AE2135 II Vibrations Reader

Aerospace Structures & Computational Mechanics

Lecture NotesVersion 2.01

AE21

35-II

-Vib

ratio

ns

Table of Contents III

Table of Contents1 Course overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 Discretisation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

2-1 Difference between statics and dynamics . . . . . . . . . . . . . . . . . . 22-2 Discretisation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

3 Free vibrations (undamped) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74 Free damped vibrations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

4-1 Case 1: Underdamped motion . . . . . . . . . . . . . . . . . . . . . . . 104-2 Case 2: Overdamped motion . . . . . . . . . . . . . . . . . . . . . . . . 124-3 Case 3: Critically damped motion . . . . . . . . . . . . . . . . . . . . . 12

5 Forced undamped vibrations (harmonic) . . . . . . . . . . . . . . . . . . . . . . 145-1 General solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145-2 Solution at resonance . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

6 Forced damped vibrations (harmonic) . . . . . . . . . . . . . . . . . . . . . . . 176-1 Solution method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176-2 Initial conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

7 Response to arbitrary excitation . . . . . . . . . . . . . . . . . . . . . . . . . . . 187-1 Unit impulse response . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187-2 Unit step response . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207-3 Standard solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217-4 Alternative solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

8 Multiple degree of freedom systems . . . . . . . . . . . . . . . . . . . . . . . . . 258-1 Solving the MDOF problem . . . . . . . . . . . . . . . . . . . . . . . . . 258-2 Modal analysis of MDOF systems . . . . . . . . . . . . . . . . . . . . . 288-3 Forced vibrations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

A Simplification standard solution . . . . . . . . . . . . . . . . . . . . . . . . . . . 30B Energy methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

B-1 Energy analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32B-2 Euler-Lagrange equation . . . . . . . . . . . . . . . . . . . . . . . . . . 33

C Coulomb damping . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34D Logarithmic decrement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37E Base excitation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38F Laplace transforms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39G General solution strategy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

G-1 Step 1: DOFs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40G-2 Step 2: Free-body diagram . . . . . . . . . . . . . . . . . . . . . . . . . 40G-3 Step 3: Equation of motion . . . . . . . . . . . . . . . . . . . . . . . . . 41G-4 Step 4: Constitutive relations . . . . . . . . . . . . . . . . . . . . . . . . 41G-5 Step 5: Kinematic relations . . . . . . . . . . . . . . . . . . . . . . . . . 41G-6 Step 6: Force and moment equilibrium . . . . . . . . . . . . . . . . . . . 42G-7 Step 7: Combining information into one relation . . . . . . . . . . . . . . 42G-8 Step 8: Solve for the motion . . . . . . . . . . . . . . . . . . . . . . . . 42

AE2135-II - Vibrations Lecture Notes

IV Table of Contents

Lecture Notes AE2135-II - Vibrations

1 Course overview 1

1 Course overview

These notes present a short summary of the content of the course AE2135-II - Vibrationsby Roeland De Breuker. These notes are ultimately based on the book Engineering Vi-brations, by Daniel J. Inman. Please feel free to report any mistakes or suggestions [email protected].

In table 1 you can find the course map for the course, which will bring you to the specifiedsection when clicked upon.

Table 1: Course map for single degree of freedom systems

Free Vibration Forced VibrationForce + base excitationHarmonic Arbitrary

Undamped Motion Section 3 Section 5 Section 7

Damped MotionSection 4

Section 6 Section 7Coulomb friction: C

A discussion on multiple degree of freedom systems is given in section 8.


mailto:[email protected]

2 2 Discretisation

2 Discretisation

2-1 Difference between statics and dynamics

There is a difference in the balance of a mass like the one below in a static and a dynamiccase:

Statics: ΣF = 0Dynamics: ΣF = mu

F1

F2

F3

u

m

Figure 1: Balance of a mass

2-2 Discretisation

Continuous structures need to be discretised in order to analyse their dynamic propertieswith the tools presented in this course. A possible discretisation of the continuous structurein figure 2 can be seen in figure 3. The question remains: What are the values of k in thediscretised version? Six different discretisation options are given in the next sections.

L

w(x)

q

x, u

w, z

Figure 2: Continuous beam deformed by a distributed load

F F2

w1 w2k1 k2 k3

/L 3

2 /L 3

m1 m2

1

Figure 3: A discretised version of figure 2


2 Discretisation 3

A: Bending stiffness

E, I, L

F

δ

=

F

k

Figure 4: Discretising a cantilever beam

In the case of a cantilever beam loaded by a transverse load, the bending stiffness of the beamis discretised as can be seen in figure 4. For the deflection:

δ = FL3

3EI

Hence, in terms of a force-displacement equation, the above can be rewritten as:

F = k δ

F = 3EIL3 δ

B: Bending stiffness with rotational spring

E, I, L

F

δ

=

F

kϕ

kr

Figure 5: Discretising a cantilever beam with a rotational spring

For the case the cantilever beam is attached via a rotational spring (see figure 5), the deflectionis described as follows:

δ = FL3

3EI + ϕL

Through the moment equilibrium at the root of the beam, the relation between the angle ofrotation ϕ and the applied load F can be found:{

M = krϕ

M = FL−→ ϕ = FL

kr

Hence:


4 2 Discretisation

F = k δ

F = 3EIkrL3kr + 3L2EI

δ

This equation can be checked by taking the limit of kr →∞. In that case k → 3EIL3 , which is

the same answer as in the previous case, validating the answer above.

C: Axial stiffness

F

δ=

F

kE, A, L

Figure 6: Discretising an axially loaded beam

For the axial deflection of the axially loaded beam in figure 6:

δ = FL

EA

Hence:

F = k δ

F = EA

Lδ

D: Axial stiffness of a multi-material rod

F

E , A , L1

E , A , L2

1

2

1

2

F1F2

F

Figure 7: Discretising a parallel multi-material rod

In figure 7, a axially loaded rod composed of two different materials is shown. Its deflectionis calculated using displacement compatibility:

δ1 = δ2 = δ


2 Discretisation 5

The right side of the figure shows force equilibrium:

F = F1 + F2

So for the individual internal loads the following can be written:

δ1 = F1L

E1A1−→ F1 = E1A1

Lδ

δ2 = F2L

E2A2−→ F2 = E2A2

Lδ

Hence, the force-displacement relation becomes the following:

F = k δ

F =(E1A1L

+ E2A2L

)δ

The relation above can again be checked by a limit case, in this case when one of the Young’smoduli goes to zero. Clearly, the relation can be rewritten to contain the stiffness of the twoindividual bars:

F = (k1 + k2)δ

E: Axial stiffness of two rods in series

F

E , A , L1

E , A , L2

1

2

1

2

Figure 8: Discretising two rods in series

There can also be a case where two different types of rod are stacked on top of one anotheras displayed in figure 8. In this situation the displacement compatibility as in the previousexample does not count, but is a summation rather:

δ = δ1 + δ2

Now the force is constant throughout the rod, and the individual displacements can be de-scribed as:

δ1 = FL1E1A1

δ2 = FL2E2A2


6 2 Discretisation

and the total displacement becomes:

δ = F

(L1E1A1

+ L2E2A2

)making the force displacement relation as follows:

F = k δ

F =(

1L1E1A1

+ L2E2A2

)δ

which can be rewritten using the spring constants as in case C (ki = EiAiLi

) as below:

F =(

11k1

+ 1k2

)δ

Check the limit case if k1 or k2 →∞

F: Rotational stiffness

ϕT,

Figure 9: Discretising for rotational stiffness

For the fixed beam loaded with a torque as displayed in figure 9:

ϕ = TL

GJ

So:

T = kr ϕ

T = GJ

Lϕ


3 Free vibrations (undamped) 7

3 Free vibrations (undamped)

The discretised version of the simplified view of a wing as shown in figure 10 is shown on theright side of the same figure. It is assumed that the mass of the wing is negligible comparedto m.

k

L

E, I =m

mFuselage

Fuel pot

Figure 10: Discretising a simplified wing

The spring constant has the following value:

k = 3EIL3

Solution

The free-body diagram (FBD) of the mass is shown in figure 11.

m

mgFS

+y

Figure 11: Free-body diagram of the mass (the spring is shown for clarity)

The next step is to write down the equation(s) of motion (EOM) and constitutive equation(s)(CE). The latter are equations that relate forces to displacements.{

EOM my = Fs +mg (3.1)CE Fs = −ky (3.2)

Combining (3.1) and (3.2) yields an inhomogeneous differential equation:

my + ky = mg

To solve this equation, assume a particular solution yp and a homogeneous solution yh:

y = yp + yh = δstatic + δdynamic = δs + x


8 3 Free vibrations (undamped)

ϕ

t

Regular sine wave

Figure 12: Phase shift of a sine wave

then:y = x

mx+ k (δs + x) = mg

−→ mx+ kx = mg − kδsFrom statics it is known that δs = mg

k which yields

mx+ kx = 0 (3.3)

x is expected to be harmonic, so assume a solution of the form

x = A sin (ωnt+ ϕ) (3.4)

where A is the amplitude of the oscillation and ϕ is the phase shift which describes howmuch the sine wave is shifted from x = 0 at t = 0, see figure 12. The standard solutionto the harmonic equation is x = eλt. How to get from eλt to A sin (ωnt+ ϕ) is shown inappendix A. The two unknowns A and ϕ can be found from the initial conditions. This isa second-order homogeneous differential equation, so two initial conditions are needed: theinitial displacement x0 and the initial velocity x0. Starting from (3.4), the initial conditionsbecome: {

x(0) = x0 = A sin(ϕ) (3.5)x(0) = x0 = ωnA cos(ϕ) (3.6)

1: Squaring (3.5) and (3.6), then adding yields

x20 +

(x0ωn

)2= A2

(sin2(ϕ) + cos2(ϕ)

)︸︷︷︸

=1

−→ A =

√x2

0 +(x0ωn

)2

2: Dividing (3.5) by (3.6) gives

x0x0

= 1ωn

tan(ϕ) −→ ϕ = arctan(x0ωnx0

)


3 Free vibrations (undamped) 9

Natural frequencies

Each mass-spring system oscillates at its own natural frequency when excited. This frequencyis ωn, and is calculated by substituting x = A sin (ωnt+ ϕ) into the homogeneous differentialequation (3.3):

−mω2nA sin (ωnt+ ϕ) + kA sin (ωnt+ ϕ) = 0

mω2n = k

−→ ωn =

√k

m

Total spring force

Fs = −ky= −k (δs + x(t))

= −k(mg

k+A sin (ωnt+ ϕ)

)The equation above is maximum for dFs

dt = 0:

−→ −kAωn cos (ωnt+ ϕ) = 0

Irrespective of the phase shift ϕ, the maximum stays the same, so:

−→ −kAωn cos (ωnt) = 0

which holds when cos (ωnt) = 0, so: ωnt = π2 + kπ, k ∈ Z

−→ t = π

ωn

(12 + k

)so

|Fsmax | = mg + kA


10 4 Free damped vibrations

4 Free damped vibrations

k c

m

x

Figure 13: A damped mass-spring system

Consider the mass-damper-spring system as shown in figure 13. It has the following equationof motion:

mx+ cx+ kx = 0To solve for the motion x(t), substitute x = xeλt:

m�xλ2��eλt + c�xλ��e

λt + k�x��eλt = 0

−→ mλ2 + cλ+ k = 0

Solve this with the quadratic formula:

λ = −c±√c2 − 4mk

2mThe discriminant c2 − 4mk makes the difference whether λ is real or complex. The criticalfactor is when it is zero:

c2 = 4mk −→ ccrit = 2√mk

Define the damping ratio ζ:ζ = c

ccrit= c

2√mk

= c

2mωnThen the equation of motion can be rewritten:

x+ 2ζωnx+ ω2nx = 0

4-1 Case 1: Underdamped motion

In this case: 0 < ζ < 1:

λ1,2 = −2ζωn ±√

4ζ2ω2n − 4ω2

n

2= −ζωn ± ωn

√ζ2 − 1

= −ζωn ± i ωn√

1− ζ2︸︷︷︸ωd

so:

λ1 = −ζωn − iωdλ2 = −ζωn + iωd


4 Free damped vibrations 11

which leads to the following solution (see also figure:

x = e−ζωnt(a1e

iωdt + a2e−iωdt

)= Ae−ζωnt︸︷︷︸

decay

sin (ωdt+ ϕ)︸︷︷︸frequency content

(4.1)

t

x

Figure 14: Decaying sinusoidal response (the dashed lines indicate ±e−ζωnt)

The amplitude A and phaseshift ϕ can be found from the initial conditions x(0) = x0 andx(0) = x0:

x(0) = A sinϕ = x0 −→ A = x0sin(ϕ) (4.2)

From equation 4.1 the velocity can be deduced:

x = −ζωnAe−ζωnt sin(ωdt+ ϕ) +Ae−ζωntωd cos(ωdt+ ϕ)

So the initial velocity condition can be written as:

x0 = −ζωnA sin(ϕ) +Aωd cos(ϕ) (4.3)

Substituting equation 4.2 in this equation yields:

x0 = −ζωnx0 + x0ωdtan(ϕ)

Hence:−→ ϕ = arctan

(x0ωd

x0 + ζωnx0

)Also, from equation 4.2 (see figure 15):

sin(ϕ) = x0A

= x0ωdP

−→ P = Aωd

and using the Pythagorean theorem:

−→ A =

√x2

0ω2d + (x0 + ζωnx0)2

ωd


12 4 Free damped vibrations

ϕ

Figure 15: Obtaining P through angle formulae

4-2 Case 2: Overdamped motion

Now: ζ > 1In this case the roots of the equation of motion are real:

λ1 = −ζωn − ωn√ζ2 − 1

λ2 = −ζωn + ωn

√ζ2 − 1

and the displacement becomes non-oscillatory (see also figure 16):

t

x

Figure 16: An overdamped motion with x(0) = 1 and x(0) = 0

4-3 Case 3: Critically damped motion

For this type of motion: ζ = 1Now the roots are equal:

λ1 = λ2 = −ωnFor repeated roots, the solution takes the form:

x = (a1 + a2t)e−ωnt

where

a1 = x0

a2 = x0 + ωnx0

A critically damped motion can be seen in figure


4 Free damped vibrations 13

t

x

Figure 17: A critically damped motion with x(0) = 1 and x(0) = −1


14 5 Forced undamped vibrations (harmonic)

5 Forced undamped vibrations (harmonic)

m

x

Fe

Figure 18: A forced mass-spring system

5-1 General solution

EOM:mx+ kx = F e(t) = F cos(ωxt)

The displacement x of the forced mass-spring system in figure 18 is composed of a transientpart, xh, and a steady-state part, xp, i.e.:

x = xh + xp

The following is assumed for x:

x = A sin (ωnt+ ϕ)︸︷︷︸xh

+Ap cos(ωxt)︸︷︷︸xp

Solving the steady-state, or particular part of the displacement:

−mApω2x cos(ωxt) +Apk cos(ωxt) = F cos(ωxt)

so:Ap = f

ω2n − ω2

x

with:f = F

m

The amplitude and phase shift are displayed in figure 19.


5 Forced undamped vibrations (harmonic) 15

θ

Figure 19: Amplitude of undamped forced motion and phase difference θ between loading andresponse. After ωx = ωn, the load and response are in opposite phase

Suppose: {x0 = 0x0 = 0

then:

A sin(ϕ) +Ap = 0

Aωn cos(ϕ)−Apωx · 0 = 0 −→ ϕ = π

2 + 2kπ

−→ A = −Ap

−→ x = Ap (cos(ωxt)− cos(ωnt))

5-2 Solution at resonance

x+ ω2nx = f cos (ωnt)

The normal assumption to reach a solution would be xp = Ap cos (ωnt), but in this case this


16 5 Forced undamped vibrations (harmonic)

is not possible as it would lead to Ap →∞. Hence, in this case, we try:

xp = Apt sin (ωnt+ ϕ)xp = ωnA

pt cos (ωnt+ ϕ) +Ap sin (ωnt+ ϕ)xp = −ω2

nApt sin (ωnt+ ϕ) + ωnA

p cos (ωnt+ ϕ) + ωnAp cos (ωnt+ ϕ)

= 2ωnAp cos (ωnt+ ϕ)− ω2nA

pt sin (ωnt+ ϕ)

Substituting this in the equation above yields:

2ωnAp cos (ωnt+ ϕ)−((((((((

((ω2nA

pt sin (ωnt+ ϕ) +(((((((((

(ω2nA

pt sin (ωnt+ ϕ) = f cos (ωnt)−→ 2ωnAp cos (ωnt+ ϕ) = f cos (ωnt)

Hence:

ϕ = 0 + 2kπ

Ap = f

2ωn

So:−→ xp = f

2ωnt sin (ωnt+ ϕ) = f

2ωnt sin (ωnt)

which can be seen in figure 20.

t

Figure 20: Sinusoidal response, with an amplitude linearly increasing by f t2ωn


6 Forced damped vibrations (harmonic) 17

6 Forced damped vibrations (harmonic)

Let’s look at the harmonic forced vibrations of an underdamped system, this has the followingequation of motion:

x+ 2ζωnx+ ω2nx = f cos (ωxt)

6-1 Solution method

To solve for the displacement, xp = Re(Apeiωxt

)is used, because: eiωt = cos (ωt) + i sin (ωt),

so cos (ωt) = Re(eiωt

). Entering this into the EOM gives:

Re((−ω2

x + 2ζωniωx + ω2n

)Apeiωxt

)= Re

(f eiωxt

)−→ Ap = Re

(f

ω2n − ω2

x + 2ζωniωx

)The denominator can be rewritten as a complex number u = v + iw, with:

v = ω2n − ω2

x

w = 2ζωnωxNow, denoting u as the complex conjugate of u:

Ap = Re(f

u

)= Re

(fu

uu

)= Re

(fu

v2 + w2

)Also:

u = v − iw = |u| (cos(θ)− i sin(θ)) = |u| (cos(−θ) + i sin(−θ)) = |u| e−iθ

with θ = arctan(wv

). Hence:

Ap = Re(f

√v2 + w2e−iθ

v2 + w2

)= Re

(f e−iθ√v2 + w2

)= Re

f e−iθ√(ω2n − ω2

x)2 + (2ζωnωx)2

So the particular solution becomes:

xp = Re(Apeiωxt

)= Re

f ei(ωxt−θ)√(ω2n − ω2

x)2 + (2ζωnωx)2

= f cos (ωxt− θ)√(ω2n − ω2

x)2 + (2ζωnωx)2

6-2 Initial conditions

x = Ahe−ζωnt sin (ωdt+ ϕ) +Ap cos (ωxt− θ)

x = Ah(−ζω−ζωnt

n sin (ωdt+ ϕ) + e−ζωntωd cos (ωdt+ ϕ))−Apωx sin (ωxt− θ)

So when x(0) = 0 and x(0) = 0, then:{Ah sin (ϕ) +Ap cos (θ) = 0Ah (−ζωn sin (ϕ) + ωd cos (ϕ)) +Apωx sin (θ)

from which ϕ and Ah can be derived as a function of the known Ap and θ.


18 7 Response to arbitrary excitation

7 Response to arbitrary excitation

To be able to solve the response to an arbitrary excitation, first, the solution methods fora unit impulse and step response need to be known. They are treated first, after which thestandard solution to an arbitrary excitation is treated.

7-1 Unit impulse response

To calculate the response of a system to a unit impulse, first, the unit impulse function δneeds to be defined. This function is called the Dirac delta function and has the followingdefinition (see also figure 21):

δ(t− a) = 0 for t 6= a

δ(t− a) 6= 0 for t = a∞∫−∞

δ(t− a)dt = 1

δ has the unit of 1/s.

t

ε1

a

area = 1

ε

Figure 21: A visualisation of the Dirac delta function

From the definition, it follows that:∞∫−∞

f(t)δ(t− a)dt = f(a)∞∫−∞

δ(t− a)dt = f(a)

Let’s have a look at the impulse response of an underdamped system:

k

c x

m

Figure 22: A forced damped mass-spring system

The EOM is:mx+ cx+ kx = P δ(t)


7 Response to arbitrary excitation 19

and the IC are: {x(0) = 0x(0) = 0

To solve the impulse response problem, the impulse is converted to the initial conditions. Tothis end, integrate both sides from 0 to ε:

ε∫0

(mx+ cx+ kx)dt =ε∫

0

P δ(t)dt

This can be split into different integrals, looking at each individual part:

−→ limε→0

ε∫0

mxdt = limε→0

mx∣∣∣ε0

= limε→0

(mx(ε)−mx(0)) = mx(0+)

−→ limε→0

ε∫0

cxdt = limε→0

cx∣∣∣ε0

= limε→0

(cx(ε)− cx(0)) = cx(0+)

−→ limε→0

ε∫0

kxdt = limε→0

kxt∣∣∣ε0

= limε→0

(kx(ε)ε− kx(0)0) = 0

−→ limε→0

ε∫0

P δ(t)dt = P

x(0+) = 0 from the definition of impulse. The impulse can thus be written as an initialvelocity and the equation can be simplified to:

mx(0+) = P −→ x(0+) = P

m

So the EOM becomes:mx+ cx+ kx = 0

with IC: {x(0) = 0x(0) = P

m

Remember, for an underdamped system (see section 4):

x = Ae−ζωnt sin (ωd + ϕ)

with

A =

√x2

0ω2d + (x0 + ζωnx0)2

ωd

ϕ = arctan(

x0ωdx0 + ζωnx0

)



which for this case become:

ϕ = 0 + 2kπA = x0

ωd= P

mωd

−→ x(t) =

Pmωd

e−ζωnt sin (ωdt) t ≥ 00 t < 0

= P g(t)

where the displacement function to a unit impulse has been labelled g(x) for later conve-nience.

7-2 Unit step response

The unit step function H, or Heaviside step function, is defined as (see also figure 23):

H(t− a) ={

0 for t < a

1 for t ≥ a

H

t

1

a0

Figure 23: A visualisation of the Heaviside step function

The Heaviside step function is mathematically linked to the Dirac delta function:

δ(t− a) = dH(t− a)dt

H(t− a) =t∫

−∞

δ(τ − a)dτ

The response of an underdamped mass-spring system to a unit step at t = a can thus befound by integrating the response to a unit impulse at t = a. In this case a step starting at



t = 0 is regarded, so g(x) from section 7-1 can be inserted in the integral:

G(t) =t∫

−∞

g(τ)dτ

=t∫

−∞

1mωd

e−ζωnτ sin (ωdτ)dτ τ ≥ 0

=t∫

0

1mωd

e−ζωnτ sin (ωdτ)H(τ)dτ

where the lower limit has been set to zero because of the inclusion of the Heaviside functionand we are only interested in the displacement after t = 0. Substitute sin(ωdt) = eiωdt−e−iωdt

2i :

G(t) =t∫

0

12imωd

e−ζωnτ(eiωdτ − e−iωdτ

)H(τ)dτ

= 12imωd

[e−(ζωn−iωd)τ

−(ζωn − iωd)− e−(ζωn+iωd)τ

−(ζωn + iωd)

]t0

The denominator has to be equal to add fractions. Multiplying both fractions with the otherdenominator yields a common denominator: −(ζωn − iωd) · −(ζωn + iωd) = ζ2ω2

n + ω2d = ω2

n.So:

G(t) = 12imωdω2

n

[(ζωn + iωd)e−(ζωn−iωd)τ + (ζωn − iωd)e−(ζωn+iωd)τ

]t0

= 12imωdω2

n

[(ζωn + iωd)

(e−(ζωn−iωd)t − 1

)+ (ζωn − iωd)

(e−(ζωn+iωd)t − 1

)]= 1

2imωdω2n

[−ζωne−ζωnteiωdt +��ζωn + iωd − iωde−ζωnteiωdt + ζωne

−ζωnte−iωdt��− ζωn

+ iωd − iωde−ζωnte−iωdt]

= 12imωdω2

n

[2iωd − e−ζωnt

(ζωn

(eiωdt − e−iωdt

)+ iωd

(eiωdt + e−iωdt

))]= 2iωd

2imωdω2n

[1− e−ζωnt

(ζωn2iωd

2i sin(ωdt) + 122 cos(ωdt)

)]= 1k

[1− e−ζωnt

(cos(ωdt) + ζωn

ωdsin(ωdt)

)]which is only valid for t ≥ 0.

7-3 Standard solution

The response to an arbitrary excitation can be found by dividing the excitation force intoinfinitesimally short impulses, see figure 24, and summing all the responses to these impulsesup:



t

Figure 24: An arbitrary load can be seen as a collection of impulses

P (τ) = f(τ)∆τF (τ) = P (τ)δ(t− τ) Impulsive force

= f(τ)∆τδ(t− τ)

linearsystem

Figure 25

x(τ) = f(τ)∆τg(t− τ)

x(t) = lim∆τ→0

∑τ

x(τ) = lim∆τ→0

∑τ

f(τ)∆τg(t− τ) =t∫

0

f(τ)g(t− τ)dτ

It does not matter which of the two functions (f or g) is function of τ which one is functionof (t− τ), as can be seen below:

t− τ = λ, so: τ = t− λ and: dτ = −dλ

also:

τ = 0 −→ λ = t

τ = t −→ λ = 0

x(t) = −0∫t

f(t− λ)g(λ)dλ =t∫

0

f(t− λ)g(λ)dλ

The response to an arbitrary excitation can be written as:

x =t∫

0

f(τ)g(t− τ)dτ



where g(t) is the impulse response function.

Integrals like these are conveniently solved in the Laplace domain, see also appendix F, wherethe integral vanishes into a straightforward multiplication and thus the solution is easilyfound. The challenge of this method usually only lies in transforming the solution back fromLaplace domain to the time domain. In the Laplace domain, the equation above becomes:

X = F (s)g(s)

so in case of the following initial conditions:

x(0) = x0

x(0) = x0

the equation of motion mx+ cx+ kx = f(t) becomes in the Laplace domain:

m(s2X − sx0 − x0

)+ c (sX − x0) + kX = F (s)

and solving the equation in the Laplace domain can be done by simply isolating X from theequation:

X = F (s)ms2 + cs+ k

+ (sm+ c)x0 +mx0ms2 + cs+ k

= F (s)m (s2 + 2ζωns+ ω2

n) + m [(s+ 2ζωn)x0 + x0]m (s2 + 2ζωns+ ω2

n)= F (s)g(s) +m [(s+ 2ζωn)x0 + x0] g(s)

where:g(s) = 1

m (s2 + 2ζωns+ ω2n)

This expression can be split up in partial fractions and transformed back to the time domainusing a table of standard Laplace transformations (Note: such a table will be provided at theexam).

7-4 Alternative solution

Instead of using the impulse response function, one can also use the step response functionin the manner as described below. This approach, though slightly more complex, is relevant,since it is often easier to get step response solutions from numerical codes than impulseresponse solutions.

∆f = ∆f∆τ ∆τH(t− τ)

−→ x(t) = lim∆τ→0

∑τ

∆f∆τ ∆τG(t− τ) =

t∫0

dfdτ G(t− τ)dτ



t

Figure 26: An arbitrary load can also be seen as a collection of steps


8 Multiple degree of freedom systems 25

8 Multiple degree of freedom systems

An example of a multiple degree of freedom (DOF) system is a wing with two masses, seefigure 27.

Fuselage

m1k1 k2m2

Figure 27: A wing with two masses

The wing can basically be discretised in two ways, see figures 28 and 29 below.

x2

x1

m1k1

k2 m2

Figure 28: Discretising the wing

x2x1

m1k1k2

m2

Figure 29: Discretising the wing in equivalent way

Note that both approaches are equivalent: although for figure 29 the direction of movementsappears to have changed, still wing bending is considered and hence it is just an easier wayto draw the discretised version of the wing.

8-1 Solving the MDOF problem

Upon solving the discretisation method as displayed in figure 29, the loads are as displayedin figure 30.


26 8 Multiple degree of freedom systems

x2x1

m1k1k2

m2F F21 F2

Figure 30: Finding the loads in the discretised wing

The loads have the following values:{F1 = k1x1

F2 = k2 (x2 − x1)

The equations of motion for the masses are:{m1x1 = −F1 + F2

m2x2 = −F2

Combining the above: {m1x1 = −k1x1 + k2x2 − k2x1

m2x2 = −k2x2 + k2x1

or: {m1x1 + (k1 + k2)x1 − k2x2 = 0m2x2 − k2x1 + k2x2 = 0

This equation can be written in matrix-vector format:[m1 00 m2

]{x1x2

}+[k1 + k2 −k2−k2 k2

]{x1x2

}={

00

}or, in short:

M x+ Kx = 0When handwriting matrix symbols, it is more common to use a double underline, becausewriting in bold is not possible.

1. M and K are always symmetric

2. The system is coupled, so you cannot solve x1 without solving x2 and vice versa

The solution can be found by assuming a harmonic displacement again:

x = xeiωnt

which stands for: {x1 = x1e

iωnt

x2 = x2eiωnt

Entering the assumed solution into the matrix equation yields:{[−ω2

nm1 00 −ω2

nm2

]+[k1 + k2 −k2−k2 k2

]}︸︷︷︸

A

x = 0



where A is the system matrix. The trivial solution is x = 0. A nontrivial solution occurswhen det(A) = 0. As an example, consider:

m1 = 4 · 103 kgm2 = 2 · 103 kgk1 = 8 kN/mk2 = 4 kN/m

which yields for the nontrivial solution:(−4ω2

n + 12) (−2ω2

n + 4)− 16 = 0

or:8(ω2n − 4

) (ω2n − 1

)= 0

so the eigenfrequencies are:ω2n = 4ω2n = 1 −→ ωn = ±2

ωn = ±1

The negative eigenfrequencies can be discarded. Also note that:

ω1 = 1 6=√k1m1

ω2 = 2 6=√k2m2

Entering the value of an eigenfrequency into the equation of motion yields the correspondingeigenmode. For ω1 = 1: [

8 −4−4 2

]{x11x12

}= 0

The two lines are a scalar multiple of each other, so no unique solution exists. Choose anequation and choose a value for e.g. x11. The first equation yields:

−→ 8x11 − 4x12 = 0

Choosing x11 = 1 gives x12 = 2 and so the first eigenmode is:

x1 ={

12

}

or any scalar multiple of this. For ω2 = 2:[−4 −4−4 −4

]{x21x22

}= 0

so:x21 = −x22


28 8 Multiple degree of freedom systems

1 2

(a) The first eigenmode

1-1

(b) The second eigenmode

Figure 31

and choosing x21 = 1 yields x22 = −1, making the second eigenmode:

x2 ={

1−1

}

or any scalar multiple of this. The two eigenmodes are visualised in figure 31 below. Thereare as many eigenmodes and eigenfrequencies as there are degrees of freedom. Consideringthe following initial conditions:{

x1(0) = x10 x2(0) = x20//x1(0) = x10 x2(0) = x20

the complete motion can be written in terms of the two eigenmodes:

x = A1 sin (ω1t+ ϕ1) x1 +A2 sin (ω2t+ ϕ2) x2

so: x1(t) = A1 sin (ω1t+ ϕ1) x11 +A2 sin (ω2t+ ϕ2) x21

x1(t) = A1ω1 cos (ω1t+ ϕ1) x11 +A2ω2 cos (ω2t+ ϕ2) x21

x2(t) = A1 sin (ω1t+ ϕ1) x12 +A2 sin (ω2t+ ϕ2) x22

x2(t) = A1ω1 cos (ω1t+ ϕ1) x12 +A2ω2 cos (ω2t+ ϕ2) x22

and thus, using the initial conditions, there are four equations to solve for the four unknowns:A1, A2, ϕ1 and ϕ2.

8-2 Modal analysis of MDOF systems

A popular way of solving MDOF systems is by modal analysis. The basic idea is that thedisplacements of the MDOF system are replaced by coordinates of the modes (modal coordi-nates) of the MDOF system. These modes are comparable to the eigenvectors of the MDOFsystem as discussed in the previous section. Using the modified eigenvectors of the MDOFsystem will result in an uncoupled system of equations which can be solved using the singledegree-of-freedom techniques from the previous chapters in this reader. Obviously from aphysical point-of-view the MDOF system remains coupled, but the coupling is moved for thesystem of equations to the modal coordinates.

The procedure is as follows:

1. Mass normalise the displacements x to a new coordinate q.

2. Tranform the mass normalised coordinates q to mass normalised modal coordinates r.



The mass normalisation is carried out by transforming x = M− 12 q. Substituting this into the

equations of motion yields:

MM− 12 q + KM− 1

2 q = 0

Premultiplying the above equation by M− 12 yields:

M− 12 MM− 1

2 q + M− 12 KM− 1

2 q = 0

This results in the following equation:

q + Kq = 0

In this equation, K is the mass normalised stiffness matrix, which is still symmetric, but full.It is obvious that the eigenvalues of this matrix are the squares of the eigenfrequencies of theMDOF system.

The next step is to tranform the mass normalised displacements q to modal coordinates r.This is done by applying the transformation q = P r. Matrix P contains the orthonormaleigenvectors of K. This transformation results in the following equation of motion:

P r + KP r = 0

Premultiplying the equation by P T yields:

P TP r + P T KP r = 0

or:

r + Λr = 0

Matrix Λ is a diagonal matrix which contains the eigenvalues of K. This system of equa-tions is a decoupled system of which the modal amplitudes can be solved individually. Thedisplacement vector can be retrieved by using x = M− 1

2 P r.

8-3 Forced vibrations

Solving MDOF systems with forced vibrations, whether they are harmonic or arbitrary be-comes rather straightforward is the decoupled system in modal coordinates is used. The samemethodologies from the previous chapters on single degree-of-freedom systems can be used toobtain the forced response of r, and the coordinate transformation back to x can be used toobtain the physical displacements. Obviously the forcing terms have to be modified to:

r + Λr = P TM− 12F


30 A Simplification standard solution

A Simplification standard solution

This section shows how to get from xeλt to A sinωt+ φ

The formal solution to the second order homogeneous differential equation mx+ kx = 0 is

x = xeλt

Substitute this into the differential equation: mλ2��xeλt + k��

�xeλt = 0

λ2 = − km

−→ λ = ±iωn

Therefore the solution becomes

x = a1eiωnt + a2e

−iωnt

a1 and a2 need to be complex conjugates for x to be real, so:

a1 = a+ ib

a2 = a− ib

In this case we first aim for a solution of the form x = A1 sin (ωnt) + A2 cos (ωnt). Thefollowing steps are then to be taken after defining the complex conjugates a1 and a2:

x = a1eiωnt + a2e

−iωnt

= (a+ ib)eiωnt + (a− ib)e−iωnt

= (a+ ib) (cos (ωnt) + i sin (ωnt)) + (a− ib) (cos (−ωnt) + i sin (−ωnt))= (a+ ib) (cos (ωnt) + i sin (ωnt)) + (a− ib) (cos (ωnt)− i sin (ωnt))= 2a cos (ωnt) + (−2b) sin (ωnt)

−→ x = A1 cos (ωnt) +A2 sin (ωnt)

where A1 = 2a and A2 = −2b. This solution is equivalent to a sine function with a phaseshift. To prove that this is indeed the case, the following identity needs to be used:

eiωnt = cos (ωnt) + i sin (ωnt)ieiωnt = i cos (ωnt)− sin (ωnt)

so:

cos (ωnt) = Im(ieiωnt

)sin (ωnt) = Im

(eiωnt

)and thus the equation for x can be rewritten to:

x = A1Im(ieiωnt

)+A2Im

(eiωnt

)= Im

((A1i+A2) eiωnt

)Lecture Notes AE2135-II - Vibrations

A Simplification standard solution 31

Figure 32: The position of an imaginary number on the imaginary plane

From figure 32 it can be deduced that for an imaginary number A1i + A2 the following isvalid:

ϕ = arctan(A1A2

)so

−→ A1 =√A2

1 +A22 sin(ϕ)

A2 =√A2

1 +A22 cos(ϕ)

and

(A1i+A2) =√A2

1 +A22 (cos(ϕ) + i sin(ϕ))

= Aeiϕ

with A =√A2

1 +A22. So now:

−→ x = Im(Aeiϕeiωnt

)= Im

(Aei(ωnt+ϕ)

)−→ x = A sin (ωnt+ ϕ)


32 B Energy methods

B Energy methods

Vibration problems can also be solved considering the energy in the system.

B-1 Energy analysis

Equate the potential energy U and kinetic energy T of a system:

∆U = ∆T−→ T + U = constant−→ Tmax = Umax

The kinetic and potential energy are always exchanged to one another, causing the oscillationaround the equilibrium position. Consider the suspended mass-spring system of figure 33:

k

m

x

0∆

Figure 33: A suspended mass-spring system

In this system:

Uspring = 12k (∆ + x)2

Ugravity = −mgx

T = 12mx

2

Now:T + U = constant = 1

2k (∆ + x)2 −mgx+ 12mx

2 (B.1)

Taking the derivative to time yields:

k (∆ + x) x−mgx+mxx = 0−→ (k∆−mg)︸︷︷︸

=0

x+ (mx+ kx)︸︷︷︸=0

x = 0

where in the last equation, coefficients have been grouped according to their dependency onx. To find ωn, a solution of the form x = A sin (ωt+ ϕ) is suggested. In this case:

Tmax = 12m (ωnA)2 = Umax = 1

2kA2

−→ ωn =

√k

m


B Energy methods 33

B-2 Euler-Lagrange equation

The Euler-Lagrange equation can be used to find equations of motion in a very effective way,and reads the following:

ddt

(∂L

∂xi

)− ∂L

∂xi+ ∂D

∂xi= Q∗i

The Lagrangian L is equal to the kinetic energy minus the potential energy: L = T − P ,D is Rayleigh’s Dissipation Function, and Q∗i is a generalised force that is not derivablefrom a potential or dissipation function, else it would have been included in one of the otherterms. The potential energy P consists of internal potential energy (U) like strain energy andexternal potential energy (V ), for example due to gravity. For the system in figure 33, thereis no dissipation or any applied force, so the Euler-Lagrange equation becomes:

ddt

(∂L

∂xi

)− ∂L

∂xi= 0

Using the previously derived kinetic and potential energy for the system under consideration,the Lagrangian becomes:

L = T − P = 12k (∆ + x)2 −

(−mgx+ 1

2mx2)

Entering this in the Euler-Lagrange equation gives:

ddt (mx) + k (∆ + x)−mg = 0

−→ mx+ kx = mg − k∆


34 C Coulomb damping

C Coulomb damping

Coulomb damping is based on friction instead of the viscous damping as treated e.g. in section4, see figure 34.

FS

N

m µ

Fd*m

x

Figure 34: A mass-spring system with friction, the FBD is shown below

The amount of damping is now determined by the friction coefficient µ. The damping forcedepends on the amount of normal force on the surface: Fd = µ ·N . The damping coefficientusually differs for moving or static cases. When there is no movement, friction is governedby the static friction coefficient µs, which is normally higher than the (dynamic) frictioncoefficient µ. For now, however, the friction coefficient is assumed to have only a singleconstant value. The equation of motion for the situation in figure 34 consists of two cases:Suppose the following initial conditions: x(0) = x0, x(0) = 0.

1 : x > 0 : FS Fdm

xmx+ kx = −Fd

2 : x < 0 : FS Fdm

xmx+ kx = Fd

Phase 1: x < 0 −→ mx+ kx = FdThe motion becomes the following:

x = A sin (ωnt+ ϕ) + Fdk

and the initial conditions can be written as:

x0 = A sin(ϕ) + Fdk

0 = Aωn cos(ϕ) −→ ϕ = π

2 + 2kπ

So the amplitude becomes:

x0 = A+ Fdk

−→ A = x0 −Fdk


C Coulomb damping 35

and thus the displacement x(t) can be written as:

x =(x0 −

Fdk

)sin(ωnt+ π

2

)+ Fd

k

=(x0 −

Fdk

)cos (ωnt) + Fd

k

This solution is only valid for 0 < t < t1, of which t1 is the point in time where x changessign. To find t1 we have to find the first instance where x = 0:

x = −ωn(x0 −

Fdk

)sin (ωnt) = 0

so t1 = πωn

, at which x = 2Fdk − x0 = x1, which is a new initial condition for the following

phase.

Phase 2: x > 0 −→ mx+ kx = −FdNow the sign of the damping force changes:

x = A sin (ωnt+ ϕ)− Fdk

Applying the initial condition: x(πωn

)= x1:

x1 = 2Fdk− x0 = A sin (π + ϕ)− Fd

k

For the velocity we can deduce the phase shift, because x is again zero at t = t1:

x = ωnA cos(π + ϕ) = 0 −→ ϕ = −π2 + 2kπ

and thus the amplitude can be derived:

3Fdk− x0 = A

making the displacement for t1 < t < t2:

x =(x0 −

3Fdk

)cos(ωnt)−

Fdk

To calculate the time t2 at which the velocity is zero again, we equate the derivative to zeroagain:

0 = ωn

(3Fdk− x0

)sin(ωnt) = 0 −→ t2 = 2π

ωn

Finally: x(t2) = x0 − 4Fdk . The displacement can be seen in figure 35. The motion stops as

soon as the amplitude becomes low enough for the spring force not to overcome the frictionalforce.


36 C Coulomb damping

t

Figure 35: The displacement of a mass-spring system with coulomb damping


D Logarithmic decrement 37

D Logarithmic decrement

Measure x at time t and time t+ T with T = 2πωd. Then the logarithmic decrement is defined

as:δ = ln

(x(t)

x(t+ T )

)So for a generic damped displacement x = Ae−ζωnt sin (ωdt+ ϕ):

δ = ln

Ae−ζωnt sin (ωdt+ ϕ)

Ae−ζωn(t+T ) sin

ωdt+ ωdT︸︷︷︸2π

+ϕ

Hence:

δ = ln(eζωnT

)= ζωnT

with T = 2πωd, so:

δ = ζ��ωn2π

��ωn√

1− ζ2 = 2πζ√1− ζ2

−→ ζ = δ√4π2 + δ2


38 E Base excitation

E Base excitation

mFS

x2

x1

m

Figure 36: A base-excited mass-spring system and the corresponding free-body diagram

The displacement of the mass spring system in figure 36 is as follows:

x2 = A2 cos(ωxt)

where, in this case, ωx stands for the frequency of the base excitation. The spring force inthe mass-spring system depends on both x1 and x2, where logically the spring force increaseswhen the spring increases in length:

Fs = k(x2 − x1)

The equation of motion thus reads:

mx1 + kx1 = kx2

or:mx1 + kx1 = kA2 cos(ωxt)

The amplitude Ap2 of the particular solution to this differential equation is:

Ap2 = ω2nx2

ω2n − ω2

x

= x2

1− (ωx/ωn)2


F Laplace transforms 39

F Laplace transforms

The definition of the Laplace transform L of a function f(x) is the following:

F (s) =∞∫0

f(t)e−stdt

= L (f(t))

The first and second derivative are thus as given below:

L(f(t)

)=∞∫0

f(t)e−stdt

=[f(t)e−st

]∞0

+ s

∞∫0

f(t)e−stdt

= −f(0) + sF (s)

L(f(t)

)=∞∫0

f(t)e−stdt

=[f(t)e−st

]∞0

+ s

∞∫0

f(t)e−stdt

= −f(0)− sf(0) + s2F (s)

and also:L (mx+ cx+ kx) = m

(s2X − sx0 − x0

)+ c (sX − x0) + kX


40 G General solution strategy

G General solution strategy

This section shows the solution steps to be taken that normally solve a generic problem. Asan example, the structure in figure 37 will be used.

+

m

k1

k2c 1

c 2

ab

L

Figure 37: An example of a vibrations problem

G-1 Step 1: DOFs

Define the independent degrees of freedom and set positive directions. In this case, there isonly one degree of freedom, θ, and a clockwise rotation is set as positive, as indicated alreadyin figure 37.

G-2 Step 2: Free-body diagram

Draw a free-body diagram in which all parts that were originally connected are separated anddefine the internal and reaction forces, see figure 38.Hint: if you cut something, like a spring, draw the resulting reaction forces as if they wantto bring the two parts back together, i.e. define tensile as positive. Moreover, draw thedisplacements belonging to each internal load and define a positive direction (x1, x2 and x3in figure 38).


G General solution strategy 41

m

x2x1 x3

FS1

FS2Fd1

Fd2RA

Fm

+ + +

Figure 38: Free-body diagram of figure 37

G-3 Step 3: Equation of motion

Set up the equation(s) of motion, describing the acceleration of each mass as a result of theforces that act on it. In this case there is one EOM:

mx3 = Fd2 − Fm

G-4 Step 4: Constitutive relations

The next step is to construct the constitutive relations, i.e. the connection between forcesand (derivatives of) displacements, and enter them into the equation of motion. Since in thiscase there are four parts that cause a load on the structure due to displacements (two springsand two dampers), four constitutive relations are to be formulated:

Fd2 = c2 (x(t)− x3)FS1 = −k1x1

Fd1 = c1x1

FS2 = k2x2

Hint: When setting up constitutive relations, be sure to check for a positive displacementwhether the force in the spring/damper becomes positive or negative.

G-5 Step 5: Kinematic relations

Write the displacements defined in the FBD as a function of the degrees of freedom of thesystem. These equations are called kinematic relations and there should be one for each


42 G General solution strategy

displacement defined in the FBD. In this case, they are (assuming small angles):

x1 = θa

x2 = θb

x3 = θL

G-6 Step 6: Force and moment equilibrium

To write Fm as a function of the displacements, force and moment equilibrium around A isto be satisfied. The moment is taken positive in positive θ-direction, so:

FS1a− Fd1a− FS2b+ FmL = 0

−→ Fm = −FS1a

L+ Fd1

a

L+ FS2

b

L

= k1x1a

L+ c1x1

a

L+ k2x2

b

L

G-7 Step 7: Combining information into one relation

The next step is to combine the four types (i.e. motion, constitutive, kinematic and equilib-rium equations) into one relation. Starting from the basic EOM in step 3 and replacing Fd2

and Fm by their equations yields:

mx3 = c2 (x(t)− x3)− k1a

Lx1 − c1

a

Lx1 − k2

b

Lx2

Entering now the kinematic relations into the EOM as well gives:

mθL = c2(x(t)− θL

)− k1

a2

Lθ − c1

a2

Lθ − k2

b2

Lθ

or, upon rearranging and multiplying by L for convenience:

mL2︸︷︷︸m∗

θ +(c2L

2 + c1a2)

︸︷︷︸c∗

θ +(k1a

2 + k2b2)

︸︷︷︸k∗

θ = c2Lx(t)

where m∗ is the equivalent mass, c∗ the equivalent damping and k∗ the equivalent stiffness ofthe system.

G-8 Step 8: Solve for the motion

As soon as this point is reached, the problem becomes purely mathematical. The aim is touse any method suitable to find the solution for the motion, in this case that is θ(t). Identifythe type of equation and check whether it is consistent with the problem. In this case theequation describes a forced damped motion because of the θ and forcing term, which seemscorrect because figure 37 shows an applied load and dampers. Now the suitable solutionmethod can be applied, which in this case is described in section 6.


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