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# Weight Distribution Over Six Points6

## Weight Distribution Over Six Points

(OP)
Hi Everyone,

I have a structure supported by six legs that are bolted to the floor. The top view of this setup is shown below. The center of gravity of this structure is marked with a green X, the fixed support legs are shown with red points. I would like to work out the force exerted on the floor by each leg.
x is bigger than a but smaller than b.
What is the best way to approach this problem? I have tried to derive an equation for this but as you can guess, I didn't have any luck.
I understand that I could go down on the FEA route but what I would like to get out this is an equation system that I can use in Excel when the numbers change.

### RE: Weight Distribution Over Six Points

Force inversely proportional to distance ?

### RE: Weight Distribution Over Six Points

Are we to assume that the plate on top of the legs is sufficiently rigid to distribute the loading to the legs? Otherwise it would be a true function of actual load distribution and direction of the framing members for the plate.

for example, if there are joists that frame the plate (platform) and they run up-down on the page then you could have the bulk of the load going to the middle two posts. However if they run left-right on the page, then almost 100% of the load would be resisted by outer 4 posts.

Really there's not enough information about the problem at hand to provide good guidance.

### RE: Weight Distribution Over Six Points

(OP)
We are assuming that the plate is a rigid structure in this case.

### RE: Weight Distribution Over Six Points

2

#### Quote (Mernok1 We are assuming that the plate is a rigid structure in this case.)

If the plate may be assumed rigid and the legs are identical, you may find the force exerted on legs with linear distribution.

Locate the neutral axis of six legs X-X and Y-Y , calculate the moment of inertia of six legs Ixx , I yy and shift the force ( W) to the center with moments Mx= W*ey and My = W*ex

Axial load for any leg Pn= W/L +,- Mx*yn/Ixx +,- My*xn/Iyy

Pls look page 18 of the following doc.

### RE: Weight Distribution Over Six Points

P1 = P*k/d1

d1 = distance from load to pt1 = sqrt(x^2+y^2)

then sum all 6 reactions P*k*(1/d1+1/d2+...) = P
and k = 1/(1/d1+1/d2+...)

### RE: Weight Distribution Over Six Points

I don't know that considering as a bolt group is right. bolt group loading tends to be load proportional to distance. here I think it's reasonable for the nearer supports to react more load.

### RE: Weight Distribution Over Six Points

the correct way is to write out the equations of equilibrium, only three are effective (FZ, MX, MY)
so the problem is triply redundant.

one way to solve is with matrix math, a least squares approach, see attachment.
another is to assume linear displacements, which I think becomes load inversely proportional to distance.

### RE: Weight Distribution Over Six Points

I think the stiffness of the blue gray beams/plate/structure is not infinitely stiff.
If it is very stiff That may result in the CG not acting like a point mass where you've indicated.
I think when installing the six red support legs there is a better than middlin' chance that some will not take their fair share, one or two may stand so proud they take nearly all the load, or some may not even touch the floor at all.

Thus one possibility is the load from the CG may be distributed between the four right hand support leg dots. And maybe even on just 3, or even TWO of those 4.
2 or 3 or 4 simple sum-of-the moment / force distribution calcs using differing four support legs will define a few possible loadings for the support legs. then just pick the worst case for each leg.
You did not say what you intend to do with the support loads, but I'd design all the supports and their footings/pads the same, and for the very worst load. I predict Asking the steel fabricator or contractor to keep track of six or even two different supports/footing is asking for trouble.

### RE: Weight Distribution Over Six Points

This is just a pile cap problem (rigid distribution to defined support points). HTURKAK is correct.

### RE: Weight Distribution Over Six Points

I've designed/analyzed quite a few equipment anchorage/isolation setups and these are the formulae i've used to determine anchor forces (these assume EQUAL spacing between anchorages - sorry, OP). It is as HTURKAK describes. The calculations assume a rigid body attached to rigid supports. The force calculations may or may not be applicable to your problem as they include baked-in ASCE-7 seismic load combinations. ### RE: Weight Distribution Over Six Points

Like the others, is start with an elastic method load analysis (Alex T's bolt group or CRSI's pile group spreadsheets being convenient tools).

From that starting point, I agree that some modification of the results for plate rigidity or leg length are likely appropriate. I'd accomplish that by engineering judgement, relying on some info that has not yet been presented here.

----
just call me Lo.

### RE: Weight Distribution Over Six Points

(OP)
Thank you for response guys. The requirements for this project has changed (as always) since my first message and I ended up generating and FE model in ANSYS.
This model is a bit more representative to the reality. You will find several load cases if you open the file. My reaction forces are short of 1500N roughly unfortunately. I have tried to run the simulation with different load cases but there was always a significant amount of reaction force missing. I am not entirely sure what’s causing this. Would someone be able to take a look at to see find the problem with the simulation? ### RE: Weight Distribution Over Six Points

"This model is a bit more representative to the reality." ... no it isn't. there is "clearly" a problem with a model where reaction .NE. applied load. Clearly, since the FEA balances (it Has too) there is a seventh support reacting some applied load. You are better off, IMHO, in solving this by hand (since it is QED), and then (if you want the practice) solving with FEA.

### RE: Weight Distribution Over Six Points

(OP)
Well. I messaged in my previous comment that there is a problem with it as you pointed out. Otherwise, I wouldn't have shared it on this forum.

### RE: Weight Distribution Over Six Points

and this is a "bad" way to approach the generalised situation (which is what I thought you were after when you used variables for dimensions).
this is "easy" to put into excel ... define the geometry (positions of the 6 legs, position of the load) and excel can calculate the results.
I would start with a s/sht that had 4 reactions ('cause it's easier to check) then make a 2nd sheet for 6 reactions.

I suspect you have an errant constraint (maybe AUTOSPC ?). Which FEA are you using ? There should be a table that shows the sum of the input loads and the sum of All the reactions and their difference, which Has to be zero (or round-off error).

### RE: Weight Distribution Over Six Points

Hi Mernok1

I have a method in mind using compatibility but I need the dimensions of the structure, centre of gravity and load applied, I will share when I get down to calculating the reactions and making sure that what I am proposing is reasonably correct.

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

### RE: Weight Distribution Over Six Points

Hi Mernok1
I would expect any decent FEM software to solve this with a difference between allied load and sum of reactions < 1%, probably < 0.1 %. If you are using Ansys you are not using it properly. Maybe, as rb1957 suggests, some AUTOSPC is "stealing" the load.

If you neeed a quick result I suggest doing it by hand, as already mentioned. But if you want to use FEA, by all means. The model you have may be more geometrically correct but the results are off. Before you use any reults you need to find the missing load and ensure that the load and boundary conditions are correct. There is no way around that problem.

Good Luck

Thomas

### RE: Weight Distribution Over Six Points

The plate cannot be assumed rigid and there is no unique solution to the problem.

BA

### RE: Weight Distribution Over Six Points

well ... anything can be assumed to be anything. you "can" assume the plate is rigid, then check simply by raising one corner and seeing if "plane section remain plane". you can assume rigid then use scales to measure to loads and compare with predictions.

### RE: Weight Distribution Over Six Points

If the plate is assumed to be rigid, the assumption is wrong! That is because the plate is not rigid. I'm somewhat surprised that this needs to be said.

BA

### RE: Weight Distribution Over Six Points

Let us know if it works.

https://dakutecyfuxew.sinoppazari.com/analysis-of-...#

A black swan to a turkey is a white swan to the butcher ... and to Boeing.

### RE: Weight Distribution Over Six Points

I do not believe there is any general solution to your problem. The answers will depend on the exact loads applied at exact locations and to the exact stiffness of every part of the structure.

This problem is very similar to how do you calculate the loads in rigging when lifting large heavy objects. Levers and fulcrums are used to remove the unknowns in cable lengths structural stiffness. You cannot do that for your problem. Your answers are only as accurate as your assumptions. In this case your assumptions are not likely to be very accurate.

### RE: Weight Distribution Over Six Points

I'd concentrate on coming up with a conservative solution to the problem rather than numerical accuracy.

A black swan to a turkey is a white swan to the butcher ... and to Boeing.

### RE: Weight Distribution Over Six Points

Reactions of a three legged table can be determined by statics, whether the plate is rigid or flexible and whether the legs are precisely the same length or not. The same is not true for a four legged table and certainly not for a six legged table.

A four legged table with load W placed dead center might resist W/4 per leg, but if one leg is slightly shorter than the rest, two legs could carry W/2 each while the others carry no load. Any attempt at calculating the actual reactions is futile.

BA

### RE: Weight Distribution Over Six Points

I'm alittle surprised by that post.

If the legs are of different lengths then the table will adjust itself until 3 legs are on the ground, and the loads are still statically determinate. I've never seen a table balance itself on two legs (despite what calcs say could happen).

In my experience any reasonably practical table top is near enough "rigid", and would use this assumption ("plane sections remain plane") to determine the loads for the 6 legs and be confident that I'm well within 10%. As said above, a test would confirm the validity of the assumption.

### RE: Weight Distribution Over Six Points

#### Quote (b1957)

I'm alittle surprised by that post.

If the legs are of different lengths then the table will adjust itself until 3 legs are on the ground, and the loads are still statically determinate. I've never seen a table balance itself on two legs (despite what calcs say could happen).
We are both a little surprised.

Have you ever sat at a table where one of the legs was short, or the floor was uneven, and the table kept rocking back and forth as people on opposite sides of the table leaned on their side?

You have never seen a table balance itself on two legs because its equilibrium is unstable when the c.g. of load is aligned with two diagonally opposite legs. One leg is off the floor and has zero reaction. One leg may be on the floor but has no reaction. The entire load is carried by only two legs.

If the c.g. of load is eccentric, but within the triangle joining the three loaded legs, the load is supported by three legs, not four. As stated earlier, tripod reactions can be calculated by statics.

For a four legged table, individual reactions can vary between 0 and W, depending on the load location.

BA

### RE: Weight Distribution Over Six Points

of course I've sat at a wobbly table. you wrote "two legs could carry W/2 each", which is the correct solution of an ideal situation (akin to spherical chickens in a vacuum).

your characterisation that no table can be rigid is similarly correct in a theoretical sense. In a practical sense, any (well, nearly any) usable table will be stiff enough to be considered rigid. As I said, assuming rigid will IMHO give reactions that are within 10% of the actuals.

Yes, the reactions change as the location (and magnitude) of the load changes ... but nothing that can't be worked out in a s/sht (assuming a rigid table). With more work and problem definition you could work with a flexible table, but then FEA would probably be the better route.

### RE: Weight Distribution Over Six Points

Hi, I thought I’d just contribute to the original request, namely what type of hand calc approach could be applied to the problem. These equations may be of use.
The location of the applied forces are to be given in a global coordinate system, as well as the location of the six supporting legs. The legs can be viewed as a bolt group, placed in the x,y plane. The x,y coordinates of the bolt group centre (BGC) is first found. This is based on bolt x and y distances. In addition, to represent stiffness, a weighting factor can be included (section area or shear / axial stiffness). Knowing the BGC, moments about the x, y and z axis are calculated.
To react out Mz, the assumption that each bolt reaction force is proportional to radial distance from the BGC is made. Based on this, the following is applied.
Pr(i) = Rz.r(i), i for individual bolt number.
If weighted, use Pr(i) = Rz.r(i).A(i), if area. Note that Pr(i) acts at the bolt position and is normal to the radial arm of the individual bolt.
Mz(i) = Pr(i).r(i)=Rz.A(i)r(i)^2
Applied moment Mz = Rz.Sum[A(i).r(i)^2]
Sum[A(i).r(i)^2] = Sum[A(i).[(x(i)-x’)^2+(y(i)-y’)^2]], where x’ and y’ are bolt group centre (BGC) coordinates. Rz can be found and used to find Pr(i). Components in the x and y coordinate system will be needed. The applied forces in the x and y directions are distributed to each bolt, incorporating any weighting.
Forces in the z direction, due to moments about x and y, can be found using the following.
Pz(i) = Rx.(x(i)-x’)-Ry.(y(i)-y’)
Note, Pz(i) can include a weighting factor (area, axial stiffness), and x’,y’ are the BGC coordinates.
Mx(i)=Pz(i).(x(i)-x’) and
Mx(i)=Rx.(x(i)-x’)^2-Ry.(x(i)-x’).(y(i)-y’)
My(i)=Ry.(y(i)-y’)^2-Rx.(x(i)-x’).(y(i)-y’)
You need to sum the individual bolt Mx and My moments, giving Ixx, Iyy and Ixy terms. Rx and Ry can be found by matrix methods. The two equations used are where the total moments are equated. The applied Pz force is distributed to the bolts and the moment / axial components combined.
I hope you find this helpful. Check the calcs to make sure you’re happy with the approach and if it is applicable to your case.

### RE: Weight Distribution Over Six Points

#### Quote (Mernok1)

I have a structure supported by six legs that are bolted to the floor. The top view of this setup is shown below. The center of gravity of this structure is marked with a green X, the fixed support legs are shown with red points. I would like to work out the force exerted on the floor by each leg.
x is bigger than a but smaller than b.
What is the best way to approach this problem?

There is no unique solution. It depends on the stiffness of the supported structure as well as the distribution of load, not just its center of gravity.

The solution proposed by Stress_Eng is similar to the instantaneous center method for a bolt group. In that case, there is a known moment about the BGC and the plate is properly deemed to be rigid. In the current case, Mx, My and Mz are all zero; there is no applied moment. The plate is not rigid as it can bend between supports. If it were truly rigid, why use six supports? Three would be enough, but four would make for better symmetry, one at each corner.

With three supports, reactions can be determined by statics alone. With four, assume the load is carried by three supports and make the fourth the same as the worst case (a little conservatism never hurts).

BA

### RE: Weight Distribution Over Six Points

Hi, just wanted to give some food for thought, for yet another possible hand calc approach. The plate could be visualised as six springs (joined at cg and individually fixed at legs), each representing load path regions of the plate. Each will have a x and y direction spring stiffness, based on their combined axial and shear deflection capabilities. If needed, a z stiffness could be calculated, based on cantilever bending (fixed at cg) and possibly shear (short beam). A load split can be determined, based on equal x,y,z displacements.

### RE: Weight Distribution Over Six Points

If the only forces under consideration are gravitational, z stiffness is the only one relevant. Without some information about the stiffness of the structure being supported and the distribution of load, there can be no unique solution.

BA

### RE: Weight Distribution Over Six Points

Hi, just wanted to expand a bit on the moments generated when viewing the plate as a bolt group. If we are only considering a gravity force acting at the cg, the BGC will be offset to the cg (cg location relative to BGC), thus Mx and My moments will be generated about the BGC. If forces were to be applied in the x and y directions, they would produce an Mz moment, and, if placed away from the surface (having a z location) then they would also generate Mx and My moments. Just clarifying what needs to be taken into account when calculating your moments about the BGC.

### RE: Weight Distribution Over Six Points

Hi, I have one other possible hand calc approach you may consider. If the structure around the plate profile is considered to give a simply supported condition, then you could consider applying the principles of rectangular plate bending theory. The approach can accommodate numerous types of loading configurations (uniform/ tapered pressure distributions, singular / numerous variably placed point loads). I would suggest referring to references such as Timoshenko. The principle would permit you to determine, under your loading condition, the distributed shear about the profile of the plate, which you could sum up between the legs, find the centroid of the load between the legs and proportion relative to distance.
You could go one step further. Permitting the plate to bend will allow the plate to shorten in length, in both the x and y directions. This change in length can be found, and used in conjunction with the leg displacement stiffnesses to determine leg attachment reaction forces in the x and y directions.

### RE: Weight Distribution Over Six Points

Four legs arranged in a square pattern, loaded with gravity force W at the center, i.e. at the intersection of both diagonals, is a perfectly symmetrical structure. You might think that each reaction would be W/4, but that is not necessarily true. If all legs are precisely the same length, it would be true, but if one leg is a micron shorter, no load will be carried by it and no load will be carried by the leg on the other end of the same diagonal until the two legs which carry all of the load shorten by one half micron.

Installers do not work with such extreme precision. One leg might be one or two mm shorter than the rest, in which case, two legs carry W/2 each and two legs carry no load. To be prudent, one should design all four legs for W/2 because the location of the short leg is unknown in advance.

In the case of four legs arranged in a square pattern, loaded at the center, each leg should be capable of supporting W/2. If the load is on a diagonal but not centered, one leg could carry more than W/2. Some conservatism is necessary when calculating reactions.

BA

### RE: Weight Distribution Over Six Points

Has Mernok1 got the luxury of being able to alter the structural configuration, and reconfigure the plate structure to make the cg coincide with the BGC? If so, 4 legs would make the situation so much easier!

The original post states 6 legs are all bolted to the ground. This therefore implies that all 6 legs are active in reacting to the applied load. The post also implies that the cg is not in the centre of the plate. If the legs are of different lengths, the act of bolting them to the floor would introduce build stresses into the structure, due to forced displacements. Can shimming be introduced, to reduce any build misalignment? If manufacturing tolerances are to be included in the analysis, the forces required to displace the plate, to allow the legs to close any gaps would need to be determined. Once these forces have been derived, reactions due to applied loading would need to be superimposed.

### RE: Weight Distribution Over Six Points

I agree with BARetired. His particular case of a table with 4 legs, for which 1 has a smaller length and 2 carry together all the load, show that the result is very sensitive to the boundary conditions and the theoretical model described above is not precise.

### RE: Weight Distribution Over Six Points

#### Quote (Have you ever sat at a table where one of the legs was short)

The expression, "As true as a trivet", comes to mind.

Rather than think climate change and the corona virus as science, think of it as the wrath of God. Do you feel any better?

-Dik

### RE: Weight Distribution Over Six Points

#### Quote (Stress_Eng)

Has Mernok1 got the luxury of being able to alter the structural configuration, and reconfigure the plate structure to make the cg coincide with the BGC? If so, 4 legs would make the situation so much easier!

The OP does not mention a plate structure. He says this:

#### Quote (Mernok1)

I have a structure supported by six legs that are bolted to the floor.

Could be any kind of structure. Possibly the six points are required to satisfy the attachment points of a mechanical unit, such as a pump.

Four legs does not make it any easier to calculate reactions, as explained in my last post.

#### Quote (Stress_Eng, BA in red)

The original post states 6 legs are all bolted to the ground. This therefore implies that all 6 legs are active in reacting to the applied load. The OP may wish that to be the case, but it is unachievable by average workmen.

The post also implies that the cg is not in the centre of the plate. It does not imply that; it states that.

If the legs are of different lengths, (as they are bound to be) the act of bolting them to the floor would introduce build stresses into the structure, due to forced displacements. Displacements are not forced if the structure is rigid.

Can shimming be introduced, to reduce any build misalignment? Certainly it can, but it won't be enough to permit precise calculation of the six reactions.

If manufacturing tolerances are to be included in the analysis, the forces required to displace the plate, to allow the legs to close any gaps would need to be determined. Once these forces have been derived, reactions due to applied loading would need to be superimposed. It's easier to overestimate the reactions and make all legs the same (a little conservative, but sensible).

BA

### RE: Weight Distribution Over Six Points

#### Quote (Mernok1)

What is the best way to approach this problem?

I suggest a procedure similar to the one below: BA

### RE: Weight Distribution Over Six Points

that looks elegant (and obtuse) but "c" and "d" are indeterminate.

there are, of course, a multitude of ways to address this indeterminacy. But none are much closer to a solution.

we are, IMHO, making this problem needlessly difficult (when we talk of 1 leg being 1 micron less than the others ... why not then also talk of a surface not being perfectly flat ?). The simplest assumption is to assume the table rigid (and the surfaces of the table and the ground being flat) and solve assuming "plane surfaces remain plane". Yes, we are told nothing about the table (nor the ground), but we can still assume simplifications, and state our assumptions.

### RE: Weight Distribution Over Six Points

#### Quote (rb1957)

that looks elegant (and obtuse) but "c" and "d" are indeterminate. I am assuming that the load "W" is the sum of two separate units, one each side of the central legs. That being the case, "c" and "d" would be known. If that is not the case, why use six legs? Four would make more sense.

there are, of course, a multitude of ways to address this indeterminacy. But none are much closer to a solution. If you are expecting an exact solution to the magnitude of each reaction, you will be disappointed. The indeterminacy remains a problem because of the inability of installers to avoid having legs of slightly variable length.

we are, IMHO, making this problem needlessly difficult (when we talk of 1 leg being 1 micron less than the others ... why not then also talk of a surface not being perfectly flat ?). Mechanical units resting on a surface which is not flat would result in precisely the same problem, I agree completely.

The simplest assumption is to assume the table rigid (and the surfaces of the table and the ground being flat) and solve assuming "plane surfaces remain plane". Yes, we are told nothing about the table (nor the ground), but we can still assume simplifications, and state our assumptions. Your assumptions are invalid and fail to find a conservative value for all six reactions. It is only necessary to find the worst case, then use the same leg at each of the six locations. Installing six different legs is impractical, so an exact solution isn't required.

BA

### RE: Weight Distribution Over Six Points

so if your applied loads are within your triangles ... why the "c" and "d" thing ? distribute each load within the triangle.
I assumed you were starting with "W" since that's how the problem was stated, and dividing this into two loads.

how do you know the legs aren't adjustable ? Oh, "not stated", so you assume something (facts not in evidence) that makes the problem unsolvable.

### RE: Weight Distribution Over Six Points

Baretired and rb, I would say you are both correct.

It's staticslly indeterminate. So stiffness of the table will influence the result as BA suggests, hence an FE model would be beneficial in resolving the loads in the supports... Except, if the supporting substrate is several orders of magnitude softer than the table, in which case the table may be assumed to behave as a rigid body, and a hand analysis similar comoact bolt group could be used to give an approximate solution, a bit like the one described above.

### RE: Weight Distribution Over Six Points

#### Quote (NG2020)

It's staticslly indeterminate. So stiffness of the table will influence the result as BA suggests, hence an FE model would be beneficial in resolving the loads in the supports...

I disagree. I feel BA is spot on. FEA is not the tool for this job and will likely yield un-conservative results.

### RE: Weight Distribution Over Six Points

really ? FEA is the sledgehammer that'll answer the questions asked of it.

With FEA you can model different lengths of different legs, ad nauseum.
With FEA you can model the stiffness of the table, the un-flatness of the ground, the lengths of the legs, until you run out of electrons.

Why would FEA not be the tool for this job ? other than because you can simplify the problem and arrive at a reasonable (though not perfectly correct) answer without it ?

Why would FEA be likely to yield unconservative results ?
If you want conservative results, then let each leg react "W", or "2*W", or ...

### RE: Weight Distribution Over Six Points

I think Mernok1 has gotten disillusioned with this (endless) discussion ...

### RE: Weight Distribution Over Six Points

We know nothing about the "supported structure". If the load consists of a single concentrated load, the structure could consist of simple span beams B1, B2 and B3 as shown below. In that case, all reactions may be obtained using statics.
We know nothing about the load to the left of leg #2. If we did, a similar system of simple span beams could be used and again, we could easily determine reactions at each leg.

If we assume, as has been suggested in earlier posts, that the supported structure is a rigid body, then differences in elevation at bearing points need to be considered. The magnitude of reactions cannot be calculated without some idea of the supported structure and the distribution of load.

While it may be a useful exercise to determine each reaction, why does it matter? It seems more likely that all six legs would be identical rather than designing each for its particular reaction. Is that in dispute? BA

### RE: Weight Distribution Over Six Points

#### Quote (RB1957)

Why would FEA not be the tool for this job ? other than because you can simplify the problem and arrive at a reasonable (though not perfectly correct) answer without it ?

Exactly. FEA won't be perfect either for all the reasons BAR mentioned.

### RE: Weight Distribution Over Six Points

but of course all of those things will be known before you make a FEM.

Is this a wind up ?

### RE: Weight Distribution Over Six Points

Just suggesting, an interesting problem has been posted and it has been asked if any methods are known that can be used to determine 6 leg reactions to a given loading condition. It is apparent that information is lacking, but proposals have been given. How about listing those KNOWN methods that COULD be used (statics, elasticity and FEA methods). Each category can be broken down. For example, within the statics category, simplified 3 leg approach, 6 leg bolt group approach (pros and cons, assumptions made)? Elasticity - springs, plates, etc. FEA - any suggestions / comments on how the task should be approached? All of the methods, if not already, can be added to the analysis toolkit!

### RE: Weight Distribution Over Six Points

have at it ! My first item would be to fully define the problem.

### RE: Weight Distribution Over Six Points

#### Quote (Stress_Eng)

How about listing those KNOWN methods that COULD be used (statics, elasticity and FEA methods).

Okay, let me give it a try.

Assumptions:

1. supported structure is rigid
2. attachment points are at the same elevation
3. supported structure is braced laterally to prevent translation and rotation about vertical axis.
4. each leg is fixed at the base and either pinned, fixed or rigidly attached to the structure above.
5. each leg is precisely the same length.
6. rb1957 is on holiday, far, far away. Calculation:

σxy = P(1/A + y/Ix + x/Iy)
where σxy is stress at eccentricity x, y from the c.g. of leg group
A, Ix and Iy are area and moment of inertia about X and Y axis respectively for the leg group.

Rn = σxy*An where R is reaction and subscript n refers to leg number.

BA

### RE: Weight Distribution Over Six Points

so now when we state all the assumptions, I think we've solved it using the method in the first reply ?
or is it load is inversely proportional to distance ^2 ?

### RE: Weight Distribution Over Six Points

The only problem is the assumptions are unachievable, so the solution is wishful thinking.

EDIT: But who cares? It doesn't really matter a tinker's dam (whatever that is) what the reactions are, provided the legs are adequate to carry the worst case. When the worst case is not precisely known, engineering judgment must be used. Hopefully it is available
BA

### RE: Weight Distribution Over Six Points

What can be mentioned is that we could have numerous load cases to consider (or 1). In addition, we could also conclude that we have a possible variation in structural support conditions. By considering both load and support variations, we develop an enveloping approach (a table or matrix of cases), whilst using the same method of analysis. An example could be that we consider a case where 1 leg has failed, and 5 active. Next case, change the failed leg, and so on.

### RE: Weight Distribution Over Six Points

@Stress_Eng,

What I am envisioning from the OP's description, is six short steel columns (the "legs") carrying a structure of some kind on top. If the weight of supported structure is "W" and its c.g. lies within the rectangle enclosing the legs, the maximum reaction cannot exceed W. If the typical leg is selected to safely sustain a load of W at a specified height, that would seem to be pretty conservative and, unless W is huge, would not entail a significant cost.

The engineering cost to conduct complex calculations would far exceed any saving in material cost.

BA

### RE: Weight Distribution Over Six Points

You are absolutely right, we could just say each leg is designed to take the total load W. That would solve the problem.

Great!

BA

### RE: Weight Distribution Over Six Points

Looks a lot like some chiller dunnage I designed not long ago. Turns out 4ft tall steel columns that are large enough to actually connect to something can carry about 17x the entire weight of the chiller by themselves, so it didn't really matter.

### RE: Weight Distribution Over Six Points

Thanks phamENG, I imagine that would be the usual case. If we were dealing with a very heavy unit, or perhaps taller columns, it might be worthwhile to sharpen the pencil a little bit, but for most cases, this method should be sufficient.

BA

### RE: Weight Distribution Over Six Points

With what has been suggested in the numerous posts given, I would think that relevant bits of information can be extracted and used for the original problem.

### RE: Weight Distribution Over Six Points

"tinker's dam" ... the expression is "tinker's damn" ... tinkers do a lot of swearing (at least according to custom).

### RE: Weight Distribution Over Six Points

#### Quote (rb1957)

Why would FEA be likely to yield unconservative results ?

Because you have to make assumptions about the rigidity of the top and the exact length of the legs, which will never reflect reality. There are literally infinite possibilities of those conditions, and no matter how many different ways you model it, there's likely one that you're going to miss.

#### Quote (rb1957)

If you want conservative results, then let each leg react "W", or "2*W", or ...

If the position of the c.g. shown is accurate, all that needs to be done for an acceptably conservative solution is to assume that legs 2,3, and 5 support the load, and calculate the reaction at 5. If the c.g. is at another location, the more general solution is to calculate the reactions assuming all possible combinations of 3 legs supporting the load, and use the largest reaction at any leg. That's just my 'brute force' approach.

Rod Smith, P.E., The artist formerly known as HotRod10

### RE: Weight Distribution Over Six Points

#### Quote (BridgeSmith)

Because you have to make assumptions about the rigidity of the top and the exact length of the legs, which will never reflect reality. There are literally infinite possibilities of those conditions, and no matter how many different ways you model it, there's likely one that you're going to miss.

I don't know about "infinite possibilities", but there are many and that comment is precisely the concern.

#### Quote (BridgeSmith)

If the position of the c.g. shown is accurate, all that needs to be done for an acceptably conservative solution is to assume that legs 2,3, and 5 support the load, and calculate the reaction at 5.

That assumption would lead to a conservative result, but without the cooperation of the installers, the assumption could be wrong. If they install leg 4 slightly higher than leg 3, the load is carried by 2, 4 and 5.

Even worse is if they install legs 1, 2 and 6 a little higher than 3, 4 and 5 because that puts leg 1 in tension (T1) so that legs 2 and 6 are sharing a load of T1 + W. It may even be possible to find a combination where one leg carries more than W.

BA

### RE: Weight Distribution Over Six Points

It’s interesting to hear how structures can be viewed. Aerospace structure and system components are usually certified by analysis and test. The limit load being the highest single load seen by the object in its life. In some instances an item is permitted some permanent deformation, as long as its function is not impeded. Most of the time no yielding is seen. Items are expected to sustain an ultimate load of 1.5 x limit for more than 3 seconds. Structures that have an ultimate safety factor of more than 2.5 would probably be included in a weight saving programme.

### RE: Weight Distribution Over Six Points

"It may even be possible to find a combination where one leg carries more than W." ... no, if the load is within the envelop of the reactions then the maximum reaction is equal to the load (when the load is above one of the legs). Only if the load goes outside the envelop of the reactions can a reaction exceed W.

### RE: Weight Distribution Over Six Points

Another interesting aspect is manufacturing tolerances. Most drawings I see have a general tolerance of +/- 0.010” (0.25mm). Key datum features have much finer tolerances.

### RE: Weight Distribution Over Six Points

#### Quote (rb1957)

"It may even be possible to find a combination where one leg carries more than W." ... no, if the load is within the envelop of the reactions then the maximum reaction is equal to the load (when the load is above one of the legs). Only if the load goes outside the envelop of the reactions can a reaction exceed W.

If legs 1, 2 and 6 are installed a little higher than the other three, then the load is outside those three columns; and they are the only ones that matter. In a different arrangement of legs, it is possible to place three legs in a position such that one leg supports a load greater than W, even if the c.g. of load is inside the envelope of all columns.

BA

### RE: Weight Distribution Over Six Points

If leg 5 is moved to 5a, or a new leg is added, legs 1 and 6 will both be in tension while leg 5a will sustain a load greater than W, notwithstanding the fact that the c.g. of load is within the envelope of columns. In effect, the structure becomes a propped cantilever. BA

### RE: Weight Distribution Over Six Points

#### Quote (Stress_Eng)

Another interesting aspect is manufacturing tolerances. Most drawings I see have a general tolerance of +/- 0.010” (0.25mm). Key datum features have much finer tolerances.

In this thread, we have the manufacturing tolerance of the supported structure; we also have the tolerance of the column installers to consider. It is not possible to calculate reactions with any confidence whatever.

BA

### RE: Weight Distribution Over Six Points

#### Quote (BAretired)

Even worse is if they install legs 1, 2 and 6 a little higher than 3, 4 and 5 because that puts leg 1 in tension (T1) so that legs 2 and 6 are sharing a load of T1 + W. It may even be possible to find a combination where one leg carries more than W.

You are correct. That is a possibility I overlooked.

#### Quote (rb1957)

"It may even be possible to find a combination where one leg carries more than W." ... no, if the load is within the envelop of the reactions then the maximum reaction is equal to the load (when the load is above one of the legs). Only if the load goes outside the envelop of the reactions can a reaction exceed W.

Incorrect. As BAretired pointed out, depending on the load location, the load on a leg could exceed 2*W. For instance, if the load is placed along the edge between 1 and 2, closer to 1, and Leg 1 is short and doesn't support any of the load, then span 2-6 becomes a back span of a cantilever.

Rod Smith, P.E., The artist formerly known as HotRod10

### RE: Weight Distribution Over Six Points

"Assumptions:

1. supported structure is rigid
2. attachment points are at the same elevation
3. supported structure is braced laterally to prevent translation and rotation about vertical axis.
4. each leg is fixed at the base and either pinned, fixed or rigidly attached to the structure above.
5. each leg is precisely the same length."

I will now apply assumption 6 ...
6. rb1957 is on holiday, far, far away.
and will not open this thread again. You guys have turned a simple problem into something unsolvable ... even when you make assumptions to make it solvable.

### RE: Weight Distribution Over Six Points

Have a happy holiday, rb1957! Where to, Mexico, Hawaii, Galapagos Islands...?  It was never a simple problem; it really is unsolvable. With assumptions, it can be made soluble, but of what value is that if the assumptions are unrealistic? Installers cannot be assumed to be magicians.

BA

### RE: Weight Distribution Over Six Points

With all of the responses given to the posted problem, it is hoped that some of the information given is helpful to the original problem. Out of all of the possible scenarios that have been highlighted, I can add just one more to the list. Here goes …
1) All detail, sub / final and the all important installation drawings are to hand. All have the required dimensional information and manufacturing / assembly tolerances, in particular those depicting interface requirements.
2) From all the drawings, calculate the following…
2a) The minimum possible final build lengths for legs 4 and 6 (to installation drawing requirements).
2b) The maximum possible final build length for leg 5 (to installation drawing requirements).
3) Assume leg 2 to be at nominal length.
3) Ignore legs 1 and 3.
4) Based on the build lengths of legs 2, 4, 5 and 6, calculate compression load on leg 5.
5) Ignoring leg 6, distribute W between legs 2, 4 and 5.
The installed structure is expected to pass final inspection. If the structure is found to be discrepant to drawing / installation requirements, a concession would need to be raised, describing all deviations. An assessment would conclude either acceptance, any alterations (repairs) or rejection.

### RE: Weight Distribution Over Six Points

@Stress_Eng,

In item 5) on your list, you cannot be certain that legs 2, 4 and 5 will carry any load at all. Perhaps legs 1, 3 and 6 carry all of the load.

Any three points in space describe a plane unless they are aligned, in which case they describe a line. If there are only three legs, three bearing points describe a plane. If there are six legs of slightly varying height, it is impossible to know which three will describe the highest plane. It could be 2, 4 and 5, but it could be any other combination (with the exception of 1, 2 and 3 or 3, 4 and 5 which align).

With six legs, the only possible combinations are as follows:
124, 125, 126, 134, 135, 136, 145, 146, 234, 235, 236, 245, 246, 345, 346, and 356.
If I haven't missed any, that is a total of 16 combinations. Any one of those could form the critical plane, leaving three legs unloaded in all 16 cases.

Discrepancies can occur due to (a) uneven level in bottom slab, (b) curvature of the bottom slab, (c)errors in length of leg, (d) workmanship errors by installers and (e) errors in fabrication of supported structure. Some errors may be compensating; others may not, but there will be some discrepancies.

If columns have identical mechanical properties, and each is designed to sustain a weight of W, or perhaps more, then most of them will have low axial stress. Loaded legs may share their load with others, but only after they strain enough to make contact. The strain in any column is Pn*L/AE where n refers to column number.

BA

### RE: Weight Distribution Over Six Points

Using only three columns, each reaction can be accurately calculated, even if the workmanship is not precise. BA

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