135B Midterm 2007

135B Midterm 2007 - 135B Midterm Avérage 76.35 Problem 1 ;...

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Unformatted text preview: 135B Midterm Avérage 76.35 Problem 1 ; 26.13 Problem 2 ; 17.58 Problem 3 ; 21.63 Problem 4 ; 11.03 I 00-96 95-91 90-86 85-8] 80-76 75 -7 I 70-66 65-6 I 6'0— Total Zoo-had N-hCOG) 40 CEE 135B xam February 14, 2007 a L. P. Felton Use the displacement method of analysis for problems 1-3. (3 O) 1.'~'All bars of the truss shown have the same value of EA except for bar CD, which is defined as axially rigid. There are vertical forces F applied at joints B and D, as shown. Write down all matrices required to find the bar forces using an exact (not approximate) formulation. Solve for the force in bar CD. (20) 2. The two forces F are removed from the truss in Problem 1, but the support at C moves horizontally to the right a known small distance 60. Write down all matrices required to find the bar forces. Do not solve. ” (30) 3. The frame structure below is composed of four axially rigid beams and is symmetric about a vertical axis through C. There is a horizontal force F applied at joint C, and forces F applied perpendicular to beams BC and CD, as shoWn. Note the directions of these forces. Use symmetry to analyze half the structure and write down all matrices required to find the beam end moments on that half of the structure. Do not solve. F F .9 1 ’. E / le— 412 Luz—q (20) 4. Analysis of the structure in Problem 3 leads to an end moment MAB = -3.0649F£ (counterclockwise). Use this information to find a_ll end forces on each of the four beams. Sketch free body diagrams of all beams and of joint C showing magnitudes of Q forces acting on each free body. ‘1; M?" QED/g VF“ ’1 fix ’1: AA} 0 W7; \Mq qué F AC 0 0 $3 w "/5 I 92:] i ’ O a?) 7‘ [3&2 V 0 “3 $33 _ a0, $2 :55,» ¥ M — AD Pa) 9 ‘NL 50> \ 5? Pg; [Ca DBC/ 0' P, - 9—. r. 3 “KM £53m :éfigfipmfiefiw mm? r, ‘L 6D :: —E_+—O.(gfé>: \P C F 3 ED : — \A’fi‘zg? Nokc Cog/Mitre; awvmx.ma,4a éffiwkérvx 019+ ,' 3 % o ( g“; 0 0 o o l i— V,2— -l O EK/ZL O O 0 AF \ I3: 0 ;_ K: o 0 0 o E: $ _ a E I 1‘2 \ W52 O O l O O ‘ O i 1 0 , \o o o o I ' _ 0 116667711 \ 3.37151. 0 _ —1 _ EA k _ u 73570239. J k ' c 71Q692L ‘ 2.18381E‘L _ EA —O.436762F\ 0.”??"5‘14 _ 3.63968FL \ ‘—/d—‘:\L‘t— —O.l85728 F _ EA - -.2 26‘ . _ _ u— [_ 15,8567”) A— _‘__5_‘b_E_L_ P_ 1.26266E‘ EA 0. E21 3.63968FL \—O.606614F 3E“ . o o g 5L 0 O 0 O _ _l_ _ 1 l O 9; O O O V? V? V? 3 Vs L 0 ELI“. [3 = 1 0 0 K = 0 O T O 0 F = —E‘ _ 1 1 zoom/'2' EA ‘F V2 0 o o o ‘ 3v; 0 EA 0 o 1 o o o o p; lflg 633A _ 1176.395}. _ 0 1178515?- 3 7601351. 0 160034 L "I L L L EA E111 k _ _ 11'? SEER 1178.63 " __ 0 117351'“ k—i _ O “600’4 1. 0 760781. 0 " J. L L ‘ EA EA _ 0 117351 SF _ " 11755133; 0.3565183? 0 502511. 0 5027231; 3 L L L \ ET: ‘33 —O.436791 F ‘v —O.185646 F u: —~ A: — P: _ 3. I _ 3.960678763FL —l.59986E‘ 9.6399254 \V—O.6O6654F, 3% add—Lax } W“? D @ Hcflé‘é \ P POP : 0‘ OW? flga/K we, °D LUZ CW Aw (Kawjyméw ow ébféfi’EéiiWng'L ‘5‘53Cw4'0‘ru 04 M (LAO—T, Ev'é 7x253 5am“- 50 0 {$6 2 0 m4, : Sb Untitled—4 ProblemZ: O \ _ "fl _ 5 _ k-1 TK _ —O.25126860\ I33- 1 115% 5‘) u” '(B' ‘fls'u‘)‘(—1.244650 I O 0 5.0305213“, ‘—O.15076153‘ {'———L—i 0 359815 5, 0.08489935A6Q A=[3.u+[33.us= 0.1199385D p: 9.119958m59 0' O. ‘0‘251268 5'0 _ 0.04187792Ao'9 135B W07 Exam1.1&2.nb Problem 2, approximate solution (reference only): o o 3— 5 .._1__ __1_ J— 2 2 V? I’M—7) = B= 1 o o ; _-1_ L o «J? «[2- 0 0 1 E o o o o 51. 0 EA 0 o o 3‘\/Z_I. In[2}.-= K: 0 0 5:;- 0 0 5041on 0 0 0 T— 0 EA 0 0 0 0 E In[3]:= k = Transpose[B] .K.B; MatrixForm[N[k]] 1179.63EA _ 1178.39EA _ 0.1178518A L L L _ 1178.39EA 1178.63EA _ 0.117551EA L L L _ 0.117851EA _ 0.117851m 0.3565185A L L L In[4] .-= kinv = Inverse [k] ; MatrixFonn [N [kinv] ] 0.760135L 0.760034L 0.50251L EA EA EA 0.760034); 0.76078L 0.5027231. EA EA EA 0.50251L 0.5027231. 3.1372L - EA EA EA ( l l 0 o In[5]:= Bs: 0 , 1/Sqrt[2] s o In[6]:= Matrix‘E'orm[Bs] Ou C [5] //Ma tri xForm= O O O 1 3 V O mg’71:= us = (60 ); In[8]:= u = —kinv. (Transpose[B] .K.Bs.us); ~43! 135B WO7 Exam]. I&2.rzb Inz'9} : = MatrixForm[N[u]] Out[9]//Ma triXForm= 0.119932 (50 [—0.880017 60 —O.251255 60 In[10j:= A = B.u+Bs.us; In!11]:= MatrixForm[N[A]] Out[ll]//MatrixForm= —0.150753 60 0.359797 60 0.119932 60 0. 0000359797 60 —0.251255 60 In[12]:= P =K.A; MatrixForm [N [P] ] 0.0301506 EAéo L 0 .0848049 EArSO L 0 . 119932 EAéo L 0.0848049 EAéO L _ 0.0418758 EAéD L AW; Ox 0 ’Vl'f gbaé‘r ’_ ‘3 ~0 'qu {JR/Z} ’Aaa ’ Q E? avg )5”? g9 4 £9 U} a F MO fly 9 c f L 9: (MEL £3 + N m -* “:9” \w * "’7 my, ABC Héé-‘AQ {k F ‘ /\ s. L D 9: ~ Mgfigfi (“fl/U (5%) m@ O:VI\MLB.ABC+ Wank : A bu} /Q—> 5’1 I $9 MO»; 2‘21? Untitled—2 1 0 0 ‘TL 1 3: l 0 _4L 1 0 O O l O 1.831 0.4m L L _ 0.431 0.8EI 1“ L L _o.3;2551 O_ 3.45969FL2 EI u: _2.65577E‘L2 EI 12.7861E‘L3 EI D1 H m blH Fl ooN _ 0.37531 L2 0. O. 1875 EI L3 ' EI ——2L 0 O o 0 o o A o 0 O 4 E1 231 P = _ 10 FL F = '5'? SL 27 E 20 FL 10 231 4 RI 0 T *5T 27 1.17647L __ 0.588235L 2.35294 L2 BI EI 51 k—1 = _ 0.588235L 1.544121, _ 1.1754713 E1 El E1 2.35294 L2 _ 1.17647 LE 10.0392L3 BI EI EI _ _ 3.19651FL2 a: ~3. 06492 FL 2 #016331” P -l.33508 FL 3.45969 FLZ _ 1.33508 FL E‘.I _ 2.65577E‘L2 0' EI E / x QMM‘ l2 Qefimé fill; AEEQAM ‘ "The bar LP TC 1253712! . "7 TM W” CD only Cm YoTa‘fe M —> I s dmdhm Es res/Femoliculw Ty fiw {w GD AAA = + M» M» U 3 E i E i i i z 1 i z i i i g i I i 3: C W m __ M W A m m , I y . M .M was_§§i§§i N _ . A! if a / ILF . o __ ,,,,,,,,,,,,,,,, ; (1332;545:3323 = M A s. M a CI V W M A A 35A» A 0 k _ w M \esz; 2 {ix A [*7 ,I: 5% WW» wlisii‘ésiséii‘szllie/ V. m M a, y, w. m risigm ,,,,,,,,, i,m_§..Maa.M. D, 155i: C M n. W F F— a m , .F l _ .d \1! [TD “W W; W ,,_, 7 .,, .....i A l I; /_ru hum .. a . Y I y (2%,. W M .M MAI _ W0], .Wn! V .om ._ . w W M. (\ (Illllllux. Ella vagiiugiggigij -Mr. s p M w, w 0M flu D; P! DI: W C! Cr P1.nb H Probleml ** < 3 Matrix > 3/5 c 0 4/15 ,3: 1 o *1.0 1/5 1 o o {{0.6, O}, {0, —0.707107}, {1., 0}, {0.707107, 1.}, {0, 0}} < KMatrix > 1/5 0 o 0 0 o 1 o 0 0 K=EA/L* o 0 1/6 0 0 o o o 1/3/V3 0 o o o o 1000 “51:, o, o, o, o}, {0, %, 0, 0, 0}, {0, O, o, o}, {0, o, 0, high, 0}, {0, 0, o. o, 1005MH M1 = (Transpose[B] .K.;3) 0.356518 EA 0.166667 EA 0.166667 EA 0.735702 EA {{—T——' —““L—*}' {*—L“' —T“—}} u = Inverse [M1] .Fm 3.63968 FL 1.78567 FL {{‘T}' {TH A=B.u 2.18381 FL 1.26266FL 3.63968FL 0.787975FL {{"T}’ {“T}' {“T}' {"T}' {0}} P=K.A {{—0.436762 F}, {—1.26266 F}, {—0.606614 F}, {—0.185728 F}, {0}} Wfifiexz -—> ' Maggy JE II , i) S pnmo‘m o I 6‘— Rm‘ofivn Forte, = —f— The {1an T6 In: ezmibbvmm a}: O "’ Wm E W mm»? "" Tau Jamil} NM £71 sol/re fie 57mm (we, T i I i 9 /I // Aw’fifimmefm é Wain/1191M #0 are Mia/7 myrr/ _) My. My and ' W90)» formng Dis /acemw€(,&) _ WP?!) “'1 AAfi/(jaf—fl-f’7LL—f‘+%; g?) = Mas-(gig) ’7 ’ PAM {'1’} +21% AA; =0 7 (31.» z ’37P! ” // r/\ 125? 1 1 W!) = F‘ Ami—I) + if!» m i , 11-7“ W5“) 5 * W4 ‘AAB 4 Pugh-‘1 )7 A " YW‘AAB ‘5’ FAA; =0 7 Wm ‘ :‘f—Ffl ll [6 #7J/fl” 1) g i E I g! 'I 1’ W9( Ha - W’fa’fe F 1 , z é ' F {M1 M6 Mead/Tim! 5 m , . xv a a. if 2% a 0 m7 m_.,7 x.) _ . I an i . + \ (Ix X \g ( I J... 0 J T113}: L «.A .8 1M, .3 M! 4Vst H {Wigwamzzi , ‘ _ 3.5? ” ar— / I: I _ 0‘ 10:9}? , x [J A “ L? I? i /? LT; . . a W e, T. ( in giggiiiiiigl; W a rigs?) m 0» a; ,0 .0 27 = fish!§¥i\ Til-Gilli} , , fr: + IN 0 0 Tlllllullulllli} :, w w m w (iii A D A A .. Jm/ Tiligk \ as.) «09/ w W/firv “Pr... — rHasHQaHJ a? o 0 V F” M M M. ‘ if z. i riiiik _. . . - / +1 a a 0 ,w i K N m V w = % NP ‘ .0 I! I; 0 .fi/ LTT/T, M FL) 0 0 e .§ 2 :13ka M (Illllrlilillm. “,1 _ w, . . “ xfixw a 0 MT w . M [ii 0 # m . F W W 5 = + e H; {52%; E r: 2:} .3: .3?» put V WW w, A w a m A w a ,, 1m «QM/m AD AV «1/; AC M H M M\ E ................. \ggifgiiiigiiii‘lgiiilI!igiggiigiiigigga .. 4. fif. 7I—3flé5‘7F/g ‘F‘qu? VA ‘ ’9' I mmmmwwaw é Axis of Antisymmetry T 1.5421F l 1.3351FL I 1.54le g 1.54le : <— 0.7421F i 0.74le 1.1F i 1.1F 1.3351FL i 1.3351FL 1-1F W : LEW 0 7421F E 0.7421]? Force & Moment are antisymmertic ...
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This note was uploaded on 03/11/2008 for the course CEE 135b taught by Professor Felton during the Spring '08 term at UCLA.

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135B Midterm 2007 - 135B Midterm Avérage 76.35 Problem 1 ;...

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