ACU-T: 5301 Ship Hull Dynamics

The problem to be addressed in this tutorial is shown below. It is a mid-section of a Wigley Ship model. Wigley hulls have been widely used as test cases for evaluating hydrodynamic behavior of ships. The present tutorial demonstrates the simulation of gravity waves hitting a freely floating Wigley hull and evaluating the position of the hull with time. This tutorial is similar to the Ship Hull Static tutorial, except that the ship hull is freely floating as a rigid body compared to a static ship hull.

Since the motion considered in this tutorial is perpendicular to the length of the ship, an analysis of a 2D section of the ship hull would be appropriate with lesser computation time without compromising on accuracy. The mid-section dimensions of the Wigley hull is a function of total ship length and the model used in this tutorial is the mid-section of Wigley hull whose ship length is 1m.

Figure 1. Schematic of a Ship Hull

Generation of Surface Gravity Waves

Wind-generated gravity waves on the free surface of the sea are generated using a UDF (User-Defined Function).

Figure 2. Gravity Waves

A linear solution of surface gravity wave propagation would result in the following equation for horizontal velocity of wave.

$u=U_0+U\frac{\cosh k\left(D-z\right)}{\sinh \left(kD\right)}\text{cos}\left(\omega t-kx\right)$

Where,

$u$ is the horizontal particle velocity of wave

$U_0$ is the speed of the wave

U is the velocity amplitude of disturbance

$k$ is wave number = $2\pi /\lambda$

$\lambda$ is the wave length of the wave

$D$ is the depth of the water

$\omega$ is the frequency of the wave = $\frac{2\pi }{T}$

$T$ is the time period of the wave

t is the time

In the present simulation, the following values are used for the variables of above equation:

U = 0.1256 m/s

$T$ = 1.0 sec

$k$ = 12.566 m^-1

$U_0$ = 0.01 m/s

$D$ = 0.5 m

In the present tutorial, at the inlet you will generate the wave for 2 seconds and simulate the motion of the wave for 5 seconds. A UDF (wave.c) written in C language is used for this purpose. For the details of the functions used in the wave.c, refer to the AcuSolve User-Defined Functions Manual.

Two Dimensional Simulations in AcuSolve

AcuSolve does not support a CFD analysis on a 2D surface mesh. However, a 2D analysis can be simulated on a volume mesh by having a single element extrusion along the perpendicular direction of 2D surface of interest and then having identical boundary conditions of symmetry or slip on both sides of extrusion. This way you can ensure that the solution does not vary along the thickness (extrusion), which is essentially a 2D representation of the problem. The present tutorial uses this approach.

Modeling the Solid Ship Hull

For the present tutorial the following properties are assumed for the ship hull.

Material = Aluminum

Density = 2702 kg/m³

Body Center = (1.12971, 0.0, 0.0082984) m

Body Mass = 0.2038 kg

Body Weight = 0.2038 × 9.81 = 2.0 N

Components of Moments of Inertia (in kg m²):

$I_{xx}$ = 0.00113969

$I_{yy}$ = 0.0051392

$I_{zz}$ = 0.004006

$I_{xy}=I_{yx}=0$

$I_{yz}=I_{zy}=0$

$I_{zx}=I_{xz}=0$

Rigid Body Dynamic Analysis

A local coordinate system is used to simplify the definition of the rigid body model and the solution of the equations of motion. The translational and rotational equations of motion are:

$ma+C v+K x =F_I+F_E$

$I \alpha +D \omega +L \theta =R_I+R_E$

Where,

$x, v, a$ are the translational displacement, velocity and acceleration vectors, respectively as:

$x=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$, $v=\left[\begin{array}{c}{v}_x\\ v_y\\ v_z\end{array}\right]$, $a=\left[\begin{array}{c}{a}_x\\ a_y\\ a_z\end{array}\right]$

$\theta , \omega , \alpha$ are the angular displacement, angular velocity and angular acceleration vectors, respectively as:

$\theta =\left[\begin{array}{c}{\theta }_x\\ {\theta }_y\\ {\theta }_z\end{array}\right]$,$\omega =\left[\begin{array}{c}{\omega }_x\\ {\omega }_y\\ {\omega }_z\end{array}\right]$, $\alpha =\left[\begin{array}{c}{\alpha }_x\\ {\alpha }_y\\ {\alpha }_z\end{array}\right]$

$m$ is the mass of the body.

$I=\left[\begin{array}{ccc}{I}_{xx}& I_{xy}& I_{xz}\\ I_{yx}& I_{yy}& I_{yz}\\ I_{zx}& I_{zy}& I_{zz}\end{array}\right]$ is the dyadic (moment of inertia) matrix of the body.

$C, D$ are the translational and rotational damping matrices, respectively.

$K, L$ are the translational and rotational stiffness matrices, respectively.

$F_I, R_I$ are the internal Forces and Moments from the fluid.

$F_E, R_E$ are the external Forces and Moments you specify.

Rigid Body Dynamics Analysis for a 2D Problem

For the present tutorial, a single element extrusion will be made along the y-axis. Based on the above assumptions, you arrive with:

$x=\left[\begin{array}{c}x\\ 0\\ z\end{array}\right]$, $v=\left[\begin{array}{c}{v}_x\\ 0\\ v_z\end{array}\right]$, $a=\left[\begin{array}{c}{a}_x\\ 0\\ a_z\end{array}\right]$

$\theta =\left[\begin{array}{c}0\\ {\theta }_y\\ 0\end{array}\right]$, $\omega =\left[\begin{array}{c}0\\ {\omega }_y\\ 0\end{array}\right]$, $\alpha =\left[\begin{array}{c}0\\ {\alpha }_y\\ 0\end{array}\right]$

The translational equations of the motion will be:

$m a =F_I+F_E$

$m\left[\begin{array}{c}{a}_x\\ 0\\ a_z\end{array}\right]=\left[\begin{array}{c}{F}_I{}_x\\ F_I{}_y\\ F_I{}_z+F_g\end{array}\right]$

The rotational equations of the motion will be:

$I \alpha =R_I+R_E$

$\left[\begin{array}{ccc}{I}_{xx}& I_{xy}& I_{xz}\\ I_{yx}& I_{yy}& I_{yz}\\ I_{zx}& I_{zy}& I_{zz}\end{array}\right]\times \left[\begin{array}{c}0\\ {\alpha }_y\\ 0\end{array}\right]=\left[\begin{array}{c}0\\ R_I{}_y\\ 0\end{array}\right]$

$\left[\begin{array}{c}{I}_{xy}{\alpha }_y\\ I_{yy}{\alpha }_y\\ I_{zy}{\alpha }_y\end{array}\right]= \left[\begin{array}{c}0\\ R_I{}_y\\ 0\end{array}\right]$

${\alpha }_y =\frac{R_I{}_y}{I_{yy}}$

The only critical component of moment of inertia for the present tutorial is $I_{yy}$.