Homework #14

Homework #14 - H5 filiZa’US _ I ‘__ HW #14 Key...

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Unformatted text preview: H5 filiZa’US _ I ‘__ HW #14 Key Tahlu-Gmup I 6.3 Member AH nfthe [mas is subjecled to a IflflU—1b tensile force. Determine the weight W and the axial force in mcmhcrxtfl‘. r- 1"" x ‘ ___ __L__ “A! I Anal-if A _I' 1 195$ '1. . 'H _..---'——-_-__' i H. ’ LL60 in—fijnfil} in a . I‘ + hulln “cal #21:; 2: O 2-— —Mdam}zgg: : 1 ’ o i.B»'-' ' ;‘-(/'§- . ‘by/ EACH} 4.: U W ER T5 if = = "-fUOO whims? 7 “ A; 9:44‘? '- W fl 9% AC- =fll2.é»4*.cr: 1269£k¢lb WMFF. W : 447.; {g 6.18 The lengths oftlu: members of the lruss arc slmwn. The mass urlhc suspended crutc is UGO kg. Determine [he axial] forum in the mcmhcrfi. __________.. 6&2 WWW-(9 rem/“32’ch EMM— (chfl 0L1.wa un‘f’ f5 :14 ha), “Si/“p.10 : fi(l$mzz§3) ufiflfifdm) _ “ ‘ffiflff. 3’ Egg}: :3 Hmfiw) _,_- "B: | L60 m )mL-jf Cl; *r-F'fiF- 55’ '— “EvCLng’DH Cbmglio FEE—pa x #17553 5“; '17.}? 9:: E: (15% w} '\ “3/ +1£5=0= fideM‘DFC-Dfi-LL‘F‘D 51:. r‘ -- 4o": / iL—Li flint-E C. p- cp= film; : inn-'1’;- w «mph Mi 1;. _______. .19 . flqu-b Fit =C‘) 1:. ’01-5-[3 5M (I‘D " 5:14.55 “two i:' dgb C455- 2?- an} / WW w M 6 (21%: p -I- 4.21;; = 0 = H [1.60 Hasn- Les- #0 +9.75 mm: sue: 56.1.2153) f. O :3 0 Chad; Am‘dtb. a, E Fx=0=H-lbo+l.sl «um-27.3.3 # no.2; mam “.[afl 46a —;- 2F]:O:grg3+hmmn}8 «9-13.2«440 1m Matlab Ex. 1. See the file "Matlab 2" in the Matlab Documents folder. This time ttse the approach we discussed itt class in which you draw ALL the FBDs for all thejoints first. showing all unknown member tbrces as tensile and showing the unknown estemal reactions in the directiotts assumed on the FBI") in the documettt. Use either of the 3 Matiab methods discussed in tltc document for solving a set oflittcar equations. Print out your code attd your results from the cotnpttter. Make sure you get the sattte results as for tltejoint—hy—joint approaelt we ttsed in the LEX itt class. First draw all the FBDs for all the joints at once assunting all intemal forces are tensile and any directions you wislt tor tlte estcntal reactions. Then organize all your equilibrium equations itt matrix form and solve for all the unknowns. A negative sign will simply indicate the assented directions are wrong. and specifically for the member forces. a negative sign will indicate the member is irt compression. ln lvl'l‘l [SC sou. toast of you will do [some already ltavc done) a tniss analysis in which you solved a large set of linear simultaneous equations using lvtatlab. The method there is tnorc complex than our simple equilibrium approach. There you assign a “stiffness” to each member that measures how much force it takes to stretch or eontpress it a cenain distance (actually a gag distance}. You then minimise the total potential energy in all the members of the structure and due to the external loadittg. The minimization process creates for you the set of goventing equations rather than doittg as we do itt ME 2:31 by applying Newton‘s Laws directly. The main difference is that. ifyour minimisation process is done with respect to small changes in the positions of the joints {as the members stretch or compress), you will first solve for these movements {deflections} and back- substitute into member stiffnesses to find the member forces. I believe this is the approach used itt your MT] [SC sou course. You will also use this approach lit a simpler way itt ME Elli artdr'or 3'36 and you will call it lC‘astigliano‘s Principle. if your truss is statically determinate. your resulting forces will be identical to those we would fittd using Newton‘s Laws directly. But with the energy minimization approach. you canjust as easily solve statically indeterminate trusses and in these cases. the gov-eating equations will account [hr the relative "stiffness" of the different members. in such stmctures. you will find that sonte member forces change its you change the stiffness ofsonte mentber relative to others. It. the structure is “estemally determinate", these members with changed stiffness must be those that cause the truss to become internally statically indeterminate. As the stiffness of such a member increases. its force will also increase. Conversely. as that member’s stiffness decreases. its force will decrease. If all members that cause the truss to be internally statically indeterminate go to zero. the truss revens back to a statically determinate truss. However. if you let a member stiffness go to zero that is not one of those that are causing the truss to be statically indetemtinate. the truss will collapse and your solution will indicate that the governing equations are “singular”. i.c. cart get a solution. As we know itt ME 201. this is also saying that the static truss is "unstable" or has entered ittto a dynamic condition and its collapse would need to be modeled by a dynamic analysis method. 0a approaclt to such a dynamic formulation is to tttiriimixc tltc total mechanical energy. i.c. the sum of the potential energy and the kinetic energy oftlte system. We will actually do this type of formulation later in ME 20] . but not for such complex assemblies as a collapsing truss. For tltosc ofyou who find these topics interesting and want to estcod your abilities to solve more comples static and dynamic problems. c.g. to solve problems wltcrc we do not have to make so many simplifying assumptions as we do in truss analysis, you will want to consider taking ME 413 (Finite Element Analysis in ME Design) as one of your technical electives. This will build on lvtF. 2m. ME Sill. ME 31116. M't'ltSC see and all your olltcr math cottrscs to allow you to solve very realistic problems using computer methods. There you will be able to tnodel a tmss but not make the assumption that thejoints are pinned as we have do in these 3tltl-Icvel courses so that you could investigate the effects that realistie welded or bolted connections at the joints have on the beltavior of the truss including the deflections of thejoints. the member forces. and the strength of the truss. You cant even investigate what happens when a connection fails or a tnentber fails. Does the truss collapse (the recent disaster in Minnesota is but one oftnatty examples -- check out the site litigation.wiitipedia.orgrfwiki.-”[—?l5"l.'vr Mississi i River bridge} or does the trussjust readjust its internal forces. increase its deflections. and “hang in there" until some other failures occur. Hopefully the benefits of statieal ittdetemtinacy are becoming evident to you. Example 2 .513 a hmncwork problem due on Friday. afar. solve the Imss problem. EXAMPLE 5.? using MHI'LAH. Th} 11:11 use [he suppnrl mucliunx gix'cn in the example. Instead apply [11¢ Inclhfld ul'juinls ElT. each 01' [hr I'uur pins and win: I'm' all uighl unknuwm uh'im:I MATIAH. 115:; Ilu: tuuurnf'unkncm-rm :15: linlluws: fr,” My. .13, .H). BI]. HE ('D. If]. (1.}, when: [he dircclinnsnfthc suppm‘l rcnctinns an: defined in the FED and lhe bar Ihrcer'. are Elfifiun'li‘d IL] he in tension. . )9 Ag, .5 emf A : é -——_ —_"5 c}, T A” Pl 1 _ a; 3’ —a “CA m 1?— Fy” a = A fl’rflfixéfi'mij (E) J'JVG'": A 51' “D 3:: 3D A? Afr x" F__q_ @ PF? :65 E:fl :(fl : afificfiszS-JS ffipméfliflfii‘m: f:— F/ QEEEDGQ flQfiJIEJFffiggrw m‘flx c.1115-3al} $59.6» GIN—49 mxi-ffl—TW Wtflflr” Hull-m III“.- _H«:2 a—wflflr 5w “La. MfiTLfirbfi.opacJ f§1 A1, A65, AD, 6515,, M3, cal)ch (2/ ow?wa lbw NIH-4'3 t): KH‘E. CD .QDA’f? Ha) : O (3;) Afrua‘ M: :0 +IQDEZI> :0 6’) "‘g% fl'SED : 406 C9 - Er; FCF =0 @ —CJ:> '57 : :33 “AD misfits =~£poo CS? .3635 +d> = O '{a I O D C) If) D g l A? *9 0 0 0 c» 0 fire 0 g "C; fl *6: I o o 0 H9 0 _. 0 --.'z? o D 0 0 " 40D {3' D E) o _I C, _I 0 6c '— D N O O O a ‘3’ "i 0 fl: 51> c: O b -| #:50 D a O 0 C“ EEG? O a o .E o I o 0 6/ O Here is the III-file used: %Input E equations as shown an previeus page to solve for the eight $unknewns Ay, AB, AD, ED, EC, CD, Cx, Cy. The E unknowns are stored in $urder in the vector x. Nate the directions assumed for the external %reaetinns are as shown on the FED and all her ferees have been assumed to %be tensile. Therefore, positive bar forces indicate tension. negative %indjcate compression. All units in N. ' a=[EI.61|I|-DDDIDI;1 .BDOflGCID;D-.I5IJ-ElDDD;D-.filfl-.EGDDD;D GUU-lfl-lU;UEUGfl-lfl-1:UD-l-.EDUUU:UUU.EGIUU] h:[U;fl;fl;4UU;D;fl}-Eflfl;fl] suinutal'h x=afib x=rref{[a bl] Here an: the results from MA'I'IAB: a _ Fnlumns 1 thruugh fi [1 {}_t5{}[} I .[lElEIfI E] [J [I I .flflfl'fl flfiflflfl 11} i] [II [I D —{}.fififlfl fl flflflflfl I .Uflflfl G U -U.Eflflfl fl —U.Eflflfl fl 0 L} U {1 D —l .flflflfl (II U H H H H -].UflUfl II} [I -1.flflflfl -fl.fiflflfl l] 0 If} [1' Cl' 0.3000 U l .flflflfl Columns 7 through E U U '3' 1'] fl [1 f} -I .flflflfl fl 0 MHMJUUD -T5E}.UUUU 45flflflflfl ESUflflflfl -fi[}U.flflflfl —?.U{J.Uflflfl em} . [Jenn lflfiflflfifi x: fiflflflflflfl —T5fl.flflflfl dfiflflflflfl Efiflflflflfl -EJUI|.flflf]If} -2flfl.flf}f}f} fiflflflflflfl Eflfl'flflfifl Jr: Iflflflflflflflfiflfl UlflflflUUU-TSU Ufllflflflflfldfifl U fl {3 1 [l [1 1'] CI 25!? UHflfllflflfl—fiflfl' fiflfllflfllflfl-ZUU Uflflflflflifl'fiflfl [J U fl l] E] [II D l2flfl Se we see the results are the same for all 3 methods and they also check with the results obtained by hand in Example 5.2 all yuur text. 6.4] Tho Pratt hridgo truss supports. five I'orcos F L Ctr-Hi kit. The dimension L — ‘3 In. Use Lho method ot‘sootions to dotomiino the axial toroo in mombor JK and also .IK and U. 'I'hon douhio tho hoight oftho tl'llSS to height - 2L and rosolvo to: .IK. L1, and Al. Conunont on the effect of tho increased depth of the truss. Now let the height of the truss bu.- doubled. f ,9 2M5: 5): F0.) +5Lfi2‘9-Kmkt.) i= is if i= r F":- {33 M333 = IJCLL) nfsrofl.) up. ZEWS 1:3“: 4'15“ w mm, .7‘) ease-T 4' 135 m J; A: Jrze’. F2: 1 *3 :4Lh‘v—A’I a! i J‘F'zf far: : 4-13 ILA] mat-G“ F5- 6.54 The truss supports Inads at N. P. and R. Detennine 1he axial forces in members IL and KM and I'M. 1m .111: 2m 2m 2m "7‘5" 1m _ . : J_ . : 'L Ar- . h. +'. a, Ikxlkxlkx EL ‘L (L L Mm =0 = {mm #0 )(L) ~(aX4) We») IL: — “5’ 1‘”? PM afimFr‘ f1 III ...
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Homework #14 - H5 filiZa’US _ I ‘__ HW #14 Key...

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