E. v. v. Ramanamurthy, S. Ramachandran-Design of Transmission Systems-Air Walk Publ. (2015)(1)

For UG

978-93-84893-67-5

Chapter - 1

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.

1.1 BELT DRIVES 1.1.1 Introduction Belt drive is a mechanical drive made up of flexible material used to transmit power from one shaft (driving shaft) to another shaft (driven shaft) which are parallel to each other and run at same (or) different speeds. The selection of belt drive depends on some important factors which include, the speed of driving and driven shaft, power transmitted, speed reduction ratio, centre distance between the two shafts, space available and so on. 1.1.2 Types of Belt drives The belt drives are classified based on their specific applications. They are (a) Light duty belt drives These are used to transmit less power (approx 5 kW) and at belt speeds upto 10 m/s. The main applications of these type of drives are in agricultural purposes (pumps, blowers, fans, etc.,) (b) Medium duty belt drives These type of belt drives are used to transmit medium powers (approx. 5 kW to 20 kW) and speed varies from 10 m/s to 20 m/s. The main applications of these type of belt drives include, machine tools, generators, etc.,

1.2

Design of Transmission Systems

(c) Heavy duty belt drives These type of belt drives are used for transmitting heavy power (ie) above 20 kW. The main applications of these type of belt drives include, crushers, bucket elevators, marine engines, etc., 1.1.3 Types of Belts The belts can be classified based on cross-section and represented in Fig. 1.1. They are (i) Flat belts (ii) V - belts (iii) Circular belts (iv) Toothed belts Types of Belts (a) Flat belt (b) V - belt (c) Circular belt F la t b elt

V -b elt . . .. . . . . .. . ..

(a) Flat belt

(b) V-b elt

C ircula r b elt . .. .............. ..

(c) Circular b elt

Fig. 1.1. Types of Belts.

(i) Flat belts The Flat belt as shown in Fig.1.1(a) is mostly used in farming, mining and logging applications. The Flat belt is a simple system of power transmission. It can deliver high power at high speeds (370 kw at 50 m/s).

Design of Flexible Elements

1.3

(ii) V - belts The V - belt as shown in Fig.1.1(b) are generally endless and their cross-section shape is trapezoidal. The V - shape of belt tracks in a mating groove in the shaft, with the result that the belt cannot slip off. (iii) Circular belts The Circular belt as shown in Fig.1.1(c) are circular cross section belt to run in a pulley with a 60 V - groove round belts are used in case of low torque requirements. (iv) Toothed belt The toothed belt, is made as a flexible belt with teeth moulded onto its inner surface. It runs over matching toothed pulleys (or) sprockets. Since toothed belts can also deliver more power that a friction drive belt, they are also used for high power transmissions. They include primary drive of some motor cycles. Materials used for belts The materials used for belts has to be strong, flexible and durable. They should have high coefficient of friction. They are cotton fabrics, leather, rubber, silk, etc., 1.1.5 Flat Belt Drives Flat belt drives can be used for transmitting large amount of power and there is no upper limit of distance between the pulleys. These drives are efficient at high speeds and they offer noiseless running. Flat belts are available for a wide range of width, thickness, weight and material. 1.1.6 Advantages of Flat-belt drive 1. Different velocity ratios can be obtained by using a stepped cone pulley.

1.4


2. A belt drive can be used as a clutch, by shifting the belt from fast pulley to loose pulley. 3. Design of flat belt drive is simple. 4. Flat belt drive is relatively cheap and easy to maintain. 5. Flat belt protection.

drives

are

flexible,

which

gives

6. Close casing is not required, like a gear box. 7. Flat belt drives can be used for long centre distances (upto 15 metres) 1.1.7 Disadvantages 1. Since velocity ratio is not constant, flat belt drive is not a positive drive 2. Flat belt drives have larger dimensions and occupy more space. 3. Flat belt drive is not suitable for smaller centre distance (less than 1 metre). 1.1.8 Types of Flat belt drives 1. Open belt drive (Fig. 1.2) 2. Crossed or twist belt drive (Fig. 1.3) 3. Quarter turn belt drive (Fig. 1.4) 4. Compound belt drive (Fig. 1.5) 5. Belt drive with idler pulleys (Fig. 1.6) 6. Stepped or cone pulley drive (Fig. 1.7) 7. Fast and loose pulley drive (Fig. 1.8)


D rive r Slack side

D rive n

+

e T ig h t s id

Fig.1.2. Open Belt Drive.

D rive r S la

ck s id e

D rive n

+

+

T ig

id e ht s

Fig.1.3 Crossed or Twist Belt Drive.

1.5

1.6


D rive r

D rive n

G uide pulle y +

(a) Q uarter Turn Belt Drive. Fig.1.4

D rive r

(b) Q uarter Turn Belt Drive with Guide Pulley.

3 1

2

+

+

+

3

1

4 Fo llow er

4

2

Fig. 1.5. C om pou nd Belt Drive.


1.7

+ + Idler pulleys

D rive r

+

+

+ + + Fig.1.6 (a). Belt D rive with M any I dler Pu lleys.

+

+ +

Idler pulley Fig.1.6 (b). Belt Drive with Single Idler Pulley

D rive n Pu lleys


1.8

D riving pu lley

L ine s ha ft

M ain o f D riving Sh aft C o ne pu lley

L oo se p ulley Fa s t p ulley

D rive sha ft

M ac hine s ha ft

Fig.1 .8 Fa st and lo os e pu lle y drive

Fig.1 .7 Stepped o r Cone Pu lley Drive .

1.1.9 Design Based on Basic Equations 



D

d N1

N2

(D river)  C

(D riven)

Fig.1.9

The Fig. 1.9 represents the open flat belt drive. Let, d  diameter of smaller pulley (Driver) D  diameter of larger pulley (Driven) N 1  speed of driver N 2  speed of driven C  center distance between two pulleys


1.9

Step 1 Velocity ratio 



N2

N1



d D

dt Dt

t  thickness of belt

(If thickness is considered) 

dt  S  1 100  D  t 

Speed ratio  N1 N2



(where S  % of slip)

1 Velocity ratio D d

Step 2 Ratio of tensions T1  Tc  e T2  Tc

where T 1  tension on tightside in Newtons T 2  tension on slackside in Newtons T c  centrifugal tension in Newtons

  coefficient of friction   angle of contact in radians     2  deg 

 180

Rr where   sin  1    C  R  larger pulley radius

1.10


  ra dia ns

r  smaller pulley radius C  centre distance

Always consider  for smaller pulley Note:  is used for cross belt drive  is used for open belt drive

Step 3: Calculation of T c Tc  Centrifugal tension  m  v2 in Newton.

where m  mass of the belt in kg f/meter length. where,   density of    volume t bel material     density in kg/m 3    Area  length

A  Area in m 2  b  t

   A  l kgf/m length

length  1 m assume b  width of belt in m

thickness of belt in m

t

If  is given in kg/cm 3, then assume A in cm 2 and length  100 cm. then, m    A  100 kg/m length v  vel. of belt (or) belt speed 

dN 1 60

or

DN 2 60

m/sec

(d and D must be in meters)


1.11

Step 4 Max. tension  T  S tress  Area    b  t Newton where   stress in the belt N/m 2 b  w idth of belt in m eter t  thickness of belt in meter  T  T1  Tc

Step 5 Power transmitted by the belt Also P  T 1  T 2 v in watts Note: For Max. Power Transmission Tc 

T 3

mv2 

T 3

v

where T  Max. tension in Newtons.

 

T [velocity of belt for m ax. pow er.] 3m

Step 6 Initial tension T0 T0  T0 

T1  T2 2 T1  T2  2T c 2

(if Tc is neglected) (if Tc is considered)

1.12


Step 7 Length of Open belt drive. L

 D  d2 D  d  2C  2 4C

Length of Cross belt drive L

 D  d D  d  2C  2 4C

2

D  larger pulley dia. d  smaller pulley dia. C  Centre distance

1.1.10 Design Based on Manufacturer’s Data Step 1: Diameter of Driver (or) Driven pulley Velocity ratio 

d N2  D N1

(or) Speed ratio



D N1  d N2

Find the unknown parameters by using the given parameters From P.S.G Data book, Pg.No. 7.54, Take the standard value of pulley diameters and the tolerances. Step 2: Velocity of the belt v v

 d N1  D N2 (or) in m/s 60 60

Step 3: Load correction factor K s From P.S.G Data book, Pg.No. 7.53, According to given application, take the Load correction factor K s


1.13

Step 4: Arc of contact  From PSG Data book, Pg.No. 7.54 Arc of contact   180  

D  d  60 C

where D  Diameter of larger pulley d  Diameter of smaller pulley C  Centre distance

Step 5: Correction factor for Arc of contact K  From PSG Data book, Pg.No. 7.54, corresponding to Arc of contact, correction factor K  can be determined. Step 6: Corrected power Corrected power or Design power

    



Given pow er in kW  K s K

Step 7: Corrected belt rating From PSG Data book, Pg.No. 7.54, corresponding to Load rating per mm per ply at 180 arc of contact at 10 m/s belt speed, (select for either Fort 949 g (or) Hi-speed 878), in kW From PSG Data book, Pg.No. 7.52, corresponding to minimum pulley diameter and maximum belt speed select the number of plies. Corrected belt rating of selected corresponding arc of contact and belt speed 

= Load rating 

belt

 v   number of plies 10 180

with

1.14


Step 8: Width of the belt From PSG Data book Pg.No. 7.54, Millimeter plies of belt 

corrected load or corrected power load rating per mm per ply at be lt spe ed

(or) Width of belt 

corrected load load rating/mm

From PSG Data book, Pg.No. 7.52, The standard belt width, corresponding to ply was found out Step 9: Length of the belt From PSG Data book, Pg.No. 7.53, the Length of the belt can be calculated. for open drive L  2C 

D  d2  D  d  4C 2

for cross drive L  2C 

D  d2  D  d  4C 2

for Quarter turn drive L

 D  d  2

d2 C2  D C2      2  

Step 10: Width of pulley From PSG Data book, Pg.No. 7.54, corresponding to the Belt width, the pulleys to be wider than the belt width by given “mm”


1.15

From PSG Data book, Pg.No. 7.54, the recommended series of pulley diameters and tolerances are determined. Problem 1.1: Design a suitable flat belt drive to transmit 10 kW at 1500 rpm to a line shaft to run at 500 rpm Approximate centre distance is 2.0 m. The diameter of larger pulley is around 750 mm.

(Oct. 2001)

P  10 kW  10  10 3 watt; N 1  1500 rpm; N 2  500 rpm ; C  2 metre  2  10 3 mm ; D  750 mm

Velocity ratio 

N2

N1



d D

750  500 500 d  d  250 mm 1500 750 1500

Velocity of belt  v 



dN 1 60

  250  1500  19.63 m  sec 60  1000

Load correction factor K s According to load classification, refer PSG data book, Page No.7.53 and take the value of KS. K s  1.3....

for line shafts

Correction arc of contact: K , refer PSG data book, Page No.7.54.

1.16


Arc of contact  180   180 

Dd  60 C 750  250  60 2000

 165

From the PSG data book, page No.7.54 Correction factor for  at   165 Arc of contact,  K   1.06

Note: Find K  by using interpolation between 160 and 170   K 160  1.08 5   0.02 165  

1.06

Calculation of Corrected Power corrected power 



K s  Given powe r in k W K

1.3  10  12.26 kW 1.06

Refer PSG data book, page No. 7.52, According to the minimum pulley diameter and the maximum belt speed, assume the no. of plies, from table at v  19.63 m  sec and d  250 mm ;


1.17

Take, n  n o. of plies  5 Calculation of Load rating: Select high speed belt, The load rating per mm width per ply at 10 m/sec  0.023 kW  mm  ply Load rating at belt speed at v  19.63 m  sec 

0.023  19.63 10

 0.0451 kW  mm  ply

Calculation of width of the belt Refer PSG data book, Page No.7.54 Millimeter plies of belt 

Width  no. of plies 

Corrected load or Corrected power Load rating  m m  ply

12.26  271.54 0.04515

 Width of the belt 

271.54  54.3 mm 5 Since n  no. of plies  5

Refer PSG data book, Page No.7.52, For 5 ply belt, the standard width of belt  76 mm Calculation of length of the belt L: Refer PSG data book, Page No.7.53

1.18


L

 D  d2 D  d  2C  4C 2

L

 750  2502 750  250  2  2000  2 4  2000

 5602.04 m m

Width of Pulley Refer PSG data book, Page No. 7.54 up to including 125 mm belt width, pulley is greater than the belt width by 13 mm  width of pulley  76  13  89 mm

Refer PSG data book, page No.7.54 The recommended pulley nominal diameter  90 mm ; with tolerance of nominal diameter as  1.2. mm Problem 1.2: A leather belt 9 mm  250 mm is used to drive a castiron pulley 90 cm in diameter at 338 rpm. If the active arc on the smaller pulley is 120 and the stress in the tightside is 20 kg  cm2, find the horse power capacity of the belt which weights 0.00098 kg  cm3. The coefficient of friction of leather on cast iron is 0.35.

(Apr. ’99 and Apr. 2000)

Given Data t  9 mm  0.9 cm ; b  250 mm  25 cm ; d  90 cm N  338 rpm;   120 

  2.094 radian 180

Stress in the tight side    20 kg  cm 2 ;   0.00098 kg  cm3;   0.35


1.19

Max. tension on the tightside T    A rea of belt  20  0.9  25  450 kg

Mass of the belt per meter length  m    b  t  l m  0.00098  25  0.9  100  2.205 kg  meter length

where l  100 cm (assume 1 meter) Centrifugal tension  T c  mv2  2.205  15.928 2  55.940 N  55.94 kg v 

 dN 60   90  338 100  60

v  15.9278 m /sec

Tension on the tight side of the belt  T1  T  Tc  450  55.94  394.05 kg

1   Power capacity of the belt  P  T1  Tc  1  v e   1    394.05  55.94  1   15.9278 0.35  2.094  e  

1.20




27987.69 kgf  m/sec 75

75 kgf  m  sec  1 HP

P  37.30 HP Problem 1.3: Two parallel shafts whose centre lines are 4.8 m apart, are connected by an open belt drive. The diameter of the larger pulley is 1.5 m and that of smaller pulley 1.05 m. The initial tension in the belt when stationary is 3 kN. The mass of the belt is 1.5 kg  m length. The coefficient of friction between the belt and the pulley is 0.3; Taking centrifugal tension into account, calculate the horse power transmitted, when the smaller pulley rotates at 400 rpm.

(Oct. ’99)

C  4.8 m  4800 m m; D  1.5 m  1500 mm;   0.3; d  1.05 m  1050 mm; N 1  400 rpm ; T0  3 kN  3  10 3 N

Initial tension T 0 

T 1  T 2  2Tc 2

 3  10 3

Centrifugal tension  T C  mv2  1.5  21.99 2  725.41 N T 1  T 2  3  10 3  2  2 725.41  T 1  T 2  4549.18 N

 1

From open belt drive Angle of contact      2 Dd  1 1500  1050  Where   sin  1   2.68   sin 2C 2  4800  


     2  2.68  174.64  

1.21

 180

,n   3.048 radia Velocity of the belt  v  

d N 1 60  100   1.05  400 60

 21.99 m  s ec T1 T2

 e;

T1 T2

 e0.3  3.048  2.495

 T1  2.495 T 2

Substitute the value of T1 in equation (1) 2.495 T2  T2  4549.18 T1 

4549.18  1301.6 N 3.495

 T 1  2.495  1301.6  3247.49 N

Power Transmitted  P  T1  T2  v  3247.5  1301.6   21.99 P  42790.34 Watts  42.790 kW Problem 1.4 Design a fabric belt to transmit 10 kW at 450 rpm from an engine to a line shaft at 1200 rpm. The diameter of the engine pulley is 300 mm and the distance of the shaft from the engine 2 m. Take coefficient of friction as 0.2. [April 2002]

1.22


Given Data Power P  10 kW; Speed of driver N 1  1200 rpm Speed of driven N 2  450 rpm Diameter of driver d   300 mm  0.3 m Centre distance C   2 m Solution Step 1: Diameter of driven D  Velocity ratio 

D



D N2  d N1 N2 N1

d

1200  300 450

 800 mm

From P.S.G Data book, Pg.No. 7.54, The standard, value of pulley diameter, D  800 mm

Step 2: Velocity of the belt Since v  

 d N1

60   0.3  1200 60

 18.84 m/s


1.23

Step 3: Load correction factor K s From P.S.G Data book, Pg.No. 7.53, corresponding to line shaft, K s  1.3

Step 4: Arc of contact From PSG Data book, Pg.No. 7.54 Dd Arc of contact   180     60  c   0.8  0.3   180     60 2  

  165 Step 5: Correction factor for Arc of contact From PSG Data book, Pg.No. 7.54, corresponding to Arc of contact, the correction factor K   1.06 (by interpolation)

Step 6: Corrected power Corrected power 



Given po wer  K s K 10  1.3 1.06

 12.26 kW

Step 7: Corrected belt rating From PSG Data book, Pg.No. 7.54, corresponding to load rating per mm per ply at 180 arc of contact at 10 m/s belt speed,

1.24


Assuming HI-SPEED 878 g duck belting, Load rating = 0.023 kW/mm/ply From PSG Data book, Pg.No. 7.52, corresponding to minimum pulley diameter and maximum belt speed (ie) corresponding to d  0.3 m  300 mm;  20 m/s v  18.84 m/s ~

and

The number of plies = 6  Corrected Belt rating of selected corresponding arc of contact and belt speed  L oad rating 

 0.023 

belt

with

 v  number of plies  10 180

18.84 165  6 10 180

 0.2383 kW/mm

Step 8: Width of the belt From P.S.G Data book, Pg.No. 7.54 Width of belt  

Corre cted Load Load rating/mm 12.26 0.2383

 Width of belt = 51.44 mm

From P.S.G Data book, Pg.No. 7.52, the standard belt width, corresponding to ply, = 100 mm


1.25

Step 9: Length of the Belt From PSG Data book, Pg.No. 7.53, L  2C 

D  d2  D  d  2 2C

22

 0.8  0.32 0.8  0.3  2 22

 5.79 m  5790 mm

Step 10: Width of the pulley From PSG Data book, Pg.No. 7.54, corresponding to belt width Pulley width  100  13  113 mm From PSG Data book, Pg.No. 7.52, the standard belt width corresponding to ply is = 125 mm Problem 1.5 A pulley of 900 m diameter revolving at 200 rpm is to transmit 7.5 kW find the width of a leather belt if the maximum tension is not to exceed 145 N in 10 mm width. The tension on the tight side is twice that on the slack side. Determine the diameter of shaft and the dimensions of the various parts of the pulley assuming it to have six arms. Maximum shear stress is not to exceed 63 MN/m2. [April 2010]

Given Data Diameter of pulley D  900 mm  0.9 m Speed of pulley

N  200 rpm

Power of

P  7.5 kW  7500 W

Maximum Tension T  145 N in 10 mm width Allowable shear stress,   63 MN/m 2  63  10 6 N/m 2

1.26


Solution We know that, Power transmitted by belt P  T1  T2 v where T 1  Tension in the Tight side T 2  Tension in slack side v  Velocity of the belt 

 DN 60



  0.9  200 60

v  9.42 m/s  7500  T1  T2 9.42

7500  T1  T2  9.42  T1  T2  796.17 N

Given that, Tension on the tight side is twice that on the slack side  T 1  2T2  2T2  T 2  796.17  T 2  796.17 N

and T 1  1592.35 N


1.27

But width of the belt b  

Tension which is maximum B elt ra ting



1592.35 145 /10

. .  . 145 N in 10 mm width    14.5 / mm width  

 109.81 mm

From PSG Data book, Pg.No. 7.52, Standard width of belt = 112 mm Diameter of shaft ds Since, power 

2 NT 60

 Torque T  

60P 2 N



60  7500 2  200

 358 Nm

But diameter of shaft d3 

16T 

 16  358  d 6     63  10  Dimensions of pulley (a) Dimensions of rim Width of pulley, B mm

1/3

 0.030702 m

1.28


Thickness of pulley, t mm From PSG Data book, Pg.No. 7.54, Belt width, upto and including 125 mm Pulley to be wider than the belt width by 13 mm  Width of pulley B  b  13 mm  112  13  125 mm

From PSG Data book, Pg.No. 7.57 Thickness of pulley rim t 

D  3 mm 200

[for single belt]  Thickness of pulley rim t 

900 3 200

 7.5 mm

(b) Dimensions of arm 

Number of arms (n)

From PSG Data book, Pg.No. 7.56, number of arms = 6 (for diameter of pulley over 450 mm) cross section of arm is elliptical 

Thickness of arm

(i) Thickness of arm near the boss (b) From PSG Data book, Pg.No. 7.56 b  2.94

aD

4n  3

for single belt


1.29

where a  width of pulley  125 mm D  diameter o f pulley  900 mm n  number of arms in pulley  6  b  2.94

125  900  49.20 mm 46

  3

Design of Flat Belt Drives (Problems) Problem 1.6: An open flat belt drive connects two parallel shafts 1.2 m apart. The driving and driven shafts rotate at 350 rpm and 140 rpm respectively and the driven pulley is 400 mm in diameter. The power to be transmitted is 1.1 kW. Design the drive.

Given P  1.1 kW  1.1  10 3 W C  1.2 m  1.2  10 3 mm N 1  350 rpm N 2  140 rpm D  400 mm

Solution Step 1: Diameter of driven pulley Velocity ratio  

N2

N1



d D

d 140  400 350 d  160 mm

1.30


Take standard value of pulley diameter from PSG Data book Pg.No. 7.54  d  160 mm

Step 2: Velocity of the belt Since v  

 D N2  d N1 or 60 60   160  350 60  1000

v  2.93 m/s

Step 3: Load correction factor K s From PSG Data book, Pg.No. 7.53, Assuming steady load K s  1.2

Step 4: Arc of contact From PSG Data book, Pg.No. 7.54 Dd Arc of contact   180      60   C   400  160     180     60  1200  

  168 Step 5: Correction factor for Arc of contact K  From PSG Data book, Pg.No. 7.54, corresponding to ~ 170   168 K   1.04


1.31

Step 6: Corrected power Since corrected power 



K s  Given power in kW

K 1.2  1.1 1.04

 1.26 kW

Step 7: Corrected belt rating From PSG Data book, Pg.No. 7.54, corresponding to Hi-speed 878 of duck belting and for 10 m/s belt speed belt rating = 0.023 kW/mm/ply  corrected belt rating for v  2.93 m/s 

0.023  2.93 168  10 180

 0.006289 kW/mm /ply

Step 8: Width of belt From PSG Data book, Pg.No. 7.54 Millimeter piles of belt 

C orrected power Load rating/mm/ply

 Width  Number of piles 

Corrected power Load rating/mm/ply

 Width 

1.26 0.006289  Number of piles

 Width 

186.97 Number of piles

From PSG Data book, Pg.No. 7.52, corresponding belt speed v  2.93 m/s, minimum pulley diameter d  160 mm

1.32


Assume number of piles = 4  Width 

186.97  46.74 mm 4

From PSG Data book, Pg.No. 7.52 standard width w  50 mm Step 9: Length of Belt From PSG Data book, Pg.No. 7.53 for open drive L  2C 

 D  d2 D  d  2 4C

 2  1200 

400  1602  400  160  2 4  1200

 3291.6 mm L  3300 mm

Step 10: Width of pulley From PSG Data book, Pg.No. 7.54, belt width upto and including 125 mm, pulleys to be wider than the width by 13 mm  Width of the pulley  50  13 mm Wid th of pulley  63 mm

From PSG Data book, Pg.No. 7.54 The recommended series of pulley diameter and tolerances is 63  0.8 mm


1.33

Problem 1.7: Two pulleys, one 430 mm diameter and the other 180 mm diameter are on parallel shafts 1.90 m apart are connected by a cross belt, find the length of the belt required and the angle of contact between the belt and each pulley. What power can be transmitted by the belt when the larger pulley rotates at 210 rpm, if the maximum permissible tension in the belt is 1 kN and the coefficient of friction between the belt and pulley is 0.25?

Given D  430 mm  0.43 m  R  0.215 m d  180 mm  0.180 m  r  0.09 m C  1.90 m ; N 1  210 rp m ; T 1  1 kN  1000 N;   0.25

Solution Since length of the flat belt is given from P.S.G Data book Pg.No. 7.53 for cross drive L  2C 

 D  d2 D  d  4C 2

 2  1.9 

0.43  0.182  0.43  0.18   2 4  1.9

 3.8  0.95818  0.04896 L  4.8071 m  4807.1 mm

We know that, for cross belt drive

1.34


Angle of contact      2 Rr where   sin  1    C 

 0.215  0.09   sin  1   1.9    9.237     180  2  9.237 

 198   198 

  3.464 rad 180

Power transmitted by belt drive P  T 1  T2 v

But v 

 DN 1 60



  0.43  210 60

 4.72 m/s

But



T1 T2

 e  and T1  1000 N

1000  e0.25  3.464 T2  T 2  420.63 N

 Power Transmitted P   1000  420.63   4.72  2734.62 W P  2.734 kW


1.35

Problem 1.8: An electric motor drives an exhaust fan. The pulley diameters of the motor and fan are 40 cm and 160 cm respectively. The angle of contact between belt and pulleys of motor and fan are 2.5 radians and 3.78 radians respectively. The coefficient of friction between the belt and motor and fan pulleys are 0.3 and 0.25 respectively. The speed of the driver pulley is 700 rpm. Power transmitted by the electric motor is 30 hp. Calculate the width of 5 mm thick flat belt. Take permissible stress for the belt material as 23 kgf/cm2.

Given d  40 c m  0.4 m; D  160 cm  1.6 m; 1  2.5 radians, 2  3.78 radians, 1  0.3, 2  0.25 N 1  700 rpm; P  30 HP; t  5 mm;   23 kgf/cm 2

To find Since Velocity of the belt v 

 d N1 60

v

(or)

 DN 2 60

  0.4  700 60

v  14.7 m /s

But the power transmitted P

30 

T1  T 2 75

T1  T 2 75

v

 14.7

1.36


 T1  T2  153.06 kgf

...(1)

But we know that T1 T2 

T1 T2

 e 1  1

 2.11

...(2)

From equations (1) and (2) 2.11 T 2  T2  153.06  T 2  137.89 kg f

and T1  290.95 kg f But know that Mass of the belt/metre length = density  Area  length Density may be taken as 1 gm/cm 3 (assume) 

1  b  0.5  100 1000

 0.05 b kg/m

Since the velocity of the belt is more than km/s, therefore centrifugal tension must be taken into consideration.


Tc  

1.37

w  v2 g

0.05 b  14.71 2 9.81

Tc  1.1b kgf

But the maximum tension in the belt; T  T1  Tc  stress  Area T  T1  Tc   bt  290.95  1.1b  23  b  0.5  b  27.88 cm ~  28 cm Problem 1.9: Design a flat belt drive to transmit 110 kW for a system consisting of two pulleys of diameters 0.9 m and 1.2 m respectively, for a centre distance of 3.6 m, belt speed of 20 m/s and coefficient of friction  0.3. There is a slip of 1.2% at each pulley and 5% friction loss at each shaft with 20% over load.

Solution

1.38


Given: P  110 kW  150 HP, d  0.9 m  90 cm ,  r1  0.45 m,  1.2 m, D  120 cm,  r2  0.6 m ; C  x  3.6 m ; v  20 m/s ;   0.3 ; S 1  S 2  1.2%

Let

N 1  Speed of the smaller or driving pulley in rpm N 2  Speed of the larger or driven pulley in rpm

We know that speed of the belt v v

20 

 d 1N 1  S1  1  60  100    0.9 N1  1.2   1  100  60  

 N 1  430 rpm

and peripheral velocity of driven pulley  d 2N 2 60

S2   v1  100  

  1.2 N 2 1.2    20  1   60 100   

N 2  315 rpm

We know that the torque acting on the driven shaft 

Power transmitte d  4500 2 N2



150  4500 2  315

 341 kg f  m


1.39

Since there is a 5% friction loss at each shaft, therefore the torque acting on the belt  1.05  341  358 kg f  m

Since belt is to be designed for 20% overload, therefore the design torque,  1.2  358  430 kg f  m

Let T1  Tension on the tight side of the belt T 2  Tension on the slack side of the belt

We know that the torque exerted on the driven pulley.  T 1  T2 r2  T1  T2 0.6  0.6 T 1  T 2 Kg f  m

Equating this to the design torque, we have 0.6 T1  T2  430 430  717 kg f  T 1  T 2  0.6 

T 1  T 2  717 kgf

...(1)

Now let us find out the angle of contact of the belt on the smaller or driving pulley. From the geometry of the figure, we find that sin   

O 2M O 1O 2

  2.4 



r2  r1 60  45   0.0417 x 360

1.40


   180  2  180  2  2.4  175.2   175.2 

  3.06 rad 180

We know that  T1   T e  2  e0.3  3.06  T1  0.918 e  T2  



T1

or

T2

 2.51

...(2)

From equations (1) and (2), we have T1  1192 kgf and T 2  475 kgf Assuming

f  safe stress for the belt = 25 kgf/cm2 t  thickness of the belt  1.5 cm b  Width of the belt.

Since the belt speed is more than 10 m/s, therefore centrifugal tension must be taken into consideration. Assuming a leather belt for which the density may be taken as 1 gm /cm 2  Weight of the belt permetre length w  Area  length  density  b  1.5  1000  1  0.15 b kg/m


1.41

and centrifugal tension Tc  

w  v2 g 0.15b 202 9.81

 6.12b kgf We know that maximum tension in the belt, T  T1  Tc  f.b.t 1192  6.12b  25  b  1.5  37.5 b  37.5b  6.12 b  1192 b  37.98 cm



From Design data book, the standard width of the belt b is 40 cm. From Design data book, Pg.No. 7.53 for open drive d 2  d 12  L  2x  d 2  d1  4x [x  C  centre distance] 2  2  360 

120  902  120  90  2 4  360

 1050.6 c m L  10.506 m Problem 1.10: Design a belt drive to transmit 20 kN at 780 rpm to an rolling machine, the speed ratio being 3.0, the distance between the pulleys is 2.8 m. Diameter of rolling machine pulley is 14 m.

1.42


Given Data P  20 kW; N 1  780 rpm ; i  3; C  2.8 m ; D  1.4 m

Solution Step 1: Diameter of Driver pulley We know that Speed ratio i  d

D N1  d N2

D 1.4   0.466 m i 3

Step 2: Velocity of belt in m/sec r 

 d N1 60

or

 D N2 60

  0.466  780 60

 19.03 m/s

Step 3: Load correction factor K s From PSG Data book Pg.No. 7.53; Load correction factor k s  1.5 (for rolling machine) Step 4: Arc of contact From PSG Data book, Pg.No. 7.54 Dd Arc of contact,   180    60   C   1.4  0.466     180    2.8    160  159.98 ~    160 


1.43

From PSG Data book, Page No. 7.54, Correction factor, K   1.08 Step 5: Corrected power Since corrected power  

K s  Given pow er in k W K 1.5  20 1.08

corrected power  27.7 kW

Step 6: Corrected belt rating From PSG Data book, Pg No. 7.54, Load rating per mm width per ply at 180 arc of contact, at 10 m/s belt speed. Assuming Fort 949 g duck belting, The load rating = 0.0289 kW per mm per ply. From PSG Databook Pg.No. 7.52, corresponding to the minimum pulley diameter, d  0.466 m  466 mm velocity = 19.03 m/s Assuming No. of piles = 6  Belt capacity of selected belt with corresponding

arc of contact 160 and belt speed of 19.03 m/s  0.0289 

19.03 160  6 10 180

 0.2933 kW /mm width

Step 7: Width of belt From PSG Data book, Pg.No. 7.54

1.44


Millimeter piles of belt 

Corrected load loa d rating/mm/ply at belt spe ed

(or) Width of belt b 

Corrected load 27.7  load ra ting/mm 0.2933

b  94.44 mm

From PSG Data book, Pg.No. 7.52, The standard belt width corresponding to 6 ply = 152 mm Step 8: Length of the belt From PSG Data book, Pg.No. 7.53, Assuming open drive belt, Belt length L  2C 

 L  2  2.8 

 D  d2 D  d  4C 2

1.4  0.466 2  1.4  0.466   2 4  2.8

 8.608 m

Step 9: Width of pulley From PSG Data book, Pg.No. 7.54, corresponding to the belt width above 125 upto and including 250 mm, pulleys to be wider than the belt width by 25 mm.  Width of the pulley  152  25 mm b  177 mm

From PSG Data book, Pg.No. 7.54, the recommended pulley diameter  180  2.0 mm


1.45

1.2. DESIGN OF V-BELT DRIVE 1.2.1 Introduction V-Belt is a type of flexible connector used for transmitting power from one pulley to another pulley having a centre distance upto 3 metres. V-Belts are used with electric nylon to drive different equipments like blowers, compressor, machine tool, industries machinery, etc. The belts are operated on grooved pulley called sheaves. The sheaves have V - shaped grooves or two inclined sides with flat bottom. The belt makes contact with the sheaves on the sides and clearance at the bottom. Usually V - Belts are endless ie., each belt is made in a circular form with various cross section which may be differentiated by different grades. It is made in trapezoidal section and the power is transmitted by the wedging action between the belt and the V -groove of the pulley or sheave. The cross-section of V - belt is discussed in article 1.2.4. A properly installed V - belt should fit tightly against the sides of the pulley grooves without making any projection beyond the rim and should have efficient clearance bottom of the groove. The materials used for V-belts are cotton fabric and cards moulded in rubber and coined with fabric and rubber (fig. 1.10). 1.2.2 Advantages 1. High velocity ratio (upto 7 and in some cases 10 also) 2. Smaller centre distance.

1.46


3. Reliability of the drive, in any position; even with vertical shafts. 4. Replacement is easy, because V-belts are available in standard sizes. 5. Smooth operation. 1.2.3 Disadvantages 1. Design of V-belt drive is more complicated. 2. Cannot be used for larger centre distance. 1.2.4 The Cross-section of V-belt Standard trapezoidal.

section

of

belt: Cross-section

2  Groove angle

x

w

C.S. Area 

a

Fa bric and R u bb er C over Fa bric

 40 general

To find cross-sectional area:

is

C o rds

T

R u bb er b 

1 [W  b] T 2

Fig.1.10.Cross - Section of a V -Belt

To find (b): tan 20 

x T

x  T  tan 20  b  W  2x  W  2 T  tan 20

1.2.5 Types of V - belts According to BIS (IS - 2494 - 1974), V-belts are classified as A, B, C, D and E types.


1.47

For various dimension of standard V-belt. Refer PSG Data book Pg.No. 7.70. DESIGN OF V-BELTS I. Based on Basic equations: T1 T2

 e/sin 

  Semi groove an gle   Angle of contact in radians T 1  Tension on tight side T 2  Tension on slack side

T1  Tc  e/sin  T2  Tc

T c  C entrifugal tension  m v2 v  Speed of the belt in m/sec.

1   Power transmitted per beltP  T 1  T c  1  v /sin   e   n  Number of V belts 

Total power transmitted Power trans mitted per b elt

II. Design Based On Manufacturer’s Data Refer data book, Page No: (From 7.58 to 7.69) *Step 1 From the given data, select the cross-section of the belt depending on the power to be transmitted. From PSG data book Page No. 7.58 Note the corresponding values, Belt cross section, W, T, mass of belt, and minimum pulley dia;

1.48


*Step 2 Select smaller pulley dia. from the table P.No. 7.58 of PSG Data book and find larger pulley dia. by using speed ratio. D  Larger pulley dia; d  Smaller pulley dia; N 1  Speed of driver pulley ; N 2  Speed of driven pulley ;

*(Take R20 series for pulley diameter) V-belt No. 7.61.

designation

refer

PSG

data

book

Page

C 3048/120 1S:2494 means, C - represents - Cross section 3048 - represents nominal inside length. Step 3 Calculate nominal pitch length (L) (P.No. 7.53) L

D  d2  D  d   2C 4C 2

From PSG Data book Page No.7.58, 7.59 and 7.60, according to the Cross-section of the belt; select the nearest nominal pitch length and take the corresponding nominal inside length. *Represent the belt length in terms of nominal inside length along with cross-section as Designation of V-belt. *Step 4 Calculation of design power Refer PSG data book, Page No. 7.62


1.49

According to the cross-section, from PSG databook, Calculate the Power Ex: Let the C.S. of belt is (B) 

Power/belt  0.79 S  0.09 

where S  belt speed 

50.8  1.32  10  4S2 S de

 d1 N 1 m/sec 60

de  equivalent pitch diameter  dp  Fb d 1  dia. of smaller pulley in m . N 1  Speed of smaller pulley in m

where d p  Smaller pulley diameter mm F b  From PSG Data book table, P. No. 7.62 according to speed ratio D/d

Take value of F b at

D ratio. d

 N 1  driven dia D  Speed ratio    N driver dia d  2

*To Find No. of belts Refer PSG Data book Page No. 7.70 No. of belts  n 

P  Fa kW  F c  fd

where P  given power in kW F a  Co rrection factor 7.69  refer PSG data boo k Page No. 7.69 kW  Power at the corresponding crosssection .

1.50


F c  Correction factor for length Refer data book Page No. 7.59 and 7.60 F d  Correction factor for arc of contact Refer data book Page No. 7.68

Dd     180    60     C  

C  given ce ntre distance.

at  value, take F d * Calculation of new centre distance (From PSG Data book Page No. 7.61) CA

where

A2  B  

Dd L A   8   4 B

D  d2 8

*Calculation of T 1 and T 2 T1  Tc  e /sin  T2  Tc

... (1)

  SemiGroove a ngle Dd   Arc of contact  180     60   C 

  C alculate in radians. Centrifugal tension Tc  m v2 where m    b  t  1 kg/m length   density of belt material kg /m 3 b  t  C.S. area of Vbelt


1.51

m    C .S. area  1 kg/m length

(C.S. must be in m 2) v

 d N1 m/sec 60

d  smaller pulley dia. in m .

(or) Mass can be read from PSG data book Pg No. 7.58 according to cross-section area. Take mass/m length. Tc  m v2 Newton. Power transmitted/belt  T1  T2  v

... (2)

Solving eq (1) and eq (2) and Calculate T1 and T 2 *Calculation of stress: fb Max. permissible tension  T  fb  area of belt fb  permissible stress in the belt

But, T  T1  Tc  T 1  T c  fb  area of belt

Calculate fb:  T1  T2  * Initial tension T o    2  

1.52


Problem 1.11: A 30 kW, 1440 rpm, motor is to drive a compressor by means of V-belts. The diameter of pulleys are 220 mm and 750 mm; The centre distance between the compressor and motor is 1440 mm. Design a suitable drive. (Apr. ’97)

Given data: P  30  103 Watts

Driver speed  N1  1440 rpm ; Driver pulley dia.  d  220 mm Driven pulley dia.  D  750 mm; Centre distance between the compressor and motor C  1440 mm  1.44 m

Step 1 From PSG data book, Pg. No. 7.58. Select Cross-section of the belt. Select either (C), (D) or (E) Select (C) type Belt (Since minimum pulley pitch dia = 220 mm given) Load of drive

 P  22 kW to 150 kW

Min. pulley pitch diameter  200 mm Nominal top width

W  22 mm

Nominal thickness

T  14 mm ;

Weight/meter

 0.343 kg f/m length


C.S. area of belt  

1.53

1 W  b T 2 1 [22  11.81 ]  14 2

 236.67 mm2

Step 2 Nominal pitch length  L  

 D  d2 D  d   2C 4C 2

750  2202   2  1440 750  220  2 4  1440

 4452.439 mm  4.452 . m

Take the nearest nominal pitch length from PSG data book. Refer PSG data book P.No. 7.60, Take standard nominal pitch length (Nearest) The corresponding  4394 mm .

nominal

 4450 mm ;

inside

length

The Designation of V-belt C 4394  1S2494 Step 3 Calculation of Design Power P in kW Refer PSG Data book Page No. 7.62 kW  1.47S  0.09 

142.7  2.34  10  4S2 S de

[S  v  velocity]

1.54


S

dN 1 60



  0.22  1440  16.587 m /sec 60

de  Equiva le nt pitch diam eter  dp  Fb  220  1.14  250.8 m.m dp  pitch dia. of smaller pulley  220 mm D 750   3.4 d 220 F b  1.14 (Refer PSG data book Pg. No. 7.62) 142.7   0.09 4 2   2.34  10  16.587  16.587  kW   1.47 16.587  250.8  

P  8.4312 kW

Step 4 To find the no. of belts (n) Refer PSG Data book page No. 7.70 No. of belts  n 

P  Fa kW  F c  F d

where P  given po wer kW F a  co rrection factor from PSG data book Page No. 7.69

Let the time period is upto 10 hr.  F a  1 (for compressor) kW  Power at the corresponding C.S. ie at C  Cross section


1.55

F c  C orrection factor for length P.No.7.60  1.04 F d  Correction factor for arc of contact Dd   180    60  C   750  220   180    60  157.9  1440  

 158 Refer PSG Data book page No. 160  0.95 7.68  for 158, F d  0.94 157 0.94  0.01 3 1 

0.01  1 3  0.0033

 158  0.94  0.0033

n  no. of belts 

30  1 8.4312  1.04  0.94

 3.63

 0.9433

 4 belts  No. of belts required  n  4

Calculation of new centre distance CA

A2  B  

A

Dd L   4  8 

B

D  d2 8

(from PSG Data book P.No. 7.61) L  nominal pitch length  4450 m m

1.56


A

 750  220  4450    731.58 mm 8 4  

B

750  2202  35112.5 mm 8

C  731.58   731.58 2  35112.5 

New centre distance C  1438.75 mm Calculation of T 1 and T 2 T1  Tc

T2  Tc

 e /sin  ... (1)

2  40

Semi-groove angle   20 Let   0.25   158 

 180

 2.75 radians Tc  mv2  0.343  16.5872  94.36 N

m  0.343 kgf/m length v  S  16.587 m /sec

1   Power/belt  [T 1  T c]  1  v /sin   e   1   8.3412  10 3  T1  94.36   1   16.587 0.25  2.75 / sin 20  e   8.4312  10 3  T1  94.36  14.364  T 1  681.3274 N


1.57

T1  Tc  e/sin  T2  Tc 681.3274  94.36  e0.25  2.75/sin 20 T 2  94.36 5869619  7.464 T2  94.36 T 2  78.638  94.36 T 2  172.998 N

To Find Stress fb T  T1  T2  fb  area of belt 681.3274  94.36  fb  236.66

Permissible stress in the belt material fb  3.28 N/mm 2 Problem 1.12: Design

a

V-belt

drive

to

the

following

specifications Power to be transmitted  75 kW Speed of the driving wheel N1  1440 rpm; Speed of the driven wheel N2  400 rpm; Diameter of the driving wheel d  300 mm; Centre distance  2500 mm Service  16 hours/day Assume any other relevant data if necessary.

Given Data: P  75  10 3 Watts

(Oct. 2007)

1.58


Driving Wheel Speed N 1  1440 rpm ; Driven Wheel Speed N 2  400 rpm Driving wheel diameter d  300 mm Centre distance C  2500 mm ; Service  16 hr/day . Step 1 Selection of Cross section; From PSG data book, Page No. 7.58 Since minimum pulley dia. is 300 mm Select (C) cross section Nominal Width

 W  22 mm

Nominal thickness t  14 mm Mass/kg  m  0.343 kg/m length 1 W  b t 2 1  [22  11.808]  14  236.656 mm 2 2

C.S. area 

Step 2 Nominal Pitch length  L 



 D  d2  2C D  d   4C 2

1050  3002  1050  300   2  2500 2 4  2500

 7176.82 mm L  7.176 m


N1 N2 D



1.59

D d

N1 N2

d 

1440  300  1050 m m 400

Take the nearest value of nominal pitch length (from PSG data book 7.60) L  6863 mm The Corresponding nominal inside length = 6807 mm Step 3 Calculation of Design Power From PSG Data book page No. 7.62 : for (C) C.S. of the belt 142.7    2.34  10 4S2  S kW   1.47S  0.09  de   S

dN 1 60



  0.3  1440  22.61 m/sec . 60

d e  equivalent pitch diameter  dp  Fb  300  1.14  342 mm

where dp  pitch dia. of smaller pulley refer PSG Data book page no. 7.62 at

D  3.6 the value d

of F b  1.14 But the max. value of d e in the formula is 300 mm for (C) cross-section.  Take de  300 mm 142.7    2.34  10  422.612  22.61 kW   1.47 22.61  0.09  300  

1.60


Max. power transmitted by belt  11.643 kW Step 4 To find the no. of belts (Refer PSG Data book page No. 7.70) No. of belts  n 

P  Fa kW  Fc  F d

P  75 kW F a  16 hrs/day

(Page No. 7.6 PSG databook)

Select medium duty [Since specific application is not given]  F a  1.2

P  Pow er  11.643 kW F c  1.14 - refer PSG Data book Page No. 7.60 at

Cross-section C Refer PSG databook, Page No. 7.68. F d  0.95

Dd   180     60  C   1080  300   180    60  2500    161.28  Take   160;

 n  no . o f belts required 

75  1.2 11.643  1.14  0.95

 7.137


1.61

= n  8 belts 8 belts are required. Calculation of New Centre Distance Refer PSG databook Page No. 7.61 CA

A2  B  

A

L D  d  4 8

B

D  d2 8

A

6863  1080  300  8 4

 1173.8252 B

1080  3002 8

 76050  New centre distance  C  1173.8252 

1173.8252 2  76050  

C  2314.796 mm Problem 1.13: A 20 kW, 1440 rpm motor is to drive a compressor by V-belt drive with a speed ratio of 3. Design the drive completely for a centre distance of about 1.5 meter. (Oct. 2000, Nov/Dec 2014)

Given data: P  20 kW  2  10 3 Watts; N1  1440 rpm

1.62


Speed radio  i  3 

N1 N2

Centre distance, C  1.5 m  1500 mm N 2  Speed of driven pulley,

N1 3



1440  480 rpm 3

Selection of Standard V-belt section Refer PSG data book, page No.7.58 Select cross-section ‘C’ and its minimum pulley pitch diameter d  200 mm But, i 

D d

 D  Pitch diameter of larger pulley  3  200  600 mm

Take the values of W, t and weight per meter length W  22 mm; t  14 mm; m  0.343 kg  m length

Calculation of Nominal Pitch length (L): Refer PSG data book Page No. 7.53 L



D  d2  D  d  2C  4C 2  600  200 2 600  200  2  1500  2 4  1500

 4283.30 mm

Nominal pitch length L  4283.30 mm From PSG data book, page No.7.50


1.63

A belt of ‘C ’ cross-section with nominal pitch length 4450 mm is selected (next higher value of the calculated pitch length L). The corresponding Nominal inside length  4394 mm Designation A V-belt cross-section C and of nominal inside Length of 4394 mm shall be designated as C  4394 1S:2494 Calculation of Design Power The maximum power in kW. Which the V-belts of section A, B, C, D and E can transmit shall be calculated from the following equations. Refer PSG data book Page No. 7.62 For C-section 142.7  2.34  10  4 S 2 S kW  1.47 S  0.09  de kW  [1.47 15.079  0.09 

142.7 228

 2.3  10  4  15.0792] 15.079  1.151  0.6258  0.05229  15.079  7.117 kW

Where S  belt speed 

d N 1 60

  200  1440  15.079 m/sec 60  1000 de  eq uivalent pitch diameter  d p  Fb

1.64


dp  pitch dia. of smaller pulley  200 mm Fb 

small diameter factor (refer PSG data book

P.No. 7.62) for

D 600   3; Fb  1.14 d 200

de  200  1.14  228 mm

To Find No. of Belts n Refer PSG data book, Page No. 7.70

No. of belts  n 

P  fa kW  F c  Fd

F a  1.1 (Assume medium duty and upto 10 hr.)

To Find F d:   Arc of contact angle Dd  180     60  C 

 600  200   180    60    1500  164

where P  given power in kW. F a  correction factor (refer PSG data book P.No.

7.69)


1.65

F c  Correction factor for length refer PSG data book

P.No. 7.58 F c  1.04 F d  Correction factor for arc of contact (refer PSG

data book P.No. 7.68) at   164; the value of F d  0.965 (approx)  n  No. of belts 

20  1.1 7.117  1.04  0.965

n  3.08

Take n  4 belts. Calculation of New centre distance (Refer PSG data book Page No. 7.61) CA A

A2  B  

L  D  d  4 8

L  N ominal P itch length  4450 mm D  600 mm

4450  600  200   8 4  798.34 mm 

2

600  200 D  d  8 8  20,000 mm

d  200 mm 2

B

C  New centre distance  A 

A2  B  

 798.34 

798.34 2  20.000  

 1584.05 mm

Take, C  New centre distance  1584 mm

1.66


Problem 1.14 Design a V-belt drive to transmit 50 kW at 1440 rpm from an electric motor to a textile machine running 24 hours a day. The speed of the machine shaft is 480 rpm.

Solution Given data P  50 kW  50  10 3 watts

Driver speed

 N 1  1440 rpm

Driven speed  N 2  480 rpm Service

 24 hours a day

we know Speed ratio  i 

N1 N2



1440 3 480

from P.S.G Data book, Pg.No. 7.58 for the power 50 kW, ‘C’ type of belt may be selected. For this belt, the minimum pulley pitch diameter is dmin  200 mm

from P.S.G Data book, Pg.No. 7.62 equivalent pitch diameter de

max

 dpmax  F b max from

P.S.G Data book Pg.No. 7.62, corresponding to Belt  300 mm cross-section ‘C’ de max

and corresponding to speed ratio range F b max  1.14  dp max 

de max F b max



300  263.15 mm 1.14


1.67

Hence we should select the diameter of smaller pulley d  between 200 to 263.15 mm Let us select d  250 mm  from P.S.G Data book, Pg.No. 7.61

Diameter of larger pulley Dd

N1 N2

 250 

 1440  0.98 480

D  735 mm

[   0.98 a ssum ed]

from P.S.G Data book, Pg.No. 7.61, corresponding to i  3, C  1.0 D  C  1  735  C  D  735 mm where c  centre distance

from P.S.G Data book, Pg.No. 7.61 C min  0.55 D  d  T

where T  N ominal T hickness  14 mm (from P.S.G Data book Pg. No. 7.58, corresponding to cross section of belt)  C min  0.55 735  250  14  556 mm

from P.S.G Data book, Pg.No. 7.61

1.68


C max  2 D  d  2 735  250  1970 mm

Since, calculated ‘C’ value is in between C m in and C m ax our selection is correct. From P.S.G Data book,

Pg.No. 7.58 corresponding to cross-section symbol ‘C’ Nominal top width W  22 mm Weight per metre

 0.343 kgf

But C.S.A of belt 

1 W  b T 2

but x  T tan 20  14  tan 20  5.0955 mm b  W  2x  22  2 5.0955   11.808 mm  C.S.A of belt 

1 [ 22  11.808 ]  14 2

 236.66 mm 2

from P.S.G Data book, Pg.No. 7.61 Nominal pitch length of belt


L



 D  d2 D  d   2C 2 4C

 738  250 2 725  250    2  735 2 4  735

 3097 mm

from P.S.G Data book, Pg.No. 7.60 Take standard Nominal pitch length  3104 mm Nominal inside length

 3048 mm

from P.S.G Data book, Pg.No. 7.62 Design power kW  1.47 S 0.09 

But S 

142.7  2.34  10  4 S 2 S de

d N 1   0.250  1440  60 60  18.85 m/s

and de  dp  F b  250  1.14  285 mm  Design power  [1.47  18.85 

 0.09



142.7 4 2  2.34  10  18.85  ]  18.85  285

 10.26 kW

from P.S.G Data book, Pg.No. 7.70 No. of belts  n 


1.69

1.70


where P  given power in kW from P.S.G Data book, Pg.No. 7.69 F a  1.4 for electric motor for textile machinery

from P.S.G Data book, Pg.No. 7.60 F c  0.97 corresponding to Nominal pitch length and

cross-section of belt from P.S.G Data book, Pg.No. 7.68 Dd   180    60  C   735  250   180    60   140.4  735    Fd  0.845 , from P.S.G Data book, Pg.No. 7.68  n  number of belts 

50  1.4 10.26  0.97  0.845

 8.32 ~  9 belts

Calculation of new centre distance From P.S.G Data book Pg.No. 7.61 CA

But A  

A2  B  

Dd L   4  8   735  250  3104   8 4  

 389.19


B

1.71

D  d2 735  2502  29403.1  8 8

 C  389.19 

389.19 2  29403.1   

New centre distance C  738.56 mm Problem 1.15 Select a suitable V-Belt drive to connect a 7.5 kW, 1440 rpm, induction motor to run a fan at approximately 480 rpm for a service of 16 hr per day. The space available for centre distance is 1 m.

Solution Given N 1  1440 rpm, N 2  480 rpm P  7.5 kW, Service  16 hrs/day C1m

Since the diameters of Driving pulley and Driven pulley are not given, they may be assumed. Let d  Diameter of smaller pulley i.e. driving pulley D  Diameter of bigger pulley i.e. driven pulley

For the power of 7.5 kW, ‘B’ type belt may be selected. For ‘B’ type belt dmin  125 mm , and dmax is calculated as follows. Equivalent pitch diameter [From P.S.G. Data book, Pg.No. 7.62] de max  dp max  F b max

(ie)

dp max 

de max F b max



175  153.5 mm 1.14

1.72


Since F b max  1.14 from P.S.G. Data book, Pg.No. 7.62 corresponding to speed ratio range. D/d. and de max is from P.S.G. Data book, Pg.No. 7.62, from corresponding to cross-section ‘B’. Hence “d” should be in between 125 to 153.5 mm. Step 1 Let diameter of driving pulley d  140 mm D

N1 N2

d

1440  140  0.98 480

 D  412 mm

here  is assumed as 0.98 from P.S.G. Data book, Pg.No. 7.61 i

D N1  3 d N2

Design power 

Rated power  service factor Length facto r  Arc of contact factor

Service factor  1.1 for light duty a nd 16 hrs/day

(from P.S.G. Data book, Pg.No. 769) Step 2 Pitch length of the belt L  2C 

 D  d2 D  d  4C 2

 2 1000  

(from Pg.No. 7.61)

412  1402  412  140  2 4  1000

 2885.4 mm


1.73

Take the nearest value of Nominal pitch length (from P.S.G. Data book, Pg.No. 7.60); L  2888 mm The corresponding nominal inside length  2845 mm (from P.S.G. Data book, Pg.No. 7.60) Step 3 Calculation of Design power from P.S.G. Data book, Pg.No. 7.62 kW  0.79S  0.09 

S 

50.8  1.32  10  4 S2 S de

 d N1 60   0.140  1440  10.555 m /sec 60

de  equivalent pitch diameter  dp  F b dp  Pitch dia of sma ller pulley  140  1.14  159.6 mm ~  160 mm 

 kW  0.79 10.555 

 0.09





50.8 4 2  1.32  10 10.55  10.55 160 

 0.63902  0.3175  0.0147  10.55  0.3068  10.555  3.2368 kW

1.74


Step 4 To find the No. of belts from P.S.G. Data book, Pg.No. 7.70 number of belts  n 


P  75 kW , F a  1.1 (for light duty application from P.S.G. Data

book, Pg.No. 7.69) kW  3.2368 kW F c  1.05

(from P.S.G. Data book, Pg.No.

7.60,

corresponding to cross section ‘B’ and nominal Inside length) from P.S.G. Data Book Pg.No. 7.68 Dd Arc of contact angle  2 cos  1    2c  Dd  180  60    c   412  140   180  60   1000    163.68  ~  163 F d  0.96

(from P.S.G. Data book, Pg.No. 7.68,

corresponding to Arc of contact on smaller pulley)


1.75

 Number of belts required 

75  1.1  26 belts  25.285 ~ 3.2368  1.05  0.96

Hence, 26 belts are required Calculation of New centre distance [from P.S.G. Data book, Pg.No. 7.61] CA A2  B 

where A 

L   D  d 4 8 2885.4   412  140 8 4

 504.58 B

D  d2  9,248 8

 New centre distance C  504.58   504.58 2  9248   C  999.91 mm

Initial tension  0.5 to 1% L [from P.S.G. Data book, Pg.No. 7.61] Take 1% of L  2888 

1  28 mm 100

 Final centre distance  999.91  28  1027 mm

1.76


width of pulley  n  1 e  2f f  E dge of pulley to first groove centre

where

n  total numb er of belts e  centre to centre distance of grooves

 26  1 19  2 12.5 (from P.S.G. Data book, Pg.No. 7.70  500 mm

Specifications Type of Belt: B 2845/116 IS 2494 (V Belt) Number of Belts: 26 Pitch diameter of smaller pulley: 140 mm Pitch diameter of bigger pulley: 412 mm Centre distance: 1027 mm Problem 1.16 A motor driven blower is to run at 600 rpm driven by an electric motor of 8.5 kW at 1900 rpm. Design v-belts.

Given data Power P  8.5 kW Speed of driver N 1  1900 rpm Speed of driven N 2  600 rpm Solution Step 1: Selection of cross section of belt Since the diameters of the pulleys are not given, they can be assumed. From PSG Data book, Pg.No. 7.58 for the power 8.5 kW, ‘C ’ type belt may be selected. For ‘C ’ type belt, dmin  200 mm


1.77

Nominal top width W   22 mm, Thickness t  14 mm, Weight per metre  0.343 kg f Step 2: Selection of diameters From PSG Data book, Pg.No. 7.62, Equivalent pitch diameter d e  dp  F b

(or) de

max

 dpmax  F b max

From PSG Data book, Pg.No. 7.62, corresponding to ‘C ’ type belt de

max

 300 mm

and corresponding to D/ d ratio, F b max  1.14  dpmax 

de

max

F b max



300  263.15 mm 1.14

Therefore Diameter of smaller pulley d  (or) driver pulley d  230 mm From PSG Data book, Pg.No. 7.61 Diameter of larger pulley D   d 

N1 N2



Where N 1  rpm of smaller pulley (Driver pulley)

1.78


and N 2  rpm of Larger pulley (driven pulley)

and   0.98 (assumed)  D  230 

1900  0.98 600

 713.76 mm

Step 3: Determination of centre distance From P.S.G Data book, Pg.No. 7.61, i

D 713.76   3.1 d 230

From P.S.G Data book, Pg.No. 7.61, corresponding to i  3.1 ~  4 C  0.95 D  C  0.95  713.76  C  678.072 mm

From PSG Data book, Pg.No. 7.61, C min  0.55 D  d    0.55 713.76  230  14  533.06 mm C max  2 D  d  2 713.76  230  1887.52 mm


C min

1.79

Since the calculated a  678.072 mm is between and C max, The calculated centre distance is correct.

Step 4: Nominal pitch length of belt L  From PSG Data book, Pg.No. 7.61 L  2C 

 D  d2 D  d   4C 2

 2  678.072 

713.76  230 2  713.76  230  2 4  678.072

 2819.35 mm

From P.S.G Data book, Pg.No. 7.60, corresponding to cross-section C Nominal pitch length L  2901 mm The corresponding Nominal Inside length = 284.5 mm Step 5: Calculation of Design power From P.S.G Data book, Pg.No. 7.62, corresponding to belt cross-section symbol ‘C’, 142.7    2.34  10  4 S2 S  kW   1.47 S  0.09  de  

since S  belt speed  

 N1 60   0.230  1900 60

 22.8 m/s

From P.S.G Data book, Pg.No. 7.62,

1.80


de  dp  Fb (or) d  F b  230  1.14  300  de  262.2  de max 142.7   0.09 4 2   kW   1.47 22.8  2.34  10 22.8    22.8  262.2  

kW power  10.113 kW

Step 6: Number of belts From P.S.G Databook, Pg.No. 7.70 Number of belts 

P  Fa kW  Fc  F d

Power P  8.5 kW (given) From P.S.G Data book, Pg.No. 7.69, Assuming medium duty service and over 16 hr and continuous service F a  1.3

From P.S.G Data book, Pg.No. 7.60, corresponding to Belt cross section ‘C’ and Nominal inside length = 2845 mm F c  0.95

From P.S.G Data book, Pg.No. 7.68, Dd Arc of contact   180     60   C   713.76  230   180      60 678.072    137 


1.81

From P.S.G Data book, Pg.No. 7.68, corresponding to Arc of contact,  and V - Flat. F d  0.85  Number of belts n 

8.5  1.3 10.113  0.95  0.85

 1.353 ~  2 belts

Step 7: Calculation of New centre distance From PSG Data book Pg.No. 7.61, centre distance ‘C ’ CA

A2  B  

where A 

D  d L  A 8 713.76  230 2901  8 4

 354.6 mm B



D  d2 8 713.76  2302 8

 29252.96 mm2  New centre distance ‘C ’  354.6 

354.62  29252.96  

 665.22 mm

1.82


Problem 1.17: Design a suitable V - Belt and design the drive for a wet grinder. The power available from a 0.75 kW motor running at 780 rpm. Drum speed is to be about 120 rpm. Drive is to be compact.

Given Data P  0.75 kW; N1  780 rpm; N 2  120 rpm

Solution Step 1: Selection of cross-section of belt Since the diameters of pulleys are not given they can be assumed. From P.S.G Data book, Pg.No. 7.58, for the power 0.75 kW, “A” type belt may be selected. For “A” type belt, minimum pulley pitch diameter dmin = 75 mm; Normal top width w  = 13 mm; Nominal thickness T  = 8 mm; weight of metre = 0.106 kg f. Step 2: Selection of diameters From PSG Data book, Pg.No.7.62, Equivalent pitch diameter de  dp  F b (or) de max  dp max  F b max

From P.S.G Data book, Pg.No. 7.62, corresponding to “A” type belt de

max

 125 mm

and corresponding to

D ratio, d


1.83

F b max  1.14 de 125 max   109.64 mm  dp max  Fb max 1.14

Therefore diameter of smaller pulley d  or driver pulley should be between 75 mm to 109.64 mm Hence let the diameter of driver pulley d = 100 mm From P.S.G Data book, Pg.No. 7.61, Diameter of larger pulley D   d 

N1 N2



where N 1  rpm of smaller pulley (Driver pulley)

and N 2  rpm of larger pulley (Driven pulley)

and   0.98 (assumed)  D  100 

780  0.98 120

 637 mm

Step 3: Determination of centre distance From P.S.G Data book, Pg.No. 7.61 i

D 637   6.37 d 100

From P.S.G Data book, Pg.No. 7.61, corresponding to C  6.37,

1.84


C  0.85 D  C  0.85  637  C  541.45 mm

From P.S.G Data book, Pg.No. 7.58, C min  0.55 D  d  t  0.55 637  100   8  413.35 mm C max  2 D  d  2  637  100  1474 mm

Since the calculated ‘C ’ = 541.45 mm, is between C m in and C max, the calculated centre distance is correct. Step 4: Nominal pitch length of belt L  From P.S.G Data book, Pg.No. 7.61, L  2C 

D  d2  D  d  4C 2

 2  541.45  

637  1002  637  100  2 4  541.45

 2373.72 mm

From P.S.G Data book, Pg.No. 7.60, corresponding to cross-section “A” Nominal pitch length L  2703 mm, The corresponding nominal Inside length = 2667 mm.


1.85

Step 5: Calculation of Design power From P.S.G Data book, Pg.No. 7.62, corresponding to belt cross-section symbol “A”. 19.62    0.765  10  4 S 2  S kw   0.45 S  0.09  de  

Since S  belt speed  

 d N1

60   0.100  780 60

 4.084 m/s

From P.S.G Data book, Pg.No. 7.62 de  dp  Fb  100  1.14  114 mm  de max  125 mm 19.62   0.09 4 2    0.765  10 4.084   4.084  kw   0.45  4.084 114  

 0.911 kW

Step 6: Number of belts From P.S.G Data book, Pg.No. 7.70, Number of belts  Power

P  Fa kw  F c  F d

P  0.75 kW given

From P.S.G Data book, Pg.No. 7.69 Assuming medium duty service and upto 10 hr service F a  1.1

1.86


From P.S.G Data book, Pg.No. 7.60, corresponding to belt cross-section “A” and Nominal inside length = 2667 mm. F c  1.1

From P.S.G Data book, Pg.No. 7.68, Dd Arc of contact   180     C

   60 

 637  100   180      60  541.45   120.49

From P.S.G Data book, Pg.No. 7.68, corresponding to Arc of contact,  and V - Flat F d  0.83  Number of belts n 

0.75  1.1 0.911  1.1  0.83

 0.9918 ~  1 belt

Step 7: Calculation of New centre distance From PSG Data book, Pg.No. 7.61, A2  B Centre distance ‘C ’  A   

where A  

L  D  d  8 4 2703  637  100  8 4

 386.33 mm


and B  

1.87

D  d2 8 637  100 2 8

 36046.12 mm 2  New centre distance ‘C’  386.33 

386.332  36046.12   

 722.78 mm Problem 1.8: Design a V-belt from an electric motor to the flywheel of a forging press from the following data. Motor power 80 kW. During operation the motor is overloaded, the overload factor being 1.5. Rated speed of motor is 720 rpm and The required speed of the press flywheel is 300 rpm. The flywheel with diameter of 1500 mm is preferably used as a driven pulley. The distance between the areas of the flywheel and motor pulley can be about 1650 mm.

[May 2011]

Given data P  80 kW; N 1  720 rp m; N2  300 rpm ; D  1500 mm ; C  1650 mm, Over load factor F a  1.5

Solution Step 1: Selection of cross-section of belt From PSG Data book, Pg.No. 7.58, for the power 80 kW, “D” type belt may be selected. For “D” type belt, minimum pulley pitch diameter d min  355 mm; Nominal top width w   32 mm; Nominal thickness T  19 mm; weight per metre = 0.596 kg f.

1.88


Step 2: Calculation of diameters From P.S.G Data book, Pg.No. 7.61, Diameter of larger pulley D  d 

N1 N2



where n 1  rpm of smaller pulley (Driver pulley) n 2  rpm of larger pulley (Driven pulley)

and   0.98 (assumed) dD

N2 1 N1 

 1500 

300 1  720 0.98

 637.75 mm

Step 3: Nominal pitch length of belt L From P.S.G Data book, Pg.No. 7.61, L  2C 

D  d2  D  d   4C 2

 2  1650 

1500  637.752  1500  637.75   2 4  1650

L  6770.6 mm

From P.S.G Data book, Pg.No. 7.60, corresponding to belt cross-section symbol “D ” Nominal pitch length L  6886 mm The corresponding nominal inside length = 6807 mm


1.89

Step 4: Calculation of design power From P.S.G Data book, Pg.No.7.62, corresponding to belt cross-section symbol “D ” 506.7    4.78  10  4 S 2  S kw   3.22 S  0.09  de  

Since S  belt speed  

 dN 1 60   0.6377  720 60

 24.04 m/s

From P.S.G Data book, Pg.No. 7.62 de  dp  Fb  637.75  1.13  720.657 mm D . . [ . F b  1.13 corresponding to ratio from PSG Data d book Pg.No. 7.62] 506.7   0.09 4 2   4.78  10 [24.04 ]   24.04  kw   3.22  24.04   720.657  

 34.6 kW

Step 5: Number of belts From PSG Data book, Pg.No. 7.70, Number of belts n  


Power P   80 kW (given)

1.90


From P.S.G Data book, Pg.No. 7.60, corresponding to belt cross-section “A” and Nominal inside length 6807 mm. F c  1.1

From P.S.G Data book, Pg.No. 7.68 Dd Arc of contact   180    60   C   1500  637.75   180     60 1650    148.64

From P.S.G Data book, Pg.No. 7.68, corresponding to Arc of contact,  and V - Flat F d  0.82  Number of belts =

80  1.5 34.6  1.1  0.82

 4 belts  3.84 ~

Step 7: Calculation of new centre distance From P.S.G Data book, Pg.No. 7.61, A2  B Centre distance ‘C’  A   

where

A 

L  D  d  A 8 6886  1500  637.5   8 4

 882.2 mm

and

D  d 2 B 8




1.91

1500  637.5 2 8

 92934.3 mm 2  C  882   882 2  92934.3    1709.6 mm

1.3. DESIGN OF WIRE ROPES 1.3.1 WIRE ROPES When a large amount of power is to be transmitted over a long distance (up to 150 meters) from one pulley to another pulley, then wire ropes are used. Wire ropes are used for hoisting and hauling purposes, example: elevators, mine hoist, cranes, conveyors, etc. Wire ropes are made from cold drawn wires to increase strength and durability. The various materials used for wire ropes are wrought iron, cast steel and alloy steel. 1.3.2 Advantages of wire ropes 1. Wire ropes are light in weight 2. They offer silent operation. 3. They can withstand shock loads 4. They are more reliable 5. They do not fail suddenly 6. They are more durable, low cost and high efficient when compared to fibre ropes.

1.92


1.3.3 Disadvantages of wire ropes 1. Not suitable for low power transmission. 2. It cannot be employed for short centre distance. 3. High manufacturing cost. 4. High Maintenance cost. 5. Should be kept away from electrical power, to prevent current leakage through them. 6. It should be kept away from corrosive environment. 1.3.4 Construction of Wire rope The wires are initially heat treated and cold drawn during stage in order to have high strength and durability. The wires are first twisted into a strand and then a number of strands are twisted about a core to form the rope as shown in the Fig. 1.11 (a), (b). dw

d

S tran d

Fig. 1.11(a)

C o re 6x7 w ire rop e R

R o pe

R1

R1

2 Fig. 11.11(b) C ross-sectio n o f a rop e.


1.93

The number of wires in each strand is generally 7, 19 or 37 and number of strands is usually 6. 1.3.5 Designation of wire ropes 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. 1.3.6 Classification of wire ropes Wire ropes are classified based on (a) Number of strand and number of wire in each strands (i) 6  7 ropes (ii) 6  19 ropes (iii) 6  37 ropes (iv) 8  19 ropes (b) Direction of twist of wire and strand as (i) Cross-lay ropes (ii) Parallel-lay ropes (iii) Composite-lay ropes (c) Direction of strand bending (i) Right hand lay (ii) Left hand lay (d) Usage (i) Running ropes (ii) Stationary ropes

1.94


Left Hand Lay

Parallel Lay Ropes

Cross Lay Ropes

Righ t Hand Lay

Steel Wire Ropes of Various Constructions

Com posite Lay Ropes Fig.1.12. Types o f W ire - R opes.

(iii) Track ropes (iv) Wire ropes slings Stress in Wire Ropes (i) Direct stress The stress induced due to the weight of the load to be lifted, weight of the rope etc. d 

W  Wr A

Where W  weight of the load to be lifted

... (1)


1.95

W r  weight of the rope A  cross  sectio na l area of the ropes

(ii) Bending stress The stress induced in the rope as it passes over a sheave or drum. It is directly proportional to the wire diameter and inversely proportional to the diameter of the sheave. dw  b  Er D

... (2)

Where, E r  Young’s modulus of the entire rope  0.8  10 5 N  mm 2 or Er 

3 E 8

E  Young’s modulus of the wire material dw  wire diameter D  s heave diameter

(iii) Stress due to acceleration The stress induced due to the acceleration which is changed due to speed of hoising. It is not due to starting or stopping acceleration.  W  Wr  a a    g  A

where a  acceleration of rope and load during hoisting 

V2  V1 t

... (3)

1.96


(Speed of the rope changes from V 1 to V2 in t sections) (iv) Stresses during starting During starting, the load is given some acceleration which induces addition at stress in the rope. s  2  d  4 

(If there is no slack in the rope)

If there is slack in the rope before starting, which result impact load in the rope then,  W  Wr  1 s   A 

 2as h E r   1  d l  g 

 

... (5)

where, h  slack during starting l  length of the rope. as  acceleration during starting  stopping

(v) Effective stress The sum of direct stress  d and the bending stress  b is called the effective stress in the rope.

The effective stress in the rope during normal working  ne  d  b The effective stress during the acceleratio n of the load  ae   d   b   a The effective stress during starting  se   s   b


1.97

1.3.7 Failures of wire rope Wire rope failure may be caused by the following reasons (a) Work at overloads. (b) Wear and corrosion due to poor maintenance. (c) Due correct selection of drum and rope diameter. (d) Incorrect winding of drum. (e) High dynamic impact. 1.3.8 Design procedure for wire rope 1. Selection of suitable wire rope Select a suitable wire rope based upon it usage from the table. 6  17 rope

used in wires, transmission

tramways,

power

6  19 rope

used in cranes, derricks, dredges, elevators

6  37 rope

used in steel mill ladles, cranes, high speed elevators

8  19 rope

used as extra flexible hoisting rope

2. Calculation of design load Design load = load to be lifted  Assumed factor of safety n Refer PSG data book Pg.No. 9.1 Assume factor of safety 1 to 1.6 times the factor of safety given in data book, it is because of duty factor. Refer PSG data book Pg.No. 9.2

1.98


3. Selection of wire rope diameter d  Select a suitable diameter from the table. Refer PSG Data book Pg.No. 9.4, 9.5 and 9.6 4. Calculation of shear diameter D  Refer PSG data book Pg.No. 9.1 Always larger shear diameter is preferred. 5. Selection of Area of useful cross-section of rope A  Refer Pg.No. 9.1 (PSG data book) A  0.4

T 2 d 4

6. Calculation of wire diameter d w  dw 

d i 1.5 

where i  Number of wire in the rope  Number of strands



Number of wire in each

strand 7. Selection of weight of rope w r Refer PSG data book Pg.No. 9.4, 9.5 and 9.6 8. Calculation of various loads (i) Direct load  W d  W  W r (ii) Bending load  W b  b  A  Er 

dw A D


1.99

(iii) Acceleration load W a  W  Wr  Wa   a g  

where a 

v2  v1 2

(when speed of rope changes from v1 to v2 in t seconds) (iv) Starting on sloping load (a) When no slack in rope starting load W st  2  W d  2 W  W r

(b) When there is slack starting load W st  st  A  W  W r 

 1 

2ash  E r   d  l  g 

 

9. Calculation of effective load (i) Effective load during normal work Wen  W d  W b (ii) Effective load during starting Wse  W b  W s (iii) Effective load during acceleration Wae  W d  W b  Wa 10. Calculation of working factor of safety FOS w  FOSw 

Breaking load Effective load durin g a cceleration

1.100


For Breaking load, refer PSG data book Pg.No. 9.4, 9.5 and 9.6 11. Checking of safe design 12. Calculation of Number of ropes Number of ropes 

Re commended factor of safety n Working factor of safety FOSw 

Problem 1.19: A workshop crane is required to lift a load of 30 kN using a wire rope and hook. The hook weighs 15 kN. The diameter of rope drum is 30 times than that of the wire rope. The load is to be lifted with an acceleration of 1m  sec2. Calculate the diameter of rope. Factor of safety is 6. Young’s modulus of wire rope is 80 kN  mm2. Ultimate stress is 180 kN  cm2. The cross sectional area of the wire rope is 0.4 times the square of the wire rope diameter. (Oct. ’96, May/June 2013) W 1  weight of the load to be lifted  30 kN W 2  weight of the crane hook  15 kN

Direct load or Dead weight on the rope  W d  W1  W 2  30  15  45 kN  45  103 N

Ultimate stress 180 kN  cm2 

180  10 3 2

10

 1800 N  mm2


1.101

Young modulus  E r  80 kN  mm2  80  10 3 N  mm2 D  diameter of rope drum  30  d

where d  d ia o f w ire rope a  1 m  sec 2; FOS  6; as  0.4 d2

Design load  Factor of safety  Dead weight on the rope  6  45  270 kN

Since the rope is used in crane, select 6  19 rope dW 

d 1.5

i 



d 1.5

6  19  

 0.0624 d

Bending load on the rope  W b  

Er  dW D

A

80  10 3  0.0624 d  0.4d 2 30 d

 66.56 d2 N Load on the rope due to acceleration  W a  

Wd g

a

45  10 3 1 9.81

 4587.15 N  Total load acting on the rope  W d  W b  W a  45  103  66.56 d 2  4597.15

... (1)

1.102


Total load on the rope  Area of wire rope  Allowable stress A

u

FOS

 0.4 d2 

1800 6

 120 d2

... (2)

Equate the equations (1) and (2) 45  10 3  66.56 d2  4587  120d2

120 d 2  66.56 d 2  49,587 d2 

49,587 53.44

d  30.46 mm From table, Refer PSG data book Page No. 9.5 standard diameter of wire  32 mm Problem 1.20: Select a wire rope for a vertical mine hoist to lift a load of 55 kN from a depth of 350 meters. A rope speed of 600 m/min is to be attained in 20 seconds.

Solution Given data: W  55 kN  55  10 3 N ; depth  350 metre ; v  600 m  min; t  20 seconds

1. Selection of wire rope for a vertical mine hoist. From 6  19 ro pe;

PSG

databook,

Page

No.

9.1

select

2. Factor of safety  n  n  duty facto r

Refer PSG Data book Pg. No.9.1


1.103

For mining installations n  10 and duty factor  1.6

refer P.No.9.2

(assume mechanism class 4)  n  10  1.6  16  Design load for the wire rope  16  55  10 3  880  103 N  89.73 tonnes

From PSG databook P.No. 9.5. For nominal breaking strength of 90.4 tonnes and tensile strength of 160 to 175 kg f  mm 2, the diameter of rope is 41 mm.  Select diameter of rope as  d  41 mm

Approximate weight w  6.37 kg f  m ; (Assume) d ep th  350 m  6.37  350  2229.5 kg f w  22295 N dw  dia of wire 

d 1.5

 i



41 1.5

6  19  

3. Diameter of sheave: (D) Refer PSG data book page No. 9.1, Min

D  100 d

 D  100  d  100  41  4100 mm

 2.56 mm

1.104


 Bending stress 

E r  dW 0.8  10 5   2.56  4100 D

b  49.95 N  mm 2 Take E r  0.8  10 5 N  mm2 (Assume) Equivalent bending load on the rope  W b  b  A  49.95  528.1  26.378 kN A  0.4   0.4 

  d2 4   41 2 4

 528.1 mm 2

4. Acceleration of the rope and load: a

600 V  60t 60 20 

 0.5 m  sec 2

Additional load due to acceleration  W a  

Ww a g

55  10 3  22295  0.5 9.81

 3939.6 N

5. Effective load on the rope during acceleration of the load  W  w  Wb  Wa  55  10 3  22295  26.378  10 3  3939.6  107.6126  10 3 N


1.105

 Actual factor of safety during acceleration of the

load 

880  10 3 107.6126  10 3

 8.177  16 Design factor of safety

6. Effective load on the rope during normal working  W  w  Wb  55  10 3  22295  26.378  10 3  103.673  10 3 N

Actual factor of safety during normal working 

880  10 3 103.673  10 3

 8.488  16

 8.488  16 des ign fa ctor of safety

Therefore, the design is satisfactory. Problem 1.21: Select a wire rope from a 6  37 group to lift a maximum load of 10 kN through a height of 60 m. The weight of bucket is 2 kN. Maximum lifting speed is 2 m/s which is attained in 3 seconds. Drum diameter is 30 times the rope diameter. Factor of safety is 6. (Anna University, May 2010)

Solution Given W 1  weight of load to be lifted = 10 kN W 2  weight of bucket = 2 kN

Direct load (or) Dead weight on the rope W d

1.106


W d  W1  W 2 W d  10 kN  2 kN  12 kN  12  10 3 N D  Diameter of rope drum  30  d F os  6 (since the load applied on the rope during

starting and acceleration is more than two times the weight) Rope Type  6  37 ropes Maximum speed v  2 m/s Time = 3 seconds i  6  37 type

1. Design load P d  F os  3  Dead weight on rope P d  3  6  12 P d  3  72 kN P d  216  10 3 N

Ultimate stress  1700 N/mm2 (Assumed) 2. Area of rope A  

Design load Ultima te stress 216  10 3 N 1700 N/mm 2

A  127 mm 2

Diameter of rope is determined by using A  0.4 d2


d



1.107

 

A 0.4

0.4 

127

 17.8 mm

Next standard diameter of rope d  18 (Refer PSG data book Pg.No. 9.4, for 6  37 ropes) Approximate weight w  6.37 k gf/m w  6.37 kg f/m  63.7 N/m

dia of wire  d w  

d 1.5

i  18

1.5

6  37  

 0.8053 mm

3. Diameter of sheave Refer PSG data book Pg.No. 9.1 D  100 d  D  100  d  100  18  1800 mm

Take E r  0.8  10 5 N/mm 2 (Assumed)  Bending stress 

Er  dw D



0.8  10 5  0.8053 1800

1.108


b  35.79 N/mm2 Equivalent bending load on the rope wb  b  A

w.k.t A  0.4 

 0.4 

  d2 4   182 4

 101.78 mm 2  w b  35.79  101.78 N  3642.7 N  3.6427 kN

4. Acceleration of rope and load a

v 2  t 3  0.66 m/s 2

load due to acceleration w a wa 

Ww a g 

10  10 3  63.7  0.66 9.81

w a  677.06 N


1.109

5. Effective load on rope during acceleration of load  W  w  wb  w a  10  10 3  63.7  3.64  10 3  677.06  14382.7 N  14.3827  10 3 N  Actual factor of safety during acceleration of load 

216  10 3 N

14.3827  10 3 N

 15.01  18

(Design factor of safety)

6. Effective load on the rope during normal working  W  w  wb  10  10 3  63.7  3.642  10 3  13.7057  10 3 N

Actual factor of safety during normal working 

216  10 3 N

13.7057  10 3 N

 15.75  18

(design factor of safety)

 The design is satisfactory. Problem 1.22: A crane is lifting a load of 18 kN through a wire rope and a hook. The weight of the hook etc., is 10 kN. The load is to be lifted with an acceleration of 1 m/sec2. Calculate the diameter of the wire rope. The rope drum diameter may be taken as 30 times the diameter of rope. Take a factor of safety of 6 and young’s modulus for the wire rope


1.110

0.8  105 N/mm2.

The

ultimate

stress

may

be

taken

2

as

1800 N/mm . The cross-sectional area of the wire rope may be taken as 0.38 times the square of the wire rope diameter.

Solution Given W 1  Weight of the load to be lifted  18 kN W 2  Weight of crane hook  10 kN

Direct load (or) Dead weight on the rope Wd  W1  W2  18  10  28 kN  28  10 3 N

Ultimate stress  u  1800 N/mm 2 Young Modulus E r  0.8  10 5  80  10 3 N/mm 2 D  diameter of rope drum

 30  d

where d  diameter of wire rope a  1 m /s 2; F.O.S  6; as  0.38 d2

where as  crosssectional area of wire rope a  acceleration

Design Load Pd  Factor of safety  Dead weight on th e rope  6  28  168 kN


1.111

Since the rope is used in crane, from P.S.G. Data book, Pg.No. 9.1, select 6  19 rope The diameter of wire dW 

d d  0.0624 d   i 1.5  1.5    6  19

Bending load on rope W b  Wb 

E r  dW  as D

80  10 3  0.0624 d  0.38 d 2 30 d

 63.232 d2 N

Load on the rope due to acceleration Wa 

Wd g

a

28  103 1 9.81

 2854.23 N  Total load acting on the rope  Wd  Wb  Wa  28  10 3  63.232 d2  2854.23

... (1)

Total load on the rope  Area of wire rope  Allowable stress u 1800  0.38 d 2   as  FOS 6  114 d 2

Equate the equations (1) and (2)

... (2)

1.112


28  10 3  63.271 d2  2854.23  114 d2 30,854.23  50.729 d2  d  24.66 mm

From table, Refer P.S.G. Data book, Pg.No. 9.5, standard diameter of rope  25 mm Problem 1.23: A 6  19 wire with a tensile designation 1120 is used to raise a load. The nominal diameter of the wire rope and the sheave diameter are 15 and 400 mm respectively. Assuming long life on the basis of fatigue consideration determine the maximum load that the wire rope can carry.

Solution Given that i  6  19 wire rope

Tensile designation 1120 means  v  1120 N/mm2 d  15 mm D  400 mm W max to be found out.

w.k.t

p  0.0012 v

Refer PSG data book Pg.No. 9.2 (Graph)  P  0.0012  v P  0.0012  1120 P  1.344 N /mm 2


Also w.k.t P 

1.113

2Wmax dD

(Refer PSG Data book Pg.No. 9.2) W max 

P  d  D 1.344  15  400  2 2

W max  4032 N Ans.

1.3.9 Design of wire rope sheaves Performance of wire rope in a machine is affected by the design of its sheaves and drum. The Bending stress induced in the wire rope.  b  Er 

dw D

So, the bending stress induced in the rope is inversely proportional to the diameter of the sheave.  Its diameter should be large enough to reduce the

bending stress in the rope. It also provides better and more economical service. Narrowing sheaves will crush the rope and damage. G ro ove D ia m e te r R o pe D ia m eter

Sh eave G roo ve w ith Too M uch C learanc e

G ro ove D ia m e te r R o pe D ia m eter

G ro ove D ia m e te r

Sh eave G roo ve w ith Too Little C learanc e

Sh eave G roo ve w ith Corre ct C learanc e

Fig.1.13

R o pe D ia m eter

1.114


Sheave materials Cast iron sheaves are employed in light and medium service where as steel castings are often employed for heavy service. The cross-section of standard sheave grooves for wire ropes are given in the data book (Refer PSG Data book Pg.No. 9.10) For various proportion of sheave groove, Refer PSG data book Pg.No. 9.10. 1.3.10 Design of wire rope drums In industries and other practical field, more than one wire rope is being used, therefore one has to use rope drum to enable the rope to be wound in multiple layers. The drum is selected similarly the sheaves. Drums for steel wire ropes are generally made up of cast iron. The drum should be provided with helical grooves so that the rope winds up uniformly and is less subjected to wear. The helical grooves should be large enough to prevent jamming of the rope. 1.3.11. For dimensions of the drum groove for wire rope, refer PSG data book Pg.No. 9.9. Problem 1.24: A 6  19 steel wire rope is used for the elevator in the building. The weight is being lifted with a maximum speed of 200 m/min. The weight of the elevator cage is 10 kN and the weight of passengers is 9 kN. The lifting sheaves are of the traction type. Design a suitable wire rope sheave. Take FOS as 10.


1.115

Solution Given data Wire Type  6  19 wire rope v  200 m/min W elevator  10 kN W passenger  9 kN

FOS = 10 1. Calculate the rope diameter To design a wire sheave, we have to find the rope diameter first. Total load  W elevator  W passenger  10 kN  9 kN  19 kN  Design load  Total load  FOS  19  10  190 kN

Taking the design load as breaking strength refering PSG data book Pg.No. 9.6 The diameter of the rope is selected as 25 mm.  d  25 mm for v  1100  1250 N/mm2

and breaking strength = 190 kN Calculation of sheave diameter Refer PSG data book Pg.No. 9.1 For velocity upto 50 m/min

1.116


D min/d  27

But lifting speed is 200 m/min, this ratio has to be modified. For every additional speed of 50 m/min, D min/d ratio has to be increased by 8%  Modified

D min

d

 27  1.08 4  1  34

 Diameter of sheave D  34  25

  . . 200  . 50  4   

 850 mm

Design of wire rope sheave Refer PSG Data book Pg.No. 9.10 Knowing the diameter and refering data book, the proportions of sheave are given below (i) a  65 mm;

(ii) b  50 mm;

(iii) c  10 mm;

(iv) e  1.5 mm;

(v) h  37.5 mm;

(vi) l  18 mm;

(vii) r  14.5 mm;

(viii) r1  5 mm;

(ix) r2  5 mm;

(x) r3  20 mm;

(xi) r4  15 mm;


1.117

ADDITIONAL PROBLEMS APRIL 1998: Design a belt drive to transmit 22.5 kW at 740 rpm to an aluminium-rolling machine. The speed ratio being 3.0 and the distance between the pulleys is 3 meter. Diameter of the rolling machine pulleys is 1.2 meter. Take fabric belt and cast iron pulley. Assume Co-efficient of friction   0.2, Permissible tension for 8 ply fabric belt = 12.5 kg/cm width. Take the length of the belt, 1% less to give initial tension. APRIL 2002: Design a fabric belt to transmit 10 kW at 450 rpm from an engine to a line shaft at 1200 rpm. The diameter of the engine pulley is 600 mm and the distance of the shaft from the engine is 2 meter. OCTOBER 1996: Design a chain drive to transmit 5 kW at 900 rpm of sprocket pinion. The speed reduction is to be 2.5:1. The driving motor is mounted on an adjustable base. Assume that the load is steady, the drive is horizontal and the service is 16 hours per day. OCTOBER 2000: Select a suitable deep groove ball bearing for a drilling machine spindle of 40 mm diameter and suggest necessary tolerances on the shaft and housing. Radial load is 2.0 kN. Thrust is 1.5 kN. Spindle speed is 3000 rpm. Desired life is 3000 hours.

1.118


1.4. DESIGN OF CHAIN DRIVE 1.4.1 Chain Drives A chain drive consists of two sprockets and an endless chain. The chain is wrapped around two sprockets. The chain consists of number of links made of plates, which are connected by pin joints. The sprockets are made of special profile toothed wheels. D rive n S procket D rive r S pro cket

C e ntre D istan ce Fig.1.14. A. Typical Ch ain Drive.

1.4.2 Advantages: 1. Chain drives can be used for long or short center distances. 2. From a single driving sprocket, a number of shafts can be driven in the same or opposite direction by means of a chain. 3. Chain drives are more compact. 4. The efficiency of chain drives is approximately 98% (higher than belt drives).


1.119

5. There is no slip between sprocket and chain. (It is a positive drive compared to the belt drive). 6. Chain drives require less maintenance, when compared to belt drives. 1.4.3 Disadvantages 1. Periodic maintenance is required, particularly lubrication and slack adjustment. 2. Noisy operation, when compared to belt drives. 3. Design is more complicated (sprocket design). 4. Chain drive is unsuitable where precise motions is required due to polygonal effect and wear in the joints. 1.4.4 Applications: 1. Transportation industry (Bicycles, motor cycles and automobile vehicles, etc.) 2. Textile industry. 3. Building construction and material handling. 1.4.5 Types of chains: The common types of chains are (a) Roller chain (b) Bushed chain (c) Silent chain (or) inverted tooth chain w

w Db

G

Pt

Pt

Pt

Dr

DP

B ush C hain

R oller C hain

A1 W Sim p lex C h ain Fig.1.15

A2 D up lex C hain

A3 Triplex C hain

1.120


1.4.6 Components of Chain Drive 1. Chain (a) Roller chain It is a type of chain in which an alternate inner and outer links are connected by hinged joints. Each joint consists of a pin of diameter ‘D p’ pressed into the outer plates and the dash is secured in the holes of the inner plates. The roller employed in the chain safeguards the sprockets teeth against wear. The projection area of the contact surface for such a joint is given by A  Dp  b

where b  width of inner plates (b) Bush chains Bush chains differ from the roller chains in that they have no rollers in it. Therefore they are lighter in weight, but the absence of rollers increases the wear of teeth on the sprockets because of the sliding friction between the teeth and bushes. (c) Silent or inverted chains It is similar to the roller chain, with the sprockets engaged by projection on the link side bars and series of toothed plates are pinned together in rows across the width of the chain. Its structure is more intricate, it is more expensive manufacturing and it requires good maintenance.


1.121

2. Sprockets A sprocket is a profiled wheel with teeth that mesh with a chain. The name sprockets applies generally to any wheel upon which are radial projections that engage a chain passing over it. The operating capacity of a chain drive depends upon the quality of the sprockets. Generally sprockets are made up of cast iron or hardened steel. Sprocket with small radius is termed as pinion-sprocket and with large radius is called wheel sprocket. d  Pitch circle diameter of sprocket  

D rive D irection

p sin  /Z p sin 180/ Z

O utside Diam eter Pitch Circle D riving Diam eter C learance for Belt U nderside

+

Fig.1.16 S procket

1.122


P = P itch of C hain

Z = N um ber of Te eth P

Dr  1 80 2 = Z r  2

=

3 60 Z

 2

o

 d 2

o

d = Pitch C ircle D ia m eter of Sp rocket =

p sin ( /2) p o sin (1 80 /z)

Fig.1.17. Action of C hain E ngaging Spro cket Teeth.

The Fig. 1.17 represents the action of chain engaging sprocket teeth. 3. Chain housing It is a pointing made up of thin sheet, provided to prevent the chain drive from dust and dirt. It also dampers the noise of the drive and pressure grease. 4. Slack adjuster It is a movable bearing to regulate the chain slag and maintain required tension in the drive. The bearings are in the form of slides in which sprocket shaft is installed. 1.4.7 Design Procedure Step 1 Calculate transmission ratio i

Z2

Z1



n1 n2


1.123

Z 1  No. of teeth on sprocket pinion (Driver) Z 2  No. of teeth on sprocket wheel (Driven) n1  Speed of rotation of pinion in rpm. n2  Speed of rotation of wheel in rpm.

Step 2 (a) Select the no. of teeth on sprocket pinion Z 1 from databook Page transmission ratio (i)

No.

7.74

based

on

(b) Calculate the No. of teeth on sprocket wheel Z 2 Since i 

Z2 Z1

 Z 2  iZ 1 Z min  7 teeth Z max  100 to 120 tee th

For uniform wear, select odd no. of teeth on driver and even no. of teeth on driven. Step 3 Calculate of pitch p Refer databook Page No. 7.74 Optimum centre distance  a  30 to 50 p i.e.,

pmax 

a 30

1.124


pmin 

a 50

Select the standard pitch from databook 7.71, 7.72 & 7.73 The

selected

standard

pitch

must

be

within

pmax and pmin .

The standard pitches are

6.0 mm 8.0 mm

25.40 mm 31.75 mm

9.525 mm

38.10 mm

12.7 mm

44.45 mm

15.875 mm

50.8 mm

19.05 mm, Step 4 Select chain No., according to the selected standard pitch. (PSG Data book page No. 7.71, 7.72 and 7.73) Take the following values according to the selected chain No. Simplex - R Duplex - DR Triplex - TR (1) Bearing area in cm 2 A (2) Weight per meter in kg f W (3) Breaking load in kg f Q


1.125

Step 5 Refer PSG databook, Page No. 7.77. Checking of breaking load Q in kg f Power transmitted on the basis of Breaking load N

Qv in kw 102  n  K s

N  given power in kW. Q  induced breaking load (to be calculated) v  chain velocity in m/sec. 

ZpN m /sec. 60  1000

p  pitch in m m N 1  N  Speed o f pinion in rpm z  Z  no. of teeth on pinion n  factor of safety allowable

Refer databook Page No. 7.77 The value of n depends on 1. speed of rotation of pinion 2. pitch in mm ks  Service factor

Refer PSG databook Page No. 7.76 and 7.77  k1  k2  k3  k4  k5  k6

1.126


The

service

factor

depends on six factors

ks

k 1 to k6

According to the given conditions, take the values of k 1, k2, k 3, k4, k 5 and k6 and Calculate K s.

Note: If conditions are not given, select K s  1 (by assuming K1  K2  K3  K4  K5  K6  1 Substitute all the values in the equation N

Qv Calculate Q 102 n K s

Q  [Q] then design is safe.

If Q  [Q] then change the chain no. and read the values of A, W and [Q] and check again Q breaking load. Step 6 Check the Actual factor of safety. Refer PSG data book Page No. 7.78. [n] 

Q P

[n]  actual Fos  n allowable factor o f safety Q  Breaking load of chain in kgf

P  P t  P c  Ps P t  Tangential force due to power transm ission Pt 

102 N kg f V


1.127

P t  C entrifugal tension Pc 

W  v2 kg f g

N  P ower in kW given v  C hain velocity in m/sec W  W eight/m length of chain kgf g  9.81 m/sec 2 P s  Tension due to sagging of chain: Ps  K  W  a

K  Coefficient of sag

(refer PSG Data book page No. 7.78) value of K depends on position of chain drive W  Weight/m length of chain kgf a  Centre distance in meter

To Find ‘a’: Centre distance 1. To find length of chain Refer PSG Data book Page No. 7.75

lp  2ap 

Z1  Z2  2

 Z2  Z1    2   ap

2

(no. of links)

(to be corrected to even no.) Actual length of chain  l  lp  p

1.128


where ap 

ao p

ap  Approximate centre distance ao  Initially assum ed centre dis tance p  Pitc h

2. To find exact centre distance e a  where e  lp 

 Z2  Z1  m   2 

e2  8m  

Z1  Z2 2

2

4

 p 

The value of m can be read directly from the PSG data book. Page No.7.76 The value of m depends on Z 2  Z 1 value  Calculate Z 2  Z 1 and take the corresponding

value of ‘m’. Calculate exact centre distance ‘a’. Ps  K  W  a

where a  centre dista nce in meters K  Coefficient. of sag W  Weight/m length

Calculate P and [n] 

Q P


1.129

[n]  n

Then the design is satisfactory. * If [n]  n, then change the chain No. and repeat the calculations and check again [n] and n Step 7 Checking of bearing stress  Refer PSG Data book, Page No. 7.77 The allowable bearing pressure [] depends on 1. Pitch 2. Chain velocity Read the value of [] in kgf/mm 2 and convert in kg f /cm 2

kgf/mm 2  10 2  kg f/cm2 Power transmitted on the basis of allowable bearing

stress  N 

 A  v 102 K s

where N  given power in kW   Induced bearing stress in kgf  cm 2 A  bearing Area  cm 2 v  Chain speed  m/sec K s  service factor

Calculate   induced stress in kgf  cm 2   []

Then the design is satisfactory.

1.130


Step 8 Diameters of pinion and wheel d and D Refer PSG data book Page No. 7.78 d

D

p 180 sin Z1

d  dia. of smaller sprocket

p 180 sin Z2

D  dia. of larger sprocket

1.4.8 Chordal action of a roller chain Sprocket with less number of teeth can affect the smooth running of a chain drive. This unsmooth running condition is termed as chordal action of the chain. When a chain roller approaches the sprocket and has just seated, it has radius R c, also known as chordal radius. When the roller passes through half of the pitch angle , the roller has radius R which is the pitch radius.  R  Rc

At N rpm, the chain velocity for pitch radius R  is 2 RN and for chordal radius is 2 R c N.

Since the velocity R  Rc, it is obvious that it is undergoing from minimum to maximum velocity for every rotation, which results in unsmooth running. Chordal rise  R is given by  R  R  Rc


 R  R cos

1.131

 2

   R  1  cos  2  

The rise can be reduced by increasing the number of teeth z in the sprocket, because 

360 z

1.4.9 DESIGN OF CHAIN DRIVES Problem 1.25: Design a chain drive to operate a compressor from a 15 kW electric motor at 900 rpm; The compressor is to run at a speed of 300 rpm; The minimum centre distance should be 550 mm.

Solution Given Data: Power of Motor P  15 kW Speed of motor N 1  900 rpm Speed of the compressorN 2  300 rpm ; Minimum centre distance ao  550 mm;

Step 1: Calculation of transmission ratio Tra ns mission ratio 

Z2 Z1

i  Transmission ratio 



N1 N2

N1 N2



900 3 300

Step 2: From PSG Design data book, refer Page No. 7.74.

1.132


For i  2 to 3 ; Z 1  25 to 27 i  3 to 4 ;

Z 1  23 to 25

Take Z 1  23 to 27 (select any odd no. of teeth) Select Z 1  27 teeth (no. of teeth on sprocket pinion) Z 2  i Z1  3  27  81  82 (no. of teeth on sprocket wheel)

Step 3: From PSG design data book, P.No.7.74 Optimum centre distance  a  30 to 50  p where a  approximate centre distance 550  18.33 mm pitchmax  pmax  30 550  11 mm pitchmin  pmin  50

Select standard pitch from PSG data book, Take any standard pitch between 11 to 18.33 mm  Select Pitch  p  15.875 mm

Step 4: Selection of Chain No. Select roller chain from PSG Data book Page No. 7.72 The available chain No. are 10A and 10B *

Select 10A-2 Duplex Chain.

Pitch  p  15.875 mm Corresponding to chain No. selected, take the values of


1.133

A  Bearing area  1.4 c m2 W  Weight per m length  1.78 kgf Q  Breaking load  4440 kgf

Step 5: Calculate Power transmitted based on breaking load: From data book, Pg. No. 7.77 N

Qv kW 102n  Ks

From the above equation calculate ‘Q’ breaking load by considering N  given power N  15 kW v

Z1 p N1 60  1000



27  15.875  900 60  1000  6.429 m/sec

n  Minimum value of factor of safety K s  1 ; Z1  15 to 30 from PSG data book, Pg. No. 7.77,

Select

n  11

for

a

pitch

15.875

mm

and

n 1  1000 rpm

Since the specific conditions are not given in the problem, assume K 1  K 2  K 3  K 4  K5  K 6  1  Ks  1 15 

Q  6.429 102  11  1

1.134


Q

15  102  11  1  2617.825 kgf 6.429

Breaking load Q  2617.825 kgf which is less than the selected chain Breaking load 4440 kgf The selection of chain no. is satisfactory based on breaking load. Step 6 (a) Calculation of Length of chain (b) Final centre distance. (a) Length of continuous chain in multiples of pitches

lp  2ap 

where ap  

Z1  Z 2 2



 Z2  Z1    2  

2

ap

ao p 550  34.64 15.875

 34.64 ao  initially assumed centre distance in mm  550 mm p  pitch  15.875 mm

No. of teeth on sprocket wheel Z 2  82 ; No. of teeth on sprocket pinion Z 1  27 Refer PSG data book Page No. 7.76,


1.135

2

 Z2  Z1   m   2  Read the value of ‘m’ directly from PSG data book Pg.No. 7.63 Z 2  Z 1  82  27  55; the value of m  76.6 lp  234.64  

27  82 76.6  2 34.64

 125.99  126 approximate d to 126 

Length of chain  l  lp  p  126  15.875  2000.25 mm

Take,

l  2000 mm

(b) Final centre distance a

e

e2  8m   4

p m  76.6 (from distance)

Z1  Z2 2 27  82  126  2

e  lp 

 126 

109  71.5 m.m 2

Final centre distance a

71.5 

71.5 2  8  76.6  

a  549.98 a  550 mm

4

 15.875

1.136


Step 7: Check the actual factor of safety. From PSG databook, Refer P.No. 7.78 Actual factor of safety  [n] 

Q P

Q  Breaking load of the chain  4440 kg

P  P t  Ps  Pc P t  Tangential force due to power Transmission  Pt 

102  15 6.429

N  15 kW v  6.429 m/sec

 237.98 kg f P c  Centrifugal fus ion  

W  v2 g 1.78  6.429 2 9.81

 7.499 kg f W  1.78 kg f P s  tension due to slagging  kWa metres  6  1.78 

550 1000

 5.874 kg K  Coefficient of Sag from PSG data book 7.78  6 (Horizontal) a  550 mm  0.55 m P  237.98  7.499  5.874  251.353 kgf

102 N v


[n] 

1.137

Q 4440  P 251.353

 17.66  11

which is greater than allowable factor of safety  The design is safe.

Step 8: Checking of Allowable Bearing Stress From PSG data book, Pg.No. 7.77 The allowable bearing stress    2.24 kg f/mm2 (for a pitch of 15.875 and speed  1000 rpm ) Refer PSG data book, from Pg. no. 7.77 Power transmitted on the basis of allowable bearing stress N

15 

 Av kW 102 Ks

  1.4  6.429 102  1

  169.98 kgf/cm 2

N  15 kW or

A  1.4 cm 2

  1.69 kgf/mm 2

v  6.429 m/sec

  []

Ks  1

 2.24 kg f/mm 2

Therefore the design is safe.

1.138


Step 9: Pitch dia. of small sprocket  d 



p 180  sin Z1 15.875 180 sin 27

 136.74 mm

Pitch dia. of large sprocket

D



p 180 sin Z2 15.875 180 sin 82

 414.46 mm Problem 1.26: A 600 rpm blower is to be driven by a 10 kW 1440 rpm motor approximately 750 mm away. Design a suitable chain drive.

(May/June 2011)

Given Data: Power  10 kW  10  10 3 W atts: Speed of motor  N 1  1440 rpm; speed of blower  600 rpm Approximate centre distance  750 mm Step 1: Calculation of transmission ratio (i) Transmission ratio  i 

i

Z2 Z1



N1 N2

1440  2.4 600


1.139

Step 2: From PSG Design data book, Page No. 7.74 For i  2 to 3; Z 1  no. of teeth on pinion  25 to 27 Take Z 1  27 teeth  Z 2  no.of teeth on wheel  i Z 1 Z 2  2.4  27  64.8 Take Z 2  66 teeth

Step 3: From PSG design data book Page No.7.74 Optimum centre distance  a  30 p to 50 p where p  pitch 750  25 mm Pitchmax  P max  30 Pitchmax  P min 

750  15 mm 50

Select standard pitch from PSG databook, Pg.No. 7.72 Take standard pitch from PSG databook, between 15 mm to 25 mm  Select Pitch 19.05 mm

Step 4: Selection of chain No. Select roller chain (from Pg. No. 7.72) from the available chain No. Select chain No. 12A-1 (simpler roller chain) p  pitch  19.05 mm

Bearing Area  A  1.05 cm 2;

1.140


weight per meter  W  1.47 kg f

Breaking load  Q  3200 kg f Step 5: Calculate power transmitted on breaking load from databook, Page No. 7.77 N

Q.V kW 102. n ks

From the above equation, calculate ‘Q’ breaking load by using N  given power in kW. N  10 kW; V  

p  Z 1 N1 60  1000 19.05  27  1440  12.34 m  sec 60  1000

n  Minimum value of factor of safety

From PSG databook, Page No. 7.77 Select n  13.2 Since the specific conditions are not given, assume K1  K2  K3  K4  K5  K6  1  KS  1 10  Q

Q  12.34 102  13.2  1

102  13.2  1  10  1091.08 kgf 12.34

Breaking load Q  1091.08 kgf, which is less then the selected chain, breaking load of 3200 kg f.


1.141

Step 6 (a) Calculation of Length of chain, and (b) Final centre distance  Z2  Z1     Z Z 2 a L ength of co ntinuous chain  1 2       l 2a p p ap in multiples of pitches  2

 lp  2 39.37  

66  27 38.5  39.37 2

 126.217 m  128 mm approximate links

Where ap 

ao p

ao  initially assumed center distance  750 mm  ap 

750  39.37 19.05

(b) Finalc entredistanc e  a 

a

e e2  8m   p 4

81.5 2  8  38.5  81.5    4

 767.18 mm e  lp 

Z1  Z2

 128 

2 66  27  81.5 2

 19.05

2

1.142


Take a  final centre distance  770 mm (rounded to nearest value) Step 7: Check the actual factor of safety From PSG data book, Page No. 7.78 Actual factor of safety  [n] 

Q P

Q  breaking load of the chain  3200 kgf

P  P t  P S  P C P t  tangential force due to power transmission  

102 N V

102  15  123.98 kg f 12.34

P C  Centrifugal tension  

W  v2 g

1.47  12.342 9.81

 22.818 kg f P S  tension due to slagging  k.W.a  6  1.47  0.77  6.7914 kg f

P  123.98  22.818  6.7914  153.5894 kgf K  coefficient of sa g from PSG databook, P.No.7.78. K  6 (Horizontal)


1.143

a  770 mm  0.77 metre  Actual factor of safety  [n] 

3200 Q  P 153.5894

[n]  20.83  13.2 (Greater than allowable factor of safety) Step 8: Checking of Allowable bearing stress: Refer data, Page No. 7.77 The allowable bearing stress    1.85 kg f  mm 2 Power transmitted on the basis of allowable bearing Av Stress  N  kW 102 KS 10 

   1.05  12.34 102  1

N  10 kW

  78.7219 kg f  cm 2 or   0.78721 9 kgf  mm

A  1.05 cm 2 2

v  12.34 m  sec KS  1

Which is less than allowable stress of 1.85 kg f  mm 2  The design is satisfactory. Problem 1.27: A truck equipped with a 8.5 kW engine uses a roller chain as the final drive to the axle. The driving sprocket runs at 900 rpm and the driven sprocket at 380 rpm, with a centre distance of approximately 550 mm. Select roller chain.

1.144


Solution Given data N  8.5 kW N 1  900 rpm N 2  380 rpm ao  550 mm

Step 1 Calculation of transmission ratio (i) Transmission ratio i  i

Z2 Z1



N1 N2

900 380

 2.368

Step 2 From PSG Data book Pg No. 7.74 For i  2 to 3 Z 1  no o f teeth o n pinio n = 25 to 27

Take Z 1  27 teeth  Z2  Z1  i  27  2.368  64.4

Take Z 2  66 teeth


1.145

Step 3 From PSG design data book Pg.No. 7.74 Optimum centre distance  a  30 p to 50 p where p  pitch P max 

a 550   18.33 30 30

P min 

a 550   11 50 50

Select standard pitch for design data book Pg.No. 7.72 Take any standard pitch between 11 to 18.33 mm  select pitch p  15.875 mm

Step 4 Selection of chain Select roller chain from PSG Data book Pg.No. 7.72 The available chain No are 10 A and 10 B. Select 10 A - Z Duplex chain. Now corresponding to chain No selected, take the value of (PSG data book Pg.No. 7.72) A  Bearing area  1.4 c m2 W  Weight per m length  1.78 kgf Q  Breaking load  440 kgf

Step 5 Calculate power transmitted based on breaking load.

1.146


From PSG Data book Pg.No. 7.77 N

Q.v kW 102. n.K s

From the above equation calculate ‘Q’ by considering N (given power) = 8.5 kW v 

Z1 p N1 60  1000 27  15.875  900 60  1000

 6.4293 m/sec

From PSG Data book Pg.No. 7.71 Select n  11 fo r pitc h = 15.875 and N 1  1000 rpm K s  1 (Assume) . . since the specific condition is not given in .

problem. K s  K 1  K 2  K3  K 4  K5  K 6  1  8.5  Q

Q  6.4293 102  11  1

8.5  102  11  1 6.4293

Q  1483.36 kg f  Breaking load Q  1483.36 kg f which is less than

the selected chain breaking load Q  4440 kg f  The Design is satisfactory based on breaking


1.147

Step 6 (a) Calculation of length of chain (b) Final centre distance (a) Length of continuous chains in multiplies of pitches  Z2  Z1   Z 1  Z2  2   L p  2ap  ap 2 where ap 

ao p



2

550 15.875

 34.64

No of teeth on sprocket wheel Z 2  64 No of teeth on sprocket pinion Z 1  27 From PSG data book Pg.No. 7.76, take the value of m for Z 2  Z 1  64  27  37  m  34.7 2

 Z2  Z1  w.k.t    m  34.7  2  Lp  2 34.62  

64  27 34.7  2 34.64

 115.74 links  116 approximated to 116 link s  Length of chain  l  lp  p l  116  15.875

1.148


l  1841.5 mm

Take,

l  1842 mm

(b) Final centre distance a 

w here e  lp 

e

e2  8m   4

p

Z1  Z 2 2

 116 

64  27 2

 70.5 a

70.5 

70.52  8  34.7   4

 15.875

a  551.66 mm

Take a  552 mm Step 7 Check actual factor of safety From PSG Data book Pg.No. 7.78 Actual factor of safety  n 

Q P

Q  Breaking load of the chain = 4440 kg

 P  Pt  Ps  Pc P t  Tangential force due to power transmission 

102 N v


Pt 

1.149

102  8.5 6.4293

P t  134.851 N  8.5 kW v  6.4293 m/sec P c  Centrifugal function 

W  v2 g W  1.78 kg f/m length



1.78  6.42932 9.81

 7.5 kg f P s  tension due to slag = kWa meter k  6 (Horizontal) from Pg.No. 7.78 PSG Data book P s  6  1.78 

550 1000

P s  5.874 kg f

 P  Pt  Pc  Ps  134.851  7.5  5.874 kg f  148.225 kg f  [n]  

Q P

4440 148.225

 29.95  1 allowable FOS  The design is satisfactory.

1.150


Step 8 Checking of Allowable Bearing stress From PSG Data Book P.No. 7.77 The allowable stress for pitch of 15.875 and speed < 1000 rpm.   2.42 kg f/mm 2 Power transmitted on the basis of allowable bearing stress N

8.5 



 AV kW 102 Ks   1.4  6.4293 102  1

8.5  102  1 1.4  6.4293

  96.322 kg f/cm 2   0.9632 kg f/mm 2    0.9632  []  2.24 kg f/mm 2

Therefore the design is safe. Step 9 Pitch dia of small sprocket  d 

d

p 180  sin Z1 15.875 180 sin 27

 136.74 mm


Pitch dia of large sprocket D 

D

1.151

p 180 sin Z2 15.875 180 sin 64

D  323.53 mm Problem 1.28: Design a chain drive to actuate a compressor from 15 kW electric motor running at 1000 rpm, the compressor speed being 350 rpm. The minimum centre distance is 500 mm. The compressor operates 15 hours per day. The chain tension may be adjusted by shifting the motor.

Solution Power transmission N  15 kW Motor speed

N 1  1000 rpm

Compressor speed N 2  350 rpm Centre distance A  500 mm Service

 15 hours/day

Let the operating chain may be a roller chain. Since the optimum centre distance is 30 to 50 pitches. pmax 

500  16.66 30

pmin 

500  10 50

(ie) p 

a 500   14.3 mm 35 35

1.152


The standard pitch value is p  15.875 mm (from P.S.G. Data book, Page No. 7.72 for pitch between 10 and 16.66 ) Transmission Ratio, i 

N1 N2



1000  2.857 350

for i  2.85 , the number of teeth on pinion sprocket, from P.S.G. Data book, Page No. 7.74. Z 1  25 (assumed)

then the number of teeth on wheel sprocket.  72 Z 2  i Z 1  2.85  25  71.42 ~

The power transmitted on the basis of breaking load is given by, PN

Q .v kW 102n Ks

(from P.S.G. Data book, Pg.No. 7.77) N  Power to b e transmitted  11 kW Q  Minimu m b reaking load in N v  Chain velocity in m/s n   F.O.S to be assumed k s  Service facto r

Service factor ks  k1  k2  k3  k4  k5  k6 from P.S.G. Data book, (Pg.No. 7.76, 7.77)


1.153

k 1  1.5 load with heavy sh ock k 2  1.0 Adjustable supports k 3  1.0 a p  30 to 50 p k 4  1.0 Assume th e line joining the centres of the sprockets is inclined up to 60 to the horizontal k 5  1.0 Drop lubrication, Assumed k 6  1.25 Double shift of 16 hours a day K s  1.5  1.0  1.0  1.0  1.0  1.25  1.875

from the P.S.G. Data book Pg.No. 7.77 Let the minimum F.O.S n   11 for k s  1 and Z 1  25 n   n  ks

(or) n   11  1.875  20.625

Chain velocity, v 

Z1 N1 p 60  1000



25  1000  15.875 60  1000

 6.61 m/s

Hence minimum breaking load Q

N n  K s 15  20.625  1.875  6.61 v

 87.75 kg f

1.154


Since minimum breaking load ((ie) developed tangential force to break the chain) is 87.75 kgf. We should select a chain, having higher breaking strength than the calculated value. For pitch value of 15.875 mm, DR Rolon chain may be selected whose breaking strength is 4440.0 kg f from P.S.G. Data book, Pg.No. 7.72. Now the chain is checked for its actual F.O.S and for allowable stresses. Checking for actual F.O.S, [n ] 

Q P

where  P  P t  P c  P s Now P t  Tangential force du e to power transmission 

102N kgf V



102  15  231.467 kg f 6.61

P c  centrifugal tens io n in kgf

From

P.S.G.

Data

book,

Pg.No.

7.72,

W

weight/metre length  17.8 N PC 

WV2 17.8  6.612   79.27 kg f g 9.81 P S  Tension due to sagging of chain in Newtons  kWa, where k  Sagging coefficient

from P.S.G. Data Book, Pg.No. 778, position of chain inclined up to 40 is assumed


1.155

hence k  4, a  0.5 m given, W  17.8  P s  4  17.8  0.5  35.6 kg f   P  231.467  79.27  35.6  346.337

Hence [n ] 

Q 4440   12.81  11  P 346.337

Since the actual F.O.S is more than adopted minimum value, chain selection is correct. ie., n  n Checking for induced bearing stress The induced bearing stress over the projected area of the chain is given by 

N  Ks AV

A  Bearing area

From P.S.G. Data book, Pg.No. 7.72, corresponding to DR 50, Bearing area  140 mm2 Hence  

15  1.875 1.4  6.61

 3.039 kgf/cm 2  0.03039 kg f/mm

This is less than the allowable bearing stress, whose value for k s  1.875, Z 1  25 and N 1  1000, fro m P.S.G. Data bo o k, Pg. 7.77 is 42 N/mm 2 Hence selection of chain is correct

1.156


Now length of chain interms of pitches is obtained as  Z2  Z 1   Z 1  Z 2  2    l p  2a p  2 ap

2

from P.S.G. Data

book, Pg.No. 7.75 2

. .  Z2  Z1  .   m 2x  

From PSG Data book Pg. No. 7.76 take m  56 But ap 

ao p



500  31.5 15.875

Hence  25  72  56 lp  2  31.5     2 31.5    113.27 mm; T ake 114

Length may be corrected to 112 pitches  Actual length of chain  114  15.875  1809 mm

Now corrected centre distance

7.75 But

e a 

e2  8 m   4

 Z1  Z 2  e  lp    2  

  from P.S.G. Data book, Pg. No. 


1.157

 25  72   114     65.5 2  

From P.S.G. Data book, Pg.No. 7.76, m  54 , for Z 2  Z 1  48  65.5  Hence a   

65.5 2  8  54   

4

   15.875 

 506.47 mm

Centre distance (decrement) accommodate initial chain sag

allowance

  0.5 F

But F  0.02 a    0.5  0.02 506.47   5.0647 mm

Hence required centre distance  506.47  5.06  501.48 mm

Now, pitch diameter of pinion sprocket d d

15.875 P   126  127 mm  180   180  sin sin   25  Z     1 

pitch diameter of wheel sprocket D D

15.875 P   363.9  364 mm 180    180  sin sin   72  Z     2 

Specifications Type of chain: 10 A  2 DR 50 Ro ller c hain Centre distanc e  501.40 mm

to

1.158


Number of teeth of sprocket pinion  25 Number of teeth of sprocket wheel  72 Length of chain  1809 mm Pitch diameter of pinion sprocket  127 mm Pitch diameter of wheel sprocket  364 mm Problem 1.29: A 15 kW squirrel cage motor with a speed of, 1250 rpm, is driving a centrifugal pump at 550 rpm. The centrifugal pump is located at 700 mm from the motor. Design a chain drive.

Solution Given Data Power of motor  N  15 kW Speed o f mo to r  N 1  1250 rpm Speed of centrifugal pump  N 2  550 rpm Minimum centre distance

 700 mm

Step 1 Calculation of transmission ratio Transmission ratio 

Z2

Z1



i  Transmission ratio 

N1 N2

N1 N2



1250  2.27 550

Step 2 From P.S.G. Data Book, Pg.No. 7.74 For i  2 to 3; Z 1  25 to 27 Take Z 1  27 teeth (No. of teeth on sprocket pinion)


1.159

Z 2  i Z 1  2.27  27  62

Step 3 From P.S.G. data book, Pg.No. 7.74 Optimum centre distance  a  (30 to 50) P where a  approximate centre distance p  pitch of chain

a 700 pitchmax  b max   23.33 mm  30 30 a 700  14 mm pitchmin  b min   50 50 Select standard pitch from P.S.G. data book, from Pg.No. 7.72 Take any standard pitch between 14 to 23.33 mm.  select pitch p  15.875 mm

Step 4 Selection of chain number Select roller chain from P.S.G. Data book, Pg.No. 7.72 The available chain Number are 10 A and 10 B Select 10 A - 2 Duplex chain Pitch p  15.875 mm Corresponding to chain number selected, take the values of A  Bearing area  1.4 c m2 W  Weight per metre length  1.78 kg f Q  Breaking load  4440 kgf

1.160


Step 5 Calculate power transmitted based on breaking load From P.S.G. Data book, Pg.No. 7.77 N

Qv  kW 102 n k s

From the above equation calculate ‘Q’ breaking load by considering N  given power  15 kW v

Z1 p N1 60  1000

n



27  15.875  1250  8.929 m/ s 60  1000

Minimum

value

of

factor

of

safety

(Ks  1; Z 1  15 to 30) from P.S.G. data book; Pg.No.7.77; Select

n  13.2

for

a

pitch

15.875

mm

and

N 1  1600 rpm .

Since the specific conditions are not given, in the problem, assume K1  K2  K3  K4  K5  K6  1  K5  K 1  K2  K3  K 4  K 5  K 6  1  15 

Q  8.929 102  13.2  1

 Q  2261.66 kg f which is less than the selected

chain Breaking load 4440 kg f. The selected chain Number is satisfactory based on breaking load. Step 6 (a) Calculation of Length of chain (b) Final centre distance


1.161

(a) Length of continuous chain in multiples of pitches 2

 Z2  Z1   Z1  Z2  2    lp  2ap  ap 2

from

P.S.G.

Data

book, Pg.No. 7.75 where ap 

ao p

where ao is initially assumed centre distance p is pitch in mm  ap 

700  44.09 15.875

No. of teeth on sprocket wheel Z 2  62 No. of teeth on sprocket pinion Z 1  27 Read the value of “m” directly from P.S.G. data book at Z 2  Z 1 from Pg.No. 7.76  the value of m corresponding to Z 2  Z1  35 is 31  62.27  62  27  2    lp  2  44.09   44.09 2

 133.38 Length of chain  l  lp  p  133.38  15.875  2117.46 mm  2117.5 mm

2

1.162


(b) Final centre distance (from P.S.G. Data book, Pg.No. 7.75) a

e

e2  8 m  

e  lp 

But

4

p

Z1  Z2 2

 133.38 

27  62  88.88 mm 2

 Final centre distance a

88.88 2  8 31 88.88     4

 15.875

 699.90 a~  700 mm

Step 7 Check the actual factor of safety From P.S.G. Data book, Pg.No. 7.78 Actual factor of safety [n] 

Q P

where Q  Breakin g loa d of the ch ain  4440 kg

 P  Pt  P s  P c P t  Tangential force due to power transmission 

102 N v


1.163

where N in kW and v in m/s 

102  15 8.929

 171.35 kg f P c  Centrifugal tension 

W  v2 g

where W  Weight per meter length. From P.S.G. Data book, Pg.No. 7.72, corresponding to p  15.875 and chain number.  W  1.78 kg f/m length  Pc 

1.78  8.929 2  14.46 kg f 9.81

P s  tension due to sagging  k.W.a

From P.S.G. Data book, Pg.No. 7.78; k - coefficient for sag. k  6 for horizontal position of chain drive. and a is the centre distance W is weight per meter length  P s  6  1.78 

700 1000

 7.476 kg f   P  171.35  14.46  7.476  193.286 kg f

1.164


 [n] 

Q 4440   22.97  13.2  P 193.286

which is greater than allowable factor of safety.  The design is safe.

Step 8 Checking of Allowable Bearing stress. From P.S.G. data book, Pg.No. 7.77, corresponding to speed of rotation of sprocket  1600 rpm , and pitch 15.875. Allowable Bearing stress []  1.85 kgf/mm 2 From P.S.G. Data book, Pg.No. 7.77 Power transmitted on the basis of allowable bearing stress. N

 15 

Av 102 Ks   1.4  8.929    122.39 kg f/cm 2 102  1

Or   1.22 kg f/mm2   []   1.85 kg f/mm2  The design is safe


1.165

Step 9 Pitch dia of small sprocket d

d

p 180 sin Z1 15.875  136.74 mm 180 sin 27

Pitch dia of large sprocket D D



p 180 sin Z2

15.875 180 sin 62

D  313.43 mm Problem 1.30: Design a chain drive to run a compressor from a 11 kW electric motor running at 970 rpm, the compressor speed being 330 rpm. The compressor operates 16 hours/day. The centre distance should be approximately 500 mm. The chain tension can be adjusted by shifting the motor on slides.

Solution Power transmitted N  11 kW Motor speed

N 1  970 rpm

Compressor speed N 2  330 rpm Centre distance

a  500 mm

1.166


Service  16 hours/day Let the operating chain may be a roller chain. Since the optimum centre distance is 30 to 50 pitches. assume a  30  50 p where p  pitch of chain (from P.S.G. Data book, Pg.No. 7.74) (ie) pmax 

a a 500 500   23.33 mm; pmin   10 mm  30 30 50 50

The next standard pitch value, p  15.875 mm (from P.S.G. Data book, Pg.No. 7.72) Transmission ratio, i 

N1 N2



970  2.93 330

for i  2.93 , the number of teeth on pinion sprocket, from the P.S.G. Data book, Page No. 7.74 Z 1  25

then the number of teeth on pinion-sprocket Z 2  iZ 1  2.93  25  73

(from P.S.G. Data book, Pg.No. 7.77) the power transmitted on the basis of breaking load is given by N

Q .v kW 102n Ks

N  P ower to be transmitted  11 kW Q  Minimu m b reaking load in N v  Chain velocity in m/s


1.167

n   F.O.S to be assumed K s  Service factor

(from P.S.G. Data book, Pg.No. 7.77) Service factor ks  k1  k2  k3  k4  k5  k6 (from P.S.G. Data book, Pg.No. 7.76, 7.77) k 1  1.5 load with heavy sh ock k 2  1.0 Adjustable supports k 3  1.0 a p  30 to 50 p k 4  1.0 Assume th e line joining the centres of the sprockets is inclined up to 60 to the horizontal k 5  1.0 Drop lubrication, Assumed k 6  1.25 Double shift of 16 hours aday k S  1.5  1.0  1.0  1.0  1.0  1.25  1.875

From the P.S.G. Data book Pg.No. 7.77 Let the minimum F.O.S n  11 for k s  1 and Z 1  25

(or) n   11  1.875 for Ks  1.875 , Z 1  25  20.625

1.168


Chain velocity, v 

Z1 N1 p 25  970  15.875  60  1000 60  1000

 6.41 m/s

Hence minimum breaking load Q 

N n  Ks v 11  20.625  1.875 6.41

 66.36 kg f

Since minimum breaking load ((ie) developed tangential force to break the chain) is 66.36 kg f. We should select a chain, having higher breaking strength than the calculated value. For pitch value of 15.875 mm. DR 50 Rolon chain may be selected whose breaking strength is 4440 kg f from P.S.G. Data book, Pg.No. 7.72 Now the chain is checked for its actual F.O.S and for allowable stresses. (from P.S.G. Data book, Pg.No. 7.78) Actual F.O.S, [n ] 

Q P

where  P  P t  P c  P s Now P t  Tangential force du e to power transmision Pt 

102 N 102  11  v 6.41


1.169

 175.03 kgf P c  centrifugal tens io n in Ne wtons 

W v2 g



17.8  6.412 9.81

 74.55 kgf  W  W eight per metre   length  17.8 N from P.S.G.   Data book, Pg.No. 7.72

    

Ps  Tension due to sagging of chain in Newtons  k.W.a , where, k  sagging coefficient

from P.S.G. Data book Pg.No. 7.78, position of chain inclined up to 40 assumed hence k  4 a  0.5 m given W  17.8  P s  4  17.8  0.5  35.6 N   P  175.03  74.55  35.6  285.18

1.170


Hence [n ] 

Q 4440   15.56  P 288.18

Since the actual F.O.S is more than minimum value, chain selection is correct.

adopted

ie 15.56  11 Checking for induced bearing stress (from P.S.G. Data book, Pg.No. 7.77) The Induced bearing stress over the projected area of the chain is given by 

N  ks Av

A  Bearing area

from P.S.G. Data book Pg.No. 7.72, corresponding to DR 50, Bearing area  140 mm 2 This is less than the allowable bearing stress whose value for k s  1.875 , Z 1  25 and N 1  970 (from P.S.G. Data book, Pg.No. 7.77) is 42 N/mm 2 Hence selection of chain is correct. Now length of chain in terms of pitches is obtained as  Z2  Z1    Z 1  Z 2 2   lp  2ap  ap 2

2

from P.S.G. Data book Pg.No. 7.75


1.171

But ap 

a0 p



500  31.5 from P.S.G. Data book Pg.No. 7.7 5 15.875

Hence  73  25     25  73   2   lp  2  31.5    2 31.5  

2

 112.24

Length may be corrected to 112 pitches  Actual length of chain (from P.S.G. Data book, Pg.No. 7.75)  112  15.875  1778 mm

Now corrected centre distance e a  Pg.No.7.75)

e2  8 m   4

 p 

(from

P.S.G.

Data

book,

 Z1  Z2  But e  lp   2    25  73   112     63 2   m  58.4 , for Z 2  Z1  73  25  48 from P.S.G. Data

book, Pg.No. 7.76

1.172


Hence    a   63  

 63  8  58.4    15.875 4 

  2

 484.8 mm

Centre distance (decrement) allowance to accommodate initial chain sag (from P.S.G. Data book, Pg.No. 7.75)   0.5 F

But

F  0.02 a    0.5  0.02 484.8 

 4.848 mm Hence, Required centre distance  484.8  4.848  479.95 mm

(from P.S.G. Data book, Pg.No. 7.78) Now, pitch diameter of pinion sprocket d d



p  180  sin  Z1    15.875  127 mm  180  sin    25 

Pitch diameter of wheel sprocket D


D

1.173

p 15.875   369 mm  180   180  sin  sin   Z2   73   

Specifications Type of chain: 10 A-2 DR 50 Rolon chain Centre distance: 479.95 mm Number of teeth of sprocket pinion: 25 Number of teeth of sprocket wheel: 73 Length of the chain: 1778 mm Pitch diameter of pinion sprocket: 127 mm Pitch diameter of wheel sprocket: 369 mm Problem 1.31: A blower is to run at 600 rpm. Power to the blower is available from a motor rated 8 kW at 1500 rpm. Design chain drive for the system if the centre distance is to be 800 mm.

Solution Given data Power N  8 kW Speed of motor N 1  1500 rpm Speed of blower N 2  600 rpm Centre distance a   800 mm Step 1 Calculate of transmission ratio i i

Z2 Z2



N1 N2



1500  2.5 600

1.174


Step 2 From P.S.G data book, Pg.No. 7.74 for i  2 to 3, Z 1  No. of teeth on pinion  25 to 27 Take Z 1  27 teeth Since Z 2  No. of teeth on wheel  i Z1  Z 2  2.5  27  68 teeth

Step 3 Fro m P.S.G Data bo o k, Pg.No. 7.74 Optimum c entre distanc e  a  30 p to 50 p where p  pitch 800  26.66 Pitchmax  pmax  30 800 Pitchmin  pmin   16 50

Select standard Pg.No. 7.72

pitch

from

P.S.G

Data

book,

Take standard pitch, between 16 mm to 27 mm  select pitch 19.05 mm  p  19.05 mm

Step 4 Selection of chain No. Select roller chain from P.S.G Data book, Pg.No. 7.72 from the available chain No. Select chain No. 12A - 1 (simpler roller chain) p  pitch  19.05 mm


1.175

Bearing area  A  1.05 cm 2 Weight per meter  W  1.47 kgf Breaking load  Q  3200 kgf Step 5 From P.S.G Data book, transmitted on breaking load N

Pg.No.

7.77,

power

Qv 102  N 1  k s

N  8 kW V

p  Z1 N1 19.05  27  1500  60  1000 60  1000



19.05  27  1500  12.85 m/s 60  1000

n  minimum value of Factor of safety from P.S.G

Data book, Pg.No. 7.77 Select n  13.2 corresponding to speed of rotation of small sprocket, rpm Since specific conditions are not given, assume K 1  K2  K 3  K 4  K 5  K 6  1  KS  K1 K 2 K3 K 4 K5 K 6  1 8

Q  12.85 102  13.2  1

 Q  838.22 kgf which is less than the selected

Breaking load of 3200 kg f.

1.176


Step 6 From P.S.G Data book, Pg.No. 7.75 length of continuous chain in multiples of pitches  Z2  Z1   Z1  Z2  2   lp  2 a p   ap 2

where ap 

2

ao p

a o  initially assu me d centre distance in mm  800 mm 800  41.99  ap  19.05

 68  27    2  27  68    lp  2 41.99   41.99 2

2

 131.48 mm  132 mm

Length of chain  l  lp  p  132  19.05  2514.6 mm

From P.S.G Data book, Pg.No. corresponding to Z 2  Z1  68  27  41  m  42.6

From P.S.G Data book, Pg.No. 7.75

7.76,

take

m


final centre distance ‘a’  but e  lp 

e2  8 m e  p 4

Z1  Z 2

 132 

a

1.177

2 27  68  84.5 2

84.5   84.52  8 42.6    19.05 4

 803.9 mm  804 mm

Step 7 Check the actual factor of safety from P.S.G Data book, Pg.No. 7.78 Actual factor of safety  [n] 

Q P

Q  breaking load of the cha in  3200 kg f

 P  Pt  P s  P c where P t  tangential force due to pow er transmission 

102 N 102  8   63.5 kg f v 12.85

P c  centrifugal tension 



W v2 g 1.47  12.85 2 9.81

 24.74 kg f

1.178


P s  tension due to sagging  k  W  a

from P.S.G Data book, Pg.No. 7.78 k  coefficient of sag  6 Horizontal a  796 mm  0.796 m  P s  6  1.47  0.796  7.02 kg f   P  63.5  24.74  7.02  95.26 kg f  Actual factor of safety [n]  

Q P

3200 95.26

 33.59  13.2 greater than allowable FOS

Step 8 Checking of Allowable bearing stress from P.S.G Data book, Pg.No. 7.77 allowable bearing stress    1.85 kg f/mm 2 Power transmitted on basis of allowable bearing [] Av stress  N  102 k s 8

[]  1.05  12.85  []  60.47 kgf/cm 2 102  1

or []  0.6047 kgf / mm 2  1.85 kgf / mm 2


1.179

(i.e.) allowable bearing stress  The design is satisfactory. Problem 1.32: A compressor is to be actuated from a 10 kW electric motor. The speed of the motor shaft is 1000 rpm and the compressor speed being 350 rpm. The minimum centre distance is 500 mm. The compressor operates 16 hours per day. Design a suitable chain drive.

Solution Given P  10 kW ; N1  1000 rp m, N 2  350 rp m , a  500 mm  0.5 m, servic e  16 hours/day.

Let the operating chain may be roller chain from P.S.G Data book, Page No. 7.74, the optimum centre distance is 30 to 50 pitches. p is the pitch of the chain.

(ie) pmin 

pmax 

a 500   16.36 mm 30 30

500  11 50

select standard pitch betn (11 to 16.6) From P.S.G Data book, Page No. 7.72, The next standard pitch value, p  15.875 m From P.S.G Data book, Page No. 7.74 Transmission ratio i 

N1

N2



1000  2.857 350

1.180


for i  2.857 , the number of teeth on pinion-sprocket, from P.S.G Data book, Page No. 7.74, Z 1  25 As su med

then the number of teeth on wheel-sprocket, Z 2  iZ 1  2.857  25  71.425 ~  72

The load applied on the chain due to transmitted power is given by. Q

N  ks  n v

where N  power to be transmitted  10 kW  10,000 watts v  chain velocity in m/s n  Factor of safety to be assumed k s  Service factor

Now service factor ks  k1  k2  k3  k4  k5  k6

From P.S.G Data book, Page No. 7.76 k 1  1.5

(Load with heavy shock)

k 2  1.0

(Adjustable supports)

k 3  1.0

ap  30 to 50p


1.181

k 4  1.0

(Assume the line joining the centres of the sprockets is inclined upto 60 to the horizontal)

k 5  1.0

(Assuming drop lubrication)

k 6  1.25

(16 hours/day service) k s  1.5  1  1  1  1  1.25



 1.875

From P.S.G Data book, Page No. 7.77 Let the minimum factor of safety, n  11 for ks  1 and Z 1  25  for

ks  1.875 and Z 1  25 n  11  1.875  20.625

Chain velocity, V  

Z 1N 1p 60  1000 25  1000  15.875 60  1000

 6.614 m/s

Hence the minimum breaking load, Q

N  n  ks 10,000  20.625  1.875  v 6.614  58,469.72 N

Since the minimum breaking load (ie., developed tangential force to break the chain) is 58,469.72 N , we

1.182


should select a chain having higher breaking strength than the calculated value. From P.S.G Data book, Page No. 7.72 for the pitch value of 15.875 mm, DR 50 Rolon chain may be selected whose breaking strength is 66600 N Now the chain is checked for its actual factor of safety and for allowable stress. Checking for actual factor of safety: From P.S.G Data book, Page No. 7.78 Actual factor of safety, [n]  where and

Q P

P  P t  P c  P s Q  Breaking strength of selected chain

Now P t  Tangential force due to power transmission N   Newtons  v  

10000  1575 Newton 6.35

F c  Centrifugal tension in Newtons 

Wv 2 g

From P.S.G Data book, Page No. 7.72, W  weigth per metre length  30.2 N


 Pc 

1.183

30.2  6.352 9.81

 124.132 N P s  Tension due to sa gging of chain in newtons kWa

From P.S.G Data book, Page No. 7.78 where k  Sagging coefficient  4 (position of chain drive inclined upto 40)  P s  4  30.2  0.5  60.4 N

P  1575  124.132  60.4



 1759.53 N

Hence

[K n]  

Q P 66600  37.85 1759.53

Sinc e the ac tual fac to r o f safety is mo re than ado pted minimum value,c hain selec tio n is c o rrec t. Chec kingfo rinduc ed bearing stress: The induc ed bearing stress o ver the pro jec ted area o f the c hain is given by 

N  ks Av

from P.S.G Data book, Page No. 7.72

1.184


A  B earing area  210 mm 2 



10,000  1.875 210  6.35

 14.06 N/mm 2

This is less than the allowable bearing stress whose value is 28.7 N/mm 2 from P.S.G Data book, Page No. 7.77. Hence the chain selection is correct. From P.S.G Data book, Page No. 7.75 Now the length of chain interms of pitches is obtained as,  Z2  Z1   Z1  Z 2  2   L p  2ap  ap 2 But

ap 

a0 p



2

500  31.5 15.875

Hence,  72  25   2  25  72    L p  2  31.5  2 31.5

2

 113.2

Length may be corrected to even number such as L p  114 pitches.  Actual length of chain l  lp  p


1.185

 114  15.875  l  1810 mm

From P.S.G Data book, Page No. 7.75 Now the correct centre distance, e a 

e2  8m   4

 p 

where  Z1  Z2  e  lp    2    25  72   114    2    65.5 ~  66

From P.S.G Data book, Page No. 7.76 Corresponding to Z 2  Z1  72  25   47 m  56

Hence  66  a 

662  8   56   4

   15.875 

 510 mm

from P.S.G Data book, Page No. 7.75 Centre distance (decrement) accommodate initial chain sag, But F  0.02 a

allowance

to

1.186


  0.5F  0.5  0.02  a  0.5  0.02  510  5.1 mm

Hence required centre distance  510  51  505 mm

From P.S.G Data book, Page No. 7.78 The pitch diameter of pinion sprocket, d d



15.875 p   180   180   sin  sin   Z1   25   

d  127 mm

and The pitch diameter of wheel sprocket, D D



15.875 p   180   180   sin  sin   Z2   72   

D  364 mm

Specifications Type of chain  10 A  2 DR 50 Rolon chain Centre distanc e  505 mm number of teeth of sprocket pinion  25 number of teeth of sprocket wheel  72


1.187

Length of chain  1810 mm Pitch diameter of pinion sprocket  127 mm Pitch diameter of wheel sprocket  364 mm Problem 1.33: Design a chain drive to actuate a compressor from a 10 kW electric motor at 960 rpm. The Compressor speed is to be 350 rpm. Minimum center distance should be 0.5 m. Compressor is to work for 8 hours/day.

Given data Power of motor N  10 kW Speed of motor N 1  960 rpm Speed of the compressor N 2  350 rpm Minimum centre distance a  500 mm Solution Step 1: Calculation of transmission ratio From PSG Data book, Pg.No: 7.74 Transmission ratio i 

 i 

Z2

Z1 N1

N2





N1 N2 960  2.74 350

Step 2: Number of Teeth From PSG Data book, Pg.No. 7.74 For i  2 to 3; Z 1  25 to 27 i  3 to 4; Z 1  23 to 25

Take Z 1  25 to 27 (select any odd number of teeth)

1.188


Select Z 1  27 teeth (number of teeth on sprocket pinion) Z 2  i Z 1  2.74  27  74  Z 2  73.98 ~  Z 2  74 teeth

Step 3: Selection of pitch From PSG Design data book, Pg.No: 7.74 Optimum centre distance a  (30 to 50) p Pitchmax 

500  16.66 30

Pitchmin 

500  10 50

Select standard Pg.No. 7.74

pitch

from

PSG

Data

book

Take any standard pitch between 10 mm and 16.66 mm  select pitch p  15.875 mm

Step 4: Selection of chain number from PSG Data book, Pg.No. 7.72 The available chain number are 10 A and 10 B Select 10 A  2 Duplex chain from PSG Data book, Pg.No. 7.72 corresponding to 10 A  2 and Pitch = 15.875 mm A  Bearing area  1.4 c m2


1.189

W  Weight per m etre length  1.78 kgf Q  Breaking load  4440 kgf

Step 5: Selection of Breaking load From PSG Data book, Pg.No. 7.77, Power transmitted on the basis of breaking load N

Q.v 102n K s

We know that v 

Z1 p N1 27  15.875  960  60  1000 60  1000

v  6.858 m/s

Since specific conditions are not given in the problem, assume K1  K2  K3  K4  K5  K6  1  Ks  1

from PSG Data book, Pg.No. 7.77, corresponding to ks  1, Z 1  15 to 30, pitch p  15.875 , N 1  960 rpm  1000 rpm  n  11 allowable factor of sa fety Q

102n k s N 102  11  1  10  6.858 u

 Q  1636.04 kg f 

selected chain Breaking load

4440 kg f

Step 6 (a) calculation of length of chain (b) final centre distance

1.190


(a) Calculation of length of chain From PSG Data book, Pg.No. 7.75, Length of continuous chain in multiples of pitches lp  Z2  Z1   Z1  Z2  2    2ap  2 ap

where ap 

2

ao p

500  ap  15.875  ap  31.49  74  27    27  74   2   lp  2  31.49     31.49 2  

2

 62.98  50.5  1.776  115.25  116 approximated to 116   from PSG Data book,

Length of chain l  lp  p  116  15.875  1841.5 mm

Take

l  1840 mm

(b) Final centre distance from PSG Data book, Pg.No. 7.75 Final centre distance corrected to even number of pitches ‘a’


a

e

e2  8 m  4

1.191

p

 z1  z2   27  74  e  lp     116    2 2    

where

e  65.5 mm 2

 z2  z1   74  27  m   2   2  

and

2

m  55.95  Final centre distance ‘a’ 

65.5 

65.52  8  55.95   4

 15.875  505.97

 a~  506 mm

Step 7 Check the actual factor of safety From PSG Data book, Pg.No. 7.78 Actual factor of safety [n] 

Q P

where Q  breaking load of ch ain  4440 kg and  P  Pt  P s  P c where

Pt 

tangential

force

transmission from PSG Data book, Pg.No. 7.78 Pt 

102 N v

due

to

power

1.192


 Pt 

102  10 6.858

 148.73 kg f

and

P c  centrifugal tension

from PSG Data book, Pg.No. 7.78, Pc 

2 W v2 1.78  6.858   8.533 kgf  g 9.81

and P s  tension due to sagging of chain from PSG Data book, Pg.No. 7.78 Ps  k W a

where k  coefficient of sag from PSG Data book, Pg.No. 7.78, corresponding to position of chain drive as horizontal k6 506  5.404 kgf  P c  6  1.78  1000   P  148.73  8.533  5.404  162.66 kg f  [n] 

Q 4440   27.29  11  P 162.66

which is greater than allowable factor of safety  The design is safe.

Step 8 Checking of Allowable bearing stress From PSG Data book, Pg.No. 7.77, corresponding to pitch = 15.875 mm, speed less than 1000 rpm


1.193

[]  2.24 kgf/mm 2

From PSG Data book, Pg.No. 7.77, Power transmitted on the basis of allowable bearing stress N



 Av 102 k s 102N k s 102  10  1  AV 1.4  6.858

 106.236 kgf/cm 2  1.062 kg f/mm 2  2.24 kgf/mm2

  []  The design is safe

Step 9 Pitch diameter of small sprocket d d

15.875 p  180  180  sin sin Z1 27

d  136.74 mm

Pitch diameter of large sprocket D D

15.875 p  180 180  sin sin 74 Z2

D  374 mm

Chapter 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.

2.1 INTRODUCTION Gears are defined as toothed wheels, that transmit power and motion from one shaft to another due to the successive engagement of teeth. 2.1.1 Advantages of Gear drives 1. It is a positive drive and the velocity ratio remains constant. 2. The drive is compact, since the centre distance between the shafts is relatively small. 3. It can transmit more power. 4. The efficiency of gear drive is very high (99% in case of spur gears) 5. In the gear box, shifting of gears is possible. Hence, velocity ratio can be changed over a wide range.

2.2


2.1.2 Disadvantages 1. Gear drives are costly. 2. Maintenance cost is high. 3. Manufacturing complicated.

processes

for

gears

are

4. Gear drives require careful attention. 5. Accurate alignment of shafts is required.

2.2 CLASSIFICATION OF GEARS Gears are broadly classified into four groups; 1. Spur gears 2. Helical gears 3. Bevel gears and 4. Worm & Worm wheel 1. Spur gears: The teeth are cut parallel to the axis of the shaft. Spur gears are used, when the shafts are parallel. 2. Helical gears: The teeth of these gears are cut at an angle  with the axis of the shaft. Helical gears impose radial and thrust loads on the Fig:2 .1 S pur G ea r shafts. To eliminate the end thrust load, double helical gears are used. These gears are called as Herringbone gears. The construction results in equal and opposite thrust reactions, balancing each other and there is no thrust load on the shaft.

Spur Gears & Parallel Axis Helical Gears

2.3

Fig 2.2 (a) Helical Fig 2.2 (b) D ouble Helical gear

Helical and Herringbone gears are used, when the shafts are parallel. 3. Bevel gear: Bevel gears are used for shafts whose axes intersect each other. The angle can be 90, less than 90 or more than 90. The tooth of the bevel gears can be cut straight or spiral.

Fig. 2.3 Bevel gear

2.4


4. Worm and Worm wheel: Worm and Worm wheel are used for shafts, whose axes do not intersect and are perpendicular to each other. Worm gears are mainly used for high speed reduction ratio. The worm is in the form of a threaded screw, which meshes with a wheel.

Fig 2.4 W orm and Wo rm wheel

2.3 TERMINOLOGY OF SPUR GEAR 1. Pitch Circle: It is an imaginary circle which allows pure rolling action without slip. 2. Pitch Circle diameter (PCD): (d) It is the diameter of the pitch circle. The size of the gear is usually specified by pitch circle diameter (d). 3. Pitch point: It is a common point of contact between two pitch circles. 4. Base circle: It is an imaginary circle from which the involute curve of the tooth profile is generated. 5. Pressure angle or angle of obliquity: : It is the angle between the common normal to two gear teeth at the point of contact and the common tangent at the pitch point. The standard pressure angles are 14 12  and 20. 6. Circular Pitch: Pc It is the distance measured along the pitch circle between two similar points on adjacent teeth.


D ede nd um

ce F a th w id

A dde ndu m

2.5

A dde ndu m circle Top la nd P itch su rfa ce Fa

Tota l de pth W orking D epth

F

ce

P itch circle

k la n

Too th spa ce C ircu lar p itch Too th th ickne ss R oot or D ede nd um circle

C lea ran ce C lea ran ce or w orking d epth circle

Fig: 2.5.Term s used in gears

Pc  

d Z

7. Diametral pitch: P d It is the ratio of the number of teeth to the pitch circle diameter. Pd 

Z d

Pc  Pd  

8. Module: m: It is defined as the reciprocal of diametral pitch. (or) It is the ratio between pitch diameter and the number of teeth. m

1 1 d    P d Z  Z  d

Module is represented in millimeter.

2.6


9. Gear ratio i: It is the ratio of number of teeth on the gear to the number of teeth on pinion.  Z 1  No . of teeth o n pinio n  Z 2 N1  Z 2  No. of teeth on gear  i Z 1 N2   N 1  Speed o f pinion  N  Speed of gear  2 10. Transmission ratio: It is the ratio of the angular speed of the first driving gear to the angular speed of the last driven gear in a gear train.

2.4 GEAR TOOTH FAILURES There are two basic modes of gear tooth failure 1. Breakage of the tooth due to the static and dynamic loads 2. Surface distruction or tooth wear. The breakage of the tooth can be avoided by adjusting the parameters in the gear design, such as module and face width. The various types of gear tooth wear are as follows: 1. Abrasive wear: The dust particles in the lubricant, dirt, rust, metallic debris can scratch the tooth surface. It can be prevented by providing oil filters, increasing surface hardness and use of high viscosity oil. 2. Corrosive wear: The corrosion of the tooth surface is caused by extreme pressure of additives present in the lubricating oil and due to external contamination. It can be prevented by providing complete enclosure of gears from external contamination.


2.7

3. Initial pitting: It is caused due to errors in the tooth profile, surface irregularities and misalignment. It can be prevented by precision machining of gears, adjusting the correct alignment of gears and reducing the dynamic loads. 4. Destructive pitting: It is a surface fatigue failure which occurs when the load on gear tooth exceeds the surface endurance strength of material. It can be prevented by designing the gears so that the wear strength of the gear tooth is more than the sum of static and dynamic loads. 5. Scoring: This is due to excessive surface pressure, high surface speed and inadequate supply of lubricant. Due to the break down of the oil film, excessive frictional heat leads to overheating of the meshing teeth. Scoring can be avoided by selecting the parameters, such as surface speed, surface pressure and the flow of lubricant in such a way that the resulting temperature at the contacting surfaces is within permissible limits.

2.5 LAW OF GEARING For a constant velocity ratio, the profile of the teeth must be such that the common normal to the profiles at any point of contact always intersects the line of centres at the same point, which is called the pitch point, which is the tangency of their pitch circles. Forms of Teeth 1. Involute teeth 2. Cycloidal teeth. 

An Involute of a circle is a plane curve generated by a point on a tangent, which rolls on the circle


2.8

V c1 1

O1

on m m al o C rm No

V c2 N C

M

2 O2

Fig: 2.7 Law o f gearing

without slipping. (Normal at any point of an involute is a tangent to the circle). 

A cycloid is the curve traced by a point on the circumference of a circle which rolls without slipping on a fixed straight line.

2.5.1 Advantages of Involute gears 1. The centre distance for a pair of involute gears can be varied within limits without changing the velocity ratio. But incase of cycloidal gears, the exact centre distance should be maintained. 2. In involute gears, the pressure angle, from the start of the engagement of teeth to the end of the engagement remains constant. Due to this condition, the drive is smooth and also the wear of gears is less. In cycloidal gears, the pressure angle is maximum


2.9

at the beginning of the engagement, reduces to zero at pitch point, starts increasing and again becomes maximum at the end of the engagement. This results in less smooth running of gears. 3. The face and flank of involute teeth are generated by a single curve. Therefore, the manufacturing of involute teeth are easy. In case of cycloidal teeth, double curves (Epi cycloid for face and Hypo-cycloid for flank) are required. Therefore, the manufacturing of cycloidal teeth are difficult. 2.5.2 Disadvantages of Involute Teeth 

The main disadvantage of the involute teeth is that the interference occurs with pinions having smaller number of teeth. This may be avoided by altering the heights of addendum and dedendum of the mating teeth. (Interference: When the tip of the tooth of pinion undercuts the root on its mating gear, it is known as Interference)



Interference may be prevented, if the addendum circles of the two mating gears cut the common tangent to the base circles between the points of tangency. (OR) The point of contact between the two teeth is always on the involute profiles of both the teeth.

2.5.3 Advantages of Cycloidal Gears 1. The cycloidal teeth have wider flanks, therefore for the same pitch, cycloidal gears are stronger.

2.10


2. Cycloidal gears have less wear, because the contact takes place between a convex flank and concave surface. 3. In cycloidal gears, the interference does not occur at all. 2.5.4 Disadvantages 

Manufacturing of cycloidal gear is difficult.



Therefore, due to simplicity during manufacturing and flexibility, the involute gears are generally used.

2.6 SPEED RATIO AND NUMBER OF TEETH Speed ratio is defined as the ratio of the speed of the faster gear (smaller size gear) to the slower gear (larger size gear). It is also called as the gear ratio (i), which is the ratio of number of teeth of large gear to smaller gear. It should be always greater than unity (1) Z 2 =6 0 To oth G ea r

d1 Z 1 =3 6 To oth G ea r

1 (Pinion )

Fig 2.8

d 2 (ge ar)

d2


i

where

N1 N2



Z2 Z1



2.11

d2 d1

Z 1  number. of teeth on smaller pinion Z 2  number. of teeth on larger gear N 1  speed of small pinion gear in rpm. N 2  speed of large gear in rpm.

2.7 MINIMUM NUMBER OF TEETH The minimum number of teeth on pinion to avoid interference is given by Z min 

2 sin 2 

  20 [full depth involute system]

Assume number of teeth as 18 to 20 for safe design.

2.8 FORCE ANALYSIS [SPUR GEAR] The power is transmitted by means of a force exerted by the tooth of the driving gear on the engage (meshing) tooth of the driven gear. According to the law of gearing, this resultant force F must always act on the pressure line. The pressure line is the tangent line which is tangent to the base circles of the gear and pinion. Since the direction of resultant force F is inclined at an angle  called as pressure angle, the resultant force can be resolved at the pitch point into two components such as.

2.12


1. Tangential component of force F t 2. Radial component of force F r 1. Tangential component of force Ft F t is the useful force, because it is employed to

transmit power. The transmitted power and torque can be calculated by using F t. Mt

2 Mt

Tangential Force,

Ft 

Radial Force

F r  F t  tan 

Resultant Force

F 

d/2



d

F 2r F 2t    

(Or) F  

Ft cos  O1

Pre ssure line

p in ion d river F Pitch po int 

G ear d riven

Fig 2.9 (a)

O2


Power transmitted P   F t  v

Pinion d river

v  pitch line velocity v

O1

 dN ; 60

Ft F

2. Radial component of force F r Radial component of force F r, is a separating

Pitch p oin t





Fr Fr

F

Pitch p oin t

Ft G ea r d riven

force, since it is always directed towards the center of gear, tries to separate the gears. The F r does not do

O2

any useful work. This force F r produces bending moment on the gear shaft, so it is also called as bending force.

Fig:2.9. (b)

where P  Power transmitted M t  Torque transmitted d  Pitch circle diameter of driving gear N  Speed of driving gear in rpm

  Pressure angle v  Linear velocity

Torque Transmitted M t 

60 P 2 N

2.13

2.14


[Note: The tangential force should be same for both pinion and gear wheel] The analysis of gear tooth is based on the following assumptions 1. The effect of dynamic load is neglected. 2. Friction losses in the bearings and gears are negligible. 3. The gears mesh at the pitch circles. 4. The shafts for pinion and gear are parallel.

2.9 TOOTH STRESSES (THE LEWIS BENDING EQUATION) Wilfred Lewis introduced an equation for estimating the bending stress in gear tooth. This equation formulated in 1892 still remains the basis for most gear design today. In the Lewis analysis, the gear tooth is treated as a cantilever beam and the tangential component F t causes

b

t

Ft ............................ . . .................................................................................. .... ............................................. .............................................. .. ................ . ...... ...... .................. ...... ................... ..... .... h

Fig. 2.10. G ear tooth as cantilever

the bending moment about the base of the tooth. The Lewis equation is based on the following assumptions. (i) The effect of radial component F r which induces compressive stress is neglected. (ii) It is assumed that the tangential component F t is uniformly distributed all over the face width of


2.15

the gear (this is possible when the gears are rigid and accurately machined). (iii) The effect of stress concentration is neglected. (iv) It is assumed that at any time, only one pair of teeth is in contact and takes total load. It is observed that the cross section of the tooth varies from free end to fixed end. Therefore, a parabola is constructed within the tooth profile and shown in Fig. 2.11 as dotted lines. Gear tooth as parabolic beam Fr

The advantage of parabolic out line is that it is a beam of uniform strength, and the stress at any cross section is uniform.

O

Ft h

X

X

We know

t

Bending Moment, M b  F t  h...(1)

N Se ctio n at XX

bt 3 t I and y 1  12 2

Fig:2.11

b A

y1  maximum distance between center line and

the extreme fibre.  b 

Mb y I



Mb  I  y   1



bt 3 bt 2 12  Z m  Section modulus  t 6 2

Mb Zm

2.16


b  Permissible bending stress (N / mm2)   bt 2   b   Ft   6h  

6F t  h bt 2

...(2)

Multiplying the numerator and denominator of the right hand side by P c (circular pitch)  t2  F t  P c b b   6hP c  

The bracketed quantity depends on the form of the tooth and is termed as Lewis factor or form stress factor y. Let

y

t2 6hP c

Then the equation can be rewritten as F t  P c b b y ...(3) Where, y is called as Lewis form factor where h  length of the tooth b  maximum bending stress I  moment of Inertia b  face width t  thickness of tooth P c  circular pitch y  Lewis form factor


2.17

When the stress reaches the permissible magnitude of bending stresses, the corresponding force F t is called the beam strength F s F s  [ b] P c by  [ b] by  m

where,

. .  . P c    m Lewis Equation

F s  beam strength of gear tooth (N) (or) kg f [b]  Permissible bending stress N/mm2 (or)

Allowable static stress kg f / cm2

The above equation is known as LEWIS EQUATION From PSG Data book Page: 8.50 y  0.124 

0.684 for 14 12 full depth Involute Z

 0.154 

0.912 for 20  Involute Z

 0.175 

0.95 for 20  Involute subteeth Z

In order to avoid the breakage of gear tooth due to bending, the beam strength should be more than the effective force between the meshing teeth. In design of gears, it is required to decide the weaker force between pinion and gear. When the same material is used for pinion and gear, the pinion is always weaker than the gear . .  . Pc    m 

We know that F s  [b] b y P c



It can be observed that ‘m’ and ‘b’ are same for pinion as well as gear in a gear pair,

2.18




When different materials are used, the product  b  y decides the weaker one between the pinion and gear



The Lewis form factor y ia always less for pinion compared to gear



Therefore, when the same material is used for pinion and gear, the pinion is always weaker than the gear and hence design should be based on pinion.

Effective load-Calculation To determine the tangential component resultant force between two meshing teeth,

of

the

This component can be calculated by using (I) M t 

(II) F t 

P  60 , and 2 N 1 2M t d1

The value of the tangential component, depends upon rated power and rated speed. We know that, [b] is permissible static bending stress which is

modified to C v b where, C v is the velocity factor considered in fatigue loading. This velocity factor C v developed by Carl. G. Barth, can be expressed in terms of Pitch line velocity.


2.19

The values of velocity factor are as given below (i) C v 

3  vm 3

, For ordinary and commercially cut

gears (made with form cutters) and vm  10 m/sec (ii) C v 

6  vm 6

, For accurately hobbed and generated

gears, vm  5  20 m/sec. (iii) C v 

5.5  vm1/2

, For precision gears 5.5 shaving grinding and lapping and vm  20 m/sec (iv) C v 

1  vm 1  0.25 vm 

with

, For non-metallic gears.

Where, vm  Pitch line velocity (m/sec) 

 dN 60  10 3

d in mm

(The velocity factor is an empirical relationship developed by past experience).

2.10 DYNAMIC EFFECTS (DYNAMIC TOOTH LOAD) When gears rotate at very low speed, the tangential tooth load F t can be considered to be the actual force present between two meshing teeth. However, the most of the cases, the gears rotates at appreciable speed and it becomes necessary to consider the dynamic force resulting from impact between mating teeth.

2.20


The Dynamic force is induced due to the following factors. 1. Inaccuracies of the tooth profile, 2. Errors in tooth spacings, 3. Misalignment between bearings, 4. Elasticity of parts, and 5. Inertia of rotating masses. These are two methods to account for Dynamic load. I. Approximate estimation by the velocity factor in the preliminary stages of gear design II. Precise estimation by Buckingham’s equation in the final stages of gear design. Note: Approximate estimation, using velocity factor C v developed by C.G. Barth was discussed earlier. In the final stages of gear design, when gear dimensions are known, errors specified and quality of gears determined, the Dynamic load is calculated by equation derived by Buckingham. Where, F d  Dyn amic load  Ft  Fi

Where F t Tangential tooth load F i  Increment load due to dynamic action   0.164 V m cb  F t Fd  Ft     cb  F t   0.164 V m  1.485 

Where Vm  Pitch line Velocity (m/min)


2.21

c  Dynamic factor kg f/cm 2 depending upon

machining errors e  measured error in action between gears in cm b  face width of tooth (cm) F t  tangential force due to rated torque kg f

The Dynamic factor c, depends upon modulus of elasticity of materials for pinion and gear and the form tooth or pressure angle and it is given by c

e 1   1  K E1 E2   

Where, K  Constant depending upon the form of teeth. E 1  Modulus of elasticity of pinion material

kgf/cm2

E2 

Modulus of elasticity of gear material

kgf/cm2

The error, e in the dynamic load equation is measuring error in action between gears in (cm) This error depends upon the quality of gear and manufacturing methods.

2.11 WEAR TOOTH LOAD (OR) FATIGUE STRENGTH OF GEAR WEAR For gears, wear is defined as loss of material from contacting surfaces of teeth.

2.22


It is further classified as 

Normal wear



Moderate wear



Destructive wear



Abrasive wear



Scratching wear, etc.



Generally, normal wear (Polishing in) does not constitute failure because it involves loss of metal at a rate too slow to affect performance



Moderate wear refers to loss of metal more rapid than normal wear.



This need not necessarily be destructive and may develop on heavily loaded gear teeth.



Destructive wear usually results from loading that is excessive for the lubricant D riving to oth employed.



The effect of destructive wear on the tooth profile of an involute gear is depicted in the Figure 2.12.

P itch circle

P itch circle

D rive n to oth Fig:2.12

PITTING Pitting is the principal mode of failure of rolling surfaces. The details of the process vary with the material and operating conditions, but in all cases, it manifests itself by the initiation and propagation of cracks in the near


2.23

surface layer until microscope pieces detach themselves to form a pit or a spall. In spur gears, surface pitting has long been recognised as one of the failure modes. This is often referred to as “Pitch line Pitting” The main factors affecting pitting type of failure, 

Contact stress.



Material pouring and hardness.



Surface finish and lubrication

Contact stress was originally conceived By “HERTZ” in 1896 whose name it is often referred to as Hertz Contact Stress. The failure of the gear tooth due to pitting occurs when the contact stress between two meshing teeth exceeds the surface endurance strength of the material. In order to avoid this type of failure, the proportion of the gear tooth and surface properties such as surface hardness should be selected in such a way that the wear strength of the gear tooth is more than the effective load between the meshing teeth. The Hertz stress is based on the assumptions of electric and isometric material behaviours, load is compressive and normal to the contacting surfaces which are stationary and the size of contacting area whose dimensions are relatively smaller compared with the curvature radius of the contacting bodies The Figure 2.13 illustrates the contact area and corresponding stress distribution between two cylinders.

2.24


z

R2

R1

pO x

p r

L b

Y

Fig:2.13

Here the area of contact stress which is theoretically rectangular with one dimension being the cylinder length L . (i.e. corresponding to face width of the gear) The distribution of pressure is represented by a semi elliptical prism and the maximum contact pressure P 0 exists on the load axis. The gear design practice is to estimate the contact stress at the pitch point of the teeth by assuming line contact between two cylinders whose radius of contact depends on the gear geometry at the pitch point. The analysis of wear strength was done by Earle Buckingham and was accepted by AGMA (American Gear Manufacturing Association) in 1926. This Buckingham equation gives the wear strength of the gear tooth based on Hertz theory of contact stress. Hence, the maximum tooth load from wear consideration as evaluated from Hertz contact stress equation applied for pitch point contact is given by


2.25

F w  d 1 bQk

Where d1  Pitch circle diameter of pinion in (cm) b  Face width of the pinion in (cm) Q  Ratio factor 

2i i1

[ for external  for internal]



2Z 2 Z2  Z1

Load stress factor also known as material combination factor in kg f/cm2 k

This load stress factor depends upon the maximum fatigue limit of compressive stress, the pressure angle, and the modulus of elasticity of the materials of the gear. According to Buckingham theory, this load stress factor is given by k

c2 sin   1 1  E E  1.4 2   1

Where,  c  Surface endurance limit of a gear pair in

kg f/cm2.

2.12 FACTOR OF SAFETY [F.O.S] Factor of safety is defined as the ratio of the ultimate stress to working stress F.O.S 

Ultimate stre ss Working stress

2.26


F.O.S 

Material strength D esign Load

2.12.1 (i) Factor of safety for bending F.O.Sb 

Beam strength of gear tooth F s Dynamic load on gear tooth Fd

F.O.S for bending is in between 1.5 to 2. 2.12.2 (i) Factor of safety for pitting F.O.Spitting 

Wear strength of gear tooth F w Dynamic load on gear tooth F d

F.O.S for pitting is 2.

2.13 GEAR MATERIALS The gears are manufactured from both metallic and non-metallic materials. Some of the materials are given here. 2.13.1 Metallic materials 1. Cast Iron 2. Steel 3. Bronze, etc. 2.13.2 Non-metallic materials 1. Wood 2. Rawhide 3. Compressed paper 4. Synthetic resins like nylon etc.


2.27

2.13.3 Selection of gear materials The gear materials should have following properties. 1. High tensile strength to prevent failure against static loads. 2. High endulance strength to withstand Dynamic loads. 3. Low co-efficient of friction. 4. Good manufacturability. PSG Design Data book Page no. 1.40 MATERIALS FOR GEARS

Material

IS Specification

Tensile strength

Brinell

N/mm 2

Hardness number

Grey Cast Iron

Grade 25

250

197

Phosphor Bronze

Sand Cast Chill cast

160 240

60 70

Centrifugally cast

270

90

Cast Steel

Grade 1

550

145

St 58

580-680

160-190

C 60

600-700

180-200

Carbon steel

C 45

700

145 Core

for surface

C 55

720

460 Case 200 Core

Constructional steel Plain Carbon Steel

hardening Carbon steel for case hardening

520 Case C 14

500-750

650 Case

2.28


Material

IS Specification

Tensile strength

Brinell Hardness

N/mm 2

number

Direct

40 Ni 2 Cr 1

1500

44

hardening Alloy steels

Mo 28 30 Ni 4 Cr 1

1000-1150 1150

400-500 400-500

35 Ni 1 Cr 60 40 Ni 3

750-1050

400-500

Alloy Stells for case

17 Mn 1 Cr 95 15 i2 Cr 1 Mo

800-1100 1040

650 Case 630 Case

hardening

15 13 13 Ni 3 Cr

900-1200 1200-1500

600-620 Case 600-650 Case

80 15 Ni 4 Cr 1

600-850

650 Case

>660

750 - 800 Case

15 Cr 65 Nitriding Alloy Steel

40 Cr 2 A1 1 Mo 18

2.14 DESIGN OF STRAIGHT TOOTH SPUR GEAR BASED ON STRENGTH AND WEAR CONSIDERATION [LEWIS AND BUCKINGHAM EQUATION METHOD] Lewis equation is used to find beam strength of gear, whereas the Buckingham equation is used to determine the maximum dynamic load and wear strength of the gears. Step 1: For a given problem, note down the given parameters like power, speed, gear ratio, etc. Step 2: Selection of material If material is not given, select a suitable material for pinion and gear wheel based on - gear ratio (i) [From PSG data book pg no: 8.4]


2.29

Choose both pinion and gear as same material. (If material is not given). Pinion is the weekest materials. So design the pinion gear alone, because Lewis equation is applicable for weaker one of the two gears. If both pinion and gear are made up of different materials, determine the gear and pinion strength, separately. Strength of gear  b2  y2 Strength of pinion  b1  y1 where y1 and y2 are the Lewis form factor of pinion and gear respectively b1 and b2 are the bending stress of pinion and gear respectively. [Based on the weaker material, Design the gear drive] [From PSG Data book Pg: 8.5 table: 7] Step 3: Calculation of Tangential load F t F t  HP 

75 kgf vm

HP – Horse power vm  mean velocity in m/s vm 

 d1 N 1 m/s 60  1000

d1  mm ; N 1  rpm

2.30


 750 Watts ; 1 kg f ~  10 N (Newton)] [Note: 1 HP ~

Step 4: Calculation of Dynamic load F d F d  F t  Cv

in kg f

Already we have calculated the value of the F t and vm C v  velocity Factor (Refer PSG Data book Pg.No.8.51) vm  10 m/s wheels for commercially cut

 3  vm  Cv   ; 3  

wheels

 6  vm  Cv   ; 6  

vm  5  20 m/s for carefully cut wheels 1

5.5   vm  2 v  20 m/s for precision wheels m   Cv  5.5 Cv 

1  vm 1  0.25 vm

For non-metallic gears.

Step 5: Calculation of strength of gear tooth (or) beam strength F s F s  [ b] by P c

. . [ . P c    m]

 [ b ]  b  y    m m  module ; b  face width y  Lewis form factor based on P c


P c  circular pitch 

 d1 Z

2.31

(cm)

If b is not given, then assume b  3 to 4 times of P c

. . [ . b  3P c o r b  4P c]

where, [ b] is the design bending stress in kg f/cm2

but P c   m ;  3 m to 4  m b~ d  where, m  module    Z d1  Diameter of pinion (weaker gear) Z  no: of teeth on pinion. [If Z is not given, then assume Z  17 to 25 teeth]

If [ b] is not given, choose the approximate value from the PSG data book based on the material. [Table: 7, pg: 8.5] Step 6: Calculation of module: m For safe Design; F s  F d By Equating this two equations, determine the value of module. [Choose the standard module value from PSG Data book: Table 1, pg no: 8.2]

2.32


Step 7: Calculate the value of face width b b  3 Pc 3m  10 m b~

where, m  module Substitute the value of m, which is calculated from previous step 6. Step 8: Calculate pitch circle diameter d 1 d1  Z1 m

Step 9: Calculate pitch line velocity in m/min V

where,

 d1 N1 m/min m  1000 [V m in m/min and vm in m/s]

N 1  speed of pinion in rpm d1  pitch circle dia. of pinion in mm

Step 10: Recalculate the beam strength Fs (or) strength of gear tooth F s  [b] by P c  [ b ]  b  y    m

Substitute the newly calculated m value Step 11: Calculate by Buckingham’s Dynamic load F d Refer PSG Data book P.No. 8.51


2.33

 0.164 V m cb  F t  Fd  Ft    cb  F t    0.164 Vm  1.485 

where, b  face width in cm c  Deformation factor F t  Tangential load (or) transmitted load in kg f V m in m/min

[Choose the “c” value from PSG Data book page no: 8.53 from table 41 and 42. According to quality and module, select e - expected error in tooth profiles - mm from Table 42 and then table 41]. Step 12: Check for Beam Strength For safe, F s  Fd Compare the both values If F s  F d, the design is not satisfactory, then the values of “m ” and “b” should be increased to increase the strength of gear tooth. Step 13: Calculation of wear load Fw F w  d1 b Qk Q is Ratio factor Q

2i for external gear i1

Q

2i for internal gear i1

1   1  [2c ] sin   E1 E2    k 1.4

2.34


k

–

Load stress factor kgf/cm2

[c]



Design contact stress in kg f/cm 2

E1

–

Young’s modulus of pinion (kgf/cm2)

E2

–

Young’s modulus of gear (kgf/cm2)

[Choose E 1 and E 2 from the PSG Data book pg: 8.14 from the table: 9 based on the material]   Pressure angle. If  is not given; take   15  to 20  To Calculate [c] Select material and heat treatment process (surface hardness) Refer PSG Data book P.No.8.16 assume HRC value, take C R - value, [c]  CR  HRC  k cl

To find kcl assume life of drive 10,000 hrs to 20,000 hrs. Life   _____  hrs  rpm  60  _____ cycles Represent life  N  _____  10 7 cycles From P.No.8.17, according to material, and life in No. of cycles (consider surface hardness) take kcl value. Find [c] in kgf  cm2 Step 14: Check for wear strength For safe Design, F w  F d If F w  F d; Design is not safe. Step 15: Calculate the Basic Dimensions of spur gear


2.35

Dimensions of Externally Meshing Gears: Suffix 1 for Pinion and 2 for Wheel Nomenclature

Notation Units

Module

m

mm

Normal module

mn

mm

Transverse module

mt

mm

Centre distance

a

mm

Height factor

f0

–

Spur Gears m

2a Z1  Z2

am

Z1  Z2 2

f0  1 (in special f0  0.8, stub teeth)

Bottom clearance

c

mm

c  0.25m (For stub teeth, h  0.3m)

Tooth depth

h

mm

h  2.25 m (For stub teeth

Pitch diameter or Reference diameter

d

mm

d1  mZ1 ; d1  mZ2

Tip diameter

dc

mm

da1  Z1  2f0 m da2  Z2  2f0 m

Root diameter

df

mm

df1  Z1  2f0 m  2c df2  Z2  2f0 m  2c

Helix angle



deg

Number of teeth

Z

–

Virtual number of

Zv

teeth

Z1 

2a Z  iZ1 m i  1 2

2.36


Step 16: Draw the neat sketch of spur gear p itch circle L in e o f a ction O utside D ia Pre ssure a ng le

Too th profile (involute)

Ba se circle Pitch circle W h ole depth

W orking d ep th

C ircu lar too th thickness

C le aran ce

Ad de nd um

D e de nd um

C e nte r d istan ce

R o ot (to oth ) fille t

R o ot D ia

C ircu lar p itch Fig: 2.14

Note: The Above Design procedure is same for Helical gears, Worm and worm wheels and Bevel gears with slight modifications. Problem 2.1: Design and draw a straight spur gear drive to transmit 35 kW. The pinion speed is 720 rpm and the speed ratio is 2. Both the Gears are made up of the same surface Hardened Carbon steel with 55 RC and core hardness less than 350 BHN. Ultimate strength is 720 N/mm2 and yield strength is 360 N/mm2.

[A.U. May/June: 2007; Dec 2010]

Step 1: Given: Power P  35 kW  35  10 3 W


2.37

Speed ratio i  2 Gear material: Pinion and Gear are same material (carbon steel) with 55 RC and core Hardness < 350 BHN. Ultimate (stress) strength  u  720 N/mm2 Yield strength (stress)

 y  360 N/mm 2

Pinion speed

N 1  720 rpm

Step 2: Selection of Material Pinion –

Carbon steel surface hardened with 55 RC with core

Gear

Carbon steel Hardness less than 350 BHN

–

Both the gears are made up of same material, so pinion is the weaker one. Design the pinion wheel. Step 3: Calculation of tangential load F t F t  HP 

i

N1 N2



75 vm

Z2 Z1

Assume Z 1  20 2

Z2 20

Z 2  40 vm 

 d1 N 1 60  1000



d2 d1 d1  Z1 m

d1 . . . m Z1 d1  mZ1 . .  . d1  20  m

2.38




3.14  20  m  720 60  1000

vm  0.7539  m m/s vm ~  0.7  m m /s} P ower 

35  10 3 HP 750

[ 1 H.P ~  750 watts]

 46.666 ~  46.7  Ft  Ft 

46.7  75 0.7m

5003.57 kg f m

Step 4: Calculation of Dynamic load F d F d  F t  C v kg f

Assume vm  12 m/s for determining the velocity factor C v Cv 

6  vm [vm  5  120 6

m/s,

for

wheels, Refer PSG Data book Pg.No. 8.51] Cv 

6  12 6

Cv  3 ; Fd 

5003.57  3 m

Fd 

15070.71 m

carefully

cut


2.39

Step 5: Calculation of strength of gear tooth F s  [b]  b  y  P c

. .  . P c   m b  3 P c or

 [b]  b  y    m

b  10  m

[For carbon steel [b]  3200 kg f/cm 2 from PSG Data book Pg: 8.5 table: 7] Z1 

2 . .  . Z 1  20 

sin 2 

sin 2  

2 20

sin 2  

1 10

  sin

1

  

 1  10 

 

  18.43 ~  20 

then Lewis factor y  0.154 

0.912 Z1

For 20 involute (from PSG Data book pg: 8.50)

 0.154 

0.912  0.1084 20

y  0.1084 and [ b]  3200 kg f/cm 2 F s  3200  10  m  0.1084    m F s  10892.0320  m 2 kgf

2.40


Step 6: Calculation of module m  For safe Design F s  F d 10892.0320  m  m 

15070.71 m

m 3  1.378 m  1.112 cm m  11.12 mm

(1 cm  10 mm)

(Choose module with nearest standard value from Table 1 P.S.G Data book pg no: 8.2]  12 mm m~ m  12 mm

Step 7: Calculate the value of face width b b  10  m  10  12 b  120 mm

Step 8: Calculate the Diameter d 1 d1  Z1 m  20  12  240 mm

Step 9: Calculate pitch line velocity vm  v1 

 d1 N 1 60  1000




2.41

3.14  240  720 60  1000

v1  9.0432 m/s

Step 10: Recalculate the beam strength Fs F s  [ b]  b  y  P c  12   12   3200  10     0.1084     10  10      15692.4815 kg f

12 . . [ . Pc    m    10 y  0.1084 ; b  10  m in cm ]

Step 11: Calculate the Buckingham’s Dynamic load Fd   0.164 Vm cb  F t Fd  Ft    cb  F t    0.164 V m  1.485  v  9.0432 m/s V m  9.0432  60 m/min  542.592 m/min b  120 mm  12 c m Ft 

5003.57 kg f m



5003.57 kg f 12

 416.96 kgf  417 kg f Ft ~

2.42


Choose the “c” value from PSG Data book pg no: 8.53 table: 41 and 42. Both pinion and gear are carbon steel   20 For 20  full depth; steel and steel, c  11860 e. e  0.064

(m  12 and vm  9.0432 , in which vm is less than 10 m/s. For commercially cut wheels) [From table: 42; Data book pg: 8.53] c  11860  e  11860  0.064  759.04   0.164 V m cb  F t Fd  Ft    cb  F t    0.164 V m  1.485    0.164  542.592 759.04  12  417 F d  417    759.04  12  417    0.164  542.592  1.485   8476   417     3623.5 kg f  233.92   F d  3623.5 kg f

Step 12: Checking For Beam Strength F d  3623.5 kg f F s  15692.481 kg f

By comparing F d and F s, F s  F d, so Design is safe.


2.43

Step 13: Calculation of wear load Fw F w  d 1 Qkb d 1  240 mm  24 c m ; Q 

Q

2i i1

(For external gear)

22  1.333 21

b  120 mm  12 cm ;

1   1  [2c ] sin   E1 E2    k 1.4   20 (Assume)

Take E 1 and E 2 from PSG Data book pg: 8.14 Table: 9 For steel both pinion and gear wheel. E 1  2.15  10 6 kg f/cm 2 (pinion) E 2  2.15  10 6 kg f/cm 2 (gear wheel) [c]  9500 kg f/cm 2 is taken from the PSG Data book pg: 8.4 table 7 gear material (for steel).

k

1 1    9500 2  sin 20   6 6  2.15  10   2.15  10

 21 k  20.509 ~

1.4

2.44


F w  d1 Q kb F w  24  1.333  21  12  8061.98 kg f

Step 14: Checking for wear strength F w  8061.98 kgf F d  3623.5 kg f

By comparing F w and F d ; F w  F d, the Design is satisfactory. Step 15: Calculate the Basic Dimensions of spur gear. From the PSG Data book pg: 8.22 table: 26 1. Module m   12 mm  Z1  Z2  2. Center Distance a  m   2    20  40   12   2    360 mm

3. Height factor f0  1 4. Bottom clearance c  0.25  m  0.25  12  3 mm

5. Tooth Depth h  2.25  m  2.25  12  27 mm

6. Pitch Diameters d d 1  Z 1 m  20  12  240 mm

Z2 Z1

i

Z 2  2  20  40


2.45

d2  Z2 m  40  12  480 mm

7. Tip Diameter dc : d c1  Z 1  2f0  20  2  1  12  264 mm d c2  Z 2  2f0 m  40  2  1 12  504 mm

8. Root Diameter df : df 1  Z 1  2f0 m  2c  20  2 12  2  2  212 mm df 2  Z 2  2f0 m  2c  40  2 12  2  2  452 mm

9. No: of Teeth: Z 1  20 ; Z 2  40 Step 16: Draw the neat sketch of spur gear. [Refer the spur gear Fig 2.14 - Pg 2.36 Design procedure step 16:] Problem 2.2: Design a spur gear drive to transmit 22 kW at 1000 rpm ; Speed reduction is 2.5. The centre distance between the gear shafts is approximately 350 mm. The materials used are: pinion – C45, gear wheel – CI Grade 30. Design the Drive [use Lewis and Buckingham equations].

2.46


Step 1 Given data  22 kW  22  10 3 W

Power P  Pinion speed

N 1  1000 rpm

Speed ratio (i)

 2.5

Center distance (a)  350 mm Pinion material

 C45 steel

Gear wheel material  C.I Grade 30 Step 2: Selection of materials Pinion – C45 steel Gear wheel – C.I Grade 30 Step 3: Calculation of tangential load F t F t  HP 

 Power 

75 kgf vm

7 . . . 1 watt  H.P 750

22  10 3 H.P 750

Power  29.333 H.P vm 

 d1 N 1 m/s 60  1000

Centre Distance a 

d1  d2 2

350  2  d 1  d2 d 1  d 2  700 mm  d2  700  d1


Speed ratio (i) 



d2 d1

2.47

 2.5

700  d1  2.5 d1 700  d 1  2.5 d1 700  3.5 d 1

Pitch circle dia. of pinion

 d 1  200 mm

d 2  2.5  200 d 2  500 mm

vm 

  200  1000  10.466 m/s 60  1000

F t  HP 

75 vm

F t  29.333 

75 10.466 F t  210.18 kg f

. .  10 N] [ . 1 kgf ~

F t  2101.8 N

Step 4: Calculation of Dynamic load F d F d  F t  C v kg f Cv 

6  vm 6

Cv 

6  10.466 6

For vm  5  20 m/s Here vm  10.466 m/s

C v  2.744 F d  210.18  2.744

2.48


F d  576.733 kg f

Step 5: Calculation of strength of gear tooth Fs The strength of gear tooth is calculated for weaker gears. Here, the pinion and gear wheel are made up of different materials, so calculate for weaker one. Strength of gear wheel  b2  y2 From PSG Data book pg no: 8.5 table: 7 For Cast Iron Grade 30; In the PSG Data book; C.I Grade 35 is available so take that value  b2  800 kg f/cm 2

For pinion: C45  b1  1400 kg f/cm 2

Lewis form factor From the PSG Data book pg: 8.50

For

y  0.154 

0.912 Z

y1  0.154 

0.912 Z1

 0.154 

0.912 20

 0.1084 pinion

For

y2  0.154 

0.912 Z2

 0.154 

0.912 50

 0.1357 gear

For 20 involute (Assume) [Assume Z 1  20 teeth]



Z2 Z1

i

Z 2  2.5  20 Z 2  50


2.49

Strength of gear wheel   b2  y2  800  0.1357  108.608 kg f/c m2

Strength of pinion  b1  y1  1400  0.1084  151.76 kg f/c m2

By comparing gear and pinion; gear is the weaker material. So we have to design for the gear. F s gear  [ b2] b  y2  Pc  800  3    m   0.1357    m . .  . b  3Pc

 3214.33 m  m

Step 6: Calculation of module m

3m

For safe Design F s  F d ;

 3 m

3214.33  m  m  576.733

Pc    m

m2 

576.733 3214.33

m  module

m  0.179425 m  0.4235 cm m  4.2 mm

The obtained module is 4.2 mm. Choose the standard value of module from the PSG data book pg: 8.2 table: 1 m  5 mm (standard value)

2.50


Step 7: Recalculate the face width (b) b  3P c 3m 35  47.12 ~  50 mm

Step 8: Recalculate Diameter d2 (gear)

. . . Z2  50

d2  Z2  m d 2  50  5 d 2  250 mm

Step 9: Recalculate the pitch line velocity (gear) v2  

 d2 N 2 60  1000

. . N1 i . N2

  250  400 60  1000

1000  i  N 2 1000  N2 2.5

 5.233 m/s

N 2  400 rpm

Step 10: Recalculate the strength of gear tooth F s  [ b2] b  y2  P c

Substitute P c and m in cm.

 50   5   0.1257      800      10   10  F s  852.196 kgf


2.51

Step 11: Calculate the Buckingham’s Dynamic load F d   0.164 V m cb  F t Fd  Ft    cb  Ft    0.164 V m  1.485  b  5 cm ; F t  210.18 kg f

Choose c value from PSG Data book table 41 and 42 pg no: 8.53 Pinion and gear material   20 (Assume) Steel and Cast Iron c  8150 e e  0.025 (for carefully cut gear) in module “5”. c  8150 e  8150  0.025  203.75 vm  10.466 m/s

Substitute V m in m/min. b in cm; F t in kg f V m  10.466  60 m/min  627.96 m/min   0.164 V m cb  F t F d  Ft    cb  F t    0.164 V m  1.485    0.164  627.96 203.75  5  210.18  Fd  210.18      205.75        627.96 1.485 5 0.164 210.18  

 210.18 

126561.8968 155.043

 210.18  816.29 F d  1026.47 kgf

2.52


Step 12: Check for Beam Strength For safe Design F s  F d F s  852.196 kg f ; Fd  1026.47 kg f F s  F d ; so the Design is not satisfactory.

For safe Design, increase the value of m and b. m  6 mm b  10  m b  60 mm

Recalculate

F s and F d using module substitute m and b in cm in equation.

(m  6

mm)

F s  [ b2]  b  y  P c  6   800  6  0.1357       10  F s  1227.162 kg f

(1)

  0.164 V m cb  F t Fd  Ft    cb  F t    0.164 V m  1.485    0.164  627.96 203.75  6  210.18   210.18    203.75  6  210.18     0.164  627.96  1.485   210.18 

147545.1802 159.19

F d  1137.02 kg f

(2)

By comparing F s and F d, Fs  Fd

So, the Design is safe.


2.53

Step 13: Calculation of wear load F w  F w  d1 Qkb b  6 cm d1  200 mm  20 cm Q

2i i1

(For external gear) Mostly all the sums are external gears only. Q

2  2.5 2.5  1

 1.4285

1   1  [2c ] sin   E 1 E2    k 1.4 From the PSG Data book pg no: 8.14, Table: 9 E 1 for C45 steel  2.15  10 6 kg f/cm2 E 2 C.I Grade 30  1.46  10 6 kg f/cm 2

  20 Assum e c  5000 kgf/c m2 [From PSG Data Book pg: 8.5 table:7; for C45 steel] 1 1    5000 2  sin 20   6  6 1.46  10   2.15  10 k 1.4

2.54




5000 2  sin 20  1.15  10  6 9.8334   7.023 kgf/cm2 1.4 1.4 F w  d 1 Qkb F w  20  1.4285  7.023  6  1205.304 kg f

Step 14: Checking for wear strength For satisfactory Design F w  F d F w  1205.304  1137.02 F d

So the Design is safe. Step 15: Calculate the Basic dimensions of spur gear. From the PSG Data book pg no: 8.22. Table: 26 1. Module  6 mm 2. Centre Distance (a)  m

Z1  Z2 2

 20  50   6 2  

Z 1  20 Z 2  50

 210 mm

where suffix 1 refers to pinion and suffix 2 refers to gear wheel. 3. Height factor f0  1 4. Bottom clearance (c)  0.25  m  0.25  6  1.5 mm 5. Pitch or Reference d Diameter  d 1  m  Z 1 d1  6  20


2.55

d1  120 mm d2  m  Z 2  6  50 d2  300 mm

6. Root Diameter df  df1  Z 1  2f0 m  2c d f1  20  2  1 6  2  1.5  105 mm d f2  Z 2  2f0 m  2c  50  2 6  2  1.5  285 mm

7. Tip Diameter dc  d c1  Z 1  2f0 m  20  2  1 6  132 mm d c2  Z 2  2f0 m  50  2 6  312 mm Problem 2.3: A Reciprocating compressor is to be connected to an Electric motor with the help of spur gears. The distance between the shafts is to be 500 mm. The speed of the electric motor is 900 rpm and the speed of the compressor shaft is desired to be 200 rpm. The torque to be transmitted is 500 N-m. Taking starting Torque as 25% more than the Normal torque, Determine. (a) Module and face width of the gears using 20 degree  stub teeth. (b) Number of teeth and pitch circle diameter of each gear. Assume suitable values of velocity factor and Lewis factor.

2.56


Step 1 Given data: Distance Distance) a  500 mm

between

the

shafts

(Centre

N 1 (pinion)  900 rpm (always pinion has maximum

rpm when compared to gear wheel) N 2 gear  200 rpm TorqueNormal  500 N  m

So, starting Torque as 25% more than the normal torque. T max  1.25  T mean or normal T max  1.25  500  625 Nm  625  10 3 N  mm

[For this type of problems, it is not needed to design a spur gear, Just find what is asked in Question] Step 2: Selection of material Material is not given in the problem. So, choose the material from the PSG Data book pg: 8.4 Table: 5 based on gear ratio (i) i

From the table: 5

N2

N1



900  4.5 200


2.57

i  4 ; pinion (material)  C 45

gear wheel (material)  cast steel To calculate module value, we have to find the F s and F d. F s will be calculated by using Lewis equation. [Lewis equation is applicable for weaker gear]. So, we have to

determine, which is weaker one. Pinion strength   b  y1 1 From PSG Data book pg: 8.5 table: 7;  b  1400 kg f/cm 2 [for C 45 material], b  3200 kg f/cm 2 [for 1 2

steel] y 1  0.154  Z min or Z 1 



0.912 Z1

(For 20 involute) . . ( .   20 given)

2 sin2  2 2

sin 20

 17.09 ~  18

y 1  0.1033

Pinion strength  1400  0.1033  144.66 kg f/cm 2

Gear strength

  b  y2 2

y2  0.154 

0.192 Z2

(for 20 involute)

2.58


Z2 Z1

i

Z 2  4.5  18 Z 2  81 y 2  0.154 

0.192 81

y 2  0.1516

Gear strength  3200  0.1516  485.12 kg f/cm 2

By comparing pinion and gear strength, pinion is the weaker material. So design for pinion. Step 3: Calculate tangential load F t F t  HP 

vm 

 d1 N1 60  1000

a

d1  d2 2

500 

d1  d2 2

d1  d2  1000 d2  1000  d 1

P 

75 kg f vm 2  N1 T max 60 2    900  625 60

P  58875 Watts

P

58875 H.P 750

 78.5 H.P


d2

d1

 i  4.5 

1000  d1 d1

2.59

 4.5

1000  d1  4.5 d1 1000  4.5 d1  d1 5.5d1  1000 d1  181.818 mm d2  1000  d 1  1000  181.818 d2  818.1818 mm vm 

 d 1 N1 60

vm 

  181.818  900 60  1000

vm  8.5636 m/s F t  78.5 

75 8.5636

F t  687.7 kg f

Step 4: Calculation of Dynamic load Fd F d  F t  C v kg f Cv 

3  vm 3

C v  3.8545

[vm  8.5636 ]

2.60


[vm  10 m/s; commercially cut wheel from Data book pg: 8.51] F d  687.5  3.845  2643.44 kg f

Step 5: Calculation of strength of gear tooth F s  [ b ] b  y  P c

b  1400 kg f/cm 2 [From PSG Data 1

book pg: 8.4 Table: 7] F s  1400  10  m  0.1033    m

b  3 to 4P c b  3Pc b3m

 4541.068 m  m

Step 6: Calculation of module m For safe Design, F s  F d 4541.068 m  m  2643.44 m 2  0.5821

b  9.39 m  10 m b~ m  module Pc    m

m  0.7629 cm

y1  lewis form

m  7.629 mm

factor for pinion y1  0.1033

 8 mm (Standard module from Data book pg: 8.2; m~

table: 1). We can also calculate module using the formula m

2  500 2a   10.1010 (for standard value) Z1  Z2 18  81

Take the “m” value largest of two values


m  10 mm

Step 7: Calculate face width (b) b  10  m  10  10 b  100 mm

Step 8: Calculate Diameter of gear and pinion: d1  Z1  m  18  10 d1  180 mm d2  Z2  m  81  10 d2 810 mm

Answer (a) Module m   10 mm Face width b  100 mm (b) No. of Teeth on pinion Z 1  18 No. of teeth on gear

Z 2  81

Pitch circle Diameter of pinion d 1  180 mm Pitch circle Diameter of gear

d 2  810 mm

2.61

2.62


Problem 2.4: The following particulars of a single reduction spur gear are given. Gear ratio 10:1, Distance between the centres 660 mm approximately. Pinion transmits 50 kW at 1800 rpm. Involute teeth of standard proportions: Addendum  1 module with pressure angle of 22.5; Permissible normal pressure between teeth  175 N per mm of width. Find 1. Standard module if no interference is to occur 2. No. of teeth on each wheel 3. The necessary width of the pinion 4. The load on the bearing of the wheels due to power transmitted.

In this type of problem, it is not needed to design a gear drive, Just find the values asked. Given Gear ratio (or) speed ratio (i)



10  10 1

Distance between the centres a  660 mm P  50 kW  50  10 3 W

Power Speed of pinion

N 1  1800 rpm   22.5

Pressure angle

Permissible normal pressure between teeth  175 N/mm of width. 1. Find the module (m) m

d1

Z1

Z 1  minimum no: of teeth on pinion. Calculate the

minimum no: of teeth on pinion to avoid interference.


Z1  

2.63

2

sin 2 

  22.5

2 2

sin 22.5 

 13.6 ~  14 Z 1  14 teeth a

d1  d2 2

660 

d1  d2 2

2  660  d1  d2 d1  1320  d 2 i 10 

d2 d1 d2 1320  d 2

13200  10d2  d2 13200  11d2 d2  1200 mm d1  1320  1200 d1  120 mm m

120  8.57 ~  8 14

m  8 mm (Standard module from Data book pg: 8.2

Table: 1]

2.64


2. No: of teeth on each wheel: Z 1 and Z 2. Corrected no: of teeth on pinion. Z1 

d1 m



120 8

Z 1  15 Z2 Z1

 i ; Z 2  10  15

Z 2  150

3. The necessary width of the pinion If permissible normal pressure between teeth  175 N/mm (unit load) of width is given, then to calculate face width, use the following formula only. b

Fn 

Total Normal load F n unit load

Ft cos 

F t  tangential load

To find Ft F t  HP 

vm 

75 vm

 d1 N 1   120  1800  60  1000 60  1000

vm  11.3040 m/s

Power

 50 kW 

50000  H.P 750


Power

2.65

 66.666 H.P Ft 

66.666  75 11.3040

F t  442.32130 kg f F t  4423.2130 N (Substitute F t in Newton) Fn  

Ft

cos 

4423.2130 cos 22.5 

Fn  4787.6512 N b 

Fn un it load 4787.6512 175

b  27.358 mm

4. The load transmitted.

on

the

bearings

where

power

is

The radial component of the normal load acting on the tooth is the load on the bearings of the wheels. So, Radial Road F r  F n sin   4787.6512  sin 22.5  1832.154 N  1.832 kN Fr  1.832 kN

2.66


Problem 2.5: A motor shaft rotating at 1500 rpm has to transmit 20 HP to a low speed shaft with a speed reduction of 3. The teeth are 14 12 involute with 25 teeth on the pinion. Both the pinion and gear are made of steel with a maximum safe stress of 2000 kgf  cm2. A safe stress of 400 kgf  cm2 may be taken for the shaft on which the gear is mounted and for the key. Design a spur gear drive to suit the above conditions. Assume starting torque to be 25% higher than the running torque.

[AU Nov/Dec 2007, Apr/May 2009]

Given Data: Speed of the pinion N 1  1500 rpm Power P  20 HP Speed reduction i  3 Pressure angle   14 12 invo lute No . o f teeth on pinion Z1  25 teeth

Pinion and gear are made of same material Therefore, the design is based on pinion. Safe static stress  [ b]  2000 kgf  cm 2 [Note: The remaining data is used for designing the shaft] Step 1: Find module m Beam strength  F S  [b] b  y P c Assume, b  Face width  3P c  3m ; where m  module in cm.

. . [ . P c   m]


2.67

 Beam strength  F S  [b] y 3 m   m  [b]  3  2  m 2 y y  Lewis form factor  0.124 

0.684 0.684  0.124  Z1 25

[ 14

1 involute] 2

y  0.09664  Beam strength  F S  2000  3 2 m 2  0.09664 FS  5722.79 m2 kg f

... (1)

* Dynamic Load  F d  F t  C v F t  Transmitted load  HP 

75 vm

vm  mean velo c ity

75  20  19.634 m



 76.398    kg f  m 



 d1 N 1 60  100



mZ 1 N 1 60  100

  m  25  1500 60  100

vm  19.634 m m  sec  d1  z1 m

Cv 

6  vm  when vm  20 m  sec (carefully cut gears) 6

 6  19.634 m     6    Dynamic Load  Fd  F t  C v  Fd 

76.398  6  19.634 m     6 m  

12.733 6  19.634 m m

... (2)

2.68


To find module, equate the above equations (1) and (2) F s  F d o r F s  Fd 5722.79 m 2 

12.733 6  19.634 m m

449.44 m3  6  19.634 m

Calculate m  method

module in cm, by trial and error

Trial 1: m  0.5 cm LHS  56.18 LHS  RHS

RHS  15.817  Satisfactory but too big m

Trial (2): m  0.4 c m RHS = 13.8536

LHS  28.764 LHS  RHS:

 Satisfactory but too big m

Trail (3): m  0.3 c m LHS  12.134 LHS  RHS Trial (4): m  0.29 c m

RHS  11.69  Satisfactory

LHS  10.96 LHS  RHS

RHS  11.69  Not Satisfactory

Therefore, the value of module (m) is in between 0.29 and 0.3. Standard module 3 mm. (P.No. 8.2 table 2) Take m  0.3cm  3 mm module m  3 mm Face width b  3 m  3    3 b  28.274 mm


2.69

CHECK FOR BEAM STRENGTH AND WEAR STRENGTH Step 2: Checking for Beam strength Buckingham’s dynamic Load   0.164 V m cb  F t  Fd  Ft     cb  F t   0.164 V m  1.485  76.398 76.398   254.66 kg f m 0.3 V m  19.634  0.3   60  353.412 m  min b  face width  2.8274 cm carefully c ut gears e  0.025 mm [Ref PSG Data book c  11440  e  11440  0.025  286 kg f  cm Pg.No 8.53 Table 4.2] Ft 

F d  254.66 

0.164  353.412 286  2.8274  254.66  0.164  353.412  1.485

 254.66 

286  2.8274  254.66  

61628.2  254.66  579.31 106.382

Fd  833.97 kg f F s  5722.79 m 2  5722.79  0.3 2  515.05 Fs 515.05 kg f Fs  Fd,

Here F s is not greater than F d. Therefore, the design is not satisfactory based on beam strength. So m can be increased to 0.4 cm and found not satisfied and increase to 0.5 and now satisfied

2.70


Ft 

76.398 76.398   152.796 kgf m 0.5

V m  19.634  0.5 60  589.02 m /m in b  3 m  3  0.5  4.712 Fd  152.796 

0.164  589.02 286  4.712  152.796  0.164  589.05  1.485

 190.995 

286  4.712  152.796   

144940  1131.43 kg f 154.12

F s  5222.79  0.5 2  1430.69 kg f

Here F s  F d satisfied. Step 3 Check for wear strength Wear Load  FW  d1 Q  k  b d1  m Z 1  0.5  25

[ c]  C R HRC K Cl

 12.5 mm  1.25 c m Q

23 6 2i    1.25 i1 31 4

b  4.712 c m k  [c]2

sin   1 1   1.4  E 1 E 2   

Assume alloy steel, case hardened C R  280; HRC  55 to 63 take HRC  60

assume life  10,000 hrs. Life N  10,000  60  1500  90  10 7 cycles

Pressure angle   14.5 

K Cl  0.585

Pinion and wheel are made of [c]  280  60  0.585 same material  9828 kgf  cm2 E 1  E 2  2.15  10 6 kg f  cm 2


2.71

[From PSG Data book Pg. 8.14, Table 9] k  9828 2

sin 14.5   1 1     6 6 1.4 2.15  10   2.15  10

 16.069 kgf  cm 2

Wear load F W  d 1 Qkb Wear load  F W  12.5  1.5  16.069  4.712 F W  1419.69 kgf

Since, F W 1419.69   Fd (1131.43) Therefore, the design is satisfactory based on wear strength  Final values:

m  module  5 mm  0.5 c m b  Face width  4.712 c m

Problem 2.6: Design a spur gear drive, from consideration of dynamic load and wear, to transmit 120 HP at 600 rpm to shaft running at 150 rpm. Assume 10 mm module and 20 full depth involute system. The pinion is to have 20 teeth and is

made

of

6

2  10 kgf/cm

heat 2

treated

steel

having

value

of 2

E

as

and safe static stress of 2000 kg/c m . Assume

suitable material for gear.

Given Data: Power

 P  120 HP

Pinion speed

 N 1  600 rpm

Gear speed (Shaft speed)  N 2  150 rpm

2.72


 m  10 mm  1 cm

Module

Pressure angle    20 No. of teeth on pinion  Z 1  20 teeth E  2  10 6 kgf/cm 2

Safe static stress  [  b ]  2000 kg f /cm 2 From PSG Databook, (Page No. 8.50 and 8.51) Beam strength  FS  [ b ]  b  y  Pc

Assume

 [ b ]  b  y  m

b  Face width

 FS  2000  9.424  0.1084    1

 3 to 4 Pc and Pc  m

2

b  3Pc  3m

FS  6418.66 kgf/cm

 3    1  9.424 c m b  9.424 cm y  0.154   0.154 

0.912 Z1 0.912

y  0.1084

Step 2 Transmitted load  F t  HP   120 

75 vm

75 6.283

 1432.43 kg f F t  1432.43 kgf

20


2.73

vm  me an velocity in m/sec 

But d1  mZ 1



d 1N 1 60  100



Z 1mN 1 60  100

  20  1  600 60  100 vm  6.283 m/sec or V m  371.99 m/min

b  9.424 cm V m  6.288  60  371.99 m/min

Step 3 : Check for beam strength Buckingham’s dynamic load   0.164 V mcb  F t  Fd  Ft    cb  Ft    0.164 Vm  1.485  c  11860 e  c  11860  0.09

; e  0.09  1067.4 kg/cm

Refer PSG Data book Pg No. 8.53 table 41 0.164  371.99 1067.4  9.424  1432.43    F d  1432.43     1067.4  9.424  1432.43   0.164  371.99  1.485 

 1432.43 

701061.15 220.196

 1432.43  3183.80  F d 4616.235 kg f

2.74


But F S  6418.66 kg f  FS 6418.66   Fd 4616.235 

Therefore, the design based on beam strength is satisfactory. Step 4 Check for Wear load.

d1  mZ 1  10  20  200 mm  20 cm

Wear load F w  d1Q  k  b sin   1 1   k  [  c ]2 1.4  E 1 E 2   

Assume both pinion and gear are made of same material. E 1  E 2  2  10 6 kg f/cm 2

To calculate [ c ] Select material, heat treatment process (Refer PSG Data book page 8.16) Assume alloy steel, case hardened HRC  55 to 63

C R  280

Take HRC  60 [ c ]  C RHRC  Kcl

Assume life  10,000 hrs N  10,000  600  60

 280  60  0.585  9828 kg f/cm 2

 36  10 7 cycle From PSG Data book table

(17) P.No. 8.17 K cl  0.585   pressure angle  20


2.75

sin   1 1    k  [  c ]2  1.4 E2  E  1   9828 2 

sin 20  1 1   1.4  2  10 6 2  106    k  23.59 kg /cm2

Wear load  F w  d 1  Q  k  b  20  1.6  23.59  9.424  7113.989 kgf

But

F d  4616.235 kg f FW 113.9   Fd 4616.235 

Therefore, the design is satisfactory based on wear strength. Since F S  F d and F W  F d, Therefore, the design is satisfactory based on beam strength and also based on Wear strength. Problem 2.7: An electric motor is to be connected to a reciprocating pump through a gear pair. The gears are overhanging in their shafts. Motor speed  1440 rpm. Speed reduction ratio  5. Motor Power  36.8 kW. The gears are to have 20 pressure angle. Design a spur gear drive. [April/May: 2010] (Nov-Dec-2009)

[Similar to problem 2-1]

2.76


Problem 2.8: Design and draw spur gear drive transmitting 30 kW at 400 r.p.m. to another shaft running approximately at 1000 r.p.m. The load is steady and continuous. The material for the pinion is cast steel and for gear is cast iron. Take module as 10 mm. Also check the design for dynamic load and wear. [May/June: 2014]

[Similar to problem 2-1] Problem 2.9: A motor shaft rotating at 1440 rpm has to transmit 15 kW power to a low speed shaft rotating at 500 rpm. A 20 pressure angle involute tooth gear – pinion is used. The pinion has 25 teeth. Both gear and pinion are made of cast iron having allowable strength of 55 N/mm2. Design a suitable gear drive. [Nov-Dec: 2007]

[Similar to problem 2.1] Problem 2.10: A motor shaft rotating at 1440 rpm has to transmit 15 kW to a low speed shaft rotating at 500 rpm. The teeth are 20 involute with 25 teeth on the pinion. Both the pinion and gear are made of cast iron with a maximum safe stress of 56 MPa. A safe stress of 35 MPa may be taken for the shaft on which the gear is mounted. Design and sketch the spur gear drive to suit the above conditions. The starting torque may be assumed as 1.25 times the running torque. [Nov-Dec: 2009]

[Similar to problem 2.5]


2.77

2.15 DESIGN OF SPUR GEAR [HERTZ METHOD] [Note: Always solve the problem by using Lewis and Buckinghams equation method. If the life of gears is given in the problem (or) Design the gear drive based on contact stresses, then use this Hertz method. Design Procedure Step 1: Selection of material (Refer PSG Data book Page No 8.4) Note (a): If the material for both pinion and wheel are same, then design for pinion alone. (because the pinion is weaker than Gear). Note (b): If the material for pinion and wheel are different, pinion is to be designed first and then, Check the design stresses for both pinion and wheel. From PSG Data book Page No 8.4: Selection of material based on gear ratio (i) From PSG Data book Page No 8.16: Selection of material based on Heat treatment process. If case hardness or surface hardness  350 BHN then its core hardness  350 BHN If surface hardness is greater than 350 BHN, then represent in HRC. Step 2: Calculate the Life in number of cycles; (N) Note (a): If life required is not given in the problem, assume N  10,000 hrs to 20,000 hrs. Note (b): If the life required is given in hrs.

2.78


Life in cycles  life in hrs   60  rpm  ...... cycles

and write in the form of N  .....  10 7 cycles

Step 3: Calculation of Equivalent Young’s modulus. From PSG Data book Page No. 8.14 

Always pinion is made of steel with E  2.15  10 6 kg f/c m2



Wheel is made of steel, CI, Bronze or Nylon.

Take the equivalent Young’s modulus  E eq in kgf /cm 2 Step 4: Calculation of Design bending stress [ b] From PSG Data book Page No. 8.18 Assume rotation in one direction only. [b] 

1.4 kbl  1 n k

kbl  Life factor for bending - From PSG Data book

Page No. 8.20; Table: 22 Take the value of kbl, based on 1. Material 2. Surface hardness (always  350 3. Life in No. of cycles N  n  factor of safety - From PSG Data book Page No.

8.19; Table: 20


2.79

Take the value of n, based on 1. Material 2. Mode of Manufacture 3. Heat treatment. k  fillet stress concentration fa ctor -

From PSG Page No. 8.19 Table: 21 Take the value of k, based on 1. Material and heat treatment 2. Addendum Modification always 0  X  0.1 value)

Coefficient,

X

(take

 1  Endurance limit stress in bending for complete

reversal of stresses. kg f/cm 2 From PSG Data book Page No. 8.19; Table: 19 Take the equation based on material of mating gear In the above equation take the values of u  ultimate stress in kgf/cm 2 and y  yield stress in kg f/cm 2

From PSG Data book Page No. 1.9 (for C15, C20, C15 Mn 35, C40, C45) From PSG Data book Page No. 1.13 to 1.15 (for alloy steels)

u

From PSG Data book Page No. 1.4 (for cast iron) (only value)

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Substitute all the values and calculate the design bending stress [ b] in kg f/cm 2 Step 5: Calculation of Design contact stress [c] From PSG Data book Page No 8.16 [ c]  C B  HB  kcl  1 If hardness represented in BHN  (or) [ c]  C R  HRC  kcl  2 If hardness represented in HRC

The value of C R depends on 1. Wheel material 2. Heat treatment process 3. Surface hardness. HB - Brinell hardness number HRC - Rockwell ‘C’ hardness number C B and C R - Coefficient depending on the surface hardness. kcl - Life factor - From PSG Data book Page No. 8.17

Table: 17

Take the value of kcl , based on Ka 

 

10 2 ; N

N -no. of cycles

1. Material 2. Surface hardness (consider surface hardness generally  350 HB 3. Life in number of cycles N


2.81

Substitute all the values in the above equation and calculate the design surface (contact compressive) stress [c] in kg f/cm 2

Step 6: Calculation of design twisting moment [M t] From PSG Data book Page No. 8.15 [M t]  Mt  k  kd M t  Nominal twisting moment transmitted by the pinion kgf  cm  97420

hp kW  71620 n n

where kW or hp  given power n  N 1  Speed of rotation of pinion, rpm.

Initially assume, k  kd  1.3 (Since k and k d values are not known).

Step 7: Calculation of approximate centre distance a From PSG Data book Page No. 8.13

a  i  1

  3

2  0.74  E  [M t]  [ ]  i c  

i  Gear ratio E  Equivalent young’s modulus  kgf/cm 2 [M t]  Design twisting moment  kg fcm [c]  Design surfac e contact compressive  stress  kg f/cm 2

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  ratio between facew idth and centre distance 

b a

initially assume   0.3 Step 8: Calculation of Z 1 and Z 2 Z 1  no. of teeth on pinion Z 2  no. of teeth on wheel

Assume Z 1  17 to 25 teeth (if Z 1 is not given in the problem). Since, i 

Z2 Z1

,

(From PSG Data Book Page No. 8.1) Calculate Z 1 Step 9: Calculation of module. (From PSG Data book Page No. 8.22) m

2a Z1  Z2

After calculation, take the nearest standard module from PSG databook, Page No. 8.2 (Take always nearest higher value). Represent module ‘m’ in ‘millimeter’. Step 10: Recalculate ‘a’ centre distance a

m Z 1  Z 2 2

Represent centre distance ‘a’ in R10 SERIES in m.m.


2.83

Step 11: Calculation of facewidth b In Step 7,  

b  0.3 assumed. a

 Face width  b  a 0.3  in m.m

Represent

 1 mo d ule m  2 centre distance a   3 face width b 

in millimeters

Step 12: Calculation of pitch line velocity V in m/sec.

v

 d1 N 1 60

in metres

where

d1  pitch circle diam eter of pin io n in m eters m N 1  speed of the pinion in rpm

But m  module 

d1 Z1

 Calculate d1 in mm and convert into m.

Select Quality of Gears - From PSG Data book Page No. 8.3 IS Quality depends on (1) type of gear

Cylindrical Straight bevel Spiral bevel

(2) Speed of gears in m/sec. Select the IS Quality (Preferred Quality). Step 13: To find the constants k and k d. Load concentration factor k.

2.84


Page No. 8.15, Table: 13 k values depends on  P 

b d1

Calculate  p and read the corresponding value of ‘k’ (assume bearings close to Gears and symmetrical) Dynamic load factor kd – From PSG Data book Page No. 8.16; Table: 15 kd value depends on

1. IS Quality 2. Pinion surface hardness 3. Pitch line velocity 4. Spur, helical, straight bevel and Spiral bevel Read the value of k d. Step 14: Recalculate the design twisting moment [M t] [M t]  M t k  k d

The value of M t is similar to step (6). Substitute the values of k and kd. Calculate the design twisting moment [M t] in kg f cm. Step 15: Checking of bending stress i1 [M t]  [b] From PSG Data book amby Page No. 8.13 b 

where i  gear ratio a  centre distance in c m


2.85

m  module in cm b  face width in cm y  form factor  Re fer P SG Data book Page No. 8 .18

Value of y depends on 1. Z 1  no. of teeth on pinion 2. X  0 addendum modification coefficient. Note If  b  [b] then change the module, and recalculate a, b and [M t]

Check again bending stress  b Step 16: Checking of surface (contact compressive) stress  c; Refer PSG Data book Page No. 8.13 c  0.74

i1 a

i1

E [M ]  [ ] ib   t

c

If  b  [ b] and c  [c] then the design is satisfactory.

Step 17: Draw the neat sketch of spur gear. Refer Pg 2.36 Fig. 2.14 Problem 2.11: Design a spur gear to transmit 22.5 kW at 900 rpm; speed reduction is 2.5; Material for pinion and wheel are C15 steel and castiron grade 30 respectively. Take pressure angle of 20 and working life of the gears as 10,000 hrs.

2.86


Solution Material Given: pinion  C15



Steel

Wheel - Cast iron grade 30 Speed reduction  i  2.5 i.e. Gear ratio  i 

Z2 Z1

 2.5

Power  P  22.5 kW Pinion Speed  n  N 1  900 rpm; Life

 N  10,000 hrs

Step 1: Material properties Refer PSG Data book P.No. 8.16; Table 16 Pinion: C15 (Given) Assume C15 - Case hardened HRC  55 to 63, assume HRC  60

i.e. Surface hardness  350 BHN  core hardness  350 BHN

Wheel: Cast Iron - GRADE 30 HB  200 to 260

Assume surface hardness HB  250 (Brinell hardness)  Core hardness  350

Step 2: Life required  10,000 hrs  10000  60  900 cycles

Life required  N  54  10 7 cycles .


2.87

Step 3: Calculation of Equivalent Young’s Modulus Refer PSG Data book P.No. 8.14; Table 9 Pinion material - Steel - E 1  2.15  10 6 kg f /cm 2 Gear (or) Wheel material - C.I. Grade 30 E 2  1.4  10 6 kg f /cm 2

Therefore, equivalent young’s modulus  E  1.7  10 6 kg f /cm 2

Since u  28 kg f/mm 2 For CI Grade 30, the ultimate stress 2

u  30 kg f /mm

kg f  cm 2 10



N mm 2

(Convert into N/mm 2) E  1.7  105 N /mm2

Step 4: Calculation of design bending stress [ b] (Refer PSG Data book P.No. 8.18) 1.4 kbl  1 Design bending stress [ b]  n k

(assume rotation in one direction) kbl  Life factor for bending  1

From PSG Data book P.No. 8.20, Table 22 (Consider core hardness)

2.88


K bl  1 (for N  10 7 cycles and BHN  350  n  Factor of safety

From PSG Data book Table (20) P.No. 8.19 (for case hardend)

n2

k  fillet stress concentration factor

Refer PSG Data book P.No. 8.19 Table (21)  1.2

 1  Endurance limit stress in bending for complete reversal of stresses Refer PSG Data book P.No. 8.19; table 19 Assume C15 as forged  1  0.25 u  y  500

steel,

for

forged

steel,

From PSG Data book P.No. 1.9 for C15 material. u  37 to 49 kgf/mm 2 y  24 kg f/mm 2 select u  40 kg f/mm 2  40  100 kg f/cm 2 y  24  100 kgf/cm 2    1  0.25 40  100  24  100  500  2100 kg f /cm 2


2.89

 Design bending stress 1.4 kbl   [b]  n  k  1 

1.4  1  2100  1225 kg f/cm 2 2  1.2

(Convert into N/mm 2) [b]  122.5 N/mm 2

Step 5: Calculation of Design surface Contact compressive stress [ c] Refer PSG Rate book P.No. 8.16 Design surface contact compressive stress  [ c]  CR  HRC  kcl  kgf/c m2 C R  Coeff. depending on surface hardness

From PSG Data book P.No. 8.16 Table 16 C R  220 for C15, Case hardened  HRC  60 assumed  kcl  Life factor Consider surface hard ness

(Table 17: P.No. 8.17 of PSG Data book) kcl  0.585 for N  10 7 c yc les and surface hardness  350 BHN [c]  220  60  0.585 [c]  7722 kgf/c m2 (Convert into N/mm 2

2.90


Design surface (Contact compressive) stress  [c]  772.2 N/mm2

Step 6: Calculation of design torque [M t] From PSG Data book P.No. 8.15 [M t]  Mt  kd  k M t  97420 

kW n

97420  22.5 900

 2435.5 kg f  cm

Initially assume, for symmetry scheme kd  k  1.3 [M t]  2435.5  1.3  3166.15 kg fcm

(Convert

into

N-mm) [M t]  316615 Nmm

Step 7: Calculation of Approximate centre distance ‘a’. From PSG Data book P.No.8.13

  3

Centre distance  a  i  1

 0.74   [ ]  c  

i  2.5 [c]  772.2 N/mm2

from Step (5)

E  1.7  10 5 N/mm 2 from Step (3)

2

E[M t] i


2.91

[M t]  316615 N mm from Step (6) 

b a

Assume  

b  0.3 from P.No. 8.14 table (10); a

  3

a  2.5  1

2

 0.74     772.2 

1.7  10 6  316615 2.5  0.3

a  142.74 mm;

Step 8: Calculation of Z 1 and Z 2 Assume Z 1  No. of teeth on pinion  25 teeth . (Refer PSG Data book P.No. 8.1) Speed reduction i 

Z2

Z1

Z 2  No. of teeth on wheel  i  Z 1  2.5  25  62.5 teeth

Take Z 2  63 teeth Step 9: Calculation of module Refer PSG data book (P.No. 8.22) Take, Module  m 



2a Z1  Z2

2  142.74 285.48   3.24 m m 88 25  63

2.92


Take Standard module, from PSG Data book P.No. 8.2 Table (1) module  4 mm

Step 10: Recalculate centre distance ‘a’ and rounded to R10 series. a

m Z1  Z2 2



4 25  63 2

Centre distance  a  176 mm  Standard R10 series a  200 mm

Step 11: Calculation of face width b  value already assumed as 0.3 but  

face width

b  0.3 a

 b  0.3  200  60 mm

Step 12: Calculation of PCD d1 and PCD d 2 PCD of pinion  PCD 1  d1  m  Z 1  4  25  100 mm

PCD of wheel  PCD 2  d2  m  Z 2  4  64  252 mm

Step 13: Calculate of pitch line velocity: v v

 d1 N 1 60

m/sec

   0.1 

900  4.712 m/sec 60


2.93

Step 14: Selection of Quality of Gear. From P.No. 8.3, table (2); at v  upto 8 m/sec Select the Quality of Gear - IS Quality 8 Step 15: Find load concentration factor k for IS Quality 8 Gear, Refer PSG Data book P.No. 8.15; table (14) k depends on  p 

b value d1



60  0.6 100

at  p  0.6; the value o f k  1.03 Step 16: Find Dynamic load factor k d Refer PSG Data book P.No. 8.16; Table (15) For pitch line velocity up to 8 m /sec , cylindrical gears, surface hardness  350 BHN kd  1.4

Step 17: Recalculate, [M t] design twisting moment, [M t]  M t k d k  2435.5  1.4  1.03 [M t]  3511.99 kg f cm [M t]  351199 Nmm .

M t  2435.5 kg fc m

[Refer PSG Data book Pg. 8.15, Table 13]

2.94


Step 18: Check  b induced bending stress Refer PSG Data book P.No. 8.13; table (8) b 

i1 [M t]  [ b] amby

[M t]  351199 N mm i  2.5 a  200 mm m  4 mm b  60 mm y  form factor

Refer PSG Data book P.No. 8.18 table (18) y value depends on X and Z value)

for Z 1  25; X  0 ; y  0.4205 b 

2.5  1 351199 200 X 4 60  0.4205

 60.89 N/mm 2  [b] 1225 N/mm 2

Step 19: Check  c induced surface (contact compressive) stress Refer PSG Data book P.No. 8.13, Table (8) c  0.74

i1 ib

E [M ]   a i1

t


2.5  1 200

 0.74

2.95

 1.7  10  351199   2.5  60

2.5  1

5

 483.34 N /mm 2  [ c] 772.2 N/mm 2  The design of pinion is satisfactory.

Step 20: Check the stresses in the wheel: (a) Gear Material Given - cast iron - Grade (30) Since contact area is same, c

wheel

 c  483.34 N /mm 2 (Already calculated) pinion

Life of Wheel  N Wheel 

N pinion 2.5



54  10 7  21.6  10 7 cycles 2.5

From PSG Data book P.No. 8.16, [c]  design surface contact compressive stress  C B  HB  k cl

For CI - Grade (30) HB = 200 to 260 BHN take HB = 260 BHN, C B  23 ; HB  260 kcl  from PS G Data b ook P.No. 8.17, table (17)



  6

10 7  N

  6

10 7

21.6  10 7

 0.5992

2.96


[c]  23  260  0.5992  3583.216 kgf /cm 2

Convert into N/mm 2  358.3216 N/mm 2 c 483.34   [ c] 358.3216 

The design of wheel is not satisfactory. Therefore, change the recalculate [c] and  c  [c]

material

of

wheel

and

Select another material, From PSG Data book P.No. 8.16; Table 16 Select either C40 or C45 Steel; k cl  Surface > 350 hardness Heat treatment - Surface  N  25  10 7 hardened. Select C40 - Surface hardened HRC  40 to 55

Assume HRC  50 ; C R  230 ;



 

10 7 N

= 0.5992

kcl  0.5992

Design surface [ c]  C R HRC kcl Contact Compressive stress [ c]  230  50  0.5992  6890.8 kg f /cm 2 [c]  689.08 N/mm 2


2.97

  c 483.34   [c] 689.08

Therefore the selection of material C40, Surface hardened is satisfactory. Checkup of b induced bending stress; b y1  b y2 1

2

[b ]2 - design bending stress of wheel 2 

1.4 k bl k  n

b  [b]2 2 wheel

Find [b]2

 1

at Z  63 ; x  0 by interpolation.

wheel 

Design bending stress 1.4 k bl  [ b]wheel  1 k n kbl  1 From PSG Data book P.No. 8.20 table (22)  350 BHN core hardness; N  10 7 n  2.5 From PSG Data book P.No. 8.19 table (20) k  1.5 From PSG Data book P.No. 8.19 table (21) [b] wheel 

1.4  1  2825 2.5  1.5

[b] wheel  105.46 N/mm 2

2.98


Pinion: b  77.49 N/mm 2 1

y1  0.4205

Wheel b  Induced ben ding stres s 2

(to be calculated) y2  0.4927 b y 1   b y 2 1

2

b  2

77.49  0.4205 0.4927

 66.13 N/mm 2

b 66.13   [b]2 105.46 N/mm2 2

Gear design is satisfactory. Problem 2.13: In a spur gear for a rock crusher, the gears are made of case hardened steel. The pinion is transmitting 18 kW. at 1200 rpm, with a gear ratio of 3.5; The Gear is to work 8 hrs/day for 3 years. Design the drive. [May, June 2009],(Oct. ’96)

Given Data: Power to be transmitted  18 kW Speed  1200 rpm Gear ratio  i  3.5 Life required  8 hr  day for 3 years  8  365  3  8760 hrs


2.99

Step 1 Selection of material Assume pinion and wheel are made of same material. 

Design only pinion

(Because pinion is weaker than wheel)

From PSG Data book - P.No. 8.16, Table 16 Select Alloy Steel - Case Hardened Surface hardness  55 to 63 HRC  55 to 63 Assume HRC  60

 i.e, case hardness  30 BHN  There fore core hardness  350 BHN  Step 2 Required Life  8760 hrs given  N  8760  60  rpm  8760  60  1200  63.07  10 7 cycles

Step 3 Refer PSG Data book pg. 8.14 Table (9) for steel (both pinion and gear) E eq.  2.15  10 6 kg f  cm 2

Step 4 Calculation of Design bending stress [b] Refer PSG Data book P.No. 8.18 Assume, rotation in one direction only.

2.100


[b] 

1.4

kbl

n. k

 1

kbl  Life factor for bending   From PSG Data book  P.No . 8.20 Table 22   

(Always consider core hardness)

Life in cycles  Hrs  60  rpm  8760  60  1200  63.07  10 7 cycles

 350 BHN for

   350 BHN   7 and life  10 cycles 

surface hardness

The value of kbl  1

n  factor of safety, from PSG Data book P.No. 8.19,

Table 20 (for steel and case hardened) n2 k  Fillet stress concentration factor

From PSG Data book P.No. 8.19, table (21) Consider always Addentum modification coefficient. 0    0.1  for case hardened steel.

the value of k  1.2  1  Endurance limit in bending kg f  cm 2


2.101

From PSG Data book P.No. 8.19, table 19 for alloy steel,  1  0.35 u  1200 where u  ultimate tensile stress kg f  cm 2 For u value, refer PSG Data book P.No. 8.5 table No.7 u  155 kgf  mm 2 assume u  160 kg f  mm 2 convert into kgf  cm 2  160  102 kgf  cm 2

 1  0.35  160  10 2  1200  6800 kg f  cm 2

Design bending stress 1.4 kbl 1.4  1  1   6800  [b]  n  k 2  1.2 [b]  3966.66 kg f  cm 2

Step 5 Calculation of design contact compressive stress [ c] P.No. 8.16 [c]  C R HRC  kcl C r  coefficient depe nding on su rfa ce hardness

From PSG Data book P.No.8.16, table (16)

2.102


for alloy steel and case hardened  280 HRC  60 assumed in the step 1 kcl  Life factor  steel From PSG Data book P.No.8.17, table 17 N  63.07  10 7 cycles} For su rface hardness  350 and N  25  10 7} kcl  0.585 [ c]  D esign surface contact compressive stress  C R  HRC . k cl  280  60  0.585 [ c]  9828 kg f  cm2

Step 6 Calculation of Design torque (or) twisting moment) Refer PSG Data book P.No.8.15 [M t]  Design twisting moment  M t kd  k Mt 

nominal twisting moment transmitted by the

pinion - kg f  cm  97420

kW n


 97420 

2.103

18  1461.3 kg f  cm 1200

Step 7 Calculation of Approximate centre distance (a)

a  i  1

  3

2  0.74  E [M t]  [ ]  i   c  

i  35 [c]  9828 kg f  cm 2

(+) for external

E  2.15  10 6 kgf  cm 2

 fo r Internal

[Mt]  1899.69 kg f  cm   0.3, Assume PSG Data book

a 

table 10, P.No. 8.14

  3

 0.74   9828   

2

 12.61 cm

2.15  10 6  1899.69 3.5  0.3

 125 mm

(From R10 series)

Step 8 Assume Z 1  20 teeth ,

a  125 mm

From PSG Data book Page No. 8.1 Z2 Z1

i

Z 2  20  3.5  70 teeth

2.104


Step 9 Calculation of module (Refer PSG Data book P.No. 8.22) Module  m 

2  125 2a   2.77 mm Z 1  Z2 20  70

Take standard module, m  3 mm From PSG Data book P.No. 8.2 table No. 1 Step 10: Recalculate ‘a’ centre distance a 

m Z 1  Z 2 2 3 20  70  135 mm 2

Take R10 series, Centre distance  a  135 mm Step 11: Calculation of ‘b’ face width Refer PSG Data book P.No.8.1, 

b  0.3 a

 are given in Pg. 8.14, table 11

Face w idth  b  0.3  a  0.3  135  40.5 mm

Step 12: Calculation of P.C.D. d1 From PSG Data book P.No. 8.22 PCD of pinion d1  mZ 1  3  20  60 mm


2.105

Step 13: Calculation of pitch line velocity From PSG Data book P.No: 8.15 v

 d1 N1   60  1200   3.769 m/sec. 60  1000 60  1000

Step 14: Select preferred Quality From PSG Data book Table 2: Page No. 8.3 Above 1 m/sec to 8 m/sec Preferred Quality IS 8 From the table 2, based on peripheral speed (or) pitch line velocity. Step 15: Calculation of kd and k Cal. of p 

b d1

*In the previous step, kd and k values were assumed as 1.3; Now cal. kd and k value. p 

40.5  0.675  0.8 60

Step 16: Take the value of k. (Load concentration factor) From PSG Data book P.No: 8.15 table 14 IS 8 Quality, at p  0.8 the value of k  1.06 (assume any one condition) k  1.06

2.106


Step 17: Select kd: Dynamic Load factor From PSG Data book Table 15; P.No: 8.16 for IS 8 Quality, Cylindrical gears, Pinion surface hardness  350 ; Pitch line velocity up to 8 m/sec Dynamic load factor  k d  1.4 Step 18: Recalculate Design twisting moment [M t] [M t]  M t  kd  k

M t  1461.3 kgcm

 1461.3  1.4  1.06

k d  1.4

Refer PSG Data book Pg 8.15, table 13  2168.5692 kg fcm

k  1.06

Step 19: Checking of bending stress From PSG Data book P.No. 8.13 table 8 b 

i1 [M t]  [ b] amby

[M t]  2168.5692 kgfcm i  3.5 ; a  135 mm  13.5 cm m  3 mm  0.3 c m b  40.5 mm  4.05 cm y  form factor

- From PSG Data book P.No. 8.18

table (18) addendum  modification coefficient x  0


2.107

for Z  20, and x  0 y  0.389 b 

3.5  1 [2168.5692 ] 16  0.3  4.05  0.389

1088.8 kg f/cm 2  [ b] 3966.66 kg f/cm 2  Design based on bending stress is satisfactory. Problem 2.13: Desin a spur gear pair to transmit 1.5 kW at 1440 rpm from an electric motor to an air compressor running at 720 rpm. Take the working life as 10,000 hrs. Material to be used is cast iron grade 25 for both pinion and wheel. [Nov/Dec: 2010]

[Similar to problem: 2.12] Problem 2.14: Design a straight spur gear drive. Transmitted power 8 kW. Pinion speed 764 rpm. Speed ratio is 2. The gears are to be made of C45 steel. Life is to be 10,000 hours. [May/June 2013]

Given Data Power P  8 kW ; Pinion speed n  N 1  764 rpm; Speed reduction i  2; N  10,000 hrs Solution Since the gears are of same material, design only pinion Step 1: Material properties Since material of gears are C45 steel from PSG Data book, Pg.No. 8.16, Table 16 For C45 steel, Assume surface hardened

2.108


HRC  40 to 55

Assume HRC  55 ; C R  230 Step 2 Life required  10,000 Hrs  10,000  60  764 Life required N  45.84  10 7 cycles Step 3 Calculation of Equivalent Young’s Modulus From PSG Data book, Pg.No. 8.14, Table 9, E  2.15  10 6 kg f/c m2  2.15  10 5 N/mm2

Step 4 Calculation of design bending stress [ b] From PSG Data book, Pg.No. 8.18 [ b] 

1.4 kbl 1 n  k

[assume rotation is in one direction only] where k bl  Life factor for bending From PSG Data book, Pg.No. 8.20, table 22 kbl  1

From PSG Data book, Pg.No. 8.19, Table 21 k  stress concentration factor

Assume surface hardened steel, modification cofficient ‘X’  0  X  0.1  k   1.5

Addendum


2.109

From PSG Data book, Pg.No. 8.19, Table 19  1  Endurance limit stress in bending for complete reversal of stresses For forged steels  1  0.25  u   y  500 From PSG Data book, Pg.No. 1.9, corresponding to C45 steel u  63 to 71 kg f/mm2  71 kgf/mm 2 assume  71  100 kg f/cm 2  7100 kgf/cm 2

y  36 kgf/mm 2  36  100 kgf/cm 2  3600 kgf/cm 2    1  0.25 7100  3600   500  3175 kg f/c m2

From PSG Data book, Pg.No. 8.19, Table corresponding to steel, forged, surface hardened

20

n  2.5  Design bending stress [b] 

1.4  1  3175 kg f/cm 2  1185.33 kgf/cm 2 2.5  1.5

[ b]  118.53 N /mm 2

Step 5 Calculation of Design compressive) stress [ c]

surface

From PSG Data book, Pg.No. 8.16,

(contact

2.110


[c]  C R HRC k cl

From PSG Data book, Pg.No. 8.16, Table 16 corresponding to C45 steel, surface hardened C R  230, HRC  55

From PSG Data book, Pg.No. 8.17, corresponding to steel, surface hardness 7350

Table

17

kcl  life factor  0.585  [ c]  230  55  0.585  7400.25 kgf/cm 2 [ c]  740.02 N/mm 2

Step 6 Calculation of design torque [M t] From PSG Data book, Pg.No: 8.15 [M t]  M t k d  k

where M t  97420

8 kW  97420   1020 kg f  cm 764 n

From PSG Data book, Pg.No. 8.15 Initially assume k d  k  1.3  [M t]  1020  1.3  1326 kg f  cm [M t]  132600 Nmm


2.111

Step 7: Calculation of Aproximate centre distance ‘a’ From PSG Data book Pg.No. 8.13

a  i  1

Here  

 

2  0.74  E [M t]  [ ]  i c  

3

b a

From PSG Data book, Pg.No. 8.14, Table 10 Assume   0.3

a  2  1

  3

2

2.15  10 6  1326  0.74   7400.2   2  0.3  

a  10.865 cm  a  108 mm

Step 8: Calculation of Z 1 and Z 2 Assume Z 1  number of teeth on pinion  25 teeth From PSG Data book, Pg.No. 8.1 Speed reduction i 

Z2 Z1

where Z 2  number of teeth on wheel  i Z1  2  25  50 teeth  Take Z 2  50 teeth

2.112


Step 9: Calculation of module From PSG Data book, Pg.No. 8.22 Take module  m 

2  108 2a  2.88 mm  Z 1  Z 2 25  50

From PSG Data book, Pg.No. 8.2, Table 1 Take standard module m = 3 mm Step 10: Recalculate centre distance ‘a’ Since a 

m Z 1  Z 2 3 25  50   112.5 mm 2 2

From PSG Data book, Pg.No. 7.20, rounding the centre distance to R10 series a  125 mm

Step 11: Calculation of face width b Since  is assumed as 0.3, but  

b  0.3 a

 b  0.3  125  37.5 mm

Step 12: Calculation of PCD d 1 and PCD d 2 Pitch circle Diameter d 1  m  Z 1  3  25  d 1  75 mm  0.075 m

Pitch circle diameter d 2  m  Z 2  3  50  d 2  150 mm  0.150 m


2.113

Step 13: Calculation of pitch line velocity v  Since v

 d1 N 1 60



  0.075  764  3 m /s 60

Step 14: Selection of Quality of gear From PSG Data book, Pg.No. corresponding to velocity upto 8 m/s

8.3,

table

2,

Select the quality of the gear, IS quality 8

Step 15: Load concentration factor ‘k’ From PSG Data book, Pg.No. 8.15, table 14 k depends upon  p 

b value d1

37.5  0.5  p  75

Corresponding to p  0.5 k  1.03

Step 16: Dynamic load factor “k d” From PSG Data book, Pg.No. 8.16, table 15, for pitch line velocity upto 8 m/s, cylindrical gears, surface hardness  350 BHN kd  1.4

Step 17: Recalculate [M t] design twisting moment, From PSG Data book, Pg.No. 8.15 [M t]  Mt  kd  k  1020  1.03  1.4

2.114


 1470.84 kgf  cm  147084 Nmm Step 18: Checking induced bending stress b From PSG Data book, Pg.No. 8.13, Table 8, b 

i1 [M t]  [ b] amby

From PSG Data book, Pg.No. 8.18, table 18, y depends on X and Z value. For Z 1, X  0 y  0.4205 21  140784  b  125  3  37.5  0.4205  71.42 N/mm 2  [b] 118.53 N/mm 2

Step 19: Check induced surface (contact compressive) stress c From PSG Data book, Pg.No. 8.13, Table 8,

c  0.74

i1  a

 0.74 

21 125

 

i1  E  [M t] ib

21  2.15  10 5  140784    2  37.5

 617.97 N/mm 2  [c] 740.02 N/mm 2

The design is satisfactory.


2.115

Problem 2.15: A motor shaft rotating at 1500 rpm has to transmit 15 kW to a low speed shaft with a speed reduction of 3:1. Assume starting torque to be 25% higher than the running torque. The teeth are 20 involute with 25 teeth on pinion. Both the pinion and gear are made of C45 steel. Design a spur gear to suit the above conditions and check for compressive and bending stresses and plastic deformation also sketch the spur gear drive. [Design based on contact stress]. [Apr/May 2008]

Solution Given data Speed of driver  N 1  1500 rpm Po wer P  15 kW Gear Ratio i  3 No of teeth on pinion Z1  25 teeth

Step 1 Selection of material (from P.S.G Data book Pg.No. 8.5) C45, surface hardened to 55 RC core hardness  350 BHN u  720 N/mm2,  y  360 N/mm 2 [b]  1400 kg f/cm 2  140 N/mm 2 [c]  5000 kgf/cm 2  500 N/mm 2

2.116


Step 2 Calculation of design torque [M t] From P.S.G Data book, Pg.No. 8.15 [M t]  M t  kd  k  1.25

since starting torque is 25% higher than the running torque But M t  97420

kW N1

 97420 

15 1500

 974.2 kg f cm

Initially assume, for symmetry scheme kd  k  1.3  [M t]  974.2  1.3  1.25  1583.07 kgfcm  [M t]  158307 Nmm

Step 3 Calculation of Approximate centre distance ‘a’ from P.S.G Data book, Pg.No. 8.13

a  i  1

  3

2  0.74  E [M t]  [ ]  i c  

From P.S.G Data book, Pg.No. 8.13


2.117

E  2.15  106 kgf / cm2  2.15  10 5 N/mm 2

Assume  

b  0.3 (from P.S.G Data book, Pg.No. a

 a  3  1

 

8.14)

3

2

 0.74  2.15  10 6  15383.07   3  0.3  500 

a  17.436 cm a  174.36 mm

Step 4 Calculation of Z 1 and Z 2 Z1  25 assume

since

i3

From P.S.G Data book, Pg.No. 8.1 speed reduction i 

Z2 Z1

 Z 2  No. of teeth on wheel  i  Z 1  3  25  75 teeth

Step 5 Calculation of module From P.S.G Data book, Pg.No. 8.22


2.118

module  m 



2a Z1  Z2 2  174.36 25  75

 3.48 mm

Take standard module, from P.S.G Data book, Pg.No. 8.2  Module m  4 mm

Step 6 Recalculate centre distance ‘a’ and rounded to R10 series a 

m Z 1  Z2 2

4 25  75 2

a  200 mm

Step 7 Calculation of face width b  value already assumed as 0.3

but  

b  0.3 a

 face width  b  0.3  200  60 mm

Step 8 Calculation of PCD d 1 and PCD d2


2.119

PCD of pinion  PCD 1  d1  m  Z 1  4  25  100 mm PCD of wheel

 PCD2  d2  m  Z 2  4  75  300 mm

Step 9 Calculation of pitch line velocity v v 

d1 N 1 m/s 60   0.1  1500  7.85 m/s 60

Step 10 Find load concentration factor k from P.S.G Data book, Pg.No. 8.15 b k depends on  p  value d1 

60  0.6 100

at  P  0.6, the value of k  1.03 Step 11 Dynamic load factor kd from P.S.G Data book, Pg.No. 8.16 For pitch line velocity up to 8 m/s, cylindrical gears, surface harndess  350 BHN

2.120


kd  1.4

Step 12 Recalculate, [M t], design hoisting moment [M t]  M t  kd  k  1.25

 974.2  1.4  1.03  1.25  1755.99 kg f cm  M t  175599 Nmm

Step 13 Check b induced bending stress from P.S.G Data book, Pg.No. 8.13 b  

i1 [Mt]  [ b] amby 31 [175599 ] 200  4  60  0.4205

 1400 kg f/cm 2  139.19 N /mm2  140 N/mm2 ~

[y  Form factor from P.S.G Data book, Pg.No. 8.18 y depends upon X and Z value  y  0.4205 ]

Step 14 Check c, induced surface stress From P.S.G Data book, Pg.No. 8.13 c  0.74

i1 a

i1

E [M ]  [ ] b   t

c


2.121

E  2.15  10 6 kg f/cm 2  c 0.74 

4 200

 2.15  10  1755.99   3  60 4

6

 4286.82 kgf/cm 2  [ c] 5000 kgf  cm 2  Design is safe

Step 15 Check for plastic Deformation: M t - Rated Torque  1755.99 kg f  c m  175.599 N  m

Starting Torque 25% higher than starting Torque (125%) [M t]ma x 

125%  Mt 100

 1.25  175.599  219.49875 N  m  2194.9875 kg f  cm

Step 16 Check for Bending  b max  b 

[M t]max Mt

 1400 

2194.9875 1755.99

 1750 kgf  cm

(Or) b max  1.25   b  1.25  1400  1750 kgf  cm

2.122


[b]max  0.8   y  0.8  3600  2880 kgf/cm 2

For safe Design, [b]max  b max By comparing satisfactory.

this

two

values,

the

Design

is

Step 17 Check for wear strength c max   c  1.25  4286.82  1.25  5358.525 kg f/c m2 [c]max  3.1  y  3600  3.1  11160 kg f/c m2 [c]max  c  The Design is safe. Problem 2.16: Design a spur gear drive to transmit 10 kW at 1440 rpm, speed reduction is 3. Take pressure angle as 20 and working life of the gears as 15,000 hrs. Assume the materials for pinion and wheel as heat treated cast steel and high grade cast iron respectively.

[Nov; Dec: 2012]

[Similar to the problem: 2.12] Problem 2.17: Design a spur drive to transmit 5 kW at 900 rpm of forged steel pinion 120 mm diameter to a cast steel gear to run at 300 rpm. Take pressure angle of 20 and working life of the gears as 10000 hrs.

[Similar to the problem 2.12]

[May/June 2012]


2.123

2.16 HELICAL GEARS For high pitch line velocities and heavy loads, helical gears are used. Two types of gearing are (i) Single helical gears. (ii) Double helical gears. (Herringbone gears) When straight tooth spur gears begin to engage, the contact theoretically extends across the entire tooth on a line parallel to the axis of the gear. This sudden application of load results in noisy operation and high impact stress. In helical gearing, contact begins at one end of the entering tooth and gradually extends along a diagonal line across the tooth face as the gear rotates. The gradual engagement of load application reduces the noise and the dynamic load. Therefore, the operation is silent and hence higher pitch line velocities can be employed. (Generally pitch line velocities upto 20 to 35 m/sec) Applications: Automobile Gears, turbines. Herringbone Gears pitch line velocity upto 60 m/sec. Since the teeth are inclined to the axis of rotation, helical gear is subjected to an axial thrust. To reduce axial thrust, another helical gear of opposite hand is to be

(b) D o ub le helica l (a) Single he lical g ea r g ea r Fig: 2.15

2.124


provided. This arrangement is called double helical Gear (or) Herringbone gear.  mn cos   m m

Pc

PN



mn Fig:2.16. Helical gear (no menclature)

cos

where   helix angle (Real pitch) P  Circular pitch along the plane perpendicular to the axis. Pm

Normal pitch P n 

Circular pitch along the plane

perpendicular to the teeth. Pn   m n.

where   helix angle  8  25 for helical gears  25  40 for herringbone gears. where m n  normal module in m.m.

2.17 BEAM STRENGTH OF HELICAL GEARS (LEWIS EQUATION FOR HELICAL GEARS) As discussed earlier in this chapter (spur gear), The formative gear is an Imaginary spur gear in a plane perpendicular to the tooth element. It is understood that the beam strength of virtual (or) equivalent or formative


2.125

gear is nothing but the beam strength of Helical gears. Therefore a modified lewis equation is used to determine the beam strength of a helical gear. F s  b [b] y Pc

where

Pc    mn

F s  Beam strength of Helical gear m n  normal module b  face width P n  normal pitch [b]  Permissible (or) Allowable static stress y  Tooth form factor (or) lewis form factor.

y  0.124 

0.684 for 14 12 full depth Involute Zv

 0.154 

0.912 for 20  Involute Zv

 0.175 

0.95 for 20  Involute subteeth Zv

Z v  virtual number of teeth.

The beam strength F s indicates the maximum value of tangential force that the tooth can transmit without bending failure.

2.126


2.18 DYNAMIC LOAD ON HELICAL GEAR TOOTH (EFFECTIVE LOAD ON GEAR TOOTH) As discussed in previous chapter, in addition to the static load due to power transmission, there is dynamic load between the meshing teeth. In order to account for dynamic loads, the following two methds are used. 1. Approximate estimation of dynamic load using the velocity factor, which is used in the initial stages of gear design, and 2. Accurate estimation of dynamic load using the Buckingham’s equation, which is used in the final stages of gear design. 1. Calculation of initial dynamic load F d The preliminary value of dynamic load can be calculated by using the relation where

F t  tangential load in kg f Fd  Ft  Cv

Ft 

where

Power in Horse power  75 vm P  Power transmitted in Horse power vm  mean velocity 

 dN m/s 60

C v  velocity factor (C v values are the same as for spur gears) 

3  vm , for vm  10 m/s, commercially cut gears 3




6  vm 6

2.127

, for vm  5 to 20 m/s, carefully cut gears



5.5  vm , for vm  20 m/s, precision gears 5.5



1  vm , for non-metallic gears. 1  0.25 vm

2. Buckingham’s equation for dynamic load Buckingham’s equation, used for accurate estimation of dynamic load, is given by Dynamic load  0.164 V cb cos3   F  cos  m t F d  F t   cb c o s3   F t   0.164 V m  1.485  

where

    ...(6.15) 

F t  Tangential load neglecting service factor 

P  75 vm

vm  Pitch line velocity (or) mean velocity m/sec. b  Face width c  Deformation or dynamic factor in kg f/cm

  Helix angle

2.19 WEAR STRENGTH OF HELICAL GEARS (WEAR TOOTH LOAD) The wear strength equation of the spur gear is modified to suit helical gears. Thus the modified wear strength equation for helical gears is given by

2.128


Wear load, F w  where

d1  b  Q  k c o s2 

(6.16)

d1  Pitch circle diameter of pinion (use pinion

diameter irrespective of whether pinion or gear is designed) b  Face width, Q  Ratio factor, 

2Z2 2i  , for external gears i  1 Z1  Z2 

2Z2 2i  , for internal gears i  1 Z2  Z1

where i  velocity ratio 

Z2 Z1

k  Load stress factor (also known as material

combination factor), in kg f/cm 2  [ ]2  sin   1 1  c n      E 14 E   p g 

where

[c]  Surface endurance limit in kg f/cm 2

n  Normal pressure angle, and E p and Eg  Young’s modulus of pinion and gear respectively.

  Helix angle.


2.129

Note: In the design of helical gears, the normal module m n should be selected from standards. For the recommended series of modules, From PSG Data book pg:8.2; table: 1.

2.20 PRESSURE ANGLE IN THE NORMAL AND TRANSVERSE PLANE 2.20.1 Pressure angle  Pressure Angle is the angle formed by the line of action with common tangent to the pitch circles of mating gears.

Pitch p oint

For Involute system of gears, the pressure angle is constant and it may be 1 (or) 20 14 2  14

Pre ssu re An gle Velocity o f Pitch P oint

Axis o f Tra nsm ission

Co mm on Tan ge nt Fig: 2.17(a)

1 or 20  2

2.20.2 Transverse Pressure Angle t The pressure angle measured in transverse plane is known as transverse pressure angle  t 2.20.3 Normal Pressure Angle n  The pressure angle measured in the normal plane is known as Normal pressure angle  n 

where

tan  n tan  t

  Helix angle

2.130


Pn

n



t

Pt Fig:2.17(b).Helical gear nom enclature

Formative or Equivalent Number of teeth for Helical gears The formative or equivalent number of teeth for a helical gear may be defined as the number of teeth that can be generated on the surface of a cylinder having a radius of curvature at a point at the tip of the minor axis of an ellipse obtained by taking a section of the gear in the normal plane. Mathematically, formative or equivalent number of teeth on a helical gear. Zv 

Z cos 3 


2.131

  Helix angle Z  Actual number of teeth on a helical gear.

It is also called as virtual number of teeth.

2.21 FORCE ANALYSIS OF HELICAL GEARS In the case of spur gears, the normal force acting on the plane of rotation of gears and inclined at an angle  to the common tangent, to the pitch circle diameters of mating gears, at the point of contact, whereas for helical gears, the force will be inclined to the plane of rotation by an angle , known as helix angle, and to the tangent by an angle  , the pressure Angle. Hence the Resultant normal force F n acting on the helical gear tooth can be derived into three components such as tangential force F t, Radial component of force F r and axial component of force F a .

Derivations for F t, F r and F a Fn 

Resultant (normal) force acting on the tooth.

Ft 

Tangential (or) Transmitted force

Fr 

Radial component of force

Fa 

Axial component of force



Pressure angle



Helix angle



Normal pressure angle

t 

Transverse pressure angle

2.132


From the Fig. 2.18 (A) A

E Fn

D

B 

C

Ft

H

cen tre lines o f teeth

F



Fr

Fa

G  Pitch C ylinder

n h ere (A)

A

B F n cos a

Fn

Fr

Fa

n B

G

Ft

C

(B)

 Ft (C )

Fig:2.18.

A G  Normal force acting on the tooth  F A D  Axial component of force  F a AE 

Tangential component of force  F t

AB 

Radial component of force  F r

From triangle ABG sin  

Opposite side Hypotenuse

G


2.133

 AB sin    AG  AB  F r  AG  F n  BG  F t   AB  AG sin   Fr  Fn sin  cos  

(1)

Adj Hyp

 BG cos    AG   BG  AG cos   Ft  F n cos 

From Triangle BCG  BC sin    BG

(2)

. .  . BG  Fn c o s 

  BC  BG sin   F a  F n cos  sin  (3)  . .   . AD  BC  F a  CG cos     BG . .   . CG  AE  F t   CG  BG cos   F t  Fn cos  c o s  (4)

The radial and axial components can expressed in terms of tangential components Divide the Equation (1) by equation (4)

also

be

2.134


1  4 Fr Ft



Fn sin  F n cos  cos 

Fr

 sin    1   F t  c o s    cos    tan   Fr  Ft     cos  

Divide the Equation (3) by Equation (4) 3  4 Fa Ft Fa Ft Fa Ft

 F t sin 





F n cos  sin  F n c o s  cos  sin  cos 

Fa  F t tan 

. . sin     . tan   cos   

The tangential force can be derived from the torque Mt Torque  Tangential Force  Radial Distance Mt  F t  r

Ft  Ft 

Mt r

2Mt d

d  r 2   


Power

P   Mt 

where

2.135

2 N M t 60 60  P 2 N

P  Power; N  Speed in rpm d  Pitch circle diameter.

2.22 DESIGN OF STRAIGHT TOOTH HELICAL AND DOUBLE HELICAL GEAR BASED ON STRENGTH AND WEAR CONSIDERATIONS The design of helical gear is almost similar to spur gear drive design with slight modification in Lewis and Buckingham Equations due to presence of Helix angle  in Helical gear. . . ( .   Helix angle)

Step 1 Similar to spur gear Design Step 2, Step 3 and Step 4 Similar to spur gear Design Step 5 Calculation of strength of gear tooth F s F s  [ b] b y v Pn

yv  0.124   0.154 

Pn   mn m n  normal pitch

0.684 for 14 12 Involute Zv 0.912 for 20  Involute Zv

2.136


 0.175 

0.95 for 20  Involute stub teeth Zv

From PSG Data book pg: 8.50 and 8.51 b  10  m n [b] is the Design bending stress in kg f/cm2. Choose

the [b] value from Data book pg: 8.5 Table: 7. Step 6 Calculation of Normal module m n For Safe Design F s  F d By Equating this two equations, determine the value of normal module. [Choose the nearest standard normal module value from PSG Data book pg: 8.2 Table: 1] Step 7 Calculate the value of face width (b) b  10  m n

Step 8 Calculate the pitch circle Diameter d1 d1 

mn cos 

 Z1

  Helix Angle Step 9 Calculation of pitch line velocity


v

 d1 N 1 60  1000

2.137

 v for m/sec V for m/min

N 1  rpm d 1  mm

Step 10 Recalculate the stength of gear tooth F s F s  [b] b yv   m n

Substitute m n which is calculated in previous step 5. Step 11 Calculation of Buckingham’s Dynamic load F d  0.164 V cb cos 3   F  c o s  t m F d  F t   cb cos 3   F t  0.164 V m  1.485   

    

C, b and V m are similar to spur gear Design. The above Equation F d is called modified Buckingham’s

Dynamic load for Helical gears. Step 12 Check for Beam strength For safe Design F s  Fd If F s  F d, then the Design is not safe, the value of m  should be increased to increase the strength of gear tooth.

2.138


Step 13 Calculation wear load F w :

Fw 

d 1 Q kb cos 2 

1   1  [2c ] sin  n  E1 E2    ; k 1.4

n - normal pressure Angle c, E 1 and E 2 are similar to spur gear design. Step 14 Checking for wear strength For safe Design F w  Fd If F w  F d, then the Design is not safe. The value of k is increased and then Recalculate the value of F w again. k is the load stress factor.

Step 15 Calculate the Basic Dimensions of Helical gear From PSG Data book PN 8.22; Table: 26


2.139

Basic Dimensions of Externally Meshing Standard Cylindrical Gears Suffix 1 for Pinion and 2 for Wheel Nomenclature

Notation

Units

Module

m

mm

Normal module

mn

mm

Transverse

mt

mm

module Centre distance Height factor

a f0

mm 

Helical and Herringbone Gears mn 

m1  a

2a cos  Z1  Z2 mn

; mt 

cos 

2a Z1  Z2

ma Z1  Z2 2

c o s

fo  1 (in special cases fo  0.8, stub teeth)

Bottom clearance

c

mm

c  0.25 mn (For stub teeth, c  0.3 mn

Tooth depth

h

mm

h  2.25 mn (For stub teeth, h  1.9 mn

Pitch diameter

d

mm

or Reference diameter Tip diameter

dc

mm

d1 

m Z ; cos  1

d2 

cos 

mn

 da1     da2   

Z2

Z1

  2f0  mn cos   Z2   2f0  mn cos  

2.140


Nomenclature

Helical and

Notation

Units

Root diameter

df

mm

 df1     df2   

Helix angle



deg.

For helical,   8 to 25

Herringbone Gears Z1

  2f0  mn  2c cos   Z2   2f0  mn  2c cos  

For herringbone,   8 to 25 cos   Number of teeth

Z

Virtual number

Zv



Z1 

2a

2a cos  ; Z2  iZ1 mn i  1

Zv1 

of teeth

mn Z1  Z2

Z1 3

cos 

Step 16 Draw the neat sketch of Helical gear

H e lix Too th

H e lix a ng le A xis

Fig: 2.17

; Zv2 

Z3 cos3 


2.141

Problem 2.18: Design a helical gear drive to transmit a power of 15 kW at 1400 rpm to the following specifications speed reduction is 3, pressure angle is 20, Helix angle is 15, The material of both gears is C45 steel, Allowable static stress

is

180 N/mm2

Young’s

modulus

of

the

material

 2  105 N/mm2.

Step 1 Given data power P  15 kW  15  10 3 W Pinion speed N 1  1400 rpm Speed reduction (i)  3 Pressure angle   20 Helix angle   15 Material for pinion  C45 steel Material for gear Static stress

 C45 steel

[b]  180 N/mm 2

Young’s modulus E 1  E 2  2  10 5 N/mm 2 Step 2 Selection of material Pinion – C45 steel Gear – C45 steel Both pinion and gear are made up of same material (C45 steel). Pinion is weaker, therefore design for pinion.

2.142


Step 3 Calculation of tangential load F t F t  HP 

power

P   15 kW

75 vm

15  10 3 H.P 750

 20 H.P vm 

 d1 N 1 60  1000



3.14  d 1  1400 60  1000



3.14  20.705  m n  1400 60  1000

vm  1.517  m n m/s Ft 

20  75 988.7936  kg f mn 1.517  m n

Assume Z 1  20 d1 



mn Z1 cos  20  m n cos 15

d1  20.705  mn

where

m n  normal module.


2.143

Step 4 Calculation of Dynamic load F d Fd  Ft  Cv Cv 

3  vm ; 3

Assume vm  10 m/s, commerically cut wheels. Cv 

3  1.517 m n 3

 3  1.517 m n  F d  988.7775  mn    kg f 3   Step 5 Calculation of strength of gear tooth F s F s  [b] b yv P cn

b  10  m n

 [b]  b  yv    m n y v  0.154 

0.192 For 20 Involute teeth Zv

(yv, from PSG Data book pg: 8.50) Zv  

Z1 cos 3  20 3

cos 15

 22.974 ~  23

2.144


y v  0.154 

0.912 23

[ b]  180 N/mm2 [ b], convert into kgf/c m2

 0.1143

[b]  18  10 2 kgf/c m2

2

F s  18  10  10  m n

. .  . 1N ~  10 kgf

 0.1143    m n F s  6460.236 m 2n

Step 6 Calculation of Normal module m n For Safe Design, F s  F d  3  1.517 m n  6460.236 m 2n  988.755   3   6460.236 m 2n  329.5916 [3  1.1517 m n] 6460.236 m2n  379.590 m n  988.775   6460.236 m 2n  0.05875 m n  0.153055 m 2n  0.05875 m n  0.153055

By solving this Equation, we get m n  0.4217;

m n   0.3629

(- Not possible) m n  0.4217 c m

 5 mm mn ~


2.145

Step 7 Calculate the value of face width b b  10  m n  10  5  50 mm

Step 8 Calculate the pitch circle diameter d1  

mn cos 

 Z1

5  20 c o s 15

d 1  103.52 ~  104 mm

Step 9 Calculation of pitch line velocity vm  

 d1 N 1 60  1000 3.14  104  1400 60  1000

vm  7.619 m/s

Step 10 Calculate the strength of gear tooth F s F s  [ b]  b  yv    m n

[ b]  18  10 2 kg f/c m2  50   5   18  10 2    0.1143  3.14      10   10   1615.059 kgf

2.146


[b]  18  10 2 kg f/cm2

Substitute m n and b in cm. Step 11 Calculation of Dynamic (Buckingham’s) load F d Fd  Ft  FI  0.164 V cb cos 3   F  cos  m t  F t   cb cos 3  F t  0.164 Vm  1.485   

    

V m  7.619  60 m/min  477.714 m/m in b  50 mm  5 cm   15

Choose “c” From the PSG Data book pg: 8.53 table 41 and 42. For   20 ; Steel and Steel. c  11860  e from table: 41 from table: 42 module 5 mm; (carefully cut wheels) e  0.025 c  11860  0.025 c  296.5 Ft 

988.7936 mn



988.7936 5

 197.758 kg f


2.147

3  0.164  477.714 296.5  5 cos 15  197.758  F d  197.758    3  0.164  477.714  1.485 296.5  5  cos 15  197.758 

 

F d  197.759 

116600.4643 135.63

 197.758  859.66 F d  1057.427 kg f

Step 12 Checking for Beam Strength For Safe Design F s  F d F s  1615.059 kg f and F d  1057.427 kgf

By comparing both values F s 1615.059   1057.427 F d

therefore the Design is satisfactory. Step 13 Calculation of wear load F w  Fw 

d 1 bQk cos 2 

d 1  104 mm  10.4 cm b  50 mm  5 cm   15  Q

23 2i   1.5 i1 31

    

2.148


1   1  [c]2 sin   E2  E  1  k 1.4 E 1  E 2  2  10 5 N/mm 2  2  10 6 kgf/cm 2 given

  20 [c]  3000 kgf/c m2

From PSG Data book pg: 8.5 table: 7 for steel 1 1    3000 2  sin 20  6 6  2  10   2  10 k 1.4 k

3.078 1.4

k  2.1987 Fw  

10.4  1.5  2.1987  5 cos 15  171.4986

cos 2 15

 183.8116 kg f k  Load stress factor

Step 14 Checking for wear strength For safe Design F w  F d F w  183.8116 kg f and Fd  1057.427 kg f


2.149

F w and F d ; F d  F w therefore the Design is not satisfactory. So we have to increase the “k ”

By

comparing

value. k  2.1987 ; It would be increased to increase the value

of F w . Assume k  21.987  ; Assume the k value; The assumed value of k should increase the wear strength and also the wear strength F w  should be larger value, when compared with Dynamic load F d Recalculation of wear load Fw 

 

d1 Qkb cos 2  10.4  1.5  21.987  5

cos 2 15 1714.986 0.9330

 1838.116 kgf k  21.987 Assu me

Now, F w  F s therefore the Design is satisfactory. Step 15 Calculate the Basic Dimensions of Helical gear 1. module m n  5 mm (Normal module) 2. Transverse module m t 

mn cos 



5  5.176 mm cos 15

2.150


3. Centre Distance a   Z2

mn  Z1  Z2    2 cos     20  60  5    207.05 mm 2 c o s 15    i Z 1  20

Z1

Z 2  3  20  60

4. Height factor f0  1 5. Bottom clearance c  2.25 m n  2.25  5  11.25 mm

6. Tooth Depth h  2.25 mm  2.25  5  11.25 mm 7. Pitch Diameters: d1 

mn

cos 

 Z1 

5  20 c o s 15

 103.5 mm ~  104 mm d2 

mn

5  60  Z2   310.5 cos  c o s 15 ~  311 mm

 Z1   Z6  mn 8. Tip Diameter dc : d c   1  cos     20 215  cos 15    113.52 ~  114 mm


  Z2  Z6  m n dc   2   cos    60 25  c o s 15    320.58 ~  321 mm  Z1   2f0  m n  2 c 9. Root Diameter df : df   1  cos   20    2  5  2  11.25  15 cos    71.02 mm ~  71 mm  Z2  df    2f0  m n  2c 2  cos    60    2  5  2  11.25 c o s 15    278.08 ~  278 mm

10. Helix angle   15 11. No: of teeth: Z 1  20 Z 2  60 12. Virtual no: of teeth Z v : Z v  1

Z1 3

cos 

 22.97 ~  23



20 cos 3 15

2.151

2.152


Zv  2



Z2 cos 3  60 cos 15

 68.92 ~  69 Problem 2.19: A pair of helical gears with 30 helix angle is used to transmit 15 kW at 10,000 rpm. The velocity ratio is 4:1; Both the gears are made of case hardened steel with a static strength of 100 N/mm2. The gears are 20 stub and the pinion is to have 24 teeth. The face width is 14 times the module. Find the module and face width for strength. (Oct ’2006)

Given Data: Helix angle    30 P  15 kW  15  10 3 Watts  20.38 HP

(HP = 736 Watts) i  4 :1 N  10,000 rpm

Both the gears are made of case hardened steel. Static strength  [b]  100 N /mm2  10  1000 kgf/cm 2

[Convert into kgf/cm 2 i.e., N/mm2  10  1 kg f/cm 2]


Pressure angle  20 involute stub. No. of teeth on pinion  24 Face width  b  14 tim es module b  14  m n

Step 1 F s  Beam strength  [ b] b yv  m n F s  1000  14 m n  0.1492    mn  6562.15 m 2n  k gf ... (1)

Transmitted load  F t  HP 

75 vm

When vm  20 m/sec ,  vm 5.5   (precision wheels) 5.5

Cv 

F t  20.38  

75 125.66 m n

12.16  mn

yv  0.175  Zv 

z 3

cos 

yv  0.1492

0.95 Zv 

24 c o s3 30

 36.95

2.153

2.154


vm 

 Z1 m n N1 m/sec 100  60

m n is in cm 

  24  m n  10,000 100  60

vm  125.66 m n

m/sec

V m  7539.82 m n m/min

Dynamic load Cv 

 Fd  Ft  Cv

  125.66 mn 5.5   vm 5.5    5.5 5.5

125.66 mn   12.16  5.5    Fd    5.5 mn   

To

2.2115  5.5    125.66 mn  mn

 2

find

equate

normal

module

m n,

the

equations (1) and (2) 6562.15 m2n 

2.2115 5.5    125.66 m n  mn

 2967.28 m3n  5.5   125.66 mn  2967.28 m3n  5.5   125.66 mn

Calculate m n, by using trail and error method.

above


Trail 1 Assume m n  0.5 cm LHS  371.1175 ; RHS  13.42 LHS  RHS  Not satisfactory.

Trail 2 m n  0.3 cm L HS  80.162 ; RH S  11.639 LHS  RHS  Not satisfactory.

Trail 3 m n  0.2 LHS  23.75 and RHS  10.513 LHS  RHS  Not satisfactory

Trail 4 m n  0.15 c m LHS  10.01; RHS  9.84 LHS  RHS

Therefore,

m n  Normal module  0.15 cm mn  1.5 mm

2.155

2.156


m

mn cos 



1.5  1.732 mm cos 30

Base width  b  14 m  14  1.732 b  24.24 mm Problem

2.20: A helical cast steel gear with 30 helix angle

has to transmit 45 HP at 1500 rpm; If the gear has 24 teeth, determine the necessary module, pitch diameter and face width for 20 full depth teeth. The static stress for cast steel may be taken as 560 kg  cm2. The width of face may be taken as 3 times the normal pitch. What could be the end thrust on the gear? The tooth factor for 20 full depth involute gear may be taken as 0.154 

0.912 ; where TE represents the equivalent TE

number of teeth.

(Apr. 2009)

Given Data: Helix angle    30 P  45 HP; N 1  1500 rpm No.of teeth on pinion  Z 1  24 teeth  T1 Pressure angle   20 full depth involute. Static stress [ b]  560 kg  cm 2 Face width  b  3 normal pitch  3  pn  3 m n

yv  0.154   0.1293

0.912 0.912  0.154  TE 37

TE 

T1 3

cos 



24 cos 3 30 

 36.95  37 teeth


2.157

Beam strength  FS  [ b]  b  yv pn  [ b] 3 m n yv   m n  560  3  m n  0.1293  m n  2143.91 m n2 kg

... (1)

Dynamic Load  F d  F t  C v

F t  HP 

75 vm

vm 

 d1 N1   Z1 m n  N1  60  100 60  100

 F t  45 

60  100

 18.84 m n m  sec

75 18.84 m n

 179.14   kg mn   

  24  m n  1500

Cv 

6  vm 6  18.84 m n  6 6

(Carefully cut gears)

 179.14   6  18.84 m n  Dynamic Load  F d     mn   6    Fd 

29.856 6  18.84 mn m n

... (2)

To find normal module (mm), Equate the equations (1) and (2) 2143.91 m 2n 

29.856 6  18.84 m n mn

71.808 m 3n  6  18.84 m n

2.158


Calculate m n by trail and error method. Trail (1): m n  0.5 LHS  17.952

RHS  15.42

Therefore it is not satisfactory

LHS  RHS

Trail (2): m n  0.4 LHS  4.595

RHS  13.536

LHS  RHS

Therefore it is not satisfactory

The value of m n is in between 0.4 and 0.5 Trail m n  0.45 LHS  14.5411

RHS  14.478

 LHS  RHS  Select normal module m n  0.45 cm  4.5 mm

From PSG Data book P.No. 8.2 Table (1)  Take standard normal module  5 mm normal mo d ule  m n  5 mm face width  b  3 m n 35  47.12 mm

Pitch diameter  d1  mn Z 1  5  24  120 mm End thrust on the Gear Axial force or    F a  F t tan  End thrust 


or F t 

179.14 mn

d  120 mm



179.14 0.5

 12 cm

 358.28 kg

2.159

 30 helix angle 

Tangential force F t 

2M t d

d 12 Torque on   358.28    mt  Ft  pinio n 2 2   2149.68 kg  cm  2148.6 kg.cm

Axial force F a  358.28 tan 30 Fa  206.74 kg or End Thrust  Problem 2.22: A pair of helical gears subjected to moderate shock loading is to transmit 30 kW at 1500 rpm of the pinion. The speed reduction ratio is 4 and the helix angle is 20. The service is continuous and the teeth are 20 FD in the normal plane. For the gear life of 10,000 hours, design the gear drive. [Nov/Dec 2007]

Step 1 Selection of material From PSG Data book Pg.No. 1.15, Select Alloy Steel 40 Ni 2 Cr 1 Mo 28, hardened and tempered. Surface hardness 55 RC and core hardness  350 BHN . and u  1550 N/mm 2

2.160


Materials for pinion and gear are assumed to be same and hence only the pinion is designed. Step 2 Calculation of design stresses From PSG Data book Pg.No. 8.18 [b] 

1.4 kbl n  k

  1

From PSG Data book Pg.No. 8.19, for Alloy steels  1  0.35  u  120  0.35  1550  120  662.5 N /mm 2

Life is 10,000 hrs  N  10,000  60  1500  90  10 7 cycles kb  0.7 from PSG Data book Pg.No. 8.20 k  1.5

from

PSG

Data

book

Pg.No.

8.19

corresponding to steel normalised, surface hardened steel normalised, surface hardened. from PSG Data corresponding to surface hardened. n  2.5

 [ b ] 

book

Pg.No.

1.4  0.7  662.5  173.13 N/mm2 2.5  1.5

From PSG Data book Pg.No. 8.16 [c]  C R HRC k cl

8.19


2.161

From PSG Data book Pg.No. 8.16, corresponding to Alloy steels, Hardened and tempered C R  26.5 HRC  55

From PSG Data book Pg.No. 8.17, corresponding to surface hardness  350 kcl  0.585  [c]  26.5  55  0.585  852.6 N/mm2

Step 3 Calculation of centre distance

a  i  1

2  0.7  E [M t]  [ ]  i  c 

  3

from PSG Data book

Pg.No. 8.13 From PSG Data book Pg.No. 8.15, design twisting moment [M t]  M t  ko k d  k

where ko  1.25 for moderate shock k  kd  1.3 Assume

and M t  97420 

kW from PS G Da ta book Pg.No. 8.15 N1

2.162


 97420 

30 1500

 1948.4 kg f  c m  194.84 Nm  [M t]  1.25  1.3  194.84  316.615 Nm

Assume   0.5 from PSG Data b ook Pg.No. 8.14  E  2.1  10 5 N/mm 2 from PSG Data book Pg.N o. 8.14 

 a  4  1

2

  3

2.1  10 5  316.615  10 3  0.7   852.6   4  0.5  

 140.96 ~  141 mm

Assuming Z 1  20 teeth  Z2  i Z1  4  20  80 teeth

From PSG Data book Pg.No.8.22 mn 



2a cos  Z1  Z2

2  141  cos 20 20  80

 2.65 mm


2.163

 Standard module m n  3 is chosen (from PSG Data book Pg.No. 8.2)

Now revise ‘a’ But d1  

m n Z1

cos 

from PSG Data book P g.N o. 8.22

3  20  63.85 mm cos 20

and d2  

mn Z2 cos 

from PSG Data book P g.N o. 8.22

3  80 cos 20

 255.4 mm a 

d1  d2

2

from PSG Data book Pg.No. 8.22

63.85  255.4 2

 159.63 mm

Step 4 Calculation of b, P a, [Mt]   d1    Pt Z1    Pa  tan  tan 

   63.85    20   27.55 mm   tan 20


2.164

but b   a from PS G D ata book Pg.No . 8.14  0.5  159.63  80 mm

we find b  P a  Revise design torque [M t]  k o  k kd  M t

We know that  p  b/d1 from PSG Data book Pg.No. 8.15  80  p   1.25 63.85  k (Load concentration factor  1.14 (from PSG Data book Pg.No. 8.15)

and v  pitch line velocity 

 d1 N 1   63.85  1500   5.01 m/s 60  1000 60  1000

From PSG Data book Pg.No. 8.16, corresponding to pitch line velocity kd  1.2 for helical gear, IS quality 8  [M t]  1.25  1.14  1.2  194.84  333.176 Nm

Step 5 Calculation of induced stresses b 

0.7 i  1 [Mt] ab mn yv

from PSG Data book Pg.No. 8.13 A


2.165

where yv is obtained for Z eq , from PSG Data book Pg.No. 8.22

but Z eq 

Z1 3

cos 



20 ~ 24 cos 3 20

From PSG Data book Pg.No. 8.18 yv  0.414 for 24 teeth  b 

0.7 4  1 333.176  10 3  73.52 N/mm 2 159.63  80  3  0.414

b 73.52   [ b] 173.1 N/mm 2 From PSG Data book, Pg.No. 8.13 c  0.7 

 0.7 

i1 a

i1 E [M t] ib

 

4  1 159.63

 2.1  10  333.176  10   4  80 41

5

3

 724.95  [c] 852.6 N /mm 2

Hence the design is safe Problem 2.22: A pair of helical gears subjected to heavy shock loading is to transmit 37.5 kW at 1750 rpm of the pinion. The speed reduction ratio is 4 and the helix angle is 15. The service is continuous and the teeth are 20 full depth in the normal plane. Select suitable material and design the gears. Check for working stresses and sketch the drive. [May, June 2009]

Refer Problem 2.19

2.166


Problem 2.23: Deduce the expressions for forces acting on a gear tooth of a helical gear with the aid of a schematic diagram.

[Nov, Dec: 2012]

Refer Article 2.22, Pg.No.2.135 Problem 24: Design a pair of helical gears for the following data. Power  7.5 kW Speed of pinion  1400 rp m Speed reduction  3 Pressure angle  20 Helix angle  10 Select the materials and heat treatment.

Given Data: P  7.5 kW  7.5  10 3 Watts ; n  1400 rpm ; i  3 ;   20  ;   10

Note (If life of the gear drive is not given, then select life  N  10,000 hrs). Step 1: Selection of material and heat treatment (Refer PSG databook, Page No. 8.16) Assume same material for both pinion and wheel. Select Alloy steel with case hardened heat treatment (Refer PSG Data book P.No. 1.14) - 40 Cr 1 MO 28 Material,  u  70  85 kgf /mm2 Take  u  80 kgf /mm2 HRC  55 to 63 and CR  280


2.167

 Surface hard ness 350 BHN  Take HRC  60    Core hardness  350 BHN 

Step 2: Life in cycles. Assume life of the gear drive  N  10,000 hrs  N  10,000  60  RPM  10,000  60  1400  84  10 7 cycles

Step 3: Equivalent young’s modulus (refer PSG databook P.No. 8.14) Since the material for the pinion and wheel is same, E eq  2.15  10 6 kgf /c m2

Step 4: Calculation of design bending stress [b] (Refer PSG databook P.No. 8.18) Design bending stress 1.4 k bl   1 [ b ]  n k

(From PSG Data book P.No. 8.20) (Consider core hardness ) k bl  Life factor for bending 1 (From PSG Data book P.No.

8.19)  1  Endurance limit stress  [b] 

1.4  1  4000  2  1.5

 1866.66 kgf /cm 2

 0.35  u  1200  0.35 8000   1200  4000 kgf /cm 2

Take [b]  1867 kgf /cm 2

(From PSG Data book P.No. 8.19) n  factor of safety 2

2.168


(From PSG Data book P.No. 8.19) k   Fillet stress concentration

 1.5

Step 5: Calculation of design surface (contact compressive) stress. (Refer PSG databook, P.No. 8.16) Design surface (contact compressive) stress  [c]  C R HRC k cl [c]  280  60  0.585 [c]  9828 kg f /cm

2

C R  280 HRC  60 kcl  Life factor

(From

PSG

Data book P. No. 8.17) (Consider surface hardness)  0.585 Step 6: Calculation of Design twisting moment [M t] (Refer PSG databook, P. No. 8.15) Design twisting moment  [M t]  M t  k d  k  521.89  1.3 [M t]  678.46 kgf  cm

Initially assume

M t  97420 

kW n

97420  7.5 1400

 521.89 kg f  c m Step 7: Calculation of approximate centre distance a (Refer PSG databook, Page No. 8.13)


  3

Centre distance a  i  1

a  3  1

2.169

2  0.7  E [M t]  [ ]  i  c 

  3

2

 0.7  2.15  106  678.46   3  0.3  9828 

assume  

b a

 0.3 a  8.07 cm

(Open type gearing) (Refer PSG Data book P.No. 8.14) i3

(Centre distance should be rounded to R10 series) Refer PSG databook, Page No. 7.20 Take, Centre distance  a  80 mm Step 8: Calculation of Z 1 and Z2 Assume, Z 1  No. of teeth on pinion  20 No. of teeth on wheel  Z2  i  Z2  3  20  60 teeth Step 9: Calculation of Normal module m n (refer PSG Data book page No. 8.22) m n  Normal module  mn 

2a cos  Z1  Z2

2  80  cos 10  1.9696 mm 20  60

2.170


Select recommended value of module, (from PSG Data book P.No. 8.2) Select, module  m n  2.5 mm (For safe design, always select m n slightly greater than required) Step 10: Recalculate the centre distance a a



m n Z 1  Z 2 2 cos  2.5 20  60 2 cos 10

a  101.54 mm  Take, a  centre distance 100 mm (nearest value)

(from R10 series) Step 11: Calculation of face width ‘b’ 

b  0.3 a

 b  0.3  a  0.3  100  30 mm Face width  b  30 mm

Step 12: Calculation of pitch line velocity, v in m/sec. Refer PSG Data book P. No. 8.22


Pitch line velocity  d1 n 1 v m/sec 60  1000

2.171

d1  pitch circle dia of pinion d1 

 3.7216 m/sec 

mn cos 

Z1

2.5  20 cos 10

 50.77 mm

Selection of Quality of gears: (Refer PSG Data book page No. 8.3) Select IS Quality - Medium Preferred Quality - 8 Step 13: To find the constants k k  Load correctionfactor  1.03 at   0.6 kd  Dynamic load fac to r  1.3 (For IS-8 Quality,



and k d.

b 30   0.59 d 1 50.77

  0.6 velocity upto  350 BHN hardness)

3

m/sec, surface

Step 14: Recalculation of [M t] [M t]  Mt  kd  k

 521.89  1.03  1.3  698.81 kg fcm Step 15: Checking of bending stress. Refer PSG Data book Page No. 8.13

yv  form factor Refer PSG Data book P.No. 8.18

2.172


b  0.7

i1

[Mt]  [b]

a  b  mn  yv

Zv 

Z1 cos3 

(Refer Data

book P. No. 8.22) 3  1  0.7 698.81   3 0.25  0.3955 10



20  20.93  21 teeth cos3 10

 659.64 kgf /cm2  [b] 1867 kgf/ cm2

 Design is satisfactory based yv  0.3955at Zv  21 teeth

on bending stress Step 16: Checking of Surface (contact compressive) stress refer PSG databook, P. No. 8.13

c  0.7

i1 a

 0.7

31 10

i1 E [M t]  [c] ib

 

31  2.15  10 6  698.81 33

 

 7235.45 kg f /cm 2  [c] 9828 kgf/cm 2

Therefore, the design is satisfactory based on surface (contact compressive) stress. Problem 2.25: Design a helical gear to transmit 7.5 kW at 1400 rpm; to the following specifications: Speed reduction is 3;

Pressure angle 20; Helix angle 10;

Design compressive stress 9020 kg/cm2; Design bending stress 2100 kg/cm2.

Modulus

 2.05  106 kg/cm2.

of

elasticity

of

materials

(Nov/Dec 2014)


2.173

Given Data: P  7.5 kW ; N 1  1400 rpm;  Z2  Speed reduction  i  3    Z  1

Pressure angle    20 Helix angle    10  Design surface compressive stress  9020 kg/cm 2 [c]  902 N/mm 2

Design bending stress  2100 kg/cm 2 [b]  210 N/mm 2 E  2.05  10 5 N/mm2

Step 1 Calculation of design twisting moment [M t] [M t]  M t  kd  k. M t  Nominal twisting m oment transmitted by the pinion  kgf c m  97420 

kW n

 97420 

7.5  521.89. kg fc m 1400

M t  521.89 .kgfcm


2.174

For symmetric scheme, initially assume k  k d  1.3 [M t]  521.89  1.3

 678.46 kg fc m  67846 Nmm Step 2 Calculation of approximate centre distance.

a  i  1

   0.7   [ ]   c 

3

2

E [Mt]

i

i  Gear ratio  3; [c]  902 N/mm 2; E  2.05  10 5 N/mm 2 [M t]  67846 Nmm 

b  0.3 a

a  3  1

assume, (for open type of Gearing)

   0.7     902 

3

2

2.05  10 5  67846 3  0.3

a  84.139 mm

Step 3 Assume Z 1  no. of teeth on pinion  20 Gear ratio i 

Z2 Z1


3

2.175

Z2 20

Z 2  60 teeth

Step 4 Calculation of normal module (m n. From PSG Data book page 8.22; table (26) mn 

a  84.138 mm

2a  cos  Z1  Z2

  10 Z 1  20

2  84.139  cos 10  20  60

Z 2  60

 2.07 mm  Represent normal mo dule  standard   m n  3 mm  boo 8.2 table 2  k P.No. from PSG Data 

Step 5 Re-calculate centre distance ‘a’ From PSG Data book P.No.8.23; table (26) Centre distance  a  

mn  Z1  Z2    2 cos     20  60  3   2 cos 10   

 121.85 mm

(rounded to R10 series) From R10 series, take centre distance a  125 mm

2.176


Step 6 Calculation of face width ‘b’ Already  is assumed as 0.3 But  

b a

ba

 125  0.3  37.5 mm Face width  b  37.5 mm

Step 7 Calculate pitch circle dia. d1 and d 2 Refer PSG Data book P.No. 8.22; table (26) mn d1  pitch circle dia   Z1  of p inio n  cos 

3  20 cos 10

d1  60.92 mm

mn d2  pitch circle dia   Z2  of wheel  cos 

3  60 cos 10

d2  182.77 mm

Step 8 Calculation of pitch line velocity v Pitch line velocity v 

d1N 1

60

m  sec




2.177

  60.92  1400 1000  60

 4.46 m  sec

Step 9 Selection of Quality, preferred Quality from PSG Data book P.No.8.3; table (2) above 1m/sec; up to 8m/sec, Preferred Quality- 1S6 Step 10 Calculate  p and k; Refer table (14); P.No. 8.15 of PSG Data book p 

37.5 b   0.61556 d1 60.92

From PSG Data book  p  0.6 : The value of k  1.03

table (14): P.No.8.15 at

Step 11 Read the value of k d : From PSG Data book Table (15): P.No. 8.16 for quality IS  8; pitch line velocity. upto 8 m/sec. kd  1.3

Step 12 Recalculate the design twisting mount [M t] [M t]  M t  kd  k  521.89  1.3  1.03

 698.81 kg f  cm  69881 N  mm

2.178


Step 13 Calculation of  b and checking of ab Refer PSG Data book P.No.8.13; Table 8 b  0.7

i1 [M t] a  b  m n  yv

i  3; a  125 mm m n  3 mm; b  36.555 mm yv  form factor From PSG Data book P .No. 8.18 and P.No.822  0.39485 [M t]  69881 N mm b  0.7

3  1 69881  36.2 125  36.5  3  0.39485

b 36.20   [b]  210 N  mm 2 b 36.2  [ b] 210   The design based on beam strength is satisfactory.

Step 14 i1 E [M t] ib

c  0.7

i1 a

 

 0.7

31 125

 

31  2.05  10 5 69881  3  36.555

 512.035 N  mm 2  [c] 902 N  mm2  The design based on wear strength is satisfactory.


2.179

Problem 2.26: A pair of helical gears with 30 helix angle is used to transmit 15 kW. at 10,000 rpm of pinion. The velocity ratio is 4 : 1. Both the gears are made of case hardened steel with a static strength of 10 N  mm2. The gear is at 20 stub and the pinion is to have 24 teeth. The face width is 14 times the module. Find the module and face width for strength. (May/June 2013)

Given Data: Helix angle    30; P  15 kW; n  N  10,000 rp m; i  4 [b]  100 N  mm 2 Z 1  24 teeth b  14 module

Refer PSG Data book P.No. 8.13: For designing,

m n  1.15 cos 

  3

[M t]

y v  [b]  m Z 1

[M t]  M t  kd  k (Refer PSG Data book P.No.8.15) M t  97420

kW 97420  15  10,000 n

 146.13 kg f cm

(Assume k d  k  1.3 

M t  14613 N  mm

Design twisting moment  [M t]  14613  1.3  18996.9 N  mm yv  form factor Z v  yv

2.180


40  0.465 35  0.452 5 _ 0.013 2  ?

Zv 

2  0.013   5.2  10  3 5



for 35  0.452

Z1 cos 3

24 cos 3 30

 36.95

for 2  5.2  10  3

 37

37  0.4572 yv  At 37 and X  0

Form factor yv  0.4572 Assume m  10 [From PSG Data book P.No. 8.14 Table (12)] Z 1  24 teeth given

m n  1.15 cos 30

 1.15 cos 30

  3

[M t ]

yv  [b] m Z1

  3

18996.9 0.4572  100  10  24

 1.195 mm take m n  1.195 mm Module (m) 

mn cos 



1.195  1.37 mm c o s 30


2.181

Take standard module  m  1.5 mm b  face width  14 m  14  1.5  21 mm Problem 2.27: A pair of helical gears subjected to moderate shock loading is to transmit 37.5 kW at 1750 r.p.m. of the pinion. The speed reduction ratio is 4.25 and the helix angle is 15. The service is continuous and the teeth are 20 FD in the normal plane. Design the gears, assuming a life of 10,000 hours.

[April/May: 2010]

[Similar to the problem: 2.24] Problem 2.28: A pair of helical gears for a turbine has a transmission ratio of 10:1. The pinion rotates at 5000 rpm and made of carbon steel and the gear wheel is made of high grade cast iron. Power transmitted  90 kW. The gear pair is required to last for atleast 12000 hrs. Select suitable gear materials. [May/June: 2012]

[Similar to the problem: 2.24] Problem 2.29: A pair of helical gears is to transmit 14 kW. The teeth are 20 sutb and helix angle is 45. Pinion runs at 10,000 rpm and has 80 mm PCD. Wheel has 320 mm PCD. Both gears are made of cast steel. Design the gear pair and obtain the basic dimensionals assuming a life of 10000 hours. [April/May: 2010]

[Similar to the problem: 2.24]

2.182


Problem 2.30: Design a pair of helical gears to transmit 10 kW at 1000 rpm of the Pinion. Reduction ratio of 5 is required. Give details of the drive in a tabular form. Life of gear is 10  103 hrs.

Given Data P  7.5 kW ; N  1000 rpm ; i  5

Solution Step 1 Selection of material and heat treatment Assume same material for both pinion and wheel. Select Alloy steel with case hardened, 40Cr 1MO 28 material

From PSG Data book Pg.No. 1.14, properties of the selected material, u  70  85 kg f/mm 2; Take u  80 kg f/mm 2 HRC  55 to 63, C R  280

 surface hardness  350 BHN  Take HRC  60,     core hardness  350 BHN Step 2: Life in cycles life of gear drive  N  10,000 hrs  N  10,000  60  RPM  10,000  60  1000 N  60  107 cycles

Step 3: Equivalent Young’s modulus From PSG Data book, Pg.No. 8.14, since the material for the pinion and wheel is same.


2.183

E eq  2.15  10 6 kg f/cm 2

Step 4: Calculation of design bending stress [b] From PSG Data book, Pg.No. 8.18 Design bending stress [b] 

1.4 kbl n  k

  1

From PSG Data book, Pg.No. 8.20 consider core hardness kbl  1

From PSG Data book, Pg.No. 8.19,  1  Endurance limit stress  0.35 u  1200  0.35  8000  1200  4000 kg f/cm 2

From PSG Data book, Pg.No. 8.19 n  F actor of safety  2 k   Fillet stress concentration  1.5  [b] 

1.4  1  4000   1866.66 kgf/cm2 2  1.5

Step 5: Design surface contact stress From PSG Data book, Pg.No. 8.16 Design surface (contact compressive) stress  [ c]  C R HRC kcl

From PSG Data book, Pg.No. 8.17,

2.184


kcl  Life factor, consider surface hardness  0.585  [c]  280  60  0.585 [ c]  9828 kgf/cm 2

Step 6: Calculation of Design twisting moment [M t] From PSG Data book, Pg.No. 8.15, Design twisting moment [M t]  M t  kd  k

Initially Assume kd  k  1.3 and M t  97420

kW 10  97420  1000 n

M t  974.2 kgf  cm  [M t]  974.2  1.3  1266.46 kgf  cm

Step 7: Calculation of approximate centre distance From PSG Data book, Pg.No. 8.13

a  i  1

2  0.7  E [M t]  [ ]  i  c 

  3

From PSG Data book, Pg.No. 8.14, corresponding to open type gearing 

b  0.3 a


 a  5  1

2.185

  3

2

6

 0.7  2.15  10  1266.46    5  0.3  9828 

a  12.57 c m

Centre distance should be rounded to R10 series. From PSG Data book, Pg.No. 7.20, Centre distance a  16 cm  160 mm Step 8: Calculation of Z 1 and Z 2 Assume, Z 1  Number of T eeth on pinion  20 teeth Number of teeth on wheel Z 2  i Z 1  5  20  100 teeth

Step 9: Calculation of Normal module m n From PSG Data book, Pg.No. 8.22 m n  Normal module 

2a cos  Z 1  Z2

Assuming Helix angle   10  

2  160  cos 10  2.626 mm 20  100

From PSG Data book, Pg.No. 8.2, the recommended value of module m n  3 mm

2.186


Step 10: Recalculate the centre distance Since a

m n Z 1  Z 2  20  100  3     182.77 m 2 2 cos  cos 10  

 Take a  centre distance = 200 mm, (nearest value

is R10 series) Step 11: Calculation of face width ‘b’ Since  

b  0.3 a

 b  0.3  a  0.3  200

Face width ‘b’  60 mm Step 12: Calculation of pitch line velocity, v in m/s From PSG Data book, Pg.No. 8.22, d1  

mn cos 

Z 1  pitch circle dia of pinion

3  20  60.92 mm cos 10

 Pitch line velocity v 

 d1 n 60



  60.92  10  3  1000 60

 v  3.189 m/s

Step 13: Selection of Quality of gears From PSG Data book, Pg.No: 8.3, select IS Quality - medium preferred Quality - 8


2.187

Step 14: To find the constants k and k d From PSG Data book, Pg.No. 8.15, p 

b 60   0.984 d1 60.92

From PSG Data book, Pg.No. 8.15, corresponding to p  0.984 k1

From PSG Data book, Pg.No. 8:16, corresponding to pitch line velocity, up to 8 m/s and IS Quality - 8 and surface Hardness  350 kd  1.2

Step 15: Recalculate value [M t] [M t]  M t  kd  k  974.2  1.2  1  1169.04 kgf  cm

Step 16: Checking the bending stress From PSG Data book, Pg.No. 8.22 Virtual Number of Teeth Zv 

Z1 3

cos 



20 c o s3 10

 20.93 ~  21 teeth

From PSG Data book, Pg.No. 8.18, corresponding to Z v  21 teeth , X  0 yv  0.3955

From PSG Data book, Pg.No. 8.13 A, b  0.7

i1 [M t]  [b] amby

2.188


 0.7 

5  1  1169.04 20  0.3  6  0.3955

 344.8 kg f/cm 2  [b]

Step 17: Checking for surface stress from PSG Data book, Pg.No. 8.13 c  0.7

i1 a

 0.7 

E [M ]  [ ] ib   i1

5  1 20

t

c

2.15  10  1169.04   56 51

6

 4708.3 kg f/cm 2  [c]  The design is satisfactory based on bending stress

and surface stress Problem 2.31: A helical gear with 30 helix angle has to transmit 35 kW at 1500 rpm, with a speed reduction ratio 2.5. If the pinion has 24 teeth, determine necessary module, pitch diameter, and face width for 20 full depth teeth. Assume 15 Ni 2 Cr 1 Mo 15 material for both pinion and wheel. [Assume Lg  10,000 hrs]

Solution Step 1 Material selection Given that material used is 15 Ni 2 Cr 1 Mo 15 From P.S.G Data book, Pg.No. 8.5


2.189

[b]  320 N/mm2, [ c]  950 N/mm 2 L g  10,000 hrs

Step 2 Calculation of centre distance from P.S.G Data book, Pg.No. 8.13

a  i  1

  3

2  0.7  E [M t]  [ ]  i  c 

i  2.5 given; [c]  950 N/mm2 E  2.15  105 N/mm2 assume for s te el

From P.S.G Data book, Pg.No. 8.14 

b  0.3 (assume) a

But P.S.G Data book, from Pg.No. 8.15 [M t]  k  k d M t

But from P.S.G Data book, Pg.No. 8.15 M t  97420   97420 

kW N1 35 1500

 2273 kg f cm

Initially assume k  k d  1.3  [M t]  1.3  2273

2.190


 2954.9 kg f cm  [M t]  295490 Nmm

 a  2.5  1

  3

2

5

 0.7  2.15  10  295490    2.5  0.3  950 

 125.39 ~  126 mm

Step 3 Calculation of module Z1  24 given  Z 2  i Z 1  2.5  24  60 teeth

From P.S.G Data book, Pg.No. 8.22 mn 



2a cos  Z1  Z2

2  126  cos 30 24  60

m n  2.59

The standard module m n  3 is chosen Now revise ‘a’ d1 

mn Z1 3  24   83.13 mm cos  cos 30

d2  i d1  2.5  88.13  207.82 mm


a

2.191

d1  d2 88.13  207.82 a 2 2  148 mm  147.975 mm ~

Step 4 Calculation of face width ‘b’ ba  0.3  147.975  44.39 mm ~  45 mm

ADDITIONAL PROBLEMS April 1998 It is desired to determine the proportions of a spur gear drive to transmit 8 kW from a shaft rotating at 1200 rpm to a low speed shaft, with a reduction of 3:1. Assume that the teeth are 20 full depth involute, with 24 teeth of the pinion. The pinion is made to be of 40 C 8 normalized and gear of 30 C 8 normalized steel. Assume that the starting torque is 130 percent of the rated torque. April 1998 A compressor running at 360 rpm is driven by 150 kW, 1440 rpm motor through a pair of 20 full depth helical gears having helix angle of 25. The center distance is approximately 400 mm. The motor pinion is to be forged steel and the driven gear is to be cast steel. Assume medium shock conditions. Design the gear pair.

2.192


October 2001 A pair of helical gears is to transmit 10 kW at 1200 rpm with a speed reduction of 4. Design the gear drive assuming suitable material. October 2000 Design a pair of spur gears to transmit 10 kW 1500 rpm with a gear ratio of 3. Assume same material for both wheels. Check for plastic deformation assuming instantaneous torque is 50% higher than mean torque. October 1998 Design a set of helical gears to transmit 80 HP to effect a speed reduction of 2.5:1 with pinion connected to motor rotating at 1440 rpm. The teeth of gears are of full depth 20 pressure angle to the normal plane. Peripheral pitch line velocity may be limited to 600 meter per minute. Assume gear and pinion are to be made of forged steel having ultimate tensile strength of 6000 kg/cm 2. The pinion is properly heat treated to obtain hardness of 240 BHN. Assume service factor of 2 and number of teeth in pinion  16 .

Chapter – 3

DESIGN OF 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 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.

3.1 INTRODUCTION Bevel gears are used to transmit power between two intersecting shafts. There are two types of bevel gears Straight and spiral. Incase of straigth bevel gears, the teeth are straight, which converge into a common apex. Incase of spiral bevel gears, the teeth are curved. Straight bevel gears are easy to design and manufacture. These gears produce noise at high speed conditions. Spiral bevel gears are difficult to design and manufacture. These gears facilitate a quiet operations, even at high speeds. In general, the angle between the axes of intersecting shafts is 90. i.e., the axes of the shafts are at right angles.

3.2 TYPES OF BEVEL GEARS Bevel gears are used to connect two intersecting shafts.


dum d en

De

Add

end

um

3.2

b

R

PC D ( ,n sid e) R oot D iam eter

O PC D (O u ts ide)

R = C on e D is ta nce b =Fa ce W idth Se m i C o ne A ng le ( o r P itc h Angle) a Ad dend um Angle f D edend um A ng le



Tip D iam eter

a

f Fig. 3.1. P arameter of bevel gear

1. Straight bevel gears 1. Axes intersect 2. Zerol bevel gears

2. Axes intersect

3. Sprial bevel gears

3. Axes intersect and teeth are curved and oblique (less noisy)

4. Face gears

4. Spur gear pinion meshes with a ring gear having conjugate teeth cut into the face of it.

5. Skew bevel gears

5. Axes are non-parallel and non-intersecting. Teeth are parallel

Design of Bevel,Worm and Cross Helical Gears 3.3





o

< 90 (A cu te)

o

= 9 0 (b)

(a)

(R igh t)

z2  

o

> 9 0 (O b tu se ) (c)

6. Hypoid gears

Fig. 3.2.

z1 M itre B evel o G ea rs = 90 (d) z 1 = z 2 = no o f te e th in ea ch g e ar.

6. Axes are non-parallel, non-intersecting. Teeth are curved. Used in Automobile differential units

3.4


3.3 BEVEL GEAR NOMENCLATURE = 90

o

M

f a

  a 

d1

1

f1 a1 a2

2

R

f2

b

mt mt a

mm d2

s = ab

b

ha hf

d a2 Fig. 3.3. N o m e n clature of s tan d ard b evel G e ars

1. Pitch cone: It is an imaginary cone that rolls without slipping on a pitch surface of another gear. 2. Cone Centre: It is the point where the axes of two mating gears intersect each other. It may be also defined as the apex of the pitch cone. 3. Pitch Angle: It is the angle made by the pitch line of a gear with the gear axes. It is denoted by  4. Cone Distance: It is the length of the pitch cone element. It is also known as pitch cone radius and is denoted by R R

d2/2 Pitch Radius d 1/2   sin  sin 1 sin 2


R

1 2

d 21  d 22  

For Right angle gears R

1 m 2 t

Z 21   Z 22  

where d 1 & d2 

Pitch circle diameter of pinion and gear.

Z1 & Z2 

Number of teeth of pinion and gear.

1 & 2 

Pitch angle of pinion and gear.

mt 

Transverse module

5. Addendum Angle  a: It is the angle subtended by the addendum of the tooth at the cone centre.  ha   addendum  a  tan  1  tan  1     R   cone distan ce 

6. Dedendum Angle  f: It is the angle substended by the dedendum of the tooth at the cone centre.  hf   dedendum   tan  1  f  tan  1     R   cone distance 

7. Tip Angle (Face Angle): It is the angle substended by the root of the tooth at the cone centre. It is denoted by a a  Pitc h Angle  Addendum Angle    a

3.6


8. Root Angle: It is the angle subtended by the root of the tooth at the cone centre. It is denoted by f f  Pitch Angle  Dedendum Angle    f

9. Back Cone: It is an imaginary cone, perpendicular to the pitch cone at the end of the tooth. It is also known as normal cone. 10. Back cone distance: It is the length of the back cone. It is also known as back cone radius. 11. Backing: It is the distance of the pitch point from the back of the boss, parallel to the axis of the gear. 12. Mounting height: It is the distance of the back of the boss from the cone centre. 13. Pitch diameter: It is the diameter of the largest pitch circle. 14. Outside (or) addendum cone diameter: It is the maximum diameter of the teeth of the gear. It is equal to the diameter of the blank from which the gear can be cut. Outside Diameter  Pitch diameter  2  h ac o s  15. Inside (or) dedendum cone diameter: Inside diameter  Pitch diameter  2hf  cos  Where ha  Addendum hf  Dedendum


3.4 FORMATIVE (OR) EQUIVALENT (OR) VIRTUAL NUMBER OF TEETH FOR BEVEL GEARS Bevel gears are replaced by equivalent spur gears, to simplify the design calculation and analysis. An Imaginary spur gear considered in a plane perpendicular to the tooth at the largest end, is called as virtual spur gear. The pitch cone radius ‘R’ of the bevel gear is equal to the pitch circle radius of the virtual spur gear. The number of teeth on this imaginary spur gear is called virtual number of teeth. Mathematically, For pinion Zv 

Z cos 

Where Z1 & Z2

Z v1 

Z1

cos 1

For gear Z v2 

Z2

cos 2

are the actual number of teeth on the pinion and gear of the bevel gear respectively.

Z v1 & Z v2 are the virtual number of teeth on the pinion and gear of the bevel gear respectively.

3.5 TOOTH STRESS (OR) STRENGTH OF BEVEL GEARS Lewis Equation The strength of a bevel gears tooth is obtained in the similar way as that of the spur gear. The modified form of the Lewis equation for the tangential tooth load is given here. Rb F s  [b] b    m  yv    R 

3.8


[b]

–

Allowable Static Stress

b

–

Face width

m

–

Module

yv

–

Lewis Factor (or) tooth form factor for the equivalent number of teeth.

R

–

Cone distance

R

  d21 2



d22 2

d1

–

Pitch Circle Diameter of pinion

d2

–

Pitch Circle Diameter of gear

Rb  R  is a bevel factor.  

3.6 DYNAMIC LOAD FOR BEVEL GEARS The dynamic load for bevel gears may be obtained in the similar way as that of spur gear. Buckingham’s Dynamic load From PSG Data book Pg No. 8.50 Fd  F t  F I  0.164 V m  cb  F t   Ft    cb  F t    0.164Vm  1.485 

All the notations are similar to spur gear deign.

3.7 WEAR STRENGTH FOR BEVEL GEARS The maximum (or) limiting load of wear strength of bevel gear is given here.


Fw 

d1 b Q v kw cos

Qv

–

Ratio factor based on the virtual number of teeth.



–

Pitch Angle

b

–

Face width

kw

–

Load stress factor Qv 

2 Z v2 Z v1  Z v2

Note: According to Buckingham’s the wear load should be greater than dynamic load for safe design.

3.8 FORCE ACTING ON THE BEVEL GEAR The normal force F N on the tooth is perpendicular to the tooth profile and thus makes an angle equal to the pressure Angle  to the pitch circle. Thus the normal force can be resolved into two components. (i) Tangential component F T (ii) Radial component F R The bearing reactions are produced by tangential component (or) tangential tooth load. The end thrust in the shaft is produced by the radial component. cos  

FT

FN

FT  FN c o s 

3.10


FN FR



FR

FT

b

R

F R V= F R

F RH= F A

G ea r S ha ft

Rm 

o

=p re ssu re an gle = 2 0 P itch co ne a ng le fo r p in io n b =fac e w id th R = con e distan ce

O ver H an g D ista n ce Fig. 3.4 .

sin  

Fo rc es a ctin g o n B ev el G ea r

FR FN

F R  F N sin  tan  

FR FT

F R  F T tan 

These Forces are considered to act at the mean radius R m. From the Fig. 3.4. d1   R m  R  b/2 sin 1 ;  sin 1   2R  


R m  R  b/2

d1

FR

2R

The radial force F R  acting at the

V

Fig. 3.5

FR FR

H

mean radius Rm  may be further resolved into two components. F RH and F RV in the axial and radial directions respectively are shown in Fig. 3.5.

The axial force acting on the pinion shaft FR H sin 1  FR

. .  . FR  F T tan 

F R  FR sin 1 H

 F T tan  sin 1 The radial force acting on the pinion shaft FR V cos 1  FR F R  FR c o s 1 V

. . ( . FR  FT tan 

 F T tan  cos 1

It shows that the axial force on the pinion shaft is equal to the radial force on the gear shaft but in opposite directions, likewise, the radial force on the pinion shaft is equal to the axial force on the gear shaft, but in opposite directions.

3.12


3.9 BASIC DIMENSIONS OF BEVEL GEAR PAIR Nomenclature Transverse

Formulae for straight and spiral bevel gears

Symbol mt

module

y  0.5  mm  mt  m m y 

Mean Module

mm

Normal Module

mn

Cone Distance

R

b sin  Z

mn  mt cos  R  0.5 mt Zi i2   1

Z21  Z22  0.5mt    Reference

2 sin 1



mt Z2

2 sin 2

d

d1  mt Z1 ; d2  mt Z2.

da

da1  mt Z1  2 cos 1 ;

Diameter Tip Diameter

mt Z1

da2  mt Z1  2 cos 2 Face Width

b

b ~ 0.3 R or 10 mt whichever is smaller For spiral bevel, see chart, p.8.40

Number of Teeth

Z

Number of teeth

Zc w

on crown wheel Minimum number of teeth on pinion to avoid

ZU

Z1, Z2 ; Z1  iZ2 Zcw  ZU 

2R mt 2X cos  sin2

; X  1,

for uncorrected gears)

undercutting Pressure Angle Mean spiral angle



  20 usually

m

m usually 30 to 35 ; 35 preferred.   0 for ZEROL BEVEL

y

y 

R b


Nomenclature

Formulae for straight and

Symbol

spiral bevel gears

Reference angle



tan   i ; 1  90  2

Addendeum angle

a

Dedendum angle

f

Height factor

f0

f0  1 ; standards

Clearance

C

C  0.2

Tip angle

a

a    a

Root angle

f

f    f

tan a1  tan a2  tan f1  tan f2 

m t f0 R

mt f0  c R

Straight Bevel Tooth height

h

h  ha  hf

Spiral Bevel h  1.88 mt for   20

Working depth

hw

hw  2mt

hw  1.7 mt for   20

Addendum

ha

ha  mt

ha1  hw  ha2 ha2  mt Ca Select Ca from Table 31a.

dedendum

hf

hf  1.1236 mt (Reinecker)  1.67mt (AEG)

hf  h  ha1 1

 1.88 mt (Gleason) hf  h  ha2 2

Circular tooth thickness

S

S1   mt  S2

S1   mt  S2

3.14

Design of Transmission Systems Straight Bevel S2 

 mt 2



Zv



Corrected gear

2



 Kmt

tan   Kamt

1

Face contact ratio

 mt

1.22 ha1  ha2

Zv  Zv

S2 

ha1  ha2 tan 

Select K from table 31 b Virtual number of teeth

Spiral Bevel

Z cos 

min

Select K2 from chart, p.8.42 Zv 

Z cos  cos2 m

 18 for   20 and f0  1

t

f

bRtan m  mt R  0.5b

usually S correction is applied. To avoid undercutting the correction factor Xou  1  2f0 cos1 Z1 sin2 

Nomenclature of Standard Bevel Gears From PSG Data book P.No. 8.38 Table (31) R  Cone radius b  Face width   90  (Angle between two shafts)


Angles Pinion

Wheel

a  addendum angle

a  addendum angle

f  dedendum angle

f  dedendum angle

a  tip angle

a  tip angle

f  root angle

f  root angle

1  semi-cone angle of

2  semi-cone angle of

1

1

1

2

2

2

1

2

pinion (1)

wheel (2)

From PSG Data book P.No. 8.38 Table (31)

1. Transverse module  m t  m m 

b  sin 1 Z1

where m m  mean module or average m odule b  face width 1  semi cone angle for pin ion Z 1  no. of teeth on pinion

1  2  90 But, tan 2  i 

Z2 Z1

Calculate 2, then

1  90  2

where 2  semi cone angle of wheel.

3.16


2. Normal module  mn  m t  cos m (only for spiral bevel gear) where m  mean spiral angle  30 to 35 Gen eral

3. Cone distance (R) R  0.5 m t  Z 1  i2 1  0.5 m t  

Z21   Z22  

m t  Z2 mt  Z1  2 sin 1 2 sin 2

4. Pitch

circle

diameter

(outside)

of

pinion

 d1  m t  Z 1

Pitch circle diameter (outside) of gear or wheel  d2  m t  Z2 5. Tip dia. of pinion  d a  m tZ1  2 cos 1 1

6. Tip dia. of wheel or gear  d a  m tZ 2  2 cos 2 2

7. Face width b  0.3R or 10 m t. Consider whichever is smaller. y 

R b

From PSG Data book table (13); P.No. 8.15, for i  1 to 4   y  3 for i  4 to 6   y  4


In general, take

y  3 3

R b

b

R 3

8. Addendum angle  a tan  a  tan a  1 2

m t  fo R

where fo  height factor 9. Dedendum angle   f tan f  tan f  2 1

m tfo  c R

Where c  clearance 10. Virtual number of teeth Zv Zv 

Zv  2

Z ; cos 

Zv  1

Z1 cos 1

Z2

cos 2

11. Tip angle a  1  a (pinion) a  2  a (wheel) 1 1 2 2 12. Root angle f  1  f pinion) f  2   f (wheel) 1

1

2

2

3.18


3.11 DESIGN OF STRAIGHT BEVEL GEARS USING LEWIS AND BUCKINGHAM EQUATIONS Step 1: For a given problem, note down the given parameters like, power, speed, gear ratio (i) etc... Step 2: Selection of Material To determine the strength of gear and pinion, use procedure similar to the spur gear design. Step 3: Calculation of Pitch Angles 1 & 2 1  2  90 ; tan 2  i 2  tan  1 i

Step 4: Calculation of virtual number of teeth Z v1  Z2

Z1

Z2 ; Z v2  cos 2 cos 1 Z1

 i if Z 1 is n ot given, Assume Z 1  20 

Step 5: Calculation of tangential load F t Ft 

HP  75 kg f vm  10 Newton] [1 kgf of ~

vm  Pitch line velocity m/sec. Ft 

HP  75  10 N vm

The above step is similar to spur gear.


Step 6: Calculation of Dynamic Load F d Fd Cv Nsf km Ft

From the data book Pg.8.52

N sf  survival factor [N sf  1 to 2]

[C v 

 vm 3.5   ] [when vm  5 m /s] 3.5

km  load distribution factor.

Choose k m value From PSG pg data book Pg.8.52 Step 7: Calculation of strength of gear tooth

Fs 

[ b] b yv 1 

b R

Pd

From PSG data book Pg no.8.52 Diametrical Pitch P d Pd 

 Pc

Pc 

Pc  Circular Pitch

d Z

b  Face width in cm R  Cone Distance in cm yv  Lewis form factor based on virtual number of

teeth.

3.20


b  0.3 R (or)

10 Pd

m t  transverse module in cm

From PSG Data book Pg 8.52 Step 8: Calculation of Transverse Module m t We know that, For satisfactory design F s  Fd

By equating this two factors, we can calculate the module m t Step 9: Calculation of Face Width (b) b  0.3 R (or)

10 Pd

1

Z1    P d1  d  1  

Step 10: Calculation of Pitch Circle Diameter

d 1

d1  m t  Z1 Step 11: Calculation of Pitch Line Velocity V

 d 1N 1 m/s 60  1000

d1  in mm N1  rpm

Step 12: Recalculation of Strength of gear tooth (Fs   b  [ b ] b y v  1       R  Fs  Pd   b  Fs  [ b] b yt   1      R


Step 13: Calculation of Buckingham’s Dynamic Load Fd   0.164 V m cb  F t Fd  F t    cb  Ft    0.164 Vm  1.485 

The calculation of dynamic load is similar to spur gear, where Vm should be taken for largest pitch circle. Step 14: Checking for beam strength For safe design Fs  Fd This step is similar to spur gear Step 15: Calculation of Wear Load Fw  Fw 

d ib Q v kw cos 1

Qv  Ratio factor based on virtual number of teeth. kw  Load stress factor 1   1 [c] sin    E 1 E2    kw  1.4 1 . . . If  is not given, assume  14 2

Step 16: Checking of Wear Strength For safe design F w  F d Step 17: Calculate the Basic Dimensions of Straight Bevel Gear From PSG data book pg 8.39

3.22


Step 18: Draw the neat sketch of straight bevel gear. DESIGN OF BEVEL GEARS (Simple steps) (Refer PSG Data book Page Nos. 8.50 to 8.53) By using Lewis and Buckingham’s equation. Step 1 b   [ b] b  Yv  1   R   F s  Beam strength  Pd

But P d  diametrical pitch 

... (1)

  1   P c  m t mt

P c   m t

where m t  Transeverese module  in cm. Y v    yv

where

yv 

virtual

form factor based on virtual no. of teeth. Z v  yv  0.154   0.124   0.175 

0.912 for 20  involute Zv 0.684 for 1412 involute Zv

0.95 for 20 involute stub Zv

Zv 

Z1 c o s 1

1  2  90  tan 2  i

Calculate

2

and

for

20

then 1 Z v  18

involute teeth


Replace Y v and P d in the above equation (1) then b   Beam strength  Fs  [ b] b  yv  m t  1   R 

... (2)

[b]  allowable static strength in kg f  cm 2

where

m t  transverse module in cm. b  face width in cm.

Note If b is not given in the problem, take from PSG databook. P.No. 8.52 b  0.3 R or

10 Pd

B ut P d 

   Pc  mt

 0.3 R to 10 m t R

cone distanc eor     0.5 m t cone length 

Z21  Z22 

in cm

Dynamic load  F d  F t  C v in kgf For straight bevel. C v 

3.5 

vm 

3.5

[when vm  5 m  sec ]

where, vm  mean velocity in m  sec  75   HP  Tangential tooth load v m 

F t  Transmitted load or

Refer PSG Data book Page. No. 8.51 Where vm  mean velocity 

 d 1 av N 1 m  sec 60  100

3.24


 R  0.5b  d 1 av  m t Z 1   R  

Calculate vm, F t and C v and substitute the values in the equation of F d  Dynamic load  F d  F t  C v

... (3)

For satisfactory design, F s  F d. To find transverse module

(m t,

Equate the

equations (2) and (3) and calculate m t in cm. Take standard value of mt in mm from PSG Data book P.No. 8.2 table (1). Step 2 Buckingham’s Dynamic load from PSG Data book Pg No 8.50  0.164 Vm cb  F t Fd  Ft    cb  F t   0.164 V m  1.485  

    

(Similar to equation for the spur gear) V m  mean velocity in m/min C  calculation is similar to spur gear b  face width in cm F t  transmitted load in k gf Fs  Fd

Then the design is satisfactory based on the Beam strength.


Note If F s  F d, change the transverse module m t, and recalculate d1 av, vm, F t, F S, b and R. Calculate again F d and check in the 2nd trail, If F S  F d then the design is satisfactory.

Step 3 Wear load  F w 

Note:

b  d1  Q  k

cos 1

d 1  PCD

of

pinion.  Z 1  m t  cm

Fw  Fd

Then the design tan 2  i is satisfactory     90 2 i based on wear Cal. 1 strength

If F w  F d

then change the k  calculate as material, Similar to spur recalculate k  Gear Design procedure.

and calcuculate again F w In the 2nd   trail 

Q

2i i1

If F w  F d

Then the design is satisfactory. LEWIS METHOD Problem 3.1 A pair of cast iron bevel gears connect two shafts at right angles. The pitch diameter of the pinion and gear are 80 mm and 100 mm respectively. The tooth profiles of the gears are of 14

1 composite form. The allowable static stress for both 2

3.26


the gears is 55 MPa. If the pinion transmits 2.75 kW at 1100 rpm, find the module and number of teeth on each gear and check the design for wear strength. Take surface endurance limit as 630 MPa. and modulus of elasticity for cast iron as 84 kN/mm2

LEWIS METHOD Solution: Step 1: Given Data: Material for pinion and gear cast iron Pitch diameter of pinion d1  80 mm Pitch diameter of gear Pressure Angle

d2  100 mm   14

1 2

Allowable Static Stress [b]  55 MPa  55  106 N/m 2  55 N/mm 2 . . [ . 1 N /m 2  10  6N/mm 2]

Power P  2.75 kW  2.75  10 3 W Speed of Pinion N 1  1100 rpm Surface endurance  e  630 MPa  630  10 6 N/m 2  630 N/mm 2 Modulus of Elasticity E 1  E 2  84 kN/mm 2  84  10 3 N/mm2


Step 2: Calculation of pitch angles 2  tan  1 i

i

d2 d1



100  1.25 80

2  tan  11.25 



2  51.34  1  2  90  1  90  51.34 1  38.65 

Step 3: Selection of Materials Pinion - Cast Iron Gear - Cast Iron Step 4: Calculation of virtual number of teeth Z v1 

Z1

In this problem, don’t assume Z 1 value cos 1 initially, because, we have to determine the accurate value of Z 1. Z1 

Z v1 

d1 m t c o s 1



d1 mt 80 m t cos 38.65 

3.28




Z v2  

102.43 mt d2 m t cos 2



100 m t cos 51.34 

160.07 mt


Power



HP  75 kgf vm 2.75  10 3  3.6666 H.P 750

d 1  mm N  rp m

vm  Ft 

 d1 N1 60  1000



  80  1100  4.6076 m/s 60  1000

3.6666  75  59.682 4.6076

F t  59.682 kg f  596.82 N

 10 Newton) (1kg f ~

Step 6: Calculation of Dynamic load Fd F d  C v Nsf km F t km  1 from PSG data book pg 8.52 [Both gears

straddle mounted]. N sf  1 (Assume) From PSG data book pg 8.52


Cv 

 3.5   vm 3.5

Fo r straight bevel gears   For commercially cut wheels. vm  5 m/s 

From PSG Data Book pg 8.52. Cv 

3.5   4.6076  1.6132 3.5

F d  C v  N sf k m F t  1.6132  1  1  596.82 F d  962.8466 N

Step 7: Calculation of strength of gear tooth F s b   F s  [ b] by v   m t  1   R   yv  0.124 

 0.124 

b  0.3R (Or)

10 Pd

1 0.684 F or 14 involute 2 Zv1 0.684 mt 102.43

 0.124  6.6777  10 3m t b   F s  [b]   m t  b  yv  1   R   0.3R    55    m t  b  0.124  6.6777  10  3 m t  1 R    55    m t  0.3  64  0.124  6.677  10  3 m t 1  0.3  F s  287.96 m t  15.507 m 2t

3.30


[R  

1 2

1 2

d 21  d 22  

80 2  100   2

 64.03 ~  64 ]

Step 8: Calculation of transverse module m t For safe design; F s  Fd 287.96 m t  15.507 m 2t  962.8466

By solving this equation, m t  4.37 or 14.19 14.19 is to o high, so neglected m t  4.37 mm

From PSG Data book, the standard module mt  5 mm pg 8.2 Step 9: Calculation of Face Width b  20 mm b  0.3  R  0.3  64  19.2 ~

Step 10: Calculation of teeth Z1 & Z 2  d1  m t  Z1  d 2  m t  Z2   d1 d2  Z1  Z2  mt mt   80 100     5 5   Z 1  16 Z 2  20  


Step 11: Calculation of pitch line velocity v m  or v 1 v1  4.6076 m/s

Step 12: Recalculation of strength of gear tooth Fs F s  287.96 m t  15.507m 2t mt  5 F s  287.96  5  15.507 52  1052.125 N

Step 13: Calculation of wear load F w Fw 

d 1 b Q v kw c o s 1

Zv1 

102.43 102.43   20.486 mt 5

Zv2 

160.07 160.07   32.014 mt 5

Qv 

2 Z v2 2  32.014   1.295 Z v1  Zv2 20.486  32.014 

1   1  [ 2e ] sin   E1 E2    kw  1.4

3.32


[c]  630 N/mm 2   14

1 2

E 1  E 2  84  103 N /mm2 [630 2]  sin 14 kw  

1  1 1    2  84  10 3 84  10 3    1.4

2.36609 1.4

b  20 mm

kw  1.69 cos 1  cos 38.65   0.7809 Fw 

d 1 b Q v kw cos 1



80  20  1.2195  1.69 0.7809



3297.528 0.7809

F w  4222.72 N

Step 14: Checking for Wear Fw  Fd 4222.72 N  962.84 N  The design is safe based on wear condition.

Answer Module m  5 mm Pinion Z1  16 Gear

Z2  20


Problem 3.2: Design a right angle bevel gear drive with speed of pinion shaft  300 rpm and that of the gear shaft  150 rpm; pinion is to have 20 teeth of involute profile with module of 20 mm and a pressure angle of 20 and is to be of suitable material. Gear is of forged steel having allowable stress of 1500 kg/cm2, kW at gear shaft  56; Assume service factor  2;

(May / June - 2008)

LEWIS METHOD Given Data: Speed of pinion shaft  N 1  300 rpm Speed of gear shaft

 N 2  150 rpm

No. of teeth on pinion  Z 1  20 teeth Module

 m t  20 mm

Pressure angle

   20

Gear is of forged steel with allowable static stress  [  b ]gear  1500 kgf/cm 2

Power at gear shaft

 56 kW  56  10 3 watts

Service factor

2

i  speed reduction  Z 1  20 teeth  Z 2  40 teeth

Z2

Z1



N1 N2



300 2 150

3.34


R  cone distance (or) cone length  0.5 mt   Z 21   Z 22 

20 2  40 2  0.5  20    447.21 mm

Assume pinion is also made of same material, forged steel, [b]gea r  1500 kg f/c m2

The design is based on gear. [Because, the power given is the power at gear shaft.] Assume face width  b  0.3R  0.3  447.21  134.163 mm  13.4163 cm

Step 1 b   Beam strength  F s  [ b ] b    yv  mt  1   R   ... (1) From PSG Databook, Pg No. 8.52 F S  1500  13.4163  13.4163     0.1438  2   1  44.721    12,728 kg f/cm

2

 Beam Strength FS  12,728 kg f/cm 2

ZV 

Z1 cos 1



Z2 cos 2

But 1  2  90 ; tan 2  i  2 2  63.43 


ZV 

40 cos 63.43 

1  90  2  26.57  89.42   90 teeth y v  0.154   0.154 

0.912 Z V gear 0.912 90

 0.1438

Step 2 75 Transmitted load (or) tangential tooth load  F t  H P  vm

Power transmitted by gear shaft  56 kW  76.08 HP vm  mean velocity 



d2

d1 av  N1 60  100 d2

av

 N2

60  100

m/sec

m /sec

 R  0.5b   m tZ 2   av R    447.21  0.5  134.163   20  40   447.21    680 mm  68 cm

vm 

  68  150  5.34 m /sec 100  60

or

3.36


Transmitted load  F t 

76.08  75 5.34

 1068.53 kgf/cm 2

Buckingham’s dynamic load   0.164 Vm cb  F t Fd  Ft     cb  F t   0.164 V m  1.485  F t  1068.53 kg f/cm2 V m  5.34  60  320.4 m/min b  13.416 cm

For (c) value, from table (42) for 10 mm module e  0.09 mm Commercially cut gears for 20 mm module also, take same value. From table (41), 20 involute teeth form, c  11860  0.09  1067.4

Buckingham Dynamic load  0.164 Vmcb  F t   Fd  Ft    cb  F t    0.164V m  1.485  0.164  320.4 1067.4  13.416  1068.53    F d  1068.53     1067.4  13.416  1068.53   0.164  320.4  1.485

 52.54515388.7684    1068.53     52.545  184.216   1068.53  3415.27  4483.8 kgf/cm 2


Beam strength F S  12,728 kg f/cm2 ; F d  4483.8 kgf/cm 2  FS  Fd  The design is satisfactory based on beam strength.

Step 3 To check for wear strength, Fw 

b  d 1  Qv  k w cos 1

b  face width  13.4163 cm d1  m tZ 1  20  20  400 mm  40 cm Qv 

22 4 2i   1.333  i1 21 3

1  26.57  k  [  c ]2

sin   1 1   1.4  E 1 E 2   

[ c ]  C R  H RC  k cl

P.No. 8.16 PSG Data book

 220  60  0.6177  8153.83 kg f/cm 2 . . . Forged steel, HRC  55 to 63

Take HRC  60 C R  220 k cl assume life 10,000 hrs Refer PSG Data book

(P.No. 8.17)

3.38


N  300  60  10,000  18  10 7 cycles kcl 

  6

107  N

  6

10 7

18  10 7

 0.6177

Take E 1  E 2  2.15  10 6 kgf/cm 2  k w  8153.83 2

P.No. 8.14 PSG Data book

sin 20  1 1   1.4  2.15  10 6 2.15  10 6   

 15,109 kg/cm2 Fw 

13.4163  40  1.33  15.109 cos 26.57 

 12,057 kg f/cm 2,

But F d  4483.8 kgf/cm 2

[ Fw  Fd]

Therefore, the design is satisfactory based on wear strength. Problem

3.3: A pair of 20 full depth involute teeth bevel

gears connect two shafts at right angles having a velocity ratio of 3:1. The gear is made of cast steel with permissible stress of 100 N/mm2. The pinion transmits 37.5 kW at 750 rpm. Determine (1) module and face width and (2) pitch diameters. Assume face width 

1  Length of Pitch cone. 3 (Nov/Dec 2007)


LEWIS METHOD Given Data: Pressure angle  20 i3

Permissible stress  [ b]  100 N /mm 2  10  1000 kgf/c m2 P  37.5 kW  50.95 HP . 1 HP  736 Watts

Speed of pinion  N  750 rpm Face width

b 

1  Length of pitch cone 3 1 R 3

(Pitch cone length (or) cone distance)

From PSG data book, F s 

[b] b Yv [1 

b ] R

Pd

(Replace Y v & P d b   F s  beam strength  [b] b  yv mt  1   R  

Yv   yv   Pd   Pc  m t 

1 mt

3.40


 1000 

1 1  R    0.1125  m t  1   3 3 

 78.539 m t  R kg f 1  2  90

Zv 

Z1



20  21.08  22 cos 18.53

cos 1 (assume Z 1  20 teeth tan 2  i  3

yv  0.154 

2  71.56

0.912 0.912  0.154  Zv 22

 0.1125

1  18.43, Beam strength  F s  78.53 m t R kg f

... (1)

Dynamic load  F d  F t  C v F t  HP 

75 vm

But

Power (P)  37.5 kW  50.95 HP vm 

Z 1  20 teeth Z 2  i  Z 1  3  20  60 teeth

R  0.5 m t

Z 22 Z 21    

R  0.5 m t

20 2  60 2  

 31.62 m t cm vm 

 d1av N 1 100  60

d1

 d av N 1 m/sec 100  60

 R  0.5b  mt Z1  av R  0.5R   R 3  mt Z 1  R 

     .  

 R  0.166 R   mt   R  




  16.6797 mt  750 60  100

 6.55 mt m/sec  F t  T ransmitted load  50.95 

75 6.55 m t

 583.396    kgf mt   Cv 

6.55 m t  vm 3.5   3.5    3.5 3.5

Dynamic load  Fd  F t  C v 

 t 6.55 m 583.396 3.5    mt 3.5

... (2)

Substitute the value of R in Eq. (1) Beam Strength F s  78.53 m t R  78.53  m t  31.62 m t  2483.1186 m t2  kgf

... (1)

Equate eq (1) and eq (2); Fs  Fd then check Fw  Fd m t2 

 t 6.55 m 583.396 3.5    mt 3.5

Calculate m t by trial and error method,

3.42


Trail 1 Let m t  0.5 cm LHS  620.779

RHS  1770.08

LHS  RHS

Not Satisfactory Trail 2 Let m t  0.6 cm LHS  893.922

RHS  1523.05

LHS  RHS  Not satisfactory.

Trail 3 Let m t  0.65 cm LHS  1049.1176

RHS  1426.65

LHS  RHS  Not satisfactory

Trail 4 Let m t  0.62 cm LHS  954.51 LHS  RHS  Not satisfactory.

; RHS  1482


Trail 5 Let m t  0.7 LHS  1216.72 ; RHS  1343 LHS  RHS  Not satisfactory

Trail 6 Let m t  0.72 LHS  1287.24

RHS  1313

LHS  RHS  Not satisfactory

Trail 7 Let m t  0.725 LHS  1305.18 ;

RHS  1305.69

LHS  RHS

Therefore, select m t  T rans verse m odu le  0.725 mm  7.25 cm According to the standard value of normal module mt Refer PSG Data book P. No. 8.2 and Table 1 The standard normal module mt  8 mm  Cone distance

 R  31.62  m t  31.62  0.8  25.296 c m

3.44


Cone distance  152.96 mm

(or)

Face width:  b 

Length

of

pitch

cone

(R)

R 252.96   84.32 mm 3 3

Pitch diameters: Pitch dia. of pinion d1  m tZ 1  8  20  160 mm Pitch dia. of wheel d2  m tZ 2  8  60  480 mm. Here check F w  F d Problem 3.4 Design a bevel gear to transmit a power of 9 kW at 200 rpm of the pinion. Gear Ratio is 3, material used is C20. Ultimate tensile strength u is 500 N/mm2, Yield strength y is 260 N/mm2

(May/June 2013)

LEWIS METHOD Solution: Step 1: Given Data: Power  9 kW  9  10 3 W Pinion Speed N 1  200 rpm Gear Ratio

i  3

Material used for pinion and gear  C20 u (ultimate strength)  500 N/mm 2 y (yield strength)  260 N/mm 2


Step 2: Selection of Materials Pinion  C20 ; Gear  C20 Step 3: Calculation of pitch angles 1 & 2 tan 2  i 2  tan  1 3 2  71.56  ; 1  2  90 1  90  71.56  1  18.43

Step 4: Calculation of virtual number of teeth Z v1   Z2 Z1

Z1 cos 1

Z 1  20 (Assume)

20  22 21.08 ~ cos 18.43 

i

Z 2  3  22 Z 2  66 Z v2 

Z2 cos 2



66 cos 71.56 

3.46


 209 Z v2  2086 ~


Power 

HP  75 kgf vm 9  10 3 HP 750

d 1  m tZ 1 d 1  20m t

 12 HP v1 

 d 1N 1   20 mt  200  60 60  1000

v1 

209.439 mt m/s 100

v1  0.2094 mt m/s Ft 

12  75 0.2094  mt Ft 

4297.99 kg f mt

Step 6: Calculation of Dynamic load Fd F d  C v N sf k m Ft Cv 

 vm 3.5   3.5

From PSG data book pg No. 8.52


N sf  1 ; k m  1 (both gears straddle mounted)

Assume vm  5 m /s Cv 

5 3.5   3.5

 1.6388 Fd 

1.6388  1  1  4297.99 mt

Fd 

70435.53 7043.553 kgf  N mt mt

Step 7: Calculation of strength of gear tooth F s b  F s  [ b] b  yv    m t 1   R   [ b]  b

u 3



500  166.66 N/mm 2 3

R~  0.33 R 3

b  0.33 R b  10.43 mt R1  

mt 2

mt

2

Z 21   Z22  

20 2  60 2  

R  31.62 mt

3.48


yv  0.154 

0.912 Zv 1

 0.154 

For 20 Involute   20 (Assume)

0.912 22

yv  0.1125 F s  166.66  10.43 mt  0.1125    m t 1  1/3  409.56  m 2t N

Step 8: Calculation of transverse module m t We know that Fs  Fd 409.56 m 2t 

70435.53 mt

m 3t 

70435.53 409.56

 m t  3 171.978 m t  5.56 mm  6 mm mt ~

(From psg Data Book Pg. 8.2 1 Table 1. Step 9: Calculation of face width (b) b

R~  10  mt  10  6 3 b  60 mm


Step 10: Calculation of pitch circle diameter d 1 d 1  m 1Z1  6  20 d 1  120

Step 11: Calculation of pitch line velocity v 1 v1  0.2094  m t m/s  1.2564 m/s

Step 12: Recalculation of strength of gear tooth Fs F s  409.56 m 2t N  409.56  36   14744.16 N

Step 13: Calculation of Buckingham’s Dynamic Load Fd   0.164 Vm cb  F t Fd  Ft    cb  Ft    0.164 V m  1.468  Ft 

4297.99 kgf mt

3.50




4297.99 kgf 6

 7163.3 N  716.33 kg f ~ V m  V 1  1.2564 m/s  60 m/m in  6 cm b  60 mm ~ c  11860 e for steel e  0.015 m  6 mm  Precis ion gears c  11860  0.015 c  177.9 kg f/cm Fd  Ft 

0.164 V m cb  F t

 0.164 V m   1.468 cb  F t

substitute F t in kg f b  in cm V m  m/min vm  m/s c  kg f/cm

then we obtain F d in kg f   0.164  1.2564  60177.9  6  716.33   716.33     177.9  6  616.33   0.164  1.2564  60  1.468 

 716.33 

22052.2118 74.36

 716.33  296.548  10128.7 N F d  1012.87 kgf ~


Step 14: Checking for beam strength Fs  Fd 14744.6  10128.7  The design is safe.

Step 15: Calculation of Wear Load Fw  Fw 

d 1bQvkw

cos 1

d 1  120 mm b  60 mm Qv 

2Zv2 Z v1  Z v2



2  209  1.809 22  209

cos 1  cos 18.43   0.9487

Fw 

Assume kw  1 N /mm 2 kw -load stress factor.

120  60  1.809  1 cos 1

 13729.102 N

Step 16: Checking for Wear Strength For satisfactory Design. Fw  Fd 13729.102  10128.7 N  The design is safe based on wear strength.

3.52


Step 17: Calculation of Basic Dimensions of Straight Bevel Gear. From PSG Data book pg no. 8.39. 1. Reference (or) Lead Angle 1  18.43 2  71.56  2. Addendum Angle  a  mtf0  a1   a2  tan  1    R  61  tan  1    190  a1   a2  1.8

R  31.62 m t  31.62  6 ~  189.72  190 mm

3. Dedendum Angle f  f0  c  f1   f2  tan  1  mt  R    6 1  0.2    tan  1   190    tan  1 0.03789  f1   f2  2.17

4. Height Factor 0 0  1


5. Tip Angle a a1  1  a1  a2  2  a2   18.43  1.8   71.56  1.8  20.23    73.36  

6. Root Angle t t1  1   f1    18.43  2.17  f1  16.26  

t2  2  f2  71.56  2.17

t2  69.39 

7. Virtual number of teeth Z v2  22 ; Z v2  209 Problem 3.5 Design a pair of bevel gears to transmit 15 kW at the pinion speed of 1600 rpm, and the gear ratio is 4. Material for gears 15 Ni 2 cr Mo 15 / steel. The tooth profiles of the gears are of 20 composite form.

LEWIS METHOD Solution Step 1 Given Data P  15 kW  15  10 3 W N 1 pinion  1600 rpm

Gear ratio (i)  4 Material for pinion and gear  15 Ni 2Cr Mo 15 / steel. Pressure angle   20

3.54


Step 2: Selection of Material Pinion - 15 Ni 2 Cr 1 Mo 15 / steel Gear - 15 Ni 2 Cr 1 Mo 15 / steel Step 3: Calculation of Pitch Angles 1 & 2 tan 2  i

1  2  90

2  tan 14

1  90  75.96

2  75.96

1  14.03

Step 4: Calculation of virtual number of teeth Z v1   Z v2 

Z1

Assume Z 1  20

c o s 1

20  20.6 ~  21 c o s 14.03  Z2 cos 2



Z2 Z1

i

Z 2  4  20 Z 2  80

80  330  329.76 ~ cos 75.96 

Steps 5: Calculation of tangential load F t Ft 

HP  75 kg f vm

P

15  10 3 H.P 750

P  20 H.P

d1  m t  Z1 d1  20m t


vm 

Ft 

 d 1N 1   20m t  1600  1.6755 m t  60  1000 60  1000 HP  75 kgf Vm



20  75 1.6755 m t



895.255 kg f mt

Step 6: Calculation of Dynamic Load F d F d  Cv N sf k m F t Cv 

 3.5   vm 3.5

 5 m/s Ass ume vm ~

[For bevel gears vm  5 m/s] Choose k m & N sf From PSG data book pg 8.52 N sf  1 ; k m  1

[  both gears are straddle mounted (Assume) ]

Cv 

5 3.5    1.6388 3.5

Fd 

1.6388  1  1  895.255 1467.212  kg f mt mt

Fd 

14672.12 N mt

3.56


Step 7: Calculation of strength of gear tooth b  F s  [ b] byv 1      m t R  

Choose [ b] From PSG Data book pg.8.5 (For 15 Ni 2Cr 1 Mo 15 / steel.) (m t upto 6) [ b]  3200 kgf/cm 2 1 kgf/cm 2 

 320 N/mm 2

1 N/mm 2 10

 320 N /mm2

Always Assume b 

R



1 R 3 mt 2 mt 2

Z 21   Z 22   20 2  80 2  

R  41.23 mt yv1  0.154 

0.912 For 20 Involute, From PSG data Z v1

book pg 8.50 yv  0.154 

0.912  yv  0.1105 21


b F s  [ b] b y v1 1    m t R R  1    320   R   0.1105  1     mt 3 3R      41.23   320    m t  0.1105 0.666     m t  3  F s  1016.78 m 2t

Step 8: Calculation of Transverse module m t We know that Fs  Fd 1016.78 m 2t 

14672.12 mt

m 3t 

14672.12 1016.78  2.43 mt ~

From PSG Data book Pg. 8.2; Table 1 standard module mt  3 mm

Step 9: Calculation of face width (b) b  0.3  R  0.3  123.69  37.107  38 mm b~

R  41.22  m t  41.23  3  123.69  124 mm R~

3.58


Step 10: Calculation of pitch circle diameter d 1 d1  mt Z 1  3  20 d1  60 mm

Step 11: Calculation of pitch line velocity v1  d 1  1600   60  100  vm  v1l  60  1000 60  1000  5.0265 m/s

Step 12: Recalculation of strength of gear tooth Fs F s  1016.78 m 2t  1016.78  9  9151.02 N

Step 13: Calculation of Buckingham’s Dynamic Load Fd  0.164 V m cb  F t Fd  Ft    cb  F t  0.164 V m  1.485    Ft 

895.255 kgf mt

 298.418 kgf

    


To determine vm for bevel gear, take largest pitch circle dia of gear. d2

d1  60 mm

d1 N1 N2

d2  4  60

i

N 2  N 1/i 

i

d2  240 mm

1600 4

N 2  400 rpm vm 

 d2 N 2   240  400  60  1000 60  1000

vm  5.0265 m/s

From PSG data book pg 8.53 ; Table 41 & Table 42. For 20 Full depth steel and steel  11860 e e  0.0125 (Precision gears)  38 mm  3.8 cm b~ c  1160  0.0125  148.25 kgf /c m

To calculate F d Substitute V m in m/min ; b  in cm F t  kg f

Then obtain F d in kg f.

b  38 mm  3.8 cm

3.60


V m  5.0265  60 m/min  301.5 m/m in   0.164 V m cb  F t Fd  Ft    cb  Ft    0.164 V m  1.468    0.164  301.59 148.25  3.8  298.418   298.418    148.25  3.8  298.418    0.164  301.59  1.468 

 298.418 

42623.70 236.11

 298.418  180.52 F d  478.9 kgf ~  4789 N

Step 14: Checking for beam strength Fs  Fd 9151.02  4789  The design is safe.

Step 15: Calculation of Wear Load Fw  Fw 

d 1b Q v kw cos 1

d 1  60 mm ; b  38 mm Qv 

2Z v2 Z v1  Z v2



2  330  1.8803 21  320


cos 1  cos 14.03   0.97 1   1  [2e ] sin   E2  E  1  kw  1.4

  20 From Data book pg 8.14 ; Table 9. (steel & steel) E 1  E 2  2.15  106 kg f/cm 2  2.15  10 5 N /mm2 kw  load Stress Factor

e  55 kg f/mm2 (steel) From PSG Data book pg No.85; Table 7 e  550 N/mm2 1 1   5502  sin20    5 5  2.15  10   2.15  10 kw  1.4  0.687 N/mm2 Fw 

d1b Q v kw 60  38  1.8803  0.687  0.97 cos 1

F w  3038.3 N

Step 16: Checking for wear strength F w  3038.3 N ; F d  4789.04 N

3.62


Fd  Fw  The design is not safe.

To increase the wear load, we have to increase the load stress factor k w kW  0.687 N/mm 2 ; New kW  2.5 N/mm 2 Assum e Fw  

d 1 b Q vkw cos 1 60  38  1.8803  2.5 cos 14.03 

 11047.26 N

Now F w  F d  Now the design is safe.

Step 17: Calculate the Basic Dimension of Straight Bevel Gear From PSG Data book pg 8.39 1. Reference (or) lead Angle tan 2  i

2  75.96  1  14.03

2. Addendum Angle  a tan  a1 

m t fo  tan a2 R 31 a 1  tan 1    124 

fo  1


a 2  a1  1.3859

3. Dedendum Angle f tan f1  tan f2  

m t f0  c

c  0.2

R

3 1  0.2  f1   f2  tan  1 29.67  10 3 124

f1   f2  1.704 

4. Height Factor f0 f0  1

5. Tip Angle a a1  1  a1    14.03  1.385    15.415  

a2  2  a2 a2  75.96  1.385 a2  77.345

6. Root Angle t    14.03   1.704     12.326  

t2  2  t2

t1  1   t1

 75.96  1.704 

7. Virtual number of teeth Z v1  21 ; Z v2  330

 74.256 

3.64


3.11.(a) DESIGN PROCEDURE FOR BEVEL GEAR (BASED ON HERTZ METHOD) Step 1 Selection of material: P.No. 8.16 Table (16) or Table (5) P.No. 8.4 of PSG Data book. Similar to Spur and helical gear. If pinion and gear are made of same material, design for only pinion (because pinion is weaker than gear) Select material, surface hardness. If surface hardness  350 BHN o r 55 to 63 RC or 40 to 55 RC The core hardness  350 BHN

Step 2 Calculate equivalent Young’s modulus. From PSG Data book P.No. 8.14 table (9) Read

young’s

modulus

E 1, E2

according

material. Find the equivalent Young’s modulus E eq  kg /cm2 N  f   to convert  10 mm2   Step 3 Assume life 10,000 hrs to 20,000 hrs.

to

the


If life is given in the problem, Calculate life N in terms of cycles. i.e., N  Life required  Given life in Hrs  RPM  60 C ycles Represent life interms of  ......   10 7 cycles Step 4 Calculation of Design bending stress [ b ] From PSG Data book P.No. 8.18 (similar to spur and helical gears) Design bending stress  [ b ] 

1.4kbl  1 n  k

kbl  Life factor for bending From PSG Data book P.No. 8.20; table (22) (consider core hardness). It depends on material, Surface hardness and life in cycles.  1  Endurance limit stress in bending for complete reversal stress in kg f/cm 2 From PSG Data book P.No. 8.19; Table (19). According to material of mating gear, take u and  y values from From PSG Data book P. No. 8.5 Table (7) in kgf/cm 2; u  ultimate stress in kg f/cm 2(Refer P.No. 1.9 of PSG Data book) y  yield stress in kg f/cm 2 n  Factor of safety

Table (20); P.No. 8.19 of PSG Data book

3.66


According to material, mode of manufacture and heat treatment. k  Fillet stress concentration factor From PSG

Data book P.No. 8.19 Table (21). Take addendum modification coefficient. 0  X  0.1 value and according to material and heat treatment.

Calculate design bending stress [ b ] in kg f/cm 2 Convert into N/mm 2

kgf/cm2 10

 N/mm 2

Step 5 Calculation of Design surface contact compressive stress. Fron PSG Data book P.No. 8.16 [ c ]  C RHRC  kcl kg f/cm2 C R  Coefficient depending on the surface

hardness. From PSG Data book Table (16) P.No. 8.16. Depends on material, heat treatment and hardness. HRC  Rockwell ‘C’ hardness number Table (16)

P.No. 8.16 kcl  Life factor From PSG Data Table (17),

P.No. 8.17 (Consider surface hardness) It depends on material, surface hardness and life in number of cycles.


Calculate [ c ] in kg f/cm 2 Step 6 Calculation of Design torque of twisting moment [ M t ] P.No. 8.15 PSG Data book Design twisting moment  [ M t ]  Mt  k d  k M t  Nominal twisting moment transmitted by the

pinion kg f  cm . From PSG Data book Pg No. 8.150 

97420 kW N

kW  nominal power transmitted in kW N  speed of rotation of pinion  rpm k  load correction factor. Table (14) (P.No. 8.15) of

PSG Data book. kd  dynamic load factor. Table (15) (P.No. 8.16) of

PSG Data book. k value depends on IS quality and b/d 1 ratio kd value depends on IS quality and pitch line velocity

Since the values of b, d1 pitch line velocity, and IS quality are not known, initially assume k d  k  1.3 for symmetric scheme. Calculate M t in kg f  cm . And [ M t ]  Mt  k d  k in kg f  cm . [ kg f  cm  100  Nmm ]

3.68


Step 7 Calculate Cone distance ‘R’ P.No. 8.13 table (8) of PSB Data book

R   y i 1  2

y 

  3

2 0.72   E[ M t ] cm    0.5  [  ]  i c   y

R table (13) From PSG Data book P.No. 8.15 b

Assume y  3 (or) 4 (based on i value) i  speed ratio 

Z2 Z1

[ c ]  Design surface (contact compressive) stress kg /cm 2 E  Equivalent Young’s modulus kg/cm 2 [ M t ]  Design twisting moment kg-cm.

Step 8 Calculation of m t  transverse module From PSG Databok P.No. 8.38 table (31) R  0.5 m t Z 1

i2  1 

(or) R  0.5 m t  Z21   Z22  Assume Z 1  20 to 30

Represent cone distance R in mm. Calculate m t in mm


Take standard module m t i  speed ratio 

Z2 Z1

From PSG Data book P.No. 8.2 table (1) Represent transverse module mt in mm Step 9 b

Calculate

d1av

to find (k)

k  Load concentration factor:

To find d 1av m m  mav  m t 

b  sin 1 Z1

From PSG Data book P.No. 8.38 table 31

Cal. v  pitch line velocity 

 d1 av N 1 60  1000

m  sec

Cal. d1av in mm  Calculate

b d1 av

d1 av    mm  Z  1  

ratio .

From PSG Data book Table (14): P.No. 8.15 Take the value of k k value depends on

b value and surface hardness. d1 av

3.70


Step 10 Selection of IS Quality (From PSG Data book P.No.8.3) Table 2 Calculate pitch line velocity v 

 d1av  N 1 60  1000

m  sec

Corresponding to velocity (m/sec) select the preferred quality. Step 11 Read the value of kd   From PSG Data book   P.No. 8.16 table 15 

It depends on Pitch line velocity (m/sec)

Read the value of k d

Corresponding to pitch line

kd  dynamic load factor

velocity, Surface hardness and preferred quality. Step 12 Recalculate [M t]- Design twisting moment. [M t]  M t  kd  k M t from step 6 kgf cm kd from step 11

kgfcm  100  N  mm

k  from ste p 9

Represent [Mt] in kg  cm Step 13 Calculation of induced bending stress b


From PSG Data book P.No. 8.13A; table (8) b 

R

i2  1 

[M t]

2

R  0.56 b  mt yv

R  Cone distance

1  [b] cos 

  pressure angle ;

(cm);

 20

i  Speed ratio;

[M t]  Design to rque  kg  cm ;

m t  Transverse module cm yv  form factor Z1 Z eq  Z v  cos 1

Refer PSG Data book table (18) P.No. 8.18 at X  0 (Addendum modification coefficient) and Z v Read yv value either directly (or) by interpolation. Note: If b  [b] Increase the value of m t Recalculate:

(R)

;

Pitch line velocity. v

(b)

;

kd, k

m av 

;

[M t ]

d1a

v

Calculate b (induced bending stress) Check again  b If  b  [b] Then the satisfactory.

design

based

on

beam

strength

is

3.72


Step 14 Calculate induced surface contact compressive stress c From PSG DAta book P.No. 8.13 table (8)

0.72 c  R  0.5b

  i2  13   ib

E [M t]  [ c]

Note: If c, b are less than design stresses [ c] and [b] respectively, then the design is satisfactory. If pinion and wheel are made of different materials. First design pinion and then check the stresses in the wheel. Procedure Step (1) For wheel material, Calculate design bending stress [b ]  [ b ]w heel 2

and design surface contac t     c    C  compressive stress 2      w heel

  similar to step 4  and step 5  

Assume c

pinion

 wheel

c

wheel

 [c]W heel

Life for wheel  N  cycles

If c

wheel

 [c]w heel,

Life of pinion i


Change the material and recalculate

[ c]wheel and

check again. If  c  [c]wheel , then the design is satisfactory. To check b,  b y v  b y v 1

1

2

y v  form factor for pinion

2

1

b  induced bending 1

stress for pinion

    

Already calculated in step 13

Zv  1

Z1 cos 1

y v  form factor for wheel 2

Zv  2

Z2 cos 2

Calculate b and check b 2

2

b  [b ] i.e. [ b]wheel 2

2

FORMULAE FOR GEAR FORCES From PSG Data book P.No. 857 table 46 (a) Spur Gear Tangential force  F t 

2 Mt d1

M t  twisting moment  97420

kW kgf  cm N1 where N 1  pinio n speed in rpm

d 1  pinion pitch circle diameter

Radial force F r  F t tan  (where   pressure angle  20 

3.74


(b) Helical Gear Tangential force Ft  Radial force F r  F t 

2M t d1

tan  cos 

(c) Bevel Gears: (Straight bevel) 2 Mt 2M t     Tangential force   F tav  dav b    d  1 0.5  1 R     Radial force  F r   F t  tan   cos 1, for pinion av    F t  tan  cos 2 fo r wheel av  Axia l fo rce  F a  F tav. tan   sin   F tav tan sin 1 for pinion  F tav tan  sin 2 for w heel

(d) Bevel (Spiral) Gear Tangential force F t:

F tav 

2M t d av



Where b    Pres sure angle  20  d1  1  0.5 R    helix angle  2M t

 18 to 25


Radial force Fr 

F tav c o s av

tan   cos   sin av sin 

Axial force  F a 

F tav cos av

tan  sin   sin av  c o s 

av  mean spiral a ngle 30 to 35  35 preferred

  Pressure angle  20   reference angle 1  semico ne angle of pinion 2  semicon e angle of wheel (a) For right angle (when   90 Z2 tan 2  i  Z1 1  90   2

(b) For acute angle when   90 sin tan 1  i  cos 

(c) Objective angle, when   90 tan 1 

sin 180   i  c o s 180  

1  2  

3.76


Problem 3.6 Design a bevel gear drive to transmit 7 kW at 1600 rpm for the following data.

Gear ratio : Material for pinion and gear : Life :

3 C45 steel 10,000 hrs. (May/June 2013)

HERTZ METHOD Given data P  7 kW ; N  1600 rpm ; i  3 ; N  10,000 hrs

Solution Selection of material Since the given material for pinion and wheel are made of same material (i.e) C45 steel, therefore, design only pinion. From PSG Data book, Pg.No. 8.16, Table 16 corresponding to C45 steel. Assume surface hardened heat treatment, HRC  40 to 55  HRC  50 (Assume)

(ie) sufarce hardness > 350 BHN core hardness < 350 BHN Step 2: Life of gears N  10,000 hrs  10,000  60  1600  96  10 7 cycles

Step 3: Equivalent Young’s modulus From PSG Data book, From PSG Data book Pg.No. 8.14, Table offer E  2.15  10 6 kg f/cm 2  2.15  10 5 N/mm 2


Step 4: Calculation of design bending stress [b] From PSG Data book, Pg.No. 8.18, [b] 

1.4 kbl  1 n  k

From PSG Data book Pg.No. 8.20, Tabl;e 22. Assume rotation in one direction only, consider core hardness  350 BHN; N  10 7 cycles kbl  1

From PSG data Blook Pg.No. 8.19, Table 19, for forged steel  1  0.25  u   y  500 kgf/cm2 From PSG Data book Pg.No. 1.9, for C45 steel, u  63  71 kgf/mm 2; y  36 kgf/mm 2 Assume u  70 kg f/mm 2  70  100 kg f/cm 2    1  0.25 70  100  36  100  500  3150 kg f/cm 2

From, PSG Data book Pg.No. 8.19, Table 20 for forged steel and surface hardened n  2.5

From PSG Data Book Pg.No. 8.19, Table 21 for steel, surface hardened at 0  X  0.1 k  1.5

3.78


 Design bending stress [b] 

1.4  1  3150  1176 kg f/c m2  117.6 N/mm 2 2.5  1.5

Step 5: Calculation of compressive stress [c]

design

surface

contact

From PSG Data book, Pg.No. 8.16 [c]  C R HRC  kcl

From PSG Data book, Pg.No. 8.16, Table 16 for C45 - surface hardened steel C R  230 HRC  50

From PSG Data book, Pg.No. 8.17, Table 17, Assuming steel, surface hardness > 350, N  25  10 7 kcl  0.585  [c]  50  230  0.585  6727.5 kgf/c m2  672.75 N/mm2

Step 6: Calculation of design torque From PSG Data book Pg.No. 8.15, [M t]  M t  kd  k

But

M t  97420

kW 7  97420   426.21 kgf  cm n 1600

Assume initially, k  kd  1.3 (for symmetric)  [M t]  426.21  1.3  554.07 kgf  cm  554.07  10 2 Nmm


Step 7: Calculation of cone distance ‘R’ From PSG Data book Pg.No. 8.13, Table 8 R  y

i2  1  

  3

2 0.72   E [Mt]    0.5 [ ]  i c   y

From PSG Data book, Pg.No. 8.15, Table 13 for i  3, Assume  y  3

R3

2 3 1  

  3

2

5

2.15  10  554.07  10 0.72    3  0.5 672.75    3  

2

 R  97.67 mm

Step 8: Calculation of Transverse module ‘m t’ From PSG data book Pg.No. 8.38, Table 31, R  0.5 m t Z1

i2  1 

Assume Z 1  20 teeth  97.67  0.5  m t  20

32 1  

 m t  3.08 mm

Represent traverse module terms of standard module, From PSG Data book Pg.No. 8.2, Table 1,  m t  5 mm

Note: If we hoose m t  4 mm , then the design will be not safe, To avoid recalculation, always choose higher standard module m t.

3.80


Step 9: Recalculate “R ” and calculate “b”  R  0.5 m t Z 1

i2  1  

 0.5  5  20 

32 1  

 158.113 m m

From PSG Data book, Pg.No. 8.15, Table 13 y  b

R 3 b R 158.113  3 3

 b  52.7 mm

Step 10: Calcualtion of k and

b d 1av

From PSG Data book, Pg.No. 8.15, Table 14 Average module sin 1

m av  m t  b

Z1

but tan 2  i  3  2  tan  1 3  71.56  1  90  71.56   18.43   m av  5 

52.7  sin 18.43 20

 m av  4.16 mm

But d1av  Z 1 m av  20  4.16  83.33 mm




52.7 b   0.6323 d 1av 83.33

From PSG Data book, Pg.No. 8.15, Table 14 for surface hardness > 350 for both the gears k  1.6

Step 11: Selection of IS Quality From PSG Data book, Pg.No. 8.3, Table 2. Pitch line velocity v  v

 d1av  N 60  1000

  83.33  1600  6.98 m/s 60  1000

According to velocity, straight bevel, select IS quality - 8 steel Step 12: Read the value of k d From PSG Data book, Pg.No. 8.16, Table No: 15, corresponding to IS Quality - 8, surface hardness > 350 and v upto 8 m/s kd  1.4

Step 13: Recalculate design twisting moment [M t] [M t]  M t  kd  k  426.21  1.4  1.6  954.71 kg f  cm  [M t]  95471 Nmm

Step 14: Calculation of induced bending stress b From PSG Data book, Pg.No. 8.13 A, Table 8 b 

R

i2  1   2

[Mt]

R  0.5b b mt yv



1  [ b] cos 


3.82

From PSG Data book, Pg.No. 818, Table 18, Zv 

Z1

cos 1



20  21.08 ~  21 teeth cos 18.43

Take average value of 20 and 22 teeth y v  0.3955

  pressure a ngle  20 assume  b 

158.113

32 1  

 95471 

2

158.113  0.5  52.7   52.7  5  0.3955



1 cos 20 

 28.076 N/mm 2  [b] 117.6 N /mm 2  The design of pinion based on beam strength is

satisfactory. Step 15: Calculation of induced surface stress From PSG Data book, Pg.No: 813, Table 8

c 



0.72 R  0.5b

  i2  13   ib

0.72 158.113  0.5  52.7 

 E [M t]

 2.15  10  95471   3  52.7 3 2  13  

5

 350.128 N/mm2  [c]672.75 N /mm 2  The design based on wear strength is satisfactory. Problem 3.7 A 25 kW motor running at 1200 rpm drives a compressor

at

780

rpm,

through

a

90

bevel

gearing

arrangement. The pinion has 30 teeth. The pressure angle of teeth is 20. Both the pinion and gear are made of heat treated


cast Iron grade 35. Determine the cone distance, average module and face width of gears.

HERTZ METHOD Solution Given data Power  P  25 kW Pinion speed  N 1  1200 rpm Wheel speed  N 2  780 rpm Z 1  30 teeth i

N1 N2

 1.54

Step 1 Selection of material Material given is CI grade 35 (heat treated) From P.S.G Data book, Pg.No. 8.5 [b]  800 kgf / c m2  80 N /mm2 [ c]  7500 kgf / c m2  750 N /mm2

From P.S.G Data book, Pg.No. 8.14 E  1.7  10 6 kgf/c m2  1.7  10 5 N/mm2

Step 2 Calculation of Design torque [M t] From P.S.G Data book, Pg.No. 8.15 [M t]  M t  kd  k

3.84


But M t  97420   97420 

kW N1 25 1200

 2029.58 kg fcm

From P.S.G Data book, Pg.No. 8.15 kd  k  1.3 assume  [M t]  2029.58  1.3  2638.45 kg fcm  263845 Nmm

Step 3 Calculate cone distance ‘R’ From P.S.G Data book, Pg.No. 8.13

R  y

i 1   2

  3

2 0.72   E [M t]     0.5 [ ]   i c  y 

From P.S.G Data book, Pg.No. 8.15 y 

R  3 correspon ding to i  1 to 4 b i  1.54

R3 1.54 1  2

 89.54 mm

  3

2

5

0.72   1.7  10  263845  3  0.5  [750]   1.54  


Step 4 Calculation of transverse module “m t” From P.S.G Data book, Pg.No. 8.38 R  0.5 m t Z 1

i2  1  

 89.54  0.5  m t  30  1.542 1   m t  3.25 mm

Take standard module m t  4 mm (From P.S.G Data book, Pg.No. 8.21) Step 5 Recalculate ‘R’ and calculate ‘b’ R  0.5 m t Z 1

i2  1 

 0.5  4  30 

1.542 1  

 110.17 mm

But y  b

R  3 assumed b R 110.17   36.72 mm 3 3

Step 6 Calculation of load concentration factor ‘k’ From P.S.G Data book, Pg.No. 8.15, b  The average module

b d 1av

3.86


m av  m t  b

sin 1

Z1

we know that tan 2  i

(i.e) tan 2  1.54  2  tan  1 1.54  57  1  90  57   33  m av  4 

36.72  sin 33  30

 3.33 mm

but d1av  Z 1 m av  30  3.33  100 mm 

b diav



36.72  0.3672 100

 From P.S.G Data book, Pg.No. 8.15 k  1.6

Step 7 Calculate pitch line velocity v From P.S.G Data book, Pg.No. 8.3 v

 d 1av N 1 60  1000



  100  1200  6.283 m/s 60  1000

Step 8 Calculate the value of k d


From P.S.G Data book, Pg.No. 8.16 kd  1.4 for v up to 8 m/s

Step 9 Recalculate design twisting moment [M t]  Mt  kd  k

 2029.58  1.4  16  4546.25 kgfcm  454625 N  mm Step 10 Calculation of Induced bending stress b From P.S.G Data book, Pg.No. 8.13 (A) b 

R

i2  1   2

[M t]

R  0.5b b m t yv



1  [ b] cos 

From P.S.G Data book, Pg.No. 8.18 take yv corresponding to Z v Zv 

Z1

cos 1



30 30  35.77  cos 1 c o s 33 

~  36 teeth  y v  0.4546 at value X  0, Z v  36  b  

110.17

1.542 1   2

 45462.5

110.17  0.5  36.72   86.72  4  0.4546



1 cos 20

91,967,564.61  173.89 N /mm 2  [ b] 80 N/mm 2 528,880.904

3.88


Hence the design is not safe. Therefore increase the value of transverse module m t

Trail 1 Assume m t  5 Repeat the steps 5 to 10 Step 5 R  0.5 m t Z 1 b

i2  1 

 137.71 mm

R 137.71   45.90 mm 3 3

Step 6 m av  mt  b

sin 1 z1

 5  45.90 

sin 33 30

m av  4.166 mm

but d1av  Z1 m av  30  4.166  120 mm 

b 45.90   0.375 d1av 120  From P.S.G Data book, Pg.No. 8.15 k  1.6


Step 7 v

 d1av  N 1   120  1200  60  1000 60  1000

 7.539 m/s

Step 8 kd  1.4 for v upto 8 m/s

Step 9 [M t]  M t  kd  k

 2029.58  1.4  1.6  4546.25 kg fcm  454625 Nmm Step 10 b 



i2  1 [M t] R  2

R  0.5 b b mt yv 137.71



1 cos 

1.54 2 1  

 454625

2

137.71  0.5  45.9   45.9  5  0.4546

b  89.03 N/mm2  [ b] 80 N/mm 2

The design is not safe Trail 2 Assume m t  6 Repeat the steps 5 to 10 Step 5 R  0.5 m t Z 1

i2  1 



1 cos 20

3.90


 0.5  6  30

1.54 2 1 

 165.25 mm b

R 165.25   55.08 mm 3 3

Step 6 m av  mt  b   6  55.08 

sin 1 Z1 sin 33 30

 5 mm

but d 1av  Z 1 m av  30  5  150 mm 

b 55.08   0.3672 d 1av 150

 k  1.6 From P.S.G Data book, Pg.No. 8.15

Step 7 v 

 d 1av N 1 60  1000   150  1200  9.42 m/s 60  1000

Step 8 From P.S.G Data book, Pg.No. 8.16 kd  1.4 for v upto 12 m/s


Step 9 [M t]  M t  kd  k

 2029.58  1.4  1.6  4546.25 kg fcm  454625 Nmm Step 10 b 



i2  1 [M t] R 

1 cos 

2



165.25

1.54 2 1  

R  0.56   b m t yv

 454625

2

165.25  0.5  55.08   55.08  6  0.4546



1 cos 20

 51.52 N/mm2  [b] 80 N/mm2

Hence design is safe. Step 11 Calculation of surface compressive stress From P.S.G Data book, Pg.No. 8.13

0.72 c  R  0.5 b

0.72  165.25  0.5  55.08 

    

 i2  1  

3

ib

    

1.542 1   1.54  55.08

 E [M t]

3

    1.7  10 5  454625

 392.67 N/mm2  750 N/mm 2

since  c  [ c] the design is safe.

3.92


Problem 3.8 Design a bevel gear to transmit a power of 9 kW at 200 rpm of the pinion. Gear Ratio is 3, material used is C20. Ultimate tensile strength u is 500 N/mm2, Yield strength y is 260 N/mm2

(May/June 2013)

HERTZ METHOD Given data: Power  9 kW  9  10 3 W Pinion speed N 1  200 rpm Gear ratio

i3

Material used Pinion-Gear  C20 Ultimate strength  u  500 N/mm 2 Yield strength

 y  260 N/mm 2

Step 1: Selection of Material Material for pinion and gear - C20 (Given) Refer PSG Data book Pg No.8.4. Since the pinion and gear are made up of same material, design pinion as it is weaker than gear. Step 2: Calculate equivalent Young’s Modulus From PSG Data book Pg.No. 8.14 table 9 Select E eq  2.15  10 6 kg f/c m2 E eq  2.15  10 5 N/mm 2


Step 3: Assume life  10000 hrs  N  10000  60  200 N  12  10 7 cycles

Step 4: Calculation of design bending stress[b] From PSG Data book Pg. No. 8.18 [b] 

1.4 kbl  1 n k

From PSG Data book Pg. No.8.20 table 22 (consider core hardness) kbl  1

 1  Endurance limit stress in bending for complete reversal stress in kgf/cm 2 From PSG Data book Pg No.8.19 table 19  1  0.22 u  y  500

(For cast steel)

 0.22 500  260  500 N/mm2  667.2 N/mm 2  6672 kg f/cm 2 n  Factor of safety

From PSG Data book Pg No. 8.19 table 20 n2 k  Fillet stress concentration factor

3.94


From PSG Data book Pg No. 8.19 table (21) Take 0  x  0.1

addendum

modification

co-efficient

k   1.2  [b]  

1.4 kbl  1 n k

1.4  1  6672 2  1.2

 3892 kg f/c m2 [b]  389.2 N/mm2

Step 5: Calculate design surface contact compressive stress [ c] From PSG Data book Pg 8.16 [ c]  C B HB kcl kg f/cm 2

Where C B  20 HB  170 to 200

From PSG Data book Table 16 Pg. No.8.16 Take HB  170 kcl  life factor

From PSG Data book Pg 8.17 table 17 kcl  1

as


 [c]  20  170  1  3400 kgf/cm 2 [ c]  340 N/mm 2

Step 6: Calculation of design twisting moment [M t] From PSG Data book Pg no. 8.15 Design twisting moment [M t]  M t k d k Where M t  Nominal twisting moment Mt 

97420 kW N

Where N  S peed in rpm  200 rpm Given kW  n ominal power in kW  9 kW Given  Mt 

97420  9 200

 4383.9 kg f  cm k  load correction facto r k d  dynamic load factor

From PSG Data book Pg No.81.5 k d  k  1.3  [M t]  M t kd  k  4383.9  1.3  5699.07 kg f  cm

3.96


 1 kgf  cm  100 N  mm

Step 7: Calculate cone distance ‘R ’ From PSG Data book Pg No.8.13 table (8)

 

R  y  i2  1 

y 

3

2 0.72   E [M t] cm    0.5  [ ]  i c   y

R From PSG Data book table 13 Pg no. 8.15 b

Take y  3 (based on i  3 Given) E  equivalent Young’s Module

R3

32 1  

10  R3

  3

2

6 0.72   2.15  10  5699.07  3  0.5 [3400]  3  

7.1750  10  9  4.0843  10 9   3

 R3 10  3.08304 R  29.24 cm R  30 cm R  300 mm

Step 8: Calculation of m t  transverse module From PSG Data book Pg No.8.38 table 31 R  0.5 mt Z 1  i2  1 


Assume Z 1  20  mt 

mt 

mt 

R 0.5  Z 1  i2  1  300

0.5  20 

32 1  

300 31.6227

m t  9.486 mm

Take standard module From PSG Data book Pg no.8.2 table 1 Take m t  10 mm Step 9: Calculate

b d 1av

to find k

k  load concentration factor

To find d 1

av

mm  m t 

b  sin 1 Z1

From PSG Data book Pg No.8.38 table 31 b  0.3 R  0.3  300  90 mm Z 1  20

3.98


m m  10 

90  sin 71.56  20

tan   i 1  tan  1 i

 5.73 mm

1  tan  1 3 d1

av

 m m  Z1

 71.56 

 5.73  20  114.6 mm Now calculate pitch line velocity v v v

 d1 N 1 av

60  1000

m/sec

  114.6  200 60  1000

v  1.2 m/sec

Calculate b d1

ratio 

av

90  0.785 114.6

From PSG Data book table (14) Pg No.8.15. Take the value of k k  1.06

Step 10: Selection of IS

Quality

Corresponding to pitch line velocity v Select the preferred quality From PSG Data book Pg No.8.3 table 2 Select IS quality 10


Step 11: Read the value of dynamic load factor kd

From, PSG Data book Pg No.8.16 table 15 The value of kd  1.1 Step 12: Recalculate [M t]-Design twisting moment [M t]  M t kd k M t  4383.9 kg f  cm from step 6 kd  1.1 k  1.06  [M t]  4383.9  1.1  1.06 [M t]  5111.6274 kgf  cm [M t]  511162.74 N  mm

Step 13: Calculation of induced bending stress b From PSG Data book Pg No.8.13 A table 8. b 

R

i2  1 

[ M t]

1  [b] R  0.5 b b m t  yv c o s  2

Where R

Cone distance  30 cm



Pressure angle  20

i

Speed ratio  3

[M t]  Design torque  5111.6274 kgf  cm b  face width  9 c m

3.100


m t  transverse module  1 cm yv  form factor

Form factor is based on equivalent no of teeth on the virtual cylinder Z v Zv  

Z1 cos 1

20 c o s 71.56 

Z v  63.22

Read the value of yv corresponding to Z v From PSG Data book Pg No.8.18 table 18 at x  0 y v  0.490  b 





30

R

i2  1  

[M t]

1 R  0.5  b b m t yv cos  2

32 1  

5111.6274

2

30  0.5  9  9  1  0.490



1 c o s 20

484931.554 2694.664

b  175.95 kg f  c m  [ b]  3892 kg f  cm  The design is safe based on beam strength.

Step

14: Calculate

induced

surface

contact

compressive stress (c) From PSG Data book Pg No. 8.13 table (8)


 

i2  13  

c 

0.72 R  0.5b

c 

0.72 30  0.5  9



ib

Eeq [M t]  [c]

  3 2  13   39

2.15  10 6  5111.6274

0.72  113453.25 25.5

 3203.3 kgf  c m   c  3203.3  [ c]  3400 kgf  c m  The design is satisfactory. Problem 3.9 Design a pair of bevel gears to transmit 15 kW at the pinion speed of 1600 rpm, and the gear ratio is 4. Material for gears 15 Ni 2 Cr Mo 15 / steel. The tooth profiles of the gears are of 20 composite form.

HERTZ METHOD Solution: Given data P  15 kW  15  10 3 W N 1 pinion  1600 rpm

Gear ratio

i  4

Pressure angle   20 Material used for pinion and gear  15 Ni 2 Cr Mo15/ steel

3.102


Steel 1: Selection of Material Material for pinion and gear

 15 Ni 2 Cr Mo 15 /

steel Refer PSG Data book Pg No.8.4 Since the pinion and gear are made up of same material, design pinion as it is weaker than gear. Step 2: Calcuatle Equivalent Young’s Modulus From PSG Data book Pg No.8.14 table 9 select E eq  2.15  10 6 kgf/cm 2 E eq  2.15  10 5 N/mm 2

Step 3: Life is not given in the question, so Assume life N   10000 hrs  N  10000  60  1600  96  10 7 cycl es

Step 4: Calculation of Design bending stress [b] From PSG Data book Pg No.8.18 [b] 

1.4 kbl  1 n k

From PSG Data book Pg No.8.20 table 22, take the value of life factor for bending kbl  k bl  0.7


From PSG Data book Pg no.8.19 table 19  1  Endurance limit stress in bending for complete reversal of stress. For alloy steel  1  0.35 u  1200 Where u  Ultimate tensile stress According to material of gear, take the value of u From PSG Data book Pg No. 8.5 table 7 in kg f/cm 2 u  90 Take u  90 kg f/mm 2  9000 kg f/cm 2    1  0.35  9000  1200  4350 kg f/cm 2 n  Factor of safety

From PSG Data book Pg 8.19 table 20 n  2.0 k  Fillet stress concentration factor.

From PSG Data book Pg 8.19 table 21 (Take addendum modification co-efficient 0  X  0.1) k   1.2

3.104


[b] 

1.4 kbl  1 n k

[b] 

1.4  0.7  4350 2  1.2

[b]  1776.25 kg f/cm 2 [b]  177.625 N/mm 2

Step 5: Calculate design surface contact compressive stress [c] From PSG Data book Pg 8.16 [ c]  C R HRC kcl kg f/cm2

From PSG Data book Pg 8.16 table 16 For Alloy steel case hardened. C R  280 HRC  55

From PSG Data book Pg 8.17 table 17 K c1  0.585 [ c]  280  55  0.585  9009 kgf/cm 2 [ c]  900.9 N/mm 2

Step 6: Calculation of design twisting moment [M t] From PSG Data book Pg No.8.15


Twisting moment [M t]  M t k d k Where M t  Nominal twisting mo ment Mt 

97420 kW N

Where N

Speed in rpm  1600 rpm

kW  Nominal power in kW  15 kW

 Mt 

97420  15 1600

M t  913.3125 kg f  c m k

Load correction factor

kd 

Dynamic load factor

From PSG Data book Pg No.8.15 k d  k  1.3  [M t]  m t kd  k [M t]  913.3125  1.3 [M t]  1187.306 kg f  cm [M t]  11873.06 N  mn

STep 7: Calculate cone distance R From PSG Data book Pg no.8.13 table 8

3.106


R   y i2  1 

y 

  3

2 0.72  E [M t]  cm    0.5 [ ]  i c   y

R From PSG Data book, table 13 Pg no.8.15 b

Take y  3 (based on i  4) E  Equivalent Youong’s Modulus (found in step 2)

R3

42 1  

 3 17

  3

2

6 0.72   2.15  10  11873.06  3  0.5 9009  4  

1.4193  10  9  6.3817  10 9 3  

 25.78 cm R  26 cm R  260 mm

Step 8: Calculation of m t transverse module From PSG Data book Pg No.8.38 table 31 R  0.5 mt Z 1  i2  1 

Assume Z 1  20  mt 

R 2

i 1 0.5 Z1  



260

0.5  20

M t  6.38 mm

42 1  


Take standard module From PSG Data book Pg No.8.2 table Take M t  8 mm (Standard value) Step 9: Calculate

b d1

to find k

av

k  load concentration facto r

To find d 1 av m m  mt 

b sin 1 Z1

From PSG Data book Pg No.8.38 table 31 b  0.3 R  0.3  260 b  78 mm Z 1  20 tan 1  i 1  tan 1 4 1  75.96 mm  8 

78  sin 75.96  20

m m  4.21 mm d1

av

 m m  Z1

 4.21  20

3.108


 84.2 mm Now calculate pitch line velocity v v v

 d1 N 1 av

60  1000

m/sec

  84.2  1600 60  1000

v  7.05 m/sec

Calculate b d1

ratio 

av

78  0.92 84.2

From PSG Data book table (14) Pg No.8.15 Take the value of k k  1.1

Corresponding to

b  1.0 d1 av

Step 10: Selection of IS Quality Corresponding to pitch line velocity v Select the preferred quality. From PSG Data book Pg No.8.3 table 2 Select IS quality 8 w.r.t to v  7.05 m/s Step 11: Read the value of dynamic load factor kd

From PSG Data book Pg No.8.16 table 15 The value of kd  1.4


Step 12: Recallcualte [M t] Design twisting moment [M t]  M t kd k

 913.3125  1.4  1.1  1406.50125 kg f  cm  140650.10 N  mm Step 13: Calculation of induced bending stress b From PSG Data book Pg No.8.13A table 8 b 

R

i2  1  

[Mt]

1  [b] R  0.5b b m t yv cos  2

where R




Pressure angle  20

i

Speed ratio  4

Mt 

Design torque  1406.50 kg f  c m

b

Face width  7.8 c m

Mt 

Transverse module  0.8 cm

yv 

Form factor

From factor is based on equivalent no of teeth on virtual cylinder Z v Zv  Zv 

Z1 cos 1

20 cos 75.96

Z v  82.44

Read the value of yv corresponding to Z v

3.110


From PSG Data book Pg No.8.18 table 18 at x  0 y v  0.499  b 

R

i2  1  

[ M t]

1 R  0.5b b mt yv cos  2

26

4 2 1 

 1406.50



1 c o s 20 26  0.5  7.8  7.8  0.8  0.499



150777.8496 1429.07

2

b  105.507 kg f/cm 2   b  105.507  [b] 1776.25 k gf /cm 2

The design is satisfactory based on beam strength. Step

14:

Calculate

induced

surface

contact

compressive stress c From PSG Data book Pg No.8.13 table (8)

c 

0.72 R  0.5b

  i2  13  

0.72  26  0.5  7.8 

ib

E eq [M t]  [t]

  42  13   4  7.8

2.15  10 6  1406.50125

0.72  82423.05  2685.27 kg f/cm 2 22.1

c 2685.27 kg f/cm 2  [ c] 9009 kg f/c m2 The design is satisfactory.


Problem 3.10 Design a bevel gear drive to transmit 3.5 kW with the following specifications: speed ratio  4; driving shaft speed  200 rpm; drive is non reversible, material for pinion is steel; material for wheel is cast iron; and life 25,000 hours.

HERTZ METHOD Solution Step 1 Selection of material (From P.S.G. Data book, Pg.No. 1.12) Pinion: C 45, surface hardened to 45 RC and core hardness  350 BHN u  700 N/mm2;  y  360 N/mm 2 Wheel: CI grade 30; u  300 N/mm2 Step 2 calculation of design stresses [c]pinion  C R HRC kcl (From P.S.G. Data book, Pg. No. 8.16) C R  23; HRC  45

for life 25,000 hrs N  25000  60  200  30  10 7 cycles

From P.S.G. Data book, Pg.No. 8.17 life factor for surface strength kcl  0.585 for HB  350 and N  25  10 7 cycles  [ c]pinion  23  45  0.585  605.5 N/mm 2

3.112


[b]pinion 

1.4 k bl    1 (From P.S.G. Data book, n  k

Pg. No. 8.18) From P.S.G. Data book, Pg.No. 8.19 for forged steels,  1  0.25  u   y  50  0.25 700  360  50  315 N/mm2

From P.S.G. Data book, Pg.No. 8.20 kbl  for HB  350 and N  10 7 cycles

From P.S.G. Data book, Pg.No. 8.19 k  for steel normalised, surface hardened n  2.5 for steel, surface hardened  [b]pinion 

1.4  1  315 2.5  1.5

 117.6 N/mm2 [c]wheel  C B  H B  k cl ( From P.S.G. Data book, Pg. No. 8.16)

From P.S.G. Data book, Pg.No. 8.16, for CI, Grade 30 C B  2.3; HB  220 assumed

we know that N wheel 

30  10 7 30  10 7   7.5  10 7 cycles i 4


From P.S.G. Data book, Pg.No. 8.17; for CI

kcl 



  6

10 7 N wheel

  10 7

6

7.5  10 7

 0.714

 [c]wheel  2.3  220  0.714  361.3 N/mm 2 [ b]wheel 

1.4 k cl 1 n  ka

From P.S.G. Data book, Pg.No. 8.19; for CI   1  0.45  u  0.45  300  135 N/mm 2

From P.S.G. Data book, Pg.No. 8.20, for CI

kbl 



  9

10 7 N wheel

  9

10 7

7.5  10 7

 0.799

From P.S.G. Data book, Pg.No. 8.19, for CI, No heat treatment


3.114

n  2.5 k   1.2  [b]wheel 

1.4  0.799  135  5.03 N/mm2 2.5  1.2

Step 3 Determination of R (Design of pinion) From P.S.G. Data book, Pg.No. 8.13

R  y  i 1  2

  3

0.72      0.5  [ ]  c   y

2



E [Mt] i

i  4 given E eq  1.7  10 5 N/mm 2 (From P.S.G. Data book, Pg.

No. 8.14)  y  3 Assume (From P.S.G. Data book, Pg. No

8.15) [M t]  k o  k  k d  M t k o  1 assumed k  kd  1.3 assume From P.S.G. Data bo ok, Pg. No. 8.15 

From P.S.G. Data book, Pg. No. 8.15 M t  97420

kW N

 97420 

3.5 200

 1704.85 kg fcm


 170.48 Nm  [M t]  1  1.3  170.48  221.63 Nm

3

R   16 1

  3

2

5

0.72   1.7  10  221.63  10  3  0.5 605.5   4  

3

 159.17 mm

Step 4 Determination of m t R  0.5 m t  i2  1  Z1 (From P.S.G. Data book, Pg.  No. 8.38)

 159.17  0.5 m t  16 1  18 assume Z1  18  mt  4.28 mm

Take m t  5 (standard) Step 5 Recalculate R and find b R  0.5  5   16 1  18  185.54 mm b

R 185.54   61.8 mm 3 3

Step 6 Calculate m av , d 1av and pitch line velocity, From P.S.G. Data book, Pg.No. 8.15

3.116


 R  0.5 b  m av  m t   R  

 185.54  0.5  61.8   5  185.54    4.17 mm d1av  Z 1 m av  18  4.17  75 mm

Pitch line velocity, v 

 d 1av N 1 (from P.S.G. Data 60  1000

book, Pg. No. 8.15) v

  75  200  0.8 m/s 60  1000

Step 7 Revise k, kd and [M t] Since

61.8 b   0.824 d1av 75

From P.S.G. Data book, Pg.No. 8.15, for

b d1av

1

and surface hardness is less than 350 BHN for wheel  k  1.1

From P.S.G. Data book, Pg.No. 8.16, for class 6 gear, v  1 m/s  kd  1  [M t]  1  1.1  1  170.48  187.53 Nm


Step 8 Calculation of Induced stresses (i) Stresses in pinion (From P.S.G. Data book, Pg. No. 8.13)   0.72  c     R  0.5 b  

i2  13   ib

 0.72      185.54  0.5  61.8   

  E [M t]  

42  13   4  61.8

1/2

1/ 2



 1.7  10  187.5  10

 442.67 N/mm2  [ c] 605.5 N /mm2

From P.S.G. Data book, Pg.No. 8.13 A b 

0.7 R

i2  1  

R  0.5 b2  b m t yv

 [M t]

From P.S.G. Data book, Pg.No. 8.39 tan 2  i  4  2  tan  1 4  76   1  90  76  14

From P.S.G. Data book, Pg.No. 8.39 Zv 

Z1 cos 1



18  19 cos 14

 yv  0.383 (from Pg. No. 8.18)  b 

 0.7  185.54  16 1  187.53  10 3 185.54  0.5  61.82  61.8  5  0.383

3



3.118


 35.48 N/mm 2  [b] 117.6 N/mm 2

(ii) Check for stresses in the wheel c

wheel

 442.67 N/mm 2  [c]wheel  361.3 N/mm 2  c pinion  Wheel does not have adequate wear strength.

Using following relation to calculate the induced bending stress b wheel  ywheel   b pinion  ypinion where ywheel is the form factor for the virtual number of

teeth of the wheel we know that Z 2  i Z 1  4  18  72

But Zv  

Z2 cos 2

From P.S.G. Data book, Pg.No. 8.39

72  297.6 ~  300 c o s 76

From P.S.G. Data book, Pg.No. 8.18 yv  0.521 correspondin g to Z v  300 teeth  wheel  0.521  35.48  0.383

wheel  26.08 N/mm 2  [ b]wheel 50.3 N/mm 2 Hence design of pinion and wheel is satisfactory


Problem 3.11 A 90 degree straight bevel gear set is needed to give a 3:1 reduction. Determine the pitch cone angle, pitch diameter and gear forces if the 25 pressure angle, pinion has 15 teeth of pitch circle diameter 4 and the transmitted power is 8 HP at 550 pinion rpm.

HERTZ METHOD Solution i  3,   25, Z1  15 teeth, P  8 HP  5.888 kW

Assume the material for pinion and gear C45 surface hardened. Since C45  350 BHN

hardened

to

55

RC,

core

hardness

u  720 N/mm2,  y  360 N/mm 2

Assume life as 10,000 hrs  10,000  60  550  33  10 7 cycles

Design compressive contact stress (From P.S.G. Data book, Pg.No. 8.16) [c]  C R  HRC  kcl  23  55  0.585  740 N/mm2

since CR  23 ; HRC  55 ; kcl  0.585 Design bending stress (From P.S.G. Data book, Pg.No. 8.18) [b] 

1.4 kbl n  k

1

3.120


But   1  0.25 u  y  50 From PSG Data bookPg. No. 8.19  0.25 720  360   50  320 N/mm2

since k bl  1, n  2.5, k   1.5  [ b ] 

1.4  1  320 2.5  1.5

 119.5 N/mm2

Cone Distance ‘R’ (From P.S.G. Data book, Pg.No. 8.13)

R  y

i2  1  

Assume y 

  3

2 0.72   E [M t]    0.5 [ ]  i c  y 

R  3 (From P.S.G. Data book, Pg.No. b

8.15) i3 [M t]  k o  k  k d  M t from P.S.G. Data book, P g. No. 8.15 ko  1 assumed k  k d  1.3 assumed  Mt 

5.888  10 3  60  102.229 Nm 2  550

 [M t]  1  1.3  102.229  132.89 Nm


E  2.1  10 5 N/mm2 assumed

R

 

  3

3

2

3 1

2

5

0.72   2.1  10  132.89  10  3  0.5  [740]  3  

3

 106.355 mm i2  1 But R  0.5 m t Z 1   10 106.355  0.5 m t  15   m t  4.48  5 mm standard from P.S.G. Data book, Pg.No. 8.2  R  0.5  5  15   10  118.58 mm b

R 118.58   39.528 mm 3 3

But m av  mt 

b sin 1 Z1

(From PSG Data book Pg.No. 8.38) tan 2  i  3  2  tan 1 3  71.5 

1  90  71.5   18.5  m av  5 

39.528 sin 18.5  4.16 mm 15

v  pitch lin e velocity 

 d1av N 1 60  1000

m/s


3.122

But d1av  Z 1 m av  15  4.16  62.45 mm v

  62.45  550  1.79 m/s 60  1000

IS Quality 6 bevel gear is assumed k  1.6 for b/d1av  0.64  from P.S.G. Data book, Pg.No. 8.15 

k  1.4  [M t]  ko  k  kd  mt  1  1.6  1.4  102.229  10 3  228.99  10 3 Nmm

(From P.S.G. Data book, Pg.No. 8.13) 1

 0.72  c  R  0.5 b  

i  1   2

3

ib

2 E [M t]  

0.72 118.58  0.5  39.528 1

3 2  32  1    5 3   2.1  10  132.89  10    3  39.528

 628.55 N/mm 2  [c] 740 N/mm2 i  1 [M t] R  2

b 

2

R  0.5 b  b mt yv

from P.S.G. Data book, Pg.No. 8.13

Z1 15   15.81  16 Z v  Z eq  cos 1 cos 18.5 Y v  0.355 From PSG Data book Pg.No. 8.18 corresponding to 2 V


 b 

118.58 

3 2 1 

 228.99  10 3

118.58  0.5  39.528 2  39.528  5  0.355

 118.22  [ b] 119.5 N/mm2

Now the design is satisfactory The pitch cone angle 1  18.5 Pitch Circle  d1  Z1 m t

Diameter

(PCD)

(outside)

of

 75 mm

PCD (outside) of gear  d2  Z 2 m t Z2 Z1

i3

Z 2  3  15  45  d 2  Z 2 m t  25 mm

PCD (average) of pinion  d1av  Z 1 m av  15  4.16  62.4 mm

PCD (average) of gear  d 2av  Z 2 m av  45  4.16  187.2 mm

Gear forces Tangential force F t 

2 M t 2  102.229  10 3  62.4 d1av  3276.5 N

pinion

3.124


Radial force F r  F t  tan   cos   3276.5  tan 25  cos 18.5   1448.93 N

Axial force F a  Ft  tan  sin  484.796 N Problem 3.12 Design a pair of bevel gears for two shafts whose axis are at right angles the power transmitted is 25 kW. The speed of the pinion is 300 rpm and the gear is rotating at 120 rpm. Take material used as C45 for both pinion and gear.

HERTZ METHOD Given data Power transmitted P   25 kW N 1 Pinion speed  300 rpm N 2 Gear speed  120 rpm

Material used

pinion  C45 gear 

w.k.t i

N1 N2



300  1.5 120

Solution: 1. Selection of Material Refer PSG Data book Pg No.1.12


Given C45, surface hardened to 45 RC and core hardness ~ 350 BHN for pinion and wheel v  700 N /mm2 ; y  360 N/mm 2

Since the material used for pinion of gear are same design only pinion as it is weaker than gear. Step 2: Calculate Equivalent Young’s Modulus From PSG Data book Pg in 8.14 table 9 Select E eq  2.15  10 6 kg f/cm 2 E eq  2.15  10 5 N/mm 2

Step 3: Life is not given in the question, so Assume N  10000 hrs N  10000  60  300

 18  10 7 cycles Step 4: Calculation of design bending stress [b] From PSG Data book Pg no 8.18 [b] 

1.4 kbl  1 nk

From PSG Data book Pg no.8.20 table 22 take the value of life factor for bending kbl kbl  1

3.126


From PSG data book Pg 8.19 table 19  1  Endurance limit stress in bending for complete reversal of stress  1  0.22 u  y  500  1  0.22 700  360  500  1  733.2 N/mm 2  1  7332 kg f/cm 2 n  factor of safety

From PSG Data book Pg 8.19 table 20 n  2.0 k  Fillet stress concentration factor

From PSG Data book Pg 8.19 table 21 Take addendum modification co-efficient 0   0.1 k   1.2 [b] 



1.4 kbl  1 n k 1.4  1  7332 2  1.2

[y]  4277 kgf/cm 2 [b]  427.7 N/mm 2


Step 5: Calculate design surface contact compressive stress [c] From PSG Data book Pg No.8.16  c  C R HRc kcl kg f/cm2 C R  230 HR c  45

life factor kcl from PSG Data book Pg No.8.17 table 17 kcl  0.585  [c]  230  45  0.585 kg f/cm 2  6054.75 kgf/cm 2 [ c]  605.47 N/mm2

Step 6: Calculation of design twisting moment [M t] From PSG Data book Pg No.8.15 Design twisting moment [M t]  M tk dk Where M t  Nominal twisting moment Mt 

97420 kW N

3.128


where N

Speed in rpm  300 rpm.

kW 

Nominal power in kW  25 kW

Mt 

97420  25 300

 8118.33 kg f  cm kd 

dynamic load factor

k

load concentration factor

From PSG Data book Pg No.8.15 k d  k  1.3 [M t]  M t  kd  k [M t]  8118.33  1.3 [M t]  10553.829 kg f  cm [M t]  1055382.9 N  mm

Step 7: Calculate cone distance ‘R ’ From PSG Data book Pg No.8.13 table (8)

R  y

y 

i2  1  

2 0.72   E [M t ] cm    0.5 [ ]  i c   y

  3

R From PSG Data book table 13 Pg no 8.15 b

Take y  3 (base on i  1.5 E  equivalent Young’s Module


 

R3 1.52 1 

3

2

6 0.72   2.15  10  10553.829  3  0.5 6054.75  1.5  

 R  5.408  34.2254 R  5.408  5.850 R  31.63 R  32 cm R  320 mm

Step 8: Calculation of m t  transverse module From PSG Data book Pg No.8.38 table 31 R  0.5 m t Z 1

i2  1 

Assume Z 1  20 mt 

mt 

R 0.5  Z 1

i2  1   320

0.5  20 

1.52 1  

m t  17.75 mm

Take standard module From PSG Data book Pg no.8.2 table 1 Take m t  18 mm

3.130


Step 9: Calculate

b d1

to find k

av

k  load concentration factor

To find d 1

av

mm  m t 

b sin 1 Z1

From PSG Data book Pg no.8.38 table 31 b  0.3 R b  0.3  320  96 mm Z 1  20 tan 1  i 1  tan 1 1.5 1  56.30 m m  18 

96  sin 56.30 20

 14.006 mm ~  14 mm d1

av

 mm  Z1

 14  20  280 mm Now calculate pitch line velocity v


v 

 d1 N1 av 60  1000

m/s ec

  280  300 60  1000

v  4.398 m/sec

Calculate b d1

av

ratio 

96  0.342 280

From PSG Data book table (14) Pg No.8.15 Take the value of k k1

Step 10: Selection of IS quality Corresponding to pitch line velocity v Select the preferred quality From PSG Data book Pg No.8.3 table 2 Select IS quality  10 Step 11: Read the value of dynamic load factor kd

From PSG Data book Pg No. 8.16 table 15 The value of kd  1.1 Step 12: Recalculate [M t] Design twisting moment [M t]  M t k d  k M t  10533.829 kg f  cm

3.132


k d  1.1 k1 [M t]  10533.839  1.1  1 [M t]  11587.2119 kg f  cm [M t]  1158721.19 N  mm

Step 13: Calculation of induced bending stress From PSG Data book Pg No. 8.13 A table 8 b 

R

i2  1  

[M t]

1  [b] R  0.5b b m t yv cos  2

where R




Pressure angle  20 (Assume).

i

Speed ratio  1.5

[M t] 

Design torque  11587.2119 kgf  cm

b

Face width  9.6 cm

Mt 

transverse module = 1.8 cm

yv 

Form factor

Form factor is based on equivalent no of teeth on the Z1 virtual cylinder Z v  cos 1 

20 c o s 56.30 

 36.04


Read the value of yv corresponding to Z v From PSG Data book Pg No.8.18 table 18 at X  20 y v  0.458  b 





R

i2  1  

[M t]

1  [ b] c o s R  0.5b b m t y v 2

32

1.5 2 1  

 11587.2119

2

32  0.59.6   9.6  1.8  0.458



1 cos 20 

668452.5863 1  5855.271 c o s 20

b  121.48   b  121.5 kgf  cm  [b] 4277 kg f  c m

The design is satisfactory based on beam strength. Step

14: Calculate

induced

surface

contact

compressive stress c From PSG Data book Pg No. 8.13 table (8)

  i2  1 3  

c 

0.72 R  0.5b

c 

0.72 32  0.5  9.6

ib

Eeq [M t]  [ c]

  1.52  13   1.5  9.6

 2.15  10 6  11587.2119

3.134




0.72  100679.2518 27.2

c  2665.039 kg f  c m c  2665.039 kg f  c m  [c] 6054.75 kg f  cm  The design is satisfactory. Problem 3.13 Design a bevel - gear drive to transmit 7.5 kW. at 1440 rpm. Gear ratio is 3; pinion and gear are made of C45 steel; Life of gears 10,000 hrs.

(Apr./May 2009)

HERTZ METHOD Given Data: Power  P  7.5 kW Pinion speed  N 1  1440 rpm

i

Z2 Z1

3

Gear ratio  i  3 Life  N  10,000 hrs Life  N  10,000  60  1440  86.4  107 cycles

Step 1 Selection of Material: From PSG Data book P.No. 8.16; table (16) Material Given C45 Pinion and wheel are made of same material. Therefore, design only pinion Assume surface hardened heat treatment C45  surface hardened HRC  40 to 55

Assume HRC  50


i.e, surface hardness > 350 BHN core hardness < 350 BHN Step 2 Life of gears. (N) N  10,000 hrs  10,000  60 1440 Life  N  86.4  10 7 Cycles of Gears 

Step 3 Equivalent Young’s modules ‘E’ From PSG Data book P.No. 8.14 table (9) For steel pinion and steel wheel E  2.15  10 6 kgf  cm2 E  2.15  10 5 N  mm2

Step 4 Calculation of design bending stress [b] From PSG Data book P.No. 8.18 Design bending 1.4 kbl  1 stress    b   n  k   (Assume rotation in one direction only) (From PSG Data book P.No. 8.20, table (22) (consider core hardness)  350 BHS; N  10 7 cycles

3.136


kbl

1

 1  0.22 u  y  500

kgf  cm 2

(From PSG Data book P.No.8.19 table (19) (Forged Steel) from PSG Data book pg no.1.9 for C45    63  71 kgf  mm2 material  u

and y  36 kgf  mm2

take u  70 kgf  mm2

 36  100 kgf  c m2

 70  100 kgf  cm2

From PSG Data book P.No.8.19 table (20)  1  0.25 70  100  36  100  500  3150 kg f  c m2

[For forged and surface hardened] n  2.5 k   From PSG Data book table 21; P.No.8.19

(Steel - surface hardened - at 0    0.1 k   1.5

Design bending stress [b] [ b] 

1.4  1  3150 [Pg.No. 8.18] table.18 2.5  1.5

 1176 kg f  cm 2 [b]  117.6 N  mm 2


Step 5 Calculation of Design surface contact compressive stress [ c] [c]  C R HRC  kCl kg f  cm 2 From PSG Data book Table 16; P.No. 8.16 C R  230 for C45  surface hardened steel

HRC  50 assumed HRC  50 k Cl  From PSG Data book P.No.8.17; P.No.17 

steel, N  25  10

Consider surface hardness which is  350 BHN kcl  0.585 [ C ]  50  230  0.585  6727.5 kg f  cm 2 [ c]  672.75 N  mm2

Step 6 Calculation of Design torque [M t] From PSG Data book P.No.8.15 Design twisting moment [M t]  M t  k d  k M t  97420 

kW N1

kg f  cm

7

3.138


 97420 

7.5 1440

 507.39 kg fcm

Assume initially, k  kd  1.3 (for symmetric scheme) [M t]  507.39  1.3  659.61 kg fcm [M t]  659.61  100 Nmm

Step 7 Calculate Cone distance ‘R’ From

R  y

y 

PSG Data book P.No. 8.13 and table (8)

i2  1  

 

2 0.72   E[M t]    0.5 [ ]  i c   y

R b

From PSG Data book table (13) P.No. 8.15 Assume y  3 (for i  1 to 4) i  Gear ratio  3 [c]  672.75 N  mm 2 E  2.15  10 5 N  mm 2 [M t]  659.61  100 N  mm


R3

3 2 1  

  2

5

0.72  2.15  10  659.61  100   3  0.5 672.75   3  

3

R  90.43 mm

Step 8 Calculation of m t  transverse module. From PSG Data book P.No. 8.38 table (31) R  0.5 mt Z 1

i2  1 

90.43  0.5 m t  20 

32 1  

assume Z 1  20 teeth gea r ratio  i  3

mt  2.859 mm

Represent transverse module in terms of standard module From PSG Data book P.No. 8.2 table (1) take n earest higher value

 m t  3mm;

Step 8 (a): Recalculate (R) and Calculate (b) R  0.5 m t Z 1

i2  1  

 0.5  3  20 R  94.86 mm

y 

R 3 b

3 2 1  

Revised

cone distance ‘R’

3.140


b

R 94.86   31.62 mm 3 3

Face width b  31.62 mm

Step 9 To find k ; Calculate b  d1 av From PSG Data book P.g. No. 8.15 Table 14 k  Load concentration factor

To find d 1 ; av

Average (or) Module

m av  m t  b 3

sin 1 Z1

31.62  sin 18.43  20

(or)

av

 Z 1 m av

d1

 20  2.5  50 mm



31.62 b   0.06324 50 d 1av

2  71.56 1  90  71.56 

m av  2.5 mm

d1

tan 2  i  3

av

 18.43  R  0.5 b   m tZ 1   R  

 94.86  0.5 31.62   3  20   94.86  

 50 mm

From PSG Data book table 14; P.No.(8.15)  The value of k  1.6

Step 10 Selection of IS Quality


From PSG Data book P.No. 8.3 table (2) Calculate pitch line velocity  v  v

d 1av  N 1 60  1000   50  1440 60  1000

v  3.769 m/sec

According to velocity of straight bevel, Select IS quality 6 steel; Step 11 Read the value kd From PSG Data book P.No. 8.16 Page No. (15) kd  1.4 for IS quality (6) surface hardness  350 and v  upto 8 m/sec

Step 12 Recalculate [ M t ] - design twisting moment. [ M t ]  M t  kd  k  507.39  1.6  1.4  1136.55 kg fcm [ M t ]  113655 Nmm

Step 13 Calculation of induced bending stress b, From PSG Data book P.No. 8.13 table (8)

3.142


b 

R i2  1 [ M t ] 

1  [ b ] cos  R  0.5b b  m tyv 2

R  94.86 mm ; i  3

To find yv: From PSG Data book P.No. 8.18; Table (18)

[Mt]  113655 Nmm

Zv 

b  31.62 mm ; m t  3 mm

Z1 cos 1



20 cos 18.43

 21.08  21 teeth

Take average for 20 and 22 and take yv value at X  0. yv  form factor  0.3955

  pressure angle  20  Induced bending stress [ b] b 



R i2  1 [ M t ] 

1  [ b ] c o s R  0.5b b  m t  y v 2

32 1 113655  94.86   2

94.86  0.5  31.62   31.62  3  0.3955



1 cos 20

 b  154.75 N/mm 2  [ b ] 117.6 N/mm 2

Design of pinion based on beam strength is not satisfactory. Therefore, increase the value of transverse module m t i.e., Revise m t.


Trail 2 Assume transverse module  m t  4 mm Repeat steps 8 (a) to (13) Step 8 (a) R  0.5m t  Z 1 i2  1   0.54  20  32 1  R  126.49 mm b

R 126.49   42.16 mm 3 3

Step 9 m av  m t  b 

sin 1 Z1

1  18.43 

sin 18.43     4  42.16  20   m a v  3.33 mm d1

av

 Z1  mav  20  3.33  66.67 mm

b d1



av

42.16  y 66.67

 0.6323

From table (14); PSG Data book P.No. 8.15 k  1.6

3.144


Step 10 v 

  d1  N1 av 60  1000   66.67  1440 60  1000

 5.026 m/sec

Select IS quality (6) gear. Step 11 kd  1.4

Step 12 [ M t ]  Mt  k d  k

(Similar to previous step)

 507.39  1.6  1.4  1136.55 kg fcm [ M t ]  113655 Nmm y v  0.3955 

(Similar to previous step) Step 13 b 



R i2  1 [ M t ] 

1 c o s R  0.5b b  m t  yv 2

126.49  3 2 1 113655  2

126.49  0.5  42.16  42.16  4  0.3955

b  65.28 N/mm 2  [ b ] 117.6 N/mm 2



1 c o s 20


Therefore, the design of pinion based on beam strength is satisfactory. Note: To avoid re-calculation from step 8(a) to 13, always take module mt (standard) greater value i.e., instead of selecting 3 mm, take 4 mm in the beginning itself. Step 14 Calculation of induced surface contact compressive stress c; ( Notes use only recalculated values)

0.72 c  R  0.5b

  i2  13   ib

E[ M t ]  [ c ]

R  126.49 mm ; b  42.16 m m ; i  3 E  2.15  10 5 N/mm 2 ; [ M t ]  113655 Nmm

0.72 c  126.49  0.5  42.16

  2 3 3   1  

3  42.16

2.15  105  113655

 c  533.89 N /mm 2  [  c ] 672.75 N/mm 2  The design based on wear strength is satisfactory.

To Find Angles From PSG Data book P.No. 8.39 Table (31) [f o  1 ]

1. Addendum angle  a tan  a  tan  a  1

2

m t  fo R

3.146


 41  a   a  tan  1  1 2  126.49  a   a  1.81 1

2

2. Dedendum angle  f f   f  tan  1 1 2  tan  1

m t fo  c R 4 1  0.2 126.49

f   f  2.17 1

2

3. Tip angle  a  1  a 1

1

(Pinion)  18.43   1.81 a  20.24  1

4. Root angle  f  1  f 1

1

(Pinion)  18.43   2.17 f  16.26  1

5. Tip dia. of pinion  da  mt Z 1  2 cos 1 1

 420  2 c o s 18.43  da  87.58 mm 1

6. Tip dia. of wheel  d a  m tZ 2  2 c o s 2 2

 420  2 c o s 71.57 


da  82.52 mm 2

7. Pitch circle dia. (outside) of pinion  d 1  m t  Z 1  4  20 d1  80 mm

8. Pitch circle dia. (outside) of wheel (or) gear  d2  m t  Z2  4  60 d2  240 mm Problem

3.12 A pair of bevel gear is to be used to transmit

12 kW from a pinion rotating at 360 rpm to a gear mounted on a shaft which intersect the pinion shafts at the angle of 70; Assuming the pinion is to have an outside pitch diameter of 200 m.m, a pressure angle of 20; a face width of 40 mm: and the gear shaft is to rotate at 120 rpm; Determine (a) the pitch angle for the gear (b) the forces on the gear (c) the torque produced about the shaft axis (d) Calculate dia. of pinion shafts, if   45 N  mm2 and if pinion shaft overhangs by 120 mm.

(Nov. 2007)

Given Data: P  12 kW P in ion s peed  N 1  360 rpm Angle between     70  two shafts  Pitch circle dia. of pinion    d 1  200 mm Outside 

3.148


Pressure     20  angle 

Face width

 b  40 mm

Gear shaft speed  120 rpm i

N1 360  3; N 2 120

Assume Z 1  20 teeth Z 2  No. of teeth    i  Z 1 Z 1  no.of teeth on pinion on Gear  Z 2  3  20  60 teeth

Let m t  transverse module; d1  m t Z1 mt 

d1 Z1



200  10 mm 20 (which is equal to standard module)

i2  1 Cone distance  R  0.5 m t Z 1    0.5  10  20

3 2 1  

R  316.22 mm

Let 1  semi cone angle of pinion 2  semicone angle of w heel or Gear   1  2  70 tan 1 

sin  i  cos 

Acute angle


sin 70 tan 1  3  cos 70 1  tan 1 0.2811 1  15.7 2  70  15.7 2  54.3.

Addendum Angle: a tan  a  1

m t  f0 R

 m t f0  Addendum angle of pinion  a  tan 1   1  R   10  1   tan  1    316.22   a  1 48   a 1

2

 m t f0  Dedudum angle of pinion  f  tan  1   1  R   10  1   tan 1    316.22  f  210  f 1 2 Tip angle of pinion   a  1  a 1

1

 1542  1 48 a  17.5  17 30 1

3.150


Tip angle of wheel  a  2   a 2

2

 54 18  148 a  566 2

Root angle   f  1  f 1 1 of pinion   1542  210  1332 Root angle   f  2  f 2 2 of pinion   5418  210  528

(b) The forces on the Gear (a) Tangential force  F tav 

M t  97420

2 Mt dav



2 Mt b   d 1  1  0.5  R  

kW N1

 97420 

12 kg fcm  3247.33 kg fcm 360

Torque  M t  324733 N  mm F tav 

2  324733

40   200  1  0.5  316.22  

Ftav  3.466 kN

 3466.58 Newton


  20

Radial force  Fr  F tav tan  c o s 1  3.466  tan 20 c o s 1542

1  1542 

 1.2016 kN Axial force  F a  F tav tan  sin 1  3.466  tan20  sin 1542 Fa  0.3413 kN

(c) Torque produced about the shaft axis Assume

over hang  120 mm

Fa and Fr induces BM;

Rm  dav  2

M1  Fr  overhang  Fa  Rm

 R  0.56  dav  mt Z1   R 1  

 1.214  120  0.3413  93.67

 316.22  0.5  40   316.22  

 10  20 

 113.7 kNmm

 187.35 mm Rm  93.67 mm

FT induces also bending moment. M 2  F t  overhang  3.466  120  415.92 kN mm

3.152


Re sultan t BM  

M21  M22  

113.7 2  415.92 2  

E quivalent   Torq ue on  Te  shaft  Te 

 431.18 kNmm

M2  T 2  

[T  Mt  324733 N  mm]

431.18  10 32  3247332  

 539785.75 N  mm

If   45N  mm 2 for shaft material Te 

   d3  539785.75 N  mm 16 S

Shaft dia. d  39.38 mm

approx 40 mm

Pinion shaft diameter  d  40 mm


3.12 DESIGN OF WORM AND WORM WHEEL 3.12.1 Introduction When the shafts are non-parallel and non-intersecting, worm and worm wheel drive is used. Worm drive can be treated as screw and nut pair. A segment of nut is coiled to form a wheel. The thread of a worm has trapezoidal profile in the axial direction. Materials: Since the sliding occurs, the materials used should have low coefficient of friction. Worm: Worms are made of steel. The threads are grounded and polished to reduce the surface roughness as low as possible. Worm wheel: Worm wheels are made of bronze and cast iron.

LW l

Pa



a

n D W D OW

x D OG

DT DG

Rf b Fig.3 .6. W orm and W orm G ear

Rr

3.154


3.13 TERMS USED IN WORM GEARING 1. Axial Pitch: Pa: It is the distance measured axially (i.e, parallel to the axis of worm) from a point on one thread to the corresponding point on the adjacent thread on the work P a  mx

where m x  axial module of the worm

2. Lead: It is the linear distance through which a point on a thread moves ahead in one revolution of the worm. Lead  P a Z1

where, P a  axial pitch Z 1  numbe r of starts

If one turn of a worm thread in unwound, it will form an inclined plane whose base will be equal to the pitch circumference of the worm and altitude equal to lead of the worm, as shown in Fig. 3.7.

LN

= P a Z1

3. Lead angle: : It is the angle between the tangent to the thread helix on the pitch cylinder and the plane normal to the axis of the worm.

 Fig. 3.7

d 1

From the Fig. 3.7 tan  



Lead of the Worm Pitch circumference of the worm Pa Z1 Pc Z1 l    d2  d1  d1

. .  . P a  P c


tan  



 m x Z1  d1

mx Z 1 d1

The lead angle () may vary from 9 to 45. A lead angle less than 9 results in rapid wear and the safe value of  is 12

1 2

Z 1  no. of starts on the w orm Z 2  no. of teeth on the worm wh ee l i  gear ratio 

Z2 Z1

m x  axial m odule d 1  Pitch circle diameter of worm d 2  pitch circle diameter of worm wheel q  diam eter factor 

d1 mx

a  centre distance  Z1    lead angle  tan  1    q  4. Tooth Pressure Angle: It is measured in a plane

containing the axis of the worm and equal to one half the thread profile angle. The pressure angle of 30  is recommended to obtain a high efficiency and to permit overhauling, for automotive applications.

3.156


5. Normal Pitch: The distance measured along the normal to the thread between two corresponding points on two adjacent threads of the worm. P N  P a cos 

P N  Normal pitch

Normal diametrical pitch, Pd 

Z1 d 1 sin 



Z2 d2 sin 



 PN

Note: The reciprocal of the normal diametrical pitch is known as Normal Module m n mn 

d1 sin  d2 sin  P N 1    Pd Z1 Z2 

Normal lead L N  l cos  6. Helix Angle : It is the angle between the tangent to the thread helix on the pitch cylinder and the axis of the worm.     90   90  

7. Velocity Ratio (i): The ratio of speed of the worm N to the speed of the worm gear N 2 in rpm. i

N1 N2



Z2 Z1

. . N1 Z2  . N2 Z1


3.14 EFFICIENCY OF WORM GEARING The ratio of work done by the worm gear to the work done by the worm is known as efficiency of worm gearing. It’s given by  

tan  cos    tan   c o s  tan   

where, 

Normal pressure angle



Co-efficient of friction



lead angle The maximum efficiency, when 1  2   tan     For square threads,   0 and tan    

tan  tan   

  Friction Angle

3.15 STRENGTH OF WORM GEAR TOOTH The Lewis equation of beam strength F s   m x b [ b] y F s  m x  b  [ b ] Y

where mx

–

Axial module

b [b]

– –

Face width Allowable Static Stress

y

–

Form factor for worm wheel

y 

Y 

3.158


3.16 DYNAMIC LOAD ON WORM GEAR TOOTH  6  vmg  Fd  Ft   6  

Where, Ft

–

Tangential load

Fd

–

Dynamic load

vmg

–

Pitch line velocity of worm gear

3.17 WEAR STRENGTH OF WORM GEARS F w  d2  b  kw d2

–

Pitch diameter of worm wheel

b

–

Face width

kw

–

Wear Factor (or) load stress factor

3.18 THERMAL RATING OF WORM GEARING The heat generated due to friction must be dissipated in order to avoid over heating of the drive and lubricating oil. The quantity of heat generated (H g) is given by H g  P 1   P

–

Power in watts



–

Efficiency of worm gearing

The heat generated must be dissipated through the lubricating oil to the gear box housing and then to the atmosphere.


The heat dissipating capacity depends upon the following factors. 1. Area of Housing (A); 2. Thermal conductivity of material (K) and 3. Temperature difference between the housing surface and 4, Surrounding air T 2  T 1 H d  A T2  T1 K

Where H d  Heat dissipated Hg  Hd P 1    A T2  T1 K

3.19 FORCES ACTING ON WORM GEARS When the worm gear is transmitting power. The forces acting on the worm are similar to power screw. The force acting on the worm is shown in Fig. 3.8. The forces on the worm and worm gear are equal in magnitude, but opposite in direction.

W orm G ea r

W orm G ea r

W orm G ea r S h aft

+ FR

FR FA

W orm

FT W orm S h aft

W orm

Fig.3.8. F orces A cting on W orm Teeth .

3.160


1. Tangential force on the Worm FT 

2  Tw d1

where, Tw

–

Torque on worm.

d1

–

Pitch circle diameter of worm.

3.20 MERITS AND DEMERITS OF WORM GEAR DRIVES 3.20.1 Merits of Worm Gear Drives 1. It is used for very high velocity ratio of about 100. 2. Very smooth and noiseless operation. 3. Self locking facility is available. 4. It is very compact when compared with equivalent spur or helical gears for the same speed reduction. 3.20.2 Demerits of Worm Gear Drives 1. Low Efficiency 2. More heat will be produced and hence this drive can only be operated inside an oil reservoir (or) an extra cooling fan is required in order to dissipate the heat from the drive. 3. Low power transmission (up to 100 kW) 4. Cost is very high, when compared with other gears drive.


3.21 BASIC DIMENSIONS OF WORM GEAR From PSG data book pg.8.43 Table 35 Basic Dimensions of Worm Gear Pair Nomenclature

Symbol

Formulae

Centre distance

a

a  0.5 mx q  Z2  2x

Face width of wheel

b

From PSG Data book table 38 Pg 8.48

Length of worm

L

From PSG Data book table 39 Pg 8.48

Axial module

mx

mx 

2a q  Z2 2x

Number of starts

Z

Determined by calculation

Number of teeth on wheel

Z2

Z2  iZ1

Gear ratio

Diameter factor

i

q

i q

Z2 Z1 d1 mx

Height factor

fo

Normally fo  1

Axial pitch

Px

Px   mx

Pitch of the helix, lead

Pz

Pz  ZPx

Pressure Angle in axial



  20

section (Archimedian worm) Lead angle on the



reference cylinder Lead angle on the pitch cylinder



tan  

Z q

tan  

Z q  2x

3.162


Bottom clearance

c

c  0.2 to 0.3 mx

Addendum modification

x

x

coefficient

a  0.5 q  Z2 mx

Worm Reference diameter

d1

d1  qmx

Worm Tip diameter

da1

da1  d1  2fomx

Worm Root diameter

df1

df1  d1  2fomx  2c

Worm Pitch diameter

d1

d1  mx q  2x

(In axial section) Reference diameter

d2

d2  Z2mx

Worm wheel Tip diameter

da2

da2  Z2  2fo  2x mx

Worm wheel Root

df2

df2  Z2  2fo mx  2c

d2

d2  d2

WORM WHEEL

diameter Worm wheel Pitch diameter Throat tip radius

R1

R1 

d1 2

 fomx

Worm wheel Throat root radius

R2

Worm Tip relief radius

r1

r1  0.1 mx

Worm Root relief radius

r2

r2  0.2 mx

Worm Nominal tooth thickness on reference

S

R2 

d1 2

 fomx  c

 mx 2

diameter in axial section. Worm Nominal tooth thickness on reference diameter in normal section

Sn

Sn 

 m cos  2 x


Note: In PSG Data Book Z 2 is denoted by z & Z 1 by Z . 3.21.1 Forces on Worm: Refer PSG databook Page No. 8.57 1. Tangential force: F t 

2M t d2

2. Radial force: F r  F t tan  3. Axial force: F a  F t tan    M t  twisting moment

  pressure angle   friction angle

3.22 DESIGN OF WORM AND WORM GEAR USING LEWIS AND BUCKINGHAM’S EQUATION Step 1: For a given problem note down the given parameters like Power, gear ratio (i), speed, etc. Step 2: Selection of Material Refer PSG Data book Pg. No. 8.45 Selection of material is similar to spur gear and take the vlaues of [c]. – Design surface stress and [ b] - Design bending stress. Step 3: Calculation of lead Angle Diameter factor q 

11 

d1 mx

d1 mx

[If q is not given Assume q  11 ] m x  axial module

3.164


 Z1  Lead Angle   tan  1    q  To determine Z 1 (no. of starts of gear); It is based on the efficiency of worm gear drive. choose the appropriate value From PSG data book pg. 8.46, Table 37. Step 4: Calculation of Tangential load Ft acting on the wheel in terms of axial module mx Tangential load F t 

HP  75 vmg

vmg pitch line velocity of worm gear vmg 

 d2 N2 60  1000

d 2  in mm N 2  rpm

Power in HP Step 5: Calculation of Dynamic Load Fd  6  vmg  Fd    Ft 6  

Step 6: Calculation of strength of gear tooth Fs 

[b] y bP cn 



[ b ] y b P dn

P cn  Normal circular pitch P dn  Normal Diametrical pitch

 [b] yb   mx  [b] Y b  mx


Step 7: Calculation of axial module mx By equating F s  F d ; Determine the m x Step 8: Calculation of face width (b) b  0.75 d 1 (If Z 1  2or 3  d1  11 m x  8.25 m x [From Data book Pg. 8.48, Table: 38] b~

Step 9: Calculation of pitch circle diameter d2 d2  Z2  m x

Step 10: Calculation of pitch line velocity of worm wheel vgm  vgm 

 d 2N 2 60  1000

d2  in mm

N2  rpm

Step 11: Recalculation of beam strength F s  [b] y b  m x

Step 12: Recalculation of Dynamic Load  6  vmg  Fd    Ft 6  

3.166


Step 13: Checking for beam strength F d  F s for safe design.

Step 14: Calculation of wear load Fw  F w  d2  b  kw

Step 15: Checking for wear strength F d  F w For safe design.

Step 16: Checking for efficiency actual  design then the design is safe.   tan  actual     tan    

[From PSG Data book pg no. 8.49]   Friction Angle  tan 1  If  is not given Assume   0.03   co-efficient of friction. Step 17: Calculation the power loss (or) Heat Generated H g  1  a  Input hp  4500 kg f m/min

(Or) a   actual


H g  1  a   P Watts

(power in Watts)

Step 18: Calculate the area required to dissipate the heat Heat generated  Heat dissipated to the atmosphere Heat Dissipated H d  h cr A  t kg f m/min hcr

–

heat transfer co-efficient.

t

–

change in temperature T2  T1

A

–

Area required to dissipate the heat. (Or)

substitute A in m 2, K in W/m 2 H d  A T 2  T 1   K H d in watts.

Step 19: Calculate the basic dimensions of worm and worm wheel. From the PSG Data book pg 8.43 table 35. Step 20: Draw the neat sketch of worm and worm wheel. Problem 3.13 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 atleast 82% in desired, and also determine the heat generated, required area of cooling to dissipate the generates heat. Temperature rise restricted to 40C and heat transfer co-efficient 10 W/m2C

3.168


Solution: Step 1: Given Data Worm

N1  1440 rpm

Power

P   12 kW  12  10 3 W

Worm wheel

N 2  60  3% rpm

desired  82%

Material (Worm) - Hardened steel Material (Worm wheel) - Phosphor bronze Change in temperature t  40C Heat transfer co-efficient, Hcr  10 W/m 2 C Step 2: Selection of Material Worm - Hardened Steel Worm Wheel - Phosphor bronze Step 3: Calculation of lead Angle  Diameter factor q  11 

d1 mx d1 mx

d1  11 mx

  tan  1 Z/q  Z1   tan 1    11 

[Assume q  11 If not given]


From PSG Data book Pg. 8.46 ; Table: 37   82% [  0.80 to 0.92] Z1  3   tan  1 3 /11   15.25 

Step 4: Calculation of tangential load Ft Ft 

V mg 

i

hp  75 V mg

[vmg  worm gear velocity]

 d2 N 2 60  1000 N1 N2

1440  3%  i 60

Z2

i  24  0.72 ; i ~  24 vmg 

  m x  72  60 60  1000 3

P

12  10 hp 750

P  16 hp Ft 



hp  75 kg f vmg

16  75 0.226 m x

d2  m x  Z2

Z1  0.226 m x

i

Z 2  24  3 Z 2  72

3.170




5309.73 53097.3 N kg f  mx mx

Step 5: Calculation of Dynamic Load Fd  6  vmg  Fd    Ft 6  

(Assume vmg  5 m/s)

 6  5  53097.3 N   mx  6  

97345.05 N mx

Step 6: Calculation of strength of gear tooth Fs

Y  y  

F s  [b]  y    m x  [ b] b  Y  m x Y  0.392 Assu me   20

From PSG Data book Pg. No.8.52. [b]  780 kgf/cm 2  78 N/mm 2

From PSG Data book pg. 8.45, Table 33. (For bronze, rotation in only one direction) b  0.75 d1 [From PSG Data book Pg.No.848 Table38]  0.75  11m x  8.25 m x F s  78  8.25 m x  0.392  mx F s  252.252 m 2x


Step 7: Calculation of axial module (mx) For safe design Fs  Fd 252.252 m 2x 

97345.05 mx

m 3x  385.903 m x  7.28  8 mm mx ~

From the PSG Data book pg. no.8.2, Table 1 (Choose the largest standard module from table). Step 8: Calculation of face width (b) b  8.25 m x  8.25  8 b  66 mm

Step 9: Calculation of pitch axile diameter (d2) d2  Z2  m x  72  8  576 mm

3.172


Step 10: Pitch line velocity of worm wheel V2 V 2  0.226 m x V 2  1.808 m/min

Step 11: Recalculation of Dynamic Load Fd Fd 

97345.05 N mx

F d  12168.13125 N

Step 12: Recalculation of Strength of Gear Tooth Fs F s  252.252 m 2x  16144.128 N

Step 13: Checking for Beam Strength Fs F s  F d for satis factory design 16144.128  12168.14  Fd  FS

The design is safe, based on the beam strength. Step 14: Calculation of wear load Fw  Fw  d2 b kw d 2  576 mm


b  66 mm k w  5.6 kg f/c m2  0.56 N/mm 2 [  20, (Assume) Hardened steel & Bronze, From

PSG data book pg no. 8.54, Table 43] Fw  d2 b kw  576  66  0.56 F w  21288.96 N

Step 15: Checking for Wear Strength For satisfactory design Fw  Fd F w  21288. 96 N ; Fd  12168.14

By comparing these values, F w  F d, so the design is safe based on wear strength. Step 16: Checking for efficiency   tan  actual       tan    

  tan 1    1.718   15.25  From PSG Data book Pg. 8.49 (Assume   0.03 )

3.174


actual 

tan  tan   



tan 15.25  tan 15.25  1.718 



0.27262 0.30512

 0.8934  100

desired  82% actual  89.34% actual   desired (for safe design)  So the design is safe, because the actual efficiency

is greater than desired efficiency. 89.3%  82% 

Step 17: Calculate the power loss (or) Heat Generated H g  1  a  Input hp  4500 kg f/min H g  1  0.893416  4500 kg f/min  7675.2 kg f/min

Step 18: Calculate the area required to dissipate the Heat. Heat atmosphere.

generated

H d 

Heat

dissipation

to

the


hcr  10 W/m2C  10

Nm Sm 2C

Nm  6 kgfm/min S  7675.2 kgf/min  h cr  t  A

10  kgfm  60 60  10 m 2C min

1 w/m 2C  6 kg fm/minC

hcr  60 kgfm/min C kg f m  A  40 7675.2 kg fm/min  60 min  C m n A  3.198 m2

Step 19: Calculate the basic dimension of worm and worm wheel. 1. Axial module m x  8 mm 2. Number of starts (worm) Z 1  3 3. Number of teeth on the worm wheel Z 2  72 4. Face Width b  66 mm 5. Length of worm L  12.5  0.09 Z 2 m x From PSG Data book pg. No. 8.48 Table 39 L  [12.5  0.09 72 ]  8  151.84  152 mm L~

6. Centre Distance (a)

3.176


a  0.5 m x q  Z  2x  0.5  8 11  72  332 mm

(Assume x  0)

7. Height Factor fo  1 8. Bottom Clearance c  0.2 m x to 0.3 m x

[Take c  0.3 m x]

c  0.3  8  2.4 mm

9. Pitch Diameter d1  mx q  2x  mx q  8  11  88 mm d 2  Z 2m x  72  8  576 mm

10. Tip diameters da d a1  d1  2fom x  88  2  1  8  104 mm d a2  Z 2  2f0  2n m x  72  2  1  0 8  592 mm

11. Root Diameter df

n  0


df  d 1  2fom x  2C 1  88  2  1  8  2 2.4   68 mm  67.2 ~ df  Z 2  2fo m x  2C 2  72  2  1 8  2  2.4  56 mm  555.2 ~ Problem 3.14 Design a worm gear drive to transmit 12 kW at 1200 rpm. Speed reduction desired is 30 : 1. The worm is made of hardened steel and the wheel of phosphor bronze. Check the heating capacity of gears and determine the efficiency.

Solution: Step 1: Given Data Power P  12 kW  12  10 3 W WormN 1  1200 rpm Speed ratio i  30 Worm - Hardened steel Work wheel - phosphor bronze To Find Heating capacity of gears H g Efficiency ( actual)

[May/June 2012]

3.178


Step 2: Selection of Material Worm - Hardened steel Worm wheel - Phosphor Bronze Step 3: Calculation of lead Angle Diameter factor q 

d1 mx

 q  11(Assume

If not given) 11 

d1 mx

d 1  11 m x

  tan  1

Z1 q

(Assume Z 1  3 if not given).

1

 tan

3/11 

  15.25 

Step 4: Calculation of tangential load Ft

Power

Ft 

hp  75 kg f V mg

P  

12  10 3 hp 750

 16 hp


vmg 



Ft 

 d2 N2 60  1000

d2  m x  Z2

  90 mx  40

Z1

Z2

60  1000

Z 2  30  3

 0.18849 m x

Z 2  90

hp  75 vmg

d2  90 m x N1 N2

16  75  0.18849 m x Ft 

i

 30

N2 

6366.385 653663.85 kgf  N mx mn

1200 30

N 2  40 rpm

Step 5: Calculation of Dynamic Load Fd  6  vmg  Fd    Ft 6  

 Fd 

(Assume vmg  5 m/s

6  5 63663.85  6 mx 116717.05 mx

Step 6: Calculation of strength of gear tooth Fs F s  [ b] b Y m x [n]  780 kg f/cm 2  78 N/mm 2

3.180


[From PSG Data book pg. No.8.45. Table 33.] For bronze rotation in one direction only b  0.74d1 Z 1  3 ; Table 38]

From PSG Data book pg No. 8.48; b  0.75  11 m x

Y  0.392   20)

(Assume) From PSG data book pg.No.8.52] b  8.25 m x F s  78  8.25 m x  0.92  m x F s  252.252 m 2x

Step 7: Calculation of axial module mx For safe design Fs  Fd 252.252 m 2x 

116717.05 mx

m 3x 

116717.05 252.252

 462.700 m x  7.73 m x~  8 mm

From PSG Data book pg no.8.2, Table 1 the standard module in chosen mx  8 mm


Step 8: Calculation of face width (b) b  8.25 m x  8.25  8 b  66 mm

Step 9: Calculation of pitch circle diameter d 2 d2  Z2 m x  90  8 d 2  720 mm

Step 10: Pitch line velocity of worm wheel V2 V 2   0.18849 m x  0.18849  8  60  90.5 m/min

Step 11: Recalculation of dynamic load Fd Fd 

116717.05 116717.05  mx 8

F d  14589.63 N

Step 12: Recalculation of strength of gear tooth Fs F s  252.252 m 2x  16144.128 N

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Step 13: Checking for Beam Strength For satisfactory design, Fs  Fd 16144.128  14589.63  The design is safe based on the strength of gear.

Step 14: Calculation of Wear Load (Fw F w  d2  b  kw d2  720 mm ; b  66 kw  5.6 kg f/cm 2  0.56 N/mm2

[Hardened steel Bronze, from psg Data book pg. 8.54, Table 43] Fw  d2 b kw  720  66  0.56  26611.2 N

Step 15: Checking for Wear strength For satisfactory design Fw  Fd 26611.2 F w   14589.63 F d  The design is safe based on the wear strength of

gear.


Step 16: Efficiency actual 

tan  tan   

  tan 1 

  0.03 (Assume)

  1.718   15.25 actual 

tan 15.25  tan 15.25  1.718 

actual  0.8935  100%  89.35%

Step 17: Heating capacity of great (H g ) (or) power loss (or) Heat generated. H g  1  a   Inpu t hp  4500 kg f m/min  1  0.8935   16  4500 kgfm/min  7668 kg f m/min

7668 Nm /s 6 H g  1278 Watts

 1 kgfm 1 Nm     6 s   mi n

  . . 1 Nm W  . s  

Step 18: Calculate the basic dimensions of worm and worm wheel. 1. Axial module m x  8 mm

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2. Number of starts (worm) Z1  3 3. Number of teeth on the worm wheel Z 2  90 4. Face Width b  66 mm 5. Length of Worm L  12.5  0.09 Z2 m x Choose L From PSG Data book pg. No. 8.48, Table 39 L  12.5  0.09  90 8  164.8 L~  165 mm

6. Centre Distance a  0.5 m x q  Z 2  2x  0.5  8 11  90

(Assume x  0 )

 404 mm

7. High Factor f0  1 8. Bottom Clearance c  0.2 to 0.3 m x c  0.3 m x

[Take c  0.3 m x]

 0.3  8  2.4 mm 9. Pitch Diameter (d) d1  m x q  2n    8 11  0    88 mm 

(Assume n  0) d2  Z 2m x  90  8  720 mm


10. Tip diameter d a

(Assume n  0)

da1  d1  2fo m x  d a2  Z 2  2fo  2nm x   88  2  1  8   90  2  0 8   104 mm da2  736 mm 

    

Root diameter (d f) df1  d 1  2fom x  2c  88  2  8  2  1  2.4  67.2  68 mm df2  Z 2  2fo m x  2c  90  2  8  2  2.4  699.2  700 mm df2 ~ Problem 3.15 Design a worm gear drive to transmit a power of 22.5 kW. The worm speed is 1440 rpm, and the speed of the wheel is 60 rpm. The drive should have a minimum efficiency of 80% and above. Select suitable materials for worm and wheel and decide upon the dimensions of the drive. (May/June 2013)

Solution: Step 1: Given Data Power P   22.5 kW  22.5  10 3 W Worm N 1  1440 rpm Worm Wheel N 2  60 rpm Minimum Efficiency (desired  80%


3.186

Step 2: Selection of Material Worm  Always Select (steel) Worm wheel  Choose bronze From PSG Data book Pg No.8.45; Table 32. Step 3: Calculation of lead angle  Diameter Factor q 

d1 mx

(Take q  11 )

d 1  11 m x  Z1    tan  1    q 

Z 1  3   0.8  0.9

d  0.8  tan 1 3/11

  15.25  Step 4: Calculation of tangential load Ft Ft 

vmg 



Z2

hp  75 vmg

Z1

 d 2 N2 60  1000

  m x72  60 60  1000

i [d 2  72 m x]

i

i N1 N2 1440 60

i  24

 0.226 mx m/s

Z 2  24  3 Z 2  72


Power P  

22.5  10 3 hp 750

 30 hp Ft  

Ft 

hp  75 kgf V mg

30  75 9955.75  kg f mx 0.226 m x 99557.5 N mx

Step 5: Calculation of Dynamic Load F d  6  vmg  Fd    Ft 6  

Assume vmg  5 m/s

 6  5  99557.5   m x  6  Fd 

182522.08 mx

Step 6: Calculation of strength of gear tooth F s F s  [ b] b Y m x Y  0.392

(Assume   20 From PSG Data book

pg. no. 8.52 [Y  0.392 in chosen] b  0.75 d 1  8.25 mx

[b]  80 N/mm 2 for bronze.

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F s  80  8.25 m x  0.392  m x  258.72 m 2x

Step 7: Calculation of axial module m x For safe design F s  Fd 258.72 m 2x 

182522.08 mx

m 3x  705.481 m x  8.90  10 mm mx ~

A standard module chosen, From PSG Data book pg. No.8.2 Table 1. Step 8: Calculation of face width b b  8.25 m x  83 mm  8.25  10  82.5 ~

Step 9: Calculation of pitch circle diameter d 2 d2  Z2  m x  72  10 d2  720 mm


Step 10: Pitch line velocity of worm wheel V2 v2  vmg  0.226 m x  0.226  10  2.26 m/s

Step 11: Recalculation of Dynamic Load Fd Fd 

182522.08 182522.08  mx 10

F d  18252.208 N

Step 12: Recalculation of strength of great tooth Fs F s  258.72 m 2x  258.72  10 2 F s  25872 N

Step 13: Check for Beam Strength F s  F d For satisfactory design. F s 25872   18252.208 F d  The design is safe based on beam strength of gear.

Step 14: Calculation of Wear Load Fw  F w  d 2  b  kw

3.190


d 2  720 mm ; b  83 mm kW  5.6 kg f/cm 2  0.56 N/mm 2 [Assum e   20

For steel and Bronze] From PSG Data book pg no.8.54 Table 43. F w  720  83  0.56  33465.6 N

Step 15: Checking for Wear Strength For safe design F w  F d F w  33465.6 F d  182522.08 N

By comparing two values F w  F d,  The design is safe based on wear strength of gear. Step 16: Checking for efficiency actual 

tan  tan   

  tan 1 

Assume   0.03

  tan 1 0.03   1.718   15.25


actual 

tan 15.25  tan 15.25  1.718 

desided  80%

actual  0.8934 actual  89.34% The actual efficiency is greater than that of desired efficiency,  The design is safe [a  d] Step 17: Calculate the basic dimensions of worm and worm wheel. 1. Axial module m x  10 mm 2. Number of starts Z 1  3 3. Number of teeth on the worm wheel Z 2  72 4. Face width b   83 mm 5. Length of Worm L  12.5 0.09Z 2 mx L  12.5  0.09  72   10  189.2 ~  190 mm

6. Centre Distance a  0.5 m x q  Z 2  2x (Assume x  0 )  0.5  10 11 1 72   415 mm

7. Height Factor fo  1 8. Bottom Clearance c  0.2 m x to 0.3 m x [Take c  0.3m ] x c  0.3 m x  0.3  10  3 mm

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9. Pitch diameter d1  m x q  2n   d1  10 11  0    110 mm 

d 2  Z 2m x  72  10  720 mm

(Assume n  0 )

10. Tip diameter d a da1  d 1  2fom x  d a2  Z 2  2fo  2n m x   72  2  1  0  10 99  2  1  10    119 mm  740 mm 

11. Root diameter d f df1  d 1  2fom x  2c    99  2  10  2  3   df1  73 mm 

df2  Z2  2fo mx  2c  72  2 10  2  3  694 d f2  694 mm

3.23 DESIGN OF WORM AND WORM WHEEL [Based on Hertz Method] Refer PSG Data book Pg. No. 8.43 to 8.49 Step 1: Selection of Material Refer PSG Data book Page No. 8.45 Worm: Steel Worm wheel: Cast iron (or) Bronze Take the values of [ c]  Design surface stress and   b   Design bending stress in kgf  cm2  


For [c] value Initially assume sliding velocity vs as 3 m  sec and select wheel material as bronze. Take the value of [c] From PSG Data book P.No. 8.45. For [b] value, The value depends on material, methods of casting of wheel, u and type of rotation. Take the value of [ b] From PSG Data book P.No., 8.45 Step 2 Select no. of starts  Z 1 Refer PSG Data book Page No. 8.46 Take the value of Z 1, based on efficiency. If efficiency is not given, assume efficiency and take Z 1. Step 3 (Refer PSG Data book P.No. 8.44) Calculation of number of teeth on wheel Z2  i  Z1

Normally Z 2  30 to 70

(For compact reduction unit choose Z 2 towards the lower limit; For higher transmission efficiency choose Z 2 towards the higher limit)

In general, For hp  20 ; Z 2  60 to 70

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for small hp, Z 2  30 to 50 Step 4 Diameter factor: (q) q

d1

d 1  pitch diameter of Worm in mm

mx

m x  axial module - mm

Initially assume q  11 Step 5 Calculation of design twisting moment [M t] Refer PSG Data book Page No. 8.44 Mt 

Nominal twisting moment transmitted by worm wheel in kg fcm hp  71620 i   kgf cm N

Design twisting moment  [M t]  M t  k  k d

N 1  speed of worm in rpm k  load concentration factor  1 when load is almost con stan t kd  dynamic load factor  1 for V s  3 m  sec

Step 6: Calculation of centre distance (a) Refer PSG Data book Page No. 8.44 Substitute c in kg f/cm 2 and M t in kg f  cm then we get (a) in cm.


  Z2 1 a   q

  3

2  540  [M t]  Z  10 cm  2   q [c]   

(Or)  Z2  1  a   q 

  3

 540 Z  2  q [c] 

2  [M t]  10 cm   

Substitute  c in N/mm 2 [M t] in N-mm, then we get ‘a’ in mm. Step 7: Calculation of axial module m x: Refer PSG Data book Page No.8.43 mx 

2a q  Z 2  2x

assume x  0 x  addendum modification coefficient

Refer data book, Refer PSG Data book Page No. 8.2, take the standard value of module. Step 8: Recalculate centre distance ‘a’ According to the standard value of module m x recalculate the centre distance a a

m x q  Z 2 2

Step 9: Calculation of sliding velocity vs  m  sec d1  Pitch circle diameter of worm  q  m x

3.196


Sliding velocity  vs 

V1 cos 

tan  

Z1 q

where   lead angle Check vs: If vs  3 m  sec, then reduce [ c] design

 Z1    tan 1    q  surface stress by 30% of initial value.

ie., [c]  0.7  Initially assumed value of [c]

Calculation of pitch line velocity of wheel V2: v2 

d 2 N 2 60

d2  mxZ

m  sec

N 2  wheel speed in rpm

for v2  3 m  sec kd  1;

Since the value of k and kd  1 There is no change in the value of moment. [M t]

design twisting

Step 10: Checking of design surface stress [c], Refer PSG Data book Page No.8.44 Substitute a in cm, M t in kg f/cm , We get, c in kg/cm 2

540 C  Z2  q

3

 

 Z 2/q  1   [M t]  a  

 [C ] in kgf/c m


(Or)

c 

540  Z2     q 

3

    Z2   1  q   a 

  [m t ]   in N/mn2  10 

Substitute a in mm; [M t] in N-mm; then get  c in N/mm 2

Step 11: Checking of design bending stress [b] Refer PSG Data book Page No. 8.44

b 

1.9 [Mt]

m3x

 q  Z 2  yv

 [ b ]

y v  form factor

Cal. Z v or Z eq 

Z1 cos 3 

at the value of Z v take value of yv From PSG Databook Page No.8.18 at X  0 Step 12: Calculation of efficiency Refer data book Page No. 8.49



tan  tan   

Where   lead angle in degree   friction angle in degree

Refer PSG Data book Pg NO.8.49

3.198


From Graph, at sliding velocity and according to material take the value of  where coefficient

  frictional

But   tan  Step 13: Calculate the Basic Dimension of Worm & Worm Wheel 1. Axial module m x 

2a q  Z2  2x

(Assume x  0 2. Number of starts Z 1 3. Number of teeth on the worm wheel Z 2  i  Z 1 4. Face width (b) [From table 38, from PSG Data book pg. 8.48] 5. Length of worm L  [Choose L formula from Data book pg.8.48; Table: 39; Assume n  0] 6. Centre Distance a  0.5 m xq  Z 2  2x 7. Height factor fo  1 8. Bottom clearance c  0.2 m x  0.3 m x 9. Pitch diameter d d1  m x q  2x  d2  Z 2 m x  Assume x  0 


10. Tip diameter d a da1  d 1  2fom x  d a2  Z2  2fo  2x mx   Assum e x  0

11. Root diameter d f df1  d 1  2fom x  2C  d f2  Z 2  2fo mx  2C   Problem 3.16 Design a worm gear drive a transmit 22.5 kW at a worm speed of 1440 rpm. Velocity ratio is 24:1. An efficiency of atleast 85% is desired. The temperature rise should be restricted to 40C. Determine the required cooling area.

Given data P  22.5 kW ;

N  1440 rpm ;

i  24 ;

T o  T a   T  40C ;

  85% Solution Step 1: Selection of material Assuming worm material as steel and wheel material as Bronze (sand cast) From PSG Data book Pg.No. 8.45 u  39 kg f/mm 2  390 N /mm2 [b]  50 N/mm2 (Assume rotation in one direction

only) Step 2: Calculation of centre distance From PSG Data book Pg.No. 8.46, Table 37 corresponding to efficiency   85% number of starts Z 1  3

3.200


since

Z2 Z1

i

 Z2  i Z 1  24  3  72

Assume q  diameter factor 

d1 mx

 11 (initially)

Assume sliding velocity vs  3 m/s (initially) From PSG Data book corresponding to sliding velocity

Pg.

8.45,

Table

[c]  1590 kg f/cm 2  159 N/mm 2

Assume intially k  k d  1 From PSG Data book, Pg.No. 8.44, M t  71620

hp i n

since the power is in kW P

2 N Mt 60

P  60 22.5  103  60   3581 N m  Mt   1440  2 N2 2     24  From PSG Data book Pg.No. 8.44 [M t]  M t  kd  k  3581  10 3  1  3581  10 3 Nmm

From PSG Data book Pg.No. 8.44 Substitute  c is N/mm 2; M t in N-mm.

32,


  Z2 1 a   q

 72   1 11  

  3

540  Z  2    q  [ c ]  

2  [M t]  10   

  3

540   72      11   159  

2

3581  10 3    345.9  10   

 346 mm a~

Step 3: Calculation of axial module Refer PSG Data book Pg.No. 8.43 Axial module m x 

2a q  Z 2  2x

Assume x  0 2  346 2a   mx  q  Z 2 11  72 m x  8.4 mm

From PSG Data book, Pg.No. 8.2, select standard module  axial module m x  10 mm

Step 4: Recalculate a and determine d 1 a

m x q  Z 2 10 72  11   415 mm 2 2

From PSG Data book Pg.No. 8.43 Pitch circle diameter of worm d 1  q m x

3.202


 d 1  11  10  110 mm

Step 5: From PSG Data book Pg.No. 8.44 Sliding velocity vs 

V1 cos 

 Z1  1 3  But   tan  1    tan  11  q       15.25 

and v1 

 d 1 N1   110  1440  60  1000 60  1000

v1  8.3 m/s 8.3  vs  cos 15.25 vs  8.6 m/s

Step 6: Modification of [c] Since vs  3 m/s (initially assumed), the [c] value has to be modified.  [ c] is reduced by 20% to account for the increased speed.  [ c]  0.8  159  127 N/mm 2

Step 7: Determination of Induced stresses From PSG Data book Pg.No. 8.44


c 

540  Z2     q 

   Z2   1   q    a  

 

3

[M t] 10

3



540 72 /11

 

   72  1 3     3581  10   11    415 10  

 121 N/mm 2  [c]  The design is satisfactory based on surface stresses.

From PSG Data book Pg.No. 8.44 b 

Zv 

1.9 [M t] m 3x qZ 2 yv Z2 3

cos 



72 3

cos 15.25 

 80

From PSG data book Pg.No. 8.18 corresponding to Z v  80 y v  0.499  b 

1.9  3581  10 3 3

10  11  72  0.499

 17.22 N/mm2  [b]

 The design is satisfactory based on bending stresses.

Step 8: Efficiency From PSG Data book Pg.No. 8.49

3.204




tan  tan   

From PSG Data book Pg.No. 8.49, from graph at vs  3.5 m/s   0.03    tan     tan  1   tan  1 0.03    1.718  

tan 15.25  tan 15.25   1.718 

  0.8935  89.35% Step 10: Heat removal analysis 1    Input power  kt To  Ta  A 1  0.8935   22500  10 40  A

[Assume kt  10 w /m 2  C ] A  5.99 m 2  Cooling Area  5.99 m 2 Problem 3.17 A 2 kW power is applied to a worm shaft at 720 rpm. The worm is of quadruple start type 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 in friction.


Solution P  2 kW N 1  720 rpm Z 1  4 since quadruple start d 1  50 mm Z 2  40 teeth m  5 mm i  V.R 

Z2 Z1



40  10 4

Step 1 Selection of material From PSG Data book Pg.No. 8.45 Worm Wheel

– –

steel Bronze, sand, chill

[b]  50 N/mm2 [c]  159 N/mm 2 at Vs  3 m/s

Step 2 Calculation of centre distance From PSG Data book Pg.No. 8.44   Z2  a 1  q 

  3

2  540  [M t]  Z  10  2  [ ]  q c   

3.206


q  10 (diameter factor) From PSG Data book

Pg.No. 8.46; Table 36. Initially assume k  kd  1 Mt 

Po wer  60 2  10 3  60  720 2  rpm of wh eel 2  10  265.25 Nm  265.25  10 3 Nmm

 [M t]  M t  k  k d  265.25  10 3  1  265.25  10 3 Nmm

  40 1 a 10  

  3

540   40   10  159 

2

265.25  10 3    10   

 133.7 mm

Step 3 Lead angle  From PSG Data book Page No. 8.43 tan   

Z1 q

4  0.4 10

tan   0.4


  tan 1 0.4  21.8 

Step 4 Efficiency of worm gearing From PSG Data book Pg.No. 8.49 

tan  tan   

We know sliding velocity vs  But v1  v1 

v1 cos 

 d1 N 1 60  1000   50  720  1.884 m/s 60  1000

1.884  2.03 m/s  vs  cos 21.8

From PSG Data book Pg.No. 8.49 from the graph, corresponding to sliding velocity and plot of bronze worm wheel   0.045   tan 

But

 tan   0.045    tan 1 0.045  2.576 

3.208




tan 21.8 tan 21.8  2.576 

 0.8827  88.27%

Step 5 Power lost in friction Q Q  P 1    2  103 1  0.8827   234.6 watts  0.234 kW Problem 3.18 A worm drive transmits 15 kW at 2000 rpm to a machine carriage at 75 rpm. The worm is triple threaded and has 65 mm Pitch diameter. The worm gear has 90 teeth of 6 mm module. The tooth form is to be 20 full depth involute coefficient of friction is 0.1; Calculate (i) Tangential force acting on the worm (ii) axial thrust and separating force on worm. and (iii) efficiency of worm drive.

Solution Given Data: P  15 kW ;

N 1  2000 rpm;

N 2  75

rpm;

Z 1  3;

d 2  65 mm; Z 2  90 ; m x  6 mm   20;   0.1

1. Tangential force acting on the worm (refer PSG Data book P.No. 8.57)


Ft 

2M t

M t  97420

d2

2  730.65 6.5  224.81 kg

kW N1

 97420 



15 2000

 730.65 kg f  cm

2. Axial thrust on Worm: (refer PSG Data book Page No. 8.57) F a  F t tan   

tan  

 224.81 tan 15.47  5.71  87.10 kg

d2

Z1 q

65  10.83 6 3   tan  1 10.83 Lead angle    15.47

q

mx



3. Separating force (or) Radial force  F t  tan 

  Frictiona l coefficient

 tan   0.1

 224.81 tan 20  81.824 kg

   tan 1 0.1

friction angle    5.71   pressure angle  20

4. Efficiency   

tan  (refer PSG Data book Pg. tan   

No. 8.49) 

tan 15.47  0.7142 tan 15.47  5.71

  71.42%

3.210


Problem 3.19 Determine the efficiency of worm gear drive to transmit 20 hp from a worm at 1440 rpm to the worm wheel the speed of the worm wheel should 40  2% rpm. (May /June 2008)

Solution Given Data: Power  P  20 hp ; Speed of worm  N  1440 rpm Speed of Worm wheel  40  2% 

Gear ratio  i 

1440 speed of worm   36  2% speed of worm wheel 40 i.e ,



% deviation of i  2%

Selection of material and design stresses refer Databook, Refer PSG Data book Page No. 8.45

Worm - steel; Worm wheel 

Bronze

(sand

u  39 kgf  mm

chill

cast)

-

2

(assume rotation in one direction only) Take [b]  and [c] 



Design bending stress  500 kgf  mm 2 Design surface stress  1590 kgf  cm 2 (assume initially, the sliding velocity vs  3 m  sec

Since the efficiency is not given, assume efficiency as 80% Refer PSG Data book page No. 8.46

For efficiency   0.70 to 0.82 ,


The value of Z 1  no. of starts  2 But, i  gear ratio 

Z2 Z1

; where Z 2  no. of teeth on

worm wheel  Z 2  iZ 1  36  2  72 teeth 

Assume q  diameter factor 

d  11 (initially) mx

where d  pitch diameter of worm  mm  m x  axial module  mm 

Design twisting moment [M t]

Refer PSG Data book Page No. 8.44 Design torque  [M t]  M t kd  k  25464.88  1  1 [M t]  25464.88 kgfcm

M t  71620

hp i n

 71620 

20  32  0.8 1440

 25464.88 kg f  c m

Assume k  kd  1 initially 

Centre distance, a refer PSD Databook, Page No. 8.44

Centre distance (a)   Z2 1 a   q

  3

2

540    Z  q [ ]  [M t] C   2

3.212


 72  1 a  11 

  3

540     72  1590   11   

2

  25464.88   

a  30.88 cm  388.8 mm

Take centre distance  a  309 mm Calculation of axial module, mx Refer PSG Data book Page No. 8.43 axial module  m x  

2a q  Z 2  2x

assume x  0

2  309  7.53 mm 11  72  0

From PSD data book, refer page No. 8.2, select standard module  axial module  m x  10 mm

Recalculate, centre distance (a) a

m x q  Z 2

2



10 11  72   415 mm 2

Refer PSG Data book Page No. 8.43 Reference diameter of worm  d 1  q  m x  11  10  110 mm

Refer PSG Data P.No. 8.44 Sliding velocity V2  vs  cos 

book

Z1

2  q 11  lea d angle    10.3  tan  




8.293 cos 10.3 

v1 

d 1 N 60

 8.429 m /sec

  0.11  1440 60



 8.293 m  sec

Since vs  3m  sec (initially assumed) Therefore, reduce [c] by 30% Now, The design surface stress  [ c]  0.7  1590  1113 kg f  cm2

Checking of surface stress, c Refer PSG Data book Page No.8.44

540 C  Z 2  q



 

540 72  11 

3

 Z 2  q  1   [Mt]  a  

 [ c]

3

    72    11   1    415 

    25464.88  

 32.276 kg f  cm 2  [ c] 1113 kg f  cm 2  The design is satisfactory based on surface stress. 

Checking of bending stress, b

Refer PSG Data book Page No.8.44

Zv 

Z2 3

cos 



72 3

c os 10.3

3.214

b  


1.9 [Mt] m3x

 q Z2  yv

 73.59

 [b]

Take approximately Zv  80 teeth

1.9  25464.88 13  11  72  0.499

 122.42 kgf  cm2  500 kgf  cm2

Refer PSG Data book P.No. 8.18 take yv  0.499 (at

Zv  80 teeth) mx  10 mm  1 c m

 The design is satisfactory based on bending stress 

To find efficiency of drive

Refer PSG Data Page No. 8.49 tan   tan    tan 10.3  tan 10.3  1.718    0.85  85% 

book   friction angle from graph (P.No. 8.49) at vs  3.5 m  sec,   0.03    tan   0.03    tan  1 0.03  1.718 

CROSSED HELICAL GEARS 3.24 Screw or Skew gears (crossed helical gears) These helical gears are also known as spiral gears. They are true helical gears and only differ in their application for interconnecting skew shafts shown in Fig. 3.9. Screw gears can be designed to connect shafts at any angle, but in most applications the shafts are at right angles.


1. Helical gears of the same hand operate at right angles. 2. Helical gears of opposite hand operate on parallel shafts.

D rive r

D rive r

D rive r

D rive r

L eft-H an d

D rive r

D rive r

Fig.3.9.Types o f H elical Gear M eshes.

Helical angles 1 and 2 need not be the same. However, their sum must be equal to shaft angle. 1  2  

Where 1 and 2 are the respective helix angles of the two gears and  is the shaft angle. Module: Because of the possibility of different helix angles for the gear pair, the radial modulus may not be the same. However, the normal modulus must always be identical.


3.216

Let a pair of screw gears have the shaft angle  and helix angles 1 and 2, fig 3.10. If they have the same hands, then   1  2

If they have the opposite hands, then   1  2 or   2  1

Crossed helical gears are noiseless in operation due to smooth engagement and disengagement. These gears are not suitable for high transmission of power, because of point contact between teeth. The efficiency of the drive is less due to relative sliding of teeth along the common tangent. G ea r 1 (R igh t-H a nd (L eft-H a nd ) 

2 2

1 1



G ea r 2 (R igh t-H a nd ) Fig. 3.10. Screw Gea rs o f N o nparallel and N on inters ecting A xes.

Fig. 3.10 Screw Gears of Non parallel and Non-intersecting Axes


3.24.1 Design procedure (Refer PSG data book Page No. 8.55) Let

P n  Circular pitch, P d  Diameteral pitch

suffix n refers to normal plane suffixes 1, 2 refer to driving and driven wheels respectively P dn  P cn   P cn  P c cos 1  P c cos 2 1 2 P dn 

P d1 cos 1



Pd 2 c o s 2

D  pitch diameter of the gear D 1  pitch diameter of driving wheel 

Z 1 P cn

 c o s 1

Similarly, D 2  pitch diameter of driven wheel 

i  velocity ratio    1  2 ;

Z2 Z1



Z 2 P cn  c o s 2

D 2 cos 2 D 1 cos 1

 when both helicals are of the same hand

when both helicals are of the opposite hand 

Z v  virtual number of te eth

3.218




Z1 cos 3 1



Z2 cos3 2

1. To calculate strength of the gear tooth Let

F s  strength of the gear tooth kgf [b]  design bending stress kgf/cm 2 b  face width – cm Y  form factor   y C  wear and lubrication factor

(Refer table 44, PSG design data book P.No. 8.55) FS 

[b]  b  Y P dn  C w

kgf

Modified as, FS 

[b]  b   y     P  C w  cm 



[b]  b  P cn  y kgf Cw

2. Calculation of Dynamic load F d Let

F d  Dynamic load kgf Fd  Ft  Cv

where

F t  Transmitted load 

hp  75 Vm

hp is the given power V m is the mean velocity in m/sec


Velocity factor C v 

4.5  V m 4.5

Equate (1) and (2) to calculate P cn and then find the value of module. 3. Calculation of wear load Let

F w  wear load F w  kQ  D 21

Where

k  Load stress factor

[refer Table 45, PSG Design book P.No. 8.55]  2D 2  Q  Ratio factor    D  D2   1

2

4. Calculation of lead l l1  lead of the driving wheel   D 1 tan 2 l2  lead of the driven wheel   D 2 tan 1

5. Calculation of center distance a  centre distance 

Z1 2P dn f c o s 1



Z2 2Pdn cos 2

3.220


3.26 The Equations

for a

Screw

Gear Pair on

Nonparallel and Non-interacting Axes in the Normal System No.

Item

Symbol

Formula

Example Pinion

Gear

1

Normal module

ma

3

2

Normal Pressure Angle

n

20

3

Helix Angle



20

30

4

Number of Teeth & Helical Hand

Z1, Z2

15 R

24 L

5

Number of Teeth

zv

18.0773

36.9504

of an Equivalent Spur Gear

Z 3

cos 

6

Radial Pressure Angle

1

7

Normal Coefficient of

xn

8

Involute

 tan n  21.1728 22.7959 tan 1    cos  

0.4

0.2

Profile Shift Function  n

inv  n

2 tan n

0.0228415

 Xn1  Xn2     Zy1  Zy2   inv n

9

Normal Working Pressure Angle

 n

Find from

22.9338

Involute Function Table

10

Radial Working Pressure Angle

xt

tan 1  tan  n     c o s 

24.2404 26.0386

Design of Bevel,Worm and Cross Helical Gears 3.221 No. 11

Item Center Distance

Symbol

Formula

Example

Y

1 Z  Zv2 2 v1

0.55977

Increment Factor

 cos n  1   cos wn 

12

Center Distance

x

z1    2 cos 1  z2  y 2 cos 2  mn

13

Pitch Diameter

d

Zmn

67.1925

47.8880

83.1384

44.6553

76.6445

49.1155

85.2695

c os  14

Base Diameter

db

15

Working Pitch

dx1

d cos t 2ax

Diameter dx2 16

Working Helix Angle

n

17

Shaft Angle



2ax

d1

d1  d2 d2 d1  d2

 dx  20.4706 30.6319 tan 1  tan    d 

w1  w2

51.1025

or w1  w2 18

Addendum

1  y 

ha1

xn2 mn ha2

1  y  xn1

4.0793

3.4793

mn

19

Whole Depth

h

[2.25  y  xn1  6.6293 xn2] mn

20

Outside Diameter

da

d  2ha

56.0466

90.0970

21

Root Diameter

dt

da  2h

42.7880

76.8384

Chapter 4

DESIGN OF SPEED REDUCERS (Gear Boxes) 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.

4.1 GEAR BOX A gearbox is a mechanical method of transferring energy from one device to another and is used to increase torque while reducing speed. A group of gears put together in a manner to increase the torque of an engine and reduce the speed is known as gear box. Generally gear boxes are also known as speed reducers, or gear reducers.

4.2 TYPES OF GEAR BOXES 

According to the type of drive used: Spur, helical, bevel and worm gears.



According to the No. of stages:

(i) Single stage Gear box: It means one pair of gear wheels, such as one pinion and one wheel can be operated to reduce the speed. (ii) Multi-stage Gear box: It means two or more pairs of gears can be operated to reduce the speed.

4.2


In case of machine tools, the gear box is used to reduce the cutting speed. The cutting speed depends on material of the job, cutting tool used and nature of operation. The same machine tool may be used for different working conditions. Therefore, it is necessary to provide a gear box between motor and spindle for changing the spindle speeds. There are variety of gear boxes used in industries such as aerospace, paper, mining, agriculture, etc... The toothed gears are used as independent units to reduce or increase the speed and they are enclosed in rigid closed housings. The housings provide 1. Support for the shafts 2. Hold lubricant inside 3. Protect the gear from dust and moisture 4. Give necessary cooling surface to dissipate the heat generated. When the unit is used as a speed reducing device, it is called as “speed reducer.” 4.2.1 Applications Speed reducers are widely used for reduction of speed in turbine generator set; from motor to machine tool spindles; in rolling mills; from engine to road wheels in automobiles, etc. Speed reducers are widely used for reduction of speed in conveyors, crushers, cranes, elevators, feeders, small and large ball mills, mixers, towers and coal pulverising units.

Design of Speed Reducers (Gear Boxes) 4.3



Major Manufacturing Industries – Sugar, cement, elevator, paper, plastic and leather, rubber, steel industry, power plant, mines and minerals, waste water treatment and extraction.

In speed reducers, gears are arranged in different forms: 1.

Single reduction

2.

Double reduction

with spur worm and with spur worm and

gear, worm gear, worm

bevel gear and wheel bevel gear and wheel

4.2.2 Requirements of Speed Gear Boxes 

It should provide silent operation of the power transmission.



To reduce / increase the rpm.



Change the direction of rotation (Clockwise / Anticlockwise)



Shift the axis of rotation (linear or angular)



To transmit the required power to the spindle.

4.2.3 Method of changing speed in gear boxes Mainly two important methods are used, (i) Sliding mesh gear box (ii) Constant mesh gear box. 4.2.4 Requirements to obtain optimum design To reduce large diameter of the gear wheels and also to limit the pitch line velocity of gear drive, the following principles are to be followed. 1. Number of gears on the last stage should be minimum.

4.4


2. Number of gears on shafts should not be more than three. (But in some cases it may be four) 3. It is necessary to have, N maximum  N input  N minimum in all stages, except in the first stage. 4. The transmission ratio between driver and driven shaft should be the maximum. N minimum 1 N maximum  2 and  (For all stages, N minimum 4 Ninput except for first stage) 4.2.5 Main components of gear box 

Shafts for mounting gears.



Oil seals for lubrication.



Bearings to support shafts.



Gears for getting different speeds.



Gear box housing used for covering all inner components.

4.2.6 Speed Changing 

In the conventional gear box, speed changing is obtained by using sliding gears.

Compound gears are mounted on splined shaft and are moved by means of forced lever pivoted at the casing.

4.3 SPEED DIAGRAM (OR) RAY DIAGRAM 

The speeds of the spindle should be in the geometric series, in order that the number of gears employed to be minimum and also to reduce speed loss. Preferred Nos. are in Geometrical progression.


The speed of driving shaft, driven shaft and intermediate shafts can be represented easily in one diagram. That particular diagram is called ‘Speed diagram’ (or) Ray diagram.

( D rive r) H ig h Sp eed

H ig h Sp eed (D rive r)

L ow Sp eed ( D rive n)

(a) Sp ur Gear

L ow S peed ( D rive n) (b) Bevel G ear

L ow speed D riven

L

L ow Sp eed D riven

H ig h Sp eed D river (c) Worm & Worm Wh eel

H ig h S peed D river Fig. 4.1 Single Red uction Sp eed R edu cers

4.6


H H

S pur G ears W orm W he el I

I

I

I L

L

I

H

L W orm

(a) Spur Gears

(b) Spu r An d (c) W orm & Wo rm Wh eel Bevel G ears and Spu r G ear H - H igh Spe ed I - In term ed iate S pe ed Fig. 4.2 Do uble R eduction Speed Reducers L - L ow S pe e d

4.3.1 Geometric progression and standard step ratio Step 1 Calculate step ratio  Let N max  Maximum speed required. N min  Minimum speed required Z  No. of speeds required. N max N min

 Z  1

Calculate step ratio ‘ ’ There are three methods to find the required speeds.


1st Method The standard step ratio values are available in (Refer PSG databook, Page No. 7.20) R5 – 1.6; R10 – 1.25; R20 – 1.12; R40 – 1.06. The standard step ratio values are given in Geometric progression in the above series. If the calculated value of  is coinciding with the standard step ratio, then select that series and write the speeds from N min , N 2, N 3  N max , continuously. For example N max  335 rpm ; N min  100 rpm; Z  12 speeds N max

Nmin

 Z  1

355  12  1 ;   1.12 100

which is standard step ratio of R20 series. The value of  obtained is exactly coinciding with R20 series. Therefore take R20 series and write 12 speeds continuously. N 1  100; N 2  112; N 3  125; N 4  140 ; N 5  160;

N6  180; N 7  200; N 8  224; N 9  250; N 10  280 ; N 11  315; N 12  355

2nd Method If the calculated value of  is non-standard step ratio, then try the multiples of standard step ratios and try the series.

4.8


For example N max  1600 rpm; N min  500 rpm; Z  6 speeds N max N min

 Z  1

1600   6  1;   1.26 500

which is a non-standard step ratio. Trail 1 Take R40 series step ratio 1.06 and try multiples.  1.06   1.06  1.06  1.06  1.26

Skip 3 speeds. By multiplying R40 series step ratio by 4 times, the value obtained is similar to the required step ratio. Therefore take R40 series by skipping three speeds ie., N 1  500; N 2  630; N 3  800; N 4  1000 ; N 5  1250 ; N 6  1600

Since the value of N 6  1600 is coinciding with the required value of N max , therefore the selection of this trail is satisfactory. Note: Try any standard step ratio multiples (i.e., R5 or R10 or R20 or R40) If the multiples are four means, skip three speeds (i.e., multiples minus one) If the multiples are three - skip two speeds.


3rd Method If the  value is not coinciding with standard step ratio. (or) If the multiples of standard step ratio are also not coinciding, then find the speeds by using this method. Note Before using this method, first try method (1) and method (2). For example: N max  600 rp m; N min  23 rpm; Z  12 speeds N max N min

 Z  1

600  12  1    1.345 23

(which is a non-standard step ratio) Trail 1 Take R20 series multiples   1.12   1.12  1.12  1.12  1.404

which is greater than required;  not satisfactory. Trail 2 Take   1.06 (R40 series) multiples  1.06   1.06  1.06  1.06  1.06  1.338

Skip 4 speeds. which is approximately equal to the required  value. Therefore, try R40 series by skipping 4 speeds.

4.10


N 1  23.6; N 2  31.5; N 3  42.5 N 4  56; N 5  75; N 6  100; N 7  132; N 8  180; N 9  236 N 10  315; N 11  425; N 12  560;

The maximum speed obtained is 560 rpm. But the required maximum speed is 600 rpm. Therefore the above trail is not satisfactory. To obtain speeds (Approximate Method) N 1  23 rp m N 2  N 1    23  1.345  30.935  31 rpm N 3  31  1.345  41.495  42 rpm Similarly N 4  56 rpm; N 5  75 rpm; N6  101 rpm; N 7  136 rp m; N 8  183 rpm; N 9  246 rpm; N 10  331 rp m N 11  445 rpm; N 12  599 rpm

The maximum speed obtained is almost equal to the required maximum speed.


4.4 STRUCTURAL FORMULA Preferred Structural formula Write the No. of speeds requried in multiples of 2 and 3 as given table 4.1 Table 4.1 6 speeds

3  2 P1 P 2

9 speeds

3  3 P1 P 2

12 speeds

or

2  3 P1 P2

3  2  2 or  2  3  2 or 2  2  3

P1 P 2 P 3

18 speeds

P1 P2 P3

P1 P2 P3

3  3  2 or 2  3  3 or 3  2  3

P1 P 2 P 3

16 speeds

P1 P2 P3

P1 P2 P3

2  2  2  2 P1 P 2 P 3 P 4

Structural formula Z  P 1 X 1

P 2 X 2

,

1 st stage

, P 3 X 3

2 nd stage

3rd stage

For example: 12 speed gear box Speed 3  2  2 P1 P2 P3

12 

P1  3

X 1  1 always

P2  2

X2  P1  3

P3  2

X3  P1  P2  3  2  6

3 1 1

st

Space

stage

2 3

, 2

nd

stage

2 6

, 3

rd

stage

4.12


In the ray diagram, 1st stage:

3 speeds required with one space interval between each speed

2nd stage:

2 speeds required with three space interval between each speed

3rd stage:

2 speeds required with six space interval between each speed.

4.5 CONSTRUCTION OF RAY DIAGRAM Let u  No. of stages. Draw u  1 vertical lines (with some distance) The 1st vertical line from left represents the driver shaft in the 1st stage and 2nd line represents the driven shaft for the 1st stage and driver shaft for the 2nd stage and so on. That means, in each stage consists of one driver shaft and one driven shaft. Draw a series of parallel horizontal lines with equal space according to number of speeds required. Selection of N input in each stage To obtain optimum design of gear box, Check, for each stage, except for the 1st stage N max N input

 2 and

Nmin N input



1 4

(or)

N max  N input  N min

If this is not possible to satisfy, then try


N max N input



N input N min



Always start the ray diagram from the minimum speed.



At the end, all speeds should be completely filled.

4.6 KINEMATIC LAYOUT To draw the kinematic arrangement, find the total number of gears required. Approximate method to find No. of gears required. Table 4.2 No. of speeds

Factors

Approximate No. of gears

2

2  1

2 2  4

3

3  1

2 3  6

do not consider 1 4

22

2 2  2  8

6

23

2 2  3  10

8

222

2 2  2  2  12

9

33

2 3  3  12

12

232

2 2  3  2  14

16

2222

2 2  2  2  2  16

18

332

2 3  3  2  16

4.14


Draw the approximate kinematic diagram according to the required input and output speeds according to the stages. 

In case of 2 speeds, there are two pairs of gears. One pair gives the maximum speed and the other pair gives the minimum speed.



In case of 3 speeds, there are three pairs of gears. One pair gives the maximum speed. One pair gives the minimum speed and the other pair gives the intermediate speed.



In each stage, the minimum No. of teeth on one gear can be assumed between 17 to 20; Maximum No. of teeth should not exceed to 120 teeth.



In case of 3 pairs of gears, to avoid interference, check the difference between adjacent gear No. of teeth. The value should be greater than or equal to 4.



To find the No. of teeth on each gear, use velocity ratio,

V.R 

Speed of driven No. of teeth on driver  Speed of driver No. of teeth on driven

... (1)

In case of 2 speeds, let Z 1 and Z 3 are driver gears and Z 2 and Z 4 are driven gears. Assume centre distance is constant, then Z 1  Z 2  Z3  Z4

... (2)

By using relations (1) and (2), calculate No. of teeth on each gear. By solving numerical understand the concepts.

problem,

we

can

easily


Basic Series of Preferred Numbers (From PSG D.B7.20) R 5

R 10

R 20

R 40

  1.6 1.00

  1.25 1.00

  1.12 1.00

  1.06 1.00 1.06 1.12 1.18 1.25 1.32 1.40 1.50 1.60 1.70 1.80 1.90 2.00 2.12 2.24 2.36 2.50 2.65 2.80 3.00 3.15 3.35 3.55 3.75 4.00 4.25 4.50 4.75 5.00 5.30 5.60 6.00 6.30 6.70 7.10 7.50 8.00 8.50 9.00 9.50 10.00

1.12 1.25

1.25 1.40

1.60

1.60

1.60 1.80

2.00

2.00 2.24

2.50

2.50

2.50 2.80

3.15

3.15 3.55

4.00

4.00

4.00 4.50

5.00

5.00 5.60

6.30

6.30

6.30 7.10

8.00

8.00 9.00

10.00

10.00

10.00

4.16


Problem 4.1: Design a 12 Speed Gear Box The speed range required is 100 to 355 rpm (a) Draw the Ray diagram, (b) Draw the Kinematic arrangement, (c) Calculate the No. of teeth on each gear.

(Oct. 2007)

Solution: N max N min

 Z  1

N max  355 rpm

355   12  1 100

N min  100 rpm

 11  3.55

Z  No. of speeds  12

  1.122

From PSG Data book Page No. 7.20 The progression ratio   1.122 Coincides with R20 series standard Geometric progression ratio. Therefore select R20 series speeds N 1  100 rpm; N 4  140 rpm; N 7  200 rpm;

N 10  280 rpm

N 2  112 rpm; N 5  160 rpm; N 8  224 rpm;

N 11  315 rpm

N 3  125 rpm; N 6  180 rpm; N 9  250 rpm;

N 12  355 rpm


Structural Formula Stage

Speed

Space

Z322

I

P1  3

X1  1

12  P 1, P 2, P 3

II

P2  2

X 2  P1  3

P3  2

X 3  P 1  P2

 P 1X 1, P 2X 2, P3X3 12  3 1, st

1 stage 2

2 3, nd

stage 3

III

326

2 6 rd

stage

u  No. of stages  3  D raw u  1  3  1  4 vertical

lines. Since 12 speeds, draw 12 horizontal lines.

Procedure for the construction of Ray diagram 3rd Stage: 2 speeds, 6 spaces. At 100 rpm (min), Locate Point (A) Leave six intervals. At 200 rpm - mark point (B). i.e., Point (B) represents N max  200 rpm , Point (A) represents N min  100 rpm Locate input speed at Point (C) and check: N max N input

2

or N min N input

Select

N input  160

rpm

and

mark it as (C) 200  1.25 which is less than 2 160



1  0.25 4

Join CA and CB

 Selection of N input  160 rpm

is satisfactory.

4.18


3 (1)

2 (3) RAY DIA GR AM

2 (6)


2nd Stage: 2 speeds, 3 spaces. Locate Point (D) above Point (C) after 3 spaces. Join DA 3 and DB 3 To find N input for 2nd stage. N max  224 rpm N min  160 rpm

Select N input  200 rpm N max N input N min N input

224  1.12 200

2



1 160  0.8 ; 4 200

which

is

less

than 2  Selection of N input  200 is satisfactory.

Locate Point (E) at 200 rpm. (Note: Selection of N input  180 rpm is also correct.) 1st Stage: There are 3 speeds with 1 space interval.  Mark (F) and (G) with 1 space interval from E .

Locate (H) as Input of Motor For first stage, it is not necessary to satisfy the condition.   N max 2    Ninput

Join H and F From (F), draw (2) speeds

4.20


FC 1 and FD 1 180 rpm 250 rpm (Maintain again 3 space intervals in II stage) From C1 draw 2 speeds. (112 rpm, and 224 rpm) C1A 1 and C 1B 1

112 rpm

(Maintain again 6 space intervals in III stage)

224 rpm

Join H and G From (G) draw 2 speeds (200 rpm and 280 rpm) GC 2 and GD 2 280 rpm 200 rpm

(Maintain again 3 space intervals in II stage) From C2 draw 2 speeds (125 rpm and 250 rpm)

C2A 2 and C 2B 2

125 rpm


250 rpm

From D: Draw (2) speeds (140 rpm and 280 rpm) DA3 and DB 3 280 rpm 140 rpm


From D1: Draw (2) speeds (160 rpm and 315 rpm) D1A4 and D 1B 4

160 rpm

315 rpm

(6 space intervals)

From D2: Draw (2) speeds (180 rpm and 355 rpm)


D2A5 and D 2B 5

180 rpm

(6 space intervals)

355 rpm

Kinematic Diagram No. of Gears required are  23  2  2  27

For details, refer page 4.13. Table 4.2 Calculation of Number of Teeth: Refer kinematic arrangement III Stage  14 Gears

1st Pair (Reducing speed) Speed of driven N 1 Z 2   Speed of driver N2 Z1 Speed of driver  N  160 rpm 2

Let Z 2  20 teeth

Speed of driven  N1  100 rpm (Assume above 17 teeth.)

100 20 1st Pair:  Z1  32 teeth  160 Z 1 Z2  20 teeth 20  160 Z1   32 teeth 100

2nd Pair: Increasing speed N 3  driven speed  200 rpm N 4  driver speed  160 rpm Speed of driven N 3 Z 4   Speed of driver N 4 Z 3

... (1)

Z 1  Z 2  Z 3  Z 4 because centre distance is same. 32  20  Z 3  Z 4 Z3  Z4  52

... (2)

4.22


3 (1)

2 (3) RAY DIA GR AM

2 (6)


4.24


200 Z 4  Z 4  1.25 Z 3  160 Z 3 Z 3  1.25 Z3  52 Z3 

52  23 2.25

Take Z3  23 teeth

2nd Pair Z3  23 teeth

Z 4  52  23  29 teeth

Z4  29 teeth

II Stage 1st Pair: Reducing speed Speed of driven N 7 Z8   Speed of driver N 8 Z 7 N8  driver speed  200 rpm

160 20  200 Z 7 Z7 

N7  driven speed  160 rpm

Assume Z 8  20 teeth

20  200 ; Z 7  25 teeth 160

1st Pair: Z7  25 teeth; Z8  20 teeth 2nd Pair: Increasing speed.

N 6  driver speed  200 rpm N 5  driven speed  224 rpm

Speed of driven N 5 Z 6   Speed of driver N 6 Z 5 Z 7  Z 8  Z 5  Z 6, b ecau se centre dis tance is same 25  20  Z 5  Z 6 Z5  Z6  45

... (1)


224 Z 6  200 Z 5 Z 6  Z 5  1.12 Z 5  1.12Z 5  45 Z5 

45  21.22 2.12

Take Z 5  21 teeth

2nd Pair: Z5  21 teeth

Z 6  45  21

Z6  24 teeth

 24 teeth

I Stage In this stage, there are 3 speeds; and 3 pairs of gears are required. 

To avoid interference of gear of one shaft with the gear of the other shaft while shifting, Z 1  Z 3  9  4 also Z7  Z5  4 Z 4  Z 2  9  4; Z 6  Z8  4;

1st Pair: Reducing speed

condition satisfied

N 10  driver speed  355 rpm N 9  driven speed  200 rpm

N9 Z 10 Speed of driven   Z9 Speed of driver N 10 

Assume Z 10  20 teeth

200 20  355 Z 9 Z9 

Take Z9  36 teeth

355  20  35.5 200

4.26


1st Pair: Z9  36 teeth Z10  20 teeth

2nd Pair: Intermediate speed Spee d of driven N 13 Z 14   Speed of driver N 14 Z 13 224 Z 14  355 Z13

Since centre distance is same, Z 9  Z 10  Z 13  Z 14  36  20  56 Z14  Z13  0.63  Z 13  0.63Z 13  56 Z 13 

56  34 teeth 1.63

Z 13  34 teeth Z 14  56  34  22 teeth

3rd Pair: Increasing speed N 11  driven speed  250 rpm N 12  driver speed  355 rpm Speed of driven N 11 Z 12   Speed of driver N 12 Z 11 250 Z 12  355 Z 11 Z12  Z11  0.704


Since centre sitance is same, Z 11  Z12  Z 13  Z 14  56 Z 11  Z12  56 Z 11  0.704 Z 11  56 Z 11  32.85  T ake Z 11  33 teeth

3rd Pair: Z11  33 teeth Z12  23 teeth

Z 12  56  33  23 teeth  No. of teeth on all the gears are given here:

1st stage Z 9  36 teeth 

Z 12  23 teeth 

Z 10  20 teeth 

Z 13  34 teeth 

Z 11  33 teeth 

Z 14  22 teeth 

2nd Stage

3rd Stage

Z 5  21 teeth

Z 1  32 teeth

Z 6  24 teeth

Z 2  20 teeth

Z 7  25 teeth

Z 3  23 teeth

Z 8  20 teeth

Z 4  29 teeth

Problem 4.2: The maximum and minimum speeds of a 6 speed gear box are to be 1600 rpm and 500 rpm respectively. Construct the speed diagram and the kinematic arrangement of the gear box.

(Apr. 2008)

4.28


Given Data: No. of speeds  Z  6 Max. speed

 N max  1600 rpm

Min. speed

 N min  500 rpm

N max N min

 Z  1

1600  6  1 500  1600     500 

1 5

 1.2619

  1.2619 is not a standard step ratio.

Therefore, Trail (1): for R40 series,   1.06   1.06   1.06  1.06  1.06  1.2624 Skip 3  speeds

Skip (3) speeds and try R40 series. N 1  500 rpm;

N 4  1000 rpm

N 2  630 rpm;

N 5  1250 rpm

N 3  800 rpm;

N 6  1600 rpm

Since N 6  1600 rpm is the max. required speed,  Select the above speeds. (i.e., R 40 series

skipping 3 speeds).

by



4.30

Structural Formula Z32 P1

P2

Z  P 1X 1, P 2X2

 3 1, 1

st

Stage

2 3

nd

2

Stage

P1  3

X1  1

P2  2

X 2  P1  3

st

In 1 stage 3 represents - No. of speeds 1 represents - space between speeds In the 2nd stage, 2 represents - No. of speeds 3 represents space between speeds

Ray Diagram: No. of stages 2; so 3 vertical lines No. of speeds 6; so 6 horizontal lines Mark N 1  500 to N 6  1000 rpm in Ray diagram. Take N max  1000 rpm and mark it as (B) N min  500 rpm and mark it as (A)

Here space between A and B is 3 and no. of speeds is 2 ie A and B. Either

Nmax Nip

 2 or

Nmin Nip



1 4

For 1st stage, the above condition need not be checked. Select N input as N max for 1st stage. For 2nd Stage Select N input  800 rpm N max  1000 rpm


1000  1.25  2 800  satisfies condition of mark N input  800 rpm

as C. Mark D and E so that CD and DE has 1 space each. Join DA 1 parallel to CA and Join DB 1 parallel to CB Similarly, join EA 2 parallel to CA and join EB 2 parallel to CB Similarly A1, A 2, B 1, B 2 are also marked. Approximate No. of Gears Required 6 speeds - (3  2 ) = 23  2  10 gears For 1st stage No. of speeds required are ‘3’  No. of gears required are ‘6’.

For 2nd stage

Refer Pg.No.4.13 for finding no. of gear

No. of speeds required are ‘2’  No. of gears required are ‘4’.

Calculation of No. of Teeth on Gear Refer Kinematic Diagram 2nd Stage 1st pair Z 1 and Z 2 - Reducing speed

4.32


4


8

7 6

*

5

* *

9

10

1

2

*

3

*

4.34


2nd pair Z 3 and Z 4 - Increasing speed Assume Z 1  20 teeth driver  1st Pair: (Reducing speed) Z 1  No. of teeth on driver  20 teeth N 1  Speed of driver  800 rpm Z 2  No. of teeth on driven N 2  Speed of driven  500 rpm Z2

Z1



N1

N2

No. of teeth on driven Sp eed o f driver  No. of teeth o n driver Speed of driven Z2 20



800 500

Z 2  32 teeth Z1  20 teeth Z2  32 teeth

2nd Pair: (Increasing speed) Z 3  No. of teeth on driver N 3  Speed of driver  800 rpm Z 4  No. of teeth on driven N 4  Speed of driven  1000 rpm

Since centre distance is same,


Z 1  Z 2  Z 3  Z 4  20  32  52  Z 3  Z 4  52

... (1)

No. of teeth on driven Speed of driver  No. of teeth on driver Speed of driven Z4 Z3



800  0.8 1000

Z 4  0.8Z 3

... (2)

Substitute the value of Z 4 in eq. (1) Z 3  0.8Z 3  52 Z3 

52  28.88 1.8

Z 3  29 teeth

But Z 3  Z 4  52  Z 4  52  29

N14  driver speed  355 rpm N13  driven speed  224 rpm

 23 teeth Z3  29 teeth Z4  23 teeth

1st Stage 1st Pair: Z 5 and Z 6 - Reducing speed 2nd Pair: Z 7 and Z 8 - Increasing speed 3rd Pair: Z 9 and Z 10  - Intermediate speed.

4.36


1st Pair: Reducing speed. Z 5  No. of teeth on driver  20 (Assume) N 5  Speed of driver  1600 rpm Z 6  No. of teeth on driven N 6  Speed of driven  800 rpm Z6

Z5



N5

N6

Z 6  20 

1600  40 800

Z5  20 teeth Z6  40 teeth

2nd Pair: Increasing speed. Z 7  No. of teeth on driver Z 8  No. of teeth on driven N 7  Speed of driver  1600 rpm N 8  Speed of driven  1250 rpm

Since centre distance is same, Z 5  Z 6  Z 7  Z 8  60  Z 7  Z 8  60 Z8 Z7



N7 N8

... (3)


 Z8 

1600 Z 1250 7

 1.28Z 7

... (4)

Substitute the value of Z 8 in eq. (3) Z 7  1.28 Z7  60 Z7 

60  26.31 2.28

Z7  27 teeth

Take

But Z 7  Z 8  60  Z 8  60  27  33 teeth Z7  27 teeth Z8  33 teeth

3rd Pair: Intermediate speed Z 9  No. of teeth on driver Z 10  No. of teeth on driven N 9  Speed of driver  1600 rpm N 11  Speed of driven  1000 rpm

Since centre distance is same, Z 5  Z 6  Z7  Z8  Z 9  Z 10  60  Z 9  Z 10  60

But,

Z 10 Z9



N9 N 10

1600 Z  Z 10  1000 9

... (5)

4.38


 1.6 Z 9

Substitute the value of Z 10 in eq. (5) Z 9  1.6 Z 9  60 Z9 

Take

60  23.07 2.6

Z 9  23 teeth

 Z 10  60  23  37 teeth Z9  23 teeth Z10  37 teeth

No. of teeth on each gear II Stage

I Stage

Z 1  20

Z 5  20

Z 2  32

Z 6  40

Z 3  29

Z 7  27

Z 4  23

Z 8  33 Z 9  23 Z 10  37

Check To avoid interference Z 7  Z 5 and Z 7  Z9  4 27  20  7 and 27  23  4

... (6)


The condition is satisfied. Therefore, the calculation of No. of teeth on each gear is satisfactory. Problem 4.3: Design a 9 speed gear box for a milling machine with speeds ranging from 56 - 900 rpm. The input speed is 720 rpm; Make a neat sketch of the gear box. Indicate the No. of teeth on all the gears and their speeds. (Apr. 2008)

Given Data: No. of speeds  Z  9 Max. speed  N max  900 rpm Min. speed  N min  56 rpm Input speed  720 rpm N max N min

 Z  1

900  9  1 56 1 8

 900    56   1.415

  1.415 is not a standard step ratio.

Trail 1 Consider R20 series, which is having   Step ratio  1.12.  1.12   1.12  1.12  1.40 Skip 2 speeds

Try R20 series by skipping (2) speeds.

4.40


C2

N 9 = 9 00 rp m

In pu t (N ) =7 20 rp m C1

F

N 7 = 4 50 rp m

C

E

B2

D

B1

B

A2

A1

3 (1)

3 (3) Ray D iagram fo r Nine S peed Gear Box

N 8 = 6 30 rp m

A

N 6 = 3 15 rp m

N 5 = 2 24 rp m

N 4 = 1 60 rp m

N 3 = 112 rpm

N 2 = 8 0 rpm

N 1 = 5 6 rpm

10

12

11

9

7

8

6

5

4

3

1

2



4.42

N 1  56 rpm;

N 4  160 rpm;

N 7  450 rpm

N 2  80 rpm;

N 5  224 rpm;

N 8  630 rpm

N 3  112 rpm;

N 6  315 rpm;

N 9  900 rpm

Since N 9  900 rpm is the maximum required speed, therefore select the above speeds from N 1 to N9, according to R20 series by skipping 2 speeds. Structural Formula Z33

P1  3

P1 P2

Z  P 1X 1, P 2X2

 31, 1

st

Stage

33 2

nd

Stage

st

P2  3

In 1 stage, 3 represents 1 represents In 2nd stage, 3 represents 3 represents

X1  1 X2  P1  3

No. of speeds space between speeds no of speeds space between speeds

Ray Diagram Mark N 1 to N 9 in Ray diagram. No. of stages is 2, so draw 3 vertical lines. No. of speeds is 9, so draw 9 horizontal lines. For 2nd stage, Mark A in N 1  56 rpm. Mark B in N 4  160 rpm after leaving 3 spaces. Similarly mark C in N 7  450 rpm after leaving 3 spaces.


Select input speed

N max N i p

Select N ip  224 rpm



2

and

450 2 224

N min N ip



1 4

56  0.25 224

Since condition is satisfying, Select N ip  224 rpm. So mark D accordingly. Join AD, BD and CD For 1st stage, no need to check the condition. Then mark E and F after leaving 1 space in between D, E and F.

Join D, E and F with input speed 720 rpm. Joint EA 1 (Parallel to DA ) EC 1 (Parallel to DC ) - Join FA 2 (Parallel to DA ) and FC 2 (parallel to DC )

Approximate No. of Gears Required Refer Page 4.13 for details 9 speed  3  3  23  3  12 gears For 1st stage, No. of speeds required are 3.  No. of gears required are 6.

For 2nd stage, No. of speeds required are 3.  No. of gears required are 6.

4.44


Calculation of No. of Teeth on Gear Refer kinematic diagram.. (Here after words, to calculate no. of teeth on all gears, the kinematic diagram should be referred. II Stage 1st Pair - Z 1 and Z 2 - Reducing speed 2nd Pair - Z 3 and Z 4 - Increasing speed 3rd Pair - Z 5 and Z 6 - Intermediate speed. 1st Pair: Reducing speed Z 1 and Z 2 Z 1  No. of teeth on driver  20 teeth (Assume) Z 2  No. of teeth on driven N 1  Speed of the driver  224 rpm N 2  Speed of driven  56 rpm No. of teeth on driven Speed of the driver  No. of teeth on driver Speed of the driven Z2

Z1



N1 N2

Z 2  20 

224  80 56

 Z1  20 teeth Z2  80 teeth

2nd Pair: Increasing speed Z 3 and Z 4 Z 3  No. of teeth on driver


Z 4  No. of teeth on driven N 3  Speed of driver  224 rpm N 4  Speed of driven  450 rpm Z4

Z3

 

N3

N4 224  0.4977 450

Z 4  0.4977Z 3

Since centre distance is same, Z 1  Z 2  Z3  Z 4  20  80  100 Z 3  Z 4  100 Z 3  0.4977 Z 3  100 Z3 

100 1.4977

 66.76

Take

Z 3  67 teeth  Z 4  100  67  33 teeth Z3  67 teeth Z4  33 teeth

3rd Pair: Intermediate speed. Z 5 and Z 6 Z 5  No. of teeth on driver Z 6  No. of teeth on driven

... (1)

4.46


N 5  Speed of driver  224 rpm N 6  Speed of driven  160 rpm Z6

Z5 Z6

Z5



N5 N6



224 160

 1.4

Z 6  1.4Z 5

 Z 5  Z 6  100

 3

 4

Since the centre distance is same, Z 1  Z 2  Z 3  Z 4  Z 5  Z 6  20  80  100

From eq. (3) and (4) Z 5  1.4Z 5  100 Z5 

Take

100  41.66 2.4

Z 5  42 teeth  Z 6  100  42  58 teeth Z5  42 teeth Z6  58 teeth

2nd Stage Z 1  20 teeth

To check interference Z 1, Z 3, Z5 are driver.

Z 2  80 teeth

Z 3  Z1 

Z 3  67 teeth

and 4 Z 3  Z5 

Z 4  33 teeth

 


Z 5  42 teeth

67  20  47  4 67  42  25 

Z 6  58 teeth

 There is no interference.

1st Stage 1st Pair - Z 7 and Z 8 - Reducing speed 2nd Pair - Z 9 and Z 10  - Increasing speed 3rd Pair - Z 11 and Z 12  - Intermediate speed. 1st Pair: Z 7 and Z 8 - Reducing speed. Z 7  No. of teeth on driver  20 teeth (Assume) Z 8  No. of teeth on driven N 7  Speed of driver  720 rpm N 8  Speed of driven  224 rpm No. of te eth on driven Speed of driver  No . of teeth on driver Speed of driven Z8

Z7



N7 N8

Z 8  20 

720  64.28 224

Z 8  65 teeth Z7  20 teeth Z8  65 teeth

4.48


2nd Pair: Z 9 and Z 10  - increasing speed. Z 9  No. of teeth on driver Z 10  No. of teeth on driven N 9  Speed of driver  720 rpm N 10  Speed of driven  450 rpm

Since the centre distance is same, Z 7  Z 8  Z 9  Z 10  20  65  85  Z 9  Z 10  85 Z 10

But,

Z9



N9 N 10



720  1.6 450

 Z 10  1.6 Z 9

From eq. (5) and (6) Z 9  1.6Z 9  85 Z9 

Take

... (5)

85  32.69 2.6

Z 9  33 teeth

 Z 10  85  33  52 teeth Z9  33 teeth Z10  52 teeth

3rd Pair: Z 11 and Z 12  - intermediate speed. Z 11  No. of teeth on driver Z 12  No. of teeth on driven N 11  Speed of driver  720 rpm N 12  Speed of driven  315 rpm

... (6)


Since centre distance is same, Z 7  Z 8  Z 9  Z 10  Z 11  Z 12  85  Z 11  Z 12  85

But,

Z 12 Z 11



N 11 N 12



... (7)

720  2.285 315

Z 12  2.285 Z 11

... (8)

From eq. (7) and (8) Z 11  2.285 Z 11  85 Z 11 

Take

85  25.87 3.285

Z 11  26 teeth  Z 12  85  26  59 teeth Z11  26 teeth Z12  59 teeth

1st Stage Z 7  20 teeth Z 8  65 teeth

Similarly, Z 7, Z 9, Z11 are driver gears.

Z 9  32 teeth Z 10  52 teeth Z 11  26 teeth Z 12  59 teeth

Check for interferen ce, Z9  Z7  4 Z 9  Z 11  32  20  12   4 Since both 32  26  6  are greater than four.  There is no interference.

4.50


Problem 4.4: A gear box is to give 18 speeds for a spindle of a milling machine. The drive is from an electric motor of 3.75 kW at 1440 rpm. Maximum and minimum speeds of the spindle are to be around 650 rpm and 35 rpm respectively. Find the speed ratios which will give the desired speeds and draw the structural diagram and kinematic arrangement of the drive.

(Apr. 2007)

Solution: N max N min

 Z  1

650  18  1 35  650  Step ratio      35 

1 17

N max  650 rpm N min  35 rp m Z  18

 1.1875

Trail 1 R40 series   1.06  1.06   1.06  1.06

skip 2 speeds

 1.191

which is approximately equal to the required step ratio.  Skip two speeds


N 1  35.5 rpm

N 9  140 rpm

N 17  560 rpm

N 2  42.5 rpm

N 10  170 rpm

N 18  670 rpm

N 3  50 rpm

N 11  200 rpm

Since N 18 is closer to the

N 4  60 rpm

N 12  236 rpm

required N max ,

N 5  71 rpm

N 13  280 rpm

N 6  85 rpm

N 14  335 rpm

N 7  100 rpm

N 15  400 rpm

N 8  118 rpm

N 16  475 rpm

Structural formula, No. of speeds required  Z  18 332 P1

P2

P3

Speeds

Space intervals

P1  3

X1  1

P2  3

X 2  P1  3

P3  2

X 3  P1  P 2  3  3  9

 Structural formula  P 1X 1, P 2X 2, P 3X3 

33,

31 , 1

st

Stage

nd

2

Construction of Ray Diagram For Third Stage To find N input , Let N ip  100 rpm

and

2 9

rd

Stage 3

N min N i/p



Stage

1 4

35.5  0.355  0.25 100

4.52

I


In pu t

N 1 8 = 67 0 N 1 7 = 56 0 N 1 6 = 47 5 N 1 5 = 40 0 N 1 4 = 33 5 E

N 1 3 = 28 0

H

N 1 2 = 23 6

G

N 11 = 200

F

B N 1 0 = 17 0

D

N 9 = 1 40 N 8 = 118 N 7 = 1 00

C

N6 = 85 N5 = 71 N4 = 60 N3 = 50 N 2 = 4 2.5 1 st Sta ge . 3 (1)

2 nd S tage . 3 (3)

3 rd Sta ge . 2 (9)

Ray D iagram fo r 18 Speed Gear Box.

A

N 1 = 3 5.5

(236 rpm )

Z 16 =1 08

14

13 11

Z 1 4 =10 4 (2 80 rpm )

15

8

9 7

5 6

D riven

D river (200 rpm )

Z 6 = 40 (100 rp m)

Z 7 =35 Z 5 =2 0

Z 8 = 25 Z 10 = 32 (280 rpm ) (170 rpm )

10

Z 9 =28

II S tage

Z1 = 20

Z 4 = 28 (170 rpm )

D rive n

Z 3 =4 9

Z 2 = 57 (35.5 rpm )

D river (100 rpm )

III Stag e

* *

Kinem atic Arrangement of 18 Speed Gear Bo x.

Z 1 2 = 114 (236 rpm )

12

Z 11 =3 4

* * *

D riven

16

Z 13 =44

* * *

I Stage

D rive r (67 0 rpm )

Z 15 =40


4.54


Nmax

Nip N max

Nip

2



170  1.7  2 100

From 100 rpm, leave 9 spaces and mark N max  170 rpm

 Select N ip as 100 rpm; and mark it as (C)

Take 9th speed as N max  170 rpm and mark as (B)

Take the 1st speed N min  35.5 rpm and mark as (A)

For Second Stage To find N input , Let N ip  200 rpm N max

Ni  p



280  1.4  2 200

and

N min N i/p

100  0.5  0.25 200

Mark N max as (E)  Select N ip as 200 rpm; and mark it as (F). Similarly mark (D) at the midpoint of (E) and (C). (or)

Mark D after leaving 3 spaces from C and mark E after leaving 3 spaces from D . Approximate No. of Gears required 18 speeds 3  3  2  23  3  2  16 gears For 1st Stage No. of speeds required are 3,


 No. of gears required  6; [i.e. 2  3  6]

For details, refer page 4.13 table 4.2 For 2nd Stage No. of speeds required are 3,  No. of gears required  6; [i.e. 2  3  6]

For 3rd Stage No. of speeds required are 2,  No. of gears required  4; [i.e. 2  2  4]

Calculation of No. of Teeth on Each Gear Refer Kinematic Arrangement For III Stage 1st Pair Z 1 and Z 2 - Reducing speed. 2nd Pair Z 3 and Z 4 - Increasing speed. 1st Pair: Reducing speed. Z 1 and Z 2 Z 1  No. of teeth on driver  20 teeth (Assume) N 1  speed of driver  100 rpm Z 2  No. of teeth on driven N 2  speed of driven  35.5 rpm Speed ratio 

Z2 Z1



Z 2  20 

N1 N2 100  56.3 35.5

Z1  20 teeth

4.56


 Take Z 2  57 tee th.

Z2  57 teeth

2nd Pair: Increasing speed Z3 and Z4 Z 3  No. of teeth on driver Z 4  No. of teeth on driven N 3  Speed of driver  100 rpm N 4  Speed of driven  170 rpm

Since the centre distance is same, Z 1  Z 2  Z 3  Z4  57  20 Z 3  Z 4  77

But, speed ratio  Z4 Z3



Z4 Z3

... (1) 

N3 N4

100  0.5882 170

Z 4  0.5882 Z 3

From eq. (1) and (2) Z 3  0.5882 Z 3  77 Z3 

77  48.48 1.5882

Take Z 3  49 teeth  Z 4  77  49 28 Z3  49 teeth Z4  28 teeth

... (2)


For II Stage 3 speeds - 3 pairs of gears. 1st Pair - Z 5 and Z 6 - Reducing speed 2nd Pair - Z 7 and Z 8 - Increasing speed 3rd Pair - Z 9 and Z 10 - Intermediate speed. 1st Pair Z5 and Z6: Reducing speed. Z 5  No. of teeth on driver  20 teeth (Assume) Z 6  No. of teeth on driven N 5  Speed of driver  200 rpm N 6  Speed on driven  100 rpm. Speed ratio 

Z6

Z5



N5 N6



200 2 100

Z 6  20  2  40 teeth Z5  20 teeth  Z6  40 teeth

2nd Pair Z7 and Z8: Increasing speed. Z 7  No. of teeth on driver Z 8  No. of teeth on driven N 7  Speed of driver  200 rpm N 8  Speed of driven  280 rpm

Since the centre distance is same. Z 5  Z 6  Z 7  Z8  20  40  60 Z 7  Z 8  60

... (3)

4.58


But, speed ratio 

Z8 Z7



N7 N8



200  0.7142 280

Z 8  0.7142 Z 7

... (4)

From eq. (3) and (4) Z 7  0.7142 Z 7  60 Z7 

Take

60  35.001 1.7142

Z 7  35 teeth Z 8  60  35  25 Z7  35 teeth Z8  25 teeth

3rd Pair Z9 and Z10: Intermediate speed. Z 9  No. of teeth on driver Z 10  No. of teeth on driven N 9  Speed of driver  200 rpm N 10  Speed of driven  170 rpm

Since the centre distance is same, Z 9  Z 10  Z 5  Z 6  Z 7  Z 8  20  40  60 Z 9  Z 10  60

Speed ratio,

... (5) Z 10 Z9



N9 N 10



200  1.176 170

Z 10  1.176 Z9

From eq. (5) and (6)

... (6)


Z 9  1.176 Z9  60 Z9 

Take,

60  27.57 2.176

Z 9  28 teeth Z 10  60  28  32 teeth Z9  28 teeth Z10  32 teeth

To avoid interference, Z 5, Z 7 and Z 9 are driver gears with Z 5  20 teeth Z 7  35 teeth and Z 9  28 teeth

Condition Z7  Z5  4 and Z 7  Z 9  4

35  20  15 35  28  7

Both are greater than 4.

 There is no interference between gears. Z5  20 teeth;

Z7  35 teeth;

Z9  28 teeth

Z6  40 teeth;

Z8  25 teeth;

Z10  32 teeth

I Stage In first stage, the No. of speeds required are 3. Therefore, six gears are required. i.e., 3 pairs of gears. 1st Pair: Z 11 and Z 12 - Reducing speed 2nd Pair: Z 13 and Z 14 - Increasing speed

4.60


3rd Pair: Z 15 and Z 16 - Intermediate speed 1st Pair: Z 11 and Z 12 - Reducing speed. Z 11  No. of teeth on driver  17 teeth (Assume) Z 12  No. of teeth on driven N 11  speed of driver  670 rpm N 12  speed of driven  200 rpm

Speed ratio 

Z 12 Z 11



N 11 N 12



670 200

Z 12  17 

670  56.95  57 teeth 200

Take Z12  57 teeth and Z11  17 teeth

2nd Pair: Z 13 and Z 14 - Increasing speed. Z 13  No. of teeth on driver Z 14  No. of teeth on driven N 13  Speed of driver  670 rpm N 14  Speed of driven  280 rpm

Since the centre distance is same, Z 13  Z 14  Z 11  Z 12  17  57  74 Z 13  Z 14  74


... (7) Z 14 Z 13



N13 N 14



670 280

Z 14  2.3928Z 13

... (8)


From eq. (7) and (8) Z 13  2.3928Z 13  74 Z 13 

74  21.81 3.3928

Take

Z 13  22 teeth

and

Z 14  74  22  52 teeth Z13  22 teeth Z14  52 teeth

3rd Pair: Z 15 and Z 16 - Intermediate speed. Z 15  No. of teeth on driver Z 16  No. of teeth on driven N 15  Speed of driver  670 rpm N 16  Speed of driven  236 rpm

Since the centre distance is same, Z 15  Z 16  Z 11  Z 12  Z 13  Z 14  74 Z 15  Z 16  74


... (9) Z 16 Z 15



N15 N 16



670 236

Z 16  2.8389Z 15

From eq. (9) and (10) Z 15  2.8389Z 15  74

... (10)

4.62


Z 15 

Take

74  19.27 3.8389

Z 15  20 teeth

From equation (1) Z 16  74  20  54 teeth Z15  20 teeth Z16  54 teeth

To Avoid Interference Z 13  Z 11  4; and Z 13  Z 15  4;

i.e., 22  17  5 i.e.,22  20  2

Since the difference between No. of teeth on Gear 13 and Gear 15 are less than 4, then multiply all the gears in the 1st stage by 2. Modify

the

No.

of

teeth

Z 11 to Z16), multiplying by 2 Z 11  17  2  34 teeth Z 12  57  2  114 teeth Z 13  22  2  44 teeth Z 14  52  2  104 teeth Z 15  20  2  40 teeth Z 16  54  2  108 teeth

Check again,

on

each

gear

(from


Z 13  Z 11  44  34  10 Z 13  Z 15  44  40  4

Since

the

difference

between

Z 13 and Z 11 , Z 13 and Z 15 is greater than 4, our calculation is correct.

Therefore, the No. of teeth on each gear are given here. 3rd Stage Z 1  20 teeth Z 2  57 teeth Z 3  49 teeth Z 4  28 teeth

2nd Stage Z 5  20 teeth Z 6  40 teeth Z 7  35 teeth

1st Stage Z 11  34 teeth Z 12  104 teeth Z 13  44 teeth

Z 8  25 teeth

Z 14  104 teeth

Z 9  28 teeth

Z 15  40 teeth

Z 10  32 teeth

Z 16  108 teeth

Problem 4.5: Design a 9 speed gear box to give output speeds between 280 and 1800 rpm. The input power is 5.5 kW at 1400 rpm. Draw the kinematic layout diagram and the speed diagram. Determine the number of teeth on all gears.

4.64


Given Data N min  280 rpm ;

N max  1800 rpm ;

P ower  5.5 kW;

Z  9 speed

Solution We know that N max N min 

 Z  1

1800  9  1 280    1.26

Since the obtained  is not a standard speed ratio take   1.06 R40 series  1.06  1.06  1.06  1.06  1.26 Skip 3 speeds, which is approximately equal to the required  value.  Try R40 series by skipping 3 speeds; From Pg.No. 7.20 N 1  280 rpm; N 2  355 rpm, N 3  450 rpm; N 4  560 rpm; N 5  710 rpm, N 6  900 rpm; N 7  1120 rpm; N 8  1400 rpm; N 9  1800 rpm

Structural formula Z33 ;

P1  3 ;

X1  1

9  P 1, P 2 ;

P2  3 ;

X2  P1  3

 P1 X 1, P 2 X 2 9  3 1, 3 3

Since number of stages u  2 ,  Draw u  1  2  1  3 vertic al lines

Sinc e 9 speeds, draw 9 ho rizo ntal lines


Procedure for constructing speed diagram At 280 rpm (min), Locate A. Leave 3 intervals at 560 rpm mark B, again Leave 3 intervals at 1120 rpm, mark C Locate input speed at point D N max N input

2

Selecting N input  710 rpm N max N input 



N m ax N input

1120  1.57  2 710 

280 1  0.39  710 4

After locating point ‘D’, at 900 mark E and again interval at 1120, mark F. Select Input speed as 1400 rpm. Calculation of number of Teeth II stage I pair Speed of driven N 11 Z 12   Speed of driver N12 Z 11

Assume

Z11  20 teeth Z 12



20



710  52 teeth  Z12  50.71 ~ 280

II stage II pair Since

Z7 Z8



N8 N7



560  0.788 710

4.66


Speed D iag ram or Ray D iagram of 9 speed gear bo x


4.68


also Z 7  0.788 Z8 Since Z 7  Z 8  Z 11  Z12 0.788 Z 8  Z 8  20  52  Z8  40 teeth  Z7  32 teeth

II stage, III pair Z9

Since Also

Z 10



N 10 N9



Z9 1120  1.57 Z 9  1.57 Z 10  Z 10 710

Z 9  Z 10  Z11  Z12 1.57 Z 10  Z 10  20  52 Z10  28 teeth and Z9  44 teeth

I stage, I pair Assume Z5  20 teeth Since

Z5 Z6



N6 N5



710  0.507  Z6  40 teeth 1400



900  0.64  Z1  0.64 Z 2 1400

I stage II pair Since also

Z1 Z2



N1

N2

Z 1  Z 2  Z 5  Z6 0.64 Z 2  Z 2  20  40 ; Z2  37 teeth

and

Z1  24 teeth


I stage III pair Since

Z3

Z4



N4 N3



1120  Z3  0.8 Z 4 1400

But

Z 3  Z 4  Z 5  Z6 0.8 Z 4  Z 4  20  40 Z4  34 teeth a nd Z3  28 teeth

Problem 4.6: The maximum and minimum speeds of nine speed gear box are to be 600 rpm and 100 rpm respectively. The drive is from an electric motor giving 3 kW at 1440 rpm. Design the gear box. Construct the speed diagram and sketch the arrangement of gear box.

Solution N max  600 rpm, Nmin  100 rpm; P  3 kW ; Z  9, N input  1440 rpm

We know that N max N min

 Z  1

600  9  1 100 8  6    6 1/8   1.25

The progression ratio   1.25 , coincides with the R10 series standard progression ratio

4.70


Therefore select R10 series speeds N 1  100 rpm; N 2  125 rpm; N 3  160 rpm; N 4  200 rpm; N 5  250 rpm; N 6  315 rpm; N 7  400 rpm; N 8  500 rpm; N 9  600 rpm;

Structural formula Z33 9  P 1, P 2  P 1 X 1, P 2 X2 P 1  3, X 1  1 P 2  3, X 2  3  9  3 1, 3 3 u  no . o f stages  2  draw u  1  2  1  3 vertical lines

since 9 speeds, draw 9 horizontal lines Procedure for the construction of ray diagram 2nd Stage: 3 speeds, 3 spaces At 100 rpm (min), locate point A Leave 3 intervals, at 200 rpm, mark point B Again leave 3 intervals, at 400 rpm, mark point C Point

“C”

represents

represents N min  100 rpm

N max  400 rpm ,

Point

“A”


Let N input  250 rpm. N max N input

Select

N input  250 

and

2

400  1.6  2 250

N min N input 



1  0.25 4

100  0.4  0.25 250

 Selection of N input  250 is satisfactory and mark it as D , similarly mark E and F above

Kinematic diagram: No. of gears required  2 3  3  12 Calculation of No. of Teeth on gear II stage 1st pair - (Z 1 and Z 2) - Reducing speed 2nd pair - (Z 3 and Z 4) - Increasing speed 3nd pair - (Z 5 and Z 6) - Intermediate speed II stage 1st pair reducing speed (Z 1 and Z 2) Let Z 1  No of teeth on driver  20 teeth Assume

4.72


Driver 600rpm

F ig. Kin em atic A rrang em ent

D riven

Driven

D river 250rpm


4.74


Z 2  No of teeth on driven N 1  Speed of the driver  250 rpm N 2  Speed of the driven  100 rpm No. of teeth on driven Speed of the driver  No. of teeth on driver Speed of the driven Z2 Z1



 Z2  

N1 N2 N1 N2

 Z1

250  20 100

Z 2  50  Z1  20 teeth, Z 2  50 teeth

2nd pair Increasing speed (Z 3 and Z 4) Z 3  no. of teeth on driver Z 4  no. of teeth on driven N 3  speed of driver  250 N 4  speed of driven  400 Z4 Z3



Z4 

N3 N4 N3 N4

 Z3


Z4 

250  Z  0.625 Z 3 400 3

Z 4  0.625 Z3

Since centre distance is same Z1  Z2  Z3  Z4  Z 3  Z 4  20  50  70 Z 3  Z 4  70  Z 3  0.625 Z 3  70  Z3  43.07  44 teeth

and Z4  26 teeth 3rd pair: Intermediate speed (Z 5 and Z 6) 

Z6 Z5



N5 N6



250 200

 Z 6  1.25 Z 5

Since centre distance is same Z 1  Z 2  Z3  Z 4  Z 5  Z 6  20  50  70  Z 5  Z 6  70  Z 5  1.25 Z 5  70 Z5  31.11  32 teeth Z 6  70  32  Z 6  38 teeth

4.76


II stage Z 1  20 teeth; Z 4  26 teeth Z 2  50 teeth; Z 5  32 teeth Z 3  44 teeth; Z 6  38 teeth

To check interference: Z 1, Z 3, Z 5 are driver

 Z3  Z1    4  a nd Z Z  5  3  44  20    and   4     44  32 

 There is no interference I stage 1st pair - (Z7 and Z8) - Reducing speed 2nd pair - (Z9 and Z10 ) - Increasing speed 3rd pair - (Z11 and {Z 12 ) -Intermediate speed 1st pair Z7  No . of teeth on driver  20 teeth Assume Z8  No. of teeth on driven N7  Speed of driver  600 rpm N8  Speed of driven  250 rpm


Z8 Z7



N7

N8

 Z8  20 

600  48 teeth 250

 Z7  20 teeth Z8  48 teeth 2nd pair Z9  No. of teeth on driver Z10  No. of teeth on driven N9  Speed of driver  600 rpm N10  Speed of driven  400 rpm Since the centre distance is same Z7  Z 8  Z 9  Z 10  20  48  68 teeth  Z 9  Z 10  68 teeth But Z10 Z9



 Z 10 

N9 N 10 600  Z 9  1.5 Z 9 400

Z10  1.5 Z 9  Z 9  1.5 Z9  68  Z9  27.2  28 teeth and Z10  40 teeth

4.78


3rd pair Z11  No. of teeth on driver Z12  No. of teeth on driven N11  Speed of driver  600 N12  Speed of driven  315 Since the centre distance is same Z7  Z 8  Z 9  Z 10  Z 11  Z 12  68  Z 11  Z 12  68 But Z12 Z 11



N 11 N 12



600 315

 Z 12  1.9 Z 11  Z 11  1.9 Z 11  68  Z11  23.44  24 teeth and

Z12  44 teeth

Checking of interference  Z9  Z7   and Z Z 11  9

  4  

 28  20   28  24   4    There is no interference


Problem 4.7: A machine tool gear box is to have 9 speeds. The gear box is driven by an electric motor whose shaft rotational speed is 1400 rpm. The gear box is connected to the motor by a belt drive. The maximum and minimum speeds required at the gear box output are 1000 rpm and 200 rpm respectively. Suitable speed reduction can also be provided in the belt drive. What is the step ratio and what are the values of 9 speeds? Sketch the arrangement., Obtain the number of teeth on each gear and also the actual output speeds.

Given Data No. of speeds  Z  9 Maximum speed  N max  1000 rpm Minimum speed  N min  200 rpm Solution We know N max N min

 Z  1

1000    1.22  8  200 which is not a standard step ratio Trail 1 Take R40 series multiples   1.06  1.06  1.06  1.06  1.06  1.26

which is greater than required  not satisfactory

(Nov/Dec 2007)

4.80


 To obtain speeds (approximate method) N 1  200 rpm N 2  N 1    200  1.22  244 rpm  298 rp m N 3  N 2    244  1.22  297.68 ~  364 rp m N 4  N 3    298  1.22  363.56 ~  444 rp m N 5  N 4    364  1.22  444.08 ~  542 rp m N 6  N 5    444  1.22  541.68 ~  662 rp m N 7  N 6    542  1.22  661.24 ~ N 8  N 7  662  1.22  807.64 ~  808 rpm N 9  N 8    808  1.22  985.76 ~  986 rp m

Structural formula Z33  P1  3 ; P2  3

and X1  1 ; X 2  P 1  3 Z  P 1 X 1, P 2 X 2  3 1, 3 3

Ray diagram Select input speed;

N max N input

Select N input  444 rp m

 2 and

N min

Ninput

4




N min 200 1 Nmax 662     2 and N inp ut 444 N input 444 4

since condition is satisfying Mark N input  444 rpm as D, and similarly mark E and F . Approximate number of gears required 9 speed  3  3  2 3  3  12 gears Calculation of No. of teeth on gear Let Z1  No. of teeth on driver  20 teeth Assume Z2  No. of teeth on driven N 1  Speed of the drive r  444 rpm N 2  Speed of driven  200 rpm

We know that Z2 Z1



N1 N2

N1 444  45 teeth  Z2  Z 1   20  N2 200  Z1  20 teeth, Z2  45 teeth

Similarly,

Z4 Z3



N3

N4

 444   Z4  Z 3    662   Z4  0.671 Z 3

Since centre distance is same

4.82


G

1400 rpm

986 rpm C2

808 rpm C1

F

C

B2 E

662 rpm

542 rpm

B1

D

444 rpm

364 rpm B A2 298 rpm

A1 244 rpm

3(1)

Ray D iag ram

3(3)

A

200 rpm

D riven

D riven


4.84


Z 1  Z 2  Z3  Z 4  Z 3  Z 4  20  45  Z 3  Z 4  65  Z 3  0.671 Z 3  65  Z3  39 teeth

and

Z4  26 teeth

Similarly, Z6 Z5



N5 N6



444 364

 Z6  Z 5 1.2197 

Since the centre distance is same Z 1  Z 2  Z 3  Z4  Z 5  Z6  20  45  65  Z 5  Z 6  65  Z 5  1.2197 Z 5  65  Z5  30 teeth

and Z 6  65  30  35 Z6  35 teeth

To check for interference Z 1, Z 3, Z 5 are driver


Z3  Z1   and 4 Z3  Z5  

 39  20   39  30   4    There is no interference Similarly, Z7  N o . of teeth on driver  20 teeth Assume Z8  No. of teeth on driven N7  Speed of driver  1400 rpm N8  Speed of driven  444 rpm 

Z8 Z7



 Z8 

N7 N8 N7 N8

 Z7  20 

1400 444

 64 teeth  Z7  20 teeth, Z8  64 teeth

Since the centre distance is same Z 7  Z 8  Z9  Z 10  20  64  84  Z 9  Z 10  84

But Z 10 Z9



N9 N 10



1400  2.11 662

... (1)

4.86


 Z 10  2.11 Z 9

... (2)

 Z 9  2.11 Z 9  84  Z9  27 teeth

and

Z10  57 teeth

Since centre distance is same Z 7  Z 8  Z 9  Z10  Z 11  Z 12  85  Z 11  Z 12  85 Z 12

But,

Z 11



N 11 N 12



1400  2.583 542

 Z 12  2.583 Z 11  Z 11  2.583 Z 11  84  Z11  23 teeth

and

Z12  61 teeth

Check of interference  Z9  Z7  4   Z 11   Z9  27  20   27  23   4    Satisfactory, there is no interference Problem 4.8: Design a nine-speed gear box for a machine to provide speeds ranging from 100 to 1500 rpm. The input is from a motor of 5 kW at 1440 rpm. Assume any alloy steel for the gear.

(May/June 2007)


Given Data No. of speeds  Z  9 N max  1500 rpm N min  100 rpm P  5 kW

Solution N max N min

 Z  1

1500  Z  1     1.40 100   1.40 is not a standa rd s tep ratio

Trial 1: Consider R20 series, which is having   step ratio  1.12  1.12  1.12  1.12  1.40

skip 2 speeds

(From P.S.G. Data book, Pg.No. 7.20) Try R20 series by skipping (2) speeds N 1  100 rpm; N 4  280 rpm; N 7  800 N 2  140 rpm; N 5  400 rpm; N 8  1000 N 3  200 rpm; N 6  560 rpm; N 9  1250

The obtained N 9  1250  N max hence the above trail is not satisfactory.  To obtain speeds (Approximate method)

4.88


N 1  100 rpm N 2  N 1   140 rpm N 3  N 2   196 rpm N 4  N 3   274 rpm N 5  N 4   384 rpm N 6  N 5   538 rpm N 7  N 6   753.2 rpm N 8  N 7   1054 rpm  1500 rpm N 9  N 8   1476 ~

Approximate method is satisfactory Structural formula Z  3  3 ; P1  3 ; X1  1 P2  3 ; X2  P1  3 Z  P 1 X 1, P 2 X 2  3 1,

3 3

For second stage Check for condition

N max N input

Select Input  384 rpm where, N min  100 rpm N max  753 rpm

 2 and

N min N input



1 4


753  1.96  2 384 N min N i/p



100  0.26  0.25 384

Satisfies the c o nditio n and mark it as D.  N inp ut  384

Approximate Number of Gears 9 speed  3  3 and 2 3  3  12 gears For I stage No. of speeds required are 3  No. of gears required are 6

For II stage No. of speeds required are 3  No. of gears required are 6

Calculation of Number of Teeth on gear II stage N1 N2



Z2 Z1

N 1  Speed of driver shaft  384 N 2  Speed of driven shaft  100

Assume Z 1  20 teeth 384  20  78 teeth  Z2  100  Z1  20, Z2  78 teeth

4.90


1 44 0 rpm

1 50 0 rpm

1 0 5 4 rpm

7 53 rp m C

F

5 38 rp m E

3 84 rp m D

2 74 rp m B

1 96 rp m

1 40 rp m

1 00 rp m 3 (1)

3 (3) Ray D iag ram

A

4

10 12

11

9

7

8

Kinem atic Diagram

6

5

3

1

2



4.92

Z3  Z4  Z1  Z2  Z 3  Z 4  98 N3 N4 N3 N4



Z4 Z3



384 753

 0.509

 Z4  0.509 Z 3  Z3  0.509 Z 3  98  Z3  65 teeth Z4  33 teeths Z5 Z6 



N6 N5

274 Z 5   Z 5  0.7135 Z6 384 Z 6

Also Z 5  Z 6  Z3  Z4 Z 5  Z 6  98  0.7135 Z 6  Z 6  98  Z6  58 teeth

and Z5  40 teeth To check interference Z 1, Z3, Z 5 are driver


 Z3  Z1    4  and Z Z  5  3 there is no interferenc e I stage N8  Speed of driver shaft  1440 rpm N7  Speed of driven shaft  384 rpm Z7 Z8

Assume

N8



N7

Z7  20 teeths  Z 8  20 

N7 N8

 20 

1440  75 teeth 384

and Z7  20, Z8  75 teeth Z9  Z 10  Z 7  Z 8   Z 9  Z 10  95 teeth and Z9 Z 10



N 10

N 10  Z 9  Z10  N9 N9 753  0.5229 Z 10  Z 10  1440

 0.5229 Z 10  Z10  95  Z10  63 teeth

and

4.94


Z9  32 teeth Z 11 Z 12



N12 N 11



Z 11

Z12



538  Z 11  0.3736 Z 12 1440

and Z 11  Z 12  Z 9  Z 10 Z 11  Z 12  95  Z12 

95  70 teeth 1.3736

Z11  95  70  25

Check for interference Z 7, Z 9, Z11 are driver gears

 Z9  Z7  4   Z 9  Z 11   There is no interference Problem 4.9: Design the layout of a 12 speed gear box for a lathe. The minimum and maximum speeds are 100 and 1200 rpm. Power is 5 kW from 1440 rpm. Draw the speed and kinematic diagram. Also calculate the number of teeth on all gears.

(May/June 2013)

Given data Z  12 ; Nmin  100 rpm ; Nmax  1200 rpm ; P  5 kW Solution Since

Nmax N min

 Z  1

Design of Speed Reducers (Gear Boxes) 4.95 1

 1200  11    100 

   1.25 The progression ratio   1.25 , coincides R10 series. From PSG Data book, Pg.No. 7.20 N1  100 rpm ;

N2  125 rpm;

N 3  160 rpm ;

N4  200 rpm;

N 5  250 rpm;

N 6  315 rpm;

N7  400 rpm;

N 8  500 rpm;

N 9  630 rpm

N10  800 rpm;

N 11  1000 rpm;

N 12  1250 rpm

Structural formula Z322

P1  3 X1  1

12  P 1, P 2, P 3

P2  2 X2  P 1  3

 P1 X 1, P 2 X 2, P 3 X 3

P3  2 X3  P1  P2  3  2  6

 3 1, 2 3, 2 6

Since Number of stages = 3  draw u  1  3  1  4 vertical lines

Since 12 speeds, draw 12 horizontal lines three spaces III stage, I pair Since

Z 13 Z 14



N 14 N 13

Procedure for constructing speed diagram At 100 rpm (min) Locate A, at sixth intervals at 400 rpm and mark as B.

4.96


STAG E 1

STAG E 2

STAGE 3


4.98


Let N input  250 Sinc e

and

N min N input

N max N input





100 1  0.4  250 4

400  1.6  2 250

Select input 250 rpm and mark C. III stage, I pair Assume  Z13  20 teeth N 13 250  20   Z 14  Z 13   Z14  50 teeth N 14 100

III stage, II pair Z 11 Z 12



N 12



N 11

Z 11 Z 12



400  Z 11  1.6 Z 12 250

Also Z 11  Z 12  70  1.6 Z 12  Z 12  70  Z12  27 teeth

and Z11  43 teeth Mark N max  500 in II stage as (D) Input speed in II stage (E) 400 rpm N min  C

II stage, I pair Since also

Z7 Z8



N8 N7



500 400

 Z7  1.25 Z 8

Z 7  Z 8  Z9  Z10  52


1.25 Z 8  Z 8  52  Z8  24 teeth and Z7  28 teeth

III stage, II pair Since

Z9 Z 10



N 10 N9

Assume Z9  20 teeth N9 400  Z 10  Z 9   Z10  32 teeth  20  250 N 10 I stage, III pair Since

Z5 Z6



N6 N5

Assume Z5  20 teeth

N5 1000  20   Z6  Z5  N6 400  Z6  50 teeth

I stage, II pair Since

Z3 Z4



N4

N3



500  0.5 1000

 Z3  0.5 Z 4

Also Z 3  Z 4  Z 5  Z6  70 0.5 Z 4  Z 4  70

 Z4  47 teeth and Z3  23 teeth

Mark in 1st stage Max speed as (G) Min speed as (E) Input at 1000 rpm as H

4.100


Z 1  Z 2  70 Z1 Z2



N2 N1



630  0.63 1000

Z 1  0.63 Z2 0.63 Z 2  Z2  70 Z 2  43 Z 1  70.43  27 Z1  27 teeth Z2  43 teeth Problem 4.10: Design a 12 speed gear for a headstock of a lathe. The maximum and minimum speeds are 600 rpm and 25 rpm respectively. The drive is from a electric motor giving 2.25 kW at 1440 rpm. Construct the speed diagram and sketch the arrangement of the gear box.

(Nov/Dec 2009)

Solution Z  12, N max  600 rpm, N min  25 rpm , P  2.25 kW ,

We know that N max N min

 Z  1

600  12  1 25   1.334

First check for the step ratio whether it coincides with series value.


Since the obtained step ratio does not coincide with the standard step ratio and also the multiples of obtained step ratio does not coincide with the standard step ratio, therefore to obtain speeds, by Approximate method N 1  25 rpm N 2  N 1    25  1.334  33.35 rpm  34 rp m N 3  N 2    34  1.334  45.356 rpm  46 rpm N 4  N 3    46  1.334  61.364 rpm  62 rpm N 5  N 4    62  1.334  82.7 rpm  83 rpm N 6  N 5    83  1.334  110.72 rpm  111 rpm N 7  N 6    111  1.334  148 rpm N 8  N 7    148  1.334  197.432 rpm  198 rpm N 9  N 8    198  1.334  264.132  265 rp m N 10  N 9    265  1.334  353.5  354 rpm N 11  N 10    354  1.334  472.23  473 rpm N 12  N 11    473  1.334  630 rpm

Structural formula Z322 12  P 1, P 2, P 3  P 1 X1, P 2 X 2, P 3 X 3 P 1  3; X 1  1 P 2  2; X 2  3  P 1


4.102

P 3  2; X 3  P 1  P 2  6  12  3 1, 2 3, 2 6

Let u  no. of stages  3  Draw u  1  4 vertical lines

Since, 12 speeds draw 12 horizontal lines Procedure for construction of ray diagram: 3rd stage: 2 speed, 6 spaces At 25 rpm (min), locate point “A ” Leave six intervals at 148 rpm, mark point “B ” Point “B ” represents N max  148 rpm Point “A ” represents N min  25 rpm Locate input speed, point “C ” N max N input

 2 or

N min N input



1 or 0.25 4

Select N input  83 rpm 

25 148  1.78  2 and  0.30  0.25 83 83

 Selection of N input  83 is satisfactory

2nd stage, 2 speeds, 3 space Locate point “D ” above point “C ” after 3 spaces  N max  198, N min  83

Select N input  148




198 83  1.3  2 and  0.56  0.25 148 148

 Selection of N input  148 is satisfactory

Locate point “E ” at 148 rpm 1st stage 3 speeds, 1 space  Mark “F ” and “G ”

Locate “H ” as input of motor For first stage, it is not necessay to satisfy the condition   N max 2    N input

Kinematic diagram No. of gears required  2 3  2  2  2 7   14 gears

Calculation of Number of Teeth III stage 1st pair (Reducing speed) Speed of driver  N 2  83 rpm Speed of driven  N 1  25 rpm But N1 N2

Let



Z2 Z1

Z 2  20 teeth 25 20   Z 1  66.4  67 teeth 83 Z 1

4.104


K inem atic Layou t


4.106


 Z1  67 teeth; Z2  20 teeth

2nd pair, Increasing speed N 4  driver speed  83 rpm N 3  driven speed  148 rpm

But N3 N4



Z4 Z3

and Z1  Z2  Z3  Z4  Z 3  Z 4  67  20 Z 3  Z 4  87

and Z4 

148  Z3 83

Z 4  1.78 Z3  Z 3  1.78 Z 3  87  Z3  31.29  32 teeth

and Z4  55 teeth II stage 1st pair: Reducing speed N 8  Speed of driver  148 rpm N 7  Speed of driven  83 rpm

But


N7 N8



 Z7 

Z8 Z7 Z 8  N8 N7



148  Z8 83

Z 7  1.783 Z 8

Assume

Z8  20 teeth  Z7  35.66  36 teeth

2nd pair: Increasing speed N 6  Speed of driver  148 rpm N 5  Speed of driven  198 rpm

But N5 N6



Z6 Z5

and Z7  Z8  Z5  Z6 36  20  Z 5  Z 6  Z 5  Z 6  56 teeth

and Z6 

198  Z5 148

Z 6  1.33 Z5  Z 5  1.33 Z 5  56  Z5  24 teeth

4.108


Z6  31.92  32 teeth

I stage In this stage, there are 3 speeds, and 3 pairs of gears are required. To avoid interference of gear of one shaft with the gear of the other shaft while shifting: Condition to avoid interference; Z 12  Z 10   4 and Z 12  Z 14   4

1st pair: Reducing speed Assume: Z10  20 teeth N 10  Speed of driver  600 rpm N 9  Speed of driven  148 rpm

But

N9 N 10



Z 10 Z9

148 20  600 Z 9  Z9  81.08  82 teeth

2nd pair: Intermediate speed Let N 14  Speed of driver  600 rpm N 13  Speed of driven  198 rpm

But


N 13 N 14



Z 14 Z 13

198 Z 14  600 Z 13  Z 14  0.33 Z 13

and Z 9  Z 10  Z13  Z 14  Z 13  Z 14  82  20 Z 13  Z 14  102  Z 13  0.33 Z 13  102  Z13  76.69  77 teeth

and Z14  25 teeth

3rd pair: Increasing speed N 11  speed of driven  265 rpm N 12  speed of driver  600 rpm

But N 11 N 12 



Z 12 N 11

265 Z 12  600 Z 11

 Z 12  0.44 Z 11


4.110

But Z 11  Z12  Z 13  Z14  Z 11  Z 12  77  25  Z 11  Z 12  102

Hence Z 11  0.44 Z 11  102  Z11  71 teeth


Check for Interference Z 12  Z 10  31  20  4 Z 12  Z 14  31  25  4

Hence, there is no interference and the design is safe. Problem 4.11: The spindle of a pillar drill is to run at 12 different speeds in the range of 100 rpm to 355 rpm. Design a three stage gear box with a standard step ratio. Sketch the layout of the gear box, indicating the number of teeth on each gear. The gear box receives 5 kW from an electric motor running at 360 rpm. Sketch also the speed diagram. (April/May 2008)

Solution Given P  5 kW, N max  355 rpm, Nmin  100 rpm Z  12


N max N m in

 Z  1

355   12  1 100    1.122

The progression ratio   1.122 , coincides with R20 series standard progression ratio. Therefore select R20 series speeds. (From P.S.G. Data book, Pg.No. 7.20) N 1  100 rpm , N2  112 rpm, N 3  125 rpm, N 4  140 rpm N 5  160 rpm, N 6  180 rpm, N 7  200 rpm, N 8  224 rpm N 9  250 rpm, N 10  280 rpm, N 11  315 rpm, N 12  355 rpm

Structural formula Z322 12  P 1, P 2, P 3  P 1 X1, P2 X2, P 3 X 3  3 1, 2 3, 2 6

1st stage 2nd stage 3rd stage P 1  3;

P 2  2;

P3  2

X1  1  1 X 2  P1  3 X 3  P1  P2 326

4.112


360

H

335

315

280

250

G D

224

F

200

E

B 180

160

C

140

125

112

100 I Stage 3(1)

II Stage 2(3) Ray D iag ram

III Stage 2(6)

A

11

14 12 10 9

To a void interfere nce, a ll ge ars in this stage are m o dife d to: Z 9 = 14 4; Z 1 0 = 80 ; Z 11 = 1 32 ; Z 1 2 = 92; Z 1 3 =1 36 ; Z 1 4 = 88

D riven

13

6 5

Kinematic D iagram

Driven

Driver 200 rp m

8 7

Driven

4 3

2 1

10 0 rpm


4.114


u  No. of stages  3  Draw u  1  3  1  4 vertical lines

Since 12 speeds, draw 12 horizontal lines Procedure for the construction of ray diagram 3rd stage : 2 speeds, 6 spaces At 100 rpm (min) locate point A Leave six intervals at 200 rpm mark point B i.e., Point B represents N max  200 rpm, Point A represents N min  100 rpm

Locate input speed, Point C N max N input

or

 2 select N input  160

200  1.25 which is less than 2 160

N min N input



1 or 0.25 4

Locate Point C at 160 rpm. 2nd stage: 2 speeds, 3 spaces. Locate point D above point C after 3 spaces. To find N input for 2nd stage N max  224 rpm N min  160 rpm


select N input  200 rpm N max N inp ut

2

224  1.12 w hich is less than 2 220 N min N input

 1/4;

160  0.8  Selection of N input  200 is satisfactory. 200

Locate point E at 200 rpm. (Note: selection of 180 rpm is also correct) 1st stage: There are 3 speeds with 1 space interval.  Mark F and G

Locate H as input of motor. For first stage it is not necessary to satisfy the condition.   N max 2 N   input

Join H, F From F draw (2) speeds F C 1 and F D 1

180 rpm, 250 rpm (maintain again 3 space intervals) From C 1, draw 2 speeds. (112 rpm and 224 rpm) C 1 A 1 and C 1 B 1

112 rpm, 224 rpm

4.116


Join H, G From G draw 2 speeds (200 rpm and 280 rpm) G C 2 and C D 2

200 rpm, 280 rpm From C 2 draw 2 speeds (125 rpm and 250 rpm) C 2 A 2 and C 2 B 2

140 rpm, 315 rpm From D 2 draw (2) speeds (180 rpm and 355 rpm) D 2 A 5 and

D 2B 5

180 rpm

355 rpm

Kinematic Diagram No. of Gears required are  2 3  2  2   2 7  14 Gears

Calculation of Number of Teeth III stage 1st pair (Reducing speed) speed of driven N 1 Z2   spe ed of driver N 2 Z1

let

Z 2  20 teeth 100 20  160 21 Z1 

20  160  32 teeth 100


speed of driver  N 2  160 rpm speed of driven  N 1  100 rpm Assume above 17 teeth 1st pair  Z1  20 teeth Z2  32 teeth N 3  drive n s peed  200 rpm N 4  driver speed  160 rpm

2nd pair: Increasing speed speed of driven N 3 Z4   speed of driver N 4 Z3

... (1)

Z1  Z2  Z3  Z4 20  32  Z 3  Z 4 Z 3  Z 4  52 200 Z 4  160 Z 3 Z 4  1.25 Z3 Z 3  1.25 Z 3  52 Z3 

52  23 2.25

Take Z3  23 teeth Z4  52  23  29 teeth

2nd pair Z3  23 teeth Z4  29 teeth

... (2)

4.118


II stage 1st pair: Reducing speed speed of driven N 7 Z8   speed of driver N 8 Z7 N 8  driver speed  200 rpm N 7  driven speed  160 rpm 160 20  200 Z7 Z7 

20  200 160

Z7  25 teeth

Assume Z8  20 teeth 1st pair: Z7  25 teeth Z8  20 teeth

2nd pair: Increasing speed speed of driven N 5 Z6   speed of driver N 6 Z5 N 6  driver speed  200 rpm N 5  driven speed  224 rpm Z7  Z8  Z5  Z6 25  20  Z 5  Z 6 Z 5  Z 6  45

... (1)


224 Z 6  200 Z 5 Z 6  Z 5  1.12 Z 5  1.127 5  45 Z5 

45  21.22 2.12

Take Z 5  21 teeth Z 6  45  21  24 teeth

2nd pair Z5  21 teeth Z6  24 teeth

I stage In this stage there are 3 speeds; and 3 pairs of gears are required. To avoid interference of gear of one shaft with the gear of the other shaft while shifting: Condition to avoid interference Z 12  Z 10   4 Z 12  Z 14   4

1st pair: Reducing speed N 10  driver speed  335 rpm N 9  driven speed  200 rpm N9 N 10



Z 10 Z9

; Assume Z 10  20

4.120


 Z 9  35.5  36 teeth N 13 N 14



Z14

224  Z14  Z 13   0.63 Z13 Z 13 355

But Z 9  Z 10  Z 13  Z 14 56  34 teeth  Z 13  Z 14  36  20  56  Z 13  1.63 Z13  34 teeth Z14  22 teeth N 11 N 12



Z 12 Z 11



250 Z 12  355 Z 11

 Z 12  Z 11  0.704 Z 11  Z 12  Z 13  Z 14  56 Z 11  0.704 Z 11  56  Z11  33 teeth Z12  23 teeth

Check the No. of teeth on Z 9, Z 10 , Z 11 , Z 12 , Z13 and Z 14 Z 12  Z 10  23  20  3 which is less than 4 Z 12  Z 14  23  22  1 which is less than 4  Interference will occur between the gears. To avoid

Interference, proportionately increase the No. of teeth on all the gears.  Multiply by 4 Z 12  4  23  92, Z10  80, Z 12  Z 10  92  80  12  4


 Satisfactory

Similarly Z 9  144 teeth, Z 10  80 teeth, Z 11  132 teeth, Z 12  92 teeth, Z 13  136 teeth, Z 14  88 teeth, Problem 4.12: Design a layout of a 12 speed gear box for a milling machine having an output of speeds ranging from 180 to 2000 rpm. Power is applied to gear box from a 6 kW induction motor at 1440 rpm. Choose standard step ratio and construct the speed diagram. Decide upon the various reduction ratios and number of teeth on each gear wheel. Sketch the arrangement of the gear box.

(April/May 2008)

Solution N max

Nmin

 Z  1

N max  2000 rpm, N m in  180 rpm 

2000   12  1 180    1.244

It is not a standard step ratio  Try second method

(i.e.) Try the multiples of standard ratio and try the series Take R20 series and try multiples  1.12  1.12  1.256 which is greater than required.  Not satisfactory

4.122


 To obtain speeds (Approximately) by Approximate

method N 1  180 rpm N 2  N 1    224 rpm N 3  N 2    278.65  279 rpm N 4  N 3    347 rpm N 5  N 4    431.66  432 rpm N 6  N 5    536 rpm N 7  N 6    667 rpm N 8  N 7    830 rpm N 9  N 8    1033 rpm N 10  N 9    1285 rpm N 11  N 10    1599 rpm N 12  N 11    1990 rpm is almost equal to N max Z322 12  P 1, P 2, P 3 P 1  3; X 1  1 P 2  2; X 2  P 2  3 P 3  2; X 3  P 1  P 2  3  2  6

3rd stage Locate N input


Mark

N min  180 rpm as (A) N max  667 rpm as (B) N max

Ninput

2

Let us select N input  432 rp m 667 2 432 N min N input



1 4

180 1  432 4  Selection of Input is satisfactory and mark it as (C)

2nd stage Mark N max  830 rpm as (B) N max N input

Let



836  1.25  2 667

N input  667 N min N input



432 1  0.647  667 4

Selection of Input 667 is satisfactory and mark it as (E) I stage 1st pair Let

Z2  20 teeth

4.124


N 12 = 1 9 9 0 rp m

N 11 = 1 5 9 9 rp m H 1440 rp m

N 10 = 1 2 8 5 rp m

N 9 = 1 0 3 3 rp m

N 8 = 8 3 0 rp m

D

N 7 = 6 6 7 rp m

E

B N 6 = 5 3 6 rp m

C

N 5 = 4 3 2 rp m

N 4 = 3 4 7 rp m

N 3 = 2 7 9 rp m

N 2 = 2 2 4 rp m

I S ta g e 3 (1 )

II S ta g e 2 (3 ) Fig. Ray Diag ram

N 1 = 1 8 0 rp m III S ta g e 2 (6 ) A

Z 1 3 =40 (83 0 rpm )

Driven

Driver 1440 rp m

Z 1 4 =23

Z 11 = 36 (10 33 rpm )

Z 1 2 = 27

Z 9 =4 3 (66 7rp m )

Kinem atic D iagram

Z 7 =3 1 (432 rpm )

Z 8 = 20

Z 5 = 23 (830 rpm )

Driven

Driver 667 rpm

Z 6 = 28

Driven

Z 2 =2 0

Z 3 =27 (6 67rpm )

Driver 432 rp m

Z 4 =41

Z 1 =48 (18 0 rpm )


4.126


N1

Since

N2



Z2 Z1

N2 432  Z 1  20   48 teeth  Z1  Z 2  N1 180 N3

Similarly

N4



Z4 Z3

and Z 1  Z 2  Z 3  Z4  Z 3  Z 4  68 teeth  Z4  Z3 

N3 N4

 Z 4  1.54 Z3

 Z 3  1.54  Z 3  68  Z 3  27 teeth

and Z 4  68  27  41  Z 4  41 teeth

II stage N7 N8

Take



Z8

Z7

Z8  20 teeth

N8  Z7  31 teeth  Z7  Z8  N7


and Z6 Z5

N5



N6

Z7  Z8  Z5  Z6  Z 5  Z 6  51 teeth

and Z6  Z5 

N5 N6

 Z 6  1.24 Z 5

 1.24 Z 5  Z 5  51  Z 5  23 teeth

and Z 6  28 teeth

I stage N9 N 10



Z 10 Z9

Take Z10  20 teeth  Z 9  Z 10  N 13 N 14



N10 N9

 Z 9  43 teeth

Z 14 Z 13

N13  Z 14  Z 13   Z 14  0.576 Z 13 N 14 But

4.128


Z 9  Z 10  Z 13  Z14  Z 13  Z 14  63 teeth  Z 13  0.576 Z 13  63  Z13  40 teeth


Similarly N 11 N 12



Z 12 Z 11

Z 12  Z 11  0.717

  N 11 1033   0.717     N 12 1440

But Z11  Z 12  Z 13  Z14  63  Z 11  Z 12  63  Z 11  0.717 Z 11  63  Z11  36 teeth and Z12  63  36  27 Z12  27 teeth Check for Interference Z 12  Z10  4; Z 12  Z 14  4 27  20  4; 27  23  4  satisfactory


Design of Gear Box 1. Selection of spindle speed Progression ratio  n  1 

N max N min

2. Construct the Ray diagram. 3. Construct the kinematic arrangement. 4. Calculate the No. of teeth of all the gears. 5. Choose the suitable material (i) C45 – 30 (ii) 15 Ni2 Crl M015 – 80 (iii) 40 Ni2 Crl M028 – 100 and for permissible shear stress N/mm 2 (i) C45  30 (ii) C14  25 (iii) 40 Ni2 Crl M0 28 – 55 (iv) Low carbon alloy steel – 40 6. Calculation of module: (i) Find the torque for the gear T

P  60 2 N

(ii) Calculate the tangential force, Ft 

T 2T  r Zm


4.130

(iii) Find the value of module,

m

where, m 

  Ft

m  

b  10 m

(M value take from point 4) 7. Find the value of centre distance by using formula,  Zx  Zy  a m 2  

8. Calculate the face width: b  10  m 9. Calculate the distance between the bearing using formula, L  25  10  4b  20  7b  10  25 

Assume, width of the bearing  25 mm



Clearance between gear and bearing  10 mm



Distance between the adjacent groups of gears  20 mm



4b is two pairs gear group and 7b is three pairs

gear group from the total length. 10. Design of shafts (i) Maximum bending moment M   F n  Normal load

Fn  L

4


Fn 

Ft cos 

(ii) Torque eq T eq    M 2  T2  T

P  60 2 N low

T  Torque on the spindle in system.

(iii) Find the diameter of the spindle  16  Teq  ds        

where,



1/3

16T eq

 d3s

(From point 4) 11. Design the diameter of the input and intermediate shaft. T  0.2d3s [] Problem 4.13: Design a 12 speed gear box for an all geared head stock of the lathe. Maximum and Minimum speeds are 600 rpm and 25 rpm respectively. The drive is from an electric motor giving 3.25 kW at 1500 rpm. (Assume length of shaft  860 mm).

Given Data: n  12 ; N max  600 rpm, N min  25 rpm ; P  3.25 kW ; N input  1500 rpm


4.132

To Design a gear box. Solution 1. Selection of spindle speeds. n  1 

N max N min

 600     25 

1/11

 1.338

From PSG data book Page No. 7.20 1.06  1.06  1.06  1.06  1.06   1.334

Since 1.334 is very close to 1.338, we can take R 40 series. N 1  25, N 2  33.5 , N 3  45, N 4  60, N 5  80 , N 6  106, N 7  140, N 8  190, N 9  250, N 10  335 N 11  450, N 12  600

2. Draw the Ray diagram: Formula: 3(1), 2(3) 2(6) Stage 1 N max

Ninput N min

Ninput



140  1.75  2 80



25  0.31  0.25 80



190  1.36  2 40

Stage 2 N max

Ninput


N min

Ninput



80  0.57  0.25 40



140  0.311  0.25 150



250  0.56  2 450

Stage 1 N min

Ninput N max

Ninput

3. Kinematic Arrangement 4. Calculate the No. of teeth of all the gears Stage 3 1st pair: Maximum reduction from 80 rpm corresponding gears are 13 and 14 on shaft 25 rpm. Assume Z13  20 (driver) Z 13 Z 14



N14 N13

20 25  ; Z14  64 Z 14 80

2nd pair: Speed Increases from 80 to 140 rpm. (Corresponding gears are 11 and 12). Z 11 12



N 12 N 11



140 80

Z 11  1.75 Z 12

where,

Z 11  Z 12  Z 13  Z14

1.75 Z 12  Z 12  20  64


4.134

3 (1)

2 (3)

2 (6)

6 00

H

4 50

3 35

2 50

G D

1 90

F

1 40

E

1 06

80

C

60

Stag e 1

Stag e 2

Sh aft-4

S haft-3

S haft-2

Sh a ft-1

45

3 3.5

Stag e 3

Fig. Ray Diagram for 12 Speed G ear B ox

25 A

I S tag e

Fig. Kinem atic Arrang em en t fo r 12 Speed G ear B ox.

II S tage

III Stag e


4.136


Z12  31 Z11  1.75 31  54

Stage 2 1st pair: Maximum speed reduction from 140 to 80 rpm (corresponding gears are 9 and 10) Assume Z9  20 (driver) Z9 Z 10



N 10

N9

20 80  Z 10 140  Z10  35

2nd pair: Speed Increase from 140 to 190 rpm. (Corresponding gears are 7 and 8). Z7 Z8



N8 N7



190 140

Z 7  1.357 Z 8

where,

Z 7  Z 8  Z 9  Z10

1.357 Z8  Z 8  20  35 Z8  24 Z7  1.357 24  32

Stage 1 1st pair: Maximum reduction from 450 to 140 rpm (corresponding gears are 5 and 10)


Assume Z5  20 (driver) Z5

Z6



N6 N5

20 140  Z 6 450 Z 6  65

2nd pair: Speed Reduction from 450 to 190 rpm. (Corresponding gears are 3 & 4). Z3 Z4



N4 N5



190 450

Z 3  0.422 Z 4

where,

Z 3  Z 4  Z 5  Z6

0.422 Z4  Z 4  20  65 Z4  60 Z3  0.422 60  25

3rd pair: Speed Reduction from 450 to 250 rpm. (Corresponding gears are 1 and 2). Z1

Z2



N2 N1



250 450

Z 1  0.555 Z 2

where,

Z 1  Z 2  60  25  85  Z2  55

 Z1  0.555 55  30

5. Material Selection: C 45

4.138


6. Calculation of module: Torque at lowest speed 25 rpm by meshing gears 13 and 14. T 14 

60  3.25  10 3 2  25

 1241.40 N  m

To Calculate tangential force on gear 14, Ft 

module



2  1241.40  10 3 64  m



38793.75 m

m 

m 

T 2  T 14  r Z 14  m

. . Zm   . r 2   

 

F t

m  

b  10 ,   30 [for C 45 ] m

m

 

m

m 

 38793.75 /m    10  30  

129.31


m2 

129.31 m

m 3  129.31 m  5.056 mm

Nearest value is  5 mm (From PSG Data book Pg No. 8.2) 7. Calculate the centre distance (i) Centre distance for  Z1  Z2  Stage 1,  a1   m 2    30  55  a1   5 2   a1  212.5 mm

(ii) Centre distance for stage 2,  Z7  Z8  a2   m 2    32  24   5 2    140 mm

(iii) Centre distance for stage 3,  Z11  Z 12  a3   m 2    54  31   5 2  

4.140


 212.5 mm

8. Find the face width bm b  10  5  50 mm

9. Length of shaft L  860 mm (given)

10. Design of shafts (i) Maximum bending moment M M Fn 

Fn  L 4 Ft cos 



38793.75 /m cos

Assume,   20 

38793.75 /5 7758.75  0.939 cos 20 Fn  8256.6 N

Maximum Bending moment, M

8256.6  860  1.77  10 6 N  mm 4

(ii) EquivalentTo rqueTeq T eq 

1.77  10 62  1241.4 2 M 2  T214       1.77  10 6 N  mm

(iii) Find the diameter of the spindle,

Design of Speed Reducers (Gear Boxes) 4.141 1/3

 16  T eq  ds       

where,

  30 N/mm 2 from the part 4  16  1.7  10 6      30  

1/3

 66.08 mm

Using R 40 series, Rounded value of the diameter is 67 mm. 11. Design of the other shaft (i) Diameter of shaft: Input speed  450 rpm Torque



P  60 3.25  10 3  60  2  450 2 N

T  69 N  m T  0.2 d 3s  1 69  10 3  0.2d3s 30 1 ds  22.57 mm or 23 mm 1

(ii) Diameter of shaft 2 Minimum speed  140 rpm Torque



P  60 3.35  10 3  60  2  140 2N

T  228.5 N  m

wkt,

T  0.2d3s  2

4.142


22.5  10 3  0.2d3s 30 2 ds2  33.64 mm

(iii) Diameter of shaft 3: Minimum speed  80 rpm Torque



P  60 3.35  103  60  2 N 2  80

 399.87 N  m 399.87  10 3  0.2 d 3s 30  3 ds  40.5 mm 3

Problem 4.14: Design a gear drive, to give 18 speeds for a spindle of a milling machine. The drive is from an electric motor of 6.75 kW at 1500 rpm. Maximum and minimum speeds of the spindle are to be around 35 and 650 rpm respectively.

Given Data: Z  18 ; P  6.75 kW; N motor  150 rpm; N max  650 rpm; N min  35 rpm.

To: Design a Gear box Solution 1. Selection of spindle speed n  1 



17

N max N min

650  650      35  35 

  1.1875

1/17

4.143


The step ratio is not standard, so multiply the standard ratio to the calculated step ratio. where, 1.06  1.06  1.06  1.191 The value is very close to the calculated step ratio 1.18 of R40 series. (skip 2 speeds from the R40 series) It’s given by, N 1  35.5, N 2  42.5 , N 3  50, N 4  60, N 5  71 , N 6  85, N 7  100, N 8  118, N 9  140, N 10  170 , N 11  200, N 12  236, N13  280, N14  335, N 15  400, N 16  475, N 17  560, N18  670 N max is very close to the N 18 , therefore speed

selection is satisfactory. 2. Ray diagram Formula: 2(1) 3(2) 3(6) Stage 3 N max

Ninput N min

Ninput



280 2 140



35.5  0.253  0.25 140



280  1.186  2 236

Stage 2 N max

Ninput

Design of Speed Reducers (Gear Boxes) 4.144 2(1)

3(6)

3(2)

670 560 475

,

400 335

H

C 280

F

236

G

200

E

170 140

D

118 B 100 85 71 60 50 42.5 Stage 1

Stage 2

Stage 3

RAY DIAG RAM for 18 Speed G ear Box

A

35.5

10 m m

4b

3

20 m m

236 rpm Z 4 =4 0

4

200 rp m

Z 6 =29

6

7 8

9

140 rpm

Stage 2

7b

Driven

Z 1 0 =34

10

Driver 236 rp m

Z 7 = 30 Z 9 =20

Z 8 =25 280 rpm

5

Z 5 = 25

20 mm

Kinematic Arrangem ent for 18 Speed G ear B ox

Stage 1

Z 2 =38 280 rpm

Driven

2

1

Z 1 =2 3 Z = 20 3

Driver 475 rpm

Z 1 2 =5 8 100 rpm

12

14

13

15

D riven

10m m

35.5 rpm Z 1 6 = 79

16

Driver 140 rpm

Z 1 5 =20

Z 1 4 = 33 280 rpm

11

Z 11 =4 1

Z 13 =6 6

7b a 1= 91 .5m m a 2= 8 1m m

Stage 3

a 3= 148.5m m

4.145 Design of Transmission Systems


N min

Ninput



140  0.59  0.25 236



280  0.59  2 475



236  0.49  0.25 475

Stage 1 N max

Ninput N min

Ninput 

All conditions are satisfied.

3. Kinematic Arragement 4. Calculate the No. of teeth of all the gears. State 3 1st pair: Maximum reduction from 140 to 35.5 rpm (Corresponding gears are 15 and 16) Assume, Z15  20 driver 

Z 15 Z 16



N 16 N 15

20 35.5  Z 16 140 Z16  79

2nd pair: Speed Reduction from 140 to 100 rpm. (Corresponding gears are 11 and 12) Z 11

Z12



N 12

N11


4.147

Z 11

Z12



100 140

Z 11  0.714 Z 12  Z 11  Z 12  Z 15  Z 16 0.714 Z 12   Z 12  20  79  Z12  58  Z11  0.714 58  41 rd

3

pair: Speed Increases from 140 to 280 rpm.

(Corresponding gears are 13 and 14). Z 13 Z 14 Z 13 Z 14





N 14 N 13 280 140

 280  Z 13    Z14  Z 13  2Z 14  140 

where,

Z 13  Z14  Z 15  Z 16 2Z 14  Z 14  20  79 Z14  33 Z 13  2 33 Z13  66

Stage 2 I pair: Maximum reduction from 236 to 140 rpm (Corresponding gears are 9 and 10) Assume, Z9  20 (driver)


Z9 Z 10



N 10 N9

140 20  Z 10 236  Z10  34

2nd pair: Speed Reduction from 236 to 200 rpm. (Corresponding gears are 7 and 8). Z5 Z6



N6 N5

 200  Z5    Z6  236  Z 5  0.847 Z 6

where,

Z 5  Z 6  Z 9  Z 10 0.847 Z 6  Z 6  20  34  Z6  29  Z5  0.847 29  25

3rd pair: Speed Increases from 236 to 280 rpm. (Corresponding gears are 7 and 8). Z7 Z8



N8 N7

 280  Z7    Z 8  Z 7  1.186 Z 8  236  Z 7  Z 8  Z 9  Z10

4.149


1.186 Z 8  Z 8  20  34 Z8  25 Z7  1.186 25  30

Stage 1 1st pair: Maximum Speed Reduction from 475 to 236 rpm (Corresponding gears are 3 and 4) Assume, Z3  20 (driver) Z3 Z4



N4 N3

20 236  Z 4 475 Z4  40

2nd pair: Speed Reduction from 475 to 280 rpm. (Corresponding gears are 1 and 2). Z1 Z2



N2 N1

 280  Z1    Z 2  Z1  0.59 Z2  475 

where, Z 1  Z2  Z 2  Z4 0.59Z 2  Z 2  20  41 Z2  38 Z 1  0.59 38 Z1  23


5. Material Selection Select 40 Ni2 Cr1 M0 28 6. Find the value of module: Torque: gear 16 has the lowest speed of 35.5 rpm. (i) T 16 

P  60 6.75  10 3  60  2 N 2  35.5 T 16  1815.71 N  m

(ii) Tangential Force, Ft

wkt,

16



T 2  T16  r Z 16  m



2  1815.71  10 3 45967.34  m 79  m

m

m 

  Ft

m  

b  10 ;   100 m

Material constant  100 m

45967.34 /m  10  100

    m2 

45.96 m

m 3  45.96 m  3.58 mm

45.96 m


4.151

(or) m  4 mm (Standard module) From PSG Data book Pg No 8.2 

Centre distance for Stage 1,  Z1  Z2  a1   m 2    23  38    4  122 mm 2  

For stage 2  Z5  Z 6  a2    2    25  29    4  108 mm 2  

For stage 3  Z11  Z 12  a3   m 2    42  58    4  200 mm 2  

8. Calculation of face width b    m  10  4  40 mm

9. Calculation of length of shaft L  25  10  46  20  7b  20  7b  10  25  110  18b  110  18 40  830 mm


10. Design of shafts (i) Maximum bending moment, M

Fn 

Ft cos 



Fn  L

4 45967.34 /m cos 20

Assume   20 

45967.34 /4  12.22  10 3 N cos 20 M

12.22  10 3  830 4

M  2.53  10 6 N  mm

(ii) To find the equivalent Torque, Teq 

M 2  T216  



2.53  10 62  1815.71  10 32 



6.40  10 12  3.29  10 12  

Teq  3.11  10 6 N  mm

(iii) Find the diameter of spindle  16  T eq  ds       

1/ 3

where,   55 N/mm 2 from point 4

4.153


 16  3.11  10 6     55   

1/3

 66.03  66 mm R 40 series 

11. (i) Diameter of shaft 1 Input speed  475 rpm Torque



P  60 6.75  10 3  60  2 N 2  475

T  135.70 N  m T  0.2  d 3s   1  135.70  10 3  ds    1   0.2  55

1/3

ds  23.10 mm 1

 25 mm R 40 series

(ii) Diameter of shaft 2 Minimum speed  236 rpm Torque 

P  60 6.75  10 3  60  2  236 2 N  273.12 N  m

273.12  10 3  0.2  d3s  55 2 d s  29.17  30 mm R 40 series 2


(iii) Diameter of shaft 3 Minimum speed  140 rpm Torque 

P  60 6.75  10 3  60  2  140 2 N  460.41 N  m

460.41  10 3  0.2  d3s  55 3

ds  34.72 mm  35 mm R40 series  3 Problem 4.15: A six speed gear box is required to provide output speeds in the range of 125 to 400 rpm, with a step ratio of 1.25 and transmit a power of 5 kW at 710 rpm. Draw the speed diagram and kinematic diagram. Determine the number of teeth module and face width of all gears, assuming suitable materials for gears. Determine the length of the gear box along the axis of the gear shaft.

(Nov/Dec 2007)

Solution No. of speeds  Z  6 Max. speed

 N max  400 rpm

Minimum speed  N min  125 rpm N max N min

 Z  1

400  6  1 125    1.262

The  is not an standard progression ratio. Therefore, for R40 series from P.S.G. Data book, Pg.No. 7.20

4.155


  1.06 skip 3  speeds

  1.06  1.06  1.06  1.06  1.262

Skip (3) speeds and try R40 series N 1  125 rpm N 2  160 rpm N 3  200 rpm N 4  250 rpm N 5  315 rpm N 6  400 rpm

Since N 6  400 rpm is the maximum required speed  Select the above speeds

Structural formula Z32

P1  3

X1  1

P1  P2

P2  2

X2  P2  3

 Z  P 1 X 1, P 2 X 2  3 1, 2 3 

Check Either

N max N input

 2 or

N min N input



1 4

Fo r 1st stage, the abo ve c o nditio n need no t be c hec ked.Selec t N input as 710 rpm. Mark N max  250 a s B N min  125 as A


D =710 rpm

3 (1)

2 (3 ) 710 rpm

400 rpm

315 rpm

B

250 rpm

200 rpm

C

160 rpm

125 rpm 3(1)

Ray D iag ram

2(3)

A

Driven

5 8

7

Z 7 = 28 Z 5 =20

9 10

Z 9 = 24

Z 8 = 63 Z 1 0 =6 7 Z 6 = 71 (3 15 rpm ) (2 50 rp m ) (200 rp m )

6

Driver 710 rpm

STAG E I

3 4

Z 3 =29

Z 4 =23 (250 rpm )

Driven

Driver 200 rp m

S TAG E II

1

Z 1 =2 0

2

Z 2 = 32 (125 rpm )



For 2nd stage Check

N max N input

 2 and

N min N input



1 4

Select N input  200 rp m Max. speed  N max  250 rpm Min. speed  N min  125 rpm 

125 1 250   2 and 200 4 200

 Selection of N input 200 rpm is satisfactory and mark it as (C)

Approximate No. of gears required 6 speeds  3  2  2 3  2  10 gears Calculation of No. of teeth on gear Assume Z 1  20 teeth driver N 1  200 rpm  Speed of driver N 2  125 rpm  Speed of driven Z 2  N o . of teeth on driven 

Z2 Z1



N1 N2

N1  Z2  Z1  N2  20 

200  32 teeth 125

4.159


 Z1  20 teeth, Z2  32 teeth

Since centre distance is same Z 1  Z 2  Z3  Z 4  20  32  52 Z4

and

Z3 

Z4 Z3





N3

N4 200  0.8 250

Z 4  0.8 Z3  Z 3  0.8 Z3  52  Z3  29 teeth

But

Z 3  Z 4  52  Z4  23 teeth

Similarly, Assume

Z6 Z5

N5



N6

Z5  20 teeth

N5  Z6  Z5  N6  20 

710 200

 71 teeth  Z5  20 teeth, Z6  71 teeth

Since centre distance is same Z 5  Z 6  Z7  Z 8  91


 Z 7  Z 8  91 Z8

and

Z7 

Z8 Z7





N7

N8 710 315

 Z8  2.253  Z 7  Z7  2.253 Z 7  91  Z7  28 teeth

and

Z8  63 teeth

Since centre distance is same Z 5  Z 6  Z 7  Z 8  Z 9  Z10  91  Z 9  Z 10  91

But

Z 10 Z9



N9 N 10

 Z 10 

710 Z 250 9

 Z 10  2.84 Z9  Z 9  2.84 Z 9  91  Z9  24 teeth

and

Z 10  91  24  67 Z10  67 teeth

4.161


To avoid interference Z 7  Z 5 and Z 7  Z 9  4 28  20 and 28  24  4

The condition satisfies, therefore the calculation of number of teeth on each gear is satisfactory. Module calculation The lowest speed is 125 rpm  Calculate the torque at 125 rpm T125 rpm 

5  10 3  60 P  60  2  125 2 N

 381.97 Nm But Torque  F t  r2 2

where r2 is the pitch circle radius of gear 2 But r2 

Z2  m 2



32  m  16 m 2

where m refers to module F t  tangential force on gear 2 2

 381.97  10 3  F t  16 m 2

 Ft  2

But

m

381.97  10 3 23,873.12  N m 16 m

   M Ft

m


where m 

b  10 (assume) from P.S.G. Data book, Pg.No. m

8.14 a1  Material constant  100, for 40 Ni2 Cr1 Mo 28  80, for 15 Ni2 Cr2 Mo15  30, for C 45

m

   23873  12  m 

m2 

 /10  100 

23873.12 m  1000

 m 3  23.873 m  2.9 mm ~  3 mm (standard) from P.S.G. Data book, Pg.No. 8.2

we know b  10  3 m  b  10  3  30 mm

From P.S.G. Data book, Pg.No. 8.22 a1  centre distance in stage 1 

Z 1  Z2 m 2



20  32 3 2

 78 mm a2 

Z 5  Z 6 m 20  71 3   136.5 mm 2 2

4.163


Length of shafts Assume Distance between the gear and gear box wall is to be about 10 mm. Distance between the adjacent groups of gears is to be about 20 mm. Total length requirement for 2 pairs group and 3 pairs group are to be 4b and 7b respectively. Thus the length of the shaft is obtained as L  30  10  4b  20  7b  10  30

(First 30 and last 30 mm are for bearings) Put b  30 mm we get L  430 mm  Length of shafts is L  430 mm Problem 4.16: Sketch the arrangement of a six speed gear box. The minimum and maximum speeds required are around 450 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.

Solution No. of speeds  Z  6 Maximum speed  N max  1400 rp m Minimum speed  N min  450 rpm N max N min 

 Z  1

1400  6  1    1.254 450

It is not a standard ratio


Therefore, Trail (1) for R20 series (from P.S.G Data book, Pg.No. 7.20)   1.12   1.12  1.12  1.254  Skipping one speed, we get N 1  450 rpm N 2  560 rpm N 3  710 rpm N 4  900 rpm N 5  1120 rpm N 6  1400 rpm

Structural Formula Z3 P1 P 1  3;



2 P2 X1  1 P 2  2;

X2  P1  3

Z  P 1 X 1, P 2 X 2  3 1, 2 3

For second stage Mark N min as A N max as B

4.165


N max

Check

N input

Let

2

N input  710 rpm 

900 2 710

N min N input   N input



1 4

450  0.25 710

selection is satisfactory, mark it as C

Approximate No. of gears required 6 speeds  3  2  2 3  2  10 gears For 1st stage No. of speeds required are 3  No of gears required are 6

For 2nd stage No. of speeds required are 2  No. of gears required are 4

Kinematic diagram Let Z1  No . of Teeth of Driver  20 teeth N 1  710 rpm N 2  540 rpm

But

Z2 Z1



N1 N2

 Z2 

N1 N2

 Z1 

710  20 450


Input D

1440 rpm

,,

1400 rpm B2

1120 rp m B1

E

900 rpm B

D

710 rpm A2

C

560 rpm A1

2 nd S tage 2(3 )

1 s t Stage 3(1) Ray D iag ram

450 rpm A

D riven

Z 6 = 40 7 10 rp m

6

Driver 1440 rpm

8

7

Z 7 =27

Z 8 = 33 1120 rpm

5

Z 5 = 20

9

Z 10 =37 900 rp m

10

Z 9 =2 3

4

3

1

900 rp m

Driven

Driver 710 rpm

Z 3 =2 9 Z 1 = 20

Z 2 = 32 (450 rp m )

2



 32 teeth  Z 1  20 tee th, Z2  32 teeth N3  710 rpm N4  900 rpm

Since centre distance is same Z 1  Z 2  Z3  Z 4  20  32  52  Z 3  Z 4  52

But

Z4 Z3

N3



N4

Z 4 

710 Z  Z 4  0.788 Z 3 900 3  0.8 Z 3

 Z 3  0.8 Z 3  52  1.8 Z3  52  Z3  29 teeth

But Z 3  Z 4  52  Z 4  52  29  23 teeth Z3  29 teeth, Z4  23 teeth

Similarly, Z6 Z5



 Z6 

N5 N6 N5 N6

 Z5

4.169


Assume Z5  20 teeth 1440  Z6   20  40 teeth 710

and hence Z5  20 teeth, Z6  40 teeth

Similarly Z 5  Z 6  Z 7  Z 8  60  Z 7  Z 8  60 Z8 Z7



 Z8 

N7 N8 1440  Z7 1120

 Z8  1.28 Z7  Z7  1.28 Z7  60  Z7  27 teeth

But Z 7  Z 8  60  Z 8  33 teeth  Z7  27 teeth, Z8  33 teeth

Since centre distance is same Z 5  Z 6  Z7  Z8  Z9  Z10  60  Z 9  Z 10  60

But

Z 10 Z9

 Z 10 



N9 N 10

1440 Z  Z 10  1.6 Z9 900 9


 Z 9  1.6 Z9  60  Z9  23 teeth

and

Z10  37 teeth

Check to avoid interference Z 7  Z 5 and Z 7  Z 9  4 27  20  7 and 27  23  4

The condition is satisfied. Therefore, the calculation of No. of teeth on each gear is satisfactory. Problem 4.17: The following data refers to an overlapping Speed Gear Box: No. of speeds required  14 Minimum speed

 20 rpm

Maximum speed

 400 rpm

Motor power

 11 kW at 970 rpm

Draw speed diagram, and kinematic arrangement. Find the actual output speeds.

Solution N max  400 rpm N min  20 rpm Z  No . o f speeds  14 N max N min

 Z  1

 400     20 

1

14

1

 1.259

Motor speed I  970 rpm.

4.171


Trail 1 The above step ratio is nearly equal to R10 series. Therefore, Try R10 series; N 1  20 rp m

N 8  100 rpm

N 2  25 rpm

N 9  125 rpm

N 3  31.5 rpm

N 10  160 rpm

N 4  40 rpm

N 11  200 rpm

N 5  50 rpm

N 12  250 rpm

N 6  63 rpm

N 13  315 rpm

N 7  80 rpm

N 14  400 rpm

The last speed is coinciding with the N max .  Selection of speeds is satisfactory.

The kinematic arrangement will be similar to that of 18 speed gear box with 4 overlapping speeds. For 18 speeds structural formula, 18  3  3  2

P1  3

X1  1

Z  P 1X 1, P 2X 2, P 3X 3

P2  3

X 2  P1  3

P3  2

X 3  P 1P 2  9

P1

P2 P 3

 31, 33, 29

For 14 speed gear box, subtract 4 from the last stage character. Now the structural formula is modified as,


Z  3 1, 3 3, 2 9  4  3 1, 3 3, 2 5

I Stage - 3 speeds with one space interval. II Stage - 3 speeds with three space intervals. III Stage - 2 speeds with 5 space intervals. Draw the speed diagram similar to previous method in which four speeds are overlapping i.e., repeating Selection of Ninput in each stage 3rd Stage N max  63 rpm N min  20 rpm

Let N input  40 rpm N max N input

Check

2 N max N input



63  1.575 40

Take N input  40 rpm 2nd Stage N max  160 rp m N min  40 rpm Try, N input  100 rpm

which is less than 2  satisfactory,

4.173


N max N input



160  1.6 100

which is less than two

Take Ninput  100 rpm

Ist Stage For 1st stage, there is no need to check the condition   N max  2 . Therefore select Ninput as 400 rpm.  N input   Kinematic Arrangement Draw the Kinematic arrangement for 18 speed gear box. In 3rd stage, 2 speeds are required – Two pairs of gears required. In 2nd stage, 3 speeds are required – Three pairs of gears are required. In 1st stage, 3 speeds are required – Three pairs of gears are required. To Find number of teeth on each gear. III Stage I Pair: Reducing speed N2  speed of the driver  40 rpm ; Z2  No. of teeth on driver; N1  speed of the driven  20 rpm ; Z1  No. of teeth on driven;


(O u tp ut)

15

13

11

7 9

* **

10

8

6

5

4 3

2

1


* 16 14 12

4.175


V.R 

Speed of the driven N 1 Z 2   Speed of the driver N2 Z 1

Take Z 2  20 teeth 20 20  40 Z 1 Z1 

40  20  40 20


II Pair: Increasing speed N 4  Speed of the driver  40 rpm ;

Z 4  driver teeth;

N 3  Speed of the driven  63 rpm ;

Z 3  driven teeth;

V.R 

Speed of the driven N 3 Z 4   Speed of the driver N4 Z 3

Z 1  Z 2  40  20  60 63 Z 4  40 Z 3 Z 4  1.575 Z 3

Z 4  60  24  36 Z3  24 teeth Z4  36 teeth

Z 3  Z4  60 Z 3  1.575 Z 3  60 60 Z3   23.30 teeth 2.575 take Z 3  24 teeth

4.177


II Stage 1st Pair: Reducing speed. N 5  speed of the driven  40 rpm ; Z 5  No. of teeth on driven; N 6  speed of the driver  100 rpm ; Z 6  No. of teeth on driver;

V.R 

Speed o f the d riven N 5 Z 6   Speed o f the driver N6 Z 5

Z6 40  100 Z5 Take Z 6  20 teeth Z5 

100  20  50 teeth 40

1st Pair Z5  50 teeth Z6  20 teeth

II Pair: Increasing speed. N 8  speed of the driver  100 rpm ; Z 8  No. of teeth on driver; N 7  speed of the driven  100 rpm ; Z 7  No. of teeth on driven;

V.R 

Speed of the driven N 7 Z 8   Speed of the driver N 8 Z 7

Z 5  Z 6  Z 7  Z 8  50  20  70 Z7  Z8  70

... (1)


160 Z 8  100 Z 7 Z 8  1.6 Z 7

... (2)

Z 7  1.6 Z 7  70 Z7 

70  26.92 2.6

Take Z 7  27 teeth Z 8  70  27  43 Z7  27 teeth Z8  43 teeth

III Pair: Intermediate speed N 9  Speed of the driven  80 rpm ; Z 9  No. of teeth on driven; N 10  Speed of the driver  100 rpm ; Z 10  No. of teeth on driver; Velocity ratio 

N9 N 10



Z 10 Z9

Z 10 80  100 Z9

Z 9  Z 10  Z 5  Z 6  50  20 Z 9  Z 10  70

1

Z 10  0.8 Z 9

From equation (1) and (2) Z 9  0.8 Z 9  70 Z9 

70 1.8

2

4.179


Z 9  38.88 ;

Z 10  70  39  31


To avoid interference, check Z 8  Z 6  43  20  23 Z 8  Z 10  43  31  12

Since (1) Z 8  Z 6  4 and (2) Z 8  Z 10  4  Calculation of number of teeth on each gear is

satisfactory. Z5  50 teeth

Z8  43 teeth

Z6  20 teeth

Z9  39 teeth

Z7  27 teeth

Z10  31 teeth

I Stage I Pair: Reducing speed. N 11  Speed of the driven  100 rpm ; Z 11  No. of teeth on driven; N 12  Speed of the driver  400 rpm ; Z 12  No. of teeth on driver; V.R 

Speed of the d river N 11 Z12   Speed of the driver N 12 Z11


Assume

Z 12  20 teeth ; Z 11 

100 Z 12  400 Z 11

20  400  80 Teeth 100

Z11  80 Teeth Z12  20 Teeth

II Pair: Increasing Speed N 13  Speed of the driven  160 rpm ; Z 13  No. of teeth on driven; N 14  Speed of the driver  400 rpm ; Z 14  No. of teeth on driver; V.R 

N 13 N 14



Z 14

Z 11  Z 12  80  20  100

Z 13

Z 13  Z 14  100

160 Z 14  400 Z 13 Z 14  0.4 Z 13

2

Z 13  0.4Z 13  100 Z 13  Z 14  100 Z 13 

100  71.42  72 1.4

Z 13  72 teeth Z 14  100  72  28 teeth

4.181


II Pair Z13  72 teeth Z14  28 teeth

III Pair: Intermediate speed N 15  speed of the driven  125 rpm

Z 15  driven teeth;

N 16  speed of the driver  400 rpm

Z 16  driver teeth;

V.R 

driven speed N15 Z 16   driver speed N 16 Z15

125 Z16  400 Z 15

Z 15  Z 16  Z 13  Z 14  Z 11  Z 12  77  23  100

Z 16  0.3125  Z 15, Z 15  0.3125 Z 15  100 Z 15  76.19  77 Z15  77 teeth Z16  23 teeth

To avoid interference, Check 1

Z 14  Z12  4

2

Z 14  Z16  4

1. 28  20  8 which is greater than (4) 2. 28  23  5 which is greater than (4) Selection of Gears is satisfactory. Therefore, calculation of No. of teeth on each gear is satisfactory.


To find actual output speeds Output speed (1)  input speed 

Output speed (1)  400 

 400 

Z 14 Z 13

No. of teeth on driver No. of teeth on driven 

Z8

Z7



Z4 Z3

28 43 36   72 27 24

 371.60 rpm


 400 

Z 14 Z 13



... (1) Z8 Z7



Z2 Z1

28 43 20   72 27 40

 123.86 rpm


 400 

Z 14 Z 13



... (2) Z6 Z5



Z4 Z3

28 20 36   72 50 24

 93.33 rpm


 400 

Z 14 Z 13

... (3) 

Z6 Z5



Z2 Z1

28 20 20   72 50 40

 31.11 rpm

... (4)

4.183


Output speed (5)  400   400 

Z 14 Z 13



Z10 Z9



Z4 Z3

28 31 36   72 39 24

 185.47 rpm

Output speed (6)  400   400 

Z 14 Z 13



... (5) Z10 Z9



Z2 Z1

28 31 20   72 39 40

 61.82 rpm

Output speed (7)  400   400 

Z 16 Z 15

... (6) 

Z8 Z7



Z4 Z3

23 43 36   77 27 24

 285.42 rpm

Output speed (8)  400   400 

Z 16 Z 15



... (7) Z8 Z7



Z2 Z1

23 43 20   77 27 40

 95.14 rpm

Output speed (9)  400   400 

Z 16 Z 15

... (8) 

Z10 Z9



Z4 Z3

23 31 36   77 39 24

 142.45 rpm

... (9)


Output speed (10) 400 

 400 

Z 16 Z 15



Z10 Z9



Z2 Z1

23 31 20   77 39 40

 47.48 rpm


 400 

Z 16 Z 15

... (10) 

Z6 Z5



Z4 Z3

23 20 36   77 50 24

 71.688 rpm


 400 

Z 16 Z 15



... (11) Z6 Z5



Z2 Z1

23 20 20   23.89 rpm  77 50 40 ... (12)


 400 

Z 12 Z 11



Z8 Z7



Z4 Z3

20 43 36   80 27 24

 238.88 rpm


 400 

Z 12 Z 11



... (13) Z8 Z7



Z2 Z1

20 43 20   80 27 40

 79.62 rpm

... (14)

4.185



 400 

Z 12 Z 11



Z10 Z9



Z4 Z3

20 31 36   80 39 24

 119.23 rpm


 400 

Z 12 Z 11



... (15) Z10 Z9



Z2 Z1

20 31 20   80 39 40

 39.74 rpm


 400 

Z 12 Z 11

... (16) 

Z6 Z5



Z4 Z3

20 20 36   80 50 24

 60 rpm


 400 

Z 12 Z 11

... (17) 

Z6 Z5



Z2 Z1

20 20 20   80 50 40

 20 rpm

... (18)

Problem 4.18: Design a Gear box, for the following data: No. of speeds required  16; Min. speed  100 rpm; Max. speed  560 rpm; Draw neat diagram and Kinematic arrangement. (May/June 2013)


Given:

N max N input

N max  560 rpm;

  Z  1

560   16  1 100

N min  100 rpm;

No. of speeds  Z  16

  1.12

which is standard ratio for R20 series. N 1  100 rpm;

N 9  250 rpm;

N 2  112 rpm;

N 10  280 rpm;

N 3  125 rpm;

N 11  315 rpm;

N 4  140 rpm;

N 12  355 rpm;

N 5  160 rpm;

N 13  400 rpm;

N 6  180 rpm;

N 14  450 rpm;

N 7  200 rpm;

N 15  500 rpm;

N 8  224 rpm;

N 16  560 rpm;

Structural formula Z  16  4  2  2

P1  4

X1  1

3 stages required.

P2  4

X2  P1  4

P3  2

X3  P1 P2

P1

P2 P3

Z  P 1X 1 , P 2X 2 , P 3X 3 st

1 Stage

16 

41  st

1 Stage

2

nd

,

Stage

24 2

nd

3

rd

Stage

, 28 rd

Stage 3

Stage

428

4.187


3rd Stage To select N input, Select N input  200 rpm N max N input



250  1.25 200

which is  2  Satisfactory. N min N input



100  0.5 200

which is  0.25  Satisfactory. Select Ninput  200 rpm - 3rd Stage Locate (C) 2nd Stage Select N input  280 rpm N max N input



315  1.125 280

which is less than (2)  Satisfactory N min N input



200  0.7142 280

which is greater than (0.25)  Satisfactory Select Ninput  280 rpm - 2nd Stage. Locate (E) 1st Stage In this stage, N input  560 rpm (No need to check condition)


I

Input

D3 D2

H D1

G

F

D

E

C3

C

B7

560

B6

500

B5

450

B4

400

B3

355

B2

315

B1

280

B 250 2

C1 C

A7

224

A6

200

A 5 180

st

1 Stage 4 (1)

nd

2 S tag e 2(4 )

rd

3 Stag e 2 (8 )

A4

160

A3

140

A2

125

A1

112

A

100

1

6 9 15

13

11

12 10

7

8

5

(1 00)

3

4

2


16 1 4

4.189


Locate I:   

In the 3rd stage, above point (C); mark (D) after four space intervals. i.e. at 315 rpm. In the 2nd stage, above point (E); mark (F), (G), (H) with one space interval. Join CA;

CB

100 rpm 250 rpm

From C 1 Join IF, FC 1

C 1A 1; 112 rpm

224

 C 1B 1 280 rpm

Join IG From G GC 2  250

GD 2  400

From C 2 C 2A 2  125

C 2B 2  315

Join I H From H HC 3  280

From C 3 C 3A 3  140

C 3B 3  355

HD 3  450


4.191

Repeat the same process; And complete the speed diagram. Kinematic Arrangement No. of speeds required are 16 The no. of stages are (3): 4  2  2  The no. of gears required  24  2  2  28  16

Z  P 1X 1, P 2X2, P 3X 3 16  41, 1

st

Stage

24, 2

nd

Stage

28 3

rd

Stage

Calculation of No. of Teeth on each gear III Stage I Pair: Increasing Speed VR

N1 N2



Z2 Z1

Z2

250  ; Z 2  31.25 200 25 Z 2  32 teeth Z1  25 teeth Z2  32 teeth

N1  Speed of driven  250 rpm N2  Speed of driver  200 rpm Z1  no. of teeth on driven  25 assume Z2  No . of teeth on driver


II Pair : Decreasing Speed VR 

N3 N4



Z4 Z3

;

100 Z4  200 Z3

N3  Speed of driven  100 rpm N4  Speed of driver  200 rpm Z3  No . of teeth on driven Z4  No . of teeth on driver

Since centre distance is constant, Z 1  Z 2  Z 3  Z4  25  32  57 57  3.8 teeth Z 3  Z 4  57; Z 3  0.5 Z 3  57; Z3  1.5 Z 4  19 teeth Z3  38 teeth and Z4  57  38  19 teeth

II Stage I Pair : Increasing speed: VR

N5 N4



Z6 Z5

315 Z 6  ; Z 6  28.1  28 teeth 280 25


N7  Speed of driven  200 rpm N8  Speed of driver  280 rpm Z7  No . of teeth on driven Z8  No . of teeth on driver


4.193

II Pair: Decreasing Speed VR

N7 N8



N5  Speed of driven  315 rpm

Z8

N6  Speed of driver  280 rpm

Z7

200 Z 8  ; Z 8  0.714 Z7 280 Z 7

Z5  No . of teeth on driven  25 assume Z6  No . of teeth on driver

But Z 7  Z 8  Z5  Z 6  25  28  53 Z 7  0.714 Z 7  53 Z 7  30.9  31 teeth and Z 8  53  31  22 Z7  31 teeth Z8  22 teeth N9  Speed of driven  315 rpm

I Stage I Pair: Increasing Speed VR

N9 N 10



Z 10

N10  Speed of driver  560 rpm Z9  No. of teeth on driven  40 assume Z10  No. of teeth on driver

Z9

315 Z 10  ; Z 10  22.5 40 560 Z9  40 teeth and

Note: In any calculation of number of teeth on the gear wheel, the value must be greater than 17; Because of this reason, Z 9 is assumed as 40 teeth.

Z10  23 teeth

II Pair : Decreasing Speed VR 

N 11 N 12 Z 12



Z 12 Z 11


;

N12  Speed of driver  560 rpm Z11  No. of teeth on driven

280 Z12  No. of teeth on driver  ; Z 12  0.5 Z 11 560 Z 11


63  42 teeth Z 11  0.5 Z 11  63; Z 11  1.5

and Z 12  6  42  21 teeth Z11  42 teeth Z12  21 teeth

III Pair: Increasing speed: VR

N 13 N 14



Z 14 Z 13

Z 14  Z 13  63 0.71428 Z 13  Z 13  63 Z 13 

63 ~ 37 1.71428

Z 14  63  37  26 Z13  37 teeth Z14  26 teeth N 13  S pe ed of driven  400 rpm N 14  S pe ed of driver  560 rpm Z 13  N o. of teeth on driven  37 teeth assume Z 14  N o. of teeth on driver

IV Pair: Decreasing Speed VR 

N 15 N 16




Z 16

N16  Speed of driver  280 rpm

Z 15

Z15  No. of teeth on driven

Z16  No. of teeth on driver 355 Z 16  ; Z 16  0.633 Z 15 560 Z 15


4.195

But Z 15  Z16  Z13  Z 14  26  19  63 Z 15  0.633 Z 15  63 ; Z 15  39 Z 15  39 teeth and Z 16  63 39  24  Z15  39 teeth Z16  24 teeth Problem 4.19: Design the layout of a gear box for a milling machine having an output of speeds ranging from 180 to 2000 rpm. Power is supplied to the gear box from a 6 kW induction motor at 1440 rpm. Choose standard step ratio and construct the speed diagram. Decide upon the various reduction ratios and the number of teeth at each gear wheel. Sketch the arrangement of gear box.

(April/May 2008)

Solution Given N max  2000 rpm N min  180 rpm P  6 kW Z  8 assum ed since not given N max N min

 Z  1

2000  8  1 180    1.41


Consider R20 series, which is   step ratio  1.12  1.12   1.12  1.12  1.40

skip 2 speeds

Try R20 series by skipping (2) speeds (From P.S.G. Data book, Pg.No. 7.20) N 1  180 rpm, N 2  250 , N 3  400, N 4  560 N 5  800 rpm, N 6  1000 , N 7  1400 , N 8  2000

Structural formula Z222 8  P 1, P 2, P 3  P 1 X 1, P 2 X 2, P 3 X 3  2 1, 2 2, 2 4 P1  2 ; P2  2 ; P3  2 X1  1  1 X2  P1  2 X3  P1  P2  2  2  4 U  N o. of stages  3  draw u  1  3  1  4 vertical lines

Since 8 speeds, draw 8 horizontal lines Procedure for the construction of ray diagram 3rd stage: 2 speeds, 4 spaces Locate input speed (i.e.) point C Point A is minimum speed and B is the maximum speed in stage 3

4.197


N max N input

Select

2

N input  560 rpm 800 2 560 N min N input



1  0.25 4

180 1   0.25 560 4  Selection of N input  560 is satisfactory.  Locate point ‘C’.

Similarly mark all the points and complete the ray diagram. Kinematic diagram No. of gears required are  2 2  2  2  2 6  12 Gears

Calculation of number of teeth III stage Speed of driven N 1 Z2   Speed of driver N 2 Z1

(i.e.) N 1  180, N 2  560 Assume above 17 teeth Let Z2  20 teeth


H

2 00 0

1 40 0

1 00 0

D

F

8 00 B

E

5 60

C

4 00

2 50

I S tag e 2 (1)

II S tag e 2 (2) R ay D iag ram

III S tag e 2 (4)

1 80 A

11

12

I STA G E

Z 11 = 29 10 00 rpm

Driven

D river 1440 rpm

Z 1 2 = 20

10

7

I I S TA G E

Z 7 =2 9 5 60 rpm

D riven

8

D river 800 rpm

Z 8 =2 0

Z 5 = 22 10 00 rpm

5

6

Z 6 =2 7

Kinem atic Diagram

Z 9 = 31 80 0 rpm

9

Z 1 0 =1 8

3

4

2

1 80 rpm Z 1 = 62

Driven

Z 2 = 20

III STA G E

Z 3 = 34 800 rp m

D river 560 rp m

Z 4 = 48

1





180 20  560 Z 1

 Z1  62 teeth

Similarly

N3 N4



Z4

Z3

Z1  Z2  Z3  Z4 20  62  Z 3  Z 4  Z 3  Z 4  82 800 Z 4   Z 4  1.42 Z 3 560 Z 3  Z 3  1.42 Z 3  82  Z3  34 teeth

and

Z4  48 teeth

II stage N7 N8



Z8 Z7

Assume Z8  20 560 20  800 Z 7  Z7  29 teeth

Similarly N5 N6



Z6

Z5

Z 7  Z 8  Z5  Z6 29  20  Z 5  Z 6

4.201


 Z 5  Z 6  49

also 1000 Z 6  Z 6  1.25 Z 5  800 Z5

Hence Z5  22 teeth Z6  27 teeth

I stage Z11 Z12



N 12 N 11

Assume Z12  20 1440 Z 11   1000 20 Z11  29 teeth

Similarly Z 9  Z 10  Z11  Z 12

(i.e.)

Z 9  Z 10  49

and Z9 Z 10



N 10 N9



Z9 Z 10

 Z 9  1.8 Z 10 1.8Z 10  Z 10  49  Z10  18 teeth Z 9  49  18 Z9  31 teeth



1440 800


Problem 4.20: A three speed reduction gear box is to have the following speed ratios as nearly as possible. Bottom gear 5:1, second gear 3:1 and top gear 1.5:1. Output shaft and input shafts are to be in line and the centre distance which is horizontal between them and the lay shaft is 144 mm. All the gears are to be 4 mm module and none should have less than 20 teeth of 20 pressure angle. Determine the suitable No. of teeth and pitch circle diameter of all the gear wheels. Find the actual speeds obtained. 

(Oct. 2007)

Gears (1) and (2) Z 1 and Z 2 are the constant mesh gears.



Gears (4), (6), (8) Z 4, Z 6 and Z 8 are sliding gears on lay shaft.



Gears (3), (5), (7) Z 3, Z 5 and Z 7 are fixed gears on output shaft.



Gears (3) and (4) Z 3 and Z 4 - Form Top gear



Gears (5) and (6) Z 5 and Z 6 - Form Second gear



Gears (7) and (8) Z 7 and Z 8 - Form Bottom gear

Given I. Bottom Gear Speed ratio 



Speed of driver No. of teeth on driven  Spee d of driven No. of teeth on driver N1 N2



N8

N7





Z2 Z1

Z2 Z1





Z7

Z8

Z7 Z8 5

5

... (1)

4.203


*

*

*

*

*

II. Second Gear Speed ratio 

N1 N2 



N6 N5

Z2 Z1



Z2



Z1

Z5 Z6



Z5 Z6

3

3

... (2)

III. Top Gear Speed ratio 



N1 N2 Z2 Z1





N4 N3 Z3 Z4



Z2 Z1



Z3 Z4

 1.5

 1.5

... (3)

Since the centre distance is same for all the gears, Z 1  Z 2  Z 3  Z 4  Z5  Z6  Z7  Z 8


But, distance between output shaft and lay shaft  144 mm given i.e., PCRGear 1  PCRGear 2  144  PCD   2  Gear 

 PC D    144 2  1  Gear 2

But, Module  m 

PCD Z



D Z

where Z  No. of teeth. D or PCD  pitch circle diameter.

DmZ

 mo dule  Z 1    2 Gear  module  m  4 mm 

4 Z1 2



4 Z2 2

 module  Z2   144   2 Gear 2 1  given 

 144

 Z 1  Z 2  72 teeth  Z3  Z4  Z5  Z6  Z7  Z8

Let Z 8  20 teeth Z 7  Z 8  72  Z 7  72  20  52 teeth

4.205



Bottom Gear From eq. (1), Z2 Z1



Z7

Z8 Z2 Z1

5 5

Z8 Z7

5

20  1.923 52

 Z2  1.923 Z 1

... (4)

Z 1  Z 2  72

But

... (5)

From eq. (4) and (5) Z 1  1.923 Z 1  72 72  24.63  Z1  2.923

Take

Z 1  25 teeth ;

Z 2  72  25  47 teeth


Second Gear From eq. (2) Z2 Z1



Z5 Z6

3

47 Z5  3 25 Z 6


Z 5  1.596 Z 6

... (6)

Z 5  Z 6  72

... (7)

From eq. (6) and (7) 1.596 Z 6  Z 6  72 Z6 

72  27.73 2.596

 Take Z 6  28 teeth and Z5  72  28  44 Z5  44 teeth Z6  28 teeth

Top Gear From eq. (3) Z2 Z1



Z3 Z4

 1.5

47 Z3   1.5 25 Z 4 Z 3  0.7978 Z 4 Z 3  Z 4  72

From eq. (8) and (9) 0.7978 Z 4  Z 4  72 Z4   Take

72  40.04 1.7978 Z 4  40 teeth

... (8) ... (9)

4.207


Z 3  72  40  32 Z3  32 teeth Z4  40 teeth

Actual Speeds Speed of driven N o. of teeth on driver  Spe ed of driver No. of teeth on driven

Top Gear Speed of driven (output speed)  Speed of driver 

Z1

Z2



Z4 Z3

(Assume speed of driver  2000 rpm ) Output speed for Top gear  2000 

25 40  47 32

 1329.78 rpm

Second Gear Output speed for 2nd gear  2000   2000 

Z1

Z2



Z6 Z5

25 28  47 44

 676.98 rpm

Bottom Gear Output speed for bottom gear

 2000   2000 

Z1 Z2



Z8 Z7

25 20  47 52

 408.16 rpm


Pitch circle diameter of gears D 1  mZ 1  4  25  100 mm D 2  mZ 2  4  47  188 mm D 3  mZ 3  4  32  128 mm D 4  mZ 4  4  40  160 mm D 5  mZ 5  4  44  176 mm D 6  mZ 6  4  28  112 mm

driven  N1  100 rpm driver  N2  160 rpm (input)

D 7  mZ 7  4  52  208 mm D 8  mZ 8  4  20  80 mm

4.7 CONSTANT MESH GEAR BOX It is a type of transmission used in motor vehicle application and also known as manual gear box. It uses a driver-operated clutch which is engaged and disengaged by a foot pedal or hand lever for regulating torque transfer from the engine to the transmission system. All transmissions donot have 4 speeds forward and all donot have the same gear reductions at the various speeds. For example: Passenger cars usually have only three forward speeds and one reverse speed. We must remember that, the gear reduction in the transmission is only between the engine and the propeller shaft. In this type of transmission, certain countershaft gears are constantly in mesh with the main shaft gears. Constant mesh gear box is an example for the application of ordinary gear train in automobiles. In this type, the main shaft gear wheels and the lay shaft gear wheels, giving different ratios on engaging with each other,

4.209


Sh iftin g leve r G ear Se lector Fo rk 3

4

Fro m En gine

5

R

D og D 3

2

Dog D1

1

Dog D2

En gagem en t d og s

To D iffere ntial M ain shaft

A Prim ary sha ft

D

C B

D1 F

E

H

D2 J

L D3

G I K L aysh aft Fig.4.3 Con stant Mesh Gear B ox

P

Idle r N G ear

M

are always in mesh (i.e), all the gear wheels are always being meshed with their pair. The general layout of constant mesh type is similar to that of the sliding mesh, however it differs in the terms of gear wheel type and wheel teeth design. Constant mesh gear box uses double helical gears for its transmission, the main benefit of double helical (or) stronger helical gear is that it helps in “quieter (less noisy) operation” The helix arrangement on the teeth (or) the gear tooth is always in contact with its pair which reduces the chattering effect which is quiet common in case of straight cut teeth (or) spur gear type.


In constant mesh gear box set up, the main shaft revolves at different speed and the connection between any one of the wheel and main shaft is obtained by “Dog clutch members”. The forward speed of the gear box works on constant mesh principle and reverse speed on sliding mesh principle. 4.7.1 Constant Mesh Gear Box Gear Shifting 1. First Gear: Drive is through ABCD gear pairs and D1 engages with D. The engagement dog D 1 is shifted to the first pair that is towards the primary shaft, the first gear gets locked to the main shaft thus the main shaft starts rotating in the first gear. (i.e. low speed but high torque). 2. Second Gear: Speed Gear Drive is through ABEF and D1 engages with F. The clutch pedal is pressed thus disconnecting the transmission, now the dog D 1 is shifted from first pair to the second pair that is away from the primary shaft. Now the clutch is released and the main shaft is at gear 2. 3. Third Gear: Drive is through ABGH and D2 engages with H The transmission is disconnected by pressing the clutch. The engagement dog D 2 is shifted towards the primary shaft to the Gear pair (3), thus the main shaft is locked to gear (3) and the clutch is brought to engaged position.


4.211

4. Fourth Gear: Drive is through ABIJ and D2 engages with J The engagement dog D2 is shifted away from the primary shaft to gear pair (4). 5. Fifth Gear or Top Gear: Drive is through ABKL and D3 engages with L The engagement dog D 3 is brought towards the primary shaft to the gear pair (5) with the highest gear ratio. 6. Reverse Gear: Drive is through ABMNP and D3 engages with P The dog D3 is shifted to the idler gear which is in constant mesh. This changes the direction of drive and the reverse gear is obtained.

4.8 SPEED REDUCER UNIT 

Worm Reducer



Helical Reducer



Miler Bevel gear reducer

4.8.1 Worm Reducer There are two types of worm gear reducers. Cylindrical type (worm power drive) and drum shaped type (Troi drive). A worm gear reducer is a reduction gear box which consists of a helical output gear and a worm input gear where a worm (which is a gear in the form of a screw) meshed with a worm gear (which is similar in appearance to a spur gear).


4.8.2 Helical Reducer It has high power and is a compact with standard right angle and parallel shaft types, including motor types. The helical power drive uses a modular housing to accomodate any mounting position. 4.8.3 Bevel Reducer Bevel gears are gears where the axis of the two shafts intersect and the tooth bearing faces of the gears themselves are conically shaped. There are four types of bevel gears. 

Straight bevel gears



Spiral bevel gears



Zerol bevel gears



Hypoid bevel gears

4.9 VARIABLE SPEED GEAR BOX Fluid couplings and torque convertes are used as variable speed gear box to suit the varying engine speed. 4.9.1 Fluid coupling (or) Fluid fly wheel It is a hydro dynamic device which is used to transmit rotating mechanical power. It has been used in automobile transmission system as an alternative to a mechanical cluth. It also has widespread applications in marine and industrial machine drives, where variable speed operations and controlled start-up without shock loading of the power transmission system are essential. It is also called as fluid flywheel. A fluid flywheel is a hydraulic device. In 1905 Dr. Hermann Fottinger, a chief engineer of the Vulcan Werke

4.213

Design of Transmission Systems Sh ell

Vanes

C o re C o re R u nn er H o using

,nput Sh aft

O utput S haft

,m peller or R u nn er

O il C ircula tio n ,m peller

Fig.4.4 Fluid Flywheel.

AG in Germany invented the fluid flywheel. The fluid flywheel is now a fundamental part of modern car design and is used in other applications including diesel locomotives. Fluid flywheel is a kind of fluid coupling in which the flywheel is the driving rotor. Fluid flywheel is a fluid device utilizing recirculated fluid within a sealed housing to maintain speed or torque applied to a mechanical apparatus without relying on inertia of weight. A fluid-flywheel consists of two concentric housings a main circulation chamber and an angular recirculation chamber surrounding the main circulation chamber. A shaft is mounted on the bearing of the outer housing and the extension of the shaft has a series of blades mounted thereon. Upon rotation of the shaft, fluid is drawn by the blades into the inlet end of the main circulation chamber.


The fluid is propelled from one series of blades to the next until it is discharged from the outlet end. The fluid is then redirected around the inner housing and re-circulated through the recirculation chamber. The fluid, being constantly re-circulated within the device will increase in speed or maintain the existing speed with relatively minimum force being applied to rotate the extending shaft. Toroidal shaped flow deflectors at opposing ends are useful for smooth transition of fluid from the main circulation chamber to the recirculation chamber and vice versa. Van es

E n gine C ra nk S h aft H u bs ,m pe ller

S h ell

C lutch S ha ft R o tor

Fig.4.5. Tw o H alve s of F luid F lywh eel.

The driving unit is called impeller and the driven limit is called as runner. When the efficiency of the liquid coupling is highest, the impeller and the runner almost runs at same speed. But usually the runner speed is less than the impeller. This speed lag of runner is called slip. The slip is greatest when the vehicle is at rest that is when the runner is stationary. Fluid coupling is used with conventional clutch and transmission to enable the driver to use clutch and gear


4.215

with less skill and fatigue as compared to the mechanical linkage. Applications 

Fluid couplings are used in many industrial applications involving rotational power in machine drives.



Rail transportation and Hydraulic braking systems.



Automobile and Aviation systems.

4.9.2 Torque Convertor A torque converter is a hydraulic fluid coupling that is used to transmit power from one or more engines or motors to a driveshaft or other output shaft. It takes the place of a mechanical clutch, and within certain operating speed ranges, multiplies input torque, providing the output equivalent of a reduction gears. Torque converters are commonly found in automotive transmission system and also used in marine systems and other various machine tools. We have read about manual transmission, system in which an engine is connected to a transmission by way of a clutch. Without this connection, a car would not be able to come to a complete stop without stopping the engine. But cars with an automatic transmission have no clutch that disconnects the transmission from the engine. Instead, they use an amazing device called torque convertor. 4.9.2.1 Torque Converter Gear Box A torque convertor is a fluid coupling or clutch that disengages when the vehicle is stopped and transfers power as engine rpm increases. When the vehicle is in motion


and not under acceleration, the torque converter slips and transfers power as vehicle speed decreases. A torque converter, by its very nature as a fluid coupling also dampens engine combustion pulses to achieve smoother operation. “Torque converters have never been considered efficient” because they tend to lose more power than they transfer by their nature as “fluid coupling”. 4.9.2.2 Torque converter principles A fluid coupling is an assembly of two element drive i.e. the impeller and the turbine which is incapable of multiplying torque, while a torque converter has one extra element (the stator) in addition to impeller and turbine which alters the drive’s characteristics during periods of high slippage, producing an increase in output torque. The purpose of Torque converter is to transmit engine torque to the driving wheels. It also increases or decreases torque to suit varying road and operating conditions. In a manual transmission system, the driver manually selects the appropriate gear ratio to suit. In an automatic transmission, it is done by both the transmission control system, automatically selecting the gearing according to load and speed, and by the torque converter. The torque converter is fitted on the engine in the same place as a clutch, and does the same job as that of clutch-transmitting engine torque to the input shaft of the transmission. It also multiplies torque according to operating conditions.

4.217


It is also called as a fluid coupling, since the fluid is acting as the driving medium and none of the converter components are physically connected to the others. It acts as an automatic clutch. At engine idling speeds, the vehicle is kept stationary by allowing the engine to operate without transmitting power to wheels. Tu rbine

S tato r

,m pe ller

Fig.4.6. A Ty pical To rq ue C o nverte r.

Torque converter has basically three elements: 1. The Impeller, 2. Turbine, and 3. Stator. All the components - impeller, turbine and stator have curved vanes and are assembled in a single casing but are separated from each other by thrust bearings. The casing is filled up with full of fluid like oil. The impeller is keyed and attached to the converter casing and both the impeller and casing rotate as a single member. In simple words, the impeller has a large number of vanes rotating along with the casing as the engine rotates. Each vane has a slight curvature and is set radially


outward to the casing. The impeller is acting as a centrifugal pump. The fluid enters at center and leaves radially outward and then it enters the turbine and rotates the turbine. The turbine has more vanes, with a greater curvature and is set radially inward. The direction of curvature of the turbine vanes is opposite to that of the impeller vanes. The turbine has no direct mechanical link with casing. i.e., the turbine is free to rotate in the casing and a central spline of the turbine mates with a spline on the input shaft. When the fluid leaving the impeller rotates the turbine, the input shaft splined with turbine also rotates. There is a chance of fluid coming out of the turbine flows in the opposite direction to the impeller and hence slows down the impeller. Fig 4.7 (a). To avoid this, a stator is placed in between the turbine and impeller and this stator changes the direction of flow from turbine and provides the positive flow to the impeller. Fig 4.7 (b). The S tator Va nes

P um p

Fig.4.7(a). Torque Con verterOil is Throw n A gainst Pump Van es.

Turb in e H elping to R ota te Pum p

Turbin e S to pp ing P um p

P um p

P um p

Pu m p

Turbine

Turbine

Fig.4.7. (b ). Torque Con verterOil Path is Changed by Stator to Aid R otation of Pum p.

4.219

Design of Transmission Systems Torque C onve rte r H ousin g (C o nn ects to F lyw hee l a nd im pello r)

Turbine O u tpu t S ha ft (C o nn ects to Transm issio n)

Flyw heel (C o nn ects to En gine)

Stator O utp ut Shaft (C o nn ects to F ixed in Tra nsm issio n) Turbine

Pu m p(im pe llo r) ( Fixe d to H ou sing ) Stator Fig.4.8 Torque Con verter Cut Away(Assemb ly)

stator is free to rotate and no mechanical link with turbine, impeller and casing. The stator has a small set of curved blades attached to a central hub. The central hub is mounted on a one-way clutch to allow the stator to rotate only in the same direction as the impeller. 4.9.2.3 Advantages. 1. Provides noise free operation. 2. Fuel consumption is reduced. 3. Provides simplified driving control. 4. Wear and tear is greatly reduced. 5. Shocks and vibrations are greatly reduced. 6. Despite the efficiency loss, moderate slippage the torque converter provides a smoother, more even


flow of power by absorbing engine and power train’s vibrations rather than transmitting to the output shaft or surrounding equipment. 4.9.2.4 Applications 

Automatic transmissions on automobiles such as cars, buses and trucks.



Forwarders and other heavy duty vehicles.



Marine propulsion systems.



Industrial power transmissions such as conveyor drives, drilling rigs, construction equipments and railway locomotives.

Chapter 5

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.

5.1 CAMS: (CAM AND FOLLOWER) 5.1.1 Introduction In a cam drive, there are two main components, cam and follower. The cam acts as a driving member and is made to have miscellaneous boundary (usually curved shape) surface in order to transmit desired motion to the corresponding follower (the driven member) by a direct contact. The cam shape is like a disc with some projection on its face. The cam transmits the desired motion (linear or oscillatory) to the follower.

Cams are widely employed in most machineries like I.C engines, printing machines and other instruments because of its accurate and efficient transmission of motion, velocity and acceleration.

5.2


To an d fro mo tion R o ller

To an d fro mo tion



 (b)

(a) plate cam (b) w edge cam (or) Radial cam w ith Translating follower. (a)

Fig.5.1

5.1.2 Classification of Cam Mechanisms Cam Mechanisms can be classified into various types. (a) Based on input of cam and output of follower they are of four types. 1. Rotating cam-translating follower mechanisms. 2. Rotating cam-oscillating follower mechanism. 3. Translating cam-translating follower mechanism. 4. Stationary cam-rotating follower mechanism. (b) Based on the motion of follower it is of three types. 1. R-R-R (Rise-Return-Rise) Mechanism. 2. D-R-R-D (Dwell-Rise-Return-Dwell) Mechanism. 3. D-R-D (Dwell-Rise-Dwell) Mechanism.

Cams, Clutches and Brakes

5.3

5.1.3 Types of Cam 1. According to the shape (a) Wedge or flat cam. (b) Radial or disc cam. (c) Cylindrical or drum cam. (d) Spherical (e) Conical 2. According

to

the

method

of

provision

of

constraint (a) Spring loaded (b) Positive drive. 3. According the motion of follower (a) Uniform velocity. (b) Simple Harmonic Motion (SHM). (c) Cycloidal Motion. (d) Uniform acceleration and retardation. 5.1.4 Type of Cam followers 1. According to the construction of contact surface (a) Knife edge (b) Flat faced follower (c) Roller follower 2. According to the type of moment (a) Translating (b) Oscillating

5.4


3. According to location of follower (a) Radial (b) Off-set 

Radial: When the motion of the follower is along the axis passing through the centre of the cam.



Off-set: When the motion of the follower is along the axis away from the axis of cam.

5.1.5 Nomenclature of a Cam-Drive 1. Base Circle The smallest circle drawn to the cam profile from the centre of rotation of cam. The base circle decides the size of the cam drive. 2. Cam Profile This is the boundary curve of the contact surface of the cam with the follower. 3. Trace Point It is the point on the follower for the purpose of tracing the cam profile. 4. Prime Circle It is the smallest circle drawn to the pitch curve from the centre of rotation of the cam. 5. Pitch Curve The imaginary curve obtained when the follower is assumed to rotate the cam, keeping the cam fixed. It is the path of the trace point.


m axim um p ressure a ng le D irection o f m otion

N o rm al

5.5

D irection of m otion Follow er

Pitch po int

Tra ce point Pitch circle

nt Ta n g e

Cam profile Ra

Prim e circle Rb

Pitch curve

rp

Ba se circle

Fig. 5.2. C am Nom enclatu re.

6. Cam Angle  The angle of rotation of the cam for a definite displacement of the follower. 7. Pitch Point It is the point on the cam pitch curve having maximum pressure angle.

5.6


8. Pressure Angle () The angle at any point on the pitch curve included between the normal to the pitch curve and line of motion of the follower at that instant. 9. Stroke The maximum displacement of the follower from the base circle of the cam. 10. Pitch Circle It is the circle obtained when the centre of cam axis is taken as its centre and its radius such that it passes through the pitch point. 11. Transition Point The point on the cam at which the follower has its maximum velocity. 12. Pulse or Jerk It is used to define the instantaneous time rate of change of acceleration in order to minimise vibration. 5.1.6 Basic Curves Refer PSG Data book P.No. 7.110 for characteristic equations of basis curves. 1. Simple polynomial: y  c  n When n  1;

straight line y  c 

When n  2;

parabolic (or) constant acceleration.

y  c

2

When n  3;

cubic (or) constant pulse y  c 3


5.7

2. Trigonometric curves 1. Simple harmonic 2. Double harmonic 3. Cycloidal. Advantages of Trigonometric curves, over simple polynomial 1. Smaller cam size; 2. Easier layout; 3. Lower follower side thrust; 4. Low manufacturing cost; 5. Easier duplication. Problem 5.1: A polynomial cam having n  2.4 rotating at 60 rpm has positive acceleration (i.e., its direction is that of the positive displacement), part of its rise of 10 mm in 40 of cam rotation. Find the velocity and acceleration value at the end of 30 of cam rotation.

(April/May 2008)

Solution Given Data: n  2.4

When   40 ; the rise  10 mm y 

2N 60 2  60  6.283 rad/sec  60

   40 

  0.6981 rad . 180

Basic equation, y  c n

N  60 rpm

... (1)

5.8


10  c  0.6981 2.4 c 

10 0.6981 2.4

 23.69

differentiate the eq. (1) w.r.t. t Velocity 

dy d  c n dt d t

[Where    t]

v  c    n  n  1

 2

Velocity at the end of 30 of Cam rotation    v  23.69  6.283  2.4   30   180  

2.4  1

v  144.39 mm/sec

Differentiate the eq. (2) w.r.t. t Acceleration 

d 2y dt 2

 c   2  n n  1 n  2

    23.69  6.283  2.42.4  1   30   180   2

2.4  2

 Acceleration at the end of 30 of Cam rotation  2420 mm/sec 2

Cam Curvature 

If radius of roller  radius of curvature of pitch surface, then the profile is satisfactory. (i.e. No under cutting). Rr  Pk No und er cutting 




5.9

If radius of roller  radius of curvature of pitch surface, then the cam profile results under cutting. Rr  Pk U nder cutting results



If roller radius  radius of curvature of pitch surface, a pointed cam profile results.



A pointed cam will not give desired motion. (Refer PSG Databook Page No. 7.114)

To Avoid Under cutting Flat follower  P c

 1 d 2y   Rb   y  2  2  min  dt   0

to avoid under cutting

Roller follower

 Pk  min

  1 dy   R p  y2        dt 

  

32

2  1 d 2y   1 dy   R b  y  2  R p  y  2     dt dt 2    

 Rr 

2

2

to avoid under cutting

For y, at point of maximum negative acceleration.

where P c  radius of curvature of the cam profile P k  radius of curvature of the pitch curve R b  base circle radius;

R p  prime circle radius

5.10


y  follower displacement;



2N ; 60

R r  radius of the roller

dy d 2y  Velocity;  A cceleration dt dt 2

[For velocity and acceleration equations, refer PSG databook (Page No. 7.110) according to the type of curve] Under cutting in Cams When a cam has a sharp point (corner) the follower does not undergo an abrupt change of motion, but gradually assumes a different velocity. If the roller radius is greater than the theoretical pitch curve at the convex portion of the cam, the roller undercuts, i.e. moves itself over the same path twice. This phenomenon is called undercutting. Problem 5.2: A radial cam rotating at 150 rpm is driving 15 mm diameter translating roller follower to produce the following motion, rise of 25 mm with SHM in 150 degrees of cam rotation, dwell for 60, fall 25 mm with SHM in 120 of cam rotation and dwell for remaining 30. Construct the cam contour. (a) with pressure angle limited to approximately 20. (b) with the pressure angle limited to exactly 20 (April 2007)

Solution: (a) Approximate pressure angle: For SHM, the position of transition point is at

 150 i.e. 2 2

 At 75, locate transition point and pitch point


5’

6’

f

e

f’ e’

d

4’

d’

c

25 3’ mm

c’

b

2’

b’

a 1’

0

5.11

a’

1

2

3

4

o =15 0

5

P 6’ 5’ 4’ 3’ 2’ 1’ 0’

6

o

d 1

R

d 2

Displacem en t Diagram

Pitch circle radius  R p   Rp 

Take

pitch

fh 

4.315  25  41.2 mm. 5  6

circle

h  rise  25 mm ;   150 

radius

as

42

mm

;

 5   rad 180 6

CAM PROFILE (Procedure): 1. Draw pitch circle with a radius of 42 mm with Q as centre 2. Locate Transition Point and Pitch point (P) 3. Divide cam angle of rise 150 into equal number of parts (say 6) on either side of the point P. (3 equal parts on each side of P. 4. Draw the radial lines Q0, Q 1, Q 2, Q 3, Q4, Q 5 and Q6. 5. With P as centre and 25 mm (rise of follower) as radius, draw a semi-circle.


5.12

6. Divide the semi circle into same number of equal parts (i.e., six). Mark the point as 0, 1, 2, 3, 4, 5 and 6. 7. Take Q 0 as radius and Q as centre, draw arc on the radial line Q 0 and get 0. Similarly Q1 radius, Q as centre draw an arc on the radial line Q1

and get a similarly. 8. Similarly, draw arcs, with radius Q2, Q3, Q4, Q5 and Q 6, interest the respective radial lines, Q 2, Q3, Q4, Q5, and Q 6, to get b, c, d, e and f respectively.

R o ller Follow er

5’

3’

4’

4 d M otio n o f Follow er

Tra nsitio n a nd Pitch P oint (P) 2’

6 5 3 P 2

e

c

b

1

1’

0

6’

1 50 f 75

60

Cam Pro file

75

o

Q o

a

o

0’

o

o 30

o

0” 1 20

a’

o

Pitch C u rve

b’

6”

c’

f’ e’ 5”

Pitch C ircle

o’



d’

1” 2”

3”

4” o (a) Pressure An gle Limited to Appro ximately 20


5.13

9. Draw a smooth pitch curve passing through 0, a, b, c, d, e and f to get pitch curve. 10. With centers 0, a, b, c, d, e and f, draw circles with radius equal to roller radius. 11. Draw the cam profile (Tangential curve to the roller circles). 12. Repeat the same procedure during fall and dwell periods. (b) Exact pressure angle method: For SHM, R p  Pitch circle radius

Rp

and

P   h  sin       p 2  cos 

2

m  20   150  150 

p    m  tan 1   cot       p  20  tan  1  cot     150   150  180 180  tan 20  1.2 cot 1.2  p

cot 1.2  p 

0.36397  0.3033 1.2

1  0.3033 tan 1.2 p tan 1.2 p  3.29697 1.2 p  tan 1 3.29697

    

 ra d 180

5.14


1.2 p  73.12   p  60.03  61

 Rp 

  60.93  25  sin   150   180 

    

   60.93 2cos     150  180 

    

2

 39.149 mm

h  25 mm

Take R p  40 mm y 

  h  1  cos   2      61 25   1  cos 2   150   180 

   8.88 mm   

Take y  9 mm CAM PROFILE Procedure 1. With Q as centre draw a pitch circle with a radius of 40 mm R p 2. Locate pitch point P. 3. Draw a radial line Q 0 at an angle of 61   p ^ i.e. P Q 0  61. 4. Mark 150 cam angle with Q 0 as reference and divide into 6 equal number of parts and draw radial lines. Name the points as 1, 2, 3, 4, 5, and 6.


5.15

5. Locate y  9 mm below the pitch point at M. i.e. (PM  9 mm ). 6. With Q as centre Q M as radius, draw an arc on line Q3, which intersects at 0. 7. From 0 mark the rise  25 mm . Draw a semicircle with radius 25 mm, divide the semicircle into same number of equal parts (six), name the points as 0, 1, 2, 3, 4, 5 and 6 and project on the Q3 line.

m =2 0

3’

4

Follow er M otio n 5’

P

d

R o ller Follow er

3

Pitch P oint (P) 1’

P

2 1 0 M

a o

6’

30

f

0’ 0

15

0

9 mm

b

61

o

e

c

Tra nsitio n P oint

2’

6 5

4’

o

o

Pitch C ircle R P = 40 m m O”

 60

o

Pitch C u rve 1”

a’ o

120

Cam Pro file

b’ 2”

c’ f’ e’

3”

d’

6”

4” 5” (b) Pressure Angle Lim ited to E xactly 20

o

5.16


8. With Q as centre and radius as Q0, draw an arc on the radial line Q 0. Name this point as 0. 9. Similarly locate a , b , c , d, e and f 10. With 0, a, b, c , d, e and f, as centers, draw circles with radius equal to roller radius. 11. Draw a smooth pitch curve passing through 0, a , b, c , d, e and f . 12. Draw the cam profile, i.e. tangential curve to the roller circles 13. Repeat the same procedure during fall and dwell periods. 5.14 Simple Harmonic Motion SHM 2

a 

when   0;

 h   cos 2    

Acceleration is maximum positive

when     Acceleration is minimum negative when  

  Acceleration is zero. 2

Cycloidal: When  

 Acceleration is maximum. 4

Parabolic: When  

 Acceleration is maximum. 2

Cams of Specified Contours For accuracy of profile, and cheapness of manufacture, profiles should be formed of circular arcs and straight lines, then the motion of the follower can be calculated. This type of cams are known as “cams of specified contours”.


5.17

I. Circular Cam with Flat Follower When the flank of the cam are connecting the base circle and nose arc of convex circular arc, then the cam is known as Circular Arc Cam. II. Tangent Cam with Roller Follower When the flank of the cam are straight and tangential to the base circle and nose circle, then the cam is known as a Tangent Cam. (These cams are used for operating the inlet and exhaust valves of internal combustion engines). Circular Arc Cam with Flat Follower r 1  Base circle radius; r 3  Flank radius; r 4  Nose radius;

  Cam angle for displacement ‘y’ 1  Cam angle for acceleration 2  Cam angle for lift.

Distance OD  radius r 5: Refer PSG data book, Page No. 7.118 0    1 A to B Flank

Displacement y  r 3  r 1 1  cos  Acceleration y  a   2r 3  r 1 c o s  when   0; a is max. a max   2r3  r1


5.18

Flank

B

N ose

r4

D

C2 r3

r5

A  2

r1 O

Base circle

1

Fig. 5.3 Circular Cam w ith Flat Follow er. C1

1    2 B to D  Nose

y  r 4  r 1  r 5  r 4 cos  2   a   2r 5  r 4 c o s 2  

Put   2 a min   2r5  r4


5.19

Line diagram B

A r4

D r3

C2

r

r5

3

O D =r 5

O

1

Fig. 5.4

C1

a y

v

N o se

Flank  = cam a ng le for position



Fig. 5.5

5.20


B r4 c2

y y

r4

A r1 

c2

D

2 

r3

r3

r1

o

D

o

1

1 B A

 (a)

D C2

c1

(b)

o

c1

r 3 =C B CA

L r5 (=O D )

C 1 (c) Fig. 5.6 Circular A rc Cam with Flat F aced Fo llo wer.

r2

B1

A1

S

y A

B D

r4



B1

D1 A1

B

c2



A

r1

L rs (=O D ) o

s

y

r2

r1

D1 D

r4

c2

  

r 5 =O D o

(a) Fig. 5.7 Tangent Cam with Ro ller F ollower

(b)


F r2 s

B1

r4



A1

r1

D1 D

B

y A

5.21

c2

 

r3 r5

o 

Fig. 5.8 C1 C ircular A rc C am w ith R o lle r Fo llo w er

Problem 5.3: The following data refers to a circular cam which operates flat faced follower; Least radius  50 mm. Total lift  25 mm. Flank radius  150 mm. Angle of lift  70 degrees; Calculate the principal dimensions of the cam and draw the cam profile. r 1  Least radius  50 mm r 3  Flank radius  150 mm

Lift  25 mm 2  Angle of lift  70

From the diagram, C 1C 2  r 3  r 4  150  r 4

... (1)

5.22


O C1  r 3  r 1  150  50  100

... (2)

O C 2  r 1  lift  r 4  50  25  r 4  75  r 4 AOC 2  70, Therefore, C 2O C 1  110

... (3)

Using Cosine Rule C 1C 22  OC 21  O C 22  2O C 1  O C 2cos 110 150  r 42  100 2  75  r 42  2100 75  r 4 c o s 110 150  r 42  75  r 42  100 2  68.4 75  r 4 150 2  300r 4  r 24  75 2  r 24  150r 4  100 2  68.475  r 4 16875  150r 4  100 2  5130  68.4r 4 16875  15130  150 r 4  68.4r 4 1745  81.6 r 4 r4 

1745  21.38 mm 81.6

 Nose radius  r 4  21.38 mm  r 5  O C2  r 4  53.615  21.38  74.99 mm

r5  75 mm O D

To find 1, Apply sine rule,

From eq. (3) OC2  75  21.38  53.615 mm

 54 mm


OC 2 sin  1



C 1C 2

5.23

C 1C 2  r 3  r 4

sin 110 

= 150 - 21.38

128.62 53.615  sin 110 sin 1

= 128.62 mm

25

B

D

Li

ft

=

r4

r3

=

15

C2

0

70

- 

r5 E o

=7 2 r1

=

0

o

50 O

11O

A



C1

5.24


53.615  sin 110   0.3917 sin  1  128.62  1  sin 10.3917  23 

1  23 The principal dimensions are, r 4  21.38 mm;

r 5  75 mm

1  23  Cam Profile: (With Dimensions)

r1 = 5 0



=

O 11 0

70

o

o

r3 =

r4 = 15

0

47

21

B

o

C

2

D



=

C

1

23

o


5.25

Problem 5.4: The following data refers to a cam operating the suction valve of a 4-stroke petrol engine, least radius  18 mm; Lift  10 mm; Nose radius  3 mm; Crank angle when the suction valve opens before TDC  4; Crank angle when the suction valve closes after BDC  50; Cam shaft speed  80 rpm. The cam is a circular type with circular nose and flanks. Calculate: 1. Maximum acceleration and retardation of the valve. 2. Minimum force to be exerted by the spring to overcome inertia of the valve parts whose mass is 250 grams. (Nov 2013)

Given Data: 4-stroke petrol engine; r 1  L east radius  18 mm ; Lift  10 mm ; r 4  Nose radius  3 mm

Crank angle when suction valve opens before TDC  4 Crank angle, when suction valve closes after BD C  50 Cam shaft speed  80 rpm Refer PSG Data book Page No. 7.118 Circular arc cam - with flat speed follower A to B

- Acceleration  a   2r 3  r 1 cos  when   0 ; acceleration is maximum.  a max   2r 3  r 1 [Refer Fig.]

But r 3  r 1  OC 1 Calculate OC 1


5.26

A

r1

O

=5 8 .5 o

(18 0-5 8.5 )

r3 r5 B r4

C2 D

1 C1

B to D

put   2

- Retardation    2r 5  r 4 cos 2   But r 5  r 4  O C 2

a    2r 5  r 4

Calculate OC 2

Step 1. Find OC 1 and OC 2 OC 2  r 1  lift  r 4  18  10  3  25 mm C 1C 2  r 3  r 4  OC 1  r 1  r 4  O C 1  18  3  O C 1  15

Cams, Clutches and Brakes Su ctio n Valve O pen s

4

5.27

TD C

o

1 80 +5 0+ 4= 23 4 o

50

o

Su ctio n Valve C loses

BD C

Angle of crank rotation during suction  4  180  50  234

For 4-stroke cycle engine, Speed of cam shaft 

1 speed of crank shaft. 2

(Therefore cam angle is also 22 

1 angle of crankshaft) 2

234  117 2

Cam angle for lift   2 

117  58.5  2

5.28


Consider triangle O C 1C 2, apply cosine law C 1C 22  OC 21  OC 22  2O C 1  O C 2  cos 180  58.5 15  O C 12  O C 21  25 2  2  O C 1  25  cos 121.5 225  30O C 1  O C 21  O C 21  625  2OC 1  25  cos 121.5 

Simplify,

r 1  18 mm

O C 1  103.36 mm

r 4  3 mm

 C 1C 2  O C 1  15  103.36  15

lift  10 mm

 118.36 mm

Step 2. To Find 1 Apply sine rule, OC 2 sin  1



C 1C 2 sin 121.5 

25 118.36  sin  1 sin 121.5 

1  10.4  Step 3. Acceleration is maximum a max   2r 3  r 1

 103.36  18  121.36 mm

2

 2  80   121.36  18      60 1000      7.254 m/sec

r 3  O C 1  18

2

N  80 rpm



2N 60


5.29

Step 4. Retardation 2

a min

 2  80  25    O C2      1000  60  2

  1.754 m /sec 2

Step 5. Minimum value of force  F min  m  accelerationmin m  250 gram  0.25  1.754  0.44 N

5.1.7 Designing CAM 

Consideration The following points should be considered. 1. The rise should be kept as small as possible and achieved in the lowest possible time. 2. Backlash should be kept to a minimum. 3. The moving parts should be as rigid and light as possible. 4. The pressure angle should be kept as low as possible (upto 30). In order to reduce the pressure angle, off-set may be provided between the cam and follower. 5. The accuracy of the cam is important to avoid the vibration effects and therefore care should be taken to produce cam profile.

5.30


Procedure Step 1. From the given data, note down all the given information like the load to be lifted, torque supplied, etc, and then find the type of cam, follower and its motion based on the lift, velocity and acceleration required. Also select proper material to be used for cam and follower. Step 2. By assuming proper pressure angle, calculate the prime circle radius of the cam. Step 3. Calculate the minimum radius of curvature of the pitch curve by using formula (or) design chart and finalise the radius of follower and base circle of cam. Step 4. Using lift and cam angles draw the displacement angle diagram and then draw the cam profile. Step 5 Find the width of the face of cam based on the contact strength of cam material. Step 6. Find the dimensions of the follower with respect to their withstanding strengths, space limitation, etc. Step 7. Evaluate the spring parameter such as stiffness, spring index and length by using suitable formulae.


5.31

5.1.8 Cam Profile Layout The cam profile can be drawn by graphical method if the displacements for different cam angles are clearly specified. To draw a cam profile, the following points are to be considered. 1. The displacement time diagram is always given for a point on the follower which is utilized to draw the follower in its correct position at any instant. 2. The cam profile will always be tangent to the follower and will thus be determined by the envelope of the follower profiles drawn in successive position. 3. Cam profile has to be drawn on the basis of principle of inversion i.e, cam should be kept fixed and the frame carrying the follower is to be rotated in the opposite direction to that of cam. 5.1.9 Pressure angle : (Refer Fig. 5.9) Pressure angle is the angle between the follower motion and a normal to the pitch curve. This is very important angle for cam design, as its maximum value establishes the cam size, torque loads, side thrust wear and acceleration of follower, etc. Due to the angle between the follower motion and a normal to the pitch curve, the normal force F n may be resolved

into

two

components

F n cos 

and

F n sin  .

F n cos  is used to lift the follower and F n sin  to produce a side thrust on the guide of the follower.


5.32

Ca m Fo rce N o rm al N 2  Pre ssure A ngle

Su rfa ce/Pitch C u rve

y

A Rr

c le

Rp

P it

ch

C ir

h

Ca m Fo rce F n



Rb



R



F orce requ ire d to lift th e follo w er Fn Cos 



Rp 

In itital Po int

Ba se C ircle

Cam p rofile

Fig.5.9 (a)

Fo rce on th e guide, F n Sin 

Fig.5.9 (b)

The allowable value of  is influenced by the speed of operation and weight of connected part. The pressure angle can be determined by using the formula. tan  

Where,

R

1 dy R d

Distance of trace point from the centre of cam.

The trace point displacement R is given by R  Rp  y  Rb  Rr  y Where,

Rp 

Prime circle radius

Rb 

Base circle radius

Rr 

Radius of the roller of follower

y

Follower option.


For parabolid, harmonic and cycloidal cams, the prime circle radius can be evaluated using the formula assuming maximum pressure angle as  dy   dt  max  tan max  R p  y  

Suppose if offset is provided for the follower (Fig. 5.10) then pressure angle is given by tan  

Where y 

5.33

Pitch C u rve y’



Rp

e

o ffset

Fig. 5.10

y  e R 2p  e 2  y   dy d

and e  offset distance

5.1.10 Under Cutting When a cam has a sharp point, the follower does not undergo an abrupt change of motion, but gradually assumes a different velocity. If the roller radius is greater than the theoretical pitch curve at the convex portion of the cam, the roller undercuts, i.e., move itself over the same path twice, this phenomenon is called under cutting. Therefore, as long as the minimum radius of the curvature of the pitch curve P k is greater than the radius

5.34


of the roller R r the profile will be satisfactory and the desired displacement may be obtained. P k  R r (No under cutting)

If P k min  R r, then the profile will be a pointed one. If P k min  R r, then the cam will have an undercut which should be avoided. To avoid under cut For Flat Follower  1 d 2y  P C min  R b   y  2    dt 2    0 (to avoid under cutting)

Where P C min  radius of curvature of cam profile For Roller Follower 2

  1 dy   R p  y2        dt  P k min  2  1 dy  2  R b  R p  y  2   dt   

  

3/2

 1 d 2y  y  2   dt 2  

 R r to avoid under cutting 

For y at point of maximum negative acceleration.

where 

2  N dy  velocity , dt 60

and

d 2y dt 2

 Acceleration.


5.35

[For velocity and displacement equation refer PSG data book Pg No. (7.110) 5.1.11 Contact Stress A compressive stress is induced in the contact line during the force transmission between cam and follower and this stress is called contact stress. Hence the materials selected for the manufacturing of cam and follower should have sufficient endurance limit and hardness. The induced contact stress ‘c’ between cam and follower (roller) is given by the equation.

c  0.591

 

F n E 1 E2  1 1    b [E 1  E 2] R r P Cmin   

Note: Minus sign is used when the follower is in contact with concave portion of the cam profile. where, Fn 

Normal force at the point of contact.

E 1, E 2 

Moduli of follower

b

Face width of cam

P C min 

Minimum radius of curvature of cam profile

P C min  P Kmin  R r

elasticity

of

cam

and

5.36


W

al

N 1

R ising

5.1.12 Force Determination

No

rm

N1

B

F n cos 

Fn

A

N 2



Fn

N2 F n sin  L ifting F orce =F n cos 



Side Thrust =F n sin 

Fig:5.11

In the Fig. 5.11, W is the external force, which is the sum of inertia force and spring force that acts against the direction of motion of the follower. Therefore the power supplied should be in such a manner that the torque produced should develop the lifting force F , which should be greater than the external force. The torque required to drive the cam is given by T

Fv F n  cos   v   

and the normal force F n is obtained by

(here F   


Fn 

5.37

W  2A  B  cos    h   sin  B  

Where 

Co-efficient of friction.

A

Overhang length.

B

Follower bearing length.

Note: Parameters like spring dimensions determined based on the given condition.

are

Problem 5.5: A cycloidal cam with a central roller follower has a rise of 25 mm in angle of 70. Base circle radius is 80 mm and the follower roller radius is 20 mm. Speed of rotation of the cam is 5000 rpm. Mass of the follower is 0.5 kg. Find the

maximum

value

of

acceleration

of

the

follower,

corresponding pressure angle, stiffness of the spring used with the follower and maximum cam force. The function between the follower guide may be ignored.

(Nov/Dec 2007 - AU)

Given data h  25 mm   70  70 

Base circle radius r b  80 mm Follower radius r r  20 mm Speed

N   5000 rpm m  0.5 kg

 rad 180

5.38


Solution From P.S.G. Data book Pg. No.7.110 for cycloidal cam, the rise ‘y ’ of the cam is given by. y

h    1 sin 2      2     

Differentiating we get the velocity (From P.S.G data book Pg.No.7.110) h  2  dy v  1  cos      d

w.k.t for cycloidal cam, acceleration is maximum (Refer PSG Data book P.G.No. 7.110) when   0.25  y

 25   1    2 sin 2      

1    y  25  0.25  sin 2  0.25   2    y  25 0.25  0.00435  y  25 0.2456  y  6.14 mm

and v

 h dy   1  cos 2      d


we know that angular velocity is given by 



2N 60 2   5000 60

 523.59 ra d/sec 

25  523.59 dy  [1  cos 2  0.25   d   70  180    

13089.75 [1  0.2484] 1.221

 10720.91  0.7516  8057.53 mm/sec

From P.S.G Data book Pg. No.7.113  dy  d  Pressure angle tan    Rp  y

But

Rp  y  r  tan  

1 dy  r d

or  1 dy    tan  1    r d

W.k.t

5.39

5.40


R p  R b  Rr  80  20  100 mm  r  100  6.14 mm r  106.14 mm 1      tan  1   8057.53  106.14    89.2

Now maximum value of follower acceleration by the equation (Refer P.S.G Data book Pg.No.7.110)  d 2y  2  h 2 2 sin  A fmax   2  2    d  m ax  A f max 

2   25  10  3  523.3 2 1.22172

 sin 2  0.25

 28819.88  sin 2  0.25   28819.88  0.0273  788.53 m/s2

But inertia force F 1  m f A f max  0.5  788.53  394.265 N

The inertia force is maximum 394.265 N but acts upward. If the spring does not exert a force greater than


5.41

394.265 N, then contact between roller and cam will be lost, therefore spring force has to be set atleast 394.265 N. w.k.t the compression of spring for the follower rise  from lower most portion to the present level at  0.75   y  25 [0.75 

1 sin 2   0.75] 2

 25 [0.75  0.1591 sin 4.712 ]  25 0.7369  18.42 mm  Required minimum stiffness of the spring k  

F1 y 394.265 18.42

 21.40 N/mm

Cam forces At any instant, both inertia force and spring force are acting on the cam. During the phase cam rotation when  varies from 0.5 to 1, the total force on the follower, F is  the difference between the inertia force (upward) and the spring force (downward). Hence the maximum force during this phase is due to spring force and this occurs, when the  follower reaches the highest portion i.e  1 

5.42


F n max  F s  h  k  25  21.40  535 N

When

 from downward these two forces donot reach 

their maximum at the same time. Therefore it is necessary to calculate both the forces and obtain their sum for  different values of  1.

  0 ; y  0 ; Fs  F1  0 ;   0 

2.

  0.25 ; y  6.14 mm  F s  6.14  21.40  131.39 N

and F 1  394.265 N and

  89.2  F  394.265  131.39  525.661 N

3.

  0.375 ,   18.15   1   y  25  0.375  sin 2   0.375    2    6.56 m


5.43

F s  21.40  6.56 m  140.384 N F1  m f 



d 2y

2    sin 2    m h 2 f 2 2  d 

0.5  2   25  10  3  523.3 2  sin 2   0.375  1.22172

 592.4 N F  592.41  140.384  732.794 N

4.

 d 2y  0.5, y  12.5 mm , 0  d F s  12.5  21.46  267.5 N F1  0 F  F s  F 1  0  267.5 N F  267.5 N dy 25  [1  c o s 2   0.5] 1.2217 d  20.46 [0.5030]  10.29 mm/ra d r  100  12.5  112.5

5.44


 1   10.29    tan 1   112.5 

 5.22  The normal force acting between follower and guide 1. F n  9 for

 0 

2. F n 

 525.661 N  37648.9 N for  0.75  cos 89.2 

3. F n 

732.794 N cos 18.15 

 15128 for

4. F n 

  0.375 

267.5 N cos 5.22

 268.61 N for

  0.5 

Thus we find F n max  37648.9 N Problem 5.6: A radial cam rotates at 1200 rpm with the follower rising 20 mm with SHM is 150 of cam rotation. The roller is 32 mm in diameter and the prime circle is 80 mm in diameter. Check whether under cutting will occur.

Solution W.k.t To avoid under cutting

Cams, Clutches and Brakes 2

  1 dy   R p  y2        dt  P K min  2  1 dy  2 R p  y  2     R p    dt 

Also

1  d 2y   2  dt 2

  

5.45

3/2

 1 d 2y  y  2   dt 2  

2

 h      2  max at    

150   180

 2.62 radian

At

  , y  20 mm, a nd

 P K min 

1 dy 0  dt

 40  202  02   

3/2

40  202  2 02  40  20  14.38

Assume R p  40 mm R r  16 m  P K min  48.4 mm

We find P K min  R r Hence there is no under cutting. Problem 5.7: A cycloidal motion cam has the following data h  25 mm,   60, max  38. Calculate the minimum radius of the base circle if the radius of the roller is 6 mm. Also

5.46


calculate the amount of offset that is necessary to reduce the maximum pressure angle to 25

Solution Calculate the minimum base circle radius R b Rb  Rp  Rr

Given data Radius of roller R r  6 mm w.k.t  dy   dt  max  2h tan  max   h  R p  y    Rp  2     2  25 tan  max  25    R p  2  60     60 

 tan 38 

tan 30 

 3

  180   3  60    2  25

25     Rp  2  3   150 R p  12.5 

 R p  48.6 mm  R b  R p  R r  48.6 mm  6 mm  R b  42.6 mm


5.47

Now find the offset (e) required to reduce the pressure angle.

tan  

tan 25 

tan 25 

0.4663 

2h e 

e   R 2p

2



dy e d R 2p  e 2  

y

h 2

2  25 e 

48.62  e 2  



25 2

47.75  e 2361.9  e 2  

 12.5

solving by trail and error, we get e  22 mm, and   24.75 

5.2 CLUTCHES 5.2.1 Definition A clutch is a device used to engage and disengage the driving shaft to the driver shaft according to the requirement. In case of automobiles, the clutch is required during changing of gears. By operating a lever, the clutch engages and disengages the driver and driven shafts. While changing gears, the driven shaft should be disengaged from the driving shaft.

5.48


5.2.2 Classification of Clutch Clutches are classified in two ways, based on 1. Method of energy transfer (actuation). 2. Method of engagement. According to the method of energy transfer, the clutches are categorised into 1. Mechanically actuated clutches. 2. Hydraulically actuated clutches. 3. Pneumatically actuated clutches. 4. Electrically actuated clutches. According to the method of engagement, clutches are categorised into 1. Friction clutches. 2. Positive contact clutches. 3. Over running clutches. 4. Magnetic clutches. In this book, mechanically operated especially friction clutches are discussed.

clutches,

The actuating force or the force required to engage the clutches is supplied mainly by springs. 5.2.3 Working principle of friction clutch The principle of transmission of power is by contacting friction surfaces, to develop frictional force and this frictional force is used to start the driven shaft from rest and gradually increases the speed without slipping. In case of automobiles, the friction clutch is used to engage the engine shaft to the driven shaft.


5.49

5.2.4 Material for friction surface The materials used for lining of friction surfaces should have the following properties. 1. The material should have high coefficient of friction. 2. The material should withstand high temperatures caused during operation. 3. The material should have high heat conductivity, high resistance to wear and should not be effected by moisture or oil. 5.2.5 Types of friction clutches 1. Disc or plate clutches. (a) Single disc clutch (b) Multiple disc clutch 2. Cone clutches. 3. Centrifugal clutches. 5.2.6 Advantages clutch.

and

Disadvantages

of

friction

Advantages of friction clutches. 1. They have a very little shock during engagement as they can slip relative to each other. 2. It can be used for high speed engagement applications. Disadvantages of friction clutches 1. They are not suitable for application that require positive transmission because they do slip. 2. Replacement of friction material is often required as they wear out.

5.50


3. External cooling is required as they generate heat during engagement. 5.27 Design considerations for friction clutches 1. Sufficient torque transmitting capacity. 2. Light weight. 3. Smooth engagement without disengagement with drag.

shock

and

fast

4. Proper friction material (high co-efficient of friction) that should not be affected by moisture and oil, etc. 5. Provision for taking up wear of the contact surfaces. 6. Provision for transmission of heat which is generated during operation. 5.28 Design of a Disc (or) plate clutch Consider two friction surfaces Single d isc o r plate dx p r1 r2

r2

x

W r1 p

Frictio n surfaces Fig.5.12(a)

Fig.5.12(b) Forces on a D isc Clutch.


5.51

Let, W

Axial thrust,,

p

Intensity of axial pressure,

r1 

External radius of friction disc

r2 

Internal radius of friction disc.

r

Mean radius of friction disc



Coefficient of friction

Ft 

Frictional torque transmitted by the clutch

For more details, refer Dynamics of machines by Dr. S. Ramachandran Consider the following two conditions 1. Uniform pressure distributed over the entire area of the friction surface. 2. Uniform axial wear due to the sliding friction. 1. By considering uniform pressure Intensity of pressure  p 

W 

r21

 r22

 r3  r3  2 1 2  Total frictional torque  Ft   W  2 2  3  r1  r2    Ft   W r

where r = mean radius of the friction surface 2 r 3

    

r31  r32    r21  r22  

... (1)

5.52


2. By considering uniform wear Let p  Intensity of pressure x  radial distance from axis of the clutch

Intensity of pressure varies inversely with the radial distance x ; p

1 or p. x  constant x

(a) The intensity of pressure is maximum at inner radius r2  pmax r2  C  or pmax 

C r2

(b) The intensity of pressure is minimum at outer radius r1 pmin r1  C

or pmin 

C r1

(c) The average pressure is given by, paverage 

W  

r21

 r22  

1 Total frictional torque  Ft   W r1  r2 2 Wr


5.53

where r  mean radius of the friction surface  r1  r2    2  

(d) Axial thrust  W  2 C r1  r2 In general The total frictional torque acting on the clutch F t Ft  n  W r

w here n  no. of pairs of contact surfaces   coefficient of friction

r1  outer radius

W  axial load

r2  inner radius

2 r  mean radius  3 

    

r31  r32    r21  r22  

for uniform pressure

1  r  r2  ... for uniform wear 2 1 

Note: 1. For a single plate or disc clutch, n  2 (both sides of the plate are effective) 2. For a multiple disc clutch n  n1  n2  1

where n 1  no. of discs on driving shaft n2  no. of discs on driven shaft. Problem 5.8 A single dry plate clutch is to be designed to transmit 10 HP. at 900 rpm. Find (a) diameter of the shaft, (b) mean radius and face width of the friction lining assuming

5.54


the ratio of the mean radius to the face width as 4, (c) outer and inner radii of the clutch plate. Take y for the shaft material as 38 kg/mm2

(Nov/Dec - 2009)

Power  10 HP  7.36 kW Speed  N  900 rpm Refer P.S.G Data book, page No. 7.89 (a) Diameter of shaft  d  where

  3

495000  kW  kw N []

kW 

rated power  7.36 kW

N

rpm of the drive N  900 rpm

kw 

factor based on working conditions

 k1  k 2  k 3  k 4 [] 

Allowed shear stress

Since working conditions are not given, Assume k 1, k2, k 3, and k4 values From P.S.G Data book, Page No. 7.90 take k1  0.5 ; k2  1.25 ; k 3  0.32 and k4  0.75 kw  k1  k2  k3  k4 kw  0.5  1.25  0.32  0.75  2.82

Assume, material of shaft as C50; y  38 kgf/mm 2 ; FOS  2 []  Allowable shear stress


 

y

2



5.55

1 FOS

38 1   9.5 kg f/mm 2 2 2

[]  950 kgf/cm2

Diameter of clutch shaft d

  3

495000  7.36  2.82  2.29 cm o r 2.3 c m 900  950

d  23 mm

(b) Let b  face width rm  mean radius b

rm

4

(given)

Refer PSG data book Page No 7.89 im 

[M t] 2 P b b  r2m

W.k.t For single plate clutch im  2 where, im  Number of friction surfaces From PSG data book, Page No. 7.89, allowable pressure  Pa  k. P b where k  speed fac to r  1 (assume) and P b  basic pressure running)

 2.5 to 3 kg f/cm 2 ( for dry

5.56


Take P b  3 kg f/cm 2   coefficien t of friction  0.25 Assume From PSG Databook, Page No. 7.8 71620 HP 71620  10  795.77 kg f  c m  N 900

Mt 

 [M t]  M t  k w 795.77  2.82  2244.07

Design torque

 2244 kg f  cm im 

2

 Mt    2 Pb b  r2m

2244 rm  0.25  r2m 23 4

r3m 

2244  4  rm  2   3  0.25  2

952.38117   

3

rm  9.8 c m or 98 mm

Mean radius

 rm  98 mm

(c) To find ri  inner radius and ro  outer radius b  face width of the plate  ro  r i  ro  ri 

rm 4

98  24.5 4 b  24.5 mm

... (1)


But rm 

5.57

ro  ri  98 2

 ro  ri  98  2  196

... (2)

By solving eq. 1 and eq. 2 we get ro  110.25 mm and ri  85.75 mm

Inner radius of the clutch plate ri  85.75 mm Outer radius of the clutch plate  ro  110.25 mm Problem 5.9: Design a suitable clutch for scooter of 150 C.C. Axial thrust is to be provided by suitable helical springs. At maximum torque, the power developed is 4 kW at 3200 rpm. Space restricts the outer diameter of the clutch to be 150 mm [May/June - 2008]

At maximum torque developed, Power  P  4kW and Speed  N  3200 rpm The outer diameter  do  150 mm The outer radius  r0 

d0 2

 75 mm

From PSG databook, Page No. 7.94 according to the maximum torque, the ratio of inside diameter is varying from 0.72 to 0.84. Assume inner radius  0.8  outside radius ri  0.8 r0

Refer PSG data book, Page No 7.89 Transmitted torque  M t 

97400 kW N

5.58




97400  4 3200

 121.75 kg f c m

Number of friction surfaces i  m1  m 2  1

where, m 1  no. of driving plates. m 2  no. of driven plates imin 

[M t] 2 P b b  r2m

where [M t]  Design torque  kw M t kw  Factor based on working conditions kw  k1  k2  k3  k4

Since, working conditions are not given, assume k 1, k2, k 3, and k4 from data book Refer PSG databook, Page No 7.90 and 7.91 k1  0.5 ; k2  1.25 ; k3  k 4  0.9 kw  0.5  1.25  0.43  0.9  3.08 [M t]  3.08  121.75  374.99 kgf cm [M t]  Design tw isting moment  375 kgf cm

Refer PSG data book, Page No. 7.89 P b  3 kg f/cm 2

and

k  1 spee d factor


5.59

P a  k P b  3  1  3 kgf/cm 2

Face width of plate b  r0  ri  75  60  15 mm  1.5 cm

Assume   0.3 imin 



ri  0.8 r0

[M t ] 2  Pb b 

ro  75 mm (Given) r2m

and ri  60 mm

375 2   3  1.5   0.3  6.75 2 rm 

 0.97



r1  r0 2

75  60  67.5 mm 2

 6.75 c m

 Take imin  1

 No. of plates  m 1  m 2  i  1  1  1  2

Axial thrust  W  2  C r0  ri (Maximum pressure occurs at inner radius)  C  pmax  ri  3  6

 18 kg/cm  W  2   18 7.5  6.0  169.64 kg

Axial thrust  169.64 kg Problem 5.10: A single plate clutch, effective on both sides, is required to transmit 30 kW at 1400 rpm. Determine the inner and outer diameter of friction surface if the co-efficient of friction is 0.25, ratio of diameter is 1.5 and the maximum pressure is not to exceed 0.2 N/mm2. Also determine the axial

5.60


thrust to be provided by springs. Assume the theory of uniform wear.

Given data Power 30 kW P  30 kW  30  10 3 W N  1400 rpm

Speed

  0.25 pmax  0.2 N/mm2 Do Di

 1.5

Solution Let ro & ri be the outer and inner radii respectively.  ro  1.5 ri

Now find Torque to be transmitted (T) T

60  30000 60 P  2  N 2   1400  204.62 N  m  204.62  10 3 N  mm

w.k.t T  n  W rm

Here,


5.61

n

Number of sides of friction surface.

n2

(both sides of the plate being effective)

  0.25

(given)

W

Axial thrust applied by the spring.

W  2  pmax ri ro  ri  2   0.2  ri 1.5 ri  ri W  0.63 r2i N

Mean radius

rm 

 ri  r o  rm     2 

 ri  1.5 ri    2   rm  1.25 ri  T  n  W rm

We have 204.62  10 3  2  0.25  0.63 r2i  1.25 ri 204.62  10 3  0.3937 r3i  204.62  10 3   ri    0.3937  

1/3

ri  [519735.839 ]1/3  80.400 mm  80 mm   Di  160 mm ri ~

5.62


And ro  1.5  ri  1.5  80 ro  120 mm  D o  240 mm  The axial thrust applied by the spring W W  0.63 r2i  0.63  802 W  4032 N Problem 5.11 A multiplate clutch is to be designed to transmit a power of 50 kW at 500 rpm. Assuming suitable materials and data, determine the main dimensions of friction lining used in that clutch.

Solution Given data Power P  50 kW ; Speed N  500 rpm Assume service factor  1.25 Let

P d  Design Power

Design power  Rated power  service factor i.e.,

P d  P  Service factor  50  10 3  1.25  62500 W

Now calculate design torque that is to be transmitted. T

60  P d 2N



60  62500 2  500

 1193.66 N  m  1193.66  10 3 N  mm


5.63

Let, 

Permissible shear stress for shaft material.

  60 N/mm2 (Assume)

and

d

diameter of clutch shaft.

Then  16T  d    

1/3

 16  1933660  d    60    30938560  d   188.49 

1/3

1/3

d  [164134.15 ]1/3 d  54.75  d  55 mm

Let ri  Inner radius ; ro  Outer radius ri  2d  2  55  110 mm ro  1.5 ri  1.5  110  165 mm

5.64


n1  Number of driving plates  5 (Assume) n2  Number of driven plates  6 (Assume)

W.k.t.  n  n 1  n 2  1  5  6  1  10

Let the friction material be asbestos based material. Assume,   0.35 Allowable pressure  0.21 to 0.56 N/mm 2 Now T  n Wrm  rm  W

ri  ro 2



110  165  137.5 mm 2

T 1193660  n  rm 10  0.35  137.5

W  2480.33 N

Now pmax 



W 2  r i r o  r i  2480.33 2    110 165  110

 0.06 N/mm 2  allowable pr. 0.21 N/mm 2

Since the pressure is less than allowable pressure, our design is safe.


5.65

Now find face width b  ro  ri  165  110 b  55 mm

Thickness t  5 mm (Assumed)

Hence, the main dimensions of friction lining are 1. Inner radius  110 mm 2. Outer radius  165 mm 3. Face width

 55 mm

4. Thickness

 5 mm

Problem 5.12: A mult-disc clutch is to transmit 30 kW at 1200 rpm and consists of 4 discs on driving shaft and three discs on the driven shaft. The inside diameter of the contact surface is 150 mm. The maximum pressure between the friction surfaces is limited to 0.2 N/mm2. The coefficient of friction is 0.25; Consider the uniform wear condition and find the outside diameter of the contact surface. P  30 kW ; N  1200 rpm m 1  no. of discs on driving shaft  4 m 2  no. of discs on driven shaft  3 150  75 mm pmax  0.2 N/mm 2 ;   0.25 ; rmin  2

Let rmax  outside diameter Refer PSG databook, Page No. 7.89

5.66


Mt 

97400 kW 97400  30  N 1200

 2435 kg f  cm Tran smitted   M t  2435 kgfc m (or) M t  243500 N  mm torque 

For uniform wear condition, P max  rmin  C  C  0.2  75  15 N /mm

Axial force on each friction surface (W) W  2C  rmax  rmin    W  2  15  rmax  75   94.24  rmax  75     

(1)

But, rm  mean radius 

rmax  rmin rmax  75  0.5 rmax  37.5   2 2

Number of friction surfaces  i  m 1  m2  1 i4316

Torque transmitted  [M t]  i    W  rm 243500  6  0.25  94.24  rmax  75   0.5 rmax  37.5     243500  6  0.25  94.24  0.5 r2max  2812.5    0.5 r2max  2812.5  1722.55 r2max  9070  Outside radius of disc  rmax  95.23 mm


5.67

Problem 5.13: A multi disk clutch consists of five steel plates and four bronze plates. The inner and outer diameters of friction disks are 75 mm and 150 mm respectively. The coefficient of friction is 0.1 and the intensity of pressure is limited to 0.3 N/mm2. Assuming the uniform wear theory, calculate

(i) the

required

operating

transmitting capacity at 750 rpm.

force,

(ii) power

(May/June 2007)

Solution Given data D o  150 mm, ro  75 mm, D i  75 mm, ri  37.5 mm ,   0.1 pmax  0.3 N /mm2, N  750 rpm, m 1  5, m 2  4

Required operating force for uniform wear condition pmax  ri  C  C  0.3  37.5  11.25 N/mm

Hence required operating force W  2  C [ro  ri]  2   11.25 75  37.5   2650.7 N

Power Transmitting Capacity at 750 rpm For uniform wear condition Normal Torque  T  i  W rm [i  no. of frictional surfaces]

We know that rm 

ro  ri 75  37.5  2 2

5.68


and No. of friction surfaces  i  m 1  m 2  1  5  4  1  8  75  37.5   T  0.1  2650.7   8 2    119.281  10 3 N  mm  119.2 N  m

Power  

2  NT 60 2    750  119.2 60

 9361.94 W

Power  9.36 kW Problem 5.14: Find the torque that two surface of dry disc clutch can transmit if the outside and inside lining diameter are 120 mm and 70 mm respectively and the applied axial force is 10 kN. Assume uniform wear and   0.4 (Apr/May - 2008 AU)

Solution For uniform wear, W  2  Pmax ri ro  ri

Given that W  10 kN, ri  35 mm ; ro  60 mm W  pmax  2  ri ro  ri

[P max ri  C ]




5.69

10 2   35 60  35

 1.8189  10  3 kN/mm2  pmax  1.8189 N/mm 2

For uniform wear condition p  ri  C (constant)

Since the intensity of pressure is maximum at the inner radius ri and mean radius rm 

ro  ri 60  35   47.5 mm 2 2

and Torque Transmitted T  n p  W rm [np  i  no.of frictional surfaces] np  2 (since both sides of plate effective)  T  2  0.4  10,000  0.0475  380 Nm

5.2.9 CONE CLUTCH Cone clutch is used in automobiles. It consists of driver which is keyed to the driving shaft and has an inside conical surface which fits exactly into the outside conical surface which is keyed to the driven shaft. Hence it consists of one pair of friction surface. The cone angle , the radii of friction surface and the face width are the important design parameters.

5.70


b

C o nical F riction Su rfa ce Sp rin g

r1

B

r2



Wa D riving sha ft D rive n sha ft

D rive n sha ft

D rive n

Fig.5.13 C one Clutch.

5.2.9.1 Design of cone clutch: Consider a pair of friction surfaces of cone clutch. Let pn  Normal pressure between the contact surfaces r1  Outer radius of friction surface r2  Inner radidus of friction surface rm  Mean radius of friction surface 

r1  r 2

  S em i cone angle   coefficient of friction between the contact surfaces Consider the following two conditions 1. Uniform pressure 2. Uniform wear

2


5.71

1. Uniform pressure 2  pn  r31  r32    Frictional torque  F t  3 sin   

(1)

Frictio n surfac e

b

F= W n r

r2



r1 Wn

 Wa Fig. 5.1 4 Forces o n a F riction S urface

Let W a  axial force; b  face width; r  mean radius Normal face W n W n  Normal pressure  Area of contact surface  pn  2r  b

Axial force  W a  Wn  sin   pn  2  rb sin 

But, sin  

r1  r2 b

r1  r2  r1  r2  (or) b    and r  2  sin  

Substitute the values of b and r in equation (2)  r1  r2  r1  r2  sin  W a  pn  2   2   sin 

(2)

5.72


W a pn    r21  r22    pn 

Wa   r21  r22   

(3)

Now, substitute the value of pn in eq (1)  r3  r3  Wa 2  1 2    Ft  3 sin    r2  r2    1 2   Wa 2   3 sin 

 r3  r3 2  1  2 2  r1  r2 

    

But

Wa sin 

 Wn

 r3  r3  2 2  1  Ft   W n   2 2  3  r1  r2   

2. Uniform wear Incase of uniform wear, the intensity of pressure varies inversely with the distance (similar to the axial clutch) Px 

1 x

Px 

C x

Total frictional torque  F t 

(C  a constant) 2 2 2   C  r1  r2    2 sin   

Normal force on the  2C r1  r2  contact surface  sin  Axial force  W a  W n  sin   2 C  r1  r2

(4)


C

5.73

Wa 2 r1  r2

Substitute the value of C in eq. (4)  r2  r 2  Wa 2  1 2   Frictional torque  F t  2 sin  2 r1  r2   

but W n 

 W a  r1  r2    2 sin   

Wa sin 

 r1  r2   Ft   W n    2   Since, W n  pn  2r  b and

r  r2 2

r

Therefore the above equation can be written as Friction torque  F t    pn  2r  b  r F t    pn  2r2  b

Axial force required to engage the clutch  W a engage  W n  sin    W n cos  Problem 5.15: An engine is developing 50 kW at 1200 rpm by using a cone clutch. The cone angle is 30. The mean radius is 300 mm; The coefficient of friction is 0.25; The normal pressure should not exceed 0.2 N/mm2; Determine (1) face width of the cone (2) axial force required to engage the clutch.

5.74


P  50 kW ; N  1200 rpm ; 2  30;   15 pn  0.2 N/mm 2 ;   0.25 ; r  300 mm Let; b  face width

Assume uniform wear, Refer PSG data book Page.No. 7.89 M t  T ran smitted torque  

(o r)

9400 kW N

97400  50  4058.33 kg f  c m 1200

M t  405833 N  mm

Frictional torque  F t  2  pn  b  r2 405833  2  0.25  0.2  b  300 2 F ace width  b  14.35 mm

Let Waengage  Axial force required to engage the clutch W n  Normal force on the contact surface  pn  2r  b  0.2  2    300  14.35  5409.82 N  Axial force required    Waengage  W n sin    cos  to engage the clutch   5409.82  sin 15   0.25 cos 15]  F aengage  2707.17 N Problem 5.16: A power of 20 kW, is to be transmitted through a cone clutch at 500 rpm. For uniform wear condition find the main dimensions of clutch and shaft. Also determine the axial


5.75

force required to engage the clutch. Assume co-efficient of friction as 0.25 and the maximum normal pressure on the friction surface is not to exceed 0.08 MPa and design stress for the shaft material as 40 MPa.

Solution Given data P  20000 Watts ; N  500 rp m ;   0.25 [pmax]  0.08 MPa  0.08 MN /m 2  0.08 N/mm 2 s  40 MPa  40 MN/m 2  40 N/mm 2

Torque to be transmitted T

60 P 60  20000  2  N 2    500  382 N  m  382  10 3 N  mm

Diameter of clutch shaft  16T  d  s   

1/3

 16  382  10 3  d    40   d  36.5 mm

Take d  40 mm

1/3

5.76


Let rm  Mean radius rm  4d assumed  rm  4  40  160 mm

Now b  face width of friction lining b

160  80 mm 2

Assume   (angle of friction surface with axis clutch)   20 ro  Outer radius of friction lining b ro  rm  sin  2  160  40 sin 20  174 mm ri  Inner radius of friction lining. ri  r m 

b sin  2

 160  40 sin 20  146 mm


T  pmax 



5.77

  pmax ri r2o  r2i  sin  T sin    ri r2o  r2i  382  10 3  sin 20   0.25  146 174 2  1462

 0.1271 N/mm 2  allow able 0.08 N/mm 2 pr

Since the pressure developed is more than the allowable pressure, therefore the design is not safe with this dimension. So let us change the dimension. rm  5d  200 mm b  100 mm   25

then ro  200  50 sin 25  221 mm ri  200  50 sin 25  179 mm  pmax 

382  10 3  sin 25

  0.25  179 221 2  179 2

 0.068 N/mm 2  [pmax] 0.08 N/mm 2

Now the pressure developed is less than the allowable pressure i.e. (0.08 N/mm 2)  The design is satisfactory.

5.78


 Axial force required to engage the clutch W a T   W n rm  Wa 



 W a rm sin 

T  sin   rm 382  10 3  sin 25 0.25  200

 3229 N

5.2.10 CENTRIFUGAL CLUTCHES The disc clutches and cone clutches are manually operated mechanical clutches whereas the centrifugal clutches are operated by an electric motor (or) I.C engine. These clutches are employed to engage the driven shaft automatically after the driving member attains a particular

Fig.5.15


5.79

speed. The centrifugal clutch permits the driver motor or engine to start, warm up and accelerate the operating speed without load. The clutch then automatically engages and the driven shaft is smoothly brought up to operating speed. This type of clutches are generally used in Motor cycles, Go-cart, Mopeds, Scooty, etc. 5.2.10.1 Designing of centrifugal clutch Torque transmitted T  Ff  R  n

where Ff 

Frictional force

R

Inside radius of drum

n

No of shoes 


Fa 

Actuating force)

force

(outward

Fc 

Centrifugal shoe

force

on

Fs 

Inward force (Spring force)

F c  2

r

Distance of centre of gravity

F c  m  2r



Angular speed of drum at running speed.

F s  m  21 r

1 

Angular speed of drum at the time of engagement

F f    Fa

Now Fa  Fc  Fs

each

5.80


 T  Ff  R  n    Fa  R  n

Also the force acting on each shoe  Net outward force F a Here, the intensity pressure exerted on each shoe due to this force. p



Fa Net outward force  Area of conta ct of shoe lb Fa Rb . . . lR

5.2.10.2 Internal expanding rim clutches The internal expanding rim clutch is often used in textile machinery, machine tools where the clutch may be located within the driving pully. It benefits from centrifugal effects and transmits high torque even at low speeds and requires both positive engagement and high release force. It is a type of centrifugal clutch. 5.2.11 Electromagnetic clutch Electromagnetic clutches are operated electrically but transmission of torque is done mechanically. This is why it is also called as electro-mechanical clutches. Electromagnetic clutch is best suited for remote operations, because it doesnot require linkages to control its engagement. It has fast and smooth operation. However there is a risk of overheating, because energy dissipates as heat in the electromagnetic actuator each time when the


5.81

clutch is engaged. Therefore the maximum operating temperature of the clutch is limited by the temperature rating of the insulation of the electromagnet, Another drawback is its high initial cost.

Arm ature Field C oil

H ub

Fig. 5.16 Electro M agnetic Clutch .

Working: Engagement: When the clutch is required to actuate, current flows through the electromagnet, which produces a magnetic field. The monitor portion of the clutch becomes magnetized and set up a magnetic loop that attracts the armature. The armature is pulled against the rotor and a frictional force is generated at contact within a relatively short time, the load is accelerated to match the speed of the motor, there by engaging the armature and output hub

5.82


of the clutch. In the meanwhile the motor is constantly rotating with the input all the time. Disengagement: When current is removed from the shaft, the armature is free to turn with the shaft. In most designs, spring holds the armature away from the motor surface when power is released creating a small air gap. Cycling: It is achieved by interrupting the current through the electromagnet. Slippage normally occurs only during acceleration. When the clutch is fully engaged, there is no relative slip. Applications Electromagnetic clutches are widely used because of its compact nature; Some of the areas of applications are. 1. Machinery like lawnmover, conveyor drive etc.

copy

machines,

2. Automobile: Helps the driver to shift gear, while the vechile is in motion or running engine. 3. Locomotives: It is used in diesel locomotives.


5.83

5.3 BRAKES 5.3.1 Introduction: Brake is a device which is used to stop the motion of a machine or retard its motion. It is due to the friction between contact surfaces. The factors required in the design are: 1. Coefficient surfaces.

of

friction

between

the

braking

2. Unit pressure between braking surfaces. 3. The projected area of the braking surfaces. 4. Heat dissipating capacity. 5.3.2 Difference between clutch and brake Clutch is used to rotate the driven member along with driving member. Where as, brake is used to control or stop the moving member. In otherwords, clutch is used to connect two moving members of the machine, i.e driven member along with driving members. Whereas brake connects a moving member to a stationary member. Therefore we can say that, if one member of the clutch is kept fixed (stationary), then the clutch becomes a brake. 5.3.3 Brake friction materials The brake lining material should have the following properties. 1. It should have high coefficient of friction. 2. It should have high wear and heat resistance. 3. It should have high heat dissipation. 4. It should not react with moisture and oil.

5.84


5.3.4 Types of Brakes 1. Simple block shoe brake 2. Pivoted block shoe brake 3. Double block shoe brake 4. Simple band brake 5. Differential band brake 6. Band and block brake 7. Internal expanding brake 5.3.5 Single Block Shoe Brake It consists of a block which is pressed against the rotating wheel. The friction between the block and wheel, it causes a tangential braking force on the wheel. This tangential braking force retards or stops the rotation of the wheel. The block is attached to a lever, which is pivoted at one end and the force is applied on the other end. Let , P 

Applied force

RN 

Normal Reaction force

r

Wheel radius

2

Angle of contact between block and wheel




Ft 

Tangential braking force (or) frictional force

l

Length of the lever; x  distance between centre of wheel to the

fulcrum. It is assumed that, the normal pressure is uniform between the block and the wheel when the angle of contact 2  60  Then, frictional braking force F t  R N


5.85

. .  Braking torque = T B  F tr  R Nr [ . F t   R N]

There are three conditions in the arrangement of contact surface and position of pivot. Note: Direction of rotation is equal to direction of F t break. Case I When the tangential braking force F t passes through the fulcrum of the lever N

N

x

x Ft

O Fulcru m

P

RN r

O Ft Fulcru m

2

P RN 2

(a) Clock - wise Ro tation

r

(b) Anti Clo ck - wise Rotation Fig.5.17

(a) Clockwise rotation of wheel. Taking moments about ‘O’

(b) Anticlockwise rotation of wheel. Taking moments about ‘O’

RN  x  P  l

RN  x  P  l

Note:

RN 

Pl x

RN 

Pl x

In clockwise rotation of wheel, F t is acting right side. Because, we design the brake shoe, not rotating wheel. Hence, with respect to brake shoe, F t should be right side. Similarly, in case of anticlockwise rotation of wheel, F t should be left side.


5.86

But, braking torque T B  RN  r  

Braking torque TB   RN r

Pl r x

TB



Plr x

TB 

Pl r x

Plr x

Case II When the tangential force F t passes through a distance ‘a’ below the fulcrum.

x

a

x

O

P RN 2

a

Ft

O

P Ft

r

RN 2

(a) Clock - w ise Ro tation

r

(b) Anti Clock - wise Rotatio n Fig.5.18

(a) When wheel rotates (b) When wheel rotates in in clockwise direction. anti-clockwise direction Taking moment about ‘O’

Taking moment about ‘O’

F t  a  R N x  Pl

Ft  a  P  l  R N  x

But F t  RN

But F t  RN

RN a  R Nx  Pl

RN a  Pl  RN x

RN 

Pl a  x

R N x   RN a  Pl


Braking torque

R N x  a  Pl

TB  R N r

TB 

5.87

Plr a  x

RN 

Pl x  a

TB 

Plr x  a

Case III When the tangential force F t passes through a distance ‘a’ above the fulcrum.

x

x Ft

Ft

a O

P O

2 R N r

(a) Clock - w ise Ro tation

a 2 R N

r

P

(b) Anti Clock - wise Rotatio n Fig.5.19

(a) When wheel rotates in (b) When wheel rotates in clockwise direction anti-clockwise direction. Taking moment about ‘O’

Taking moment about ‘O’

F t  a  P  l  RN  x

F t  a  R N x  Pl

But F t  RN

But F t  RN

RN a  Pl  RN x

RN a  R Nx  Pl

R N x   R N a  Pl

RN 

Pl a  x

5.88


R N x  a  Pl RN 

Pl x  a

TB 

Plr x  a

Braking torque TB  R N r

TB 

Plr a  x

Case II (b) and Case III (a) are same i.e., RN x   Pl R N a moment of frictional force In this type of brakes, frictional force adds to the moment of force (Pl) i.e. friction helps to apply the force. These brakes are self-energizing brakes. Case II (a), and case III (b) are same When the frictional force is greater enough to apply the brake without any external force, then the brake is called self-locking brake. From the above equations, condition for self locking if x  a ; the P will be –ve or equal to zero.

No external force is required to apply the brake. The self locking brake is used only in back - stop applications. In order to avoid self locking, x  a 

The brake should be self energizing but not self locking




5.89

P b  bearing pressure on the block or shoe 

RN Ab

A b  width of shoe  projecting length of shoe  w 2rsin 

Angle of contact It is assumed that the normal reaction and frictional force act at the mid point of the block; Usually the angle of contact 2  is small i.e less than 60. When the value of

2

is

more

than

60  2  60,

then

equivalent

co-efficient of friction w is used in torque equation.  Equivalent co-efficient of friction

  eq 

4  sin  2   sin 2 

  actual coefficient of friction

where

2   Angle of conta ct Problem 5.17: A single block brake shown in the diagram has the diameter 300 mm. The angle of contact is 120. The coefficient of friction is 0.3; If the torque transmitted by the brake is 80 N – m, find the force P required. Assume length of the lever l  300 mm and distance between centre of wheel to fulcrum x  150 mm

Given: 2  120; x  150 mm

5.90


T B  80 N m

x

 80  10 3 N mm

Diameter of drum d  300 mm (or) radius of

Ft

O Fu lcru m

P

RN 2

drum  150mm Let Force P required to operate the brake

(a ) Clo ck - W ise R otatio n

Taking moments about O RN x  P  l RN 

Pl P  300   2P x 150

RN  2 P

Since angle of contact is greater than (or) equal to 60   







4  sin  2   sin 2 

4  0.3  sin 60 120   sin 120 4  0.3  sin 60   sin 120  1.5 1.03923 2.96042

  0.351


5.91

Braking torque TB  R Nr 80  10 3  0.351  [2 P]  150 P 

80  103  759.7 N 105.3

Problem 5.18: The block brake is shown in the diagram; Brake drum diameter provides a braking torque of 360 Nm, and coefficient of friction is 0.3 Find (1) the force P to be applied at the end of the lever for the clockwise and counter clockwise rotation of the brake drum. 2. The location of the pivot or fulcrum to make the self locking for the clockwise direction. 2 00

6 00

P 50 O

1 00

Solution 1. (a) Consider clockwise direction TB  360 N m  360  10 3 Nmm r

300  150 mm 2

Let P is the force applied at the end of the lever

5.92


6 00

2 00

Ft

50 O

RN 1 00

Fig.5.3.4.

Braking torque TB  Ft  r  360  10 3  F t  150 F t  2400 N.

But

F t  R N

 2400  0.3  R N R N  8000 N

Taking moments about ‘O’ P  800  R N  200  Ft  50 P

8000  200  2400  50 800

P  1850 N

(b) Consider counter clockwise direction Taking moments about ‘O’ P  800  F t  50  RN  200 P  2150 N

P


200

5.93

600

P Ft

50 100

O

RN

2. Location of pivot or fulcrum to make self-locking (when anti clockwise rotation). Let x be the distance of the pivot from the line of action. Taking moments about ‘O’ P  600  200  F t x  R N  200

In order to make the brake self-locking F t x must be equal to RN  200 (i.e. P must be zero)

Ft

x 100

O

RN

 F t x  RN  200

P

5.94


x

8000  200 2400

F t  2400 N R N  8000 N

 667 mm Problem 5.19: In a single block brake, the diameter of the drum is 250 mm and the angle of contact is 90. The operating force of 700 N is applied at the end of lever which is at 250 mm from the centre of the brake block. The coefficient of friction between the drum and the lining is 0.35. Determine the torque that may be transmitted. Fulcrum is at 200 mm from the centre of brake block with an offset of 50 mm from the surface of contact.

(a) Consider clockwise direction Taking moments at ‘O’ P  450  F t  50  RN  200

But P  700 N  700  450  F t  50  R N  200 3,15,000  50 F t  R N  200 2 00 m m

2 50 m m

P 7 00 N

50 mm O

Fig.5.3.13.


200 m m

5.95

250 m m

Ft

50

P 700 m m

RN

But F t   RN we know

  0.35

 F t  0.35 R N

Substituting in above equation 3,15,000  50  0.35 R N  200 R N  R N  1726 N  F t  604.109 N

But Braking Torque T B  F t  r  604.109 

250  75513 Nmm 2

 75.513 Nm

(b) Anticlockwise direction Taking moments about ‘O’ P  450  F t  50  RN  200 3,15,000  0.35 R N  50  RN  200

5.96


250 m m

200 m m

Ft

50

P= 70 0N

RN

R N  1448.27 N  F t  506.89 N

But Braking Torque TB  F t  r  506.89 

250  63.36 Nm 2

5.3.6 Simple band brake

a

a

F

O T2

A

L

A

T1 L

T1 

O

B

B ra ke d ru m

B

T2 

B ra ke d ru m

S h aft B a nd

F

S h aft B a nd

Fig.5.2 0(a) C lockw is e D irectio n

In a simple band brake, one end of the band is attached to the fixed pin or fulcrum O of the lever and the


5.97

other end is attached to the lever at a distance ‘a’ from the fulcrum. Let F

Force applied at the end of the lever

T1 

Tension on tight side of the band

T2 

Tension on the slack side of the band



Angle of lap




r

Brake drum radius

Ratio of tensions

T1 T2

  e

Braking force on the drum  T 1  T 2  Braking torque on the drum  TB   T 1  T 2  r  

(Neglecting thickness of band) Let

t

thickness of the band

then

re 

effective radius of the drum  t  r  2

Braking torque on the drum  TB   T1  T2  re   (Considering thickness of band) Consider clockwise rotation, taking moments about fulcrum ‘O’

5.98


F  l  T1 a

Then,

F

T1 a

l

Consider anti-clockwise rotation, taking moments about fulcrum ‘O’ Then, F  l  T2 a  T2 a  F   l 

where a  perpendicular distance from O to the line of action l  length of the lever

Note 1 Let

P

Power absorbed in watts.

N

Speed of the drum in rpm

TB 

Braking torque in N  m P

2 NT B 60

P  60  T B  Braking torque  2 N

Note 2 To find width of steel band Let

b

width of the band – mm

t

thickness of band – mm


tensile

t 

N/mm

stress

in

the

band

5.99

material

in

2

maximum tension in Newtons

T1 

Then, T 1   t b t Newtons (a) If stress and thickness are given, find b (b) If cross-section of band t are given, find  stress. 5.3.7 Differential Band brake A a

T2

b O

A

O

B

F

C

T2

L

M 

b a

C T1

B T1

L

F

M Ba nd Bra ke drum Sh aft

(a) Clockwise R otation of the Drum



Ba nd Bra ke drum Sh aft

(b) Anti Clo ckwise Ro tation of the Drum

Fig.5.21 Differential Ban d Brake

In case of differential band brake, the ends of the band are attached at A and B to a lever AOC as shown in the diagram. The lever is pivoted on a fixed pin or fulcrum at ‘O’. Let T1  Tension on tight side T2  Tension on slackside

5.100


F  Applied force l  length of lever (from O to C) a  distance from fulcrum to other end (O to A) b  distance from O to B (from O to B)

Consider clockwise rotation, taking moments about ‘O’ F  l  T1 b  T 2  a Fl  T 2a  T 1b

... (1)

For anti-clockwise rotation, F  l  T2 b  T 1a F  l  T1a  T2b

... (2)

When the frictional force helps to apply the brake it is called self energizing brake. In

the

above

two

equations

the

moment

T 1b and T2 b helps in applying brake.

Note 1 When the force F is negative or zero, then the brake is self locking Therefore, for clockwise rotation of the drum, the condition for self-locking is, T 2 a  T 1 b or T 1 b  T2 a

(or)


T2 T1



5.101

b a

For anticlockwise rotation of the drum, the condition for self locking is, T1 a  T2 b or T2 b  T 1 a

(or)

T1 T2



b a

Note 2 If length of OB is greater than OA, then the force F must act in the upward direction to apply brake. Note 3 If the length OB is less than OA, then force F must act in the downward direction in order to apply brake. T 1 and T2 remain same. Problem 5.20: A differential band brake shown in the diagram has a width of 80 mm and thickness of 2 mm. The permissible tensile stress in the band material is limited to 60 N/mm2. The Coefficient of friction is 0.25 between drum and the friction lining. The brake

drum

diameter

is

a 50

for

F

b 2 00 O

T2

500 mm. Calculate

A

B

C

T1

clockwise

direction (a) Tension in the band

2 20

o

B ra ke d rum

(b) Actuating force

B and

(c) Braking torque (d) Whether

the

self-locking or not.

brake

is

Fig.

5.102


Given data: b  80 mm ; t  2 mm ; t  60 N/mm 2 d  500 mm (or) A

r  250 mm

T2

F

a 50

T2

b O

2 00

C

B

T1

8 00

Since b  a therefore F will act upwards Max tension  T 1   b t  60  80  2  9600 N T1 T2

 e

9600  e0.25  3.839 T2

  220 

 3.839 radian

T2  3676 N

Taking moments about ‘O’ F  800  T 1  200  T 2  50 F

9600  200  3676  50 800

 2629.75 N

Braking torque  T B  T 1  T 2 r  9600  3676  250  1481000 N m m  1481 N m

 180


5.103

Checking for self locking Clockwise rotation Since, b  a, the c o nditio n fo r self lo c king T2 T1 T2 T1



b a



3676  0.3829 9600

b 200  4 a 50

Since

T2 T1



b a

0.3829  4

The brake is self-locking Problem 5.21: Design a differential band brake for a crane lifting a load of 50 kN through a rope wound round a barrel of 550 mm diameter, as shown in Fig. The brake drum keyed to the same shaft be 650 mm in diameter and the angle of lap of the brake band over the drum is 240. Operating arms of the brake are 45 mm and 210 mm. Operating lever is 1.5 m long. Take

  0.25.

Solution: Given data Load  50 kN  50  10 3 N Barrel Diameter  550 mm

5.104


2 40

o

6 50 mm

5 50 m m

T1

L oa d

2 10

T2

F

O

45

Fig.

1 50 0

Barrel radius  225 mm Brake drum diameter  650 mm Brake drum radius  325 mm Angle of lap   240  240 

 rad 180

 4.188 rad

Length of lever  1.5 m   0.25

Designing differential band brake by following steps. Step 1 Find braking torque T B  Load  Barrel radius T B  50  10 3 N  225 mm T B  11250  10 3 Nmm T B  11250 N.m


5.105

Step 2 Find Tension T1 & T 2 Tension ratio



T1 T2

T1 T2

 e 

 T 1  Tight side tension  T  Slack side tension  2

 e0.25  4.188  2.849

o r T1  2.849 T2

...(1)

Now, TB  T1  T2  r 11250 N m  T1  T2  0.325 m T 1  T 2  34,615.4 N

...(2)

Comparing and solving both equation 1 and 2, we get T 1  53336.5 N and T 2  18721.14 N

Step 3 Thickness of band brake t  0.005  D  0.005  650 t  3.25 mm

Now find band width w Assume Induced tensile stress  t  60 N/mm2

5.106


t  

T1 wt

53,336.5  60 N/mm 2 w  3.25

Band Width w  273.5 mm w  273.5 mm

Step 4 Calculate the bearing pressure P max  

T1 w .r 53,336.5 273.5  325

P m ax  0.599 N/mm 2

[P ] for Metal drum  1.5 N/mm 2 Since P max  [P],  The design is satisfactory.

Step 5 Force applied at the end of lever. From the figure. Taking moments about O, we get F  1500  T1  45  T2  210 F  1500  53336.5  45  18721.14  210 F  1020.85 N


5.107

Problem 5.22: A differential band brake is operated by lever as shown in fig. The brake drum has a diameter of 600 mm and the maximum torque on the drum is 900 N-m. If the co-efficient of friction between the brake lining and drum is 0.3, find the operating force.

Solution: Given data Diameter  600 mm or  Radius

 300 mm  0.3 m

T B  900 N  m

  0.3 10

0

m

a

m

D

5 00 m m

F

B O

T1

b 8 0m m

2 40

C T2

o

Fig.

5.108


OB  l  500 mm  0.5 m OC  b  80 mm  80  10  3m OD  a   100 mm  0.1 m   240  240 

 180

 4.188 rad

Step 1 Braking torque is given  900 N  m Step 2 Tension of tight and slack side. Tension ratio is given by T1 T2 T1 T2

 e   e0.3  4.188  3.5136

T1  3.5136 T2

...(1)

Also TB  T1  T2 r 900 N  m  T1  T2  0.3 m T1  T2  3000

...(2)

Comparing and solving equation (1) and (2) we get T 1  4193.48 N T 2  1193.50 N


5.109

Step 3 Taking moments about the fulcrum O, we get F  0.500  T1  0.100  T 2  80  10  3  F  0.5  4193.48  0.1  1193.50  80  10  3  F  0.5  323.868  Operating force F  647.736 N Problem 5.23: A band and block brake having 12 blocks, each of which subtends at an angle of 16 at the centre, is applied to a rotating drum of diameter 600 mm. The blocks are 75 mm thick. The drum and the flywheel. mounted on the same shaft have a mass of 1800 kg and have a combined radius of gyration of 600 mm. The two ends of the band are attached to pins on the opposite sides of the brake fulcrum at distances of 40 mm and 150 mm from the fulcrum. If a force of 250 N is applied at a distance of 900 mm from the fulcrum, find: (i) The maximum braking torque, (ii) The angular retardation of the drum, and (iii) The time taken by the system to be stationary from the rated speed of 300 r.p.m. Take coefficient of friction between the blocks and the drum as 0.3.

Solution Given data Angle of contact  2  16   8 d  600 mm  0.6 m

or

r  300 mm  0.3 m

5.110


D

N

900 m m

40

T1

a mm

A

O

B

b 150 m m

T2 F = 2 50 N

r

D rum Fig.

t  75 mm  0.075 m

Mass

m  1800 kg

Radius of gyration k  600 mm  0.6 m Force F  250 N (a) Number of blocks n  12 From the figure OD  a  40 mm  0.04 m OA  b  150 mm  0.15 m OB  l  900 mm  0.9 m

Step 1 Calculate Maximum braking torque T B  The braking torque will be maximum when all the following conditions are satisfied.


5.111

1. b > a. 2. The applied force acts upwards. 3. Anticlockwise rotation of brake drum. From the fig we know that band attached to A is tighter T 1 and end of band attached to B is slack side tension T2.  Tension ratio is given by T1

 1   tan    T2  1   tan  

n

T1

 1  0.3 tan 8   T2  1  03 tan 8  T1 T2

12

 2.7524

T1  2.752 T2

...(1)

Now taking moments about O. F  l  T 1  a  T2  b 250  0.9  T1  0.04  T2  0.15 225  0.04 T1  0.15 T 2

...(2)

Comparing and solving equation (1) and (2) we get T1  5636 N and T2  2048 N

5.112


Now Maximum braking torque TB   d  2t  T B  T 1  T 2    2 

 0.6  2  0.075   5636  2048   2    1345.5 N  m II. Find Angular retardation of drum  T B  I   mk 2  1345.5  1800  0.62     2.07 rad/s2

III. Time taken by the system to come to rest from the speed of 300 rpm. Initial angular speed i  

2N 60 2   300  31.4 rad/s 60

Final angular speed  f  0 rad/s Now w.k.t f  i    t 0  31.4  2.07 t t

31.4 2.07

t  15.2 sec


5.113

5.3.8 Double shoe brake In case of single block shoe brake, an additional load is acting on the shaft and its bearings. In order to overcome this additional load, double block shoe brake is used. Note When angle of contact is more than 60, then the equivalent coefficient of friction  

4 sin  2  sin 2 

TB   Ft  Ft  r t   1 2  

where F t and F t 1

2

are braking forces on two blocks and

r  radius of the wheel Problem 5.24: A double shoe brake shown in the diagram, is capable of absorbing a torque of 1500 N-m. Brake drum dia = 400 mm; Angle of contact for each shoe 110;   0.4 Find: (a) Spring force necessary to set the brake (b) Width of the shoe brake, if the bearing pressure on the lining material is not to exceed 0.5 N/mm2

Solution Given data TB  1500  10 3 N mm d  400 mm;

r  200 mm;

2  110  110 

  1.919 rad 180

5.114


S

Ft

S

3 00 2

110 RN

o

110

o

R N1

2

2 50

Ft

O2

O1

40

40

Fig.

  0.4 ; P b  0.5 N/mm 2

Let S  Spring force necessary to set the brake. Since 2  60 ; Therefore  

4 sin  equivalent coefficient   of friction  2  sin 2 

4  0.4  sin 55 1.919  sin 110 

 0.4584


Taking moments about O 1 S  550  R N  250  F t 200  40 1 1 RN 

But

1

Ft

1

 S  550  545.37 F t 160 1  705.37 F t 1

 F t  0.78  S Newtons 1

Similarly, S  550  F t  200  40  RN  250 2

2

Ft 2  250  545.37 F t S  550  160 F t  2 2 0.4584 550 S  385.37 F t 2 F t  1.4272  S Newtons 2

Braking torque  T B  F t  F t  r 1

2

1500  10 3  [0.78 S  1.4272  S]  200 S  3398 Newtons  F t  0.78  S  0.78  3398  2650.4 N 1

and F t  1.4272 S  1.4272 3398  4849.6 N 2

To find width of shoe brake: Let

b  width of the brake shoe.

5.115

5.116


RN  1

RN  2

Ft

1

 Ft

2





2650.4  5782 N 0.4584

 10579.4 N

Projected bearing area of one shoe  Ab  b  2rsin   Pressure 

 0.5  b

RN Ab 10579.4 b 2r sin 

R N  RN 1 2

11515 0.5 2  200  sin 55

b  64.6 mm

Max. normal force is on the left hand side of the shoe Problem 5.25: The layout of double block brake is shown in Fig. The brake is rated at 250 N-m at 650 rpm. The drum diameter is 250 mm. Assuming coefficient of friction to be 0.3 and for conditions of service. A pressure value of 1000 kPa may be assumed. Determine (i) Spring force ‘S’ required to set the brake (ii) Width of shoes Which shoe will have greater wear and what will be the ratio of rates of wear of the two shoes.

Solution 1. Relation between friction and Normal forces


S

S

FL2

110 RN

1 60 m m

o

RN 110

2

1 o

F L1

O1

O2

60

60

RN 

But e 

Fig.

Ft e

4  sin  2  sin 2

since   0.3, 2  110 

  1.919 180

 110  4  0.3 sin    2   0.3437  e  1.919  sin 110 

Ft 1  RN  1 0.3437

1 60 m m

5.117

5.118


R N  2.91 F t 1 1

Similarly RN  2.91 F t 2

2

2. Calculation of spring force Considering LH yoke, take moment about O 1, S  320  F t  125  60   R N  160  2.91 Ft  160 2 2 2

(ie) 320S  65 F t  465.5 F t 2

2

320S  400.5 F t 2

 Ft  0.799 S 2

Considering RH yoke, take moment O2 S  320  F t  125  60  RN  160 1

2

320 S  65 F t  2.91 F t  160 1

1

320 S  473 F r  F t  0.6 S 1

Now Braking Torque  F t  F t  r  250  10 3 Nmm (given) 1 2  0.6 S  0.799 S  125  250  10 3  S  1426.4 N

3. Wear depends upon the friction force  Wear ratio is

Ft

2

Ft

1



Fl

Fr


Ft

2

Ft

1



5.119

Fl 0.799 S  1.33  0.6 S Fr

As F l  F r, LH side shoe will have greater wear Determination of width of shoe Determine the greater normal force and obtain bearing pressure in terms of the unknown width of the shoes; by knowing pressure value, b can be found out R N  R nl  2.91 F l  2.91  0.799 S  2.91  0.799  1426.4 2

 3316.5 N R N  R nr  2.91 F r 1

 2.91  0.6 S  2.91  0.6  1426.4  2490.5 N We find Rnl  R nr , and hence R nl will be used in calculating the maximum bearing pressure. Projected area of shoe A  2r sin   b

 2  125  sin 55  b  204.78 b mm 2 P  Unit bearing pressure 

R nl A



3316.5 204.78 b



16.19 N /mm 2 b



16.19  10 6 N/m2 b

5.120


V  rubbing velocity



 Dn   250  10  3  650  60 60

 8.508 m/s Pressure  1000 kPa given 16.19  10 6  8.508  1000000 b  b  137.74 mm

5.3.8 Internal Expanding Brake Internal expanding brake is commonly used in automobiles. It consists of two brake shoes S 1 and S2 and outer surfaces are lined with friction material. These shoes are pivoted at one end about O 1 and O2. The other end of the shoes are made to contact a cam. When the cam rotates, the brake shoes are pushed outwards. The friction between the brake shoes and drum produces the braking torque, which reduces the speed of the drum. The brake shoes are held in their original position by means of a spring. When the drum rotates in the anti-clockwise direction, lefthand shoe is known as leading or primary shoe and the right hand shoe is known as trailing or secondary shoe. Refer PSG data book page No 7.99 Let r  Internal radius of the brake drum b  Face width of the shoe


5.121

R o tation F

F

r D ru m  

L in ing

c

a

 O1

O2

Fig. 5.22

a  Distance from the centre of the drum to the shoe pivot F  Actuating force p  pressure at angle ‘’  1  centre angle of heel from pivot  2  centre angle of toe from pivot  m  centre angle of point of maximum pressure from pivot. M n  moment of normal force on the right shoe. M n  moment of normal force on the left shoe. M f  moment of the frictional force on the left shoe pm  maximum pressure on the right shoe pm   maximum pressure on the left shoe

  coefficient of friction J  polar moment of inertia of the brake drum  Kg  m.sec 2 t


5.122

tr  temperature rise of the brake drum C p  sp. heat of drum material  0.13 for cast iro n  0.116 fo r steel

Total moment of frictional force 2

 pm  b  r  Mf  sin  m



 pm  b  sin m



sin  r  a cos  d

1

r a   r cos 1  cos  2  4 cos 22  cos 21   

Total moment of normal force

Mn 



pm b r

sin m

2



sin 2 d

1

1 pm b r a  1     1  sin 2   sin 22  2 2 sin m  2 

For right shoe, F

Mn  Mf C

For left shoe F

M n  M f C

M n 

M npm pm

and Mf 

M f pm pm


5.123

Braking torque  cos 1  cos 2   M t   b r2    pm  pm  sin m    Note 1

max  90 when 2  90 max  2 when  2  90 Note 2 If M f  M n, then the brake becomes self locking. Problem 5.26: An automotive type internal – expanding shoe is shown in the diagram. The face width of the friction lining is 60 mm and the coefficient of friction is 0.35. The maximum intensity of pressure is limited to 1.2 N/mm2. Assume angle

1 can be zero. Calculate (1) the actuating force ‘F’ and (2) The torque capacity of the brake.

30

30

o

F

1 20

o

o

F

1 50

R

1 20

1 77

o

1 25 R

30

o

30

o

Fig.

5.124


Solution Assume that the maximum normal pressure will occur between the lining on right hand shoe and the brake drum. Therefore, for right hand shoe

Since,

1  0;

max  90  when   90

2  120

max   2É when 2  90

2 is greater than 90 m  max  90

From the given diagram a  125 mm C  177 mm r  150 mm

Given data b  60 mm   0.35 p  1.2 N/mm 2

Total moment of frictional force Mf 



 pm b r  a   r  cos 1  c o s 2   4  cos 22  cos 21   sin m     

0.35  1.2  60  150  150  cos 0  cos 120  sin 90 

125  cos 2120  cos 20  4


5.125

 3780  150  1  0.5   31.25   0.5  1   3780  225  46.875   673312.5 N mm

Total moment of normal force Mn 

pm b  r

1     2   1    sin 21  sin 22    2 sin  m     2 

1.2  60  150  125 2  sin 90     1  120  180  2  sin 20  sin 2 120    

 675000 2.09 

1 [0    0.866 ] 2

 1706000 N mm F

Mn  Mf C



1706000  673312.5 177

 5834.39 N

Torque on the right hand shoe  r2 pmax b  cos 1  c o s  2     M t   sin m  R 

0.35  150 2  1.2  60 c o s 0  c o s 120  sin 90

M tR  283500 Nmm or 283.5 N m

For identical shoes, the expressions for M n and M f are proportional to pmax .

5.126


For Left shoe, let pm is maximum pressure on the left shoe. Mf 

M f  pm 673312.5  pm  pmax 1.2 

 561093.75 pm

Similarly, MN 

M n pm 1706000  pm  pmax 1.2

 1421666.66 pm

For left shoe,

F

9037.78 

M n  M f C

1421666.66 pm  561093.75 pm 177

pm  1.858 N/mm 2

Since, shoes are identical p m Torq ue on th e    M tL  MtR  le ft hand shoe  pmax



283500  1.858 1.2

 438952.5 N mm  438.9525 N mm Therefore,  The to taltorque  283500  438952.5 capacity of the brace 

M t  722452.5 N mm  722.4525 Nm


5.127

Alternately Refer PSG data book, page No 7.99  c o s 1  cos  2  Braking torque   b r2    p m  p m  sin m   

 c o s 0  c o s 120    0.35  60  150 2    1.2  1.858   sin 90    472500 0.5 3.058  722452.5 N mm M t  722.452 Nm Problem 5.27: An automotive type internal expanding double shoe shown in fig. is 300 mm in diameter and is actuated by a mechanism that exerts the same force (F) on each shoe. The shoes are identical and have a face width of 32 mm. The lining is a moulded asbestos having a coefficient of friction of 0.32 and a pressure limitation of 1000 k.Pa. 1. Determine the actuating force P, and 2. Find the braking capacity (i.e torque absorbing capacity of the brake).

Solution Given data Diameter  300 mm  0.3 m b  32 mm  0.032 m   0.32 pmax  1000 kPa  1000  10 3 N /m 2

5.128


27

A

F

o

F

1 00 15

0

O

1 26

o

112 O1

O2 B

50

50 27

Fig.

o

(i) Actuating force (F) The force F is found on the basics that the maximum pressure occured on shoe, as it is a right hand shoe (self energizing) From the fig:  1  0  2  126  126 

 180

 2.199 rad   90


5.129

and sin   sin 90  1 From the fig, by using phythagorous theorem, we get OO 1 

OB 2  BO 21 



112 2  50  2 

 122.65 mm

Now, the total moment of the normal forces about the fulcrum O, equal to Mn  

1 pmax b  r  OO1  1  2  1  sin 2  1  sin 2 2   2 2 sin   

1  0.3   1000  10 3  32  10  3     0.12265 2  2 

[2.199  0 

1 sin 0  sin 2  126] 2

 787.27 N  m

Again find moment of friction force m f about O 1 Mf 

  pmax  b  r  OO1  c os 2 2  cos 2 1   r cos 1  cos 2  4 sin   

 0.3  0.32  1000  10 3  32  10  3     2   1   0.3   0.12265  [cos 2  126  c o s 0]  [cos 0  c o s 126    2 4   

 304.174 N  m

5.130


 The actuating force F F 

M n  Mf l 787.27  304.174 0.112  0.100 

 2278.75 N

(ii) Braking capacity T B The braking torque T B  for right hand shoe is given by T B   pmax  b  r2 cos  1  cos 2 2

 0.3   0.32  1000  10 3  32  10  3    cos 0  cos 126  2   365.83 N  m

Now find T B for left hand shoe. For identical shoes, the expressions of M n and M f are both proportional to pmax.  The maximum intensity of pressure for left hand

shoe  pmax  M 1N 

787.27 p max pmax



787.27 p max 1000  10 3

 7.813  10  3 pmax

and


Mf 

5.131

304.174 pmax 304.174  pmax  pmax 1000  10 3

 3.042  10  4 pmax  F for left hand shoe equal to F

2278.75 

M N   M f l

8.813  10  4 pmax  3.042  10  4 pm ax 0.112  0.100

or pmax  445.04  10 3 N/m2 Torque required by left hand shoe TB   pmax  b  r2 c o s 1  cos 2 3

 0.32  445.04  10  32  10

3

2

 0.3    c o s 0  cos 126  2 

 162.8 N  m  The braking torque T is the total torque produced by both right and left hand shoes.  T  T B  TB  365.83  162.8  528.64 N  m Problem

5.28: An

automotive

type

internal-expanding

double-shoe brake is shown in figure. The face width of friction lining is 40 mm and the intensity of normal pressure is limited to 1 N/mm2. The coefficient of friction is 0.32. The angle 1

5.132


can be assumed to be zero. Calculate (i) the actuating force P, (ii) the torque-absorbing capacity of the brake. (Apr/May 2008) 30

F

1 00 .9

o

F

1 25 R

1 20

o

8 6.6

50 Fig.

50 30

o

Solution Assume that the maximum normal pressure will occur between the lining on right hand shoe and the brake drum.  For right hand shoe

1  0;  max  90, when   90 2  120;  max  2 when 2  90 


5.133

Since 2 is greater than 90 m   max  90 From given diagram a  100 mm ; c  187.5 mm ; r  125 mm

Also given b  40 mm ;   0.32 pmax  1 N/mm 2

Total moment of frictional force Mf   Mf 

 pm br  a  r c o s 1  c o s 2  cos 2  2  cos 2 1   4 sin m   0.32  1  40  125  125 cos 0  c o s 120  sin 90 

100 cos 2 120  c o s 2 0  ] 4

 1600 [125 1  0.5  25  0.5  1]  240000 Nmm

Total moment of Normal force Mn  

pm b r a  1  2  1  sin 2  1  sin 2 2   2 2 sin  m  

  1  40  125  100  1  120  180  2 sin 2 0  sin 2 120  2 sin 90  

1    250000  2.09 0    0.866  2    630750 Nmm

5.134


F F

Mn  Mf C 630750  240000 187.5

 2084 N

Torque absorbing capacity of brake from P.S.G Data book, Pg.No. 7.99 Braking Torque  cos 1  cos  2    br 2   pm  pm sin m    cos 0  cos 120   0.32  40  1252   1  pm sin 90   pm is unknown (i.e.) Max. pressure on left shoe

Hence Mf 

M f  pm 240000 pm  240000 pm  pmax 1

Similarly Mn 

M n pm P max

 630750 pm

for left shoe F 2084 

Mn  Mf C 630750 pm  240000 pm 187.5

 pm  1


5.135

 Braking Torque

 cos 0  cos 120   0.32  40  1252   i  1 sin 90    600000 Nmm  600 Nm 5.3.10 Energy considerations When the rotating members of a machine are allowed to stop by using a brake, the kinetic energy of rotation must be absorbed by the brake. When the members of a machine which are initially at rest and it is required to rotate upto a speed, slipping occurs between driver and driven until the driven member rotates with the same speed of driver. In case of clutch or brake the kinetic energy is absorbed during slippage. In the design of mechanical brake, the braking torque depends on the amount of energy absorbed by the brake. Let m  mass of the mechan ic al system V 1  Initial velocity V 2  final velocity

The kinetic energy 1  KE  m  V 21  V 22    2

absorbed

by

the

brake

Similarly, in case of rotating body, The Kinetic energy absorbed  KE 

1  2 I    22   2  1

where I  mass moment of inertia of rotating body  m .k2 k  radius of gyration

5.136


 1  Initial angular velocity  2  final angular velocity

The energy absorbed by the brake  E  M t   Jo ules where M t  Braking torque N m   angle through which the brake drum rotates during the braking period in radian  1  2  Q t 2   t  time in seconds

5.3.11 Temperature Rise The energy absorbed by the brake is converted into heat. The heat produced between rubbing surfaces increases the temperature of friction materials. When the temperature rises the coefficient of friction reduces, which in turn reduces the braking torque. At higher temperature the wear of friction lining is high, which reduces the life of lining material. In order to increase the life of tech lining materials, the temperature should be kept within the permissible value. In general, it is very difficult to find the temperature rise. But according to practical considerations, the given equation is used for preliminary design. Refer PSG databook, page No. 7.99 Temperature rise of brake tr 

Hg W Cp

where Hg  heat generated per operation

2 n 2 J k.cal 1800  427


J

5.137

polar moment of inertia of the brake drum

assembly. kg f m. sec 2 W  w eight o f the brake drum, kg f C p  Specific heat of drum material  0.13 for cast iron  0.116 for steel n  speed of the drum, rpm

In SI units: Temperature rise  t 

E mC

where E  Total energy absorbed by the brake (J) m  Mass of the drum assembly (kg) C  Specific heat of the brake drum material  460 J/kg C for cast iron

5.3.12 Friction Materials A brake or clutch friction material should have the following characteristics. 1. A high and uniform coefficient of friction 2. High resistance to wear 3. It should withstand high temperatures 4. It should have high heat conductivity For the characteristics of friction materials, Refer PSG databook, page No. 7.97

5.138


5.4 GENEVA MECHANISM Refer P.S.G Data book (Page No. 7.81, 7.82, 7.83, 7.84) Geneva mechanism consists of a Geneva wheel with a number of slots and a driving disc (Crank) whose projected pin enters the slot and causes the intermittent rotation of the “Geneva wheel.” This mechanism is used in rotary tables and indexing turrets in machine tools and in automatic machines. The number of slots in the mechanism ranges from three to eight depending on the applications. Z  No. of slots on the driven disc. OP  e  Centre distance OB  r  Radius of the driving crank R  Radius of Geneva wheel

  Semi-indexing angle (driven)   Semi-indexing angle (driver)   Angular velocity of driving crank (Assumed to be constant) d  Angular velocity of driven disc.  OBP  90 (P. No. 7.81 of PSG Data book)

    90  

 2

Z  2   2Z 2


s c l) D i ee n Wh e iv a Dr nev e (G

5.139

R

wd P  e

B 

90

o

3 O D riving D isc (or) C rank

r

Fig.5.23 External Gen eva Mechanisam

1 

  R  sin   cos   cos     1  2  e Z  

where R  Radius of Geneva wheel. 

Length of the slot h hrRe  e sin

   e cos  e Z 2

    h  e  sin  c o s  1  Z Z  



r e

re r  e sin   e  sin

 1 Z

R  e  cos

 2 Z

5.140


   h    sin  cos  1  e  Z Z 

(or)

The pin enters the Geneva wheel tangentially with no shock. i.e., The centre lines of the slot and crank must be mutually perpendicular at engagement and at disengagement. The circular segment attached to the crank effectively locks the wheel against rotation when the roller is not in engagement, and also positions the wheel for correct engagement of the roller with the next slot. 

For entry without shock, 

 r  sin   sin Z e

  Z  2   2Z 2  

t  time taken for one complete rotation of driving

crank, in seconds. 

60 n

where n  speed of rotation of crank in rpm; 

tr

t



Z2 2Z w here tr  resting time, seconds.



Angular acceleration of driven disc, at the start and finish of indexing.  do   2tan

 Z


where,  

2N rad /sec 60

Z  No. of slots

P

P

d



R

 

e

A

90 o

A

r

B

 e  



r

Fig.5.24

tan  

 sin  1   cos

 d  Angular velocity of driven disc. 

cos   

 1  2 cos   2

 d  Angular acceleration of driven disc.

5.141

5.142






1  2 sin  2 2

1  2 cos    

2

    dmax    1 To avoid the wheel overhanging the bearing, dW e

    2  1  sin  Z  

Non-interference of Geneva wheel shaft-with driving disc (or) The shaft of the driving disc does not interfere with the geneva disc, the following condition has to be satisfied.    d  2 1  cos  e Z 

where dW  diameter of wheel shaft. Problem 5.29: The fig. shows a Geneva mechanism used to obtain intermittent rotary motion from continuous rotation. The pin A is carried in a constant speed shaft which is rotating at

100

rpm,

and

it

engages

OP  125 mm; OA  88 mm;

with

a

POA  20;

slotted

member.

Find

the

instantaneous angular velocity and angular acceleration of the slotted member.

(May/June 2010)

Crank angle    20 N  100 rpm



2N 2  100   10.47 rad/sec 60 60



r 88  e 125

  0.704


O



20

88

o



1 25

A 

P

Fig.5.25

OA  r  Radius of the driving crank

 88 mm

OP  e  Centre distance

 125 mm

Angular velocity of driven disc (or) slotted member  d 

c o s   

 1  2 cos   2

5.143


5.144

d  

0.704 cos 20   0.704 1  2  0.704  cos 20  0.7042

 10.47

1.7372  10.06 rad/sec 0.1725

Angular acceleration of driven disc  d  

d   

1  2 sin 

2 1  2 c o s   2 0.704 1  0.704 2 sin 20 2 2

[ 1  20.704 cos 20  0.704 ]

 10.47 2

13.313  448.25 rad/sec 2 0.0297

Problem 5.30: A Geneva wheel is to have six slots. Driving crank radius is 50 mm. Determine various dimensions of geneva wheel.

(October 2000)

Solution: Refer PSG Data book page No. 7.81 Given data: Z = number of slots on the driven disc  6 r  Driving crank radius  50 mm

For entry without shock, 

 r  sin   sin e Z

where e  Centre distance. 

 r  sin Z e


e

5.145

r sin

 Z

e  centre distance 

50  100 mm 180 sin 6

e  100 mm

But, sin  

r 50   0.5 e 100

where,   semi – indexing angle (driven)   sin  1 0.5   30   radius ra tio   R  r c o t   50 

R  cot r

1 50  86.6025 mm  tan tan 30

Radius of Geneva wheel  R  86.6025 mm h  Slot length  r  R  e  50  86.6025  100 h  36.6025 mm

To avoid the wheel overhanging the bearing dw e

    2  1  sin  Z  

180   dw  2  100  1  sin 6  

5.146


dw  100 mm

Angle of locking action 

 180 Z  2  6  2  240 6 Z

Indexing time ratio   

 

where   Semi – indexing angle driver  

      30  60  2 2 60  0.333 180

Note: If n  100 rpm, (speed of rotation of crank is given) then, t  time taken for one complete rotation of the driving

crank in seconds 

60 60  0.6  100 n

Time for indexing  ti   t  0.333  0.6  0.199 sec Resting time

Z2 62  tn  t  0.6  0.4 sec Z2 62


5.147

5.5 RATCHET AND PAWL MECHANISM Ratchet and Pawl mechanism is mainly used in hoisting machines. It allows rotation in one direction and prevents rotation in other direction. In hoisting machines, this mechanism allows to raise the load, but prevents the load falling down due to gravity. The mechanism is shown in Fig. 5.5.1. This mechanism consists of (1) Ratchet (2) Pawl. Ratchet is fitted to the driver shaft and the Pawl is fixed to the machine frame. The pawl engaging the ratchet locks the clockwise rotation of the driver shaft. When the rotation is reversed, the pawl gets disengaged and allows the ratchet to rotate in counter clockwise direction. Ratchet and Pawl Mechanism (7.85 and 7.86) Refer PSG Data book. 1. Material: Ratchet - Grey cast iron Grade 15 - 30, Cast steel, C30 and C45. Pawl - C45, 40 Cr 1 Ratchet and pawl are surface hardened to 50 HRC. 2. No. of teeth on ratchet wheel Z  6 to 8  12 to 20  16 to 25

3. Values of p - linear unit pressure - P/b kgf/cm p

P b

where P  Peripheral force (or)

5.148


P G

P



o

D

90

t

b

a1

(a)

(b)

b P

e1

h P

a

x

(c)

b

(d)

Fig.5.26

P p

Steel pawl and C.I. ratchet - 50  100 kgf/cm Steel pawl and steel ratchet - 150  300 kgf/cm   b/m  1.5 to 3.0



2M t Zm

Z  no. of teeth on ratchet m  D/Z (module) D

tip

circle

dia.

ratchet.  m Z M t  torque transmitted

of


4. Module: m  2

5.149

  Mt

3

Z  [ b ]

where [ b ]  300 to 500 kg f/cm 2 (for C45 steel) Other Dimensions a  m;

h  0.75m ;

bm

5. Bending moment M b  P  h Design of Pawl M b  P  e1 1  6M b P  1  Total stress     2 xb bx 

Dia. of the pawl pin  d  2.71

a    2 [  ]  2  3

P

b

b

1



where a1  thickness of collar on pawl pin (from table 7.86) [ b ] for C 45 steel 30 to 50 N/mm2 Problem 5.31: Design a ratchet and pawl mechanism for the following data: Braking torque = 180 N-m Brake drum diameter = 400 mm

Solution: Given data: M t  180 N  m  1800 kg f  cm

(Convert in to kg f  cm

5.150


Let Z  No. of teeth on ratchet  18 assume Step 1: Calculate module (m) Refer PSG databook, Page No. 7.85

2

2

assume   2.5

  Mt

3

and [b]  450 kgf/c m2

Z  [b]

for ratchet

  3

1800  0.8925 c m 18  2.5  450

Take m  1 cm or 10 mm From data book, a  m  1 cm; h  0.75 m  0.75 c m b  m  2.5  1  2.5 cm

Step 2: Peripheral force  P  

2M t D

But D  Zm

2  1800  200 kg f 18  1

Step 3: Linear unit pressure 

P 200   80 kg f/cm 2 b 2.5


5.151

Step 4: Refer PSG databook Page No. 7.85 Calculation of diameter of pawl pin

d  2.71

 

d  2.71 

[b] 

P b   a1   2[b]  2 

FOS for C45 steel (Pawl) y  36 kgf/mm 2

  3

y

200  25  6  2 2  7.2  

Take FOS  5 36  [ b]  5

 17.228 mm

Take d  18 mm

 7.2 kgf/ mm 2

Refer table (P.No. 7.86) of PSG Data book a1  6 mm b  2.5 cm = 25 mm

Step 5: Checking stress for pawl Refer PSG databook page No. 7.85  

6M b1 bx

2



P  [b] x.b

6  2000

25  10

2



200 10  25

 4.8  0.8  5.6 kgf/ mm 2  7.2 kgf/ mm

2

 Design is safe

M 1  Pe1 e1  a  1 cm  10 mm (assume) M b  20010 1

 2000 kg f  mm x  a  1 cm  10 mmassume  b  2.5 c m  25 mm

Two Marks Question and Answers

TM.1

Two Marks Question and Answers Chapter - I 1.1 Name the types of transmission belts. Based on the belt transmission belts are:

cross-section,

the

types

of

1. Flat belt 2. V-belt 3. Round belt. 1.2 What are the commonly used belt materials? 1. Leather 2. Cotton or Canvas 3. Rubber 4. Nylon core 5. Balata 6. Fabric. 1.3 State Law of Belting The law of belting states that the center line of the belt as it approaches the pulley must lie on a plane perpendicular to the axis of that pulley, or must lie in the plane of the pulley, otherwise the belt will run off from the pulley. 1.4 Explain the term Crowning of Pulleys Pulleys are provided the belt from running off crowning of pulleys. The rounded. The height of the speed and length of belt.

with a slight dwell to prevent the pulley. This is known as crowning maybe tapered (or) crown depends upon the width,

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1.5 What are the different types of flat belt drives? The different types of flat belt drives are: 1. Open belt drive 2. Crossed belt drive 3. Quarter-turn belt drive 4. Compound belt drive 5. Belt drive with idler pulley 6. Stepped or cone pulley 7. Fast and loose pulley drive. 1.6 What are the advantages of flat belt drive? The advantages of flat belt drive are: 1. They are more flexible and absorb shock loads and vibration. 2. They are relatively cheap and easy to maintain. 3. They can be atmosphere.

used

in

dusty

or

abrasive

4. The design of flat belt drive is simple and inexpensive. 5. They can be used for long distance, even upto 15 m. 6. They are quiet and efficient. 1.7 Why is the face of a pulley crowned? (Nov/Dec 2009)

To keep the belt running on the pulley face, it is necessary that the face has to be crowned.


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1.8 Define maximum tension in a belt? (Arp/May 2008) Maximum Tension in a belt T    b  t where   m amimum safe stre ss b  width of the belt t  th ickness of the belt

1.9 What is slip in a flat belt drive? In a flat belt drive, when the frictional grip between the pulley and belt is insufficient, the driver will have some forward motion without belt, which is known as slip. 1.10 What is creep in a flat belt drive? Due to the difference in tensions on tight side and slack side, the belt undergoes a change in strain as it passes around the pulley. To compensate for these different strains the belt moves relative to the surface of the pulley. This relative movement of the belt is called creep. 1.11 How does creep effect the belt design? The creep in belt leads to slip, loss of power and velocity ratio. So, elastic nature of the belt also taken into account in the belt design. 1.12 What is centrifugal tension in a belt drive? When the belt runs over the pulley, some centrifugal force is created due to the belt velocity, which increases the belt tension on both tight as well as the slack side. The increase of tension due to centrifugal force is called centrifugal tension.

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1.13 What are the factors affecting the coefficient of friction between a belt and pulley? The coefficient of friction between the belt and the pulley depends upon the factors like material of belt, material of pulley, slip of belt, and speed of belt. The value of coefficient of friction lies between 0.25 to 0.30. 1.14 What are the factors to be considered in design of a belt drive? The factors to be considered in design of a belt drive are power to be transmitted, input and output speeds, the centre distance, peak load, nature of load, effect of centrifugal force, the ratio of tight side of slack side tension, coefficient of friction, etc. 1.15 Sketch the cross section of a V-belt and label its (Nov/Dec 2009) important parts.

2  = 40 x

o

Fa bric and ru bb er co ve r

W

Fa bric C o rds

T

R u bb er b

2

Cross-section of a V-b elt


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1.16 Give the relationship of ratio of tensions in a V-belt drive. (Apr/May 2008) T1 T2



e

sin 

where, T 1  Ten sions on tight side T 2  Tensions on slack side

  A ngle of conta ct in radians   S emi groove angle 1.17 State reasons why V-belt drive is preferred over flat belt drive. (a) Replacement of V-belt is easy, because V-belts are available according to standards. (b) Power transmitted by the V-belts is more. (c) For smaller centre distances V-belts are more suitable. (d) Efficiency is high, when compared to flat belt. 1.18 How will you determine the no. of required in the design of V-belt drives? No. of belts  n 

P  Fa k W  Fc  F d

P  given power in kW F a  service factor k W  power transmitted per belt F c  correction factor for length F d  correction factor for arc of contact.

belts

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1.19 What are the factors affecting the belt life? The belt life is very much affected by the following factors. Decrease in pulley diameter, increase in belt thickness, increase in tight side tension, too large centre distance, service and environmental conditions, deflection in load, etc. 1.20 Give few disadvantages of V belt drive over flat belt drive. The disadvantages of V belt drive over flat belt drive are: 1. V belts cannot be used for large centre distances. 2. Design of V belts are more complicated. 3. V belts are not so durable. 4. V belts are costly. 5. The centre distance should be exact for V belts. 1.21. What are the advantages of wire rope drives? (a) High strength to weight ratio. (b) Silent operation at high speeds. (c) Greater reliability. (d) High efficiency. (e) Low cost. (f) Ability to withstand shock loads. 1.22 Name some applications of wire rope drive. Wire rope drive is best suited for the applications where large amount of power is to be transmitted over a long distance. Wire ropes are used in elevators, mine hoists, cranes, conveyors, hauling devices, suspension bridges, etc.


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1.23 Where are the stresses induced in a wire rope? A wire rope is generally subjected to the following stresses. 1. Direct stress due to axial loads. 2. Bending stresses due to bending of ropes over sheaves. 3. Stresses due to acceleration or retardation of moving masses. 4. Stresses during starting. 1.24 How are the wire ropes designated? The wire ropes are designated by a number specifying the number of strands and number of wires in each strand. The wire diameter, type of construction and wire material grade are also included in the designation. 1.25 What is 6  12/6/1?

the

meaning

of

wire

designation

6 - refers to the number of strands 12/6/1 - refers number of wires in a strand are 12  6  1  19. (i.e. 19 wires are formed by 12 wires over 6 wires with 1 core wire) 1.26 What is a chain drive? The chain drive consists of an endless chain running over two or more sprockets to transmit power. The chain consists of large number of links connected by pin joints and the sprockets are toothed wheels with a special tooth profile.

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1.27 What are the advantages of chain drive? The advantages of chain drive are: 1. Compact drive when compared to flat-belt drives. 2. No slip between chain and sprocket. 3. High efficiency of the drive. 4. Possibility of driving several sprockets by using a single chain. 1.28 What is a silent chain? Silent or inverted tooth chain consists of a series of links formed form laminated plates. A special profile is used to reduce noise level for chain links and sprocket. The links themselves engage with the sprocket teeth directly to reduce noise. They can be operated at high speeds and has more reliability. 1.29 How are chains designated? According to BIS (IS: 2403-1967) the roller chains are designated as per roller diameter and number of stands in the chains. For example, 5.0 S IS: 2403-1967 represents single strand chain with roller diameter 5 mm. 10.0 D IS: 2403-1967 represents double strand chain with roller diameter 12.17 mm. 24.0 T IS: 2403-1967 represents triple strand chain with roller diameter 25.5 mm. 1.30 Why is odd number of teeth for sprocket preferred? In order to distribute wear evenly on all the chain links, it is the usual practice to have odd numbers of teeth


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on the smaller sprocket and an even number of pitches in the chain. 1.31 What are the types of failure of chain drive? The various failures in chain drives are: 1. Elongation: It is caused by increased pitch due to joint wear under the action of tension and dynamic loads during operation. 2. Failure of Joints and Plates Improper erection causes misalignment of sprockets and wear occurs on the plate of the chain, which causes the failure of joints. 3. Wear of the sprocket teeth Wear of sprocket occur due to the relative motion between bush and the teeth. 1.32 What are the main classification of chains? The main classification of chains are: 1. Roller Chains 2. Bushed Chains 3. Silent (inverted tooth) Chains. Again the roller and bushed chains are of three types. (a) Simpler (b) duplex (c) Triplex.

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1.33 List the applications of the chain drives. Chain drives are widely used in machine tools, material handle equipments, automobiles, bicycles, wood working machinery, construction machinery, agricultural machinery, rolling mills, coal cutters, etc. 1.34 Explain the chordal action in chain drive. When the chain passes over a sprocket, it moves as a series of chords instead of a continuous arc as in case of the belt drive. Thus the centre line of the chain does not have uniform radius. 1.35 What are the limiting values on number of teeth on sprockets? The minimum number of teeth recommended for small sprocket is 17 and the maximum number of teeth on larger sprocket can be between 100 and 120.

Chapter - II 2.1 State the advantages of the gear drives over other type of drives. The advantages of the gear drives over other types of drives are: 1. Gear drives are positive drives. 2. Suitable for shorter centre distance. 3. The gear drives are better than others in case of amount of power transmission and efficiency. 4. Change of speed is very easy in gear drives.


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2.2 State the fundamental law of gearing? The fundamental law of gear drive states that the common normal to the tooth profile at the point of contact should always pass through a fixed point, called the pitch point, in order to obtain a constant velocity ratio. 2.3 How is gear classified? 1. Based-on peripheral velocity (i) Low velocity (0 to 3 m/s) (ii) Medium velocity (3 to 15 m/s) (iii) High velocity (15 m/s and above). 2. Based on mutual position of shafts (i) Parallel shafts-spur, helical, herringbone gears. (ii) Intersecting shafts-straight bevel, spiral bevel gears. (iii) Non-Parallel, non-intersecting helical, spiral, hypnoid gears.

shafts-crossed

3. Based on the relative motion of shafts (i) Row gears-(relatively motion of shafts are fixed). (ii) Planetary and differential gears. 4. Based on contact area (i) External gears (ii) Internal gears. 5. Based on teeth on wheel (i) Straight teeth-spur gears (ii) Helical teeth-helical, herringbone gears (iii) Inclined teeth-bevel gears (iv) Curved teeth-spiral gears.

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2.4 Define the following terms of a gear. (a) pressure angle ; (b) diameteral pitch D p Pressure angle : It is the angle between the common normal of two gear teeth profiles at the point of contact and the common tangent at the pitch point. Diametral pitch D p: It is the ratio of number of teeth of the pitch circle diameter in inches. 2.5 Define ‘module’ of a gear. The module is a basic parameter, by which a gear tooth is specified and is equal to the ratio between pitch circle diameter in millimeters and number of teeth. 2.6 What is a pitch point? The intersecting point of the line of action and the line joining the axes of two mating gears is called as pitch point. The circle drawn through pitch point, from gear centre is called pitch circle. 2.7 What is interference in gears? In a pair of gears in mesh, the possibility of the tip of one tooth digging into the portion of the flank of the mating tooth gear can occur in certain conditions. This phenomenon is termed as interference. 2.8 How can interference in gears be avoided? The interference in gears can be avoided by ensuring that the point of contact between the two teeth always remains on the involute profiles of the both the teeth. This also can be avoided by, (1) reducing the tooth height, (2) increasing the pressure angle, and (3) the radial flank of pinion teeth be cut back.


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2.9 What is the minimum number of teeth on pinion to avoid interference? The minimum number of teeth on the pinion in order to avoid interference are given in the following table. Pressure angle 14.5

20

Z min (Theoretical)

32

17

Z min (Practical)

27

14

2.10 Explain undercutting in gears. While generating gear teeth, if there is interference of the cutter with the rotating job, then a recess is cut at the root of the gear tooth which is known as undercutting. This happens when the cutter extends beyond the base circle of the pinions having small number of teeth. 2.11 What is backlash? Backlash is defined as the play between a mating pair of gear teeth in assembled condition. It is the amount by which the width of the tooth space exceeds the thickness of the meshing tooth measured on pitch circle. 2.12 What are skew gears and list its application? When the helical gears are mounted on non-parallel, non-intersecting shafts, and mesh with each other, they form a skew gearing (or) crossed helical gearing. Skew gears are used to drive the cam shafts and auxiliaries on small internal combustion engines, feed

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mechanisms on machine automobile engines, etc.

tools,

distributor

drive

of

2.13 Give the requirements of a gear materials. The main requirements for a gear materials are low cost, easy to manufacture, high strength, high wear resistance, low coefficient of friction and noiseless operation. 2.14 What are the materials commonly used for gears? Cast iron, steel, brass, Al-Zinc alloy, phosphor bronze, nonmetallic materials like wood, fibre reinforced plastics, backlite, mica, vulcanized rubber, composites, lead/graphite, pressed paper, etc. are commonly used as gear materials. 2.15 What are the main types of gear tooth failure? (May/June 2013)

(i) Tooth breakage (due to static and dynamic loads) (ii) Tooth wear (a) Abrasion (b) Pitting (c) Scoring or seizure 2.16 What are the common profiles used for gear tooth? The common types of profiles used for gear tooth are: 1. Involute 2. Cycloidal 3. Combination of the above two (composite system)


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2.17 Write any two properties of involute curve. Involute curve has the following properties. 1. The shape of the involute is dependent only upon the size of the base circle. 2. If one involute, rotating at a uniform rate of motion, acts against another involute, it will transmit a uniform angular motion to the second, regardless of the distance between the centers of the two base circles. 2.18 What are the characteristics of cycloidal gears? Cycloidal gears do not have interference. A cycloidal tooth is stronger than involute tooth because it has spreading flanks in contrast to the radial flanks of an involute tooth. Also cycloidal gear teeth have less sliding and therefore less wear, better contact and lower contact stress. 2.19 Compare involute and cycloid profiles.

Pressure angle

Involute profile Constant

Cycloid profile Varies from zero

Running Manufacturing Velocity ratio

Smooth Easy Small variation

to maximum. Less smooth Complicate This does not

in centre distance does not affect

permit any variation in

velocity ratio

centre distance.

2.20 What are the disadvantages of using cycloidal profile for a gear tooth? The disadvantages of using cycloidal profile for a gear tooth are:

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1. The cycloidal profile cutters are formed with greater difficulty and with less accuracy. 2. The cost of machining cycloidal profile is very high. 3. Accurate centre distance setting of cutter is required. 4. The cycloidal profile does not permits even a small variation in centre distance of gears without undercutting or variation in velocity ratio. 5. In the cycloidal profile system, the pressure angle varies from zero at pitch line to maximum at the tip of the teeth. 2.21 What are the general characteristics of spur gearing? The general characteristics of spur gearing are: 1. Transmit torque between parallel shafts. 2. Teeth are parallel to axis. 3. Line contact takes place during mating. 4. Can be used as sliding gears in gear boxes. 5. Easy to design and manufacture. 2.22 What are the applications of spur gears? The spur gears are used in sliding gearboxes, machine tools, automobiles, material handling equipments, rolling mills, marine power plants, etc. 2.23 Why are spur applications?

gears

preferred

for

certain

Spur gears are preferred for certain applications due to following reasons.


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1. Provision can be made for shifting of gears 2. Higher efficiency (99%) 3. The maintenance of drive is inexpensive and life is long 4. Design and economical.

fabrication

is

easier

and

more

2.24 What is interchangeability of gears? In a gear drive the ability of replacing a failed gear by a similar gear (same tooth form) without affecting the functionality of the drive is called interchangeability of gears. This interchangeability is available in spur and helical gears. 2.25 What is meant by cluster gears?

If two or more gears are cut on same blank with gap between them and they are functioning as a single unit, then they are called as cluster gears. The cluster gears are widely used in speed gearboxes as sliding gears. They are manufactured either by die forging, smith forging or by turning form bar stock and the teeth are machined.

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2.26 How teeth does contact occur in spur and in helical gears? In spur gears, the contact between meshing teeth occurs along the entire face width of the tooth. In helical gears, the contact between meshing teeth begins with a point on the leading edge of the tooth and gradually extends along the diagonal line across the tooth. 2.27 What are the advantages of helical gears over straight spur gears? The advantages of helical gears over straight spur gears are: 1. High peripheral speed (20 to 35 m/sec). 2. Low noise. 3. High power transmission. 4. More smooth engagement, etc. 2.28 What are the demerits of a helical gear? The demerits of a helical gear are: 1. The axial thrust is produced (thrust bearings are required) 2. Complicated design (design is based on equivalent number of teeth) 3. Machining of gear is difficult and costly 4. The transmission efficiency is slightly less compared to the spur gear. 2.29 Differentiate between the normal pitch and axial pitch in helical gears. The axial pitch is the distance parallel to axis between similar faces of adjacent teeth, whereas normal pitch is the distance between similar faces of adjacent teeth along a helix on the pitch cylinders normal to the teeth.


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2.30 Show the circular pitch, normal pitch in a helical gear using a sketch.

2.31 What is the virtual number of teeth in a helical gear? The virtual number of teeth or equivalent number of teeth of a helical gear is the number of teeth on an assumed equivalent spur gear whose pitch circle is in a plane perpendicular to helical tooth. The virtual number of teeth is given by, Zv 

Z cos 

Where ; Z  number of teeth.

2.32 What are the ranges of helix angle used for helical gears? Helical gears have helix angle between 8 and 25. Herringbone gears used in industrial gears have helix angle between 20 and 30 and herringbone gears used in turbines have helix angle upto 45.

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2.33 What is a herringbone gear? A gear fabricated such that half of its width is cut with helix in one direction and other half of the teeth is cut in opposite direction is called herringbone gear. The herringbone gears are cut with help of gear shaping machine. They do not induce any axial component of force as in case of helical gears. 2.34 When do we prefer herringbone gears? Herringbone gears are preferred to transmit large power, without any noise and axial thrust. They are applied in rolling mills, drive of reciprocating machineries, drive for hoisting machineries, drive for machines tools, drives for alternators from steam turbines, etc. 2.35 What are the herringbone gear?

merits

and

demerits

of

The merits of a herringbone gears are silent operation, absence of vibration and axial thrust, higher efficiency, higher velocity ratio and high load carrying capacity with high speeds and the demerits are high cost and requirement of large space.

Chapter - III 3.1 State the use of bevel gears.

(Apr/May 2008)

The bevel gears are used for transmitting power at a constant velocity ratio between two shafts whose axes intersect at a certain angle.


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3.2 What are the various forces acting on bevel gears? The normal force F n acting on bevel gear perpendicular to tooth making an angle  circle. This normal resolved into component F t radial

tooth is profile and with pitch force is tangential component

(F r ) p

Pinion

(F a ) p (F r ) g

(F a ) g G ear

F r and axial component F a. F r  Fttan  c o s  F a  F ttan  sin 

where   pressure angle and   pitch angle.  2M t  Ft   ; d  av  M t  torque on bevel gear dav  pitch circle and of bend gear.

3.3 What are Mitre gears. When the angle between two shafts is 90 and no. of teeth on pinion is equal to no. of teeth on wheel (i.e. Z 1  Z2 and i  1 ), then the arrangement of bevel gears are called mitre gears. 3.4 What are the crown gears. When the bevel gears connect two shafts whose axes intersect at an angle greater than a right angle and one of the bevel gear has a pitch angle of 90, then it is known

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as a crown gear. The crown gear corresponds to a rack in spur gearing. 3.5 Classify the bevel gears. The bevel gears are classified as 1. 3. 5.

Straight bevel gears Zerol bevel gears Spiroid gears

2. 4. 6.

Spiral bevel gears Hypnoid gears Mitre gear.

3.6 What are the advantages of spiral bevel gears over straight teeth bevel gears? The advantages of the curved tooth spiral bevel gears over the straight ones are in general the same as those of helical gears over spur gears, namely lesser noise, greater contact ratio, gradual engagement of the mating teeth, greater load carrying capacity (15 to 25%), lesser vibrations, etc. 3.7 Define the following gears. (a) Zerol bevel gear (b) Hypnoid gear. Zerol bevel gear is a patented bevel gear having curve teeth but with a zerol spiral angle. Hypnoid gear is similar to spiral bevel gear but with the axis of the pinion is offset from the axis of the gear. In hypnoid gears pitch surfaces are hyperboloids of revolution. The tooth action between such gear is a combination of rolling and sliding. 3.8 What are the advantages of hypnoid gears over bevel gears? The advantages of hypnoid gears are comparatively smoother action and possibility of extending the shafts past


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each other. So the bearings can be used on both sides of the gear and the pinion. This makes the drive more rigid and hence higher power can be transmitted than bevel gears. 3.9 Why is hypnoid gear used in automobiles? The hypnoid gear is used in automobile gear box because the drive can be placed beneath the floor level. 3.10 Differentiate between the spiral bevel gears and hypnoid gears. The spiral and hypnoid gears are similar and the most important difference being that the pinion axis of hypnoid gear set is offset above or below the gear axis. 3.11 How are bevel gears classified based on pitch angle? Based on pitch angle the bevel gears are classified as External gears 1  2  90 Internal gears

1  2  90

Crown gears

1 or 2  90

Miter gears

1  2  45

Where 1, 2 are pitch angles of pinion and gears respectively. 3.12 When do you prefer worm and worm wheel drive? We prefer the worm and worm wheel drive in following circumstances: 1. In large reduction in velocity ratio is required

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2. Self locking specially is required 3. To connect non-intersecting perpendicular shafts. 3.13 List various application of worm gear drive. The worm gear drives are used to connect non-intersecting shafts whose axes are right angles to each other. They are used in cranes, lifting machines, automobile steering mechanism, indexing head, table drive for hobbling machine, auto feed drive in lathe carriage, reduction gear boxes, etc. 3.14 Give the advantages of worm gear drive. The advantages of the worm gear drive are: 1. Transmission ratio upto 500:1 with comparatively low weight. 2. Smaller overall requirement.

dimensions

and

space

3. Smooth and noiseless operation. 4. Irreversibility or self locking ability. 3.15 What are the disadvantages of worm gear drive? The disadvantages of worm gear drive are: 1. Wear is more due to sliding. 2. Transmission efficiency is less (only 65 to 70%). 3. Not suitable for high power applications. 4. Drive is costlier because use of materials like phosphor bronze. 5. High amount lubricating oil.

heat

to

be

dissipated

using


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3.16 What are the factors controlling the efficiency of the worm drive? The factors controlling the efficiency of the worm drive are lead angle, number of starts in worm, surface finish, type and amount of lubrication and type of design of the system. The efficiency of the worm gearing also depends on the material of the worm and worm wheel and the rubbing velocity. 3.17 How efficiency of worm drive can be increased? Worm drive efficiency can be increased by 1. Increasing the lead angle. 2. Using multistart worm with small diameter. 3. Improving finish, lubrication, etc. 4. Using rigid, non-yielding worms with smooth, ground or polished flanks. 3.18 Define terms of worm gear drive. (a) helix angle, and (b) lead angle. Helix angle: It is the angle between the tangent to the thread helix on the pitch cylinder and the axis of the worm. Helix angle of worm wheel + lead angle of worm = 90 Lead angle: It is the angle between the tangent to the pitch helix and the plane of rotation. 3.19 In which gear drive self-locking is available? Self-locking is available in worm and worm wheel drive. Condition for self-locking is the friction angle of the surfaces in contact is greater than the lead angle of the worm.

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3.20 What is irreversibility in worm gears and how is it obtained? The worm drive is so designed that the gear can not turn the worm although the worm can turn the gear. If the worm wheel is unable to drive the worm, the phenomenon is called irreversibility (ability to transmit motion in one direction only). The irreversibility is obtained by having the friction angle of worm greater then the lead angle. It is also known as self locking ability of the gear. 3.21 Write short notes on worm and worm wheel? The relative sliding of the profiles of the mating members in a worm gear drive leads to seizure due to wear. Therefore the allowable (or) designed contact stresses are assigned depending on the combination of contacting materials and the velocity of sliding. 3.22 Name the materials used for the manufacture of worm and worm wheel? Materials for Worm Worms are made of steel. Low carbon alloy steels case hardened to 55 to 63 RC Medium carbon alloy steels surface hardened to 50 to 55 RC. Materials for Worm Wheel Cast iron used for low speed, light duty drives velocity upto 2 m/sec. Tintless bronzes and brasses are used for velocities upto 8 m/sec. Cast tin, bronzes are used for velocities upto 25 m/sec.


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3.23 Usually worm is made of hard material and worm gear is made of softer material - Justify. (Nov/Dec 2009)

Because of relatively high sliding velocities the worm material is harder than the worm wheel material. The worm goes through many more contact stress cycles than the worm wheel, so it is advantageous to use harder, more durable material for the worm. 3.24 Why is the crossed helical gear drive not used for power transmission? (Nov/Dec 2007) When helical gears are mounted on non-parallel shafts (i.e.) crossed helical gears, the gears may be of same or opposite hand. Crossed helical gears are essentially non-enveloping worm gears. The teeth have point contact with each other. For this reason they will carry only very small loads and hence they are for instrumental applications and not for transmission of power. 3.25 What is Lewis (tooth) form factor? (Apr/May 2008) F t  bm b Y

where the Y   y, is called the Lewis from factor; whose value depends on the number of teeth and pressure angle of the gear drive. The empirical relations for the Lewis form factor are for a 20 involute full-depth tooth 0.912   Y    0.154  Z  

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for a 20 sub tooth 0.95   Y    0.175  Z  

for a 14.5  tooth 0.684   Y    0.124  Z  

3.26 What is the main disadvantage of a helical gear drive? How is it eliminated? The helical gears drive will produce axial thrusts. The axial thrusts can be eliminated by the use of double helical gears. Both hands produce axial thrusts which are equal and opposite to each other, so they nullify each other. 3.27 What is helix angle? A helix angle is the angle between any helix and an axial line on its right. 3.28 What is the helix angle of a herringbone gears? The helix angle of a herringbone gear ranges from 25 to 40. 3.29 What are the various materials used for the production of gears? The materials used for the production of gears are: Metallic Materials 1. Plain carbon steels 2. Alloy Steels 3. Cast iron 4. Bronze (Worm wheel)


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Non - Metallic Materials: 1. Wood 2. Synthetic resins 3. Compressed papers 3.30 What are the various losses in the worm gear? Meshing occurs with sliding action, which is greater than that of a toothed gears, because of this the friction and intensive heating of the drive takes place. Therefore, it results in power loss. 3.31 What is a power screw? Power screw is power transmission device which is used to convert rotary motion into reciprocating motion with help of a screw and nut mechanism. 3.32 What are screws?

the

main

application

of

power

The power screws are used to 1. Raise the load. 2. Obtain an accurate motion in machine tools, e.g. Leadscrew of lathe. 3. Clamp the workpieces, e.g. Vices. 4. Load the specimen, e.g. Universal testing machines. 3.33 What are advantages of power screws? The main advantages of power screws are: 1. Large load carrying capacity with small overall dimensions. 2. Simple design and easy to manufacture. 3. Provides large mechanical advantages. 4. Smooth and noiseless operation.

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3.34 What are the various types of screw threads used in power screws? The various types of threads used in power screws are: 1. Square threads: high efficiency and difficult to machine. 2. Acme threads: wear can be taken care of by use of split nut. 3. Butters threads: transmission, stronger threads.

unidirectional power than other types of

4. Ball round threads (ball screws): no backlash, hither efficiency. 3.35 How is self locking obtained in power screws? In order to make a screw as a self locking one, the helix angle chosen should be low to introduce intentionally a frictional force that is sufficient to prevent reverse motion. In other words a screw will be self locking, if the helix angle is less than friction angle.

Chapter - IV 4.1 Define the progression ratio  of a gear box. If the speed steps in a gear box are arranged in a manner to have the ratio adjacent speeds is constant, then the progression ratio (step ratio) is called geometric progression ratio . N 1  first speed, N2  N1   N3  N2   


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4.2 What are the various components of automation speed reducers? Casing shafts, lay shaft, spine shaft, bearings, gears, keys, spacers. (To maintain space between gear and bearings) etc... 4.3 Where do you use speed reducers. A speed reducer is a gear mechanism, which is used to reduce the angular speed of the output shaft as compared with that of input shaft. In a multi-speed gear box, the speed of the output shaft is varied in discrete steps, these may be increased or decreased speeds, from the input speed. 4.4 What type of gear will you recommend for silent operation of gear boxes? The gears recommend for silent operation of gear boxes are helical gears. 4.5. Name the types of speed reducers? 1.

According to the type of drive (Spur, helical, bevel and Worm gears)

2.

According to the no. of stages (single-stage, double-stage)

3.

According to the arrangement (Horizontal, Parallel, Vertical, Intersecting).

of

shafts Co-axial,

4.6 Why are spacers applied to gear box? The spacers in the gear box are used to maintain distance between the gears and the bearings.

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4.7 What is the structural diagram of a gear box? The structural diagram represents the arrangement of shafts, and the arrangement of movable and fixed gears. 4.8 What does indicate?

the

‘ray-diagram’

of

gear

box

The ray diagram of a gear box, represents the no. of output speeds, no. of stages required and also the no. of shafts required. 4.9 What are the important components of a gear box? 1. Proper housing to support the bearings, to hold lubricants and to protect the unit from dirt and dust. 2. Shafts for mounting gears. 3. Suitable bearings to support the shafts. 4. Spacers to maintain distance between the gears and the bearings. 4.10. Draw the structural diagram of a Gear box. For A 6-speed gear box Structural formula  Z  P1X1, P2X2, P3X3 P1, P2 X1  1, X2  P1  3, X3  P1 P2  3.2  6 Z  3 1,

1st

2 3

x

1 60 0

2nd

E

The No (3) indicates No. of speeds. The No. inside the bracket indicates No. of spaces in speed diagram.

C

D

1 25 0 1 00 0

B 8 00 6 30 A 5 00 1 St S tag e

Second Sta ge


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4.11 What is a multi-speed gear box? Explain? When the no. of output speeds are more than four, then the gear box is known as multispeed gear box. The multi speed gear box is used in machine tools to obtain more output speeds such as 6,9,12 speeds. 4.12 What is a gear box? A gear box is a device employed for stepped regulation of the input speed and to give required output speed. It consists of gears, shafts, levers, keys, bearings, etc. assembled in a housing. 4.13. What is the use of gear box in a machine tool? The gear box is used in a machine tool to get 1. Required torque and speed. 2. Multispeed in more number of steps. 3. Change in the direction of rotation. 4.14 Give the applications of the speed reducers. The speed reducers are used in turbine generators, between motor and machine tools, in rolling mills, between engine and road wheels in automobiles and in many industrial processing like cement factories, paper plants, chemical industries, etc. 4.15 What is step ratio? The ratio between the adjacent multispeed gearbox is called step ratio.

speeds

in

a

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4.16 Why are the steps of speed arranged in geometric progression in a machine tool gear box? The steps of speeds of a gear box of machine tools are arranged in geometric progression because it provides, 1. Constant loss of economic cutting speed in whole rpm range 2. Constant loss of productivity in the whole rpm range 3. Better design feature 4. Compact design, etc. 4.17 What are the considerations to select a particular transmission group in a gear box? The transmission ratio must lie in the following range. 1. The maximum reduction of speed is limited to four times to keep the radial dimensions of gear box 1  within the reasonable limits.  i.e. imin   4  2. Maximum increase of speed in a stage is restricted to two times due to limitations of the pitch line velocity.

In a transmission group.  i.e. imax  2 .   4.18 What are the informations structural diagram?

given

Structural diagram gives informations about 1. Number of shafts in a gear box. 2. Number of gears in shaft.

by

a


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3. The order of changing transmission in individual groups to get a desired speed. 4. Transmission range and characteristic of each transmission group. 4.19 Distinguish between structural diagram and ray (speed) diagram. The structural diagram shows the general arrangement of the gear transmission, where as the ray diagram gives the gear arrangement for the required output speed. 4.20 What are preferred numbers?

(May/June 2013)

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. 4.21 What are the possible arrangements to achieve 12 speeds from a gear box? (May/June 2013) (i) 3  2  2 scheme (ii) 2  3  2 scheme (iii) 2  2  3 scheme 4.22 Which type of gear is used in constant mesh gear box? Justify. (Nov/Dec 2009) Planetary reduction gears are remarkable for compact layout and higher gear ratios. These reduction gears are widely employed in aircraft, helicopters and winches and other machines. The weight of planetary reduction gears can be decreased two to three times as compared with ordinary toothed gears of same power and gear ratios.

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4.23 Compare sliding mesh and synchromesh gear box. (Nov/Dec 2009) Sliding mesh gear box is simplest type of gear box. The clutch gear is rigidly fixed to clutch shaft. It remains always connected to the drive gear of the lay shaft. The required speed is obtained by shifting the gears in counter shaft by selective mechanism. Synchromesh gear box is used to synchronize the rotation of gears that are about to be meshed. Due to friction, the rotary pinion is made to rotate at the same speed as the synchromesh unit. 4.24 List six standard speeds starting from 18 rpm with a step ratio 1.4. (Apr/May 2008) Consider R20   step ratio  1.12  1.12 

series,

which

1.12  1.12  1.4 skip 2 speeds

Take R20 series by skipping (2) speeds. N 1  18 rpm N 2  25 rpm N 3  35.5 rpm N 4  50 rpm N 5  71 rpm N 6  100 rpm

is

having


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4.25 Sketch the kinematic layout of gears for 3 (Apr/May 2008) speeds between two shafts. Z4 Z2

Z6

D rive r

D rive n

Z3 Z5

Z1

4.26 List out the Basic rules to be followed for optimum gear box design. To avoid excessively large diameter of the wheel and also to limit the pitch line velocity of the gears. 1. Number of gears on the last (spindle) should be minimum. 2. Number of gears on the shafts should not be more than 3, though in exceptional cases it may be 4. 3. It is preferable to have N max  N input  N min in all stages except in first stage. 4. The transmission ratio between the spindle and the shaft preceding it should be the maximum possible (i.e.) speed reduction should be maximum possible.

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Follow for all stages N max N input

 2 and

N min N input



1 4

If it is not possible keep

N max N input



N input N min

(Or) keep

N max N input



N input N min

4.27 Explain why the discrete speeds are specified in geometric series for any machine tools? (Apr/May 2008)

The speeds of the spindle should be in the geometric series, in order that the number of gears employed is to be minimum and also to reduce speed loss. Preferred No. are in geometrical progression. 4.28 Why gear boxes are provided with housings? Gear boxes are provided with housings because of the following reasons. 1. Supports for the shafts. 2. Hold lubricant inside. 3. Protects the gears from dust and moisture. 4. Gives necessary cooling surface to dissipate the heat generated.


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4.29 What is preferred structural formulae? The no. of speeds required when given in the multiples of 2 and 3 is known as preferred structural formulae. 4.30 Where are speed reducers applied? Speed reducers are widely used for reduction of speed is turbine generator set, from motor to machine tool spindles, in rolling mills and in automobiles.

Chapter - V 5.1 What is a cam? Cam is a mechanical member having miscellaneous contour surface which transmits a desired motion to a follower by direct contact. With help of cam, it is very easy, accurate and efficient to produce a motion with required velocity and acceleration. 5.2 Give the list of types of cams used in industries. The common types of cams used in industries are : (1) Wedge cam (2) Radial cam (3) Globoloidial cam (4) Cylindrical cam (5) Spherical cam (6) 3-dimensional cam (3D) and (7) Inverse cam. 5.3 Give some applications of cams. Cams are used in wide variety of machines such as, packaging machines, can making machines, wire forming machines, IC engines, machine tools, computing mechanisms, mechanical and electronic computers, etc.

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5.4 Why are cam mechanisms preferred? The cam mechanisms are preferred over other types because the use of cams makes it possible to obtain an unlimited variety of motions and the cams perform satisfactorily year after year. The mechanical movements from cams are not readily obtained by other types of mechanisms. 5.5 Define following terms of a cam (a) trace point, (b) stroke. Trace Point: This is a reference point on the follower which is used in laying out the theoretical curve or pitch curve of the cam. Stroke: This is the maximum movement of the follower away from the cam shaft from its initial position. This also known rise or life or throw of the cam. 5.6 Define the following (a) pitch point (b) jerk Pitch point: It is the point on the cam pitch having the highest pressure angle. Jerk or pulse: The instantaneous time rate of change of acceleration is called as jerk or pulse (i.e. sudden change in follower acceleration). 5.7 What is the pressure angle? It is the angle between normal to cam profile and line of action of cam follower at the point of contact. The pressure angle represents the steepness of cam profile. The pressure angle is limited to 30 for smooth cam-follower action.


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5.8 What is the importance of pressure angle in cam design? The size of pressure angle is important because 1. Increasing pressure angle increases the side thrust and this increase the forces exerted on cam and follower. 2. If the pressure angle is too large, jamming of follower takes place, 3. Reducing the pressure angle increase the cam size. 5.9 Define the following (a) base circle (b) offset Base circle: It is defined as the smallest circle which is drawn tangent to the physical cam surface. Offset: It is defined as the perpendicular distance between the follower’s axis of motion and the centre of the cam. 5.10 What are the common types of cam followers? The common types of cam followers are: According to shape: Knife edge follower, flat faced follower, spherical faced follower and roller follower. According to oscillating follower.

motion:

reciprocating

follower

and

5.11. Write short notes on side thrust on: Flat, Knife edge and Roller follower. Knife edge follower is subjected to excessive wear due to sliding motion. If the load on the follower is small, this follower can be used. The roller follower has rolling contact, so wear is less.

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The roller and knife-edge subjected to side thrust.

follower

guides

are

Incase of flat face follower, there is no side thrust. 5.12 What is undercutting in cams? When a cam has a sharp point or corner, the follower does not undergo an abrupt change of motion, but gradually a different velocity. If the roller radius is greater than the theoretical pitch curve at the convex portion of the cam, the roller undercuts. i.e., moves itself over the same path twice. This phenomenon is called under cutting. 5.13 Why a roller follower is preferred to that of an knife-edge follower? Knife edge follower is subjected to excessive wear due to sliding motion. Roller follower has rolling contact, hence the wear is less. If the load on the follower is small, knife-edge follower can be used. 5.14 What is a clutch and why is it necessary? A clutch is a machine member used to engage or disengage a driving member with driven member. The clutch is required for the following circumstances. 1. To engage and disengage driving and driven member will without stopping the engine. 2. To avoid overloading of driving member. 5.15 What are the various types of friction clutches? The friction clutches are classified according to: 1. Use: Connecting directional clutches.

clutches,

starting

clutches

and


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2. Construction: Shoe and cone clutches, disc clutches and band clutches. 3. Friction pairing and lubrication: Dry and lubricated clutches with or without special friction lining. 4. Operation: Mechanical clutches, magnetic clutches, hydraulic or pneumatic operated clutches. 5.16 What are clutch?

the

characteristic

of

a

positive

The characteristics of a positive clutch are: 1. They do not slip; so no heat is generated. 2. They can not be engaged at high speeds. 3. Engagements shocks.

at

any

speed

accompanied

by

5.17 Why are the friction clutches preferred? The friction clutches are preferred, because: 1. They transmit torque at the desired rate. 2. They permit smooth engagement at all speeds. 3. They permit variable output and regulation in output. 4. They are simpler, smaller and invariably much cheaper than fluid or electrodynamic clutches. 5.18 What are the required qualities for a friction material used in clutches? The required qualities for a friction material used in clutches are: 1. High and uniform coefficient friction 2. Unaffected by environmental conditions

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3. Ability to withstand high temperature together with good heat conductivity 4. Good resilience, high wear resistance, scoring and galling. 5.19. What is the axial force required at the engagement and disengagement of cone clutch? (May/June 2013) W  W n 1   c o t 

5.20 Why is it necessary to dissipate the heat generated during clutch operation?(Nov/Dec 2009) In order to save the friction plates and lining materials from melting by the heat produced during operation, the generated heat should be dissipated. 5.21 Clutches are usually designed on the basis of uniform of wear. Why? (Apr/May 2008) In clutch design, generally two cases are considered. In one case it is assumed that the intensity of the pressure on friction surfaces is constant. This assumption is valid only when the discs are relatively flexible. On the other hand, if the discs are rigid, wearing of the friction surface is approximately uniform after initial wearing - in has taken place. In practical situations, neither of these assumptions (uniform pressure or uniform wear) is correct. So the designer has to choose a hypothesis which is more close to the actual situation. Alternately, it is better to assume a uniform wear rate because it is more conservative than assuming uniform pressure but it results into lower torque transmission capacity.


5.22 Sketch a cone clutch.

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(May/June 2007)

5.23 Define a brake? Brake is a device which is used to stop the motion of a moving body or retard its motion. It is due to the friction between the contact surface. 5.24 What are the factors required to design a brake? The factors required to design a brake are: 1. Coefficient between the braking surfaces. 2. Unit pressure between braking surfaces. 3. The projected area of the breaking surface. 4. Heat dissipating capacity. 5.25 What are the different types of brakes? The different types of brakes are: 1. Simple block (or shoe) brake 2. Pivoted block (or shoe) brake 3. Double block (or shoe) brake

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4. Simple band brake 5. Differential band brake 6. Band and block brake 7. Internal expanding brake. 5.26 List few materials?

properties

of

a

brake

friction

The brake lining materials should have the following properties. 1. It should have high coefficient of friction. 2. It should have high wear and heat resistance. 3. It should have high heat dissipation capacity. 4. It should not react with moisture. 5.27 What is the difference between clutch and brake? Clutch is used to rotate the driven member along with driving member where as brake is used to control or stop the moving member. 5.28 How does Geneva mechanism function? The Geneva mechanism consists a Geneva wheel having radial slots and a crank. The crank has a pin which enters into a radial slot and causes the Geneva wheel to turn through a portion of revolution. When the pin leaves the slot, the Geneva wheel remains stationary until the pin enters the next slot. This way the continuous motion of the crank is converted into intermittent motion of the Geneva wheel.


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5.29 List the applications of the Geneva mechanism. The Geneva mechanism is used in rotary tables, indexing mechanisms of turret and capstan lathes, automatic machines, watches, motion-picture projectors, etc. 5.30 What is the requirement to obtain a smooth motion from a Geneva wheel? The slot in the Geneva wheel must be tangential to the path of the pin upon engagement in order to reduce shock during engagement. In other words, the centre lines of slot and crank must mutually be perpendicular at the engagement and at the disengagement. 5.31 State a few applications of ratchet and pawl mechanism. Shaping machine table movement Planing machine Hoisting machine Bearing machine Some types of screw jack. 5.32 What do Mechanism’

you

understand

by

‘Geneva

In machine tools, the mechanism used in rotary tables and indexing turrets is known as Geneva mechanism. Geneva mechanism consists of a Geneva Wheel with a no. of slots and a driving disc (Crank) whose projected pin enters the slot and causes the intermittent rotation of the Geneva Wheel.

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5.33 What is an intermittent mechanism? It is a mechanism that consists a sequence of motions and dwells. A dwell is a periods in which the output link or links remains stationary while the input link continues to move. The intermittent motions may be circular, reciprocating or oscillating. 5.34 How ratchet and pawl mechanism works? In a ratchet and pawl mechanism, the arm pivots about the centre of the toothed ratchet wheel and is moved back and forth to index the wheel. The driving pawl rotates the ratchet wheel in one direction and does no work on the return trip. The locking pawl prevents the ratchet from reversing direction, while the driving pawl returns. Both pawls are usually spring loaded. 5.35 What is the function of pawl in the ratchet and pawl mechanism? Pawl is the machine member used to interrupt the further rotation of the driver (ratchet), by engaging in the ratchet tooth. The pawl may be activated by its own weight or by a spring. When the direction of rotation is reversed the pawl gets disengaged and there is no interference of driver motion. One or more pawls are used in a ratchet mechanism.

1


Index B Based on Hertz Method, 3.64, 3.192 Basic Curves, 5.6 Bevel Reducer, 4.212 Bevel Gear Nomenclature, 3.4 Brake friction materials, 5.83 Brakes, 5.83 Bush chain, 1.120

C Cam Profile Layout, 5.31

Differential Band brake, 5.99 Double shoe brake, 5.113 Dynamic Effects (Dynamic Tooth Load), 2.19

E Efficiency of Worm Gearing, 3.157 Electromagnetic clutch, 5.80

F Factor of Safety, 2.25

Cam, 5.1

Failures of wire rope, 1.97

Centrifugal Clutches, 5.78

Fatigue Strength of Gear, 2.21

Chain, 1.120

Flat belts, 1.2

Chain Drives, 1.118

Flat Belt Drives, 1.3

Chain housing, 1.122

Fluid coupling, 4.212

Circular belts, 1.3

Fluid fly wheel, 4.212

Clutches, 5.47

Follower, 5.1

Cone Clutch, 5.69

Force Determination, 5.36

Constant Mesh Gear Box, 4.208

Friction Materials, 5.137

Contact Stress, 5.35 Crossed Helical Gears, 3.214

D Design of chain drive, 1.118 Design of V-Belt Drive, 1.45 Design of Wire Ropes, 1.91 Design of Gear Box, 4.129 Design of Spur Gear, 2.77

G Gear Tooth Failures, 2.6 Gear Box, 4.1 Gear Materials, 2.26 Geneva Mechanism , 5.138 Geometric progression, 4.6

H

Design of V-Belts, 1.47

Helical Gears, 2.123

Design of Worm, 3.192, 3.153

Helical Reducer, 4.212

Designing CAM, 5.29

Hertz Method, 2.77

Index

I

Single Block Shoe Brake, 5.84

Internal Expanding Brake, 5.120

Slack adjuster, 1.122

Inverted chains, 1.120

Speed Diagram, 4.4

K Kinematic Layout, 4.13

L Law of Gearing, 2.7

M Metallic materials, 2.26

N Nomenclature of a Cam-Drive, 5.4 Non-metallic materials, 2.26 Normal Pressure Angle n, 2.129 Number of Teeth, 2.10

P Plate clutch, 5.50 Pressure angle , 2.129, 5.31

R Ratchet and Pawl Mechanism, 5.147 Ray Diagram, 4.4 Roller chain, 1.120

S

2

Speed Reducer Unit, 4.211 Speed Ratio, 2.10 Sprockets, 1.121 Standard step ratio, 4.6 Strength of Bevel Gears, 3.7 Structural Formula, 4.11

T Tooth Stress, 3.7 Tooth Stresses (The Lewis Bending Equation), 2.14 Toothed belt, 1.3 Torque Convertor, 4.215 Torque Converter Gear Box, 4.215 Transverse Pressure Angle t, 2.129

U Under Cutting, 5.33

V V - belts, 1.3 Variable Speed Gear Box, 4.212

W

Screw or Skew gears, 3.214

Wear Tooth Load, 2.21, 2.127

Sheave materials, 1.114

Wear Strength of Helical Gears,

Silent chains, 1.120

2.127

Simple band brake, 5.96

Wire Ropes, 1.91

Simple Harmonic Motion SHM, 5.16

Worm Reducer, 4.211 Worm Wheel, 3.153, 3.192

E. v. v. Ramanamurthy, S. Ramachandran-Design of Transmission Systems-Air Walk Publ. (2015)(1)

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