Aerospace Structures & Computational Mechanics
Lecture NotesVersion 2.01
AE21
35-II
-Vib
ratio
ns
Copyright c© Aerospace Structures and Materials (ASM)All rights reserved.
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.
AE2135-II - Vibrations Lecture Notes
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
Lecture Notes AE2135-II - Vibrations
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:
AE2135-II - Vibrations Lecture Notes
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
Lecture Notes AE2135-II - Vibrations
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
AE2135-II - Vibrations Lecture Notes
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
Lecture Notes AE2135-II - Vibrations
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
AE2135-II - Vibrations Lecture Notes
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
)
Lecture Notes AE2135-II - Vibrations
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
AE2135-II - Vibrations Lecture Notes
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
Lecture Notes AE2135-II - Vibrations
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
AE2135-II - Vibrations Lecture Notes
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
Lecture Notes AE2135-II - Vibrations
4 Free damped vibrations 13
t
x
Figure 17: A critically damped motion with x(0) = 1 and x(0) = −1
AE2135-II - Vibrations Lecture Notes
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.
Lecture Notes AE2135-II - Vibrations
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
AE2135-II - Vibrations Lecture Notes
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
Lecture Notes AE2135-II - Vibrations
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.
AE2135-II - Vibrations Lecture Notes
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)
Lecture Notes AE2135-II - Vibrations
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
)
AE2135-II - Vibrations Lecture Notes
20 7 Response to arbitrary excitation
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
Lecture Notes AE2135-II - Vibrations
7 Response to arbitrary excitation 21
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:
AE2135-II - Vibrations Lecture Notes
22 7 Response to arbitrary excitation
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
Lecture Notes AE2135-II - Vibrations
7 Response to arbitrary excitation 23
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
AE2135-II - Vibrations Lecture Notes
24 7 Response to arbitrary excitation
t
Figure 26: An arbitrary load can also be seen as a collection of steps
Lecture Notes AE2135-II - Vibrations
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.
AE2135-II - Vibrations Lecture Notes
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
Lecture Notes AE2135-II - Vibrations
8 Multiple degree of freedom systems 27
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
AE2135-II - Vibrations Lecture Notes
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.
Lecture Notes AE2135-II - Vibrations
8 Multiple degree of freedom systems 29
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
AE2135-II - Vibrations Lecture Notes
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)
AE2135-II - Vibrations Lecture Notes
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
Lecture Notes AE2135-II - Vibrations
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∆
AE2135-II - Vibrations Lecture Notes
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
Lecture Notes AE2135-II - Vibrations
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.
AE2135-II - Vibrations Lecture Notes
36 C Coulomb damping
t
Figure 35: The displacement of a mass-spring system with coulomb damping
Lecture Notes AE2135-II - Vibrations
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
AE2135-II - Vibrations Lecture Notes
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
Lecture Notes AE2135-II - Vibrations
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
AE2135-II - Vibrations Lecture Notes
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).
Lecture Notes AE2135-II - Vibrations
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
AE2135-II - Vibrations Lecture Notes
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.
Lecture Notes AE2135-II - Vibrations
Top Related