二、分配系数设A点有力矩M,求MB、M和MAD A Br如用位移法求解 于是可得 AD iab MAB=4iB0A-=SABE MR=5ABMI AC MAC=LC0=S4C0A ∑S C 4D=3i4DE4=SADe M MA==ACM AD M LAB ∑ m=0 M=(SB+SC+SAD)6 ∑ A A AC M M M -AD Sn+sts AD AD ∑ A MA=·M ∑=1 分配系数 2021/2/21 42021/2/21 分配系数 4 SAB = 4i 1 SAB= 3i 1 1 SAB= i 二、分配系数 设A点有力矩M，求MAB、MAC和MAD C D A B iAB iAC iAD  A M 如用位移法求解： AB AB A SAB A M = 4i  =  AC AC A SAC A M = i  =  AD AD A SAD A M = 3i  =   M MAB MAC MAD  = 0 A m M SAB SAC SAD  A = ( + + )  = + + = A AB AC AD A S M S S S M  M S S M A AD AD  = 于是可得 M S S M A AB AB  = M S S M A AC AC  =  M Aj =  Aj M  = A Aj Aj S S    =1