3.2 – Vergleich verschiedener Diskretisierungen

 

Auszug aus dem Skript der Vorlesung Finite Elemente bei Dr.-Ing Philipp Höfer an der UniBw München.

Wir betrachten als Beispiel einen Zug-Druck-Stab mit konstanter Belastung und EA = \operatorname{const}:

zug-druck-stab-konstante-belastung-verschiebung-finite-elemente

3.2.1 Exakte Lösung durch DGL

Die Differentialgleichung und ihre Lösung lauten:

{N^\prime }+q = 0

EA{u^{\prime \prime }}+q = 0

\quad \Rightarrow \quad {u^{\prime \prime }} = -\frac{q}{{EA}}

\quad \Rightarrow \quad u\left( x \right) = -\frac{q}{{2EA}}{x^2}+{C_1}x+{C_2}

Randbedingungen:

u\left( 0 \right) = 0\quad \Rightarrow \quad {C_2} = 0

N\left( L \right) = EA{u^\prime }\left( L \right) = -qL+EA{C_1}\mathop = \limits^! 0\quad \Rightarrow \quad {C_1} = \frac{{qL}}{{EA}}

Die Gleichung für u lautet also:

u\left( x \right) = \frac{{q{L^2}}}{{2EA}}\left( {2\frac{x}{L}-{{\left( {\frac{x}{L}} \right)}^2}} \right)

Die Gleichung für die Normalkraft lautet:

N\left( x \right) = qL\left( {1-\frac{x}{L}} \right)

3.2.2 Approximation durch ein lineares Element

approximation-ein-finites-element-zug-druck-stab

Es ergibt sich:

\underbrace {\left[ {{m^e}} \right]\left\{{\ddot{\hat{u}}} \right\}}_{ = 0}+\left[ {{k^e}} \right]\left\{{\hat u} \right\} = \left\{ r \right\}

\left[ {{k^e}} \right] = \frac{{EA}}{L}\left[ {\begin{array}{*{20}{c}} 1&{-1} \\ {-1}&1 \end{array}} \right]

\left\{ r \right\} = \left\{{{F^*}} \right\}+\int_0^l {{{\left[ H \right]}^T}\left[ H \right]dx} \left\{{\hat q} \right\}

\quad = \left\{{\begin{array}{*{20}{c}}{F_1^*} \\ {F_2^*} \end{array}} \right\}+\int_0^L {\left[ {\begin{array}{*{20}{c}}{1-\frac{x}{l}} \\ {\frac{x}{l}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}}{1-\frac{x}{l}}&{\frac{x}{l}} \end{array}} \right]dx} \left\{{\hat q} \right\}

\quad = \left\{{\begin{array}{*{20}{c}}{F_1^*} \\ {F_2^*} \end{array}} \right\}+\frac{l}{6}\left\{{\begin{array}{*{20}{c}}{2{q_1}+{q_2}} \\ {{q_1}+2{q_2}} \end{array}} \right\}\mathop = \limits^{{q_1} = {q_2} = q} \left\{{\begin{array}{*{20}{c}}{F_1^*} \\ {F_2^*} \end{array}} \right\}+\frac{{qL}}{2}\left\{{\begin{array}{*{20}{c}} 1 \\ 1 \end{array}} \right\}

\quad \Rightarrow \quad \frac{{EA}}{L}\left[ {\begin{array}{*{20}{c}} 1&{-1} \\ {-1}&1 \end{array}} \right]\left\{{\begin{array}{*{20}{c}}{{u_1}} \\ {{u_2}} \end{array}} \right\} = \left\{{\begin{array}{*{20}{c}}{F_1^*} \\ {F_2^*} \end{array}} \right\}+\frac{{qL}}{2}\left\{{\begin{array}{*{20}{c}} 1 \\ 1 \end{array}} \right\}

Randbedingungen:

{u_1} = 0,\quad F_2^* = 0

\quad \Rightarrow \quad \frac{{EA}}{L}\left[ {\begin{array}{*{20}{c}} 1&{-1} \\ {-1}&1 \end{array}} \right]\left\{{\begin{array}{*{20}{c}} 0 \\ {{u_2}} \end{array}} \right\} = \left\{{\begin{array}{*{20}{c}}{F_1^*} \\ 0 \end{array}} \right\}+\frac{{qL}}{2}\left\{{\begin{array}{*{20}{c}} 1 \\ 1 \end{array}} \right\}\quad \quad \begin{array}{*{20}{c}}{\left( 1 \right)} \\ {\left( 2 \right)} \end{array}

\quad \mathop \Rightarrow \limits^{\left( 2 \right)} \quad \frac{{EA}}{L}{u_2} = \frac{{qL}}{2}\quad \Rightarrow \quad {u_2} = \frac{{{L^2}q}}{{2EA}}

\quad \mathop \Rightarrow \limits^{\left( 1 \right)} \quad -\frac{{EA}}{L}{u_2} = F_1^*+\frac{{qL}}{2}\quad \Rightarrow \quad F_1^* = -\frac{{qL}}{2}-\frac{{EA}}{L}\frac{{{L^2}q}}{{2EA}} = -\frac{{qL}}{2}-\frac{{Lq}}{2} = -qL

{u^{\left( 1 \right)}}\left( x \right) = \left[ H \right]\left\{{\hat u} \right\} = \left[ {\begin{array}{*{20}{c}}{1-\frac{x}{L}}&{\frac{x}{L}} \end{array}} \right]\left\{{\begin{array}{*{20}{c}}{{u_1}} \\ {{u_2}} \end{array}} \right\} = \frac{{{L^2}q}}{{2EA}}\frac{x}{L}

{N^{\left( 1 \right)}}\left( x \right) = EA\left[ B \right]\left\{{\hat u} \right\} = EA\left[ {\begin{array}{*{20}{c}}{-\frac{1}{L}}&{\frac{1}{L}} \end{array}} \right]\left\{{\begin{array}{*{20}{c}}{{u_1}} \\ {{u_2}} \end{array}} \right\} = \frac{{qL}}{2}

3.2.3 Approximation durch zwei lineare Elemente

approximation-zwei-finite-elemente-zug-druck-stab

Randbedingungen:

{u_1} = 0,\quad F_2^* = F_3^* = 0

Die Elementsteifigkeitsmatrizen der beiden Elemente sind identisch:

\left[ {k_{\left( 1 \right)}^e} \right] = \left[ {k_{\left( 2 \right)}^e} \right] = \frac{{2EA}}{L}\left[ {\begin{array}{*{20}{c}} 1&{-1} \\ {-1}&1 \end{array}} \right]

Die Flächenlasten werden gleichmäßig auf die Knoten verteilt. Auf den mittleren Knoten wirken daher die Anteile des linken und des rechten Stabes. Es ergeben sich die Gesamtsteifigkeitsmatrix und die Bewegungsgleichungen:

\frac{{2EA}}{L}\left[ {\begin{array}{*{20}{c}} 1&{-1}&0 \\ {-1}&2&{-1} \\ 0&{-1}&1 \end{array}} \right]\left\{{\begin{array}{*{20}{c}}{{u_1}} \\ {{u_2}} \\ {{u_3}} \end{array}} \right\} = \left\{{\begin{array}{*{20}{c}}{F_1^*} \\ {F_2^*} \\ {F_3^*} \end{array}} \right\}+\left\{{\begin{array}{*{20}{c}}{\frac{{qL}}{4}} \\ {\frac{{qL}}{4}+\frac{{qL}}{4}} \\ {\frac{{qL}}{4}} \end{array}} \right\}

Einsetzen der Randbedingungen liefert die Verschiebungen {u_2} und {u_3}:

\frac{{2EA}}{L}\left[ {\begin{array}{*{20}{c}} 1&{-1}&0 \\ {-1}&2&{-1} \\ 0&{-1}&1 \end{array}} \right]\left\{{\begin{array}{*{20}{c}} 0 \\ {{u_2}} \\ {{u_3}} \end{array}} \right\} = \left\{{\begin{array}{*{20}{c}}{F_1^*} \\ 0 \\ 0 \end{array}} \right\}+\left\{{\begin{array}{*{20}{c}}{\frac{{qL}}{4}} \\ {\frac{{qL}}{4}+\frac{{qL}}{4}} \\ {\frac{{qL}}{4}} \end{array}} \right\}\quad \quad \quad \begin{array}{*{20}{c}}{\left( 1 \right)} \\ {\left( 2 \right)} \\ {\left( 3 \right)} \end{array}

\quad \mathop \Rightarrow \limits^{\left( 2 \right)} \quad \frac{{2EA}}{L}\left( {2{u_2}-{u_3}} \right) = \frac{{qL}}{2}\quad \Rightarrow \quad \frac{{2EA}}{L}2{u_2}-\frac{{qL}}{2} = \frac{{2EA}}{L}{u_3}

\quad \Rightarrow \quad {u_3} = 2{u_2}-\frac{{q{L^2}}}{{4EA}}

\quad \mathop \Rightarrow \limits^{\left( 3 \right)} \quad \frac{{2EA}}{L}\left( {-{u_2}+{u_3}} \right) = \frac{{qL}}{4}\quad \Rightarrow \quad {u_2} = -\frac{{q{L^2}}}{{8EA}}+{u_3}

\quad \Rightarrow \quad {u_2} = -\frac{{q{L^2}}}{{8EA}}+{u_3} = -\frac{{q{L^2}}}{{8EA}}+2{u_2}-\frac{{q{L^2}}}{{4EA}}\quad \Rightarrow \quad {u_2} = \frac{{q{L^2}+2q{L^2}}}{{8EA}} = \frac{{3q{L^2}}}{{8EA}}

\quad \Rightarrow \quad {u_3} = 2{u_2}-\frac{{q{L^2}}}{{4EA}} = \frac{{6q{L^2}}}{{8EA}}-\frac{{q{L^2}}}{{4EA}} = \frac{{3q{L^2}}}{{4EA}}-\frac{{q{L^2}}}{{4EA}} = \frac{{q{L^2}}}{{2EA}}

Für das erste Element ergibt sich also das Verschiebungsfeld:

{u^{\left( 1 \right)}}\left( x \right) = \left[ H \right]\left\{{\begin{array}{*{20}{c}}{{u_1}} \\ {{u_2}} \end{array}} \right\} = \left[ {\begin{array}{*{20}{c}}{1-\frac{x}{{\frac{L}{2}}}}&{\frac{x}{{\frac{L}{2}}}} \end{array}} \right]\left\{{\begin{array}{*{20}{c}} 0 \\ {\frac{3}{8}} \end{array}} \right\}\frac{{q{L^2}}}{{EA}} = \frac{3}{4}\frac{{q{L^2}}}{{EA}}\frac{x}{L}

Und für das zweite Element:

{u^{\left( 2 \right)}}\left( x \right) = \left[ {\begin{array}{*{20}{c}}{1-\frac{{2x}}{L}}&{\frac{{2x}}{L}} \end{array}} \right]\left\{{\begin{array}{*{20}{c}}{\frac{3}{8}} \\ {\frac{1}{2}} \end{array}} \right\}\frac{{q{L^2}}}{{EA}} = \frac{{q{L^2}}}{{EA}}\left( {\frac{3}{8}-\frac{3}{4}\frac{x}{L}+\frac{x}{L}} \right) = \frac{{q{L^2}}}{{EA}}\left( {\frac{3}{8}+\frac{x}{{4L}}} \right)

Im nächsten Schritt berechnen wir die Normalkräfte. Für das erste Element gilt:

{N^{\left( 1 \right)}}\left( x \right) = EA\left[ B \right]\left\{{\begin{array}{*{20}{c}}{{u_1}} \\ {{u_2}} \end{array}} \right\} = EA\left[ {\begin{array}{*{20}{c}}{-\frac{2}{L}}&{\frac{2}{L}} \end{array}} \right]\left\{{\begin{array}{*{20}{c}} 0 \\ {\frac{3}{8}} \end{array}} \right\}\frac{{q{L^2}}}{{EA}} = \frac{3}{4}qL

Und für das zweite Element:

{N^{\left( 2 \right)}}\left( x \right) = EA\left[ B \right]\left\{{\begin{array}{*{20}{c}}{{u_2}} \\ {{u_3}} \end{array}} \right\} = EA\left[ {\begin{array}{*{20}{c}}{-\frac{2}{L}}&{\frac{2}{L}} \end{array}} \right]\left\{{\begin{array}{*{20}{c}}{\frac{3}{8}} \\ {\frac{1}{2}} \end{array}} \right\}\frac{{q{L^2}}}{{EA}} = \frac{1}{4}qL

Zuletzt berechnen wir noch die Reaktionskräfte:

\quad \mathop \Rightarrow \limits^{\left( 1 \right)} \quad -\frac{{2EA}}{L}{u_2} = F_1^*+\frac{{qL}}{4}\quad \Rightarrow \quad F_1^* = -\frac{{2EA}}{L}{u_2}-\frac{{qL}}{4} = -\frac{{3qL}}{4}-\frac{{qL}}{4} = -qL

3.2.4 Approximation durch drei lineare Elemente

approximation-drei-finite-elemente-zug-druck-stab

Randbedingungen:

{u_1} = 0,\quad F_2^* = F_3^* = F_4^* = 0

Elementsteifigkeitsmatrizen:

\left[ {k_{\left( 1 \right)}^e} \right] = \left[ {k_{\left( 2 \right)}^e} \right] = \left[ {k_{\left( 3 \right)}^e} \right] = \frac{{3EA}}{L}\left[ {\begin{array}{*{20}{c}} 1&{-1} \\ {-1}&1 \end{array}} \right]

Indextafel:

\begin{array}{*{20}{c}}{Element\:Nr.}&\vline & {u_1^*}&\vline & {u_2^*} \\ \hline 1&\vline & 1&\vline & 2 \\ 2&\vline & 2&\vline & 3 \\ 3&\vline & 3&\vline & 4 \end{array}

Zusammensetzen der Gesamtsteifigkeitsmatrix über die Indextafel:

\left[ K \right] = \frac{{3EA}}{L}\left[ {\begin{array}{*{20}{c}} 1&{-1}&0&0 \\ {-1}&{1+1}&{-1}&0 \\ 0&{-1}&{1+1}&{-1} \\ 0&0&{-1}&1 \end{array}} \right] = \frac{{3EA}}{L}\left[ {\begin{array}{*{20}{c}} 1&{-1}&{}&{} \\ {-1}&2&{-1}&{} \\ {}&{-1}&2&{-1} \\ {}&{}&{-1}&1 \end{array}} \right]

Bewegungsgleichungen:

\underbrace {\left[ M \right]\left\{{\ddot{\hat{u}}} \right\}}_{ = 0}+\left[ K \right]\left\{{\hat u} \right\} = \left\{ R \right\}

Wir brauchen noch das \left\{ R \right\}. Dafür müssen wir die Kraft q auf die Knoten verteilen. Jedes der drei Elemente bekommt 1/3 der Kraft. Jeder der beiden Knoten eines Elements bekommt davon die Hälfte, also 1/6. Die beiden mittleren Knoten bekommen so Anteile von zwei angrenzenden Elementen. Es ergibt sich:

\left\{ R \right\} = \left\{{\begin{array}{*{20}{c}}{F_1^*} \\ {F_2^*} \\ {F_3^*} \\ {F_4^*} \end{array}} \right\}+\left\{{\begin{array}{*{20}{c}}{\frac{{qL}}{6}} \\ {\frac{{qL}}{6}+\frac{{qL}}{6}} \\ {\frac{{qL}}{6}+\frac{{qL}}{6}} \\ {\frac{{qL}}{6}} \end{array}} \right\} = \left\{{\begin{array}{*{20}{c}}{F_1^*} \\ {F_2^*} \\ {F_3^*} \\ {F_4^*} \end{array}} \right\}+\frac{{qL}}{6}\left\{{\begin{array}{*{20}{c}} 1 \\ 2 \\ 2 \\ 1 \end{array}} \right\}

Einsetzen in die Bewegungsgleichungen:

\left[ K \right]\left\{{\hat u} \right\} = \left\{ R \right\}

\quad \Rightarrow \quad \frac{{3EA}}{L}\left[ {\begin{array}{*{20}{c}} 1&{-1}&{}&{} \\ {-1}&2&{-1}&{} \\ {}&{-1}&2&{-1} \\ {}&{}&{-1}&1 \end{array}} \right]\left\{{\begin{array}{*{20}{c}}{{u_1}} \\ {{u_2}} \\ {{u_3}} \\ {{u_4}} \end{array}} \right\} = \left\{{\begin{array}{*{20}{c}}{F_1^*} \\ {F_2^*} \\ {F_3^*} \\ {F_4^*} \end{array}} \right\}+\frac{{qL}}{6}\left\{{\begin{array}{*{20}{c}} 1 \\ 2 \\ 2 \\ 1 \end{array}} \right\}

Einsetzen der Randbedingungen:

\frac{{3EA}}{L}\left[ {\begin{array}{*{20}{c}} 1&{-1}&{}&{} \\ {-1}&2&{-1}&{} \\ {}&{-1}&2&{-1} \\ {}&{}&{-1}&1 \end{array}} \right]\left\{{\begin{array}{*{20}{c}} 0 \\ {{u_2}} \\ {{u_3}} \\ {{u_4}} \end{array}} \right\} = \left\{{\begin{array}{*{20}{c}}{F_1^*} \\ 0 \\ 0 \\ 0 \end{array}} \right\}+\frac{{qL}}{6}\left\{{\begin{array}{*{20}{c}} 1 \\ 2 \\ 2 \\ 1 \end{array}} \right\}\quad \quad \quad \begin{array}{*{20}{c}}{\left( 1 \right)} \\ {\left( 2 \right)} \\ {\left( 3 \right)} \\ {\left( 4 \right)} \end{array}

Wir könnten nun Gleichungen (2) bis (4) miteinander verrechnen, um die drei Verschiebungen zu bestimmen. Bei größeren Gleichungssystemen wird dies unübersichtlich, daher wollen wir in diesem Beispiel lösen, indem wir das reduzierte Gleichungssystem in Matrixschreibweise aufschreiben und die reduzierte Elementsteifigkeitsmatrix invertieren:

\frac{{3EA}}{L}\left[ {\begin{array}{*{20}{c}} 2&{-1}&0 \\ {-1}&2&{-1} \\ 0&{-1}&1 \end{array}} \right]\left\{{\begin{array}{*{20}{c}}{{u_2}} \\ {{u_3}} \\ {{u_4}} \end{array}} \right\} = \frac{{qL}}{6}\left\{{\begin{array}{*{20}{c}} 2 \\ 2 \\ 1 \end{array}} \right\}

\quad \Rightarrow \quad \underbrace {\left[ {\begin{array}{*{20}{c}} 2&{-1}&0 \\ {-1}&2&{-1} \\ 0&{-1}&1 \end{array}} \right]}_{\left[ {{K_R}} \right]}\left\{{\begin{array}{*{20}{c}}{{u_2}} \\ {{u_3}} \\ {{u_4}} \end{array}} \right\} = \frac{{q{L^2}}}{{18EA}}\left\{{\begin{array}{*{20}{c}} 2 \\ 2 \\ 1 \end{array}} \right\}

{\left[ {{K_R}} \right]^{-1}} = \frac{1}{{4-2-1}}{\left[ {\begin{array}{*{20}{c}}{+\left( {2-1} \right)}&{-\left( {-1} \right)}&{+1} \\ {-\left( {-1} \right)}&{+2}&{-\left( {-2} \right)} \\ {+1}&{-\left( {-2} \right)}&{+\left( {4-1} \right)} \end{array}} \right]^T} = \left[ {\begin{array}{*{20}{c}} 1&1&1 \\ 1&2&2 \\ 1&2&3 \end{array}} \right]

\quad \Rightarrow \quad \left\{{\begin{array}{*{20}{c}}{{u_2}} \\ {{u_3}} \\ {{u_4}} \end{array}} \right\} = \frac{{q{L^2}}}{{18EA}}\left[ {\begin{array}{*{20}{c}} 1&1&1 \\ 1&2&2 \\ 1&2&3 \end{array}} \right]\left\{{\begin{array}{*{20}{c}} 2 \\ 2 \\ 1 \end{array}} \right\}

\quad \Rightarrow \quad {u_2} = \frac{{5q{L^2}}}{{18EA}},\quad {u_3} = \frac{{4q{L^2}}}{{9EA}},\quad {u_4} = \frac{{q{L^2}}}{{2EA}}

Nun fehlt nur noch die Reaktionskraft. Diese erhalten wir aus Gleichung (1):

\frac{{3EA}}{L} \cdot \left( {-{u_2}} \right) = F_1^*+\frac{{qL}}{6}

\quad \Rightarrow \quad F_1^* = -\frac{{qL}}{6}-\frac{{3EA}}{L}{u_2} = -\frac{{qL}}{6}-\frac{{3EA}}{L}\frac{{5q{L^2}}}{{18EA}} = -qL

Jetzt bestimmen wir die Verschiebungsfelder:

{u^{\left( 1 \right)}}\left( x \right) = \left[ H \right]\left\{{\begin{array}{*{20}{c}}{{u_1}} \\ {{u_2}} \end{array}} \right\} = \left[ {\begin{array}{*{20}{c}}{1-\frac{x}{{\frac{L}{3}}}}&{\frac{x}{{\frac{L}{3}}}} \end{array}} \right]\left\{{\begin{array}{*{20}{c}} 0 \\ {\frac{5}{{18}}} \end{array}} \right\}\frac{{q{L^2}}}{{EA}} = \frac{5}{6}\frac{{qL}}{{EA}}x,\quad 0 \leq x \leq \frac{L}{3}

{u^{\left( 2 \right)}}\left( x \right) = \left[ H \right]\left\{{\begin{array}{*{20}{c}}{{u_2}} \\ {{u_3}} \end{array}} \right\} = \left[ {\begin{array}{*{20}{c}}{1-\frac{x}{{\frac{L}{3}}}}&{\frac{x}{{\frac{L}{3}}}} \end{array}} \right]\left\{{\begin{array}{*{20}{c}}{\frac{5}{{18}}} \\ {\frac{4}{9}} \end{array}} \right\}\frac{{q{L^2}}}{{EA}}

= \left[ {\left( {1-\frac{x}{{\frac{L}{3}}}} \right)\frac{5}{{18}}+\frac{x}{{\frac{L}{3}}}\frac{4}{9}} \right]\frac{{q{L^2}}}{{EA}}

\quad \quad = \left[ {\frac{5}{{18}}-\frac{{5x}}{{6L}}+\frac{{4x}}{{3L}}} \right]\frac{{q{L^2}}}{{EA}} = \left[ {\frac{5}{{18}}-\frac{{13x}}{{6L}}} \right]\frac{{q{L^2}}}{{EA}},\quad \quad 0 \leq x \leq \frac{L}{3}

{u^{\left( 3 \right)}}\left( x \right) = \left[ H \right]\left\{{\begin{array}{*{20}{c}}{{u_3}} \\ {{u_4}} \end{array}} \right\} = \left[ {\begin{array}{*{20}{c}}{1-\frac{x}{{\frac{L}{3}}}}&{\frac{x}{{\frac{L}{3}}}} \end{array}} \right]\left\{{\begin{array}{*{20}{c}}{\frac{4}{9}} \\ {\frac{1}{2}} \end{array}} \right\}\frac{{q{L^2}}}{{EA}}

= \left[ {\left( {1-\frac{x}{{\frac{L}{3}}}} \right)\frac{4}{9}+\frac{x}{{\frac{L}{3}}}\frac{1}{2}} \right]\frac{{q{L^2}}}{{EA}}

\quad \quad = \left[ {\frac{4}{9}-\frac{{4x}}{{3L}}+\frac{{3x}}{{2L}}} \right]\frac{{q{L^2}}}{{EA}} = \left[ {\frac{4}{9}+\frac{{1x}}{{6L}}} \right]\frac{{q{L^2}}}{{EA}},\quad \quad 0 \leq x \leq \frac{L}{3}

Für die Normalkraftfelder in den Elementen folgt:

{N^{\left( 1 \right)}}\left( x \right) = EA\left[ B \right]\left\{{\begin{array}{*{20}{c}}{{u_1}} \\ {{u_2}} \end{array}} \right\} = EA\left[ {\begin{array}{*{20}{c}}{-\frac{3}{L}}&{\frac{3}{L}} \end{array}} \right]\left\{{\begin{array}{*{20}{c}} 0 \\ {\frac{5}{{18}}} \end{array}} \right\}\frac{{q{L^2}}}{{EA}} = \frac{5}{6}qL

{N^{\left( 2 \right)}}\left( x \right) = EA\left[ B \right]\left\{{\begin{array}{*{20}{c}}{{u_2}} \\ {{u_3}} \end{array}} \right\} = EA\left[ {\begin{array}{*{20}{c}}{-\frac{3}{L}}&{\frac{3}{L}} \end{array}} \right]\left\{{\begin{array}{*{20}{c}}{\frac{5}{{18}}} \\ {\frac{4}{9}} \end{array}} \right\}\frac{{q{L^2}}}{{EA}} = \frac{1}{2}qL

{N^{\left( 3 \right)}}\left( x \right) = EA\left[ B \right]\left\{{\begin{array}{*{20}{c}}{{u_3}} \\ {{u_4}} \end{array}} \right\} = EA\left[ {\begin{array}{*{20}{c}}{-\frac{3}{L}}&{\frac{3}{L}} \end{array}} \right]\left\{{\begin{array}{*{20}{c}}{\frac{4}{9}} \\ {\frac{1}{2}} \end{array}} \right\}\frac{{q{L^2}}}{{EA}} = \frac{1}{6}qL