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In den folgenden Darstellungen haben die \glqq dicken\grqq\ Federn eine Federkonstante von $\SI{4.0}{\Npm}$ und die \glqq dünnen\grqq\ Federn eine solche von $\SI{12}{\Npm}$. Berechne jeweils die Ersatzfederkonstante für das gesamte Gebilde! \begin{center} \begin{tikzpicture} \begin{scope} \draw[ultra thick] (0,0)--(0,1); \draw[decoration={aspect=0.3, segment length=1mm, amplitude=3mm,coil},decorate] (0,0)--(2,0); \draw[decoration={aspect=0.3, segment length=1mm, amplitude=3mm,coil},decorate] (0,1)--(2,1); \draw[ultra thick] (2,0)--(2,1); \draw[decoration={aspect=0.4, segment length=.5mm, amplitude=1mm,coil},decorate] (2,.5)--(3,.5); \end{scope} \begin{scope}[xshift=4cm] \draw[ultra thick] (0,0)--(0,1); \draw[decoration={aspect=0.3, segment length=1mm, amplitude=3mm,coil},decorate] (0,0)--(3,0); \draw[decoration={aspect=0.3, segment length=1mm, amplitude=3mm,coil},decorate] (0,1)--(1.5,1); \draw[ultra thick] (3,0)--(3,1); \draw[decoration={aspect=0.4, segment length=.5mm, amplitude=1mm,coil},decorate] (1.5,1)--(3,1); \end{scope} \begin{scope}[xshift=2cm,yshift=-2cm] \draw[ultra thick] (0,0)--(0,1); \draw[decoration={aspect=0.4, segment length=.5mm, amplitude=1mm,coil},decorate] (0,0)--(1.5,0); \draw[ultra thick] (3,0)--(3,1); \draw[decoration={aspect=0.3, segment length=1mm, amplitude=3mm,coil},decorate] (1.5,0)--(3,0); \draw[decoration={aspect=0.3, segment length=1mm, amplitude=3mm,coil},decorate] (0,1)--(1.5,1); \draw[ultra thick] (3,0)--(3,1); \draw[decoration={aspect=0.4, segment length=.5mm, amplitude=1mm,coil},decorate] (1.5,1)--(3,1); \end{scope} \end{tikzpicture} \end{center}
\begin{abcliste} \abc Die Ersatzfederkonstante ist \begin{align} \Ders &= \left(\frac{1}{D_1+D_1}+\frac{1}{D_2}\right)^{-1}\\ &= \SI{4.8}{\newton\per\meter}. \end{align} \abc Die Ersatzfederkonstante ist \begin{align} \Ders &= \left(\frac{1}{D_1}+\frac{1}{D_2}\right)^{-1}+D_1\\ &= \SI{7.0}{\newton\per\meter}. \end{align} \abc Die Ersatzfederkonstante ist \begin{align} \Ders &= \left(\frac{1}{D_1}+\frac{1}{D_2}\right)^{-1}+\left(\frac{1}{D_1}+\frac{1}{D_2}\right)^{-1}\\ &= \SI{6.0}{\newton\per\meter}. \end{align} \end{abcliste}
11:14, 17. Nov. 2019 | Initial Version. | Patrik Weber (patrik) | Current Version |