习题课 D P E E C XB 解ⅡBAC:2Mn=0,YC=0.5P AEC ∑ MA=O NDE Yc:4tan22.5。=0 tan 22. 5. nde=2P tan 22.5 2P tan2 0U4×20×103×tan422.5。×2000 2 EA P210×10文(7.5)×π =0.127mm习题课 解Ⅱ BAC： ∑M B = 0, YC = 0.5P AEC：∑ M 1 A = 0, NDE ⋅ − YC ⋅ 4 tan 22.5 = 0 tan 22.5 N DE = 2 P tan 22.5 2  (2P tan U= 22.5 2 EA 2 2)l DE ∂U 4 × 20 ×103 × tan 4 22.5 × 2000 ⇒δA = = = 0.127mm ∂P 210 ×10 ×3 2 (7.5) × π