\[ R_{pl}^{+} = 12{,}5 \ \frac{\text{kN}}{\text{cm}^2} \]

Odcinek \( AC \):

\[ J_0 = \frac{\pi}{2} \cdot 8^4 = 6434 \ \text{cm}^4 \]

\[ W_s^0 = \frac{J_0}{r} = \frac{J_0}{4} = 804 \ \text{cm}^3 \]

\[ M_s^0 = W_s \cdot R_{pl}^{+} = 804 \cdot 125 = 100500 \ \text{kNcm} \]

\[ W_{pl}^0 = \frac{2}{3} \pi \cdot 8^3 = 1072 \ \text{cm}^3 \]

\[ M_n^0 = W_{pl}^0 \cdot R_{pl}^{+} = 1072 \cdot 12{,}5 = 13400 \ \text{kNcm} \]

Odcinek \( CF \):

\[ J_0 = \frac{\pi}{2} \left( 8^4 - 5^4 \right) = 4398 \ \text{cm}^4 \]

\[ W_s^0 = \frac{J_0}{r} = 550 \ \text{cm}^3 \]

\[ M_s^0 = 550 \cdot 12{,}5 = 6872 \ \text{kNcm} \]

\[ W_{pl}^0 = \frac{2}{3} \pi \left( 8^3 - 6^3 \right) = 620 \ \text{cm}^3 \]

\[ M_n^0 = 620 \cdot 12{,}5 = 7750 \ \text{kNcm} \]

a)

\[ 2M = 2M_n^0 \Rightarrow M = M_n^0 = 77{,}5 \ \text{kNm} \]

b)

\[ 4M = 2M_n^0 \Rightarrow M = \frac{M_n^0}{2} = 67 \ \text{kNm} \]

c)

\[ 3M = M_n^0 + M_n^0 \Rightarrow M = \frac{M_{n1} + M_{n2}}{3} \] \[ M = 70{,}5 \ \text{kNm} \]

Wniosek: Tylko \( M_n = 67 \ \text{kNm} \) jest statycznie dopuszczalne.

Rozwiązanie sprężyste

Warunek geometryczny:

\[ \frac{(3M - x)a}{J_{o1}G} + \frac{-(M + x)a}{J_{o1}G} + \frac{(2M - x)a}{J_{o2}G} + \frac{-x a}{J_{o2} G} = 0 \]

\[ 2M = M_{s1}^0 \Rightarrow M = \frac{100{,}5}{2} = 50{,}25 \ \text{kNm} = M_s \]

\[ M = M_{s2}^0 \Rightarrow M = 68{,}7 \ \text{kNm} \]

\[ \frac{M_n}{M_s} = {1{,}333} \]

Nośność graniczna

Przykład 0