Dreaming of and designing my own pool table

[El Jefe]

I'm not a mechanic, but I am an engineer, and can analyze just about any design that can be imagined.

Let me start with a concern that was expressed in post #2: "...a big, heavy slate is only directly supported by a couple of dozens of small wood pieces, totalling maybe 20 square inches of contact between the slate and the frame. This is a lot weight for such a small surface to handle."

How much weight is that? And what are the limits of what can be supported on a small surface?

From several internet references, the slate for a 9' table measures 107" by 57". From another reference, the density of slate is 2.691 g/cm3 (a number I find absurdly precise, most sources give a range of densities ranging up to as much as 3.3 g/cm3). If it turns out that our slate is heavier, we can simply adjust our dimensions.

For slate 25mm thick, that works out to 269 kg (593 lbs).

I have not seen slate thicker than that used in Toulet's carom tables at 80mm. For sake of computation, we'll assume 80mm slate for pool tables can also be had. At 107" x 57", that works out to 847.2 kg or 1868 lbs.

What load can wood bear? Well, that depends on the wood, of course. For sake of calculation, I chose Sugar Pine, likely the weakest species one is likely to encounter in the lowest price category in most US lumberyards. There are weaker woods, but they are more expensive and are likely purchased for reasons other than structural use (appearance, light weight, etc.) Your plywood will be stronger than this, but because it is of 2-way construction rather than the 1-way construction of lumber "as the tree grows", it is unlikely to be significantly stronger...perhaps on par with better grades of pine sold in the US, or some hardwoods. More importantly, if a design works with Sugar Pine, it will work with anything better.

We are speaking of wood in compression, being crushed by the weight of the slate, and wood is weakest in compression perpendicular to the grain. That is to say, a cube of wood will bear more weight if the grain runs up and down than if the grain runs from one side to another.

Sugar Pine in compression, perpendicular to the grain will bear 4,460 pounds to the square inch.

If we take 4 pieces of sugar pine, each at one corner bearing an equal load, then the weight of a 25 mm slate can be carried by 4 pieces of 1/4" by 1/4", the four pieces together totaling 0.25 square inches. There will be considerable bearing capacity left over.

When life safety is an issue, we often multiply the load by a factor of 5. If four 1/4" by 1/4" pieces are a lab curiosity, then four 1/2" by 1/2" pieces totaling one square inch of area is a practical, everyday design.

For the 80 mm slate and a safety factor of 5, four pieces of 3/4" by 3/4" pine totaling 2.25 square inches is more than adequate.

Now, I'm not suggesting that you change your design to six spindly little 3/4" by 3/4" legs, or even that this would be safe. Other factors come into play, such as the slenderness of the components. But if you're at all worried about the amount of table supported by the adjusting pads, rest easy. There is good reason why those millions of the pool tables were able to get by with less.

In general, our materials are much, much stronger than we think, and things usually fail, or fail to function properly for other reasons. I would suggest you think more about the stiffness of your design, how well it resists the deflection of its parts, than in how much overall strength it has.

Give me a day or so, and I'll post some things on stiffness that may either reassure you, or prompt some late change if you don't have too much of it built already.

 

[MamboFats]

I'm not a mechanic, but I am an engineer, and can analyze just about any design that can be imagined.

Let me start with a concern that was expressed in post #2: "...a big, heavy slate is only directly supported by a couple of dozens of small wood pieces, totalling maybe 20 square inches of contact between the slate and the frame. This is a lot weight for such a small surface to handle."

How much weight is that? And what are the limits of what can be supported on a small surface?

From several internet references, the slate for a 9' table measures 107" by 57". From another reference, the density of slate is 2.691 g/cm3 (a number I find absurdly precise, most sources give a range of densities ranging up to as much as 3.3 g/cm3). If it turns out that our slate is heavier, we can simply adjust our dimensions.

For slate 25mm thick, that works out to 269 kg (593 lbs).

I have not seen slate thicker than that used in Toulet's carom tables at 80mm. For sake of computation, we'll assume 80mm slate for pool tables can also be had. At 107" x 57", that works out to 847.2 kg or 1868 lbs.

What load can wood bear? Well, that depends on the wood, of course. For sake of calculation, I chose Sugar Pine, likely the weakest species one is likely to encounter in the lowest price category in most US lumberyards. There are weaker woods, but they are more expensive and are likely purchased for reasons other than structural use (appearance, light weight, etc.) Your plywood will be stronger than this, but because it is of 2-way construction rather than the 1-way construction of lumber "as the tree grows", it is unlikely to be significantly stronger...perhaps on par with better grades of pine sold in the US, or some hardwoods. More importantly, if a design works with Sugar Pine, it will work with anything better.

We are speaking of wood in compression, being crushed by the weight of the slate, and wood is weakest in compression perpendicular to the grain. That is to say, a cube of wood will bear more weight if the grain runs up and down than if the grain runs from one side to another.

Sugar Pine in compression, perpendicular to the grain will bear 4,460 pounds to the square inch.

If we take 4 pieces of sugar pine, each at one corner bearing an equal load, then the weight of a 25 mm slate can be carried by 4 pieces of 1/4" by 1/4", the four pieces together totaling 0.25 square inches. There will be considerable bearing capacity left over.

When life safety is an issue, we often multiply the load by a factor of 5. If four 1/4" by 1/4" pieces are a lab curiosity, then four 1/2" by 1/2" pieces totaling one square inch of area is a practical, everyday design.

For the 80 mm slate and a safety factor of 5, four pieces of 3/4" by 3/4" pine totaling 2.25 square inches is more than adequate.

Now, I'm not suggesting that you change your design to six spindly little 3/4" by 3/4" legs, or even that this would be safe. Other factors come into play, such as the slenderness of the components. But if you're at all worried about the amount of table supported by the adjusting pads, rest easy. There is good reason why those millions of the pool tables were able to get by with less.

In general, our materials are much, much stronger than we think, and things usually fail, or fail to function properly for other reasons. I would suggest you think more about the stiffness of your design, how well it resists the deflection of its parts, than in how much overall strength it has.

Give me a day or so, and I'll post some things on stiffness that may either reassure you, or prompt some late change if you don't have too much of it built already.

thanks, as a college engineer drop-out (due to partying and other non-educational stuff) I still love to know the physics behind things and stuff. I find this very interesting.
I'm looking forward to learn some more on stiffness.
thx,

 

[El Jefe]

Well, then let's start with members in compression, because that's easier. Every material has a modulus of elasticity, which is the ratio of stress to strain, or how far it moves per pound of force. For sugar pine parallel to the grain, that's given as 1,190,000 pounds per square inch.

That can seem very confusing, but what it means is this. If you make a 1" by 1" cube of sugar pine, and put one pound on it, it will move 1 / 1,190,000 inches. Not very much! The crushing strength parallel to the grain is probably higher than the crushing strength perpendicular to the grain, but let's see what happens if we apply that much force to our little cube. It squishes down 4460 / 1190000 inches, or 0.00375 inches. Go much more than that, say maybe 5 or 6 thousandths, and it starts to look like this:

Up until it starts to crush, it's a fairly linear relationship. It's very linear (nice straight line on a graph) for many metals, maybe a little less so for wood. But 1,190,000 is a good ballpark estimate. Quality structural wood (denser yellow pines and some hardwoods) will go up to 1,900,000, and some engineered wood products go up to 2,100,000 -- they're stiffer and deform less under pressure. Your plywood is probably around 1,300,000 - 1,400,000 or so, nearly as soft as my pine.

Now, let's think about 2 of those 1-inch cubes stacked on top of each other. Push down on one with 4460 pounds, and it pushes down on the bottom one with 4460 pounds, also. So, they both shrink about 0.00375 inches, or about 0.0075 inches in total. Place them side by side, and let them share the load, and each one only takes 2230 pounds, and only shrinks 0.00187 inches. So, that gives us the formula for other shapes.

Deflection (we use the lowercase Greek letter delta) = (Force x Length) / (Elastic Modulus x Area).

To keep the units straight, force is in pounds, length in inches, area in square inches, and Modulus in pounds per square inch. We could also use metric units, if so Modulus is usually in MegaPascals or GigaPascals.

So, let's look at some pool table loads. If we have 4 legs, and our leg layout is symmetrical from end to end and side to side (like 99% of them are), it's easy. Each leg gets 1/4 of the load. Let me use two 2x3s glued (and screwed) together for each leg. Each one is 1.5" x 2.5", and there's two of them, so the area is 7.5 square inches. I'll put my slate exactly at the WPA maximum, 31 inches, deduct 1 inch for my 25.4 mm slate, and go from there.

Deflection = ((593 / 4) x 30") / (1,190,000 x 7.5) = 0.000498"

That was deliberate. If it's less than 0.001", I pretty much don't have to worry about it.

If you try the same thing with the 80mm slate (1868 lbs) and three 2x6s glued together (24.75 square inches), you'll get a similar result.

It's important for you that this be negligible, since you're building a six-legged table. Every wonder just how much weight each leg would be taking? It's a little more complex with 6, because it matters where each one is.

My understanding is that you have one leg way out at each corner, and the center legs exactly in the middle. So, here's how the weight of the slate divides between them: The dividing line is exactly halfway between each support. If a molecule of table slate is closer to one leg than the other, that's the leg that bears its weight. So, the two end legs get half of one half of the weight, or 1/4, the center legs get half of the "north" half" (1/4), plus half of the "south" half (another 1/4) and the other two end legs get the remaining 1/4.

Yep, that's right, the center pair of legs carries twice the weight of either of the end pairs. And from our formula...

Deflection = (Force x Length) / (Modulus x Area)

If they have double the force, are the same length, are made out of the same stuff, and are the same area, they should settle twice as much. Is that going to cause the center of your table to sag?

Let's see what happens if you use three 2x4s (area = 15.75 sq. in.) for your legs:

End legs, heavy slate: ((1868 / (4 x 2 legs) x 30") / (1,190,000 x 15.75) = 0.000374" No problem there!

Center legs, heavy slate: ((1868 / (2 x 2 legs) x 30") / (1,190,000 x 15.75) = 0.000747"

So it will sag...a little less than 4 ten-thousandths! I think those balls would still roll fairly straight.

So now you can look at the cross sectional area of your legs, and determine if you have a problem or not. If it's anything near 16 square inches per leg, you don't have to worry. If it's less than that, you can use thinner slate (really hard to find 80mm with pool table cutouts, anyway), or make your middle legs a little bigger (probably not the aesthetic you were looking for), move the outside legs in so that they take some of the load from the center legs, etc. Even doubling plywood where you had planned for singling it would fix this.

One last thing before we leave compression members, and that's the leg adjusters. I think you can see that you want as much area as you can get. Well, nearly every coin-operated bar table in the US has 1/2" screws with 13 threads per inch, and I think Diamond uses them, too. I found out that the Brunswick Gold Crowns use 3/4" with 16 threads per inch, which gives you infinitesimally less sag and a slightly finer touch on leveling the legs. If you put 3/4" 16 tpi nuts in your legs, you can buy Gold Crown feet off the internet and use those.

I don't know what load they're rated at, but I found these:

They're rather pricey but note that they are rated at 4,650 lbs per leg. As I said before, most materials...including little 3/4" screws...are stronger than you think.

 

[rexus31]

I'm not an engineer so take this for what it is worth. Even though lessor gauge materials can handle the weight doesn't mean it is the best choice for pool table construction. Total mass is your friend in terms of stability and playability. One of things that separates say, a Gold Crown from a furniture grade table is the construction and engineering. While the "streamlined" frame and pedestals of a furniture grade table can effectively support the weight of the slate, the robust Gold Crown frame and pedestals not only provides said support but also adds mass to the equation providing stability for the sum of the whole, thus providing superior playbility. IMO, bigger (heavier) is better when it comes to pool table engineering. Total weight of a 9' Gold Crown I is 1,270 lbs. so 54% of its total mass is derived from materials other than slate. IMO, this makes a superior playing table.

 

[MamboFats]

if in college these subjects ands lessons were given in relation to pooltables, my interest and attention span would have been much higher ...

 

[realkingcobra]

Well, then let's start with members in compression, because that's easier. Every material has a modulus of elasticity, which is the ratio of stress to strain, or how far it moves per pound of force. For sugar pine parallel to the grain, that's given as 1,190,000 pounds per square inch.

That can seem very confusing, but what it means is this. If you make a 1" by 1" cube of sugar pine, and put one pound on it, it will move 1 / 1,190,000 inches. Not very much! The crushing strength parallel to the grain is probably higher than the crushing strength perpendicular to the grain, but let's see what happens if we apply that much force to our little cube. It squishes down 4460 / 1190000 inches, or 0.00375 inches. Go much more than that, say maybe 5 or 6 thousandths, and it starts to look like this:

View attachment 632325

Up until it starts to crush, it's a fairly linear relationship. It's very linear (nice straight line on a graph) for many metals, maybe a little less so for wood. But 1,190,000 is a good ballpark estimate. Quality structural wood (denser yellow pines and some hardwoods) will go up to 1,900,000, and some engineered wood products go up to 2,100,000 -- they're stiffer and deform less under pressure. Your plywood is probably around 1,300,000 - 1,400,000 or so, nearly as soft as my pine.

Now, let's think about 2 of those 1-inch cubes stacked on top of each other. Push down on one with 4460 pounds, and it pushes down on the bottom one with 4460 pounds, also. So, they both shrink about 0.00375 inches, or about 0.0075 inches in total. Place them side by side, and let them share the load, and each one only takes 2230 pounds, and only shrinks 0.00187 inches. So, that gives us the formula for other shapes.

Deflection (we use the lowercase Greek letter delta) = (Force x Length) / (Elastic Modulus x Area).

To keep the units straight, force is in pounds, length in inches, area in square inches, and Modulus in pounds per square inch. We could also use metric units, if so Modulus is usually in MegaPascals or GigaPascals.

So, let's look at some pool table loads. If we have 4 legs, and our leg layout is symmetrical from end to end and side to side (like 99% of them are), it's easy. Each leg gets 1/4 of the load. Let me use two 2x3s glued (and screwed) together for each leg. Each one is 1.5" x 2.5", and there's two of them, so the area is 7.5 square inches. I'll put my slate exactly at the WPA maximum, 31 inches, deduct 1 inch for my 25.4 mm slate, and go from there.

Deflection = ((593 / 4) x 30") / (1,190,000 x 7.5) = 0.000498"

That was deliberate. If it's less than 0.001", I pretty much don't have to worry about it.

If you try the same thing with the 80mm slate (1868 lbs) and three 2x6s glued together (24.75 square inches), you'll get a similar result.

It's important for you that this be negligible, since you're building a six-legged table. Every wonder just how much weight each leg would be taking? It's a little more complex with 6, because it matters where each one is.

My understanding is that you have one leg way out at each corner, and the center legs exactly in the middle. So, here's how the weight of the slate divides between them: The dividing line is exactly halfway between each support. If a molecule of table slate is closer to one leg than the other, that's the leg that bears its weight. So, the two end legs get half of one half of the weight, or 1/4, the center legs get half of the "north" half" (1/4), plus half of the "south" half (another 1/4) and the other two end legs get the remaining 1/4.

Yep, that's right, the center pair of legs carries twice the weight of either of the end pairs. And from our formula...

Deflection = (Force x Length) / (Modulus x Area)

If they have double the force, are the same length, are made out of the same stuff, and are the same area, they should settle twice as much. Is that going to cause the center of your table to sag?

Let's see what happens if you use three 2x4s (area = 15.75 sq. in.) for your legs:

End legs, heavy slate: ((1868 / (4 x 2 legs) x 30") / (1,190,000 x 15.75) = 0.000374" No problem there!

Center legs, heavy slate: ((1868 / (2 x 2 legs) x 30") / (1,190,000 x 15.75) = 0.000747"

So it will sag...a little less than 4 ten-thousandths! I think those balls would still roll fairly straight.

So now you can look at the cross sectional area of your legs, and determine if you have a problem or not. If it's anything near 16 square inches per leg, you don't have to worry. If it's less than that, you can use thinner slate (really hard to find 80mm with pool table cutouts, anyway), or make your middle legs a little bigger (probably not the aesthetic you were looking for), move the outside legs in so that they take some of the load from the center legs, etc. Even doubling plywood where you had planned for singling it would fix this.

One last thing before we leave compression members, and that's the leg adjusters. I think you can see that you want as much area as you can get. Well, nearly every coin-operated bar table in the US has 1/2" screws with 13 threads per inch, and I think Diamond uses them, too. I found out that the Brunswick Gold Crowns use 3/4" with 16 threads per inch, which gives you infinitesimally less sag and a slightly finer touch on leveling the legs. If you put 3/4" 16 tpi nuts in your legs, you can buy Gold Crown feet off the internet and use those.

I don't know what load they're rated at, but I found these:

View attachment 632326

They're rather pricey but note that they are rated at 4,650 lbs per leg. As I said before, most materials...including little 3/4" screws...are stronger than you think.

Everything you just mentioned is nothing but word salad!
Explain why the 6 leg 1912 Regina 10' billiards table i set up in Billings, MT had almost NO weight on the center legs!

Slates were 1 1/2" thick.

 

[realkingcobra]

Well, then let's start with members in compression, because that's easier. Every material has a modulus of elasticity, which is the ratio of stress to strain, or how far it moves per pound of force. For sugar pine parallel to the grain, that's given as 1,190,000 pounds per square inch.

That can seem very confusing, but what it means is this. If you make a 1" by 1" cube of sugar pine, and put one pound on it, it will move 1 / 1,190,000 inches. Not very much! The crushing strength parallel to the grain is probably higher than the crushing strength perpendicular to the grain, but let's see what happens if we apply that much force to our little cube. It squishes down 4460 / 1190000 inches, or 0.00375 inches. Go much more than that, say maybe 5 or 6 thousandths, and it starts to look like this:

View attachment 632325

Up until it starts to crush, it's a fairly linear relationship. It's very linear (nice straight line on a graph) for many metals, maybe a little less so for wood. But 1,190,000 is a good ballpark estimate. Quality structural wood (denser yellow pines and some hardwoods) will go up to 1,900,000, and some engineered wood products go up to 2,100,000 -- they're stiffer and deform less under pressure. Your plywood is probably around 1,300,000 - 1,400,000 or so, nearly as soft as my pine.

Now, let's think about 2 of those 1-inch cubes stacked on top of each other. Push down on one with 4460 pounds, and it pushes down on the bottom one with 4460 pounds, also. So, they both shrink about 0.00375 inches, or about 0.0075 inches in total. Place them side by side, and let them share the load, and each one only takes 2230 pounds, and only shrinks 0.00187 inches. So, that gives us the formula for other shapes.

Deflection (we use the lowercase Greek letter delta) = (Force x Length) / (Elastic Modulus x Area).

To keep the units straight, force is in pounds, length in inches, area in square inches, and Modulus in pounds per square inch. We could also use metric units, if so Modulus is usually in MegaPascals or GigaPascals.

So, let's look at some pool table loads. If we have 4 legs, and our leg layout is symmetrical from end to end and side to side (like 99% of them are), it's easy. Each leg gets 1/4 of the load. Let me use two 2x3s glued (and screwed) together for each leg. Each one is 1.5" x 2.5", and there's two of them, so the area is 7.5 square inches. I'll put my slate exactly at the WPA maximum, 31 inches, deduct 1 inch for my 25.4 mm slate, and go from there.

Deflection = ((593 / 4) x 30") / (1,190,000 x 7.5) = 0.000498"

That was deliberate. If it's less than 0.001", I pretty much don't have to worry about it.

If you try the same thing with the 80mm slate (1868 lbs) and three 2x6s glued together (24.75 square inches), you'll get a similar result.

It's important for you that this be negligible, since you're building a six-legged table. Every wonder just how much weight each leg would be taking? It's a little more complex with 6, because it matters where each one is.

My understanding is that you have one leg way out at each corner, and the center legs exactly in the middle. So, here's how the weight of the slate divides between them: The dividing line is exactly halfway between each support. If a molecule of table slate is closer to one leg than the other, that's the leg that bears its weight. So, the two end legs get half of one half of the weight, or 1/4, the center legs get half of the "north" half" (1/4), plus half of the "south" half (another 1/4) and the other two end legs get the remaining 1/4.

Yep, that's right, the center pair of legs carries twice the weight of either of the end pairs. And from our formula...

Deflection = (Force x Length) / (Modulus x Area)

If they have double the force, are the same length, are made out of the same stuff, and are the same area, they should settle twice as much. Is that going to cause the center of your table to sag?

Let's see what happens if you use three 2x4s (area = 15.75 sq. in.) for your legs:

End legs, heavy slate: ((1868 / (4 x 2 legs) x 30") / (1,190,000 x 15.75) = 0.000374" No problem there!

Center legs, heavy slate: ((1868 / (2 x 2 legs) x 30") / (1,190,000 x 15.75) = 0.000747"

So it will sag...a little less than 4 ten-thousandths! I think those balls would still roll fairly straight.

So now you can look at the cross sectional area of your legs, and determine if you have a problem or not. If it's anything near 16 square inches per leg, you don't have to worry. If it's less than that, you can use thinner slate (really hard to find 80mm with pool table cutouts, anyway), or make your middle legs a little bigger (probably not the aesthetic you were looking for), move the outside legs in so that they take some of the load from the center legs, etc. Even doubling plywood where you had planned for singling it would fix this.

One last thing before we leave compression members, and that's the leg adjusters. I think you can see that you want as much area as you can get. Well, nearly every coin-operated bar table in the US has 1/2" screws with 13 threads per inch, and I think Diamond uses them, too. I found out that the Brunswick Gold Crowns use 3/4" with 16 threads per inch, which gives you infinitesimally less sag and a slightly finer touch on leveling the legs. If you put 3/4" 16 tpi nuts in your legs, you can buy Gold Crown feet off the internet and use those.

I don't know what load they're rated at, but I found these:

View attachment 632326

They're rather pricey but note that they are rated at 4,650 lbs per leg. As I said before, most materials...including little 3/4" screws...are stronger than you think.

Furthermore, pool table legs don't sag, the frame BETWEEN the legs does. Your whole focus is on the stress and compression of the legs, who cares? We ship 7ft Diamond tables stacked 4 high on CARDBOARD stickers under the bottom table, and between the others. The 2 cardboard stackers under the bottom table have 3,400lbs on the 2 of them.

The frame the slates sit on go through compression and stress, which in turn can cause a slate to sag 3/4" and even more in the middle of the frame, yet the legs have NO problem supporting the overall weight of everything they're holding up, so where's your explanation to the frame of the table and its abilities to handle the weight of the slate? The legs are really irrelevant to the table maintaining level!

 

[El Jefe]

Everything you just mentioned is nothing but word salad!
Explain why the 6 leg 1912 Regina 10' billiards table i set up in Billings, MT had almost NO weight on the center legs!

Slates were 1 1/2" thick.

Well, if the corner legs were an inch taller than the center legs, then the center legs would be taking zero weight. Even if they were all of equal length, if the center slate was bowed (or if the frame supporting it was bowed) upward, then the center legs would take less weight.

Everything I wrote assumes that the legs were equal length and the bed and frame are flat. If you can adjust the height of the legs, then you can adjust how much weight they carry. And, as I showed, even an adjustment of just a few thousandths can make a difference.

This is all part of the discussion as to why 6 (or 8, or 10) legs aren't always "better".

Now, as to the legs, that was just introductory. I'm well aware that the compression of the legs is, if not infinitesimal, at least very, very small compared to other things. But the math is easier, the concepts are easier, and that's a good place to start. More importantly, I wanted to show MamboFats just how much he's probably overthought that part of it in comparison to some of the other parts.

Today, I'll start talking about the frames, which I suspect will be the bulk of my contribution here. My biggest worry is that I'll talk about things like section properties and modulus to the point of turning off anyone who isn't geeking out about those things; I'll try to keep it brief and appropriate to the interests here.

Meanwhile, I had a question for you, but I think you may have already answered it. 3/4" is a lot of sag, without a lot of work I'd guess that table was just about unplayable. I was going to ask just how well does superglue "weld" pieces of slate together, but after poking around a little bit and finding examples of people who could lift an entire 3-piece slate from one end as if it was one piece (and no, I don't recommend that anybody do that!), I think I have an answer.

I suppose I do have a question for you before I leave the subject of legs. Do you know of any better leg adjusters than the Gold Crown ones I mentioned? From what I was able to find, everyone either uses the 1/2" or the 3/4" size. Beyond those, it seems to be so much cramming shims under the feet.

 

[El Jefe]

And that brings me to today's installment...beams. Beams are very important. If you really understand how beams work, then you can understand pool table frames a lot better.

A beam is any long member that carries a load over an empty space beneath it. In our case, it will be carrying the weight of the slate, the rails, anything that sits on top of it. But beams are everywhere around us, from flooring joists in our houses, to the floor pan of our automobiles, to the bookshelf in our study. I can't give you a comprehensive method to analyze any beam, that would take a college course and probably a lot more math than most of you want to be involved with. But I can give you some tools and ideas that will help you think about this more clearly, and maybe help you decide between different options.

Most beam analysis looks at three properties:
1. How much load can the beam carry in shear?
2. What bending moment can the beam safely carry?
3. How much will the beam deflect under load?

For the most part, I'm going to skip over the first two items, shear and bending moment. That's because I doubt that they'll apply much to pool tables, unless you get a wild idea to build one out of 1/4" by 1/4" furring strips. They are of tremendous importance to an engineer designing a skyscraper or a ship, for example. If a beam doesn't satisfy the first two criteria, it will simply collapse. But how many times have you heard of a pool table breaking in half and dumping the bed slates on the floor, vs. one that simply sagged so much that the balls rolled crooked? As we saw from the leg example, it only took a slim 3/4" stick to (barely) carry the load without breaking but ensuring that deflection was kept within bound required something at least as substantial as a 2x4. So it will be with the horizontal members as well.

Analyzing a beam for shear is usually simple and something that could probably be taught to most high school seniors. For something called a "statically determinate" beam, bending moment isn't much harder, and anyone familiar with algebra could do that. But deflection and analysis of "statically indeterminate" beams requires calculus and is typically taught in about the second year of engineering school. I'm not interested in solving a new calculus problem for every example I want to show you, and unless you want to learn a bunch of math, you're probably not interested in seeing that. Fortunately, we have the magic of the internet to do the heavy (mental) lifting for us.

I'll be using this calculator here (https://clearcalcs.com/freetools/beam-analysis/us) to do the calculations. Go ahead and take a quick look at it, you might find it more convenient to put it in a new tab (right click and select new tab in most browsers).

When you first go to that page, it has a default example. That's a 10' steel beam (I can tell by the Young's Modulus, 29,000,000) sitting on two supports, carrying an evenly distributed load that starts at 100 lbs/foot at the left end and ramps up to 1000 lbs/ft at the right end. I can also tell by looking up in a table that it's a steel I-beam that is 14" tall and weighs 68 lb/ft (680 lbs overall)...far stouter, stronger, and more expensive than anyone in their right mind would use to build a pool table out of, but just the ticket for building a floor in an office building.

As for how it behaves, I can also tell from looking at the right hand side of the page that it carries the most shear load right at the right support, the most bending moment about 5.6' from the left support, and sags 0.00592", just shy of six thousandths, at a point 5.12' from the left support.

All told, a very useful calculator. Now, let me conclude today with a more practical example that illustrates something about what we call "end conditions", or how a beam is attached to something.

First, let me change the beam to a 10x10 carrying an even 1000 lbs/ft, let's say 5 tons of sandbags evenly distributed along its length. Young's modulus will become 1,300,000, I'm going to ignore Area because it doesn't affect the result, "Second moment of inertia" becomes 610. The top part of the left hand side should look like this:

And the bottom left side should look like this:

On the right hand side, we're interested in this graph:

If you mouse over that graph (on the beam calculator page, not on this post), you'll see that maximum deflection is 0.284" at the midpoint of the span, 5' from the left support. So, 5 tons of sandbags would make even a mighty 10x10 sag over 1/4".

An engineer would call this a "simply supported beam", as if it was just lying on two (very strong) sawhorses, spaced 10' apart. Let's look at what happens if I remove one of the sawhorses, and clamp one end of the beam so that it doesn't just fall over.

Change the left support to "fixed", and delete the right support (just click on the trash can symbol to delete it).

Now, look at the deflection curve.

If you mouse over this graph on the beam calculator page, you'll see that the maximum sag is 2.72"...nearly 10x as much!

In fact, if you shorten the beam to 5':

And redistribute the same 5 ton load over 5' instead of 10' (2000 pounds per foot instead of 1000 pounds per foot):

You'll see that the deflection is still worse at 5' than at the midpoint of the simply supported span, 0.34" vs 0.284".

And that's today's installment. When we dangle something off the edge like that, we call it a cantilever. Cantilevers are all around you as well, from those nifty tables that wheel in over your hospital bed when you're not feeling so well, to this particular house:

Now, while it is certainly possible to design a pool table that is completely cantilevered from one end (sounds like a fashion statement!) that plays rock solid, I hope you can appreciate how it is much easier and cheaper to do so with one that is simply supported...you know, with legs at each end.

In the next installment, I hope to show how this might be made even easier with a combination of cantilevers and simple spans.

 

[garczar]

And that brings me to today's installment...beams. Beams are very important. If you really understand how beams work, then you can understand pool table frames a lot better.

A beam is any long member that carries a load over an empty space beneath it. In our case, it will be carrying the weight of the slate, the rails, anything that sits on top of it. But beams are everywhere around us, from flooring joists in our houses, to the floor pan of our automobiles, to the bookshelf in our study. I can't give you a comprehensive method to analyze any beam, that would take a college course and probably a lot more math than most of you want to be involved with. But I can give you some tools and ideas that will help you think about this more clearly, and maybe help you decide between different options.

Most beam analysis looks at three properties:
1. How much load can the beam carry in shear?
2. What bending moment can the beam safely carry?
3. How much will the beam deflect under load?

For the most part, I'm going to skip over the first two items, shear and bending moment. That's because I doubt that they'll apply much to pool tables, unless you get a wild idea to build one out of 1/4" by 1/4" furring strips. They are of tremendous importance to an engineer designing a skyscraper or a ship, for example. If a beam doesn't satisfy the first two criteria, it will simply collapse. But how many times have you heard of a pool table breaking in half and dumping the bed slates on the floor, vs. one that simply sagged so much that the balls rolled crooked? As we saw from the leg example, it only took a slim 3/4" stick to (barely) carry the load without breaking but ensuring that deflection was kept within bound required something at least as substantial as a 2x4. So it will be with the horizontal members as well.

Analyzing a beam for shear is usually simple and something that could probably be taught to most high school seniors. For something called a "statically determinate" beam, bending moment isn't much harder, and anyone familiar with algebra could do that. But deflection and analysis of "statically indeterminate" beams requires calculus and is typically taught in about the second year of engineering school. I'm not interested in solving a new calculus problem for every example I want to show you, and unless you want to learn a bunch of math, you're probably not interested in seeing that. Fortunately, we have the magic of the internet to do the heavy (mental) lifting for us.

I'll be using this calculator here (https://clearcalcs.com/freetools/beam-analysis/us) to do the calculations. Go ahead and take a quick look at it, you might find it more convenient to put it in a new tab (right click and select new tab in most browsers).

When you first go to that page, it has a default example. That's a 10' steel beam (I can tell by the Young's Modulus, 29,000,000) sitting on two supports, carrying an evenly distributed load that starts at 100 lbs/foot at the left end and ramps up to 1000 lbs/ft at the right end. I can also tell by looking up in a table that it's a steel I-beam that is 14" tall and weighs 68 lb/ft (680 lbs overall)...far stouter, stronger, and more expensive than anyone in their right mind would use to build a pool table out of, but just the ticket for building a floor in an office building.

As for how it behaves, I can also tell from looking at the right hand side of the page that it carries the most shear load right at the right support, the most bending moment about 5.6' from the left support, and sags 0.00592", just shy of six thousandths, at a point 5.12' from the left support.

All told, a very useful calculator. Now, let me conclude today with a more practical example that illustrates something about what we call "end conditions", or how a beam is attached to something.

First, let me change the beam to a 10x10 carrying an even 1000 lbs/ft, let's say 5 tons of sandbags evenly distributed along its length. Young's modulus will become 1,300,000, I'm going to ignore Area because it doesn't affect the result, "Second moment of inertia" becomes 610. The top part of the left hand side should look like this:

View attachment 632506

And the bottom left side should look like this:

View attachment 632507

On the right hand side, we're interested in this graph:

View attachment 632508

If you mouse over that graph (on the beam calculator page, not on this post), you'll see that maximum deflection is 0.284" at the midpoint of the span, 5' from the left support. So, 5 tons of sandbags would make even a mighty 10x10 sag over 1/4".

An engineer would call this a "simply supported beam", as if it was just lying on two (very strong) sawhorses, spaced 10' apart. Let's look at what happens if I remove one of the sawhorses, and clamp one end of the beam so that it doesn't just fall over.

Change the left support to "fixed", and delete the right support (just click on the trash can symbol to delete it).

View attachment 632509

Now, look at the deflection curve.

View attachment 632510

If you mouse over this graph on the beam calculator page, you'll see that the maximum sag is 2.72"...nearly 10x as much!

In fact, if you shorten the beam to 5':

View attachment 632511
And redistribute the same 5 ton load over 5' instead of 10' (2000 pounds per foot instead of 1000 pounds per foot):

View attachment 632512

You'll see that the deflection is still worse at 5' than at the midpoint of the simply supported span, 0.34" vs 0.284".

View attachment 632513

And that's today's installment. When we dangle something off the edge like that, we call it a cantilever. Cantilevers are all around you as well, from those nifty tables that wheel in over your hospital bed when you're not feeling so well, to this particular house:

View attachment 632515

Now, while it is certainly possible to design a pool table that is completely cantilevered from one end (sounds like a fashion statement!) that plays rock solid, I hope you can appreciate how it is much easier and cheaper to do so with one that is simply supported...you know, with legs at each end.

In the next installment, I hope to show how this might be made even easier with a combination of cantilevers and simple spans.

Please no. Put it all in a book and sell it. Please.

 

[rexus31]

Please no. Put it all in a book and sell it. Please.

Nominee for longest post in AZB history.

 

[garczar]

Nominee for longest post in AZB history.

Poster-boy for my recent poll.

 

[realkingcobra]

And that brings me to today's installment...beams. Beams are very important. If you really understand how beams work, then you can understand pool table frames a lot better.

A beam is any long member that carries a load over an empty space beneath it. In our case, it will be carrying the weight of the slate, the rails, anything that sits on top of it. But beams are everywhere around us, from flooring joists in our houses, to the floor pan of our automobiles, to the bookshelf in our study. I can't give you a comprehensive method to analyze any beam, that would take a college course and probably a lot more math than most of you want to be involved with. But I can give you some tools and ideas that will help you think about this more clearly, and maybe help you decide between different options.

Most beam analysis looks at three properties:
1. How much load can the beam carry in shear?
2. What bending moment can the beam safely carry?
3. How much will the beam deflect under load?

For the most part, I'm going to skip over the first two items, shear and bending moment. That's because I doubt that they'll apply much to pool tables, unless you get a wild idea to build one out of 1/4" by 1/4" furring strips. They are of tremendous importance to an engineer designing a skyscraper or a ship, for example. If a beam doesn't satisfy the first two criteria, it will simply collapse. But how many times have you heard of a pool table breaking in half and dumping the bed slates on the floor, vs. one that simply sagged so much that the balls rolled crooked? As we saw from the leg example, it only took a slim 3/4" stick to (barely) carry the load without breaking but ensuring that deflection was kept within bound required something at least as substantial as a 2x4. So it will be with the horizontal members as well.

Analyzing a beam for shear is usually simple and something that could probably be taught to most high school seniors. For something called a "statically determinate" beam, bending moment isn't much harder, and anyone familiar with algebra could do that. But deflection and analysis of "statically indeterminate" beams requires calculus and is typically taught in about the second year of engineering school. I'm not interested in solving a new calculus problem for every example I want to show you, and unless you want to learn a bunch of math, you're probably not interested in seeing that. Fortunately, we have the magic of the internet to do the heavy (mental) lifting for us.

I'll be using this calculator here (https://clearcalcs.com/freetools/beam-analysis/us) to do the calculations. Go ahead and take a quick look at it, you might find it more convenient to put it in a new tab (right click and select new tab in most browsers).

When you first go to that page, it has a default example. That's a 10' steel beam (I can tell by the Young's Modulus, 29,000,000) sitting on two supports, carrying an evenly distributed load that starts at 100 lbs/foot at the left end and ramps up to 1000 lbs/ft at the right end. I can also tell by looking up in a table that it's a steel I-beam that is 14" tall and weighs 68 lb/ft (680 lbs overall)...far stouter, stronger, and more expensive than anyone in their right mind would use to build a pool table out of, but just the ticket for building a floor in an office building.

As for how it behaves, I can also tell from looking at the right hand side of the page that it carries the most shear load right at the right support, the most bending moment about 5.6' from the left support, and sags 0.00592", just shy of six thousandths, at a point 5.12' from the left support.

All told, a very useful calculator. Now, let me conclude today with a more practical example that illustrates something about what we call "end conditions", or how a beam is attached to something.

First, let me change the beam to a 10x10 carrying an even 1000 lbs/ft, let's say 5 tons of sandbags evenly distributed along its length. Young's modulus will become 1,300,000, I'm going to ignore Area because it doesn't affect the result, "Second moment of inertia" becomes 610. The top part of the left hand side should look like this:

View attachment 632506

And the bottom left side should look like this:

View attachment 632507

On the right hand side, we're interested in this graph:

View attachment 632508

If you mouse over that graph (on the beam calculator page, not on this post), you'll see that maximum deflection is 0.284" at the midpoint of the span, 5' from the left support. So, 5 tons of sandbags would make even a mighty 10x10 sag over 1/4".

An engineer would call this a "simply supported beam", as if it was just lying on two (very strong) sawhorses, spaced 10' apart. Let's look at what happens if I remove one of the sawhorses, and clamp one end of the beam so that it doesn't just fall over.

Change the left support to "fixed", and delete the right support (just click on the trash can symbol to delete it).

View attachment 632509

Now, look at the deflection curve.

View attachment 632510

If you mouse over this graph on the beam calculator page, you'll see that the maximum sag is 2.72"...nearly 10x as much!

In fact, if you shorten the beam to 5':

View attachment 632511
And redistribute the same 5 ton load over 5' instead of 10' (2000 pounds per foot instead of 1000 pounds per foot):

View attachment 632512

You'll see that the deflection is still worse at 5' than at the midpoint of the simply supported span, 0.34" vs 0.284".

View attachment 632513

And that's today's installment. When we dangle something off the edge like that, we call it a cantilever. Cantilevers are all around you as well, from those nifty tables that wheel in over your hospital bed when you're not feeling so well, to this particular house:

View attachment 632515

Now, while it is certainly possible to design a pool table that is completely cantilevered from one end (sounds like a fashion statement!) that plays rock solid, I hope you can appreciate how it is much easier and cheaper to do so with one that is simply supported...you know, with legs at each end.

In the next installment, I hope to show how this might be made even easier with a combination of cantilevers and simple spans.

And again all your words add up to a bowl of word salad! What about the floor not being level for you to set your perfect pool table legs on? How about the fact that most pool tables are built with side panels and NO beams for support?

 

[realkingcobra]

And that brings me to today's installment...beams. Beams are very important. If you really understand how beams work, then you can understand pool table frames a lot better.

A beam is any long member that carries a load over an empty space beneath it. In our case, it will be carrying the weight of the slate, the rails, anything that sits on top of it. But beams are everywhere around us, from flooring joists in our houses, to the floor pan of our automobiles, to the bookshelf in our study. I can't give you a comprehensive method to analyze any beam, that would take a college course and probably a lot more math than most of you want to be involved with. But I can give you some tools and ideas that will help you think about this more clearly, and maybe help you decide between different options.

Most beam analysis looks at three properties:
1. How much load can the beam carry in shear?
2. What bending moment can the beam safely carry?
3. How much will the beam deflect under load?

For the most part, I'm going to skip over the first two items, shear and bending moment. That's because I doubt that they'll apply much to pool tables, unless you get a wild idea to build one out of 1/4" by 1/4" furring strips. They are of tremendous importance to an engineer designing a skyscraper or a ship, for example. If a beam doesn't satisfy the first two criteria, it will simply collapse. But how many times have you heard of a pool table breaking in half and dumping the bed slates on the floor, vs. one that simply sagged so much that the balls rolled crooked? As we saw from the leg example, it only took a slim 3/4" stick to (barely) carry the load without breaking but ensuring that deflection was kept within bound required something at least as substantial as a 2x4. So it will be with the horizontal members as well.

Analyzing a beam for shear is usually simple and something that could probably be taught to most high school seniors. For something called a "statically determinate" beam, bending moment isn't much harder, and anyone familiar with algebra could do that. But deflection and analysis of "statically indeterminate" beams requires calculus and is typically taught in about the second year of engineering school. I'm not interested in solving a new calculus problem for every example I want to show you, and unless you want to learn a bunch of math, you're probably not interested in seeing that. Fortunately, we have the magic of the internet to do the heavy (mental) lifting for us.

I'll be using this calculator here (https://clearcalcs.com/freetools/beam-analysis/us) to do the calculations. Go ahead and take a quick look at it, you might find it more convenient to put it in a new tab (right click and select new tab in most browsers).

When you first go to that page, it has a default example. That's a 10' steel beam (I can tell by the Young's Modulus, 29,000,000) sitting on two supports, carrying an evenly distributed load that starts at 100 lbs/foot at the left end and ramps up to 1000 lbs/ft at the right end. I can also tell by looking up in a table that it's a steel I-beam that is 14" tall and weighs 68 lb/ft (680 lbs overall)...far stouter, stronger, and more expensive than anyone in their right mind would use to build a pool table out of, but just the ticket for building a floor in an office building.

As for how it behaves, I can also tell from looking at the right hand side of the page that it carries the most shear load right at the right support, the most bending moment about 5.6' from the left support, and sags 0.00592", just shy of six thousandths, at a point 5.12' from the left support.

All told, a very useful calculator. Now, let me conclude today with a more practical example that illustrates something about what we call "end conditions", or how a beam is attached to something.

First, let me change the beam to a 10x10 carrying an even 1000 lbs/ft, let's say 5 tons of sandbags evenly distributed along its length. Young's modulus will become 1,300,000, I'm going to ignore Area because it doesn't affect the result, "Second moment of inertia" becomes 610. The top part of the left hand side should look like this:

View attachment 632506

And the bottom left side should look like this:

View attachment 632507

On the right hand side, we're interested in this graph:

View attachment 632508

If you mouse over that graph (on the beam calculator page, not on this post), you'll see that maximum deflection is 0.284" at the midpoint of the span, 5' from the left support. So, 5 tons of sandbags would make even a mighty 10x10 sag over 1/4".

An engineer would call this a "simply supported beam", as if it was just lying on two (very strong) sawhorses, spaced 10' apart. Let's look at what happens if I remove one of the sawhorses, and clamp one end of the beam so that it doesn't just fall over.

Change the left support to "fixed", and delete the right support (just click on the trash can symbol to delete it).

View attachment 632509

Now, look at the deflection curve.

View attachment 632510

If you mouse over this graph on the beam calculator page, you'll see that the maximum sag is 2.72"...nearly 10x as much!

In fact, if you shorten the beam to 5':

View attachment 632511
And redistribute the same 5 ton load over 5' instead of 10' (2000 pounds per foot instead of 1000 pounds per foot):

View attachment 632512

You'll see that the deflection is still worse at 5' than at the midpoint of the simply supported span, 0.34" vs 0.284".

View attachment 632513

And that's today's installment. When we dangle something off the edge like that, we call it a cantilever. Cantilevers are all around you as well, from those nifty tables that wheel in over your hospital bed when you're not feeling so well, to this particular house:

View attachment 632515

Now, while it is certainly possible to design a pool table that is completely cantilevered from one end (sounds like a fashion statement!) that plays rock solid, I hope you can appreciate how it is much easier and cheaper to do so with one that is simply supported...you know, with legs at each end.

In the next installment, I hope to show how this might be made even easier with a combination of cantilevers and simple spans.

The problem with you, is you can draw up a pool table all day long, but when it comes to actually building it, you'd never get finished!

And for what its worth, I HAVE built over 200 pool tables, so I kind of know what I'm talking about!

 

[MamboFats]

The problem with you, is you can draw up a pool table all day long, but when it comes to actually building it, you'd never get finished!

And for what its worth, I HAVE built over 200 pool tables, so I kind of know what I'm talking about!

I appreciate El Jefe's contribution very much, since it makes a lot of things more clear on the theoretical side of my project?

I encourage everyone, and especially you as a very experienced mechanic, to educate me more: on all subjects and views. Theoretical, practical, financial, build-wise, and any other matter that help me obtain my goal.

 

[realkingcobra]

I appreciate El Jefe's contribution very much, since it makes a lot of things more clear on the theoretical side of my project?

I encourage everyone, and especially you as a very experienced mechanic, to educate me more: on all subjects and views. Theoretical, practical, financial, build-wise, and any other matter that help me obtain my goal.

If it was that simple, I wouldn't have to correct manufacturers right?

 

[realkingcobra]

Even the cushion manufacturers are no help in designing the rails to play at the best they can using their own cushions in doing so, all they can do is take how thick your rails are and put them on a cad program and tell you what the bevel is you need to get the nose height to what ever it needs to be, in their opinion, instead of telling you our cushions play at their best if you make your rails like xyz!!!

 

[realkingcobra]

I appreciate El Jefe's contribution very much, since it makes a lot of things more clear on the theoretical side of my project?

I encourage everyone, and especially you as a very experienced mechanic, to educate me more: on all subjects and views. Theoretical, practical, financial, build-wise, and any other matter that help me obtain my goal.

Ask anyone what the nose height is suppose to be and they'll tell you its 63 1/2% +/- 1% of the ball height, so 1 25/64ths to 1 29/64ths of an inch, or for simple purposes, 1 7/16", but they fail to understand that rule only applys if the rails are 1 11/16" thick, not 1 5/8" to 1 1/2" thick, those don't play right with a 1 7/16" nose height!

 

[realkingcobra]

I appreciate El Jefe's contribution very much, since it makes a lot of things more clear on the theoretical side of my project?

I encourage everyone, and especially you as a very experienced mechanic, to educate me more: on all subjects and views. Theoretical, practical, financial, build-wise, and any other matter that help me obtain my goal.

No one on this forum has the experience of building pool tables to tell you what to do to design and build your own table, nor would they take the time to walk you through all the steps to do so. Its your labor of love, not anyone else's. As far as that goes, I don't even know what tools you have to work with either, as far as I know you might only have a drill, hammer, and a skillsaw.

 

[trentfromtoledo]

And that brings me to today's installment...beams. Beams are very important. If you really understand how beams work, then you can understand pool table frames a lot better.

A beam is any long member that carries a load over an empty space beneath it. In our case, it will be carrying the weight of the slate, the rails, anything that sits on top of it. But beams are everywhere around us, from flooring joists in our houses, to the floor pan of our automobiles, to the bookshelf in our study. I can't give you a comprehensive method to analyze any beam, that would take a college course and probably a lot more math than most of you want to be involved with. But I can give you some tools and ideas that will help you think about this more clearly, and maybe help you decide between different options.

Most beam analysis looks at three properties:
1. How much load can the beam carry in shear?
2. What bending moment can the beam safely carry?
3. How much will the beam deflect under load?

For the most part, I'm going to skip over the first two items, shear and bending moment. That's because I doubt that they'll apply much to pool tables, unless you get a wild idea to build one out of 1/4" by 1/4" furring strips. They are of tremendous importance to an engineer designing a skyscraper or a ship, for example. If a beam doesn't satisfy the first two criteria, it will simply collapse. But how many times have you heard of a pool table breaking in half and dumping the bed slates on the floor, vs. one that simply sagged so much that the balls rolled crooked? As we saw from the leg example, it only took a slim 3/4" stick to (barely) carry the load without breaking but ensuring that deflection was kept within bound required something at least as substantial as a 2x4. So it will be with the horizontal members as well.

Analyzing a beam for shear is usually simple and something that could probably be taught to most high school seniors. For something called a "statically determinate" beam, bending moment isn't much harder, and anyone familiar with algebra could do that. But deflection and analysis of "statically indeterminate" beams requires calculus and is typically taught in about the second year of engineering school. I'm not interested in solving a new calculus problem for every example I want to show you, and unless you want to learn a bunch of math, you're probably not interested in seeing that. Fortunately, we have the magic of the internet to do the heavy (mental) lifting for us.

I'll be using this calculator here (https://clearcalcs.com/freetools/beam-analysis/us) to do the calculations. Go ahead and take a quick look at it, you might find it more convenient to put it in a new tab (right click and select new tab in most browsers).

When you first go to that page, it has a default example. That's a 10' steel beam (I can tell by the Young's Modulus, 29,000,000) sitting on two supports, carrying an evenly distributed load that starts at 100 lbs/foot at the left end and ramps up to 1000 lbs/ft at the right end. I can also tell by looking up in a table that it's a steel I-beam that is 14" tall and weighs 68 lb/ft (680 lbs overall)...far stouter, stronger, and more expensive than anyone in their right mind would use to build a pool table out of, but just the ticket for building a floor in an office building.

As for how it behaves, I can also tell from looking at the right hand side of the page that it carries the most shear load right at the right support, the most bending moment about 5.6' from the left support, and sags 0.00592", just shy of six thousandths, at a point 5.12' from the left support.

All told, a very useful calculator. Now, let me conclude today with a more practical example that illustrates something about what we call "end conditions", or how a beam is attached to something.

First, let me change the beam to a 10x10 carrying an even 1000 lbs/ft, let's say 5 tons of sandbags evenly distributed along its length. Young's modulus will become 1,300,000, I'm going to ignore Area because it doesn't affect the result, "Second moment of inertia" becomes 610. The top part of the left hand side should look like this:

View attachment 632506

And the bottom left side should look like this:

View attachment 632507

On the right hand side, we're interested in this graph:

View attachment 632508

If you mouse over that graph (on the beam calculator page, not on this post), you'll see that maximum deflection is 0.284" at the midpoint of the span, 5' from the left support. So, 5 tons of sandbags would make even a mighty 10x10 sag over 1/4".

An engineer would call this a "simply supported beam", as if it was just lying on two (very strong) sawhorses, spaced 10' apart. Let's look at what happens if I remove one of the sawhorses, and clamp one end of the beam so that it doesn't just fall over.

Change the left support to "fixed", and delete the right support (just click on the trash can symbol to delete it).

View attachment 632509

Now, look at the deflection curve.

View attachment 632510

If you mouse over this graph on the beam calculator page, you'll see that the maximum sag is 2.72"...nearly 10x as much!

In fact, if you shorten the beam to 5':

View attachment 632511
And redistribute the same 5 ton load over 5' instead of 10' (2000 pounds per foot instead of 1000 pounds per foot):

View attachment 632512

You'll see that the deflection is still worse at 5' than at the midpoint of the simply supported span, 0.34" vs 0.284".

View attachment 632513

And that's today's installment. When we dangle something off the edge like that, we call it a cantilever. Cantilevers are all around you as well, from those nifty tables that wheel in over your hospital bed when you're not feeling so well, to this particular house:

View attachment 632515

Now, while it is certainly possible to design a pool table that is completely cantilevered from one end (sounds like a fashion statement!) that plays rock solid, I hope you can appreciate how it is much easier and cheaper to do so with one that is simply supported...you know, with legs at each end.

In the next installment, I hope to show how this might be made even easier with a combination of cantilevers and simple spans.

You really went all out on this! I bet it sure rattles some peoples cages!
Great Post!

TFT