In an earlier article we determined the required size for a Glulam roof beam carrying a heavy snow load. The determination was based on the assumption that `bending stress’ governed. We also had to guess on the weight of the beam itself. In this article we will check on our guess of the `self-weight’, as well as the other design considerations of deflection (beam sag) and shear.
From the previous article the following information is obtained:
Applied `line’ load: 2450 pounds per linear foot (plf);
Assumed self weight: 50 plf;
Assumed beam species and grade: Douglas Fir, `24F’;
Acceptable Sizes (based on bending):
6-3/4 inches (in.) wide x 45 in. deep;
8-3/4 in. x 39 in.; and
10-3/4 in. x 36 in.
First let’s check the self-weight assumption.
Douglas Fir beams in service weigh from 32 pounds per cubic foot (pcf) to 35 pcf, or more, or less, depending on the moisture content of the wood. The beam in question will be in service, indoors, in the intermountain west, for which moisture content values tend to be low. Hence, 32 pcf will be used in our check.
The weight of a beam per linear foot, ω self weight, may be obtained using the following equation:
ω self weight = γ x A,
γ is the weight per cubic foot, and
A is the cross-sectional area of the beam.
For the 8-3/4 x 39 beam,
ω self weight = 32 pcf x (8.75/12 ft x 39/12 ft) = 76 plf
(where, of course, the cross section dimensions have been divided by 12 for the calculation to be dimensionally consistent).
The weight is a bit more than assumed, hence, the earlier beam sizes should be checked again to assure they are still adequate.
For example, the 8-3/4 x 39 beam will carry at total of 2450 plf + 76 plf = 2526 plf.
The bending moment will be (drawing on the equations and values from the previous article),
M = 2526 plf (36 ft)2/8 = 409,212 lb-ft = 4,910,544 lb-in.
The bending stress will be,
fb = M/S = 4,910,544 lb-in. / 2218 in.3 = 2214 psi.
The Allowable stress for the 8-3/4 x 39 is (see previous article)
Fb’ = 2400 psi x 1.15 x 0.798 = 2202 psi.
DANG! … we are `barely over’ (over-stressed). Strictly speaking a larger beam section should be selected. Many designers would say `close enough’ (after all, they could both be rounded to the nearest hundred at 2200 psi). Alternately, one could state that the design stress is only … 2214/2202 = 1.005 … ½ of 1 % over. Perhaps a better way to justify use of the 39 in. deep section (if we really wanted it) would be to evaluate the design loads to see if they were robust, or lean. In the example of the previous article a roof dead load of 25 psf was used. This value was an assumed `robust’ value for wood roof construction. If push comes to shove, the actual roof material weights could be identified and be shown to be, perhaps, significantly lighter … perhaps, so much so that the 39 in. deep section would be suitable. The other thing we did in the previous article was to neglect the counter-balancing effect of the beam overhang to carry roof rake. Taking into account the counter-balance of the overhang load might also bring the `exact’ calculations back to `good’.
Or just go with the bigger size beam, and be a bit over-safe.
Continuing with the 8-3/4 x 39 beam …
To investigating deflection we will consider the `live’ load only, in this case the snow load. The deflection, Δ, of a `simple’ single span beam uniformly loaded is,
Δ = (5/384) (ω L4 / E I),
ω is the line load, again,
L is the beam length,
E is the beam Modulus of Elasticity, and
I is the beam Moment of Inertia.
The 24F Douglas Fir beam is commonly available in the `1.8E’ stress class meaning it has a Modulus of Elasticity of 1,800,000 psi.
From Table 1C if the National Design Specification for Wood Construction – Supplement, we obtain a value of I = 43,250 in.4 for the 8-3/4 x 39 section.
Getting ready to plug in the numbers (again borrowing from the previous article) …
ω = 150 psf x 14 ft = 2100 pounds per linear foot =2100 / 12 = 175 pounds per inch.
L = 36 ft = 36 x 12 = 432 in.
Plugging in the numbers gives,
Δ = (5/384) [ 175 lb/in. (432 in.)4 / (1,800,000 psi x 43,250 in.4)] = 1.01 in.
A typical deflection `limit’ for a wood roof (with no crackable materials such as sheetrock) is 1/240th of the span for the live load deflection.
In our case, then, the limiting deflection would be 432 in. / 240 = 1.8 in. Since the deflection under applied load is less than the limit, the deflection `checks’ (meaning the sag in the beam is deemed not to be excessive).
Now let’s check shear in the beam.
The shear force in the beam, V, greatest at the ends, and is
V = ω L / 2; in our case,
V = 2526 plf x 36 ft / 2 = 45,468 lb.
The corresponding shear stress, fv, is (for a rectangular shape section),
fv = 3 V / 2 A,
where, again, A is the cross section area.
Thus, for the 8-3/4 x 39 beam,
fv = 3 x 45,468 lb / (2 x 8.75 in. x 39 in.) = 200 psi.
The Allowable shear stress for the beam, Fv’, is the Design Value for Shear Parallel to Grain, Fv, multiplied by the adjustment factors appropriate for the conditions.
From the National Design Specification – Supplement, Table 5A, Fv = 265 psi, for the 24F-1.8 stress class.
The only applicable adjustment is for load duration, 1.15 for `snow’.
Thus, the Allowable shear stress in our case is,
Fv’ = 265 psi x 1.15 = 305 psi.
Since the applied stress is only 200 psi, the Shear Stress check is `good’.
One thing remains. Let’s also determine the required `bearing length’ so the wood doesn’t get crushed at the ends of the beam. That will be the subject of the next article.
Designing a Glulam Beam to Carry a Heavy Snow Load, Jeff Filler, YAHOO Contributor Network (Submitted).
National Design Specification for Wood Construction and Supplement – Design Values for Wood Construction, 2005, American Wood Council, Washington, D.C.