shafts &couplings.pdf

7/30/2019 shafts &couplings.pdf

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I L LU S T RAT E D S O U R C E 80 0 of M E C H A N I C A L C 0 M P 0 N E N T S

S E C T I O N 4SHAFTSOverview of Shaft& CouplingsBeams &Stepped ShaftsShafts: 5 Steps Find Strength& DefectionShaft Torque: Charts Find Equivalent SectionsCritical Speeds of End Supported Bare ShaftsTorsional Strength of ShaftsBearing Loads on Geared Shafts7Ways to limit Shaft RotationFriction for Damping15 Ways to Fasten Gears to Shafts14 Ways to Fasten Hubs to ShaftsAttaching Hubless Gears to Shafts10Different Types of Splined ConnectionsTypical Methods of Coupling Rotating Shafts ITypical Methods of Coupling Rotating Shafts IITypical Designs of Flexible Couplings ITypical Designs of Flexible Couplings IITypical Designs of Flexible Couplings 111low Cost Methods of Coupling Small Diameter ShaftsCoupling of Parallel ShaftsNovel linkage for CouplingOffsetShaftsTorque of Slip CouplingsHigh-speed Power CouplingsNovel Coupling Shifts Shafts

4-24-84-134-154-174-184-204-224-244-294-344-364-384-404-424-444-464-484-504-524-544-564-584-70


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4-8

Beams & Stepped ShaftsEquations simplify analysis. Two general equations can now handleany type of loading and number of stepped contours. The methodi s exact and avoids integration.William Griffe1

N practice, it is quite common f or beams o r shafts toI be subjected to a combination of concentrated loads,distributed loads, and moments. Also, the beams mayhave cross-sections that vary by distinct steps.To find the deflection and slope at any point on abeam with, say, only three loads, as shown in Fig 1, pagesof calculations are needed, and you will certainly turnfor help from a computer, if one is available. Thus, thevalue of the simplified method given here becomes appar-ent. It can handle any num ber and combination of loads,and any number of steps in the contour. The methodis similar to the one developed by W. H. Macaulay ofGreat Britain (see Note on the Deflection of Beams,

Symbolsa, , c =dis tance a long a beam loca t ingpo i n t s uf applicat ion of momentM, oncent ra ted load P , a n d t h e

po i n t w he re t he un i fo rm l oa d wbeginsC, = con stants of integrat io nE =mo dulu s of e last ic i ty, psiI =m o m e n t of ine r t ia , in.*x =po i n t on t he be a m whe re t he de -flection is calcula ted, in.XI, z, e t c = specific valu es of x for segmentsof t he be a m, i n.yl, yz, e t c =deflections at points xi, 2, etc.,in.yo =deflect ion a t th e origin of t he co-ordina te sys tem, in .w =uni form dis t r ibuted load, Ib / in .8 = s lope of the bea m a t point of in-te res t , r , r a d i a ns8 0 = s lope at the or igin of the coordi -na t e sys te m, r a d i a nsUpward def lect ions and c lockwise mom entst o t h e l e f t of a sec t ion a re cons idered posi-t ive . See il lus t ra t ion on fac ing page .

Messenger Math, Vol 4 8 , C a n d d g e , E n g l a n d ) , b u t itsapplication to stepped beams is new.The method makes use of two design equations, onefor deflection and the other for slope. There is nointegration-simple algebr aic calculatio ns ar e all thatsneeded. There is no approximation involved: The methodis exact. But it has its limitations in that it is applicableonly to cases where the reactions are known-staticallyindeterminate beams are beyond the scope of the meth-o d s . Most beams and cantilever problems, however, arestatically determinate.The derivation of these two general equations (Eq 7and 8) which follows gives a clearer picture of the con-cepts underlining the method. But if you wish, you mayskip directly to the sample problems starting on p 84to se e how easily the equations can produce beam form-ulas (starting with simple loadings), and to the stepped-shaft problem, p 86.DESIGN EQUATIONS FOR UNIFORM BEAMS

Although integration is employed in deriving themethod, it is not required in the actual solution of prob-lems.The usual method of determining the curved line of abeam under load is to start with the functional equation:

and integrate it twice, whereby the equation is obtainedof the deflection curve y = f ( x ) . (Most of the currentmethods, such as the moment-area method and theconjugate-beam method, are based on carrying out insome way this double integration.) Th e constants ofintegration, C and D, are easily determined from theboundary conditions. As a rule:C =EI&D = E l y o

where yo an d BO are the deflection and angle of rotationat the origin of coordinate system (see list of symbols).For example, for a cantilever beam with a uniformlydistributed load w, f th e origin of the coordinates isplaced at the free end, then Fq 1 ca n be written as:


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Integrating twicedY 2 3El- -w 7+Cax (3)

F o r x =0, E q 3 and 4 becomeEIyo =D (5 )EZ&=C (6)

where upward deflection and clockwise movement to th eleft of a section are considered positive.Note that the relationships in Eq 5 an d 6 are valid onthe condition that the same origin be used for the xcoordinate no matter what portion of the beam is con-sidered.For the cantilever beam with all three types of load-ing, Fig 1, we denote the four segments of the beam asA-B, A-C, A-D, A-E, as defined by their coordinates,X I , x 2 , x 3 , x4. Each of these segments has its own differen-tial equation describing the elastic line, and each equationif solved separately, introduces two constants of integra-tion. Thus, eight constants of integration must be deter-mined from appropriate boundary conditions. This re-quires an involved procedure, but the number of integra-tion constants can be reduced to two, regardless of th enitmher of segments, by following these steps:I . Obtain the equation for the bending moment at adesired section by considering the external forces lyingbetween that section and the origin.2. Integrate the moment equation without opening thebrackets.3. Introduce a beam factor ( x - )" , for convenience.Because this factor is raised to the zero power and hencemust be equal to unity, it will not affect the equations,as will be seen.For segment A-B (where Coordinates of a point are xl,yl):

Double integration gives

EIyi =Cix1+DIFor segment A-C (coordinates XZ, yz)Multiplying by the beam factor

Integrating twice gives

Shafts &Couplings 4-9

A XI I 1I

Note-Moments and loads ar e positive in the directionsindicated

For segment A-D; (x3, y3)

For segment A-E; (x4, y4)d2y4 ( 2 4 - )Z2I __x42 =M(x4 - )O +P(x4 - ) +w

+e4p (x4 - 12 (x4 - l 32 6+C4 ~ 4+ 4( 5 4 - 14

24The eight constzints of integration are now determinedfrom the boundary conditions of the segments. For ex-ample, when x1 = x 2 , then both are equal to a, and thedeflections yl and yz will be equal to each other. Follow-ing this type of reasoning we getW h e n XI = x2 = a t h e n yl = y2 and 81 = 02

~2 = 2 3 = b e2 = 0,x3 = x4 = c e3 = e4x4 =L y4 =0 e4 =o

92 = y39 3 = y4

These values for y an d 0 are substituted into the pre-vious equations for each sector to yield:c1 =cz =c3 =c4 =cDI =D2 =D3 =Dp =D

Thus, the substitution of (x - )O caused all constantsto be equal to each other, except two, C and D. These


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4-10

two can be determined from the fourth condition whichwas already calculated (E q 5 an d 6) :C =EIOo, D =EIyo

Now, considering a case where there may be a multi-plicity of forces and moments, we turn to segment A-E.and from its equation obtain a general formula goodfor any arrangement of forces M , P, w :Design equation for deflection

M ( z - >aE I y =EIyo +EIOox +1(7)

Design equation for slope (by differentiating Eq 7)E10 =ED,, + M ( x - )E

+1 b )e + Is (8)Because the reactions of the supports are included inthe external forces and moments, Eq 7 an d 8 are appli-cable to beams with any type of support.The term that contains the distributed load w is validonly if the distributed load continues at least to the sec-tion where y and O are to be determined. If the loadbreaks off before the section of interest, it should be con-tinued to the section and an equivalent load of oppositedirection should be added (example 6 illustrates this).Thus, one equation is all that is needed to calculatethe deflection of a beam with any supports, any type ofloading and number of applied loads.Example 1-Fixed-end with conc entrated load

c 5M M

Determine the maximum deflection for a beam withboth ends fixed subjected to a center load.SolutionYou can place the origin of the coordinates at eitherthe left or right end of the beam. In both cases, x o =0an d yo =0. Thus, in the equation for deflection, Eq 7,the first two term s dro p out. Also, because of symmetry, itis best to calculate ymaz a t x =%. Include in Eq 7 allforces and moments lying between the origin and the sec-tion under consideration (which is at x = L / 2 ) . Thismeans that the reactive force R = P/2 and reactivemoment M =% X L / 2 x P/2=P L / 8 at the built-inend should be used too. Because the direction of thereactive mom ent is counter-clockwise, it will enter in the

general equation with a minus sign. Also a =b =c =0.Thus ymaZ, fo r x = L/2, s

PL3or y,,, =- ___L3192 192EI- __From the sign convention, the minus sign designatesa downward deflection.

Example 2-Fixed-end with dis trib uted loadYt

Here, too, the location of the origin is arbitrary.Maximum deflection y m a Z s a t x =L 2 and a =b =c =0. Thus f rom E q 7:

W L ( ~ 1 2 )* ~ ( ~ 1 214- -2 6 24=- __L4 or ym0,=- __L4384EI84Note that the downward load w was entered with aminus sign.Example 3F an t i le v er w i th d is tr ibuted load

In this case find the maximum deflection and slope(F ig 4).Sdut ionIt is best to place the origin at the built-in end. ThenY O =0 and Bo =0, and maximum deflection and slope,ymaz nd e,,,,, are at n =L. With a =0, b =0, c =0,obtain from Eq 7:


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4-11

r 6 IOOIb.IO /b/in.

wL483--From Eq 8:

(L- ) +LeEA?&?&a--2wL(L - 0.y - w ( L - )a

2 6

Example &-Simple beam with fixed endsYI 2I P

X

Find ymor.e for the beam in Fig 5.SolutionBecause Eq 7 contains both yo and Bo , choose thelocation of the origin such that either of the two terms(or both) are equal to zero. By placing the origin at theleft end, you can determine Bo by making use of thecondition that at x = L the deflection is zero. HenceP ( L - ) s - P ( L - /2)3 =6 6IyZ=z =EIB& +

and

The maximum deflection is at x = L / 2 :PL348

force P / 2 is zero, but to the applied load P t is L / 2 .

L2 16EIy, , - -- -

Note that the distance from the origin to the reactive

Example 5-Beam with complex loadingFind the deflection at point D, and the slope at pointA , of the beam loaded as shown in Fig 6.SolutionPlace the origin at the left support where yo =0. The

Shafts &Couplings

angle of rotation Bo then will be determined from th econdition that y =0 at n =20 in.loo(20 - 10)' 20(20 - 10)s2 + 6Iy,zo =EIB020 +

i o ( 2 0 - 014 i o ( 2 0 - 0 14 =,24- 24225or eo =- __E l

As was pointed out before, the design equations recog-nize only a distributed load that does not break off be-fore the section where y or 0 are to be determined.Because the uniform load in Fig 6 breaks off at pointB, you must extend the load to point C. To compensatefor this addition, add a uniform load of opposite direc-tion (upward) in the portion BC of the beam. The lastmember in the equation below reflects this change.Deflection at point D ( x =30 in .) :lOO(30 - 0) '2I ~ D - 25(30) +

ZO(30 - )' 180(30 - 0)866 +

In the previous cases of beams with moment of inertiaconstant all along the length of the beam, only M wasintegrated as a function of x . But when the moment ofinertia Z also varies with x , Eq 1 can be written as:

where Z is arbitrarily ch osen as a basis t o which we referthe varying l o . Integrating the M / ( Z o / Z ) function givesthe deflection line.Because l o / l s dimensionless, M / ( Z , J l ) has the di-mension of bending moment and represents as a functionof x , a corrected bending-moment for the beam w ithuniform moment of inertia I. Hence, it follows that the $ 4moment diagram on the beam of constant inertia willresult in the same deflection curve as the M / ( I o / Z ) mo-ment diagram on the beam of varying moment of inertia.In other words, a certain P / ( Z o / Z ) loading system, whichoriginates the M / Z o / l ) diagram, will produce exactlythe same deflection curve on the beam of constant cross


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Stepped beam 500/b 7

. .I

(B) Moment diagramI(C) Shear diag ram

212 /b 500/b 288/b

I288 /b 28811,I I ' ID) Shaft segmentsI2/2/b 500/b 288/b5;7/b 5;7/b 703/b 7dJIb(E) Change to con stant diameterI I I

7b3 b(F) Final shaft of uniform rigidityI

section as the original P loading (which gave the Mdiagram) would do on the beam of variable moment ofinertia.Thus, if you replace a beam or a portion of a beamwith a moment of inertia Io , by a beam with a momentof inertia I =KIo, and at the same time change all theloads and reactions acting upon the beam by a factorof K =U I o , the elastic curve of the two b ea m will bethe same. Therefore, to apply the general formulas, Eq7 and 8, to a beam whose cross section varies by steps,you must replace the beam of variable rigidi ty by abeam of constant rigidity in the manner described above.This is illustrated in the following numerical example:

Determine the end slope BO, and the deflection underthe 500-lb load of a steel stepped shaft, Fig 7A . Theshaft has a 1 in. dia with IO = 0.049 in.* and 1% in.dia with I =0.120 in.*. E =30 x IO6 psi.Solution

Construct a bending moment diagram, Fig 7B, whereM = 212 x 15 = 3180 Ib-in., followed by a sheardiagram, Fig 7C. Fig 7D shows the shaft with momentsand shears at the discontinuities.In Fig 7E the stepped shaft is replaced by a shaftwith a constant diameter of 1% in. Now transform theloads by a factor of I / I o =2.444. The loads acting uponthe middle portion of the shaft remain unchanged, be-cause we have not changed the moment of inertia ofthis portion. Finally, in Fig 7F, the shaft of uniformrigidity is schematically shown for which the generalequations, Eq 7 and 8 , a p p l y . Thus, the effect of thechange in the cross section has been established as iden-tical to that of a certain force system, and it is possibleto find the deflection in a beam with any number ofsteps and discontinuities by a simple expedient of usingone equation.Continuing the numerical computations, assume theorigin of coordinates at the left support. Thus, at x =26 in. y o =0, and Eq 7 yields:

Example 6-Stepped shaft

2755(26 - )z2Zy,zs =EI&26 -3742(26 - 17)' 517(26 - )'+---2 6305(26 - 9)s 500(26 - 15)'-- 6- 6415(26 - 7)' =- 6Hence

eo =- _ _ -21948 =-0.0092 radiansEZNow again, from Eq 7, but at a new position ( x =15), and with inclusion of the Bo value:

2755(15 - 9)'2Zy,l, =- 2,948(15) -305(15 - 9)8 517(15 - )'

' 6- 6=- 64,518


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Shafts:5 Steps Find Strength & DefectionCharles Tiplitz

B e n d i n g or torsional stresses, or a combillation, maybe limiting requirements in shaf t design. Shaft stressdiagrams usually deal with combined stress only. Th re esupplementary graphs presented here, together with astep-by-step example, bring quick aid to difficult shaftdesign problems.EXAMPLE

Design a shaft that will transmit at 1200 rpm, 0.1 h pthrough a 4-in.-dia, 20" gear, centrally mounted betweenself-aligning bearings 2 in. apart. Limiting combined

i.Force vs r a d i u s a n d shaft d ia top r o d u c e i0,OOO psi shear stress

Dia, in.

shear stress t o be 10,000 psi, deflections must not exceed0.05" or 0.0002 in.Step 1.Find:hp X 63,025 - 0.1 X 63,025 =5.25 in . Ib- 1,200orque = *Pm

Tangential force a t the gear pitch-radius (2 in.) is,thStep 1 5.25/2 =2.63 Ib. From chart 1, a 2-in. radiusintcrpolated for 2.6 lb gives D =0.14 in. for 10,000 psishear. (CONTINUED)

2.Deflection an d l oad of uniformlyl oaded s tee l shaft.(For 10,000ps i stress, an d E E 30 x 1 0 6 p s i )

os 1.0 5.0Dio, in .


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4-14

Dia, in.

Step 2Since the bearings are self aligning and there is 3central load, correction factor is 0. 5 (see table, simplesupp orts). Since a 20" gear-tooth is transmitting thcpower, the bending force on the shaft will be greater thanthe tangential force. Therefore,uncorrected force to stress the shaft to 10,000 psi willbe 2.8/0.5 = 5.6 Ib. Therefore, enter chart 2 a t 5.6Ib, at which point D =0.11'' (solid lin e).Step 3Check combined stress on the 0.14 shaft dia foundin step 1. Bending capacity on chart 2 for this dia isapproximately 11 Ib (dotted line ). The load correctionfactor is 0.5 so actual (corrected) load to produce 10,000psi is 5 . 5 Ib. Load is proportional to stress, hence a 2.8 Ibload will cause a bending stress of 5,100 psi. Combinedstress

Resolving: 2.63 x secant 20" = 2.8 Ib.

I

~

=d ( actuaastress +(max stress)Z3.Angular deflection an d torsionalelasticity-steel shaft.

0.1 io

8 =d(%j)'+10,OOOz =10,600 psiRead deflection of 0.0026 in. at junction of 2 in.length and 0.11 in. dia. Deflection factor from tableis 03 , therefore correct value is (0 .8) (0.0026) =0.0022

in .Step 4Target deflection is 0.0002 in.; check for deflectionand torsion instead of adjusting for combined stressrequirement. Desired deflection is 0.000 2/0.00 22 =1 901 the deflection of the 0.11 in.-dia shaft. Interpo latebetween lines for 1000 Ib/in. and 10,000 Ib/in. foran elasticity of abo ut 2000 Ib /in . Elasticity factor is0.625 so actual elasticity is 1250 Ibi in. Valu e ondiagram should be 9 x 2000 lb/in or 18,000 Ib/in. toreduce deflection correctly. Interpo late between 10'Ib/in. and IO5 Ib/in.: shaft dia is 0.20 in .Step 5 Angulardeflection is based on a stress of 10,00 0 psi; chart 1showed that a shaft 0.14 in. dia would be stressed thismuch by the design torque. A 2 in. length will betwisted approximately 1.3' according to interpolationof chart 3 (solid lin e). Desired deflection is 0.05" .Estimate torsional elasticity of 5 .5 in. lb/deg by inter-polating between 1 in.lb/deg on chart 3 . Elasticitymust be greater by the factor 1.3/0.05 = 26. Calcu-late: 26 x 5.5. =143. Interpolate betweeh 1P an d10' in.lb/deg. (dotted line) to find a dia of 0.034 in .This is the correct shaft dia to use.

Obtain torsional deflection from chart 3.

CORRECTION FACTORS

Load Deflection Elasticity1 I I Ib/in.Condit ionSimple supports, uniformlyloaded (the charts are basedon this condition)

~

Simple support s, one concen-trated load a t center

1 .o--0 . 5

1 .o0 . 8

1 .o

0 . 6 2 5

Fixed ends, uniformly dirtrib- I 1 .5 I 0.3uted load I 5 . 0Fixed ends, one concentratedload at centerCantilever, one end fixed,other unsupported, one concen.trated lo ad at unsupported end

1 .o0 .125

0.4

3.22 . 5

0.0391

Square shafts, length of sidesame as shaft dia.


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-3-ExOmpk? I finds D for T =I< 300 n. lb andmax perm;ssible shear stress =18,000p51:Example 2 finds dl. for equivalent -strength,

- -----

Shaft Torque:Charts Find Equivalent SectionsAn easy way to convert solid circular shafts to equivalent-strengthshafts of hollow circular, ell iptical, square, and rectangular sections.Dr. Biswa Nath Ghosh

1-ROUND and ELLIPTICAL SHAFTS

5 0 2.04 E 0.53.0j4.0Location of Torque formulas:maxshear T =

Outerf iber

Ends Of rd,d:fminor Ix i s

T in.-lb2,00q000

l,000,000

100,000

10,000

0.7

0.5i


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4-16

?-SQUARE and RECTANGULAR SHAFTSTorque, T, in.-lb

2,000,000~i,000,000 Exomple 4 indsS fur squareshoft thoi will fronsmif/6300in.4 orque ot14OOOpsi sheor stress.Exomple 5 findsA for rectungulorshoft forrutio A M = /.20k00*000IFI0,Ooo

200E

Square side, S, in.5.0

4.0f

Max shear stress, f, psi

0.7

0.5

I I I 1

Ih a f t Locotion of Torque formulos:section max shear 1 T =It 1 I IMiddle 0.208s fof sides

t I I 1ofidpainajor A 2 B 2 f

3A t 1.8Bsides 50,000

10,000


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Shafts 8t Couplings 4-17

Critical Smeeds ofEnd Supported Bare ShaftsL. Morgan PorterTHIS OMOGRAM solves the equation for the critical speedof a bare steel shaft that is hinged at the bearings. Forone bearing fixed and the other hinged rnuItiply the criticalspeed by 1.56. For both bearings fixed, multiply the critical

speed by 2.27. The scales for critical speed and length ofsh aft are folded; the right han d sides, or the left hand sides,of each are used together. Th e chart is valid for both hollowand solid shafts. For solid shafts, D2=0.

8,000

7,000

6,000

50,000

40,000

5,00030,000

g ,0003,O 00

20,000 E?0z

70c.-

BO

40 100 *

503E x am p l e ; -

DI=6.3 n., D2=5.8 n., -=8.56L =130in. Nc=2,375p r n

For Aluminum multiply uolues o f N byLOO26 ,For Mognosium m u l t i p b vofues of Nc by 0.9879

\.


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4-18

Torsional Strength of ShaftsFormulas and charts for horsepower capacity of shafts from 1/2 to 2 1/2 inchdiameter and 100 to 1000 rpm.DouglasC. Greenwood

For a maximum torsionaldcflection of 0.08" per foot,shaf t Icng th , d iameter andhorsepower capacity are rc-latcd ind =4.644%

wherc d = shaft diamctcr ,in.; h p = horscpowcr; R =sha f t spccd , rpm. Th is dc-f lcction is rccommendcd bymany authorit ies as being asafe genera l maximum. T hctwo charts arc plotted fromthis formula, providing arapid means of chcckingtransmission-shaft s t r e n g t hfor usual industrial speeds upto 20 hp. Although shaftsunder 1-in. dia are n ot trans-mission shafts, strictly apcak-ing, lower sizes %ave been in -cluded.W h cn sh a f t d e s ig n isbased on strength alone, thed iameter can be smaller thanvalues plot ted here. In suchc a m u sc th c f or mula

2.50

2.25

2 oo

I .75

i0Q 1.50._._t

0v)-

I .25

I .oo

0.75

0.50 I I I I II O 0 200 300 no0 500 600S h a f t speed, rpm


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LOADING CONDITION _IjHead shafts subjed to heavy strains.and slow speeds, clutches or g earing carr ied)

(Intermittent loads(2600 psi)

The value of k varies from1 25 to X ccording ro allow-able stress used. Thc figurca x o u n s f or m mbcrs tha tintrodu ce bcndii ig loads, suchas gears, clutch cs and pullcys.B,ut bcndin g loads are n o t asreaclily determined as tor-sional strcss. Thercfore, toalIow for combined bendingand torsional stresses, it isusual to assume simple tor-sion and usc a lower designstress for thc shaft dcpendingupon h ow i t is loaded. Fo reuamplc, 12 5 represents astress of approxiniatcly 2600psi, which is very lo w andshould thus insure a strong-cnough shaf t . Ot her valuesdit ions arc shown in thetable .W h e n bending strcss isnot cons idercd, lower k val-ues can bc used, hut a valueof 38. should be regarded ast h e m i n i m u m .

of k fo r differcnt loadin," con-

__- __Lineshafts 75-100 ft long, heavily loaded. Bearings(3200 psi) 8 ft apart.Lineshafts 50-75 ft long, medium laad, bearings 8 ft

loo I90(3550psi) 1 apart.14300 psi) apart.

______75 Lineshafts 20-50 ft long, lightly loaded, bearings 6 ft

2.oo

I 75

I 50

c'0

.-1.25+c

0cvl

I .oo

0.75

0.506

I I I 1

0 700 800 900 1000Shof f speed, rpm


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Bearing Loads on Geared ShaftsSimple, fast and accurate graphical method of calculating both direction andmagnitude of bearing loads.Zbigniew J ania

To calculatc thc bearing loads resulting froiii gear action, both tlic magnitudc anddirection of the tooth reaction mu st bc known. Th is reaction is thc forcc at the pitchcircle excrted by thc tooth in the direction peiyciidicular to, and away froni the toothsurface. Th us , the too th reaction of a gear is always in the sanie geiicral direction a5i ts mot ion .Most techniques for evaluating bearing loads scparate thc total foroe acting onthc gear into tangential and separating components. Th is tends to com plicate thesolution. 'The me tho d described herein uses th e total force directly.I t T is the torque transmitted by a gear , the tangential tooth force is

FT = 2 T / D ,Also, from Fig. 1 ,F =FT w c Q,F =FT sec (spur gears),(helical gears).

Sincc a forcc can be replaced by an equal force acting at a different point, plus acouple, the total gear force can be considered as acting at th e intersection of the shaftcenterline and a line passing through the mid-face of the gear, if the appropriate coupleis included. For example, in Fig. 2 the total force on gear B s equivalent to a force F Bapplied a t poin t X plus the couple b x F B . In establishing the couples for the othergears, a sign convention must be-adopted todistinguish between clockwise and counter.clockwise moments.If a vector diagram is now drawn for allcouples acting on the shaft , the closing l inewill be equal ( to scale) to the couple result-ing from the reaction at bearing 11. Know-ing the distance between the two bearings,the load on bearing I1 can be found , thedirection being the same as that of thecouple caused by it.The load on bearing I is found in thesame manner by drawing a force vector dia-gram for all the forces acting a t X ncludingthe load on bearing I1 found f rom thecouple diagram.

Tooth reocf ionIi


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Shafts &Couplings

The construction of both diagrams is illustrated on page 2 13. Referring to Fig. 2 t h egiven data arePitch Dia. of GesrB, in.A ................ 2 . 0 0B ................. . 5 0c................. 4 . 0 0

Driver. ........... 1 . 7 5Moment A r m, in .

a . . . . . . . . . . . . . . . . . . . 5 0b .................. . 5 0e ..................5 . 0 0d .................. . 0 0Ang l e , deg Torque on driver. ........... lo 0 Ib-in.

a. .... 5 5f l ...... 487 . . .... 45T h e n ,

Torque delivered by A ............ 4 0 pe r cent of torque on center shaftTorque delivered by B ............ 60per cent of torque on center shaftPressure angle of all gears, +. ...... 20 de gTangential force of driver =200/1.75 = 114 lbTorque on center shaft =2 X 114 =228 Ib-in.

Gear loads areOe 4 228 aec 20 deg =97 Ib2.00.4 =

F B =Fc = 114 seo 20 deg = 121.5 lb

:c 228 aec 20 deg = 19 5 lbBefore drawing the diagrams, i t is convenient to collect all the data as in T able I.The n, the couple d iagram, Fig. 3, is drawn. I t is impor tant to note khat:( 3 ) Vectors representing negative couples are drawn in the same direction but inopposite sense to the forces causing them;( b ) The direction of the closing line of th e diagram should be such as to makethe sum of all couples equal to zero. Thus, th e direction of 7 P,r is the direction ofbearing reaction. T h e bearing load has the same direction but is of opposite sense.

4-2 1

- seporor ing f o r ce - Toto1 geor lood (51


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4-22

7 Ways to Limit Shaft RotationTraveling nuts, clutch plates, gear fingers, and pinning members are thebases of these ingenious mechanisms.I.M.Abeles

M ec h an ic al stops are often required in automatic machinery and servomech-anisms to limit shaft rotation to a given number of turns. Tw o problems toguard against, however, are: Excessive forces caused by abrupt stops; largetorque requirements when rotation is reversed after being stopped.

7ravefing nut

\frome

Troveling nut, Stop pih? finger, Shaft,finge

Rubber A Metal grommet

?top pin

Section A - A

TRAVELING N U T moves (1) alongthreaded shaft until frame preventsfurther rotation. A simple device, butnu t jams so tight that a large torqueis required to move the shaft from its

CLUTCH PLATES tighten and stoprotation as the rotating shaft movesthe nut against the washer. When rota-tion is reversed, the clutch plates canturn with the shaft from A to B. Duringthis movement comparatively lowtorque is required to free the nu t fromthe clutch plates. Thereafter, subse-quent movement is free of clutch fric-tion until the action is repeated atother end of the shaft. Device is recom-mended for large torques becauseclutch plates absorb energy well.

stopped position. This fault is over- than the thread pitch so pin can clearcome a t the expense of increased finger on th e first reverseturn. Thelength by providing a stop pin in the rubber ring and grommet lessen im-traveling nut (2). Eng agem ent between pact, provide a sliding surface. Thepin and rotati ng finger mus t be sh ort er gromm et can be oil-impregnated metal.

Clutch plotes Clutch plateskeyed to shaft4ith proeconTi-ove/ing nut P-B Section 8-B


23/70

I \Output lnput snort

Shafts &?Couplings 4-23

SHAFT FINGER on output shaft hits re-silient stop after making less than onerevolution. Force on stop depends upongear ratio. Device is, therefore, limited tolow ratios and few turns unless a worm-gear setup is used.

TWO FINGERS butt together at initial and final positions, preventrotation beyond these limits. Rubber shock-mount absorbs impactload. Gear ratio of almost 1:l ensures that fingers will be out ofphase with one another until they meet on the ana l turn. Example:Gears with 30 to 32 eeth limit shaf t rotation to 25 turns. Space issaved here but gears are costly.

Gear makes less thun one revolufion

Po i 4 1huff, N fingers ro fote on shuft

finger fixefo ffume

LARGE GEAR RAT IO limits idler gear to less than one turn.Sometimes stop fingers can be added to already existing; gearsin a train, making this design simplest of all. Inpu t gear, how-ever, is limited to a maximum of about 5 turns.PINNED FINQERS limit shaft turns toapproximately N +1 revolutions ineith er direction. Resilient pin-bushingswould help reduce impact force.


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4-24

Friction for DampingWhen shaft vibrations are serious, t ry this simple technique ofadding a sleeve to the shaft can keep vibrations to a minimum.Heres how to design one and predict it s effect.Burt Z i m m e r m a n

H E N B OOS T I N G T HE OP E R A T I N G SPEED of any ma-W chine, the most formidable obstacle to successfuloperation that the designer faces is structural vibration.There is always some vibration in a system, and as th espeeds are increased the vibration amplitudes becomelarge (relatively speaking, for they may still be too smallto be seen).These amplitudes drastically reduce life by causingfatigue failures and also damage the bearings, gears, andother components of the machine. It is not over-simplify-in g the case to say that the easiest way to prevent vibra-tion damage is to damp the vibration amplitudes.An interesting but little-known technique for vibrationdamping is to apply a small amount of dry friction(coulomb friction) at key places of the structure. Thisproduces a greater amount of damping than one wouldnormally expect, and the technique is used with success

by some product designers and structural engineers but,i t seems, only after the machine or structure has beenbuilt. There seems to have been little attempt to applythis concept to initial design or to develop the equationsnecessary for the proper location of the friction points.We will apply this concept here to the solution oftorsional vibrations of shafts, as this is a serious problemin both industrial machinery and in military systems suchas submarines, missiles, and planes. The necessary designformulas are developed and put to work to solve a typicalshaf,t problem from industry.How t h e technique works

Vibration amplitudes in a shaft become a problemwhen the shaft length to the thickness ratio, L 1 / D 1 , e-comes large. One can of course make the shaft thicker.But this would greatly add to its weight.

Symbolsa = b / L zC = Dz/D1D 1 = Diame te r o f sh a f tD z = Diam eter o f s leeveG =S h e a r m o d u l u s of elasticityH = Thickness o f the sleeve wallJ = P o l a r m o m e n t of iner t ia ( fo r the shaf t :

T DI4/32)J E G nD13H/4L I =Length o f sh a f tLz = Len g th of sleevem =D , / 8 H C 3 = ra t io of torsional st iffness

of the s h a f t t o t h a t o f t h e sleeve

r = l + mR = D a m p i n g r a t i oT =Appl ied to rque on t h e sh a f tT, = Resis t ing f r ic t ional to rque appl ied byU = Residual in ternal energy of shaf t and

VI = In ternal energy of the shaf tU,= In ternal energy of the s leeveW =Energy d iss ipa ted in a half oscil lation

the sleevesleeve

h = T,/T6 =Angular d isp lacement o f the shaf t6, =Angular d isp lacement o f the s leeve


25/70


u

I

1. Thin sleeve added to rotating shaft greatly reduces torsional vibrations. The,disk is rigidly attached to the shaft and ha s a snug fi t with the sleeve. Extendingthe sleeve over the entire length provides the most effective damping condition.

To apply the friction-damping technique to a shaft,Fig I . a sleeve is added which is attached to the shaftat one cnd ( A ) . The sleeve is extended along the shaftlength and makes contact with some point on the shaft.In this particular design, a disk is rigidly attached to theshaft (by welding it or tightly pressing it on), and thereis a snug fi t between the disk and the sleeve.The exact amount of fit is not too important, but itmust be neither too loose nor too tight: If the fit is tootight, the shaft and sleeve will tend to move together as aunit and there will be no damping (just an increase in themoment of inertia) ; f to o loose, with a clearance betweendisk and sleeve, again there will be no damping.The frictional forces in question occur at the contactbetween the inside surface of the sleeve and the edgeof the disk, and their magnitude depends on the coeffi-cient of friction and on the pressure between the surfaces.

The most effective damping condition is when thesleeve extends the entire length of the shaft, but theremay be cases, depending on the product design andapplication, where this is impossible. Therefore, the gen-eral case where the Sleeve length is variable is consideredhere.To avoid corrosion or; fretting at the interface, try alayer of viscoelastic s tripping (elastomer) a t the edgeof the disk.Analysis of concept

When a shaft is rotating, a resisting torq ue is developedin the shaft which varies along the length of the shaft.Because the angular displacement is a function of thisresisting torque, the surface fibers of the shaft willundergo different angular displacements which dependon the distance of the specific fiber from the point of

2. Energy present in arotating shaft throughone complete cycle ofvibration with dampingtaking place for one halfof the cycle. At ti, th eenergy in the shaft i sequal to its residual in-ternal energy, U, p lusthe energy dissipatedduring half t h e cycle,W. A t t,, the energy isequal to U, which indi-cates that the energydissipated through dryfriction damping i s W.W ( d i s s i p o l e d enerqy ldI I


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4-26

m :D, / 8HC33. Design chart for different values of the dimensional constant, rn. The frictionalamount of energy dissipated per cycle is a function of the sleeve-shaft length ratio.Critical damping is the amount of damping above which the sleeve-disk interfacewill stick. The curv e fo r th e amplitude-damping ratio (which is read a t the right scale )can be used for most design problems, as illustrated in the numerical example.

the applied torque. The magnitude of the torsional vibra-tion is measured by the difference of displacements alongthe shaft length.The torque difference OF the shaft (applied torque, T,minus the torque at the disk, T,,,) is greater than thecorresponding torque difference along the length of thesleeve. Therefore, there will bc an angular difference be-tween the sleeve and the di5k. Because the inside edgeof the sleeve and the outside surface of the disk havea pressure contact (however slight) this tends to resistrelative motion, hence, torsional vibration damping. Onecan see that as the sleeve diameter approaches infinityan d as the length of the sleeve approaches the length ofthe shaft, the damping becomes more and more eflicient.(The point to remember here is that i t is not the contactpressure which causes damping but rather the frictionaltorque, T, , which opposes the direction of the appliedtorque on the shaft.)Because a shaft is usually driving a load a t its end,it is safe to assume (to simplify the equations withoutmuch error) that the system consists of two rotatingniasses connected by a shaft whose inertia is negligiblea5 compared to the end masses. So we can say that theapplied torque is a constant along the shaft length.If the angular displacement is assumed to be zero atthe end ( A ) of the shaft, the displacement at the disk(a t EG) is in the form (see list of symbols):

1'Le = - - JG

And that the displacement at the end B in the formT ( h - L2)aD14GG/32s = errc +

The angular displacement of the sleeve at EG (withzero displacement at the fixed end A ) is

The strain energy of the shaft caused by vibration isu =u1+ uz ( 3 )u, = ; [ T ( I- ) e EG+u e n - LCc)i (6 )Uz =!zXT(es)~c (7)

(8)

The maximum strain energy of the shaft can be derivedasThe corresponding energy in the sleeve isThe difference between the energies in the shaft andth sleeve must be the energy dissipated by friction:

11 =~ X T [ ~ L ' Ge s ) ~ ( i ]Chenea and Hansen (Ref 1 ) proved that the reduc-tion in amplitude (and hence dissipated energy) is 5 0 %of the maximum strain for a half cycle. Therefore, for


27/70

d to shaffJ

4. Application of the friction-damping techniqueto dampen torsional vibrations in an engine f l ywheel system. Both flywheels are free to rotateon bushings and are driven by a crankshaftthrough friction disks. The flywheels are pressedagainst the disks by mea ns of loading springsand adjustable nuts. When, due to resonance,large deflections (vibrations) of t h e shaft andhub occur, the inertia of the. flywheel preventsthem from duplicating t h e vibrations; there isrelative motion between the hub and t h e flywheels. As a result, friction energy (o f vibration)is dissipated. The change of total system energyf rom a torsional deflection results in a decreasein the amplitudes of vibrations.

each full cycle of damping, the amplitude is recfuced bya factor of 4 or, in other words, the energy dissipatedis raised by a fac tor of 4. This accounts for the factor2 in Eq 8.Note at this point that when the relative displacement~ K c ; - ( ~ . ) ~ ; cs zero there is no relative motion, and henceno damping action.Determination of dampingpressed in terms of a ratio (and is shown in Fig 3) :The amount of damping in any system can be ex-

Mngiiitutlr of cncrgy after d a m p i n gMagnitude of energy beforc d a m p i n g=

Uu + f7=-\ Y I l ( T ( ? u = LI +u 2

I f the damping action is to be a maximum, the ratio>f R must be so chosen as to make U/(U - W ) a mini-mum or W/(U+ W ) , which is the percentage of en-ergy dissipated, a maximum.I17 1-=L +1v L1 V + l

The ratio U / must be a minimum, hence W / mustbe a maximum. Using the previous equations (Eq 1 to 8)


Differentiating with respect to T J T , and equating theresult to zero, results in a value for T , / T which is theoptimum value.

where X = T s / T .Solving the quadratic for h , results in:

x = a [ l -1/1 - (l/ar)] (11)where r = 1 +D1/8HC3=1+mc = Dz/D1a = L l / L 2The ratio in , which is equal to D 1 / 8 H C 3 , s the ratioof the torsional stiffness of the shaft to that of the sleeve.The corresponding fractional value of the energydissipated per oscillation at optimum A is equal to 1-R.The key to the design chart, Fig 3, is the fact that thefractional energy curve is not in direct proportion to theratio L z / L I of the sleeve length to the shaft length. Thisallows the designer a choice between a full-length sleeveand a stiffer sleeve placed over part of the shaft length.The chart shows that for the same damping capacity,a sleeve 1 / 3 or 1/5 the length of the shaft must be manytimes stiffer than one covering the entire length of theshaft.Damping vs forced vibration

Suppose a cyclic forcing function is imposed upon theshaft, causing a vibration at its fundamental naturalfrequency. The resulting increment per half oscillationof torsional displacement in the absence of damping,is equal to AB. As a result of introducing dry frictiondamping, this displacement will become zero when thelosses due to the energy dissipated are equal to the gainfrom the forcing function. It is desirable to have theenergy dissipated at the smallest possible torsional dis-placement; in other words when

B/AO =minimum (12)This can only be true when h is equal to 1-R. here-fore, the inverse of Eq 12, AWB, is a ratio of energydissipated, and nele = 1- R (13)

or A0e=--- 1 - RThus, if we know the increment of amplitude ABproduced by the forcing function (assuming the forcingfrequency equals the natural frequency and that dampingis zero), we can calculate the torsional displacement towhich the system can be limited for any value of thedamping ratio, R.Application to an engine

Actually, this procedure could be used for any appli-cation of rotating parts where space and weight con-siderations are critical. The general effect of torsionalvibrations is to decrease the allowable stresses on atransmission shaft.One of the earliest applications of coulomb frictionto reduce torsional vibrations is found in gasoline anddiesel engines and is called the Lanchester Damper.


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4-28

Synchronous =otornduction motor Gear b o x

It is shown in Fig 4 (see Vibration Problems in Engi-neering, 3rd Edition, Van Nostrand Co, pp 265-268).Other design considerations

If weight is a primary objective, make the dampingsleeve diameter as large as possible to gain the largestweight saving.If weight is not important, it is probably best to goto a sleeve diameter only slightly larger than the shaftitself.You have a choice fo r the length of sleeve, rangingfro m a full-length sleeve to one of one-tenth the totallength. In the latter case. make s ure that you design intothe sleeve sufficient rigidity and stiffness.Reduce the wall thickness at the end of the sleevein contact with the disk so that the contact pressure willnot induce large stresses in the sleeve. Make sure thatthis contact pressure is uniform around the peripheryof the friction end of the sleeve.Frictional torque depends on the coefficient of frictionand the normal pressure exerted by the sleeve. It is noteasy to measure the coefficient of friction under dynamicconditions, but there are values tabulated by manyauthors. You can vary the pressure by using a variable-diameter disk. In this way, the optimum value of damp-ing can be empirically determined.Dont worry too much about fretting corrosion at thefriction surfaces because: 1 ) the friction torque is low(relative to shaft torque); 2) the normal force is dis-tributed over a large area so as to limit the pressure tolow values; 3) even i f fretting occurs to some slightdegree, it will not affect the torque-carrying shaft; 4 ) thefriction surfaces need not be metallic (an elastomer orany viscoelastic material works well)Numerical problem

A sleeve is to be designed for a shaft transmittingpower to an air compressor for a supersonic windtunnel, Fig 5. The shaft has a diameter of D I = 7.5 in.and a length of L1 = 8 ft =96 in. Thus L, nd D1 arefixed and L2, H and D 2 are values which must be deter-mined. As will be seen in this problem, L? an d Dz areselected on a trial and error basis, and H, which is thethickness of the sleeve wall and thus the importantparameter which influences the total weight of the shaft,is determined from the chart in Fig 3 and its abscissaequation.Solution

It is generally accepted that with most dry-frictiondamping there will be approximately 3% of dampingtaking place per cycle. If the forcing torque were re-

Compressor

moved for one cycle, the strain energy would drop to97% of its maximum value and the angular displacementwould likewise drop to 0.97 8. Therefore, the forcingtorque must be such as to increase the angular displace-ment by an amount. or ( i n the absence of damping):A0 =0.038 per cycle =0.0150 or half cycle

We would like to limit the A t value of torsional vibra-tion to 10% of the steady displacement which is a resultof the mean torque in the shaft.Substituting in Eq 14 and solving for R,

e -H -

From Fig 3, this value of R (damping ratio) requiresan m-value equal to 5.2 and a damping/critical dampingvalue of 2.6%. Thusm =D1/811C3= 5.2

Since D , =7.5 in.HC S =0.1S02We can now choose how the produce HCR is to bemad e up. If we pick D, to be twice D n, then C =D 1 / D p = 2, and H = 0.1802/8 = 0.0225.This provides a sleeve thickness of about 24 page,which has only 2.7 per cent of the weight of the shaft.Thus we obtained a 10:1 reduction in the amplitude orvibration at the cost of very little extra weight. Tocompute the resisting friction torque, from Eq 11 weobtain - -

L

x =.- T, = 1 1 1 - J l -1 1 =0.09T 2)~ -From the engine characteristics, it is known that T =800,000 Ib-in. Therefore. the frictional torque is

T,=800,000(0.09)= 72,000 Ib-in.With a diameter of 15 in. on the sleeve, the frictionforce is equal to 1528 Ib/in. If we further assume amaterial will be used with a coefficient of friction equalto 0.6, the normal force per inch of periphery is 2,547 Ib.This amount of pressure is small compared with thekind of pressure usually associated with fretting fatigue.References1. P. F. Chenea and H. M. Hansen, Mechanics of Vibra-tion, John Wiley and Sons nc, 1952, pp 319-324.2. D. Williams, Damping of Torsional Vibrations by DryFriction, Royal Aircraft Establishment, 1960 (Fig 3).

5. Transmission system designed fo r frictiondamping. Numerical ex-ample below shows how the addition of a very thin sleeve with a wallthickness of 0.023 in. reduces the amplitude of vibration by 10 to 1.


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15 Ways to FastenGears to ShaftsSo you've designed or selected a good set of gears foryour unit-now how do you fasten them to their shafts?Here's a roundup of methods-some old, some new-witha comparison table to help make the choice.1 M. Rich

1 PINNINGPin nin g of gears to shafts is still considered on e of th e m ost posi-t ive methods. Various types can be used: dowel, taper, grooved, rollpin or spiral pin. These pins cross through shaft (A) or are parallel(B) . Latter method requires shoulder and retaining r ing to preventen d play, b u t allows quick removal. Pin can be designed to shearwhen gear is overloaded.Main drawbacks to pinning are: Pinning reduces the shaft cross-section; difficulty in reorienting the gear once it is pinned; problemof drilling the pin holes if gears are hardened.Recommended practices are:Fo r good concentr icity keep a maximum clearance of 0.0002

Use steel pins regardless of gear material. Ho ld gear in place on*Pin d ia should never be larger than 8 the shaft-recommended

Simplified formula for torque capacity T of a pinned gear is:

to 0.0003 in . between bore and shaft .shaft by a setscrew during machining.size is 0.20 D to 0.25 D.

T =0.757 @Ilwhere S is safe shear stress and d is pin mean diameter .

( A )Shoulder


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4-30

2 CLAMPS AND COLLETS1 Hub clamp Sloffed hub

Slighf clearance

Clamping is popular with instrument-gear users becausethese gears can be purchased or manufactured with clamp-type hubs that are: machined integral ly as part of the gear( A ) , or pressed in to th e gear bore. Gears arc a lso availablewith a colle t-hub asscmbly ( B ) . Clamps can be obtainedas a separatc i tem.Clam ps of onc-picce construct ion can brcak u ndcrexcessivc clamping pressure; hence the preference for thetwo-piece c lamp ( C ) . Th is p laces the s tress onto th escrcw threads which hold the clamp together, avoidingpossible fracturc of th e clamp i tself. H u b of th e gearshould be slot ted into three or four equal segments, witha thin wall sect ion to reduce the size of the clamp. Hard-

3 PRESS FITSPress-fit gears to shafts wh en shafts are too sinall forkeyways and w here to rqu e transmission is relatively low.Method i ,s inexpensive but impractical where adjustmentsor disassemblies are expected.Torque capacity is:

T =0.785 f D l LeE [ - ($Result ing tensi le stress in the gear bore is:

S = e E j D ,where f =coefficient of friction (generally varies between0.1 and 0.2 for small metal assemblies), D , is shaft dia ,D , is OD of gear, L is gear width, e is press fit (differencein d imension be tween bore and shaf t ) , and E is modulusof elasticity.Similar metals (usu ally stainless steel wh en used ininstruments) are recommended to avoid difficul t ies aris-ing from changes in tem perature . Press-fi t pressures be-tween s tee l hu b and shaf t a re shown in char t a t right (fromMarks' IIand boo k). Curves are a lso applicablc to hollowshafts, providing d is not over 0.25 D.

ened gears can be sui tably fas tened wi th c lamps , but hubof the gear should b e s lo t ted pr ior to ha rdening.Ot her recommenda t ions are : M ake gear hub approxi-mately same length as for a pinned gear; slot throughto the gear face at approximately 90" spac ing. Whi leclamps can fasten a gear on a spl ined shaft , results arebest if bo th shaft an d bore are smooth. If both sp l ined,clainp then keeps gear from moving la teral ly.Materia l of c lamp should be same as for the gear, espe-cially in military eq ui pm en t because of specifications ondissimilarity of metals. However, if weight is a factor,a l un~ i nun~ -a l l oylamps are effective. Cost of the clampand sl i t t ing the gear hu b are relat ively low.

L I I

Allowonce per inch of shaft diam., e


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Shafts & Couplings 4-3 1

ComparisonMethod

-Pinning-Clamping-Press flts-1octite--Setscrews--Splining-Integral shaft-Knurling-Keying-Staking-Spring washer-Tapered shaft-Taperad rings-Tapered bushing-Die-cast assembly

o f Gear-Fastening Methods

ExcellentGoodFairGoodFairExcellentExcellentGoodExcellentPoorPoorExcellentGoodExcellentGood

PoorExcellentFairGoodExcellentExcellentPoorPoorExcellentFairExcellentExcellentExcellentExcellentPoor

ExcellentFairGoodGoodPoorExcellentExcellentGoodExcellentPoorGoodExcellentGoodExcellentGood

ExcellentFairFoirExcellentGoodFairGoodPoorPoorPoorFairGoodExcellentGoodExcellent

ExcellentGoodGoodExcellentFairExcellentExcellentGoodExcellentGoodGoodExcellentGoodGoodGood

y15.s E.- .s gL 5

HighModerateModerateLittleModerateHighHighModerateHighModerateModerateHighModerateModerateLittle

Poor HighExcellent MediumExcellent MediumExcellent LowGood LowExcellent HighExcellent HighPoor MediumExcellent HighPoor LowExcellent MediumExcellent HighExcellent MediumExcellent HighFair Low

4 RETAINING COMPOUNDSSeveral different compounds can fasten the gear ontothe shaft-one in particular is "Loctite," m anufa ctured byAmerican Sealants C o . This material remains l iquid aslong as it is exposed to air , but hardens when confinedbetween closely fitting metal parts, such as with close fitsof bolts threaded into nuts. (Mili tary spec MIL-S-40083

approves the use of retaining com po und s) .Loctite sealant is supplied in several grades of shearstrength . T h e grade, coupIed with the contact area,de termines t he to rque t ha t can be t ransmit ted . For exam-ple: with a gear 3 in . long on a A -in.-dia shaft , th e bonde d'area is 0.22 in ." Using Loctite A with a shear strength

5 Setscrews

of 1000 psi, the retaining force is 20 in.-lb.Loct i te will wick into a space 0,0001 in . or less and filla clearance up t o 0.010 in . I t requires about 6 h r t oharden, 10 min. with activator or 2 min. if heat i9 applied.Sometimes a setscrew in the hub is needed to posit ionthe gear accurately and permanently until the sealant hasbeen completely cured.

Gears can be easily removed from a shaft or adjustedo n t h e s h a f t by forcibly breaking the bo nd and thenreapplying th e sealant af ter th e new posit ion is determin ed.I t will hold any metal to any other me tal . Co st is lowin comparison to other methods because extra machiningand tolerances can be eased.

6 GEARS INTEGRALWITH SHAFTFabricating a gear and shaft fromthe same material is sometimes eco-

nomical with small gears where costof machining shaft from OD of gearis not prohibitivc. Method is alsoused when die-cast blanks are feasibleor when space limitations are severeand there is no room for gear hubs.No l imi t to the amount of to rquewh ich can be resisted-usualIy gearteeth will shear before any other dam-age takes place.

Tw o se tscrews a t 90" or 120" to each other are usuallysufficient to hold a gear firmly to a shaft. M ore securityresults with a flat on the shaft , which prevents the shaftfrom being marred. Flats give added to rque capacity andare helpfu l for freq uen t disassembly. Sealants applied onsetscrews preve nt loosening d urin g vibration.


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4-32

0' / z3/4

7 SPLINED SHAFTS

4-spline 6-splineW WI _ _

0.120 0.1250 I8 0. I88

Ideal where gear must sl ide in la t-eral direct ion during rotation. Squaresplines often used, b ut involute spl inesare self-centering and stronger. Non-sliding gears are pinned or held bythreaded n ut o r re taining ring.

78 [ 0.211I 0241

I - '% 0301

Torque s t rength i s h igh and de-pe nde n t on numbe r of splines em-ployed. Use these recomm ended di-mensions for width of square toothfor 4-spline and 6-spline systems; al-

0 2190.2500 3 1 3

though other spl ine systems are sometimes used. Stainless steel shafts andgears are recom men ded. Avoid dis-similar metals or alum inu m. Relativecost is high.

8 KNURLING A knurled shaft can be pressed into th e gear bore, to doits own broaching, thu s keying itself in to a close-fittinghole. T h i s avoids need for supplementary locking devicesuch as lock rings and threaded nuts.not weaken or distort parts by the machining of grooveor holes. I t is inexpensive and requires no extra parts.Knurling increases shaft dia by 0.002 to 0.005 in . I t isrecomm ended that a chip groove be cu t a t th e trai l ing edgeof t he knurl . Ti gh t tolerances on shaft and bore dia arenot needed unless good concentric i ty is a requirementThe unit can be designed to sl ip under a specific load-hence acting as a safety device.

T h e meth od i s applied to shafts $ in. or under and does

9 KEYING

Generally employed with large gears, but occasionallyconsidered for small gears in instrum ents. Feath er key(A) al lows axial movement but keying must be mil led toen d of shaft. For blind keyway ( B ) , use setscrew againstthe key, but m etho d permits locat ing the gear anywherealong length of shaft.

Keyed gears can withstand high torque, much morethan the pinned or knurled shaft and, a t times, lnorc thanthe splined shafts bccausc thc key extends wcll into both

the shaft and gear bore. Torque capacity is comparablewi th tha t of the integral gear and shaft . Main tenance iseasy because the key can be removed while the gearremains in the system.Materials for gear, shaft and key should be similarpreferably steel. Larger gears can be either cast or forgedand t he key either hot- or cold-rolled steel. How ever, ininstrument gears, stainless steel is required for mostapplicat ions. Avoid alum inum gears and keys,


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10 STAKING 11 SPRING WASHER

8 I 0020

Torque

0.002000020

I

It is difficult to predic t the stre ngth of a staked joint-but it is a quick andeconomical method w hen t he gear is posit ioned a t the end of the shaft .Results from five tests we made on gears staked on 0.375-in. hubs are shownhere with typical notations for specifying staking on an assembly drawing.Staking was done with a 0.062-in. punch having a 15" bevel. Variables inthe test were: depth of stake, number of stakes, and clearance between huband gear. Breakaway torque ranged from 20 to 52 in.-lb.Replacing a gear is not simple with this method because the shaft is muti -rated by the staking. Bu t production costs are low.12 TAPERED SHAFT

Tapered sha ft and m atching taperin gear bore need key to providehigh torque resistance, and threadednut to t ighten gear onto taper . Ex-pensive but suitable for larger gearapplications where rigidity, concen-tricity and easy disassembly are impor-t an t . A larger clia shaft is neecledthan wi th o ther methods. Space canbe problem because of protrudingt! ireadcd end. Keep nu t t igh t.

13 TAPERED RINGS

Tliese inter lock and expand whentightened to lock gear on shaft . Apurchased item, the rings are quickand easy to use, and d o no t need c losetolerance on bore or shaft . No specialmachining is required and torque ca-pacity is fairly high. If lock washer isemploycd, tlic gear can bc adjusted toslip at predetermined torque.

15 DIE-CAST HUBDie-casting machines are available, which automaticallyassemble and position gear on shaft, then die-cast a metal

h u b o n b o th s id es of gear for retention. Method canreplace staked assembly. Gears are fed by hopper, shaftsby magazine. Method maintains good tolerances on gearwobble, concentrici ty and location. For high-procluctionapplications. Cost s are low once dies are made.

Assembly consists of lockn ut, springwasher, flat washer and gear. Thelocknut is adjusted to apply a pre-determined retaining force to thegear . This permits the gear to sl ipwhen overloaded-hence avoidinggear breakage or protecting the drivemotor from overheating.Construction is simple and costsless than if a slip clutch is employed.Popular in breadboard models.

14 TAPERED BUSHINGS

This, too, is a purchased item-but generally restricted to shaft di-ameters t in. and over. Adaptersavailable for untapered bores of gears.Unthreaded half-holes in bushingalign with threaded half-holes in gearbore. Screw pulls bushing into bore,also prevents rotational slippage ofgear under load.


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4-34

14 Ways to Fasten Hubs to ShaftsM. Levine

Shoddei may bei n f h r o o g h s h a f f

1 Cuppoint setscrew.. .in hub (A ) bears against flat on shaft. Fastening suitable forfractional horsepower drives with low shock loads. Unsuitablewhen frequent removal and assembly necessary. Key with set-screw (B) prevents shaft marring from frequent removal andassembly. Not suitable where high concentricity is required.

Can withstand high shock loads. Two keys 120" aport (C) rans-mit extra heavy loads. Straight or tapered pin ( 0 ) preventsend ploy. For experimental setups expanding pin is positiveyet easy to remove. Gear-pinning machines are available. Taperpin (E) parallel to shaft may require shoulder an shaft. Can beused when gear or pulley has no hub.

Stroight -s ided 4-spl inef l n v o l u i e splines may

4 Splined shafts ...are frequently used when gear must slide. Square splines canbe ground to close minor diameter fits but involute splines arewlf-centering and stronger. Nan-sliding gears may be pinnedto shaft if provided with hub.

7 Interlocking.. .tapered rings hold hub tightly to shaft whennut is tightened. Coarse tolerance machiningof hub and shaft does not effect concentricityas in pinned and keyed auernblier. Shoulderis required (A) for end-of-shaft mounting; endplates ond four bolts (B) allow hub to bemountad anywhere on shaft.


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r I

2 Tapered shaft.. . 3 Feather key . . .with key ond threoded end provides rigid, (A) allows axial movement. Keyway must be milled to end of shaft. For blind keywayconcentric assembly. Suitable for heovy-duty (B) hub ond key must be drilled and tapped, but design ollows gear to be mounted any-opplications, yet con be easily dissasembled. where on shaft with only a short keyway.

-Re ta in ing r ing

5 Retaining ring ... 6 Stamped gear.. .allows quick removal in light load applications. Shoulder on ond formed wire shaft used mostly in toys. Lugs stomped onshaft necessary. Pin securing gear to shaft can be shear-pin Bend radii of shaftif protection against excessive load required. both legs of wire to prevent disouembly.shou!d be small enough to allow gear to seat.

8 Split bushing .

o f ho / & th readed .B u s h in g h a l f o fhole &t threaded.

for removing f roms h a f t : B us h ing ha f fof ho l e t h readed .Hub half o f holen 2 f t h r e a d e d .

ISllght clearancehas tapered outer diameter.bushing half i s un-topped.hub as screw is screwed into hub.by a reverse procedure.IO-in. dia shafts. Adapters are available for untapered hubs. shaft. Ideal for experimental set-ups.

Split holes in bushing align withScrew therefore pulls bushing into

Bushing is iocked from hubSizes of bushings ovaliable for %- to

split holes in hub. For tightening, hub half of hole is topped, 9 Split hub.. .of stock precision gear is clomped onto shaft with separate hubclamp. Manufacturers list correctly dimensioned hubs and clompsso that efficient fastening can be made based on precision ground


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4-36

Attaching HublessGears to-ShaftsThin gears and cams save space-but how to fastenthem to their shafts? These illustrated methods givesimple, effective answers.L. Kasper

-.---__--_/------

/-Plafe \< _- -I=--+,

3 P L A T E gives greater resistance to shear whenratli,il loads ;ire likely to be hea vy. W1ir.n th e ge ar ismounted, the plate becomes the driver; the centers c r e w merely ar t s as a retainer .

1 C O U N T E R B O R E with close fit un shaf t ensuresconcentr ic mounting. Torque is t ransmit ted by pins ;posit ive fastening is provided by f lathead screw.

2 T I G H T - F I T T I N G washer in counterbored holecarries the rad ia l load ; i t s shear area is large enoughto ensure aniple strength.


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Shafts & Couplings 4-37

4 K E Y A N D F L A T T E D T A PE R- PI N should not pro-'trude above surface of gear; pin length should beslightly shorter tha n gear width. Note that this at -t a ch ment i s n o t p o s i t i v e g e a r r e t en t io n i s b y f ri c ti o nI nly.6 T A P E R E D P L U G is another fr iction holding de-vice. This type mounting should be used so t h a t t h eradial load will tend to t ighten ra the r than loosen theI hread. For added security, thread can be lef thandto reduce tampering r isk.I

I

5 D-PLATE keys gear to shaf t ; op t imum slo tdepth in shaft will depend upon torque forces andstop-an d-start requirements-low, con stan t torq uerequires only minimum depth and groove length;heavy-duty operation requires enough depth t o pro-vide longer bearing surface.

W7 TW O FRICTION D ISKS, tapered to about 5" in -cluded angle on their r ims, are bored to fit t h e sh a f t .Flathead screws provide clamping force, which canbe quickly eased to allow axial or rad ia l ad justmentof gear.

8 T W O P IN S in radia l hole of shaft provide posit ivedr ive tha t can be easi ly d isassembled . Pins w i thconical end are forced t igh tly togethe r by f latheadscrew s. Slot leng th should be sufficient to allow pins:o be withdrawn while ge ar is in place if backside3f gear is " t ight" against housing.


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4-38

10 Different Typesof Splined ConnectionsW. W. Heath

C Y L I N D R I C A L T Y PE S

Tooth Propor t ions

SQUARE SPLINES make a simple connection1 and are used mainly for applications of lightloads, where accurate positioning is not important.This type is commonly used on machine tools; a capscrew is necessary to hold the enveloping member.

SERRATIONS of small size are used mostly for applications of2 light loads. Forcing this shaft into a hole of softer material makesan inexpensive connection. Originally straight-sided and limited tosmall pitches, 45 deg serrations have been standardized (SAE) withlarge pitches up to 10 in. &a. For tight fits, serrations are tapered.

INVOLUTE-FORM splines are used where high loads are width or side positioning has the advantage of a full fillet radius5 to be transmitted. Tooth proportions are based on a 30 deg at the roots. Splines may be parallel or helical. Contact stressesstub tooth form. (A) Splined members may be positioned either of 4,000psi are used for accurate, hardened splines. Diametralby close fitting major or minor diameters. (B) Use of the tooth pitch above is the ratio of teeth to the pitch diameter.

Addendum-- --F A C E T Y P E S

MILLED SLOTS in hubs or shafts make an inexpensive connection.8 This type is limited to moderate loads and requires a lockingdevice to maintain positive engagement. Pin and sleeve method is usedfor light torques and where accurate positioning is not required.

9 RADIAL SERRATIONS by milling or shaping theteeth make a simple connection. (A) Tooth propor-tions decrease radially. (B) Teeth may be straight-sided(castellated) or inclined; a 30 deg angle is common.


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STRAIGHT-SIDED splines have been widely used in the auto-3 motive field. Such splines are often used for sliding members. Thesharp corner at the root limits the torque capacity to pressures of ap-proximately 1,000 psi on the spline projecred area. For different appli-cations, tooth height is altered as shown in the table above.

4 MACHINE-TOOL spline has a wide gap betweensplines to permit accurate cylindrical grinding of

the lands-for precise positioning. Internal parts canbe ground readily so that they will fit closely with thelands of the external member.

Snap ring holds ARBER-COLMAN CO1-ossembly togetherSPECIAL INVOLUTE splines are made by using gear6 tooth proportions. With full depth teeth, greater contact

area is possible. A compound pinion is shown made by croppingthe smaller pinion teeth and internally splining the larger pinion.

TAPER-ROOT splines are for drives which require positivepositioning. This method holds mating parts securely.

With a 30 deg involute stub tooth, this type is stronger thanparallel root splines and can be hobbed with a range of tapers.

GLEASON GAR WORKS10 CIJRVIC COUPLING teeth are machined by a face-millWhen hardened parts are used which

require accurate positioning, th e teeth can be ground. (A) Thisprocess produces teeth with uniform depth and can be cut at

any pressure angle, although 30 deg is most common. (B) Dueto the cutting action, the shape of the teeth will be concave(hour-glass) on one member and convex on th e other-themember with which it will be assembled.

type of cutter.


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4-40

Typical Methods ofCoupling Rotating Shafts IMethods of coupling rotating shafts vary from simple bolted flangeconstructions to complex spring and synthetic rubber mechanisms.Some types incorporating chain, belts, splines, bands, and rollersare described and illustrated below.

She/ grid fransmifs flo ati ng s/eeve, carryingaskef befween housingf/anges rrfuihs /ubricanf, /power a d bsorbs genera ed nf ernu / spfines Loch sef o f pl ines in mesharound enfire circumfevence.I fhe s/eeve permanem'& en- unit. 1a t e r a / andangu/arI page fhe spfinesof ach hub, P/V a/ow dbe fwtm SPfine

I I shock and vibmfion afm& end. 7he spf inesd Assembly revdvesas oneII

//// I faces-

II ---_/'fau./irclte epration I,/ ofceni+Uanges---ubs ar e pressedon andkeyedbeach .$'ewejack ho/es fo 7

nufa in ffanges

'Oi/ f i jm befween s p h e se/linirxr3 mefa/fc-mata/con adThe Falk Corp. oi/ o immersesphnesBartlett- HoywordDiv., Y o p p m Co., Inc.IG . 2

Oi/ho/ewith sat&& -nod ing housing she//-Nus---_ ~ wf wi'fh nferna/geurs Tapered bores do no+run comp/ete& fhrou9hhubs',II

Ienerafeedspherical Iqears on hubs, I

Double - fapered jb w shel d by kevseafs in e n dwunterbored

', Jaws machihedon h n e r s u r f a c e rb', ad/us es s f han shaR Shcrffgr//;opdfogefher by bolts'\bU/bws when f lange s a re dmwnGaskef befweenflanges to ensure / LC/earuncespace befween Io,/iighhf>ea/------ J hubs fo a//ow i r e n d p / a y / FIG. 4/80ffs draw ffanqedhubs togefher W. H Nicho l~on nd CoIG . 3 Barcus Engineering CaJnc.


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Shafts &Couplings 4-4 1

- -

FIG. 5Removable uccess p/u& iosprings

,-Mess kofher be/f,'acedfiro& a/-

on each ri m

I \/ '.

I

I

/ II I/ IWith ro//ers /ocafedin/ I

II

l 4 . r F I G . 8..-,-t o ah&


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4-42

Typical Methods ofCoupling Rotating Shafts IIShafts couplings that uti lize internal and external gears, balls, pinsand non-metallic parts to transmit torque are shown herewith.

Me fa / hous ing over T f l a n g eneoprene bc;cuifs I

F/ex ib /e d isk /onga ' ted hdesM e f a / facin p / a t e s o f v u /c a ni z edi v e f e d t o J e x i b l e r u b b er e d f ab ri c?disk iDrevenfrece ive boss f rommating f / a n y e

Neoprene cenrer designe d for unifor m stress,/,near deflection and abs orpt ton of yibration

Sha ft keyedt o f lange7

.JM e f a / / i c _,,'scre en core--&' ,'

Oufer fabr ic r i n i mpregna fed w i thneoprene, provides support fo r center sectionF I G . 1 Morse Chain Cc

Compensat ing member r -F /ang ed outerpro9ides connect ion )I s/eeve bofteabetween hub and outer I d i rec t /y tosleeve---, ,' f / a n g e d hubF/ a n s e d h u b ', IIk e y g d t oshaf t - . _. .Tapped hofesf a c i / i t a t eassemblyan d d;s-assemb/y---

Generatede x t e r n a la n d n f er n o/g e a r f e e t h --

I I- - _ _xcessive wear----. -.

disk andente,

-- Boston Gear W o r k , IncFIG. 2---Large numberSetscrews o f e e t h

secure hubs pro du ce veryt o shaff ---- large bearin gsur face

FIG.3 Boston Gear W o rks . Inc.

- W i l e f a ceof in ternalyea r feefhperm i isf u l l e n df / o a f wi fh -ou t di i -engagemenf

'- Gearedhub keye df o s h a f t

L o n g yea r teefb in s/eeve r-Two fap pe d ho/es io e a c h h u bp r e v e n t h o b f r o m ; fac i / i fa te assembly an dr em add i s e n g a g h g - - - - - ;Clearance between \, ,' o i l / e a k a g e# ,askef pre ven ts

INexib/e,oi/-resis fant pac king retmmso i l inside the coupling and excludesdir t , grit a n d mo is tu re

Farrel Birmingham CO.,InC.IG .4 hr r t l -8 i rmmghom b . , I n c . FlG.5


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Shafts & Couplings 4-43

4ub ber vutcanize dfo stee/p/ates,,;'I'// /

Oufer fubric ring Me al l i ' c r - -Trunn ion pins f i f i e dimpregnu e d wi ih sc reen ,' info ou f e r d i am e t e r,' o f hub und we/ dede o p r e n e y videss u p p o r t or cenfer /ore-\Plates bo / iedio flanges------

F10.6

- wo-piece housingc l om ps aroundneoprene biscui s .f a c e ofhous inga t t a c h e s tos t a n d a r d f l a ng e

Hubs secured fo

& /I bearing inserf perm&unrestr,cte d end p / a - --,F I G . 8 B o s t o n Gear Works, In c

f / e x i 6 / e l a m i n u f e d p m u ni tscompensafe for misa/ ignmentof connected s h a f f s . One endheld by s p r ih g r e h n i n g nng,other end moves lafera//y inbushjng . I ri ny holds f/exib/e F I G . 9 Boston Geur Works, ~ Lr -Spr ing re f a i n i ng1 pm u n i t s r n /angef / a n g e hubssecured foshaf t w i thsetscrewsand keys ---

Sf e1aminat;onssw i ve l oncross p i nsin d o t fkeeper

Cross pin

wax ~brol:. impregnu ed , se /f ;ubriccrt ingm e bushings

F I G . IO John Waldron Corp

No lubricafibn is req wre d on this c o u p h g,4Ba//s con t u i ned i npockefs in f l angePrcyect ion onr e t a i n i n g cove rholds bo l l s inp ro p e r posifion --

,Huh keyed for,* 5haf i .sFacep / u erefo ics bu/ l s -,

---Smal/ h&s in f langefo a i d i n r em ov in gPower frons bu//am i f f e d h v six :ubber b L I k - - - /

Crocker-Wheeler D i v ,FIG . 11 J os hua Hendy I r o n Works


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4-44

Typical Designs ofFlexible Couplings ICyril Donaldson

H u b , ,Hub

FIG.5

Fig . 1-A rubber hose clamped to two shafts. Fo r applicatiotlswhere the torque is low and slippage unimportant. It is easilyassembled and disconnected without disturbing either machine ele-ment. Adaptable to changes in longitudinal distance betweenmachines. This coupling absorbs shocks, is not damaged by over-loads, does not set up end thrusts, requires no lubrication andcompensates for both angular and offset misalignment.Fig. 2-Similar to Fig. 1, but positive drive is assured by bolt-ing hose to shafts. Has Same advantages as type in Fig. 1, exceptthere is no overload protection other than the rup ture of the hose.Fig. &The use of a coiled spring fastened t o shaft s gives th esame action as a hose. Ha s excellent shock absorbing qualities,but torsional'vibra tions are possible. Will allow end play in

shafts, but sets up end thrust in so doing. Othe r advantages aresame as in types shown in Figs. 1 and 2. Compensates for mis-alignment in any direction.Fig. 4-A simple and effective coupling for low torques andunidirectional rotation. Stranded cable provides a positive drivewith desirable elasticity. Inertia of rotating parts is low. Easilyassembled and disconnected without disturbing either shaf t. Cablecan be encased and length extended to allow for right angle bendssuch as used on dental drills and speedometer drives. End s ofcable are soldered or bound with wire to prevent unraveling.Fig. 5-A type of Fa lk coupling that operates on the sameprinciple as design shown in Fig. 6, but has a single flat springin place of a series of coiled springs . High degree of flexibilityobtained by use of tapered slots in hubs. Smooth operation, ismaintained by inclosing the working parts and packing with grease.Fig . &Tw flanges and a series of coiled springs give a high

degree of flexibility. Used only where the shafts have n o free endplay. Needs no lubrication, absorbs shocks and provides protec-


45/70


t i o n again>t overloads. but will set up torsional vibrations. Springscan be o f round or square wire with varying sizes and pitches toallow fo r any degree of flexibility.

Fig. 7-1s similar to Fig. 6, except that rubber tubing, re-inforced by bolts, is used instead of coiled springs. Is of sturdierconstruction but more limited in flexibility. Ha s no.over load pro-tection other than shearing of the bolts. Good anti-vibrationpropert ies if thick rubber tubing is used. Can absorb minorshocks.Fig. %--A series of pins engage rubber bushings cemented intoflange. Coupling is easy to install. Flanges being accurately ma-chined and of identical size makes accurate lining-up with spiritlevel possible. Will allow minor end play in shafts, and providesa positive drive with good flexibility in all direction.Fig. 9--1 Foote Gear Works flexible coupling which has shearpins in a separate set of bushings to provide overload protection.Construction of studs, rubber bushings and self-lubricating bronze

bearings is in principle similar to that shown in Fig. 10. Replace-able shear pins are made of softer material than the shear pinbushings.Fig. l&A design made by the Ajax Flexible Coupling Com-

pany. Stu ds are firmly anchored with nuts and lock washers andbear in self-lubricating bronze bushings spaced alternately in bothflanges. Thick rubber bushings cemented in flanges are forcedover the bronze bushings. Life of coupling said to be considerablyincreased because of self-lubricated bushings.Flexibility isobtained by solid conically-shaped pins of metal or fiber. Thistype of pin is said t o provide a positive drive of sturdy construc-tion with flexibility in all directions.

Fig . 12-In this Smith & Serrell coupling a high degree of flex-ibility is obtained by laminated pins built-up of tempered springsteel leaves. Sprin g leaves secured to holder by keeper pin. Phos-phor bronze bearing strips are welded to outer spring leaves andbear in rectangular holes of hardened steel bushings fastened inflange. Pins ar e free to slide endwise in one flange, but are lockedin the other flange by a spring retaining ring. Th is type is usedfo r severe duty in both marine and land service.

Connection can be quickly disassembled.

Fig. 1 -Another Foote Gear Works coupling.

1 FIG.10


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4-46

Typical Designs oFlexible Couplings IICyril Donaldson

i er or f iber disk

I FJexibIe d isks .

F g . 13-111 this B r o w n Engineerillg Coinpany couplillg1kxihiIit)- is increased by addition of 1)ufler-slots in the lami-nated leather. These slots also aid in the absorption of shockloads and torsional vibration. Under parallel misalignmentor shock loads, buffer slots will close over their entire width,but under angular misalignment buffer slots will close on!>-on one side.F ig . 14--Flexibility is provided by resilience of a rubber ,

leather, or fiber disk in this w'. A. Jones Foundry & Ma-chine Company coupling. Degree of flexibility is limitedto clearance between pins and holes in the disk plus theresilience of the disk. Ha s good shock absorbing properties,allows fo r end play and needs no lubrication.Fig. 15-A coupling made by Altlrich I'ump Company,similar to Fig . 14, except bolts are used instead of pins. Thiscoupling permits only slight endwise movement of the shaftand allows machines to he temporarily disconnected with-out disturbing the flanges. Driving and driven membersare flanged for protection against projecting bolts.Fig. l c l am in at ed metal disks are used in this couplingmade by Thom as Flexible Coupling Company. Th e disks

are bolted to each flange and connected to each other bymeans of pins supported by a steel center disk. Th e springaction of the center ring allows torsional flexibility and thetwo side rings compensate for angular and offset misalign-ment. This type of coupling provides a positive drive ineither direction without setting up backlash. No lubricationi s required.

F ig . 17-A design made by Palmer-Bee Company fo rheavy torques. Ead i flange carries two studs upon whichare mounted square metal blocks. The Mocks slide i n theslots of the center metal disk.

Me fa / block -- - -- -

Section A-A


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Shafts & Couplings

* R ~ v e fmembers

4-47

Fig. l G 1 n this Charles 13ond Company coupling a leatherdisk Hoats between two identical flanges. Drive is throughfo ur laminated leather lugs cemented and riveted to theleathcr disk. Compensates for misaligninent in all directionsand sets up no end thrusts. Th e flanges are made of castiron and the driving lug slots are cored. .--

Fi g. 19-The principle of the T. B. Wood Br Sons Com-pany coupling is the same as Fig. 18, but the dr iving lugsare cast integrally with the metal flanges. Th e laminatedleather disk is punched out to accommodate the metal drivinglugs of each flange. This coupling has flexibility in all direc-tions and does not require lubrication.

Fig . 2 L A n o t h e r design made by Charles Bond Company.T h e flanges have square recesses into which a built-up leathercube fits. Endwi se movement is prevented by through bolts5et at right angles. Th e coupling operates quietly and isused where low torque loads are to be transmitted. Die-castings can be used for the flanges.

Fi g. 21-Similar to Fig. 20, being quiet in operation andused fo r low torques. This is also a design of Charles BondCompany. Th e floating member is made of laininated leatherand is shaped like a cross. Th e ends of the intermediatemember engage the two cored slots of each flange. Th ecoupling will withstand a limited amount of end play.

Fi g. 22-Pins mounted in flanges are connected by leather,canvas, or rubber hands. Coupling is used for temporaryconnections where large torques are transmitted, such a s thedriving- of dynamometers by test engines. Allows for a largeamount of flexibility in all directions, absorbs shocks hutrequires frequent inspection. Machines can be quickly dis-connected, especially when belt fasteners are used on thebands. Driven member lags behind drive r when under load.

Fi g. 2.3-This Bruce-Macbeth Engine Company couplingis similar to that of Fig. 22, except that six endless wirecable links ar e used, made of plow-steel wire rope. Th elinks engage small metal spools mounted on eccentric hush-ings. By turni ng these bushings the links ar e adjusted to theproper tension. Th e load is transmitted from one flange tothe other by direct pull on the cable links. Thi s type of coup-ling is used for severe service.

+A ,----Leafher disk,

F IG . 2 1


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4-48

Typical Designs ofFlexible Couplings 111Cyril Donaldson

Pins

IO L r o t h r r r n k 0FIG . 26

Fig. 2 6 T h i s Wehster Manufacturing Company couplinguses a single endless leather belt instead of a series of bands,as in Fig. 22. The belt is looped over alternate pins in bothflanges. Has good shock resisting properties because of beltstretch and the tendency of the pins to settle back into theloops of the belt.

Fig. 2S-This coupling made by the Welle r Manufact ur-ing Company is similar to the design in Fig. 24, but insteadof a leather belt uses hemp rope. made endless bv splicing.Th e action under load is the same as in the endless belt type.

Fi g. 26-This Bruce-Macbeth design uses leat her links in -stead of endless wire cables, as shown in Fig. 23. The loadis transmitted from one flange to the other by direct pull ofthe links, which at the same time allows for the proper flcx-ibility. Inten ded fo r permanent installations reqn iring a min-imum of supervision.

Fi g. 27-The Oldhani form of coupling made by W. A.Jones Foundry and Machine Company is of the two-jawtype with a metal disk. Is used for transmitting heavy loadsat low speed.

Fi g. 28-The Char les Bond Company sta r coupling is sim-ilar to the cross type shown i n Fig. 21. The star-shapedAoating member is made of laminated leather. Ha s threejaws in each flange. Torque capacity is thus increased overthe two-jaw or cros s type. Couplirip takes care of limitedend play.

Fi g. 29-Combination rubber and canvas disk is bolted totwo metal spiders. Extensively used for low torques wherecompensation for only slight angular misalignment is re-quired. Is quiet in operation and needs no lubrication orother attention. Offset niisalignnient shortens disk life.


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Shafts & Couplings

Ruhberlred f n b r i c dmks_..._- .

Fig . 30-A metal block ab a floating center is used in thisAmerican Flexilile Coupling Conipany design. Quiet opera-tion is secured by lacing the block with removable fiber stri psantl packing the center with grease. Th e coupling sets up noend thrusts, is easy to assemble and does not depend on flex-ible material f or the driving action. Can he built in smallsizes by using hardwood block without facings, for the float-ing member.

Fig. 31-This Westinghouse Nuttall Company coupling isan all-metal type having excellent torsional flexibility. Th eeiglit compression springs conipensate for angular an d offsetmisalignment. Allows for some free endwise float of the.hafts . Will transmit high torques in either direction. N oluhication is needed.

Fi g. 32-Similar to Fig. 29, but will withstand offset mis-alignment by addition of the extra disk. In this instance thecenter spider is free to float. By use of two rubber-canvasdisks. as shown, coupling will withstand a considerable angu-lar niisalignment.

F ig . 33-111 this Siiiith & Serrcll coupling a flexible crossmade of laniinatetl steel strips floats between two spiders.I he Iaminatetl S ~ C J ~ ~ S ,etained by lour segmental shoes,engage lugs integral with the flanges. Coupling is intendedfor th e transmission o i light loads only.I .

Fi g. 34-This coupling niade 113Brown Engineering Con- L Apaiiv is useful fo r iiiiiirovi5iiiq connections between ani)aratns.~ I .in laboratories ;rind siiiiilar temporary installations. Co n-pciisates for n~isalignnient n all directions. Will absclrhwryinc degreps of torsional shocks by changing the size ofthe springs. Springs are retained by threaded pins engagingthe coils. Overload protection is possilile by the slippage orhrrakage of replacable springs.Fi g. 35-In another drsizn hy Hri ~wnE n g i n c e r i n ~Com-pany, a series of laiiiinatetl spokes transinit power betweentlic two flanges without setting up em 1 thrusts. This t ypeallows free eiitl play. Among other advantages are al)sorp-tion of torsional sliocks, has no exposed nioving parts. antl iswell I)alancetl at all slxetls. \Veariiig parts are replacableand ~ v n r k i n g )arts are protected froni dust.

Spring re fa in iny p h - ~

Section A-A

4-49


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4-50

l o w costCoupling Small DiameterSixteen types of low cost couplings, including flexible and non-flexible types. Most arefor small diameter, lightly loaded shafts, but a few of them can also be adapted to heavyduty shafts. Some of them are currently available as standard commercial parts.

Rubber sleeve.

Rubber base adhrsiviFig I-Rubber sleeve has insidediameter smaller than shaft diameters.Using rubber-base adhesive will in-crease the torque capacity.

Fig 2-Slit sleeve of

shafts &couplings.pdf

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