Module 2- GEARS Lecture 9 - SPUR GEAR DESIGN Contents 9.1 Problem 1 Analysis 9.2 Problem 2 Spur gear 9.1 PROBLEM 1 SPUR GEAR DESIGN In a conveyor system a step-down gear drive is used. The input pinion is made of 18 teeth, 2.5 mm module, 20 o full depth teeth of hardness 330Bhn and runs at 1720 rpm. The driven gear is of hardness 280Bhn and runs with moderate shock at 860 rpm. Face width of wheels is 35 mm. The gears are supported on less rigid mountings, less accurate gears and contact across full face may be assumed. The ultimate tensile strength of pinion and gear materials is 420 and 385MPa respectively. The gears are made by hobbing process. Find the tooth bending strength of both wheels and the maximum power that can be transmitted by the drive with a factor of safety 1.5. The layout diagram is shown in the Fig 9.1. Fig 9.1 Conveyor drive layout
Solution: The bending fatigue stress is found from AGMA equation as, Ft K v Ko K (9.1) m bmj We know that, Z 2 = Z 1 x (N 1 /N 2 ) Substituting values from table 1, Z 2 = 18 X (1720/860) = 36 Table 9.1 Data given for gear and pinion N Z m d b Pinion 1720rpm 18 2.5 mm 45 mm 35 mm Gear 860 rpm 36 2.5 mm 90 mm 35 mm Using the values from Table 9.1, V = π dn/60000 = π x 45 x 1720/60000 = 4.051m/s We know that K v 0.5 50 (200V) (9.2) 50 Table 9.2 J values for pinion and gear Z J (sharing) K v K o K m Pinion 18 0.338 1.569 1.25 1.6 Gear 36 0.385 1.569 1.25 1.6 The J value is obtained from Fig. 9.2 for sharing teeth as in practice. K o and K m values are obtained from Tables 9.3 and 9.4 for the given conditions.
Fig.9.2 - Geometric Factor J SPUR GEAR TOOTH BENDING STRESS (AGMA) Table 9.3 - Overload factor K o Driven Machinery Source of power Uniform Moderate Shock Heavy Shock Uniform 1.00 1.25 1.75 Light shock 1.25 1.50 2.00 Medium shock 1.50 1.75 2.25
Table 9.4 - Load distribution factor K m Face width ( mm) Characteristics of Support 0-50 150 225 400 up Accurate mountings, small bearing 1.3 1.4 1.5 1.8 clearances, minimum deflection, precision gears Less rigid mountings, less accurate gears, 1.6 1.7 1.8 2.2 contact across the full face Accuracy and mounting such that less than Over Over Over Over full-face contact exists 2.2 2.2 2.2 2.2 For pinion: Ft K bmj t K K v o m Ft x1.569x1.25x1.6 35x 2.5x 0.338 = 0.1061 F (9. 3) And for Gear: Ft σ = Kv Ko Km bmj Ft = x1.569x1.25x1.6 35x 2.5x0.385 = 0.0932 F t (9.4) Fatigue strength of the material is given by, σ e = σ e k L k v k s k r k T k f k m (9.5) Table 9.5 Properties of pinion and gear Prop. σ ut MPa σ e =0.5σ ut MPa k L K v k s Pinion 420 210 1 1 0.8 Gear 385 187.5 1 1 0.8
SPUR GEAR PERMISSIBLE TOOTH BENDING STRESS (AGMA) Endurance limit of the material is given by: σ e = σ e k L k v k s k r k T k f k m (9.6) Where, σ e is the endurance limit of rotating-beam specimen From table 9.5, k L = load factor = 1.0 for bending loads k v = size factor = 1.0 for m < 5 mm and = 0.85 for m > 5 mm k s = surface factor, is taken from Fig.9.3 based on the ultimate tensile strength of the material for cut, shaved, and ground gears. k r = reliability factor given in Table 9.5. k T = temperature factor = 1 for T 350 o C = 0.5 for 350 < T 500 o C Fig.9.3 Surface factor K s Reliability of 90%, working temperature <150 o C and reversible is assumed. k f = 1.0 since it is taken in J factor.
k m = 1.0 for reverse bending assumed here Table 9.6 K terms of pinion and gear Prop. k r k T k f k m Pinion 0.897 1.0 1.0 1.0 Gear 0.897 1.0 1.0 1.0 Table 9.7 Reliability factor R Reliability factor R 0.50 0.90 0.95 0.99 0.999 0.9999 Factor K r 1.000 0.897 0.868 0.814 0.753 0.702 Permissible bending stress e [ ] n ( (9.7) Hence the design equation from bending consideration is, σ [ σ ] (9.8) Factor of safety required = 1.5 ```````````````````````Table 9.8 Strength values of pinion and gear Prop. σ e MPa [σ]= σ e / s MPa σ MPa F T N Pinion 150.7 100.5 0.1061 F t 947 Gear 134.6 89.7 0.0932 F t 962 Table 9.8 shows that the pinion is weaker than gear. And maximum tangential force that can be transmitted is: F t = 947 N So, the maximum power that can be transmitted is: W = F t v / 1000 = 947 x 4.051 /1000 = 3.84 kw -----------------
9.2 PROBLEM 2 SPUR GEAR DESIGN In a conveyor system a step-down gear drive is used. The input pinion is made of 18 teeth, 2.5 mm module, 20 o full depth teeth of hardness 340Bhn and runs at 1720rpm. The driven gear is of hardness 280Bhn and runs with moderate shock at 860 rpm. Face width of wheels is 35mm. The gears are supported on less rigid mountings, less accurate gears and contact across full face may be assumed. The ultimate tensile strength of pinion and gear materials is 420 and 385MPa respectively. The gears are made by hobbing process. From surface durability consideration, find the maximum power that can be transmitted by the drive with a factor of safety 1.2 for a life of 10 8 cycles. Drive layout is shown in the Fig 9.4. Fig. 9.4 Conveyor drive Layout diagram Data given: i = n 1 /n 2 = 1720/860 = 2 Z 2 = Z 1 x i = 18 X 2 = 36
Table 9.9 Data given for pinion and gear n Z m d = mz b Pinion 1720rpm 18 2.5 mm 45 mm 35 mm Gear 860 rpm 36 2.5 mm 90 mm 35 mm Table 9.10 Properties of gear and pinion Bhn Ø Reliability Life Temp Pinion 340 20 o 99 % 10 8 <120 o C Gear 280 20 o 99 % - <120 o C Solution: The induced dynamic contact stress is given by equation below, Ft H Cp KVKoK (9.9) m bd I 1 When both pinion and gear material are made up of steel, from Table 9.11, C p = 191 (9.10) MPa SPUR GEAR CONTACT STRESS Table 9.11 Elastic coefficient Cp for spur gears in MPa Pinion Material(μ=0.3 in all cases) Gear material Steel Cast iron Al Bronze Tin Bronze Steel, E=207Gpa 191 166 162 158 Cast iron, E=131Gpa 166 149 149 145 Al Bronze, E=121Gpa 162 149 145 141 Tin Bronze, E=110Gpa 158 145 141 137
sin cos i I (9.11) 2 i 1 Substituting the values from table 10, o o sin 20 cos20 2 I 0.1071 2 2 1 SPUR GEAR SURFACE DURABILITY From table 3 and 4, V = π dn/60000 = π x 45 x 1720/60000 = 4.051m/s For hobbed gear, 50 (200V) K v 50 0.5 (9.12) Table 9.14 K Values of pinion and gear Z K v K o K m Pinion 18 1.569 1.25 1.6 Gear 36 1.569 1.25 1.6 Substituting values from Table 14, we have, C F t H p V o m bd1 I t 191 1.569x1.25x1.6 t K K K F 35x45x0.1071 26.051 F MPa Surface fatigue strength of the material is given by, σ sf = σ sf K L K r K T (9.13) From table 10, for steel life is 10 7 cycles & reliability 99% and from Table 9.15,
σ sf = 28(Bhn) 69 = 2.8x340 69 = 954MPa K L = 0.9 for 10 8 cycles from Fig.9.2 K R = 1.0. for 99% reliability from Table 9.10 SPUR GEAR SURFACE FATIGUE STRENGTH Table 9.15 Surafce fatigue strength σ sf for metallic spur gears (10 7 cycle life 99% reliability and temperature <120 0 C) Material Steel Nodular iron σ sf (MPa) 2.8 (Bhn)-69MPa 0.95 (2.8(Bhn)-69MPa) Cast iron, grade 20 379 Cast iron, grade 30 482 Cast iron, grade 40 551 Tin Bronze, AGMA 2C (11% Sn) 207 Aluminium Bronze (ASTM 148 52) 448 (Alloy 9C H.T.) Fig. 9.5 Life factor K l
SPUR GEAR ENDURANCE LIMIT Table 9.16 Reliability factor K R Reliability (%) K R 50 1.25 99 1.00 99.9 0.80 SPUR GEAR ALLOWABLE SURFACE FATIGUE STRESS (AGMA) We know that, [ σ H ] = σ Sf / f s = 954/1.2 = 795MPa For factor of safety f s = 1.2 Design equation is, σ H [ σ H ] 26.051 F t = 795 F t = 931 N Maximum Power that can be transmitted is, W = F t V/1000 = 931x4.051/1000 = 3.51kW -----------------------