Why this loop is executing 4 times?

% For plane slider: H = Ho + a(1-x)
Ho = 1;
alpha = 0.1;
eps = 0.1;
a = 1.0;
lbar = 0.1;
SIGMA = 0:0.05:0.15;
sigma = zeros(1,length(SIGMA));
for j = 1:length(SIGMA)
sigma = SIGMA(j);
H = @(x) Ho + a*(1 - x); % For plane slider: H = Ho + a(1-x)
G1 = @(x) H(x).^3 + 3 .* H(x).^2 .* alpha + 3 .* H(x) .* alpha^2 + 3 .* H(x) .* sigma^2 + eps + 3*sigma^2*alpha + alpha^3 - 12*lbar^2 .* (H(x) + alpha);
G2 = @(x) 24 * lbar^3 .* tanh(H(x)./(2*lbar));
G3 = @(x) (12*lbar^2*alpha - eps - alpha^3 - 3*sigma^2*alpha) .* (1 - (tanh(H(x)./(2*lbar))).^2);
G = @(x) G1(x) + G2(x) + G3(x);
Hm1 = @(x) H(x).* (1 ./ G(x));
Hm2 = @(x) (1 ./ G(x));
IntHm1 = integral(Hm1,0,1);
IntHm2 = integral(Hm2,0,1);
Hm = IntHm1 / IntHm2;
P1 = @(x) 6 .* (1 ./ G(x)) .* (H(x) - Hm);
P2 = @(x) integral(P1,0,x);
% Calculating dimensionless load carrying capacity(W):
W(j) = integral(P2,0,1, 'ArrayValued', true)
% Calculating non dimensional Frictional Force(F):
F1 = @(x) (H(x).* P1(x)) ./2 + (1 ./ H(x));
F(j) = integral(F1,0,1)
% Calculating coefficient of friction(f):
f(j) = F(j)/W(j)
% Calculating non dimensional temperature rise(deltaT):
deltaT(j) = F(j)/Hm
% Calculating the center of pressure(Xbar):
Xbar1 = @(x) P2(x) .* x;
Xbar(j) = integral(Xbar1, 0, 1, 'ArrayValued', true)/W(j)
end

 Accepted Answer

SIGMA is of size 1-by-4, and the loop goes from 1 to length(SIGMA) (which is 4), so what else would you expect?
SIGMA = 0:0.05:0.15 % 1-by-4
SIGMA = 1×4
0 0.0500 0.1000 0.1500
for j = 1:length(SIGMA) % 1:4
disp(j)
end
1 2 3 4

4 Comments

Thank you for clarification and is there any way that I can get only the final result?
If, for instance, when the code is complete, you want Xbar to contain just the final element of Xbar that's calculated inside the loop, you can do this after the loop:
Xbar = Xbar(end);
Is that what you mean by "final result"?
Yes, thank you!
You're welcome!

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Asked:

on 7 Jun 2022

Commented:

on 7 Jun 2022

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