Metoda 3 momentów

SSN = 5 - 3 = 2

\[ M_A = 20 \]

\[ \sum M_C = 0 \] \[ 20 \cdot 3 - R_D \cdot 5 = 0 \] \[ R_D = 12 \]

Belki wtórne

\[ \begin{aligned} \text{I:} \quad & \sum M_A = 0 \quad \overline{lR_B} = 0 \\ & \\ \text{II :} \quad & \sum M_B = 0 \quad 6{,}67 \cdot 1 - \overline{lR_C} \cdot 2 = 0 \quad \overline{lR_C} = 3{,}335 \\ & \sum y = 0 \quad \overline{pR_B} + \overline{lR_C} - 6{,}67 = 0 \quad \overline{pR_B} = 3{,}335\\ & \\ \text{III:} \quad & \sum M_C = 0 \quad 36 \cdot 2 + 24\left(3 + \frac{2}{3}\right) - \overline{R_D} \cdot 5 = 0 \quad \overline{R_D} = 32 \\ & \sum y = 0 \quad \overline{pR_C} - 36 - 24 + \overline{R_D} = 0 \quad \overline{pR_C} = 28 \end{aligned} \]

\[ \textcolor{red}{M_{k-1} \cdot l_{k-1} + 2M_k(l_{k-1} + l_k) + M_{k+1} \cdot l_k = \textcolor{red}{= -6\left( \overline{lR_k} + \overline{pR_k} \right)}} \]

\[ M_A \cdot 3 + 2 \cdot M_B(3 + 2) + M_C \cdot 2 = -6(0 + 3{,}335) \]

\[ \begin{aligned} (1) \quad 10M_B + 2M_C &= -20{,}01 - 60 \\ &= -80{,}01 \\ & \\ M_B \cdot 2 + 2M_C \cdot (2 + 5) + M_D \cdot 5 &= -6(3{,}335 + 28) \end{aligned} \]

\[ \begin{aligned} (2) \quad 2M_B + 14M_C &= -188{,}01 |\cdot (-5) \\ & \\ 10M_B + 2M_C &= -80{,}01 \\ +|-10M_B - 70M_C &= 940{,}05 \\ \hline -68M_C &= 860{,}04 \\ \\ M_C &= -12{,}648\, \text{kNm} \\ M_B &= \frac{-80{,}01 - 2M_C}{10} \\ M_B &= -5{,}471\, \text{kNm} \end{aligned} \]

\[ \begin{aligned} \text{I:} \quad & \sum M_A = 0 \quad 20 + 5{,}471 - \overline{lR_B} \cdot 3 = 0 \\ & \overline{lR_B} = 8{,}49 \\ & \\ & \sum y = 0 \quad R_A + \overline{lR_B} = 0 \\ & R_A = -8{,}49 \\ \\ \text{II:} \quad & \sum M_B = 0 \quad -5{,}471 + 10 \cdot 2 \cdot 1 + 12{,}648 - \overline{lR_C} \cdot 2 = 0 \\ & \overline{lR_C} = 13{,}589 \\ & \\ & \sum y = 0 \quad pR_B + lR_C - 10 \cdot 2 = 0 \\ & pR_B = 6{,}411 \\ \\ \text{III:} \quad & \sum M_C = 0 \quad -12{,}648 + 20 \cdot 3 - \overline{R_D} \cdot 5 = 0 \\ & \overline{R_D} = 9{,}47 \\ & \\ & \sum y = 0 \quad pR_C - 20 + R_D = 0 \\ & pR_C = 10{,}53 \end{aligned} \]

\[ \text{Ekstremum:} \quad \frac{6{,}411}{x} = \frac{13{,}589}{2-x} \] \[ 13{,}589x = 6{,}411(2-x) \] \[ 13{,}589x = 12{,}822 - 6{,}411x \] \[ 20x = 12{,}822 \] \[ x = 0{,}6411 \]

\[ \text{Mextr:} \] \[ \text{II:} \quad pR_B \cdot 0{,}6411 - 5{,}471 - 10 \cdot 0{,}6411 \cdot \frac{0{,}6411}{2} = -3{,}416 \]

\[ \begin{aligned} SSN = 4 - 3 - 0 = 1\\ \\ \text{I:} \quad & \sum M_A = 0 \quad 83{,}33 \cdot 2{,}5 - \overline{lR_B} \cdot 5 = 0 \\ & \overline{lR_B} = 41{,}665 \\ \\ \text{II:} \quad & \sum M_C = 0 \quad -20 \cdot \frac{2}{3} \cdot 2 - 20 \cdot \left(2 + \frac{1}{3} \cdot 2\right) + \overline{pR_B} \cdot 4 = 0 \\ & \overline{pR_B} = 20 \end{aligned} \]

\[ \begin{aligned} M_A = 0\\ M_C = 0\\ \\ M_A \cdot 5 + 2 \cdot M_B(5 + 4) + M_C \cdot 4 &= -6(41{,}665 + 20) \\ M_A \cdot l_1 + 2 \cdot M_B(l_1 + l_2) + M_C \cdot l_2 &= -6(\overline{lR_B} + \overline{pR_B}) \end{aligned} \]

\[ \begin{aligned} 18M_B &= -369{,}99 \\ M_B &= -20{,}555 \end{aligned} \]

\[ \begin{aligned} \text{I:} \quad & \sum M_A = 0 \quad 8 \cdot 5 \cdot 2 \cdot 5 + 20{,}555 - \overline{lR_B} \cdot 5 = 0 \\ & \overline{lR_B} = 24{,}111 \\ & \\ & \sum y = 0 \quad R_A - 8 \cdot 5 + lR_B = 0 \\ & R_A = 15{,}889 \\ \\ \text{II:} \quad & \sum M_B = 0 \quad -20{,}555 + 20 \cdot 2 - R_C \cdot 4 = 0 \\ & R_C = 4{,}861 \\ & \\ & \sum y = 0 \quad pR_B - 20 + R_C = 0 \\ & pR_0 = 15{,}139 \end{aligned} \]

\[ \text{Ekstremum:} \quad \frac{15{,}889}{x} = \frac{24{,}111}{5-x} \quad \Rightarrow \quad x = 1{,}986 \]

\[ \text{Mextr} = R_A \cdot 1{,}986 - 8\cdot1{,}986 \cdot \frac{1{,}986}{2} = 15{,}779 \]

Metody energetyczne - Układy statycznie niewyznaczalne

Wytrzymałość materiałów - Metody energetyczne - Układy statycznie niewyznaczalne - Przykład 1

Przykład 1