I was planning to make a wind/privacy screen with horizontal slats between two posts at the end of the patio. The distance between the posts is 2.75m and I don't plan to support it in the middle on the patio floor, just attach it at the ends to the posts. What lumber dimensions are required for the slats so they don't sag in the middle like a "hammock"? In other words: Are there any guidelines for the relationship between lumber dimension and distance for beams and with different loads? In my case, the load would be equal to its own weight. I was thinking of slats that might be 44x21 (on edge), 44x44, or somewhere around there.
/L-O
/L-O
Deformation due to long-term load is difficult to address without oversizing. If you use thin slats, they weigh less and warp more easily. Slightly thicker slats are more stable, but they too can twist/bend. They also impact the horizontal beam more due to their greater weight.
Much of how the slats will behave under the influence of weather and wind depends on how they are cut from the log. Much of how the beams will bend down depends on the quality of the wood, the better the quality, the greater the strength of the wood. And here too it is crucial how the beam is taken from the log and mounted in place.
And of course, everything depends on the dimension you choose. The ‘taller’ the beam, the more resistant it is. And the best is, of course, a plank in the standard dimension 45x220 in the highest wood class.
Luckily, there are some tricks to avoid having to support with planks for a span of 2.75 m.
The first is to use K-plywood. It is dimensionally stable because the veneer is laid crosswise in multiple layers. If you take 2-3 layers of 13 mm K-plywood in strips of 15 cm from a full sheet of 3 meters length, you can be fairly sure that the 'beam' (glued and nailed/screwed together) will not sag.
The second is to place a tensioning beam under the load-bearing beams. You make it from a thin wire or equally thick steel wire that you tension up on the underside of the beam at both ends. Then insert a spacer in the middle between the wire/cable and the underside of the beam so that the wire/cable is tensioned like a bowstring, causing the beam to bend slightly upwards, allowing room for the beam to settle over time.
Unfortunately, this last method can also cause the beam to crack sideways, either towards the side where the slats are or the opposite.
Therefore, method one is better. However, you need to seal the cut edge of the plywood with a thin wooden strip to make it water-resistant, and the same goes for the lower edge. Nail-glue it in place. Then paint over everything, including the slats, and you have a fence that will last for many years.
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The Builder
Much of how the slats will behave under the influence of weather and wind depends on how they are cut from the log. Much of how the beams will bend down depends on the quality of the wood, the better the quality, the greater the strength of the wood. And here too it is crucial how the beam is taken from the log and mounted in place.
And of course, everything depends on the dimension you choose. The ‘taller’ the beam, the more resistant it is. And the best is, of course, a plank in the standard dimension 45x220 in the highest wood class.
Luckily, there are some tricks to avoid having to support with planks for a span of 2.75 m.
The first is to use K-plywood. It is dimensionally stable because the veneer is laid crosswise in multiple layers. If you take 2-3 layers of 13 mm K-plywood in strips of 15 cm from a full sheet of 3 meters length, you can be fairly sure that the 'beam' (glued and nailed/screwed together) will not sag.
The second is to place a tensioning beam under the load-bearing beams. You make it from a thin wire or equally thick steel wire that you tension up on the underside of the beam at both ends. Then insert a spacer in the middle between the wire/cable and the underside of the beam so that the wire/cable is tensioned like a bowstring, causing the beam to bend slightly upwards, allowing room for the beam to settle over time.
Unfortunately, this last method can also cause the beam to crack sideways, either towards the side where the slats are or the opposite.
Therefore, method one is better. However, you need to seal the cut edge of the plywood with a thin wooden strip to make it water-resistant, and the same goes for the lower edge. Nail-glue it in place. Then paint over everything, including the slats, and you have a fence that will last for many years.
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The Builder
Another trick is to insert a couple of crosses in the middle of the construction. In this way, you get some diagonal braces that take up a lot of tension/compression loads, and in this way, you minimize the deflection.
I don't know if I understand what you mean.imported_Byggaren said:
To begin with, it's this kind of 180 cm high "fence" I've been thinking about, but planning to make a variant of myself:
Romb-series - Jabo
or something like this:
Funkis-series - Jabo
The problem is that these ready-made panels aren't available in suitable measurements, only up to 180 cm. But perhaps that's understandable.
Regarding the plywood, if I understood you right, I should "manufacture" my own "beams" by laminating three layers of plywood myself? It might become a bit cluttered if you want "beams" that are about 2" high...
K-plywood - what type is that?
The wire you mentioned, wouldn't that be visible?
And would it then require a wire under each slat?
Meaning, I don't want to add any diagonal braces either, even though it would naturally improve the situation.mats_o said:
Quick responses, by the way!
There is hardly any point in manufacturing a 2" K-plywood (construction plywood) rule. It helps as little as a 2" wood beam to keep the nail rule straight for the load. You increase in dimension because the height of the rule is what is needed to withstand the weight represented by the slats.
Otherwise, you should choose a rectangular hollow profile of iron instead. There, a KKR 60x40x4 can handle it but will be difficult to nail into.
The thread should not be under the slats at all, but under the beams. Since it is thin, it is not visible from a few meters away.
Addition:
I now see from your links that you are considering horizontal slats. I had assumed standing = horizontal support beams in my answer. Therefore, forget what I've written so far.
For horizontally laying slats of the type you refer to, dimensioned for their own weight only, it is clear that the thinner the slat, the less it weighs and the easier it turns/twists.
At 2.75 m span, you must consider at least 1/2" thickness and a height of at least 12" for each slat if they are to bear their own weight without bending downwards. I have then calculated with a density of 500kN/m3 for pine/spruce, which is the weight it has.
So, insert cross braces, a center post, or abandon the project.
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Byggaren
Otherwise, you should choose a rectangular hollow profile of iron instead. There, a KKR 60x40x4 can handle it but will be difficult to nail into.
The thread should not be under the slats at all, but under the beams. Since it is thin, it is not visible from a few meters away.
Addition:
I now see from your links that you are considering horizontal slats. I had assumed standing = horizontal support beams in my answer. Therefore, forget what I've written so far.
For horizontally laying slats of the type you refer to, dimensioned for their own weight only, it is clear that the thinner the slat, the less it weighs and the easier it turns/twists.
At 2.75 m span, you must consider at least 1/2" thickness and a height of at least 12" for each slat if they are to bear their own weight without bending downwards. I have then calculated with a density of 500kN/m3 for pine/spruce, which is the weight it has.
So, insert cross braces, a center post, or abandon the project.
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Byggaren
The brain is a wonderful 'invention.' If it gets a problem to work on, at least mine gladly solves them at night when I sleep.
There is actually a possibility that you can get what you want with 2" horizontal slats without a middle post, but it would be expensive and ineffective and assumes one thing: that plexiglass exists in sheets as large as your fence is long and tall. You can probably guess the rest. Otherwise:
A plexiglass sheet the same size as the fence's length and height. Mounted firmly between the posts. Slats are mounted on the plexiglass with KH screws drilled from the plexiglass into the battens.
The downside? A solid fence provides less wind protection than a perforated one.
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Byggaren
There is actually a possibility that you can get what you want with 2" horizontal slats without a middle post, but it would be expensive and ineffective and assumes one thing: that plexiglass exists in sheets as large as your fence is long and tall. You can probably guess the rest. Otherwise:
A plexiglass sheet the same size as the fence's length and height. Mounted firmly between the posts. Slats are mounted on the plexiglass with KH screws drilled from the plexiglass into the battens.
The downside? A solid fence provides less wind protection than a perforated one.
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Byggaren
Could one not consider a discreet support leg in the middle under each section?
If you look at the "Romb series," it also has a vertical slat in the middle – let it take some load as well.
Best regards,
Ronnie
If you look at the "Romb series," it also has a vertical slat in the middle – let it take some load as well.
Best regards,
Ronnie
Or just a narrow strip of plexiglass or glass in the middle as support? Should be invisible.
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Byggaren
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Byggaren
I actually have another option that I was planning to avoid for aesthetic reasons: Between the posts and above the windbreak - there is a support beam made of laminated wood on which the rafters rest. I could actually hang a middle post from it. This would halve the distance to about 1.35. The reason I haven't planned to attach it to the deck is that it sits loosely like a piece of cake on several tiles on the ground (floating). The posts, on the other hand, are on foundations and are also connected to the house via the roof structure. So I thought that the floor should be able to move freely without causing strange tensions and weakness in the windbreak.
But when you Imported_Byggaren say that a timber dimension of 12" in height is required to support its own weight, I was a bit surprised. Does the requirement for dimension decrease so drastically to 180cm length? Because that length is what Jabo's factory-made sections have. And in the Romb series, all cross bars including the slats are 45mm. Or would Jabo's sections also sag?
But when you Imported_Byggaren say that a timber dimension of 12" in height is required to support its own weight, I was a bit surprised. Does the requirement for dimension decrease so drastically to 180cm length? Because that length is what Jabo's factory-made sections have. And in the Romb series, all cross bars including the slats are 45mm. Or would Jabo's sections also sag?
Yep!
The length is crucial for determining the size needed to avoid getting a hammock effect.
The formula for a simply supported 2-span beam (as we're discussing here) is qxl2/8, where the length is squared for load alone. But if deflection consideration is necessary, the formula is 5xqxl4/384xExI where the length is raised to the fourth power. (384 is a constant, E is the modulus of elasticity for wood, and I is the moment of inertia of the surface/cross-section.) The larger the I, the stiffer the profile.
If you have 2.75 m in spv, the square of it becomes 7.5625, but only 3.24 for 1.8 m. Similarly, you get a factor of 57.19140625 for I to the fourth, compared to 10.4976 for 1.8 m. Thus, more than a doubling in the first case for load alone and almost a sixfold increase in the second case for deflection. This requires a sixfold increase in I as well to achieve the same resistance (moment of inertia) against deflection. And since the moment of inertia for a rectangular cross-section is bxh3/12, where the height is cubed, it's easy to see that, given the same width, a significantly greater height is required to achieve a sixfold increase in the moment of inertia.
(qxl can also be expressed as Q, i.e., the total load instead of load/m beam. But you cannot use total load in these calculations; you must use load/m.)
Perhaps a bit too theoretical and simplified for the uninitiated to understand, but that's how it is according to Hooke's law and the theory of elasticity, which underpins all strength calculations.
(Must try to figure out how to write a formula/exponent in this new forum)
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The Builder
The length is crucial for determining the size needed to avoid getting a hammock effect.
The formula for a simply supported 2-span beam (as we're discussing here) is qxl2/8, where the length is squared for load alone. But if deflection consideration is necessary, the formula is 5xqxl4/384xExI where the length is raised to the fourth power. (384 is a constant, E is the modulus of elasticity for wood, and I is the moment of inertia of the surface/cross-section.) The larger the I, the stiffer the profile.
If you have 2.75 m in spv, the square of it becomes 7.5625, but only 3.24 for 1.8 m. Similarly, you get a factor of 57.19140625 for I to the fourth, compared to 10.4976 for 1.8 m. Thus, more than a doubling in the first case for load alone and almost a sixfold increase in the second case for deflection. This requires a sixfold increase in I as well to achieve the same resistance (moment of inertia) against deflection. And since the moment of inertia for a rectangular cross-section is bxh3/12, where the height is cubed, it's easy to see that, given the same width, a significantly greater height is required to achieve a sixfold increase in the moment of inertia.
(qxl can also be expressed as Q, i.e., the total load instead of load/m beam. But you cannot use total load in these calculations; you must use load/m.)
Perhaps a bit too theoretical and simplified for the uninitiated to understand, but that's how it is according to Hooke's law and the theory of elasticity, which underpins all strength calculations.
(Must try to figure out how to write a formula/exponent in this new forum
)_____________________
The Builder
I guess that's how it will be!mats_o said:Hang on to the glulam beam and it will definitely be much better
Thanks for the tips!
Imported_byggaren:
Just wish for a clarification:
What (which magnitude) do you calculate with the formula you mentioned at the beginning - qxl2/8?
q = N or kN per meter62_an said:
l = m (squared)
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Byggaren