\begin{aligned} &\sigma_{r}(r)=A-\frac{B}{r^{2}}-\frac{3+\nu}{8 g} \gamma \cdot \omega^{2} r^{2} \\ &\sigma_{\varphi}(r)=A+\frac{B}{r^{2}}-\frac{1+3 \nu}{8 g} \gamma \cdot \omega^{2} r^{2} \end{aligned}

Warunki brzegowe:

\begin{aligned} &\sigma_{r}(0)=\sigma_{\varphi}(0) \\ &\sigma_{r}(b)=p_{z} \end{aligned}

Używając pierwszego warunku:

\begin{aligned} &\sigma_{r}(0)=\sigma_{\varphi}(0) \\ &A-\frac{B}{r^{2}}-\frac{3+\nu}{8 g} \gamma \cdot \omega^{2} r^{2}=A+\frac{B}{r^{2}}-\frac{1+3 \nu}{8 g} \gamma \cdot \omega^{2} r^{2} \\ &-\frac{3+\nu}{8 g} \gamma \cdot \omega^{2} r^{2}=2 \frac{B}{r^{2}}-\frac{1+3 \nu}{8 g} \gamma \cdot \omega^{2} r^{2} \\ &B=\frac{(\nu-1) \cdot \gamma \cdot \omega^{2} \cdot r^{4}}{8 \cdot g} \\ & r=0 \\ &B=0 \end{aligned}

Używając drugiego warunku:

\begin{aligned} &\sigma_{r}(b)=p_{z} \\ &\sigma_{r}(r)=A-\frac{B}{r^{2}}-\frac{3+\nu}{8 g} \gamma \cdot \omega^{2} r^{2} \\ &p_{z}=A-\frac{3+\nu}{8 g} \gamma \cdot \omega^{2} b^{2} \\ &A=p_{z}+\frac{3+\nu}{8 g} \gamma \cdot \omega^{2} b^{2} \end{aligned}

Naprężenia zredukowane:

\begin{aligned} \sigma_{r e d}=\sqrt{\sigma_{r}^{2}-\sigma_{r} \cdot \sigma_{\varphi}+\sigma_{\varphi}^{2}} \end{aligned}

Wartość maksymalna dla r=0:

\begin{aligned} &r:=0\\ &\sigma_{r}=\sigma_{\varphi}=A\\ &A=p_{z}+\frac{3+\nu}{8 g} \gamma \cdot \omega^{2} b^{2}\\ &\sigma_{r e d}=\sqrt{A^{2}-A \cdot A+A^{2}}=A\\ \end{aligned}

Warunek:

\begin{aligned} &\sigma_{r e d}\leq k_{r} \\ &\frac{1}{3} k_{r}+\frac{3+\nu}{8 g} \gamma \cdot \omega^{2} b^{2}=k_{r} \\ &\frac{3+\nu}{8 g} \gamma \cdot \omega^{2} b^{2}=\frac{2}{3} k_{r} \\ &\omega=\sqrt{\frac{k_{r} \cdot 16}{3(3+\nu) \cdot \gamma \cdot b^{2}}} \end{aligned}