发布网友 发布时间:2024-10-23 21:55
共1个回答
热心网友 时间:2024-10-26 02:00
解(1) 梁 CD,受力如图所示
Σ MC= 0 ,
− q/2 ×(2m)^2 −M + FD×4m = 0
F = (M + 2q) / 4 = 15 kN
ΣFy = 0 ,FC + FD − q × 2 m = 0 ,FC = 5 kN
(2)梁AC,受力如图所示
Σ MA= 0 ,FB × 2 m − FC ' × 4 m − 2 m⋅ q ×3 m = 0
FB = (4FC '+ 6q) / 2 = 40 kN
ΣFy = 0 ,FA + FB − FC ' − q × 2 m = 0 ,FA = −15 kN