Tài liệu Handbook of Mechanical Engineering Calculations P21 docx

21.1
SECTION 21
TRANSMISSIONS, CLUTCHES,
ROLLER-SCREW ACTUATORS,
COUPLINGS, AND SPEED
CONTROL
Constructing Mathematical Models for
Analyzing Hydrostatic Transmissions
21.1
Selecting a Clutch for a Given Load
21.12
Clutch Selection for Shaft Drive
21.13
Sizing Planetary Linear-Actuator Roller
Screws
21.16
Designing a Rolling-Contact Translation
Screw
21.21
Selection of a Rigid Flange-Type Shaft
Coupling
21.30
Selection of Flexible-Coupling for a
Shaft
21.32
Selection of a Shaft Coupling for Torque
and Thrust Loads
21.34
High-Speed Power-Coupling
Characteristics
21.35

Selection of Roller and Inverted-Tooth
(Silent) Chain Drives
21.38
Cam Clutch Selection and Analysis
21.42
Timing-Belt Drive Selection and Analysis
21.43
Geared Speed Reducer Selection and
Application
21.47
Power Transmission for a Variable-
Speed Drive
21.48
CONSTRUCTING MATHEMATICAL MODELS FOR
ANALYZING HYDROSTATIC TRANSMISSIONS
Construct a mathematical model of vehicle performance for a construction vehicle
powered by a hydrostatic transmission when the vehicle is driven by a 45-hp (33.6-
kW) engine at 2400 rpm, with a high idle-speed of 2600 rpm. The vehicle has a
supercharge pump rated at 2 hp (1.5 kW). Other vehicle data are: loaded radius, r
L
ϭ
14.5 in (36.8 cm); gross vehicle weight, W
g
ϭ
8500 lb (3825 kg); weight on
drive wheels, W
w
ϭ
5150 lb (2338 kg); final drive ratio, R
fd

ϭ
40:1; coefficient of
slip, C
s
ϭ
0.8; and coefficient of rolling resistance, C
r
ϭ
60 lb/1000 lb (27.2 kg/
454 kg) of gross vehicle weight. The vehicle is powered by a hydrostatic trans-
mission with a 2.5-in
3
/rev (41-mL/rev) displacement pump, rated at 5000 lb / in
2
(34.5-MPa). Compare the performance produced by using a 2.5-in
3
/rev (41-mL /
rev) displacement fixed-displacement motor and a 2.5-in
3
/rev (41-mL/rev) dis-
placement variable-displacement motor with an 11-degree displacement stop. Other
pump and motor data are given on performance curves available from the pump
manufacturer.
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Source: HANDBOOK OF MECHANICAL ENGINEERING CALCULATIONS
21.2
DESIGN ENGINEERING
Calculation Procedure:

1. Determine the vehicle speed at maximum tractive effort
The theoretical pump displacement required to absorb the input horsepower from
the engine, using the nomenclature at the end of this procedure, is:
396,000 H
p
D
ϭ
pt
PN
pp
Substituting,
396,000 (45
Ϫ
2)
D
ϭ
pt
5,000 (2,400)
3
ϭ
1.42 in /rev (23.3 mL / rev)
Next, find the horsepower-limited displacement from
D
ϭ
DE
ppttp
Substituting,
D
ϭ
1.42 (0.92)

p
3
ϭ
1.31 in /rev (21.5 mL / rev)
Now we must find the pump flow from
DNE
pp
v
p
Q
ϭ
p
231
Substituting,
1.31 (2,400) (0.88)
Q
ϭ
p
231
ϭ
12 gal/min (0.76 L/s)
Using the motor torque curve from the manufacturer for the pump being con-
sidered, similar to Fig. 1, these data give a motor torque, T
m
ϭ
1800 lb / in (203.3
Nm) at a motor speed of 960 rpm.
Maximum tractive effort is given by
TREn
m ƒd ƒdm

T
ϭ
emax
r
L
Substituting,
1,800 (40) (0.9)
T
ϭ
emax
14.5
ϭ
4,469 lb (2029 kg)
The vehicle speed at this tractive effort is given by
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TRANSMISSIONS, CLUTCHES, ROLLER-SCREW ACTUATORS, COUPLINGS, AND SPEED CONTROL
TRANSMISSIONS, CLUTCHES, ETC.
21.3
100
99
98
97
96
95
94
80
70
60

50
40
30
20
10
0
40
35
30
25
20
15
10
5
0
100
90
80
70
60
50
Volumetric efficiency, E
rv
(%)
Outlet flow, Q
p
(gpm)
Input horsepower, H
p
(hp)

Pump speed, N
p
(rpm)
(a)
Overall efficiency, E
pod
(%)
FIGURE 1 (a) Typical pump performance curves relate input horsepower, fluid outlet flow
rate, speed, and pressure to volumetric and overall efficiency. (b) Motor performance curves
relate output horsepower, fluid inlet flow rate, speed, and pressure to volumetric and overall
efficiency. (Machine Design.)
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1000
500
0
60
50
40
30
20
10
0
30
25
20
15
10

5
0
100
80
60
40
20
0
Output torque, T
m
(lb-in.)
Output horsepower (hp)
Inlet flow, Q
p
(gpm)
Overall efficiency, E
moa
(%)
Motor speed, N
m
(rpm)
(b)
FIGURE 1 Continued.
SI values for Fig. 1a and 1b:
gpm L / sec lb-in. Nm
00 0 0
5 0.32 500 56.4
10 0.63 1000 112.9
15 0.95
20 1.26 psi MPa

25 1.58 200 1.4
30 1.89 250 1.72
35 2.2 500 3.4
40 2.5 1000 6.89
2000 13.8
3000 20.7
4000 27.6
5000 34.5
1.5 cu in / rev (24.6 mL
/ revf)
2.5 cu in / rev (41 mL /
rev)
hp kW
00
10 7.46
20 14.9
30 22.4
40 29.8
50 37.3
60 44.8
70 52.2
80 59.7
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TRANSMISSIONS, CLUTCHES, ROLLER-SCREW ACTUATORS, COUPLINGS, AND SPEED CONTROL
TRANSMISSIONS, CLUTCHES, ETC.
21.5
FIGURE 2 Performance curve for vehicle analyzed in calculation procedure. (Ma-
chine Design.)

Fig. 2 SI
lb kg mph m / sec
00
1000 454 2 0.89
2000 908 4 1.78
3000 1362 6 2.68
4000 1816 8 3.58
5000 2270 10 4.47
Nr
mL
N
ϭ
v
168 RE
ƒd ƒd
Substituting,
960 (14.5)
N
ϭ
v
168 (40)
ϭ
2.1 mi/h (0.939 m/s)
Plot these computed values as point A on the tractive-effort vs. vehicle-speed
curve, Fig. 2.
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21.6

DESIGN ENGINEERING
2. Determine the tractive effort at the maximum vehicle speed
From the pump performance curve obtained from the manufacturer, the maximum
pump flow is 25.2 gal/min (1.59 L / s) at 2700 lb/in
2
18.6 (MPa) and 2.5 in
3
/rev
(41 mL / rev). From these data, the motor torque curves for the fixed-displacement
motor give N
m
ϭ
2240 rpm and T
m
ϭ
1000 lb /in (112.9 Nm).
The maximum vehicle speed produced by the fixed-displacement motor is
Nr
mL
N
ϭ
v
168 RE
ƒd ƒd
Substituting,
2,240 (14.5)
N
ϭ
v
168 (40)

ϭ
4.8 mi/h (2.15 m/s)
The tractive effort at this speed is
TREn
m ƒd ƒdm
T
ϭ
emax
r
L
Substituting,
1,000 (40) (0.9)
T
ϭ
e
14.5
ϭ
2,482 lb (1127 kg)
Plot these values as point B on Fig. 2.
From the curves for the variable-displacement motor, N
m
ϭ
3580 rpm and T
m
ϭ
560 lb/in (63.2 Nm). Therefore, as before, maximum vehicle speed produced by
the variable-displacement motor is
3,580 (14.5)
N
ϭ

v
168 (40)
ϭ
7.7 mi/h (3.44 m/s)
And the tractive effort, as before, is:
560 (40) (0.9)
T
ϭ
e
14.5
ϭ
1,390 lb (631 kg)
Plot these values as point C on Fig. 2.
3. Find intermediate points on the tractive-effort vs. vehicle-speed curve
To plot an intermediate point on the curve, Fig. 2, a pump flow of 21 gal/min
(1.325 L/s) is chosen arbitrarily. For the fixed-displacement motor, this flow gives
N
m
ϭ
1800 rpm and T
m
ϭ
1200 lb / in (135.5 Nm). Therefore, vehicle speed and
tractive effort are
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TRANSMISSIONS, CLUTCHES, ETC.
21.7

1,800 (14.5)
N
ϭ
v
168 (40)
ϭ
3.8 mi/h (1.698 m/s)
800 (40) (0.9)
T
ϭ
e
14.5
ϭ
2,979 lb (1352 kg)
which are plotted as point D on Fig. 2.
For the variable-speed motor, N
m
ϭ
2600 rpm and T
m
ϭ
800 lb / in (90.3 Nm).
Therefore, the vehicle speed and tractive effort are:
2,600 (14.5)
N
ϭ
v
168 (40)
ϭ
5.6 mi/h (2.5 m/s)

800 (40) (0.9)
T
ϭ
e
14.5
ϭ
1,986 lb (901.6 kg)
Plot these values as point E on Fig. 2.
4. Find the maximum theoretical speed for each motor type
The final point needed to construct the performance curve is the maximum theo-
retical speed of the vehicle for each type of motor. For the fixed-displacement
motor, maximum motor speed is given by
NDEE
pmax p
v
p
v
m
N
ϭ
mmax
nD
mm
Substituting,
2,600 (2.5) (0.95) (0.95)
N
ϭ
mmax
2.5
ϭ

2,346 rpm
The maximum theoretical vehicle speed is found from
Nr
mmax L
N
ϭ
v
max
168 R
jd
Substituting,
2,346 (14.5)
N
ϭ
v
max
168 (40)
ϭ
5.1 mi/h (2.5 m/s)
which is plotted as point F on Fig. 2.
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TRANSMISSIONS, CLUTCHES, ROLLER-SCREW ACTUATORS, COUPLINGS, AND SPEED CONTROL
21.8
DESIGN ENGINEERING
For the variable-displacement motor, maximum motor speed is
2,600 (2.5) (0.95) (0.95)
N
ϭ

mmax
1.5
ϭ
3,911 rpm
and the maximum theoretical vehicle speed is
3,911 (14.5)
N
ϭ
v
max
169 (40)
ϭ
8.4 mi/h (3.75 m/s)
which is plotted as point G on Fig. 2.
5. Refine the curve with rolling resistance and tractive effort at wheel slip
To refine the curve, rolling resistance, tractive effort at wheel slip, and gradability
must be determined. Rolling resistance is found from
R
ϭ
WC
rg
v
r
Substituting,
8,900 (60)
R
ϭ
r
1,000
ϭ

510 lb (231.5 kg)
Tractive effort at wheel slip is given by
T
ϭ
WC
ews
Single-path
T
ϭ
0.6 WC
ews
Dual-path
Substituting,
T
ϭ
5,150 (0.8)
e
ϭ
4,120 lb (1870 kg)
The gradability at slip is given by
T
Ϫ
R
er
Ϫ
1
G
ϭ
tan sin 100
ͫͩ ͪͬ

W
g
v
Substituting,
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TRANSMISSIONS, CLUTCHES, ETC.
21.9
4,120
Ϫ
510
Ϫ
1
G
ϭ
tan sin 100
ͫͩ ͪͬ
8,500
ϭ
47%
for the fixed-displacement motor.
The maximum gradability for the variable-displacement motor, limited by 5000
lb/in
2
(34.45 MPa), is
2,482
Ϫ
510

Ϫ
1
G
ϭ
tan sin 100
ͫͩ ͪͬ
8,500
ϭ
24%
These data are also shown on the tractive-effort curve, Fig. 2, which now gives a
complete picture of vehicle performance.
Related Calculations. The analytical technique presented here allows the hy-
drostatic transmission to be evaluated on paper, and necessary changes made before
the unit is actually built. The procedure uses a series of calculations that gradually
define transmission and vehicle data. With these data, a curve can be constructed,
Fig. 2, so that vehicle performance can be predicted for the entire operating range.
The first step in the analysis is to compare the ‘‘application values’’ of the vehicle
and the available transmissions. This comparison provides a simple way to match
vehicle requirements to transmission capabilities.
The vehicle application value expresses vehicle requirements and depends on
required vehicle speed and maximum tractive effort. For single-path applications
only one transmission is used. For dual-path applications, where two transmissions
are used, the transmission on each side of the vehicle must be treated as if it were
a single-path system. In such a case, a normal assumption is that 60 percent of the
total weight on the drive wheels transfers to one side of the vehicle when it ne-
gotiates a slope or turn. The transmission application value expresses transmission
capabilities and depends on motor torque and speed.
If the transmission application value is greater than that for the vehicle, the
proposed transmission is viable, and calculations to size properly the transmission
can be made. If the vehicle application value is greater, consideration must be given

to increasing pump speed, using variable-displacement motors, increasing pump
displacement, lowering maximum vehicle speed, or accepting a lower vehicle trac-
tive effort.
If a comparison of application values indicates that a proposed transmission is
adequate, a more refined procedure must be used to size the transmission. This
ensures that the transmission is applied within its horsepower rating, and that it
meets vehicle power requirements.
The input power available to the transmission is the net engine flywheel horse-
power (kW) at full-load governed speed, less the horsepower (kW) required for
supercharge and auxiliary no-load losses. For a transmission to operate satisfacto-
rily, input horsepower (kW) must not exceed its rated horsepower (kW). In dual-
path applications, power is split between two pumps and 80 percent of the total
input horsepower (kW) should be used in this calculation. Thus, up to 80 percent
of the total input horsepower (kW), is assumed to be directed to one pump.
Next, the transmission’s ability to produce the required tractive effort and pro-
pelling speed must be checked. For these calculations, the pump and motor per-
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21.10
DESIGN ENGINEERING
TABLE 1
Coefficients of Rolling Resistance
Surface Rubber tire kg /454 kg Crawler kg/454 kg
Concrete 10–20 4.5–9.1 40 18.2
Asphalt 12–22 5.5–9.9 40 18.2
Packed gravel 15–40 6.8–18.2 40 18.2
Soil 25–40 11.4–18.2 40 18.2
Mud 37–170 16.8–77.2 — —

Loose sand 60–150 27.2–68.1 100 45.4
Snow 25–50 9.1–22.7 — —
Units are lb / 1,000 lb (kg / 454 kg) gross vehicle weight.
formance curve for the proposed transmission must be available, usually from the
manufacturer.
The maximum tractive effort is limited by the pump relief-valve setting or wheel
slip. For multiple-path systems, maximum tractive effort must be divided by the
number of motors and multiplied by 0.6 before the comparison is made.
The final calculation required to determine whether a transmission meets vehicle
requirements is to check maximum vehicle speed. If the transmission can produce
the required tractive effort and speed, it is sized properly. However, if speed is too
low and tractive effort acceptable, consideration should be given to increasing pump
speed, using a variable-displacement motor, or decreasing the ratio of the final drive.
If speed is acceptable but tractive effort too low, give consideration to increasing
the final drive ratio. The resultant loss in maximum speed can be recovered by
increasing pump speed or by using a variable-displacement motor.
Once the transmission is sized to meet vehicle requirements, a mathematical
model can be generated to predict system performance. The calculations necessary
to produce the model take into account such factors as pump speed, pump and
motor displacement, and pump and motor efficiency. The expected vehicle perform-
ance is represented by a tractive-effort vs. speed curve.
The first two steps in generating the math model are to define the upper and
lower limits on the curve. The upper limit is the vehicle speed produced at the
maximum tractive effort; the lower limit is the tractive effort produced at maximum
vehicle speed.
Typically, four intermediate points on the performance curve are sufficient to
provide a rough approximation of vehicle performance. Six to eight points may be
required for a complete analysis. These points are calculated as shown here, except
that motor torque and speed are determined for pump displacements between max-
imum displacement and displacement at maximum tractive effort.

To complete the analysis, a number of factors must be calculated to determine
how they affect vehicle performance. One factor that must be considered is rolling
resistance, the portion of tractive effort required to overcome friction and move the
vehicle. For the vehicle to move, available tractive effort must be greater than the
rolling resistance. If the actual coefficient of rolling resistance is not known, the
values in Table 1 can be used.
Few vehicles operate only on level ground; so the slope or grade it can climb
must be determined. This factor, called gradability, is calculated as shown above.
Gradability can be determined for any point along the tractive-effort vs. speed curve,
up to the slip-limited effort. Gradability at wheel slip is usually the upper limit.
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TRANSMISSIONS, CLUTCHES, ETC.
21.11
TABLE 2
Coefficients of Slip
Surface Rubber tire Crawler
Concrete or asphalt 0.8–1.0 0.5
Dry clay 0.5–0.7 0.9
Sand & Gravel 0.3–0.6 0.4
Firm soil 0.5–0.6 0.9
Loose soil 0.4–0.5 0.6
The final factor to be considered is the tractive effort required to slip the wheels.
If the actual coefficient of slip is not known, the values in Table 2 can be used.
This procedure is valid for a variety of off-the-road vehicles: tractors, draggers,
bulldozers, rippers, scrapers, excavators, loaders, trenchers, hauling units, etc. It is
the work of Charles Griesel, Sperry Vickers, as reported in Machine Design Mag-
azine. SI values were added by the handbook editor.

Nomenclature
A
t
ϭ
Transmission application value
A
r
ϭ
Vehicle application value
C
r
ϭ
Coefficient of rolling resistance, lb /1,000 lb (kg / 454 kg)
C
s
ϭ
Coefficient of slip
D
m
ϭ
Motor displacement, in
3
/rev (mL/rev)
D
p
ϭ
Pump displacement in
3
/rev (mL/rev)
D

pt
ϭ
Theoretical pump displacement, in
3
/rev (mL/rev)
E
ƒd
ϭ
Final drive efficiency, %
E
poa
ϭ
Overall pump efficiency, %
E
tm
, E
tp
ϭ
Pump or motor torque efficiency, %
E
v
m
, E
v
p
ϭ
Pump or motor volumetric efficiency, %
G
ϭ
Gradability, %

H
p
ϭ
Pump input horsepower, hp (kW)
H
r
ϭ
Pump rated horsepower, hp (kW)
N
m
, N
p
ϭ
Motor or pump speed, rpm
N
mmax
ϭ
Maximum motor speed, rpm
N
pmax
ϭ
Maximum pump speed, rpm
N
pr
ϭ
Rated pump speed, rpm
N
v
ϭ
Vehicle speed, mi/ h (m / s)

N
v
max
ϭ
Maximum vehicle speed, mi/ h (m / s)
n
m
ϭ
Number of motors
P
p
ϭ
Pump pressure, lb/ in
2
(kPa)
Q
p
ϭ
Pump output flow, gal/min (L / s)
R
ƒd
ϭ
Final drive output ratio
R
r
ϭ
Rolling resistance, lb (kg)
r
L
ϭ

Loaded radius, in (cm)
T
e
ϭ
Tractive effort, lb (kg)
T
emax
ϭ
Maximum tractive effort, lb (kg)
T
m
ϭ
Motor torque, lb/in (Nm)
W
g
v
ϭ
Gross vehicle weight, lb (kg)
W
w
ϭ
Weight on drive wheels, lb (kg)
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21.12
DESIGN ENGINEERING
FIGURE 3 Basic friction clutch. Adjustable spring tension
holds the two friction surfaces together and sets the overload

limit. As soon as the overload is removed the clutch reengages.
(Product Engineering.)
SELECTING A CLUTCH FOR A GIVEN LOAD
Choose a clutch for a lathe designed for automatic operation. There will be no gear
shifting in the headstock. All speed changes will be made using hydraulically op-
erated clutches to connect the proper gear train to the output shaft. Determine the
number of plates and the operating force required for the clutch if it is to transmit
a torque of 300 lb / in (33.9 Nm) under normal operating conditions. Design the
clutch to slip under 300 percent of rated torque to protect the gears and other parts
of the drive. Space limitations dictate an upper limit of 4 in (10.2 cm) and a lower
limit of 2.5 in (6.35 cm) for the diameters of the friction surfaces. The clutch will
operate in an oil atmosphere.
Calculation Procedure:
1. Choose the type of clutch to use
Based on the proposed application, choose a wet clutch with hardened-steel plates.
(Since the clutch is operating in an oil atmosphere, use of a dry clutch could lead
to operational problems.)
2. Compute the number of friction plates needed for this clutch
Use the general clutch relation
D
ϩ
d
T
ϭ
N
␮
P
4
where T
ϭ

torque transmitted by clutch, lb / in (Nm); N
ϭ
number of friction plates
in the clutch;
␮
ϭ
coefficient of friction for the clutch; P
ϭ
total operating force
on the clutch, lb (kg); D
ϭ
maximum space limitation, in (cm); d
ϭ
minimum
space limitation, in (cm).
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TRANSMISSIONS, CLUTCHES, ETC.
21.13
Using a pressure valve, p, of 100 lb /in
2
(689 kPa) for long wear of this clutch
and its plates, the total operating force for this clutch will be P
ϭ
pA
ϭ
100
ϫ

␲
ϫ
[(D
2
Ϫ
d
2
)/4]
ϭ
100
ϫ
␲
ϫ
[(4
2
Ϫ
2.5
2
)/ 4)]
ϭ
766 lb (347.8 kg).
Substituting in the general clutch relation, with the torque at 300 percent of
normal operating conditions, or 3
ϫ
300 lb / in
ϭ
900 lb/ in (101.7 Nm), and a
coefficient of friction of 0.1, 900
ϭ
N

ϫ
0.1
ϫ
766
ϫ
[(4
ϩ
2.5)/ 4]. Solving for
N,wefindN
ϭ
7.23. The next larger even whole number of friction plates is 8.
Therefore, eight friction planes and nine plates will be specified.
3. Determine if the chosen number of plates is optimum for this clutch
Once we’ve chosen the number of plates we have the option of either reducing the
operating force, P, and thus the pressure on the plates, by the ratio 7.23 /8 or
keeping the pressure between the plates at 100 lb/ in
2
(689 kPa) and reducing the
outer diameter of the plates. Since space is important in the design of this clutch,
we will determine the outer diameter required when p
ϭ
100 lb/in
2
(689 kPa) and
N
ϭ
8.
Substituting pA
ϭ
p(

␲
)[D
2
Ϫ
d
2
)/4] for P in the general clutch relation gives
22
D
Ϫ
dD
ϩ
d
T
ϭ
N
␮
p
␲
44
22
D
Ϫ
2.5 D
ϩ
2.5
900
ϭ
8
ϫ

0.1
ϫ
100
44
Solving the resulting cubic equation for D,wefindD
ϭ
3.90 in (9.906 cm). Solving
for P,wefindP
ϭ
704 lb (319.6 kg).
Our specifications for this clutch will be: Plates 9 (eight friction planes), hard-
ened steel, outer diameter of friction surface
ϭ
3.90 in (9.906 cm); inner diameter
friction surface
ϭ
2.50 in (6.35 cm); operating force
ϭ
704 lb (319.6 kg).
Related Calculations. Use this general procedure to choose either wet or dry
clutches. The relations given here can be applied to either type of clutch. A wet
clutch is chosen wherever the atmosphere in the clutch operating area is such that
oil or moisture are present and cannot be conveniently removed. Using a wet clutch
saves the cost of seals and other devices needed to seal the clutch from the atmos-
pheric moisture.
Dry-plate clutches are used where there is no danger of oil or moisture getting
on the plates. Most such clutches use either natural or forced convection for cooling.
The drive material is a manmade composition in contact with cast iron, bronze, or
steel plates.
This procedure is the work of Richard M. Phelan, Associate Professor of Me-

chanical Engineering, Cornell University. SI values were added by the handbook
editor.
CLUTCH SELECTION FOR SHAFT DRIVE
Choose a clutch to connect a 50-hp (37.30 kW) internal-combustion engine to a
300-r/ min single-acting reciprocating pump. Determine the general dimensions of
the clutch.
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21.14
DESIGN ENGINEERING
TABLE 3
Clutch Characteristics
Calculation Procedure:
1. Choose the type of clutch for the load
Table 3 shows a typical applications for the major types of clutches. Where econ-
omy is the prime consideration, a positive-engagement or a cone-type friction clutch
would be chosen. Since a reciprocating pump runs at a slightly varying speed, a
centrifugal clutch is not suitable. For greater dependability, a disk or plate friction
clutch is more desirable than a cone clutch. Assume that dependability is more
important than economy, and choose a disk-type friction clutch.
2. Determine the required clutch torque at starting capacity
A clutch must start its load from a stopped condition. Under these circumstances
the instantaneous torque may be two, three, or four times the running torque. There-
fore, the usual clutch is chosen so it has a torque capacity of at least twice the
running torque. For internal-combustion engine drives, a starting torque of three to
four times the running torque is generally used. Assume 3.5 time is used for this
engine and pump combination. This is termed the clutch starting factor.
Since T

ϭ
63,000hp/R, where T
ϭ
torque, lb
⅐
in; hp
ϭ
horsepower transmitted;
R
ϭ
shaft rpm; T
ϭ
63,000(50)/ 300
ϭ
10,500 lb
⅐
in (1186.3 Nm). This is the
required starting torque capacity of the clutch.
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TRANSMISSIONS, CLUTCHES, ETC.
21.15
TABLE 4
Clutch Service Factors
3. Determine the total required clutch torque capacity
In addition to the clutch starting factor, a service factor is also usually applied.
Table 4 lists typical clutch service factors. This tabulation shows that the service
factor for a single-reciprocating pump is 2.0. Hence, the total required clutch torque

capacity
ϭ
required starting torque capacity
ϫ
service factor
ϭ
10,500
ϫ
2.0
ϭ
21,000-lb
⅐
in (2372.7 Nm) torque capacity.
4. Choose a suitable clutch for the load
Consult a manufacturer’s engineering data sheet listing clutch torque capacities for
clutches of the type chosen in step 1 of this procedure. Choose a clutch having a
rated torque equal to or greater than that computed in step 3. Table 5 shows a
portion of a typical engineering data sheet. A size 6 clutch would be chosen for
this drive.
Related Calculations. Use the general method given here to select clutches for
industrial, commercial, marine, automotive, tractor, and similar applications. Note
that engineering data sheets often list the clutch rating in terms of torque, lb
⅐
in,
and hp /(100 r/min).
Friction clutches depend, for their load-carrying ability, on the friction and pres-
sure between two mating surfaces. Usual coefficients of friction for friction clutches
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TRANSMISSIONS, CLUTCHES, ROLLER-SCREW ACTUATORS, COUPLINGS, AND SPEED CONTROL
21.16
DESIGN ENGINEERING
TABLE 5
Clutch Ratings
FIGURE 4 Recirculating roller screw which has high positioning accuracy is well-
suited for precise work, such as refocusing lenses for laser beams. (Machine Design.)
range between 0.15 and 0.50 for dry surfaces, 0.05 and 0.30 for greasy surfaces,
and 0.05 and 0.25 for lubricated surfaces. The allowable pressure between the
surfaces ranges from a low of 8 lb / in
2
(55.2 kPa) to a high of 300 lb/ in
2
(2068.5
kPa).
SIZING PLANETARY LINEAR-ACTUATOR ROLLER
SCREWS
A high-speed industrial robot requires a linear actuator with a 1.2-m (3.94-ft) stroke
to advance a load averaging 5700 N (1281 lb) at 20 m/min (65.6 ft/ min). The load
to reposition the arm is 1000 N (225 lb). Positioning should be within 1 mm
(0.03937 in). Find the mean load, expected life, life in million revolutions, and
maximum speed of a roller screw for this application. Suggest a type of lubrication,
estimate screw efficiency, and calculate the power required to drive the roller screw,
Fig. 4.
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TRANSMISSIONS, CLUTCHES, ETC.
21.17

Calculation Procedure:
1. Choose the type of screw to use
Roller screws are best suited for loads exceeding 100,000 N (22,482 lb) or speeds
over 6 m/min (19.7 ft /min). A roller screw without preload has a backlash of about
0.02 to 0.04 mm (0.00079 to 0.00157 in). This backlash adds to the thread inac-
curacy of the roller screw. Depending on the accuracy class, positioning is usually
within 0.1 to 0.25 mm (0.00384 to 0.0098 in) over a travel of 4000 mm (157.5 in).
Preloading the screw eliminates backlash, and overall positioning accuracy is not
significantly affected by a variable external load. Because more than 18 mm / min
(0.709 in) velocity is required, a roller screw will be used for this application, with
single support bearings at each end. The machine served by this actuator will make
300 load cycles /h, operate 16 h /day, 240 days / yr, and function for 5 years.
2. Determine the life expectancy of this roller screw actuator
Because length and load are not excessive, a medium-duty roller screw will be
chosen. To accommodate linear speed, shaft speed, and life, an initial selection of
a roller screw has a 44-mm (1.73-in) diameter, a 12-mm (0.472-in) lead, and a
108,200 N (24,326-lb) dynamic nut capacity, based on manufacturer’s catalog data.
Manufacturers often list the dynamic load of a nut or screw for an L
10
life of 1
million revolutions. Total life of the screw is:
l
L
ϭ
lttt
ͩͪ
rev hhdy
s
where L
ϭ

life in 10
6
revolutions; I
ϭ
stroke, mm (in); s
ϭ
screw lead, mm/ rev
(in/ rev); I
h
ϭ
strokes/ h; t
h
ϭ
operating hours /day; t
y
ϭ
years of service. Another
factor may be included to account for variations in load alignment, acceleration,
and lubrication.
Substituting for this roller screw,
1,200
ϫ
2
L
ϭ
(300) 16 (240) 5
rev
12
6
ϭ

1,152
ϫ
10 rev
3. Find the mean load on the screw
For the advance and retract loads, the mean load on the screw is found from:
33
F
ϩ
F
3
ar
F
ϭ
m
Ί
2
where F
a
ϭ
load during advance, N (lb); F
r
ϭ
load during retraction, N (lb).
Substituting,
33
5,700
ϩ
1,000
3
F

ϭ
m
Ί
2
ϭ
4,532 N (1019 lb)
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21.18
DESIGN ENGINEERING
4. Determine the L
10
life of this roller-screw actuator
The L
10
value for a roller screw is found from
3
C
6
L
ϭϫ
10 rev
ͩͪ
10
F
m
where L
10

ϭ
life corresponding to a 10 percent probability of failure, and C
ϭ
dynamic nut capacity, N (lb). Substituting,
3
108,000
6
L
ϭϫ
10
ͩͪ
10
4,532
9
ϭ
13.5
ϫ
10 rev
5. Find the rotational speed and type of lubrication needed
With a linear speed of 20 m/mm (65.6 ft/min), the rotational speed in rpm will be
l/ s, where I
ϭ
stroke length, mm (in); s
ϭ
screw lead, mm / rev (in/ rev). Substi-
tuting, rpm
ϭ
n
ϭ
(20 m

ϫ
1000 mm /m)/ 12)
ϭ
1667 rpm is required.
Knowing the rotational speed, we can compute the nD value, where D
ϭ
nom-
inal screw diameter, mm. Substituting, nD
ϭ
1667(44)
ϭ
73,348.
Lubrication type defines the roller screw speed limit. This limit is given by the
nD value computed above, or
v
D/s, where
v ϭ
linear nut speed, mm/s (in / s); other
symbols as before. Oil lubrication allows nD values as high as 140,000, while
grease permits nD values to 93,000. For rolled-thread ball screws, nD values are
about 64 percent of these. Since this roller screw has an nD value of 73,348, grease
lubrication is acceptable.
If lubrication of the roller screw is not regular or old lubricant is used, the life
figure can be modified by a factor of 0.5 to 0.66. Further, if the lubricant is likely
to be contaminated, an adjustment factor of 0.33 to 0.5 can be used.
6. Compute the maximum speed of the screw shaft
The maximum permissible speed of the screw shaft is 80 percent of the first critical
speed and is given by:
5
0.8(392) (10 ) ad

o
␻
ϭ
max
2
l
where
␻
max
ϭ
maximum permissible speed, rpm; a
ϭ
screw support factor from
Table 3; d
o
ϭ
screw shaft root diameter, mm (in); l
ϭ
distance between centers of
screw shaft support bearings, mm (in). Substituting,
5
0.8(392
ϫ
10 ) 2.47(42)
␻
ϭ
max
2
1,200
ϭ

2,259 rpm
7. Calculate the theoretical efficiency of this roller screw
Efficiency of converting rotary motion to linear motion is estimated with
s
e
ϭ
s
ϩ
Kd
o
where e
ϭ
efficiency; K
ϭ
friction angle factor, 0.0375 for heavy duty and 0.0325
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TRANSMISSIONS, CLUTCHES, ETC.
21.19
for medium-duty designs. At low loads, less than 10 percent of dynamic capacity,
e is within 10 percent of the calculated value. As load increases to the dynamic
capacity, efficiency estimates are less certain, usually within 25 percent of the cal-
culated value. Substituting,
12
e
ϭ
12
ϩ

0.0325(42)
ϭ
0.90
8. Find the input power to the roller screw
Input power to drive the load at a given speed is found from
Fs
␻
m
P
ϭ
60,000e
where P
ϭ
power, W (hp); other symbols as before. Substituting,
4,532(12) 1,667
P
ϭ
60,000 (0.9)
ϭ
1,679 W (2.25 hp)
9. Calculate the screw root diameter required to avoid buckling
Buckling of roller screws becomes a problem when the screw shaft is loaded in
compression. To avoid buckling, the screw root diameter must exceed the critical
or minimum screw diameter for the load, or
Fl
4
m
d
Ͼ
0

Ί
34,000b
where b
ϭ
screw support factor from Table 6. Substituting,
4,532(1,200)
4
d
Ͼ
0
Ί
34,000(1)
d
Ͼ
3.56 mm (0.14 in.)
0
The root diameter of the selected shaft is d
o
ϭ
44 mm (1.74 in). Since this is greater
than the 3.56 mm (0.14 in), the shaft is unlikely to buckle.
The limiting factor in this application appears to be the maximum shaft speed.
If the bearings are encased at one end of the shaft, allowing a
ϭ
3.85 and b
ϭ
2,
a 36-mm (1.41-in) diameter screw with 66,000 N (14,838 lb) dynamic capacity and
the same lead will also perform adequately.
Related Calculations. Roller screws are cost-effective alternatives to ball

screws in applications requiring high speed, long life, and high load capacity. The
load-bearing advantage of roller screws lies in their contact area. Ball screws trans-
fer load through point contact on the balls. Consequently, load is carried through
a discrete number of points. Roller screws, by contrast, transmit load through con-
tacts on each roller. Unlike a ball, contact on a roller can be ground precisely for
a duty requirement.
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21.20
DESIGN ENGINEERING
TABLE 6
Screw Speed and Compression Load Support Factors***
Support bearings on shaft Speed factor, a Compression factor, b
0.88 0.25
2.47 1
3.85 2
5.6 4
*= simply supported shaft end
**= encased end
***Machine Design
Roller screws have other advantages. Because the roller and nut threads have
the same helix angles, the rollers do not move axially as they roll inside the nut.
Hence, no recirculation is required. Further, a planetary gear in each housing end
turns the rollers during movement so that if something hinders rolling motion, they
are driven past the problem area. Rolling also minimizes the friction penalty of
area contact.
The absence of recirculation has the added benefit of allowing greater speeds
than ball screws. Because the rollers are in constant contact with the screw, as

opposed to being lifted and repositioned, roller screws can be driven at higher
rotational linear speeds than ball screws. Nonrecirculation also means the rollers
are not subjected to cyclic stressing. This further improves fatigue life.
Load is a good parameter for starting the sizing process for a roller screw. While
the load often fluctuates and reverses with each cycle, once a mean load is calcu-
lated, a unit can be selected by using the nut’s dynamic capacity. Constant mean
load is given by
33 3
FL
ϩ
FL
ϩ ⅐⅐⅐ ϩ
FL
3
11 22 nn
F
ϭ
m
Ί
L
ϩ
L
ϩ ⅐⅐⅐
L
12 n
where F
m
ϭ
constant mean load, N (lb); F
1

through F
n
ϭ
constant loads encountered
during operation, N (lb). These loads correspond to L
1
through L
n
, which are the
number of revolutions under a particular constant load.
If the loads are fairly constant as the screw advances or retracts, mean load is
found from
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