Shear Design Check for a Pitched Roof Joist

In a previous lesson / article we determined the appropriate Snow load(s) for a gable roof in Moscow, Idaho. The joists on each side of the ridge will be designed for the unbalanced load of I p_g = 64 psf (where I is the Importance factor assigned to the structure and p_g is the ground snow load.) The roof Dead load was given to be 15 psf. The Snow load is taken to be that acting on a horizontal projection of the roof, and the roof Dead load is that of the actual (inclined) roof surface, in this example sloped at 5/12 (22.6°). We will chiefly be considering the loads acting perpendicular to the joists, so we must determine what components of the above loads to use, and the appropriate span of the members. The joists under consideration are 2 x 12 DFL No. 2 spaced at 16 in. o.c.

Joist Spans

Not considering wall and ridge beam thicknesses, the joists span 16 ft from eave to ridge in plan. The slope length of these joists (for structural consideration) is,

L _slope = L / cos 22.6° = 16 ft / cos 22.6° = 17.3 ft = 208 in.

For now we will not take into consideration the eaves for analysis of the main spans of the joists. In reality short eaves will tend of `off-load' the main spans a bit, and increase the shear at the eave wall a bit. Thus for flexure and deflection we will be conservative (by not considering the eaves), and non-conservative for shear. But, also, we will see that in longer span joists, shear generally does not control.

Loads on the Joists

DEAD

Now for the loads on the joists: the Dead load is the weight of the roof per square foot of roof surface. That portion of the load acting perpendicular to the surface (and the joists) is,

σ _perp = σ cos 22.6° = 15 cos 22.6° = 14 psf (approximately) = σ _{DL perp}

For joists spaced 16 in. o.c., the Dead line load on the joists is,

ω _{DL perp} = σ _{DL perp} trib width = 14 psf × 16/12 ft = 19 plf = 1.6 pli (lb/in.).

In this particular case the 15 psf Dead load has included the weight of the joists.

SNOW

The Snow load is different than the Dead load. For every square foot of roof plan there is 1/cos 22.6° = 1.08 square foot of roof surface. So, the Snow load spread out upon the surface of the roof is,

64 lb/sq ft of plan × 1 sq ft of plan / 1.08 sq ft of roof = 59 psf.

Now, this 59 psf is acting downward. The component acting perpendicular to the roof and thus the joists is,

59 psf cos 22.6° = 55 psf.

The Snow line load on each joist is thus,

ω _{SL perp} = 55 psf × 16/12 ft = 73 plf = 6.1 pli (lb/in.).

Summarizing, the line loads acting perpendicular to the joists are:

ω _{DL perp} = 19 plf, and

ω _{SL perp} = 73 plf.

Design Checks

The design checks we will perform on the joists will be: 1) shear parallel to grain, 2) bending, 3) Live load deflection, and 4) Total load deflection. Shear will be considered here; bending and deflection will be considered in a continuation article.

SHEAR PARALLEL TO GRAIN

For members that are loaded at the compression edge and supported from the other edge, producing a `compressive' reaction, the uniform loads within a distance `d' from the end may be neglected in shear design, where d is the depth of the member.

Hence, the shear load under consideration is,

V _{@ d} = ω L/2 - ω d = ω (L/2 - d) = (19 + 73) plf (17.3 /2 ft - 11.25/12 ft) = 92 plf (7.7 ft) = about 710 lb.

The shear stress is (shear parallel to grain at neutral axis),

... fv = (3/2) (V/A) = (3/2) ((710 lb / (1.5 in. x 11.25 in.) ) = 63 psi.

This is checked against the Allowable stress, which is the shear parallel to grain `Design Value' multiplied by all applicable adjustment factors.

The Design Value for DFL No. 2 is found in the National Design Specification (NDS) Supplement, and the adjustment factors are found in the NDS itself.

In `formula' form,

... F_v' = F_v C_D C_M C_t ...

Where

... F_v' = Allowable Stress,

F_v = Design Value = 180 psi (from the Supplement),

C_D = 1.15 for Snow load duration,

C_M = 1.0 (dry service)

C_t = 1.0 (normal temperatures), and so on.

Thus,

F_v' = 180 psi (1.15) (1.0) ... = 207 psi.

Our design check for shear becomes, ...

Is f_v = 63 psi ≤ F_v' = 207 psi? ...

And the answer is `yes', so this design check is ... `GOOD'.

Conclusion

So far our design check is good; but we have only looked at shear, and the span is pretty long. Shear generally governs in only relatively short spans. Next we will look at bending and deflection.

References

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.