We are building a living roof. on loadbearing straw walls.
The (dead&live) load for the beams is 185kg per m2
The span for the round larch beams is 4meters
Round larch beams between 7" and 9'' diameter.
They look the part but how do we prove it with numbers?

Ben Law's book has some of the calcs his SE did for his roundwood building.

Well the official way would be to get them visually graded to the softwood rules and then treat the cross section as square (losing the rounded bits). Calc as normal.

<blockquote><cite>Posted By: SteamyTea</cite><blockquote><cite>Posted By: Timber</cite>treat the cross section as square (losing the rounded bits). Calc as normal.</blockquote>Makes sense to me. There will only be a small difference in strength. Be different if it was hollow or being used as a shaft to transmit power.</blockquote>

The section modulus for a round log is (pi*r^3)/4 or 0.785r^3.
Trim the log square and the modulus drops to (2.82*r^3)/6 which is 0.47 r^3
The optimum rectangular shape is about 1.15r wide and 1.64 r high, giving 0.513 so that's lost a third of the round pole strength. Surprising eh?

The max bending moment is wl^2/8 or 480*16/8 = 960kg.m
Max stress is that divided by the section modulus which makes it about 250,000 kg/m^2
Larch has a working tensile strength of about 70,000 kg/m^2 so you’d need at least 4 logs per metre.

Max shear load is 480*2 = 960 kg/m
Cross section area of 4 logs is 4*pi *0.09^2 or about 0.102 m^2
Max shear stress is 960/.102 = 9400kg/m^ which is well below working shear strength of larch of about 80,000kg/m^2

About.. you might want to check this, figures used for larch etc.

At one point I was interested in using Douglas fir for my posts and beams but was put off by the fact that it wasn't structurally graded. I know of some houses which have been built with it (near Loch Eribol, from the mill I was talking to) but when I asked the timber guy about that he didn't know how it was approved.

Is there just some default strength which can be assumed for non-graded timber? Per species?

Or can you just load test it? Using what safety factor? Every log?

Is fatigue ever taken into account with timber structures?

Not often the maximum load on a house is reached, but there are many small loads, over many decades.

I believe timber is generally believed to have much better fatigue properties than other, manufactured materials. That is one of its attractions in boat and aircraft building (strength for weight being the other great advantage).

Google brings up some interesting points; according to:

http://link.springer.com/article/10.1007%2FBF02265220

"The extreme lack of a satisfactory data base and the persistent neglect of fatigue characteristics in wood material design is emphasized." December 1980

However http://www.byg.dtu.dk/-/media/Institutter/Byg/publikationer/byg_rapporter/byg_r038.ashx has:

"Wood fatigue is of interest in regular fatigue loaded structures such as wood bridges but
wood fatigue also covers low cycle fatigue encountered in ordinary structures. The study
concentrates entirely on fatigue in tension perpendicular to the grain as this strength
parameter often is strength limiter in critical details." 2002

and http://eu.wiley.com/WileyCDA/WileyTitle/productCd-0471487082.html looks like it might be the dogs danglies if you have the dosh. Though there might have been some advance in the last decade or so.

From what I understand, fatigue is cyclical loading, usually below the elastic limit.
It is also affected by temperature, and with semi porous materials, probably moisture/humidity levels.

It is hard to fatigue a car tyre, but easy to do a bit of mild steel.
I suppose it is not much of a problem as we would have heard about it. But I have been in some creaky boats that have got me worried in the past.

Moving on from that a bit. I think timber fails in a more 'gentle' way than masonry and steel.
I know that some of the composite chassis cars I was involved with has some serious problems. Got around some of the problems by laminating in composite tubes as they failed in a more predicable way.

Ah that's pig fumes does that

Posted By: gyrogearthe exact same phenomenon (of load-reversal) occurs on windows...

As I said, it occurs even more significantly, for same wind-buffet reasons, on airtight (or any) membranes/tape. Explains a lot about their notorious failure rate.

Posted By: Carol hunterThe algebraic formula's looking quite esoteric

It does look it, but all it is saying is, how much does it move when I put a load on it, why you end up with that odd looking m^4 unit.
All that is saying is for a bit of material with dimensions m x m x m (which is m^3), how many m's does it move when loaded, so it becomes m x m x m x m (m^4).
Or just jump on a bit and see if it feel strong enough.

In fact it totally dominates - no such thing as a 'lightweight' structure as far as calcs concerned!

Posted By: Carol hunter

Posted By: mike7
About.. you might want to check this, figures used for larch etc.

Erblushmumble bit of an error here - correct figure for tensile strength is more like 500,000kg/m^2 or as it might be given now 5mPa.
Assuming no other errors that'd mean 1 log per 2m, which seems like not a lot.

I calculated the roof beam section once for a shed down my garden. It didn't break, but it did sag quite a lot

Hi Mike7,

Tried to reply to your original post last night, but a similar gremlin deleted the post... Think I just took too long writing it, so here goes again!

Thanks so much for these figures, but I think we need a dummy’s guide, as tried to work through them yesterday and there was lots that we just couldn’t understand, not having done any calculations like this before. Hope you don’t mind, but you may need to hold our hands! These might sound like silly questions:

What does wl mean?

What exactly is max stress? 250,000kg/m^2 seems a lot when our total load (including snow!) is only 185kg/m^2.

Again (at the risk of sounding simple!), 500,000kg/m^2 seems like a very big number- can you please explain tensile strength? The total load of the materials above the beams is 18,053kg.

Is the shear load figure the load that can be applied before shearing occurs?

How did you arrive at the figure of one beam per 2m? Or beams are actually about 1.4m apart.

Is ‘sag’ the same as deflection?

Also, I don’t know if this will have any bearing on how you’re thinking about it, but thought a bit of extra background on the structure might be handy, as our roof isn’t a conventional one! The single-storey house is only 8m x 4m internal, with walls 0.5m thick. We have 7 beams in total, 2 sitting on top of the 5m walls, and 5 spanning the gap. The beams are equidistant and are sitting on composite posts creating a 10 degree pitch. Some of the beams had a slight curve in them, so we placed them with a high point in the middle to help balance out deflection.

Any more help would be much appreciated!

Many thanks :-)