§9-2剪力和弯矩及其方程 2F Me=Fa F 5F SE A ℃ B E D 分析右段得到: Me=Fa B ∑F=0 F+f=0 SE E D F F SE By M=0 M=F +F E 3FC 2 目录FBy FAy FBy FSE ME O 3 F FBy = 3 5F FAy = 分析右段得到： FSE ME O Fy = 0 FSE + FBy = 0 3 F FSE = −FBy = − Mo = 0 Fa a ME = FBy  + 2 3 2 3Fa ME = ◼§9-2 剪力和弯矩及其方程 目录