建设网站挣钱,网站推广实施计划,如何制作app软件要多少钱,wordpress 注册小工具文章目录 1.2.2 基于最小势能原理的线性有限元一般格式1.2.2.1 离散化1.2.2.2 位移插值1.2.2.3 单元应变1.2.2.4 单元应力1.2.2.5 单元刚度矩阵1.2.2.6 整体刚度矩阵1.2.2.7 处理约束1.2.2.8 求解节点载荷列阵1.2.2.9 求解位移列阵1.2.2.10 计算应力矩阵等 1.2.2 基于最小势能原… 文章目录 1.2.2 基于最小势能原理的线性有限元一般格式1.2.2.1 离散化1.2.2.2 位移插值1.2.2.3 单元应变1.2.2.4 单元应力1.2.2.5 单元刚度矩阵1.2.2.6 整体刚度矩阵1.2.2.7 处理约束1.2.2.8 求解节点载荷列阵1.2.2.9 求解位移列阵1.2.2.10 计算应力矩阵等 1.2.2 基于最小势能原理的线性有限元一般格式
现在市场上的有限元分析软件普遍都是基于位移法进行求解的其数学原理就是上节的最小势能原理变分提法。回顾上节推导过程首先需要根据位移条件确定可能位移的范围其次根据假设的可能位移代入几何方程和本构方程得到基于位移的应力应变的表达式最后代入 δ I I 0 \delta II0 δII0的方程求解相关待定参数。线性有限元求解的一般格式基本上也是一样的除此之外线性有限元会多一些其他步骤比如结构的离散化、位移模式假定等。下图为线性有限元求解的一般格式。 加图
1.2.2.1 离散化
将结构进行离散化建模是有限元分析工作中的第一步某典型的结构如下图所示其中结构为一个圆盘将其离散成若干四边形网格单元如下图所示。结构离散化合适与否一定程度上决定使用有限元分析方法解决工程问题的精度。
1.2.2.2 位移插值
a位移模式 在最小势能原理中首先需要根据位移条件确定可能位移的范围但是在通常的分析中我们为了方便处理往往将可能位移用多项式来逼近这个过程就是单元的位移模式选择。 我们以一阶二维三角形单元为例二维三角形单元有6各位移自由度。用多项式拟合可能位移如下式所示 u ( x , y ) a ‾ 0 a ‾ 1 x a ‾ 2 y v ( x , y ) b ‾ 0 b ‾ 1 x b ‾ 2 y (1-54) \begin{aligned} u(x,y)\overline a_{0}\overline a_{1}x\overline a_{2}y\\ v(x,y)\overline b_{0}\overline b_{1}x\overline b_{2}y \end{aligned}\tag{1-54} u(x,y)a0a1xa2yv(x,y)b0b1xb2y(1-54) b形函数 多项式位移模式有六个待定参数二维三角形单元有六个节点位移变量所以六个待定参数可以表示成六个节点位移变量的形式如下式待定系数满足下式。 [ u 1 u 2 u 3 ] [ 1 x 1 y 1 1 x 2 y 2 1 x 3 y 3 ] [ a ‾ 0 a ‾ 1 a ‾ 2 ] , [ v 1 v 2 v 3 ] [ 1 x 1 y 1 1 x 2 y 2 1 x 3 y 3 ] [ b ‾ 0 b ‾ 1 b ‾ 2 ] (1-55) \begin{aligned} \begin{bmatrix} u_1\\ u_2\\ u_3\\ \end{bmatrix} \begin{bmatrix} 1 x_1 y_1\\ 1 x_2 y_2\\ 1 x_3 y_3\\ \end{bmatrix} \begin{bmatrix} \overline a_0\\ \overline a_1\\ \overline a_2\\ \end{bmatrix}, \begin{bmatrix} v_1\\ v_2\\ v_3\\ \end{bmatrix} \begin{bmatrix} 1 x_1 y_1\\ 1 x_2 y_2\\ 1 x_3 y_3\\ \end{bmatrix} \begin{bmatrix} \overline b_0\\ \overline b_1\\ \overline b_2\\ \end{bmatrix} \end{aligned}\tag{1-55} u1u2u3 111x1x2x3y1y2y3 a0a1a2 , v1v2v3 111x1x2x3y1y2y3 b0b1b2 (1-55) 那么系数为 [ a ‾ 0 a ‾ 1 a ‾ 2 ] [ 1 x 1 y 1 1 x 2 y 2 1 x 3 y 3 ] − 1 [ u 1 u 2 u 3 ] , [ b ‾ 0 b ‾ 1 b ‾ 2 ] [ 1 x 1 y 1 1 x 2 y 2 1 x 3 y 3 ] − 1 [ v 1 v 2 v 3 ] (1-56) \begin{aligned} \begin{bmatrix} \overline a_0\\ \overline a_1\\ \overline a_2\\ \end{bmatrix} \begin{bmatrix} 1 x_1 y_1\\ 1 x_2 y_2\\ 1 x_3 y_3\\ \end{bmatrix}^{-1} \begin{bmatrix} u_1\\u_2\\u_3\\ \end{bmatrix}, \begin{bmatrix} \overline b_0\\ \overline b_1\\ \overline b_2\\ \end{bmatrix} \begin{bmatrix} 1 x_1 y_1\\ 1 x_2 y_2\\ 1 x_3 y_3\\ \end{bmatrix}^{-1} \begin{bmatrix} v_1\\ v_2\\ v_3\\ \end{bmatrix} \end{aligned}\tag{1-56} a0a1a2 111x1x2x3y1y2y3 −1 u1u2u3 , b0b1b2 111x1x2x3y1y2y3 −1 v1v2v3 (1-56) 其中逆矩阵的求解如下 [ 1 x 1 y 1 1 x 2 y 2 1 x 3 y 3 ] − 1 [ M i j ⋅ ( − 1 ) i j ] T A (1-57) \begin{bmatrix} 1 x_1 y_1\\ 1 x_2 y_2\\ 1 x_3 y_3\\ \end{bmatrix}^{-1}\frac{[M_{ij}\cdot (-1)^{ij}]^T}{A}\tag{1-57} 111x1x2x3y1y2y3 −1A[Mij⋅(−1)ij]T(1-57) 其中 M i j M_{ij} Mij为原矩阵的余子式典型如下其乘上 ( − 1 ) i j (-1)^{ij} (−1)ij称为代数余子式矩阵的逆是其代数余子式组成的伴随矩阵除以该矩阵的行列式。 经过求解系数矩阵的逆矩阵如下式 [ 1 x 1 y 1 1 x 2 y 2 1 x 3 y 3 ] − 1 ∣ 1 x 1 y 1 1 x 2 y 2 1 x 3 y 3 ∣ − 1 ⋅ [ ∣ x 2 y 2 x 3 y 3 ∣ − ∣ x 1 y 1 x 3 y 3 ∣ ∣ x 1 y 1 x 2 y 2 ∣ − ∣ 1 y 2 1 y 3 ∣ ∣ 1 y 1 1 y 3 ∣ − ∣ 1 y 1 1 y 2 ∣ ∣ 1 x 2 1 x 3 ∣ − ∣ 1 x 1 1 x 3 ∣ ∣ 1 x 1 1 x 2 ∣ ] 1 A [ a 1 a 2 a 3 b 1 b 2 b 3 c 1 c 2 c 3 ] (1-58) \begin{aligned} \begin{bmatrix} 1 x_1 y_1\\ 1 x_2 y_2\\ 1 x_3 y_3\\ \end{bmatrix}^{-1} \begin{vmatrix} 1 x_1 y_1\\ 1 x_2 y_2\\ 1 x_3 y_3\\ \end{vmatrix}^{-1}\cdot \begin{bmatrix} \begin{vmatrix} x_2 y_2\\ x_3 y_3\\ \end{vmatrix} -\begin{vmatrix} x_1 y_1\\ x_3 y_3\\ \end{vmatrix} \begin{vmatrix} x_1 y_1\\ x_2 y_2\\ \end{vmatrix}\\ \\ -\begin{vmatrix} 1 y_2\\ 1 y_3\\ \end{vmatrix} \begin{vmatrix} 1 y_1\\ 1 y_3\\ \end{vmatrix} -\begin{vmatrix} 1 y_1\\ 1 y_2\\ \end{vmatrix}\\ \\ \begin{vmatrix} 1 x_2\\ 1 x_3\\ \end{vmatrix} -\begin{vmatrix} 1 x_1\\ 1 x_3\\ \end{vmatrix} \begin{vmatrix} 1 x_1\\ 1 x_2\\ \end{vmatrix}\\ \end{bmatrix} \\\frac{1}{A}\begin{bmatrix} a_1 a_2 a_3\\ b_1 b_2 b_3\\ c_1 c_2 c_3\\ \end{bmatrix} \end{aligned}\tag{1-58} 111x1x2x3y1y2y3 −1A1 a1b1c1a2b2c2a3b3c3 111x1x2x3y1y2y3 −1⋅ x2x3y2y3 − 11y2y3 11x2x3 − x1x3y1y3 11y1y3 − 11x1x3 x1x2y1y2 − 11y1y2 11x1x2 (1-58) 那么系数可以表达成 a ‾ 0 1 A ( a 1 u 1 a 2 u 2 a 3 u 3 ) a ‾ 1 1 A ( b 1 u 1 b 2 u 2 b 3 u 3 ) a ‾ 2 1 A ( c 1 u 1 c 2 u 2 c 3 u 3 ) (1-59) \begin{aligned} \overline a_0\frac{1}{A}(a_1u_1a_2u_2a_3u_3)\\ \overline a_1\frac{1}{A}(b_1u_1b_2u_2b_3u_3)\\ \overline a_2\frac{1}{A}(c_1u_1c_2u_2c_3u_3) \end{aligned}\tag{1-59} a0A1(a1u1a2u2a3u3)a1A1(b1u1b2u2b3u3)a2A1(c1u1c2u2c3u3)(1-59) 上式代入位移模式可得 u ( x , y ) a ‾ 0 a ‾ 1 x a ‾ 2 y 1 A ( a 1 u 1 a 2 u 2 a 3 u 3 ) 1 A ( b 1 u 1 b 2 u 2 b 3 u 3 ) x 1 A ( c 1 u 1 c 2 u 2 c 3 u 3 ) y 1 A ( a 1 b 1 x c 1 y ) u 1 1 A ( a 2 b 2 x c 2 y ) u 2 1 A ( a 3 b 3 x c 3 y ) u 3 N 1 ( x , y ) u 1 N 2 ( x , y ) u 2 N 3 ( x , y ) u 3 (1-60) \begin{aligned} u(x,y)\overline a_{0}\overline a_{1}x\overline a_{2}y\\ \frac{1}{A}(a_1u_1a_2u_2a_3u_3)\frac{1}{A}(b_1u_1b_2u_2b_3u_3)x\frac{1}{A}(c_1u_1c_2u_2c_3u_3)y\\ \frac{1}{A}(a_1b_1xc_1y)u_1\frac{1}{A}(a_2b_2xc_2y)u_2\frac{1}{A}(a_3b_3xc_3y)u_3\\ N_1(x,y)u_1N_2(x,y)u_2N_3(x,y)u_3 \end{aligned}\tag{1-60} u(x,y)a0a1xa2yA1(a1u1a2u2a3u3)A1(b1u1b2u2b3u3)xA1(c1u1c2u2c3u3)yA1(a1b1xc1y)u1A1(a2b2xc2y)u2A1(a3b3xc3y)u3N1(x,y)u1N2(x,y)u2N3(x,y)u3(1-60) 同理可得 v ( x , y ) a ‾ 0 a ‾ 1 x a ‾ 2 y 1 A ( a 1 v 1 a 2 v 2 a 3 v 3 ) 1 A ( b 1 v 1 b 2 v 2 b 3 v 3 ) x 1 A ( c 1 v 1 c 2 v 2 c 3 v 3 ) y 1 A ( a 1 b 1 x c 1 y ) v 1 1 A ( a 2 b 2 x c 2 y ) v 2 1 A ( a 3 b 3 x c 3 y ) v 3 N 1 ( x , y ) v 1 N 2 ( x , y ) v 2 N 3 ( x , y ) v 3 (1-61) \begin{aligned} v(x,y)\overline a_{0}\overline a_{1}x\overline a_{2}y\\ \frac{1}{A}(a_1v_1a_2v_2a_3v_3)\frac{1}{A}(b_1v_1b_2v_2b_3v_3)x\frac{1}{A}(c_1v_1c_2v_2c_3v_3)y\\ \frac{1}{A}(a_1b_1xc_1y)v_1\frac{1}{A}(a_2b_2xc_2y)v_2\frac{1}{A}(a_3b_3xc_3y)v_3\\ N_1(x,y)v_1N_2(x,y)v_2N_3(x,y)v_3 \end{aligned}\tag{1-61} v(x,y)a0a1xa2yA1(a1v1a2v2a3v3)A1(b1v1b2v2b3v3)xA1(c1v1c2v2c3v3)yA1(a1b1xc1y)v1A1(a2b2xc2y)v2A1(a3b3xc3y)v3N1(x,y)v1N2(x,y)v2N3(x,y)v3(1-61) 可以将二维三角形单元内任意一点的位移写成矩阵的形式如下 u ( x , y ) [ u ( x , y ) v ( x , y ) ] [ N 1 ( x , y ) 0 N 2 ( x , y ) 0 N 3 ( x , y ) 0 0 N 1 ( x , y ) 0 N 2 ( x , y ) 0 N 3 ( x , y ) ] [ u 1 v 1 u 2 v 2 u 3 v 3 ] N ( x , y ) ⋅ q e (1-62) \begin{aligned} \mathbf u(x,y) \begin{bmatrix} u(x,y)\\ v(x,y)\\ \end{bmatrix} \begin{bmatrix} N_1(x,y) 0 N_2(x,y) 0 N_3(x,y) 0\\ 0 N_1(x,y) 0 N_2(x,y) 0 N_3(x,y)\\ \end{bmatrix} \begin{bmatrix} u_1\\v_1\\u_2\\v_2\\u_3\\v_3 \end{bmatrix}\\\mathbf N(x,y)\cdot \mathbf q^e \end{aligned}\tag{1-62} u(x,y)[u(x,y)v(x,y)][N1(x,y)00N1(x,y)N2(x,y)00N2(x,y)N3(x,y)00N3(x,y)] u1v1u2v2u3v3 N(x,y)⋅qe(1-62) 我们称 N ( x , y ) \mathbf N(x,y) N(x,y)为形状函数矩阵而 q e \mathbf q^e qe为单元的节点位移列阵。形状函数矩阵作用 就是用节点位移列阵插值得到任意一点位移的插值函数。
1.2.2.3 单元应变
应变和位移的关系由几何方程确定在本文二维的例子中应变仅有三个分量那么应变与位移的关系可以如下式所示。 ε ( x , y ) [ ε x x ε y y γ x y ] [ ∂ u ∂ x ∂ v ∂ y ∂ u ∂ y ∂ v ∂ x ] [ ∂ ∂ x 0 0 ∂ ∂ y ∂ ∂ y ∂ ∂ x ] ⋅ [ u ( x , y ) v ( x , y ) ] [ ∂ ∂ x 0 0 ∂ ∂ y ∂ ∂ y ∂ ∂ x ] ⋅ N ( x , y ) ⋅ q e B ( x , y ) ⋅ q e (1-63) \begin{aligned} \boldsymbol{\varepsilon}(x,y) \begin{bmatrix} \varepsilon_{xx}\\ \varepsilon_{yy}\\ \gamma_{xy} \end{bmatrix} \begin{bmatrix} \frac{\partial u}{\partial x}\\ \frac{\partial v}{\partial y}\\ \frac{\partial u}{\partial y}\frac{\partial v}{\partial x} \end{bmatrix} \begin{bmatrix} \frac{\partial }{\partial x} 0\\ 0 \frac{\partial }{\partial y}\\ \frac{\partial }{\partial y} \frac{\partial }{\partial x} \end{bmatrix}\cdot \begin{bmatrix} u(x,y)\\ v(x,y) \end{bmatrix}\\\begin{bmatrix} \frac{\partial }{\partial x} 0\\ 0 \frac{\partial }{\partial y}\\ \frac{\partial }{\partial y} \frac{\partial }{\partial x} \end{bmatrix}\cdot\mathbf N(x,y)\cdot \mathbf q^e\\\mathbf B(x,y)\cdot \mathbf q^e \end{aligned}\tag{1-63} ε(x,y) εxxεyyγxy ∂x∂u∂y∂v∂y∂u∂x∂v ∂x∂0∂y∂0∂y∂∂x∂ ⋅[u(x,y)v(x,y)] ∂x∂0∂y∂0∂y∂∂x∂ ⋅N(x,y)⋅qeB(x,y)⋅qe(1-63)
1.2.2.4 单元应力
应力应变关系又成本构方程在线性范围内应力与应变的关系写成矩阵如下(本例中为二维三角形单元应力分量只有三个) σ ( x , y ) [ σ x x σ y y τ x y ] E 1 − μ 2 [ 1 μ 0 μ 1 0 0 0 1 − μ 2 ] [ ε x x ε y y γ x y ] D ( x , y ) ⋅ ε ( x , y ) D ( x , y ) ⋅ B ( x , y ) ⋅ q e S ( x , y ) ⋅ q e (1-64) \begin{aligned} \boldsymbol{\sigma}(x,y) \begin{bmatrix} \sigma_{xx}\\ \sigma_{yy}\\ \tau_{xy} \end{bmatrix} \frac{E}{1-\mu^2}\begin{bmatrix} 1 \mu 0\\ \mu 1 0\\ 0 0 \frac{1-\mu}{2} \end{bmatrix}\begin{bmatrix} \varepsilon_{xx}\\ \varepsilon_{yy}\\ \gamma_{xy} \end{bmatrix}\\ \mathbf D(x,y)\cdot \boldsymbol{\varepsilon}(x,y)\\ \mathbf D(x,y)\cdot \mathbf B(x,y)\cdot \mathbf q^e\\ \mathbf S(x,y)\cdot \mathbf q^e \end{aligned}\tag{1-64} σ(x,y) σxxσyyτxy 1−μ2E 1μ0μ100021−μ εxxεyyγxy D(x,y)⋅ε(x,y)D(x,y)⋅B(x,y)⋅qeS(x,y)⋅qe(1-64)
1.2.2.5 单元刚度矩阵
回顾物体的总势能表达式并将其写成矩阵的形式 I I 1 2 ∫ Ω σ i j ε i j d Ω − ∫ Ω b i u i d Ω − ∫ A p i u i d A 1 2 ∫ Ω σ T ε d Ω − ∫ Ω b T u d Ω − ∫ A p T u d A (1-65) \begin{aligned} II\frac{1}{2}\int_{\Omega}\sigma_{ij}\varepsilon_{ij}d\Omega-\int_{\Omega} b_{i}u_{i}d\Omega -\int_{A}p_{i}u_{i}dA\\ \frac{1}{2}\int_{\Omega}\boldsymbol{\sigma}^T\boldsymbol{\varepsilon}\mathbf d\Omega-\int_{\Omega} \mathbf b^T \mathbf u \mathbf d\Omega -\int_{A}\mathbf p^T\mathbf u \mathbf dA \end{aligned}\tag{1-65} II21∫ΩσijεijdΩ−∫ΩbiuidΩ−∫ApiuidA21∫ΩσTεdΩ−∫ΩbTudΩ−∫ApTudA(1-65) 将前述单元的应力、应变矩阵等代入上式建立单元的总势能表达式如下所示 I I 1 2 ∫ Ω σ T ε d Ω − ∫ Ω b T u d Ω − ∫ A p T u d A 1 2 ∫ Ω q e T B T D T B q e d Ω − ∫ Ω b T N q e d Ω − ∫ A p T N q e d A 1 2 q e T ( ∫ Ω B T D T B d Ω ) q e − ( ∫ Ω b T N d Ω ∫ A p T N d A ) q e 1 2 q e T K e q e − P e T q e (1-66) \begin{aligned} II \frac{1}{2}\int_{\Omega}\boldsymbol{\sigma}^T\boldsymbol{\varepsilon}\mathbf d\Omega-\int_{\Omega} \mathbf b^T \mathbf u \mathbf d\Omega -\int_{A}\mathbf p^T\mathbf u \mathbf dA\\ \frac{1}{2}\int_{\Omega}\boldsymbol{q^e}^T\boldsymbol{B^TD^TBq^e}\mathbf d\Omega-\int_{\Omega} \mathbf b^T \boldsymbol{ Nq^e d}\Omega -\int_{A}\mathbf p^T\boldsymbol{ Nq^e d}A\\ \frac{1}{2}\boldsymbol{q^e}^T(\int_{\Omega}\boldsymbol{B^TD^TB d}\Omega)\boldsymbol{q^e}-(\int_{\Omega} \mathbf b^T \boldsymbol{ Nd}\Omega \int_{A}\mathbf p^T\boldsymbol{ Nd}A)\boldsymbol{q^e}\\ \frac{1}{2}\boldsymbol{q^e}^T\boldsymbol{K^e}\boldsymbol{q^e}-\boldsymbol{P^e}^T\boldsymbol{q^e} \end{aligned}\tag{1-66} II21∫ΩσTεdΩ−∫ΩbTudΩ−∫ApTudA21∫ΩqeTBTDTBqedΩ−∫ΩbTNqedΩ−∫ApTNqedA21qeT(∫ΩBTDTBdΩ)qe−(∫ΩbTNdΩ∫ApTNdA)qe21qeTKeqe−PeTqe(1-66) 回顾之前最小势能原理真实位移是使得势能取得最小值上式是单元的总势能那么真实的位移使得势能取到最小值也就是说总势能一阶变分为零即 δ I I ∂ I I ∂ q e δ q e ( K e q e − P e ) δ q e 0 (1-67) \delta II \frac{\partial II}{\partial q^e}\delta q^e(\boldsymbol{K^e}\boldsymbol{q^e}-\boldsymbol{P^e})\delta q^e0 \tag{1-67} δII∂qe∂IIδqe(Keqe−Pe)δqe0(1-67) 由于 δ q e \delta q^e δqe的任意性可以得到 K e q e − P e 0 (1-68) \boldsymbol{K^e}\boldsymbol{q^e}-\boldsymbol{P^e}0\tag{1-68} Keqe−Pe0(1-68) 其中 K e \boldsymbol{K^e} Ke为单元刚度矩阵由上式中的量纲分析 K e \boldsymbol{K^e} Ke为刚度的量纲 P e \boldsymbol{P^e} Pe为节点等效载荷。
1.2.2.6 整体刚度矩阵
对于物体来说一般有多个单元那么就可以建立多个单元平衡方程如下所示 K ( i ) e q ( i ) e − P ( i ) e 0 (1-69) \boldsymbol{K^e_{(i)}}\boldsymbol{q^e_{(i)}}-\boldsymbol{P^e_{(i)}}0\tag{1-69} K(i)eq(i)e−P(i)e0(1-69) 按照顺序将所有的单元的节点位移组成总体节点位移列阵相应的节点等效载荷和单元刚度矩阵组成总体节点等效载荷列阵和整体刚度矩阵以下图为例下图总共有两个单元得到两个单元平衡方程 [ k 11 1 k 12 1 k 13 1 k 14 1 k 15 1 k 16 1 k 21 1 k 22 1 k 23 1 k 24 1 k 25 1 k 26 1 k 31 1 k 32 1 k 33 1 k 34 1 k 35 1 k 36 1 k 41 1 k 42 1 k 43 1 k 44 1 k 45 1 k 46 1 k 51 1 k 52 1 k 53 1 k 54 1 k 55 1 k 56 1 k 61 1 k 62 1 k 63 1 k 64 1 k 65 1 k 66 1 ] [ u 1 v 1 u 2 v 2 u 3 v 3 ] [ p x 1 1 p y 1 1 p x 2 1 p y 2 1 p x 3 1 p y 3 1 ] [ k 11 2 k 12 2 k 13 2 k 14 2 k 15 2 k 16 2 k 21 2 k 22 2 k 23 2 k 24 2 k 25 2 k 26 2 k 31 2 k 32 2 k 33 2 k 34 2 k 35 2 k 36 2 k 41 2 k 42 2 k 43 2 k 44 2 k 45 2 k 46 2 k 51 2 k 52 2 k 53 2 k 54 2 k 55 2 k 56 2 k 61 2 k 62 2 k 63 2 k 64 2 k 65 2 k 66 2 ] [ u 2 v 2 u 3 v 3 u 4 v 4 ] [ p x 2 2 p y 2 2 p x 3 2 p y 3 2 p x 4 2 p y 4 2 ] (1-70) \begin{aligned} \begin{bmatrix} k^1_{11} k^1_{12} k^1_{13} k^1_{14} k^1_{15} k^1_{16}\\ k^1_{21} k^1_{22} k^1_{23} k^1_{24} k^1_{25} k^1_{26}\\ k^1_{31} k^1_{32} k^1_{33} k^1_{34} k^1_{35} k^1_{36}\\ k^1_{41} k^1_{42} k^1_{43} k^1_{44} k^1_{45} k^1_{46}\\ k^1_{51} k^1_{52} k^1_{53} k^1_{54} k^1_{55} k^1_{56}\\ k^1_{61} k^1_{62} k^1_{63} k^1_{64} k^1_{65} k^1_{66} \end{bmatrix}\begin{bmatrix} u_{1}\\ v_{1}\\u_{2}\\ v_{2}\\u_{3}\\ v_{3}\\ \end{bmatrix}\begin{bmatrix} p^1_{x1}\\ p^1_{y1}\\p^1_{x2}\\ p^1_{y2}\\p^1_{x3}\\ p^1_{y3}\\ \end{bmatrix}\\ \begin{bmatrix} k^2_{11} k^2_{12} k^2_{13} k^2_{14} k^2_{15} k^2_{16}\\ k^2_{21} k^2_{22} k^2_{23} k^2_{24} k^2_{25} k^2_{26}\\ k^2_{31} k^2_{32} k^2_{33} k^2_{34} k^2_{35} k^2_{36}\\ k^2_{41} k^2_{42} k^2_{43} k^2_{44} k^2_{45} k^2_{46}\\ k^2_{51} k^2_{52} k^2_{53} k^2_{54} k^2_{55} k^2_{56}\\ k^2_{61} k^2_{62} k^2_{63} k^2_{64} k^2_{65} k^2_{66} \end{bmatrix}\begin{bmatrix} u_{2}\\ v_{2}\\u_{3}\\ v_{3}\\u_{4}\\ v_{4}\\ \end{bmatrix}\begin{bmatrix} p^2_{x2}\\ p^2_{y2}\\p^2_{x3}\\ p^2_{y3}\\p^2_{x4}\\ p^2_{y4}\\ \end{bmatrix}\end{aligned}\tag{1-70} k111k211k311k411k511k611k121k221k321k421k521k621k131k231k331k431k531k631k141k241k341k441k541k641k151k251k351k451k551k651k161k261k361k461k561k661 u1v1u2v2u3v3 px11py11px21py21px31py31 k112k212k312k412k512k612k122k222k322k422k522k622k132k232k332k432k532k632k142k242k342k442k542k642k152k252k352k452k552k652k162k262k362k462k562k662 u2v2u3v3u4v4 px22py22px32py32px42py42 (1-70) 叠加整理后整体平衡方程为 [ k 11 1 k 12 1 k 13 1 k 14 1 k 15 1 k 16 1 0 0 k 21 1 k 22 1 k 23 1 k 24 1 k 25 1 k 26 1 0 0 k 31 1 k 32 1 k 33 1 k 11 2 k 34 1 k 12 2 k 35 1 k 13 2 k 36 1 k 14 2 k 15 2 k 16 2 k 41 1 k 42 1 k 43 1 k 21 2 k 44 1 k 22 2 k 45 1 k 23 2 k 46 1 k 24 2 k 25 2 k 26 2 k 51 1 k 52 1 k 53 1 k 31 2 k 54 1 k 32 2 k 55 1 k 33 2 k 56 1 k 34 2 k 35 2 k 36 2 k 61 1 k 62 1 k 63 1 k 41 2 k 64 1 k 42 2 k 65 1 k 43 2 k 66 1 k 44 2 k 45 2 k 46 2 0 0 k 51 2 k 52 2 k 53 2 k 54 2 k 55 2 k 56 2 0 0 k 61 2 k 62 2 k 63 2 k 64 2 k 65 2 k 66 2 ] [ u 1 v 1 u 2 v 2 u 3 v 3 u 4 v 4 ] [ p x 1 1 p y 1 1 p x 2 1 p x 2 2 p y 2 1 p y 2 2 p x 3 1 p x 3 2 p y 3 1 p y 3 2 p x 4 2 p y 4 2 ] (1-71) \begin{bmatrix} k^1_{11} k^1_{12} k^1_{13} k^1_{14} k^1_{15} k^1_{16} 0 0\\ k^1_{21} k^1_{22} k^1_{23} k^1_{24} k^1_{25} k^1_{26} 0 0\\ k^1_{31} k^1_{32} k^1_{33}k^2_{11} k^1_{34}k^2_{12} k^1_{35}k^2_{13} k^1_{36}k^2_{14} k^2_{15} k^2_{16}\\ k^1_{41} k^1_{42} k^1_{43}k^2_{21} k^1_{44}k^2_{22} k^1_{45} k^2_{23} k^1_{46}k^2_{24} k^2_{25} k^2_{26}\\ k^1_{51} k^1_{52} k^1_{53}k^2_{31} k^1_{54}k^2_{32} k^1_{55}k^2_{33} k^1_{56}k^2_{34} k^2_{35} k^2_{36}\\ k^1_{61} k^1_{62} k^1_{63}k^2_{41} k^1_{64}k^2_{42} k^1_{65}k^2_{43} k^1_{66}k^2_{44} k^2_{45} k^2_{46}\\ 0 0 k^2_{51} k^2_{52} k^2_{53} k^2_{54} k^2_{55} k^2_{56}\\ 0 0 k^2_{61} k^2_{62} k^2_{63} k^2_{64} k^2_{65} k^2_{66} \end{bmatrix}\begin{bmatrix} u_{1}\\ v_{1}\\u_{2}\\ v_{2}\\u_{3}\\ v_{3}\\u_{4}\\ v_{4}\\ \end{bmatrix}\begin{bmatrix} p^1_{x1}\\ p^1_{y1}\\p^1_{x2}p^2_{x2}\\ p^1_{y2}p^2_{y2}\\p^1_{x3}p^2_{x3}\\ p^1_{y3}p^2_{y3}\\p^2_{x4}\\ p^2_{y4}\\ \end{bmatrix}\tag{1-71} k111k211k311k411k511k61100k121k221k321k421k521k62100k131k231k331k112k431k212k531k312k631k412k512k612k141k241k341k122k441k222k541k322k641k422k522k622k151k251k351k132k451k232k551k332k651k432k532k632k161k261k361k142k461k242k561k342k661k442k542k64200k152k252k352k452k552k65200k162k262k362k462k562k662 u1v1u2v2u3v3u4v4 px11py11px21px22py21py22px31px32py31py32px42py42 (1-71)
1.2.2.7 处理约束
如果在上例中假设节点1为约束那么 u 1 v 1 0 u_1v_10 u1v10上式变为 [ k 11 1 k 12 1 k 13 1 k 14 1 k 15 1 k 16 1 0 0 k 21 1 k 22 1 k 23 1 k 24 1 k 25 1 k 26 1 0 0 k 31 1 k 32 1 k 33 1 k 11 2 k 34 1 k 12 2 k 35 1 k 13 2 k 36 1 k 14 2 k 15 2 k 16 2 k 41 1 k 42 1 k 43 1 k 21 2 k 44 1 k 22 2 k 45 1 k 23 2 k 46 1 k 24 2 k 25 2 k 26 2 k 51 1 k 52 1 k 53 1 k 31 2 k 54 1 k 32 2 k 55 1 k 33 2 k 56 1 k 34 2 k 35 2 k 36 2 k 61 1 k 62 1 k 63 1 k 41 2 k 64 1 k 42 2 k 65 1 k 43 2 k 66 1 k 44 2 k 45 2 k 46 2 0 0 k 51 2 k 52 2 k 53 2 k 54 2 k 55 2 k 56 2 0 0 k 61 2 k 62 2 k 63 2 k 64 2 k 65 2 k 66 2 ] [ 0 0 u 2 v 2 u 3 v 3 u 4 v 4 ] [ R x 1 R y 1 p x 2 1 p x 2 2 p y 2 1 p y 2 2 p x 3 1 p x 3 2 p y 3 1 p y 3 2 p x 4 2 p y 4 2 ] (1-72) \begin{bmatrix} k^1_{11} k^1_{12} k^1_{13} k^1_{14} k^1_{15} k^1_{16} 0 0\\ k^1_{21} k^1_{22} k^1_{23} k^1_{24} k^1_{25} k^1_{26} 0 0\\ k^1_{31} k^1_{32} k^1_{33}k^2_{11} k^1_{34}k^2_{12} k^1_{35}k^2_{13} k^1_{36}k^2_{14} k^2_{15} k^2_{16}\\ k^1_{41} k^1_{42} k^1_{43}k^2_{21} k^1_{44}k^2_{22} k^1_{45} k^2_{23} k^1_{46}k^2_{24} k^2_{25} k^2_{26}\\ k^1_{51} k^1_{52} k^1_{53}k^2_{31} k^1_{54}k^2_{32} k^1_{55}k^2_{33} k^1_{56}k^2_{34} k^2_{35} k^2_{36}\\ k^1_{61} k^1_{62} k^1_{63}k^2_{41} k^1_{64}k^2_{42} k^1_{65}k^2_{43} k^1_{66}k^2_{44} k^2_{45} k^2_{46}\\ 0 0 k^2_{51} k^2_{52} k^2_{53} k^2_{54} k^2_{55} k^2_{56}\\ 0 0 k^2_{61} k^2_{62} k^2_{63} k^2_{64} k^2_{65} k^2_{66} \end{bmatrix}\begin{bmatrix} 0\\ 0\\u_{2}\\ v_{2}\\u_{3}\\ v_{3}\\u_{4}\\ v_{4}\\ \end{bmatrix}\begin{bmatrix} R_{x1}\\ R_{y1}\\p^1_{x2}p^2_{x2}\\ p^1_{y2}p^2_{y2}\\p^1_{x3}p^2_{x3}\\ p^1_{y3}p^2_{y3}\\p^2_{x4}\\ p^2_{y4}\\ \end{bmatrix}\tag{1-72} k111k211k311k411k511k61100k121k221k321k421k521k62100k131k231k331k112k431k212k531k312k631k412k512k612k141k241k341k122k441k222k541k322k641k422k522k622k151k251k351k132k451k232k551k332k651k432k532k632k161k261k361k142k461k242k561k342k661k442k542k64200k152k252k352k452k552k65200k162k262k362k462k562k662 00u2v2u3v3u4v4 Rx1Ry1px21px22py21py22px31px32py31py32px42py42 (1-72) 其中 R x 1 、 R y 1 R_{x1}、R_{y1} Rx1、Ry1为支反力将上式化为两个方程 [ k 33 1 k 11 2 k 34 1 k 12 2 k 35 1 k 13 2 k 36 1 k 14 2 k 15 2 k 16 2 k 43 1 k 21 2 k 44 1 k 22 2 k 45 1 k 23 2 k 46 1 k 24 2 k 25 2 k 26 2 k 53 1 k 31 2 k 54 1 k 32 2 k 55 1 k 33 2 k 56 1 k 34 2 k 35 2 k 36 2 k 63 1 k 41 2 k 64 1 k 42 2 k 65 1 k 43 2 k 66 1 k 44 2 k 45 2 k 46 2 k 51 2 k 52 2 k 53 2 k 54 2 k 55 2 k 56 2 k 61 2 k 62 2 k 63 2 k 64 2 k 65 2 k 66 2 ] [ u 2 v 2 u 3 v 3 u 4 v 4 ] [ p x 2 1 p x 2 2 p y 2 1 p y 2 2 p x 3 1 p x 3 2 p y 3 1 p y 3 2 p x 4 2 p y 4 2 ] K ‾ q ‾ e P ‾ (1-73) \begin{aligned}\begin{bmatrix} k^1_{33}k^2_{11} k^1_{34}k^2_{12} k^1_{35}k^2_{13} k^1_{36}k^2_{14} k^2_{15} k^2_{16}\\ k^1_{43}k^2_{21} k^1_{44}k^2_{22} k^1_{45} k^2_{23} k^1_{46}k^2_{24} k^2_{25} k^2_{26}\\ k^1_{53}k^2_{31} k^1_{54}k^2_{32} k^1_{55}k^2_{33} k^1_{56}k^2_{34} k^2_{35} k^2_{36}\\ k^1_{63}k^2_{41} k^1_{64}k^2_{42} k^1_{65}k^2_{43} k^1_{66}k^2_{44} k^2_{45} k^2_{46}\\ k^2_{51} k^2_{52} k^2_{53} k^2_{54} k^2_{55} k^2_{56}\\ k^2_{61} k^2_{62} k^2_{63} k^2_{64} k^2_{65} k^2_{66} \end{bmatrix}\begin{bmatrix} u_{2}\\ v_{2}\\u_{3}\\ v_{3}\\u_{4}\\ v_{4}\\ \end{bmatrix}\begin{bmatrix} p^1_{x2}p^2_{x2}\\ p^1_{y2}p^2_{y2}\\p^1_{x3}p^2_{x3}\\ p^1_{y3}p^2_{y3}\\p^2_{x4}\\ p^2_{y4}\\ \end{bmatrix}\\ \overline K \overline q^e\overline P \end{aligned}\tag{1-73} k331k112k431k212k531k312k631k412k512k612k341k122k441k222k541k322k641k422k522k622k351k132k451k232k551k332k651k432k532k632k361k142k461k242k561k342k661k442k542k642k152k252k352k452k552k652k162k262k362k462k562k662 u2v2u3v3u4v4 Kqe px21px22py21py22px31px32py31py32px42py42 P(1-73) [ k 13 1 k 14 1 k 15 1 k 16 1 k 23 1 k 24 1 k 25 1 k 26 1 ] [ u 2 v 2 u 3 v 3 ] [ R x 1 R y 1 ] (1-74) \begin{bmatrix} k^1_{13} k^1_{14} k^1_{15} k^1_{16}\\ k^1_{23} k^1_{24} k^1_{25} k^1_{26}\\ \end{bmatrix}\begin{bmatrix} u_{2}\\ v_{2}\\u_{3}\\ v_{3} \end{bmatrix}\begin{bmatrix} R_{x1}\\ R_{y1}\\ \end{bmatrix}\tag{1-74} [k131k231k141k241k151k251k161k261] u2v2u3v3 [Rx1Ry1](1-74)
1.2.2.8 求解节点载荷列阵
1.2.2.9 求解位移列阵
最终通过上式求得位移列阵 q ‾ e K ‾ − 1 ⋅ P ‾ \overline q^e\overline K^{-1}\cdot \overline P qeK−1⋅P再将与 [ u 1 , v 1 ] T [ 0 , 0 ] T [u_1,v_1]^T[0,0]^T [u1,v1]T[0,0]T组装成最终的节点位移列阵 q e \boldsymbol{q^e} qe。
1.2.2.10 计算应力矩阵等
计算得到最终的节点位移列阵 q e \boldsymbol{q^e} qe后通过一系列的回代得到单元内任意位置的位移、应变、应力矩阵而式1-74用来求解约束支反力。 u ( x , y ) N ( x , y ) ⋅ q e ε ( x , y ) B ( x , y ) ⋅ q e σ ( x , y ) S ( x , y ) ⋅ q e (1-75) \begin{aligned} \mathbf u(x,y) \mathbf N(x,y)\cdot \mathbf q^e\\ \boldsymbol{\varepsilon}(x,y) \mathbf B(x,y)\cdot \mathbf q^e\\ \boldsymbol{\sigma}(x,y) \mathbf S(x,y)\cdot \mathbf q^e \end{aligned}\tag{1-75} u(x,y)N(x,y)⋅qeε(x,y)B(x,y)⋅qeσ(x,y)S(x,y)⋅qe(1-75)
