C B 111111111 1/2 立B f (c) XA 1/2 1/2 B XB 对O点取矩即得： YA YB (b) ∑M。=0 ,24a马=0 X4=Xa-q功 ∑Mc=0 X2×f-X0 X=-4 X 多跨刚架支座反力计算见 于是X4=一4 P44例4.2XA 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