4 Steps for Gear Drive Design: From design requirements, identify speed of pinion, nP, desired output speed of gear, nG, and power to be transmitted, P.Choose type of material for the gears (steel, cast iron, bronze, etc.)Determine overload factor, Ko, using table 9-5Calculated Pdes = KoP and calculate a trial value for the diametral pitch, Pd (for steel use Figure 9-27). Note diametral pitch must be a standard size (see Table 8-2).Note, as Pd decreases, tooth size increases thus bringing down St and Sc. But….. As Pd increases, # teeth increases and gear train runs smoother and quiter and the drive gets smaller as well!
5 Steps for Gear Drive Design: Specify Np and NG to meet VR requirement. Calculate center distance, D, OD to make sure there aren’t any interference issues.Specify face width using recommended range: 8/Pd < F < 16/Pd.Remember, increasing face width reduces St and Sc but consider alignment factor. Face width is normally less than 2X Dp.Compute transmitted load, Wt, pitch line speed, vt, quality number, Qv, and other factors required for calculating bending stress and contact stress.Calculate St and required Sat. Does material in 2 meet Sat #? No – then select new material or define new geometry (step 4). If yes, continue to 9.Calculate Sc and required Sac. Does material in 2 meet Sac? No – then select new material to meet Sac and Sat or define new geometry (step 4). If yes, continue to 10.Summarize design
7 Problem # 9.61A gear pair is to be a part of the drive for a milling machine requiring 20 hp with the pinion speed at 550 rpm and the gear speed to be between 180 and 190 rpm.Given:Driven = Milling MachinePower = 20 hpPinion Speed = 550 rpmOutput Speed = rpm ≈ 185 rpmContinuous Use = 30,000 hoursFind:Compact Gear Design
12 Geometry Factors Pinion: JP = .36 Gear: JG = .415 What is it? Fucntion of stress concentration, tooth geometry, number of teeth, etc. As J decreases, St increases. Don’t really need to calculate Jg really this and KB only difference between pinion and gear.Pinion: JP = .36Gear: JG = .415Figure 9-17, Page 387
14 Load Distribution Factor Equation 9-16, Page 390; Equation is solved on next slidePageWhat is it? Alignment factor, takes into account misalighnment of meshing teeth which impacts load transfer – worst case vibration or loaded like helical tooth.. Note as F/D increases, Cpf increases because further away more difficult to control perpendicularity.
15 Load Distribution Factor Cont… Size FactorPage 389As diametral pitch decreases, tooth size increases and therefore size factor increases.ks = 1.0 since Pd ≥ 5
16 Rim Thickness Factor Page Draw FBD of tooth with rim. Really only need to calculate this if not solid. Since M = F*d, where d will increase if flexible ring gear.For this problem, specify a solid gear blank KB = 1.00KB = for mB = 1.2 or larger
17 Rim Thickness Factor Cont… We are assuming a solid gear blank for this problem, but if not then use:Min Rim Thickness = (1.2)(.45 in) = .54 inMin back-up ratio
18 Safety Factor Hardness Ratio Reliability SF = 1.25 (Mid-Range) CH =1.00 for early trials until materials have been specified. Then adjust CH if significantdifferences exist in the hardness of the pinion and the gear.ReliabilityPage 396KR = 1.5 (for 1 in 10,000 failures)
19 Dynamic Factor KvPage 393What is it? Accounts for quality of gear manufacture which leads to tolerance which impact how load is transferred to tooth = impact + Wt = total force. If Qc decreases, tolerance increases and gears will vibrate, not smooth X of torque, aggravated at high pitch line speeds. In other words, Kv increases with increasing pitch line speed and increases as Qv gets worse!
20 Dynamic Factor Cont… Kv Qv comes from Figure 9-21Kv can be calculated like in above equations or taken from Figure 9-21.Equations are more accurate.
21 Design Life Ncp = (60)(L)(n)(q) L = 30,000 hours from Table 9-7 n = 550 rpmq = 1 contactsNcp = (60)(30,000 hours)(550 rpm)(1 contact) = 9.9x108 cyclesNcG = (60)(30,000 hours)( rpm)(1 contact) = x108 cycles
28 Hardness Numbers BENDING (Grade 1) Page 379Pinion BendingSatp = 34,324.5 psi = HB 270Gear BendingThese stresses are OKSatG = 28,741.9 psi = HB 215Go to appendix A3 or A4 and spec out material that meets this hardness requirement! Example AISI 1040, Temper at 900 F
29 Hardness Numbers CONTACT (Grade 1) Page 380Pinion ContactSacp = 261,178 psiGear ContactThese Stresses are WAY too HIGH!Values are off table!SacG = 245,609 psi
30 Summary of ProblemContact stresses are too High. Must iterate until stress are low enough untila usable material can be found.NOTE: Contact Stress generally controls. If material cannot be found for bending,contact stress is too high!Iterate! Decrease Pd and increase FExcel is a GREAT tool to use for these Iterations.This problem solved after third iteration using Excel
31 Guidelines for Adjustments in Successive Iterations. Decreasing the numerical value of the diametral pitch results in larger teeth and generally lower stresses. Also, the lower value of the pitch usually means a larger face width, which decreases stress and increases surface durability.Increase the diameter of the pinion decreases the transmitted load, generally lowers the stresses and improves surface durability.Increase the face width lowers the stress and improves surface durability but less impact than either the pitch or pitch diameter.Gears with more and smaller teeth tend to run more smoothly and quietly than gears with fewer and larger teeth.Standard values of diametral pitch should be used for ease of manufacture and lower cost (See table 8-2).Use high alloy steels with high surface hardness – results in the most compact system but the cost is higher.Use gears with high quality number, Qv – adds cost but lowers load distribution factor, Km.The number of teeth in the pinion should be as small as possible to make the system compact. But the possibility of interference is greater with fewer teeth. Check Table 8-6 to ensure no interference will occur.