Page 261 - Aircraft Stuctures for Engineering Student

242 Airworthiness and airframe loads

whence
N = 124.6 kN
Now resolving forces perpendicular to the axis of the fuselage
S - rnlusin 10" - 4.5~0s 10" = 0
i.e.

S - 13.5 sin lo" - 4.5 cos 10" = 0
so that
S = 6.8kN
Note that, in addition to the axial load and shear load at the section AA, there will
also be a bending moment.
Finally, from elementary dynamics
v2 = vi + 2as

where vo is the touchdown speed, v the final speed (= 0) and s the length of deck
covered. Then
2
210 = -2us
i.e.
252 = -2(-3 x 9.81)s
which gives
s = 10.6m

Example 8.2
An aircraft having a weight of 250 kN and a tricycle undercarriage lands at a vertical
velocity of 3.7m/s, such that the vertical and horizontal reactions on the main wheels
are 1200 kN and 400 kN respectively; at this instant the nose wheel is 1 .Om from the
ground, as shown in Fig. 8.7. If the moment of inertia of the aircraft about its CG is
5.65 x lo8 N s2 mm determine the inertia forces on the aircraft, the time taken for its
vertical velocity to become zero and its angular velocity at this instant.

Nose wheel

1.0m @ 400kN
' I 25OkNv Tl200kN
r
I. 5.0m 4 /10m

Fig. 8.7 Geometry of the aircraft of Example 8.2.