Module 2- GEARS. Lecture 16 WORM GEARS WORKED OUT PROBLEMS

Transcription

1 Module 2- GEARS Lecture 16 WORM GEARS WORKED OUT PROBLEMS Contents Worm gears force analysis problem Worm gears- design problem Gearbox design procedure 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, 20 o 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 Fig.1 General arrangement of the worm drive

2 Data: Fig.16.2 Isometric view of the worm gears in mesh W = 2 kw, n 1 = 2950 rpm, Z 1 = 2, Z2 = 32, m = 4mm, φ = 20 o, 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 = mm Lead: L = p Z 1 = x 2 = mm Lead angle: λ = tan -1 ( L / π d 1 ) = tan -1 (25.12 /π x 50) = 9.09 o

3 Solution: (b) V 1 = V m = (πd 1 n 1 /60000) = π x 50 x 2950 / = 7.72 m/s n 2 = n 1 / i = {n 1 (Z 2 /Z 1 )} = 2950 / (32/2) = rpm V 2 = (πd 2 n 2 /60000) = πx128x184.38/60000 =1.24m/s V S = V 1 /cos λ = 7.72 / cos9.09 o = 7.82 m/s For V S = 7.82 m/s and the given materials f = from Fig Since the helix angle of the gear is the same as the lead angle of the worm, Φ n = tan -1 ( tanφ 1 cos ψ) = tan -1 (tan20 o cos 9.09 o ) = o 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. F t Fn cosφ n sinλ fcosλ 259 cos19.77 sin cos9.09 o o o 1503N F r1 = F y = F n sinφ n = 1503 sin19.77 o = 508 N F a1 = F z = F n (cos φ n cosλ - f sin λ) = 1503 (cos o cos 9.09 o sin 9.09 o ) = 1391 N

4 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 Fa2 Fr2 = F r1 = 508 N i.e., F t2 Fig.16.5 Sketch showing the forces acting on worm gear shaft

5 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 y B x 105 F a2 x 64 F r2 x 40 = 0 y i.e., 105 F B x x 40 = 0 F y B = 351 N F y = 0, from which F y a = = 157 N By taking moment about y axis through A, we have F t2 x 40 F z B x 105 = 0 i.e., 1391 x F z B = 0 F z B = 530 N F z = 0 from which F z A = = 861 N T = F t2 x r 2 = 1391 x 64 x10-3 = Nm Fig.16.6 Sketch showing the calculated value of forces acting on worm gear shaft

6 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 x C x 80 F t1 x40 = 0, 80 F x C 259x40 =0 F x c = N F x D = N since F x = 0 Taking moment about x (horizontal) axis through D, F y c x 80 F a1 x 25 - F r1 x 40 = 0 80 F y c x x 40 = 0 F y C = 689 N From F y = 0, F y D = -181 N

7 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 cosn - f tan cos f cot n o cos tan9.09 o cos cot o o

8 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: 20 o normal pressure angle worm gear is assumed for which the lead angle should not exceed 25 o (Table 1) and Z 2 minimum is 21 (Table 2). Allowing 6 o 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 Φ n (Degrees) Maximum Lead Angle λ (degrees) Lewis form factor y Modified Lewis form factor Y From the worm gears design guidelines we have, Table 16.2 Minimum number of teeth in the worm gear Pressure angle φ n 14.5 o 17.5 o 20 o 22.5 o 25 o 27.5 o 30 o Z 2 minimum

9 i = n 1 / n 2 = 1500 /75 = 20 = Z 2 / Z 1 ω 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. C C d d 1 C /3 = /3 42 mm and d 1 C /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 / 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 = mm tan λ = L / π d 1 = / 3.14x72 = λ = o = ψ ω 2 = (2πn 2 /60) = (2x3.14x75/60) = 7.85 rad/s V 2 = ω 2 r 2 = 7.85 x (0.5 x 480) x 10-3 = m/s F t = 1000W/ V = 1000 x 12/ = 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 = from Table 1 for φ n = 20 o

10 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 (Degrees) Maximum Lead Angle λ (degrees) Lewis form factor y Modified Lewis form factor Y F d =F 6.1+V t 6370x 8133N 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 F = [ b 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 [σ b ] 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

11 F b ( 8489) > F d ( 8133) Hence the design is safe from bending fatigue consideration. Check for the wear strength. F =d bk 480x 45x N w 2 w K w = for steel worm vs bronze worm gear with λ < 25 o from Table F w (11189) > F d (8133), the design is safe from wear strength consideration. Table 16.4 Worm Gear Wear Factors K w Material K w (MPa) Worm Gear <10 <25 >25 Steel, 250 BHN Bronze Hardened steel (Surface 500 BHN) Bronze Chill-cast Bronze Cast iron Bronze AGMA recommendation for the axial length of the Worm is, L w Z Lw p a(4.5 ) 18.84x(4.5 ) 115mm Worm Velocity V 1 = ω 1 r 1 = 157 x 0.04= 6.28 m/s V V 6.28 cos λ cos s o 6.62 m/ s From the Fig worm gear friction characteristics, for V s = 6.62 m/s f = 0.025

12 Fig 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 F 6370 F 7145N t2 n o o cosφn cos λ cos20 cos18.43 cos n - f tan cos f cot n o o cos tan18.43 o o cos cot Heat generated during operation: H g = (1-η)W = ( )12000= 984 Nm Surface area A for conventional housing designs may be roughly estimated from the equation: 1.7 A=14.75 C Where A is in m 2 and C (the distance between the shafts) is in m. 1.7 A=14.75 x m Heat generated during operation: H g 2 From Fig the C H = 32 Nm/s/m 2 / o C for n 1 =1500rpm

13 H d = C H A (T o T a ) assuming T a = 35 o C = 32 x 1.694x (T o 35 o ) = (T o 35 o ) H g = 984 Nm T 0 = 53.2 o C < 93 o C permissible for oil. Hence the design is OK from thermal considerations. Fig Influence of worm speed on heat transfer

14 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 d 1min = 2 x bore 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 t min between the keyway and the root circle shall be: Z tmin m (16.2) 5 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 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 hrs. operations per day life= 40,000-50,000 hrs 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.

15 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 = ( ) L t litres or Q = ( ) W litres (16.3) For spray lubrication rate of supply should be Q e = 30 ( L t /ΔT) lpm (16.4) where L t - 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 o C. Oil circulation time Q t ' min. (16.5) Q e t = 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) mm (16.6)

16 and that of the cover from: s c = 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: d f =(2T) 1/3 12 mm (16.9) The thickness of the foundation flange should be: s ff 1.5 d f. (16.10) The width of the flanges at the base and at the two halves of the housing should be: w f = (2.1 to 2.5 )d (16.11) Tab le 16.5 Alternate approach for wall thickness s in mm for the gearboxes Non-case hardened Case hardened gears gears CI castings 0.007L + 6 mm 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. Wall thickness of the load carrying upper half of the housing is taken as 0.8s, and that for the non-load carrying upper half of the housing 0.5s. Flange thickness is taken as 1.5s for castings and 2s for welded construction. Outside dimension of the bearing housing is kept 1.2 times outside diameter of the bearing. Flange screw/bolt diameter = 1.2s casting & 1.5s for welded construction. Bolt spacing 6-10 times the bolt diameter.

17 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 = L t + L ch + L b (16.12) L t - power loss at tooth engagement. L ch - churning power losses & L b - bearing power losses 1 1 Lt 2.3f W kw (16.13) Z1 Z2 or Lt W kw (16.14) Zcosψ 1 V 2 f - is the coefficient of friction between the teeth Z 1 and Z 2 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 20 o pressure angle gears V m/s f Vμ 3 Lch cbv x10 kw (16.15 Z1 Z2 Where V - peripheral speed ( m/s) )

18 b - face width of the gear ( mm ) c - factor equal to for splash lubrication, for stream lubrication - viscosity of oil at the operating temperature ( cp ) L b = 5.23 x 10-8 F f b d n kw (16.16) where F - radial load on the bearing ( N ) f b - coefficient of friction at the bearing reduced to the shaft diameter 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 = k t (T o T a ) 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) m 2. T o & T a - temperature of the oil and the surrounding air, o C k t - heat transfer coefficient, equal to 10 to 16 kcal / (m 2. o C.h), larger values are used under favourable conditions of air circulation; in new standard reducing gears, k t 2 o =14 kcal/(m. C.h)

19 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, H g = L / J Where J is the mechanical equivalent of heat J = 4270 Nm / kcal And L is the total power loss in kw H g = 843 L (16.18) H d H g (16.19) And T o 93 o C, otherwise provide heat exchanger for cooling the oil