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I am quite confused with the definition of Load factor and how it applies to the wing structures of an aircraft.

For example, you have an Aircraft being subjected to a Load Factor of 2 and whose total mass is 5000kg and wings + fuel mass is 1500kg, and are asked to calculate the Inertia Loading of Said wings.

Would you multiply the weight of the wings by 2? Or does Load factor only affect the lift and not the weight?

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  • $\begingroup$ If it affects the weight it should affect the lift as well, right? For the flight condition to remain stationary. It sounds like you are almost at the answer. $\endgroup$ – AEhere May 2 at 11:16
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    $\begingroup$ Related: aviation.stackexchange.com/questions/46287/… $\endgroup$ – AEhere May 2 at 11:18
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https://en.m.wikipedia.org/wiki/Load_factor

As a start for an aircraft the load factor is the ratio of total lift to total Weight

An easy way to imagine this concept is to understand the idea that when the aircraft is flying at high speed during cruise with a small angle of attack the load factor is very close to 1, this corresponds to the first case on the left of the picture here below

Now what is lift, the picture here below is self explanatory, so it is important to understand that the lift is not always an upward vertical force; when the aircraft is under a coordinated bank angle the lift has two components ; horizontal and vertical.

enter image description here

Now to simplify we shall consider the lift to be totally produced by the wings, while in real flight it is the result of the wing, the body, and the tail.

Since the Aircraft is subjected to a load factor of 2 this means the aircraft as a complete body is subjected to a total lift( not vertical) equal to twice the total weight of the Aircraft

The lift produced is therefore: $2\cdot(5000 + 1500)\,\mathrm{kg}\cdot g = 13000\,\mathrm{kg}\cdot g = 128\,\mathrm{kN}$

If we consider we are in the steep banked turn indicated on the right of the picture above:

The weight of the aircraft body without the wings is a downward force on the root of the wings

Also the wings have their own distributed weights, that produces a torque to the root of the wings

Also the distributed fuel in the wings produces a torque to the wings roots

Also each wing is subject to a distributed lift of $\ 6500\,\mathrm{kg}\cdot g = 63.8\,\mathrm{kN}$

The exact computation are difficult because

  • we don’t know the bare weight of the wing nor the centre of gravity of the wing

  • we don’t know the distribution of the fuel in the wing, so we don’t know the centre of gravity of the fuel of each wing.

Therefore it is impossible to calculation the shear forces and the torques at the wings roots.

Nevertheless I imagine the main question concerns the relationship between lift and load factor, And here the most important point is to understand that the lift is not always a vertical force

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  • $\begingroup$ It appears that this answer doesn't actually answer the main question being asked which was "For example, you have an Aircraft being subjected to a Load Factor of 2... and are asked to calculate the Inertial Loading of Said wings. Would you multiply the weight of the wings by 2?" $\endgroup$ – quiet flyer Jun 24 at 18:42
  • $\begingroup$ +1 for making the effort to use Newtons $\endgroup$ – bogl Jun 24 at 18:42
  • $\begingroup$ @quiet flier, what applies to the body applies to the wings. Indeed the weight is vertically down, but the lift is not vertical, it is 60° apart, so yes any weight is neutralized by an oblique force of double magnitude. $\endgroup$ – user40476 Jun 24 at 19:52
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The load factor is not the lift force. It is the total force vector in the aircraft + Z-axis (pointing downwards), in units [force from earth gravity]. A picture earlier used in this answer:

enter image description here

In a horizontal steady turn of 60°, the resultant vector of gravity and centrifugal force has magnitude 2 * normal gravity force. So due to some physical circumstance, all elements of the aeroplane experience a force twice that of gravity. The wing and the fuel as well.

In order to maintain the steady turn, lift must also be twice that of normal steady flight. So the inertial loading is twice that of gravity, and lift loading is twice that of steady flight.

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For example, you have an Aircraft being subjected to a Load Factor of 2 and whose total mass is 5000kg and wings + fuel mass is 1500kg, and are asked to calculate the Inertia Loading of Said wings.

Would you multiply the weight of the wings by 2?

Yes.

Note that, strictly speaking, kg are not a measure of weight. You would need to multiply the weight of the wings by two.

In more detail:

In the most general terms, "load factor" could be said to be equal total aerodynamic force divided by weight. The usual convention is to use "load factor" to mean lift force divided by weight. The difference between the two results could be very significant in a steep steady-state climb, or in a sideslip.

For the remainder of this answer, "load factor" will be used to mean lift force divided by weight.

As other answers have noted, just knowing the "load factor" is not enough to predict the stresses and strains at the wing root etc-- the load distribution matters as well.

It may help to note the following:

  • "Centrifugal force" is not a real force, in the sense that it is often applied to an analysis of the flight of a turning aircraft.

  • Gravitational force (weight) acts with an equal force per unit mass on every single particle of the aircraft, and thus exerts no stresses or strains on the aircraft structure.

However, it is not uncommon to hear the "inertial loading" acting on the aircraft as a whole being described as the combined vector sum of "centrifugal force" and gravitational force. Whether this is truly "accurate" or "inaccurate" may to some extent be a matter of opinion or reference frame; at any rate the net result is always exactly equal in magnitude (but opposite in direction) to the net aerodynamic force being generated by the aircraft. In the more narrow case where all aerodynamic forces other than lift add up to zero, the sum of "centrifugal force" and gravitational force will be equal in magnitude and opposite in direction to the lift vector. In other words, in the case where the net aerodynamic force is equal to the lift vector, the "inertial loading" acting on the aircraft as a whole is exactly equal in magnitude to the lift vector, which is exactly equal in magnitude to the "load factor" times the aircraft weight.

(Note that this conception of "inertial loading" is a little limited-- it considers effects related to turning but not effects related to rotation-- the difference will rarely be significant, except in unusual cases like a flat spin with a high rotation rate-- these nuances only apply to a discussion of the inertial load acting on a given element of the aircraft, not on the aircraft as a whole.)

It would seem that the original question is using the phrase "inertial loading of the wings" to describe the portion of the wing's lift force that is absorbed by accelerating the mass of each wing, and therefore does not contribute to the upward bending moment acting at the wing root. This would be equal to the weight of each wing, times the "load factor", with the "load factor" being defined as total lift force divided by total weight.

Since we don't know that the center of lift of each wing is the same as the center of mass of each wing, it's only a rough approximation to say that the lift "absorbed" by the "inertial loading" of each wing, calculated as we've stated above, makes no contribution to the bending moment at the wing roots.

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  • $\begingroup$ An alternate definition of "inertial loading" could construe it to only encompass the ADDITIONAL apparent weight caused by maneuvering. This would seem to be an inferior definition and also would seem to be not the usual one, but ultimately it just comes down to a matter of convention. $\endgroup$ – quiet flyer Jun 24 at 18:14
  • $\begingroup$ Related-- sort of-- physicsforums.com/threads/what-is-an-inertia-load.771167 $\endgroup$ – quiet flyer Jun 24 at 18:17
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As a start for an aircraft the load factor is the ratio of total lift to total weight

Since the Aircraft is subjected to a load factor of 2 this means the aircraft as a complete body is subjected to a total lift equal to twice the total weight of the Aircraft

The lift produced is therefore 2x(5000 + 1500)= 13000kg

The downward force that opposes the lift is the total weight of the aircraft that is 6500kg

The net upward force on the wings is 13000-6500=6500kg

To appreciate intuitively what is load factor, please note that when the load factor is equal to « 1 » the aircraft is not subject to any vertical acceleration, it’s vario is constant

An easy way to imagine this concept is to understand the idea that when the aircraft is standing in the runway or parked, it's load factor value is 1.

Note: For completeness please note the answer is dealing with forces therefore all above Kg units refer to kilogram-force, otherwise you may multiply all the results by « g » to obtain Newton’s.

On the other hand the notion of load factor applies too to satellites therefore « g » should be theoretically considered as a variable function of altitude.

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    $\begingroup$ The unit of force is N, not kg! $\endgroup$ – bogl Jun 19 at 11:06
  • $\begingroup$ Yes indeed, I meant kilogram-force, and I preferred to use kilogram-force rather than multiplying every thing by g because of the definition of the simplified definition of load factor which refers to the aircraft weight (even though the weight in this definition represents the vertical force produced by the weight), I will add a note for completeness $\endgroup$ – user40476 Jun 19 at 12:15
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    $\begingroup$ No please, there is no such thing as kilogram-force! If you want to be lazy, at least use the word "equivalent" when mixing forces with masses. (E.g. "The net upward force on the wings is equivalent to 6500 kg.") $\endgroup$ – bogl Jun 19 at 13:57
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    $\begingroup$ Did you read the part where it says "Kilogram-force is a non-standard unit and is classified in SI Metric System as a unit that is unacceptable for use with SI.[4]"? $\endgroup$ – bogl Jun 19 at 14:33
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    $\begingroup$ Your force units are still wrong though. $\endgroup$ – AEhere Jun 19 at 17:00

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