Timber beams are inherently fire safe by virtue of their size. Think of a big tree trunk in a forest fire. The trunk is the `last to go'. Actually, most tree trunks survive forest fires. It is the same thing with big timbers in structures. While other materials `go up in smoke' ... beams, posts, and girders last a long time. As such, major collapse of a structure is prevented or delayed until adequate time to allow occupants, if any, to flee.
Just `how safe' depends on just `how big' the member is. This bigness is approached a couple, actually three, ways.
The first way is to evaluate the size of the timbers with respect to `Heavy Timber' (Type) Construction. In Heavy Timber (HT) construction timber members above certain sizes are deemed adequately (fire) safe. I'll save the HT sizes discussion for another post.
The other approaches to `bigness' deal with: 1) calculating how long the timber (cross) section endures a fire, in number of minutes, and 2) calculating the strength of the structural member at the end of a certain amount of time.
How long a timber retains sufficient strength can be calculated using the procedure in Technical Note 7 by the American Institute of Timber Construction (AITC), the method of which is also in the International Building Code, Chapter 7. I provide more discussion and examples of this procedure in the following articles. Both Tech Note 7 and the method in Ch. 7 of the IBC are in the context of `1-HOUR' fire resistance, and whether (or not) the timber has enough strength for a `1-Hour' fire.
( ... for a Glulam Beam) Fire Resistance Calculation for a Glulam Beam
( ... for a Glulam Column) Fire Resistance Calculate for a Glulam Column
The AITC Tech Note 7 method is nice in that its basics can be taught to non-engineers!
The drawback of the method is that for glulam beams the method is restricted to optimized glulam beams with a modified `1-hour' layup, or to uniform layup beams. Modified `1-hour' beams are not a stock product, and thus must be ordered as such. Uniform layup timbers are not commonly used for beams. Uniform layup timbers are commonly used for columns; thus the procedure is applicable to a stock or unmodified column glued laminated timber.
The other way to address fire safety is to evaluate its capacity after a certain amount of fire (1 hour, etc.). This method is provided in the National Design Specification for Wood Construction, Chapter 16, and referenced in the IBC. With regard to glulam beams it is restricted to those beams with the modified layup. The AITC Timber Construction Manual provides a procedure that deals with unmodified beams. The following article follows the method of the NDS and TCM including dealing with unmodified beams.
( ... for Glulam Beams, modified and not ... HERE)
This latter method is better digested by structural engineers.
Tuesday, April 23, 2013
Monday, March 11, 2013
Here is a mis-match in widths from beam to support column. The beam is 6-3/4 x 33 and the column is 6-3/4 x 6 ... but the 6-3/4 got rotated 90 deg. Thus, trim (and siding) had to be ripped above. Better framing would be to make the 6-3/4 in. dimensions flush. In fact, I think that is how it was originally designed. Rotating the column gave it a different reveal.
In a previous post we tackled the design of the framing elements (decking, joists, support beams, and posts) for a 12’ x 32’ exterior deck for a mountain cabin. In that design we came up with four 3-ply 2 x 12 Douglas Fir – Larch (DF) beams, each 8 feet (ft) long; the inner beams spanning 8 ft `simply’, and the outer beams 7 ft with 1 ft overhangs. In this article we examine the use of the same 2 x 12s, except that instead of four 8-ft beams, we will use two 16-ft beams. These two 16-ft beams will span 8 ft, 7 ft, and then over the outer supports to overhang 1 ft. The advantage to using two longer beams (instead of four) will be better framing `continuity’ and reduced beam deflection (sag).
The original design using the 8-ft simple spanning beams was `easy’; we used a `Wood Beam Capacity Calculator’ that we found on line. That Calculator, however, does not handle continuously spanning beams. This article covers the more cumbersome `hand calculations’ that we will need instead.
We will start by determining the load on the beam, in pounds per linear foot (plf). Then we will calculate the `internal forces’ (Bending Moment and Shear) and the beam deflections. From the internal forces we will calculate the internal stresses (bending stress and shear stress). Then we will check the stresses due to load with regard to the `Allowable’ stresses for the beams (based on species, grade, and service conditions). And we will make sure the deflection under load is not excessive. We will start with the same 3-ply 2 x 12 built-up beam, Select Structural grade lumber, and then add or subtract plies, change lumber size, if necessary, until everything `checks out’.
To get the internal forces we will model the each 8 + 7 + 1 ft beam as just having two 8 ft continuous spans. Item 29a of Table B.1 from the Timber Construction Manual gives the following:
Bending Moment, M = w L2/8,
Shear, V = 5 w L / 8, and
Deflection, Δ = w L4 / 185 EI,
w = line load on the beam,
L = span length (support to support),
E = Modulus of Elasticity of the wood, and
I = Moment of Inertia of the beam section.
The beams will carry approximately half of the 11 ft joist spans plus the 1 ft joist overhangs, for a tributary width of ½ of 11 + 1 = 6.5 ft. Thus, the line load on the breams will be,
w = 260 psf x 6.5 ft = 1690 plf.
Let’s add 10 plf for the weight of the beams themselves, or
w (total carried by beams) = (total carried by beams) = 1700 plf.
From the National Design Specification for Wood Construction – Supplement – Design Values for Wood Construction, we obtain the following Design Values and section properties for Douglas Fir Select Structural Dimension Lumber:
E = 1,900,000 psi, and
Fv = 180 psi. (Yes, we will also check Shear.)
For a `3 ply’ we will have:
A = 3 x 16.88 = 50.6 in.2,
S = 3 x 31.64 = 95 in.3, and
I = 178 x 3 = 534 in.4.
The bending moment is,
The bending stress, fb, is,
The `allowable’ bending stress (Fb’) is the Fb multiplied by appropriate adjustment factors;
Fb’ = Fb (from the Supplement) x 1.15 (Load Duration factor for Snow load) x 1.0 Size Factor (see Supplement) x (0.9 presumed Stability factor) x 1.15 (Repetitive Member factor, 3 or more members side-by-side, see Supplement), or
Fb’ = 1500 psi x 1.15 x 0.9 x 1.15 = 1785 psi.
Since the bending stress is 1718 psi, the 3 ply 2 x 12 is strong enough with regard to bending stress.
Now we check Shear in the beam.
The Shear stress in the (rectangular shape) beam is,
fv = 3 V / 2 A,
V = 5 w L / 8 = 5 (1700 plf)(8 ft) / 8 = 8500 lb.
fv = 3 (8500 lb) / [2 x 50.6 in.2] = 252 psi.
The `allowable’ shear stress is the published value from the Supplement multiplied by the appropriate adjustment factors for shear,
Fv’ = 180 psi x 1.15 (for Snow) = 207 psi.
This is a bit scary in that the design stress ends up being more than the allowable stress. Not good! However, let’s check one thing. The National Design Specification (NDS) allows `uniform’ loads within a distance `d’ of the supports to be omitted in the shear design check, where d is the depth of the member. In our case, then, we may deduct 11.25 twelfths of the 1700 plf for V. Expressed in equation form,
V (design) = V (@ support) – w d, or,
V (design) = 8500 lb – 1700 plf (11.25/12 ft) = 8500 lb – 1594 lb = 6906 lb.
The correspond stress is,
fv = 3 (6906 lb) / [2 x 50.6 in.2] = 205 psi.
Δ = (1700 / 12 pounds per inch) (8 x 12 in.)4 / [185 x 1,900,000 psi x 534 in.4] = 0.064 in. ( ... about 1/16th of an inch).
With respect to the span, this corresponds to 0.06 in. / (8 x 12 in.) = 0.00067 or about 1/1500.
Going back to our deflection limit of `span / 240’for a deck covered with snow (original article), we can see that the 3-ply 2 x 12 beams will be plenty stiff enough.
The 3 – ply 2 x 12 built-up beams WORK!!!
Wood Beam Capacity Calculator by J. Ochshorn, Cornell University, Ithaca, New York.
National Design Specification for Wood Construction and Supplement – Design Values for Wood Construction, 2005, American Wood Council, Washington, D.C.
Consider the deck illustrated. It is 32’ x 12’ and will be attached to the gable end of a proposed mountain cabin. The cabin is in an area of heavy snowfall, making the design of such, including the deck, outside the scope of `Conventional’ construction. This article covers the selection of the main structural elements of the deck, specifically, the decking, joists, beams, and support posts. The steps, though structural, will not be considered here, nor other non-structural aspects, such as the railing for the deck.
The cabin is to be located in Valley County, Idaho, in a location for which the following Design Criteria are provided by the local building authority: Snow load = 150 psf (pounds per square foot), and Percentage of Snow for Seismic = 35%. In this area the following materials are presumed to be economically available: Douglas fir (DF) Glued Laminated Timber (Glulam) beams; Douglas fir or Alaska Yellow Cedar decking; and Douglas fir dimension lumber (joists and built up beams).
The design Snow load provided by the local building authority is presumed to be a `roof’ design snow load (since not indicated otherwise). In general, snow accumulations on elevated roof surfaces, particularly those that are heated, are about 70% of what occurs naturally on the `ground’. To design the deck, a `ground’ value of, say, 250 psf will be used, presumed to account for what might accumulate on the unheated deck near the ground, and also accommodating some drifting off the gable roof above.
( ... yes, there is a deck under here, and this isn't even the design snow load)
The joists will be `designed’ first, as they drive the decking design. Spacing of joists for floor applications is generally at 16 inches (in.) on-center (o.c.), or less; for decks, 16 in. o.c., or perhaps more. However, due to the robust snow load, it may be necessary to keep the joists pretty close. We shall see!
The joists planned will span 11 feet (ft) from ledger to support beam and then overhang 1 ft more to get the full deck width. To simplify the design of the joists they will be treated as simple, single span joists of 11 ft; the 1 ft overhangs will provide a bit of counter bending, making the single 11 ft span assumption conservative. Using the American Wood Council `Maximum Span Calculator for Joists and Rafters’ (Span Options tab), 11 ft span, 250 psf Snow load, 16 in. o.c. spacing, and an assumed 5 psf Dead load, the following design options are revealed: Douglas fir, Select Structural, 2 x 12. This design assumes that the Wet Service factor is not applicable (moisture content of wood less than 19%), and that the joists have not been incised for Preservative treatment. In situations where high moisture content in wood is expected, during the design loading condition, it would be appropriate to apply the Wet Service factor (`Yes’ on the Wet Service option). Note further that a deflection limit of L/240 was used. For floors in general a value of L/360 is set, primarily for walking comfort and to protect crack-able materials. Since comfort will not be an issue for a deck with 12 to 15 feet of snow on it (unless it is some other kind of comfort), and since the deck will not be constructed of crack-able materials such as concrete or gypsum, the deflection limit was relaxed (to that of a wood roof).
(Indeed by playing with the `Calculator’ it is seen that changing the deflection limit back to L/360, or further to 180, doesn’t control the determination of the deck joists.)
Once the joist span is obtained the required decking thickness may be determined. The cabin roof will be constructed of Douglas fir `Select’ grade heavy timber decking. Let us use this material for the attached deck also. However, the decking for the roof will have a tongue and groove profile, whereas the decking for the `deck’ will be S-4-S, meaning `surfaced (straight) four sides’. In fact, the decking for the deck should be specified to be installed with ¼ in. gaps between courses (adjacent pieces) to allow for drainage (rain, snowmelt, various forest litter).
Using the American Institute of Timber Construction Standard for Tongue-and-Groove Roof Decking, AITC 112-93, Table 3, design values for Douglas Fir (Douglas Fir – Larch) Select are: Bending Stress (Fb) = 2000 psi (pounds per square inch) and Modulus of Elasticity (E) = 1,800,000 psi. Pages 11 and 13 of the Standard provide adjustment values for the Bending design value as follows: multiply by 1.15 for Snow and 1.10 for 2 in. thickness or 1.04 for 3 in. thickness. (The latter adjustment is based on the published design values being for 4 in. decking). Thus, assuming we will use 2 in. decking, Fb (adjusted) = 2000 psi x 1.15 x 1.10 = 2530 psi, and E = 1,800,000 psi (doesn’t get adjusted). (Hot stuff, indeed!)
The load and span tables provided in the Standard aren’t going to work for our deck design as our spans are way shorter than those for which allowable loads are published. Hence, we are going to have to do some design-check calculations.
A single piece (strip) of decking will carry a `line’ load of,
w = σ x S = (250 psf Snow + 5 psf Dead + 5 more Dead for decking itself) x (5-1/2 in. actual width + ¼ in. gap) / 12 in. per foot = 260 psf x (5.75/12 ft) = 125 pounds per linear foot (plf), where, of course, σ is the load in psf, and S is the width (`swath’) of deck that each piece supports.
Assuming any particular piece will span 4 or more joist spaces, the associated bending moment, M, and deflections, Δ, are (Item 42, Table B.1, American Institute of Timber Construction Timber Construction Manual):
M = 0.107 w L2 = 0.107 (125 plf) (16/12 ft)2 = 23.8 pound-ft (lb-ft) or 285 lb-in. (multiplying by 12 ... we’ll need M in lb-in. later). Obviously, L here is the structural span length of the decking ... equal to the joist spacing.
fb = M/S,
Making sure we take into account the decking is oriented flat,
S = (5.5 in.)(1.5 in.)2/6 = 2.06 in.3. (Of course the actual thickness of the `2 X’ decking is 1-1/2 in.).
fb = M/S = 285 lb-in. / 2.06 in.3 = 138 psi.
The so-called bending design check is, then, ... (is) fb = 138 psi ≤ Fb (adjusted) = 2530 psi (?). Yes! Way yes! The 2 x 6 DF decking is way strong enough to carry all that snow.
Checking deflection; Δ = 0.0065 w L4/EI (also from Item 42, Table B.1, Timber Construction Manual), where I is the Moment of Inertia, bh3/12.
Again remembering that the decking is flat,
First dividing w by 12 to get it into a pounds per inch line load, and leaving L in inches for span,
Δ = 0.0065 (125 / 12 pounds per inch) (16 in.)4 / [(1,800,000 psi) (1.55 in.4)] = 0.0016 in.
Wow, that is only about 1/600th of an inch.
In terms of a ratio with the span, 0.0016 in. / 16 in. = .0001 = 1/10,000.
If we follow what we did for the joists, our limit is as much as 1/240, or 16 in. / 240 = 0.066 in. The 2 x 6 DF Select decking is strong enough ... and plenty stiff enough.
The beams will carry approximately half of the 11 ft spans plus the 1 ft overhang for a tributary width of about ½ of 11 + 1 = 6.5 ft. Let’s use DF-L again for species, also Select Structural grade. Further, let us use 4 beams, 8’-0” long each, with the two `interior’ spans being `simple’, and the two end spans with the supports brought in 1 ft (for looks), with 1’ overhangs. We will model all four beams as 8’ simple span, presuming the 7 ft + 1 ft to be a bit conservation (as we did with the joists), but (just) bringing in a bit of counter-flexure. (Besides, the Ochshorn doesn’t handle an overhang.)
So, using the Calculator with Span = 8 ft, Spacing = 6.5 ft (the tributary width), Snow load duration, not wet service, Select Structural grade, Live load deflection limit L/240, we get 4 x 14 or 3-ply 2 x 12. Since the deck is pretty close to the ground and the support beams not real visible, a solid beam-like look is not essential; let’s use the 3-ply 2 x 12 (same lumber as for the joists).
Now let’s get the posts, quickly!
Let’s do a `one-size-fits-all’ approach for the support posts. From Item 29a of Table B.1 of the TCM the maximum support reaction is,
R = 10 w L / 8, or, in our case,
R = 10 (1700 plf ) (8 ft) / 8 = 17,000 lb.
Decking: 2 x 6 `Select’ Grade, S4S.
Joists: 2 x 12 DF Select Structural Grade @ 16 in. o.c.
Outer Support Beams: four 3 – Ply 2 x 12 Select Structural, 8 ft length each (outer beams 7 ft + 1 ft overhang).
Support posts: 3 ply 2 x 6 DF No. 2.
1) we will deal with the fasteners, connections, and ledger later (another article); and
2) instead of four 8-ft outboard beams, two 16-ft continuous spanning beams could arguably be better. The calculations for two 16-ft beams will be handled in another post.
Standard for Tongue-and-Groove Roof Decking, AITC 112-93, and Errata, American Institute of Timber Construction, West Coast Lumber Inspection Bureau (Custodian), Portland, Oregon.
`Maximum Span Calculator for Joists and Rafters’, American Wood Council, Washington, D.C.
`Wood Beam Capacity Calculator’ and `Wood Column Capacity Calculator’, J. Ochshorn, Cornell University, Ithaca, New York.
Thursday, February 28, 2013
Consider a shallow slope roof subject to a Snow load of 50 pounds per square foot (psf) and applied Dead (weight) load of 5 psf. Let’s find, and specify, heavy timber decking to carry the required loads while spanning 6 feet (ft) rafter-to-rafter. By case of example let’s assume that Ponderosa Pine, `Commercial’ grade decking is available, and in lengths up to 12 or 16 ft. Further, the decking will be exposed on the underside and also be the `ceiling’ material. We will use the American Institute of Timber Construction AITC 112-93, Standard for Tongue-and-Groove Heavy Timber Decking, to determine the required thickness.
According to the Standard, a number of `layups’ are typically used for heavy timber decking construction. Some layups are stronger and stiffer than others; while others are more (or less) affordable. Often the `Controlled Random Layup’ is the most affordable (per piece), as it allows the manufacturer to provide decking pieces of various (convenient) lengths. On the other hand, it may result in more trim waste than layups of pieces ordered to `exact lengths’.
Typical heavy timber thicknesses are 2 inch (in.) nominal (1-1/2 in. net), 3 in. nominal (2-1/2 in. net), 4 in. nominal (3-1/2 in. net). And the typical width is 6 in. nominal (5-1/2 in. net), though other thicknesses and widths are available from some manufacturers.
In this example we will investigate using 2 in. nominal (nom.) Commercial Grade Ponderosa Pine with a Controlled Random Layup.
Page 10 of the Standard gives us weights of various decking species. For Ponderosa Pine we are given:
2 in. nominal weighs ... 4.1 psf
3 in. nom. weighs ... 6.9 psf, and
4 in. nom. weighs ... 9.6 psf.
Total load of ... 50 + 9 = 59 psf (Snow plus Dead plus `self’ weight).
(Wood is 15% `stronger’ under snow load than `normal’.)
For Ponderosa Pine, Commercial Grade,
Fb (Design Bending stress) = 1250 psi (pounds per square in.), and
E (Modulus of Elasticity) = 1,100,000 psi.
for 2 in. nominal ... 1.10.
(The way decking is `graded’ the 2-in. and 3-in. thicknesses have a bit more unit strength. The values in Table 3 on Page 13 are based on 4 in. thickness.)
Thus, for our 2 in. `P. Pine’ subject to Snow load, the Bending Stress adjusted for size and load duration, for our application, is
The Modulus of Elasticity, E, doesn’t get adjusted; it stays at 1,100,000 psi.
To deal with `strong enough’ we look at Page 14, Table 4, ... ALLOWABLE ROOF LOAD LIMITED BY BENDING.
Scrolling down the left hand side of the Table to a Bending Stress of 1581 psi ... wait! ... we have 1550 and 1600 ... let’s go to 1550 psi (and be a bit conservative) ... for Controlled Random Layup, Span 6 ft, we get an Allowable load of 108 psf.
l/180 ... 66 psf
l/240 ... 50 psf.
Now let’s check.
The l/180 deals with total load; our total load is 59, and the Allowable is 66: good!
The l/240 deals with (the effect of) Live (Snow) load; our Snow load is 50, and the Allowable is 50. Whoa, good, barely. Perfect! (Close!)
The 2 in. decking works!
Here is our specification: 2 in. Ponderosa Pine Decking, Commercial Grade, Controlled Random Layup.
The Standard goes on to dictate how the pieces must be laid, as well as fastening requirements.
2. use a stiffer or stronger WOOD (different species or grade of decking),
3. try THICKER decking,
4. change the rafter spacing (???), or
5. take into consideration actual roof slope (Page 12 of the Standard).
See also ... http://voices.yahoo.com/weight-board-foot-quantity-calculations-decking-12006629.html?cat=6
Monday, February 25, 2013
... here is the cabin header solution in `SCL'.
Note that we couldn't use a span table since our loads are so huge. That's why I switched over to a `member properties' approach.
Note that we couldn't use a span table since our loads are so huge. That's why I switched over to a `member properties' approach.