3.单元刚度矩阵和单元等效结点荷载 用虚位移原理或势能原理可推得A=(1m)5+(+355)n 2 6 kS=F+FE A,=un, 5 25/ 时=[Dt A3=75+ k A4=,( )7 (1+,mm)k 24 2a3 kk !时+时两 2 43 对于平面应力问题 算例 Kk]=abI B[DB]d5dn见教98页例题52、95页例题53 E|A4423.单元刚度矩阵和单元等效结点荷载 用虚位移原理或势能原理可推得       E  e e k  = F + F        − − = 1 1 1 1 k B D B abdd e T                                             = 4 1 4 2 4 3 4 4 3 1 3 2 3 3 3 4 2 1 2 2 2 3 2 4 1 1 1 2 1 3 1 4 k k k k k k k k k k k k k k k k 对于平面应力问题        − − = 1 1 1 1 k ab B D B j dd T ij i       − = 3 4 1 2 2 4(1 ) A A E A A  i j i j i j i j b a a b A        ) 3 1 (1 2 1 ) 3 1 (1 1 + − = + + A j i  j i    2 1 2 − = + A i j  i  j    2 1 3 − = + i j i j i j i j a b b a A        ) 3 1 (1 2 1 ) 3 1 (1 4 + − = + + F  ab N F d d N F ds L s T b T E    = + − −    1 1 1 1 三.算例 见教材98页例题5-2、99页例题5-3