Pile Supported Wall - Foundation Engieering by Peck Hanson Thornburn 4

[CWEngineer]

I am using the Pile Supported Retaining Wall Example on Page 436 as a reference for a problem that I am working on.

Here are the loads I have about point A:
Sum V = 13.16 k
Sum H = 9.39 k
Sum M About Point A = 176.55 k-ft clockwise, (I have water pressures). With this type of loading my vertical resultant is near the edge of the toe, but still within the footing.

I am following the same pile layout right now, except I am using 5.5 ft spacing instead of 3 feet.

When I calculate the Vertical Components of Pile Reaction, I get a positive value for Row 1, a positive value for Row 2. BUT I get a negative value for Row 3.

Do you guys know what this negative value means? Also do you guys have an idea of what I can do to make this value positive?

THANKS

 

[BigH]

The pile is in tension.

 

[CWEngineer]

Do you have any advice on what I can do so that, that pile will not be in tension. I added another pile in row 2, but I still get a tension on pile 3. Should I make the footing longer? Or do you have any other suggestions?

Appreciate your help.

 

[kxa]

I don't have the book but you are referencing but make a suggestion: Can you move the stem (wall) so that it is closer to piles in row three?

 

[CWEngineer]

Yes, I can do that. If I do that I will be decreasing my heel lenght and increasing my toe length. Do you still recommend I do that?

TAHNKS

 

[CWEngineer]

I also, can also move both my batter piles about 2.5 towards the driving side. Or I can move my middle pile 2.5 away from the driving side. Should oen of these approaches help me out?

THANKS

 

[UcfSE]

Really the best way to figure out what will help you is to do both kxa's suggestion and your own and see what happens to your results in each case. Then you will see how changing something like stem wall location affects the resulant forces and pile forces and so on.

 

[PSlem]

I typically switch heel and toe dimensions on pile supported walls from standard practice with two rows of piles but still have one in tension. There is a resultant bearing point on the footing. If you want them all in compression keep them on the toe side of that point, but a tension pile design is the most economical.

 

[BigH]

One important thing to remember is that just because the pile is in tension doesn't mean that it has no capacity. All piles in tension have the capacity of the adhesion/skin friction that must be overcome before they offer no resistance. It is typically taken, though, that the capacity in tension is much smaller than the capacity in compression (see Tomlinson's 'Foundation Design and Construction' - PH&T should have some comments on uplift as well). I would see what the magnitude of tension is, then determine if this would exceed the capacity of my pile in tension uplift - say use a FS of 6 for tension rather than 3 for skin friction/adhesion. Do you have enough?

 

[kxa]

BigH, I agree with your posting regarding the uplift capacity of the piles. The uplift load may not even overcome the friction capacity + Wt. of pile. The footing attachment to the pile however will have to be designed for uplift.

Can you explain the safety factors a bit more. I was under the impression that we use FS=2. If we test a pile to the failing point, the pile capacity becomes the load at failure divided by 2. Can you please elaborate on the 2, 3 and the 6 safety factors.
Thanks

 

[CWEngineer]

Thanks for the help, just wanted to confirm some calcs. - if someone has this book.

On page 437 the following equation is being used for the Vertical Components of Pile ReactionsSum V/n) +/- (Sum M *d/ Sum d^2). Is the second term positive when "d" is to the left of C.G. and negative when "d" is to the right? I think there is more to it, but I can figuere it out since a vertical load is being calculated.

Appreciate Your Help

 

[BigH]

Quoting from Tomlinson (my appointed guru on piles)
"Various published test results have indicated that the skin frictional resistance of piles to uplift loads is appreciably lower than that mobilied in resistance to compression loading. A reduction of 50% has been suggested for granular soils and the same order of reduction for long-term sustained loading for clays. . . . the appropriate safety factor should be applied to the skin friction values calculated by the methods described earlier . . . and if the uplift load on piles necessitates mobilization of skin friction approaching the ultimate values, it will be advisable to carry out uplift tests on selected full-scale piles to ensure that there is an adequate safety factor. . . jetty and wharf structures are subjected to lateral forces from berthing ships and from wave action. If these forces are transferred to supporting piles, the resutling lateral movement of the upper par tof the embedded length of the piles may destroy most of the skin friction . . . a generous factor of safety on the ultimate poull-out load should be adoped for piles carrying both lateral and uplift loads."
Usually one uses a SF of 3 (or 2.5 on piles) so my "suggestion" of 6 is the halfing of the skin friction loading based on skin friction/adhesion reduction. Of course, you can do tension tests to get more appropriate values. I've done this before and on timber piles we had 19 tons (vs 40 tons in compression - although the later would have included some base mobilization and was not loaded to failure).

 

[chichuck]

gman1,

I have that reference. Your statement is correct, but there's really not much more to it than that.

The equation used is simply P/A +/- Mc/I. This is simple statics. Looking at a vertical load,

Sum(V)/n is the P/A term.

Sum(M) * d/ Sum(d^2) is the Mc/I term; for piles, "I" is SUm(d^2), "c" is d.

To apply this equation, you must remember to keep your signs straight; a downward vertical load == positive (= compression in the pile) and upward load == negative (= tension in pile). When you define signs for the moment, counterclockwise moments == positive. the example uses distances defined as left of cg of pile group. Keep the correct sign on d as well.

Do the algebra, keep track of signs & it works. You'll note that for all rows, M is ccw = positive.

for row 1 M is positive, d is positive (row is left of cg of pile group), so Mc/I is positive, this is compression in pile.
for row 2, M is positive, d is negative row is right of cg of pile group), so Mc/I is negativel, this is tension in pile.
for row 3, M is positive, d is negative row is right of cg of pile group), so Mc/I is negativel, this is tension in pile.

Is this some help?

Regards,

chichuck

 

[CWEngineer]

Thanks chichuck,

That was very helpfull. Just one more quick question.

In the last section you mentioned:

for row 1, Mc/I is positive, this is compression in pile.
for row 2, Mc/I is negative, this is tension in pile.
for row 3, Mc/I is negative, this is tension in pile.

But the final results are:
Row 1: +22.7 k
Row 2: +21.9 k
Row 3: +21.0 k

So this means that all the piles in Row 1,2 and 3 are in compression?

THANKS

 

[chichuck]

gman1,

I meant that only the Mc/I portion is positive or negative, meaning compression or tension in the pile.

the P/A part (for piles, P/npiles) is always positive = compression.

P/A +/- Mc/I = total force in the piles. These are the numbers you refer to in your last reply; they are all positive so that meand all are compressive.

BTW, I looked over that example closely. I think the author has made an error. At the bottom of Sht.1 he calculates the resultant V, H and M loads on the pile group. H is correct. V is incorrect because it omits the value of Pv (the soil force acting vertically through point A; reference his example DP 26-1). M is incorrect because it only includes the eccentricity of the vertical load acting through the point at 0.83' from the cg of piles (and that is off by a little, as I said), but does not include the moment effect of the Ph acting at a height 5.33' above the piles. Correcting this, you would have a larger M than is shown in the example. Thus, the Mc/I term is larger for each pile. Some of the piles might actually have tension on them.

regards,

chichuck

 

[chichuck]

gman1,

I apologize for making two errors in my most recent reply.

1) the value of Ph from the figure given in HTP is actually zero, for flat backfill. His value of V is correcat.

2) he does in fact account for the fact that Pv is acting at 5.33' above pile group. He does this when he calculates the resultant of the vertical forces. So his value for M is also correct.


I've crafted a spreadsheet with the calculations necessary, including the ones that the book example omits with the words "similar to DP 26-1". that might make it more clear. I will post again with my website address when I have it up and available.

Sorry for the error.


Regards,


chichuck

 

[CWEngineer]

Thanks for the clarification to my questions, you have been very helpfull.

Look forward to seeing to spreadsheet to check my hand calcs.

THANKS
gman1

 

[chichuck]

gman1,

I got it posted and it appears to work now.

my website address is:

http://home.mn.rr.com/gbbadger/

download the spreadsheet hansen thornburn and peck.xls

regards,

chichuck

 

[CWEngineer]

Thanks,

Got it. It was very helpfull. I had another quick question. I am looking at the possibility of using "one" big pile to support the loading, Is the approach in this example still appropriate, or is their other things I need to consinder.

THANKS

 

[chichuck]

gman1,

I think this example is not applicable to one big pile. With one pile, the entire load is resisted by that one pile, so the Mc/I portion of the calculation is not correct. I think one pile will resist the moment through some lateral soil resistance along the pile length. That is an entirely different case. I'm not sure how to analyze tha though.

Perhaps someone else on the forum can help.


Regards,


chichuck