DYNAMIKA

Drgania harmoniczne układu prętowego z masami skupionymi.

Dane:

m₁ = 280kg

m₂ = 120kg

P(t) = P • sinpt

P=100kN

$$p = 40\frac{\text{rad}}{s}$$

E=205GPa

Przekrój: HD 320 o I=30820cm⁴

EI = 205 • 10⁶ • 30820 • 10⁻⁸ = =63181kNm²

$$tg\alpha = \frac{2}{6}\ \rightarrow \ \alpha = 18,43494882$$

Układ posiada dwa stopnie swobody dynamicznej (SSD=2): q₁ oraz q₂.

CZĘSTOŚCI KOŁOWE DRGAŃ WŁASNYCH UKŁADU

Siły bezwładności:

$$\left\{ \begin{matrix} B_{1}\left( t \right) = - m_{1}{\ddot{q}}_{1}(t) \\ B_{2}\left( t \right) = - m_{2}{\ddot{q}}_{2}(t) \\ \end{matrix} \right.\ $$

Przyjęto masę porównawczą m=100kg oraz m₁ = 2, 8m oraz m₂ = 1, 2m.

$$\left\{ \begin{matrix} B_{1}\left( t \right) = - 2,8m{\ddot{q}}_{1}(t) \\ B_{2}\left( t \right) = - 1,2m{\ddot{q}}_{2}(t) \\ \end{matrix} \right.\ $$

Równania ruchu:

$$\left\{ \begin{matrix} q_{1}(t) = \delta_{11}B_{1}\left( t \right) + \delta_{12}B_{2}\left( t \right) \\ q_{2}(t) = \delta_{21}B_{1}\left( t \right) + \delta_{22}B_{2}\left( t \right) \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} q_{1}(t) = \delta_{11}( - 2,8m{\ddot{q}}_{1}\left( t \right)) + \delta_{12}( - 1,2m{\ddot{q}}_{2}\left( t \right)) \\ q_{2}(t) = \delta_{21}( - 2,8m{\ddot{q}}_{1}\left( t \right)) + \delta_{22}( - 1,2m{\ddot{q}}_{2}\left( t \right)) \\ \end{matrix} \right.\ $$

Przyjęto: q₁ = A₁sinωt stąd: ${\ddot{q}}_{1} = - \omega^{2}A_{1}\text{sinωt}$

q₂ = A₂sinωt ${\ddot{q}}_{2} = - \omega^{2}A_{2}\text{sinωt}$

$$\left\{ \begin{matrix} A_{1}sin\omega t = \delta_{11}2,8m\omega^{2}A_{1}sin\omega t + \delta_{12}1,2m\omega^{2}A_{2}\text{sinωt} \\ A_{2}sin\omega t = \delta_{21}2,8m\omega^{2}A_{1}sin\omega t + \delta_{22}1,2m\omega^{2}A_{2}\text{sinωt} \\ \end{matrix} \right.\ \ \ /:sin\omega t$$

$$\left\{ \begin{matrix} A_{1} = \delta_{11}2,8m\omega^{2}A_{1} + \delta_{12}1,2m\omega^{2}A_{2} \\ A_{2} = \delta_{21}2,8m\omega^{2}A_{1} + \delta_{22}1,2m\omega^{2}A_{2} \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} \delta_{11}2,8m\omega^{2}A_{1} - A_{1} + \delta_{12}1,2m\omega^{2}A_{2} = 0 \\ \delta_{21}2,8m\omega^{2}A_{1} + \delta_{22}1,2m\omega^{2}A_{2} - A_{2} = 0 \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} (\delta_{11}2,8m\omega^{2} - 1)A_{1} + \delta_{12}1,2m\omega^{2}A_{2} = 0 \\ \delta_{21}2,8m\omega^{2}A_{1} + (\delta_{22}1,2m\omega^{2} - 1)A_{2} = 0 \\ \end{matrix} \right.\ $$

Ustalenie wartości δ_ik wykonano metodą sił.

$${\tilde{\delta}}_{11} \bullet X_{1}^{i} + {\tilde{}}_{1P}^{i} = 0$$

Stan X₁ = 1

$${\tilde{\delta}}_{11} = \int_{s}^{}{\frac{M_{1} \bullet M_{1}}{\text{EI}}\text{ds}} = \frac{1}{\text{EI}}\left\lbrack 4 \bullet 6 \bullet 6 + \frac{1}{2} \bullet 6 \bullet \frac{2}{\text{sinα}} \bullet \frac{2}{3} \bullet 6 \right\rbrack = 219,8946639\frac{1}{\text{EI}}\lbrack m^{3}\rbrack$$

Stan P₁ = 1

$${\tilde{}}_{1P}^{1} = \int_{s}^{}{\frac{M_{1} \bullet M_{P}^{1}}{\text{EI}}\text{ds}} = \frac{1}{\text{EI}}\left\lbrack - \frac{1}{2} \bullet 4 \bullet 4 \bullet 6 \right\rbrack = - 48\frac{1}{\text{EI}}\ \lbrack m^{3}\rbrack$$

$$X_{1}^{1} = - \frac{{\tilde{}}_{1P}^{1}}{{\tilde{\delta}}_{11}} = \frac{48}{\text{EI}} \bullet \frac{\text{EI}}{219,8946639} = 0,218286334$$

Stan P₂ = 1

$${\tilde{}}_{1P}^{2} = \int_{s}^{}{\frac{M_{1} \bullet M_{P}^{2}}{\text{EI}}\text{ds}} = \frac{1}{\text{EI}}\left\lbrack - \frac{1}{2} \bullet 4 \bullet 4,427188725 \bullet 6 - \frac{1}{2} \bullet 4 \bullet 3,162277661 \bullet 6 - \right.\ $$

$$\left. \ - \frac{1}{2} \bullet \frac{1}{\text{sinα}} \bullet 3,162277661 \bullet \left( \frac{2}{3} \bullet 6 + \frac{1}{3} \bullet 3 \right) \right\rbrack = - 116,07359661\frac{1}{\text{EI}}\ \left\lbrack m^{3} \right\rbrack$$

$$X_{1}^{2} = - \frac{{\tilde{}}_{1P}^{2}}{{\tilde{\delta}}_{11}} = \frac{116,07359661}{\text{EI}} \bullet \frac{\text{EI}}{219,8946639} = 0,527859997$$

Wykresy momentów od siły P₁ = 1 dla układy statycznie niewyznaczalnego:

$$\delta_{11} = \int_{s}^{}{\frac{M_{1}^{D} \bullet M_{1}^{D}}{\text{EI}}\text{ds}} = \frac{1}{\text{EI}}\left\lbrack \frac{1}{2} \bullet 4 \bullet 2,690281996 \bullet \left( \frac{2}{3} \bullet 2,690281996 - \frac{1}{3} \bullet 1,309718004 \right) + \right.\ $$

$$+ \left. \ \frac{1}{2} \bullet 4 \bullet 1,309718004 \bullet \left( \frac{2}{3} \bullet 1,309718004 - \frac{1}{3} \bullet 2,690281996 \right) \right\rbrack +$$

$$+ \frac{1}{\text{EI}}\left\lbrack \frac{1}{2} \bullet \frac{1}{\text{sinα}} \bullet 1,309718004 \bullet \left( \frac{2}{3} \bullet 1,309718004 - \frac{1}{3} \bullet 0,654859002 \right) + \right.\ $$

$$+ \frac{1}{2} \bullet \frac{1}{\text{sinα}} \bullet 0,654859002 \bullet \left( \frac{2}{3} \bullet 0,654859002 + \frac{1}{3} \bullet 1,309718004 \right) +$$

$$+ \left. \ \frac{1}{2} \bullet \frac{1}{\text{sinα}} \bullet 0,654859009 \bullet \frac{2}{3} \bullet 0,654859002 \right\rbrack = \mathbf{10,85558931}\frac{\mathbf{1}}{\mathbf{\text{EI}}}\mathbf{\ }\left\lbrack \mathbf{m}^{\mathbf{3}} \right\rbrack$$

Korzystając z twierdzenia redukcyjnego uzyskano:

$$\delta_{11} = \int_{s}^{}{\frac{M_{P}^{1} \bullet M_{1}^{D}}{\text{EI}}\text{ds}} = \frac{1}{\text{EI}}\left\lbrack \frac{1}{2} \bullet 4 \bullet 4 \bullet \left( \frac{2}{3} \bullet 2,690281996 - \frac{1}{3} \bullet 1,309718004 \right) \right\rbrack = \mathbf{10,8555893}\frac{\mathbf{1}}{\mathbf{\text{EI}}}$$

Wykresy momentów od siły P₂ = 1 dla układy statycznie niewyznaczalnego:

$$\delta_{22} = \int_{s}^{}{\frac{M_{2}^{D} \bullet M_{2}^{D}}{\text{EI}}\text{ds}} = \frac{1}{\text{EI}}\left\lbrack \frac{1}{2} \bullet 4 \bullet 1,260028742 \bullet \left( \frac{2}{3} \bullet 1,260028742 - \frac{1}{3} \bullet 0,00488216 \right) + \right.\ $$

$$+ \left. \ \frac{1}{2} \bullet 4 \bullet 0,00488216 \bullet \left( \frac{2}{3} \bullet 0,00488216 - \frac{1}{3} \bullet 1,260028742 \right) \right\rbrack +$$

$$+ \left\lbrack \frac{1}{2} \bullet 0,00488216 \right.\ \bullet \frac{1}{\text{sinα}} \bullet \left( \frac{2}{3} \bullet 0,00488216 + \frac{1}{3} \bullet 1,58357991 \right) +$$

$$+ \frac{1}{2} \bullet \frac{1}{\text{sinα}} \bullet 1,58357991 \bullet \left( \frac{2}{3} \bullet 1,58357991 + \frac{1}{3} \bullet 0,00488216 \right) +$$

$$+ \left. \ \frac{1}{2} \bullet \frac{1}{\text{sinα}} \bullet 1,58357991 \bullet \frac{2}{3} \bullet 1,58357991 \right\rbrack = \mathbf{7,403649956}\frac{\mathbf{1}}{\mathbf{\text{EI}}}\mathbf{\ }\left\lbrack \mathbf{m}^{\mathbf{3}} \right\rbrack$$

Korzystając z twierdzenia redukcyjnego uzyskano:

$$\delta_{22} = \int_{s}^{}{\frac{M_{P}^{2} \bullet M_{2}^{D}}{\text{EI}}\text{ds}} = \frac{1}{\text{EI}}\left\lbrack \frac{1}{2} \bullet 4,427188725 \bullet 4 \bullet \left( \frac{2}{3} \bullet 1,260028742 - \frac{1}{3} \bullet 0,00488216 \right) \right.\ +$$

$$+ \frac{1}{2} \bullet 3,162277661 \bullet 4 \bullet \left( \frac{1}{3} \bullet 1,260028742 - \frac{2}{3} \bullet 0,00488216 \right) +$$

$$+ \left. \ \frac{1}{2} \bullet \frac{1}{\text{sinα}} \bullet 3,162277661 \bullet \left( - \frac{2}{3} \bullet 0,00488216 - \frac{1}{3} \bullet 1,58357991 \right) \right\rbrack = \mathbf{7,403652341}\frac{\mathbf{1}}{\mathbf{\text{EI}}}\mathbf{\ }\left\lbrack \mathbf{m}^{\mathbf{3}} \right\rbrack$$

$$\delta_{12} = \delta_{21} = \int_{s}^{}{\frac{M_{1}^{D} \bullet M_{2}^{D}}{\text{EI}}\text{ds}} = \frac{1}{\text{EI}}\left\lbrack \frac{1}{2} \bullet 1,260028742 \bullet 4 \bullet \left( \frac{2}{3} \bullet 2,690281996 - \frac{1}{3} \bullet 1,309718004 \right) \right.\ +$$

$$+ \left. \ \frac{1}{2} \bullet 4 \bullet 0,00488216 \bullet \left( \frac{2}{3} \bullet 1,309718004 - \frac{1}{3} \bullet 2,690281996 \right) \right\rbrack +$$

$$+ \frac{1}{\text{EI}}\left\lbrack \frac{1}{2} \bullet \frac{1}{\text{sinα}} \bullet 0,00488216 \bullet \left( \frac{2}{3} \bullet 1,309718004 + \frac{1}{3} \bullet 0,654859002 \right) \right.\ +$$

$$+ \frac{1}{2} \bullet \frac{1}{\text{sinα}} \bullet 1,58357991 \bullet \left( \frac{2}{3} \bullet 0,654859002 + \frac{1}{3} \bullet 1,309718004 \right) +$$

$$+ \left. \ \frac{1}{2} \bullet \frac{1}{\text{sinα}} \bullet 1,58357991 \bullet \frac{2}{3} \bullet 0,654859002 \right\rbrack = \mathbf{6,707133324}\frac{\mathbf{1}}{\mathbf{\text{EI}}}\left\lbrack \mathbf{m}^{\mathbf{3}} \right\rbrack$$

$$\left\{ \begin{matrix} \delta_{11} = 10,85558931\frac{1}{\text{EI}} \\ \delta_{12} = \delta_{21} = 6,707133324\frac{1}{\text{EI}} \\ \delta_{22} = 7,403649956\frac{1}{\text{EI}} \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} (\delta_{11}2,8m\omega^{2} - 1)A_{1} + \delta_{12}1,2m\omega^{2}A_{2} = 0 \\ \delta_{21}2,8m\omega^{2}A_{1} + (\delta_{22}1,2m\omega^{2} - 1)A_{2} = 0 \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} (10,85558931\frac{1}{\text{EI}}2,8m\omega^{2} - 1)A_{1} + 6,707133324\frac{1}{\text{EI}}1,2m\omega^{2}A_{2} = 0 \\ 6,707133324\frac{1}{\text{EI}}2,8m\omega^{2}A_{1} + (7,403649956\frac{1}{\text{EI}}1,2m\omega^{2} - 1)A_{2} = 0 \\ \end{matrix} \right.\ $$

Przyjęto parametr:

$$\lambda = \frac{m\omega^{2}}{\text{EI}}$$

$$\left\{ \begin{matrix} (30,39565007\lambda - 1)A_{1} + 8,048559989\lambda A_{2} = 0 \\ 18,77997331\lambda A_{1} + (8,884379947\lambda - 1)A_{2} = 0 \\ \end{matrix} \right.\ $$

$$\det\begin{bmatrix} (30,39565007\lambda - 1) & 8,048559989\lambda \\ 18,77997331\lambda & (8,884379947\lambda - 1) \\ \end{bmatrix} = 0$$

(30,39565007λ−1) • (8,884379947λ−1) − 18, 77997331λ • 8, 048559989λ = 0

270, 046504λ² − 8, 884379947λ − 30, 39565007λ + 1 − 151, 1517418λ² = 0

118, 8947622λ² − 39, 28003002λ + 1 = 0

Δ = b² − 4ac = ( − 39, 28003002)² − 4 • 118, 8947622 • 1 = 1067, 341709

$$\sqrt{} = 32,67019604$$

$$\lambda_{1} = \frac{- b - \sqrt{\Delta}}{2a} = \frac{- \left( - 39,28003002 \right) - 32,67019604}{2 \bullet 118,8947622} = 0,027796994$$

$$\lambda_{2} = \frac{- b + \sqrt{\Delta}}{2a} = \frac{- \left( - 39,28003002 \right) + 32,67019604}{2 \bullet 118,8947622} = 0,302579461$$

Częstości kołowe drgań własnych:

$$\omega = \sqrt{\frac{\text{λEI}}{m}}$$

$$\omega_{1} = \sqrt{\frac{\lambda_{1}\text{EI}}{m}} = \sqrt{\frac{0,027796994 \bullet 63181000}{100}} = \mathbf{132,5232764\ }\left\lbrack \frac{\mathbf{\text{rad}}}{\mathbf{s}} \right\rbrack$$

$$\omega_{2} = \sqrt{\frac{\lambda_{2}\text{EI}}{m}} = \sqrt{\frac{0,302579461 \bullet 63181000}{100}} = \mathbf{437,2330377\ }\left\lbrack \frac{\mathbf{\text{rad}}}{\mathbf{s}} \right\rbrack$$

AMPLITUDY DRGAŃ WYMUSZONYCH W PUNKTACH MAS SKUPIONYCH WYWOŁANE DZIAŁANIEM SIŁY WYMUSZAJĄCEJ P(t)=sin(pt)

Siły bezwładności:

$$\left\{ \begin{matrix} B_{1}\left( t \right) = - m_{1}{\ddot{q}}_{1}(t) \\ B_{2}\left( t \right) = - m_{2}{\ddot{q}}_{2}\left( t \right) \\ P\left( t \right) = P \bullet sin(pt) \\ \end{matrix} \right.\ $$

Przyjęto masę porównawczą m=100kg oraz m₁ = 2, 8m oraz m₂ = 1, 2m.

$$\left\{ \begin{matrix} B_{1}\left( t \right) = - 2,8m{\ddot{q}}_{1}(t) \\ B_{2}\left( t \right) = - 1,2m{\ddot{q}}_{2}\left( t \right) \\ P\left( t \right) = P \bullet sin(pt) \\ \end{matrix} \right.\ $$

Równania ruchu:

$$\left\{ \begin{matrix} q_{1}\left( t \right) = \delta_{11}B_{1}\left( t \right) + \delta_{12}B_{2}\left( t \right) + \delta_{13}P(t) \\ q_{2}(t) = \delta_{21}B_{1}\left( t \right) + \delta_{22}B_{2}\left( t \right) + \delta_{23}P(t) \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} q_{1}\left( t \right) = \delta_{11}\left( - 2,8m{\ddot{q}}_{1}\left( t \right) \right) + \delta_{12}\left( - 1,2m{\ddot{q}}_{2}\left( t \right) \right) + \delta_{13}P \bullet sinpt \\ q_{2}(t) = \delta_{21}( - 2,8m{\ddot{q}}_{1}\left( t \right)) + \delta_{22}( - 1,2m{\ddot{q}}_{2}\left( t \right)) + \delta_{23}P \bullet sinpt \\ \end{matrix} \right.\ $$

Przyjęto: q₁ = A₁sinpt stąd: ${\ddot{q}}_{1} = - p^{2}A_{1}\text{sinp}t$

q₂ = A₂sinpt ${\ddot{q}}_{2} = - p^{2}A_{2}\text{sinp}t$

$$\left\{ \begin{matrix} A_{1}\text{sinp}t = \delta_{11}2,8mp^{2}A_{1}\text{sinp}t + \delta_{12}1,2mp^{2}A_{2}\text{sinp}t + \delta_{13}P \bullet sinpt \\ A_{2}\text{sinp}t = \delta_{21}2,8mp^{2}A_{1}\text{sinp}t + \delta_{22}1,2mp^{2}A_{2}\text{sinp}t + \delta_{23}P \bullet sinpt \\ \end{matrix} \right.\ \ \ /:\text{sinp}t$$

$$\left\{ \begin{matrix} A_{1} = \delta_{11}2,8mp^{2}A_{1} + \delta_{12}1,2mp^{2}A_{2} + \delta_{13}P \\ A_{2} = \delta_{21}2,8mp^{2}A_{1} + \delta_{22}1,2mp^{2}A_{2} + \delta_{23}P \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} A_{1} - \delta_{11}2,8mp^{2}A_{1} - \delta_{12}1,2mp^{2}A_{2} = \delta_{13}P \\ - \delta_{21}2,8mp^{2}A_{1} + A_{2} - \delta_{22}1,2mp^{2}A_{2} = \delta_{23}P \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} A_{1}\left( 1 - \delta_{11}2,8mp^{2} \right) - \delta_{12}1,2mp^{2}A_{2} = \delta_{13}P \\ - \delta_{21}2,8mp^{2}A_{1} + A_{2}(1 - \delta_{22}1,2mp^{2}) = \delta_{23}P \\ \end{matrix} \right.\ $$

Ustalenie wartości δ_ik wykonano metodą sił.

Stan X₁ = 1

$${\tilde{\delta}}_{11} \bullet X_{1}^{i} + {\tilde{}}_{1P}^{i} = 0$$

$${\tilde{\delta}}_{11} = \int_{s}^{}{\frac{M_{1} \bullet M_{1}}{\text{EI}}\text{ds}} = \frac{1}{\text{EI}}\left\lbrack 4 \bullet 6 \bullet 6 + \frac{1}{2} \bullet 6 \bullet \frac{2}{\text{sinα}} \bullet \frac{2}{3} \bullet 6 \right\rbrack = 219,8946639\frac{1}{\text{EI}}\lbrack m^{3}\rbrack$$

Stan P₃ = 1

$${\tilde{}}_{1P}^{3} = \int_{s}^{}{\frac{M_{1} \bullet M_{P}^{3}}{\text{EI}}\text{ds}} = \frac{1}{\text{EI}}\left\lbrack - \frac{1}{2} \bullet 4 \bullet 6 \bullet 6 - \frac{1}{2} \bullet 4 \bullet 2 \bullet 6 \right.\ \left. \ - \frac{1}{2} \bullet \frac{2}{\text{sinα}} \bullet 2 \bullet \frac{2}{3} \bullet 6 \right\rbrack = - 121,2982213\frac{1}{\text{EI}}\ $$

$$X_{1}^{3} = - \frac{{\tilde{}}_{1P}^{3}}{{\tilde{\delta}}_{11}} = \frac{121,2982213}{\text{EI}} \bullet \frac{\text{EI}}{219,8946639} = 0,551619667$$

Wykresy momentów od siły P₁ = 1 dla układy statycznie niewyznaczalnego:

$$\delta_{13} = \int_{s}^{}{\frac{M_{1}^{D} \bullet M_{3}^{D}}{\text{EI}}\text{ds}} = \frac{1}{\text{EI}}\left\lbrack \frac{1}{2} \bullet 4 \bullet 2,690281996 \bullet \left( \frac{2}{3} \bullet 2,690281998 - \frac{1}{3} \bullet 1,309718002 \right) \right.\ +$$

$$+ \left. \ \frac{1}{2} \bullet 4 \bullet 1,309718004 \bullet \left( \frac{2}{3} \bullet 1,309718002 - \frac{1}{3} \bullet 2,690281998 \right) \right\rbrack +$$

$$+ \frac{1}{\text{EI}}\left\lbrack \frac{1}{2} \bullet 1,309718004 \bullet \frac{1}{\text{sinα}} \bullet \left( \frac{2}{3} \bullet 1,309718002 + \frac{1}{3} \bullet 0,654859001 \right) \right.\ +$$

$$+ \frac{1}{2} \bullet \frac{1}{\text{sinα}} \bullet 0,654859002 \bullet \left( \frac{2}{3} \bullet 0,654859001 + \frac{1}{3} \bullet 1,309718002 \right) +$$

$$+ \left. \ \frac{1}{2} \bullet \frac{1}{\text{sinα}} \bullet 0,654859002 \bullet \frac{2}{3} \bullet 0,654859001 \right\rbrack = \mathbf{10,85558931}\frac{\mathbf{1}}{\mathbf{\text{EI}}}\mathbf{\ }\left\lbrack \mathbf{m}^{\mathbf{3}} \right\rbrack$$

$$\delta_{23} = \int_{s}^{}{\frac{M_{2}^{D} \bullet M_{3}^{D}}{\text{EI}}\text{ds}} = \frac{1}{\text{EI}}\left\lbrack \frac{1}{2} \bullet 4 \bullet 1,260028742 \bullet \left( \frac{2}{3} \bullet 2,690281998 - \frac{1}{3} \bullet 1,309718002 \right) \right.\ +$$

$$+ \left. \ \frac{1}{2} \bullet 4 \bullet 0,00488216 \bullet \left( \frac{2}{3} \bullet 1,309718002 - \frac{1}{3} \bullet 2,690281998 \right) \right\rbrack +$$

$$+ \frac{1}{\text{EI}}\left\lbrack \frac{1}{2} \bullet \frac{1}{\text{sinα}} \bullet 0,00488216 \bullet \left( \frac{2}{3} \bullet 1,309718002 + \frac{1}{3} \bullet 0,654859001 \right) \right.\ +$$

$$+ \frac{1}{2} \bullet \frac{1}{\text{sinα}} \bullet 1,58357991 \bullet \left( \frac{2}{3} \bullet 0,654859001 + \frac{1}{3} \bullet 1,309718002 \right) +$$

$$+ \left. \ \frac{1}{2} \bullet \frac{1}{\text{sinα}} \bullet 1,58357991 \bullet \frac{2}{3} \bullet 0,654859001 \right\rbrack = \mathbf{6,707133324}\frac{\mathbf{1}}{\mathbf{\text{EI}}}\mathbf{\ }\left\lbrack \mathbf{m}^{\mathbf{3}} \right\rbrack$$

$$\begin{matrix} \delta_{11} = 10,85558931\frac{1}{\text{EI}} & \delta_{12} = 6,707133324\frac{1}{\text{EI}} & \delta_{13} = 10,85558931\frac{1}{\text{EI}} \\ \delta_{21} = 6,707133324\frac{1}{\text{EI}} & \delta_{22} = 7,403649956\frac{1}{\text{EI}} & \delta_{23} = 6,707133324\frac{1}{\text{EI}} \\ \end{matrix}$$

$$\left\{ \begin{matrix} A_{1}\left( 1 - \frac{10,85558931}{\text{EI}}2,8mp^{2} \right) - \frac{6,707133324}{\text{EI}}1,2mp^{2}A_{2} = \frac{10,85558931}{\text{EI}}P \\ - \frac{6,707133324}{\text{EI}}2,8mp^{2}A_{1} + A_{2}(1 - \frac{7,403649956}{\text{EI}}1,2mp^{2}) = \frac{6,707133324}{\text{EI}}P \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} A_{1}\left( 1 - \frac{10,85558931}{63181000}2,8 \bullet 100 \bullet 40^{2} \right) - \frac{6,707133324}{63181000}1,2 \bullet 100 \bullet 40^{2}A_{2} = \frac{10,85558931}{63181000}100000 \\ - \frac{6,707133324}{63181000}2,8 \bullet 100 \bullet 40^{2}A_{1} + A_{2}(1 - \frac{7,403649956}{63181000}1,2 \bullet 100 \bullet 40^{2}) = \frac{6,707133324}{63181000}100000 \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} 0,923025846A_{1} - 0,02032229A_{2} = 0,017181731 \\ - 0,047558534A_{1} + 0,977501135A_{2} = 0,010615744 \\ \end{matrix} \right.\ $$

0, 923025846A₁ − 0, 02032229A₂ = 0, 017181731

−0, 047558534A₁ + 0, 977501135A₂ = 0, 010615744

$$\overline{0,875467312A_{1} + 0,957178845A_{2} = 0,027797475}$$

A₁ = −1, 093334762A₂ + 0, 031751585

0, 923025846(−1,093334762A₂+0,031751585) − 0, 02032229A₂ = 0, 017181731

A₂ = 0, 011778359

A₁ = −1, 093334762 • 0, 011778359 + 0, 031751585

A₁ = 0, 018873896

Sprawdzenie:

0, 923025846 • 0, 018873896 − 0, 02032229 • 0, 011778359 = 0, 017181731

0, 017181731 = 0, 017181731

−0, 047558534 • 0, 018873896 + 0, 977501135 • 0, 011778359 = 0, 010615744

0, 010615744 = 0, 010615744

Amplitudy drgań wymuszonych:

$$\left\{ \begin{matrix} \mathbf{A}_{\mathbf{1}}\mathbf{= 0,018873896} \\ \mathbf{A}_{\mathbf{2}}\mathbf{= 0,011778359} \\ \end{matrix} \right.\ $$

WYKRESY EKSTREMALNYCH MOMENTÓW ZGINAJĄCYCH

Siły bezwładności dla sin pt = 1,0:

$$\left\{ \begin{matrix} B_{1}\left( t \right) = - m_{1}{\ddot{q}}_{1}\left( t \right) = m_{1}p^{2}A_{1}sinpt = 2,8mp^{2}A_{1}\text{sinpt} \\ B_{2}\left( t \right) = - m_{2}{\ddot{q}}_{2}\left( t \right) = m_{2}p^{2}A_{2}sinpt = 1,2mp^{2}A_{2}\text{sinpt} \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} B_{1}\left( t \right) = 2,8 \bullet 100 \bullet 40^{2} \bullet 0,018873896 \bullet 1,0 = 8455,505408\ \lbrack N\rbrack \\ B_{2}\left( t \right) = 1,2 \bullet 100 \bullet 40^{2} \bullet 0,011778359 \bullet 1,0 = 2261,444928\ \lbrack N\rbrack \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} \mathbf{B}_{\mathbf{1}}\left( \mathbf{t} \right)\mathbf{= 8,455505408\ \lbrack kN\rbrack} \\ \mathbf{B}_{\mathbf{2}}\left( \mathbf{t} \right)\mathbf{= 2,261444928\ \lbrack kN\rbrack} \\ \end{matrix} \right.\ $$

Obciążenie statyczne:

Q₁ = m₁ • g = 280 • 9, 81 = 2746, 8 [N] =  2, 7468 [kN]

Q₂ = m₂ • g = 120 • 9, 81 = 1177, 2 [N] =  1, 1772 [kN]

Wykresy momentów zginających od obciążenia statycznego:

$${\tilde{\delta}}_{11} \bullet X_{1}^{i} + {\tilde{}}_{1P}^{i} = 0$$

Stan X₁ = 1

$${\tilde{\delta}}_{11} = \int_{s}^{}{\frac{M_{1} \bullet M_{1}}{\text{EI}}\text{ds}} = \frac{1}{\text{EI}}\left\lbrack 4 \bullet 6 \bullet 6 + \frac{1}{2} \bullet 6 \bullet \frac{2}{\text{sinα}} \bullet \frac{2}{3} \bullet 6 \right\rbrack = 219,8946639\frac{1}{\text{EI}}\lbrack m^{3}\rbrack$$

Stan „P”

$${\tilde{}}_{1P}^{S} = \int_{s}^{}{\frac{M_{1} \bullet M_{P}^{S}}{\text{EI}}\text{ds}} = \frac{1}{\text{EI}}\left\lbrack \frac{1}{2} \bullet \frac{1}{\text{sinα}} \bullet 3,5316 \bullet \left( - \frac{2}{3} \bullet 6 - \frac{1}{3} \bullet 3 \right) + 4 \bullet 3,5316 \bullet \left( - 6 \right) \right\rbrack$$

$$\ {\tilde{}}_{1P}^{S} = - 112,6781495\frac{1}{\text{EI}}\ \left\lbrack \text{kN}m^{3} \right\rbrack$$

$$X_{1}^{S} = - \frac{{\tilde{}}_{1P}^{S}}{{\tilde{\delta}}_{11}} = \frac{112,6781495}{\text{EI}} \bullet \frac{\text{EI}}{219,8946639} = 0,512418753\ \left\lbrack \text{kN} \right\rbrack$$

Wykresy momentów zginających od obciążenia dynamicznego:

Stan X₁ = 1

$${\tilde{\delta}}_{11} \bullet X_{1}^{i} + {\tilde{}}_{1P}^{i} = 0$$

$${\tilde{\delta}}_{11} = \int_{s}^{}{\frac{M_{1} \bullet M_{1}}{\text{EI}}\text{ds}} = \frac{1}{\text{EI}}\left\lbrack 4 \bullet 6 \bullet 6 + \frac{1}{2} \bullet 6 \bullet \frac{2}{\text{sinα}} \bullet \frac{2}{3} \bullet 6 \right\rbrack = 219,8946639\frac{1}{\text{EI}}\lbrack m^{3}\rbrack$$

Stan „P”

$${\tilde{}}_{1P}^{D} = \int_{s}^{}{\frac{M_{1} \bullet M_{P}^{D}}{\text{EI}}\text{ds}} = \frac{1}{\text{EI}}\left\lbrack 6 \bullet 4 \bullet \left( - \frac{1}{2} \bullet 643,8338651 - \frac{1}{2} \bullet 207,1513168 \right) \right.\ +$$

$$+ \frac{1}{2} \bullet \frac{1}{\text{sinα}} \bullet 207,1513168 \bullet \left( - \frac{2}{3} \bullet 6 - \frac{1}{3} \bullet 3 \right) + \frac{1}{2} \bullet 100 \bullet \frac{1}{\text{sinα}} \bullet \left( - \frac{1}{3} \bullet 6 - \frac{2}{3} \bullet 3 \right) +$$

$$\left. \ + \frac{1}{2} \bullet 100 \bullet \frac{1}{\text{sinα}} \bullet \frac{2}{3} \bullet 3 \right\rbrack = - 12165,7249\frac{1}{\text{EI}}\ \left\lbrack \text{kN}m^{3} \right\rbrack$$

$$X_{1}^{D} = - \frac{{\tilde{}}_{1P}^{D}}{{\tilde{\delta}}_{11}} = \frac{12165,7249}{\text{EI}} \bullet \frac{\text{EI}}{219,8946639} = 55,3252393\ \left\lbrack \text{kN} \right\rbrack$$