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Re: Foundation A.B. Capacity

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The reason you are getting bad results is not so much in the math, but in
the approach.  You are double penalizing your condition.

Try calculating the relative capacity of each embedment without truncating
the cone, and then apply the proportional reduction(s).

Since you are looking at a stem wall, you are dealing with an extreme
condition.  If you were looking at single 4 and 6 inch nelson studs with
only one edge reduction the intended methodology would be more apparent.
Notice that there is no mention of calculating a reduced cone area due to
edge distance in the definition of Ap.

There are some excellent references available regarding the shear cone
approach.  One of the best used to be available for free from TRW with their
nelson stud catalog.

Hopefully I am communicating clearly, it is the end of a very long day :-)

Paul Feather

----- Original Message -----
From: Jake Watson <jwatson(--nospam--at)>
To: <seaint(--nospam--at)>
Sent: Tuesday, July 20, 1999 7:01 PM
Subject: Re: Foundation A.B. Capacity

> Seaintonln(--nospam--at) wrote:
> >
> > Jake, here is some comments:
> >
> > 1. Typically, the concrete will fail before the steel does. I am not
> > what type of foundation you have, but the deeper the embedment, the
> > the cone (concrete cone failure). If you are considering a stem wall for
> > raised foundation, your capacity should not (in my opinion) consider the
> > as helping to resist the tension. I would develop the capacity in the
> I agree with you, but here is the exact problem:
> 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)*(4/12)*(4/12) = 1.54 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).
> Therefore, phiPc = (0.65)*(1)*(4)*(251.32)*(2500^(1/2)) = 32.67 kips
> Edge distance reduction, (32.67)*(4/30)*(4/30) = 0.58 kips
> See the dilema?  Note the edge distance reduction seems to overpower the
> increase in depth.  Because it is on two sides, the reduction must be
> taken twice.  As a side note. This also happens with only 1 truncated
> edge, but not to the same degree.
> > 2. This generally means designing the footing of the all to resist
bending so
> > as to distribute enough load into a rectangular foundations to allow for
> > sufficient concrete to act as a "deadman" before failing as a beam.
> > this is starting to get a bit beyond my talents (since I would have
created a
> > deeper deadman type resistance for the anchor) I would assume that you
> > use the stem of the foundation wall and model the foundation as a "T"
> > in bending.
> I have designed the footing / foundation wall as a grade beam to
> compensate for the uplift.  The real problem is getting the forces into
> the foundation in the first place.
> Can anyone see a flaw in my logic (or bad math?)
> Thanks again,
> Jake Watson, E.I.T.
> Salt Lake City, UT