# int function return ans in 'int' itself . I got the expression J and L in terms of g but the integration of expression U1 return as in int function. Thanks in advance

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arvind thakur on 25 Sep 2021
Answered: arvind thakur on 9 Oct 2021
clc
clear
syms x g
assume(g >= 0)
Ef=427000*10^6; % in MPa
Em=89000*10^6;
Ec=184000*10^6;
mu=0.2825;
ao=0.001;
a=0.00189;
w=0.00512;
x1=0;
%x1=0.001
%R=.0725;
S=198*10^6;
%tow=20;
m1=0.6147+17.1844*(a^2/w^2)+8.7822*(a^6/w^6);
m2=0.2502+3.2889*(a^2/w^2)+70.0444*(a^6/w^6);
%m1=2.7552;
%m2=0.7889;
H=(1+m1*((g-x)/g)+m2*((g-x)/g)^2);
H1=(1+m1*((g-x1)/g)+m2*((g-x1)/g)^2);
% for remote stress S=198 MPa
c=S*((w/(w-ao))+6*w*ao*((0.5*(w-ao)-(x-ao))/(w-ao)^3));
j=int(((S*H)/(sqrt(g-x))),x,[0,ao]);
L=int(((S-c)*H)/(sqrt(g-x)),x, [ao,a]);
R=(H1*j+H1*L);
R1=(R/(sqrt(g-x1)))
U1=int(R1,g,0, a);

Walter Roberson on 25 Sep 2021
Edited: Walter Roberson on 25 Sep 2021
clc
clear
syms x g
assume(g >= 0)
Ef=427000*10^6; % in MPa
Em=89000*10^6;
Ec=184000*10^6;
mu=0.2825;
ao=0.001;
a=0.00189;
w=0.00512;
x1=0;
%x1=0.001
%R=.0725;
S=198*10^6;
%tow=20;
m1=0.6147+17.1844*(a^2/w^2)+8.7822*(a^6/w^6);
m2=0.2502+3.2889*(a^2/w^2)+70.0444*(a^6/w^6);
%m1=2.7552;
%m2=0.7889;
H=(1+m1*((g-x)/g)+m2*((g-x)/g)^2);
H1=(1+m1*((g-x1)/g)+m2*((g-x1)/g)^2);
% for remote stress S=198 MPa
c=S*((w/(w-ao))+6*w*ao*((0.5*(w-ao)-(x-ao))/(w-ao)^3));
j=int(((S*H)/(sqrt(g-x))),x,[0,ao]);
L=int(((S-c)*H)/(sqrt(g-x)),x, [ao,a]);
R=(H1*j+H1*L);
R1=(R/(sqrt(g-x1)))
R1 = sR1 = simplify(R1)
sR1 = limit(sR1, g, 0)
ans =
NaN
If you look at the pretty() output, you will see that there is a division by . The numerator has g to a variety of powers. The result is an expression whos limit cannot be taken at 0, and therefore cannot be integrated.
digits(100)
vpa(subs(sR1, g, sym(10)^(-20)))
ans = vpa(subs(sR1, g, sym(10)^(-50)))
ans = You can see from those last couple of evaluations that the real part is steady near 0, but that the imaginary part is exploding. I suspect that the denominator has a root in the range of integration.
Walter Roberson on 9 Oct 2021
inmy case u(x) and c(x) dependent on each other
No they do not. u(x) depends upon c(x) but c(x) does not depend upon u(x) . An iterative approach is not called for.

arvind thakur on 9 Oct 2021 here is the c(x), ignore previous c(x) and all other value and equation is same.

R2019a

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