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

 Risposta accettata

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 Commenti

Thank you for clarification and is there any way that I can get only the final result?
Voss
Voss il 7 Giu 2022
Modificato: Voss il 7 Giu 2022
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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