HW11_sol - Problem 14.35 was 5on Pmb 14.9 using Moll“ aim...

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Unformatted text preview: - Problem 14.35 was 5on Pmb. 14.9, using Moll“: aim; 14.19T‘wplamsofuniformmlmfion IOHBflmmwddadmgudlu-uahm Knowing that cent-it: lN-kNfixuumappliudmtheweldnd plateslndlhltthein- planclhcu'ing aims: pnfldmmewddhlflMPmMmhe{a)fllemglefl.(b)flle manning nomalmmmdiaflartolhcwcld. was 10‘ 4. - "' 1: z: WW) - ‘IQEIII‘O PG. 125”?“- =0 me Huh-'5 41‘th (an sin 23:- 43; = ones (9* “La" i {In 6: 52.5+C3.5Gu5 233 «.- “7.3 ”Pa. Poi-I4: P? X: Y: (6}, 15.. ) = (-33:15, a) c: Q: m 6-: 1133 WWI-1.11m Mobil-'5 circle. parallel to diagram (5) themnml stress pajamdiwllrtomegrain, Sir—SHEA %:—l.8 HP“ 113: 6...: 2.551 = -2.'+MPa. 0w ruf- Hok‘l’h‘s Lift-age. (Sh-Tu} . 03er a} (6'...) D 3 1 (-2." H950) _ '5' 29' ~30' 12 -E-X'Sr'n$0 = - R Ila 30' '-' -O.6 sin 30' f - 0.3 MPa. 0.6 H‘Pa. ' 3 fin-fi'mw =' -2.."'l-O.G 60:30, -‘ —2.92 “Pd. 14.lelLlimpainofnwodmmmfmmufleaflffithfln untied. Fmfllcmofmmmhc{ajmein-planesbauingm O 6' (Him I “JIM Imammmmmmmwmmm Problem 14"” mkofplmmmulfinzfiomdwwflfloflhcmmfim shown. ’1 —-— m / 1 . II! + m léJTwofigidbarsACandBCareconnectedasshowntoaspringofconsmtk. Knowing that the spring can act in eidm’ tension or campmsiom demfine the critical load 1-"“ for the system. Let} X be ‘Hu fw‘l‘fif‘fa! Jcpleakw aPPof-n‘l C- x =' i1. sine fink)!” 5'“. nine Fm. Iain+ C:+1{F‘§ zozificch-Fmoasefi'o IF.” 1;. $§Fx Fae 'ans 5;“9 . 'iil'k" sine '-' (LB —-(?~.+ it‘kL‘IsSnQ '-' 0 {-1.3 “1%“. Join+ A7 +1ZFJ=Oi ”P"- quse = O P=-Fucase 3 'nljldlusé VHL 9-50 PW=15LL - Problem 16.17 15.17 Knowing that? = 5.2 1:14. duel-nine the factor ofsafety for theme shown, Use 1‘: = 209 Oh and consider only buckling in the plane offl'le mature. Jain'f B : From ~90“: ‘I'r-t'amjlc Far. .5: 2 fibula. 5% HS“ 3.:o1q an (“up 2.55: klu (cowl lie—fink“ As: I»: %(§)" = 3&3)” r 5.15'3wa’m“ : 51153::0“ m' Fl-‘J"= 13% = .fl‘ 2w'loll'1115'l53 H0". “'2 105363th N '= 1096 k“ I. .. EL. E5. =- MEMLM BC: I“ = E’ Ln: = L2H Lat . 2.35 w.‘ ' z 1 "' F“. = V‘EI-L _ 7F 1‘00““? (”34'9er J: Ismsm’w = 7.3513 In: L.“ ‘ 2.33 .. Fa“. _ 7.35:3 _ ES. - E— - 15.51 - 3.13 Sat-unfit!" RS. awn-n5. Es. '-' 2.27 ludTw columuueusedmsuppmahlack weighing 3.25 tips in achofdwefour Problem 18.24 ways shown, (:1) Knowing that the mhnm ufFig. (I) is made 01°me with :1 1.25411, dim, damn-nine the factor of safety with rcspoct to buckling fur the loading sham (b) Danube the dimofuch ofthe other aollmms for which the fictornfsafety isflicmnsfllcmufaafetyohninedinpuia. USES-29! Io‘psi. Tl" I = o. H‘NUZ in“ 843' ' FG lino- F‘E; LI 1 11' ”I"? .IHS‘H) -. 3'32: )1: = 3-722 kip. JFar- on: column, am. Lac-fen r Lanai». Problem 16.50 1650mm barADis mathedmtwasprings orwmnndis inoquilibfimin the positim shown. Knowing that thc equal and opposite loads P and I}! will horizourai, data-mine the magnitude Pu. fifth: critical load for the swim. L25 2“ and y: L: Hue JJ-Peakw: a; Pei-11's 3 mm! C, 901(th upward. Then FB = Lk‘ys PEI-k}: +fEF330§ FE-i- FL: 0 Fc=-F. y: 2 -y' Fa. final FL “cur-M A courJg 9 LB+ 9 Lo. 'Hue Ith-pe alumnae: y5=-J$_ = iQJ'ffl-e J 3: fish-:6 PM Pl ¥oflh a. cow)?! a GP armada-f P5 th'o; kKa'ga-siné)a_ use - PE Sine = O P: 12%:- case) LenL e - o p 2 1‘ .... ...
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