Eccentrically Loaded area supported on 4 piers 1

[Pete1919]

Hello all,
I need help please. I have 6'(say, x-axis) x 4'(Y-axis) area with a point load of 100 tons acting 2' from short neutral axis (ex = 2') and 1' from long neutral axis (ey=1). The area will be supported on four (4) piers. What will be the load distribution on 4 piers? Weight of the pad (area) will be ignored.

I calculated moment and calculated stresses at 4 corners using p/A+/-Mx/Ix*y +/-My/Iy*x. I interpolated stresses at mid-points between corners. Stress at center would be 100/24=4.17 tsf. Then I assigned tributary areas (four quadrants) to each pier. Took "average" stress for the quadrants multiplied by quadrant area of 6 sq.ft to get pier loads. The pier loads were 69, 31, -19 and 19 tons.

Using different method of 25+/-My/(6*2) +/- Mx/(4*2), I came up with the pier loads of 54, 29, -4, and 21 tons.

I am trying to find which method and answer is correct. Any comments or calculation will be appreciated. Thanks.

 

[Guastavino]

Have you tried to create a pseudo moment of inertia for the Ix and Iy, where Ix=4*3^2=36piers*ft^2 and Iy=4*2^2=16piers^ft^2
Then the "stress" P/4piers+My/Ix+My/Iy=200k/4piers+(200k*2')*3'/(36piers*ft^2)+(200k*1')*2'/(16piers*ft^2)=83.33k (41.667tons)

Forgive me if my math is off, late at night here, but point is, I would do P/A+MY/I concept but my "A" is a pseudo Area, by that I mean the units are in piers, and my "I" is a pseudo moment of inertia, where my units are piers*ft^2

Hope that makes sense. This also assumes rigid rotation, etc.

 

[hokie66]

So are you saying the plate will be supported at the four corners with the load 1' in from one end, and 1' in from one side? If so, I get the reactions to be 62.5, 20.8, 12.5, 4.2.

 

[Pete1919]

I don't think this is correct. At least one pier load has to be negative since the applied 100T point load is outside the kernel of the section. I wonder what was your approach to arrive at 62.5, 20.8, 12.54.2. Please clarify.

 

[hokie66]

It is "kern", not "kernel", but that only applies when there is contact on the entire base. When you have discrete support points, and the supports do not impart fixity, it is just a simple beam in one direction, and a simple beam in the other direction. 100 x 5/6 x 3/4 = 62.5

 

[Pete1919]

niltzwe,
Yes, the math is off. Correcting the math, your answers are same as my second set (54,29,-4,21 tons) even though your approach is different. I do not understand pseudo M of I concept. Could you please explain.

Besides, I would like to find out what is wrong in my first approach (69, 31,-19, 19 tons). The concept seems to be correct, I think. So both the answers should match, but they don't.

Thanks.

 

[Pete1919]

Hokie66,
Thanks for the answer. Also thanks for clarifying that Kern (or kernel or core or whatever) applies when there is contact on the entire base. If that is true, my entire first set of answer (69,31,-19,19) and underlying concept are incorrect. Is it true?

Now let us extend your approach to 100T point load application at the corner itself at 3' from the y-axis, and 2' from the x-axis. Then, the pier just below the corner will have a load of 100T and all other piers will have zero loads. Is this your understanding?

Thanks for your time and help.

 

[hokie66]

Yes, exactly. What you have is just a table with four legs. You can't get uplift on a leg unless the load is outside the top of the table.

 

[Pete1919]

hokie66,
That sure makes sense. What do you think of the other set of answers: 54, 29, -4, and 21 tons, which are also the answers of njlutzwe. Any comments?
Thanks.

 

[hokie66]

Not really. I didn't try to go through it. Just seemed to me that you were making too much of a simple force distribution.

 

[Hokie93]

Using the same approach I would use for distribution of an eccentric load to piles in a pile cap (njlutzwe's approach), I get 20.83 tons, 54.17 tons, -4.17 tons, and 29.17 tons for the leg reactions. This assumes the slab is perfectly rigid and that each leg is located in the corner of the 4'-0 x 6'-0 area. These results also ignore the weight of the slab. With the load location as defined, the system is stable (it won't overturn) but that does not preclude one leg from being in tension.

 

[hokie66]

Two Hokies disagreeing. Hokie93, do you agree that if all the load is placed directly over the corner column, that column would take the entire 100 tons? Or if all the load was placed at a point along one side, the two columns on that side would share all the load?

 

[Hokie93]

Hokie66 - Using the same theory, placing the 100 ton load directly above one of the piles (legs) would result in a 75 ton reaction at that location, 50 tons each in the next two closest piles (legs), and -25 tons in the most remote pile (leg). In that particular case I would likely make some effort to assess the stiffness of the slab/cap in question and possibly design the pile (leg) immediately under the load for a reaction greater than 75 tons. For the other case, I agree that the two piles (legs) on the same side as the load would equally share the load.

 

[hokie66]

By the "same theory" you mean the way you are doing it? If so, that is obviously wrong, as the forces goes straight down. As well, that sums to 150 tons, while you only have 100 tons to distribute. If I am distributing the forces incorrectly, I would like someone to explain why.

 

[Hokie93]

I should have said 50 tons total in each of the two legs closest to the load (25 tons each). Sorry about that. So that is 75 tons in the leg under the load, 25 tons each in the two legs closest to the load, and -25 tons in the most remote leg.

 

[hokie66]

So do you agree that that theory is wrong?

 

[Hokie93]

No. I believe P/A + Mc/I is the correct theory to apply in this case assuming the slab in question is sufficiently stiff and each of the legs is of equal stiffness.

 

[hokie66]

That is for stress, not distribution of forces. Carried to the extreme per my example of the full force applied at one corner, the only answer that give static equilibrium is a 100 ton reaction at that point.

 

[Hokie93]

My results are in static equilibrium. The sum of the reactions (75 tons + 25 tons + 25 tons - 25 tons = 100 tons) equals the applied load (100 tons). Using P/A + Mc/I to distribute loads within a pile group is a commonly accepted practice, to the best of my knowledge. It is certainly done that way in all of my foundations books.

 

[hokie66]

To be in equilibrium, not only do the forces have to add up, but also the moments. Try summing the moments about a corner.