AASHTO 2012 Side resistance for drilled shafts into rock

[eit09]

To resist uplift to a footing that sits directly on solid granite my customer plans on drilling a 5.5" diameter hole, 10 feet deep with 2 #8 rebars embedded into the hole with 4000 psi concrete.I came accross equation 10.8.3.5.4b-1 on page 10-137 of 2012 AASHTO book to calculate to calculate the side friction. I found an examples within a presentation that raised a few questions. I would appreciate any comments or feedback on the 3 items listed below.


qs=0.65*∝[sub]E[/sub]*p[sub]a[/sub]*(q[sub]u[/sub]/p[sub]a[/sub])[sup]1/2[/sup] < 7.8*p[sub]a[/sub](f[sup]'[/sup][sub]c[/sub]/p[sub]a[/sub])[sup]1/2[/sup] ______Formula 10.8.3.5.4b-1

I am using ∝[sub]E[/sub]=0.45 from table 10.8.3.5.4b-1 to be conservative
For compressive strength of granite I am using q[sub]u[/sub]=2736ksf
Atmospheric pressure p[sub]a[/sub]=2.12ksf


1. Should I use the lower value of f[sup]'[/sup][sub]c[/sub] and compressive strength of the rock for q[sub]u[/sub] within the formula?
2. The formula doesnt mention it but do I also need to multiple by the resistance factor of 0.4 from table 10.5.5.2.4-1?
3. is the 7.8 in the 7.8*p[sub]a[/sub](f[sup]'[/sup][sub]c[/sub]/p[sub]a[/sub])[sup]1/2[/sup] formula a conversion and truly need to use ksi units for f[sup]'[/sup][sub]c[/sub] & ksf units for p[sub]a[/sub]?

 

[BigH]

I cannot address AASHTO - don't have it. I have included something that might help you . . . see attached.

As a matter of curiosity, how did you ever come up with qu = 2736 ksf? That seems unduly accurate.

 

[eit09]

BigH,

The q[sub]u[/sub]is actually just = 19000 psi. Thanks for the attachment! What reference does the attachment come from?

 

[BigH]

See Tomlinson's Pile Design and Construction Book (7th Ed for example). Also see Seidel and Collingwood's "A New Socket Roughness Factor for Prediction of Rock Socket Shaft resistance" Canadian Geotechnical Journal, 2001 (pp 138-153).

I was making a jab that your "exact" approach to the qu value needs to be tempered. 19,000 is to 2 significant figures meaning that, if there is no "fuzziness" in the value - it could range from 18,500 to 19,500. You have, in ksf, given the accuracy to 4 significant figures which, in my opinion, is a bit unreasonable - almost bizarre! The accuracy of these values are no where close to that precise.

 

[eit09]

BigH,

Thanks for the references! There was no relevance for me to post the q[sub]u[/sub]or p[sub]a[/sub] values not sure what I was thinking!