AME 3353: Design of Mechanical Components

Class Notes
© 2002
Dr. J.D. Baldwin
School of Aerospace & Mechanical Engineering
University of Oklahoma
Norman, Oklahoma
OVID - The Oklahoma Virtual Interactive Design System

AME 3353: Design of Mechanical Components Lecture Notes

To illustrate the topic of spur gear stress analysis, consider the following problem: We have an
application requiring two gears to mesh and deliver 180 horsepower between their respective
shafts. A steel pinion (gear 2) rotates at 1000 rpm and is to drive a steel gear (gear 3) at 400 rpm.
The steel pinion and gear are AGMA Class A3, through hardened to 300HB. The teeth are to have
standard 20 degree involute profiles. Determine the proper diametral pitch, number of teeth and
face width of these gears from the standpoint of bending strength and surface strength. Use
application factors (safety factors) of 2.0.
It should be noted that this speed reduction (1000 RPM to 400 RPM) is quite large to be
accomplished in a single stage. In practice, a more complicated gear system would be used to
achieve the speed reduction. These numbers are chosen for illustration purposes only. To begin
our illustration of the AGMA gear performance analysis, we consider the bending stress and
contact stress separately.

Pinion (gear 2) Gear (gear 3)

n2 1000 RPM n3 400 RPM

S fbp2 36000 psi S fbp3 36000 psi

S fcp2 120000 psi S fcp3 120000 psi

HB 2 300 BHN HB 3 300 BHN

H 180 HP
φ 20 degrees

Because we will need the elastic moduli and Poisson ratios for the gear materials in the contact
stress analysis that follows, we list them here (Norton, pg. 994)
6 6
E2 30. 10 E3 30. 10

ν2 0.292 ν3 0.292

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At this point, we do not yet know the number of teeth on each gear, the diametral pitch or the face
width of the gears. All of these quantities must be estimated before we can proceed. Therefore, we
will assume values for these parameters, compute the stresses, compare them with the allowables
and decide whether the design is adequate. If not, we must iterate until we achieve a suitable
design. We begin by assuming that the pinion has 18 teeth (a practical minimum for 20 degree
involute profiles, e.g., see Table 11-4 in Norton)

Input the number of pinion teeth, N2: N2 20

n2
N 3c N 2. N 3c = 50 (calculated value)
n3

Based on this result, we choose to use a value of N2 of 112 teeth. With this number, we can also
confirm that the speed ratio is preserved.

Input the number of gear teeth, N3: N3 50

N2
n 3c .n n 3c = 400 RPM (calculated output speed - OK)
2
N3

We now must choose a value for the diametral pitch pd for the gear set. Remember that pd gives a
measure of the number of teeth per unit diameter and must be the same for both mating gears.
Based on Table 11-2 in Norton, we will choose a diametral pitch pd = 4 from the relatively easy to
manufacture coarse pitch series. This is a good starting point for calculating the gear diameters.

Input the diametral pitch, pd: pd 6

N2
Pinion diameter: d2 d 2 = 3.333 inch
pd

N3
Gear diameter: d3 d 3 = 8.333 inch
pd

It should be clear from these diameters why the required speed reduction will, in practice, be
accomplished by a different arrangement of gears. If we attempt the reduction in more than one
step, smaller diameter gears are required, reducing the size of the resulting gearbox enclosure.

2
 From these values, we can now compute the pitch line velocity for the gear set. Note that either
d2 and n2 or d3 and n3 can be used here; they give the same answers if all is correct.

π. d 2 . n 2
V V = 872.7 fpm
12

The last geometric characteristic for this gear set we must choose is the face width. Norton states
that current design practice is to have face widths in the range
8/pd < F < 16/pd
with a nominal value of 12/pd. Here, we will calculate the range and select an appropriate value
within that range.

8
F1 F 1 = 1.333
pd

16
F2 F 2 = 2.667
pd

Input a face width value between F1 and F2: F 2.5 inch

We now compute the load transmitted to the tooth as in a cantilever beam. The transmitted load,
W t, can be computed either in terms of the transmitted torque, T, or the horsepower, H.

H. 6600 4
T T = 1.134 10 in lb
2.π
n 2.
60

T
Wt 2. W t = 6806.7 lb
N2
pd

The alternate formulation is in terms of the horsepower directly

33000 . H lb
Wt W t = 6806.7
V

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AGMA Bending Stress Analysis
It should be quickly apparent that the AGMA gear analysis is very similar to the Lewis analysis
when we consider the AGMA bending stress relationship, given by
σb = (Wt pd/FJ) (Ka Km /Kv) Ks KB KI
where the various factors are
σb = AGMA bending stress
W t = transmitted tangential load
pd = diametral pitch
F = gear tooth face width
J = bending strength geometry factor
Ka = application factor
Km = load distribution factor
Kv = dynamic factor
Ks = size factor
KB = rim thickness factor
KI = idler factor

We will consider each of these factors individually below.

Bending Strength Geometry Factor, J:

Input the pinion geometry factor, J2: J2 0.3

Input the gear geometry factor, J3: J3 0.4

Application Factor, Ka: This term functions as the safety factor in the calculation. It can be used
to take account of fluctuating loads and heavy impacts. The default value is 1.0.

Input the application factor, Ka: Ka 2.0

Load Distribution Factor, Km : This factor is intended to to account for variable load transition
across the width of the mating teeth. Wider teeth promote an uneven force distribution, while
narrower teeth experience relatively even loading. If the tooth face width is within AGMA
guidelines given above, the default value is 1.0.

Input the load distribution factor, Km: Km 1.0

4
Dynamic Effects Factor, Kv: Like in the Lewis analysis, we will account for speed effects in the
bending stress. Specifically, higher speeds tend to promote shorter lives so we apply a factor to
the bending stress to drive the allowable transmitted load down. This should extend the life
accordingly. The dynamic factor requires the specification of a gear quality index, Qv, which is a
measure of the accuracy with which the gear was manufactured. In general, a higher quality index
indicates a more accurate gear. In these calculations, we use the quality index for the lowest
quality gear in the mesh pair.

Input the controlling quality factor, Qv: Qv 7

2
3
12 Qv
B B = 0.731
4

A 50 56. ( 1 B) A = 65.064

B
A
Kv K v = 0.761
A V

Size Factor, Ks: With this factor, we can account for large gears that are expected to behave
differently from the relatively small gears used in fatigue tests. Usually, we will take this factor to
have a value of 1.0, but values up to 1.5 can be used to introduce some conservatism into the
calculation for larger gears.

Input the size factor, Ks: KS 1.0

Rim Thickness Factor, KB : This factor is intended to for use when spoked gears are specified.
For solid gears, take the factor to be 1.0.

Input the rim thickness factor, KB : KB 1.0

Idler Factor, KI : It is possible that one of the gears in a gearset is an idler, i.e., it serves only to
reverse the direction of rotation of the shaft. In this case, it is more heavily loaded than a non-idler
gear and we use KI = 1.42; otherwise we use 1.0.

Input the idler factor, KI : KI 1.0

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With all of the controlling parameters fixed, we can now compute the AGMA bending stress from
the formula

W t. p d K a . K m 5
σ b2 . .K .K .K
S B I σ b2 = 1.432 10
F. J 2 Kv

W t. p d K a . K m 5
σ b3 . .K .K .K
S B I σ b3 = 1.074 10
F. J 3 Kv

In order to assess the suitability of this design, we will compare the bending stress with the AGMA
allowable stress value, given by
Sfb = (KL/KT KR) Sfb'
where
Sfb = AGMA allowable fatigue-rated stress
KL = life factor
KT = temperature factor
KR = reliability factor
Sfb' = baseline fatigue strength

Life Factor, KL: The factor allows us to design our gears for specified life targets. The factor is
defined in Figure 11-24 (Norton) for various hardness values. In order to compute the life factor, we
need to specify the target life and the gear's hardness value.

6
Input the target life, in cycles, N: N 10 cycles

Hardness Condition
* 400 HB (1)
* 250 HB (2)
* 160 HB (3)
* Case Carburized (4)

Input the life factor, KL: KL 1

Temperature Factor, KT:

Input the lubricant temperature, TF: TF 200 deg. F

KT 1.0 if T F 250
KT =1
460 TF
if T F > 250
620

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Reliability Factor, KR:

Input the target reliability, R: R 0.99

KR 0.90 if R 0.85 KR =1
1.00 if R 0.99
1.25 if R 0.999
1.50 if R 0.9999

Baseline AGMA Bending-Fatigue Strengths, Sfb': Unlike the classical Lewis analysis, the AGMA
method gives us allowable stresses for various materials. By convention, these are given at a life of
107 cycles. Table 11-20 (Norton) lists representative strength values for common steel and iron
gear materials For reference, we reiterate the allowable stresses input initially.

4
The pinion's baseline AGMA allowable strength, Sfb2': S fbp2 = 3.6 10 psi

4
The gear's baseline AGMA allowable strength, Sfb3' S fbp3 = 3.6 10 psi

With all of the required parameters established, we can now compute the AGMA allowable
stresses for the pinion and gear and compare it with the bending stress computed above.

Pinion:

KL 4
S fb2 .S S fb2 = 3.6 10 psi
fbp2
K T. K R

S fb2
η b2 η b2 = 0.251
σ b2

Gear:

KL 4
S fb3 .S S fb3 = 3.6 10 psi
fbp3
K T. K R

S fb3
η b3 η b3 = 0.335
σ b3

7
AGMA Surface Stress Analysis
Just like in the AGMA bending stress analysis, the surface stresses are computed from an
expression reminiscent of the classical, Hertz-based, contact stress expression. The AGMA
surface stress is given by
σc = Cp [(Wt/F I d) (Ca Cm/Cv) Cs Cf]^1/2
where
σc = AGMA surface stress
C p = elastic coefficient (same as in the Lewis analysis)
Wt = transmitted load
F = face width
I = geometry factor
d = pitch diameter
C a = application factor, equal to Ka
C m = load distribution factor, equal to Km
Cv = dynamic effects factor, equal to Kv
Cs = size factor, equal to Ks
Cf = surface finish factor

Elastic Coefficient, Cp: As in the Lewis analysis, we again use the AGMA-recommended elastic
coefficient, Cp, to summarize the stiffness properties of the individual gears. This coefficient is
given by

1
Cp C p = 2284.7
2 2
1 ν2 1 ν3
π.
E2 E3

Surface Geometry Factor, I: Unlike the AGMA bending case where we pick the relevant geometry
factor from a table or a figure, we use an equation to compute I. In this case we assume that our
gears have full-depth tooth profiles, i.e., xp = 0.

Input the pinion addendum coefficient, xp: xp 0

Also, we need the pitch radii for the pinion and gear, computed from the pitch diameters above.

d2
rp r p = 1.667 inch
2

d3
rg r g = 4.167 inch
2
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 The shaft center distance between the pinion and gear is the sum of the two pitch radii, given by

C rp rg C = 5.833 inch

The contact radii of the two teeth are based on the pressure angle, φ, the diametral pitch, pd, and
the individual gear diameters; these radii are given by

2 2
1 xp φ π . φ
ρp rp r p . cos π. cos π. ρ p = 0.461 inch
pd 180 pd 180

φ
ρg C. sin π. ρp ρ g = 1.534 inch
180

Finally, the geometry factor is given by

φ
cos π.
180
I I = 0.1
1 1 .
d2
ρp ρg

Application Factor, Ca: Ca Ka C a =2

Load Distribution Factor, Cm: Cm Km Cm=1

Dynamic Effects Factor, Cv: Cv Kv C v = 0.761

Size Factor, Cs: CS KS C S =1

Surface Finish Factor, Cf: Unless we have unusually rough gear tooth surfaces, we will take this
factor to be 1.0.

Input the surface finish factor, Cf: Cf 1

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The AGMA surface stress in each of the two gears is computed below:

Wt C a. C m 5
σ c2 C p. . .C .C
S f σ c2 = 3.349 10 psi
F. I. d 2 Cv

Wt C a. C m 5
σ c3 C p. . .C .C
S f σ c3 = 2.118 10 psi
F. I. d 3 Cv

The AGMA-allowable surface stress is computed (just like the allowable bending stress) from a
baseline value modified by several performance factors. The AGMA surface stress equation is
Sfc = (CL CH/CT CR) Sfc'
where
Sfc = AGMA allowable surface-fatigue strength
CL = surface life factor
CH = hardness ratio factor, applied only to the gear
CT = temperature factor, equal to KT
CR = reliability factor, equal to KR

Baseline Surface Fatigue Strength, Sfc': This value is established by AGMA for various common
gear materials in Table 11-21 in Norton. These strengths are defined on the basis of a 107 cycle life
for the gears.

Input the baseline surface-fatigue strength, Sfc': S fcp 100000 psi

Surface Life Factor, C L: This factor accounts for design lives other than 107 cycles. For lives
below 104 cycles, C L = 2.5. Between 104 and 107 cycles, the factor is defined by the equation
CL = 2.466 N -0.056
At lives greater than 107,
we can choose between critical service conditions (and use the above
relationship) and standard, commercial use, with the life factor defined by
CL = 1.4488 N -0.023
These regions are shown graphically in Figure 11-26 in Norton. This calculation presumes we have
standard, commercial gears.

8
Input the design life of the gears, N: N 10 cycles

10
 4
CL 2.5 if N 10 C L = 0.948
0.056 4 7
2.466. N if 10 < N 10
0.023 7
1.4488. N if N > 10

Hardness Ratio Factor, CH: Here, we will account for differences in hardness between the pinion
and gear. In order to make the calculation, we use the Brinell hardness values for both gears input
at the beginning of the analysis. We will assume that both the pinion and the gear are
through-hardened and calculate the factor using the relation
C H = 1 + A(mG -1)
where mG is the gear tooth ratio

N3
mG m G = 2.5
N2

HB 2
The hardness ratio, HB2/HB3 is =1
HB 3

and the parameter A is given by

HB 2
A 0 if < 1.2
HB 3

HB 2 HB 2
0.00898 . 0.00829 if 1.2 1.7 A =0
HB 3 HB 3

HB 2
0.00698 if > 1.7
HB 3

CH 1 A. m G 1 CH=1

Temperature Factor, CT: CT KT C T =1

Reliability Factor, CR: CR KR CR =1

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 Pinion:

CL 5
S fc2 .S S fc2 = 1.138 10
fcp2
C T. C R

S fc2
η c2 η c2 = 0.34
σ c2

Gear

C L. C H 5
S fc3 .S S fc3 = 1.138 10
fcp3
C T. C R

S fc3
η c3 η c3 = 0.537
σ c3

Summary: The safety factors computed by this analysis are as follows:

Bending Compression

Pinion η b2 = 0.251 η c2 = 0.34

Gear η b3 = 0.335 η c3 = 0.537

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