桿件 AB 與 CD 具相同的楊氏模數 $E$ 與斷面積 $A$，則剛性桿件 AC 之傾斜角為何？

由圖：A點接AB桿(長L，距中點a/2處)，C點接CD桿(長L/2，距中點a/2處)。設AB受力F_AB、CD受力F_CD。靜力平衡：F_AB·(a/2) + F_CD·(a/2) = P·0（對A點取矩）→ F_AB + F_CD = P（垂直平衡），且力矩平衡：F_AB×(a/2)×(-1) + F_CD×(a/2) = 0方向需重新整理。依圖P在中點正下方，A在左端，C在右端，距離各為a/2。對A點取矩：P×(a/2) = F_CD×a → F_CD = P/2；再由垂直平衡：F_AB = P/2。AB桿長L，變形δ_AB = F_AB·L/(EA) = PL/(2EA)（伸長）；CD桿長L/2，變形δ_CD = F_CD·(L/2)/(EA) = PL/(4EA)（縮短，即向上）。A點下移δ_AB，C點上移δ_CD，AC總相對位移 = δ_AB + δ_CD = PL/(2EA) + PL/(4EA) = 3PL/(4EA)。傾斜角θ = 總位移差/AC桿長 = [3PL/(4EA)]/(3a) ... 重新計算：A端下移 δ_A = PL/(2EA)，C端下移 δ_C = -PL/(4EA)（CD縮短代表C上移即負）。相對位移 = δ_A - δ_C = PL/(2EA) + PL/(4EA) = 3PL/(4EA)，AC長度為a（A至C = a/2+a/2 = a），θ = (3PL/4EA)/a... 再核對：對C取矩：F_AB × a = P × (a/2) → F_AB = P/2；對A取矩：F_CD × a = P × (a/2) → F_CD = P/2。δ_A(下) = F_AB×L/(EA) = PL/(2EA)；δ_C(上縮) = F_CD×(L/2)/(EA) = PL/(4EA)。AC傾斜角 = (δ_A + δ_C)/a = [PL/(2EA) + PL/(4EA)]/a = 3PL/(4aEA)... 此為C選項。重新審視幾何：AB長L，CD長L/2（由圖D在中點L/2處）；A在AC左端，C在AC右端，間距為a。P作用在AC中點(距A為a/2)。力矩平衡對A：P×(a/2) = F_CD×a → F_CD = P/2；對C：P×(a/2) = F_AB×a → F_AB = P/2。δ_B = PL/(2EA)(A端下移)；δ_D = (P/2)(L/2)/(EA) = PL/(4EA)(C端下移，CD縮短)。θ = |δ_B - δ_D|/a = |PL/(2EA) - PL/(4EA)|/a = PL/(4aEA)。答案B正確！