Aerospace Structures & Computational Mechanics

Lecture NotesVersion 1.6

AE21

35-I I

-Vib

ratio

ns

Discretisation

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 w2k 1 k 2 k 3

/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