1.3 Heavy Springs

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A heavy spring has its mass and flexibility properties continuously distributed throughout its body.

In that sense it has an infinite number of DoF(s), and a single coordinate cannot represent its motion.

However, for many practical purposes, a lumped-parameter approximation with just one lumped mass to

represent the inertial characteristics of the spring may be sufficient. Such an approximation may be

obtained by using the energy approach. Here, we represent the spring by a lumped-parameter “model”

such that the original spring and the model have the same net kinetic energy and potential energy. This

energy equivalence is used in deriving a lumped mass parameter for the model. Even though damping

(energy dissipation) is neglected in the present analysis, it is not difficult to incorporate that as well in

the model.

Box 1.1

APPROACHES FOR DEVELOPING EQUATIONS OF MOTION

1. Conservative Systems (No Nonconservative Forces/No Energy Dissipation):

Kinetic energy ¼ T

Potential energy ¼ V

Conservation of energy: T þ V ¼ const:

Differentiate with respect to time t

2. Lagrange’s Equations:

Lagrangian L ¼ T 2 V

d

dt

›L

›q_i

2

›L

›qi ¼ Qi for i ¼ 1; 2; …; n

n ¼ number of DoFs

Qi ¼ generalized force corresponding to generalized coordinate qi:

Find Qi using:

dW ¼

X

Qidqi

where dW ¼ work done by nonconservative forces in a

general incremental motion ðdq1; dq2; …; dqnÞ:

3. Newtonian Approach:

X

Forces ¼

d

dt

X

Linear Momentum

X

Torques ¼

d

dt

X

Angular Momentum

(About centroid or a fixed point)

1-12 Vibration and Shock Handbook

© 2005 by Taylor & Francis Group, LLC

1.3.1 Kinetic Energy Equivalence

Consider the uniform, heavy spring shown in

Figure 1.9, with one end fixed and the other end

moving at velocity v:

Note that:

k ¼ stiffness of spring

ms ¼ mass of spring

l ¼ length of spring

Local speed of element dx of the spring is given by ðx=l Þv: Element mass ¼ ðms=l Þdx: Hence,

Element KE ¼

1

2

ms

l

dx

x

l

v

􀀏 􀀐2

In the limit, we have dx ! dx: Then,

Total KE ¼

ðl

0

1

2

ms

l

dx

x

l

v

􀀏 􀀐2

¼

1

2

msv2

l3

ðl

0

x2 dx ¼

1

2

msv2

3 ð1:45Þ

Hence, the equivalent lumped mass concentrated at the free end ¼ ð1=3Þ £ spring mass:

Note: This derivation assumes that one end of the spring is fixed. Furthermore, the conditions are

assumed to be uniform along the spring.

An example of utilizing this result is shown in Figure 1.10. Here, a system with a heavy spring and a

lumped mass is approximated by a light spring (having the same stiffness) and a lumped mass.

Another example is shown in Figure 1.11. In this case, it is not immediately clear which of the

approximations shown on the right-hand side is most appropriate.

Example 1.2

A uniform heavy spring of mass ms and stiffness k is attached at one end to a mass m that is free to roll on

a frictionless horizontal plane. The other end is anchored to a vertical post. A schematic diagram of this

arrangement is shown in Figure 1.12.

l

x δx

k, ms

v

FIGURE 1.9 A uniform heavy spring.

m

k, ms

= +

3

m m s

k

FIGURE 1.10 Lumped-parameter approximation for an oscillator with heavy spring.

m2

k, ms

=

k

m1 3

ms m 1 +

k

6

ms m 1+

6

ms m2 +

m2

k

3

ms m1 m2 +

etc.

FIGURE 1.11 An example where lumped-parameter approximation for spring is ambiguous.

Time-Domain Analysis 1-13

© 2005 by Taylor & Francis Group, LLC

The unstretched length of the spring is l:

Assume that, when the velocity of the connected

mass is v; the velocity distribution along the spring

is given by

vsðxÞ ¼ v sin

px

2l

where x is the distance of a point along the

spring, as measured from the fixed end. We will

determine an equivalent lumped mass located

at the moving end of the spring (i.e., at the

moving mass m) to represent the inertia effects of

the spring.

Consider an element of length dx at location x

of the spring. Since the spring is uniform, we have

element mass ¼ ðms=l Þdx: Also, according to the

given assumption, element velocity ¼

vsinðpx=2l Þ: Hence, kinetic energy of the spring is

ðl

0

1

2

v sin

px

2l

􀀏 􀀐2 ms

l

dx ¼

1

2

ms

l

v2

ðl

0

sin2 px

2l

dx ¼

1

4

ms

l

v2

ðl

0

1 2 cos

px

l

􀀒 􀀓

dx

¼

1

4

ms

l

v2 x 2

l

p

sin

px

l

􀀒 􀀓l

0¼

1

4

ms

l

v2l ¼

1

2

ms

2

v2

It follows that the equivalent lumped mass to be located at the moving end of the spring is ms=2: This

result is valid only for the assumed velocity distribution, and corresponds to the first mode of motion

only. In fact, a linear velocity distribution would be more realistic in this low frequency (quasi-static

motion) region, which will give an equivalent lumped mass of ð1=3Þms; as we have seen before. Such

approximations will not be valid for high frequencies

􀀍

say, higher than

ffiffiffiffiffiffi

k=ms

p 􀀎

.

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