homework 9 - PM?“ Thecamilevcr bumshouminfigfiflhas

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Unformatted text preview: PM?“ Thecamilevcr bumshouminfigfiflhas adownwarddjstfibmgdloadthatvariesfinurlyfiromw,” x-Otowzatx-Ljheeqfiationofdfisbum'sdcflecfim mi: ' vm-53:2[-&>’-s&)'+w<-zr-so (9'1 (aJDewrnfinecxpreafioufinrw.andw3;md(b)demmim cxpreuiomfatheremimmndfil‘. («j flerrmbar 24x w; “Jazz , fry/"‘71.: J = 19 (x) r ,- x ,t I, z ' - V“! m = M [4%] {(2) m/f») —w/-:-) ] ‘7' /Z0 aw gait—{(522%) i w/gx') ‘- mm] m; = EN" = 54—1 z['ZO/Z&J"GO/Z&)z*/w/£' )”°”] /&0 W) = Ely” = j‘jf f”“‘£fw”= We — 2’!) w] M=—f{g):u/‘) {’6’} flan/IBM)“ re; fins/Mt, A; .- 17(0) = fZ-Vm (=0 : Mag/1450]“: El?” _ = mfg/312 x-o 6 ug-yn’th 2w; [-60 $2.5) 5/20 {-5} I W] Prat-.136. The cantilever beamAB in Fig. P736 supports a triangle distributed load of maximum intensity m. (3) Determine an expression for the deflection ma, 1:0), for Lhisbemand (b)detelmineexpressions formeslope, 0., and the deflection. a". at and B. 501?: by integrating the second-order difi'erenfial equation for deflection. 03W "'1 W” 4%” E: : beam sad-.10 .0:- x t L —w.[L- ‘ 3| X H n L .(L-xii] Wimfi can/1w m. X) -x} f 2 EIT”: *WeEl.“XJ§ Mm -—x + wntL—XV' FHA law—int: +0.. V00 5 wa + Co: + Ca WW .4 ' “(OFF—ff?” O = [253+ CZ ][ + 8 Cl: 74; O = 13%? 1' C| C2: #6 Wx\=“—*I2%El§:: -(L—L"5-5&‘Z + I] rum-mu. Apemnmdingatend Coradiuingbom exerts a domd'l'eree P. as shown, in Fig. P1342. (:1) Determine expressions for the deflection curve ultx) in seg- ment AB and the deflectidn m 03(x) in the section BC. Solve by integrating the second-order differential equation: for deflection. (b) Determine the deflectinnlac where the aim is standing. and (e) determine the maximum (upward) deflection. a... - max 19.0)]. in segment AB. Egg: board mam oexe L I A A Mod X ET“? If f ‘ ZWX 1%sz = 0: M. m= - 2m '-—-—x~—»l \an EIVI. = 'H Em: board 5585mm Le x e 51. X abx w +J(2M)x=0; M‘zbd = '" WfBL—fl EIV” WX‘SWL MM -—-x 1 2 C1 ‘3 EIQ’EW—fa‘gmgc +D. um EIVWEJ-—zl+nx+bz ' ‘ :mohwnflamko ‘ ' ' ' - mm: wave 2 XML— O=EQWL5 Ci: 3 o=_ 3 3+C4L =3, C1“0 2 0“ ms + BLED; I D.=_WL5 -WL2+CI'= ig—+D. D2=T (ConiHJ 1.54 nztan-JJ W<x1=%%i§[’{ M[ (1 won = tag: L W b) 4+ La \ngL)” Lug: [mt—am + luvs) - 5] - E1 5 .5: MEL“: 33%? 3&2 WLS —-'| l Vi'fXJ‘EEE—L['3@Z+I 1: “(7L)” +75] Xfi’fi = fix]; 23 ms <55..=v:(x,3 = max M7346» AWlGXfiDwide-flangcstealcanfilembm AB carries a uniformly disuibmed load of we - 2.7 Idpsm unaspanofL-Sfi(seeFig.P13-26).E#3OX10’ ksL(a)DctermineIhemxim1md:fleai0n.5_.and (b) determine the maximum flexm'al stress in the beam. (See Table D1 oprpendifoorthe unss-seflionalproperfics ofthebeam.) a) . i I I, mm) o A E,_._x Rom-Fable, DJ, 'I ‘58 in4 6M1 d= rum in. P—"fi 1— —“—-—’4 Tabl LI 'La 1 ¥ro_m Wolf" _ (IE—IPHGFfS/S )(‘IFL/pginjmla‘mf _ (Em: 8E1 ' ‘8 one 1 51'8m “MM BMW beam ElectionJ 0 ex éL. -W.?(2 a; It“: "9(2ML'0:M051' M2} ()0 M / H—L—>k~?$.fl—al 5m”: W; 2) 2. 2 2d W30 :_ 353i— _(2.‘1‘-kiEs/fl’}{‘IL, in.}zUL.,o11nJ — 11%, 5:8 m“ =Mzzk51. Umax = No.02 ksi M1328. The simplympportedbeminfig. FIB-28 issnbjmdwadimibmedloadofinmnmyfix)= 1 ~w,(§);mbu;‘mmadamdowwu¢mmm finnxignintbeuprusionimpmJ (a}Usethefom1h- ammmwdomminethsuqmfimofthedeflufion mu(x}:and(b)demminethemsfimnmdeflmfim,d_ .mh(x)],otthisbcan(c)netemfineexprusiomtor themaionsA,and-B,. Infix‘t Wm: A B X b“: 5"“ 2S 2 — L —fl MM: 4300* 'W.(L) (I _ 02—1 {HEW ' WI? , -w.1c‘ V= (EEK) ' 61);: + C» —W. M= El'xr" Tel? + Cm. * C.L _“W-=C" CELL“ EIT" flat-13+ 921* CEKK+ C; — " I‘aL 2 EIxr= fi?‘ Ln a. “braved WM: Wch" \rm'o 3 M(o\= MCDSO w.L (3‘= CH. C s [2 0: 5% + L; + T+C5L c} C;"'O .5 «-w.L 0 C5 ‘ fio 0= ‘2 +CIL Cq=0 m = 5%[‘ (“3" + 5%? 4%)] 2 &*fi[-L(fl5+ 15?“) 4] ‘ \r'ixJ" 3w 1:” - - o *Evawolvu, x: 0.5521l9153L wme gjfiqfi ['(flh 59513441)?” = "5.232%0‘5] OL‘I Gingham}! = ‘ 5.28001) ET (J reactions at A anal B. V: —Wax’+ ELL A,=V'{OJ* 3’51 Av m J B: T "5'1! :2 wgfi Bf-VUJ’ 7r Pub. 7“ The uniformly loaded beam in Fig. P144 is completely fixed at end: A and B. [3) Use the second-order integmt'mn method to determine the reaetbns RA and M4 and to determine an expression [orthe deflection m 9(1). (b) Sketch the shear'diagmm. WI). and the moment dia- gram. Hts)- rnu, I'M-35, ml Ive-36 0.1me film) .439 I. . l I : A/leu B . . I‘e-———‘— —‘—-——‘—‘—)f 2 I EBB: beam Scolc'ion, ofix’, e L ‘II H1er misfifio: MOO: M1: Rm“ 22321: 13" EIv'EMHR‘x'rflfs wé Hth Maw Hewmefiflem R, 2 2 Wm EIV' rmf+fll¥+%§*alx*Cz WW: rte) ' Wm} = WJ " WM 7 0 “C22 C110 C Prob. 14-12. Solve Prob. 7.4-1 using the fourth—order inte- gration method. 69 I- 5’V00 WW ANKLE/ALE B I‘— L——*-~—>4 MM: 4300: "Wu (EIWY‘ “w. V “ {EIHT ‘ "Wax + C. —w.x2 M: EIVJ: 7; +C1X2+Ca EIv’ = '“J‘ + 9%— +sz +C; —w.x.‘* (Inc; Q2? EL” 24 * la * 2. +C5X+Cn| : we}: NHL} ‘ T’IL} = Mfo‘r = o C1 iamL (:sz 2 + CSL. a; C2: 0 2 -V&L§ T 4 + C3 C5 ‘5 zig C4: 0 SWQL LBW Fromm, inward-[man and bomdan} Condflfions in problm seal-jam (a), W,% 1%): 59:95- [a] ...
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This homework help was uploaded on 04/06/2008 for the course AME 20241 taught by Professor Wagner during the Spring '08 term at Notre Dame.

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homework 9 - PM?“ Thecamilevcr bumshouminfigfiflhas

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