MECHANICAL VIBRATIONS:, Summaries of Aerospace Engineering

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MECHANICAL VIBRATIONS:

LECTURE NOTES FOR COURSE EML 4220

ANIL V. RAO

University of Florida

Spring 2009

ii

Anil V. Rao earned his B.S. in mechanical engineering and A.B. in mathematics from Cornell University, his M.S.E. in aerospace engineering from the University of Michi- gan, and his M.A. and Ph.D. in mechanical and aerospace engineering from Princeton University. After earning his Ph.D., Dr. Rao joined the Flight Mechanics Department at The Aerospace Corporation in Los Angeles, where he was involved in mission sup- port for U.S. Air Force launch vehicle programs and trajectory optimization software development. Subsequently, Dr. Rao joined The Charles Stark Draper Laboratory, Inc., in Cambridge, Massachusetts. As a Draper employee, Dr. Rao led numerous projects related to trajectory optimization, guidance, and navigation of both space flight and atmospheric flight vehicles. Concurrently, from 2001 to 2006 Dr. Rao was an Ad- junct Professor of Aerospace and Mechanical Engineering at Boston University where he taught the core undergraduate engineering dynamics course. While at Boston Uni- versity, Dr. Rao was voted the 2002 and 2006 Mechanical and Aerospace Engineering Faculty Member of the Year and was voted 2004 College of Engineering Professor of the Year for outstanding teaching.

Vakratunda Mahaakaaya Soorya Koti Samaprabha Nirvighnam Kuru Mein Deva Sarva Kaaryashu Sarvadaa

vi

viii Contents

Chapter 1

Response of Single Degree-of-Freedom

Systems to Initial Conditions

In this chapter we begin the study of vibrations of mechanical systems. Generally speaking a vibration is a periodic or oscillatory motion of an object or a set of objects. Vibrating systems are ubiquitous in engineering and thus the study of vibrations is extremely important. The most basic problem of interest is the study of the vibration of a one degree-of-freedom (i.e., a system whose motion can be described using a single scalar second-order ordinary dif- ferential equation). The generic model for a one degree-of-freedom system is a mass connected to a linear spring and a linear viscous damper (i.e., a mass-spring-damper system). Because of its mathematical form, the mass-spring-damper system will be used as the baseline for analysis of a one degree-of-freedom system. In particular, the differential equation of motion will be derived for the mass-spring-damper system. It will then be shown that the time response of this system is the sum of the zero input response and the zero initial condition response. In this chapter we will focus attention on the zero input response, i.e., the response of the system to a given set of initial conditions. Several examples of single degree-of-freedom systems will then be given. In each of these examples the differential equation will be derived and will be shown to have the same mathematical form as the generic mass-spring-damper system.

1.1 Mass-Spring-Damper System

The most basic system that is used as a model for vibrational analysis is a block of mass m connected to a linear spring (with spring constant K and unstretched length ℓ 0 ) and a viscous damper (with damping coefficient c). In addition, an external force P(t) is applied to the block and the displacement of the block is measured from the inertially fixed point O, where O is the point where the spring is unstretched. Finally, the spring and damper are both attached at the inertially fixed point Q. This system is shown in Fig. 1–1 Denoting unit vector in the direction from O to Q as Ex and the inertial reference frame of the ground by F , the inertial acceleration of the block is given as Fa = x¨Ex (1–1)

Next, the forces exerted by the spring and damper are given, respectively, as

Fs = −K(ℓ − ℓ 0 )us (1–2) Ff = −cvrel (1–3)

First, because the spring is attached at point Q, we have

ℓ = ‖r − rQ‖ (1–4)

1.2 General Solution of a Second-Order LTI Differential Equation 3

Now historically it has been the case that the differential equation has been written in a form that is normalized by the mass, i.e., we divide Eq. (1–15) by m to obtain

x¨ +

c m

x˙ +

K

m

x =

P

m

= p(t) (1–16)

where p(t) = P (t)/m. Furthermore, it is common practice to define the quantities K/m and c/m as follows:

ω^2 n =

K

m 2 ζωn =

c m

The quantities ωn and ζ are called the natural frequency and damping ratio of the system, respectively. In terms of the natural frequency and damping ratio, the differential equation of motion for the mass-spring-damper system can be written in the so called standard form as

x¨ + 2 ζωn x˙ + ω^2 nx = p(t) (1–17)

It is seen that Eq. (1–17) is a second-order linear constant coefficient ordinary differential equa- tion. Often, the term “constant coefficient” is replaced with the term time-invariant , i.e., we say that Eq. (1–17) is a called a second-order linear time-invariant (LTI) ordinary differential equation. The terminology “time invariant” stems from the fact that, for a given input p(t) and a given set of initial conditions (x(t 0 ), x(t˙ 0 ) = (x 0 , x˙ 0 ) at the initial time t = t 0 is the same as the solution to the input p(t + τ) for the initial conditions (x(t 0 + τ), x(t˙ 0 + τ) = (x 0 , x˙ 0 ) at the (shifted) initial time t = t 0 + τ. Because of this fact associated with an LTI system, without loss of generality we can assume that the initial time is zero, i.e., t 0 = 0. Thus, when studying the zero input response of an LTI system we can restrict our attention to initial conditions (x( 0 ), ˙x( 0 ) = (x 0 , x˙ 0 ).

1.2 General Solution of a Second-Order LTI Differential Equation

Eq. (1–17) can be written as d^2 x dt^2

• 2 ζωn

dx dt

• ω^2 nx = p(t) (1–18)

which can be further written as ( d^2 dt^2

• 2 ζωn

d dt

• ω^2 n

x = p(t) (1–19)

Now let

L =

d^2 dt^2

• 2 ζωn

d dt

• ω^2 n (1–20)

Then we can view the system of Eq. (1–17) as a system of the form

Lx = f (1–21)

It is seen that the operator L defined in Eq. (1–20) is linear because

L(αx 1 + βx 2 ) = αL(x 1 ) + βL(x 2 ) (1–22)

for all constants α and β. Then it is seen that Eq. (1–21) is a linear system whose general solution is of then form Eq. (1–17) is given as

x(t) = xh(t) + xp (t) (1–23)

4 Chapter 1. Response of Single Degree-of-Freedom Systems to Initial Conditions

here xh(t) is the homogeneous solution (i.e., the solution for a particular set of initial conditions (x(t 0 ), x(t˙ 0 ) = (x 0 , x˙ 0 ) with a zero input function p(t) ≡ 0) while xp (t) is the particular solution (i.e., the solution for zero initial conditions (x(t 0 ), x(t˙ 0 ) = ( 0 , 0 ) and an arbitrary input function p(t) ≠ 0). The homogeneous solution and particular solutions are also called the zero input response and zero initial condition response , respectively. The general solution x(t) to a second-order LTI system is then given as the sum of the zero input response and the zero initial condition response. Because the zero input response satisfies Eq. (1–17) when p(t) ≡ 0, we have x¨h + 2 ζωn x˙h + ω^2 nxh = 0 (1–24)

Contrariwise, because the zero initial condition response satisfies Eq. (1–17) when p(t) ≠ 0 and the initial conditions are zero, we have

x¨p + 2 ζωn x˙p + ω^2 nxp = p(t) (1–25)

From the preceding discussion, it is seen that studying the general response of a second- order LTI system amounts to studying independently the zero input response and the zero initial condition response. Consequently, the study of single degree-of-freedom vibrations amounts to quantifying the zero input response and the zero initial condition response. In this remainder of this chapter we study in detail the zero input response of a second-order LTI system that arises in the study of mechanical vibrations.

1.3 General Solution to Second-Order Homogeneous LTI System

We now focus on the zero input response of the second-order LTI system of Eq. (1–17), i.e., we focus on the system x¨h + 2 ζωn x˙h + ω^2 nxh = 0 (1–26)

Suppose that we guess the solution to Eq. (1–26) as

xh(t) = eλt^ (1–27)

where λ is constant that has yet to be determined. Differentiating the assumed solution of Eq. (1–27) twice, we have

x˙h(t) = λeλt^ (1–28) x¨h(t) = λ^2 eλt^ (1–29)

Substituting the results of Eqs. (1–28) and (1–29) into (1–26), we obtain

λ^2 eλt^ + 2 ζωnλeλt^ + ω^2 neλt^ = 0 (1–30)

Then, because eλt^ is not zero as a function of time, it can be dropped from Eq. (1–30) to give

λ^2 + 2 ζωn + ω^2 n = 0 (1–31)

Equation (1–31) is called the characteristic equation whose roots give the behavior of the zero input response of Eq. (1–17). Using the quadratic formula, the roots of Eq. (1–31) are given as

λ 1 , 2 = −ζωn ±

4 ζ^2 ω^2 n − 4 ω^2 n = −ζωn ± ωn

ζ^2 − 1 (1–32)

It can be seen that the types of roots admitted by Eq. (1–31) depend upon the value of ζ. In particular, the types of roots are governed by the quantity ζ^2 − 1. We have three cases to consider: (1) 0 ≤ ζ < 1, (2) ζ = 1, and (3) ζ > 1. We now consider each of these cases in turn.

6 Chapter 1. Response of Single Degree-of-Freedom Systems to Initial Conditions

x

(t)h

t

ζ = 0. 05 ζ = 0. 1 ζ = 0. 2 ζ = 0. 5

Figure 1–2 Schematic of the zero input response of an underdamped second-order linear time-invariant system.

Case 2: ζ = 1 (Critical Damping)

When ζ = 1 the zero input response is said to be critically damped. For critically damped

system the quantity ζ^2 − 1 = 0 which implies that

ζ^2 − 1 = 0. The roots of the characteristic equation for an underdamped system are then given as

λ 1 , 2 = −ζωn = −ωn (1–42)

It is seen from Eq. (1–42) that the roots of the characteristic equation for a critically damped system are real and repeated (i.e., the two roots are the same). Furthermore, the general zero input response for a critically damped system is given as

xh(t) = e−ωnt^ (c 1 + c 2 t) (1–43)

The constants c 1 and c 2 can be solved for by using the initial conditions (x( 0 ), x(˙ 0 )) = (x 0 , x˙ 0 ) as follows. First, applying the initial condition x( 0 ) = x 0 into Eq. (1–43), we have

xh( 0 ) = x 0 = c 1 (1–44)

Next, differentiating Eq. (1–43), we obtain

x˙h(t) = −ωne−ωnt^ (c 1 + c 2 t) + c 2 e−ωnt^ (1–45)

Applying the initial condition ˙x( 0 ) = x˙ 0 , we obtain

x˙h( 0 ) = x˙ 0 = −ωnc 1 + c 2 (1–46)

1.3 General Solution to Second-Order Homogeneous LTI System 7

Substituting the result for c 1 from Eq. (1–44), we have

x˙h( 0 ) = x˙ 0 = −ωnx 0 + c 2 (1–47)

Solving Eq. (1–47) for c 2 gives c 2 = x˙ 0 + ωnx 0 (1–48) The zero input response for an critically damped system is then given as

xh(t) = e−ωnt^ [x 0 + ( x˙ 0 + ωnx 0 )t] (1–49)

A schematic of a critically damped zero input response is shown in Fig. 1–3.

Action

x

h

(t)

t

Figure 1–3 Schematic of the zero input response of a critically damped second-order linear time-invariant system.

Case 3: ζ > 1 (Overdamping)

When ζ > 1 the zero input response is said to be overdamped. For an overdamped system the quantity ζ^2 − 1 > 0 which implies that

ζ^2 − 1 > 0. The roots of the characteristic equation for an underdamped system are then given as

λ 1 , 2 = −ζωn ± ωn

ζ^2 − 1 (1–50)

It is seen from Eq. (1–50) that the roots of an overdamped system are real and distinct. Further- more, the general zero input response for an overdamped system is given as

xh(t) = c 1 eλ^1 t^ + c 2 eλ^2 t^ (1–51)

Chapter 2

Forced Response of Single

Degree-of-Freedom Systems

2.1 Response of Single Degree-of-Freedom Systems to Nonperiodic In-

puts

2.2 Physics of Impulsive Motion

Recall from dynamics that the principle of impulse and momentum for a particle states that

ˆF = N^ G′^ − N^ G (2–1)

where N^ G is the linear momentum of the particle as viewed by an observer in an inertial ref- erence frame N. Suppose now that we consider the following system. A block of mass m is connected to a linear spring with spring constant K and unstretched length ℓ 0 and a viscous linear damper with damping coefficient c as shown in Fig. 2–1. The block is initially at rest (i.e., its initial velocity is zero) at its static equilibrium position (i.e., the spring is initially un- stressed) when a horizontal impulse ˆP is applied. We are interested here in determining the velocity of the block immediately after the application of the impulse ˆP.

c

g

m ˆP

x

K

Q

Figure 2–1 Block of mass m connected to linear spring and linear damper struck by

horizontal impulse ˆP.

The solution of the above problem is found as follows. First, let F be the ground. Then,

10 Chapter 2. Forced Response of Single Degree-of-Freedom Systems

choose the following coordinate system fixed in F :

Origin at block when x = 0 Ex = To the left Ez = Into page Ey = Ez × Ex

Then, the position of the block is given in terms of the displacement x as

r = xEx (2–2)

Because {Ex , Ey , Ez } is a fixed basis, the velocity of the block in reference frame F is given as

F (^) v =

Fdr

dt

= x˙Ex = vEx (2–3)

Now because we are going to apply the principle of linear impulse and momentum to this problem, we do not need the acceleration of the block. Instead, we know that neither the spring nor the damper can apply an instantaneous impulse. Therefore, the only impulse applied to the system at t = 0 is that due to ˆP. Consequently, the external impulse acting on the system at t = 0 is ˆF = ˆP = Pˆ Ex (2–4)

Furthermore, the linear momentum of the block the instant before the impulse is applied is zero (i.e., the block is initially at rest) while the linear momentum of the block the instant after the impulse is applied is given as

FG′ (^) = mFv′ (^) = mv′E x (2–5)

Setting ˆF equal to FG′, we obtain

Pˆ = mv′^ ≡ mv(t = 0 +) (2–6)

Solving for v(t = 0 +), we obtain

v(t = 0 +) =

Pˆ

m

The result of this analysis shows that the response of a resting second-order linear system to an impulsive force ˆF is equivalent to giving the system the initial velocity shown in Eq. (2–7).

2.3 Impulse Response of Second-Order Linear System

Suppose now that we consider the general motion of the system in Fig. 2–1, i.e., we consider motion to a general force F(t). Then, recalling the result from earlier, the differential equation of motion is given as m x¨ + c x˙ + Kx = F (t) + Kℓ 0 (2–8)

It is noted that the equilibrium point of the system in Eq. (2–8) is xeq = ℓ 0 , we can define the variable y = x − xeq and rewrite Eq. (2–8) in terms of y to give

m y¨ + c y˙ + Ky = F (t) (2–9)

Now suppose that F (t) is the following function:

F (t) = F δ(t)ˆ (2–10)

where δ(t) is defined as follows:

δ(t − a) =

∞ , t = a 0 , t ≠ τ (2–11)

12 Chapter 2. Forced Response of Single Degree-of-Freedom Systems

where ωn =

k/m is the natural frequency, ζ is the damping ratio, and ωd = ωn

1 − ζ^2 is the damped natural frequency. Differentiating this last equation, we have

g(t)˙ = −ζωne−ζωnt^ (A cos ωdt + B sin ωdt) + e−ζωnt^ (−Aωd sin ωdt + Bωd cos ωdt) (2–24)

Noting that g( 0 ) = 0 and that ˙g( 0 +) = 1 /m, we have

A = 0 (2–25)

B =

mωd

Therefore, the response of the system to a unit impulse at t = 0 is given as

g(t) =

mωd e

−ζωnt (^) sin ωdt , t > 0 0 , t ≤ 0

It is seen that, for an underdamped system, the impulse response is a decaying sinusoid with a zero phase (i.e., the applied impulse did not result in a nonzero phase shift). A schematic of the impulse response is shown in Fig. 2–2.

t

g(t)

Figure 2–2 Schematic of Impulse Response of Underdamped Second-Order Linear System.

2.4 Step Response of Second-Order Linear System

After the unit impulse function, the next fundamental function of importance in the analysis of vibratory systems is the unit step function. The unit step function, denoted u(t), is defined

2.4 Step Response of Second-Order Linear System 13

as

u(t − a) =

0 , t ≤ a 1 , t > a (2–28)

Recalling the unit impulse function δ(t) from Eq. (2–11), it is seen that u(t) is related to δ(t) as follows:

u(t − a) =

∫ (^) t

−∞

δ(τ − a)dτ (2–29)

where τ is a dummy variable of integration. Now suppose we want to determine the response, s(t), of the system of Eq. (2–9) to a unit step input at t = 0. The function s(t) is called the step response and, from Eq. (2–9), satisfies

m¨s + c˙s + Ks = u(t) (2–30)

It is noted that Eq. (2–30) can be written as

m

d^2 s dt^2

• c

ds dt

• Ks = u(t) (2–31)

We can obtain s(t) as follows. Consider again the relationship that holds between the unit impulse and the impulse response, i.e.,

m g¨ + c g˙ + Kg = δ(t) (2–32)

Then, from Eq. (2–29), we have du(t − a) dt

= δ(t − a) (2–33)

Therefore, for a unit step function at t = 0, we have

m g¨ + c g˙ + Kg =

du dt

Integrating both sides of Eq. (2–34) gives

∫ (^) t

−∞

[

d^2 g dτ^2

• c

dg dτ

• Kg

]

dτ =

∫ (^) t

−∞

du(a) da

da = u(t) (2–35)

Now from the fundamental theorem of calculus we have ∫ (^) t

−∞

d^2 g dτ^2

d^2 dt^2

∫ (^) t

−∞

g(τ)dτ (2–36) ∫ (^) t

−∞

dg dτ

d dt

∫ (^) t

−∞

gdτ (2–37)

Therefore, Eq. (2–35) can be rewritten as [ d^2 dτ^2

• c

d dτ

+ K

] ∫ (^) t

−∞

g(τ)dτ = u(t) =

∫ (^) t

−∞

δ(τ)dτ (2–38)

Now if we compare Eq. (2–38) to Eq. (2–31), it is seen that

s(t) =

∫ (^) t

−∞

g(τ)dτ (2–39)

In other words, the response of the system of Eq. (2–9) to a unit step function is the integral of the response of the system to a unit impulse^1. We can then use the result of Eq. (2–39) and the

(^1) More generally, it is the case that the response of any linear time-invariant system to the integral of a

function f (t) is equal to the integral of the response to the original function f (t).