1. Jun 10, 2014

### laurens

I try to build a loading diagram in the lateral direction of an aircraft, however have not succeeded yet.

For example, in the z-direction (direction where weights and lift of the aircraft point). I have made a simple loading diagram. I first assumed a simple beam, which is hinged at the wing and the horizontal tail surface. Then adding all the weights to the aircraft, I could balance them when I assumed that the aircraft was flying in a straight and level flight. An example of this can be seen in the attached files.

Now this was assumed to be just a static case. However now I want to calculate the loads in lateral direction, for example when a gust hits the vertical fin. In the attached files I put a top-down view of such a load case. In this picture I also drew the shear force and moment, as I though they would act on the aircraft. Since there is no static equilibrium anymore I first assumed I could maybe simplify it as this. However when calculating with "real" values, I got incredibly high stresses at the nose of the aircraft.

So although this is not a static problem, is there any possibility to draw a shear force and moment diagram as I would have done in the static case?

Can I maybe assumed that I cut somewhere in the tail and then assume from the cut to the rear of the fuselage as beam. Which is then clamped at the cut? As an example I drew the third picture. So that in the rest of the fuselage the bending moment is constant? Or is this really not allowed to use as a first estimation of the fuselage loading? If this is not allowed, maybe someone has a suggestion to build such loading diagrams?

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2. Jun 10, 2014

### AlephZero

The correct way to do this is to treat it as a dynamics problem. Restraining an arbitrary point on the structure will give the wrong answers, if there are any reaction forces at the restraint.

If you apply a side force to the tail, there will be a sideways acceleration of the center of mass of the aircraft, and an angular acceleration about the COM. If you assume the aircraft is rigid, you can find the corresponding inertia forces (mass x acceleration) distributed along the length of the fuselage. Now, if you apply MINUS those forces to the structure as well as the force on the tail fin, the resultant load on the whole aircraft is zero, and you can restrain it anywhere convenient without getting a reaction force.

That is equivalent to working in a coordinate system fixed to the aircraft. The "fictitious inertia forces" are caused by the fact that the coordinate system is accelerating.

You can find more on this by googling for "inertia relief".

3. Jun 11, 2014

### laurens

Thanks for the quick answer AlpehZero, it seems pretty logical how you explained it. I think I can make it work now :).

4. Jun 11, 2014

### laurens

Okay, so I tried a to make a sample calculation, but I have the feeling somewhere it goes horribly wrong. The resultant force on the aircraft is not zero. But when calculating the acceleration with the final resultant force I also get an acceleration a factor 10x smaller than with the initial external force. Could someone maybe take a look at the sample calculation? As provided in the attachments?

EDIT: The red bars are basically the masses, so it is not the "white space" which is 10kg, but the red bar.

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Last edited: Jun 11, 2014
5. Jun 16, 2014

### laurens

Well for those wondering (including me), I found the error. The sideways acceleration is 60 m/s^2 and not 66.667 m/s^2, which was a calculation mistake. Using 60 m/s^2, the resultant load leads to zero again, as AlephZero explained.