2
$\begingroup$

How does one calculate the compressive force that the inner section of a strutted or braced wing (from the strut attach point to the wing root) experiences?

I know how to calculate the tension in the struts (or flying wires), which is simply

$$ \mathrm{Tension} = \cfrac{\mathrm{Lift}}{\sin \theta} $$

but I don't know to calculate the compressive force that a strutted or braced wing experiences.

enter image description here

$\endgroup$
  • $\begingroup$ Are you sure about that tension equation? I don't think it is correct, as it disregards the cantilevered wing. For $\theta$ approaching $\pi/2$ it would imply that the flying wire would take on near all the lift, which is patently untrue. $\endgroup$ – AEhere supports Monica Jun 13 '19 at 13:27
  • $\begingroup$ Are you really worried about the compressive force on the lower area? I would rather be more worried about the stretching force on the upper area. $\endgroup$ – user40476 Jun 13 '19 at 14:32
  • $\begingroup$ @user40476 buckling can kill a beam in an instant. $\endgroup$ – AEhere supports Monica Jun 13 '19 at 15:16
  • $\begingroup$ @AEhere, correct, but the split or the crack starts on the stretched surface $\endgroup$ – user40476 Jun 13 '19 at 16:29
  • 1
    $\begingroup$ Good thing you included the image, it's much clearer now. Although what's the text left of "strut"? $\endgroup$ – Sanchises Jun 13 '19 at 16:58
1
$\begingroup$

These pictures hopefully will improve the question and serve to validate Sanchises answer.

If I could find a flying wire to take on near all the lift, I would patent it!

enter image description hereenter image description here

| improve this answer | |
$\endgroup$
  • $\begingroup$ Note that I purposefully did not express the compressive forces as a function of lift, but as a function of force in the strut. There can be a coincidental geometry where the vertical force exerted by the strut is equal to lift, but it's definitely not true in general. $\endgroup$ – Sanchises Jun 20 '19 at 17:34
  • $\begingroup$ (also, just placing the strut exactly under the center of lift is hardly patentable, and a waste of bending strength of the wing...) $\endgroup$ – Sanchises Jun 20 '19 at 17:41
  • $\begingroup$ @Sanchises Hey, I can't believe they handed me the check mark either (thanks to you). After reviewing this design I can see why they went cantilever. Would love to see a flitched wooden spar and plywood stressed skinned wing. Now that would strong. $\endgroup$ – Robert DiGiovanni Jun 20 '19 at 19:04
3
$\begingroup$

It's very easy. Consider that the point at which the strut is attached does not accelerate, i.e., the forces in all directions exactly even out.

The force in the spanwise direction exerted by the strut is

$$F_{strut}^{spanwise}=\cos\theta\cdot F_{strut}$$

This force must be balanced by the wing in compression, so the above expression is also valid for the amount of compression on the wing.

This assumes the strut only transfers longitudinal forces, while in reality it may have some bending stiffness. Also, I'm not sure where you got your equation for the tension in the strut, but it's not necessarily true (there's a lot more structural members exerting a vertical force on the wing, so the force- and moment balance is not as simple).

| improve this answer | |
$\endgroup$
2
$\begingroup$

For dimensioning purposes, consider the intersection points to exert no moments and to behave as hinges.

enter image description here

Force equilibrium in point 1:

  • $F_{13} \cdot sinψ = ½L => F_{13} = \frac{L}{2sinψ}$
  • $F_{12} = F_{13} \cdot cosψ = \frac{L}{2tanψ}$
| improve this answer | |
$\endgroup$
  • $\begingroup$ Yes the forces create stretch. Strut 1-3 is the longest strut with the highest force, and will stretch more than the other members. The triangle will be slightly deformed, but because of the hinge modelling all of the members will still be loaded by compression/tension forces only. $\endgroup$ – Koyovis Jun 18 '19 at 11:46
  • $\begingroup$ Would it be correct to start with the lift and tension vectors at point 1 only, then decompose the tension vector of the strut to vertical (opposing lift) and horizontal (compressing inner wing spar)? $\endgroup$ – Robert DiGiovanni Jun 18 '19 at 16:47
  • $\begingroup$ To balance forces F12 could be coming from the other strutted wing! $\endgroup$ – Robert DiGiovanni Jun 18 '19 at 16:54
1
$\begingroup$

Analytically, using beam theories, like Timoshenko or Bernoulli, maybe, see below.

Numerically, via FEM.


This is a pretty classic exam question involving a statically indeterminate (or over-constrained) system. To solve it you will need to solve the beam equations with the correct boundary conditions, which may or may not be possible (I do not recall at present whether the "cantilever and wire" has an analytical solution, I'll update this later).

Alternatively you can obtain an approximate solution by relaxing your boundary conditions, for example exchanging the cantilever for a point support with a torsional spring of suitable stiffness.


An anonymous comment pointed out that strut braced wings are almost invariably pin-jointed structures to maximise structural efficiency, the system is statically determinate. In this case the structure is analytically solvable without the need to muck around with boundary conditions and beam equations:

Reactions and internal loads for the 2D free body shown can be resolved by basic statics. The strut is a two force member, experiencing axial loads only. The inboard portion of the spar which experiences compressive loads in reacting the spanwise component of strut tensile load is usually idealised as a beam-column.


| improve this answer | |
$\endgroup$
  • $\begingroup$ It most certainly has an analytic solution as a Bernoulli beam (probably Timoshenko too but that seems overkill for such a slender member as a wing) $\endgroup$ – Sanchises Jun 13 '19 at 17:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.