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Final Exam Solution

Final Exam Solution - CE 573 Structural Dynamics Final Exam...

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Unformatted text preview: CE 573: Structural Dynamics Final Exam Problem #1 (25 pts.) K— ‘18" ——jl Sinai-we A 5+fuclure 5 Given: The structures shown are modeled with truss bar elements, which behave similarly to axial bar elements except that they allow both axial and transverse displacements at each node. In Structure A, it is assumed that the lower right comer is supported by a roller that permits horizontal motion. In Structure B, it is assumed that this same corner is supported by a pin that does not permit any motion. The assembled mass and stiffness matrices for each structure in the appropriate global coordinates are given below: 7.2 0 2.0 45.36: r1.sg__—1s.36_ Structure A: [ML = 0 6.4 0 basal-{KL = 11.52 g 48.64 —11.52 “06%. 2.0 0 6.4 —15.36 R —11.52 15.36 6.4 0 . 48.64 —11.52 1 Structure B: [M],3 2|: O 64]”? ;[K]H =[_]152 15.36 ]x106%. Reguired: (a) (15 pts.) Using static condensation, determine the global transformation matrix that eliminates global DOF #1 from Structure A. Calculate the mass and stiffness matrices for the reduced system. (b) (7 pts.) Determine the natural frequencies associated with the reduced vibration problem for Structure A. (Do r1_ot calculate the mode shapes.) (c) (3 pts.) The natural frequencies associated with the vibration problem for Structure B are 601 =1.3556fl and (02 = 2.8570m. Based on your results, what effect (if any) would you attribute to see see the presence of the roller in Structure A versus the pin in Structure B? Selma: (at) E7 w, (fl = 4393]" MP1 um7 at. War W Afldwncuon—el Lb-(KLAQ flat [14331: [4936]}H06A [KW] : [11.52 nasal woe “43; ["52 45.36] = {—13%}, 333$]. 77% flow tWivW 5T15[fi1 )m CT} new) 33% it [13M 1 o O l I. KER]: CTTEMECT11EJWQ I o] 7.1 0 2‘0 -,2§m.33éé AYEQ o l o M c) 1 0 2‘0 a 6ft (3 z = 4.3288 (A ~o.so&o «2m we r ': . if “’. [#8701 0 7.0772 I O 486 5 H272 . O I 4.1272 857% m _. r 7 ,_.__¢ ‘ [K]: [(3 [MEL .1 : «‘25m 1 Olffsge (Ls: -$S.7e][:zsm .338é [We 0 1 H52 fist—n52 l o 4934. rma ($.36 Cs i 2 O '7.6L ’- 73M ‘ ’ = . ( - . 2 [0 v7.61 lows] 1 O 187 76 #06 Q l —7.61 10/6 I...— {M M[Cfl-wlifll}0 :7 4a “ISW— é.X€q5w‘:—7.6Z+I.IZ7ZNI 1:6 ~7L62“ L/Z72Lol' I -8qu an (497! —5.86qu/o.ze~857m) ._ (_7-62*/J27Zw1)z C O L :7 S76§wq'L/LILI.7S‘.,01-uf0533=c 'r_ _ . - , - _ 2‘ W“ = Lommofifi w ‘ M rwwwamW' :7 w“; 2.02q’7k WM! LOL52,88074_¢, (.lqloxfiyffivuflww. FthLMUA-QWW (C) TMMWMW imam ' 'w (awchleg (Klgmwmfia) LTLQMUM Wéxmrm W MM. Writ-7% M14351“ WW WMfl‘l A W aimajmfm 5.144%. Problem #2 (25 pts.) 0.150 0 o [M]: 0 0.100 0 kip-secZ/in; 0 0 0.050 “I ‘50 —20 0 K‘N-x, - meow-m [K]: —20 30 -10 kip/in; m‘alsoib-m’ym 0 -10 “1 7.735 1.000 1.000 1.000 {m}= 16.15] gins]: 2.051 0.544 4.345. 22.642 2.927 -l.787 0.860 Given: The structure shown, with the mass matrix, stiffness matrix, and modal properties indicated, is subjected to a base acceleration given by a(t), which has a peak value of 06g. Plots of the Newmark— Hall design spectra for earthquakes with this peak value are found on the last page of this exam. The modal damping factors for modes #1 and #3 are 51 = 0.03 and £3 = 0.06, and you may assume Rayleigh damping. Reguired: (a) (20 pts.) Determine the equation of motion for each mode in terms of a(t) and other appropriate terms. (b) (5 pts.) Give tw_o reasons why you would expect mode #1 to contribute the most to the displacement response of the lowest story. (Note: you need n_ot solve for the displacement response, just give reasons why mode #1 makes the largest contribution.) Hint: You may find the following information helpful — l a b “ p 1 d —b c d _ad—bc —c a I I . . 41 Sclu‘i’luvl; (a) 6'7 WW [£1 “444—9 2 i "/o. mil egtgng :71 0.1243 7.735 94?: 0‘07 2 iv“); to; i r; l oat/L12 22.672 {5 ooé x ._ 2 22.542 »7.73§1§O«03§ : EO-MH :7 §fl<;fl 13an 0.1143 006 0-0060 , “1‘92 4 W1} [0146“ 2 0005051615“ = 0.0955 " $2 ’1wz 16m ‘MM‘ ta W {J W 20 do MW a 04500.“) “415,134 [Kliuh =—[M] 317‘“) = * 0,004”) 0 .050 a“) M} i.- m‘ ii 4 E13? : ’ 07 +Z§;w;z‘—;4w: 3;: ' =-T‘;a[) ' wwwg;wa;/mwm,wtamw WWW/n MW- EMF-rim? Hernia“ T F : gaETEMJH-S : ‘ (Ho 50W 241:"? ) 0.05 ‘ W 3/150 322% $3322.: '4‘“ fi . 1 L000 T 0“; Ma. n: if??? t :41? W a M Love 7 0.19 “a L5 “1 {3:25 {3125:4272 "3" M‘s: R: mfmm : Egfiflé'ié? .0939 W -— —— {7570 $06: 011%, .o‘ggggffggfw .m E 2. + Mew? + 94-23%, : -. 7016? m) 21 + 1.4577 =2; 260.55 = aflqza [+) Z; * 2.7170? 92.56 g r 15404 n) (5) EM, wwgw i; WK ,W ml P; 36’; AWMMWWM.M)%memWJ €— I .9 :J 9_V\ ‘j m '9 Problem #3 (30 pts.) Rigid 75 0 0 1750 —750 0 k‘ [M]: 0 75. 0 1b-sec2/1n;[1<}= L750 1250 —5001b/in. 2‘ 0 0 50 0 —500 500 —" 1.64 0.2748 0.5350 -O.7356 M = l(XlOlD/1n “mm {50}: 3.95 [(1)]: 0.5672 0.4133 0.6152 . k3" xvi—rsoulh/m 5.63 0.7764 0.7369 -0.2835 Given: The mass matrix, stiffness matrix, and modal properties of the following structure are as shown. The structure is subjected to external forcing for the form I —2.00 {Fm} : +1.00 fo), -0.90 where f (l) is as shown above, t, =l.90 seconds, and F“ : 2000 lb. The assumed plot of dynamic load factors for this function f (t) is shown above. Reguired: (a) (20 pts.) Estimate the maximum displacement of the second story of this structure. Use the SRSS rule to combine the appropriate quantities. (b) (5 pts.) Which single mode contributes most to the displacement of the second story? Justify your answer. (0) (5 pts.) Using only the mode you identified in part (b), estimate the maximum base shear force at ground level. Sakai/911a) F; M Mia, 11% WW 404M270“ we W [A M Sagflu/ufllm: 38w; - TW/ffl Mtg M flawwe MM 7230 defgjgmmmfcut Mm - [Uzlmvx]; : 0;; (M/MSJWMJ; Essie-c .' Maxth ‘ (Ea/ith *5/56 _.Lfl}3 ,73fi *( W: WM: [@JTFFM: “.7793 502 076%] W 17354 M51 —.2333 1‘? :7 mm? 9 {gig W KW : 26812,(200m5)= 'l362flflr 235mg PM“; = 10068 (2000119): 13 fit PM” = 2.3% (mm = #633,216 NWAJWW : _WWL412WW K/fl: w:- «m K : Half. TNWMWW M 751% I an: g : T 7g 27%? 7 [5-41 ' 0‘ ' 357% 73 gm : 5619: /w .7791 75 ,7764 N T ML'?aSzL—M1;agg_: 25350 T 7; 4.33m [ LU ‘HZ; ‘.‘((33 [ 7s “,‘(l'53 : 5 ' IBM $0 57360! M —' a T M (1—5 : — 7 .' SQ A “9,41 ‘ m —.zi1§ fi283§ VH3 (‘94-?3‘55‘30MW ) R:/é/aW-3‘)(3fi§%f) K3=K7Z‘4Q'*:*)(mz~“) : WM L7“ = fiSMW»; = 2313.96 ‘9‘; . 21: _ tr — : .M x v 2 “$5136” =~q3v~uce=se7z M7! K IMMLHVL. %’ J z: 0(th ‘~ (5672) (0016\03-‘43‘43 = “0.07% _ UK 43 I MMSH‘Z “ TL: Essex: W544ng ‘7 tr/Ez I. H5 '7! lusklmz - P r [3 15 - . Z- . ' fr—WL: .- *aO/qw 3 a 3 30/403 M“ K mum 2” ‘If Lt.”wa (“OAD’SMO51mm).qu = 000%; Mnb’r}? 7;: ’2’" : ’J’éQJ—av-JA57 {577—3 L7 - awfiu/ 963"? M W W Ju/Umf : , 20; 2 qémw = 2.01%; MP “an? 231mm; :7 [ufimjst (0,6.sz)(.zo)(zozqm = .262»; '1“ UAW : (1078)“ (.oo’JSV—t (12ml , n (M m.M_w#3mmmM. Mmew (Maw QM_ m,m*z MQWMMMW, W (c) Maggi; M47: P — 2 — 3 W§> §U§>— '7SO O ‘,73fé ‘(lohaozq 450 I250 500 .mya Q) o ~Soc) 900 33335 Problem #4 (20 pts.) Given: You are asked to formulate a new type of beam element in which both .the mass per unit length and flexural rigidity of the beam vary linearly along the element; the relevant functions are x ‘_ __ i E1(x)— E1” [1+Zj,m(x)—mn(1+ L]. 2 2 3 3 2 Two of the shape functions for this beam element are N 2 (x) = x — % + 2—2 and N 4 (x) = % — XI. Required: (3) (12 pts) Determine the stiffnesses k44 and k14 for this element in terms of E1” and Le (b) (6 pts.) Determine the mass m24 for this element using the consistent mass formulation in terms of fin and L. (c) (2 pts.) Determine the mass mM for this element using the lumped mass formulation in terms of mo and 1.. Hint: Here are some potentially useful algebraic results —— 4x3 6x“ 4x5 x6 , ,, 16 48x 36x2 Nl(x)*N2(x):x2~T+—L2 —?+F;N2(x)*N2(x)=L—Z——L3 + L4 . )53 3x‘1 3x5 x6 ,, ,, 8 36x 36x2 N2(x)*N4(X)=—?+7—?+F;N2(X)*N4(x)=F-?+ L4 . 2:“ 2x5 x" ,, ,, 4 24x 36x2 N4(X)*N495)ZF—7+F;N4(X)*N4(X)=F— £3 + L4 . ' ‘ z, SDI“JW‘M I (q) k 7' L ” I x g_qu 34k,— —— w mm w) w - Wm t a u o O L L2 L3 L4 - L L qu=5 M1[‘F\NU(‘€)-=LX ‘= ([+E\(_g4?fifii€<4fi)&( ° 0 L 2, 3 IM if} 2 q XL 1 ' XL‘ 2X; Z 1 K = o “—— +_,_X —2 ..’S§.[ =M° —ng4§],1_"%+8£5 O L- Lz L“ L" ‘ 71. o _. 3 7 _ '5 m mtg—W“ .ifgtflzhgj :7 Mug—Swot [‘f 5 7 3 Zgo (L) 1:579:57 MWGWWWq’mfiqM TWW, ...
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