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.
Engineers use an iterative process to fine tune the various elements into final structural element designs. Think of this as repetitive in nature working toward the ultimate goal of an efficient design that meets the variety of requirements the structure’s configuration places on the path that the applied load will need to take to get to the ground. The engineer starts with a broad understanding of the loads on individual elements and narrows the focus until each element and ultimately the entire structure is designed to safely transfer all loads, meet code requirements and provide an acceptable solution that can be signed and sealed. Through this process the load paths are accurate, specific and reliable. With the accurate load paths, drafting can be completed with fully detailed structural plans available for construction. 
Table 1C specifies an overall length of 5″ and 3″ of thread length. Allowable fastener shear is 1235# which by Footnote 4, “Allowable shear strength values apply only to shearing in the unthreaded shank portion of the fastener”. This would be fastener failure itself. This however is not our limiting value. 
