Using different equations for a single line plot

% In the code below i need calculate tmax value for 2 different cases. A
% indepent variable matrix named: "hu" starts from the value of "0" and
% ends at "3" when the "hu" value is lower than 0.436 the tmax1 formula
% needs to be used. For the values greater than 0.436 the tmax2 formula
% must be used.
%the resulting calculation must be plotted as a single line 
%I would hihgly appriciate your feedback on how this could be achieved,
%thank you in advance

close all ; %hertzian contact calculator for a cylinder
clc;
format long
F= 5000 ;
v1=0.3;
v2=0.3;
E1=700000;
E2=700000;
d1=10 ;
d2=100 ;
L= 25;
b=(1/d1)+(1/d2);
m1=(1-v1^2)/E1;
m2=(1-v2^2)/E2;

a=0.173;

z=[];
U=[];
Q=[];
OK=[];
Pmax=[];
qxy=[];
Pmax=(2*F)/(pi*a*L)
z= 0.786.*a;
ZA=(1+(z^2/a^2))^(1/2)

qx=(-2*v1*Pmax)*(ZA-abs(z/a));
qy=-Pmax*(((1+2*(z^2/a^2))/ZA)-2*abs(z/a));
qz=-Pmax/ZA;
tx=(qx-qz)/2;
ty=(qy-qz)/2

%Plot creation
tmax2=[];
tmaxP=[];
zi=[];
% The issue is under this section
hu=[0:0.25:3]; %the used  values
zi=a*hu;


ZI=(1+(zi.^2/a.^2)).^(1/2)

qx2=(-2*v1*Pmax)*(ZI-abs(zi./a));
qy2=-Pmax*(((1+2.*(zi.^2/a^2))./ZI)-2*abs(zi./a))
qz2=-Pmax./ZI
tmax1=(qx2-qz2)./2 %equation to be used when "hu" variable is lower than 0.436
tmax2=(qy2-qz2)./2 %equation to be used while "hu" variable is above 0.436


tmaxP=tmax2./Pmax;
plotx=zi./a;
plot(plotx,tmaxP)
legend('ty')
title('Stress distribution along cylindrical bodies')

Answer 1

[Voss]

close all ; %hertzian contact calculator for a cylinder
clc;
format long
F= 5000 ;
v1=0.3;
v2=0.3;
E1=700000;
E2=700000;
d1=10 ;
d2=100 ;
L= 25;
b=(1/d1)+(1/d2);
m1=(1-v1^2)/E1;
m2=(1-v2^2)/E2;

a=0.173;

Pmax=(2*F)/(pi*a*L);
z= 0.786.*a;
ZA=(1+(z^2/a^2))^(1/2);

qx=(-2*v1*Pmax)*(ZA-abs(z/a));
qy=-Pmax*(((1+2*(z^2/a^2))/ZA)-2*abs(z/a));
qz=-Pmax/ZA;
tx=(qx-qz)/2;
ty=(qy-qz)/2;

%Plot creation

hu=[0:0.05:3]; %the used  values

ZI=(1+hu.^2).^(1/2);

qx2=(-2*v1*Pmax)*(ZI-hu);
qy2=-Pmax*(((1+2.*hu.^2)./ZI)-2*hu);
qz2=-Pmax./ZI;
tmax1=(qx2-qz2)./2; %equation to be used when "hu" variable is lower than 0.436
tmax2=(qy2-qz2)./2; %equation to be used while "hu" variable is above 0.436

tmax = zeros(size(hu));
tmax(hu < 0.436) = tmax1(hu < 0.436)/Pmax;
tmax(hu >= 0.436) = tmax2(hu >= 0.436)/Pmax;
plot(hu,tmax)

legend('ty')
title('Stress distribution along cylindrical bodies')

[Erdem Turan]

Thank you sir!