(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, a*nd 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

]]>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

]]>

“Let’s take a walk.” (We’re walking toward *my* shop.)

As it happens, the trusses for my shop span 40 feet, the roof slope is maybe close to 4/12, the members are sawn lumber, the connections are steel plates, and we’re in the same `locale’. Looks like 2 x 6 bottom and top chords, 2 x 4 webs, metal plates both sides of connections. The trusses are spaced 4 feet on center.

“Easy answer? (now we’re in the shop, and as I look up, my body language beckons that he do the same) … build your trusses just like these, except *double* them at the supports.”

“If a single truss can carry 4 feet of roof, all things equal, then two of them (doubled together) can carry 8! … just make 2-ply trusses; each ply like the ones you can see here! Set them on the posts! Done.”

Miles: “I want a single-ply truss to drop between the laminations of a 3-ply post.”

(I tried to talk him into the 2-ply idea, without success.)

“Okay, my guess is … 2 x 8 rough sawn top and bottom chords, and more metal (probably *twice as much*) for the plates.”

Let’s do it.

(Selah)

Following are hand calcs to `solve’ Miles’ truss. I start with basic engineering mechanics … Method of `Pins’, or `Joints’, as taught in a typical 200-level college engineering course. The `connections’ of the lumber pieces of the truss are modeled as though smooth pins, or single through-bolts, with no resistance to the pieces rotating (in the plane of the truss) at the joints. Each joint (or pin) is analyzed for equilibrium under the actions of the adjoining lumber pieces pulling or pushing on the pin. Solution of the forces acting on each pin thus, by Newton’s Third Law, solves for the forces in the wood members … `pulling’ being tension, and `pushing’ being compression. I am taking you through a hand calculation for several reasons: 1) for my own practice; 2) for your illustration (especially if you’re a college student looking on); and 3) so I can use it to check more elaborate computer solutions of the same truss, or variants, later on.

Some notes getting started …

I am going to start with truss spacing at 4 feet, like in my shop, so that I can compare results on paper, with what is `real life’ over my head.

I am also going to use loading conditions presumably similar to those used (by someone, not me) for the original design of the shop trusses, specifically:

Downward (gravity) load = 20 pounds per square foot (psf), minimum roof construction live load, plus 5 psf dead load (weight of trusses and roofing materials … a light roof) = 25 psf.

This (25 psf) number is also a good starting place for a wind section uplift load … say 30 psf of pure wind uplift (suction) as a 90 mile-per-hour wind races over the roof, and the roof wants to fly away like an airfoil … *minus* the weight of the roof (5 psf) … giving 25 psf. Thus, for a `starting place’, the solution using 25 psf downward will also give me the solution for uplift, just with everything in reverse (members in tension under download will be in compression with uplift).

Pages 1 and 2 …

First I look at the loads as they apply to the truss. For the Method of Pins, the loads must be applied (also) at the pins. This might indeed be the case if the purlins holding up the actual roof material attach at the joints, or connections; otherwise applying the loads at the joints is a good (and necessary) first approximation. Each pin will carry the load (25 psf) times the tributary area (plan view) of roof carried by the pin (4 ft x 2 x 1/2 of 10 ft or 40 square feet) = 25 psf x 40 sq ft = 1000 pounds … say `P = 1000 lb’.

I look at the truss as a whole … the whole truss carries 1/2 P at one end, P at the three interior joints, and another 1/2 P at the other end … or 4P or 4000 lb. The truss is symmetric; the loading is symmetric, and the support conditions (walls at ends) are symmetric … so I could argue that the support (reactions) will be equal and up to 4P, or be 2P each … or I can go ahead and prove it (as in the calcs).

And then I start in on the Joints …

Note that I show force AD `pushing’ on Pin A. Thus, Pin A will be pushing on Member AD. This makes sense. A truss carrying download will tend to have the top (top chords or members) in `compression’, and bottom chords (bottom of truss) in tension. Had I computed a negative number, it would have immediately `raised a red flag`. The `(C)’ stands for `Compression’.

On across the truss …

Note that I (you) can only solve the truss by attacking, one at a time, joints that have no more than two unknown members. To solve a joint with, say, three members, you (I) need to first get the force in at least one of the three members, before I (you) can use the joint to get the other two.

Note at Pin D I use a `rotated’ set of axes, to get DE directly. A bit more geometry, but at least I don’t have to solve two equations for two unknowns. Just my preference.

Yikes … on Page 4 you see some `red’. That’s because, on the first time through (`blue’), by the time I got to Page 5, things weren’t coming out quite right … my original calcs (on my cell phone) were a bit sloppy … the `red` numbers are corrected.

And on Page 6 you’ll see I do a check using the Method of Sections …

Now that I have proved to myself that I can still do the Method of Joints/Pins, by hand (and using the calculator on my cell phone, if necessary), I will leave it up to you to agree, or not, that the rest of the truss is a mirror image of the side I have now solved for.

(Rest)

Oh, let’s see if we’re in the ballpark with stresses in the lumber. The bottom chord of the truss being examined carries 4500 lb. The bottom chord of the truss in my shop will also need to carry, presumably, 4500 lb, and it’s a 2×6, dimension lumber.

The tension stress (f_{ t}) in the bottom chord will be the tension force (T) divided by the cross section area (A) …

where A = 1.5 in. x 5.5 in. = 8.25 sq in. (where 1.5 and 5.5 are the actual dimensions of the 2×6 nominal)

… f_{ t} = T / A = 4500 lb / 8.25 sq. in. = 545 pounds per square inch (psi).

The number (545 psi) appears reasonable for a working tensile stress of typical sawn dimension lumber.

(Stresses in lumber to be discussed in greater detail … MUCH GREATER DETAIL … in other posts.)

NEXT …

I decided to put together a quick spreadsheet, to be able to `tweak’ things, such as roof slope, truss spacing, and so on. The spreadsheet follows the hand calcs, so one checks the other.

Here it is for the trusses spaced 4 feet apart, with roof slope 4/12.

And, so now what’s cool, is that I can change things … for example, the truss spacing.

I don’t show the spreadsheet here, but, as one might guess, with the truss spacing doubled, everything else doubles.

I get a tension load in the bottom chords of … 9000 lb. If we guess that 2×8 dim lum would be required …

… f_{ t} = T / A = 9000 lb / (1.5 x 7.25 sq. in.) = 828 psi … that’s probably `workable’.

If Miles wants to use 2 x 6 rough …

… f_{ t} = T / A = 9000 lb / (2 x 6 sq. in.) = 750 psi … that’s probably doable.

If Miles wants to use 2 x 8 rough …

… f_{ t} = T / A = 9000 lb / (2 x 8 sq. in.) = 563 psi … that’s doable, but a heavier truss.

(Pause)

Now, what’s interesting, is that I went out and measured the rise/run for the trusses in my shop.

Turns out they are more like 3/12 … (56/240-ish) … and thus *shallower.*

The shallower truss will place a higher demand on the top and bottom chords.

Running through the spreadsheet … yeah, the bottom chord, instead of carrying 4500 lb under design load, carries 6429 lb.

The corresponding stress is …

… f_{ t} = T / A = 6429 lb / (1.5 x 5.5 sq. in.) = 779 psi … probably still `workable’.

Spreadsheet calcs follows:

So, at this juncture, it looks like Miles can probably use 2×6 or 2×8 rough sawn for the top and bottom chords … depending on the `grade’ of the lumber.

… the loads in the members are …

Bottom Chords AC, CE, EG, and BG … all 4500 lb, TENSION

Top Chords AD and BH, … 4743 lb, COMPRESSION

Top Chords DF and FH … 3162 lb, COMPRESSION

King Post EF … 1000 lb, TENSION

Webs DE and EH … 1581 lb, COMPRESSION

Webs/Posts CD and GH … 0 lb (under the method of pins, simplification)

More to follow.

JRF

]]>

Pictured: Fitsum Peak (Main Block), Central Idaho, (c) Jeff R. Filler

]]>