worm gear design2

Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram

Indian Institute of Technology Madras

Module 2- GEARS

Lecture 16 – WORM GEARS WORKED OUT PROBLEMS Contents 16.1. Worm gears – force analysis problem

16.2. Worm gears- design problem

16.3. Gearbox design procedure.

16.1 WORM GEARS- PROBLEM 1 A two tooth right hand worm transmits 2 kW at 2950 rpm to a 32 tooth worm gear. The

worm gear is of 4 mm module, 20o pressure and a face width of 30 mm. The worm is of

pitch diameter of 50 mm with a face width of 65 mm. The worm is made of steel case

carburized OQ and T and ground. The worm gear is made of phosphor bronze.

(a) Find the centre distance, the lead and the lead angle.

(b) Find the bearing reactions on the worm gear and worm shaft and the torque

output.

(c) Find the efficiency.

The general arrangement and isometric views are shown in Fig.16.1 and Fig.16.2. Fig.1 General arrangement of the worm drive



Fig.16.2 Isometric view of the worm gears in mesh Data:

W = 2 kW, n1 = 2950 rpm, Z1 = 2, Z2 = 32, m = 4mm, φ = 20o, d1= 50 mm, b1= 65 mm,

b2= 28 mm. Pinion material case carburized steel and Gear material phosphor bronze.

Q – (a) C =?, L=?, λ = ?, Q-(b) bearing reactions? T2=? and Q (c) η = ?

Solution: (a)

d2 = m Z2 = 4 x 32 = 128 mm

Centre distance C= 0.5(d1+ d2) = 0.5(50+128) =79mm

Axial pitch: p = πm = 3.14x4 = 12.56 mm

Lead: L = p Z1 = 12.56 x 2 = 25.12 mm

Lead angle: λ = tan-1( L / π d1) = tan-1(25.12 /π x 50) = 9.09o



Solution: (b)

V1 = Vm = (πd1n1/60000) = π x 50 x 2950 / 60000 = 7.72 m/s

n2 = n1/ i = {n1 (Z2/Z1)} = 2950 / (32/2) = 184.38 rpm

V2 = (πd2n2/60000) = πx128x184.38/60000 =1.24m/s

VS = V1 /cos λ = 7.72 / cos9.09o = 7.82 m/s

For VS = 7.82 m/s and the given materials f = 0.024 from Fig.16.3.

Since the helix angle of the gear is the same as the lead angle of the worm,

Φn = tan-1( tanφ1cos ψ) = tan-1(tan20o cos 9.09o) = 19.77o

Ft1 = W / V1 = 2000/ 7.72 = 259 N

Fig.16.3 Friction of well lubricated worm gears, A for cast iron worm and worm gear and B for case hardened steel worm and phosphor bronze worm gear.

tn

n

o o o

FF

cosφ sinλ f cosλ

2591503N

cos19.77 sin9.09 0.024cos9.09

Fr1= Fy = Fn sinφn = 1503 sin19.77o = 508 N

Fa1= Fz = Fn (cos φn cosλ - f sin λ) = 1503 (cos 19.77o cos 9.09o - 0.024 sin 9.09o) =

1391 N



Worm Gears – Force Analysis

Fig.16.4 Forces on the worm gear tooth on the pitch cylinder.

ng to the Fig.16.4, we can now write down the forces acting on the worm gear oth.

t2 = Fa1 Fr2

a2 = Ft1 F

r2 = Fr1 = 508 N i.e.,

Referrito F = 1391 N F = 259 N a2

Ft2

F

Fig.16.5 Sketch showing the fo es acting on worm gear shaft rc



Since Bearing B takes the entire thrust load,

FBx = Fa2 = 259 N

A, we get

- 259 x 64 – 508 x 40 = 0

Fy = 0, from which Fay = 508-351 = 157 N

y axis through A, we have

FBz = 0

∑ Fz = 0 from which FAz = 1391 – 530 = 861 N

T = Ft2 x r2 = 1391 x 64 x10-3 = 89.02 Nm

Taking moment about z axis through

FBy x 105 – Fa2 x 64 – Fr2 x 40 = 0

i.e., 105 FBy

FBy = 351 N

∑

By taking moment about

Ft2 x 40 – FBz x 105 = 0

i.e., 1391 x 40 – 105

FBz = 530 N

Fig.16.6 Sketch showing the calculated value of forces acting on worm gear shaft



Fig.16.7 Forces acting on pinion shaft and bearing reactions. Since the bearing at C takes the entire thrust,

FCz = Fa1 = 1391 N.

Taking moment about y (vertical) axis through D,

FCx x 80 – Ft1x40 = 0, 80 FC

x – 259x40 =0

Fcx = 129.5 N

FDx = 129.5 N since ∑Fx = 0

Taking moment about x (horizontal) axis through D,

Fcy x 80 – Fa1 x 25 - Fr1 x 40 = 0

80 Fcy - 1391 x 25 – 508 x 40 = 0

FCy = 689 N From ∑Fy = 0, FD

y = -181 N



Fig.16.8 Calculated values of forces acting on pinion shaft and bearing reactions. Solution: (C) Efficiency of the gearbox The efficiency of the gearbox is given by η

n

n

o o

o o

cos - f tan

cos f cot

cos19.77 - 0.024 tan9.09

cos19.77 0.024cot9.09

0.859

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16.2 WORM GEARS- PROBLEM 2 Design a worm gear set to deliver 12 kW from a shaft rotating at 1500 rpm to another

rotating at 75 rpm.

Solution:

20o normal pressure angle worm gear is assumed for which the lead angle should not

exceed 25o (Table 1) and Z2 minimum is 21 (Table 2). Allowing 6o lead per thread of the

worm, the worm could have 4 or less teeth. Z1 = 4 or quadruple threaded worm is

assumed

Table 16.1 Maximum Worm Lead Angle and Worm Gear Lewis Form Factor for Various Pressure angles

Pressure Angle Φn

(Degrees)

Maximum

Lead Angle λ

(degrees)

Lewis form

factor y

Modified

Lewis form

factor Y

14.5

15

0.100

0.314

20

25

0.125

0.393

25

35

0.150

0.473

30

45

0.175

0.550

From the worm gears–design guidelines we have,

Table 16.2 Minimum number of teeth in the worm gear

Pressure angle φn

14.5o

17.5o

20o

22.5o

25o

27.5o

30o

Z2 minimum

40

27

21

17

14

12

10



i = n1 / n2 = 1500 /75 = 20 = Z2 / Z1

ω1 = 2πn1 /60 = 2x3.14x1500 /60 =157 rad/s

Z2 = i x Z1 = 20 x 4 = 80

A centre distance of 250 mm (as per R10 series) is assumed.

1

0.875 0.875C Cd

3.0 1.7

d1 ≥C 0.875 /3 = 2500.875 /3 ≥ 42 mm and

d1≤ C 0.875 /1.7=75 mm.

d1 = 72 mm is taken.

Since d1≈ 4p2 or circular pitch,

p2 = d1/4 = 72 / 4= 18 mm

m = p /π = 18/3.14 = 5.73 mm take standard module of 6mm.

Hence, d2 = m Z2 = 6 x 80 = 480 mm.

Actual centre distance: C = 0.5 (d1+ d2)

= 0.5(72+480) = 276 mm.

Check for d1 ≤ C 0.875/1.7 ≤ 80.4 mm, d1 = 80 mm is taken.

C = 0.5(d1 + d2) = 0.5(80+480) = 280 mm

Lead = Ntw x pa = 4 x 18.84 = 75.36 mm

tan λ = L / π d1 = 75.36 / 3.14x72 = 0.3333

λ = 18.43o = ψ

ω2 = (2πn2 /60) = (2x3.14x75/60) = 7.85 rad/s

V2 = ω2 r2 = 7.85 x (0.5 x 480) x 10-3 = 1.884 m/s

Ft = 1000W/ V = 1000 x 12/ 1.884 = 6370 N

b ≤ 0.5 da1 , b ≤ 0.5(d1 + 2m) ≤ 0.5 x (80+2x6) ≤ 46

b = 45 mm is assumed.

Y = 0.393 from Table 1 for φn = 20o



Table 16.1 Maximum Worm Lead Angle and Worm Gear Lewis Form Factor for Various Pressure angles which is reproduced below for convenience of selection.

Pressure Angle Φn (Degrees)

Maximum Lead Angle λ

(degrees)

Lewis form factor y

Modified Lewis form

factor Y

14.5

15

0.100

0.314

20

25

0.125

0.393

25

35

0.150

0.473

30

45

0.175

0.550

2d 2t 8133N

6.1+V 6.1+1.884F =F 6370x

6.1 6.1

Choosing phosphor bronze for the gear and heat treated C45 steel for the ground worm,

[σb ] = 80 MPa from Table 16.3

Beam strength of the worm gear

b bF = [ ] bmY = 80x45x6x0.393 = 8489 N

Worm gears – bending and surface fatigue strengths are given Table 16.3

Table 16.3 Permissible stress in bending fatigue

Material of the gear

[σb] MPa

Centrifugally cast Cu-Sn bronze

23.5

Phosphor bronze

80

Aluminium alloys Al-Si

11.3

Zn Alloy

7.5

Cast iron

11.8



Fb ( 8489) > Fd ( 8133) Hence the design is safe from bending fatigue consideration.

Check for the wear strength.

w w2F =d bK 480x 45x0.518 11189 N

Kw = 0.518 for steel worm vs bronze worm gear with λ < 25o from Table 16.4.

Fw (11189) > Fd (8133), the design is safe from wear strength consideration.

Table 16.4 Worm Gear Wear Factors Kw

Material

Kw (MPa)

Worm

Gear

<10

<25

>25

Steel, 250 BHN

Bronze

0.414

0.518

0.621

Bronze

0.552

0.690

0.828

Hardened steel (Surface 500 BHN) Chill-cast

Bronze

0.828

1.036

1.243

Cast iron

Bronze

1.036

1.277

1.553

AGMA recommendation for the axial length of the Worm is, Lw

2w a

Z 80L p (4.5 ) 18.84x(4.5 ) 115mm

50 50

Worm Velocity V1 = ω1 r1 = 157 x 0.04= 6.28 m/s

1

s o

V 6.28V 6.62 m/ s

cosλ cos18.43

From the Fig. 16.9 worm gear friction characteristics, for Vs = 6.62 m/s f = 0.025



Fig. 16.9 Friction of well lubricated worm gears, A for cast iron worm and gear and B for case hardened steel worm and phosphor bronze worm gear

t2n o o

n

F 6370F 7145N

cosφ cosλ cos20 cos18.43

n

n

o o

o o

cos - f tan

cos f cot

cos 20 - 0.025 tan18.43

cos 20 0.025 cot 18.430.918

Heat generated during operation: Hg = (1-η)W = (1-0.918)12000= 984 Nm Surface area A for conventional housing designs may be roughly estimated from the equation: 1.7A=14.75 C Where A is in m2 and C (the distance between the shafts) is in m.

21.7A=14.75 x0.28 1.694m Heat generated during operation: Hg From Fig. 16.10 the CH = 32 Nm/s/m2/oC for n1=1500rpm



Hd= CH A (To – Ta) assuming Ta = 35oC

= 32 x 1.694x (To – 35o) = 54.21 (To – 35o)

Hg = 984 Nm

T0 = 53.2o C < 93o C permissible for oil.

Hence the design is OK from thermal considerations.

Fig.16.10 Influence of worm speed on heat transfer.

--------------------------



16.3 GEARBOX DESIGN 1. Design of gears is based on beam strength, pitting and scuffing (high speed gears)

considerations. The minimum pitch diameter of the pinion should be

d1min = 2 x bore + 0.25 m (16.1)

where d is the bore diameter and m is the module expressed in mm.

2. As per Nuttall Works of the Westinghouse Electric and Manufacturing company, the

minimum thickness of metal tmin between the keyway and the root circle shall be:

min

Zt m (16

5.2)

3. The outside diameter of the hubs in larger gears should be 1.8 times the bore for

steel, 2 times for CI and 1.65 times for the forged steel. The hub length should be at

least 1.25 times the bore and never less than the width of the gear.

4. Design of the shaft is based on fatigue strength and rigidity considerations. (At the

contact region the deflection of the shaft should be less than 0.01 module, the slope of

the shaft at the radial bearing should be less than 0.008 radians and for self-aligning

bearings it should be less than 0.05 radians.

5. Bearings selection is based on 90% reliability for the following life:

8 hrs. operation per day life = 20,000 to 30,000 hrs.

8-16 hrs. operations per day life= 40,000-50,000 hrs.

16-24 hrs. operations per day life = 50,000-60,000 hrs.

6. Selection of lubricant is based on peripheral velocity, load, type of application and

operating temperature etc.

SAE 30, 40, 50, 60, 80, 90 are being recommended.

For low friction and high temperature operation synthetic oils are used.



7. Selection of method of lubrication is based on the peripheral speed. Up to a

peripheral speed of 15 m/s oil bath (oil immersion / splash) lubrication is used. Higher

depth of immersion is recommended for slow speed application. The maximum depth

should not exceed 100 mm. At higher speed the depth of immersion is reduced to cut

the churning losses. 0.7 tooth height or a minimum of 1 module is taken. However, the

depth of immersion should not be less than 10 mm. Generally recommended depth of

immersion is 3 to 4 times the module and a maximum of 6 modules. Above a peripheral

velocity of 15 m/s stream (jet) lubrication is used. Oil is delivered by a pump through a

filter and if necessary through a cooler directly to the teeth of gears as they are coming

out of mesh.

8. Quantity of oil required is given below by the thumb rule:

For splash lubrication Q = ( 2.5 - 8 ) Lt litres or

Q = ( 0.35 - 0.7 ) W litres (16.3)

For spray lubrication rate of supply should be

Qe = 30 ( Lt /ΔT) lpm (16.4)

where

Lt - loss of power at the teeth contact ( kW )

W - is the power transmitted ( kW )

ΔT is the difference in oil temperature at the outlet and inlet in oC .

Oil circulation time e

Qt ' min. (16.5)

Q

t’ = 1 - 2.5 for splash lubrication with no external circulation where lack of space is

there in compact design.

t’ = 4 - 30 for oil circulation with cooling or reservoir.

9. Gear box housing design is based on thumb rule and thermal consideration.

The wall thickness of the CI housing can be found from the empirical formula:

s = 2 (0.1 T)0.25 ≥ 6 mm (16.6)



and that of the cover from: sc = 0.9 s (16.7)

where the T is torque on the slow speed shaft in Nm.

The diameter “d” of the bolts for securing the cover should be: d = (T)1/3 ≥ 10 mm

(16.8)

and that of foundation bolts:

df =(2T)1/3 ≥ 12 mm (16.9)

The thickness of the foundation flange should be: sff ≥ 1.5 df.

(16.10)

The width of the flanges at the base and at the two halves of the housing should be:

wf = (2.1 to 2.5 )d (16.11)

le 16.5 lternate appr ess ‘s’ i s

Non-case ened Case hardened gears

Tab A oach for wall thickn

ard

n mm for the gearboxe

hgears

CI castings 0.007L + 6 mm 0.010 L + 6 mm

Steel castings

0.005L + 4 mm

0.007L + 4 mm

Welded construction 0.004L + 4 mm 0.005L + 4 mm

where L is the largest dimension of the housing in mm.

ing is taken as 0.8s, and that

r the non-load carrying upper half of the housing 0.5s.

lange thickness is taken as 1.5s for castings and 2s for welded construction.

imension of the bearing housing is kept 1.2 times outside diameter of the

1.5s for welded construction.

Bolt spacing 6-10 times the bolt diameter.

Wall thickness of the load carrying upper half of the hous

fo

F

Outside d

bearing.

Flange screw/bolt diameter = 1.2s casting &



In designing reducing gear housing simple geometric shapes are to be preferred with

the outside as plain as possible. In order to reduce the air draft noises, the gap between

the gear and the side wall should be at least 15 mm.

Losses in gear boxes :

Total power loss L = Lt + Lch + Lb (16.12)

Lt - power loss at tooth engagement.

Lch - churning power losses &

Lb - bearing power losses

or

t1 2

1 1L 2.3 f W kW (16.13)

Z Z

t

1

0.1 0.3L W kW (16.14)

Z cosψ V 2

f - is the coefficient of friction between the teeth

Z1 and Z2 are number of teeth on the pinion and the gear,

V - peripheral velocity m/s and

ψ - helix angle

Table 16.6 Coefficient of friction for 20o pressure angle gears

V m/s 0.7 7 14 21 42

f 0.08 0.06 0.05 0.04 0.025

0.5

3ch

1 2

200VμL cb V x10 kW (16.15

Z Z ) Where

V - peripheral speed ( m/s)



b - face width of the gear ( mm )

c - factor equal to 0.009 for splash lubrication, 0.006 for stream lubrication

- viscosity of oil at the operating temperature ( cP )

Lb = 5.23 x 10-8 F fb d n kW (16.16)

where

F - radial load on the bearing ( N )

fb - coefficient of friction at the bearing reduced to the shaft diameter 0.005 - 0.01

d - shaft diameter ( mm )

n - shaft speed ( rpm )

The heat generated by the total power loss will raise the temperature of the oil and the

housing. The housing will dissipate heat H by radiation, convection and by conduction

through the foundation plate or the frame. When equilibrium conditions are set in the

heat generated and the heat dissipated will be the same. This equilibrium temperature

should be less that the maximum operating temperature for the oil otherwise the oil will

be getting oxidized. If the temperature exceeds then additional heat has to be dissipated

by separate cooling arrangement.

H = kt (To – Ta) A (1+U) kcal/h (16.17)

where A - free surface of the housing from which heat is removed to cool the drive

(included is the 50% of the surface of the fins) m2.

To & Ta - temperature of the oil and the surrounding air, oC

kt - heat transfer coefficient, equal to 10 to 16 kcal / (m2 .oC .h) , larger values are

used under favourable conditions of air circulation; in new standard reducing gears, kt

=14 kcal/(m . C

.h) 2 o



U - factor taking into account heat transfer to the foundation plate or frame of the

machine and amounting up to 0.3 when the housing seating surface is large.

Heat generated per hour, Hg = 3600000 L / J

Where J is the mechanical equivalent of heat

J = 4270 Nm / kcal

And L is the total power loss in kW

Hg = 843 L (16.18)

Hd ≥ Hg (16.19)

And To ≤ 93o C, otherwise provide heat exchanger for cooling the oil.

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worm gear design2

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