# 16.3: Elastic Behaviour

- Page ID
- 7887

To start let us assume that the predominant contribution to the elastic strain comes from the axial compression of the vertical struts, as shown below.

We can estimate the magnitude of this strain as the cross-sectional area of solid material, A_{V}, in a cut across just the vertical faces is less than that if the material were completely solid by the ratio of the cell wall thickness, t, to the horizontal distance across each cell, 2 l cos θ.

As t << l cos θ, this is given by

\[A_{\mathrm{V}}=t / 2 l \cos \theta\]

Using the measured values of t ( = 0.09 mm) and l ( = 6.30 mm), and taking θ = 30° as the cells are hexagonal gives A_{V} as 0.008. Taking the Young modulus of aluminium as 70 GPa, this predicts the Young modulus of the honeycomb to be 560 MPa. This is greater than the observed value of 435 kPa by more than 3 orders of magnitude and shows that axial compression of the vertical faces makes a negligible contribution to the elastic strain.

If the axially loaded faces do not deform, then clearly it must be those at an angle to the loading axis that do, the diagonal faces. And because they are at an angle to the loading axis they will bend.

The bending of each face beams must be symmetrical about the mid-point of the face and can be estimated using beam bending theory. To do this each face is described as two beams, cantilevered at the vertices of the hexagonal cell and loaded at the centre point. Note that one beam (i.e. a half cell wall) is pushed upward, the other downward.

This [derivation] gives the Young modulus of the honeycomb as

\[E=\frac{4}{\sqrt{3}} E_{\mathrm{S}}\left(\frac{t}{l}\right)^{3}\]

Using the measured values of t( = 0.09 mm) and l ( = 6.30 mm) and taking E_{S} as 70 GPa, the elastic modulus is predicted to be 471 kPa. This is within 10% of the measured value for the sample.

The predominant contribution to the Young modulus is therefore from the bending of the diagonal faces.