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1、第3章部分习题第3章部分习题答案3－6FAB=-65kN,FBC=-50kN,FCD=-30kN,FDd=Fcd=FCc=Fbc=0kN,FBb=-15kN,Fab=30kN,FBa=21.213kN,FCa=22.361kN,FDb=33.541kN3－7(a)FCB=-qa,FBC=-qa,FBA=0,FAB=-1.5qa,222MCB=0,MBC=-qa,MBA=qa,MAB=2.125qa(b)MAB=Fpa,MBA=Fpa,MBC=FpaMBD=-2Fpa23－8FAB=0.433ql,FBA=-0.433ql,跨中弯矩M=0.433ql3－9(a)FA
2、C=-32kN,FAB=41.6kN,FBA=-38.4kN,FBD=12kNMAC=-32kN•m,MAB=-32kN•m,MBA=-24kN•m,MBD=-32kN•m,(b)FAB=9kN,FBA=-9kN,FBC=12kN,AB跨中剪力F=-1kNMBA=122kN•mAB跨中弯矩M=-18kN•m3－10(a)FBA=0.5ql,FBC=0.5ql,FBD=-0.125ql222MBA=ql,MBC=-0.125ql,MBD=1.125ql(b)FAB=-0.5ql,FAC=ql,FAE=ql,FAD=-0.5ql2222MAB=0.5ql,MAC=ql
3、,MAE=-0.5ql,MAD=ql3－11FCD=16kN,FDC=-112kN,FDB=40kNMDC=480kN•m,MDB=480kN•m3－12(a)FN1=-56.568kN,FN2=-40kN,FN3=-22.361kN,FN4=60kN(b)FN1=-55.902kN,FN2=17.678kN,FN3=-17.678kN,FN4=50kN3－13(a)FDA=0.5kN,FDC=2kN,FEC=-2kN,FEB=0.5kNMDA=-2kN•m,MDC=-2kN•m,MEC=-2kN•m,MEB=-2kN•m(b)FDA=-3.409kN,FDC=8
4、.030kN,FEC=-1.030kN,FEB=1.288kNMDA=-13.637kN•m,MDC=-13.637kN•m,MEC=-5.152kN•m,MEB=-5.152kN•m3－14(a)FDA=0.125ql,FDC=0,FEC=-ql,FEB=0.375ql2222MDA=0.125ql,MDC=0.125ql,MEC=-0.375ql,MEB=-0.375ql(b)FDA=-0.333Fp,FDC=0.667Fp,FEC=-0.333Fp,FEB=0.333FpMDA=-0.667Fpa,MDC=-0.667Fpa,MEC=-0.333Fpa,ME
5、B=-0.333Fpa3－17FN1＝4kN,FN2=13.333kN,FN3=-8.944kN,FN4=-48.333kN3－18FNBC=FNAC=-0.335ql,FNFD=FNGE=0.15ql,FNBE=FNAD=0.335ql,FNDE=0.3ql3－19(a)FAB=2kNFBA=-2kN,FBC=-2kN,FCB=-6kN,FCD=4kNMCB=-2kN•m,MCD=-8kN•m,MED=8kN•mAB跨中弯矩M＝4kN•m，BC跨中弯矩M＝－2kN•m(b)FAB=-6.667kN,FBA=-6.667kN,FBC=24kN,FCB=-24kNF
6、CD=27.333kN,FDC=20.667kNMBA=-20kN•m,MBC=-20kN•m,MCB=-20kN•m,MCD=-20kN•mBC跨中弯矩M＝16kN•mCD跨中弯矩M＝26kN•m3－21(a)MDA=-Fpa,MDE=-Fpa,MED=-Fpa,EF=-Fpa,MFE=-Fpa,MCF=Fpa2(b)MGD=Fpa,MGF=0.5Fpa,MHF=-0.5Fpa,MHE=0.5Fpa3－22FNAC=-60kN,FNCB=-45kN,FNBD=64.616kN,FNCD=-47.019kNMDC=20kN•m,,MDE=-20kN•m3－24(a
7、)几何可变(b)几何不变，无多余约束
