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Homework6Solution - 4‘ sin(a At 4 = to Art" ’...

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Unformatted text preview: 4‘, . sin (a) At.) 4 = . to Art + . " ’ Problem 5.1 u’“ u’ COS( " J a”: a)” i f I(‘Dver the time interval 1,. S t S 1m the excrtation + %_ [1 _ cos (0),, At,- )1 one ionis W) = h 0 S T 5 At,- (a) am = ui [— can sin(a)nAti)] + dicos(a)nAti) where Ari a ’m - ti and the equation to be solved is + it. a)" sin (50,1139) (d2, " + k = ". b m“ u p’ ( ) Substituting 13,. = (pi + pm)/2 gives , subject to the initial conditions u(T = 0) = ui and . Q 12(r : O) z u “m : Aui + B“; + CPI + DPM (3-11 The response 14(1) over 0 S 2' S Ari is the sum of “”1 z A u‘ + Bu‘ + Cpi + DP‘“ ((2.2 two parts: (1) free vibration due to initial displacement ui where and velocity :21 at r = O; and (2) response to step force [3,. A = cos (mum!) A’ z __ a)” sin (wnAIi) (£1 with zero initial conditions: . a. A f). B = $111 (w"At‘) B’ = cos(m,,A1i)‘ (£2 u(T) = uicoswnr + 451nm”? + 7‘0 ~ coswnz') a)" -, (on ; (cl) C = 1 — cozslfwnAri) C’ = (0n sm (kwnAti) (f3 it”) =—u-sina)T+—&-cosa)’r+ £1311”)? (c2) -2 a)“ 1 n U)" n k n v D : l. - COZSIEWHAIi) D, = (on Slnszwnmi) (£4 Evaluate Eqs. (c) at 2' 2 Ali on = ti + Ari: C:\Documents and Settings\Jerome Peter Lynch\My Documents\CEE Classes Teaching\CF March 2, 2004 7511 DynamicsPage 10rk\r 9:04:14 AM 0\° 0\° 0\° 0\° 0\° o\° o\° o\° (c) Jerome Peter Lynch, (all rights reserved) University of Michigan February 14, 2004 This program implements a Newmark solution to Problem 2 and 3 in Homework #6. In particular, the Newmark Average Acceleration method (gamma = 1/2, beta 2 1/4) % PROBLEM #2 % O 6Establish Structural Properties 0 /:__--...2::._....___..______________.________ O m = 1; % in kips-SAZ/in k = 100; % in kips/in c = 0.05*2*sqrt(k*m); % in kips—s/in wn = sqrt(k/m); % in rad/sec Tn = 2*pi/wn; % in sec fn = wn/(2*pi); % in Hz %First Setup Applied Loading dt = 0.1; tmax = 5; for 1:124:49 f(i,1) = 0; f(i+l,l) = 386.4; f(i+2,1) = 0; f(i+3,l) = —386.4; end %Apply Newmark Method gamma = 1/2, beta = 1/4, [X,t] = newmark(m,c,k,f,dt,gamma,beta,0,0); plot(t,X,'r-O'); xlabel(‘Time (sec)'); ylabel('Response (in)'); C:\Documents and Settings\Jerome Peter Lynch\My Documents\CEE Classes Teaching\CE3511 DynamicsPage Zork\r March 2, 2004 9:04 14 AM 9 ____________ 9 o ———————————— o O 0 6 PROBLEM #3 6 9 ____________ 9 o ———————————— o wn — 2*pi*fn; % in rad/sec m = 1; % in kips—s‘Z/in k = (wn*2)*m; % in kips/in c = 0.0S*2*sqrt(k*m); % in kips~s/in %Apply Newmark Method x n u newmark(m,c,k,f,dt,gamma,beta,0,0); %Find Response Spectra fi:==::========:=:====: count = count + l; Sd(count) = max(abs(X)); F(count) = fn; end %Plot Response Spectra O ____________ fizz: __________________ figure; plot(F,Sd,‘r~o‘); ylabel('Max Displacement (in)'); xlabel('Frequency (Hz)'); C:\Documents and Settings\Jerome Peter Lynch\My Documents\CEE Classes Teaching\CEE511 DynamicsPage lork\x March 2, 2004 9:04:28 AM newmark(M,C,K,f,dt,gamma,beta,Xi,Xdi) function [X,t] newmark(M,C,K,f,dt,gamma,beta,Xi,Xdi) m r: s 0 H P. O :5 N n n 0\° 0\° 0\° (c) Jerome Peter Lynch, (all rights reserved) University of Michigan February 14, 2004 o\° o\° o\° o\° This function is intended to perform the numerical integration of a structural system subjected to an external dynamic excitation such as a wind or earthquake. The structural model is assumed to be a lumped mass shear model. The integration scheme utilized for this analysis is the newmark alpha-beta method. The newmark alpha~beta method is an implicit time steping scheme so stability of the system need not be considered. o\O o\° o\o o\o o\° o\° o\° o\° % Input Variables: % [M] = Mass Matrix (nxn) % [C] = Damping Matrix (nxn) % [K] = Stiffness Matrix (nxn) % {f} = Excitation Vector (mxl) % dt = Time Stepping Increment % beta: Newmark Const (1/6 or 1/4 usually) % gam = Newmark Const (1/2) % Xi = Initial Displacement Vector (nxl) Xdi 2 Initial Velocity Vector (nxl) o\° o\° Output Variables: {t} = Time Vector (mxl) Response Matrix (mxn) o\° o\° o\° N L—J | I o\° I I I | | | | I I I I I I I I I I I I I I | I | | I | | | I I I I I I I I I I I I I I | n = size(M,l); fdimc = size(f,2); fdimr = sizeIf,l); o\° Check Input Excitation if (fdimc == I f=f‘; end; m = size(f,2); o\° % Coefficients c0 = 1/(beta*dt*dt) ; cl = gamma/(beta*dt) ; C:\Documents and Settings\Jerome Peter Lynch\My Documents\CEE Classes Teaching\CEE511 DynamicsPage 20rk\1 March 2, 2004 9:04:28 AM c2 = l/(beta*dt) ; c3 = l/(beta*2) — 1 ; c4 = gamma/beta — l ; c5 = O.5*dt*(gamma/beta - 2 ) ; C6 = dt*(l — gamma ) ; C7 = dt* gamma ; o\o Initialize Stiffness Matricies Keff = CO*M + cl*C + K ; Kinv = inv(Keff) ; o\° o\° Initial Acceleration 0\0 I l I I l I l I o\° W m H H1 0 H B W H . H U) (T m ('3' m U f(:,l) = f(:,l) + M*(c0*Xi+c2*Xdi+c3*dei) +C*(Cl*Xi+c4*Xdi+c5*dei) ; X(:,l) = KinV*f(:,l) ; o\° de( ,1): cO*(X(:,l)—Xi) — c2*Xdi — c3*dei ; Xd( ,1) = Xdi + c6*dei + c7*de(:,l) ; % Perform Subsequent Steps o\° | H) H- + H f(:,i+l) + M * ... (CO*X(:,i)+c2*Xd(:,i)+c3*de(:,i)) .. + C*(c1*X(:,i)+C4*Xd(:,i)+c5*de(:,i)) ; X(:,i+l) = Kinv*f(:,i+l) ; de(:,i+l)= cO*(X(:,i+l)—X(:,i))—c2*Xd(:,i)-C3*de(:,i) ; Xd(:,i+l) Xd(:,i)+c6*de(:,i)+c7*de(:,i+l) ; end; o\° Strip Off Padded Response Zeros o\° o\° D (D 5 m H 9 (.r m (‘1‘ {3‘ m H H . E (D < m 0 (‘1' O H o\° C:\Documents and Settings\Jerome Peter Lynch\My Documents\C3E Classes Teaching\CEE511 DynamicsPage 30rk\z March 2, 2004 9:04:28 AM 0!?" I AJ: 0 0 0 0 0 “ fl 0 0 (I fl 0 0 1D 0 0 Time (sec) Ac: mwcoammm Max Displacement (in) 16 .x O I Frequency (Hz) Response (in) cl k 0 9 % 2% fi 4 [ i 7 I I n o 0 0 o ” .1 fi 0 0 n ’o "a ” u ”9 0 0 u o o _. 0 ‘p (D “ 0 0 0 u “ u u u u _ Cb u u u " on “ u u “ o —1 ~ . n _ " u 0 a ‘ u u o u 0 u 0 l‘ n “ -2 _ 0 u 0 ‘ «V 0 u o -3 _ u _ -4 ,— M -5 1 l 1 J O 2 3 4 5 6 Time (sec) ““57"" (:23? 16 14—- Max Displacement (in) I 2 Frequency (Hz) 2.5 Response (in) I I I I ””T‘“ 7 .J v? «v 0 f; a: m «a . a - " r“ 0 cm ‘ 0 e, o 0 'n 0' I‘ ’0 "c; Q‘ 0 0 u 0 (D o “ u u n “ «a n 0 u o "I " u 0 ., 0 I! u u u 0 “ 0 0 0 0 u _ n 0 u u u “ o 0 0 0 n “ " 0 0 u 0 “ “ u 0 0 , 0 ‘4 0 " fl “0 0 «v " a " 0 "fl, ') “ J (m “ u "5» ‘I I '5 0 .‘L’ a) I I I I I 1 2 3 4 5 6 Time (sec) 16 14 12 .3 0 Max Displacement (in) 00 Frequency (Hz) ...
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