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How do I calculate the lift force that the the bolt that join the wings to the fuselage of a strutted or braced airplane carry?

I know in a cantilever airplane,the bolts have to carry all of the lift load on a wing but how about a strutted airplane - how do the wing attach bolts share the lift with the wing braces or struts, i.e. is the lift load shared between them, the bolts carrying half and the bracing wires or struts carrying the other half?

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    $\begingroup$ The wings carry themselves, any engines attached to them and the fuel inside. Only the remaining load e.g. the weight of the fuselage is carried through the connection of the fuselage and wing. $\endgroup$ – Jan Jun 13 at 13:12
  • $\begingroup$ This only applies to aircraft with conventional spar designs. Some aircraft, like the homebuilt BD-5, have a spar that slides over a center carry-through spar and uses a small taper bolt to prevent movement in torsion. Works well as long you remember to inspect the hardware on a regular basis. $\endgroup$ – Juan Jimenez Jun 13 at 14:40
  • $\begingroup$ On an aircraft with flexible wire-based wings and significant washout, I've seen the lower wires go completely slack during high-speed flight, due to the torque generated by the downlifting wingtips during high-speed flight. Food for thought when formulating the answers-- $\endgroup$ – quiet flyer Jun 13 at 19:27
  • $\begingroup$ Instead of the lift loads being carried by the bolts you might consider bolted flanges. You basicly make this part stronger than the wing and spar, not weaker. Like Peter K said, it is also important to know the strength properties of the components of the structure. Find the weakest point. That is where improvement is possible, provided weight increase is not too extreme. $\endgroup$ – Robert DiGiovanni Jun 15 at 0:42
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The answer depends on the elastic deformation of wing and strut. Extreme example: Make the strut a rubber band and all lift needs to be carried by the wing and its fittings. Other extreme: Make the wing spar (or attachment points - this works either way) from rubber - now the strut needs to take all the loads.

In real life, stiffness attracts tension and compression, and in order to know the load distribution between wing spar and strut you need to look at them with a slight deformation from loading. How much resistance does each show (in isolation) against that deformation? This will tell you in what ratio the forces will split between wing and strut. Make sure to use the stiffness of the real structure - the material property called Young's modulus is not what I mean here.

If you want to be safe, you need to consider any single point failure and still have a flyable airplane. Normally, you design your structure with a safety factor (usually 1.5, but higher for fittings). Now assume that in turn each of the structural members is missing and calculate the loads on the remaining structure with a safety factor of 1.0. Dimension each part at least for those loads and you should be pretty sure that the intact structure will work out fine.

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    $\begingroup$ Anonymous down voter: If you don't understand structural mechanics please ask in the comments or your own question. Sprinkling downvotes on answers you don't understand isn't helping anybody. $\endgroup$ – Peter Kämpf Jun 13 at 18:40
  • $\begingroup$ Deformation is caused by the forces acting on the structural members. Determine forces first, then deformation follows from the material properties. I wasn't the downvoted, by the way. $\endgroup$ – Koyovis Jun 14 at 8:08
  • $\begingroup$ @Koyovis And I didn't downvote your answer, either. The deformation is to determine stiffness of the part. The ratio in stiffness is inverse to the ratio of the forces in an overdetermined structure. The forces first approach only works if there are enough degrees of freedom (Cremona plan etc.). $\endgroup$ – Peter Kämpf Jun 14 at 10:32
  • $\begingroup$ The diagonal strut transforms the inner wing construction to a kinematically and statically determined system, with the triangle experiencing compression and tension forces. Within the triangle, both upper and lower skin experience stress in the same direction (tension or compression). Only outside the strut is the wing still a cantilever beam, with upper skin under compression and lower skin under tension. Point is: adding the strut does not create a statically overdetermined structure, which your answer addresses. $\endgroup$ – Koyovis Jun 19 at 7:53
  • $\begingroup$ @Koyovis: Depends entirely on the wing root mount. If that is a hinge, you are correct. If it has a central spar running all the span (which is how it is mostly done), it is indeed an overdetermined structure. $\endgroup$ – Peter Kämpf Jun 19 at 9:07
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In flight, the fuselage hangs off of the wing. For dimensioning purposes, consider:

  • the intersection points to exert no moments and to behave as hinges;
  • weight W of the fuselage to be concentrated in the Centre of Gravity;
  • lift of each wing to be concentrated in its Centre of Lift.

enter image description here

If we dimension the construction in this way, we over-dimension which is never a bad idea with primary construction bolts. In reality the following factors alleviate the loads:

  • The bolts are modelled as hinges which cannot exert a moment, but actually they do exert torque.
  • The wing lift is distributed loading, with most of it near the wing root.

Bolts 1, 2 and 3 experience lift and gravity forces from the construction, and exert equal and opposite forces in order for everything to stay in one piece. Fuselage weight is transferred to bar 2-3 which distributes the load evenly over bolts 2 and 3. Remove the bolts and the fuselage falls from the assembly.

Force equilibrium in point 1 from this answer:

  • $F_{13} \cdot sinψ = ½L => F_{13} = \frac{L}{2sinψ}$
  • $F_{12} = F_{13} \cdot cosψ = \frac{L}{2tanψ}$

In point 2, bolt reaction force in green:

enter image description here * $F_{V} = ¼W$ $\tag{Vertical}$ * $F_{H} = \frac{L}{2tanψ}$ $\tag{Horizontal}$

In point 3, bolt reaction force in green:

enter image description here * $F_{V} = ¼W - F_{13} \cdot sinψ = ¼W - ½L$ $\tag{Vertical}$ * $F_{H} = \frac{L}{2tanψ}$ $\tag{Horizontal}$

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  • $\begingroup$ The forces on the strut are not in equilibrium $\endgroup$ – Gypaets Jun 14 at 8:26
  • $\begingroup$ @Gypaets Strut 2 - 3 can be removed, all forces to be absorbed by the bolt supports. $\endgroup$ – Koyovis Jun 14 at 8:29
  • $\begingroup$ I mean the vertical forces on strut 1-3. Vertical reaction force on the upper hinge (2) should also be 0. $\endgroup$ – Gypaets Jun 14 at 8:30
  • $\begingroup$ This appears to ignore that a portion of created lift must lifts the wing itself, not the fuselage. $\endgroup$ – J Walters Jun 14 at 12:44
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    $\begingroup$ +1 just for the picture. Try making a bookshelf without the "strut". Shown here is the positive G case. The bolts 1 and 3 would have to be torn from their mountings. $\endgroup$ – Robert DiGiovanni Jun 15 at 0:15

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