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.

]]>It may be a bit premature, but let’s also look at Section 28.3.5 `Minimum Design Wind Load’. It’s not clear whether we should consider this at `each frame’, or the entire structure as a whole, or maybe even `piece-by-piece’. Anyway, we’ll look at it now.

Section 28.3.5 requires that the minimum design wind pressure used in the design of the MWFRS … be not less than the wall area multiplied by 16 psf and a pressure of 8 psf applied to the roof area projected onto a vertical surface. (Perhaps the use of singular `MWFRS’ answered the question above.) Let’s do it.

Wall area: 8 ft wide x 8.5 ft tall = 68 sf

Horizontal force from wall = F = p x A = 16 psf x 68 sf = 1,088 lb.

Roof area: 8 ft wide x ½ of 20 ft x 4/12 = 8 x 6.67 sf = 53 sf.

Horizontal force from roof = 8 psf x 53 sf = 427 lb.

__Total horizontal force =__ 1,088 + 427 = 1,515 lb, say __1,52____0 lb.__

Note that the contribution of horizontal force from the wall(s) didn’t change much.

The big difference is the contribution from the roof … another 500 lb.

My `guess’ is this: the pressure coefficients in 28.3-1 (Load Case 1) capture the effects of the wind blowing over the structure with the structure acting as a (rather strong) air foil … so much so, that, roof alone, by *calculation*, the horizontal effect is actually negative.

Two Different Air Flows Across Shop

Section 28.3.3 makes sure that we don’t have a negative effect. Section 28.3.6, I suggest, captures the fact that in the turbulence (chaos) of the design wind event, there may be, or are, gusts that exert positive pressure on the windward roof slope … the wind separates’ from the roof at the roof, instead of the windward eave.

*****

The question remains, at least in my mind, should I have done this calculation at this stage (per `frame’, or `swath’), or on the entire structure?

]]>

Here goes…

Windward eave pressure: p = qz Kd (GCpf – CGp) = 18.5 psf (0.85) (-0.69 – 0.7) = 18.5 psf (0.85) (-1.39) = __21.8 psf__ (the minus sign dropped as we know it’s acting upward and outward).

Leeward eave pressure: 18.5 psf (0.85) (-0.47 – 0.7) = … __18.4 psf__ (upward and outward).

For an 18-inch overhang:

Windward upward force on eave, per longitudinal foot of wall: F per foot = p x A per foot -21.8 psf x 18/12 ft x 1 ft = 33 plf (rounded). For an 8-ft `swath’: 33 plf x 8 ft = 260 lb.

For the leeward eave: F = 28 plf (rounded) or 220 lb, for the 8-ft swath.

Note again that the windward upward force magnitude is greater than the leeward … causing a `negative drag’, and that the horizontal component of which must be ignored, for the MWFRS, per Section 28.3.3.

But the added vertical forces are:

Windward: 260 lb cos (18.4°) + 220 cos (18.4°) = 460 lb … (8-ft swath).

If the shop is 100 ft long, we’re adding … another 460 x 100 / 8 = 5600 lb. What does a car weigh? Apparently the average small car weights about 2500 lb. Maybe we need to park a couple (more cars on the roof, one on the windward eave, and one on the leeward, to keep it from flying away.)

]]>In previous examples (here and here) we looked at the horizontal and vertical wind forces on a `shop’. The shop is/was a `pretend’ structure in a real location, Latah County, Idaho. It derives from an actual shop of the same dimensions, but in a very different location. The intended use in the example was, as with the real one, `uninhabited’ occasional (hobby) farm use. As such, we assigned in a Risk Category of I, per Chapters 1 and 26 of ASCE 7-22.

Here’s a summary of relevant information from the shop example:

Basic Wind Speed: 96.5 miles per hour (mph)

Exposure C

Kd = 0.85

Kzt = 1.00

Ke = 0.915 (Elevation 2500 ft)

Roof Slope: 18.4°

Mean Roof Height: 12.3 ft (0.5 ft stemwall + 8.5 ft framed wall + ½ of the roof height of 6.67 ft)

Kz, Kh = 0.85 (Table 26.10-1, from the Exposure Condition and Mean Roof Height)

(Total Roof Height: 15.7 ft, in case we want it later)

We found with that example that the design horizontal and vertical wind forces, for an 8-foot wide transverse `swath’ of building *not* near the ends, were 1,000 lb and 5300 lb, respectively. The vertical (uplift) wind force on the shop (swath) is over 5 times than that of the horizontal force. The shop wants to act like an airfoil and `fly’ away in the design wind.

Consider another pretend structure, a `house’, near the shop (also located in Latah County, Idaho). The house width is 30 feet (ft), one story, with 9-ft framed walls on 12-in. (deep) framed floor system on 6-inch stemwalls, and with roof pitch of 12/12. (Note the steepness of the roof … 45° slope.) The total height of the house is 0.5 ft + 1 ft + 9 ft + ½ of 32 ft x 12/12, or 26.5 ft; it is significantly taller `than’ the shop. Let’s calculate the total horizontal and vertical design wind forces on the house.

To determine the wind forces on the house we will, as with the shop, use ASCE 7-22 Chapters 26 General Requirements, and Chapter 28, Envelope Procedure. The steps are given in Table 28.2-1. Here goes:

Step 1: Determine the Risk Category: easy; it’s Risk Category II (ASCE 7-22 Table 1.5-1).

Step 2: Determine the Basic wind speed, V. Here’s where it gets a bit interesting. Even though the house is potentially `right next to’ the shop, we end up using a different wind speed. Why? It’s because we are assigning different `risk’ to house, versus shop. We’re willing to risk an 86.5-mph wind hitting the shop, and possibly `blowing it over’, but we’re not willing to risk that same wind blowing over the house. That’s how the `code’ parses it. For the sake of example, we could determine the wind forces on the shop based on Risk Category II, as, say, for business use or, at least, regular use by persons. In fact, the *real* shop, upon which the examples derive, was at times used for a business. However, in that case, I would assume that, for the wind event, the big garage doors would be *closed*; then I would have smaller internal pressure coefficients, and * lesser* total wind uplift forces. I’ll stick with Risk `Categ’ I and open garage door (partially enclosed structure).

Basic wind speed, for the house, Risk Category II, ASCE 7-22, Figure 26.5-1B, __102.5 mph__.

Step 3: Determine the wind load parameters:

Wind directionality factor … same as for the shop; Kd = 0.85.

Exposure Category … same as for the shop; C.

Topographic factor … same, Kzt = 1.0.

Ground elevation factor … same, Ke = 0.915 (2500 ft elevation).

Enclosure classification: Section 26:12; `Enclosed’.

Internal pressure coefficient, GCpi, from Section 26.13 / Table 26.13l-1; *GCpi = +/- 0.18*.

Step 4: Determine the velocity pressure exposure coefficient, Kz or Kh, from Table 26.10-1. First we need the mean roof height; it is, 0.5 ft + 1.0 ft + 9 ft + ½ of 32 ft x 12/12 x ½ = 18.5 ft. From Table 26.10-1 I get __Kz, Kh = 0.885__ (interpolated).

Step 5: Determine the velocity pressure, qz or qh, using Equation 26.10-1. Here goes:

qz = 0.00256 Kz Kzt KeV^{2} = 0.00256 (0.885)(1.0)(0.915)(102.5^{2}) = __21.8____ psf__.

(Recall for the shop we had qz = 18.5 psf.)

Depending on the actual pressure coefficients, this higher qz may, or may not, result in higher pressures on wall and roof surfaces; let’s see.

Step 6: Determine the external pressure coefficients, GCpf. For our house we will, for comparison with the shop. again look at … the windward wall (Zone 1), windward roof (Zone 2), leeward roof (Zone 3), and leeward wall (Zone 4), Case 1, not near the end of the building. From Figure 28.3-1 we get the following external pressure coefficients (Load Case 1) … (45° roof slope) …

GCpf windward wall (Zone 1) = 0.56

GCpf windward roof (Zone 2) = + 0.21 (notice the `positive’)

GCpf leeward roof (Zone 3) = -0.43

GCpf leeward wall (Zone 4) = -0.37

We’ll look at positive internal pressures, for now, so GCpi = 0.18 (enclosed).

The resulting pressures are:

p, windward wall = qz Kd [(GCpf) – (GCpi)] = 21.8 psf (0.85) (0.56 – 0.18) = 7.0 psf

p, windward roof = 21.8 (0.85) (0.21 – 0.18) = 0.56 psf (external and internal pressures just about cancel)

p, leeward roof = 21.8 (0.85) (-0.43 – 0.18) = – 11.3 psf (negative meaning suction)

p, leeward wall = 21.8 (0.85) (-0.37 – 0.18) = – 10.2 psf (again, suction).

We’ll again consider an 8-ft `swath’ of building. And considering the `swath’ as a whole, the internal pressure coefficient (GCpi) acting on the inside faces of the (outside) walls cancel. (Yeah, the building is symmetric about its longitudinal axis.)

The wall areas are 8 ft wide x 10 ft tall = 80 sf.

The roof slope length (neglecting eaves) is … 16 ft √2 = 22.6 ft, each side, ridge to wall line.

The slope area, each side, for the 8-ft swath, 8 ft x 22.6 ft = 181 sf.

The wind forces are …

windward wall, F = p x A = 7.0 psf x 80 sf = 560 lb (horizontal)

windward roof, 0.56 psf x 181 sf = 101 lb (pushing in @ 45°)

leeward roof, 11.3 psf x 181 sf = 2045 lb ( … dropped the negative sign but understood to be sucking up and away)

leeward wall, 10.2 psf x 80 sf = 816 lb (suction).

Per the 45 degree roof slope, the horizontal and vertical forces are:

Fx = 563 lb + 101 lb (cos 45°) + 2045 (cos 45°) + 815 = 2900 lb (rounded a bit)

Fy = – 101 lb (sin 45°) + 2045 (sin 45°) = 1400 lb (rounded).

This is interesting. The horizontal wind load is greater than the vertical … and nearly twice so. Recall the vertical wind load on the shop was much greater than the horizontal wind load on the shop. The shop will tend to fly up and away in a big wind. This particular house would have a greater tendency to be pushed/pried off its foundation, and slide along the ground.

Provided neither started to get ripped apart first (Components and Cladding … stay tuned!).

]]>Now let’s look at the total horizontal and vertical wind forces, going clear down to, but excluding, the foundation. The total horizontal force will be easy, since we will again end up neglecting the roof; we end up using the force(s) on the whole walls, 1,000 lb (see the previous post).

Now let’s get the vertical force(s).

Both the windward and leeward roof planes span 20 ft, and rise 4/12 or 6.67 ft. The slope length, bottom to top, of each, is √(20^{2} + 6.67^{2}) = 21.1 ft. Both roof surfaces, then, for an 8-ft swath, have areas of 8 ft x 21.1 ft = 169 square feet (sf). Again we’ll omit the eaves, addressing them in another post, or posts.

In the earlier example we considered the shop `partially enclosed’, allowing for a large door open at one end, and the other end closed. The corresponding internal pressure coefficient, GCpi, is +/- 0.55. Since we’re looking at vertical forces, particularly `uplift’, we’ll use the +0.55; the internal pressure is pushing upward on the underside of the roof, inflating the building. This internal pressure will `add’ to the suction of the external pressures, helping it `explode’.

From the previous example the external pressure coefficients are GCpf (windward roof, Zone 2) = -0.69 and GCpf (leeward roof, Zone 3) = -0.47.

The Velocity pressure and Directionality factor, from that example, are, respectively, qz = 18.5 psf, and Kd = 0.85.

The pressures on the roof surfaces, p = qz Kd [(GCpf) – (GCpi)], are:

p (windward slope) = 18.5 psf (0.85) (-0.69 – 0.55) = 18.5 (psf 0.85) (-1.24) = -19.5 psf, and

p (leeward slope) = 18.5 (0.85) (-0.47 – 0.55) = 18.5 psf (0.85) (-1.02) = – 16.0 psf.

The resulting forces (F = p x A; 8-ft swath) are, thus,

F (windward slope) = 19.5 psf x 169 sf = 3296 lb, say 3300 lb, and I leave off the minus sign knowing it’s suction (outward and upward), and

F (leeward slope) = 16.0 psf x 169 sf = 2704 lb, say 2700 lb, also outward and upward.

The actual vertical components of these forces are,

Fy (windward slope) = 3300 lb cos (18.4°) = 3130 lb; and

Fy (leeward slope) = 2700 cos (18.4°) = 2570 lb.

The total vertical forces is … 3130 lb + 2570 lb = 5300 lb.

Whoa! The vertical (uplift) force is over five times that of the horizontal force. All other things equal, this building will have a greater tendency to lift off the ground and `fly away’, versus just being pushed along the ground from the horizontal components of the wind action. The shop acts more like an `airfoil’ than a `blunt’ obstacle to the wind.

*****

The walls, per these calculations, do not contribute directly to the (purely) vertical wind load, but we can talk more about that, later.

TAGS: wind, force, load, wind load, ASCE 7-22, Envelope Procedure, horizontal, vertical, gust, coefficient, internal, external, GCpf, GCpi, wall, roof, zone, enclosed, partially enclosed

]]>Δ = P L^{3} / (3 E I),

where

Δ = how far the top end moves (deflects), in this case, horizontally,

P = the horizontal force, applied at the top,

L = length (height) of the post,

3 = the number 3 (`three’),

E = the modulus of elasticity of the post material, and

I = the moment of inertia of the post (a function of the post cross section and orientation).

Suppose the shop has walls 9 feet tall, the posts are in the walls, and that trusses are attached at ends to the tops of the posts, the trusses span 40 feet, and the roof slope is 4/12.

The trusses, and supporting posts, are spaced 8 feet apart.

Further suppose that the force involved is a wind force hitting the shop from the side. Let’s look at the wind force involved trying to push over an 8-foot-wide `swath’ of the shop, including windward wall, leeward wall, and 8-foot-wide swath of roof, being kept from blowing over by the `frame’ action created by the truss and pair of posts at ends on opposing walls.

Let’s assume the framing of the structure takes the wind pressure forces acting on the bottom halves of the windward and leeward walls to the ground, circumventing the posts. And let’s assume that the wind pressure forces acting on the top halves of the walls (via girts, or whatever), and swath of roof, are delivered to the tops of the posts (at the roof/wall line).

The total horizontal force delivered to the `swath’ walls and roof, and, in this case, delivered to the cantilever posts, is, say, 400 pounds (lb), which we will assume divides equally, to 200 lb, to each of two posts (windward and leeward).

(See here and here for an example calculation on a similar structure, but with different wind design parameters.)

Now let’s say we are using a 6 x 6 Douglas-fir wood post, Grade No. 2, for which E = 1,300,000 pounds per square inches (psi), and I = 76.3 in.^{4} (both values from the *National Design Specification® for Wood Construction Supplement*).

Before we drop the above `ingredients’ into our `standard formula’, let’s get the post height into `inches’;

L = 9 feet x 12 inches / foot = 108 inches (in.).

Δ = P L^{3} / (3 E I) = 200 lb (108 in.)^{3} / (3 x 1,300,000 psi x 76.3 in.^{4} ;

__Δ____ = ____0.8____5____ inches__. (Almost 1 inch.)

If you were sitting on the eave line, and could ignore the wind that would be causing this force, you would certainly feel this movement.

If the shop is really long, the whole building would essentially experience this lateral movement, sway.

If the building is not really long, then the roof would act as a `diaphragm’ … a long, flat, thin, horizontal beam, and as the building starts to sway, the top of the post would engage the stiffness of the roof, and some of the force would be carried through the plane(s) of the roof to the end walls. The amount of wind force carried by the posts, versus the amount carried to the end walls, relates to the relative stiffness of the roof (as a diaphragm) and the (stiffness of) the posts. `Load goes to stiffness.’ Really stiff posts – the wind load will be carried by the posts. Really stiff roof, and end walls – the lateral load will be carried by the roof and end walls. Most real buildings fall somewhere in between. We’ll discuss this in another post.

But there’s more to the sway of the post than the pure `bending’ of the post itself. If the post is embedded in a concrete pier, the concrete pier probably holds the post pretty `rigid’ (as long as the pier itself doesn’t move). If the column is attached to the pier, or other type of foundation, with some kind of metal `bracket’, we need to look at how `rigid’ the bracket is. And the answer probably is, `not very’. Typically, wood connections, being wood-to-wood or wood-to-metal, are not relied upon to provide rigidity, though the attempt at making such brackets is not new, and is in fact, ongoing. The only real `tried-and-true’ rigid wood connection is the so-called `Moment Splice’; see Chapter 15 of the *Timber Construction Manual*, Sixth edition, by the American Institute of Timber Construction. This connection is not what you would see at the base of a post for a shop, or barn.

Supposing, though, hurdles cleared, a rigid (`moment’) connection is found for the 6 x 6 post, for installation in concrete, and that it has a stiffness of (I’m just picking a number, but not terribly dissimilar from what you might find `published’ by a manufacturer) 3,000 lb-ft / degree. This means that a `moment’ of 3,000 lb-ft exerted by the post on/in the bracket, will cause the post to rotate, in the bracket, 1 degree of angular measure. In the example above, we’re looking at a moment of 200 lb at the top of a 9-ft post, or, about 1,800 lb-ft. Thus, the bracket will rotate … 0.6 degrees. The post comes out of the bracket at an angle! The angle is not a big one, but watch! This angle is magnified by the length (height) of the post. The top of the post, just per the angular rotation, moves, laterally, 0.6 degrees times π rad per 180 degrees … times 108 inches, or 1.13 inches. The total movement of the top of the post is the sum of that produced by rotation in the bracket, plus the bending of the post itself, … 0.85 inches plus 1.13 inches, or 1.98 total inches … __almost ____2____ inches__. (Pick your bracket carefully!)

It is interesting that, in this example, the lateral movement of the top of the post is due more *to the bracket* than the bending of the post itself.

And, again, note that in any `real’ building, the top of the post will have engaged the `diaphragm action’ (resistance) of the roof system itself, long before this calculated deflection, or lateral movement, meaning that the post will not take on the full 200-lb load, and nor will the top end deflect the full 2 inches. How much, or how not much, depends on the stiffness of the roof.

+++++

Rotation of the post in the bracket might result from any number of effects, such as crushing of wood by the fasteners and in bearing with the bracket metal, oversize fastener holes, gaps between wood and metal due to poor fit and/or shrinkage, and metal deformation.

+++++

Also note that this assumes no movement (rotation) of the foundation itself, where it is assumed that the bracket is in some way attached to or embedded in concrete. If the concrete is a concrete pier, then attention should be paid to how much the *concrete pier *might rotate in the founding or backfilled soil. Seriously.

Step 6: Determine the external pressure coefficients, GCpf … Section 28.3.2.

For Case 1 we get … roof slope tan^{-1} 4/12 = 18.4° … interpolating between 5° and 20° …

interp |
|||

GCpf |
per |
||

slope, deg |
actual |
||

Zone |
5 |
20 |
18.4 |

1 |
0.4 |
0.53 |
0.52 |

2 |
-0.69 |
-0.69 |
-0.69 |

3 |
-0.37 |
-0.48 |
-0.47 |

4 |
-0.29 |
-0.43 |
-0.42 |

Note that the positive coefficient for Zone (1) and the negative coefficient for (4), are, as far as the frame (bent) is concerned, `additive’; both tend to push (`flop’) the frame over.

The coefficient for Zone (2), the windward roof plane, is interesting. The negative sign means suction (`away’ from the surface). This pressure actually counteracts the lateral (horizontal) wind effects on the wall. It begs the thought of a `negative’ wind drag force. In fact, if one looks at the pressure distribution on an airfoil (this low-rise shop is kinda like an airfoil), there is a zone on the upper front of the foil where the pressure does indeed act `upstream’! But, for both the airfoil, and any building, any `upstream’ effect is presumably wiped out by the other, `downstream’, effects, such as on the rest of the surfaces.

It is also interesting to note the coefficients for Zones (1) and (4) for Load Case 2, wind hitting the end walls. In Case 2 the side walls experience negative pressure (suction), and the building wants to `explode’. We are not looking at that effect in this post; just the flopping over from net force pushing on the windward (side) wall, wind going over the roof like a foil, and sucking on the leeward (side) wall.

Note further that we are not looking at the ends of the side walls and roof (Zones 1E, 2E, 3E, etc.); we can look at those later. Right now we are getting the lateral force(s) for the interior (between-the-ends) transverse frames (bents).

Step 7: Calculate the wind pressure, p, from Equation 28.3-1 …

p = qz Kd [(GCpf) – (GCpi)] … for each surface.

Without proof here, and per the symmetry of the frame (bent), I state the following: as far as the frame (bent) is concerned (an 8-ft wide swath through the building), the internal pressures cancel out for the walls, and, for the walls, we will look at the `sum’ of the external pressures. For the roof surfaces, I only need to look the `sum’ of the roof pressures as acting on a vertical projection of the roof volume. Here goes:

Wall(s): p = qz Kd GCpf windward (Zone 1) – qz Kd GCpf leeward (Zone 4) =

p = qz Kd (GCpf windward – GCpf leeward) =

p = qz Kd (0.52 – (-0.42)] =

p = qz Kd (0.94).

Aside from the Kd term, the net horizontal pressure is (94%) that of `stagnation’ pressure.

p = 18.5 psf (0.85)(0.94) = __14.8 psf__.

This force acts on an 8-ft wide swath of the windward and leeward walls, with (projected) area equal to:

A = 8 ft x 8.5 ft = 68 square feet (sf).

The total (net) force acting on the walls (pushing on the windward and sucking on the leeward) is, thus,

F = p A = 14.7 psf x 68 sf = 1,000 lb.

Oh, wait! The walls are no doubt attached to some kind of sill, at the ground (foundation); so let’s say that half of the wall force goes up to the eave, and half goes to the ground. Therefore, the force from wind action on the walls, acting at the top of the wall, is

F = 1,000 lb / 2 = 500 lb.

NOW FOR THE ROOF …

p = qz Kd (GCpf Zone 2 – GCpf Zone 3) =

= qz Kd [(-0.69) – (-0.47)] =

= qz Kd (-0.22) … whoa, yeah, the `negative drag’ effect …

= 18.4 psf (0.85) (-0.22) = -3.4 psf … again, I wish I didn’t have 0.85 for two different effects.

The vertical projected area of the roof is … 8 ft wide x 6.67 ft tall, or 53.4 sf.

F = p A = -3.44 psf x 53.4 sf = -183 lb.

The total horizontal force on the frame (bent) is …

Fx = 500 lb – 183 lb = 317 lb.

But, we need to look at Section 28.3.3. It requires that the total horizontal force not be less than that determined without the pressures on the roof. So, while this `negative drag’ on the roof might be in some sense real, it should not be counted on, and, in other cases, perhaps *doesn’t* exist … these provisions and requirements, remember, are part of the `envelope procedure’, intended to capture a whole range of possibilities and geometries. We’ll talk more about this `negative drag’ in another post.

So, the total horizontal force is … 500 lb.

Let’s divide this equally to the ends of the trusses/tops of the posts; the lateral (horizontal) force at the top of each post is thus,

Fx, top of post = 500 lb / 2 = 250 lb, each.

We’ll see that this is not very much compared to the vertical (uplift) forces on the roof (and posts).

(Stay tuned for Part 3 … the vertical (uplift) wind load on the shop … *in progress.*)

*****

And on roof overhangs (28.3.5) … here (forthcoming)

And on `Minimum Design Wind Loads’ (Sec. 28.3.6) … here (forthcoming).

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