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FW: Foundation A.B. Capacity

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Not to pile on, but I have a suggestion for your use in this problem.
Sometimes applying coefficients to formulas leads to a loss of understanding
of what is actually represented.  Ignoring the footing below, it might be
best to visualize the truncated pullout cone, calculate it's projected area
and then apply stresses to it.

For a 45 degree cone, and 12" embed,

Ap= 8*24=192 in^2    (ignoring radius at each end and area of rod head/nut)
phi*Pc=.65 * 4 * sqrt(2500)* 192/1000= 25.0 kips

for 30" embed

Ap= 8*60=480 in^2
phi*Pc= .13 ksi * 480 = 62.4 kips

Then, as others have suggested, start thinking about other actions involved
such as stem wall bending and adequate dead load to offset uplift.  And if
the stem wall in now bending, you've attached to a tension zone and the
special extra load factors from UBC 1925.2 apply.

I hope this has helped.

Curt La Count
Jacobs Engineering
Portland, OR

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

        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

        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