UKŁAD PRZESTRZENNY

1. Układ zadany – obciążenia mechaniczne

1. Wyznaczenie stopnia statycznej niewyznaczalności

e = 6 + 1 = 7

t = 1

n_h = e − 6t = 7 − 6 • 1 = 1 - jednokrotnie statycznie niewyznaczalny

1. Układ podstawowy

1. Postać ogólna równania dla metody sił

$$\delta_{11} \bullet x_{1} + \delta_{1F} = \delta_{1rz} = - \frac{x_{1}}{k}$$

1. Rozwiązanie układu podstawowego od obciążenia mechanicznego

$${M_{AB,y}}^{F} = - 10\frac{\text{kN}}{m} \bullet 10m \bullet 3m + 15kNm - 20kN \bullet 11m = - 505kNm$$

$${M_{BA,y}}^{F} = 10\frac{\text{kN}}{m} \bullet 10m \bullet 3m + 15kNm - 20kN \bullet 5m = 215kNm$$

$${M_{BC,y}}^{F} = - 10\frac{\text{kN}}{m} \bullet 10m \bullet 5m + 20kN \bullet 3m = - 440\ kNm$$

$${M_{\left( L = 0,5 \right)\ CB,y}}^{F} = - 10\frac{\text{kN}}{m} \bullet 5m \bullet 2,5m \bullet 20kN \bullet 8m = 35kNm$$

M_CB, y^F = 20kN • 13m = 260kNm

M_CD, y^F = −20kN • 11m = −220kNm

M_DC, y^F = 0

M_DE, y^F = −20kN • 8m = −160kNm

M_ED, y^F = 0

$${M_{AB,x}}^{F} = - 10\frac{\text{kN}}{m} \bullet 10m \bullet 4m = - 400kNm$$

$${M_{BA,x}}^{F} = - 10\frac{\text{kN}}{m} \bullet 10m \bullet 4m = - 400kNm$$

M_BC, x^F = 20kN • 4m = 80kNm

M_{(L=0,5) CB, x}^F = 20kN • 4m = 80kNm

M_CB, x^F = 20kN • 4m = 80kNm

M_CD, x^F = 20kN • 8m = 160kNm

M_DC, x^F = 20kN • 8m = 160kNm

M_DE, x^F = 0

M_ED, x^F = 0

$${V_{AB,z}}^{F} = 20kN + 10\frac{\text{kN}}{m} \bullet 10m = 120kN$$

$${V_{BA,z}}^{F} = 20kN + 10\frac{\text{kN}}{m} \bullet 10m = 120kN$$

$${V_{BC,z}}^{F}20kN + 10\frac{\text{kN}}{m} \bullet 10m = 120kN$$

$${V_{\left( x = 0,5L \right)\ CB,z}}^{F} = 20kN + 10\frac{\text{kN}}{m} \bullet 5m = 70kNm$$

V_CB, z^F = 20kN

V_CD, z^F = 20kN

V_DC, z^F = 20kN

V_DE, z^F = 20kN

V_ED, z^F = 20kN

1. Rozwiązanie układu podstawowego od obciążenia X1=1

M_AB, y¹ = 1 • 11m = 11m

M_BA, y¹ = 1 • 5m = 5m

M_BC, y¹ = −1 • 3m = −3m

M_{(L=0,5) CB, y}¹ = −1 • 8m = −8m

M_CB, y¹ = −1 • 13m = −13m

M_CD, y¹ = 1 • 11m = 11m

M_DC, y¹ = 0

M_DE, y¹ = 1 • 8m = 8m

M_ED, y¹ = 0

M_AB, x¹ = 0

M_BA, x¹ = 0

M_BC, x¹ = −1 • 4m = −4m

M_{(L=0,5) CB, x}¹ = −1 • 4m = −4m

M_CB, x¹ = −1 • 4m = −4m

M_CD, x¹ = −1 • 8m = −8m

M_DC, x¹ = −1 • 8m = −8m

M_DE, x¹ = 0

M_ED, x¹ = 0

V_AB, z¹ = −1

V_BA, z¹ = −1

V_BC, z¹ = −1

V_{(x=0,5L) CB, z}¹ = −1

V_CB, z¹ = −1

V_CD, z¹ = −1

V_DC, z¹ = −1

V_DE, z¹ = −1

V_ED, z¹ = −1

1. Obliczenie współczynników układu równań

$$\delta_{11} = \sum_{P}^{}{\int_{}^{}\frac{{M_{x}}^{1} \bullet {M_{x}}^{1}}{GI_{x}}\text{dx}} + \sum_{P}^{}{\int_{}^{}\frac{{M_{y}}^{1} \bullet {M_{y}}^{1}}{EI_{y}}dy = \frac{6m}{6EI_{y}}\left( 11^{2}m^{2} + 4 \bullet 8^{2}m^{2} + 5^{2}m^{2} \right)} +$$

$$+ \frac{10m}{6EI_{y}}\left( {( - 3)}^{2}m^{2} + 4 \bullet {( - 8)}^{2}m^{2} + {( - 13)}^{2}m^{2} \right) + \frac{11m}{6EI_{y}}\left( 11^{2}m^{2} + 4 \bullet {5,5}^{2}m^{2} + 0 \right) +$$

$$+ \frac{8m}{6EI_{y}}\left( 8^{2}m^{2} + 4 \bullet 4^{2}m^{2} + 0 \right) + 0 + \frac{10m}{6 \bullet 0,8EI_{y}}\left( {( - 4)}^{2}m^{2} + 4 \bullet {( - 4)}^{2}m^{2} + {( - 4)}^{2}m^{2} \right) +$$

$$+ \frac{11m}{6 \bullet 0,8EI_{y}}\left( {( - 8)}^{2}m^{2} + 4 \bullet {( - 8)}^{2}m^{2} + {( - 8)}^{2}m^{2} \right) + 0 = \left( 402 + 723,333 + 443,667 + 170,667 + 200 + 880 \right)\frac{m^{3}}{EI_{y}} = 2819,667\frac{m^{3}}{EI_{y}}\ $$

$$\delta_{1F} = \sum_{P}^{}{\int_{}^{}\frac{{M_{x}}^{F} \bullet {M_{x}}^{1}}{GI_{x}}\text{dx}} + \sum_{P}^{}{\int_{}^{}\frac{{M_{y}}^{F} \bullet {M_{y}}^{1}}{EI_{y}}dy =}$$

$$= \frac{6m}{6EI_{y}}\left( - 505kNm \bullet 11m + 4 \bullet \left( - 145kNm \right) \bullet 8m + 215kNm \bullet 5m \right) +$$

$${+ \frac{10m}{6EI_{y}}\left( \left( - 440 \right)kNm \bullet \left( - 3 \right)m + 4 \bullet 35kNm \bullet \left( - 8 \right)m + 260kNm \bullet \left( - 13 \right)m \right) + \backslash n}{+ \frac{11m}{6EI_{y}}\left( ( - 220)kNm \bullet 11m + 4 \bullet \left( - 110 \right)kNm \bullet 5,5m + 0 \right) +}$$

$${+ \frac{8m}{6EI_{y}}\left( \left( - 160 \right)kNm \bullet 8m + 4 \bullet \left( - 80 \right)kNm \bullet 4m + 0 \right) + 0 + \backslash n}{+ \frac{10m}{6 \bullet 0,8EI_{y}}\left( 80kNm \bullet \left( - 4 \right)m + 4 \bullet 80kNm \bullet \left( - 4 \right)m + 80kNm \bullet \left( - 4 \right)m \right) +}$$

$${+ \frac{11m}{6 \bullet 0,8EI_{y}}\left( 160kNm \bullet \left( - 8 \right)m + 4 \bullet 160kNm \bullet \left( - 8 \right)m + 160kNm \bullet \left( - 8 \right)m \right) + 0 = \backslash n}{= \left( - 9120 - 5300 - 8873,333 - 3413,333 - 4000 - 17600 \right)\frac{m^{3}}{EI_{y}} = - 48306,666\frac{\text{kN}m^{3}}{EI_{y}}}$$

$$\delta_{11} \bullet x_{1} + \delta_{1F} = - \frac{x_{1}}{6\ \frac{EI_{y}}{m^{3}}}$$

$$\left( 2819,667 + \frac{1}{6} \right)\frac{m^{3}}{EI_{y}} \bullet x_{1} - 48306,666\frac{\text{kN}m^{3}}{EI_{y}} = 0$$

x₁ = 17, 132 kN

1. Obliczenie rzeczywistych sił przekrojowych od obciążeń mechanicznych dla układu zadanego

M_ij, y^RZ = M_ij, y^F + M_ij, y¹ • x₁

M_ij, x^RZ = M_ij, x^F + M_ij, x¹ • x₁

V_ij, z^RZ = V_ij, z^F + V_ij, z¹ • x₁

M_AB, y^RZ = −505kNm + 11m • 17, 132kN = −316, 548 kNm

M_BA, y^RZ = 215kNm + 5m • 17, 132kN = 300, 660 kNm

M_BC, y^RZ = −440 kNm − 3m • 17, 132kN = −491, 396 kNm

M_{(L=0,5) CB, y}^RZ = 35kNm − 8m • 17, 132kN = −102, 056 kNm

M_CB, y^RZ = 260kNm − 13m • 17, 132kN = 37, 284 kNm

M_CD, y^RZ = −220kNm + 11m • 17, 132kN = −31, 548 kNm

M_DC, y^RZ = 0

M_DE, y^RZ = −160kNm + 8m • 17, 132kN = −22, 944 kNm

M_ED, y^RZ = 0

M_AB, x^RZ = −400kNm

M_BA, x^RZ = −400kNm

M_BC, x^RZ = 80kNm − 4m • 17, 132kN = 11, 472 kNm

M_{(L=0,5) CB, x}^RZ = 80kNm − 4m • 17, 132kN = 11, 472 kNm

M_CB, x^RZ = 80kNm − 4m • 17, 132kN = 11, 472 kNm

M_CD, x^RZ = 160kNm − 8m • 17, 132kN = 22, 944 kNm

M_DC, x^RZ = 160kNm − 8m • 17, 132kN = 22, 944 kNm

M_DE, x^RZ = 0

M_ED, x^RZ = 0

V_AB, z^RZ = 120kNm − 1 • 17, 132kN = 102, 868 kN

V_BA, z^RZ = 120kNm − 1 • 17, 132kN = 102, 868 kN

V_BC, z^RZ = 120kNm − 1 • 17, 132kN = 102, 868 kN

V_{(x=0,5L) CB, z}^RZ = 70kNm − 1 • 17, 132kN = 52, 868 kN

V_CB, z^RZ = 20kN − 1 • 17, 132kN = 2, 868 kN

V_CD, z^RZ = 20kN − 1 • 17, 132kN = 2, 868 kN

V_DC, z^RZ = 20kN − 1 • 17, 132kN = 2, 868 kN

V_DE, z^RZ = 20kN − 1 • 17, 132kN = 2, 868 kN

V_ED, z^RZ = 20kN − 1 • 17, 132kN = 2, 868 kN

1. Układ zadany – obciążenia niemechaniczne

1. Układ podstawowy

1. Postać ogólna równania dla metody sił

$$\delta_{11} \bullet x_{1} + \delta_{1T} = \delta_{1rz} = - \frac{x_{1}}{k}$$

1. Obliczenie współczynników układu równań

PRZYJETO PRZEKRÓJ I₂₀₀ o EI=4387 kNm²

$$\delta_{11} = \sum_{P}^{}{\int_{}^{}\frac{{M_{x}}^{1} \bullet {M_{x}}^{1}}{GI_{x}}\text{dx}} + \sum_{P}^{}{\int_{}^{}\frac{{M_{y}}^{1} \bullet {M_{y}}^{1}}{EI_{y}}dy = \frac{6m}{6EI_{y}}\left( 11^{2}m^{2} + 4 \bullet 8^{2}m^{2} + 5^{2}m^{2} \right)} +$$

$$+ \frac{10m}{6EI_{y}}\left( {( - 3)}^{2}m^{2} + 4 \bullet {( - 8)}^{2}m^{2} + {( - 13)}^{2}m^{2} \right) + \frac{11m}{6EI_{y}}\left( 11^{2}m^{2} + 4 \bullet {5,5}^{2}m^{2} + 0 \right) +$$

$$+ \frac{8m}{6EI_{y}}\left( 8^{2}m^{2} + 4 \bullet 4^{2}m^{2} + 0 \right) + 0 + \frac{10m}{6 \bullet 0,8EI_{y}}\left( {( - 4)}^{2}m^{2} + 4 \bullet {( - 4)}^{2}m^{2} + {( - 4)}^{2}m^{2} \right) +$$

$$+ \frac{11m}{6 \bullet 0,8EI_{y}}\left( {( - 8)}^{2}m^{2} + 4 \bullet {( - 8)}^{2}m^{2} + {( - 8)}^{2}m^{2} \right) + 0 = \left( 402 + 723,333 + 443,667 + 170,667 + 200 + 880 \right)\frac{m^{3}}{EI_{y}} = 2819,667\frac{m^{3}}{EI_{y}}\ $$

$$\delta_{1T} = \sum_{P}^{}{\left( \ \alpha_{T}\frac{T_{d,z} - T_{g,z}}{h_{z}} \bullet \Omega_{\text{My}^{1}} \right) =} = 1,2 \bullet 10^{- 5}\frac{1}{}\left( \frac{- 12 - 12}{0,2m} \bullet \frac{11m + 5m}{2} \bullet 6m + \frac{10 - \left( - 10 \right)}{0,2m} \bullet \frac{- 3m + \left( - 13 \right)m}{2} \bullet 10m \right) = 1,2 \bullet 10^{- 5}\frac{1}{}\left( - 120\frac{}{m} \bullet 48m^{2} + 100\frac{}{m} \bullet \left( - 80 \right)m^{2} \right) = - 0,16512\ m$$

$$\left( 2819,667 + \frac{1}{6} \right)\frac{m^{3}}{EI_{y}} \bullet x_{1} - 0,16512\ m = 0$$

$$x_{1} = 5,856 \bullet 10^{- 5}\frac{EI_{y}}{m^{2}}$$

1. Obliczenie rzeczywistych sił przekrojowych od obciążeń niemechanicznych dla układu zadanego

M_ij, y^RZ, T = M_ij, y^F + M_ij, y¹ • x₁

M_ij, x^RZ, T = M_ij, x^F + M_ij, x¹ • x₁

V_ij, z^RZ = V_ij, z^F + V_ij, z¹ • x₁

$${M_{AB,y}}^{RZ,T} = 11m \bullet 5,856 \bullet 10^{- 5}\frac{EI_{y}}{m^{2}} = 6,442 \bullet 10^{- 4}\frac{EI_{y}}{m}$$

$${M_{BA,y}}^{RZ,T} = 5m \bullet 5,856 \bullet 10^{- 5}\frac{EI_{y}}{m^{2}} = 2,928 \bullet 10^{- 4}\frac{EI_{y}}{m}$$

$${M_{BC,y}}^{RZ,T} = - 3m \bullet 5,856 \bullet 10^{- 5}\frac{EI_{y}}{m^{2}} = - 1,757 \bullet 10^{- 4}\frac{EI_{y}}{m}$$

$${M_{\left( L = 0,5 \right)\ CB,y}}^{RZ,T} = - 8m \bullet 5,856 \bullet 10^{- 5}\frac{EI_{y}}{m^{2}} = - 4,685 \bullet 10^{- 4}\frac{EI_{y}}{m}$$

$${M_{CB,y}}^{RZ,T} = - 13m \bullet 5,856 \bullet 10^{- 5}\frac{EI_{y}}{m^{2}} = - 7,613 \bullet 10^{- 4}\frac{EI_{y}}{m}$$

$${M_{CD,y}}^{RZ,T} = 11m \bullet 5,856 \bullet 10^{- 5}\frac{EI_{y}}{m^{2}} = 6,442 \bullet 10^{- 4}\frac{EI_{y}}{m}$$

M_DC, y^RZ, T = 0

$${M_{DE,y}}^{RZ,T} = 8m \bullet 5,856 \bullet 10^{- 5}\frac{EI_{y}}{m^{2}} = 4,685 \bullet 10^{- 4}\frac{EI_{y}}{m}$$

M_ED, y^RZ, T = 0

M_AB, x^RZ, T = 0

M_BA, x^RZ, T = 0

$${M_{BC,x}}^{RZ,T} = - 4m \bullet 5,856 \bullet 10^{- 5}\frac{EI_{y}}{m^{2}} = - 2,342 \bullet 10^{- 4}\frac{EI_{y}}{m}$$

$${M_{\left( L = 0,5 \right)\ CB,x}}^{RZ,T} = - 4m \bullet 5,856 \bullet 10^{- 5}\frac{EI_{y}}{m^{2}} = - 2,342 \bullet 10^{- 4}\frac{EI_{y}}{m}$$

$${M_{CB,x}}^{RZ,T} = - 4m \bullet 5,856 \bullet 10^{- 5}\frac{EI_{y}}{m^{2}} = - 2,342 \bullet 10^{- 4}\frac{EI_{y}}{m}$$

$${M_{CD,x}}^{RZ,T} = - 8m \bullet 5,856 \bullet 10^{- 5}\frac{EI_{y}}{m^{2}} = - 4,685 \bullet 10^{- 4}\frac{EI_{y}}{m}$$

$${M_{DC,x}}^{RZ,T} = - 8m \bullet 5,856 \bullet 10^{- 5}\frac{EI_{y}}{m^{2}} = - 4,685\frac{EI_{y}}{m}$$

M_DE, x^RZ, T = 0

M_ED, x^RZ, T = 0

$${V_{AB,z}}^{RZ,T} = - 1 \bullet 5,856 \bullet 10^{- 5}\frac{EI_{y}}{m^{2}} = - 5,856 \bullet 10^{- 5}\frac{EI_{y}}{m^{2}}$$

$${V_{BA,z}}^{RZ,T} = - 1 \bullet 5,856 \bullet 10^{- 5}\frac{EI_{y}}{m^{2}} = - 5,856 \bullet 10^{- 5}\frac{EI_{y}}{m^{2}}$$

$${V_{BC,z}}^{RZ,T} = - 1 \bullet 5,856 \bullet 10^{- 5}\frac{EI_{y}}{m^{2}} = - 5,856 \bullet 10^{- 5}\frac{EI_{y}}{m^{2}}$$

$${V_{\left( x = 0,5L \right)\ CB,z}}^{RZ,T} = - 1 \bullet 5,856 \bullet 10^{- 5}\frac{EI_{y}}{m^{2}} = - 5,856 \bullet 10^{- 5}\frac{EI_{y}}{m^{2}}$$

$${V_{CB,z}}^{RZ,T} = - 1 \bullet 5,856 \bullet 10^{- 5}\frac{EI_{y}}{m^{2}} = - 5,856 \bullet 10^{- 5}\frac{EI_{y}}{m^{2}}$$

$${V_{CD,z}}^{RZ,T} = - 1 \bullet 5,856 \bullet 10^{- 5}\frac{EI_{y}}{m^{2}} = - 5,856 \bullet 10^{- 5}\frac{EI_{y}}{m^{2}}$$

$${V_{DC,z}}^{RZ,T} = - 1 \bullet 5,856 \bullet 10^{- 5}\frac{EI_{y}}{m^{2}} = - 5,856 \bullet 10^{- 5}\frac{EI_{y}}{m^{2}}$$

$${V_{DE,z}}^{RZ,T} = - 1 \bullet 5,856 \bullet 10^{- 5}\frac{EI_{y}}{m^{2}} = - 5,856 \bullet 10^{- 5}\frac{EI_{y}}{m^{2}}$$

$${V_{ED,z}}^{RZ,T} = - 1 \bullet 5,856 \bullet 10^{- 5}\frac{EI_{y}}{m^{2}} = - 5,856 \bullet 10^{- 5}\frac{EI_{y}}{m^{2}}$$

1. Kontrola rozwiązania układu zadanego od obciążeń mechanicznych

KONTROLA RÓWNAŃ RÓWNOWAGI

Dla pręta AB

$$\sum_{}^{}M_{y} = - M_{AB,y} + M_{BA,y} - V_{BA,z} \bullet 6m = - \left( - 316,55kNm \right) + 300,66kNm - 102,87kN \bullet 6m = - 0,01 \approx 0$$

$$\sum_{}^{}M_{x} = - M_{AB,x} + M_{BA,x} = - \left( - 400kNm \right) + 400kNm = 0$$

$$\sum_{}^{}V_{z} = - V_{AB,z} + V_{BA,z} = - 102,87kN + 102,87kN = 0$$

Dla pręta BC

$$\sum_{}^{}M_{y} = - M_{BC,y} + M_{CB,y} - V_{CB,z} \bullet 10m - q \bullet \frac{10^{2}m^{2}}{2} = - \left( - 491,40kNm \right) + 37,28kNm - 2,87kN \bullet 10m - 10\frac{\text{kN}}{m} \bullet \frac{10^{2}m^{2}}{2} = - 0,02 \approx 0$$

$$\sum_{}^{}M_{x} = - M_{BC,x} + M_{CB,x} = - \left( - 11,47kNm \right) + 11,47kNm = 0$$

$$\sum_{}^{}V_{z} = - V_{BC,z} + V_{CB,z} + q \bullet 10m = - 102,87kN + 2,87kN + 10\frac{\text{kN}}{m} \bullet 10m = 0$$

Dla pręta CD

$$\sum_{}^{}M_{y} = - M_{CD,y} + M_{DC,y} - V_{DC,z} \bullet 11m = - \left( - 31,55kNm \right) + 0 - 2,87kN \bullet 11m = - 0,02 \approx 0$$

$$\sum_{}^{}M_{x} = - M_{CD,x} + M_{DC,x} = - \left( - 22,94kNm \right) + 22,94kNm = 0$$

$$\sum_{}^{}V_{z} = - V_{CD,z} + V_{DC,z} = - 2,87kN + 2,87kN = 0$$

Dla pręta DE

$$\sum_{}^{}M_{y} = - M_{DE,y} + M_{ED,y} - V_{ED,z} \bullet 8m = - \left( - 22,94kNm \right) + 0 - 2,87kN \bullet 8m = - 0,02 \approx 0$$

$$\sum_{}^{}M_{x} = - M_{DE,x} + M_{ED,x} = 0$$

$$\sum_{}^{}V_{z} = - V_{DE,z} + V_{ED,z} = - 2,87kN + 2,87kN = 0$$

KONTROLA ZGODNOŚCI PRZEMIESZCZEŃ

$$\delta_{1RZ} = \sum_{P}^{}{\int_{}^{}\frac{{M_{x}}^{\text{RZ}} \bullet {M_{x}}^{1}}{GI_{x}}\text{dx}} + \sum_{P}^{}{\int_{}^{}\frac{{M_{y}}^{\text{RZ}} \bullet {M_{y}}^{1}}{EI_{y}}dy =}$$

$$= \frac{6m}{6EI_{y}}\left( 316,548kNm \bullet 11m + 4 \bullet \left( - 7,944kNm \right) \bullet 8m + 300,660kNm \bullet 5m \right) +$$

$${+ \frac{10m}{6EI_{y}}\left( \left( - 491,396 \right)kNm \bullet \left( - 3 \right)m + 4 \bullet ( - 102,056)kNm \bullet \left( - 8 \right)m + 37,284kNm \bullet \left( - 13 \right)m \right) + \backslash n}{+ \frac{11m}{6EI_{y}}\left( ( - 31,548)kNm \bullet 11m + 4 \bullet \left( - 15,774 \right)kNm \bullet 5,5m + 0 \right) +}$$

$${+ \frac{8m}{6EI_{y}}\left( \left( - 22,944 \right)kNm \bullet 8m + 4 \bullet \left( - 11,472 \right)kNm \bullet 4m + 0 \right) + 0 + \backslash n}{+ \frac{10m}{6 \bullet 0,8EI_{y}}\left( 11,472kNm \bullet \left( - 4 \right)m + 4 \bullet 11,472kNm \bullet \left( - 4 \right)m + 11,472kNm \bullet \left( - 4 \right)m \right) +}$$

$${+ \frac{11m}{6 \bullet 0,8EI_{y}}\left( 22,944kNm \bullet \left( - 8 \right)m + 4 \bullet 22,944kNm \bullet \left( - 8 \right)m + 22,944kNm \bullet \left( - 8 \right)m \right) + 0 = \backslash n}{= \left( - 2233,936 + 7092,247 - 1272,836 - 490,772 - 573,60 - 2523,84 \right)\frac{\text{kNm}^{3}}{EI_{y}} = - 2,737\frac{\text{kN}m^{3}}{EI_{y}} \approx - 2,855\frac{\text{kN}m^{3}}{EI_{y}} = - \frac{{S_{s}}^{\text{RZ}}}{k_{s}} = - \frac{17,132kN}{6\frac{EI_{y}}{m^{3}}}}$$

1. Kontrola rozwiązania układu zadanego od obciążeń niemechanicznych

KONTROLA RÓWNAŃ RÓWNOWAGI

Dla pręta AB

$$\sum_{}^{}M_{y} = - M_{AB,y} + M_{BA,y} - V_{BA,z} \bullet 6m = - \left( 6,442 \bullet 10^{- 4}\frac{EI_{y}}{m} \right) + 2,928 \bullet 10^{- 4}\frac{EI_{y}}{m} - \left( - 5,856 \bullet 10^{- 5}\frac{EI_{y}}{m^{2}} \right) \bullet 6m = - 4 \bullet 10^{- 8} \approx 0$$

$$\sum_{}^{}M_{x} = - M_{AB,x} + M_{BA,x} = 0$$

$$\sum_{}^{}V_{z} = - V_{AB,z} + V_{BA,z} = - \left( - 5,856 \bullet 10^{- 5}\frac{EI_{y}}{m^{2}} \right) - 5,856 \bullet 10^{- 5}\frac{EI_{y}}{m^{2}} = 0$$

Dla pręta BC

$$\sum_{}^{}M_{y} = - M_{BC,y} + M_{CB,y} - V_{CB,z} \bullet 10m = - \left( - 1,757 \bullet 10^{- 4}\frac{EI_{y}}{m} \right) - 7,613 \bullet 10^{- 4}\frac{EI_{y}}{m} - \left( - 5,856 \bullet 10^{- 5}\frac{EI_{y}}{m^{2}} \right) \bullet 10m = 0$$

$$\sum_{}^{}M_{x} = - M_{BC,x} + M_{CB,x} = - \left( - 2,342 \bullet 10^{- 4}\frac{EI_{y}}{m} \right) - 2,342 \bullet 10^{- 4}\frac{EI_{y}}{m} = 0$$

$$\sum_{}^{}V_{z} = - V_{BC,z} + V_{CB,z} = - \left( - 5,856 \bullet 10^{- 5}\frac{EI_{y}}{m^{2}} \right) - 5,856 \bullet 10^{- 5}\frac{EI_{y}}{m^{2}} = 0$$

Dla pręta CD

$$\sum_{}^{}M_{y} = - M_{CD,y} + M_{DC,y} - V_{DC,z} \bullet 11m = - \left( 6,442 \bullet 10^{- 4}\frac{EI_{y}}{m} \right) + 0 - \left( - 5,856 \bullet 10^{- 5}\frac{EI_{y}}{m^{2}} \right) \bullet 11m = 0$$

$$\sum_{}^{}M_{x} = - M_{CD,x} + M_{DC,x} = - \left( - 4,685 \bullet 10^{- 4}\frac{EI_{y}}{m} \right) - 4,685 \bullet 10^{- 4}\frac{EI_{y}}{m} = 0$$

$$\sum_{}^{}V_{z} = - V_{CD,z} + V_{DC,z} = - \left( - 5,856 \bullet 10^{- 5}\frac{EI_{y}}{m^{2}} \right) - 5,856 \bullet 10^{- 5}\frac{EI_{y}}{m^{2}} = 0$$

Dla pręta DE

$$\sum_{}^{}M_{y} = - M_{DE,y} + M_{ED,y} - V_{ED,z} \bullet 8m = - \left( 4,685 \bullet 10^{- 4}\frac{EI_{y}}{m} \right) + 0 - \left( - 5,856 \bullet 10^{- 5}\frac{EI_{y}}{m^{2}} \right) \bullet 8m = - 2 \bullet 10^{- 8} \approx 0$$

$$\sum_{}^{}M_{x} = - M_{DE,x} + M_{ED,x} = 0$$

$$\sum_{}^{}V_{z} = - V_{DE,z} + V_{ED,z} = - \left( - 5,856 \bullet 10^{- 5}\frac{EI_{y}}{m^{2}} \right) - 5,856 \bullet 10^{- 5}\frac{EI_{y}}{m^{2}} = 0$$

KONTROLA ZGODNOŚCI PRZEMIESZCZEŃ

$$\delta_{1T} = \sum_{P}^{}{\int_{}^{}\frac{{M_{x}}^{T} \bullet {M_{x}}^{1}}{GI_{x}}\text{dx}} + \sum_{P}^{}{\int_{}^{}\frac{{M_{y}}^{T} \bullet {M_{y}}^{1}}{EI_{y}}\text{dy}} =$$

$$= \frac{6m}{6EI_{y}}\left( 6,442 \bullet 10^{- 4}\frac{EI_{y}}{m} \bullet 11m + 4 \bullet 4,682 \bullet 10^{- 4}\frac{EI_{y}}{m} \bullet 8m + 2,928 \bullet 10^{- 4}\frac{EI_{y}}{m} \bullet 5m \right) +$$

$${+ \frac{10m}{6EI_{y}}\left( \left( - 1,757 \bullet 10^{- 4}\frac{EI_{y}}{m} \right) \bullet \left( - 3 \right)m + 4 \bullet \left( - 4,685 \bullet 10^{- 4}\frac{EI_{y}}{m} \right) \bullet \left( - 8 \right)m - 7,613 \bullet 10^{- 4}\frac{EI_{y}}{m} \bullet \left( - 13 \right)m \right) + \frac{11m}{6EI_{y}}\left( 6,442 \bullet 10^{- 4}\frac{EI_{y}}{m} \bullet 11m + 4 \bullet 3,221 \bullet 10^{- 4}\frac{EI_{y}}{m} \bullet 5,5m + 0 \right) + \backslash n}{+ \frac{8m}{6EI_{y}}\left( 4,685 \bullet 10^{- 4}\frac{EI_{y}}{m} \bullet 8m + 4 \bullet 2,343 \bullet 10^{- 4}\frac{EI_{y}}{m} \bullet 4m + 0 \right) + 0 + \backslash n}{+ \frac{10m}{6 \bullet 0,8EI_{y}}\left( - 2,343 \bullet 10^{- 4}\frac{EI_{y}}{m} \bullet \left( - 4 \right)m + 4 \bullet \left( - 2,343 \bullet 10^{- 4}\frac{EI_{y}}{m} \right) \bullet \left( - 4 \right)m + \left( - 2,343 \bullet 10^{- 4}\frac{EI_{y}}{m} \right) \bullet \left( - 4 \right)m \right) + \frac{11m}{6 \bullet 0,8EI_{y}}\left( - 4,685 \bullet 10^{- 4}\frac{EI_{y}}{m} \bullet \left( - 8 \right)m + 4 \bullet \left( - 4,685 \bullet 10^{- 4}\frac{EI_{y}}{m} \right) \bullet \left( - 8 \right)m + \left( - 4,685 \bullet 10^{- 4}\frac{EI_{y}}{m} \right) \bullet \left( - 8 \right)m \right) + 0 = \backslash n}$$

$$= \left( 235,476 + 423,600 + 259,827 + 99,957 + 117,15 + 515,35 \right) \bullet 10^{- 4}m = 0,16514m \approx 0,16512m = - \sum_{P}^{}\left( \ \alpha_{T}\frac{T_{d,z} - T_{g,z}}{h_{z}} \bullet \Omega_{\text{My}^{1}} \right) = - ( - 0,16512m)$$

1. Wyznaczenie zadanego przemieszczenia od obciążeń mechanicznych

$$\delta_{\text{DRZ}} = \sum_{P}^{}{\int_{}^{}\frac{{M_{x}}^{\text{RZ}} \bullet {M_{x}}^{1}}{GI_{x}}\text{dx}} + \sum_{P}^{}{\int_{}^{}\frac{{M_{y}}^{\text{RZ}} \bullet {M_{y}}^{1}}{EI_{y}}dy =}$$

$$= \frac{6m}{6EI_{y}}\left( 316,548kNm \bullet \left( - 11m \right) + 4 \bullet \left( - 7,944kNm \right) \bullet \left( - 8m \right) + 300,660kNm \bullet ( - 5m) \right) +$$

$${+ \frac{10m}{6EI_{y}}\left( \left( - 491,396 \right)kNm \bullet \left( - 3,4 \right)m + 4 \bullet \left( - 102,056 \right)kNm \bullet 1,6m + 37,284kNm \bullet 6,6m \right) + \backslash n}{+ \frac{11m}{6EI_{y}}\left( \left( - 31,548 \right)kNm \bullet \left( - 11m \right) + 4 \bullet \left( - 15,774 \right)kNm \bullet \left( - 5,5m \right) + 0 \right) +}$$

$$+ \frac{8m}{6EI_{y}}\left( ( - 22,944)kNm \bullet 8m + 4 \bullet \left( - 11,472 \right)kNm \bullet 4m + 0 \right) +$$

$${+ \frac{6m}{6 \bullet 0,8EI_{y}}\left( \left( - 400 \right)kNm \bullet ( - 8m) + 4 \bullet \left( - 400 \right)kNm \bullet ( - 8m) + \left( - 400 \right)kNm \bullet ( - 8m) \right) + \backslash n}{+ \frac{10m}{6 \bullet 0,8EI_{y}}\left( 11,472kNm \bullet 8,8m + 4 \bullet 11,472kNm \bullet 8,8m + 11,472kNm \bullet 8,8m \right) + 0 + 0 =}$$

$$= \left( 2233,936 + 2106,104 + 1272,436 - 489,472 + 24000 + 1261,92 \right)\frac{\text{kNm}^{3}}{EI_{y}} = 30384,924\frac{\text{kN}m^{3}}{EI_{y}}$$