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:搜搜考试网作业帮 分类:物理作业 时间:2024/07/31 07:48:53
这种外伸梁A,B的支座反力怎么计算!请尽量详细点,
![](http://img.wesiedu.com/upload/2/22/2225aa67899c497ecf021b8f6f0dc814.jpg)
![](http://img.wesiedu.com/upload/2/22/2225aa67899c497ecf021b8f6f0dc814.jpg)
![这种外伸梁A,B的支座反力怎么计算!请尽量详细点,](/uploads/image/z/3127739-59-9.jpg?t=%E8%BF%99%E7%A7%8D%E5%A4%96%E4%BC%B8%E6%A2%81A%2CB%E7%9A%84%E6%94%AF%E5%BA%A7%E5%8F%8D%E5%8A%9B%E6%80%8E%E4%B9%88%E8%AE%A1%E7%AE%97%21%E8%AF%B7%E5%B0%BD%E9%87%8F%E8%AF%A6%E7%BB%86%E7%82%B9%2C)
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.