Chapter = For one gear straddle-mounted, the load-distribution factor is:

Transcription

1 Chaper iven: Uncrowned, hrough-hardened 300 Brinell core and case, rade 1, N C 10 9 rev of pinion a R 0.999, N 0 eeh, N 60 eeh, Qv 6, d 6 eeh/in, normal pressure angle 0, shaf angle 90, n p 900 rev/min, F 1.5 in, S F S 1, K o 1. Fig. 15-7: J 0.49, J 0.16 Mesh Eq. (15-7): d 0/ in, d 60/ in v π(3.333)(900/1) f/min Eq. (15-6): B 0.5(1 6) / A ( ) Eq. (15-5): K v Eq. (15-8): v,max [ (6 3)] 3940 f/min Since < 3904, Kv is valid. The size facor for bending is: Eq. (15-10): K s / For one gear sraddle-mouned, he load-disribuion facor is: Eq. (15-11): K m (1.5) For K L, he more conservaive criical equaion will be used here. Though i is accepable, according o AMA, o use he less conservaive general equaion for general applicaions. Eq. (15-15): (K L ) (10 9 ) (K L ) (10 9 / 3) Eq. (15-14): (C L ) 3.48(10 9 ) (C L ) 3.48(10 9 / 3) Eq. (15-19): K R log( ) 1.5 (or Table 15-3) C K R R Shigley s MED, 10 h ediion Chaper 15 Soluions, age 1/0

2 Bending Fig : 0.99S s 44(300) psi Eq. (15-4): Eq. (15-3): a sakl (0.86) all) sw psi SFKT KR 1(1)(1.5) all) FK xj K K K K 1 d o v s m (1.5)(1)(0.49) 690 lbf 6(1)(1.374)(0.5)(1.106) 690(785.3) 16.4 hp Eq. (15-4): (0.893) all) psi 1(1)(1.5) (1.5)(1)(0.16) 60 lbf 6(1)(1.374)(0.5)(1.106) 60(785.3) 14.8 hp Ans. The gear conrols he bending raing. 15- Refer o rob for he gearse specificaions. ear Fig. 15-1: s ac 341(300) psi For he pinion, C 1. From rob. 15-1, C R Thus, from Eq. (15-): sac( CL) C c,all) S KTCR 15 90(1)(1) c,all) psi 1(1)(1.118) For he gear, from Eq. (15-16), B (300 / 300) C (3 1) From rob. 15-1, (C L ) Equaion (15-) hus gives Shigley s MED, 10 h ediion Chaper 15 Soluions, age /0

3 sac( CL) C c,all) S KTCR 15 90(1.0685)( ) c,all) psi 1(1)(1.118) For seel: C p 90 psi Eq. (15-9): C s 0.15(1.5) Fig. 15-6: I Eq. (15-1): C xc Eq. (15-1): ) FdI C K KvK C C c,all p o m s xc (3.333)(0.083) 90 1(1.374)(1.106)(0.5937)() 464 lbf 464(785.3) 11.0 hp (3.333)(0.083) 90 1(1.374)(1.106)( )() 531 lbf 531(785.3) 1.6 hp The pinion conrols wear: 11.0 hp Ans. The power raing of he mesh, considering he power raings found in rob. 15-1, is min(16.4, 14.8, 11.0, 1.6) 11.0 hp Ans AMA 003-B97 does no fully address cas iron gears. owever, approximae comparisons can be useful. This problem is similar o rob. 15-1, bu no idenical. e will organize he mehod. A follow-up could consis of compleing robs and 15- wih idenical pinions, and cas iron gears. iven: Uncrowned, sraigh eeh, d 6 eeh/in, N 30 eeh, N 60 eeh, ASTM 30 cas iron, maerial rade 1, shaf angle 90, F 1.5, n 900 rev/min, φ n 0, one gear sraddle-mouned, K o 1, S F, S. Fig. 15-7: J 0.68, J 0.8 Shigley s MED, 10 h ediion Chaper 15 Soluions, age 3/0

4 Mesh d 30/ in, d 60/ in v π (5)(900 / 1) 1178 f/min Se N L 10 7 cycles for he pinion. For R 0.99, Table 15-7: Table 15-5: Eq. (15-4): s a 4500 psi s ac psi sak L 4500(1) sw S K K (1)(1) F T R 50 psi The velociy facor Kv represens sress augmenaion due o mislocaion of ooh profiles along he pich surface and he resuling falling of eeh ino engagemen. Equaion (5-67) shows ha he induced bending momen in a canilever (ooh) varies direcly wih E of he ooh maerial. If only he maerial varies (cas iron vs. seel) in he same geomery, I is he same. From he Lewis equaion of Secion 14-1, M Kv σ I / c FY e expec he raio σ CI /σ seel o be σ σ CI ( K ) ( K ) v CI E E CI seel v seel seel In he case of ASTM class 30, from Table A-4(a) (E CI ) av ( )/ 14.7 kpsi ( K ) 14.7 ( K ) 0.7( K ) 30 Then, v CI v seel v seel Our modeling is rough, bu i convinces us ha (Kv) CI < (Kv) seel, bu we are no sure of he value of (Kv) CI. e will use Kv for seel as a basis for a conservaive raing. Eq. (15-6): B 0.5(1 6) / A ( ) Eq. (15-5): inion bending K v (σ all ) sw 50 psi From rob. 15-1, K x 1, K m 1.106, K s 0.5 Shigley s MED, 10 h ediion Chaper 15 Soluions, age 4/0

5 Eq. (15-3): ear bending ) FK J all x d KoKvKsKm 50(1.5)(1)(0.68) lbf 6(1)(1.454)(0.5)(1.106) 149.6(1178) hp J lbf J (1178) 4.54 hp The gear conrols in bending faigue hp Ans Coninuing rob. 15-3, Table 15-5: Eq. (15-1): s ac psi sw σ c,all psi σ FdI C K KvK C C c,all p o m s xc Fig. 15-6: I 0.86 From robs and 15-: C s , K s 0.5, K m 1.106, C xc From Table 14-8: C p 1960 psi (5.000)(0.086) Thus, (1.454)(1.106)( )() 91.6(1178) hp Ans. Raing Side noe: Based on resuls of robs and 15-4, 91.6 lbf min(5.34, 4.54, 3.7, 3.7) 3.7 hp The mesh is weakes in wear faigue. Shigley s MED, 10 h ediion Chaper 15 Soluions, age 5/0

6 15-5 Uncrowned, hrough-hardened o 180 Brinell (core and case), rade 1, 10 9 rev of pinion a R 0.999, N z 1 eeh, N z 4 eeh, Qv 5, m e 4 mm, φ n 0, shaf angle 90, n rev/min, S F 1, S S 1, F b 5 mm, K o K A K T K θ 1 and C p 190 Ma. Fig. 15-7: J Y J1 0.3, J Y J 0.05 Mesh d d e1 mz 1 4() 88 mm, d m e z 4(4) 96 mm F Eq. (15-7): v e 5.36(10 5 )(88)(1800) 8.9 m/s Eq. (15-6): B 0.5(1 5) / A ( ) (8.9) Eq. (15-5): Kv Eq. (15-10): K s Y x (4) 0.50 Eq. (15-11): wih K mb 1 (boh sraddle-mouned), K m K β (10 6 )(5 ) From Fig. 15-8, ( CL) ( ZNT ) 3.48(10 ) ( C ) ( Z ) 3.48[10 ( / 4)] L NT Eq. (15-1): C xc Z xc (uncrowned) Eq. (15-19): K R Y Z log ( ) 1.5 CR ZZ YZ From Fig , C Zw 1 Eq. (15-9): Z x (5) ear of inion Fig. 15-1: σ lim.35 B (180) Ma Fig. 15-6: I Z I Eq. (15-): ) ( Z ) Z lim NT ) S K θ Z Z Shigley s MED, 10 h ediion Chaper 15 Soluions, age 6/0

7 585.9(1)(1) 54.1 Ma 1(1)(1.118) Eq. (15-1): σ bd Z The consan 1000 expresses in kn. e1 I C p 1000K AKvK βz xz xc (88)(0.066) (1)(1.663)(1.0035)(0.56)() π dn 1 π (88)(1800)(0.591) Eq. (13-36): k kn ear of ear σ lim Ma 585.9(1.0054) ) 56.9 Ma 1(1)(1.118) 4 ) kn ) 54.1 π (88)(1800)(0.594) 4.93 k Thus in wear, he pinion conrols he power raing; 4.90 k Ans. e will rae he gear se afer solving rob Refer o rob for erms no defined below. Bending of inion ( KL) ( YNT ) (10 ) KL YNT ( ) ( ) [10 ( / 4)] Fig : σ F lim 0.30 B (180) Ma Eq. (15-13): K x Y β 1 From rob. 15-5: Y Z 1.5, v e 8.9 m/s, K A 1, Kv 1.663, Kθ 1, Y 0.5, K , Y 0.3 x β J1 Shigley s MED, 10 h ediion Chaper 15 Soluions, age 7/0

8 Eq. (5-4): σ F limynt 68.5(0.86) F ) 47. Ma SFK θ YZ 1(1)(1.5) Eq. (5-3): F ) bmeyβ YJ K K Y K Bending of ear 1 A v x β 47.(5)(4)(1)(0.3) 1.5 kn 1000(1)(1.663)(0.5)(1.0035) π ( 88)( 1800)( 1.5) k σ F lim 68.5 Ma 68.5(0.864) F ) 47.3 Ma 1(1)(1.5) 47.3(5)(4)(1)(0.05) 1.1 kn 1000(1)(1.663)(0.5)(1.0035) π ( 88)( 1800)( 1.1) 9.9 k Raing of mesh is raing min(10.37, 9.9, 4.90, 4.93) 4.90 k Ans. wih pinion wear conrolling σ σ ( S ) ( S ) all all (a) F F σ σ ( sak L / KT KR) ( sakl / KT KR) ( K K K K / FK J ) ( K K K K / FK J ) d o v s m x d o v s m x All erms cancel excep for s a, K L, and J, (s a ) (K L ) J (s a ) (K L ) J From which ( sa) ( KL) J J ( s ) ( s ) m ( K ) J J β a a L where β or β as appropriae. This equaion is he same as Eq. (14-44). Ans. (b) In bending Shigley s MED, 10 h ediion Chaper 15 Soluions, age 8/0

9 In wear σ FK J s K FK J S K K K K S K K K K K K all x a L x F d o v s m 11 F T R d o v s m 11 1/ ac L U KoKv KmCsCxc C p T R Fd I s C C S K C (1) Squaring and solving for gives s C C Fd I ac L S KTCRC K okvkmcscxc () Equaing he righ-hand sides of Eqs. (1) and () and canceling erms, and recognizing ha CR K R and d d N, we obain ( s ) C S ( s ) ( K ) K J K C C p a 11 L 11 x 11 T s xc ac ( CL) SF C NKsI For equal in bending and wear S S F ( S ) F 1 S F So we ge (c) Cp ( sa) ( KL) J KxKTCsCxc ( sac) Ans. ( C ) C N IK ( S ) ( ) L s σ σ c,all σ c,all S σ c c Subsiuing in he righ-hand equaliy gives [ saccl / ( CRKT )] [ sacclc / ( CRKT )] C p KoK KmCsCxc / ( FdI) C p KoK KmCsCxc / ( FdI) v v Denominaors cancel, leaving Solving for (s ac ) gives, (s ac ) (C L ) (s ac ) (C L ) C Shigley s MED, 10 h ediion Chaper 15 Soluions, age 9/0

10 ( CL) ( sac) ( sac) C (1) ( C ) From Eq. (15-14), ( ) ( ) ( ) Thus, L C 3.48 N and C 3.48 N / m. L L L L ( ) ( ) ( 1 ) ( ) ac ac ac s s m C s m C Ans. This equaion is he ranspose of Eq. (14-45) iven ( B ) Brinell Eq. (15-3): (s a ) 44(300) psi J 0.49 ( ) ( sa) ( sa) m psi J ( B) Brinell Ans (0.86)(0.49)(1)( )() ( s ac) (1) 0(0.086)(0.5) psi ( B) 345 Brinell Ans ( s ) ( s ) m C (3 ) psi ac ac ( B) Brinell Ans. 341 Core Case inion ( B ) ( B ) Ans. ear ( B ) ( B ) 345 ( ) 15-9 inion core ( ) ( 0.86) 1( 1)( 1.5) ( 1.5)( 0.49) ( )( )( )( ) ( s ) psi a all ) psi lbf Shigley s MED, 10 h ediion Chaper 15 Soluions, age 10/0

11 ear core inion case ear case ( s ) 44(339) psi a (0.893) all) psi 1(1)(1.5) 1 156(1.5)(0.16) lbf 6(1)(1.374)(0.5)(1.106) ( s ) 341(373) psi ac (1) c,all) psi 1(1)(1.118) (3.333)(0.086) lbf 90 1(1.374)(1.106)( )() ( s ac) 341(345) psi (1.0685)(1) c,all) psi 1(1)(1.118) (3.333)(0.086) lbf 90 1(1.1374)(1.106)( )() All four ransmied loads are essenially equal, so he equaions developed wihin rob are effecive. Ans The caalog raing is 5. hp a 100 rev/min for a sraigh bevel gearse. Also given: N 0 eeh, N 40 eeh, φ n 0, F 0.71 in, J 0.41, J 0.01, d 10 eeh/in, hrough-hardened o 300 Brinell-eneral Indusrial Service, and Qv 5 uncrowned. Mesh d 0 / in, d 40 / in π dn π ()(100) v 68.3 f/min 1 1 K 1, S 1, S 1 o F Eq. (15-6): B 0.5(1 5) / A ( ) Eq. (15-5): Kv Eq. (15-10): K s / Shigley s MED, 10 h ediion Chaper 15 Soluions, age 11/0

12 Eq. (15-11): K m (0.71) 1.5, where K mb 1.5 Eq. (15-15): (K L ) ( ) (K L ) ( /) Eq. (15-14): (C L ) 3.48( ) (C L ) 3.48( /) Eq. (15-19): K R log(1 0.99) 1.0 C K 1.0 R R Bending inion: Eq. (15-3): Eq. (15-4): Eq. (15-3): ear: Eq. (15-4): Eq. (15-3): ear inion: Eq. (15-): Eq. (15-): (s a ) 44(300) psi (1.040) ( s w ) psi 1(1)(1) ( sw ) FK x J K K K K 1 d o v s m 15 91(0.71)(1)(0.41) 303. lbf 10(1)(1.41)(0.508)(1.5) 303.(68.3) 5.8 hp (s a ) psi (1.053) ( s w ) psi 1(1)(1) (0.71)(1)(0.01) 56.0 lbf 10(1)(1.41)(0.508)(1.5) 56.0(68.3) 4.9 hp ( C ) 1, I 0.078, C p 90 psi, Cxc C 0.15(0.71) s (s ac ) 341(300) psi 15 90(1.419)(1) c,all) psi 1(1)(1) Shigley s MED, 10 h ediion Chaper 15 Soluions, age 1/0

13 Eq. (15-1): ) Fd I c,all C p KoK KmCsCxc v (.000)(0.078) 90 1(1.41)(1.5)(0.56 5)() 36.4 lbf ear: Raing: 36.4(68.3) hp ( s ac) psi 15 90(1.479)(1) c,all) psi 1(1)(1) (.000)(0.078) 90 1(1.41)(1.5)(0.56 5)() 393.7(68.3) hp min(5.8, 4.9, 6.9, 7.5) 4.9 hp lbf inion wear conrols he power raing. The caalog raing of 5. hp is slighly higher han prediced here, bu seems reasonable. Ans From Ex. 15-1, he core hardness of boh he pinion and gear is 180 Brinell. So ( B ) 11 and ( B ) 1 are 180 Brinell and he bending sress numbers are: ( sa) 44(180) psi ( s ) psi a The conac srengh of he gear case, based upon he equaion derived in rob. 15-7, is C p S ( sa) ( KL) KxJ KTCsCxc ( sac) ( C ) C S N IK L F s Subsiuing (s a ) from above and he values of he remaining erms from Ex. 15-1, Shigley s MED, 10 h ediion Chaper 15 Soluions, age 13/0

14 (1)(1)(0.16)(1)(0.575)() ( sac) 1.3(1) 1.5 5(0.065)(0.59) psi ( B) 66 Brinell 341 The pinion conac srengh is found using he relaion from rob. 15-7: ( sac) ( sac) m C (1) (1) psi ( B) 1 66 Brinell 341 Realizaion of hardnesses Core Case inion ear The response of sudens o his par of he quesion would be a funcion of he exen o which hea-reamen procedures were covered in heir maerials and manufacuring prerequisies, and how quaniaive i was. The mos imporan hing is o have he suden hink abou i. The insrucor can commen in class when sudens curiosiy is heighened. Opions ha will surface may include: (a) Selec a hrough-hardening seel which will mee or exceed core hardness in he ho-rolled condiion, hen hea-reaing o gain he addiional 86 poins of Brinell hardness by bah-quenching, hen empering, hen generaing he eeh in he blank. (b) Flame or inducion hardening are possibiliies. (c) The hardness goal for he case is sufficienly modes ha carburizing and case hardening may be oo cosly. In his case he maerial selecion will be differen. (d)the iniial sep in a niriding process brings he core hardness o Rockwell C-scale (abou Brinell), which is oo much Compuer programs will vary A design program would ask he user o make he a priori decisions, as indicaed in Sec. 15-5, p. 798, of he ex. The decision se can be organized as follows: Shigley s MED, 10 h ediion Chaper 15 Soluions, age 14/0

15 A priori decisions: Funcion:, K o, rpm, m, emp., N L, R Design facor: n d (S F n d, S nd ) Tooh sysem: Involue, Sraigh Teeh, Crowning, φ n Sraddling: K mb Tooh coun: N (N m N ) Design decisions: ich and Face: d, F Qualiy number: Qv inion hardness: ( B ) 1, ( B ) 3 ear hardness: ( B ), ( B ) 4 Firs, gaher all of he equaions one needs, hen arrange hem before coding. Find he required hardnesses, express he consequences of he chosen hardnesses, and allow for revisions as appropriae. Shigley s MED, 10 h ediion Chaper 15 Soluions, age 15/0

16 Load-induced sress (Allowable sress) Tabulaed srengh Associaed hardness Chosen hardness New abulaed srengh Facor of safey s inion Bending ear Bending inion ear ear ear KoK KmKs K v okv KmKs s11 s s FK xj 1 FK xj ( s ) a s S K K s S K K 1/ KoK CsC v xc σ c C p s FdI s S K C 1 s s 1 s S K C 11 F T R 1 F T R 1 T R ( sa) ( s ) ( K L) ( KL) T R ac ( sac) ( CL) ( C ) ( CL) ( C ) ( sa) Bhn Bhn ( sa) ( s ) a1 n ( sa) Bhn ( sa) ( sac) Bhn ( sac) ( sac) ( sac) ( B ) 11 ( B ) 1 ( B ) 1 ( B ) 44( B) ( B) σ ( s ) ( K ) ( s ) a1 44( B) ( B) ( s ) ( K ) all a1 L a1 L 11 n1 σ s11k T KR s1kt KR ( s ) ac1 n 341( B) ( B) ( s ) ( C ) ( C ) ac1 L 1 s1k TCR ( s ) ac1 n 341( B) ( B) ( s ) ( C ) ( C ) ac1 L sktcr Noe: S n, S S F d F Shigley s MED, 10 h ediion Chaper 15 Soluions, age 16/0

17 15-14 N 1, N 56, 8 eeh/in, d 1.5 in, o 1hp, φ n 0, a 70 F, K a 1.5, n d 1, F e in, A 850 in (a) Eq. (15-39): Eq. (15-40): Eq. (15-41): Eq. (15-4): Eq. (15-43): Eq. (15-44): Eq. (15-45): Eq. (15-46): m N /N 56, d N / 56/8 7.0 in p x π / in, C in a p x / π / π 0.15 in b p x in h p x in d o (0.15) 1.75 in d r 3 (0.1446).711 in D 7 + (0.15) 7.5 in D r 7 (0.1446) in c in Eq. (15-47): ( )( ) Eq. (13-7): Eq. (13-8): Eq. (15-6): ( F ) in max V π(1.5)(175/1) f/min π(7)(175/ 56) V f/min 1 L p N in x o λ an π (1.5) 8 n 8.08 cos λ cos π pn in n π(1.5)(175) V s f/min 1cos (b) Eq. (15-38): f exp 0.110(679.8) Eq. (15-54): cosφn f an λ cos an e cosφ + f co λ cos co n Ans. Shigley s MED, 10 h ediion Chaper 15 Soluions, age 17/0

18 Eq. (15-58): Eq. (15-57): n K (1)(1)(1.5) 966 lbf Ans. V e 56.45(0.7563) d o a n n cosφ sin λ + f cos λ cosφ cos λ f sin λ cos 0 sin cos cos 0 cos sin lbf Ans. (c) Eq. (15-33): C s log Eq. (15-36): C + + m (56) Eq. (15-37): C v 0.659exp[ (679.8)] 0.31 Eq. (15-38): ( ) all 787(7) 0.8 ()(0.767)(0.31) 1787 lbf Since < ( ), he mesh will survive a leas h. all Eq. (15-61): Eq. (15-63): 0.05(966) f 9.5 lbf 0.05sin cos 0 cos (679.8) f hp 106.4(677.4).18 hp 966(56.45) 1.65 hp The mesh is sufficien Ans. λ p n n o / cos 8 / cos π / in 966 σ psi (0.5)(0.15) The sress is high. A he raed horsepower, 1 σ psi accepable 1.65 Shigley s MED, 10 h ediion Chaper 15 Soluions, age 18/0

19 (d) Eq. (15-5): A min 43.(8.5) in < 1700 in Eq. (15-49): loss ( )(.18) f lbf/min Assuming a fan exiss on he worm shaf, 175 o Eq. (15-50): h CR f lbf/(min in F) o Eq. (15-51): s F Ans (1700) Shigley s MED, 10 h ediion Chaper 15 Soluions, age 19/0

20 15-15 o 15- roblem saemen values of 5 hp, 115 rev/min, m 10, K a 1.5, n d 1.1, φ n 0, a 70 F are no referenced in he able. The firs four parameers lised in he able were seleced as design decisions p x d F A FAN FAN f N N K C s C m Cv V f e ( ) n C-o-C s L λ σ d Shigley s MED, 10 h ediion Chaper 15 Soluions, age 0/0