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我的计算结果,FB亦是向上。 计算过程如下:
.
(1)取梁CDE段为隔离体:
ΣMC =0, FD.3m -(2KN/m).4m.2m =0
FD =5.3KN(向上)
ΣMD =0, 10KN.3m -FC.3m +(2KN/m).4m.1m =0
FC =12.7KN(向上)
.
(2)取梁ABC段为隔离体:
FC的反作用力FC' =(38/3)KN(向下)
ΣMB =0, 6KN.m -FAy.4m -FC'.2m =0
6KN.m -FAy.4m -12.7KN.2m =0
FAy = -4.85KN(向下)
ΣFx =0, FAx =0
ΣMA =0, 6KN.m +FB.4m -FC'.6m =0
6KN.m +FB.4m -12.7KN.6m =0
. FB = 17.55KN(向上)
验算:
ΣFy =0, FAy +FB -FC' =0
-4.85KN +FB -12.7KN =0
FB = 17.55KN(向上),与用ΣMA =0方程计算结果相同
.
(1)取梁CDE段为隔离体:
ΣMC =0, FD.3m -(2KN/m).4m.2m =0
FD =5.3KN(向上)
ΣMD =0, 10KN.3m -FC.3m +(2KN/m).4m.1m =0
FC =12.7KN(向上)
.
(2)取梁ABC段为隔离体:
FC的反作用力FC' =(38/3)KN(向下)
ΣMB =0, 6KN.m -FAy.4m -FC'.2m =0
6KN.m -FAy.4m -12.7KN.2m =0
FAy = -4.85KN(向下)
ΣFx =0, FAx =0
ΣMA =0, 6KN.m +FB.4m -FC'.6m =0
6KN.m +FB.4m -12.7KN.6m =0
. FB = 17.55KN(向上)
验算:
ΣFy =0, FAy +FB -FC' =0
-4.85KN +FB -12.7KN =0
FB = 17.55KN(向上),与用ΣMA =0方程计算结果相同
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