# A Floor Raising Exercise: I Joists

For some obscure reason people planning new buildings tend to scrimp on height. In most instances, designing a new fully engineered post frame building – whether for a barndominium, shop house (shouse), garage, shop, etc., just a little bit taller is a relatively inexpensive proposition and can save many more dollars and mental anguish than having to alter after construction.

Reader CHRIS in SNOHOMISH writes:

“I have a pole barn with a center door for Rv, above is an additional living space, the width is 12’6” depth of 41’ height of 13’, I need to shorten the truss’s so I can gain 6” height , current truss are HY floor joists, question is can I put 2×6” spaced every 8”’s and have the same weight carrying capacity?”

Mike the Pole Barn Guru writes:

HY floor joists are wood prefabricated I joists. Let’s take a look at Chris’ proposed design solution (please keep in mind, any structural design solution should be reviewed by your building’s engineer to confirm structural adequacy):

FLOOR JOIST DESIGN

Assumptions:

Joist span 12.5-ft.
joist spacing = 8″ o.c.

joist  live load = 40 psf

Fb: allowable bending pressure
Fb‘ = Fb * CD * CM * Ct * CL * CF * Cfu * Ci * Cr
CD: load duration factor
CD = 1 NDS 2.3.2
CM: wet service factor
CM = 1 because floor joists 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 = 1.3 NDS Supplement table 4A
Cfu: flat use factor
Cfu = 1 NDS Supplement table 4A
Ci: incising factor
Ci = 1 NDS 4.3.8
Cr: repetitive member factor
Cr = 1.15 NDS 4.3.9
Fb = 900 psi NDS Supplement Table 4-A
Fb‘ = 900 psi * 1 * 1 * 1 * 1 * 1.3 * 1 * 1 * 1.15
Fb‘ = 1345.5 psi

fb = live + dead load * joist spacing / 12 / 12 * (sf * 12 – 1.5)2 / 8 / (b * d2 / 6)
fb = 50 psf * 8″ / 12 in./ft. / 12 in./ft. * (12.5′ * 12 in./ft. – 1.5″)2 / 8 / (1.5″ * 5.5″2 / 6)
fb = 1012.5 psi <= 1345.5 psi so okay in bending

Δallow: allowable deflection
Δallow = sf * 12 / 360 IBC 1604.3
Δallow = 12.5′ * 12 in./ft. / 360
Δallow = 0.4167″

Δmax: maximum deflection
Δmax = 5 * live load * joist spacing / 12 / 12 * (sf * 12 – 1.5)4 / 384 / 1600000 / b / d3 * 12 from http://www.awc.org/pdf/DA6-BeamFormulas.pdf p.4
Δmax = 5 * 40 psf * 8″ / 12 in./ft. / 12 in./ft. * (12.5′ * 12 in./ft. – 1.5)4 / 384 / 1600000 / 1.5″ / 5.5″3 * 12
Δmax = 0.423″ > 0.4167:
2×6 #2 DougFir joists will not work at 8″ on center due to not meeting deflection criteria

Chris’ options are to buy #1 or Select Structural graded 2×6 DougFir or go to 6″ on center spacing.

# Feedback Needed From RDP’s and Building Officials

I am asking for feedback from RDP’s and Building Officials because:

There is a method to my madness. Seriously. I want to make sure we are doing things 100% correctly. In my humble opinion there are currently numerous post frame buildings being constructed where wall girts do not meet Code or acceptable engineering practice.

I have developed a professional respect for a builder based in Northern Idaho. Recently I visited his website and saw some photographs leading me to ask about how he solves “barn style” wall girt design issues. He was right on top of it – his photos were of older buildings and he switched to all bookshelf style wall girts years ago, I applaud him for doing so!

Lots of architects, engineers and building officials read my articles, thank you! Your wisdom is appreciated. Attached is an example set of wall girt calculations. If there is an error in any direction, or something missed, your feedback would be more than appreciated. Thank you in advance.

Code is 2015 IBC (International Building Code)

Building Summary

 Building Footprint Width 40′ Building Footprint Length 60′ Building Footprint Height 17′ Square Footage (area contained by embedded poles) 2400 ft2 Total Roof Area 2745 ft2 Total Wall Area 3191 ft2 Building Eave Height 17′ Roof Style GABLE Slope 4/12 Roof Height 20.33′ Building Conditioned Yes

Wind Summary

 Vult 110 mph Vasd 85 mph Risk Category I Wind Exposure B Applicable Internal Pressure Coefficient 0.18 Components and Cladding Design Wind Pressure Zone 1 -19.78 Zone 2 -32.985 Zone 3 -49.217 Zone 4 -23.522 Zone 5 -27.936 Zone 1 Positive 11.826 Zone 2 Positive 11.826 Zone 3 Positive 11.826 Zone 4 Positive 21.188 Zone 5 Positive 21.188 Duration of Load for Wind 1.6 Structure type Enclosed

wall girt size: 2″X6″
spacing between girts = 22.5″

girt span = 139.875″
supported by 2×4 blocking every 139.875″

Fb: allowable girt pressure
Fb‘ = Fb * CD * CM * Ct * CL * CF * Cfu * Ci * Cr NDS 4.3
CD: load duration factor
CD = 1.6 NDS 4.3
CM: wet service factor
CM = 1 because girts are protected from moisture by building envelope
Ct: temperature factor
Ct = 1 NDS 4.3
Cfu: flat use factor
Cfu = 1 NDS 4.3
Ci: incising factor
Ci = 1 NDS 4.3
Emin: reference adjusted modulus of elasticity
Emin = 470000 psi NDS Supplement
Cr: repetitive member factor
Cr = 1.15 NDS 4.3
lu: laterally unsupported span length
lu = 139.875″
le: effective length
le = 1.63 * lu NDS table 3.3.3
le = 244.496″
CF: size factor
CF = 1.3 NDS 4.3
CL: beam stability factor
CL = 1 NDS 3.3.3
Fb‘ = 850 psi * 1.6 * 1 * 1 * 1 * 1.3 * 1 * 1 * 1.15
Fb‘ = 2033.2 psi

fb: girt test pressure
fb = 6 * 0.6wall_wind_force / 144 * girtSpacing * span2 / 8 / (b * d2) NDS 3.3
fb = 6 * 17.389 psf / 144 in.2/ft.2 * 24″ * 139.875″2 / 8 / (1.5″ * 5.5″2)
fb = 937.255 psi
937.255 ≤ 2033.2 stressed to 46% 6″X2″ #2 OK in bending

Fv‘: allowable shear pressure
Fv = 135 NDS Supplement Table 4-A
Fv‘ = Fv * CD * CM * Ct * Ci NDS 4.3
Fv‘ = 135 psi * 1.6 * 1 * 1 * 1
Fv‘ = 216 psi NDS Supplement

fv: shear girt pressure
fv = 3 * (0.6wall_wind_force / 144 * girtSpacing * span / 2) / (2 * b * d) NDS 3.4
fv = 3 * (17.389 psf / 144 in.2/ft.2 * 24″ * 139.875″ / 2) / (2 * 1.5″ * 5.5″)
fv = 36.854 psi

36.854 ≤ 216 stressed to 17% 6″X2″ #2 OK in shear

Deflection

Δallow: allowable deflection
l = 139.875″
Δallow = 139.875″ / 90
Δallow = 1.5542″
Δmax: maximum deflection
Δmax = 5 * 0.6W * spacing * span4 / 384 / E / I from http://www.awc.org/pdf/DA6-BeamFormulas.pdf p.4
E: Modulus of Elasticity
E = 1300000 psi NDS Supplement
I: moment of inertia
I = b * d3 / 12
I = 1.5″ * 5.5″3 / 12
I = 20.796875 in.4
Δmax = 5 * 12.173 psf / 144 psi/psf * 24″ * 139.875″4 / 384 / 1300000 psi / 20.796875 in.4 components and cladding reduced by .7 per footnote f of IBC table 1604.3
Δmax = 0.37401″ ≤ 1.5542″