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Re: More confusion with concrete

[Subject Prev][Subject Next][Thread Prev][Thread Next] Okay. Set c1 to h/1.5 and let Av exceed nAvo. I do not have my copy of the 2000 IBC handy, but believe they had a reference to thin members which is not part of Appendix D. This is what I think caused the problem. I spoke with Basile Rabbat ( PCA) who directed me to Hilti who, in turn, recommended the Concrete International article. Everybody's guessing was somewhat entertaining. Sorry you let this degenerate into a political discussion. Elias Hahn wrote:

I'm trying to use 1 1/2" anchors spaced at 6" (not my idea, I've requested
the spacing be doubled to 12").  They have an hef of 12" (8d) and I'm trying
to get away with a 18" footing, but I think I'll bump that up.  I am
currently using a 10d edge distance (15".)  This is for the foundation of a
prefab metal structure (thus the spacing that I don't like.)  Ummm,
f'c=4000... I think that's about it.  But I've sort of given up on 6"
spacing, and I've requested that the building manufacture increase the size
of the baseplate they are using, so we'll see if that flies.

-----Original Message-----
From: Scott Maxwell [mailto:smaxwell(--nospam--at)engin.umich.edu] Sent: Wednesday, June 08, 2005 3:46 PM
To: seaint(--nospam--at)seaint.org
Subject: RE: More confusion with concrete

Elias,

What is your particular case (give the specific example with numbers) that
you are trying to look at?  I believe you gave it before, but I deleted
that message.

Regards,

Scott
Adrian, MI


On Wed, 8 Jun 2005, Elias Hahn wrote:

Ahhh see, so there it is.  I didn't catch/understand/pay attention to the
doubling factor.  It still seems a little silly, and for my case with a
relatively big couple of anchors trying to get away with a relatively thin
footing, I'm still losing capacity by increasing my edge distance, which
is
not how it should work.  Or can I double it because the force is in fact
parallel to an edge?  (c1=c2, so either way it is both parallel and
perpendicular to an edge.)

-----Original Message-----
From: Scott Maxwell [mailto:smaxwell(--nospam--at)engin.umich.edu]
Sent: Wednesday, June 08, 2005 3:27 PM
To: seaint(--nospam--at)seaint.org
Subject: RE: More confusion with concrete

Elias,

Yes...you are correct in what you wrote below...but your are forgeting
that there is another c1 in the overall equation for the shear breakout
strength.  Remember the overall breakout strength is Av/Av0*(a bunch of
factors depending on either single anchor or group of anchors)*Vb.  The
key is that Vb is a function of c1^1.5.

So, your case of the long beam in the "big" c1 direction...Av will be
(c2+c2)*h (unless you want to consider a REALLY deep beam such that
h>1.5c1). and Av0 will be 4.5*c1^2.  Thus, Av/Av0 will nominally be 1/c1^2
if you consider c2 and h to be negligble compared to c1.  But then Vb is
some value*c1^1.5.  So, in fact, you end up with Vcb being proportional to
1/c1^.5.  Now, keep in mind that you have some "constant"/value there as
well, so it is not just 1/c1^.5 but rather a/c1^.5.  And then beyond that
section (c) of D.6.2.1 states: "for shear force parallel to an edge, Vcb
or Vcbg shall be permitted to be twice the value for shear force
determined from Eq. (D-20) or (D-21), repectively, with psi6 taken equal
to 1"...and your "big" c1 direction is such a case.  Thus, the end result
is that I am willing to bet in most cases the Vcb of the "big" c1
direction will still be bigger than the Vcb of the "small" c1 direction.

Let's do an example with some numbers.  Say it is a 24"x42" concrete beam
that spans 24 ft.  Put a 1" dia. anchor in the middle of the beam with an
embedment of say 24".  OK.

For the "big" c1 case: c1 is 144".  c2 is 12".  h is 42".  d0 is 1".
l is 8*d0 or 8".  Let's make f'c equal to 3000 psi.  psi7 will be the same
for either case.  psi6 will be 1.  So, Av will be (c2+c2)*h or 1008 sq in.
Av0 will be 4.5c1^2 or 93312 sq in.  So, Av/Av0 is .0108.  Vb is
7*(l/d0)^0.2*d0^0.5*f'c^0.5*c1^1.5 or 1004200.4 lbs.  So, Vcb will be
10845.4 lbs (assuming psi7 is 1.0), but section (C) let's us double that
to 21690.7 lbs.

For the "small" c1 case: c1 is 12".  c2 is 144".  h is 42".  d0 is 1".  l
is 8".  Using f'c of 3000 psi again.  psi6 will be 0.7.  So, Av will be
1.5*c1*(1.5*c1+1.5*c1) or 4.5c1^2.  Av0 will be 4.5*c1^2.  So, Av/Av0 will
be 1.  Vb will be 24157.3 lbs.  Vcb will end up being 16910.1 lbs (again
assuming that psi7 is 1.0).

Thus, here you have a situation that you describe in basics terms but this
time with some specific numbers.  You can see that the "big" c1 direction
does produce a higher breakout condition than the "small" c1 direction.

Now, one could argue that this may still not fit the intuitive sense,
which for me would want to say that the "big" c1 direction should be
SIGNIFICANTLY larger than the "small" c1 direction is such a case and
while it is bigger, I would say not "big enough" to match the intuitive
feel...and I would agree with this arguement.  But, from a structural
point of view in checking the anchor, while it might not be completely
accurate, it does produce the expected result of the "small" c1 direction
being the "weak link".

HTH,

Scott
Adrian, MI


On Wed, 8 Jun 2005, Elias Hahn wrote:

Agreed, the problem I have with formulas is that if you increase c1
until
it
is "big" (ie, 1.5c1 is greater than c2 and/or h), Av/Avo gets small. (or
conversely, have a thin slab, or a corner condition where c1 is roughly
equal to c2.)

Ok, now my long hypothetical - feel free to not read:
The problem in my mind is if you take a long beam, and put an anchor in
it,
and look at the calculated strength of that anchor (side-face blow out)
in
both directions.  Now, in one direction, c1 is "big", much bigger than
either c2 or h, while in the other direction c1 is "small."  Now on the
case
were c1 is "big" Avo is great, but Av is small, (assume c2 and h are
negligible in that direction, Av/Avo tends to 1/(3*c1), which makes the
overall equation roughly (a number less than one)*(Square root(c1)).
Now
if
you look at the other direction, were Av/Avo tends to one, you now end
up
with an overall equation of (a number greater than one)*(c2)^1.5)).  It
is
easy to see that the direction without much edge distance could have a
larger concrete side-face blow out strength...


-----Original Message-----
From: Scott Maxwell [mailto:smaxwell(--nospam--at)engin.umich.edu]
Sent: Wednesday, June 08, 2005 10:21 AM
To: seaint(--nospam--at)seaint.org
Subject: Re: More confusion with concrete

Elias,

First, I would say that you need to be a little clearer about what kind
of
condition you are looking at.  The edge distance, the effective
thickness
of the concrete (h), and the number of anchors can affect the "view" of
things.

Take a simple basic condition...a single anchor in a concrete element
that
is MUCH thicker than the depth of the anchor and is only at a side, not
a
corner.  Kind of like what is shown in Fig. RD.6.2.1(a) and RD.6.2.1(c)
in
the 2002 ACI 318.  In such a case, c2 and h don't enter the picture.

In such a case, then Av is equal to Av0.  Both have a value of 4.5*c1^2.
Av/Av0 will be 1.

Ok, now make it two anchors that are spaced 3*c1 apart.  This will be
that
same as above but Av will be twice as big because the failure surface
(Av)
will still be bottom of two FULL half pyramids with a base of 3*c1 by
1.5*c1.  This is because you now have two anchors with FULL failure
surfaces that do not interesect (remember Av is the failure surface for
the entire group of anchors).  So, Av/Av0 will be 2.

OK, now decrease that space of the two anchors.  The failure "half
pyramids" will now overlap.  Av will become 1.5*c1 (depth of failure
plane) times 1.5*c1+s+1.5*c1 where s is less than 3*c1.  The result is
that Av/Av0 will be less than 2 and approaching 1 as s decreases toward
zero.

I will leave it to you to look at more complitcated, "less perfect"
cases
such as when concrete thickness is not much greater than anchor depth
(i.e. h<1.5*c1) or corners or multiple rows of anchors parallel to the
edge.

This is as it should be.  Av0 is the shear breakout of ONE anchor
assuming
"perfect" edge conditions (i.e. no other anchors to overlap the failure
surface, at just a plain edge not corner, and concrete thickness much
greater than the anchor depth).  Av is the shear breakout of the entire
anchor group, which could in fact be single anchor in the group [if you
have an isolated anchor] or an actual group of anchors, under
potentially
less than ideal conditions (i.e. close spacing, at a corner, and/or
concrete thickness less than 1.5*c1).  Thus, things like spacing,
another
edge (i.e. at a corner), or "thin" concrete will affect Av.  The point
is
that under "perfect" conditions (i.e. spacing greater than 3*c1, edges
not
corners, concrete thickness much greater than anchor depth [i.e.
h>1.5*c1], anchors all in one parallel row/line that is parallel to the
edge) Av will be a integer number multiple of Av0, where the integer
multiple is the number of anchors.  This multiple becomes a non-integer
and less than the number of anchors as the conditions become less than
perfect.

Now, I can only hope that I am remembering/explaining this right...I
am little scatter-brained today.

HTH,

Scott
Adrian, MI


On Wed, 8 Jun 2005, Elias Hahn wrote:

So, I come again to this list with a question stemming primarily from
ACI
318.



I'm "designing" some anchor bolts, and I'm confused about the formula
for
concrete breakout strength of anchor bolts.  Specifically about
Av/Avo.
Because Avo is a function of 1.5*c1, and Av is a function of 1.5*c1,
c2,
and
h - it seems like the larger c1 gets, the smaller Av/Avo gets, which
seems
counter-intuitive.



The reason it gets so small is because as c1 gets large, 1.5*c1 will
be
much
larger than h and c2, so that Av becomes much smaller than Avo.



My confusion is why is Avo a function of c1 if it supposed to
approximate
the "surface area of full breakout prism. unaffected by edge
distance"?

Thank you,

Elias Hahn

Evergreen Engineering, LLC

phone 503.502.0698




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