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Question and answers

Examination: 2017 SUMMER
Que.No Marks
Q 1 i )

2

Question:

Write down the formula of length of belt for open belt drive and cross belt drive.


Answer:

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Q 1 i )

2

Question:

List the methods to reduce the slip in belt and pulley.


Answer:

Methods to reduce the slip in belt and pulley:

1. Vertical belt drive should be avoided.

2. In horizontal belt drive the upper side should be kept as loose side.

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Q 1 k )

2

Question:

Define law of gearing.


Answer:

Law of Gearing: The law of gearing states that the angular velocity ratio of all gears of a meshed gear system must remain constant also the common normal at the point of contact must pass through the pitch point.

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Q 2 d )

4

Question:

Explain condition for maximum power transmission.


Answer:

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Q 2 e )

4

Question:

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


Answer:

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Q 3 d )

4

Question:

Find the width of the belt, necessary to transmit 7.5 kW to a pulley 300 mm diameter, if the pulley makes 1600 rpm and the co-efficient of friction between the belt and pulley is 0.3. Assume the angle of contact as 180o and the maximum tension in the belt is not to exceed 8 N/mm width.


Answer:

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Q 4 c )

4

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 5 c )

8

Question:

A leather belt is required to transmit 7.5 kW from a pulley 1.2 m in diameter running at 250 rpm. The angle of contact is 165o and the co-efficient of friction between the belt and the pulley is 0.35. If the safe working stress for the leather belt is 2 MPa, density of leather is 1050 kg/m3 and the thickness of belt is 10 mm, determine the width of belt, taking centrifugal tension into account.


Answer:

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Examination: 2017 WINTER
Que.No Marks
Q 1a)(c)

2

Question:

State law of gearing.


Answer:

Gearing law (Law of gearing) :

Gearing law states that, "The law of gearing states that the angular velocity ratio of all Gears of a meshed gear system must remain constant also the common normal at the point of contact must pass through the pitch point."

gearing law

Gearing law illustration

As illustrated in above animation the common normal at the point of contact passes through the pitch point. Gearing law must be followed in order to two gears transmit motion form one to another.

In order to have a constant angular velocity ratio for all positions of the wheels, it is must that the point P must be the fixed point (called pitch point) for the two wheels. In other words it can be said that , the common normal at the point of contact between a pair of teeth should always pass through the pitch point for  proper working.

This is the fundamental condition which must be satisfied while designing the profiles for the teeth of gear wheels, it is also known as the law of gearing.

Gearing law explination with diagram

 

 

 

 

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Q 1a)(d)

2

Question:

State the types of chains & sprockets.


Answer:

Types of Chains & Sprockets: The chains, on the basis of their use, are classified into the following three groups : 1. Hoisting and hauling (or crane) chains, 2. Conveyor (or tractive) chains, and 3. Power transmitting (or driving) chains. Sprockets: 1. Taper lock sprockets 2.Pilot bore sprocket 3.Platewheel sprocket

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Q 2 f )

4

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 3 c )

4

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 4 a )

4

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 5 c )

8

Question:

Two pulley, one 450 mm diameter and the other 200 mm diameter are on parallel shafts 1.95 m apart and are connected by a crossed belt. Find the length of the belt required and the angle of contact between the belt and each pulley. What power can be transmitted by the belt when the larger pulley rotates at 200 rpm, if the maximum permissible tension in the belt is 1 kN and the co-efficient of friction between the belt and pulley is 0.25 ?


Answer:

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Q 6a)(i)

8

Question:

State types of gear train and explain any one.


Answer:

i) Types of gear trains 1) Simple gear train 2) Compound gear train 2) Epicyclic gear train 4) Inverted gear train Simple gear train. When there is only one gear on each shaft, it is known as simple gear train. The gears are represented by their pitch circles. When the distance between the two shafts is small, the two gears are made to mesh with each other to transmit motion from one shaft to the other Epicyclic gear train: A simple epicyclic gear train is shown in Fig. where a gear A and the arm C have a common axis at O1about 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 vice- versa, 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 move upon and around another member are known as epicyclic gear trains.

Compound Gear Train When there are more than one gear on a shaft, it is called a compound train of gear. Whenever the distance between the driver and the driven or follower has to be bridged over by intermediate gears and at the same time a great (or much less) speed ratio is required, then the advantage of intermediate gears is intensified by providing compound gears on intermediate shafts

 

Reverted Gear Train When the axes of the first gear (i.e. first driver) and the last gear (i.e. last driven or follower) are co-axial, then the gear train is known as reverted gear train. We see that gear 1 (i.e. first driver) drives the gear 2 (i.e. first driven or follower) in the opposite direction.

 

 

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Examination: 2016 SUMMER
Que.No Marks
Q 1a)(iii)

2

Question:

How are drives classified?


Answer:

Classification of drives:
(i) Belt drives.
(ii) Chain drives.
(iii) Rope.
(iv) Gear drives.

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Q 1a)(v)

2

Question:

Write any two disadvantages of chain drive.


Answer:

Disadvantages of chain drives:
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 1b)(iii)

4

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:

 

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Q 2 f )

4

Question:

A shaft runs at 80 rpm & drives another shaft at 150 rpm through belt drive. The diameter of the driving pulley is 600 mm. Determine the diameter of the driven pulley in the following cases: (i) Taking belt thickness as 5 mm. (ii) Assuming for belt thickness 5 mm and total slip of 4%.


Answer:

Ans.: Given data; N1 = 80 rpm. N2 =150 rpm. D1= 600 mm. S = 4 % To find; D2 =?;

(i) Case I: Taking t = 5 mm. Velocity ratio, (V.R.) N2/N1 = (D1 + t)/ (D2 + t)

150/80 = (600 + 5)/ (D2 + 5)

Therefore, diameter of driven pulley D2 = 317.66 mm ~ 318mm

(ii) Case II: Assuming for belt thickness 5 mm and total slip of 4%. Velocity ratio, (V.R.) N2/N1 = {(D1 + t)/ (D2 + t)} × {1- (S/100)}

 150/80 = {(600 + 5) / (D2 + 5)} × {1- (4/100)

Therefore, diameter of driven pulley D2 = 304.76 mm ~ 305 mm

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Q 3 c )

4

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

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Q 5 a )

4

Question:

State and explain Law of Gearing.


Answer:

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 for all positions of mating gear. This law forms the basis for the gear profile design. This is a must condition for the two gears to perform properly.

Law of gearing Proof

Consider the portions of two gear teeth in mesh.o1 and o2 are centre points,

Let K= point of contact

TT = COmmon tangent at the point of contact K

N'N' = common tangent at the pont of contact K

O1m and O2N are perpendicular to common normal N'N'.

 

law of gearinglaw of gearing

Law of gearing animation

As shown in the diagram below the common normat at the point of contact between a pair of teeth must always pass through the pitch point for all positions of mating gears.

This is the fundamental condition which must be satisfied while designing the profiles of teeth for gears.

 

This law is must for a gearing pair to perform properly. The animation clearly demonstrates the blue line which traces the path of the point of contact.

 

Link to other topics of Theory of machines is given below.

 

 

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Q 5 c )

8

Question:

Two parallel shafts, connected by a crossed belt, are provided with pulleys 480 mm and 640 mm in diameters. The distance between the centre lines of the shafts is 3 m. Find by how much the length of the belt should be changed if it is desired to alter the direction of rotation of the driven shaft.​


Answer:

                   

 

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Q 6a)(i)

4

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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Examination: 2016 WINTER
Que.No Marks
Q 1a)(iv)

2

Question:

Define slip and creep in case of belt drive.


Answer:

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 1a)(v)

2

Question:

Give four advantages of chain drive over belt drive.


Answer:

Advantages of chain drive over belt drive 

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

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Q 1a)(vi)

2

Question:

State the effect of centrifugal tension on power transmission.


Answer:

Effect of centrifugal tension on power transmission:

As the belt passes over the pulley with high velocity, centrifugal force is produced on the belt, which tends to act on the belt. This force tries to move the belt away from the pulley.

This force is given by,

TC = m x V2

There is no effect of centrifugal tension on power transmitted.

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Q 1b)(iii)

4

Question:

The central distance two shaft is 4m having two pulleys with diameter having 500mm and 700mm respectively find the length of belt required - (1) for open belt drive (2) for cross belt drive


Answer:

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Q 2 f )

4

Question:

A pulley is driven by the flat belt running at speed of 600m/min. and transmit 4 kW. The coefficient of friction between belt and pulley is 0.3 and angle of lap is 160°. Find maximum tension in the belt.


Answer:

Flat belt speed = V = 600 m/min = 600/60 m/sec = 10 m/sec;

Power transmitted = P = 4 kW ;

Coefficient of friction =µ = 0.3;

Angle of lap = θ =1600

Belt tension ratio = T1/ T2 = eµθ = e 0.3(160x π/180) = 2.31; T1/ T2 = 2.31;

T1= T2 x 2.311--------------------------------(1)

P = ( T1 - T2) x V ; --------------------------------(2)

P = ( T2 x 2.31- T2)x 10; Putting value of power

P = 4 kW 4 x1000 = ( T2 x 2.31 - T2)x 10;

T2 = 305.34 N

T1 = 705.34N

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Q 3 c )

4

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:

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Q 3 f )

4

Question:

Explain with neat sketch working principle of epicyclic gear train.


Answer:

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Q 4 a )

4

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.

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Examination: 2015 SUMMER
Que.No Marks
Q 1a)(c)

2

Question:

Define angle of lap and slip in belt drive.


Answer:

Slip of belt : The motion of belts and shafts assuming a firm frictional grip between the belts and the shafts. But sometimes, the frictional grip becomes insufficient. This may cause some forward motion of the driver without carrying the belt with it. This may also cause some forward motion of the belt without carrying the driven pulley with it. This is called slip of the belt and is generally expressed as a percentage.

Angle of Lap : The angle of lap is defined as the angle subtended by the portion of the belt which is in contact at the pulley surface of the pulley.

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Q 1a)(d)

2

Question:

State four conditions under which the ‘V’ belt drive is selected.


Answer:

Conditions for ‘V’ Belt drive selection :

 1. Great amount of Power to be transmitted,

2. Requirement of the high velocity ratio (maximum 10).

3. Small Centre distance between the shafts

4. Positive drive requirement

5. Compact Space

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Q 1b)(c)

4

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 :

 

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Q 2 b )

4

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.

 

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Q 2 f )

4

Question:

The central distance between two shaft is 4 m having two pulleys with diameter having 500 mm and 700 mm respectively. Find length of belt required (i) for open belt drive (ii) for cross belt drive


Answer:

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Q 3 d )

4

Question:

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


Answer:

Types of power transmission chains :

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Q 4 a )

4

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.

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Q 5 c )

8

Question:

A belt is required to transmit 10 kW from a motor running at 600 rpm. The belt is 12 mm thick and has a mass density 0.001 gm/mm3 . Safe stress in the belt is not to exceed 2.5 N/mm2 , diameter of the driving pulley is 250 mm whereas the speed of the driven pulley is 200 rpm. The two shafts are 1.25 m apart. The coefficient of friction is 0.25, determine (1) Angle of contact at driving pulley (2) The width of the belt


Answer:

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Q 6a)(i)

4

Question:

(i) Define ‘Gear Train’. State its purpose and types of gear train.  


Answer:

Definition: When two or more gears are made to mesh with each other to transmit power from one shaft to another. Such a combination is called gear train

Purpose: The purpose of the train used is

To obtain correct & required velocity ratio between driver & driven shafts.

To decide upon the relative position of the axes of shafts.

To decide upon amount of power to be transmitted between shafts

Types: Following are the different types of gear trains, depending upon the arrangement of wheels :

1. Simple gear train,

2. Compound gear train,

3. Reverted gear train, and

4. Epicyclic gear train.

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Examination: 2015 WINTER
Que.No Marks
Q 1a)(iii)

2

Question:

Define slip and creep in the belt.


Answer:

Slip : When the frictional grip becomes insufficient. This may cause some forward motion of the driver without carrying the belt with it. This may also cause some forward motion of the belt without carrying the driven pulley with it. This is called slip of the belt and is generally expressed as a percentage. Creep : When the belt passes from the slack side to the tight side, a certain portion of the belt extends and it contracts again when the belt passes from the tight side to slack side. Due to these changes of length, there is a relative motion between the belt and the pulley surfaces. This relative motion is termed as creep.

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Q 1a)(iv)

2

Question:

State any two advantages of V belt drive over flat belt drive.


Answer:

Advantages -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 of 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 1b)(iii)

4

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.

 

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Q 3 c )

4

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 :

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Q 4 a )

4

Question:

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


Answer:

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Q 5 c )

8

Question:

Two parallel shafts whose centre line are 4.8 m apart, are connected by open belt drive. The diameter of larger pulley is 1.5 m and that of smaller pulley 1 m. The initial tension in the belt when stationary is 3 kN. The mass of the belt is 1.5 kg/m length. The coefficient of friction between the belt and pulley is 0.3 Taking centrifugal tension into account, calculate the power transmitted when the smaller pulley rotates at 400 rpm.


Answer:

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Q 6a)(i)

4

Question:

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


Answer:

             

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