这种外伸梁A,B的支座反力怎么计算!请尽量详细点,
来源:学生作业帮 编辑:神马作文网作业帮 分类:物理作业 时间:2024/11/11 11:23:05
这种外伸梁A,B的支座反力怎么计算!请尽量详细点,
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.
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.