1. ROZWIĄZANIE RAMY METODĄ PRZEMIESZCZEŃ

1. WYZNACZANIE STOPNIA STATYCZNEJ NIEWYZNACZALNOŚCI

   1. WYZNACZENIE LICZBY STOPNI SWOBODY OBROTU WĘZŁÓW

Uwzględniając, że φ_A1 = 0 , φ_1A = φ₁₃ = φ₁₂ = φ₁ , φ₃₁ = φ₃₂, φ₂₁ = φ₂₃ = φ_2B = φ₂, φ_B2 = 0, stwierdzam, że wszystkie kąty obrotu końców prętów określone są przez kąty obrotu 2 węzłów φ₁ oraz φ₂, więc

- liczba stopni swobody obrotu węzłów wynosi n_φ=2

1.1.2 WYZNACZENIE LICZBY STOPNI SWOBODY PRZESUWU WĘZŁÓW

MODEL PRZEGUBOWY UKŁADU

w = 7;    p = 7;    r = 6 ∖ nn_δ ≥ 2w − p − r = 2 • 7 − 7 − 6 = 1

Należy dodać co najmniej 1 więź.

Dodając więź  II_δ przekształciłem model przegubowy układu w układ geometrycznie niezmienny. Zatem liczba stopni swobody przesuwu układu danego wynosi n_δ=1.

Stopień geometrycznej niewyznaczalności wynosi : n_g=n_δ+n_φ=1 + 2 = 3

1. UKŁAD PODSTAWOWY

1.2.1 OGÓLNA POSTAĆ UKŁADU RÓWNAŃ METODY PRZEMIESZCZEŃ

k₁₁φ₁ + k₁₂φ₂ + k_1IIδ_II + k₁₀ + k_1(0T) = 0

k₂₁φ₁ + k₂₂φ₂ + k_2IIδ_II + k₂₀ + k_2(0T) = 0

k_II1φ₁ + k_II2φ₂ + k_{II II}δ_II + k_II0 + k_II(0T) = 0

1. WYZNACZENIE WSPÓLCZYNNIKÓW UKŁADU RÓWNAŃ OD OBCIĄŻEŃ MECHANICZNYCH

1.3.1 ROZWIAZANIE UKŁADU PODSTAWOWEGO OD OBCIĄZENIA DANEGO (φ₁ = φ₂ = δ_II = 0)

$${M_{A1}}^{0} = - \frac{p \bullet L^{2}}{12} = - \frac{4 \bullet \frac{16}{25} \bullet 25}{12} = - 5.3333kNm$$

$${M_{1A}}^{0} = + \frac{p \bullet L^{2}}{12} = + \frac{4 \bullet \frac{16}{25} \bullet 5}{12} = + 5.3333kNm$$

M₃₁⁰ = 0

M₁₃⁰ = 0

M₂₁⁰ = 0

M₁₂⁰ = 0

M₂₃⁰ = 0

M₃₂⁰ = 0

$${M_{2B}}^{0} = - \frac{P \bullet a}{2} \bullet \left( 2 - \frac{a}{L} \right) = - \frac{24 \bullet 3}{2} \bullet \left( 2 - \frac{3}{4} \right) = - 45.00kNm$$

$${M_{B2}}^{0} = - \frac{P \bullet a^{2}}{2 \bullet L} = - \frac{24 \bullet 3^{2}}{2 \bullet 4} = - 27.00kNm$$

M₂⁰ = −20kNm

Siły równoważne mają wartości:

$$P_{1} = P_{2} = p \bullet \frac{L}{2} = \frac{4 \bullet \frac{16}{25} \bullet 5}{2} = 6.40kN$$

$$P_{3} = \frac{P}{L} = \frac{24}{4} = 6.00kN$$

$$P_{4} = \frac{a}{L} \bullet P = \frac{3}{4} \bullet 24 = 18.00kN$$

1.3.2 ROZWIAZANIE UKŁADU PODSTAWOWEGO OD φ₁ = 1

(F = 0         φ₂ = δ_II = 0)

W tym stanie obciążenia do wzorów transformacyjnych podstawiamy:

φ_1A¹ = φ₁₂¹ = φ₁₃¹ = φ₁ = 1 pozostałe φ_ij¹ = 0 ψ_ij¹ = 0 M_ij¹ = 0

Momenty brzegowe wynoszą:

$${M_{A1}}^{1} = \frac{2EI}{5} \bullet 2 = 0.8\frac{\text{EI}}{m}\ $$

$${M_{1A}}^{1} = \frac{2EI}{5} \bullet 4 = 1.6\frac{\text{EI}}{m}\ $$

M₃₁¹ = 0 

$${M_{13}}^{1} = \frac{\text{EI}}{4} \bullet 3 = 0.75\frac{\text{EI}}{m}\ $$

$${M_{21}}^{1} = \frac{2EI}{4 \bullet \sqrt{2}} \bullet 2 = 0.7071\frac{\text{EI}}{m}$$

$${M_{12}}^{1} = \frac{2EI}{4 \bullet \sqrt{2}} \bullet 4 = 1.4142\frac{\text{EI}}{m}\ $$

M₂₃¹ = 0 

M₃₂¹ = 0 

M_2B¹ = 0 

M_B2¹ = 0 

1.3.3 ROZWIAZANIE UKŁADU PODSTAWOWEGO OD φ₂ = 1

(F = 0         φ₁ = δ_II = 0)

W tym stanie obciążenia do wzorów transformacyjnych podstawiamy:

φ₂₁¹ = φ_2B¹ = φ₂₃¹ = φ₂ = 1 pozostałe φ_ij¹ = 0 ψ_ij¹ = 0 M_ij¹ = 0

Momenty brzegowe wynoszą:

M_A1² = 0

M_1A² = 0 

M₃₁² = 0 

M₁₃² = 0

$${M_{21}}^{2} = \frac{2EI}{4 \bullet \sqrt{2}} \bullet 4 = 1.4142\frac{\text{EI}}{m}$$

$${M_{12}}^{2} = \frac{2EI}{4 \bullet \sqrt{2}} \bullet 3 = 0.7071\frac{\text{EI}}{m}\ $$

$${M_{23}}^{2} = \frac{2EI}{4} \bullet 3 = 1.50\ \frac{\text{EI}}{m}$$

$${M_{32}}^{2} = \frac{2EI}{4} \bullet 0 = 0$$

$${M_{2B}}^{2} = \frac{\text{EI}}{4} \bullet 1 = 0.25\frac{\text{EI}}{m}$$

$${M_{B2}}^{2} = \frac{\text{EI}}{4} \bullet - 1 = - 0.25\frac{\text{EI}}{m}$$

1.3.4 ROZWIAZANIE UKŁADU PODSTAWOWEGO OD δ_II=1

(F = 0         φ₁ = φ₂ = 0)

W tym stanie obciążenia do wzorów transformacyjnych podstawiamy: φ_ij^II = 0

$$\frac{a}{x} = \frac{4}{3}\ \ \ \ \ \ \ \ \ \ \ \rightarrow \ \ \ \ \ \ \ \ \ a = \frac{4}{3}x$$

$$\frac{a}{1} = \text{tg}\left( \text{beta} \right) = \frac{4}{7}\text{\ \ \ \ } \rightarrow \text{\ \ \ \ \ \ }x = \frac{3}{7};\ \ a = \frac{4}{7} = c\ ;\ d = \frac{5}{4}a = \frac{5}{7};b = \sqrt{2} \bullet \frac{4}{7}$$

$${\psi_{A1}}^{\text{II}} = \frac{{_{A"1"}}^{\text{II}}}{L_{A1}} = \frac{d}{5} = 0.14285\frac{1}{m}\ $$

$${\psi_{13}}^{\text{II}} = \frac{{_{1"3"}}^{\text{II}}}{L_{13}} = \frac{c}{4} = 0.14285\frac{1}{m}$$

$${\psi_{23}}^{I} = \frac{{_{2"3"}}^{\text{II}}}{L_{13}} = \frac{a}{4} = 0.14285\frac{1}{m}$$

$${\psi_{12}}^{\text{II}} = \frac{{_{1"2"}}^{\text{II}}}{L_{12}} = \frac{d}{4\sqrt{2}} = 0.14285\frac{1}{m}$$

Momenty brzegowe wynoszą:

$${M_{A1}}^{\text{II}} = \frac{2EI}{5} \bullet \left( - 6 \right) \bullet 0.14285 = - 0.3429\frac{\text{EI}}{m^{2}}$$

$${M_{1A}}^{\text{II}} = \frac{2EI}{5} \bullet \left( - 6 \right) \bullet 0.14285 = - 0.3429\frac{\text{EI}}{m^{2}}\ $$

$${M_{31}}^{\text{II}} = \frac{\text{EI}}{4} \bullet \left( 0 \right) \bullet 0.14285 = 0\ $$

$${M_{13}}^{\text{II}} = \frac{\text{EI}}{4} \bullet \left( - 3 \right) \bullet 0.14285 = - 0.1071\frac{\text{EI}}{m^{2}}$$

$${M_{21}}^{\text{II}} = \frac{2EI}{4 \bullet \sqrt{2}} \bullet \left( - 6 \right) \bullet 0.14285 = - 0.3030\frac{\text{EI}}{m^{2}}$$

$${M_{12}}^{\text{II}} = \frac{2EI}{4 \bullet \sqrt{2}} \bullet \left( - 6 \right) \bullet 0.14285 = - 0.3030\frac{\text{EI}}{m^{2}}\ $$

$${M_{23}}^{\text{II}} = \frac{2EI}{4} \bullet \left( - 3 \right) \bullet 0.14285 = - 0.2143\ \frac{\text{EI}}{m^{2}}$$

$${M_{32}}^{\text{II}} = \frac{2EI}{4} \bullet 0 = 0$$

$${M_{2B}}^{\text{II}} = \frac{\text{EI}}{4} \bullet 0 = 0$$

$${M_{B2}}^{\text{II}} = \frac{\text{EI}}{4} \bullet 0 = 0$$

δ_s2^II = cos45 • −1 = −0.7071

δ_P1^II = 0

δ_P2^II = 0.7143

δ_P3^II = 1

δ_P4^II = 1

1.3.5 OBLICZENIE WSPÓLCZYNNIKÓW UKŁADU RÓWNAŃ

$$k_{11} = \sum_{j}^{}{M_{1j}}^{1} + k_{1}^{\varphi} = 1.6\frac{\text{EI}}{m} + 0.75\frac{\text{EI}}{m} + 1.4142\frac{\text{EI}}{m} + 6\frac{\text{EI}}{m} = 9.7642\frac{\text{EI}}{m}$$

$$k_{12} = {M_{12}}^{2} = 0.7071\frac{\text{EI}}{m}$$

$$k_{1II} = \sum_{j}^{}{M_{1j}}^{\text{II}} = \left( - 0.3429\frac{\text{EI}}{m^{2}} \right) + \left( - 0.1071\frac{\text{EI}}{m^{2}} \right) + \left( - 0.3030\frac{\text{EI}}{m^{2}} \right) = - 0.7530\frac{\text{EI}}{m^{2}}$$

$$k_{10} = \sum_{j}^{}{M_{1j}}^{0} - {M_{1}}^{0} = 5.3333kNm$$

$$k_{21} = {M_{21}}^{1} = 0.7071\frac{\text{EI}}{m} = k_{12}$$

$$k_{22} = \sum_{j}^{}{M_{2j}}^{2} = 1.4142\frac{\text{EI}}{m} + 1.50\ \frac{\text{EI}}{m} + 0.25\frac{\text{EI}}{m} = 3.1642\frac{\text{EI}}{m}$$

$$k_{2II} = \sum_{j}^{}{M_{2j}}^{\text{II}} = \left( - 0.3030\frac{\text{EI}}{m^{2}} \right) + \left( - 0.2143\ \frac{\text{EI}}{m^{2}} \right) = - 0.5173\frac{\text{EI}}{m^{2}}$$

$$k_{20} = \sum_{j}^{}{M_{2j}}^{0} - {M_{2}}^{0} = - 45.00kNm - \left( - 20kNm \right) = - 25kNm$$

$$k_{II1} = - \sum_{\text{ij}}^{}{{(M_{\text{ij}}}^{1} + {M_{\text{ji}}}^{1}}) \bullet {\psi_{\text{ij}}}^{\text{II}} = = - \left\lbrack \left( 0.8\frac{\text{EI}}{m} + 1.6\frac{\text{EI}}{m} \right) \bullet 0.14285\frac{1}{m} + \left( 0.75\frac{\text{EI}}{m} + 0 \right) \bullet 0.14285\frac{1}{m} + \left( 0.7071\frac{\text{EI}}{m} + 1.4142\frac{\text{EI}}{m} \right) \bullet 0.14285\frac{1}{m} \right\rbrack = - 0.7530\frac{\text{EI}}{m^{2}} = k_{1II}$$

$$k_{II2} = - \sum_{\text{ij}}^{}{{(M_{\text{ij}}}^{2} + {M_{\text{ji}}}^{2}}) \bullet {\psi_{\text{ij}}}^{\text{II}} = = - \left\lbrack \left( 0.7071\frac{\text{EI}}{m} + 1.4142\frac{\text{EI}}{m} \right) \bullet 0.14285\frac{1}{m} + \left( 1.50\frac{\text{EI}}{m}\ + 0 \right) \bullet 0.14285\frac{1}{m} \right\rbrack = - 0.5173\frac{\text{EI}}{m^{2}} = k_{2II}$$

$$k_{\text{II\ II}} = - \sum_{\text{ij}}^{}{{(M_{\text{ij}}}^{\text{II}} + {M_{\text{ji}}}^{\text{II}}}) \bullet {\psi_{\text{ij}}}^{\text{II}} + {k_{2}}^{\delta} \bullet {\delta_{s2}}^{\text{II}} \bullet {\delta_{s2}}^{\text{II}} = - \left\lbrack \left( - 0.3429\frac{\text{EI}}{m^{2}} - 0.3429\frac{\text{EI}}{m^{2}} \right) \bullet 0.14285\frac{1}{m} + \left( - 0.1071\frac{\text{EI}}{m^{2}}\ + 0 \right) \bullet 0.14285\frac{1}{m} + \left( - 0.3030\frac{\text{EI}}{m^{2}}\ - 0.3030\frac{\text{EI}}{m^{2}} \right) \bullet 0.14285\frac{1}{m} + \left( - 0.2143\ \frac{\text{EI}}{m^{2}}\ + 0 \right) \bullet 0.14285\frac{1}{m} \right\rbrack + 4\frac{\text{EI}}{m^{3}} \bullet \left( - 0.7071 \right) \bullet \left( - 0.7071 \right) = 0.2305\frac{\text{EI}}{m^{3}} + 2\frac{\text{EI}}{m^{3}} = 2.2305\frac{\text{EI}}{m^{3}}$$

$$k_{II0} = - \sum_{\text{ij}}^{}{{(M_{\text{ij}}}^{0} + {M_{\text{ji}}}^{0}}) \bullet {\psi_{\text{ij}}}^{\text{II}} - \sum_{P}^{}{P_{P} \bullet}{\delta_{P}}^{\text{II}} = = - \left\lbrack \left( - 5.3333kNm + 5.3333kNm \right) \bullet 0.14285\frac{1}{m} \right\rbrack - \left\lbrack 6.40kN \bullet 0.7143 + 6.00kN \bullet 1 + 18.00kN \bullet 1 \right\rbrack = - 28.5715kN$$

1. WYZNACZENIE WSPÓLCZYNNIKÓW UKŁADU RÓWNAŃ OD OBCIĄŻEŃ NIEMECHANICZNYCH

1.4.1 WYZNACZENIE MOMENTÓW BRZEGOWYCH W UKŁADZIE PODSTAWOWYM OD ZMIAN TEMPERATURY PO WYSOKOŚCI PRZEKROJU PRĘTA

PRZYJETO PRZEKRÓJ I₂₀₀ oraz 2I₂₀₀ o EI=4387 kNm²

Momenty brzegowe wynoszą:

$${M_{A1}}^{\varphi T} = - \frac{2EI}{0.2} \bullet 1.2 \bullet 10^{- 5} \bullet ( - 16 - 16) = 0.00384\frac{\text{EI}}{m}$$

$${M_{1A}}^{\varphi T} = \frac{2EI}{0.2} \bullet 1.2 \bullet 10^{- 5} \bullet \left( - 16 - 16 \right) = - 0.00384\frac{\text{EI}}{m}\ $$

M₃₁^φT = 0 

M₁₃^φT = 0

$${M_{21}}^{\varphi T} = \frac{2EI}{0.2} \bullet 1.2 \bullet 10^{- 5} \bullet ( - 20 - 20) = - 0.0048\frac{\text{EI}}{m}$$

$${M_{12}}^{\varphi T} = - \frac{2EI}{0.2} \bullet 1.2 \bullet 10^{- 5} \bullet ( - 20 - 20) = 0.0048\frac{\text{EI}}{m}\ $$

M₂₃^φT = 0

M₃₂^φT = 0

$${M_{2B}}^{\varphi T} = - \frac{\text{EI}}{0.2} \bullet 1.2 \bullet 10^{- 5} \bullet \left( 10 + 10 \right) = - 0.0012\frac{\text{EI}}{m}$$

$${M_{B2}}^{\varphi T} = \frac{\text{EI}}{0.2} \bullet 1.2 \bullet 10^{- 5} \bullet (10 + 10) = 0.0012\frac{\text{EI}}{m}$$

1.4.2 OBLICZENIE WSPÓLCZYNNIKÓW UKŁADU RÓWNAŃ

$$k_{1(OT)} = \sum_{j}^{}{M_{1j}}^{(OT)} = {M_{1A}}^{\varphi T} + {M_{13}}^{\varphi T} + {M_{12}}^{\varphi T} = - 0.00384\frac{\text{EI}}{m} + 0 + 0.0048\frac{\text{EI}}{m} = 9.6 \bullet 10^{- 4\ }\frac{\text{EI}}{m}\ $$

$$k_{2(OT)} = \sum_{j}^{}{M_{2j}}^{(OT)} = {M_{21}}^{\varphi T} + {M_{23}}^{\varphi T} + {M_{2B}}^{\varphi T} = - 0.0048\frac{\text{EI}}{m} + 0 - 0.0012\frac{\text{EI}}{m} = - 6 \bullet 10^{- 3\ }\frac{\text{EI}}{m}\ $$

$$k_{II(OT)} = - \sum_{\text{ij}}^{}{{(M_{\text{ij}}}^{(0T)} + {M_{\text{ji}}}^{(0T)}}) \bullet {\psi_{\text{ij}}}^{\text{II}} = 0$$

1. POSTAĆ SZCZEGÓŁOWA UKŁADU RÓWNAŃ OD OBCIĄŻEŃ MECHANICZNYCH

$$9.7642\frac{\text{EI}}{m}\varphi_{1} + 0.707\frac{\text{EI}}{m}1\varphi_{2} - 0.7530\frac{\text{EI}}{m^{2}}\delta_{\text{II}} + 5.3333kNm = 0$$

$$0.7071\frac{\text{EI}}{m}\varphi_{1} + 3.1642\frac{\text{EI}}{m}\varphi_{2} - 0.5173\frac{\text{EI}}{m^{2}}\delta_{\text{II}} - 25kNm = 0$$

$$- 0.7530\frac{\text{EI}}{m^{2}}\varphi_{1} - 0.5173\frac{\text{EI}}{m^{2}}\varphi_{2} + 2.2305\frac{\text{EI}}{m^{3}}\delta_{\text{II}} - 28.5715kNm = 0$$

1. ROZWIĄZANIE UKŁADU RÓWNAŃ OD OBCIĄŻEŃ MECHANICZNYCH
   

$$\varphi_{1} = - 0.1295\frac{\text{kN}m^{2}}{\text{EI}}$$

$$\varphi_{2} = 10.4116\frac{\text{kN}m^{2}}{\text{EI}}$$

$$\delta_{\text{II}} = 15.1804\frac{\text{kN}m^{3}}{\text{EI}}$$

1.5.2 MOMENTY BRZEGOWE I SIŁY W WIĘZIACH SPRĘŻYSTYCH

	-0.1295	f1	10.4116	f2	15.1804	dII
	M ^1		M ^2		M ^II		M^o		M ^rz
M A1	0.8000	-0.1038	0.0000	0.0000	-0.3429	-5.2055	-5.3333		-10.6426
M 1A	1.6000	-0.2075	0.0000	0.0000	-0.3429	-5.2055	5.3333		-0.0797
M 31	0.0000	0.0000	0.0000	0.0000	0.0000	0.0000	0.0000		0.0000
M 13	0.7500	-0.0973	0.0000	0.0000	-0.1071	-1.6259	0.0000		-1.7231
M 21	0.7071	-0.0917	1.4142	14.7269	-0.3030	-4.5998	0.0000		10.0354
M 12	1.4142	-0.1834	0.7071	7.3635	-0.3030	-4.5998	0.0000		2.5803
M 23	0.0000	0.0000	1.5000	15.6204	-0.2143	-3.2532	0.0000		12.3672
M 32	0.0000	0.0000	0.0000	0.0000	0.0000	0.0000	0.0000		0.0000
M 2B	0.0000	0.0000	0.2500	2.6034	0.0000	0.0000	-45.0000		-42.3966
M B2	0.0000	0.0000	-0.2500	-2.6034	0.0000	0.0000	-27.0000		-29.6034

$$S_{\varphi 1} = k^{\varphi} \bullet \varphi_{1} = 6\frac{\text{EI}}{m} \bullet \left( - 0.1295\frac{\text{kN}m^{2}}{\text{EI}} \right) = - 0.7770\ kNm$$

$$S_{\delta 2} = k^{\delta 2} \bullet \left( {\delta_{s2}}^{\text{II}} \bullet \delta_{\text{II}} \right) = 4\frac{\text{EI}}{m^{3}}\mathbf{\bullet}\left( - 0.7071 \bullet 15.1804\frac{\text{kN}m^{3}}{\text{EI}} \right) = - 42.9362\ kN$$

1. BRZEGOWE SIŁY TNĄCE I OSIOWE

Pręt A-1

$$\sum_{}^{}M_{A} \rightarrow \text{\ \ \ V}_{1A} = - \frac{\left( M_{A1} + M_{1A} \right)}{5m} - 4 \bullet \frac{4}{5} \bullet \frac{4}{5} \bullet \frac{5^{2}}{2} = - \frac{\left( - 10.6426 - 0.0797 \right)}{5m} - 2.56 \bullet \frac{5}{2} = - 4,2555\ kN$$

$$\sum_{}^{}M_{1} \rightarrow \text{\ \ \ V}_{A1} = - \frac{\left( M_{A1} + M_{1A} \right)}{5m} + 4 \bullet \frac{4}{5} \bullet \frac{4}{5} \bullet \frac{5^{2}}{2} = - \frac{\left( - 10.6426 - 0.0797 \right)}{5m} + 2.56 \bullet \frac{5}{2} = 8.5445\ kN$$

$$\sum_{}^{}X \rightarrow \ \ \ \ N_{1A} + 4 \bullet \frac{4}{5} \bullet \frac{3}{5} \bullet 5 - N_{A1} = 0\ \ \ \ \ N_{1A} = N_{A1} - 9.6$$

Pręt 1-3

$$\sum_{}^{}M_{1} \rightarrow \text{\ \ \ V}_{31} = - \frac{\left( M_{13} \right)}{4m} = - \frac{\left( - 1.7231 \right)}{4m} = 0.4308\ kN$$

$$\sum_{}^{}M_{3} \rightarrow \text{\ \ \ V}_{13} = - \frac{\left( M_{13} \right)}{4m} = - \frac{\left( - 1.7231 \right)}{4m} = 0.4308\ kN$$

$$\sum_{}^{}X \rightarrow \text{\ \ \ \ \ \ N}_{13}{= N}_{31}\text{\ \ \ \ }$$

Pręt 1-2

$$\sum_{}^{}M_{1} \rightarrow \text{\ \ \ V}_{21} = - \frac{\left( M_{12} + M_{21} \right)}{4 \bullet \sqrt{2}m} = - \frac{\left( 2.5803 + 10.0354 \right)}{4 \bullet \sqrt{2}m} = - 2.2302\ kN$$

$$\sum_{}^{}M_{2} \rightarrow \text{\ \ \ V}_{12} = - \frac{\left( M_{12} + M_{21} \right)}{4 \bullet \sqrt{2}m} = - \frac{\left( 2.5803 + 10.0354 \right)}{4 \bullet \sqrt{2}m} = - 2.2302\ kN$$

$$\sum_{}^{}X \rightarrow \text{\ \ \ \ \ \ N}_{12}{= N}_{21}\text{\ \ \ \ }$$

Pręt 2-3

$$\sum_{}^{}M_{2} \rightarrow \text{\ \ \ V}_{32} = - \frac{\left( M_{23} \right)}{4m} = - \frac{\left( 12.3672 \right)}{4m} = - 3.0918\ kN$$

$$\sum_{}^{}M_{3} \rightarrow \text{\ \ \ V}_{23} = - \frac{\left( M_{23} \right)}{4m} = - \frac{\left( 12.3672 \right)}{4m} = - 3.0918\ kN$$

$$\sum_{}^{}X \rightarrow \text{\ \ \ \ \ \ N}_{23}{= N}_{32}\text{\ \ \ \ }$$

Pręt 2-B

$$\sum_{}^{}Y \rightarrow \text{\ \ \ V}_{2B} = 24.00\ kN$$

$$\sum_{}^{}X \rightarrow \text{\ \ \ \ \ \ N}_{2B}{= N}_{B2}\text{\ \ \ \ }$$

WĘZEŁ 3

$$\sum_{}^{}M_{3} = 0 - \text{Spe}l\text{nione\ to}z\text{samo}s\text{ciowo}$$

$$\sum_{}^{}Y = 0\ \ \ \ \ \ N_{32} = N_{23} = V_{31} = 0.4308\ kN$$

$$\sum_{}^{}X = 0\ \ \ \ \ N_{31} = N_{13} = - V_{32} = 3.0918\ kN\ $$

WĘZEŁ 2

$$\sum_{}^{}M_{2} = - M_{21} - M_{23} - M_{2B} - 20 = - \lbrack 10.0354 + 12.3672 - 42.3966 + 20\rbrack = - 0.006 \approx 0 - \text{Spe}l\text{nione\ to}z\text{samo}s\text{ciowo}$$

$$\sum_{}^{}Y = N_{23} - V_{2B} + 0.7071V_{21} + 0.7071N_{21} - 0.7071S_{\delta 2} = 0$$

$$N_{21} = N_{12} = - \frac{1}{0.7071} \bullet \left\lbrack 0.4308 - 24 + 0.7071 \bullet \left( - 2.2302 \right) - 0.7071 \bullet \left( - 42.9362 \right) \right\rbrack = - 7,3738\ kN\ $$

$$\sum_{}^{}X = N_{2B} + + 0.7071S_{\delta 2} + V_{23} + 0.7071V_{21} - 0.7071N_{21} = 0$$

N_2B = N_B2 = −[0.7071 • (−42.9362) − 3.0918 + 0.7071 • (−2.2302) − 0.7071 • (−7.3738)]=30.1904 kN

WĘZEŁ 1

$$\sum_{}^{}M_{1} = - M_{1A} - M_{13} - M_{12} - S_{\varphi 1} = - ( - 0.0797) - ( - 1.7231) - 2.5803 - \left( - 0.7770 \right) = - 0.0005 \approx 0 - \text{Spe}l\text{nione\ to}z\text{samo}s\text{ciowo}$$

$$\sum_{}^{}Y = - \frac{4}{5} \bullet N_{1A} + \frac{3}{5} \bullet V_{1A} - 0.7071 \bullet V_{12} - 0.7071N_{12} - V_{13} = 0$$

N_1A = 1.25 • [0.6•(−4,2555)−0.7071•(−2.2302)−0.7071•(−7,3738)−0.4308 ] = 4.7586 kN

N_1A = N_A1 − 9.6           N_A1 = 9.6 + N_1A = 9.6 + 4, 7586 = 14.3586 kN 

WĘZEŁ PODPOROWY B

$$\sum_{}^{}M_{B} = M_{B} - M_{B2} = 0\ \ \ \ \ \ M_{B} = - 29.6034\ kNm$$

$$\sum_{}^{}Y = {0\ \ \ \ \ \ \ \ V}_{B} = 0$$

$$\sum_{}^{}X = N_{B} - N_{B2} = 0\ \ \ \ \ \ \ \ \ \ \ \ N_{B} = 30.1904\ kN\ $$

WĘZEŁ PODPOROWY A

$$\sum_{}^{}M_{A} = M_{A} - M_{A1} = 0\ \ \ \ \ M_{A} = - 10.6426kNm\ $$

$$\sum_{}^{}X = \frac{3}{5}N_{A1} + \frac{4}{5} \bullet V_{1A} + N_{A} = 0$$

N_A = −[0.6•14.3586+0.8•8.5445] = −15.4508 kN

$$\sum_{}^{}Y = \frac{4}{5}N_{A1} - \frac{3}{5} \bullet V_{1A} + V_{A} = 0$$

V_A = −0.8 • 14.3586 + 0.6 • 8.5445 = −6.3602 kN

1. MOMENTY ZGINAJĄCE

Pręt A-1

$$M\left( 2.5 \right) = M_{A} + \text{\ \ \ V}_{A1} \bullet 2.5 - 2.56 \bullet \frac{{2.5}^{2}}{2} = - 10.6426 + 8.5445 \bullet 2.5 - 8 = 2.7187\ kNm$$

V(2.5) = V_A1 − 2.56 • 2.5 = 8.5445 − 2.56 • 2.5 = 2.1445 kN

9. SPRAWDZENIE

Dla X₆ = 1

$$\sum_{}^{}M_{A} = 0$$

M_A + x₆ = 0

M_A = −x₆ = −1

$${_{6}}^{o} = \sum_{p}^{}{\int_{}^{}M_{6} \bullet \frac{M^{\text{rz}}}{\text{EI}}}dx + \frac{S_{\varphi 1}S_{6}}{k^{\varphi}} = 0$$

$$\frac{5}{6 \bullet 2EI} \bullet \left\lbrack \left( - 10.6426 \right) \bullet \left( - 1 \right) + 4 \bullet 2.7187 \bullet \left( - 1 \right) + 0.0797 \bullet \left( - 1 \right) \right\rbrack + \frac{- 1 \bullet - 0.7770}{6 \bullet 4387} = 9.4978 \bullet 10^{- 5} \bullet \left( - 0.3119 \right) + \frac{- 1 \bullet - 0.7770}{6 \bullet 4387} = - 1.05 \bullet 10^{- 7} \approx 0\ \ \ \ OK!\ $$

Dla X₅ = 1

$${_{5}}^{o} = \sum_{p}^{}{\int_{}^{}M_{5} \bullet \frac{M^{\text{rz}}}{\text{EI}}}dx = 0$$

$$\frac{4 \bullet \sqrt{2}}{6 \bullet 2EI} \bullet \left\lbrack 2.5803 \bullet 1 + 4 \bullet ( - 3.5926) \bullet 0.5 + ( - 10.0354) \bullet 0 \right\rbrack + \frac{4}{6 \bullet EI} \bullet \left\lbrack \left( - 1.7231 \right) \bullet \left( - 1 \right) + 4 \bullet ( - 0.8616) \bullet \left( - 0.5 \right) + 0 \bullet 0 \right\rbrack = 1.0745 \bullet 10^{- 4} \bullet - 4.6049 + 1.5196 \bullet 10^{- 4} \bullet 3.4443 = 0.000028599\ \approx 0\ \ \ \ \ \ \ OK!\ $$

9. POSTAĆ SZCZEGÓŁOWA UKŁADU RÓWNAŃ OD OBCIĄŻEŃ NIEMECHANICZNYCH

$$9.7642\frac{\text{EI}}{m}\varphi_{1} + 0.707\frac{\text{EI}}{m}1\varphi_{2} - 0.7530\frac{\text{EI}}{m^{2}}\delta_{\text{II}} + 9.6 \bullet 10^{- 4\ }\frac{\text{EI}}{m} = 0$$

$$0.7071\frac{\text{EI}}{m}\varphi_{1} + 3.1642\frac{\text{EI}}{m}\varphi_{2} - 0.5173\frac{\text{EI}}{m^{2}}\delta_{\text{II}} - 6 \bullet 10^{- 3\ }\frac{\text{EI}}{m} = 0$$

$$- 0.7530\frac{\text{EI}}{m^{2}}\varphi_{1} - 0.5173\frac{\text{EI}}{m^{2}}\varphi_{2} + 2.2305\frac{\text{EI}}{m^{3}}\delta_{\text{II}} + 0 = 0$$

φ₁ = −0.00021

φ₂ = 0.002008

δ_II = 0.000394 m

1.10.2 MOMENTY BRZEGOWE I SIŁY W WIĘZIACH SPRĘŻYSTYCH

	-0.00021	f1	0.002008	f2	0.000394	dII			[EI/m]	[kNm]
	M ^1		M ^2		M ^II		M^(OT)		M ^rz	M ^rz
M A1	0.8000	-0.0002	0	0.0000	-0.3429	-0.0001	0.00384		0.0035	15.3545
M 1A	1.6000	-0.0003	0	0.0000	-0.3429	-0.0001	-0.00384		-0.0043	-18.8641
M 31	0.0000	0.0000	0	0.0000	0.0000	0.0000	0.00000		0.0000	0.0000
M 13	0.7500	-0.0002	0	0.0000	-0.1071	0.0000	0.00000		-0.0002	-0.8774
M 21	0.7071	-0.0002	1.4142	0.0028	-0.3030	-0.0001	-0.00480		-0.0022	-9.6514
M 12	1.4142	-0.0003	0.7071	0.0014	-0.3030	-0.0001	0.00480		0.0058	25.4446
M 23	0.0000	0.0000	1.5	0.0030	-0.2143	-0.0001	0.00000		0.0029	12.7223
M 32	0.0000	0.0000	0	0.0000	0.0000	0.0000	0.00000		0.0000	0.0000
M 2B	0.0000	0.0000	0.25	0.0005	0.0000	0.0000	-0.00120		-0.0007	-3.0709
M B2	0.0000	0.0000	-0.25	-0.0005	0.0000	0.0000	0.00120		0.0007	3.0709

$${S_{\varphi 1}}^{T} = k^{\varphi} \bullet \varphi_{1} = 6\frac{\text{EI}}{m} \bullet \left( - 0.00021 \right) = - 0.0013\frac{\text{EI}}{m} = - 5.7031\ kNm$$

$${S_{\delta 2}}^{T} = k^{\delta 2} \bullet \left( {\delta_{s2}}^{\text{II}} \bullet \delta_{\text{II}} \right) = 4\frac{\text{EI}}{m^{3}} \bullet \left( - 0.7071 \bullet 0.000394\ m \right) = - 0.001144\frac{\text{EI}}{m^{2}} = - 4.8889\ kN$$

Pręt A-1

$$\sum_{}^{}M_{A} \rightarrow \text{\ \ \ V}_{1A} = - \frac{\left( M_{A1} + M_{1A} \right)}{5m} = - \frac{\left( 15.3545 - 18.8641 \right)}{5m} = 0.7019\ kN$$

$$\sum_{}^{}M_{1} \rightarrow \text{\ \ \ V}_{A1} = - \frac{\left( M_{A1} + M_{1A} \right)}{5m} = - \frac{\left( 15.3545 - 18.8641 \right)}{5m} = 0.7019\ kN$$

$$\sum_{}^{}X \rightarrow \ \ \ \ N_{1A} - N_{A1} = 0\ \ \ \ \ N_{1A} = N_{A1}$$

Pręt 1-3

$$\sum_{}^{}M_{1} \rightarrow \text{\ \ \ V}_{31} = - \frac{\left( M_{13} \right)}{4m} = - \frac{\left( - 0.8774 \right)}{4m} = 0.2194\ kN$$

$$\sum_{}^{}M_{3} \rightarrow \text{\ \ \ V}_{13} = - \frac{\left( M_{13} \right)}{4m} = - \frac{\left( - 0.8774 \right)}{4m} = 0.2194\ kN$$

$$\sum_{}^{}X \rightarrow \text{\ \ \ \ \ \ N}_{13}{= N}_{31}\text{\ \ \ \ }$$

Pręt 1-2

$$\sum_{}^{}M_{1} \rightarrow \text{\ \ \ V}_{21} = - \frac{\left( M_{12} + M_{21} \right)}{4 \bullet \sqrt{2}m} = - \frac{\left( 25.4446 - 9.6514 \right)}{4 \bullet \sqrt{2}m} = - 2.7917\ kN$$

$$\sum_{}^{}M_{2} \rightarrow \text{\ \ \ V}_{12} = - \frac{\left( M_{12} + M_{21} \right)}{4 \bullet \sqrt{2}m} = - \frac{\left( 25.4446 - 9.6514 \right)}{4 \bullet \sqrt{2}m} = - 2.7917\ kN$$

$$\sum_{}^{}X \rightarrow \text{\ \ \ \ \ \ N}_{12}{= N}_{21}\text{\ \ \ \ }$$

Pręt 2-3

$$\sum_{}^{}M_{2} \rightarrow \text{\ \ \ V}_{32} = - \frac{\left( M_{23} \right)}{4m} = - \frac{\left( 12.7223 \right)}{4m} = - 3.1806\ kN$$

$$\sum_{}^{}M_{3} \rightarrow \text{\ \ \ V}_{23} = - \frac{\left( M_{23} \right)}{4m} = - \frac{\left( 12.7223 \right)}{4m} = - 3.1806\ kN$$

$$\sum_{}^{}X \rightarrow \text{\ \ \ \ \ \ N}_{23}{= N}_{32}\text{\ \ \ \ }$$

Pręt 2-B

$$\sum_{}^{}M_{B} \rightarrow \text{\ \ \ V}_{2B} = - \frac{\left( M_{B2} + M_{2B} \right)}{4m} = - \frac{\left( 3.0709 - 3.0709 \right)}{4m} = 0\ \ $$

$$\sum_{}^{}X \rightarrow \text{\ \ \ \ \ \ N}_{2B}{= N}_{B2}\text{\ \ \ \ }$$

WĘZEŁ 3

$$\sum_{}^{}M_{3} = 0 - \text{Spe}l\text{nione\ to}z\text{samo}s\text{ciowo}$$

$$\sum_{}^{}Y = 0\ \ \ \ \ \ N_{32} = N_{23} = V_{31} = 0.2194\ \ kN$$

$$\sum_{}^{}X = 0\ \ \ \ \ N_{31} = N_{13} = - V_{32} = 3.1806\ kN\ $$

WĘZEŁ 2

$$\sum_{}^{}M_{2} = - M_{21} - M_{23} - M_{2B} = - \lbrack - 9.6514 + 12.7223 - 3.0709\rbrack = 0 - \text{Spe}l\text{nione\ to}z\text{samo}s\text{ciowo}$$

$$\sum_{}^{}Y = N_{23} - V_{2B} + 0.7071V_{21} + 0.7071N_{21} - 0.7071{S_{\delta 2}}^{T} = 0$$

$$N_{21} = N_{12} = - \frac{1}{0.7071} \bullet \left\lbrack 0.7019 - 0 + 0.7071 \bullet \left( - 2.7917 \right) - 0.7071 \bullet \left( - 4.8889 \right) \right\rbrack = - 3.0898\ kN\ $$

$$\sum_{}^{}X = N_{2B} + + 0.7071S_{\delta 2} + V_{23} + 0.7071V_{21} - 0.7071N_{21} = 0$$

N_2B = N_B2 = −[0.7071 • (−4.8889) − 3.1806 + 0.7071 • (−2.7917) − 0.7071 • (−3.0898)]=6.4268 kN

WĘZEŁ 1

$$\sum_{}^{}M_{1} = - M_{1A} - M_{13} - M_{12} - {S_{\varphi 1}}^{T} = - ( - 18.8641) - ( - 0.8774) - 25.4446 - \left( - 5.7031 \right) = 0 - \text{Spe}l\text{nione\ to}z\text{samo}s\text{ciowo}$$

$$\sum_{}^{}Y = - \frac{4}{5} \bullet N_{1A} + \frac{3}{5} \bullet V_{1A} - 0.7071 \bullet V_{12} - 0.7071N_{12} - V_{13} = 0$$

N_1A = 1.25 • [0.6•(0.7019)−0.7071•(−2.7917)−0.7071•(−3.0898)−0.2194 ] = 5.4507 kN

N_1A = N_A1 = 5.4507kN 

WĘZEŁ PODPOROWY B

$$\sum_{}^{}M_{B} = M_{B} - M_{B2} = 0\ \ \ \ \ \ M_{B} = 3.0709\ kNm$$

$$\sum_{}^{}Y = {0\ \ \ \ \ \ \ \ V}_{B} = 0$$

$$\sum_{}^{}X = N_{B} - N_{B2} = 0\ \ \ \ \ \ \ \ \ \ \ \ N_{B} = 6.4268\ kN\ $$

WĘZEŁ PODPOROWY A

$$\sum_{}^{}M_{A} = M_{A} - M_{A1} = 0\ \ \ \ \ M_{A} = 15.3545\ kNm\ $$

$$\sum_{}^{}X = \frac{3}{5}N_{A1} + \frac{4}{5} \bullet V_{1A} + N_{A} = 0$$

N_A = −[0.6•5.4507+0.8•0.7019] = −3.8319 kN

$$\sum_{}^{}Y = \frac{4}{5}N_{A1} - \frac{3}{5} \bullet V_{1A} + V_{A} = 0$$

V_A = −0.8 • 5.4507 + 0.6 • 0.7019 = −3.9394 kN

9. MOMENTY ZGINAJĄCE

Pręt A-1

M(2.5) = M_A +  V_A1 • 2.5 = 15.3545 + 0.7019  • 2.5 = 17.1093 kNm

13. SPRAWDZENIE

Dla X₆ = 1

$$\sum_{}^{}M_{A} = 0$$

M_A + x₆ = 0

M_A = −x₆ = −1

$${_{6}}^{o} = \sum_{p}^{}{\int_{}^{}M_{6} \bullet \frac{M^{T}}{\text{EI}}}dx + \sum_{p}^{}{\int_{}^{}M_{6}d\varphi_{1}} + \sum_{S}^{}\frac{{S_{\varphi 1}}^{T}S_{6}}{k^{\varphi}} = 0$$

$$\frac{5}{6 \bullet 2EI} \bullet \left\lbrack \left( 15.3545 \right) \bullet \left( - 1 \right) + 4 \bullet 17.1093 \bullet \left( - 1 \right) + 18.8641 \bullet \left( - 1 \right) \right\rbrack + \left( - 5 \right) \bullet 1.2 \bullet 10^{- 5} \bullet \frac{- 32}{0.2} + \frac{- 1 \bullet - 5.7031}{6 \bullet \text{EI}} = - 9.75 \bullet 10^{- 3} + 9.6 \bullet 10^{- 3} + 2.1667 \bullet 10^{- 4} = 6.667 \bullet 10^{- 5} \approx 0\ \ \ \ OK!\ $$

Dla X₅ = 1

$${_{5}}^{o} = \sum_{p}^{}{\int_{}^{}M_{6} \bullet \frac{M^{T}}{\text{EI}}}dx + \sum_{p}^{}{\int_{}^{}M_{6}d\varphi_{1}} = 0$$

$$\frac{4 \bullet \sqrt{2}}{6 \bullet 2EI} \bullet \left\lbrack - 9.6514 \bullet 1 + 4 \bullet ( - 17.5475) \bullet 0.5 + ( - 25.446) \bullet 0 \right\rbrack + \frac{4}{6 \bullet EI} \bullet \left\lbrack \left( - 0.8774 \right) \bullet \left( - 1 \right) + 4 \bullet ( - 0.4387) \bullet \left( - 0.5 \right) + 0 \bullet 0 \right\rbrack + \left( \frac{4\sqrt{2}}{2} \right) \bullet 1.2 \bullet 10^{- 5} \bullet \frac{- 40}{0.2} = 1.0745 \bullet 10^{- 4} \bullet - 44.7464 + 1.5196 \bullet 10^{- 4} \bullet 1.7548 - 6.7883 \bullet 10^{- 3} = - 0.0113 \approx 0\ \ \ \ \ \ OK!\ $$