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prex · May 8, 2007

hello2006,
in the first site below there is a sheet for calculating those reactions under Plates -> Simple bending -> Rectangular -> 4 corners -> Conc.load. Keep in mind however that the solution is approximate, the approximation being not very good if the load is on one side.

JStephen,
my answer to the question would have been exactly the same as yours. However, by thinking a bit more to this problem, it appears evident that the reactions do not really depend on the elastic properties of the plate. This is simple to experiment with the sheet above, but is also evident: why should the distribution of reactions change as the plate becomes more and more rigid, and towards what values would the reactions tend when the plate becomes substantially rigid? The intuitive answer is that they would remain the same.
By playing a bit around, I found a simple rule to which the reactions calculated by the sheet above are in quite a good approximation, and that gives the exact answer when there is a symmetry that allows for calculating the reactions by equilibrium.
It is very simple: divide the rectamgle into 4 rectangles by tracing two lines parallel to the sides and intersecting at the load. Now each reaction is proportional to the area ratio of the partial rectangle containing the opposite vertex to that of the full rectangle.
So, hello2006, this would be a simple answer to your question, as this is a supplemental and independent equation with respect to the three equilibrium equations.

Is there someone that can give a proof that this equation is incorrect, or that indeed it is correct?

prex

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BigInch · May 8, 2007

Prex, If your method gives a correct solution in all cases, then there would be no need for thick plates, right?

With a thin plate, shear and beam deformation must be considered. With a thick plate, deformations remain small enough to neglect.

The reactions change according to the rigidity of the plate, since shear deformation and excessive bending takes place in a thin plate. Beam theory is based on relativly small deflections and plane sections remaining relatively plane. Given a 4 point reaction with both an Mx & My acting on the plate, the reactions would not vary directly with distance, but vary in accordance with the bending resistance and differential rotation of an effective beam between the loading point and the reaction point under consideration. A thick rigid plate's reaction at a corner would be quite high in relation to that of a thin plate, where the effective beam "strip" running between load and reaction point is long and thin in depth and can deflect and rotate along its length. A far corner could much more easily deflect and rotate than a close corner, thus reducing the far corner's effective contributing resistance to the applied load in relation to a corner having a shorter effective length and subsequent high stiffness of its effective beam.

Some possible reaction distributions depending on plate stiffness,

BigInch

-born in the trenches.

prex · May 8, 2007

BigInch, thanks for considering that: your doubts are mine too.
However I can't follow your way of reasoning.
I think we can assume the plate is thick enough to make its behavior linear, and also that shear deflections may be neglected, just as in the ordinary theory of beams and flat plates subject to simple bending.
What I found by very simple numerical tests is that the values of the 4 reactions do not change by changing the stiffness of the plate (varying either the thickness or the elastic modulus or both).
Said in other words: if you calculate two plates with the same geometry, except that one has a maximum deflection of, say, half the thickness (hence a relatively large, though still small deflection) and the other one has a totally negligible deflection (for example because it is done with a very stiff material), then you'll find exactly the same corner reactions (for whatever load position, but of course the same in the two plates).
This is a fact.
Unfortunately I cannot explain why that simple rule of area ratios gives results that are exact in some simple cases, and closely match those obtained by the calc sheet in xcalcs, at least in the few cases that I tested.

prex

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http://www.amazon.com/Design-Welded-Structures-Omer-Blodgett/dp/9998474922

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BigInch · May 8, 2007

If you assume that the plate is "thick enough" in both cases, there will never be any difference.

Its not my way of thinking. I confess I learned it from, Blodgett's "Design of Welded Structures" about 30 years ago.

here's another good one,

http://www.amazon.com/Roarks-Formul...007072542X/ref=pd_sim_b_1/002-1169651-1760853

BigInch

-born in the trenches.

prex · May 9, 2007

BigInch,
Roark doesn't cover rectangular plates supported at four corners, don't know about Blodgett.
As you seem to know the answer, can you state your solution to the original question: formula to calculate the four corner reactions in a rectangular plate supported at four corners only, with a concentrated load at a defined off center location.
Note that your formula above P/4 + Mx/2dx + My/2dy is only valid for loads at center; also discard the moments, they are zero.

prex

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BigInch · May 9, 2007

The thick plate formula is valid for loads at any location on the plate. When the load is offcenter there is moment created by the eccentricity of the load, so its not zero; simply substitute Mx = P * Ex and My = P * Ey,

where Ex, Ey = eccentricity in x direction and/or y direction

BigInch [worm]

-born in the trenches.

prex · May 9, 2007

OK BigInch, this takes us back to the post by JStephen: you can write two moment equations plus one resultant equation, but you have four unknowns. Try a practical example with both nonzero eccentricities and you'll see the point.

prex

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BigInch · May 9, 2007

Look this is just simple normal bolt load calculations for plain vanilla thick plate bolted connections. It can also be used to find individual bolts loads for 1, 2 bolts or 32 anchor bolts on a vessel skirt, or n bolts. Just find the centroid of the bolt pattern, the neutral axis, calculate distanct to each bolt from the neutral axis, assume load is proportional to that distance and calculate each bolt loads.

Here's four more equations for you, + n more

R1/d1 = R2/d2 = R3/d3 = R4/d4 = Rn/dn

BigInch [worm]

-born in the trenches.

prex · May 9, 2007

Sorry to insist, but we seem to be connected to different wavelengths: what you state assumes there is an axis of symmetry.
And I'm still without answers to my questions.
Will you please take this very simple example:
-square plate a x a=1000x1000 mm
-a single load of say P=10 kN at Ex=300 and Ey=200.
Counting the vertices ccw from the first quadrant, my formula gives:
R1=5.6 kN
R2=1.4 kN
R3=0.6 kN
R4=2.4 kN
It's easy to check that these values satisfy the resultant equation ([Σ]R=10 kN) and the two moment equations:
P[·]Ex=(R1+R4)[·]a/2-(R2+R3)[·]a/2
P[·]Ey=(R1+R2)[·]a/2-(R3+R4)[·]a/2

Will you please make a similar calculation with your method and give the results?

prex

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BigInch · May 9, 2007

OK. I'll make this calc and, since you're structural, you do one for me showing the force (or i'll accept stress) in each of 8 vertical 25 mm diameter reinforcing bars and the stress distribution in the concrete of a vertical reinforced circular column of 300 mm diameter with a 10 kn upward directional load applied 200 mm off the X centerline and 300 mm of the Y centerline of the column's axis. Assume the concrete in the column is cracked all the way through and will not transmit tension.

Deal or no deal?

BigInch [worm]

-born in the trenches.

prex · May 9, 2007

That's quite easy, as you have two axes of symmetry here and the concrete is not contributing.
The point of application of the load is at E=[√](200²+300²)=360 mm from centerline.
Let's assume the bars are equispaced and placed on a circle D=240 mm dia., and also (as a minor detail) that 4 bars cross the axes of symmetry.
Calling R1, R2, R3, R4, R5 the unknown reactions (because of symmetry R6=R4, R7=R3, R8=R2), with R1 farthest from the load, we simply have:
R1=P/8-PE/(2D)=-6.25 kN
R2=P/8-PE/(2[√]2D)=-4.05 kN
R3=P/8=1.25 kN
R4=P/8+PE/(2[√]2D)= 6.55 kN
R5=P/8+PE/(2D)= 8.75 kN

It's your turn now!
Note however that I assumed above, as you probably expected, the behavior of a section in a bent beam (coming from the assumption that the bent sections remain plane) with linear distribution of stresses. This is a rough approximation for example for the base ring of a column (though it is customarily used), but is not valid for a plate, even if infinitely rigid.
To be convinced of that think to another example: a continuous beam with uniform section over for example 3 supports. With your way of reasoning you would expect there too, I suppose, a linear distribution of the support reactions, but this is far from being true, and for whatever stiffness: try a sample case to be convinced of that.

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BigInch · May 9, 2007

Non numeric solution, since numeric solutions should account for stiffness and flexibility. They will always show the same proportional reactions with the same proportional stiffnesses, since it is true, stiffness or rigidity makes no difference to them. This is a pure brain power only solution, where rigid remains rigid and proportionality holds at all costs.

Now reduce one bar by 1mm^2. Or add 1mm^2 of concrete to the south side of the column. Does that invalidate your answer, or is it still (approximately) a good solution? At what point does the solution become invalid?

Here's my simple statics solution. Not the exact problem you gave me, and its solved backwards, so turn the solution around... OK, but then I gave you one way too easy. I have no doubt that it can be worked in reverse... just no time to do it... and I don't like structural calculations. I'd rather make stress diagrams and take the integral of the stress over any given area to arrive at an equivalent force when such is necessary.

BigInch

-born in the trenches.

prex · May 10, 2007

OK BigInch, now we can come back to the original question.
You found the following corner reactions:
R1=600, R2=0, R3=500, R4=1200
You might check that also the following combinations satisfy the moment and resultant equations:
R1=470, R2=130, R3=370, R4=1330
R1=340, R2=260, R3=240, R4=1460
R1=1100, R2=-500, R3=1000, R4=700
and infinite others may be obtained.
Of course the last one is quite physically unrealistic, but look at the first one: is perfectly realistic and is safer than yours, as the maximum reaction is higher.

My point remains the same: you assume a linear distribution of the reactions around a so called neutral axis, but there is nothing that supports such an assumption in a general situation. This assumption is more or less valid only when we have something that can be assimilated to the section of a beam: there the constraint of the parts of the beam adjacent to our section will force the section to remain plane after deformation, and this is the basis for assuming a linear distribution of reactions.

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BigInch · May 10, 2007

In the case of a base plate or pipe flange where there are stresses in the "column" member that are entering the "flange", crossing through and then finally reaching the bolts, what would cause the flange to behave differently from one side to the other (given enough bolts)? Let's consider eccentricity? Than start with a balanced load and stress distribution on the column side and slowly move the loads away from balanced towards increasing eccentricity. What would suddenly cause the plate to behave in an entirely different manner, as the load moved from one dE to another? Can you support some theory as to why some type of bifurcation would suddenly appear?

I may be wrong, but I also think minimum internal energy is going to be expended by the first group of (my) reactions, as it appears that the other solutions seem to create a greater sum of moments on various similar elements. I think that is the basis of the theory I'm using, but I can't say that I'm sure about that. It would be interesting to confirm, but I'm not going to volunteer. Would you?

prex · May 10, 2007

Well yes, I did that by the calc sheet in the first site below, that uses the principle of minimum energy to calculate deflections and reactions in a rectangular plate. The results for your example are:
R1=489 R2=126 R3=374 R4=1310
As you see this is not very different from my first set above, that was hand calculated by the simple rule of areae. Of course we are speaking of an elastic plate of whatever stiffness here, as in the original question, not of a pipe flange, that sure behaves much more like you calculate.

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BigInch · May 10, 2007

I knew there should be some semblence to reality there somewhere. I understand the need to analyze these eccentric things using indeterminate methods, but was under the mistaken impression that it was conservative to use that stress transition method. I'm also glad I avoid eccentricities whenever possible. And I guess I'll have to give more respect to eccentrically loaded plates with corner reactions. OK, I'm happy. I appreciate the time you put into this.

Wonder how the OP is doing? Appologies for the thread hijack.