Tài liệu Handbook of Mechanical Engineering Calculations P20 pptx

20.1
SECTION 20
GEAR DESIGN AND
APPLICATION
Analyzing Gears for Dynamic Loads
20.1
Helical-Gear Layout Analysis
20.12
Analyzing Shaft Speed in Epicyclic Gear
Trains
20.14
Speeds of Gears and Gear Trains
20.17
Selection of Gear Size and Type
20.18
Gear Selection for Light Loads
20.21
Selection of Gear Dimensions
20.25
Horsepower Rating of Gears
20.26
Moment of Inertia of a Gear Drive
20.28
Bearing Loads in Geared Drives
20.29
Force Ratio of Geared Drives
20.30
Determination of Gear Bore Diameter
20.31
Transmission Gear Ratio for a Geared
Drive

20.32
Epicyclic Gear Train Speeds
20.33
Planetary-Gear-System Speed Ratio
20.34
ANALYZING GEARS FOR DYNAMIC LOADS
A two-stage, step-up gearbox drives a compressor and has a lubrication pump
mounted on one of the gear shafts, Fig. 1. Tables 1, 2, and 3 show the spur-gear
data, tolerances for tooth errors, and polar moments of inertia for the masses in the
compressor drive. All gears in the drive have 20
Њ
pressure angles. The gears are
made of steel; the compressor is made of cast iron. A 50-hp (37.3-kW) motor drives
the gearbox at 3550 rpm. What are the dynamic loads on this gearbox?
Calculation Procedure:
1. Determine the pitchline velocity, V, applied load, W, at each mesh, and
shaft speed, n
In the equations that follow, subscripts identify the shaft, gear, or mesh under con-
sideration. For example, the subscript A denotes Shafts A
1
and A
2
; B, Shaft B; and
C, Shaft C. Also, subscripts 1, 2, 3, 4 refer respectively to Gears 1, 2, 3, and 4 in
the gearbox. The subscripts a and b denote the mesh of Gears 1 and 2 and the
mesh of Gears 3 and 4, respectively. Finally, subscripts r and n refer to the driver
gear and driven gear. Full nomenclature appears at the end of this procedure.
For Shaft A
1
and A

2
,
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Source: HANDBOOK OF MECHANICAL ENGINEERING CALCULATIONS
20.2
DESIGN ENGINEERING
SI Values
0.95 in (22.7 mm) 4.5 in (114.3 mm) 3.2 in (81.3 mm)
1.0 in (25.4 mm) 5.8 in (147.3 mm) 8 in (203.2 mm)
1.25 in (31.8 mm) 6.65 in (168.9 mm)
1.5 in (38.1 mm) 14 in (355.6 mm)
1.57 in (39.9 mm) 16 in (406.4 mm)
2 in (50.8 mm)
2.35 in (59.7 mm)
2.5 in (63.5 mm)
FIGURE 1 Two-stage step-up gearbox driving a compressor (Machine Design.)
TABLE 1
Spur Gear Data
Number
of teeth,
N
Pitch radius
R (in) mm
Face width
F (in) mm
Lewis
form
factor, y

Gear 1 133 3.325 84.5 1.25 31.8 0.276
Gear 2 47 1.175 29.8 1.25 31.8 0.256
Gear 3 116 2.90 73.7 1.00 25.4 0.275
Gear 4 64 1.60 40.6 1.00 25.4 0.264
n
ϭ
input speed
A
ϭ
3550 r/min
P
ϭ
input power
A
ϭ
50 hp (37.3 kW)
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GEAR DESIGN AND APPLICATION
GEAR DESIGN AND APPLICATION
20.3
TABLE 2
Tolerances for Tooth Errors
Tolerance for tooth profile
error e
p
in mm
Tolerance for tooth spacing
error e

s
in mm
Gear 1 0.0010 0.0254 0.0007 0.0178
Gear 2 0.0009 0.0229 0.00065 0.0165
Gear 3 0.0010 0.0254 0.0007 0.0178
Gear 4 0.0009 0.0229 0.00065 0.0165
TABLE 3
Polar Moment of Inertia of Cylindrical Masses
Element
Diameter, D
in mm
Length, L
in mm
Polar moment of inertia, I
o
in
2
⅐
lb
⅐
s
2
/ft cm
2
⅐
kg
⅐
s
2
/m

Motor 14 355.6 16 406.4 534.751 5138.78
Shaft A
1
0.95 24.1 2.5 63.5 0.002 0.0192
Shaft A
2
1.57 39.88 4.0 101.6 0.021 0.2018
Gear 1 6.65 168.9 1.25 31.75 2.127 20.439
I
oA
ϭ ͚
I
o
ϭ
536.901 5159.44
Gear 2 2.35 59.69 1.25 31.75 0.033 0.3171
Shaft B 2.0 50.8 12.543 318.59 0.175 1.6816
Gear 3 5.8 147.3 1.0 25.4 0.985 9.465
I
oB
ϭ ͚
I
o
ϭ
1.193 11.463
Gear 4 3.2 81.28 1.0 25.4 0.091 0.8744
Shaft C 1.5 38.1 4.5 114.3 0.020 0.1921
Compressor 8.0 203.2 1.5 38.1 4.915 47.231
I
oC

ϭ ͚
I
o
ϭ
5.026 48.297
Tabulated values do not include the polar moment of inertia for the lubrication pump.
A material mass factor B
ϭ
0.00087 lb
⅐
s
2
/in
3
⅐
ft was assumed for steel; B
ϭ
0.00080 lb
⅐
sec
2
/in
3
⅐
ft
(0.000073 kg
⅐
s
2
/cm

3
⅐
m) cast iron.
The polar moment of inertia for cylindrical masses was calcualted as, I
o
ϭ
BD
4
L.
T
ϭ
63,025 P /n
AAA
ϭ
63,025 (50)/3550
ϭ
888 lb
⅐
in (100.3 N
⅐
m)
W
ϭ
T / R
aA1
ϭ
888/3.325
ϭ
267 lb (121.2 kg)
V

ϭ
0.5236 Rn
a 1 A
ϭ
0.5236 (3.325) (3550)
ϭ
6180 ft/min (1883.7 m/min)
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GEAR DESIGN AND APPLICATION
20.4
DESIGN ENGINEERING
Therefore, for Shafts B and C, the following values for speed, torque, load and
pitchline velocity are obtained.
n
ϭ
n (N /N )
BA12
ϭ
3550 (133/47)
ϭ
10,046 r/min
T
ϭ
T (N /N )
BA21
ϭ
888 (47/133)
ϭ

314 lb
⅐
in (35.5 N
⅐
m)
n
ϭ
n (N /N )
CB34
ϭ
10,046 (116/64)
ϭ
18,208 r/min
T
ϭ
T (N /N )
CB43
ϭ
314 (64/116)
ϭ
173 lb
⅐
in (19.5 N
⅐
m)
W
ϭ
T /R
bB3
ϭ

314/2.90
ϭ
108 lb (49 kg)
V
ϭ
0.5236 Rn
b 3 B
ϭ
0.5236 (2.9) (10,046)
ϭ
15,254 ft/min (4649.4 m/min)
From these values, dynamic loads can be calculated.
Find the Total Effective Mass at Mesh a and Mesh b
Total effective mass m
t
must be determined at each gear mesh. The parameter m
t
can be calculated from
1/m
ϭ
1/m
ϩ
1/m
trn
where m
r
is the effective mass at the driver gear and m
n
is the effective mass at the
driven gear.

For rotating components such as gears, an effective mass m at the pitch radius
R can be calculated from
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GEAR DESIGN AND APPLICATION
GEAR DESIGN AND APPLICATION
20.5
2
m
ϭ ͚
I /R
o
The gears in the compressor drive mesh at two points, for which the total ef-
fective mass m
t
must be determined.
At Mesh a and Mesh b, m
t
is
1/m
ϭ
1/m
ϩ
1/m
ta ra na
1/m
ϭ
1/m
ϩ

1/m
tb rb nb
However, a lubrication pump is directly coupled to Shaft A
2
. For Mesh a, the
effective mass at the driver gear m
ra
is infinite because of the incompressible fluid
in the pump that prevents instantaneous accelerations. Also at Mesh b, the effective
mass at the driver gear m
rb
is infinite because the reflected inertia of the pump is
also infinite.
Consequently, the equations for m
t
become
1/m
ϭ
1/
ϱϩ
1/m
ta na
m
ϭ
m
ta na
and
1/m
ϭ
1/

ϱϩ
1/m
tb nb
m
ϭ
m
tb nb
Finally, the total effective mass m
t
at Mesh a and Mesh b can be calculated as
2
I
ϩ
I (R / R )
oB oC 34
m
ϭ
m
ϭ
ta na
2
R
2
2
1.193
ϩ
5.026(2.90/160)
ϭ
2
1.175

22
ϭ
12.82 lb
⅐
s /ft (19.1 kg
⅐
s/m)
2
m
ϭ
m
ϭ
I / R
tb nb oC 4
2
ϭ
5.026/1.6
22
ϭ
1.963 lb
⅐
s /ft (2.924 kg
⅐
s/m)
3. Compute the acceleration force for the spur gears
For spur gears, the acceleration force ƒ
1
is
2
ƒ

ϭ
Hm V
1 t
where H
ϭ
A
1
(1/R
r
ϩ
1/R
n
). The calculation factor A
1
ϭ
0.00086 for 14.5
Њ
teeth,
0.00120 for 20
Њ
teeth, and 0.00153 for 25
Њ
teeth. The parameters R
r
and R
n
are the
pitch radii of driver and driven gears, respectively.
Therefore at Mesh a,
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GEAR DESIGN AND APPLICATION
20.6
DESIGN ENGINEERING
H
ϭ
0.00120(1R
ϩ
1/R )
arn
ϭ
0.00120(1/3.325
ϩ
1/1.175)
ϭ
0.00138
V
ϭ
6180 ft/min (1883.7 m/min)
a
22
m
ϭ
12.82 lb
⅐
s /ft (19.1 kg
⅐
s/m)
ta

2
ƒ
ϭ
Hm V
1a a ta a
2
ϭ
(0.00138)(12.82)(6180)
ϭ
675,685 lb (306,761 kg)
And at Mesh b,
H
ϭ
0.00120(1/R
ϩ
1/R )
brn
ϭ
0.00120(1/2.90
ϩ
1/1.60)
ϭ
0.00116
V
ϭ
15,254 ft
⅐
min (4649.4 m/min)
b
22

m
ϭ
1.963 lb
⅐
s /ft (2.924 kg
⅐
s/m)
tb
2
ƒ
ϭ
Hm V
1b b tb b
2
ϭ
(0.00116)(1.963)(15,254)
ϭ
529,841 lb (240,548 kg)
4. Calculate the deflection force
The deflection force ƒ
2
is
1/ƒ
ϭ
1/C
ϩ
(1/C
ϩ
1/C
ϩ ⅐⅐⅐

1/C )
212x
where C accounts for deflection from bending and compressive loads in the gear
teeth, and C
x
accounts for deflection from torsional loads in shafts, flexible cou-
plings, and other components. C is the load required to deflect gear teeth by an
amount equal to the error in action. C
x
is the load required at the pitch radius to
deflect a shaft or coupling by an amount equal to the error in action. Previously,
calculations of ƒ
2
only considered the parameter C. With the addition of C
x
to the
equation for ƒ
2
, the value of ƒ
2
will always be less than the smallest value of C,
C
1
, C
2
...C
x
.
The parameter C is
C

ϭ
W
ϩ
1000 eFA
where A is the load required to deflect teeth by 0.001 in (0.0254 mm).
If the Lewis form factor y is known, A can be calculated from
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GEAR DESIGN AND APPLICATION
GEAR DESIGN AND APPLICATION
20.7
EZEZ
rrnn
A
ϭ
1000(EZ
ϩ
EZ)
rr nn
where for the driver and driven gears, E is the modulus of elasticity, and Z is the
calculation factor for A. Thus calculation factor Z is
Z
ϭ
y/(0.242
ϩ
7.25y)
The table below summarizes the results of calculations for A. For both Mesh a and
Mesh b, A
ϭ

1837 lb (834 kg).
Summary of Calculations for Deflection Load
Lewis form Calculation Deflection
factor, y factor, Z load, A (lb)
Gear 1 0.276 0.123 At Mesh a,
Gear 2 0.256 0.122 A
ϭ
1837 (834 kg)
Gear 3 0.275 0.123 At Mesh b,
Gear 4 0.264 0.122 A
ϭ
1837 (834 kg)
Previously, the error in action e had to be assumed for a given class of gears,
based on recommendations tabulated in handbooks. However, this error can be
approximated as
22
e
ϭ ͙
(e
ϩ
e )
ϩ
(e
ϩ
e )
pr sr pn sn
where for the driver and driven gears, e
p
is the tooth profile error and e
s

is the tooth
spacing error. If actual measurements of e
p
and e
s
are not available, then the tol-
erances for allowable e
p
and e
s
can be used for calculations instead. Therefore, for
both Mesh a and Mesh b, the approximate error in action e is
22
e
ϭ ͙
(0.0010
ϩ
0.0007)
ϩ
(0.0009
ϩ
0.00065)
ϭ ͙
0.0000053
ϭ
0.0023 in (0.058 mm)
Finally, C can be calculated for both meshes. At Mesh a,
C
ϭ
W

ϩ
1000eFA
aa aaa
ϭ
267
ϩ
1000(0.0023)(1.25)(1837)
ϭ
5548 lb (2519 kg)
and at Mesh b,
C
ϭ
W
ϩ
1000eFA
bb bbb
ϭ
108
ϩ
1000(0.0023)(1.00)(1837)
ϭ
4333 lb (1967 kg)
For steel shafts, the parameter C
x
is calculated from
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GEAR DESIGN AND APPLICATION
20.8

DESIGN ENGINEERING
42
C
ϭ
1,080,000eD /RL
xss
where D
s
is the shaft diameter, and L
s
is the shaft length. Generally, for other
components such as flexible couplings, C
x
has to be determined experimentally or
from information supplied by the component manufacturer.
In the gear box, the portion of Shaft A
2
between the motor and Gear 1 deflects
interdependently with Shaft A
1
. On the other hand, the portion of Shaft A
2
between
the lubrication pump and Gear 1 deflects independently of the shaft elements to
the right of Gear 1. The table below summarizes the calculations of C
x
for all shaft
elements affecting each mesh.
Summary of Calculations for Shaft Deflection Load
Shaft

elements
for Mesh a
Shaft
diameter,
D
s
(in) mm
Shaft
length,
L
s
(in) mm
Pitch
radius,
R (in) mm
Shaft
deflection
load, C
x
(lb) kg
Shaft A
1
0.95 24.1 2.5 63.5 3.325 84.5 73 33.1
Shaft A
2
1.57 39.9 2.5 63.5 3.325 84.5 546 247.9
Shaft A
2
1.57 39.9 1.5 38.1 3.325 84.5 910 413.1
Shaft B 2.00 50.8 8.543 216.9 1.175 29.8 3370 1529.9

Shaft
elements
for Mesh b
Shaft B 2.00 50.8 8.543 216.9 2.90 73.7 553 251
Shaft C 1.5 38.1 4.5 114.3 1.60 40.6 1091 495.3
When shafts have steps of different diameter, the loads required to deflect each
step torsionally by the error in action must be determined. Then the loads for each
step must be combined as an inverse of the sum of the reciprocals of the loads.
For example, Shaft A
2
has a different diameter than Shaft A
1
has. Consequently,
the effective value for C
x
between the motor and pump is
C
ϭ
C
ϭ
C
ϩ
1/(1C
ϩ
1/C )
x 43 2 1
ϭ
910
ϩ
1/(1/546

ϩ
1/73)
ϭ
974 lb (442.2 kg)
Note that values of C
x
for each shaft element in the table are differentiated from
each other by the use of a subscript x, enumerated as x
ϭ
1, 2, 3, 4 . . . This
subscript should not be confused with numerical subscripts used to denote Gears
1, 2, 3, 4.
For Mesh a,ƒ
2
is calculated as
1/ƒ
ϭ
1/C
ϩ
1/C
ϩ
1/C
2aa45
ϭ
1/5548
ϩ
1/974
ϩ
1/3370
F

ϭ
665 lb (301.9 kg)
2a
and for Mesh b,ƒ
2
is
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GEAR DESIGN AND APPLICATION
GEAR DESIGN AND APPLICATION
20.9
1/ƒ
ϭ
1/C
ϩ
1/C
ϩ
1/C
2bb67
ϭ
1/4333
ϩ
1/553
ϩ
1/1091
ƒ
ϭ
338 lb (153.5 kg)
2b

5. Find the resultant force and dynamic load at each mesh
At Mesh a,
1/ƒ
ϭ
1/ƒ
ϩ
1/ƒ
aa 1a 2a
ϭ
1/675,685
ϩ
1/665
ƒ
ϭ
664 lb
aa
W
ϭ
W
ϩ ͙
ƒ (2ƒ
Ϫ
ƒ )
da a aa 2aaa
ϭ
267
ϩ ͙
664[2(665)
Ϫ
664]

ϭ
932 lb (423.1 kg)
and at Mesh b,
1/ƒ
ϭ
1/ƒ
ϩ
1/ƒ
ab 1b 2b
ϭ
1/529,841
ϩ
1/338
ƒ
ϭ
338 lb
ab
W
ϭ
W
ϩ ͙
ƒ (2ƒ
Ϫ
ƒ )
db b ab 2bab
ϭ
108
ϩ ͙
338[2(338)
Ϫ

338]
ϭ
446 lb (202.5 kg)
At Mesh a and Mesh b, W
d
/W
ϭ
3.49 and 4.13 respectively. Because the
W
d
/W ratio for both meshes is greater than two, there will be free impact between
the gear teeth. Also, the drive will be noisy and may wear rapidly. The dynamic
load W
d
can de decreased by placing a flexible coupling between the motor and
Shaft A
1
. This will effectively isolate the motor mass. Also, the lubrication pump
can be isolated by driving it with a quill shaft. Finally, the diameter of Shaft B can
be minimized to meet the required torque capacity while increasing shaft resilience
to lower the magnitude of W
d
.
Related Calculations. Gears in mesh never operate under a smooth, continuous
load. Factors such as manufacturing errors in tooth profile and spacing, tooth de-
flections under load, and imbalance all interact to create a dynamic load on gear
teeth. The resulting action is similar to that of a variable load superimposed on a
steady load.
Consider, for example, how tooth loads can fluctuate from manufacturing errors.
The maximum, instantaneous tooth load occurs at the maximum error in action.

The average load is the applied load on the teeth at the pitchline of the gears.
As each pair of teeth moves through its duration of contact, errors in action
create periods of sudden acceleration that momentarily separate mating gear teeth.
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GEAR DESIGN AND APPLICATION
20.10
DESIGN ENGINEERING
Impact occurs as each pair of teeth returns to mesh to complete its contact duration.
The impact loads can be significantly greater than the applied load at the pitchline.
The maximum magnitude of the dynamic load depends on gear and pinion masses,
connected component masses, operating speed, and material elasticity. Elastic de-
flections in gear teeth and drive components, such as shafts and flexible couplings,
help reduce dynamic loads.
Generally, gears should be designed so that their bending and wear capacities
are equal to or greater than the maximum, instantaneous dynamic load. However,
the exact magnitude of the maximum dynamic load is seldom known. Although
many gear studies have been conducted, there is no full agreement on the single
best method for determining dynamic loads.
One of the most widely accepted methods for calculating dynamic loads con-
siders the maximum dynamic load W
d
, resulting from elastic impact, to be
W
ϭ
W
ϩ
W
di

The term W is the applied load at the pitchline of a gear. For gears, the incremental
load W
i
is
W
ϭ ͙
ƒ (2ƒ
Ϫ
ƒ )
ia2 a
In the above equation, ƒ
a
is the resultant force required to accelerate masses in
a system of elastic bodies. The resultant acceleration force ƒ
a
is
1/ƒ
ϭ
1/ƒ
ϩ
1/ƒ
a 12
The term ƒ
1
is the force required to accelerate masses in a system of rigid bodies.
The term ƒ
2
is the force required to deflect the system elastically by the amount of
error in action. As mentioned previously, free impact loads occur in gears when
the teeth separate and return to mesh suddenly during the contact interval. For free

impact, the value of the ratio W
d
/W will always be greater than two.
If forces ƒ
1
, ƒ
2
, and ƒ
a
are plotted as functions of pitchline velocity, force ƒ
2
is
seen to be an asymptote of the incremental load W
i
. The equations that define ƒ
1
and ƒ
2
depend on the type of gears being analyzed for dynamic loads. The example
presented in this procedure provides equations defining ƒ
1
and ƒ
2
for spur gear
applications.
Sometimes this method for calculating W
d
gives values for dynamic load that
are conservative. These high dynamic load estimates can lead to overdesigned gears.
Conservative calculations of W

d
can be minimized if ƒ
2
can be calcualted more
accurately. A less conservative calculation of ƒ
2
effectively reduces W
d
because ƒ
2
is an asymptote for the incremental load W
i
.
The method presented here has been refined to give more accurate values for
ƒ
2
. Previously, only the elastic deflection in the gear teeth was considered in the
calculation of ƒ
2
. Now, elastic deflections in other mechanical components, such as
flexible couplings and shafts, are also considered for their effects on ƒ
2
through
inclusion of a parameter C
x
.
Also, the calculation of ƒ
2
is further refined by a more accurate approximation
of the error in action. Previously, the error in action was assumed for a given class

of gears, based on recommendations commonly tabulated in gear design handbooks.
Now the error in action can be calculated from tooth profile errors and tooth-to-
tooth spacing errors. The equation for approximating error in action results from
tests where gears were measured for error in action, profile error, spacing error, and
runout error.
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GEAR DESIGN AND APPLICATION
GEAR DESIGN AND APPLICATION
20.11
This procedure shows how the refinements in the dynamic load analysis apply
in spur gear applications. However, the principles can be applied to other forms of
gearing as well. Furthermore, the example describes how resilience in mechanical
components, such as shafts, can be applied to reduce dynamic loads on gears.
Reduced dynamic loads result in lower operating noise and longer component life.
In all new applications where there is no design experience, gear drives should
be analyzed for the possibility of dynamic loading. Also, all gear drives operating
at peripheral speeds of about 1000 ft / min (304.8 m/ min) and higher should be
checked.
At low operating speeds, usually less than 1000 ft/min (304.8 m / min), gear
drives follow the dynamics of rigid bodies. Consequently, elastic deflections in
components have little effect on dynamic loads. Inertial forces generated by mo-
mentum variations appear to be directly proportional to mass magnitude and the
square of mass velocity. At intermediate operating speeds, elastic deflections help
reduce instantaneous dynamic loads. In this range, inertial forces appear to be di-
rectly proportional to mass velocity and the square root of mass magnitude. At
higher operating speeds, elastic deflections have a pronounced influence on dynamic
loads. The deflections tend to limit inertial forces to an asymptotic value. Conse-
quently, inertial forces appear to be independent of mass magnitude and mass ve-

locity.
Some applications must be carefully analyzed for dynamic loading. Examples
include:
•
Drives where large masses are directly connected, or connected with short shafts,
to gear drives;
•
Gear drives connected with solid couplings;
•
Drives with fluid pumps connected directly to gears or gear shafts;
•
Long gear trains, such as those in printing presses, paper machines, and process
machinery;
•
Auxiliary drives with small gears connected to high power systems;
•
Drives with multiple power inputs;
•
High-speed drives that include servodrives;
•
Any gear drive where gears shown signs of distress or operate with high noise
for no apparent reason.
This procedure is the work of Eliot K. Buckingham, President, Buckingham As-
sociates, Inc., as reported in Machine Design magazine.
Nomenclature
A
ϭ
Load which deflects gear teeth by 0.001 in, lb (0.0254 mm, kg)
A
1

ϭ
Calculation factor for H
B
ϭ
Material mass factor, lb
⅐
s
2
/in
3
⅐
ft (kg
⅐
s
2
/cm
3
-m)
C
ϭ
Load which deflects gear teeth by an amount equal to the error in action, lb (kg)
C
1,2,3
...
r
ϭ
Load at the pitch radius which deflects a shaft by an amount equal to the error in
action, lb (kg)
D
ϭ

Cylinder diameter, in (mm)
D
s
ϭ
Shaft diameter, in (mm)
E
ϭ
Modulus of elasticity, lb / in
2
(kPa)
e
ϭ
Error in action, in (mm)
e
p
ϭ
Tooth profile error, in (mm)
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GEAR DESIGN AND APPLICATION
20.12
DESIGN ENGINEERING
e
s
ϭ
Tooth spacing error, in (mm)
ƒ
ϭ
Face width, in (mm)

ƒ
a
ϭ
Resultant force required to accelerate masses in a system of elastic bodies, lb (kg)
ƒ
1
ϭ
Force required to accelerate masses in a system of rigid bodies, lb (kg)
ƒ
2
ϭ
Force required to deflect elastically the system by the amount of error in action, lb
(kg)
H
ϭ
Calculation factor for ƒ
1
I
a
ϭ
Polar moment of inertia, in
2
⅐
lb
⅐
s
2
/ ft (cm
2
⅐

kg
⅐
s
2
/m)
L
ϭ
Cylinder length, in (mm)
L
s
ϭ
Shaft length, in (mm)
m
ϭ
Effective mass, lb
⅐
s
2
/ ft (kg
⅐
s
2
/m)
m
n
ϭ
Effective mass at the driven gear, lb
⅐
s
2

/ ft (kg
⅐
s
2
/m)
m
r
ϭ
Effective mass at the driver gear, lb
⅐
s
2
/ ft (kg
⅐
s
2
/m)
m
t
ϭ
Total effective mass at the mesh, lb
⅐
s
2
/ ft (kg
⅐
s
2
/m)
N

ϭ
Number of gear teeth
n
ϭ
Shaft speed, rpm
P
ϭ
Power, hp (kW)
R
ϭ
Pitch radius, in (mm)
T
ϭ
Torque, lb
⅐
in (N
⅐
m)
V
ϭ
Pitchline velocity, ft / min (m / min)
W
ϭ
Applied load at the pitch radius, lb (kg)
W
d
ϭ
Dynamic load, lb (kg)
W
i

ϭ
Incremental load, lb (kg)
y
ϭ
Lewis form factor
Z
ϭ
Calculation factor for C
Subscripts
A, B, C
ϭ
Shaft A, B, and C
1, 2, 3, 4
ϭ
Gear 1, 2, 3, and 4
a, b
ϭ
Mesh a and b
n, r
ϭ
Driven gear, driver gear
HELICAL-GEAR LAYOUT ANALYSIS
Helical gears are to be designed for shafts 4-in (10.16-cm) apart and at right angles,
Fig. 2. The normal dimeteral pitch, P
n
ϭ
20; the number of teeth, N, in the gear
or worm wheel
ϭ
64; the number of teeth, n, in the pinion or threads in the

worm
ϭ
32. Determine the suitable helix angle, or angles,
␣
, of the gear or worm.
Calculation Procedure:
1. Determine if the selected variables for this gearset are suitable
In the preliminary layout of gear drives, the designer must determine if the dia-
meteral pitch, gear ratio, and center distance are suitable for the proposed layout.
With this settled, the designer can proceed to calculate the specific helix angle.
To check a proposed layout, use the relation,
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GEAR DESIGN AND APPLICATION
GEAR DESIGN AND APPLICATION
20.13
FIGURE 2 Helical-gear layout.
(Product Engineering.)
2CP R 1
n
ϭϩ
N sin a cos a
where
C
ϭ
center distance, in (mm)
P
n
ϭ

normal diameteral pitch
N
ϭ
number of teeth in gear or worm wheel
n
ϭ
number of teeth in pinion or threads in worm
R
ϭ
ratio, n / N
a
ϭ
helix angle of gear or worm wheel, deg.
Find the value of the lefthand portion of this equation for this gear layout from:
2(4)(20)/64
ϭ
2.5. Then, R
ϭ
32/64
ϭ
0.5.
2. Find the suitable helix angle or angles
Knowing the value of the lefthand side of the equation and the value of R we can
assume suitable values for alpha and solve the equation. If the chosen helix angle
is correct, the two sides of the equation will be equal.
As a first choice, assume a helix angle of 20
Њ
. Substituting, using values from
standard trigonometric tables, we have 2.5
ϭ

(0.5/0.3584)
ϩ
(1/0.9336)
ϭ
2.466.
This helix angle is not suitable because the two sides of the equation are not equal.
Try 20
Њ
30
Ј
. Or, 2.5
ϭ
(0.5/0.3486)
ϩ
(1/0.9373)
ϭ
2.4968. This is much closer
than the first try. Another trial calculation shows that 20
Њ
25
Ј
is the correct helix
angle for this layout.
Another angle, namely 58
Њ
30
Ј
is also suitable, based on the equation above. For
greater efficiency and less wear, the gear and pinion helix angles should be nearly
equal. Thus, the gear angle chosen could be 58

Њ
30
Ј
and the pinion angle becomes
31
Њ
and 30
Ј
.
Related Calculations. Use this equation and general approach for 90 deg shaft
angles for helical gears in any service. Note that the trial helix angles chosen should
be in the general range for the type of drive being considered. This will save time
in the computation.
This procedure is the work of Wayne A. Ring, Barber-Colman Company, as
reported in Product Engineering magazine.
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GEAR DESIGN AND APPLICATION
20.14
DESIGN ENGINEERING
FIGURE 3 Epicyclic gear with drive and driven shafts rotating in same direction. (Product
Engineering.)
ANALYZING SHAFT SPEED IN EPICYCLIC GEAR
TRAINS
Analyze the input and output shaft rpm for the six epicyclic gear trains shown in
Figs. 3 through 8. The gear train in Fig. 3 has arm A integral with the righthand
shaft. Gears C and D are keyed to a short length of shaft which is mounted in a
bearing in arm A. Gear C meshes with internal gear E which is keyed to the lefthand
shaft.

Calculation Procedure:
1. Find the ratio of the shaft speeds
To find the ratio of the speed of shaft E to the speed of shaft A, proceed thus: Let
N
b
be the number of teeth in gear B, N
c
the number in gear C, and so on. Let arm
A, which was originally in a vertical position, be given an angular displace-
ment,
␪
.
In giving arm A the angular displacement, gear C will traverse through arc ab
on gear B. Since angles are inversely proportion to radii, or to the number of teeth,
gears C and D with turn through angle (
␪
)N
b
/N
c
.
While the foregoing was taking place, gears D and E were rotating on each other
through the equal arcs ed and ef. Gear E will have been turned in the reverse
direction through angle (
␪
)(N
b
/N
c
)(N

d
/N
e
).
The net effect of these two operations is to move the point of gear E, which
was originally vertical at g, over to location ƒ. Gear E has thus rotated through
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GEAR DESIGN AND APPLICATION
GEAR DESIGN AND APPLICATION
20.15
FIGURE 4 Drive and driven shafts rotate in opposite directions. (Product Engineering.)
FIGURE 5 Drive and driven shafts rotate in same direction. (Product Engineering.)
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GEAR DESIGN AND APPLICATION