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文章目錄
- 1.2.2 基于最小勢(shì)能原理的線性有限元一般格式
- 1.2.2.1 離散化
- 1.2.2.2 位移插值
- 1.2.2.3 單元應(yīng)變
- 1.2.2.4 單元應(yīng)力
- 1.2.2.5 單元?jiǎng)偠染仃?/li>
- 1.2.2.6 整體剛度矩陣
- 1.2.2.7 處理約束
- 1.2.2.8 求解節(jié)點(diǎn)載荷列陣
- 1.2.2.9 求解位移列陣
- 1.2.2.10 計(jì)算應(yīng)力矩陣等
1.2.2 基于最小勢(shì)能原理的線性有限元一般格式
現(xiàn)在市場(chǎng)上的有限元分析軟件普遍都是基于位移法進(jìn)行求解的，其數(shù)學(xué)原理就是上節(jié)的最小勢(shì)能原理變分提法。回顧上節(jié)推導(dǎo)過(guò)程，首先需要根據(jù)位移條件確定可能位移的范圍，其次根據(jù)假設(shè)的可能位移代入幾何方程和本構(gòu)方程，得到基于位移的應(yīng)力應(yīng)變的表達(dá)式，最后代入 δ I I = 0 \delta II=0 δII=0的方程求解相關(guān)待定參數(shù)。線性有限元求解的一般格式基本上也是一樣的，除此之外線性有限元會(huì)多一些其他步驟，比如結(jié)構(gòu)的離散化、位移模式假定等。下圖為線性有限元求解的一般格式。
（加圖）
1.2.2.1 離散化
將結(jié)構(gòu)進(jìn)行離散化建模，是有限元分析工作中的第一步，某典型的結(jié)構(gòu)如下圖所示，其中結(jié)構(gòu)為一個(gè)圓盤，將其離散成若干四邊形網(wǎng)格單元，如下圖所示。結(jié)構(gòu)離散化合適與否一定程度上決定使用有限元分析方法解決工程問(wèn)題的精度。
1.2.2.2 位移插值
（a）位移模式
在最小勢(shì)能原理中首先需要根據(jù)位移條件確定可能位移的范圍，但是在通常的分析中，我們?yōu)榱朔奖闾幚?#xff0c;往往將可能位移用多項(xiàng)式來(lái)逼近，這個(gè)過(guò)程就是單元的位移模式選擇。
我們以一階二維三角形單元為例，二維三角形單元有6各位移自由度。用多項(xiàng)式擬合可能位移，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)=a0?+a1?x+a2?yv(x,y)=b0?+b1?x+b2?y?(1-54)
（b）形函數(shù)
多項(xiàng)式位移模式有六個(gè)待定參數(shù)，二維三角形單元有六個(gè)節(jié)點(diǎn)位移變量，所以六個(gè)待定參數(shù)可以表示成六個(gè)節(jié)點(diǎn)位移變量的形式，如下式，待定系數(shù)滿足下式。
[ 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} ?u1?u2?u3?? ?= ?111?x1?x2?x3??y1?y2?y3?? ? ?a0?a1?a2?? ?, ?v1?v2?v3?? ?= ?111?x1?x2?x3??y1?y2?y3?? ? ?b0?b1?b2?? ??(1-55)
那么系數(shù)為
[ 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} ?a0?a1?a2?? ?= ?111?x1?x2?x3??y1?y2?y3?? ??1 ?u1?u2?u3?? ?, ?b0?b1?b2?? ?= ?111?x1?x2?x3??y1?y2?y3?? ??1 ?v1?v2?v3?? ??(1-56)
其中逆矩陣的求解如下
[ 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)^{i+j}]^T}{A}\tag{1-57} ?111?x1?x2?x3??y1?y2?y3?? ??1=A[Mij??(?1)i+j]T?(1-57)
其中 M i j M_{ij} Mij?為原矩陣的余子式，典型如下，其乘上 ( ? 1 ) i + j (-1)^{i+j} (?1)i+j稱為代數(shù)余子式，矩陣的逆是其代數(shù)余子式組成的伴隨矩陣除以該矩陣的行列式。
經(jīng)過(guò)求解，系數(shù)矩陣的逆矩陣如下式
[ 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} ?111?x1?x2?x3??y1?y2?y3?? ??1=A1? ?a1?b1?c1??a2?b2?c2??a3?b3?c3?? ??= ?111?x1?x2?x3??y1?y2?y3?? ??1? ? ?x2?x3??y2?y3?? ?? ?11?y2?y3?? ? ?11?x2?x3?? ??? ?x1?x3??y1?y3?? ? ?11?y1?y3?? ?? ?11?x1?x3?? ?? ?x1?x2??y1?y2?? ?? ?11?y1?y2?? ? ?11?x1?x2?? ?? ??(1-58)
那么，系數(shù)可以表達(dá)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_1+a_2u_2+a_3u_3)\\ \overline a_1=\frac{1}{A}(b_1u_1+b_2u_2+b_3u_3)\\ \overline a_2=\frac{1}{A}(c_1u_1+c_2u_2+c_3u_3) \end{aligned}\tag{1-59} a0?=A1?(a1?u1?+a2?u2?+a3?u3?)a1?=A1?(b1?u1?+b2?u2?+b3?u3?)a2?=A1?(c1?u1?+c2?u2?+c3?u3?)?(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_1+a_2u_2+a_3u_3)+\frac{1}{A}(b_1u_1+b_2u_2+b_3u_3)x+\frac{1}{A}(c_1u_1+c_2u_2+c_3u_3)y\\ &=\frac{1}{A}(a_1+b_1x+c_1y)u_1+\frac{1}{A}(a_2+b_2x+c_2y)u_2+\frac{1}{A}(a_3+b_3x+c_3y)u_3\\ &=N_1(x,y)u_1+N_2(x,y)u_2+N_3(x,y)u_3 \end{aligned}\tag{1-60} u(x,y)?=a0?+a1?x+a2?y=A1?(a1?u1?+a2?u2?+a3?u3?)+A1?(b1?u1?+b2?u2?+b3?u3?)x+A1?(c1?u1?+c2?u2?+c3?u3?)y=A1?(a1?+b1?x+c1?y)u1?+A1?(a2?+b2?x+c2?y)u2?+A1?(a3?+b3?x+c3?y)u3?=N1?(x,y)u1?+N2?(x,y)u2?+N3?(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_1+a_2v_2+a_3v_3)+\frac{1}{A}(b_1v_1+b_2v_2+b_3v_3)x+\frac{1}{A}(c_1v_1+c_2v_2+c_3v_3)y\\ &=\frac{1}{A}(a_1+b_1x+c_1y)v_1+\frac{1}{A}(a_2+b_2x+c_2y)v_2+\frac{1}{A}(a_3+b_3x+c_3y)v_3\\ &=N_1(x,y)v_1+N_2(x,y)v_2+N_3(x,y)v_3 \end{aligned}\tag{1-61} v(x,y)?=a0?+a1?x+a2?y=A1?(a1?v1?+a2?v2?+a3?v3?)+A1?(b1?v1?+b2?v2?+b3?v3?)x+A1?(c1?v1?+c2?v2?+c3?v3?)y=A1?(a1?+b1?x+c1?y)v1?+A1?(a2?+b2?x+c2?y)v2?+A1?(a3?+b3?x+c3?y)v3?=N1?(x,y)v1?+N2?(x,y)v2?+N3?(x,y)v3??(1-61)
可以將二維三角形單元內(nèi)任意一點(diǎn)的位移寫成矩陣的形式，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)0?0N1?(x,y)?N2?(x,y)0?0N2?(x,y)?N3?(x,y)0?0N3?(x,y)?] ?u1?v1?u2?v2?u3?v3?? ?=N(x,y)?qe?(1-62)
我們稱 N ( x , y ) \mathbf N(x,y) N(x,y)為形狀函數(shù)矩陣，而 q e \mathbf q^e qe為單元的節(jié)點(diǎn)位移列陣。形狀函數(shù)矩陣作用 就是用節(jié)點(diǎn)位移列陣插值得到任意一點(diǎn)位移的插值函數(shù)。
1.2.2.3 單元應(yīng)變
應(yīng)變和位移的關(guān)系由幾何方程確定，在本文二維的例子中，應(yīng)變僅有三個(gè)分量，那么應(yīng)變與位移的關(guān)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)= ?εxx?εyy?γxy?? ?= ??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)?qe=B(x,y)?qe?(1-63)
1.2.2.4 單元應(yīng)力
應(yīng)力應(yīng)變關(guān)系又成本構(gòu)方程，在線性范圍內(nèi)，應(yīng)力與應(yīng)變的關(guān)系寫成矩陣如下(本例中為二維三角形單元，應(yīng)力分量只有三個(gè))
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)= ?σxx?σyy?τxy?? ??=1?μ2E? ?1μ0?μ10?0021?μ?? ? ?εxx?εyy?γxy?? ?=D(x,y)?ε(x,y)=D(x,y)?B(x,y)?qe=S(x,y)?qe?(1-64)
1.2.2.5 單元?jiǎng)偠染仃?/h5>
回顧物體的總勢(shì)能表達(dá)式，并將其寫成矩陣的形式
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} II?=21?∫Ω?σij?εij?dΩ?∫Ω?bi?ui?dΩ?∫A?pi?ui?dA=21?∫Ω?σTεdΩ?∫Ω?bTudΩ?∫A?pTudA?(1-65)
將前述單元的應(yīng)力、應(yīng)變矩陣等代入上式，建立單元的總勢(shì)能表達(dá)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} II?=21?∫Ω?σTεdΩ?∫Ω?bTudΩ?∫A?pTudA=21?∫Ω?qeTBTDTBqedΩ?∫Ω?bTNqedΩ?∫A?pTNqedA=21?qeT(∫Ω?BTDTBdΩ)qe?(∫Ω?bTNdΩ+∫A?pTNdA)qe=21?qeTKeqe?PeTqe?(1-66)
回顧之前最小勢(shì)能原理，真實(shí)位移是使得勢(shì)能取得最小值，上式是單元的總勢(shì)能，那么真實(shí)的位移使得勢(shì)能取到最小值，也就是說(shuō)總勢(shì)能一階變分為零，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^e=0 \tag{1-67} δII=?qe?II?δqe=(Keqe?Pe)δqe=0(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?Pe=0(1-68)
其中 K e \boldsymbol{K^e} Ke為單元?jiǎng)偠染仃?#xff08;由上式中的量綱分析， K e \boldsymbol{K^e} Ke為剛度的量綱）， P e \boldsymbol{P^e} Pe為節(jié)點(diǎn)等效載荷。
1.2.2.6 整體剛度矩陣
對(duì)于物體來(lái)說(shuō)，一般有多個(gè)單元，那么就可以建立多個(gè)單元平衡方程，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)e?q(i)e??P(i)e?=0(1-69)
按照順序?qū)⑺械膯卧墓?jié)點(diǎn)位移組成總體節(jié)點(diǎn)位移列陣，相應(yīng)的節(jié)點(diǎn)等效載荷和單元?jiǎng)偠染仃嚱M成總體節(jié)點(diǎn)等效載荷列陣和整體剛度矩陣，以下圖為例，下圖總共有兩個(gè)單元，得到兩個(gè)單元平衡方程
[ 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} ?k111?k211?k311?k411?k511?k611??k121?k221?k321?k421?k521?k621??k131?k231?k331?k431?k531?k631??k141?k241?k341?k441?k541?k641??k151?k251?k351?k451?k551?k651??k161?k261?k361?k461?k561?k661?? ? ?u1?v1?u2?v2?u3?v3?? ?= ?px11?py11?px21?py21?px31?py31?? ? ?k112?k212?k312?k412?k512?k612??k122?k222?k322?k422?k522?k622??k132?k232?k332?k432?k532?k632??k142?k242?k342?k442?k542?k642??k152?k252?k352?k452?k552?k652??k162?k262?k362?k462?k562?k662?? ? ?u2?v2?u3?v3?u4?v4?? ?= ?px22?py22?px32?py32?px42?py42?? ??(1-70)
疊加整理后，整體平衡方程為
[ 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} ?k111?k211?k311?k411?k511?k611?00?k121?k221?k321?k421?k521?k621?00?k131?k231?k331?+k112?k431?+k212?k531?+k312?k631?+k412?k512?k612??k141?k241?k341?+k122?k441?+k222?k541?+k322?k641?+k422?k522?k622??k151?k251?k351?+k132?k451?+k232?k551?+k332?k651?+k432?k532?k632??k161?k261?k361?+k142?k461?+k242?k561?+k342?k661?+k442?k542?k642??00k152?k252?k352?k452?k552?k652??00k162?k262?k362?k462?k562?k662?? ? ?u1?v1?u2?v2?u3?v3?u4?v4?? ?= ?px11?py11?px21?+px22?py21?+py22?px31?+px32?py31?+py32?px42?py42?? ?(1-71)
1.2.2.7 處理約束
如果在上例中，假設(shè)節(jié)點(diǎn)1為約束，那么 u 1 = v 1 = 0 u_1=v_1=0 u1?=v1?=0，上式變?yōu)?br /> [ 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} ?k111?k211?k311?k411?k511?k611?00?k121?k221?k321?k421?k521?k621?00?k131?k231?k331?+k112?k431?+k212?k531?+k312?k631?+k412?k512?k612??k141?k241?k341?+k122?k441?+k222?k541?+k322?k641?+k422?k522?k622??k151?k251?k351?+k132?k451?+k232?k551?+k332?k651?+k432?k532?k632??k161?k261?k361?+k142?k461?+k242?k561?+k342?k661?+k442?k542?k642??00k152?k252?k352?k452?k552?k652??00k162?k262?k362?k462?k562?k662?? ? ?00u2?v2?u3?v3?u4?v4?? ?= ?Rx1?Ry1?px21?+px22?py21?+py22?px31?+px32?py31?+py32?px42?py42?? ?(1-72)
其中 R x 1 、 R y 1 R_{x1}、R_{y1} Rx1?、Ry1?為支反力，將上式化為兩個(gè)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} ?k331?+k112?k431?+k212?k531?+k312?k631?+k412?k512?k612??k341?+k122?k441?+k222?k541?+k322?k641?+k422?k522?k622??k351?+k132?k451?+k232?k551?+k332?k651?+k432?k532?k632??k361?+k142?k461?+k242?k561?+k342?k661?+k442?k542?k642??k152?k252?k352?k452?k552?k652??k162?k262?k362?k462?k562?k662?? ? ?u2?v2?u3?v3?u4?v4?? ?Kq?e?= ?px21?+px22?py21?+py22?px31?+px32?py31?+py32?px42?py42?? ?=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} [k131?k231??k141?k241??k151?k251??k161?k261??] ?u2?v2?u3?v3?? ?=[Rx1?Ry1??](1-74)
1.2.2.8 求解節(jié)點(diǎn)載荷列陣
1.2.2.9 求解位移列陣
最終通過(guò)上式求得位移列陣 q ￣ e = K ￣ ? 1 ? P ￣ \overline q^e=\overline K^{-1}\cdot \overline P q?e=K?1?P，再將與 [ u 1 , v 1 ] T = [ 0 , 0 ] T [u_1,v_1]^T=[0,0]^T [u1?,v1?]T=[0,0]T組裝成最終的節(jié)點(diǎn)位移列陣 q e \boldsymbol{q^e} qe。
1.2.2.10 計(jì)算應(yīng)力矩陣等
計(jì)算得到最終的節(jié)點(diǎn)位移列陣 q e \boldsymbol{q^e} qe后通過(guò)一系列的回代得到單元內(nèi)任意位置的位移、應(yīng)變、應(yīng)力矩陣，而式（1-74）用來(lái)求解約束支反力。
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)