e.x.1-4,(a),e.x.1-5,e.x.补,M,e.x.2-1,e.x.2-9,解:一、A:,X=0 Q+SAcos45=0, SA=- Q,解: CD:,二、B:,SA=SB=- Q,Y=0,SB+Rcos30=0, R=1.633Q,解: AB:,RA,m=0 m-SCcos45(0.12+0.3)=0, SC= RA= 269.4N,e.x.4-7,mA=0 qa /2+Pa+2aNB 3Qa=0 NB= 21kN,2,Y=0 , NB+NA qa=0 NA=15kN,e.x.4-16,解:整体:,my=0, Gr+2a(T2 T1)R=0, T1=2T2= 10kN,mx=0, 100ZB+ (T1 T2)sin3060-G30=0 ZB= 1.5kN,Z=0, ZB+ ZA+ (T1T2)sin30-G=0 ZA= 6kN,mz=0, -100XB-(T1+T2)cos3060=0 XB=-7.794kN,X=0, XB+ XA+ (T1+T2)cos30=0 XA=-5.196kN,e.x.2-2,解：铰链A+物体:,受力如图,y,x,z,P,A,B,C,D,SC,SB,SA,
