# How Far Can a 2×6 Purlin Span?

How Far Can a 2×6 Purlin Span?

How far can a 2×6 purlin on a 6:12 sloped roof span?”

The following describes 2×6 SYP #2 purlins spanning a 14′ bay, with an on-center spacing of 24″ (sf).

Purlins are recessed between rafters with their top edges flush with rafter top edges. Purlins are mounted to rafters with Simpson Strong-Tie LU-26 joist hangers at both ends.

Effective simple beam span length (le) will be taken as 165.

Dpurlin = purlin density × ((b × d × le) / (sf × l))
Purlin density found via NDS Supplement 2015 Section 3.1.3:
density = 62.4 × (G / (1 + (G × 0.009 × moisture content))) × (1 + (moisture content / 100))
moisture content = 19%
density = 62.4 × (0.55 / (1 + (0.55 × 0.009 × 0.19))) × (1 + (0.19 / 100))
density = 34.56 pcf
Dpurlin = 34.56 pcf × ( ( 1.5″ × 5.5″ × 165″ ) / ( 24″ × 168″ ) ) × 1/12 in/ft
Dpurlin = 0.966 psf

Roof designed for 29g corrugated steel
Dead load from weight of steel (Dsteel) based on values from the American Building Components catalogue:
Dsteel = 0.63 psf

D = Dpurlin + Dsteel
D = 0.966 psf psf + 0.63 psf psf
D = 1.596 psf

Project load to a vector acting perpendicular to the roof plane:
D = D × cos(Θ)
D = 1.596 psf × cos(0.464)
D = 1.428 psf

A conversion from psf to psi will be made for ease of calculation:
D = 1.428 psf × 1/144 psi/psf
D = 0.01 psi

Lr = 18 psf

Project load to a vector acting perpendicular to the roof plane:
Lr = Lr × cos(Θ) × cos(Θ)
Lr = 18 psf × cos(0.464) × cos(0.464)
Lr = 14.4 psf

A conversion from psf to psi will be made for ease of calculation:
Lr = 14.4 psf × 1/144 psi/psf
Lr = 0.1 psi

S = 13.267 psf

Project load to a vector acting perpendicular to the roof plane:
S = S × cos(Θ) × cos(Θ)
S = 13.267 psf × cos(0.464) × cos(0.464)
S = 10.614 psf

A conversion from psf to psi will be made for ease of calculation:
S = 10.614 psf × 1/144 psi/psf
S = 0.074 psi

W = 9.6 psf

A conversion from psf to psi will be made for ease of calculation:
W = 9.6 psf × 1/144 psi/psf
W = 0.067 psi

Wu = -11.763 psf

A conversion from psf to psi will be made for ease of calculation:
Wu = -11.763 psf × 1/144 psi/psf
Wu = -0.082 psi

Lr ≥ S, so roof live loads will dictate in load combinations.

Bending test (fb / Fb′ ≤ 1.0)

Fb: allowable bending pressure
Fb′ = Fb × CD × CM × Ct × CL × CF × Cfu × Ci × Cr
CL = 1
CM = 1 because purlins are protected from moisture by roof
Ct = 1 NDS 2.3.3
CF = 1 NDS Supplement
Ci = 1 NDS 4.3.8
Cr = 1 NDS 4.3.9

S: section modulus
S = (b × d2) / 6
S = (1.5″ × (5.5″)2) / 6
S = 7.563 in3

w: pounds force exerted per linear inch of beam length
M: maximum moment
fb: maximum bending stress

1. D

CD = 0.9
Cfu = 1
Fb′ = 1000 psi × 0.9 × 1 × 1 × 1 × 1 × 1 × 1 × 1
Fb′ = 900 psi

w = (D) × sf
w = 0.008 psi × 24″
w = 0.186 pli

M = (w × l2) / 8
M = ( 0.18559992381479 pli × (165″)2 ) / 8
M = 654.797 in-lbs

fb = M / S
fb = 654.797 in-lbs / 7.563 in3
fb = 86.585 psi

fb / Fb′ ≤ 1.0
86.585 psi / 900 psi ≤ 1.0
0.096 ≤ 1.0

1. D + Lr

CD = 1.25
Cfu = 1
Fb′ = 1000 psi × 1.25 × 1 × 1 × 1 × 1 × 1 × 1 × 1
Fb′ = 1250 psi

w = (D + Lr) × sf
w = 0.108 psi × 24″
w = 2.586 pli

M = (w × l2) / 8
M = ( 2.5855999238148 pli × (165″)2 ) / 8
M = 9121.997 in-lbs

fb = M / S
fb = 9121.997 in-lbs / 7.563 in3
fb = 1206.214 psi

fb / Fb′ ≤ 1.0
1206.214 psi / 1250 psi ≤ 1.0
0.965 ≤ 1.0

1. D + W

CD = 1.6
Cfu = 1
Fb′ = 1000 psi × 1.6 × 1 × 1 × 1 × 1 × 1 × 1 × 1
Fb′ = 1600 psi

w = (D + W) × sf
w = 0.074 psi × 24″
w = 1.786 pli

M = (w × l2) / 8
M = ( 1.7855999238148 pli × (165″)2 ) / 8
M = 6299.597 in-lbs

fb = M / S
fb = 6299.597 in-lbs / 7.563 in3
fb = 833.004 psi

fb / Fb′ ≤ 1.0
833.004 psi / 1600 psi ≤ 1.0
0.521 ≤ 1.0

1. D + Wu

CD = 1.6
Cfu = 1
Fb′ = 1000 psi × 1.6 × 1 × 1 × 1 × 1 × 1 × 1 × 1
Fb′ = 1600 psi

w = (D + Wu) × sf
w = -0.074 psi × 24″
w = -1.775 pli

M = (w × l2) / 8
M = ( -1.7748379435325 pli × (165″)2 ) / 8
M = -6261.628 in-lbs

fb = M / S
fb = -6261.628 in-lbs / 7.563 in3
fb = -827.984 psi

fb / Fb′ ≤ 1.0
-827.984 psi / 1600 psi ≤ 1.0
-0.517 ≤ 1.0

1. D + 0.75Lr + 0.75W

CD = 1.6
Cfu = 1
Fb′ = 1000 psi × 1.6 × 1 × 1 × 1 × 1 × 1 × 1 × 1
Fb′ = 1600 psi

w = (D + 0.75Lr + 0.75W) × sf
w = 0.133 psi × 24″
w = 3.186 pli

M = (w × l2) / 8
M = ( 3.1855999238148 pli × (165″)2 ) / 8
M = 11238.797 in-lbs

fb = M / S
fb = 11238.797 in-lbs / 7.563 in3
fb = 1486.122 psi

fb / Fb′ ≤ 1.0
1486.122 psi / 1600 psi ≤ 1.0
0.929 ≤ 1.0

1. D + 0.75Lr + 0.75Wu

CD = 1.6
Cfu = 1
Fb′ = 1000 psi × 1.6 × 1 × 1 × 1 × 1 × 1 × 1 × 1
Fb′ = 1600 psi

w = (D + 0.75Lr + 0.75Wu) × sf
w = 0.021 psi × 24″
w = 0.515 pli

M = (w × l2) / 8
M = ( 0.51527152330431 pli × (165″)2 ) / 8
M = 1817.878 in-lbs

fb = M / S
fb = 1817.878 in-lbs / 7.563 in3
fb = 240.381 psi

fb / Fb′ ≤ 1.0
240.381 psi / 1600 psi ≤ 1.0
0.15 ≤ 1.0

Purlin stressed in bending to a maximum of 96.5%

Shear test (fv / Fv′ ≤ 1.0)

Fv: allowable shear pressure
Fv′ = Fv × CD × CM × Ct × Ci
CM = 1 because purlins are protected from moisture by roof
Ct = 1 NDS 2.3.3
Ci = 1 NDS 4.3.8
V: max shear force
fv: max shear stress

1. D

CD = 0.9
Fv‘ = 175 psi × 0.9 × 1 × 1 × 1
Fv‘ = 157.5 psi

V = w × (le – (2 × d)) / 2
V = 0.186 pli × ( 165″ – (2 × 5.5″) ) / 2
V = 14.57 lbs

fv = (3 × V) / (2 × b × d)
fv = (3 × 14.57 lbs) / ( 2 × 1.5″ × 5.5″ )
fv = 2.649 psi

fv / Fv′ ≤ 1.0
2.649 psi / 157.5 psi ≤ 1.0
0.017 ≤ 1.0

1. D + Lr

CD = 1.25
Fv‘ = 175 psi × 1.25 × 1 × 1 × 1
Fv‘ = 218.75 psi

V = w × (le – (2 × d)) / 2
V = 2.586 pli × ( 165″ – (2 × 5.5″) ) / 2
V = 202.97 lbs

fv = (3 × V) / (2 × b × d)
fv = (3 × 202.97 lbs) / ( 2 × 1.5″ × 5.5″ )
fv = 36.904 psi

fv / Fv′ ≤ 1.0
36.904 psi / 218.75 psi ≤ 1.0
0.169 ≤ 1.0

1. D + W

CD = 1.6
Fv‘ = 175 psi × 1.6 × 1 × 1 × 1
Fv‘ = 280 psi

V = w × (le – (2 × d)) / 2
V = 1.786 pli × ( 165″ – (2 × 5.5″) ) / 2
V = 140.17 lbs

fv = (3 × V) / (2 × b × d)
fv = (3 × 140.17 lbs) / ( 2 × 1.5″ × 5.5″ )
fv = 25.485 psi

fv / Fv′ ≤ 1.0
25.485 psi / 280 psi ≤ 1.0
0.091 ≤ 1.0

1. D + Wu

CD = 1.6
Fv‘ = 175 psi × 1.6 × 1 × 1 × 1
Fv‘ = 280 psi

V = w × (le – (2 × d)) / 2
V = -1.775 pli × ( 165″ – (2 × 5.5″) ) / 2
V = -139.325 lbs

fv = (3 × V) / (2 × b × d)
fv = (3 × -139.325 lbs) / ( 2 × 1.5″ × 5.5″ )
fv = -25.332 psi

fv / Fv′ ≤ 1.0
-25.332 psi / 280 psi ≤ 1.0
-0.09 ≤ 1.0

1. D + 0.75Lr + 0.75W

CD = 1.6
Fv‘ = 175 psi × 1.6 × 1 × 1 × 1
Fv‘ = 280 psi

V = w × (le – (2 × d)) / 2
V = 3.186 pli × ( 165″ – (2 × 5.5″) ) / 2
V = 250.07 lbs

fv = (3 × V) / (2 × b × d)
fv = (3 × 250.07 lbs) / ( 2 × 1.5″ × 5.5″ )
fv = 45.467 psi

fv / Fv′ ≤ 1.0
45.467 psi / 280 psi ≤ 1.0
0.162 ≤ 1.0

1. D + 0.75Lr + 0.75Wu

CD = 1.6
Fv‘ = 175 psi × 1.6 × 1 × 1 × 1
Fv‘ = 280 psi

V = w × (le – (2 × d)) / 2
V = 0.515 pli × ( 165″ – (2 × 5.5″) ) / 2
V = 40.449 lbs

fv = (3 × V) / (2 × b × d)
fv = (3 × 40.449 lbs) / ( 2 × 1.5″ × 5.5″ )
fv = 7.354 psi

fv / Fv′ ≤ 1.0
7.354 psi / 280 psi ≤ 1.0
0.026 ≤ 1.0

Purlin stressed in shear to a maximum of 16.9%

Deflection test (Δmax / Δallow ≤ 1.0)

I: moment of inertia
I = b × d3 / 12 NDS 3.3.2
I = ( 1.5″ × (5.5″)3 ) / 12
I = 20.797 in4

E: modulus of elasticity
E′ = E × CD × CM × Ct × Ci
CM = 1 because purlins are protected from moisture by roof
Ct = 1 NDS 2.3.3
Ci = 1 NDS 4.3.8

Δallow: allowable deflection
Δmax: maximum deflection

1. D + Lr

CD = 1.25
E′ = 1400000 × 1.25 × 1 × 1 × 1
E′ = 1750000 psi

Per IBC 1604.3 footnote d, dead load may be taken as 0.5D.
w = ((0.5 × D) + Lr) × sf
w = ( (0.5 × 0.01 psi) + 0.1 psi ) × 24″
w = 0.104 pli

Δallow = l / 150 IBC 1604.3
Δallow = 165″ / 150
Δallow = 1.1″

Δmax = (5 × w × l4) / (384 × E′ × I)
Δmax = ( 5 × 2.493 pli × (165″)4 ) / ( 384 × 1750000 psi × 20.797 in4 )
Δmax = 0.661″

Δmax / Δallow ≤ 1.0
0.661″ / 1.1″ ≤ 1.0
0.601 ≤ 1.0

1. D + W

CD = 1.6
E′ = 1400000 × 1.6 × 1 × 1 × 1
E′ = 2240000 psi

Δallow = l / 150 IBC 1604.3
Δallow = 165″ / 150
Δallow = 1.1″

Δmax = (5 × w × l4) / (384 × E′ × I)
Δmax = ( 5 × 1.786 pli × (165″)4 ) / ( 384 × 2240000 psi × 20.797 in4 )
Δmax = 0.37″

Δmax / Δallow ≤ 1.0
0.37″ / 1.1″ ≤ 1.0
0.336 ≤ 1.0

1. D + Wu

CD = 1.6
E′ = 1400000 × 1.6 × 1 × 1 × 1
E′ = 2240000 psi

Δallow = l / 150 IBC 1604.3
Δallow = 165″ / 150
Δallow = 1.1″

Δmax = (5 × w × l4) / (384 × E′ × I)
Δmax = ( 5 × -1.775 pli × (165″)4 ) / ( 384 × 2240000 psi × 20.797 in4 )
Δmax = -0.368″

Δmax / Δallow ≤ 1.0
-0.368″ / 1.1″ ≤ 1.0
-0.334 ≤ 1.0

1. D + 0.75Lr + 0.75W

CD = 1.6
E′ = 1400000 × 1.6 × 1 × 1 × 1
E′ = 2240000 psi

Δallow = l / 150 IBC 1604.3
Δallow = 165″ / 150
Δallow = 1.1″

Δmax = (5 × w × l4) / (384 × E′ × I)
Δmax = ( 5 × 3.186 pli × (165″)4 ) / ( 384 × 2240000 psi × 20.797 in4 )
Δmax = 0.66″

Δmax / Δallow ≤ 1.0
0.66″ / 1.1″ ≤ 1.0
0.6 ≤ 1.0

1. D + 0.75Lr + 0.75Wu

CD = 1.6
E′ = 1400000 × 1.6 × 1 × 1 × 1
E′ = 2240000 psi

Δallow = l / 150 IBC 1604.3
Δallow = 165″ / 150
Δallow = 1.1″

Δmax = (5 × w × l4) / (384 × E′ × I)
Δmax = ( 5 × 0.515 pli × (165″)4 ) / ( 384 × 2240000 psi × 20.797 in4 )
Δmax = 0.107″

Δmax / Δallow ≤ 1.0
0.107″ / 1.1″ ≤ 1.0
0.097 ≤ 1.0

Purlin stressed in deflection to a maximum of 60.1%

# Solving Steel Roofing Leaks

Solving Steel Roofing Leaks

Nothing much more frustrating than finding your post frame building’s new steel roofing has a leak (or two) – however they can be solved!

Reader (and client) BRYAN from MECHANICSVILLE writes:

“I’m really loving my pole barn.

With all of the rain we’ve had, I’ve noticed two leaks, I believe they are coming through roof screw holes, what is your recommended way of correcting the problem? Should I put a sealer around the screws, replace the screws or something else?

Thank you for your time and help!”

Quite pleased you are loving your new pole barn! In case anyone has wondered why we do what we do, this would be why – there are such great rewards in being able to aid clients in their new building journeys!

If you have been able to narrow a water issue down to these two locations, you have solved a most challenging puzzle piece. Often, with use of a Radiant Reflective Barrier under the roof steel, water will enter in one place and leak out in a different location. This can cause leak searching to be a true chore.

Under no circumstance should a sealant be used around or over top of screws. Eventually this sealant works away from the screw and the leak is back! Now it becomes diagnosing why there a leak exists and how to best fix it.

Most often leaks are caused by a predrilled screw hole not having a screw in it. Although this sounds obvious – it does occur. Easiest fix, put a screw in hole. Next up would be a screw has missed a purlin. If this happens to be your case, chances are you can see a “shiner” (galvanized screw shank) alongside, or poking out a side of, a purlin. Provided screw holes were pre-drilled, this can be resolved by removing offending screw, as well as its adjacent neighbors for several feet in each direction along the purlin. Have someone push that purlin uphill or pull it downhill until a screw can be replaced into solid wood.

In event of a random miss, for whatever reason, have someone hold a block of 2×4 under the screw hole and drive a screw through roofing and into the block.

Next possible culprit would be a screw not properly seated. If under driven, (EPDM gasket not compressed) screw can normally be driven in further. If over driven (gasket smashed) screw should be replaced with either a larger diameter and longer length screw (say #14 x 2″) or by driving a wooden match stick or other slender piece of solid wood into screw hole, then use an originally sized screw back in original hole. If a screw was driven in other than perpendicular to roofing, it may be possible to remove the screw and drive it straight in.