取AB梁為受力分析對象：
ΣMA =0, Fcd.2m -F.3m =0
Fcd =(3/2)F =(3/2)(30kN) =45kN（拉力）
.
σ =Fcd /[(π/4)d^2] ≦ [σ]
d^2 ≧ (4/π)Fcd /[σ] =(4/3.1416)(45000N) /120Mpa
d ≧ 21.85mm
答：取CD桿直徑為 22mm