Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram. Module 2 - GEARS

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Charity Snow
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1 Module 2 - GEARS Lecture 10 - SPUR GEAR DESIGN Contents 10.1 Problem 3 Spur gear design Buckingham Approach AGMA Approach 10.1 Problem 3 - Design of Spur gear A pair of gears is to be designed to transmit 30kW power from a pinion running at 960rpm to a gear running at 320rpm. Design the gears so that they can last for 10 8 cycles. Assume 20 o full depth involute spur gear for the system. Motor shaft diameter is 30mm. Data given: W = 30 kw; n 1 = 960 rpm; n 2 = 320 rpm; Life = 10 8 cycles; 20 o full depth involute spur gear. Solution: i = n 1 / n 2 = 960 / 320 = 3 In order to keep the size small and meet the centre distance, Z 1 = 17 chosen Z 2 = i Z 1 = 3 x 17 = 51 n xx rad/s Torque is given by, w 30x1000 T Nm

2 rom Lewis equation for pinion we have, 2T t 1 p [ ] 2 p bym byz1m (10.1) SAE 1050 hardened by OQT with permissible bending strength of 532 and hardness of 223Bhn is selected for pinion and SAE 1045 hardened by WQT with permissible bending strength of 487 and hardness of 215 Bhn is selected for the gear from Table ace width b = 10m is chosen for both wheels. Table 10.1 Safe static stresses for use in the Lewis equation Material [σ ] MPa BHN Gray cast iron ASTM 25 ASTM 35 ASTM 50 Cast steel(low carbon) 0.2% C not heat treated 0.2% C WQT orged carbon steel SAE 1020 case hardened and WQT SAE 1030 not heat treated SAE 1035 not heat treated SAE 1040 not heat treated SAE 1045 hardened by WQT SAE 1045 hardened by WQT SAE 1050 hardened by OQT Alloy steel SAE 2320 case hardened and WQT SAE 2345 hardened by OQT SAE 3115 case hardened and OQT SAE 3145 hardened by OQT SAE 3245 hardened by OQT SAE 4340 hardened by OQT SAE 4640 hardened by OQT SAE 6145 hardened by OQT

3 Buckingham approach: The preliminary dimensions are found from Lewis equation and then they are checked for dynamic loads by Buckingham equation. rom equation (10.1) substituting the value of b = 10m, we have, T 1 p 3 5YZ1m [ ] ( p (10.2) rom Table 10.2, for the pinion Y = for Z 1 = 17 or the gear, Y = , for Z 2 = 51 or gear, Y[σ] g = x 487 = or pinion, Y[σ] p = x 542 = Table 10.2 Values of the Lewis form factor Y Number of teeth Φ=20 a=0.8m* b=m Φ=20 a=m b=1.25m Φ=25 a=m b=1.25m

4 Hence, for the same face width pinion will be weaker and consideration for the design is, T x YZ m 5x x17m m p MPa (10.3) m = 2.93 mm. Since motor shaft diameter is 30 mm, to get sufficiently large pinion m = 4 mm is taken. Table 10.3 Data for pinion and gear Wheel Z m b=10m d V =wrv Material Hardness Pinion 17 4mm 40 mm 68mm 3.42 m/s SAE Gear 51 4mm 40 mm 204mm 3.42 m/s SAE We will now use Buckingham dynamic load approach for the design. t = T 1 /r 1 = /0.034 = 8781 N Buckingham dynamic load is given by, i 9.84V(Cb+ t ) 9.84V Cb+t (10.4) or V=3.42 m/s permissible error is e= mm from ig rom Table 10.4, if we choose I class commercial cut gears, expected error is for m=4mm. In order to keep the dynamic load low precision cut gears are chosen. So, e =

5 ig Permissible error Table 10.4 Expected error in tooth profile Gear quality and expected error e Module irst class commercial gears Carefully cut gears Precision gears Up to Table 10.5 Value of C Tooth form Material of pinion and gear 14.5 o Cast iron and cast iron steel and cast iron steel and steel 20 o ull depth Cast iron and cast iron steel and cast iron steel and steel 20 o Stub tooth Cast iron and cast iron steel and cast iron steel and steel C 5720 e 7850 e e 5930 e 8150 e e 6150 e 8450 e e rom Table 10.5, if material for both gear and pinion are steel, then,

6 C = 11860e = x = Substituting the values t = 8781 N, C = , V=3.42 m/s, b= 40mm in eqn. (10.4), Buckingham dynamic load is given by, 9.84x3.42(148.25x ) i 5464N 9.84x x (10.5) d = t + i = = N Beam strength of the pinion is given by, tp = bym [σ] p = 40x x4x542 = N Since tp (22381)> d (14245) the design is safe from tooth bending failure consideration. Wear strength of the pinion is given by, [ ] H bd I C p ts 1 2 (10.6) rom Table for steel vs steel, pinion and gear C p = 191 MPa 0.5 and substituting i = 3, Ф=20 0 we o o sin cos i sin 20 cos20 3 I i Table 10.6 Elastic coefficient C p for spur gears in MPa 0.5 Pinion Material (µ=0.3 in all cases) Gear Material Steel Cast iron Al Bronze Tin Bronze Steel, E=207 GPa Cast iron, E=131 GPa Al Bronze, E=121 GPa Tin Bronze, E=110 GPa

7 Surface fatigue strength of the pinion material is σ sf = σ sf K L K R K T σ sf = 2.8(Bhn) 69MPa = 2.8 x = 555.4MPa K L = 0.9 for 10 8 cycles life from graph1 K R = 1.0 taken for 99 reliability K T = 1.0 for operating temperature <120 C (assumed) o Table 10.7 Surface fatigue strength σ Pa) for metallic spur gears (10 7 sf (M cycle life 99% reliability and temperature <120 o C) Material Steel Nodular iron σ sf (MPa) 2.8 (Bhn)-69MPa 0.95(2.8(Bhn)-69MPa) Cast iron, grade Cast iron, grade Cast iron, grade Tin Bronze, AGMA 2C (11% Sn) 207 Aluminium Bronze (ASTM ) (Alloy 9C H.T.) 448 ig.10.2 Life factor K L

8 Table 10.8 Reliability factor K R Reliability (%) K R Surface fatigue strength of the pinion material is σ sf = σ sf K L K R K T = 555.4x0.9x1x1 = 500MPa Assuming, factor of safety, s = 1.1 [σ H ] = σ sf /s = 500/1.1 = 455MPa Wear strength of the pinion is: 2 2 [σ H] 455 ts = bd1 I = 40x68x =1860 N C p 191 Since ts (1860) << d (14245), the design is not safe. Revision is necessary. As the SAE1050 can attain a hardness of 800 VPN(~750 Bhn) after oil quenching, increase the hardness to 475 Bhn and increase the b to 13m = 13 x 4 = 52 mm. rom Table 10.7, we know that, σ sf = 2.8(Bhn) 69MPa = 2.8 x = 1261MPa K L = 0.9 for 10 8 cycles life from ig.10.1 K R = 1.0 taken for 99 reliability K T = 1.0 for operating temperature <120 o C Assumed. Surface fatigue strength of the pinion material is σ sf = σ sf K L K R K T = 1261x0.9x1x1 = 1135MPa

1205 =12439 N C p 191 Since ts (12439)< d (14245), still it is not safe.

9 ig.10.3 Effect of carbon content on the hardness of fully hardened steel Assuming factor of safety, s = 1.1 [σh] = σ sf /s = 1135 /1.1 = 1032MPa 2 2 [σ H ] 1032 ts = bd1i =52x68x =12439 N C p 191 Since ts (12439)< d (14245), still it is not safe. Hence increase the module to 5mm. Table 10.9 Properties of pinion an d gear Wheel Z m b=13m d V =wrv Material Hardness Pinion 17 5mm 65 mm 85mm 4.27 m/s C Gear 51 5mm 65 mm 255mm 4.27 m/s C

10 With new dimensions d = N ts = N. Since ts > d, the revised design is safe from surface fatigue (pitting) considerations. If b = 50 mm, d = N ts = N, ace width of 50 mm is adequate AGMA Approach Data given: W = 30 kw; n 1 = 960 rpm; n 2 = 320 rpm; Life = 10 8 cycles; 20 o full depth involute spur gear. Solution: i = n 1 / n 2 = 960 / 320 = 3 In order to keep the size of gears small and avoid interference, Z 1 = 17 is chosen. Z2 = i Z 1 = 3 x 17 = 51 2n 2x = = rad / s 1000W 1000x30 T Nm AGMA equation for tooth bending stress is, t bmj K v Ko Km d 1 = m Z 1 2T1 2 bz m J 1 K K K [ ] v o m ace width, b= 10 to 13 m.

11 b = 10 m is assumed for the first trial. J = for pinion Z 1 = 17 mating with gear Z 2 =51 or gear J = These values are obtained from the table Table AGMA geometry factor J for teeth having Ø = 20 o, a=1m, b=1.25m and r f =0.300m Number Number of teeth in mating gear of teeth Rack The tooth bending stress is given by, (200V) Kv is assumed.

12 K o = 1.25 is taken assuming uniform power source and moderate shock load from the table 7 K m = 1.3 assuming accurat e mounting and precision cut gears for face width of about 50mm. Table Ov erload factor K o Driven Machinery Source of power Uniform Moderate Shock Heavy Shock Uniform Light shock Medium shock Table Load distribution factor K m ace width ( mm) Characteristics of Support up Accurate mountings, small bearing clearances, minimum deflection, precision gears Less rigid mountings, less accurate gears, contact across the full face Accuracy and mounting such that less than Over Over Over Over full-face contact exists Substituting these values in the equation, 2T K K K x1.15 x1.25 x1.3 10m x17xm x v o m 2 bz1 m J m

13 σ e = σ e k L k v k s k r k T k f k m The pinion is of steel C50 OQT with 223Bhn hardness and tensile strength of 660MPa and the gear is of C45 OQT with hardness 210Bhn and tensile strength of 465MPa. or pinion σ e = 0.5 σ ut = 0.5 x 660 = 330MPa k L = 1 for bending, k V = 1 assumed expecting m to be <5mm; k S = 0.73 from the ig.10.4 for σut= 660MPa, k r = for 90% reliability ig Surface factor k s Table Reliabili ty factor Kr Reliability factor R actor K r k T =1 assumed based on operating temperature <120 o C k f = 1.- and k m = 1.33 for σ ut = 660MPa ( Ultimate tensile strength = 660 MPa for SAE 1050 OQT condition) σ e = σ e k L k v k s k r k T k f k m = 330x1x1x0.73x0.897x1x1x1.33 = 287.4MPa actor of safety on bending of 1.5 assumed

14 [σ] = σ e / s = / 1.5 =191.6MPa ig Miscellaneous effects factor k m rom tooth bending fatigue considerations, m [ ] Solving the equation we get m = 3.68mm Now take m=4 mm as the next standard value. rom this module, the dimensions calculated are given in Table Table Dimensions of pinion and gear Wheel Z m b=10m d V =wrv Material Hardness Pinion 17 4mm 40 mm 68mm 3.42 m/s C Gear 51 4mm 40 mm 204mm 3.42 m/s C t = T 1 / r 1 = 29857/34 = 8781N The tooth has to be checked from surface durability considerations now. The contact stress equation of AGMA is given below: C K K K t H p V o m bd1i C p = 191 MPa 0.5 from the table for steel vs steel

15 0 Substituting i =3, Ф=20 we get I= o o sin cos i sin 20 cos20 3 I i (200V) 78 (200x3.42) Kv rom Table and 10.12, Ko = 1.25 and K m = 1.3 assumed as in the case of bending stress calculation C K K K 191 t H p V bd1i o m 8781x1.15x1.25x1.3 40x68x σh = 1209MPa The surface fatigue strength of the pinion material is given by, σ sf = σ sf K L K R K T Where σ sf = 2.8(Bhn) 69MPa = 2.8 x = 555.4MPa K L = 0.9 for 10 8 cycles life from graph1 K R = 1.0 taken for 99% reliability K o T = 1.0 for operating temperature <120 C assumed. Substituting the values in the equation, σ sf = σ sf K L K R K T = 555.4x0.9x1x1 = 500MPa Assuming a factor of safety, s = 1.1 rom ig and Table 10.8, we get, [σ H ] = σ sf /s = 500/1.1 = 455MPa σ H = 1209MPa

16 Since σ H (1209) >> [σ H ] (455), the design is not safe and surface fatigue failure will occur. Solution: Increase the surface hardness of the material to 475Bhn and also increase the b to 13m = 13 x 4 = 52 mm rom ig. 3 we get, Surface fatigue strength of the pinion material as σsf = σ sf K L K R K T where σ sf = 2.8(Bhn) 69MPa = 2.8 x = 1261MPa K L = 0.9 for 10 8 cycles life from graph1 KR = 1.0 taken for 99% reliability K <120 o T = 1.0 for operating temperature C Assumed. Substituting these values we get, σ sf = σ sf K L K R K T = 1261x0.9x1x1 = 1135MPa Assuming a factor of safety s = 1.1 [σh] = σ sf /s = 1135 /1.1 = 1032MPa C K K K 191 t H p V o m bd1 I 8781x1.15x1.25x1.3 52x68x As σ H (1185) > [σ H ] (1032) the design is not safe from surface durability considerations. Hence increase the module to 5mm and take b=10m C K K K 191 t H p V o m bd1 I 7025x1.17x1.25x1.3 50x85x0.1205

17 σ H =975MPa < [σ H ] (1032MPa). Hence the design is safe from surface durability consideration. inal specification of the pinion and gear are given in the Table and Table Values for gear and pinion Wheel Z m b=10m d Pinion 17 5mm 50 mm 85mm Gear 51 5mm 50 mm 255mm Table Specification of gear and pinion Wheel Material Steel Hardness Manufacturing quality Pinion SAE1050 OQT 475Bhn Precision cut Gear SAE 1045 OQT 450Bhn Precision cut

Spur Gear Design. Pinion. Gear

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