AASHTO 2012 Side resistance for drilled shafts into rock
AASHTO 2012 Side resistance for drilled shafts into rock
(OP)
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*∝E*pa*(qu/pa)1/2 < 7.8*pa(f'c/pa)1/2 ______Formula 10.8.3.5.4b-1
I am using ∝E=0.45 from table 10.8.3.5.4b-1 to be conservative
For compressive strength of granite I am using qu=2736ksf
Atmospheric pressure pa=2.12ksf
1. Should I use the lower value of f'c and compressive strength of the rock for qu 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*pa(f'c/pa)1/2 formula a conversion and truly need to use ksi units for f'c & ksf units for pa?
qs=0.65*∝E*pa*(qu/pa)1/2 < 7.8*pa(f'c/pa)1/2 ______Formula 10.8.3.5.4b-1
I am using ∝E=0.45 from table 10.8.3.5.4b-1 to be conservative
For compressive strength of granite I am using qu=2736ksf
Atmospheric pressure pa=2.12ksf
1. Should I use the lower value of f'c and compressive strength of the rock for qu 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*pa(f'c/pa)1/2 formula a conversion and truly need to use ksi units for f'c & ksf units for pa?





RE: AASHTO 2012 Side resistance for drilled shafts into rock
As a matter of curiosity, how did you ever come up with qu = 2736 ksf? That seems unduly accurate.
RE: AASHTO 2012 Side resistance for drilled shafts into rock
The quis actually just = 19000 psi. Thanks for the attachment! What reference does the attachment come from?
RE: AASHTO 2012 Side resistance for drilled shafts into rock
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.
RE: AASHTO 2012 Side resistance for drilled shafts into rock
Thanks for the references! There was no relevance for me to post the quor pa values not sure what I was thinking!