专业方向系列-00-Python与有限元初探

案例1

给出4个弹簧的劲度系数，离散后，求其总的刚度矩阵。
**代码：**
```
import numpy as np
k1, k2, k3, k4 = 500, 250, 2000, 1000
ki = np.array([k1, k2, k3, k4])
p = 1000
K = np.matrix(np.zeros((4, 4)))
eindex = [K[0:2, 0:2], K[1:3, 1:3], K[0:3:2, 0:3:2], K[2:, 2:]]
for i in range(4):
ke = ki[i] * np.array([[1, -1], [-1, 1]])
eindex[i] += np.matrix(ke)
print(K)
```
**结果如下：**
[[ 2500. -500. -2000. 0.]
[ -500. 750. -250. 0.]
[-2000. -250. 3250. -1000.]
[ 0. 0. -1000. 1000.]]

案例2

两根杆的实例。
**代码分析：**
```
import numpy as np
# 系统的初始参数如下：
A1, A2, L1, L2, E1, E2 = 2400, 600, 300, 400, 7e4, 2e5
# 计算每根杆的刚度
k1 = (E1*A1/L1)*np.array([[1, -1], [-1, 1]])
print('k1', k1)
k2 = (E2*A2/L2)*np.array([[1, -1], [-1, 1]])
print('k2', k2)
ki = [k1, k2]
# 初始化总的刚度矩阵、力以及位移
K, F, d = np.zeros((3, 3)), np.zeros((3, 1)), np.zeros((3, 1))
print(K)
print(F)
print(d)
# 组装总的刚度矩阵
eindex = [K[0:2, 0:2], K[1:, 1:]]
for i in range(2):
eindex[i] += ki[i]
print(K)
# 给出力向量
F[1] = 200000
print(F)
# 求解过程
KK = K[1, 1]
FF = F[1]
dd = FF/KK
d[1] = dd
print(d)
```
**计算结果：**
1. 每根杆的刚度矩阵
k1 [[ 560000. -560000.]
[-560000. 560000.]]
k2 [[ 300000. -300000.]
[-300000. 300000.]]
2. 总的刚度矩阵
[[ 560000. -560000. 0.]
[-560000. 860000. -300000.]
[ 0. -300000. 300000.]]
3. 力向量
[[ 0.]
[200000.]
[ 0.]]
4. 求解的位移向量
[[0. ]
[0.23255814]
[0. ]]