Division by zero error

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Elton Sim
Elton Sim 2020-9-30
评论： Elton Sim 2020-9-30
I have the following functions in which I am doing a Ritz Method analysis for a tapered cantilever beam, under clamped free conditions.
I am repeatedly getting a zero division error even though I have checked that there are no zeros in the denominators.
Can I know how do I fix this?
I have attached the file code and pasted the lines for reference.
Thanks in advance
>> %Constants
E = 2*10^12; %Young's Modulus
rho = 300; %Density
L = 29; %Length
syms i j y %Declare i, j, y variables
%Variable Equations
o = y/L; %Position Ratios
A = 4.1*(1.005-o); %Area
I = 0.03*(1.005-0.5*o-0.5*o^2); %Moment of Inertia
%Miscellenous
n = 7; %Iterations
%Basis Functions Cosine
phis_i = cos((i-3)*pi*o);
phis_j = cos((j-3)*pi*o);
%For Mc Matrix
sig_fun = rho*A*phis_i*phis_j; %Function for integral a
sigma = int (sig_fun, y, 0, L); %Integral a for y from 0 to L
%Matrices Cosine Series
Mc = zeros (n);
for si = 1:n
for sj = 1:n
Mc(si,sj) = subs (sigma,{i,j},{si,sj});
end
end

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采纳的回答

Walter Roberson
Walter Roberson 2020-9-30
%Mc(si,sj) = subs (sigma,{i,j},{si,sj});
t1 = limit(sigma, i, si);
t2 = limit(t1, j, sj);
Mc(si, sj) = t2

更多回答（1 个）

Ameer Hamza
Ameer Hamza 2020-9-30
Division by zero occurs in your equation. Following image show expression for sigma from the live editor
As you can see, there are several conditions when 0 can come in the denominator. One such condition is i==j. Therefore, when the loop starts, you have i=1, j=1, and i-j=0, and MATLAB throws an error.
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Elton Sim
Elton Sim 2020-9-30
Hi Walter, I can't do that because although it resolves the problem, the results will not be the one I want. There must be cases where i=j for my numerical solution.
Ameer Hamza
Ameer Hamza 2020-9-30
It is just one case. You will also get zero in the denominator at j=3, and also at i+j-6=0. I don't think there is an easy way to avoid this other than adding all these conditions in for-loop. For example,
for si = 1:n
for sj = 1:n
if ~((i==j) | (j==3) | ((i+j-6)==0))
Mc(si, sj) = subs(sigma, {i,j}, {si,sj});
end
end
end

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