材力课件+习题答案 习题解答1练习册P61-P64.ppt

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1、P6134-1(1)FF2FFFF11FF2FFN111F2FFN22FFN3FFN11122223333FFFN2FFFFN3＋＋＋4F3F2FFN图F－＋FFN图Fy=0FN1=4FFy=0FN2=3FFy=0FN3=2FFy=0FN1=FFy=0FN2=0Fy=0FN3=－FP6134-1(4)六要素(阴影线方向)RFN3截面法分段相等正向假设P6134-2ABCDF4F5F2F1F3E＋＋＋＋500N920N640N240NF1=500NF2=420NF3=280NF4=400NF5=240NFN图s=AFNsAB=1.25MPasBC=2.3MPasC

2、D=1.6MPasDE=0.6MPa应力单位：MPaA=400mm2与点有关P6234-3F3F1F2ABCD112233FN1F3Fx=0FN1=F3F3F2FN2F3F1F2FN3=30kNFx=0FN2=F3+F2=40kNFx=0FN3=F3+F2–F1=30kN＋＋＋30kN40kN30kNFN图s1=A1FN1A1=4×10–2m2=0.75MPas2=A2FN2A2=2×10–2m2=2MPas3=A3FN3A3=4×10–2m2=0.75MPaP6234-4FFsa=scos2a2sta=sin2aa=0ºs=AFNFN=–F=–5kN=–50MPaa

3、=30ºa=45ºa=60ºa=90º–50MPa0–37.5MPa–21.65MPa–25MPa–25MPa–12.5MPa–21.65MPa00ansatassss力的单位：kN正负P6335-1FAFBFDFCABCDaaaFNiEAFNiaDli=Dli=SDli–4kN1kN3kN–0.05mm0.0125mm0.0375mm0e呢?e与Dl区别?Se?讨论：P6335-2l2l2hhABCDEH123FN1FN2F=FN3FN1=–80kNFN2=–40kNFN3=120kNE1A1FN12hDl1==–0.4mmE2A2FN22hDl2==–0.4mmE3A3

4、FN3hDl3==0.2mmdCdHDl1Dl2dC=│Dl1│dH=dC+Dl3=0.4mm=0.6mm(1)(2)dCdHDl1Dl2Dl1=–0.2mmDl2=–0.4mmDl3=0.2mmdH=dC+Dl3=0.267mm=0.467mmdC=│Dl1│+│Dl2│3132位移分析!P6435-3ABCl=4mh=3m5mq=10kN/mSMA=0FBCFAyFAxFBC=kN3100FN=(1)斜杆用钢丝索A=n4pd2s=AFN≤[s]∴取n=67或(2)斜杆用两根等边角钢A=2A1s=AFN≤[s]∴取∟20×3(A1=1.13cm2)结论!n≥66.366A

5、1≥1.044cm2平衡计算用型号表示P6435-4FACB①②45º60ºFN2FN1SFx=0SFy=0FN1=0.897FFN2=0.732Fs1=A1FN1≤[s]1F≤107kNs2=A2FN2≤[s]2F≤123kN则[F]=107kN两杆一般不会同时达到容许应力！