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So, of course, the rafters need to `bear’ on (be supported by) something. For Linda’s Goat Barn the rafters will rest on `headers’. The design of the barn, remember, is `pole barn’, e.g., rafters supported by headers supported by posts that act as `poles’. The poles embedded in the ground provide the lateral stability (keep the barn from flopping over)^{1}. In terms of construction, I want three things of the support conditions: 1) I need enough `length of contact’ so that the end of the rafter doesn’t `accidentally’ slide off the header; 2) I need some `anchorage’ to make sure the rafter doesn’t slide or get pulled off the header … during construction and afterward; and, 3) I need enough contact area (rafter on header), so that the wood isn’t crushed. I like to start with at least 1-1/2 inches `bearing length’^{2} … which generally keeps the rafter from sliding or falling off the support, and keeps the wood from crushing..
One and one-half inches `seems’ to provide enough length in most cases to keep the rafter (joist, beam, whatever) from accidentally sliding off the support during construction. Then to make sure it doesn’t slide off, we `anchor’ the rafter with, say, some nails. A typical code requirement for anchoring a rafter to a wall is … (3) 10d common nails in toenail (slant) configuration (two on one side and one on the other). Each end of rafter. Yup! That will anchor the ends of the rafter. Note that at this stage I’m not really worried about wind. If a wind comes along, construction stops, and I wait till it’s over. Besides, without the roofing material on yet, there’s not much to `catch sail’. Once I’m done with construction, I also need to make sure that the rafters don’t get pulled off the supports/headers if a wind comes along. For Linda’s Barn, I will `assume’ that the toenails used for anchorage also provide the necessary resistance to wind uplift^{3}. For buildings occupied by humans, a tie-down design would be required, likely resulting in `hurricane ties’ being required to hold the rafters to the walls.
This idea of `sliding off’ is even more important with sloped rafters. Linda’s Barn’s roof is `1 in 12’ … not too steep, but not flat, either. I’ll definitely anchor the ends of the rafters before I `get up on them.’
Let’s see if we crush the wood … in this case the construction worker is not out there at mid-span … but closer to one end, so that nearly the whole weight of the worker goes through one end … 300 lb … a bit more if we count the added weight of the roof. Using 300 lb … and a (true) 2-inch wide rafter, bearing on at least 1.5 inch of header …
300 lb (man) divided by 2 inch (wide rafter) x 1.5 inch (length along header)… equals 300 lb / 3 square inches = 100 psi … any decent available wood should be able to carry 500 psi or more in bearing … Good!
Does this mean we could use less than 1.5 inches of bearing? NO!
________________
1As construction continues, the boards on the sides will also provide lateral support … kind of like `shear walls’ … relieving the posts, acting like vertical cantilever columns. from providing all the lateral support.
2Minimum of 1.5 inches bearing length Code references … (see, e.g., 2018 IBC 2308.7.3.1).
3Let’s do a quick calc … 100 mph wind … say 25 psf wind uplift (roof acting like an airfoil) … each end of rafter holds down (½ of 12 ft + 1 ft eave times 2 feet per rafter … 14 square foot of roof for each rafter end… 14 square feet times 25 pounds per square foot … gives … 350 pounds! I can probably get that with (3) 10d common toe nails. I’m actually going to use 8d and some 16d sinkers. I can probably get 100 pounds each out of the 8-d nails … that would correspond to, say, 90 mph wind. Fine. I haven’t accounted for the weight of the roof itself (though relatively light) in helping to hold the roof down. TBH, if we get a 100 mph wind, it will probably be accompanying a tornado, and I suspect the whole barn will be gone, with the cows, if still alive, standing in the rain, and the goats, if still alive, hiding under anything they can find to get out of the rain.
Now for the headers …
Starting from the top … we found that we’re going to use 2×6 rough sawn Southern pine rafters … spanning 12 feet, and overhanging 1 foot each end. The design was `governed’ by a concentrated load (construction worker) at mid-span of the rafter. The 2×6 rafter also accommodates a uniform construction load of 20 psf, as well as 25 or so psf of wind uplift. The barn has two 12-ft x 12-ft bays, and is `pole-barn’ construction, meaning the gravity loads (and lateral loads) are resisted (carried down … to the ground) by the posts/columns. `Headers’ (beams/girders/girts … whatever you want to call them) will support the rafters and transfer the loads to the posts/columns.
Each header will need to carry one set of ends from the rafters. In addition to the rafters spanning 12 feet, and overhanging 1 foot each, I need to have some kind of rake, so I’ll assume the headers also span 12 feet (the other direction, capturing the ends of the rafters), and (also) overhang 1 foot. Otherwise I’ll need to come up with some kind of outrigger system.
Let’s start out with the headers simply spanning 12 feet between posts/columns. This will be slightly conservative with respect to `bending’, since it ignores the counter flexure from the rakes. I’ll assume the roof weighs 5 psf (relatively light). In terms of a uniform, or line load, the headers will need to carry (between the posts) …
Dead Load …
w DL = (½ of 12 ft^{1} + 1 ft overhang) x 5 psf = 7 ft x 5 psf = 35 plf …
or
W = w x l = 35 plf x 12 ft = 420 lb … (dead weight of roof on the header span).
Live Load …
A single construction worker would contribute 300 lb live load … (probably won’t govern) …
A construction crew/operation will deliver … 7 ft x 20 psf = 140 plf … or 140 plf x 12 ft or 1680 lb^{2}. (Yup, the whole crew governs over a single worker.)
Let’s look at the total … roof weight plus construction crew = 420 lb + 1680 lb = 2100 lb total load, uniform.
Check … total uniform `area’ load … 5 + 20 = 25 psf.
Total load to each header = uniform load x area supported by each header = 25 psf x [(½ of 12 ft + 1 ft) x 12 ft] = 25 psf x 84 sf = 2100 lb. Check.
Assuming the design of the header is `controlled’ by bending (stress) … let’s find the minimum Section Modulus needed.
In equation form …
Let fb = applied stress under load = Fb’ = allowable stress, where
fb = M / S,
and
assuming an allowable bending stress, Fb’, of … say, 1000 psi …
M = bending moment in header under design (applied) load = W L / 8 …
W = 2100 lb;
L = 12 ft;
8 = 8 …
M = 2100 lb x 12 ft / 8 = 3150 lb-ft.
Let’s get this into lb-in. …
3150 lb-ft x 12 in./ft = 37,800 lb-in.
Doing some rearranging …
fb = Fb’ = M/S … gives S needed …
S needed = M / fb = M / Fb’ = 37,800 lb-in / 1000 psi = 37.8 in.^3.
What sawn lumber (Southern pine) sections give me 37.8 in.^3???
Looking the the National Design Specification ® for Wood Construction (NDS) Supplement …
2 x 12 dimension lumber (1.5 x 11.25) … S = 31.64 … not enough.
Try 2 x 12 `actual’ … S = b h^2/ 6 = 2 x 12^2 / 6 = 48 … ENOUGH!
I will ask my son-in-law to cut me a 14-ft 2 x 12 (actual dims) header!
WAIT!
Several problems.
First, my son-in-law can cut sawn lumber in dimensions (actual) up to 6 inches, easily … 12 inches, more difficultly.
Also, I have been carrying the 2 x 6 rafters around … 14 ft long … and they are about all I can handle by myself … they are remarkably heavy; I think it’s because they are `GREEN’.
What are my other options?
2 x 10 actual … S = 33.3 … not enough.
4 x 10 actual … 67 … way enough.
4 x 8 actual … 43.7 … enough!
But that is still going to be heavy …
The `line weight’ of a wood member goes like … w = γ x A, where γ = specific weight, and A = cross section area. I could compare line weights, but I don’t know the specific weights of this `green lumber’ here^{3} … so let’s compare cross section area …
2 x 6 = 12 square inches (actual) … (I can lift into place)
4 x 8 = 48 square inches … that’s four times as much … will be four times as heavy … no way! I’m not going to ask my son-in-law to come over with his tractor to lift each header … if I can avoid doing so.
Hmmmm.
Try 6 x 6 … S = 36 in.^3. Enough! Actually, not quite … BUT … I’ll bet I’ll be able to split hairs, and make it work.^{4}
Turns out A = 36 in.^2. Same number; different property.
At least my son-in-law can more easily cut the 6 x 6.
Or I can laminate three 2 x 6s side by side.
But still way to heavy to heave up above my head.
OMG! I don’t have to nail laminate the header together on the ground … this is what I’ll do … I’ll lift each 2×6 ply into place, one at a time … something I can do … and nail laminate them together once they’re up in place!!!^{5}
*****
This `three’ is very interesting. Remember that the rafters span 12 feet … and that the headers themselves span 12 feet … same support condition … 12 foot span. And same load condition, … roof dead plus live load. The rafters are spaced 2 feet apart … so each header carries the ends of 12 ft / 2 ft per rafter = 6 rafters. Since the rafters are supported by headers at each end … we could say that each header carries half of 6 rafters, or the same as three rafters! A 3-ply 2×6 header will carry the ends of three 2 x 6 headers.
Interesting.^{6}
Selah
The rest is easy … especially if we look at it in the context of a 3-ply 2×6 carrying 3 – 2×6’s. Deflection should be good. Shear should be good. Bearing should be good, as long as each 2 x 6 ply of the header has the bearing required for each of the 2 x 6 rafters … or the header as a whole has enough bearing … and so far, we’re way good in bearing. Let’s look!
BEARING!
Bearing (rafter on header) …
The roof is sloped, so I’ll bevel one of the lams of the 2×6 header to match the slope of the rafter. That means that the rafters will only actually bear on one ply of the header … but we already determined that would be fine … we assumed that we would have bearing area of at least 2 in. (width of rafter) x 1.5 in. (code min) … but the beveled lam of the header is 2 inches wide, … we’ll have 2 in. (width of header) times 2 in. (width of lam) = 4 sq. in. … way good!
JRF
___________________________
1 … in the direction of the rafter.
2Note that we don’t combine the concentrated single construction worker and the uniform construction crew … it’s either or … or … as we crunch along down through the structure, the worst of the two. It (the concentrated load) governed the bending stress limit state for the rafter … but will not govern the bending stress limit for the header (the uniform load representing the crew, materials, etc.).
3 … … plus the green weight is changing day by day as the lumber dries out.
4At the very least, I’ll be able to employ the repetitive member factor, Cr, since I have 3 plies … Cr is (at least) 1.15 for this condition … 1.15 x 1000 = 1150 psi. S needed = 37,800 / 1150 = 32.9 in.^3. I’ll have 36. Yay! (We’ll talk more about `allowable stress’ later.)
5This (3-ply) solution is not the most conservative with regard to wood mass … but just think! … I can proceed without employing additional labor (son-in-law and his sons), as well as machinery (his tractor). I’m thrilled that I can proceed with `just’ 2x6s … at my disposal. Plus, I am surrounded by Southern pine …
6We could also look at it like this … each rafter carries 2 ft x 12 ft = 24 ft of roof, at 12-ft span … and each header carries ½ x 12 x 12 = 72 sf of roof (okay, not counting overhangs) … 72 = 3 x 24 … if one 2 x 6 carries 24 sf of roof at 12 ft span, then 3 of them can carry 72 sf at 12 ft span. (This only plays out with this particular framing layout.)
(c) Jeff R. Filler, 2021
… last year I built a chicken `palace’ (huge coop) for my wife, actually for her chickens. I made sure I found out how high she wanted the roof. It would slope downward, front to back, with the lowest clearance in back. She wanted to be able to stand, in back, without ducking, so I `designed’ the walls, foundation, and so on to `just give her head clearance in back. I did not plan on caring for the chickens, so my head clearance was not relevant.
Or so I thought.
The problem was, I built it. During the construction process I discovered it would have been nice if the roof had been high enough for me to move about underneath, while building it.
So this year she ordered a goat barn/shelter. I would not make the same mistake. It would also slope downward front to back, and she still wanted head clearance in back, but this time I framed it so that I would also have clearance … again, not because I planned to be in the shelter working with goats, but because I had to build the thing. I made sure the rafters, and the supporting beams and headers, were all above my head.
About halfway through the project I realized my genius! I was able to flit about unhindered! And because it was a post/pole-frame design, I could move from inside to outside, so easily, without having to `duck’ , or otherwise be `bonked’.
Only later did the `bonking’ begin, as my wife also wanted interior walls, that required various bracing schemes during their construction. The braces required horizontal pieces below the roof rafter volume, and without headroom. After getting bonked several times (both she and I), I flagged and/or orange-painted the low-hanging braces. It helped. Mostly.
Just as there are `code minimums’ for headroom in finished structures, I am sure there are also flagging or warning or barricading requirements for structures under construction, to keep people from getting `bonked’ in the head. I am not a tall person; I have never had to deal much with bonking my head on a construction project, or in a finished building. But now I am building stuff for my wife, and she’s less tall than I am
]]>The dead weight deflection of a beam, rafter, etc. is the amount the beam deflects, bends, sags, due to the `dead weight’ of the things it is carrying, including itself. Going back to the example of the rafters for my wife’s barn, we have 2 x 6 rough (actual dim) Southern Pine rafters spaced 24 inches on center, supporting 1 x 2 purling, and metal. So far I have supposed the whole assembly, rafters included, to weigh 5 pounds per square foot. Of course this is an average. Out there between the rafters is just the weight of the purlins and (thin) sheet meal roofing. And out there between the purlins is only the (thin) sheet metal. Out `there’ the weight of the roof is much less. But, if we add up the weight of the rafters, purlins, metal roofing, screws, and nails (yes, the Oxford comma), and smear it out over a per-square-foot basis, we’ll probably get (not more than) 5 psf.
Let’s look at the rafters.
They are 2 in. x 6 in., `actual’.
Now, Southern Pine (SP) lumber is `assigned’ a specific gravity of 0.55. This means that (on the average, of course, since wood is variable), (that) Southern Pine lumber weighs 55% times that of water, or 0.55 times 62.4 pounds per cubic foot (look it up if you want to) equals 34 pounds per cubic foot (pcf). BUT, the 0.55 relates to dry, in fact, `oven dry’ lumber (dried in an oven so that there’s NO water in it). In `service’, or `real life’, in-place wood will have some water (moisture) in it, and weight more than that determined using the assigned specific gravity. If we look at Page 47 in the 6^{th} edition Timber Construction Manual (TCM), Table 2.2-1, we see a specific weight for SP, at 12% moisture content (another conversation), of 36 pcf. We could use this/that number for how much our rafter weighs. Instead, I’m going to use 50 pcf, since I’m constructing this barn with `green’ lumber.
Here’s the equation …
w = γ A
where
w = the weight per foot of rafter, and
A = cross section area.
Let’s do it …
w = 50 lb per cubic ft x (2 / 12 ft x 6 / 12 ft)
w = 4.17 lb/ft.
Does this make sense? Could a 2 x 6 rafter, green (just cut from the tree), weigh 4 pounds per foot of length?
I would question 0.4 pounds, or 40 pounds … per foot, but maybe 4 is reasonable. Let’s see, … a 12 foot rafter would weigh … 12 x 4 or 48 to 50 pounds? Yeah, maybe.
Let’s calc how much it bends under its own weight.
The formula is …
Δ = 5 w L^{4} / (384 E I)
where 5 and 384 are just numbers … if you take a course in Mechanics of Materials, and pay attention, you’ll probably prove it …
Δ = deflection at midspan (that’s where it will be sagged the most)
w = weight per length, or `load’, in this case the weight is the load
L = length of the beam (rafter, etc.)
E = Modulus of Elasticity of the wood, and
I = MOI or Moment of Inertia of the wood `section’.
Sometimes (e.g., Page 595 for that same TCM) you’ll see deflection cast in terms of the `Whole’ load, W … W = w L … or …
Δ = 5 W L^{3} / (384 E I).
This is kind of nice since the `Whole load’ might be easier to get our mind around.
Let’s do it.
The `Whole load’ due to the weight of the rafter is … (at 50 pcf) …
W = w L = 4.17 lb/ft x 12 ft = 50 lb.
E = 1,600,000 psi, from before … (our concentrated load deflection calc), and
I = 36 in.^{4} …
So,
Δ = 5 W L^{3} / (384 E I) = 5 (50 lb) (144 in.)^{3} / (384 x 1,600,000 psi x 36 in.^{4} )
Δ = I get … 0.036 inches.
Hmmmm.
Gosh, that would be hard to even see! Well, yeah, … when I put that rafter up on the headers, girders, and it’s only carrying itself, I can’t see it sag … so maybe that’s correct?!!!
What if the rafter is carrying the design roof load, of 20 psf? It’s the same equation! But we need to do a bit of crunching first …
w = σ x s
w = 20 psf x 2 ft = 40 lb/ft … hmmm … about 10 times the weight of the rafter itself …
W = 40 x 12 = 480 lb.
Δ = 0.036 in. (40 / 4.2) = 0.34 in.
Seems reasonable. In other words, the sag due to the weigh of the rafter itself is only about a 30^{th} of an inch … but if we load the rafter up to the design roof load (probably a re-roofing project, with workers, shingles, etc.) … the rafter would sag some 1/3rd of an inch.
Let’s look at the deflection due to the whole `light roof’ construction.
w = 5 psf x 2 ft = 10 lb per foot of rafter length
W = w L = 12 ft x 10 lb/ft = 120 lb.
Δ = 5 W L^{3} / (384 E I) = 5 (120 lb) (144 in.)^{3} / (384 x 1,600,000 psi x 36 in.^{4} )
Δ = 0.086 inches … almost 1/10th of an inch.
Still not much. Might be hard to see. The rafter might even warp more than that!
…
Let’s check our work …
Swinging over to … http://www.timbertoolbox.com/Calcs/beamcalc.htm
(Again, I’m not vouching, necessarily, for the content on this site; I don’t know these people … I’m just seeing if they come up with the same thing I do. If they do, I feel comfy with my work. The chances of us both being the same wrong is small. If they don’t come up with what I do, then at least one of us is wrong.)
Gotta do a few calcs first …
Uniform Load … Load on Beam … notice it says `pounds’ … that means we need to use the `Whole’ load … in this case the `Whole’ uniform dead load on the beam (rafter) … W ( = w L) … 120 lb from above.
Span … 12 feet is 144 inches.
Width … 2 in.
Depth … 6 in.
Max allowable fiber stress … we’re not checking fiber stress here, but I’ll just put in, say, 999 (psi)
Modulus of Elasticity … well, to check the above hand calc, I better use the same E … 1,500,000 psi.
Maximum allowable horizontal shear … I’m not checking shear now, but I’ll just put in 180 (psi).
Show Result!
… over to Deflection … 0.0864. Yup; that’s what we got above.
Yay!
…
Oh, let’s talk a bit about … `CREEP’
`Creep’ is long-term deflection.
So, in the example above, I put up the rafter, and it `immediately’ sags a whole 0.036 inches. But if I leave it there, which I will, as time progresses, it will sag (deflect) some more. Let’s say in the mean time I get the rest of the roof up, and, under the (assumed) 5 psf roof dead load, the rafter(s) immediately sags, from above, 0.086 inches. Section 3.5 of the National Design Specification® for Wood Construction (NDS) tells us, essentially, how to handle this.
The long term deflection of a sustained (long-term) load is equal to the short term deflection of that load multiplied by a factor, K_{cr}. K_{cr }for seasoned lumber is 1.5; for unseasoned is 2.0. So, since I’m using unseasoned lumber, the long-term deflection of the rafter, carrying the `light’ roof, can be expected to be 2.0 x 0.086 in. = 0.17 in. … somewhere between 1/8th and ¼ of an inch. Maybe I could see it, but no big deal.
If I wait a while, probable at least several months, for this `creep’ thing to take effect … and climb up on the roof (assuming I weight 300 lb – the design concentrated load), and stand on a single rafter, midspan, then the total (midspan) deflection will be the long-term deflection due to the sustained effect of the dead weight of the roof, plus the immediate effect of my weight … from above … 0.17 in. plus 0.35 inches, giving 0.52 inches … about half an inch.
And if I stay up there for a long time (which I won’t!) … the rafter would creep some more, due to the sustained effect of my weight!
JRF
]]>So far we’ve been dodging the `deflection’ issue. `Deflection’ of a beam, rafter, girder, header … is how much the beam `bows’, or bends. For a beam or rafter supported at it’s ends, the deflection is generally measured at midspan, e.g., how much does the beam sag, at the middle of the span. (See sketch.)
We look at the deflection of beams, etc., generally, for the following reasons:
1) if the beam bends (deforms) too much, stuff attached to it, like a `sheetrock’ ceiling, may be damaged (cracked)
2) the beam may `look’, or `feel’, unsafe, even though it may be nowhere near the condition of actually `breaking’
3) for flat, or near-flat roofs, we don’t want to cause `ponding’. `Ponding’, or the so-called `ponding instability’ is a condition where the roof rafter, beam, whatever starts to sag, and form a `dish’, or `valley’. Then rain (or melting snow) comes along, and `fills the dish (pond)’. But rain (or melted snow) is HEAVY. So the beam sags some more. This makes a deeper bend or depression in the roof, capable of capturing more water. If the beam (rafter, etc.) is not stiff enough, the `ponding’ situation gets away from us, and eventually leads to collapse (rupture of the beam) … NOT GOOD. Easy way to deal with this … put some slope on the roof. Less easy way … do some ponding instability calcs.
4) limiting deflection can be an indirect way of mitigating potential `bounce’ (vibration) of the, typically, floor. (This goes back to 2), above.) … bounce, or vibration, of a floor, as we walk across it, makes us feel unsafe, or in the case of vibration, at least feels `weird’. Easy way to mitigate bounce or vibration … make sure the beam is really stiff … select a recommended and very restrictive deflection limit. Less easy way … do a vibration analysis.
5) other effects … we won’t get into here.
So far I haven’t been worried about deflection of the roof rafters for my wife’s barn, since 1) it will have a (flexible) metal roof, that I don’t need to worry about cracking; 2) I anticipate that the deflections of the rafters are not going to look unsafe, and I doubt they will even feel unsafe. Besides, my `customer’ on this project is my wife … no one will be up on the roof while she’s in the barn taking care of her animals; 3) I’m gonna build it with a 1/12 slope … no ponding for this roof (even though we get tons of rain); 4) I’ll either like the bounce, or not, when I’m up there; and, 5) not going to worry about other effects.
So, even though I’m not worried about deflection for this roof, let’s look anyway! …
The formula for the deflection (sag) at midspan of a beam, subject to a concentrated load at midspan (me, up there working, standing, sitting) is, without proof …
Δ = P L^{3} / (48 E I)
where
Δ = deflection (sag) at midspan (where concentrated load acts)
P = the concentrated load
L = rafter span (length)
48 = a number, dependent on the location/distribution of the load, support conditions, etc.
E = the so-called Modulus of Elasticity (of the wood), and
I = the Moment of Inertia (MOI) of the wood section.
For a rectangular section, without proof,
I = b h ^{3 }/ 12
The Modulus of Elasticity is a measure of the material (wood) `stiffness’.
Doing the calc,
P = 300 lb
L = 12 ft or 144 inches (L in inches works out better below)
E = let’s use 1,500,000 psi … (mid-range for Southern Pine lumber)
For a 2 x 6 sawn (actual dimension), `on edge’
I = 2 in. (6 in.) ^{3 }/ 12
I = 36 in.^{4}
Δ = 300 lb (144 in.)^{3} / (48 x 1,500,000 psi x 36 in.^{4} )
Δ= 0.3456 in.
Let’s look at this …
This is a tiny bit cumbersome of a calculation … so let’s first see if it makes sense. This says that when I stand out there at the middle of the 2×6 rafter, spanning 12 feet, that it will sag about a third of an inch, under my weight (assuming it doesn’t break first, which we determined it probably wouldn’t) … does that make sense? Well, thinking of the lumber, my weight, the span, … it doesn’t seem Un-reasonable. If the number came out to be 3.4 inches … I would worry about my calculations being wrong … did I put in the wrong weight? … do have the correct MOI? … did I take the length to the correct power, and so on. If I had calculated 0.03456 inches … that’s a pretty stiff beam … hmmm … only sagging 1/30th of an inch under 300 lb???
NOTE: I show 0.3456 inches … four significant figures. None of the input information is that exact, so I really can’t argue that exactness in the deflection calculation. Perhaps especially with the value for E. Lumber is a highly variable material. When I grab a (published) value of 1,500,000 psi, first of all, it’s published at only one or at most two significant figures. Plus, it’s the average of a range of E values for that species and grade of lumber. All this said … deflection calculations are not exact, even though they might look it.
JRF
]]>Let’s go over to another favorite webpage of mine. I don’t know if I have met this guy or not. Maybe, or maybe we have only passed each other in the halls of some meeting somewhere. Anyway, here’s the website:
https://jonochshorn.com/scholarship/calculators-st/example8.1/index.html
I’m going to have to `trick’ the program to give me a single load at midspan (midspan gives me the worst `bending’ condition for a concentrated load). Looking at Fig. 1, plan (b), the framing drops a single load at midspan, delivering a uniform load from the shaded area. If our rafter is 12 feet long, and they are spaced 2 ft apart, the shaded area is ½ of 12 ft, times 2 times ½ of 2 ft, equals 12 square feet. I want to model a 300-lb concentrated load, so, to use the program, I’ll spread this out of the shaded area … 300 lb / 12 sq ft = 25 psf.
Span 12 ft
Spacing 24 in.
Construction Load
no (wet service)
Southern Pine
live load 25 psf
dead load 5 psf
Grade No. 2
allow live load deflection 180
allow dead load deflection 180
Let’s look …
for b) (load at midspan) we see … Mmax 900 lb-ft … yeah, that’s what we had in the hand calcs!
Vmax = 150 lb … we haven’t talked about `V’ yet (shear) … but, basically, a concentrated load of 300 lb will tend to shear the beam with ½ of 300 lb = 150 lb toward each support.
And it gives us 2 x 6 (No. 2).
Good!
(I wish I didn’t have to trick the program to give me a concentrated load … generally the simplest solution to any problem is the best … once we start getting fancy, more complicated, or try playing tricks, we’re asking for trouble.)
Now, before we leave the page, let’s put in 20 psf live (construction) load, and 0 dead load, we get the same M = 720 lb-ft as our hand calcs. Good. If we add 5 psf dead load, we should get the same thing as AWC did … No. 2, No. 1, and SS 2×6. Good.
REALLY GOOD.
We’ll talk about shear, bearing, lateral support … in other posts.
(c) 2021, Jeff R. Filler
]]>The American Wood Council (AWC) has put together a free, online, span calculator. I know a bunch of the folks at the AWC, and they are some of my favorite people. I also like the AWC `stuff’ … (use it all the time).
The Span Calculator is available here:
https://awc.org/codes-standards/calculators-software/spancalc
I’m used to using the `2012 Version’ (the one on the left) …
Here goes …
Species: Southern Pine
Required Horizontal Span: 12 ft 0 in.
Member Type: Rafter (Roof Live Load)
Deflection Limit: L/180 … we haven’t talked about deflection yet … I will set it to the least stringent condition … L/180 means that the calculator will select a rafter size/grade that deflects (bends, bows) not more than the length/180 under the design roof load … (L/180 = 12 ft x 12 in./ft / 180 = … in this case … 0.80 in.)
Spacing: 24 (in.)
Live Load: 20 psf
Dead Load: 5 psf
Wet Service: No (we’ll talk about later)
Incised: No
Calculate Span Options:
Gives … SP Select Structural 2×6, No. 1 same 2×6, and No. 2 same 2×6.
That’s marvelous!
That’s what I (we) came up with earlier … the Select Structural (SS), No. 1, and No. 2 are relatively high grades, which is what I (we) expected with the dim lum sections.
NOTE: it does not appear that the Span Calculator took into account the 300-lb concentrated load, which, we showed earlier, `governs’ the design. It could be argued that the 300-lb load need not be entirely carried by a single rafter. I could also argue that the structure in question is `unregulated’, so why the worry??? I’ll tell you the worry! … I’m gonna be up there building the thing, and at times I will be sitting, or standing on, a single rafter!!! I’ll snoop around and find some other (independent, and free, online, for kicks and giggles) means of checking the concentrated load condition (other post).
JRF
]]>(c) Jeff R. Filler, 2021
The use of building products in construction that are not described in the building code, or not described completely in the building code, generally requires the specific approval of the building authority (department) having jurisdiction over the building project. The approval of such products is greatly facilitated with a code evaluation report of the product. A code evaluation report (code report or evaluation report) is generally provided by an agency qualified to evaluate such products (such as the International Code Council Evaluation Service, ICC-ES). The evaluation report is particularly valuable as a single evaluation report of the product issued by the evaluation agency can generally be presented for product acceptance to any building department functioning under the building codes to which the product was evaluated. The evaluating agency generally relies on evidence provided by the manufacturer, for review by the agency, with respect to compliance with the code. The evaluating agency generally charges fees for such evaluations, and the cost of obtaining the evidence needed for review, are carried by the product manufacturer. The time and expense involved in the evaluation process are outweighed by the benefit of obtaining a product evaluation report that may be used for multiple projects under multiple jurisdictions. To avoid unnecessary delay and expenses with the evaluation process, seekers of code evaluation reports can benefit from the following so as to have the evaluation process, ending in successful completion (publication) of a code report, go as smoothly, and swiftly, as possible.
1. Engage the evaluation agency
2. Execute necessary administrative actions
3. Test and evaluation plan
4. Utilize accredited test laboratories (where product testing is required)
5. Retain competent professionals (where analyses, computations, or third party conclusions are advised or required)
6. Provide a complete and well-organized submittal of needed documentation
The evaluation process may be punctuated with time required for product testing, and obtaining documentation from any necessary third parties. The evaluation may also be drawn out if acceptance criteria need to be developed, or existing criteria revised. As such, it is crucial that, as early as possible, the manufacturer (or product sponsor), and evaluation agency, agree upon what acceptance criteria are applicable, and the plan in which the requirements of the acceptance criteria will be fulfilled.
JRF
]]>Linda wants a barn-shelter for her goats. Okay – I’ll do it. `Twelve by twenty-four.’ `Let’s go walk the site. It’s got some slope. I’ll make a stepped-barn, each section will be twelve by twelve’.
We have rough-sawn 2×6 pine all over the place; let’s see if I can do it (the ROOF) with 2×6 rough sawn!
What are the design conditions?
The roof will need to hold up construction workers (me!, alone, putting up the framing, and, later, a team, putting on the final metal roof).
The roof will also need to withstand wind loads … typically an uplift load … the air coming into the barn (open front) and trying to pry off the roof!
Let’s do it!
The 2×6 is actual dimensions … 2 in. thick x 6 in. deep.
Species – (southern) pine!
Let’s space the roof rafters 24 inches apart (`on center’ … or `o.c.’)
Span will be 12 feet, actually a bit less, as the support posts are 12 feet apart, outside to outside … but we’ll use 12.
We’ll have 12 in. overhangs at both ends. We’ll ignore the overhangs (counter-bending) for our `bending’ calcs.
The Code-required construction loads are … a 300-lb construction worker, concentrated load, or 20 pounds per square foot (psf) uniform load.
Plus, of course, the roof rafters have to carry the weight of the roof itself.
First let’s see what `governs’ for the two construction loads above …
Case 1 – Single Construction Worker – Concentrated Load
The `worst’ bending effect of a construction worker will be if he/she is at the middle of the 12-ft span. In this case the bending effect (`bending moment’) in the beam will be …
M = P L / 4
where
M = bending moment (effect)
P = the concentrated (`person’) load (Code says 300 pounds – a worker and stuff)
L = beam span-rated, and
4 = the number 4, shown here without proof.
So,
M = 300 lb (12 ft) / 4 = 900 lb-ft … this is the bending moment, or effect, in the beam, at mid-span, due to a 300-lb load (person) at mid-span.
Case 2 – Uniform Load of 20 psf
If the rafters are 2 feet apart, then every foot of rafter (length) carries 2 feet of roof (one foot on either side, or halfway to adjacent rafters). In equation form …
w = σ x s
where
w is the `line load’
σ is the `area load’ or pressure (in this case 20 psf of construction activity),
x is the multiplication symbol
s is the rafter spacing …
So,
w = 20 psf x 2 ft = 40 lb / ft … every foot of length of beam carries 40 pounds …
(Think on this for a bit … if this doesn’t make sense, there is no need for you to go further.)
The `Whole’ load on the rafter, W, is …
W = w x L …
W = 40 lb/ft x 12 ft = 480 pounds.
Let’s make sure you follow this … every rafter carries 2 ft x 12 ft of roof … or 24 square foot. If each square foot of roof is loaded with 40 pounds … then each rafter carries 24 x 20 = 480 lb.
The bending effect of a `uniform’ (uniform along the length) load is …
M = W L / 8,
In this case …
M = 480 lb x 12 ft / 8 = 720 lb-ft.
Sometimes you’ll see the bending moment expressed as …
M = w L^{2} / 8 ….
(Little) w vs. (big) W.
Gives the same thing.
Again I show the `8’ here without proof … but you’ll see it in engineering textbooks, design aids, etc.
NOTE: even though the `whole’ load, 480 lb, is more than the `person’ load, 300 lb, the 300 has a greater bending effect (900 vs 720) because it is `concentrated’ at the middle of the span! For other effects, such as shear and bearing, we need to come back to the `480’.
When we design wood members (beams, rafters, joists, etc.) we generally look at the `stress’ the effect, in this case the bending effect, has on the wood fibers. For a member in bending, the `fibers’ on the bottom side of the beam are stressed in tension, and the fibers on the top are stressed in compression. In terms of numbers, for a rectangular-section member, the bottom and top stresses are the same (number) … in this case …
f _{b} = M / S
where
f _{b} = bending stress (tension on one face, bottom; and compression on the other, top)
M = bending moment, from above
S = the so-called Section Modulus.
Without proof … for a rectangular section
S = b h^{ 2} / 6
where
b = rafter thickness (typically the narrow dimension, rafter on edge)
d = depth, and
6 shown (again, without proof).
I like to think of S in terms of the cross section area (A) of the rafter (beam, girder, joist, whatever);
A = b x h
So,
S = (b h ) x h / 6
S = A d / 6.
In the case of a 2 x 6 on edge (actual dimensions) …
S = (2 in. x 6 in.) x 6 in. / 6
S = 12 in.^{3}.
The bending stress is …
But, wait … you’ll see our `units’ get messed up if we don’t first get the moment into (units of) lb-in. …
M = 900 lb-ft x 12 in. / ft = 10,800 lb-in.
Now, then …
f _{b} = M / S = 10,800 lb-in. / 12 in.^{3}
f _{b} = 900 lb / in.^{2 } … (900 psi).
YIKES! … I wish the numbers had come out a bit different … the `900’ lb-ft only gives `900’ psi because we happened to multiply by 12 feet, and then divide by 12 in.^{3.}. A rafter with a different Section Modulus would give a different stress.
(Let’s say we use 2×6 dimension lumber, from `Lowes’ (or Home Depot, or local lumber yard). For dimension lumber, the 2×6 are `nominal’ dimensions; the actual dimension are … 1.5 in. x 5.5 in. … maybe even a tiny bit less. The Section Mod for these dim lum rafters would be … S = 1.5 x 5.5 x 5.5 / 6 = 7.563 in.^{3}. Less wood … less Section Modulus … trying to carry the same load … more stress … let’s see: f _{b} = M / S = 10,800 lb-in. / 7.5625 in.^{3 }= 1428 psi … yup.)
I need rough sawn 2×6 that can carry at least 900 psi … (or dim lum 2×6 that can carry 1428 psi).
…
But what about the weight of the roof?
This is going to be a `light’ roof … without going through all the fuss … it will probably weigh 5 psf or less, including the rafters. I guess I should add that in. We’ll treat it as an area load, crunched into a line load, like we did above with the `20’. If we got a bending moment of 720 lb-ft from 20 psf, then 5 psf would give us … 720 (5/20) = 180 lb-ft … should be added in.
BUT! … I’m the only construction worker who’s gonna be up there, and I don’t weigh 300 lb. I’m not going to tell you what I weigh, but let’s say me, plus hammer, lot of nails, boots, clothes, etc. … takes me up to … 250 lb. The bending moment due to `me’ at midspan will be …
M = P L / 4 = 250 (12) / 4 = 750 lb-ft … down from 900 an amount of … 900 – 750 = 150 … about a wash! … 180 lb-ft more vs 150 lb-ft less … and the 5 psf and the 250 lb were just `guestimates’. If I really want to split hairs, I can calculate the finish roof weight, and I can also strip down and carry fewer nails with me up on the roof.
So, yeah, I need 2×6 rough sawn (actual dims) rafters that can carry 900 psi. This is doable.
We’ll `make sure’ we get 900 psi on another post. (Or, if we want to use dim lum, that will also be on another post.)
…..
We’ll also look at wind, later.
JRF
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