给我一个用python语言写的三角形常应变单元的有限元代码
时间: 2024-03-17 12:43:25 浏览: 57
以下是一个用Python语言编写的三角形常应变单元的有限元代码:
```python
import numpy as np
# 定义三角形节点坐标和单元节点关系
nodes = np.array([[0, 0], [0, 1], [1, 0]])
elems = np.array([[0, 1, 2]])
# 定义材料的弹性模量和泊松比
E = 210e9
nu = 0.3
# 计算材料的弹性矩阵
D = E / (1 - nu ** 2) * np.array([[1, nu, 0], [nu, 1, 0], [0, 0, (1 - nu) / 2]])
# 构造三角形常应变单元的刚度矩阵
def get_element_stiffness_matrix(elem):
x1, y1 = nodes[elem[0]]
x2, y2 = nodes[elem[1]]
x3, y3 = nodes[elem[2]]
A = 0.5 * abs(x1 * y2 + x2 * y3 + x3 * y1 - x1 * y3 - x2 * y1 - x3 * y2)
B = np.array([[y2 - y3, 0, y3 - y1, 0, y1 - y2, 0],
[0, x3 - x2, 0, x1 - x3, 0, x2 - x1],
[x3 - x2, y2 - y3, x1 - x3, y3 - y1, x2 - x1, y1 - y2]])
return A * D @ B.T @ np.linalg.inv(B @ D @ B.T)
# 构造整体刚度矩阵
num_nodes = nodes.shape[0]
num_elems = elems.shape[0]
K = np.zeros((2 * num_nodes, 2 * num_nodes))
for i in range(num_elems):
elem = elems[i]
ke = get_element_stiffness_matrix(elem)
for r in range(3):
for c in range(3):
K[2 * elem[r], 2 * elem[c]] += ke[2 * r, 2 * c]
K[2 * elem[r], 2 * elem[c] + 1] += ke[2 * r, 2 * c + 1]
K[2 * elem[r] + 1, 2 * elem[c]] += ke[2 * r + 1, 2 * c]
K[2 * elem[r] + 1, 2 * elem[c] + 1] += ke[2 * r + 1, 2 * c + 1]
# 定义边界条件
fixed_nodes = [0]
fixed_dofs = [2 * i for i in fixed_nodes]
free_dofs = [i for i in range(2 * num_nodes) if i not in fixed_dofs]
# 定义外力
F = np.zeros(2 * num_nodes)
F[2] = -5000
# 求解位移场
K_ff = K[np.ix_(free_dofs, free_dofs)]
F_f = F[free_dofs]
u_f = np.linalg.solve(K_ff, F_f)
u = np.zeros(2 * num_nodes)
u[free_dofs] = u_f
# 输出结果
print("位移场:")
print(u.reshape(-1, 2))
print("应力场:")
for i in range(num_elems):
elem = elems[i]
u_e = u[2 * elem]
B_e = np.array([[y2 - y3, 0, y3 - y1, 0, y1 - y2, 0],
[0, x3 - x2, 0, x1 - x3, 0, x2 - x1],
[x3 - x2, y2 - y3, x1 - x3, y3 - y1, x2 - x1, y1 - y2]])
epsilon_e = B_e @ u_e
sigma_e = D @ epsilon_e
print(sigma_e)
```
该代码中,首先定义了三角形的节点坐标和单元节点关系,在此基础上计算了材料的弹性矩阵,并利用它构造了三角形常应变单元的刚度矩阵。然后,通过遍历所有单元,将它们的刚度矩阵叠加到整体刚度矩阵中。接着,定义了边界条件和外力,并利用求解线性方程组的方法求解了位移场。最后,利用位移场计算了应变场和应力场,并输出了结果。
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