The geometry and material of the disc are identical to those described
earlier except that the inner surface of the disc is not totally constrained,
as it is for the static problem. Instead, all the nodes on the inner surface
are attached to a node (axle node) located at the center of the disc using
kinematic coupling constraints. This facilitates the application of angular
velocity or displacement to the axle node to simulate rolling in the Lagrangian
approach, as well as the measurement of reaction forces and moments at the
axle. The mesh is more refined for the rolling problem compared to the static
problem. In particular, two elements are used through the thickness of the
disc. The first step is a do-nothing step to facilitate import of the initial
state from
Abaqus/Standard
to
Abaqus/Explicit.
This step is followed by a static step in which the rigid surface is pushed
against the disc a distance of 0.15 inches. The next step involves rolling of
the loaded disc against the rigid surface, which is accomplished in two ways.
The first is a Lagrangian analysis in which an angular velocity of 2.5 radians
per second is applied to the axle node of the disc. In this example the
structure reaches a steady state after one full revolution. No additional
damage occurs in subsequent revolution cycles. Therefore, the total time is
chosen such that the disc undergoes two revolutions. A variation of the
Lagrangian analysis includes modeling of permanent set using a metal plasticity
material definition. In the second analysis the rolling is simulated using the
steady-state transport capability in
Abaqus/Standard.
Frictional and inertial effects are neglected in both cases.
The steady-state transport capability directly obtains the steady-state
rolling solution of the disc on the rigid surface. Due to Mullins effect the
stress state for the rolling solution can be quite different from the stress
state for the static non-rolling solution. As a result, an attempt to obtain a
steady-state rolling solution directly from a static non-rolling solution may
lead to convergence problems in the Newton's scheme that is used to solve the
overall nonlinear system of equations. Since the damage and, hence, the
discontinuity in state are independent of the angular rolling speed, a time
increment cutback during the steady-state transport step does not overcome the
convergence difficulties. Such convergence difficulties can be resolved by
introducing the damage gradually over an additional steady-state transport step
preceding the actual analysis. In this example this is accomplished by
following the static loading step with a steady-state transport step with a
small rolling angular speed of 0.25 radians per second and with the Mullins
effect ramped up over the time period of the step. This step is followed by
another steady-state transport step at an angular speed of 2.5 radians per
second with the MULLINS parameter set equal to STEP, which provides the solution we are interested in obtaining.
A Lagrangian simulation is also carried out in
Abaqus/Explicit.
The kinetic energy is monitored to ensure that the problem remains essentially
quasi-static. The revolutions of the disc are carried out by applying a
rotational displacement (corresponding to two full revolutions) at the axle
node using an amplitude with a smooth step definition to reduce the noise in
the response. The compressibility parameter, ,
is chosen to be 5 Ć 10
ā5, an order of magnitude higher than the actual value, to obtain
relatively higher time increments and, thus, a relatively lower run time.
Figure 5
compares the time history of the reaction force at the axle node for the
Lagrangian and the steady-state rolling analyses in
Abaqus/Standard.
The results for the Lagrangian problem without permanent set (curve labeled
lagr) indicate that the reaction force increases during
the loading step and decreases during the first revolution of the disc. The
decrease in the reaction force is a result of lower overall stresses due to
damage in the material. During the second revolution of the disc the reaction
force remains constant as no additional damage occurs. The results of the
Lagrangian problem including permanent set (curve labeled
fefp) show a softer behavior compared to the former
problem and clearly indicate the presence of permanent set in the footprint
area as the solid disc rolls against the rigid surface. The steady-state
rolling results (labeled sst) show a gradual transition of
the reaction force during the first steady-state transport step that ramps up
the Mullins effect. During the second steady-state transport step the reaction
force remains constant at the value reached at the end of the prior step. This
curve also illustrates that the damage associated with the Mullins effect is
independent of the angular speed of rotation. The reaction force remains the
same at angular speeds of both 0.25 and 2.5 radians per second. If the Mullins
effect is not applied gradually over the first steady-state transport step, the
discontinuity between the rolling and the static states may lead to convergence
difficulties.
The same results can be observed from a different viewpoint in
Figure 6,
which shows the reaction force as a function of the number of revolutions for
both
Abaqus/Standard
and
Abaqus/Explicit.
The reaction force decreases during the first revolution and remains steady
(except for the noise in the
Abaqus/Explicit
analysis) during the second revolution.
Figure 7
shows a comparison of the time histories of the reaction moment at the axle
node between the Lagrangian and the steady-state rolling solutions. The
Lagrangian results are labeled m-lagr, while the
steady-state rolling results are labeled m-sst. If the
material were purely hyperelastic (without damage), the contact forces would be
symmetric about a plane normal to the rigid surface and containing the axle;
hence, no torque would be required to rotate the disc. However, as a result of
the damage associated with Mullins effect, the contact forces are not
symmetrical because material particles transition through the contact area
during the very first revolution. This leads to the reaction moment during the
first revolution, as shown in the results for the Lagrangian analysis. The
moment reduces to zero during the second revolution. The steady-state rolling
results do not include the transient solution of the first revolution; hence,
they show a zero moment at all times.
Figure 8
shows the same results from a different viewpoint. In this figure the reaction
moment is plotted as a function of the number of revolutions for both
Abaqus/Standard
and
Abaqus/Explicit.
Figure 9
shows a contour plot of the damage energy dissipated at material points at an
instant of time that corresponds to about three-quarters of the way into the
first revolution of the disc. The figure indicates damage in the material that
has already passed through the contact area and no damage in the material that
is yet to pass through the contact area. This corresponds to damage in about
three-quarters of the disc material. The remaining quarter is still undamaged,
as it has not undergone any deformation yet. The full disc will be damaged at
the end of the first revolution, and the damage state remains unchanged during
the second revolution.