lecture notes ae2135 ii vibrations - appendices

Aerospace Structures & Computational Mechanics

Lecture NotesVersion 1.6

AE21

35-II

-Vib

ratio

ns

Appendices

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∆


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

k 2c 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.


lecture notes ae2135 ii vibrations - appendices

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