Weird Roof Plan

[Spaski]

Greetings All,

After building a deck around a tree my pole configuration (4 poles) for the roof is rombus shaped. I had originally planned to shade sail this area but want to build a gazebo style roof (4m X 3m) with shingles or slate. Will this present any unique challenges given that the poles are not square other than to have a weird shape ?

Cheers.

[journeyman Mick]

Spaski
If the pokes form a true rhombus the the "top" two will be in a line parrallel with a line running through the "bottom" two. Thus if you run a beam on each pair and have the two beams the same length it will give you a rectangle to build on.

Mick

[joe greiner]

I don't see anything parallel to anything in the pic. I'd suggest placing the peak directly above the intersection of the quadrilateral's diagonals. This should give you a fighting chance of making the ridge members right. You should expect a lot of fiddling to get all the bevels fitted. This will be easier if you build the roof (framing at least) on the deck. Might have to loosen the corner fixings to lift it. If the poles aren't plumb, you should be able to make them plumb with a few hand winches (aka come-alongs).

For a true rectangular roof, use beams between two pairs of poles, and build a rectangluar frame (all four side) to span between those.

Good thing you didn't include the tree within the covered area!

Joe

[journeyman Mick]

I just had a look at the pic and it isn't a rhombus, so scrap my idea.

Mick

[Spaski]

Thanks Joe. Appreciate your ideas.

[pawnhead]

I would pitch it to a point at the geometric centre of the rhombus.

There is a difference between this point and the intersection of the diagonals as shown: -



In a more exaggerated rhombus, the difference becomes more apparent : -



I can show you how to do the calculations for cutting the hips, or I can do them for you if you post your dimensions for the rhombus, the roof pitch, and at least one angle.

If that front corner is 90 degrees, then I don't need an angle.

And I'd build it in place. You could do it single handed.

[journeyman Mick]

I thought a parallelogram was like a square that had been pushed in one corner to make it out of square, ie all sides equal lengths, oppposite sides parallel, but corners not 90 deg. Whereas a rhombus had four sides, two of which were parallel to each other. Of course it's been a few years since high school......

Mick

[pawnhead]

Yes, it's not a rhombus. I suppose you'd just call it an irregular polygon, and I'm not totally happy with my method of finding it's centre.
It gets you a lot closer than joining the diagonals, but it's not quite correct I've found.

I've posed the question on another forum that has a lot of geeks (no offense if you read this Assasinator_2. Thanks for the help ), but I suspect that the solution will be a quite complex method of averages, getting closer to the point, the more calculations that you do.

I'd imagine that it's something like halving all the angles. Drawing a polygon from their intersections, then halving the angles again, etc.

I should get a definitive answer soon.

[Spaski]

Thanks John,

Outstanding. I will get back to you on the pitch and angle. Not yet decided. Dimensions are 2750 X 2900 X 4300 X 3200 with diagonals 4500 and 4850. It bares an uncanny resemblance to your diagram.

Cheers,

Brad

[pawnhead]

What you'll need are the dimensions to the outside corners of your beams. You can house the beams in flush, at the appropriate height by setting your power saw to the depth of the timber thickness. It looks like a 190X45 should do the trick at those spans. (Hmmm, perhaps a 240X45) scrub that. I've just noticed you've said that it's 4300mm. That would require some beam (if you want to stick to the code). Span table time.

Measure it at deck level using straightedges or stringlines as I've shown in this diagram : -



And check that the front corner is square using Pythagoras (3,4,5 or calculate a diagonal)

[pawnhead]

Well here it is, and as I suspected it's not all that simple. I think I could get my head around it with some concentration, but I think that my method would get you near enough.

Young Einstein, the physics guru has offered to crunch the numbers anyway, but I'm going to try to understand myself : -

So you're wanting to work out the centroid of the object.

That's easily doable, and shredder hinted at it. It's first year statics so not exactly my everyday dealings-with but I have covered it a few years ago and can hopefully do it with a bit of help from lecture notes. Give me a minute to look stuff up and I'll edit it in after this.

Edit:

H'ok, so...

Bascially, the way I was taught to find centroids of objects was to break them down to a bunch of basic, easy objects and then sum the centroids of those respective elementary objects together to find the overall centroid. With this one, it can be done using right-angle triangles quite nicely, as the centroid for a right angle triangle is located both 1/3 of the side length inwards from the right-angle corner on both the x and y directions (imagine, baiscally, that those arrows go for 1/3 of the length they run along, starting where they form T shaped ends).

Observe how I've broken it down:


Now, the formula states that you can either add bits on (like I have) and then subtract them in the formula, or break the overall complex object up internally and sum the parts. I've chosen to add bits on then subtract them.

From this, we need some data:

Ai : The area of object i, where in this example you can see i will be 1, 2, 3 and 4 according to the diagram. Note that 1 is the overall thing including the areas I've added, so as to make 1 a simple triangle, rather than some ugly shape that sucks to calculate.
xi : The x-location of the centre of object i, RELATIVE TO A GLOBAL COORDINATE AXIS. I suggest you take the bottom-left corner (though it doesn't matter) as your global point, and measure the centre relative to that.
yi : The y-location of the centre of object i, again relative to a global axis.

We also then need to compute, from that. Ai*xi for each, as well as Ai*yi for each.

Once you have doen that, the formula that calculates it all is:

Basically, sum together all the Ai*xi that you calculated, then divide that by the sum of the areas (in this case, you'll need to subtract the areas, as you've added parts on, so go A1-A2-A3-A4, as A2, A3 and A4 are effectively 'negative areas' of A1 if you want to get your original object). Just remember that in this example, you will be doing (A1*x1)-(A2*x2)-(A3*x3)-(A4*x4) as I've made 1 the overall object, including the parts i've 'added on', and then i'm subtracting them afterwards. Because Ai is an area and thus in m^2, and xi is a distance, in m, that gives m^3. the overall area is m^2, so m^3/m^2 gives you a distance, m. This will then give you the overall x and y points for the overall object, as distances from your global axis.

I've made your example use right-angle triangles, because it makes everything nice and easy. You could use other shapes and, given you do everything right, it should come out the same, but this is the easiest, most logical way.

If you give me the measurements (or even fake ones that are easy) I'll punch them in, but I'm out of time for now (I have a family lunch at the pub in 10 mins) so I'll disappear to that and if you leave me fake measurements that are nice and neat and easy, I'll run through an example if you feel you need the extra help. Hopefully I've explained how the entire thing works, though. It's not an overly hard or complex idea.

[ChipperChip]

I know this isnt going to be the perfect solution, but there is another way to get the location of the centroid.

Make a scaled drawing of the shape of the roof (plan view) and cut the shape out of some even-density material, like cardboard. Put a pin through the cardboard somewhere near a corner (doesn't matter exactly where) and hang a plumb-line from this pin. Allow both to freely hang from the pin until they settle. Mark the line of the string on the cardboard.

Repeat this at another corner, and for a cross-check do it at a third location. The lines should all cross at the same point, and they will do so at the centroid of the object concerned.

For this kind of purpose the location found will be more than accurate enough. Measure it, scale it back up, and use trig from that point on.

That said, what you're really wanting to achieve is to spread the load evenly among the roof members. I assume you want to build this without 'ceiling' joists (ie, the bottom of your rafters aren't tied together) so the load coupling will be done by your fascias. You will have fascias in tension and with a bending load (good), and hip rafters in compression and with a bending load (potentially bad). The bending loads in the hip rafters are what you want to minimise and by shifting the centre of your roof you can vary those loads. As a punt, I'd have thought you could do this by minimsing the total length of all your hip rafters, but I'm not an engineer. Regardless, my question is whether the centroid gives you the minimum bending loads. I expect it DOES, but I couldn't say for sure. Pawnhead, any chance of taking that question back to your geeks?

Oh, and you'll want to tie your fascias and hip rafters together VERY well.

[pawnhead]

Your method looks alright (a bit 'howyagoin' though ) so long as it's a big enough piece of cardboard. it is the centre of gravity that we're looking for to give the wonky roof the most symetrical look possible.
The intersection of the diagonals would look quite wonky if the polygon was extremely irregular.

I think I've got my head around that formula, but I'll wait for confirmation of the dimensions and angles before I tackle it. I can post it in the other forums for Assasinator_2 to double check my calcs.

Pawnhead, any chance of taking that question back to your geeks?

Oh, and you'll want to tie your fascias and hip rafters together VERY well.

If the perimeter beams are bolted off well to the posts, then it should be alright without a bottom chord (ceiling joist), but they will have a tendency to bow out as well as sagging. You'll need a pretty decent sized beam to span over four metres though.

The thread I started, asking the question is here, but it's a private forum so you'd have to sign up (free).
It's a computer geeks forum, much larger than these forums (second largest in Australia behind Whirlpool forums), but I have no real interest in turbocharging my PC for gaming. I mainly go there for the heated debates about current events, the science forum, and the internet humour.

It's quite an interesting place if you can get over all the nerds and geeks (they're not everywhere there). There's a lot of very clever and successful people there.

I posted this there going off the dimensions Spaski has already given: -

Yeh, well I think I get it, but I'll try it out on some dimensions that he's posted later (I'm busy myself for the rest of the day). I'm not sure that I've got his dimensions in the right place. He's going to double check them all, and check the front corner to see if it's square. He'll also give me the pitch of the roof so I can work out the angle and pitch of the hip rafters. So if he does that by tonight I'll post up what he's measured on site instead of mucking around with these dimensions that might be wrong.


Not to scale

And someone else plugged it into a program and came up with this : -

According to Rhinoceros 3D, this is the centre. (Multiply the dimensions by 100)

I don't know how to work it out though.....

Cheers

[ChipperChip]

Your method looks alright (a bit 'howyagoin' though ) so long as it's a big enough piece of cardboard. it is the centre of gravity that we're looking for to give the wonky roof the most symetrical look possible.
The intersection of the diagonals would look quite wonky if the polygon was extremely irregular.

Yeah, a bit howyagoin, but my Dad (an engineer) taught me that technique when I was about 8. It works every time, and as I said, for this application it's more than adequate. Bush engineering

If the perimeter beams are bolted off well to the posts, then it should be alright without a bottom chord (ceiling joist), but they will have a tendency to bow out as well as sagging.

I doubt there'd be much lateral deflection, actually - the hip rafters would have to buckle for that to happen. Your rafters should essentially just be transferring a dead load to your hip rafters and the perimeter beams. Lateral deflection would only happen if you had a ridge. Remember, hip ends are self-bracing, and this is like two back-to-back hip ends.

You'll need a pretty decent sized beam to span over four metres though.

Where would you look it up? Isn't it structurally the same as a trimmer? I have a feeling that AS1684 only goes up to 3m spans for trimmers (going completely from memory) before calling for it to be 'designed according to engineering principles'.

It's a computer geeks forum, ... It's quite an interesting place if you can get over all the nerds and geeks (they're not everywhere there).

Ah, I'd be at home there I was a geek long before I became a carpenter. I googled the nickname as soon as I saw it and figured you must have been talking about the overclockers. Those blokes do some totally insane stuff. Years back I saw a double-barreled liquid nitrogen filled copper heat sink about a foot high to run a 1GHz cpu at 3.5GHz. It's like hot rods for geeks. I just used to build home-brew computers that had freaky CPUs and only ran home-cooked versions of Minix

But back to the issue.. I'm still not completely convinced that the centroid will give you the best case for hip rafter compressive loads. It'd be dead easy to calculate the centroid for that roof though, if you do have one right angle. The shape is a rectangle with two right triangles attached. Use the right angle as your reference point (coordinates 0,0 ) and do the sums. Easy areas to calculate and easy 'sub-centroids' to plug in. What your geek described is just a weighted average.

Or you could get a piece of cardboard

[pawnhead]

I doubt there'd be much lateral deflection, actually - the hip rafters would have to buckle for that to happen. Your rafters should essentially just be transferring a dead load to your hip rafters and the perimeter beams. Lateral deflection would only happen if you had a ridge. Remember, hip ends are self-bracing, and this is like two back-to-back hip ends.

Yeh, you're right there.

Where would you look it up? Isn't it structurally the same as a trimmer? I have a feeling that AS1684 only goes up to 3m spans for trimmers (going completely from memory) before calling for it to be 'designed according to engineering principles'.

Yeh they wouldn’t.Hyspan goes that far but you couldn’t use them outside. He could put in a cabletruss but it wouldn’t look all that crash hot. If it’s just a lightweight roof, then I’d just whack in a 300X50 chunk of HWD, and whack a cabletruss on it if it sags too much.

Those blokes do some totally insane stuff.

There’s some mega geek building a rail gun there at the moment.

But back to the issue.. I'm still not completely convinced that the centroid will give you the best case for hip rafter compressive loads. It'd be dead easy to calculate the centroid for that roof though, if you do have one right angle. The shape is a rectangle with two right triangles attached. Use the right angle as your reference point (coordinates 0,0 ) and do the sums. Easy areas to calculate and easy 'sub-centroids' to plug in. What your geek described is just a weighted average.

Or you could get a piece of cardboard

Probably something like this : -



That's still not quite right though, because it doesn't take into account the overlapping of the triangles.

I do believe that the centre of gravity would be the best place for distributing hip loads evenly though.

edit : - No that's not right at all, the more that I think about it.

That formula above is the way to go I reckon. It makes sense when you think about it.