The maximum loads on the components of an aircraft's structure generally occur when the aircraft is undergoing some form of acceleration or deceleration, such as in landings, take-offs and manoeuvres within the flight and gust envelopes. Thus, before a structural component can be designed, the inertia loads corresponding to these accelerations and decelerations must be calculated. For these purposes we shall suppose that an aircraft is a rigid body and represent it by a rigid mass. in. as shown in Fig. 8.3. We shall also, at this stage, consider motion in the plane of the mass which would correspond to pitching of the aircraft without roll or yaw. We shall also suppose that the centre of gravity (CG) of the mass has coordinates .v, V referred to .v and y axes having an arbitrary origin O; the mass is rotating about an axis through O perpendicular to the at plane with a constant angular velocity uj.

The acceleration of any point, a distance /■ from O, is ui'r and is directed towards O. Thus, the inertia force acting on the clement, dm. is ui'rbm in a direction opposite to the acceleration, as shown in Fig. 8.3. The components of this inertia force, parallel to the a and v axes, are wrbni cos 0 and x'rkm sin 0 respectively, or, in terms of a and y. and The resultant inertia forces. Fy and /•',. are then given by

id in

Fig. 8.3 Inertia forces on a rigid mass having a constant angular velocity.

in which we note that the angular velocity w is constant and may therefore be taken outside the integral sign. In the above expressions j x dm and j y dm are the moments of the mass, m, about the y and x axes respectively, so that

If the CG lies on the x axis, y = 0 and Fv = 0. Similarly, if the CG lies on the v axis, Fx = 0. Clearly, if O coincides with the CG, * = y = 0 and Fx = Fy = 0.

Suppose now that the rigid body is subjected to an angular acceleration (or deceleration) a in addition to the constant angular velocity, w, as shown in Fig. 8.4. An additional inertia force, arSm, acts on the element 6m in a direction perpendicular to r and in the opposite sense to the angular acceleration. This inertia force has components arSm cos 6 and ar6m sin 9, i.e. ax6m and ay6m, in the y and x directions respectively. Thus, the resultant inertia forces, Fx and Fy, are given by

Fig. 8.4 Inertia forces on a rigid mass subjected to an angular acceleration

Fy = — | ax dm = —,a Jjc dm for a in the direction shown. Then, as before

Also, if the CG lies on the x axis, y = 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 8m is given by

Thus, for the complete mass

The integral term in this expression is the moment of inertia, 70, of the mass about the axis of rotation. Thus

Equation (8.5) may be rewritten in terms of 7Cg, 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

IQ = m(rf + /CG where r is the distance between O and the CG. Then

Example 8.1

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 touchdown 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

Fig. 8.5 Forces on the aircraft of Example 8.1.
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