Stopping a loop at the maximum value of y and finding the x from the maximum y

1 vue (au cours des 30 derniers jours)
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]')

Réponse acceptée

David Hill
David Hill le 23 Nov 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 commentaires
David Hill
David Hill le 23 Nov 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
Shimon Katzman
Shimon Katzman le 23 Nov 2019
doesnt work
Attempted to access M(0); index must be a positive integer or logical.
Error in advencedconcrete32a (line 22)
if M(i)<M(i-1)

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