how to triple Integrate this?

1 Ansicht (letzte 30 Tage)
Saurabh Agarwal
Saurabh Agarwal am 7 Aug. 2021
Beantwortet: Walter Roberson am 8 Aug. 2021
clear
clc
D = 200000;
syms xi eta zeta
N(1) = (1/8)*(1-xi)*(1-eta)*(1-zeta);
N(2) = (1/8)*(1+xi)*(1-eta)*(1-zeta);
N(3) = (1/8)*(1+xi)*(1+eta)*(1-zeta);
N(4) = (1/8)*(1-xi)*(1+eta)*(1-zeta);
N(5) = (1/8)*(1-xi)*(1-eta)*(1+zeta);
N(6) = (1/8)*(1+xi)*(1-eta)*(1+zeta);
N(7) = (1/8)*(1+xi)*(1+eta)*(1+zeta);
N(8) = (1/8)*(1-xi)*(1+eta)*(1+zeta);
for i = 1:8
N_xi(i) = diff(N(i),xi);
N_eta(i) = diff(N(i),eta);
N_zeta(i) = diff(N(i),zeta);
end
% Nodes xi eta zeta
% 1 -1 -1 -1
% 2 1 -1 -1
% 3 1 1 -1
% 4 -1 1 -1
% 5 -1 -1 1
% 6 1 -1 1
% 7 1 1 1
% 8 -1 1 1
% x1_xi = (N_xi(1)*-1) + (N_xi(2)*1) + (N_xi(3)*1) + (N_xi(4)*-1) + (N_xi(5)*-1) + (N_xi(6)*1) + (N_xi(7)*1) + (N_xi(8)*-1);
% x1_eta = (N_eta(1)*-1) + (N_eta(2)*1) + (N_eta(3)*1) + (N_eta(4)*-1) + (N_eta(5)*-1) + (N_eta(6)*1) + (N_eta(7)*1) + (N_eta(8)*-1);
% x1_zeta = (N_zeta(1)*-1) + (N_zeta(2)*1) + (N_zeta(3)*1) + (N_zeta(4)*-1) + (N_zeta(5)*-1) + (N_zeta(6)*1) + (N_zeta(7)*1) + (N_zeta(8)*-1);
% x2_xi = (N_xi(1)*-1) + (N_xi(2)*-1) + (N_xi(3)*1) + (N_xi(4)*1) + (N_xi(5)*-1) + (N_xi(6)*-1) + (N_xi(7)*1) + (N_xi(8)*1);
% x2_eta = (N_eta(1)*-1) + (N_eta(2)*-1) + (N_eta(3)*1) + (N_eta(4)*1) + (N_eta(5)*-1) + (N_eta(6)*-1) + (N_eta(7)*1) + (N_eta(8)*1);
% x2_zeta = (N_zeta(1)*-1) + (N_zeta(2)*-1) + (N_zeta(3)*1) + (N_zeta(4)*1) + (N_zeta(5)*-1) + (N_zeta(6)*-1) + (N_zeta(7)*1) + (N_zeta(8)*1);
% x3_xi = (N_xi(1)*-1) + (N_xi(2)*-1) + (N_xi(3)*-1) + (N_xi(4)*-1) + (N_xi(5)*1) + (N_xi(6)*1) + (N_xi(7)*1) + (N_xi(8)*1);
% x3_eta = (N_eta(1)*-1) + (N_eta(2)*-1) + (N_eta(3)*-1) + (N_eta(4)*-1) + (N_eta(5)*1) + (N_eta(6)*1) + (N_eta(7)*1) + (N_eta(8)*1);
% x3_zeta = (N_zeta(1)*-1) + (N_zeta(2)*-1) + (N_zeta(3)*-1) + (N_zeta(4)*-1) + (N_zeta(5)*1) + (N_zeta(6)*1) + (N_zeta(7)*1) + (N_zeta(8)*1);
%
% J = [x1_xi x1_eta x1_zeta;x2_xi x2_eta x2_zeta;x3_xi x3_eta x3_zeta];
J = [1 0 0;0 1 0;0 0 1];
Inv_J = inv(J);
Det_J = det(J);
N_x = [Inv_J(1,1) * N_xi; Inv_J(2,2) * N_eta; Inv_J(3,3) * N_zeta];
N_x = [N_x(1,1) 0 0 N_x(1,2) 0 0 N_x(1,3) 0 0 N_x(1,4) 0 0 N_x(1,5) 0 0 N_x(1,6) 0 0 N_x(1,7) 0 0 N_x(1,8) 0 0;...
0 N_x(2,1) 0 0 N_x(2,2) 0 0 N_x(2,3) 0 0 N_x(2,4) 0 0 N_x(2,5) 0 0 N_x(2,6) 0 0 N_x(2,7) 0 0 N_x(2,8) 0;...
0 0 N_x(3,1) 0 0 N_x(3,2) 0 0 N_x(3,3) 0 0 N_x(3,4) 0 0 N_x(3,5) 0 0 N_x(3,6) 0 0 N_x(3,7) 0 0 N_x(3,8)];
K_ma = N_x.' * N_x;
for i = 1:24
for j = 1:24
K_mat(i,j) = @(xi,eta,zeta) K_ma(i,j);
K(i,j) = integral3(K_ma(i,j),-1,1,-1,1,-1,1);
end
end

Melden Sie sich an, um diese Frage zu beantworten.

Akzeptierte Antwort

Walter Roberson
Walter Roberson am 8 Aug. 2021
Ran in:
clear
clc
D = 200000;
syms xi eta zeta
N(1) = (1/8)*(1-xi)*(1-eta)*(1-zeta);
N(2) = (1/8)*(1+xi)*(1-eta)*(1-zeta);
N(3) = (1/8)*(1+xi)*(1+eta)*(1-zeta);
N(4) = (1/8)*(1-xi)*(1+eta)*(1-zeta);
N(5) = (1/8)*(1-xi)*(1-eta)*(1+zeta);
N(6) = (1/8)*(1+xi)*(1-eta)*(1+zeta);
N(7) = (1/8)*(1+xi)*(1+eta)*(1+zeta);
N(8) = (1/8)*(1-xi)*(1+eta)*(1+zeta);
for i = 1:8
N_xi(i) = diff(N(i),xi);
N_eta(i) = diff(N(i),eta);
N_zeta(i) = diff(N(i),zeta);
end
% Nodes xi eta zeta
% 1 -1 -1 -1
% 2 1 -1 -1
% 3 1 1 -1
% 4 -1 1 -1
% 5 -1 -1 1
% 6 1 -1 1
% 7 1 1 1
% 8 -1 1 1
% x1_xi = (N_xi(1)*-1) + (N_xi(2)*1) + (N_xi(3)*1) + (N_xi(4)*-1) + (N_xi(5)*-1) + (N_xi(6)*1) + (N_xi(7)*1) + (N_xi(8)*-1);
% x1_eta = (N_eta(1)*-1) + (N_eta(2)*1) + (N_eta(3)*1) + (N_eta(4)*-1) + (N_eta(5)*-1) + (N_eta(6)*1) + (N_eta(7)*1) + (N_eta(8)*-1);
% x1_zeta = (N_zeta(1)*-1) + (N_zeta(2)*1) + (N_zeta(3)*1) + (N_zeta(4)*-1) + (N_zeta(5)*-1) + (N_zeta(6)*1) + (N_zeta(7)*1) + (N_zeta(8)*-1);
% x2_xi = (N_xi(1)*-1) + (N_xi(2)*-1) + (N_xi(3)*1) + (N_xi(4)*1) + (N_xi(5)*-1) + (N_xi(6)*-1) + (N_xi(7)*1) + (N_xi(8)*1);
% x2_eta = (N_eta(1)*-1) + (N_eta(2)*-1) + (N_eta(3)*1) + (N_eta(4)*1) + (N_eta(5)*-1) + (N_eta(6)*-1) + (N_eta(7)*1) + (N_eta(8)*1);
% x2_zeta = (N_zeta(1)*-1) + (N_zeta(2)*-1) + (N_zeta(3)*1) + (N_zeta(4)*1) + (N_zeta(5)*-1) + (N_zeta(6)*-1) + (N_zeta(7)*1) + (N_zeta(8)*1);
% x3_xi = (N_xi(1)*-1) + (N_xi(2)*-1) + (N_xi(3)*-1) + (N_xi(4)*-1) + (N_xi(5)*1) + (N_xi(6)*1) + (N_xi(7)*1) + (N_xi(8)*1);
% x3_eta = (N_eta(1)*-1) + (N_eta(2)*-1) + (N_eta(3)*-1) + (N_eta(4)*-1) + (N_eta(5)*1) + (N_eta(6)*1) + (N_eta(7)*1) + (N_eta(8)*1);
% x3_zeta = (N_zeta(1)*-1) + (N_zeta(2)*-1) + (N_zeta(3)*-1) + (N_zeta(4)*-1) + (N_zeta(5)*1) + (N_zeta(6)*1) + (N_zeta(7)*1) + (N_zeta(8)*1);
%
% J = [x1_xi x1_eta x1_zeta;x2_xi x2_eta x2_zeta;x3_xi x3_eta x3_zeta];
J = [1 0 0;0 1 0;0 0 1];
Inv_J = inv(J);
Det_J = det(J);
N_x = [Inv_J(1,1) * N_xi; Inv_J(2,2) * N_eta; Inv_J(3,3) * N_zeta];
N_x = [N_x(1,1) 0 0 N_x(1,2) 0 0 N_x(1,3) 0 0 N_x(1,4) 0 0 N_x(1,5) 0 0 N_x(1,6) 0 0 N_x(1,7) 0 0 N_x(1,8) 0 0;...
0 N_x(2,1) 0 0 N_x(2,2) 0 0 N_x(2,3) 0 0 N_x(2,4) 0 0 N_x(2,5) 0 0 N_x(2,6) 0 0 N_x(2,7) 0 0 N_x(2,8) 0;...
0 0 N_x(3,1) 0 0 N_x(3,2) 0 0 N_x(3,3) 0 0 N_x(3,4) 0 0 N_x(3,5) 0 0 N_x(3,6) 0 0 N_x(3,7) 0 0 N_x(3,8)];
K_ma = N_x.' * N_x;
K = int(int(int(K_ma,xi,-1,1),eta,-1,1),zeta,-1,1)
K = 

Weitere Antworten (0)

Kategorien

Mehr zu Linear Least Squares finden Sie in Help Center und File Exchange

Tags

Community Treasure Hunt

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

Start Hunting!

Translated by