9. SHAFTING ‘Shaft—a rotating member, usally of circular section, used to transmit power or motion. Is primary load is torsion “Axle —a non-rotating member which carries no torque ‘and is used to support wheels, pulleys, andthe like. It is «primarily laded with bending. Spindle—a short shaft or axle. ‘Terms such as lineshaft, headshaft, transmission shat, ‘machinery shaft, countershaft, and flexible shaft are names, associated with special usage. Most machine design books generally use the term shaft to mean all of the above terms. © SHAFT UNDER PURE TORSION saidstatt ¥ Wat shakt*T Maximum torsional stress 5, = 1 or 55,4: (fora solid shat) m0 Sp ere Seg: Cat a hollow shaft) or ‘ or Ot LENS cere" Angle of twist (orsonal deformation) @= 7g! imradians where: J=qgPt: for solid shafts J=ZO'-a"}: forhollow shafts * SHAFT UNDER PURE BENDING M ay, Pye Sold shaft —_ shat ‘Machine Design I: BASAEN, RV. ‘Maximum Flexural stress 32M D’ $5, =e > Sy: (solid shaft) a “Case 3) Shaft has fixed supports with load P at midspan u2_|P y., [conte ———— ‘Two General Classifications of Shafts 1. Transmission shaft— shaft hat connects a prime ‘mover to a machine. 2. Machinery shaft—shaft that is an integral part of a machine ASME, Formulas ‘Shaft Design Stresses (Only S, and Sy are known) Shear design stress: Sa~ 03S, whichever is smaller: Sa 0.18 Sy Mfshafh has a kevway, ‘consider a strength reduction factor of 0.75 Normal design stress: Se=0.6 S, whichever is smaller: S2= 0.368, [shaft has a keyway, considera strength reduction factor of 0.75: 14Allowable deformations Angle of Twist ‘Transmission shaft: Onion = 1° ina length of 20 diameters 0.08" to 1° per foot of length ‘between supports “Transverse Deflection Transmission shaft: Yeuow = 0.01 inch per foot of length between supports Machinery shaft: ‘Yetiow = 0.002 inch per foot of length between supports ‘© COMBINED STRESSES (cx fea PPP) 05, and, from the Max. Normal Stress equation Sex = 164 arn(t +B?) = go (=m) | Rae arpa + BA] + (cee + [ia PPE* EY) 9 5, where a = factor for column action K. = combined shock and fatigue factor for shear Ky = combined shock ane fatigue factor for bending B= ratio of inside to outside diameters of hollow shaft Note For combined torsion and bending on a solid sha, above formulas can be reduced to 16 Soma = ps VET + Bea? > Sn [Machine Desin I: BASAEN, RV. and Snax = 5s KM AV RTI CMF = Si --» then solve for the diameter D from each ofthe formulas, and use whichever is larger! If Ka and K, are not specified, use Km = K, = 1.0. Note: ~ Af the shaft material is ductile, use tho first equation. A ductile material for a shaft is more ‘common than a brittle material, - LF uncertainty arises as to the shaft material, use the more conservative result of first and second ‘equations (e.g, forthe shaft diameter, D, use the ‘greater diameter from the equations). ~ Ifthe axial load is tensile a = 10. = Ifloading conditions are not specified, the factors K, 8 Ky, become insignificant. K,= Ky =10 ~ If some toads are Considerably low, they may become negligible, thus, the first and second equations may be simplified. - From the possible loads on the shaft, bending ‘moment M is oftentimes the mos difficult load to compute. * Transmission shafts from the PSME Code, Machinery’s handbook, Kent's M.E. handbook, ete. Main Power Transmiting ‘Su Line Jo Machine ‘hat Driven cal Machine A Deiven Machine 8 Main Power Transmitting Shafts DN 30 Line Shafts Carrying Pulleys 1s‘Small, Short Shaft and Countershafts “oi “38 ‘whore P= horsepower transmitted N D = angular speed in sp. “= shaft diameter in inches Note: : ‘Above formulas were based on design stresses of = 4 ksi for main power transmitting shafts = Gksifor line shafis, and = 85 ksi forsmall /short shafts, and countershafis ‘¢ Empirical Formalas from Machinery Handbook Required Shaft diameter: 8) foran allowable angle of twist not exceeding, 0.08° per foot of length D=029 VF )_foranallowable angle of twist not exceeding I ina length of 20 diameters D=01 VT ©) for short, solid shaft subjected only to heavy transverse shear hav 5 D = shaft diameter in inches T = torque in intb V_ = maximum transverse shear load, Ib S, = maximum torsional shearing stress, psi Shaft Supported length VS. Shaft Diameter (Deflection is limited to 0.01 inch per foot of length) a)» Shafting subjected to no bending action except its own weight L=895 VD subjected to bending action due to L=52 VDF L = maximum distance between bearings, feet D = shaft diameter in inches ¢ Standard Sizes of Shafts For standard sizes of shafts in inches: refer to p. 269 of DME Faires Machine Design I: BASAEN, RV For standard sizes of shafts in mm Tnerement 4-10 Imm 12,15,17 20-110 Smm 110-200 10mm 200-280 20mm ® Cases of Shaft Problems (Stress Based) 1. 54g 07 Sais given (5,, 5,,cuul N) For Pure‘Torsion, 5-12 45, = ps Sea (solid shaft) . 11D 5207S aot ay > Sse N SOP Sus S.= aps pp) Sa (hollow shaft) where B For Pure Bending So at = aF> (solid shaft) 5 = M0 _ SyorSy 1" 3D a ~ aN 32m Syors, Si aD \i-# N (hollow shaft) 2. Only S,,and S, are given but no factor of safety,V Use the ASME Code for Shaft Desig Shear design stress 018 8, ‘whichever is smaller Note: = Above Szq and Sq are based on the ASME Code (p. 278 DME Faires shows estimation formulas) ~ If the shaft is deseribed to have a keyway, multiply above design stresses by 0.75. 1163. No Su, Sy, or N given, The shaft is described to bbe any of the following transmission shafts, MPTS —main power tansmitting shafts LS line shaft carrying pulleys SS ~short, small shaft or counter shaft For each of the above shafts, design handbooks give the following shear design stresses, Su=4 ksi (MPTS) Sar 6 ksi (L5) Sue 85 ksi (SS) 4. Shaft is under combined stresses Use the derived formulas for Sines 2nd Snes Combined Stresses in Shafts ‘Axial & Flexural Loads ‘Axial & Torsional Loads Flexural & Torsional Loads ‘Axial, Torsional, & Flexural Loads eesP Note: ‘These combinations are treated with the aid of ‘maximum stress equations in the previous chapters. = Maximum Shear Stress Uqution, The caution is more gencrally applicable fo shaft made of ductile instal Po ee |e (oye Smmar= |5:2+ (5) S04 = Maximum Normal Suess Equation. The equation is more applicable o bitle materials BS eles ey inex = 3+ [8+ (5) > Sa = Af uncertainly arises as to the material, use the more conservative results from both equations above. Conversion from Solid to Hollow Shaft ‘+ Percent Saving Weight Nate: “Applicable to equal length and same material. ‘© Percent Reduction in Torque Capacity 1,-T, RTC Where, T, = torque thatthe solid shaft can transmit Tj, = torque thatthe hollow shaft can transmit Machine Design 1: BASAEN, BV Example 1: ‘A 76mm solid shaft is to be replaced by a hollow shaft of equal length and equal torsional strength, but its ‘weight should only be one half that of the solid shaft Determine the diameter D and dof the hollow shaft. Solution: T Hollowshatt Solid shart Conditions given for the shafts include equal length, equal torsional strength and that hollow shaft should weigh only half as much a the solid shaft. Its then assumed that the shaft are made of the same material ‘© Equal torsional strength, Sag or Seoonay = Spy 7 Sut ter ora) St Ths, 1610 (D+ 4) ‘nd, with equal power being transmited, torque is also constant. Sor 16TD D*—d* =76°D + (1) nae) ma) + 50% weight reduction we=y (Goa): ee a GO? -a)1 = 05 [y Gee) ] D? — d? = 288 d? = D* — 2888 + (2) = 7G@?-@)L Substituting eqn. (2) into equation (1) Dt —[D? — 2888]? = 76*D: D = 91.74mm And, from eqn. (2), d = 74,35mm. Thus, asa replacement shaft, the hollow shaft should have the following diamet TA mam 7Example 2: ‘A shaft is made of steel whose S, = 300 MPa and S, = 400MPa, it is to transmit 300 kW at 600 rpm. (a) Calculate the diameter ofthe shaft if the load is gradually applied. Consider the effect of keyway on the shaft. (b) The same as a), except that the load is suddenly applied. Solution: T solidshart Sy + 800 10a Su +400 MPa T+ 300, nto) 1 T# 4774 500 Nem 408 c4-549 «10% (A052 Ssd + 0-3 Sy » 0°3 (308) =90 Ssd* 0-18 Su + 0-18 (4001 = 72 BUT, W/ KEYWAY= Sse" * 0-75 (72) THEN, os 46L ‘ $5 ks IEE, Sod TABLE at *S]enaoonusy aomico (1-0) (46 = 4 73a 500) (0.79972) [De 36-65 mm * 80 mm (B)D=9 (SUDDENLY APPLIED LOAD) eee 210 210 015 Keqee * 125 (125)(1es 4.772 500) + 0-75.(72) [D+ 62-87 mm = 85am THUS, Example 3: 2) What maximum horsepower can a 3-inch diameter short shaft transmit at 600 rpm? 'b) What diameter in mm is required by a main power transmitting shaft to transmit 500 kW at 200 ppm? Machine BASAEN, RV Solucon: 3) Maximum Power P=? Chip) £ SHORT SHAFT? FROM PSNE PDBN , P+ hp ee OF in No rem b) Min diameter required for a main power transiting shaft From the PSME_ Code formulas for transmission shatts: 0+? (mm) { MPTS} FROM PON; Po hp 0 0+ in Nov epm B00 RN a thar] 168-6 rim ‘© Bending Lond Calculation for Shaft J. Bending Load jrom Belts and Pulley usMachine Design 1: BASAEN, RV, Useful Relations: F, = tight side tension in the bet BM, =0; T=@- (2) Fz = slack side tension in the belt fa a Fp = bending load on shaft F.-F= >): WetBeltPall) Fay = chain tension where, T = torque transmitted by the shaft F,= tight side tension in the belt = slack side tension in the belt Fy ~ bending load on shaft 3. Bending Load from Spur Gears Belt tension ratio: a E shen! where f = pressure angle of the gear teeth 4.5°, 20°or 25°) Fe = tangential force on the gear F, ~ radial force on the gear Fa ~ resultant force on the gear @ = conlact angle between belt and pulley (based on the small pulley, where belt slip is ‘more likely to take place) 6 =180— 2y From Machine Elements 2, For tangential load, F y= sin where, Fp=ot Dyas Ppa = play diameter D, C= center distance between the pulleys (tangential load of the gear tooth) ‘Thus, the bending load is, Forradal lod Fe Fp = (Fi + Fa) cosy Fem cong (ending load on the shaft) ae From the set-up as shown, the bending moment M is (adial load or separating fore between gears) For the resultant load, Fe Fa= J Fet+F,? 2, Bending Load Chains and Sprockets (esultant tooth load) ‘This resultant tooth load Fe becomes the bending load on the shaf, Fo=Fr Note: If the shaft has complex bending loads, the bending, moment M is generally determined with the aid of shear and moment diagrams. Other gear types (helical, bevel, worm gears, etc.) Note: result ins more complicated loading on the gear and on the For chains the slack side tension Fis pmetically vero. Supporting shaft ‘The bending lad then becomes, Fa= Fi= Fea EM, = Of: 8 a es = Fea =5- 9Example 4: a) Determine the bending load Fy at the shaft from the given set up. The driven pulley is 400mm and is ‘800 mm away from the driving palley. 'b) The same as a. except that the pulleys are replaced by spur gears (g} = 20°) and the diameter of the driving gear is 100mm. Also compute for the bending ‘moment on the shaft ® simply supported and the gear is replaced midway( between the bearings that are 100 mm pat. ae Solution: 8) Bending load fiom belt and pulley, F Fox (Pir Fe) cos wee: Fi-ras 2p e Fiote + 200 cassanoe 205 Fi-Fa= 3183-—-© Brae? Fete 180-28 Beso [oasp4) a Bein fage 200 2¢600) Bese 2165-6" fe, (oss-622) i, i.e ie Pre 2.330 F2 —» @ SOLVING © & © + SimivL Fi 54895 N F2+ 2306-5 N THs, Fe (5469-5 +2306-5) cos(t-181") Fer 7734-6 N] ‘Machine Design I: BASAEN, RV b) Bending load from spur gear, Fy 1591 500 Feeat + 21060) asa 108] 160 Fes @366N Fre Ft tang 4, + 6366 tan zo") or Fre 2317N THs, Fre [TRPe cree = [Teaeay® (2a) FR: 6774 SN > Fo BENOING, Ms Fe CL) = 6774-5 ¢1comm) 4 + Me 169 362 Nome] Example 5: ‘A.150mm diameter spur pinion transmits 20 KW at 600 pm. Considering a pressure angle of 20° and the distance between bearing supports as shown, recommend the standard shaft diameter necessary if S..~ 60 MPa. T H 180mm r *yo0mm 150m ‘Solution: “+ TORAUE , (7) Ts 20. (4.549.2108 EB (5494109) T+ 318 300 N-mm + BENOING MOMENT, ¢M) Fes 27 = £(316300) ; Fee4aaan 0 150 Fre Fe tang = 4244 tan 20 Pee [FP VF + (Canaay®s (rea Fes 45ioN Ms Fr (a)(b) » C4516) (400) (160) atb 250 Mz 270 460 N-mm 120Ds? FROM Ssmax Ssraax * [5st+ (FPP Ssd Ssr1ot ; 5+ 32M 105 703 Sewn “18 ({FFEME) = Ssd sh ([Fo18 3cor* + cero a8cF ) 9 60 Bram = 35mm) Example 6: ‘The shaft of a heavy tractor transmits 120 kW at 600 pm. It acts as a cantilever supporting a load of 2800 N located 60 em from the supporting bearing. The allowable stresses are 137 MPa for normal stress and 62 MPa for shear. If axial load is negligible, find the shaft diameter. Solution: ‘Sketch of shaft loading from the description: FINO 0:7 *TORAUE ,¢T) T ato# Pawtes. + 128 (4.549 «1087 T+ 1909 800 N-mm *BENOING MOMENT. (IM) MF CL) + 2600 (600) Mr 1.680.000 N-mm *USING Ssmax =» Ssd Ssmax * J5e*+(E)" => Sed=62 Mia Seek; S2aaM 0s Oe Sinan 8 [TR ] 62 rea + Ceeaouatt + 54.34 nm Machine Design 1: BASAEN.RV. USING Smax > Sd Suac* $4 [SFECE). op Sa = 137 are ane [FRE Suan +88, [four [Teme f] 06 187 0258-45 mm qwus, use [Desacbden = 60mm LARGEST O Veta: aia Example 7: ‘A centrifugal pump is directly coupled to a motor. The ‘pumping rate is 3600 li/min, against a total head of 8 meters of water. The pump efficiency is 65% at a shaft speed of | about 500 rpm. Determine the motor shaft diameter if the shear design stress of the shaft is 60 MPa. Neglect the bending load from the centrifugal force due to the coupling, © wor Bo Bn = NK Solution: FIND 0? Pane GBH & Bb mt (961AK) com ‘ Prag * 4-709 ko Papo Puone., Np > Prise iv Puork Me 657 Paoroe #-4:704. | Pome + 7-245 KW. 0-65 Ts Porn * 7-245. tot Pugne + 1-248. (9.949 10°) T+ 188 365 Nom THs, Ssmax Ssd*60 MPa Sistas +16 T mp Sed: 60 od 16618365). 66 08 D= 22-35 cam = 250mm 121THEN, ; as $6 (88)(167-57 » Miso! feel ce Fis 2.405 F2 —+@ SOLVING © & O+ O- reo 1987-5 O— Fi 2.405 Fo Frets e212 Fas 5663-7N THus, 76-215" FA= (Prefs) cos 8 FAD IV 171-6 N Fax © Fa Cos 60 Fay + Fa sin 60" + 16603-N @ POINT. (@)- +1991 800 N-mm 2T +2(1541500) OF 186 1220 Fe F Fre FE tan® * 21220 tan 20° Fre 7723-4N ays fe *SHAFT's LOAD ae one 4c 2 ae ey o> pee 22088 og 2 a7 = = 300-—F i504 Ray Rey bso. MOMENT: Mey (17 1921 (180) © 2 578 800, Mey + (70717 (160) = 4 180 650 X- FORCES 9566 4 iso. Roe MOMENT? Max + (4586) (150) + 1.487 400 21220) (180) 3 183 000 ‘RESULTANT MOMENTS , C0) Ms [Mate Mato. moo rinay? + [traara00 Mg 482.586 Nomn Me fracct> wey? * (3163 oc0)** G1 T9050 Me*3394 411 Nem ® Mex 578 800)" OY + 181 CRITERION: Ssmax T+ 1591 500 N-mm Ms 3 394 911 Nem Ssmax + Ssd 54 MP; KEELE jKme tS Scrone “36> [TRETI+ (Rm MF => Sse 054 2+ 80:05 mm. + 2N° CRITERION * Smax Staaz “> Sd*108 MPR Stax #18 [iin Me [CRS TIE + (Km sts [hm Me [ERSTE + Chim MOF JS: 0579.19 0m THUS, USE THE LaKcEst D D> 86-08 » Bm, sea diaExample 8: ‘A transmission shaft ‘x’ carries a pulley and a pinion at points A and D, respectively. The shaft is supported by " bearings at B and C. Dimensions given are in mm, Reconunend the diamcter of this shaft ‘x’. Use 5, = ee 1.5 and coefficient of belt friction f = 350 300150 6s pulley toarhen 4 machine 300] = shaft ou a a 1D} | 150 13 kg pinion wee 72N SoKW 600 rpm > [200 souW 600 rpm Solution: Shaft ‘x’ is under combined torsion and bending. Machine Design Ic BASAEN, BV + DESIGN STRESSES: Ssd + 0-8 Sy +03 (360) = 90 Mita Ssd= 048 Su = 048 (400) = 72 MPa wy KEYWAY: Sed" s 0-75 0172) 54 MPa Sd20-G Sy = 180 MPa Sd 0-36 Sus 144 MPa Sur- 0-75 (1449 top Me SHAFT *x’ LOAD! + TORBUE, (1) 1? Pov ca Tex Sq * Tn Dpcm Tx Ns (400) + (606) (200) Tix © 3006 9m Tso x08 $b; (4549 x10") T2141 500_N-mm_ “BENDING MOMENT, (M1): @ POINT AY Tes Ts 1841 800 LS Orhng Patey [Maso]: ri-ra © 20 + 261501500) oY Fras ras75—+@ M50, contact tole @ smal plty a : €*180-2% = Sema enuey Bre" fos | 00 «sin go" sco B22 + sino Cd+ 423-7 men we Ee] Se eaist 10 ~ 2¢6-215) 8167.5 122