HomeMy WebLinkAbout1827 W 16th St Technical - Building
TECHNICAL
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LJANCE ENGINEERIN
PROJECT No.
150408 Rev. 2
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POST FRAME BUILDING
STRUCTURAL CALCULATION
(This structure has been analyzed and designed for structural adequacy only.)
BUILDING OWNER I LOCATION:
Dale Seward FILE
1827 W 16th St ~
Port Angeles, WA 98363 S3
<t.
CLIENT: ,
Sound Building Systems, Inc. -
t:r
3546 Thorndyke Rd 1-
Port Ludlow, WA 98365 RECEIVED ~
JUL 1 4 200B
ENGINEER: CITY OF PORT ANGELES 12..:00
BUILDING DIVISION tJ
~ "o/'j
~- ~~.-(
, ~XP!P'ES: '/1--::
"~i~lI""'''''''''\_"", ..l ,
Property of Alliance Engineering ar , authorized duplication prohibited.
Copyright @ Alliance Eng. eerin of Oregon, Inc.
2700 Market Street N.E. Alliance Engineering of Oregon, Inc. Phone: (503) 589-1727
Salem, OR 97301 www.aeoregon.com Fax: (503) 589-1728
,--
I
7/3/2008 150408R2 (Seward) 28x36x18.xmcd 1
POST FRAME BUILDING SUMMARY:
This is a post-frame building with wooden trusses or rafters and preservately treated posts that
are pressure treated for burial. Post size, post embedment depth. post hot.! di~~ter and
backfill is given in the body of the calculation. The building will depend on the diaphragm
action of the roof and wall sheathing for lateral stability. 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 will be imposed on the windward and
leeward sides of the building simultaneously. .If 1here.is .no concrete floor, the concrete backfill
will provide lateral constraint in the windward and leeward direction. If a concrete floor is
used, lateral restraint for the post will be provided at the ground line by the concrete floor.
REFERENCES:
1. 2006 Edition of the InternationalBuikiin~ Code
2. ASCE 7..()S - Minimum Design Loads for Buildings and Other Structures
American Society of Civil Engineers, 2006
3. 2005 Edition, National Design SpecnlCation (NOS) Supplement For Wood
Construction, American Wood Counsel
7/3/2008 150408R2 (Seward) 28x36x18.xmcd 2
DESIGN INPUT VALUES:
Building Dimensions
Width of Building
Wb1dg:= 24 ft
4,ldg := 30 ft
Hb1dg:= 18 ft
Length of Building
Eave Height of Building
Overhang:= 18 in Length of Eave Overhang
Rpitch := 4 /12 Roof pitch
Bay:= 10 ft Greatest nominal spacing between eavewall posts
W gableopenings:= 16 ft Total width of openings in one gable wall
Weaveopenings:= 0 ft Total width of openings in one eave wall
DesiQn Loads for Buildina:
Wind DesiQn Values:
Fastest wind speed (3 second gust)
V wind:= 100 MPH
Wind Exposure:
ExposlU'C:= "cn
Roof Load DesiQn Values:
Pg := 25
Pd:= 3
Pd2:= 0
Ibs Ground snow load
Ibs Roof dead load
Ibs Additional truss bottom chord dead load (if applicable)
Seismic Deskin Values:
Ss:= 145.6 Mapped spectral acceleration for short period
Sl := 60.0 Mapped spectral acceleration for 1 second period
In = 1.00 Importance factor
Rs := 7 Response modification factor
7/3/2008 150408R2 (Seward) 28X?.A)x18.xmcd 3
DESIGN INPUT VALUES (Continued):
Structural Members for BuildinQ:
Post Properties:
P width:= 6
P depth:= 8
in Post width y-:axis
in Post depth x-axis
POST SIZE
(Solid rough-:sawn Hem-Fir post
unless otherwise specified)
Grade := "2" Grade of Post ( 2, 1, or SS" Select Structural)
PurlinProoerties:
P urlin _spacing:= 24 in
GirtProoerties:
Girt_spacing := 48 in
Spurlin:= Sx26
Sgirt;= 2 Sy26 + Sx26
F purlin:= F bDF2dim
Fgirt:= FbHF2dim
Footinq and Post HoleDesKID Values:
qsoil:= 1500 pst Assumed soil vertical bearing capacity
Ssoil = 150 pst Assumed soH lateral bearing capacity
<!jaJooting := 2 ft Main truss post footing diameter
Slab and backfill infonnation
Concrete_slab = "Required for post cOllStraint"
Concrete backfill = "No" Backfill in main posts
(GO TO LAST PAGE FOR SUMMARY OFRESUL TS)
7/3/2008 150408R2 (8evyard) 28x36x18.xmcd 4
SNOW lOAD ANALYSIS:
Design per ASeE 7-05
For roof slopes greater than 5 degrees, and less than 70 d~grees.
Pg = 25 psf Ground Snow Load (from above)
Ce:= 1.0 Exposure factor
Ct:= 1.0 Thermal Factor
C. = 1.00 Roof slope factor
I,. = 1.00 Importance factor
Pt= Flat roof snow load, psf (see analysis below)
Ps= Sloped roof snow load, psf (see analysis below)
1. Detennine Pf and Ps
Pr:= .7-Ce-Ct-I,.-Pg pr= 17.5 psf
Flat roof snow load
Note: This is NOT the snow
Ps = 17.5 psf Sloped (balanced) roof snow load load Ut Sbedttfor desf ign-See
Psu a 0 om 0 page.
2. Detenninethe unbalanced snow load
Ps := PrCs
Wh1d!l
W rid"~:= ----::..
=' J
Wridgc = 12
ft Horizontal distance from eave to ridge
Note: If Wridge < 20', use Method 1 to ~etermine unbalanced snow load, otherwise use Method 2
Method 1
Psul := I,..Pg
Psul = 25
psf Unbalanced snow load for buildings with Wridge < 20'
Method 2
The unbalanced snow load will occur from the ridge to a distance Is. and intensity, Psu2 as follows:
h.t = 1.56 ft Height of drifted snow
y = 17.25 pcf Snow density
S = 3 ft Run in roof for a rise of 1
Is = 7.2 ft Distance of unbalanced snow from ridge (if applicable-see below)
Psu2 = 33 psf Unbalanced snow load for buildings with Wridge> 20'
Final unbalanced snow load
Psu = 25 psf Final (roof) snow load used for desiQn of stmctural members
and connections as required per Chapter 7 of ASCE 7-05
Application of snow load to buildina
The snow load, Psu' was calculated using Method = I ,therefore the final roof snow load used for desiqn
shall be Distributed = "across entire building width"
If Method 2 is used. the remainder of roof shall be designed using no less than Ps = 20 pst snow load
7/312008 150408R2 (Seward) 28x36x18.xmcd 5
WIND ANALYSIS:
Design per ASCE 7-05
Methoq 2 - Analytical Procedure
V wind = 100
Basic Wind Speed
k.J := .85
WlOd Directionality Factor
kn = 1.0
Topographic Factor
kz = 0.902
Wind Exposure Factor
Iw = 1.00
Importance factor
2
%:= .00256.kz'kn'k.J. V wind .Iw
Velocity Pressure
qb = 19.63 pst
Calculated Wind Pressures:
Wmdward Eave Wall:
Leeward Eave Wall:
qww:= %.GCpfWw
qlw := qh.GCpflw
qww = 10.14 pst
qlw = -8.15
pst
Wmdward Gable Wall:
Leeward Gable Wall:
qwwg:= qh.GCpfwwg
qlwg:= %.GCpflwg
qwwg = 7.85 pst
qlwg = -5.69 pst
Wmdward Roof:
Leeward Roof:
qwr:= qh'GCpfWr
<Ilr:= <Ib.GCpflr
%- = -9.20 pst
qwr = -13.54 pst
Wall Elements:
Roof Elements:
qwe:= qh.GCpfw
qwe = -20.02 pst
qr:= ~.GCpfr
q, = -29.05 pst
Internal Wind Pressure (+/-):
qj:= %.GCpj
qj = 3.53 pst
~--
7/3/2008 150408R2 (Seward) 28x"...6x18.xmcd 6
BUILDING MODEL:
STEP 1: DETERMINE THE SHEAR STIFFNESS OF THE TE~T PANEL
This procedure relies on tests conducted by the National Frame Builders Association.
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 Iblin
STEP 2: CAlCULATED ROOF DIAPHRAGM STIFFNESS OF THE TEST PANEL
c' = (E X t) I (2 X (1+V) X (gJp) + (K2/ (b' X t)^2))
Where: E =
t=
v=
g/p ;;
b'=
27 .5x1 0^6 psi (modulus of elasticity for steel)
0.017" (thickness of 29 gauge steel)
0.3 (Poisson's Ratio for steel)
1 .139 ratio of sheathing corrugation length to corrugation pitch
144" (12'-0" length attest pan el)
STEP 2.1
This equation was set equal to the stiffness of the test panel (2166 Iblin) and the unknown value
(K2) was solved for.
K2 = 1215in4 sheet edge purlin fastening constant
STEP 2.2:
Use new building width to determine stiffness of new roof diaphragm ( Ch
bn.:w:=
Wb1dg.12
2
cos((':') )
K.,;= 1175
Ibf I ft
t:= 0.017
in
E> = 18.435 deg
(Angle of roof pitch from horizontal)
bnew = 152 in
E := 27500000
E.!
.c:=
K"}
C = 2404
Ibf / in
2.961 +
.,
(b""".tt
STEP 2.3 & 2.4:
Calculate the equivalent horizontal roof stiffness ( cJJ for the full roof:
Since Ch is for the full roof, the roof length must be ratioed by the aspect ratio of the roof panel (b / a)
where "aN is the truss spacing in inches.
a := Bay" 12
( )2 bn.:"
cll := 2.c.cos (~) .-
n
a = 120 in
Ch;;= 5474
lbf / in
71312008 150408R2 (8eward) 28JC>v6x18.xmcd 7
STEP 3: DETERMINE 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. The
stiffness of the post frame can be calculated from the amount of force required to deflect the system one
inch. The spring constant (k) in pounds per inch of deflection results directly.
k = 72 Ibffm
STEP 4: DETERMINE THE TOTAL SIDE SWAY FORCE (R):
Apply wind loads to the walls to determine the moment, fiber stress and end reaction at prop point R.
Calculate Total Wind Pressure:
qc:= if(qww - qlw ~ 10, lO,qww - ~w)
qc = 18.29 psf
'( '\
a
q\\\\vosl:= q". 12.12)
qwwposl = 15.24 pH
..,
Lposl_hndg ..
MWind := q""wp,,sr _
X
Mwind = 79289 in-Ibf
Mwind
t~'-ind :=
Sxea"epost
fwind = 619 psi
. LposI_lwdg
R := .' .tlm'lJOsl" 8
R = 1166 Ibs
STEP 5: DETERMINE THE RATIO OF THE FRAME STIFFNESS TO THE ROOF STIFFNESS:
This ratio (kJ ctJ will be used to determine the side sway force modifiers.
k
- = 0.013
ch
STEP 6: DETERMINE SIDE SWAY RESISTANCE FORCE:
mD = 0.99
STEP 7: DETERMINE THE ROOF DIAPHRAGM SIDE SWAY RESISTANCE FORCE:
Q:= mD.R Q = 1151 Ibf
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. The distributed load that is not resisted by the roof diaphragm will apply
additional moment and fiber stress to the post.
Mdfl = 4193 in-lbf
fdfl = 33 psi
Calculate the total moment and the total fiber stress in the post
Mtat := mD. Mwind + ~
Mtot = 82434 in-Ibf
ftot := mD.fwind + fdfl
ftot = 644 psi
7/3/2008 150408R2 (Seward) 28x36x18.xmcd 8
MAIN POST DESIGN:
Calculate allowable unit compression stress, Fcc'
Fcl = 575 psi
F c := F cl' 1.15
Fc = 661 psi
Allowable compression stress including load factors
~_Dndg = 204 in Bending length of post ~ = 8 in Minimum unbraced dimension of post
K.: := 0.8
c:= 0.8
Ewood = 400000 psi
Ie:= K.:.I1IosUllldg
Ie = i63.2in
.882. Ewo<m
FeE :=
(~r
FcE = 848
CalculateCohlmn Stability FactQr. ep:
~( FeE J ( FeE J:2 . ..F'cE.'.l
1+- 1+- -
. '_ Fe Fe 2
Cp ._ . . . ... - ..... ..0 .... ,-..
2.c 2.c c
Cp = 0.77
Fcc := Fc'Cp
Fcc = 509 psi Allowable compression stress on the post
Wroof = 28 psf Total roof loading
Psnowpost = 3375 fbs Axial loading per post due to roof snowload
Pdeadpost = 405 Ibs Axial loading per post due to roof dead load
Fb := I-b}" 1.6
Fb = 920 psi Allowable bending stress per post including load factors
7/3/2008 150408R2 (Seward) 28x"..6x18.xmcd 9
Check Load Cases:
Load Case 1: Dead Load + .75 * Wind Load +.75 * Snow Load
fbI := .75ftot
fbI = 483
psi Actual bending stress on post
. '_ .75 P snoul"lSl + P d""dl")sl
f.:.-
1\ost
f~=61
psi Actual compression stress per post
CCF ALl I := l(~)2 +
F.:.:
f"J
( t~)
F". I - -
\. F.:E J
1
CCFALI1 = 0.58
Load Case 2: Dead Load -+ Wind Load
fbI := ftot
fl.} = 644
psi (Actual bending stress on post)
. P d"..dl"lSt
t.: :=
Apost
fe= 8
psi (Actual compression stress per post)
r.., j
CCFAU2= (:J + ('b' ()
Ft>. 1--
F.:E
CCFALI2 = 0.71
Load Case 3: Dead Load + Snow Load
fbI:= 0
fbI = 0
psi (Actual bending stress on post)
, P snowpost + P deadpost
fe :=
Apoo
f~ = 79
psi (Actual compression stress per post)
l 7 j
f - f
CCFALI3= (F:) +, ( bI ( I
FI,I--
F.:E)
CCFALI3 = 0.02
CCFALI = 0.71 Less than or equal to 1.00 thus OK
7/3/2008 150408R2 (Seward) 28x36x18.xmcd 10
SEISMIC CALCULATIONS:
Design per A8CE 7.:.05
Ss = 145.6 Mapped spectral acceleration for short periods (from above)
SI = 60 Mapped spectral acceleration for 1-second period (from above)
IE = 1.0 Importance factor
W = Dead load of building
Rs = 7 Response modification factor (from above)
1. Determine the Seismic Design Category
a. Calculate Sos and SDl
For 8DS:
For SD1:
For Ss = 1.46
For SI = 0.60
Fa == 1.00
Fv == 1.50
SMS:== Ss.Fa
S~H := SrFv
SMS '" 1.46
Sr.u '" 0.9
(")\
Sps:== i )"S!\IS
SDJ := (f }SHI
Sns = 0.97
Sm = 0.60
Seismic _ Design_Category = ''0''
2. Determinetlte building parameters
Building dead load weight, W:
W:== [(Wb1dg'4Idg),(pr.2)J + [(Wbldg'4Idg) + [2.(Wb1dg + 4Idg)' H;dg]'Pd
W = 5076 Ibf
Building area, Ab:
At, ;= Lbldg' Wb1dg
At, == 720 ft2
7/312008 150408R2 (Seward) 28x36x18.xmcd 11
3. Determine the shear force to be applied
a. Determine the structural period, T
Ta:= .02. (Hb1dg+ H.oof).75 T:= Ta
b. Detemine the Seismic Response Coefficient, Cs:
Cs is calculated as:
'. ,_ SDS
C." ,- ,
'- Rs
T = 0.20
Cs2 = 0.139
IE
But shall not be less than:
C.l := .044.Sos.IE
Csl = 0.043
But need not exceed:
Cs3 :=
SDl
Cs3 = 0.422
('RS)'
T-
IE
Cs::=' 0.139
c. Detemine the Seismic Base Shear:
Vhase_shear:= Cs.W
V base_shear = 704
Ibf
4. Detennine the seismic load on the building:
Per ASCE 7-05 Section 12.3.4.1 & 12.3.4.2, for Seismic Design Category's A, S, andC, p =1.0; for
Seismic Design Category 0, E, or F, p shall 1.3.
Since Seismic_Design _Category = "D", p = 1.3
E:= P,Vbase_shear
E = 915
Ibf Seismic load on building
- ----,
7/3/2008 150408R2 (Seward) 28x36x18.xmcd 12
DETERMINE GABLE WALL SHEAR LOADS:
1. Determine the wind load on the eave wall to be resisted by the gable wall in shear:
4e == 18.3 pst Eave wall wind pressure from above
. (O.37S.mD.l4.l<Ig.Lhklg.q,,) + (I4m,rLbl<lg'(L-oof)
Veuye wmd:==
- ~
Veave wind == 1827 Ibf
2. Determine the seismic load to be resisted by the gable wall in shear:
.. E
Vea,'c seISllllC:== -
- ...
Veave seismic == 458 Ibt
3. Determine the controlling load to be resisted by the gable wall in shean
The controlling load = "Vcavc_wind" . Therefore, V gable shear = 1827 Ibf
V gable_shear is the shear load that is transmitted through the roof diaphragm to each gable wall.
Normalize the load to a per foot basis.
Ygabl,,\\all :~
Whldg - 12
V gahl,,_ sh"ar
v gablcwall = 152
pit
The gable wall diaphragms can resist the shear loads as follows:
v gablewall < 11 0 plf
Vgablewall < 188 plf
Use 29 gauge metal sheathing. Install per the
Typical Screw Schedule as shown on the Standard
Details drawing in the engineered drawing package.
Use 29 gauge metal sheathing. Install per the Alternate
Screw Schedule as shown on the Standard Details drawing in
the engineered drawing package.
Determine the lateral load that is transmitted to the gablewall with the -large openings that will be
resisted by the gable wall posts in bending. Check the bending stress in these posts.
OpcningJlCight:= 120
Mgablcwall :== V gablc_shcar" Opening_height
fvlg;Jbl"wall
Fxgahk\\a/l := 4'~fi8
Fxgablcwall :::: 857 psi
Fxallow:== 1.6.575
Fxallow = 920 psi
Since F xgablewall < F xallow this is ok.
7/312008 150408R2 (Seward) 28x36x18.xmcd 13
DETERMINE EAVE ViALL SHEAR LOADS:
1. Detennine the wind load on the gable wan to be resisted by the eave wall in shear:
qg := if( qwwg - qlwg ~ 10, 10, qwwg - qlwJ
qg = 13.5 psf Gable wall wind pressure
H.-oof = 4 ft
O.375.mD. Hhldl!' W"ldl!'ql! + O.5.II.-oof' W"ldl!'q!!
V gable \\'ind:= ~ ~ ~ - ~
- 2
VgabJe_wind;::= 1407 Ibf
2. Determine the seismic toad to be resisted by the eave wall in shear:
.. E
V gable selsmIC:= -
- 1
Vgable_seismic = 458 Ibf
3. Determine the controlling load to be n:-sisted by the eave wall in shear:
The controlling load = "V gable_wind" . Therefore, V cave _ sbwr = 1407 Ibf
Veave_shear is the shear load that is transmitted through the roof diaphragm to each eave wall.
Normalize the load to a per foot basis.
Veave shear
veavewall:= T. w-
'-1>ldg - eavcopeningll
veavewall = 47
pit
The eave wall diaphragms can resist the shear loads as follows:
Veavewall < 110 pIt
Use 29 gauge metal sheathing. Install per the
Typical Screw Schedule as shown on the Standard
Details drawing in the engineered drawing package.
- -- -----,
7/3/2008 150408R2 (Se'Nard) 28x36x18.xmcd 14
EMBEDMENT FOR MAIN POST:
Calculate the minimum required post embedment depth tor lateral loading tor the main posts. The
backfill may be gravel, natural or concrete backfill as specified on page 3.
Post_is = "constrained by a concrete slab"
Concrete backfill = "No" (Input from page 3)
Va = 1037 Ibf Lateral shear load at thegroundline
Ma = 6869 ft-Ibt Moment at the groundline
djaJooting = 2 ft. Main post footing diameter
Ssoil = 150 pst Lateral capacity of soil
Trial depth:: 1.5 ft.- The starting depth of the post hole depth. The final post hole depth is determined
by iterating to a final depth, per ASAE EP486.1, as allowed per 2006 IBC.
depth Jl<lS1 = 3
ft. This is the minimum required post embedment depth for lateral loading
Gable walt uplift due to shear loading on gable wall shear panel:
Calculate uplift pullout of the gable wall posts due to shear loads on the gable walls.
Veave wind = 1827 Ibf
Calculated from above
Veaye _wind. Ht.ldg
Cpo.<:l :=
Wh1dg - 11
Cpos! = 2741 Ibf This is the uplift load on one gable wall post
Assume a dead load weight of roof and wall area to be 2.0 pst. The area of the roof and wall that
will tend to keep the gable wall post in the ground will be as follows:
B,,\ .
Roof:= ---=-- Wh1d,z-2
., .
..
Roof = 240 Ibs Dead load of roof
Gablc_"(tll := [Ht.IJ{(Wbldg - 12) + (HlOofOWbldg) + (Hbldg" 2o:-!ay)J2
Gable wall = 984 Ibf Dead load of gable wall <\.1llh.gabl..:]ootllg = 5.0 ft gable post embedment depth
Posts:= (Hb1dg + <\:plIUlable Jooting)' W post
Posts = 268 Ibs Weight of post
dja .$olble Jooting = 2
ft Diameter of gable wall posthole footing
Concrete backfill in the gable end posts is = "required"
to resist gable wall panel uplift.
Backfill = 21051bs Gable post backfill weight if gable end post hole is backfilled with
concrete (0 if granular or native soil backfill. Concrete backfill mayor
may not be required to resist gable wall panel uplift).
Wttot := Gable_wall + Roof + Posts + Backfill
Wttot = 3597 Ibf
Total resistance for gable wall panel uplift. Since ~ is greater than
the gable wall panel uplift, Cpost> the gable wall footing is adequate.
,-
7/3/2008 150408H2 (Se-.vard) 28x36x18.xmcd 15
FOOTING DESIGN FOR MAIN POST:
Determine the footing size and depth for vertical bearing for the main posts.
( 2J'
~a Jooting
Afooting:= 1t. 4
Atooting = 3.14 ft2 Footing area
qsoil = 1500 psf S9i1 bearing capacity for f99ting
~a~tOoting = 2 ft Footing diameter
P oslo.depth = 5 ft Minimum required post embedment depth
Pfooting:= Acooting.qsoil.dfactor
Pfooting = 8482 Ibf End bearing capacity of footing
p snow = 3780 1M
Total footing load
Note that the end bearing capacity (Pfooting) is greater than the snow load (P snow)' This is OK.
7/3/2008 150408R2 (Seward) 28x36x18.xmcd 16
GIRT DESIGN:
The girts will simple span between posts and loaded horizontally for wind. Calculate bending
stress due to wind loading and determine the adequacy of the girts.
Girt _SP.1Cing
qw~girt := q"int! _ltirt. 12. 12
qwegirt = 7.85 pli
Lgirt_span = 138 in Orientation = "Strongback"
.,
Lgirt _span'"
M Ilirt : = q,,~ltirt.
- - 8
, Mgirt
tbllil1 := -
- Sgirt
Mgirt = 18687 in-Ibf
fbgirt = 1600 psi
Stress applied to the girt
Determine the allowable member stress including load factors.
LDFwind := 1.6
Cfugirt = U5
CFgirt = 1.00
Cc:= 1.15
F girt = 850 psi
Fbgirt := LDFwind.Cfugirt.CFgirt,Cc.Fgirt
Fbgirt = 1799 psi > fbgirt This is OK.
PURLlN DESIGN:
The purlins simply span between pairs of trusses or rafters. Determine the adequacy of the purl ins.
Purlin = "2x6"
Purlin_spacing = 24 in O.C.
l:'pllrlin_~ = III in Bending length of purl in
"'purlin = 4.43
pli Distributed snow load along top edge of purl in
')
wpllrliu' Lpurlin --"p.,n-
Mpurlin := :;:
M.-Jin = 6818 in-lbf Bending moment in the purlin
M "in
..... ~':1"'"
lbpurlin := -
Spurlin
tb"unm = 902 psi Bending stress applied to the purlin
Determine the allowable member stress including load factors
LDFsnow:= 1.15
CFpurI;n = 1.30 Cc:= 1.15
Cfupurlin = 1.00
F purI;n = 900 psi
F bpurlin := LDF snow,CFpurlin ,Cc,Cfupurlin' Fpurlin
Fbpurlin = 1547 psi> fbpurlin This is OK
I
7/312008 150408R2 (Seward) 28x"..6x18.xmcd 17
MAIN POST CORBEL BLOCK DESIGN:
Determine the required number and size of bolts required in the main post corbel block.
A$sum~ full snow load and dead load on the. roof.
Allowable fastener shear capacities
PbolU8 := 1590 Ibf Shear capacity for 5/8" dia. bolts
Pbolt 34:= 2190 Ibf Shear capacity for 3/4" dia. bolts
Pbolt 10 := 3600 Ibf Shear capacity for 1" dia. bolts
PI6d:= 122 Ibf Shear capacity for 16d nails
P20d:= 147 1M Shear capacity for 20d nails
p snow = 3780 Ibf Combined snow and dead load on corbels
If 5/8 dia. bolts ale used:
Nbol\s58 = 2.1
Number of 5/8" dia. bolts required in the corbel block
If 3/4 dia. bolts are used:
Nbo1ts34 = 1.5
Number of 3/4" dia. bolts required in the corbel block
If 1 dia. bolts are used:
NbollslO = 0.9
Number of 1" dia. bolts required in the corbel block
If 20d nails are to be used:
N ail.2Od = i1.2
number of 20dnails required in each corbel block.
If 16d nails are to be used:
Nailsl6d = 135
number of 16d nails required in each corbel block.
7/312008 150408R2 (Seward) 28x?v6x18.xmcd 18
SUMMARY OF RESULTS:
Building Dimensions
Building Des",n Loads
Wb1dg = 24 ft (Width of Building)
Wind_speed = 100 MPH Growld_snow_load = 25 psf
4,ldg = 30
ft (Length of Building)
Wind_exposure = "C"
Roof snow load = 25 pst
Roof dead load = 3 psf
Hb1dg = 18 ft (Eave Height of Building)
Seismic_Design_Category = "0"
Overhang = 18 in (Length of Eave Overhang)
Rpitch = 4 /12 (Roof pitch)
FootinQ Details:
Post Details
Post size = "6x8"
Postjs = "constrained by a concrete slab"
Post_grade = "No.2 Hem-Fir"
Usage;;::: 71 % (C9mbinedstre.ss usage !Jfpost)
Shear Wall Details:
Postdepth = 5.0 ft (Design Post Depth)
!1jaJooting"" 2
ft(Design Footing Diameter)
Footingusage = 45 % (Stress usage of footing)
Vgablewall = 152 plf (Max. shear in gable wall)
Veavew:lIl = 47 plf (Max. shear in eave wall)
Girt Details:
Girt_usage = 89 % (Stress usage of wall girt)
Orientation = "Strongback"
Purtio Details:
Purlin_usage = 58 % (Stress usage of roof purlin for snow loading)
Corbel Block Bolts:
Nbolts58 = 2.1 Number of 5/8" dia. bolts required in the corbel block if used.
NboIts34 = 1.5 Number of 314" dia. bolts required in the corbel block if used.
NboltslO = 0.9 Number of 1" dia. bolts required in the corbel block if used.
N ails20d = 11.2 Number of 20d nails required in each corbel block if used.
Nailsl6d = 13.5 Number of 16d nails required in each corbel block if used.
SPECIAL NOTE:
The drawings attendant to this calculation shall not be modified by the builder unless authorized in
writing by the engineer. No special inspections are required. No structural observation by the
design engineer is required.