将单元矩阵组装成6*6的矩阵,且梁的参数如下
时间: 2024-03-27 11:40:45 浏览: 23
好的,下面是将单元矩阵组装成6*6的矩阵的示例代码,假设梁的几何和材料参数如下:
```matlab
E = 2.1e11; % 弹性模量
G = 0.8e11; % 剪切模量
rho = 7800; % 密度
L = 2; % 梁的长度
d = 0.1; % 梁的直径
A = pi*(d/2)^2; % 梁的横截面积
I = pi*(d/2)^4/4; % 梁的截面惯性矩
```
我们可以先计算出单元刚度矩阵和质量矩阵,然后将其组装成6*6的矩阵。这里假设每个单元有2个节点,每个节点有3个自由度(x、y方向的位移和转角),因此单元刚度矩阵和质量矩阵的大小均为6*6。我们使用类似上面的循环遍历方法,将单元矩阵组装成6*6的全局矩阵。
```matlab
% 计算单元刚度矩阵和质量矩阵
k11 = E*A/L;
k12 = 0;
k13 = 0;
k14 = -E*A/L;
k15 = 0;
k16 = 0;
k22 = 12*E*I/L^3;
k23 = 6*E*I/L^2;
k24 = 0;
k25 = -12*E*I/L^3;
k26 = 6*E*I/L^2;
k33 = 4*G*A/L;
k34 = 0;
k35 = 0;
k36 = -4*G*A/L;
k44 = E*A/L;
k45 = 0;
k46 = 0;
k55 = 12*E*I/L^3;
k56 = -6*E*I/L^2;
k66 = 4*G*A/L;
K_e = [k11, k12, k13, k14, k15, k16;
k12, k22, k23, k24, k25, k26;
k13, k23, k33, k34, k35, k36;
k14, k24, k34, k44, k45, k46;
k15, k25, k35, k45, k55, k56;
k16, k26, k36, k46, k56, k66];
m11 = rho*A*L/3;
m12 = 0;
m13 = 0;
m14 = rho*A*L/6;
m15 = 0;
m16 = 0;
m22 = rho*I*L/15;
m23 = rho*I*L/30;
m24 = 0;
m25 = -rho*I*L/15;
m26 = rho*I*L/30;
m33 = rho*A*L/3;
m34 = 0;
m35 = 0;
m36 = rho*A*L/6;
m44 = rho*A*L/3;
m45 = 0;
m46 = 0;
m55 = rho*I*L/15;
m56 = -rho*I*L/30;
m66 = rho*A*L/3;
M_e = [m11, m12, m13, m14, m15, m16;
m12, m22, m23, m24, m25, m26;
m13, m23, m33, m34, m35, m36;
m14, m24, m34, m44, m45, m46;
m15, m25, m35, m45, m55, m56;
m16, m26, m36, m46, m56, m66];
% 定义全局刚度矩阵和质量矩阵的大小
N = 2; % 假设有2个节点
K = zeros(3*N, 3*N); % 初始化全局刚度矩阵
M = zeros(3*N, 3*N); % 初始化全局质量矩阵
% 循环遍历每个单元,将其贡献添加到全局刚度矩阵和质量矩阵中
for i = 1:N-1
K_i = K_e; % 第i个单元的刚度矩阵
M_i = M_e; % 第i个单元的质量矩阵
% 将第i个单元的刚度矩阵和质量矩阵添加到全局刚度矩阵和质量矩阵中
K(3*i-2:3*i+3, 3*i-2:3*i+3) = K(3*i-2:3*i+3, 3*i-2:3*i+3) + K_i;
M(3*i-2:3*i+3, 3*i-2:3*i+3) = M(3*i-2:3*i+3, 3*i-2:3*i+3) + M_i;
end
```
在上面的代码中,我们首先计算出单元的刚度矩阵和质量矩阵,然后定义全局刚度矩阵和质量矩阵的大小,并初始化为零矩阵。最后,我们循环遍历每个单元,将其贡献添加到全局刚度矩阵和质量矩阵中。这样,我们就可以得到组装后的6*6全局刚度矩阵和质量矩阵了。