同济大学 材料力学 习题解答4(练习册P75-P82).ppt

材料力学,习题解答(四),P7842-2-1,M图(M0),M图,M图(F),FS图(M0),FS图,FS图(F),叠加法,(过程),(单个荷载),P7842-2-2,M图(M0),M图,M图(q),FS图(M0),FS图,FS图(q),(抛物线,(抛物线-中点或极值点),P7541-1-1,qa,形状：,FS图,M图,斜直线,斜直线,FSA=0,FSB=qa,FSC=0,FSB=qa,MA=0,MD=qa2,MB左=qa2,MB右=-qa2,ME=-qa2,MC=0,D,E,形状：,抛物线,抛物线,简便方法,微分关系,(突变),(特殊点),(形状.方向),a=l,斜直线,P7541-1-2,qa,qa2,D,FS图,形状：,水平线,斜直线,FSB左=-qa,FB=qa,FC=qa,mB=0,Fy=0,FSB右=qa,FSC=-qa,M图,形状：,抛物线,MA=0,MD=qa2,MB=-qa2,MC=0,MB=-qa2,(突变),(特殊点),抛物线,E,斜直线,斜直线,P7541-1-3,FD=qa,FA=qa,mD=0,Fy=0,FS图,形状：,水平线,水平线,斜直线,FSB左=qa,FSB右=-qa,FSC=-qa,FSD=-qa,MA=0,MC=qa2,MB=qa2,MD=0,MB=qa2,M图,形状：,MC=qa2,ME=qa2,(突变),(按比例),P8043-1,C,D,FD=20kN,FC=10kN,mD=0,Fy=0,M1,=FC1.5,=15kNm,=8103cm4,yA=-5cm,yB=10cm,=-9.375MPa,=18.75MPa,-,反力,指定截面,形心主轴,P8043-2,E,G,A,B,C,D,I32a,FE=10kN,FG=50kN,mE=0,Fy=0,M1,=10kNm,Iz=11080cm4,h=32cm,t=1.5cm,查表,yA=-16cm,=-yD,yB=-14.5cm,=-yC,=-14.44MPa,sD=14.44MPa,=-13.09MPa,sC=13.09MPa,指定截面,：导学篇,附录B-3P380中,I32a,P8143-3,A,B,I20a,C,D,M图,mA=0,Fy=0,Mmax,Wz=236.9cm3,查表,s,F56.9kN,则,：导学篇,附录B-3P380中,I20a,F=57kN,P8143-4,A,B,M图,Mmax,=80kNm,s,=41.6cm,取,=28cm,判定,h=42cm,P8244-1,C,D,A,B,FD=20kN,FC=10kN,mD=0,Fy=0,FS1=10kN,=8103cm4,b=12cm,SzA=450cm3,SzB=0,=0.469MPa,tB=0,*,*,*,剪力方向,t分布,指定截面,MPa,P8244-2,21.15MPa,14.39MPa,=85mm,负面积法,C,=3752cm4,组合法,Szmax,*,=264.5cm3,tw=4cm,=21.15MPa,*,分布,(腹板)：,*,*,抛物线分布,20mm,115mm,D,=14.39MPa,tD,t分布规律,(tmax)位置,P8445-1,A,B,D,C,M图,MA,=20kNm,20kNm,20kNm,MB,=-20kNm,Mmax,=20kNm,s,=210-3m3,(1),(2),(3),28.8cm,h,取h=288mm,b=144mm,22.9cm,a,取a=229mm,27.3cm,d,取d=273mm,三种截面形状的材料用量比:,A1:A2:A3,=1:1.26:1.41,危险截面：,P8445-2,A,B,C,D,M图,12kNm,8kNm,sCmax,sCmax,stmax,stmax,B截面,C截面,FB=36kN,FA=12kN,B截面,C截面,MC,=12kNm,MB,=-8kNm,几何性质：,=147mm,yC,Iz,=2901cm4,正