单自由度受动力荷载的newmark程序

单自由度体系受动力荷载Newmark法小程序

W = 200 kip; k = 12.2 k/in;po = 36.6 kip; t1 = 0.4 sec

.△t=0.05s

Repeat the analysis assuming zero damping.

Include both cases (zero and 2% damping) on same plot

Solution

The computer program is as followed:

p0 = 0; p1 = 36.6; p2 = 0;

t0 = 0; t1 = 0.4;

u0 = 0; v0 = 0;

dt = 0.05; % Solution time step

n = t1/dt/2;

nn = 10; % Dynamic load cycle Number

p = [p0:p1/n:p1, p1-p1/n:-p1/n:p2]; % One cycle input load

for ii = 1:72

a(1,ii) = 0;

end

p=[p,a];

t = [t0:dt:t1*nn]'; % Total time

m = 200/386; k = 12.2; zeta = 0.05; % System mass, stiffness and damping ratio

gamma = 0.5; beta = 0.25; % Define the method

omega = sqrt(k/m);

c = 2*m*omega*zeta; % Damping

[u1,v1,a1] = dynresp(m,k,c,p,dt,0,0,gamma,beta);

[u2,v2,a2] = dynresp(m,k,0,p,dt,0,0,gamma,beta);

figure

plot(t,u1,t,u2)

figure

plot(t,v1,t,v2)

figure

plot(t,a1,t,a2)

function [u,v,a] = dynresp(m,k,c,p,dt,u0,v0,gamma,beta)

% solve the dynamic response of SDOF

% u = displacement of the structural response

% v = velocity

% a = acceleration

% m = mass of the system

% k = stiffness of the system

% c = damping

% p = input dynamic load for the system

% gamma, beta = Newmark's Method selection

% t = input load duration

% u0, v0 = initial displacement and velocity

n = length(p);

% dt = t/(n-1);

u(1) = u0;

v(1) = v0;

a(1) = (p(1)-c*v0-k*u0)/m;

k1 = k+gamma*c/(beta*dt)+m/(beta*dt^2);

pm1 = m/(beta*dt)+gamma*c/beta;

pm2 = m/2/beta+dt*(gamma/2/beta-1)*c;

dp = p(2:n)-p(1:n-1);

for ii = 1:n-1

dp1(ii) = dp(ii)+pm1*v(ii)+pm2*a(ii);

du(ii) = dp1(ii)/k1;

dv(ii) = gamma*du(ii)/beta/dt-gamma*v(ii)/beta+dt*(1-gamma/2/beta)*a(ii);

da(ii) = du(ii)/beta/dt^2-v(ii)/beta/dt-a(ii)/2/beta;

u(ii+1) = u(ii)+du(ii);

v(ii+1) = v(ii)+dv(ii);

a(ii+1) = a(ii)+da(ii);

end

u = u';

v = v';

a = a';

The result: