Page 579 - Aircraft Stuctures for Engineering Student

560 Elementary aeroelasticity

Inserting the values of m, r, I and EI we have
1435 x 4 x 0.763 1435 x 0.1522 x 3 x 0.762
w2e = o
(l -9.81 x 3 x 1.44 x lo6 w2)v- 9.81 x 2 x 1.44 x lo6 (viii)
1435 x 3 x 0.762 1435 x 0.1522 x 2 x 0.76
- 8 = 0 (ix)
9.81 x 2 x 1.44 x lo6 "+ (l - 9.81 x 1.44 x lo6 w2)
or

(1 - 6 x 10-5w2)v - 0.203 x 10-5w28 = 0 (4
-8.8 x lO-'w%+ (1 - 0.36 x 10-5w2)8 = 0 (4
Solving Eqs (x) and (xi) as before gives
w= 122 or 1300
from which the natural frequencies are

61 650
h=--, h=-
7r 7r
From Eq. (x)
v - 0.203 x lOP5w2
-
8 - 1 - 6 x 10-5w2
which is positive at the lowest natural frequency, corresponding to w = 122, and
negative for w = 1300. The modes of vibration are therefore as shown in Fig. 13.14.
So far we have restricted our discussion to weightless beams supporting concen-
trated, or otherwise, masses. We shall now investigate methods of determining
normal modes and frequencies of vibration of beams possessing weight and therefore
inertia. The equations of motion of such beams are derived on the assumption that
vibration occurs in one of the principal planes of the beam and that the effects of
rotary inertia and shear displacements may be neglected.
Figure 13.15(a) shows a uniform beam of cross-sectional area A vibrating in a
principal plane about some axis Oz. The displacement of an element 6z of the
beam at any instant of time t is v and the moments and forces acting on the element
are shown in Fig. 13.15(b). Taking moments about the vertical centre line of the

Fig. 13.14 The first two natural modes of vibration of the beadmass system of Fig. 13.13.