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What kind of load does the interplane strut(s) carry,is it compression, tension or shear? In order words,what's their function and how do you calculate the force that the interplane strut carries?enter image description here

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    $\begingroup$ As Koyovis's answer shows, you simply have to calculate the vertical lift load acting at the interplane connection point to the lower wing panel to determine the compression load on the interplane strut, which will be somewhere around half or a bit more or less of the panel's total lifting force depending on the spanwise location. $\endgroup$ – John K Jul 11 at 14:36
  • $\begingroup$ Could you crop that image so it doesn't waste so much space above and below? $\endgroup$ – AEhere Jul 12 at 7:03
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The wing strut prevents the lower wing from bending upwards due to the lift forces.

enter image description here

The biplane is a truss structure: rectangular shapes are "crossed out" by diagonal structural members, in order to create triangular shapes with only tension and compression in their structural members, no bending. If the diagonal is a rod, it can transfer both compression and tension forces and only one diagonal is required. If it is a cable, it can only transfer tension forces, and the other diagonal needs to be crossed out by a cable as well.

In flight, lift wants to bend the wings upward and lengthen the diagonal, which becomes tension loaded. The upper wing is then prevented from bending upwards and becomes compression loaded.

The interplane strut connects the lower wing with the upper wing, and prevents the lower wing from bending upwards. It is compression loaded: the diagonal exerts a downward force, lower wing lift exerts an upward force.

The method of how to compute the force is in this answer. In a truss structure the intersection points are modelled as hinges, and the structure members are dimensioned to carry the resulting tension/compression loads.

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    $\begingroup$ Minor quibble, but on biplanes with wire bracing it's a pin joint structure so the dotted lines should really be straight. The curve implies root fittings able to take bending, like on a Fokker DR-1. $\endgroup$ – John K Jul 11 at 14:31
  • $\begingroup$ @JohnK the dotted lines are what the wings would do if there was no diagonal structural member. $\endgroup$ – Koyovis Jul 11 at 14:36
  • $\begingroup$ the root fittings on biplanes are pin joints that can't take any bending. If you took out the diagonal member, it would fold up like a parallelogram. $\endgroup$ – John K Jul 11 at 14:39
  • $\begingroup$ @JohnK I know. It's a truss structure. $\endgroup$ – Koyovis Jul 11 at 14:46
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    $\begingroup$ This feels like a good reminder that wing structures are complex with many different fine details in various design options. - Vertical struts may at times carry a dynamic load as the end of a box-beam structure in some cases, transferring both compression and tension loads. [Wing loading can get complex, especially if a structure is approaching a 'rapid unplanned disassembly' state due to 'unfavourable conditions'] $\endgroup$ – TheLuckless Jul 11 at 22:35
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It carries compression. The compression force is the share of the lower panel's lift being borne by the interplane strut's connection to the lower wing. So to calculate the compression load for, say 1G flight at gross weight, you simply need to determine the lift force present at the interplane strut's attachment based on the wing panel's loading and the attachment's location in the lift distribution of the panel. It's going to be somewhere in the range of half of the total lift being produced by the lower panel.

The ONLY member in significant tension in this structure is the flying wire (the red shows what happens if the flying wire breaks), and in most biplanes with center sections and single bolt jointed structures at the fuselage attachments there are no bending loads at all, except airloads pushing up on the spar beam along its length.

enter image description here

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