Page 260 - Aircraft Stuctures for Engineering Student

8.3 Aircraft inertia loads 241

A

\", , ..
Wheel Arrester /
reaction R
hook
Fig. 8.5 Forces on the aircraft of Example 8.1.

i.e.

which gives
T = 137.1 kN
Now resolving forces vertically
R- W-TsinlO"=O
i.e.
R = 45 + 131.1 sin 10" = 68.8 kN
Assuming two undercarriage struts, the load in each strut will be (R/2)/cos2Oo =
36.6 kN.
Let N and S be the axial and shear loads at the section AA, as shown in Fig. 8.6.
The inertia load acting at the centre of gravity of the fuselage aft of AA is mla, where
ml is the mass of the fuselage aft of AA. Thus
4.5
mla=-3g= 13.5kN
g
Resolving forces parallel to the axis of the fuselage
N - T + mlacos 10" - 4.5 sin 10" = 0
1.e.
N- 137.1 + 13.5~0~10~-4.5sin1O0=O

4.5 kN
Fig. 8.6 Shear and axial loads at the section AA of the aircraft of Example 8.1.