work than the requirement. This excess work (equal to the area BbC) is stored in the flywheel and hence the speed of the flywheel increases while the crank moves from p to q.

Similarly, when the crank moves from q to r, more work is taken from the engine than is developed. This loss of work is represented by the area C c D. To supply this loss, the flywheel gives up some of its energy and thus the speed decreases while the crank moves from q to r. As the crank moves from r to s, excess energy is again developed given by the area D d E and the speed again increases. As the piston moves from s to e, again there is a loss of work and the speed decreases. The variations of energy above and below the mean resisting torque line are called fluctuations of energy. The areas BbC, CcD, DdE, etc. represent fluctuations of energy.

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Q 4 e )

Question:

Explain the working of internal expanding brake with neat sketch.

Answer:

Internal Expanding shoe brake:

An internal expanding brake consists of two shoes S1 and S2. The outer surface of the shoes are lined with some friction material (usually with Ferodo) to increase the coefficient of friction and to prevent wearing away of the metal. Each shoe is pivoted at one end about a fixed fulcrum O1and O2 and made to contact a cam at the other end. When the cam rotates, the shoes are pushed outwards against the rim of the drum. The friction between the shoes and the drum produces the braking torque and hence reduces the speed of the drum. The shoes are normally held in off position by a spring . The drum encloses the entire mechanism to keep out dust and moisture. This type of brake is commonly used in motor cars and light trucks.

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Q 6 a )

Question:

Draw a neat sketch of Oldham’s coupling and explain the working of it.

Answer:

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Q 6 b )

Question:

Define following terms :

Fluctuation of energy, co-efficient of fluctuation of energy, co-efficient of fluctuation speed, maximum fluctuation of energy.

Answer:

Fluctuations of energy: The variations of energy above and below the mean resisting torque line are called fluctuations of energy.

Coefficient of fluctuation of energy: It may be defined as the ratio of the maximum fluctuation of energy to the work done per cycle. Mathematically, Coefficient of fluctuation of energy, E = Maximum fluctuation of energy/Work done per cycle

Coefficient of fluctuation of speed: The difference between the maximum and minimum speeds during a cycle is called the maximum fluctuation of speed. The ratio of the maximum fluctuation of speed to the mean speed is called the coefficient of fluctuation of speed.

Maximum fluctuation of energy: Δ E = Maximum energy – Minimum energy = (E + a1) – (E + a1 – a2 + a3 – a4) = a2 – a3 + a4

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Q 6 c )

Question:

Explain the working of rope brake dynamometer with neat sketch

Answer:

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Q 6 d )

Question:

Explain the working of single plate clutch with neat diagram.

Answer:

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Q 6 e )

Question:

State reasons for balancing of rotating elements of machine. Explain balancing concept.

Answer:

Reasons for balancing of rotating elements of machine: The balancing of the moving parts both rotating and reciprocating of such machine is having greater importance. Because, if these parts are not balanced properly then the unbalanced dynamic forces can cause serious consequences, which are harmful to the life of the machinery itself, the human beings and all the property around them. These unbalanced forces not only increase the load on the bearings and stresses in various members, but also produces unpleasant and dangerous vibrations in them.

Concept of balancing: When a mass moves in circular pitch, it experience a centripetal acceleration which generates a force acting towards the center of rotation. An equal and opposite force which is acting radially outwards which is called centrifugal force. This force is the disturbing force for the system. The magnitude of this force remains constant but the direction goes on changing with the rotation of mass. The centrifugal force , on a rotating machine can be expressed mathematically as follows:

Fc = m. ω².r Newton

Where, m = Mass of rotating part in kg,

Ω = angular speed of this part in rad/sec, and r = Distance of the center of gravity of mass from the axis of rotation of part in m.

For the balance of rotating masses, it is the centrifugal force which is to be balanced. This type of problem is very common in steam turbine rotors, engine crank shafts, rotory compressors and centrifugal pumps.

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Examination: 2017 WINTER
Que.No Question/Problem marks Link
Q 1 b )

Question:

State any four types of friction clutch, along with its application each.

Answer:

(Types of clutches: Two marks, applications Two marks) Types of clutches: a) Single plate clutch b) Multi plate clutch c) Cone clutch d) Centrifugal clutch Applications: a) Single plate clutch: Heavy vehicles, four-wheeler such as car, truck, bus b) Multi plate clutch: Two wheelers, mopeds, scooters, bikes c) Cone clutch: Machine tools, automobiles, press work d) Centrifugal clutch: mopeds, Luna

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Q 1b)(a)

Question:

Define completely constrained motion and successfully constrained motion with neat sketch. State one example of each.

Answer:

a) 1. Completely constrained motion: When the motion between a pair is limited to a definite direction irrespective of the direction of force applied, then the motion is said to be a completely constrained motion. For example, the piston and cylinder (in a steam engine) form a pair and the motion of the piston is limited to a definite direction (i.e.it will only reciprocate) relative to the cylinder irrespective of the direction of motion of the crank. Examples: 1. The motion of a square bar in a square hole 2. the motion of a shaft with collars at each end in a circular hole,

2. Successfully constrained motion: When the motion between the elements, forming a pair, is such that the constrained motion is not completed by itself, but by some other means, then the motion is said to be successfully constrained motion. Consider a shaft in a foot-step bearing as shown in Fig. The shaft may rotate in a bearing or it may move upwards. This is a case of incompletely con-strained motion. But if the load is placed on the shaft to prevent axial upward movement of the shaft, then the motion of the pair is said to be successfully constrained motion. Examples:1. The motion of an I.C. engine valve (these are kept on their seat by a spring) 2. The piston reciprocating inside an engine cylinder 3. Shaft in a foot step bearing

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Q 1b)(b)

Question:

State function of clutch. Explain working principle of clutch.

Answer:

b) Function of the Clutch 1. Function of transmitting the torque from the engine to the drive train. 2. Smoothly deliver the power from the engine to enable smooth vehicle movement. 3. Perform quietly and to reduce drive-related vibration. WORKING PRINCIPLE OF CLUTCH It operates on the principle of friction. When two surfaces are brought in contact and are held against each other due to friction between them, they can be used to transmit power. If one is rotated, then other also rotates. One surface is connected to engine and other to the transmission system of automobile. Thus, clutch is nothing but a combination of two friction surfaces

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Q 2 a )

Question:

What is a machine ? Differentiate between a machine and a structure.

Answer:

Sl. No
Machine
Structure

1
All parts / links have relative motion
No relative motion between the links

2
It transforms the available energy into some useful work
No energy transformations

3
The kinematic link of a machine may transmit both power and motion
The member of the structure transmit forces only

4
Examples: I.C. Engine, Machine tools, steam engine, type writer, etc.
Example: Truss of roof, frame of machine, truss of bridge

5
Studied under 'Dynamics'
Studied under 'Statics'

 

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Q 2 a )

Question:

Differentiate between machine and structure.

Answer:

Sl. No
Machine
Structure

1
All parts / links have relative motion
No relative motion between the links

2
It transforms the available energy into some useful work
No energy transformations

3
The kinematic link of a machine may transmit both power and
The member of the structure transmit forces only

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Q 2 b )

Question:

Describe with neat sketch the working of scotch yoke mechanism.

Answer:

Crank and slotted lever quick return motion mechanism. This mechanism is mostly used in shaping machines, slotting machines and in rotary internal combustion engines. In this mechanism, the link AC (i.e. link 3) forming the turning pair is fixed, as shown in fig. The link 3 corresponds to the connecting rod of a reciprocating steam engine. The driving crank CB revolves with uniform angular speed about the fixed centre C. A sliding block attached to the crank pin at B slides along the slotted bar AP and thus causes AP to oscillate about the pivoted point A. A short link PR transmits the motion from AP to the ram which carries the tool and reciprocates along the line of stroke R1R2. The line of stroke of the ram (i.e. R1R2) is perpendicular to AC produced

 

In the extreme positions, AP1 and AP2 are tangential to the circle and the cutting tool is at the end of the stroke. The forward or cutting stroke occurs when the crank rotates from the position CB1 to CB2 (or through an angle β) in the clockwise direction. The return stroke occurs when the

 

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Q 2 b )

Question:

Explain with the neat sketch working of crank and slotted lever quick return mechanism.

Answer:

Crank and slotted lever quick return motion mechanism: This mechanism is mostly used in shaping machines, slotting machines and in rotary internal combustion engines. In this mechanism, the link AC (i.e. link 3) forming the turning pair is fixed, as shown in fig. The link 3 corresponds to the connecting rod of a reciprocating steam engine. The driving crank CB revolves with uniform angular speed about the fixed centre C. A sliding block attached to the crank pin at B slides along the slotted bar AP and thus causes AP to oscillate about the pivoted point A. A short link PR transmits the motion from AP to the ram which carries the tool and reciprocates along the line of stroke R1R2. The line of stroke of the ram (i.e. R1R2) is perpendicular to AC produced

In the extreme positions, AP1 and AP2 are tangential to the circle and the cutting tool is at the end of the stroke. The forward or cutting stroke occurs when the crank rotates from the position CB1 to CB2 (or through an angle β) in the clockwise direction. The return stroke occurs when the crank rotates from the position CB2 to CB1 (or through angle α) in the clockwise direction. Since the crank has uniform angular speed,

 

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Q 2 c )

Question:

Explain the inter-relation between linear and angular velocity, linear and angular acceleration with suitable example.

Answer:

 

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Q 2 c )

Question:

Define linear velocity, angular velocity, absolute velocity and state the relation between linear velocity and angular velocity

Answer:

Relation between linear and angular velocity: V = ω.r

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Q 2 d )

Question:

Explain the Klein’s construction to determine velocity and acceleration of single slider crank mechanism

Answer:

 

If ωAO is the angular velocity of the crank, then Linear velocity’s of the links is given byVAO = ωAO x AO, VAP = ωAO x AM, VPO = ωAO x MO Acceleration of the links is given bya r AO = ω 2 AO x AO, a r AP = ω 2 AO x AC, a t AP = ω 2 AO x CN, aPO = ω 2 AO x NO

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Q 2 d )

Question:

Explain the Klein’s construction to determine velocity and acceleration of a link in an I.C. engine mechanism.

Answer:

Let OC be the crank and PC the connecting rod of a reciprocating steam engine, as shown in Fig. Let the crank makes an angle θ with the line of stroke PO and rotates with uniform angular velocity ω rad/s in a clockwise direction. The Klien’s velocity and acceleration diagrams are drawn as discussed below:

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Q 2 e )

Question:

Draw the labelled displacement, velocity and acceleration diagrams for a follower when it moves with uniform velocity.

Answer:

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Q 2 e )

Question:

Draw the labelled displacement, velocity and acceleration diagrams for a follower when it moves with simple harmonic motion.

Answer:

Roller follower is preferred over knife edge follower  Knife-edge of the follower will cause the wear of the cam.  Higher load on the small contact area the follower likely to cause wear at the tip of Knifeedge due to more stresses.  Knife-edge follower practically not feasible for higher torque / load applications.  More friction due to sliding motion of the knife-edge follower and hence, more maintenance.  Roller follower on the other hand produces smooth operation with less wear and tear of both cam and follower.  Pure rotational motion of roller follower causes less friction and less loss of power.  Considerable side thrust exists between knife-edge follower and the guide.

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Q 2 f )

Question:

A flat belt drive is required to transmit 35 kW from a pulley of 1.5 m effective diameter running at speed of 300 rpm. The angle of contact is spread over 11/24 of the circumference co-efficient of friction for the surface is 0.3. Determine the maximum tension in the belt. 

Answer:

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Q 2 f )

Question:

A pulley rotating at 50 m/s transmits 40 kW. The safe pull in belt is 400 N/cm width of belt. The angle of lap is 170º. If coefficient of friction is 0.24, find required width of belt.

Answer:

Data: Initial tension, To = 2000 N, coefficient of friction, µ = 0.3, Angle of lap, θ = 1500 = 1500 x П / 180 = 2.618 rad, Smaller pulley radius, R = 200 mm, hence, D = 400 mm, Speed of smaller pulley, N = 500 r.p.m. We know that the velocity of the belt, v = П = П = 10.47 m/sec (01 mark) Let T1 = Tension in the belt on the tight side, N Let T2 = Tension in the belt on the slack side, N We know that, T0 = Hence, 2000 = (T1 + T2) / 2 Thus, (T1 + T2) = 4000 N ....................... (1) We also know that, = therefore, = or = 2.2 ............. (2) From equations 1 and 2, T1 = 2750 N and T2 = 1250 N (02 marks) Power transmitted by belt, P = [T1 - T2] v = [2750 - 1250] 10.47 = 15700 watts = 15.7 kW

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Q 3 a )

Question:

In a four bar chain ABCD, AD is fixed and is 150 mm long. The crank AB is 40 mm long and rotates at 120 r.p.m. clockwise, while the link CD = 80 mm oscillates about D. BC and AB are of equal length. Find the angular velocity of link CD when angle BAD = 60.

Answer:

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Q 3 b )

Question:

In a slider crank mechanism, the length of crank OB and connecting rod AB are 125 mm and 500 mm respectively. The centre of gravity G of the connecting rod is 275 mm from the slider. The crank speed is 600 rpm clockwise. When the crank has turned 45 from the inner dead centre position, determine :

(i) Velocity of slider ‘A’,

(ii) Velocity of the point ‘G’ graphically.

Answer:

 

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Q 3 c )

Question:

Explain slip and creep phenomenon in belts.

Answer:

Define slip and creep in the belt drive Slip --- Slip is defined as insufficient frictional grip between pulley (driver/driven) and belt. Slip is the difference between the linear velocities of pulley (driver/driven) and belt. Creep ----- Uneven extensions and contractions of the belt when it passes from tight side to slack side. There is relative motion between belt and pulley surface, this phenomenon is called creep of belt.

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Q 3 d )

Question:

Draw the neat sketch of diaphragm clutch and explain its working.

Answer:

 

 

Diaphragm Spring Type Single Plate Clutch A diaphragm spring type clutch is shown in fig. where shows the clutch in the engaged position and in the disengaged position. It is seen from the above figures that the diaphragm spring is supported on a fulcrum retaining ring so that any section through the spring can be regarded as a simple lever. The pressure plate E is movable axially, but it is fixed radically with respect to the cover. This is done by providing a series of equally spaced lugs cast upon the back surface of the pressure plate. The drive from the engine flywheel is transmitted through the cover, pressure plate and the friction plate to the gear box input shaft. The clutch is disengaged by pressing the clutch pedal which actuates the release fingers by means of a release ring. This pivots the spring about its fulcrum, relieving the spring load on the outside diameter, thereby disconnecting the drive.

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Q 3 e )

Question:

Write the procedure for balancing of a single rotating mass by single masses rotating in the same plane.

Answer:

Procedure :Balancing of a Single Rotating Mass By a Single Mass Rotating in the Same Plane Consider a disturbing mass m1 attached to a shaft rotating at ω rad/s as shown in Fig. Let r1 be the radius of rotation of the mass m1 (i.e. distance between the axis of rotation of the shaft and the centre of gravity of the mass m1). We know that the centrifugal force exerted by the mass m1 on the shaft, FCl= m1.ω2 . r1 . . . (i) This centrifugal force acts radially outwards and thus produces bending moment on the shaft. In order to counteract the effect of this force, a balancing mass (m2) may be attached in the same plane of rotation as that of disturbing mass (m1) such that the centrifugal forces due to the two masses are equal and opposite.

 

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Q 3 f )

Question:

Give detailed classification of followers.

Answer:

Types of followers The followers may be classified as discussed below:

1. According to the surface in contact. (a)Knife edge follower. When the contacting end of the follower has a sharp knife edge, it is called a knife edge follower. (b) Roller follower. When the contacting end of the follower is a roller, it is called a roller follower. (c) Flat faced or mushroom follower. When the contacting end of the follower is a perfectly flat face, it is called a flat faced follower and when the flat faced follower is circular, it is then called a mushroom follower. (d) Spherical faced follower. When the contacting end of the follower is of spherical shape, it is called a spherical faced follower.

2.According to the motion of the follower. (a) Reciprocating or translating follower. When the follower reciprocates in guides as the cam rotates uniformly, it is known as reciprocating or translating follower. (b) Oscillating or rotating follower. When the uniform rotary motion of the cam is converted into predetermined oscillatory motion of the follower, it is called oscillating or rotating follower.

3. According to the path of motion of the follower. (a) Radial follower. When the motion of the follower is along an axis passing through the centre of the cam, it is known as radial follower (b) Off-set follower. When the motion of the follower is along an axis away from the axis of the cam centre, it is called off-set follower.

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Q 4 a )

Question:

State advantages and disadvantages of chain drive over belt drive

Answer:

Advantages of chain drive over belt drive (Any four)

a) No slip takes place in chain drive as in belt drive there is slip.

b) Occupy less space as compare to belt drive.

c) High transmission efficiency.

d) More power transmission than belts drive.

e) Operated at adverse temperature and atmospheric conditions.

f) Higher velocity ratio.

g) Used for both long as well as short distances.

Disadvantages of chain drive: 1. Manufacturing cost of chains is relatively high

2. The chain drive needs accurate mounting and careful maintenance

3. High velocity fluctuations especially when unduly stretched

4. Chain operations are noisy as compared to belts

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Q 4 b )

Question:

Justify that slider crank mechanism is a modification of the basic four bar mechanism with neat sketch.

Answer:

Justification for a single slider crank mechanism is a modification of four bar chain mechanism is as given below. 1) Single slider mechanism has four kinematic links – crank, connecting rod, frame and slider and four bar mechanism has crank, coupler, frame and a follower. 2) A follower in four bar mechanism is replaced by a slider. 3) A four bar mechanism has 4 turning pairs and single slider crank mechanism has also four pairs, but one of the turning pairs is replaced by a sliding pairs. 4) A four bar mechanism rotary motion of the crank into oscillating motion of the follower whereas in single slider motion is converted in sliding motion of the piston

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Q 4 c )

Question:

Compare flywheel and governor.

Answer:

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Q 4 d )

Question:

Explain with neat sketch construction and working of eddy current dynamometer.

Answer:

Eddy Current Dynamometer : It consists of a stator on which are fitted a number of electromagnets and a rotor disc made of copper or steel and coupled to the output shaft of the engine. When the rotor rotates, eddy currents are produced in the stator due to magnetic flux set up by the passage of field current in the electromagnets. These eddy currents oppose the motion of the rotor thus loading the engine. The eddy currents are dissipated in producing heat so that this type of dynamometer also requires some cooling arrangements. The torque is measured similar to absorption dynamometers i.e. with the help of moment arm. The load is controlled by regulating the current in the electromagnets.

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Q 4 e )

Question:

A flat foot step bearing 225 mm in diameter supports a load of 7500 N. If the co-efficient of friction is 0.09 and the shaft rotates at 600 rpm, calculate the power lost in friction.

Answer:

Problem on Foot step bearing D = 225 mm = 0.225 m W = 7500 N µ = 0.09 N = 600 rpm ω = 2 π N / 60 =62.83 rad/sec Uniform pressure condition Frictional torque T = 2/3 µ W R = 50.625 Nm Power lost in friction = T x ω = 50.625 x 62.83 = 3180.8 W --------Ans Uniform wear condition Frictional torque T = 1/2 µ W R = 37.98 Nm Power lost in friction = T x ω = 37.98 x 62.83 = 2385.57 W -------- Ans

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Q 4 f )

Question:

Four masses attached to a shaft and their respective radii of rotation are given as :
m 1 = 180 kg m 2 = 300 kg m 3 = 230 kg m 4 = 260 kg
r 1 = 0.2 m r 2 = 0.15 m r 3 = 0.25 m r 4 = 0.3 m
The angles between successive masses are 45, 75 and 135. Find the
position and magnitude of the balance mass required, it its radius of rotation is
0.2 m. The masses revolve in same plane.

Answer:

Given : m1 = 180 kg, m2 = 300 kg, m3 = 230 kg, m4 = 260 kg r1 = 0.2 m, r2 = 0.15 m, r3 = 0.25 m, r4 = 0.3 m ϴ1 = 45, ϴ2 = 75, ϴ = 135 The centrifugal forces are given by - m1r1 = 36, m2r2 = 45, m3r3 = 57.5, m4r4 = 78

 

 

 

From vector diagram the resultant force is at 60 to the mass m1 and is represented by ar ar = 12 kg m Therefore mb * rb = 12 kgm Balancing mass mb = 12/0.2 = 60 kg at an angle of 2400 with the direction of m1 mass

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Examination: 2016 SUMMER
Que.No Question/Problem marks Link
Q a)(ii)

Question:

Explain single cylinder 4-stroke I.C. engine using turning moment diagram.

Answer:

A turning moment diagram for a four stroke cycle internal combustion engine, we know that in a four stroke cycle internal combustion engine, there is one working stroke after a crank has turned through two revolution i.e.7200 . Since the pressure inside the engine cylinder is less than the atmospheric pressure during suction stroke therefore a negative loop is formed. During the compression stroke, the work is done on gases, therefore a higher negative loop is obtained.

During the expansion or working stroke, the fuel burns and the gases expand, therefore a positive loop is obtained. In this stroke the work done is by the gases. During exhaust stroke, the work is done on the gases, therefore negative loop is formed. It may be noted that effect of inertia forces on the piston is taken is account.

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Q 1b)(i)

Question:

State inversions of double slider crank chain. Explain Oldham's coupling with neat sketch

Answer:

Inversions of double slider crank chain:
i.Scotch Yoke mechanism.
ii.Oldham’s coupling.
iii. Elliptical trammel.

Oldham’s coupling:
An Oldham’s coupling is used for connecting two parallel shafts whose axes are at a
small distance apart. The shafts are coupled in such a way that if one shaft rotates, the other shaft
also rotates at the same speed. This inversion is obtained by fixing the link 2, as shown in Fig.
The shafts to be connected have two flanges (link 1 and link 3) rigidly fastened at their ends by
forging. The link 1 and link 3 form turning pairs with link 2. These flanges have diametrical slots
cut in their inner faces, as shown in Fig. The intermediate piece (link 4) which is a circular disc,
have two tongues (i.e. diametrical projections) T1 and T2 on each face at right angles to each
other. The tongues on the link 4 closely fit into the slots in the two flanges (link 1 and link 3).
The link 4 can slide or reciprocate in the slots in the flanges.

When the driving shaft A is rotated, the flange C (link 1) causes the intermediate piece
(link 4) to rotate at the same angle through\ which the flange has rotated, and it further rotates the
flange D (link 3) at the same angle and thus the shaft B rotates. Hence links 1, 3 and 4 have the
same angular velocity at every instant. A little consideration will show that there is a sliding
motion between the link 4 and each of the other links 1 and 3.

4

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Q 1b)(ii)

Question:

Explain: (i) Uniform pressure theory. (ii) Uniform wear theory in clutches and bearing.

Answer:

(i) Uniform pressure theory:
 When the mating component in clutch, bearing are new, then the contact between
surfaces may be good over the whole surface.
It means that the pressure over the rubbing surfaces is uniform distributed.
 This condition is not valid for old clutches, bearings because mating surfaces may
have uneven friction.
The condition assumes that intensity of pressure is same.
P = W/A =Constant; where, W= load, A= area
(ii) Uniform wear theory in clutches and bearings:

When clutch, bearing become old after being used for a given period, then all
parts of the rubbing surfaces will not move with the same velocity.
The velocity of rubbing surface increases with the distance from the axis of the
rotating element.
It means that wear may be different at different radii and rate of wear depends
upon the intensity of pressure (P) and the velocity of rubbing surfaces (V).
 It is assumed that the rate of wear is proportional to the product of intensity of
pressure and velocity of rubbing surfaces.
This condition assumes that rate of wear is uniform;
P*r = Constant; where, P = intensity of pressure, r = radius of rotation.

4

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Q 1b)(iii)

Question:

Compare cross belt drive and open belt drive on the basis of:

(i) Velocity ratio.

(ii) Direction of driven pulley.

(iii) Length of belt drives

(iv) Application.

Answer:

 

4

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Q 2 a )

Question:

Draw a labeled sketch of quick return mechanism of shaper and explain its working

Answer:

Crank and slotted lever quick return motion mechanism. This mechanism is mostly used in shaping machines, slotting machines and in rotary internal combustion engines. In this mechanism, the link AC (i.e. link 3) forming the turning pair is fixed, as shown in fig. The link 3 corresponds to the connecting rod of a reciprocating steam engine. The driving crank CB revolves with uniform angular speed about the fixed centre C. A sliding block attached to the crank pin at B slides along the slotted bar AP and thus causes AP to oscillate about the pivoted point A. A short link PR transmits the motion from AP to the ram which carries the tool and reciprocates along the line of stroke R1R2. The line of stroke of the ram (i.e. R1R2) is perpendicular to AC produced.

 

n the extreme positions, AP1 and AP2 are tangential to the circle and the cutting tool is at the end of the stroke. The forward or cutting stroke occurs when the crank rotates from the position CB1 to CB2 (or through an angle β) in the clockwise direction. The return stroke occurs when the crank rotates from the position CB2 to CB1 (or through angle α) in the clockwise direction. Since the crank has uniform angular speed,

 

4

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Q 2 b )

Question:

What are the types of kinematic pair ? Give its examples.

Answer:

4

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Q 2 c )

Question:

Define linear velocity, angular velocity, absolute velocity and state the relation between linear velocity and angular velocity.

Answer:

Linear Velocity: It may be defined as the rate of change of linear displacement of a body with respect to the time. Since velocity is always expressed in a particular direction, therefore it is a vector quantity. Mathematically, linear velocity, v = ds/dt

Angular Velocity: It may be defined as the rate of change of angular displacement with respect to time. It is usually expressed by a Greek letter ɷ (omega). Mathematically, angular velocity, ɷ = dƟ /dt

Absolute Velocity: It is defined as the velocity of any point on a kinematic link with respect to fixed point. Relation between v and ɷ: V = r. ɷ Where V = Linear velocity. ɷ = angular velocity. r = radius of rotation.

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Q 2 d )

Question:

Explain the Klein's construction to determine velocity and acceleration of single slider crank mechanism.

Answer:

We have already discussed that the velocity diagram for given configuration is a triangle OCP as shown in Fig. If this triangle is rotated through 90°, it will be a triangle oc1 p1, in which oc1 represents VCO (i.e. velocity of C with respect to O or velocity of crank pin C) and is parallel to OC, op1 represents VPO (i.e. velocity of P with respect to O or velocity of cross-head or piston P) and is perpendicular to OP, and c1p1 represents VPC (i.e. velocity of P with respect to C) and is parallel to CP. A little consideration will show that the triangles oc1p1 and OCM are similar. Therefore,

Thus, we see that by drawing the Klein’s velocity diagram, the velocities of various points may be obtained without drawing a separate velocity diagram. Klien’s acceleration diagram: The Klien’s acceleration diagram is drawn as discussed below: 1. First of all, draw a circle with C as centre and CM as radius. 2. Draw another circle with PC as diameter. Let this circle intersect the previous circle at K and L. 3. Join KL and produce it to intersect PO at N. Let KL intersect PC at Q. This forms the quadrilateral CQNO, which is known as Klien’s acceleration diagram. We have already discussed that the acceleration diagram for the given configuration is as shown in Fig. We know that (i) o'c' represents CO ar (i.e. radial component of the acceleration of crank pin C with respect to O ) and is parallel to CO; (ii) c'x represents PC ar (i.e. radial component of the acceleration of crosshead or piston P with respect to crank pin C) and is parallel to CP or CQ; (iii) xp' represents PC at (i.e. tangential component of the acceleration of P with respect to C ) and is parallel to QN (because QN is perpendicular to CQ); and (iv) o'p' represents aPO (i.e. acceleration of P with respect to O or the acceleration of piston P) and is parallel to PO or NO. A little consideration will show that the quadrilateral o'c'x p' is similar to quadrilateral CQNO . Therefore,

4

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Q 2 e )

Question:

 Draw neat sketch of radial cam with follower and show on it  (i) Base circle. (ii) Pitch point. (iii) Prime Circle. (iv) Cam profile

Answer:

4

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Q 3 c )

Question:

Explain epicyclic gear train with neat sketch.

Answer:

In case of Epicyclic Gear train, the axis of shafts on which gears are mounted may have a relative motion between them, unlike other gear trains. This gives advantage that, very high or low velocity ratio can be obtained compared to simple and compound gear trains; in the small space. In above sketch, if gears A and B are rotating and arm RS is fixed, then it behaves like simple gear train. However, when Arm C rotates and gear A is fixed, then train becomes epicyclic. It is also known as planetary gear train. Applications- Differential gears of the automobiles, back gear of lathe, hoists, pulley blocks

4

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Q 3 d )

Question:

Draw a labelled sketch of multiplate clutch and state its applications.

Answer:

Applications- Automobiles like scooters, motorcycles, textile and paper industries, machine tools

4

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Q 3 e )

Question:

Write the procedure of balancing single rotating mass when it balance mass is rotating in the same plane as that of disturbing mass.

Answer:

4

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Q 3 f )

Question:

What are the different types of follower motion ? Also draw displacement diagram for uniform velocity.

Answer:

Different types of follower motions –

The follower during its travel may have one of the following motions:-

Uniform velocity, Simple harmonic motion, Uniform acceleration and retardation, Cycloidal motion.

Displacement Diagram of Uniform Velocity:

 

 

 

 

 

 

 

 

4

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Q 4 b )

Question:

Justify with neat sketch elliptical trammel as an inversion of double slider crank chain.

Answer:

Elliptical trammel :

Since Elliptical trammel consist of two turning pairs and two sliding pairs, it is inversion of double slider crank chain. This instrument is used for drawing ellipses. This inversion is obtained by fixing a slotted plate (link 4) as shown in fig. It has got two right angled grooves cut into it. 1-2 is turning pair                                     2-3 is turning pair                                                                           1-4 is sliding pair                                      3-4 is sliding pair.................................. [2 M]

 

 

 

 

 

 

4

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Q 4 c )

Question:

Differentiate between flywheel and governor.

Answer:

Difference between Flywheel and Governor :

4

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Q 4 d )

Question:

Explain construction and working of eddy current dynamometer.

Answer:

Construction and Working of Eddy current dynamometer :

4

view
Q 5 a )

Question:

State and explain Law of Gearing.

Answer:

Law of Gearing :

 

 

 

 

 

 

4

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Q 6a)(i)

Question:

Explain steep and creep phenomenon in belts.

Answer:

Slip of Belt:- Ans:- When driver pulley rotates firm grip between its surface and the belt. This firm grip between pulley and belt is because of friction and known as frictional grip. If this frictional grip becomes insufficient to transmit the motion of pulley to belt. Then there will be. 1) Forward motion of driver pulley without carrying belt called as slip on driving side.

2) Some forward motion of belt without carrying driven pulley this is called as slip on driver side. The difference between linear speed of rim of pulley and belt on the pulley is known as slip of belt. The velocity ratio considering slip is given by:-

Creep of Belt:- The belt moves from driving pulley is known as Tight side and belt moves from driving pulley to driver pulley as slack side.

Tension on both i.e. on tight sides and slack side is not equal ( T1> T2 ) . The belt material is elastic material which elongates more on Tight side than the slack side resulting in unequal stretching on both sides of drive. A certain portion of belt when passes from slack side to tight side extends and certain portion of belt when contracts, passes from tight side to slack side because of relative motion. The relative motion between belt and pulley surface due to unequal stretching of two sides of drives is known as creep.

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Q 6a)(ii)

Question:

Explain single cylinder 4-stroke I.C. engine using turning moment diagram.

Answer:

A turning moment diagram for a four stroke cycle internal combustion engine, we know that in a four stroke cycle internal combustion engine, there is one working stroke after a crank has turned through two revolution i.e.7200 . Since the pressure inside the engine cylinder is less than the atmospheric pressure during suction stroke therefore a negative loop is formed. During the compression stroke, the work is done on gases, therefore a higher negative loop is obtained.

During the expansion or working stroke, the fuel burns and the gases expand, therefore a positive loop is obtained. In this stroke the work done is by the gases. During exhaust stroke, the work is done on the gases, therefore negative loop is formed. It may be noted that effect of inertia forces on the piston is taken is account.

4

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Examination: 2016 WINTER
Que.No Question/Problem marks Link
Q 1b)(i)

Question:

State any four inversions of single slider crane chain. Describe any one with neat sketch.

Answer:

1.Reciprocating engine, Reciprocating compressor;

2. Whitworth quick return mechanism, Rotary engine,

3.Slotted crank mechanism, Oscillatory engine.

4.Hand pump, pendulum pump.

1.Reciprocating engine, Reciprocating compressor link 1 is fixed

2.Whitworth quick return mechanism, Rotary engine​ link 2 is fixed,

3.- Oscillatory engine. Slotted crank mechanism. link 3 is fixed

4. Hand pump, pendulum pump. link 4 is fixed

 

 

 

 

4

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Q 1b)(ii)

Question:

Compare multiplate clutch with cone clutch on the following basis.

(1) Power Transmission

(2) Size

Answer:

Comparison of multiplate clutch and Cone clutch:

4

view
Q 2 a )

Question:

Explain a scotch yoke mechanism with a neat sketch.

Answer:

4

view
Q 2 b )

Question:

What is machine ? Differentiate between a machine and a structure.

Answer:

4

view
Q 2 c )

Question:

Explain Klein’s construction to determine velocity and acceleration of different links in single slider crank mechanism.

Answer:

Klein’s construction

a) For velocity of different links

b) For acceleration of different links

4

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Q 2 d )

Question:

Define the terms:

(i) Linear velocity

(ii) Angular velocity

(iii) Absolute velocity

(iv) Relative velocity

Answer:

4

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Q 2 d )

Question:

Explain with neat sketch different types of follower.

Answer:

4

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Q 3 a )

Question:

Discuss the following motion of the follower by drawing the displacement velocity and acceleration diagram.

(i) Uniform Velocity

(ii) Simple Harmonic Motion

(iii) Uniform acceleration and retardation

Answer:

4

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Q 3 b )

Question:

The crank and connecting rod of steam engine are 0.5m and 2m long respectively. The crank makes 180r.p.m. in clockwise direction. When it has turned through 45° from I.D.C. Find the velocity of piston and angular velocity of connecting rod by relative velocity method.

Answer:

4

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Q 3 c )

Question:

Compare cross belt drive and open belt drive on the basis of -

(i) Velocity ratio

(ii) Direction of driven pulley

(iii) Application

(iv) Length of belt drive

Answer:

4

view
Q 3 d )

Question:

State the applications of :

(i) Band brake

(ii) Disc brake

(iii) Internal expanding shoe brake

(iv) External shoe brake

Answer:

4

view
Q 3 f )

Question:

Explain with neat sketch working principle of epicyclic gear train.

Answer:

4

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Q 4 a )

Question:

Generally, the lower side is kept “Tight side” and upper side is kept as “Slack side” with the belt drives having small driving pulley and big driven pulley. Why ?

Answer:

Power transmission in belt drive depends on angle of lap and frictional grip between belt and pulley. As slack side is at upper side angle of lap and grip increases.

4

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Q 4 b )

Question:

Describe with neat sketch the working of Oldham’s coupling.

Answer:

4

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Q 4 c )

Question:

Distinguish between flywheel and governor.

Answer:

4

view
Q 4 d )

Question:

Discuss the working of Rope brake dynamometer with the help of a neat sketch.

Answer:

4

view
Q 4 e )

Question:

Explain the working of internal expanding shoe brake with the help of neat sketch.

Answer:

4

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Q 4 f )

Question:

Explain the process of balancing of single rotating mass by a single mass rotating in the same plane.

Answer:

m = Mass attached to shafts,

r = Distance of CG from axis of rotation.

Consider mass ‘m’ is attached to rotating shaft at a radius are then the centrifugal force exerted by mass ‘M’ on the shaft is

Fc = Mw2R Where,

W = Angular velocity of shaft

R = Distance of CG from axis of rotation

M = Mass attached to shaft.

Due to continuous rotation of shaft the centrifugal force developed will be continuously changing its direction. It will cause bending moment on shaft. To counter act the effect of centrifugal force the balance weight may be introduced in same plane of rotation. This balance weight should be attached it will result in exactly equal but opposite centrifugal force to that of disturbing weight ‘M’.

The balanced centrifugal force is given by Fb = mbw2Rb For balancing the shaft – Mw2R = mbw2Rb.

 

 

 

 

 

 

 

4

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Q 6a)(i)

Question:

Define the following terms as applied to cam with neat sketch.

(1) Pitch circle

(2) Pressure angle

(3) Stroke of follower

(4) Module

Answer:

(1) Pitch circle- Circle drawn from centre of cam through pitch points.

(2) Pressure angle- Angle between direction of follower motion and normal to pitch curve.

(3) Stroke- Maximum travel of follower from its lowest position to top most position.

(4) Module –(Gears) – Ratio of pitch circle diameter in mm to No. of teeth on gear.

4

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Q 6a)(ii)

Question:

Differentiate between disc brake and internally expanding brake.

Answer:

4

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Examination: 2015 SUMMER
Que.No Question/Problem marks Link
Q 1b)(a)

Question:

Define completely constrained motion and successfully constrained motion with neat sketch. State one example of each.

Answer:

4

view
Q 1b)(b)

Question:

Explain working principle of clutch. State its location in transmission system of an automobile.

Answer:

A friction clutch has its principal application in the transmission of power of shafts and machines, which must be started and stopped frequently. The force of friction is used to start the driven shaft from rest and gradually brings it up to the proper speed without excessive slipping of the friction surfaces. In automobiles, friction clutch is used to connect the engine to the driven shaft. In operating such a clutch, care should be taken so that the friction surfaces engage easily and gradually brings the driven shaft up to proper speed.

                                                                                   Location: Between the engine and gear box.

 

4

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Q 1b)(c)

Question:

Compare cross belt drive and open belt drive on the basis of

(i) velocity ratio

(ii) application

(iii) direction of driven pulley

(iv) length of belt drive

Answer:

             Comparison between cross belt drive and open belt drive :

 

4

view
Q 2 a )

Question:

Differentiate machine and structure on any four points.

Answer:

4

view
Q 2 b )

Question:

Explain with neat sketch working principle of Oldham’s coupling.

Answer:

When the driving shaft A is rotated, the flange C (link 1) causes the intermediate piece (link 4) to rotate at the same angle through which the flange has rotated, and it further rotates the flange D (link 3) at the same angle and thus the shaft B rotates. Hence links 1, 3 and 4 have the same angular velocity at every instant. A little consideration will show that there is a sliding motion between the link 4 and each of the other links 1 and 3.

 

4

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Q 2 c )

Question:

Define linear velocity, angular velocity, absolute velocity and state the relation between linear velocity and angular velocity.

Answer:

4

view
Q 2 d )

Question:

Describe stepwise procedure for determination of velocity and acceleration by Klein’s construction with suitable data.

Answer:

Steps in Klein’s construction :

 Klein’s construction is a simpler construction to get velocity and acceleration diagrams. For example : for reciprocating engine mechanism OPC. draw a circle with PC as diameter as shown. and obtain velocity diagram OCM ie. produce PC to cut perpendicular to line of stroke in ‘M’ . Draw another circle with ‘C’ as center and “CM” as radius cutting the first circle in points K and L. Join “KL” which is the chord common to both the circles. Let it cuts PC and OP in “Q” and “N” respectively. Then “OCQN” is the required quadrilateral which is similar to acceleration diagram.

4

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Q 2 e )

Question:

Draw a neat sketch of radial cam with roller follower and show the following on it :

(i) Pitch point                      (ii) Pressure angle

(iii) Prime circle                 (iv) Trace point

Answer:

4

view
Q 3 a )

Question:

Draw a neat labelled sketch of “Multiplate Clutch”.

Answer:

4

view
Q 3 b )

Question:

Why roller follower is preferred over a knife follower ? State two advantages and application of roller follower.

Answer:

In case of knife edge follower there is sliding motion between the contacting surface of cam and follower. Because of small contact area, there is excessive wear; therefore it is not frequently used. Whereas in roller follower there is rolling motion between contacting surfacing and more contact area, therefore rate of wear is greatly reduced.

Advantages: i) Less wear, more life ii) Less side thrust as compared to knife edge follower.

Application: Used in stationary oil and gas engines

4

view
Q 3 c )

Question:

Write the procedure for balancing of a single rotating mass by single masses rotating in the same plane.

Answer:

4

view
Q 3 d )

Question:

State the type of power transmission chains. Describe any one with its sketch.

Answer:

Types of power transmission chains :

4

view
Q 4 a )

Question:

Explain the phenomenon of slip and creep in a belt drive. State its effect on velocity ratio.

Answer:

Slip of the belt:  A firm frictional grip between belt and shaft is essential. But sometimes it becomes insufficient. This may cause some forward motion of the belt without carrying the driven pulley with it. This called as slip of the belt. It is expressed as a percentage.

Effect on velocity ratio: Result of belt slipping is to reduce the velocity ratio of the system.

Creep in belt drive : When the belt passes from slack side to tight side, a certain portion of the belt extends and it contracts again when the belt passes from tight sight to slack side. Due to these changes in length, there is a relative motion between the belt and the pulley surfaces. This relative motion is called as creep.

Effect on velocity ratio: The total effect of creep is to reduce slightly the speed of the driven pulley or follower.

4

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Q 4 b )

Question:

Explain with the diagram working of crank and slotted lever quick return mechanism.

Answer:

Crank and slotted lever quick return motion mechanism:

 This mechanism is mostly used in shaping machines, slotting machines and in rotary internal combustion engines. In this mechanism, the link AC (i.e. link 3) forming the turning pair is fixed, as shown in fig. The link 3 corresponds to the connecting rod of a reciprocating steam engine. The driving crank CB revolves with uniform angular speed about the fixed centre C. A sliding block attached to the crank pin at B slides along the slotted bar AP and thus causes AP to oscillate about the pivoted point A. A short link PR transmits the motion from AP to the ram which carries the tool and reciprocates along the line of stroke R1R2. The line of stroke of the ram (i.e. R1R2) is perpendicular to AC produced.

 

4

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Q 4 c )

Question:

Explain with sketch working of hartnell governor.

Answer:

4

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Q 4 d )

Question:

Explain working of hydraulic brake dynamometer with sketch.

Answer:

Hydraulic dynamometer is also called as water brake absorber. Invented by British engineer William Froude in 1877 in response to a request by the Admiralty to produce a machine capable of absorbing and measuring the power of large naval engines, water brake absorbers are relatively common today. The schematic shows the most common type of water brake, known as the "variable level" type. Water is added until the engine is held at a steady RPM against the load, with the water then kept at that level and replaced by constant draining and refilling (which is needed to carry away the heat created by absorbing the horsepower). The housing attempts to rotate in response to the torque produced, but are restrained by the scale or torque metering cell that measures the torque.

4

view
Q 6a)(ii)

Question:

Explain the concept of fluctuation of energy related with turning moment diagram with sketch.

Answer:

4

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Examination: 2015 WINTER
Que.No Question/Problem marks Link
Q 1b)(iii)

Question:

Draw the neat sketch of epicyclic gear train and explain how it works.

Answer:

In an epicyclic gear train, the axes of the shafts, over which the gears are mounted, may move relative to a fixed axis. A simple epicyclic gear train is shown in Fig. where a gear A and the arm C have a common axis at 1 about which they can rotate. The gear B meshes with gear A and has its axis on the arm at O2, about which the gear B can rotate. If the arm is fixed, the gear train is simple and gear A can drive gear B or viceversa, but if gear A is fixed and the arm is rotated about the axis of gear A (i.e. O1), then the gear B is forced to rotate upon and around gear A. Such a motion is called epicyclic and the gear trains arranged in such a manner that one or more of their members moves upon and around another member are known as epicyclic gear trains (epi - means upon and cyclic means around). The epicyclic gear trains may be simple or compound.

The epicyclic gear trains are useful for transmitting high velocity ratios with gears of moderate size in a comparatively lesser space. The epicyclic gear trains are used in the back gear of lathe, differential gears of the automobiles, hoists, pulley blocks, wrist watches etc.

 

4

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Q 2 a )

Question:

State and explain various types of constrained motions with suitable examples.

Answer:

Types of Constrained Motions :

Following are the three types of constrained motions:

1. Completely constrained motion: When the motion between a pair is limited to a definite direction irrespective of the direction of force applied, then the motion is said to be a completely constrained motion. For example, the piston and cylinder (in a steam engine) form a pair and the motion of the piston is limited to a definite direction (i.e. it will only reciprocate) relative to the cylinder irrespective of the direction of motion of the crank.

2. Incompletely constrained motion: When the motion between a pair can take place in more than one direction, then the motion is called an incompletely constrained motion. The change in the direction of impressed force may alter the direction of relative motion between the pair. A circular bar or shaft in a circular hole is an example of an incompletely constrained motion as it may either rotate or slide in a hole. These both motions have no relationship with the other.

3. Successfully constrained motion: When the motion between the elements, forming a pair, is such that the constrained motion is not completed by itself, but by some other means, then the motion is said to be successfully constrained motion. Consider a shaft in a foot-step bearing. The shaft may rotate in a bearing or it may move upwards. This is a case of incompletely constrained motion. But if the load is placed on the shaft to prevent axial upward movement of the shaft, then the motion of the pair is said to be successfully constrained motion.

4

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Q 2 b )

Question:

Draw the neat labeled sketch of Oldham’s coupling. State its applications.

Answer:

Applications:

An Oldham's coupling is used for connecting two parallel shafts whose axes are at a small distance apart.

Used to transmit motion and power.

4

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Q 2 c )

Question:

Define the terms linear velocity, relative velocity, angular velocity and angular acceleration.

Answer:

4

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Q 2 d )

Question:

For a single slider crank mechanism , state the formulae to calculate by analytical method – Also state the meaning of each term.

Answer:

i) Velocity of slider:

                                Vp = w.r [sinθ + sin2θ/2n ]

              where,

                  Vp - velocity of slider

w- angular velocity

θ – angle of crank to line of stroke ‘PO’

n- l/r = ratio of length of connecting rod to crank radius.

ii) Acceleration of slider:

               fp = w2 r(cos θ + cos2θ/n)

                 where, fp – acceleration of slider

iii) Angular velocity of connecting rod.:

            wpc = w cos θ / ( n2 – sin2θ)1/2

Where, wpc is angular velocity of connecting rod

iv) Angular acceleration of connecting rod.:

           αpc =  -w2sinθ (n2 -1)/ ( n2 -sin2θ) 3/2

Where, αpc is angular acceleration of connecting rod.

 

 

 

4

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Q 2 e )

Question:

Define the following terms related to cams.

Answer:

i) Trace point : It is a reference point on the follower and is used to generate the pitch curve. In case of knife edge follower, the knife edge represents the trace point and the pitch curve corresponds to the cam profile. In a roller follower, the centre of the roller represents the trace point.

ii) Pitch curve: It is the curve generated by the trace point as the follower moves relative to the cam. For a knife edge follower, the pitch curve and the cam profile are same whereas for a roller follower, they are separated by the radius of the roller.

iii) Prime circle: It is the smallest circle that can be drawn from the centre of the cam and tangent to the pitch curve. For a knife edge and a flat face follower, the prime circle and the base circle are identical. For a roller follower, the prime circle is larger than the base circle by the radius of the roller.

iv) Lift of stroke: It is the maximum travel of the follower from its lowest position to the topmost position.

4

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Q 3 a )

Question:

Space diagram 01 Mark, Velocity Diagram 02 marks , Calculations 01 Mark

Note In QP length BC & AB are equal. Read length AD = length BC = 150 mm

Answer:

4

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Q 3 b )

Question:

In a single slider crank mechanism, crank AB = 20 mm and connecting rod BC = 80 mm. Crank AB rotates with uniform speed of 1000 rpm in anticlockwise direction. Find

(i) angular velocity of connecting rod BC and

(ii) Velocity of slider C when crank AB makes angle of 60° with the horizontal.​

Answer:

Given: Crank AB = 20 mm = 0.02 m, C. R. BC = 80 mm = 0.08 m

N = 1000 rpm, ωBA= 2πN/60 = 2π x 1000/60 = 104.7 rad/sec

VBA = ωBA x AB = 104.7 x 0.02 = 2.09 m/s

From velocty diagram: Velocity of C w.r.t. B -

VCB = vector cb = 1.15 m/s

Angular velocity of Connecting rod ‘BC’ ωCB = VCB / CB = 1.15/0.08 = 14.375 rad /sec

Velocity of slider ‘C’

VC= vector ac = 2 m/sec

4

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Q 3 c )

Question:

State the formulae to calculate the length of open belt drive and cross belt drive. State the meaning of each term by drawing suitable diagrams in both cases.

Answer:

Formulae to calculate the length of open belt drive :

4

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Q 3 d )

Question:

Draw the neat sketch of single plate clutch and explain its working.

Answer:

4

view
Q 3 e )

Question:

State the procedure of balancing single rotating mass when its balancing mass is rotating in the same plane as that of disturbing mass.

Answer:

4

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Q 3 f )

Question:

Give detailed classification of followers.

Answer:

4

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Q 4 a )

Question:

What is centrifugal tension ? State its formula. Explain its effect on power transmitted by a belt drive.

Answer:

4

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Q 4 b )

Question:

State the meaning of sliding pair, turning pair, rolling pair and spherical pair with one example each.

Answer:

4

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Q 4 c )

Question:

Draw turning moment diagram for single cylinder four stroke I.C. Engine. Label all parts.

Answer:

4

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Q 4 d )

Question:

Explain the working of rope brake dynamometer with neat sketch.

Answer:

4

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Q 4 e )

Question:

A vertical shaft 150 mm in diameter and rotating at 100 rpm rests on a flat end footstep bearing. The shaft carries vertical load of 20 kN. Assuming uniform pressure distribution and coefficient of friction equal to 0.05, estimate power lost in friction​

Answer:

4

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Q 4 f )

Question:

Four masses m1, m2, m3 and m4 are 200 kg, 300 kg, 240 kg, and 260 kg respectively. The corresponding radii of rotation are 0.2 m, 0.15 m, 0.25 m and 0.3 m respectively and the angles between successive masses are 45°, 75° and 135°. Find the position and magnitude of balance mass required, if its radius of rotation is 0.2 m.

Answer:

4

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Q 6a)(i)

Question:

State and explain law of gearing with the help of suitable sketch.

Answer:

             

4

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Q 6a)(ii)

Question:

Compare flywheel and governor.

Answer:

4

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