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Mechanical Engineering

### Transcript of Machine Design

SECTION 7 SHAFT DESIGN Page 1 of 76 471.A short stub shaft, made of SAE 1035, as rolled, receivers 30 hp at 300 rpm via a 12-in.spurgear,thepowerbeingdeliveredtoanothershaftthroughaflexible coupling. The gear is keyed (profile keyway) midway between the bearings. The pressureangleofthegearteeth o20 = ;5 . 1 = N basedontheoctahedralshear stress theory with varying stresses. (a) Neglecting the radial componentRof the toothloadW ,determinetheshaftdiameter.(b)Consideringboththetangential andtheradialcomponents,computetheshaftdiameters.(c)Isthedifferencein the results of the parts (a) and (b) enough to change your choice of the shaft size? Problem 471. Solution: For SAE 1035, as rolled ksi sy55 =ksi su85 =( ) ksi s su n5 . 42 85 5 . 0 5 . 0 = = = cos W A =( )lb innhpT = = = 630030030 000 , 63 000 , 63 2ADT=( )2126300A=lb A 1050 = cos W A =20 cos 1050 W =lb W 1118 =Shear stress ( )3 36300 16 16d dTss = =3800 , 100ds sms s= =0 =ass SECTION 7 SHAFT DESIGN Page 2 of 76 bending stress From Table AT 2 4FLM=(a) NegligibleR : ( )( )lb inALM = = = 4200416 10504 ( )3 3 3400 , 134 4200 32 32d d dMs = = =0 =ms3400 , 134ds sa= =SFs Kssssa fmyne+ =For profile keyway 0 . 2 =fK6 . 1 =fsK85 . 0 = SF( )( )( )( )3 3661 , 10085 . 0400 , 134 0 . 2d d SFs Ksa fe= = = SFs Kssssas fsmsysnses+ =294 . 11555 . 42= = =ynysnsssss 3 3796 , 24 800 , 100294 . 11d dssssmsysnses=|||

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\|= = Octahedral-shear theory 212 2577 . 01(((

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\|=nesnessssN ( )212323500 , 42 577 . 0796 , 24500 , 42661 , 1005 . 11(((

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\|=d d in d 569 . 1 =SECTION 7 SHAFT DESIGN Page 3 of 76 usein d16111 = (b) Considering bothradial and tangential component. ( )( )lb inWLM = = = 4472416 11184 ( )3 3 3104 , 143 4472 32 32d d dMs = = =0 =ms3104 , 143ds sa= =( )( )( )( )3 3180 , 10785 . 0104 , 143 0 . 2d d SFs Ksa fe= = = 212 2577 . 01(((

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\|=nesnessssN ( )212323500 , 42 577 . 0796 , 24500 , 42180 , 1075 . 11(((

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\|=d d in d 597 . 1 =usein d16111 = (c) The difference in the results of the parts (a) and (b) is not enough to change the choice of the shaft size. 472.A cold-finished shaft, AISI 1141, is to transmit power that varies from 200 to 100 andbackto200hpineachrevolutionataspeedof600rpm.Thepoweris receivedbya20-in.spurgearAanddeliveredbya10-in.spurgearC.The tangentialforceshaveeachbeenconvertedinto a force ( A and C shown) and a couple (not shown). The radial componentRof the tooth load is to be ignored in theinitialdesign.Let2 and,consideringvaryingstresseswiththemaximum shear theory, compute the shaft diameter. Problems 472 474 SECTION 7 SHAFT DESIGN Page 4 of 76 Solution: For AISI 1141, cold-finished ksi sy90 =ksi sn50 =8 . 11=ynss 85 . 0 = SFnhpT000 , 63=( )lb in T = = 000 , 21600200 000 , 63max ( )lb in T = = 500 , 10600100 000 , 63min ( ) ( ) lb in T T Tm = + = + = 750 , 15 500 , 10 000 , 212121min max ( ) ( ) lb in T T Ta = = = 250 , 5 500 , 10 000 , 212121min max 316dTss=( )3 3000 , 252 750 , 15 16d dsms = =( )3 3000 , 24 5250 16d dsas = =SFs Kssssas fsmsysnses+ =For profile keyway 0 . 2 =fK6 . 1 =fsK8 . 11= =ynysnsssss ( )( )3 3 3894 , 9485 . 0000 , 84 6 . 1 000 , 2528 . 11d d dses= +|||

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\|= Bending stress, negligible radial load lb in T = 000 , 21at 200 hp For A: T A = ||

\|220 ( ) 000 , 21 10 = ASECTION 7 SHAFT DESIGN Page 5 of 76 lb A 2100 =at 200 hp For C: T C = ||

\|210 ( ) 000 , 21 5 = Clb C 4200 =at 200 hp [ ]= 0BM ( ) ( ) ( ) 15 25 10 C D A = +at 200 hp ( )( ) ( ) ( )( ) 15 4200 25 10 2100 = + Dlb D 1680 =[ ]= 0VFD B C A + = +at 200 hp 1680 4200 2100 + = + Blb B 4620 =At 200 hp:lb A 2100 = ,lb B 4620 = ,lb C 4200 = ,lb D 1680 =Shear Diagram Maximum moment at B ( )( ) lb in M = = 000 , 21 10 2100( )3 3 3000 , 672 000 , 21 32 32d d dMs = = =0 =ms 3000 , 672ds sa= =( )( )3 3304 , 50385 . 0000 , 672 0 . 20d d SFs Kssssa fmyne= + = + = SECTION 7 SHAFT DESIGN Page 6 of 76 3894 , 94dses=Maximum Shear Theory 212 25 . 01(((

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\|=nesnessssN ( )212323000 , 50 5 . 0894 , 94000 , 50304 , 50321(((

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\|=d d in d 78 . 2 =usein d432 = 475.A shaft S, of cold-drawn AISI 1137, is to transmit power received from shaft W, which turns at 2000 rpm, through the 5-in. gear E and 15-in. gear A. The power is delivered by the 10-in. gear C to gear G, and it varies from 10 hp to 100 hp and backto10hpduringeachrevolutionofS.Thedesignistoaccountforthe varyingstresses,withcalculationsbasedontheoctahedralshearstresstheory. Let8 . 1 = N andcomputetheshaftdiameter,usingonlythetangentialdriving loads for the first design. Problem 475 477 Solution. For AISI 1137, cold drawn ksi sy93 =ksi su103 =( ) ksi s su n5 . 51 103 5 . 0 5 . 0 = = =806 . 11935 . 51= = =ysnsynssss nhpT000 , 63=( ) rpm rpmA inE inn 667 2000. 15. 5= =SECTION 7 SHAFT DESIGN Page 7 of 76 ( )lb in T = = 9450667100 000 , 63max ( )lb in T = = 94566710 000 , 63min ( ) ( ) lb in T T Tm = + = + = 5 . 5197 945 94502121min max ( ) ( ) lb in T T Ta = = = 5 . 4252 945 94502121min max 316dTss=( )3 3160 , 83 5 . 5197 16d dsms = =( )3 3040 , 68 5 . 4252 16d dsas = =SFs Kssssas fsmsysnses+ =For profile keyway 0 . 2 =fK6 . 1 =fsK85 . 0 = SF( )( )3 3 3425 , 5585 . 0040 , 68 6 . 1 160 , 83806 . 11d d dses= +|||

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\|= Bending stress, using only tangential loads For 100 hp:lb in T = 9450T A = ||

\|215 ( ) 9450 5 . 7 = Alb A 1260 = For C: T C = ||

\|210 SECTION 7 SHAFT DESIGN Page 8 of 76 ( ) 9450 5 = Clb C 1890 = [ ]= 0BM C D A 14 20 6 = +( ) ( ) 1890 14 20 1260 6 = + Dlb D 945 =[ ]= 0VFD B C A + = +945 1890 1260 + = + Blb B 2205 =Shear diagram Maximum moment at B ( )( ) lb in M = = 7560 6 1260( )3 3 3920 , 241 7560 32 32d d dMs = = =0 =ms 3920 , 241ds sa= =( )( )3 3189 , 18185 . 0920 , 241 0 . 2d d SFs Kssssa fmyne= = + = 3425 , 55dses=Octahedral Shear Theory 212 2577 . 01(((

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\|=nesnessssN ( )212323500 , 51 577 . 0425 , 55500 , 51189 , 18121(((

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\|=d d in d 997 . 1 =usein d 2 = SECTION 7 SHAFT DESIGN Page 9 of 76 478.AshaftmadeofAISI1137,coldrolled,foraforageharvesterisshown. Power is supplied to the shaft by a vertical flat belt on the pulley A. At B, the rollerchaintothecutterexertsaforceverticallyupwards,andtheV-beltto thebloweratCexertsaforceverticallyupwards.Atmaximumoperating conditions, the flat belt supplies 35 hp at 425 rpm, of which 25 hp is delivered to the cutter and 10 hp to the blower. The two sections of the shaft are joined byaflexiblecouplingatDandthevariouswheelsarekeyed(sled-runner keyway)totheshafts.Allowingforthevaryingstressesonthebasisofthe von Mises-Hencky theory of failure, decide upon the diameters of theshafts. Choose a design factor that would include an allowance for rough loading. Problem 478. Solution: For AISI 1137, cold rolled ksi sy93 =ksi su103 =( ) ksi s su n5 . 51 103 5 . 0 5 . 0 = = =806 . 11935 . 51= = =ysnsynssss Pulley,( )lb innhpTA = = = 518842535 000 , 63 000 , 63 For flat-belt ( )( )lbDTF F F F FAAA692305188 4 22 21 2 2 1= =|||

\|= = + =Sprocket, ( )lb innhpTB = = = 370642525 000 , 63 000 , 63 For chain, ( )lbDTFBBB741103706 2 2= = =Sheave,( )lb innhpTC = = = 148242510 000 , 63 000 , 63 SECTION 7 SHAFT DESIGN Page 10 of 76 For V-belt, ( )( )lbDTF F F F FCCC445101482 3 25 . 1 5 . 11 2 2 1= =|||

\|= = + =Consider shaft ABD. 35 hp Shaft ABD [ ]= 0' DM( ) ( )B AF A F 4 ' 4 8 4 8 6 + + = + +( ) ( ) 741 4 ' 12 692 18 + = Alb A 791 ' = [ ]= 0VFA F D FB A + = +791 741 692 + = + Dlb D 840 = ShearDiagram Maximum M at A. ( )( ) . 4152 692 6 lb in M = =( )3 3 3864 , 132 4152 32 32d d dMs = = = 0 =ms 3864 , 132ds sa= =SECTION 7 SHAFT DESIGN Page 11 of 76 SFs Kssssa fmyne+ =For sled-runner keyway (Table AT 13) 6 . 1 =fK6 . 1 =fsK85 . 0 = SF( )( )3 3610 , 7985 . 0864 , 132 60 . 10d d SFs Kssssa fmyne= + = + = at Alb in T TA = = 5188( )3 3 3008 , 83 5188 16 16d d dTss = = =s mss s = 0 =assSFs Kssssas fsmsysnses+ =3 3630 , 14 000 , 83806 . 11d dses=|||

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\|= Choose a design factor of 2.0 0 . 2 = Nvon Mises-Hencky theory of failure (Octahedral shear theory) 212 2577 . 01(((

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\|=nesnessssN ( )212323500 , 51 577 . 0630 , 14500 , 51610 , 7921(((

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\|=d d in d 48 . 1 =usein d211 = Consider shaft D-C ( )lb innhpTC = = = 148242510 000 , 63 000 , 63 For V-belt, ( )( )lbDTF F F F FCCC445101482 3 25 . 1 5 . 11 2 2 1= =|||

\|= = + =SECTION 7 SHAFT DESIGN Page 12 of 76 [ ]= 0' CMCF D 3 8 = ( ) 445 3 8 = Dlb D 167 = [ ]= 0VFCF D C + = 445 167 + = Clb C 612 = Shear Diagram ( )( ) lb in M = = 1336 8 167( )3 3 3752 , 42 1336 32 32d d dMs = = =0 =ms ,s sa=( )( )3 3616 , 2585 . 0752 , 42 60 . 10d d SFs Kssssa fmyne= + = + = at C,lb in TC =1482( )3 3 3712 , 23 1482 16 16d d dTss = = =s mss s = 0 =assSFs Kssssas fsmsysnses+ =3 341800712 , 23806 . 11d dses= +|||

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\|= 212 2577 . 01(((

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\|=nesnessssN ( )212323500 , 51 577 . 04180500 , 51616 , 2521(((

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\|=d d SECTION 7 SHAFT DESIGN Page 13 of 76 in d 011 . 1 =usein d 1 = 479.A shaft for a punch press is supported by bearings D and E (with L = 24 in.) and receives 25 hp while rotating at 250 rpm, from a flat-belt drive on a 44-in.pulleyatB,thebeltbeingat45owiththevertical.An8-in.gearatA delivers the power horizontally to the right for punching operation. A 1500-lb flywheelatChasaradiusofgyrationof18in.Duringpunching,theshaft slowsandenergyforpunchingcomesfromthelossofkineticenergyofthe flywheelinadditiontothe25hpconstantlyreceivedviathebelt.A reasonableassumptionfordesignpurposeswouldbethatthepowertoA doublesduringpunching,25hpfromthebelt,25hp from the flywheel. The phaserelationsaresuchthataparticularpointinthesectionwherethe maximummomentoccursissubjectedtoalte