Calculate normal and shear stress of a beam where values of depth and width are under certain conditions
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clear
clc
b = 1:200;
d = 1:200;
A = 2640;
F = 300e3;
d_max = 125;
b_min = 20;
NL = length(b);
sigma_max = zeros(1,NL);
tau_max = zeros(1,NL);
for i = 1
if (b(i)>=d(i)) && (d(i)<=d_max) && (b(i)>=b_min) && (b(i)*d(i)==A)
My = F*(b(i)/2);
Iyy = (b(i)*d(i)^3)/12;
sigma_max(i) = (My*Iyy)*(d(i)/2);
Qy = (b(i)*d(i)^2)/8;
tau_max(i) = (Vz/(Iyy*b(i)))*Qy;
end
end
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Accepted Answer
VBBV
on 31 Mar 2023
Edited: VBBV
on 31 Mar 2023
close all
clear
clc
b = 1:200;
d = 1:200;
A = 2640;
F = 300e3;
d_max = 125;
b_min = 20;
NL = length(b);
sigma_max = zeros(1,NL);
tau_max = zeros(1,NL);
for i = 1:NL
if ((b(i)>=d(i)) | (d(i)<=d_max)) & ((b(i)>=b_min) | (b(i)*d(i)==A))
My(i) = F*(b(i)/2);
Iyy(i) = (b(i)*d(i)^3)/12;
sigma_max(i) = (My(i)/Iyy(i))*(d(i)/2);
Qy(i) = (d(i)*b(i)^2)/6;
tau_max(i) = (My(i)/Iyy(i))*(d(i)/2)*Qy(i);
end
end
% normal stress
figure
bar(b,sigma_max)
xlim([min(b) max(b)])
%shear stress
figure
bar(b,tau_max)
xlim([min(b) max(b)])
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