Extracting Data from Figures and Vectors

Dear all,

I have the following code in which I want to extracxt y values (displacements) for 12ft, 24ft and 36ft along a 48ft beam.

clear, clc, close all

EI = 29000*13400;  % Flexural rigidity
L = 48;  % Length of Beam
P = 1;  % Loading
R1y = 0.75;  % Support reaction at support 1
R2y = 0.25;  % Support reaction at support 2
x1 = 0:0.1:12;  % X values for 0 < x < 12
dx1 = x1(2) - x1(1);  % Step size
x2 = 12:0.1:48;   % X values for 0 < x < 12
Mx1 = R1y*x1/(EI);   % Moment function for x1
dx2 = x2(2) - x2(1);  % Step size
Mx2 = (R1y*x2 - P*(x2-12))/(EI);  % Moment function for x2
y1(1) = 0;  % Initial condition for y1
b = 36;
z1(1) = P*b*(L^2 - b^2)/(6*L*EI); % Initial condition for z1 (slope)

% Compute displacement for 0 < x1 < 12

for i = 1:length(x1)-1
   y1(i+1) = y1(i) + z1(i)*dx1;
   z1(i+1) = z1(i) - Mx1(i)*dx1;
end

y2(1) = y1(end)  % Initial condition for y2
z2(1) = z1(end)  % Initial condition for z2 (slope)

% Compute displacement for 12 < x2 < 48

for i = 1:length(x2)-1
   y2(i+1) = y2(i) + z2(i)*dx2;
   z2(i+1) = z2(i) - Mx2(i)*dx2;
end

% Plot results

plot(x1,y1,x2,y2)
grid on

My y1 vector is 121 interations long and my y2 vector is 361 iterations long. How do I extract from the data the values at 12ft, 24ft and 36ft of the beam in the most efficient way? If for example I put y1(12) I do not get the value at 12ft along the 48ft beam, because I have made a step size 0.1 and not 1.

Many thanks in advance,

Scott

Answer 1

[Star Strider]

The easiest way is to use the interp1 function —

clear, clc, close all

EI = 29000*13400;  % Flexural rigidity
L = 48;  % Length of Beam
P = 1;  % Loading
R1y = 0.75;  % Support reaction at support 1
R2y = 0.25;  % Support reaction at support 2
x1 = 0:0.1:12;  % X values for 0 < x < 12
dx1 = x1(2) - x1(1);  % Step size
x2 = 12:0.1:48;   % X values for 0 < x < 12
Mx1 = R1y*x1/(EI);   % Moment function for x1
dx2 = x2(2) - x2(1);  % Step size
Mx2 = (R1y*x2 - P*(x2-12))/(EI);  % Moment function for x2
y1(1) = 0;  % Initial condition for y1
b = 36;
z1(1) = P*b*(L^2 - b^2)/(6*L*EI); % Initial condition for z1 (slope)

% Compute displacement for 0 < x1 < 12

for i = 1:length(x1)-1
   y1(i+1) = y1(i) + z1(i)*dx1;
   z1(i+1) = z1(i) - Mx1(i)*dx1;
end

y2(1) = y1(end)  % Initial condition for y2

y2 = 3.3489e-06

z2(1) = z1(end)  % Initial condition for z2 (slope)

z2 = 1.8644e-07

% Compute displacement for 12 < x2 < 48

for i = 1:length(x2)-1
   y2(i+1) = y2(i) + z2(i)*dx2;
   z2(i+1) = z2(i) - Mx2(i)*dx2;
end

x2_vals = [12 24 36];                                   % Independent Variable Values
y2_vals = interp1(x2, y2, x2_vals)                      % Interpolated Dependent Variable Values

y2_vals = 1×3

1.0e-05 * 0.3349 0.4113 0.2645

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% Plot results

plot(x1,y1,x2,y2)
hold on
plot(x2_vals, y2_vals, 'rs')
hold off
grid on

●

.

[Scott Banks]

Briliant! Thank you, Star Strider.

[Star Strider]

As always, my pleasure!

[Scott Banks]

Hi Star Strider, I have come into a bit of difficulty with the Interp1 function.

I have the two vectors for displacement (y1 and y2) and I have combined them into one vector called "Y".

clear, clc, close all

EI = 29000*13400;  % Flexural rigidity
L = 48;  % Length of Beam
P = 1;  % Loading
R1y = 0.75;  % Support reaction at support 1
R2y = 0.25;  % Support reaction at support 2
x1 = 0:0.1:12;  % X values for 0 < x < 12
dx1 = x1(2) - x1(1);  % Step size
x2 = 12:0.1:48;   % X values for 0 < x < 12
Mx1 = R1y*x1/(EI);   % Moment function for x1
dx2 = x2(2) - x2(1);  % Step size
Mx2 = (R1y*x2 - P*(x2-12))/(EI);  % Moment function for x2
y1(1) = 0;  % Initial condition for y1
b = 36;
z1(1) = P*b*(L^2 - b^2)/(6*L*EI); % Initial condition for z1 (slope)

% Compute displacement for 0 < x1 < 12

for i = 1:length(x1)-1
   y1(i+1) = y1(i) + z1(i)*dx1;
   z1(i+1) = z1(i) - Mx1(i)*dx1;
end

y2(1) = y1(end);  % Initial condition for y2
z2(1) = z1(end);  % Initial condition for z2 (slope)

% Compute displacement for 12 < x2 < 48

for i = 1:length(x2)-1
   y2(i+1) = y2(i) + z2(i)*dx2;
   z2(i+1) = z2(i) - Mx2(i)*dx2;
end

X = [x1 x2];
Y = [y1 y2];


X_vals = [12,24,36];
Y_vals = interp1(X,Y,X_vals)

I keep getting an error message for the Interp1 function and I don't know why!

[Star Strider]

Apparently, you have duplicated values in ‘X’.

Using unique seems to have solved it, although you need to check those results to be sure.

Try this —

clear, clc, close all

EI = 29000*13400;  % Flexural rigidity
L = 48;  % Length of Beam
P = 1;  % Loading
R1y = 0.75;  % Support reaction at support 1
R2y = 0.25;  % Support reaction at support 2
x1 = 0:0.1:12;  % X values for 0 < x < 12
dx1 = x1(2) - x1(1);  % Step size
x2 = 12:0.1:48;   % X values for 0 < x < 12
Mx1 = R1y*x1/(EI);   % Moment function for x1
dx2 = x2(2) - x2(1);  % Step size
Mx2 = (R1y*x2 - P*(x2-12))/(EI);  % Moment function for x2
y1(1) = 0;  % Initial condition for y1
b = 36;
z1(1) = P*b*(L^2 - b^2)/(6*L*EI); % Initial condition for z1 (slope)

% Compute displacement for 0 < x1 < 12

for i = 1:length(x1)-1
   y1(i+1) = y1(i) + z1(i)*dx1;
   z1(i+1) = z1(i) - Mx1(i)*dx1;
end

y2(1) = y1(end);  % Initial condition for y2
z2(1) = z1(end);  % Initial condition for z2 (slope)

% Compute displacement for 12 < x2 < 48

for i = 1:length(x2)-1
   y2(i+1) = y2(i) + z2(i)*dx2;
   z2(i+1) = z2(i) - Mx2(i)*dx2;
end

X = [x1 x2];
Y = [y1 y2];

[Xu,ix] = unique(X);
Yu = Y(ix);

X_vals = [12,24,36];
Y_vals = interp1(Xu,Yu,X_vals)

Y_vals = 1×3

1.0e-05 * 0.3349 0.4113 0.2645

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Results = table(X_vals(:), Y_vals(:), VariableNames=["X Vals", "Y Vals"])

Results = 3x2 table
    X Vals      Y Vals  
    ______    __________

      12      3.3489e-06
      24      4.1132e-06
      36      2.6449e-06

.

[Scott Banks]

That's fine. Thanks for that, Star Strider.

[Star Strider]

As always, my pleasure!