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求图示梁支座的约束反力.已知 :F=2KN,a=2m,求A及B的支座反力
人气:392 ℃ 时间:2020-06-04 13:21:03
解答
题目未给出支座B处的斜面倾角,无法计算支座反力.呃 忘了说了 斜面倾角是30° 谢谢ΣMA =0,FBy.6m -2KN.2m -2KN.4m =0
FBy = 2KN (向上)
FBx = (FBy)tang30 =(2KN).(0.577) =1.154KN(向左)

ΣFx =0, FAx -FBx =0
FAx -1.154KN =0
FAx =1.154KN(向右)

ΣFy =0, FAy -2F +FBy =0
FAy -4KN +2KN =0
FAy =2KN(向上)
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