Pile end bearing capacity on rock

[oldestguy]

I'd like to hear opinions on this.

For years I've been estimating pile lengths by combining side friction and end bearing estimates, generally mostly for highway bridges. Usually I can hit final driven lengths within 5 feet of this estimate.

For piles on rock, generally I've taken the driven pile capacity as being governed by stresses during driving, and for length estimates assuming near 100 percent as end bearing, depending on rock type and soundness.

Now comes a railroad bridge designer that wants pile "bearing capacity" of a rather sound sandstone. I'd have given it a near 100 percent of load as the end bearing part. My experience with piles of 60 tons or less is that the field capacity is determiend by driving resistance or possibly a load test. I've given little thought to rock bearing "safety factor" or "end bearing capacity" as a number based upon "strength of the rock". Never had any settle later.

One of my references, "Pile Foundations" by Chellis would indicate there is a significant safety factor for almost any rock that is confined. That is why I don't get involved there.

Apparently the designer would like to use some formula, such as those suitable for soil. He isn't satisfied with something like "100 percent of design load". I suppose I could pull a number out of the air, like 100 tons per sq. ft. I'd want to be high enough so that his driving stresses would then govern.

What would you do?

 

[Mickney]

I am an engineer who practices in the Piedmont. Some of my work involves bridge foundations where rock is relatively shallow.

I am not sure if you have this resource, but it is a very valuable book for any geotechnical engineer...Foundation Design and Construction by MJ Tomlinson. Section 7.11.1 "The carrying capacity of piles founded on rock - driven piles" should answer your questions.

I have extracted some sections verbatum for your use.

"The ultimate bearing capacity of a rock mass with open or clay filled joints is no more than the unconfined compression strength of the intact rock. Where the joints are widely spaced...the ultimate end-bearing resistance may be calculated from the equation qub=2*Nphi*quc. Nphi is tan squared of 45 + phi/2. quc is the uniaxial compression strength of the rock.

 

[Panars]

When driving piles to refusal on sound rock, the structural capacity of the pile is what controls. The pile length is simply going to be the depth to rock. It sounds to me like the designer is worrying too much.

I believe the piedmont rock is usually pretty weathered (I could be wrong on this however). So the text from Tomlinson about rock with open or clay filled joints would apply for that region. But for a sound bedrock (one that is intact, RQD>95) without any open joints, the equation quoted doesn't apply. The rock is going to have more capacity than the load any driven pile can apply to it.

 

[fattdad]

I think you can use qsubp=9Su (or 4.5 Uc). What you need though is the unconfined compressive strength of the rock mass (rather than just the intact rock. You also need a factor of safety. So, if you use a safety factor of 2.5 (for single piles) you'd end up with 1.8 Uc for a design tip bearing capacity. If you're interested in an overview of rock mass properties, you can check my thesis:

http://home.comcast.net/~fatt-dad/thesis.pdf

Hope this helps. I'm just looking at my old college notes from J. M. Duncan.

f-d

¡papá gordo ain’t no madre flaca!