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Wood Diaphragm Nail Slip Maximum Load

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Jerehmy

Structural
Aug 23, 2013
415
If you look at table C4.2.2D in ANSI/AF&PA SDPWS-2008 ( for 10d common nails the maximum fastener load for calculating nail slip is 260lb (this is for deflection, so ASD loads).

If you calculate the fastener strength for a 10d nail connecting plywood into framing (assuming the plywood's specific gravity is 0.5 and framing also), you get the following load from Table 11Q in NDS 2005 multiplied by 1.6 wind factor and 1.1 diaphragm factor:

15/32" plywood : 88*1.6*1.1 = 155#
1" plywood : 118*1.6*1.1 = 208#
1-1/4" plywood : 118*1.6*1.1 = 208#


What am I missing? Why would they allow for a 260# maximum fastener load when in no situations is that plausible? I must be missing something.

Thanks.
 
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I guess if I had higher density framing and sheathing you could maybe achieve the 260#. I will have to check at work tomorrow
 
Check out Table 4.3A on page 31 for the where the fastener loads come from.

Garth Dreger PE - AZ Phoenix area
As EOR's we should take the responsibility to design our structures to support the components we allow in our design per that industry standards.
 
Look at Table 4.3A, the last column, 5th row down. 2435plf nominal shear capacity for 15/32" panels, 10d nails.

ASD load for this is 2435plf/2 = 1217.5 plf

The edge of the diaphragm has fasteners every 2in. So the load per fastener is 203# and the allowable load is 155#.

So why would they design a diaphragm that has 203# per fastener when the allowable for the conditions listed is 155#. I find nothing about this in the commentary or any diaphragm design guides. I've designed diaphragms using this before I just never noticed this before. What am I missing.
 
Can't answer that, but I will say I would never design a building with diaphragm loads that high - that's ridiculous to have fasteners 2" o.c.
 
The reasoning for me doing it is we are tearing down a building(building A) that has another building (building B) "leaned" up against it (brick walls). As in, building B has no 4th wall and there is literally no connection.

We're going to connect to the existing floor diaphragm of building B to the wall but how dependable the existing diaphragm is a judgement call (its 58.5ft x 58.5ft building so even if the diaphragm has barely any capacity, the wind load is only 142plf at the roof it should be ok).

So just to give us some piece of mind we're going to heavily reinforce the diaphragm where we connect to the wall to make a "mini" diaphragm capable of resisting all the wind load.
 
First, the calculation is only for seismic (column A) see eq 4.3-1 where Gn = from column A. The 2435plf is in column B wind.
Second, why are you going to the commentary?

Garth Dreger PE - AZ Phoenix area
As EOR's we should take the responsibility to design our structures to support the components we allow in our design per that industry standards.
 
Woodman88

I use equation C4.2.2-1 on page 54 which has nail slip as a separate calculation, 0.188Len which is also the same as the IBC equation.

en = (Vn/769)^3.276 Where Vn is fastener load (Table C4.2.2D on npage 57).

If I'm calculating wind deflection why would I use seismic loads. Also, if you look at design guides (ex: they use the wind load for Vn when calculating deflection.
 
Per this link ( on page 14, it is the interior nail spacing not the edge nailing spacing that is used for deflection calculations.

This may be the reason for the 260 lb/nail requirement.

Garth Dreger PE - AZ Phoenix area
As EOR's we should take the responsibility to design our structures to support the components we allow in our design per that industry standards.
 
If that is the case, would you say that the load being higher than would be allowed for a connection is just for deflection purposes and that the nail isn't actually seeing that high of a load?
 
That is the way I see it.

Garth Dreger PE - AZ Phoenix area
As EOR's we should take the responsibility to design our structures to support the components we allow in our design per that industry standards.
 
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