Theory of Machines Pune University MCQs

This set of Theory of Machines Multiple Choice Questions & Answers  focuses on “Planar, Spherical and Spatial Motion”.

1. If the relative motions of the rigid bodies are in one plane or in parallel planes, then the mechanism is called

a) spherical mechanism

b) planar mechanism

c) spatial mechanism

d) flexure mechanism

Answer: b

Explanation: In a planar mechanism, all particles describe plane curves in space and all the curves lie in a single or parallel planes. The relative motions are constrained to a single plane. Hence, the planar mechanism is a 2 – D mechanism.

2. Which mechanism is shown in the given diagram?

theory-machines-questions-answers-planar-spherical-spatial-motion-q2

a) spatial

b) spherical

c) planar

d) flexure

Answer: c

Explanation: The given diagram is of a plane four bar linkage which is included under planar mechanisms as the motions are constrained in a single plane. Thus, it is planar.

3. Which of the following are not included under planar mechanisms?

a) Hooke’s universal joint

b) Plate cam and follower

c) Plane four bar linkage

d) Slide crank mechanism

Answer: a

Explanation: Hooke’s universal joint is an example of spherical mechanism as the relative motion of the particles is three dimensional, whereas the rest are examples of planar mechanisms 

4. Coplanar motion is the motion transformation of any planar mechanism. True or false?

a) True

b) False

Answer: a

Explanation: As all the relative motions of the rigid bodies are in a single plane or in parallel plane, the motion transformation is coplanar. Thus, the statement is true.

5. In a mechanical system, if the bodies move in a way such that the trajectories of points in the system lie on concentric sphere, then the mechanism is called

a) spatial mechanism

b) flexure mechanism

c) planar mechanism

d) spherical mechanism

Answer: d

Explanation: In a spherical mechanism, the locus of every point is a curve contained in a spherical surface. These spherical surfaces are concentric.

6. Hooke’s universal joint is an example of spatial mechanism. True or false?

a) True

b) False

Answer: b

Explanation: Hooke’s universal joint is a joint in a rigid rod which allows the rod to bend in any direction. It is included under spherical mechanisms. Thus, the given statement is false.

7. A mechanism where there are no restrictions on the relative motion between the particles, the mechanism is called

a) planar mechanism

b) flexure mechanism

c) spherical mechanism

d) spatial mechanism

Answer: d

Explanation: The relative motion can be random in a spatial mechanism. It is not necessary for the motion transformation to be coplanar or concentric. Spatial mechanism can move in any of the three dimensions.

8. Planar mechanisms and spherical mechanisms are included under spatial mechanisms. True or false?

a) True

b) False

Answer: a

Explanation: The relative motion between particles can be the same as that of planar mechanisms  or spherical mechanism . Thus, planar and spherical mechanisms are included under spatial mechanism.

9. Planar mechanisms are ___________ dimensional whereas spatial mechanisms are ___________ dimensional.

a) 1, 2

b) 2, 1

c) 2, 3

d) 3, 2

Answer: c

Explanation: Planar mechanisms are restricted in a single plane and spatial mechanisms have no such restriction. The relative motion between particles in a spatial mechanism is not in the same or parallel planes. Thus, planar mechanisms are 2 dimensional whereas spatial mechanisms are 3 dimensional.

10. A linkage containing a screw pair is an example of

a) Planar mechanism

b) Spatial mechanism

c) Flexure mechanism

d) Spherical mechanism

Answer: b

Explanation: Relative motion within a screw pair is helical. It is not constrained to one plane. Hence, a linkage containing a screw pair is an example of spatial mechanism.

This set of Theory of Machines Multiple Choice Questions & Answers  focuses on “Mobility”.

1. Mobility is different than the number of degrees of freedom. True or false?

a) True

b) False

Answer: b

Explanation: The mobility or number of degrees of freedom of a rigid body is defined as the number of independent movement the body has. The number of parameters that must be controlled independently to get the device in a particular position is mobility. Mobility and number of degrees of freedom are the same thing.

2. According to Kutzbach criterion, the mobility of a rigid body is given by the formula:

a) 3-2j 1 -j 2

b) 3+2j 1 +j 2

c) 3+2j 1 +j 2

d) 3-2j 1 -j 2

Answer: d

Explanation: The Kutzbach criterion is used to determine the number of degree of freedoms of a kinematic chain or a linkage. According to Kutzbach criterion, mobility  = 3-2j 1 -j 2 ; where n = number of links, j 1 = number of lower pairs, j 2 = number of higher pairs.

3. What is the equation for Grubler’s criterion for plane mechanisms with constrained motion?

a) 3n-2j 1 -4=0

b) 3n-3j 1 -4=0

c) 3n+2j 1 +4=0

d) 3n+3j 1 +4=0

Answer: a

Explanation: Equation for Grubler’s criterion for plane mechanisms with constrained motion can be derived by putting m = 1 and j 2 = 0 in the equation of mobility according to Kutzbach criterion.

1 = 3-2j 1 -0

Therefore, 3n-2j 1 -4=0

4. State the three dimensional form of Kutzbach criterion.

a) 6+5j 1 +4j 2 +3j 3 +2j 4 +j 5

b) 6-5j 1 -4j 2 -3j 3 -2j 4 -j 5

c) 6+5j 1 +4j 2 +3j 3 +2j 4 +j 5

d) 6-5j 1 -4j 2 -3j 3 -2j 4 -j 5

Answer: b

Explanation: According to the three dimensional form of Kutzbach criterion, mobility  = 6-5j 1 -4j 2 -3j 3 -2j 4 -j 5 . It is also called as Kutzbach criterion to determine the degree of freedom of a spatial mechanism.

5. Find the mobility of the following three bar mechanism.

theory-machines-questions-answers-mobility-q5

a) 3

b) 2

c) 0

d) 1

Answer: c

Explanation: Here, n = 3, j 1 = 3, j 2 = 0.

Hence, m = 3--0 = 0

As the degree of freedom is 0, there is no relative motion between the links and the mechanism forms a structure.

6. Find the mobility of the given mechanism.

theory-machines-questions-answers-mobility-q6

a) 0

b) 1

c) 2

d) 3

Answer: b

Explanation: Here, n = 4, j 1 = 4, j 2 = 0.

Hence, m = 3--0 = 1

As the degree of freedom is 1, the mechanism can be brought into motion by a single input motion.

7. Find the mobility of the given mechanism.

theory-machines-questions-answers-mobility-q7

a) 0

b) 1

c) 2

d) 3

Answer: c

Explanation: Here, n = 5, j 1 = 5, j 2 = 0.

Hence, m = 3--0 = 2

As the degree of freedom is 2, two separate input motions are required to produce constrained motion for the mechanism.

8. If there are redundant constraints in the chain and it forms a statically indeterminate structure, what is the degree of freedom or the mobility of this structure?

a) 0

b) 1

c) 2

d) Less than 0

Answer: d

Explanation: If n ≤-1, then there are redundant constraints in the chain and a statically indeterminate structure is formed. The structure is called a preloaded structure as some stress is created within the link so that it has negative mobility.

9. Find the mobility of the given mechanism

theory-machines-questions-answers-mobility-q9

a) -2

b) 2

c) -1

d) 1

Answer: c

Explanation: Here, n = 6, j 1 = 8, j 2 = 0.

Hence, m = 3--0 = -1

The degree of freedom is -1, thus there are redundant constraints in the chain and a statically indeterminate structure is formed.

10. Find the mobility of the following mechanism with a higher pair.

theory-machines-questions-answers-mobility-q10

a) 3

b) -1

c) 1

d) 2

Answer: d

Explanation: Here, n = 4, j 1 = 3, j 2 = 1.

Hence, m = 3--1 = 2

The degree of freedom or mobility is 2.

11. Find the mobility of the given mechanism

theory-machines-questions-answers-mobility-q11

a) 1

b) 3

c) 2

d) 0

Answer: a

Explanation: Here, 1 and 3 form ternary joints.

Hence, n = 6, j 1 = 7, j 2 = 0.

Hence, m = 3--0 = 1

The degree of freedom or mobility is 1.

12. State the equation for the three dimensional form of Grubler’s criterion.

a) 6n-5j 1 -7=0

b) 6n-6j 1 -6=0

c) 6n+5j 1 +7=0

d) 6n+6j 1 +6=0

Answer: a

Explanation: Equation for three dimensional form of Grubler’s criterionfor plane mechanisms with constrained motion can be derived by putting m = 1 and j 2 = j 3 = j 4 = j 5 = 0 in the equation of mobility according to three dimensional form of Kutzbach criterion.

1 = 6-5j 1 ----0

Therefore,6n-5j 1 -7=0.

This set of Theory of Machines Multiple Choice Questions & Answers  focuses on “Kinematic Inversion”.

1. The method to obtain different types of mechanisms by fixing different links in a kinematic chain is known as inversion of mechanism. True or False.

a) True

b) False

Answer: a

Explanation: Many mechanisms can be obtained as the number of links in a kinematic chain by fixing, different links in a kinematic chain one by one. Here the relative motions of the links of the mechanisms do not change. Fixing different links would result in different types of motions. This method is called as inversion of mechanism.

2. Which of these is an inversion of a single slider crank chain?

a) Coupling rod of a locomotive

b) Beam engine

c) Watt’s indicator mechanism

d) Pendulum pump

Answer: d

Explanation: Beam engine, coupling rod of a locomotive, watt’s indicator mechanism are all inversions of a four bar chain whereas the pendulum pump is an inversion of a single slider crank chain.

3. In the given inversion of four bar chain, the link 1 is fixed. Which one is it?

theory-machines-questions-answers-kinematic-inversion-q3

a) Coupling rod of a locomotive

b) Beam engine

c) Bull engine

d) Rotary internal combustion engine

Answer: a

Explanation: CD is a coupling rod and link AB is fixed so as to maintain a constant centre to centre distance between them. This mechanism is used to transmit rotary motion from one wheel to the other wheel. When a wheel rotates, the connecting rod starts reciprocating. The other end of the connecting rod is fixed with the second circle which also starts moving.

4. What is the use of beam engine?

a) To convert rotary motion into oscillatory motion

b) To convert rotary motion into reciprocating motion

c) To convert reciprocating motion into rotary motion

d) To convert oscillatory motion into rotary motion

Answer: b

Explanation: In this, due to the rotation of the crank about the fixed centre A, the lever oscillates about the point D. Point E of the lever CDE is connected to a piston rod which reciprocates due to the rotation of the crank i.e. this mechanism converts rotary motion into reciprocating motion.

5. In the given inversion of a four bar chain rotary motion is converted into reciprocating motion. Which one is it?

theory-machines-questions-answers-kinematic-inversion-q5

a) Rotary internal combustion engine

b) Watt’s indicator mechanism

c) Beam engine

d) Bull engine

Answer: d

Explanation: In this, due to the rotation of the crank about the fixed centre A, the lever oscillates about centre D. E is connected to a piston rod which reciprocates due to rotation of the crank. In other words, the purpose of this mechanism is to convert rotary motion of the crank into reciprocating motion. It is a beam engine or a crank lever mechanism.

6. What is the purpose of the oscillating cylinder engine?

a) To convert rotary motion into oscillatory motion

b) To convert reciprocating motion into rotary motion

c) To convert rotary motion into reciprocating motion

d) To convert oscillatory motion into rotary motion

Answer: b

Explanation: The oscillating cylinder engine is used to convert reciprocating motion into rotary motion. The cylinder oscillates, as the piston is moved by the crank shaft. Thus, due to the reciprocating motion of the crank shaft, the cylinder oscillates resulting in rotary motion.

7. The following mechanism is an inversion of a single slider crank chain obtained by fixing link 3. What is the name of the mechanism?

theory-machines-questions-answers-kinematic-inversion-q7

a) Pendulum pump

b) Gnome engine

c) Oscillating cylinder engine

d) Watt’s indicator mechanism

Answer: c

Explanation: It is used in converting reciprocating motion to rotary motion. In this, link 3  is fixed. When the crank  rotates, the piston attached to piston rod reciprocates and the cylinder  oscillates about a pin pivoted to the fixed link at A. Nowadays, oscillating cylinder engine are used in small toys and models but they were previously used in ships and small stationary engines.

8. The pressure of gas or steam which acts on the indicator plunger can be measured using a ______________

a) Rotary internal combustion engine

b) Watt’s indicator mechanism

c) Bull engine

c) Oscillating cylinder engine

Answer: b

Explanation: The displacement of the link is proportional to the pressure of gas or steam which acts on the piston rod. Thus, by calculating the amount of displacement, the pressure of the gas or steam can be calculated. This mechanism was used in Watt’s engines.

9. Watt’s indicator mechanism is an inversion of a single slider crank chain. True or false.

a) True

b) False

Answer: b

Explanation: Watt’s indicator mechanism is not an inversion of a slider crank chain but is an inversion of a four bar chain. Thus, the given statement is false. In Watt’s mechanisms, link 1 is fixed whereas links 3 and 4 act as lever.

10. Identify the mechanism:

theory-machines-questions-answers-kinematic-inversion-q10

a) Watt’s indicator mechanism

b) Bull engine

c) Beam engine

d) Rotary internal combustion engine

Answer: b

Explanation: The given diagram is of a bull engine or pendulum pump. Here, link 4 is fixed at A. In this inversion the slotted link shape is given to the slider to get the desired motion.

11. Rotary internal combustion engine is an inversion of a four bar chain. True or false?

a) True

b) False

Answer: b

Explanation: Rotary internal combustion engine or a gnome engine is an inversion of single slider crank chain and not a four bar chain. Thus, the statement is false. In the rotary I.C. engine the crank  is fixed. The pistons, connecting rod and the fixed crank form the inversion of the slider crank chain.

12. The relative motions between different links changes due to the process of inversion. True or false?

a) True

b) False

Answer: b

Explanation: The relative motions between the various links do not change through the process of inversion, but the motions measured with respect to the fixed link change. Thus, the given statement is false.

This set of Theory of Machines Multiple Choice Questions & Answers  focuses on “Grashof’s Law”.

1. What is the condition involved in Grashof’s law?

a) s+l≥p+q

b) s-l ≥ p-q

c) s+l≤p+q

d) s-l ≤ p-q

Answer: c

Explanation: According to Grashof’s law, in a four bar mechanism, the sum of the lengths of the shortest and the longest link should be less than or equal to the sum of the lengths of two remaining links for continuous relative motion between them.

Therefore, s+l≤p+q.

2. If Grashof’s law is not satisfied, no link will make a complete revolution relative to one another. True or False?

a) False

b) True

Answer: b

Explanation: According to Grashof’s law, in a four bar mechanism, the sum of the lengths of the shortest and the longest link should be less than or equal to the sum of the lengths of two remaining links for continuous relative motion between them. Thus, the given statement is false.

3. If the link next to the shortest link is fixed, what type of mechanism is obtained?

a) Double crank mechanism

b) Linkage is not planar

c) Double rocker mechanism

d) Crank rocker mechanism

Answer: d

Explanation: When the link next to the shortest link is fixed, the short link is able to rotate continuously and is called crank. The link opposite to it can only oscillate between limits and is called the rocker. Thus, the mechanism obtained is the crank rocker mechanism.

4. If the shortest link is fixed, what type of mechanism is obtained?

a) Double crank mechanism

b) Double rocker mechanism

c) Linkage is not planar

d) Crank rocker mechanism

Answer: a

Explanation: When the shortest link is fixed, the links adjacent to it rotate continuously and can be described as proper cranks. Thus, this mechanism is called as double crank mechanism.

5. If the link opposite to the shortest link is fixed, what type of mechanism is obtained?

a) Double crank mechanism

b)Linkage is not planar

c) Double rocker mechanism

d) Crank rocker mechanism

Answer: c

Explanation: When the link opposite to the shortest link if fixed, the shortest link is able to make a complete revolution but the two links adjacent to it oscillate between limits and are called rockers.

Thus, this mechanism is called as double rocker mechanism.

6. If the s+l=p+q, what is the type of linkage?

a) Trapezium

b) Parallelogram

c) Circular

d) Six bar

Answer: b

Explanation: According to the given condition, s+l=p+q i.e. the sum of the lengths of the adjacent sides are equal. Thus, the linkage has to be a parallelogram.

7. What type of mechanism is shown in the given diagram?

theory-machines-questions-answers-grashofs-law-q7

a) Double crank mechanism

b) Linkage is not planar

c) Double rocker mechanism

d) Crank rocker mechanism

Answer: d

Explanation: When the link next to the shortest link is fixed, the short link is able to rotate continuously and is called crank. The link opposite to it can only oscillate between limits and is called the rocker. In the given diagram, the link adjacent to the shortest link is fixed, thus crank rocker mechanism is obtained.

8. What type of mechanism is shown in the diagram below?

theory-machines-questions-answers-grashofs-law-q8

a) Double rocker mechanism

b) Crank rocker mechanism

c) Double crank mechanism

d) Linkage is not planar

Answer: b

Explanation: When the link next to the shortest link is fixed, the short link is able to rotate continuously and is called crank. The link opposite to it can only oscillate between limits and is called the rocker. In the given diagram, the link adjacent to the shortest link is fixed, thus crank rocker mechanism is obtained.

9. What type of mechanism is shown in the diagram given below?

theory-machines-questions-answers-grashofs-law-q9

a) Double crank mechanism

b) Crank rocker mechanism

c) Double rocker mechanism

d) Linkage is not planar

Answer: a

Explanation: When the shortest link is fixed, the links adjacent to it rotate continuously and can be described as proper cranks. In the given diagram, the shortest link is fixed, thus double crank mechanism is obtained.

10. What type of mechanism is shown in the diagram given below?

theory-machines-questions-answers-grashofs-law-q10

a) Crank rocker mechanism

b) Double crank mechanism

c) Double rocker mechanism

d) Linkage is not planar

Answer: b

Explanation: When the shortest link is fixed, the links adjacent to it rotate continuously and can be described as proper cranks. In the given diagram, the shortest link is fixed, thus double crank mechanism is obtained.

11. What type of mechanism is shown in the diagram given below?

theory-machines-questions-answers-grashofs-law-q11

a) Double rocker mechanism

b) Double crank mechanism

c) Crank rocker mechanism

d) Linkage is not planar

Answer: c

Explanation: When the link next to the shortest link is fixed, the short link is able to rotate continuously and is called crank. The link opposite to it can only oscillate between limits and is called the rocker. In the given diagram, the link adjacent to the shortest link is fixed, thus crank rocker mechanism is obtained.

12. What type of mechanism is shown in the diagram given below?

theory-machines-questions-answers-grashofs-law-q12

a) Double rocker mechanism

b) Double crank mechanism

c) Crank rocker mechanism

d) Linkage is not planar

Answer: c

Explanation: When the link next to the shortest link is fixed, the short link is able to rotate continuously and is called crank. The link opposite to it can only oscillate between limits and is called the rocker. In the given diagram, the link adjacent to the shortest link is fixed, thus crank rocker mechanism is obtained.

13. What type of mechanism is shown in the diagram given below?

theory-machines-questions-answers-grashofs-law-q13

a) Double rocker mechanism

b) Double crank mechanism

c) Crank rocker mechanism

d) Linkage is not planar

Answer: d

Explanation: When s+l is not less than or equal to p+q, Grashof’s law is not satisfied and the linkage isn’t planar. According to the given figure, the Grashof’s law is not satisfied. Hence, the four bar linkage is not planar.

14. What type of mechanism is shown in the diagram given below?

theory-machines-questions-answers-grashofs-law-q14

a) Double rocker mechanism

b) Double crank mechanism

c) Crank rocker mechanism

d) Linkage is not planar

Answer: a

Explanation: When the link opposite to the shortest link if fixed, the shortest link is able to make a complete revolution but the two links adjacent to it oscillate between limits and are called rockers.

Here, the link opposite to the shortest side is fixed. Thus, the mechanism obtained is double rocker mechanism.

15. What type of mechanism is shown in the diagram given below?

theory-machines-questions-answers-grashofs-law-q15

a) Double rocker mechanism

b) Linkage is not planar

c) Crank rocker mechanism

d) Double crank mechanism

Answer: b

Explanation: When s+l is not less than or equal to p+q, Grashof’s law is not satisfied and the linkage isn’t planar. According to the given figure, the Grashof’s law is not satisfied. Hence, the four bar linkage is not planar.

This set of Theory of Machines Multiple Choice Questions & Answers  focuses on “Straight Line Mechanisms”.

1. Identify the given straight line mechanism.

theory-machines-questions-answers-straight-line-mechanisms-q1

a) Hart’s mechanism

b) Peaucellier mechanism

c) Tchebicheff’s mechanism

d) Watt’s mechanism

Answer: b

Explanation: The given straight line mechanism is called Peaucellier mechanism. Peaucellier mechanism is used for converting an input circular motion to an exact straight line motion. This mechanism has 8 members and 6 joints. Two links having equal length are connected at opposite ends to a rhombus.

2. Identify the given straight line mechanism.

theory-machines-questions-answers-straight-line-mechanisms-q2

a) Hart’s mechanism

b) Tchebicheff’s mechanism

c) Grasshopper mechanism

d) Watt’s mechanism

Answer: d

Explanation: The given straight line mechanism is called Watt’s mechanism. It consists of three links. Two of them are of equal length whereas one is shorter. Due to the rotation motion of the longer links, the centre of the shorter link traces an approximate straight line. Hence, this mechanism is classified under approximate straight line mechanisms.

3. Identify the given straight line mechanism.

theory-machines-questions-answers-straight-line-mechanisms-q3

a) Hart’s mechanism

b) Tchebicheff’s mechanism

c) Scott Russell’s mechanism

d) Robert’s mechanism

Answer: c

Explanation: The given straight line mechanism is called Scott Russell’s mechanism. It consists of one short and a long link. The length of the shorter link is approximately equal to the half of the longer link. The bottom part of the longer link traces a straight line.

4. Which of these is an approximate straight line motion mechanism?

a) Scott Russell’s mechanism

b) Hart’s mechanism

c) Peaucellier mechanism

d) Watt’s mechanism

Answer: d

Explanation: Watt’s consists of three links. Two of them are of equal length whereas one is shorter. Due to the rotation motion of the longer links, the centre of the shorter link traces an approximate straight line. Out of the following mechanism, Watt’s mechanism is an approximate straight line mechanism whereas the rest are exact straight line mechanisms.

5. Identify the given straight line mechanism.

theory-machines-questions-answers-straight-line-mechanisms-q5

a) Hart’s mechanism

b) Modified Scott Russel mechanism

c) Peaucellier mechanism

d) Tchebicheff’s mechanism

Answer: b

Explanation: The given straight line mechanism is called Modified Scott Russel mechanism. It is similar to Scott Russell’s mechanism. The working of it is similar to Scott Russel mechanism.

6. Identify the given straight line mechanism.

theory-machines-questions-answers-straight-line-mechanisms-q6

a) Grasshopper mechanism

b) Hart’s mechanism

c) Peaucellier mechanism

d) Tchebicheff’s mechanism

Answer: a

Explanation: The given straight line mechanism is called Grasshopper mechanism. It is an improvement over modified Scott Russel mechanism. The point P does not trace an exact straight line. It is an approximate straight line motion.

7. Which of the given mechanisms is an exact straight line motion mechanism?

a) Grasshopper mechanism

b) Robert’s mechanism

c) Peaucellier mechanism

d) Tchebicheff’s mechanism

Answer: c

Explanation: Peaucellier mechanism is used for converting an input circular motion to an exact straight line motion. This mechanism has 8 members and 6 joints. Two links having equal length are connected at opposite ends to a rhombus. Peaucellier mechanism is an exact straight line motion mechanism whereas the rest of them are approximate straight line motion mechanism.

8. Identify the given straight line mechanism.

theory-machines-questions-answers-straight-line-mechanisms-q8

a) Tchebicheff’s mechanism

b) Hart’s mechanism

c) Robert’s mechanism

d) Peaucellier mechanism

Answer: c

Explanation: The given straight line mechanism is called Robert’s mechanism. This mechanism has the form of a trapezium. The point P traces an approximate straight line. Thus, Robert’s mechanism is classified under approximate straight line mechanism.

9. Identify the given straight line mechanism.

theory-machines-questions-answers-straight-line-mechanisms-q9

a) Chebychev mechanism

b) Tchebicheff’s mechanism

c) Robert’s mechanism

d) Peaucellier mechanism

Answer: a

Explanation: The given straight line mechanism is called Chebychev mechanism. This mechanism is used for converting rotary motion to an approximate straight line motion. The point P traces an approximate straight line mechanism.

10. Hart’s mechanism requires only eight links whereas the Peaucellier mechanism requires six links. True or false?

a) True

b) False

Answer: b

Explanation: Hart’s mechanism requires only six links whereas the Peaucellier mechanism requires eight links and contains 6 joints. Thus, the statement is false.

11. __________ mechanism is a crossed four bar chain mechanism in early steam engines to guide the piston rod in a cylinder to have an approximate straight line motion.

a) Peaucellier’s

b) Chebychev’s

c) Grasshopper

d) Watt’s

Answer: d

Explanation: Watt’s mechanism is a four bar chain mechanism and was used by Watt in steam engines so as to guide the piston rod into a cylinder to have a straight line motion. Due to the rotation motion of the longer links, the centre of the shorter link traces an approximate straight line.

This set of Theory of Machines Multiple Choice Questions & Answers  focuses on “Quick Return Mechanisms”.

1. For what purpose are the quick return mechanisms used?

a) To convert reciprocating motion into oscillatory motion

b) To convert oscillatory motion into reciprocating motion

c) To convert reciprocating motion into rotary motion

d) To convert rotary motion into reciprocating motion

Answer: d

Explanation: Due to the rotation motion of the disc, the ram moves forward and backwards. During half rotation, the ram moves forward whereas during the other half rotation. The ram quickly returns. Thus it converts rotary motion into reciprocating motion.

2. In a quick return mechanism, the forward reciprocating motion is faster rate than the backward stroke. True or false?

a) True

b) False

Answer: b

Explanation: In a quick return mechanism, the forward reciprocating motion is slower rate as compared to the backward stroke. That is why it is called a quick return mechanism.

Thus, the statement is false.

3. For a crank and slotted lever quick return mechanism, α = 150°. Find the ratio of time of cutting stroke to time of return stroke.

a) 1.2

b) 1.3

c) 1.4

d) 1.5

Answer: c

Explanation: Ratio of the time of cutting stroke to the time of return stroke for a crank and slotted lever quick return mechanism = /α = /150 = 1.4.

4. For a crank and slotted lever quick return mechanism,β = 260°. Find the ratio of time of cutting stroke to time of quick return stroke.

a) 2.6

b) 1.6

c) 0.2

d) 0.4

Answer: a

Explanation: : Ratio of time of cutting stroke to time of return stroke for a crank and slotted lever quick return mechanism = β/ = 260/ = 2.6.

5. For a Whitworth quick return motion mechanism α = 200°. Find the ratio of time of cutting stroke to time of return stroke.

a) 1.25

b) 1.35

c) 1.30

d) 1.40

Answer: a

Explanation: : Ratio of time of cutting stroke to time of return stroke for a Whitworth quick return motion mechanism= α/ = 200/ = 1.25.

6. For a Whitworth quick return motion mechanism β = 110°. Find the ratio of time of cutting stroke to time of return stroke.

a) 0.44

b) 2.27

c) 2.37

d) 0.42

Answer: b

Explanation: Ratio of time of cutting stroke to time of return stroke for a Whitworth quick return motion mechanism = / β = /110 = 2.27.

7. A crank and slotted lever quick return mechanism has a centre distance of 1000 mm between the centre of oscillation of slotted lever and centre of rotation of the crank. Radius of the crank is 420 mm. Find the ratio of time of cutting stroke to time of return stroke.

theory-machines-questions-answers-quick-return-mechanisms-q7

a) 1.677

b) 6.901

c) 6.248

d) 1.762

Answer: d

Explanation: AC = 1000 mm and BC = 420 mm

sin = BC/AC = 0.42

90- α/2 = 24.834

α = 130.33°

Ratio of time of cutting stroke to time of return stroke = /α = /130.33

= 1.762.

8. In a crank and slotted lever quick return mechanism, the distance between the fixed centres is 400 mm and the length of the crank is 250 mm. If the length of the slotted bar is 2000 mm, find the length of the stroke if the line of stroke passes through the extreme positions of the free side of the lever.

theory-machines-questions-answers-quick-return-mechanisms-q8

a) 1500 mm

b) 2000 mm

c) 1000 mm

d) 2500 mm

Answer: d

Explanation: AC = 400 mm, CB = 250 mm and AB = 2000 mm

sin = BC/AC = 0.625

α = 102.63°

Length of stroke = 2 x AB sin = 2500 mm.

9. In a Whitworth quick return mechanism, distance between the fixed centres is 30mm and the length of the driving crank is 60 mm. The length of the slotted lever is 100 mm and the length of the connecting rod is 85 mm. Find the ratio of time of cutting stroke to time of return stroke.

theory-machines-questions-answers-quick-return-mechanisms-q9

a) 1.5

b) 2.5

c) 2

d) 1

Answer: c

Explanation: CD = 30 mm, CA = 60 mm, PA = 100 mm, PR = 85 mm

cos β/2 = 0.5

β = 120°

Ratio of time of cutting stroke to time of return stroke = / β = /120 = 2.

This set of Theory of Machines Multiple Choice Questions & Answers  focuses on “Single Plate Clutch”.

1. Which of these is not a type of clutch?

a) Single disc

b) Conical

c) Centrifugal

d) Cylindrical

Answer: d

Explanation: Single disc, conical and centrifugal are types of clutches. Clutches are used for the transmission of power in shafts and machines which are started and stopped frequently. It is also used to deliver power to the machine which are partially or fully loaded.

2. Identify the clutch.

theory-machines-questions-answers-single-plate-clutch-q2

a) Single plate clutch

b) Conical clutch

c) Centrifugal clutch

d) Multi plate clutch

Answer: a

Explanation: The given diagram is of a single plate clutch. It consists of a clutch plate whose both sides are faced with Ferrodo . It is mounted on the hub which moves along the splines of the shaft which is driven.

3. In a single plate clutch, considering uniform pressure, T = nµWR. What is R equal to?

a) 2(r 1 3 + r 2 3 ) / 3(r 1 2 + r 2 2 )

b) 2(r 1 3 – r 2 3 ) / 3(r 1 2 – r 2 2 )

c) (r 1 – r 2 )/2

d) (r 1 + r 2 )/2

Answer: b

Explanation: For uniform pressure theory, T = nµW(2(r 1 3 – r 2 3 ) / 3(r 1 2 – r 2 2 ))

Therefore, considering uniform pressure theory in a single plate clutch, T = nµWR where R = 2(r 1 3 – r 2 3 ) / 3(r 1 2 – r 2 2 ).

4. In a single plate clutch, considering uniform wear, T = nµWR. What is R equal to?

a) 2(r 1 3 + r 2 3 ) / 3(r 1 2 + r 2 2 )

b) 2(r 1 3 – r 2 3 ) / 3(r 1 2 – r 2 2 )

c) (r 1 – r 2 )/2

d) (r 1 + r 2 )/2

Answer: d

Explanation: For uniform wear, T = nµW((r 1 + r 2 )/2)

Therefore, considering uniform pressure theory in a single plate clutch, T = nµWR where R = (r 1 + r 2 )/2.

5. The intensity of pressure is minimum at the outer radius that is r 1 . True or false?

a) True

b) False

Answer: a

Explanation: The intensity of pressure is minimum at the outer radius that is r 1 . It is given by C/r 1 . The intensity of pressure is maximum at the inner radius which is r 2 .

6. In a single plate clutch, the pressure is uniformly distributed. If the outer and inner radii are 100 mm and 70 mm respectively, find the value of R.

a) 76.39 mm

b) 23.48 mm

c) 85.88 mm

d) 34.98 mm

Answer: c

Explanation: Considering uniform pressure theory in a single plate clutch, T = nµWR where R = 2(r 1 3 + r 2 3 ) / 3(r 1 2 + r 2 2 )

Thus, here, R = 2(100 3 – 70 3 ) / 3(100 2 – 70 2 ) = 85.88 mm.

7. Maximum intensity of pressure is given by ____

a) C/R

b) C/R 2

c) C/r 2

d) C/r 1

Answer: c

Explanation: We know that, pmax x r 2 = C.

Thus, pmax = C/r 2

Similarly, pmin = C/r 1

8. When the pressure is uniformly distributed over entire area, then the intensity of pressure is _______________

a) W/ (π(r 1 2 – r 2 2 )

b) W/ (π(r 1 2 + r 2 2 )

c) W/ (π(r 2 2 – r 1 2 ))

d) W/ (π(r 2 2 x r 1 2 ))

Answer: a

Explanation: Pressure = Force / Area

Therefore, the intensity of pressure considering uniform pressure is W/ (π(r 1 2 – r 2 2 ).

9. For uniform wear, the axial thrust W = ______________

a) 2 π C (r 2 + r 1 )

b) 2 π C (r 1 – r 2 )

c) 2 π C (r 2 – r 1 )

d) 2 π C (r 1 x r 2 )

Answer: b

Explanation: In uniform wear theory, C = W/(2 π (r 1 – r 2 ))

Thus, W = 2 π C (r 1 – r 2 ).

10. Find the axial thrust to be provided by the springs if the maximum intensity of pressure in a single plate clutch should not exceed 0.2 N/mm 2 . The outer and the inner radius is 75 mm and 35 mm. Assume theory of uniform wear.

a) 1792.01 N

b) 1789.21 N

c) 1723.41 N

d) 1759.29 N

Answer: d

Explanation: Given: pmax = 0.2 N/mm 2 , r 1 = 75 mm and r 2 = 35 mm

Thus, W = 2 π C (r 1 – r 2 ) = 2 π C (r 1 – r 2 ) = 2 π x pmax x r 2 (r 1 – r 2 ) = 2 π x 0.2 x 35  = 1759.29 N.

11. Find the average pressure in a single plate clutch if the axial force is 5 kN. The inside radius and the outer radius is 30 mm and 70 mm respectively. Assume uniform wear.

a) 398 N/mm 2

b) 0.398 N/m 2

c) 0.398 N/mm 2

d) 398 N/m 2

Answer: c

Explanation: Given: W = 5 kN = 5 x 10 3 N, r 1 = 70 mm and r 2 = 30 mm

Average pressure = Total normal force on the surface/ Cross sectional area of those surfaces = W/ (π(r 1 2 – r 2 2 )) = 5 x 103/(π(70 2 – 30 2 )) = 0.398 N/mm 2 .

12. In a single plate clutch if the axial force is equal to 8 kN, find the minimum and maximum intensity of pressure. The outer radius = 100 mm and inner radius = 65 mm.

a) 0.36 N/mm 2 , 0.56 N/mm 2

b) 0.56 N/mm 2 , 0.36 N/mm 2

c) 0.36 N/m 2 , 0.56 N/m 2

d) 0.56 N/m 2 , 0.36 N/m 2

Answer: a

Explanation: Given : W = 8 kN = 8 x 10 3 N, r 1 = 100 mm and r 2 = 65 mm

Maximum intensity of pressure = p min = C/r 1

Thus, C = 100 x p min

W = 2 π C (r 1 – r 2 )

Thus,

C = W/(2 π (r 1 – r 2 )) = 8 x 103/ )

C = 100 x p min = 36.378

p min = 0.36 N/mm 2

Maximum intensity of pressure = pmax = C/r 2

Thus, C = 65 x p max

W = 2 π C (r 1 – r 2 )

Thus,

C = W/(2 π (r 1 – r 2 )) = 8 x 103/ )

C = 65 x pmax = 36.378

pmax = 0.56 N/mm 2 .

13. A single plate clutch, having n = 2, has outer and inner radii 150 mm and 100 mm respectively. The maximum intensity of pressure at any point is 0.1 N/mm 2 . If the µ is 0.3, determine the power transmitted by a clutch at a speed 3000 r.p.m.

a) 74031 kW

b) 740.31 kW

c) 74.031 kW

d) 706.95 kW

Answer: c

Explanation: Given : n = 2, r 1 = 150 mm ; r 2 = 100 mm ; p = 0.1 N/mm 2 ; µ = 0.3 ; N = 3000 r.p.m. or ω = 2π × 3000/60 = 314.16 rad/s.

Since the intensity of pressure  is maximum at the inner radius (r 2 ), considering uniform wear, p max x r 2 = C or C = 0.1 × 100 = 10 N/mm.

Axial thrust, W = 2 π C (r 1 – r 2 ) = 2π × 10  = 3142 N

Mean radius = R = (r 1 + r 2 )/2 = 125 mm = 0.125 m.

We know that torque transmitted,

T = n.µ.W.R = 2 × 0.3 × 3142 × 0.125 = 235.65 N-m

Thus, power transmitted,

P = T.ω = 235.65 × 314.16 = 74031 W = 74.031 kW.

14. A single plate clutch, having n = 2, has outer and inner radii 200 mm and 175 mm respectively. The maximum intensity of pressure at any point is 0.05 N/mm 2 . If the µ is 0.4 and the power generated by the clutch is equal to 32.384 kW, determine the speed of the clutch.

a) 209.44 r.p.m.

b) 2000 r.p.m.

c) 157.08 r.p.m.

d) 1500 r.p.m.

Answer: d

Explanation: Given : n = 2, r 1 = 200 mm ; r 2 = 175 mm ; p = 0.05 N/mm 2 ; µ = 0.4 ; P = 32.384 kW = 32384 W

Since the intensity of pressure  is maximum at the inner radius (r 2 ), considering uniform wear, p max x r 2 = C or C = 0.05 × 175 = 8.75 N/mm

Axial thrust, W = 2 π C (r 1 – r 2 ) = 2π × 8.75  = 1374.44 N

Mean radius = R = (r 1 + r 2 )/2 = 187.5 mm = 0.1875 m.

We know that torque transmitted,

T = n.µ.W.R = 2 × 0.4 × 1374.44 × 0.1875 = 206.16 N-m

Thus, power transmitted,

P = T.ω

ω = P/T = 32384/206.16 = 157.08 rad/s

N = ω x 60 /  = 1500 r.p.m.

15. A single plate clutch, where n = 2, is required to transmit 30 kW at 3000 r.p.m. Determine the outer and inner radii if the coefficient of friction is 0.3, the ratio of radii is 1.5. pmax = 0.1 N/mm 2 .

a) 0.111 mm, 0.074 mm

b) 111 mm, 74 mm

c) 74 mm, 111 mm

d) 0.074 mm, 0.111 mm

Answer: b

Explanation: Given: n = 2 ; P = 30 kW = 30 × 103 W ; N = 3000 r.p.m. or ω = 2π × 3000/60 = 314.2 rad/s ; µ = 0.3 ; r 1 /r 2 = 1.5 ; p = 0.1 N/mm 2 .

Since the ratio of radii (r 1 /r 2 ) is 1.5, therefore r 1 = 1.5r 2

We know that the power transmitted , 30 × 10 3 = T.ω = T × 314.2 ∴ T = 30 × 103/314.2 = 95.48 N-m = 95.48 × 10 3 N-mm

The intensity of pressure is maximum at inner radius (r 2 ),

pmax x r 2 = C or C = 0.1 x r 2 N/mm

W = 2 π C (r 1 – r 2 ) = 2 π × 0.1 r 2 (1.5r 2 – r 2 ) = 0.314 (r 2 ) 2

Mean radius for uniform wear = R = (r 1 + r 2 )/2 = 1.25 r 2

Torque transmitted , 95.48 × 10 3 = n.µ.W.R = 2 × 0.3 × 0.314 (r 2 )2 × 1.25 r 2 = 0.2355 (r 2 ) 3

(r 2 ) 3 = 95.48 × 10 3 /0.2355 = 405.43 × 10 3 or r 2 = 74 mm

r 1 = 1.5 r 2 = 1.5 × 74 = 111 mm.

This set of Theory of Machines Multiple Choice Questions & Answers  focuses on “Multi plate Clutch”.

1. In a multi plate clutch, the formula for T is given by ______

a) n.µ.W.R

b) n.µ.W.r 1

c) n.µ.W.r 2

d) n.µ.W.(r 1 +r 2 )

Answer:a

Explanation: In a multi plate clutch, the formula for T is given by n.µ.W.R

The formula is the same for torque in a single plate clutch. But for large amount of torque to be transmitted, multi plate clutch is used. Multi plate clutches are used in motor vehicles and machine tools.

2. In a multi plate clutch, considering uniform pressure, T = nµWR. What is R equal to?

a) 2(r 1 3 + r 2 3 ) / 3(r 1 2 + r 2 2 )

b) 2(r 1 3 – r 2 3 ) / 3(r 1 2 – r 2 2 )

c) (r 1 – r 2 )/2

d) (r 1 + r 2 )/2

Answer: b

Explanation: Considering uniform pressure theory in a multi plate clutch, T = nµWR where R = 2(r 1 3 – r 2 3 ) / 3(r 1 2 – r 2 2 ). This value is the same as that for the single plate clutch considering uniformly distributed pressure.

3. In a multi plate clutch, considering uniform wear, T = nµWR. What is R equal to?

a) 2(r 1 3 + r 2 3 ) / 3(r 1 2 + r 2 2 )

b) 2(r 1 3 – r 2 3 ) / 3(r 1 2 – r 2 2 )

c) (r 1 – r 2 )/2

d) (r 1 + r 2 )/2

Answer: d

Explanation: Considering uniform pressure theory in a multi plate clutch, T = nµWR where R = (r 1 + r 2 )/2. This value is the same as that for the single plate clutch considering uniform wear.

4. In a multi plate clutch, T = 150 N-m, n = 4, µ = 0.3 and R = 0.1 m. Find the axial thrust.

a) 18

b) 1800

c) 1250

d) 200

Answer: c

Explanation: T = nµWR

150 = 4 x 0.3 x W x 0.1

W = 1250 N

Thus, the axial thrust = 1250 N.

5. Maximum intensity of pressure for multi plate clutch is given by ____

a) C/R

b) C/R 2

c) C/r 2

d) C/r 1

Answer: c

Explanation: We know that, pmax x r 2 = C.

Thus, pmax = C/r 2

Similarly, pmin = C/r 1 .

6. For a single plate clutch, n = 2, whereas for a multi plate clutch, n can take any values ≥ 2. True or false?

a) True

b) False

Answer: a

Explanation: A single plate clutch has both sides effective, thus n = 2. In a multi plate clutch, the value of n can be equal or greater than 2, which helps in transmitting large amounts of torque.

7. In a multi plate clutch, number of discs on the driving shaft is given by n 1 . True or false?

a) True

b) False

Answer: a

Explanation: In a multi plate clutch, number of discs on the driving shaft is given by n 1 , whereas number of discs on the driven shaft is given by n 2 . With the help of these two variables we can find out the number of pairs of contact surfaces.

8. In a multi plate clutch, number of pairs of contact surfaces  = ________

a) n 1 – n 2 + 1

b) n 1 – n 2 – 1

c) n 1 + n 2 – 1

d) n 1 + n 2 + 1

Answer: c

Explanation: In a multi plate clutch, number of pairs of contact surfaces  is equal to n 1 + n 2 – 1. Due to the high number of pairs of contact surfaces, the torque transmitted is higher.

n is always a whole number.

9. A multi-disc clutch has 6 discs on the driving shaft and 4 on the driven shaft. Find the number of pairs of contact surfaces.

a) 3

b) 10

c) 9

d) 11

Answer: c

Explanation: In a multi plate clutch, number of pairs of contact surfaces  is equal to n 1 + n 2 – 1 = 6+4-1 = 9

Thus, T = n.µ.W.R = 9.µ.W.R

10. Calculate the maximum and minimum intensity of pressure in multi plate clutch when the axial force is 5 kN assuming uniform wear. Inner radius = 20 mm and outer radius = 40 mm.

a) 1.989 N/mm 2 , 0.994 N/mm 2

b) 0.994 N/mm 2 , 1.989 N/mm 2

c) 1.989 N/m 2 , 0.994 N/m 2

d) 0.994 N/m 2 , 1.989 N/m 2

Answer: a

Explanation: We know that, p max x r 2 = C

p max x 20 = C

Now, W = 2 π C (r 1 – r 2 )

5 x 10 3 = 2 π x p max x 20 x 

p max = 1.989 N/mm 2

We know that, p min x r 1 = C

p min x 40 = C

Now, W = 2 π C (r 1 – r 2 )

5 x 10 3 = 2 π x p max x 40 x 

p max = 0.994 N/mm 2 .

11. A multi-disc clutch has 3 discs on the driving shaft and 2 on the driven shaft. The outer radius is 120 mm and inside radius 60 mm. Considering uniform wear and µ = 0.3, find the Find the value of W for transmitting 25 kW at 1500r.p.m.

a) 651 N/m 2

b) 651 N/mm 2

c) 0.0651 N/m 2

d) 0.0651 N/mm 2

Answer: d

Explanation: Given : n 1 = 3 ; n 2 = 2 ; r 1 = 120 mm ; r 2 = 60 mm ; µ = 0.3 ; P = 25 kW = 25 × 10 3 W ; N = 1500 r.p.m. or ω = 2 π × 1500/60 = 157.08 rad/s

Power transmitted P = 25 × 10 3 = T.ω = T × 157.08

T = 25 × 10 3 /157.08 = 159.15 N-m

n = n 1 + n 2 – 1 = 4

Mean radius = R = (r 1 + r 2 )/2 = 90 mm = 0.09 m

159.15 = n.µ.W.R = 4 × 0.3 × W × 0.09 = 0.108 W

W = 1473.65 N.

12. A multi-disc clutch has 3 discs on the driving shaft and 2 on the driven shaft. The outer radius is 100 mm and inside radius 40 mm. Considering uniform wear and µ = 0.25, find the maximum axial intensity of pressure between the discs for transmitting 30 kW at 1000 r.p.m.

a) 0.4071 N/m 2

b) 40.71 N/mm 2

c) 407.1 N/m 2

d) 0.4071 N/mm 2

Answer: d

Explanation: Explanation: Given : n 1 = 3 ; n 2 = 2 ; r 1 = 100 mm ; r 2 = 40 mm ; µ = 0.25 ; P = 30 kW = 30 × 10 3 W ; N = 1000 r.p.m. or ω = 2 π × 1000/60 = 104.71 rad/s

Power transmitted P = 30 × 10 3 = T.ω = T × 104.71

T = 30 × 10 3 /157.08 = 286.48 N-m

n = n 1 + n 2 – 1 = 4

Mean radius = R = (r 1 + r 2 )/2 = 70 mm = 0.07 m

286.48 = n.µ.W.R = 4 × 0.25 × W × 0.07

W = 4092.55 N

W = 4092.55 = 2 π C (r 1 – r 2 ) = 80 π C

C = 16.283 = p max x 40

p max = 0.4071 N/mm 2 .

13. A single plate clutch, having n 1 = 3 and n 2 = 2, has outer and inner radii 200 mm and 175 mm respectively. The maximum intensity of pressure at any point is 0.05 N/mm 2 . If the µ is 0.4, determine the power transmitted by a clutch at a speed 1500 r.p.m.

a) 32384 x 10 3 W

b) 64.768 W

c) 64768 kW

d) 64768 W

Answer: d

Explanation: Given : n = 4, r 1 = 200 mm ; r 2 = 175 mm ; p = 0.05 N/mm 2 ; µ = 0.4 ; N = 1500 r.p.m. or ω = 2π × 1500/60 = 157.08 rad/s.

Since the intensity of pressure  is maximum at the inner radius (r 2 ), considering uniform wear, p max x r 2 = C or C = 0.05 × 175 = 8.75 N/mm

Axial thrust, W = 2 π C (r 1 – r 2 ) = 2π × 8.75  = 1374.44 N

Mean radius = R = (r 1 + r 2 )/2 = 187.5 mm = 0.1875 m.

We know that torque transmitted,

T = 4.µ.W.R = 4 × 0.4 × 1374.44 × 0.1875 = 412.32 N-m

Thus, power transmitted,

P = T.ω = 412.32 × 157.08 = 64768 W = 64.368 kW.

14. A plate clutch has three discs on driving shaft and two discs on driven shaft. The outer radius of the contact surfaces is 100 mm and inner radius 50 mm. Assuming uniform pressure and µ = 0.3; find the total spring load pressing plates together to transmit 35 kW at 1600r.p.m.

a) 2372.3 N

b) 2260.7 N

c) 2401.8 N

d) 2131.4 N

Answer: b

Explanation: Given : n 1 = 3 ; n 2 = 2 ; n = 4 ; r 1 = 100 mm ; r 2 = 50 mm ; µ = 0.3 ; P = 35 kW = 35 × 10 3 W ; N = 1600 r.p.m. or ω = 2 π × 1600/60 = 167.55 rad/s

We know that power transmitted , 35 × 10 3 = T.ω = T × 167.55

T = 35 × 10 3 /167.55 = 208.89 N-m

Mean radius of the contact surface, for uniform pressure,R = 2(r 1 3 – r 2 3 ) / 3(r 1 2 – r 2 2 ) = 77.77 mm = 0.077 m

208.89 = n.µ.W.R = 4 × 0.3 W × 0.077

W = 2260.7 N.

15. A multi plate clutch, having n 1 = 2 and n 2 = 1, has outer and inner radii 1000 mm and 600 mm respectively. The maximum intensity of pressure at any point is 0.2 N/mm 2 . If the µ is 0.25, determine the speed of the clutch if the power generated = 31582.807 kW.

a) 2500 r.p.m.

b) 261.8 r.p.m.

c) 2000 r.p.m.

d) 209.44 r.p.m.

Answer: a

Explanation: Given : n = 2, r 1 = 1000 mm ; r 2 = 600 mm ; p = 0.2 N/mm 2 ; µ = 0.25 ; P = 31582.807 kW = 31582.807 x 10 3 W

Since the intensity of pressure  is maximum at the inner radius (r 2 ), considering uniform wear, p max x r 2 = C or C = 0.2 × 600 = 120 N/mm

Axial thrust, W = 2 π C (r 1 – r 2 ) = 2π × 120  = 301592 N

Mean radius = R = (r 1 + r 2 )/2 = 800 mm = 0.8 m.

We know that torque transmitted,

T = n.µ.W.R = 2 × 0.25 × 301592 × 0.8 = 120637 N-m

Thus, power transmitted,

P = T.ω

ω = P/T = 261.8 rad/s

N = ω x 60 /  = 2500 r.p.m.

This set of Theory of Machines Multiple Choice Questions & Answers  focuses on “Conical Clutch”.

1. Identify the given clutch.

theory-machines-questions-answers-conical-clutch-q1

a) Single plate

b) Multi plate

c) Conical

d) Centrifugal

Answer: c

Explanation: The given diagram is that of a conical clutch. Conical clutch was used in automobiles earlier but nowadays single plate and multi plate clutches are used instead of it.

2. In a conical clutch, pn = ______________

a) W/(π(r 1 2 – r 2 2 ))

b) W/(π(r 1 2 + r 2 2 ))

c) W/(π(r 2 2 – r 1 2 ))

d) W/(π(r 2 2 x r 1 2 ))

Answer: a

Explanation: Pressure = Force/ Area.

Thus, here, intensity of pressure considering uniform pressure pn is W/(π(r 1 2 – r 2 2 )).

3. In a conical clutch, the formula for T is given by ______

a) n.µ.W.R

b) n.µ.W.r 1

c) n.µ.W.r 2

d) n.µ.W.(r 1 +r 2 )

Answer: a

Explanation: In a multi plate clutch, the formula for T is given by n.µ.W.R

It is same as that of single plate and multi plate clutch but the value of R is different for conical clutch.

4. In a conical clutch, considering uniform pressure, T = nµWR. What is R equal to?

a) {2(r 1 3 + r 2 3 ) / 3(r 1 2 + r 2 2 )} x cosec α

b) {2(r 1 3 – r 2 3 ) / 3(r 1 2 – r 2 2 )} x cosec α

c) {(r 1 – r 2 )/2} x cosec α

d) {(r 1 + r 2 )/2} x cosec α

Answer: b

Explanation: Considering uniform pressure theory in a single plate clutch, T = nµWR where R = {2(r 1 3 – r 2 3 ) / 3(r 1 2 – r 2 2 )} x cosec α

α = the semi cone angle.

5. In a conical clutch, considering uniform wear, T = nµWR. What is R equal to?

a) {2(r 1 3 + r 2 3 ) / 3(r 1 2 + r 2 2 )} x cosec α

b) {2(r 1 3 – r 2 3 ) / 3(r 1 2 – r 2 2 )} x cosec α

c) {(r 1 – r 2 )/2} x cosec α

d) {(r 1 + r 2 )/2} x cosec α

Answer: d

Explanation: Considering uniform pressure theory in a single plate clutch, T = nµWR where R = {(r 1 + r 2 )/2} x cosec α.

6. In a conical clutch, what is the axial force required for engaging the clutch ?

a) W n µ

b) W n µ

c) W n µ

d) W n µ

Answer: a

Explanation: W e = W + µ.W n cos α = W n sinα + µ.W n cosα = W n µ. If the semi cone angle of the clutch decreases, the torque produced by the clutch increases which in turn reduces the axial force W.

7. In a conical clutch, what is the axial force required for disengaging the clutch ?

a) W n µ

b) W n µ

c) W n µ

d) W n µ

Answer: b

Explanation: The axial force required for disengaging the clutch i.e. W d = W n µ. The axial force for disengagement is only required if the value of tan α is less than µ.

8. In a conical clutch, the breadth of the contact surface is given by ______

a) (r 1 + r 2 )/sinα

b) (r 2 -r 1 )/sinα

c) (r 1 – r 2 )/sinα

d) (r 1 2 + r 2 2 )/sinα

Answer: c

Explanation: In a conical clutch b sinα = r 1 -r 2

Thus, b = (r 1 -r 2 )/sinα

9. In a conical clutch, the mean radius of the bearing surface is 300 mm whereas the breadth is 20 mm. Find the inner and outer radii. The semi cone angle is 30°.

a) 145 mm, 155 mm

b) 140 mm, 160 mm

c) 160 mm, 140 mm

d) 155 mm, 145 mm

Answer: d

Explanation: In a conical clutch b sinα = r 1 – r 2

r 1 – r 2 = 10 and r 1 + r 2 = 300

r 1 = 155 mm and r 2 = 145 mm.

10. If the outer and inner radius of the contact surfaces are 100 mm and 75 mm respectively and the semi cone angle is 22.5°, find the value of the face width required.

a) 89.43 mm

b) 78.94 mm

c) 65.33 mm

d) 23.87 mm

Answer: c

Explanation: In a conical clutch b sinα = r 1 – r 2

b x sin 22.5° = 100 – 75

b = 65.33 mm.

11. A conical friction clutch is used to transmit 75 kW at 1500 r.p.m. The semi cone angle is 20° and the coefficient of friction is 0.3. If the mean diameter of the bearing surface is 500 mm and the intensity of normal pressure is not to exceed 0.1 N/mm 2 , find the dimensions of the conical bearing surface.

a) 118.07 mm, 131.93 mm

b) 131.93 mm, 118.07 mm

c) 121.72 mm, 128.28 mm

d) 128.28 mm, 121.72 mm

Answer: b

Explanation: Given : P = 75 kW = 75 × 10 3 W ; N = 1500 r.p.m. or ω = 2 π × 1400/60 = 157.08 rad/s ; α = 20° ; µ = 0.3 ; D = 500 mm or R = 250 mm ; pn = 0.1 N/mm 2

Power transmitted  = 75 × 10 3 = T.ω = T × 157.08

T = 75× 10 3 /157.08 = 477.4 N-m = 477.4× 10 3 N-mm

Torque transmitted  = 477.4 × 10 3 = 2 π µ p n .R 2 .b = 2π × 0.3 × 0.1  2 b = 11780.97 b

b = 40.52 mm

W e know that, r 1 +r 2 = 250 and r 1 -r 2 = 40.52 sin 20 = 13.86

Therefore, r 1 = 131.93 mm, r 2 = 118.07 mm.

12. A conical friction clutch is used to transmit 50 kW at 1000 r.p.m. The semi cone angle is 15° and the coefficient of friction is 0.2. If the mean diameter of the bearing surface is 450 mm and the intensity of normal pressure is not to exceed 0.15 N/mm 2 , find the axial spring force necessary to engage the clutch.

a) 4796 N

b) 4774 N

c) 4785 N

d) 4742 N

Answer: a

Explanation: Given : P = 50 kW = 50 × 10 3 W ; N = 1000 r.p.m. or ω = 2 π × 1000/60 = 104.72 rad/s ; α = 15° ; µ = 0.2 ; D = 450 mm or R = 225 mm ; pn = 0.15 N/mm 2

Power transmitted  = 50 × 10 3 = T.ω = T × 104.72

T = 50× 10 3 /104.72 = 477.4 N-m

477.4 = µ.W n .R = 0.2 × W n × 0.225

W n = 10610.33 N

Axial spring force = W e = W n µ = 10610.33  = 4796 N.

13. A conical friction clutch is used to transmit 40 kW at 1300 r.p.m. The semi cone angle is 12.5° and the coefficient of friction is 0.25. If the mean diameter of the bearing surface is 400 mm and the intensity of normal pressure is not to exceed 0.2 N/mm 2 , find the axial spring force necessary to disengage the clutch.

a) 154.28 N

b) 121.45 N

c) 201.78 N

d) 162.39 N

Answer: d

Explanation: Given : P = 40 kW = 40 × 10 3 W ; N = 1300 r.p.m. or ω = 2 π × 1300/60 = 136.13 rad/s ; α = 12.5° ; µ = 0.25 ; D = 400 mm or R = 200 mm ; pn = 0.2 N/mm 2

Power transmitted  = 40 × 10 3 = T.ω = T × 136.13

T = 40× 10 3 /136.13 = 293.82 N-m

293.82 = µ.W n .R = 0.25 × W n × 0.2

W n = 5876.5 N

Axial spring force = W d = W n µ = 5876.5 = 162.39 N.

This set of Theory of Machines Multiple Choice Questions & Answers  focuses on “Centrifugal Clutch”.

1. Which of the following clutches include shoes and spider inside the rim of the pulley?

a) Centrifugal clutch

b) Cone clutch

c) Multi plate clutch

d) Single plate clutch

Answer: a

Explanation: Centrifugal clutches are incorporated into motor pulleys which consist of a number of shoes on inside of the rim of the pulley. These clutches are generally incorporated into the motor pulleys. The outer surfaces of the shoe are covered with a layer of a friction material.

2. In a centrifugal clutch, what is ω?

a) Angular acceleration of the pulley

b) Angular running speed of the pulley

c) Angular acceleration at which the engagement begins to take place

d) Angular running speed at which the engagement takes place

Answer: b

Explanation: In a centrifugal clutch, ω is the angular running speed of the body. ω1 is the angular speed at which the engagement begins to take place.

3. What is the formula for the total frictional torque transmitted?

a) µ(P c – P s )R × n

b) µ(P c – P s )R

c)µ(P c + P s )R

d)µ(P c + P s )R x n

Answer: a

Explanation: Frictional torque acting on each shoe = µ(P c – P s )R. Therefore, total frictional torque transmitted = µ(P c – P s )R × n; where n is the number of shoes.

4. In a centrifugal clutch, total frictional torque transmitted = µ(P c – P s )R × n; where P c is the ____________________ acting on each shoe and is given by the formula P c = _______

a) centrifugal force, mω 1 2 r

b) inward force, mω 2 r

c) centrifugal force, mω 2 r

d) inward force, mω 1 2 r

Answer: c

Explanation: P c is the centrifugal force acting on each shoe at the running speed; P c = mω 2 r.

A little consideration will show that P c is greater than P s . The net outward radial force of the shoe is given as P c – P s .

5. In a centrifugal clutch, total frictional torque transmitted = µ(P c – P s )R × n; where P s is the ____________________ acting on each shoe and is given by the formula P s = _______

a) centrifugal force, mω 1 2 r

b) inward force, mω 2 r

c) centrifugal force, mω 2 r

d) inward force, mω 1 2 r

Answer: d

Explanation: P s is the inward force acting on each shoe at the running speed; P s = mω 1 2 r. A little consideration will show that P c is greater than P s . The net outward radial force of the shoe is given as P c – P s .

6. If the radial clearance  is specified and is not negligible, then what is the operating radius of the mass centre of the show from the axis of the clutch?

a) r 1 = r – c

b) r 1 = c – r

c) r 1 = r + c

d) r 1 = r x c

Answer: c

Explanation: If the radial clearance  is specified and is not negligible, then the operating radius of the mass centre of the show from the axis of the clutch is r 1 = r + c. So now, P c = mω2r 1 .

7. In a centrifugal clutch, if l and b are the contact length and the width of the shoes respectively and p is the intensity of pressure exerted in the shoe, l.b.p = P c – P s . True or false?

a) True

b) False

Answer: a

Explanation: l.b.p = Area x p; which is also equal to the net outward force with which the show presses against the rim at running speed. Thus, l.b.p = P c – P s .

8. If the inside radius of the pulley rim is equal to 150 mm and the angle subtended by shoes at the centre of the spider is 60°, find the net outward radial force of the centrifugal clutch. The intensity of pressure is equal to 0.1 N/mm 2 and the width is given to be 100 mm.

a) 1570.8 N

b) 1823.7 N

c) 1289.4 N

d) 1438.9 N

Answer: a

Explanation: θ = 60° = π/3 rad, R = 150 mm, b = 100 mm and p = 0.1 N/mm 2

l = θ R = 157.08 mm

Net outward radial force of the centrifugal clutch = l.b.p = P c – P s

P c – P s = 157.08 x 100 x 0.1 = 1570.8 N.

9. A centrifugal clutch is to transmit 25 kW at 1200 r.p.m. There are four shoes. The speed at which the engagement begins is 0.5 times the running speed. The inside radius of the pulley rim is 100 mm and the centre of gravity of the shoe lies at 75 mm from the centre of the spider. Coefficient of friction may be taken as 0.3. Determine mass of the shoes.

a) 1.955 kg

b) 1.866 kg

c) 2.272 kg

d) 2.139 kg

Answer: b

Explanation: Given : P = 25 kW = 25 × 10 3 W ; N = 1200 r.p.m. or ω = 2 π × 1200/60 = 125.66 rad/s ; n = 4 ; R = 100 mm = 0.1 m ; r = 75 mm = 0.075 m ; µ = 0.3 ; ω1 = 0.5 ω.

ω1 = 0.5 ω = 0.5 x 125.66 = 62.83 rad/s

Power transmitted = Tω = T x 125.66

T = 198.95 N-m

P c = mω 2 r = m x 125.66 2 x 0.075 = 1184.28 m N

P s = mω1 2 r = m x 62.83 2 x 0.075 = 296.07 m N

Frictional force acting tangentially on each shoe, F = µ(P c –P s ) = 0.3  = 266.46m N

Torque transmitted , 198.95 = n.F.R = 4 × 266.46 m × 0.1 = 106.584 m

m = 1.866 kg.

10. A centrifugal clutch is to transmit 22.5 kW at 1250 r.p.m. There are four shoes. The speed at which the engagement begins is 0.5 times the running speed. The inside radius of the pulley rim is 175 mm and the centre of gravity of the shoe lies at 120 mm from the centre of the spider. Coefficient of friction may be taken as 0.3. Determine the width of the shoe if the angle subtended by the shoes at the centre of the spider is 60° and the intensity of pressure exerted on the shoes is 0.05 N/mm 2 .

a) 104.493 mm

b) 89.198 mm

c) 52. 837 mm

d) 113.983 mm

Answer: b

Explanation: Given : P = 22.5 kW = 22.5 × 103 W ; N = 1250 r.p.m. or ω = 2 π × 1250/60 = 130.9 rad/s ; n = 4 ; R = 175 mm = 0.175 m ; r = 120 mm = 0.12 m ; µ = 0.3 ; ω1 = 0.5ω ; θ = 60° = π/3 rad ; p = 0.05 N/mm 2

ω1 = 0.5ω = 0.5 x 130.9 = 65.45 rad/s

Power transmitted = Tω = T x 130.9

T = 171.887 N-m

P c = mω 2 r = m x 130.9 2 x 0.12 = 2056.17 m N

P s = mω 1 2 r = m x 65.45 2 x 0.12 = 514.04 m N

Frictional force acting tangentially on each shoe, F = µ(P c – P s ) = 0.3  = 462.64 m N

Torque transmitted , 171.887 = n.F.R = 4 × 462.64 m × 0.175 = 323.847 m

m = 0.53 kg

l = θ x R = π/3 x 175 = 183.26 mm

l.b.p = P c – P s = 1542.13 m

183.26 x b x 0.05 = 1542.13 x 0.53

b = 89.198 mm.

11. A centrifugal clutch has four shoes. When the clutch is at rest, each shoe is pulled against a stop by a spring so that it leaves a radial clearance of 10 mm between the shoe and the rim. The pull exerted by the spring is then 700 N. The mass centre of the shoe is 150 mm from the axis of the clutch. If the internal diameter of the rim is 500 mm, the mass of each shoe is 6 kg, the stiffness of each spring is 75 N/mm and the coefficient of friction between the shoe and the rim is 0.35; find the power transmitted by the clutch at 400 r.p.m.

a) 3.439 kW

b) 34.39 kW

c) 3439 kW

d) 3.439 W

Answer: a

Explanation: Given : n = 4 ; c = 10 mm ; S = 700 N ; r = 150 mm ; D = 500 mm or R = 250 mm = 0.25 m ; m = 6 kg ; s = 75 N/mm ; µ = 0.35 ; N = 400 r.p.m. or ω = 2 π × 400/60 = 41.89 rad/s

r 1 = r + c = 150 + 10 = 160 mm = 0.16 m

P c = m.ω2.r 1 = 6 x  2 × 0.16 = 1684.58 N

P s = S + c.s = 700 + 10 x 75 = 1450 N

F = µ (P¬c – P s ) = 0.35  = 82.1 N

T = n.F.R = 4 × 82.1 × 0.25 = 82.1 N-m

Power transmitted  = Tω = 82.1 x 41.89 = 3439.31 W = 3.439 kW.

12. A centrifugal clutch has four shoes. When the clutch is at rest, each shoe is pulled against a stop by a spring so that it leaves a radial clearance of 7.5 mm between the shoe and the rim. The pull exerted by the spring is then 600 N. The mass centre of the shoe is 120 mm from the axis of the clutch. If the internal diameter of the rim is 350 mm, the mass of each shoe is 5 kg, the stiffness of each spring is 60 N/mm and the coefficient of friction between the shoe and the rim is 0.2; find the speed of the centrifugal clutch, if the power transmitted is 40.168 kW.

a) 83.779 r.p.m.

b) 800 r.p.m.

c) 104.66 r.p.m.

d) 1000 r.p.m.

Answer: b

Explanation: Given : n = 4 ; c = 7.5 mm ; S = 600 N ; r = 120 mm ; D = 350 mm or R = 175 mm = 0.175 m ; m = 5 kg ; s = 60 N/mm ; µ = 0.2; P = 40.168 kW = 40.168 x 103 W

r 1 = r + c = 120 + 7.5 = 127.5 mm = 0.1275 m

P c = m.ω2.r 1 = 5 x  2 × 0.1275 = 4474.67 N

P s = S + c.s = 600 + 7.5 x 60 = 1050 N

F = µ (P c – P s ) = 0.2  = 684.93 N

T = n.F.R = 4 × 684.93 × 0.175 = 479.45 N-m

Power transmitted  = Tω = 479.45 x ω

40.168 x 10 3 W = 479.45 x ω

ω = 83.779 rad/s

Therefore, N = 60 x ω /  = 800 r.p.m.

13. A centrifugal clutch is to transmit 35 kW at 1500 r.p.m. There are four shoes. The speed at which the engagement begins is 0.7 times the running speed. The inside radius of the pulley rim is 120 mm and the centre of gravity of the shoe lies at 80 mm from the centre of the spider. Coefficient of friction may be taken as 0.35. Determine intensity of pressure exerted on the shoes if the angle subtended by the shoes at the centre of the spider is 60° and the width of the shoes is 105.52 mm.

a) 109.32 mm

b) 186.45 mm

c) 150.57 mm

d) 105.52 mm

Answer: d

Explanation: Given : P = 35 kW = 35 × 103 W ; N = 1500 r.p.m. or ω = 2 π × 1500/60 = 157.08 rad/s ; n = 4 ; R = 120 mm = 0.12 m ; r = 80 mm = 0.08 m ; µ = 0.35 ; ω1 = 0.7 ω ; θ = 60° = π/3 rad ; b = 105.52 mm. ω1 = 0.7 ω = 0.75 x 157.08 = 109.95 rad/s

Power transmitted = Tω = T x 104.72

T = 222.816 N-m

P c = mω 2 r = m x 157.08 2 x 0.08 = 1973.93 m N

P s = mω 1 2 r = m x 109.95 2 x 0.08 = 967.12 m N

Frictional force acting tangentially on each shoe, F = µ(P c – P s ) = 0.35  = 352.38 m N

Torque transmitted , 190.98 = n.F.R = 4 × 352.38 m × 0.12 = 169.144 m

m = 1.317 kg

l = θ x R = π/3 x 120 = 125.66 mm

l.b.p = P c – P s = 1006.81 m

125.66 x 105.52 x p = 1006.81 x 1.317

p = 0.1 N/mm 2 .

This set of Theory of Machines Multiple Choice Questions & Answers  focuses on “Gear Terminology”.

1. The smaller and generally the driving gear of a pair of mated gears is called _________

a) pitch

b) module

c) rack

d) pinion

Answer: d

Explanation: Pinion is the smaller and the driving gear of a pair of mated gears. It is used in gear train drives and is usually the smaller gear. The pinion engages the larger gear or rack that helps in the turning of the wheels of the vehicle.

2. The distance measured along the circumference of the pitch circle from a point on one tooth to the same point on the adjacent tooth is called ________________

a) pitch diameter

b) circular pitch

c) pitch point

d) line of centres

Answer: b

Explanation: Circular pitch is the distance measured along the circumference of the pitch circle from a point on one tooth to the same point on the adjacent tooth. So basically, it is the distance between corresponding points of consecutive gear teeth.

p = d π/ N; where, p is the circular pitch, d is the standard pitch diameter and N is the number of teeth.

3. The number of teeth per unit length of the pitch circle diameter is called ________________

a) diametral pitch

b) module

c) clearance

d) gear ratio

Answer: a

Explanation: Diametral pitch is the number of teeth per unit length of the pitch circle diameter.

It is used to classify different types and sizes of gears. It is a function of the pitch circle.

P = T/d.

4. The ratio of the number of teeth on the gear to the number of teeth on the pinion is called ______

a) velocity ratio

b) module

c) gear ratio

d) pitch point

Answer: c

Explanation: Gear ratio is given by T/t and is defined as the ratio of the number of teeth on the gear to the number of teeth on the pinion. Velocity ratio is the inverse of gear ratio.

Gear ratio is inversely proportional to the radius of the pitch circle and also the number of teeth on the driving gear.

5. The ratio of pitch diameter to the number of teeth is called _________

a) module

b) addendum

c) dedendum

d) backlash

Answer: a

Explanation: Module is the ratio of pitch diameter to the number of teeth.

Module indicates the size of the gear and classifies whether the gear is big or small as module is a unit of size.

m = d/T or m = p/π.

6. The circle passing through the upper tips of the teeth is called _________________

a) pitch circle

b) circular pitch

c) dedendum circle

d) addendum circle

Answer: d

Explanation: Addendum circle is the circle passing through the upper tips of the teeth. The addendum circle lies on the outside cylinder for external gears whereas the addendum circle lies on the internal cylinder for internal gears. The diameter for the addendum circle of an internal gear is called as inside diameter.

6. The radial height of a tooth above the pitch circle is called ____________

a) addendum

b) dedendum

c) rack

d) line of centres

Answer: a

Explanation: Addendum is the radial height of a tooth above the pitch circle. Basically, it is the radial distance between pitch diameter and diameter of the outside circle. Addendum angle is the angle between the face cone and the pitch cone.

7. The circle passing through the roots of the teeth is called ________________

a) addendum

b) addendum circle

c) dedendum circle

d) rack

Answer: c

Explanation: Dedendum circle is the circle passing through the roots of the teeth. It touches the bottom of the spaces between the teeth of a gear.

8. The radial depth of a tooth below the pitch circle is called _______________

a) dedendum

b) addendum

c) rack

d) line of centres

Answer: a

Explanation: Dedendum is the radial depth of a tooth below the pitch circle. It is the radial distance between the pitch circle and the root circle. Dedendum angle is the angle between the root cone and the pitch cone.

9. The radial difference between the addendum and the dedendum of the tooth is called _________

a) clearance

b) rack

c) line of action

d) line of centres

Answer: a

Explanation: Clearance is the radial difference between addendum and dedendum of the tooth. Clearance circle is a tangent to the addendum circle on the gear. It has a clearance which equal to the distance between the dedendum of the gear and the addendum of the mating gear.

10. The locus of the point of contact on two mating teeth from beginning of engagement to end of engagement is known as the _______________

a) angle of action

b) arc of contact

c) path of contact

d) line of action

Answer: c

Explanation: Path of contact is the locus of the point of contact on two mating teeth from beginning of engagement to end of engagement. It is also called as contact length. It is the tangent to both the base circles and passes through the pitch point.

11. The portion of path of contact from the beginning of contact to the pitch point is called _______

a) arc of recess

b) arc of approach

c) path of recess

d) path of approach

Answer: d

Explanation: Path of approach is the portion of path of contact from the beginning of contact to the pitch point. It is given by the formula 0.5 – Rsinφ; where Ra is the radius of addendum of the larger gear, R is the radius of the larger gear and φ is the pressure angle between the two gears.

12. The portion of the path of contact from the pitch point to the end of contact is called _________

a) arc of recess

b) arc of approach

c) path of recess

d) path of approach

Answer: c

Explanation: Path of recess is the portion of the path of contact from the pitch point to the end of contact. It is given by the formula (r a 2 – r 2 cos 2 φ) 0.5 – rsinφ; where r a is the radius of addendum of the smaller gear, r is the radius of the smaller gear and φ is the pressure angle between the two gears. Path of approach + Path of recess = Path of contact.

13. Arc of contact = ___________________ Complete the equation.

a) Path of contact / cos θ

b) Path of contact / sin θ

c) Path of contact x cos θ

d) Path of contact x sin θ

Answer: a

Explanation: Arc of contact = Path of contact / cos θ; θ = the pressure angle between the two gears.

14. The locus of a point on the pitch circle from the beginning to end of engagement of two mating gears is called as ___________

a) Arc of contact

b) Path of contact

c) Path of approach

d) Arc of approach

Answer: a

Explanation: Arc of contact is defined as the locus of a point on the pitch circle from beginning to end of engagement of two mating gears. It is subdivided into two sections namely arc of approach and arc of recess. Arc of approach + arc of recess = arc of contact.

15. The ratio of angle of action to the pitch angle is called ___________

a) space width

b) angle of recess

c) angle of approach

d) contact ratio

Answer: d

Explanation: Contact ratio is the ratio of angle of action to the pitch angle. It is also defined as the number of pairs of teeth in contact. Contact ratio = arc of contact / circular pitch.

16. The angle turned by the gear from the beginning to the end of engagement of a pair of teeth is called ______________

a) angle of approach

b) angle of recess

c) angle of action

d) angle of contact

Answer: d

Explanation: Angle of action is the angle turned by the gear from the beginning to the end of engagement of a pair of teeth. Angle of action = Arc of contact / r; where r is the radius of the pinion.

This set of Theory of Machines Multiple Choice Questions & Answers  focuses on “Path of Contact”.

1. In the given diagram identify the path of contact.

theory-machines-questions-answers-path-contact-q1

a) CD

b) AB

c) FE

d) BP

Answer: a

Explanation: Path of contact is the locus of the point of contact on two mating teeth from the beginning of engagement to the end of it. Thus CD is the path of contact. It is also called as contact length. It is the tangent to both the base circles and passes through the pitch point.

2. The formula to calculate path of contact is ______________________

a) (R a 2 + R 2 cos 2 φ) 0.5 + (r a 2 + r 2 cos 2 φ) 0.5 – sin φ

b) (R a 2 – R 2 cos 2 φ) 0.5 + (r a 2 – r 2 cos 2 φ) 0.5 + sin φ

c) (R a 2 – R 2 cos 2 φ) 0.5 + (r a 2 – r 2 cos 2 φ) 0.5 – sin φ

d) (R a 2 – R 2 cos 2 φ) 0.5 + (r a 2 – r 2 cos 2 φ) 0.5 – sin φ

Answer: c

Explanation:

theory-machines-questions-answers-path-contact-q1

From the diagram of two gears in a mesh, we can clearly see that, CD = CP + PD =  +  =(R a 2 – R 2 cos 2 φ) 0.5 + (r a 2 – r 2 cos 2 φ) 0.5 – sin φ.

3. The condition which must be fulfilled by two gear tooth profiles to maintain a constant angular velocity ratio between them is called __________________

a) arc of contact

b) path of contact

c) law of gearing

d) interference

Answer: c

Explanation: Law of gearing is the condition which must be fulfilled by two gear tooth profiles to maintain a constant angular velocity ratio between them. In order to maintain this constant angular velocity, the common normal of the tooth profiles should always pass through a fixed point on the line of centres, called fixed point.

4. Find the path of contact if r = 20 mm, r a = 25 mm, R = 50 mm, R a = 55 mm and φ = 20°.

a) 11.836 mm

b) 21.135 mm

c) 69.018 mm

d) 36.046 mm

Answer: b

Explanation: Path of contact = (R a 2 – R 2 cos 2 φ) 0.5 + (r a 2 – r 2 cos 2 φ) 0.5 – sin φ = 21.135 mm.

5. What is the formula for the path of approach?

a) (R a 2 – R 2 cos 2 φ) 0.5 – Rsinφ

b) (R a 2 + R 2 cos 2 φ) 0.5 + Rcosφ

c) (R a 2 + R 2 cos 2 φ) 0.5 + Rsinφ

d) (R a 2 + R 2 cos 2 φ) 0.5 – Rcosφ

Answer: a

Explanation: The path of approach is given by the formula (R a 2 + R 2 cos 2 φ) 0.5 – Rsinφ; where R a = The addendum radius of the larger gear, R = The radius of the larger gear and φ is the pressure angle. Path of approach is the portion of path of contact from the beginning of contact to the pitch point. Similarly, the formula for the path of recess is (R a 2 + R 2 cos 2 φ) 0.5 – rsinφ.

6. To avoid interference what is the maximum length of the path of approach?

a) Rsinφ

b) rsinφ

c) Rcosφ

d) rcosφ

Answer: b

Explanation: The maximum length of the path of approach is equal to rsinφ. Similarly, the maximum length of the path of recess is equal to Rsinφ; where r is the radius of the smaller gear, R is the radius of the larger gear and φ is the pressure angle between these two gears.

7. Two gears in a mesh have 45 teeth each and the module is 5 mm. The pressure angle is given to be φ = 15°. The addendum is equal to 1 module. Find the path of contact.

a) 31.159 mm

b) 48.201 mm

c) 34. 356 mm

d) 42.543 mm

Answer: a

Explanation: φ = 15°, t = T = 45 and m = 5 mm. Addendum = 1 module.

R = r = mT/2 = 112.5 mm and R a = r a = 112.5 + 5 = 117.5 mm

Path of contact = (R a 2 – R 2 cos 2 φ) 0.5 + (R a 2 – R 2 cos 2 φ) 0.5 – sin φ= 31.159 mm.

8. The pressure angle of two gears in a mesh is φ = 20°. The gear ratio is 3 and the number of teeth on the pinion is 20. The module is given to be 8 mm. The addendum is one module. Find the path of contact of these two gears.

a) 21.324 mm

b) 12.543 mm

c) 56.343 mm

d) 39.458 mm

Answer: d

Explanation: φ = 20°, t = 20. Hence, T = 20 x 3 = 60. Addendum = 1 module

r = mt/2 = 80 mm, R = mT/2 = 240 mm

r a = r + a = 88 mm, R a = r + a = 248 mm

Path of contact = (R a 2 – R 2 cos 2 φ) 0.5 + (R a 2 – R 2 cos 2 φ) 0.5 – sin φ = 39.458 mm.

9. The pressure angle of two gears in a mesh is φ = 25°. The number of teeth on the pinion is 25 and the number of teeth on the gear is 70. The module is given to be 10 mm. The addendum is 1.1 times the module. Find the path of approach and path of contact of these gears.

a) 24.421 mm, 22.417 mm

b) 22.417 mm, 24.421 mm

c) 26.407 mm, 20.431 mm

d) 20.431 mm, 26.407 mm

Answer: a

Explanation: φ = 25°, t = 25, T = 70, m = 10 mm and addendum = 1.1 module = 11 mm.

r = mt/2 = 125 mm, R = mT/2 = 350 mm

r a = r + a = 136 mm, R a = R + a = 361 mm

Path of approach = (R a 2 – R 2 cos 2 φ) 0.5 – Rsinφ = 24.421 mm

Path of recess = (r a 2 – R 2 cos 2 φ) 0.5 – rsinφ = 22.417 mm.

10. The pressure angle of two gears in a mesh is φ = 15°. The number of teeth on the larger gear is 50. The module is 4 mm. The addendum is equal to be 1.25 module. Find the path of approach of these gears.

a) 39.211 mm

b) 15.287 mm

c) 31.092 mm

d) 21.122 mm

Answer: b

Explanation: φ = 15°, T = 50, m = 4 mm and addendum = 1.25 x module = 5 mm

R = mT/2 = 100 mm, R a = R + a = 105 mm

Path of approach = (R a 2 – R 2 cos 2 φ) 0.5 – Rsinφ = 15.287 mm.

11. The pressure angle of two gears in a mesh is φ = 35°. The number of teeth on the pinion is 20. The module is 10 mm and the addendum is 1 module. Find the path of recess of these gears.

a) 34.213 mm

b) 23.328 mm

c) 16.058 mm

d) 32.091 mm

Answer: c

Explanation: φ = 35°, t = 20, m = 10 mm and addendum = 1 module = 10 mm.

r = mt/2 = 100 mm and r a = r + a = 110 mm

Path of recess = (r a 2 – r 2 cos 2 φ) 0.5 – rsinφ= 16.058 mm.

This set of Theory of Machines Multiple Choice Questions & Answers  focuses on “Velocity of Sliding”.

1. Velocity of sliding at the beginning of engagement of the two gears = (ω p + ω g ) x ______________

a) path of contact

b) arc of contact

c) arc of approach

d) path of approach

Answer: d

Explanation: Velocity of sliding at the beginning of engagement = (ω p + ω g ) x path of approach. Similarly, the velocity of sliding at the end of engagement = (ω p + ω g ) x Path of recess.

2. Velocity of sliding at the pitch point = (ω p + ω g ) x _____________

a) path of contact

b) 0

c) arc of approach

d) path of recess

Answer: b

Explanation: At the pitch point, there is no relative motion between the gears. Thus, the velocity of sliding is 0. Velocity of sliding at the pitch point = (ω p + ω g ) x 0 = 0.

3. The angular velocity of the pinion is 25 rad/s and that of the gear is 10 rad/s. The path of approach is equal to 15 mm. Find the velocity of sliding at the beginning of contact.

a) 394 mm/s

b) 525 mm/s

c) 134 mm/s

d) 348 mm/s

Answer: b

Explanation: ω p = 25 rad/s and ω g = 10 rad/s. Path of approach = 15 mm.

Velocity of sliding = (ω p + ω g ) x Path of approach =  x 15 = 525 mm/s.

4. The angular velocity of the pinion is 40 rad/s and that of the gear is 12 rad/s. The path of recess is equal to 16 mm. Find the velocity of sliding at the end of the contact.

a) 832 mm/s

b) 213 mm/s

c) 1029 mm/s

d) 720 mm/s

Answer: a

Explanation: ω p = 40 rad/s and ω g = 12 rad/s. Path of recess = 16 mm.

Velocity of sliding = (ω p + ω g ) x Path of recess =  x 16 = 832 mm/s.

5. The angular velocity of the pinion is 100 rpm and that of the gear is 30 rpm. The path of recess is equal to 50 mm and the path of contact = 100 mm. Find the velocity of sliding at the beginning of the contact.

a) 1033.29 mm/s

b) 902.93 mm/s

c) 394.02 mm/s

d) 680.58 mm/s

Answer: d

Explanation: N p = 100 rpm. Therefore, ω p = 10.47 rad/s

N g = 30 rpm. Therefore, ω g = 3.14 rad/s

Path of approach = Path of contact – Path of recess = 100 mm – 50 mm = 50 mm.

Velocity of sliding = (ω p + ω g ) x Path of approach =  x 50 = 680.58 mm/s.

6. The angular velocity of the pinion is 200 rpm and that of the gear is 50 rpm. The maximum addendum radius of the larger wheel is 85 mm and the radius of the gear is equal to 75 mm. The pressure angle between the gears is equal to 30°. Find the velocity of sliding at the beginning of the contact.

a) 1289.129 mm/s

b) 732.324 mm/s

c) 453.681 mm/s

d) 239.238 mm/s

Answer: c

Explanation: N p = 200 rpm. Therefore, ω p = 20.944 rad/s

N g = 50 rpm. Therefore, ω g = 5.236 rad/s

R a = 85 mm and R = 75 mm; φ = 30°

Path of approach = (R a 2 – R 2 cos 2 φ) 0.5 – Rsinφ = 17.33 mm

Velocity of sliding = (ω p + ω g ) x Path of approach =  x 17.33 = 453.68 mm/s.

7. The angular velocity of the pinion is 325 rpm and that of the gear is 155 rpm. The number of teeth on the larger wheel is 60 and the module is equal to 7 mm. Addendum is 1.1 times the module. The pressure angle is 20°. Find the velocity of sliding at the beginning of the contact.

a) 1010.858 mm/s

b) 1317.858 mm/s

c) 1217.858 mm/s

d) 1117.858 mm/s

Answer: a

Explanation: N p = 325 rpm. Therefore, ω p = 34.034 rad/s

N g = 155 rpm. Therefore, ω g = 16.231 rad/s

m = 7 mm, T = 60 mm and R = mT/2 = 210 mm

R a = R + a = 210 + 7.7 = 217.7 mm

Path of approach= (R a 2 – R 2 cos 2 φ) 0.5 – Rsinφ = 20.11 mm

Velocity of sliding = (ω p + ω g ) x Path of recess =  x 20.11 = 1010.858 mm/s.

8. The angular velocity of the pinion is 75 rpm and that of the gear is 25 rpm. The addendum radius is 45 mm for the pinion and the radius of the pinion is 40 mm. The pressure angle between the gears is 25°. Find the velocity of sliding at the end of the contact.

a) 89.92 mm/s

b) 91.01 mm/s

c) 94.29 mm/s

d) 97.77 mm/s

Answer: d

Explanation: N p = 75 rpm. Therefore, ω p = 7.854 rad/s

N g = 25 rpm. Therefore, ω g = 2.168 rad/s

r = 40 mm, ra = 45 mm, φ = 25°

Path of recess = (r a 2 – r 2 cos 2 φ) 0.5 – rsinφ = 9.755 mm

Velocity of sliding = (ω p + ω g ) x Path of recess =  x 9.755 = 97.77 mm/s.

9. The angular velocity of the pinion is 150 rpm and that of the gear is 75 rpm. The number of teeth on the pinion is equal to 20 and the module = 6 mm. The value of addendum is equal to 1 module. The pressure angle between the two gears is 20°. Find the velocity of sliding at the end of the contact.

a) 591.29 mm/s

b) 601.12 mm/s

c) 324.87mm/s

d) 512.02 mm/s

Answer: c

Explanation: N p = 150 rpm. Therefore, ω p = 15.708 rad/s

N g = 75 rpm. Therefore, ω g = 7.854 rad/s

t = 20 and m = 6 mm.

r = mt/2 = 60 mm and ra = r + a = 66 mm; φ = 20°

Path of recess = (r a 2 – r 2 cos 2 φ) 0.5 – rsinφ = 13.79 mm

Velocity of sliding = (ω p + ω g ) x Path of recess =  x 13.79 = 324.87 mm/s.

10. What is the formula for rolling velocity?

a) ω p x r

b) ω g x R

c) 0

d) ω g x r

Answer: a

Explanation: We know that v = r. ω. The rolling velocity is also represented by the same formula ω p x r; where ω p is the angular velocity of the pinion and r is the radius of the smaller gear. Rolling velocity is also called as pitch line velocity.

11. Two involute gears in a mesh have path of approach = 29.83 mm and path of recess = 21.09 mm. The larger gear has 40 teeth and the pinion has 12 teeth. Find the ratio of the sliding to rolling velocity at the:

i) beginning of contact

ii) end of contact

iii) pitch point

r = 72 mm

a) 0.38, 0.54, 0

b) 0.54, 0.38, 0

c) 0.24, 0.63, 0

d) 0.63, 0.24, 0

Answer: b

Explanation: i) Ratio of the sliding to rolling velocity = ((ω p + ω g ) x Path of approach)/ (ω p x r) = (ω p + ω p ) x 29.83/ω p x 72 = 0.54

ii) Ratio of the sliding to rolling velocity = ((ω p + ω g ) x Path of recess)/ ω p x r = (ω p + ω p ) x 21.09/(ω p x 72) = 0.38

iii) Ratio of the sliding to rolling velocity = ((ω p + ω g ) x 0)/ ω p x r = 0.

This set of Theory of Machines Multiple Choice Questions & Answers  focuses on “Arc of Contact and Contact Ratio”.

1. The distance travelled by a point on either pitch circle of the two wheels during the period of contact of a pair of teeth is called ____________________

a) path of contact

b) arc of contact

c) contact ratio

d) angle of action

Answer: b

Explanation: The arc of contact is defined as the distance travelled by a point on either pitch circle of the two wheels during the period of contact of a pair of teeth. Simply, it is traced out along the pitch circle while one pair of the teeth is in contact.

2. The contact ratio is the ratio of arc of contact to the ____________

a) circular pitch

b) dedendum

c) circular pitch

d) module

Answer: c

Explanation: Contact ratio is the of the arc od contact to the circular pitch. The number of teeth in contact = arc of contact / circular pitch = path of contact / .

3. If the contact ratio is 2.7, two pairs of teeth are always in contact and three pairs of teeth are in contact for _________ % of the time.

a) 100

b) 50

c) 45

d) 70

Answer: d

Explanation: If the contact ratio is 2.7, two pairs of teeth are always in contact and three pairs of teeth are in contact for 70 % of the time.

4. For continuous transmission of motion, the contact ratio i.e. n must always be _______

a) equal to 1

b) equal to 0

c) more than unity

d) less than unity

Answer: c

Explanation: For a continuous transmission of motion, at least one tooth of the wheel must be in contact with another tooth of the second wheel. Hence, n must be greater than unity. If the contact ratio is less than unity, then it implies that the even a single tooth of the two gears are not in contact with each other.

5. Two involute gears have path of contact 40 mm and the pressure angle is 50°. Find the arc of contact.

a) 62.23 mm

b) 25.71 mm

c) 30.64 mm

d) 52.21 mm

Answer: a

Explanation: Arc of contact = Path of contact/cos φ = 40/cos 50° = 62.23 mm.

6. Two gears in a mesh have the arc of contact = 27 mm. The pressure angle is 15°. Find the path of contact.

a) 104.32 mm

b) 6.98 mm

c) 27.95 mm

d) 26.08 mm

Answer: d

Explanation: Path of contact = Arc of contact x cos φ = 27 x cos 15° = 26.08 mm.

7. Two involute gears in a mesh have a circular pitch of 25 mm. The arc of contact is 55 mm. Find the number of pairs of teeth in contact.

a) 2.2

b) 0.45

c) 13.75

d) 1.3

Answer: a

Explanation: Contact ratio = Arc of contact/ Circular pitch = 55/25 = 2.2.

8. Two involute gear have a module of 5 mm, the arc of contact is 25 mm. Find the contact ratio of these two gears.

a) 5

b) 1.59

c) 0.2

d) 3

Answer: b

Explanation: Contact ratio = Arc of contact/ Circular pitch = Arc of contact / πm = 25/ = 1.59.

9. The path of contact of two gears in a mesh is 50 mm and the pressure angle is 45°. The module is 8 mm. Find the contact ratio of these two gears.

a) 4.2

b) 1.9

c) 3.7

d) 2.8

Answer: d

Explanation: Arc of contact = Path of contact/cos φ = 50/cos 45° = 70.71 mm

Contact ratio = Arc of contact/ Circular pitch = Arc of contact / πm = 70.71/ = 2.8.

10. The path of approach = 13.92 mm and path of recess = 11.56 mm. The pressure angle is 22.5°. The module is 4 mm. Find the contact ratio of these two gears.

a) 2.19

b) 3.54

c) 3.12

d) 2.53

Answer: a

Explanation: Path of contact = Path of approach + Path of recess = 13.92 + 11.56 = 25.48 mm

Arc of contact = Path of contact/cos φ = 25.48/cos 22.5° = 27.58 mm

Contact ratio = Arc of contact/ Circular pitch = Arc of contact / πm = 27.58/ = 2.19.

11. Two involute gears in a mesh have a module of 10 mm and the pressure angle is 35°. The larger gear has 45 teeth whereas the pinion has 15 teeth. The addendum is equal to one module. Find the arc of contact.

a) 36. 832 mm

b) 32.460 mm

c) 39.626 mm

d) 27.239 mm

Answer: c

Explanation: φ = 35°, t = 15, T = 45, m = 10 mm and addendum = 1 module = 10 mm.

r = mt/2 = 75 mm, R = mT/2 = 225 mm

r a = r + a = 85 mm, R a = R + a = 235 mm

Path of contact = (R a 2 – R 2 cos 2 φ) 0.5 + (r a 2 – r 2 cos 2 φ) 0.5 – sinφ = 32.46 mm

Arc of contact = Path of contact/cos φ = 32.46/cos 35° = 39.626 mm.

12. The pressure angle of two gears in a mesh is φ = 22.5°. The number of teeth on the pinion is 25 and the gear ratio is 2. The module is 7 mm and addendum = 1 module. Find the angle of action of these two gears.

a) 32.056°

b) 34.697°

c) 22.72°

d) 28.531°

Answer: c

Explanation: φ = 22.5°, t = 25 and T = 25 x 2 = 50, m = 7 mm and addendum = 7 mm.

r = mt/2 = 87.5 mm and r a = r + a = 94.5 mm

R = mT/2 = 175 mm and R a = R + a = 182 mm

Path of contact = (R a 2 – R 2 cos 2 φ) 0.5 + (r a 2 – r 2 cos 2 φ) 0.5 – sinφ = 32.056 mm

Arc of contact = Path of contact/cos φ = 32.056/cos 22.5° = 34.697 mm.

Angle of action = Arc of contact / r = 0.3965 rad = 22.72°.

13. The pressure angle of two gears in a mesh is φ = 25°. The number of teeth on the pinion is 45 and the gear ratio is 2. The module is 6 mm and addendum = 1.1 module. Find the contact ratio of these two gears.

a) 1.7

b) 2.3

c) 4.2

d) 3.5

Answer: a

Explanation: φ = 25°, t = 45 and T = 45 x 2 = 90, m = 6 mm and addendum = 6.6 mm.

r = mt/2 = 135 mm and r a = r + a = 141.6 mm

R = mT/2 = 270 mm and R a = R + a = 276.6 mm

Path of contact = (R a 2 – R 2 cos 2 φ) 0.5 + (r a 2 – r 2 cos 2 φ) 0.5 – sinφ = 29.068 mm

Arc of contact = Path of contact/cos φ = 29.068/cos 25° = 32.073 mm.

Contact ratio = Arc of contact / Circular pitch = 32.073 / πm = 32.073 /  = 1.7.

This set of Theory of Machines Multiple Choice Questions & Answers  focuses on “Minimum Number of Teeth and Interference”.

1. In the given diagram, identify the maximum value of addendum radius of the larger gear or the wheel to avoid interference.

theory-machines-questions-answers-minimum-number-teeth-interference-q1

a) AD

b) BE

c) BC

d) BA

Answer: BE

Explanation: BE is the maximum value of addendum radius of the wheel or the larger gear to avoid interference of the gears. It is given by R a max. When two gears are in mesh at one instant there is a chance to mate involute portion with non involute portion of mating gear. This phenomenon is described as interference. In case of a pinion, AF is the maximum value of the addendum radius of the pinion.

2. The maximum radius of addendum to avoid interference is given by the formula ____________

a)  2 +  2 ) 0.5

b)  2 +  2 ) 0.5

c)  2 –  2 ) 0.5

d)  2 –  2 ) 0.5

Answer: a

Explanation:

theory-machines-questions-answers-minimum-number-teeth-interference-q1

We know that the maximum value of addendum radius is equal to BE. From the given diagram it is evident that BE =  2 +  2 ) 0.5 =  2 +  2 ) 0.5 . If the actual value of the addendum radius is less than the maximum value of the addendum radius, then interference does not occur.

3. What is the formula for calculating the minimum number of teeth on the wheel ?

a) T = 2a w /+2)sin 2 φ) 0.5 – 1)

b) T = a w /+2)sin 2 φ) 0.5 – 1)

c) T = 2a w /+2)sin 2 φ) 0.5

d) T = 2a w /+2)sin 2 φ) 0.5 + 1)

Answer: a

Explanation: T ≥ 2a w /+2)sin 2 φ) 0.5 – 1)

This implies that the minimum teeth must always be greater than the value obtained using this formula.

In the limit, T = (2a w /+2)sin 2 φ) 0.5 – 1). The minimum number of teeth on the pinion is given by t = T/G.

4. What is the maximum value of the addendum of the pinion?

a) mtsin 2 φ) 0.5 +1)/2

b) mtsin 2 φ) 0.5 -1)/2

c) mtsin 2 φ) 0.5 +1)/2

d) mtsin 2 φ) 0.5 -1)/2

Answer: d

Explanation: Since, a w max = mT+2)sin 2 φ) 0.5 – 1)/2 is the maximum value of addendum for the wheel, similarly, for pinion, the maximum value of addendum is given by mtsin 2 φ) 0.5 -1)/2. If the maximum value of addendum is less than the actual value of addendum, then interference occurs.

5. If the maximum addendum radius of the wheel is given as 260 mm and the actual value of the addendum radius is found out to be 255 mm, the interference will occur. True or false?

a) True

b) False

Answer: b

Explanation: As the value of Ra obtained is less than the value of the maximum addendum radius, the interference does not occur. It only occurs if the value of the addendum radius obtained is more than the maximum addendum radius. Thus, the given statement is false.

6. Two involute gears have a pressure angle of 20°. The gear ratio is given to be 3. The module is 5 mm and the value of addendum is equal to 1 module. Determine the minimum number of teeth on each wheel to avoid interference, if the pinion rotates at 100 rpm.

a) T = 51, t= 17

b) T = 45, t =15

c) T = 60, t =20

d) T = 54, t = 18

Answer: b

Explanation: Given: φ = 20°, G = T/t = 3, m = 5 mm and addendum = 1 module.

Now, T = (2a w /+2)sin 2 φ) 0.5 – 1) is the minimum number of teeth on the larger gear.

T = /+2)sin 2 ) 0.5 – 1) = 44.94 ~ 45.

Therefore, T = 45 and t = 45/3 = 15.

7. Find the maximum value of addendum radius of the larger gear if the pressure angle between two teeth is equal to 25°. T = 50 and t = 25. The gears have a module of 8 mm and the addendum is equal to one module.

a) 211. 202 mm

b) 231. 202 mm

c) 221. 202 mm

d) 201. 202 mm

Answer: c

Explanation: Given: φ = 25°, T = 50 and t = 25, m = 8 mm and addendum = 1 module = 8 mm

R = mT/2 = 200 and r = mt/2 = 100

R a = 200 + 8 = 208 mm.

R a max =  2 +  2 ) 0.5 = 221. 202 mm.

8. The number of teeth on a pinion is equal to 20 and the gear ratio is 3. The pressure angle is equal to 15°. If the module is equal to 6 mm, find the path of contact when interference is just avoided.

a) 62.116 mm

b) 73.324 mm

c) 42.831 mm

d) 51.483 mm

Answer: a

Explanation: Given: φ = 15°, t = 20 and G = T/20 = 3, T = 20 x 3 = 60, m = 6 mm

r = mt/2 = 60 mm and R = mT/2 = 180 mm

Path of contact when interference is just avoided = maximum path of approach + maximum path of recess = r sinφ + R sinφ = 60 sin15° + 180 sin15° = 62.116 mm.

9. Find the maximum value of addendum radius of the pinion if the pinion has 25 teeth, the larger gear has 100 teeth and the pressure angle is equal to 20°. Take the value of module as 6 mm.

a) 78.943 mm

b) 71.345 mm

c) 78.431 mm

d) 74.459 mm

Answer: b

Explanation: Given: φ = 20°, t = 25 and T = 100, G = T/t = 100/25 = 4, m = 6 mm

r = mt/2 = 75 mm

a p max = rsin 2 φ) 0.5 -1) = 71.345 mm.

10. Find the maximum value of addendum radius of the larger gear if the pinion has 17 teeth, the larger gear has 51 teeth and the pressure angle is equal to 15°. Take the value of module as 8 mm.

a) 9.092 mm

b) 8.923 mm

c) 5.247 mm

d) 4.389 mm

Answer: c

Explanation: : φ = 15°, t = 17 and T = 51, G = T/t = 51/17 = 3, m = 8 mm

R = mT/2 = 204 mm

a w max = R+2)sin 2 φ) 0.5 – 1) = 5.247 mm.

11. Two 25° involute spur gears have a module of 7 mm. The addendum is equal to one module. If the larger gear has 60 teeth and the pinion has 30 teeth, they will interfere with each other. True or false?

a) True

b) False

Answer: b

Explanation: Given: φ = 25°, t = 30, T = 60 and m = 7 mm

r = mt/2 = 105 mm and R = mT/2 = 210 mm

R a = R + a = 210 + 7 = 217 mm

R a max =  2 +  2 ) 0.5 = 232.26 mm

As the value of R a obtained is less than the value of the maximum addendum radius, the interference does not occur. Hence, the given statement is false.

12. Two 20° involute spur gears have a module of 12 mm. The addendum is equal to 1.5 module. If the larger gear has 40 teeth and the pinion has 20 teeth, find the pressure angle at which interference can be avoided.

a) 20.66°

b) 21.96°

c) 22.23°

d) 21.24°

Answer: a

Explanation: Given: φ = 20°, t = 20, T = 40 and m = 12 mm, addendum = 1.5 x m = 18 mm

r = mt/2 = 120 mm and R = mT/2 = 240 mm

R a = R + a = 240 + 18 = 258 mm

R a max =  2 +  2 ) 0.5 = 256.95mm

As the calculated value of the addendum radius exceeds the maximum value of the addendum radius, interference occurs.

The new value of φ can be calculated by taking R¬a max equal to Ra.

258 =  2 +  2 ) 0.5

66564 =  2 +  2

66564 = 57600 cos 2 φ +129600 (1-cos 2 φ)

66564 = 129600 – 72000 cos 2 φ

cos 2 φ = 0.8755

cos φ = 0.935

φ = 20.66°

If the pressure angle is increased to 20.66°, the interference can be avoided.

This set of Theory of Machines Multiple Choice Questions & Answers  focuses on “Helical Gears”.

1. Identify the given gear.

theory-machines-questions-answers-helical-gears-q1

a) Spur gear

b) Helical gear

c) Worm and worm gear

d) Bevel gear

Answer: b

Explanation: The given diagram is of a helical gear. In helical gears, the teeth are inclined to the axis of the gear. The gears can be either left handed or can be right handed depending on the direction in which the helix slopes when viewed. Here, 1 is left handed gear and 2 is a right handed gear.

2. The angle at which the teeth of the gear are inclined to the axis of a gear is called as __________

a) pitch angle

b) normal angle

c) helix angle

d) gear angle

Answer: c

Explanation: Helix angle is the angle at which the teeth are inclined to the axis of a gear.

theory-machines-questions-answers-helical-gears-q1

In the diagram above, Ψ is called as helix angle. Depending upon the direction of this angle, the gear can be either left handed or right handed.

3. The distance between the corresponding points on adjacent teeth measured on the pitch circle is called ______________

a) helical pitch

b) normal pitch

c) gear pitch

d) circular pitch

Answer: d

Explanation: Circular pitch is the distance between the corresponding points on adjacent teeth measured on the pitch circle.

theory-machines-questions-answers-helical-gears-q3

Here, the distance p is called as circular pitch.

p = πm, where m is the module.

4. The shortest distance measured along the normal to the helix between corresponding points on the adjacent teeth is called ____________

a) gear pitch

b) helical pitch

c) circular pitch

d) normal circular pitch

Answer: d

Explanation: Normal circular pitch or normal pitch is the shortest distance measured along the normal to the helix between corresponding points on the adjacent teeth.

theory-machines-questions-answers-helical-gears-q4

In the given figure, pn is the normal circular pitch.

pn = p cos Ψ

Therefore, mn = m cos Ψ.

5. Two spiral gears have a normal module of 10 mm and the angle between the skew shaft axes is 50°. The driver has 20 teeth and the helix angle of it is 30°. If the velocity ratio is 1/3 and the driver as well as the follower are both left handed, find the centre distance between the shafts.

a) 434.72 mm

b) 452.83 mm

c) 582.19 mm

d) 523.39 mm

Answer: a

Explanation: Ψ 1 = 30°, mn = 10 mm and Ψ 2 = 50° – 30° = 20°, T 1 = 20.

VR = 1/3 = T 1 /T 2

T 2 = 3 x T 1 = 3 x 20 = 60

Centre distance = C =  ((T 1 /cos Ψ 1 ) + (T 2 /cos Ψ 2 )) = 434.72 mm.

6. The centre distance between two meshing gears is 250 mm and the angle between the shafts is 60°. The normal circular pitch is 10 mm and the gear ratio is 2. Ψ 1 = 35°. Find the value of number of teeth on both the wheels.

a) 46,92

b) 45, 90

c) 54,108

d) 62, 124

Answer: a

Explanation: Given: Ψ 1 = 35°, Ψ 2 = 60° – 35° = 25°, G = T 2 /T 1 = 2, p n = 10 mm and C = 250 mm

Centre distance = C = 250 =  ((T 1 /cos Ψ 1 ) + (T 2 /cos Ψ 2 )) = (p n /2π) ((T 1 /cos Ψ 1 ) + (T 2 /cos Ψ 2 )) = 1.59 x (1.22T 1 + 2.207T 1 ) = 5.45 T 1

T 1 = 250/5.45 = 45.88 ~ 46

T 2 = 2 x T 1 = 92.

7. Find the exact centre distance for the above problem.

a) 248.992 mm

b) 250.934 mm

c) 251.831 mm

d) 251.029 mm

Answer: b

Explanation: Ψ 1 = 35°, Ψ 2 = 60° – 35° = 25°, p n = 10 mm

We found out that T 1 = 46 and T 2 = 92.

Centre distance = (p n /2π) ((T 1 /cos Ψ 1 ) + (T 2 /cos Ψ 2 )) = 250.934 mm.

8. Find the efficiency in the above problem if the friction angle is 6°.

a) 89.23 %

b) 91.02 %

c) 88.58 %

d) 85.69 %

Answer: c

Explanation: Given: φ = 6°, Ψ 1 = 35°, Ψ 2 = 60° – 35° = 25°

Efficiency = η = (cos (Ψ 2 + φ) x cos Ψ 1 ¬)/ (cos (Ψ 1 – φ) x cos Ψ 2 ¬) = 0.8858 = 88.58 %.

9. The shaft angle of two helical gears is 60°. The normal module is 6 mm. The larger gear has 50 teeth and the gear ratio is 2. The centre distance is 300 mm. Find the helix angles of the two gears.

a) Ψ 1 = 11.68°, Ψ 2 = 48.32°

b) Ψ 1 = 13.02°, Ψ 2 = 46.98°

c) Ψ 1 = 10.11°, Ψ 2 = 49.89°

d) Ψ 1 = 12.21°, Ψ 2 = 47.79°

Answer: d

Explanation: Given: Ψ 1 + Ψ 2 = 60°; Ψ 1 = 60° – Ψ 2 ; mn = 6 mm; T 2 = 50 and T 1 = T 2 /G = 25. C = 300 mm

C = 300 =  ((T 1 /cos Ψ 1 ) + (T 2 /cos Ψ 2 )) = 3((25/cos(60° – Ψ 2 ))+(50/cos Ψ 2 )

100 = ((25/cos(60° – Ψ 2 ))+(50/cos Ψ 2 )

By trial and error method, we get Ψ 2 = 47.79°, Ψ 1 = 12.21°.

10. Find the spiral angles of the two helical gears for maximum efficiency if the shaft angle is 100° and the friction angle is 10°.

a) 55°,55°

b) 50°,50°

c) 55°,45°

d) 65°,45°

Answer: c

Explanation: For maximum efficiency of the helical gears, Ψ 1 = /2 = /2 = 55°

Ψ 2 = 100° – 55° = 45°.

11. Find the normal circular pitch as well as the normal module if the larger gear has 40 teeth and the gear ratio is 4. The helix angle is 20° and the shaft angle is 50°. Assume the centre distance between the gears to be 200 mm.

a) 22.12 mm, 7.04 mm

b) 21.31 mm, 6.78 mm

c) 20.37 mm, 6.48 mm

d) 21.98 mm, 6.996 mm

Answer: a

Explanation: Given: T 2 = 40 and T 1 = T 2 /G = 40/4 = 10, Ψ 1 + Ψ 2 = 50°; Ψ 1 = 20°; Therefore, Ψ 2 = 30°; C = 200 mm

C = 200 =  ((T 1 /cos Ψ 1 ) + (T 2 /cos Ψ 2 ))

400 = mn x 

mn = 7.04 mm and thus p n = πmn = 22.12 mm.

12. The normal circular pitch of two helical gears in a mesh is 7 mm. If the number of teeth on the smaller gear is 22 and the gear ratio is 2, find the pitch diameters of the gears. Ψ 1 = 40° and Ψ 2 = 25°.

a) 83.01 mm, 109.93 mm

b) 63.99 mm, 108.17 mm

c) 85.34 mm, 96.32 mm

d) 79.12 mm, 100.13 mm

Answer: b

Explanation: T 1 = 22 and T 2 = T 1 x G = 22 x 2 = 44, p n = 7 mm, Ψ 1 = 40° and Ψ 2 = 25°

d1 = P1T 1 /π = p n T 1 /(cos Ψ 1 x π) = 63.99 mm

d2 = P2T 2 /π = p n T 2 /(cos Ψ 2 x π) = 108.17 mm.

13. For two gears having shaft angle 90° and friction angle 6°, find the maximum efficiency of the helical gears.

a) 89.23%

b) 85.64%

c) 79.02%

d) 81.07%

Answer: d

Explanation: θ = 90°, φ = 6°

ηmax = +1)/ +1) = 0.8107 = 81.07%.

This set of Theory of Machines Multiple Choice Questions & Answers  focuses on “Worm Gears”.

1. When large gear reductions are needed _________ gears are used.

a) helical

b) spur

c) worm

d) bevel

Answer: c

Explanation: Worm gears are used where large speed reductions are needed. The horizontal portion of the gear is called as a worm and the assembly of it with the gear is known as a worm gear. Worm can easily turn a gear but a gear cannot turn a worm because of the shallow angle on the worm.

2. The driven gear in the worm gear is a helical gear. True or false?

a) True

b) False

Answer: a

Explanation: The helical gear is driven in the worm gear and the driving element is the screw or the worm. Worm gears are used in transmission of power between two non-parallel and non-intersecting shafts.

3. Which is of these is an advantage of worm gear?

a) It is expensive

b) Has high power losses and low transmission efficiency

c) Produce a lot of heat

d) Used for reducing speed and increasing torque

Answer: d

Explanation: Reduction of speed and increasing the torque is an advantage of worm gear. The rest are the disadvantages of the worm gear. Worm gears are used in gate control mechanisms, hoisting machines, automobile steering mechanisms, lifts, conveyors and presses.

4. The distance between corresponding points on adjacent teeth measured along the direction of the axis is called ____________

a) joint line

b) normal link

c) axial pitch

d) lead

Answer: c

Explanation: Axial pitch is the distance between corresponding points on adjacent teeth measured along the direction of the axis. The axial pitch of the worm gear is the same thing as the circular pitch of the helical gear.

5. The distance by which a helix advances along the axis of the gear for one turn around is called _____________

a) joint line

b) normal link

c) axial pitch

d) lead

Answer: d

Explanation: Lead is the distance by which a helix advances along the axis of the gear for one turn around. The axial pitch is equal to the lead in a single helix and the axial pitch is one half of the lead in a double helix and so on.

6. The angle at which the teeth are inclined to the normal of the axis of rotation is called _______________

a) pitch angle

b) lead angle

c) normal angle

d) joint angle

Answer: b

Explanation: Lead angle is the angle at which the teeth are inclined to the normal of the axis of rotation. Lead angle of the worm gear is same as the helix angle of the helical gear. Thus, Ψ2 = λ1.

7. What is the velocity ratio of worm gears?

a) /d 2

b) (πd 2 )/l

c) l/(πd 2 )

d) d 2 /

Answer: c

Explanation: Velocity ratio in worm gears is given as the ratio of the angle turned by the gear to the angle turned by the worm.

Thus, velocity ratio = (2l/d 2 )/ = l/(πd 2 ).

8. What is the centre distance for the worm gear?

a) (m n /2)(T 1 cotλ 1 – T 2 )

b) (m n /2)(T 2 cotλ 1 + T 1 )

c) (m n /2)(T 2 cotλ 1 – T 1 )

d) (m n /2)(T 1 cotλ 1 + T 2 )

Answer: d

Explanation: C = (m n /2)(T 1 cotλ 1 + T 2 )

This equation can be derived by using the formula for centre distance of a helical gear which is given as C = (m n /2) ((T 1 /cos Ψ 1 ) + (T 2 /cos Ψ 2 )

As, Ψ 2 = λ 1 , Ψ 1 = 90° – λ 1 .

9. What is the formula to calculate maximum efficiency of a worm gear?

a) ø/ø

b) ø/ø

c) (tan(λ 1 -ø))/tan λ 1

d) (tan(λ 1 +ø))/tan λ 1

Answer: b

Explanation: The maximum efficiency of the worm gear is given to be ø/ø, whereas the efficiency of the worm gear is given as (tan(λ 1 -ø))/tan λ 1

10. Find the helix angle of the worm if the pitch of the worm gear is 12 mm and the pitch diameter is 50 mm.

a) 8.687°

b) 11.231°

c) 9.212°

d) 10.319°

Answer: a

Explanation: tan λ 1 = Lead / Pitch circumference = 2p/πd1 = 24/50π = 0.1528

λ 1 = 8.687°.

11. Find the speed of the gear if the worm is a three start worm rotating at 500 rpm. The gear has 20 teeth.

a) 125 rpm

b) 100 rpm

c) 75 rpm

d) 50 rpm

Answer: c

Explanation: N 1 /N 2 = T 2 /T 1

500/N 2 = 20/3

N 2 = 75 rpm

Thus, the gear rotates at a speed of 75 rpm.

12. For a two start worm gear having a pitch of 20 mm and a lead angle 12°, find the centre distance if the larger gear has 25 teeth.

a) 148.22 mm

b) 124.93 mm

c) 121.19 mm

d) 109.53 mm

Answer: d

Explanation: C = (m n /2)(T 1 cotλ 1 + T 2 )

Therefore, C = (p n /2π)(T 1 cotλ 1 + T 2 ) = 109.53 mm.

13. Calculate the lead angle of the worm gear for maximum efficiency if θ = 90° and the coefficient of friction is 0.05.

a) 48.21°

b) 42.23°

c) 43.57°

d) 46.43°

Answer: c

Explanation: µ = 0.05°; ø = tan -1  = 2.862°; θ = 90°

For maximum efficiency, Ψ 1 = ø/2 = 92.862/2 = 46.43°

Ψ 1 = 90° – λ 1 = 46.43°

λ 1 = 90° – 46.43° = 43.57°.

14. Find the maximum efficiency if the lead angle is given to be 10° and the coefficient of friction is 0.07.

a) 79.82%

b) 72.23%

c) 76.29%

d) 70.72%

Answer: d

Explanation: λ 1 = 10°, ø = tan -1  = 4°

Efficiency = tan(λ 1 )/ tan(λ 1 + ø) = 0.7072 = 70.72%.

15. Calculate the maximum efficiency of the worm gears which have a friction angle of 0.06.

a) 88.71%

b) 83.23%

c) 89.91%

d) 86.49%

Answer: a

Explanation: ø = tan -1  = 3.43°

Maximum efficiency = ø/ø = 0.8871 = 88.71%.

This set of Theory of Machines Multiple Choice Questions & Answers  focuses on “Bevel Gears”.

1. Bevel gears are used to have a gear drive between two intersecting shafts. True or false?

a) True

b) False

Answer: a

Explanation: To have a gear drive between two intersecting shafts, bevel gears are used. Bevel wheels are used to work a gear to another gear. They are generally mounted on shafts which are perpendicular to each other but can be designed to work at other angles too. Thus, the statement is true.

2. Bevel gears are equivalent to rolling ________

a) cubes

b) cones

c) spheres

d) cuboids

Answer: b

Explanation: By geometry of bevel gears, it is evident that they are equivalent to rolling cones. Their pitch surface is a cone. Bevel gears exist as pairs which can be either straight or inclined at different angles.

3. Identify the given gear.

theory-machines-questions-answers-bevel-gears-q3

a) Spur

b) Helical

c) Worm

d) Bevel

Answer: d

Explanation: The given gear is bevel gear. Here, ƴ g , ƴ p = pitch angles of gear and pinion respectively.

r g , r p = pitch radii of gear and pinion respectively.

4. What is the formula to calculate the pitch angle of gear of a bevel gear using the radius?

a) y g = tan -1 /((r g /r p )+sin θ))

b) y g = tan -1 /((r p /r g )+sin θ))

c) y g = tan -1 /((r p /r g )+cos θ))

d) y g = tan -1 /((r g /r p )+cos θ))

Answer: c

Explanation: tan y g = /((r p /r g )+cos θ))

Therefore, y g = tan -1 /((r p /r g )+cos θ))

Similarly, tan y p = /((r g /r p )+cos θ))

Therefore, y p = tan -1 /((r g /r p )+cos θ)).

5. What is the formula to calculate the pitch angle of pinion of a bevel gear using the angular velocity?

a) y p = tan -1 /((w g /w p )+sin θ))

b) y p = tan -1 /((w p /w g )+sin θ))

c) y p = tan -1 /((w g /w p )+cos θ))

d) y p = tan -1 /((w p /w g )+cos θ))

Answer: d

Explanation: tan y p = /((w p /w g )+cos θ))

Therefore, y p = tan -1 /((w p /w g )+cos θ))

Similarly, tan y g = /((w g /w p )+cos θ))

Therefore, y g = tan -1 /((w g /w p )+cos θ)).

6. Which of the gears has the highest contact ratio?

a) Helical

b) Spur

c) Bevel

d) Worm

Answer: c

Explanation: Bevel gears increase the contact ratio. It avoids interference and results in stronger pinion teeth. The gear ratio can be determined from the number of teeth, pitch diameters or pitch cone angles.

7. What is the working depth of the gear and the pinion of the bevel gears?

a) 2 m, 0.7 m

b) 0.7 m, 2 m

c) 3 m, 0.4 m

d) 0.4 m, 3 m

Answer: a

Explanation: The working depth of the teeth is usually 2 m which is the same as for standard spur and helical gears, but the bevel pinion is designed with the larger addendum and has 0.7 m working depth. The gear addendum varies according to the gear ratio and is 1 m for a gear ratio 1.

8. A pair of bevel gears is mounted on two intersecting shafts who have a shaft angle of 70°. The velocity ratio of the gears is 2. Find the pitch angles.

a) 42.21°, 27.79°

b) 48.14°, 21.86°

c) 39.29°, 30.71°

d) 46.95°, 23.05°

Answer: b

Explanation: θ = 70°, w g /w p = 1/2

tan y g = /((w g /w p )+cos θ)) = /+cos 70°)) = 1.16

y g = 48.14°

tan y p = /((w p /w g )+cos θ)) = /+cos 70°)) = 0.4

y p = 21.86°

We can confirm this by checking that y g + y p = θ = 70°.

9. A pair of bevel gears is mounted on two intersecting shafts who have a shaft angle of 80°. The gear ratio of the gears is 1/3. Find the pitch angles.

a) 62.76°, 17.24°

b) 49.31°, 30.69°

c) 56.97°, 23.03°

d) 64.63°, 15.37°

Answer: a

Explanation: θ = 80°, w g /w p = 1/3

tan y g = /((w g /w p )+cos θ)) = /+cos 80°)) = 1.94

y g = 62.76°

tan y p = /((w p /w g )+cos θ)) = /+cos 80°)) = 0.31

y p = 17.24°

We can confirm this by checking that y g + y p = θ = 80°.

10. The gear ratio of bevel gears = tan y g = cot y p . True or false?

a) True

b) False

Answer: a

Explanation: The gear ratio of bevel gears can be given from the number of teeth, the pitch diameters or the pitch cone angles as i = ω 1 /ω 2 = n 1 /n 2 =d 2 /d 1 = tan y g = cot y p . Thus, the statement is true.