Cast in-situ BORED pile with working load of 15000 KN 1

[Kpaudel]

Dear friends,

Please give yours opinion regarding possibility of cast in-situ bored pile with working load of 15000 KN in soft claystone to mudstone having SPT N value less than 30 around 25-27 and undrained shear strength of about 150 KPa.


Give me very rough idea regarding length and size of cast in-situ bored pile based on your experience...

 

[jayrod12]

You are going to need a gigantic pile to resist loading like that. I'm not even sure how you can get that much load going into a single pile. The framing members above must be ridiculous. 15000kN or 3 million pounds just does not seem likely.

 

[hokie66]

What you are describing is a very large "caisson". If the rock had a bearing capacity of 1 MPa, which this mudstone doesn't, the bearing area required would be 15 square metres. If stronger material is not accessible further down, you may have to use a group of driven piles.

 

[Kpaudel]

Dear friends

My structure engineer provided this much of load on each pile...probably due to 45 m long PSC girder beam bridge having pier height of about 30 m from ground...they proposed with 4 nos of pile in each pier....but I am not sure each pile can resist such a huge load or not, There is hard rock in about 160 m below the proposed pile cap level...only possibility is cast in-situ pile due to very fragile nature of slope and unavailability of machinery...driven pile is almost impossible...

 

[civeng80]

As jayrod said that is one mighty load.

Do some rough computations e.g. look up my post (Capacity of a bored Pier in clay) and you will see that one bored pile taking such a load would be mega huge.

 

[hokie66]

I don't see why there would be much difference in accessibility for driving piles vs. boring. The boring rig is going to be large and cumbersome.

 

[AndreBC]

Kpaudel :

I will give you the rough idea if a single pile were installed :
For end bearing capacity, you got :
Qp = 9 Su Ap = 9*150*Ap = 1350 Ap (in kN) with Ap = cross sectional area of the pile = 0.25*3.14*diameter^2
For friction capacity, assuming that Su would be 150 kPa from top to bottom, you got :
Qs = adhesion*Su*As = 0.385*150*As = 57.75 As with As = sleeves area of the pile = 3.14*diameter*length of pile

Your ultimate capacity would be : Qu = (1059.75*diameter^2) + (181.335*diameter*length of pile)
Assuming safety factor of 2.5, for Qu/SF should be larger than load 15000 kN, you got :
15000 kN < (423.9*diameter^2) + (72.534*diameter*length of pile)
Try to input the diameter and length of the pile by yourself to get the idea, then you'll realized that it doesn't make sense, unless you use a group of piles (many piles, not just 4 piles).

For the idea of group piles, first input the diameter and length of pile that you think make sense for a single pile:
for example D = 1 meter and length of 30 meters. You'll get allowable capacity for a single pile. Divide the load of 15000 kN, with the allowable capacity, then you get number of piles required (theoretically, it should be reduced with factor of group efficiency, but I ignore it for simplicity).

I'm just curious, is the load of 15000 kN came from a gravity load nad live load, or already combined with seismic load ?
If it came from seismic load, I suspect that your engineer use static load method to calculate seismic load, which is very conservative.

 

[Kpaudel]

Dear Andrew sir,

Very thank you for your idea... I am sorry for short information...

Actually,

After all Cu is about 150 Kpa from laboratory test result… I am bit confuse with information provided by Structure Engineer… He provided Service load of 15000 KN and Ultimate load 16480 KN… Then I had made some geo-technical calculation with 2.5 m dia. pile… with Cu 150 Kpa…minimum alpha 0.3 and K of 0.5…

Only 85 m long pile have Ultimate Bearing Capacity 37500 KN and Allowable Bearing capacity is 15000 KN…..with FoS 2.5.

Now my question is how can I compare geo-technical ultimate and allowable capacity with ultimate and service load provided by Structure Engineer???

But same time his designated allowable settlement is only 1 inch (i.e 25 mm), But as per my calculation settlement of 85 m long, 2.5 m dia. pile with allowable load of 15000 KN will be more than 400 mm….

What is the best solution for this foundation? As he proposed 4 piles @ 3D spacing already pile cap is around 10.0 m wide (Square). Pile Steam have radius of only 3.0 m so there is no option for more piles…even client is not interested on others type of foundation than pile.

 

[AndreBC]

Kpaudel :

Structural engineer usually used load combinations based on code such as IBC :
a.DL (dead load) + LL (live load)
b.1.2 DL + 1.6 LL
c.0.9 D + E (earthquake load), etc.
Noted that some of the combination is factored with SF (1.2 for dead load, 1.6 for live load for examples).
They usually use this factored load combination to calculate structural reinforcement (beams & columns), and the maximum forces result from all of the combinations is the ultimate load. While the service load is the result from DL and LL only (not factored).
I don't know about the others, but I usually use Qu / SF with SF = 3.00 for service load, and Qu / SF with SF = 1.50 for ultimate load. My reason is the ultimate load only occur temporary within a short time (such as wind and earthquake), so I could take SF as much as half SF value for permanent load.

For the best solution, I suggest you should contact a geotechnical consultant expert in your area. Don't take it too lightly, foundation is not a minor issue, it could lead to a disaster especially with an important structure such as bridges.

 

[Kpaudel]

Dear friends, Is there any limiting value for ultimate skin friction and tip resistance, based on IS code limiting ultimate value for Tip resistance is 11Mpa and limiting ultimate skin friction should not be more than 70 Kpa…is that justifiable??