# Help with a tiny project ( building a graph)

2 ビュー (過去 30 日間)
Netanel Malihi 2021 年 11 月 17 日

I will try again, if anyone can help I would be very grateful...
I was wondering if someone could help me with a problem I have been given. It is about deflection of beams in a course in college called numerical analysis.
Given: 𝑞0 = 1𝑁𝑚𝑚, E = 140MPa, L = 1000mm, v = -0.7mm.
Draw a graph of x in front of I for 100∙ 10^6 mm^4 ≤ I ≤ 450∙10^6 mm^4. What conclusion can .be physically deduced from the graph?
Now, here is the code for x, I found it numerically (function beam2).I used the newton raphson method.
function X_coordinate = beam2(x0,q0,v,L,I,E)
epsilon = 1e-4;
flag = 0;
counter = 0;
% v=-0.5
% x0=500;
% q0=0.5
% L=1000
% I=351*10^6
% E= 100
x = x0;
while flag == 0
%Check for infinite loop
counter = counter + 1
if counter > 10000;
error('Too many iterations');
end
F = -(q0*L)/(3*pi^4*E*I)*(48*L^3*cos((pi*x)/(2*L))-48*L^3+3*pi^3*L*x^2-pi^3*x^3)-v
F_d = -(q0*L)/(3*pi^4*E*I)*(-48*L^3*(pi/(2*L))*sin((pi*x)/(2*L))+6*pi^3*L*x-3*pi^3*x^2)
x = x-F/F_d %solution
if abs(F/F_d) < epsilon || abs(F) < epsilon
flag = 1;
end
end
X_coordinate = x;
if X_coordinate <=0 || X_coordinate > 1000
error('No solution found,Please enter valid value')
else
X_coordinate = x
end
The code down is the code for buliding the graph. I need to use the solution of the beam2 function for that, numerically of course, and I have no idea how to proceed from here.
%function x_I = MomentI(x0,q0,L,v,E)
epsilon = 1e-4;
flag = 0;
counter = 0;
v=-0.7
x0=500;
q0=1
L=1000
% I=351*10^6
E= 140
x = x0
% I_x = 1:10
% x_x = 1:10
% while flag == 0
%Check for infinite loop
% counter = counter + 1
% if counter > 10;
% error('Too many iterations');
% end
if I_x >= 100*10^6 & I_x<=450*10^6
for x=1:1000
I_x(x) = -(q0*L)/(3*pi.^4*E*v)*(48*L^3*cos((pi*x)/(2*L))-48*L.^3+3*pi.^3*L*x.^2-pi^3*x.^3)
I_d(x) = -(q0*L)/(3*pi.^4*E*v)*(-48*L.^3*(pi/(2*L))*sin((pi*x)./(2*L))+6*pi^3*L*x-3*pi.^3*x.^2)
x_x(x) = x- I_x(x)./I_d(x) %solution
end
end
% if abs(I_x(x)./I_d(x)) < epsilon | abs(I_x(x)) < epsilon
% flag = 1;
% end
%end
% end
x_I = x;
if x_I <=0
error('No solution found,Please enter valid value')
else
x_I = x
end
x = 1:1000;
hold on
title ('I as a function of x')
xlabel('x[mm]')
ylabel('I[mm]')
plot(x_x,I_x)
hold off
Any suggestions to improve the current code would be great . thanks ...

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### 採用された回答

Chris 2021 年 11 月 17 日
If I understand correctly, you only wish to plot I as a function of x.
I have modified your function slightly (at the end of the code), including some semicolons to make it less talkative.
% Make a vector of I values
I_x = linspace(100e6,450e6,1000);
% Set constants
v=-0.7;
x0=500;
q0=1;
L=1000;
E=140;
% Preallocate x vector for speed
x_x = zeros(size(I_x));
% Repeatedly call the function, assigning the result to an element of the
% x_x vector
for idx = 1:numel(I_x)
x_x(idx) = beam2(x0,q0,v,L, I_x(idx) ,E);
end
% Plot
hold on
title ('I as a function of x')
xlabel('x[mm]')
ylabel('I[mm]')
plot(x_x,I_x)
hold off function X_coordinate = beam2(x0,q0,v,L,I,E)
epsilon = 1e-4;
flag = 0;
counter = 0;
x=x0;
while flag == 0
%Check for infinite loop
counter = counter + 1;
if counter > 10000
error('Too many iterations');
end
F = -(q0*L)/(3*pi^4*E*I)*(48*L^3*cos((pi*x)/(2*L))-48*L^3+3*pi^3*L*x^2-pi^3*x^3)-v;
F_d = -(q0*L)/(3*pi^4*E*I)*(-48*L^3*(pi/(2*L))*sin((pi*x)/(2*L))+6*pi^3*L*x-3*pi^3*x^2);
x = x-F/F_d; %solution
if abs(F/F_d) < epsilon || abs(F) < epsilon
flag = 1;
end
end
X_coordinate = x;
if X_coordinate <=0 || X_coordinate > 1000
error('No solution found,Please enter valid value')
else
X_coordinate = x;
end
end

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R2019b

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