# Evaluate values of a vector with values of variable p

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ABG on 23 Oct 2022
Commented: ABG on 23 Oct 2022
Hi guys!
I am having a bit of a problem with a code. In this case I have calculated values of exx and eyy, which depend on p. After having computed this two vectors, I want to obtain the values of vector uc which depends on the values of exx and eyy related with the particular value of p.
However, when I try to substitute all the exx and eyy values into uc, I do not obtain the correct substitution as I am only obtaining the same result, which also keeps depending on p (I do not know why if I am giving p values to the two previous vectors). I do not know if I have to do it with another FOR value, which allows me to store a certain value of p and pair it to the values of exx and eyy.
Thanks a lot!
syms E ex ey r v t p
Pa=(E*ex)/((r/2*t)-v*r/t);
Po=(E*ey)/(-v*(r/2*t)+r/t);
%Value of all derivatives
%Calculation of the derivatives
dPa1=diff(Pa,E);
dPa2=diff(Pa,ex);
dPa3=diff(Pa,r);
dPa4=diff(Pa,v);
dPa5=diff(Pa,t);
dPo1=diff(Po,E);
dPo2=diff(Po,ey);
dPo3=diff(Po,r);
dPo4=diff(Po,v);
dPo5=diff(Po,t);
%Values of u
u1=sqrt((0.05*70*10^3)^2/3); % u for E coefficient
u2=sqrt((0.05*0.3)^2/3); % u for v Poisson's coefficient
u3=sqrt(1^2/3); % u for r
u4=sqrt(0.01^2/3); % u for t
u5=sqrt((5*10^-6)^2/3); % u for strain
%Now we give values to all the variables
E=70*10^3;
v=0.3;
r=33;
t=0.1;
exx=[];
eyy=[];
ex=(((p*r)/(2*t))-v*((p*r)/t))*1/E;
ey=(-v*((p*r)/(2*t))+((p*r)/t))*1/E;
uc=[];
ucc=[];
for p=0:0.001:0.2
exx=[exx eval(ex)];
eyy=[eyy eval(ey)];
uc=[uc sqrt((eval(dPa1)*u1)^2+(eval(dPa2)*u2)^2+(eval(dPa3)*u3)^2+(eval(dPa4)*u4)^2+(eval(dPa5)*u5)^2+(eval(dPo1)*u1)^2+(eval(dPo2)*u2)^2+(eval(dPo3)*u3)^2+(eval(dPo4)*u4)^2+(eval(dPo5)*u5)^2)]; %uc in terms of p
end

Walter Roberson on 23 Oct 2022
eval() of a symbolic expression is not defined. Use subs() and double()
##### 2 CommentsShowHide 1 older comment
ABG on 23 Oct 2022
Thanks Walter!

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