parallel cord truss

I was wondering if someone can help me design a parallel cord truss, warren truss. the span it needs to cover is 26' and was trying to figure out how to size the members.

I know you can use method of section to determin the load on each member but how do you load it (for a 26' roof section) and how do you get a beam size from it

Thanks in advance for your help

Yes, I have access to 26' long timber beams.

Is that necessary though since their will be node at 52" if i split it into 6 sections wouldn't i be more worried about the connection or would it be better to cut a 26' piece.

tietje_3

I would use the full length timber if available

You may want to check out this site http://www.dot.state.oh.us/se/coveredbridges/truss_types.htm

Lateral stability, especially during construction, would be a big reason to go long if possible.

If the truss is uniformly loaded, to size for gravity loads, determine the tributary width, multiply the live and dead loads by the square footage bearing on the truss (trib width x 26' x (LL+DL)). Distribute the load uniformly on the nodes of the loaded edge. The members are solved as either columns or columns with bending.

"Wood Engineering and Construction Handbook" Mcgraw Hill, has a good description of sizing a 50' span heavy timber parallel chord truss. Interlibrary loan might be able to get it, chapter 6.

Don answered part of your question - the forces are normally unifor - so you have a continuous load across the truss top chord. If you are confident that the span from panel point to panel point is small enough to result in very low bending stresses, then you can collect the load at each panel point.

Example:

4' on center trusses
20psf Dead Load, 40psf Live Load (residential floor)
52" from panel point to panel point

4' x [20D, 40L] x 52" = 1040lbs per panel point

Now you've got a load at each panel point at the top chord, and you can do your frame analysis.

Once you've got the forces in the members, you can size your timbers.

Take the force in each member and divide it by the allowable tension or compression stress from the NDS tables. Now you have the area of your cross section and you can choose a size. There are lots of other factors to consider, and if you're not familiar, I would stick to the following:

Your members should be 2x the area you calculate above
No member should be longer than 16x its smallest dimension
No cross section should be more than a 2:1 ratio of long:short side

You will also need to connect the ends of every member properly - if you're using 1" oak dowels, count 400lbs for each one in the joint (probably conservative...I can give you the actual equation if you want it). Use 200lbs if you use 1/2" steel bolts. Make sure you have proper end and edge clearances for the pegs, though.

This is a pretty complex thing you're trying, and it sounds like this is either a ridge beam (in which case your loads will be much higher than my floor example), or you're doing a flat roof, in which case you've got a lot of calculation work ahead of you to determine - and limit - the deflection of the truss to prevent ponding and progressive collapse.

It's worth a shot, but you might want to pay to have a second set of eyes review the calculations (or do them independently) to make sure you don't miss a decimal somewhere.

sorry for the late reply. Been kind of busy

to answer some of the questions asked species is Ponderosa pine,
it would be for design of a ridge beam, snow load is 53psf ground snow load not reduced, connections will be bolts and steel plates. Steel plates will be hidden in for example (2x) 6x12 top ans bottom cords and steel washers on the outside.
I would probably have to split the cords into 2 sections and find where which section has the minimum bending momend and still comply with the IRC and the ASD/NDS. I think the steel plates will really help the strength and connection situation

I'm designing it myself and sending it to the PE for review (trying to save cost)

I'm open to any other ideas and any comment is welcome

the building code (IRC 2003 in my county) uses L/180= (26*12)/180 = 1.73 inches for deflection would you go with that number or would you go with all other structural members at L/240= 1.3 inches.