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However, Equation 10.122 gives the displacement response of generalized displacements rather than actual nodaldisplacements, owing to the transformation described by Equation 10.112. As themodal matrix is known, conversion of the generalized displacements to actualdisplacements requires only multiplication by the normalized modal matrix.Hutton: Fundamentals ofFinite Element Analysis10. Structural DynamicsText10.8 Harmonic Response Using Mode Superposition© The McGraw−HillCompanies, 2004423EXAMPLE 10.8Again consider the 3 degrees-of-freedom system of Example 10.3 and determine thesteady state response when a downward force F = F0 sin ␻ f t is applied to mass 2.■ SolutionFor the given conditions, the applied nodal force vector is0{F(t)} = F0 sin ␻ f t0and the generalized forces are 0.2956 0.42090.6064 0  0.4209  F sin ␻ t10f[A]T {F} = √  0.6575 0.5618 −0.3550  F0 sin ␻ f t =√0.5618 mm0−0.71240.6930 −0.7124 0.0782The equations of motion for the generalized coordinates are thenp̈1 + ␻12 p1 =0.4209F0 sin ␻ f t√mp̈2 + ␻22 p2 =0.5618F0 sin ␻ f t√mp̈3 + ␻32 p3 =−0.7124F0 sin ␻ f t√mfor which the solutions are0.4209F0 sin ␻ f t√p1 (t) = 2␻1 − ␻2f m0.5618F0 sin ␻ f t√p2 (t) = 2␻2 − ␻2f mp3 (t) =−0.7124F0 sin ␻ f t 2√␻3 − ␻2f mThe actual displacements, x (t ) = q (t ) in this case, are obtained by application of Equation 10.112:0.4209 22 ␻−␻f 10.2956 0.65750.6930 0.5618  F0 sin ␻ f t1 {x} = [A]{ p} = √√0.4209 0.5618 −0.7124  ␻22 − ␻2f mm0.6064 −0.3550 0.0782 −0.7124 ␻2 − ␻2 f3Hutton: Fundamentals ofFinite Element Analysis42410.

Structural DynamicsC H A P T E R 10Text© The McGraw−HillCompanies, 2004Structural DynamicsExpanding, the steady-state displacements are given by$0.1244x 1 (t ) =␻12−␻ 2f+0.3694␻22−␻ 2f+␻32−%0.1772␻12 − ␻ 2f+$x 3 (t ) =0.2552␻12−␻ 2f+0.3156␻22 − ␻ 2f−0.1994␻22−␻ 2f++0.5075␻32 − ␻ 2f−0.0557␻32−F0 sin ␻ f tm␻ 2f$x 2 (t ) =%−0.4937␻ 2fF0 sin ␻ f tm%F0 sin ␻ f tmA few observations need to be made regarding the displacements calculated in thisexample:1.2.3.4.The displacement of each mass is a sinusoidal oscillation about the equilibriumposition, and the circular frequency of the oscillation is the same as the frequencyof the forcing function.The characterstics of the principal modes of vibration are reflected in the solutions,owing to the effects of the natural circular frequencies and modal amplitude vectorsin determining the forced oscillation amplitudes.The displacement solutions represent only the forced motion of each mass; inaddition, free vibration may also exist in superposition with the forced response.Energy dissipation mechanisms are not incorporated into the model.The mode superposition method may seem quite complicated and, when attempting to obtain solutions by hand, the method is indeed tedious.

However, therequired computations are readily amenable to digital computer techniques andquite easily programmed. Additional ramifications of computer techniques forthe method will be discussed in a following article.10.9 ENERGY DISSIPATION:STRUCTURAL DAMPINGTo this point, the dynamic analysis techniques dealt only with structural systemsin which there is no mechanism for energy dissipation. As stated earlier, all realsystems exhibit such dissipation and, unlike the simple models presented, do notoscillate forever, as predicted by the ideal model solutions.

In structural systems,the phenomenon of energy dissipation is referred to as damping. Damping maytake on many physical forms, including devices specifically designed for the purpose (passive and active damping devices), sliding friction, and the internal dissipation characteristics of materials subjected to cyclic loading. In this section,we begin with an idealized model of damping for the simple harmonic oscillatorand extend the damping concept to full-scale structural models.Hutton: Fundamentals ofFinite Element Analysis10.

Structural DynamicsText© The McGraw−HillCompanies, 200410.9 Energy Dissipation: Structural Dampingkccx•kxm⫹x(a)(b)(c)Figure 10.13(a) A spring-mass system with damping. (b) The schematicrepresentation of a dashpot piston. (c) A free-body diagramof a mass with the damping force included.Figure 10.13a depicts a simple harmonic oscillator to which has been added adashpot.

A dashpot is a damping device that utilizes a piston moving through aviscous fluid to remove energy via shear stress in the fluid and associated heat generation. The piston typically has small holes to allow the fluid to pass through butis otherwise sealed on its periphery, as schematically depicted in Figure 10.13b.The force exerted by such a device is known to be directly proportional to thevelocity of the piston asf d = −c ẋ(10.123)where f d is the damping force, c is the damping coefficient of the device, and ẋis velocity of the mass assumed to be directly and rigidly connected to the pistonof the damper.

The dynamic free-body diagram of Figure 10.13c represents asituation at an arbitrary time with the system in motion. As in the undamped caseconsidered earlier, we assume that displacement is measured from the equilibrium position. Under the conditions stated, the equation of motion of the mass ism ẍ + c ẋ + k x = 0(10.124)Owing to the form of Equation 10.124, the solution is assumed in exponentialform asx (t ) = C e st(10.125)where C and s are constants to be determined. Substitution of the assumed solution yields(ms 2 + cs + k)C e st = 0(10.126)As we seek nontrivial solutions valid for all values of time, we conclude thatms 2 + cs + k = 0(10.127)must hold if we are to obtain a general solution.

Equation 10.127 is the characteristic equation (also the frequency equation) for the damped single degree-offreedom system. From analyses of undamped vibration, we know that the natural425Hutton: Fundamentals ofFinite Element Analysis42610. Structural DynamicsC H A P T E R 10Text© The McGraw−HillCompanies, 2004Structural Dynamicsfrequency given by ␻ 2 = k/m is an important property of the system, so we modify the characteristic equation tos2 +cs + ␻2 = 0m(10.128)Solving Equation 10.128 by the quadratic formula yields two roots, as expected,given by 21 c 2cs1 =−− 4␻2 (10.129a)2mm 21 c 2cs2 =+− 4␻2 2mm(10.129b)The most important characteristic of the roots is the value of (c/m) 2 − 4␻2 , andthere are three cases of importance:1.

If (c/m) 2 − 4␻ 2 > 0, the roots are real, distinct, and negative; and thedisplacement response is the sum of decaying exponentials.2. If (c/m) 2 − 4␻ 2 = 0, we have a case of repeated roots; for this situation,the displacement is also shown to be a decaying exponential. It isconvenient to define this as a critical case and let the value of the dampingcoefficient c correspond to the so-called critical damping coefficient.Hence, cc2 = 4␻ 2 m 2 or cc = 2m␻.3.

If (c/m) 2 − 4␻ 2 < 0, the roots of the characteristic equation areimaginary; this case can be shown [2] to represent decaying sinusoidaloscillations.Regardless of the amount of damping present, the free-vibration response, asshown by the preceding analysis, is an exponentially decaying function in time.This gives more credence to our previous discussion of harmonic response, inwhich we ignored the free vibrations. In general, a system response is definedprimarily by the applied forcing functions, as the natural (free, principal) vibrations die out with damping.

The response of a damped spring-mass system corresponding to each of the three cases of damping is depicted in Figure 10.14.We now define the damping ratio as ␨ = c/2m␻ and note that, if ␨ > 1 , wehave what is known as overdamped motion; if ␨ = 1, the motion is said to becritically damped; and if ␨ < 1, the motion is underdamped. As most structuralsystems are underdamped, we focus on the case of ␨ < 1. For this situation, it isreadily shown [2] that the response of a damped harmonic oscillator is describedbyx (t ) = e −␨ ␻t ( A sin ␻d t + B cos ␻d t )(10.130)Hutton: Fundamentals ofFinite Element Analysis10.

Structural DynamicsText© The McGraw−HillCompanies, 200410.9 Energy Dissipation: Structural Dampingx(t)Xe⫺␨␻tt␨⬍1(a)x(t)x(t)tt␨⫽1␨ ⬎1(b)(c)Figure 10.14 Characteristic damped motions: (a) Underdamped.(b) Critically damped. (c) Overdamped.where ␻d is the damped natural circular frequency, given by␻d2 = (1 − ␨ 2 )km(10.131)and the coefficients are determined by the initial conditions.While we demonstrate the effect of damping via the simple harmonic oscillator, several points can be made that are applicable to any structural system:1. The natural frequencies of vibration of a system are reduced by the effectof damping, per Equation 10.131.2. The free vibrations decay exponentially to zero because of the effects ofdamping.3. In light of point 2, in the case of forced vibration, the steady-state solutionis driven only by the forcing functions.4.

Damping is assumed to be linearly proportional to nodal velocities.10.9.1 General Structural DampingAn elastic structure subjected to dynamic loading does not, in general, have specific damping elements attached. Instead, the energy dissipation characteristicsof the structure are inherent to its mechanical properties. How does, for example,a cantilevered beam, when “tweaked” at one end, finally stop vibrating? (If thereader has a flexible ruler at hand, many experiments can be performed to exhibitthe change in fundamental frequency as a function of beam length as well as the427Hutton: Fundamentals ofFinite Element Analysis42810. Structural DynamicsC H A P T E R 10Text© The McGraw−HillCompanies, 2004Structural Dynamicsc12kFigure 10.15 A model of abar element with damping.decay of the motion.) The answer to the damping question is complex. Forexample, structures are subjected to the atmosphere, so that air resistance is afactor.

Air resistance is, in general, proportional to velocity squared, so this effectis nonlinear. Fortunately, air resistance in most cases is negligible. On the otherhand, the internal friction of a material is not negligible and must be considered.If we incorporate the concepts of damping as applied to the simple harmonicoscillator, the equations of motion of a finite element model of a structure become[M ]{q̈ } + [C ]{q̇ } + [K ]{q } = {F (t )}(10.132)where [C] is the system viscous damping matrix assembled by the usual rules.

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