UNIT 6 KEYS AND COUPLINGS Keys And Couplings
UNIT 6 KEYS AND COUPLINGSKeys and s of Keys6.2.1Sunk Keys6.2.2Saddle Keys6.2.3Tangent Keys6.2.4Round Keys6.2.5Splines6.3Forces Acting on a Sunk Key6.4Strength of a Sunk Key6.5Effect of Keyways6.6Couplings6.7Summary6.8Key Words6.9Answers to SAQs6.1 INTRODUCTIONA key is a piece of steel inserted between the shaft and hub or boss of the pulley toconnect these together in order to prevent relative motion between them. It is alwaysinserted parallel to the axis of the shaft. Keys are used as temporary fastenings and aresubjected to considerable crushing and shearing stresses. A keyway is a slot or recess ina shaft and hub of the pulley to accommodate a key.ObjectivesAfter studying this unit, you should be able to identify keys and their application, calculate forces on keys, and design keys.6.2 TYPES OF KEYSThe following types of keys are important from the subject point of view :(a)Shunk keys,(b)Saddle keys,(c)Tangent keys,(d)Round keys, and(e)Splines.We shall now discuss the above types of keys, in detail, in the following sections.6.2.1 Sunk KeysThe sunk keys are provided half in the keyway of the shaft and half in the keyway of thehub or boss of the pulley or gear. The sunk keys are of the following types :145
Machine DesignRectangular Sunk KeyA rectangular sunk key is shown in Figure 6.1. The usual proportions of this keyare :Width of key, w whered2w d; and thickness of key, t 364d Diameter of the shaft or diameter of the hole in the hub.The key has taper 1 in 100 on the top side only.Taper 1: 100wtdFigure 6.1 : Rectangular Sunk KeySquare Sunk KeyThe only difference between a rectangular sunk key and a square sunk key is thatits width and thickness are equal, i.e.w t d4Parallel Sunk KeyThe parallel sunk keys may be of rectangular or square section uniform in widthand thickness throughout. It may be noted that a parallel key is a taperless and isused where the pulley, gear or other mating part is required to slide along theshaft.Gib-head KeyIt is a rectangular sunk key with a head at one end known as gib head. It isusually provided to facilitate the removal of key. A gib head key is shown inFigure 6.2(a) and its use in shown in Figure 6.2(b).Taper 1:100Gib head key1.75 ttShaft1.5 tw(a)(b)Figure 6.2 : Gib-head KeyThe usual proportions of the gib head key are :Width,w d;4and thickness at large end, t 2w d .36Feather KeyA key attached to one member of a pair and which permits relative axialmovement of the other is known as feather key. It is a special key of parallel type146
which transmits a turning moment and also permits axial movement. It is fastenedeither to the shaft or hub, the key being a sliding fit in the key way of the movingpiece.Keys and CouplingsThe feather key may be screwed to the shaft as shown in Figure 6.3(a) or it mayhave double gib heads as shown in Figure 6.3(b). The various proportions of afeather key are same as those of rectangular sunk key and gib head key.Feather keysSet screw(a)(b)Figure 6.3 : Feather KeyThe following Table 6.1 shows the proportions of standard parallel, tapered andgib head keys, according to IS : 2292 and 2293-1974 (Reaffirmed 1992).Table 6.1 : Proportions of Standard Parallel, Tapered and Gib Head KeyShaft Diameter(mm) upto andIncludingKey Cross-sectionWidth(mm)Thickness(mm)6228310Shaft Diameter(mm) upto andIncludingKey 40652012380904575221444010050Woodruff KeyThe woodruff key is an easily adjustable key. It is a piece from a cylindrical dischaving segmental cross-section in front view as shown in Figure 6.4. A woodruffkey is capable of tilting in a recess milled out in the shaft by a cutter having thesame curvature as the disc from which the key is made. This key is largely used inmachine tool and automobile construction.147
Machine DesignRFigure 6.4 : Woodruff KeyThe main advantages of a woodruff key are as follows :(a)It accommodates itself to any taper in the hub or boss of the matingpiece.(b)It is useful on tapering shaft ends. Its extra depth in the shaft preventsany tendency to turn over in its keyway.The disadvantages are :(a)The depth of the keyway weakens the shaft.(b)It can not be used as a feather.6.2.2 Saddle KeysThe saddle keys are of the following two types :(a)Flat saddle key, and(b)Hollow saddle key.A flat saddle key is a taper key which fits in a keyway in the hub and is flat on the shaftas shown in Figure 6.5. It is likely to slip round the shaft under load. Therefore, it is usedfor comparatively light loads.Hollow saddle keywtwdtFigure 6.5 : Saddle KeyA hollow saddle key is a taper key which fits in a keyway in the hub and the bottom ofthe key is shaped to fit the curved surface of the shaft. Since hollow saddle keys hold onby friction, therefore, these are suitable for light loads. It is usually used as a temporaryfastening in fixing and setting eccentrics, cams, etc.6.2.3 Tangent KeysThe tangent keys are fitted in pair at right angles as shown in Figure 6.6. Each key is towithstand torsion in one direction only. These are used in large heavy duty shafts.Figure 6.6 : Tangent Keys148
Keys and Couplings6.2.4 Round KeysThe round keys, as shown in Figure 6.7(a), are circular in section and fit into holesdrilled partly in the shaft and partly in the hub. They have the advantage ofmanufacturing as their keyways may be drilled and reamed after the mating parts havebeen assembled. Round keys are usually considered to be most appropriate for lowpower drives.Round key(a)Tapered pin(b)Figure 6.7 : Round KeysSometimes the tapered pin, as shown in Figure 6.7(b), is held in place by the frictionbetween the pin and the reamed tapered holes.6.2.5 SplinesSometimes, keys are made integral with the shaft which fit in the keyways broached inthe hub. Such shafts are known as splined shafts as shown in Figure 6.8. These shaftsusually have four, six, ten or sixteen splines. The splined shafts are relatively strongerthan shafts having a single keyway.The splined shafts are used when the force to be transmitted is large in proportion to thesize of the shaft as in automobile transmission and sliding gear transmissions. By usingsplined shafts, we obtain axial movement as well as positive drive.bdDFigure 6.8 : Splines6.3 FORCE ACTING ON A SUNK KEYWhen a key is used in transmitting torque from a shaft to a rotor or hub, the followingtwo types of forces act on the key :(a)Forces (F1) due to fit of the key in its keyway, as in a tight fitting straightkey or in a tapered key driven in place. These forces produce compressivestresses in the key which are difficult to determine in magnitude.(b)Forces (F) due to the torque transmitted by the shaft. These forces produceshearing and compressive (or crushing) stresses in the key.149
Machine DesignThe distribution of the forces along the length of the key is not uniform because theforces are concentrated near the torque-input end. The non-uniformity of distribution iscaused by the twisting of the shaft within the hub.The forces acting on a key for a clockwise torque being transmitted from a shaft to a hubare shown in Figure 6.9.ShaftF1tFFF1dFigure 6.9 : Forces Acting on a Shunk KeyIn designing a key, forces due to fit of the key are neglected and it is assumed that thedistribution of forces along the length of key is uniform.6.4 STRENGTH OF A SUNK KEYA key connecting the shaft and hub is shown in Figure 6.9.LetT Torque transmitted by the shaft,F Tangential force acting at the circumference of the shaft,d Diameter of shaft,l Length of key,w Width of key,t Thickness of key, and and c Shear and crushing stresses for the material of key.A little consideration will show that due to the power transmitted by the shaft, the keymay fail due to shearing or crushing.Considering shearing of the key, the tangential shearing force acting at the circumferenceof the shaft,F Area resisting shearing Shearing stress l w Torque transmitted by the shaft,T F dd l w 22. . . (6.1)Considering crushing of the key, the tangential crushing force acting at thecircumference of the shaft,F Area resisting crushing Crushing stress l t c2 Torque transmitted by the shaft,T F dtd l c 222. . . (6.2)The key is equally strong in shearing and crushing, ifl w 150dtd l c 222[Equating Eqs. (6.1) and (6.2)]
w c t2 orKeys and Couplings. . . (6.3)The permissible crushing stress for the usual key material is atleast twice the permissibleshearing stress. Therefore, from Eq. (6.3), we have w t. In other words, a square key isequally strong in shearing and crushing.In order to find the length of the key to transmit full power of the shaft, the shearingstrength of the key is equal to the torsional shear strength of the shaft.The torque transmitted by the key,T l w d2. . . (6.4)and the torque transmitted by the shaft,T 1 d 316. . . (6.5)(Taking 1 Shear stress for the shaft material)From Eqs. (6.4) and (6.5), we havel w l d 1 d 32 16 1 d 2 d 1 1.571d 18 w 2 (Taking w d)4. . . (6.6)When the key material is same as that of the shaft, then 1. l 1.571 d[From Eq. (6.6)]Example 6.1Design the rectangular key for a shaft of 50 mm diameter. The shearing andcrushing stresses for the key material are 42 MP and 70 MPa.SolutionGiven d 50 mm; 42 MPa 42 N/mm2; 70 MPa 70 N/mm2.The rectangular key is designed for a shaft of 50 mm diameter,Width of key, w d,4w 12.5 mmand thickness of key asd6t 8.3 mmThe length of key is obtained by considering the key in shearing and crushing.Letl Length of key.Considering shearing of the key. We know that shearing strength (or torquetransmitted) of the key,T l w d50 l 17 42 13125 l N-mm22. . . (6.7)and torsional shearing strength (or torque transmitted) of the shaft,T d3 42 (50)3 1.03 106 N-mm1616. . . (6.8)From Eqs. (6.7) and (6.8), we havel 1.03 106 79.25 mm13125151
Machine DesignNow considering crushing of the key. We know that shearing strength (or torquetransmitted) of the key,T l td8.350 c l 70 7262.5 l N-mm2222. . . (6.9)From Eqs. (6.8) and (6.9), we have106 141.8 mm8750l 1.03 Taking larger of the two values, we have length of key,l 141.8 say 142 mm.Example 6.2A 45 mm diameter shaft is made of steel with a yield strength of 400 MPa. Aparallel key of size 14 mm width and 9 mm thickness made of steel with a yieldstrength of 340 MPa is to be used. Find the required length of key, if the shaft isloaded to transmit the maximum permissible torque. Use maximum shear stresstheory and assume a factor of safety of 2.SolutionGiven d 45 mm; for shaft 400 MPa 400 N/mm2; w 14 mm; t 9 mm; yt for key 340 MPa 340 N/mm2.Letl Length of key.According to maximum shear stress theory, the maximum shear stress in the shaft, max 2 F . S. 400 100 N/mm22 2and maximum shear stress for the key, k 2 F . S. 340 85 N/mm22 2(Note : Yield strength for shaft and key materials are different).We know the maximum torque transmitted by the shaft and key,T max d 3 100 (45)3 1.8 106 N-mm1616First of all, let us consider the failure of key due to shearing. We know that themaximum torque transmitted (T),1.8 106 l w k l d45 l 14 85 26775 l221.8 106 67.2 mm26775Now considering the failure of key due to crushing. We know that the maximumtorque transmitted by the shaft and key (T),1.8 106 l l td9 340 45 ck l 17213 l Taking ck F . S. 22222 1.8 106 104.6 mm17213Taking the larger of the two value, we have152l 104.6 say 105 mm.
Keys and Couplings6.5 EFFECT OF KEYWAYSA little consideration will show that the keyway cut in the shaft reduces the load carryingcapacity of the shaft. This is due to the stress concentration near the corners of thekeyway and reduction in the cross-sectional area of the shaft. In other words, thetorsional strength of the shaft is reduced. The following relation for the weakening effectof the keyway is based on the experimental results by H. F. Moore. w h e 1 0.2 1.1 d d where ke Shaft strength reduction factor. It is the ratio of the strength of the shaft withkeyway to the strength of the same shaft without keyway.w Width of keyway,d Diameter of shaft, andh Depth of keyway Thickness of key (t ).2It is usually assumed that the strength of the keyed shaft is 75% of the solid shaft, whichis somewhat higher than the value obtained by the above relation.In case the keyway is too long and the key is of sliding type, then the angle of twist isincreased in the ratio k as given by the following relation : w h k 1 0.4 0.7 d d where k Reduction factor for angular twist.Example 5.3A 15 kW, 960 rpm motor has a mild steel shaft of 40 mm diameter and theextension being 75 mm. The permissible shear and crushing stresses for the mildsteel key are 56 MPa and 112 MPa respectively. Design the keyway in the motorshaft extension. Check the shear strength of the key against the normal strength ofthe shaft.SolutionGiven P 15 kW 15 103 W; N 960 rpm; d 40 mm; l 75 mm; 56 MPa 56 N/mm2; c 112 MPa 112 N/mm2.We know that the torque transmitted by the motor,T P 60 15 103 60 149 N-m 149 103 N-mm2 N2 960Let w Width of keyway or key.Considering the key in shearing. We know that the torque transmitted (T).Assuming that length of the key is equal to length of the shaft (i.e. extension)149 103 l w w 149 10384 103d40 75 w 56 84 103 w22 1.8 mmThis width of keyway is too small. The width of keyway should be at least w d 40 10 mm44d.4153
Machine DesignSince c 2 , therefore, a square key of w 10 mm and t 10 mm is adopted.According to H. F. Moore, the shaft strength factor,t w h w t ke 1 0.2 1.1 1 0.2 1.1 (because h )2 d d d 2d 10 10 1 0.2 0.8125 20 2 40 Strength of the shaft with keyway, d 3 75 10 56 (40)3 0.8125 571 844 N16and shear strength of the key, i.e. torque carrying capacity l w d40 75 10 56 840 000 N22Shear strength of the key840 000 1.47Normal strength of the shaft 571 8446.6 COUPLINGSIn engineering applications there arise several cases where two shafts have to beconnected so that power from driving shaft is transmitted to driven shaft without anychange of speed. Such shafts are normally coaxial with slight or no misalignment andcan be connected through devices known as couplings. Permanent couplings, oftenreferred to as couplings, are the connectors of coaxial shafts and cannot be disengagedwhen shafts are running. On the other hand, those couplings which can be readilyengaged or disengaged when driving shaft is running are termed as clutches. The poweris transmitted when a clutch is engaged and not transmitted when clutch is disengaged.In this unit only permanent couplings will be considered. Figure 6.10 shows one suchcoupling connecting the shaft of an electric motor with the shaft of a worm and wormwheel reducer.CouplingFigure 6.10 : A Permanent Coupling Connecting Coaxial Shafts of an Electric Motorand a Worm and Worm Wheel ReducerSeveral types of couplings are used in practice. A few are described here. Muff or sleevecoupling is shown in Figure 6.11. It is the simplest form of a permanent coupling,consisting of a steel or cast iron sleeve fitted on the ends of shaft to be connected. The154
sleeve is connected to the shaft by means of keys. The length of sleeve can be taken as(3.5 to 4) diameter of the shaft while the outer diameter of the muff or sleeve, D, isgiven byKeys and CouplingsD d 2 D 1.67 d 20 mm. . . (6.10)where d is the diameter of shaft in mm, , the thickness of the muff (Figure 6.11).However, the shear stress in the muss must be checked by treating it as a hollow shat ofinternal diameter d and external diameter D. The muff or sleeve coupling has theadvantage of simple design and easy manufacture. However, need of perfect alignmentof shafts is apparent and if not present the connection through a sleeve will inducebending stresses in the shafts. Yet another disadvantage is that while removing the sleevemust move on one of the shafts at least over a distance equal to half its length. Thisrequires the shaft to be longer by this much amount.dD3.5 - 4dd 1cm3Figure 6.11 : A Sleeve CouplingIn case of split muff coupling, the sleeve is made to have two halves which are heldtogether on two coaxial shafts by bolts. This coupling also known as clamp coupling isshown in Figure 6.12. When the bolts are tightened a compression is induced betweenthe inner surface of sleeve and outer surface of shaft. This compressive force causesfriction between the muff and the shaft which transmit the torque form one shaft to theother. In addition, a key is also used to connect the split muff with the two shafts.(a) Split Muff with BoltsDQ(b) Split Muff Tightened on Two Coaxial ShaftsFigure 6.12Split muff coupling has a distinct advantage over ordinary muff coupling as it can beremoved or disassembled without disturbing the shafts.155
Machine DesignThe outer diameter of the muff, D, the length of the muff, L, and the bolt diameter db arethe dimensions required to be determined for split muff coupling. These dimensions canbe calculated from following empirical relations with shaft diameter, d.D 2.5 dorL 1.5 DorD 2 d 13 mmL 3.5 d. . . (6.11)db 0.2 d 10 mmThe dimensions of the key can be calculated by strength consideration or selected fromstandards. Such standards will be described later in this unit. Even if the bolt diameter insplit muff coupling is calculated from last of Eq. (6.11) it will be worthwhile to checkcompression force and consequent frictional torque which results from tightening ofthese bolts.6.6.1 Flange CouplingFlange coupling, as was mentioned earlier is used to connect two strictly coaxial shafts.One such coupling is shown in Figure 6.10 and details are shown in Figure 6.13. The twoflanges are usually made in cast iron. These flanges are separately keyed to driving anddriven shafts.BoltKeydDDc DoDrShaftFlangettLFigure 6.13 : Flange CouplingThe two flanges are identical in all respects except that one has a circular projection andother has a corresponding recess to make a register. When the two faces of flanges arebrought in contact the projection fits into recess ensuring condition of coaxiality. Theflanges are further connected through bolts placed near the periphery of the flanges. Thefaces of flanges are machine finished true right angled to the axis of shafts. The powermay be transmitted by friction between the flange faces or by bolts in which case boltswill be subjected to shearing stress.Flange couplings are often employed to transmit great torque and are largely dependableconnections for shafts ranging in diameter between 18 mm to 200 mm. They are easilydesigned and manufactured.Flange coupling normally refers to unprotected types as shown in Figure 6.13. The bolthead and nut, in this case are fully exposed and may present risk to operators. The boltheads and nuts are often protected by providing cover in the flange on them as shown inFigure 6.14. This coupling is known as protected flange coupling.156While designing, the shaft diameter is calculated for transmission of torque, designatedas d. The hub diameter of the flange may be calculated by treating the hub as hollowshaft but hub diameter D 2d is often adopted and is found safe. The thickness of theflange may be calculated by considering it to be in shear along the circumference whereit joins the hub. However, this thickness, t, is often taken as slightly greater thandiameter of the bolt.
Keys and CouplingsD dD1DcFlangeShafttfCover in Flange(a) Cut View(b) Sectional ViewFigure 6.14 : Protected Flange CouplingThe number of bolts which are placed symmetrically in a circle is determined in advanceby an empirical formulad. . . (6.12) 350where d is the shaft diameter in mm. The number of bolts normally varies between4 to 8.The diameter of bolt, d1, is determined by yet another empirical formula to obtainapproximate value d1.n d1 d. . . (6.13)2 nwhere d and d1 are in mm.The pitch circle diameter, Dc, is then determined from,Dc 2d 2d1 12 mm. . . (6.14)The diameter of bolt is then accurately determined by taking it in single shear at theinterface of two flanges.Mt nD 2d1 s1 c42. . . (6.15)where s1 is the permissible shearing stress in bolt and Mt is the torque transmitted. Thefactor of safety for the bolt is higher as compared to other parts because it is subjected tosudden load at the start.The keys in the coupling are designed in the normal manner and its depth is selected onthe basis of shaft diameter which is calculated for transmission of torque only. The keydimensions for rectangular section defined w h (width height) can be chosen fromTable 6.2.Table 6.2 : Standard Key Section Dimensionsw h (mm2)8 710 812 814 916 1018 1120 1224 1428 1632 1836 20d (mm)28, 3032, 34, 3536, 37, 38, 4042, 44, 45, 46, 4748, 50, 5255, 58, 6065, 68, 70, 7264, 76, 78, 80, 82, 85, 8890, 92, 95, 98, 100105, 110, 115120, 125157
Machine DesignExample 6.4A driving shaft is joined with coaxial driven shaft through a muff coupling. Theshaft transmits 60 kW of power at 150 rpm. Design the shaft, key and muff.Assume a factor of safety of 5 with following ultimate strength values.Ultimate shear strength for shaft 300 N/mm2Ultimate shear strength for key 200 N/mm2Ultimate shear strength for muff 50 N/mm2Ultimate compressive strength for key 500 N/mm2SolutionIf torque transmitted by the shaft is Mt Nm, power transmitted is H Watt andangular velocity is rad/s,H 60 103 M t M t Mt 2 1506060 103 3819.7 Nm5 The stress caused by torque at outer surface of shaft of diameter, d s 16 M t d3 16 3819.7 103 d3This stresses is not to exceed permissible value s 300 60 N/mm2 .51 19.45 106 3d 68.7 mm 60 This diameter is increased by 25% to take care of weakening by key so thatd 85.9 mm say 86 mm.From Table 6.2 choose a key with w 24 mm and h 14 mm. See Figure 6.15below.h wFigure 6.15 : KeyThe length l of key is calculated from shear force on it.The shear force F M t 3819.7 3 10 88.83 103 Nd8622The shear area w . l 24 l mm2The permissible shear stress 15888.83 103 4024 l200 40 N/mm25
orl Keys and Couplings2.22 103 92.53 mm24The key has to be slightly less than the half muff length. The muff length 3.5 dto 4 d, i.e. 301 mm to 344 mm. Let’s take muff length 301 mm, half of which is150.5 mm hence, key length of 140 mm is safe.We check height of the key against crushing under same force that causesshearing. c h l F 88.83 103 N2 c 88.83 103 90.6 N/mm27 140Permissible compressive stress 500 100 N/mm25Thus, key is safe in crushing.The muff is designed as hollow shaft with internal diameter as the diameter of theshaft. The muff will transmit same power or torque as shaft.With D as outside diameterJ ( D4 d 4 )32M 3819.7 103 t DJ ( D4 d 4 )232The permissible shear stress in muff 50 10 N/mm25D4 d 4 16 3819.7 103 1945.26 103D10 D4 864 1945.36 103 DD4 1.95 106 D 54.7 106. . . (i)This equation can be solved by trial and error and to get an idea of starting pointtake D 2.5 d 215 mm. With this value, the term on right hand side can beneglected resulting in D 13 3(1950 10 ) 125 mm .Choose value of 130, 140, 150, 160 mm for D. Then for D 140 mm,D4 38.4 107 Left hand side of (i) 384 106 273 106 31.5 106For D 130 mm, D4 285 106 Left hand side of (i) 285 106 253.5 106 31.5 106For D 135 mm, LHS 332 106 263 106 69 106For D 132 mm, LHS 303.6 106 257.4 106 46.2 106159
Machine DesignFor D 133 mm, LHS 313 106 259.4 106 53.6 106D 133 mm comes closest to solution of (i).The above trial and error method has been given to make reader familiar with suchmethod. We would rather select the outer diameter from empirical formula.D 2d 13 mm 2 86 13 185 mmExample 6.5A shaft transmitting 150 kW is to be connected to a coaxial shaft through cast ironflange coupling. The shaft runs at 120 rpm. The key and shaft are to be made ofsame material for which permissible shearing stress is 60 N/mm2 and compressivestrength is 120 N/mm2. The steel bolts may be subjected to maximum shearingstress of 26 N/mm2. Design protected type flange coupling.SolutionShaft Diameter dH M t or 150 103 M t Mt 2 12060150 103 12 103 Nm 12 106 Nmm12.57For the shaft 16 M t d3 60 N/mm21or 16 12 106 3d 100.6 mm 60 Increase diameter by 25% to take care of keyway. d 125 mm. . . (i)Bolt Diameter d1Let there be n bolts clamping two flanges and let each bolt be subjected toshearing stress 1. The force produced tangential to pitch circle of bolts(The diameter of pitch circle is DC from Figure 6.13)F n 2d1 14The torque produced by F must be equal to torque transmitted by the shaft. Mt FDcD n d12 1 c242From Eq. (6.13)n d125 3 3 5.5 say 65050Also from Eq. (8.24)d1 d2 n 125 25.5 mm2 6Dc can be obtained form Eq. (6.14)Dc 2d 2d1 12 mm 2 125 2 25.5 12 313 mm160. . . (ii)
We calculate RH side of (ii) by using values of d1, Dc and 1 26 N/mm2.Mt 6 Keys and Couplings 313(25.5)2 26 12.47 106 Nmm42Since this value is greater than torque transmitted, 12 106 Nmm,n 6, d1 25.5 mm, Dc 313 mm. . . (iii)are acceptable values.Hub Diameter DThe hub diameter can be taken as 2d, with internal diameter d. Thentreating hub as hollow shaft under torque Mt, the shear stress should be lessthan 6.6 N/mm2 (shear stress in C.I). ( D4 d 4 ) Mt 2 with D 2d 250 mmD 32 2 2 16 12 106 250 (250 125 )44 15.3 10936.6 108 4.18 N/mm2This stress is less then 6.6 N/mm2, hence, D 250 mm is safe. D 250 mm. . . (iv)Length of Hub, LLength of the hub is equal to length of the key.From Table 6.2 for shaft diameter of 125 mm, find w 36 mm, h 20 mm.dYou may also choose a square key with w h 31.25 mm .4Shear stress in key is same as in shaft, 60 N/mm2. Mt l w l d l 36 60 62.5, M t 12 106 Nmm212 106 89 mm36 60 62.5. . . (v)Thickness of FlangeThere is possibility of failure by shear along the circumference where flangejoint the hub. If t is the thickness of the flange, the area over which shearmay occur is D t. The shear force will be D t 3, 3 being the permissibleshearing stress in cast iron flange. This, will cause the torque equal to thetorque transmitted by the shaft D t 3 t D Mt22 12 106 (250)2 6.6 18.5 mmThe bolts in holes of flange may be crushed. Of course the hole surface mayalso be crushed but if bolts are safe then the hole surface will be safe sincethe CI is stronger than steel in compression.161
The area resisting crushing is d1 t and force in n bolts is n d1 t c at a radiusDof c . Thus, the torque is2Machine Designn d1 t 3Dc313 6 25.5 18.5 120 53 106 Nmm22This torque is much larger than 12 106 Nmm, and hence dimensions aresafe.Thus,t 18.5 mm. . . (vi)Other DimensionsThe outer diameter of flange is calculated fromDo 2Dc D 2 313 250 376 mmThe diameter of register, Dr Do 188 mm2. . . (vii). . . (viii)Thickness of the protective cover on the top of the flangetf dor t , Choose tf t 18.5 mm4. . . (ix)The extension of protection should be 5 mm greater than nut height on bothflanges.Summary of ResultsShaft diameterd 125 mm. . . (i)Bolt diameterd1 25.5 mm. . . (ii)Number of boltsn 6. . . (iii)Pitch circle diameter of boltsDc 313 mm. . . (iv)Hub diameterD 250 mm. . . (v)Length of hubL 89 mm. . . (vi)Key dimensionsw 36 mm, h 20 mm, L 89 mm . . . (vii)Thickness flanget 18.5 mm. . . (viii)Outer diameter of flangeDo 376 mm. . . (ix)Diameter of registerDr 188 mm. . . (x)Thickness of protective cover tf 18.5 mm. . . (xi)SAQ 1162(a)Sketch a muff coupling and identify its advantages and disadvantages.(b)Sketch a flange coupling and mention how strength of bolts and thickness ofthe flange can be calculated.(c)Mention materials for shaft, flange, keys and bolt.(d)Show register in flange. What purpose does it serve?
(e)Design and draw a flange coupling, to connect two coaxial shafts of anelectric motor and worm and worm wheel reducer. The shafts transmit 7 kWof power at 300 rpm. The permissible stresses are :Keys and CouplingsShearing stress in shaft 50 N/mm2Shearing stress in key 25 N/mm2Shearing stress in coupling 3 N/mm2Shearing stress in bolt 25 N/mm2The results must consist of shaft diameter (d), which has to be increased by25% to take care of keyway, number of bolts (n), diameter of bolts (d1),pitch circle diameter of bolts (Dc), diameter of hub (D), length of hub (L),dassume square key of size , thickness of flange (t), outside flange4diameter (Do).6.7 SUMMARYShaft is an important machine element and transmits power. The keyways becomesessential feature of shafts because some part like gear or pulley has to be attached on itto transmit power. The keys are standardised and can be selected from relevant table.1There is yet simpler method to use a square key of depthof diameter of shaft.4Couplings connect coaxial shafts. They are formed by two discs attached to shaftsthrough key and jointed by bolts, parallel to shaft axis. The discs are made as flangesintegral with the hub. The flanges are often made in cast iron. Muff couplings are thickcylinders which could be used as sleeves or split to be bolted around the shafts. Thedriving force in muff coupling is friction between the inner surface o
(a) Shunk keys, (b) Saddle keys, (c) Tangent keys, (d) Round keys, and (e) Splines. We shall now discuss the above types of keys, in detail, in the following sections. 6.2.1 Sunk Keys The sunk keys are provided half in the keyway of the shaft and hal
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with a keyway of the appropriate size and style for the key to be used. Shaft keys are available in a variety of types and include Parallel Keys, Square Keys, Flat Keys, Rectangular Keys, Woodruff Keys, Plain Taper Keys, Gib-Head Taper Keys, Perpendicular Pins, and the Feather Key. Parallel keys are inexpensive, readily
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Coupling Preselection and Options 3 Torsionally Rigid Gear Couplings ZAPEX ZW 4 ZAPEX ZI 5 Torsionally Rigid All-Steel Couplings ARPEX 6 Flexible Couplings N-EUPEX 7 RUPEX 8 BIPEX 9 Highly Flexible Couplings ELPEX-B 10 ELPEX-S 11 ELPEX 12 Fluid Couplings FLUDEX 13 Taper Clamping Bushes 14 Ap
Quick Start: Downloads and keys Microsoft Volume Licensing 5 Export all keys To export all the keys for an account, do the following: 1. Select EXPORT ALL KEYS. You can print a list of all keys or export the file in CSV format. 2. To print a list of all keys, select PRINT KEYS. 3. To export the file for later use, select EXPORT AS A CSV.
PGP makes use of four types of keys: one-time session symmetric keys, public keys, private keys, passphrase-based symmetric keys. 1 Unpredictable session keys must be generated. 2 PGP allows users to have multiple public/private key pairs. There is not a one-one correspondence between users and public keys. 3 Each entity must maintain a file of
PGP makes use of four types of keys: one-time session symmetric keys, public keys, private keys, passphrase-based symmetric keys. 1 Unpredictable session keys must be generated. 2 PGP allows users to have multiple public/private key pairs. There is not a one-one correspondence between users and public keys.
Anatomy 2-5 Indications 5 Contra-indications 5 General preparation 6 Landmarks 6-7 Performing the block 7-8 Complications 8 Trouble shooting 9 Summary 9 References 10 Appendix 1 11. 6/10/2016 Fascia Iliaca Compartment Block: Landmark Approach 2 FASCIA ILIACA COMPARTMENT BLOCK: LANDMARK APPROACH INTRODUCTION Neck of femur fracture affect an estimated 65,000 patients per annum in England in .
