ERROR : "Input arguments must be convertible to floating-point numbers"

Hi everyone, im getting this error:

Error using sym/min (line 98)

Input arguments must be convertible to floating-point numbers.

Error in advencedconcrete32a (line 16)

tension= min(Es*(d-c)/c*epscm*As/1000,fy*As/1000);

i guess the problem is that i use "c" before i solved it, how else can i find "c" from the equation compression= tension?

please help.

thank you very much.

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;

epscm=1.5/1000;
syms c
funC=@(epsc) (k*epsc/epsc1-(epsc/epsc1).^2)./(1+(k-2)*epsc/epsc1);
compression= b*fc*c./epscm*integral(funC,0,epscm)/1000;
tension= min(Es*(d-c)/c*epscm*As/1000,fy*As/1000);
solc= compression==tension;
c=solve(solc,c)

[Shimon Katzman]

Anybody please?

Answer 1

[Walter Roberson]

Replace

tension= min(Es*(d-c)/c*epscm*As/1000,fy*As/1000);

with

tension = piecewise(Es*(d-c)/c*epscm*As/1000<=fy*As/1000,Es*(d-c)/c*epscm*As/1000,fy*As/1000)

Replace

funM=@(epsc)(k*epsc/epsc1-(epsc/epsc1).^2)./(1+(k-2)*epsc/epsc1)*(d-c+(c/epscm)*epsc);

with

funM=@(epsc)(k*epsc./epsc1-(epsc./epsc1).^2)./(1+(k-2)*epsc./epsc1).*(d-c+(c./epscm).*epsc);

[Shimon Katzman]

Walter hi and thank you.

i replaced funM and it works well but without the loop.

once i add the loop, again it Errors.

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+(c./epscm).*epsc); 
    M(i)=b*fc*c/epscm*integral(funM,0,epscm)/1000000;
end


plot(epscmv, M)
grid on
xlabel('epscm')
ylabel('M')

Error:

Error using  .* 
Matrix dimensions must agree.

Error in @(epsc)(k*epsc./epsc1-(epsc./epsc1).^2)./(1+(k-2)*epsc./epsc1).*(d-c+(c./epscm).*epsc)


Error in integralCalc/iterateScalarValued (line 314)
                fx = FUN(t);

Error in integralCalc/vadapt (line 132)
            [q,errbnd] = iterateScalarValued(u,tinterval,pathlen);

Error in integralCalc (line 75)
        [q,errbnd] = vadapt(@AtoBInvTransform,interval);

Error in integral (line 88)
Q = integralCalc(fun,a,b,opstruct);

Error in advencedconcrete32a (line 20)
    M(i)=b*fc*c/epscm*integral(funM,0,epscm)/1000000;
    
    

but without the loop:

    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;


epscm = 1.5/1000
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=fsolve(@(c) compression(c)-tension(c),1);
funM=@(epsc)(k*epsc./epsc1-(epsc./epsc1).^2)./(1+(k-2)*epsc./epsc1).*(d-c+(c./epscm).*epsc); 
M=b*fc*c/epscm*integral(funM,0,epscm)/1000000

it works fine:

M =

  331.2854

[Walter Roberson]

Change

    funM=@(epsc)(k*epsc./epsc1-(epsc./epsc1).^2)./(1+(k-2)*epsc./epsc1).*(d-c+(c./epscm).*epsc); 
    M(i)=b*fc*c/epscm*integral(funM,0,epscm)/1000000;

to

    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;

[Shimon Katzman]

THANK YOU VERY MUCH!!!

another question, how do i stop the loop when M(i) gets to the maximum and how can i know what value is the maximum from the graph?

Answer 2

[Star Strider]

The min function is not compatible with symbolic objects, for obvious reasons.

Try this instead:

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;
epscm=1.5/1000;

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=fsolve(@(c) compression(c)-tension(c), 1 )

producing (with this initial parameter estimate):

c =
   154.2368

[Star Strider]

Shimon Katzman’s Answer moved here —

Hi star, Thanks for your help. What is the solution if id like to make a loop that changes the epscm... The purpouse is to create a graph which c is the y And epscm is the x. Thank you

[Star Strider]

My pleasure.

One approach:

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(1, 2, 25)*1E-3;                           % Vector Of ‘epscm’ Values

for k = 1:numel(epscmv)
    epscm = epscmv(k);
    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(k) = fsolve(@(c) compression(c)-tension(c), 1 );
end

figure
plot(epscmv, c)
grid
xlabel('epscm')
ylabel('c')

I have no idea what values ‘epscm’ could range over, or how many you want to solve for.

Experiment to get the result you want.

[Shimon Katzman]

Hi star and thank you very much again.

Now i am trying to add another function but again it continues to ERROR.

the purpose now is to plot a graph like earlier but now M is the y and epcm is the x

epcm starts from 0.05/1000 to 3.5/1000 and the incriments are 0.05/1000.

This example i didnt even created the loop because i wanted to see if my code is OK but unfortunettly it errors :((

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;

epscm=1.5/1000;
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=fsolve(@(c) compression(c)-tension(c),1)
funM=@(epsc)(k*epsc/epsc1-(epsc/epsc1).^2)./(1+(k-2)*epsc/epsc1)*(d-c+(c/epscm)*epsc);
M=b*fc*c/epscm*integral(funM,0,epscm)/1000

This is the ERROR i get:

c =

  154.2368

Error using  * 
Inner matrix dimensions must agree.

Error in @(epsc)(k*epsc/epsc1-(epsc/epsc1).^2)./(1+(k-2)*epsc/epsc1)*(d-c+(c/epscm)*epsc)


Error in integralCalc/iterateScalarValued (line 314)
                fx = FUN(t);

Error in integralCalc/vadapt (line 132)
            [q,errbnd] = iterateScalarValued(u,tinterval,pathlen);

Error in integralCalc (line 75)
        [q,errbnd] = vadapt(@AtoBInvTransform,interval);

Error in integral (line 88)
Q = integralCalc(fun,a,b,opstruct);

Error in Untitled2 (line 18)
M=b*fc*c/epscm*integral(funM,0,epscm)/1000

THANK YOU VERY MUCH FOR YOUR HELP.

[Star Strider]

I was sleeping when you posted that (night here), and just now saw it.

Oh, well ...

[Shimon Katzman]

Thanks star,

maybe you could help me with this?

https://www.mathworks.com/matlabcentral/answers/492749-stopping-a-loop-on-the-peak

[Star Strider]

My pleasure.

I appreciate the reference. I have posted my Answer.