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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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 2 Design of Transmission System
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:
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 3 Design of Transmission System
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 4 Design of Transmission System
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 5 Design of Transmission System
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)
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 6 Design of Transmission System
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 7 Design of Transmission System
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 8 Design of Transmission System
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.
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 9 Design of Transmission System
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 -
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 10 Design of Transmission System
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 11 Design of Transmission System
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.
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 12 Design of Transmission System
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 13 Design of Transmission System
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.
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 14 Design of Transmission System
(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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 15 Design of Transmission System
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 16 Design of Transmission System
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 17 Design of Transmission System
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 18 Design of Transmission System
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 19 Design of Transmission System
=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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 20 Design of Transmission System
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 21 Design of Transmission System
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 22 Design of Transmission System
(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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 23 Design of Transmission System
(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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 24 Design of Transmission System
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 25 Design of Transmission System
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 26 Design of Transmission System
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 27 Design of Transmission System
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 28 Design of Transmission System
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 29 Design of Transmission System
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 30 Design of Transmission System
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 31 Design of Transmission System
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,
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 32 Design of Transmission System
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 33 Design of Transmission System
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 34 Design of Transmission System
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 35 Design of Transmission System
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 36 Design of Transmission System
=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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 37 Design of Transmission System
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 38 Design of Transmission System
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 39 Design of Transmission System
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 40 Design of Transmission System
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 41 Design of Transmission System
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 42 Design of Transmission System
=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]
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 43 Design of Transmission System
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)
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 44 Design of Transmission System
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.
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 45 Design of Transmission System
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 46 Design of Transmission System
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,
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 47 Design of Transmission System
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 48 Design of Transmission System
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 49 Design of Transmission System
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 50 Design of Transmission System
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 51 Design of Transmission System
Step 8: length of roller(lr)
lr=0.9bs-0.15
=0.9x(11.11)-0.15
lr=9.849mm
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 52 Design of Transmission System
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.
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 53 Design of Transmission System
(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.
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 54 Design of Transmission System
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.
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 55 Design of Transmission System
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 56 Design of Transmission System
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.
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 57 Design of Transmission System
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 58 Design of Transmission System
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 59 Design of Transmission System
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 60 Design of Transmission System
= 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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 61 Design of Transmission System
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)
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 62 Design of Transmission System
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 63 Design of Transmission System
=
= 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
5. Calculation of b,d, and v
Face width (b): b=10 X 10 =100mm Pitch circle diameter (d1)= mz1=10 X 18 =180mm
Pitch line Velocity (v) =
=
= 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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 64 Design of Transmission System
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 65 Design of Transmission System
= 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.
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 66 Design of Transmission System
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=
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 67 Design of Transmission System
= 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
Pitch line Velocity (v) =
=
= 11.31 m/s
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 68 Design of Transmission System
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.
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 69 Design of Transmission System
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:
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 70 Design of Transmission System
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 71 Design of Transmission System
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
8. Calculation of module (m)
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
Pitch line Velocity (v) =
=
= 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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 72 Design of Transmission System
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.
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 73 Design of Transmission System
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
5. Calculation of E eg [ ] and [ ]
(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.
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 74 Design of Transmission System
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
8. Calculation of module (m)
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 75 Design of Transmission System
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 76 Design of Transmission System
= 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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 77 Design of Transmission System
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.
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 78 Design of Transmission System
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=
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 79 Design of Transmission System
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.
7. Calculation of b,d, and v
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 80 Design of Transmission System
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 81 Design of Transmission System
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]
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 82 Design of Transmission System
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 83 Design of Transmission System
2. Z1=20; z2=80
3. Calculation of module
a=(
)X(
) = 7.25 mm
4. Calculation of b,d, and v
(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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 84 Design of Transmission System
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 85 Design of Transmission System
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
Virtual Number of teeth
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]
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 86 Design of Transmission System
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 87 Design of Transmission System
5. Calculation of E eg [ ] and [ ]
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) √(
)
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 88 Design of Transmission System
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 89 Design of Transmission System
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
Virtual Number of teeth
Zv1= Zv2=
=
=78
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 90 Design of Transmission System
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.
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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.
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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
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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.
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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
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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
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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
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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
Substituting all the values,
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
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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
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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]:-
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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,
Substituting all the values,
[ 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.
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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
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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
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[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.
15. Calculation of basic dimensions:-
From DD: 8.38,
Mt=3 mm
Z1=20
Z2=40
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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:-
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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.)
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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,
Substituting all the values,
[ 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,
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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
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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
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c=
[ ( )] [
√( )
] 1/2
c = 466.06 N/mm2
Since, c [ c ] design is satisfactory.
15. Calculation of basic dimensions:-
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
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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
SOLVE THIS PROBLEM SAME AS” PROBLEM NO:-3”
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]
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Solution:-
Since the material of pinion and gear are different, we have to design the pinion first
and check the gear, and
SOLVE THIS PROBLEM SAME AS” PROBLEM NO:-3”
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]
SOLVE THIS PROBLEM SAME AS” PROBLEM NO:-3”
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
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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 =
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=
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
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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
13. Check for wear:-
Since Fd Fw .the design is safe.
14. Check for efficiency:-
act = 0.95x
( )
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Mechanical Engineering Department 115 Design of Transmission System
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 116 Design of Transmission System
=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)
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 117 Design of Transmission System
=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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 118 Design of Transmission System
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 119 Design of Transmission System
=596.83 Nm
3. Selection of z1 and z2:-
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 120 Design of Transmission System
6. Calculation of axial module (mx):-
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
Lead angle: - =tan-1(z1/q)
=tan-1(3/11)
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 121 Design of Transmission System
=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
11. Checking for bending:-
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.
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 122 Design of Transmission System
12. Check for wear:-
c =
(
)√*
( )
+3
[ ]
= 118.59N/mm2
Since c [ c] the design is satisfactory
13. Check for efficiency:-
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
15. Calculation of basic dimensions for worm and worm gear:-
From DD: 8.43,
Axial module Mx =10 mm
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 123 Design of Transmission System
Number of startsZ1=3
Number of teeth on worm wheel Z2 =30
Length of worm (from DD: 8.48), L (12.5+0.09xZ2) x mx
(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
Tip diameter da1=d1+2f0mx
=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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 124 Design of Transmission System
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
3. Selection of z1 and z2:-
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 125 Design of Transmission System
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
6. Calculation of axial module (mx):-
Mx=2a/ (q + z2)
=2x345.96 / (11+72)
=8.34 mm
Therefore std.axial module=mx=10 mm
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 126 Design of Transmission System
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
Lead angle: - =tan-1(z1/q)
=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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 127 Design of Transmission System
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
11. Checking for bending:-
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.
12. Check for wear:-
c =
(
)√*
( )
+3
[ ]
=
(
)√*
( )
+3
= 121.03 N/mm2
Since c [ c] the design is satisfactory
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 128 Design of Transmission System
13. Check for efficiency:-
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
15. Calculation of basic dimensions for worm and worm gear:-
From DD: 8.43,
Axial module mx =10 mm, Number of startsZ1=3, Number of teeth on worm wheel Z2
=72
Length of worm (from DD: 8.48), L (12.5+0.09xZ2) x mx
(12.5+0.09x72)10 = 189.8 mm
Centre distance =415 mm
Face width b=0.75d1
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 129 Design of Transmission System
=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
Tip diameter da1=d1+2f0mx
=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]
Solution:-Solve this problem same as the “PROBLEM NUMBER: 10
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 130 Design of Transmission System
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]
Solution:-Solve this problem same as the “PROBLEM NUMBER: 10
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 131 Design of Transmission System
(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
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Mechanical Engineering Department 132 Design of Transmission System
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 133 Design of Transmission System
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 134 Design of Transmission System
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,
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 135 Design of Transmission System
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,
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 136 Design of Transmission System
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,
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 137 Design of Transmission System
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.
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 138 Design of Transmission System
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.
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 139 Design of Transmission System
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 140 Design of Transmission System
(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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 141 Design of Transmission System
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 142 Design of Transmission System
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 143 Design of Transmission System
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 144 Design of Transmission System
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 145 Design of Transmission System
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)
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 146 Design of Transmission System
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]
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Mechanical Engineering Department 147 Design of Transmission System
(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.
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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 %
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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.
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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.
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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
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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
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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
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Mechanical Engineering Department 154 Design of Transmission System
= 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
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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
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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
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Mechanical Engineering Department 157 Design of Transmission System
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
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Mechanical Engineering Department 158 Design of Transmission System
=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.
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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
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Mechanical Engineering Department 160 Design of Transmission System
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:-
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Mechanical Engineering Department 161 Design of Transmission System
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.
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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)
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Mechanical Engineering Department 164 Design of Transmission System
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Mechanical Engineering Department 165 Design of Transmission System
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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:-
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Mechanical Engineering Department 168 Design of Transmission System
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.
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Mechanical Engineering Department 169 Design of Transmission System
4. Kinematic arrangements:-
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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:-
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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
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Mechanical Engineering Department 173 Design of Transmission System
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.
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Mechanical Engineering Department 174 Design of Transmission System
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.
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Mechanical Engineering Department 175 Design of Transmission System
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.
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Mechanical Engineering Department 176 Design of Transmission System
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.
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Mechanical Engineering Department 177 Design of Transmission System
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 178 Design of Transmission System
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]
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 179 Design of Transmission System
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 180 Design of Transmission System
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 181 Design of Transmission System
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 182 Design of Transmission System
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.
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 183 Design of Transmission System
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)
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 184 Design of Transmission System
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
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]
S.K.P. Engineering College, Tiruvannamalai VI SEM
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
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]
S.K.P. Engineering College, Tiruvannamalai VI SEM
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.
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 195 Design of Transmission System
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)
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 196 Design of Transmission System
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 197 Design of Transmission System
=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.
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 198 Design of Transmission System
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:-
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 199 Design of Transmission System
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.
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 200 Design of Transmission System
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 201 Design of Transmission System
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]
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Mechanical Engineering Department 202 Design of Transmission System
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 203 Design of Transmission System
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 ,
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 204 Design of Transmission System
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
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 205 Design of Transmission System
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
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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
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Mechanical Engineering Department 207 Design of Transmission System
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]
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Mechanical Engineering Department 208 Design of Transmission System
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Mechanical Engineering Department 209 Design of Transmission System
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Mechanical Engineering Department 210 Design of Transmission System
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]
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 211 Design of Transmission System
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
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Mechanical Engineering Department 213 Design of Transmission System
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
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Total number of plates : n1 + n2 = 617 = 13
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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
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(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
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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
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Mechanical Engineering Department 218 Design of Transmission System
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:-
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(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
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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]
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Mechanical Engineering Department 222 Design of Transmission System
157500 = F 2[20/0.409 - 13.5]
F2 = 4449.2 N
Considered right hand side brake shoes:-
Taking moment about O1,
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=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:-
S.K.P. Engineering College, Tiruvannamalai VI SEM
Mechanical Engineering Department 224 Design of Transmission System
(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
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Mechanical Engineering Department 226 Design of Transmission System
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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]
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Mechanical Engineering Department 228 Design of Transmission System
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Mechanical Engineering Department 229 Design of Transmission System
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]
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fig.4
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Mechanical Engineering Department 231 Design of Transmission System
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.
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Fig 5
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Mechanical Engineering Department 234 Design of Transmission System
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.
S.K.P. Engineering College, Tiruvannamalai VI SEM
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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
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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,
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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
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Mechanical Engineering Department 239 Design of Transmission System
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Mechanical Engineering Department 240 Design of Transmission System
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Mechanical Engineering Department 241 Design of Transmission System
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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]
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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]
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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.
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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
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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
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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)
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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
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Mechanical Engineering Department 249 Design of Transmission System
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.
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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,
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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.
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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.
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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,
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