Ridge beam question?

[phalynx]

If I built, say, a 20x20 building and did not want to have any ceiling joists or beams across the ceiling, I assume that I would need to have a ridge beam that was structural.  I "assume" that I would need to build support in the wall to carry the ridge beam as well as appropriate foundation under the wall support.  The question I have is, I guess, architecture 101.  If you put a structural ridge beam for your roof, how do you calculate the beam requirements?  I am assuming that roof area and loading per square foot are in the calculation.  I am thinking that the beam needs to carry the entire load.  Am I on track?  Is this how it is done?   I know John won't answer as "an engineer" and that's fine, an answer like "an opinion" is fine too.  I will do the math, I just want to know what I am supposed to calculate.

[MountainDon]

An engineer is qualified to do the calculations. The rest of us can only guess.

There are various types of trusses that might fulfill your wants. The truss company engineer will do the calculations for a price estimate, usually at no charge, in my experience. The price for the trusses will include the engineered, and stamped, drawings for the permitting process. I think it's worth a look to see what the costs would be. Also check out the insulation amounts that could be achieved with trusses vs beam and rafter.

Back to the ridge beam.... It does not carry the entire load. Some of the load is carried by the side walls. That should be half to the ridge beam and one fourth to each side wall. You are correct in that the end walls must supply adequate support for the beam and the foundation must carry the weight from all that.

[NM_Shooter]

Scissor trusses will do what you want, and will be easy to work with.  What sort of pitch do you want on your ceiling?  Steep pitch ceilings might need to be done with a beam, as scissor trusses start to get real "noodly" above 5:12 or so.  Keep in mind that for some reason, the pitch looks "steeper" when you are viewing it from inside. 

[phalynx]

I understand the scissor trusses.  I am actually trying to learn the math of the ridge beam to determine how it is calculated.  I like the absolutely clear look of a catherdral ceiling.  The pitch I am looking is 12/12. 

[NM_Shooter]

Google Euler–Bernoulli.  That's what you'll need to use. 

It's not algebra. 

[phalynx]

Wow,  its really NOT algebra... hahaha.  Thanks, I have some reading to do...   

[Don_P]

It's nothing to guess at and it's pretty simple. This used to be known by every competent builder and was allowed to be figured by the builder... the strength tables are in my old codebooks. I had to hire an engineer this past week to do the math and got into a discussion with the building official over this. The old book is on his shelf and I pointed to it and quoted its content. He replied that the new IRC does not allow me to do the math. I correctly replied that it does not disallow it and quoted again. People rise or fall to our expectations. We are dumbing our population down...don't fall for it. Know how and stand there tapping your foot with arms crossed and a glare  . This ain't rocket science, it's 9th grade math, if that.

MD described the tributary width... 1/2 the building width is bearing on the ridge, 1/4 is bearing on each sidewall. So the trib area in the example is 200 sf. Typically for a conventional roof use 10 pounds per square foot for the dead load (the weight of the roof materials). Your snow load is typically the live load. Multiply the total load per square foot times the trib area to get the total uniformly distributed load on the beam.

This beam is called a simply supported uniformly loaded beam, a beam on 2 supports.

Now you need to know the various strength properties of the material you wish to use.
Fb- allowable extreme fiberstress in bending. The extreme fiber is the outermost (bottom) strap of fibers in this beam... how much bending stress is allowed before they tear. Bend a popsicle stick, where it breaks in the middle is a fiber failure.
Modulus of elasticity- How stiff is the material. A rubber band is incredibly strong but would sag badly as a beam. Deflection is normally the limiting factor in horizontal members.
Fv- horizontal shear. Bend a phone book, watch the pages slip past each other as the beam bends. If the beam bends and slits lengthwise into 2 beams it has become 2 shallower beams which wioll fail individually at much lower load than the full depth beam. This is a shear failure. It normally occurs in short very heavily loaded beams. Longer members would fail in one of the other two modes before getting here.
Fcperp- Fiberstress in compression. The posts at the end are driving into the side(perpendicular) grain of the beam. There must be enough bearing surface not to crush into the beam.

There's one side of the equation, gotta go unload a trailer before dark, more in a bit.

[phalynx]

Don_P,  thanks for chiming in.  So, if I follow the start of what you are saying, it calculates as follows:
20x20 building has a 400 sq ft roof area (not sure if you have to calculate the pitch in there)
1/2 of the 400 sq ft is carried by the beam so we use 200 sq ft in the calculation.
The dead load is 10psf and we'll say the snow load is 30psf for a total of 40psf
40psf X 200 sq ft roof area carried by the beam = 8000lbs.

Is the next part, dividing 8000lbs by the length of the beam?  If so, that would be 400 lbs uniform load per linear foot?

So to follow the dumbing it down method, we look at a chart for a beam capable of carrying 400lbs per linear foot at 20' length?

Or then go to the next section of math?

[Don_P]

You got it, you're there if you want to use a table. Use the horizontal measurement. If you want to get picky the DL is the actual (sloped measure) and snow is the horizontal projection. Most folks don't get that fine for lightweight residential. I was typing while waiting for some pdf's to download and missed your response... this is a bit different than your example but gives an overview of the longhand. The tables are developed from the formulas that follow.

I'm going to use 10psf dead load(DL)+ 40psf live load(LL)(snow)=50psf
first check bending:
Calculate maximum bending moment;
WL/8
W= 50 psf x 200 sf = 10,000 lbs total load on the ridge
L=20'
Max Moment= 25,000 ft lbs
There's the bending force we need to safely resist.
Design values are in psi so we'll convert to inch lbs
25000 ft-lbs X 12= 300,000 in-lbs
Lets use the design values of a typical LVL... FB=2900 psi (sawn lumber runs from ~600-~2000psi)
300,000/2900=103.448"³...there is the minimum section modulus we are looking for.

Section modulus is determined by the formula bd²/6...breadth x depth squared/6 (remember the oft repeated phrase "deeper is cheaper", depth is squared here going deeper builds strength fastest)

Lets try a double 14" LVL
1.75" X 14²/6 =114.33

Checks in bending.

I got a call, need to sketch some stuff for a client, I'll give deflection and shear a whirl tomorrow if you like.