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# Weight Distribution Problem

Hello All,

I've been noodling over this for several days and can't figure it out. Â Here is my problem. Â Suppose you have a table with four legs. Â If you set a weight on the table somewhere inside the area bounded by the legs, what is the distribution of that weight on each leg? Â I realize this is a statically indeterminate problem so solving with classical statics (without introducing deflections) produces garbage. Â I keep thinking there must be a way to ratio the weight on each leg using the distance from the weight to each leg. Â Or maybe there is an iterative approach that will work?

Additional Notes:

1. Â There can be no "inside" legs. Â That is to say the location of one leg cannot be inside the boundary created by the other three. Â

2. Â Can the solution be expanded to tables with more than 4 legs

3. Â What if the weight is outside the area bounded by the legs? Â It would be nice if the solution would begin to report negative weight distributions upon this occurrence.

Thank you in advance!

Kyle

I've been noodling over this for several days and can't figure it out. Â Here is my problem. Â Suppose you have a table with four legs. Â If you set a weight on the table somewhere inside the area bounded by the legs, what is the distribution of that weight on each leg? Â I realize this is a statically indeterminate problem so solving with classical statics (without introducing deflections) produces garbage. Â I keep thinking there must be a way to ratio the weight on each leg using the distance from the weight to each leg. Â Or maybe there is an iterative approach that will work?

Additional Notes:

1. Â There can be no "inside" legs. Â That is to say the location of one leg cannot be inside the boundary created by the other three. Â

2. Â Can the solution be expanded to tables with more than 4 legs

3. Â What if the weight is outside the area bounded by the legs? Â It would be nice if the solution would begin to report negative weight distributions upon this occurrence.

Thank you in advance!

Kyle

ASKER

d-glitch,

Thank you for your response. Â I have tried summing moments and forces to no avail. Â There are simply not enough independent equations available. Â There are only 3 possible equations of equilibrium (in this 2-D problem) and we have four unknowns here.

Thank you for your response. Â I have tried summing moments and forces to no avail. Â There are simply not enough independent equations available. Â There are only 3 possible equations of equilibrium (in this 2-D problem) and we have four unknowns here.

If you have four legs of equal length, then you have four moment equations and one force equation. Â You

The moment equation for each leg depends on the distance of the applied force from the top of the leg and the length of the leg. Â The applied force will have radial and tangential components.

The moment equation for each leg depends on the distance of the applied force from the top of the leg and the length of the leg. Â The applied force will have radial and tangential components.

ASKER

I apologize, I forgot to mention the legs are vertical. Â No need for all those ugly sines and cosines. Â

I'm not sure what you mean by "four moment equations"? Â Sure, you can multiply the reaction variables for each leg with the distance to the origin but then what do you set that equal to?

I'm not sure what you mean by "four moment equations"? Â Sure, you can multiply the reaction variables for each leg with the distance to the origin but then what do you set that equal to?

ASKER

Perhaps if we ran through an example.

Let's say our table has legs at the following coordinates

(10,2)

(-11,4)

(-6,-16)

(5,-14)

a force of 100 lbs is applied at (2,-8)

How would you go about finding the reaction forces on the legs?

Let's say our table has legs at the following coordinates

(10,2)

(-11,4)

(-6,-16)

(5,-14)

a force of 100 lbs is applied at (2,-8)

How would you go about finding the reaction forces on the legs?

Turn the table upside down and balance it at (2,-8).

Distribute 100 lbs at (10,2) (-11,4) (-6,-16) and (5,-14) so that the table balances.

I have to commute now. Â

I will think about it some more on the train.

I will be surprised if it isn't solved by the time I get home.

Distribute 100 lbs at (10,2) (-11,4) (-6,-16) and (5,-14) so that the table balances.

I have to commute now. Â

I will think about it some more on the train.

I will be surprised if it isn't solved by the time I get home.

ASKER

Hmmm....fascinating way of looking at it. Â I think you're onto something here. Â Placing the "balance point" at (2,-8) I summed moments about the x-axis and Â y-axes, summed forces, and summed moments about another line passing through (2,-8) at a 30 deg angle. Â I got results of:

R1 = 12.139

R2 = 22.543

R3 = 0

R4 = 65.318

I then changed the angle of the arbitrary line from 30 deg to 80 deg and got these results:

R1 = 29.670

R2 = 9.890

R3 = 26.374

R4 = 34.066

Both sets of results sum to 100 so that's good but why are they different. Â You should be able to sum moments about any line you want as long as it passes though the "origin" right? Â All you need is another equation that describes the relationship between the four reactions. Â Why the different numbers?

I have to get home as well. Â Thanks for your help thus far. Â I think we're almost there.

Kyle

R1 = 12.139

R2 = 22.543

R3 = 0

R4 = 65.318

I then changed the angle of the arbitrary line from 30 deg to 80 deg and got these results:

R1 = 29.670

R2 = 9.890

R3 = 26.374

R4 = 34.066

Both sets of results sum to 100 so that's good but why are they different. Â You should be able to sum moments about any line you want as long as it passes though the "origin" right? Â All you need is another equation that describes the relationship between the four reactions. Â Why the different numbers?

I have to get home as well. Â Thanks for your help thus far. Â I think we're almost there.

Kyle

>Perhaps if we ran through an example.

Let's say our table has legs at the following coordinates

(10,2)

(-11,4)

(-6,-16)

(5,-14)

a force of 100 lbs is applied at (2,-8)

>How would you go about finding the reaction forces on the legs?

Sorry but that is a nonsense example. The weight is not even on the table. Just draw it out on a piece of paper and you'll see what I mean. If the weight was at (2,2) then fair enough.

I have charted it using Excel - see screenshot below.

tableleg-loading-01.bmp

Let's say our table has legs at the following coordinates

(10,2)

(-11,4)

(-6,-16)

(5,-14)

a force of 100 lbs is applied at (2,-8)

>How would you go about finding the reaction forces on the legs?

Sorry but that is a nonsense example. The weight is not even on the table. Just draw it out on a piece of paper and you'll see what I mean. If the weight was at (2,2) then fair enough.

I have charted it using Excel - see screenshot below.

tableleg-loading-01.bmp

ASKER

You plotted -5,14. Â It should be 5,-14. Â Now the example makes sense.

OK - you're quitte right - my mistake.

My solution is in the attached file. I have no doubt that it's wrong as frankly I can't remember how to do it. I'm sure Brad (byundt) will provide the correct solution in due course. Screenshot attached.

tableleg-loading-02.bmp

tabeleg-loading02.xls

My solution is in the attached file. I have no doubt that it's wrong as frankly I can't remember how to do it. I'm sure Brad (byundt) will provide the correct solution in due course. Screenshot attached.

tableleg-loading-02.bmp

tabeleg-loading02.xls

ASKER

Patrick,

Thank you very much for your efforts. Â Let's hope Brad can shed some light on this for us. Â If he can't figure it out the solution most likely does not exist. Â Thanks again.

Kyle

Thank you very much for your efforts. Â Let's hope Brad can shed some light on this for us. Â If he can't figure it out the solution most likely does not exist. Â Thanks again.

Kyle

ASKER

After reviewing some of my old text books I don't believe a theoretical solution exists. Â Consider the definition of the balance point location from multiple objects.

X = (1/MTotal)*(m1*x1 + m2*x2 + m3*x3 + m4*x4)

Y = (1/MTotal)*(m1*y1 + m2*y2 + m3*y3 + m4*y4)

Obviously to solve for the four individual masses you need four equations. Â One additional equation can be written since we know the sum of all the masses but I do not believe there is a way to come up with the fourth and final equation. Â Perhaps there is still a way to solve this problem iteratively.

Kyle

X = (1/MTotal)*(m1*x1 + m2*x2 + m3*x3 + m4*x4)

Y = (1/MTotal)*(m1*y1 + m2*y2 + m3*y3 + m4*y4)

Obviously to solve for the four individual masses you need four equations. Â One additional equation can be written since we know the sum of all the masses but I do not believe there is a way to come up with the fourth and final equation. Â Perhaps there is still a way to solve this problem iteratively.

Kyle

Kyle,

If you put the coordinates of the legs to make a square table (10,10), (-10,10), (10,-10), (-10,-10) and place the weight in the centre (0,0) my worksheet produced 25% for each of the legs, which of course is correct. However if the weight's location is changed to be at the same as the location of a leg I would expect it to be 100% for that leg - it isn't.

Likewise if the weight is halfway between two legs I'd expect the result to be 50% for each of the legs - it isn't. Perhaps there are problems with zeros - I dunno.

Anyhow try my workbook with different data and see what happens.

Patrick

If you put the coordinates of the legs to make a square table (10,10), (-10,10), (10,-10), (-10,-10) and place the weight in the centre (0,0) my worksheet produced 25% for each of the legs, which of course is correct. However if the weight's location is changed to be at the same as the location of a leg I would expect it to be 100% for that leg - it isn't.

Likewise if the weight is halfway between two legs I'd expect the result to be 50% for each of the legs - it isn't. Perhaps there are problems with zeros - I dunno.

Anyhow try my workbook with different data and see what happens.

Patrick

ASKER

Patrick,

I played around with your workbook a little bit last night and found the same inconsistencies. Â I will take another look and try to determine exactly how you are calculating you percentages. Â Thanks

Kyle

I played around with your workbook a little bit last night and found the same inconsistencies. Â I will take another look and try to determine exactly how you are calculating you percentages. Â Thanks

Kyle

Hello Kyle,

I'm back. Â I was halfway home when I discovered that you were way ahead of me.

You are correct, that here is not a unique solution.

There is an infinite number of solutions, and you can find them all.

Behold the lowly tripod.

You have four (or N) legs. Â Look at them in sets of three.

You decompose your 4-leg table into four 3-leg/triangular Â tables.

In general, two of them will support the load, and two of them won't.

[If the load is on a boundary between tables, then you may have to consider

three or all four sub-tables.]

The nice thing about a 3-leg table, is that it does have a unique solution.

It will give you a weight distribution for the three included legs with zero

load on the fourth.

After you find the solutions for each of the relevant 3-leg tables, you can distribute the weight among them any way you want, i.e. linear combinations.

My linear algebra is long ago and far, but I think we are talking about eigen vectors.

In a physical table, the location of the load relative to the centroids of the component tables and the deflection characteristics of the top and legs will determine the relative weighting of the component vectors.

I think this sort decomposition will work for more legs as well, including interior ones.

I'm back. Â I was halfway home when I discovered that you were way ahead of me.

You are correct, that here is not a unique solution.

There is an infinite number of solutions, and you can find them all.

Behold the lowly tripod.

You have four (or N) legs. Â Look at them in sets of three.

You decompose your 4-leg table into four 3-leg/triangular Â tables.

In general, two of them will support the load, and two of them won't.

[If the load is on a boundary between tables, then you may have to consider

three or all four sub-tables.]

The nice thing about a 3-leg table, is that it does have a unique solution.

It will give you a weight distribution for the three included legs with zero

load on the fourth.

After you find the solutions for each of the relevant 3-leg tables, you can distribute the weight among them any way you want, i.e. linear combinations.

My linear algebra is long ago and far, but I think we are talking about eigen vectors.

In a physical table, the location of the load relative to the centroids of the component tables and the deflection characteristics of the top and legs will determine the relative weighting of the component vectors.

I think this sort decomposition will work for more legs as well, including interior ones.

You can't 'break-down' a 4 legged table into 3-legged components. The reason is that whilst the weight will be in two of the set of 4 overlapping sectors (unless it's in the centre) it is then not possible to resolve for the other two overlapping sectors. The reason is that the weight is not in those sectors at all. So decomposing tables into 3-legged sectors is somewhat unlikely to be the route to a successful solution.

Lastly there are not an infinite number of solutions for a given table and a given weight in a specific location. In those circumstances there are just four figures for the loading on each leg - for a 4-legged table.

Lastly there are not an infinite number of solutions for a given table and a given weight in a specific location. In those circumstances there are just four figures for the loading on each leg - for a 4-legged table.

Take your square table with legs at:

Â Â Â A=(10,10), B=(-10,10), C=(10,-10), D=(-10,-10)

Put the load at (5,0). Â

CASE 1: Â Take away B. Â The load is supported uniquely on ACD.

Â Â Â Â Â Â Â This is also a solution for the 4-leg table with load on B=0

CASE 2: Â Take away D. Â The load is supported uniquely on ABC.

Â Â Â Â Â Â Â This is also a solution for the 4-leg table with load on D=0.

Slit up the load between CASES 1 and 2 anyway you like and you will valid solutions.

Â Â Â A=(10,10), B=(-10,10), C=(10,-10), D=(-10,-10)

Put the load at (5,0). Â

CASE 1: Â Take away B. Â The load is supported uniquely on ACD.

Â Â Â Â Â Â Â This is also a solution for the 4-leg table with load on B=0

CASE 2: Â Take away D. Â The load is supported uniquely on ABC.

Â Â Â Â Â Â Â This is also a solution for the 4-leg table with load on D=0.

Slit up the load between CASES 1 and 2 anyway you like and you will valid solutions.

The thoery is probably correct but...

The problem with looking at in such a simple way is that it ignores reality. The reality is that if you sit on a table the increase in load will be felt on all 4 legs. That's assuming of course that you are sitting inside the line joining any two adjacent legs. The reason is that the table flexes and so the load is shared by all the legs inversely proportional to the distance from the load.

So in the case of the table Â A=(10,10), B=(10,-10), C=(-10,-10), D= (-10,10), with the weight as you suggest at (5,0) the load is shared in pairs of legs as follows:

A &Â B - 10/20 together ie. 50%/50%

C &Â D - 5/20 together ie. 25%/75%

As the weight is at (5,0) then the load is shared so:

A = 25%

B = 25%

C = 12.5% (25% of 50%)

D = 37.5% (75% of 50%)

Note: I have changed the name locations of the legs to make it easier to visualise.

The problem with looking at in such a simple way is that it ignores reality. The reality is that if you sit on a table the increase in load will be felt on all 4 legs. That's assuming of course that you are sitting inside the line joining any two adjacent legs. The reason is that the table flexes and so the load is shared by all the legs inversely proportional to the distance from the load.

So in the case of the table Â A=(10,10), B=(10,-10), C=(-10,-10), D= (-10,10), with the weight as you suggest at (5,0) the load is shared in pairs of legs as follows:

A &Â B - 10/20 together ie. 50%/50%

C &Â D - 5/20 together ie. 25%/75%

As the weight is at (5,0) then the load is shared so:

A = 25%

B = 25%

C = 12.5% (25% of 50%)

D = 37.5% (75% of 50%)

Note: I have changed the name locations of the legs to make it easier to visualise.

A simpler example would be a one-dimensional table supported by 3 legs. Â With a simple rigid statics model it's just not possible to determine the load on each leg. Â That's like trying to find a unique solution for (a + b = 3).

If you use a nonrigid physics model, and take sagging of the table structure and compression of the legs into account, then you get a unique answer.

a.png

If you use a nonrigid physics model, and take sagging of the table structure and compression of the legs into account, then you get a unique answer.

a.png

ASKER

All,

Have a look at this workbook. Â It follows the following steps in arriving at a solution.

1. Â Break the 4 leg table down into 4 - 3 leg tables (4 loops)

2. Â Solve for the three leg reactions for each sub-table using sum of forces and sum of moments

3. Â Average the three loadings found for each leg and ratio accordingly so the sum of the three averages equals the total weight

The only problem that I see with this method is a more complex way of finding the end solutions is necessary. Â The averaging trick falls short when the weight is placed close to one of the legs. Â For example, with our 10x10 square table, if you place the load at (10,10) you end up with leg reactions of R1 = 75, R2 = 25, R3 = -25, and R4 = 25. Â For a perfectly rigid structure it should be R1 = 100, R2 = 0, R3 = 0, R4 = 0

I think we're getting closer though:)

kyle

Approx-Pad-Reactions.xls

Have a look at this workbook. Â It follows the following steps in arriving at a solution.

1. Â Break the 4 leg table down into 4 - 3 leg tables (4 loops)

2. Â Solve for the three leg reactions for each sub-table using sum of forces and sum of moments

3. Â Average the three loadings found for each leg and ratio accordingly so the sum of the three averages equals the total weight

The only problem that I see with this method is a more complex way of finding the end solutions is necessary. Â The averaging trick falls short when the weight is placed close to one of the legs. Â For example, with our 10x10 square table, if you place the load at (10,10) you end up with leg reactions of R1 = 75, R2 = 25, R3 = -25, and R4 = 25. Â For a perfectly rigid structure it should be R1 = 100, R2 = 0, R3 = 0, R4 = 0

I think we're getting closer though:)

kyle

Approx-Pad-Reactions.xls

I believe in the flexible real world if the loads are resolved vertically and horizontally in pairs of legs the apportionment of load per leg is simple in a square/rectangular table - as above.

In your workbook example, only sub-tables 124 and 134 can support the load.

If you put the load exactly on leg 4, then you have to consider three subtables:

Â Â Â 124, 134, and 234.

But I believe the only solution for each sub-table places the full load at 4.

So any linear combination of these solutions would also place the full load at 4.

If you put the load exactly on leg 4, then you have to consider three subtables:

Â Â Â 124, 134, and 234.

But I believe the only solution for each sub-table places the full load at 4.

So any linear combination of these solutions would also place the full load at 4.

I am pretty happy with this solution.

I used Matlab for the calculations.

I used Matlab for the calculations.

```
I took your table and translated it so that the load is applied
at (0,0) rather than (2,-8).
The leg positions become A=(8,10) B=(-13,12) ...
The matrix equation for the table is
| 8 -13 -8 3 | | A | | 0 | X moment
| 10 12 -8 -6 | x | B | = | 0 | Y Moment
| 1 1 1 1 | | C | |100| Total Load
| D |
Where A B C D are the unknown loads on each leg.
You have three eqations and four unknowns, but you
can force any one of the unknowns to 0 and solve for
the other three. Equivalent to taking subtables.
Case 1 Sub-Table BCD is not valid
0.0000
31.3043
-18.2609
86.9565
Case 2 Sub-Table ACD is a solution
43.3735
0.0000
46.9880
9.6386
Case 3 Sub-Table ABD is a solution
12.1387
22.5434
0.0000
65.3179
Case 4 Sub-Table ABD is not valid
48.7805
-3.9024
55.1220
0.0000
```

ASKER

d-glitch,

Our numbers agree. Â I get exactly the same thing as you using my excel workbook. Â So, what would be your proposed answer to the loads on legs A, B, C, and D? Â I'm assuming you're saying it can be either the answers from sub-table ACD or ABD.

The more I work on this the more I believe that is impossible to approximate the loadings on a four legged table without introducing deflection in either the legs or the table top.

Kyle

Our numbers agree. Â I get exactly the same thing as you using my excel workbook. Â So, what would be your proposed answer to the loads on legs A, B, C, and D? Â I'm assuming you're saying it can be either the answers from sub-table ACD or ABD.

The more I work on this the more I believe that is impossible to approximate the loadings on a four legged table without introducing deflection in either the legs or the table top.

Kyle

ASKER CERTIFIED SOLUTION

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Like I said above, you can see this principle more clearly with a one-dimensional 3-legged table. Â It's like trying to find a unique solution for (a + b = 3).

Doing the least squares calculation:

Â k = 0.554 Â Â [ A B C D ] Â = Â [ 28.83 Â 10.49 Â 25.11 Â 35.56 ]

Â k = 0.554 Â Â [ A B C D ] Â = Â [ 28.83 Â 10.49 Â 25.11 Â 35.56 ]

ASKER

All,

I haven't forgotten about this Q. Â I am currently working with one of my colleagues to try and come up with a fourth equation that takes deflection of the legs into account. Â We are essentially assuming that the legs are springs with some constant 'k'. Â Since we are assuming the table top to be perfectly rigid all the legs will remain in the same plane as as the system deflects. Â This should provide us with enough information to write one more independent equation that relates the deflections (and thus forces) of all the legs to each other. Â I'm guessing that as the spring constant approaches infinity the reactions in the legs will approach the theoretical solution. Â When we figure it out I'll reply with our solution.

Kyle

I haven't forgotten about this Q. Â I am currently working with one of my colleagues to try and come up with a fourth equation that takes deflection of the legs into account. Â We are essentially assuming that the legs are springs with some constant 'k'. Â Since we are assuming the table top to be perfectly rigid all the legs will remain in the same plane as as the system deflects. Â This should provide us with enough information to write one more independent equation that relates the deflections (and thus forces) of all the legs to each other. Â I'm guessing that as the spring constant approaches infinity the reactions in the legs will approach the theoretical solution. Â When we figure it out I'll reply with our solution.

Kyle

ASKER

Ok,

Here is our solution. Â It's different than I originally anticipated but it seems to work pretty well. Â The steps are as follows.

1. Â Find the area centroid of all the pad reactions. Â To do this you'll need to assume a shape for each of the table legs. Â For the purposes of this discussion lets assume we have square table legs that measure 1x1.

X coordinate of centroid: Â Xc = (A1*X1 + A2*X2 + A3*X3 + A4*X4)/(A1+A2+A3+A4)

X coordinate of centroid: Â Yc = (A1*Y1 + A2*Y2 + A3*Y3 + A4*Y4)/(A1+A2+A3+A4)

2. Â Now shift all the leg locations and the load location so the original origin sits on top of the area centroid.

Pad 1: Â X1' = X1 - Xc Â Â Y1' = Y1 - Yc

Pad 2: Â X2' = X2 - Xc Â Â Y2' = Y2 - Yc

Pad 3: Â X3' = X3 - Xc Â Â Y3' = Y3 - Yc

Pad 4: Â X4' = X4 - Xc Â Â Y4' = Y4 - Yc

Load: Â XL' = XL - Xc Â Â YL' = YL = Yc

3. Â Find the moment of inertia for each leg

MOI for rectangle: Â I = (b*d^3)/12

4. Â Use the parallel axis theorem to find the moment of inertia of each pad reaction about the area centroid. Â Since all the legs are the same size I = I1 = I2 = I3 = I4.

Pad 1: Â I1X = I1 + A1*X1' Â Â I1Y = I1 + A1*Y1'

Pad 2: Â I2X = I1 + A2*X2' Â Â I2Y = I2 + A2*Y2'

Pad 3: Â I3X = I1 + A3*X3' Â Â I3Y = I3 + A3*Y3'

Pad 4: Â I4X = I1 + A4*X4' Â Â I4Y = I4 + A4*Y4'

5. Â Sum the X- and Y-components to achieve the total moment of inertia about the X' and Y' axes

IX' = I1X + I2X + I3X + I4X

IY' = I1Y + I2Y + I3Y + I4Y

Note: Â We have just created a "virtual beam" with polar moments of inertia IX' and IY'.

6. Â The "leg reactions" can be now be found by calculating the total stress at each of the leg coordinates. Â The stress is caused by the moment of the load about the X' and Y' axes.

Axial stress: Â SigmaA = Load/(A1 + A2 + A3 + A4)

Bending Stress about X' axis: Â SigmaBx = (Mx*Cy)/IX'

Bend Stress about Y' axis: Â SigmaBy = (My*Cx)/IY'

MX' = Load*YL'

MY' = Load*YX'

Pad 1: Â SigmaBx1 = (MX'*Y1')/IX' Â Â SigmaBy1 = (MY'*X1')/IY'

Pad 2: Â SigmaBx2 = (MX'*Y2')/IX' Â Â SigmaBy2 = (MY'*X2')/IY'

Pad 3: Â SigmaBx3 = (MX'*Y3')/IX' Â Â SigmaBy3 = (MY'*X3')/IY'

Pad 4: Â SigmaBx4 = (MX'*Y4')/IX' Â Â SigmaBy4 = (MY'*X4')/IY'

7. Â Calculate leg loadings

R1 = SigmaA + SigmaBx1 + SigmaBy1

R2 = SigmaA + SigmaBx2 + SigmaBy2

R3 = SigmaA + SigmaBx3 + SigmaBy3

R4 = SigmaA + SigmaBx4 + SigmaBy4

Using the pad coordinates of our example and a load of 100 lbs at (2,-8) the following leg loadings are calculated using this method

R1 = 29.46

R2 = 9.58

R3 = 26.20

R4 = 34.76

d-glitch,

Here are the percent differences between our two approaches:

R1 = 2.1%

R2 = 9.5%

R3 = 4.2%

R4 = 2.3%

Here is our solution. Â It's different than I originally anticipated but it seems to work pretty well. Â The steps are as follows.

1. Â Find the area centroid of all the pad reactions. Â To do this you'll need to assume a shape for each of the table legs. Â For the purposes of this discussion lets assume we have square table legs that measure 1x1.

X coordinate of centroid: Â Xc = (A1*X1 + A2*X2 + A3*X3 + A4*X4)/(A1+A2+A3+A4)

X coordinate of centroid: Â Yc = (A1*Y1 + A2*Y2 + A3*Y3 + A4*Y4)/(A1+A2+A3+A4)

2. Â Now shift all the leg locations and the load location so the original origin sits on top of the area centroid.

Pad 1: Â X1' = X1 - Xc Â Â Y1' = Y1 - Yc

Pad 2: Â X2' = X2 - Xc Â Â Y2' = Y2 - Yc

Pad 3: Â X3' = X3 - Xc Â Â Y3' = Y3 - Yc

Pad 4: Â X4' = X4 - Xc Â Â Y4' = Y4 - Yc

Load: Â XL' = XL - Xc Â Â YL' = YL = Yc

3. Â Find the moment of inertia for each leg

MOI for rectangle: Â I = (b*d^3)/12

4. Â Use the parallel axis theorem to find the moment of inertia of each pad reaction about the area centroid. Â Since all the legs are the same size I = I1 = I2 = I3 = I4.

Pad 1: Â I1X = I1 + A1*X1' Â Â I1Y = I1 + A1*Y1'

Pad 2: Â I2X = I1 + A2*X2' Â Â I2Y = I2 + A2*Y2'

Pad 3: Â I3X = I1 + A3*X3' Â Â I3Y = I3 + A3*Y3'

Pad 4: Â I4X = I1 + A4*X4' Â Â I4Y = I4 + A4*Y4'

5. Â Sum the X- and Y-components to achieve the total moment of inertia about the X' and Y' axes

IX' = I1X + I2X + I3X + I4X

IY' = I1Y + I2Y + I3Y + I4Y

Note: Â We have just created a "virtual beam" with polar moments of inertia IX' and IY'.

6. Â The "leg reactions" can be now be found by calculating the total stress at each of the leg coordinates. Â The stress is caused by the moment of the load about the X' and Y' axes.

Axial stress: Â SigmaA = Load/(A1 + A2 + A3 + A4)

Bending Stress about X' axis: Â SigmaBx = (Mx*Cy)/IX'

Bend Stress about Y' axis: Â SigmaBy = (My*Cx)/IY'

MX' = Load*YL'

MY' = Load*YX'

Pad 1: Â SigmaBx1 = (MX'*Y1')/IX' Â Â SigmaBy1 = (MY'*X1')/IY'

Pad 2: Â SigmaBx2 = (MX'*Y2')/IX' Â Â SigmaBy2 = (MY'*X2')/IY'

Pad 3: Â SigmaBx3 = (MX'*Y3')/IX' Â Â SigmaBy3 = (MY'*X3')/IY'

Pad 4: Â SigmaBx4 = (MX'*Y4')/IX' Â Â SigmaBy4 = (MY'*X4')/IY'

7. Â Calculate leg loadings

R1 = SigmaA + SigmaBx1 + SigmaBy1

R2 = SigmaA + SigmaBx2 + SigmaBy2

R3 = SigmaA + SigmaBx3 + SigmaBy3

R4 = SigmaA + SigmaBx4 + SigmaBy4

Using the pad coordinates of our example and a load of 100 lbs at (2,-8) the following leg loadings are calculated using this method

R1 = 29.46

R2 = 9.58

R3 = 26.20

R4 = 34.76

d-glitch,

Here are the percent differences between our two approaches:

R1 = 2.1%

R2 = 9.5%

R3 = 4.2%

R4 = 2.3%

ASKER

Correction:

Pad 1: Â I1X = I1 + A1*X1' Â Â I1Y = I1 + A1*Y1'

Pad 2: Â I2X = I1 + A2*X2' Â Â I2Y = I2 + A2*Y2'

Pad 3: Â I3X = I1 + A3*X3' Â Â I3Y = I3 + A3*Y3'

Pad 4: Â I4X = I1 + A4*X4' Â Â I4Y = I4 + A4*Y4'

Should be...

Pad 1: Â I1X = I + A1*X1' Â Â I1Y = I + A1*Y1'

Pad 2: Â I2X = I + A2*X2' Â Â I2Y = I + A2*Y2'

Pad 3: Â I3X = I + A3*X3' Â Â I3Y = I + A3*Y3'

Pad 4: Â I4X = I + A4*X4' Â Â I4Y = I + A4*Y4'

Pad 1: Â I1X = I1 + A1*X1' Â Â I1Y = I1 + A1*Y1'

Pad 2: Â I2X = I1 + A2*X2' Â Â I2Y = I2 + A2*Y2'

Pad 3: Â I3X = I1 + A3*X3' Â Â I3Y = I3 + A3*Y3'

Pad 4: Â I4X = I1 + A4*X4' Â Â I4Y = I4 + A4*Y4'

Should be...

Pad 1: Â I1X = I + A1*X1' Â Â I1Y = I + A1*Y1'

Pad 2: Â I2X = I + A2*X2' Â Â I2Y = I + A2*Y2'

Pad 3: Â I3X = I + A3*X3' Â Â I3Y = I + A3*Y3'

Pad 4: Â I4X = I + A4*X4' Â Â I4Y = I + A4*Y4'

ASKER

Thank you for all your help. Â This question dragged on for a few days. Â Thanks for staying engaged!

Your solution appears to be a linear combination of

Case 2 and Case 3 with Â k = 0.5545

Happy to see that.

I think my compliance criteria corresponds to a fairly soft spring on each leg.

Don't know that there can be a final resolution, but it is an interesting problem.

Case 2 and Case 3 with Â k = 0.5545

Happy to see that.

I think my compliance criteria corresponds to a fairly soft spring on each leg.

Don't know that there can be a final resolution, but it is an interesting problem.

Hi,

just implemented this method in a software. However had to correct several issues. Took me some time, so I share what I found out:

1.

Â "A" stands for area of the table leg surface

2.

MOI for rectangle: Â I = (b*d^3)/12

explanation: b,d - dimensions of the table leg surface

3.

MX' = Load*YL'

MY' = Load*YX'

should be

MX' = Load*XL'

MY' = Load*YL'

4.

Pad 1: Â I1X = I1 + A1*X1' Â Â I1Y = I1 + A1*Y1'

was corrected to

Pad 1: Â I1X = I + A1*X1' Â Â I1Y = I + A1*Y1'

but according to parallel axis theorem it should be

Pad 1: Â I1X = I + A1*(X1'^2) Â Â I1Y = I + A1*(Y1'^2)

5.

Pad 1: Â SigmaBx1 = (MX'*Y1')/IX' Â Â SigmaBy1 = (MY'*X1')/IY'

should be

Pad 1: Â SigmaBx1 = (MX'*X1')/IX' Â Â SigmaBy1 = (MY'*Y1')/IY'

Then it works well and gives the results as indicated.

just implemented this method in a software. However had to correct several issues. Took me some time, so I share what I found out:

1.

Â "A" stands for area of the table leg surface

2.

MOI for rectangle: Â I = (b*d^3)/12

explanation: b,d - dimensions of the table leg surface

3.

MX' = Load*YL'

MY' = Load*YX'

should be

MX' = Load*XL'

MY' = Load*YL'

4.

Pad 1: Â I1X = I1 + A1*X1' Â Â I1Y = I1 + A1*Y1'

was corrected to

Pad 1: Â I1X = I + A1*X1' Â Â I1Y = I + A1*Y1'

but according to parallel axis theorem it should be

Pad 1: Â I1X = I + A1*(X1'^2) Â Â I1Y = I + A1*(Y1'^2)

5.

Pad 1: Â SigmaBx1 = (MX'*Y1')/IX' Â Â SigmaBy1 = (MY'*X1')/IY'

should be

Pad 1: Â SigmaBx1 = (MX'*X1')/IX' Â Â SigmaBy1 = (MY'*Y1')/IY'

Then it works well and gives the results as indicated.

This technique should work for any convex polygon.

This is certainly a solved problem: Â The touch screen and electronic bathroom scale

folks have figured it out.

If the applied force is large enough and outside of the area bounded by the legs,

the table will start to tip over on the one or more likely two legs that define the

nearest side.

If the force is smaller, you will take weight off all but the nearest legs.

Torque is still the way to go.