Stopping a loop at the maximum value of y and finding the x from the maximum y
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Shimon Katzman
el 23 de Nov. de 2019
Comentada: Shimon Katzman
el 23 de Nov. de 2019
Hi everyone,
How can i stop the loop when i get the maximum M(i) and how do i know what is the value of the x (phi) for the maximum value of M?
Thank You.
b=300; %mm
d=400; %mm
fc=40; %Mpa
Ecm=22*(fc/10)^0.3*10^3; %Mpa
Es=200000; %Mpa
As=2400; %mm^2
fy=400; %Mpa
epsc1=min(2.8/1000,0.7*fc^0.31/1000);
epscu=3.5/1000;
k=1.05*Ecm*epsc1/fc;
epscmv = linspace(0.05, 3.5, 50)*1E-3;
for i=1:50(epscmv);
epscm = epscmv(i);
funC=@(epsc) (k*epsc/epsc1-(epsc/epsc1).^2)./(1+(k-2)*epsc/epsc1);
compression=@(c) b*fc*c/epscm*integral(funC,0,epscm)/1000;
tension=@(c) min(Es*(d-c)/c*epscm*As/1000,fy*As/1000);
c(i)=fsolve(@(c) compression(c)-tension(c),1);
funM=@(epsc)(k*epsc./epsc1-(epsc./epsc1).^2)./(1+(k-2)*epsc./epsc1).*(d-c(i)+(c(i)./epscm).*epsc);
M(i)=b*fc*c(i)/epscm*integral(funM,0,epscm)/1000000;
phi(i)=epscm/c(i);
end
Mmax=max(M(i))
plot(phi, M)
grid on
xlabel('phi [1/mm]')
ylabel('Moment [kNm]')
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Respuesta aceptada
David Hill
el 23 de Nov. de 2019
b=300; %mm
d=400; %mm
fc=40; %Mpa
Ecm=22*(fc/10)^0.3*10^3; %Mpa
Es=200000; %Mpa
As=2400; %mm^2
fy=400; %Mpa
epsc1=min(2.8/1000,0.7*fc^0.31/1000);
epscu=3.5/1000;
k=1.05*Ecm*epsc1/fc;
epscmv = linspace(0.05, 3.5, 50)*1E-3;
for i=1:50(epscmv);
epscm = epscmv(i);
funC=@(epsc) (k*epsc/epsc1-(epsc/epsc1).^2)./(1+(k-2)*epsc/epsc1);
compression=@(c) b*fc*c/epscm*integral(funC,0,epscm)/1000;
tension=@(c) min(Es*(d-c)/c*epscm*As/1000,fy*As/1000);
c(i)=fsolve(@(c) compression(c)-tension(c),1);
funM=@(epsc)(k*epsc./epsc1-(epsc./epsc1).^2)./(1+(k-2)*epsc./epsc1).*(d-c(i)+(c(i)./epscm).*epsc);
M(i)=b*fc*c(i)/epscm*integral(funM,0,epscm)/1000000;
phi(i)=epscm/c(i);
end
[Mmax,idx]=max(M);%this is how you would get the max of M and the index at the maximum
phiAtmax=phi(idx);%using the index at maximum you can get the phi at the maximum
plot(phi, M)
grid on
xlabel('phi [1/mm]')
ylabel('Moment [kNm]')
3 comentarios
David Hill
el 23 de Nov. de 2019
If your function is smooth and has only one peak, then you can just check when the value is less than the previous value and break.
for i=1:50(epscmv);
epscm = epscmv(i);
funC=@(epsc) (k*epsc/epsc1-(epsc/epsc1).^2)./(1+(k-2)*epsc/epsc1);
compression=@(c) b*fc*c/epscm*integral(funC,0,epscm)/1000;
tension=@(c) min(Es*(d-c)/c*epscm*As/1000,fy*As/1000);
c(i)=fsolve(@(c) compression(c)-tension(c),1);
funM=@(epsc)(k*epsc./epsc1-(epsc./epsc1).^2)./(1+(k-2)*epsc./epsc1).*(d-c(i)+(c(i)./epscm).*epsc);
M(i)=b*fc*c(i)/epscm*integral(funM,0,epscm)/1000000;
phi(i)=epscm/c(i);
if M(i)<M(i-1)
break;
end
end
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