Espressioni in forma chiusa di $K0$ e $K1$

I valori di $K_{0}$ e $K_{1}$ possono essere calcolati utilizzando le seguenti formule ¹ :

$$
K_0 = 2 \lambda b \frac{1}{\sinh^2(2\cdot\lambda b) \sin(2\cdot\lambda b)} [a’\cdot A + b’\cdot(B_1 + B_2)]
$$
con:
$$
\begin{cases}
a’ = 2 \cosh(\lambda (y+b)) \cos(\lambda (y+b)) \\
A = \sinh(2\cdot\lambda b) \cos (\lambda (b+e)) \cosh (\lambda (b-e)) – [\sin(2\cdot\lambda b) \cosh(\lambda (b+e)) \cos(\lambda (b-e))] \\
b’ = \cosh(\lambda (y+b)) \sin(\lambda (y+b)) + \sinh(\lambda (y+b)) \cos(\lambda (y+b)) \\
B_1 = \sinh(2\cdot\lambda b) [\sin(\lambda (b+e)) \cosh(\lambda (b-e))] – [\cos(\lambda(b+e)) \sinh(\lambda(b-e))] \\
B_2 = \sin(2\cdot\lambda b) [\sinh(\lambda(b+e))\cos(\lambda(b-e))] – [\cosh(\lambda(b+e))\sin(\lambda(b-e))] \\
\lambda=\dfrac{\pi}{L\sqrt{2}}\sqrt[4]{\dfrac{\rho_P}{\rho_E}}
\end{cases}
$$

$$
K_1 =\frac{\sigma}{2\sinh^2(\sigma)} [ C – D + E + F ]
$$
con:
$$
\begin{cases}
C = \cosh (\theta \cdot \chi) ( \sigma \cosh(\sigma) + \sinh(\sigma)) \\
D = \theta \chi \sinh( \sigma \sinh( \sinh(\theta \chi)) \\
E =\dfrac{R_\beta \cdot R_\psi}{3 \sinh (\sigma) \cosh(\sigma) – \sigma} \\
F =\dfrac{Q_\beta \cdot Q_\psi}{3 \sinh(\sigma )\cosh(\sigma)+ \sigma}
\end{cases}
$$
$$
\begin{cases}
R_u = \cosh(\theta u) ( \sigma \cosh(\sigma) – sh(\sigma)) – \theta u sh(\sigma) sh (\theta u) \\
Q_u = \sinh(\theta u) (2 \sinh(\sigma) + \sigma \cosh(\sigma)) – \theta u \sinh(\sigma) \cosh(\theta u)
\end{cases}
$$
$u$ = indice da sostituire con $\beta$ e $\psi$

$$
\theta=\frac{b}{L}\sqrt[4]{\frac{\rho_P}{\rho_E}}
$$

$$
\psi=\frac{\pi e}{b}
$$
$$
\beta=\frac{\pi y}{b}
$$
$$
\sigma=\theta\pi
$$
$$
\chi=\pi-|\beta-\psi|
$$
Nota importante: Per il calcolo di $K_{0}$, quando $e \le y$ cambiare $y$ in $-y$ ed $e$ in $-e$.