**
8.2 DYNAMIC MODELS
**

It
is often convenient in dynamic analysis to create a simplified model of a
complicated part. These models are sometimes considered to be a collection of
point masses connected by massless rods, referred to as a lumped-parameter
model, or just lumped model. For a lumped model of a rigid body to be
dynamically equivalent to the original body, three things must be true:

1 The mass of the model must equal that of the
original body.

2 The center of gravity must be in the same location
as that of the original body.

3 The mass moment of inertia must equal that of the
original body.

**
**

**
8.3 MASS
**

Mass
is not weight. Mass is an invariant property of a rigid body. The weight of the
same body varies depending on the gravitational system in which it sits. We
will assume the mass of our parts to be constant over time in our calculations.

When
designing cam-follower systems (or any machinery), we must first do a complete
kinematic analysis of our design in order to obtain information about the rigid
body accelerations of the moving parts. We can then use Newton’s second law to
calculate the dynamic forces. But to do so, we need to know the masses of all
the moving parts that have these known accelerations. If we have a design of
the follower train done in a CAD program that will calculate masses and mass
moments of inertia, then we are in good shape, as the data needed for dynamic
calculations are available. Lacking that luxury, we will have to calculate
estimates of the mass properties of the follower train to do the dynamics
calculations.

Absent
a solids-modeler representation of your design, a first estimate of your parts’
masses can be obtained by assuming some reasonable shapes and sizes for all the
parts and choosing appropriate materials. Then calculate the volume of each
part and multiply its volume by the material’s
**
mass density
**
(not weight density) to obtain a first approximation of its mass.
These mass values can then be used in Newton’s equation to

estimate
the dynamic forces.

[Portions of this chapter were adapted from R. L.
Norton,
*
Design
of Machinery
*
2ed,
McGraw-Hill, 2001, with permission.]

How
will we know whether our chosen sizes and shapes of links are even acceptable,
let alone optimal? Unfortunately, we will not know until we have carried the
computations all the way through a complete stress and deflection analysis of
the parts. It is often the case, especially with long, thin elements such as
shafts or slender links, that the deflections of the parts under their dynamic
loads will limit the design even at low stress levels. In other cases the
stresses at design loads will be excessive.

If
we discover that the parts fail or deflect excessively under the dynamic
forces, then we will have to go back to our original assumptions about the shapes,
sizes, and materials of these parts, redesign them, and repeat the force,
stress, and deflection analyses.

Design
is, unavoidably, an
**
iterative
process
**
. We need the dynamic
forces to do the stress analyses on our parts. (Stress analysis is addressed in
a later chapter.) It is also worth noting that, unlike a static force situation
in which a failed design might be fixed by adding more mass to the part to
strengthen it, to do so in a dynamic-force situation can have a deleterious
effect. More mass with the same acceleration will generate even higher forces
and thus higher stresses. The machine designer often needs to remove mass (in
the right places) from parts in order to reduce the stresses and deflections
due to
**
F
**
=
*
m
*
**
a.
**
The
designer needs to have a good understanding of both material properties and
stress and deflection analysis to properly shape and size parts for minimum
mass while maximizing strength and stiffness to withstand dynamic forces.

**
8.4 MASS MOMENT AND CENTER OF
GRAVITY
**

When
the mass of an object is distributed over some dimensions, it will possess a
moment with respect to any axis of choice. Figure 8-1 shows a mass of general
shape in an
*
xyz
*
axis system. A differential element of mass is
also shown. The
**
mass moment
(first moment of mass)
**
of the
differential element is equal to the
**
product
of its mass and its distance
**
along
the axis of interest. With respect to the
*
x
*
,
*
y,
*
and
*
z
*
axes these are:

*
dM
*
*
x
*
=
*
r
*
*
x
*
1
*
dm
*
(8.2a)

*
dM
*
*
y
*
=
*
r
*
*
y
*
1
*
dm
*
(8.2b)

*
dM
*
*
z
*
=
*
r
*
*
z
*
1
*
dm
*
(8.2c)

The
radius from the axis of interest to the differential element is shown with an
exponent of 1 to emphasize the reason for this property being called the first
moment of mass. To obtain the mass moment of the entire body we integrate each
of these expressions.

*
M
*
*
x
*
=
∫
*
r
*
*
x
*
*
dm
*
(8.3a)

*
M
*
*
y
*
=
∫
*
r
*
*
y
*
*
dm
*
(8.3b)

*
M
*
*
z
*
=
∫
*
r
*
*
z
*
*
dm
*
(8.3c)

If
the mass moment with respect to a particular axis is numerically zero, then
that axis passes through the
**
center
of mass (
**
**
***
CM
*
**
)
**
of
the object, which for earthbound systems is coincident with its
**
center of gravity (
**
**
***
CG
*
**
)
**
.
By definition, the summation of first moments about all axes through the center
of gravity is zero. We will need to locate the

*
CG
*
of all moving
bodies in our designs because the linear acceleration component of each body is
calculated as acting at that point.

It
is often convenient to model a complicated shape as several interconnected
simple shapes whose individual geometries allow easy computation of their
masses and the locations of their local
*
CGs
*
. The global
*
CG
*
can
then be found from the summation of the first moments of these simple shapes
set equal to zero. Appendix C contains formulas for the volumes and locations
of centers of gravity of some common shapes. Of course, if the system is
designed in a solids modeling CAD package, then the mass and other properties
can be automatically calculated.

Figure
8-2 shows a simple model of a mallet broken into two cylindrical parts, the
handle and the head, which have masses
*
m
*
*
h
*
and
*
m
*
*
d
*
, respectively. The individual centers of
gravity of the two parts are at
*
l
*
*
d
*
and
*
l
*
*
h
*
/2, respectively, with respect to the axis
*
ZZ
*
.
We want to find the location of the composite center of gravity of the mallet
with respect to
*
ZZ
*
. Summing the first moments of the individual
components about
*
ZZ
*
and setting them equal to the moment of the entire
mass about
*
ZZ
*
gives.

This
equation can be solved for the distance
*
d
*
along the
*
x
*
axis,
which, in this symmetrical example, is the only dimension of the composite
*
CG
*
not
discernible by inspection. The
*
y
*
and
*
z
*
components
of the composite
*
CG
*
are both zero.

*
*

Dynamic models, composite center of gravity, and
radius of gyration of a mallet