Can you share your program. Then I can try to suggest you.

// Beam on an Elastic foundation.
clear
// Example: Aluminum I-beam sitting on
// loose sand
E=72E9; // in N/m^2
J=2.45E-6; // in m^4
b=0.075; // width of beam in m
k0=5E6; // Modulus of foundation in N/m^3
k=b*k0; // in N/m^2
L=6.8; // in m
P0=1E3; // in N
ke=k/(E*J);
N=20;
dx=L/N;
dx2=dx*dx;
for i=1:N-1
x(i)=i*dx;
P(i)=P0*(0.25-(x(i)/L-0.5)^2); // The loading function
end
x=[0;x;L];
// Matrix A setup
A=0;
A(1,1)=-2;
A(1,2)=1;
for i=2:N-2
A(i,i-1)=1;
A(i,i)=-2;
A(i,i+1)=1;
end
A(N-1,N-2)=1;
A(N-1,N-1)=-2;
// Matrix B setup
B=[A,-dx2*eye(N-1,N-1);dx2*ke*eye(N-1,N-1),A];
// RHS vector R setup
for i=1:N-1
R(i)=0;
end
R=[R;dx2*P/(E*J)];
// Solve system BW=R
W=B\R;
// Augment of output vector W to include endvalues
for i=1:N-1
y(i)=W(i);
z(i)=W(i+N-1);
end
y=[0;y;0];
z=[0;z;0];
// plot output
//plot(x,y,"*",x,z,"+")
// Let's compare this to the beam without the foundation
// All we need to do is set ke=0 and recalculate the matrix B
// Designate the no foundation case as y0 and z0 for y and z
ke=0;
B=[A,-dx2*eye(N-1,N-1);dx2*ke*eye(N-1,N-1),A];
W=B\R;
for i=1:N-1
y0(i)=W(i);
z0(i)=W(i+N-1);
end
y0=[0;y0;0]/100;
z0=[0;z0;0]/100;
plot(x,y,"*",x,z,"+",x,y0,"^",x,z0,"o")

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