DESIGN OF A GEAR DRIVE
CLASS NOTES FOR
MECHANICAL ENGINEERING
Gear is one of the most important part
of each machinery. It controls different
speeds during power transmission. There
are different types of gears. Each is
for a specific service. Most simple gear is
spur gear. Gear design guidelines results in
a sound and safe design of a gear.
A gear drive consists of two shafts,
one gear and one pinion. There are
many types of gears. These all
transmit power from the motor to
the machine.
(i) DESIGN OF A GEAR DRIVE depends on the material of the pinion and gear
If material selected is same, pinion is the basis of design.
(ii) If material is different, Calculate Ft for both pinion and gear. Lesser Ft OR weaker of the two becomes the basis of design.
(iii)Decide number of teeth on the pinion if not given. Then find the number of teeth on the gear from the velocity ratio.
Zp min = minimum Number of teeth on pinion
(i) For 14 ½0, Zp min = 18
(ii) For 200, Zp min = 18
(iii) For 200 Stub involutes Zp min = 14
(iv) Circular pitch, pc = Pitch circle circumference/number of teeth = πD / Z
(v) Diametric pitch, pd = no. of teeth / pitch diameter = Z/ D
(vi) Module m = D/Z=1/pd
(vii) Assume face width, b in any one way of the followings:
(a) ‘b’ between 10 m to 20 modules
(b) b = 3 to 5 pc
(c) b as pd
When pd > 2 and < 16
(d) b=dp = pinion diameter
(viii) a =Addendum =1/pd
(ix) d = Dedendum = 1.25/ pd
(x) Depth = a+ d
(xi) Clearance = d–a
(xii) Outside diameter Do= Dp + 2a
(xiii) Root diameter Di= Dp– 2d
(xiv) Normally Face width ‘b’ is same for gear and pinion whether of same or different materials.
(xv) Circular pitch is also same for pinion and gear.
(xvi) Base circle radius = pitch circle radius* cosφ Where ‘φ’ is the pressure angle. Hence base circle is smaller than the pitch circle.
(xvii) Assume service is intermittent or continuous if not given.
(xviii) If service is intermittent, do the design on strength basis. Use Lewis’ Equation Ft = k σo Y b / pd
(xix) If service is continuous , do the design on wear basis.
Use Buckingham equation FW = Dp b q W
CALCULATION OF DYNAMIC LOAD
Use under shock and impact loads due to surface irregularities. Such load on gears is the Dynamic Load.
Fd =[ 0.05V(bC + Ft)/(0.05V(bC+Ft)0.5)]+ Ft
Where
(a) Fd = Dynamic load
(b) Ft = tangential load
(c) V = pitch line velocity
(d) b = face width
(e) C is constant depending on material, tooth form, accuracy of transmission.
(f) Find the value of C from the standard table given below:
Material of pinion and gear
|
Tooth form
|
Error in tooth action ‘e’ in inches
|
0.0005
|
0.001
|
0.002
|
0.003
|
Gray iron and gray iron
|
14 ½ involute
|
400
|
800
|
1660
|
2400
|
Gray iron and gray iron
|
20 involutes
|
415
|
830
|
1660
|
2490
|
Gray iron and gray iron
|
200 Stub involutes
|
430
|
860
|
1720
|
2580
|
Gray iron and steel
|
14 ½ involutes
|
550
|
1100
|
2200
|
3300
|
Gray iron and steel
|
20 involutes
|
570
|
1140
|
2280
|
3420
|
Gray iron and steel
|
200 Stub involutes
|
590
|
1180
|
2360
|
3540
|
Steel and steel
|
14 ½ involutes
|
800
|
1600
|
3200
|
4800
|
Steel and steel
|
20 involutes
|
830
|
1680
|
3320
|
4980
|
Steel and steel
|
200 Stub involutes
|
860
|
1720
|
3340
|
5160
|
(i) Selection of Spur gears
(a) When shaft axis are parallel
(b) Vm <3000 m/min
(c) speed ratio is up to 10:1.
(ii) Selection Bevel gears
(a) When the shaft axis are at angle and are intersecting
(b) Vm < 3000 m/min

Fig. Bevel Gears
(iii) Selection Helical spur gears
(a) When Vm >3000 m /min
(b) speed ratio up to 10:1
(c) More power transmission
NOTE: Helical gears are spur gears with twisted teeth across the face in the form of a helix about the axis of rotation.
(iv) Standard helix angles are 7030’, 150 and 230
(v) Minimum theoretical face width for helical gears bmin=(π/pd).tanα where ‘α’ is helix angle
(vi) Pitch diameter of a helical gear based upon normal pitch i.e. Dg= Zg/(pn. cos α)
(vii) For helical gears , velocity factor k = 4000/4000+Vm
(viii) Wear load for helical gear, Fwear = Dp b q W/ cos2α
(v) Select Double Helical gears Or Herringbone gears if V > 4000 m/min
(x) Spur gears and Bevel gears are of INVOLUTE profile with pressure angles of
14 1/20, 200 full and 200 stub involutes
(xi) Use gears of cycloid profile with skew shafts. Worm gears are of Cycloid profile.
(xii) Various parts of a gear—Rim, arms and hub
(xiii) NUMBER OF ARMS
(a) Four or five, diameter < 500 mm
(b) Six for >500 mm & < 1500 mm
(c) Eight for diameter >1500 mm and < 2400 mm
(d) 10 or 12 if diameter > 2400 mm
(xiv) SHAPE OF CROSS SECTION OF THE ARM
(a) Elliptical for light load or for less power
(b) H or I shape for heavy load or for more horse power
Standard proportions adopted by many manufacturers
Type of service
|
Diameter of hub
|
Length of Hub, L
|
Cast Iron (CI)
|
Steel
|
Light load without shock
|
1 d+ 3 mm
|
1 d+ 6 mm
|
1.75 to 2.25 d
|
Medium load medium shock
|
1 d+ 3 mm
|
1 d +5 mm
|
1.75 to 2.25 d
|
Heavy load heavy shock
|
2 d
|
1 d + 3 mm
|
1.75 to 2.25 d
|
Useful Data For Spur And Helical Gears
Useful data for both spur and helical gears, including double helical and internal gears, used in power transmission
Module: m 1.5 to 25 mm
Pitch Circle: d 25 to 3200 mm
Linear Speed: v less than 25 m/sec
Rotating Speed: N less than 3600 rpm
Numerical Data on Gears
Type of gear
|
Efficiency
|
Speed ratio
|
Peripheral velocity
|
Spur
|
97 to 99 %
|
1:1 to 10: 1
|
< 3000 m/min
|
Bevel
|
96 to 98 %
|
5:1 to 400 :1
|
< 3000 m/min
|
Worm
|
50 to 90 %
|
5:1 to 400 :1
|
< 7600 m/min
|
Helical
|
96 to 98 %
|
1:1 to 8:1
|
> 3000 m/min
|
–
Design of a gear drive includes the followings:
(i) Design of a pinion
(ii) Design of a gear
(iii) Design of arms – number of arms and section of the arm
(iv) Design of hubs
(v) Design of shafts and keys
The design is carried on the basis of the following considerations.
-
LEWIS EQUATION –DESIGN ON THE BASIS OF STRENGTH FOR INTERMITTENT SERVICE
Design load Ft is found from Lewis equation Ft = k σo Y b / pd
-
BUCKINGHAM EQUATION—DESIGN ON THE BASIS OF WEAR FOR CONTINUOUS SERVICE
Fw= Dp b q W
-
BUCKINGHAM EQUATION—DESIGN ON THE BASIS OF DYNAMIC LOAD UNDER SHOCK/IMPACT LOADS
Fd =[ 0.05V(bC + Ft)/(0.05V(bC+Ft)0.5)]+ Ft
-
DESIGN ON THE BASIS OF EFFECTIVE LOAD
Feff = ks Ft/kv
Design of gears is hit and trial process as some assumptions are made during design and then checked as follows :
(i) Ft > Fd
(ii) (Ft –Fd) should not be > !0 % of Ft
(iii) Fw > Fd
(iv) (Fw – Fd) should not be > !0 % of Fw
(v) Fw > FOS. Feff
(vi) Ft > FOS. Feff
NOTE: IF CHECKS DO NOT WORK, CHANGE THE ASSUMPTIONS AND REDESIGN AND GO ON DOING IT TILL THE ASSUMPTIONS ARE VERIFIED BY THE VARIOUS CHECKS. THIS MAKES THE DESIGN OF GEARS AS AN ITERATIVE PROCEDURE.
Q. VARIOUS TYPES OF FAILURES IN GEARS
(i) Fatigue failure due to bending
(ii) Pitting failure due to improper lubrication
(iii) Scoring failure due to improper lubrication under heavy load, lubrication film breaks and metal to metal contact occurs, very high temperatures are developed, welding takes place at the contact points
(iv) Abrasive failure due to foreign material, lubricating oil must be filtered at regular intervals
(v) Corrosive wear due to the chemical action like rusting
SCORING, ABRASIVE WEAR AND CORROSIVE WEAR ARE DUE TO IMPROPER LUBRICATION.
SURFACE FATIGUE FAILURE DUE TO REPEATED CONTACT LOAD.
https://www.mesubjects.net/wp-admin/post.php?post=1677&action=edit Gears Introduction