Układ wyjściowy:
Układ podstawowy:
Układ równań kanonicznych:
$$\left\{ \begin{matrix}
r_{11} \bullet Z_{1} + r_{12} \bullet Z_{2} + R_{1p} = 0 \\
r_{21} \bullet Z_{1} + r_{22} \bullet Z_{2} + R_{2p} = 0 \\
\end{matrix} \right.\ $$
Stan Z1:
Wartości przywęzłowych momentów podporowych:
$$M_{12} = - \frac{1}{L_{12}}EI = - \frac{1}{4}\text{EI}\text{\ \ \ \ \ \ \ \ \ \ }M_{21} = \frac{1}{L_{21}}EI = \frac{1}{4}EI$$
$$M_{23} = \frac{3}{L_{23}}EI = \frac{3}{5}\text{EI}\text{\ \ \ \ \ \ \ \ \ \ }M_{32} = 0$$
$$M_{24} = \frac{3}{L_{24}}EI = \frac{3}{6}2EI = EI\text{\ \ \ \ \ \ \ \ \ \ }M_{42} = 0$$
M45 = 0 M54 = 0
Wykres momentów stanu Z1:
$$\mathbf{r}_{\mathbf{11}} = \frac{1}{4}EI + EI + \frac{3}{5}EI = \frac{\mathbf{37}}{\mathbf{20}}\mathbf{\text{EI}}$$
Stan wirtualny – węzłowi „4” nadano przesuw poziomy Z2=1:
Wartości kątów obrotu prętów obliczono z równań łańcucha kinematycznego:
↓12
4 • ψ12 = 0 ⇒ ψ12 = 0
→54
$$5 \bullet \psi_{45} = 1\ \ \ \ \ \Longrightarrow \ \ \ \ \ \psi_{45} = \frac{1}{5}$$
↓542
0 • ψ45 − 6 • ψ24 = 0 ⇒ ψ24 = 0
→3245
$$4 \bullet \psi_{23} + 0 \bullet \psi_{24} - 5 \bullet \psi_{45} = 0\ \ \ \ \ \Longrightarrow \ \ \ \ \ \psi_{23} = \frac{1}{4}$$
Stan Z2:
Wartości przywęzłowych momentów podporowych:
M12 = 0 M21 = 0
$$M_{23} = - \frac{3}{L_{23}}EI \bullet \psi_{23} = - \frac{3}{5}EI \bullet \frac{1}{4} = - \frac{3}{20}\text{EI\ \ \ \ \ \ \ \ \ \ }M_{32} = 0$$
M24 = 0 M42 = 0
$$M_{45} = 0\ \ \ \ \ \ \ \ \ \ M_{54} = - \frac{3}{L_{54}}EI \bullet \psi_{45} = - \frac{3}{5}EI \bullet \frac{1}{5} = - \frac{3}{25}\text{EI\ \ \ \ \ \ \ \ \ \ }$$
Wykres momentów stanu Z1:
$$\mathbf{r}_{\mathbf{21}}\mathbf{= -}\frac{\mathbf{3}}{\mathbf{20}}\mathbf{\text{EI}}$$
Wartości r12 i r22 obliczono wykorzystując równanie pracy wirtualnej. W tym celu węzłowi „4” nadano wirtualny przesuw poziomy ${\overset{\overline{}}{}}_{4} = \overset{\overline{}}{1}$ zgodny z kierunkiem niewiadomych wartości. Wirtualne wartości ${\overset{\overline{}}{\psi}}_{\text{ik}}$ określają kąty obrotu prętów obliczone dla stanu wirtualnego.
$$r_{12} \bullet \overset{\overline{}}{1} + \left( M_{32} + M_{23} \right) \bullet {\overset{\overline{}}{\psi}}_{23} + \left( M_{12} + M_{21} \right) \bullet {\overset{\overline{}}{\psi}}_{12} + \left( M_{24} + M_{42} \right) \bullet {\overset{\overline{}}{\psi}}_{24} + \left( M_{45} + M_{54} \right) \bullet {\overset{\overline{}}{\psi}}_{45} = 0$$
$$r_{12} = - \left( 0 + \frac{3}{5}\text{EI} \right) \bullet \frac{\overset{\overline{}}{1}}{4} - \left( - \frac{1}{4}EI + \frac{1}{4}\text{EI} \right) \bullet 0 \bullet \overset{\overline{}}{1} - \left( EI + 0 \right) \bullet 0 \bullet \overset{\overline{}}{1} - (0 + 0) \bullet \frac{\overset{\overline{}}{1}}{5}$$
$$\mathbf{r}_{\mathbf{12}}\mathbf{= -}\frac{\mathbf{3}}{\mathbf{5}}\mathbf{\text{EI}}$$
$$r_{22} \bullet \overset{\overline{}}{1} + \left( M_{32} + M_{23} \right) \bullet {\overset{\overline{}}{\psi}}_{23} + \left( M_{12} + M_{21} \right) \bullet {\overset{\overline{}}{\psi}}_{12} + \left( M_{24} + M_{42} \right) \bullet {\overset{\overline{}}{\psi}}_{24} + \left( M_{45} + M_{54} \right) \bullet {\overset{\overline{}}{\psi}}_{45} = 0$$
$$r_{22} = - \left( 0 - \frac{3}{20}\text{EI} \right) \bullet \frac{\overset{\overline{}}{1}}{4} - \left( 0 + 0 \right) \bullet 0 \bullet \overset{\overline{}}{1} - \left( 0 + 0 \right) \bullet 0 \bullet \overset{\overline{}}{1} - (0 - \frac{3}{25}EI) \bullet \frac{\overset{\overline{}}{1}}{5}$$
$$\mathbf{r}_{\mathbf{22}} = \frac{3}{80}\text{EI} + \frac{3}{125}EI = \frac{\mathbf{123}}{\mathbf{2000}}\mathbf{\text{EI}}$$
Stan „P”:
Pręt 1-2
$$M_{12} = - \frac{ql^{2}}{6} = - \frac{6 \bullet 4^{2}}{6} = - 16\ \lbrack kNm\rbrack$$
$$M_{21} = - \frac{ql^{2}}{3} = - \frac{6 \bullet 4^{2}}{3} = - 32\ \lbrack kNm\rbrack$$
Pręt 2-3
$$M_{23} = \frac{q{l_{y}}^{2}}{8} = \frac{6 \bullet 4^{2}}{8} = 12\ \lbrack kNm\rbrack$$
M21 = 0
Pręt 2-4
$$M_{24} = - \frac{3Pl}{16} = - \frac{3 \bullet 90 \bullet 6^{2}}{16} = - 101,25\ \lbrack kNm\rbrack$$
M42 = 0 [kNm]
$$M_{4'} = \frac{5Pl}{32} = \frac{5 \bullet 90 \bullet 6}{32} = 84,375\ \lbrack kNm\rbrack$$
Wykres momentów stanu „P”:
R1p = −32 − 101, 25 + 12 = −121, 25 [kNm]
Wyznaczenie wartości przemieszczeń wirtualnych sił QA, QB oraz P zakładając wirtualny przesuw poziomy węzła „4” o 4=1:
QA = QB = 6 • 4 = 24kN
$$\left\{ \begin{matrix}
{\overset{\overline{}}{}}_{A} = \overset{\overline{}}{1} \\
{\overset{\overline{}}{}}_{B} = 2 \bullet {\overset{\overline{}}{\psi}}_{23} = 2 \bullet \frac{\overset{\overline{}}{1}}{4} = \frac{\overset{\overline{}}{1}}{2} \\
{\overset{\overline{}}{}}_{C} \\
\end{matrix} \right.\ $$
Wartość R2p wyznaczono wykorzystując równanie pracy wirtualnej:
$$R_{2p} \bullet \overset{\overline{}}{1} + \left( M_{32} + M_{23} \right) \bullet {\overset{\overline{}}{\psi}}_{23} + \left( M_{12} + M_{21} \right) \bullet {\overset{\overline{}}{\psi}}_{12} + \left( M_{24} + M_{42} \right) \bullet {\overset{\overline{}}{\psi}}_{24} + \left( M_{45} + M_{54} \right) \bullet {\overset{\overline{}}{\psi}}_{45} +$$
+QB•B + QA•A + P•C = 0
$$R_{2p} \bullet \overset{\overline{}}{1} = - \left( 0 + 12 \right) \bullet \frac{\overset{\overline{}}{1}}{4} - ( - 16 - 32) \bullet 0 \bullet \overset{\overline{}}{1} - ( - 101,25 + 0) \bullet 0 \bullet \overset{\overline{}}{1} + \left( 0 + 0 \right) \bullet \frac{\overset{\overline{}}{1}}{5} -$$
$$- 24 \bullet \frac{\overset{\overline{}}{1}}{2} - 24 \bullet \overset{\overline{}}{1} - 90 \bullet 0$$
R2p = −3 − 24 − 12 = −39 [kN]
Rozwiązanie układu równań kanonicznych:
$$\left\{ \begin{matrix}
r_{11} \bullet Z_{1} + r_{12} \bullet Z_{2} + R_{1p} = 0 \\
r_{21} \bullet Z_{1} + r_{22} \bullet Z_{2} + R_{2p} = 0 \\
\end{matrix} \right.\ $$
$$\left\{ \begin{matrix}
\frac{37}{20}EI \bullet Z_{1} - \frac{3}{20}EI \bullet Z_{2} - 121,25 = 0 \\
- \frac{3}{20}EI \bullet Z_{1} + \frac{123}{2000}EI \bullet Z_{2} - 39 = 0 \\
\end{matrix} \right.\ $$
Po rozwiązaniu układu równań kanonicznych otrzymano:
$$\left\{ \begin{matrix}
Z_{1} = 145,788825\frac{1}{\text{EI}} \\
Z_{2} = 989,7288414\frac{1}{\text{EI}} \\
\end{matrix} \right.\ $$
Sprawdzenie poprawności obliczeń:
$$\frac{37}{20}EI \bullet 145,788825\frac{1}{\text{EI}} - \frac{3}{20}EI \bullet 989,7288414\frac{1}{\text{EI}} - 121,25 = 0$$
0, 000 = 0
$$- \frac{3}{20}EI \bullet 145,788825\frac{1}{\text{EI}} + \frac{123}{2000}EI \bullet 989,7288414\frac{1}{\text{EI}} - 39 = 0$$
0, 000 = 0
Obliczenia momentów przywęzłowych z wykorzystaniem superpozycji:
Mikn = Mik0 + Mik1 • Z1 + Mik2 • Z2
M12n = M120 + M121 • Z1 + M122 • Z2
$$M_{12}^{n} = - 16 + \left( - \frac{1}{4}\text{EI} \right) \bullet 145,788825\frac{1}{\text{EI}} + 0 = - 52,44720625\ \lbrack kNm\rbrack$$
M21n = M210 + M211 • Z1 + M212 • Z2
$$M_{21}^{n} = - 32 + \frac{1}{4}EI \bullet 145,788825\frac{1}{\text{EI}} + 0 = 4,44720625\ \lbrack kNm\rbrack$$
M23n = M230 + M231 • Z1 + M232 • Z2
$$M_{23}^{n} = 12 + \frac{3}{5}EI \bullet 145,788825\frac{1}{\text{EI}} - \frac{3}{20}EI \bullet 989,7288414\frac{1}{\text{EI}} = - 48,98603121\ \lbrack kNm\rbrack$$
M32n = M320 + M321 • Z1 + M322 • Z2
M32n = 0 + 0 + 0 = 0 [kNm]
M24n = M240 + M241 • Z1 + M242 • Z2
$$M_{24}^{n} = - 101,25 + EI \bullet 145,788825\frac{1}{\text{EI}} + 0 = 44,538825\ \lbrack kNm\rbrack$$
M4′n = M4′0 + M4′1 • Z1 + M4′2 • Z2
$$M_{4'}^{n} = 84,375 + \frac{1}{2}EI \bullet 145,788825\frac{1}{\text{EI}} + 0 = 157,2694125\ \lbrack kNm\rbrack$$
M42n = M420 + M421 • Z1 + M422 • Z2
M42n = 0 + 0 + 0 = 0 [kNm]
M45n = M450 + M451 • Z1 + M452 • Z2
M45n = 0 + 0 + 0 = 0 [kNm]
M54n = M540 + M541 • Z1 + M542 • Z2
$$M_{54}^{n} = 0 + 0 - \frac{3}{25}EI \bullet 989,7288414\frac{1}{\text{EI}} = - 118,767461\ \lbrack kNm\rbrack$$
Ostateczny wykres momentów:
ΣM2 = 0
48, 98603121 − 4, 44720625−
−44, 538825 = 0
0, 000 = 0
Sprawdzenie kinematyczne:
$$\overset{\overline{}}{1} \bullet_{3} = \sum_{}^{}{\int_{s}^{}{\frac{M_{p}^{n} \bullet \overset{\overline{}}{M}}{\text{EI}}\text{ds}}}$$
$$_{3} = \frac{1}{\text{EI}}\left\lbrack \frac{1}{2} \bullet 5 \bullet 48,98603121 \bullet \frac{2}{3} \bullet \left( - 4 \right) + \frac{2}{3} \bullet \frac{6 \bullet 4^{2}}{8} \bullet 5 \bullet \frac{1}{2} \bullet \left( - 4 \right) \right\rbrack +$$
$$+ \frac{1}{2EI}\left\lbrack \frac{1}{2} \bullet 3 \bullet 44,538825 \bullet \left( \frac{2}{3} \bullet \left( - 4 \right) + \frac{1}{3} \bullet \left( - 2 \right) \right) + \right.\ $$
$$+ \frac{1}{2} \bullet 3 \bullet 157,2694125 \bullet \left( \frac{1}{3} \bullet \left( - 4 \right) + \frac{2}{3} \bullet \left( - 2 \right) \right) + \left. \ \frac{1}{2} \bullet 157,2694125 \bullet 3 \bullet \frac{2}{3} \bullet \left( - 2 \right) \right\rbrack +$$
$$+ \frac{1}{\text{EI}}\left\lbrack \frac{1}{2} \bullet 5 \bullet 118,767461 \bullet \frac{2}{3} \bullet 5 \right\rbrack = \frac{1}{\text{EI}}\left\lbrack - 406,5735414 \right\rbrack +$$
$$+ \frac{1}{2EI}\left\lbrack - 222,694125 - 629,07765 - 314,538825 \right\rbrack + \frac{1}{\text{EI}}\left\lbrack 989,7288417 \right\rbrack =$$
=0, 0000003 ≅ 0, 000
Obliczenia sił poprzecznych:
Pręt 2-3
ΣM2 = 0
$$- M_{23} + T_{32} \bullet 5 - \frac{6 \bullet 4^{2}}{2} = 0$$
$$T_{32} \bullet 5 = M_{23} + \frac{6 \bullet 4^{2}}{2}$$
$$T_{32} \bullet 5 = 48,98603121 + \frac{6 \bullet 4^{2}}{2}$$
T32 = 19, 39720624 [kN]
ΣM3 = 0
$$- M_{23} - T_{23} \bullet 5 + \frac{6 \bullet 4^{2}}{2} = 0$$
$$T_{23} \bullet 5 = - M_{23} + \frac{6 \bullet 4^{2}}{2}$$
$$T_{23} \bullet 5 = - 48,98603121 + \frac{6 \bullet 4^{2}}{2}$$
T23 = −0, 197206242 [kN]
Pręt 1-2
ΣM1 = 0
$$- M_{12} + M_{21} + T_{21} \bullet 4 - \frac{6 \bullet 4^{2}}{2} = 0$$
$$T_{21} \bullet 4 = M_{12} - M_{21} + \frac{6 \bullet 4^{2}}{2}$$
T21 • 4 = 52, 44720625 − 4, 44720625+
$$+ \frac{6 \bullet 4^{2}}{2}$$
T21 = 24 [kN]
Pręt 2-4
ΣM2 = 0
M24 − T42 • 6 + 90 • 3 = 0
T42 • 6 = M24 + 270
T42 • 6 = 44, 538825 + 270
T42 = 52, 4231375 [kN]
ΣM4 = 0
M24 + T24 • 6 − 90 • 3 = 0
T24 • 6 = −M24 + 270
T24 • 6 = −44, 538825 + 270
T24 = 37, 5768625 [kN]
Pręt 4-5
ΣM4 = 0
−M54 − T54 • 5 = 0
T54 • 5 = −M24
T54 • 5 = −118, 767461
T54 = −23, 7534922 [kN]
ΣM5 = 0
−M54 + T45 • 5 = 0
T45 • 5 = M54
T45 • 5 = 118, 767461
T45 = 23, 7534922 [kN]
Wykres sił poprzecznych:
Obliczenia sił normalnych:
Węzeł „4”
ΣX = 0
−N42 + T45 = 0
N42 = T45 = 23, 7534922 [kN]
N42 = N24
ΣY = 0
−N45 + T42 = 0
N45 = T42 = 52, 4231375 [kN]
N45 = N54
Węzeł „2”
sinβ = 0, 8 cosβ = 0, 6
ΣX = 0
−N23 • cosβ + T23 • sinβ − T21 + N24 = 0
$$N_{23} = \frac{T_{23} \bullet sin\beta - T_{21} + N_{24}}{\text{cosβ}} =$$
$$= \frac{0,197206242 \bullet 0,8 - 24 + 23,7534922}{0,6}$$
N23 = −0, 147904677 [kN]
ΣY = 0
−N23 • sinβ − T23 • cosβ − N21 + T24 = 0
N21 = T23 • cosβ + N23 • sinβ − T24=
N21 = (−0,147904677) • 0, 8 + 0, 197206242 • 0, 6 − 37, 5768625
N21 = −37, 5768625 [kN]
N21 = N12
Pręt 2-3
sinβ = 0, 8 cosβ = 0, 6
ΣX = 0
−N32 + N23 + q • cosβ • 4 = 0
N32 = N23 + q • cosβ • 4
N32 = 0, 14, 54904677 + 6 • 0, 6 • 4
N32 = 14, 54904677 [kN]
Wykres sił normalnych: