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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