4.6 Damped Continuous Systems

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All practical mechanical systems have some form of energy dissipation (damping). When the level of

dissipation is small, damping is neglected, as we have done thus far in this chapter. Yet, some effects of

damping (e.g., the fact that at steady state the natural [modal] vibration components decay to zero

leaving only the steady forcing component) are tacitly assumed even in undamped analysis.

The natural behavior of a system is expected to change due to the presence of damping. In particular,

the system’s natural frequencies will decrease (and be called damped natural frequencies) as a result of

damping. Furthermore, it is quite possible that a damped system would not possess “real” modes in

which it could independently vibrate. Mathematically, in that case, the modes will become complex (as

opposed to real) and, physically, all the points of the system will not move, maintaining a constant phase

at a given damped natural frequency. In other words, a real solution that is separable in space ðxÞ and time

ðtÞ may not be possible for the free vibration problem of a damped system. Also, node points of an

undamped system may vary with time as a result of damping. With light damping, of course, such effects

of damping will be negligible.

Since there are damped systems that do not possess real natural modes of vibration, care should be

exercised when extending the results of modal analysis from an undamped system to a damped one.

However, in some cases, the mode shapes will remain the same after including damping (even though the

natural frequencies will change). This is analogous to the case of proportional damping, which was

discussed in the section on lumped-parameter (multi-degree-of-freedom) vibrating systems. The modal

analysis of a damped system will become significantly easier if we assume that the mode shapes will remain

the same as those for the undamped system. Even when the actual type of damping in the system results in

complex modes, for analytical convenience, an equivalent damping model that gives real modes is used in

simplified analysis. This is analogous to the use of proportional damping in lumped-parameter systems.

4.6.1 Modal Analysis of Damped Beams

Consider the problem of free damped transverse vibration of a thin beam, given by

›2

›x2 EI

›2v

›x2 þ L

›v

›t

􀀏 􀀐

þ rA

›2v

›t2 ¼ 0 ð4:216Þ

where L is a spatial differential operator (in x). Consider the following two possible models of damping:

1: L ¼

›2

dx2 EpI

›2

›x2 ð4:217Þ

2: L ¼ c ð4:218Þ

Model 1 corresponds to the Kelvin – Voigt model of material (internal) damping given by the stress –

strain relation

s ¼ E1 þ E p

›1

›t ð4:219Þ

where Ep is the damping parameter of the beam material. Hence, we obtain the damped beam equation

simply by replacing E in the undamped beam equation by E þ Ep ð›=›tÞ: Also, Ep is independent of the

frequency of vibration for the viscoelastic damping model, but will be frequency dependent for the

hysteretic damping model. Modal analysis is done regardless of any frequency dependence of Ep and, in

the final modal result for a particular modal frequency vi; the appropriate frequency function for EpðvÞ

is used with v ¼ vi if the damping is of the hysteretic type. It can be easily verified that the mode shapes

of the damped system with model 4.217 are identical to those of the undamped system, regardless of

whether the beam cross section is uniform or not.

In Model 2 (Equation 4.218), the operator is a constant c: This corresponds to external damping of the

linear viscous type, distributed along the beam length. For example, imagine a beam resting on a

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foundation of viscous damping material. For Model 2, it can be shown that the damped mode shapes are

identical to the undamped ones, assuming that the beam cross section is uniform. If the beam is

nonuniform, the damped and the undamped mode shapes are identical if we assume that the damping

constant c varies along the beam in proportion to the area of cross section AðxÞ of the beam. We shall

show this in the example given below.

Example 4.8

Perform the modal analysis for transverse vibration of a thin nonuniform beam with linear viscous

damping distributed along its length and satisfying the beam equation

›2

›x2 EIðxÞ

›2v

›x2 þ rAðxÞb

›v

›t þ rAðxÞ

›2v

›t2 ¼ 0 ð4:220Þ

Determine damped natural frequencies, modal damping ratios, and the response vðx; tÞ as a modal series

expansion, given vðx; 0Þ ¼ dðxÞ and v_ðx; 0Þ ¼ sðxÞ:

Solution

Substitute the separable solution

vðx; tÞ ¼ Y ðxÞqðtÞ ð4:221Þ

into Equation 4.220. We obtain

d2

dx2 EI

d2Y ðxÞ

dx2 qðtÞ þrAðxÞbYðxÞq_ðtÞ þrAðxÞYðxÞq€ðtÞ ¼ 0

Group the functions of x and t separately and equate to the same constant v2; as usual:

d2

dx2 EI

d2Y ðxÞ

dx2

rAðxÞY ðxÞ ¼ 2

q€ðtÞ þ bq_ðtÞ

qðtÞ ¼ v2 ð4:222Þ

We have

d2

dx2 EI

d2Y ðxÞ

dx2 2 v2rAY ðxÞ ¼ 0 ð4:223Þ

and

q€ðtÞ þ bq_ðtÞ þv2qðtÞ ¼ 0 ð4:224Þ

Note that Equation 4.223 is identical to that for the undamped beam. Hence, with known BCs, we will

obtain the same mode shapes YiðxÞ and the same undamped natural frequencies vi in the usual manner.

However, the equation of modal generalized coordinates qðtÞ given by Equation 4.224 is different from

that for the undamped case ðb ¼ 0Þ: We write, for mode i

q€iðtÞ þ 2ziviq_iðtÞ þv2i

qiðtÞ ¼ 0 ð4:225Þ

where

zi ¼

b

2vi ð4:226aÞ

is the modal damping ratio for mode i: Damped natural frequencies are

vdi ¼

ffiffiffiffiffiffiffiffi

1 2 z2i

q

vi ð4:226bÞ

Equation 4.225 can be solved in the usual manner, with initial conditions qið0Þ and q_ið0Þ determined

a priori, using known vðx; 0Þ and v_ðx; 0Þ:

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The modal series solution is

vðx; tÞ ¼

X

YiðxÞqiðtÞ ð4:227Þ

The initial conditions are

X

YiðxÞqið0Þ ¼ dðxÞ ð4:228Þ

X

YiðxÞq_ið0Þ ¼ sðxÞ ð4:229Þ

Multiply Equation 4.228 and Equation 4.229 by rAðxÞYjðxÞ and integrate from x ¼ 0 to l using the

orthogonality condition

ðl

0

rAðxÞYiðxÞYjðxÞdx ¼

0 for i – j

aj for i ¼ j

(

ð4:230Þ

We obtain

qjð0Þ ¼

1

aj

ðl

0

dðxÞrAðxÞdx ð4:231Þ

q_jð0Þ ¼

1

aj

ðl

0

sðxÞrAðxÞdx ð4:232Þ

This completes the solution for the free damped beam. The forced damped case can be analyzed in the

same manner as for the forced undamped case because the mode shapes are the same.

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