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Cable element

The formulation of the cable element is not based on a finite element point of view. No special kinematics assumption is done and it is not possible to separate the internal contribution from the external one.

Let $w_l$ denote the load per unit length, including apparent weight and hydrodynamic forces. The entire cable will lay in the plane formed by the segment joining the two extremities and the direction of $w_l$.

A local co-ordinate system is attached to the origin of the cable. The y axis is oriented in the direction opposite to $w_l$ . The x axis points towards the cable end. The local z axis is then obtained by the right hand rule.

Let $H$ and $V$ be the horizontal and vertical distances between both ends; $F_{xo}$ and $F_{yo}$ the reaction forces at the origin; $F_{xe}$ and $F_{ye}$ the reaction forces at the end. Then the catenary equation writes :

\[ H = - F_x \left [ \frac {L_o}{EA} + \frac {1}{w_l} ln \left( \frac{F_{ye}+T_e}{-F_{yo}+ T_o} \right) \right] \]

\[ V = \frac 1 {2EA w_l} (T_e^2 - T_o^2) + \frac{T_e - T_o}{w_l} \]

with $T_o$ and $T_e$ the tensions at both cable ends and $EA$ the axial stiffness. When a length $L_d$ of the cable is laying on the seabed, it comes :

\[ H = - F_x \left [ \frac {L_{susp}}{EA} + \frac {1}{w_l} ln \left( \frac{F_{ye}+T_e}{-F_{yd}+ T_d} \right) \right] \]

$$ V = \frac 1 {2EA w_l} (T_e^2 - T_d^2) + \frac{T_e - T_d}{w_l} $$ with the following additional relationships :

$H_s = H - L_d cos \theta$

$V_s = V - L_d sin \theta$

$T_D=T_I + w_l sin \theta L_d$

Where $\theta$ is the sea bed slope.

The stretching of the laying part writes : $$ e_d = \frac {T_D - T_I} {2EA} $$

In both cases, an iterative procedure is used to solve the previous couple of equations. In case of seabed contact, an another iterative process is included in order to determine the laying part. The corresponding convergence criteria are fixed with the keyword *CONVC.

As a conclusion, the contribution of a cable element to the Virtual work principle writes in the local reference :

\[ G_{internal}^{cable} = \sum_{node} \left( \vec {F_x} . \delta \vec x + \vec F _y . \delta \vec y + \vec F_z . \delta \vec z \right) \]

In dynamic, the inertia contribution of a cable element is simply taken into account with lumped masses. Each nodes accounts for half of the total mass. It comes :

$$ G_{inertia}^{cable} = \sum_{node} \left ( \frac 1 2 m \vec { \ddot X} . \delta \vec X \right) $$ where $X$ is the global displacement of a node.