# Why this loop is executing 4 times?

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AVINASH SAHU on 7 Jun 2022
Commented: Voss on 7 Jun 2022
% 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

Voss on 7 Jun 2022
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
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Voss on 7 Jun 2022
You're welcome!