Bending Strength of Edge Loaded Plywood

So I took a 32" long piece of 3/4" x 3/4" 7 ply plywood with no voids, put the ends on blocks and measured the deflection of all 4 sides when I put a fairly heavy piece of iron in the middle.

With the plies vertical the 2 sides deflected 63 and 65 mm, with the plies horizontal they deflected 55 and 58 mm, the averages being 64 and 56.5, a difference of 7.5 mm.

I think that's a difference of 11.7%, maybe 13.2%, the horizontal plies being stiffer.

 

Barry, you threw the whole book at me but failed to read it first. Here is some images to refresh your memory. The images were from book Strength of Materials by Pytel & Singer.

Wood is anisotrophic and its greatest strength is when the fibers are aligned to the load. When the load is applied at an angle to the fibers, there is a corresponding loss of strength/modulus. The remaining strength could be anywhere from 17 to 25% of the ultimate depending on the species of wood. At +45/-45 degree angle, the fibers loses much of its vertical strength but gains 100% of the shear strength because the fibers are tugging at each other, preventing displacement.

When the plywood is laid on its edge, the “I” is increased and there is greater stresses on the faces(edge) of the plywood. The fibers now is aligned at 0/90 degrees. The horizontal fibers will have 100% strength in tensile/compression region while the horizontally aligned fibers will retain about 25% in the perpendicular. Half of the strength to make it simple. The only way to achieve 100% strength is to have all fibers aligned along the load direction, that is to Cap it with a solid lumber or metal and the resulting beam is a reinforced beam.

The mid portion or the neutral axis occurring in the Web will have the greatest shear stress as the fibers tends to slide past each other. As stated, the +45/-45 fibers has the greatest strength in shear, hence it is logical to use angled fibers in the web.

In the example posted in No 3, it shows a concentrated load on a simply supported beam. Agree, the shear will greatest at he midspan but it is greatest at the support edge if the beam is uniformly loaded. This V shear depends on the load type.

This V shear (vertical shear) is reacted by by a horizontal shear, a couple to counteract the vertical shear.

When you say couple because the other fibers are carrying the load in a vertical shear this is erroneous because what is causing the fibers to bond is the glue or bondline which is usually the weakest part of the design. If the fibers are in +45/-45 orientation, the unit area is only increased as both set of fibers share the same degraded properties. In both cases, the limiting factor is the least mechanical properties of the glue or the degraded properties of wood.

When a concentrated load is applied at the center on a simply supported beam, the load path is vertical at the point of load and changes direction gradually until it reaches 45 degree at about ¼ of the span. It then reverses direction and continue to change at it approach the point of support. This is because the support is resisting and applies a counter rotating force. The top region, which was previously in compression is now in tension as it nears the fixed support point.

I don’t know about the Parralam beams, I did not design it. Maybe you know better.

 

That is flexural loading and you increase the load bearing by a factor of 4 not 2.
When you double the overall thickness (depth) you increase the "I", the stresses on the top and bottom fibers increase but the shear stress on the NA is reduced.
The formula is

Fs=4x max load x support span/(3x width x depth x depth)

 

SamSam
Interesting results. Everything I have read suggests that edge loaded plywood is stiffer than panel loaded. I wonder what the result would be for a 3/4 x 1 1/2 piece of plywood.

 

It will be stiffer (less deflection) when laid edgewise as the moment of inertia is increased. See post 48.

 

You did not read the post properly. The post which you have only partly quoted was developed to show to a contributor that stated that there is no shear flow in a beam under bending.
The example that you quoted was two INDIVIDUAL
pieces not joined, simply resting on one another.
Your come back, ie raising the area moment of inertia would be accurate if the two were in fact joined at the interface.

I then went on to explain that if they were attached and the joint would carry the shear flow, then would then act as a stronger piece.

Post 34

 

Did not mean to throw the book at you. Sorry that you took it like that. Also did not need a refresher in beam design.

Your comment that solicited my response was your comment that the 45 degree oriented strands make the beam act like a truss. A truss design produces compressive and tensile stresses in the members through out the structure while orienting alternating plys at 45 degrees do not produce the same stress directions.

But regarding your last paragraph above.
1) Your statement the load path changes from vertical to a 45 degree orientation. In about 40 years of practice, I have never ran across this theory and if you have some supporting data to back this up, it would be much appreciated.

2) ""Then the load path reverses direction and the compression area at the top of the beam changes to tension. ""
You said in this part that the beam is simply supported ( and that is what we were discussing) but then refer to the couple that results from the FIXED supports
This is confusing. We are discussing a simply supported beam where all the upper fibres are in compression not the stresses/forces in a beam with fixed end supports

Any information to clear this up for me would be appreciated

 

In explaining the shear flow my mind shortcircuited the process in explaining the simple beam to fixed beam.

This is a well understood phenomena in almost all discipline.
In Marine Composites (E Green and Associates), where the hull can be considered a simple beam. It is shown that the v hear is almost vertical at point load and how it traverses the length and slopes down towards the mid point and reverses.

In the second illustration it shows (in a uniformly distributed beam) how the v shear and the couple h shear orient itself in a beam.

In concrete beams (which can also be considered a composite beam), it explains the direction of stress as it approaches the support points (Reinforced concrete Design by Besavilla). Further from my reference article shows how the crack propagates in a concrete beam.

The LR manual on composite design shows the regions of tension and compression and how it reverses due to reaction of the fixed beam or stiffeners and that is how I explained it.

 

May I first wish everyone a happy new year

Barry

If you think of this as a composite. By that I mean a typical composite such as woven roven or carbon fibre etc. ..fibres that are laid over a formed shape to be aligned in the direction of maximum stress. Thus the material is anisotropic or at best orthotropic.

In a typical beam, say an I-beam, when the vertical load is applied how is the shear transferred? Thus, imagine if you will two flat bars that are equal in their size yet at ‘some distance’ apart. By the simple I=ah^2, the further the distance apart the greater the stiffness. But with two FBs that are at some distance there is nothing between the two….so if a load is applied what occurs…one of the FBs will merely bend on its own with its own “I” and fail rather quickly.

So if we now place a vertical thin member, a web, that joins the two FBs together, the system works as one. Thus the “I” is now significantly greater than one “I” of the single FB alone. In addition, there is now a shear path connecting the two FBs together.

Thus when a load is applied this newly introduced web can carry the shear loads.

If we look at the vertical load we need a material that has it properties aligned with the direction of the stress. If the I-beam is isotropic, like any metal, this is easy. Nothing needs to be done. But, in the case of composites, not so.

If we lay the fibres vertically, they align with the direction of the load and thus stress, in either tension, or compression, depending where the shear load is being measured along the beam.

However, since the beam is a composite (not isotropic) we must account for the shear stress horizontally. So the fibres need to be laid horizontally. Oh dear, we now have fibres being laid in two direction. The reason why this is not ideal, is that in regions where the vertical load is greatest having fibres running horizontally provides little resistance and thus will simply shear apart...similarly vice versa.

Going back to the 2 FBs at some distance apart. If the web were not a solid continuous web, but say FB or tube, that is connected vertically only, we would have a series of FBs spaced, again, at some distance apart to form an I-beam. However, if a load is applied between two of the vertical FBs connecting the two horizontal ones, what occurs to the load? There is no load path to transfer the load from the upper horizontal FB to the lower, except the horizontal FB itself. The path is the horizontal FB alone and if this is insufficient to take the load (as one would expect) it shall fail locally, just as if there were no connection between the two FBs.

So as a composite not only do we have an issue with the fibres being sheared by being orientated in the wrong axis, but also we need to ensure the load is able to be transferred via the vertical fibres under a vertical load in-between them where there are little or no fibres.

Thus the solution, is rather easy and already noted by several posters above. For the sake of simplicity we assume the load is uniform in the beam, otherwise the resultant vector of the shear, vertically and horizontally, shall have an ever changing axis of direction. Thus it is assumed to be 45 degrees as any off set will be relatively minor. Since at 45 degrees, the loads, whether horizontal or vertical can easily be transferred. And all Classification societies require webs to have their fibres arranged as such, for this reason, as noted by RX's above image too.

This can also easily be seen with regards to wood, as posted in mark’s diagrams posted here.



Which shows such behaviour and how it is indeed analysed.

 

Happy New Year to everyone!

My recollection from composite courses four decades ago is that fibers parallel the long direction of the beam ("horizontal") do not appreciably affect the shear stiffness of the web, and the shear stiffness will depend on the shear stiffness of the matrix only. Am I mistaken?

 

No.
But we are not talking about stiffness here. We, or least Mark, RX and myself, are talking about the loads and what paths they take. Nothing to do with stiffness per se.