Matlab code: deflection of simply supported beam using ode45

조회 수: 9 (최근 30일)
Camryn
Camryn 2023년 11월 7일
편집: hamawandy 2025년 3월 7일
Hi, my code keeps returning a function for a cantilever beam rather than a simply supported. How do I fix this??
My code:
clear;
clc;
x0 = 0;
Y0 = [0; 0; 0; 0];
L = 5000; % Length of the beam in mm
xRange = [x0, L];
[x, y] = ode45(@(x, y) beamDeflection(x, y, L), xRange, Y0);
plot(x, y);
xlabel('Length (mm)');
ylabel('Deflection (mm)');
title('Simply Supported Beam Deflection');
function dydx = beamDeflection(x, y, L)
E = 69000; % Young's modulus of aluminum in MPa
b = 500; % Length of cross-sectional base in mm
h = 750; % Height of cross-sectional height in mm
I = (b * h^3) / 12; % Moment of inertia about x
q = 10000; % Load in newtons per mm^2
if x == 0 || x == L
dydx = [0; 0; 0; 0];
else
dydx = [y(2); y(3); y(4); (q / (E * I))];
end
end

답변 (2개)

Torsten
Torsten 2023년 11월 7일
편집: Torsten 2023년 11월 7일
Hi, my code keeps returning a function for a cantilever beam rather than a simply supported. How do I fix this??
By using bvp4c instead of ode45. You have a boundary value problem, not an initial value problem because you try to impose boundary condition at x = 0 and x = L.
Here is a symbolic solution:
syms x y(x)
E = 69000; % Young's modulus of aluminum in MPa
b = 500; % Length of cross-sectional base in mm
h = 750; % Height of cross-sectional height in mm
I = (b * h^3) / 12; % Moment of inertia about x
q = 10000; % Load in newtons per mm^2
L = 5000;
Dy = diff(y,x);
D2y = diff(y,x,2);
D3y = diff(y,x,3);
D4y = diff(y,x,4);
eqn = D4y == q/(E*L);
conds = [y(0)==0,y(L)==0,D2y(0)==0,D2y(L)==0];
sol = dsolve(eqn,conds)
sol = 
fplot(sol,[0 L])
  댓글 수: 7
Torsten
Torsten 2025년 3월 5일
syms x y(x)
E = 69000; % Young's modulus of aluminum in MPa
b = 500; % Length of cross-sectional base in mm
h = 750; % Height of cross-sectional height in mm
I = (b * h^3) / 12; % Moment of inertia about x
q = 10000; % Load in newtons per mm^2
L = 5000;
Dy = diff(y,x);
D2y = diff(y,x,2);
D3y = diff(y,x,3);
D4y = diff(y,x,4);
eqn = D4y == q/(E*I);
conds = [y(0)==0,y(L)==0,D2y(0)==0,D2y(L)==0];
sol = dsolve(eqn,conds);
xn = 0:0.1:L;
yn = subs(sol,x,xn);
area(xn,yn)
hamawandy
hamawandy 2025년 3월 6일
Thank you so much ...... I am so grateful to you.

댓글을 달려면 로그인하십시오.


hamawandy
hamawandy 2025년 3월 6일
편집: hamawandy 2025년 3월 6일
Dear Sir
I am facing a problem in sketching the shear force diagram from the span 4m till 8m.......the line should be horizontal from 4m to 8m.
Can you help me to fix this error ?
This is the code:
R1=6;
R2=2;
X= linspace (0, 8, 100);
V= (R1-2*X).*(X>=0) - (R2)*(X>=4);
subplot (211)
plot (X, V)
area (X,V, 'FaceColor', 'r', 'LineWidth', 2);
  댓글 수: 4
Torsten
Torsten 2025년 3월 7일
Which function for M do you want to plot in a mathematical notation (not code) ?
hamawandy
hamawandy 2025년 3월 7일
편집: hamawandy 2025년 3월 7일
Thank you so much I have fixed the error just now.
The final code for both shear force and bending moment diagrams is the following:
R1=6;
R2=2;
X= linspace (0, 8, 100);
V= (R1-2*X).*(X>=0).*(X<=4) - (R2)*(X>=4);
subplot (211)
plot (X, V)
area (X,V, 'FaceColor', 'r', 'LineWidth', 2);
M=(R1*X-X.^2 ).*(X>=0).*(X<4) + (R1*X-8*(X-2)).* (X>=4).* (X<=8);
subplot(212)
plot (X,M)
area(X,M, 'FaceColor', 'b', 'LineWidth', 2);
xline(0, 'Color', 'k', 'LineWidth', 2); % Draw line for Y axis
yline(0, 'Color', 'k', 'LineWidth', 2); % Draw line for X axis

댓글을 달려면 로그인하십시오.

카테고리

Help CenterFile Exchange에서 Ordinary Differential Equations에 대해 자세히 알아보기

제품


릴리스

R2023b

Community Treasure Hunt

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

Start Hunting!

Translated by