HomeMy WebLinkAboutBLD2000-01099 Structural Calculation - BLD Engineering / Geo-tech Reports - 8/9/2000 BUILDING SUPPLY
22175 S. Highway 99E,Canby,Oregon 97013
Phone:(503)263-6953 Fax:(503)266-7102
POST FRAME BUILDING
STRUCTURAL CALCULATION
(This structure has been analyzed and designed for structural adequacy only.
The analysis is in compliance with the 1997 Uniform Building Code.)
PROJECT No.
MWO0330
OWNER:
David Spandl
276 NE Tee Lake Rd.
Tanuya, WA 98588
ENGINEER:
O�'PS qs,
8765
G�
SSj�NAL
[—EXPIRES: i/12/—p
pppl-
job Name_ : Truss ID: tMWO0330 2 Drw .COM21 UM-002
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7/31100 MWO0330 CALC 0731 (Spandl) 30x36x16.mcd 1
POST FRAME BUILDING SUMMARY:
This is a steel sided post-frame building with wooden trusses or rafters and preservately treated
posts that are pressure treated for ground contact. Post size, post embedment depth, post hole
diameter and backfill is given in the body of the calculation.
The building will depend on the diaphragm action of the sheathing for lateral stability. The
eave and gable walls and the roof structure will be modeled as shear diaphragms. The posts will
be modeled as propped cantilevers that are fixed at the base and propped by the deep beam
action of the roof.The roof structure spans horizontally between the wall diaphragms where it is
simply supported.
The post frames will be assumed to act as a unit. Wind loads(wind controls over seismic)will
be imposed on the windward and leeward sides of the building simultaneously. The trusses/
rafters will act as a drag strut that will cause the two posts to act together.
The post frame model for the building comes from Reference 2. The actual post length for
bending will be assumed to be measured from top of post hole backfill to the top of the corbel
block. The concrete backfill will provide lateral constraint in the windward and leeward direction.
REFERENCES:
1. 1997 Edition of the Uniform Building Code
2. Post-Frame Building Design
Walker and Woeste. ASAE Monograph Number 11. ASAE 1992
3. Timber Construction Manual,AITC 3rd Edition 1985
4. 1997 Edition, National Design Specification (NDS) 2nd Supplement For Wood
Construction,American Wood Counsel
5. Wood Technology and the design of Structures Hoyle and Woeste 5th Edition.
6. Strength and Stiffness of Screw-Fastened Roof Panels for Pole Buildings.
Hoagland and Bundy. Transactions of the ASAE 1983
7. Experimental Verification of Stressed Skin Design for Pole-Frame Buildings.
Johnson and Curtis. Transactions of the ASAE 1984.
8. ASAE Technical Paper 84-4511
9. ASAE Technical Paper 92-4540
10. 1996 Diaphragm Test by the National Frame Builders Association and
observed by Braun Intertec Corporation Canby, Oregon.
11. ASAE Technical Paper EP484.1
12. 1989 Shear Panel Test with Val-Rib Cladding by Devco Engineering Inc.for Valley
Rolling Mills, Inc.of Salem Oregon
13. Shear and Flexibility Factors for BHP Steel Building Products Nor Clad Metal Cladding
System By Capital Engineering Laboratories,July 1996,CYS Job NO.94115-02
7/31/00 MW00330 CALC 0731 (Spandl) 30x36xl6.mcd 2
SECTION PROPERTIES OF WOOD MEMBERS:
Ix24 =5.36 5x24 :=3.06 Sv24 :=1.31
Where:
Ix26 :=20.8 Sx26 :=7.56 Sy26 :=2.06 1 = Moment of Inertia
I x28 :=47.64 S x28 :=13.14 S y28 :=2.72 S = Section Modulus
A= Cross sectional area
Ix210 :=98.93 Sx210 :=21.39 Sy210:=3.47 W= Weight per foot
b= Diagonal of member
I x212 :=177.98 Sx212 :=31.64 Sy212 :=4.22
I x46 :=48.53 S x46 :=17.65 S y46 :=11.23 A 46:=19.25 W 46:=4.68 b 46 :=7.21
1x66 76.26 Sx66 :=27.73 Sy66 :=27.73 A66:=30.25 W66:=7.35 b66 :=8.49
I x68 :=193.36 S x68 :=51.56 S y68 :=37.81 A 68:=41.25 W 68:=10.03 b 68 :=10.0
I x610 :=392.96 S x610 :=82.73 S y610:=47.90 A 610:=52.25 W 610:=12.70 b 610:=11.66
I x612 :=697.07 S x612 :=121.23 Sy6 12 :=57.98 A 612:=63.25 W 612 :=15.37 b 612 :=13.42
Design Values of Visually Graded Timbers (5" x 5" and larger)
(NDS Table 4-D, WCLIB):
F b1]F2post:=575 psi (bending stress for No. 2 Hem-Fir)
F b11F1post:=975 psi (bending stress for No. 1 Hem-Fir)
F bHFSSpost:=1200 psi (bending stress for Structural Select Hem-Fir)
F cHF2post:=575 psi (compression stress parallel to grain for No. 2 Hem-Fir)
F cHFIpost:=850 psi (compression stress parallel to grain for No. 1 Hem-Fir)
F c1]FSSpost:=975 psi (compression stress parallel to grain for Structural Select
Hem-Fir)
E FIF2post:=1100000 psi (modulus of elasticity for No. 2 Hem-Fir)
E HFlpost:=1300000 psi (modulus of elasticity for No. 1 Hem-Fir)
E HFSSpost:=1300000 psi (modulus of elasticity for Structural Select Hem-Fir)
7/31/00 MWO0330 CALC 0731 (Spandl) 30x36xl6.mcd 3
BEAM & STRINGERS DESIGN VALUES (NDS Table 4-D, WCLIB):
FbMbeam :=675 psi (bending stress for No. 2 Hem-Fir)
F bHFlbeam =1050 psi (bending stress for No. 1 Hem-Fir)
F bHFSSbeam :=1300 psi (bending stress for Structural Select Hem-Fir)
F cHF2beam:=500 psi (compression stress parallel to grain for No. 2 Hem-Fir)
F cHFlbeam :=750 psi (compression stress parallel to grain for No. 1 Hem-Fir)
F cHFSSbeam :=925 psi (compression stress parallel to grain for Structural Select
Hem-Fir)
E HF2beam :=1100000 psi (modulus of elasticity for No. 2 Hem-Fir)
E HFlbeam:=1300000 psi (modulus of elasticity for No. 1 Hem-Fir)
E HFSSbeam :=1300000 psi (modulus of elasticity for Structural Select Hem-Fir)
VISUALLY GRADED DIMENSION LUMBER DESIGN VALUES
(NDS Table 4-A, WCLIB and WWPA):
FbHF2dim :=850 psi (bending stress for No. 2 Hem-Fir)
FbDF2dim :=900 psi (bending stress for No. 2 Doug-Fir)
F bDFIdim :=1000 Psi (bending stress for No. 1 Doug-Fir)
F bDFIBdim:=1200 psi (bending stress for No. 1 and better Doug-Fir)
F bDFSSd m :=1500 psi (bending stress for Structural Select Doug-Fir)
F cHF2dim :=1300 psi (compression stress for No. 2 Hem-Fir)
E HF2dim :=1300000 psi (modulus of elasticity for No. 2 Hem-Fir)
MECHANICALLY GRADED DIMENSION LUMBER DESIGN VALUES
(NDS Table 4-C, WCLIB and WWPA):
F bMSR1650 :=1650 psi (bending stress for MSR1650f-1.3E)
F bMSR2100 :=2100 psi (bending stress for MSR2100f-1.8E)
F bMSR2400 :=2400 psi (bending stress for MSR2400f-2.OE)
E MSR1650 :=1300000 psi (modulus of elasticity for MSR1650f-1.3E)
E MSR2100:=1800000 psi (modulus of elasticity for MSR2100f-1.8E)
E MSR2400 :=2000000 psi (modulus of elasticity for MSR2400f-2.OE)
7/31/00 MWO0330 CALC 0731 (Spandl) 30x36xl6.mcd 4
SUMMARY OF DESIGN VALUES:
Building Dimensions
W bldg:=30 [ft] Width of Building)
Lbldg:=36 [ft] (Length of Building)
H bld8:=16 [ft] (Eave Height of Building)
R pitch 3 / 12 [roof pitch]
B av :=13 [ft] (Spacing between eavewall posts)
W 8ableopenin8s :=3 [ft] (Total width of openings in one gable wall)
W eaveopenings:=22 [ft] (Total width of openings in one eave wall)
Truss heel:=12 [in] (Depth of truss heel)
Post Properties:
Pwidtha =6 [in] (Post width y-axis)
POST SIZE
Pdeptha =8 [in] (Post depth x-axis)
P width P widtha— .125
P depth :=P depWa— .125
This accounts for the rough sawn dimensions of the post)
P width'P depth 3 I S x
_P depth'P width _P width'P depth 2
1 xpost'= 12 YPost'- 12 3 post'- 6
1 xpost= 239.1 [in^4] Iypost= 133.07 [in^4] S xpost= 60.72 [in^3]
5 A post
b post �P width 2+P depth 2 A post:=P width'P depth W post 144 35
b post=9.83 [in] A post=46.27 [in^2] W post= 11.25 [plf]
1' c I '=F cHF2post
I' cl = 575 [psi] (Allowable compression stress for the
posts)
F b 1 :=F bHF2post
F b l = 575 [psi] (Allowable bending stress for the
posts)
E wood:=E HF2post
E wood= 1100000 [psi] (Allowable modulus of elasticity for
posts)
7/31/00 MW00330 CALC 0731 (Spandl) 30x36xl6.mcd 5
SUMMARY OF DESIGN VALUES (Continued):
Purlin Properties:
Purlin_spacing 24 [in]
S purlin '=S x26 F purlin :=F bMSR1650
S purlin- 7.56 [in A3] F purlin= 1650 [psi] (Allowable bending stress for the purlins)
Girt Properties:
G irt_spacmg :=24 [in]
S girt :=S y26 F girt :=F bMSR1650
S girt=2.06 [in A3] F girt= 1650 [psi] (Allowable bending stress for the girts)
Footing and Post Hole Desiqn Values:
q soil =1500 [psf] (Assumed soil vertical bearing capacity, 1997 UBC Table 18-IA pg 2-49)
S sod :=150 [psf] (Assumed soil lateral bearing capacity, 1997 UBC Table 18-IA pg 2-49)
dia footing:=2 ft [diameter of footing]
d epth postembed :=4 ft [Trial depth of posthole]
Design Loads for Building:
Wind Design Values:
Fastest wind speed: 80 MPH
q s :=16.4 (Wind stagnation pressure at standard height of 33ft (1997 UBC Table 16-F)
Wind Exposure: B
C e :=0.64 (Combined height, exposure and gust factor coefficient- 1997 UBC Table 16-G
p2-2)
Roof Load Design Values:
W snow:=25 [Ibs] (Ground snow load)
W dead:=5 [Ibs] (Roof dead load)
7/31/00 MWO0330 CALC 0731 (Spandl) 30x36xl6.mcd 6
WIND ANALYSIS:
UBC conditions: Method 1 (Normal Force Method)
Calculate the Design Wind Pressure by Equation 20-1, page 2-7 in the UBC
q = CeXCgXgsXI
Where:
Ce = Combined height, exposure and gust factor coefficient (UBC Table 16-G p 2-28)
Cq = Pressure coefficient (UBC Table 16-H p2-29, use various)
qs = Wind stagnation pressure at standard height of 33 feet (UBC Table 16-F p 2-28)
Iw = Importance Factor(UBC Table 16-K, p 2-30)
Ce=0.64 q s= 16.4 Iw:=1.0
C qww:=0.8 (Pressure coefficient for windward wall)
C qlw :=-0.5 (Pressure coefficient for leeward wall)
C qwr:=0.3 (Pressure coefficient for windward roof)
C qlr:=-0.7 (Pressure coefficient for leeward roof)
C qwe :=1.2 (Pressure coefficient for wall elements)
Calculated Wind Pressures:
Windward Wall: Leeward Wall:
q ww :=C e•C gww'q s•1 w q lw '=C e'C glw'q s'1 w
q ww= 8.4 [Psf] q Iw=—5.25 [Psf]
Windward Roof: Leeward Roof:
q wr:=C e•C gwr.q s•I w q lr:=C e'C glr'q s'1 w
q wr= 3.15 [psf] q lr=—7.35 [psf]
Wall Elements:
q we:=C e•(C qwe—0.1 •q s•Iw (Note: Subtract 0.1 due to footnote 2, UBC Table 16-H, page
2-29)
q we= 11.55 [psq
7/31/00 MWO0330 CALC 0731 (Spandl) 30x36xl6.mcd 7
BUILDING MODEL:
STEP 1: CALCULATE THE SHEAR STIFFNESS OF THE TEST PANEL
This procedure relies on tests conducted by the National Frame Builders Association
(reference 10), the outlines given in ASAE technical paper(reference 11), and chapter 4 of
reference 2.
The test was conducted using 29 gauge ribbed steel panels. These ribbed steel
panels are similar to Strongpanel, Norclad, and Delta-Rib which are in common use by builders
in this area. The material and section properties for the test panels are thus reasonable and will
be used throughout.
The stiffness of the test panel was calculated to be: c= 2166 lb/in (Table 5, Reference 10)
STEP 2: CALCULATED ROOF DIAPHRAGM STIFFNESS OF THE TEST PANEL
c' = (E X t)/ (2 X (1+V) X (g/p) + (K 2 / (b'X t)^2))
Where: E = 27.5x10^6 psi (modulus of elasticity for steel)
t= 0.017" (thickness of 29 gauge steel)
V= 0.3 (Poisson's Ratio for steel)
g/p= 1.17 ratio of sheathing corrugation length to corrugation pitch
V = 144" 02'-0" length of test panel)
Note: 2 X (1+V)X (g/p) = 3.0420 (a constant for this type of panel)
STEP 2.1:
This equation was set equal to the stiffness of the test panel (2166 lb/in) and the unknown
value (K2)was solved for.
K2 = 1275 inA4 sheet edge purlin fastening constant
STEP 2.2:
Use new building width to determine stiffness of new roof diaphragm (c h):
W bldg-12 K 2 :=1275 [lb/ft] 'R pitch
b new '
2 t :=0.017 [in] 0 :=atan� j
12
'=
cos(@) @ = 14.036-deg (roof angle of
b new= 186 [in]
E :=27500000 incline)
c :_ (E-t)
K2 c= 3563 [lb/in]
3.042
2
�b new't J
7/31/00 MW00330 CALC 0731 (Spandl) 30x36xl6.mcd 8
STEP 2.3&2.4:
Calculate the equivalent horizontal roof stiffness(c h)for the full roof:
Since c h is for the full roof,the roof length must be ratioed by the aspect ratio of the
roof panel (b/a)where "a"is the truss spacing in inches.
a :=B ay-12 :ch'-
=2 c cos(®)2 b new
a= 156 (in) c h= 7978 [Ib/in]
STEP 3: CALCULATE THE STIFFNESS OF THE POST FRAME(k):
Since the connection between the posts and the rafters can be assumed to be a
pinned joint, the model for the post frame can be assumed to be the sum of two
cantilevers (the posts)that act in parallel. This can be accurately modeled by doubling
the moment of inertia (1) of a single post. The stiffness of the post frame can be
calculated from the amount of force (P) required to deflect the system one inch. The
spring constant(k) in pounds per inch of deflection results directly.
L post_bndg:=H bldg-12— T russ_hcel (The bending length of post is reduced by the length
of the truss heel.)
L post_bndg= 180 [in]
6 :=1 in
P:=3.6-E we 2 1 xpost P= 271 Ib
3
L post_bndg
k:=P k= 271 lb/in
STEP 4: CALCULATE TOTAL SIDE SWAY FORCE(R):
Apply wind loads to the walls to determine moment (Mwind), fiber stress (fwind)
and end reaction at prop point (R).
Calculate Total Wind Pressure:
_ a
q wwpost'-q ww- a
12-12 q lwpost '=q lw'12.12 q tot'=q wwpost- q lwpost
q wwpost=9.1 [pli] q lwpost=-5.69 [pli] q tot= 14.78 [pli]
2
M wind '-q tot'L post 8 wind=_bndg M 59867 [in-lbs)
M wind [ps
f wind := ]f wind=493 i
2-S xpost
R :=3 L post_budg R= 998 [lbs]
q tot' 8
7/31/00 MWO0330 CALC 0731 (Spandl) 30x36xl6.mcd 9
STEP 5: CALCULATE THE RATIO OF THE FRAME STIFFNESS TO THE ROOF STIFFNESS:
This ratio Nc h)will be used to enter Table 4.1 or the matrix on page 238 in reference
2 to determine the side sway force modifiers.
-=0.034
ch
STEP 6: DETERMINE SIDE SWAY RESISTANCE FORCE:
From Table 4.1 or the matrix on page 238 in reference 2.
mD:=0.97
STEP 7: CALCULATE THE ROOF DIAPHRAGM SIDE SWAY RESISTANCE FORCE:
Q :=mD•R Q=968 lb
Since not all of the total side sway force (R) is resisted by the roof diaphragm, some translation
will occur at the top of the post.
wtr'_8. Q w = 14.3 [pli]
3'L post_budg rr
Subtract w LI. from the applied wind load (q tot). This is the distributed load that is NOT resisted
by the post. Determine the moment(M dfl) and the additional fiber stress (fbdfl)that results
from this deflection.
w dfl '=g tot-w ff w df,=0.4 [pli]
2
M dfl :=w dfl L post_bndg M dfl=7184 [lb in]
2
[psi]
f�:= M f�= 59 ]i
2 S xpost
Calculate the total moment(M tot) and the total fiber stress(f tot),
M tot:=mD'M wind+M dfl M tot= 65255 [lb in]
ftot:=mD'f wind+fdfl ftot= 537 [psi]
7/31/00 MW00330 CALC 0731 (Spandl) 30x36xl6.mcd 10
POST DESIGN:
Assume the following post properties:
1. The posts will be modeled as propped cantilevers fixed at the base and propped at the
eave line by the roof diaphram. The two posts will act at each frame to resist bending.
2. The roof will act as a diaphragm and act as a simple support for the posts.
3. The roof will act as a simple horizontal beam spanning between shear walls.
4. The posts will be pressure treated for ground contact.
5. The controlling load case will be Dead Load +Wind Load + 1/2 Snow Load.
Calculate allowable unit stress(compression F cc). NDS for Wood Construction Section 3.7, p 22.
F cl = 575 [psi]
F c :=F cl•1.15 (Note: The 1.15 accounts for LDF for snow.)
F c =661 [psi] (Allowable compression stress including load factors)
L post_bndg= 180 [in] (Bending length of post)
d post '=P deptha
d post= 8 [in] (Unbraced dimension of post)
K e:=0.8 [NDS App. G Table G1 page 1621
1 e'=K e'L post_bndg c :=0.8 [sawn lumber NDS page 22]
I e= 144 [in]
K cE :=0.3 [visually graded sawn lumber NDS 221
E wood= 1100000 [psi]
FcE '=K cE' E wood F cE = 1019
le 2
�d
post
z
I +FcE ltFcE
C = Fc — Fc — F cE C =0.45
P 2-c 2-c Fc P
c
F cc :=F c C p F cc = 299 psi
7/31/00 MWO0330 CALC 0731 (Spandl) 30x36xl6.mcd 11
POST DESIGN:(Continued)
Calculate Combine Flexural and Axial Loading Index from UBC p 2-818:
W roof'=W snow+W dead
W roof= 30 [psf] (Total roof loading)
W bldg
P snow:= 2 a-W roof
P snow= 5850 [lbs] (Axial loading per post)
P snow
fc := 2 fc=63 [psi] (Actual compression stress per post)
A post
I' b :=F bl-1.6 F b=920 [psi] (Allowable bending stress per post including load factors)
(The 1.6 factor accounts for the LDF for wind - reference
NDS Table 2.3.2 page 9.)
f bl :=f tot f bl = 537 [psi] (Actual bending stress on post)
f
2
CCFALI := c fb1+
IFCCI F 1_ fc
b I FcEI
CCFALI= 0.67 less than or equal to 1.0 is OK
7/31/00 MWO0330 CALC 0731 (Spandl) 30x36xl6.mcd 12
POST EMBEDMENT FOR NON CONSTRAINED CONDITION:
Calculate the required post depth. Use UBC equation 6-1, p 2-46. This assumes full depth
backfill of post holes with concrete.
M tot 11
M tot=65255 in - Ibs
16•—
P post 2.12 16 P post=623 Ibs [lateral load applied @ mid point of
3•H bldg the truss post]
�d epth_postembed)
S 1 :=2•Ssod 3 ( g a 1.33 calculated using trial depth of embedment and note 3
S 1 = 532 from UBC Table 18-1-A)
A:=2.34 P post H bldg
S 1.d ia_footing 2
A= 1.37 [ft^2] h= 8 [ft]
depth:=2' 1+ 1 +4.36 A depth= 4.2 [ft] (required post embedment depth)
Calculate pullout of the posts due to shear loads on the end walls.
W bldg
H roof:= 2 'tan(( ) H roof= 3.75 [ft]
0.375.mD-lH bldg)'L bldg'(q ww—q lw)+ H roof).L bldg'`q wr—q lr)
V eave-wind 2
V eave wind= 2138 [Ibs] (total load transferred into each gable wall)
C post := v eave-wind'H bldg C post= 1267 Ibs [this is the uplift load on one post]
W bldg- W gableopenings
Assume a total weight of roof and wall area to be 2.0 psf. The area of the roof and wall that will
tend to keep the pole in the ground will be as follows:
E ;=H L bldg.2 R =L bldg W bldg.2 G able wall H bldg'W bldg 2
ave_wall bldg' 2 oof' 2 2 — 2
G able—wall =480 [Ibs]
E ave wall= 576 [Ibs] R oof= 540 [Ibs]
_ z
P osts:= H bldg+d e th,•W post d ia_footing _ A post
' � g P J P Post hole:=150�depth�ostembed" 3.14•
I 4 144
Posts= 227 [Ibs]
Post hole= 1691 [Ibs]
Wt tot'=E ave wall+U able wall+R oof+P osts+P ost hole
Wt tot=3514 [Ibs] (Note that Wt tot is greater than C post. Thus OK.)
7/31/00 MWO0330 CALC 0731 (Spandl) 30x36xl6.mcd 13
FOOTING DESIGN:
Check the soil bearing capacity of the punch pads. UBC Table 18-1-A, page 249.
,z
A :=3.14 j d ia_footing'I ft^2 [this is the area of the footing]
footing 1 2
q soil= 1500 [psf]
d ia_footing- 2 ft
d epth_postembed- 4 ft
- 1 [additional q per UBC Table 181-B, p 2-57, note 2]
d factor'=1+.2 id epth�ostembed
d factor= 1.6
Pfooting :=A footing'q soil Afactor Pfooting- 7536 [lbs] (end bearing capacity of footing)
P snow= 5850 [Ibs]
Note that the end bearing capacity (Pf,,ti„9) is greater than the snow load (Ps ). This is OK.
7131/00 MWO0330 CALC 0731 (Spandl) 30x36xl6.mcd 14
GABLE WALL SHEAR ANALYSIS:
Calculate the load that must be resisted by the shear capacity of the gable walls. Per
"Procedures For Calculating Roof Shear"in reference 2, page 62, use K= 3/8 for the eave wall
wind. Use the building length L bldg, the actual eave height (Hbid9) for calculation and the
projected roof height (H roof).
H bldg= 16 [ft] q ww= 8.4 psf q = 3.1 psf
H roof= 3.75 [ft] L bidg- 36 [ft] q Iw=-5.2 psf q Ir=-7.3 psf
0.375-mD- H bldg- IJ'L bldg*�qww- qlw)+ �H roof+ 1).Lbldg* gwr- qlr
V eave wind'= 2
V eave wind= 2237 [Ibs]
This is the lateral wind load that is transmitted to each gable wall. This load will be transmitted
through the roof diaphragm to the gable walls. Subtract the total maximum width of doors and
windows in one gable wall and normalize the load to a per foot basis.
V eave-wind
v gablewall'-W bldg- W gableopenings g v ablewall= 83 [Ibs/ft]
The able wall diaphragms can resist 100% of the minimum ultimate load given in reference 10.
9
This value may be increased by 1/3 for wind per UBC section 1612.3.2 on page 2-5 (110 plf X
1.33 = 142 plf).
If v gablewall < 142 plf Then no additional sheathing is required.
7/31/00 MWO0330 CALC 0731 (Spandl) 30x36xl6.mcd 15
EAVE WALL SHEAR ANALYSIS:
Calculate the load that must be resisted by the shear capacity of the eave walls. Per
"Procedures For Calculating Roof Shear" in reference 2, page 62, use K= 3/8 for the gable
wall wind. Use the building width (Wh,,,,),the actual eave height (Hhli,)for calculation and the
projected roof height (H roof)-
H roof= 3.75 [ft] H bldg= 16 [ft] q ww= 8.4 psf
W bldg= 30 [ft] L bldg= 36 [ft]
q 1w_-5.2 psf
0.375•mD.�H bldg- 1J'W bldg*�qww- glw)+0.5'�Hrooft1J'Wbldg- qww-qlw,
V gable_wind'= 2
V gable_wind= 1603 [lbs]
This is the lateral wind load that is transmitted to each eave wall. This load will be transmitted
through the roof diaphragm to the eave walls. Subtract the total maximum width of doors and
windows in one eave wall and normalize the load to a per foot basis.
_ V gable_wind
v eavewall ' L bldg-W eaveopenmgs v eavewall= 114 [Ibs/ft]
The eave wall diaphragms can resist 100% of the minimum ultimate load given in reference 10.
This value may be increased by 1/3 for wind per UBC section 1612.3.2 on page 2-5 (110 plf X
1.33= 142 plf).
If v gablewall ` 142 plf Then no additional sheathing is required.
7/31/00 MWO0330 CALC 0731 (Spandl) 30x36xl6.mcd 16
EAVE WALL TOP GIRT ANALYSIS:
The roof diaphragm can be modeled as a simple beam that spans between the gable walls.
Use the value for V gable wind to calculate forces (tension and compression) in the roof
diaphragm chords. Note that these chords are the top girts.
V able wind
L bldg
2
M max :=W.. Bldg M max= 14426 [ft Ibs]
x
Calculate the maximum fiber stress (tension) in the diaphragm chords (T max). The fiber stress
can be determined by dividing the moment by the moment arm. The moment arm is the width in
feet of the roof(Wbid9).
.l.max =M max .11max= 481 [lb]
'
W bldg
T max will be resisted by the shear strength of the nails driven through the steel strapping (if
req'd). Determine the required number of 8d nails needed to resist tension load. Use 111 lb
(1.33 X 83 per NDS Table 12.31' page 125 and NDS Section 7.3 page 43) per nail for nails
driven through 16 ga steel strapping. The steel strapping will be installed where the top girls are
butted on the truss posts. Use 20d nails in each end of the top girt. The shear capacity of the
20d nails is 147 lb. This value may also be increased by 33% due to wind per NDS Table 12.313
page 117 and NDS Section 7.3 page 43.
N ails eave'=4
Shear cave '=N ails cave-1.33.147 Shear eave= 782 [lb]
Nailsbutted :=T max— S hear eave
133 Nailsbutted=—2.3
Therefore no strapping is required.
BOTTOM GIRT CALCULATION:
The shear in the gable walls will be transferred to the posts through the pressure treated splash
board (bottom girt). The steel panels will pass a maximum of 142 pif shear to the splash
boards. Determine the adequacy of the six 20d gals nails on each end of the splash board (12
nails total per splash board) to resist this maximum shear:
V 111ax:=142.12 V max= 1704 [lb]
Use (6) 20d gals nails in each end of the splash board. Determine the adequacy:
�'allowable :=1.33-12-147 v allowable= 2346 > V max This is adequate.
7/31100 MW00330 CALC 0731 (Spandl) 30x36xl6.mcd 17
GIRT DESIGN:
The girts will simple span between posts. Calculate bending stress f bg )due to wind loading rt
q weairt) and determine the required girt size. Subtract 6"for the post width.
G irt_spacing
q wegirt :=q we, q wegirf= 1.92 [pli] L girt-spy :=a- 6
12-12
z
M girt '=q wL girt_span M 5412
[lb in)
egirt' 8 girt
M
fbgirt := S gut fbgut = 2627 psi (stress applied to the girt due to wind loading)
0
Determine the allowable member stress. Use UBC:
LDF for wind = 1.6 NDS Section 2.3 Table 2.3.2 page 9. Flat Use Factor= 1.15 NDS Table 4A
page 25. Size Factor= 1.3 NDS Table 4A page 25.
LDF wind :=1.6 C fu :=1.15 C f:=1.3 C r:=1.15 F girt= 1650 [psi NDS table 4A]
F bgirt :=LDF wind'C fu•C fC r-F girt F bgirt=4539 > fbo psi This is OK.
PURLIN DESIGN:
Assume that the purlins simply span between pairs of trusses or rafters.
Determine the required purlin size.
roof urlin_spacing'c°s(n)
purlinsnow,:= 12-12 w purlinsnow= 4.85 [pli]
Calculate the maximum moment M pure in the purlin and the maximum fiber stress �;f bpurlin)'
due to snow and dead load.
L purlin_span :=a-6-2-1.5 Lpurhn_span= 147 in The bending length of the purlin is reduce by
twice the width of the truss/ rafter.
z
_W purlinsnow-L purhn_span
M purlin g M purlin= 13102 [lb in]
f bpurlin :=M purlin f bpurlin= 1733 psi [stress applied to the purlin
S purlin due to snow and dead load]
Determine the allowable member stress. Use the UBC:
LDF for snow= 1.15 NDS Section 2.3 Table 2.3.2 page 9. Size factor= 1.3 NDS Table 4A page
25. Repetitive Member Factor= 1.15 NDS Table 4A page 25.
LDF snow:=1.15 C f:=1.3 C r:=1.15 F purlin= 1650 [psi NDS table 4A]
F bpurlin :=LDF snow'C fC r'F purlin
F bpurlin= 2837 > f bpurhn [psi] This is OK.
7/31100 MWO0330 CALC 0731 (Spandl) 30x36xl6.mcd 18
CORBEL BLOCK DESIGN:
Determine the required number and size nails and bolts required in the truss block.
Use for bolts: NDS Table 8.3A page 69. 5.5" main member and 1.5"side members. G=0.43
H-F.
Use for nails: NDS Table 12.3E page 117. 5.5" main member and 1.5"side members. G=0.43
H-F.
Assume full snow load and dead load on the roof.
Multiply the allowable loads by 1.15: Snow NDS section 7.3.2 page 43, NDS section 2.3.2 Table
2.3.2 page 9, and Table 7.3.1 page 44.
P bolt 58:=1590 [lb] P bolt 34 :=2190 [lb]
P 16d :=122 [lb] P 20d :=147 [lb]
P snow= 5850 [lb]
If nails are to be used:
N ails P snow N ails= 17.3 number of nails required in each corbel
2.1.15-P 20d block.
If bolts are to be used:
N bolts P snow N bolts= 3.2 number of bolts required through the corbel
1.15'P bolt_58 blocks.
SPECIAL NOTE:
The analysis shown here is in compliance with accepted engineering techniques. The material
referenced is recognized by professional societies and municipalities as accurately modeling the
load cases required by the Uniform Building Code. The drawings and construction notes
attendant to this calculation shall not be modified by the builder unless authorized in writing by
the engineer. The materials and construction techniques described herein all have the required
safety factors included. This construction package which includes this calculation, several
sheets of drawings, and construction notes is in compliance with the 1997 Edition of the Uniform
Building Code. No special inspections are required. No structural observation by the design
engineer is required.
i