how to correctly get values for stress for Y >82 values should be close to zero at Y = 88.25

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p = -100;
y = 0:0.25:88.25;
l = 1:1:199;
ybar = 63.17;
i = 790200.11;
t = 88.25;
[L,Y,P] = meshgrid(l,y,p);
V = p.*L;
Q1 = -3.125.*(Y.^2) + 394.81.*Y;
Q2 = -82.8.*(((Y.^2)/2)-63.17.*Y)-139163.72;
if Y >= 82
stress = abs(V.*Q2)/(i*t);
else
stress = abs(V.*Q1)/(i*t);
end
contourf(L,Y,stress);
  5 Comments
VBBV
VBBV on 7 Oct 2020
Edited: VBBV on 7 Oct 2020
Bcoz you were not using a loop for matrix Y. Use a for loop instead instead to step thru each value of Y as
% if true
% code
%end
for i = 1: length(Y)
for j = 1: length(Y)
if Y(i,j) >= 82
...
else
...
end
end
end
contourf(...)

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Accepted Answer

Asad (Mehrzad) Khoddam
Asad (Mehrzad) Khoddam on 7 Oct 2020
You can use vectorized if statement:
Q = Q2 .* (Y >= 82) + Q1 .* (Y < 82);
stress = abs(V.*Q)/(i*t);
  4 Comments
Joel Rebello
Joel Rebello on 7 Oct 2020
thanks vasishta, the analysis is with regards to the shear stress on a T-beam, i checked the results from the code you presented and not sure why but the contour does not represent the behaviour of the stress caused by shear (V) acting on the beam. For shear stress the stress is maximum stress should be around the neutral axix (ybar) where Y =63.17, but your results indicate the maximum value is around Y = 5.
Appreciate your help

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More Answers (1)

KSSV
KSSV on 7 Oct 2020
You have to use logical indexing to achieve this.
clc; clear all ;
p = -100;
y = 0:0.25:88.25;
l = 1:1:199;
ybar = 63.17;
i = 790200.11;
t = 88.25;
[L,Y,P] = meshgrid(l,y,p);
V = p.*L;
Q1 = -3.125.*(Y.^2) + 394.81.*Y;
Q2 = -82.8.*(((Y.^2)/2)-63.17.*Y)-139163.72;
stress = zeros(size(L)) ;
stress(Y>=82) = abs(V(Y>=82).*Q2(Y>=82))/(i*t);
stress(Y<82) = abs(V(Y<82).*Q2(Y<82))/(i*t);
contourf(L,Y,stress);

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