Log In

Come Join Us!

Are you an
Engineering professional?
Join Eng-Tips Forums!
  • Talk With Other Members
  • Be Notified Of Responses
    To Your Posts
  • Keyword Search
  • One-Click Access To Your
    Favorite Forums
  • Automated Signatures
    On Your Posts
  • Best Of All, It's Free!

*Eng-Tips's functionality depends on members receiving e-mail. By joining you are opting in to receive e-mail.

Posting Guidelines

Promoting, selling, recruiting, coursework and thesis posting is forbidden.


AASHTO 2012 Side resistance for drilled shafts into rock

AASHTO 2012 Side resistance for drilled shafts into rock

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 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

I am using ∝E=0.45 from table 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
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


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

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.

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!

Red Flag This Post

Please let us know here why this post is inappropriate. Reasons such as off-topic, duplicates, flames, illegal, vulgar, or students posting their homework.

Red Flag Submitted

Thank you for helping keep Eng-Tips Forums free from inappropriate posts.
The Eng-Tips staff will check this out and take appropriate action.

Reply To This Thread

Posting in the Eng-Tips forums is a member-only feature.

Click Here to join Eng-Tips and talk with other members!


Close Box

Join Eng-Tips® Today!

Join your peers on the Internet's largest technical engineering professional community.
It's easy to join and it's free.

Here's Why Members Love Eng-Tips Forums:

Register now while it's still free!

Already a member? Close this window and log in.

Join Us             Close