同济大学 材料力学 习题解答7(练习册P94-P104)ppt课件

材料力学,习题解答(七),P9450-2,sx=0,=sy,a=30,Mn=-T0,=-2.5kNm,=42.41cm3,=-58.95MPa,s30=-tsin(60),s-60=-tsin(-120),=-51.05MPa,=51.05MPa,=319106,广义虎克定律：,sz=0,而,正应变,（线应变）,单元体,无量纲,P9550-3,sy=0,=tx,sx,a=-30,sz=0,=160MPa,=540106,F=sxA,=50kN,则,而,广义虎克定律：,P9550-4,I25b,A,B,C,D,E,危险截面：,E,A,C左,(D右),危险点：,1,2,3,Iz=5278cm4,h=250mm,t=13mm,查表,Wz=422.72cm3,b=118mm,tw=10mm,E截面：,A截面：,C左截面：,Mmax,=45kNm,s=,=106.5MPa,FS,=208kN,=98.0MPa,t,s,M,=41.8kNm,FSmax,=210kN,=72.35MPa,t=,=88.70MPa,s1=,=129.2MPa,s3=,=-40.5MPa,=210kN=,(B),=169.7MPa,=153.5MPa,s,全面校核,内力图,不满足强度条件,：导学篇附录B-3,P9651-1,A,C,B,Fy,=Fcosj,Fz,=Fsinj,=9.66kN,=2.59kN,=7.24kNm,=1.94kNm,=1104cm4,=5625cm4,(1),s=0,令,=0.476,a=25.5,=1103cm3,=750cm3,stmax=scmax,=9.83MPa,(2),=5.43mm,=2.59mm,=6.02mm,=tana,b=a=25.5,跨中截面：,中性轴,stmax,scmax,中性轴,拉压区,P9651-2,A,C,B,查表：,b=20cm,z0=5.69cm,Iz=4554.55cm4,Iy=1180.04cm4,L20020,stmax在B点：,危险截面：,跨中,Fy=Fz,=17.68kN,My=,=17.68kNm,s=0,令,=3.86,a=75.5,yB=0,=-80.47mm,sB=120.6MPa,sA=-146.2MPa,scmax在A点：,=60.95mm,=141.4mm,中性轴,弯曲中心,形心主惯性平面,P9751-3,A,B,C,D,FN=2F,=20kN,=7.46kNm,=5.6kNm,A=bh,=72cm2,sA=2.78+51.81+77.78,sB=2.78+51.81-77.78,sC=2.78-51.81-77.78,sD=2.78-51.81+77.78,=132.4MPa,=-23.2MPa,=-126.8MPa,=28.8MPa,=144cm3,=72cm3,斜弯曲偏心受压,P9751-4,50,250,FN=F,A=bh,M=Fe,=625kN,=25mm,讨论：,(1),smax0,smin,0,0,整个截面：,lP,=2.32cm,=216,=99.35,属细长杆,=42.3MPa,或,=178kN,Fcr=scrA,=178kN,球铰,失稳范围,P10254-1,(1),(2),失稳方向,失稳范围,A=24cm2,=32cm4,=72cm4,=1.155cm,=1.732cm,my=0.5,mz=1,lzly,lzlP,=132.8,=99.6,:绕z轴失稳,=99.35,属细长杆,=269kN,失稳的可能性相同：,l相同,scr相同,P10254-2,A,B,C,D,a,a,a,a,FN2=F,C,B,(压),(拉),(四杆),(中间杆),Q235钢：,scr=304-1.12l,直线公式,=99.35,=57.1,l,稳定问题,强度问题,AB段先失稳,nw=2.5,F=160kN,P10455-1,B,C,A,SMC=0,=158.7kN,从稳定角度考虑,=1cm,=80,安全系数法：,(压),=99.35,=57.1,l,lP,lS,scr=304-1.12l,直线公式：,Fcr=Ascr,=214.4MPa,=269.4kN,=1.70,nw,=2,所以,不安全,P10455-2,A,B,C,D,SFx=0,FBD=0,SMA=0,SFy=0,FBC=2.25q,FAD=6.75q,(