I am trying to test a part of my code but it shows me this message ! how can i solve it , Help plz !

I am trying to test a part of my code but it shows me this message ! how can i solve it , Help plz !

Cannot find an exact (case-sensitive) match for 'Material'

The closest match is: material in D:\toolbox\matlab\graph3d\material.m

[Jorg Woehl]

Are you trying to execute code that you have saved to a file called Material.m?

If that's the case, make sure to first navigate to the folder containing Material.m before calling Material from the command line.

[Sura Hatem Basir Al-Ani]

%Initialisation---------------------------------------
    clc;clear;format short g;format compact;
    %-------------------------------------------------
    %file name: Mat_Data
    %thickness and cost factor 
    
    %INPUTS-1: search parameters
    
    material(1).Sy=170;     material(1).t=[2,4,6,8,10,20,30];                   material(1).CF=1.00;
    material(2).Sy=280;     material(2).t=[1,2,3,4,6,8,10,12,14,16,18,20];      material(2).CF=1.37;
    material(3).Sy=180;     material(3).t=[2,4,6,8,10,20];                      material(3).CF=0.88;
    material(4).Sy=300;     material(4).t=[2,4,6,8,10,20];                      material(4).CF=1.47;
    material(5).Sy=190;     material(5).t=[2,4,6,8,10,20];                      material(5).CF=0.93;
    material(6).Sy=320;     material(6).t=[2,4,6,8,10,20];                      material(6).CF=1.57;
    material(7).Sy=220;     material(7).t=[2,4,6,8,10,20];                      material(7).CF=1.08;
    material(8).Sy=370;     material(8).t=[1,2,3,4,6,7,8,9,10,12,14,16];        material(8).CF=1.81;
    material(9).Sy=210;     material(9).t=[2,4,6,8,10,14,16,20,30,40];          material(9).CF=1.03;
    material(10).Sy=390;    material(10).t=[1,2,3,4,6,8,10,12,14,16,18,20];     material(10).CF=1.91;
    material(11).Sy=260;    material(11).t=[ 2,4,6,8,10,12,16,20,30,40,50];     material(11).CF=1.27;
    material(12).Sy=440;    material(12).t=[2,4,6,8,10];                        material(12).CF=2.16; 
    material(13).Sy=270;    material(13).t=[1,2,3,4,5,6,7,8,9,10];              material(13).CF=1.32;
    material(14).Sy=460;    material(14).t=[1,2,3,4,5,6,7,8,9,10];              material(14).CF=2.25;
    material(15).Sy=290;    material(15).t=[2,4,6,8,10,12,18,20,30];            material(15).CF=1.42;
    material(16).Sy=490;    material(16).t=[1,2,3,4,5,6,7,8,9,10];              material(16).CF=2.40;
    material(17).Sy=310;    material(17).t=[1,2,3,4,5,6,7,8,9,10];              material(17).CF=1.52;
    material(18).Sy=530;    material(18).t=[10,20,30];                          material(18).CF=2.60;
    material(19).Sy=340;    material(19).t=[4,5,8,14,20];                       material(19).CF=1.67;
    material(20).Sy=580;    material(20).t=[2,4,6,8,10];                        material(20).CF=2.84;
    material(21).Sy=370;    material(21).t=[2,4,6,8,10,15,18,20];               material(21).CF=1.81;
    material(22).Sy=420;    material(22).t=[2,4,6,8,10,12,14];                  material(22).CF=2.06;
    material(23).Sy=460;    material(23).t=[1,2,4,6,8,10,12]; 
    save ('Materials_table.mat', 'material')
    
    %INPUTS-2: Physical inputs
    P_bar=str2double(get(findobj('tag','Pressure'),'string')); %Internal Pressure of pressure vessel (Bar)
    C=str2double(get(findobj('tag','Capacity'),'string')); %Capacity (Litres)
    n_d=str2double(get(findobj('tag','Safety Factor'),'string')); %safety factor
    L_t=str2double(get(findobj('tag','Length Constraint'),'string')); %length of pressure vessel (m)
    H=str2double(get(findobj('tag','Height Constraint'),'string')); %Height of pressure vessel (m)
    step=str2double(get(findobj('tag','step'),'string')); %step for loop
    
    %conversion
    P=P_bar*10^5; %conversion of Mpa to Pa
    C=C*10^-3;    %conversion of litres to m^3
    
    %------------------------------------------------------------------------------------
    
    %Errors
    
    sol=[];
    
    if P_bar <= 0
       set(findobj('tag','errors'),'string','Un valid value of internal pressure')
       
       clc
       
       return 
    elseif isempty (P_bar) == true
        set(findobj('tag','errors'),'string','Un valid value of internal pressure')
        
        clc
        
        return
    end
    
     if L_t <= 0
       set(findobj('tag','errors'),'string','Un valid value of the lenght')
    
    
     elseif isempty (L_t) == true
        set(findobj('tag','errors'),'string','Un valid value of the lenght')
    
        clc
        return
     end
     
     if n_d <= 0
       set(findobj('tag','errors'),'string','Un valid value of safety factor')
    
    
     elseif isempty (n_d) == true
        set(findobj('tag','errors'),'string','Un valid value of safety factor')
        
        clc
        return
     end
     if C <= 0
       set(findobj('tag','errors'),'string','Un valid value of the capacity')
    
    
     elseif isempty (C) == true
        set(findobj('tag','errors'),'string','Un valid value of the capacity')
        
        clc
        return
     end
     
     if H <= 0
       set(findobj('tag','errors'),'string','Un valid value of the height')
    
    
     elseif isempty (H) == true
        set(findobj('tag','errors'),'string','Un valid value of the height')
     
        clc
        return
     end
     
     if step <= 0
       set(findobj('tag','errors'),'string','Un valid value of the step radius')
    
    
     elseif isempty (step) == true
        set(findobj('tag','errors'),'string','Un valid value of the step radius')
    
        clc
        return
     elseif isempty (step) > 1
        set(findobj('tag','errors'),'string','Un valid value of the step radius')
        
        clc
        return
     end
     
     
     %Calculations
     
    for  k=1:1:lenght(Material)
         CF=material(k).CF;
         S_y=material(k).S_y*10e6;
         allow_stress=S_y/n_d;
   
    end
    
     for  x=1:3:lenght(material(x).t)
          t=material(x).t(x)/1000;
          z=H/2-t;
     end
     
     
     for  r_i=0:step:z
          r_o=r_i+t;
          L_t=L+2*r_o;
     
     end
     
          
     if   t/r_i >= 0.05
          stress_t=((ri^2*P)/(ro^2-ri^2))*(1+(ro^2/ri^2));
          stress_r=((ri^2*P)/(ro^2-ri^2))*(1-(ro^2/ri^2));
          stress_l=((ri^2*P)/(ro^2-ri^2));
          type=thick;
          
     else
         stress_t=P*ri/t;
         stress_r=0;
         stress_l=P*ri/(2*t);    
         type=thin;
         
     end
     
stress_VM= sqrt(((stress_t-stress_r)^2+(stress_t-stress_l)^2+(stress_l-stress_r)^2)/2);

if stress_VM <= allow_stress && L>0 
    V_mat= 2*pi*r_i*t*L+4*pi*r_i^2*t;
    CI=V_mat*CF;
     stress_alldisp=stress_all/10e6;
     stress_VMdisp=stress_VM/10e6;
     L_tDisp=L+2*ro;
     Hdisp=2*ro;
    sol=[sol;r_i,r_o,t,stress_alldisp,stress_VMdisp,V_mat,CI,L_tDisp,Hdisp,type];
end
    
 sol_sorted=sortrows(sol,8);
 opt_sol=sol_sorted(1,:);
     
 
 %outputs
 
 set(findobj('tag','Inner radius'),'string', (sol_sorted(1,1)))
 set(findobj('tag','Material number'),'string', (sol_sorted(1,2)))
 set(findobj('tag','Thickness'),'string', (sol_sorted(1,3)))
 set(findobj('tag','Total lenght'),'string', (sol_sorted(1,4)))
 set(findobj('tag','Allowable Stress'),'string', (sol_sorted(1,5)))
 set(findobj('tag','Von Mises Stress'),'string', (sol_sorted(1,6)))
 set(findobj('tag','Vessel material volume'),'string', (sol_sorted(1,7)))
 set(findobj('tag','Vessel Cost Index'),'string', (sol_sorted(1,8)))
 set(findobj('tag','Height'),'string', (sol_sorted(1,9)))
 set(findobj('tag','Outer radius'),'string', (sol_sorted(1,10)))
 set(findobj('tag','Thickness type'),'string', (sol_sorted(1,11)))

this is my code

Answer 1

[KALYAN ACHARJYA]

This is due to a name mismatch between the upper case or lower case in the custom function name. Matlab is case-sensitive. Modify either in the function name or the main script.

[Walter Roberson]

... or it could mean that you really did mean Material with a capital but that there is no function named Material in the MATLAB path.

... or it could mean that you were use a matrix named Material but that matrix is not defined at that point in the code.

[Sura Hatem Basir Al-Ani]

%Initialisation---------------------------------------
    clc;clear;format short g;format compact;
    %-------------------------------------------------
    %file name: Mat_Data
    %thickness and cost factor 
    
    %INPUTS-1: search parameters
    
    material(1).Sy=170;     material(1).t=[2,4,6,8,10,20,30];                   material(1).CF=1.00;
    material(2).Sy=280;     material(2).t=[1,2,3,4,6,8,10,12,14,16,18,20];      material(2).CF=1.37;
    material(3).Sy=180;     material(3).t=[2,4,6,8,10,20];                      material(3).CF=0.88;
    material(4).Sy=300;     material(4).t=[2,4,6,8,10,20];                      material(4).CF=1.47;
    material(5).Sy=190;     material(5).t=[2,4,6,8,10,20];                      material(5).CF=0.93;
    material(6).Sy=320;     material(6).t=[2,4,6,8,10,20];                      material(6).CF=1.57;
    material(7).Sy=220;     material(7).t=[2,4,6,8,10,20];                      material(7).CF=1.08;
    material(8).Sy=370;     material(8).t=[1,2,3,4,6,7,8,9,10,12,14,16];        material(8).CF=1.81;
    material(9).Sy=210;     material(9).t=[2,4,6,8,10,14,16,20,30,40];          material(9).CF=1.03;
    material(10).Sy=390;    material(10).t=[1,2,3,4,6,8,10,12,14,16,18,20];     material(10).CF=1.91;
    material(11).Sy=260;    material(11).t=[ 2,4,6,8,10,12,16,20,30,40,50];     material(11).CF=1.27;
    material(12).Sy=440;    material(12).t=[2,4,6,8,10];                        material(12).CF=2.16; 
    material(13).Sy=270;    material(13).t=[1,2,3,4,5,6,7,8,9,10];              material(13).CF=1.32;
    material(14).Sy=460;    material(14).t=[1,2,3,4,5,6,7,8,9,10];              material(14).CF=2.25;
    material(15).Sy=290;    material(15).t=[2,4,6,8,10,12,18,20,30];            material(15).CF=1.42;
    material(16).Sy=490;    material(16).t=[1,2,3,4,5,6,7,8,9,10];              material(16).CF=2.40;
    material(17).Sy=310;    material(17).t=[1,2,3,4,5,6,7,8,9,10];              material(17).CF=1.52;
    material(18).Sy=530;    material(18).t=[10,20,30];                          material(18).CF=2.60;
    material(19).Sy=340;    material(19).t=[4,5,8,14,20];                       material(19).CF=1.67;
    material(20).Sy=580;    material(20).t=[2,4,6,8,10];                        material(20).CF=2.84;
    material(21).Sy=370;    material(21).t=[2,4,6,8,10,15,18,20];               material(21).CF=1.81;
    material(22).Sy=420;    material(22).t=[2,4,6,8,10,12,14];                  material(22).CF=2.06;
    material(23).Sy=460;    material(23).t=[1,2,4,6,8,10,12]; 
    save ('Materials_table.mat', 'material')
    
    %INPUTS-2: Physical inputs
    P_bar=str2double(get(findobj('tag','Pressure'),'string')); %Internal Pressure of pressure vessel (Bar)
    C=str2double(get(findobj('tag','Capacity'),'string')); %Capacity (Litres)
    n_d=str2double(get(findobj('tag','Safety Factor'),'string')); %safety factor
    L_t=str2double(get(findobj('tag','Length Constraint'),'string')); %length of pressure vessel (m)
    H=str2double(get(findobj('tag','Height Constraint'),'string')); %Height of pressure vessel (m)
    step=str2double(get(findobj('tag','step'),'string')); %step for loop
    
    %conversion
    P=P_bar*10^5; %conversion of Mpa to Pa
    C=C*10^-3;    %conversion of litres to m^3
    
    %------------------------------------------------------------------------------------
    
    %Errors
    
    sol=[];
    
    if P_bar <= 0
       set(findobj('tag','errors'),'string','Un valid value of internal pressure')
       
       clc
       
       return 
    elseif isempty (P_bar) == true
        set(findobj('tag','errors'),'string','Un valid value of internal pressure')
        
        clc
        
        return
    end
    
     if L_t <= 0
       set(findobj('tag','errors'),'string','Un valid value of the lenght')
    
    
     elseif isempty (L_t) == true
        set(findobj('tag','errors'),'string','Un valid value of the lenght')
    
        clc
        return
     end
     
     if n_d <= 0
       set(findobj('tag','errors'),'string','Un valid value of safety factor')
    
    
     elseif isempty (n_d) == true
        set(findobj('tag','errors'),'string','Un valid value of safety factor')
        
        clc
        return
     end
     if C <= 0
       set(findobj('tag','errors'),'string','Un valid value of the capacity')
    
    
     elseif isempty (C) == true
        set(findobj('tag','errors'),'string','Un valid value of the capacity')
        
        clc
        return
     end
     
     if H <= 0
       set(findobj('tag','errors'),'string','Un valid value of the height')
    
    
     elseif isempty (H) == true
        set(findobj('tag','errors'),'string','Un valid value of the height')
     
        clc
        return
     end
     
     if step <= 0
       set(findobj('tag','errors'),'string','Un valid value of the step radius')
    
    
     elseif isempty (step) == true
        set(findobj('tag','errors'),'string','Un valid value of the step radius')
    
        clc
        return
     elseif isempty (step) > 1
        set(findobj('tag','errors'),'string','Un valid value of the step radius')
        
        clc
        return
     end
     
     
     %Calculations
     
    for  k=1:1:lenght(Material)
         CF=material(k).CF;
         S_y=material(k).S_y*10e6;
         allow_stress=S_y/n_d;
   
    end
    
     for  x=1:3:lenght(material(x).t)
          t=material(x).t(x)/1000;
          z=H/2-t;
     end
     
     
     for  r_i=0:step:z
          r_o=r_i+t;
          L_t=L+2*r_o;
     
     end
     
          
     if   t/r_i >= 0.05
          stress_t=((ri^2*P)/(ro^2-ri^2))*(1+(ro^2/ri^2));
          stress_r=((ri^2*P)/(ro^2-ri^2))*(1-(ro^2/ri^2));
          stress_l=((ri^2*P)/(ro^2-ri^2));
          type=thick;
          
     else
         stress_t=P*ri/t;
         stress_r=0;
         stress_l=P*ri/(2*t);    
         type=thin;
         
     end
     
stress_VM= sqrt(((stress_t-stress_r)^2+(stress_t-stress_l)^2+(stress_l-stress_r)^2)/2);

if stress_VM <= allow_stress && L>0 
    V_mat= 2*pi*r_i*t*L+4*pi*r_i^2*t;
    CI=V_mat*CF;
     stress_alldisp=stress_all/10e6;
     stress_VMdisp=stress_VM/10e6;
     L_tDisp=L+2*ro;
     Hdisp=2*ro;
    sol=[sol;r_i,r_o,t,stress_alldisp,stress_VMdisp,V_mat,CI,L_tDisp,Hdisp,type];
end
    
 sol_sorted=sortrows(sol,8);
 opt_sol=sol_sorted(1,:);
     
 
 %outputs
 
 set(findobj('tag','Inner radius'),'string', (sol_sorted(1,1)))
 set(findobj('tag','Material number'),'string', (sol_sorted(1,2)))
 set(findobj('tag','Thickness'),'string', (sol_sorted(1,3)))
 set(findobj('tag','Total lenght'),'string', (sol_sorted(1,4)))
 set(findobj('tag','Allowable Stress'),'string', (sol_sorted(1,5)))
 set(findobj('tag','Von Mises Stress'),'string', (sol_sorted(1,6)))
 set(findobj('tag','Vessel material volume'),'string', (sol_sorted(1,7)))
 set(findobj('tag','Vessel Cost Index'),'string', (sol_sorted(1,8)))
 set(findobj('tag','Height'),'string', (sol_sorted(1,9)))
 set(findobj('tag','Outer radius'),'string', (sol_sorted(1,10)))
 set(findobj('tag','Thickness type'),'string', (sol_sorted(1,11)))

this is my code

Answer 2

[Walter Roberson]

    for  k=1:1:lenght(Material)

You have not defined any variable named Material but you have defined a variable named material . As you are looping over the number of elements of something and inside that loop you access material() indexed by the loop variable, it would be most natural to take the length of material rather than of whatever Material is.

         CF=material(k).CF;
         S_y=material(k).S_y*10e6;
         allow_stress=S_y/n_d;
   
    end
    
     for  x=1:3:lenght(material(x).t)
          t=material(x).t(x)/1000;
          z=H/2-t;
     end

Please also pay attention to the fact that in both of those for loops you use a function named lenght when you probably intended length -- ht compared to th

[Sura Hatem Basir Al-Ani]

now it shows me this:

Index exceeds the number of array elements (7).

what does that mean?

[Rik]

That means that you are trying to get the 8th element from a list that is 7 elements long.

[Walter Roberson]

     for  nux=1:3:lenght(material(x).t)
          t=material(x).t(x)/1000;
          z=H/2-t;
     end

You are taking whatever value if x is left over in memory and using it to index the material structure, and you find the length of the t field there. You use that length as the upper limit in a for x loop. You then use that to index the material structure and pull out the t field and then you use the same index to index the t vector. This is sort of like going down the diagonal

For example if the value of x in memory happened to be 2 then you would have taken length of material(2).t which is 12. You would then be doing for x = 1:3:12. You would access materials(1).t(1) then materials(4).t(4) then materials(7).t(7) but that would fail because materials(7).t is only 6 elements long.

Note also that when you succeed in indexing the t vector that you calculate z. However you do not save the value and it will be overwritten by the next for x iteration.