弯矩分配法基本思想 4i+3i M M IP i1=E1/1 El2/ L2B Z1=M(4i1+3i2) Mc4=M×4i1(4i1+32) B MCB=M×3i24i1+3i2) M AC M×2i,/4 CA M =MCg×0/3 由此可得到什么结论呢?④ BC 1P弯矩分配法基本思想 R1P C M 11 4 1 3 2 r = i + i R1P = −M /(4 3 ) 1 1 2 Z = M i + i 4 /(4 3 ) 1 1 2 M M i i i CA =  + 3 /(4 3 ) 2 1 2 M M i i i CB =  + 由此可得到什么结论呢？ M A B 1 l 2 l 1 1 1 i = EI / l 2 2 2 i = EI / l C Z1 =1 A B 1 i 2 C i 1 4i 2 3i 1 2i 11 r 1 C 4i 2 3i 1 4 1 M M 2i / i AC = CA  3 2 M M 0/ i BC = CB 