Aerospace Structures & Computational Mechanics

Lecture NotesVersion 1.6

AE21

35-I I

-Vib

ratio

ns

Free Vibrations (Undamped)

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