hw-3-solution

hw-3-solution - $313!? assent ECH s“ Homework 3, problem...

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Unformatted text preview: $313!? assent ECH s“ Homework 3, problem 1 Solve for the approximate deflection of a simply—supported prestressecl cable with triangular static distributed load using the finite element Galerkin method. Use a mesh of four equal—length finite elements, and number the nodes and the elements left to right. (a) Compare the midpoint deflection computed analyt— ically and approximately. (b) Compute the transverse force analytically and compare with the finite element result. Homework 3, problem 2 Find the natural frequencies and normal mode shapes for the roller—pin sup~ ported cable with uniform mass density using a mesh of three finite elements of equal length. Number the nodes and the elements left to right. Integrate ele— ment Wise quantities exactly. (a) Plot the normal mode Shapes. (b) Calculate the relative error of the natural frequencies. P.” epfl®fi®fle a? l;_L__.m HOmework 3, problem 3 Consider the free vibration of a pin—pin supported cable. Construct the “Solid- Works Simulation” finite element model to compute the seventh natural fre- quency of free Vibration using the cable-uniaxial rod analogy. Use both draft— and high—quality tetrahedra. Find the uniform mesh size necessary for attaining accuracy better than 1% in angular frequency. Compare the lnumerical results to the analytical formulas for the prestressed cable free vibration. Produce a brief report. ‘ yrs» mm; (EEC “224mm Pu’fl-fl) flax Mia: Mam-T“ WM Law ,[K]{w_} + {L} :50} Wyammmm ‘ (I: v v f v (73% , 7‘ 117%!)33 {flit/V2) fig? 35 [,1 [l A UL M MA at ojflmfl z. W Ia *- iéfl/‘A~©) fi ~ij [I L L/ A L i V I [kw ‘3 M Wm)? 70 pyjmww V , D ‘1 . i, I l I UL. r [Mzflm NMNMXW WWW zépgwfydx ‘D L L. » a t: :1 Mail" 9"»; V WWW 0 X2; 3 3 2/???) Y Kata) i j/ q £44 + gm {:3 / _‘ 0/“ 273%! w’: ~3§)&ZFG 7,0 L l m. 2 ,wcr i)£%fi¥+€a fl. 3 jammy My} 2 X31; 3%? 31% Li, w/rfié): {L2 :2 "~' flak 3W Lira? '29}? v g 5;» x3 H. (@Mflflié” 5am paws ,4?" JV 3*" 5/1 ,mea gm,- M Tmmg Wg’ iii} 3 L2 flflk’géwvmfifi” 3&5??an («/3 '3 law 3‘" 5.3/2" #32 P I2 P V W; Mm wmmaflemmmwmmrmWrmwmmwmg « memummw /7¢22”AWWMME fifiiWfifiW lg" mag?" 7" 1/ 3 V21] _ WmeWWWMWumstwwmwmmu Wtwmmzmmwmwwwmmmmmm _ -' .mwxg“ W65” WMij Wm WEXHfir “Mm-f i5- Mérfiwa m. W} L, a}; ,1 / / {bf/"5] 69W”; #7ng +— yszWZU wow/Or @510 : 71/1ij Wgaww mu [5)? my; ' 8 0/9" VJ 19“) 3 Wg’waea» fifiijbg 2P/*4E)(;;;z)%"+%)(fiz)$7 ‘ W Pu’wzgié WMWMME 631“ [VH1 646 Force [qL] 0.1 0.2 0.3 0.4 0.5 x [L] 0.6 0.7 0.8 0.9 7 x. fix" '4‘ 1;: Eg‘.‘ ' Nam: $X/L ~ 3 Nmtx’) =1 “lax/HE fig/w: @ WW,Wmewiwwimw ‘ ‘ é ;( {fig in mm v Key, v D] chmw awe? WWW 5/ 19" if} 3 I. j V § “22 3 f'l'ifiméaém‘i‘ ' L via 5’13"”; M fifya Wk m wwfifir‘L; 3,“) D “r; 0 MM “W i L m w 2: [a w J «:2 W I 1 3% “N23 5 \4 *" (/31 M L, % :13 fl“? V. m m [\AmrmfiéK W) (was; vwf‘osmca @3081 Q $47.2, Q 10133 D‘QQQO “G “153157” 0 370$ mama 02%! a <99?“ D: o 0%} C) o 0 055000 Q O O \eflsg >4: 00"“)?- _ 31:03 1 $33: Lé‘fié wzm 2 02915888 3;, I t m; u; L war—«SOSWWSTF 1, wzztlflflfl ‘fL IL 43%? 41 E ()3. 1:; {33% ")x 5‘: 30m AL wwmflHC/WE Wm” M @390: fixgfiu '1‘ f: ‘ “§f‘§;}w¢af:&mémk @E—Z’Qu £34 ii. WW5 QM Wat) '2 E exvflfit) S FATE/1k Rd 3‘. : [g P WAVE, affmm CAL clear r_w close 2;: L=l;q:l;F1=11*q*L/48;F2=5*q*L/48; xx1=[O:L/200:L/2]; f=@(X,L)(—q*X“2/L+q*L/4); XX2=[L/2:L/200:L]; xx=Lxxl xx2(2:end)];n=length(xx); f0: i=1:length(xxl) yy(i)=f(xxl(i),L>; yy(n—i+l)=-yy(i); grid ,. :;hold plot<xx,yy,“~fl',?1"cwigvi*,2); plot([0 L/4],[Fl Fl], #k“',‘,irfifiif";’,2); plot([3*L/4 L],-[F1 Fl],vw§:‘,f 'T& Lfi:35,2); plot([L/4 L/21,[F2 F2],*~,og,§l;mfriqu3,2); plot([L/2 3*L/4],-[F2 F2],§~.:2,‘TLVfi1 mfih¥,2); ‘,. “0 var“, A, i]; 171 ,»,y ;>; xlabel(vf ’»;’);ylabel(“a/“v syms 5 i ; u “ Nl=—3*X/L+l;N2L=3*X/L;N2R=—3*X/L+2;N3L=3*X/L-l;N3R=‘3*X/L+3;N4=3*X/L‘2; kll=int(diff(Nl,X)*P*diff(Nl,X),X,O,L/3); mll:int(Nl*u*Nl,X,O,4/3); m12=int(Nl*u*N2L,X,O,L/3); m22=int(N2L*u*N2L,X,O,L/3)+int(N2R*u*N2R,X,L/3,2*L/3); m23=in:(N2R*u*N3L,X,;/3,2*L/3); m33=in1(N3L*u*N3L,X,L/3,2*L/3)+int(N3R*u*N3R,X,2*L/3,L); K=3*P/;*[l —l O;-l 2 *1;O —l 2]; M=[m11 ml2 O;ml2 m22 m23;0 m23 m33]; L- ' Mn:[2 1 O;l 4 1;O 1 4]; , Kn=[l -l O;—l 2 —1;O -l 2]; [Vn Dn]:eig(Kn,Mn); x=[0;1/3;2/3;1]; modefil:[Vn(:,l);O];modefil=mode_1/mode_l(1,1); mode;2:[Vn(:,2);O];mode:2=mode:2/mode:2(1,1); mode_3=[Vn(:,3);O];modefl3:mode_3/mode_3(1,1); xx:[0:l/lOOO:1]; far i=l:length(xx) Modefll(i)=~sin(l*pi*xx(i)/2/l—pi/2); ModeflZ(i)=—sin(3*pi*xx(i)/2/l—pi/2); Mode_3(i)=—sin(5*pi*xx(i)/2/l—pi/2); figure;hold “4; p..o:(x,mode_l,‘fiwfl,E p;0d(x,mode_2,“i Lg," plot(x,mode~3,‘~*“‘,‘ V” legend(’3 L» ifi,‘::,w ’“ ‘ * , b ’), plo:(xx,Mode‘l,‘ww ,*,i_ax _, plot(xx,Mode)2,V' 3, ;; av*':,“ plot(xx,Mode!3,bfiiiW,S 377<W‘” Xlabel($a [Q]');ylabel(7‘ '5); title('$5ri“ 31*-: “3?? 9';N ...
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hw-3-solution - $313!? assent ECH s“ Homework 3, problem...

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