Subscript indices must either be real positive integers or logicals.

Why do i get " Subscript indices must either be real positive integers or logicals." when i run my code and how can i fix it

%% problem parameters

P = 0.5*9.81; % [load N]

L = 0.5; % [Length M]

E = 69*10^6;

I = ((0.0032^3)*(0.0088))/12;

%%

x = (0:0.025:L);

v = zeros(length(x));

for i = 0.025:length(a)

a = 1/(E*I);

b = 1/12;

c = 1/16;

v(i) = a*((b*P*(i^3))-(1*c*P*(L^2)));

end

Answer 1

[KALYAN ACHARJYA]

The error is

 v(i) 
%..^

In Matlab indexing, the i value must be positive interger, like 1,2,3,,,,

v(1)>> Allow

v(0)>> Not Allow

v(0.25)>> Not Allow

v(-3)>> Not Allow

v(7)>> Allow

Hope you get the sufficients hints. In your case i starts from 0.25, hence v(0.25) Not allowed.

% Define a
% ????

%% problem parameters
P = 0.5*9.81;      % [load N]
L = 0.5;            % [Length M]
E = 69*10^6;
I = ((0.0032^3)*(0.0088))/12;
%%
x = (0:0.025:L);
v = zeros(length(x));
i=0.025:length(a) 
for j =1:length(i) 
    a = 1/(E*I);
    b = 1/12;
    c = 1/16;
v(j)= a*((b*P*(i(j)^3))-(1*c*P*(L^2)));
end

Without Loop:

% Define a
% ????

%% problem parameters
P = 0.5*9.81;      % [load N]
L = 0.5;            % [Length M]
E = 69*10^6;
I = ((0.0032^3)*(0.0088))/12;
%%
x = (0:0.025:L);
v = zeros(length(x));
i=0.025:length(a) 
a=1/(E*I);
b=1/12;
c=1/16;  
v=a*((b*P.*(i.^3))-(1*c*P*(L^2)));

Any issue let me know?

[nathan Short]

this hasnt produced what i wanted, i am after an array that stores all the values for when 0 to 0.5 is entered in incriments of 0.025

[KALYAN ACHARJYA]

Please define variable a?

[nathan Short]

i think it would be 20?? would a = 0.5/0.025

[KALYAN ACHARJYA]

If a=20,

the what does the following means

i = 0.025:length(a)

it return i as a single element

i=0.025

Are you sure? No Need of loop.The v becomes

v =
  -46.2192 

If no, What exactly you are trying to do?

[nathan Short]

i want to return v as an array for each input of i (0;0.025;0.5). then plot the values against each other as it should display a deflection curve

Answer 2

[Image Analyst]

The FAQ explains it pretty well: Click here for the FAQ on that error