Gelöste Aufgaben/Bike

Aufgabenstellung

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Gesucht ist "SOME EXPLANATION"

Lösung mit Maxima

Lorem Ipsum ....

$\underset{\_}{\tilde{Q}}=\left(\begin{array}{c}{W}_0\\ {\mathrm{\Phi }}_0\\ W_1\\ {\mathrm{\Phi }}_1\\ W_2\\ {\mathrm{\Phi }}_2,\\ {\mathrm{\Psi }}_1,\\ {\mathrm{\Psi }}_2\end{array}\right)$

$\underset{\_}{\delta \tilde{Q}}=\left(\begin{array}{c}\delta W_0\\ \delta {\mathrm{\Phi }}_0\\ \delta W_1\\ \delta {\mathrm{\Phi }}_1\\ \delta W_2\\ \delta {\mathrm{\Phi }}_2,\\ \delta {\mathrm{\Psi }}_1,\\ \delta {\mathrm{\Psi }}_2\end{array}\right)$

$\underset{\_}{Y}=\left[\begin{array}{c}\underset{\_}{Q}\\ \underset{\_}{\dot{Q}}\end{array}\right]$

$\begin{array}{ll}\underset{\_}{\dot{Y}}& =\underset{\_}{f}\left(\underset{\_}{Y}\right)\\ & =\left[\begin{array}{l}\underset{\_}{\dot{Q}}\\ -{\underset{\_}{\underset{\_}{M}}}^{-1}\cdot \underset{\_}{\underset{\_}{K}}\cdot \underset{\_}{Q}+{\underset{\_}{\underset{\_}{M}}}^{-1}\cdot \underset{\_}{P}\end{array}\right]\end{array}$

$\underset{\_}{\underset{\_}{M}}\cdot \underset{\_}{\ddot{Q}}+\underset{\_}{\underset{\_}{K}}\cdot \underset{\_}{Q}=\underset{\_}{P}(\underset{\_}{Q},\underset{\_}{\dot{Q}})$

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

$\delta \mathrm{\Pi }=\delta {\mathrm{\Pi }}_G+\delta {\mathrm{\Pi }}_S$

$\delta {\mathrm{\Pi }}_S=K_S\cdot ({\mathrm{\Psi }}_2-{\mathrm{\Psi }}_1)\cdot (\delta {\mathrm{\Psi }}_2-\delta {\mathrm{\Psi }}_1)$

$\delta {\mathrm{\Pi }}_G=\delta {\mathrm{\Pi }}_{G1}+\delta {\mathrm{\Pi }}_{G2}$

$\delta {\mathrm{\Pi }}_{Gi}=\underset{0}{\overset{{\ell }_i}{\int }}{M}_i(x)\cdot \delta {w_i}^{″}(x)dx$

$M_i(x)=E\cdot I(x)\cdot {w_i}^{″}(x)$

$w_i(x)={\underset{\_}{Q}}_i^T\cdot \underset{\_}{\phi }$

$\underset{\_}{\phi }=\left[\begin{array}{c}(\xi -1{)}^2\cdot (2\cdot \xi +1)\\ {\ell }_i\cdot \xi \cdot (\xi -1{)}^2\\ -{\xi }^2\cdot (2\xi -3)\\ {\ell }_i\cdot {\xi }^2\cdot (\xi -1)\end{array}\right]$

${\underset{\_}{\tilde{Q}}}_i=\left(\begin{array}{c}{W}_{i-1}\\ {\mathrm{\Phi }}_{i-q}\\ W_i\\ {\mathrm{\Phi }}_i\end{array}\right)$

$\delta {\mathrm{\Pi }}_{Gi}=\underset{0}{\overset{{\ell }_i}{\int }}E\cdot I_i(x)\cdot (W_{i-1}{{\phi }_1}^{″}\cdot \delta W_{i-1}{{\phi }_1}^{″}+{\mathrm{\Phi }}_{i-1}{{\phi }_2}^{″}\cdot \delta W_{i-1}{{\phi }_1}^{″}+W_i{{\phi }_3}^{″}\cdot \delta W_{i-1}{{\phi }_1}^{″}+\dots +{\mathrm{\Phi }}_i{{\phi }_4}^{″}\cdot \delta {\mathrm{\Phi }}_i{{\phi }_4}^{″})$

$(.):=\frac{d(.)}{dx}$

$\delta {\mathrm{\Pi }}_{Gi}=\delta Q_i^T\cdot {\underset{\_}{\underset{\_}{K}}}_i\cdot Q_i$

$k_{i,jk}=\underset{0}{\overset{{\ell }_i}{\int }}E\cdot I_i(x){{\phi }_j}^{″}\cdot {{\phi }_k}^{″}dx$

$I_i(x)=\frac{\pi }{64}(D_i(x{)}^4-d_i(x{)}^4)$

$\begin{array}{ccccc}{D}_i(x)& =& D_{i-1}\cdot ({\xi }_1-1)& +& D_i\cdot {\xi }_1\\ d_i(x)& =& d_{i-1}\cdot ({\xi }_1-1)& +& d_i\cdot {\xi }_1\end{array}$

${\xi }_1=\frac{x}{{\ell }_1}$

${\xi }_2=\frac{(x-H)}{{\ell }_2}$

$\frac{d(.)}{dx}=\frac{d(.)}{d{\xi }_i}\cdot \frac{1}{{\ell }_i}$

$\underset{\_}{\underset{\_}{K}}=\left(\begin{array}{cccccccc}{k}_{1,11}& k_{1,12}& k_{1,13}& k_{1,14}& 0& 0& 0& 0\\ k_{1,12}& k_{1,22}& k_{1,23}& k_{1,24}& 0& 0& 0& 0\\ k_{1,13}& k_{1,23}& k_{1,33}+k_{2,11}& k_{1,34}+k_{2,12}& k_{2,13}& k_{2,14}& 0& 0\\ k_{1,14}& k_{1,24}& k_{1,34}+k_{2,12}& k_{1,44}+k_{2,22}& k_{2,23}& k_{2,24}& 0& 0\\ 0& 0& k_{2,13}& k_{2,23}& k_{2,33}& k_{2,34}& 0& 0& \\ 0& 0& k_{2,14}& k_{2,24}& k_{2,34}& k_{2,44}& 0& 0& \\ 0& 0& 0& 0& 0& 0& K_S& -K_S\\ 0& 0& 0& 0& 0& 0& -K_S& K_S\\ \end{array}\right)$

${\underset{\_}{r}}_{B,1}=\left(\begin{array}{c}H-h-b\cdot \sin ({\mathrm{\Phi }}_1(t))\\ -v_{0}t+W_1(t)+b\cdot \cos ({\mathrm{\Phi }}_1(t))\end{array}\right)$

${\underset{\_}{r}}_{B,2}=\left(\begin{array}{c}H-r_2\cdot \cos (\mathrm{\Omega }t+{\mathrm{\Psi }}_2(t)+{\mathrm{\Theta }}_2)\\ -v_{0}t+W_2(t)-r_2\cdot \sin (\mathrm{\Omega }t+{\mathrm{\Psi }}_2(t)+{\mathrm{\Theta }}_2)\end{array}\right)$

${\underset{\_}{r}}_A=\left(\begin{array}{c}H-R_1\cdot \cos (\mathrm{\Omega }t+{\mathrm{\Psi }}_1(t)+{\mathrm{\Theta }}_1)\\ -v_{0}t+W_1(t)-R_1\cdot \sin (\mathrm{\Omega }t+{\mathrm{\Psi }}_1(t)+{\mathrm{\Theta }}_1)\end{array}\right)$

${\underset{\_}{v}}_{B,rel}={\underset{\_}{\dot{r}}}_{B,1}-{\underset{\_}{\dot{r}}}_{B,2}$

$\begin{array}{lll}{\underset{\_}{v}}_{A,rel}& =& {\underset{\_}{\dot{r}}}_A\\ & =& \left(\begin{array}{c}0\\ -v_0+R_1\cdot (\mathrm{\Omega }+{\dot{\mathrm{\Psi }}}_1(t))+{\dot{W}}_2(t)\end{array}\right)\end{array}$

$\dot{(.)}:=\frac{d(.)}{dt}$

$\delta W^a=\delta W_A^a+\delta W_B^a+\delta W_{d^{\prime }Alembert}^a$

$\delta {\underset{\_}{r}}_{B,1}=\left(\begin{array}{c}-\delta {\mathrm{\Phi }}_{1}b\\ \delta W_1\end{array}\right)$

$\delta {\underset{\_}{r}}_{B2}=\left(\begin{array}{c}{r}_2\cdot \delta {\mathrm{\Psi }}_2\sin (\psi )\\ \delta W_2-r_2\cdot \delta {\mathrm{\Psi }}_2\cos (\psi )\end{array}\right)$

$\delta {\underset{\_}{r}}_A=\left(\begin{array}{c}0\\ \delta W_2+R_1\cdot \delta {\mathrm{\Psi }}_1\end{array}\right)$

$\delta W_A^a={\underset{\_}{F}}_A^T\cdot \delta {\underset{\_}{r}}_A$

$\delta W_B^a={\underset{\_}{F}}_B^T\cdot \delta {\underset{\_}{r}}_{B,1}-{\underset{\_}{F}}_B^T\cdot \delta {\underset{\_}{r}}_{B,2}$

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