Chapter / rev/min Ans. C / in. C mm Ans teeth Ans. C / mm Ans.

Transcription

1 Chaper dp 17 / 8 15 in N 110 dg dp in N3 544 NG PdG eeh C / 35 in Ans Ans ng / rev/min Ans p m 3 mm Ans C mm Ans N G eeh Ans d N m mm Ans G G d N m mm Ans P P C / 40 mm Ans 13-4 Mesh: a1/ P1/ in Ans b15 / P15 / in Ans cba00834 in Ans p / P / in Ans p/ 1047 / 053 in Ans Pinion Base-Circle: d N / P1/3 7 in 1 1 d1 b 7 cos in Ans Gear Base-Circle: d N / P 8 / in db 9333cos in Ans Base pich: p p cos / 3 cos in Ans b c Conac Raio: mc Lab / pb 153 / Ans See he following figure for a drawing of he gears and he arc lenghs Chaper 13, Page 1/35

2 13-5 (a) 14 / 6 3 / 6 A in Ans (b) 1 an 14 / Ans 1 an 3 / Ans 1/ (c) d 14 / in d P G Ans 3 / in Ans Chaper 13, Page /35

3 (d) From Table 13-3, 03A 0 = 03(910) = 0873 in and 10/P = 10/6 = < 167 F 0873 in Ans 13-6 (a) pn / P n / in p pn / cos / cos in p p / an / an in x (b) Eq (13-7): p p cos 07854cos in Ans nb n n (c) p Pncos 4cos eeh/in 1 an an / cos an 1 n (an 5 / cos 30 ) 83 Ans (d) Table 13-4: a1/ in Ans b15 / in Ans 0 dp 4cos in Ans 36 dg 4cos in Ans 13-7 N 19 eeh, N 57 eeh, 0, m 5 mm P G n n (a) p m mm Ans n n pn 7854 p cos cos mm Ans p 9069 px an an mm Ans (b) mn 5 m cos cos mm Ans Chaper 13, Page 3/35

4 1 an 0 an 80 Ans cos30 (c) a m 5 mm Ans n b15m mm Ans d P dg n N Nm =5485 mm Ans P mm Ans 13-8 (a) Using Eq (13-11) wih k = 1, = 0º, and m =, k NP m m 1msin 1 m sin 1 1 sin eeh 1 sin 0 Round up for he minimum ineger number of eeh N P = 15 eeh Ans (b) (c) (d) Repeaing (a) wih m = 3, N P = 1498 eeh Rounding up, N P = 15 eeh Ans Repeaing (a) wih m = 4, N P = 1544 eeh Rounding up, N P = 16 eeh Ans Repeaing (a) wih m = 5, N P = 1574 eeh Rounding up, N P = 16 eeh Ans Alernaively, a useful able can be generaed o deermine he larges gear ha can mesh wih a specified pinion, and hus also he maximum gear raio wih a specified pinion The Max N G column was generaed using Eq (13-1) wih k = 1, = 0º, and rounding up o he nex ineger Min N P Max N G Max m = Max N G / Min N P unlimied unlimied Wih his able, we can readily see ha gear raios up o 3 can be obained wih a minimum N P of 15 eeh, and gear raios up o 631 can be obained wih a minimum N P of 16 eeh This is consisen wih he resuls previously obained Chaper 13, Page 4/35

5 13-9 Repeaing he process shown in he soluion o Prob 13-8, excep wih = 5º, we obain he following resuls (a) For m =, N P = 943 eeh Rounding up, N P = 10 eeh Ans (b) For m = 3, N P = 99 eeh Rounding up, N P = 10 eeh Ans (c) For m = 4, N P = 100 eeh Rounding up, N P = 11 eeh Ans (d) For m = 5, N P = 1038 eeh Rounding up, N P = 11 eeh Ans For convenien reference, we will also generae he able from Eq (13-1) for = 5º Min N P Max N G Max m = Max N G / Min N P unlimied unlimied (a) The smalles pinion ooh coun ha will run wih iself is found from Eq (13-10) N P k 1 13sin 3sin 1 3sin sin eeh Ans (b) The smalles pinion ha will mesh wih a gear raio of m G = 5, from Eq (13-11) is k N P m m 1 msin 1 m sin sin eeh Ans sin 0 The larges gear-ooh coun possible o mesh wih his pinion, from Eq (13-1) is N G N sin 4k 4k sin P NP 15 sin sin eeh Ans Chaper 13, Page 5/35

6 (c) The smalles pinion ha will mesh wih a rack, from Eq (13-13), N P k 1 sin sin eeh Ans , 30 n 1 From Eq (13-19), an an 0 / cos30 80 (a) The smalles pinion ooh coun ha will run wih iself, from Eq (13-1) is N P k cos 1 13sin 3sin 1cos sin 80 3sin eeh Ans (b) The smalles pinion ha will mesh wih a gear raio of m = 5, from Eq (13-) is 1cos30 N P sin sin eeh Ans The larges gear-ooh coun possible o mesh wih his pinion, from Eq (13-3) is N G N sin 4k cos 4 cos sin P k NP 10 sin cos cos 30 0 sin eeh Ans (c) The smalles pinion ha will mesh wih a rack, from Eq (13-4) is N P k cos 1cos30 sin sin eeh Ans Chaper 13, Page 6/35

7 1ann 1an From Eq (13-19), an an 796 cos cos 30 Program Eq (13-3) on a compuer using a spreadshee or code, and incremen N P The firs value of N P ha can be doubled is N P = 10 eeh, where N G 601 eeh So N G = 0 eeh will work Higher ooh couns will work also, for example 11:, 1:4, ec Use N P = 10 eeh, N G = 0 eeh Ans Noe ha he given diameral pich (ooh size) is no relevan o he inerference problem 1ann 1an From Eq (13-19), an an 736 cos cos 45 Program Eq (13-3) on a compuer using a spreadshee or code, and incremen N P The firs value of N P ha can be doubled is N P = 6 eeh, where N G 176 eeh So N G = 1 eeh will work Higher ooh couns will work also, for example 7:14, 8:16, ec Use N P = 6 eeh, N G = 1 eeh Ans The smalles pinion ha will operae wih a rack wihou inerference is given by Eq (13-13) k NP sin Seing k = 1 for full deph eeh, N P = 9 eeh, and solving for, 1 k 1 1 sin sin 816 Ans NP (a) Eq (13-3): p m 3 mm Ans n n Eq (13-16): p p / cos 3 / cos mm Ans n Eq (13-17): p p / an 1040 / an 5 30 mm Ans x (b) Eq (13-3): m p / 1040 / 3310 mm Ans Chaper 13, Page 7/35

8 Eq (13-19): an an 0 cos cos 5 1 n 1 an an 188 Ans (c) Eq (13-): d p = m N p = 3310 (18) = 5958 mm Ans Eq (13-): d G = m N G = 3310 (3) = 1059 mm Ans (a) Skeches of he figures are shown o deermine he axial forces by inspecion The axial force of gear on shaf a is in he negaive z-direcion The axial force of gear 3 on shaf b is in he posiive z-direcion Ans The axial force of gear 4 on shaf b is in he posiive z-direcion The axial force of gear 5 on shaf c is in he negaive z-direcion Ans 1 16 nc n rev/min ccw Ans dp 1 / 1cos in (b) (c) dg3 48 / 1cos in in Ans d P4 16 / 8cos 5 07 in Cab dg5 36 / 8cos in Cbc 3586 in Ans e nd rev/min cw Ans e n rev/min cw Ans 304 Chaper 13, Page 8/35

9 1 1 c rev/min cw abou x Ans 40 1 d 1 / 8cos in (a) n (b) P G d 40 / 8cos in d P d G 3531 in Ans 3 (c) d 8 in a he large end of he eeh Ans Applying Eq (13-30), e = (N / N 3 ) (N 4 / N 5 ) = 45 For an exac raio, we will choose o facor he rain value ino inegers, such ha N / N 3 = 9 (1) N 4 / N 5 = 5 () Assuming a consan diameral pich in boh sages, he geomery condiion o saisfy he in-line requiremen of he compound revered configuraion is N + N 3 = N 4 + N 5 (3) Wih hree equaions and four unknowns, one free choice is available I is necessary ha all of he unknowns be inegers We will use a normalized approach o find he minimum free choice o guaranee inegers; ha is, se he smalles gear of he larges sage o uniy, hus N 3 = 1 From (1), N = 9 From (3), N + N 3 = = 10 = N 4 + N 5 Subsiuing N 4 = 5 N 5 from () gives 10 = 5 N 5 + N 5 = 6 N 5 N 5 = 10 / 6 = 5 / 3 To eliminae his fracion, we need o muliply he original free choice by a muliple of 3 In addiion, he smalles gear needs o have sufficien eeh o avoid inerference From Eq (13-11) wih k = 1, = 0, and m = 9, he minimum number of eeh on he pinion o avoid inerference is 17 Therefore, he smalles muliple of 3 greaer han 17 is 18 Seing N 3 = 18 and repeaing he soluion of equaions (1), (), and (3) yields N = 16 eeh N 3 = 18 eeh N 4 = 150 eeh N 5 = 30 eeh Ans Chaper 13, Page 9/35

10 13-1 The soluion o Prob 13-0 applies up o he poin of deermining he minimum number of eeh o avoid inerference From Eq (13-11), wih k = 1, = 5, and m = 9, he minimum number of eeh on he pinion o avoid inerference is 11 Therefore, he smalles muliple of 3 greaer han 11 is 1 Seing N 3 = 1 and repeaing he soluion of equaions (1), (), and (3) yields N = 108 eeh N 3 = 1 eeh N 4 = 100 eeh N 5 = 0 eeh Ans 13- Applying Eq (13-30), e = (N / N 3 ) (N 4 / N 5 ) = 30 For an exac raio, we will choose o facor he rain value ino inegers, such ha N / N 3 = 6 (1) N 4 / N 5 = 5 () Assuming a consan diameral pich in boh sages, he geomery condiion o saisfy he in-line requiremen of he compound revered configuraion is N + N 3 = N 4 + N 5 (3) Wih hree equaions and four unknowns, one free choice is available I is necessary ha all of he unknowns be inegers We will use a normalized approach o find he minimum free choice o guaranee inegers; ha is, se he smalles gear of he larges sage o uniy, hus N 3 = 1 From (1), N = 6 From (3), N + N 3 = = 7 = N 4 + N 5 Subsiuing N 4 = 5 N 5 from () gives 7 = 5 N 5 + N 5 = 6 N 5 N 5 = 7 / 6 To eliminae his fracion, we need o muliply he original free choice by a muliple of 6 In addiion, he smalles gear needs o have sufficien eeh o avoid inerference From Eq (13-11) wih k = 1, = 0, and m = 6, he minimum number of eeh on he pinion o avoid inerference is 16 Therefore, he smalles muliple of 3 greaer han 16 is 18 Seing N 3 = 18 and repeaing he soluion of equaions (1), (), and (3) yields N = 108 eeh N 3 = 18 eeh N 4 = 105 eeh N 5 = 1 eeh Ans Chaper 13, Page 10/35

11 13-3 Applying Eq (13-30), e = (N / N 3 ) (N 4 / N 5 ) = 45 For an approximae raio, we will choose o facor he rain value ino wo equal sages, such ha N / N N / N If we choose idenical pinions such ha inerference is avoided, boh sages will be idenical and he in-line geomery condiion will auomaically be saisfied From Eq (13-11) wih k = 1, = 0, and m 45, he minimum number of eeh on he pinions o avoid inerference is 17 Seing N 3 = N 5 = 17, we ge N N eeh Rounding o he neares ineger, we obain N = N 4 = 114 eeh N 3 = N 5 = 17 eeh Ans Checking, he overall rain value is e = (114 / 17) (114 / 17) = H = 5 hp, i = 500 rev/min Le ω o = 300 rev/min for minimal gear raio o minimize gear size o i o 1 N N 8333 N N i N N4 1 1 Le N N From Eq (13-11) wih k = 1, = 0, and m = 887, he minimum number of eeh on he pinions o avoid inerference is 15 Le N = N 4 = 15 eeh N 3 = N 5 = 887(15) = 4331 eeh Try N 3 = N 5 = 43 eeh o Too big Try N 3 = N 5 = 44 Chaper 13, Page 11/35

12 o rev/min N = N 4 = 15 eeh, N 3 = N 5 = 44 eeh Ans 13-5 (a) The plane gears ac as keys and he wheel speeds are he same as ha of he ring gear Thus, na n / rev/min Ans (b) nf n5 0, nl n6, e1 n n6 300 n6 600 rev/min Ans (c) The wheel spins freely on icy surfaces, leaving no racion for he oher wheel The car is salled Ans 13-6 (a) The moive power is divided equally among four wheels insead of wo (b) Locking he cener differenial causes 50 percen of he power o be applied o he rear wheels and 50 percen o he fron wheels If one of he rear wheels ress on a slippery surface such as ice, he oher rear wheel has no racion Bu he fron wheels sill provide racion, and so you have wo-wheel drive However, if he rear differenial is locked, you have 3-wheel drive because he rear-wheel power is now disribued Le gear be firs, hen n F = n = 0 Le gear 6 be las, hen n L = n 6 = 1 rev/min n e n n 16 L F n n 0 A 1 na 51 1 na 1749 rev/min (negaive indicaes cw) Ans 35 / Le gear be firs, hen n F = n = 0 rev/min Le gear 6 be las, hen n L = n 6 = 85 rev/min A A Chaper 13, Page 1/35

13 nl na e nf na 16 0nA 85 na na na n A na 139 rev/min The posiive sign indicaes he same direcion as n 6 na 139 rev/min ccw Ans 13-9 The geomery condiion is d5 / d /d3 d4 Since all he gears are meshed, hey will all have he same diameral pich Applying d = N / P, N /( P) N /( P) N / PN 4 / P N N N N eeh Ans Le gear be firs, n F = n = 30 rev/min Le gear 5 be las, n L = n 5 = 0 rev/min n e n na 0 na 3 n rev/min A 14 The negaive sign indicaes opposie of n na 6857 rev/min cw Ans Le n F = n, hen n L = n 7 = 0 L F n n A A n e n L F n n n5 10 n Chaper 13, Page 13/35

14 n n n n n n n5 n b 348 urns in same direcion (a) n /60 H T Tn/ 60 ( T in N m, H in W) So 3 60 H T 10 n 9550 H / n ( H in kw, n in rev/min) Ta 398 N m 1800 mn 517 r 45 mm So F3 Ta kn r 45 F3b Fb kn in he posiive x-direcion Ans (b) mn r4 175 mm Tc N m ccw T4 c 1193 N m cw Ans Noe: The soluion is independen of he pressure angle Chaper 13, Page 14/35

15 13-3 N N d P 6 d 4 in, d 4 in, d 6 in, d 4 in e 1/ nf n 1000 rev/min n n L 6 0 nl na 0nA 1 e n n 1000 n 6 F A A n 00 rev/min A Noing ha power equals orque imes angular velociy, he inpu orque is T H 5 hp 550 lbf f/s 60 s 1 rev 1 in 1576 lbf in n 1000 rev/min hp min rad f For 100 percen gear efficiency, he oupu power equals he inpu power, so T arm H 5 hp 550 lbf f/s 60 s 1 rev 1 in 7878 lbf in na 00 rev/min hp min rad f Nex, we ll confirm he oupu orque as we work hrough he force analysis and complee he free body diagrams Gear 1576 W 788 lbf r F3 788 an 0 87 lbf Gear 4 F 4 W lbf A Chaper 13, Page 15/35

16 Gear 5 Arm Tou lbf in Ans Given: m = 1 mm, n P = 1800 rev/min cw, N = 18T, N 3 = 3T, N 4 = 18T, N 5 = 48T Pich Diameers: d = 18(1) = 16 mm, d 3 = 3(1) = 384 mm, d 4 = 18(1) = 16 mm, d 5 = 48(1) = 576 mm Gear From Eq (13-36), 60000H W 7368 kn dn d 16 Ta W N m r W 7368 an 0 68 kn Gears 3 and W 7368 W 1310 kn W r 1310 an kn Ans Chaper 13, Page 16/35

17 13-34 Given: P = 5 eeh/in, N = 18T, N 3 = 45T, 0, H = 3 hp, n = 1800 rev/min Gear n Tin 110 lbf in dp 3600 in 5 45 dg 9000 in W3 6 lbf 36 / r W3 6 an 0 6 lbf r r F W 6 lbf, F W 6 lbf a 3 a 3 1/ Fa lbf Each bearing on shaf a has he same radial load of R A = R B = 66/ = 331 lbf Gear 3 W3 W3 6 lbf r r W3 W3 6 lbf Fb3 Fb 66 lbf R R 66 / 331 lbf C D Each bearing on shaf b has he same radial load which is equal o he radial load of bearings A and B Thus, all four bearings have he same radial load of 331 lbf Ans Given: P = 4 eeh/in, 0, NP = 0T, n n = 900 rev/min N P 0 d 5000 in P Tin 40 lbf in 900 W T / d / 40 / 5 / 1681 lbf 3 in r W an 0 61 lbf Chaper 13, Page 17/35

18 The moor moun resiss he equivalen forces and orque The radial force due o orque is r 40 F 150 lbf 14 Forces reverse wih roaional sense as orque reverses The compressive loads a A and D are absorbed by he base plae, no he bols For he ensions in C and D are F M F 1109 lbf AB W, 3 Chaper 13, Page 18/35

19 If W3 reverses, 155 in changes o 135 in, 4815 in changes o 875 in, and he forces change direcion For A and B, r For W, 3 F F 184 lbf 1 1 M / 646 lbf in a 14 / 115 / 898 in 646 F 179 lbf 4898 A C and D, he shear forces are: FS / / 898 A A and B, he shear forces are: FS / / lbf The shear forces are independen of he roaional sense The bol ensions and he shear forces for cw roaion are, Chaper 13, Page 19/35

20 For ccw roaion, (a) N = N 4 = 15 eeh, N 3 = N 5 = 44 eeh N N P d d P 15 d d4 5 in 6 Ans 44 d3 d5 733 in 6 Ans (b) (c) 5500 dn Vi V V f/min Ans 1 1 dn / Vo V4 V5 558 f/min Ans 1 1 Inpu gears: H Wi lbf 504 lbf Ans V 1636 i Wri Wi an 5043 an lbf Ans Wi 5043 Wi 537 lbf Ans cos cos 0 Oupu gears: H Wo lbf Ans V 558 o Wro Wo an 1478 an lbf Ans Wo 1478 Wo 1573 lbf Ans cos 0 cos 0 d 5 (d) Ti Wi lbf in Ans Chaper 13, Page 0/35

21 44 44 (e) To Ti lbf in Ans H 35 hp, n 100 rev/min, =0 i N N 16 eeh, N N 48 eeh, P 10 eeh/in (a) n n n n n inermediae 3 4 N 3 N 16 i rev/min Ans N 48 N rev/min 4 o ni N3 N Ans (b) (c) N N P d d P 16 d d4 16 in Ans d3 d5 48 in Ans 10 dn Vi V V3 507 f/min Ans 1 1 dn Vo V4 V f/min Ans 1 1 W i H lbf lbf Ans V 507 i Wri Wi an 98 an lbf Ans Wi 98 Wi 445 lbf Ans cos cos 0 W o H lbf Ans V 1676 o Wro Wo an 6891an lbf Ans Wo 6891 Wo 7333 lbf Ans cos 0 cos 0 d 16 (d) Ti Wi lbf in Ans (e) To Ti lbf in Ans Chaper 13, Page 1/35

22 o (a) For, from Eq (13-11), wih m =, k = 1, 0 1 i 1 NP 1 sin 0 So N 15 Ans min P 1 sin (b) N 15 P 1875 eeh/in Ans d 8 (c) To ransmi he same power wih no change in pich diameers, he speed and ransmied force mus remain he same For A, wih = 0, W A = F A cos0 = 300 cos0 = 819 lbf For A, wih = 5, same ransmied load, F A = W A /cos5 = 819/cos5 = 3110 lbf Ans Summing he orque abou he shaf axis, da db WA WB d / A d A WB WA WA lbf db / db 8 WB FB 7776 lbf Ans cos 5 cos 5 o (a) For, from Eq (13-11), wih m = 5, k = 1, 0 1 i 1 NP 1 5 sin 5 So N 11 Ans min P sin (b) d 300 m 73 mm/ooh Ans N 11 (d) To ransmi he same power wih no change in pich diameers, he speed and ransmied force mus remain he same Chaper 13, Page /35

23 For A, wih = 0, W A = F A cos0 = 11 cos0 = 1033 kn For A, wih = 5, same ransmied load, F A = W A /cos5 = 1033 / cos 5 = 1140 kn Ans Summing he orque abou he shaf axis, da db WA WB d / A d A WB WA WA 80 kn db / db 300 WB 80 FB 516 kn Ans cos 5 cos (a) Using Eq (13-11) wih k = 1, = 0º, and m =, k NP m m 1msin 1 m sin 1 1 sin eeh 1 sin 0 Round up for he minimum ineger number of eeh N F = 15 eeh, N C = 30 eeh Ans (b) d 15 m 833 mm/ooh Ans N 15 (c) T H kw 1000 W rev 60 s 100 N m 191 rev/min kw rad min (d) From Eq (13-36), W H kn 1600 N dn Ans Or, we could have obained W direcly from he orque and radius, Chaper 13, Page 3/35

24 W T N d / 015 / W W an 1600 an N Ans r W 1600 W 1700 N Ans cos cos (a) Using Eq (13-11) wih k = 1, = 0º, and m =, k NP m m 1msin 1 m sin 1 1 sin eeh 1 sin 0 Round up for he minimum ineger number of eeh N C = 15 eeh, N F = 30 eeh Ans (b) (c) N 30 P 3 eeh/in Ans d 10 H 1 hp 550 lbf f/s 1 in rev 60 s T 70 rev/min hp f rad min T 900 lbf in Ans (d) From Eqs (13-34) and (13-35), 1070 dn V 1833 f/min 1 1 H W lbf Ans V 1833 W W an 180 an lbf Ans r W 180 W cos cos 0 19 lbf Ans 13-4 (a) Eq (13-14): 1 N P 1d P 1130 an an an 185 NG dg 388 Ans (b) Eq (13-34): dn V 4084 f/min 1 1 Ans Chaper 13, Page 4/35

25 (c) Eq (13-35): Eq (13-38): Eq (13-38): H 10 W lbf V 4084 W W ancos 808 an 0 cos lbf r W W ansin 808 an 0 sin lbf a Ans Ans Ans The angenial and axial forces agree wih Prob 3-74, bu he radial force given in Prob 3-74 is shown here o be incorrec Ans T H n in / / lbf in W T / r 6565 / 383 lbf R AG = i + 517j, R AB = 5j M R W R F T0 Solving gives z x RAB FB 5FB i5fb k R W AG 1697 i 6566 j 4459 k 1 an / an 4 / cos 6565 a / 67 in r W 383an 0 cos lbf a W 383an 0 sin lbf W = 1069i 534j + 383k lbf 4 AG + AB B So So z x i FB 5FB T j k i k j 0 z FB 1697 / lbf T 6566 lbf in F 4459 / lbf x B FB lbf Ans F A = (F B + W) = ( 1788i 6788k i 534j + 383k) = 715i + 534j k 1/ Chaper 13, Page 5/35

26 A 1/ F (radial) lbf Ans FA (hrus) 534 lbf Ans d 18 /10 18 in, d3 30 /10 30 in 1d / 109 an an 3096 d3 / DE 05cos in 16 W 5 lbf W r 5an 0 cos lbf W a 5 an 0 sin lbf W = 4681i 7803j +5k R DG = 08197j + 15i R DC = 065j MD RDGWRDCFC T0 R W DG 049 i 315 j 5917 k z R F 065F i065f k x DC C C C z x i C 065 C j k F i F k Tj 0 T 315 lbf in Ans F 947i 38 k lbf Ans C FC 1/ lbf Ans F 0 F D 479 i 780 j 578 k lbf FD (radial) lbf Ans a FD (hrus) W 780 lbf Ans 1/ Chaper 13, Page 6/35

27 13-45 P Pncos 4cos eeh/in 1 ann 1 an 0 an an 80 cos cos dp 5196 in 3464 The forces on he shafs will be equal and opposie of he forces ransmied o he gears hrough he meshing eeh Pinion (Gear ) r W W an 800 an lbf a W W an 800 an lbf W 336i46j800 k lbf Ans W lbf Ans Gear 3 W 336i46j800 k lbf Ans W 983 lbf Ans 3 dg 938 in 3464 TG W r lbf in 1/ Chaper 13, Page 7/35

28 13-46 From Prob soluion, Noice ha he idler shaf reacion conains a couple ending o urn he shaf end-overend Also he idler eeh are ben boh ways Idlers are more severely loaded han oher gears, belying heir name Thus, be cauious Gear 3: P Pncos 7 cos eeh/in an 0 an 0403, 8 cos30 54 d in 606 W 500 lbf a W 500 an lbf r W 500 an 8 10 lbf W3 10i887j500 k lbf Ans Gear 4: 14 d4 309 in W lbf 309 a W 199 an lbf r W 199 an lbf W4 811i1114j199 k lbf Ans P 6cos eeh/in 4 d in Chaper 13, Page 8/35

29 16 d 3079 in T 916 lbf in 170 T 916 W 595 lbf r 3079 / a W 595 an lbf r W 595 an 8 50 lbf W 344i50j595 k lbf R 6 i, R 3i404j DC DG MD RDCFC RDGWT0 R W DG 404 i 1785 j 140 k z y R F 6F j 6F k DC C C C (1) Subsiuing and solving Eq (1) gives T404 i lbf in z FC 975 lbf 3657 lbf y FC FF F W 0 D C Subsiuing and solving gives x FC y FD z FD 344 lbf 1067 lbf 975 lbf FC 344i3567j975 k lbf Ans FD 1067j975 k lbf Ans Chaper 13, Page 9/35

30 13-49 Since he ransverse pressure angle is specified, we will assume he given module is also in erms of he ransverse orienaion d mn mm d3 mn mm d4 mn mm H 6 kw 1000 W rev 60 s T 3581 N m 1600 rev/min kw rad min T 3581 W 1119 N d / 0064 / Chaper 13, Page 30/35

31 r W W an 1119 an N a W W an 1119 an N F a 1119i4073j998 k N Ans F3 b i j 7117i7117 jn Ans F4 c 4073i1119j998 k N Ans d d N in, d 5196 in 3 Pn cos 8cos 30 8cos in, d in 5cos15 5cos For gears and 3: n 1 For gears 4 and 5: an an 0 / cos an an / cos an an 0 / cos30 8 F T / r 100 / 01/ 1188 lbf 5196 F lbf 3106 r F3 F3 an 1188 an lbf r F an lbf a F F an 1188 an lbf 3 3 a an15 53 lbf F Nex, designae he poins of acion on gears 4 and 3, respecively, as poins G and H, as shown Posiion vecors are Chaper 13, Page 31/35

32 R 1553 j 3 k CG R 598j65k CH R 85k CD Force vecors are F i748j53k F i500j686k F F if j x y C C C x y z FD FDiFDj FDk Now, a summaion of momens abou bearing C gives MC RCGF54 RCH F3 RCDFD 0 The erms for his equaion are found o be R F CG i 5961 j3086k R F CH i 77 j 3086 k y x R F 85F i85f j CD D D D When hese erms are placed back ino he momen equaion, he k erms, represening he shaf orque, cancel The i and j erms give F y D x FD Nex, we sum he forces o zero Subsiuing, gives lbf 85 Ans lbf 85 Ans FF F F FD 0 C 54 3 Solving gives x y F F Ci j1987i 746j53k 1188i499j 686k C z 1610i 45j F D k0 x FC lbf Ans y FC lbf Ans z FD lbf Ans Chaper 13, Page 3/35

33 13-51 V W W W dwn W m/s H N V W L p N mm x an W 1 L d an 4550 lead angle W WW W cosn sin f cos VW V S 315 m/s cos cos 4550 In f/min: V S = 38(315) = 1033 f/s = 60 f/min Use f = 0043 from curve A of Fig 13-4 Then, from he firs of Eq (13-43) 637 W 533 N cos145 sin cos 455 n y W W sin 533sin N z 533cos145 cos in 455 W s 5119 N The force acing agains he worm is W 637i1333j5119 k N Thus, A is he hrus bearing Ans R 005 j 010 k, R 00k AG AB M R WR F T0 A AG AB B R W AG 16 i 637 j 3185 k y x R F 0F i0f j AB B B B Subsiuing and solving gives T 3185 N m Ans x y F 3185 N, F 613 N B So F 3185i613 jn Ans B B Chaper 13, Page 33/35

34 Or 1/ FB N radial FFA WRB 0 FA WFB 637i1333j5119k3185i613j 3185i1946j5119 k Ans Radial F r 3185 i 1946 j N A 1/ r F A N a Thrus FA 5119 N 13-5 From Prob 13-51, W G 637i 1333 j 5119 k N p px 485 G So d N px G 38 mm Bearing D akes he hrus load MD RDGWG RDCFC T0 R DG 0075 i 0191 j R 01075i DC The posiion vecors are in meers RDG WG 9777i3711j 50k z y R F 01075F j 01075F k DC C C C Puing i ogeher and solving, T 9777 N m Ans FC 33j3450 k N, FC 3460 N Ans FFC WG FD 0 F F W i j k N D C G Ans Radial F r 1566 j 1669 k N Or D 1/ F N (oal radial) r D F D 637 i N (hrus) Chaper 13, Page 34/35

35 V W 357 f/min 1 x W W W 1050 lbf 357 p px 0397 in 8 L in an W 5153 lbf cos 0 sin cos946 y W 5153sin lbf z 5153 cos 0 cos946 W 005sin lbf So W 105i176j473 k lbf Ans T lbf in Ans Compuer programs will vary Chaper 13, Page 35/35