Page 301 - Fluid-Structure Interactions Slender Structure and Axial Flow (Volume 1)

282 SLENDER STRUCTURES AND AXIAL FLOW

quantities for the enclosed fluid), and where rL and t~. represent, respectively, the position
vector and the tangential unit vector at the end of the pipe [Figure 5.1(c)].
However, before proceeding with the derivation of the equations of motion, some order-
of-magnitude considerations are necessary. The lateral displacement of the pipe may be
considered to be small relative to the length of the pipe, i.e.

z = w - 6(E), E << 1. (5.1 1)

Large motions imply that terms of higher order than the linear ones have to be kept in the
equation of motions. Because of the symmetry of the system, the nonlinear equations will
necessarily be of odd order, and the derivation here will give a set of equations correct
to 6'(c3). However, the variational technique always requires a formulation correct to one
order higher than that of the equation sought, so that all expressions under the integrand
in statement (S.10) have to be at least of 0(c4). Finally, by considering the inextensibility
condition, one can easily see that the longitudinal displacement u is

u - S(€2>, (5.12)

i.e. one order higher than w.
The total kinetic energy of the system is the sum of the kinetic energy of the pipe, 9,
and the kinetic energy of the fluid, q, defined by

(5.13)

V, and V, being the corresponding velocities.
The potential energy comprises gravitational and strain energy components. In general,
the gravitational energy depends on the distribution of mass (Fung 1969), and is written
as % = J p#(()dQ, where r$ is the gravitational potential per unit mass; in a uniform
gravitational field it becomes Yi = J pgcdQ, where g is the gravitational acceleration, 6
is a distance measured from a reference plane in a direction opposite to the gravitational
field, and d"lr is an elemental volume. Consequently, with the notation used here,

(5.14)

It is very important to define an exact form of the strain energy in the case of large
deflections, correct to 0(c4). This problem is solved by Stoker (1968), with only one
major (but not drastic) assumption: the strain is small even though the deflection can be
large. His analysis finally leads to

L
.(r = + EZ(1 + E)~K~] dr0, (5.15)

where no represents the Lagrangian coordinate, A the cross-sectional area, I the moment
of inertia and E the axial strain.