python三维有限元代码

三维有限元方法是非常复杂的数值计算方法，需要涉及许多数学和物理概念。在 Python 中，我们可以使用许多库来实现三维有限元代码，其中最常用的是 NumPy、SciPy、Matplotlib 等。下面是一个简单的示例代码：

import numpy as np
from scipy.sparse import csr_matrix
from scipy.sparse.linalg import spsolve


class FEM3D:
    def __init__(self, nodes, elems, loads, E, nu):
        self.nodes = nodes  # 节点矩阵
        self.elems = elems  # 单元矩阵
        self.loads = loads  # 荷载矩阵
        self.E = E  # 弹性模量
        self.nu = nu  # 泊松比

        self.nnodes = len(nodes)  # 节点数
        self.nelems = len(elems)  # 单元数
        self.ndof = 3  # 自由度数
        self.edof = np.arange(self.ndof * 4).reshape((4, self.ndof))  # 单元自由度矩阵

        self.K = np.zeros((self.nnodes * self.ndof, self.nnodes * self.ndof))  # 刚度矩阵
        self.F = np.zeros((self.nnodes * self.ndof, 1))  # 荷载矩阵
        self.U = np.zeros((self.nnodes * self.ndof, 1))  # 位移矩阵

    def _get_elastic_matrix(self):
        D = np.zeros((6, 6))
        D[0, 0] = D[1, 1] = D[2, 2] = 1 - self.nu
        D[0, 1] = D[0, 2] = D[1, 0] = D[1, 2] = D[2, 0] = D[2, 1] = self.nu
        D[3, 3] = D[4, 4] = D[5, 5] = (1 - 2 * self.nu) / 2
        D = self.E / (1 + self.nu) / (1 - 2 * self.nu) * D
        return D

    def _get_shape_function(self, x, y, z):
        N = np.zeros((4, 1))
        N[0] = (1 - x) * (1 - y) * (1 - z) / 8
        N[1] = (1 + x) * (1 - y) * (1 - z) / 8
        N[2] = (1 + x) * (1 + y) * (1 - z) / 8
        N[3] = (1 - x) * (1 + y) * (1 - z) / 8
        return N

    def _get_jacobian(self, elem):
        x = self.nodes[elem, 0::3]
        y = self.nodes[elem, 1::3]
        z = self.nodes[elem, 2::3]
        J = np.zeros((3, 3))
        J[0, :] = (y[1:] - y[0]) / 2
        J[1, :] = (x[2] - x[0], y[2] - y[0], z[2] - z[0])
        J[2, :] = (y[2] - y[0]) / 2
        return J

    def _get_B_matrix(self, elem):
        J = self._get_jacobian(elem)
        B = np.zeros((6, 12))
        for i in range(4):
            B[0, i * 3:i * 3 + 3] = J[1, :] @ self.edof[i, :]
            B[1, i * 3:i * 3 + 3] = J[2, :] @ self.edof[i, :]
            B[2, i * 3:i * 3 + 3] = J[0, 0] @ self.edof[i, :]
            B[3, i * 3:i * 3 + 3] = J[1, :] @ self.edof[i, :]
            B[4, i * 3:i * 3 + 3] = J[0, 1] @ self.edof[i, :]
            B[5, i * 3:i * 3 + 3] = J[2, :] @ self.edof[i, :]
        B = 1 / np.linalg.det(J) * B
        return B

    def _get_element_stiffness(self, elem):
        D = self._get_elastic_matrix()
        B = self._get_B_matrix(elem)
        Ke = np.transpose(B) @ D @ B * np.linalg.det(self._get_jacobian(elem))
        return Ke

    def _assemble_stiffness_matrix(self):
        for i in range(self.nelems):
            Ke = self._get_element_stiffness(i)
            edof = np.repeat(self.edof[np.newaxis, :], 4, axis=0).reshape((16, self.ndof))
            edof += np.repeat(self.ndof * self.elems[i, np.newaxis], 4, axis=0)
            edof = edof.flatten()
            self.K[np.ix_(edof, edof)] += Ke

    def _assemble_loads(self):
        for i in range(len(self.loads)):
            node = self.loads[i][0]
            dof = self.loads[i][1]
            value = self.loads[i][2]
            self.F[node * self.ndof + dof] += value

    def solve(self):
        self._assemble_stiffness_matrix()
        self._assemble_loads()
        self.U = spsolve(csr_matrix(self.K), self.F)

该代码使用了稀疏矩阵求解器来求解方程组，以提高效率。您可以根据需要进行修改和扩展。