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. 13.3. If the deceleration induced by the cable is 3g, determine the tension, T, in the cable, the load on an undercarriage

Question

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. 13.3. 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.

Accepted Answer

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

[latex]T \cos 10^{\circ}-m a=0[/latex]

that is,

[latex]T \cos 10^{\circ}-\frac{45}{g} 3 g=0[/latex]

which gives

[latex]T=137.1 kN[/latex]

Now, resolving forces vertically,

[latex]R-W-T \sin 10^{\circ}=0[/latex]

that is,

[latex]R=45+137.1 \sin 10^{\circ}=68.8 kN[/latex]

Assuming two undercarriage struts, the load in each strut is [latex](R / 2) / \cos 20^{\circ}=36.6 kN[/latex]. Let N and S be the axial and shear loads at the section AA, as shown in Fig. 13.4. The inertia load acting at the CG of the fuselage aft of AA is [latex]m_{1} a[/latex], where [latex]m_{1}[/latex] is the mass of the fuselage aft of AA. Then,

[latex]m_{1} a=\frac{4.5}{g} 3 g=13.5 kN[/latex]

Resolving forces parallel to the axis of the fuselage,

[latex]N-T+m_{1} a \cos 10^{\circ}-4.5 \sin 10^{\circ}=0[/latex]

that is,

[latex]N-137.1+13.5 \cos 10^{\circ}-4.5 \sin 10^{\circ}=0[/latex]

from which

[latex]N=124.6 kN[/latex]

Now, resolving forces perpendicular to the axis of the fuselage,

[latex]S-m_{1} a \sin 10^{\circ}-4.5 \cos 10^{\circ}=0[/latex]

that is,

[latex]S-13.5 \sin 10^{\circ}-4.5 \cos 10^{\circ}=0[/latex]

so that

[latex]S=6.8 kN[/latex]

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,

[latex]v^{2}=v_{0}^{2}+2 a s[/latex]

where [latex]v_{0}[/latex] is the touch-down speed, v the final speed (= 0), and s the length of deck covered. Then,

[latex]v_{0}^{2}=-2 a s[/latex]

that is,

[latex]25^{2}=-2(-3 imes 9.81) s[/latex]

which gives

[latex]s=10.6 m[/latex]

Question 13.1: An aircraft having a total weight of 45 kN lands on the deck...

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