Page 575 - Aircraft Stuctures for Engineering Student

556 Elementary aeroelasticity

For a non-trivial solution

(1 - 16d) -15Xw’
-( 1 - 6XJ)
Expanding this determinant we have
-(1 - 16Xw2)(1 - 6XJ) + 75(Xw2)’ = 0
or
21(AJ)’ - 22xw2 + 1 = 0

Inspection of Eq. (ix) shows that
xw2= 1/21 or 1
Hence
2 3 x48EI 3 x 48EI
w= or
21m13 m13
The normal or natural frequencies of vibration are therefore

h=g-; w2-6F
m13

The system is therefore capable of vibrating at two distinct frequencies. To determine
the normal mode corresponding to each frequency we first take the lower frequencyfi
and substitute it in either Eq. (vii) or Eq. (viii). From Eq. (vii)

- 1 5Xw2 - 15 x (1/21)
-
-
~2 1 - 16Xw2 - 1 - 16 x (1/21)
which is a positive quantity. Therefore, at the lowest natural frequency the cantilever
oscillates in such a way that the displacement of both masses has the same sign at the
same instant of time. Such an oscillation would take the form shown in Fig. 13.10.
Substituting the second natural frequency in Eq. (vii) we have

v1 15Xw’ 15
-= -
--
~2 1 - 16h2 1- 16
which is negative so that the masses have displacements of opposite sign at any instant
of time as shown in Fig. 13.11.

Fig. 13.10 The first natural mode of the massheam system of Fig. 13.9.