Matlab code: deflection of simply supported beam using ode45

8 Ansichten (letzte 30 Tage)
Camryn
Camryn am 7 Nov. 2023
Bearbeitet: hamawandy am 7 Mär. 2025
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

Antworten (2)

Torsten
Torsten am 7 Nov. 2023
Bearbeitet: Torsten am 7 Nov. 2023
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])
  7 Kommentare
Torsten
Torsten am 5 Mär. 2025
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 am 6 Mär. 2025
Thank you so much ...... I am so grateful to you.

Melden Sie sich an, um zu kommentieren.


hamawandy
hamawandy am 6 Mär. 2025
Bearbeitet: hamawandy am 6 Mär. 2025
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);
  4 Kommentare
Torsten
Torsten am 7 Mär. 2025
Which function for M do you want to plot in a mathematical notation (not code) ?
hamawandy
hamawandy am 7 Mär. 2025
Bearbeitet: hamawandy am 7 Mär. 2025
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

Melden Sie sich an, um zu kommentieren.

Produkte


Version

R2023b

Community Treasure Hunt

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

Start Hunting!

Translated by