homework 10 - m 7.5-3. For the beamst in Fig. Pm,...

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Unformatted text preview: m 7.5-3. For the beamst in Fig. Pm, (a)deler-. mine dimfinuityiumim expressions [or the slope 0(1) udthcdenecfionv(i]totos:51.;and(b)evalmthe displmmtfipoinuflandc. .4301) ‘= (EIW "’ |\4’1»<-.|c>"°'"135.60"1 +4100” " gLa‘kx—Lfi' - (vi-L9) erm = [EIv'1‘= M;<x>" +R5<x>° + +3.60: - ‘% (cc—a) - <X~L> Moa= Elv" = Maw“ + Raw' + “92‘ 092- =§IE(<x-L2>’ - <x- L9) Elem: EIv' =Ma<x>' + %<x>‘ +96 <x>‘- i%_(<x~ %>“*<x—L>‘) + 5137(5)? co EL! = 2 (392+ Ef<x>5+ <70"— fikx— lig‘ (1-D?) + E]: 53:20“ boundary condif-10n5:V(LJ=O -= RA 'th—“ ‘93—]- ‘* R4 = fi: L.1 " Mm=o=m+mu%- 4%? 9* m % :29 I _ é em r 24 El[?L‘<x> "1603 + 409‘ — L (<x— é-‘J - <x—LS'fl 4 \rth = ago—"9%: [as]: (x)? — 50L «>5 + Io «>4 — t (<x— %>‘- (x— 05)] b7 ‘ £2 2. L ‘| NA? = ZHOEI [SSE * 50L(E]3* 10%)? ’ 15—72%]: :3.” 'c‘. '5 | L” HI)“ 2405; [mam # mm; + mm —4 (%3 ] = LIE-5'3? v(%1=%€& Jl “ VII)” meoé: Mew-15 through 15-18. For the clmped—clmped beam shown. use the discontinuity-function method (a) to determine expressions for the reactions at A and C, that is, RA. MA, R0. and Mg; 03) to debermine discontinuity-function expressions for the slope 30:) and the deflection 0(x) for O S x 5 L: and (e) to determine an expression for the deflection at point B. Walden: pod: (En-“Y- Mm“ aa$'-P<x—'%>"+ R; GOD-1* M: <x-L>“ Von -—- 631?)” M,“ «W raw—Pa- e: + :2;ng we a- L>" Mix1= Elr”7Ma<x>°+R;<x5*P<x—%$'+R¢<x-L9'Mc(x—L>° 0 Stew 51‘1": M. «>3 +%<x>2- % <x—%>"'+ % <x—L>2'" MC ()9 L>' + [2.3925 <x>° 2. O 9 Hr =%"<x>2+%<x>‘-E<v ‘§>’+%<x- L>‘- Mé‘ <><-L> + new» 223 W2V(m“o=RrP*Rs [24'22 'QEL a? OEMI+RALF P(%)'Mc .3 MA" .. , t _ _ BE ,2— - 1P hawdanjfiofldfizmmEIfiL‘] “O ' MAJ 2m '5 R5 :3,- ammo=%(L1*+%(L)‘~E[%F MB- 22% P —4 1E ~;EL R??? Ma'% REF 2? Me' 2? 65.5mm in” 11:13 Efingfmns _ From “the in—chfafiomldo'sure. awaiions,md bowdfiu‘j CondIEions creed-Lon (a) P L 2 and" EEL'SMX; + moo2 ' 2‘4 <x— §>] p \rbCF #quEileLéc‘f moods - 24 <29 %>‘] C‘dtflznimu _ I a I. 3 v(%)=ié%i[‘lzL * 20 " {33% (g) : —%PL5 \J' 5 2'13? E1 M7546 fllmugh 15-18. For the clamped-clamped been shown. use the discontinuity-Won method (a) to determine expressiomsfordreteaelionsatdmdcmmigflmMAJOand Me: {13} to determine dimfinnity—fumfinn expressions for the slopefix)mddnedeflecfionv(x)fur05x5£;md(c)to demufinemexpminnforflndeflecdonapoimfl. a) . $=WM Mm 5| 9 51V“ Ma ' M" a l 77:3 E g L. ‘3 l. a), x A" B G K Wan. 4300‘ (EIHT’ = Mu¢+a<p§l + 10066 " gflw‘ '<x‘ 159%“ Rea-ISLMA» LSz Wfl=fEIv”Y*MA<x>'1+Ri<x>’ was; - w? (602* -%§}+au-L§- Max-L9 Mod= Elr”= M.<x>°+ aw +102" <20“ $105 ‘<x‘ %>‘)+R;<x—b' * MJX- L.>° o EMF F,va 54.45% ®3+¥f<x>§—fl(¢-<x- [39% %<X»L>?" MAX-DI flfimi Elv = M2¢O<>2+%<x>3+ “33¢; — fib‘dd—a- %§)+ %<x~t§- h—gfiq—Lfiafiéafi W2V{M‘OERA+F°L‘%[L2“(L5Y]*R: @5131? M(U)‘0=MA+ RAL+%(L]2‘§&[L5'(1§F]_M¢ .) Mn: gage; L WEE-1L1: o = MM %(LY“+ %(Q‘-%_[E‘-(§3;]] Re“ "432 3+ — 5 EIth= og%(02*% (ll % (LTL %[f-(2 — .L Re" —g|0_ fifiE‘I‘bBLz-ai — rosL <x>2+ Moo «>1 38%?" <x- v00 = H%5‘ET[23 Deaf" ?2L<x§ 1' ‘30 (x; ‘ %(<X>S- <X’ 15$] cgdegecfim E. L a L a L 5 U1 HE): Heal“ EI[25L?(2]z—?2L(‘2’) +%o(‘2‘ "Ca 1: ' M7514. Ans-in. (nominal) mdudmflpipemas aunfilmrbumthnshppomms-flplmdsfiomhanm Sin. rods as shown in Fig. PMS-14. (See Table 13.7 [or the m Hm mainnalpropertiesofthepipa) Dem-mineth “mud vintsecaiunflanducuthefightenddatfid=29xlfl'kxi. NPG fleffl'cfi'anr V, and fl. I . Skips Slaps ly,vm “a ".644 "1.5-9 % fi'B—% X - («1 C——* Lazy, P: {My P p Mm fl/é x27, [var/5* I}, m ' from am am}, J i = P1: m A B c x og‘ 245! 3 6m. a.“ 0k ’ 5791'. 4351' P 5/92.? {C} 4 B are): C K age = a}: , m" a; 35! M m? 4%" vac-4044')? r 1 ~ — : - -——. _ w I” _ ' 3 a“ dé‘ {45! 4%! #55! ' fiflflm’hf 72.575!) ' ‘M’wm- L’ 3 3 . a Vr—d‘ -Jr :3”? -fl - 7x05 , -7@44«J//4mJ . C [5 If ‘ 3 'fi/067/fl. M7566. Use superposition of beam-deflation solutions from Table 1-3.2 to solve Prob. 1.4-4. a) . .. ILjinx) “Pa ng. (M A B ' +-—— L——~l ‘JJW X 2:. Era 9:: A 8 Fig.5!) k—L~——>l MsL , QQ_ n Haw-e» "Garza: 3&1 ‘- , _ - 0L5 v; =eab+ 93¢ + 653.; = 553?ij van " vgm + vzm + V300 ‘ ‘E [(flq- 269‘ Em Table. 5.20%“ '5), 9m Gab: fifi min = 3% 151sz + w) hmmm I32 (Hm: am. Bu = ‘Ef‘e‘i esc'M‘L'LEI V200 " LLEI{2L2 - 5Lx 40(2) xvom Table 5-2 #0, T MEL. = em 19E; 93a 35:: v; m -' L, LEE [2L2 - 5H m + (L—xfl (L2 -x‘3 a 'MX = LI L lotIz a, MA 1’22 LL Prat?“ The cantilever beamACinFig. PMS-4311118: mammofinerfial= 50 x 10mm‘andissnpponedby rodCD,whosocmss-secfiomlamaisA- 200mm’.Let E...- Efl.A_mnvnodmdP-25kfiaap§fiedm thematfifl‘herodcbisiomficepdonbdpiifléafion dtheloadflbetermincuuwmimindmdindeD. EanaEL? 5.! (item MN, éch: L—Ef (EL-5) x A c. :ESCD P*25kr~1 film-(n T :I‘Dm #5‘ Fl' .k] I 2 I A C, $66 |<—'——"—L‘bm ~———-‘r1 -Ih __ P 2 TL!» w: AE = L‘E‘fiSL‘dj “' 13—51 BE .11. L_"’- "' T 3 Eiflkfl’mfiflw 521‘ a tum“ " a m = @5000“ mm [3[bm3‘4m][200rnm2 +{5350tmocfimm“] "‘ @5323 ...
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homework 10 - m 7.5-3. For the beamst in Fig. Pm,...

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