Chances are good if you have to ask a structural design question, then you are in over your head.

Reader LARRY in DITTMER writes:

*“Can you 2 by 4 flat on an 8 foot span Truss”*

A few years ago, one of my neighbors bought a pole building kit from someone other than Hansen Pole Buildings. It was for a garage and sidewall columns and single roof trusses were placed every eight feet. Now I am relatively certain this building’s roof purlins were supposed to be 2×8 on edge between trusses – however for some obscure reason, they got installed flat wise! I am unsure as to how they were even able to get roofing installed without falling through.

This is just one of many reasons why post frame buildings should be designed by a Registered Professional Engineer.

When it comes to designing whether a roof purlin can achieve a given span, it takes a lot of calculations – both for live or snow loads, as well as wind loads. In high wind areas, wind will fail purlins (or their connections) rather than snow! I have condensed calculations down to just bending and deflection and will use minimum snow loads in this example:

**ROOF PURLIN DESIGN – Main Building (Balanced snow load)**

Assumptions:

Roof slope = 4:12 (18.435° roof angle)

Trusses spaced 8-ft. o.c.

Purlin span = 8-ft.

Purlin spacing = 24 in.

Purlin size 2″ x 4″ #2 Southern Pine

Roof steel dead load = 0.63 psf steel American Building Components catalogue

Roof lumber dead load = 0.587 psf

Total purlin dead load = 1.217 psf

**Check for gravity loads**

**Bending Stresses**

Fb: allowable bending pressure

Fb‘ = Fb * CD * CM * Ct * CL * CF * Cfu * Ci * Cr

CD: load duration factor

CD = 1.15 NDS 2.3.2

CM: wet service factor

CM = 1 because purlins are protected from moisture by roof

Ct: temperature factor

Ct = 1 NDS 2.3.3

CL: beam stability factor

CL = 1 NDS 4.4.1

CF: size factor

CF = 1 NDS Supplement table 4B

Cfu: flat use factor

Cfu = 1.1 NDS Supplement table 4B

Ci: incising factor

Ci = 1 NDS 4.3.8

Cr: repetitive member factor

Cr = 1.15 NDS 4.3.9

Fb = 1100 psi NDS Supplement Table 4B

Fb‘ = 1100 psi * 1.15 * 1 * 1 * 1 * 1 * 1.1 * 1 * 1.15

Fb‘ = 1600 psi

fb: bending stress from snow/dead loads

fb = (purlin_dead_load + S) * spacing / 12 * cos(θ) / 12 * (sf * 12 – 3)2 / 8 * 6 / b / d2 * cos(θ)

S = 21.217 psf using the appropriate load calculated above

fb = 21.217 psf * 24″ / 12 in./ft. * cos(18.435) / 12 in./ft. * (8′ * 12 in./ft.)2 / 8 * 6 / 3.5″ / 1.5″2 * cos(18.435)

fb = 2961.59 psi > 1600 psi; stressed to 185.1%

**Deflection**

Δallow: allowable deflection

Δallow = l / 180 IBC table 1604.3

l = 96″

Δallow = 96″ / 180

Δallow = 0.533″

Δmax: maximum deflection

Δmax = S * spacing * cos(θ * π / 180) * (sf * 12)4 / 185 / E / I from http://www.awc.org/pdf/DA6-BeamFormulas.pdf p.18

E: Modulus of Elasticity

E = 1400000 psi NDS Supplement

I: moment of inertia

I = b * d3 / 12

I = 3.5″ * 1.5″3 / 12

I = 0.984375 in.4

Δmax = 21.217 psf / 144 psi/psf * 24″ * cos(18.435° * 3.14159 / 180) * (8′ * 12 in./ft.)4 / 185 / 1400000 psi / 0.984375 in.4

Δmax = 1.118″ > 0.533″; 209.68% overstressed in deflection

These calculations are based upon purlins every 24 inches on center. If you were to reduce spacing to say 11 inches on center then flatwise 2×4 #2 Southern Pine with a 20 psf roof snow load would be adequate.

If you were able to somehow acquire 2850f Machine Stress Rated 2×4 with a E value of 2300000 psi (very high grade material used by some truss manufacturers) spacing could be 18 inches on center.

Again – remember these equations are just for checking for bending due to a minimal snow load, wind conditions may dictate. Please consult with a Registered Professional Engineer for actual designs.