结构力学一—第九章力矩分配法与近似法 W 海南大学土木建筑工程学院 (2)在结点B加上附加刚臂,计算固端弯矩 M=92=15x4 =-20kN.m 12 12 M1=9_15x4 =20kN.m 12 12 M泰=-33x80x4.-60kNm 16 16 则结点B的不平衡力矩MB=∑MF=M4+M6c=20-60=-40kN.m (3)放松附加刚臂,计算分配弯矩和传递弯矩 Mg4=4gM×(MB)=0.4×40=16kN.m MHe=4Bc×(-MB)=0.6×40=24kN.m MGB CBaMB =0.5x16=8 kN.m MCn CBCMBC =0 12 12 结构力学—第九章 力矩分配法与近似法 土木建筑工程学院 (2)在结点B加上附加刚臂,计算固端弯矩 20 12 15 4 12 2 2 = − = − = − ql M F AB kN.m 20 12 15 4 12 2 2 = = = ql M F BA kN.m 60 16 3 80 4 16 3 = − = − = − F l M F P BC kN.m 则结点B的不平衡力矩 = = + = 20 − 60 = −40 F BC F BA F MB M M M kN.m (3)放松附加刚臂,计算分配弯矩和传递弯矩 = = 0 BC BC C M AB C = CBAMBA = 0.516 = 8 kN.m MCB C M M BC = BC (−M B ) = 0.6 40 = 24 kN.m MBA = BA (−MB ) = 0.4 40 =16 kN.m