Section modulus in case of stacking the plates 5

[kaffy]

Hello guys,

I am working on a project where layers of plywood were stacked on top of each other and nailed. My question is in bending, whenever flat plates(steel, plywood, planks)are stacked together, what is the section modulus?

For example,
Two plates with same thickness
Is the new section modulus same as what I am showing in attached or
are there any method to check those cases?

Thank you
Newbie

 

[271828]

Your calculation is correct for the elastic section modulus.

To make the two plates behave in this manner, you'll need adequate connection between them to transfer the horizontal shear. Compute the required horizontal shear strength using VQ/I.

 

[phamENG]

The calculation is correct ONLY if you fasten them together to satisfy shear flow requirements as 271828 indicated.

If you do not fasten them together, the elastic section modulus becomes roughly 2*(b*d^2)/6 (assuming EI of each plate is equal).

I looked at that a little too quickly. Your moment of inertia is correct only if they are fastened together for shear flow. If not, it becomes I=2(b*t^3)/12. Your section modulus is not correct. If you're going for elastic section modulus, it would be [b*(2t)^2]/6 if fastened for shear flow and plastic section modulus would be [b*(2t)^2]/4. If not fastened for shear flow, they would be 2[b*(t)^2]/6 and 2[b*(t)^2]/4, respectively.

Unless I'm missing something in your calc, you end up with Z (whether that's supposed to be elastic or plastic depends on where you are in the world - in the US it's standard notation for plastic section modulus) being independent of b, the width of the plate.

 

[rb1957]

"are there any method to check those cases?" ... you could make a simple test ... put a 1ft wide strip between to "simple" supports (two work benches ?) and apply a load in the middle. measure deflection, compare with calc. I suspect you'll find more "error" from E, so you may want to test that too.

another day in paradise, or is paradise one day closer ?

 

[HTURKAK]

Dear kaffy ,

I looked at your calculation quickly.

Apparently you missed (b).. how did you cancel the width ? Pls check your calculation..

If the combined beam has dimensions b, 2t

I = b*(2t)**3/12 and elastic section modulus (in my zone we use notation W for section modulus) Z = ( b*(2t)**3/12 )/t =4b*t**2/6= 2bt**2/3

When the two layers connected ( with glue, nail, bolt etc ) , between them to resist horizontal shear , the two plates behave as a combined section.

 

[Agent666]

Look into the GAMMA METHOD from eurocode 5, what you will find is that the slip that occurs in the fasteners is crucial, and that the behaviour will be more like two separate sheets instead of 1 semi-composite sheet. This method explicitly allows for the slip that will occur between the layers enabling you to work out an effective stiffness and the stresses in each layer due to the limited composite action you create via the fasteners between the layers.

Just providing for the strength in the fixings as suggested by others is not adequate to ensure "composite action" when you are joining using a fastener that will slip in the plywood/timber under load.

What you will find is you'll need a massive amount of fixings to get anywhere close to maybe even 50% of the fully composite stiffness. It quickly becomes impractical, there will however be some improvement over and above just taking two separate sheets and adding them. But usually if you follow through the calculation on this sort of thing it is only marginally better than simply taking the capacities & stiffnesses of the two sheets and adding them.

If you are after true composite action between the sheets, then gluing/laminating the sheets together would be the practical option. As there is then no slip possible between the two layers. Or simply specify a thicker plywood to start with as that is what you're effectively wanting to create.

https://engineervsheep.com

 

[kaffy]

Thank you very much guys. Really Helpful. I forgot to put B in the attached screenshot.
As I am working with flat bar and the maximum horizontal shear stress = 3V/2A.
I just have to make sure that allowable shear stress in nail/bolt/glue/weld is more than 3V/2A.
Is that right?

 

[271828]

kaffy, please clarify if you're after the elastic section modulus, S, or plastic section modulus, Z. For plywood, I would assume you're after S.

 

[kaffy]

Yes. I am only working with elastic section modulus.

As I am working with flat bar and the maximum horizontal shear stress = 3V/2A.
I just have to make sure that allowable shear stress in nail/bolt/glue/weld is more than 3V/2A.
Is that right?

 

[BAretired]

Correct!

BA

 

[phamENG]

I'm not quite getting the same thing, but I'm going to assume I'm making a simple algebra error somewhere on my overcrowded post-it note as I trust BA. (3V/8A...)

Regardless, designing the fasteners will be simpler with a shear flow calculation as 271828 posted in the very first response. q=VQ/I. That will give you lb/in (or kN/mm, or whatever units you're using), which can then be multiplied by your fastener spacing to get the load in each fastener.

If you're dealing with wood, heed Agent666's words of warning. Everyone else is talking general shear flow and how to use it in composite shapes as it is frequently employed for built-up steel shapes. Wood does not do composite action well. Things like plywood and glulams are thoroughly tested and are glued together using highly specialized adhesives in controlled factory settings. About the only thing I use this composite action for in field built assemblies is for serviceability considerations of wood floor systems. I don't count on them for strength.

 

[BAretired]

Thanks for the compliment, phamENG. I'm flattered.

Shear stress is VQ/Ib.
Q = (bd/2)d/4 = bd^2/8
A = bd
I = bd^3/12
VQ/Ib = V(bd^2/8)/bd^3/12b = 3V/2A

BA

 

[kaffy]

Thank u very much guys...