(not the actual house … but close!)

Woeste and Dolan did a paper on floor bounce … wait, no, on floor vibration (copy here). They provided an equation for determining the fundamental frequency of a floor joist, as follows:

f = 1.57 √ (386 EI/WL^3).

Further, their paper says that occupants are the most sensitive to vibrations in the 7 to 10 hz range.

Plugging in the old-house joist dimensions, and span, presumed modulus of elasticity, number of joists involved at any one time, and various weights of movie-set `occupants’ and stuff (for mass) … yeah, it was pretty easy to `land’ in the 7-10 range.

I advised that the floor was indeed safe, despite the bounce … “as long as you don’t put a crazy huge amount of stuff up there … if you want to shoot your movie up there, you’ll have some bounce.”

…

I tried to reconcile the bounce I felt due to the sheer effect of walking across the floor. I `calculated’ that my walking was, say, 1 to 2 hz (steps per second) … not exactly in the 7 to 10 range … but maybe a harmonic of my stepping. Or, each one of my steps engaged more than one joist!

…

To date, that’s been the extent of my movie-set engineering, except for some quick advice on what factor of safety to use for stage rigging.

Over the years I never did see very many 2 x 8 floors. Maybe `bounce’ is why.

I you have a movie for me (engineer, cast, or crew) … let me know!

]]>Consider a conventionally-framed house facing a remodel project. The house is, for the most part, a simple, two-story `box’, with stick-framed roof and interior bearing wall. The interior bearing wall supports both roof and main story floor. Two walls are to be removed per the remodel; one wall is a segment of the interior bearing wall, and the other a non-bearing wall. The interior bearing wall runs longitudinally through the house, roughly down the `middle’. Ten or so feet of the wall are to be removed, opening up the dining room to the kitchen. The segment to be removed, being on the main floor, supports only the roof and ceiling. The non-bearing wall to be removed is between present dining room and living room. Non-bearing walls will be discussed separately. The idea here is to see how we can determine what kind of beam, or header, is needed, to carry the roof and ceiling over this new space, created by removing a part of the bearing wall. A cross-section sketch of the house framing is shown below.

The `easiest’, and perhaps most expensive, way to determine what we need, is to `hire an engineer’. I am such a person. But why? I’m retired (mostly), and since the house was originally designed without an engineer, as are many, why hire an engineer to help with a remodel?!

First, let me say, that you may *have* to hire an engineer. Second, you may *want* to hire an engineer, whether or not you `have’ or `need’ to. Let me explain. A bearing wall is a `structural’ part of the home; it holds part of the home up. Do it wrong and your house may collapse, or look like it’s collapsing. You may *want* to hire an engineer, just for peace of mind. You may *have* to hire an engineer … if the building regulations for your locale require it! This would typically come about via the permit process. You may want, or have to, get a building *permit* to do the remodel. And to be granted the permit, if dictated by the building official, you may have to hire an engineer. In some cases the local building official himself/herself might determine what kind, size, etc. header is needed, or might have design aids for such, but don’t count on this^{1}.

The contractor you choose to do the work (if not yourself) might be willing to determine, or `size’, the header for you. Careful with this one, unless he or she plans to hire an engineer, and pass the associated fees as part of the remodeling contract. While good contractors can do amazing things with buildings and building materials, they typically do not have the science background to properly `engineer’ a solution. And by this I mean a long-term solution … a header that will hold up the roof, not only for the duration of the construction project, but for the `life’ of the structure itself. But, again, if we can come up with a solution without hiring an engineer, let’s do it. And if a permit is required, perhaps your`non-engineered’ solution will make enough sense to the building official, that he / she agrees, and issues the permit. (Yay!) After all, all the headers in the existing building got past having to involve an engineer; why not one or two more!?

Let’s look at some ways to do so.

1. Provisions in the *International Residential Code* (IRC) … online version here. I will be referring to the 2018 edition. Further, the IRC was likely the basis of the building code governing your original structure when first built anyway (if governed at all).

**R602.7 Headers.** For header spans, see Tables R602.7(1), R602.7(2) and R602.7(3).

Table R602.7(1) is for exterior walls … R602.7(2) is for interior bearing walls … and R602.7(3) is for open porches.

The problem with Table R602.7(2) is that it doesn’t seem to accommodate any roof load.

Let’s look at Table R602.7(1) … and … R602.4 …

**R602.4 Interior load-bearing walls.** Interior load-bearing walls shall be constructed, framed and fireblocked as specified for exterior load-bearing walls.

So, let’s use Table R602.7(1) to come up with our header. I suggest that the braces direct the roof load in such a way that the interior bearing wall carries *more* than half of the roof … all of the roof between the braces, and half the spans from braces to exterior walls, or, 5 feet between the braces + ½ of ½ of (30 – 5) on the left and ½ of ½ of (30 – 5) feet on the right, or … 5 + 2 x ½ x 12.5 = 17.5 feet (ft) (of building width). This would correspond to an equivalent (more simple) building width of … 2 x 17.5 ft = 35 ft. We’ll use the first set of `spans’ in Table R602.7(1) … `headers supporting roof and ceiling’. The house is in a region of `no snow’, so, per footnote e to the Table, we’ll use the 30 psf Ground Snow Load. Recall that we want an opening (span) of (about) 10 ft. For a building width of 36 ft we get … 4 – 2 x 12s give us 9-10 (9 feet 10 inches). We should also look at the footnotes, but, before we go any farther … note the `4’. This means FOUR 2 x 12s, each of which is 1.5 inches thick … or a header of 4 x 1.5 = 6 inches. The resulting header will be 6 inches wide (thick) by almost 12 inches deep. That will be hard to hide in an interior bearing wall. Yeah, our building `equivalent’ width is 35 ft, not 36, so maybe we could split some hairs and go a bit smaller (by `interpolating’, per the footnotes), but we won’t be able to reduce it (the header requirement) by much. Even if our building were only 24 feet wide, the requirement (without the weird braces) would be 3 plies of 2 x 12s. Yikes. (That’s why big openings in bearing walls in conventional construction are rare.)

2. Let’s look at provisions in the *International Building Code* (IBC) … online version here.

Chapter 2308 of the IBC addresses conventional light-frame construction … (our stick-framed house) … but we end up in the same dilemma, exactly … 4-ply 2 x 12s. See Table 2308.4.1.1(1). Ugh.

3. Try the *Wood Frame Construction Manual* (WFCM), which is referenced in the IRC, as applicable to (usable for) residential construction (per Section R301.1.1).

Same issues … we don’t see anything directly giving us a solution for an interior bearing wall carrying roof/ceiling only. Tables 3.22A1 and A2 give us options for exterior wall, roof/ceiling only, with the difference in the Tables being dropped header, or raised. A dropped header has a top somewhat below the wall top plates, necessitating cripple studs, which can’t be counted on as providing so-called lateral stability for the top of the header (I’ll explain elsewhere). A raised header is mounted right up under the roof/ceiling construction, stabilizing the top of the header, and is the case of the remodel considered. So we’ll go with A2 … and we see `solutions’ such as 4 – 2 x 12s, roof live load 20 psf, can span 10-8, for building width of 36 feet. We also see 4 – 2 x 10s can span 9-1, and so on … not a whole lot different than what we got with the IRC and IBC.

Big opening … big header required.

Oh, the *Wood Frame Construction Manual* is also available online … here. (I link to the 2018 version, but newer is available, as per the IRC and IBC.)

Let’s look at WFCM Chapter 2. Chapter 2 gives us loads and stresses, instead of specifying the actual header makeup. Look at Table 2.11, upper right. There’s an interior bearing wall holding up roof only! Yay! For a roof live load of 20 psf, building width of 36 feet, the Table gives `720’. This is the unit load on the header or beam or girder. For a building width of 24 feet, we get 480. Going back to the 36 width, then, the 720 means that each foot of header (length), no matter how long, carries 720 pounds. If the header is 10 feet long, the total amount of roof/ceiling load is 720 x 10 or 7,200 pounds (lb). This 720 unit load is also called a `line load’, or pounds per lineal/linear feet, or plf, etc. This doesn’t help us much directly, but can be tremendously useful as we look to some further tools. This number enables us to `engineer’ a header.

Note that if we had a shorter amount of wall removed, we’d be done … try 6 feet in the examples above. Go back and look at Table 3.22A2 in the WFCM … a 2-ply 2 x 10 would carry an opening of 6-3 in a 36-ft wide house. Looking at IBC 2308.4.1.1(1) we see a 2-ply 2 x 10 can carry 5-9, and 2-ply 2×12 … 6-10. Perhaps the difference is that the 2308.4.1.1(1) uses 30 psf snow load for both it and 20 psf live load. And perhaps an eave is considered. Or perhaps there are other effects mixed in (wind?) We’d look closer at the line items in the Manual (or code), and perhaps Commentary(s), if the 6 foot opening is really what we are designing. The IRC comes up with the same thing as the IBC.

4. Design aids provided by industry / manufacturers.

Already you might have realized, justifiably, that dimension lumber isn’t going to work, unless we want a fat header. So let’s look to `engineered’ wood. There are several options. We can go with structural glued-laminated timber (glulam), or structural composite lumber (SCL). And there are others. But, sticking with either glulam or SCL, let’s see what we can find. Let’s see what the glulam people have to offer … the American Institute of Timber Construction (AITC) (on www.aitc-glulam.org). (You’ll get redirected to www.plib.org/aitc as AITC was absorbed by the Pacific Lumber Inspection Bureau (PLIB), and the West Coast Lumber Inspection Bureau (WCLIB) along the way.) Look under Resources, then Publications … and then Southern Pine, and then Roof Beam Construction Load. (Construction load governs where snow loads don’t … consider it the Roof Live Load.) And let’s open up SP 26F – 1.9E, and we have options. Looking at a span of 10 feet … and something that carries (at least) 720 plf, we see … 3-1/8 x 8-1/4 carries 768 B. The `B’ means bending (stress) controls. Whoa, nice … the 3-1/8 width certainly fits in the 2×4 interior bearing wall dimension. And 8 inches is pretty modest. But, we’re not done; we need to be a bit careful. The Table says that the deflection limit is Span/180 for total load. Deflection is a different conversation. But, for now, let’s look back at IRC Section R301.7 and Table R301.7. The Table indicates a deflection limit of L/240 for members supporting flexible finishes, including gypsum, and R301.7 says the limit is based on the … live load of Section R301.6. There is a disconnect … L/180 for `total’ from AITC, and L/240 for `live’ (our construction load) from IRC. We can reconcile the two using Table 1604.3 of the IBC … which relates the L/240 for live load and L/180 for live load plus dead (equaling total).

So, a 3-1/8 x 8-1/4 SP 26F-1.9E (Southern Pine) glulam beam is adequate for the header! (Show that to the building official.)

Now note! … these answers are bare-bones minimums! The AITC Table indicates that the `solution’ is based on bending stress (the `B’). But if we look one column to the right, for the same beam, and an 11-foot opening, the allowable load is 618, and `D’ … that means it’s deflection-controlled. I would argue that the 10-foot span is `almost’ deflection controlled. That would mean that, under the 768 plf, the beam would be expected to have a total deflection of `Span divided by 180’, or 10 feet x 12 inches per foot / 180 = 0.67 inches … about 5/8 of an inch. It might `meet code’, but is that really what you want (to see) above the opening between the kitchen and dining rooms? You might want it to be less, maybe way less! What if we want to drive it down to, say, ¼ inch total. Well, a wood beam is like a spring. We would need a heavier (stronger) spring. Look at the 3-1/8 x 11; it can carry a load of 1365 plf, and is controlled by bending stress. That suggests that, under the 1365, the deflection is not yet at the (Span /180) limit. Maybe significantly under the limit. Since the 3-1/8 x 11 also acts like a spring, under only 720 plf instead of 1365, it would have a total deflection of not greater than 0.67 x 720 / 1365 = 0.35 inches. (Half the load … half the deflection!) Actually less (since bending controls the 1365, not deflection). Maybe way less. (The actual engineering `calc’ not shown.) Turns out to be almost exactly 0.25 inches. Sweet. Generally it’s not a problem going stiffer and stronger than the code minimum(s). Obviously the 11-inch beam will cost more (and weigh more). In a remodel the only issue that we might run into with a deeper beam is `headroom’ … make sure there’s enough clearance under the beam / header to `meet code’ (and your needs … maybe you’re a tall person!).

Let’s try a different product … say Laminated Veneer Lumber (LVL), a type of SCL.

I found that a Roseburg 2-ply 1-3/4 x 9-1/4 2.1E will carry 1307 total (non-snow) load over 10 feet. The 2-ply 1-3/4 is sweet, as the total thickness will be 3.5 inches … perfect for an interior wall (typically 2×4 nominal). Engineered lumber is typically stronger, and stiffer … alas the smaller dimensions. (In the project considered, 2-ply 1-3/4 x 11-1/4 2.1E LVLs were used, and the opening enlarged a bit, to 11 feet). Note that when selecting engineered lumber, you need to make sure it’s `available’ in your locale. Actually, it’s the same with dimension lumber, when dealing with specific species and grade. Southern pine lumber, and engineered lumber made from Southern pine, is generally available on the `South’ (east); Douglas fir, and products made from Douglas fir, are available in the West. And/but there are other lumber species … Hemlock, other firs, cedars, and so on, and some lumbers are `mixed’, e.g., hem-fir, which is a mix of Western Hemlock, and firs. (Just make sure what you want is available, or, find out what’s available, and design within.)

5. Some straightforward calculations and common sense.

Go back to the discussion above where we jump into Chapter 2 of the WFCM, specifically, Table 2.11A. Note the Table heading says Roof Assembly DL (Dead Load) = 20 psf. This is the assumed weight of the roof and ceiling materials (including insulation, whatever). Under RLL (Roof Live Load) of 20 psf, and following the 36-foot width over to the upper right, where it shows a header (or beam) carrying the middle of the roof, we got (get) 720 plf. If we go up to 24-foot building width, we get 480 plf. Now let me show you something. Maybe you already see it. We could say this header (the one in the WFCM graphic) carries half the roof. Or we could say that it carries half the way from the peak to the exterior wall, both directions. Same thing. For a building width of 36 feet, the interior header (or wall) carries an 18-foot wide `swath’ of roof (9 feet each side of the ridge). Each foot of wall (or header) length thus carries 1 ft x 18 ft or 18 square feet of roof. If the roof weighs 20 pounds per square foot (psf), then the wall (or header) carries 18 x 20 = 360 pounds of roof, per foot of wall (or header). Similarly, if the live load on the roof is 20 psf, each foot of wall (or header) carries 360 pounds. Add the two together, and the header (or wall) carries 720 plf, under the design condition of dead weight (or course), and live (probably a re-roofing crew). Yeah, it’s that easy! Similarly, for a building width of 24 ft, the interior bearing wall (and any headers therein) carries ½ of 24 ft x (20 + 20) psf = 480 plf. Determining weights of the material of the roof (dead loads) is not particularly difficult, and the design loads (live, snow, etc.) are typically provided by the local building code … so sizing any header, now, unless the building is unduly complicated, is … doable!

Let’s say we have a building that is 32 feet wide, and the roof is constructed of pre-manufactured trusses (allowed in the IRC, IBC, etc.), that span wall-to-wall. Let’s say we have a design snow load of 60 psf. And we determine that the roof weighs no more than 15 psf. Oh, and let’s add 12-inch eaves. Each exterior wall carries ½ of 32, plus 1 ft, or 17 ft. The total load on the exterior wall (and any headers therein) is 17 x (60 + 15), or 1275 psf. Let’s go back to the AITC Beam Capacity Tables, but this time find one for beams made of Douglas fir, and let’s say we have an opening of 12 feet. Table AITC DF-26 Roof Beams Snow Loads … the 3-1/8 x 12 will carry 1497 plf for 12 feet. Alternately, the 5-1/8 x 12 will carry 1572 plf for 12 feet. Either would work. The 3-1/8 x 15 weighs 11.4 pounds per foot, and the 5-1/8 x 12 weighs 14.9 pounds per foot. Other things equal, the heavier beam will likely cost more. On the other hand, if we’re fighting 60 psf of snow, we’re probably also dealing with 6-inch exterior walls; the 5-1/8 inch wide beam might fit better. Ask the manufacturer; they may well offer widths of 5.5 inches, fitting even better in 2×6 walls. Notice again that the deflection limit is L/180; we could go wider or deeper to get an even `bigger’ (stiffer) beam with less deflection. Notice also that both sizes are controlled by bending stress; thus we know that the deflections will be *less* than the L/180 … how much less we don’t know without some more work. Finally, and importantly, the footnotes indicate that the beams must be supported laterally along the top. This is typically by attachment to a double top plate to which the trusses are attached … holding the top of the beam in place, from buckling out of the plane of the wall. This is the `raised’ header. If the header is lower, a `dropped’ header, with cripple studs from header to the top of the wall, the beam is, strictly speaking, not laterally supported, and there will be lessened bending capacity. A `bigger’ beam may be required. But the bigger beam will give us more deflection control. Some design aids accommodate for the dropped header, either by providing additional tables, such as we saw in the WFCM Chapter 3 Tables, and some aids provide adjustment factors for the tabular values. Otherwise more difficult (`engineering’) calculations may be necessary to deal with the unbraced beam condition. And, of course, make sure this beam is even available in your locale.

So, with this start, you can see some ways to `design’ a header for an interior load-bearing wall. The idea was to avoid hiring an engineer for what are typically non-engineered buildings (residential construction). And we could take what we did here as a start into a plethora of additional ways to design headers … using various online resources, calculators, software, and so on. We could have switched over to steel, and specified a steel beam for the header. Switching to steel might be the only way to deal with high loads and long spans. And, interestingly, though not surprisingly, if you think about it, various manufacturers have design aids by which you can `switch’ from one material to another (typically* their *material) … say a parallel strand lumber (PSL) beam (a type of SCL), to glulam. Or steel to glulam, or … the information out there is endless. No doubt, somewhere, is a table trading dimension lumber for SCL.

At some point, yeah, hire an engineer.

1If it is necessary to involve the building department, make the building official your friend. Who knows what he / she might come up with to assist you on your project.

Aha!, yes, design values for E are average values. Modulus of elasticity is a measure of wood *stiffness.* Wood stiffness is related to how much wood deforms when subject to load, commonly, how much a beam (or joist or rafter) will `sag’. Engineers call this sag `deflection’. We have fancy equations for deflection. The deflection (sag) of the midspan (relative to the ends) of a simply supported beam (or joist, rafter, girder, or similar) is given by the equation:

Δ = 5 W L3 / (384 E I ),

where,

W is the `whole’ load on the beam (joist, rafter, etc.),

L = beam length,

E = the modulus of elasticity of the beam (more to say here),

I = the moment of inertia of the beam, and

5 and 384 are just pure numbers, related to how the load is distributed and the beam supported.

The load W is generally `predicted’, the result of some future event, say a `once-in-a-lifetime’ snow storm. For us engineers this future event is typically predicted by others, and we work off of some amount (e.g., of snow) associated with the event. The number we work with is a prediction, even though it may look like it’s exact.

The beam length, L, and moment of inertia, on the other hand, are more exact. We know pretty darn close how long the beam is (or is to be). The moment of inertia is based on the beam (or rafter, or …) width and depth. We can actually measure these with a tape measure, pretty close, or we can just rely of the dimensions stated (by the manufacturer), plus or minus a bit of tolerance (allowed in manufacturing).

E likewise might very well appear to be exact, such as, E = 1,800,000 psi (pounds per square inch).

Actually, all the zeros betray that, perhaps, only the 1 and the 8 are significant. But, importantly, remember, E is average. Actual E values, board to board, beam to beam, rafter to rafter, etc. … vary. Even for boards, beams rafters of the same species, and grade, and even from the same manufacturer, or even from the same day of* manufacturing.* A common way to describe this variability is through the coefficient of variation, C.O.V. (or COV). It is defined as the standard deviation divided by the average. For a certain beam with a design value for E of 1,800,000 psi, and COV or 0.10, the standard deviation (measure of variability) of the E’s of individual beams of this species and grade is … 0.10 x 1,800,000 psi = 180,000 psi. (As AITC 119 Section 4.1.6 points out) … in a normal distribution, 2/3 of the actual E values `in population’ would be within 1,800,000 plus 180,000 and 1,800,000 minus 180,000 psi. Or, 5/6 of the E’s would be greater than 1,800,000 – 180,00 psi = 1,620,000 psi. Or, half of the E’s would be less than 1,800,000 psi, and 1/6 would be less than 1,620,000 psi. Is that a big deal?

Well, in most cases not. Since E is in the `denominator’ of the deflection equation, a lower E (in an individual, or specific, beam), albeit only a bit lower, will result in a bit greater deflection, or sag (notice I didn’t use the word `higher’). So, if the calculated sag for a beam is ½ inch, there’s a 1/6 chance it could be as much as … 10 percent greater, or 0.505 inches. Sure we could measure the difference; *seeing* it might be more difficult. Only in a case where 0.50 inches (0.500) would be critical, might we worry about this variability. Where specific deflection (sag) amounts are critical, one solution would be to order a stiffer beam (higher E). We could order a beam with E = 2,000,000 psi (if one exists), and calculate the deflection, and its variability. There would still be some `chance’ of the beam having an unacceptably low E, though less chance. If we really want to make sure we have an acceptable E for a particular beam, we need to `measure’ E for that particular beam, though this is typically not done in practice, say, beam by beam, on a jobsite. You* might* be able to get a manufacturer to *measure* the E of individual beams for a particular order … I’ve never done it.

Pictured is a screen grab from the Course, showing an example of conventional roof framing, including the use of a purlin and brace.

The term `conventional’ means `in accordance with general agreement’, `established practice’, and so on. One might also say, with regard to conventional framing … `normal’ construction … `the way we’ve been doing roofs for years’. And this goes two ways in real life: 1) builders who build roofs unencumbered by codes, rules, engineers, etc., … framing them the way they `always have’, and 2) builders who build roofs according to the Building Codes. Those following the Building Codes have two options: one being Conventional, a.k.a. prescriptive, a.k.a. stick framing, and the other, engineered design. Conventional framing is covered by the International Residential Code (IRC), the Wood Frame Construction Manual (WFCM), and Section 2308 of the International Building Code (IBC). Again, the idea being, if one follows the `prescriptions’ in these codes, the employment of an engineer is not necessary. These requirements of these codes seem pretty restrictive, pretty `complicated’, but, I argue, they are generous in that they: 1) relieve you of having to hire an engineer (my contact information is here, somewhere, on the website), and 2) if you follow the requirements, you will not only be following the Code (probably also the law in the area you are building), but you will also be building a `safe’ roof. The requirements of the prescriptive codes are not trivial; nor are they intended to be unnecessarily burdensome or troublesome; they are the product of people coming together to come up with requirements that are both safe and workable in the modern world.

I tell my students, friends, relatives, etc. that are interested in building, or designing buildings, to learn where and how to apply conventional / prescriptive design; so you’ll be able to avoid hiring an engineer. (I am trying to retire.) Conventional design is generally applicable to residential structures. But learn it well, as not all residential structures fall within the limitations of the conventional provisions, and not all `parts’ of some residential structures can be designed (and built) `conventionally’. A bad-case scenario for a designer, or architect, is to embark on a project assumed to be covered nicely by conventional framing requirements, only to find out that an engineer must needs be employed, as the structure, or a part, or parts, fall outside the limitations of use of the prescriptive (`conventional’) provisions.

Another beauty of the conventional framing relates to the term `stick framing’. The roof, however large, can be made up of … `sticks!’ Yeah, pieces of wood that a single person can manage. So, not only may you avoid having to hire an engineer to design a `stick-built’ structure, you can avoid renting a crane, and hiring a crane operator, to put it together. Roof rafters, ties, ceiling joists, and so on, are of manageable size, and also typically available at the local building supply store. Further, the framing pieces are typically attached with nails, as opposed to bolts, or heavy hardware. (Neither do you need a welder!) Where bigger building elements are needed (beams, headers, girders), the prescriptive codes provide for their construction to be of `sticks’, e.g., by nail-laminating them together. (I have found, in fact, that when I have been framing alone, instead of nailing them together on the ground, and having to lift the (heavy) finished beam or header, the header can actually be installed piece by piece. Yay! No crane, no crying for the help of others in the moment!

(So, while I make a living as an engineer designing structures that fall outside of conventional framing, I actually do my living, making structures out of `sticks’.)

(Maybe I’ll pick apart my `conventionally framed’ house in another post.)

Oh, the IRC does provide for the use of roof trusses. The trusses are made up of multiple pieces of lumber, connected, typically, with metal plates (metal-plate connected wood trusses). They typically span from exterior wall to exterior wall, or from intermediate girder trusses to wall, and thus are `bigger’ than single rafters and joists. They are designed by a `truss manufacturer’, and are typically installed with the aid of cranes. You don’t have to hire an engineer to design manufactured roof trusses; the truss manufacturer does. The truss manufacturer typically provides site-specific design and layout, as well as the actual trusses. And the cost of the `engineering’ is part of the `truss package’.

]]>In new construction, wood framing more or less starts from the ground up; wood members are placed on top of one another, and are fastened into place. Just about everything is `statically determinate’, and unless something is cut grossly wrong, things `fit into place’. In remodeling, on the other hand, walls, etc. are removed, surrounding or supporting framing is more or less supported (`shored’), and new construction (beams, etc.) put (sometimes `jammed’) into place. Even with the best intentions, things don’t always `fit’. And when they don’t fit, the common remedy is the hammer, big or small. A common situation is removing a wall placing a header or beam over the new space; the header supporting the framing above, floor or roof/ceiling. While the header (or beam) can be ordered and cut to easily fit, it’s the supports at the ends of the new opening, typically studs (jack studs), that often don’t quite fit.

So let’s look at a recent wall-removed-header-placed `situation’. Roof and ceiling framing were shored using temporary walls on both sides of the existing wall, adjacent to planned opening. The wall framing, from sole plate up to but excluding the double top plate, was removed, and the new header `lifted’ into place (human labor). Some of the jack studs, intended to support the ends of the headers, cut beforehand, didn’t quite fit. Alas the hammer. After a lot of smashing, the jack studs were in place, supporting the header, readying for the removal of the temporary walls. Not quite!

Let’s run some numbers.

The header was designed to have very little `sag’ … `deflection’ in engineer language (as well, of course, be strong enough). The existing wall held up ceiling and roof framing above; the new header would take over the role of the wall above the new opening. The clear span for the header is 10’-9”. Two jack studs were used for support each end. In the design headers, beams, etc., the `structural’ span for the header is taken to extend half way into each support, so, 10’-9” + 2 x ½ of 2 x 1.5” = 10’-9” + 3” = 11.0 ft. The header is a 2-ply 1.75-in. x 11.875-in. 2.1E LVL. The `design’ of the header accommodated the design load by several hundred percent, with, importantly, very little anticipated deflection, or `sag’ (very stiff header). The intent was that the top of the opening be visually `flat’.

The jack stud rammed into place at one end caused the header to be `bent’, with respect to a straight line end to end, about 1/16th of an inch. (Lots of ramming). This `bent-ness’ was before the shoring was removed. Properly shored, and assuming a flat (straight) ceiling/roof above, the header initially should `fit’ under the framing with no deflection; it’s not carrying any load! Only as shoring is removed should the header take on load, and begin to bend. Structurally, the action of forcing the header to be bent into a curved shape, however `slight’, produces, in the wood, the same effect as if the header was simply supported at ends (just `resting’ on supports), and being loaded downward along the span, enough to bend it. The load comes about as jamming the jack stud(s) in place tends to `lift’ the header and framing on top of it … the framing above resisting the lifting action, pushing down.

Assuming this `lifting’ effect is greatest at the end where the stud is jammed in place, and tapers off to … nothing, or little, at the other end, we’ll model the situation using the so-called `SIMPLE BEAM-LOAD INCREASING UNIFORMLY TO ONE END’, found in typical engineering design manuals. (See graphic.) Let’s look at the load `required’ to produce a 0.10-inch deflection.

Δ = 0.01304 W L^3 / EI,

where,

Δ = deflection (sag, out-of-straightness),

W = the `whole’ load (`weight’) or downward force on the beam (header),

L = header span, 11 ft or 132 in.

E = the modulus of elasticity of the header, 2,100,000 psi (the `2.1E’), and

I = area header moment of inertia of the header , n x (b h^3/12) = 1.75 (11.875)^3 / 12,

where

n = number of plies,

b = thickness of each ply, and

d = depth of the header.

We get, for I, 488 in.4.

Solving for W, we get, … W = 2,137 lb (call it 2,100 lb).

Hmmmm. The total dead load (roof and ceiling weight) above the header is about 2,300 lb, and more or less distributed uniformly. By smashing the jack stud in place we are just about starting to `lift’ the roof. No. Take the jack stud(s) out, take off a sixteenth of an inch or so, so that it goes in with less `smashing’.

*Can’t we just leave the header as is, since it, and the jack studs, are in place, and nothing `happened’? After all, the header was intended to carry the roof weight anyway!*

Maybe, but why? What if the out-of-straightness is 1/10th of an inch? … or 1/8th of an inch? At 1/8th of an inch we’d be well overcoming the weight of the roof … things would start to `give’ that we don’t want to give. Additional to the noise from the banging of jack studs in place, we might be able to hear creaking and groaning as the framing (wood members and fasteners and other materials) above and adjacent to the header start crushing, yielding, cracking. At some point we could, theoretically, `break’ the header itself (if we haven’t already broken something else … maybe the jack stud). Besides, we’d be carrying the roof wrong … by the end of the header, not equally along the entire span.

Are you SURE nothing happened? Is the roof plane still straight? Have you cracked any gypsum?

The header is now pre-bent, and pre-loaded. Is the out-of-straightness visible? So far the header is only carrying the weight of the roof itself, plus some more. What happens when the header experiences its `design load’ (a future re-roofing project)? In this particular case, re-running the design calculations with some extra load (and deflection) might turn out to be `okay’ (as there was way more strength, and stiffness in the header than required), but, again, why? Take the studs out and do it right. The header should rest snug, tight, on top of the studs, but not so jammed as to deflect the beam.

…

Where a header or beam is more deeply embedded in a structure, and, particularly, if the structure has continuous framing elements or inflexible systems, providing little `give’, it’s possible, by `jamming’ studs into place, to induce a load in the header that exceeds the design load. Not good.

Don’t ruthlessly smash studs or columns into place; cut them so they fit. Tight, yes! And with a bit of pounding; but you don’t want the header ends to `settle’ (move) when the shoring is removed.

…

A NOTE ON NUMBERS

The numbers in the calculations above look rather `exact’. Actually, many are not. For example, the modulus of elasticity value, E, for the header is (published) 2.1E, meaning 2,100,000 psi. This is an average value. A bunch of LVL specimens were measured in a test lab for E, and the average value was used (published). Some of the LVLs that arrive on a project (or lumber yard) have higher E, some lesser. That’s the way wood is, variations in properties, between species, between grades in species, and between `engineered lumber’ (e.g., LVL) made up of wood. (LVL tends to have less `variation’ than other wood products, due to how it is manufactured … that is another conversation.) So, when I dump and un-exact number E into an equation that looks exact (and, in fact, might very well be), I get an un-exact number out. The applied force to bend an 11-foot long 2-ply 1.75 x 11.875 2.1E LVL header 1/16th of an inch may take somewhat more, or somewhat less, than the one I calculated; in fact, it would be highly unlikely that it would take, exactly, 2,137 lb to bend the header 1/16th inch, exactly.

Further, the deflection equation used does not account for `shear deflection’. Shear deflection is the relatively small amount of additional `sag’ that a beam, header, joist, girder, whatever, experiences due to the `shearing’ of the wood fibers. The equation used above only determines the deflection (sag) that comes about from bending stresses (compressing the top of the beam and elongating the bottom). By neglecting the shear-deflection term I have been a bit `conservative’ – a lesser amount of force (jamming) could indeed produce the same effect. (The header is a tiny bit less stiff than I calculated.) But, this `shear deflection’ may very well be (more than) eclipsed by the variation in E, already discussed above (and talked about more in some of my other posts).

Don’t worry so much about the exactness in E; worry about cutting the studs so that they don’t need to be ruthlessly smashed into place.

…

]]>Δ = 5 w L^{4 }/ (384 E I),

where

w = the uniformly applied load, i.e., such as in pounds (lb) per foot (ft) (of beam length), or plf,

L = beam length, i.e., in ft, or inches (in.),

E = the modulus of elasticity of the wood, typically psi (psi per inch of deflection per inch of wood),

I = the area moment of inertia of the wood section, i.e., in in.^{4.}, and

5 and 384 are just numbers (pure, exact numbers), based on the support and load distribution on the beam.

We call this `case’, or condition, the simple beam – uniform load, or `uniformly loaded simple beam’.

The equation is dimensionally consistent; it can be used in any `system of units’, as long as done so correctly (e.g., get all the input into inches and pounds).

I tend to like a slightly different form of the equation, commonly used in the structural steel industry, but it still works for wood. Here it is …

Δ = 5 W L^{3 }/ (384 E I),

where

W = the `whole’ load on the beam.

(The `whole’ load on the beam is simply the load per foot (or whatever unit) of length, times the whole length … W = w x L.)

Let’s do an example. Let’s say we intend to use a 2 x 10 dimension lumber joist to support a floor. The floor is estimated to weight 15 pounds per square foot (psf), including joists, and is intended to carry a design occupancy load of 40 psf. The joists will be spaced 16 inches apart.

The joist must carry the floor itself, plus the occupancy load (people, furniture, etc.); thus the total load applied to the joist will be 40 psf + 15 psf = 55 psf. Further, the joist, obviously, must carry itself. Sometimes the weight of the joist itself is `included’ in the weight of the floor (e.g., floor system). Or it may be examined separately. Often for the floor system, or volume, the estimated (or assumed) floor weight includes the joists, as the joists are part of the system (i.e., enclosed by floor sheathing above, gypsum ceiling below, and perhaps insulation between); but when examining, for example, a beam *carrying* the joists, the beam weight will be explicitly (additionally) examined. Back to our example.

The total load is 55 pounds per square feet. The area (square feet) supported by each joist is … 16 inches / 12 inches per foot times 10 feet equals 13.33 square feet (sf or sq ft). Thus the total load on the joist is … 55 psf times 13.33 sq ft = 733 lb (W = 733 lb).

Before we drop this number into our equation, we need E and I.

We generally `look up’ E. Perhaps the best place to get it is the* National Design Specification® for Wood Construction* (NDS) by the American Wood Council. In its *Supplement* are provided design values, including E values, for commonly used wood products. Let’s, for our example, use Douglas-fir / Larch Grade No. 2 joists. In the *Supplement* we find E = 1,600,000 psi.

(Strictly speaking, modulus of elasticity is psi per inch per inch, … the amount of stress required to produce a unit deflection … but the inch per inch is typically not shown.)

The moment of inertia, I, is a function of the joist cross section, and how oriented. By formula, for a rectangular cross section,

I = b h^{3} / 12,

where

b = dimension of the side, or edge, perpendicular to the load direction, and

d = dimension of the side parallel to the load direction.

For a floor, the load is downward; b is the horizontal top or bottom dimension, and d is the dimension of the vertical sides of the joist. Typically we orient the joist `on edge’, so b is the narrow edge, and d is the wider face, or depth.

For a dimension lumber 2 x 10, the actual dimensions are 1.5 in. x 9.25 in.

For our joist,

I = b h^{3} / 12 = 1.5 in. (9.25 in.)^{3} / 12 = 98.9 in.^{4}.

Here we go …

The calculated deflection for the intended loading condition is … (10 ft is 120 inches) …

Δ = 5 W L^{3 }/ 384 E I = 5 (733 lb) (120 in.)^{3} / [ 384 (1,600,000 psi) (98.9 in.^{4}) ] = __0.104 in.__

A typical limitation on floor deflection is that the sag under live load be not greater than the span divided by 360, in this case, 120 inches / 360 or 0.33 inches.

Since the deflection calculated using the total load (live and dead) is 0.104, of which the deflection due to live load would be just a fraction, the deflection is acceptable.

But there’s more we need to talk about.

For starters, E is not a `true’ E. In the lumber world, published values for E, though often not obvious, are `apparent’ values. They arise from actually bending lumber pieces under load, in a `lab’, and back- calculating the E, using an equation(s) `like’ the one above. The problem is, the equation above, and others like it, are based on bending (beam curvature) arising solely from the tension and compression stresses in the lumber, where tension on the bottom half of the beam stretches the wood fibers, and compression on the top half of the beam compress the fibers, resulting in a curved shape. In real life, when a beam is loaded, it bends (sags) a bit *more* than that caused by the tension and compression stresses described above; as the wood also `shears’ (has shear distortion). The value obtained by this bending test, nevertheless, is `published’ … but, is usable for deflection calculations of other lumber, suffering the same phenomenon; that the resulting deflection includes the effect of `pure’ bending (tension and compression), and shear.

You can read more about this in Appendix F of the NDS.

The effect of shear on sag is not huge; for sawn lumber it is 3 percent; the ratio of true (pure, or `shear-free’) E to apparent E is 1.03. If you want the true (shear-free) E, take the published value and multiply it by 1.03. In the case above, one `gets’ 1,600,000 psi x 1.03 = 1,648,000 psi. (If we round it to the nearest hundred-thousand psi, we still get 1,600,000. I’ll let you decide what you want to do with rounding. But I’ll also say, look at the Coefficient of Variation of sawn lumber, in that same NDS Appendix F, and don’t get too carried away `splitting hairs’.)

While the difference between true and apparent E values for sawn lumber is not huge, the difference is bigger for some other structural wood products. In Appendix F we see that for glulam the ratio is 1.05 … 5% difference. Further, for glulam (Appendix F), the Coefficient of Variation (of E) is less, so we might say that the difference between the two E’s is less moot than as for sawn lumber (not as smeared by the overall variability of E).

Products such as prefabricated wood I-joists have even greater ratios of E true to E app. Important to note, and so I’ll do it here, the relative effect of shear deflection is a function on the member *length. * To accurately determine deflections of such products, deflection equations typically have two terms (components), the first term using true (shear-free) E to calculate the component of deflection due to tension and compression stresses, and a second term to address shear. Design equations and E values are provided by the I-joist manufacturer!

It is the growing tendency in the structural wood industry, engineered wood products, now even glulam, to show the `pure’ (pure, shear-free) E values … after all, they’re `higher’.

But for now, for sawn lumber, it’s just E, and it’s actually apparent E.

]]>And, further, when off-loaded, the beam does not go back to its un-deflected shape.

Example

Let’s look at an example. Consider 2 x 10 floor joists, spaced 16 inches on center, spanning 12 feet (ft) `simply’, to be used in the main floor of a residential structure. The International Building Code (IBC) requires that the joists be able to carry (be designed for) an occupancy load of 40 pounds per square feet (psf). The joists, ultimately, will need to carry: 1) this occupancy load (people, whatever, for the most part coming and going); 2) the weight of the floor system; and 3) the weight of the joists themselves. In the case of joists, their weight is often included in the weight of the floor system (i.e., floor made up of joists, sheathing, finish floor materials, gypsum ceiling, perhaps thermal or acoustic insulation). On the other hand, the weights of beams, girders, etc. carrying joists, should be specifically considered. The dead weight (load) of the a typical wood-framed floor system seldom weighs more than 10 to 15 psf. If the floor construction includes tile, or concrete, then, obviously … more.

Live Load Deflection

The live (occupancy) load that each joist is required to support can be calculated two ways. I like to look at the `whole’ live load. The `whole’ load is the `area load’, in this case 40 psf, times the `whole area’ (floor area) carried by the joist (not the whole floor, just each joist). Floor joists are spaced 16 inches (16/12 or 1.33 ft); so, each joist will need to carry (support) … 1.33 ft x 12 ft = 16 square feet (sf). The `whole’ live load is, thus, 40 psf x 16 sf or 640 lb. You could think of this as several people ending up standing on the joist at one time.

The general formula for calculating deflection of a simple beam supporting a uniform load, in terms of `whole load’ is …

Δ = 5 W L3 / (384 E I),

where

Δ = the calculated deflection,

W = whole load,

L = rafter span,

E = the moment of inertia of the wood joist, and

I = the moment of inertia of the wood section, and

5 and 384 are `numbers’ particular to the beam loading and support condition, in this case uniformly distributed load on simply supported beam (joist).

Another common form of this equation is …

Δ = 5 w L4 / (384 E I),

where

w (`lower case’) = the load in terms of load per length of joist.

It’s pretty easy to see that W = w L.

This `line load’ w can (also) be calculated as,

w = σ x s,

where

σ = area load (in psf),

s = tributary width for the joist (joist spacing), and, of course,

x is the multiplication symbol, in this case.

(Let’s check: w = 16/12 ft x 40 psf = 53.3 lb/ft (plf); W = w L = 53.3 lb/ft x 12 ft = 640 lb. Yay!)

Continuing,

(So, W = 640 lb.)

Let’s get L into inches … 12 ft x 12 in./ft = 144 in.

For E we need to pick a wood species, and `grade’ within the species. This will depend on availability/cost. Availability depends on location. Around here (Southeastern USA) Southern Pine is typically available. A good starting place for cost is `Grade No. 2’. A good place to get the E for wood products used in construction is the Supplement to the American Wood Council National Design Specification® for Wood Construction (NDS). You can view it online at awc.org. It’s a good place (good source) because the NDS is referenced by model building codes, and thus likely acceptable to the building department with jurisdiction over `your’ project. (Plus you can view it `free’.) From Table 4B in the 2018 NDS Supplement we get … E = 1,400,000 psi.

For I we use the formula for the area moment of inertia for a rectangular section,

I = b d3 /12.

Assuming the joists will be installed `on edge’ and from Table 1A in the (same) Supplement, b = 1.5 in., d = 9.25 in., and, thus,

I = 1.5 in. (9.25 in.)3 / 12 = 98.9 in.4.

(We could have just looked I up in Table 1B of the `Supp’.)

Here goes, … the calculated deflection for the 40 psf live load is …

Δ = 5 W L3 / (384 E I) = 5 (640 lb) (144 in.)3 / [ 384 (1,400,000 psi) (98.9 in.4) ] = 0.180 in.

The `bare bones’ minimum deflection required for live load deflection of a floor joist, per Section 1604.3 of the IBC, is that the deflection due to live load shall not exceed the span divided by 360. By `bare bones’ I mean that this is a minimum requirement. The owner, designer, floor material manufacturer, etc. may require or recommend something more stringent … say L/480, or L/600, or an actual (fraction of an inch) amount. The manufacturer of a brittle floor material may, for example marble, recommend a more stringent deflection. (Not to mention it might very well weigh more, motivating a more deliberate calculation of the floor material weights.) But don’t try going any less stringent, or an occupant might be able to feel the beam deflect; or an observer see the floor deflect; or more deflection might cause a problem with floor or ceiling coverings or materials. In the case of our example,

L / 360 = 144 in. / 360 = 0.40 in.; 0.18 in. ≤ 0.40 in.; by this criteria we are okay.

Dead Load Deflection

The IBC places a limit of L/240 on the additional deflection due to dead load, plus that due to live load. And the additional deflection due to dead load may be taken as the immediate (direct) deflection of 0.50 times the dead load (0.5D) (for dry lumber in dry condition of use). It’s another way of saying that the additional deflection due to the long-term (dead) load is 1.5 times the immediate, or 50% more.

Let’s do it.

Our `whole’ dead load is …

W = 15 psf x 16 sf = 240 lb.

Δ DL = 5 (240 lb) (144 in.)3 / [ 384 (1,400,000 psi) (98.9 in.4) ] = 0.067 in. (CHECK THIS!)

The deflection of `0.5D’ would be half that, or … 0.033 in.

Added to the live load deflection of 0.180 in., gives 0.213 in.

Comparing this with L/240 = 144 in. / 240 = 0.60 inches … we’re good!

The TOTAL dead load deflection, long term, is 0.066 in. + 0.033 in. = 0.099 in., which `persists’. Then the occupants arrive, and we get 0.180 inches more, or 0.099 + 0.180 or 0.279 total inches. The occupants leave, and the deflection `goes back’ to 0.099 … (assuming the occupants don’t stick around `long term’).

]]>`Components and Cladding’ … or C & C or `see and see’. In earlier posts we looked at the Main Wind Force Resisting System (MWFRS) wind forces on a couple buildings, specifically, on a relatively wide, low, flat `shop’, and a not-as-wide, and taller, `house’. The MWFRS forces are the `big-picture’ forces acting on the structure, or overall-building forces. These forces tend to lift the structure off the ground, or push, slide it along the ground, lift the roof off the walls, or pry or tip the building over. These forces are used to design the main `force-resisting systems’ … the main frames, or framing system(s). Typical systems might be actual structural frames within the building, or roof-diaphragm-shear-wall systems, or `cantilever-post-and-roof-cladding’ systems. And the system(s) that attach (anchor / tie down) the building to the foundation, or ground.

Winds act on structures via air `pressures’ they exert on building surfaces. Really strong winds also `throw stuff’ at structures … parts of other buildings already torn apart by the wind, tree branches, kitchen sinks, cows, etc. We won’t talk about these `projectiles’ here … maybe later.

Winds contain `swirls’ of air. The swirls are not necessarily visible, unless made so by water vapor, smoke, other markers, or by `waves’ as they move across fields. These swirls are characteristic of turbulent flow. Water in rivers is also turbulent, and we might say that some rivers are more turbulent than others. Serious white-water rafters like really turbulent flow. These swirls of air are in all different sizes – look at clouds – small swirls within bigger ones. As these swirls hit and move across, or are blocked by, a building, there are parts of the swirls that exert higher (or lower) pressures than others. Some parts of the swirls have higher (or lower) speed air particles than others. The highest and lowest pressures tend to act on relatively smaller amounts of building surface area. Alternately speaking, the pressures are smeared out over larger areas and as the swirls move across the building surfaces. These areas of relatively high or low pressure are smaller than the amounts of surface we looked at for the MWFRS wind forces … they are on the scale of individual `components’ of the building surface … thus we call these pressures `Components and Cladding’ pressures. Whereas before we looked at the whole building lifting off the ground, or the whole roof, or the whole house being slid across the ground, now we are going to look at the pressures and forces that tend to rip pieces off the structure. It is important, of course, for the components and cladding sake that we address these pressures and resulting forces, to keep them from coming off the building, resulting in local damage, or flying debris, or allowing water penetration to the building. It is also important to keep these components intact so that the wind doesn’t penetrate the building, causing it to `inflate’, and then explode, or cause to roof to start, and continue, to unzip, allowing the roof to peel off.

Okay, let’s do it! … ASCE 7-22 Chapter 30. And let’s return to the `shop’ we have been examining. We’re dealing with a low-rise building, partially enclosed (garage door open at one end), and so on; the Steps are provided in Figure 30.3-1.

Step 1: Risk category … I

Step 2: Basic wind speed … V = 96.5 mph, for the location of interest, Risk Category I

Step 3: Determine the wind load parameters:

… Kd = 0.85 … from before

… Exposure C, ditto

… Kzt = 1.00, ditto

… Ground elevation factor 0.915, ditto

… Enclosure classification … partially enclosed, from before

… Internal pressure coefficient(s) … +/- 0.55

Step 4: Kh = 0.85, from before

Step 5: Determine the velocity pressure, qh … from before, 18.5 psf

… and now …

Step 6: Determine the external pressure coefficient, GCp … for gable roofs … we use Figure 30.3-2B … and / but it points also to Section 30.7.

For the top surface we use Figure 30.3-2B.

We’ll assume that the bottom surface of the roof (plywood sheathing) is open to below (as per the actual structure that inspired these posts). The bottom of the plywood outside the exterior wall will be subject to the wall pressure, pw, of Figure 30.3-1, and the bottom of the plywood within the exterior wall will be subject to the internal pressure, GCpi.

Consider a 4 ft x 4 ft piece of plywood, installed at a corner.

The area of the component is 4 ft x 4 ft = 16 ft^{2}.

Going back and getting h, … h = 12.3 ft

And a = 4 ft ( … from our exam of `Corners’)

This 4 x 4 piece falls entirely in Zone 2 …

From Fig. 30.3-2B … GCp = ~ -2.4.

And let’s use a rake of 1.5 ft, equal to the eave overhang.

From Figure 3.3-1, pw = + 1, -1.4 (Zone 4 and 5).

From Equation 30.3-1 the total pressure on the `rake’ and `eave’ portions of the plywood component is … p = qh Kd (GCp – GCpi) = 18.5 psf (0.85)( -2.4 – 1.0) = __53 psf__. Whoa!

The total pressure for the plywood component within the exterior wall boundary is …

p = 18.5 psf (0.85)( -2.4 – (0.55)) = __46 psf__.

The area of the plywood exposed to the interior of the building below is (4 ft – 1.5 ft) x (4 ft – 1.5 ft) = 9.75 ft^{2}. The area of plywood exposed underneath to the `outside air’ is … 16 ft^{2} – 9.75 ft^{2} = 6.25 ft^{2}.

The total uplift force on the 4 x 4 piece of plywood is … 53 psf x 6.75 ft^{2} + 46 psf x 9.75 ft^{2} = 331 lb + 449 lb = 780 lb. CRAZY! This would be no place for the fasteners attaching the plywood to the framing below to *miss* the framing! It also is a good place to point out that the framing members will also need to be attached (held down, tied down) … as they will hold down individual components as well as larger areas of the rest of the roof.

Note that 53 psf is equivalent to the weight of almost a foot of standing water, except upward! … or 6 or 8 feet of snow! These wind forces are very significant! In a particular windstorm (or tornado, or hurricane), it might very well be that just a single component or cladding that gets ripped off the building, … or, often, more and bigger pieces follow, or the whole roof, or the whole structure.

]]>In earlier posts we looked at the `main’, or `overall’, wind forces on a rather low, wide `shop’ (here and here), and on a relatively narrower and taller `house’ (here). Specifically, we calculated the Main Wind Force Resisting System (MWFRS) forces hitting the structures from the `side’ (long dimension). We calculated the forces per 8-foot `swath’ of building, as the shop was framed in 8-foot column-truss-column bents. We found that, overall, the wind forces tend to `lift’ a low, wide building off the ground (or roof off the wall), with perhaps a bit of pushing and prying due to the horizontal component of wind load, while for a narrower, taller (and steeper-roof) residence, the wind tends to push the house off its foundation … the horizontal components of wind force being greater than the vertical (upward) components. The comparison is often exaggerated as the roofs of `shop’ type buildings are generally light (small roof dead load) … there’s little dead weight helping hold the roof, or building, down.

We might very well design the MWFRS elements (bents of columns-trusses-columns) of the shop, and the main wind-force resisting system of the house, utilizing the forces we obtained in the MWFRS analysis, but before we do so, let’s look at the `corners’. ASCE 7-22 Figure 28.3-1 shows us what MWFRS pressure coefficients to use for corners. A quick look shows them to be roughly 50 percent more than `away’ from the corners. The amount of the walls and roof surfaces upon which these pressures act are denoted by *a* and *2a*, where *a* is the lesser of 10% of the building width and 40% of the mean roof height, but not less than 4% of the building width (least horizontal dimension) or 3 feet (ft), and so on. For our 40 ft by 100 ft shop with 4/12 roof, a is the lesser of 0.1 x 40 ft = 4 ft and 0.4 x ~ 12 ft = 5 ft, but not less than 0.04 x 40 ft = 1.6 ft or 3 ft … or … 4 feet. 2*a* = 8 ft. So, the end 8 feet of 100 feet of side wall is affected by the higher pressures.* *The MWFRS forces were good when looking at the MWFRS away-from-corner bents. The end bents, and the next ones in, have higher pressures. * What about the other end (corner)?* Yeah, the other corner (other end) needs to be considered, but not simultaneous with the first. Going back to Figure 28.3-1, there are 2 Load Cases. Case 1 is with the wind hitting the side, and at some angle. Case 2 is the wind hitting the end, and at some angle. Each Case has 4 corners, one at a time. So, yeah, there are 8 overall building loading conditions to examine. It’s nice if the building has a lot of symmetry, especially if you’re doing the calculations by hand. At any one time, only one corner (8 percent of the side wall in the example) experiences the `corner’ pressures.

Go look at structures damaged by wind. The damage is at, or starts at, the corners.

]]>