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Unformatted text preview: CE 573: Structural Dynamics
Final Exam Problem #1 (25 pts.) K— ‘18" ——jl
Sinaiwe 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[Cﬂwliﬂl}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) ._ (_762*/J27Zw1)z C O L :7 S76§wq'L/LILI.7S‘.,01uf0533=c 'r_ _ .  ,  _ 2‘ W“ = Lommoﬁﬁ
w ‘ M rwwwamW'
:7 w“; 2.02q’7k WM! LOL52,88074_¢, (.lqloxﬁyfﬁvuﬂww.
FthLMUAQWW (C) TMMWMW imam ' 'w
(awchleg
(Klgmwmﬁa) LTLQMUM Wéxmrm W MM.
Writ7% M14351“ WW WMﬂ‘l A W aimajmfm
5.144%. Problem #2 (25 pts.) 0.150 0 o
[M]: 0 0.100 0 kipsecZ/in;
0 0 0.050 “I ‘50 —20 0 K‘Nx,  meowm [K]: —20 30 10 kip/in;
m‘alsoibm’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§ : EOMH
:7 §ﬂ<;ﬂ 13an 0.1143 006 00060
, “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 EMFrim? Hernia“ T F : gaETEMJHS : ‘ (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? + 9423%, : . 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 1bsec2/1n;[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; Essiec .' 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 [541
' 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/aW3‘)(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—avJA57 {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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 Fall '05
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