Tranmissions,Cluthes,Roller-Screw Actuators,CouplingsiAnd Speed Control

21.1

SECTION 21TRANSMISSIONS, CLUTCHES,ROLLER-SCREW ACTUATORS,

COUPLINGS, AND SPEEDCONTROL

Constructing Mathematical Models forAnalyzing Hydrostatic Transmissions21.1

Selecting a Clutch for a Given Load21.12

Clutch Selection for Shaft Drive 21.13

Sizing Planetary Linear-Actuator RollerScrews 21.16

Designing a Rolling-Contact TranslationScrew 21.21

Selection of a Rigid Flange-Type ShaftCoupling 21.30

Selection of Flexible-Coupling for aShaft 21.32

Selection of a Shaft Coupling for Torqueand Thrust Loads 21.34

High-Speed Power-CouplingCharacteristics 21.35

Selection of Roller and Inverted-Tooth(Silent) Chain Drives 21.38

Cam Clutch Selection and Analysis21.42

Timing-Belt Drive Selection and Analysis21.43

Geared Speed Reducer Selection andApplication 21.47

Power Transmission for a Variable-Speed Drive 21.48

CONSTRUCTING MATHEMATICAL MODELS FORANALYZING HYDROSTATIC TRANSMISSIONS

Construct a mathematical model of vehicle performance for a construction vehiclepowered 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 asupercharge pump rated at 2 hp (1.5 kW). Other vehicle data are: loaded radius, rL

� 14.5 in (36.8 cm); gross vehicle weight, Wg � 8500 lb (3825 kg); weight ondrive wheels, Ww � 5150 lb (2338 kg); final drive ratio, Rfd � 40:1; coefficient ofslip, Cs � 0.8; and coefficient of rolling resistance, Cr � 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-in3 / rev (41-mL/rev) displacement pump, rated at 5000 lb/ in2

(34.5-MPa). Compare the performance produced by using a 2.5-in3 / rev (41-mL/rev) displacement fixed-displacement motor and a 2.5-in3 / rev (41-mL/rev) dis-placement variable-displacement motor with an 11-degree displacement stop. Otherpump and motor data are given on performance curves available from the pumpmanufacturer.

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Source: HANDBOOK OF MECHANICAL ENGINEERING CALCULATIONS

21.2 DESIGN ENGINEERING

Calculation Procedure:

1. Determine the vehicle speed at maximum tractive effortThe theoretical pump displacement required to absorb the input horsepower fromthe engine, using the nomenclature at the end of this procedure, is:

396,000 HpD �pt P Np p

Substituting,

396,000 (45 � 2)D �pt 5,000 (2,400)

3� 1.42 in / rev (23.3 mL/rev)

Next, find the horsepower-limited displacement from

D � D Ep pt tp

Substituting,

D � 1.42 (0.92)p

3� 1.31 in / rev (21.5 mL/rev)

Now we must find the pump flow from

D N Ep p vpQ �p 231

Substituting,

1.31 (2,400) (0.88)Q �p 231

� 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, Tm � 1800 lb/ in (203.3Nm) at a motor speed of 960 rpm.

Maximum tractive effort is given by

T R E nm ƒd ƒd mT �emax rL

Substituting,

1,800 (40) (0.9)T �emax 14.5

� 4,469 lb (2029 kg)

The vehicle speed at this tractive effort is given by



TRANSMISSIONS, CLUTCHES, ROLLER-SCREW ACTUATORS, COUPLINGS, AND SPEED CONTROL

TRANSMISSIONS, CLUTCHES, ETC. 21.3

100

99

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umet

ric e

ffici

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, Erv

(%)

Out

let f

low

, Qp(

gpm

)

Inpu

t hor

sepo

wer

, Hp(

hp)

Pump speed, Np (rpm)

(a)

Ove

rall

effic

ienc

y, E

pod(

%)

FIGURE 1 (a) Typical pump performance curves relate input horsepower, fluid outlet flowrate, speed, and pressure to volumetric and overall efficiency. (b) Motor performance curvesrelate output horsepower, fluid inlet flow rate, speed, and pressure to volumetric and overallefficiency. (Machine Design.)




1000

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put t

orqu

e, T

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.)O

utpu

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sepo

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(hp

)

Inle

t flo

w, Q

p(gp

m)

Ove

rall

effic

ienc

y, E

moa

(%)

Motor speed, Nm (rpm)

(b)FIGURE 1 Continued.

SI values for Fig. 1a and 1b:gpm L/sec lb-in. Nm

0 0 0 05 0.32 500 56.410 0.63 1000 112.915 0.9520 1.26 psi MPa25 1.58 200 1.430 1.89 250 1.7235 2.2 500 3.440 2.5 1000 6.89

2000 13.83000 20.74000 27.65000 34.5

1.5 cu in / rev (24.6 mL/ revf)

2.5 cu in / rev (41 mL/rev)

hp kW0 010 7.4620 14.930 22.440 29.850 37.360 44.870 52.280 59.7





FIGURE 2 Performance curve for vehicle analyzed in calculation procedure. (Ma-chine Design.)

Fig. 2 SIlb kg mph m/sec

0 01000 454 2 0.892000 908 4 1.783000 1362 6 2.684000 1816 8 3.585000 2270 10 4.47

N rm LN �v 168 R Eƒd ƒd

Substituting,

960 (14.5)N �v 168 (40)

� 2.1 mi/h (0.939 m/s)

Plot these computed values as point A on the tractive-effort vs. vehicle-speedcurve, Fig. 2.





2. Determine the tractive effort at the maximum vehicle speedFrom the pump performance curve obtained from the manufacturer, the maximumpump flow is 25.2 gal /min (1.59 L/s) at 2700 lb/ in2 18.6 (MPa) and 2.5 in3 / rev(41 mL/rev). From these data, the motor torque curves for the fixed-displacementmotor give Nm � 2240 rpm and Tm � 1000 lb/ in (112.9 Nm).

The maximum vehicle speed produced by the fixed-displacement motor is

N rm LN �v 168 R Eƒd ƒd

Substituting,

2,240 (14.5)N �v 168 (40)

� 4.8 mi/h (2.15 m/s)

The tractive effort at this speed is

T R E nm ƒd ƒd mT �emax rL

Substituting,

1,000 (40) (0.9)T �e 14.5

� 2,482 lb (1127 kg)

Plot these values as point B on Fig. 2.From the curves for the variable-displacement motor, Nm � 3580 rpm and Tm

� 560 lb/ in (63.2 Nm). Therefore, as before, maximum vehicle speed produced bythe variable-displacement motor is

3,580 (14.5)N �v 168 (40)

� 7.7 mi/h (3.44 m/s)

And the tractive effort, as before, is:

560 (40) (0.9)T �e 14.5

� 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 curveTo 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 givesNm � 1800 rpm and Tm � 1200 lb/ in (135.5 Nm). Therefore, vehicle speed andtractive effort are





1,800 (14.5)N �v 168 (40)

� 3.8 mi/h (1.698 m/s)

800 (40) (0.9)T �e 14.5

� 2,979 lb (1352 kg)

which are plotted as point D on Fig. 2.For the variable-speed motor, Nm � 2600 rpm and Tm � 800 lb/ in (90.3 Nm).

Therefore, the vehicle speed and tractive effort are:

2,600 (14.5)N �v 168 (40)

� 5.6 mi/h (2.5 m/s)

800 (40) (0.9)T �e 14.5

� 1,986 lb (901.6 kg)

Plot these values as point E on Fig. 2.

4. Find the maximum theoretical speed for each motor typeThe 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-displacementmotor, maximum motor speed is given by

N D E Epmax p vp vmN �mmax n Dm m

Substituting,

2,600 (2.5) (0.95) (0.95)N �mmax 2.5

� 2,346 rpm

The maximum theoretical vehicle speed is found from

N rmmax LN �vmax 168 Rjd

Substituting,

2,346 (14.5)N �vmax 168 (40)

� 5.1 mi/h (2.5 m/s)

which is plotted as point F on Fig. 2.





For the variable-displacement motor, maximum motor speed is

2,600 (2.5) (0.95) (0.95)N �mmax 1.5

� 3,911 rpm

and the maximum theoretical vehicle speed is

3,911 (14.5)N �vmax 169 (40)

� 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 slipTo refine the curve, rolling resistance, tractive effort at wheel slip, and gradabilitymust be determined. Rolling resistance is found from

R � W Cr gv r

Substituting,

8,900 (60)R �r 1,000

� 510 lb (231.5 kg)

Tractive effort at wheel slip is given by

T � W Ce w s

Single-path

T � 0.6 W Ce w s

Dual-path

Substituting,

T � 5,150 (0.8)e

� 4,120 lb (1870 kg)

The gradability at slip is given by

T � Re r�1G � tan sin 100� � ��Wgv

Substituting,





4,120 � 510�1G � tan sin 100� � ��8,500

� 47%

for the fixed-displacement motor.The maximum gradability for the variable-displacement motor, limited by 5000

lb/ in2 (34.45 MPa), is

2,482 � 510�1G � tan sin 100� � ��8,500

� 24%

These data are also shown on the tractive-effort curve, Fig. 2, which now gives acomplete 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 beforethe unit is actually built. The procedure uses a series of calculations that graduallydefine 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 vehicleand the available transmissions. This comparison provides a simple way to matchvehicle requirements to transmission capabilities.

The vehicle application value expresses vehicle requirements and depends onrequired vehicle speed and maximum tractive effort. For single-path applicationsonly one transmission is used. For dual-path applications, where two transmissionsare used, the transmission on each side of the vehicle must be treated as if it werea single-path system. In such a case, a normal assumption is that 60 percent of thetotal 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 transmissioncapabilities and depends on motor torque and speed.

If the transmission application value is greater than that for the vehicle, theproposed transmission is viable, and calculations to size properly the transmissioncan be made. If the vehicle application value is greater, consideration must be givento increasing pump speed, using variable-displacement motors, increasing pumpdisplacement, lowering maximum vehicle speed, or accepting a lower vehicle trac-tive effort.

If a comparison of application values indicates that a proposed transmission isadequate, a more refined procedure must be used to size the transmission. Thisensures that the transmission is applied within its horsepower rating, and that itmeets 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 forsupercharge 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 totalinput horsepower (kW) should be used in this calculation. Thus, up to 80 percentof 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-





TABLE 1 Coefficients of Rolling Resistance

Surface Rubber tire kg /454 kg Crawler kg /454 kg

Concrete 10–20 4.5–9.1 40 18.2Asphalt 12–22 5.5–9.9 40 18.2Packed gravel 15–40 6.8–18.2 40 18.2Soil 25–40 11.4–18.2 40 18.2Mud 37–170 16.8–77.2 — —Loose sand 60–150 27.2–68.1 100 45.4Snow 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 themanufacturer.

The maximum tractive effort is limited by the pump relief-valve setting or wheelslip. For multiple-path systems, maximum tractive effort must be divided by thenumber of motors and multiplied by 0.6 before the comparison is made.

The final calculation required to determine whether a transmission meets vehiclerequirements is to check maximum vehicle speed. If the transmission can producethe required tractive effort and speed, it is sized properly. However, if speed is toolow and tractive effort acceptable, consideration should be given to increasing pumpspeed, 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 increasingthe final drive ratio. The resultant loss in maximum speed can be recovered byincreasing pump speed or by using a variable-displacement motor.

Once the transmission is sized to meet vehicle requirements, a mathematicalmodel can be generated to predict system performance. The calculations necessaryto produce the model take into account such factors as pump speed, pump andmotor 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 andlower limits on the curve. The upper limit is the vehicle speed produced at themaximum tractive effort; the lower limit is the tractive effort produced at maximumvehicle speed.

Typically, four intermediate points on the performance curve are sufficient toprovide a rough approximation of vehicle performance. Six to eight points may berequired for a complete analysis. These points are calculated as shown here, exceptthat 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 determinehow they affect vehicle performance. One factor that must be considered is rollingresistance, the portion of tractive effort required to overcome friction and move thevehicle. For the vehicle to move, available tractive effort must be greater than therolling resistance. If the actual coefficient of rolling resistance is not known, thevalues in Table 1 can be used.

Few vehicles operate only on level ground; so the slope or grade it can climbmust 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.





TABLE 2 Coefficients of Slip

Surface Rubber tire Crawler

Concrete or asphalt 0.8–1.0 0.5Dry clay 0.5–0.7 0.9Sand & Gravel 0.3–0.6 0.4Firm soil 0.5–0.6 0.9Loose 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 isthe work of Charles Griesel, Sperry Vickers, as reported in Machine Design Mag-azine. SI values were added by the handbook editor.

NomenclatureAt � Transmission application valueAr � Vehicle application valueCr � Coefficient of rolling resistance, lb /1,000 lb (kg/454 kg)Cs � Coefficient of slip

Dm � Motor displacement, in3 / rev (mL/rev)Dp � Pump displacement in3 / rev (mL/rev)Dpt � Theoretical pump displacement, in3 / rev (mL/rev)Eƒd � Final drive efficiency, %

Epoa � Overall pump efficiency, %Etm, Etp � Pump or motor torque efficiency, %

Evm, Evp � Pump or motor volumetric efficiency, %G � Gradability, %

Hp � Pump input horsepower, hp (kW)Hr � Pump rated horsepower, hp (kW)

Nm, Np � Motor or pump speed, rpmNmmax � Maximum motor speed, rpmNpmax � Maximum pump speed, rpm

Npr � Rated pump speed, rpmNv � Vehicle speed, mi/h (m/s)

Nvmax � Maximum vehicle speed, mi/h (m/s)nm � Number of motorsPp � Pump pressure, lb / in2 (kPa)Qp � Pump output flow, gal /min (L/s)Rƒd � Final drive output ratioRr � Rolling resistance, lb (kg)rL � Loaded radius, in (cm)Te � Tractive effort, lb (kg)

Temax � Maximum tractive effort, lb (kg)Tm � Motor torque, lb / in (Nm)

Wgv � Gross vehicle weight, lb (kg)Ww � Weight on drive wheels, lb (kg)





FIGURE 3 Basic friction clutch. Adjustable spring tensionholds the two friction surfaces together and sets the overloadlimit. 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 gearshifting 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 thenumber of plates and the operating force required for the clutch if it is to transmita torque of 300 lb/ in (33.9 Nm) under normal operating conditions. Design theclutch to slip under 300 percent of rated torque to protect the gears and other partsof the drive. Space limitations dictate an upper limit of 4 in (10.2 cm) and a lowerlimit of 2.5 in (6.35 cm) for the diameters of the friction surfaces. The clutch willoperate in an oil atmosphere.


1. Choose the type of clutch to useBased 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 leadto operational problems.)

2. Compute the number of friction plates needed for this clutchUse the general clutch relation

D � dT � N�P

4

where T � torque transmitted by clutch, lb / in (Nm); N � number of friction platesin the clutch; � � coefficient of friction for the clutch; P � total operating forceon the clutch, lb (kg); D � maximum space limitation, in (cm); d � minimumspace limitation, in (cm).





Using a pressure valve, p, of 100 lb/ in2 (689 kPa) for long wear of this clutchand its plates, the total operating force for this clutch will be P � pA � 100 � �� [(D2 � d2) /4] � 100 � � � [(42 � 2.52) /4)] � 766 lb (347.8 kg).

Substituting in the general clutch relation, with the torque at 300 percent ofnormal operating conditions, or 3 � 300 lb/ in � 900 lb/ in (101.7 Nm), and acoefficient of friction of 0.1, 900 � N � 0.1 � 766 � [(4 � 2.5) /4]. Solving forN, we find N � 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 clutchOnce we’ve chosen the number of plates we have the option of either reducing theoperating force, P, and thus the pressure on the plates, by the ratio 7.23/8 orkeeping the pressure between the plates at 100 lb/ in2 (689 kPa) and reducing theouter diameter of the plates. Since space is important in the design of this clutch,we will determine the outer diameter required when p � 100 lb/ in2 (689 kPa) andN � 8.

Substituting pA � p(�)[D2 � d 2) /4] for P in the general clutch relation gives

2 2D � d D � dT � N �p�

4 4

2 2D � 2.5 D � 2.5900 � 8 � 0.1 � 100

4 4

Solving the resulting cubic equation for D, we find D � 3.90 in (9.906 cm). Solvingfor P, we find P � 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 � 3.90 in (9.906 cm); inner diameterfriction surface � 2.50 in (6.35 cm); operating force � 704 lb (319.6 kg).

Related Calculations. Use this general procedure to choose either wet or dryclutches. The relations given here can be applied to either type of clutch. A wetclutch is chosen wherever the atmosphere in the clutch operating area is such thatoil or moisture are present and cannot be conveniently removed. Using a wet clutchsaves 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 gettingon 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, orsteel plates.

This procedure is the work of Richard M. Phelan, Associate Professor of Me-chanical Engineering, Cornell University. SI values were added by the handbookeditor.

CLUTCH SELECTION FOR SHAFT DRIVE

Choose a clutch to connect a 50-hp (37.30 kW) internal-combustion engine to a300-r /min single-acting reciprocating pump. Determine the general dimensions ofthe clutch.





TABLE 3 Clutch Characteristics


1. Choose the type of clutch for the loadTable 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 clutchwould be chosen. Since a reciprocating pump runs at a slightly varying speed, acentrifugal clutch is not suitable. For greater dependability, a disk or plate frictionclutch is more desirable than a cone clutch. Assume that dependability is moreimportant than economy, and choose a disk-type friction clutch.

2. Determine the required clutch torque at starting capacityA clutch must start its load from a stopped condition. Under these circumstancesthe 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 therunning torque. For internal-combustion engine drives, a starting torque of three tofour times the running torque is generally used. Assume 3.5 time is used for thisengine and pump combination. This is termed the clutch starting factor.

Since T � 63,000hp/R, where T � torque, lb � in; hp � horsepower transmitted;R � shaft rpm; T � 63,000(50) /300 � 10,500 lb � in (1186.3 Nm). This is therequired starting torque capacity of the clutch.





TABLE 4 Clutch Service Factors

3. Determine the total required clutch torque capacityIn 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 servicefactor for a single-reciprocating pump is 2.0. Hence, the total required clutch torquecapacity � required starting torque capacity � service factor � 10,500 � 2.0 �21,000-lb � in (2372.7 Nm) torque capacity.

4. Choose a suitable clutch for the loadConsult a manufacturer’s engineering data sheet listing clutch torque capacities forclutches of the type chosen in step 1 of this procedure. Choose a clutch having arated torque equal to or greater than that computed in step 3. Table 5 shows aportion of a typical engineering data sheet. A size 6 clutch would be chosen forthis drive.

Related Calculations. Use the general method given here to select clutches forindustrial, commercial, marine, automotive, tractor, and similar applications. Notethat engineering data sheets often list the clutch rating in terms of torque, lb � 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





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 thesurfaces ranges from a low of 8 lb/ in2 (55.2 kPa) to a high of 300 lb/ in2 (2068.5kPa).

SIZING PLANETARY LINEAR-ACTUATOR ROLLERSCREWS

A high-speed industrial robot requires a linear actuator with a 1.2-m (3.94-ft) stroketo advance a load averaging 5700 N (1281 lb) at 20 m/min (65.6 ft /min). The loadto 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, andmaximum 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.






1. Choose the type of screw to useRoller screws are best suited for loads exceeding 100,000 N (22,482 lb) or speedsover 6 m/min (19.7 ft /min). A roller screw without preload has a backlash of about0.02 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 usuallywithin 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 notsignificantly 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, withsingle support bearings at each end. The machine served by this actuator will make300 load cycles /h, operate 16 h/day, 240 days/yr, and function for 5 years.

2. Determine the life expectancy of this roller screw actuatorBecause length and load are not excessive, a medium-duty roller screw will bechosen. To accommodate linear speed, shaft speed, and life, an initial selection ofa roller screw has a 44-mm (1.73-in) diameter, a 12-mm (0.472-in) lead, and a108,200 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 L10 life of 1million revolutions. Total life of the screw is:

lL � l t t t� �rev h h d ys

where L � life in 106 revolutions; I � stroke, mm (in); s � screw lead, mm/rev(in/ rev); Ih � strokes/h; th � operating hours /day; ty � years of service. Anotherfactor may be included to account for variations in load alignment, acceleration,and lubrication.

Substituting for this roller screw,

1,200�2L � (300) 16 (240) 5rev 12

6� 1,152 � 10 rev

3. Find the mean load on the screwFor the advance and retract loads, the mean load on the screw is found from:

3 3F � F3 a rF �m � 2

where Fa � load during advance, N (lb); Fr � load during retraction, N (lb).Substituting,

3 35,700 � 1,0003F �m � 2

� 4,532 N (1019 lb)





4. Determine the L10 life of this roller-screw actuatorThe L10 value for a roller screw is found from

3C 6L � � 10 rev� �10 Fm

where L10 � life corresponding to a 10 percent probability of failure, and C �dynamic nut capacity, N (lb). Substituting,

3108,000 6L � � 10� �10 4,532

9� 13.5 � 10 rev

5. Find the rotational speed and type of lubrication neededWith a linear speed of 20 m/mm (65.6 ft /min), the rotational speed in rpm will bel /s, where I � stroke length, mm (in); s � screw lead, mm/rev (in/ rev). Substi-tuting, rpm � n � (20 m � 1000 mm/m)/12) � 1667 rpm is required.

Knowing the rotational speed, we can compute the nD value, where D � nom-inal screw diameter, mm. Substituting, nD � 1667(44) � 73,348.

Lubrication type defines the roller screw speed limit. This limit is given by thenD value computed above, or vD/s, where v � linear nut speed, mm/s (in /s); othersymbols as before. Oil lubrication allows nD values as high as 140,000, whilegrease permits nD values to 93,000. For rolled-thread ball screws, nD values areabout 64 percent of these. Since this roller screw has an nD value of 73,348, greaselubrication is acceptable.

If lubrication of the roller screw is not regular or old lubricant is used, the lifefigure can be modified by a factor of 0.5 to 0.66. Further, if the lubricant is likelyto be contaminated, an adjustment factor of 0.33 to 0.5 can be used.

6. Compute the maximum speed of the screw shaftThe maximum permissible speed of the screw shaft is 80 percent of the first criticalspeed and is given by:

50.8(392) (10 ) ado� �max 2l

where �max � maximum permissible speed, rpm; a � screw support factor fromTable 3; do � screw shaft root diameter, mm (in); l � distance between centers ofscrew shaft support bearings, mm (in). Substituting,

50.8(392�10 ) 2.47(42)� �max 21,200

� 2,259 rpm

7. Calculate the theoretical efficiency of this roller screwEfficiency of converting rotary motion to linear motion is estimated with

se �

s�Kdo

where e � efficiency; K � friction angle factor, 0.0375 for heavy duty and 0.0325





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 dynamiccapacity, efficiency estimates are less certain, usually within 25 percent of the cal-culated value. Substituting,

12e �

12�0.0325(42)

� 0.90

8. Find the input power to the roller screwInput power to drive the load at a given speed is found from

F s �mP �60,000e

where P � power, W (hp); other symbols as before. Substituting,

4,532(12) 1,667P �

60,000 (0.9)

� 1,679 W (2.25 hp)

9. Calculate the screw root diameter required to avoid bucklingBuckling of roller screws becomes a problem when the screw shaft is loaded incompression. To avoid buckling, the screw root diameter must exceed the criticalor minimum screw diameter for the load, or

F l4 md �0 �34,000b

where b � screw support factor from Table 6. Substituting,

4,532(1,200)4d �0 � 34,000(1)

d � 3.56 mm (0.14 in.)0

The root diameter of the selected shaft is do � 44 mm (1.74 in). Since this is greaterthan 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 � 3.85 and b � 2,a 36-mm (1.41-in) diameter screw with 66,000 N (14,838 lb) dynamic capacity andthe same lead will also perform adequately.

Related Calculations. Roller screws are cost-effective alternatives to ballscrews in applications requiring high speed, long life, and high load capacity. Theload-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 througha 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 fora duty requirement.





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 havethe 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 endturns the rollers during movement so that if something hinders rolling motion, theyare driven past the problem area. Rolling also minimizes the friction penalty ofarea contact.

The absence of recirculation has the added benefit of allowing greater speedsthan ball screws. Because the rollers are in constant contact with the screw, asopposed to being lifted and repositioned, roller screws can be driven at higherrotational linear speeds than ball screws. Nonrecirculation also means the rollersare not subjected to cyclic stressing. This further improves fatigue life.

Load is a good parameter for starting the sizing process for a roller screw. Whilethe 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 meanload is given by

3 3 3F L � F L � � � � � F L3 1 1 2 2 n nF �m � L � L � � � � L1 2 n

where Fm � constant mean load, N (lb); F1 through Fn � constant loads encounteredduring operation, N (lb). These loads correspond to L1 through Ln, which are thenumber of revolutions under a particular constant load.

If the loads are fairly constant as the screw advances or retracts, mean load isfound from





3 3F � F3 a rF �m � 2

where Fa � load during advance, N (lb); Fr � load during retraction N (lb).Lubrication of a roller screw is similar to that for a roller bearing. A circulating

lubrication system is preferred, especially in hot or dirty environments. Here, cooloil reduces frictional heat buildup during operation and flushes debris from thethreads. Oil should have a viscosity of 100 ISO or about SAE 30 during operation.With heavy load or at low speed, an EP additive is recommended to improve filmstrength.

Grease lubrication is acceptable when oil is impractical. Wipers should be usedon the nut to prevent dirty grease from entering the threads. The bearing greasesmost often used are NLGI 2. Lithium-base greases are generally suitable from �30to 110�C (�20�F to 230�F).

Alignment and shock can severely alter roller-screw bearing life. If the load isnot completely axial, the constant mean load, Fm, may be adjusted by a factor of1.05 to 1.1. If traverse loads are applied to the screw, Fm is adjusted by a factor of1.1 to 1.5.

If accelerations are low and speed variations in the application are smooth, noadjustment to the screw capacity factor is necessary. However, if the speed variesrapidly, or vibrations or high-frequency oscillations are present, an adjustment factorof 1.05 to 1.2 should be applied to Fm.

Roller screws find many applications including catapult reset after aircraft launchfrom aircraft carriers, automated arms for industrial presses, rudder control on largeaircraft, robot actuators, etc.

The data and calculation in this procedure are the work of Pierre C. Lemor,Manager, Linear Components, SKF Component Systems, as reported in MachineDesign magazine. SI units were added by the handbook editor.

DESIGNING A ROLLING-CONTACT TRANSLATIONSCREW

A rolling-contact screw-operated translation device is being designed to raise a10,000-lb (44,480-N) load a distance of 15 in (38.1 cm) as rapidly as possible. Theproposed design is shown schematically in Fig. 5. Modified square threads havebeen proposed for this power screw. The nut will be made of bronze; the screw ofAISI 3140 steel oil-quenched and tempered at 1000�F (537.8�C). All thrust andguide bearings will be rolling-contact types with negligible friction. Determine (a)the dimensions of the screw and nut for a factor of safety of 2; (b) the time requiredto raise the load; (c) the required horsepower (kW) of the electric drive motor.


1. Determine the required screw diameterConsider the screw as a column. Then, the J. B. Johnson formula applies. Checkingthread-property tables, it is found that a 1-in (2.54-cm) four-threads-per-inch (2.54cm) is the smallest modified square thread that will be satisfactory for this design.





FIGURE 5 Translation device to raise load.

However, the unsupported length of this screw must also carry the screw torque,and screw strength must be checked under the combined stress of the column andtorque loads.

2. Find the screw torque

Qd cos � tan � � �T � lb-in (Nm)

2 cos � � � tan �

where T � screw torque, lb-in (Nm); Q � screw load, lb (kg); d � screw pitchdiameter, in (mm); trigonometric functions are defined in Figs. 6 and 7; cos � �0.9962; � � 5 deg; tan � � l /� d � (1 /4) / (�)(0.875) � 0.0909 for single-threadscrew, � � 0.14 from Table 7, which gives the starting friction for high-gradematerials, superior-quality workmanship, and best running conditions. Thus,

10,000 � 0.875 0.9962 � 0.0909 � 0.14T � � 1,163 lb-in (131.4 Nm)

2 0.9962 � 0.14 � 0.0909

3. Check screw strength under combined stressMake the check by comparing the factor of safety under combined stress with thespecified factor of safety of 2. Or,

s /2yf.s. �2 2�(s /2) � (s )s

where sy � yield point, lb / in2 (kPa) from an AISI-3140 steel properties plot �132,000 lb/ in2 (909.5 MPa); s � equivalent normal stress for the column load, lb/ in2 (kPa); ss � shear stress produced by the torque, lb / in2 (kPa).




21.23

FIG

UR

E6

Thr

ead

geom

etri

cre

latio

ns,

and

colu

mn

end-

fixity

coef

ficie

nts.





FIGURE 7 Translation screw geometric relations.

To compute an equivalent normal stress for the column load we will use

Q 1S �equiv 2 2A [s (L/k) /4C� E]i s

with

Q � 10,000 lb (44,480 N)Ai � �di

2 /4 � � (0.750)2 /4 � 0.442 in2 (2.85 cm2)sy � 132,000 lb/ in2 (909.5 MPa) from a table of steel propertiesL � 15 in (38.1 cm)k � di /4 � 0.750/4 � 0.1875 in (0.48 cm)C � 3, from Fig. 6 on the basis that the screw is essentially a column with fixed

ends but that the normal clearance between the threads at the nut will give aslight decrease in the rigidity at that end

E � 30,000 lb/ in2 (206,700 MPa)

Thus,





TABLE 7 Coefficients of Friction for Screw Threads and Thrust Collars*

Steel screw and bronze or cast-iron nut Thrust-collar friction

Averagecoefficient

of friction, �

Averagecoefficient

of friction, �c

Conditions Starting Running Materials Starting Running

High-grade materialsand workmanshipand best runningcoditions . . . . . . . . . . . 0.14 0.10

Soft steel on cast iron . . 0.17 0.12

Average quality ofmaterials andworkmanship andaverage runningconditions . . . . . . . . . . 0.18 0.13

Hardened steel on castiron . . . . . . . . . . . . . . . 0.15 0.09

Poor workmanship orvery slow andinfrequent motionwith indifferentlubrication or newlymachined surfaces . . . 0.21 0.15

Soft steel on bronze . . .Hardened steel on

bronze . . . . . . . . . . . . .

0.10

0.08

0.08

0.06

*After C. W. Ham and D. G. Ryan, An Experimental Investigation of the Friction of Screw Threads, Univ.Illinois Eng. Expt. Sta. Bull. 247.

10,000 1s �equiv 2 20.442 1 � [132,000 � (15/0.1875 ] / (4 � 3 � � � 30,000,000)

2� 29,700 lb/ in (204.6 MPa)

To calculate the shear stress due to torque, we will use:

Tc ts � �s J J /c

and T � 1,163 lb-in.3 3J/c � �d /16 � [� � (0.750) ] /16 � 0.0828i

Thus,

1,163 2s � � 14,000 lb/ in (96.5 MPa)s 0.0828

Solving for the factor of safety we find:

132,000/2f.s. � � 3.23

2 2�(29,700/2) � (14,000)

Therefore, since 3.23 is greater than 2, the screw has more than adequate strengthas a column.





4. Determine the length of thread engagementThe specified length of thread engagement will be the greatest of the lengths re-quired for adequate shear strength of the screw threads, adequate shear strength ofthe nut threads, and satisfactory wear life. In this design, we are interested in raisingthe load as rapidly as possible, and the wear rate will probably be the most criticalfactor. The procedure we will follow will be to calculate the lengths of engagementrequired for adequate strength, then select a length at least equal to the larger ofthe two, and to determine the maximum operating speed of the nut for a reasonablewear rate.

The shear of the screw threads with a factor of safety of 2 and the shear stressdetermined earlier gives a length of engagement of

pQL � f.s.e,screw �d t s ,i i s screw

with

f.s. � 2p � 1⁄4 � 0.250 in (0.635 cm)Q � 10,000 lb (44,480 N)di � 0.750 in (1.91 cm)ti � tr � 0.1359 in (0.345 cm)ss � ssy � sy /2 � 132,000/2 � 66,000 lb/ in2 (454.7 MPa)

Thus,

0.250 � 10,000L � 2 � 0.237 in (0.60 cm)e,screw � � 0.750 � 0.1359 � 66,000

The shear of the nut threads is found from

pQL � f.s.e,nut �d t sp o s,nut

with all the values the same as for the screw except that the factor of safety � 4because the design will be based upon the ultimate shearing strength of the bronzenut. In the above relation, do � 1.00 in (2.54 cm); ss � ssu � 35,000 lb/ in2 (241MPa) from AISI 3140 steel property data. Thus,

0.250 � 10,000L � 4 � 0.669 in (1.699 cm)e,nut � � 1.000 � 0.1359 � 35,000

Therefore, any length of thread engagement equal to or greater than 0.669 in(1.699 cm) will result in adequate shear strength of the threads. Since wear is afunction of the product of the bearing pressure and sliding velocity, it is evidentthat the maximum speed of operation will be possible when the bearing pressureis a minimum. Hence, the length of engagement should be as large as practical.The load will not be uniformly distributed over the several threads, being carriedmostly by the first thread or two. Consequently, there is little to be gained by usinga length of engagement more than 1.0 or 1.5 times the screw diameter. Here, weshall use Le � 1.5 � 1 in (2.54 cm), calculate the bearing pressure, and see whatsliding speed will be permissible on the basis of values in Table 8, using




21.27

TA

BLE

8A

llow

able

Bea

ring

Pres

sure

sfo

rSc

rew

s

App

licat

ion

Mat

eria

l

Scre

wN

ut

Allo

wab

lebe

arin

gpr

essu

re,

psi

MPa

Slid

ing

spee

dat

thre

adpi

tch

diam

eter

m/m

in

Han

dpr

ess

Stee

lB

ronz

e2,

500–

3,50

017

.2–2

4.1

Low

spee

d,w

ell

lubr

icat

edJa

cksc

rew

Stee

lC

ast

iron

1,80

0–2,

500

12.4

–17.

2L

owsp

eed,

8fp

m2.

43Ja

cksc

rew

Stee

lB

ronz

e1,

600–

2.50

011

.0–1

7.2

Low

spee

d,10

fpm

3.04

Hoi

stin

gsc

rew

Stee

lC

ast

iron

600–

1,00

04.

1–6.

89M

ediu

msp

eed,

20–4

0fp

m6.

09–1

2.2

Hoi

stin

gsc

rew

Stee

lB

ronz

e80

0–1,

400

5.5–

9.65

Med

ium

spee

d,20

–40

fpm

6.06

–12.

2L

ead

scre

wSt

eel

Bro

nze

150–

240

1.03

–1.6

5H

igh

spee

d,50

fpm

15.2





4pQL � ine 2 2� (d � d )so i b

14 � ⁄4 � 10,0001.5 �Thus, 2 2� � (1.000 � 0.750 )sb

Solving for sb, we find

2s � 4,850 lb/ in (33.4 MPa)b

which is higher than recommended for even low-speed, well-lubricated operation.Therefore, a new approach to the design is required. Nevertheless, our efforts havenot been wasted entirely because we now know that any screw selected for thisapplication on the basis of wear will be more than adequately strong. The differenceis so great that, except for the increased resistance of the hardened surface to wear,there is little to be gained in using the heat-treated alloy steel. If conditions warrant,the material may be charged to cold-drawn low-carbon bar stock, 1020 or equiva-lent.

5. Perform any needed redesignThe design will now be based entirely on wear considerations. A reasonable com-bination of allowable bearing pressure and sliding speed is 800 lb/ in2 (5.5 MPa)at 40 ft /min (12.2 m/min) from Table 8. The size of screw that will give a bearingpressure of 800 lb/ in2 (5.5 MPa) or less, with a length of engagement of 1.5 timesthe nominal diameter, do, must be determined by trial and error as in the tablebelow.

do, in cm di, in* cm p*, in cm Le, in cm sb, psi MPa

2.000 5.08 1.5556 3.952 1 /2.25 1.1288 3.00 7.62 1194 8.2262.500 6.35 2.000 5.08 1 /2 1.27 3.75 9.525 755 5.2

*From a table of thread dimensions for translation screws

Hence, a 2.5-in (63.5-cm) two-threads-per-inch (25.4 mm) modified square threadwill be satisfactory.

6. Determine the time required to raise the loadThe pitch-line sliding velocity is:

�dnV � ft /min (m/min)

12

with these values: V � 40 ft /min (12.2 m/min); d � 2.50 � h � 2.50 � 0.250 �2.250 in (5.72 cm); n � rpm of nut. Solving for n, we find n � 67.9 rpm. Eachrevolution will raise the load a distance equal to the lead. If a single-thread screwis used, the lead will be 0.500 in (1.27 cm), and the lifting speed will be

V � nl 67.9 � 0.500 � 33.8 in/min (85.9 cm/min)load

and the time to raise the load 15 in (38.1 cm) will be:





S 15t � � � 0.444 min or 26.6 sec

V 33.8

If 26.6 s is too long, a double-thread screw will require only half the time, or 13.3sec, a triple-thread screw will required only 8.9 sec, etc.

7. Find the motor horsepower (kW) requirementsIf a lift time of 13.3 s for the double-thread screw is considered reasonable, thedrive horsepower can be determined in terms of torque and rotative speed from

Tnhp �

63,000

where T � torque, lb / in (Nm), and n � rpm.The torque calculated in the first part of the solution is no longer valid, because

both the lead and the diameter have been changed, and another calculation mustbe made. Since the motor will probably have a starting torque of 200 percent ofrated torque, there is considerable justification for determining the power require-ments under running conditions. However, there is an appreciable difference be-tween the starting and running coefficients of friction, 0.14 and 0.10, respectively,from Table 7. A slight change in lubrication conditions could increase the coeffi-cient of friction to the point where overheating of the motor might become a prob-lem. Thus, the power requirement will be based on the starting coefficient of fric-tion. The excess of the motor will produce a more rapid acceleration of the load.The required torque is found from

Qd cos � tan � � �T � lb / in (Nm)

2 cos � � � tan �

where Q � 10,000 lb (44,480 N); D � 2.25 in (6.35 cm); cos � � 0.9962; tan �� l /� (d) � (2 � 1/2) / (� � 2.25) � 0.1415; � � 0.14. Substituting,

10,000 � 2.250 0.9962 � 0.1415 � 0.14T � � 3,960 lb/ in (447.4 Nm)

2 0.9962 � 0.14 � 0.1415

Therefore, if the relatively small power losses in the thrust bearings, the guidebearing, and the gear train between the motor and the nut are neglected, the powerrequired is

3,960 � 67.9hp � � 4.27 (3.19 kW)

63,000

Thus, a 5-hp (3.2-kW) motor would be specified. Note that for the double-threadscrew, � � 0.14 and cos � � tan � � 0.1409. Hence, the screws is not self-locking.The guide bearings, gears, etc., might possibly provide enough additional frictionto hold the load at any point under starting or static friction. However, if for anyreason, such as pronounced vibration, the friction coefficient decreases to the run-ning value, the load would rapidly lower itself. The solution is to use either theself-locking single-thread screw with its slower motion (and smaller power require-ment) or to use a brake of some type, such as that in Fig. 8. In this design we will





FIGURE 8 Brake for translation screw.

assume that the saving in lift time will justify the additional complication of addinga brake.

Our specifications for this actuator will therefore be: Screw and nut: double-thread, modified-square-thread, 2.5-in (6.35-cm)-diameter, four-threads-per-inch(2.54-cm), low-carbon-steel screw and bronze nut with length of engagement (nutheight) of 3.75 in (9.52 cm). Operation time: 13.3 s (neglecting acceleration anddeceleration). Motor: 5 hp (3.19 kW).

Related Calculations. Use this general procedure to design ball and roller bear-ing translation screws. Note that various adjustments must be made to initial as-sumptions as the design work proceeds.

This procedure is the work of Richard M. Phelan, Associate Professor of Me-chanical Engineering, Cornell University. SI vales were added by the handbookeditor.

SELECTION OF A RIGID FLANGE-TYPE SHAFTCOUPLING

Choose a steel flange-type coupling to transmit a torque of 15,000 lb � in (1694.4Nm) between two 21⁄2-in (6.4-cm) diameter steel shafts. The load is uniform andfree of shocks. Determine how many bolts are needed in the coupling if the allow-able bolt shear stress is 3000 lb/ in2 (20,685.0 kPa). How thick must the couplingflange be, and how long should the coupling hub be if the allowable stress bearingfor the hub is 20,000 lb/ in2 (137,900.0 kPa) and in shear 6000 lb/ in2 (41,370.0kPa)? The allowable shear stress in the key is 12,000 lb/ in2 (82,740.0 kPa). Thereis no thrust force acting on the coupling.


1. Choose the diameter of the coupling bolt circleAssume a bolt-circle diameter for the coupling. As a first choice, assume the bolt-circle diameter is three times the shaft diameter, or 3 � 2.5 � 7.5 in (19.1 cm).This is a reasonable first assumption for most commercially available couplings.

1John H. Glover, ‘‘Planetary Gear Systems,’’ Product Engineering, Jan. 6, 1964.





2. Compute the shear force acting at the bolt circleThe shear force Fs lb acting at the bolt-circle radius rb in is Fs � T/rb, where T �torque on shaft, lb � in. Or, Fs � 15,000/(7.5/2) � 4000 lb (17,792.9 N).

3. Determine the number of coupling bolts neededWhen the allowable shear stress in the bolts is known, compute the number of boltsN required from N � 8Fs / (�d2ss), where d � diameter of each coupling bolt, in;ss � allowable shear stress in coupling bolts, lb / in2.

The usual bolt diameter in flanged, rigid couplings ranges from 1⁄4 to 2 in (0.6to 5.1 cm), depending on the torque transmitted. Assuming that 1⁄2-in (1.3-cm)diameter bolts are used in this coupling, we see that N � 8(4000) / [� (0.5)2(3000)]� 13.58, say 14 bolts.

Most flanged, rigid couplings have two to eight bolts, depending on the torquetransmitted. A coupling having 14 bolts would be a poor design. To reduce thenumber of bolts, assume a larger diameter, say 0.75 in (1.9 cm). Then N �8(4000) / [� (0.75)2(3000)] � 6.03, say eight bolts, because an odd number of boltsare seldom used in flanged couplings.

Determine the shear stress in the bolts by solving the above equation for ss �8Fs / (�d2N) � 8(4000) / [� (0.75)2(8)] � 2265 lb/ in2 (15,617.2 kPa). Thus, the boltsare not overstressed, because the allowable stress is 3000 lb/ in2 (20,685.0 kPa).

4. Compute the coupling flange thickness requiredThe flange thickness t in for an allowable bearing stress sb lb / in2 is t � 2Fs / (Ndsb)� 2(4000) / [(8)(0.75)(20,000)] � 0.0666 in (0.169 cm). This thickness is much lessthan the usual thickness used for flanged couplings manufactured for off-the-shelfuse.

5. Determine the hub length requiredThe hub length is a function of the key length required. Assuming a 3⁄4-in (1.9-cm)square key, compute the hub length l in from l � 2Fss / (tkst), where Fss � forceacting at shaft outer surface, lb; tk � thickness, in. The force Fss � T/rh, where rh

� inside radius of hub, in � shaft radius � 2.5/2 � 1.25 in (3.2 cm) for this shaft.Then Fss � 15,000/1.25 � 12,000 lb � in (1355.8 Nm). Then l � 2(12,000) /[(0.75)(20,000)] � 1.6 in (4.1 cm).

When the allowable design stress for bearing, 20,000 lb/ in2 (137,895.1 kPa)here, is less than half the allowable design stress for shear, 12,000 lb/ in2 (82,740.0kPa) here, the longest key length is obtained when the bearing stress is used. Thus,it is not necessary to compute the thickness needed to resist the shear stress forthis coupling. If it is necessary to compute this thickness, find the force acting atthe surface of the coupling hub from Fh � T/rh, where rh � hub radius, in. Thents � Fh /�dhss, where dh � hub diameter, in; ss � allowable hub shear stress, lb /in2.

Related Calculations. Couplings offered as standard parts by manufacturersare usually of sufficient thickness to prevent fatigue failure.

Since each half of the coupling transmits the total torque acting, the length ofthe key must be the same in each coupling half. The hub diameter of the couplingis usually 2 to 2.5 times the shaft diameter, and the coupling is generally made thesame thickness as the coupling flange. The procedure given here can be used forcouplings made of any metallic material.





TABLE 9 Allowable Flexible Coupling Misalignment

SELECTION OF FLEXIBLE COUPLING FOR ASHAFT

Choose a stock flexible coupling to transmit 15 hp (11.2 kW) from a 1000-r /minfour-cylinder gasoline engine to a dewatering pump turning at the same rpm. Thepump runs 8 h/day and is an uneven load because debris may enter the pump. Thepump and motor shafts are each 1.0 in (2.5 cm) in diameter. Maximum misalign-ment of the shafts will not exceed 0.5�. There is no thrust force acting on thecoupling, but the end float or play may reach 1⁄16 in (0.2 cm).


1. Choose the type of coupling to useConsult Table 9 or the engineering data published by several coupling manufactur-ers. Make a tentative choice from Table 9 of the type of coupling to use, based onthe maximum misalignment expected and the tabulated end-float capacity of thecoupling. Thus, a roller-chain-type coupling (one in which the two flanges areconnected by a double roller chain) will be chosen from Table 9 for this drivebecause it can accommodate 0.5� of misalignment and an end float of up to 1⁄16 in(0.2 cm).

2. Choose a suitable service factorTable 10 lists typical service factors for roller-chain-type flexible couplings. Thus,for a four-cylinder gasoline engine driving an uneven load, the service factor SF �2.5.

3. Apply the service factor chosenMultiply the horsepower or torque to be transmitted by the service factor to obtainthe coupling design horsepower or torque. Or, coupling design hp � (15)(2.5) �37.5 hp (28.0 kW).

4. Select the coupling to useRefer to the coupling design horsepower rating table in the manufacturer’s engi-neering data. Enter the table at the shaft rpm, and project to a design horsepowerslightly greater than the value computed in step 3. Thus, in Table 11 a typical ratingtabulation shows that a coupling design horsepower rating of 38.3 hp (28.6 kW) isthe next higher value above 37.5 hp (28.0 kW).





TABLE 10 Flexible Coupling Service Factors*

TABLE 11 Flexible Coupling hp Ratings*

5. Determine whether the coupling bore is suitableTable 11 shows that a coupling suitable for 38.3 hp (28.6 kW) will have a maximumbore diameter up to 1.75 in (4.4 cm) and a minimum bore diameter of 0.625 in(1.6 cm). Since the engine and pump shafts are each 1.0 in (2.5 cm) in diameter,the coupling is suitable.

The usual engineering data available from manufactures include the stock key-way sizes, coupling weight, and principal dimensions of the coupling. Check theoverall dimensions of the coupling to determine whether the coupling will fit theavailable space. Where the coupling bore diameter is too small to fit the shaft,choose the next larger coupling. If the dimensions of the coupling make it unsuit-able for the available space, choose a different type or make a coupling.

Related Calculations. Use the general procedure given here to select any typeof flexible coupling using flanges, springs, roller chain, preloaded biscuits, etc., totransmit torque. Be certain to apply the service factor recommended by the man-ufacturer. Note that biscuit-type couplings are rated in hp/100 r /min. Thus, a bis-cuit-type coupling rated at 1.60 hp/100 r /min (1.2 kW/100 r/min) and a maximumallowable speed of 4800 r /min could transmit a maximum of (1.80 hp)(4800/100)� 76.8 hp (57.3 kW).





FIGURE 9 Shaft-coupling characteristics.

SELECTION OF A SHAFT COUPLING FORTORQUE AND THRUST LOADS

Select a shaft coupling to transmit 500 hp (372.9 kW) and a thrust of 12,500 lb(55,602.8 N) at 100 r /min from a six-cylinder diesel engine. The load is an evenone, free of shock.


1. Compute the torque acting on the couplingUse the relation T � 5252hp/R to determine the torque, where T � torque actingon coupling, lb � ft; hp � horsepower transmitted by the coupling; R � shaft rotativespeed, r /min. For this coupling, T � (5252)(500) /100 � 26,260 lb � ft (35,603.8Nm).

2. Find the service torqueMultiply the torque T by the appropriate service factor from Table 10. This tableshows that a service factor of 1.5 is suitable for an even load, free of shock. Thus,the service torque � (26,260 lb � ft)(1.5) � 39,390 lb � ft, say 39,500 lb � ft (53,554.8Nm).

3. Choose a suitable couplingEnter Fig. 9 at the torque on the left, and project horizontally to the right. Usingthe known thrust, 12,500 lb (55,602.8 N), enter Fig. 9 at the bottom and projectvertically upward until the torque line is intersected. Choose the coupling modelrepresented by the next higher curve. This shows that a type A coupling having amaximum allowable speed of 300 r /min will be suitable. If the plotted maximumrpm is lower than the actual rpm of the coupling, use the next plotted couplingtype rated for the actual, or higher rpm.

In choosing a specific coupling, use the manufacturer’s engineering data. Thiswill resemble Fig. 9 or will be tabulation of the ranges plotted.





Related Calculations. Use this procedure to select couplings for industrial andmarine drives where both torque and thrust must be accommodated. See the MarineEngineering section of this handbook for an accurate way to compute the thrustproduced on a coupling by a marine propeller. Always check to see that the cou-pling bore is large enough to accommodate the connected shafts. Where the boreis too small, use the next larger coupling.

HIGH-SPEED POWER-COUPLINGCHARACTERISTICS

Select the type of power coupling to transmit 50 hp (37.3 kW) at 200 r /min if theangular misalignment varies from a minimum of 0 to a maximum of 45�. Determinethe effect of angular misalignment on the shaft position, speed, and acceleration atangular misalignments of 30 and 45�.


1. Determine the type of coupling to useTable 12, developed by N. B. Rothfuss, lists the operating characteristics of eighttypes of high-speed couplings. Study of this table shows that a universal joint isthe only type of coupling among those listed that can handle an angular misalign-ment of 45�. Further study shows that a universal coupling has a suitable speed andhp range for the load being considered. The other items tabulated are not factorsin this application. Therefore, a universal coupling will be suitable. Table 13 com-pares the functional characteristics of the couplings. Data shown support the choiceof the universal joint.

2. Determine the shaft position errorTable 14, developed by David A. Lee, shows the output variations caused by mis-alignment between the shafts. Thus, at 30� angular misalignment, the position erroris 4�06�42�. This means that the output shaft position shifts from �4�06�42� to�4�06�42� twice each revolution. At a 45� misalignment the position error, Table14, is 9�52�26�. The shift in position is similar to that occurring at 30� angularmisalignment.

3. Compute the output-shaft speed variationTable 14 shows that at 30� angular misalignment the output-shaft speed variationis �15.47 percent. Thus, the output-shaft speed varies between 200(1.00 � 0.1547)� 169.06 and 230.94 r /min. This speed variation also occurs twice per revolution.

For a 45� angular misalignment the speed variation, determined in the same way,is 117.16 to 282.84 r /min. This speed variation also occurs twice per revolution.

4. Determine output-shaft accelerationTable 14 lists the ratio of maximum output-shaft acceleration A to the square ofthe input speed, �2, expressed in radians. To convert r /min to rad/s, use rps �0.1047 r /min � 0.0147(200) � 20.94 rad/s.

For 30� angular misalignment, from Table 14 A /�2 � 0.294571. Thus, A ��2(0.294571) � (20.94)2(0.294571) � 129.6 rad/s2. This means that a constant




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TABLE 14 Universal Joint Output Variations*

input speed of 200 r /min produces an output acceleration ranging from � 129.6to � 129.6 rad/s2, and back, at a frequency of 2(200 r /min) � 400 cycles /min.

At a 45� angular misalignment, the acceleration range of the output shaft, de-termined in the same way, is �346 to �346 rad/s2. Thus, the acceleration rangeat the larger shaft angle misalignment is 2.67 times that at the smaller, 30�, misa-lignment.

Related Calculations. Table 12 is useful for choosing any of seven other typesof high-speed couplings. The eight couplings listed in this table are popular forhigh-horsepower applications. All are classed as rigid types, as distinguished fromentirely flexible connectors such as flexible cables.

Values listed in Table 12 are nominal ones that may be exceeded by specialdesigns. These values are guideposts rather than fixed; in borderline cases, consultthe manufacturer’s engineering data. Table 13 compares the functional character-istics of the couplings and is useful to the designer who is seeking a unit withspecific operating characteristics. Note that the values in Table 12 are maximumand not additive. In other words, a coupling cannot be operated at the maximumangular and parallel misalignment and at the maximum horsepower and speedsimultaneously—although in some cases the combination of maximum angular mis-alignment, maximum horsepower, and maximum speed would be acceptable. Whereshock loads are anticipated, apply a suitable correction factor, as given in earliercalculation procedures, to the horsepower to be transmitted before entering Table12.

SELECTION OF ROLLER AND INVERTED-TOOTH(SILENT) CHAIN DRIVES

Choose a roller chain and the sprockets to transmit 6 hp (4.5 kW) from an electricmotor to a propeller fan. The speed of the motor shaft is 1800 r /min and of thedriven shaft 900 r /min. How long will the chain be if the centerline distance be-tween the shafts is 30 in (76.2 cm)?





TABLE 15 Roller Chain Loads and Service Factors*


1. Determine, and apply, the load service factorConsult the manufacturer’s engineering data for the appropriate load service factor.Table 15 shows several typical load ratings (smooth, moderate shock, heavy shock)for various types of driven devices. Use the load rating and the type of drive todetermine the service factor. Thus, a propeller fan is rated as a heavy shock load.For this type of load and an electric-motor drive, the load service factor is 1.5,from Table 15.

Apply the load service factor by taking the product of it and the horsepowertransmitted, or (1.5)(6 hp) � 9.0 hp (6.7 kW). The roller chain and sprockets musthave enough strength to transmit this horsepower.

2. Choose the chain and number of teeth in the small sprocketUsing the manufacturer’s engineering data, enter the horsepower rating table at thesmall-sprocket rpm and project to a horsepower value equal to, or slightly greaterthan, the required rating. At this horsepower rating, read the number of teeth in thesmall sprocket, which is also listed in the table. Thus, in Table 16, which is anexcerpt from a typical horsepower rating tabulation, 0.9 hp (6.7 kW) is not listedat a speed of 1800 r /min. However, the next higher horsepower rating, 9.79 hp





TABLE 16 Roller Chain Power Rating*

(7.3 kW), will be satisfactory. The table shows that at this power rating, 16 teethare used in the small sprocket.

This sprocket is a good choice because most manufacturers recommend that atleast 16 teeth be used in the smaller sprocket, except at low speeds (100 to 500 r/min).

3. Determine the chain pitch and number of strandsEach horsepower rating table is prepared for a given chain pitch, number of chainstrands, and various types of lubrication. Thus, Table 16 is for standard single-strand 5⁄8-in (1.6-cm) pitch roller chain. The 9.79-hp (7.3-kW) rating at 1800 r /minfor this chain is with type III lubrication—oil bath or oil slinger—with the oil levelmaintained in the chain casing at a predetermined height. See the manufacturer’sengineering data for the other types of lubrication (manual, drip, and oil stream)requirements.

4. Compute the drive speed ratioFor a roller chain drive, the speed ration Sr � Rh /Rl, where Rh � rpm of high-speed shaft; Rl � rpm of low-speed shaft. For this drive, Sr � 1800/900 � 2.

5. Determine the number of teeth in the large sprocketTo find the number of teeth in the large sprocket, multiply the number of teeth inthe small sprocket, found in step 2, by the speed ratio, found in step 4. Thus, thenumber of teeth in the large sprocket � (16)(2) � 32.

6. Select the sprocketsRefer to the manufacturer’s engineering data for the dimensions of the availablesprockets. Thus, one manufacturer supplies the following sprockets for 5⁄8-in (1.6-cm) pitch single-strand roller chain: 16 teeth, OD � 3.517 in (8.9 cm), bore � 5⁄8in (1.6 cm); 32 teeth, OD � 6.721 in (17.1 cm), bore � 5⁄8 or 3⁄4 in (1.6 or 1.9cm). When choosing a sprocket, be certain to refer to data for the size and type ofchain selected in step 3, because each sprocket is made for a specific type of chain.Choose the type of hub—setscrew, keyed, or taper-lock bushing—based on thetorque that must be transmitted by the drive. See earlier calculation procedures inthis section for data on key selection.





TABLE 17 Roller Chain Length Factors*

7. Determine the length of the chainCompute the chain length in pitches Lp from Lp � 2C � (S /2) � K/C, where C� shaft center distance, in /chain pitch, in; S � sum of the number of teeth in thesmall and large sprocket; K � a constant from Table 17, obtained by entering thistable with the value D � number of teeth in large sprocket � number of teeth insmall sprocket. For this drive, C � 30/0.625 � 48; S � 16 � 32 � 48; D � 32� 16 � 16; K � 6.48 from Table 17. Then, Lp � 2(48) � 48/2 � 6.48/48 �120.135 pitches. However, a chain cannot contain a fractional pitch; therefore, usethe next higher number of pitches, or Lp � 121 pitches.

Convert the length in pitches to length in inches, Li, by taking the product ofthe chain pitch p in and Lp. Or Li � Lpp � (121)(0.625) � 75.625 in (192.1 cm).

Related Calculations. At low-speed ratios, large-diameter sprockets can beused to reduce the roller-chain pull and bearing loads. At high-speed ratios, thenumber of teeth in the high-speed sprocket may have to be kept as small as possibleto reduce the chain pull and bearing loads. The Morse Chain Company states:Ratios over 7:1 are generally not recommended for single-width roller chain drives.Very slow-speed drives (10 to 100 r /min) are often practical with as few as 9 or10 teeth in the small sprocket, allowing ratios up to 12:1. In all cases where ratiosexceed 5:1, the designer should consider the possibility of using compound driveto obtain maximum service life.

When you select standard inverted-tooth (silent) chain and high-velocity in-verted-tooth silent-chain drives, follow the same general procedures as given above,except for the following changes.

Standard inverted-tooth silent chain: (a) Use a minimum of 17 teeth, and anodd number of teeth on one sprocket, where possible. This increases the chain life.(b) To achieve minimum noise, select sprockets having 23 or more teeth. (c) Usethe proper service factor for the load, as given in the manufacturer’s engineeringdata. (d) Where a long or fixed-center drive is necessary, use a sprocket or shoeidler where the largest amount of slack occurs. (e) Do not use an idler to reducethe chain wrap on small-diameter sprockets. (ƒ) Check to see that the small-diameter sprocket bore will fit the high-speed shaft. Where the high-speed shaftdiameter exceeds the maximum bore available for the chosen smaller sprocket,increase the number of teeth in the sprocket or choose the next larger chain pitch.(This general procedure also applies to roller chain sprockets.) (g)Compute the





chain design horsepower from (drive hp)(chain service factor). (h) Select the chainpitch, number of teeth in the small sprocket, and chain width from the manufac-turer’s rating table. Thus, if the chain design horsepower � 36 hp (26.8 kW) andthe chain is rated at 4 hp/ in (1.2 kW/cm) of width, the required chain width � 36hp/(4 hp/ in) � 9 in (22.9 cm).

High-velocity inverted-tooth silent chain: (a) Use a minimum of 25 teeth andan odd number of teeth on one sprocket, where possible. This increases the chainlife. (b) To achieve minimum noise, select sprockets with 27 or more teeth. (c) Usea larger service factor than the manufacturer’s engineering data recommends, iftrouble-free drives are desired. (d) Use a wider chain than needed, if an increasedchain life is wanted. Note that the chain width is computed in the same way asdescribed in item (h) above. (e) If a longer center distance between the drive shaftsis desired, select a larger chain pitch [usual pitches are 3⁄4, 1, 11⁄2, or 2 in (1.9, 2.5,3.8, or 5.1 cm)]. (ƒ) Provide a means to adjust the centerline distance between theshafts. Such an adjustment must be provided in vertical drives. (g) Try to use aneven number of pitches in the chain to avoid an offset link.

CAM CLUTCH SELECTION AND ANALYSIS

Choose a cam-type clutch to drive a centrifugal pump. The clutch must transmit125 hp (93.2 kW) at 1800 r /min to the pump, which starts and stops 40 times perhour throughout its 12-h/day, 360-day/year operating period. The life of the pumpwill be 10 years.


1. Compute the maximum torque acting on the clutchCompute the torque acting on the clutch from T � 5252hp/R, where the symbolsare the same as in the previous calculation procedure. Thus, for this clutch, T �5252 � 125/1800 � 365 lb � ft (494.9 Nm).

2. Analyze the torque acting on the clutchFor installations free of shock loads during starting and stopping, the running torqueis the maximum torque that acts on the clutch. But if there is a shock load duringstarting and stopping, or at other times, the shock torque must be added to therunning torque to determine the total torque acting. Compute the shock torque usingthe relation in step 1 and the actual hp and speed developed by the shock load.

3. Compute the total number of load applicationsWith 40 starts and stops (cycles) per hour, a 12-h day, and 360 operating days peryear, the number of cycles per year is (40 cycles /h)(12 h/day)(360 days/year) �172,800. In 10 years, the clutch will undergo (172,800 cycles /year)(10 years) �1,728,000 cycles.

4. Choose the clutch sizeEnter Fig. 10 at the maximum torque, 365-lb � ft (494.9 Nm), on the left, and thenumber of load cycles, 1,728,000, on the bottom. Project horizontally and verticallyuntil the point of intersection is reached. Select the clutch represented by the nexthigher curve. Thus a type A clutch would be used for this load. (Note that the





FIGURE 10 Cam-type-clutch selection chart.

clutch capacity could be tabulated instead of plotted, but the results would be thesame.)

5. Check the clutch dimensionsDetermine whether the clutch bore will accommodate the shafts. If the clutch boreis too small, choose the next larger clutch size. Also check to see whether the clutchwill fit into the available space.

Related Calculations. Use this general procedure to select cam-type clutchesfor business machines, compressors, conveyors, cranes, food processing, helicop-ters, fans, aircraft, printing machinery, pumps, punch presses, speed reducers,looms, grinders, etc. When choosing a specific clutch, use the manufacturer’s en-gineering data to select the clutch size.

TIMING-BELT DRIVE SELECTION AND ANALYSIS

Choose a toothed timing belt to transmit 20 hp (14.9 kW) from an electric motorto a rotary mixer for liquids. The motor shaft turns at 1750 r /min and the mixershaft is to turn at 600 � 20 r/min. This drive will operate 12 h/day, 7 days/week.Determine the type of timing belt to use and the driving and driven pulley diametersif the shaft centerline distance is about 27 in (68.6 cm).


1. Choose the service factor for the driveTiming-belt manufacturers publish service factors in their engineering data basedon the type of prime mover, the type of driven machine (compressor, mixer, pump,etc.), type of drive (speedup), and drive conditions (continuous operation, use ofan idler, etc.).

Usual service factors for any type of driver range from 1.3 to 2.5 for varioustypes of driven machines. Correction factors for speed-up drives range from 0 to0.40; the specific value chosen is added to the machine-drive correction factor.





TABLE 19 Typical Timing-Belt Pitch*

TABLE 18 Typical Timing-Belt Service Factors*

Drive conditions, such as 24-h continuous operation or the use of an idler pulleyon the drive, cause an additional 0.2 to be added to the correction factor. Seasonalor intermittent operation reduces the machine-drive factor by 0.2.

Look up the service factor in Table 18, if the manufacturer’s engineering dataare not readily available. Table 18 gives safe data for usual timing-belt applicationsand is suitable for preliminary selection of belts. Where a final choice is beingmade, use the manufacturer’s engineering data.

For a liquid mixer shock-free load, use a service factor of 2.0 from Table 18,since there are no other features which would required a larger value.

2. Compute the design horsepower for the beltThe design horsepower hpd � hpl � SF, where hpl � load horsepower; SF � servicefactor. Thus, for this drive, hpd � (20)(2) � 40 hp (29.8 kW).

3. Compute the drive speed ratioThe drive speed ration Sr � Rh /Rl, where Rh � rpm of high-speed shaft; Rl � rpmof low-speed shaft. For this drive Sr � 1750/600 � 2.92:1, the rated rpm. If thedriven-pulley speed falls 10 r /min, Sr � 1750/580 � 3.02:1. Thus, the speed ratiomay vary between 2.92 and 3.02.

4. Choose the timing-belt pitchEnter Table 19, or the manufacturer’s engineering data, at the design horsepowerand project to the driver rpm. Where the exact value of the design horsepower isnot tabulated, use the next higher tabulated value. Thus, for this 1750-r /min drive





TABLE 21 Timing-Belt Center Distances*

TABLE 20 Minimum Number of Sprocket Teeth*

having a design horsepower of 40 (29.8 kW), Table 19 shows that a 7⁄8-in (2.2-cm)pitch belt is required. This value is found by entering Table 19 at the next higherdesign hp, 50 (37.3 kW), and projecting to the 1750-r /min column. If 40 horse-power (29.8 kW) were tabulated, the table would be entered at this value.

5. Choose the number of teeth for the high-speed sprocketEnter Table 20, or the manufacturer’s engineering data, at the timing-belt pitch andproject across to the rpm of the high-speed shaft. Opposite this value read theminimum number of sprocket teeth. Thus, for a 1750-r /min 7⁄8-in (2.2-cm) pitchtiming belt, Table 19 shows that the high-speed sprocket should have no less than24 teeth nor a pitch diameter less than 6.685 in (17.0 cm). (If a smaller diametersprocket were used, the belt service life would be reduced.)

6. Select a suitable timing beltEnter Table 21, or the manufacturer’s engineering data, at either the exact speedratio, if tabulated, or the nearest value to the speed-ratio range. For this drive, havinga ratio of 2.92:3.02, the nearest value in Table 21 is 3.00. This table shows that





TABLE 22 Belt Power Rating*

with a 24-tooth driver and a 72-tooth driven sprocket, a center distance of 27.17 in(69.0 cm) is obtainable. Since a center distance of about 27 in (68.6 cm) is desired,this belt is acceptable.

Where an exact center distance is specified, several different sprocket combi-nations may have to be tried before a belt having a suitable center distance isobtained.

7. Determine the required belt widthEach center distance listed in Table 21 corresponds to a specific pitch and type ofbelt construction. The belt construction is often termed XL, L, H, XH, and XXH.Thus, the belt chosen in step 6 is an XH construction.

Refer now to Table 22 or the manufacturer’s engineering data. Table 22 showsthat a 2-in (5.1-cm) wide belt will transmit 38 hp (28.3 kW) at 1750 r /min. Thisis too low, because the design horsepower rating of the belt is 40 hp (29.8 kW). A3-in (7.6-cm) wide belt will transmit 60 hp (44.7 kW). Therefore, a 3-in (7.6-cm)belt should be used because it can safely transmit the required horsepower.

If five, or less, teeth are in mesh when a timing belt is installed, the width ofthe belt must be increased to ensure sufficient load-carrying ability. To determinethe required belt width to carry the load, divide the belt width by the appropriatefactor given below.

Thus, a 3-in (7.6-cm) belt with four teeth in mesh would have to be widened to3/0.60 � 5.0 in (12.7 cm) to carry the desired load.

Related Calculations. Use this procedure to select timing belts for any of thesedrives: agitators, mixers, centrifuges, compressors, conveyors, fans, blowers, gen-erators (electric), exciters, hammer mills, hoists, elevators, laundry machinery, lineshafts, machine tools, paper-manufacturing machinery, printing machinery, pumps,sawmills, textile machinery, woodworking tools, etc. For exact selection of a spe-cific make of belt, consult the manufacturer’s tabulated or plotted engineering data.





GEARED SPEED REDUCER SELECTION ANDAPPLICATION

Select a speed reducer to lift a sluice gate weighing 200 lb (889.6 N) through adistance of 6 ft (1.8 m) in 5 s or less. The door must be opened and closed 12times per hour. The drive for the door lifter is a 1150-r /min electric motor thatoperates 10 h/day.


1. Choose the type of speed reducer to useThere are many types of speed reducers available for industrial drives. Thus, aroller chain with different size sprockets, a V-belt drive, or a timing-belt drive mightbe considered for a speed-reduction application because all will reduce the speedof a driven shaft. Where a load is to be raised, often geared speed reducers areselected because they provide a positive drive without slippage. Also, moderngeared drives are compact, efficient units that are easily connected to an electricmotor. For these reasons, a right-angle worm-gear speed reducer will be tentativelychosen for this drive. If upon investigation this type of drive proves unsuitable,another type will be chosen.

2. Determine the torque that the speed reducer must developA convenient way to lift a sluice door is by means of a roller chain attached to abracket on the door and driven by a sprocket keyed to the speed reducer outputshaft. As a trial, assume that a 12-in (30.5-cm) diameter sprocket is used.

The torque T lb � in developed by sprocket � T � Wr, where W � weight lifted,lb; r � sprocket radius, in. For this sprocket, by assuming that the starting frictionin the sluice-door guides produces an additional load of 50 lb (222.4 N), T � (200� 50)(6) � 1500 lb � in (169.5 Nm).

3. Compute the required rpm of the output shaftThe door must be lifted 6 ft (1.8 m) in 5 s. This is a speed of (6 ft � 60 s/min) /5 s � 72 ft /min (0.4 m/s). The circumference of the sprocket is �d � � (1.0) �3.142 ft (1.0 m). To lift the door at a speed of 72 ft /min (0.4 m/s), the output shaftmust turn at a speed of (ft /min) / (ft / r) � 72/3.142 � 22.9 r /min. Since a slightincrease in the speed of the door is not objectionable, assume that the output shaftturns at 23 r /min.

4. Apply the drive service factorThe AGMA Standard Practice for Single and Double Reduction Cylindrical Wormand Helical Worm Speed Reducers lists service factors for geared speed reducersdriven by electric motors and internal-combustion engines. These factors range froma low of 0.80 for an electric motor driving a machine producing a uniform load foroccasional 0.5-h service to a high of 2.25 for a single-cylinder internal-combustionengine driving a heavy shock load 24 h/day. The service factor for this drive,assuming a heavy shock load during opening and closing of the sluice gate, wouldbe 1.50 for 10-h/day operation. Thus, the drive must develop a torque of at least(load torque, lb � in)(service factor) � (1500)(1.5) � 2250 lb � in (254.2 Nm).





TABLE 23 Speed Reducer Torque Ratings*

5. Choose the speed reducerRefer to Table 23 or the manufacturer’s engineering data. Table 23 shows that asingle-reduction worm-gear speed reducer having an input of 1.24 hp (924.7 W)will develop 2300 lb � in (254.2 N � n) of torque at 23 r /min. This is an acceptablespeed reducer because the required output torque is 2250 lb � in (254.2 Nm) at 23r /min. Also the allowable overhung load, 1367 lb (6080.7 N), is adequate for thesluice-gate weight. A 1.5-hp (1118.5-W) motor would be chosen for this drive.

Related Calculations. Use this general procedure to select geared speed re-ducers (single- or double-reduction worm gears, single-reduction helical gears, gearmotors, and miter boxes) for machinery drives of all types, including pumps, load-ers, stokers, welding positioners, fans, blowers, and machine tools. The startingfriction load, applied to the drive considered in this procedure, is typical for appli-cations where a heavy friction load is likely to occur. In rotating machinery ofmany types, the starting friction load is usually nil, except where the drive is con-nected to a loaded member, such as a conveyor belt. Where a clutch disconnectsthe driver from the load, there is negligible starting friction.

Well-designed geared speed reducers generally will not run at temperatureshigher than 100�F (55.6�C) above the prevailing ambient temperature, measured inthe lubricant sump. At higher operating temperatures the lubricant may break down,leading to excessive wear. Fan-cooled speed reducers can carry heavier loads thannoncooled reducers without overheating.

POWER TRANSMISSION FOR A VARIABLE-SPEED DRIVE

Choose the power-transmission system for a three-wheeled contractor’s vehicle de-signed to carry a load of 1000 lb (4448.2 N) at a speed of 8 mi/h (3.6 m/s) overrough terrain. The vehicle tires will be 16 in (40.6 cm) in diameter, and the enginedriving the vehicle will operate continuously. The empty vehicle weighs 600 lb(2668.9 N), and the engine being considered has a maximum speed of 4200 r /min.


1. Compute the horsepower required to drive the vehicleCompute the required driving horsepower from hp � 1.25 Wmph /1750, where W� total weight of loaded vehicle, lb (N); mph � maximum loaded vehicle speed,





mi/h (km/h). Thus, for this vehicle, hp � 1.25(1000 � 600)(8) /1750 � 9.15 hp(6.8 kW).

2. Determine the maximum vehicle wheel speedCompute the maximum wheel rpm from rpmw � (maximum vehicle speed, mi/h)� (5280 ft /mi) /15.72 (tire rolling diameter, in). Or, rpmw � (8)(5280) /[(15.72)(16)] � 167.8 r /min.

3. Select the power transmission for the vehicleRefer to engineering data published by drive manufacturers. Choose a drive suitablefor the anticipated load. The load on a typical contractor’s vehicle is one of suddenstarts and stops. Also, the drive must be capable of transmitting the required horse-power. A 10-hp (7.5-kW) drive would be chosen for this vehicle.

Small vehicles are often belt-driven by means of an infinitely variable transmis-sion. Such a drive, having an overdrive or speed-increase ration of 1:1.5 or 1:1,would be suitable for this vehicle. From the manufacturer’s engineering data, adrive having an input rating of 10 hp (7.5 kW) will be suitable for momentaryoverloads of up to 25 percent. The operating temperature of any part of the driveshould never exceed 250�F (121.1�C). For best results, the drive should be operatedat temperatures well below this limit.

4. Compute the required output-shaft speed reductionTo obtain the maximum power output from the engine, the engine should operateat its maximum rpm when the vehicle is traveling at its highest speed. This preventslugging of the engine at lower speeds.

The transmission transmits power from the engine to the driving axle. Usually,however, the transmission cannot provide the needed speed reduction between theengine and the axle. Therefore, a speed-reduction gear is needed between the trans-mission and the axle. The transmission chosen for this drive could provide a 1:1or a 1:1.5 speed ratio. Assume that the 1:1.5 speed ratio is chosen to provide higherspeeds at the maximum vehicle load. Then the speed reduction required � (maxi-mum engine speed, r /min)(transmission ratio) / (maximum wheel rpm) �(4200)(1.5) /167.8 � 37.6.

Check the manufacturer’s engineering data for the ratios of available gearedspeed reducers. Thus, a study of one manufacturer’s data shows that a speed-reduction ratio of 38 is available by using a single-reduction worm-gear drive. Thisdrive would be suitable if it were rated at 10 hp (7.5 kW) or higher. Check to seethat the gear has a suitable horsepower rating before making the final selection.

Related Calculations. Use the general procedure given here to choose powertransmissions for small-vehicle compressors, hoists, lawn mowers, machine tools,conveyors, pumps, snow sleds, and similar equipment. For nonvehicle drives, sub-stitute the maximum rpm of the driven machine for the maximum wheel velocityin steps 2, 3, and 4.







Tranmissions,Cluthes,Roller-Screw Actuators,CouplingsiAnd Speed Control

Documents

motor speed

vehicle data

kgthe vehicle speed

selection of roller

mlrev displacement pump

gross vehicle weight

selection of flexible

motor data