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:
[email protected] www.staff.city.ac.uk/~ra600/intro.htm 1
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 2
Ahmed Kovacevic, City University London
Plan for this week Gears Gear trains Examples
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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 6
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 8
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 10
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
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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 20
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Gear materials 21
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. 23 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 24
Ahmed Kovacevic, City University London
