# Why are table legs placed at the corners?

Quietrabbit
TL;DR Summary
If I have four posts with a board on top, what is the optimal distance from the center to put them so they can support the most weight added on the board?
I have 4 tiny ~7in wooden posts that have cross bracing (let me know if I need to describe the bracing), and I place a solid wooden board on top of the posts then as much weight as possible on the board until the posts break. The posts are equidistant from the center of where the boards and weights are placed so the force is evenly distributed among them.

I am wondering where the optimal location to place the four posts is. (Ie far out to the corners of the board, like a table, somewhere in between, or no difference and I am wasting my time). For the purpose of this, I am assuming the weights are placed perfectly in the center of the four posts so I am interested in force/load (sorry, I forget the correct term) distribution based on distance for the center, not any torque or tipping of the “table”.

The posts will break long before any buckling, or the like, occurs in the board so I am assuming I don’t need to worry about that. I have considered just testing, but the materials are expensive so I was hoping for an general answer instead. Any mathematical reasoning is also appreciated.

I phrased the initial question referencing a table as that is essentially the setup, and I am assuming the same principles apply. However, I can’t visualize why so I was hoping someone could guide me through that. Thank you.

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Gold Member
no difference and I am wasting my time
This.

You've constrained the problem with so much symmetry that there are no forces described except the gravitational weight downwards, which must be equally shared between the legs. You said no tipping or buckling, so, I'm not worried about tipping or buckling.

My "mathematical reasoning" is symmetry. How can anything happen except in the downward direction?

russ_watters and Quietrabbit
Use the Airy points as a first approximation.
https://en.wikipedia.org/wiki/Airy_points
Bessel points are also explained there.

But it depends on how the table top is loaded.

berkeman and Lnewqban
Homework Helper
Gold Member
Welcome!
How will the load be distributed over the board?

Mentor
You have a table with four posts. Your statements of symmetry say that the legs are equally loaded. Therefore, the load on each leg is 1/4 of the total load. That is true regardless of the locations of the legs.

However, in the real world it is difficult to get equal loading if the legs are close together near the center. But you specify exact symmetry and equal loading, which makes this a math problem rather than a real world design problem. A real world design problem would take into account that loads are never exactly centered, and thus space the legs as far apart as practical so that the load would be shared as equally as possible between the legs. In that case, the legs go to the corners. Assuming, of course, that the top is strong enough.

russ_watters, DaveE, Quietrabbit and 1 other person
Quietrabbit
Welcome!
How will the load be distributed over the board?
We try to place the weights (the donut ones) on it as centered as possible but there is some margin of error. At most we are two inches off center in a direction. I’m not sure if I’d even understand an upper level static’s answer to this though, so assuming its symmetric should still give me a good enough idea to start with.

Homework Helper
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We try to place the weights (the donut ones) on it as centered as possible but there is some margin of error. At most we are two inches off center in a direction. I’m not sure if I’d even understand an upper level static’s answer to this though, so assuming its symmetric should still give me a good enough idea to start with.
Do you have a rough diagram to show us?
Basic dimensions with a plan view will suffice.
Sorry, I still can’t imagine the centered donuts.

Quietrabbit
Do you have a rough diagram to show us?
Basic dimensions with a plan view will suffice.
Sorry, I still can’t imagine the centered donuts.
Hopefully the attached drawling is sufficient.

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• imageForPhysicsForum.pdf
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Gold Member
I've taken the liberty of posting the OP's diagram directly. Hope that's OK, OP.

russ_watters, DaveE and Quietrabbit
Gold Member
The posts will break long before any buckling, or the like, occurs in the board so I am assuming I don’t need to worry about that.
Then the placement of posts is arbitrary.

I agree with others: you have eliminated all the usual factors that would come into play.

As you said yourself: the only mechanism of failure you haven't explicitly ruled out is the posts breaking.

Gold Member
The answer you're looking for is "Put the posts along the square's diagonals, about 0.7x the radius of the average weight disk out from the center"

But, given the top looks pretty thick ; realistically it's best to put the posts at the corners : that way you'll avoid the posts working loose from incidental torquing in various directions.

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Homework Helper
Gold Member
Those steel plates, being the exterior diameter of the biggest one about 17.7 inches, are more rigid than the top board.
Because of that, it seems to me that those could be considered a distributed load, reducing the importance of the locations of the legs.

The weakest link seems to be the bracing of the legs: an X-type bracing is much stronger than any Z-type bracing.
Also, a central post could reduce the risk of placing a heavy plate too much out of the center of the board.

Gold Member
The weakest link seems to be the bracing of the legs: an X-type bracing is much stronger than any Z-type bracing.
Indeed, the bracing as-diagrammed has almost no resistance to counterclockwise rotation, and will surely collapse immediately on its own - never mind adding weights.

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Homework Helper
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Indeed, the bracing as-diagrammed has almost no resistance to counterclockwise rotation, and will surely collapse immediately on its own - never mind adding weights.

View attachment 300464
A cylinder (like a section of 14 or 16-inch pipe) solidly attached to the board, would make a stronger more stable base.

Gold Member
A cylinder (like a section of 14 or 16-inch pipe) solidly attached to the board, would make a stronger more stable base.
Sure, if the OP were asking for that kind of advice, but they are not. Presumably, the design has been chosen for reasons beyond the scope of this question (based on how specific the OP is about what they do and do not want analyzed).

DaveE
Gold Member
Sure, if the OP were asking for that kind of advice, but they are not. Presumably, the design has been chosen for reasons beyond the scope of this question (based on how specific the OP is about what they do and do not want analyzed).
Yes, a frequent PF dilemma. Do you answer the question the OP actually asked, or do you elaborate and answer the question you think they should have asked. IDK. I do it both ways from time to time.

Richard Crane, sysprog, sophiecentaur and 2 others
Quietrabbit
Thank you for all the great feedback. I’ll be sure to change the cross bracing to an X(s).
The base has some other safety measures I didn’t include in the drawing for simplicity like a pole through all the weights so they don’t fall on someone.

Yes, a frequent PF dilemma.
“The only interesting answers are those which destroy the questions.” ― Susan Sontag.

If a structure has 4 legs, one will not carry any weight unless the table top and the floor are flexible. Two of the legs will carry half the weight each, while the other shorter diagonal pair have no load, they will simply keep the balanced table rocking, close to level.

A 3 legged table will be a stable table and will not rock, while the weight on each leg will be one third of the total.

In a rigid structure a 3 leg table will therefore carry 3/2 = 50% more load than 4 legs.
Reduce the structure to carry more load.

If you must use 4 legs, arrange the 4 legs in a triangle with two legs at the same corner.

If 4 legs must be at the corners of a rectangle, place the weight off centre, biassed towards a short leg, so the short diagonal pair will carry some real load.

Or place rubber pads under all feet to balance the load on a rigid structure.

sysprog, anorlunda, DaveE and 2 others
Gold Member
I can imagine lifting a weight onto the table and catching the side of the table top. That could knock the table over unless it is heavy (the same sort of weight as the weights placed on it). Perhaps that's excessive but it should be a consideration.
Stability would be as important as strength and, as long as you don't need to be moving the table around, it should be pretty heavy (including the legs).
A peg in the middle would be good and perhaps wheels on one pair of legs which would touch ground when the table is tilted a bit to allow you to move the table around (a sort of sack barrow idea).
You will appreciate that when I make anything it will withstand anything but a direct hit with a tactical nuclear weapon.

PS is there really any point in making the table as light as possible? If it's to be used for gym weights there should be no shortage of muscle to move it about. Steel or timber are not expensive.

Mentor
As others have said, the vertical weight of the centered weights is not the real design load here. The real design load is weights dropped onto the table (impact loading), bumping into the top of the table from the side, weights dropped off center, plus general abuse. Any table in this application must be designed for loads much larger than just the weight of the weights, or it will be destroyed in a short time.

A better design, if made of wood, would be a box. The horizontal dimensions would be slightly smaller than the diameter of the weights in order to make is easy to grab hold of the weights. The top and bottom should be about 1.5" thick (2 X 12's are 1.5" thick). The sides could be plywood at least 0.5" thick if fitted to the top and bottom using rabbet joints (search the term). If rabbet joints are used, they should be used between each side piece, and between the sides and top and bottom. All joints to be carefully fitted and well glued.

Such a box will still be damaged by weights inadvertently dropped on it, so a piece of rubber on top would be a very good idea. The rubber should be at least 1/4" to 1/2" thick. Such a box will only weigh about 10 pounds.

Lnewqban and berkeman
Gold Member
A better design...
I'd really like to hear some feedback from @Quietrabbit about the bigger picture, so we can make suggestions about perhaps better was to do what he wants.

I somehow doubt the final device is really a weight holder in a public location - because if it is, I'd say it is a very dangerous design.

sysprog
Aren't exercise weights held in racks, and not stacked on tables? ##-## and as for putting load on a pedestal-style table, my opinion is heck no, the tabletop itself is already too much load if there's ever an imbalance that tends to tilt it##\dots##

I suspect we may have been doing someones homework for a competitive laboratory exercise in a structural design class.

sophiecentaur, DaveC426913 and sysprog
Mentor
I suspect we may have been doing someones homework for a competitive laboratory exercise in a structural design class.
In which case, most of our answers are useless because they are outside the scope of the stated problem.

sophiecentaur