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 +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
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)
Displacement continuity位移连续性 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。 徐汉忠第一版2000/7 弹性力学第六章有限元
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
To obtain123求 123 +2x+3y(1) Substitution of the nodal coordinates into equation(1) yields 结点坐标代入方程(1)得: 2 i y 3 i X十 m 2 3 u=u(xi,yi u=u(xi yi um=u(xm,ym) 徐汉忠第一版2000/7 弹性力学第六章有限元 23
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
To obtain123求 123 +2x+3y(1) Substitution of the nodal coordinates into equation(1) yields 结点坐标代入方程(1)得: Xi vi 吗}=1xy u=u(xi,yi u=u(xi yi um=u(xm,ym) 徐汉忠第一版2000/7 弹性力学第六章有限元 24
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
To obtain123求 123 2 Solving eq.(3), we obtain:解方程(3)得 Xi yi 1/2A u; X y 1 um yi 1 x u x;y;|=2A X The above expression is ensured when the node iim are in an anti-clockwise order. (A- area of triangle ijm单元面积) 微綪点遊钟向编号再瘟黜。正向时,上式成立2
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