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# Re: Foundation A.B. Capacity

• To: seaint(--nospam--at)seaint.org
• Subject: Re: Foundation A.B. Capacity
• From: Jake Watson <jwatson(--nospam--at)inconnect.com>
• Date: Wed, 21 Jul 1999 06:35:03 -0600

```Charles Greenlaw wrote:

> Keep up the comments and questions. But you are going to be forever
> frustrated if you expect any tidy consistency or simple clarity in
> structural parts of the Building Code. The Structural community is so good
> it doesn't need those attributes, and can stand behind a book of rules as
> transparent and illusory as the King's robe, conceived to cover everything
> so elegantly, but an observant newcomer can see right through it.

This is a lesson I am slowly learning, unfortunately it is rather
pessimistic and apparently true. What I have learned also is that when
something doesn't make sense, an "old fashioned" approach will often
work better.

I would like to propose one one additional thought.  (sorry I am on the
topic and would like to feel good about it for future use) If you use
the table given in the UBC (table 19-D) it tells you to multiply by the
edge distance divided by the embedment length, however the embedment
length is fixed.  So what if I calculate the minimum "full cone"
embedment required to fully develope the steel and use that number as my
embedment length.  That way as I increase depth, I get the effect of the
shear cone without additional penalty. So now my calc becomes:

8" FNDN Wall
4" Edge Distance (2 sides) ed = 4"

12" Embedment (le = 12")
Projected area = 106.4 in^2 (Ap), this has taken into account
that the
wall is only 8" thick (I truncated the shear cone).

Therefore, phiPc = (0.65)*(1)*(4)*(106.4)*(2500^(1/2)) = 13.83
kips

Edge distance reduction, (13.83)*(4/16)*(4/16) = 0.86 kips

30" Embedment (le = 30")
Projected area = 251.32 in^2 (Ap), this has taken into account
that the
wall is only 8" thick (I truncated the shear cone - the PCA examples do
this, although the UBC is very vague on the issue).

Therefore, phiPc = (0.65)*(1)*(4)*(251.32)*(2500^(1/2)) = 32.67
kips

Edge distance reduction, (32.67)*(4/16)*(4/16) = 2.04 kips

Still table 19-D lists a 5/8" Dia. A.B. with 4-1/2" embedment and 3-3/4"
edge distance with a capacity of 3 kips (service loads - inspected).

Thank you all for you comments and suggestions. I appreciate the time
thought you have put into them.

Jake Watson, E.I.T.
Salt Lake City, UT

```