Derivation of Young modulus
Here we estimate the Young modulus when the load is applied to the honeycomb made of hexagonal cells with some of the faces parallel to the loading axis where the displacement results from bending of diagonal faces The thickness of the cell wall is t and its through-thickness width is w and the length of each of the hexagonal faces is l.
The displacements in the upper and lower halves of the diagonal faces must be symmetrical about the centre, with the lower half being bent downward and the upper half in the opposite direction. Each face can therefore be treated as if it were made up of two cantilevered beams. Each is length l/2, cantilevered at the end fixed to the vertical face and loaded at the other end.
From beam bending theory the displacement of the loaded end, δ, of a cantilevered beam is given by
\[\delta = \frac{1}{3}\frac{{W{L^3}}}{{EI}}\]
where W is the applied load, L is the length of the beam, E is the Young modulus and I is the second moment of area (definition).
Here the bending beam lies at an angle (90-θ)° to the axis of loading. The component of the applied force in the direction normal to the beam is therefore F cos θ. As the length of each beam is l/2, then the displacement from its original position (in the direction normal to the diagonal beam) is
\[\delta = \frac{1}{{24}}\frac{{F{l^3}}}{{{E_{\rm{S}}}I}} \cdot \cos \theta \]
ES is the modulus of the cell wall material from which the honeycomb is made and the second moment of area of the cell wall, I, is wt3/12.
Just as with the force, the direction of the displacement is also not in the direction of loading. The vertical component of the displacement (that is in the loading direction) due to the bending of the complete cell face, Δx, and taking the displacements produced by the two half-beams is
Δx = 2 δ cos θ
Substituting for δ and I gives Δx as
\[\Delta x = \frac{F}{{{E_{\rm{S}}}w}} \cdot {\left( {\frac{l}{t}} \right)^3}{\cos ^2}\theta \]
This can be expressed as a strain by dividing this downward displacement by the original vertical height of the honeycomb structure, that is two vertical half-faces and the vertical component of the diagonal face, giving the strain, ε, under an imposed load F as
\[\varepsilon = \frac{F}{{{E_{\rm{S}}}w}} \cdot {\left( {\frac{l}{t}} \right)^3}\frac{{{{\cos }^2}\theta }}{{l{\rm{ }}(1 + \sin \theta )}}\]
Now the applied force F acts over an area w l cosθ, so that the stress, σ, can be expressed as \[\sigma = \frac{F}{{w{\rm{ }}l\cos \theta }}\]
The Young modulus of the honeycomb in this direction, E (= σ/ε), is therefore \[E = {E_{\rm{S}}}{\rm{ }}{\left( {\frac{t}{l}} \right)^3}\frac{{(1 + \sin \theta )}}{{{{\cos }^3}\theta }}\]
As the cells are regular hexagons θ = 30° and this becomes \[E = \frac{4}{{\sqrt 3 }}{E_{\rm{S}}}{\rm{ }}{\left( {\frac{t}{l}} \right)^3}\]