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连续介质力学作业-11. 给定一组协变基矢量g1=0 1 1T,g2=2 0 0T,g3=1 1 0T(1) 求逆变基g1,g2,g3(2) 求gij(3) 在上述协变基下，若a的逆变分量为p q rT，求a的协变分量解：(1)g=g1 g2 g3=2g1= g2g3g =0,0,1T,g2= g3g1g =12,-12,12T,g3= g1g2g =0,1,-1T(2)gij=gigj=112-11234-1-1-12(3)a=aigi=aigijgj=ajgjaj=aigijgij=gij-1=201042122a1a2a3=201042122pqr=2p+r4q+2rp+2q+2r (#)2. 已知笛卡尔坐标系e1,e2,e3，一个新的坐标系定义为：e1e2e3=01326-1213-161213-16e1e2e3向量x=x1e1+x2e2+x3e3，给定函数fx=x12-x32(1) 求函数f的梯度grad(f)(2) 求向量x参考新坐标系的表示形式x=xiei(3) 求函数f在新坐标系下的表达形式f(x1,x2,x3)(4) 判断gradf的客观性解：(1)gradf=fx1,fx2,fx3T=2x1,0,-2x3T(2)ei=Tijej, xi=Tijxjx1x2x3=01326-1213-161213-16x1x2x3=13x2+26x3-12x1+13x2-16x312x1+13x2-16x3x=xiei=13x2+26x3e1+-12x1+13x2-16x3e2+12x1+13x2-16x3e3(3) xi=Rijxj,Rij=TijTx1x2x3=0-121213131326-16-16x1x2x3=-12x2+12x313x1+13x2+13x326x1-16x2-16x3fx=x12-x32=-12x2+12x32-26x1-16x2-16x32=-23x12+13x22+13x32+23x1x2+23x1x3-43x2x3 (4)gradfei=2x1,0,-2x3Tgradfei=fx1fx2fx3=x1x1x1x2x1x3x2x1x2x2x2x3x3x1x3x2x3x3fx1fx2fx3=Hgradfei其中H=ij，故gradf具有张量的客观性。(#)3. 二维情况下，一质点应力张量主值1=1.6,2=2.3。主方向N1=32e1-12e2，N2=12e1+32e2。应变张量主值1=1,2=2，主方向与应力张量相同。e1,e2为平面直角坐标系的单位基矢量。(1) 以N1, N2为基，计算该质点处应变能密度W(2) 求ij,使得=ijeiej(3) 求ij,使得=ijeiej(4) 以e1,e2为基，计算该质点处的应变能密度W(5) 计算的球应力张量和偏应力张量，并计算偏应力张量的主值和方向解：(1)=1N1N1+2N2N2, =1N1N1+2N2N2W=12:=1211+22=3.1(2)=1N1N1+2N2N2=132e1-12e232e1-12e2+212e1+32e212e1+32e2=ijeiejij=714073407340178(3)=1N1N1+2N2N2=132e1-12e232e1-12e2+212e1+32e212e1+32e2=ijeiejij=54343474(4)W=12:=12ijij=12714054+7340342+17874=3.1(5)ij=ij0 +sIjij0 =12trijI=3920003920,sIj=ij-ij0 =-74073407340740detsij-sij=detij-12trij+sij=0s=-12trij1s=1.6-3920=-0.35, 2s=2.3-3920=0.35ijuj=iujij-12trijuj=i-12trijujsijuj=isujN1=32e1-12e2,N2=12e1+32e2 (#)4. A,B是二阶张量，证明：A:B=trATB=trABT=trBTA=trBAT证明：将张量按照标准正交基分解有：A=Aijeiej, B=Bklekel,AT=Aijejei,BT=BklelekA:B=Aijeiej:Bklekel=AijBijtrATB=tr(Aijejei)(Bklekel)=tr(AijBilejel)=AijBijtrABT=tr(Aijeiej)(Bklelek)=tr(AijBkjeiek)=AijBijtrBTA=tr(Bklelek)(Aijeiej)=tr(AijBilelej)=AijBijtrBAT=tr(Bklekel)(Aijejei)=tr(AijBkjekei)=AijBijA:B=trATB=trABT=trBTA=trBAT=AijBij (#)5 (1) 如果二阶张量S是反对称张量，对于任意一阶张量x，证明xSx=0(2) S是二阶反对称张量，A是二阶对称张量，证明A:S=0证明：(1)S=Sijgigj=-Sjigigj,x=xigi,xSx=xkgkSijgigjxlgl=Sijxixj-xSx=x-Sx=xkgkSjigigjxlgl=Sjixixj=Sijxjxi=Sijxixj=xSx故对于任意x，均有xSx=0(2)S=Sijgigj=-Sjigigj,A=Aklgkgl=Alkgkgl,A:S=Aklgkgl:Sijgigj=AijSij-A:S=A:-S=Aklgkgl:Sjigigj=Alkgkgl:Sjigigj=AjiSji=AijSij=A:SA:S=0 (#)
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