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 F_{t} for both pinion and gear. Lesser F_{t } 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.
Z_{p min} = minimum Number of teeth on pinion
(i) For 14 ½^{0}, Z_{p min} = 18
(ii) For 20^{0}, Z_{p min} = 18
(iii) For 20^{0} Stub involutes Z_{p min} = 14
(iv) Circular pitch, p_{c } = Pitch circle circumference/number of teeth = πD / Z
(v) Diametric pitch, p_{d} = no. of teeth / pitch diameter = Z/ D
(vi) Module m = D/Z=1/p_{d }
(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 p_{c}
(c) b as p_{d }
When pd > 2 and < 16
(d) b=d_{p} = pinion diameter
(viii) a =Addendum =1/p_{d }
(ix) d = Dedendum = 1.25/ p_{d}
(x) Depth = a+ d
(xi) Clearance = d–a
(xii) Outside diameter D_{o}= D_{p }+ 2a
(xiii) Root diameter D_{i}= D_{p}– 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 F_{t} = k σ_{o }Y b / p_{d}
(xix) If service is continuous , do the design on wear basis.
Use Buckingham equation F_{W} = D_{p} 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.
F_{d} =[ 0.05V(bC + F_{t})/(0.05V(bC+F_{t})^{0.5})]+ F_{t}
Where
(a) F_{d} = Dynamic load
(b) F_{t} = 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

20^{0} 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

20^{0} 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

20^{0} Stub involutes

860

1720

3340

5160

(i) Selection of Spur gears
(a) When shaft axis are parallel
(b) V_{m} <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) V_{m} < 3000 m/min
Fig. Bevel Gears
(iii) Selection Helical spur gears
(a) When V_{m} >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 7^{0}30^{’}, 15^{0 } and 23^{0}
(v) Minimum theoretical face width for helical gears b_{min}=(π/p_{d}).tanα where ‘α’ is helix angle
(vi) Pitch diameter of a helical gear based upon normal pitch i.e. D_{g}= Z_{g}/(p_{n}. cos α)
(vii) For helical gears , velocity factor k = 4000/4000+V_{m}
(viii) Wear load for helical gear, F_{wear }= D_{p} b q W/ cos^{2}α
(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 1/2^{0}, 20^{0} full and 20^{0} 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

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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 F_{t} is found from Lewis equation F_{t} = k σ_{o }Y b / p_{d}

BUCKINGHAM EQUATION—DESIGN ON THE BASIS OF WEAR FOR CONTINUOUS SERVICE
F_{w}= D_{p} b q W

BUCKINGHAM EQUATION—DESIGN ON THE BASIS OF DYNAMIC LOAD UNDER SHOCK/IMPACT LOADS
F_{d} =[ 0.05V(bC + F_{t})/(0.05V(bC+F_{t})^{0.5})]+ F_{t}

DESIGN ON THE BASIS OF EFFECTIVE LOAD
F_{eff} = k_{s} F_{t}/k_{v}
Design of gears is hit and trial process as some assumptions are made during design and then checked as follows :
(i) F_{t} > F_{d}
(ii) (F_{t} –F_{d}) should not be > !0 % of F_{t}
(iii) F_{w} > F_{d}
(iv) (F_{w} – F_{d}) should not be > !0 % of F_{w}
(v) F_{w} > FOS. F_{eff}
(vi) Ft > FOS. F_{eff}
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/wpadmin/post.php?post=1677&action=edit Gears Introduction