Question

求支反力，如图

Answer 1

ΣMC =0, -RBy.6m +(10KN/m).8m.3m -10KN.2m =0
RBy =110/3 KN（向上）
ΣMB =0, RCy.6m -(10KN/m).8m.3m -10KN.8m =0
RCy =160/3 KN（向上）
无X方向荷载，RCx =0, RBx =0

追问

乘以3m是什么意思

追答

计算支反力时，均布荷载q可用集中荷载Q代替，
Q的大小=(10KN/m).8m,
Q的作用点到C的距离=(8m/2) =4m
Q对C取矩的力臂 =4m,
Q对B取矩的力臂 =6m-4m =2m,
以上计算的力臂搞错了，现更正如下：
.
ΣMC =0, -RBy.6m +(10KN/m).8m.4m -10KN.2m =0
RBy =50 KN（向上）
ΣMB =0, RCy.6m -(10KN/m).8m.2m -10KN.8m =0
RCy =40 KN（向上）