1.SCHEMAT STATYCZNY Z OBCIĄŻENIEM RZECZYWISTYM
2.OBLICZENIA SIŁ BIERNYCH OD OBCIĄŻENIA RZECZYWISTEGO
$$\left\{ \begin{matrix}
\Sigma M_{D} = 0 \\
\Sigma M_{C}^{L} = 0 \\
\end{matrix} \right.\ $$
$$\left\{ \begin{matrix}
9H_{A} + 9V_{B} - 6 \times 5 - 1,8 \times 9 \times 4,5 = 0 \\
3H_{A} + 4V_{B} - 1,8 \times 3 \times 1,5 = 0 \\
\end{matrix} \right.\ $$
$$\left\{ \begin{matrix}
9H_{A} + 9V_{B} - 30 - 72,9 = 0 \\
3H_{A} + 4V_{B} - 8,1 = 0 \\
\end{matrix} \right.\ $$
$$\left\{ \begin{matrix}
9H_{A} + 9V_{B} - 102,9 = 0 \\
4V_{B} = 8,1 - 3H_{A} \\
\end{matrix} \right.\ $$
$$\left\{ \begin{matrix}
9H_{A} + 9V_{B} = 102,9 \\
V_{B} = 2,025 - 0,75H_{A} \\
\end{matrix} \right.\ $$
$$\left\{ \begin{matrix}
9H_{A} + 9(2,025 - 0,75H_{A}) = 102,9 \\
V_{B} = 2,025 - 0,75H_{A} \\
\end{matrix} \right.\ $$
$$\left\{ \begin{matrix}
9H_{A} + 18,225 - 6,75H_{A} = 102,9 \\
V_{B} = 2,025 - 0,75H_{A} \\
\end{matrix} \right.\ $$
$$\left\{ \begin{matrix}
2,25H_{A} = 84,675 \\
V_{B} = 2,025 - 0,75H_{A} \\
\end{matrix} \right.\ $$
$$\left\{ \begin{matrix}
H_{A} = 37,63333333 \\
V_{B} = 2,025 - 0,75 \times 37,63333333 \\
\end{matrix} \right.\ $$
HA=37, 633kN VB= − 26, 200kN
ΣX = 0
37, 63333333 − HD − 1, 8 × 9 = 0
HD=21, 433kN
ΣY = 0
−26, 200 + VD − 6 = 0
VD=32, 200kN
3.OBLICZENIE SIŁ W PRZEGUBIE „C”
ΣX = 0
37, 63333333 − HC − 1, 8 × 3 = 0
HC=32, 233kN
ΣY = 0
−26, 200 + VC − 6 = 0
VC=32, 200kN
ΣX = 0
−21, 43333333 + HC − 1, 8 × 6 = 0
HC=32, 233kN
ΣY = 0
32, 200 + VC − 6 = 0
VC= − 26, 200kN
4.OBLICZENIA SIŁ WEWNĘTRZNYCH OD OBCIĄŻENIA RZECZYWISTEGO
PRĘT B-E
0 ≤ x ≤ 4
ΣX = 0 T = OkN
ΣY = 0 N − 26, 200 = 0 N = 26, 200kN
ΣM = 0 M = OkNm
PRĘT A-E
0 ≤ x ≤ 3
0 ≤ y ≤ 3
$$\tan \propto = \frac{3}{3} = 1\ \ \ \ \ \propto = 45$$
sin ∝ =0, 707106781
cos ∝ =0, 707106781
ΣX = 0
37, 63333333 × sin ∝ −1, 8 × y × cos ∝ +N = 0
N = −37, 63333333 × sin ∝ +1, 8 × y × cos∝
N(0)= − 26, 611kN
N(3)= − 22, 085kN
ΣY = 0
37, 63333333 × cos ∝ −1, 8 × y × sin ∝ −T = 0
T = 37, 63333333 × cos ∝ −1, 8 × y × sin∝
T(0)=26, 611kN
T(3)=22, 085kN
ΣM = 0
37, 63333333 × y − 0, 9y2 − M = 0
M = 37, 63333333 × y − 0, 9y2
M(0)=0kNm
M(3)=104, 800kNm
PRĘT C-E
0 ≤ x ≤ 4
ΣX = 0
−N − 32, 233 = 0
N = −32, 233kN
ΣY = 0
T − 6 + 32, 200 = 0
T(0)= − 26, 200kN
ΣM = 0
M + 6x − 32, 200x = 0
M = −6x + 32, 200x
M(0)=0kNm
M(4)=104, 800kNm
PRĘT C-F
0 ≤ x ≤ 5
ΣX = 0
N + 32, 233 = 0
N = −32, 233kN
ΣY = 0
−T − 6 − 26, 200 = 0
T = −32, 200kN
ΣM = 0
−M − 6x − 26, 200x = 0
M = −6x − 26, 200x
M(0)=0kNm
M(5)= − 161, 000kNm
PRĘT D-F
0 ≤ x ≤ 6
ΣX = 0
T − 1, 8x − 21, 43333333 = 0
T = 1, 8x + 21, 43333333
T(0)= − 21, 433kN
T(6)=32, 233kN
ΣY = 0
N + 32, 200 = 0
N = −32, 200kN
ΣM = 0
21, 43333333x + 0, 9x2 + M = 0
M = −21, 43333333x − 0, 9x2
M(0)=0kNm
M(6)= − 161, 000kNm
5.WYKRESY SIŁ WEWNĘTRZNYCH OD OBCIĄŻENIA RZECZYWISTEGO.
6.SCHEMAT STATYCZNY OBCIĄŻONY WIRTUALNĄ SIŁĄ SKUPIONĄ $\overset{\overline{}}{\mathbf{P}}$
$$\left\{ \begin{matrix}
\Sigma M_{D} = 0 \\
\Sigma M_{C}^{L} = 0 \\
\end{matrix} \right.\ $$
$$\left\{ \begin{matrix}
9{\overset{\overline{}}{H}}_{A} + 9{\overset{\overline{}}{V}}_{B} + 2 \times \overset{\overline{}}{1} = 0 \\
3{\overset{\overline{}}{H}}_{A} + 4{\overset{\overline{}}{V}}_{B} - 4 \times \overset{\overline{}}{1} = 0 \\
\end{matrix} \right.\ $$
$$\left\{ \begin{matrix}
{\overset{\overline{}}{H}}_{A} = - {\overset{\overline{}}{V}}_{B} - \frac{2}{9} \times \overset{\overline{}}{1} \\
{\overset{\overline{}}{V}}_{B} = - \frac{3}{4}{\overset{\overline{}}{H}}_{A} + \overset{\overline{}}{1} \\
\end{matrix} \right.\ $$
$$\left\{ \begin{matrix}
{\overset{\overline{}}{H}}_{A} = - {\overset{\overline{}}{V}}_{B} - \frac{2}{9} \times \overset{\overline{}}{1} \\
{\overset{\overline{}}{V}}_{B} = \frac{3}{4}{\overset{\overline{}}{V}}_{B} + \frac{1}{6} \times \overset{\overline{}}{1} + \overset{\overline{}}{1} \\
\end{matrix} \right.\ $$
$$\left\{ \begin{matrix}
{\overset{\overline{}}{H}}_{A} = - {\overset{\overline{}}{V}}_{B} - \frac{2}{9} \times \overset{\overline{}}{1} \\
\frac{1}{4}{\overset{\overline{}}{V}}_{B} = \frac{7}{6} \times \overset{\overline{}}{1} \\
\end{matrix} \right.\ $$
$$\left\{ \begin{matrix}
{\overset{\overline{}}{H}}_{A} = - {\overset{\overline{}}{V}}_{B} - \frac{2}{9} \times \overset{\overline{}}{1} \\
{\overset{\overline{}}{V}}_{B} = \frac{14}{3} \times \overset{\overline{}}{1} \\
\end{matrix} \right.\ $$
$$\left\{ \begin{matrix}
{\overset{\overline{}}{\mathbf{H}}}_{\mathbf{A}}\mathbf{= -}\frac{\mathbf{44}}{\mathbf{9}}\mathbf{\times}\overset{\overline{}}{\mathbf{1}} \\
{\overset{\overline{}}{\mathbf{V}}}_{\mathbf{B}}\mathbf{=}\frac{\mathbf{14}}{\mathbf{3}}\mathbf{\times}\overset{\overline{}}{\mathbf{1}} \\
\end{matrix} \right.\ $$
ΣX = 0
${\overset{\overline{}}{H}}_{A} - {\overset{\overline{}}{H}}_{D} + \overset{\overline{}}{1} = 0$
$${\overset{\overline{}}{H}}_{D} = - \frac{44}{9} \times \overset{\overline{}}{1} + \overset{\overline{}}{1} = 0$$
$${\overset{\overline{}}{\mathbf{H}}}_{\mathbf{D}}\mathbf{= -}\frac{\mathbf{35}}{\mathbf{9}}\mathbf{\times}\overset{\overline{}}{\mathbf{1}}$$
ΣY = 0
${\overset{\overline{}}{V}}_{B} + {\overset{\overline{}}{V}}_{D} = 0$
$${\overset{\overline{}}{\mathbf{V}}}_{\mathbf{D}}\mathbf{= -}\frac{\mathbf{14}}{\mathbf{3}}\mathbf{\times}\overset{\overline{}}{\mathbf{1}}$$
7.OBLICZENIE SIŁ W PRZGUBIE „C”
ΣX = 0
$$- 44 - {\overset{\overline{}}{H}}_{C} + 9 = 0$$
$${\overset{\overline{}}{\mathbf{H}}}_{\mathbf{C}}\mathbf{= - 35}$$
ΣY = 0
$$14 + {\overset{\overline{}}{V}}_{C} = 0$$
$${\overset{\overline{}}{\mathbf{V}}}_{\mathbf{C}}\mathbf{= - 14}$$
ΣX = 0
$$35 + {\overset{\overline{}}{H}}_{C} = 0$$
$${\overset{\overline{}}{\mathbf{H}}}_{\mathbf{C}}\mathbf{= - 35}$$
ΣY = 0
$${\overset{\overline{}}{V}}_{C} - 14 = 0$$
$${\overset{\overline{}}{\mathbf{V}}}_{\mathbf{C}}\mathbf{= 14}$$
8.OBLICZENIE MOMENTÓW ZGINAJĄCYCH OD OBCIĄŻENIA WIRTUALNEGO
MOMENTY W WĘŹLE „E”
$$\overset{\overline{}}{M} = - \frac{44}{3} \times \overset{\overline{}}{1}$$
$$\overset{\overline{}}{M} = - \frac{12}{3} \times \overset{\overline{}}{1}$$
$$\overset{\overline{}}{M} = - \frac{56}{3} \times \overset{\overline{}}{1}$$
MOMENTY W WĘŹLE „F”
$$\overset{\overline{}}{M} = \frac{70}{3} \times \overset{\overline{}}{1}$$
$$\overset{\overline{}}{M} = \frac{70}{3} \times \overset{\overline{}}{1}$$
9.WYKRES MOMENTÓW ZGINAJĄCYCH OD OBCIĄŻENIA WIRTUALNEGO
10.PRZYJĘCIE PRZEKROJU POPRZECZNEGO PRĘTA DLA CAŁEJ RAMY
Przyjęto dwuteownik równoległościenny IPE 400 wg PN-91/H-93419 o następujących danych:
h=40cm
A=84,5cm2
IX=23130cm4 = 0,00023130m4
IY=1320cm4
WX=1160cm3
WX=146cm3
E = 205000MPa = 205000000kN/m2
$$\sigma_{\text{dop}} = 150MPa = 15,0\frac{\text{kN}}{\text{cm}^{2}}$$
Mmax = 161kNm = 16100kNcm
$$\frac{M_{\max}}{\sigma_{\text{dop}}} = \frac{16100}{15} = 1073,333\text{cm}^{3} < W_{x} = 1160\text{cm}^{3}$$
$$\sigma_{x} = \frac{M_{\max}}{W_{x}} = \frac{16100}{1160} = 13,879\frac{\text{kN}}{\text{cm}^{2}} = 138,79MPa < \sigma_{\text{dop}} = 150MPa$$
11.OBLICZENIE PRZEMIESZCZENIA POZIOMEGO PUNKTU „K”
PRĘT A-E
$$\delta = \frac{1}{\text{EI}} \times \frac{1}{3}\left\lbrack \left( - \frac{1}{2} \times 104,800 \times 4,242641 \times \frac{2}{3} \times 44 \right) - \left( \frac{2}{3} \times 4,242641 \times \frac{1,8 \times 3^{2}}{8} \times \frac{1}{2} \times 44 \right) \right\rbrack = \frac{1}{\text{EI}} \times \frac{1}{3}\left\lbrack - 6521,22206 - 126,0064377 \right\rbrack = \frac{\mathbf{1}}{\mathbf{\text{EI}}}\mathbf{\times}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{\times}\left( \mathbf{- 6395,215622} \right)$$
$$\left\lbrack \left( kNm \times m \times m \right) - \left( m \times \frac{\text{kN}}{m} \times m^{2} \times m \right) \right\rbrack = \left\lbrack \text{kN}m^{3} \right\rbrack$$
PRĘT B-E
δ = 0 [kNm3]
PRĘT C-E
$$\delta = \frac{1}{\text{EI}} \times \frac{1}{3} \times \left( - \frac{1}{2} \times 104,800 \times 4 \times \frac{2}{3} \times 56 \right) = \frac{\mathbf{1}}{\mathbf{\text{EI}}}\mathbf{\times}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{\times}\left( \mathbf{- 7825,066667} \right)$$
[kNm×m×m] = [kNm3]
PRĘT C-F
$$\delta = \frac{1}{\text{EI}} \times \frac{1}{3} \times \left( - \frac{1}{2} \times 161,000 \times 5 \times \frac{2}{3} \times 70 \right) = \frac{\mathbf{1}}{\mathbf{\text{EI}}}\mathbf{\times}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{\times}\left( \mathbf{- 18783,33333} \right)$$
[kNm×m×m] = [kNm3]
PRĘT D-F
$$\delta = \frac{1}{\text{EI}} \times \frac{1}{3} \times \left\lbrack \left( - \frac{1}{2} \times 161,000 \times 6 \times \frac{2}{3} \times 70 \right) + \left( \frac{2}{3} \times 6 \times \frac{1,8 \times 6^{2}}{8} \times \frac{1}{2} \times 70 \right) \right\rbrack = \frac{1}{\text{EI}} \times \frac{1}{3} \times \left\lbrack - 22540 + 1134 \right\rbrack = \frac{\mathbf{1}}{\mathbf{\text{EI}}}\mathbf{\times}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{\times}\left( \mathbf{- 21406} \right)$$
$$\left\lbrack \left( kNm \times m \times m \right) - \left( m \times \frac{\text{kN}}{m} \times m^{2} \times m \right) \right\rbrack = \left\lbrack \text{kN}m^{3} \right\rbrack$$
$$\delta = \frac{1}{\text{EI}} \times \frac{1}{3} \times \left\lbrack - 6395,215622 - 7825,066667 - 18783,33333 - 21406 \right\rbrack = \frac{1}{\text{EI}} \times \frac{- 41619,18438}{3} = \frac{- 13873,06146}{205000000 \times 0,00023130} = - 0,292578774 \approx \mathbf{- 0,293}$$
$$\left\lbrack \frac{\text{kN}m^{3}}{\frac{\text{kN}}{m^{2}} \times m^{4}} \right\rbrack = \left\lbrack m \right\rbrack$$
Punkt „K” doznał przemieszczenia poziomego 0,293m.