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<tr><th>Element #4</th></tr>
<tr><th>Element #4</th></tr>
<tr><td>
<tr><td>
<math>
{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}}}\\
{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{A E}{{\ell_0}} & 0 & 0 & -\left( \frac{A E}{{\ell_0}}\right)  & 0\\
0 & \frac{A E}{{\ell_0}} & 0 & 0 & -\left( \frac{A E}{{\ell_0}}\right)  & 0\\
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0 & -\left( \frac{A E}{{\ell_0}}\right)  & 0 & 0 & \frac{A E}{{\ell_0}} & 0\\
0 & -\left( \frac{A E}{{\ell_0}}\right)  & 0 & 0 & \frac{A E}{{\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}
\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}
</math>
</td></tr>
</td></tr>
</table>
</table>

Version vom 21. Oktober 2024, 18:47 Uhr


Aufgabenstellung

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Caption

Gesucht ist "SOME EXPLANATION"


Lösung mit Maxima

Lorem Ipsum ....

(2A2Eη3ℓ00−(2A2Eηℓ02)0A2Eη3ℓ000002A2Eη3ℓ0−(A2Eη4ℓ02)−(3A2Eη4ℓ02)A2Eη6ℓ0−(A2Eη2ℓ02)0A2Eη6ℓ0−(2A2Eηℓ02)−(A2Eη4ℓ02)A2Eη+3ℓ02AE2ℓ03+8A2Eηℓ030−(2A2Eηℓ02)−(A2Eη+3ℓ02AE4ℓ03)3A2Eη−3ℓ02AE4ℓ03A2Eη4ℓ020−(3A2Eη4ℓ02)03A2Eη+ℓ02AE2ℓ03+2AEℓ0−(3A2Eη2ℓ02)3A2Eη−3ℓ02AE4ℓ03−(3A2Eη+ℓ02AE4ℓ03)−(3A2Eη4ℓ02)A2Eη3ℓ0A2Eη6ℓ0−(2A2Eηℓ02)−(3A2Eη2ℓ02)4A2Eη3ℓ0−(A2Eη4ℓ02)3A2Eη4ℓ02A2Eη6ℓ00−(A2Eη2ℓ02)−(A2Eη+3ℓ02AE4ℓ03)3A2Eη−3ℓ02AE4ℓ03−(A2Eη4ℓ02)A2Eη+3ℓ02AE4ℓ03+A2Eηℓ03−(3A2Eη−3ℓ02AE4ℓ03)−(3A2Eη4ℓ02)003A2Eη−3ℓ02AE4ℓ03−(3A2Eη+ℓ02AE4ℓ03)3A2Eη4ℓ02−(3A2Eη−3ℓ02AE4ℓ03)3A2Eη+ℓ02AE4ℓ03+AEℓ03A2Eη4ℓ020A2Eη6ℓ0A2Eη4ℓ02−(3A2Eη4ℓ02)A2Eη6ℓ0−(3A2Eη4ℓ02)3A2Eη4ℓ022A2Eη3ℓ0)⋅(W1,0U1,0Φ1,0W2,0U2,0Φ2,0W3,0U3,0Φ3,0W4,0U4,0Φ4,0)=(00000F00)

Hier kommt jetzt irgendein Text.

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Title

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1+1




Element-Steigigkeitsmatrizen mit globalen Koordinaten
Element #1

k1=(8A2Eηℓ0302A2Eηℓ02−(8A2Eηℓ03)02A2Eηℓ0202AEℓ000−(2AEℓ0)02A2Eηℓ0202A2Eη3ℓ0−(2A2Eηℓ02)0A2Eη3ℓ0−(8A2Eηℓ03)0−(2A2Eηℓ02)8A2Eηℓ030−(2A2Eηℓ02)0−(2AEℓ0)002AEℓ002A2Eηℓ020A2Eη3ℓ0−(2A2Eηℓ02)02A2Eη3ℓ0)

Element #2

k2=(A2Eη+3ℓ02AE4ℓ033A2Eη−3ℓ02AE4ℓ03A2Eη4ℓ02−(A2Eη+3ℓ02AE4ℓ03)−(3A2Eη−3ℓ02AE4ℓ03)A2Eη4ℓ023A2Eη−3ℓ02AE4ℓ033A2Eη+ℓ02AE4ℓ033A2Eη4ℓ02−(3A2Eη−3ℓ02AE4ℓ03)−(3A2Eη+ℓ02AE4ℓ03)3A2Eη4ℓ02A2Eη4ℓ023A2Eη4ℓ02A2Eη3ℓ0−(A2Eη4ℓ02)−(3A2Eη4ℓ02)A2Eη6ℓ0−(A2Eη+3ℓ02AE4ℓ03)−(3A2Eη−3ℓ02AE4ℓ03)−(A2Eη4ℓ02)A2Eη+3ℓ02AE4ℓ033A2Eη−3ℓ02AE4ℓ03−(A2Eη4ℓ02)−(3A2Eη−3ℓ02AE4ℓ03)−(3A2Eη+ℓ02AE4ℓ03)−(3A2Eη4ℓ02)3A2Eη−3ℓ02AE4ℓ033A2Eη+ℓ02AE4ℓ03−(3A2Eη4ℓ02)A2Eη4ℓ023A2Eη4ℓ02A2Eη6ℓ0−(A2Eη4ℓ02)−(3A2Eη4ℓ02)A2Eη3ℓ0)

Element #3

k3=(A2Eη+3ℓ02AE4ℓ03−(3A2Eη−3ℓ02AE4ℓ03)A2Eη4ℓ02−(A2Eη+3ℓ02AE4ℓ03)3A2Eη−3ℓ02AE4ℓ03A2Eη4ℓ02−(3A2Eη−3ℓ02AE4ℓ03)3A2Eη+ℓ02AE4ℓ03−(3A2Eη4ℓ02)3A2Eη−3ℓ02AE4ℓ03−(3A2Eη+ℓ02AE4ℓ03)−(3A2Eη4ℓ02)A2Eη4ℓ02−(3A2Eη4ℓ02)A2Eη3ℓ0−(A2Eη4ℓ02)3A2Eη4ℓ02A2Eη6ℓ0−(A2Eη+3ℓ02AE4ℓ03)3A2Eη−3ℓ02AE4ℓ03−(A2Eη4ℓ02)A2Eη+3ℓ02AE4ℓ03−(3A2Eη−3ℓ02AE4ℓ03)−(A2Eη4ℓ02)3A2Eη−3ℓ02AE4ℓ03−(3A2Eη+ℓ02AE4ℓ03)3A2Eη4ℓ02−(3A2Eη−3ℓ02AE4ℓ03)3A2Eη+ℓ02AE4ℓ033A2Eη4ℓ02A2Eη4ℓ02−(3A2Eη4ℓ02)A2Eη6ℓ0−(A2Eη4ℓ02)3A2Eη4ℓ02A2Eη3ℓ0)

Element #4

k4=(A2Eηℓ030A2Eη2ℓ02−(A2Eηℓ03)0A2Eη2ℓ020AEℓ000−(AEℓ0)0A2Eη2ℓ020A2Eη3ℓ0−(A2Eη2ℓ02)0A2Eη6ℓ0−(A2Eηℓ03)0−(A2Eη2ℓ02)A2Eηℓ030−(A2Eη2ℓ02)0−(AEℓ0)00AEℓ00A2Eη2ℓ020A2Eη6ℓ0−(A2Eη2ℓ02)0A2Eη3ℓ0)


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