Gelöste Aufgaben/StaF

Aufgabenstellung

Wir untersuchen die Belastung eines ebenen Stabwerks. Die Stäbe haben wie skizziert die Länge ℓ bzw. ℓ/2. Die Struktur wird mit der Kraft F belastet.

Caption

Gesucht ist ein Vergleich zwischen der klassischen Stabwerkstheorie und einer Herangehensweise, bei der wir eine feste Verbindung der Stäbe in den Knoten ansetzten. Grundlage des Modells ist die FEM-Lösung der Felddifferentialgleichung im Vergleich zur Lösung in Problemstellung „Stab“.

Wir stellen das Modell des Stabwerks mit dem Prinzip der virtuellen Verrückungen auf und vergleichen, wie sich diese von der Herangehensweise aus „Stab“ mit der analytischen Lösung unterscheidet.

Lösung mit Maxima

Wir nutzen das Computer-Algebra-System Maxima zur Lösung. Das macht hier Sinn, weil wir die Herangehensweise mit der aus Stab vergleichen wollen – für die wir ebenfalls Maxima eingesetzt haben.

Declarations

Wir übernehmen alle Vereinbarungen und Parameter aus der Problemformulierung „Stab“.

Gleichgewichtsbedingungen

Für die Gleichgewichtsbedingung nach dem Prinzip der virtuellen Verrückungen

${\begin{array}{ccc}\delta W&{\stackrel {!}{=}}&0\\&=&\delta \Pi -\delta W^{a}\end{array}}$

benötigen wir die virtuelle Formänderungsenergie $\delta \Pi$ und die virtuelle Arbeit der äußeren Kraft $\delta W^{a}$ der äußeren Kräfte und Momente.

Mit den Konventionen für die Knoten-Verschiebungen aus Stab ist

$\delta W^{a}=-\delta W_{4,0}\cdot F$.

Für $\delta \Pi$ gilt

$\delta \Pi =\sum _{i=0}^{4}\delta \Pi _{i}$

mit den virtuellen Formänderungsarbeiten der vier Stäbe.

Dabei haben wir Anteile der Arbeit aus der [Biegung] und der Längs-Dehnung des Stabes.

Für den Stab k mit den Knoten I und J haben wir als Koodinaten der Knoten

$U_{I,k},W_{I,k},\Phi _{I,k}$ und $U_{J,k},W_{J,k},\Phi _{J,k}$.

Damit haben wir

$\delta \Pi _{k}=\left(\delta W_{I,k},\delta \Phi _{I,k},\delta \Phi _{J,k},\delta W_{J,k}\right)\cdot {\frac {EI}{\ell _{i}^{3}}}\cdot {\begin{pmatrix}12&6\,{\ell _{i}}&-12&6\,{\ell _{i}}\\6\,{\ell _{i}}&4\,{\ell _{i}^{2}}&-6\,{\ell _{i}}&2\,{\ell _{i}^{2}}\\-12&-6\,{\ell _{i}}&12&-6\,{\ell _{i}}\\6\,{\ell _{i}}&2\,{\ell _{i}^{2}}&-6\,{\ell _{i}}&4\,{\ell _{i}^{2}}\end{pmatrix}}\cdot \left({\begin{array}{c}W_{I,k}\\\Phi _{I,k}\\W_{J,k}\\\Phi _{J,k}\end{array}}\right)+\left(\delta U_{I,k},\delta U_{J,k}\right)\cdot {\frac {EA}{\ell _{i}}}\cdot {\begin{pmatrix}1&-1\\-1&1\end{pmatrix}}\cdot \left({\begin{array}{c}U_{I,k}\\U_{J,k}\end{array}}\right)$

Für den Stab k definieren wir

${\underline {Q}}_{k,k}=\left({\begin{array}{c}U_{I,k}\\W_{I,k}\\\Phi _{I,k}\\U_{J,k}\\W_{J,k}\\\Phi _{J,k}\\\end{array}}\right)$ sowie $\delta {\underline {Q}}_{k,k}=\left({\begin{array}{c}\delta U_{I,k}\\\delta W_{I,k}\\\delta \Phi _{I,k}\\\delta U_{J,k}\\\delta W_{J,k}\\\delta \Phi _{J,k}\\\end{array}}\right)$

und finden damit

$\delta \Pi _{k}=\delta {\underline {Q}}_{k,k}^{T}\cdot {\underline {\underline {K}}}_{k,k}\cdot {\underline {Q}}_{k,k}$

mit der Element-Steifigkeitsmatrix des Elements k im k-Koordinatensystem

${\underline {\underline {K}}}_{k,k}={\frac {EI}{\ell _{i}^{3}}}\cdot {\begin{pmatrix}0&0&0&0&0&0\\0&12&6\,{\ell _{i}}&0&-12&6\,{\ell _{i}}\\0&6\,{\ell _{i}}&4\,{\ell _{i}^{2}}&0&-6\,{\ell _{i}}&2\,{\ell _{i}^{2}}\\0&0&0&0&0&0\\0&-12&-6\,{\ell _{i}}&0&12&-6\,{\ell _{i}}\\0&6\,{\ell _{i}}&2\,{\ell _{i}^{2}}&-0&6\,{\ell _{i}}&4\,{\ell _{i}^{2}}\end{pmatrix}}+{\frac {EA}{\ell _{i}}}\cdot {\begin{pmatrix}1&0&0&-1&0&0\\0&0&0&0&0&0\\0&0&0&0&0&0\\-1&0&0&1&0&0\\0&0&0&0&0&0\\0&0&0&0&0&0\end{pmatrix}}$

Transformation der Koordinaten in das globale System

In den Ausdrücken der virtuellen Formänderungsenergie stehen die Koordinaten des lokalen Koordinatensystems von k. Die müssen wir, wie in Stab mit der Euler-Drehmatrix ineinander überführen.

Dafür haben wir

$\left({\begin{array}{c}U_{I,0}\\W_{I,0}\\\Phi _{I,0}\end{array}}\right)={\underline {\underline {D}}}_{R}(\alpha )\cdot \left({\begin{array}{c}U_{I,k}\\W_{I,k}\\\Phi _{I,k}\end{array}}\right)$

mit der Transformationsmatrix

${\underline {\underline {D}}}_{R}(\alpha )=\left({\begin{array}{c}\cos(\alpha )&\sin(\alpha )&0\\-\sin(\alpha )&\cos(\alpha )&0\\0&0&1\end{array}}\right)$

Damit wir für die Elementsteifigkeitsmatrix - mit beiden Anfangs- und Endknoten des Elements - vom "0"-System ins "k"-System transformieren, brauchen wir die neue Matrix

${\underline {\underline {D}}}_{T}(\alpha )=\left({\begin{array}{c}{\underline {\underline {D}}}_{R}^{T}(\alpha )&{\underline {\underline {0}}}\\{\underline {\underline {0}}}&{\underline {\underline {D}}}_{R}^{T}(\alpha )\end{array}}\right)$

Mit diesen Transformationsmatrizen ist

$\delta \Pi =\sum _{k=1}^{4}\delta {\underline {Q}}_{k,0}^{T}\cdot \underbrace {{\underline {\underline {D}}}_{T}^{T}(\alpha _{k})\cdot {\underline {\underline {K}}}_{k,k}\cdot {\underline {\underline {D}}}_{T}(\alpha _{k})} _{:={\underline {\underline {K}}}_{k,0}}\cdot {\underline {Q}}_{k,0}$

Wir sammeln alle Koordianten der Knoten im 0-System in ${\underline {Q}}_{0}$ und schreiben die Gleichgewichtsbedingungen in der Form

$\delta {\underline {Q}}_{0}^{T}\cdot \left({\underline {\underline {K}}}_{0}\cdot {\underline {Q}}_{0}-{\underline {P}}\right)=0$

an. Die Randbedingungen arbeiten wir hier durch das Streichen der passenden Zeilen für $\delta U_{1,0},\delta W_{1,0},\delta U_{2,0},\delta W_{2,0}$ sowie der passenden Spalten für $U_{1,0},W_{1,0},U_{2,0},W_{2,0}$ ein.

Das resultierende Gleichungssystem ist dies:

${\begin{pmatrix}{\frac {2{{A}^{2}}E\eta }{3{\ell _{0}}}}&0&0&-\left({\frac {2{{A}^{2}}E\eta }{{\ell }_{0}^{2}}}\right)&{\frac {{{A}^{2}}E\eta }{3{\ell _{0}}}}&0&0&0\\0&{\frac {2{{A}^{2}}E\eta }{3{\ell _{0}}}}&-\left({\frac {{\sqrt {3}}{{A}^{2}}E\eta }{4{{\ell }_{0}^{2}}}}\right)&-\left({\frac {{{A}^{2}}E\eta }{4{{\ell }_{0}^{2}}}}\right)&{\frac {{{A}^{2}}E\eta }{6{\ell _{0}}}}&0&-\left({\frac {{{A}^{2}}E\eta }{2{{\ell }_{0}^{2}}}}\right)&{\frac {{{A}^{2}}E\eta }{6{\ell _{0}}}}\\0&-\left({\frac {{\sqrt {3}}{{A}^{2}}E\eta }{4{{\ell }_{0}^{2}}}}\right)&{\frac {3{{A}^{2}}E\eta +{{\ell }_{0}^{2}}AE}{2{{\ell }_{0}^{3}}}}+{\frac {2AE}{\ell _{0}}}&0&-\left({\frac {{\sqrt {3}}{{A}^{2}}E\eta }{2{{\ell }_{0}^{2}}}}\right)&-\left({\frac {3{{A}^{2}}E\eta +{{\ell }_{0}^{2}}AE}{4{{\ell }_{0}^{3}}}}\right)&{\frac {{\sqrt {3}}{{A}^{2}}E\eta -{\sqrt {3}}{{\ell }_{0}^{2}}AE}{4{{\ell }_{0}^{3}}}}&-\left({\frac {{\sqrt {3}}{{A}^{2}}E\eta }{4{{\ell }_{0}^{2}}}}\right)\\-\left({\frac {2{{A}^{2}}E\eta }{{\ell }_{0}^{2}}}\right)&-\left({\frac {{{A}^{2}}E\eta }{4{{\ell }_{0}^{2}}}}\right)&0&{\frac {{{A}^{2}}E\eta +3{{\ell }_{0}^{2}}AE}{2{{\ell }_{0}^{3}}}}+{\frac {8{{A}^{2}}E\eta }{{\ell }_{0}^{3}}}&-\left({\frac {2{{A}^{2}}E\eta }{{\ell }_{0}^{2}}}\right)&{\frac {{\sqrt {3}}{{A}^{2}}E\eta -{\sqrt {3}}{{\ell }_{0}^{2}}AE}{4{{\ell }_{0}^{3}}}}&-\left({\frac {{{A}^{2}}E\eta +3{{\ell }_{0}^{2}}AE}{4{{\ell }_{0}^{3}}}}\right)&{\frac {{{A}^{2}}E\eta }{4{{\ell }_{0}^{2}}}}\\{\frac {{{A}^{2}}E\eta }{3{\ell _{0}}}}&{\frac {{{A}^{2}}E\eta }{6{\ell _{0}}}}&-\left({\frac {{\sqrt {3}}{{A}^{2}}E\eta }{2{{\ell }_{0}^{2}}}}\right)&-\left({\frac {2{{A}^{2}}E\eta }{{\ell }_{0}^{2}}}\right)&{\frac {4{{A}^{2}}E\eta }{3{\ell _{0}}}}&{\frac {{\sqrt {3}}{{A}^{2}}E\eta }{4{{\ell }_{0}^{2}}}}&-\left({\frac {{{A}^{2}}E\eta }{4{{\ell }_{0}^{2}}}}\right)&{\frac {{{A}^{2}}E\eta }{6{\ell _{0}}}}\\0&0&-\left({\frac {3{{A}^{2}}E\eta +{{\ell }_{0}^{2}}AE}{4{{\ell }_{0}^{3}}}}\right)&{\frac {{\sqrt {3}}{{A}^{2}}E\eta -{\sqrt {3}}{{\ell }_{0}^{2}}AE}{4{{\ell }_{0}^{3}}}}&{\frac {{\sqrt {3}}{{A}^{2}}E\eta }{4{{\ell }_{0}^{2}}}}&{\frac {3{{A}^{2}}E\eta +{{\ell }_{0}^{2}}AE}{4{{\ell }_{0}^{3}}}}+{\frac {AE}{\ell _{0}}}&-\left({\frac {{\sqrt {3}}{{A}^{2}}E\eta -{\sqrt {3}}{{\ell }_{0}^{2}}AE}{4{{\ell }_{0}^{3}}}}\right)&{\frac {{\sqrt {3}}{{A}^{2}}E\eta }{4{{\ell }_{0}^{2}}}}\\0&-\left({\frac {{{A}^{2}}E\eta }{2{{\ell }_{0}^{2}}}}\right)&{\frac {{\sqrt {3}}{{A}^{2}}E\eta -{\sqrt {3}}{{\ell }_{0}^{2}}AE}{4{{\ell }_{0}^{3}}}}&-\left({\frac {{{A}^{2}}E\eta +3{{\ell }_{0}^{2}}AE}{4{{\ell }_{0}^{3}}}}\right)&-\left({\frac {{{A}^{2}}E\eta }{4{{\ell }_{0}^{2}}}}\right)&-\left({\frac {{\sqrt {3}}{{A}^{2}}E\eta -{\sqrt {3}}{{\ell }_{0}^{2}}AE}{4{{\ell }_{0}^{3}}}}\right)&{\frac {{{A}^{2}}E\eta +3{{\ell }_{0}^{2}}AE}{4{{\ell }_{0}^{3}}}}+{\frac {{{A}^{2}}E\eta }{{\ell }_{0}^{3}}}&-\left({\frac {3{{A}^{2}}E\eta }{4{{\ell }_{0}^{2}}}}\right)\\0&{\frac {{{A}^{2}}E\eta }{6{\ell _{0}}}}&-\left({\frac {{\sqrt {3}}{{A}^{2}}E\eta }{4{{\ell }_{0}^{2}}}}\right)&{\frac {{{A}^{2}}E\eta }{4{{\ell }_{0}^{2}}}}&{\frac {{{A}^{2}}E\eta }{6{\ell _{0}}}}&{\frac {{\sqrt {3}}{{A}^{2}}E\eta }{4{{\ell }_{0}^{2}}}}&-\left({\frac {3{{A}^{2}}E\eta }{4{{\ell }_{0}^{2}}}}\right)&{\frac {2{{A}^{2}}E\eta }{3{\ell _{0}}}}\end{pmatrix}}\cdot {\begin{pmatrix}{{\Phi }_{1,0}}\\{{\Phi }_{2,0}}\\{W_{3,0}}\\{U_{3,0}}\\{{\Phi }_{3,0}}\\{W_{4,0}}\\{U_{4,0}}\\{{\Phi }_{4,0}}\end{pmatrix}}={\begin{pmatrix}0\\0\\0\\0\\0\\F\\0\\0\end{pmatrix}}$

Hier kommt jetzt irgendein Text.

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Element-Steigigkeitsmatrizen mit globalen Koordinaten
Element #1
${k_{1}}={\begin{pmatrix}{\frac {8{{A}^{2}}E\eta }{\ell _{0}^{3}}}&0&{\frac {2{{A}^{2}}E\eta }{\ell _{0}^{2}}}&-\left({\frac {8{{A}^{2}}E\eta }{\ell _{0}^{3}}}\right)&0&{\frac {2{{A}^{2}}E\eta }{\ell _{0}^{2}}}\\0&{\frac {2AE}{\ell _{0}}}&0&0&-\left({\frac {2AE}{\ell _{0}}}\right)&0\\{\frac {2{{A}^{2}}E\eta }{\ell _{0}^{2}}}&0&{\frac {2{{A}^{2}}E\eta }{3{\ell _{0}}}}&-\left({\frac {2{{A}^{2}}E\eta }{\ell _{0}^{2}}}\right)&0&{\frac {{{A}^{2}}E\eta }{3{\ell _{0}}}}\\-\left({\frac {8{{A}^{2}}E\eta }{\ell _{0}^{3}}}\right)&0&-\left({\frac {2{{A}^{2}}E\eta }{\ell _{0}^{2}}}\right)&{\frac {8{{A}^{2}}E\eta }{\ell _{0}^{3}}}&0&-\left({\frac {2{{A}^{2}}E\eta }{\ell _{0}^{2}}}\right)\\0&-\left({\frac {2AE}{\ell _{0}}}\right)&0&0&{\frac {2AE}{\ell _{0}}}&0\\{\frac {2{{A}^{2}}E\eta }{\ell _{0}^{2}}}&0&{\frac {{{A}^{2}}E\eta }{3{\ell _{0}}}}&-\left({\frac {2{{A}^{2}}E\eta }{\ell _{0}^{2}}}\right)&0&{\frac {2{{A}^{2}}E\eta }{3{\ell _{0}}}}\end{pmatrix}}$
Element #2
${k_{2}}={\begin{pmatrix}{\frac {{{A}^{2}}E\eta +3{\ell _{0}^{2}}AE}{4{\ell _{0}^{3}}}}&{\frac {{\sqrt {3}}{{A}^{2}}E\eta -{\sqrt {3}}{\ell _{0}^{2}}AE}{4{\ell _{0}^{3}}}}&{\frac {{{A}^{2}}E\eta }{4{\ell _{0}^{2}}}}&-\left({\frac {{{A}^{2}}E\eta +3{\ell _{0}^{2}}AE}{4{\ell _{0}^{3}}}}\right)&-\left({\frac {{\sqrt {3}}{{A}^{2}}E\eta -{\sqrt {3}}{\ell _{0}^{2}}AE}{4{\ell _{0}^{3}}}}\right)&{\frac {{{A}^{2}}E\eta }{4{\ell _{0}^{2}}}}\\{\frac {{\sqrt {3}}{{A}^{2}}E\eta -{\sqrt {3}}{\ell _{0}^{2}}AE}{4{\ell _{0}^{3}}}}&{\frac {3{{A}^{2}}E\eta +{\ell _{0}^{2}}AE}{4{\ell _{0}^{3}}}}&{\frac {{\sqrt {3}}{{A}^{2}}E\eta }{4{\ell _{0}^{2}}}}&-\left({\frac {{\sqrt {3}}{{A}^{2}}E\eta -{\sqrt {3}}{\ell _{0}^{2}}AE}{4{\ell _{0}^{3}}}}\right)&-\left({\frac {3{{A}^{2}}E\eta +{\ell _{0}^{2}}AE}{4{\ell _{0}^{3}}}}\right)&{\frac {{\sqrt {3}}{{A}^{2}}E\eta }{4{\ell _{0}^{2}}}}\\{\frac {{{A}^{2}}E\eta }{4{\ell _{0}^{2}}}}&{\frac {{\sqrt {3}}{{A}^{2}}E\eta }{4{\ell _{0}^{2}}}}&{\frac {{{A}^{2}}E\eta }{3{\ell _{0}}}}&-\left({\frac {{{A}^{2}}E\eta }{4{\ell _{0}^{2}}}}\right)&-\left({\frac {{\sqrt {3}}{{A}^{2}}E\eta }{4{\ell _{0}^{2}}}}\right)&{\frac {{{A}^{2}}E\eta }{6{\ell _{0}}}}\\-\left({\frac {{{A}^{2}}E\eta +3{\ell _{0}^{2}}AE}{4{\ell _{0}^{3}}}}\right)&-\left({\frac {{\sqrt {3}}{{A}^{2}}E\eta -{\sqrt {3}}{\ell _{0}^{2}}AE}{4{\ell _{0}^{3}}}}\right)&-\left({\frac {{{A}^{2}}E\eta }{4{\ell _{0}^{2}}}}\right)&{\frac {{{A}^{2}}E\eta +3{\ell _{0}^{2}}AE}{4{\ell _{0}^{3}}}}&{\frac {{\sqrt {3}}{{A}^{2}}E\eta -{\sqrt {3}}{\ell _{0}^{2}}AE}{4{\ell _{0}^{3}}}}&-\left({\frac {{{A}^{2}}E\eta }{4{\ell _{0}^{2}}}}\right)\\-\left({\frac {{\sqrt {3}}{{A}^{2}}E\eta -{\sqrt {3}}{\ell _{0}^{2}}AE}{4{\ell _{0}^{3}}}}\right)&-\left({\frac {3{{A}^{2}}E\eta +{\ell _{0}^{2}}AE}{4{\ell _{0}^{3}}}}\right)&-\left({\frac {{\sqrt {3}}{{A}^{2}}E\eta }{4{\ell _{0}^{2}}}}\right)&{\frac {{\sqrt {3}}{{A}^{2}}E\eta -{\sqrt {3}}{\ell _{0}^{2}}AE}{4{\ell _{0}^{3}}}}&{\frac {3{{A}^{2}}E\eta +{\ell _{0}^{2}}AE}{4{\ell _{0}^{3}}}}&-\left({\frac {{\sqrt {3}}{{A}^{2}}E\eta }{4{\ell _{0}^{2}}}}\right)\\{\frac {{{A}^{2}}E\eta }{4{\ell _{0}^{2}}}}&{\frac {{\sqrt {3}}{{A}^{2}}E\eta }{4{\ell _{0}^{2}}}}&{\frac {{{A}^{2}}E\eta }{6{\ell _{0}}}}&-\left({\frac {{{A}^{2}}E\eta }{4{\ell _{0}^{2}}}}\right)&-\left({\frac {{\sqrt {3}}{{A}^{2}}E\eta }{4{\ell _{0}^{2}}}}\right)&{\frac {{{A}^{2}}E\eta }{3{\ell _{0}}}}\end{pmatrix}}$
Element #3
${k_{3}}={\begin{pmatrix}{\frac {{{A}^{2}}E\eta +3{\ell _{0}^{2}}AE}{4{\ell _{0}^{3}}}}&-\left({\frac {{\sqrt {3}}{{A}^{2}}E\eta -{\sqrt {3}}{\ell _{0}^{2}}AE}{4{\ell _{0}^{3}}}}\right)&{\frac {{{A}^{2}}E\eta }{4{\ell _{0}^{2}}}}&-\left({\frac {{{A}^{2}}E\eta +3{\ell _{0}^{2}}AE}{4{\ell _{0}^{3}}}}\right)&{\frac {{\sqrt {3}}{{A}^{2}}E\eta -{\sqrt {3}}{\ell _{0}^{2}}AE}{4{\ell _{0}^{3}}}}&{\frac {{{A}^{2}}E\eta }{4{\ell _{0}^{2}}}}\\-\left({\frac {{\sqrt {3}}{{A}^{2}}E\eta -{\sqrt {3}}{\ell _{0}^{2}}AE}{4{\ell _{0}^{3}}}}\right)&{\frac {3{{A}^{2}}E\eta +{\ell _{0}^{2}}AE}{4{\ell _{0}^{3}}}}&-\left({\frac {{\sqrt {3}}{{A}^{2}}E\eta }{4{\ell _{0}^{2}}}}\right)&{\frac {{\sqrt {3}}{{A}^{2}}E\eta -{\sqrt {3}}{\ell _{0}^{2}}AE}{4{\ell _{0}^{3}}}}&-\left({\frac {3{{A}^{2}}E\eta +{\ell _{0}^{2}}AE}{4{\ell _{0}^{3}}}}\right)&-\left({\frac {{\sqrt {3}}{{A}^{2}}E\eta }{4{\ell _{0}^{2}}}}\right)\\{\frac {{{A}^{2}}E\eta }{4{\ell _{0}^{2}}}}&-\left({\frac {{\sqrt {3}}{{A}^{2}}E\eta }{4{\ell _{0}^{2}}}}\right)&{\frac {{{A}^{2}}E\eta }{3{\ell _{0}}}}&-\left({\frac {{{A}^{2}}E\eta }{4{\ell _{0}^{2}}}}\right)&{\frac {{\sqrt {3}}{{A}^{2}}E\eta }{4{\ell _{0}^{2}}}}&{\frac {{{A}^{2}}E\eta }{6{\ell _{0}}}}\\-\left({\frac {{{A}^{2}}E\eta +3{\ell _{0}^{2}}AE}{4{\ell _{0}^{3}}}}\right)&{\frac {{\sqrt {3}}{{A}^{2}}E\eta -{\sqrt {3}}{\ell _{0}^{2}}AE}{4{\ell _{0}^{3}}}}&-\left({\frac {{{A}^{2}}E\eta }{4{\ell _{0}^{2}}}}\right)&{\frac {{{A}^{2}}E\eta +3{\ell _{0}^{2}}AE}{4{\ell _{0}^{3}}}}&-\left({\frac {{\sqrt {3}}{{A}^{2}}E\eta -{\sqrt {3}}{\ell _{0}^{2}}AE}{4{\ell _{0}^{3}}}}\right)&-\left({\frac {{{A}^{2}}E\eta }{4{\ell _{0}^{2}}}}\right)\\{\frac {{\sqrt {3}}{{A}^{2}}E\eta -{\sqrt {3}}{\ell _{0}^{2}}AE}{4{\ell _{0}^{3}}}}&-\left({\frac {3{{A}^{2}}E\eta +{\ell _{0}^{2}}AE}{4{\ell _{0}^{3}}}}\right)&{\frac {{\sqrt {3}}{{A}^{2}}E\eta }{4{\ell _{0}^{2}}}}&-\left({\frac {{\sqrt {3}}{{A}^{2}}E\eta -{\sqrt {3}}{\ell _{0}^{2}}AE}{4{\ell _{0}^{3}}}}\right)&{\frac {3{{A}^{2}}E\eta +{\ell _{0}^{2}}AE}{4{\ell _{0}^{3}}}}&{\frac {{\sqrt {3}}{{A}^{2}}E\eta }{4{\ell _{0}^{2}}}}\\{\frac {{{A}^{2}}E\eta }{4{\ell _{0}^{2}}}}&-\left({\frac {{\sqrt {3}}{{A}^{2}}E\eta }{4{\ell _{0}^{2}}}}\right)&{\frac {{{A}^{2}}E\eta }{6{\ell _{0}}}}&-\left({\frac {{{A}^{2}}E\eta }{4{\ell _{0}^{2}}}}\right)&{\frac {{\sqrt {3}}{{A}^{2}}E\eta }{4{\ell _{0}^{2}}}}&{\frac {{{A}^{2}}E\eta }{3{\ell _{0}}}}\end{pmatrix}}$
Element #4
${k_{4}}={\begin{pmatrix}{\frac {{{A}^{2}}E\eta }{\ell _{0}^{3}}}&0&{\frac {{{A}^{2}}E\eta }{2{\ell _{0}^{2}}}}&-\left({\frac {{{A}^{2}}E\eta }{\ell _{0}^{3}}}\right)&0&{\frac {{{A}^{2}}E\eta }{2{\ell _{0}^{2}}}}\\0&{\frac {AE}{\ell _{0}}}&0&0&-\left({\frac {AE}{\ell _{0}}}\right)&0\\{\frac {{{A}^{2}}E\eta }{2{\ell _{0}^{2}}}}&0&{\frac {{{A}^{2}}E\eta }{3{\ell _{0}}}}&-\left({\frac {{{A}^{2}}E\eta }{2{\ell _{0}^{2}}}}\right)&0&{\frac {{{A}^{2}}E\eta }{6{\ell _{0}}}}\\-\left({\frac {{{A}^{2}}E\eta }{\ell _{0}^{3}}}\right)&0&-\left({\frac {{{A}^{2}}E\eta }{2{\ell _{0}^{2}}}}\right)&{\frac {{{A}^{2}}E\eta }{\ell _{0}^{3}}}&0&-\left({\frac {{{A}^{2}}E\eta }{2{\ell _{0}^{2}}}}\right)\\0&-\left({\frac {AE}{\ell _{0}}}\right)&0&0&{\frac {AE}{\ell _{0}}}&0\\{\frac {{{A}^{2}}E\eta }{2{\ell _{0}^{2}}}}&0&{\frac {{{A}^{2}}E\eta }{6{\ell _{0}}}}&-\left({\frac {{{A}^{2}}E\eta }{2{\ell _{0}^{2}}}}\right)&0&{\frac {{{A}^{2}}E\eta }{3{\ell _{0}}}}\end{pmatrix}}$

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