同济大学 材料力学 习题解答6(练习册P86-P94).ppt

材料力学,习题解答(六),P8646-1,C,A,B,=FC,FB=,m=0,弯矩方程,挠曲线近似微分方程,积分,边界条件,x=0,y=0：,x=l,y=0：,挠曲线方程,转角方程,A处挠度,B处转角,x=l：,讨论：,q=0：,令,0.0625,0.06415,误差2.6,积分法,坐标,P8646-2(b),A,C,B,弯矩方程,AC,BC,M1(x)=-M,M2(x)=0,挠曲线近似微分方程,AC,BC,EIy2=C2x+D2,边界条件,x=0,q1=0：,C1=0,x=0,y1=0：,D1=0,连续条件,=qC,=yC,x=a,q2=q1,x=a,y2=y1,C2=Ma,积分,P8746-4,A,C,D,B,FA=qa,FB=qa,mA=0,Fy=0,弯矩方程,AB,BC,挠曲线近似微分方程,边界条件,x=0,y1=0：,D1=0,x=2a,y1=0：,连续条件,=qB,x=2a,q2=q1,x=2a,y2=yB,=0：,=0：,qB=0,P8847-1(b),B,C,A,P224表19-1序号1,P224表19-1序号3,yCF=yBF+qBFa,qCF=qBF,P224表19-1序号5,yCq=yBq+qBqa,qCq=qBq,yC=yCM+yCF+yCq,qC=qCM+qCF+qCq,P8847-2(a),A,C,B,P225表19-1序号8,yCM=0,P225表19-1序号9,yC=yCM+yCF,qA=qAM+qAF,=qBM,=-qBF,qB=qBM+qBF,P8947-3(c),A,B,C,P225表19-1序号11,yCq=qBqa,qCq=qBq,P226表19-1序号12,yC=yCq+yCF,qC=qCq+qCF,P9048-1,A,B,M图,Mmax,=45kNm,s,=16.5cm,查表：,=17.8cm,取,b=8.9cm,其中：,h=17.8cm,变形协调条件：,变形几何方程：,P9148-3,l,l,FC,静定基,查表：,补充方程：,多余约束力：,yC=0,平衡：,yCq+yCF=0,（简便方法）,P9249-1,Mn=2T0,Mn=T0,FN=F,危险截面,(内力分布),危险点,(应力分布),横截面位置,应力方向,P9249-2(a),50MPa,20MPa,30MPa,sx=30MPa,sy=50MPa,tx=-20MPa,a=30,=52.32MPa,=-18.66MPa,应力表示,(方向数值),P9249-2(b),20MPa,40MPa,sx=-40MPa,sy=0,tx=20MPa,a=60,=-27.32MPa,=-27.32MPa,P9349-3(a),20MPa,50MPa,sx=50MPa,sy=0,tx=-20MPa,a=30,=54.82MPa,=11.65MPa,a0=19.33,a0=109.33,=0.8,s1=57.02MPa,s2=0,s3=-7.02MPa,=1.25,a1=64.33,a1=-25.67,=25MPa,解析法,P9349-3(e),30MPa,20MPa,20MPa,sx=-20MPa,sy=30MPa,tx=20MPa,a=-30,=9.82MPa,=31.65MPa,a0=19.33,a0=109.33,=0.8,s1=37.02MPa,s2=0,s3=-27.02MPa,a1=a0+45,a1=-25.61,=5MPa,=64.33,32.02MPa,P9349-4,M,=10kNm,=100cm3,=100MPa,FS=120kN,=500cm4,=30MPa,=22.5MPa,=sd,=-sa,=2sc,s1=s2=0,s3=-100MPa,s2=0,s1=30MPa,=-s3,s1=58.6MPa,s2=0,s3=-8.6MPa,s2=s3=0,s1=100MPa,=tb,P9450-1(b),25MPa,sx=0,=sy,tx=25MPa,a=45,sC=0,R=25MPa,ta=0,sa=-25MPa,s1=25MPa,s3=-25MPa,s2=0,tmax=25MPa,a0=45,a1=0,sa1=sC=0,图解法,几何关系,对应关系,P9450-1(d),80MPa,20MPa,sx=0,sy=-80MPa,tx=20MPa,a=60,R=44.72MPa,sa=-77.32MPa,=4.72MPa,=-84.72M