Where are the bugs for this ODE finite difference problem that solve using Newton Raphson method?

조회 수: 3 (최근 30일)
Could someone provide me help to solve this Euler-Bernoulli beam equation by using finite difference method and Newton Raphson please.
With boundary value of y(0) = 0 and dy/ds(L) = 0
I am continually getting answers that are nowhere near the results from the bvp4c command.
  댓글 수: 6
Zhipeng Li
Zhipeng Li 2019년 12월 2일
편집: Zhipeng Li 2019년 12월 2일
Why should it be horizontal, at L is the maxiumum angle, because this is a cantilever beam springboard problem

댓글을 달려면 로그인하십시오.

채택된 답변

darova
darova 2019년 12월 2일
Here is my attempt
clc,clear
N = 10; % 10 Set parameters
L = 16*0.3048; %meter
b = 19.625 * 0.0254; d = 1.625 * 0.0254;
p = 26.5851; I = (b*(d^3))/12; h = L/N; F = 0.7436; E = 68.9e6;
alpha = (h^2)/(E*I);
w = p*b*d;
fode = @(s,f) [f(2); -1/E/I*(w*(L-s)+F)*cos(f(1))];
fbound= @(ya,yb) [ya(1)-0; yb(2)-0];
ss = linspace(0,5);
finit = [0 0];
solinit = bvpinit(ss,finit);
sol = bvp4c(fode,fbound,solinit);
plot(sol.x,sol.y(1,:))
  댓글 수: 7
darova
darova 2019년 12월 3일
If f(s) :
image.png
then f'(s)
Try the code
syms phi(s) s
syms E I w L F
f = 1/E/I*(w*(L-s)+F)*cos(phi);
df = diff(f,s);
pretty(df)
Zhipeng Li
Zhipeng Li 2019년 12월 3일
Yes, finaly got it !! Still not perfect, but I am happy with it.This have been struggled me for days, and again, thank you so much for your help!!Screenshot 2019-12-03 at 21.49.11.png

댓글을 달려면 로그인하십시오.

추가 답변 (1개)

Thiago Henrique Gomes Lobato
Thiago Henrique Gomes Lobato 2019년 12월 1일
I didn't check exactly your FEM implementations but one thing that I quickly noticed is that the Newton-Rapson is an iteractive approach, so maybe you get different results simply because your result did not converged. When I iterate about your code I get a very different result that converges actually fast (3 iterations):
% substitute by your images
N = 20 % Set parameters
L = 16*0.3048; %meter
b = 19.625 * 0.0254; d = 1.625 * 0.0254;
p = 26.5851; I = (b*d^3)/12; h = L/N; F = 0.7436; E = 68.9e6;
alpha = h^2/E*I;
w = p*b*d;
S = [h:h:L] ;
y = ones(N,1);
e = ones(N,1);
A = spdiags([e -2*e e],[-1 0 1],N,N);
A(N-1,N) = -2; % fictitious boundaries method
Iterations = 1000;
tol = 1e-20;
for idx =1:Iterations
function_1 = zeros(N,1);
for n = 1:N
function_1(n) = alpha*(w*(L-S(n))+F)*cos(y(n));
end
Fun = A*y + function_1;
% To create the Jacobian of F(y)
Dfdia = zeros(N,1);
for n = 1:N
Dfdia(n) = alpha*(w*(L-S(n))+F)*sin(y(n));
end
J = diag(Dfdia);
step = inv(A+J)*Fun;
y = (y-step);
if norm(step)<tol
Iter = idx
break
end
end
  댓글 수: 1
Zhipeng Li
Zhipeng Li 2019년 12월 2일
Thank you very much for answering my question. One of the main erro was that I missed a pair of brackts for 'alpha', and now is been edited, but the result is still quite off from the actual solution i got from bvp4c as shown in above comment. I have tried your version of code with bracket bug fixed and is the same, what sort of issue do you think this is? I am not even sure is it math issue or code issue now.

댓글을 달려면 로그인하십시오.

카테고리

Help CenterFile Exchange에서 Symbolic Math Toolbox에 대해 자세히 알아보기

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by