ME 452: Machine Design II Spring Semester Name of Student: Circle your Lecture Division Number: Lecture 1 Lecture 2 FINAL EXAM

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1 ME 45: Mchine Design II Spring Semester 01 Nme o Student: Circle your Lecture Division Number: Lecture 1 Lecture FINAL EXAM Thursdy, My 5th, 01 OEN BOOK AND CLOSED NOTES For ull credit you must show ll your work nd clcultions clerly nd logiclly on the sheets o pper ttched to the end o ech problem. lese use only the blnk pges or ll your work nd write on one side o the pper only. lese drw your plots nd igures (such s ree body digrms) netly nd clerly lbelled. At the end o the exm, plese stple ech problem seprtely. Stple this cover sheet nd your crib sheet in ront o roblem 1. roblem 1 roblem roblem 3 roblem 4 Totl 1

2 ME 45: Mchine Design II FINAL EXAM SRING 01 Nme o Student: Circle one: Lecture 1 Lecture roblem 1 (5 points). The spur pinion shown in Figure 1 hs dimetrl pitch o 8 teeth/in, ce width o 1.5 in, 3 teeth, nd 0 pressure ngle. The pinion rottes t 000 rpm nd trnsmits lod o 400 lbs to the meshing spur ger. The pinion nd ger re commercil qulity nd the lod is pplied t the highest point o single-tooth contct. oisson s rtio nd modulus o elsticity or the pinion nd the ger re 0.9, E 30Mpsi, G 0.11, nd EG 14.5 Mpsi, respectively. 8 For the pinion, the lie is 10 cycles, the AGMA bending strength nd wer strength re 30 kpsi nd 95 kpsi, respectively, the repetedly pplied bending strength stress cycle ctor Yn nd the pitting resistnce stress cycle ctor Z N Also, or the pinion, the AGMA geometry ctor J 0.38 nd the AGMA modiiction ctors re: KO 1, KV 1.414, KS 1, Km 1.414, KB 1, KT 1, nd KR 1 (i) Determine the horsepower trnsmitted by the gerset. (ii) Determine the AGMA bending stress nd the AGMA contct stress cting on the pinion. (iii) Determine the AGMA ctor o sety gurding ginst bending tigue ilure o the pinion. (iv) Determine the AGMA ctor o sety gurding ginst wer o the pinion. Figure 1. The spur ger nd pinion.

3 ME 45: Mchine Design II FINAL EXAM SRING 01 Nme o Student: Circle one: Lecture 1 Lecture roblem (5 points). The spring index o liner helicl compression spring is 8 nd the men coil dimeter o the spring is 4 inches. The spring mteril is ASTM No. A8 music wire. The spring hs squred nd ground ends nd is not set nd is not peened. The spring is subjected to n initil lod o 0.5 lbs nd the corresponding spring length is inches, see Figure. The spring is then subjected to mximum working lod o 405 lbs nd the corresponding spring length is 5 inches. Determine the ollowing (do not use n itertion procedure): (i) The totl number o coils in the spring. (ii) The lod required to shut the spring (tht is, the lod t the shut height o the spring). (iii) lot the lod cting on the spring ginst the length o the spring. On your plot, clerly show the vlues o: () the ree length, (b) the solid length, nd (c) the lod required to shut the spring. (iv) The sttic ctor o sety t the shut height o the spring. Is ilure predicted to hve occured? F = 0.5 lbs F = 405 lbs in 5 in F = 0.5 lbs F = 405 lbs Figure. Lods on the compression spring (Not drwn to scle). 3

4 ME 45: Mchine Design II FINAL EXAM SRING 01 Nme o Student: Circle one: Lecture 1 Lecture roblem 3 (5 points). lte A is bolted to lnge B by 3/4 in - 10 UNC-A, SAE grde 5, corse series, steel bolt, s shown in Figure 3. The stiness o the plte is 45 x 10 lbs / in, the stiness o the lnge is preloded to 5 x 10 lbs / in, nd the stiness o the conined gsket is 5 x 10 lbs / in. The bolt is 3 0 x 10 lbs nd then subjected to repeted tensile lod which hs mximum vlue o x 10 lbs. The ully corrected endurnce limit o the bolt is S e gurding ginst joint seprtion is speciied s 5. Determine the ollowing (do not use n itertion procedure): 17.5 kpsi. The ctor o sety (i) The mximum vlue o the externl lod tht is tken by the bolt. (ii) The totl stiness o the members (plte A nd lnge B). Neglect the stiness o the two wshers. (iii) The ctor o sety gurding ginst overloding the bolt. (iv) The ininite lie tigue ctor o sety or the bolt using the modiied Goodmn criterion. Assume tht the slope o the lod line is 1. lte A Flnge B A conined gsket Figure 3. A plte bolted to lnge. 4

5 ME 45: Mchine Design II FINAL EXAM SRING 01 Nme o Student: Circle one: Lecture 1 Lecture roblem 4 (5 points). An externl contrcting rim-type brke hs two symmetric shoes cting on the drum s shown in Figure 4. The brke lining mteril or both the let shoe nd the right shoe is woven cotton. The drum is rotting counterclockwise nd the rdius o the drum is inches. The ce width o ech shoe is 3 inches. The igure shows tht the limiting vlue o the ctuting orce F is horizontl nd the mgnitude is the sme on both the let shoe nd the right shoe. (i) Determine the moments o the norml orce nd the riction orce on the sel-energizing shoe. (ii) Determine the moments o the norml orce nd the riction orce on the sel-deenergizing shoe. (iii) Determine the brking cpcity o the let shoe nd the right shoe. 5 in 3 4 in. 3 4 in. 85 in QG A HB O 8 in 1 8 in. in r = in in Drum Rottion 81 in in. Let Shoe Drum Right Shoe F F Figure 4. The geometry o the rim-type drum brke. (Not drwn to scle). 5

6 Solution to roblem 1 (5 points). (i) oints. The horsepower trnsmitted by the gerset, see Exmple 14-1, pge 740, is t W V hp (1) 33,000 The pitch line velocity, see Eq. (13.34), pge 707, nd Exmple 14-1, see pge 740, cn be written s d n V () 1 The pitch circle dimeter o the pinion, see Eq. (13-1), pge 7, nd Exmple 13-1, pge 83, is d N 3 4 in (3) 8 Substituting Eq. (3) nd the pinion speed n 000 rpm into Eq. (), the pitch line velocity is Substituting Eqs. (4) nd the trnsmitted lod V t/min (4) 1 t W 400 lbs into Eq. (1), the horsepower is hp 5.39 hp (5) 33, 000 Alterntive Method: The horsepower trnsmitted by the gerset cn lso be written rom Eq. (13-33), see pge 70, s hp = T ω () The torque trnsmitted by the pinion cn be written rom Eq. (b), see pge 70, s T =(d /)W t (7) Substituting Eqs. (3) nd t W The ngulr velocity o the pinion is 400 lbs into Eq. (7), the torque trnsmitted by the pinion is T (4 / ) in-lbs (7b) n rd/s (8) 0 Substituting Eqs. (7b) nd (8b) into Eq. (), the horsepower trnsmitted by the spur gerset is hp 5.39 hp (9) 00

7 (ii) 10 oints. The AGMA bending stress or the pinion rom Eq. (14-15), see pge 74, is K K σ =W K K K (10) F J t m B o v s Substituting t W 400 lbs, KO 1, KV 1.414, KS 1, 1 8in, nd J 0.38 into Eq. (10), the AGMA bending stress or the pinion is F 1.5in, Km 1.414, B K 1, (8) ( 1.414) (1) σ = (400) (1) ( 1.414)(1) kpsi (10b) (1.5) (0.38) The AGMA contct stress or the pinion cn be written rom Eq. (14-1), see pge 74, s σ =C c K t m p W KKK o v s d F C I (11) The AGMA elstic coeicient cn be written rom Eq. (14-13), see pge 744, s C p Ep EG G (1) Substituting 0.9, E 30 Mpsi, G 0.11, nd EG 14.5 Mpsi into Eq. (1), the elstic coeicient is 1 1/ C p psi (1b) The surce condition ctor, rom Fig , see pge 77, nd Exmple 14-4, see pge 770, is C 1 (13) The geometry ctor or externl gers cn be written rom Eq. (14-3), see pge 755, s cos sin mg I m m 1 N G (14) Note tht Eq. (14) includes the lod shring rtio, however, Eqs. (b) nd (c) on pge 755 do not. The current prctice or spur gers is to ssume tht the lod shring rtio, see pge 753, is The ger rtio cn be written rom Eq. (14-), see pge 754, s m 1 (15) m N d N G G G d N (1) 7

8 The rdius o the pitch circle o the ger, see Exmple 13-1, pge 83, is Thereore, the pitch circle dimeter o the ger is r Cr in (17) G d G r x3.5 7 in (17b) G Substituting Eqs. (3) nd (17b) into Eq. (1), the ger rtio is m G 1.75 (18) 4 Substituting Eqs. (15), (18), nd the pressure ngle 0 o into Eq. (14), the geometry ctor is I o o cos 0 sin 0 (1.75) 0.10 x (19) Substituting Eqs. (1b), (18), nd (19) into Eq. (11), the AGMA contct stress or the pinion is tht is σ =1817. c c (1. 414) (1) (400)(1)(1.414)(1) psi (0) ( 4)( 1. 5) (0.10) σ psi = kpsi (0b) (iii) 4 oints. The ctor o sety gurding ginst bending tigue ilure o the pinion cn be written rom Eq. (14-41), pge 75, s St ( YN / KTKR) SF (1) Substituting St 30 kpsi, YN 0.978, KT 1, KR 1, nd Eq. (10b) into Eq. (1), the ctor o sety gurding ginst bending tigue ilure o the pinion is 30x SF.18 (1b) (iv) 5 oints. The ctor o sety gurding ginst pitting ilure o the pinion, see Eq. (14-4), pge 75, cn be written s S Z /( ) c N KTK R SH () c Substituting Sc 95 kpsi, ZN , KT 1, KR 1, nd Eq. (0b) into Eq. (), the ctor o sety gurding ginst pitting ilure o the pinion is 8

9 S H 95 x (b) 9

10 Solution to roblem (5 points). (i) points. The totl number o coils or spring with squred nd ground ends, see Tble 10-1, see pge 51, is Nt N (1) The number o ctive coils or the spring cn be written rom Eq. (10-9), see pge 50, is The spring stiness cn be written s d4 G N () 3 8 D k F k y (3) Substituting the given dt into Eq. (3), the spring stiness is k lbs/in (3b) The spring index cn be written rom Eq. (10-1), see pge 519, is Rerrnging this eqution, the wire dimeter cn be written s D C (4) d D 4 d 0.5 in (4b) C 8 The modulus o elsticity nd the modulus o rigidity (or ASTM No. A8 music wire nd wire dimeter d 0.15 inches ) rom Tble 10-5, see pge 5, re E 8.0 Mpsi nd G 11. Mpsi (5) Substituting Eqs. (4b) nd (5) into Eq. (), the number o ctive coils or the spring is 0.54 x 11. x 10 3 Note tht the recommended rnge, see pge 533, is totl number o coils in the spring is N.99 7 () 8 x 4 x N 15. Substituting Eq. () into Eq. (1), the Nt (7) In generl, the totl number o coils is speciied to the nerest 1/4 o turn, see pge 533, i.e., (ii) points. The lod required to shut the spring cn be written s Nt 9 (7b) F k ( L L ) (8) shut s

11 The ree length o the spring cn be written s L L y where the ssembled length o the spring is speciied s nd the initil delection is initil (9) L in (10) F min 0.5 yinitil 1in (10b) k 0.5 Substituting Eqs. (10) nd (10b) into Eq. (9), the ree length o the spring is L in+1in=7in (11) The shut height (or the solid length) o the spring with squred nd ground ends, see Tble 10-1, pge 51, cn be written s Ls d Nt (1) Substituting Eqs. (4b) nd (7) into Eq. (1), the shut height o the spring is Ls 0.5 x ins (1b) Substituting Eqs. (3b), (11), nd (1b) into Eq. (8), the lod required to shut the spring is F 0.5 (7 4.5) 50.5 lbs (13) shut (iii) oints. A plot o the lod F ginst the length L o the spring is shown in the igure below. Lod, F y clsh F shut = 50.5 lbs F work = 405 lbs y work F min = 0.5 lbs y initil = 1 ins 0 L s = 4.5 ins L m = 5 ins L = ins L = 7 ins Length, L The plot shows: (i) the ree length, (ii) the ssembled length, (iii) the minimum working length, (iv) the solid length, (v) the initil delection, (vi) the working delection, nd (vii) the clsh delection. The plot lso shows the lod required to shut the spring. 11

12 (iv) 7 points. The sttic ctor o sety t the shut height o the spring cn be written rom Eq. (), see pge 530, s Ssy Ns (14) The ultimte tensile strength cn be written rom Eq. (10-14), see pge 53, s Sut Ad m (15) The coeicient nd the exponent rom Tble 10-4, see pge 54, respectively, re A = 01 kpsi.in m nd m = (1) Substituting Eqs. (4b) nd (1) into Eq. (15), the ultimte tensile strength o the spring mteril is Sut kpsi (17) Note tht the dimeter o the wire must be expressed in inches in this eqution. The torsionl yield strength or not set spring cn be written rom Tble 10-, see pge 5, s S sy 0.45 S (18) Substituting Eq. (17) into Eq. (18), the torsionl yield strength or the not set spring is ut S kpsi (18b) sy Since the spring is not set then the sher stress t the shut height cn be written rom Eq. (), see pge 530, s 8 F D K shut B (19) 3 d where the Bergstrsser ctor cn be written rom Eq. (10-5), see pge 519, is K B 4C C (0) Substituting Eqs. (4b), (13), nd (0) into Eq. (19), the sher stress t the shut height is kpsi (1) Substituting Eqs. (18b) nd (1) into Eq. (14), the ctor o sety t the shut height is Ns.1 () Since the ctor o sety t the shut height is greter thn 1 then ilure is predicted not to hve occurred. 1

13 Solution to roblem 3 (5 oints). (i) 7 oints. The mximum vlue o the externl lod tht is tken by the bolt cn be written rom Eq. (d), see pge 43, s C C (1) nd the mximum vlue o the externl lod tht is tken by the members is b mx (1 ) (1 ) m C mx C (1b) The mximum externl tensile lod cting on the bolted connection is speciied s = mx 4800 lbs = 4.8 kip (1c) The ctor o sety gurding ginst joint seprtion cn be written rom Eq. (8-30), see pge 441, s F i n0 (1 C) () Rerrnging this eqution, the joint constnt cn be written s The prelod on the bolt is speciied s C F 1 i (b) n0 F 0,000 lbs (3) i nd the ctor o sety gurding ginst joint seprtion is speciied s Substituting Eqs. (1b), (3), nd (3b), into Eq. (b), the joint constnt is n0 5 (3b) 0000 C Substituting Eqs. (1c) nd (4) into Eq. (1), the mximum vlue o the externl lod tht is tken by the bolt is b lbs (5) Substituting Eqs. (1c) nd (4) into Eq. (1b), the mximum vlue o the externl lod tht is tken by the members is m lbs (5b) Check: The resultnt lod cting on the members cn be written rom Eq. (8-5), see pge 43, s Fm m Fi (ii) 4 oints. The stiness o the members cn be written rom Eq. (8-18), see pge 47, s 0 (4) 13

14 1 1 1 () k k k m m m 1 The stiness o plte A nd the stiness o lnge B re speciied, respectively, s m1 45 x 10 lbs / in k nd k 5 x 10 lbs / in (b) Substituting Eqs. (b) into Eq. (), the stiness o the members is m k m x10 lbs / in (7) Wrong Solution. It is wrong to include the stiness o the conined gsket in the stiness o the members. I the conined gsket is included in Eq. () then the stiness o the members is written s In this cse, the stiness o the members would be (8) k k k k m 1 g k m x 10 lbs / in (8b) Eqution (7) gives the correct vlue or the stiness o the members. Eqution (8b) does not give the correct vlue or the stiness o the members (i.e., it is close to the stiness o the gsket). Aside: The stiness constnt o the joint cn be written rom Eq. (), see pge 43, s C k b k k b m (9) Substituting Eqs. (3) nd () into Eq. (9) gives 0.17 k b kb 1.07 x10 (9b) Rerrnging this eqution, the stiness o the bolt is k 3.1 x 10 lbs / in (10) b Compring Eqs. (7) nd (10), the rtio o the stiness o the members to the stiness o the bolt is km kb 1.07 x 10 =5 3.1 x 10 (10b) Notes: (i) The lod to cuse joint seprtion cn be written rom Eq. (e), see pge 441, s o n (11) o 14

15 Substituting Eqs. (1c) nd (3b) into Eq. (11), the lod to cuse joint seprtion is , 000 lbs (1) o (ii) The resultnt lod cting on the members cn be written rom Eq. (8-5), see pge 43, s Fm m Fi (13) The portion o the externl lod tken by the members, s determined in Eq. (5b), is (1 0.17) lbs (14) m Substituting Eqs. (3) nd (14) into Eq. (13), the resultnt lod cting on the members is F lbs (15) m Note tht the members re in compression. (The ctor o sety is greter thn 1). Aside: A plot o the resultnt lod cting on the members F m ginst the externl lod is shown in the igure below. When the externl lod 0 then the lod in the members is Fm Fi 0,000 lbs nd when the mximum externl tensile lod mx 4,800 lbs then the lod in the members is F 1,000 lbs. Also, rom Eq. (1) the lod required to cuse joint seprtion is 0 4,000 lbs. m F m (kip) 0 mx = 4.8 kip Joint Seprtion o = 4 kip (kip) 1-C = F mx = - 1 kip F i = - 0 kip Figure. lot o the resultnt lod cting on the members F m ginst the externl lod. (iii) 7 oints. The ctor o sety gurding ginst overloding (or the lod ctor), see Eq. (8-9), pge 440, cn be written s Sp At Fi nl (1) C The proo strength or 3/4 in SAE Grde No. 5 steel bolt, see Tble 8-9, pge 433, is S 85 kpsi (17) p 15

16 The tensile stress re o 3 / 4 in-10 UNC bolt, see Tble 8-, pge 413, is The prelod orce is speciied s A in (18) t Fi 0,000 lbs = 0 kip (19) Substituting Eqs. (1c), (4), (17), (18), nd (19), into Eq. (1), the ctor o sety gurding ginst overloding is ( ) 0 n L (0) (iv) 7 oints. The ininite lie tigue ctor o sety o the bolt using the modiied Goodmn criterion cn be written rom Eq. (8-48), see pge 447, s n Se ( Sut A t Fi ) C( S S ) ut e (1) Note tht this eqution is vlid or slope o the lod line r 1 (s speciied in the problem sttement). The tensile strength or 3/4 in SAE Grde No. 5 steel bolt rom Tble 8-9, see pge 433, is S 10 kpsi () ut The ully corrected endurnce strength or SAE Grde 5 steel bolt with rolled threds rom Tble 8-17, see pge 445, is S 18. kpsi (3) e Substituting Eqs. (4), (17), (18), (19), (), nd (3) into Eq. (1), the ininite lie tigue ctor o sety o the bolt using the modiied Goodmn criterion cn be written s n x 18. (10 x ) (10 18.) (4) Thereore, the ininite lie tigue ctor o sety o the bolt using the modiied Goodmn criterion is n.7 (4b) Alterntive Approch: The ininite lie tigue ctor o sety o the bolt using the modiied Goodmn criterion cn lso be written rom Eq. (8-37), see pge 44, s n S (5) The lternting component o the strength using the modiied Goodmn criterion cn be written rom Eq. (c), see pge 445, s 1

17 S S ut S ( S ) e ut i S ( ) e m i () For the given slope o the lod line is 1, see Figure 8.0, pge 44, we cn write (7) m i Substituting Eq. (7) into Eq. (), nd rerrnging, the lternting component o the strength is Se( Sut i ) S (8) Sut Se The minimum stress in the bolt (i.e., the prelod stress), see Exmple 8-5, pge 448, cn be written s F i i (9) At Substituting Eqs. (18) nd (19) into Eq. (9), the minimum stress in the bolt is 0 i kpsi (9b) Substituting Eqs. (), (3), nd (9b) into Eq. (8), the lternting component o the strength is S 18.( ) 8.08 kpsi The lternting component o the bolt stress cn be written rom Eq. (8-35), see pge 445, s (30) C (31) A Substituting Eqs. (1c), (4), nd (18), into Eq. (31), the lternting component o the bolt stress is t kpsi Substituting Eqs. (30) nd (31b) into Eq. (5), the ininite lie tigue ctor o sety is (31b) 8.08 n.7 (3) 1. Note tht this nswer is in good greement with Eq. (4b). 17

18 Solution to roblem 4. (i) 9 oints. The x nd y xes or the let shoe nd the right shoe re s shown in Figure. x 34 in in 34 in in x QA BO 85 in 1 Drum Rottion in in c = 4 in c = 1 in y y 1 8 in in F F The ngle to the heel o ech shoe is Figure. The x nd y xes or ech shoe. The ngle to the toe o ech shoe is o 30 tn 9.44 o o o (1) (1b) The loction o the mximum pressure cting on ech shoe, see pge 834, is The ce width o ech shoe is speciied s The distnce rom the center o the drum to the hinge pins is mx 18 o 90 () b 3in (3)

19 The moment rm or the ctuting orce F is ins (4) c 1 ins (5) Note tht or the counterclockwise rottion o the drum the let shoe is sel-energizing nd the right shoe is sel-deenergizing. Bsed on the ct tht the limiting ctuting orce is cting on the let shoe nd this shoe is sel-energizing then the limiting pressure is equl to the mximum pressure o the rictionl mteril, tht is p pmx () The mximum pressure or woven cotton mteril rom Tble 1.3, see pge 8, is pmx 100 lbs/in (b) nd this is the mximum pressure cting on the let shoe. Since the right shoe is sel-deenergizing then the limiting pressure on the right shoe is less thn the mximum pressure o the rictionl mteril (cnnot use the vlue in Tble 1.3 to obtin the mximum pressure). The limiting pressure o the right shoe is unknown t this point in the nlysis. See Eq. (19b). For the let shoe (the sel-energizing shoe), the moment due to the norml orce bout the hinge pin A cn be written rom Eq. (1-3), see pge 841, is M N p br sin d (7) sin 1 where the integrl cn be written rom Eq. (1-8), see pge 83, s B ( 1) 1 sin d sin( ) (7b) Substituting Eqs. (), (3), (4), (b), nd (7b), into Eq. (7), the moment due to the norml orce bout the hinge pin A cn be written s M N lbs.ins (8) For the let shoe (the sel-energizing shoe), the moment due to the rictionl orce bout the hinge pin A cn be written rom Eq. (1-), see pge 841, s pbr M sin ( rcos ) d sin (9) The irst integrl in Eq. (9) cn be written rom Eq. (1-8), see pge 83, s 1 19

20 The second integrl in Eq. (9) cn be written s sin sin A sincos d (10) sin d cos cos cos cos (10b) C Substituting Eqs. (10) into Eq. (9), the moment due to the rictionl orce bout the hinge pin A cn be written s pbr M ( rc A) (11) sin where the integrls in Eqs. (10) re nd A (1) C (1b) nd the coeicient o riction or woven cotton, see Tble 1.3, pge 8, is 0.47 (1c) Substituting Eqs. (), (3), (4), (b), nd (1), into Eq. (11), the moment o the rictionl orce bout the hinge pin A cn be written s tht is M ( x x 0.848) lbs.ins (13) 1 3 M lbs.ins (13b) (ii) 1 oints. To determine the moments o the norml orce nd the rictionl orce bout the hinge pin B. First, the mximum pressure must be determined or the right shoe. Note tht the limiting pressure on this shoe is not the mximum pressure o woven cotton given in Tble 1.3. Recll tht the limiting ctuting orce or the let shoe (tht is, the sel-energizing shoe) cn be written rom Eq. (1-4), see pge 835, s M N M F (14) c Substituting Eqs. (8) nd (13b) into (14), the limiting ctuting orce or the let shoe is F lbs (14b) 1 Note tht the ctuting orce on the right shoe is equl to the limiting ctuting orce or the let shoe, tht is F lbs (15) 0

21 However, the right shoe is sel-deenergizing, thereore, the limiting ctuting orce on this shoe, see Eq. (1-7), pge 83, cn be written s M N M F (15b) c Rerrnging this eqution, the moment bout the hinge pin B due to the ctuting orce is Fc M M N (1) Substituting Eqs. (7b) nd (9) into Eq. (1), the moment bout the hinge pin B due to the ctuting orce on the right shoe cn be written s or s p br p br Fc sin d sin ( r cos ) d sin sin 1 1 (17) p br p br Fc B ( rc A) sin sin (17b) Rerrnging this eqution in terms o the limiting pressure gives br br Fc p B ( rca) sin sin (18) Substituting Eqs. (), (3), (4), (5), (7b), (11), (11b), nd (1) into Eq. (18), the moment bout the hinge pin B due to the ctuting orce on the right shoe (tht is, the sel-deenergizing shoe) cn be written s or s p (1.3734) ( x x 0.848) 1 1 (19) p ( ) (19b) Then rerrnging this eqution, the mximum pressure on the right shoe is p lbs/in (0) Note tht the mximum pressure on the right shoe is, indeed, less thn the mximum pressure on the let shoe, see Eqs. (b) nd (0). Substituting Eqs. (1), (), (3), (4), (7b), nd (0), into Eq. (7), the moment due to the norml orce bout the hinge pin B is M N lbs.ins (1) Substituting Eqs. (1), (), (3b), (b), (11), (1), nd (0) into Eq. (9), the moment o the rictionl orce bout the hinge pin B cn be written s 1

22 M lbs.ins () Check: The ctuting orce on the let shoe nd the right shoe is the sme nd cn be written s or MN M MN M F c c let right (3) ( M M ) ( M M ) (3b) N let N right Note tht ( M ) ( M ) nd ( M ) ( M ). tht is N let N right let right Substituting Eqs. (8), (13), (1), nd () into Eq. (3b), gives (1.1 x x10 ) let (11.53 x x10 ) right (4) 3 3 ( x10 ) let ( x10 ) right (4b) The dierence in the let side nd the right side o the eqution is due to round o error. (iii) 4 oints. The totl brking torque (generted by both shoes) cn be written s Ttotl T l Tr (5) The brking torque generted by ech shoe cn be written rom Eq. (1-), see pge 835, s T p br (cos cos ) sin 1 (5b) Substituting the mximum pressure p pmx 100 lbs/ins, see Eq. (b), into Eq. (5b), the brking torque generted by the let shoe (tht is, the sel-energizing shoe) cn be written s T l o o (cos9.44 cos19.44 ) lbs.ins () 1 Thereore, the brking torque generted by the let shoe (sel-energizing shoe) is 3 T lbs.ins (b) l Substituting Eq. (0) into Eq. (5b), the brking torque generted by the right shoe (tht is, the seldeenergizing shoe) cn be written s T r o o (cos9.44 cos19.44 ) lbs.ins (7) 1 Thereore, the brking torque generted by the right shoe (tht is, the sel-deenergizing shoe) is 3 T lbs.ins (7b) r

23 Check: Note tht the brking torque generted by the let shoe is, indeed, greter thn the brking torque generted by the right shoe, see Eqs. (b) nd (7b). The mximum brking torque must occur on the sel-energizing shoe (the let-hnd shoe). Substituting Eqs. (b) nd (7b) into Eq. (5), the totl brking torque generted by both shoes is 3 3 Ttotl ( ) lbs.ins (8) Check: Compring Eq. () with Eq. (7) we note tht the rtio o the brking torque on ech shoe is equl to the rtio o the mximum pressure on ech shoe, tht is Tr T l ( p ) right (9) ( p ) let Rerrnging this eqution, the brking torque on the right-hnd shoe cn be written s T r ( p ) right Tl (9b) ( p ) let Substituting Eqs. (b), (0), nd (b) into Eq. (9b), the brking torque on the right-hnd shoe is T 100 Note tht Eqs. (7b) nd (30) re in complete greement. 3 3 r lbs.ins (30) 3