P P/2 N CD AD 3xa / F N D AC LP/2 Y C B P A CE G K 6× M DE 1求支座反力X4=0=3PYB=3P DE NN 2取结点A DA B ∑F,=0NDx√2/2+3P-P2=0.Nm0==52P/2 ∑F=0,Nmx√2/2+Nc=0,Nc=5P/2 3取结点C N=0.N 其它杆件轴力求 2CE N=5P/2 4取结点D 法类似 求出所有轴力后 ∑ F=0,N DE N+P212=2把轴力标在杆件旁 ∑Fn=0.NE=-2P/2其它杆件轴力求 法类似 . 求出所有轴力后 , 应把轴力标在杆件旁. P / 2 P P P P P P/ 2 A F C E D B G I J H 6a L K 3aYA XA YB 1.求支座反力 XA = 0 YA = 3 P YB = 3 P 2.取结点 A C NCE NCD NCA YA A NAC NAD P/ 2 NDC D NDE NDF NDAP   F 0,NAD 2 / 2 3P P/ 2 0,NAD 5 2P/ 2 y  =  + − = = − F 0, NA D 2 / 2 NA C 0, NA C 5 P / 2 x  =  + = = 3.取结点 C NCD = 0,NCE = NCA = 5 P / 2 4.取结点 D  F = 0, ND F = ND A + P  2 / 2 = − 2 2 P   F = 0,NDE = − 2 P / 2 