5.2 – Beispiel: Kragbalken mit finiten Stabelementen

 

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

5.2.1 Ein quadratisches Element, drei Knoten

quadratisch-drei-knoten-kragbalken-finite-stabelemente

Randbedingungen:

{U_1} = 0,\quad F_2^* = F_3^* = 0

Bewegungsgleichungen für {q_{x1}} = {q_{x2}} = {q_{x3}} = q:

\left[ {{k^e}} \right]\left\{{\hat U} \right\} = \left\{ r \right\}

\Rightarrow \frac{{EA}}{{3L}}\left[ {\begin{array}{*{20}{c}} 7&{-8}&1 \\ {-8}&{16}&{-8} \\ 1&{-8}&7 \end{array}} \right]\left\{{\begin{array}{*{20}{c}}{{U_1}} \\ {{U_2}} \\ {{U_3}} \end{array}} \right\} = \left\{{{F^*}} \right\}+\frac{L}{{30}}\left[ {\begin{array}{*{20}{c}} 4&2&{-1} \\ 2&{16}&2 \\ {-1}&2&4 \end{array}} \right]\left\{{\begin{array}{*{20}{c}}{{q_{x1}}} \\ {{q_{x2}}} \\ {{q_{x3}}} \end{array}} \right\}

\Rightarrow \quad \frac{{EA}}{{3L}}\left[ {\begin{array}{*{20}{c}} 7&{-8}&1 \\ {-8}&{16}&{-8} \\ 1&{-8}&7 \end{array}} \right]\left\{{\begin{array}{*{20}{c}}{{U_1}} \\ {{U_2}} \\ {{U_3}} \end{array}} \right\} = \left\{{\begin{array}{*{20}{c}}{F_1^*+\frac{1}{6}qL} \\ {F_2^*+\frac{2}{3}qL} \\ {F_3^*+\frac{1}{6}qL} \end{array}} \right\}

Wir müssen nur noch {U_2} und {U_3} bestimmen und reduzieren daher das Gleichungssystem nach dem Einsetzen der Randbedingungen:

\Rightarrow \quad \frac{{EA}}{{3L}}\left[ {\begin{array}{*{20}{c}} 7&{-8}&1 \\ {-8}&{16}&{-8} \\ 1&{-8}&7 \end{array}} \right]\left\{{\begin{array}{*{20}{c}} 0 \\ {{U_2}} \\ {{U_3}} \end{array}} \right\} = \left\{{\begin{array}{*{20}{c}}{F_1^*+\frac{1}{6}qL} \\ {\frac{2}{3}qL} \\ {\frac{1}{6}qL} \end{array}} \right\}

\Rightarrow \quad \frac{{EA}}{{3L}}\left[ {\begin{array}{*{20}{c}}{16}&{-8} \\ {-8}&7 \end{array}} \right]\left\{{\begin{array}{*{20}{c}}{{U_2}} \\ {{U_3}} \end{array}} \right\} = \left\{{\begin{array}{*{20}{c}}{\frac{2}{3}qL} \\ {\frac{1}{6}qL} \end{array}} \right\}

Berechnung der Verschiebungen durch Invertieren der Matrix:

\Rightarrow \quad \left\{{\begin{array}{*{20}{c}}{{U_2}} \\ {{U_3}} \end{array}} \right\} = \frac{{3L}}{{EA}}{\left[ {\begin{array}{*{20}{c}}{16}&{-8} \\ {-8}&7 \end{array}} \right]^{-1}}\left\{{\begin{array}{*{20}{c}}{\frac{2}{3}qL} \\ {\frac{1}{6}qL} \end{array}} \right\}

\Rightarrow \quad \left\{{\begin{array}{*{20}{c}}{{U_2}} \\ {{U_3}} \end{array}} \right\} = \frac{L}{{16EA}}\left[ {\begin{array}{*{20}{c}} 7&8 \\ 8&{16} \end{array}} \right]\left\{{\begin{array}{*{20}{c}}{\frac{2}{3}qL} \\ {\frac{1}{6}qL} \end{array}} \right\}

\Rightarrow \quad {U_2} = \frac{3}{8}\frac{{q{L^2}}}{{EA}},\quad \quad {U_3} = \frac{1}{2}\frac{{q{L^2}}}{{EA}}

Damit können wir das Verschiebungsfeld bestimmen:

{U^{\left( 1 \right)}}\left( x \right) = \left[ {\begin{array}{*{20}{c}}{{H_1}\left( x \right)}&{{H_2}\left( x \right)}&{{H_3}\left( x \right)} \end{array}} \right]\left\{{\begin{array}{*{20}{c}} 0 \\ {{U_2}} \\ {{U_3}} \end{array}} \right\}

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

Dehnungsfeld:

{\varepsilon ^{\left( 1 \right)}}\left( x \right) = \left[ {\begin{array}{*{20}{c}}{{B_1}\left( x \right)}&{{B_2}\left( x \right)}&{{B_3}\left( x \right)} \end{array}} \right]\left\{{\begin{array}{*{20}{c}} 0 \\ {{U_2}} \\ {{U_3}} \end{array}} \right\}

= \frac{3}{8}\frac{{qL}}{{EA}}\left( {4-8\frac{x}{L}} \right)+\frac{1}{2}\frac{{qL}}{{EA}}\left( {4\frac{x}{L}-1} \right) = \frac{{qL}}{{EA}}\left( {1-\frac{x}{L}} \right)

Und schließlich die Normalkräfte:

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

5.2.2 Zwei quadratische Elemente, fünf Knoten

quadratisch-vier-knoten-kragbalken-finite-stabelemente

Randbedingungen:

{U_1} = 0,\quad \quad F_2^* = F_3^* = F_4^* = F_5^* = 0,\quad \quad l = \frac{L}{2}

Es ergeben sich die Elementsteifigkeitsmatrizen:

\left[ {{k^{e1}}} \right] = \left[ {{k^{e2}}} \right] = \frac{{EA}}{{3l}}\left[ {\begin{array}{*{20}{c}} 7&{-8}&1 \\ {-8}&{16}&{-8} \\ 1&{-8}&7 \end{array}} \right]

Daraus können wir die Gesamtsteifigkeitsmatrix aufstellen:

\left[ K \right] = \frac{{2EA}}{{3L}}\left[ {\begin{array}{*{20}{c}} 7&{-8}&1&0&0 \\ {-8}&{16}&{-8}&0&0 \\ 1&{-8}&{14}&{-8}&1 \\ 0&0&{-8}&{16}&{-8} \\ 0&0&1&{-8}&7 \end{array}} \right]

Die Belastung wird gemäß der quadratischen Ansatzfunktion auf die Elemente verteilt:

\left\{ R \right\} = \left\{{{F^*}} \right\}+\frac{l}{{30}}\left[ {\begin{array}{*{20}{c}} 4&2&{-1}&0&0 \\ 2&{16}&2&0&0 \\ {-1}&2&8&2&{-1} \\ 0&0&2&{16}&2 \\ 0&0&{-1}&2&4 \end{array}} \right]\left\{{\begin{array}{*{20}{c}}{{q_{x1}}} \\ {{q_{x2}}} \\ {{q_{x3}}} \\ {{q_{x4}}} \\ {{q_{x5}}} \end{array}} \right\}\mathop = \limits^{{q_{x1}} = \ldots = {q_{x5}} = q} \left\{{\begin{array}{*{20}{c}}{F_1^*+\frac{{qL}}{{12}}} \\ {F_2^*+\frac{{qL}}{3}} \\ {F_3^*+\frac{{qL}}{6}} \\ {F_4^*+\frac{{qL}}{3}} \\ {F_5^*+\frac{{qL}}{{12}}} \end{array}} \right\}

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

\Rightarrow \quad \frac{{2EA}}{{3L}}\left[ {\begin{array}{*{20}{c}} 7&{-8}&1&0&0 \\ {-8}&{16}&{-8}&0&0 \\ 1&{-8}&{14}&{-8}&1 \\ 0&0&{-8}&{16}&{-8} \\ 0&0&1&{-8}&7 \end{array}} \right]\left\{{\begin{array}{*{20}{c}}{{U_1}} \\ {{U_2}} \\ {{U_3}} \\ {{U_4}} \\ {{U_5}} \end{array}} \right\} = \left\{{\begin{array}{*{20}{c}}{F_1^*+\frac{{qL}}{{12}}} \\ {F_2^*+\frac{{qL}}{3}} \\ {F_3^*+\frac{{qL}}{6}} \\ {F_4^*+\frac{{qL}}{3}} \\ {F_5^*+\frac{{qL}}{{12}}} \end{array}} \right\}

Wir setzen die Randbedingungen ein und reduzieren das Gleichungssystem:

\Rightarrow \quad \frac{{2EA}}{{3L}}\left[ {\begin{array}{*{20}{c}}{16}&{-8}&0&0 \\ {-8}&{14}&{-8}&1 \\ 0&{-8}&{16}&{-8} \\ 0&1&{-8}&7 \end{array}} \right]\left\{{\begin{array}{*{20}{c}}{{U_2}} \\ {{U_3}} \\ {{U_4}} \\ {{U_5}} \end{array}} \right\} = \frac{{qL}}{{12}}\left\{{\begin{array}{*{20}{c}} 4 \\ 2 \\ 4 \\ 1 \end{array}} \right\}

Durch Invertieren der Matrix erhalten wir die gesuchten Verschiebungen:

{U_2} = \frac{7}{{32}}\frac{{q{L^2}}}{{EA}},\quad \quad {U_3} = \frac{3}{8}\frac{{q{L^2}}}{{EA}},\quad \quad {U_4} = \frac{{15}}{{32}}\frac{{q{L^2}}}{{EA}},\quad \quad {U_5} = \frac{1}{2}\frac{{q{L^2}}}{{EA}}

Es ergeben sich die Verschiebungsfelder für die beiden Elemente:

{U^{\left( 1 \right)}}\left( x \right) = \left[ {\begin{array}{*{20}{c}}{\underbrace {1-3\frac{x}{l}+2{{\left( {\frac{x}{l}} \right)}^2}}_{{H_1}\left( x \right)}}&{\underbrace {4\frac{x}{l}\left( {1-\frac{x}{l}} \right)}_{{H_2}\left( x \right)}}&{\underbrace {\frac{x}{l}\left( {2\frac{x}{l}-1} \right)}_{{H_3}\left( x \right)}} \end{array}} \right]\left\{{\begin{array}{*{20}{c}} 0 \\ {{U_2}} \\ {{U_3}} \end{array}} \right\}

= \frac{7}{{32}}\frac{{q{L^2}}}{{EA}}4\frac{{2x}}{L}\left( {1-\frac{{2x}}{L}} \right)+\frac{3}{8}\frac{{q{L^2}}}{{EA}}\frac{{2x}}{L}\left( {2\frac{{2x}}{L}-1} \right)

= \frac{{q{L^2}}}{{EA}}\left( {\frac{1}{2}\left( {\frac{{2x}}{L}} \right)-\frac{1}{8}{{\left( {\frac{{2x}}{L}} \right)}^2}} \right)

{U^{\left( 2 \right)}}\left( x \right) = \left[ {\begin{array}{*{20}{c}}{{H_1}\left( x \right)}&{{H_2}\left( x \right)}&{{H_3}\left( x \right)} \end{array}} \right]\left\{{\begin{array}{*{20}{c}}{{U_3}} \\ {{U_4}} \\ {{U_5}} \end{array}} \right\}

= \frac{3}{8}\frac{{q{L^2}}}{{EA}}\left( {1-3\frac{{2x}}{L}+2{{\left( {\frac{{2x}}{L}} \right)}^2}} \right)+\frac{{15}}{{32}}\frac{{q{L^2}}}{{EA}}\left( {4\frac{{2x}}{L}\left( {1-\frac{{2x}}{L}} \right)} \right)+\frac{1}{2}\frac{{q{L^2}}}{{EA}}\left( {\frac{{2x}}{L}\left( {2\frac{{2x}}{L}-1} \right)} \right)

= \frac{{q{L^2}}}{{EA}}\left( {\frac{3}{8}-3\frac{3}{8}\frac{{2x}}{L}+2\frac{3}{8}{{\left( {\frac{{2x}}{L}} \right)}^2}+\frac{{15}}{{32}}\left( {4\frac{{2x}}{L}-\frac{{2x}}{L}4\frac{{2x}}{L}} \right)+\frac{1}{2}\left( {\frac{{2x}}{L}2\frac{{2x}}{L}-\frac{{2x}}{L}} \right)} \right)

= \frac{{q{L^2}}}{{EA}}\left( {\frac{3}{8}+\left( {-\frac{9}{8}+\frac{{60}}{{32}}-\frac{1}{2}} \right)\left( {\frac{{2x}}{L}} \right)+\left( {\frac{6}{8}-\frac{{60}}{{32}}+1} \right){{\left( {\frac{{2x}}{L}} \right)}^2}} \right)

= \frac{{q{L^2}}}{{EA}}\left( {\frac{3}{8}+\frac{1}{4}\left( {\frac{{2x}}{L}} \right)-\frac{1}{8}{{\left( {\frac{{2x}}{L}} \right)}^2}} \right)

Für die Dehnungen folgt:

{\varepsilon ^{\left( 1 \right)}}\left( x \right) = \frac{{\partial {U^{\left( 1 \right)}}\left( x \right)}}{{\partial x}} = \left[ {{H^\prime }\left( x \right)} \right]\left\{{{{\hat U}^{\left( 1 \right)}}} \right\} = \left[ {B\left( x \right)} \right]\left\{{\hat U} \right\}

= \left[ {\begin{array}{*{20}{c}}{\underbrace {-\frac{3}{l}+4\frac{x}{{{l^2}}}}_{{B_1}\left( x \right)}}&{\underbrace {\frac{4}{l}-\frac{{8x}}{{{l^2}}}}_{{B_2}\left( x \right)}}&{\underbrace {-\frac{1}{l}+\frac{{4x}}{{{l^2}}}}_{{B_3}\left( x \right)}} \end{array}} \right]\left\{{\begin{array}{*{20}{c}} 0 \\ {{U_2}} \\ {{U_3}} \end{array}} \right\}

= \left( {\frac{4}{{\left( {\frac{L}{2}} \right)}}-\frac{{8x}}{{{{\left( {\frac{L}{2}} \right)}^2}}}} \right)\frac{7}{{32}}\frac{{q{L^2}}}{{EA}}+\left( {-\frac{1}{{\left( {\frac{L}{2}} \right)}}+\frac{{4x}}{{{{\left( {\frac{L}{2}} \right)}^2}}}} \right)\frac{3}{8}\frac{{q{L^2}}}{{EA}}

= \frac{{q{L^2}}}{{EA}}\left( {\frac{7}{{32}}\left( {\frac{8}{L}-\frac{{32x}}{{{L^2}}}} \right)+\frac{3}{8}\left( {-\frac{2}{L}+\frac{{16x}}{{{L^2}}}} \right)} \right)

= \frac{{qL}}{{EA}}\left( {1-\frac{1}{2}\left( {\frac{{2x}}{L}} \right)} \right)

{\varepsilon ^{\left( 2 \right)}}\left( x \right) = \frac{{\partial {U^{\left( 2 \right)}}\left( x \right)}}{{\partial x}} = \left[ {B\left( x \right)} \right]\left\{{\begin{array}{*{20}{c}}{{U_3}} \\ {{U_4}} \\ {{U_5}} \end{array}} \right\} = \frac{{qL}}{{EA}}\left( {\frac{1}{2}-\frac{1}{2}\left( {\frac{{2x}}{L}} \right)} \right)

Nun fehlen nur noch die Normalkraftfelder:

{N^{\left( 1 \right)}}\left( x \right) = EA{\varepsilon ^{\left( 1 \right)}}\left( x \right) = qL\left( {1-\frac{1}{2}\left( {\frac{{2x}}{L}} \right)} \right)

{N^{\left( 2 \right)}}\left( x \right) = EA{\varepsilon ^{\left( 2 \right)}}\left( x \right) = qL\left( {\frac{1}{2}-\frac{1}{2}\left( {\frac{{2x}}{L}} \right)} \right)

Ähnliche Artikel

Kommentar verfassen