SKP Engineering College - mech.skpec.edu.inmech.skpec.edu.in/wp-content/uploads/sites/8/2017/11/Design-of... · Shigley J.E and Mischke C. R., ―Mechanical Engineering Design‖,

S.K.P. Engineering College, Tiruvannamalai VI SEM

Mechanical Engineering Department 1 Design of Transmission System

SKP Engineering College

Tiruvannamalai – 606611

A Course Material

on

Design of Transmission Systems

By

Mr.C.Naveen kumar, Mr.R.Saravanan, Mr.R.Velmurugan,

Mr.R. Bharath

Assistant Professor

Mechanical Engineering Department



Quality Certificate

This is to Certify that the Electronic Study Material

Subject Code: ME 6601

Subject Name: Design of Transmission Systems

Year/Sem: III / VI

Being prepared by and it meets the knowledge requirement of the University curriculum.

Signature of the Author

Name: Mr.C.Naveenkumar, Mr.R.Saravanan,Mr.R.Velmurugan, Mr.R. Bharath

Designation: Assistant Professor

This is to certify that the course material being prepared by Mr.C.Naveen kumar, Mr.R.Saravanan, Mr.R.Velmurugan and Mr.R. Bharath is of the adequate quality. He has referred more than five books and one among them is from abroad author.

Signature of HD Signature of the Principal

Name: Dr.J.Kuberan Name: Dr.V.Subramania Bharathi

Seal: Seal:



ME6601 DESIGN OF TRANSMISSION SYSTEMS L T P C 3 0 0 3

Lecture : 3 hrs/Week Internal Assessment: 20 Marks

Tutorial : 1 hrs/week Final Examination: 80 Marks

Practical : - Credits: 4

TYPE OF COURSE: Required course

ASSESSMENT METHOD: Tutorial classes, 2 internal tests, 1 Model examination and

Course end university examination.

PREREQUISITE: Design of machine elements, Strength of material, Kinematics of machinery,Engineering and material and metallurgy, automobile engineering. COURSE OBJECTIVES:

To gain knowledge on the principles and procedure for the design of power Transmission components.

To learn to use standard data and catalogues.

COURSE OUTCOMES:

Upon completion of this course the student will be able to:

CO1 Use of standard data and catalogues.

CO2 Design the power transmission components like gears, belts.

CO3 Ability to analyze gear forces.

CO4 Analysis the gear box, speed reducers, speed diagrams.

CO5 Design the brakes and clutches.

CO-PO MAPPING

CO/PO

PO1

PO2

PO3

PO4

PO5

PO6

PO7

PO8

PO9

PO10

PO11

PO12

CO1 1 1

CO2 3 2 3 2 3 3

CO3 3 2 3 3 3

CO4 3 3 3 3

CO5 3 2 2 2 3



OBJECTIVES:

1. To gain knowledge on the principles and procedure for the design of

Mechanical power Transmission components.

2. To understand the standard procedure available for Design of Transmission of

Mechanical elements

3. To learn to use standard data and catalogues

(Use of P S G Design Data Book permitted)

SYLLABUS

UNIT I DESIGN OF FLEXIBLE ELEMENTS 9

Design of Flat belts and pulleys - Selection of V belts and pulleys – Selection of

hoisting wire ropes and pulleys – Design of Transmission chains and Sprockets.

UNIT II SPUR GEARS AND PARALLEL AXIS HELICAL GEARS 9

Speed ratios and number of teeth-Force analysis -Tooth stresses - Dynamic effects –

Fatigue strength - Factor of safety - Gear materials – Design of straight tooth spur &

helical gears based on strength and wear considerations – Pressure angle in the

normal and transverse plane- Equivalent number of teeth-forces for helical gears.

UNIT III BEVEL, WORM AND CROSS HELICAL GEARS 9

Straight bevel gear: Tooth terminology, tooth forces and stresses, equivalent number

of teeth. Estimating the dimensions of pair of straight bevel gears. Worm Gear: Merits

and demeritsterminology. Thermal capacity, materials-forces and stresses, efficiency,

estimating the size of the worm gear pair. Cross helical: Terminology-helix angles-

Estimating the size of the pair of cross helical gears.

UNIT IV GEAR BOXESES 9

Geometric progression - Standard step ratio - Ray diagram, kinematics layout -Design

of sliding mesh gear box - Design of multi speed gear box for machine tool

applications - Constant mesh gear box - Speed reducer unit. – Variable speed gear

box, Fluid Couplings, Torque Converters for automotive applications.

UNIT V CAMS, CLUTCHES AND BRAKES 9

Cam Design: Types-pressure angle and under cutting base circle determination-forces

and surface stresses. Design of plate clutches –axial clutches-cone clutches-internal

expanding rim clutches- Electromagnetic clutches. Band and Block brakes - external

shoe brakes – Internal expanding shoe brake.

TOTAL: 45 PERIODS



OUTCOMES:

1. Upon completion of this course, the students can able to successfully design transmission components used in Engine and machines

CONTENT BEYOND SYLLABUS

Direct gear design for spur and helical involute gears

Direct gear design for automotive application

Animations of gear transmission process.

LEARNINGRESOURCES:

TEXT BOOKS: 1. Shigley J.E and Mischke C. R., ―Mechanical Engineering Design‖, Sixth Edition,

Tata McGraw-Hill , 2003.

2. Sundararajamoorthy T. V, Shanmugam .N, "Machine Design", Anuradha Publications, Chennai, 2003. REFERENCES:

1. Maitra G.M., Prasad L.V., ―Hand book of Mechanical Design‖, II Edition, Tata McGraw-Hill, 1985. 2. Bhandari, V.B., ―Design of Machine Elements‖, Tata McGraw-Hill Publishing Company Ltd., 1994. 3. Prabhu. T.J., ―Design of Transmission Elements‖, Mani Offset, Chennai, 2000, 4. Hamrock B.J., Jacobson B., Schmid S.R., ―Fundamentals of Machine Elements‖, McGraw-Hill Book Co., 1999. 5. Ugural A,C, "Mechanical Design, An Integrated Approach", McGraw-Hill , 2003. WEB RESOURCES:

1. IS 2458 : 2001, Vocabulary of Gear Terms

2. IS 2467 : 2002 (ISO 701: 1998), International Gear Notation

ADDITIONAL RESOURCES :

1. NPTEL TUTORIALS (Internal Server)

2. Theory of machines Books (PDF Formats)

3. Online Objective Questions

4. Videos Materials if any (You tube)

NOTE

L1 –Rember, L2 –Understand, L3 –Apply (LOTS – Lower order thinking)

H1-Analysis, H2- Evaluate, H3 – Create ( HOTS – Higher order thinking)





CONTENTS

S.No Particulars Page

1 Unit – I 8

2 Unit – II 52

3 Unit – III 90

4 Unit – IV 144

5 Unit – V 174



UNIT I

UNIT I DESIGN OF FLEXIBLE ELEMENTS

Design of Flat belts and pulleys - Selection of V belts and pulleys – Selection of

hoisting wire ropes and pulleys – Design of Transmission chains and Sprockets.

PART-A

1. State the –“Law of Belting” [CO1 - L1 - Apr13]

The law of belting states that the centerline of the belt when it approaches the

pulley must lie in the mid plane of that pulley which should be perpendicular to the axis

of the pulley. Otherwise the belt will run off the pulley.

2. State the materials for belts. [CO1 - L2 - Apr 13]

Leather, cotton fabrics, rubber, animal's hair, silk, rayon, woolen etc.

3. Explain creep in belts. [CO1 - L2 - Nov14]

Since the tensions produced by the belt on the two sides of the pulley are not

equal, the belt moves with a very negligible velocity, due to the difference of two

tensions. This slow movement of the belt over the pulley is known as creep of belt and

it is generally neglected."

4. Centrifugal forces add to belt tension without increasing the power capacity.

Justify the statement. [CO1 - H1 - Nov15]

The tension caused by centrifugal force is called centrifugal tension. At lower

belt speeds (less than 10 m/s), the centrifugal tension is very small, but at higher belt

speeds (more than 10 m/s), its effect is considerable and thus should be taken into

account. When centrifugal tension is taken into account,

Total tension in the tight side,

Tt1 = T1 + TC

Total tension in the slack side,

Tt2 = T2 + TC

Power transmitted, P = (Tt1 – Tt2) v

= [(T1 + TC) – (T2 + TC)] v

= (T1 – T2) v

Thus we see that the centrifugal tension has no effect on the power transmitted.



5. Distinguish between open drive and cross drive of a belt drive .which is better?

[CO1 - H1]

Open bet drive: used with shafts arranged parallel and rotating in same

direction.

Cross belt drive: used with shaft arranged parallel and rotating in opposite

direction.

6. What are the various losses in the power transmission by belts? [CO1 - L2 - Nov14] The losses in a belt drive are due to:

(i) slip and creep of the belt on the pulleys,( 3% )

(ii) wind age are air resistance to the movement of belt and pulleys,

(iii) bending of the belt over the pulleys,( 1 % ) and

(iv) Friction in the bearings of pulley.( 1 % )

7. In what ways the timing belts are superior to ordinary V-belts?

[CO1 - H1 - May15]

Flat belt and V belt drives cannot provide a precise speed ratio, because

slippage occurs at the sheaves, but certain applications required an exact output to

input speed ratio. In such situations, timing belts are used.

Since the timing belts possess toothed shape in their -inner side, engagement

with toothed pulley will provide positive drive without, belt-slip where as in the case of

ordinary V-belts, chances of slip are there and hence positive drive is not possible at

all times. Hence toothed belts (I timing belts) are superior to ordinary V-belts.

8. Why tight -side of the belt should be a t the bottom side of the

pulley? [CO1 - L1]

Because of the driving pulley pulls the belt from bottom side and delivers it to

the upper side .so it is obvious that the bottom side of the belt is tight.

9. How is V-belt specified? [CO1 - L1 - May10]

V-belt is designated by a grade letter followed by its inside length in code

number, year of coding. For example, D 3048: IS 2494: 1964. M belts are designated

by the grade letter and inside length only such D - 3048. Sometimes, the inside length

may be denoted in inches as D -



10. What is slack adjuster? [CO1 - H1]

Slack adjuster is an arrangement of maintaining both tight and slack side

tensions. Slack adjusters are also known as belt tensioning devices.

11. What will be the effect on limiting ratio of tensions of a belt if the co-efficient

of fr iction between the belt and the rim of pulley is doubled while

angle of lap remains the same? [CO1 - H1 - Nov07]

The ratio of tension will be squared.

12. Define: maximum tension in a belt. [CO1 - L3 - May10]

Maximum tension in a belt = tension on tight side + centrifugal tension.

13. Give the condition for maximum power transmission in terms of centrifugal

tension in case of belt drive. [CO1 - L3 - May09]

The power transmitted shall be maximum when the centrifugal tension is one

third of the maximum belt tension.

14. Sketch the cross section of a V-belt and label its important parts. [CO1 - L2 -

Nov09]

15. Derive the expression for tension ratio of belts. [CO1 - H1 - Apr07]

T1 T1 - mv2

---- = ey0 and ------------ = ey0

T2 T2 – mv 2

16. Give the relationship of ratio of tensions in a V-belt drive. [CO1 - L2 - May10]

T1

---- = ey0 COSEC B

T2



17. State reasons for V-belt drive being preferred to flat belt drive?(or)Why slip is less in the case of v- belts when compared with flat belts? [CO1 - L2 - May13] 1. Since the V-belts are made endless and there is no joint trouble, therefore the drive is smooth. 2. The wedging action of the belt in the groove gives high value of limiting ratio of

tensions. Therefore the power transmitted by V-belts is more than flat belts for the

same coefficient of friction, arc of contact and allowable tension in the belts.

18. What is the effect of centre distance and diameter of pulley on the

life of a belt? [CO1 – L2]

The life of a belt is a function of the centre distance and diameter of driver and

driven pulleys.

The shorter the belt, the more often it will be subjected to additional bending

stress while running around the pulleys at a given speed. And also it will be destroyed

quickly due to fatigue.

Hence the increased centre distance and diameter of pulley will increase the

belt life.

19. Explain the term “Crowning of pulley”. Specify the purpose of it. [CO1 - L1 - May14]

Pulleys are provided. a -slight conical shapes (or), convex shapes in their rim's

surface in order to prevent the belt from running off the pulley due to centrifugal force.

This is known as crowning, of pulley.

Usually the crowning height t may be 1/96 of pulley face width.The

purpose of it is to prevent slipping from pulley due to centrifugal force

20. What are the five parts of roller chain? [CO1 - L3 - Apr10]

Pin link or coupling link

Roller link

Pins

Bushes and Roller.



21. Give any three application of chain drive.What are their limitations?[CO1-L3]

Chain drives are widely used in transportation industry, agricultural machinery,

metal and wood working machines.

22. What is chordal action (Polygonal action) in chain drive? Also name a

company that produces driving chains[Co1 - H1 - Apr10] [CO1 - H1 - Nov15]

[CO1 - H1 - Apr15]

When chain passes over a sprocket, it moves as a series of chords instead of a

continuous arc as in the case of a belt drive. It results in varying speed of chain drive.

This phenomenon is known as chordal action.

Anish trading co, wiperdrive engineering, keneria engineering corporation, Rockman

industries ltd, perfect chain industries.

23. What do you mean by galling of roller chains? [CO1 - H1 - Nov10]

Galling is sticking slip phenomenon between the pin and bushing. When the

load is heavy and the speed is high, the high spots (joints) of the contacting surfaces

are welded together. This is galling of roller chains.

24. How does a hoisting chain differ from a roller chain? [CO1 - L1 - Apr08]

Hoisting chains also known as link chains, are used to suspend and /or lift the

loads in hoisting machines.

Roller chains also known as transmission chains are used for transmitting

power between parallel shafts using sprockets.

25. What is a silent chain? In what situations, silent chains are preferred? [Co1 -

L3 - Nov08]

Inverted tooth chains are called silent chain because of their relatively quiet

operation.

Silent drives are prepared for high power, high speed, and smooth operation.

26. In what way silent chain is better than ordinary driving chain? [CO1 - H1 -

Apr07]

Silent chains are prepared for high power, high speed, and smooth operation



27. What is done to accommodate initial sag in chain drives? [CO1 - L1 - Apr0]

In order to accommodate initial sag in chain drive, centre distance should

decreased by the amount 0.01 a.

28. What do you understand by simplex, duplex and triplex chain? [CO1 - L3 -

Apr0]

Roller chains are available in a single row or multi row construction such a

simplex, duplex or triplex strands.

29. How is a wire rope specified? [CO1 - L1 - Apr0]

The wire ropes are designated by the number of strands and the number of

wires in each strand. For example, a wire rope having six strands and seven wires in

each strand is designated by 6 × 7 rope. Standard designations are as follows:

6 × 7 rope, 6 × 19 rope, 6 × 37 rope, 8 × 19 rope

30. What do you understand by 6x9 constructions in wire ropes? [CO1 - L3 - Apr08] [CO1 - L3 - Nov14-0] A wire rope having six strands and nine wires in each strand is designated by 6

× 9 rope.

31. Sketch and name different type of compound wire ropes. [CO1-L1]

(i)Cross or regular lay ropes

(ii) Parallel or Lang lay ropes.



(iii)Composite or reverse laid ropes.

32. What is meant by the ply of belt? [CO1 - L2 - Nov13]

Thin layers which are laminated together to form the belt thickness is known as ply of

belt.

33. Write any four wire rope applications. [CO1 - L1 - Nov13]

(i) Conveyers (ii) Hoisting purpose in mines (iii) Cranes (iv) Elevators

34.What are the factors upon the which the coefficient of friction between the belt and pulley depend? [CO1 - L1 - May14] (i) Material of belt (ii) Material of pulley (iii) Slip of belt and (iV) Speed of belt



PART-B

1. Find the width of the belt necessary to transmit 7.5KW to a pulley of 300mm

diameter, if the pulley makes 1600 rpm and the co-efficient of friction between

the belt and the pulley is 0.22. Assume the angle of contact as 210O and the

maximum tension in the belt is not to exceed 8 N/mm widths.[ CO1 - H3 - Apr11]

Given:

P=7.5KW =7500W D=300 mm =0.3m N=1600 rpm

µ=0.22, α=210o =210o xπ/180o =3.6652 rad To Find:

Width of the belt ―b‖

Selection:

Vel.of belt V=πdN/60 =πx0.3x1600/60 =25.13 m/s

WKT,

P = (T1-T2) V

7500= (T1-T2) x25.13

T1-T2=298.415 -----------------------> 1

WKT,

T1/ T2 = eµα

= e 0.22x3.6652

= 2.2396

T1 = 2.2396 T2----------------------> 2

Solving 1&2 we get,

T1 =539.15, T2 =240.73

Given that,

Maximum tension is not to exceed 8 N/mm width

T1 = Max. Tension per mm width x width T1 = 8b 539.15 = 8b

Standard width of the belt =76 mm

b=67.4mm



2. A flat belt is required to transmit 35 KW from a pulley of 1.5m effective diameter

running at 300 rpm. The angle of lap is 165o and µ=0.3. Determine taking

centrifugal tension into account, width of the belt required. It is given that the

belt thickness is 9.5 mm, density of its material is 1.1 Mg/m3 and the related

permissible working stress is 2.5 Mpa. [Co1 - H3 - Apr11]

Given:

P=25 KW =35x103 w

d= 1.5 m

N=300 rpm

α=165o =165xπ/180=2.88 rad

µ=0.3

t=9.5 mm

e=1.1 Mg/m3 =1100 kg/m 3

σ=2.5 mpa=2.5x106 N/m2

To Find:

Width of the belt ―b‖

Solution: Velocity v= πdN = πx1.5x300 = 23.56 m/s 60 60 WKT,

P = (T1-T2) V

35x103 = (T1-T2)23.56

T1-T2 = 1485.45 ------------------->1

WKT,

T1/ T2 = eµα

= e 0.3X2.88

= 2.373

T1 =2.373 T2--------------->2

Solving 1&2



T1=2568 N

T2=1082.19N

WKT,

while considering centrifugal tension (Tc),

The maximum tension T= T1+Tc ----------->3

But,

Tc=mv2

m=mass of the belt per meter length

= volume x density

= bxtx1x1100

= bx9.5x1x1100x10-6

= 0.01045 b kg/m

Tc= 0.01045bx (23.56)2

WKT,

maxi. Tension is the belt T = σ (b x t)

= 2.5x106x9.5 bx10-6

=23.75 b N

Substituting all the values in eq.3

23.75 b = 2568 + 5.8 b

Standard width of the belt=152 mm.

Tc =5.8 b N

b=143 mm



3. A belt drive is required to transmit 12KW from a motor running at 720 rpm. The belt is 12mm thick and has a mass density of 0.001 gm/mm3. Permissible stress in the belt not to exceed 2.5 N/mm2. Diameter of driving pulley is 250mm whereas the speed of the driven pulley is 240 rpm. The two shafts are 1.25m apart. Co-efficient of friction is 0.25. Determine the width of the belt. [Co1 - H3 - Apr11] Given:

P = 12 KW = 12x103 W N1 = 720 rpm t = 12 mm e = 0.001 gm/mm3 = 1000 kg/ m3

σ = 2.5 N/mm2

= 2.5x106 N/ m2

d1 = 250mm=0.25m N2 = 240 rpm C = 1.25m µ = 0.25

To find:

Width of the belt ―b‖.

Solution:

Vel. of belt V= πd1N1 = πx0.25x720 = 9.425 m/s

60 60

WKT, Speed ratio, N2/N1=d1/d2

240 =0.25

720 d2

d2=0.75m

For an open belt drive, Sin α = d2-d1/2C = 0.75-0.25 = 0.2 2x1.25 α = sin-1(0.2) α = 11.54o

Since the material is same for both the pulleys The smaller pulley governs the design. Arc of contact for smaller pulley αs = (180o-2α) π/180o



=2.74 rad WKT, T1/ T2 =eµαs =e0.25x2.74 =1.983 T1=1.983 T2--------------->1 WKT, p= (T1- T2) V 12x103= (T1- T2) x9.425 T1- T2=1273.2-------------->2 Solving 1& 2 T1=2545.42 N T2=1272.20 N WKT, T= T1 +Tc --------------->3 But,

Tc=mV2

m= mass of the belt per meter length

=Volume x density

= bx12x10-6x1x1000 = 0.012 b kg/m Tc=1.065 b N

WKT, maxi. Tension in the belt T=σ x b x t =2.5x106x12bx10-6 T=30b N Substituting all the values in eq 3

30 b=2545.42 +1.065 b b= 87.97 mm Standard belt width =90mm

4. A leather belt 125mm wide and 6mm thick transmits power from a pulley with the angle of lap 150o and µ=0.3. if the mass of 1 m3 of leather is 1Mg and the stress in the belt is not to exceed 2.75Mpa.Find the maximum power that can be transmitted and the corresponding speed of the belt. [Co1 - H3 - Apr2000] Given data:

b=125mm=0.125m t=6mm=6x10-3m

ѳ=150o = =2.62 rad



µ=0.3,e=1Mg/m3=1000kg/m3 σ=2.75Mpa=2.75x106 N/m2

To Find:

Maximum power and speed Solution:

Speed of the belt: for maximum power

WKT, v=

And T=σ b t =2.75x106x0.125x6x10-3

=2062.5N

m=b x t x l x e

= 10125x6x10-3x1x1000

=0.75 kg/m

Speed of the belt for max.power

V=√

V=30.25 m/s

Max.power transmitted:-

WKT,

Tc=T/3=2062.5 = 687.5 N

3

And,

T1=T-Tc=2062.5-687.5=1375 N

And

T1/T2=eµα =e0.3x2.62 =2.195

T2=T1/2.195=1375 = 626.53 N

2.195

Power transmitted p = (T1-T2) V

= (1375-626.53)30.28

P=22.66 KW



DESIGN OF FLAT BELT PULLEY:

5. Design an overhanging pulley for the following specification:

Power=18Kw Speed=200 rpm Angle of contact=165o

Co-efficient of friction=0.25 Overhanging length(i.e) the distance of the pulley centre line from the nearest bearing is 0.30 m. belt thickness=10mm Safe shear stress for shaft=40Mpa Safe stress for belt=2.5Mpa Safe stress for rim=4Mpa Density of the leather=1000kg/m3

[Co1 - H3 - Apr11] Given data: P=18 KW=18x103W N=200 rpm α=165o=165xπ/180=2.88 rad µ=0.25 L=0.3 m t=10mm σs for shaft=40Mpa=40x106N/m2=40 N/mm2 σbelt=2.5Mpa=2.5x106N/m2=2.5N/mm2

σrim=4Mpa=4x106N/m2=4N/mm2

σleather=1000kg/m3

To find: Design an overhanging pulley Solution: (1)Dimensions of pulley: (i)Diameter of the pulley WKT, centrifugal stress of the pulley=tensile stress in rim σc= σrim

But σc=ev2 ev2 = σrim Assume: P=density of pulley material=7200 kg/m3 for CI 7200xv2=4x106 V=23.57 m/s WKT, V=πdx200/60=23.57 D=2.25m



(ii) Width of the pulley (a): To find width of the pulley, first we need to find the width of the belt WKT, p= (T1- T2) V 18x103= (T1- T2) x23.57 (T1- T2)=763.68----------------->1 WKT, T1/T2=eµα=e0.25x2.88=2.054

T1=2.054 T2-------------->2

Solving 1&2

T1=1487.2N, T2=725.25N

(Since the velocity is more than 10m/s, we need to consider Tc also)

Tc=mv2=10bx10-6 x1000x (23.57)2

Tc=5.55b N

Maximum tension in the belt T= σbelt x b x t

=2.5x106x10bx10-b

T=25b N

T1=T-Tc

1487.62=25‖b‖-5.55‖b‖

From DD 7.52, b=76.48mm

Std width ‗b‘=90mm

From DD7.54,

(For belt width up to 125mm,we need to add 13mm with belt size, to get the pulley width).

Pulley width=90+13=103mm

From DD 7.55,

the standard pulley width=112mm



(iii)thickness of the pulley rim:(t) From DD 7.57, for single belt, t= D/200+3mm =2250/200+3=14.25mm

2. Dimensions of arm:

From DD7.56

(i) Number of arms n=6 [for D>450mm] (ii) Cross section of arms: Major axis

b=2.94

for single belt here,

b=2.94

4x6

=64.38mm

b=65mm

Minor axis = b/2 = 65/2 =32.5mm

(iii) Radius of cross sections of arm:

=3/4 x major axis

=3/4x 65 = 48.75mm

3. dimensions of hub:

WKT,

Dia of hub = 2 x dia of shaft

We need to find dia of shaft first

WKT,

P=2πNT/60



18x103=2xπx200xT

60

T=859.44Nm

WKT,

Bending moment on the shaft due to tension of the belt

M= (T1+T2+2Tc) L

= (1487.62+724.25+2x5.5x90)0.3

=960.56Nm

Equivalent twisting moment (Te)

Te=

=√( ) ( )

=1288.92x103 Nmm

WKT,

Te= πd3 σs

16

1288.92=πd3/16x40

d=54.75mm

Say d=55mm

Dia of hub=2xdia of shaft =2x55

Dia of hub=110mm,

Length of hub=2/3xa

=2/3x112=74.66mm



4. Crown height of the pulley(h)

From DD7.55,

for 2250mm pulley dia and 112mm pulley width.

The crown height h=2mm.

DESIGN OF THE FLAT BELT DRIVE USING MANUFACTURE DATA:

6. It is required to select a flat belt drive for a fan running at 360rpm. Which is driven by a 10kw, 1440 rpm motor? The belt drive is open type and space available for a centre distance of 2m approximately. The dia of driven pulley 1000 mm. [Co1 - H3 - Apr11]

Given:

N1=1440rpm

N2=360 rpm

P=10x103W

D=1000 mm

C=2m

To find:

Open flat belt drive design



Solution:

Step1: calculation of pulley diameter

WKT,

N1/N2=D/d

1440/360=1000/d

D=250mm

From DD 7.54

The std.dia of driven pulley=250 mm

Step2: Calculation of design power in Kw:

Design power (KW) = rated Kw x Ks------------------>1 Kα x Kd

From DD 7.53, Ks =1.2(for steady load) Kα=180o-(D-d/c) x60o = (180o-(1000-250/2000) x60 =157.5o From DD: 7.54, arc of contact for 157.5o is Kα=1.08 From DD: 7.62 small pulley factor Kd =0.7 Substituting all the values in eq 1 Design Kw=10x1.2 =15.873KW 1.08x0.7 Step 3: selection of belt: From DD 7.54, select H1-SPEED duck belting.It‘s capacity=0.023 KW/mm/ply Step 4: Load rating correction: Velocity of the belt=πdN1

60 =πx0.25x1440 60 =18.85 m/s



From DD 7.54, load rating at‖V‖ m/s = [load rating at 10 m/s] x V/10 Load rating at 18.85 m/s =0.023x18.85/10 =0.04335 Kw/mm/ply Step 5: determination of belt width From DD 7.52 For 250mm smaller pulley, and velocity 18.85m/s, the no.of pulleys=5 n=5 Width of the belt =design power Load rating x no of plies =15.873 0.04335x5 =73.23 mm From DD 7.52, The standard width of belt =76mm Step 6: determination of pulley width: From DD 7.54, Pulley width=belt width+13 =73.225+13 =86.225 mm From DD 7.55, standard width of pulley=90mm Step 7: calculation of length of belt: (L) From 7.53 for open belt, L=2C+π/2 (D+d) + (D-d)2

4C = 2X2000+π/2 (1000+250) + (1000-250)2

4 x 2000

=6033.8 mm



7.A Compressor is to run by a motor pulley running at 1440 rpm, Speed ratio 2.5. Choose a flat belt crossed drive. Centre distance between pulleys is 3.6m. Take belt speed as 16 m/s. Load factor is 1.3. Take a 5ply, flat Dunlop belt. Power to be transmitted is 12kw. High speed load rating is 0.0118 Kw/ply/mm width at V=5m/s. determine the width and length of the belt. [Co1 - H3 - Nov14]

Solve this problem by following the above procedure. [Prb:No:6]

8.Design a belt drive to transmit 20Kw at 720 rpm to an aluminumrolling machine, the speed ratio being 3. The distance between the pulleys in 3m. Diameter of rolling machine pulley is 1.2m [Co1 - H3 - Apr12]

Given:

N1=720 rpm Speed ratio=3,C=3m,D=1.2m

To find: Design a belt drive: Solution:

Step1: calculation of pulley diameter

WKT,

speed ratio=D/d=3

d=1200 =400 mm 3 From 7.54, The std.dia of driven pulley=400 mm



Step2: Calculation of design power in Kw:

Design power (KW) =rated Kw x Ks------------------>1 Kα x Kd

From DD 7.53, load correction factor Ks = 1.5(for rolling mill) Arc of contact Kα= 180o-(D-d/c) x60o = 180o-(1200-400/3000) x60o =164o For 164o , From DD: 7.53 , Kα=1.06 From DD: 7.62 Kd=0.8 Substituting all the values in eq 1 Design Kw=20x1.5 1.06x0.8 =35.373KW Step 3: selection of belt: From DD 7.54, FORT belt ducking is selected. It‘s capacity=0.0289 KW/mm/ply Step 4: Load rating correction:- Velocity of the belt v= πd1N1

60 = πx0.4x720 60 = 15.08 m/s From DD 7.54, load rating at‖V‖ m/s = [load rating at 10 m/s] x v/10 = 0.0289 x (15.08/10) = 0.04358 Kw/mm/ply Step 5: determination of belt width:- Width of the belt = design power Load rating x no of plies = 3.5377 0.04358 x 6 =135.29 mm



From DD 7.52, for 400mm dia , and velocity 15.08m/s, the no. of plies=6 From DD 7.52, the standard width of belt =152mm Step 6: determination of pulley width:- From DD 7.54, pulley width = belt width+25mm =152+25 =177 mm From DD 7.55, standard width of pulley=180mm Step 7: calculation of length of belt: (L) From 7.53, for open belt, L=2C+π/2(D+d) + (D-d)2

4 x c

=2x3000+π/2(1200+400) + (1200-400)2 4 x 3000

=8566.6 mm

DESIGN OF “V‟ BELT:

9.Design a “V” belt drive to the following specification:

Power to be transmitted =7.5 Kw Speed of driving wheel =1440 rpm Speed of driven wheel =400 rpm Diameter of driving wheel =300 mm Centre distance =1000 mm Service =16 hrs/days

Suggest a suitable multiple ‟V‟ belt drive for the application. Also calculate the actual belt tensions and stresses induced. [Co1 – H3 - Nov12] [Co1 - H3 - Apr13] [Co1 - H3 - Apr15] [Co1 - H3 - Nov15]

Given:

N1=1440 rpm N2=400 rpm C=1000mm=1m d=300mm=0.3m



To find: Design a ‗V‘ belt drive: Solution:

Step1: selection of the belt section

From DD 7.58,for 7.5KW power the suitable cross section is ―B‖

Step2: calculation of pulley diameter (D,d)

WKT,

N1/ N2 = D/d

1440 = D 400 315 D=1134 mm From 7.54 The std.dia of driven wheel for 300mm=315 mm From 7.54 the standard D=1250mm Step 3: selection of centre distance(c) C=1000mm (given) Step 4: determination of nominal pitch length. From 7.61 L=2C+π/2(D+d) + (D-d)2

4 C

=2x1000+π/2(1250+315) + (1250-315)2 4 x 1000

=4676.85 mm

From 7.60,

for ―B‖ section the standard nominal pitch length is 4996mm

Step 5: selection of various modification factors.

(i)Length correction factor (Fc) From DD 7.60 for ―B‖ section Fc=1.18 (ii)Correction factor for arc of contact(Fd): From DD:7.68,



Arc of contact=180o-60o(D-d/c) =180o-60o (1250-315/1000) =123.9o For 123.9o the value of Fd=0.83 (iii)Service factor(Fa): From DD 7.69, for light duty,16 hrs continuous Fα=1.3

Step 6: calculation of maximum power capacity

From DD 7.62,

for ―B‖ section,

KW=(0.79s-0.09-50.8/de-1.32x10-4 s2)s

Where s=belt speed= πdN1= πx315x1440 =23.75 m/s 60 60 de=dp x Fb

dp=pitch dia of smaller pulley=d=315 mm Fb=for 3.6 speed ratio =1.14 de=315 x 1.14=359.1mm But max.value of de =175 mm must be taken KW= [0.79x23.7 5-0.09-50.8/175 -1.35x10-4(23.75)2]23.75 = 5.445KW Step 7: determination of number of belts (nb): From DD: 7.70, nb=Px Fa/ KwxFc x Fd =7.5 x1.3/5.445 x 1.18 x0.83 =1.828 =2 belts Step 8: calculation of actual centre distance From DD: 7.61,

Cactual = A+ A= L/4-π[D+d/8] =4996/4-π[1250+315/8] =634.42 B = (D-d)2/8 =(1250-315)2/8=109278

Cactual=634.42+√( )

=1175.92 mm



Step 9: calculation of belt tensions(T1,T2) WKT, P=( T1-T2)V 7.5 x 103/2 =( T1-T2)x 23.75 T1-T2=157.89------------->1 WKT, T1-mv2/ T2-mv2=eµαcosecβ From 7.58 for‖B‖ section m=0.189 kg/m From 7.70 for ―B‖ section angle 2β=34o For β=17o From step5,we found that, arc of contact α=123.9o =123.9 x π/180 =2.162 radi From step 6: we found that v=23.75 m/s Substituting all values in the above equation T1-0.189(23.75)2/ T2-0.189(23.75)2=e0.3x2.162xcosec17o

T1-106.6/ T2 -106.6=9.193

T1-106.6=9.193 T2-979.97

T1-9.193 T2=-873.37 T1- T2 =157.89 ------------------------- -8.193 T2 =-1031.26 T2 =125.87N, and hence, T1 =283.76N Step 10: calculation of stresses induced: From DD: 7.58, For ―B‖ section Nominal width=17mm Nominal thickness=11 mm Area of cross section=187 mm2,(17x11=187) Stress induced = maximum tension (T1)/cross section area =283.76/187 N/mm2 =1.517 N/mm2



10.A V-belt drive is to transmit 45kw in a heavy duty saw mill which works in two

shift of 8hrs each. The speed of motor shaft is 1400rpm with the

approximate speed reduction of 3 in the machine shaft. Design the drive and

calculate the average stress induced in the belt. [Co1 - H3 - May14]

Solution:-

Solve this problem by following the above procedure. [Prb :No:9]

11. A centrifugal pump running at 340rpm is to be drive by a 100kw motor

running at 1440rpm. The drive is to work for atleast 20 hours every day. The

center distance between the motor shaft and pump shaft is 2000mm, suggest a

suitable multiple V-belt drive for this application. Also calculate the actual belt

tensions and stress induced. [Co1 - H3 - Nov13]

Solution:-

Solve this problem by following the above procedure.

12. Two shafts whose centers are 1m apart are connected by a “V” belt drive.

The driving pulley is supplied with 100Kw and has an effective diameter of

300mm It runs at 1000 rpm. While the driven pulley runs at 375 rpm. The angle of

groove on the pulley is 40o the permissible tension is 400m2 cross sectional

area of belt is 2.1mpa. The density of the belt is 1100Kg/m3. Taking µ =0.28,

estimate the number of belts required. Also calculate the length required of

each belt. [Co1 - H3 - Nov11]

Given:

C=1m,p=100kw=100x103w,d=300mm,N1=1000rpm,N2=375rpm,2β=40o or β=20o,

a=400 mm2 (or) 400x10-6 m2, =2.1Mpa =2.1 x106 N/m2, =1100 Kg/m3,µ=0.28

To find:

No of belt required and length of each belt.

Solution:

WKT, N1/N2=D/d

1000/375=D/300



D=800mm

v=πdN1

60

=πx0.30x1000

60

=15.71 m/s

For an open belt drive,

Sinα=D-d/2c=800-300/2x1000=0.25

α=sin-10.25=14.48o

angle of contact (180o-2α)xπ/180

=(180o-2x14.48)xπ/180

=2.636 rad

Tension ratio, T1/T2=eµ cosecβ =e0.28x2.636xcose20

T1=2.158 T2-------------->1

WKT, maximum tension T= T1+Tc

Hence,T= xa=2.1x106x400x10-6=840N

Tc=mv2

=(1100x400x10-6x1)(15.71)2

=108.59N

T1=T-Tc

=840-108.59=731.41N

Therefore T2=338.93N

Power transmitted p=( T1-T2)V

=(731.41-338.93)(15.71)

=6165.86W

No of belts=total power transmitted/power transmitted per belt

=100x103/6165.86=16.22 =17 nos

Length of each belt:

L=2C+π/2(D+d) + (D-d)2

2c



=2x1+π/2(0.8+0.3) + (0.8-0.3)2

2x1

=3.852m

13. Select a wire rope for a vertical mine hoist to lift a load of 20KN from a depth

of 60m. a rope speed of 4m/s is to be obtained in 10 sec. [Co1 - H3 - Nov09] [Co1

- H3 - Nov15]

Given data:

Height=60m

W=20KN=20x103N

V=4m/s=240m/min

T=10sec

To find:

Design a wire rope

Solution:

Step 1:

Selection of suitable wire rope

Given that wire rope is used for mines, the std designation of wire

rope=6x19rope

Step 2:

Calculation of design load:

Design load=load to be lifted x assumed factor of safely(15)

=20x15=300KN

Step:3

Selection of wire rope diameter(d)

From DD:9.5, for nominal breaking strength of rope 300KN(take

design load as breaking strength)

Diameter of rope d=25mm

And σu=1600 N/mm2 to 1250 N/mm2

And breaking strength=340 KN



Step4:

calculation of sheave diameter(D):

From DD:9.1 ratio of drum diameter to rope diameter for the size 6x

19 & class ―4‖

D/d=27

This value is for rope speed of 50m/min

But in our problem the rope speed is given as 4m/s = 240m/min

To find D/d value for this speed, modified D/d value is

D/d=27x(1.08)5-1

=36.73 say ―40‖ [240/50=5times]

D/d=40

D=40d=40x25=1000mm

Step 5:

Selection of area of useful cross section of the rope(A):

Type of rope area(A)mm2

6x7 0.38d2

6x19 0.40d2

6x37 0.40d2

So, for 6x19 size of rope A=0.4(25)2=250mm2

Step 6:

Calculation of wire diameter(dw)

dw=d/ i=25/1.5 6x19=1.56mm

[i= no of strands x no of wires in each strands]

step 7:

selection of weight of rope(Wr)

from DD: 9.5 for d=25mm the app.weight

=2.41 kgf/m

=2.41x9.81 N/m=23.6 N/m

Weight of 60m rope=23.6x60=1418N



Step 8:

Calculation of various loads:

i) Direct load Wd=W+Wr

=20000+1418=21418N

ii) Bending load Wb=σb xA

=Er dw/D xA

Assume Er=0.84x105x1.56/1000 x250

=32760 N

iii) Acceleration load Wa=(W+Wr/g)a

A=acceleration of the load=V2-V1/t1=4-0/10

=0.4 m/s2

Wa=(20000+1418/9.81)0.4

=873.23N

iv) Starting load(Wst)

Wst=2xWd=2x21418

=42832 N

Step 9:

Calculation of effective loads on the rope:

i) effective load during normal working:

Wen=Wd+Wb

=21418+32760=54178N

ii) effective load during acceleration of the load

Wea=Wd+Wb+Wa

=21418+32760+873.23

=55051.23N

iii) effective load during starting West

West=Wb+Wst

=32760+42832

=75592N



Step 10: calculation of working factor of safety(Fsw)

Fsw=breaking load/ Wea

=340000/55051.23

=6.176

Step 11:

Check for safe design

From DD:9.1, factors of safety for class 4,

Cranes and hoists=6.0

Since the working FOS>recommended FOS

Therefore the design is safe.

14.At the construction site, 1tonne of steel is to be lifted up to a height of 20m

with the help of 2 wire ropes of 6x19 size, nominal diameter 12mm, and breaking

load 78 kN. Determine the factor of saftey if the sheave diameter is 56 d and if

wire rope is suddenly stopped in 1 sec when travelling at a speed of 1.2m/s.

what is the factor of safety if bending load is neglected? [Co1 - H3 - Nov14]

GIVEN:

Height=20m

W=78KN=78x103N

V=4m/s=72m/min

T=1sec

Size=6x19

To find:

Design a wire rope

Solution:

Step 1: Selection of suitable wire rope

Given that wire rope is used for mines, the std designation of wire

rope=6x19rope



Step 2:

Calculation of design load:

Design load=load to be lifted x assumed factor of safely(15)

=78x15=1170KN

Step:3

Selection of wire rope diameter(d)

From DD:9.5, for nominal breaking strength of rope 1170KN(take design

load as breaking strength)

Diameter of rope d=12mm

And σu=1600 N/mm2 to 1250 N/mm2

And breaking strength=340 KN

Step4: calculation of sheave diameter(D):

From DD:9.1 ratio of drum diameter to rope diameter for the size 6x 19 &

class ―4‖

D/d=27

This value is for rope speed of 50m/min

But in our problem the rope speed is given as 1.2m/s = 72m/min

To find D/d value for this speed, modified D/d value is

D/d=27x(1.08)5-1

=36.73 say ―40‖ [72/50=2times]

D/d=40

D=40d=40x12=480mm

Step 5:

Selection of area of useful cross section of the rope(A):

Type of rope area(A)mm2

6x7 0.38d2

6x19 0.40d2

6x37 0.40d2

So, for 6x19 size of rope A=0.4(12)2=57.6mm2



Step 6:

Calculation of wire diameter(dw)

dw=d/ i=12/1.5 6x19=1.56mm

[i= no of strands x no of wires in each strands]

step 7: selection of weight of rope(Wr)

from DD: 9.5 for d=12mm the app.weight

=2.41 kgf/m

=2.41x9.81 N/m=23.6 N/m

Weight of 60m rope=23.6x20=472N

Step 8:

Calculation of various loads:

v) Direct load Wd=W+Wr

=78000+472=78472N

vi) Bending load Wb=σb xA

=Er dw/D xA

Assume Er=0.84x105x1.56/1000 x57.6

=7547.90 N

vii) Acceleration load Wa=(W+Wr/g)a

A=acceleration of the load=V2-V1/t1=4-0/10

=0.4 m/s2

Wa=(78000+472/9.81)0.4

=3199.6N

viii) Starting load(Wst)

Wst=2xWd=2x78472

=156944 N

Step 9:

Calculation of effective loads on the rope:

iv) effective load during normal working:

Wen=Wd+Wb



=78472+7547.90=86019N

v) effective load during acceleration of the load

Wea=Wd+Wb+Wa

=78472+7547.90+3199

=89218.9N

vi) effective load during starting West

West=Wb+Wst

=7547.90+156944

=164491N

Step 10: calculation of working factor of safety(Fsw)

Fsw=breaking load/ Wea

=340000/89218.9

=3.81

Step 11:

Check for safe design

From DD:9.1, factors of safety for class 4,

Cranes and hoists=6.0

Since the working FOS<recommended FOS

Therefore the design is unsafe

15. Select a suitable wire rope from 6x37 groups to lift a maximum load of 10KN

through a height of 60m. The weight of bucket is 2KN maximum lifting speed is

2m/s which are attained in 3 seconds. Drum diameter is 30 times the rope

diameter. Factor of safety is 6. [Co1 - H3 - May10]

DESIGN OF CHAIN DRIVE

16. Design a chain drive to actuate a compressor from a 10KW electric motor at

960 rpm. The compressor speed is to be 350 rpm. Minimum centre distance

should be 0.5m. Motor is mounted on an auxiliary bed. Compressor is to work

for 8hrs/day [Co1 - H3 - Apr09] [Co1 - H3 - Nov12] [Co1 - H3 - Apr15]



Given:

Power to be transmitted p=10KW

Motor speed N1=960 rpm

Compressor speed N2=350 rpm

Centre distance a=0.5m=500mm

Service=8hrs/day

To find: design a chain drive

Solution:

Let us assume roller chain. Since the optimum centre distance is 30 to 50

pitches.

Assume, a=35p

P=a/35=500/35=14.3mm

From 7.72, standard pitch,p=15.875mm

Transmission ratio i=N1/N2=960/350=2.74

From 7.74, for i=2.74 the number of teeth on pinion sprocket,

Z1=25(assume)

Z2=iZ1

=2.74x25

=69

The load applied on the chain due to transmitted power

Q=PKsKn/v

Ks=K1K2K3K4K5K6

K1=1.5(load with heavy shock)

K2=1(adjustable support)

K3=1(a=30-50p)

K4=1(60o to horizontal)

K5=1(Assume drop lubrication)

K6=1(8hrs/day service)



Ks=1.5x1x1x1x1x1=1.5

From DD:7.77, minimum factor of safety Kn=11 for K5=1 &z1=25.

Chain velocity=v=Z1n1p/60x1000=25x960x15.875/60x1000

=6.35m/s

Minimum breaking load

Q=10000x11x1.5/6.35=25984N

From DD:7.72 for pitch value of 15.875mm and breaking load 25984

the suitable chain is Dr50 rolon chain

Checking for actual factor of safety:

Actual factor of safety Kn=Q/∑F

∑F=Ft+Fc+Fs

Ft=tangential load=P/V=10,000/6.35=1575N

Fc=centrifugal tension =WV2/g=17.8x6.352/9.81=73.2N

Fs=tension due to sagging=KWa=4x17.8x0.5

=35.6N

(k=4=sagging co-effi)

Wi=Wt/m length of chain from 7.72DD

∑F=1575+73.2+35.6=1684N

Kn=44400 = 26.4

1684

Since actual Fos is>adopted minimum value(11) the chain selection

is correct.

Checking for induced bearing stress:

Σ=p.Ks/Av=10,000x1.5

140x6.35

[from 7.72 A=140mm2 ]

= 17 N/mm2

This is less than the allowable bearing stress(From DD:7.77

σ=22N/mm2), hence the chain selection is correct.



Length of chain: (in terms of pitches)

From 7.75, Lp=2αp+(Z1+Z2/2)+(Z2-Z1/2π)2/αp

ap=a0/p=500/15.875=31.5

Lp=(2x31.5)+(25+69/2)+(69-25/2π)/31.5

Lp=111.5 =112pitches

Actual length of chain L=Lpxp

=112x15.875

=1778 mm

Corrected centre distance:

From 7.75, a=( )

Where e=Lp-(Z1+Z2/2)

=112-(25+69/2)

E=65

M=49(for Z2-Z1=44)

a=( )

=504 mm

Decrement allowance=0.01xa=0.01x504

=5.04mm

Exact centre distance=504-5.04

=499mm =500mm

Calculation of pitch circle diameters:

i) pitch dia of pinion sprocket d1=p/sin(180/z1)

=15.875/sin(180/25)=127mm

ii) pitch dia of wheel sprocket: d2=p/sin(180/z2)

=15.875/sin(180/69)=349mm

Resulting design dimensions:

Type of chain=10A-2 DR50 Roller chain

Centre distance=500mm



No of teeth of pinion sprocket=25

No of teeth of wheel sprocket=69

Length of chain=1778mm

Pitch dia of pinion sprocket=127mm

Pitch dia of wheel sprocket=349mm

17. Design a chain drive to actuate a compressor from a 15KW electric motor at

1000 rpm. The compressor speed is to be 350 rpm. Minimum centre distance

should be 0.5m. The Compressor is to work for 15hrs/day. The Chain tention

may be adjusted by Shifting the motor. [Co1 - H3 - May14]

Solve this problem by following the above procedure. [Prb:No:16]

18. the transport of a heat treatment furnace is driven by a 4.5kw, 1440rpm

induction motor through a chain drive with a speed reduction ratio of 2.4. The

transmission is horizontal with bath type of lubrication. Rating is continuous

with 3 shifts per day. Design the complete chain drive. [Co1 - H3 - Nov13]

Given:

Power to be transmitted p=4.5KW

Motor speed N1=1440 rpm

Furnace speed N2=1440/2.4=600 rpm

Centre distance a=0.5m=500mm

Service=3 shift /day

To find: design a chain drive

Solution:

Let us assume roller chain. Since the optimum centre distance is 30 to 50

pitches.

Assume, a=35p

P=a/35=500/35=14.3 mm

From 7.72,



standard pitch, p=15.875mm

Transmission ratio i=2.4

From 7.74,

for i=2.4 the number of teeth on pinion sprocket, Z1=25(assume)

Z2=iZ1

=2. 4x25

=60

The load applied on the chain due to transmitted power

Q=P Ks Kn /v

Ks=K1K2K3K4K5K6

K1=1.5(load with heavy shock)

K2=1(adjustable support)

K3=1(a=30-50p)

K4=1(60o to horizontal)

K5=1(Assume drop lubrication)

K6=1(8hrs/day service)

Ks=1.5x1x1x1x1x1=1.5

From DD: 7.77,

Minimum factor of safety Kn=11 for K5=1 & z1=25.

Chain velocity=v=Z1n1p/60x1000

=25x1440x 4.5/60x1000 =2.7m/s

Minimum breaking load

Q=4500 x11x1.5/2.7=27500 N

From DD: 7.72

For pitch value of 15.875mm and breaking load 27500 N

The suitable chain is DR-50 roll-on chain



Checking for actual factor of safety:

Actual factor of safety Kn=Q/∑F

∑F=Ft+Fc+Fs

Ft=tangential load=P/V=4500/2.7=1666.66 N Fc=centrifugal tension =WV2/g=17.8 x 2.72/9.81=13.2N Fs=tension due to sagging=KWa=4 x 17.8 x 0.5 =35.6 N (k=4=sagging co-effi) Wi=Wt/m length of chain from 7.72DD ∑F=1666.66 +13.2+35.6=1715.46 N K n =27500/1715.46=16.0

Since actual Fos is>adopted minimum value(11) the chain selection is correct. Checking for induced bearing stress:

Σ=p.Ks/Av =4500x1.5

140 x 2.7

=17.85 N/mm 2

[From 7.72 A=140mm2 ]

= 17 N/mm2

This is less than the allowable bearing stress (From DD: 7.77

σ=22N/mm2), hence the chain selection is correct.

Length of chain: (in terms of pitches)

From 7.75, Lp=2αp+ (Z1+Z2/2) +(Z2-Z1/2π)2/αp

ap=a0/p=500/15.875=31.5

Lp=(2x31.5)+(25+60/2)+[(60-25/2π)]2/31.5

=63 +42.5 +0.986

Lp=106.49 =112pitches

Actual length of chain L=Lpx p

=112x15.875 =1778 mm



Corrected centre distance:

a = √

x p

a = √

x 15.875

= 504 mm

From 7.75,

Where , e=Lp - (Z1+Z2/2)

=112-(25+69/2)

e=65, M=49(forZ2-Z1=44)

Decrement allowance=0.01x a=0.01x504

=5.04mm

Exact centre distance=504-5.04

=499mm

=500mm

Calculation of pitch circle diameters:

i) pitch dia of pinion sprocket d1=p/sin(180/z1)

=15.875/sin (180/25) =127mm

ii) pitch dia of wheel sprocket: d2=p/sin(180/z2)

=15.875/sin (180/60) =305mm

Resulting design dimensions:

Type of chain=10A-2 DR50 Roller chain

Centre distance=500mm

No of teeth of pinion sprocket=25

No of teeth of wheel sprocket=60

Length of chain=1778mm

Pitch dia of pinion sprocket=127mm

Pitch dia of wheel sprocket=305 mm



DESING OF SILENT CHAIN

20. A 12.7mm pitch silent chain operating under steady load conditions;

transmit 30 KW from an electric motor to a centrifuge. Design the chain. [Co1 -

H3 - Apr11]

Given:

Pitch p=12.7, power=30KW

To find: design the silent chain

Solution:

Step1: roller diameter (dr):

dr=5/8 x pitch

=5/8 x12.7=7.9375

Step2: Pin diameter (dp)

dp=5/16 x pitch=5/11x12.7

dp=3.968 mm

step3: chain width (b)

bi =5/8xpitch=5/8x12.7

bi=7.9375

step4: thickness of link plate (tp)

tp=1/8xpitch=1/8x12.7

=1.5875mm

Step 5: width b/n outer plates(bs)

bs=bi+2tp

=7.933+2(1.5875)

=11.11mm

Step 6: max.length of pin link plates(ho)

ho=0.82xpitch=10.414mm

Step7:

Max.length of roller link plate(hi)

hi=0.95xpitch=0.95x12.7

=11.43mm



Step 8: length of roller(lr)

lr=0.9bs-0.15

=0.9x(11.11)-0.15

lr=9.849mm



UNIT-2

SPUR GEARS AND PARALLEL AXIS HELICAL GEARS

Speed ratios and number of teeth- Force analysis -Tooth stresses - Dynamic

effects – Fatigue strength- Factor of safety - Gear materials – Design of straight

tooth spur & helical gears based on strength and wear considerations – Pressure

angle in the normal and transverse plane- Equivalent number of teeth-forces for

helical gears.

PART A . 1,Define Module. [CO2 - L3 - Nov13 ] [C02 - L3 - Nov15]

Module (m) is the ratio of pitch circle diameter to the number d of gear teeth, and is usually represented in millimeters.

. 2,Specify the effect of increasing the pressure angle in gear design. [CO2 - H1 - Nov14]. [CO2 - H1 - May14] It is the angle formed by the line of action with the common tangent to the pitch circles

of mating gears. For involute system of gears, the pressure angle is constant and it

may be 14 ½0 (or) 200

3. Why are gear drives superior to belt drives or chain drives? The advantages of gear drives? [CO2 – L2]

1. The gear drives possess high load carrying capacity, high compact layout. 2. They can transmit power from very small values to several kilowatts.

3. Write short notes on backlash of gears. [CO2- L1 - Apr10] 4. Backlash can be defined as the play between a mating pair of gear assembled

condition. 5. How are gears classified? [CO2 - L1]

Gears are classified based on (a) Axes of gear shafts as

(i) Parallel - Eg. spur, helical, herring-bone gears. ii) Intersecting - Eg. Bevel gears. iii) Non-parallel and non-intersecting - Eg.- worm, gears, Skew gears. b) Profile of gear tooth

(i) Involute gears. ii) Cycloidal gears.



(c) Position of teeth on wheel rim. (i)Tooth parallel to axis of gear - Eg. Spur gears. (ii) Tooth inclined to axis of gear - Eg. Helical gears.

(d) Pressure angle i) Gears with 201 pressure angle. ii) Gears with 14 1/20 pressure angle. 5. IlIustrate the materials for making gears'. [CO2 - H1]

1. Ferrous metals such as carbon steels, alloy steels of nickel, chromium and vanadium.

2. Cast-iron of different grades. 3. Non-ferrous metals such as brass, bronze, etc. 4. Non-metals like phemolic resins nylon, bakelite etc.

Among them steel with proper heat treatment is extensively, employed in many of' the engineering applications. 6. Specify the types of gears-failures. [CO2 – L2]

a) Tooth breakage. b) Pitting of tooth surface. c) Abrasive- wears. d) Seizing of teeth etc.

7. What is meant by spur-gear? [CO2 - L1]

Spur-gear is the gear in which teeth are cut at the circumference of a slab called as gear-blank such that the teeth are parallel to qear-axis. 8. Define the following terms. [CO2 - L3 - Apr07] a) Tip circle. b), Root circle. c) Pitch circle. a) Tip circle or addendum circle is the circle which coincides crests or tops of all teeth. b) Root circle or addendum circle is the circle which coincides with. roots or bottoms of all teeth. c) Pitch circle is the imaginary circle in which the pair of gears rolls one over the other. This circle can be visible when the pair of gears fatly rotating. This will lie between tip circle and root circle. 9. How are the following terms defined? [CO2 - L2 - Apr14] a) Pressure angle. b) Module. a)Pressure angle (a) is the angle between the common normal two gear teeth at the point of contact and the common tangent at the pitch point. The standard pressure angle are 141/2o and 20o b) Module( m )is the ratio of pitch circle diameter to the number d of gear teeth, and is usually represented in millimeters.



10. Define the following terms. [CO2 – L2] a) Back lash b) Gear ratio

a) Back lash is the difference between tooth thickness and the space into 'which it meshes, measured along the pitch circle. If we assume the tooth thickness as t, and space width as t2 then backlash = t2 - tl .

b) Gear ratio (i) is the ratio of number of teeth of larger gear to that of smaller gear. At is also defined as the ratio of high speed to the low speed in a gear drive. Usually, the gear ratio should always be greater than one. 11.What factors influence backlash? [CO2 - H1]

The factors like errors in tooth thickness, pitch, tooth spacing, mounting misalignment, etc influence the backlash. 12. What preliminary design considerations should be, adopted, when selecting gear drive? [CO2 - L1]

All kinds of gears cannot be useful for all kinds of work. Hence following factors should be considered for selecting a specific type of gear drive.

i) The amount of. power to be transmitted. ii) Space availability. iii) Amount of gear ratio for single step. iv) Causes for gear failures and their preventing methods. v) Proper material vi) Life of gears required, usually 10,000 hours.

13. What is interference in gears? How can you overcome it? [CO2 – L3]

Gear profile usually starts, from base circle and ends with tip teeth are made in such a way that their contact is along the pro the top surface of teeth is made, flat, the tip of the teeth of one gear dig I into the bottom flank of mating gear. This action is called interference. 14. On what basis gear cutters are selected? [CO2 - L1]

Gear cutters are selected based on the following conditions. 1) Properties of materials for work piece and tools. 2) Cost of production. 3) Structure of gears such as spur gear, helical gears and so on. 4) Module of the gear.

15. How do gears fail? [CO2 – L2] a)Gears may fail due to tooth breakage by overload and misalignment of shafts. b)corrosion of teeth by improper lubricants. c) tooth wear because of insufficient lubrication. d) interference because of no under-cut.

16. What is working depth of a gear-tooth? [CO2 - H1] Working depth of gear is the radial distance between addendum circle and

clearance circle. It is equal to two times, the addendum value.



17. What stresses are induced in gear tooth? [CO2 – L3] 1. Surface compressive stress. 2. Bending stress.

1. Define form factor? [CO2 - L1] Form factor is a constant, employed in the design of gear which, design the

shape and the number of teeth. 19.Why dedendum Value is more than addendum value? [CO2 - L1] In order to get clearance between the teeth of one gear and bottom surface of mating gear so as to avoid interference, dedendum is having more value than addendum. 20. What is a helical gear? [CO2 – L3] A helical gear is a cylindrical gear similar to spur-gear except that the teeth are cut at an angle, known as helix angle 'to the axis of the gear shaft, whereas in spur-gear, teeth are cut parallel to the axis. 21.1n what ways helical gears are differed from spur gears? [CO2 - L1]

Spur gears Helical gears 1. Teeth are cut parallel to the axis. Teeth are cut inclined to the axis. 2. Entire width of tooth is Gradual engagement is obtained simultaneously engaged with full since their teeth are inclined to width of mating gear. axis. 3. Rough and noisy operation Smooth and silent operation. 4. Less power is transmitted. More power can be. transmitted. 22. What are the advantages of helical gears? [CO2 – L2] Helical gears i) transmit more power. ii) provide smooth and soundless operation. iii) used for high speed and high velocity ratio processes. 23. What is helix angle? How this angle differentiate helical gear from. [CO2 - H1]

Helix angle -is the angle between the axis of the gear and the through tooth face. For helical gear, teeth are cut at an inclined axis, specifled as'helix angle and its value ranges from 80t025' the case of spur gear, tooth-are cut parallel to the axis, the spur gear ls zero.

25. What is a herringbone gears? [Co2 - L3 - Apr07] A herring bone gear is made of two single helical gears attached other hence called as double helical gear in which the teeth of be set in the opposite direction to the teeth of another gear arrangement the axial thrust produced in one gear will be null', thrust produced in another gear, and the resultant thrust is improves the life of the gear. Sometimes, a single cylindrical block is ova p/oyod for making, herring bone,. gear. 26. Write any two applications of a skew gear-drive. (or) Where do we, use skew gears? [CO2 – L3]

The skew gears or crossed helical gears are employed in instruments, distributor drive of automobile



engine etc,where small loads are applied. 27. Differentiate axial pitch and normal pitch of the helical gear. [CO2 - H1]

Axial pitch is the distance, parallel to the axis, between similar faces of adjacent teeth. It is also defined as the circular pitch in the plane of rotation and is denoted as p, Normal pitch is the distance between similar faces of adjacent teeth along a helix on the pitch cylinders normal to the teeth, and is denoted as Pn-

29. What are the main types of gear tooth failure? [Co2 - L1 - Apr13] 1. Tooth breakage (Due to static and dynamic loads) 2. Tooth wear or surface deterioration .Ex; abrasion, pitting and scoring 30. Differintate between circular pitch and diametral pitch. [Co2 - H1 - Nov1] Circular pitch: - It is the distance along the pitch circle between corresponding points of adjacent teeth. Diametral pitch: - it is the ratio of the number of teeth on the gear to the unit length of diameter of the pitch circle. 31. Where do we use spiral gears? [Co2 - L2 - Nov13] Spiral gears are used where we need to transmit the power and motion between two shafts which are not intersecting and non parallel.

32. State the law of gearing. [Co2 - H2 - Apr15]

The law of gearing states that for obtaining a constant velocity ratio, at any instant of

teeth the common normal at each point of contact should always passes through a

pitch point (fixed point) on the line joining the centre of rotation of the pair of mating

gears.

33.Why is a gear tooth subjected to dynamic load[Co2 - L3 - Nov14] If all the teeth of gear are cut perfectly, the engagement of teeth pair of pinion and

gear will be smooth. Sometimes, due to inaccuracies of tools used for tooth cutting,

the gear teeth may not have smooth surface and this defect causes the jerky

operation of toothed gears which results the angular acceleration and decelerations of

the gear wheels even though angular velocity of driving wheel is constant.



35.What condition must be satisfied for the pair of spur gears to have a constant velocity ratio? [Co2 - L3 - May14] For the spur pair to have a constant velocity ratio, the common normal, at the point of

contact between a pair of teeth, must always pass through the pitch point. This is also

called ―Law of gearing‖.

PART - B

GEAR DESIGN USING LEWIS AND BUCKINGHAM EQUATIONS (DESIGN OF

SPUR GEAR RECOMMENDED BY AGMA)

PROBLEMS:

1. Design a spur gear drive required to transmit 45kw at a pinion speed of 800

rpm. The velocity ration is 3.5. The teeth are 20. Full depth involutes with 18

teeth. On the pinion both the pinion and gear are made of steel with a

maximum safe static stress of 180N/mm2 Assume medium shock

conditions. [Co2 - H3 - Nov15] [CO2 - H3 - Apr15]

Given: P=45kw; N1=800 rpm I=3.5

=200 Z1=18

b=180n/mm2 To find : Design a spur gear. Solution:

Since both the pinion and spur gear are made of the same material, The

pinion is weaker than the gear. So we have to design only pinion Selection of

material

Given that the pinion and gear are made of steel. Assume steal is haidened to 200 BHN.

1. Calculation of Z1 and Z2

Number of teeth on pinion z1=18



Number of teeth on gear z2= i X z1

= 3.5X18=63.

2. Calculation of F1:

Tangential load Fi=p/ v X ko (d1=m.z1 m is mm)

V=

=

=

= 0.754m.

Ko=1025 for medium shock condition from table class notes.

Ft=

X 1.25

=

3. Calculation of ignition Fd:

Initail dynamic load Fd=

cv- Velocity factor ; Assuming V=12m/s

=

for accurately hobbed and generated gear with V <20 m/s

=

= 0.333

Fd=

X

=

4. Calculation of Fs :

Beam strength Fs= Xm.b[ b]X Y Where b= Face width =10 X m Y=From factor (DDB.8.53) = 0.154-(0.912/z1) for 200 full depth system. =0.154-(0.912/18)=0.1033

Fs= XmX10mX180X0.1033 =584.15m2



5. Calculation of module:

k/kt Fs Fd

584.15m2

m3 =

m = (283.13)1/3

m 7.26mm. From DDB 8.2, The nearest higher standard module value under choice 1 is 8mm.

6. Calculation of b,d, and v

Face width (b): b=10 X 8 =80mm Pitch circle diameter (d1)= mz1=8 X 18 =144mm

Pitch line Velocity (v) =

=

= 6.03 m/s 7. Recalculation of beam strength (Fs)

Fs= [ ] = = 37385.45 N.

8. Calculation of accorate dynamic load (Fd)

Fd=Ft+ ( )

Ft=p/v =

= 7462.68N c-Dgormation factor (DDB 8.53) =11860 e, for 20o FD, steel and steel e = 0.038 for module upto 8o and carefully cut gears. C= 11860 e =11860 X 0.038 =450.68 N/mm

Fd= 7462.68 + ( )

Fd = 50908.19 N 9. Check for beam strength (or tooth breakage)

Since Fd>Fs, The design is unsatisfactory that is dynamic loud is greater then the beam strength. In order to reduce the dynamic load Fd, slect the precision gears. Therefore from table e = 0.019 for precision gears. Then the dgormation factor c= 11860 X e = 11860 X 0.019



= 225.34

Fd= 7462.68 + ( )

= 32920.46 N Now we find Fd<fs. It means the gear tooth has adequate beam strength and it will not fall by breakage. Therefore the design is Satisfactory.

10. Calculation of maximum wear load (Fw)

Fw= d1 X b X Q X kw

Q = Ration factor =

=

=1.555

Kw= load factor = 0.99919 N/mm2 for steel hardened to 250 BHN. Fw= 144 X 80 X 1.555 X 0.919 = 16462.6 N

11. Check for Wear:

Since Fd>Fw The design is unsatisfactory this is dynamic load is greater than the Wear load. N order to increase the Wear load (Fw) We have the increase the hardness(BHN) so now for steel hardned to 400 BHN kw = 2.553 N/mm2 Fw=144 X 80 X 1.555 X 2.553 = 45733.42 N. Now we find Fw > Fd It means the gear tooth has adequate. Wear capacity and it will not Wear out. Therefore the design is Satisfactory.

12. Basic dimension of pinion and gear:

Module M = 80 mm

Number of teeth = z1=18 and z3=63

Pitch Circle diameter = d2=144 mm

d2=m.Z2 =8 X63

= 504 mm.

Center distance = a = ( )

= ( )

= 324 mm



Face width b = 80mm

Height factor fo =1 for 20o full depth teeth.

Bottom clearance C = 0.25 m

=0.25 X 8 = 2 mm

Tip diameter da1=(z1+2f0)m

=(18 +2 X1) 8

=160 mm

da2=(z1+2fo)m

=(63 + 2 X 1) X 8

= 520 mm

Root diameter df1 =(z1-2f0)m-2c

=(18 – 2X 1) 8-2 X 2

=124 mm

df2=(z1-2f0)m-2c

=(63 – 2 X 1)8 -2 X 2

=484 mm.

2.Design and draw spur gear drive transmitting 30kw at 400 rpm to another

shaft running approximately at 1000 rpm .The load is steady and continuous

.The material for the pinion is cast steel and for gear is cast iron. Take

modulus as 10mm .Also check the design for dynamic load and wear. [CO2 -

H3 - Apr14]

Given: P=30kw; N1=1000 rpm N2=400rpm Z1=18 (assume)

b=180n/mm2 (assume)



To find : Design a spur gear. Solution:

Since both the pinion and spur gear are made of the same material, The

pinion is weaker than the gear. So we have to design only pinion Selection of

material

Given that the pinion and gear are made of steel. Assume steal is haidened to 200 BHN.

1. Calculation of Z1 and Z2

Number of teeth on pinion z1=18 Number of teeth on gear z2= i X z1

= 2.5X18

=45

2. Calculation of F1:

Tangential load Fi=p/ v X ko (d1=m.z1 m is mm)

V=

=

=

= 4.18m/s

Ko=1025 for medium shock condition from table class notes.

Ft=

X 1.25

=8971.29

3. Calculation of ignition Fd:

Initail dynamic load Fd=

cv- Velocity factor ; Assuming V=12m/s

=

for accurately hobbed and generated gear with V <20 m/s



=

= 0.333

Fd= 8971.29 x 0.333

= 2987.44

4. Calculation of Fs :

Beam strength Fs= Xm.b[ b]X Y Where b= Face width =10 X 10=100 Y=From factor (DDB.8.53) = 0.154-(0.912/z1) for 200 full depth system. =0.154-(0.912/18)=0.1033

Fs= X10X100X180X0.1033 =58415N


Face width (b): b=10 X 10 =100mm Pitch circle diameter (d1)= mz1=10 X 18 =180mm


=

= 9.42 m/s 6. Recalculation of beam strength (Fs)

Fs= [ ]

= = 58414.77 N.

7. Calculation of accorate dynamic load (Fd)

Fd=Ft+ ( )

Ft=p/v =

= 3184.71N c-Dgormation factor (DDB 8.53) =11860 e, for 20o FD, steel and steel e = 0.038 for module upto 8o and carefully cut gears. C= 11860 e =11860 X 0.038 =450.68 N/mm

Fd= 3184.71 + ( )

Fd = 46566 N



8. Check for beam strength (or tooth breakage)

Since Fd>Fs, The design is unsatisfactory that is dynamic loud is greater then the beam strength. In order to reduce the dynamic load Fd, slect the precision gears. Therefore from table e = 0.019 for precision gears. Then the dgormation factor c= 11860 X e = 11860 X 0.019 = 225.34

Fd= 3184.71 + ( )

= 28215.32 N Now we find Fd<fs. It means the gear tooth has adequate beam strength and it will not fall by breakage. Therefore the design is Satisfactory.

9. Calculation of maximum wear load (Fw)

Fw= d1 X b X Q X kw

Q = Ration factor =

=

=1.5

Kw= load factor = 0.99919 N/mm2 for steel hardened to 250 BHN. Fw= 180 X 100 X 1.5 X 0.919 = 24813 N

10. Check for Wear:

Since Fd>Fw The design is unsatisfactory this is dynamic load is greater than the Wear load. N order to increase the Wear load (Fw) We have the increase the hardness(BHN) so now for steel hardned to 400 BHN kw = 2.553 N/mm2 Fw=180X 100 X 1.5 X 2.553 = 68931 N. Now we find Fw > Fd It means the gear tooth has adequate. Wear capacity and it will not Wear out. Therefore the design is Satisfactory.

11. Basic dimension of pinion and gear:

Module M = 10mm

Number of teeth = z1=18 and z2=45

Pitch Circle diameter = d2=180 mm

d2=m.Z2 =10X45



= 450 mm.

Center distance = a = ( )

= ( )

= 315 mm

Face width b = 100mm

Height factor fo =1 for 20o full depth teeth.

Bottom clearance C = 0.25 m

=0.25 X 10 = 2.5 mm

Tip diameter da1=(z1+2f0)m

=(18 +2 X1) 10

=200 mm

da2=(z2+2fo)m

=(45 + 2 X 1) X 10

= 470 mm

Root diameter df1 =(z1-2f0)m-2c

=(18 – 2X 1) 10-2 X 2

=96 mm

df2=(z2-2f0)m-2c

=(45 – 2 X 1)10 -2 X 2

=258 mm.



3. A compressor running at 300 rpm is driven by a 15 kw 1200 r.p.m motor

through a 141/2 full depth spur gears. The centre distance is 375 mmm the

motor pinion is to be of c 30 forged steel hardned and tempered and the

driven gear is to be of cast iron. Assuming medium shock condition design

the gear drive completely. [CO2 - H3 - Apr11]

Given: N2=300 r.p.m P=15 kw N1=1200 rpm

=14 ½

To find:

Design the spur gear drive.

Solution:-

Since the materials of pinion and gear are different, first kle have to valuate

[ ]y1 and [

] y2 to find out the Weaker dement

Gear ratio i=

=

=4

Assume z1 =18

Z2=i X z1

=4 X 18

=72

For Pinion:

From factor y1=0.270 from DDB 8.53

Z1=18

Permissible static stress b=112 N/mm2

[ ] y1=112 X

Y1=



= 9.625

For gear:

From factor Y2=0.360; z2=72 permissible static stress[ ]=56 N/mm2

=56 X

Y1=

=6.42

We find [ ]y2 < [

] y1

Ie, the gear is weaker than the pinion therefore we have to design the gear only.

1. Material Selection:

Pinion: c30 Forged steel ; and

Gear : Cast iron

2. Calculation of module (m)

Since the centre distance (a) is give, we need not to equate Fs and Fd to find the module.Here the module can be calculated using the relation a =m(z1+z2)/2 375=m(18+72)/2 m =8.333 from nearest value higher standard module 1-10.

3. Calculation of b,d and v

Face Width (b) = 10 X m =10 X 10 =100mm Pitch circle diameter of pinion (d1) = m.z1 = 10 X 18 =180 mmm Pitch circle diameter of gear (d2) = m.z2 =10 X 72 =720 mm


=

= 11.31 m/s



4. Calculation of beam strength (Fs)

Beam strength Fs= ].y2

= [

]

=20160 N

5. Calculation of dynamic load (Fd):

Dynamic load Fd = )

Ft= p/v

=

= 1326.26 N c- Dgormation factor = 7850 e for 14o FD steel – Cast iron. e=0.022 for module upto 10 and precision gear. V= 7850 X 0.022 = 172.7 N/mm

Fd = 1326.26 + ( )

= 19911.85 N 6. Check for beam strength (or) tooth breakage

We find Fd < Fs, It means the gear tooth has adequate beam

strength and it will not fall by breakage thus the design is satisfactory.

7. Calculation of Wear load (Fw)

Maximum wear load Fw = d1 X b X Q X kw Q= Ration factor = 2i/i+1

=

= 1.6

kw= Load stress factor = 1 N/mm2 for steel (250 BHN ) Cast iron and 14o FD Fw= 180 X 100 X 1.6 X 1 =28800 N

8. Check for Wear :

We find Fw> Fd. It means the gear tooth has adequate wear capacity and it will not wear out. Therfore the design is satisfactory.



9. Basic dimensions of pinion and gear:

Module m = 10 mm Number of teeth z1 n= 180and z2 =72 Pitch circle diameter d1 = 180 mm d2= 720 mm centre distance a =375 mm Face width b= 100 mm Height factor fo =1 for 14 ½ 0 FD Bottom clearance c= 0.25 m = 0.25 X 9 = 2.25 mm Tip diameter da1 =(z1+2fo)m =(18+2X1)9 =180 mm da2=(z2 + 2fo)Xm =(72 + 2 X 1) X9 = 666 mm Root diameter df1 =(z1-2fo)m-2c = (18 -2 X 1) 9 -2 X 2.25 = 139.5 mm df2= (z2-2fo)Xm -2c =(72 – 2 X 1) 9-2 X 2.25 = 625.5 mm GEAR DESIGN BASED ON GEAR LIFE (Gear design Using Basic relation)

1. In a spur gear drive for a store cruster the gears are made of c 40 steel the

pinion is transmitting 30 kw at 120 rpm the gear ratio is 3. Gear is to work

8 hours per day. Six days a week and for 3 years. Design the drive [CO2 -

H3 - Apr15]

Given data: Pinion and gear material C 40 steel p= 30 kw N1=1200 r.pm i=3 To find: Design the spur gear drive Solution: Since the pinion and gear are made of same material (ie c 40 steel) therefore we have to do the design of pinion alone. 1. Gear ratio : i=3

2. Material Selection: Pinion and gear are made of c40 steel.

Assume surface hardness > 350 3. Gear Life:



Given that the gear is to work 8 hours per day. Six days a week, and for 3 years therefore gear life in terms of hours is given by Gear life = 80X (52 X 6) X 3 =7488 hours =449280 min. Life in numbers of cycle, N= 449280 X N1 = 449280 X 1200 = 53.9 X 10 7 cycle.

4. Calculation of intail design torque (Mt)

Design torque [Mt]= Mt.k.kd

Mt=

=

= 238.73 N -m k.kd = 1.3 [Mt]= 238.73 X 1.3 = 310.34 N-m

5. Calculation of E eg [ ] and [ ]

(i) To find E eg : c=40 steel

E eg = 2.15 X 105 N/mm2

(ii) To find[ ]

The design bending stress is given

Kb1=0.7 for HB > 350 and N N= 2 for steel tempered

K 0.35 +120 for c 40

=0.35 X 630 +120

=

X 340.5

= 111.23 N/mm2

(iii) To find ( )

=cR HRC.kcl CR = 26.5 for c 40 steel hardned and tempered. HRC = 40 to 55 for c 40

Kcl= 0.585 for HB > 350 and N 25 X 107

=26.5 X 55 X 0.585 = 852.64 N/mm2



6. Calculation of centre distance (a)

a (i+1) √( )

[ ]

X

[ ]

a (3+1) √(

)

a=155 mm

7. Selection of z1 and z2

(i) Assume z117 for 20o full depth system (ii) z2=i X z1=3 X 17 =51


WkT m =

=

( )

( )

=4.56 mm.

9. Revision of centre distance

New centre distance a= ( )

= ( )

= 170 mm 10. Calculation of b,d, v and p

Face width (b) = = Pitch diameter of pinion (d1) = d1=m.z1 =5 X 17 = 85 mm


=

= 5.34 m/s

p = b/d1 = 51/85= 0.6

11. Selection of quality of gear

Pitch line Velocity 5.34 m/s is quatlity 8 gears are selected.

12. Revision of design torque of gear (Mt)

Revise k= p =0.6 and for bearing close to gears k=1.03 Revise kd = Is quality 8 HB >350 and v=5.34 m/s Kd=1.4



Revise [Mt] = Design torque [Mt] =Mt X k X kd =238.73 X 1.03 X 1.4 =344.24 N.m

13. Check for bending:

Calculation of induced bending stress

( )

Y Form factor = 0.366 z1

( )

We find < [ ] (i.e 86.78 N/mm2<111.23 N/mm The design is satisfactory.

14. Check for Wear strength

√

[ ]

√

= 765.9n/mm2

We find 765.9 N/mm2<852.64 N/mm2 design is satisfactory. 15. Calcultaion of basic dimension of pinion and gear:

Module m=5 mm Face Width b=51 mm Height factor f0 =1

Bottom clearance c=0.25 m = 0.25 X 5 =1.25 mm Tooth depth h = 2.25 m = 2.25 X 5= 11.25 mm Pitch circle diamete d1=mz1 = 5 X 17 =85 mm d2 =mz2=5 X 51 =225 mm Tip diameter da1=(z1+2fo)m=(17 + 2 X1)5=95 mm da2=(z2+2fo)m=(51 + 2 X1)5 =265mm df1=(z1-2fo)m-2c=(17-2 X 1) 5-2 X1.25 =72.5 mm

df2=(z2-2fo)m-2c = (51-2 X1) 5-2 X 1.25

=242.5 mm.



2. Design a spur gear drive to transmit 22.5 kw at 900 rpm speed reduction

is 2.5 material for pinion and wheel are c15 steel and cast iron grade 30

respectively. Take presure angle 20oand working life of the gear as 10000

hrs. [Co2 - H3 - Apr13 ][CO2 - H3 - Nov13]

Given data: P=22.5 kw N1=900 rpm I=2.5

To find : Design a spur gear. Solution:- Since the materials for pinion and wheel are different, therefore we have to design the pinion first and check both pinion and wheel. 1. Gear ratio i= 2.5

2. Material Selection

Pinion : c15 steel case hardned to 55 Rc and core hardness <350 Wheel C.I grade 30

3. Gear Life N =10000hrs

N=10000 X 60 X 900 = 54 X 107cycles.

4. Design torque [Mt]

[Mt]=Mt.k.kd.

Mt=

=

=238.73 N-m k.kd=1.3 Design torque [Mt] =238.73 X1.3 =310.35 N-m


(i). To find Eeg Pinion steel and cast iron 280 N/mm2 equivalent Young‘s modules Eeg=1.7 X 105 N/mm2

(ii) To find [ ]

[ ] =

X

Assuming rotation in one direction only.



For steel (HB ) and N kbl =1 steel case hardned,

factor of saftey n=2. Steel case hardned, stress concenbation factor kb=1.2

Forged steel ( )

( )

=232.5

= 135.625

6. Calculation of centre distance (a)

a (i+1) √(

)

[ ]

= 0.3

a (2.5+1) √(

)

7. To find Z1 and Z2

((i). For 20o full depth system select z1=18 (ii). Z2=iX z1 = 2.5 X 18 =45


WkT m=

=

= 4.32mm

9. Revision of centre distance

A= ( )

=

( )

= 157.5 mm

10. Calculation of b,d va and

Face width b= = 0.3 X 157 .5= 47.25 mm Pitch diameter of pinion d1=m.z1 = d X 18 = 90 mm



Pitch line Velocity v=

=

=4.24 m/s

11. Revision of design torque[Mt]

k

kd: Is quality 8 v= 4.24 m/s

kd=1.4

[Mt]= Mt.k.kd

=238.73 X 1.03 X 1.4 =344.24 n-m

12. Selection of quality wear

v=4.24 m/s quality 8 gear

13. Check for bending

[ ]

( )

= 85.89 N/mm2

14. Check for wear strength

√

[ ]

=

√

=684.76 N/mm2

< design is safe and satisfactory. 15. Check for wheel

[ ] wheel and [ ]

=

=360 r.p.m

Life of wheel=10,000hrs = 10,000 X 60 X 360



= 21.6 X cycle

[ ] wheel =

X

Kb1=√

= √

= 0.918

N=2; k ;

[ ]

X 130.5

= 69.88

[ ]

[ ] wheel = CB. HB kci

CB=2.3 for cast iron grade 30

HB=200 to 260

Kci=√

= √

= 0.879

[ ] wheel=2.3 X 260 X 0.879

=525.64

Check for bending

=

and = Induced bending stresses in the pinion and wheel respectively.

Y1 and y2 = from factors



Y2=0.471 for z2 =45

85.89 N/mm2 y1=0.377

85.89 X 0.377 = X 0.471

= 68.75 N/mm2

< wheel the design is satisfactory.

Check for wear strength:

Since contact area is same

wheel = pinion =684.76 N/mm2

Decrease the induced conduct stress- Increase the face width

(b) value (or) in order to increase the design contact stress.

Increase the surface hardness say to 340 HB

Increasing the surface hardness will give [ ]=2.3 X340X0.879=684.34

< Design is satisfactory.

16. Calculation of basic dimension of pinion and wheel:

m=5mm b=47.25 mm Height factor fo =1 for full depth Bottom clearance c =0.25m =0.25 X5 =1.25mm Tooth depth h=2.25 m =2.25 X5=11.25mm Pitch circle diameter =d1=m.z1=5 X18 =90mm D2=m.z2=5 X 45 =225mm da1=(z1+2fo)m=(18+2 X1)X5=100mm da2=(z2+2fo)m=(45+1 X2)=235mm Root diameter df1=(z1-2fo)m -2c) = (18 –2X1)5-2 X 1.25 = 77.5mm df2=(z2-2fo)m -2c) =(45-2X1)-2 X 1.25

=212.5mm.



HELICALGEAR DESIGN USING LEWIS AND BUCKINGHAM‟S

EQUATIONS(Helical gear design Recommended by AGMA)

3. Design a helical gear to transmit 15 kw at 1400 rpm to the following

specification speed reduction is 3, pressure angle is 20, Helix angle is 15

the material of both the gears is c45 steal Allowable static stress 180

N/mm2 surface endurance limit is 800 N/mm2, young‟s modules of

material 200Gpa. [CO2 - H3 - Apr13][CO2 - H3 - Nov13] [CO2 - H3 - Apr15]

[CO2 - H3 - Nov15]

Given: P=15kw N1=1400rpm i=3

=20o =15o =180 N/mm2 Fes=800 N/mm2

E1=E2=2 X 105 N/mm2 To Find: Design a helical gear. Solution:- 1. Material Selection : pinion and gear c45 steel

2. Calculation of z1 and z2:

Assume z1=20 z2=i Xz1=3 X 20=60

3. Calculation of tangential load of tooth (Ft)

k/kt Ft=p/v X ko

v=

=

(

)

[ d1=

]

=

=1.518 mn m/s ko=1.25 assuming medium shock

Ft=

=

4. Calculation of initial dynamic load (Fd)

Fd=



Cv=

for v=5 to 20 m/s

=

=0.286 assume v=15 m/s

Fd=

X

=

5. Calculation of beam strength (Fs)

Fs= [ ]

b=Face width =10 mn

y1=0.154-

for 20o involute

Zeq=

=

= 22.192

y1=0.154-

= 0.1143

Fs=

= 646.62 mn2

6. Calculation of normal module (mn)

Fs

646.62 mn2

From DDb 8.2 the nearest higher standard normal module is 5 mm.


Face Width(b) =10 mm=10 X 5 =50 mm

Pitch circle diameter d1=

= 103.53mm

Pitch line velocity (v) =

=

=7.59 m/s 8. Recalculation of the beam strength (Fs)

Fs=

= 5 X = 16158.78 N



9. Calculation of accurate dynamic load (Fd)

Fd=Ft+ ( )

√

Ft=p/v=

= 1976.28 N

c- Dgormation factor - 11860 e for steel and steel 20o full depth

e= 0.025 for mn upto 5 and carefully cut gear

c=11860 X 0.25 = 296.5 N/mm

Fd=1976.28 + ( )

10. Check for beam strength (or tooth breakage)

We find Fs<fd. So the design is unsatisfactory. In order f0 reduce Fd, we try with precision gears. C=1186 X 6.0125 = 148.25

Fd=1976.28 + ( )

= 10863.26 N

Fs>Fd. It means the gear has adequate beam strength and will

not fail by breakage. The design is satisfactory.

11. Calculation of the maximum wear load(fw)

Fw=

Q= Ratio factor

=

=

=1.5

Kw= [

]

= 1.5635 N/mm2

Fw=

= 13011.8 N



12. Check for Wear:

He find Fw > Fd. It means the gear tooth has adequate wear capacity and will not wear out. The design is safe and satisfactory.

13. Calculation of basic dimension of pinion and gear:

Normal Module =mn=5mm Number of feeth = z1=20 z2 =60 Pitch circle dia d1= 103.53 mm

D2=

=

= 310.58mm Centre distance fo=1 Bottom distance c= 0.25mn =0.25 X 5 =1.25 mm Tooth depth h = 2.25 mn = 2.25 X 5 = 11.25 mm

Tip diameter da1= (

)mn

=

= 113.53 mm

da2= (

)mn

= (

)

df1=(

)mn-2c

= (

)5-2 X 1.25 = 91.03 mm

df2=(

)mn-2c

=(

)

Virtual Number of teeth

Zv1=

=

=22.192

Zv2=

=

=66.57

4. A compressor running at 360 rpm is driven y 140 kw, 1440 rpm, motor

through a pall of 20ofull depth helical gears having helix angle of 250 the

centre distance is approximately 40 mm. The motor pinion is to be forged

steel and the driven gear is to be cast steel. Assume shock conditions.

Design the gear pinion. [CO2 - H3 - Nov11]



Given data : P=40kw N2=360rpm N1=1440rpm A=400mm

=20o =25o To Find: Design thehelical gear pair. Solution:-

Since the material for pinion and gear are different, first we have to

evaluate [ ]y11, and [ ]y1

2 To find out the weaker element

Gear ratio I =N1/N2=

= 4

Z1=20 Z2=i X z1 =4 X 20 =80

Zv1=

=

Virtual number of teeth zv2=

Pinion = Forged steel Gear=Grade 1 i.e cs 65 cast steel For pinion

=112 N/mm2

Y1 =0.154-(

) for 200 full depth

=0.154-(

)

=0.1202

[ ]y11=112 X 0.1202

=13.465 N/mm2

Form factor y12=0.154-(

)

= 0.154 -(

) = 15.21 N/mm2

[ ]y11< [ ]y1

2 ie the pinion is weaker we have to design only pinion.

1. Material Selection:-

Pinion 40 Ni 2 crl M628

Year-grade 1 cast steel



2. Z1=20; z2=80

3. Calculation of module

a=(

)X(

) = 7.25 mm


(b) =10 mn=10 X 8 =80 mm

d1=

= 176.54mm

(v) =

=

= 13.31 m/s 5. Calculation of beam strength (Fs)

Fs= [ ]

= X8 X 80 X 112 X 0.1202 =27067.76 N

6. Calculation of accurate dynamic load (Fd)

Fd=Ft+ ( )

√

Ft=p/v=

= 10518.4 N

c- Dgormation factor - 11860 e for steel and steel 20o full depth

e= 0.025 for mn upto 8 and carefully all gear

c=11860 X 0.38 = 450.68 N/mm

Fd=10518 + ( )

=46865.44 N 7. Check for beam strength (or) tooth breakage:

B=10mn =10 X9 =90mm

D1=

X z1 =

X 20 = 198.61 mm

Pitch circle dia d1=

=

= 198.61 mm

Pitch line velocity v=

=

=14.97 m/s



Tangential load Ft = p/v =

= 9352 N

Expected error e = 0.0205 mm for mn

Dgormation factor c= 11860 e

= 11860 X 0.0205

=243.13 N/mm

Fd= 9352 + ( )

= 34104.29 N

Fs =

8. Calculation of the limiting wear load(fw)

Fw=

Q= Ratio factor

=

=

=1.6

Kw= load stress factor =2.533 N/mm2 for steel hardned BHN 400 = 1.5635 N/mm2

Fw=

= 88892.06 N 9. Check for wear:

We find Fw>fs It means the gear tooth has adequate wear capacity and will not wear out thus the design is safe and satisfactory.

10. Calculation of basic dimension of pinion

Normal Module =mn=9mm Face width b = 90 mm Number of feeth = z1=20 z2 =80 Pitch circle dia d1= 198.61 mm

d2=

=

= 794.43mm



Centre distance a

(

) = 496.52 mm

Height factor fo=1 Bottom distance c= 0.25mn =0.25 X 9 =2.25 mm Tooth depth h = 2.25 mn = 2.25 X 9 = 20.25 m

Tip diameter da1= (

)mn

= (

) X 9

= 216.61 mm

da2= (

)mn

= (

)

df1=(

)mn-2c

= (

)9-2 X 2.25 = 176.11 mm

df2=(

)mn-2c

=(

) =771.93 mm


Zv1=27 Zv2=

8.Design a helical gear drive to connect an electric motor to a reciprocating

pump. Gear are overhanging in their shafts. Motor speed=1440 rpm. Speed

reduction ratio=5, motor power=37kw pressure angle=200 Helix angle=250[CO3 -

H3 - Nov14]

Solve this problem by following the above procedure. [Prb:No:6&7] 9.A single stage helical gear reducer is to receive power from a 1440 rpm, 25kw induction motor. The gear tooth profile is involute full depth with 200 normal pressure angle. The helix angle is 230, number of teeth on pinion is 20 and the gear ratio is 3. Both the gears are made of steel with allowable beam stress of 90Mpa and hardness 250 BHN (i) Design the gears for 20% overload carrying capacity from the standpoint of bending strength and wear. (ii) If the incremental dynamic load of 8KN is estimated in tangential plane, what will be the safe power transmitted by the pair at the same speed? [CO2 - H3 - Apr14] Solve this problem by following the above procedure. [Prb:No:6]



HELICAL GEAR DESIGN BASED ON GEAR LIFE

(Helical gear design using basin equation)

10.For intermittent duty an elevator two cylindrical gears have to transmit 12.5

kw at a pinion speed of 1200 rpm. Design the gear pari for the following

specification gear ratio 3.5 pressure angle 20 involutes full depth, helix angle

150 gears are expected to work 6 hours a day for 10 years. [CO2 - H3 - Nov11]

Given data : P=12.5kw N1=1200rpm A=400mm

=20o FD =15o To Find: Design the helical gear. Solution:- 1. Gear Ration : i=3.5

2. Selection of material :

For both pinion and gear allow steel 40 ni Ni 2cr 1 M0 28 can be selected,

consulting data book. Since the gear are of same material, we design only

the pinion.

3. Gear life:-

Given that gear are to work 6 hours a day for 10 years Gear Life = 6 hours/day X 365 day/years X10 = 21,900 hours N=21900 X 1200 X 60 = 157 .7 X 107 cycles.

4. Calculation of initial design torque(Mt)

[Mt]=Mt.k.kd.

Mt=

=

=99.47 N-m k.k0=1.3 Design torque [Mt] =99.47 X1.3 =129.31 N-m




Eeg=2.15 X 105 N/mm2 for steel pinion and steel gear.

[ ] =

X

Assuming rotation in one direction only.

Kbl-0.7 For steel (H> ) and N k =1.5 for steel

hardned.

N=2.5

Forged steel ( )

( )

=662.5

= 173.133

CR=26.5 for alloy steel hardned HRC=40 f0 55

Kcl=0.585 for steel HB>350 and N from DDB(8.17)

= 852.6 6. Calculation of centre distance (a)

a (i+1) √(( )

[ ])

X [ ]

a (3.5+1) √(

)



7. Assume z1=20 z2=iXz1

= 3.5 X 20 =70

8. Calculation of Normal Module (mn)

mn=(

) cos

=2.576mm

From DDB 8.2

9. Revision of centre distance:-

a= (

) x (

)

=(

) x (

)

=139.76mm

10. Calculation of b,d va and

Face width b= = 0.3 X 139= 41.93 mm

Axial pitch pa=

=

= 36.4 mm b>pa

Pitch diameter of pinion d1=

X z1

=

X 20

=62.12 mm

Pitch line Velocity v=

=

=3.903 m/s

11.Selection of Quality of gear:

From DDB 83 HB>350 and v upto 8 m/s is quality 8 is selected. 12.Revision of design torque (Mt)

Mt=Mt X k X kd

k 1.045 =0.676 DDB 8.15 kd=1.2 for is quality DDB 8.16 Mt=99.47 X 1.045 X 1.2 =124.74 N-m



13.Check for bending:

[ ] = ( )( )

= Form factor on virtual number of tooth

Zv1=

=

22

= 0.402 zv1=22 DDB 8.15

[ ] = ( )( )

=55.5 N/mm2

14.Calculation of basic dimension of pinion and gear:

Normal Module =mn=3mm Number of feeth = z1=20 z2 =70 Pitch circle dia d1= 62.12 mm

d2=

=

= 217.4mm Centre distance a=139.76mm Height factor fo=1 Bottom distance c= 0.25mn =0.25 X 3 =0.75 mm Tooth depth h = 2.25 mn = 2.25 X 3= 6.75 mm

Tip diameter da1= (

)mn

= (

) X 3

= 68.12mm

da2= (

)mn

= (

)

df1=(

)mn-2c

= (

)3-2 X 0.75 = 54.6 mm

df2=(

)mn-2c

=(

) =209.91


Zv1= Zv2=

=

=78



UNIT III

BEVEL, WORM AND CROSS HELICAL GEARS

Straight bevel gear: Tooth terminology, tooth forces and stresses, equivalent number

of teeth. Estimating the dimensions of pair of straight bevel gears. Worm Gear: Merits

and demerits terminology.Thermal capacity, materials-forces and stresses, efficiency,

estimating the size of the worm gear pair. Cross helical: Terminology-helix angles-

Estimating the size of the pair of cross helical gears.

PART A

1. State the use of bevel gears. [CO3 – L1 - Nov08]

The bevel gears are used for transmitting power at a constant velocity ratio between

two shafts whose axes intersect at a certain angle.

2. When bevel gears are used? [CO3 - L1 - Apr09] When the power is to be transmitted in an angular, direction, i.e., between the shafts whose axes intersecting at an angle, bevel gears are employed.

3. What are the various forces acting on a bevel gear? [CO3 - L1- Apr13]

1. Tangential or useful component (Ft)

2. Separating force (Fs). It is resolved into two components.

(i) Axial force, (ii) Radial force.

4. For bevel gears define back cone distance. [CO3 - L1]

Back cone distance is the length of the back cone. Back cone is the

imaginary cone perpendicular to the pitch cone at the end of the tooth.

5. What is virtual number of teeth in bevel gears? [CO3 - L2 - Nov14] [CO3 - L2-

- Apr14]

The virtual number of teeth is defined as the number of teeth of an

imaginary spur gear whose radius is equal to the back cone radius R b and having

pitch of the bevel gear. This is called Tredgold‘s approximation.



6. State true or false and justify. “Miter gears are used for connecting non-

intersecting shafts” [CO3 – L2]

False:When equal bevel gears (having equal teeth and equal pitch

angle) connect two shafts whose axes intersect at right angle, then they are known as

miter gears.

\7. What are commonly used materials for worm and wheel? [CO3 - L2 - Apr07]

Worm material: steel, case hardened steel, hardened molybdenum steel.

Worm wheel material: cast iron and phosphor bronze.

8. What is the specific feature of mitre gear? [CO3 – L3]

When equal bevel gears (having equal teeth and equal pitch angle) connect two shafts

whose axes intersect at right angle, then they are known as miter gears.

9. Name the different applications of worm Gear. (or) Write some applications of

worm gear drive. [CO3 – L2]

Worm gear, drive find wide applications like milling machine indexing head, table

fan, and steering rod of automobile and so on.

10. Why the efficiency of worm gear drive is comparatively low? [CO3 – L2]

Because of power loss due to friction caused by sliding.

11. when the number of start of a worm is increased in a worm

gear drive, how it affects the other parameters and action of the drive? [CO3

- L1]

The increase in number of starts on the worm will increase the lead and lead angle of

the worm .this results in higher friction losses and hence the lower efficiency.

12. Define the following terms: a) Cone distance or pitch cone radius. b) Face

angle. [CO3 - H1]

(a) Cone distance or pitch cone radius is the slant length of pitch cone, i.e., distance

between the apex and the extreme point of tooth of bevel gear.

(b) Face angle is the angle subtended by the face of the teeth at the cone centre. It is

equal to the pitch angle plus addendum angle. It is also called as tip angle.



13. Why multi start worm more efficient than the single start one? [CO3 – L2]

The efficiency of the worm depends mainly on pressure angle (also known as pitch

angle of the worm).for a single start worm this pressure angle will be less. In a multi

start worm, this pressure angle can be increased (of the order 45 deg) that‘s why multi

start worm is more efficient.

14. Mention the reasons for irreversibility in worm gears. [CO3 - L1 - Apr10]

The drive is called irreversible if the worm gear cannot drive the worm. It is possible

only when friction force exceeds the driving force.

15. State the advantage of worm gear drive in weight lifting machines. [CO3-L2 -

Nov14]

The worm gear drives are irreversible. It means that the motion cannot be transmitted

from worm wheel to the worm. This property of irreversible is advantageous in load

hoisting applications like cranes and lifts.

16. A pair of worm gears is designated as 2/54/10/5. Find the gear ratio. [CO3 -

L1 - Nov10]

Gear ratio i=z2/z1 =54/2 =27.

17. Where do we use skew gears? [CO3 – L2]

Skew gears are used to connect and transmit motion between two non-parallel and

non intersecting gears.

18. In which gear-drive, self-locking is available? [CO3 – L3]

Self locking is available in worm-gear drive.

19. What is a crown gear? [CO3 – L2]

A crown gear is a type of bevel gear whose shaft angle is 90 degree and angle of

pinion is not equal to the pitch angle of gear. Let Shaft angle

20. What are the merits and demerits of worm gear drive? [CO3 – L3]

Merits

1) Used for very high velocity ratio of about 100



2) Smooth and noiseless operation.

3) Self-locking facility is available.

Demerits

1) Low efficiency.

2) More heat will be produced and hence this drive can be operated inside an oil

reservoir or extra cooling fan is required in order to dissipate the heat from the drive.

3) Low power transmission.

21. What are the assumptions made in deriving Lewis equation? [CO3 - L1]

Fr-negligible, Ft- is uniformly distributed across the face with. At any time only one pair

of teeth is in contact. Stress concentration and tooth sliding is negligible.

22. State the advantages of Herringbone gear. [CO3 - L3 - Apr15]

1. Silent operation, 2.absence of vibration, 3.axial thrust, 4.higher efficiency, high load

carrying capacity.

23. Why phosphor bronze is widely used for worm gears? [CO3 - L2 - Nov13]

Phosphor bronze has high antifriction properties to resist seizure. Because in worm

gear drive, the failure due to seizure is more.

24. Where do we use worm gears? [CO3 - L3 - Apr13]

Worm gears are used as speed reducer in material handling equipment,

machine tools and automobiles.

25. What is the difference between an angular gear and a miter gear?

[Co3 - H1 - Nov12] [C03 - H1 - Nov13], [CO3 - H1 - Apr15]

Miter gears are mating bevel gears with equal numbers of teeth and with axes at right

angles.

Bevel gears are gears where the axes of the two shafts intersect and the tooth-

bearing faces of the gears themselves are conically shaped. Bevel gears are most

often mounted on shafts that are 90 degrees apart, but can be designed to work at

other angles as well. The pitch surface of bevel gears is a cone.



26. Mention the main types of failure in worm gear drives. [CO3-L1 -

Nov12]Seizure-significant sliding occurs between teeth and thread of worm,

1. Pitting and rupture: - the worm wheel wears off more than worm.

27. What is zerol bevel gear? [CO3 - L2 - Apr15]

Spiral bevel gear with curved teeth but with a zero degree spiral angle is known

as zerol bevel gear.

28. What kind of contact occurred between worm and wheel? How does this

differ from other gears? [C03 - L3 - Nov15]

Worm and worm gear form a lower pair(line ) as they have sliding contact with each other.

In a worm gear drive, power is always transmitted from worm to worm wheel. Power cannot be transmitted from worm wheel to worm.

16 marks:-

1. Design a pair of bevel gear to transmit 10 k w at a pinion speed of 1440 rpm.

Required transmission ratio is 4.material for gear is 15 Ni 2 Cr 1 Mo 15/ steel.

The tooth profile of the gears are of 20 0 composite form. [CO3 - H3 - Nov11]

Given data:-P=10 kW, N1=1440 rpm, i=4 =200

To find: - design the pair of bevel gear.

Solution:-

Since the same material is used for both pinion and gear, the pinion is weaker than

the gear. Therefore we have to design only pinion.

1. Material: - 15 Ni 2 Cr 1 Mo 15

2. Calculation of z1and z2:-

Z1=20, Z2=i x Z1= 4 x 20 = 80

http://mechteacher.com/kinematic-pair/#c1



3. Calculation of pitch angle & virtual number of teeth:-

Pitch angle: - tan δ2 = i

tan δ2 =4, δ1 = 90 – δ2 =90-75.96

δ2 = tan -1(4) = 14.04 0

= 75.960

ZV1 & ZV2:-

ZV1 =

=

= 20.61 =21

ZV2 =

=330

4. Calculation of tangential load on tooth (ft)

Assume, K0 = 1.25

Ft =

K0 v=

=

x 1.25 =

( | )

= 8289.32/ m t = 1.508 m t

5. Calculation of initial dynamic load (F d)

F d = Ft/C v Cv =

=

=0.74

= 8289.32/m t x 1/0.74

= 11599.23/m t



6. Calculation of beam strength (F s):-

b=10 m t, b = 450 N / mm 2, y‘ =0.154 – 0.912 / Zv1 for 20 0 FD

=0.1106

R = 0.5 m t

= 0.5 m t

= 41.23 m t

7. Calculation of transverse module (m t)

Fs Fd 1184.38 m t 3 11599.23 m t 2.14

From D.D:8.2 std. module = 3

8. Calculation of b, d1, v:-

b =10 m t = 10 x 3=30 mm

d1=mt x Z1 =3 x 20 = 60

v =

=

=4.52 m/s

9. Recalculation of beam strength:-

Fs=1184.38 mt2=1184.38 x 32=10659 N

10. Calculation of accurate dynamic load (Fd):-

Fd = Ft + ( )

Where,

Ft=P/v=10x103/4.52=2212.4 N

C=11860 e, from DD; 8.53, e=0.0125

C=11860x0.0125 =148.25 N/mm



Substituting all the values,

Fd =8866.58 N

11. Check for beam strength:-

Fs Fd, thus the design is satisfactory.

12. Calculation of maximum wear load:-

Fw=

Q‘ =

=

= 1.88

Kw=2.553 N/mm2 for steel gears 400 BHN


Fw=

= 6679 N

13. Check for wear:-

Since Fw Fd the design is unsatisfactory.

So increase mt=5 mm and repeat from step 8 and we get,

B=50 mm, d1= 100 mm v1=7.54 m/s

Fs=29608.5 N, Ft=1326.26 N, Fd=10059.9 N, Fw=18552.89 N

Now Fs Fd, and Fw Fd so, the design is satisfactory.

14. Calculation of basic dimensions:-

Transverse module=5 mm

Number of teeth: Z1=20, Z2=80



Pitch circle diameter: d1=100 mm d2=mt x Z2 =5 x 80 =400 mm

Cone distance R=0.5 mt =206.15 mm

Face width b=10 mt=10x5=50 mm

Pitch angle: δ1 =14.040, δ2=75.960

Tip diameter: da1= mt (z1+2 ) =109.7 mm

da2=402.43 mm

Height factor f0=1

Clearance c=0.2

Addendum angle: tan a1= tan a2=

=

=0.02425

a1= a2=1.40

Dedendum angle: tan f1= tan f2= ( )

= ( )

=0.0291

f1 = f2 = 0.670

Tip angle: a1= 1+ a1

=14.040+1.40 =15.440

a2=77.360

Root angle: f1= 1- f1=14.040-1.670=12.370

f2 =74.290

Virtual number of teeth: Zv1 =21, and Zv2 = 330



2. Design a cast-iron bevel gear drive for a pillar drilling machine to transmit

1875 w at 800 rpm to a spindle at 400 rpm .the gear is to work for 40 hrs/weak for

3 years ,pressure angle is 20 0. [CO3 - H3 - Nov10]

Given data: - P=1875 W, N1=800 rpm, N2=400 rpm, =20 0

To find:-design a bevel gear drive.

Solution:-

1. Gear ratio: - i=N1/N2

=800/400 = 2

Pitch angles: - for right angle bevel gears, tan 2= i=2

2 =63.43 0

1 =90-63.43 = 26.57 0

2. Material for pinion and gear:-cast-iron grade 35, heat treated,

From DD; 8.5, u = 350 N/mm2

3. Gear life in hours:-40 x52 x 3 =6240 hrs

Gear life in cycles= 6240x60x800 = 29.952x107cycles.

4. Calculation of initial design torque [M t]:-

[M t]= M t x k x k o

M t =

=

= 22.38 Nm, k x ko = 1.3

M t = 22.38 x 1.3 = 29.095 Nm

5. Calculation of Eeq, [ b], and [ c]:-



From DD: 8.14, Eeq, =1.4 x10 5N/mm2 for CI

u 280 N/mm2

[ b] =

-1 (for rotation in one direction.)

From DD: 8.2,

Kbl=√

Kbl=√

= 0.8852

From DD: 8.19, k = 1.2 for CI, n=2, -1=0.45 u,

FromDD: 1.40, u,= 350 N/mm2 for CI,

-1 = 0.45 x 350 =157.5 N/mm2,


[ b] =

x 157.5

=81.33 N/mm2

To find [ c]:-

[ c] = CB x HB x K cl

From DD: 8.16, CB= 2.3, HB=200 to 260.



From DD: 8.17,

Kcl = √

Kcl=√

=0.833

[ c] =2.3x260x0.833

=498.08 N/mm2

6. Calculation of cone distance (R):-

R ψy √

( )[ ]

]

[ ]

y =R/b = 3 (initially assume),

R 3 √

( )[ ]

]

[ ]

R 50.2,

R = 51 mm

7. Assume z1=20, z2=iz1=2x20=40,

Zvi=z1/cos 1 zv2=40/cos 63.43

=20/cos 26.57 =90

=23

8. Calculation of transverse module (mt) Mt =

=

=2.28 mm



Standard module from DD: 8.2, mt=2.5 mm

9. Revision of cone distance:-

R = 0.5 m t

= 0.5 x 2.5

= 55.9 mm

10. Calculation of b, mav, d1av, V and, y.

b=r/ y=55.9/3 =18.63 mm,

mav = mt - b sin 1 / z1

= 2.5 – [18.63 x sin 26.5 / 20]

= 2.083 mm

d1av =mav x z1

=2.083 x 20 = 41.66 mm

V=

=

=1.745 m/s

y= b/d1av

=18.63/41.66

=0.447

11. from DD: 8.3, assume IS quality 6 bevel gear.

12. Revision of design torque [mt]:-

[Mt]=Mt x k x kd

K=1.1,for b/d1av 1,and kd=1.35 for IS quality 6 and v upto 3 m/ s



[Mt]=22.38x1.1x1.35=33.24 Nm

13. Check for bending:-

b = [ ]

(

Yv1=0.408 for Zv1=23

b = [ ]√

( )

= 100.75 N/mm2

Now b [ b] thus the design is not satisfactory.

Now increase mt=3 mm and repeating from step 9, we get:- b=58.3 N/mm2.

Thus the design is satisfactory.

14. Check for wear strength:-

c=

( ) [

√( )

[ ]] 1/2

c=

[ ( )] [

√( )

] 1/2

=439.33N/mm2

Now [ ] thus the design is satisfactory.


From DD: 8.38,

Mt=3 mm

Z1=20

Z2=40



d1=60mm,

d2=120mm

R=67.08

b=22.36, 1 =26.570, 2=63.430

da1=65.37

Da2=122.68

f0=1

c=0.2

Tan a1=tan a2=0.0447

a1= a2=2.560

Tan f1=tan f2=0.05366

f1= f2=3.07 0

a1=29.13 0 a2=65.99 0

f1=23.50 f2=60.360

3. Design a bevel gear drive to transmit 7 KW at 1600 rpm of for the following

data: Gear ratio: 3, material for pinion and gear: C 45 steel, life: 10000hrs. [Co3 -

H3 - Apr13] [Co3 - H3 - Apr15]

Given data: - P=7 K W, N1=1600rpm,

To find:-



design a bevel gear drive.

Solution:-

1. Gear ratio: - i=3

Pitch angles: - for right angle bevel gears, tan 2= i=3

2 =71.33 0

1 =90-71.33 = 18.26 0

2. Material for pinion and gear:-C 45, forged steel,

From DD; 1.4, u = 700 N/mm2

3. Gear life in hours:-10000 hrs

Gear life in cycles N = 10000 x60x1600= 96 x107cycles.

4. Calculation of [Mt]:-

[M t]= M t x k x k o

M t =

=

= 41.778 Nm, k x ko = 1.3

M t = 22.38 x 1.3 = 54.31 Nm

5. Calculation of Eeq, [ b], and [ c]:-

From DD: 8.14, Eeq, =2.15 x10 5N/mm2 for steel

[ b] =

-1 (for rotation in one direction.)



From DD: 8.2,

Kbl =

= 7

=0.602

From DD: 8.19, k = 1.2 for C45, n=2.5 , -1=0.45 u,

From DD: 1.40, u,= 700 N/mm2 for C45,

-1 = 0.45 x 700 =315 N/mm2,


[ b] =

x 315

=88.509 N/mm2

To find [ c]:-

[ c] = CRx HRC x K cl

From DD: 8.16, CR= 23, HRC=40 TO 55, K cl =1 for HB 350

[ c] =23x50 x1

=1150 N/mm2

6. Calculation of cone distance (R):-

R ψy √

( )[ ]

]

[ ]

y =R/b = 3 (initially assume),

R 117.91,



R = 120 mm

7. Assume z1=20, z2=iz1=3x20=60,

Zvi=z1/cos 1 zv2=40/cos 71.33

=20/cos 18.26 =187

=21

8. Calculation of transverse module (mt)

mt=

=3.79

Standard module from DD: 8.2, mt= 4 mm

9. Revision of cone distance:-

R=

=126.49 mm

10. Calculation of b, mav, d1av, V and, y.

b=r/ y=126.49/3 =42.16 mm,

mav = mt - b sin 1 / z1

= 4 – [42.16 x sin 18.26 / 20]

= 3.339 mm

d1av =mav x z1

=3.339 x 20 = 66.79 mm



V=

=

=5.6 m/s

y= b/d1av

=42.16/66.79

=0.631

11. from DD: 8.3, assume IS quality 6 bevel gear.

12. Revision of design torque [mt]:-

[Mt]=Mt x k x kd

From DD: 8.15, K=1.1

From DD: 8.15, KD=1.45

[Mt]=54.31 X 1.1 X 1.45

=86.624 Nm

13. Checking for bending:-

b= [ ]

( )

Yv1=0.395 for zv1=21 from DDF:8.18,

b= [ ]

[( ( )]

b = 46.81 N/mm2

Since b [ b],design is satisfactory.

14. Check for wear strength:-

c=

( ) [

√( )

[ ]] 1/2



c=

[ ( )] [

√( )

] 1/2

c = 466.06 N/mm2

Since, c [ c ] design is satisfactory.


From DD: 8.38,

Mt=4 mm

Z1=20

Z2=60

d1=80mm,

d2=180mm

R=126.49mm

b=42.16mm, 1 =18.260, 2=71.330

da1=87.91

da2=242.56

f0=1, c=0.2

Tan a1=tan a2=1.810

a1= a2=2.560

Tan f1=tan f2=0.05366

f1= f2=1.84 0

a1=20.07 0 a2=73.14 0



f1=16.420 f2=69.490

4. Design a bevel gear drive to transmit 7.36 kW at 1440 rpm for the following

data. Gear ratio=3.material for pinion and gear C 45 surface hardened. [Co3 -

H3 - Apr11]

Solution:-

Since the same material is used for both pinion and gear, the pinion is weaker than

the gear. Therefore we have to design only pinion and,

SOLVE THIS PROBLEM SAME AS” PROBLEM NO:-1”

5. Design a straight bevel gear drive between two shafts at right angles to each

others to transmit 4 kW. Speed of the pinion shaft is 300 rpm and the speed of

the gear wheel shaft is 900 rpm. Pinion is of steel and wheel of cast iron.

Assume the expected gear life as 20,000. [Co3 - H3 - Nov12]

Solution:-

Since the material of pinion and gear are different, we have to design the

pinion first and check the gear, and


6. Design a straight bevel gear drive between two shafts at right angles to each

other. Speed of the pinion shaft is 360 rpm and the speed of the gear wheel

shaft is 120 rpm. Pinion is of steel and wheel of cast iron. Each gear is expected

to work 2 hours/day for 10 years. The drive transmits 9.37 kW. [Co3- H3 -

Nov13] [Co3 - H3 - Nov15]



Solution:-

Since the material of pinion and gear are different, we have to design the pinion first

and check the gear, and


7.Design a bevel gear drive to transmit 10 KW at 1440 rpm of for the following

data: Gear ratio: 3, material for pinion and gear: C 45 steel,and minimum no of

teeth is 20, life: 10000hrs. [CO3 - H3 - Apr13], [CO3 - H3 - Apr14]


8. A hardened steel worm rotates at 1440 rpm and transmits 12 kw to a

phosphor bronze gear. The speed of the worm wheel should be 60 3% rpm.

Design the worm gear drive if an efficiency of at least 82 % is desired. [CO3 - H3

- Apr14]

Given data: - P=12kW, N1=1400 rpm, N2= 60 3% rpm, , desired =82 %,

To find:-design the worm gear drive.

Solution:- i=1440/60 3%

=24 0.72

1. Selection of material:-worm-hardened steel, wheel-phosphor bronze

2. Selection of z1 and z2:-

For =85%

z1=3(or) 4

z2=i x z1 =3x24 =72



3. Calculation of q, :-

q=d1/mx

=11 (assume)

Lead angle: - =tan-1(z1/q)

=tan-1(3/11)

=15.25 0

4. Calculation of Ft:-

Ft= P/V x K0

V= Z2mxN2 / 60 x 1000

=0.226 mx

K0=1.25

Ft =

x1.25

= 66371.68 mx

5. Calculation of dynamic load F d:-

Fd=Ft/Cv

Cv =

= 6 / (6+5), assume V‘=5

=0.545

F d =



=

6. Calculation of beam strength Fs:-

Fs= mx b [ b] y‘

From DD: 8.48, b=0.75d1

=0.75 x q mx

=0.75 x 11mx

=8.25 mx

From DD: 8.45, b = 80 N/mm2

From DD: 8.52, y‘ = 0.125, assuming =200

Fs= x mx x 8.25 mx x 80 x 0.125

=259.18 mx 2

7. Calculation of axial module (mx):-

F s F d

259.18mx2 121681.4 /mx

mx 7.77,

mx = 8 mm

8. Calculation of b.d2, v:-

b=8.25 mx=66 mm

d2=Z2mx=72x8=576 mm

v=0.226mx=0.226x8=1.808 m/s



9. Recalculation of beam strengthFs:-

Fs=259.18mx2=258.18x (8)2=16587.52 N

10. Recalculation of dynamic load F d:-

Fd = Ft/Cv

Ft =

=

=8296.46 N

Cv =

=0.768

Fd =

=10802.68 N

11. Check for beam strength:-

Since Fd Fs. the design is safe.

12. Calculation of wear load Fw:-

Fw=d2bKw

From DD: 8.45 Kw=0.56,

Fw=576x66x0.56

=21288.96 N


Since Fd Fw .the design is safe.

14. Check for efficiency:-

act = 0.95x

( )



From DD8.52, for Vs=1.432, & bronze the value of =0.03,

Then tan-1( ) = tan-1(0.03) =1.700

act = 0.95 x tan(15.25) /tan(15.25+1.70)

= 84.98%, thus we find that, act desi , the design is satisfactory

15. Calculation of basic dimensions for worm and worm gear:-

From DD: 8.43,

Axial module mx =8 mm

Number of startsZ1=3

Number of teeth on worm wheel Z2 =72

Face width b=66 mm

Length of worm (from DD: 8.48), L (12.5+0.09xZ2) x mx

(12.5+0.09x72)8= 151.84

Centre distance =0.5 mx(q+Z2)

=0.5x8(11+72)

=332 mm

Height factor f0=1

Bottom clearance c=0.25 mx=0.25x8=2 mm

Pitch diameter d1=qmx=11x8=88mm, d2=72x8=576 mm

Tip diameter da1=d1+2f0mx

=88+2x1x8



=104 mm

da2= (Z2+2f0)mx

= (72+2x1)8

=592 mm

Root diameter:

df1=d1-2f0mx-2c

=88-2x1x10 -2x2

=68 mm

df2= (Z2-2f0)mx - 2c

= (72-2x1)8-2x2

= 556 mm

9. A pair of worm gear is designated as 2/54/10/5.calculate (i) the centre

distance,(ii) the speed reduction(iii) the dimensions of worm and (iv) the

dimensions of worm wheel. (CO3-H3- Nov10)

Given data: 2/54/10/5, (z1/ z2/ q/ mx).

To find: a, i, da1, d1, df1, px, and d2, da2, df2,

Solution:-

(I) centre distance (a):-

a =0.5 mx(q+z2)

=0.5 x 5(10+54)



=160 mm

(ii) Speed reduction (i):-

i =z2 / z1

=54/2

=27

(iii) Dimensions of worm:-

d1=qmx=10x5=50 mm

da1=d1+2f0mx

=50+2x1x5

=60 mm

df1= d1-2f0mx-2C

Where C=0.25mx

=0.25 x 5 =1.25 mm

df1=50-2x1x5 – 2x1.25 =37.5 mm

Px= mx

= x 5=15.71 mm

(iv) Dimensions of worm wheel:-

d2=z2xmx=54x5=270 mm

da2= (z2+2f0) mx

= (54+2x1)5

=280 mm



df2=(z2-2f0)mx-2C

=(54-2x1)5-2x1.25 [C=0.25mx]

=257.5 mm

10. A steel worm running at 240 rpm receives 1.5 K w from it shaft. The speed

reduction is 10:1 design the drive so as to have an efficiency of 80 %. Also

determine the cooling area required, if the temperature rise is restricted to 450C.

The overall heat transfer co-efficient as 10W/m2 0C. [CO3 - H3 - Apr10]

Given data: - N1=240 rpm, P=1.5 K w, i=10, desired =80 %, t0-ta=450, kt =10 w/m2 0c

To find: - (i) design the worm gear drive (ii) the cooling area required

Solution:-

N2=N1/i

=240/10 =24

1. Selection of material:-worm- steel,

Wheel- bronze

2. Calculation of initial design wheel torque [Mt]:-

[Mt]=Mt x k x kd

Mt=60P / 2 N2

=60x15x1000 / 2x x24

=596.83 Nm.

Assume, K x kd =1

[Mt]=596.83 x 1



=596.83 Nm


From DD: 8.46, for =80%

z1=3(or) 4

z2=i x z1

=10x3

=30

4. Selection of [ b],[ c]:-

From DD: 8.45, for bronze wheel: 390 N/mm2, b=50 N/mm2

(rotation on one direction),

c=159 N/mm2-assuming Vs=3 m/s

5. Calculation of centre distance (a):-

a = *(

) + √[

(

)[ ]

]

[ ]

Initially assume q=11

a = *(

) + √[

(

)[ ]

]

[ ]

= 168.6 mm




Mx=2a/ (q + z2)

=2x168.6 / (11+30)

=8.22 mm

Therefore std.axial module=mx=10 mm

7. Revision of centre distance (a):-

a=0.5 mx(q+z2)

=0.5x10(11+30)

=205 mm

8. Calculation of d, v, and vs

Pitch diameter d1=qmx=11x10=110 mm

D2=z2xmx=30x10=300 mm

Pitch line velocity v1=

=

=1.382 m/s

N2= /60

= -3x24 /60

=0.377 m/s


=tan-1(3/11)



=15.25 0

Sliding velocity:-

Vs=v1/cos

=1.382/cos 15.25

=1.432 m/s

9. Recalculation of design contact stress[ c]:-

From DD: 8.45, for vs=1.432 m/s, the value of [ c] =172 N/mm2

10. Revision of [Mt]:-for v2 3 m/s,Kd=1,

[Mt]=Mt K Kd

=596.83x1x1

=596.83 m/s


b= [ ]

From DD:8.18,Yv=0.452,

From DD:8.52,Zv=Z/cos 3

=30/cos3 15.25

=34

b =

= 7.6 N/mm2

Thus, b [ b], the design is satisfactory.




c =

(

)√*

( )

+3

[ ]

= 118.59N/mm2

Since c [ c] the design is satisfactory


act = 0.95x

( )

From DD8.52, for Vs=1.432, & bronze the value of =0.05,

Then tan-1( ) = tan-1(0.05) =2.8620

act= 0.96 x tan(15.25) /tan(15.25+2.862)

=80%, thus we find that, act = desi ,

The design is satisfactory

14. Calculation of cooling area required (A):-

(1- )xinput power= KtxAx450

(1-0.8) x1.5x103=10xAx450

A=0.666 m2


From DD: 8.43,

Axial module Mx =10 mm



Number of startsZ1=3

Number of teeth on worm wheel Z2 =30


(12.5+0.09x30)10 = 152

Centre distance =205 mm

Face width b=0.75d1

=0.75x110=82.5 mm

Height factor f0=1

Bottom clearance c=0.25 mx=0.25x10=2.5 mm

Pitch diameter d1=110 mm, d2=300 mm


=110+2x1x10

=130 mm

da2= (Z2+2f0)mx

= (30+2x1)10

=320 mm

Root diameter:

df1=d1-2f0mx-2c

=110-2x1x10 -2x2.5

=85 mm

Similarly df2 = 275 mm



11. Design a worm gear drive to transmit 22.5 KW at a worm speed of 1440 rpm.

Velocity ratio is 24:1. An efficiency of at least 85% is desired. The temperature

rise should be restricted to 400 C. Determine the required cooling area. [Co3 - H3

- Apr13], [C03 - H3 - Apr15

Given data: - N1=1440 rpm, P=22.5 K w, i=24, desired =85 %, t0-ta=400,

Assume k t =10 w/m2 0c

To find: - (i) design the worm gear drive (ii) the cooling area required

Solution:-

N2=N1/i

=1440/24

=60 rpm

1. Selection of material:-Worm- steel,

Wheel- bronze

2. Calculation of initial design wheel torque [Mt]:-

[Mt]=Mt x k x kd

Mt=60P / 2 N2

=60x22.5 x1000 / 2x x60

=3580.98 Nm.

Assume, K x kd =1

[Mt] = 3580.98 Nm




From DD: 8.46, for =85 %

z1=3(or) 4

z2=i x z1

=24 x3

=72

4. Selection of [ b],[ c]:-

From DD: 8.45, for bronze wheel: 390 N/mm2, [ b] =50 N/mm2

(rotation on one direction),

c=159 N/mm2-assuming Vs=3 m/s

5. Calculation of centre distance (a):-

Initially assume q=11

a = [(z2 /q) + 1]√[

(

)[ ]

]2

[ ]

a = [(72 /11) + 1]√[

(

)[ ]

]2 [ ]

=345.96 mm


Mx=2a/ (q + z2)

=2x345.96 / (11+72)

=8.34 mm

Therefore std.axial module=mx=10 mm



7. Revision of centre distance (a):-

a=0.5 mx(q+z2)

=0.5x10(11+72)

=415 mm

8. Calculation of d, v, and vs

Pitch diameter d1=q mx=11x10=110 mm

D2=z2xmx=72 x10=720 mm

Pitch line velocity v1=

=

=8.294 m/s

V2 = /60

= -3x 60/60

=2.26 m/s


=tan-1(3/11)

=15.25 0

Sliding velocity:-

Vs=v1/cos

=8.294/cos 15.25

=8.596 m/s

9. Recalculation of design contact stress[ c]:-

From DD: 8.45, for vs=8.596 m/s, the value of [ c] =149 N/mm2



10. Revision of [Mt]:-for v2 3 m/s,Kd=1,

[Mt]=Mt K Kd

=3580.98 x 1 x 1

=3580.98 m/s


b= [ ]

From DD:8.18,Yv=0.505,

From DD:8.52,Zv=Z/cos 3

=72/cos3 15.25

=80.17

b =

= 17.01 N/mm2

Thus, b [ b], the design is satisfactory.


c =

(

)√*

( )

+3

[ ]

=

(

)√*

( )

+3

= 121.03 N/mm2

Since c [ c] the design is satisfactory




act = 0.95x

( )

From DD8.52, for Vs=8.596 m/s, & bron ze the value of =0.035,

Then tan-1( ) = tan-1(0.035) =2

act= 0.96 x tan(15.25) /tan(15.25+2.0)

=83.41%,

Thus we find that, act desi , The design is satisfactory

14. Calculation of cooling area required (A):-

(1- )x input power= Kt x Ax 450

(1-0.8341) x22.5x103=10 x A x 400

A=9.332 m2


From DD: 8.43,

Axial module mx =10 mm, Number of startsZ1=3, Number of teeth on worm wheel Z2

=72


(12.5+0.09x72)10 = 189.8 mm

Centre distance =415 mm

Face width b=0.75d1



=0.75x110=82.5 mm

Height factor f0=1

Bottom clearance c=0.25 mx=0.25x10=2.5 mm

Pitch diameter d1=110 mm, d2=720 mm


=110+2x1x10

=130 mm

da2= (Z2+2f0)mx

= (72+2x1)10 =740 mm

Root diameter:

df1=d1-2f0mx-2c

=110-2x1x10 - 2x2.5

=85 mm

Similarly df2 = 275 mm

12. Design a worm gear drive to transmit 10 K W at 1440 rpm with a gear ratio

Of 12.use steel worm and cast steel wheel. [CO3 - H3 - Nov12]

Solution:-Solve this problem same as the “PROBLEM NUMBER: 10

13. Design a worm drive for a speed reducer to transmit 15 kW at1440 rpm of the

worm shaft. The desired wheel speed is 60 rpm. Select suitable worm and wheel

materials. [CO3 - H3 - Apr11]




14. The input to worm gear shaft is 18 k w and 600 rpm. Speed ratio is 20. The

worm is to be of hardened steel and the wheel is made of chilled phosphor

bronze. Considering wear and strength, design worm and worm wheel. [Co3 -

H3 - Nov13] [CO3 - H3 - Nov15]


15. A 2 KW power is applied to a worm shaft at 720 mm. The worm is of

quadruple start with 50 mm as pitch circle diameter. The worm gear has 40

teeth with 5 mm module. The pressure angle in the diametral plane is 20°.

Determine (i) the lead angle of the worm, (ii) velocity ratio, and (iii) centre

distance. Also , calculate efficiency of the worm gear drive, and power lost

friction. [CO3 - H3 - Apr14]

Given data: P=2 kw Z1=1 ,z2=40, d1=50mm,α=20°, q=40 teeth

To find:

(i) the lead angle of the worm, (ii) velocity ratio, and (iii) centre distance.

Solution:

(i) Lead angle

Lead angle: γ =tan-1(z1/q)

=tan-1(1/40)

=1.43 0

(II) velocity ratio

i=z2/z1

=40/1

=40



(iii) centre distance

a=0.5mx(q+z2)

q=d1/mx

mx=d1/q=50/40=1.25

q=50/1.25=40

z2=q=40

a=0.5×1.25(40+40)=50 mm

(iv) Efficency Of worm gear

η=(cos∞-µtan∞)/(cos∞+µcot∞)

η = 38 %

(v) Power lost in fricition

= (1- η)×P

= (1-38)×200

=7400 W



15.Derive expression for determining the forces acting on a Bevel gear with

suitable illustration. [Co3 - H3 - Nov14]

Reference Diagram:

Given: (Known Values)

Pinion rotates clockwise driving gear.

Diametral pitch, P

Number of teeth on gear, Np

Number of teeth on pinion, Np

Pressure angle, phi

Face Width, F

Torque applied to pinion gear, Tp

Gear ratio, m



Required:

For pinion and gear: calculate the stress, strain and deflection of the tooth. Also calculate

the factor of safety using Distortion Energy Theory (DET) for ductile materials and Maximum

Normal Stress Theory (MNST) for Brittle Materials.

Assumptions

Analysis is specifically for a 90 degree bevel gear.

Force is uniformly distributed at the end of the tooth. (serving as a maxima)

Neglect surface wear and fatigue life.

Neglect cyclic and impact effects from applied loading.

For this analysis, bevel gear tooth dimensions will be identical to a standard spur gear.

Load application only on one tooth at any given time. (minimum req. for continuous contact, serving as a maximum



Equations

All of the following equations are from Richard G. Budyna's and J. Keith Nisbett's book, Shigley's Mechanical Engineering Design eighth edition.

Calculation of tangential tooth force

Tooth Forces Diagram

The pinion pitch diameter dp, and gear pitch diamerter dg can be found using the following equation.

Solving for the pitch diameters

The tangential force can be found using the following torque equations. Where all variables are magnitudes only. See diagram of tooth forces for directions.

where,



The tangential forces acting on the gear teeth are equal in magnitude and opposite in direction.

where,

From tangential force, can calculate radial and axial forces,

To simplify for subsequent stress analysis, the radial and axial force can be resolved into a tooth compression force,

There is also a shearing component that acts along the tooth face, but for now will be neglected. If this component were to be calculated, it would be found by,



See diagram of tooth forces for direction of forces.

Calculation of tooth geometry

Tooth Geometry Diagram Image

goes here

The pinion and gear teeth have the same following geometric properties.

Face width for bevel gears in uniform contact (from Shigley, page 697, Table 13-3)

where,

Working depth, hk (how deep the gear tooth meshes into the mated gear teeth)

Clearance, c (the distance between the bottom land of the gear and the top land of the mating gear tooth)

Tooth depth, ht, is then found by summing the clearance and working depth,



The pitch tooth thickness t_pitch, along the pitch circle can be calculated using

The innermost tooth thickness can then be found by replacing dP with the inner

diameter, di. Doing this effectively uses the principal of arc length, and that the

diametral pitch (teeth per in) along the inner radius must be larger (smaller

circumference for same number of teeth). This can be easily seen by examining the

circular pitch equation,

Which can be rewritten as,

Therefore innermost tooth thickness is,

To simplify our calculations and maintain a rectangular cross-section, we can use an average tooth thickness, t_av, to serve as a uniform tooth thickness.



For an approximate value of the tooth thickness at the base or root of the tooth assume

that the tooth thickness at the pitch circle is equal to the thickness at the root. This will give conservative values for stress, strain, and deflection. For more precise values measure the actual thickness at the root of the tooth. Bear in mind that this analysis is neglecting the

non-uniform thickness of bevel gear teeth. A common bevel gear tooth with be thinner toward the center to allow for meshing. Hence why an average thickness was determined.

The cross sectional area of the tooth at the root At, can be calculated using

Calculation of shear force at base of tooth

Positive Shear Notation Diagram

Image goes here

The shear force V, at the base of the tooth is equal to the tangential force acting on the tooth in magnitude. See positive notation diagram for sign.



Calculation of moment at base of

tooth Positive Bending Notation

Diagram Image goes here

For tangential force applied at the tip of tooth

The moment induced by the eccentric loading of the compressive force can be found by



(negative to indicate opposing

direction) The effective bending

moment is then,

Calculation of maximum normal stress due to bending

Bending Normal stress Profile

Image goes here

The maximum normal stress due to bending is located at A, B, Y, and Z.

For normal stress at point A or Y use c=t_av/2. For normal stress at point B or Z use c=- t_av/2. Use the appropriate moment for the gear being analyzed and loading location.

Where the second moment of inertia I, can be calculated using

Positive normal stress means it is in tension and negative normal stress means it is in compression

Combining the above, can write a simplified equation for bending stress on pinion,

Calculation of maximum shear stress due to bending

Shear Stress Element

Take a small element isolated from the tooth section. A shear stress in the y direction

will cause a reaction shear force in the x direction. This is because the sum of

moments acting on that small element in the tooth is equal to zero. This is why we



don't see elements of the tooth rotating under shear stress. Therefore Shear stress

contributes to the normal stress due to bending.

Shear Stress Profile

Image goes here

The maximum shear stress is in the middle. The 3/4 in the equation below accounts for the shear stress profile not being uniform and is for a rectangular cross-section.

Calculation of normal/principal stressThe normal/principal stresses can be found by

drawing a Mohr's Circle or by calculation using the following formulas

Calculation of factor safety using Distortion Energy Theory (DET) for ductile

Factor of safety n, can be calculated using

Where Sy is the yield strength of the material and sigma' can be can be calculated

using the following equation for plane stress



Calculation of factor safety using Maximum Normal Stress Theory (MNST) for

Brittle Materials

Factor of safety n, can be calculated using

Calculation of elastic strain

Elongation is equal to

Stress is related to strain by

Where E is the Young's modulus or modulus of elasticity

Therefore

Calculation of deflection

The strain energy for bending is

The strain energy for bending shear is



Where C=1.2 for rectangular cross-section

Total strain energy is

Castigliano's Theorem gives

For a cantilever beam the deflection is

Use the length corresponding to the tangential force location



UNIT-4

GEAR BOXESES

Geometric Progressions, Standard Step Ratio.- Ray Diagram- Kinematic layout- Design of

Sliding Mesh Gear Box- - Design of Multi Speed Gear Box for machine tool applications-

Constant Mesh Gear Box-speed reducer unit.-variable speed gearbox, Fluid coupling,

Torque converters for automotive applications.

PART-A

4. List any two method used for changing speed in gear box [CO4 - L1 - Nov11] Sliding mesh gear box Constant mesh gear box.

5. What are preferred numbers?[ CO4 - L3 - Apr13], [CO4 - L3 - Nov12] [CO4 - L3 -

Nov14] Preferred numbers are the conventionally rounded off values derived from geometric

series. There are five basic series, denoted as R5, R10, R20, R40 and R80 series. 6. What is step ratio? Define progression ratio.name the series in which speed of

multi speed gear box are arranged? [CO4 - L3 - Nov13], [CO4 - L3 - Apr14 ][ CO4 - L3 - Nov51]

When the spindle speeds are arranged in geometric progression, then the ratio between the two adjacent speeds is known as step ratio or progression ratio.

= n-1

Series: - R 20, R 40 etc., 7. What is kinematic arrangement as applied to gear box? [CO4 - L1] The kinematic layout shows the arrangement of gears in a gear box. It also provides

information‘s like number of speeds available at each spindle and the number of stages used.

8. What does the ray-diagram of gear box indicates? [CO4 - H1] The ray diagram is a graphical representation of the drive arrangement in general from. It serves to determine the specific values of all the transmission ratios and speeds of all the shafts in the drive. 9. What is a speed reducer? [CO4 – L2] Speed reducer is a gear mechanism with a constant speed ratio, to reduce the angular speed of output shaft as compared with that of input shaft. 10. List out the possible arrangements to achieve 16 speed gear box? [CO4 – L3]

4 x 2 x 2 scheme 2 x 4 x 2 scheme



2 x 2 x 4 scheme 12. What are the points to be considered while designing a sliding-mesh type of multi-speed gear box? [[CO4 - L1-Nov13]

The transmission ratio (i) in a gear box is limited by 1/4 < i < 2 For stable operation, the speed ratio at any stage should not be greater than

8. In other words, Nmax / Nmin <

8 In all stages except

in the first stages, Nmax > Ninput > Nmin .

The sum of teeth of mating gears in a given stage must be the same for same module in a sliding gear set.

The minimum number of teeth on smallest gear in drives should be greater than or

equal to 17.

13. Give some applications of constant mesh gear box. [CO4 - L2 - Apr10]

Constant mesh gear boxes are employed in various machine tools viz.., lathe, milling machine, etc., to provide a wide range of spindle speeds.

14. What are the main requirements of a speed gear boxes? [CO4 - L1]

It should provide the designed series of spindle speeds. It should transmit the required amount of power to the spindle. It should provide smooth silent operation of the transmission. It should have simple construction.

15. What are the possible arrangements to achieve 12 speeds from a gear box? [CO4 - L3 - Apr13]

3 X 2 X 2 scheme 2 X 3 X 2 scheme 2 X 2 X 4 scheme

16. Specify four types of gearboxes. [CO4 - L3 - Nov14] 1. Selective type (i) Sliding mesh (ii) Constant mesh (iii) Synchromesh gear box

S.No. Number of speeds

Preferred structural formula

1. 6 speeds

(i) 3(1) 2(3) (ii) 2(1) 3(2)

2. 8 speeds

(i) 2(1) 3(2) 2(4) (ii) 4(1) 2(4)

3. 9 speeds

(i) 3(1) 3(3)

4. 12 speeds

(i) 3(1) 2(3) 2(6) (ii) 2(1) 3(2) 2(6) (iii) 2(1) 2(2) 3(4)



2. Progressive type gear box

3. Epicyclic type gear box.

17. Sketch the kinematic layout of gears for 3 speeds between two shafts. [CO4 - L3 - Apr14]

18. In which gear drive, self locking is available? [CO4 - L2 - Apr15]

In worm gear drive self locking is available.

19. Draw the ray diagram for a six speed gear box. [CO4 - L2 - Apr15]

2(1) 3(2)

Stage1 Stage

2

20. Write the significance of structural formula. [CO4 - L2 - Nov15]

Let ―n‖ is the number of speeds available at the spindle. p1, p2, p3…is the stage numbers in the gear box, X1,X2,X3,… are the characteristics of the stages,

Then the structural formula is given by

n = p1 (X1). p2(X2). p3(X3)…,

The structural formula provides the following in formations required for gear box design

1. The number of speeds available at the spindle 2. The number of stages used. 3. The number of simple gear trains required to get the spindle speeds. 4. The information required to draw the kinematic diagram and ray diagram.

21. Name the types of speed reducer. [CO4 – L3]



(a) Single reduction speed reducer. (b) Multi reduction speed reducer.

22. What is the function of spacers in a gear box? [CO4 – L3]

Spacers are sleeve like components which are mounted on shafts in between gears and bearings in order to maintain the distance between them so as to avoid interruption between them.

23. What is speed diagram? What is the structural diagram of a gear box? [CO4 - L1] Speed diagram or structural diagram is the graphical representation of different speeds of outpour shaft, motor shaft and intermediate shaft. 1. What situations demand use of gear boxes? [CO4 - L1] Gear boxes are required wherever the variable spindle speed is necessary. 2. Write any two requirement of speed gear box. [CO4 - L1]

Gear box should provide the designed series of spindle speeds. Gear box should transmit the required amount of power to the spindle.

3. Why G.P series is selected for arranging the speed in gear box? [CO4 – L2] The speed loss is minimum. if G.P is used The number of gear to be employed is minimum. If G.P is used G.P. provides a more even range of spindle speed at each step.



PART-B

1. Find the progression ratio for a 12 speed gear box having speeds between 100 and 355 rpm also find the spindle speed. [CO4 - H3 - Nov11]

Given:- n = 12, Nmin = 100rpm, Nmax= 355rpm

To find:-

(i) Progression ratio ф (ii) Spindle speed

Solution:-

(i)Progression ratio ф:

WKT, Nmax / Nmin = фn-1 355 / 100 = ф12-1 (or) ф = (3.55)1/11 = 1.1222

(ii) Spindle speed:

Since the calculated ф (= 1.12) is the standard step ratio for R 20 series. Therefore the

spindle speeds from R 20 series are.

100, 112, 125, 140, 160, 180, 200, 224, 250, 280, 315, 335rpm

2. Select the spindle speeds for the following data 12 speeds between 50 and 600

rpm.

Given:- n = 12,Nmin = 50 rpm, Nmax = 600rpm, To find:-

Spindle speed:- Solution:-

= фn-1

= ф12-1

Ф = 1.253

We find the calculated Ф is a standard step ratio for R10 series. So from R10 series the

spindle speeds are 50, 63, 80, 100, 125, 160, 200, 250,400, 500 and 630 rpm.

It can be seen that the calculated Nmax = 630 rpm. This is greater than the required

maximum speed. Therefore we have to check whether the deviation is within the

permissible range or not.

Permissible deviation = ± 10 (Ф-1) % ± 10 (1.253-1) % ± 2.53 %



Actual deviation = (630-600) X

= 2.5

3. A Nine speed gear box used as a head stack gear box of a turret lathe is to provide

a speed range of 180 rpm to 1800 rpm using standard step ratio. Draw the speed

diagram and kinematic layout. Also find the fix the number of teeth on all gear. [CO4

- H3 - Nov13] [Co4 - H3 - Apr13] [CO4 - H3 - Nov15]

Given:-

n = 9 Nmin =180 rpm Nmax =1800 rpm

To find:

Construction of speed diagram and kinematic layout. Solution:-

WKT,

= фn-1

= ф9-1 (or) Ф = 1.333

We find Ф = 1.333 is not a standard ratio so let us find out whether multiple of standard ratio 1.12 or 1.06 comes close 1.333

We can write 1.12 X 1.12 = 1.2544

1.12 X 1.12 = 1.405

1.06 X (1.06 X1.06 X 1.06 X1.06) = 1.338

So ф =1.06 satisfies the requirement therefore the spindle speeds from R40 series

skipping 4 speeds are given by

180, 236, 315, 425, 560, 750, 1000, 1320, 1800 rpm

Structural formula

For 9 speed, the preferred Structural formula

= 3(1) 3(3) Speed diagram:-

The speed diagram is drawn as shown in fig. the procedure is the same as

discussed in the previous problem.



Stage (2)

=

= 0.32 > 1/4

=

= 1.78 < 2 Ratio requirements are satisfied

Kinematic Arrangement:-

Calculation of Number of teeth:-

Let Z1 , Z2 ,Z3 …….. Z12 = Number of teeth of the gears 1, 2, 3…..12 respectively.



N1, N2, N3 …….. N12 = speeds of the gear 1, 2, 3…..12 respectively.

Second stage:-

First consider the ray that gives the maximum speed reduction. From the speed

diagram. We find that the speed reduced from 560 rpm 180 rpm. (Refer 9.6) we may

assume that this speed reduction is achieved by using the gear 11, 12

WKT Zmin > 17

Z11 = 20 (driver)

=

=

(or)

=

Z12 = 62.22 63 Second Pair:-

Now consider the lay that gives the minimum speed reduction from 560 rpm to 425

rpm. This can be achieved by using the gear 7 & 8.

=

=

Z7 = 0.76 Z8

Z8 = 47.16 48 Z7 = 83 – 48 =35

Third pair:-

Now consider the ray that gives the speed increase from 560 r.p.m to 1000 r.p.m.

this can be achieved by using the gear 9 and 10.

Z9/Z10 = N10/N9 = 1000/560

Z9=1.786 Z10

Z9 + Z10 = Z11 + Z12

=20+63 (solving) =83

Z10 = 29.79 30 ,Z9 =83-30 = 53

First stage:

First pair:- Consider the maximum speed reduction from 1320 r.p.m to 560 r.p.m. this

can be achieved by gear 5 and 6



Z5 = 20(drive)

Z5/Z6 = N6/N5 =560/1320

=0.4242

(or)

Z6 = Z5/0.4242

= 20/0.4242

=47.14, 48

Second pair:- Consider a speed reduction from 1320 r.p.m . this can be achieved by

gears 1 and 2

Z1/Z2 = N2/N1

=750/1320 Z1 = 0.5 Z2

Z1+Z2 = Z5+Z6 =20+45 =68

On solving Z2=43.3 44

Z1 = 68-44=24

Third pair:- Finally consider the speed reduction from 1320 r.p.m to 1000 r.p.m. this can

be achieved by gears 3 and 4.

Z3/Z4 = N4/N3= 1000/1320

Z3=0.76 Z4

And solving the eq

Z3+Z4 = Z5+Z6=20+48=68

Z4=38.64 39

Z3 = 68-39=29



4) Design a 12 speed gear box for an all geared headstock of a lathe. Maximum and

minimum speeds are 600 r.p.m and 25 r.p.m. respectively the drive is from an

electric motor giving 2.25kw at 1440 r.p.m. [CO4 - H3 - Apr13] [CO4 - H3 - Apr15]

Given:

n =12

N max =600 r.p.m

N min =25 r.p.m

P =2.25kw, N input =1440 r.p.m

To find:-Design the 12 speed gear box

Solution:-

1) Selection of spindle speed

Øn-1 = Nmax/Nmin

Ø12-1 = 600/25

Ø = 1.335

We can write 1.06 x(1.06 x 1.06 x 1.06 x 1.06 ) = 1.330

Ø = 1.06 satisfies the requirement.

Spindle speed R40.

25,33.5,45,60,80,106,140,190,250,335,450, and 600 r.p.m.

2) Ray diagram

Structural Formula

Nmin/N input = 25/80



= 0.31>1/4

Nmax/N input = 140/80 = 1.75<2

Stage 2 = Nmin/Nip = 80/140

0.57>1/4

Nmin/N input =140/450 = 0.311>1/4

Nmax/N input = 250/450 =0.56<2

3) Kinematic arrangement

4) Number of teeth on all gears:-

Stage 3:- Zmin>/17

Z13 = 20 (drive)

Z13/Z14 = N14/N13 (or) 20/Z14 = 25/80

Second pair:-

Z11/Z12 = N12/N11 = 140/80

Z11 =1.75 Z12

Z11+Z12 = Z13+Z14

=20+64, =84



Z12 = 30.5 31 and Z11 = 84-31=53.

Stage 2:-

First pair:- Maximum reduction from 140 r.p.m to 80 r.p.m.

Z9/Z10 = N10/N9

20/Z10 = 80/140

Z10 = 35

Second pair:- Z7/Z8 = N8/N7 = 190/140

Z7 = 1.357 Z8

Z7+Z8 = Z9+Z10

=20+35, =55

Stage 1:- First pair:- maximum reduction from 450 r.p.m to 140 r.p.m.

Z5/Z6 = N6/N5

20/Z6 = 140/450

Z6 = 64.28 65

Second pair:-

Speed reduction from 450 – 190 r.p.m

Z3/Z4 =N4/N3

=190/450

Z3 = 0.422 Z4

Z3+Z4 = Z5+Z6

=20+65 =85

Z4 = 59.77 60

Z3 = 85-60 =25

Third pair:- [450-250]r.p.m

Z1/Z2 = N2/N1 = 250/450 =Z1 = 0.555 Z2



Z1+Z2 = 23+24

=60+25 =85

Z2 = 54.66 55

Z1 = 85-55=30

5.material selection:

c-45

6.calculation of module:

i)torque:

25rpm –(13£14)

Ti4=p*60/2 N

=2.25* *60/2 25

=859.44N/m

ii)Tangential force:

F14=T/r

=2* Z14*m

=2*859.44* /64*m

=26857.5/m

4m=b/m=10

m=30 -c45

89.225/m

M2=89.525/m

M=4.47m 5mm



7.centre distance:

i) 1a1 =(z1+z2//2)m

=(30+55/2) =212.5mm

ii)2a2=(z7+z8/2)m

=(31+24/2)*5 =137.5mm

iii)3a3=(z11+z12/2)*m

=(53+31/2)*5 =210mm

8.width b= *m

=10*5 =50mm

9.Length of shflt:

L=25+10+7b+20+4b+20+4b+10+25

=110+15b

=110+(15*50)

=860mm

10. Design of shaft:

Maximum bending moment

M=Fn*L/4

Fn =Ft/cos α =[26857.5/m/cosα]

=[26857.5/5/cos 20] = 5716.23N

M=5716.23/4*860

=12.29*105 N-mm.

Equalent torque (Teq)

Teq= M2+T142

=(12.29*105)2+(859.44*103)2



=1.5*106N-mm.

Dia = [16*Teq/π(µ)]1/3

=[16*1.5*106/π*30]1/3 =63.38mm, ~R40 sere 67mm

Design of the shaft:-

(a) Dia of shaft: input speed =450 rpm

T=p*60/2πN

=2.25*103*60/2π*450

=47.746 N-m.

T=0.2 ds230

=19.96 mm~- 20mm (R40)

(b) Shaft 2 = 140 rpm

=2.25*10360/2π*140

=153.47 N-m

T=0.2ds23(£)

153.47*103=0.2*ds23*30

Ds2=29.46mm~_30mm (R-40)

(c) Diameter of shaft

Maximum speed = 80 rpm

Torque =p*60/2πN

=2.25*103*60/2π*80

=268.57 N-m

268.57*103=0.2*ds33*30

=35.5mm.



5. Sketch the arrangement of a six speed gear box .the minimum and maximum speeds required are around 460 and 1400 rpm. Drive speed is 1440 rpm. Construct speed diagram of the gear box and obtain various reduction ratios. Use standard output speeds and standard step ratio. Calculate number of teeth in each gear and verify whether the actual output speeds are within =2 % of standard speed. [CO4 - H3 - Apr14]

6. 6 speed gear box design N=6

Nmax =1400 rpm

Nmin 460 rpm

Progression ratio:-

=

=

=1.25

Standard speed ratio:-450, 560, 710, 900, 1120, 1400

Structural formula:-3(1) 2(3) –two shafts



Number of teeth:-

Stage 1 :-z1= 20(assume)

Z2/Z1 =1.97

Z2 =40

Z4/Z3 =1.55 Z4=1.55 Z3

Z6/Z5 =1.25 Z6=1.25Z5

Z1+Z2 = Z3+Z4 =Z5+Z6,

60=1.55Z3+Z3

Z3=23.5 =24, Z4=38,

Similarly calculate all other teethes:-z5=27, z6=34, z7=20, z8=32, z9=29, z10=23

Actual speed, N 01 =N1 ×

×

=437.5 rpm

N 02 =N1 ×

×

=882.6 rpm

N 03 =N1 ×

×

=552.6 rpm

N 04 =N1 ×

×

=1114.8 rpm

N 05=N1 ×

×

=694.8 rpm

N 06=N1 ×

×

=1401.8 rpm

Percentage deviation:-



N01 =

= - 2.7 %

N02=

=- 1.9 %

Similarly calculate all other Percentage deviations; all are within the limit +2%.

6. Draw the ray diagram and kinematic layout of a gear a box for an all geared head stock of a lathe. The maximum and minimum speeds are to be 600 and 23 rpm respectively. Number of steps is 12 and drive is from a 3000 w electric motor running at 1440 rpm. [CO4 - H3 - Apr14]

12 speed gear box:-

n=12

N max=600rpm

N min=23 rpm

(600/23)1/11

This is a non standard step ratio; therefore the speeds are calculated as follows,

N1=23 rpm

N2=23x1.345 =30.9 rpm

N3=23x1.3452=41.6 rpm

- N4=23x1.3453=55.9 rpm

Similarly other speeds are 75.3, 101.23, 136.2, 183.1, 246.3, 331.3, 445.6, 599.3 rpm.



Speed diagram:

Kinema

tic

Arrange

ment:-

7. A sixteen speed gear box is required to furnish output speeds in the range of 100 to 560 rpm. Sketch the kinematic arrangement and draw the speed diagram. (Nov14-Co4-H3)









8. A sliding mesh gear box is to be used for „4‟ forward and „1‟ reverses speeds. First gear speed ratio is 5.5and reverse gear speed ratio is 5.8. Clutch gear on clutch shaft and gear (in constant gear) on lay shaft has speed ratio of 2. Calculate the number of teeth on all the gears. Assume that the minimum number of teeth on any gear should not be less than 18. Calculate actual gear ratios. Assume that the geometric progression for gear ratios, top gear (fourth), third gear, second and first gear is 1: x

: x2 : x3. [CO4 - H3 - Nov14] Solution:-





9. A gear box is to give 18 speeds for a spindle of a milling machine. Maximum and minimum speeds are of the spindle are to be around 650 and 35 rpm respectively. Find the speed ratio which will give the desire speeds and draw the structural diagram and kinematic arrangement of the drive. [CO4 - H3 - Nov15]

Given:-

N=18, Nmax=650rpm, Nmin=35

Solution:-

1. Spindle speeds:-

=

=1.1875

We can write, 1.06 x 1.06 x 1.06 =1.191

Skip 2 speeds

From R 40 series: the spindle speeds are given by:-

35.5,42.5,50,60,71,85,100,118,140,170,200,236,280,335,400,475,560,and 670 rpm

2. Structural diagram: - 2(1) 3(2) 3(2)

3. Ray diagram:-

Checking the conditions: -

2,

0.25

Stage 3: [280/140] =2, [35.4/140] =0.253 Stage 2:- [280/236]=1.186, [140/236]=0.59 Stage:-1 [280/475]=0.59, [236/475] = 0.479

In all the 3 stages the ratio requirements are satisfied.



4. Kinematic arrangements:-



10. Design a nine speed gear box for a machine to provide speed ranging from 100 rpm to 1500 rpm .the input is from a motor of 5 KW at 1440 rpm .Assume any alloy steel for the gears. [CO4 - H3 - Apr15]

Given data:-n=9, Nmax=1500, N min=100rpm,P=5kW,Ninput 1440

Assume material=C45

Solution:-design 9 speed gear box

1. Selection of spindle speed:- =1500/100, =1.403

From R20 series, 1.12 x (1.12 x1.12) =1.405, 2 skips

Speed:- 100,140,200,280,400,560,800,1120,1600 rpm

2. Ray diagram:-

Structural formula:-3(1) 3(3).

Ratio conditions checking: - stage2:-800/400 =2, 100/400 = 0.25—o.k

Stage 1:- 800/1120 = 0.71‘ 400/1120 =0.35 o.k

3. Kinamatic arrangement:-





4. Calculation of number of teeth:-

Stage 2:-pair (11, 12)

=

, assume Z11=20 teeth

=

Z12=80

Pair (9, 10)

=

=

= 0.7,

wkt Z9 +Z10 =Z11 +Z12 =100

0.7Z10+Z10=100

Z10=59.

Z9=41

Similarly calculating all the pairs the teeth are Z8=34,Z7=66

For stage 1:-

Z5=20,Z6=56,

Z4=51,Z3=25

Z1=31,Z2=45

5. Selection of suitable material: C45

6. Calculatoion of module:-m

T12 = 60x5x103 /2x3.14x100 =477.465 nm

F t=11936.625/m

m=√

=√

,

m=6.31

Standard module m=6.5 mm



7. Calculation of centre distance:-

Stage 1 [(Z1+Z2) /2] x m

=247 mm

Stage 2 [(Z7+Z8)/2] X m

=325 mm

8. Calculation of face width: - b= m X m =10 x 6.5 =65 mm

9. Length of shaft L=25+10+7(6.5)+20+7(605)+10+25 =181 mm

10. Design of spindle shaft:-

M=

, but Fn =

, substitute Ft value ,and assume = 20o then

Fn= 1954.24

Substitute L=181mm, then M=8.84x104 N mm

Wkt, T eq = , substituting M & T value,

We get Teq=4.855 x10 5 Nmm.

Wkt, Teq=

, from this we can find,

The diameter of the spindle shaft d=45 mm.

Using the equation T = 0.2 ds3 ( ),

We can find the diameter of shaft 1,and shaft 2 as follows.

Shaft 1 dia=20 mm.

Shaft 2 dia=30 mm.



UNIT V

CAMS, CLUTCHES AND BRAKES

Cam Design: Types-pressure angle and under cutting base circle determination-forces and

surface stresses. Design of plate clutches –axial clutches-cone clutches-internal expanding

rim clutches- Electromagnetic clutches. Band and Block brakes - external shoe brakes –

Internal expanding shoe brake.

PART A

1. State the advantage of cam mechanisms. [CO5 - L2 - Apr09]

Cams are used for transmitting desired motion to a follower by direct contact.Cam

mechanisms are used in the operation of IC engine valves.

2. Define base circle and pitch circle with respect to cam. [CO5 - L1]

(i) Base circle is the smallest circle drawn to cam profile.

(ii)Pitch circle is the circle drawn with its centre as the centre of cam axis and the radius equal to the distance between cam centre and the pitch point.

3. State the advantage of cam over other reciprocating mechanisms. [CO5 - L1]

(i) It is very easy, accurate and efficient to produce a given motion, velocity and

acceleration.

(ii) The rise period, dwell period and the return period can be increased or

decreased depending upon the shape of the cam which cannot be performed by

other reciprocating mechanisms.

4. When do we use multiple disk clutches? [CO5 – L2]

When large amount of torque is to be transmitted. In multiple clutch the number of frictional linings and the metal plates are increased which increases the capacity of the clutch to transmit torque.

5. Name the profile of cam that gives no jerk. [CO5 - L2 - Nov15]

Circle-arc cam gives no jerk. Because the derivative of acceleration of cam is zero .i.e., Jerk=∞=d3θ/dt3=0 where θ is cam rotation.



6. What is significance of pressure angle in cam design? [CO5 - L1]

The pressure angle in cam design plays a very important role as its maximum value establishes, the cam size, torque loads, side thrust, wear, acceleration of follower etc.,

7. Name four profiles normally used in cams. [CO5 – L2]

1. Uniform velocity, 2.simple harmonic motion, and 3.uniform acceleration and retardation and4.cycloidal motion.

8. Name four materials used for lining of friction surfaces in

clutches. Write the desirable properties of lining materials. [CO5 – L2]

Properties of materials commonly used for lining of friction

surfaces.

9. How the “uniform rate of wear” assumption is valid for clutches? [CO5 - L1]

In clutches, the value of normal pressure, axial load for the given clutch is

limited by the rate of wear that cam be tolerated in the brake linings. Moreover,

the assumption of uniform wear rate gives a lower calculated clutch capacity

than the assumption of uniform pressure, hence clutches are usually designed

on the basis of uniform wear

10.What are the effects of temperature rise in clutches? [CO5 – L3]

During operation of a clutch, most of the work done against frictional forces opposing the motion is liberated as heat at the interface. It has been found that at the actual point of contact, the temperature as high as 10000C is reached for short duration. ((i.e.) for

0.0001sec).Due to this, the temperature of the contact surfaces will increase and may

destroy the clutch.



11.Under what condition of clutch, uniform rate of wear assumption is more valid?

[CO5 - H1]

In clutches, the value of n normal pressure, axial load for the given clutch is limited by the

rate of wear that can be tolerated in the brake linings. Moreover, the assumption of

uniform wear rate gives a lower calculated clutch capacity than the assumption of uniform

pressure. Hence clutches are usually designed on the basis of uniform wear.

12.Why is it necessary to dissipate the heat generated during clutch operation?

[CO5 – L1]When clutch engages, most of the work done will be liberated as heat at the

interface. Consequently the temperature of the rubbing surfaces will increase. This

increased temperature may destroy the clutch. So heat dissipation is necessary in

clutches.

13.Name different types of clutch. Why positive clutch is used? [CO5 - H1]

Following are the two main types of clutches commonly used in engineering practice

1. Positive clutches,

2. Friction clutches.

14.What is fading of brakes? [CO5 - L1]

When the brake is applied continuously over a period of time, the brake becomes overheated and the coefficient of friction drops. This result in sudden fall of efficiency of the brake. This is known as fading of brake.

15.How does the function of a brake differ from that of a clutch?(or)Differentiate

between clutch and brake[CO5 - L2 - Nov10] [CO5 - L2 - Nov13]

The break connects a rotating member with a non rotating member but, a clutch connects two rotating members.

16. What is meant by a self – energizing brake? [CO5 - L2 - Apr14]

When the moment of applied force and the moment of frictional force are in the same

direction, the frictional force helps in applying the brake. This type of brake is known as

self energizing brake.

17. What is self locking in brake? [CO5 - L1 - Apr13]

When the frictional force alone is sufficient to apply the brake without any external force,

then the brake is said to be self locking.



18.Why in automobiles braking action when travelling in reverse is not as effective

as when moving forward? [CO5 - L2 - Apr08]

When an automobile moves forward the braking force acts in the opposite direction to the

direction of motion of the vehicle. Whereas in reverse travelling the braking force acting

the same direction to the direction of motion of the vehicle. So it requires more braking

force to apply brake.

19. Give the reason for left and right shoes of internal expansion brakes

having different actuating forces. [CO5 - L2 - Apr07]

Depending upon the direction of the drum rotation, one shoe would be a leading shoe and another shoe is a trailing shoe. The leading shoe is self energizing whereas the trailing shoe is not. In the leading shoe, the friction force helps the applied force and hence more

actuating force thab the trailing force.

20. List the characteristics of material used for brake lining. [CO5 - L3 - Apr10]

The material used for the brake lining should have the following characteristics:

1. It should have high coefficient of friction with minimum fading. In other words, the

coefficient of friction should remain constant over the entire surface with change in

temperature.

2. It should have low wear rate.

3. It should have high heat resistance.

4. It should have high heat dissipation capacity.

5. It should have low coefficient of thermal expansion.

6. It should have adequate mechanical strength.

7. It should not be affected by moisture and oil.

21. Classify the clutches based on the coupling methods. [CO5 - L1 - Apr14]

1. Positive contact clutches

2. Frictional clutches

3. Overrunning clutches



Sl. n

Uniform pressure theory Uniform wear theory

1 Discs used in this type of clutches are flexible.

Discs used in this type of clutches are rigid.

2. Pressure obtained on the friction surface is uniform.

Wearing of the friction surface is uniform.

3. It gives higher friction torque.

it gives less friction torque 4. This is not preferable theory. This is most preferable theory.

4. Magnetic clutches

5. Fluid couplings

22. Differentiate between uniform pressure and uniform wear theories adopted in

the design of clutches. [CO5 - L1 - Nov14]



23. In a hoisting machinery, what are the different energies absorbed by a brake

system? [C05 - L2 - Nov14]

Energy absorbed by the brake system = [change in kinetic energy due to linear

Movement

+

Change in kinetic energy due to rotational

Movement

+

Potential energy due to lifted load.]

24. What is meant by positive clutch? [CO5 - H1 - Nov15]

The positive clutches are used when a positive drive is required. The simplest type of a positive clutch is a jaw or claw clutch. The jaw clutch permits one shaft to drive another through a direct contact of interlocking jaws. It consists of two halves, one of which is permanently fastened to the driving shaft by a sunk key. The other half of the clutch is movable and it is free to slide axially on the driven shaft, but it is prevented from turning relatively to its shaft by means of feather key.

25.Why in automobiles, braking action when traveling in reverse is not as effective as when moving forward? [CO5 - L2 - APR15]

(i) When the vehicle is moving in 'reverse' (anti-clockwise rotation of the brake drum), the couple due to actuating force (P x C) and the couple due to friction force (p dN), i.e., Mf are opposite. Therefore, friction force tends to increase the actuating force and consequently, the force on the brake pedal is increased [P = (Ma + MIK]

(ii) Therefore, braking action when traveling in 'reverse' is not as effective as when traveling 'forward'.

26.If a multidisc clutch has 6 discs in driving shaft and 7 discs in driven shaft then how many number of contact surface it will have? [CO5 - L3 - APR15]

Given:-

N1=6

N2=7

Solution:-

Number of contact surfaces , n=n1+n2 – 1

= 6+7-1

=12



16 MARKS

CLUTCHES

Problems on single plate clutch:-

1. An automatic single plate clutch consists of two pairs of contacting surfaces.

The inner and outer radii of friction plate are 120mm and 250mm respectively. The

coefficient of friction is 0.25 and the total axial force is 15KN.calculate the power

transmitting capacity of the clutch plate at 500 rpm. Using, 1) Uniform wear theory

and 2) Uniform pressure theory. [CO5 - H3 - Nov10]

Given data:-

N=2, r1=250mm=0.25m,W=15KN=15*103N;r2=120mm=0.12m,N=500rpm

Solution:-

1) Using uniform wear theory

Torque transmitted on clutch is given by

T=n.µ.w(m1+m2) /2

=2*0.25*15*103(0.25+0.12)/2

=1387.5nm

Power transmitted

=2∏nt/60

=2∏*500*1387.5/60

=72.65KW

2) Using uniform pressure theory:-

Torque transmitted on clutch is given by

2



T=nµw2/3[r1-r23 /r12 -r22 ]

= 2*0.25*15*103*2/3*[(0.25)3-(0.12)3/(0.25)2-(0.12)2]

=1444.6n-m

Power transmitted = 2πNT/60

=2π*500*1444.6/60

75.64KW.

2. A single plate friction clutch, with both sides of the plate being effective,is used

to transmit power at 1440rpm. It has outer and inner radii 80mm and 60mm

respectively. The maximum intensity of pressure is limited to 10*104N/m2. If the

coefficient of friction is 0.3 determine; [CO5 - H3 - Nov10]

1) Total pressure exerted on the plate, and 2) power transmitted.

Given:-

n =2,N =1440rpm,R1=80mm=80*10-3m, R2=60mm=60*10-3m, Pmax =10*104N/m2;µ=0.3.

Solution:-

1) Total pressure exerted on the plate:-

Since the intensity of pressure(p) is maximum at the inner radius(r2),therefore for

uniform wear .

Pmax.r2=c (or)

C=10*104*60*10-

3

=6000 N/m



We know that the axial

trust, W = 2πc

(r1-r2)

= 2π*6000(80*10-3-60*10-3)

= 754N

Axial thrust or total pressure exerted on the plate =754

N.

2) power transmitted:-

Torque transmitted => T=n.µ.w.(r1+r2)/2

= 2*0.3*754(80*10-3+60*10-3)/2

= 31.67 N-m

Power transmitted,

P=2πNT/60

= 2π*1400*31.67/60

= 4.643KW.



3. A single plate clutch transmits 25kw at 900rpm. The maximum pressure

intensity between the plates is 85KN/m2. The ratio of radii is 1.25. Both the sides

of the plate are effective and the coefficient of friction is 0.25.determine

1) The inner diameter of the plate and,2)The axial force to engage the clutch.

Assume theory of uniform wear. [CO5 - H3 - Nov09]

Given:-

P =25Kw = 25*103w, N=900rpm,R1/r2=1.25,n =2,Pmax = 85KN/m2=85*103N/m2,µ =

0.25

Solution:-

1) the inner diameter of the plate:-

WKT the power transmitted = 2πNT/60

25*103=2π *900*T/60, T=265.26N-m

Since the intensity of pressure is maximum at the inner radius

(r2), Pmax .r2=c (or)

C = 85*103r2 N/mm

2. the axial thrust transmitted to the frictional surface,

W = 2πc (r1-

r2)

= 2π*85*103 r2 (1.25r2-r2)

=1.335*103(r2)



2

The mean radius for uniform wear is given

by,

R=r1+r2/2

=1.25r2+r2/2

=1.125r2.

Torque transmitted=n.µ.w.R

265.26=2*0.25*1.335*105(r2)2*1.125r2

265.26=75.104*103r 3

R2=0.1523 m (or) 152.3mm

R1=1.25r2

=1.25*152.3


185

R1=190.375mm

3) The axial force to engage the clutch:-

W= 2πc (r1-r2)

=

1.335*105(r2)

2

=

1.335*105(0.1523)

2

=

3096.57N

.

4.Determine the maximum, minimum and average pressure in a plate clutch

hen the axial force is 5000N.the outer diameter of the friction surface are

200mm and100mm respectively. Assume uniform wear. [CO5 - H3 - Nov11]

Given:-

W=5000N,d1= 200mm (or) r1=100mm=100*10-3m ,d2= 100mm (or)

r2=50mm=50*10-3m

Solution:-

1) Maximum pressure:-

Since the intensity of pressure is maximum at the inner radius (r2),

therefore


186

Pmax x r2 =c

C=50*10-3pmax

Axial force exerted on the contact surface (w) is given by

W=2πc

(r1-r2)

5000=2π*50*10-3pmax (100*10-3-50*10-3)

=0.0157pmax

Pmax=31.83*104N/m2

2) Minimum pressure:-

Since the intensity of pressure is minimum at the outer surface (r1), therefore

Pmin*r1=c (or)

=100*10-3pmin

Axial force exerted on the contact surface (w) is given by

W= 2πc (r1-

r2)pmin

5000=2π*100*10-3*(100*10-3-50*10-3)

=0.0314pmin


187

Pmin =15.92*104N/m2

3) Average pressure:-

Pav= total normal force on contact surfaces/cross sectional area of contact surface

=w/π[r12-r2

2]

=5000/π[(100*10-3)2-(50*10-3)2]

=21.22*104N/m2

5.A friction clutch is used to rotate a machine from a shaft rotating in a uniform

speed of 250rpm.the disc type clutch has both of its sides effective,the

coefficient of friction being 0.3.the outer and inner diameters of the friction plate

are 200mm and 120mm respectively. Assuming uniform wear of the clutch the

intensity of pressure is not to exceed 100kN/m2.if the moment ofinteria of the

rotating parts of the machine is 6.5 kg-m2,determine the time to attain the full

speed by the machine and the energy lost in slipping of the clutch.What will be

the intensity of pressure if the condition of uniform pressure of the clutch is

considered? Also determine the radio of power transmitted with uniform wear to

that with uniform pressure. [CO5 - H3 - Nov09]

Given:-

N=250rpm,N=2,µ=0.3,d1= 200mm (or) r1=100mm (or) 0.1m,d2= 120mm (or)

r2=60mm (or) 60*10-3m

Solution: - 1. The time to attain the full speed by the machine:-

Also T= I α, Where (α) is angular

acceleration

72.38=

6.5*α


188

α = 11.135

rad/sec2

We also know that

α =

ω/t

= 11.135

(or)

(2πN/60)*1/t = 11.135

2π*250/60*t = 11.135

t =2.35sec

2) The energy lost in slipping of the clutch:-

Angle turned by the driving shaft,

Ø1=ω t=2πN/60*t

= 2π*250/60*2.35

=61.52rad

And angle turned by the driven shaft,

Ø2=ω0t + ½

αt2

=0+1/2*11.135*(2.35)2


189

=30.75rad

Energy lost in friction = T (ø1-ø2)

=72.38*(61.52-30.75)

Energy lost in friction = 226 N-m.

3) Intensity of pressure, if the condition is uniform pressure:-

Intensity of pressure, p = w/π(r12 -r22 )

=1507.96/π[(100*10-3)2-(60*10-3)2]

=75000N/m2

=75 KN/m2.

4) Ratio of power transmitted with uniform wear to that with uniform pressure;

Power transmitted with uniform wear = 1895w, [already calculated,]

Torque transmitted with uniform pressure

2

= n. μ.w.2/3 [r13-r2

3/r12-r2

2]

=2*0.3*1507.96*2/3[(0.1)3-(0.06)3/(0.1)2-(0.06)2]


190

T= 73.89N-m.

Power transmitted with uniform pressure is given by,

P = 2π*250*73.89/60

P= 1934 w

Power transmitted with uniform wear

Power transmitted with uniform pressure = 1895/1934

=0.98


191

6. Determine the time required to accelerate a counter shaft of rotating mass500kg

and radius of gyration 200mm to the full speed of 250rpm. From rest through a

single clutch of internal and external radii 125mm and 200mm, taking µ as 0.3 and

the spring force as 600N. assume that only one side of clutch is effective. [CO5 -

H3 - Apr10]

Given:-

M=500kg,k=200mm=0.2m,N=250rpm,R2=125mm=0.125m,r1=200mm=0.2m,W=600N;

µ=0.3;n=1

Solution:-

Moment of inertia of shaft is given

by

I=mk2

=500*(0.2)2

=20kg-m2

Torque transmitted is given by,

T=I*α

N µ w (r1+r2/2) =I. α

1*0.3*600(0.2+0.125/2) =20.α (or)

Angular acceleration , α a = 1.4625rad/sec2

Wkt, angular speed = angular acceleration *time


192

ω = α .t (or) t=w/α

t= (2πN/60)*1/α

t= 2π*250/60*1/1.4625

t= 17.9 seconds.

Problems on multiplate clutch & Cone Clutch

7. A multiplate disc clutch transmits 55kw of power at 1800rpm.coefficient of friction for the friction surfaces is 0.1. Axial intensity at pressure is not to exceed

160KN/m2. The internal radius is 80mm and 0.7 times the external radius. Find the number of plates needed to transmit the required torque. [CO5 - H3 - Nov09]

Given:-

P=55kw =55*103w; N=1800rpm; µ=0.1;Pmax=160KN/m2=160*103N/m2;

r2=80mm=8*10-3m; r2=0.7r1 (or) r2/r1=0.7

To find:-

Number of plates needed to transmit the required torque


193

Solution:-

r2=0.7r1 (or) r2/r1=0.7

r1=r2/0.7 = (80*10-3)/0.7

r1=0.1143m.

Assuming uniform wear, axial force exerted is given

by w = 2πc (r1-r2)

wkt the maximum intensity of pressure (pmax) is at the inner radius (r2).

Pmax .r2=c (or)

C=160*103*80*10-

3

C=12800N/m

Then, w= 2πc (r1-r2)

=2π*12800(0.1143-0.08)

=2758.57N

Torque transmitted by a single friction surface is given by

T=µ.w. (r1+r2)/2

=0.1*2758.57*(0.1143-0.08)/2

Torque required per surface [T=26.8N-m]


194

Total torque required can be calculated as given

below, Power ,p=2πNT/60

(55*103)=2π*1800*T/60

Total torque required, T=291.78N-m

Number of friction surfaces required =total torque required

Total required per surface

=291.78/26.8

=10.887

= 11

Total number of plates ={number of pairs of contacts surface +1}

=11+1

=12 surface

Hence, there will be 12 total plates, which driving and driven shafts having six plates

each.



8. A multi plate clutch has three discs on the driving shaft and two on the driven

shaft. the outside diameter of the contact surfaces is 240mm and inside diameter

is 120mm. assume uniform wear coefficient of friction as 0.3, find the maximum

axial intensity of pressure between the discs for transmitting 25kw at 1575rpm.

[CO5 - H3 - Apr11]

Given:-

n1 = 3, n2 = 2,d1 = 240mm (or) r1 = 120mm = 0.12m,d2 = 120mm (or) r2 = 60mm

=60*10-3m,µ = 0.3 ,p = 25kw = 25*103w , N = 1775rpm

Solution:-

Number of pairs of contacts

surfaces, n = n1 + n2

– 1

=3+2-1=4

Power transmitted p = 2πNT/60

25*103 = 2π(1575)T/60

T=151.6N-m

For uniform wear, torque transmitted is given

by, T=n.µ. (r1+r2/2)

151.6 = 4*0.1*w*(0.12+0.06/2)

W =1404N (or)



The axial force exerted (w) can also be given by,

W = 2πc (r1-r2)

W = 2π * pmax * r2 (r1-r2)

1404 = 2π*pmax*0.06(0.12-0.16)

Pmax = 62.07*103N/m2

Pmax = 62KN/m2.

9. A multi disc clutch has three discs on the driving shaft and two on the driven

shaft is to be designed for a machine tool, driven by an electric motor of 22KW

running at 1440rpm. The inside diameter of the contact surface is 130mm. the

maximum pressure between the surface is limited to 0.1N/mm2. Design the clutch

take µ=0.3, n1=3, n2=2. [CO5 - H3 - Apr11]

Given:- P=22kw=22*103w; N=1440rpm;d2=130mm (or) r2=65mm,Pmax=0.1N/mm2,

=0.1*106N/m2

To find:-

Design the clutch

Solution:-

Assume uniform wear

1) Outside diameter of disc (d1)

Wkt the torque transmitted,

T = p*60/2πN

=22*103*60/2*π*1440



=145.98N-m

Design torque,(T)=T*ks

Service factor ,kg=K1+K2+K3+K4

From table 10.2, k1

10.Determine the axial force required to engage a cone clutch transmitted 25kw of

power at 750rpm average friction diameter of the cone is 400mm, semi-cone angle

10˚ and co-efficient of friction 0.25. [CO5 - H3 - Apr10]

Given:-

P=25kw = 25*103w, N=750rpm,D=400mm (or) r=200mm=0.2m,Α=10˚, µ=0.25

Solution:-

Power transmitted

P=2πNT/60

25*103=2π*750*T/60

T=318.3N-m (or)

Normal load acting on friction surface can be obtained by

T=µ. wn .r

Wn =T/µr=318.31/0.25*0.2

Wn =6366.2N (or)

The axial force required to engage the cone clutch is given by

We=wn (sinα+µcosα)

=6366.2(sin10˚+0.25cos10˚)

We=2672.85N.



11. The following data relate to a cone clutch: minimum and maximum surface

contact radii are 125mm and 150mm respectively. Semi-cone angle=20; allowable

normal pressure is 14*104N/m2;µ=0.25. find a) the axial load, and b) the power

transmitted , if the speed is 750rpm. [CO5 – H3]

Given:-

r1=150mm=0.15m, r2=125mm=0.125m,α=20˚, µ=0.25 ,pn=14*104N/m2, N=700rpm.

Solution:-

a) The axial load transmitted to the clutch:-

W = ∏pn (r12-r2

2)

= π*14*104[(0.150)2-(0.125)2] =3023.78N

b) Power transmitted :-

Torque T=2/3µw [r13 -r23 / r12 -r22 ] cosecα

=2/3*0.25*3023.78[(0.150)3-(0.125)3/(0.150)3-(0.125)2]cosec 20˚

=304.73N-m

Power transmitted

P=2∏NT/60

= 2∏*700*304.73/60

=22.34kw.

12. A torque of 350N-m is transmitted through a cone clutch having a mean

diameter of 300mm and a semi-cone angle of 15˚. The maximum normal pressure

at the mean radius is 150KN/m2. The coefficient of friction is 0.3. calculate the

width of the contact surface . also find the axial to engage the clutch. [CO5 – H3]

Given:-



T=350N-m,D= 300mm (or) R=150mm=0.15mk,α=15˚,pn

=150KN/m2=150*103N/m2,µ=0.3

Solution:-

1) width of the contact surface:-

Torque transmitted

T=µ.wn.

R

350=0.3*wn*0.15

Wn= 7778N (or)

Let ‗b‘ be the width of the contact surface wkt the normal load acting on the friction

surface,

Wn= Intensity of cross sectional pressure * area of friction surface

Wn=pn*(2πR*b)

7778=150*103*2π*0.15*b (or)

Face width, b=0.055m (or) 55mm

2) axial force to engage the clutch:-

we=wn(sinα+µcosα)

=7778(sin15˚+0.3cos15˚)

=4.267KN.



13. A leather faced conical friction clutch has a cone angle of 30˚. The intensity of

pressure between the contact surface is not to exceed 6*104N/m2 and the breadth

of the conical surface is not to be greater than 1/3 of the mean radius if µ=0.20 and

the clutch transmits 37KW at 2000rpm. Find the dimensions of the contact

surface. [CO5 – H3]

Given:-

α=30˚, pn=6*104N/m2 , b=R/3 , µ=0.2 , p=37KW=37*103w , N=2000rpm

To find:-

Dimensions of contact surface (r1and r2)

Solution:-

Power transmitted

P=2∏NT/60

37*103=2∏*2000*T/60

T=176.66N-m (or) Assuming service

factor, ks=2.5,

Design torque (T) =176.66*2.5

=441.65N-m

Torque transmitted is also given by,

T=2∏µpn.R2.b

441.65=2∏*0.2*6*104*R2(R/3)

=25132.74R3

R=0.25998m (or) 259.98mm



Face width is given by,

B=R/3=0.25998/3

B=0.08666 (or) 86.66mm

We find that

r1-r2/b =sinα

r1-r2= b sinα

=0.08666sin15˚

r1-r2 = 0.02242m ………..1

Mean radius R= r1+r2/2

R=0.25995m r1+r2=0.5199m …………..2

Solving eqns 1 and 2

1) outer radius of contact surface

r1=0.2711m (or) 271.1mm

2) Inner radius of contact surface

r2=0.2487m (or) 248.7m.

14. A centrifugal clutch has four shoes each having a mass of 5Kg and having its

center of gravity at a radius of 60mm. the diameter of the friction surface

is150mm. the clutch is to transmit 6Kw power at a speed of 1000rpm.

Determinethe force which must be exerted by each spring . if the speed of

rotation is gradually increased from rest, at what speed will the clutch begins to

transmit torque? Take µ=0.30. [CO5 – H3]



Given:-n=4 , m=5Kg , r=60mm =60*10-3m , D=150mm (or) R=75mm=75*10-3m ,

p=6Kw=6*103w , N=1000rpm ,µ=0.35.

To find:-

1) Force exerted by each spring, Fs and

2) Speed at which transmitted begin, N

Solution:-

1) Force exerted by each spring :-

P=2∏NT/60

6*103=2π*1000*T/60

T=57.29N-m (or)

Total friction torque transmitted can also be given

by, T=n(Fc-Fs).µ.R

57.29=4*(Fc-Fs)*0.35*75*103

Radial force required at each

shoe, Fc-Fs =545.67N

Centrifugal force Fc=mw2r

=5*(2π*1000/60)2*60*10-3

=3289.9N

Then the force exerted by the spring is

obtained by, Fc-Fs=545.67



Fs= Fc-545.67

=3289.9-545.67

=2744.2N

2) Then the speed at which transmission begin:-

Transmission will begin when centrifugal force is equal to spring force

When m(2πN/60)2r = Fs=2744.2

5*4*π2*N2*60*10-3/3600

=2744.2

N=913.31rpm.

15. A centrifugal clutch is to transmit 15Kw at 900rpm. The shoes are four in

number. The speed at which the engagement begins is ¾th of the running pulley

rim is 150mm and the center of gravity of the shoe lics at the center of gravity of

the spider. The shoes are lined with fexrodo for which the co effient of friction

may be taken as 0.25 determine . [CO5 – H3]

1) Mass of the shoes and

2) Size of the shoes ,if angle subtended by the shoes at the center of

the spider is 60˚ and the pressure exerted on the shoes is0.1N/mm2

Given:-

P=15kw=15*103w , N=900rpm, n=4, R=150mm=0.15m , r=120mm=0.12m ,



µ=0.25, θ=60˚, p=0.1N/mm2 = 0.1*106N/m2

To find :-

1) Mass of the shoes (m), and

2) Size of the shoes (b)

Solution:-

1) Mass of the shoes(m)

Angular speed, ώ=2πN/60=2π*900/60

=94.26 rad/s

Speed at which the engagement begins (ώ1) is ¾th of running speed (ώ)

ώ 1=3/4 ώ

3/4*94.26

= 70.7rad/s

Power transmitted ,p =2∏NT/60

15*103=2*900*T/60

T=159N-m

Centrifugal force on each shoes

, Fc=m ώ2r

=m(94.26)20.12=1066mN

Spring force on each shoe ,ie the centrifugal force at the engagement speed ώ1,

Fs=m(ώ1)2r

=m(70.7)20.12

=600mn



Frictional force acting tangential on each

shoe, F=µ(Fc-Fs)

=0.25(1066m-600m)

=116.5mN

Wkt the torque transmitted,

T=η.F.B

159=4*116.5m*0.15

=70m

M=2.27kg

2) Size of the shoes:-

Θ=60˚=60˚*∏/180=∏/3rad

Contact length of shoe= angle subtended by the shoe*contact radius of shoe

L=θ.R

=∏/3*0.15

=0.1571m

Wkt ,

Fc-Fs=l.b.p

1066m-600m=0.1571*b*0.1*106

466m=466.2.27=0.1571*b*0.1*106



B=0.0673m (or) 67.3mm.

16. An electric motor rotating at 350rpm. Drives a machine through a plate clutch

whose both sides are effective. when the clutch is engaged it takes 3 seconds for

the driven machine to attain the speed of motor . the moment of inertia of driven

machine is 4.5Kg.m2. calculate the torque produced by the motor and its power.

Also calculate the energy dissipated by the clutch. [CO5 – H3]

Given:-

N1=350rpm ώ1=2∏N1/60=2∏(350)/60=36.65rad/s t1=3sec I2=4.5Kg-m2

To find:

1) Torque produced by the motor(T)

2) Power developed (P) and

3) Energy dissipated by the clutch(E)

Solutions:-

1) Torque produced by the motor(T);

Wkt the time required for completing the clutching operation(considering only I2),

t1=I2* ώ1/T

torque (T)=I2* ώ1/t1

=45*36.65/3

T=54.98N-m



1

2) Power developed(P);

Wkt, power, p=T* ώ

=54.98*36.65

=2014.94 ώ

3) Energy dissipated by the clutch(E);

Wkt , E1=1/2I2 ώ 2

=1/2*4.5*(36.65)2

=3022.25I.

17.A cone clutch is used to connect an electric motor running at 1440 rpm with a

machine that is stationary. The machine is equivalent to a rotor of muss 150 kg

and radius of gyration as 250 mm. The machine has to be brought to the full speed

of 1440 rpm from a stationary condition in 40 s. The semi-cone angle a is 12.5".

The mean radius of the clutch is twice the face width. The coefficient of friction is

0.2 and the normal intensity of pressure between contacting surfaces should not

exceed 0.1 Nimm2. Assuming uniform wear criterion, calculate: (i) the inner and

outer diameters: (ii) the face width of friction lining: (iii) the force required to

engage the clutch: and (iv) the amount of heat generated during each engagement

of clutch. [CO5 – H3]







18.A cone clutch with asbestos friction lining transmits 30 kW power at 500 rpm.

The coefficient of friction is 0.2 and the permissible intensity of pressure is 0.35

N/mm2. The semi-cone angle a is 12.5°. The outer diameter is fixed as 300 mm

from space limitations. Assuming uniform wear theory, calculate: (i) the inner

diameter. (ii) the face width of the friction lining; and (iii) the force required to

engage the clutch. [CO5 – H3]



19.A multi disk clutch consists of live steel plates and four bronze plates. The inner and outer diameters of friction disks are 75 and 150 mm respectively. The coefficient of friction is 0.1 and the intensity of pressure on friction lining is limited to 0.3 N/mm2. Assuming uniform wear theory, calculate: (i) the required operating force; and (ii) power transmitting capacity at 750 rpm. [CO5 – H3]

Solution:-

20.A multi plate clutch of alternate bronze and steel plates is to transmit 6 kW

power at 800 rpm. The inner radius is 38 mm and outer radius is 70 mm. The

coefficient of friction is 0.1 and maximum allowable pressure is 350 kN /m2

determine (i) Axial force required (ii) Total number of discs (iii) Average pressure

and (iv) Actual maximum pressure[CO5 – H3]

Given: P = 60 kW, N = 800 rpm, R1 = 38mm, Di= 76 mm R0 = 70, D0 = 140mm, µ = 0.1,

P = 350 kN / m2 = 0.35 N / mm2





21.In a maultilate clutch radial width of the friction material is to be 0.2 of

maximum radius. The coefficient of friction is 0.25. The clutch is 60KW at 3000

rpm. Its maximum diameter is 250mm and the axial force is limited is to 600N.

Determine (i) Number of driving and driven plates (ii) mean unit pressure on each

contact surface assume uniform wear. [CO5 – H3]

Given: Radial width = 0.2 Ro, µ = 0.25, P = 60KW, N = 3000rpm, D0 = 250mm,

Ro = 125mm, Fa = 600N uniform wear condition.

Solution b = Ro- Ri 0.2 Ro =Ri Ri= 0.8 Ro = 0.8x 125 = 100mm Inner diameter

2x100 = 200mm



Total number of plates : n1 + n2 = 617 = 13



BRAKES

1 .A single black brake is shown in fig.1 The diameter of the drum is 180 mm and

the angle of contact is 600.if the operating force of 400 N is applied at the end of a

lever and the co-efficient of friction between the drum and the lining is

0.30, determine

(i) The torque that may be transmitted by the block brake.

(ii) The rate of heat generated during the braking action when the initial

brake speed is 200 rpm.

(iii) The dimensions of the block if the intensity of pressure between the block

and brake drum is 1 N/mm2.the breath of the block are twice its width. [CO5 –

H3]

Fig-1

Given:-d=180mm, r=90 mm=90x10-3,F=400 N,2⍬=600= ∏/3rad, µ=0.30,

N=300 rpm, b=2w.

To find: -

(i) the torque transmitted by the block shoe.

(ii) The rate of heat generated during the braking action



(iii) The dimensions of the block

Solution:-

Since the angle of contact is greater than 400, the equivalent co-efficient of friction is

given

By

(i) The torque transmitted by the block shoe.

Rn=normal reaction of force on the block.

Taking moment about the fulcrum O we get,

Rn = 993.3 N

(ii) The rate of heat generated during the braking action:-

Initial velocity of the drum

Final velocity of the drum V2=0



Average velocity of the drum = (V1+V2)/2

= (2.827+0)/2 =1.414 m/s

Wkt,

Rate of heat generated=friction force x average velocity.

= 0.3x993.3x1.414

= 421.36 Nm/s (or) W

(ii) The dimensions of the block

Let b and w be the breath and width of the brake shoe respectively,

wkt,

Rn=pbw

993.3=1x2wxw

w=22.285 mm

b=2w=44.57 mm

2. The diameter of the brake drum of a single block shown in fig 2 is 1 m .it

sustains 240 Nm of torque at 400 rpm. the coefficient of friction is 0.32 .determine

the required force to be applied when the rotation of the drum is (a)

clockwise,(b)counter clockwise and the angle of contact is (i) 350 and (ii) 1000



given that a=800 mm, b=150 mm, c= 25 mm. also find the new values of “C” for

self locking of the brake. [CO5 – H3]

Fig-2

Given:-d=1m (or) r=0.5 m, TB=240 Nm, a=800 mm=0.8 m, b=150 mm=0.15m,

c=25 mm =0.025 m, µ=0.32,N=400 rpm,

To find: to determine the required force

Solution:-

Braking torque is given by,

240 = 0.32 x Rn x 0.5

Rn = 1500

(i) When angle of contact 2⍬ =350:-



(a) Rotation of drum is clock wise:-.

Taking moment about O we get, F x a = µRn x c + Rn x b F x 0.8 = 0.32x1500x0.025 +1500x0.15 F = 296.25 N

(b) Rotation of drum is counter clock wise:-. Taking moment about O we get,

Rn x b = F x a + µRn x c

1500 x 0.15 = Fx0.8 +0.32x1500x0.025

F = 266.25 N

(c) The new values of ―C‖ for self locking of the brake. For self locking, the externally applied force F must be zero. This is possible for

Counterclockwise rotation of the drum. Therefore,

Rn x b = 0 + µRn x c

B = μ c

c = μ / b

= 0.15 / 0.32 = 0.468 m = 468 mm

(ii) When angle of contact 2⍬ =1000:- Since the angle of contact is greater than 400 , the equivalent co-efficient of friction is given by





3. The block brake shown in fig. 3 is set by a spring that produce a force S on

each arch equal to 3500 N .the wheel diameter is 350 mm and the angle of

contact for each block is 1200 take coefficient of friction is 0.35, determine (i)

the maximum torque that the brake is capable of absorbing, and (ii) the width

of the brake shoes if the bearing pressure on the lining material is not to

exceeded 0.3 N/mm2. [CO5 – H3]



157500 = F 2[20/0.409 - 13.5]

F2 = 4449.2 N

Considered right hand side brake shoes:-

Taking moment about O1,



=4449.2/0.409

=10878.24 N

4. Determine the capacity and main dimensions of a double block brake for the

following data:

The brake sheave is mounted on the drum shaft. The hoist with its load weights

45 KN and moves downwards with a velocity of 1.15 mps. The pitch diameter of

the hoist drum is 1.25 m. the hoist must be stopped within a distance of 3.25

m. the kinetic energy of the drum may be neglected. [Co5 - H3 - Nov12]

Given:- Load=45Kn, V=1.15 m/s, D=1.25 m, x=3.25m

To find:-capacity and main dimensions of double black brake.

Solutions:-



(1) Calculation of total energy absorbed by the brakes. The various sources of energy to be absorbed are, (a) Kinetic energy of translation = 1/2 mv2

= 1/2 m (v12-v22)

V=velocity at the time of applying brake.

V1,v2 = initial and final velocity of the loads.

(b) potential energy = weight X vertical distance.

= w x X

(c ) kinetic energy of rotation=

Neglecting the kinetic energy of the drum,

Initial velocity of the load v1=1.15 m/s

final velocity of the load v2=0







5. a simple band brake is operated by a lever of length 500 mm long .the brake

drum has a diameter of 500 mm and the brake band embraces 5/8 of the

circumference one end of the band attached to the fulcrum of the lever while

the other is attached to a pin on the lever 100 mm from the drum. If the effort

applied to the end of the ------- 1000 N and the coefficient of friction is 0.25,

then design a simple band brake. [CO5 – H3]





6. Design a differential band brake for a crane lifting a load of 50 KN rope

wound round barrel of 550 mm diameter, as shown in fig.4 the brake drum to

be keyed to the same shaft is to be 650 mm in dia and the angle of tap of the

brake band over the drum is 2400 operating arms of the brake are 45 mm and

210 mm as shown in fig. Operating lever is 1.5 m long .take µ =0.25 [CO5 - H3 -

Apr15]



fig.4



7. Fig 5 shows the arrangements of two brake shoes which act on the

internalsurface of a cylindrical brake drum. The braking force F1and F2 are applied

asshown and each shoe pivots in its fulcrum O1 and O2.the width of the brake

lining is 55 mm and the intensity of pressure at any point A is 4x105 sin⍬

N/m2.wherw ⍬ is measured as shown from either pivot. The coefficient of friction

is0.40.determine the braking torque and the magnitude of the force F1 and F2.



Fig 5





8. Describe with the help of a neat sketch the design procedure of an internal

expanding shoe brake. Also deduce the expression for the braking

torque. [CO5 - H3 - Apr13] [CO5 - H3 - Nov13] [CO5 - H3 - Nov14]

An internal expanding brake consists of two shoes S1 and S2 as shown in Fig. 6 (a).

The outer surface of the shoes are lined with some friction material (usually with Ferodo)

to

increase the coefficient of friction and to prevent wearing away of the metal. Each shoe

is

pivoted at one end about a fixed fulcrum O1 and O2 and made to contact a cam at the

other

end. When the cam rotates, the shoes are pushed outwards against the rim of the drum.

The friction between the shoes and the drum produces the braking torque and hence

reduces the speed of the drum. The shoes are normally held in off position by a spring

as

shown in Fig. 6 (a). The drum encloses the entire mechanism to keep out dust and

moisture. This type of brake is commonly used in motor cars and light trucks.



We shall now consider the forces acting on such a brake, when the drum rotates in the

anticlockwise direction as shown in Fig. 6 (b). It may be noted that for the anticlockwise

direction, the left hand shoe is known as leading or primary shoe while the right hand

shoe is known as trailing or secondary shoe.

Let r = Internal radius of the wheel rim.

b = Width of the brake lining.

p1 = Maximum intensity of normal pressure,

pN = Normal pressure,

F1 = Force exerted by the cam on the leading shoe, and

F2 = Force exerted by the cam on the trailing shoe.

Consider a small element of the brake lining AC subtending an angle at the centre. Let

OA

makes an angle with OO1 as shown in Fig. 6 (b). It is assumed that the pressure

distribution on the shoe is nearly uniform; however the friction lining wears out more at

the



free end. Since the shoe turns about O1, therefore the rate of wear of the shoe lining at

A

will be proportional to the radial displacement of that point. The rate of wear of the shoe

lining varies directly as the perpendicular distance from O1 to OA, i.e. O1B. From the

geometry of the figure,



9.The layout of a brake to be rated at 250 N-m at 600 rpm is shown in figure. The

drum diameter is 200 mm and the angle of contact of each shoe is 120°. The

coefficient of friction may be assumed as 0.3 Determine. 1. Spring force F required

to set the brake. 2. Width of the shoe if the value of pv is 2 N-m/mm2 –sec.

[CO5 – H3]

Data: Mt = 250 N-m = 250 x 103 N-mm, n = 600 rpm, d = 200 mm, r = 100 mm, 2θ

=120°, 0=60°, µ = 0.3, pv = 2 N-m/mm2 –sec











10.A plate clutch with maximum diameter 6 cm has maximum lining pressure of

350 kpa .The power to be transmitted at 400 rpm is 135 kw and μ =0.3. Find in

side diameter and spring force required to engage the clutch. Spring s with

spring index 6 and material of the spring is steel with safe shear stress 600 m

pas are used. Find the diameters of 6 springs are used. [CO5 - H3 - Apr14]



11.A hydraulically operated clutch is to be designed for an automatic lathe.

Determine the number of plates and operating force requires for the clutch to

transmit 35 nm. The clutch is to be designed to slip under 300 % of rated

tensional moment to protect the gears and other part of the drive. The limits

for the diameter of frictions surfaces due to space limitation are 100 mm and

62.5 mm. this clutch is to operate in an oily atmosphere. [Apr14-Co5-H3- Apr14]



12. Explain with a neat sketch the working of a single plate clutch. Derive an

expression for the torque to be transmitted by clutch assuming

(i) uniform pressure condition and

(ii) Uniform wear condition. [Co5 - H3 - Nov15]

This type of clutch is mostly used in motor vehicles. It consists of one clutch plate, clutch

shaft, clutch spring, pressure plate, friction lining and bearing.

The flywheel is mounted on the engine crankshaft and rotates with it. The

pressure plate is bolted to the flywheel through clutch spring. The friction linings are on

both sides of the clutch plate. Fig.10.2 shows the arrangement of single plate clutch.

Working:-

When the clutch is engaged, the clutch plate is gripped between the

flywheel and the pressure plates. Due to friction, the clutch plate and shaft revolves.



When the clutch pedal is pressed, the pressure plate moves back against the force of the

spring, and the clutch plate becomes free between the flywheel and the pressure plate.

Thus the flywheel remains rotating as long as the engine is running and the clutch

shaft speed reduces slowly and finally it stops rotating.

Now consider two friction surfaces, maintained in contact by an axial thrust W, as

shown in Fig. 10.22 (a).

T = Torque transmitted by the clutch,

p = Intensity of axial pressure with which the contact surfaces are held together

r1 and r2 = External and internal radii of friction faces, and

μ = Coefficient of friction.

Consider an elementary ring of radius r and thickness dr as shown in Fig. 10.22 (b).

We know that area of contact surface or friction surface,

= 2 π r.dr



We shall now consider the following two cases :

1. When there is a uniform pressure, and

2. When there is a uniform wear.

1.Considering uniform pressure

When the pressure is uniformly distributed over the entire area of the friction face, then the intensity of pressure,

---------------- (1) where

W = Axial thrust with which the contact or friction surfaces are held together.

We have discussed above that the frictional torque on the elementary ring of radius r and

thickness dr is

Integrating this equation within the limits from r2 to r1 for the total frictional torque



Substituting the value of p from equation (i),

2. Considering uniform wear

In Fig. 10.22, let p be the normal intensity of pressure at a distance r from the axis of the

clutch. Since the intensity of pressure varies inversely with the distance, therefore

p.r. = C (a constant) or p = C/r .................(i)



13. Design a cam for operating the exhaust valve of an oil engine. It is required to

give equal uniform acceleration and retardation during opening and closing of the

valve each of which corresponds to 60° of cam rotation. The valve must remain in

the fully open position for 20° of cam rotation. The lift of the valve is 37.5 mm and

the least radius of the cam is 40 mm. The follower is provided with a roller of

radius 20 mm and its line of stroke passes through the axis of the cam. [CO5 - H3 -

Apr15]

Construction



First of all, the displacement diagram, as shown in Fig. 20.27, is drawn as discussed in

the

following steps :

1. Draw a horizontal line ASTP such that AS represents the angular displacement of the

cam

during opening (i.e. out stroke ) of the valve (equal to 60°), to some suitable scale. The

line ST represents the dwell period of 20° i.e. the period during which the valve remains

fully open and TP represents the angular displacement during closing (i.e. return stroke)

of the valve which is equal to 60°.

2. Divide AS and TP into any number of equal even parts (say six).

3. Draw vertical lines through points 0, 1, 2, 3 etc. and equal to lift of the valve i.e.

37.5mm.

4. Divide the vertical lines 3f and 3′ f ′ into six equal parts as shown by the points a, b, c .

..

and a′, b′, c′ . . . in Fig. 20.27.



5. Since the valve moves with equal uniform acceleration and retardation, therefore the

displacement diagram for opening and closing of a valve consists of double parabola.

6. Complete the displacement diagram as shown in Fig. 20.27.

Now the profile of the cam, with a roller follower when its line of stroke passes through

the

axis of cam, as shown in Fig. 20.28,



14.A cam is to be designed for a knife edge follower with the following data : 1.

Cam lift = 40 mm during 90° of cam rotation with simple harmonic motion. 2. Dwell

for the next 30°. 3. During the next 60° of cam rotation, the follower returns to its

original position with simple harmonic motion. 4. Dwell during the remaining 180°.

Draw the profile of the cam when (a) the line of stroke of the follower passes

through the axis of the cam shaft, and (b) the line of stroke is offset 20 mm from

the axis of the cam shaft. The radius of the base circle of the cam is 40 mm.

Determine the maximum velocity and acceleration of the follower during its ascent

and descent, if the cam rotates at 240 r.p.m. [CO5 - H3 - Nov11]

Solution. Given : S = 40 mm = 0.04 m; θO = 90° = π /2 rad = 1.571 rad ; θR = 60° = π

/3 rad = 1.047 rad ; N = 240 r.p.m.

First of all, the displacement diagram, as shown in Fig 20.13, is drawn as discussed in

the following steps :

1. Draw horizontal line AX = 360° to some suitable scale. On this line, mark AS = 90° to

represent out stroke ; SR = 30° to represent dwell ; RP = 60° to represent return stroke

and PX = 180° to represent dwell.

2. Draw vertical line AY = 40 mm to represent the cam lift or stroke of the follower and

complete the rectangle as shown in Fig. 20.13.

3. Divide the angular displacement during out stroke and return stroke into any equal

number of even parts (say six) and draw vertical lines through each point.

4. Since the follower moves with simple harmonic motion, therefore draw a semicircle

with AY as diameter and divide into six equal parts.



5. From points a, b, c ... etc. draw horizontal lines intersecting the vertical lines drawn

through 1, 2, 3 ... etc. and 0′ ,1′ , 2′ ...etc. at B, C, D ... M, N, P.

6. Join the points A, B, C ... etc. with a smooth curve as shown in Fig. 20.13. This is the

required displacement diagram.

(a) Profile of the cam when the line of stroke of the follower passes through the axis of

the cam shaft The profile of the cam when the line of stroke of the follower passes

through the axis of the cam shaft, as shown in Fig. 20.14, is drawn in the similar way as

is discussed in Example 20.1.

(b) Profile of the cam when the line of stroke of the follower is offset 20 mm from the axis

of the cam shaft The profile of the cam when the line of stroke of the follower is offset 20

mm from the axis of the cam shaft, as shown in Fig. 20.15, is drawn in the similar way as

discussed in Example 20.1.



Maximum velocity of the follower during its ascent and descent

We know that angular velocity of the cam,

Maximum acceleration of the follower during its ascent and descent We know that the

maximum acceleration of the follower during its ascent,



SKP Engineering College - mech.skpec.edu.inmech.skpec.edu.in/wp-content/uploads/sites/8/2017/11/Design-of... · Shigley J.E and Mischke C. R., ―Mechanical Engineering Design‖,

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