Why is the resultant divided 2(R/2)when the downward compression way meets the shaft friction? Why is it not some other ratio which would be dependent on the value of the upward friction?
here's a little "better" explaination (straight out of my manual): as the force wave travels down and encounters the friction on the sides, two friction force waves are created at that point. one force wave is compression upward and the other is tension downward...these waves must be equal and opposite so the total friction resistance at that point is Rcomp+Rtens with each 1/2 the total friction resistance at that point. if you have say a minimal friction force, that doesn't necessary mean the input force at the top is immediately cut in half. small friction equal small resistance force at that point. big friction equal big resistance force at that point. the summation of all the friction resistance should approximately equal the input force at the top.
one thing to point out is that you can have higher tension forces say if you have hard driving passing through a hard layer in to a very soft layer. you could also have high stresses for hard toe conditions. i believe the theory says you could approach 2x the input at the toe for the hard toe condition with no/minimal side friction.
msucog, you are correct, for a fixed toe, the theory does state that you can experience 2x the impact force.
Conversely, for a free toe, you reflect a tension wave, which is why tension cracking has to be monitored during the early stages of pile driving for concrete piles.
A quick program which is good to illustrate this and many other aspects of wave propagation is the Pilewave program, which can be downloaded from
Well, those are the extreme cases, which leave only the "general case," in which the pile has finite length, and at each point along the pile there is resistance greater than zero and less than infinity. That's when the numerical solutions (WEAP, CAPWAP, Case method) come in.