Illustrated Sourcebook of Mechanical Components Part 3 docx

4-16 ?-SQUARE and RECTANGULAR SHAFTS Torque, T, in lb 2,000,000~ i 1,000,000 Exomple 4 finds S fur square shoft thoi will fronsmif /6300in. 4 torque ot 14 OOOpsi sheor stress. Exomple 5 finds A for rectungulor shoft for rutio AM = /. 20 k l00*000 I- F I l0,Ooo 200 E Square side, S, in. 5.0 4.0 f Max shear stress, f, psi 0.7 0.5 I I I 1 I Shaft Locotion of Torque formulos: section max shear 1 T= I t 1 I I Middle 0.208s f of sides t I I 1 of Midpain major t A2B2f 3A t 1.8B I I sides 50,000 1 j 60,000 Shafts 8t Couplings 4-17 Critical Smeeds of End Supported Bare Shafts L. Morgan Porter THIS NOMOGRAM solves the equation for the critical speed of a bare steel shaft that is hinged at the bearings. For one bearing fixed and the other hinged rnuItiply the critical speed by 1.56. For both bearings fixed, multiply the critical speed by 2.27. The scales for critical speed and length of shaft are folded; the right hand sides, or the left hand sides, of each are used together. The chart is valid for both hollow and solid shafts. For solid shafts, D2 = 0. 8,000 7,000 6,000 50,000 40,000 5,000 30,000 g 4,000 3,O 00 20,000 E ? 0 z 70 c BO 40 100 * 50 3 Example;- DI= 6.3 in., D2=5.8 in., -=8.56 L = 130in. Nc=2,375 rprn For Aluminum multiply uolues of N, byLOO26 , For Mognosium multipb vofues of Nc by 0.9879 \ . 4-18 Torsional Strength of Shafts Formulas and charts for horsepower capacity of shafts from 1/2 to 2 1/2 inch diameter and 100 to 1000 rpm. Douglas C. Greenwood For a maximum torsional dcflection of 0.08" per foot, shaft Icngth, diameter and horsepower capacity are rc- latcd in d = 4.644% wherc d = shaft diamctcr, in.; hp = horscpowcr; R = shaft spccd, rpm. This dc- flcction is rccommendcd by many authorities as being a safe general maximum. Thc two charts arc plotted from this formula, providing a rapid means of chccking transmission-shaft strength for usual industrial speeds up to 20 hp. Although shafts under 1-in. dia are not trans- mission shafts, strictly apcak- ing, lower sizes %ave been in- cluded. Whcn shaft design is based on strength alone, the diameter can be smaller than values plotted here. In such cam usc thc formula 2.50 2.25 2 .oo I .75 i 0 Q 1.50 ._ ._ t 0 t v) - I .25 I .oo 0.75 0.50 I I I I I IO0 200 300 no0 500 600 Shaft speed, rpm Shafts & Couplings 4-19 LOADING CONDITION _I k j Head shafts subjed to heavy strains. and slow speeds, clutches or gearing carried) (Intermittent loads (2600 psi) The value of k varies from 125 to X according ro allow- able stress used. Thc figurc a xoun ts for m cmbcrs that introduce bcndiiig loads, such as gears, clutchcs and pullcys. B,ut bcnding loads are not as reaclily determined as tor- sional strcss. Thercfore, to alIow for combined bending and torsional stresses, it is usual to assume simple tor- sion and usc a lower design stress for thc shaft dcpending upon how it is loaded. For euamplc, 12 5 represents a stress of approxiniatcly 2600 psi, which is very low and should thus insure a strong- cnough shaft. Other values ditions arc shown in the table. When bending strcss is not considercd, lower k val- ues can bc used, hut a value of 38. should be regarded as the minimum. of k for 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 I 90 (3550 psi) 1 apart. 1 (4300 psi) apart. ______ 75 Lineshafts 20-50 ft long, lightly loaded, bearings 6 ft 2 .oo I .75 I .50 c' 0 1.25 + 'c 0 c vl I .oo 0.75 0.50 6 I I I 1 0 700 800 900 1000 Shoff speed, rpm 4-20 Bearing Loads on Geared Shafts Simple, fast and accurate graphical method of calculating both direction and magnitude of bearing loads. Zbigniew Jania To calculatc thc bearing loads resulting froiii gear action, both tlic magnitudc and direction of the tooth reaction must bc known. This reaction is thc forcc at the pitch circle excrted by thc tooth in the direction peiyciidicular to, and away froni the tooth surface. Thus, the tooth reaction of a gear is always in the sanie geiicral direction a5 its motion. Most techniques for evaluating bearing loads scparate thc total foroe acting on thc gear into tangential and separating components. This tends to complicate the solution. 'The method described herein uses the total force directly. It T is the torque transmitted by a gear, the tangential tooth force is FT = 2T/D, Also, from Fig. 1, F = FT wc Q, F = FT sec (spur gears), (helical gears). Sincc a forcc can be replaced by an equal force acting at a different point, plus a couple, the total gear force can be considered as acting at the intersection of the shaft centerline and a line passing through the mid-face of the gear, if the appropriate couple is included. For example, in Fig. 2 the total force on gear B is equivalent to a force FB applied at point X plus the couple b x FB. In establishing the couples for the other gears, a sign convention must be-adopted to distinguish between clockwise and counter. clockwise moments. If a vector diagram is now drawn for all couples acting on the shaft, the closing line will 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, the direction being the same as that of the couple caused by it. The load on bearing I is found in the same manner by drawing a force vector dia- gram for all the forces acting at X including the load on bearing I1 found from the couple diagram. Tooth reocfion I i Shafts & Couplings The construction of both diagrams is illustrated on page 2 13. Referring to Fig. 2 the given data are Pitch Dia. of GesrB, in. A 2.00 B 1.50 c 4.00 Driver. .1.75 Moment Arm, in. a 1.50 b 3.50 e 5.00 d 7.00 Angle, deg Torque on driver. .lo0 Ib-in. a. .55 fl .48 7 .45 Then, Torque delivered by A .40 per cent of torque on center shaft Torque delivered by B .60 per cent of torque on center shaft Pressure angle of all gears, +. .20 deg Tangential force of driver = 200/1.75 = 114 lb Torque on center shaft = 2 X 114 = 228 Ib-in. Gear loads are Oe4 228 aec 20 deg = 97 Ib 2.00 P.4 = FB = Fc = 114 seo 20 deg = 121.5 lb :c 228 aec 20 deg = 195 lb Before drawing the diagrams, it is convenient to collect all the data as in Table I. Then, the couple diagram, Fig. 3, is drawn. It is important to note khat: (3) Vectors representing negative couples are drawn in the same direction but in opposite sense to the forces causing them; (b) The direction of the closing line of the diagram should be such as to make the sum of all couples equal to zero. Thus, the direction of 7 P,r is the direction of bearing reaction. The bearing load has the same direction but is of opposite sense. 4-2 1 -sepororing force - Toto1 geor lood (5 1 4-22 7 Ways to Limit Shaft Rotation Traveling nuts, clutch plates, gear fingers, and pinning members are the bases of these ingenious mechanisms. I. M. Abeles Mechanical stops are often required in automatic machinery and servomech- anisms to limit shaft rotation to a given number of turns. Two problems to guard against, however, are: Excessive forces caused by abrupt stops; large torque 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 NUT moves (1) along threaded shaft until frame prevents further rotation. A simple device, but nut jams so tight that a large torque is required to move the shaft from its CLUTCH PLATES tighten and stop rotation as the rotating shaft moves the nut against the washer. When rota- tion is reversed, the clutch plates can turn with the shaft from A to B. During this movement comparatively low torque is required to free the nut from the clutch plates. Thereafter, subse- quent movement is free of clutch fric- tion until the action is repeated at other end of the shaft. Device is recom- mended for large torques because clutch plates absorb energy well. stopped position. This fault is over- than the thread pitch so pin can clear come at the expense of increased finger on the first reverseturn. The length by providing a stop pin in the rubber ring and grommet lessen im- traveling nut (2). Engagement between pact, provide a sliding surface. The pin and rotating finger must be shorter grommet can be oil-impregnated metal. Clutch plotes Clutch plates keyed to shaft 4 with projection\ ‘Ti-ove/ing nut P- B Section 8-B I \ Output lnput snort Shafts &? Couplings 4-23 SHAFT FINGER on output shaft hits re- silient stop after making less than one revolution. Force on stop depends upon gear ratio. Device is, therefore, limited to low ratios and few turns unless a worm- gear setup is used. TWO FINGERS butt together at initial and final positions, prevent rotation beyond these limits. Rubber shock-mount absorbs impact load. Gear ratio of almost 1:l ensures that fingers will be out of phase with one another until they meet on the anal turn. Example: Gears with 30 to 32 teeth limit shaft rotation to 25 turns. Space is saved here but gears are costly. Gear makes less thun one revolufion Poi 4 1 Shuff ,, N fingers rofote on shuft finger fixe fo ffume LARGE GEAR RATIO limits idler gear to less than one turn. Sometimes stop fingers can be added to already existing; gears in a train, making this design simplest of all. Input gear, how- ever, is limited to a maximum of about 5 turns. PINNED FINQERS limit shaft turns to approximately N + 1 revolutions in either direction. Resilient pin-bushings would help reduce impact force. 4-24 Friction for Damping When shaft vibrations are serious, try this simple technique of adding a sleeve to the shaft can keep vibrations to a minimum. Here’s how to design one and predict its effect. Burt Zimmerman HEN BOOSTING THE OPERATING SPEED of any ma- W chine, the most formidable obstacle to successful operation that the designer faces is structural vibration. There is always some vibration in a system, and as the speeds are increased the vibration amplitudes become large (relatively speaking, for they may still be too small to be seen). These amplitudes drastically reduce life by causing fatigue failures and also damage the bearings, gears, and other components of the machine. It is not over-simplify- ing 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 vibration damping is to apply a small amount of dry friction (coulomb friction) at key places of the structure. This produces a greater amount of damping than one would normally expect, and the technique is used with success by some product designers and structural engineers but, it seems, only after the machine or structure has been built. There seems to have been little attempt to apply this concept to initial design or to develop the equations necessary for the proper location of the friction points. We will apply this concept here to the solution of torsional vibrations of shafts, as this is a serious problem in both industrial machinery and in military systems such as submarines, missiles, and planes. The necessary design formulas are developed and put to work to solve a typical shaf,t problem from industry. How the technique works Vibration amplitudes in a shaft become a problem when the shaft length to the thickness ratio, L1/D1, be- comes large. One can of course make the shaft thicker. But this would greatly add to its weight. Symbols a = b/Lz C = Dz/D1 D1 = Diameter of shaft Dz = Diameter of sleeve G = Shear modulus of elasticity H = Thickness of the sleeve wall J = Polar moment of inertia (for the shaft: T DI4/32) JEG nD13H/4 LI = Length of shaft Lz = Length of sleeve m = D,/8HC3 = ratio of torsional stiffness of the shaft to that of the sleeve r=l+m R = Dampingratio T = Applied torque on the shaft T, = Resisting frictional torque applied by U = Residual internal energy of shaft and VI = Internal energy of the shaft U, = Internal energy of the sleeve W = Energy dissipated in a half oscillation the sleeve sleeve h = T,/T 6 = Angular displacement of the shaft 6, = Angular displacement of the sleeve Shafts & Couplings 4-25 u I 1. Thin sleeve added to rotating shaft greatly reduces torsional vibrations. The ,disk is rigidly attached to the shaft and has a snug fit with the sleeve. Extending the 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 shaft at one cnd (A). The sleeve is extended along the shaft length and makes contact with some point on the shaft. In this particular design, a disk is rigidly attached to the shaft (by welding it or tightly pressing it on), and there is a snug fit between the disk and the sleeve. The exact amount of fit is not too important, but it must be neither too loose nor too tight: If the fit is too tight, the shaft and sleeve will tend to move together as a unit and there will be no damping (just an increase in the moment of inertia) ; if too loose, with a clearance between disk and sleeve, again there will be no damping. The frictional forces in question occur at the contact between the inside surface of the sleeve and the edge of 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 the sleeve extends the entire length of the shaft, but there may be cases, depending on the product design and application, where this is impossible. Therefore, the gen- eral case where the Sleeve length is variable is considered here. To avoid corrosion or; fretting at the interface, try a layer of viscoelastic stripping (elastomer) at the edge of the disk. Analysis of concept When a shaft is rotating, a resisting torque is developed in the shaft which varies along the length of the shaft. Because the angular displacement is a function of this resisting torque, the surface fibers of the shaft will undergo different angular displacements which depend on the distance of the specific fiber from the point of 2. Energy present in a rotating shaft through one complete cycle of vibration with damping taking place for one half of the cycle. At ti, the energy in the shaft is equal to its residual in- ternal energy, U, plus the energy dissipated during half the cycle, W. At t,, the energy is equal to U, which indi- cates that the energy dissipated through dry friction damping is W. LW(dissipoled enerqyld I I [...]... m :D, / 8HC3 3 Design chart for different values of the dimensional constant, rn The frictional amount 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 interface will stick The curve for the amplitude-damping ratio (which is read at the right scale) can be used for most design problems, as illustrated. .. the end ( A ) of the shaft, the displacement at the disk (at EG) is in the form (see list of symbols): e= 1'L JG And that the displacement at the end B in the form 6s = errc ( h - L2) + TaD14GG /32 The angular displacement of the sleeve at EG (with zero displacement at the fixed end A ) is The strain energy of the shaft caused by vibration is u= u1+ u z (3) The maximum strain energy of the shaft can... (11) The ratio in, which is equal to D1/8HC3,is the ratio of the torsional stiffness of the shaft to that of the sleeve The corresponding fractional value of the energy dissipated per oscillation at optimum A is equal to 1-R The key to the design chart, Fig 3, is the fact that the fractional energy curve is not in direct proportion to the ratio L z / L I of the sleeve length to the shaft length This... and error basis, and H, which is the thickness of the sleeve wall and thus the important parameter which influences the total weight of the shaft, is determined from the chart in Fig 3 and its abscissa equation Solution From Fig 3, this value of R (damping ratio) requires an m-value equal to 5.2 and a damping/critical damping value of 2.6% Thus D1/811C3= 5.2 7.5 in HCS = 0.1S02 m = D, Since = We can... Medium Low 4 -32 4-spline 0 W 6-spline W I _ '/z 0.120 3/ 4 0 I8 I 0 I88 0.211 0 219 I 0 241 0.250 I-'% 030 1 031 3 78 7 SPLINED SHAFTS Ideal where gear must slide in lateral direction during rotation Square splines often used, but involute splines are self-centering and stronger Nonsliding gears are pinned or held by threaded nut or retaining ring Torque strength is high and dependent o n number of splines... Lugs stomped on both legs of wire to prevent disouembly Bend radii of shaft shou!d be small enough to allow gear to seat of ho/& threaded Bushing half of hole &t threaded for removing from shaft: Bushing haff of hole threaded Hub half of hole n2f threaded 8 Split bushing I has tapered outer diameter Split holes i n bushing align with split holes i n hub For tightening, hub half of hole i s topped, bushing... principle as design shown in Fig 6, but has a single flat spring in place of a series of coiled springs High degree of flexibility obtained by use of tapered slots in hubs Smooth operation, is maintained by inclosing the working parts and packing with grease Fig &o T w flanges and a series of coiled springs give a high degree of flexibility Used only where the shafts have no free end play Needs no lubrication,... References 1 P F Chenea and H M Hansen, Mechanics of Vibration, John Wiley and Sons Inc, 1952, pp 31 9 -32 4 2 D Williams, Damping of Torsional Vibrations by Dry Friction, Royal Aircraft Establishment, 1960 (Fig 3) It is generally accepted that with most dry-friction damping there will be approximately 3% of damping taking place per cycle If the forcing torque were re- Gear box Induction motor e -H Synchronous... frictiondamping Numerical example below shows how t h e addition of a very thin sleeve with a wall thickness of 0.0 23 in reduces the amplitude of vibration by 10 to 1 = L Compressor Shafts & Couplings 15 Ways to Fasten Gears to Shafts So you've designed or selected a good set of gears for your unit-now how do you fasten them to their shafts? Here's a roundup of methods-some old, some new-with a comparison table... gear onto the shaft-one in particular is "Loctite," manufactured by American Sealants Co This material remains liquid as long as it is exposed to air, but hardens when confined between closely fitting metal parts, such as with close fits of bolts threaded into nuts (Military spec MIL-S-400 83 approves the use of retaining compounds) Loctite sealant is supplied in several grades of shear strength T h e . Thickness of the sleeve wall J = Polar moment of inertia (for the shaft: T DI4 /32 ) JEG nD13H/4 LI = Length of shaft Lz = Length of sleeve m = D,/8HC3 = ratio of torsional. 50,000 40,000 5,000 30 ,000 g 4,000 3, O 00 20,000 E ? 0 z 70 c BO 40 100 * 50 3 Example;- DI= 6 .3 in., D2=5.8 in., -=8.56 L = 130 in. Nc=2 ,37 5 rprn For Aluminum. Determination of damping pressed 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 damping Magnitude of energy