# Honeycomb structures vs shear stress

Honeycomb structures are made of top and bottom shells and honeycomb geometry at core. In terms of bending load, maximun normal strains take place on top and bottom hence, sheet parts can support them as expected.

However maximun shear strain take place at the center of the cross section which coincide honeycomb structure. As I know, stiffness of the honeycomb very low even negligible. So, what is the idea behind shear stiffness of a honeycomb structure?

• Shear stress along which axis? Paralell or perpendicular to the sheets? (I assume paralell)
– mart
Nov 25 '19 at 8:46
• The stiffness is not very low or negligible. It is chosen based on what it needs to be for the job. Honeycomb cores, and any core for that matter, has to be up to the loads it bears. Here's a test report Nov 19 '20 at 3:09

As I know, stiffness of the honeycomb very low even negligible.

That is only correct if all the edges of the honeycomb are free.

If the top and bottom of the honeycomb are constrained by bonding to the face plates, the honeycomb can not "squash flat" with negligible stiffness, and it uses the material very efficiently to resist shear, because applying a "global" shear strain means that the walls of the honeycomb are in either tension or compression.

Honeycomb structures can be much stiffer than solid metal with the same weight. I have used 2m x 1m sheets of aluminium honeycomb (with top and bottom sheets) which are light enough for one person to pick up and carry around easily, but stiff enough and strong enough to stand on without damaging then, if they are supported at each end and used as a 2m long beam.

A honeycomb member has usually hexagonal pattern shear members. Because it is closest to an isotropic solid member within the constrains of manufacturing.

Lets imagine we put a honeycomb Formica door flat and use it as a beam supporting 100lbs at the center, which will have 50lbs shear at support.

If the support end cuts the hexagons exposing the diagonal blades at abutment, they are under out of plane shear and torque, which they can poorly take and will rotate into a mechanism, starting thee chain of collapse in the entire door.

However as has been said by alephzero these 60 degree lose blades are contained in a solid frame such as a 2in solid wood and they won't develop the rotation mechanism and the door will act very stiff, supporting the load.

Why do we use the honeycomb slabs like concrete ceilings, parking slabs, shopping mall floors, is because it is the most efficient use of material for large span slabs. For a slab to have a large 'I', it needs most of material on top and bottom and the shear can be taken care of hollow, lightweight honeycomb.

When we talk about shear stress, the first equation comes to our mind will most likely be $$f_v = V/A$$, which, in fact, is the average shear stress, not the maximum. Then, we realize for shear on irregular shape, the best way to determine shear stress is using the shear formula, $$\tau = \dfrac{V \cdot Q}{I \cdot b}$$. For this case, you need to workout $$Q$$ and $$I$$, before you can say the honeycomb section is weak, or not, compared to the normal stresses it can resist. Good luck in finding those parameters :)