Aerospace Structures & Computational Mechanics

Lecture NotesVersion 1.6

AE21

35-I I

-Vib

ratio

ns

Multiple Degree of Freedom Systems

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

m1k 1 k 2m2

Figure 27: A wing with two masses

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

x2

x1

m1k 1

k 2 m2

Figure 28: Discretising the wing

x2x1

m1k 1k 2

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

m1k 1k 2

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