Mechanical Analysis and Design
ME 2104
Lecture 4
Mechanical Analysis
Gear trains
Prof Ahmed Kovacevic
School of Engineering and Mathematical Sciences
Room C130, Phone: 8780, E-Mail: a.kovacevic@city.ac.uk
www.staff.city.ac.uk/~ra600/intro.htm
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Ahmed Kovacevic, City University London
Plan for the analysis of
mechanical elements
Objective:
Procedures for design and selection of
mechanical elements
 Week 1 – Shafts and keyways
 Week 2 – Bearings and screws
 Week 3 – Belt and chain drives
 Week 4 – Gears and gear trains
 Week 5 – Design Project Review
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Ahmed Kovacevic, City University London
Plan for this week
Gears
 Gear trains
 Examples

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Ahmed Kovacevic, City University London
Gear types
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Ahmed Kovacevic, City University London
How gears work
Law
of Gearing:
A common normal to the tooth profiles at
their point of contact must, in all positions
of the contacting teeth, pass through a fixed
point on the line-of-centres called the pitch
point.
Any two curves or profiles engaging each
other and satisfying the law of gearing are
conjugate curves
Pitch point
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Pitch
circle
Pressure
angle
Basic
circle
Pressure
line
Ahmed Kovacevic, City University London
Module and Pitch
Diametral
pitch
P
NG N P
in 1 

DG DP

D
p    m    mm 
P
N
D
25.4
Module m 
[mm]; m 
N
P
Circular
pitch
m=6.35
m=2.11
m=0.97
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Standard modules are 0.5, Ahmed
0.8, 1,Kovacevic,
1.25, 1.5,
2.5, 3, London
4, 5, 6
City2,University
Relations between gear parameters
Pressure angle:
DP  DG
C
2 o
o
Addendum:
Dedendum:
Clearance:
a = module
b = 1.25 module
c = 0.25 module
Backlash:
clearance measured on the
pitch circle of a driving gear
Centre distance:
20
(14.5 )
Backlash is function
of module
and centre distance
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Ahmed Kovacevic, City University London
Relations between gear parameters

Pow  T  GTG  PTP 

DG
DG N G nP TG
DP

v  G
 P
GR





2
2
D
N
n
TP
P
P
G


  Gear ratio
G  nG ; P  nP
30
30 
rb  r cos 
Radius of the base circle
pb  p cos   
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D
cos 
N
Pitch of the
base circle
Ahmed Kovacevic, City University London
Contact ratio and interference
Contact ratio:
cr 
Lab
Lab

pb
p cos 
Lab  rop2  rbp2  rog2  rbg2  Cd sin 
Interference (the smallest number of teeth for contact without interference):
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Ahmed Kovacevic, City University London
Helical gears
Normal circular pitch pcn  pc cos
Normal diametral pitch pdn  pd cos
Axial Pitch pa  pc cot 
Axial load Fa  Ft tan
Radial load Fr  Ft tan 
Normal load F 
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Ft
cos  cos
N
Number of teeth for helical
3
gear
cos 
tan n
tan  
Normal pressure angle
cos
bw – gear width
D p  Dg N p  N g
cd 

2 cos
2 cos
Crt  Cr  Cra Contact ratio for helical gear
b tan
Cra  w
Axial contact ratio
pc
Nn 
pcn
sin
Ahmed Kovacevic, City University London
Example 11 –helical gear
An involute gear drives a high-speed centrifuge. The speed of the centrifuge is
18000 rpm. It is driven by a 3000 rpm electric motor through 6:1 speedup
gearbox. The pinion has 21 teeth and the gear has 126 teeth with a diametral
pitch of 14 per inch. The width of gears is 45.72 mm and the pressure angle is
20o. Power of the electric motor is 10kW.
Determine :
a) A contract ratio of a spur gear
b) A contact ratio of a helical gear with helix angle of 30o.
c) A helix angle if the contact ratio is 3
d) Axial, radial and contact (normal) force for both helix angles.
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a)
b)
c)
d)
Cr=1.712
Cr1=6.343
9.122o
Fa1=160.8 N
Fa2=44.7 N
Ahmed Kovacevic, City University London
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Ahmed Kovacevic, City University London
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Ahmed Kovacevic, City University London
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Ahmed Kovacevic, City University London
Examples of gear trains
Two stage compound
Gear train
Planetary gear train
Compound reverted
gear train
Planetary gear train
on the arm
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Ahmed Kovacevic, City University London
Example 12 – Compound gear train
A gearbox is needed to provide an exact 30:1 increase in speed, while
minimizing the overall gearbox size. The input and output shafts should be inline.
Governing equations are:
N2/N3 = 6
N4/N5 = 5
N2 + N3 = N4 + N5
Specify appropriate teeth numbers.
N2 = 108; N3 = 18; N4 = 105; N5 = 21
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Ahmed Kovacevic, City University London
Example 13 – Planetary gear train
The sun gear in the figure is the input, and it is driven clockwise at 100 rpm. The
ring gear is held stationary by being fastened to the frame.
Find the rev/min and direction of rotation
of the arm and gear 4.
n3 = - 20 rpm, n4 = 33.3 rpm,
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Ahmed Kovacevic, City University London
Gear Force Calculation

Fr

F
a
Ft
Stress based on the Force acting
The force is caused by the
transmitted torque. That force
always acts along the pressure line.
T  P  Ft R 

R
F
Ft
cos a

30 P
[N ]
nR
30 P
F
 n R cos a
Ft 
The force induces stress
concentration on gear.
We need to answer:
How much power a pair of gears in
question can transfer?

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Ahmed Kovacevic, City University London
Basic gear stress calculation

Basic analysis of gear-tooth is
based on Lewis Equation which has
following assumptions:
» The full load is applied on the tip of a
single tooth (the worse case)
» Radial component is negligible
» The load is distributed uniformly
along the teeth width
» Friction forces are negligible
» Stress concentration is negligible.
Ft
mbY
Sy
P 
n D m bY
Power transmitted B
fs
Basic stress in
teeth
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B 
Ahmed Kovacevic, City University London
Correction
factors
Basic stress must be corrected for:
- Shocks,
- Size effects,
- Uneven load distribution,
- Dynamic effects.
 B
Ka Ks Km
Kv
Ka Ks Km
P  PB
Kv
Ka – Application factor
Ks – Size factor
Km – Load distribution factor
Kv - Dynamic factor
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Gear
materials
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Ahmed Kovacevic, City University London
Example 14 – Gear stress calculation
Pinion A and gear B are shown in figure. Pinion A rotates
at 1750 rpm, driven directly by an electric motor. The
driven machine is an industrial saw consuming 20 kW.
The following conditions are given:
NP=20
m=3 mm
Qv=6
NG=70
b=38 mm
fs=1.5
np=1750 rpm
Pow=20 kW
What is the centre distance? Compute the stress due to
bending in the pinion and gear and find required Brinell
hardness for this application.
SOLUTION:
( DP  DG )
( N P  NG )
The pitch diameter of the pinion is:
 3 90  135  mm
2
2
2
DP  mN P  3  20  60[mm]  0.06  m
The pitch velocity is:
vP 
Centre distance:
Transferred load (Tangential force):
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c
m
n P DP 1750  0.06

 5.5[m / s]  1090 ft / min 
60
60
60 Pow
60  20000
Ft 

 3638  N 
 nP DP  1750  0.060
Ahmed Kovacevic, City University London
Example 14 – cont.
From the diagram:
YP=0.34
Basic bending stress is: pinion –
gear Correction factors are:
(from diagrams and tables)
and
YG=0.42
 BP 
Ft
3638

 94 106  Pa 
mbYP 0.003  0.038  0.34
 BG 
Ft
3638

 76 106  Pa 
mbYG 0.003  0.038  0.42
Application factor
Size factor
Load distribution
Dynamic factor
Ka=1.5
Ks=1.0
Km=1.2
Kv=0.68
Ka Ks Km
 2.64  94 106  248  MPa 
Kv
Corrected pinion bending stress:
 P   BP
Allowable stress required for
this application:
S  f s P  248 1.5  372  MPa 
From the diagram, any material with Brinel hardness higher than HB=400 will satisfy application
requirements.
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Ahmed Kovacevic, City University London
Coursework
Example 15 – Industrial mixer
A gear train is made up of helical gears with their shafts in a single plane, such as the
arrangement in the vertical mixer shown. The gears have a normal pressure angle of 20°
and a 30° helix angle. The middle shaft is an idler. Gears AB and CD are engaged, the
others in the illustration are not in contact. The module of all gears is 2 mm. The table
gives the number of teeth per gear. Gear A exerts a load of 1200 N onto gear B. The
shaft containing gear A is driven by the motor at a speed of 200 rpm. All gears are Grade
2 and have been
Gear
Number of Teeth
hardened to
A
20
HB 350 and have
B
50
50 mm face widths.
C
12
D
75
Find:
The normal load exerted by gear C on gear D
and the speed of gear D and the safety factor
for gear D based on bending and contact
stresses.
F=4066 N; nD=12.8 rpm; fsb=2.76; fsc=1
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Ahmed Kovacevic, City University London