132
SINGLE DEGREE OF FREEDOM STRUCTURES
équations of structural motion. Some possible responses such as jumps and
subharmonics were discussed above, based on closed form solutions to those
équations. Such possible responses greatly aid in the interpretion of computerderived numerical results, as the following example will demonstrate.
Many software packages are available to carry out numerical solutions. A
popular package is Mathematica® (1999). For solutions to nonlinear dynamics
problems, such software usually employs a step-by-step intégration method,
which was used in the following problem also. With this method, the response
was evaluated at successive incréments of time where the restraint and drag
force terms were taken as constant during each interval and then updated at
the end of each interval. Clough and Penzien (1993) and Paz (1980) discussed
this basic method in the context of nonlinear structural dynamics.
The following problem is a rather comprehensive one. It begins with a physical description of the buoy, proceeds with a careful formulation of its dynamic
model with its accompanying assumptions, and concludes with a discussion of
numerical results relative to the closed form responses derived previously in this
chapter.
Physical Description and Dynamic Model
A type of open sea mooring System used to load crude oil into tankers is the
single anchor leg mooring (SALM) buoy. One of a wide variety of such buoy
designs is shown in Figure 5.12: a buoy with a long, upright cylinder stabilized
by cables to sea floor anchors and held in place by a short bottom chain to the
pipeline end manifold (PLEM). This PLEM is anchored to the seafloor with
pin piles. When the oil is not flowing through the flexible hoses to the ship,
the tanks within the buoy are full, and the buoy has just sufficient buoyancy
to keep it afloat. Then the tension in the bottom connecting chain is very
slight. It is under these conditions that the rotational motion of the buoy is
now analyzed. The chosen excitation is that of a single, harmonie, linear water
wave. Responses to a range of excitation frequencies are sought.
The mathematical model of the upright buoy with its geometry and its
loading are defined in Figure 5.13a. The buoy is treated as a rigid body rotating
at angle 9 about a frictionless pin at its base. The free body sketch of the buoy
shown in Figure 5.13b identifies: the net cable tension force Fc at angle 0 with
the horizontal: the damping force Fd acting at the mass center G; and the
total wave load p-,i’i The center of buoyancy B is assumed to be coïncident
with G. The buoy weight and its buoyant force are equal and opposite, and the
net restoring moment due to these forces is zéro. The horizontal wave particle
\elot ity and accélération are assumed to be much larger than those for the buoy.
When équation (2.3), the équation of plane motion for rotation of a rigid body
about a fixed point 0, is applied to the free body sketch of Figure 5.13c, the
resuit for small buoy rotations 9 is
M + hcFc cos 0 - hc9Fc sin 0 + hdFd = - hwP1 (t)
(5.101)
SINGLE DEGREE OF FREEDOM STRUCTURES
équations of structural motion. Some possible responses such as jumps and
subharmonics were discussed above, based on closed form solutions to those
équations. Such possible responses greatly aid in the interpretion of computerderived numerical results, as the following example will demonstrate.
Many software packages are available to carry out numerical solutions. A
popular package is Mathematica® (1999). For solutions to nonlinear dynamics
problems, such software usually employs a step-by-step intégration method,
which was used in the following problem also. With this method, the response
was evaluated at successive incréments of time where the restraint and drag
force terms were taken as constant during each interval and then updated at
the end of each interval. Clough and Penzien (1993) and Paz (1980) discussed
this basic method in the context of nonlinear structural dynamics.
The following problem is a rather comprehensive one. It begins with a physical description of the buoy, proceeds with a careful formulation of its dynamic
model with its accompanying assumptions, and concludes with a discussion of
numerical results relative to the closed form responses derived previously in this
chapter.
Physical Description and Dynamic Model
A type of open sea mooring System used to load crude oil into tankers is the
single anchor leg mooring (SALM) buoy. One of a wide variety of such buoy
designs is shown in Figure 5.12: a buoy with a long, upright cylinder stabilized
by cables to sea floor anchors and held in place by a short bottom chain to the
pipeline end manifold (PLEM). This PLEM is anchored to the seafloor with
pin piles. When the oil is not flowing through the flexible hoses to the ship,
the tanks within the buoy are full, and the buoy has just sufficient buoyancy
to keep it afloat. Then the tension in the bottom connecting chain is very
slight. It is under these conditions that the rotational motion of the buoy is
now analyzed. The chosen excitation is that of a single, harmonie, linear water
wave. Responses to a range of excitation frequencies are sought.
The mathematical model of the upright buoy with its geometry and its
loading are defined in Figure 5.13a. The buoy is treated as a rigid body rotating
at angle 9 about a frictionless pin at its base. The free body sketch of the buoy
shown in Figure 5.13b identifies: the net cable tension force Fc at angle 0 with
the horizontal: the damping force Fd acting at the mass center G; and the
total wave load p-,i’i The center of buoyancy B is assumed to be coïncident
with G. The buoy weight and its buoyant force are equal and opposite, and the
net restoring moment due to these forces is zéro. The horizontal wave particle
\elot ity and accélération are assumed to be much larger than those for the buoy.
When équation (2.3), the équation of plane motion for rotation of a rigid body
about a fixed point 0, is applied to the free body sketch of Figure 5.13c, the
resuit for small buoy rotations 9 is
M + hcFc cos 0 - hc9Fc sin 0 + hdFd = - hwP1 (t)
(5.101)