案例1
給出4個彈簧的勁度系數，離散后，求其總的剛度矩陣。 **代碼：** ``` 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) ``` **結果如下：** [[ 2500. -500. -2000. 0.] [ -500. 750. -250. 0.] [-2000. -250. 3250. -1000.] [ 0. 0. -1000. 1000.]]
案例2
兩根桿的實例。 **代碼分析：** ``` import numpy as np # 系統的初始參數如下： A1, A2, L1, L2, E1, E2 = 2400, 600, 300, 400, 7e4, 2e5 # 計算每根桿的剛度 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] # 初始化總的剛度矩陣、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) # 組裝總的剛度矩陣 eindex = [K[0:2, 0:2], K[1:, 1:]] for i in range(2): eindex[i] += ki[i] print(K) # 給出力向量 F[1] = 200000 print(F) # 求解過程 KK = K[1, 1] FF = F[1] dd = FF/KK d[1] = dd print(d) ``` **計算結果：** 1. 每根桿的剛度矩陣 k1 [[ 560000. -560000.] [-560000. 560000.]] k2 [[ 300000. -300000.] [-300000. 300000.]] 2. 總的剛度矩陣 [[ 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. ]]
