Machine Design II Module 2-GEARS Lecture 16 – WORM GEARS WORKED OUT PROBLEMS Contents PDF

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

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Machine Design II

Prof. K.Gopinath & Prof. M.M.Mayuram

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

Indian Institute of Technology Madras

Machine Design II

Prof. K.Gopinath & Prof. M.M.Mayuram

Fig.16.2 Isometric view of the worm gears in mesh Data: W = 2 kW, n 1 = 2950 rpm, Z 1 = 2, Z2 = 32, m = 4mm, φ = 20o, d 1 = 50 mm, b 1 = 65 mm, b 2 = 28 mm. Pinion material case carburized steel and Gear material phosphor bronze. Q – (a) C =?, L=?, λ = ?, Q-(b) bearing reactions? T 2 =? and Q (c) η = ? Solution: (a)

d 2 = m Z 2 = 4 x 32 = 128 mm Centre distance C= 0.5(d 1 + d 2 ) = 0.5(50+128) =79mm Axial pitch: p = πm = 3.14x4 = 12.56 mm Lead: L = p Z 1 = 12.56 x 2 = 25.12 mm Lead angle: λ = tan-1( L / π d 1 ) = tan-1(25.12 /π x 50) = 9.09o

Indian Institute of Technology Madras

Machine Design II

Prof. K.Gopinath & Prof. M.M.Mayuram

Solution: (b) V 1 = V m = (πd 1 n 1 /60000) = π x 50 x 2950 / 60000 = 7.72 m/s n 2 = n 1 / i = {n 1 (Z 2 /Z 1 )} = 2950 / (32/2) = 184.38 rpm V 2 = (πd 2 n 2 /60000) = πx128x184.38/60000 =1.24m/s V S = V 1 /cos λ = 7.72 / cos9.09o = 7.82 m/s For V S = 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, Φ n = tan-1( tanφ 1 cos ) = tan-1(tan20o cos 9.09o) = 19.77o F t1 = W / V 1 = 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.

Fn  

Ft cos φn sin  f cos

259  1503N cos19.77 sin9.09o  0.024 cos 9.09o o

F r1 = Fy = F n sinφ n = 1503 sin19.77o = 508 N F a1 = Fz = F n (cos φ n cosλ - f sin λ) = 1503 (cos 19.77o cos 9.09o - 0.024 sin 9.09o) = 1391 N

Indian Institute of Technology Madras

Machine Design II

Prof. K.Gopinath & Prof. M.M.Mayuram

Worm Gears – Force Analysis

Fig.16.4 Forces on the worm gear tooth on the pitch cylinder. Referring to the Fig.16.4, we can now write down the forces acting on the worm gear tooth. F t2 = F a1 = 1391 N

F r2

F a2 = F t1 = 259 N F r2 = F r1 = 508 N

F a2 i.e.,

F t2

Fig.16.5 Sketch showing the forces acting on worm gear shaft

Indian Institute of Technology Madras

Machine Design II

Prof. K.Gopinath & Prof. M.M.Mayuram

Since Bearing B takes the entire thrust load, 

F B x = F a2 = 259 N

Taking moment about z axis through A, we get F B y x 105 – F a2 x 64 – F r2 x 40 = 0 i.e., 105 F B y - 259 x 64 – 508 x 40 = 0 F B y = 351 N ∑ Fy = 0, from which F a y = 508-351 = 157 N By taking moment about y axis through A, we have F t2 x 40 – F B z x 105 = 0 i.e., 1391 x 40 – 105 F B z = 0  F B z = 530 N

 ∑ Fz = 0 from which F A z = 1391 – 530 = 861 N  T = F t2 x r 2 = 1391 x 64 x10-3 = 89.02 Nm

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

Indian Institute of Technology Madras

Machine Design II

Prof. K.Gopinath & Prof. M.M.Mayuram

Fig.16.7 Forces acting on pinion shaft and bearing reactions. Since the bearing at C takes the entire thrust, F C z = F a1 = 1391 N. Taking moment about y (vertical) axis through D, F C x x 80 – F t1 x40 = 0,  F c x = 129.5 N

 F D x = 129.5 N

80 F C x – 259x40 =0 since ∑Fx = 0

Taking moment about x (horizontal) axis through D, F c y x 80 – F a1 x 25 - F r1 x 40 = 0 80 F c y - 1391 x 25 – 508 x 40 = 0  F C y = 689 N

Indian Institute of Technology Madras

From ∑Fy = 0, F D y = -181 N

Machine Design II

Prof. K.Gopinath & Prof. M.M.Mayuram

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

 

cos n - f tan  cos n  f cot 

cos19.77o - 0.024 tan9.09o cos19.77o  0.024 cot 9.09o

 0.859

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

Indian Institute of Technology Madras

Machine Design II

Prof. K.Gopinath & Prof. M.M.Mayuram

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 Z 2 minimum is 21 (Table 2). Allowing 6o lead per thread of the worm, the worm could have 4 or less teeth. Z 1 = 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 (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

Z 2 minimum

40

27

21

17

14

12

10

Indian Institute of Technology Madras

Machine Design II

Prof. K.Gopinath & Prof. M.M.Mayuram

i = n 1 / n 2 = 1500 /75 = 20 = Z 2 / Z 1 1

= 2πn 1 /60 = 2x3.14x1500 /60 =157 rad/s

Z 2 = i x Z 1 = 20 x 4 = 80 A centre distance of 250 mm (as per R10 series) is assumed.

C0.875 C0.875  d1  3.0 1.7 d 1 ≥C 0.875 /3 = 2500.875 /3 ≥ 42 mm and d 1 ≤ C 0.875 /1.7=75 mm. d 1 = 72 mm is taken. Since d 1 ≈ 4p 2 or circular pitch, p 2 = d 1 /4 = 72 / 4= 18 mm m = p /π = 18/3.14 = 5.73 mm take standard module of 6mm. Hence, d 2 = m Z 2 = 6 x 80 = 480 mm. Actual centre distance: C = 0.5 (d 1 + d 2 ) = 0.5(72+480) = 276 mm. Check for d 1 ≤ C 0.875/1.7 ≤ 80.4 mm, d 1 = 80 mm is taken. C = 0.5(d 1 + d 2 ) = 0.5(80+480) = 280 mm Lead = N tw x p a = 4 x 18.84 = 75.36 mm tan λ = L / π d 1 = 75.36 / 3.14x72 = 0.3333 λ = 18.43o = 2

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

V2 =

2

r 2 = 7.85 x (0.5 x 480) x 10-3 = 1.884 m/s

F t = 1000W/ V = 1000 x 12/ 1.884 = 6370 N b ≤ 0.5 d a1 , b ≤ 0.5(d 1 + 2m) ≤ 0.5 x (80+2x6) ≤ 46 b = 45 mm is assumed. Y = 0.393 from Table 1 for φ n = 20o

Indian Institute of Technology Madras

Machine Design II

Prof. K.Gopinath & Prof. M.M.Mayuram

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 Φ n Maximum (Degrees) Lead Angle (degrees)

Lewis form Modified factor y 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

 6.1+ V2   6.1+1.884   6370x     8133N 6.1    6.1 

Fd =F2t 

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

[σ b ] = 80 MPa from Table 16.3 Beam strength of the worm gear

Fb = [ b ] 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 Centrifugally cast Cu-Sn bronze Phosphor bronze

[σ b ] MPa 23.5 80

Aluminium alloys Al-Si

11.3

Zn Alloy

7.5

Cast iron

11.8

Indian Institute of Technology Madras

Machine Design II

Prof. K.Gopinath & Prof. M.M.Mayuram

F b ( 8489) > F d ( 8133) Hence the design is safe from bending fatigue consideration. Check for the wear strength.

Fw =d2 bK w  480x 45x 0.518  11189 N

K w = 0.518 for steel worm vs bronze worm gear with λ < 25o from Table 16.4. F w (11189) > F d (8133), the design is safe from wear strength consideration. Table 16.4 Worm Gear Wear Factors K w K w (MPa)

Material Worm

Gear

...