solutions chapt 3 - PI'IHIII 3.343 I vathalpolcofsolid...

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Unformatted text preview: PI'IHIII 3.343 I! vathalpolcofsolid cirudarms suction istwistudby hofizmialfumsP= “mlbmfingatfllcmdsol'a hu'izoutdarmAB [sac figmc}. Tll: dim from 11!: oulsith: oftll: poi: to flu: lint: ofaction offich feminist: = 5.0m. Iffllcdlowflrb all-arm“ in thpob i545“) psi, what istll: miniulm] mquimddimrbrdnfinofflcpob? Sdllhn 3.3-1: WI 9* M4 L' P=llDle a: = 5.0 in. Fllld damn Tflw = 4500 psi r = H2c+d}d = lflPt2c+d} m“ wd‘flfi «d3 (“In”)? — [15PM — 32?: = 0 Summmmm mm: UNITS: Mnds, Inclts («x-1500}? — [Jammy — 32(1100}(5.0} =0 ll «F — 124495;: —12.4495 =0 Sol“: numerical}: d =2.496 in. duh = 150m. ‘— “3.4-1 AuppedMADCcom‘limofmnfid Mumsi-bjmdwmnaMfimh oppodefiuhnassbvnhlnefim'l‘hhgasqml dllnslnl mama. =2.25'n.mdlelmhL, =30h.: Mullermnlndm¢=lj$haflmlh b=23h11¢udulimclwihshamodflu a=ux no‘pimmemunr, =2n0mb-‘n. “12:8,“,540. Calculus the following qualifies: (I) III: minim lieu an“ 1... in II! shaft. “(mmmofw‘mdrfn Mus) qudC. Smart DC 1.: = +12- = a mo b-in. = 16 1x_ l6(&ONI’-‘n.) "c :4 1:11.15 in)’ Tlclc (8.000 b-iI.X20 in) =76029I' . . was "(1) . d.=225n L. =30n. mxufpa 32 may ¢=l.15n. l¢=20im = +QOISI9'Ind G=uxufpsi =ZMIJOIHIL (a) Minimum . SesmnDChslhcunxinnmtm fi=&000l)-n. =76mpsi «— Samar” (b) Mmmnmc 7'4" = 7'2 '71: '12.th «k = d» + 4pc = (—0.013m7 +0.015197) lad 610|_16(12.000l>-h.) . l— = 5365 = = . m= _n._)’1(225 9- 4r Immoral 0.16 «— 74.11 = (-12.1100th31):‘:;) 0mm (11 xw‘ pd)(3—';)(2x if = -o.013m1 M m: M3401 AWMADCeomifimofmsofid Mmskmbjcuedwmnanifimh oppodetieahnassbwninhcfimmllgusqw d‘llnslul midlands-d, =2.25‘I|.andlu¢th, =30‘n.: MMumnumqarlsnmmm I4=mh.mmaflimlwihlhumddu a=11x1¢pd.aumemmser, =m0mb-‘n. ande=8.w0IHn. Calculus the following minim: (a) line minim m an: 1.. in lb shalt. and(b)lh¢angle of lw‘nldr ("a deyus) aendC. M18" DC Tx- = +7: = 8. m “II. _ 16 rx_16(s.ooom-m "c :4 «1.15 'n)’ Txlc (woos-mm» =76mpl' file: = 4=225in A=3on 0",)” ‘llx'¢")(3_1tz)"'75i"‘f 4=l.15‘n lg=20im =+omfi97nd G=uxufpd =z).m0I>-in mMaxnlwmmms . SegnnDChnhemax‘mumsm Tz=&000l>-n. h=7mpfi .— $611311” (1)) Mmmnmc 7.41:7: m”n - + - —oonsm7+oons197 d , F—iél- Wm-” “brag-:0; ' ’° ”= «(2.25mi p. “-0. M- '1 F 1' (-Izmobnxaoin.) M: 4011 - Gum _ m x 10‘ “(3—1;)23 in.)‘ = - 0.013(1)"! tad man Farmmcntkdtoadrah Mia-1m! mwmtmn-gmdloubbshmmiuhum wound. (A) wuewmddumwun alumni“? (b) Wistengindoufidedhmdifmeiais hollowwithninidecfimetcol 10h? “ISA-1 with" 19.000154; «moth-m 3.000 [bar (5' g g A B C D [No “HI. 7...... - lemma Tut- +IIWIb-h. 1'4. - 4000 Ibvh. 1“,. - +mlb-h. (0) Souosum --_61 h d; _ I61“... _ lflllMlb-n.) - amt, in"... 100.0111") Wd- L78 5. ‘- 8.” Iris. /'\ (b) [bun-1mm mma- l.0in. fr 7‘6) 7m'— Th- 1, I. (II M lb-u.)(;) (%)u‘ - an m “owl-inch 56.0234! t—I “)1!!de - l0.” - a r-smzsd-I -o mans» war-Lash. .. FBI-M" Thumb-it u-ihac'lulrbr duo-nin-inuulcyldfidhdehulowznhenu phrmmdiuilotkmw-btme-I i mypTwTiswuflxflWMiw nx-LAuuchGisw-tu. (I) fi-lrsb‘ol math. (11) fidimflmhimmfi fiilWltZ (c) fidxq-Hwofli-MtnhhSdfi-W, 2 3 (d) Wishwflk-i'p'mlg? (e) Dmtemb-lmfim:nx).0$xsl.)nd I‘—1 L" did-cacti (TDD fix). 0 s x s Lumpy-u “OM-M (n) Imus nun-rally L.2,+_'L 1 -3 u 1 2"”‘0 ""(7‘3) "'77" “(1.) 1 x-_ ._ x-_L .— 0) 1mm “musmml £2 '7 2 '7 r (d) Rmmnanuxvathc) T.-—R. r.-I.sr 7,-— 2 3 7 r 37 FL (dmemalnwwmnmu‘rnnl ‘3- “ ‘2.— ” 6(L) G(") l' 'r m- 26,” s s rL_ r; +1',u.-x) .fl .— 6» 6(71) 6:, "'7" — P 8 T (e)1‘MD&TDD-summam (g7)! (-5)tl-- I! Micum-Tflorokafilurxbl; i- +— mmni:lba-mnjtl.¢znja2a¢. 6" 7 5” I03 G —I, 8 (3 3 I L- +-(l.- x) Profluu 3.448 A prisnmic bar AB of basil: L and solid circuhr cross mien (d'nmcmd) is loaded by a distributed (aquc of coastal! htnsily I per unit disuncc (see figure). (a) Dctmninc 1h: maximum shearsucs rm in line but. (b)Dc1uminclh:anglcoftwist ¢bdwccn Ihccmboflh: In. (a) Mam“ SHEAR smass (b) ANGLEOF msr mi‘ 1(x)=u I =33- “ = 7km: = 32nd: Glp «04' = imalsily ofdisu'ihncd mquc L 32. 15.1.2 d=diamctcr ¢=£a=moxm=fl ‘— G = sitar modulus of clslicily mm Asdidciclhrbxofdmd= Shun “88" (see figm)hlwi§edinawaimwhhennilmcapplied mmmheslhevath=SWNvalmisvaheofmua asuflwomma45°lolbaxisofflcbugivuamunihg e = 339x104. Whaismcsllamodducoflhemial? .Inn-monounsa-mayInna-annulunuunnluumo-Inn-unnnnnuonn-"mun-n-anm-nnnmu-nluau-omn-Ia-umu-nnuunnnuuun 1' ““354 "ham-uh m Stn‘m gage: 45‘: sign sum m 3;. 3-12) = “ 161' "BOON-In) ‘nn 339x10 Tm=7=—=m.372MPI d=50|nn 1 «10050.11? T=500N-m Smmwws Simsrumtmfiq.3-33) =2=m=mom .— 1. =2c_.=618x10-° Tmmmmm “0:!”qu Tu=94541b4m 21MT ”=33.” Hahp ".1an T=M Mnmummwmmmfls 33.0“)“ Thebmalmquoccusinsegnenfl = 2n = 161;. = 161'“ . 33.00am») f" and“ "an MmMzTA =— Zuloww) _ 16117.332w-in.) _ II THE, =|mnm «7500M ' = 17.332 lb-in. 4' = 337 in- 15 Mummmowmmwm Atme:T.=fi TA =78781b-m. I - If. — TL - 3m .50 ' 32 GI, .aa' AtpoinCTc=fi TA=945HML — m mam SegmlAB: = 0.40“ ¢ _ 32ml... 4‘ u -— '°"' HomAw62¢Ac=6u+hc=fl _ numaon: - mxo ”(121nm 4' 10 1.5x Hf psi)!‘ (“‘5' _°m6nnd 1.1052 . _ L507” _ - ¢u=T .. 0.02618- J' and d—ZJSm. Segment DC: Angle of twin gowns é _ 327.05”. d=1751n '- It '3‘ _ 320450 11mm £002 mm) nuns x nomad“ ms: ”and?" an M174 Amdeliveu 275nm: [waninwlheendofadnfimee figmlesuBandCukeom 13 mdlwhnmspecfively mummmdfiwedmimemmuemmwmm’mumofmmmn madwCiinniwdlolfi‘. (AmGr— n.5x 10°de, =6LandL, =4m A L ms. 174‘ 17-10 Mont 14 man may.“ {o 275 hp 125 [p [50 hp F.-noov mm»: A L. e; L; EEC ri'csl'IMID-h. Tcsmlb-in. _ A on a 43 c - t 2 j; Til-7373531 d = «new: T.‘ = 17.332 lb-in. n=lOGJ|pm Tc=9456nrim 7M=7mwi d=damut Mac)... = 15' = (102613 no T. = 7878 lb—m. G=ll.5xlo‘psi INIBMLtoIquu Tmmounssm Tu=l73321b-'m. M334 A solid cilcuhrbarABCD with fixed suppom is ,- xbduponbytuqucsToandZTLanlrlomfionsslnwnintlrfigum “ II Obtain afamulafurlhmaximmn mghoftwist “dumber. (Him: Use Eqs. 346nm b ofEuunplc 3-9 I) oblinthc native blqucs.) 1. Salmon 3.8-1 Grub III will mo and: To ANGLE 0F TWIST AT sacnou B TA A s T, 9"» sh» L L From Eqs. (3-46: and b): 101,, = = 7;,(3UIO) = 970L 7" = L 4" ¢“’ 61, 2001, T _ To LA ANGLE or TWIST A'l' SBCI'ION C ’ L 4’ _¢ _r,,(4u10) _ 3101. APPLY THE ABOVE FORMULA: to 1m: GWEN m: C CD GI, 56], 1 4 1510 MAXIMUM ANGLE 0F TWIST = _ + _ = _ TA 70(10) 210( IO) 10 3101‘ ¢m "' ¢C " 3 6 1570 50! To = 10(E) + 210(W) = —‘0 P MOSH Huminflfi Gmun. km“: 75 _ K“ m) ToLl _ R04 mXMm) L la!) mm 12. L. - no.4 mXém m) L anm - 0.I6 P 7:- - 0.24? A“? 1"! Ammmnfimsm" 7‘ UnmP-Nem r-Nmm Fran 5150-4633“ b): ToLg C 1"- L T. 13: LA 1‘. - L mm mmmuummmm X0,” mmcminpmcaoflhifll. mu mammal-nun m 1., - r. - 0.24? swam: medal viminouiq. I y Susan M mm a: 0 W - 2(45 x I0‘Nm1X36226X mam") run-“’2’ 273-1! I 0.053: 71-. - Tu— - d (54- I) " - 6520? N1: Uthouadmem p-“z'MN'“-sz 1;... - 45X IO'NIII2 0.24m - 1— ” -%(d§-J:) _ 362.26 x lo-oma P... Z720N d-dz-Omm mas-7 Amppedsmacnishddaaimmufionam Aallflndnbmwamm nmmuMmC(uefigI-e). Mtwsmoflbmwmmmamd. uddg. ”naively. andpda ms of hath I... “(5.Mpeaiwly. 'lheshlluhngandsegmAChaslemlha. (a) RummaLwinmem'lmmMmumbemgm ‘llbahsegmnudmeshfl? (b) Fawhlmhdeillmehumlmuubemumh bothqmnoflhem(mnc Unfiqs.3-4Gaaflbof Exanple 3-9loobuhlhclmfivewques.) SummACMAJ. (“=0 or (L-aM‘=d. SammCB:d,.In l.=L-a _ 2: 4n WWMEqL3-4Sandb) Solvefa'a/l. L d‘+dg ‘— = A) = (i) (Hiram—mom 7‘ T wm+wn 7' 7" Mum» n=r. a 1.1.4.4.. (a) Mmmm or (L‘¢M=¢ln f =TAMJ2) 1 =7'lua’2) a In AC (m c‘ In Solvefu'a/L: Z=W = flufl a 4: 74¢- 10 or I“ In (Ell-l) uz=4+d «— Salaam TA lid 1', into a]. (I): UM4A=MPI¢ a' M=lfl‘ 'm In Mal 3.9-7 A mticdly indctamimr mapped shit ACE is fixed I cndsAandBIndebdbyatuqucToalpoimC(mfigln).113mb scgmhofthbrmmdcdlhcsmcmmid,“ kWh-“la. andhlvc polarmomcnuafincnialm until". Dcmmimlhclnglcofmfim ¢dlhccmsssoclionl€byu§ng sninalagy. Hint:Usc Eq. 3-511» toddcnninc Ill: mincnclgy Uhlmnsoflll: anglc¢Tlrncqulcllxminmugytolhcwkdoncbythctaqmn. Comma your ml with Eq. 3-48ofExlnplc 3-9. Section 3.8. WORKDONEIY'IHEWTO 70¢ W T EQlMTEUANnWmsomeé WC” In) 70¢ _+— =— T A. la 2 ‘60,; +1.1 ) ‘— SMNENWHROMEQ-3'5m ("us In" ":th 3-48 ofExampl $9 ' ran ayccs w' .( ) c . U = - 01,43: 01m,2 + GM,2 Section 3.8.) (=1 211 21w 219 1 LA L, Mina." MeatamdaformemmgyUofmemfilmrhushowninlhefigm muhudlmhrmmmugthLfikmhjecmbaMWuqndMtpernnidisuncej'hc blemityvaieslimlyfmnt=oadnfmeudloamxim nlIeI=loauhesuppoa 34 am: I’m sums.” ammmmum x=dismcefiumrighMendoflheba Md; Considcracilemhldmdfatdmcffmmlhe right-land ad. “F. flag}..— Jr=exmlm|caaingonthiselanem dr=mdf =~(%>« m 4: AT usnnca x "x, m!" aar— Tu)=intemalmactingonlhisebmem m)=uallotquefiumx=0lox=x w» = M15494: ,0‘2 I Smmmovumdx [1!”th I to 2 ‘U‘ 261, ‘ 261, E) ”‘1‘ = % x‘dx 81.161, Smmmovmm L 6 Lx‘ U= d0 = 4: lo 8861,10 = '3 5’) 81.16:, s 3 U: %L 0- 4061, WHISJD-Z Asoliddrcullrbuhlving dimd-bbe rcplncdby antagulttubehlving cross-seem!!! di-ums dxutotemtflanlhledfitcmsecfitmOecfigurc). Dermincllemquimdfichrssrmdthembcsotlnthe mainumslrzmusintkmhewillnotemdfinulximun slrarstrus'nthewlidbl’. “3." mm 329 Sdtlbn 310-2 h and his A. =(dx2d) = 2:? (Eq- W) = — (E4. 3-12) T 7 (El. 361) '5‘“ zinc—m? memmmwmmesmuvemr Rmotmm 167' T Ifl>thhcsharstrcssinthehlbeislusdnntlr slnarstrcssintlrbr. Problem 3.10-9 Compare the angle of twist d" for a thin-walled circular tube (see figure) calculated from the approximate theory for thin-walled bars with the angle of twist (#2 calculated from the exact theory of torsion for circular bars. (a) Express the ratio ¢|l¢g in terms of the nondimensional ratio [3 = r/!. (b) Calculate the ratio of angles of twist for fl = 5, l0, and 20. What conclusion abou the accuracy of the approximate theory do you draw from these results? Soltnlon 3.10-9 Tun-walla! the ._ APPROXIMATE THEORY TL 71. = — J = 271'r3! = ¢l G] ¢l 211013! EXACT THEORY 4,2 = E From Eq. (3-17): 1,, = “—"(4r2 + :2) 01" 2 As the tube becomes thinner and B becomes larger, the _ TL _ 2TL ratio ¢./¢._ approaches unity. Thus. the thinner the tube. 4’2 ' 6—1,, ’ ‘n’Gr!(4r2 + ,2) the more accurate the approximate theory becomes. PI'IHIII 3.343 I! vathalpolcofsolid cirudarms suction istwistudby hofizmialfumsP= “mlbmfingatfllcmdsol'a hu'izoutdarmAB [sac figmc}. Tltdimnu: from 11!: oulsith: oftll: poi: to flu: lint: ofaction offich film: in: = 5.0 in. Iffllcdlowflrb all-arm“ in thpob i545“) psi, what istll: miniulm] mquimddiambrdnfinofflcpob? Sdllhn 3.3-1: WI 9* M4 F r =Pt25+d}d= lflPt2c+d} m“ wd‘flfi «d3 [qr-gnu]? — [Ni-PmI — 32h: = 0 Summmmm mm: UNITS: Mnds, Inclts («x-1500}? — [Jammy — 32(1100}(5.0} =0 ll «F — 124495;: —12.4495 =0 Sol“: numerical}: d =2.496 in. duh = 150m. ‘— L' P= 1]!)le a: = 5.0m. Tflw =45CIIIpsi Flllddfin ...
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  • Fall '16
  • Chen-wei shin
  • M&M, vathalpolcofsolid cirudarms suction, cirudarms suction istwistudby, approximate theory, Summmmm mm

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