X B 1/2 1/2 B 对O点取矩即得: (b) ∑ XA·2f+f 3f X=XR-af ∑MC=0XB×f-B×=0 多跨刚架支座反力计算见 于是X qfP44例42 4XA q f l /2 l /2 A B C (b) YA YB XB f l /2 C (c) YB XB B XC YC  MC = 0 0 2  −  = l XB f YB 4 qf XB = 于是 X qf A 4 3 = − X X qf A = B − O 对O点取矩即得： MO = 0 0 2 3  2 +  = f X f qf A X qf A 4 3 = − 多跨刚架支座反力计算见 P44例4.2