Pullout Force Questions

[sub3marathonman]

The first posting here, so please be kind. I'm somewhat out of my area, which is why I'm hoping the experts will help!

I'm trying to calculate something I never expected to be so complex, the pullout force of an 8mm lag bolt 2" into a pine 2x4. I researched this, and found a formula, using .41 as the specific gravity, on this site:

https://docs.google.com/viewer?a=v&q=cache:uBG4voqvG74J:

http://www.fpl.fs.fed.us/documnts/f...cMC-5u&sig=AHIEtbR73qKvjJsoRWnmRTOCc70WvpMVaQ

page 8-12. Using that formula, p=8100*G^1.5*D^0.75*L, I come up with 149 lbs for 2 inches of thread penetration. Or just 75 lbs per inch. Which does seem low in my opinion, but I don't know.

However, on this forum,

http://www.eng-tips.com/viewthread.cfm?qid=183649

, the pullout strength for a #10 wood screw into southern yellow pine (SYP) was stated, by CJJS, to be 163 lb/inch. And no negative comment implied toward CJJS, it is just that the formula used to calculate the 163 lb/inch of thread penetration wasn't shown. I guess it may be common knowledge among experts, but I'm not an expert, I'm just hoping to try to learn.

So I'm wondering if somebody here could either show me the formula that CJJS used and/or show me where my formula is wrong.

The help is greatly appreciated.

 

[woodman88]

Per the NDS 2005 the equation is p=1800*G^1.5*D^0.75 in lbs per inch. Their table for withdrawal gives 163 lbs per inch for a #10 wood screw in G=.55 lumber. A 5/16" dia. lag screw in g=.41 lumber is 198 lbs per inch. The Wood Handbook which you are using is a good reference book, but I would go with the NDS for the equations and values.

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.

 

[MiketheEngineer]

Yes - the NDS is the "bible"

 

[sub3marathonman]

Thank you both for the help.

I've been doing more research, and have seen the NDS formula. I was confused by that first formula, is it possible it is a typographical error? It seemed though that the units were in inch-lbs.

Would the pullout force be the same if it was, instead of one 2x4 piece, it was two 1x2 pieces? I suppose the question is more about the repetitive factor for calculating, which with two it isn't, but I didn't know if it was worthy enough for a separate thread.

Thanks again for the help. I do appreciate it, and it is interesting to learn.

 

[sub3marathonman]

I mean two 1x4 pieces to make a 2x4 piece.

I didn't see any way to edit the previous posting.

 

[MiketheEngineer]

I would think that this is more of an engineering decision. It wouldn't be as good as a 2 x4 but if the 1x4's were well connected it wouldn't be too bad. I mights use say 2/3 - again an engineering choice.

 

[structSU10]

I would say that the value could be reduced per the minimum embedment required for the lag bolt (5/8*D maybe?) for each piece. Or at least for the peice the tip of the lag is in.

 

[chicopee]

Per Kent ME HDBK"design and production" withdrawal resistance of lag bolts
For penetration into the side of the grain:
P=1500*D^(3/4)*G^(3/2) whereby D=diameter lag bolt shank in inch; G= specific gravity of wood species.
For penetration into the end of the wood grain reduce P shown above to 75%.

 

[Tmoose]

http://www.fpl.fs.fed.us/documnts/fplgtr/fplgtr190/chapter_08.pdf

- p8-12

"Penetration of the threaded part to a distance about seven times the shank diameter in the
denser species (specific gravity greater than 0.61) and 10 to 12 times the shank diameter in the less dense species (specific gravity less than 0.42) will develop approximately the ultimate tensile strength of the lag screw. Penetrations at
intermediate densities may be found by straight-line interpolation."