HW13_Solution - Problem 13.5 1 — 1 0.3333 The modal...

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Unformatted text preview: Problem 13.5 1 — 1 0.3333 The modal expansion of 1111 is shown next: h E; E; m/2 0.6220711 0.1667m, 0.0447m m l .0773m 0 0.0773171 Mass and stiffness matrices (from Problem 9.7): = + + 1 2 -1 0 m 0.6220m 0.3333m 0.0447»: In = m 1 k = k —1 2 —1 1/2 0 «1 1 m1 132 31 $2 = w~03333m l 0 = m 0 ""2 a 05 1 -0.1667 3 1 05 0.0447 h :7: s3 = 0.0893m 1 —0.866 = m —0.0773 @ —=-u2 05 1 ' 0.0447 h - m % where k = MEI/113. Vibration properties (from Problem 10.11): ‘ Part 17 a)? = (2 _ {3—) .11; mg = 2i; 6,): = (2 .I. fl) 3 The floor displacements due to the nth mode are m m m. 11H = I",I gt)” DHU) 05 —1 05 . . . Substituting for I“,I and q)” glves 4211 = 0.866 42 z 0 43 = «0.866 1 1 1 111(1) 05 0.6220 Theinflume vectoris 112(1) = 1.244 0.866 D10) = 1.0774 131(1) 1:1 1430‘) 1 1 1.2440 The first-mode properties are computed from Eq. (13.2.3): “1(1) 4.1 0.3333 3 : — . T = 1!; 2 Z mjafl = m(0.5) + m(0.866) + fl(1) = 1.866m “3m 03333 0 D20 0 _ D2“) i=1 2 43(1) 2 1 —0.3333 3 M1 2 2 775.43I = mtotsf’ + m(0.866)2 + $0? = 1.5m 111(1) 0.5 0.0447 :1 j [41 112(1) = 0.0393 —O.866 133(1) = 40.0774 173(1) 1" = ——-— = 1.244 11 z 1 0.0893 1 M1 30 3 Similar calculations for the second and third modes give: combining the mOdal responses gives the floor h h displacements: = 405 = 0.134 I“ m L’ m ale) 4 0.6220 D30) + 0.3333 192(7) + 0.0447 03(1) M2 = 1-5’” M3 2 1-5’" 12(1) = 1.0774 131(1) — 0.0774 193(1) 1‘2 : —0.3333 13 = 0.0893 113(1) = 1.2440 121(1) — 0.3333 132(1) + 0.0893 133(1) Parta Partc Substituting I}. , 111, and £19,; in Eq‘ (13-2-4) giVES Static analysis of the frame for external floor forces 5” 1 05 0.6220 gives 14:21: 1, 2, 3: 31 x 1244’" 1 0866 = m 10773 V35: = 0.6220m V; = —0.1667m 11;; = 0.0447744 05 1 0.6220 V25; = 1.6993m V25; : —0.1667m V25; = —0.0326m Vf‘f = 2.3213m 1215; = 0.1667711 vi; = 0.0121m The total story shears are 3 Vj-(r) : E V2110 = 23‘, V)? Ana) n=§ n=1 Substituting values of gives V10) = m[2.3213A1(t) + 016672420) + 0.0121A3(t)] V20) = m[1.6993 A10) — 0.1667 A20) — 0.0326 7130)] V3(t) = m[0.6220 A10) — 0.1667 A20) + 0.0447 213(1)] Part d Static analysis of the frame for external floor forces s" gives MS; : M3; = mh[0.6220(1) + 1.0773 (2) + 0.6220(3)] = 4.6426mh M3; = mh[0.3333(1) + 0(2) — 0.1667(3)] = "0.1667172?! M3; = mh [0.0447(1)—0.0773(2)+0.0447(3)] = 0.0242762 The base overturning moment response is r 3 3 Mr) = E M220) = 2 M35. A20) (1 =1 n=1 Substituting values of ME; gives M),(:) = mh [4.6420 A10) — 0.1667 A20) + 0.0242 213(1)] 10 Problem 13.9 The floor masses, and the height of each floor above the base are m m1=m mz=m m3=-~ 2 111:}; h2=2h hjzh The natural modes and generalized masses (from Problem 13.5) are: 0.5 —1 05 61 = 0.866 62 = 0 63 = “0.866 1 1 1 M1 = 1.5m M2 = 1.5m M3 = 1.5m Substituting for mi and 0);.” inEq. (13.2.3) gives 15;: 1:: = ijajl = m[1 x 0.5 +1>< 0.866 + 0.5 x 1] j=1 =1.866m. 3 L’; = E mjgbjz = m[1><(—1)+ 0 + 0.5 x11: —0.5m i=1 3 13 = 2 611.613 = m[1 x 0.5 +1><(—0.866)+ 0.5 x1] _r'=1 = 0.134412 If: Ii mh[1x1x0.5+2x1x0.866+3x0.5x1] 3.732mh il hjquijz = mhg1x1x(—1)+ O + 3 X 0.5 x1] j: r» 0. U1.— mh $3., I} 3 E hjmfijs j=l =mh{1x1xo.5+2x1x(—0.866)+3x0.5x1] = 0.268mh The effective modal masses and effective modal heights are given by Eq. (13.2.9a}: 21 h 2 2 M = (11) = m = 2.3213m M1 15 - h 2 2 M: = (12) = "3.01 = 0.1667m M2 1.5 h 2 2 M; = (L3) = M = (10121,” M3 1.5 Mode 2 Mode 3 Mode 1 Verify Eq. (13.2.14): 3 Z M; = 2.3213m + 0.1667m + 0.0121m = 2.5m n=1 3 Z 6:114; = 2h(2.3123m) + (_ h)(0.1667m) + 26010120...) 1 I! 4.13sz Problem 13.17 _ 5 1 M1 = giftinng : (1 —1.949 1.949)m 1 —1.949 1 1.949 “3 “2 = 12597m M 1 1‘1 = ~51— : 0.397 M1 Similar caicuiations for the second and third modes gives: IQ=¢§mt=5m lg=¢§mt=0 Fig.P13.l7a M2 : ¢§m¢2 I 8292?” M3 = ¢§~m¢3 = 2m 11 L3 Parta 1‘2 = _M_ : 0603 f3 = .11? z 0 2 3 The equations of motion are given by Eqs. (13.1.1) and (13.1.2). The mass and stiffness matrices (from The effective earthquake forces are given by Eq. (13.1.4): Problem 9.13) are 5m 1 5 3m 28 6 -6 peflz) = —mtiig(t) = — m 0 113(1) m =1?! 1 k = 1 L3 6 7 3 m 0 1 O —6 3 7 5m The influence vector due to horizontal ground motion is = _ 0 its“) (from Problem 9.13) 0 1 ' Substituting 1"" , m and q)” inEq. (13.1.6) gives 1 = 0 0 5m 1 s1 = 1"1 mail = 0.397 m —1.949 The natural frequencies and modes of the system (from m 3949 Problem 10.23) are ' 1.985111 EI (01 = 0.526 “a; £02 2 1,614 _3 = —0.774m M “L 0.774m El (03 _ 1.732 m 5m 1 s2 = F2 Inge)2 = 0.603 m 1.283 4) 11949 1 ¢ 3 m —1.233 1 i - - d2 = 1.283 3 = 3.0 1.949 — 1.233 1 15m = 0.774111. Substituting for m, t and on in Eq. (13.1.5) gives the 4177 4m first-mode quantities: 5 1 5m 0 0 n=¢fmr=(1 —1.949 1.949)m 1 0 33 213111453 =0 m 1 a o l 0 m 1 O = 5m The modal expansion of the spatial distribution of effective forces is shown in Fig. 1313.171). The effective forces in the third mode are ail zero, impiying that this mode will not be excited by horizontal ground motion. 35 Part I: The modal displacements from Eq. (13.1.10) are “10) I 1 u1(t) = 5420) = T1 ¢1 D10) = 0.397 «1.949 D10) u3(t) 1 1.949 0.397 = [—- 0.774 D10) 0.774 “10) 1 uga) = age) = F2 ¢2 02m = 0.603 1.283 D20?) Lea) 2 —-1.283 0.603 ={ 0.774 D20) —0.774 u1(t) 0 £130) = 1429‘) = 1‘3 4213 D30) = O 1 D30) H30) 3 1 0 = 0 D30) 0 Combining the modal displacements gives the total displacements: aim = 0.3971313) +0.603D2(t) u2(t) = 4.7741310) + 0.774D2(t) “3(19) = 0.7741310)— 0.774D2(t) Observe that the third mode does not contribute to the total displacements, and the total displacements L120) and “3(1‘) are antisymmetric. 36 Fig. P13.17b Part c The bending moment at the base of the column clue to the nth mode is Mam} = MEL AN) Substituting the modal static responses , shown in Fig. P13.17b, and combining modal responses gives 3 MN) = Z Man) = 3.533mLA10‘) + 1.467 mL A20) n=1 The bending moment at location a of the beam due to the nth mode is Mano) = M3; Ana) ' Substituting the modal static responses M2; , shown in Fig; P131713, and combining modal responses gives 3 Mac) = Z MMU) = —0.774 mLA1(t) + 0.774mLA2(r} n=1 37 ...
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This note was uploaded on 01/04/2010 for the course CE 225 taught by Professor Chopra during the Spring '09 term at University of California, Berkeley.

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HW13_Solution - Problem 13.5 1 — 1 0.3333 The modal...

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