Chapter 9 Spur Gear Design
Pinion
Gear
Lecture Steps:
1. Quick review, gear geometry (Chapter 8)
2. Transmitted loads
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.
handout
6. Example(s) gear design!!
(overhead)
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 9-5
4. 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.
7.
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.
–
7.
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.
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 # 9.61
A 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 Machine
Power = 20 hp
Pinion Speed = 550 rpm
Output Speed = 180 -190 rpm ≈ 185 rpm
Continuous Use = 30,000 hours
Find:
Compact Gear Design
Solution:
Design Power:
Assume: Light Shock Driver and Moderate Shock Driven
Ko = 1.75 (Table 9-5, page 389)
PDesign = (Ko)(PInput)
= (1.75)(20hp)
= 35hp
Trial Size
Pick Sizes
Pd = 5 T/in
Dp = 4.80 in
DG = 14.2 in
Np = 24 teeth
NG = 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 Decisions
Quality Number, Qv = 6 (Table 9-2, Page 378)
Steel Pinion
Cp = 2300 (Table 9-9, Page 400)
Steel Gear
Softer material, more relative
deformation, therefore contact area
increases and stress decreases
More precision, higher quality number!
Geometry Factors
Pinion: JP = .36
Gear: JG = .415
Figure 9-17,
Page 387
Geometry Factors Cont…
Page 402
I = .108
Load Distribution Factor
Equation 9-16, Page 390; Equation is solved on next slide
Page
Load Distribution Factor Cont…
Size Factor
ks = 1.0 since Pd ≥ 5
Page 389
Rim Thickness Factor
Page
For this problem, specify a solid gear blank KB = 1.00
KB = 1.00 for mB = 1.2 or larger
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 Ratio
CH =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.
Reliability
Page 396
KR = 1.5 (for 1 in 10,000 failures)
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 Life
Ncp = (60)(L)(n)(q)
L = 30,000 hours from Table 9-7
n = 550 rpm
q = 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)
Page 379
Pinion Bending
Satp = 34,324.5 psi = HB 270
Gear Bending
SatG = 28,741.9 psi = HB 215
These stresses are OK
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)
Page 380
Pinion Contact
Sacp = 261,178 psi
Gear Contact
SacG = 245,609 psi
These Stresses are WAY too HIGH!
Values are off table!
Summary of Problem
Contact stresses are too High. Must iterate until stress are low enough until
a 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.
2.
3.
4.
5.
6.
7.
8.
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.