Homework8Solution - ‘ ex? Lyman; #OMEWL‘JR’K...

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Unformatted text preview: ‘ ex? Lyman; #OMEWL‘JR’K :{QLUT’B‘QUQ Problem 10.2 Part a: Determine natural frequencies and modes. The mass and stiffness matrices were determined in Problem 9.2: leO m=——~ 3 01 8 —7 k 2 16251 J 5L3 7 8 Then 162E] 8 — A —7 l k - a)2 = ——--—- a m 51,3 [—7 8-2} H where 4 ’1 z 602 486E] Substituting Eq. (a) in Eq. (10.2.6) gives the frequency equation: 22~162+15=(/1-1)(2~15)=o The solution of the frequency equation gives: [11 = l and 112 - 15. The corresponding natural frequencies are E1 I E1 601 = ml} (02 = 38.184 m (C) \e The natural modes are determined from Eq. (10.2.5) ? following the procedure shown in Example 10.1 to obtain: «m =8 $2 =11 l(’11=l ¢21=1 9’) 12:1 ,. I \x \ . . I ’ .l \ . . \ . \~V~_'_.V” ‘~Vz ll>22: —l w. = 38.184 x/EI/mL 4 Second mode (antisymmetric) ‘ w. = 9.859 JEI/mu First mode (symmetric) Part b: Verifi’ orthogonality properties. dhm=3%hlflljlfi=0 T 162E] 8 —7 1 k = 1 1 = O 1 ¢2 5L3 < > “7 8 __1 Thus the natural modes satisfy the orthogonality properties. Part c: Normalize modes so that M" = 1. L l 1 2 It 111% L 1 1 2 wig-<1 —1>[ it.) L 5 w ! Divide {ll} of Eq. ((1) by 1[2mL/3 and (2)2 of Eq. (d) by 1[2m L/3 to obtain normalized modes: .. 3 1 -—- _ i— l ‘1" = 115217 {1} ¢2 " \lsz l— 1} (e) ¢11=xl3l2mL ¢21=«13/2 mL ¢12=J3IZmL / ’1?— ‘ \ ‘ : : , ’ : -L-_J ’ \~.V”' ¢22=— «lB/ZmL First mode Second mode The modes of Eq. (e) are scalar multiples of the modes in Eq. (d); the shapes of the two sets of modes are the same. Problem 10.3 The free vibration response of the system without damping is computed using Eq. (10.8.6), 12 2 . um: Z q)" [q n (O)cosmnt+q—;‘J~(9-)-sinw "1] (a) n21 where according to Eq. (10.8.5) T T - 0 mmkfigfllqnwfl$fl9 m ¢n m¢n ¢n m¢n Using m and k from Problem 10.2, the values of q1(0) and q2(0) for the initial displacements 11(0) are —"i3I—‘—(1 1)[1 1].;(0) ——————————-—=(0.5 0.5)u(0) (c) 611(0): 113141 1)[1 32—11(14)? 1]u(0) =<o_5 ~O.5)u(0) (d) 92 (0)2W Similarly, 41(0) : (0.5 0.5) 11(0) 42(0) = (0.5 ~05) 11(0) (21) For 111(0) 2 land 112(0) = O, 1 ‘ 0 = 8 = 8 Substituting in Eqs. (c)-(d) gives mm=as wm=as mmzo em=o (6) Substituting Eq. (e) in Eq. (a) and using the modes from Problem 10.2 gives 0.5co 9.859 31-4—1 +0.5co 38.184 it {u1(t)}_ mL mL4 1420‘) 0.5co 9.859 311-: —O.5co 38.184 J-E-I—t mL mL4 (b) For u1(0) = land [42(0) : 1, (0)_1.0’0 u ~1u()—-O Substituting in Eqs. (c)-(d) gives 41(0) = 1 (12(0) = 0 (21(0) = 0 q'z(0) == 0 (5;) Substituting Eq. (g) in Eq. (a) and using the modes from Problem 10.2 gives co 9.859 Ilia—z u (1‘) mL co 9.859 ~51? mL (c) For 141(0) = land u2(0) = «1, 0"1 .0_0 u<>—_1u()—O Substituting in Eqs. (c)-(d) gives (11(0) = 0 (12(0) = 1 41(0) = 0 622(0) = 0 (1) Substituting Eq. (i) in Eq. (21) and using the modes from Problem 10.2 gives In case (a) both modes contribute to the response because the initial displacement condition has components in both modes. In case (b) the initial displacement is proportional to the first mode and therefore only this mode is excited and contributes to the response. In case (c) the initial displacement is proportional to the second mode and therefore only this mode contributes to the response. Problem 10.4 The free vibration response of the system of Problem 9.2 including damping is computed using Eq. (10.104): 2 u(3‘)= 2 (3,1 6‘50"! [qn (O)cosa)n Dt ":1 (a) +——-———————-——-sq" (O)+{”w"q” (0) incl)” pt 60,, D where (0,10 mfl/l - {2 ; a)” are given by Eq. (0) of Problem 10.2 and {n = 0.05. Therefore, ll E1 E1 to = 38.136 ~— mL4 20 de mm = 9.847 (b) For the given initial conditions u1(0) = l and u2(0) = 0, Eq. (e) of Problem 10.3 gives q,(0) = 0.5 (12(0) 3 0.5 é2(0) = 0 (0) 6M0) = 0 Substituting Eqs. (b) and (c) in Eq. (a) gives "(2‘) = 1110) + “2(1) (d) where the modal contributions are - l u] (t)=e 0493’ (0.5coscu1Dt+0.025sinw1Dt){l} (e) _ l u2(t)=e 1909‘ (0.5cosw2Dt+0.025$ina)2Dr){ l } (0 8,0) 0.5 0.866 1 o 200 132(2) = ~1 0 1 0 = 200 (b) P30) 05 ~0.866 1 200 200 The modal equations are “ P 0 .<. t s t M + K = "0 d "61.1 "61,. {O I 2 Id (0) The solution of Eq. (c) is P“ l~ coszm 0< < q (t) = K” T" x I _ td " P "a [25111314 sin2n —t- ~ 3—31 t 2 Id Kn n 711 2 71. (d) Substituting PM, K,l and T,, in Eq. (d) gives (110‘) = 15249 [1 — cos2fl—%~] l q2(t) = 0.2043(1 -— c0522???) 0 s z s rd ('6) 2 it 430) 0.1095(1 - cosZflL) T3 (11(2) = 3.0498 sin.27r[—%— ~ 0250) l 1112(1) = 4.3729 sin 27z[-T-’- - 0.687) t 2 rd (1) 2 3(t) = "0.0895811! 27: ~ 0.933] 3 W) = 2 0.4.0) (g) ‘where ¢n are known and qn(t) is given by Eq. (6) if 0 S t S ta, andbyEq. (f) ift 2 td. Problem 12.12 From Problem 12.10, (01 = 18.38 602 = 50.22 (03 = 68.60 I; = 0.3418 T2 = 0.1244 T3 = 0.0916 05 —1 I 05 61 = 0.866 ¢2 = 0 gig = -O.866 1 1 1 M1 = 0.3882 M2 = 0.3882 M3 = 0.3882; K1 = 131.16 K2 = 978.97 K3 = 1826.80 (a) The rectangular pulse force at the third floor is shown 0': in the accompanying’figure: * p, Rips 200 I, 86C \ rd: 73/2 = 0.1709 The generalized modal forces are ...
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Homework8Solution - ‘ ex? Lyman; #OMEWL‘JR’K...

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