Error - Undefined function or variable
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Hello everybody.
I am trying to make an if loop but the problem is that "epsc" is undefined veriable.
Is there a better way to make it?
Thank you very much.
b=300; %mm
d=400; %mm
fc=40; %Mpa
Ecshah=57000/145*(fc*145)^0.5; %Mpa
Es=200000; %Mpa
As=2400; %mm^2
fy=400; %Mpa
eps0=1.027*10^-7*fc*145+0.00195;
epscu=3.5/1000;
kshah=0.025*fc*10^3;
A=Ecshah*eps0/fc;
epscmv = linspace(0.05, 3.5, 500)*1E-3;
for i=1:numel(epscmv);
epscm = epscmv(i);
if epsc<=eps0
funCshah=@(epsc) fc*(1-(1-epsc./eps0).^A);
elseif epsc>eps0
funCshah=@(epsc) fc*exp(-kshah*(epsc-eps0).^1.15);
end
end
compression=@(c) b*fc*c/epscm*integral(funCshah,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);
Getting the error:
Undefined function or variable 'epsc'.
Error in Untitled (line 16)
if epsc<=eps0
4 个评论
Jim Riggs
2019-11-25
You must define the variable "epsc".
What is it? Assign it a value.
Shimon Katzman
2019-11-25
Jim Riggs
2019-11-25
Please explain. What is the value of epsc?
In the context:
funCshah=@(epsc) fc*(1-(1-epsc./eps0).^A)
and also
funCshah=@(epsc) fc*exp(-kshah*(epsc-epsc0).^1.15)
epsc is a place-holder and represents the argument of function funCshah. When function funCshah is called, a numerical value must be supplied as the argument to the function. Therefore, when function funCshah is called, internally epsc has the value of the argument to the function.
In the context
if epsc<=eps0
This is a comparison of two values, so epsc and eps0 must both be defined as variables, and their values are compared. The error message that you are getting is that, in this context, the variable epsc has no definition. It must be defined.
Shimon Katzman
2019-11-26
采纳的回答
Star Strider
2019-11-26
My apologies for not seeing this Question earlier.
I am not certain what you want to do.
Taking a guess, this runs without error:
b=300; %mm
d=400; %mm
fc=40; %Mpa
Ecshah=57000/145*(fc*145)^0.5; %Mpa
Es=200000; %Mpa
As=2400; %mm^2
fy=400; %Mpa
eps0=1.027*10^-7*fc*145+0.00195;
epscu=3.5/1000;
kshah=0.025*fc*10^3;
A=Ecshah*eps0/fc;
epscmv = linspace(0.05, 3.5, 500)*1E-3;
c = zeros(size(epscmv)); % Preallocate
for i=1:numel(epscmv)
epscm = epscmv(i);
funCshah=@(epsc) fc*(1-(1-epsc./eps0).^A) .* (epsc<=eps0) + fc*exp(-kshah*(epsc-eps0).^1.15) .* (epsc>eps0);
compression=@(c) b*fc*c/epscm*integral(funCshah,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);
end
figure
plot(epscmv, c)
grid
xlabel('\epsilon (cm)')
ylabel('c(\epsilon)')
The ‘funCshah’ function eliminates the if block (and the necessity of having to define ‘epsc’ first) by using ‘logical indexing’, so that the first part of the function:
fc*(1-(1-epsc./eps0).^A)
returns the appropriate values when ‘(epsc<=eps0)’ is true, and the second part of the function:
fc*exp(-kshah*(epsc-eps0).^1.15)
returns the appropriate values when ‘(epsc>eps0)’ is true. Since ‘epsc’ is appropriately passed as an argument, the test is done inside the function and the if block is not necessary.
If my guess is not correct, make appropriate changes to get the result you want.
2 个评论
Shimon Katzman
2019-11-26
Star Strider
2019-11-26
As always, my pleasure!
Your Questions provide interesting challenges!
更多回答（2 个）
Line 16:
if epsc<=eps0
Variable epsc is not defined.
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