# 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%

# ABC Component Shortages

Remember Hoping You Could Buy Toilet Paper?

I grew up seeing photos of people in Eastern Bloc countries standing in line for nearly every commodity. More recently we have witnessed residents of Venezuela enduring similar circumstances.

Similar situations are now occurring with building materials – as witnessed by this notification from one of our steel roofing and siding suppliers, American Building Components:

“May 26, 2021

Dear Valued Customer Partners,

With the intent to keep you as updated on product availability, please note that the supply of materials from our vendors has tightened which has limited our production capacity. In order to improve material planning and delivery schedules, the following are changes to our material allocation which will begin June 1.

• Material allocations will be established at the regional level to align with the capacity of the respective manufacturing locations.
• Upon reaching monthly allocated order volume, additional orders will not be accepted until allocations are reset at the beginning of the next month.
• Non-active and/or new customers will not be eligible to participate in the allocation program.

Your specific material allocation will be communicated to you by your respective sales representative no later than June 1.

As you plan your business for the coming months, we expect to experience extended lead times. Please anticipate that all new orders will receive ship dates aligned with the production capacity based upon the expected steel from our suppliers.

With the material indices continuing to rise since our last communication on May 3, we will be monitoring the market to provide you with pricing guidance for the third quarter in the coming weeks.

We understand the challenges each of you is facing with rising material costs, limited supply, and the availability of labor. We are actively working to increase our network capacities and will continue to proactively communicate with you on our progress.”

It has been tough enough to see double digit increases in steel costs pretty well every month. According to NAHB (National Association of Home Builders):

“Steel mill products prices climbed 18.4% in April following a 17.6% increase in March.  Prices are up 55.6%, year-to-date, and the month-over-month percentage increase set a record high for the third month in a row. Steel mill products price volatility is greater than it has been at any time since the Great Recession.”

Now, besides being increasingly expensive, product may not even be available. Hansen Pole Buildings does happen to be one of American Building Components’ largest clients, so we are hopeful we will be able to continue to fulfill orders in a timely manner, however for those who are smaller, regional roll formers, it could be they are going to run out of coil and not be able to source more.

We obviously do not own a crystal ball to forecast when prices and availability of building products will stabilize. For those of you who are looking to build, if you can borrow at low interest rates over long terms, it is yet an opportunity, as higher future interest rates will likely more than offset any lower prices.

Preston Bowen

President, Building Envelope Solutions