D Bodies Test 2 - Problem 3.141 3.141 A hollow cylindrical...

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Unformatted text preview: Problem 3.141 3.141 A hollow cylindrical shaft was designed with the cross section shown in Fig. (I) to withstand a maximum torque Tn . Defective fabrication, however, if —' ' resulted in a slight eccentricity 9 between the inner and outer cylindrical surfaces of the shaft as shown in Fig. (2). (a) Express the maximum torque T which canbe safely applied to the defective shaft in terms of To , e, and r. (b) Caimlate the percent decrease in the allowable torque for values of the ratio e/t equal to 0. 1, 0.5, and 0.9. (CL) For bo‘Hn con‘FifiuWJl’fo-ns Let t = C1~C| be "HID. 'Hm’cakhcss 6F (I) "The mint'muwt Hate/knees d? C9.) is t- e . fioV‘ I For =2;—_e)a QGCPUUFl'o-a :le +°qu£ Eipres'seal as a. '79 3.156 The long, hollow, tapered shaft AB has a uniform thickness t. Denoting by G the modulus of rigidity, show that the angle of twist at end A is ¢ _ TL cA+cB AH4R‘GI cicfi ' Problem 3.156,, From Sear-1dr?! CB'CA L3 93: YES—W TD Shir: TL3 b lily”) J: GJ 'Zfitcfi'CAV—EG" 7A 73 ZVCCB'CAY'ZG 231 YA :th ,I_ _ .L = TL: {can—cgflgcn—c.y} ficg-cnste Y? 75‘ were—c.1310 Lick! L‘Ca“ TL- _ _l_ _ (Cal—Cc) WCcs—cnte Ca“ Ca" LfTICC5*Cp)tGCAIC32‘ ; Problem 4;? 4. 7 through 4.9 Two vertical forces are applied to a beam of the cross section /I shown. Determinethemaximum tensile and compressive stresses in portion BC of the 3 in. 3 in. 3 in. beam- f Yo 6 m. s ' l5 kips IFS lips Nevin/m! unis ,pies 3 I‘VL. z _2 above "Hm base. :1 ~69 3 I, = 7:5,}sz AA,“ gem-H (mm? = 12G w I: = £131523+ Aloft TIE-(“13(1334 (183(13‘ = 73 ;,,"‘ I= 13:12:: \264-78 = 204m" 3“? T 5"“ yin-t = " 3 {H- I? M M- Pa. = 0 mi) M: Pa. = USXHo): 600 Japan. ‘ _ M _ 600) 5) _ ‘ 6;? — '- _‘L20_.+£—' - - k5! (ooumsax'onx ‘4 65"} : _ M bf : _ t 8- RSI. (+CMSI.9I‘|) "a I 20L} [. Problem 4:1}0 4.10 Two equal and opposite couples of magnitudeM = 25 kN- m are applied to theehamel-shaped beam AB. Observing that the couples cause the beam to bend in a honzontal plane, determine the stress at (a) point C, (6) point D, (0) point E.. 506?]; MC? ye: '1- H‘HJS HM '~ ‘O.0!+L}’]_5'm DID: i20~'1H.25' ‘~= YSJSMM : 63-07575 “'1 jg : 3G-‘4‘L25' : — g_2'5 mm =-O.005’25' m d1? ED- 44-2.5 - 15275 m,“ d2: LH.75:-—i8 ‘ 26.25 mm 0.3 " a‘ I. = L. = fishhfmdf = BEGUM”4860030535)? ‘= amouo‘ m4 z 3 I: = 355$? Adz '- r'afllbksé) +(H320)(2e.2s)‘ = 3-4;435x :04 W‘f -= 1.), 12+ I .- 6.51.37 xro‘ M“ 1%-36‘HHO“ m“ (m, Pam’fC: g M _g25x:o3)(-o.owm “h I = ‘3;36?9X=£6“ " 77-BWO‘ Pa : 75?.8 MPQ sq ‘ ‘ M 25 1")6.<‘). 75: ‘ “=1 PM 13- cab—ill r 43st pg 7-136”! MP4. «a to PM E: 6 = l .3 xlo‘ P F I 13.2w»; “(b-s 1* T Q = 1187 Misc» all ' PrOblem 4- 4.15 Knowing that for the extruded beam shown the allowable stress is 120 MP3. in \‘ ,1 tension and 150 MPa in compression, determine the largest couple M that can be '«fi 80 applied - in”! 5"- I 40 mm The namih‘ap (son's Pies 30 mm 6L1“)!!! 'H»: Lo‘H’ow. 31hr: 54—30 : 2'4 MM = 0.024»; X“... :' ‘30 mm ? - 0.030M I! I. = EH1} MA.2 = 7%(40165‘41‘ momma? Haw/c“ m“ Ii: blkf+A1Jfi= gizfaoxsfih iCuoxlmcs)‘ = m3. 24m“ m" If I1+Iz - 753.I'4a/o’»..." = 753.16 #04...” _ M ._ §_I ls! — {-31} 1M\ [3 I +91; +ensrom siA-e 0-02% :- 3,7708 > )01 N-vn (usouo‘)(7_5_3.rs woo-:1 bo‘H’om; compresSJ‘o-a M = O. 030 " 3‘ 3_7908 x [02 m Choose 'H-ue smalper 5.3 M." Mun!= 3.7‘303 HOS N-hn = 3.77 kN-IM " -'-'II-tlll v. I ll ....,. u.» nun:me oI rroo. 5.113 has been loaded and unloaded as described in that problem, a torque T1 of sense opposite to the original torque T is applied to the shaft. Assuming no change in the value of 1-,, determine the magnitude T1 of the torque Tl required to initiate yield in this second leading and compare it with the magnitude T, of the torque T that caused the shalt to yield in the original loading. 11% —3.G:< ks. = -3.g2x;03 P3.- Apflbv'aogle reversed, Sin/£55 ‘Fov‘ rel/evsas’ you} = ’21, 4—11.; = lexio‘ —3.czxzo3 e H 33xzosp5i d 129.1 _ J‘ 11’ 3-337% l 0‘ JG? L25 38" t Sflffxlos jLJn, i, '—' 38.” m 4 -3 new“? :23 3 “(2’22 ) = it.“ I. 4.41 and 4.42 The6 x 20-in. timberbeanhasbeenstrengthenedbyboltingtoitthe steel reinforcement shown. The modulus of elasticity for wood is 1.8 X 10‘ psi and for steel 29 X 10‘ psi. knowing that the beam is bent about a horizontal axis by a couple of moment M = 450 kip - in., determine the maximum stress in (a) the wood, (b) the steel. Us: Woes, as #12 V‘C‘FEV‘QAACQ. Mdewicu’. FOV‘ M10090 “:1 For steel) n: Es/Ew = 201/” = lam: Four “-5 J‘anncj 5cr/‘l't‘on) A 1' 3.38 inz tw tw=0.220in,) 72:30.57! in.) 1:7 .= L31M" J; .. %‘ :6 For Hoe eakpofic’fe seo'ifmA ‘HIc Ccn‘fmfa' e? +112 [ U chairian Pzes (2 + 0.2204357! - 11.54751. aJom/e ‘H-e base. yo = ILG‘l‘iin. 43°» okannej. A {"1 . 1 TV‘éMS‘FD'rMan Sec/How. > J .. 1046.353 ' 3.38 54.456 Y., = _ (24354 72 72 = 8.433 in. TL: M¢U+Nj M13 flies 3.433 in. above ‘Hte bo‘Hom. 1;, = n, I, + n,A,Af = (1e. uu)(i.32)+(54,454)0:th - 3.433% 53149 an“ 12 = Effie}; + mm”: = T'flelém3+ (70(3333 «a? = I2qo.2o m“ I = I,.+.Iz = Immeqm“ M= 450 1a.“. g=_n_Vly_ P I can Wooal: n: I ) \y = - 3.433 in. (450\(-8.433) . ‘ ‘ _ . :—————-——— r _0 k: —2_0X,|£$| ‘ 6: 1274.97 2 2 ’ a» (b) Shel): H: 19.1111 1.- 12+o.220— 3-433 = 3.727 .‘n. 6 r _ (IéJIllwlSo) (3.727) = 44.9571“.- 55=-H.e$ks; 4 I87”. 6? If" pro ble’m- 432 4.81 and 4.82 Determine the plastic moment ML of a steel beam of the cross \4 // section shown, assuming the steel to be elastoplastic with a yield strength of 240 ' j MPa 50mm 30:!“ 773%? area A = (503(90) -— (30}(30) 7 3600 my..." 10 mm L — 1800 L —_+_. 1 - MM LOmmaj L—J LlOmm EA 1300 30mm 9631-2???— 1‘36MM P. 91 —-—9- I3: 9, Pq—-‘-r Al 2 (50)(3G): "SOOmm-l) g} 1' PHI-n A13, : >610: mm: Al: (so)(m)= 700 m: y: = 7...... A31 = Lmuoz‘ m3 A3 ': (20)(3D) T- 600 hung-J ya = mm A3 3/": e 5! ID: Mina A9 ~‘ {50)(10) "' 500 km: 3* : L16, MM Any“ : 24-5.)”03 MM: AIS" +A2311 + 5:7: + A,}* = 7?.2yfos MM? : 712 x/o" ",3 Mp - 6r SAG;- '-‘ (240%10‘X7‘74xl0-‘j: 2 110033103 UH“ : l9.0l kN‘hn ‘ ll \IPrOblem 1 36 4.135 through 4.137 The couple M acts in a vertical plane and is applied to a. beam oriented as shown. Determine (a) the angle that the neutral axis forms with the horizontal, (b) the maximum tensile stress in the beam. 12:“: 2L4 [all J IJ:=6-7'1' in” 2*: 2,; = 0.8571}: 2'0 : -’+..l 0-85:7}... = —3.I‘H {In jA :- —’+lm) ya: Llfn’ Lyn r. *O-JZb—{n My, : —25 Sin #50 r — l7_678 kt'P-lh M2! = 25' 49% “1‘52" = 17.67% éome _ I, _ 2m _.,9 z (a) hm g) - +«»» 9 - mq 'l’flun( 4s) 3.175: v cp = - 72.5“ 2 3' ct = 72.5°-‘+5” v 2-7.5” .- 9h #5“ ‘3’" 6 = _ M210 + ann _ _ (. I7.c72)(—o.25) +_(—I7.(,72)(—&HH 12 IJ 21.4 6.79 = o-:zos$+ 8.238 = 3.41! ks.‘ -- 4.145 An axial load P of magnitude 50 kN is applied as shown to a short section of W150 X 24 roiled-steel member. Determine the largest distance a for which the maximum compressive stress does not exceed 90 MPa. "Problem 4.145? Add 1— war) 2“ axes. For W 'ISO =1 2‘1 roilea— s+eef sec-Ham .‘AL'= 30.30 m1 = soéoréo“w‘ 11* |3.4I106mm“ = 1%.Lmo“m” y IJ r 1.83m? mm" = 1.33:6" .J’ r d:iGOr-1MJ bF=102 ham 2 y,-—§-’=—30m zA-éf-grm p P = 50:4103 N A fialg Mz:—(50*103M75¥103)'—375x[o mm M3 = ~ POL 6A=_E‘%+MZA GA:—?o>r(o‘?a J W éi-f “1:” +1? + 613' -t x 3 _ x -3 r 3 - 58x33: E (‘3-7f3fx1f0‘85 to } + 3:103:36 + (‘¢°‘lo€)} = £ 142,333 + 16.340 ~ qogxlo‘ = - 1.2373x:03 N-M — 33 3 — a: -311, ___'_-__”_“'O = 3c_gx£osm = 36.8mm 4 P 50 xtos Problem 3.37 / M___u_u__E——F£Hfs’ 3.3"? The aluminum rod AB (G = 27 GPa) is bonded to the brass red BD (G = 39 GPa). Knowing that pertion CD of the brass red is hollow and has an inner diameter ef40 mm determine the angle of twist at A. T3 = $600 - 111.. R059 AB: 6 =- 27x10‘Pe} L= 0.400 m 36mm T 3-" 5-111 T:- 800 NM c=é&=o.o:gm A we“ 250mm J‘ : EEC": :Rofia'eVr magenta” m . C B . - . /J/375mm A @Mes: LL 344% EDD (CHOU) " 71.275Mr53mJ 400mm J (27 wtofi)(15%_8qexto'*)" Paw'r BC: (3— = sauna“ Fa. L: 0.375% c.=—;ol = 0-C30m T —= 800 + [600 = Z‘ioo N-m I :: @ch ~= E (0&303'": l.2723‘f We)"4 m’ .3 #00 0,325) _ @flicg L. :LJ—L—h : 18-1373’03M GJ ' (3%10“)0.2723HWO") Pad CD '- c:I 2 J59, =- 0.020 .m Cl T7542 = 0.030 m. L: 0-250....“ l l J‘ = g (cf—c.” : 11-} (0.0303 0.020") = LOQIOZHOA m." T _ (2¥oo)(o.25o‘) _h _3 (Pm: GJ ' —' '5~Oé3“0 W 6P+uf5+ a+ A g)” '-'-' Qua *- (page; 4- 92;“: :- {050363 mots M qfl‘ e 6.02“ .4 Knowing that G = 11.2 X105 psi and that the shafi at F is fixed, determine the lug]! . ‘ 3.43 Two shafts, each of 7/8 in. diameter are connected by the gears shown Problem 3.43.} - ‘- " through which endA rotates when a 1.2 kip - in. torque is applied at A. Cajcufi¢+:'om a“? 4Ounes¢ Civ‘cum-chw‘h‘mp Cowhgbi‘ 'PDMQ LéA’weam jean/‘5 8 MA E. FrIfifi :1??? "TE— we G F — 1287,32 kiln-In ‘7 1200 filo-f” “Er: $50200) ': Isooib-in TWI5+ in SL144?" PE: ‘1- re L: 12 he.) C=$1al '—' {2:51 Gr ILQVNJGPS.’ J: Ec‘* -: gg-E)" r 57.542210" w." cps, = lL=_—____U6°°)(m = 2?,739xlo‘” ma F GI (ll.'2«lo‘3(57.5‘t3w153) Reflexion MILE, C9,; = (PE/F = 29.733Jv/0F3m4 Twajedfj ahspjacememj a} clear c-fr‘cie 5: PECPE = Ways I" 6 v3 _ —3 Rothko“ a B - 993? écpg Tfi(za.7wvzo 3 ~ 37.71-3on m? Tum-5+ in 511.4va EA, L= 8+ c; = Hm- J= 57.598Ho'3 w." 9 -: LL—T—M—L T 2Q_Oé‘5'¥lD—3 Way! "3 63 (Ir.2wao‘3(s7.st+~gxio-3) {Zohzhou J A, (pa: 993+ cpm,5 = 65.723340”: w? q); 3.77" 4 3.113 The hollow shafi AB is made of a mild steel that is assumed to be elastoplastic with z} =_ 18 ksi and G = 11.2 x 10“ psi. The magnitude Tof the torque is slowly increased until the plastic zone first reaches the inner surface; the torque is then removed. Determine (a) the maximum residual shearing stress, (5) the permanent angle of twist. Problem 3.1 / C‘.I '-‘ 0-75 in. C2 = L25in. L: = L}? in. J = ECcJ— c1”) = 4‘} 0.26—0.75") : 3.331% in” . _ 3 . 1.25m. 0-751“. 2"." = kSI ‘ "lo P5; Loading; : when (or, mac/Les 'Hne inner Surface.) Hem f= Tr auux/ wile-PG... c2 3 ct "QM—— Zflj loz’l’rd/o = 2772? = gg—T’l’ykzg—CF) 5. t gilmhowmss- 0.7533 = 57-72%“? WW" LY‘ L’L’r _ (%B)CI‘8>¢IO"" = o_102357 w“? = 5.891330 @9044! " (0:25 )(lex 10") I T Unloauiinfi : T =' $7. 7273‘“? flavtn Epas‘i’t'c. ’El 2 3‘2 3 3 At ,a .- cz= 1.25;“. t’: -s——————___(57'727”O XL“): zigzwo F5.- 3- 337‘? At {0 = c. = 0.75 in. 1" = 4(57'727H‘33) 0'75) = 1237»: :03 rs; 3.337‘! I: TL 2 W = 0.074“? mm! = H.296?” 9" 9; (il.2v:0‘)(3.337q) RBS‘JUJ: Tr”: —’El ¢Pm: _ g) (0.) Al fun; 21,5: [Zrlofi- 21.62vlo3‘: - 3-52vyo‘ps; 21m = “3-62 ‘C‘Sl‘ Al ,0 = C, ’13,“ : lgxfoa— ;2_q7y'(03 e 505 Hg: P.“- hnmlmum Tm, = 5. 03 VS? 1! (bl (ppm: 0.:02857 — 0.074% = 0.0227353 m1 cpfer; LGH‘] ° 4 300 r' - m Problem 3.35 _ exerted on pulleys A. B. and C. Knowing that both :1), determine the angle oftwist between 40 12.00 N I 111 mm 3 3.35 The torques shown‘ire shafts are solid and made of brass (G = 39 GP (a) A and B, (15).»; and C. 400 N - m [0-) Angie SPF-wig} be+waan A so»? 13 TAB = L100 H-MJ L”: L? m r; = £0! :- 0.0mm a; = 37m)” Pa. 4 3;; Id” 79.5mm” M (p : TL .— Mfl— Ala 55 L36: x m“ l (79. 52:10") «.— 0154772 M! 9 90m = 3 37° 9 4” Col Anjfle a‘P +uJu‘s'Jl Berrween A Sung? C "rec =' 800 N-M J L“: Lg M, T " =' 1i(0.020Y'= 25I.3mx :0“ m : (flame) __ m9 9°" (39¥109)C25l.327¥lo“) “ (imam 3 99m = (Pym (Page 0.154171;.0.Hém‘a.=_0.0073:co ml?) c,='l'a. =o.ozo m, G =- 3‘?X/D‘Pa If 3.36 The electric motor exerts a 500 N - In torque on the aluminum shaft ABCD eed. Knowing that G = 27 GPa and that the torques when it is rotating at a constant 5;) exerted on pulleys B and C are as shown, determine the angle twist between ((1)3 and C(11)}? 2:11:11). 200 “In (CL) Amqlc‘ay-lwfs‘f E6+W¢gm 3 M C = T837430 N-mJ Lu— LZrn lat-0-022.“, G=27xlolPa L .‘ I: T [200]“ 1) - ZLIJSTYIO'Z’NA (PB/c: I-SSL} é talc: a? - Uol Ana}: o'F +uls+ between 3 map D. Tan 2 500 N-md LED =03.“ 3,13.“ = $0,024)": 52.1. 153x10"i w." c riot: 0.01%; G = manta (500 \(ofi‘) -3 ': _______________-——-—-—-— :_ CPL": (Twlo‘txsxl .ng Hoe) 3-980on mp clam: (payr (pm = 21:573.:6131330455 = 56.137y153r4 ' CPS/n -= 3-22“ 4 3.7 The solicl spindleAB has I]. diameter d, = 1.5 in. and is made of a steel with an allowable shearmg stress of 12 ksi, while sleeve CD is made ofa brass with an allowable shearing stress of 7 ksi. Determine the largest torque T that can be applied at A. Problem 3.7 chJ sF;nelJe AB '. . c: = 47159,, = 1710.5) 2 0.75 in. : 15c! = Ecol-Is? = oJ-H'lOl ant 7m : B. W J2.) C '1“? l O. 70! ‘Tnet = c “a: 0-15 (Hm 2' 7.9532 kip-{v1 Spear/e. co: czeéefll = '§(3.0) = 1.5;..- CI: Cid-t = LS—OJIS ‘—' Llfi' in. T T It: Jamzlh) q, C: - lob— A’pflwcfipe. chpd'e d"? 'lbv‘que. T ls +Le smdflefi. T= 7.95 kt'Yl-t‘n 4 r 19. 2:3 10‘me Problem 3.8 3.3 The solid spindle AB is made ofa steel with an allowable shearing stress of 12 ksi, and sleeve CD is made of a brass with an allowable shearing stress of 7 ksi. Determine (a) the largest torque T that can be applied at A if the allowable shearing stress is not to be exceeded in sleeve CD, ([7) the corresponding required value of the diameter d, of spindle AB. J to __ (Ll-ll77e)(7) -. . C1 - T ‘: [CL :UEJ ktfr‘cm PW" equip-lott'tmn T: l9-1lr kfilhvl "6 CED Sofit‘ol apt‘nelfle AB '. T '2 151,1l3 k.}q..‘n = LC = 2T FE J" ‘lTC3 = .13/11" 1. .13/Cmmctat3) _ . C. a W5)— .. l'fl. cl = 2c: = (230.0064) 35: 2.0] “4...... .5 Problem 4.49 4.49 A concrete beam is reinforced by three steel rods placed as shown The modulus of elasticity is 3 x 10‘ psi for the concrete and 30 x 106 psi for the steel. Using an allowable stress as 1350 psi for the concrete and 20 ksi for the steel, determine the largest allowable positive bending moment in the beam. —-in. diameter 7.1: 2m. ‘3 SXIOS L—Bm-J—r A5, = 3 $1M — SGZ-K-ET = Lsoqo if F‘ggi nA5= [8.040 of x Locall‘e heal-wall axis. axra b 14 3x35: — Usmoflmxfi -= o le1+ l3.0"fD 95' "' : O _ o W . Solve 4;)“ X X: £3-04 :22ch) ( )( M252 5; r 5,005 m lLl " X = 7179.5 in, I : J3“ :8- x3 + nASCH—xl‘ = g-{sflsoosf + (:3-oqol(7.eesl" = 173:2.4 in“ '2 n M = fl (6‘! M w Caner-«5+3: h = LOJ lJl= (3.005;:«1J lS'l ‘= LES—0 psi __(13503(I7'so.5) _. 3 _ _ . . M - — 33‘340 .«Dla'tm — 33? top-m Sled: In = to J rljl = 7. W75 J 5 = 20 was last- M : %:¥;€:Zgl;:D-g) 7" 435 “[03 Jab-iv. = 433 K'P' in aLbOS? +le SMQJJEr VCR/003 M = 335’ lap-I31 3 '32-“! k‘IP fl ...
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This note was uploaded on 04/08/2008 for the course MASE 255 taught by Professor Genin during the Spring '08 term at Washington University in St. Louis.

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D Bodies Test 2 - Problem 3.141 3.141 A hollow cylindrical...

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