# damper & equivalent-converted

```Mechanical Systems
173
and the spring is twisted in the counterclockwise direction when viewed from the right- hand side.
The potential energy stored in a torsional spring element is expressed as
Ve 
1
K2
(5.15)
2
5.1.3
Damper Elements
A spring element exerts a reaction force that is dependent on the relative displacement between
two ends of the spring. In contrast, a force that depends on the relative velocity between two
bodies is modeled by a damper element. Figure 5.6 shows a mass sliding on a fixed horizontal
surface, where the two surfaces are separated by a film of liquid. The mass is subjected to a friction
force generated between the two surfaces, and the friction caused by the liquid is called viscous
damping. The direction of the damping force is opposite to the direction of the motion, and its
magnitude depends on the nature of fluid flow between the two surfaces. The exact viscous
damping force is complex; thus, for modeling in system dynamics, we use a linear relationship
f  bv
(5.16)
where the symbol b is used to denote the viscous damping coefficient in units of N&middot;s/m or lb&middot;s/ft.
The damping force exerted on the mass in Figure 5.6 is to the left. Note that the symbol c is also
often used to denote the viscous damping coefficient. Therefore, both b and c will be used
interchangeably in this book.
The viscous friction can be modeled using a viscous damper (or a dashpot). The symbol in
Figure 5.7a is the representation of a viscous damper, which is like a piston moving through a
liquid-filled cylinder, as shown in Figure 5.7b. There are small holes in the pis- ton through which
the liquid flows as the parts move relative to each other. If v2 &gt; v1 &gt; 0, the right end of the damper
moves to the right with respect to the left end. The force
v
m
Damping
force
Liquid film
Fixed surface
FIGURE 5.6
A mass sliding on a lubricated fixed surface.
f
f
v1
b
Flow
v2
(a)
FIGURE 5.7
A viscous damper: (a) symbol and (b) physical system.
(b) Cylinder motion v1
Piston motion v2
174
Modeling and Analysis of Dynamic Systems
B
B
ı +
(a)
(b)
ı +1
ı +2
FIGURE 5.8
A rotational viscous damper with (a) one fixed end and (b) two free ends.
applied to the right end is dependent on the relative velocity vrel = v2 − v1. The force has a magnitude
of
f  bvrel  b(v2  v1)
(5.17)
and points to the right. Assume that the damper is massless or of negligible mass. Then, the forces
at the two ends of the damper are equal in magnitude but opposite in direction.
For a torsional damper, as shown in Figure 5.8a, the linear relationship between the externally
applied torque and the angular velocity is given by
  B
(5.18)
where B is the rotational viscous damping coefficient in units of N&middot;m&middot;s/rad or ft&middot;lb&middot;s/rad. The
symbol in Figure 5.8b represents a rotational viscous damper, which can be used to model the
viscous friction between two rotating surfaces separated by a film of liquid. If ω2 &gt; ω1 &gt; 0, the
magnitude of the applied torque is
  Brel  B(2  1)
(5.19)
and the direction is as shown.
Note that the damping dissipates the energy of the system. Besides viscous damping, there are
two other types of damping in engineering mechanics: Coulomb damping asso- ciated with dry
friction and structural damping. The former will be discussed in Chapter 9, and the latter is beyond
the scope of this text.
5.1.4
Equivalence
In many mechanical systems, multiple springs or dampers are used. In such cases, an equivalent
spring stiffness constant or damping coefficient can be obtained to represent the combined
elements.
Example 5.1: Springs in Parallel
Consider a system of two springs, k1 and k2, in parallel, as shown in Figure 5.9. Prove that the
system is equivalent to a single spring whose stiffness is
keq  k1  k2
Proof
Because of parallel interconnection, the bottom ends of the springs are attached to the same
fixed body, and their top ends are also attached to a common body. This implies that both
springs have the same deflection x. Assume that the forces applied
Mechanical Systems
175
f
f
f1
x
k1
k2
x
x
k1
keq
=
f2
x
k2
FIGURE 5.9
Equivalence for two springs in parallel.
to the two springs are f1 and f2, respectively. Because the system is in static equilib- rium, the
total force is given by
f  f1  f2  k 1x  k2x  (k1  k2 )x
Comparing it with the equivalent system,
f  k eq x
we obtain the equivalent spring stiffness
k eq  k1  k2
The result can be extended to n springs. For a system of n springs in parallel, the equivalent
spring stiffness keq is equal to the sum of all the individual spring stiffness coefficients ki:
k eq  k1  k2 
 kn
Example 5.2: Springs in Series
Consider a system of two springs, k1 and k2, in series, as shown in Figure 5.10. Prove that the
equivalent spring stiffness of the system is
keq 
k1k2
k1  k2
Proof
Because both springs are in static equilibrium, they are subjected to the same force f. Assume
that the two springs are deformed by x1 and x2, respectively. The total deforma- tion of the
system is given by
x  x1  x2 
f

k1
f
k2
For the equivalent system, the deformation is
x
f
k eq
f
 1
k
 1

1


k2 
176
Modeling and Analysis of Dynamic Systems
f
x1
f
k1
x1
k1
f
=
k2
x2
f
x
x2
keq
k2
FIGURE 5.10
Equivalence for two springs in series.
Thus,
1 1 1
 
k eq k1 k2
or
k1k2
k eq 
k1  k2
The result can also be extended to n springs. For a system of n springs in series, the reciprocal of
the equivalent spring stiffness keq is equal to the sum of all the reciprocals of the individual
spring stiffness coefficients ki:
1
1
1

 
k eq k1 k2

1
kn
The above-mentioned two examples show how one can derive the equivalent spring stiffness for
springs connected in parallel or in series. For a system of dampers, the equivalent damping
coefficient can be derived using the same logic and similar steps.
Springs are the most familiar elastic elements. However, many engineering applications
involving elastic elements do not contain springs but other mechanical elements, such as beams
and rods, which can be modeled as springs. The equivalent spring constants can be determined
using the results from the study of mechanics of materials [2, 12].
Example 5.3: Equivalent Spring Constant of a Cantilever Beam
Consider a uniform cantilever beam of length L, width b, and thickness h in Figure 5.11.
Assume that a force f is applied to the free end of the beam, and the corresponding deflection is
x. Derive the equivalent spring constant keq.
Mechanical Systems
177
f
L
x
b
=
keq
h
x
f
FIGURE 5.11
A beam in bending under a transverse force.
Solution
The force–deflection relation of a cantilever beam can be found in mechanics of materi- als
references. The relation is
L3
x
f
3EIA
where:
x is the deflection at the free end of the beam
f is the force applied at the free end of the beam
E is the modulus of elasticity of beam material
IA is the area moment of inertia about the beam’s longitudinal axis
For a beam having a rectangular cross-section, with width b and thickness h, the area moment
of inertia is
IA 
bh3
12
Thus, the force–deflection relation reduces to
x
4L3
f
Eb h3
For the equivalent system, the force–deflection relation is
x
f
keq
Thus, the equivalent spring stiffness is
keq 
Eb h3
4 L3
```