HomeMy WebLinkAboutBLD2005-01189 Structural Calculation - BLD Engineering / Geo-tech Reports - 6/30/2005 i BUILDING SUPPLY
22175 S.fEgffi v 99E Canbb-,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.)
PROJECT No.
MW05340
OWNER:
Rowland Binion
97 E Spencer Lake Rd
Shelton, WA 98584
ENGINEER:
tip,s CLAr
l�t � i
87 5
ssd�tdF"y��yG
EXPIRES: 1/12/
6/27/2005 MW05340 (Binion) 52x60x14.mcd 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 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 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. The actual post length for bending will be
assumed to be measured from top of the post hole backfill to the top of the corbel block. If
there 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, 2003 Edition of the International Building Code
2. ASCE 7-02 - Minimum Design Loads for Buildings and Other Structures
American Society of Civil Engineers, 2003
3. 2001 Edition, National Design Specification (NDS) Supplement For Wood
Construction, American Wood Counsel
t I
6/27/2005 MW05340 (Binion) 52x60x14.mcd 2
SUMMARY OF DESIGN VALUES:
Buildin4 Dimensions
Wbldg:= 52 ft (Width of Building)
Lbldg:= 60 ft (Length of Building)
Hbldg:= 14 ft (Tallest Eave Height of Building)
Rpitch:= 4 /12 (Roof pitch)
Bay:= 12 It (Greatest spacing between eavewall posts)
Wgableopenings:= 30 ft (Total width of openings in one gable wall)
Weaveopenings= 13 ft (Total width of openings in one eave wall)
Truss heel:= 12 in (Depth of truss/rafter heel)
Post Properties:
Pwidth := 6 in (Post width y-axis) POST SIZE
Pdepth:= 6 in (Post depth x-axis)
Grade := "2" (Grade of Post(2, 1,or SS=Select Structural))
Fb1 = 575 psi (Allowable bending stress for the posts)
Fc1 = 575 psi (Allowable compression stress for the posts)
Ew,,d = 1100000 psi (Allowable modulus of elasticity for posts)
Lpost_bndg= 156 in (Bending length of post)
Purlin Properties:
Girt Properties:
Purhn_spacing:= 24 in
dirt_spacing= 26.4 in
Spurlin:= Sx26 Sgirt= "Sy26"
Fpurlin:= 1'bMSR1650 Fgirt:= FbMSR1650
6/27/2005 MWO5340 (Binion) 52x60x14.mcd 3
SUMMARY OF DESIGN VALUES (Continued):
Footing and Post Hole Design Values:
qsoil 1500 psf (Assumed soil vertical bearing capacity)
dia_f.ting 2 ft (Diameter of footing)
Ssoil = 150 psf (Assumed soil lateral bearing capacity)
Desiqn Loads for Building:
Wind Design Values: Roof Load Design Values:
Fastest wind speed (3 second gust) 1,t :-- 25 Ibs (Ground snow load)
V,,i,,d:= 85 MPH 1,a - 3 Ibs (Roof dead load)
Wind Exposure:
Exposure '13"
Seismic Design Values:
Ss:= 122.2 Mapped spectral acceleration for short period
SI := 46.6 Mapped spectral acceleration for 1 second period
IE:= 1.0 Importance factor
W= Dead load of building (See analysis below)
Rs:= 7 Response modification factor
(GO TO LAST PAGE FOR SUMMARY OF RESULTS)
6/27/2005 MW05340 (Binion) 52x60x14.mcd 4
SNOW LOAD ANALYSIS:
Design per IBC 2003
For roof slopes greater than 5 degrees, and less than 70 degrees.
pg= 25 psf Ground Snow Load(from above)
Ce:= 1.0 Exposure factor
Ct := 1.0 Thermal Factor
Cs = 1.00 Roof slope factor
Is := 1.0 Importance factor
pf= Flat roof snow load, psf(see analysis below)
IDS= Sloped roof snow load, psf(see analysis below)
1. Determine pf
Pf .7•Ce•CYls•Pg Equation 1
Pf= 17.5 Mf
Ps PrCs Equation 2
ps = 17.5 psf This is the balanced snow load on the roof.
2. Determine the unbalanced snow load
Equation 3
Ps
Psul 1.5 —
Ce
Equation 4
NO Ps 1.2 1 + a -
2 Ce
Psu = 26 psf This is the unbalanced snow load on the leeward side of the roof.
6/27/2005 MW05340 (Binion) 52AN14.mcd 5
WIND ANALYSIS:
Design per IBC 2003
Method 2-Analytical Procedure
IW:= 1.0 Importance factor
Vwind = 85 Basic Wind Speed
kd :_ .85 Wind Directionality Factor
kzt = 1.0 Topographic Factor
kZ= 0.701 Wind Exposure Factor
2
qh:_ .00256 kZ kn•kd•Vwir,d` Iw Velocity Pressure
qh= 11.01 psf
Calculated Wind Pressures:
Windward Eave Wall: Leeward Eave Wall:
gwtiv gh•GCpfiXw qlw:= gh•GCpflw
qww= 5.69 psf q1w= —4.58 psf
Windward Gable Wall: Leeward Gable Wall:
gwwg= gh'GCpfwwg glwg= gh'GCpflwg
gwwg = 4.41 psf ql wg= —3.19 psf
Windward Roof: Leeward Roof:
qwr:= qh-GCpfwr
qlr:= gh'GCpflr
qwr= —7.60 psf
glr = —5.16 psf
Wall Elements: Roof Elements:
qwe gh•GCpfw qr= gh'GCpfr
qw,= —10.68 psf qr= —14.87 psf
Internal Wind Pressure
qi := gh•GCpi
qi = 1.98 psf
6/27/2005 MW05340 (Binion) 52x60x14.mcd 6
BUILDING MODEL:
STEP 1: CALCULATE THE SHEAR STIFFNESS OF THE TEST 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 lb/in
STEP 2: CALCULATED ROOF DIAPHRAGM STIFFNESS OF THE TEST PANEL
c'= (EXt)J(2X(1+V)X(g/p) +(K2/(b'Xt)"2))
Where: E= 27.5x10116 psi(modulus of elasticity for steel)
t= 0.017"(thickness of 29 gauge steel)
V= 0.3 (Poisson's Ratio for steel)
g/p = 1.139 ratio of sheathing corrugation length to corrugation pitch
b'= 144"(12'-0"length of test 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 in4 sheet edge purlin fastening constant
STEP 2.2:
Use new building width to determine stiffness of new roof diaphragm (c1J:
Wbldg-12 K,) := 1275 Ibf/ft
bnew t:= O.Ol7 in
cos(e) O = 18.435 deg (roof angle of incline)
bnew= 329 in E:= 27500000
(E-t)
c:_
K, c = 10686 Ibf/in
2.961 +
�bnew•t)r
6/27/2005 MWO5340 (Binion) 52x60xl4.mcd 7
STEP 2.3& 2.4:
Calculate the equivalent horizontal roof stiffness(ch)for the full roof:
Since ch is for the full roof,the ruof lengti) must be ratioed by the aspect ratio of the roof panel (b/a)
where "a"is the truss spacing in inches.
a:= Bay 12 brew
ch:= 2•c•cos�U)�
a = 144 in a
ch = 43928 IDT/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. 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= 188 Ibf/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:
qe iggwtiv— qlw< 10,10,qww— qlw) qe= 10.27 )sf
a
gwwpost= qe'
12.12 gtot:= gwwpost
gwwpost= 10.27 pli gtot = 10.27 pli
Lpost_bndg�
Mwind gtot' Mwind= 31227 in-Ibf
Mwind
wwind= fwnnd= 434 psi
2'Sspost
R== 3-gtot I post�bndg R = GOi Ibs
STEP 5: CALCULATE THE RATIO OF THE FRAME STIFFNESS TO THE ROOF STIFFNESS:
This ratio (k/ch)will be used to determine the side sway force modifiers.
k
— = 0.004
ch
6/27/2005 MW05340 (Binion) 52x60xl4.mcd 8
STEP 6: DETERMINE SIDE SWAY RESISTANCE FORCE:
mD = 0.99
STEP 7: CALCULATE THE ROOF DIAPHRAGM SIDE SWAY RESISTANCE FORCE:
Q := mD•R Q = 593 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 = 1585 in-Ibf
fd fl = 22 psi
Calculate the total moment(Mtot)and the total fiber stress(1ioo.
Mtot MD'Mwind + Mdfl Mtot = 32416 in-Ibf
ftot -D'fwind+ ldtl ftot= 450 psi
6/27/2005 MW05340 (Binion) 52x60x14.mcd 9
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.
Calculate allowable unit stress(compression FCC).
Fcl = 575 psi FC:= FC1.1.15•.80
FC= 529 psi (Allowable compression stress including load factors)
Lpost_bndg= 156 in (Bending length of post) dp,st = 6 in (Minimum unbraced dimension of post)
Ke:= 0.9 C:= 0.9 KCE:= 0.3 EN,00d= 1100000 psi
Ie K,.Lpost_bndg Ie= 124.8 in
.95•Ewo,d
FcF:= KcE- - FcE= 725
( le ..
dpost
2
FcE FcE FcE
FC Fc Fc
Cp — Cp = 0.79
2•c 2•c c
FCC:= FC.Cp FCC= 416 psi
Wroof= 29.25 psf (Total roof loading)
Psnowpost= 6300 lbs (Axial loading per post due to roof snow load)
Pdeadpost= 720 Ibs (Axial loading per post due to roof dead load)
Fb:= Fb1.1.6•.80 Fb= 736 psi (Allowable bending stress per post including load factors)
6/27/2005 MW05340 (Binion) 52x60xl4.mcd 10
Check Load Cases:
Load Case 1: Dead Load+ .75«'Rind Load+ .75` Snow Load
fbI :_ .75ftot fbl = 339 psi (Actual bending stress on post)
75Psnowpost+ Pdeadpost
fc:= fc= 151 psi (Actual compression stress per post)
Apost
CCFALII := fe + fbI
Fcc b• 1 fc CCFALII = 0.71
F — FcE)—
Load Case 2: Dead Load +Wind Load
fbI := ftot fbI = 450 psi (Actual bending stress on post)
Pdeadpost
fc:= fc,= 20 psi (Actual compression stress per post)
Apost
fc ' fbl
CCFALI2 := — +
Fcc _ fc CCFALI2 = 0.63
Fb I FcE
Load Case 3: Dead Load +Snow Load
fbI := 0 tM = 0 psi (Actual bending stress on post)
Psnowpost+ Pdeadpost
fc:= fc= 195 psi (Actual compression stress per post)
Apost
CCFALI3 := f` 2 + fbI
Fc, fc CCFALI3 = 0.22
Fb I FcE
CCFALI = 0.71 Less than or equal to 1.00 taus OK
i
6/27/2005 MW05340 (Binion) 52x60x14.mcd 11
POST EMBEDMENT FOR NON CONSTRAINED CONDITION:
Calculate the required post depth. This assumes full depth backfill of post holes with concrete.
Mtot = 32416 in-lbf
Ppost = 354 Ibs (Equivalent lateral load applied @ mid point of the truss post)
Ssoil = 150 psf (Assumed soil lateral bearing capacity)
depth_postnc = 3.5 ft (Trial depth of embedment)
S1 = 465.5 (Calculated using a trial depth of embedment)
A= 0.89 ft2(Area of footing)
h= 7 ft (Height at which equivalent point load will be applied)
depthnc= A.2 I 1 + 1 + 4.36-A depthnc= 3.1 ft (Required post embedment depth)
Calculate pullout of the gable wall posts due to shear loads on the gable walls.
Hroof Wbldg tan((-)) Hroof= 9.7 ft
0.3 75•mD•(Hbldd'Lbldg'qe
Veave wind:_
Veave_wind= 1596 Ibs (Total load transferred into each gable wall)
Veave_wind•Hbld
Cpost g Lpost= 1016 lbf (This is the uplift load on one gable wall post)
Wbldg— Wgableopenings
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 gable wall post in the ground will be as follows:
Lbldg Lbldg Wbldg Wbldg
Eave wall = Hbldg' 2 •2 Roof 2 2 •2 Gable wall Hbldg• 2
Eave wall = 940 lbf Roof= 1560 Ibs Gable wall = 728 lbf
2
Posts (Hbldg+ depthnc)•Wpost dia_footing Apost
Post_hole�= 150-depth_postnc' 3.14• —
4 144
Posts = 150 lbf Post hole= 1734 Ibs
Wttot Eave wall + Gable wall + Roof+ Posts + Post hole
Wttot = 5012 lbf (Note that Wttot is greater than Cpost• Thus OK.)
6/27/2005 MW05340 (Binion) 52x60x14.mcd 12
FOOTING DESIGN:
Check the soil bearing capacity of the punch pads. .
2
Afooting'- 3.14.(�-a_footingl ft2 This is the area of the footing)
� J
gsoil = 1500 psf
dia_footing= 2 ft
depth_postnc = 4 ft (Minimum embedment depth)
Pfooting Afooting'gsoilAactor Pfooting= 7536 Ibf (End bearing capacity of footing)
Psnow = 7020 Ibf
Note that the end bearing capacity(Pfooting) is greater than the snow load(Psno„). This is OK.
6/27/2005 MW05340 (Binion) 52x60x14.mcd 13
SEISMIC CALCULATIONS
Design per IBC 2003
SS = 122.2 Mapped spectral acceleration for short periods(from above)
SI = 46.6 Mapped spectral acceleration for 1-second period(from above)
I:= 1.0 Importance factor
W= Dead load of building
RS = 7 Response modification factor(from above)
1. Determine the Seismic Design Category
a. Calculate SDS and SDI
For SDS: For SDI:
For SS = 1.22 For SI = 0.47
Fa= 1.01 Fv= 1.53
SMS := Ss•Fa
SMI := Sl'Fv
SMS = 1.24 SM1 = 0.71
SDS C3 'SMS SDI C3 'SM1
SDS=0.82 SDI = 0.48
Seismic_Design_Category= 'D"
2. Determine the building parameters
Building dead load weight,W: r 11
W:_ 1(Wbldg�•(Lbldg— 12)•(Pf.2)] + L[(Wbldg�•(Lbldg— 12)] + 2•[Wbldg+ (Lbldg— 12)]•HZdgJ Pd
W= 11688 Ibf
Building area,Ab:
Ab Lbldg'Wbldg Ab = 3120 ft2
6/27/2005 MWO5340 (Binion) 52x60x14.mcd 14
3. Determine the shear force to be applied
a. Determine the structural period, T
T,:= Hbldg'.02 T := Ta T = 0.28
b. Detemine the Seismic Response Coefficient, Cs:
Cs is calculated as:
SDS
Cs2 :=
Rs
FCs2 = 0.118
IE
But shall not be less than:
Csl := .044•SDS•IE Csl = 0.036
But need not exceed: ft
SDI
CS3 := CO = 0.243
Rs
T —
IE
Cs = 0.118
c. Detemine the Seismic Base Shear:
Vbase shear= Cs'W
— Vbase shear= 1375 Ibf
4. Determine the seismic load on the building:
Per IBC,for Seismic Design Category's A, B, and C,p=1.0. For Seismic Design Category D, E, or F, p
shall be calculated using r,,,a,,.
4a.Determine p for Seismic Design Category D,E or F(only if required).
Determine the shortest shear panel, Lw:
Lwg Wbldg— Wgableopenings
Ltive Lbldg— Weaveopenings
LW = igLwg<Lwe,Lwg,Lwe) Lw. = 22
10 20
rmax:_ — p := 2 — p = 1.21
LW rmax= 0.45 rmax'F
p = 1.21
E := p•V base shear
E = 1667 lbf This is the seismic load on the building
6/27/2005 MW05340 (Binion) 52x60x14.mcd 15
ANALYSIS FOR GABLE WALL:
1.Check Wind Loads:
Hroof = 8.67 ft Hbldg= 14 ft qe= 10.3 psf
Lbldg= 60 ft
0.375-mD#1bld0-Lbldg'ge
Veave wind:_
_ 2
Veave wind = 1596 Ibf
2.Check Seismic Loads:
Veave seismic := E Veave seismic = 834 Ibf
The controlling load= "Veave—wind" . Therefore, Vgable_shear= 1596 Ibf
This is the lateral load that is transmitted to each gable wall. This load will be transmitted through
the roof diaphragm to the gable walls. Normalize the load to a per foot basis.
V gable_shear
vgablewall Wbldg— Wgableopenings g v ablewall = 73 plf
The gable wall diaphragms can resist the shear loads as follows:
If vgablewall" 142 plf Then no additional sheathing is required.
6/27/2005 MW05340 (Binion) 52x60x14.mcd 16
ANALYSIS FOR EAVE WALL:
1. Check Wind Loads:
Hroof = 8.67 ft libldg= 14 ft qg = 10 psf
Wbldg= 52 ft Lbldg= 60 ft
Vgable_wind:_ 0.375•mD•(11bldj-Wbldg'gg + 0.5•(Hroof)'Wbldg'gg
2
Vgable_wind= 2474 Ibf
2. Check Seismic Loads:
Vgable_seismic:= E Vgable_seismic= 834 Ibf
2
The controlling load= "Vgable_wind" . Therefore, Veave shear= 2474 :)f
This is the lateral load that is transmitted to each eave wall. This load will be transmitted through
the roof diaphragm to the eave walls. Normalize the load to a per foot basis.
Veave shear
veavewal l
Lbldg— Weaveopenings Veavewall = 53 plf
The eave wall diaphragms can resist the shear loads as follows:
If veavewall< 142 plf Then no additional sheathing is required.
6/27/2005 MW05340 (Binion) 52x60x14.mcd 17
GIRT DESIGN:
The girts will simple span between posts. Calculate bending stress (fbgirt)due to wind loading
Owegirt)and determine the required girt size.
gwegirt Girtspacing lgwe — qil' gwegirt = 2.32 pli 1-girt-span= 138 in
12.12
1-girt_span�
Mgirt= gwegirt' 8 1v1girt = 5528 in-Ibf
fbgirt•— Mgirt fbgirt = 2684 psi (Stress applied to the girt due to wind loading)
Sgirt
Determine the allowable member stress.
LDFwind 1.6 Cfugirt = 1.15 Cfgirt = 1.00 Cr:= 1.15 Fgi t = 1650 psi
Fbgirt:= LDFwind'Cfugirt'Cfgirt'Cr'Fgirt Fbgirt = 3491 > fbgirt Psi This is OK.
PURLIN DESIGN:
Assume that the purlins simply span between pairs of trusses or rafters. Determine the required purlin
size.
Lpurlin_span = 135 in (Bending length of purlin)
'purlin= 4.62 pli (Distributed snow load along top edge of purlin)
Mpurlin=
�`purlin'Lpurlin_span2
8 Mpurlin= 10536 in-Ibf
f Mpurlin f = 1394 psi Stress applied to the purlin due to
bpurlin �= bpurlin — P ( PP
Spurlin snow and dead load)
Determine the allowable member stress.
LDFsnow 1.15 Cfpurlin= 1.00 Cr:= 1.15 Cfupurlin= 1.00 Fpurlin= 1650 psi
F,bpurlin LDFsnow'Cfpurlin'Cr'Cfupurlin'Fpurlin
Fbpurlin = 2182 psi � fbpurlin This is OK.
6/27/2005 MWO5340 (Binion) 52x60x14.mcd 18
CORBEL BLOCK DESIGN:
Determine the required number and size of bolts required in the truss block.
Assume full snow load and dead load on the roof.
Pbolt 58 1590 Ibf Pbolt 34 2190 Ibf Plod:_ 147 Ibf
Psnow= 7020 Ibf
If 5/8 dia. bolts are used:
Nbo11s58 = 3.8 Number of 5/8"dia. bolts required in the corbel block
If 314 dia. bolts are used:
Nbolts34 = 2-8 Number of 3/4"dia.bolts required in the corbel block
If 20d nails are to be used:
Nails20d = 20.8 number of 20d nails required in each corbel block.
• 6/27/2005 MW05340 (Binion) 52x60x14.mcd 19
RAFTER DESIGN:
Determine the required section for rafters. The rafters will simple span between posts. It will be
assumed that both ends are pinned.
I-rafter span= 138 in
Srafter Sx212
wratler = 333 plf
Mrafter = 66056 lb-inch
Determine fiber stress:
fbrafter Mrafter fbratler = 1044 psi Note: Use single rafter on each side
2.Srafter of each post.
Frafter FbDFldim Frafter = 1000 psi (Allowable bending stress)
F := LDF •F F = 1150 si> f
brafter snow rafter brafter p (brafter)
6/27/2005 MW05340 (Binion) 52x60x14.mcd 20
SUMMARY OF RESULTS:
Building Dimensions Building Design Loads
Wbldg= 52 ft (Width of Building) Wind speed= 85 MPH Ground—snow—load = 25 psf
Lbidg= 60 ft (Length of Building) Wind exposure = "B" Roof snow load = 26 psf
Hbldg= 14 ft (Eave Height of Building) Roof_dead_load= 3 psf
Seismic_Design_Category= "D"
Rpitch = 4 / 12 (Roof pitch)
Footing Details Non-Constrained:
Post Details
Postdepth= 4.0 ft(Design Post Depth)
Post—size= "6x6"
Post grade = "No. 2 Hem-Fir" a_footing
di, 2 ft(Design Footing Diameter)
Usage = 71 %(Combined stress usage of post) Footingusage= 93 % (Stress usage of footing)
Shear Wall Details:
Vgablewall = 73 plf(Max.shear in gable wall)
veavewall = 53 plf(Max.shear in eave wall)
Girt Details:
Girt usage = 77 % (Stress usage of wall girt)
Orientation= "Flat"
Purlin Details:
Purlin_usage = 64 % (Stress usage of roof purlin for snow loading)
Corbel Block Botts:
Nbolts58 = 3.8 Number of 5/8"dia.bolts required in the corbel block if used.
Nbolts34 = 2.8 Number of 3/4"dia. bolts required in the corbel block if used.
Nails20d = 20.8 Number of 20d 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.