Ahmed Kovacevic, City University London 1 1
Mechanical Analysis Gear trains
Prof Ahmed Kovacevic
Lecture 4
School of Engineering and Mathematical Sciences
Room C130, Phone: 8780, E-Mail: [email protected]
www.staff.city.ac.uk/~ra600/intro.htm
Mechanical Analysis and Design ME 2104
Ahmed Kovacevic, City University London 2
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
Ahmed Kovacevic, City University London 3
Plan for this week
Gears
Gear trains
Examples
Ahmed Kovacevic, City University London 4
Gear types
Ahmed Kovacevic, City University London 5
How gears work
Basic
circle
Pressure
line
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
Pressure
angle
Pitch point
Pitch
circle
Ahmed Kovacevic, City University London 6
Diametral
pitch
Circular
pitch
Module
Standard modules are 0.5, 0.8, 1, 1.25, 1.5, 2, 2.5, 3, 4, 5, 6
Module and Pitch
m=6.35
m=2.11
m=0.97
1
25.4[ ];
G P
G P
N NP in
D D
Dp m mm
P N
Dm mm m
N P
Ahmed Kovacevic, City University London 7
Relations between gear parameters
Centre distance:
Pressure angle: 20o (14.5o)
Addendum: a = module
Dedendum: b = 1.25 module
Clearance: c = 0.25 module
Backlash: clearance measured on the
pitch circle of a driving gear
Backlash is function
of module
and centre distance
2
P GD DC
Ahmed Kovacevic, City University London 8
Relations between gear parameters
Gear ratio
2 2
;30 30
ow G G P P
G G G GP PG P
P P G P
G G P P
P T T T
D D N TD nv GR
D N n T
n n
Radius of the base circle
Pitch of the
base circle
cos
cos cos
b
b
r r
Dp p
N
Ahmed Kovacevic, City University London 9
Contact ratio:
Interference (the smallest number of teeth for contact without interference):
2 2 2 2
cos
sin
ab abr
b
ab op bp og bg d
L Lc
p p
L r r r r C
Contact ratio and interference
Ahmed Kovacevic, City University London 10
Helical gears
cos
cos
cotsin
cn c
dn d
cna c
p p
p p
pp p
Number of teeth for helical
gear
Normal pressure angle
bw – gear width
Contact ratio for helical gear
Axial contact ratio
3cos
tantan
cos
2cos 2cos
tan
n
n
p g p g
d
rt r ra
wra
c
NN
D D N Nc
C C C
bC
p
Normal circular pitch
Normal diametral pitch
Axial Pitch
tan
tan
cos cos
a t
r t
t
F F
F F
FF
Axial load
Radial load
Normal load
Ahmed Kovacevic, City University London 11
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.
a) Cr=1.712
b) Cr1=6.343
c) 9.122o
d) Fa1=160.8 N
Fa2=44.7 N
Ahmed Kovacevic, City University London 12
Ahmed Kovacevic, City University London 13
Ahmed Kovacevic, City University London 14
Ahmed Kovacevic, City University London 15
Examples of gear trains
Two stage compound
Gear train
Planetary gear train
Compound reverted
gear train
Planetary gear train
on the arm
Ahmed Kovacevic, City University London 16
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 in-
line.
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
Ahmed Kovacevic, City University London 17
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,
Ahmed Kovacevic, City University London 18
Gear Force Calculation
Stress based on the Force acting
The force is caused by the
transmitted torque. That force
always acts along the pressure line.
The force induces stress
concentration on gear.
We need to answer:
How much power a pair of gears in
question can transfer?
Ft
F Fr
R
a
30[ ]
30
cos cos
t t
t
PPT F R F Nn R
PFF F
n R
a a
Ahmed Kovacevic, City University London 19
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.
Basic stress in
teeth
Power transmitted
tB
y
B
s
F
mbY
SP n D mbY
f
Ahmed Kovacevic, City University London 20
Correction factors
Basic stress must be corrected for: - Shocks, - Size effects, - Uneven load distribution, - Dynamic effects.
a s mB
v
a s mB
v
K K K
K
K K KP P
K
Ka – Application factor
Ks – Size factor
Km – Load distribution factor
Kv - Dynamic factor
Ahmed Kovacevic, City University London 21
Gear materials
Ahmed Kovacevic, City University London 22
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:
Centre distance:
The pitch diameter of the pinion is: 3 20 60[ ] 0.06P PD mN mm m
The pitch velocity is: P PP
n D 1750 0.06v 5.5[m / s] 1090 ft / min
60 60
Transferred load (Tangential force): 60 60 20000
36381750 0.060
t
P P
PowF N
n D
( ) ( ) 903 1352 2 2
P G P GD D N Nc m mm
Ahmed Kovacevic, City University London 23
Example 14 – cont.
6
6
363894 10
0.003 0.038 0.34
363876 10
0.003 0.038 0.42
tBP
P
tBG
G
FPa
mbY
FPa
mbY
Basic bending stress is: pinion –
gear -
62.64 94 10 248a s mP BP
v
K K KMPa
K
Correction factors are: Application factor Ka=1.5
(from diagrams and tables) Size factor Ks=1.0
Load distribution Km=1.2
Dynamic factor Kv=0.68
Corrected pinion bending stress:
248 1.5 372s PS f MPa Allowable stress required for
this application:
From the diagram, any material with Brinel hardness higher than HB=400 will satisfy application
requirements.
From the diagram: YP=0.34 and YG=0.42
Ahmed Kovacevic, City University London 24
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
hardened to
HB 350 and have
50 mm face widths.
Gear Number of Teeth
A 20
B 50
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
Coursework