Thursday, September 26, 2013

Bearing Length ...


... this lecture was so sweet (and short), that I just have to go PUBLIC with it!


Monday, September 16, 2013

apartment building framing ...

 ... the trusses themselves are made of 2 x 4, 2 x 6, etc. ... spaced (I assume) at 24 in. o.c.  There are some cool things to note:

1.   ... note how the trusses DON'T touch the top of the framed wall.  This is intentional! ... The trusses are intended to bear end to end.  Once loaded they will deflect a bit.  A gap here allows the trusses (roof) to sag, without loading the wall.  On my blog I have a blurb about what happens when the wall is framed tight, and the roof sags everywhere else except over the wall ... makes things look not so good.
2. ... note the single strong-back brace that ties the trusses to one another.
3. ... note the gap at the peak, probably for venting.
4. ... note the staggered stud arrangement ... probably to develop sound insulation between different units.  (It's gonna be an apartment building.)
5.  ... note the double top plate, which is typical.  There is a double stud, also.  I'm not sure why that is there except perhaps it has something to do with the splice in the bottom layer of the double top plate.
I really liked the framing of this project.  The Super on the project gave me permission to take photos, use as teaching example, blog, etc.  I need to go back when he is there and make sure he has my business card.  I like working for people who do good work. 

Thursday, August 1, 2013

Unsightly Ceiling Sag ...

Here's a fun little (REAL) thing on Ceiling Sag (Deflection) ...

 ... and here are some fun calcs to go along with.

Monday, July 8, 2013

Glulam Poles

Jamie Brett sent this photo of a glulam utility pole.  Cool! 

I had heard glulam was being used for poles, but this is the first I have seen.  Oddly, I assumed they would be round in section.  LOL. 

Tuesday, April 23, 2013

Fire Safety Calculations for Timber Beams

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.


Monday, March 11, 2013

beam-post mis-match

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.


Step-by-Step Calculations for Design of a Wood Beam for a Deck

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:

Fb = 1500 psi,

E = 1,900,000 psi, and

Fv = 180 psi.  (Yes, we will also check Shear.)

We will be using the 2 x 12s on edge (strong x-x direction); the section properties for a single 2 x 12 are:

Area, A = 16.88 in.2,

Section Modulus, S = 31.64 in.3, and

Moment of Inertia, I = 178 in.4.

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,

M = 1700 plf (8 ft)2 / 8 = 13,600 lb-ft = 163,200 lb-in. 

The bending stress, fb, is,

fb = M/S = 163,200 lb-in. / 95 in.3 = 1718 psi.

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.

Sweet! ... we did it!  The stress under load, 205 psi, does not exceed the allowable, 207. (CLOSE!) 

The deflection under the full load will be,

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.

Timber Construction Manual , 6th Edition, John Wiley and Sons, 2012, Hoboken, New Jersey.

National Design Specification for Wood Construction and Supplement – Design Values for Wood Construction, 2005, American Wood Council, Washington, D.C.