Textbook (Second Semester).PDF

CHAPTE,R I
CI-IAPTER 2
CHAP'TER 3
T]HAPTER 4
CONTENTS
SPUR GEARS
HEI-ICAL GEARS
BEVEL GEARS
WORM GE,ARS
( SUPPI,EMENTARY PROBLEMS)
Dr. Aung Ko Latt (Professor)
Department of Mechanical Engineering
Mandalay Technological University
dr.aungkolat@gmail.com, 09440014010

CHAPTER I
SPUR GEARS
Cear Drives
Gears are defined as toothed wheels or n.rulti lobed cams which transrnit power
and motion fiom one shaft to another by means of successive engagement of the teeth
Gear drives offer the following advantages compared with chain or belt drives
( I ) It is a positrve drive and the velocitv ratio remains constant
(2) The center distance belu,cen the shafi is relatively. small. which results in
cornpact collstructlon.
(3) It can transmit very largc power. which is beyond the range of beit or chain
dri ves.
(4) lt can transmit rnotion at..er/ lo*'r'elocitv. which is not possible w'ith the belt
drives.
(5) T'he efficiency ofgear drives rs very high even up to 90oir in case ofspur gears.
(6) A provisron can be made in the gear box fc'r qea' .!rlftirrg, thus changing the
' elocit) rati() o cr a lr'ide ranse i,:
Gear drivers are. however, costly and their maintenance cost is aiso higher The
manuf'acturing processes for gear are complicated and highly specialized. Gear drir,'es
rer,luire carefll attention for lubrication and cleanliness fhev also reclulre precise
alignment of the shaft.
Cllassification of gears
Gears are broadly classified tnto four grollps-spur. helrcal. bevel and worrn
gears
Spur gears
Spur gears (Fig I l) pror.'ide r positire rneiins of'transr.uitting power between
parallel shalis at a constant angular vekicity latio ln tire spur sears. the teeth are cul
parallel to the axis of the shalt "J' r

7
Spur gears are simple, least expansive and the formula and method of attack (or
solution) are applicable to other types ofgear such as bevel, helical, etc.
Fig l.l Spur gears
Components of gear forces
Resultant of the gear forces can be found by the vector sum of the components.
The cornponents are used in calculating bearing reaction and shaft sizes.
Fx
=Mt
Dt2
=Ft tan0
Fr
F1
Fr

Gear tooth terminology
The principal parts of the gear teeth are denoted as shown in Fig 1.2.
D.ir.n G.rt
Add6d{d (:ir<1.
PircI Circlc
BrrrCirclc

4
Definitions
Pitch circles
Fig I 3 shows the relative positions of the teeth at several stages of'engagelnent
when the teeth two gears mesh. Throughout the engagement circle. there are two circles.
one from each gear. that remain tangent These are called pitch ctrcles
'fhe
dianreter ot'
the pitch circle of a gear is itch ciian-reter
When the two gears mesh. the srnallet gear is called the pinion and the larger is
the gear. I'he pitch circle dialneters are denoted as Dp and Do fbr the pinion and gear'
respectively
rt Cll
{i r{
Gcar I
drivcr
ir
(c, t(ll
Fig 1.3 Cycle of engagement of gear tceth
Base circle
'fhe base circle is an inragiri;ry circie from r.l'hich the inr,olute curve of the tooth
protile is generated.
Addendum circle
Acldenrlurn crrcle limits the top of the gear tooth. Addenclurn circle tliametei
(outsicle diarneter) is denoted as Do Do " D * 2 adcl
Adderrdum. add
I-he raclial distarrce front the pitch ctrcle to thc or-rtsitic ol'the tooth

-
Dedendum circle
The dedendum circle or the root circle is the curve of intersectior.r of the rool
cylinder with a plane which is perpendicular to rhe axis of the shaft. D,: D 2 ded
Dedendum. ded
The dedendunr is the radial distance between the pitch and the dedendurn circle
Circular pitch, P.
The distance fi'om a point on one tooth to the corresponding point orr an adjacenr
tooth measured on the pitch circle
P. -- n D/N
Module, m
Where D - pitch diameter. N - no. of teeth
aa
r
i
al
€
;
N,Iodule is the pitch circle diameter in mm divided by ninnber of teeth
m .-- DN [PJm -r ]
Line of action
Line of action is the line normal to a parr of rrrating tooth profiles at their poinr
of'contact.
Pressure angle. g
-l-he
angle between the line of action and the comnlon tangenr to the pitch circle.
Pitch point
Pitch point is the point of tangency of the pitch circles.
.Angular velocif-v" ratio (transmission ratio). V.R
Angular velocity ratio is the ratio of angular velocity of the pinion to the angular
r"elocity of its rnating gear lt is inversely proportlonal to the nur.nber of teeth on the two
gears. and fbr spur gears it is also rnversely proportional to the pitch diameters
' R " (rpm)pi(rpm)" - Nur'Np - De/Dr,
Fundamental lall of gearing
l-he ctttrttnott rtorrnal to the rooth profile at tlre point of contact rrrust always pass

6
through a fixed point called the pitch point in order to maintain a constant angular
velocity ratio of the two gears. The involute curve satisfies the law of gearing and is
most commonly used for gear teeth profiles. Frequently a combination of the involute
and cycloid curves is used for a gear tooth profile in order to avoid interference in the
composite form, approximately the middle third of the profile has an involute shape
while the remainder is cycloidal.
Table 1.1 Proportions of standard gear teeth
Standard module series
Preferred l, 1.25, 1.5,2,2.5,3, 4,5,6, 8, 10, 12, 16'20'25,32, 40, 50
Secondchoice 1.125,1.375,1.75,2.25,3.5,4.5"5.5"7,9,11, 14, 18,22,28,36,45
Gear design
Gear tooth design involves primarily the determination of the proper pitch and
face width for adequate strength, durability and economy of manufacture.
In a gear design it is necessary to calculate: - strength check and dynamic check.
Strengthpointof view >ftIld the smallest possible module and corresponding
Dynamic point of view
face width
+ find the endurance force Fo, w€&r force F* and dynamic
force Fo.
Required condition Fo, F*. ) Fo
=) lt'class commercial cut
10
14-
2
Composite
lo
14:
2
Full depth
involute
20n
Full depth
involute
200
Stub
Involute
Addendum m m m 0.8m
Minimum dedendum l. l57m 1.157m l. I 57m m
Whole deph 2.157nt 2.151m 2.157m l.8m
Clearance 0.157m 0. l57m 0. l57m 0.2m
Economy of manufacture

Carefully cut
Precision cut
Force analysis
Fig l.a
M: Ft x h, I : bt3l12, s : bending stress induce.
Using bending equation
sbt' sbt'
'66f,
h: C t2 : equation ofparabola.
The parabola shown in Fig 1.4 outlines a beam of uniform strength. Hence the
weakest section of the gear tooth is a section A-A where the parabola is tangent to the
tooth outline. The load is assumed to be uniformly distributed across the face of the
M=-t
Iy
F,h
=t
bt'ltz tl2
ct
aa
q5
;
;
r!
chtr ( ,:
f,=?=sbl
-
Itp.=sbyR
' 6h [6hP.,l
'
Where (C/6hP.) : y: dimensionless quanrity called form factor

8
The values of form factor y is tabulated in the Appendix I The fomr factor y is
a f'unction of the tooth shape and the number of teeth on the gear.
The equation lrr - sbyPc is known as the l-ewis equation
For ordinary design condition. the face u'idth b is limited to a maximum of 4
times the circular pitch
b:kP. wherek < 4.
Ft-.sk+P.2y:sk n2m2y
In design of gear strength, the pitch diameter is either known or unknown. In
both cases the smallest possible module rvill provide the most economical desrgn. In
general, where the diameters are known, design for the largest number of teeth possible,
where the diameters are unknown design for the snallest pitch diarneters possible. The
minimum number of teeth. N is usuallv limited to 15.
Allowable tooth stress
The allowable stress tbr gear tootlr design depends upon the selected material
and pitch line velocity, For spur gears, Barth's equation in SI unit is,
/ 1
I
: sn I .- | for V less than l0 m/s
[]+ V/
(
Allor.vable s : So I o I fur It l0 to 20 nrs
6+V/
( st')"
' s,, I --- i for V creater than 20 nr/s
['so*J','
Where su is the endurance strength of the gear rnaterial, N/mr and V is the pitch line
velocity, m/s. r
Values of sn
The su values for cast iron and bronze are 8000 psi (55 N'lNlrnl.1 and 12000 psi
(83 MNAI2) respectively. Carbon steei ranges frorrr about 70 to 350 MN/ttrr depending
upon their carbon content and degree of heat ffeatment, In general, so mav be taken as
approximately one-third of the ultimate strengith (s.) of the material.

-
9
Base design
The amount of force that can be transmitted to a tooth is a function of the su y
product (load carrying capacity) as shown by the Lewis equations. For two mating
gears, the weaker wrll have the smaller so 1l value.
When trvo mating gears are to be made of the same material, the smaller gear
(pinion) will be the weaker and control the design
Case I Known diameter case
( 1 ) skzr'
| . | = ^ (t)
m-Y /u' l,
( z
s : So I " I for V less than l0 rn/s
3+ Vl
( a )
s -s,, 1 lfbrVl0to20m/s
6+V/
( s.o )..
: su I ---** I fbr V sreater than 20 m/s
s.6+JVl
k:4 (max)
- 9550xkw
f r =i ---,
rpmxD12
Substitute these value in equation (1) and we get t + I value
m-Y /,|
Assuluey: 0 1 -:-> m:. . ., and write std: module series.
I'rvm:..
t
N = l)irrr" = ..... r) y : . .. (from the table of Appendix 1)
/'
lrl
|)I
m-v/
( t / r
('ontparel-' I andl '.-l valtre
m-Y1,,]] nl-Y./,,a

10
( r /
lfl , | .l
m-vi
J /lno
( r ) f
rfl , | >l
tm-v/
r' /tno
",,
=U.rt* is satisfied, decrease the module and try again
",,
-0".t* is satisfied, increase the module and try again.
And take the smallest module that satisfies the condition
"t f+'] a f+l
m'Y,/,,0 m-Y.i
^,'
Then reduce k value.
. (l/m'?y)ind
Kretl: I(max X --- --*
(l/m'y)all
Face width, b
b:kr"oxr xm
Case II Unknown diameter case
Sind: tY' , ...... (2
kyr'Nm'
where Mt
955oxkw
rpm
k : 4 (max), N : no. of teeth of weaker, y from table of appendix l.
Substituting these value in equation 2 and we get Sina : (..../rnt).
Assume V.F : and equation 2 becomes (sJ2) : (. . ../-t) and we get m value. Then write
std: module series.
Trym:...
D:Nm,V- nDrpm/6d
s.rtt:soxV.F
Sin,J : (.../mr)
Cornpare the value of sind and sall.
If si"a < sarr :) design is satisfied, decrease module and try again.
r)
fr)
t)
e)

11
If sina > sal + design is satisfied, increase module and try again.
Try these procedures until sino just ( sau and take the smallest module that satisfies the
condition.
Reduce k
q.
k rea: k,r"* X 3
Sall
Face width, b
b:k..aX 7[ Xm
After determining the design from strength point of view, it is necessary to
check the dynamic effect.
Dynamic effect
l. In accurate in tooth cutting.
2. Spacing ofgears.
3. Misalignment in mounting.
4. Tooth deflection.
All these causes change in velocity and produce dynamic effect
Endurance Force, Fo
Fo = s,, by nm
Wear Force, F." (Wear tooth load)
Buckingham Equation
Fw = DpbKQ
Where Dp = plt.tl diameter of the pinion, m
b : face width, m
a :2 Ne/ (Np + Ng) -'zDsl(Dp + Dr,)
K - stress fatigue factor, N/m2

LZ
K : f.')-$-Tlll5'r-?
1.4
Where ses : surface endurance limit of a gear pair, N/m2
Ser - (2.75 (BHN.,,g) - 70)MN/m2
BHNavg -= average brinell hardness number of the pinion and gear
material.
En: modulus of elasticity of the pinion material, N/m2
E* = rnodulus of elasticity of the gear material, N/rn2
Several values of K fbr various materials and tooth forms have been tabulated in Table
1.2 as tentative value recommended by Br"rckingham.
Dynamic Force, Fo
- r- , 2lv(bc + F, )
Ld:r Ft +
2lV rJbC+f,
Where C'- deir-rrmation factor, N/m
C is the function of material of the gear pair, pressure angle/ ,tooth error e.
Sorne values for C have been tabulated in Table 1.3. Curves showing the relation of
errors in footh profiles vs pitch line r,elocity and modules are shown in Fig 1.5 and 1.6.
V - ..... m/s =>permissible tooth error ep (Fig 1.5) .+choose type of cut (Fig I 6) type
of cut, rnodule :)tooth error e (Fig 1.6)
I{aterial of pinion & gear,$ ,e C (Table 1.3).
I'he required condition to satisf)' the dynamic check rs Fu, F* ) Fa.
If the design is not satisfied (F,,, [i" F,r), the follorving rer,redial action is to be
taketi.
( I ) Reduce tooth eror.
(2) lrtcrease module.
(3) Increase module.
(4) Increase the Bl"lN. ie apply heat treatment.

-
I5
Tatrle 1.2
Values for s". as used in the wear load equation depend
and pinion materials. Some values fbr various materials
upon a combination of the gear
for both Ses ord K are tabulated.
Stress Fatigue Factor
K (kNim2)
Surface
Endurance
Lrmit
S". (N{Nz'm2)
342
480
618
155
| 030
)+ /.
480
618
-)+ I
206
405
o /J
100,1
I 869
303
600
I 000
3tl
282
5-5-5
919
t-)/-
25 53
503
,114
820
t3l0
A1a
689
1420
1 960
150
200
250
150
400
-
cIP*i"" -
C.l Pinion
Materials
Pinion Gear
1
Phosphor Bronze
I-L!":Sg.'B',"ry
i C. I Gear
i ...,
I cl
l^_^
L. l Lrear
-L*--.-,,,_
Involute
Tooth
fonl
tal/-"
' /2
s49
618
I 050
I 330 I
Table 1.3 zalues of Defornration Factor c in kN/rn for d_vnamic load check
cast iron
cast iron
steel
cast iron
cast iron
steel
r+r/2"
t+r/ z
20'full
depth
20" fr,rll
depth
20u full
deptir
Tooth error -rnnr
03f -i-
l
l
l
0.02 o.o4
--r (].06
.5-5
76
I l0
57
Aa t
+/+
ll0
152
220
Il4
t58
228
220
304
440
228
3t6
156
JJO
J./"+
476
330
4-s6
660
342
684
.1,r+
486
714
440
632
| 1')
648
952
608
880
4-56
'to
;
!!
59
8i
I l9
I i8
t62
238
Average Brinell Hardness
Number of steel pinion & gear
t-50
200
250
300
400
Bnnell l.lardness Nurnber. BHN
cast tron
steel
steel
cast iron
steel
steei
cast iron
steel
steel

t4
E
e
6
9l
a
E
d
Fig 1.5
Fig 1.6
Plish LI'rt v.locity, rolr (Sgur Gcrrri
Crrci.ulty lut Gczrs
Modolt of Spor 6errt. nrrri

15
tL
(..t
€
;
!l
2
Example (l)
A pair of mating spur gears have 14
/'t"tt depth teeth of l0 module..l.he pitch
diameter of the smaller gear is l60mm. If the transmission ratio is 3 to 2. Calculate 1a;
number of teeth of each gear, (b) addendurn, (c) whole depth, (d) clearance, (e) outside
diameters, (f)root diameters, (g) dedendum.
Solution:
(a) Np: Dp/m: 160/10: 16
Np : Dp/m : 160 x t.SltO : 24
(b) Add: m: l0
(c) Whole depth : 2.157 x m:21.57mm
(d) clearance : 0.157 x m = l.57mm
(e) outside diameters, Dop: Dp + zadd: 160 | 2 x l0: lg0mm
Dog : Dq + 2add : 240 + 2 x l0: 260rnm
(f rootdiameters, Dip:Do.-2 whole depth - 160 -2x21.57: ll6.g6mm
Dig: Dg - 2 whole depth : 240 - 2 x21.57: l96.g6mm
(g) ded= l.l57xm= ll.57mm
Erample (2)
A spur pinion of cast steel (so: l40MN/m2) is to drive a spur gear of cast iron (so :
55MNh2). The transmission ration is to be 2-l13 to 1. The diameter of the pinion is to
be 105mm and 20kW will be transmitted at 90Orev/min of the pinion. 'I'he teeth are to
be 20 deg fuIl depth involute form. Design for the greatest number of teeth. Determine
the necessary module and face width of the gears for strength requirement only.
Solution: *
sop :l40MN/m2, sng:55MN/m2, rpmo:900,V.R:7/3, Dn:l05mm,20kW
D-s : 105 x 713:245mm, r?mg: 9001(713):385.714
Different material
.'. check load carrying capacity. (soy)

16
A$sume Np : 15 > Yp
: 0.092
Ng: l5 x713:35 > Yg = 0.1l9
(so y)p: I 40x0.092'- 1 2.88MN/m2
(so y)g:55 x0. I I 9:6.545MN/m?<(so y)p-
.'" Gear is weaker. Base design on gear.
Known diameter case.
:4.948nVs<l0m/s
{ I I st<n-
t.t_
I m-r' / t,
tl r/f,tl
,ii- nx245xl 0' )c385.7 I 4
-
* - ",,
*
-]-
- 55xl06x --l- .--,20.76x106N/mr
-u -"'3iV -1 r--1.948
r,- 4 F. - 21s!,.kY- : -LI9"2o- :4042
' rp nrxD/2 385 7l4fl45xl 0 '/2
( t ) zo.zo*tou x/' -:
l-l-
rA/'
=20214'1 513
[to'yJ 4042.328
Assume y:0 I
rn:7mm
328N
std module series .6, 7, 8,
Try m:7
N,, = Do/m : 24517 = 35
= Yg =' 0.1 19
(lim2,v)i,a: l/(5x10 3)2
0.119'- l71497.ll '- (lintry)orr
.'. Design of 6 cannot be used due to the required velocity ratio
Try m:5
Nc =' Dg/m = 24517 =' 35 :) Yg
: 0.1 19
(1/m:y),no : lr(5x10-3)2 0.12942 : 309a7 | 24 > (1/m2y).il

-
1"7
.'. Design is not satisfied.
.'. Take smallest module : 7mm.
Reduce k value-
kred : krnax x ( l/m2y)*6/( l/m2y)un : 4 x 171497 .17 nA27 47 .513 : 3.383
Face width, b
b: krea xrx m:3.383 xn x7 :74.4mm
Use b:75mm
Example (3)
A cast steel pinion (s,,: 103MN/m2) rotating at 900rev/min is to drive a cast iron gear
(s" : 55MN/m2) at 144 rev/min. The teeth are to have 2O degstub involute profiles and
the maximum power to be transmitted is 25kw. Determine the proper module, number
of teeth, and the face width for these gears from the stand point of strength, dynamic
load and wear. Pinion is surface hardened to BHN 250.
Solution:
s,,p: 103MN/m2,sos - 55MN/m2, rpmp:900, rpmg :144,25kW, V, R = 6-25 Different
material
.'. check load carrying capacity. (soy)
assume Np : 15, Nr * whole number.
Np =, 16 +y'u:0.115
Np: 100
=yg:0.161
(s. y)p: 103 x 0.115 : I l 845 MNlmZ
(s. y)e : 5-5 x 0. 161 : 8.855 MN/m'.(rn y)n
. . Gear is rveaker. Base design on gear.
llnknolvn diameter case. :
, tMt
5lilcl -' --.----:-.---.-,:-
kyn-Nrn'

18
M,
9550xkw 9550t5
=_ :1657986Nm
sind:
rpm 144
2x1657.986 5.217
m'
4x0.l61xa2xl00x:n 3
assume V F: y2.
55xl0" 5.217
Z =-F-:.>m:5.75mm
Std module series ......., 5, 5.5, 6,7, ........
Try m:6
Dg: Ne m: 100 x 6 = 600mm
V: (nx 600 x 10-3 x 144)160:4.524rn1s < l0m/s
srl = 5-5 x 106 x 3 : 21 93xL 06 nsi
3 r 4.524
s,,,a : -13{- - 24. r 53x l o6 psi > surr
(6x10'')'
.'. Design is not satisfied. Increase m.
Irym=7
Dc: Ng m : 100 x 7:700 mnr
V : ( rx 700 x l0-3 * 14{)160: 5.278rr,1s < t0m/s.
srrr : 55x l0t'x 3 : 19 9lx l0(' nsi
3+5278
,;n,1 :
-l{-
: 15 21" t o,, psi , s"u
(7r10')'
.'. Design is satisfred. Take smallest module = 7
Reduce k
kLert '= knrax x si,r,l/ Satl :4 x 15.21119.93:3 fl-52

-
19
Face width, b
b : k..a x ?rx m : 3.052 xnxT : 67.l3mm
Use b = 68mm
Dynamic check
Endurance force, Fo
Fo:sob ym: 55 x 106x 68 x l0'3 x0.16l xx 7 x l0-3 : 13.242kN
Wear force, F*
Fr" :DpbKQ
Dp :Npm
=
2Nr-
No +N*
:16x7:ll2
2x100
a = 1."724
l6+100
K : 1310 (Table 1.1)
F,u : l12xl0-3x 68x10-3x l3l0xl03xl.724: l7.2kN
Dynamic force, Fa
Ft = MJ(D/2) - 1657.9s6l350x10-3 : 4717.103N
Y = 5.2781s *€p :0.07 (Fig 3-5), m = 7, we should use carefully cur gear (Fig 1 6)
m = 7, carefully cut :> e : 0.035
steel pinion & C.l gear,:20 deg stub, e:0.035
=,C
= 283.5kN/m
Fa:4737 to.r *-?1'fZ8(6lg=j{1l1gL : 14 75
21>15.278 +
"Gsralls +alh.rcz
F" > Fa and 96 error = (F,r - Fo)rFo - 10.22% Fo =Fa. The clesign rvill probably be
satisfied.
Fornroreaccuracy,weincrease facewidthup tob:kmaxX n xll =4x ax 7 =g1 .96mm

20
Use 80mm
Fu : 15.579kN
F.n :20.235kN
Fa == l5.73kN
Fn -F,1, F.u > F,r >Design is satisfied.
m:7, b: 80mm, Np: 16, Ns: 100 €Ans:

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