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Author: Calvin
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2003-01-30 |
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If the calculated NSF on a square driven pile is say, 20 ton while the max. load from the column onto the pile is say, 15 ton, then should the pile be driven to 35 ton (based on Hiley formulae)?
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Follow-up:
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Author:
Bengt H. Fellenius
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2003-01-31 |
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With the letter "NSF" is usually meant "dragload" (the load accumulated from negative skin friction, n.s.f.). If a pile has a dragload of 20 tons plus a dead load on the pile head of 15 tons, then, the positive direction forces must amount to that same total amount of 35 tons and the pile will have a bearing capacity of a minimum twice this, that is, 70 tons. The factor of safety is the 70 tons divided by the dead load of 15 tons, that is, 4.7. (Now, where does the 35-ton value come from to which the pile was to be driven?) If you accept this factor of safety as indicating a completed and conservative design, perhaps you would be right, and perhaps you would not. It all depends on what the settlement is at the point of equilibrium, the neutral plane, where the maximum load of 35 tons occurs. If the soil settles here, then the pile will settle an equal amount and the design is in trouble with the serviceability.
If you have difficulty following my reasoning, go to the search function and search the Discussion Forum for "dragload". You will see a large amount of previous entries on the same question and I suggest you read them.
I am aware of the I used the term "bearing capacity" in my reply, which may appear to conflict with my other reply this evening. To avoid implying a conflict between the replies, for a pile, bearing capacity is defined not really by a formula but by the maximum shear along the pile shaft (shaft resistance has an ultimate value; a "shaft capacity" does exist) plus the resistance at the pile toe occurring at an enforced toe movement of about 5mm to 10mm, as a toe capacity does not exist.
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Subject:
NSF in Pile Driving 