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Mbinu za Uchanganuzi wa Mienendo

Sehemu hii inaeleza mbinu za kuchanganua matatizo ya mienendo kwa kutumia ujumuishaji wa moja kwa moja kwa wakati. Miundo ya mbinu fiche na mbinu wazi imeonyeshwa hapa chini.

Udiskretishaji wa Mlinganyo wa Mwendo (Mfumo wa Pamoja)

Bado haijaandikwa kikamilifu (itakamilishwa katika awamu inayofuata).

Mbinu Fiche (Mbinu ya Newmark-β)

Kwa matatizo ya mienendo, mbinu ya ujumuishaji wa moja kwa moja kwa wakati hutumika kutatua mlinganyo wa mwendo ulioonyeshwa hapa chini.

\[\begin{equation} M( t + \Delta t ) \ddot{U} (t + \Delta t) + C( t + \Delta t ) \dot{U}(t + \Delta t) + Q( t + \Delta t ) = F( t + \Delta t ) \label{eq:2.5.1} \end{equation}\]

Hapa, \(M\) ni matriki ya masi, \(C\) ni matriki ya udampishaji, \(Q\) ni vekta ya nguvu ya ndani, na \(F\) ni vekta ya nguvu ya nje. Matriki ya masi inachukuliwa kuwa thabiti bila kujali mgeuko, hata katika uchanganuzi usio mstari.

Mabadiliko ya uhamisho, kasi na mchapuko ndani ya nyongeza ya muda \(\Delta t\) yanakadiriwa kwa kutumia mbinu ya Newmark-\(\beta\), kama inavyoonyeshwa katika Mling. \(\eqref{eq:2.5.2}\) na Mling. \(\eqref{eq:2.5.3}\).

\[\begin{equation} \dot{U}(t + \Delta t) = \frac{\gamma}{\beta \Delta t} \Delta U( t + \Delta t ) - \frac{\gamma - \beta}{\beta} \dot{U}( t ) - \Delta t \frac{\gamma - 2\beta}{2\beta} \ddot{U}(t) \label{eq:2.5.2} \end{equation}\]
\[\begin{equation} \ddot{U}(t + \Delta t) = \frac{1}{\beta \Delta t^2}\Delta U(t + \Delta t) - \frac{1}{\beta \Delta t} \dot{U}(t) - \frac{1 - 2\beta}{2\beta} \ddot {U}(t) \label{eq:2.5.3} \end{equation}\]

Hapa, \(\gamma\) na \(\beta\) ni vigezo vya mbinu ya Newmark-\(\beta\).

Kama inavyojulikana, thamani zifuatazo za \(\gamma\) na \(\beta\) hulingana na mbinu ya mchapuko mstari na kanuni ya trapezoidi, mtawalia.

\(\gamma = \displaystyle \frac{1}{2}\)，\(\beta = \displaystyle \frac{1}{6}\) (mbinu ya mchapuko mstari)

\(\gamma = \displaystyle \frac{1}{2}\)，\(\beta = \displaystyle \frac{1}{4}\) (kanuni ya trapezoidi)

Kwa kuweka Mling. \(\eqref{eq:2.5.2}\) na Mling. \(\eqref{eq:2.5.3}\) katika Mling. \(\eqref{eq:2.5.1}\), mlinganyo ufuatao hupatikana.

\[\begin{align} \nonumber \left( \frac{1}{\beta \Delta t^2} \mathbf{M} + \frac{\gamma}{\beta \Delta t} C + K \right) \Delta U ( t + \Delta t ) &= F ( t + \Delta t ) - Q ( t + \Delta t ) \\\ \nonumber &+ \frac{1}{\beta \Delta t} M \dot{U} ( t ) + \frac{1 - 2\beta}{2\beta} M \ddot{U} ( t ) \\\ &+ \frac{\gamma - \beta}{\beta} C \dot{U} (t) + \Delta t \frac{\gamma - 2\beta}{2 \beta} C \ddot{U}(t) \label{eq:2.5.4} \end{align}\]

Hasa kwa tatizo la mstari, \(K_L\) ni matriki ya ugumu mstari na \(Q ( t + \Delta t ) = K_L U (t + \Delta t)\). Kwa kuweka uhusiano huu katika mlinganyo ulio juu, mlinganyo ufuatao hupatikana.

\[\begin{align} \nonumber M \left\lbrace -\frac{1}{\beta \Delta t^2} U(t) -\frac{1}{\beta \Delta t}\dot U(t) - \frac{2\beta}{1-2\beta} \ddot U(t) \right\rbrace &+ C\left\lbrace - \frac{\gamma}{\beta \Delta t} U(t) + \left(1 - \frac{\gamma}{\beta}\right) \dot U(t) + \Delta{t}\frac{ 2\beta-\gamma}{2\beta}\ddot U(t)\right\rbrace \\\ & + \frac{1}{\beta \Delta{t}^2} M + \frac{\gamma}{\beta \Delta{t}} C + K_L U(t+\Delta{t}) = F(t+\Delta{t}) \label{eq:2.5.5} \end{align}\]

Katika sehemu ambazo mchapuko umebainishwa kama sharti la mpaka la kijiometri, uhamisho hupatikana kutoka Mling. \(\eqref{eq:2.5.2}\) kama ifuatavyo.

\[\begin{equation} u_{is} (t+\Delta{t}) = u_{is} (t) + \Delta t \dot{u}(t) + \Delta t^2 \left(\frac{1}{2} -\beta \right) {\ddot{u}}_{is} (t + \Delta t) \label{eq:2.5.6} \end{equation}\]

Vivyo hivyo, katika sehemu ambazo kasi imebainishwa, uhamisho hupatikana kutoka Mling. \(\eqref{eq:2.5.6}\) kama ifuatavyo.

\[\begin{equation} u_{is}(t+\Delta{t})= u_{is}(t)+\Delta t \frac{ \gamma - \beta}{ \gamma}\dot{u_{is}}(t) +(\Delta{t})^2 \frac{ \gamma - 2\beta}{ 2\gamma} \ddot{u_{is}}(t) +\Delta t \frac{\beta}{ \gamma}\dot{u_{is}}(t+\Delta{t}) \label{eq:2.5.7} \end{equation}\]

Hapa, \(u_{is}(t+\Delta{t})\) ni uhamisho wa nodi kwa wakati \(t+\Delta{t}\), na \(\dot{u_{is}}(t+\Delta{t})\) ni kasi ya nodi kwa wakati \(t+\Delta{t}\), \(\ddot{u_{is}}(t+\Delta{t})\) ni mchapuko wa nodi kwa wakati \(t+\Delta{t}\), \(i\) ni nambari ya kiwango cha uhuru cha nodi, na \(s\) ni nambari ya nodi. Vipengele vya masi na udampishaji vinashughulikiwa kama ifuatavyo.

Ushughulikiaji wa Kipengele cha Masi

Kwa kanuni, matriki ya masi hushughulikiwa kama matriki ya masi iliyojumlishwa.

Ushughulikiaji wa Kipengele cha Udampishaji

Kipengele cha udampishaji hushughulikiwa kama udampishaji wa Rayleigh unaoelezwa na Mling. \(\eqref{eq:2.5.8}\).

\[\begin{equation} C = R_m M + R_k K_L \label{eq:2.5.8} \end{equation}\]

Hapa, \(R_m\) na \(R_k\) ni vigezo vya udampishaji wa Rayleigh.

Thamani za \(R_m\) na \(R_k\) zilizobainishwa kwenye kadi !DYNAMIC hutumika kwa usawa katika modeli nzima. Ili kutoa thamani tofauti za \(R_m\) na \(R_k\) kwa kila nyenzo, ndani ya bloku ya !MATERIAL ya nyenzo hiyo bainisha kadi !DAMPING. Kwa elementi za nyenzo ambayo !DAMPING imebainishwa, matriki ya udampishaji ya elementi hukokotolewa kutoka kwenye matriki ya masi ya elementi \(M_i\) na matriki ya ugumu wa tanjenti \(K_i\) kama \(C_i = R_m M_i + R_k K_i\), kisha huunganishwa katika matriki ya jumla ya udampishaji. Kipengele hiki hutumika tu kwa mbinu fiche.

Mbinu Wazi (Mbinu ya Tofauti ya Kati)

Mbinu wazi inategemea mlinganyo wa mwendo kwa wakati t ulioonyeshwa hapa chini.

\[\begin{equation} M \ddot{U}(t) + C (t) \dot{U}(t) + Q(t) = F(t) \label{eq:2.5.9} \end{equation}\]

Kwa kuwakilisha uhamisho kwa nyakati \(t + \Delta t\) na \(t - \Delta t\) kwa upanuzi wa Taylor kuhusu wakati \(t\) na kubakiza vipengele hadi daraja la 2 katika \(\Delta t\), milinganyo ifuatayo hupatikana.

\[\begin{equation} U(t+\Delta{t}) = U(t)+\dot{U}(t)(\Delta{t}) +\frac{1}{2!}\ddot{U}(\Delta{t})^2 \label{eq:2.5.10} \end{equation}\]
\[\begin{equation} U(t-\Delta{t})=U(t)-\dot{U}(t)(\Delta{t}) +\frac{1}{2!}\ddot{U}(\Delta{t})^2 \label{eq:2.5.11} \end{equation}\]

Kwa kuchukua tofauti na jumla ya Mling. \(\eqref{eq:2.5.3}\) na Mling. \(\eqref{eq:2.5.4}\), milinganyo ifuatayo hupatikana.

\[\begin{equation} \dot{U}(t)=\frac{1}{2\Delta{t}} (U(t+\Delta{t})-U(t-\Delta{t})) \label{eq:2.5.12} \end{equation}\]
\[\begin{equation} \ddot{U}= \frac{1}{(2\Delta{t})^2} (U(t+\Delta{t})-2U(t)+U(t-\Delta{t})) \label{eq:2.5.13} \end{equation}\]

Kwa kuweka Mling. \(\eqref{eq:2.5.12}\) na Mling. \(\eqref{eq:2.5.13}\) katika Mling. \(\eqref{eq:2.5.9}\), mlinganyo ufuatao hupatikana.

\[\begin{equation} \left( \frac{1}{\Delta t^{2}} M + \frac{1}{2\Delta t} C \right) U ( t + \Delta t ) \\\ = F(t) - Q(t) - \frac{1}{\Delta t^{2}} 2 U(t) - U( t - \Delta t) - \frac{1}{2\Delta t} C U(t - \Delta t) \label{eq:2.5.14} \end{equation}\]

Hasa kwa tatizo la mstari, \(Q(t) = K_L U(t)\), na mlinganyo ulio juu huwa

\[\begin{equation} \left( \frac{1}{\Delta t^{2}} M + \frac{1}{2\Delta t} C \right) U( t + \Delta t ) \\\ = F(t) - K_L U(t) - \frac{1}{\Delta t^{2}} M U(t) - U(t - \Delta t) - \frac{1}{2\Delta t} C U (t - \Delta t) \label{eq:2.5.15} \end{equation}\]

Ikiwa matriki ya masi \(M\) inachukuliwa kuwa matriki ya masi iliyojumlishwa na matriki ya udampishaji kuwa matriki ya udampishaji wa uwiano \(C = R_m M\), Mling. \(\eqref{eq:2.5.15}\) hauhitaji kutatua mfumo wa milinganyo sambamba.

Kwa hiyo, kutoka Mling. \(\eqref{eq:2.5.15}\), \(U(t+\Delta t)\) inaweza kupatikana kwa mlinganyo ufuatao.

\[\begin{equation} U( t + \Delta t ) \\\ = \frac{1}{( \frac{1}{\Delta t^{2}} M + \frac{1}{2\Delta t} C )} \{ F(t) - Q(t) - \frac{1}{\Delta t^{2}} M U(t) - U(t - \Delta t) - \frac{1}{2\Delta t} C U(t - \Delta t) \} \label{eq:2.5.17} \end{equation}\]

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