例1.求图示简支梁E截面的内力 求E截面的内力 解：1.确定支反力 Me=Fa ∑M=0 -F3a+2F.2a+F.a=0 1.5aC D ∑M4=0 F·3a+Fa-2F·a=0 2.用截面法研究内力 5F F 3 3 ∑F,=0 2F+FSE=FAy FSE=- F 3 ∑。-02F号+M:=E 3a ME= Fa 2
FAy 解：1. 确定支反力 FAy F By 0 ∑ MB = 3 22 0 − ⋅ + ⋅ + ⋅= F a F a Fa Ay ∑ M A = 0 3 20 F a Fa F a By ⋅ + − ⋅= 3 By F F = 5 3 Ay F 2. 用截面法研究内力 F = FSE ME ∑Fy = 0 2FF F + = SE Ay 0 ∑ MO = 3 2 2 2 E Ay a a F MF ⋅+ = ⋅ 例1.求图示简支梁 E 截面的内力 O
Me=Fa 若分析右段，得到： SE ∑F,=0FE+F,=0 F Fse=-Foy=-3 ∑M。=0 Mg=FRy'2 ME= 3Fa 2
F By FSE ME O 若分析右段，得到： ∑Fy = 0 FSE + FBy = 0 0 ∑ MO = Fa a ME = FBy ⋅ + 2 3