Can downward pressure split a 2 x 4?

Hi.

... beginning with a 6' 2x4 (110 S-DRY STUD)

... drilled a total of 30 (5/16") holes through the 2x4 (1 1/2") down the center spaced 2" apart

... so it's just a 2x4 with evenly spaced 5/16" holes down the center

... now the 2x4 is standing on its end, perpendicular to the ground

If a 5/16" bolt is pushed through one of the holes near the top, how much downward pressure on that bolt before the 2x4 splits down the center?

In other words, if that 5/16" bolt were a dumbbell, how heavy can that dumbbell be before it splits the 2x4 down the center?

The next 5/16" hole is 2 inches below (center to center).

If possible, an extremely rough estimate is fine.

Thank you.

Reply to
John Doe
Loading thread data ...

If i'm reading this right, *all* of the holes are through the same grain line, which has essentially destroyed the structural integrity of the piece for the purposes you describe.

I don't know what kind of weight it would hold, but I wouldn't trust it with anything I wanted to keep.

jc

Reply to
noonenparticular

please don't answer someone's homework problem.

Reply to
Charles Spitzer

That is correct.

Reply to
John Doe

No. Resistance against splitting is the strangth perpenticular to the fibers, so the question here is how much pressure perpenticular to the fibers will the bolt exert. If it fits snug into the hole that is rather little, so splitting the stud along the middle will probably not be the mist likely failure mode. A small bolt, however, might act like the tip of a wedge and get the thing split.

Reply to
Juergen Hannappel

I agree, the smaller the bolt, the *earlier* it would split the 2x4 (given increasing weights), but I disagree in that regardless of the size of the bolt or how snugly it fits into the hole, the method of failure would still be the longitudinal splitting of the piece especially since gravity is pulling the bolt/weight/whatever down the long axis.. If you don't agree, we'll have to agree to disagree. Agree? ;-)

jc

Reply to
noonenparticular

I agree (maybe to disagree), I can't do the calculations (because they are rather complex as crushing of the wood around the bolt has to be considered). But another possible failure mode would be for the 2x4 to buckle and break under the load. Unfortunatly I am in no position to make a test...

Reply to
Juergen Hannappel

Buy some 2x4s. Drill some holes. Apply various pressures until they split. Let us know the results. You'll have to do it a few times and get an average.

Reply to
Guess who

Well 3,000,000 lbs would be a rough estimate. However, if you run a bolt and a nut between the holes on the side 90 degrees to the side with all the holes you will add tremendous strength to the board.

Reply to
Leon

A shit load

Reply to
C&S

I would have explained in detail, but I think most members of this group would be more interested in the problem as generically stated than in my specific circumstance. Usually less writing is better when asking a question IMO.

I really need and appreciate the discussi> Path:

newssvr29.news.prodigy.net!newsdbm05.news.prodigy.com!newsdbm04.news.prodigy.com!newsdst01.news.prodigy.com!newsmst01b.news.prodigy.com!prodigy.com!newscon02.news.prodigy.com!prodigy.net!wns13feed!worldnet.att.net!209.244.4.230!newsfeed1.dallas1.level3.net!news.level3.com!transfer.stratus.com!not-for-mail

Reply to
John Doe

With a 1 s.d. 'uncertainty' of approximately +/- 2,999,900 lbs.

Reply to
Robert Bonomi

Solving this kind of problem is fundamental to structural engineering of wood construction. The guiding theory is called dowel bearing strength. The equations for this are given in section 11.3 of the National Design Specification for wood construction.

If we approximate the "dumbell" as an infinitely strong bolt being loaded in double shear by infinitely strong side members, then the equation 11.3-7 governs.

The allowable compression parallel to the grain for this scenario is given as: Z = D * lm * Fem / Rd

where D is the diameter of the bolt (5/16) lm is the thickness of the main member (1.5 inches) Fem is the dowel bearing strength for the wood species in question. If we assume Douglas Fir with a specific gravity of 0.5, the dowel bearing strength is 5600 PSI. Rd is a reduction factor. In this case of perfectly parallel to the grain loading, Rd = 4

so Z = .3125 * 1.5 * 5600 / 4 Z = 656

The wood can be relied upon to carry a weight of 656 pounds for 10 years. This can be increased by a factor of up to 1.6 for "short duration" events (approximating seismic loads or wind gusts) to 1050 pounds.

There is approximately a 3:1 safety factor in this number, so we would not expect it to fail at less than 3000 pounds in a quick test. A particularly good specimen might go to as much as 7:1 or 7350 pounds.

The proposed additional drilled holes are considered of no consequence from an engineering design perspective, since the spacing is greater than 5 times the diameter of the fastener. The wood has sufficient shear strength to transfer the compressive load to the neighboring fibers and thus around the holes below the one carrying the load. After all, this kind of loading is exactly what the trunk of a tree has to support as it is growing -- carrying the weight of the crown down past branches and woodpecker holes.

Reply to
david_papworth

I can't vouch that this is right, but it sounds damn impressive!

SteveP.

Reply to
Highland Pairos

This assumes that the load applied is insufficient to cause buckling of the column.

todd

Reply to
todd

It's not a homework problem. Homework problems have clear right answers. This one doesn't -- it depends too much on how the grain of the wood runs, for one thing, and for another thing there's no good way to _calculate_ an answer without either more guesswork than a homework problem allows, or lots of computer analysis.

Also, it includes irrelevant data, which homework problems generally don't.

- Brooks

Reply to
Brooks Moses

It does. However, the question _was_ "how much will cause it to split", not "how much will cause it to fail somehow".

- Brooks

Reply to
Brooks Moses

Thu, Jan 26, 2006, 1:45am (EST+5) snipped-for-privacy@usenet.love.invalid (John=A0Doe) doth clameth: I would have explained in detail, but I think most members of this group would be more interested in the problem as generically stated than in my specific circumstance. Usually less writing is better when asking a question IMO. I really need and appreciate the discussion.

Horse hocky. I'm thinking you just want someone to do your homework too. Either that or you're another troll. And, if you're gonna ask questions, the more details you can provide the better, if you want to get usable answer, that is.

JOAT You only need two tools: WD-40, and duct tape. If it doesn't move and it should, use WD-40. If it moves and shouldn't, use the tape.

Reply to
J T

Wed, Jan 25, 2006, 9:35pm (EST-3) snipped-for-privacy@cits1.stanford.edu (Brooks=A0Moses) doth burble: It's not a homework problem.

Really? Did I miss the original poster saying that?

JOAT You only need two tools: WD-40, and duct tape. If it doesn't move and it should, use WD-40. If it moves and shouldn't, use the tape.

Reply to
J T

That explains all the generic postings about woodworking and joinery here without references to actual projects. Have you actually read any postings in this newsgroup?

It's been my experience that you get better answers to specific questi> I would have explained in detail, but I think most members of this

Reply to
Mike Berger

HomeOwnersHub website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.