I need to do 2 plots

First plot between time (t)and consoledation settlement(Se)

Second plot between time and total settelment (Se+Scplot)

But it keeps telling me an error about unmatching matrixes in the U loop

The problem is that bothe Seplot and Se depends on time.

Note: I can put any values for time and the relation should be a curve Below is the code

clc;

disp('This code determines stress increment distribution ,immediate ');
disp('settlement ,consolidation settlelment and total settlelment for ');
disp('various area shapes');
disp('This code determines the variation of total settlement vs.time ');
disp('----------------------------------------------------------------');
disp('Codes of the loaded area shapes are ');
disp('circle=0');
disp('square=1');
disp('rectangular=2');
disp('strip=3');
disp('----------------------------------------------------------------');
%code=input('enter your loaded area code=');
code=1;

while(0)
if code<0||code>3
     disp('Code that you entered is not in the range of 0 and 3');
     disp('Run the program with appropriate codes that are between 0 and 3');
else
break
end
end
disp('----------------------------------------------------------------');  
%qa=input('insert q(all) (KN/m^2)=');
qa=210;
%p=input('insert axial load (KN) =');
p=2460;
%FS=input('range of FS is (2-4),choose yours =');
FS=3;
while FS<2 || FS>4
      FS= input('your FS is out of range ,insert FS (2-4)');
end
qu=qa*FS; 
disp('----------------------------------------------------------------');
disp('Es of sand =7000 KN/m^2');
disp('angle of friction of sand=37 deg');
disp('angle of friction of clay=23 deg');
disp('----------------------------------------------------------------');
%n=input('number of layers');
N=3;

%for i=1:n
%h(i)=input('insert depth of the layer(m)= ');
%Es(i)=input('insert modulus of elasticity of the layer(KN/m^2)=');
%uw(i)=input('insert unit weight of the layer(KN/m^3)= ');
%o(i)=input('insert angle of friction for the layer =');
%c(i)=input('cohesion of the layers  = ');
%end
h=[2.5 0.5 2.5];
Es=[7000 6000 6000];
uw=[16 17.5 16];
o=[37 37 23];
c=[0 0 15];

%df=input('insert foundation depth (m)=');
df=1;
%GWT=input('insert total ground water level');
GWT=3;
gwt=sum(h(1:N))-GWT;
%GWTf=input('insert ground water level from the foundation base(m)= ');
GWTf=0;
%y=input('insert number of layers above the foundation base from top to bottom ');
%for i=1:y

%dlaf(i)=input('insrt depth of layer above foundation from top to bottom (m) = ');
%uwf(i)=input('insert unit weight of layer above foundation from top to bottom (KN/m^3)= ');

%end

y=1;
dlaf=1;
uwf=16;
%for i=1:2
%res=sum(dlaf(i).*uwf(i));
res=dlaf*uwf;
%end
qd=res-(GWTf*9.81);
disp(['effective stress at the foundation base = ' num2str(qd) ' N/mm2 ' ]);
r=N-y;
%dlbf(i)=input('insert depth of layer below the foundation from top to bottom (m) =');
dlbf=[1.5 0.5 2.5];
B=0.3;
i=1;
er=0.02;
%end
if i==1
    while B>=0.3 && B<=dlbf(i) && er>0.001
          Bold=B;
    if code==0
       k1=1.3;
       k2=0.3;
    elseif code==1
           k1=1.3;
           k2=0.4;
    elseif code==2
           L=B;
           k1=1+(0.2*(B\L));
           k2=0.5-(0.1*(B\L));
    else
           k1=1;
           k2=0.5;
    end
    for w=1:y
        if gwt<=df
           uw(i)=uw(i)-9.81;
    elseif gwt>df && gwt<(df+B)
           uw(i)=uw(i)-9.81*(1-((gwt-df)/B));
        end
    end

      Nq=exp(pi*(tan(o(i))))*(tan(45+(o(i)/2)))^2;
      Nc=(Nq-1)*atan(o(i));
      Nuw=(Nq-1)*(tan(1.4*o(i)));
      if o(i)~=0
         Nc=(Nq-1)./tan(o(i));
      else
             Nc=5.7;
      end

      B=(qu-k1.*c(i).*Nc-qd.*Nq)./(k2.*uw(i).*Nuw);
      er=(B-Bold)./B;

end
end

B=abs(B);
i=i+1;
g=y+1; 
%layerB is the order of the layer at which the B interscts.
while B>=0.3 && B<sum(h(g:N)) && er>0.001

for i=g:N 

         Bold=B;
         if code==0
            k1=1.3;
            k2=0.3;
         elseif code==1
                k1=1.3;
                k2=0.4;
         elseif code==2
                k1=1+(0.2*(B\l));
                k2=0.5-(0.1*(B\l));
         else
                k1=1;
                k2=0.5;
         end

         for ji=g:N
             if sum(h(g:ji))>B
                 layerB=ji;
             end
         end
         m=(sum(h(g:i)))-B;
         v=h(layerB)-m;
         h(layerB)=v;
         xxx=r+1;
         hh(layerB)=xxx;

         for ij=g:layerB
            if ij<layerB 
              hh(ij)=h(ij);  
            end  
         end

         for ii=g:layerB
             w(ii)=hh(ii)/B;
         end
             c1=sum(w(g:layerB).*c(g:layerB));
             o1=sum(w(g:layerB).*o(g:layerB));
            uw1=sum(uw(g:layerB).*w(g:layerB));

         if gwt<=df
            uw1=uw1-9.81;
         elseif gwt>df && gwt<(df+B)
                uw1=uw1-9.81.*(1-((gwt-df)./B));
         end

         aq=exp(pi.*(0.75-o1./360)*tan(o1));
         Nq=sqrt(aq)./(2.*sqrt(cos(45+o1./2)));
         Nuw=(2.*(Nq+1).*tan(o1))/(1+(0.4.*sin(4.*o1)));

         if o1~=0
            Nc=(Nq-1)/tan(o1);
         else
           Nc=5.7;
         end

     B=(qu-k1.*c1.*Nc-qd.*Nq)/(k2.*uw1.*Nuw);
     er=abs((B-Bold)/B);

end
end
 disp(['B=' num2str(B) ' m ']);
disp('----------------------------------------------------------------');
disp('do you want to insert the foundation load in terms of axial');
disp('load or in terms of axial load plus weight of foundation');
disp('axil= 1 , axil+weight of foundation= 0 ');
%ss=input('insert your answer=');
ss=1;
i=1;
if ss==1
if code==1
    L=B;
      a=B*L;
      disp(['L= ' num2str(L) ' m ']);
      disp(['A= ' num2str(a) ' m^2 ']);
      wf=(p/a)-9.81*GWTf;

        m=B/sum(h(i:N));
        n=L/sum(h(i:N));
        aa=2*m*n*(m^2+n^2+1)^0.5;
        bb=m^2+n^2-m^2*n^2+1;
        cc=(m^2+n^2+2)/(m^2+n^2+1);
        dd=2*m*n*((m^2+n^2+1)^0.5);
        ee=m^2+n^2-m^2*n^2+1;

        Ir=(1/(4*pi))*((aa/bb)*cc+atan(dd/ee));
        seg=wf*Ir;

        disp(['stress increement at the last layer= ' num2str(seg) ' KN/m^2 ']);
        %t=input('insert number of years to calculate the total settelment');
        t=3;
        %nsl=input('insert number of NONE CLAY layers below foundation base');
        nsl=2;
        for i=1:nsl
            %osl(i)=input('insert orders of  NONE CLAY layers below foundation base ');
        end
        osl=[1 2];
        rr=r+1;
      xx=0;
      yy=0;

    for iii=1:length(dlbf)
        if xx==0 && yy==0
        if sum (dlbf(1:iii))>=(B/2)
            layern=iii;

            xx=sum(dlbf(1:layern))-(B/2);

            yy=dlbf(layern)-xx;

            z(layern)=sum(dlbf(1:layern-1))-yy/2;

            z(layern+1)=sum(dlbf(1:layern-1))+yy-(xx/2);
            disp(['layern= ' num2str(layern)]);
            disp(['hight of layern= ' num2str(dlbf(layern))]);
            disp(['xx = ' num2str(xx)]);      
            disp(['z(layern)=' num2str(z(layern))]);
            disp(['z(layern+1)=' num2str(z(layern+1))]);
            disp(['yy = ' num2str(yy)]);

           for i=1:rr
               if i~=layern &&i~=layern+1 
                     z(i)=sum(dlbf(1:i))-(dlbf(i)/2); 
               end
           end  

                     z(rr+1)=sum(dlbf(1:rr))-(dlbf(rr)/2);
         end
        end

    disp(['z=' num2str(z) ]);
    for i=rr+1
     if i==osl
          if z(i+1)>=0 && z(i+1)<=(B/2)
             Iz=0.1+(z(i+1)./B).*(2.*0.5-0.2);

          end
          if z(i+1)>(B/2) && z(i+1)<(2*B)
              Iz=0.667.*0.5.*(2-(z(i+1)./B));

          end

      end

    end

     disp(['Iz= ' num2str(Iz)] );
  end
  end
     C1=1-0.5.*(qd./(wf-qd));
     C2=1+0.2.*log((linspace(1,t,100)./0.1));
     C3=1.03-0.03*(L./B);
      if C3<0.73
         C3=0.73;
      end
     X=sum((z(i).*Iz(i))./Es(ocl));
     disp(['X = ' num2str(X)]);

     Se=C1.*C2.*C3.*(wf-qd).*X;
     gg=input('Flexible foundation =1,Rigid Foundation=0');
     if gg==0
         Se=(0.93).*Se;
     end
     disp(['elastic settelment (m) = ' num2str(Se)]); 
     disp('-----------------------------------------------------');

     %ocl=input('insert orders of clay layers as a vector ');
     ocl=[2 3];
     %eo=input('insert void ratio of clay layer');
     eo=[0.5 0.5];
     %Ms=input ('insert poissons ratio of clay= ');
     Ms=[0.5 0.5];
     %disp('Normally consolidated clay ----> OCR=1')
     %OCR=input('insert over consoloidation ratio of clay layer'); 
     OCR=1;
     %if OCR==1
         %Cr=input('insert recompretion index of clay layer');

         Sc=(Cr./(1+eo)).*h(ocl).*log(OCR);
     %else
         %Cc=input('insert compretion index of clay layer'); 
         %Sc=(Cc./(1+eo)).*h(ocl).*log(OCR);
     %end

      %Cv=input('insert coffetiont of consoledation Cv ');
      Cv=0.002;
      Scplot=zeros([1,9]);
  for U=0.1:0.1:0.9
      if U==0.1
          Tv=0.008;
          Scplot(1)=Sc.*U;
      end
      if U==0.2
          Tv=0.031;
          Scplot(2)=Sc.*U;
      end
      if U==0.3
          Tv=0.071;
          Scplot(3)=Sc.*U;
      end
      if U==0.4
          Tv=0.126;
          Scplot(4)=Sc.*U;
      end
      if U==0.5
          Tv=0.197;
          Scplot(5)=Sc.*U;
      end
      if U==0.6
          Tv=0.287;
          Scplot(6)=Sc.*U;
      end
      if U==0.7
          Tv=0.403;
          Scplot(7)=Sc.*U;
      end
      if U==0.8
          Tv=0.567;
          Scplot(8)=Sc.*U;
      end
      if U==0.9
          Tv=0.848;
          Scplot(9)=Sc.*U;
      end

end
    disp(['Scplot' num2str(Scplot)']);

      opl=input('insert order of nonepermiable layers');

while i<n
if ocl(i)==opl(i+1)

      HD(i)=h(i);
  else
      HD=h(i)./2;

end
i=i+1;
end
tc=(Tv.*(HD(i).^2)./Cv);
plot(t,Scplot,'r-');
xlabel('Time (s)')
ylabel('Sc(m)')
title ('Sc vs. time')

St=Sc(nsl)+Se;
disp(St);
plot(t,St,'r-',[0 t],[Sc(i) Sc(i+1)],'b--',[0 t],[Se(i) Se(i+1)],'k--');
xlabel('Time (s)')
ylabel('St(m)')
title ('St vs. time')
legend('St','Sc','Se')
end

[Image Analyst]

You can learn how to format your code here: http://www.mathworks.com/matlabcentral/answers/13205-tutorial-how-to-format-your-question-with-markup, though in this case, it's probably best just to attach the m-file with the paperclip icon since there is so much code.

Also, z never gets defined, so look at this link to discover why z never gets assigned.

Answer 1

[Walter Roberson]

Undefined function or variable 'z'.

This occurs because you only assign to z inside your for loop and multiple "if" conditions need to be true for an assignment to be made. You should get in the habit of assigning something to your variables that are written to conditionally, even if it is z=[]; (and then make sure you test the size before relying on elements being there.)

[saja Alawneh]

Do you mean that the z value is not saved in the memory , if yes then how should I do that ,by the way if u run the code z is displayed !!

[Walter Roberson]

I copied the code and ran it, and it gave me the above error message right after the output

stress increement at the last layer= 24.97 KN/m^2

Nothing is being stored in your z. If you are not getting that error then you might potentially already have a z stored in your workspace.

[pinaki adak]

PLease share the complete code without error. I think Mr.Saja has initiated a good code . Please share final one . It will help us

[Walter Roberson]

The calculated B is 3.7345

dlbf=[1.5 0.5 2.5];

Notice the first entry 1.5 is not greater than 3.7345/2 -- so the first time,

if sum (dlbf(1:iii))>=(B/2)

is false. The body of the if is not executed, so nothing has been stored into z yet.

After the end of that if there is

        disp(['z=' num2str(z) ]);

but nothing has been assigned to z yet, so that statement fails.

I cannot say what the "correct" way to repair the code is, as I am not familiar with the algorithm being used.

It seems to me, though, that if z has not been assigned to at all yet, that it does not make sense to display z, and it probably does not make sense to go on to the calculations after that, which depend upon z.

My suspicion is that the prior loop should be ended before that line that depends upon z. That way z would get assigned to if there is at least one case where the cumulative sum of dlbf is >= B/2 . Really the code should be designed to detect the case where z never gets assigned to at all (the plane is never met).