Bending Check of a Pitched Roof Rafter

Introduction

In earlier lessons we determined the appropriate loads on a pitched roof joist / rafter for an example gable roof in Moscow, Idaho. For the rafters themselves, the unbalanced snow load of I p_g = 64 psf governs, and this load is taken to be per square foot of horizontal projection (plan area). The Dead load of the roof was given to be 15 psf, and this is per square foot of inclined roof surface. So, then, to look at `bending' of the beam, in the sense of also including shear and deflection, we determined the loads perpendicular to the roof surface (perpendicular to the joists). These loads are:

ω _{DL perp} = 19 plf, and

ω _{SL perp} = 73 plf,

and the span of the joist is

L _sloped = 17.3 ft (208 in.).

Now let's look at flexure.

M = ω L² / 8 = (19 + 73) plf (17.3 ft)² / 8 = 3440 lb-ft = 41,300 lb-in.

The Design Check is,

... Is f_b ≤ F_b' ? ...

where,

f_b = M/S,

where,

S = bh2/6 = 1.5 in. (11.25 in.)² / 6 = 31.6 in.³.

(We're checking 2 x 12 DFL No. 2 joists.)

(Alternately, we can grab S from Table 1B of the National Design Specification (NDS) Supplement, which is not that crazy of an idea since we are also going to get our Design Value(s) from the Supplement. For the situation at hand, F_b = 900 psi (Table 4A).

And, where F_b' = F_b × Applicable Adjustment Factors.

So,

... f_b = 41,300 lb-in. / 31.6 in.³ = 1310 psi.

Our Adjustment Factors are ... (See Table 4.3.1 of the NDS)

C_D = 1.15 for Snow load duration,

C_M = 1.0 (dry service)

C_t = 1.0 (normal temperatures),

C_L = 1.0 (Beam Stability Factor, which is 1.0 since our roof sheathing will be attached directly to the compression edge of the joists preventing lateral displacement),

C_F = 1.0 (Size Factor, from Supplement Table 4A),

C_r = 1.15 (Repetitive Member Factor),

and C_I and C_fu are both 1.0.

So,

F_b' = 900 psi (1.15)(1.15) = 1190 psi.

Now for our Design Check ... (it's not lookin' very good)

Is f_b = 1310psi ≤ F_b' = 1190 psi ? ... NO! ... not good.

Whoa; normally we frame roofs around here @ 24 in. o.c.; we anticipated that wouldn't be adequate and dropped down to 16 in. o.c., and that still isn't adequate.

Options:

1) go with tighter joist spacing,

2) take a look at higher grade joist material, or perhaps a larger section, or

3) consider engineered lumber.

Lumber is cheap these days; let's try 12 in. o.c. joist spacing.

Our loads on the joists will be ... 12/16 of those calculated, so, likewise,

... f_b = (12/16) 1310 psi = 980 psi.

Is f_b = 980psi ≤ F_b' = 1190 psi ? ... YES! ... good.

Deflection

Now, before we go and order all that wood, let's check deflection.

The ceiling will be gypsum, so the Live load deflection criteria will be,

Δ _{Live load} ≤ L / 360.

The Live load on the joist is ... 12/16 (73 plf) = 55 plf.

We'll use L = 17.3 ft = 208 in.

For a `simple' span,

Δ = (5/384) ω L⁴ / EI,

where,

E = 1,600,000 psi (Table 4A of the NDS Supplement), adjusted with all the applicable Adjustment Factors for E, in this case all 1.0),

and,

I = 178 in.⁴.

So,

Δ = (5/384) (55 / 12 lb/in.)(208 in.)⁴ / (1,600,000 psi × 178 in.⁴) = 0.39 in.

Is 0.39 in. ≤ L/360 = 208 in. / 360 = 0.58 in.? ... Yes, good.

But let's also check the deflection due to Total load.

Per Table 1604.3 of the International Building Code (IBC), the appropriate limit is L/240 and we are permitted to use the deflection caused by L + 0.5D,

So,

that caused by L is 0.39 in.;

that caused by 0.5D will be, ... 0.5 × 19/73 × 0.39 in. = 0.05 in.;

so, then,

Δ _{L + 0.5 D} = 0.39 in. + 0.05 in. = 0.44 in.

Is it less than L/240 = 208 / 240 = 0.87 in.? ... Yes, way yes; good!

REALLY GOOD!

All our design checks check ... albeit now with the joists at 12 in. o.c.

Discussion

Interestingly, if we had calculated `bending' based on the horizontal projection length of the joists (16 ft), and the horizontal projections of the load (64 psf and 16 psf), we would get the same f_b. Here, I'll show you.

ω = (64 + 16) psf × 1 ft (trib width) = 80 plf.

M = ω L² / 8 = 80 plf (16 ft)2 / 8 = 2560 lb-ft = 30,700 lb-in.

f_b = M/S = 30,700 lb-in. / 31.6 in.³ = 970 psi.

(Well, I think I lost some accuracy in rounding, but we're close!)

That doesn't mean that using the horizontal projection is correct; it just means that for bending all the cosines cancel out top and bottom. For shear and deflection, THEY DO NOT (CANCEL OUT)!

References

Balance and Unbalanced Snow Loads on a Gable Roof, Jeff Filler, Associated Content.

Shear Design Check for a Pitched Roof Joist, Jeff Filler, Associated Content.

National Design Specification for Wood Construction and Supplement Design Values for Wood Construction, American Forest & Paper Association / American Wood Council, 1111 Nineteenth St., NW, Suite 800, Washington, D.C., 20036, www.awc.org, 2005 Edition.

International Building Code, International Code Council, 4051 West Flossmoor Road, Country Club Hills, IL 60478.