Hello everyone,

I want to apply boundary conditions In Differential Quadrature Method using Wilson’s method. So I have this condition

Clamped-Clamoed: w=0; phi= 0; a x=0 and x =L

i use this formulation and its works

d(nn+1,1)=1.0;

d(nn+2,n)=1.0;

d(nn+3,n+1)=1.0;

d(nn+4,nn)=1.0;

d(1,nn+1)=1.0;

d(n,nn+2)=1.0;

d(n+1,nn+3)=1.0;

d(nn,nn+4)=1.0;

The boundary conditions for simply-simply end are as follows: w =0; diff (phi, x) =0; ax =0; and x =L

I use this formulation and it also works.

d(nn+1,1)=1.0

d(nn+2,n)=1.0

d(nn+3:nn+3,n+1:nn)=c(1:1,1:n,1)/l

d(nn+4:nn+4,n+1:nn)=c(n:n,1:n,1)/l

d(1,nn+1)=1.0

d(n,nn+2)=1.0

for i=1:n

d(n+i,nn+3)=d(nn+3,n+i)

d(n+i,nn+4)=d(nn+4,n+i)

end

Clamped-Free i use this formulation

w=0; phi= 0 a x=0 and diff(phi,x)=0,diff(w,x)+phi=0 a x=L

d(nn+1,1)=1.0;

d(nn+2,n)=1.0;

d(nn+3,n+1)=1.0;

d(nn+4,nn)=1.0;

d(nn+1:nn+1,n+1:nn)=c(n:n,1:n,1)/l;

d(nn+2:nn+2,n+1:nn)=c(n:n,1:n,1)/l;

d(nn+3:nn+3,n+1:nn)=1.0;

d(nn+4:nn+4,n+1:nn)=1.0;

for i=1:n

d(i,nn+1)=d(nn+1,i)+d(nn+3,i);

d(i,nn+2)=d(nn+2,i)+d(nn+4,i);

d(i,nn+3)=d(nn+3,i);

d(i,nn+4)=d(nn+4,i);

end

But it doesn't work, please tell me what's wrong with this formulation.