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.

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

"John Doe" wrote in message
...
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.

please don't answer someone's homework problem.

"noonenparticular" wrote in message
. net...
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

"John Doe" wrote in message
...
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.

"noonenparticular" wrote:

If i'm reading this right, *all* of the holes are through the same
grain line,

That is correct.

"noonenparticular" writes:

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.

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.

--
Dr. Juergen Hannappel http://lisa2.physik.uni-bonn.de/~hannappe
Phone: +49 228 73 2447 FAX ... 7869
Physikalisches Institut der Uni Bonn Nussallee 12, D-53115 Bonn, Germany
CERN: Phone: +412276 76461 Fax: ..77930 Bat. 892-R-A13 CH-1211 Geneve 23

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


"Juergen Hannappel" wrote in message
...
"noonenparticular" writes:

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.

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.

--
Dr. Juergen Hannappel http://lisa2.physik.uni-bonn.de/~hannappe
Phone: +49 228 73 2447 FAX ... 7869
Physikalisches Institut der Uni Bonn Nussallee 12, D-53115 Bonn, Germany
CERN: Phone: +412276 76461 Fax: ..77930 Bat. 892-R-A13 CH-1211 Geneve 23

"noonenparticular" writes:

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? ;-)

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...

jc


"Juergen Hannappel" wrote in message
...
"noonenparticular" writes:

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.

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.

--
Dr. Juergen Hannappel http://lisa2.physik.uni-bonn.de/~hannappe
Phone: +49 228 73 2447 FAX ... 7869
Physikalisches Institut der Uni Bonn Nussallee 12, D-53115 Bonn, Germany
CERN: Phone: +412276 76461 Fax: ..77930 Bat. 892-R-A13 CH-1211 Geneve 23



--
Dr. Juergen Hannappel http://lisa2.physik.uni-bonn.de/~hannappe
Phone: +49 228 73 2447 FAX ... 7869
Physikalisches Institut der Uni Bonn Nussallee 12, D-53115 Bonn, Germany
CERN: Phone: +412276 76461 Fax: ..77930 Bat. 892-R-A13 CH-1211 Geneve 23

On Wed, 25 Jan 2006 21:05:00 GMT, John Doe
wrote:

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.

"John Doe" wrote in message
...
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.

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.

A **** load

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?

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.

"Charles Spitzer" charlie.spitzer nospam.stratus.com wrote:

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From: "Charles Spitzer" charlie.spitzer nospam.stratus.com
Newsgroups: rec.woodworking
Subject: Can downward pressure split a 2 x 4?
Date: Wed, 25 Jan 2006 14:14:09 -0700
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please don't answer someone's homework problem.

"noonenparticular" invalid nowhere.nothing wrote in message
news:INRBf.22045$F_3.10773 newssvr29.news.prodigy.net...
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

"John Doe" jdoe usenet.love.invalid wrote in message
news:Xns975699701116Cfollydom 207.115.17.102...
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.

In article ,
Leon wrote:

"John Doe" wrote in message
...
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.

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.

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

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.

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

SteveP.

wrote in message
ups.com...
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.

wrote in message
ups.com...

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.

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

todd

Charles Spitzer wrote:
please don't answer someone's homework problem.

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


--
The "bmoses-nospam" address is valid; no unmunging needed.

todd wrote:
wrote in message
ups.com...
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.
[...]

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

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

- Brooks


--
The "bmoses-nospam" address is valid; no unmunging needed.

Thu, Jan 26, 2006, 1:45am (EST+5) lid (John*Doe)
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.

Wed, Jan 25, 2006, 9:35pm (EST-3)
(Brooks*Moses) doth burble:
It's not a homework problem. snip

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.

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 questions than very general ones. The answer
to the question you originally posed is "42". But you
won't understand the answer until you truly understand
the question.

John Doe wrote:
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.

There's nothing in the question that suggests it's a homework
problem.

Maybe I should be flattered for apparently defining the problem well
enough to make the question look academic.

Troll.

Jakofalltrades webtv.net (J T) wrote:

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From: Jakofalltrades webtv.net (J T)
Newsgroups: rec.woodworking
Subject: Can downward pressure split a 2 x 4?
Date: Thu, 26 Jan 2006 10:55:18 -0500
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Thu, Jan 26, 2006, 1:45am (EST+5) jdoe usenet.love.invalid (JohnÿDoe)
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.

My question was so well defined/specified that it was well discussed
and very clearly answered. Much appreciated. I'm delighted.

A sarcastic, semantics game playing troll.

Mike Berger berger shout.net wrote:

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From: Mike Berger berger shout.net
Newsgroups: rec.woodworking
Subject: Can downward pressure split a 2 x 4?
Date: Thu, 26 Jan 2006 10:47:38 -0600
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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 questions than very general ones. The answer
to the question you originally posed is "42". But you
won't understand the answer until you truly understand
the question.

John Doe wrote:
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.

On Wed, 25 Jan 2006 21:05:00 GMT, John Doe wrote:

yes, but not right away...

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.









mac

Please remove splinters before emailing

In article ,
lid says...

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.

So what is the application anyway?

I'd guessed from your example that you were doing something with weights
(as in barbells and dumbells) and thinking serious kilogrammage.

If so, I don't know about anyone else, but I'd cross reinforce it.

Frankly, I hate questions that leave out the application.

Knowing what weight will cause a 2x4 with a pattern of holes to fail may
be interesting, but knowing, that you will use a pair of the boards to
store assembled barbells of 50 to 500 lbs up to a total of x lbs. is
important because you, or others may drop one of them onto the rack...
and cause it to fail.

Knowing if the board will be vertical or set back at an angle is also
important because the weight will be transmitted differently.

etc. etc. etc.

Wow!

A lucid response to an inane question. I am impressed.

Well done!

Regards.

Tom


On 25 Jan 2006 19:29:17 -0800, wrote:

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.

John Doe wrote:

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.


Almost sounds like you're making a pair of stilts. I don't know the
answer to your question, but if you *were* making stilts I can tell
you that TWO 5/16" bolts per foot ledge seemed to hold well for all
the kids that walked around on the various pairs of stilts that I
made.

Joe Barta

Troll

Tom Banes xleanone nospam.mail.airmail.net wrote:

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From: Tom Banes xleanone nospam.mail.airmail.net
Newsgroups: rec.woodworking
Subject: Can downward pressure split a 2 x 4?
Date: Thu, 26 Jan 2006 14:19:43 -0600
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Wow!

A lucid response to an inane question. I am impressed.

Well done!

Regards.

Tom


On 25 Jan 2006 19:29:17 -0800, david_papworth hotmail.com wrote:

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.

phorbin phorbin1 yahoo.com wrote:


Knowing what weight will cause a 2x4 with a pattern of holes to
fail may be interesting,

My favorite part of the replies is the general rule about the
consequential/important distance from hole to hole, the answer being
given as five times the diameter of the dowel hole.

Message-ID: 1138246157.771503.41270 z14g2000cwz.googlegroups.com

See you all later.











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Subject: Can downward pressure split a 2 x 4?
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"phorbin" wrote in message
...
: In article ,
: lid says...
:
: 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.
:
: So what is the application anyway?
:
: I'd guessed from your example that you were doing something
with weights
: (as in barbells and dumbells) and thinking serious
kilogrammage.
:
: If so, I don't know about anyone else, but I'd cross reinforce
it.
:
: Frankly, I hate questions that leave out the application.
:
: Knowing what weight will cause a 2x4 with a pattern of holes to
fail may
: be interesting, but knowing, that you will use a pair of the
boards to
: store assembled barbells of 50 to 500 lbs up to a total of x
lbs. is
: important because you, or others may drop one of them onto the
rack...
: and cause it to fail.
:
: Knowing if the board will be vertical or set back at an angle
is also
: important because the weight will be transmitted differently.
:
: etc. etc. etc.

And unless I missed it, not a single person has yet mentioned
that you'll very seldom find a 2 x 4 of any length where the same
grains at one end appear at the other end. When we were testing
benches, there were two split mechanisms expected: One where the
wood bowed north/south first, and the other where it'd bow
east/west first. Occasionally the first split to occur would be
longitudinally the whole length, but not often. When that
occurred, it was anyone's guess which way the bow would progress
before the break started.
Also, a sharp blow is much more likely to break the wood,
where a steadily increasing pressure reached a much higher
pressure psi. Vibrations set uip by a shapr strike are much more
damaging to the structure than constantly increasing pressure.
It's a whole science in itself.
And then you get into angular displacements, as in someone
leaning back on a bench... . Credit Hammond Organ for this part
of my experineces. It was fun, but not as much fun as the later
earthquake testing on rollers and shakers. g

Pop


Pop

Thu, Jan 26, 2006, 5:04pm (EST+5) lid (John*Doe)
doth say:
There's nothing in the question that suggests it's a homework problem.
Maybe I should be flattered for apparently defining the problem well
enough to make the question look academic. snip

Sure there is - what suggests homework is because it's pretty
similar to other questions from students wanting to get out of doing
their school work - college freshmen usually. So, don't be flattered..



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.

Thu, Jan 26, 2006, 10:43pm (EST+5) lid (John*Doe)
deigns ro respond:
Troll

Sigh. Looks like I was right. Just another sad, dorky, troll.
Why else responses like that? Sigh.



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.