Rysunek:

Liczba sztywnych węzłów (1) i (2): w = 2 → φ₁, φ₂

Liczba możliwych przesuwów: p = 3t − 2p − r = 3 • 5 − 2 • 3 − 8 = 1 → ψ

Stopień geometrycznej niewyznaczalności: g = w + p = 2 + 1 = 3

Łańcuch kinematyczny:

Kąty obrotów:

w₃ = w₅ = w

w₁ = w₂ = w₄ = 0

Układ podstawowy:

Zablokowano jeden przesuw w kierunku poziomym i dwa obroty.

Przyjęto kąty obrotu przeciwnie do ruchu wskazówek zegara:

Kąty obrotów cięciw pręta:

ψ_1A = 0;  ψ_1B = ψ;  ψ_1C = 0;  ψ₁₂ = 0;  ψ_2D = ψ

Równania pracy sił wirtualnych:

$$1) - \overset{\overline{}}{\varphi_{1}}\left( M_{1A} + M_{1B} + M_{1C} + M_{12} \right) = 0$$

$$2)\ - \overset{\overline{}}{\varphi_{2}}\left( M_{21} + M_{2D} - 5ql^{2} \right) = 0$$

$$3)\ \left( M_{1A} + M_{A1} \right) \bullet \overset{\overline{}}{\psi_{1A}} + M_{1B} \bullet \overset{\overline{}}{\psi_{1B}} + M_{1C} \bullet \overset{\overline{}}{\psi_{1C}} + \left( M_{12} + M_{21} \right) \bullet \overset{\overline{}}{\psi_{12}} + \left( M_{2D} + M_{D2} \right) \bullet \overset{\overline{}}{\psi_{2D}} + 6q \bullet 6l \bullet 3l \bullet \overset{\overline{}}{\psi_{1B}} + \frac{11}{2}\sqrt{2}ql \bullet 3\sqrt{2}ql \bullet \overset{\overline{}}{\psi_{1C}} = 0$$

Tu coś pojebałem, u mnie i tak się zeruje, ale powinno być inaczej to na czerwono.

$$3)\ M_{1B} \bullet \overset{\overline{}}{\psi_{1B}} + \left( M_{2D} + M_{D2} \right) \bullet \overset{\overline{}}{\psi_{2D}} + 6q \bullet 6l \bullet 3l \bullet \overset{\overline{}}{\psi_{1B}} = 0$$

Zakładamy, że kąty są jednostkowe:

$$\overset{\overline{}}{\varphi_{1}} = \overset{\overline{}}{1};\ \ \ \ \ \ \overset{\overline{}}{\varphi_{2}} = \overset{\overline{}}{1};\ \ \ \ \ \ \overset{\overline{}}{\psi} = \overset{\overline{}}{1}$$

1)−(M_1A+M_1B+M_1C+M₁₂) = 0     /•(−1)   ⇒   M_1A + M_1B + M_1C + M₁₂ = 0

2)  − (M₂₁+M_2D−5ql²) = 0     /•(−1)   ⇒   M₂₁ + M_2D − 5ql² = 0

3) M_1B + M_2D + M_D2 + 6q • 6l • 3l = 0   ⇒   M₂₁ + M_2D − 5ql² = 0

Wzory transformacyjne:

$M_{1B}^{0} = \frac{1}{8} \bullet 6q \bullet \left( 6l \right)^{2} = 27ql^{2}$

$$M_{1C}^{0} = - \frac{3}{16} \bullet \frac{11}{2}\sqrt{2}ql \bullet 6\sqrt{2}l = - \frac{99}{8}ql^{2}$$

$$M_{1B} = \frac{3EJ}{6l} \bullet \left( \varphi_{1} - \psi_{1B} \right) + M_{1B}^{0} = \frac{1}{2} \bullet \frac{\text{EJ}}{l} \bullet \left( \varphi_{1} - \psi \right) + 27ql^{2}$$

$$M_{1C} = \frac{3\sqrt{2}\text{EJ}}{6\sqrt{2}l} \bullet \left( \varphi_{1} - \psi_{1C} \right) + M_{1C}^{0} = \frac{1}{2} \bullet \frac{\text{EJ}}{l} \bullet \varphi_{1} - \frac{99}{8}ql^{2}$$

$$M_{12} = \frac{2EJ}{6l} \bullet \left( 2\varphi_{1} + \varphi_{2} - 3\psi_{12} \right) = \frac{1}{3} \bullet \frac{\text{EJ}}{l} \bullet \left( 2\varphi_{1} + \varphi_{2} \right)$$

$$M_{21} = \frac{2EJ}{6l} \bullet \left( 2\varphi_{2} + \varphi_{1} - 3\psi_{12} \right) = \frac{1}{3} \bullet \frac{\text{EJ}}{l} \bullet \left( 2\varphi_{2} + \varphi_{1} \right)$$

$$M_{2D} = \frac{2EJ}{6l} \bullet \left( 2\varphi_{2} + \varphi_{D} - 3\psi_{2D} \right) = \frac{1}{3} \bullet \frac{\text{EJ}}{l} \bullet \left( 2\varphi_{2} - 3\psi \right)$$

$$M_{D2} = \frac{2EJ}{6l} \bullet \left( 2\varphi_{D} + \varphi_{2} - 3\psi_{2D} \right) = \frac{1}{3} \bullet \frac{\text{EJ}}{l} \bullet \left( \varphi_{2} - 3\psi \right)$$

$$M_{1A} = \frac{2EJ}{5l} \bullet \left( 2\varphi_{1} + \varphi_{A} - 3\psi_{1A} \right) = \frac{4}{5} \bullet \frac{\text{EJ}}{l} \bullet \varphi_{1}$$

$$M_{A1} = \frac{2EJ}{5l} \bullet \left( 2\varphi_{A} + \varphi_{1} - 3\psi_{1A} \right) = \frac{2}{5} \bullet \frac{\text{EJ}}{l} \bullet \varphi_{1}$$

Momenty M_A1 oraz M_1A przyjęto jak dla utwierdzeń, ponieważ siły normalne mają pomijalny wpływ na odkształcenia ram.

1) M_1A + M_1B + M_1C + M₁₂ = 0

2) M₂₁ + M_2D − 5ql² = 0

3) M_1B + M_2D + M_D2 + 108ql² = 0

Podstawienie wartości momentów do równań:

$$1)\ \frac{4}{5} \bullet \frac{\text{EJ}}{l} \bullet \varphi_{1} + \frac{1}{2} \bullet \frac{\text{EJ}}{l} \bullet \left( \varphi_{1} - \psi \right) + 27ql^{2} + \frac{1}{2} \bullet \frac{\text{EJ}}{l} \bullet \varphi_{1} - \frac{99}{8}ql^{2} + \frac{1}{3} \bullet \frac{\text{EJ}}{l} \bullet \left( 2\varphi_{1} + \varphi_{2} \right) = 0$$

$$2)\ \frac{1}{3} \bullet \frac{\text{EJ}}{l} \bullet \left( 2\varphi_{2} + \varphi_{1} \right) + \frac{1}{3} \bullet \frac{\text{EJ}}{l} \bullet \left( 2\varphi_{2} - 3\psi \right) - 5ql^{2} = 0$$

$$3)\ \frac{1}{2} \bullet \frac{\text{EJ}}{l} \bullet \left( \varphi_{1} - \psi \right) + 27ql^{2} + \frac{1}{3} \bullet \frac{\text{EJ}}{l} \bullet \left( 2\varphi_{2} - 3\psi \right) + \frac{1}{3} \bullet \frac{\text{EJ}}{l} \bullet \left( \varphi_{2} - 3\psi \right) + 108ql^{2} = 0$$

$$1)\ \frac{37}{15} \bullet \frac{\text{EJ}}{l} \bullet \varphi_{1} + \frac{1}{3} \bullet \frac{\text{EJ}}{l} \bullet \varphi_{2} - \frac{1}{2} \bullet \frac{\text{EJ}}{l} \bullet \ \psi + \frac{117}{8}\text{\ q}l^{2} = 0$$

$$2)\ \frac{1}{3} \bullet \frac{\text{EJ}}{l} \bullet \varphi_{1} + \frac{4}{3} \bullet \frac{\text{EJ}}{l} \bullet \varphi_{2} - \frac{\text{EJ}}{l} \bullet \ \psi - 5ql^{2} = 0$$

$$3)\ \ \frac{1}{2} \bullet \frac{\text{EJ}}{l} \bullet \varphi_{1} + \frac{\text{EJ}}{l} \bullet \varphi_{2} - \frac{5}{2} \bullet \frac{\text{EJ}}{l} \bullet \ \psi + 135ql^{2} = 0$$

Poniższa macierz musi być symetryczna:

$$\begin{bmatrix} \frac{37}{15} & \frac{1}{3} & - \frac{1}{2} \\ \frac{1}{3} & \frac{4}{3} & - 1 \\ \frac{1}{2} & 1 & - \frac{5}{2} \\ \end{bmatrix}\begin{bmatrix} \varphi_{1} \\ \varphi_{2} \\ \psi \\ \end{bmatrix} + \begin{bmatrix} \frac{117}{8} \\ - 5 \\ 135 \\ \end{bmatrix} \bullet \frac{ql^{3}}{\text{EJ}} = 0$$

Mnożymy dolne wartości przez (-1):

$$\begin{bmatrix} \frac{37}{15} & \frac{1}{3} & - \frac{1}{2} \\ \frac{1}{3} & \frac{4}{3} & - 1 \\ - \frac{1}{2} & - 1 & \frac{5}{2} \\ \end{bmatrix}\begin{bmatrix} \varphi_{1} \\ \varphi_{2} \\ \psi \\ \end{bmatrix} + \begin{bmatrix} \frac{117}{8} \\ - 5 \\ - 135 \\ \end{bmatrix} \bullet \frac{ql^{3}}{\text{EJ}} = 0$$

$$\varphi_{1} = 1,681\frac{ql^{3}}{\text{EJ}};\ \ \ \ \ \varphi_{2} = 62,974\frac{ql^{3}}{\text{EJ}};\ \ \ \ \ \psi = 79,526\frac{ql^{3}}{\text{EJ}}$$

Podstawienie wartości kątów do wzorów na momenty:

$$M_{1B} = \frac{1}{2} \bullet \frac{\text{EJ}}{l} \bullet \left( 1,681\frac{ql^{3}}{\text{EJ}} - 79,526\frac{ql^{3}}{\text{EJ}} \right) + 27ql^{2} = - 11,923\ ql^{2}$$

$$M_{1C} = \frac{1}{2} \bullet \frac{\text{EJ}}{l} \bullet 1,681\frac{ql^{3}}{\text{EJ}} - \frac{99}{8}ql^{2} = - 11,535ql^{2}$$

$$M_{12} = \frac{1}{3} \bullet \frac{\text{EJ}}{l} \bullet \left( 2 \bullet 1,681\frac{ql^{3}}{\text{EJ}} + 62,974\frac{ql^{3}}{\text{EJ}} \right) = 22,112ql^{2}$$

$$M_{21} = \frac{1}{3} \bullet \frac{\text{EJ}}{l} \bullet \left( 2 \bullet 62,974\frac{ql^{3}}{\text{EJ}} + 1,681\frac{ql^{3}}{\text{EJ}} \right) = 42,543ql^{2}$$

$$M_{2D} = \frac{1}{3} \bullet \frac{\text{EJ}}{l} \bullet \left( 2 \bullet 62,974\frac{ql^{3}}{\text{EJ}} - 3 \bullet 79,526\frac{ql^{3}}{\text{EJ}} \right) = - 37,543ql^{2}$$

$$M_{D2} = \frac{1}{3} \bullet \frac{\text{EJ}}{l} \bullet \left( 62,974\frac{ql^{3}}{\text{EJ}} - 3 \bullet 79,526\frac{ql^{3}}{\text{EJ}} \right) = - 58,535ql^{2}$$

$$M_{1A} = \frac{4}{5} \bullet \frac{\text{EJ}}{l} \bullet 1,681\frac{ql^{3}}{\text{EJ}} = 1,345ql^{2}$$

$$M_{A1} = \frac{2}{5} \bullet \frac{\text{EJ}}{l} \bullet 1,681\frac{ql^{3}}{\text{EJ}} = 0,672ql^{2}$$

Siły tnące:

$$\sum_{}^{}M_{1} = 0:\ T_{A1} \bullet 5l - M_{A1} - M_{1A} = 0$$ $$T_{A1} = \frac{M_{1A} + M_{A1}}{5l} = \frac{1,345ql^{2} + 0,672ql^{2}}{5l} = 0,403ql$$ $$\sum_{}^{}y = 0: - T_{A1} + T_{1A} = 0$$ TA1 = T1A = 0, 403ql	$$\sum_{}^{}M_{1} = 0:\ T_{21} \bullet 6l - M_{21} - M_{12} = 0$$ $$T_{21} = \frac{M_{21} + M_{12}}{6l} = \frac{42,543ql^{2} + 22,112ql^{2}}{6l} = 10,776ql$$ $$\sum_{}^{}y = 0: - T_{12} + T_{21} = 0$$ T12 = T21 = 10, 776ql
$$\sum_{}^{}M_{2} = 0: - T_{D2} \bullet 6l + M_{D2} + M_{2D} = 0$$ $$T_{D2} = \frac{M_{D2} + M_{2D}}{6l} = \frac{58,535ql^{2} + 37,543ql^{2}}{6l} = 16,013ql$$ $$\sum_{}^{}x = 0: - T_{D2} + T_{2D} = 0$$ TD2 = T2D = 16, 013ql	$$\sum_{}^{}M_{B} = 0:T_{1B} \bullet 6l + M_{1B} - 6q \bullet 6l \bullet 3l = 0$$ $$T_{1B} = \frac{{- M}_{1B} + 108ql^{2}}{6l} = \frac{- 11,923ql^{2} + 108ql^{2}}{6l} = 16,013ql$$ $$\sum_{}^{}x = 0:{- T}_{1B} - T_{B1} + 36ql = 0$$ TB1 = 36ql − 16, 013ql = 19, 987ql
$$\sum_{}^{}M_{C} = 0: - T_{1C} \bullet 6\sqrt{2}l + M_{1C} + \frac{11}{2}\sqrt{2}ql \bullet 3\sqrt{2}l = 0$$ $$T_{1C} = \frac{M_{1C} + 33ql^{2}}{6\sqrt{2}l} = \frac{11,535ql^{2} + 33ql^{2}}{6\sqrt{2}l} = 5,249ql$$ $$\sum_{}^{}x = 0:T_{1C} + T_{C1} - \frac{11}{2}\sqrt{2}ql = 0$$ $$T_{C1} = \frac{11}{2}\sqrt{2}ql - 5,249ql = 2,529ql$$	Siły normalne: N1A = NA1 = 0 N2D = ND2 = T21 = 10, 776ql N12 = N12 = T2D = 16, 013ql NC1 = TC1 = 2, 529ql N1C = T1C = 5, 249ql $$N_{1B} = N_{B1} = T_{1A} - T_{1C} \bullet \sqrt{2} + T_{12}$$ $$N_{1B} = N_{B1} = 0,403ql + 5,249ql \bullet \sqrt{2} - 10,776ql = = - 2,950ql$$