Computational and theoretical developments in rod mechanics
The problem: static equilibria of elastic rods
An elastic rod can be described by
(r(s),d1(s),d2(s),d3(s)),
where:
s is arclength along the rod,
the centerline r(s)
in R3 runs through the centers of mass of the rod cross-sections,
the frames
(d1(s),d2(s),d3(s)) in SO(3)
are the principal axes of inertia of the rod cross-sections.
If the rod is inextensible, then d3(s) = r'(s), so
the rod is completely specified by r(s) (shown below as a green
tube) and a normal vector field d1(s) tracking the
rod's twist (depicted below as a blue ribbon).
The energy of a rod configuration is a functional of the form:
I have been studying the possible equilibrium configurations
of an inextensible rod subject to various boundary constraints; for example,
the twisted ring, in which the centerline closes smoothly and the ribbon
forms an angle 
at the point of closure (see below).
Mathematically, this is a calculus of variations problem: we seek critical
points of the functional E subject to prescribed boundary conditions
on (r,di). These critical points are found via the classic
Euler-Lagrange equations for E, which yield a two-point boundary value
problem for a system of ODEs. Since we want to solve this 2-point
BVP as various parameters are varied
(either or various
parameters appearing in W), our numerical computations use the
parameter-continuation package AUTO .
The energy functional
The function W incorporates all material-dependent properties (e.g.,
intrinsic curvature, bending stiffnesses, twisting stiffness).
For example, a common energy model used today is:
where the strains ui are defined by:
and the expression 
denotes the strains defined from the rod's intrinsic shape.
Thus, 
and 
describe intrinsic curvatures while
describes intrinsic
twist. The coefficients Ki are called the stiffnesses
of the rod (bend stiffnesses for i=1,2 and twist stiffness for i=3).
The computations described below can be easily extended to
more elaborate energy models, such as ones with
higher-than-quadratic terms or twist-bend coupling.
For the simplest case in which Ki are independent of s,
K1=K2, and the
are 0,
which we call the perfect rod
(physically: uniform, circular cross-section, and intrinsically straight),
the set of solutions to the twisted ring problem
is highly symmetric. As varies
the solutions sweep out the bifurcation diagram shown below. The energy
of each solution is plotted against the torque m3 at s=0 (both
scaled to be unitless). Coloring indicates stability of the solutions,
with green denoting local minima of E, red denoting saddles with one
negative direction, etc. (see section on stability below for further information).
Degenerate orbits of solutions to the perfect problem (cf.
Manning & Maddocks, 1998 )
Due to symmetries of the perfect rod, each point
on the above bifurcation diagram actually represents an entire manifold
(or orbit)
of twisted ring solutions (all with the same energy and torque). For example,
see below. Here, we take a given rod equilibrium, and in Step 1 do a rigid-body
rotation by an angle 
(about 45 degrees in this case),
and in Step 2 rotate the rod at every point about its tangent vector
by 
. In the end,
we will still be at an equilibrium of the energy (because of the isotropy
of the rod) and the boundary conditions on r and di
are the same as in the original.
On the parabolic branches in the perfect diagram, the orbit of
degenerate solutions is homeomorphic to a circle (generated by the
operation shown above for 0 
).
On the other branches, the orbit is homeomorphic to a torus (generated by the
operation shown above for 0
, plus
another operation relating to translation of the rod along its arclength).
Splitting of bifurcation diagram for imperfect problems (cf.
Manning & Maddocks, 1998 )
If we remove the symmetries of the perfect rod, say by introducing
intrinsic curvature into the rod, then the orbits of degenerate
solutions for the perfect rod yield in general a finite set of solutions,
each with different energy, to the imperfect problem. In fact, through
a perturbation computation, we have determined exactly which points
on the orbit yield solutions to the imperfect problem. Most commonly,
the parabolic branches in the perfect diagram (whose orbits are circles)
yield 2 branches in the imperfect diagram, while the remaining
branches (whose orbits are tori) yield 4 branches. See for example,
the diagram below.
We can make even stronger statements using the perturbation expansion.
For example, for infinitesimal intrinsic curvatures
, a solution on
a parabolic branch in the perfect diagram always yields two
solutions in the imperfect diagram, unless the following
two conditions hold:
For non-infinitesimal perturbations, and especially those nearly satisfying
the above conditions, the nice "generic" splitting shown above
need not occur. Thus, the perturbation expansion suggests
which intrinsic shapes will lead to more unusual imperfect diagrams.
See for example, the diagram below. To better illustrate the
unusual splitting, just the bottom portion of the diagram is shown,
and a combination of energy and bending moment m1 is
plotted on the y axis.
In the category of half-science, half-entertainment, here are a few movies
demonstrating the transition from perfect to imperfect diagrams. Be forewarned
that they're somewhat large (400 KB to 1 MB).
Movie 1: a superposition
of the perfect diagram (in red) and an imperfect diagram (in purple and yellow)
as the imperfection (which in this case intrinsic curvature of a
157-base-pair DNA) is gradually turned on. As above, energy
is plotted against m3.
Movie 2: same as Movie 1 (but without
the perfect diagram superimposed) for a larger imperfection
representing protein-bound DNA. The dots mark solutions for which
d1(0)=d1(1).
Movie 3: same
as Movie 2, except that m3 is plotted versus
.
Return to my home page
rmanning@haverford.edu