Kern distance question 1

[diarmud]

Hello,
I'm new here and have a question. concerning the kern distance and eccentricities. I understand the kern distance for a rectangle is the dimension we're working with divided by 6.
I'm working on some short circular piles that have unfortunately been loaded eccentrically.
Do I use the diameter divided by 6?

I cannot find the answer and this will drive how we fix the problems.

Hope I'm in the correct forum.
Thanks,
Dermott

 

[Lion06]

I am coming up with a value of (r^3)/4.
I did this by writing the expression P/A-M/S=0
Then to P/(pi*r^2) - 4*Pe/(pi*r^3) = 0

solving for e, I get (r^3)/4.

Hope that helps.

 

[swearingen]

Shouldn't it be e = r/4?



If you "heard" it on the internet, it's guilty until proven innocent. - DCS

 

[diarmud]

Unfortunately I don't know how the l/6 was derived but it is for a square or rectangle.

My question probably should have been how was this determined, empirical or derived.

It's shown on page 791, MacGregor & Wight, Fourth Edition, "Reinforced Concrete", Mechanics and Design

I honestly have really looked hard for this one,

Thanks for the suggestions,
Dermott

 

[Denial]

As exemplified by the formula given by StructuralEIT, the outer boundary of the kern of a section is the set of points at which an applied axial load will cause the stress distribution across that section to be zero at one extreme.

For a solid circular section Swearingen has correctly given the solution for the kern as a circle whose radius is one quarter of the radius of the section.

For a solid rectangle bent about either of its principal axes, the solution is even easier, and you already have the answer (but do try working it out from first principles). That gives you four points on the kern's perimeter. The full shape of the kern for a solid rectangle is the rhombus that has these four points as its vertices.

 

[civilperson]

No, the eccentricity being greater than the middle third puts a fraction of the bearing area in a tension zone. Piles with skin friction can generate tension strength as opposed to footings which can not glue themselves to the soil. Piles do not use triangular shape soil pressure bearing to resist moments, (as do rectangular footings), the surrounding soil resists the bending and pivoting of the pile itself.

 

[youngstructural]

How short are the piles? It's key to whether the solution offered by StructuralEIT swearingen & Denial or the approach proposed by civilperson is correct...

This is much like the question of what is a wall vs. what is a column... Except in this case you're looking at whether you have a long, thin section that behaves as a flexible pipe when you load it with moment (pile) or as a block that rotates (post hole found)

Personally I consider a pile to be anything with a base less than 1/10 of length, and a post hole found to be anything with a base greater than 1/4 of the length/depth. Anything in between is a matter of judgement, and must be based upon the amount of reinforcing provided and grade of concrete (ie: The actual stiffness of the element).

IF you DO have a post hole foundation, you need to consider a completely different behaviour regarding overturning, more akin to a retaining wall's face loading than to a footing's base loading... This is what I believe civilperson was alluding to. And furthermore, you need to differentiate between a post hole foundation that is fixed at the top and one that it free to rotate at the top. I have long-hand solutions for the formulae as well as the formulae written up; Please let me know if you need them.

Cheers,

YS

B.Eng (Carleton)
Working in New Zealand, thinking of my snow covered home...

 

[diarmud]

The are 16 at 16 inches diameter and length is no less than 5 feet and no more than 6 feet. differences in length is because someone can't measure properly.

They are spaced about 5 feet apart and carry a fairly light load.

Any comments would be helpful.
Thank you.

 

[youngstructural]

I make them 0.2222:1 width to length... I'd say they will likely behave as postholes, and civilperson's approach will be the way to go. That said, you really need to check to see if they are stiff enough to behave this way!

Okay, so presuming they do act as postholes, please see my attachment for a way to calculate the stresses on the soils involved. Do you own check of this design procedure; It's fairly simple from first principles... And last time I typed a formula in I missed terms, hence the scan!

Hope it helps,
Cheers,

YS

B.Eng (Carleton)
Working in New Zealand, thinking of my snow covered home...

 

[DaveAtkins]

What do you mean by "eccentric load?" Piles in a group only take axial load, unless there is some horizontal load causing bending in the piles. If that is the case, the piles should be checked for shear plus bending. In any case, I don't think the Kern limit has anything to do with piles.

DaveAtkins

 

[apsix]

I would guess he means a single pile, not a group.

 

[youngstructural]

I'm hoping for a bit of clarification here too, however I am guessing that there is either an ofset axial (read: compressive column-like) load, or a moment into the pile cap... Since the OP was regarding field error, it's almost certainly super structure not alligning with pile.

There is also the easy-ticket solution of grade beams between the piles... But I assumed (which I shouldn't) that this was not a feasible correction.

Best advice: Post your original detail, and a photo of the field condition.

And again, please, please, please have the courtesy to read the thread before posting!

Cheers,

YS

B.Eng (Carleton)
Working in New Zealand, thinking of my snow covered home...

 

[Lion06]

I apologize, but someone above caught it, it should be r/4.
That is simply the location of load that will cause 0 stress at one edge, except as modified by youngstructural.

The derivation for L/6 is given by:
P/A - M/S = 0
A = BL
M = Pe
S = (BL^2)/6
therefore
P/(BL) - 6Pe/(BL^2) = 0
solving for e:
e=L/6

The definition of the kern is the location from the neutral axis in which the entire section is in compression (or full bearing in the case of a footing).

 

[ARLORD]

q=P/A +/- M/S

M = Pe
A = pi x r**2 FOR A CIRCLE
S = (pi x r**3)/4 FOR A CIRCLE

q = P/(pi x R**2) +/- 4Pe/(pi x r**3)
q = P/(pi x r**2)(1 +/- 4e/r)

SET q = 0
0 = 1 +/- 4e/r

e = +/- r/4

 

[diarmud]

I apologize for not describing the problem properly. This is a 400 SF 2 storey (total of 800 SF) residential addition.

The piles were loaded eccentrically because the contractor had no idea what he was doing. They should have used footers but that is too late.

I was called in to see how to rectify the situation since the Building Inspector is concerned. He is correct, the foundation will not take the loadings.

My goal is to correct any eccentricity I can, then add at least 4 footers to bring the foundation up to code in terms of loading.

Thank you all, next time I will more clearly explain the problem.

Dermott

 

[civeng80]

For a column of circular cross section r, the kern is a circle of radius r/4.

 

[tngolfer]

Can you add some additional piles on the eccentric side?

 

[diarmud]

Unfortunately, the structure is finished.

I really should have stated the situation more clearly.

My goal is to attempt to minimize the eccentricity by adding blocking (in the center of the pile) under the pile up to the floor joists.

I also intend to add some footers/columns on the edges and run some beams between them.

Thank all of you for your help.