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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 (311
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”) =—usina)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 kipsSAZ/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,'rO');
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
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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 alphabeta 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
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 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
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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\°
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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
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 Spring '06
 Lynch

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