一个目前没找到原因的问题
用np.linalg.solve(a, b) 解方程的问题。
错误消息：
r = gufunc(a, b, signature=signature, extobj=extobj)
TypeError: No loop matching the specified signature and casting was found for ufunc solve1。
既然 Ax=b 无法解，那就求 x=A**-1 *b
tmpA = np.linalg.inv(A)
X = np.dot(tmpA , b)
这个问题太奇葩了，
import math
import numpy as np
nodeNumber = 4
KK = np.zeros((2*nodeNumber,2*nodeNumber))
"""
边界条件，U表示节点的位移向量，如果某个自由度的位移未知，则该处填写‘u_Unknown’，
F表示节点的荷载向量，如果某个自由度的荷载未知，则该处填写‘f_Unknown’
"""
U = np.array([0, 0, 'u_Unknown', 0, 'u_Unknown', 'u_Unknown', 0, 0], dtype=object)
F = np.array(['f_Unknown', 'f_Unknown', 2e4, 'f_Unknown', 0, -2.5e4, 'f_Unknown', 'f_Unknown'], dtype=object)
class Bar2D:
"""定义二维杆单元类，该类包含杆件的基本信息：
E 弹性模量，A 杆单元面积，i 单元起点的节点编号，j 单元终点的节点编号
x1 y1 起点的坐标，x2 y2 终点的坐标，
DOF 单元在总体刚度矩阵里面所在的位置,L 单元的长度，
cos sin 单元的方向余弦 方向正弦，
k 单元刚度矩阵"""
def __init__(self, E, A, x1, y1, x2, y2, i, j):
self.E = E
self.A = A
self.i = i
self.j = j
# 定义一个由单刚矩阵的自由度向总刚矩阵自由度转换的数组
self.DOF = np.array([2 * i - 2, 2 * i - 1, 2 * j - 2, 2 * j - 1], dtype=np.int16)
self.L = math.sqrt((x1 - x2) ** 2 + (y1 - y2) ** 2)
self.cos = (x2 - x1) / self.L
self.sin = (y2 - y1) / self.L
L, c, s = self.L, self.cos, self.sin
self.k = (E * A / L) * np.array([[c * c, c * s, -c * c, -c * s],
[c * s, s * s, -c * s, -s * s],
[-c * c, -c * s, c * c, c * s],
[-c * s, -s * s, c * s, s * s]])
"定义求解单元应力的函数"
def stress(self):
# 从位移矩阵U中获取该单元两个节点的1*4位移矩阵
u = U(self.DOF)
E, L, c, s = self.E, self.L, self.c, self.s
T = np.array([-c, -s, c, s])
self.bar_Stress = E / L * np.dot(T, u)
return self.bar_Stress
"定义总刚矩阵集成函数"
def Bar2D2Node_Assembly(KK, bar):
for n1 in range(4):
for n2 in range(4):
KK[bar.DOF[n1], bar.DOF[n2]] += bar.k[n1, n2]
return KK
'求解节点位移'
def node_Disaplacement(KK, U, F):
# 获取缩减后的总刚矩阵
del_row_col = np.where(U == 0)
kk_delRow = np.delete(KK, del_row_col, 0)
kk_delCol = np.delete(kk_delRow, del_row_col, 1)
kk = kk_delCol
# 获取节点位移位置对应的节点力矩阵
f = F[np.where(U == 'u_Unknown')]
invKK = np.linalg.inv(kk)
u = np.dot(invKK, f)
print("求出位置的节点位移")
print (u)
'u = np.linalg.solve(kk, f)'
U[np.where(U == 'u_Unknown')] = u
return U
'求解节点力，必须在已经求得节点位移U后才可调用本函数'
def node_Force(KK, U):
F = np.dot(KK, U)
return F
E, A, x1, y1, x2, y2, x3, y3, x4, y4 = 2.95e11, 0.0001, 0, 0, 0.4, 0, 0.4, 0.3, 0, 0.3
bar1 = Bar2D(E, A, x1, y1, x2, y2, 1, 2)
bar2 = Bar2D(E, A, x3, y3, x2, y2, 3, 2)
bar3 = Bar2D(E, A, x1, y1, x3, y3, 1, 3)
bar4 = Bar2D(E, A, x4, y4, x3, y3, 4, 3)
bars = [bar1, bar2, bar3, bar4]
for bar in bars:
Bar2D2Node_Assembly(KK, bar)
U = node_Disaplacement(KK, U, F)
print("节点位移")
print(U)
F = node_Force(KK, U)
print("节点受力")
print(F)