Extract data from countour and add two (or more) contours together.

2 次查看（过去 30 天）
William White
William White 2015-7-7
编辑： William White 2015-7-7
My problem is thus:
I have a mesh onto which I 'plot' a function.
That is, for every point x,y, I have a value Q
What I want to do is extract the value of Q for every x,y and ADD (superimpose) it to the value of Q for every x,y .
My code plots a stress field centered at at point (0,0) then it is moved a length dx and dy and an appropratiate step length. Then this is plotted.
If I have more than one point, I wish to superimpose the data.
i.e. I wish to set up a loop where the original position is kept, then another stress field is generated at different dx, dy, and then the two are added.
regards William
u = 80000; % Shear modulus, measure in Mpa nu = 0.3; % Poissions ratio of material b = .00025; % Burgers Vector measured in microns
[x,y]=meshgrid(-100:1:100,-100:1:100); % area of grid to which stress field is mapped
angle = 54.7; % angle of slip plane in degrees
theta = angle*pi/180; % angle converted to radians
steplength = 400;
interplanarspacing = 100*b; % spacing between slip planes is about 100 b
x2 = (x.*cos(theta) + y.*sin(theta));% rotates and displaces dislocation
y2 = (y.*cos(theta) - x.*sin(theta));%; rotates and displaces dislocation
QxxPRIME = -(u .* b / (2.*pi .* (1-nu))) .* ((y2 .* (3.*x2.^2 +y2.^2)) ./ ((x2.^2 + y2.^2).^2)); % Qyy stress PRIME
QyyPRIME = (u .* b / (2.*pi .* (1-nu))) .* ((y2 .* (x2.^2 -y2.^2)) ./ ((x2.^2 + y2.^2).^2)); % Qyy stress PRIME
QxyPRIME = (u .* b / (2.*pi .* (1-nu))) .* ((x2 .* (x2.^2 -y2.^2)) ./ ((x2.^2 + y2.^2).^2)); % Qxy stress PRIME
QxxPRIME(QxxPRIME<-.5) = -.5; % erase low value
QxxPRIME(QxxPRIME>.5) = .5; % erase high value
dx = steplength.*cos(theta);
dy = steplength.*sin(theta);
y = y+dy;
x = x+dx;
caxis([-0.5,0.5]); % set the legend limit
contourf(x(1,:),y(:,1)',QxxPRIME)
axis equal

请先登录，再回答此问题。

采纳的回答

Thorsten
Thorsten 2015-7-7
编辑：Thorsten 2015-7-7
The idea is to compute a new QxxPRIME for the different offsets and add them to QxxPRIME0 for zero offset:
u = 80000; % Shear modulus, measure in Mpa
nu = 0.3; % Poissions ratio of material
b = .00025; % Burgers Vector measured in microns
N = 500; % grid size
[x,y]=meshgrid(-N:1:N,-N:1:N); % area of grid to which stress field is mapped
angle = 54.7; % angle of slip plane in degrees
theta = angle*pi/180; % angle converted to radians
steplength = 400;
interplanarspacing = 100*b; % spacing between slip planes is about 100 b
Nsteps = 10;
c = -(u*b/(2*pi*(1 - nu))); % constants independent of loop variable sl
for sl = linspace(0, steplength, Nsteps)
dx = sl.*cos(theta);
dy = sl.*sin(theta);
x2 = x.*cos(theta) + y.*sin(theta) + dx;% rotates and displaces dislocation
y2 = y.*cos(theta) - x.*sin(theta) + dy;%; rotates and displaces dislocation
QxxPRIME = c*(y2.*(3*x2.^2 + y2.^2))./((x2.^2 + y2.^2).^2); % Qyy stress PRIME
QxxPRIME(QxxPRIME<-.5) = -.5; % erase low value
QxxPRIME(QxxPRIME>.5) = .5; % erase high value
subplot(1,2,1)
contourf(x,y,QxxPRIME)
caxis([-0.5,0.5])
axis square
title(['Contour at Steplength ' num2str(sl)])
if sl == 0
QxxPRIME0 = QxxPRIME;
else
QxxPRIME = QxxPRIME0 + QxxPRIME;
end
subplot(1,2,2)
contourf(x,y,QxxPRIME)
caxis([-0.5,0.5])
axis square
title(['Steplength 0 + ' num2str(sl)])
pause
end
  6 个评论
显示 4更早的评论
Thorsten
Thorsten 2015-7-7
编辑：Thorsten 2015-7-7
The field is the same because the arguments x and y are just rotated and displaced, i.e., moved in the plane. If a function(x,y) is zero for (0,0), then the function f(x+dx, y+dy) is zero at x = -dx, y = -dy.
William White
William White 2015-7-7
编辑：William White 2015-7-7
so would it be possible to just plot two fucntions; one centred at 10,10 and another at say, 12,10? So that indvidually, they look exactly the same, but superimposed, give a new field?
Also, I think that by adding dx to the rotation, it has broken the correct working of the code?
For example, if my steplength is, say, 10; then for the stress field to move correctly, it moves 10 units along a 54.7 incline. So dx is 5.7 and dy is 8.2; that is, if the steplength is zero, then the origin of the stress field is at 0,0; if the steplength is 10, the new orign is at 5.7, 8.2.
I took the liberty of removing the loop from your code to see what was going on with the movement.
kind regards
u = 80000; % Shear modulus, measure in Mpa
nu = 0.3; % Poissions ratio of material
b = .00025; % Burgers Vector measured in microns
N = 50; % grid size
[x,y]=meshgrid(-N:.1:N,-N:.1:N); % area of grid to which stress field is mapped
angle = 54.7; % angle of slip plane in degrees
theta = angle*pi/180; % angle converted to radians
steplength = -10;
interplanarspacing = 100*b; % spacing between slip planes is about 100 b
Nsteps = 30;
c = -(u*b/(2*pi*(1 - nu))); % constants independent of loop variable sl
dx = steplength*cos(theta);
dy = steplength*sin(theta);
x2 = (x.*cos(theta) + y.*sin(theta)) +dx;% rotates and displaces dislocation
y2 = (y.*cos(theta) - x.*sin(theta)) +dy;%; rotates and displaces dislocation
QxxPRIME = c*(y2.*(3*x2.^2 + y2.^2))./((x2.^2 + y2.^2).^2); % Qyy stress PRIME
QxxPRIME(QxxPRIME<-.5) = -.5; % erase low value
QxxPRIME(QxxPRIME>.5) = .5; % erase high value
contourf(x,y,QxxPRIME)
caxis([-0.5,0.5])
axis equal
title('Contour at Steplength ')
regards Willam

请先登录，再进行评论。

更多回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 Stress and Strain 的更多信息

标签

Community Treasure Hunt

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

Start Hunting!

Translated by