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Vic Dura wrote:

I think what he demonstrated must have been different than what you recall or mine is all awash...see my note to Edwin.

As a practical thought experiment, if it were as you say, you could continue to reduce the wall thickness of the tube until it was paper thin and the result wouldn't be different--that obviously wouldn't be true in reality.

Vic Dura wrote:

Sounds like an urban legend to me. Kinda like hot water freezes faster than cold water.

The wikipedia says stiffness is a measure of resistance to deformation. Solid rod resists deformation more than a hollow rod. It makes no sense that boring out a steel rod would make it stronger. How thin should I make it? The thinner the stronger? And the opposite is true too? The more I fill it, the less stiff it becomes?

I didn't write the laws of physics, I just use them like everyone else. Why are airplane wings not solid?

Well that explains a lot.

And that even more.

Say E = 1.1 million psi for Eastern hemlock...

I = bd^3/12 in^4, for a b" wide x d" deep beam. I = 2x6^3/12 = 36 in^4 for a rough-sawn (real) 2x6.

....the total load. Say W = 400 lb.

....in inches. So a 10' rough-sawn 2x6 beam would have a D = 5W(10x12)^3/(384EI) = 0.23" max deflection.

You may be confusing something like the polar moment of inertia (including mass, for dynamics) with the geometric moment of inertia about the neutral axis, eg the horizontal diameter x-axis. That's the sum of the products of each tiny area and the square of the perpendicular distance from that area to the axis. Ix = Pir^4/4 for a disk of radius r, eg Pi2^4/4 = 12.57 in^4 for a 2" radius rod, IMO.

....with the same axis, consider the rod to be a composite area and subtract I = Pi1^4/4 = 0.79 in^4 for the 1" radius bore from 12.57 to get 11.78 in^4 for a 2" radius rod with a 1" radius bore.

So a 10'x2" radius Eastern hemlock rod with a 400 pound total load would have D = 5x400(10x12)^3/(384x1.1x10^6x12.57) = 0.650" max, and the hollow version would have D = 5x400(10x12)^3/(384x1.1x10^6x11.78) = 0.695" max.

If the hemlock weighs 30 lb/ft^3 and the solid rod weighs 26.2 pounds, a 26.2 pound 4" radius rod with a 3.46" radius bore with I = 201.1-113.1 = 88 in^4 would have D = 5x400(10x12)^3/(384x1.1x10^6x88) = 0.093" max.

And a 26.2 pound 4"x8" hemlock "I beam" with 2 4"x1.6" boards bolted onto a 4"x4.9" foamboard sandwich and I = 4x8^3/12-4x4.9^3/12 = 133 in^4 might have D = 5x400(10x12)^3/(384x1.1x10^6x133) = 0.062" max, if nothing slips.

Nick

That's just not true. A (larger diameter) tube made from the same amout of steel as a solid rod would have more resistance to bending force, but a solid tube of the same diameter as a hollow tube will be stronger. If you don't believe that, try bending a lenght of 1/2" EMT over your knee, then try with a solid 1/2" steel bar.

at

special

similiar old building or a speciality lumberyard. does it have to be a 4x12? just nail two 2x12's together with some 1/2 ply sandwiched in the middle.

Maybe at a sawmill. Or a place that supplies pole barns and the like. Look into making your own composite beam to achieve what you need.

#### Site Timeline

- posted on August 18, 2005, 1:21 am

I think what he demonstrated must have been different than what you recall or mine is all awash...see my note to Edwin.

As a practical thought experiment, if it were as you say, you could continue to reduce the wall thickness of the tube until it was paper thin and the result wouldn't be different--that obviously wouldn't be true in reality.

- posted on August 18, 2005, 2:16 am

Sounds like an urban legend to me. Kinda like hot water freezes faster than cold water.

The wikipedia says stiffness is a measure of resistance to deformation. Solid rod resists deformation more than a hollow rod. It makes no sense that boring out a steel rod would make it stronger. How thin should I make it? The thinner the stronger? And the opposite is true too? The more I fill it, the less stiff it becomes?

--

Respectfully,

CL Gilbert

Respectfully,

CL Gilbert

Click to see the full signature.

- posted on August 18, 2005, 3:14 am

On Wed, 17 Aug 2005 22:16:21 -0400, "CL (dnoyeB) Gilbert"

Not a legend. Hot water DOES freeze faster. It's because it dont have as much oxygen. Dont ask me why it has less oxygen, but it does. and thus it freezes faster.

Not a legend. Hot water DOES freeze faster. It's because it dont have as much oxygen. Dont ask me why it has less oxygen, but it does. and thus it freezes faster.

- posted on August 18, 2005, 11:12 am

snipped-for-privacy@UNLISTED.com wrote:

/me shakes head...

/me shakes head...

--

Respectfully,

CL Gilbert

Respectfully,

CL Gilbert

Click to see the full signature.

- posted on August 18, 2005, 1:31 pm

"CL (dnoyeB) Gilbert" wrote:

Well, it's true that water w/ less entrained air will freeze faster than that w/ more entrained air--and, heating water will drive out some of the entrained air so there's a kernel of truth in the saw...

Well, it's true that water w/ less entrained air will freeze faster than that w/ more entrained air--and, heating water will drive out some of the entrained air so there's a kernel of truth in the saw...

- posted on August 18, 2005, 1:35 pm

"CL (dnoyeB) Gilbert" wrote:

Might look at...

http://math.ucr.edu/home/baez/physics/General/hot_water.html

Might look at...

http://math.ucr.edu/home/baez/physics/General/hot_water.html

- posted on August 17, 2005, 11:38 pm

I didn't write the laws of physics, I just use them like everyone else. Why are airplane wings not solid?

- posted on August 18, 2005, 1:19 am

Edwin Pawlowski wrote:

Weight, mostly...

I'm not sure what laws you're thinking of...let's see--if we consider a simple beam w/ uniform load w/ simple support at both ends the maximum deflection at the center is 5/384 (W*l^3)/(EI) where

E = modulus of elasticity (material property only) I = moment of inertia (dependent on geometry) W = applied load l = length

Now for a rod Irod = MR^2

where M = mass of beam and R = radius

and for a hollow tube it is Itube = M(R1^2 + R2^2) where

R1,R2 are inner/outer radii, respectively.

This superficially makes it look like Itube>Irod for R2 = R, but that doesn't include the mass which will be less for a hollow tube than for a solid rod.

Since we're after comparing two geometries of the same material, we can consider the density of the two to be the same as well as the length. On that basis, for the rod the weight/unit length is

mRrod = density***pi***R^2/4

and similarly,

mTube = density***pi***(R2^2-R1^2)

Substituting into the formulae for I the geometrical terms for each M we get that for each the moment of inertia is proportional to

iRod ~ R^4

and

iTube ~ (R2^2 - R1^2)*(R1^2 + R2^2) = R2^4 - R1^4

Thus, it can be seen that the moment of inertia for the tube section is always slightly smaller than that of the solid rod and since I is in the denominator of the deflection, the larger deflection will occur for the tube, not the rod for R2==R (the outer diameters equal).

If you figure on an equivalent weight basis, the tube will be stronger as the same amount material will be located at a farther distance from the neutral axis.

Weight, mostly...

I'm not sure what laws you're thinking of...let's see--if we consider a simple beam w/ uniform load w/ simple support at both ends the maximum deflection at the center is 5/384 (W*l^3)/(EI) where

E = modulus of elasticity (material property only) I = moment of inertia (dependent on geometry) W = applied load l = length

Now for a rod Irod = MR^2

where M = mass of beam and R = radius

and for a hollow tube it is Itube = M(R1^2 + R2^2) where

R1,R2 are inner/outer radii, respectively.

This superficially makes it look like Itube>Irod for R2 = R, but that doesn't include the mass which will be less for a hollow tube than for a solid rod.

Since we're after comparing two geometries of the same material, we can consider the density of the two to be the same as well as the length. On that basis, for the rod the weight/unit length is

mRrod = density

and similarly,

mTube = density

Substituting into the formulae for I the geometrical terms for each M we get that for each the moment of inertia is proportional to

iRod ~ R^4

and

iTube ~ (R2^2 - R1^2)*(R1^2 + R2^2) = R2^4 - R1^4

Thus, it can be seen that the moment of inertia for the tube section is always slightly smaller than that of the solid rod and since I is in the denominator of the deflection, the larger deflection will occur for the tube, not the rod for R2==R (the outer diameters equal).

If you figure on an equivalent weight basis, the tube will be stronger as the same amount material will be located at a farther distance from the neutral axis.

- posted on August 18, 2005, 7:37 am

Well that explains a lot.

And that even more.

- posted on August 18, 2005, 12:45 pm

Say E = 1.1 million psi for Eastern hemlock...

I = bd^3/12 in^4, for a b" wide x d" deep beam. I = 2x6^3/12 = 36 in^4 for a rough-sawn (real) 2x6.

....the total load. Say W = 400 lb.

....in inches. So a 10' rough-sawn 2x6 beam would have a D = 5W(10x12)^3/(384EI) = 0.23" max deflection.

You may be confusing something like the polar moment of inertia (including mass, for dynamics) with the geometric moment of inertia about the neutral axis, eg the horizontal diameter x-axis. That's the sum of the products of each tiny area and the square of the perpendicular distance from that area to the axis. Ix = Pir^4/4 for a disk of radius r, eg Pi2^4/4 = 12.57 in^4 for a 2" radius rod, IMO.

....with the same axis, consider the rod to be a composite area and subtract I = Pi1^4/4 = 0.79 in^4 for the 1" radius bore from 12.57 to get 11.78 in^4 for a 2" radius rod with a 1" radius bore.

So a 10'x2" radius Eastern hemlock rod with a 400 pound total load would have D = 5x400(10x12)^3/(384x1.1x10^6x12.57) = 0.650" max, and the hollow version would have D = 5x400(10x12)^3/(384x1.1x10^6x11.78) = 0.695" max.

If the hemlock weighs 30 lb/ft^3 and the solid rod weighs 26.2 pounds, a 26.2 pound 4" radius rod with a 3.46" radius bore with I = 201.1-113.1 = 88 in^4 would have D = 5x400(10x12)^3/(384x1.1x10^6x88) = 0.093" max.

And a 26.2 pound 4"x8" hemlock "I beam" with 2 4"x1.6" boards bolted onto a 4"x4.9" foamboard sandwich and I = 4x8^3/12-4x4.9^3/12 = 133 in^4 might have D = 5x400(10x12)^3/(384x1.1x10^6x133) = 0.062" max, if nothing slips.

Nick

- posted on August 18, 2005, 1:57 am

Edwin Pawlowski wrote:

Why are bird bones hollow?

Why are bird bones hollow?

--

Respectfully,

CL Gilbert

Respectfully,

CL Gilbert

Click to see the full signature.

- posted on August 18, 2005, 3:10 am

wrote:

I'm not going to comment about the pipe/rod issue because I was not familiar with that. However, as far as putting two (or more) timbers together, a good part of the reason is the same as plywood. A natural board has all the grain running a certain way, and that allows for weak spots, such as when you buy a board with a crack in it. (a weak spot). So, with two boards, it's highly unlikely that you will end up with the weak spot in the same place as the other board. Thus one compensates for the weak spot on the other. Then adding the plywood down the center, which is mainly just a spacer, that too adds quite a bit. If the boards were glued together they would be stronger still.

Look at the beams they make now. They are simply plywood beams made to be a 2X12 or whatever, and are 2 or 4 inches thick. They are supposed to be considerably stronger, yet each layer is less than 1/4 inch thick. Alone, the pieces would not hold up anything, but together, they are very strong. I used to think that was bogus till I read up on it. As for the particle board type beams, I still find them inferior, regardless what is said about them. Many will disagree with me, but I wont use them.

One final note. I spoke with someone that was building a huge barn at a county fairgrounds. Instead of using solid posts (uprights), like in the old days where they used a "power pole". Instead they used three 2X8's nailed together. Treated lumber below ground, non treated above, and they consist of pieces with the joints at different spots. The builder said they are stronger than single posts, and because of the height of the building, it's next to impossible to get solid poles that long.

Mark

I'm not going to comment about the pipe/rod issue because I was not familiar with that. However, as far as putting two (or more) timbers together, a good part of the reason is the same as plywood. A natural board has all the grain running a certain way, and that allows for weak spots, such as when you buy a board with a crack in it. (a weak spot). So, with two boards, it's highly unlikely that you will end up with the weak spot in the same place as the other board. Thus one compensates for the weak spot on the other. Then adding the plywood down the center, which is mainly just a spacer, that too adds quite a bit. If the boards were glued together they would be stronger still.

Look at the beams they make now. They are simply plywood beams made to be a 2X12 or whatever, and are 2 or 4 inches thick. They are supposed to be considerably stronger, yet each layer is less than 1/4 inch thick. Alone, the pieces would not hold up anything, but together, they are very strong. I used to think that was bogus till I read up on it. As for the particle board type beams, I still find them inferior, regardless what is said about them. Many will disagree with me, but I wont use them.

One final note. I spoke with someone that was building a huge barn at a county fairgrounds. Instead of using solid posts (uprights), like in the old days where they used a "power pole". Instead they used three 2X8's nailed together. Treated lumber below ground, non treated above, and they consist of pieces with the joints at different spots. The builder said they are stronger than single posts, and because of the height of the building, it's next to impossible to get solid poles that long.

Mark

- posted on August 18, 2005, 5:53 pm

That's just not true. A (larger diameter) tube made from the same amout of steel as a solid rod would have more resistance to bending force, but a solid tube of the same diameter as a hollow tube will be stronger. If you don't believe that, try bending a lenght of 1/2" EMT over your knee, then try with a solid 1/2" steel bar.

--

Larry Wasserman Baltimore, Maryland

Larry Wasserman Baltimore, Maryland

Click to see the full signature.

- posted on August 17, 2005, 1:12 pm

at

special

similiar old building or a speciality lumberyard. does it have to be a 4x12? just nail two 2x12's together with some 1/2 ply sandwiched in the middle.

- posted on August 17, 2005, 2:17 pm

Maybe at a sawmill. Or a place that supplies pole barns and the like. Look into making your own composite beam to achieve what you need.

- posted on August 17, 2005, 4:12 pm

Thanks guys. I had considered that, and that is probably what I will do is
use 2 x 12s with the plywood in between, just didn't know if 4 x 12s were
commonly available.

- posted on August 17, 2005, 9:56 am

Eric and Megan Swope wrote:

No, 4x12 is not common. Making up your own as suggested will make a far stronger and straighter header. When making it, try to pick fairly straight stock and if it is a faily long header, lay it out so the curves are opposite, i.e., if the top one curves left, the bottom one curves right. Start nailing from one end and keep pulling the other end together. That makes for a nice straight beam. You might (probably will) need clamps to draw the ends together when you get near. I have done that often with stock up to 2x10, haven't tried with 2x12.

Harry K

No, 4x12 is not common. Making up your own as suggested will make a far stronger and straighter header. When making it, try to pick fairly straight stock and if it is a faily long header, lay it out so the curves are opposite, i.e., if the top one curves left, the bottom one curves right. Start nailing from one end and keep pulling the other end together. That makes for a nice straight beam. You might (probably will) need clamps to draw the ends together when you get near. I have done that often with stock up to 2x10, haven't tried with 2x12.

Harry K

- posted on August 17, 2005, 11:52 pm

crowns are ALWAYS up.

12

find it

be

12

find it

be

- posted on August 17, 2005, 10:24 pm

I R Baboon wrote:

Except when building headers. Check out some construction manuals if you don't believe it.

Top posting corrected

Harry K

Except when building headers. Check out some construction manuals if you don't believe it.

Top posting corrected

Harry K

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