5-58. The two shafts are made of A-36 steel. Each has a diameter of 1 in., and they are supported by bearings at A,

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1 J. J.'- '-'.1J L.d 1V J L...,V./ The two shafts are made of A-36 steel. Each has a diameter of 1 in., and they are supported by bearings at A, L and C, which allow free rotation. If the support at D is J fixed, determine the angle of twist of end B when the torques are agplied to tl1e assmbly as shown. r - fts t!?c; are mde - of A-36 steel. Each has a diameter of 1 in., and they are supported by bearings at A, B, and C, which allow free rotation. If the support at D is fixed, determine the angle of twist of end A when the torques are applied to the assembly as shown , " 0 80 lb. ft 30 in. Probs. 5-58/5-59 *-60. The two A-36 steel shafts AC and FD are coupled together using the meshed-gear arrangement shown. If A is a fixed support, determine the angle of twist at end F due to the applied torsional loading. Each shaft has a diameter of,.

2 .. "" The two shafts are made of A-36 steel. Each has a diameter of 1 in., and they are supported by bearings at A, B, and C, which allow free rotation. If the support at D is fixed, determine the angle of twist of end B when the torques are applied to the assembly as shown. '0""ok -f"'"". 030" -<-. ;... 'G 10,..""" - F.!1..L2120 II, O.JU3 "12,; 4"0 " Internal Torque: As shown on FBD. Angle of Twist: TL tpe=l JC 1 =y(0.54)(11.0)(1()6) [-60.0(12)(30)+20.0(12)(10)] c:::p-- - TF=-40.0Ib.Jt 0 4<>. lb.jt, o.sfe = rad = rad 6 6 tpf = "4tPE = "4( ) = rad Since there is no torque applied between F and B then tps = tp F = rod = Ans 7;H =ZO.O IbJt../ 80 Ibj<:. t::.izoi The two shafts are made of A-36 steel. Each has a diameter of 1 in., and they are supported by bearings at A, B, and C, which allow free rotation. If the support at D is fixed, determine the angle of twist of end A when tbe torques are applied to the assembly as sbown. F=.$L= ;;]3 Internal Torque: As shown on FBD. Angle of Twist: TL tpe=l JC 1 = y(0.54)( 11.0)( 1()6) [-60.0( 12)(30) ( 12)( 10)] = rad = fad 6 6 tpf ="4tPE= "4( ) = rad TGFLcF tpaif =---;a - -40(12)(10) - y(0.54)( 11.0)( 1()6) = fad = rad tpa=tpf+tpaif = = rad = Ans 7;H =,(,0.0Ibjt../ 'ff =-4<).0Ibft. a.5}t 801b.A C) f:.j,?o 16 O.5J-t. &Jt# 184

3 80 lb. ft, ju Ill. ( Probs. 5-58/5-59 *5-60. The two A-36 steel shafts AC and FD are coupled together using the meshed-gear arrangement shown. If A is a fixed support, determine the angle of twist at end F due to e applied torsional loading. Each shaft has a diameter of.:-j mm, and the journal bearings at F, B, and G allow free rotation. 40mm I 120 mm. E 18 N.m Prob.5-60

4 Ii.. *5-60. The two A-36 steel shafts AC and FD are coupled together using the meshed-gear arrangement shown. If A is a fixed support, determine the angle of twist at end F due to the applied torsional loading. Each shaft has a diameter of 30 rom, and the journal bearings at F, B, and G a,(lw frpp rotation. Internal Torq ue: As shown on FED. Angle of Twist: "'c = 'C' TL I.JJG - JG[6(1.5) +(-6)(0.5)] = 6.ooN.m2 JG 40 nun C -il J2 I"VG/D '. Ojmj" o.5m //' F:'<,=1500N. /8N. {;' O./Zrn "'D = 0.12"'c = JG 1;;1>=-/8'OAJ.rY1 /81).", l DLED -18.0(0.5) N. m N. ro2 rfieid = :J(; = JG - JG =---xi <PE=rflD+rflEJD N. m2 = -+-=lg lg JG Since there is no torque berween E and F, then rflf= rfle=r(0.oi5)'(75.0)(109) = roo = Ans bf=/50.0"; O.O+rn F=/ O.ON Tc.J{=b.OON'tr'I 71H=-(,.00 N.m The assembly is made of A-36 steel and consists of a solid rod 15 rom in diameter connected to the inside of a tube using a rigid disk at B. Determine the angle of twist at A. The tube has an outer diameter of 30 rom and wall thickness of 3 rom. Internal Torque: As shown on FBD. Angle of Twist: TL "'A=2:- JG 50(0.3) 80(0.3) = + ;' (0.0075') 75.0( [09) ;'(0.Oi ') 75.0( 109) 50,,)"" 1:= 70.m = roo = 2.70 Ans 185

5 (C;;-76. The shaft is made of L2 tool steel, has a diameter 'of 40 mm, and is fixed at its ends A and B. If it is subjected to the couple, determine the maximum shear stress in regions AC and CB. ( 2kN 2kN ( m Prob.5-76

6 *5-76. The shaft is made of L2 tool steel, has a diameter of 40 mm, and is fixed at its ends A and B. If it is subjected to the couple, determine the maximum shear stress in regions AC and CB. '" an Equilibrium: Compatibility: 1A+1j, -2(0.1) =0 [ I] 400 '"'" 600./ '.,.. tpcia=tpcib 1A (0.4) - 1j, (0.6) JG-JG Solving Eqs. [I] and [2] yields : 1A=l.501j, [2].."'::, "" To 0'00l'? 1j, =0.080kN.m 1A=0.120kN'm Maximum Shear stress: 1Ac 0.12( IQ3)(Q.O2) MPa (AC>mu= J = r(0.024) 1j, c - O.OS( 1Q3)(Q.O2) MPa (CB)mu= J - r(0.024) Ans Ans The bronze C86100 pipe has an outer diameter of 1.5 in. and a thickness of in. The coupling on it at Cis being tightened using a wrench. If the torque developed at A is 125 Ib. in., determine the magnitude F of the couple forces. The pipe is fixed supported at end B. F '.._.-.-. Equilibrium: F(l2) -1j, -125 =0 [ I] Compatibility: E!PCIB=tpCIA 1j,(8) = 125( 10) JG JG 18 = Ib. in. From Eq. [I], F( 12) = 0 F =23.4Ib Ans, 'I 194

7 , (1/ * A steel (G = 80 GPa) shaft 2.5 m long extends through and is attached to a hollow bronze (G = 40 GPa) shaft 1.5 m long, as shown in Fig. P5-60. Both shafts are fixed at the wall. When the two couples shown are applied to the shaft, determine (a) The maximum shearingstress in each of the materials. (b) The rotation of the right end of the shaft.!f/g The torsional assembly of Fig. P5-62 consists of an alu!l)ipum alloy (G = 28 GPa) segment AB securelyconnected to a steel (G = 80GPa) segment BCD by means of a flange cupling with four bolts. The diameters of both segments are 80 mm, the cross-sectional area of each bolt is 130mm2, FIG. P5-62 and the bolts are located 65 mm from the center of the shaft. If the cross shearing stress in the bolts is limited to 55 MPa, determine (a) The maximum torque T tat can be applied at section C. (b) The maximum shearing stress in the steel.

8 '. PROBLEM!I,' """"'" f' t /,,""T j-! -' I, -(77 I IboiT C. Ii) '-+! (f) rt.. cp EN C

9 , PROBLEM NAME DATE I ' z:;ill; 10:". 4t Jy;fnzjt(hrJ ': ; I 'n. j '" 12, If 2. "j ) '1 (, I 7- t:;\"-":-- 1.' J.6 J; ', ' J' "%""1 ' ',","-"-""" ' "'j """""'..:"-""i"""""'", ",,,,"'" """"- """',," '", <:; f,..r P P A(p<J ","'" rc,( CSh <, t t 7 J/' ( I <D"!'/r), /-4- JL"""! 6l ),,"/ " ė. '2.-,, j,... - """""'"- 1:?rl 1! """"""",, CJr ;,1' - " - :3 (/ ( '&>%2 " -Eftt') 'I EN C

10 'PROBLEM II) NAME DATE,IY'/'H<>.../' F /j ) 1+ /) - [ fif ; -J :, -7:> T,7 t, t",3 4-0 " """ """" """-+-"""'--""""""""-"'-"" 0 -ri.. 'f 'h't:)'1, I L tjllf;'c<. EN C

11 PROBLEM Ih/ F - t. L II The,Plniversity COLLEGEOF 0 Toledo ENGINEERING NAME DATE Ir {" 11 _!:1...3 / i prt ( ( ) ) I, 7 : ' - 0y EN 113 9a6 2C

12 FIG. P * Two 80-mm diameter solid circular steel (G = 80 GPa) and bronze (G = 45 GPa) shafts are rigidly connected and supp6rted as shown in Fig. P5-54. A torque T is applied at the junction of the two shafts as indicated. The allowable shearing stresses are 140MPafor the steel and 50 MPa for the - bronze. Determine (a) (b) The maximum torque T that can be applied. The angle of rotation of a section 1.25 m from the left support when the torque of part (a) is applied. 150 m(l1 100 mm 50 mm FIG. P * A composite shaft consists 'Ofa hollow bronze (G = 40 GPa) cylinder with a I50-mm outside diameter and a 1O0-mminside diameter over a 50-mm diameter solid steel (G = 80 GPa) core. The two members are rigidly connected to a bar at the right end and to the wail at the left end as shown in Fig. P5-58. When the couple shown is applied to the shaft, determine (a) The rotation of bar AB. (1)) The maximum shearing stress in each of the members.

13 .". PROBLEM J-- r IIThe 7 NAME DATE Hi.nt f 8j I 011 L &>(flu; ",.', I.(s;X/;).( i ft- ". j, 1144.:x. '" c:i) - L.... T - -'- 2- v) d>l 2-- f" -,... ;(ltfiq'tl, j1tz-(l,'4r!r;, ",'t '. ( p", X ;f' 'f I i G d"7li(. ix"tj..- EN C

14 ': 1 PROBLEM III' r-s-r COLLEGE OF ENGINEERING NAME DATE A/;' "" J '; G.f J'to J"':rr'L '- f& l( :c: c:.tjiu (i) T<fi:..IZ"i: IT(.. -- p q /,If/, M ) EN C

15 PROBLEM - - rf II The Ir-;niversity COLLEGE OF 0 Toledo ENGINEERING NAME DATE I "-;!A'i:: -, -' (J.I,. L - I I i t,,",",," d Vi i.,i."t?4f ( "'Jf '- C I G1)'i i EN C

16 5-87. The A-36 steel shaft is made from two segments: AC has a diameter of 0.5 in. and CB has a diameter of 1 in. If the shaft is fixed at its ends A and B and subjected tn llnlfnrm nl,trlhntp,n tornllf>: of no 1h. in./in. along o z.. - PROBLEMS Jhe two shafts are made of A-36 steel. Each has a eter of 25 mm and they are connected using the gears fixed to their ends. Their other ends are attached to fixed supports at A and B. They are also supported by journal bearings at C and D, which allow free rotation of the shafts along their axes. If a torque of 500 N. m is applied to the gear at E as shown, determine the reactions at A and B Determine the rotation of the gear at E in Probe mm Probs. 5-85/5-86

17 PROBLEM 1./', I IIThe,r=;niversity COLLEGEOF 0 Toledo ENGINEERING NAME DATE (!) te:t.-",,/ 4't.../,; -+..1IId' --- T_";_,,,::...Ti,L,t ell' c.ll_.,'e.' J-O..-". C)...,.- M".TIi.'tI (, D,r,..,) TIJ.,.2 Tc r.. ;Yo (SJ, (: Q.." eti,' J,,'/"7 (j) If :f...: 2.)... d/r /' c:. Co G" "I i...,_._ ('It..',..'2.74'. - Ir {:' Ii 1E } l 3) ,--- t.# III '1. "" EN 173 9a6 2C

Allowable Shear Stress: Applying the torsion formula. t max = t allow = Tc T (0.75) 12 = T = 7.95 kip # in. Ans. t max = t allow = T c 12 = T (0.

Allowable Shear Stress: Applying the torsion formula. t max = t allow = Tc T (0.75) 12 = T = 7.95 kip # in. Ans. t max = t allow = T c 12 = T (0. 05 Solutions 46060 5/5/10 3:53 PM Page 14 010 Pearson Education, Inc., Uer Saddle River, N. ll rights reserved. his material is rotected under all coyright laws as they currently exist. No ortion of this

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