Page 262 - Aircraft Stuctures for Engineering Student

8.3 Aircraft inertia loads 243

The horizontal and vertical inertia forces ma, and ma, act at the CG, as shown in
Fig. 8.7; pn is the mass of the aircraft and a, and a,, its accelerations in the horizontal
and vertical directions respectively. Then, resolving forces horizontally
ma, - 400 = 0
whence
ma, = 400 kN
Now resolving forces vertically
ma, + 250 - 1200 = 0

which gives
ma, = 950 kN
Then
950 - 950
a, = - - 3.8g
m 250/g
Now taking moments about the CG
ICG~ 1200 x 1.0 - 400 x 2.5 = 0
-
from which

Ic~a 2200 m kN
=
Hence

(iii)

From Eq. (i), the aircraft has a vertical deceleration of 3.8g from an initial vertical
velocity of 3.7m/s. Therefore, from elementary dynamics, the time, f, taken for the
vertical velocity to become zero, is given by
v = vo + a,t (iv)
in which v = 0 and vo = 3.7m/s. Hence
0 = 3.7 - 3.8 x 9.81t

whenc.e
t = 0.099 s
In a similar manner to Eq. (iv) the angular velocity of the aircraft after 0.099 s is
given by

W=Wo+af
in which wo = 0 and a = 3.9 rad/s2. Hence
w = 3.9 x 0.099

i.e.
w = 0.39 rad/sec.