Page 259 - Aircraft Stuctures for Engineering Student

240 Airworthiness and airframe loads

s
and
axdm=-a xdm

for a in the direction shown. Then, as before
F, = aJm
and
Fy = aXm

Also, if the CG lies on the x axis, J = 0 and Fx = 0. Similarly, if the CG lies on the y
axis, X = 0 and Fy = 0.
The torque about the axis of rotation produced by the inertia force corresponding
to the angular acceleration on the element Sm is given by
ST^ = a46m

Thus, for the complete mass
s s
To= ar2dm=a 4dm

The integral term in this expression is the moment of inertia, Io, of the mass about the
axis of rotation. Thus

To = a10 (8.5)
Equation (8.5) may be rewritten in terms of ICG, the moment of inertia of the mass
about an axis perpendicular to the plane of the mass through the CG. Hence, using
the parallel axes theorem

10 = m(TI2 + ICG
where F is the distance between 0 and the CG. Then

10 = m[(jsl2 + (J)'] + ICG
and
+
To = m[(X)' + (J)2]a ICGQ (8.6)
Example 8. I
An aircraft having a total weight of 45 kN lands on the deck of an aircraft carrier and
is brought to rest by means of a cable engaged by an arrester hook, as shown in
Fig. 8.5. If the deceleration induced by the cable is 3g determine the tension, T, in
the cable, the load on an undercarriage strut and the shear and axial loads in the
fuselage at the section AA; the weight of the aircraft aft of AA is 4.5 kN. Calculate
also the length of deck covered by the aircraft before it is brought to rest if the touch-
down speed is 25 m/s.

The aircraft is subjected to a horizontal inertia force ma where m is the mass of the
aircraft and a its deceleration. Thus, resolving forces horizontally
T cos IO" - ma = 0