Looking to solve this integral with variable bounds

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adam puchalski 2022-7-11

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1757370-looking-to-solve-this-integral-with-variable-bounds

评论： Torsten 2022-7-12

%variables

F = .00024; %kg

E = 10^6; %mPa

I = 6.1*10^-9; %kg m^2

L = .01; %m

%solve for Theta sub 0

func = @(OO) L/sqrt((E*I)/(2*F)) - integral(@(theta)1./sqrt(cos(OO)-cos(theta)),OO,pi/2);

g = fzero(func,0);

OS = fsolve(func,.5);

ideally wanted to use fzero but not sure how to do it, the like below, OS = ... kinda works but I keep getting pi/2 and not sure why. Im using this is the follwing equation:

aa = linspace(pi/2*.99, 0, 49);

aa=aa';

%shape of rod given

xi = sqrt((2*E*I)/F)*(sqrt(cos(OS))-sqrt(cos(OS)-cos(aa)));

and not getting the values i am hoping for. any help appreciated

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adam puchalski 2022-7-11

编辑：Torsten 2022-7-11

在 MATLAB Online 中打开

ok so i edited it and ff1 should cross zero now but still does not seem to work

format long

F = .00024; %kg

E = 10^6; %mPa

I = 6.1*10^-9; %kg m^2

L = .01; %m

ep = 10^-8;

z(1) = 0;

for n = 2:10001

z(n) = z(n-1) + ((pi/2)/10000);

ff1(n) = L/sqrt((E*I)/(2*F)) - (2*(ep)^.5)/ (sin(z(n)))^.5 - integral(@(theta) 1./sqrt(cos(z(n))-cos(theta)),z(n)+ep,pi/2);

end

Warning: Minimum step size reached near x = 1.5708. There may be a singularity, or the tolerances may be too tight for this problem.

%figure(5)

plot(z,real(ff1))

func = @(OO) L/sqrt((E*I)/(2*F)) - (2*(ep)^.5)/ (sin(OO))^.5 - integral(@(theta)1./sqrt(cos(OO)-cos(theta)),OO,pi/2);

%OS = fzero(func,.5);

OS = fsolve(func,0.5)

Equation solved, inaccuracy possible. The vector of function values is near zero, as measured by the value of the function tolerance. However, the last step was ineffective.

OS =

1.570794630102631

L/sqrt((E*I)/(2*F))

ans =

0.002805147493273

pi/2

ans =

1.570796326794897

adam puchalski 2022-7-12

ah i see i see, thank you very much for your help!

Torsten 2022-7-12

So in short:

Forget the complete discussion. Your problem is ill-posed.

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