计算支反力时,均布荷载可用等效的集中荷载P代替, P的大小 =q(2m) =(2KN/m)(2m) =4KN P的作用点在均布段中点,即: P对B点取矩的力臂=1m, P对A点取矩的力臂=6m+1m =7m ΣMA =0, FBy.6m -P.7m -F.4m -M =0 FBy.6m -(4KN)(7m) -(3KN)(4m) -4KN.m
是不是得啥病了