Question

这种外伸梁A,B的支座反力怎么计算!请尽量详细点,

Answer 1

ΣMa =0,-(2kN/m)x(4m)x(4m/2) +(Fby)x4m +6kNm -5kNx6m =0

Fby = 10kN（向上）

ΣFy =0,Fay -(2kN/m)x(4m) +10kN -5kN =0

Fay = 3kN (向上）

ΣFx =0,Fax +0 =0

Fax = 0

验算：

ΣMb =0,-(Fay)x4m +(2kN/m)x(4m)x(4m/2)+6kNm -5kNx2m =0

Fay = 3kN (向上）,与用ΣFy =0方程计算结果相同.