用matlab写一段圆柱体受净应力的有限元分析

好的，以下是一个简单的圆柱体受净应力的有限元分析的Matlab代码：

% 圆柱体受净应力的有限元分析

% 定义圆柱体尺寸和材料属性
r = 0.1; % 圆柱体半径
h = 0.2; % 圆柱体高度
E = 200e9; % 弹性模量
nu = 0.3; % 泊松比
F = 1000; % 施加的力

% 定义有限元网格
nr = 10; % 圆柱体半径方向网格数
nh = 20; % 圆柱体高度方向网格数
[rr, hh] = meshgrid(linspace(r, r+h/nh, nr+1), linspace(0, h, nh+1));
nodes = [rr(:), hh(:)];
elements = delaunay(nodes);

% 计算单元刚度矩阵和全局刚度矩阵
K = zeros(2*(nr+1)*(nh+1), 2*(nr+1)*(nh+1));
for e = 1:size(elements, 1)
    % 获取单元节点编号
    n1 = elements(e, 1);
    n2 = elements(e, 2);
    n3 = elements(e, 3);
    % 获取单元节点坐标
    x1 = nodes(n1, 1);
    y1 = nodes(n1, 2);
    x2 = nodes(n2, 1);
    y2 = nodes(n2, 2);
    x3 = nodes(n3, 1);
    y3 = nodes(n3, 2);
    % 计算单元刚度矩阵
    Ke = get_element_stiffness_matrix(x1, y1, x2, y2, x3, y3, E, nu);
    % 装配单元刚度矩阵到全局刚度矩阵
    K([2*n1-1, 2*n1, 2*n2-1, 2*n2, 2*n3-1, 2*n3], [2*n1-1, 2*n1, 2*n2-1, 2*n2, 2*n3-1, 2*n3]) = ...
        K([2*n1-1, 2*n1, 2*n2-1, 2*n2, 2*n3-1, 2*n3], [2*n1-1, 2*n1, 2*n2-1, 2*n2, 2*n3-1, 2*n3]) + Ke;
end

% 施加边界条件
fixed_nodes = find(nodes(:,1)==r);
fixed_dofs = [2*fixed_nodes-1; 2*fixed_nodes];
K(fixed_dofs,:) = 0;
K(:,fixed_dofs) = 0;
K(fixed_dofs,fixed_dofs) = eye(length(fixed_dofs))*1e30;

% 计算位移和应力
u = K \ [zeros(length(K)-2*length(fixed_dofs), 1); F];
sxx = zeros(nr+1, nh+1);
syy = zeros(nr+1, nh+1);
sxy = zeros(nr+1, nh+1);
for e = 1:size(elements, 1)
    % 获取单元节点编号
    n1 = elements(e, 1);
    n2 = elements(e, 2);
    n3 = elements(e, 3);
    % 获取单元节点坐标
    x1 = nodes(n1, 1);
    y1 = nodes(n1, 2);
    x2 = nodes(n2, 1);
    y2 = nodes(n2, 2);
    x3 = nodes(n3, 1);
    y3 = nodes(n3, 2);
    % 计算单元位移
    u_e = [u(2*n1-1); u(2*n1); u(2*n2-1); u(2*n2); u(2*n3-1); u(2*n3)];
    % 计算单元应力
    [sxx_e, syy_e, sxy_e] = get_element_stress(x1, y1, x2, y2, x3, y3, E, nu, u_e);
    % 装配单元应力到全局应力
    sxx(n1, n2) = sxx(n1, n2) + sxx_e;
    syy(n1, n2) = syy(n1, n2) + syy_e;
    sxy(n1, n2) = sxy(n1, n2) + sxy_e;
    sxx(n2, n3) = sxx(n2, n3) + sxx_e;
    syy(n2, n3) = syy(n2, n3) + syy_e;
    sxy(n2, n3) = sxy(n2, n3) + sxy_e;
    sxx(n3, n1) = sxx(n3, n1) + sxx_e;
    syy(n3, n1) = syy(n3, n1) + syy_e;
    sxy(n3, n1) = sxy(n3, n1) + sxy_e;
end

% 绘制圆柱体应力分布图
figure;
surf(rr, hh, sxx);
title('Sxx');
xlabel('Radius');
ylabel('Height');
zlabel('Stress');
figure;
surf(rr, hh, syy);
title('Syy');
xlabel('Radius');
ylabel('Height');
zlabel('Stress');
figure;
surf(rr, hh, sxy);
title('Sxy');
xlabel('Radius');
ylabel('Height');
zlabel('Stress');

% 获取单元刚度矩阵函数
function Ke = get_element_stiffness_matrix(x1, y1, x2, y2, x3, y3, E, nu)
    A = 0.5 * abs(x1*y2 + x2*y3 + x3*y1 - x1*y3 - x2*y1 - x3*y2); % 计算单元面积
    b = [y2-y3; y3-y1; y1-y2] / (2*A);
    c = [x3-x2; x1-x3; x2-x1] / (2*A);
    B = [b(1), 0, b(2), 0, b(3), 0;
         0, c(1), 0, c(2), 0, c(3);
         c(1), b(1), c(2), b(2), c(3), b(3)];
    D = E / (1 - nu^2) * [1, nu, 0;
                         nu, 1, 0;
                         0, 0, (1-nu)/2];
    Ke = A * B' * D * B;
end

% 获取单元应力函数
function [sxx, syy, sxy] = get_element_stress(x1, y1, x2, y2, x3, y3, E, nu, u_e)
    A = 0.5 * abs(x1*y2 + x2*y3 + x3*y1 - x1*y3 - x2*y1 - x3*y2); % 计算单元面积
    b = [y2-y3; y3-y1; y1-y2] / (2*A);
    c = [x3-x2; x1-x3; x2-x1] / (2*A);
    B = [b(1), 0, b(2), 0, b(3), 0;
         0, c(1), 0, c(2), 0, c(3);
         c(1), b(1), c(2), b(2), c(3), b(3)];
    D = E / (1 - nu^2) * [1, nu, 0;
                         nu, 1, 0;
                         0, 0, (1-nu)/2];
    s = D * B * u_e;
    sxx = s(1);
    syy = s(2);
    sxy = s(3);
end

这段代码使用了三角剖分法生成了圆柱体的有限元网格，并计算了单元刚度矩阵和全局刚度矩阵。然后通过施加边界条件和施加力来求解位移，最终计算了圆柱体的应力分布。