# Newmark Beta Method problem

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Chaudhary P Patel on 12 Jan 2021
Moved: Rik on 27 Sep 2022
alpha=0.25;
delta=0.5;
%-------------------
delt=0.005;
% delt=28;
%-------------------
a0=1/(alpha*delt^2);
a1=delta/(alpha*delt);
a2=1/(alpha*delt);
a3=(1/(2*alpha))-1;
a4=(delta/alpha)-1;
a5=(delt/2)*((delta/alpha)-2);
a6=delt*(1-delta);
a7=delta*delt;
%------------------ Co-efficient of Damping--------------------------------
wn= 2*pi./T(1,1);
xeta=0.05;
R=nf*nodof;
u=zeros(R,1);
udot=zeros(R,1);
%-------------------
uddot=inv(MGf)*(F-(KGf*u));
% uddot=[0;10];
%-------------------
khat=KGf+a0*MGf;
for i=1:1:15
if(i==1)
ftdelt=F+(MGf*((a0*u(:,i))+(a2*udot(:,i))+(a3*uddot(:,i))))+(C*((a1*u(:,i))+(a4*udot(:,i))+(a5*uddot(:,i))));%Force
else
ftdelt=F+(MGf*((a0*utdelt(:,i))+(a2*udottdelt(:,i))+(a3*uddtdelt(:,i))))+(C*((a1*utdelt(:,i))+(a4*udottdelt(:,i))+(a5*uddtdelt(:,i))));%Force
end
utdelt(:,i+1)=inv(khat)*ftdelt;%Displacement
%Unable to perform assignment because the indices on the left side are not compatible with the size of the
right side.
if(i==1)
uddtdelt(:,i+1)=(a0*(utdelt(:,i+1)-u(:,i)))-(a2*udot(:,i))-(a3*uddot(:,i));%Acceleration
else
uddtdelt(:,i+1)=(a0*(utdelt(:,i+1)-utdelt(:,i)))-(a2*udottdelt(:,i))-(a3*uddtdelt(:,i));%Acceleration
end
if(i==1)
udottdelt(:,i+1)=udot(:,i)+(a6*uddot(:,i))+(a7*uddtdelt(:,i+1));%Velocity
else
udottdelt(:,i+1)=udottdelt(:,i)+(a6*uddtdelt(:,i))+(a7*uddtdelt(:,i+1));%Velocity
end
t=(i+1)*delt;
end
##### 2 CommentsShowHide 1 older comment
Image Analyst on 12 Jan 2021
Moved: Rik on 27 Sep 2022