弹性力学第六章有限单元法.ppt

徐汉忠第一版2000/7,弹性力学第六章有限元,1,Chapter 6 Finite Element Method for Plane Stress and Plane Strain Problems,第六章 有限单元法解平面问题,徐汉忠第一版2000/7,弹性力学第六章有限元,2,References 参考书,徐芝纶，弹性力学简明教程第六章。高等教育出版社。华东水利学院，弹性力学问题的有限单元法，水利电力出版社。卓家寿，弹性力学中的有限元法，高等教育出版社。O.C.Zienkiewicz,The Finite Element Method,Third Edition,51.818,Z66K.C.Rockey and so on,The Finite Element Method,Second Edition,51.818,R682-2,徐汉忠第一版2000/7,弹性力学第六章有限元,3,Introduction-1 导引-1,The finite element method is an extension of the analysis techniques(matrix method)of ordinary framed structures.有限元法是刚架结构分析技术的扩充。The finite element method was pioneered in the aircraft industry where there was an urgent need for accurate analysis of complex airframes.有限元法首先应用于飞机工业。,徐汉忠第一版2000/7,弹性力学第六章有限元,4,Introduction-2,The availability of automatic digital computers from 1950 onwards contributed to the rapid development of matrix methods during this period.从 1950以后 数字计算机的出现使矩阵位移法迅速发展。,徐汉忠第一版2000/7,弹性力学第六章有限元,5,Introduction-3,The finite element method was developed rapidly from 1960 onwards and known in China from 1970 onwards.从 1960以后 有限元法迅速发展。1970以后 传入我国。,徐汉忠第一版2000/7,弹性力学第六章有限元,6,Introduction-4,In a continuum structure,a corresponding natural subdivision does not exist so that the continuum has to be artificially divided into a number of elements before the matrix method of analysis can be applied.连续结构不存在自然的单元，须人为划分为单元,徐汉忠第一版2000/7,弹性力学第六章有限元,7,Introduction-5,The artificial elements,which are termed finite elements or discrete elements,are usually chosen to be either rectangular or triangular in shape.单元通常取为三角形或矩形。,徐汉忠第一版2000/7,弹性力学第六章有限元,8,6.1 Fundamental quantities and fundamental equations expressed by matrix6.1 基本量和基本方程的矩程表示,Body force 体力：p=X YT Surface force 面力：p=X YTDisplacement 位移：f=u vTStrain 应变：=x y rxy T Stress 应力：=x y xy TGeometrical equations Physical equationsvirtual work equations,徐汉忠第一版2000/7,弹性力学第六章有限元,9,Geometrical Equation 几何方程,x u/x/x 0 u=y=v/y=0/y v=Lf rxy u/y+v/x/y/x x/x 0=y L=0/y f=u vT rxy/y/x=Lf,徐汉忠第一版2000/7,弹性力学第六章有限元,10,Physical Equation for Plane Stress Problem 平面应力问题的物理方程,x x+y 1 0 x y=E/(1-2)y+x=E/(1-2)1 0 y xy rxy(1-)/2 0 0(1-)/2 rxy x 1 0 x=y D=E/(1-2)1 0=y xy 0 0(1-)/2 rxy=D,徐汉忠第一版2000/7,弹性力学第六章有限元,11,Virtual Work Equation 虚功方程,状态1：p=X YT p=X YT=x y xy T状态2：f*=u*v*T*=L f*虚功方程：f*Tpdx dy t+f*Tpds t=*Tdx dy t注：f*Tp=u*v*X=X u*+Y v*Y*T=x*y*rxy*x=x x*+y y*+xy rxy*y xy,徐汉忠第一版2000/7,弹性力学第六章有限元,12,6.2 Basic Concepts about Finite Element Method6.2 有限单元法的概念,有限单元法的计算模型1.The continuum structure is idealized as a structure consisting of a number of individual elements connected only at nodal points.连续的结构理想化为仅由在结点相连的单元组成。,徐汉忠第一版2000/7,弹性力学第六章有限元,13,2.Displacement boundary:place a bar support at the node where displacement is zero.位移边界：结点位移为零处，设置连杆.3.The system of external loads acting on the actual structure has to be replaced by an equivalent system of forces concentrated at the element nodes.This can be done by using the principle of virtual work and equating the work done by the actual loads to the work done by the equivalent nodal loads.外力按静力等效的原则移置到结点上,徐汉忠第一版2000/7,弹性力学第六章有限元,14,徐汉忠第一版2000/7,弹性力学第六章有限元,15,徐汉忠第一版2000/7,弹性力学第六章有限元,16,补充：关于离散 About Discretization,In reality Elements are connected together along their common boundaries.Here it is assumed that these elements are only interconnected at their nodes.实际：单元间相连-假定：只结点相连,徐汉忠第一版2000/7,弹性力学第六章有限元,17,关于离散-2,However,in the finite element method,the individual elements are constrained to deform in specific patterns.然而，单元变形按指定模式.,徐汉忠第一版2000/7,弹性力学第六章有限元,18,关于离散-3,Hence,although continuity is only specified at the nodal points,the choice of a suitable pattern of deflection for the elements can lead to the satisfaction of some,if not all,of the continuity requirements along the sides of adjacent elements.位移模式使相连单元位移连续得某些满足,徐汉忠第一版2000/7,弹性力学第六章有限元,19,关于离散-4,Hence,as stated by Clough,finite elements are not merely pieces cut from the original structure,but are special types of elastic elements constrained to deform in specific patterns such that the overall continuity of the assemblage tends to be maintained,徐汉忠第一版2000/7,弹性力学第六章有限元,20,6.3 Displacement pattern and convergence criteria6.3 位移模式和收敛性,Fig.1 shows the typical triangular element with nodes ijm numbered in an anti-clockwise order.Y m图1为一典型的三角形单元，i 结点 ijm 逆钟向编号-x正向到 jy正向。Fig.1 x,Element with nodes numbered 单元的结点编号,徐汉忠第一版2000/7,弹性力学第六章有限元,21,Displacement pattern 位移模式,The displacement representation is given by the two linear polynomials with six constants 位移用有6个常数的线性多项式表示 u=1+2x+3y(1)v=4+5x+6y(2),徐汉忠第一版2000/7,弹性力学第六章有限元,22,Since these displacements are both linear in x and y,displacement continuity is ensured along the interface between adjoining elements for any identical nodal displacement.因为位移在单元上均为线性，相邻单元交界面上的位移连续性因同一结点位移相同而得到保证。,Displacement continuity 位移连续性,徐汉忠第一版2000/7,弹性力学第六章有限元,23,u=1+2x+3y(1)Substitution of the nodal coordinates into equation(1)yields:结点坐标代入方程（1）得：ui=1+2 xi+3 yi uj=1+2 xj+3 yj(3)um=1+2 xm+3 ym ui=u(xi,yi)uj=u(xj,yj)um=u(xm,ym),To obtain 1 2 3 求 1 2 3-1,徐汉忠第一版2000/7,弹性力学第六章有限元,24,u=1+2x+3y(1)Substitution of the nodal coordinates into equation(1)yields:结点坐标代入方程（1）得：ui 1 xi yi 1 uj=1 xj yj 2(3)um 1 xm ym 3 ui=u(xi,yi)uj=u(xj,yj)um=u(xm,ym),To obtain 1 2 3 求 1 2 3-1,徐汉忠第一版2000/7,弹性力学第六章有限元,25,Solving eq.(3),we obtain:解方程（3）得 1 ui xi yi 1 ui yi 1 xi ui T 2=1/2A uj xj yj 1 uj yj 1 xj u j 3 um xm ym 1 um ym 1 xm um 1 xi yi 1 xj yj=2A(4)1 xm ym The above expression is ensured when the node ijm are in an anti-clockwise order.(A-area of triangle ijm 单元面积)当结点逆钟向编号 x正向到 y正向 时，上式成立,To obtain 1 2 3 求 1 2 3-2,徐汉忠第一版2000/7,弹性力学第六章有限元,26,Substitution of 1 2 3 into eq.(1)yields:将 1 2 3代入方程（1）得：,ui xi yi 1 ui yi 1 xi ui u=1/2A uj xj yj+1 uj yj x+1 xj u j y um xm ym 1 um ym 1 xm um 1 x y 1 x y 1 x y=1/2A 1 xj yj ui+1 xm ym uj+1 xi y i um 1 xm ym 1 xi yi 1 xj yj u=Ni(x,y)ui+Nj(x,y)uj+Nm(x,y)um v=Ni(x,y)vi+Nj(x,y)vj+Nm(x,y)vm,徐汉忠第一版2000/7,弹性力学第六章有限元,27,In which：1 x y 1 xi yi 其中：Ni(x,y)=1 xj yj 1 xj yj 1 xm ym 1 xm ym=(ai+bix+ciy)/(2A)(i,j,m)xj yj 1 yj ai=xm ym=xjym-xmyj bi=-1 ym=yj-ym 1 xj ci=1 xm=xm-xj(i,j,m)1 xi yi 2A=1 xj yj 1 xm ym,徐汉忠第一版2000/7,弹性力学第六章有限元,28,Ni is called element displacement function or element shape function.Ni 叫做单元位移函数或单元形函数。Ni(xi,yi)=1 Ni(xj,yj)=0 Ni(xm,ym)=0(i,j,m)1 x y 1 xi yi Ni(x,y)=1 xj yj 1 xj yj 1 xm ym 1 xm ym,徐汉忠第一版2000/7,弹性力学第六章有限元,29,u=Ni(x,y)ui+Nj(x,y)uj+Nm(x,y)um v=Ni(x,y)vi+Nj(x,y)vj+Nm(x,y)vmf=N e f=u vT nodal displacement matrix:结点位移列阵：e=ui vi uj vj um vmTshape function matrix:形函数矩阵：Ni 0 Nj 0 Nm 0 N=0 Ni 0 Nj 0 Nm 有限个自由度问题,徐汉忠第一版2000/7,弹性力学第六章有限元,30,Convergence Criteria 收敛准则-1,Criterion 1：The displacement function chosen should be such that it does not permit straining of an element to occur when the nodal displacements are caused by a rigid body displacement.准则1：位移模式必须反映单元的刚体位移。,徐汉忠第一版2000/7,弹性力学第六章有限元,31,Convergence Criteria 收敛准则-2,Criterion 2：The displacement function has to be taken so that the constant strain(constant first derivative)could be observed.准则2：位移模式必须反映单元的常量应变。,徐汉忠第一版2000/7,弹性力学第六章有限元,32,Convergence Criteria 收敛准则-3,Criterion 3：The displacement function should be so chosen that the strains at the interface between elements are finite(even though indeterminate and not equal).准则3：位移模式必须使单元公共边上的应变在不同单元中为常量。,徐汉忠第一版2000/7,弹性力学第六章有限元,33,Convergence Criteria 收敛准则-3,准则3：位移模式必须使位移处处连续.(1)单元内位移连续.(2)单元公共边上的位移连续。,徐汉忠第一版2000/7,弹性力学第六章有限元,34,Further discussion about criteria-1 准则的进一步讨论-1,Criterion 3 implies a certain continuity of displacements between elements-In the case of strains being defined by first derivative,the displacements only have to be continuous between elements.That is C0 continuity is sufficient,徐汉忠第一版2000/7,弹性力学第六章有限元,35,Further discussion about criteria-2 准则的进一步讨论-2,Criterion 3 implies a certain continuity of displacements between elements.-In the plate and shell problems,the strains are defined by second derivatives of deflections,first derivatives of deflections have to be continuous between elements.,徐汉忠第一版2000/7,弹性力学第六章有限元,36,Further discussion about criteria-1,2 准则的进一步讨论-1,2,准则3 意味着对单元间位移的连续性有一定要求。-应变是位移的一阶导数的情况，例如弹性力学平面问题，仅要求单元之间位移连续。称为C0连续性。-板和壳问题，应变是位移的二阶导数，要求位移的一阶导数在单元间连续,C1 连续性。,徐汉忠第一版2000/7,弹性力学第六章有限元,37,Further discussion about criteria-3 准则的进一步讨论-3,Criterion 1 and 2 are necessary conditions.Criterion 3 is the sufficient condition.准则1和2是收敛的必要条件，准则3是充分条件。,徐汉忠第一版2000/7,弹性力学第六章有限元,38,准则1和2是收敛的必要条件，不满足一定不收敛.在满足准则1和2的必要条件的前提下,再满足准则3,一定收敛。-协调元在满足准则1和2的必要条件的前提下,不满足准则3-可能收敛(非协调元,例薄板弯曲问题),可能不收敛.,徐汉忠第一版2000/7,弹性力学第六章有限元,39,Further discussion about criteria-4 准则的进一步讨论-4,u=1+2x+3y=1+2x-y(5-3)/2+y(5+3)/2 v=4+5x+6y=4+6y+x(5-3)/2+x(5+3)/2刚体位移 u=-y+u0 v=x+v0u0=1 v0=4=(5-3)/2 反映刚体位移x=2 y=6 rxy=5+3 反映常量应变,徐汉忠第一版2000/7,弹性力学第六章有限元,40,Further discussion about criteria-4 准则的进一步讨论-4,u=1+2x+3y=1+2x-y(5-3)/2+y(5+3)/2 v=4+5x+6y=4+6y+x(5-3)/2+x(5+3)/2位移在单元内部连续,在单元公共边上连续,满足准则3,徐汉忠第一版2000/7,弹性力学第六章有限元,41,6.4 Strain,Stress and Stiffness 应变，应力和劲度,=Lf=LNe=B e=B e/x 0 Ni 0 Nj 0 Nm 0 B=LN=0/y 0 Ni 0 Nj 0 Nm/y/x=Bi Bj Bm,A.Strain 应变,徐汉忠第一版2000/7,弹性力学第六章有限元,42,/x 0 Ni 0 Nj 0 Nm 0 B=LN=0/y 0 Ni 0 Nj 0 Nm/y/x=Bi Bj Bm Ni/x 0 bi 0Bi=0 Ni/y=0 ci/(2A)Ni/y Ni/x ci biThe B matrix is independent of the position within the element,and hence the strains are constant throughout it.应变在单元中为常量。,徐汉忠第一版2000/7,弹性力学第六章有限元,43,B.Stress 应力,=D=DB e=Se S=DB=DBi DBj DBm=Si Sj Sm 1 0 bi 0 Si=DBi=E/(1-2)1 0 1/2A 0 ci 0 0(1-)/2 ci bi bi ci=E/2A(1-2)bi ci(1-)ci/2(1-)bi/2 S-Elasticity matrix;弹性矩阵，应力转换矩阵,徐汉忠第一版2000/7,弹性力学第六章有限元,44,C.Stiffness 劲度,C.1 Element Nodal Force Matrix 单元结点力列阵 Fe=Ui Vi Uj Vj Um VmTElement nodal forces are the internal forces between elements and nodes.It is considered positive or negative according as it acts in the positive or negative direction of the coordinate axis when it acts on the element.单元结点力：单元和结点间相互作用力，作用在单元上时，沿坐标正向为正.作用在结点上时，沿坐标负向为正.,徐汉忠第一版2000/7,弹性力学第六章有限元,45,Vi2 Y1 Ui2 i m U1=Uj1+Ui2 1 X1 Uj1 Vj1 j y j Um m i Ui Vm Vi x,V1=Vj1+Vi2,单元结点力列阵Fe=Ui Vi Uj Vj Um VmT整体结点力列阵F=U1 V1 U2 V2 U3 V3 U4 V4 T 作用在结点上时，沿坐标负向为正.,徐汉忠第一版2000/7,弹性力学第六章有限元,46,C.Stiffness 劲度,C.2 The relation between the element nodal force and nodal displacementsIsolate an element from the structure.Since body forces and surface forces are moved to the nodes,only element nodal forces are the external forces acting on the element.Impose an arbitrary virtual nodal displacement.The work done by the nodal forces is equal to the work done by the stresses.,徐汉忠第一版2000/7,弹性力学第六章有限元,47,C.Stiffness 劲度,C.2 单元结点力和单元结点位移列阵的关系将单元取出作为隔离体，因为体力面力已移置到结点上，单元上结点力为外力，应力为内力。施加一个虚位移，结点力作功等于应力作功可导得 单元结点力和单元结点位移的关系,徐汉忠第一版2000/7,弹性力学第六章有限元,48,C.2-continue,We=(*e)T Fe=ui vi uj vj um vm Ui Vi Uj Vj Um Vm=Uiui+Vivi+Ujuj+Vjvj+Umum+Vmvm,徐汉忠第一版2000/7,弹性力学第六章有限元,49,C.2-continue,WI=*Tdx dy t=(*e)T BTDBdx dy t e 注：*T=x*y*rxy*x=xx*+yy*+xyrxy*y xy*=B*e*T=(*e)T BT=DBe,徐汉忠第一版2000/7,弹性力学第六章有限元,50,C.2-continue,We=(*e)T Fe WI=*Tdx dy t=(*e)T BTDBdx dy t e We=WI(*e)T Fe=(*e)T BTDBdx dy t e Since*e is arbitrary,we have Fe=BTDBdx dy t e=k e k=BTDBdx dy t=BTDBAt k-element stiffness matrix.单元劲度矩阵,徐汉忠第一版2000/7,弹性力学第六章有限元,51,C.3 The Explicit Form for Element Stiffness Matrix C.3 单元劲度矩阵的表达式,k=BTDBdx dy t=BTDBAt BiT k=BjT D Bi Bj BmAt BmT BiT D Bi BiT D Bj BiT D Bmk=At BjT D Bi BjT D Bj BjT D Bm BmT DBi BmT DBj BmT D Bm,徐汉忠第一版2000/7,弹性力学第六章有限元,52,kii kij kimk=kji kjj kjm krs=BrTDBs(r,s=i,j,m)kmi kmj kmm brbs+(1-)crcs/2 brcs+(1-)crbs/2 krs=Et/4(1-2)A crbs+(1-)brcs/2 crcs+(1-)brbs/2(r,s=i,j,m)(plane stress problem 平面应力问题)krsxx krsxykrs=krsyx krsyy,徐汉忠第一版2000/7,弹性力学第六章有限元,53,C.4 The Physical Explanation for Element Stiffness matrix C.4 单元劲度矩阵的物理意义,Fe=k e Ui kiixx kiixy kijxx kijxy kimxx kimxy ui Vi kiiyx kiiyy kijyx kijyy kimyx kimyy vi Uj=kjixx kjixy kjjxx kjjxy kjmxx kjmxy uj Vj kjiyx kjiyy kjjyx kjjyy kjmyx kjmyy vj Um kmixx kmixy kmjxx kmjxy kmmxx kmmxy um Vm kmiyx kmiyy kmjyx kmjyy kmmyx kmmyy vm kijyx-j结点x方向发生单位位移在i结点y方向的结点力,徐汉忠第一版2000/7,弹性力学第六章有限元,54,kijyx-j结点x方向发生单位位移在i结点y方向的结点力方向 kijyx局部结点号 结果 原因,徐汉忠第一版2000/7,弹性力学第六章有限元,55,C.5 The Characteristics of the Element Stiffness Matrix C.5 单元劲度矩阵的特点,1.对称矩阵2.每一行元素之和为零.Assume e=ui vi uj vj um vmT=1 1 1 1 1 1T Fe=ke=03.每一列元素之和为零.4.k为奇异矩阵。|k|的各行元素乘1后加到第一行，行列式值不变，由于第一行元素全为零，故|k|=05.k的元素的数值取决于单元形状，大小，方位和弹性常数，不随单元的平行移动或作n的转动而改变。n为正整数。,徐汉忠第一版2000/7,弹性力学第六章有限元,56,转动,k不变,m j i i j m,徐汉忠第一版2000/7,弹性力学第六章有限元,57,kii kij kimk=kji kjj kjm krs=BrTDBs(r,s=i,j,m)kmi kmj kmm brbs+(1-)crcs/2 brcs+(1-)crbs/2 krs=Et/4(1-2)A crbs+(1-)brcs/2 crcs+(1-)brbs/2(r,s=i,j,m)(plane stress problem 平面应力问题)krsxx krsxykrs=krsyx krsyy,徐汉忠第一版2000/7,弹性力学第六章有限元,58,In which：1 x y 1 xi yi 其中：Ni(x,y)=1 xj yj 1 xj yj 1 xm ym 1 xm ym=(ai+bix+ciy)/(2A)(i,j,m)xj yj 1 yj ai=xm ym=xjym-xmyj bi=-1 ym=yj-ym 1 xj ci=1 xm=xm-xj(i,j,m)1 xi yi 2A=1 xj yj 1 xm ym,徐汉忠第一版2000/7,弹性力学第六章有限元,59,6.5 Element Load Matrix 单元荷载列阵,1.Element load matrix Re=Xi Yi Xj Yj Xm YmT It is positive when it acts in the positive direction of the corresponding axis.作用在坐标正向为正。,徐汉忠第一版2000/7,弹性力学第六章有限元,60,Yi2 Y1 Xi2 i m 1 X1 j Xj1 Yj1 y j Xm m i Xi Ym Yi x,V1=Vj1+Vi2,单元结点列阵Re=Xi Yi Xj Yj Xm YmT作用在结点上整体结点荷载列阵R=X1 Y1 X2 Y2 X3 Y3 X4 Y4 TX1=Xj1+Xi2+R1 Y1=Yj1+Yi2 作用在结点上时，沿坐标正向为正.,徐汉忠第一版2000/7,弹性力学第六章有限元,61,Re=Xi Yi Xj Yj Xm YmT和P=Px PyT p=X YT p=X YT为静力等效，在虚位移f*=N*e上作功相等,2 we1=(*e)TRe we2=f*TP+f*Tpdx dy t+f*Tpds t=(*e)T(NTP+NTpdx dy t+NTpds)we1=we2(*e)TRe=(*e)T(NTP+NTpdx dyt+NTpds)Re=NTP+NTpdx dy t+NTpdsnote:f*=N*e f*T=(*e)T NT,徐汉忠第一版2000/7,弹性力学第六章有限元,62,Re=NTP+NTpdx dy t+NTpds,Xi NiPx NiXdxdy NiXds Yi NiPy NiYdxdy NiYds Xj NjPx NjXdxdy NjXds Yj=NjPy+t NjYdxdy+t NjYds Xm NmPx NmXdxdy NmXds Ym NmPy NmYdxdy NmYds,徐汉忠第一版2000/7,弹性力学第六章有限元,63,Body forces:X=0,Y=-g Re=NTpdx dy t,Xi NiXdxdy 0 Yi NiYdxdy-W/3 Xj NjXdxdy 0 Yj=t NjYdxdy=-W/3 Xm NmXdxdy 0 Ym NmYdxdy-W/3note:Nidxdy=A/3(i,j,m)W=gAt,徐汉忠第一版2000/7,弹性力学第六章有限元,64,ij边上面力 p=X=q Y=0T Re=NTpds t,Xi ijNiXds 1 Yi ij NiYds 0 Xj ij NjXds 1 Yj=t ij NjYds=0 0.5tqLij Xm ij NmXds 0 Ym ij NmYds 0note:ij Nids=Lij/2 ij Njds=Lij/2 ij Nmds=0,徐汉忠第一版2000/7,弹性力学第六章有限元,65,6.6 global analysis 结构整体分析,例：,徐汉忠第一版2000/7,弹性力学第六章有限元,66,example 1-A.结点编号和 位移列阵,结点的局部号:i j m结点的整体号:1 2 3 4element nodal displacement matrix:单元结点位移列阵：e=ui vi uj vj um vmTglobal nodal displacement matrix:整体结点位移列阵：=u1 v1 u2 v2 u3 v3 u4 v4T,徐汉忠第一版2000/7,弹性力学第六章有限元,67,example 1-B.荷载列阵,Element load matrix 单元荷载列阵 Re=Xi Yi Xj Yj Xm YmT R1=0 0 0.5q2 0 0.5q2 0T R2=0-0.5q1 0 0 0-0.5q1Tglobal load matrix 整体荷载列阵 R=X1 Y1 X2 Y2 X3 Y3 X4 Y4T R=R1+0.5q2-0.5q1 0.5q2 R2 R3x R3y-0.5q1 R4x R4yT,徐汉忠第一版2000/7,弹性力学第六章有限元,68,Element load matrix 单元荷载列阵 Re=Xi Yi Xj Yj Xm YmT R1=0 0 0.5q2 0 0.5q2 0T R2=0-0.5q1 0 0 0-0.5q1Tglobal load matrix 整体荷载列阵 X1=Xj)1+Xi)2+R1=R1+0.5q2 Y1=Yj)1+Yi)2=-0.5q1 X2=Xm)1=+0.5q2 Y2=Ym)1+R2=R2,徐汉忠第一版2000/7,弹性力学第六章有限元,69,链杆反力不放入单元结点荷载列阵中,放入整体结点荷载列阵中,徐汉忠第一版2000/7,弹性力学第六章有限元,70,C.单元劲度 brbs+(1-)crcs/2 brcs+(1-)crbs/2 krs=Et/4(1-2)A crbs+(1-)brcs/2 crcs+(1-)brbs/2(r,s=i,j,m)(plane stress problem 平面应力问题),y j m i x,bi=yj-ym=1 bj=ym-yi=0 bm=yi-yj=-1ci=-xj+xm=0 cj=-xm+xi=1 cm=-xi+xj=-1Et/4(1-2)A=18Et/35kii=18Et/35 1 0 0 5/12 kij=18Et/35 0 1/6 5/12 0 取=1/6,徐汉忠第一版20