135B_Hw-1_Solution_2008

# 135B_Hw-1_Solution_2008 - CEE 135B Homework#1 January 7...

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Unformatted text preview: CEE 135B Homework #1 January 7, 2008 1. Refer to Problem 1 on the ﬁnal exam from 135A dated Dec. 13, 2007. (a) Solve for the bar forces and deformations. (b) Find vertical and horizontal components of displacement at D. 2. Refer to Problem 3 on the ﬁnal exam from 135A dated Dec. 13, 2007 . (a) Solve for element forces and deformations. (b) Sketch the bending moment diagram for beam AB. (c) Find displacement at B. 3. Refer to Problem 4 on the ﬁnal exam from 135A dated Dec. 13, 2007. (a) Solve for element forces and deformations. (b) Find the vertical component of displacement at B. 4. For the beam shown below: (a) Find v(x) (b) Let AAB = - v’(0); express v(x) in terms of AAB rather than M. Hwy, CEE 135A Final Exam December 13, 2007 L. P. Felton (30) 1. The structure is the same as on the midterm exam, i.e., rigid beam ABCD is supported by four truss bars having identical values of EA. This time the only load is a" vertical force F applied at D. Write down all matrices required for the analysis of the bar forces and solve for the bar forces. (In order to simplify grading, please order the terms in your P matrix as follows: (10) 2. A uniform simply supported beam is loaded by an external moment Mo applied at the midpoint. Use the moment—area method to ﬁnd the rotations at each end. . M) W !<‘—-———lZ———>l (40) 3. The structure in Figure (a) below consists of a uniform axially rigid beam AB, and a bar BC with pinned ends. There is a ﬁxed support with vertical roller at end B of the beam and a ﬁxed support with horizontal roller at end A The bar Stiﬂ‘hBSS is such that E4 = 3‘1—ZEI/F. The only load is an External m‘omentM, applied at the midpoint of the beam. Write down all matrices required for the analysis of the element forces. Solution is not required. (20) 4. The load M0 is removed from the structure in Problem 3, but the supports at A and B undergo small motions of known magnitudes 51, 82 and 53, as shown in Figure (1)) above. Explain in detail all modiﬁcations to matrices necessary to analyze the element forces. Solution is not required. _——- + -—u—- “ELF ‘ ” FBF ll as "2;: "ﬁ 1“ DA? 1' H; b» ‘65 ‘1 W Wmmm‘“ mm :6» “*5 m 5?. 9L5 60m [/31 MW 25am > (NM) 3 = -N(m+£) Z s 7‘1‘X+/Da Ta? "ﬂf‘r‘J'F W IM—L/D AA? i ’an’fb ‘ be; 4.7/0 Aw “DAN/V H —> Afﬂ)’ 7W Vii/ﬁm/ ﬁne Fx = / .0 a —%> / I VWT‘M/ PW“ ‘Wf In’fevm/ b/‘vrfwn/ Poms 1ng a a “:13? ’ VI" 4/ =0 W =4 ( L) FBF ZFx=0 —-> 6*) FAKF ’FDF—f-FA.F~T:/—=o L ’ Z? =0 6 ¢ / FF)? 3: +FAF'f -/ =0 e / " :— FW 0 I J— : Fsp‘ J” Tr‘é 7% FA? / FAF’4LJJ713 / I Ex’fevm' VwTw/ Work = I . ux= Mx= .1511. 3A ﬁg 9A // inTEW, RAVEN Wax/K [(mwoxo) + (—ang—ww) + (—9.7/0)(r) 4- (— p- 83") ('33) T F)? (vi->73) (6-) = EA—(NJN) (——->) I?! ‘1“: ‘T‘Q. ZF =0 6') _.+ 4 _ FBFJ EMF +J——0 Fe; 0 FAF ,3, [SW = #0! FAF = @«SZN #7 ~ (’0‘ s ' r) I ' ‘5 l; 5 a Mn/resﬁ/‘cﬁ/ 5/ 8/0/6174 .8 fw lea/n P ' - y y _ 110.1} V'ELL'[71':(‘H°) + “N” 1L'””] ' {7‘51 ‘. I ;» Pﬂwwr‘hwv‘ M-%M-HJ = i C/mk 7%! mm w W foﬂff (T ( Vaiﬁmﬂmknam) WWW 3/ Mine umﬁ— c/ookmqe if p977?” for mm and M72 W5] 1 0 X/ + 0 = ‘ 645/6 "/ ’é Xv H0 / f < Asézmbla M143 _ I z, 0 HM - -/ ’33,: X + H0 I’M/e 0 / y 0 = A : fa FoY’ﬁyU MW 1 ‘ I _ 77: AL(W;MW7P) ' 72—A— — 9A , _LE1_ " z .1123. m 91%, Icy :: ——I— '1, 3’ MBA’ 4" 1151; ‘ H” #J 7° =A .A .1 l M 17-——-—+§ ' M/J’Mo ’0J68-g37‘ -—mx) =0 z H“) = —0\¢/J'Ha -— wk”?! '-—> MHz) = -I.J’Ho I .W Nesta, ! ZHW =0 m 69 -0v¢/J’H0 - ox/bﬁ-‘Ef-Dx +Mb -—H(X) =0 H“) = o-J'J'J'Ha —o~/é£—§{"x >0) {7M wt {8 i dime #/ gm!me ﬁrmuﬁﬁa; fw W6 fem? WW5! (NA/M X = We — NOW +3) WE —’ ENZan Virﬁm/ WK 5 41/30 _/fa/ .. 2/ rSf/aceM/W WMW/ M W Miami/M fang. \$77k} Mva W3; WM Xl=l WE = W? with X3! XI ( 1 XI —"5/ 59 . - \L 4'” WéQ=X~\$=\$ x—X—/ V 5y I 3 3 l— 4 K, . Ky 55 F “ff/ —-> {3 I X» 1/ W99 7' Xy'Sv +T5/ 3;; iv _ T W m aim/t Ala/e 7f? ﬂog/24M form —> z» -—— 2 = o .7 ct; S 7&va / Ali/re 7759 MW; gym (I‘m/5*) Z? = W; ME: .22. MM = -%T+;—[ro‘70m-f “5 p.6735v - 9‘578071 k; We = fl Age {3-1 — A578» -— \$6732] To, == %—[' mffﬂﬂ + A 7017 57 4 /\ 700 5/] 5 Au *5 (0-3/14 SJ + o~633 5y + 0.5335,, FW- WVWA/ digféwwﬂ at “‘9 4,5077%? WV/Zla/ fan-.2 FV=/ owl/6’ ' :M9A=0 M743 =4ﬂéﬂ /, 5’, F“=/ “7 @QYM/ YWTM/ Mark = M] 4— WWW! ViYTM/ mark = J; - —o.s/653+4.,5335> +0.53%) : a: is! 5XT€WI VWTWK/ Work == V‘W‘ﬁm/ Wank M7 +1 5/} ‘- 3': (’0-9/653y'f‘0‘63352r ‘f’ 0-5335!) 4 A ‘u ' o4¢>§e I —f é-Mé 5y 1/9! 2” ...
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• Spring '08
• felton
• Second moment of area, Bar association, element forces

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