Chapter 9 Spur Gear Design

Pinion

Gear

Lecture Steps:

1. Quick review, gear geometry (Chapter 8)2. Transmitted loads (overhead)

3. Review bending stress, bending stress number, St, allowable bending stress number, Sat and adjusted allowable bending stress number, S’at.

4. Review contact stress number, Sc, allowable contact stress number, Sac and adjusted allowable contact stress number S’ac.

5. Overview gear design steps.6. Example(s) gear design!! handout

Quick review:

Bending Stress No:

Required Allowable Bending Stress No:

Contact Stress No:

Required Allowable Contact Stress No:

Steps for Gear Drive Design:

1. From design requirements, identify speed of pinion, nP, desired output speed of gear, nG, and power to be transmitted, P.

2. Choose type of material for the gears (steel, cast iron, bronze, etc.)

3. Determine overload factor, Ko, using table 9-54. Calculated 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!

Steps for Gear Drive Design:6. Specify Np and NG to meet VR requirement. Calculate center

distance, D, OD to make sure there aren’t any interference issues.7. 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.

7. Compute transmitted load, Wt, pitch line speed, vt, quality number, Qv, and other factors required for calculating bending stress and contact stress.

8. 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.

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.

10. Summarize design

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 = 180 -190 rpm ≈ 185 rpmContinuous Use = 30,000 hours

Find:Compact Gear Design

Solution:

Design Power:

Assume: Light Shock Driver and Moderate Shock DrivenKo = 1.75 (Table 9-5, page 389)

PDesign = (Ko)(PInput) = (1.75)(20hp) = 35hp

Trial SizePick Sizes

Pd = 5 T/inDp = 4.80 inDG = 14.2 inNp = 24 teethNG = 71 teeth

Check physical size!!

Center Distance

Pitch Line Speed

Tangential Load

Note: Use Input Power Here as Ko is applied Later!

Face Width

Assumptions:

Design DecisionsQuality Number, Qv = 6 (Table 9-2, Page 378)Steel PinionSteel Gear

Cp = 2300 (Table 9-9, Page 400)

More precision, higher quality number!

Softer material, more relative deformation, therefore contact area increases and stress decreases

Geometry Factors

Pinion: JP = .36Gear: JG = .415Figure 9-17, Page 387

Geometry Factors Cont…

I = .108

Page 402

Load Distribution FactorEquation 9-16, Page 390; Equation is solved on next slide

Page

Load Distribution Factor Cont…

Size Factor

ks = 1.0 since Pd ≥ 5Page 389

Rim Thickness Factor

KB = 1.00 for mB = 1.2 or larger

For this problem, specify a solid gear blank KB = 1.00

Page

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 in

Min back-up ratio

Safety Factor

SF = 1.25 (Mid-Range)

Hardness RatioCH =1.00 for early trials until materials have been specified. Then adjust CH if significant

differences exist in the hardness of the pinion and the gear.

ReliabilityKR = 1.5 (for 1 in 10,000 failures)

Page 396

Dynamic Factor Kv

Page 393

Dynamic Factor Cont… Kv

Qv comes from Figure 9-21

Kv can be calculated like in above equations or taken from Figure 9-21. Equations are more accurate.

Design LifeNcp = (60)(L)(n)(q)L = 30,000 hours from Table 9-7n = 550 rpmq = 1 contacts

Ncp = (60)(30,000 hours)(550 rpm)(1 contact) = 9.9x108 cycles

NcG = (60)(30,000 hours)( 185.92 rpm)(1 contact) = 3.34656x108 cycles

Stress Cycle Factors

Page 395

Stress Cycle Factors Cont…

Page 403

Bending Stress Numbers

Pinion:

Gear:

Required Bending Stress Allowable:

Contact Stress Number

Required Contact Stress Allowable:Pinion:

Gear:

Hardness Numbers BENDING (Grade 1)

Satp = 34,324.5 psi = HB 270Pinion Bending

Gear Bending

SatG = 28,741.9 psi = HB 215These stresses are OK

Page 379

Go to appendix A3 or A4 and spec out material that meets this hardness requirement! Example AISI 1040, Temper at 900 F

Hardness Numbers CONTACT (Grade 1)

Pinion ContactSacp = 261,178 psi

Gear ContactSacG = 245,609 psi

These Stresses are WAY too HIGH!Values are off table!

Page 380

Summary of Problem

Contact 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 F

Excel is a GREAT tool to use for these Iterations. This problem solved after third iteration using Excel

Guidelines for Adjustments in Successive Iterations.

1. 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.

2. Increase the diameter of the pinion decreases the transmitted load, generally lowers the stresses and improves surface durability.

3. Increase the face width lowers the stress and improves surface durability but less impact than either the pitch or pitch diameter.

4. Gears with more and smaller teeth tend to run more smoothly and quietly than gears with fewer and larger teeth.

5. Standard values of diametral pitch should be used for ease of manufacture and lower cost (See table 8-2).

6. Use high alloy steels with high surface hardness – results in the most compact system but the cost is higher.

7. Use gears with high quality number, Qv – adds cost but lowers load distribution factor, Km.

8. 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.