Specifications .
RECEIVED
JUN 3u
South Valley Engineering CITY OF BUILDING DIVSION
4742 Liberty Rd. 5 #151 • Salem, OR. 97302
Ph. (503) 302-7020 • Fax (888) 535-6341
www.southvalleyengineering.com
OFFICE COPY
Project No.
12006036
MST202o- 00241
Calculations for
John Grant
11270 SW Fairhaven Ct.
Tigard, OR. 97223
Date
6/26/2020
Engineer
WV; ;4%
OREGON
fip <r92° Fps
44RH06
RENEWS: 6/30/21
POST FRAME BUILDING SUMMARY SHEET
Owner: John Grant Date: 6/26/2020
Building location: 11270 SW Fairhaven Ct.
Tigard,OR.97223 Project No.: 12006036
Building Description: Private shop Building Codes: 2019 OSSC,ASCE 7-16
Building dimensions: Environmental information:
Width: 22 ft. Wind speed: 100 MPH
Length: 24 ft. Wind exposure: B
Height: 12 ft. Seismic design category: D
Eave overhang: 1 ft. Ss: 0.98
Gable overhang: 1 ft. Si: 0.42
Roof pitch: 4 /12 Ground snow load: 25 psf.
Bay spacing:, 12 ft. Design Snow Load: 25 psf.
Post tributary width: 12 ft. Roof dead load: 7 psf.(inc.'.ceiling load if any)
Concrete Slab: Yes Soil bearing capacity: 1,500 psf.
Risk Category: I Per Table 1.5-1
ASCE 7-16
Post&posthole information:
Eave wall posts: Gable wall posts:
Size: 6x6 Size: 6x6
Grade: #2 H-F Grade: #2 H-F
Type: RS* Type: RS'
Posthole diameter: 24 in Posthole diameter: 24 in
Posthole depth": 4.00 ft. Posthole depth": 4.00 ft.
Post Constraint/backfill: Varies-see Post Constraint/ba�ll: Slab w/granular
calculations backfill
*Rough Sawn *Rough Sawn
**To bottom of footing "To bottom of footing
Purlin&girt information:
Purl ins Girts
Size: 2x6 Size: 2x6
Grade: MSR 1650 Grade: #2 D-F
Spacing: 24 in.o.c. Spacing: 24 in.o.c.
Orientation: Commercial
Sheathing information:
Roof: Composition Roof
Walls: All walls are 29 ga.metal only
Page 1 of 14
Snow Load Calculations
Snow load calculations per ASCE 7-16 Chapter 7
P4: 25 psf-Ground Snow Load
Ce: 1.0 Exposure Factor from ASCE Table 7-2
G: 1.2 Thermal Factor from ASCE Table 7-3
Is: 1.0 Importance Factor from ASCE Table 1.5-2
Flat Roof Snow Load,pf=0.7 x p9„Ce x Ct x l
pf: 21.0 psf-Flat Roof Snow Load
Cs: 1.00 Figure 7.4-1 based on Ct, roof slope and surface
ps: 21.0 psf-Sloped roof snow load
Pdestan: 25 psf-Design Snow Load
Page 2 of 14
Wind Pressure Calculations
Wind calculations per ASCE 7-16 Chapters 28 and 30
Roof Pitch: 4 112 Design Wind Speed,V: 100 MPH
Eave Height: 12 ft. Wind Exposure: B Risk Category
Velocity pressures qh per equation 26.10.2
qh=0.00256xKhxKnxKdxKexV2 at mean roof height h
Angle: 18.43 °
Kh: 0.70 Velocity pressure coefficient at roof ht. h from Table 26.10-1
Kn: 1.00 Topographic effect-assume no ridges or escarpments
Kd: 0.85 Wind Directionality Factor,Table 26.6-1
Ke: 1.00 Ground Elevation Factor,Table 26.9-1
Velocity Pressures: qh= 15.23 psf
Determine Velocity Pressure Coefficients&Wind Pressures per ASCE 7-16 Figure 28.3-1 for MWFRS
MWFRS
1. Windward Eave Wall Pressure 2. Leeward Eave Wall:
GCppM: 0.51 GCpfw: -0.42
qww: 7.76 psf quw: -6.33 psf
3. Windward Eave Roof Pressure 4. Leeward Eave Roof:
GCpfwr: -0.69 GCpfl: -0.47
qwr: -10.51 psf q1r -7.14 psf
5. Windward Gable Wall: 6. Leeward Gable Wall:
GCvr v: 0.40 Cpf„y: -0.29
qh„: 6.09 psf q1w: -4.42 psf
Components&Cladding
GCdt: 0.18 Internal pressure per Figure 26.13-1
7. Roof elements
GCpr: -0.82
qer: 15.26 psf Roof elements per Figure 30.3-2A thru I
8. Wall elements:
GCpw: -0.96
qer: 17.31 psf IWall elements per Figure 30.3-1
Page 3 of 14
Seismic Desiqn Parameters
Calculate seismic building loads from ASCE 7-16 Chapter's 11 & 12
Seismic Parameters
Ss= 0.98 S1= 0.42
F' 1.11 Fv= 1.88 per Tables 11.4-1 &11.4-2
SMs= 1.08 SM1= 0.80 Calculated per Section 11.4.3
Sos= 0.72 Sol= 0.53 Calculated per Section 11.4.4
Seismic Design
Category= D From Section 11.6 Importance factor: 1.00
F= 1.0 for 1 story building
Response Mod. Factor R:
Roof: 7 From Table 12.14-1,Section B-22
Left gable wall: 2.5 From Table 12.14-1,Section B-24
Right gable wall 2.5 From Table 12.14-1,Section B-24
Front eave wall 2.5 From Table 12.14-1,Section B-24
Rear eave wall 2.5 From Table 12.14-1,Section B-24
Calculate building weights,W,for seismic forces
Building width= 22 ft. Building length= 24 ft. Building height= 12 ft.
Roof area= 624 sf Gable wall area= 304 sf Eave wall area= 288 sf
Roof+ceiling DL= 7 psf Snow LL(if appliable)= 0 psf Roof W= 4,368 lbs
Loft(y/n): n Loft dead load: N/A psf Full or partial loft: N/A
Wall Areas Building dead loads Loft dead loads
Left gable wall: 304 SF Left gable wall: 3 psf Left gable wall: 0 lbs
Right gable wall: 304 SF Right gable wall: 3 psf Right gable wall: 0 lbs
Front eave wall: 288 SF Front eave wall: 3 psf Front eave wall: 0 lbs
Rear eave wall: 288 SF Rear eave wall: 3 psf Rear eave wall: 0 lbs
Calculate Seismic Base Shear,V per Section 12.14.8
V=[(FxSps)/R]xW (Eqn.12.14-12)
Total dead loads.W(incl roof, loft)
Roof: 2,184 lbs Vrcisf= 225 lbs base shear for roof diaphragm
Left gable wall: 2,641 lbs VLcyr 763 lbs base shear for wall diaphragm
Right gable wall: 2,641 lbs VR6w= 763 lbs base shear for wall diaphragm
Front eave wall: 2,616 lbs ViEw= 755 lbs base shear for wall diaphragm
Rear eave wall: 2,616 lbs VREw= 755 lbs base shear for wall diaphragm
Page 4 of 14
Diaphracm Stiffness Calculation
The diaphragm stiffness will be calculated based on the methodology from"Post Frame Building Design",
by John N.Walker and Frank E.Woeste. This method is widely accepted in the post frame industry for
determining metal diaphragm stiffness.
1. The diaphragm stiffness,c'=(Ext)/[2x(1+u)x(g/p)+ (l9/(bxt)2)
Where: c'= 3130 lbs/in=Diaphragm stiffness of the test panel(1992 Fabral Test for Grandrib Ill)
E= 2.75E+07 psi=Modulus of elasticity for metal sheathing
t= 0.017 in=Steel thickness for 29 ga metal sheathing
u= 0.3 =Poisson's ratio for steel
g/p= 1.085 =Ratio of steel corrugation pitch to steel sheet width
b= 144 in.=Length of test panel
K2= - =Sheet edge purlin fastening constant(unknown)
2.The diaphragm for the same metal for a different length b can be calculated with the above
above equation once the constant K2 is known. Solving for K2 yields:
K2=[((E)(t)x(bxt2))/c]-[2x(1+u)x(bxtlx(gfp) K2= 878 in4
3.The stiffness of the acutal panel will be calculated from equation in 1.above,based on its actual length,b'
Roof pitch= 4 /12 Building width= 22 ft 8= 18.43 °roof angle
b'= 139.14 in=length of steel roof panel at the given angle for 1/2 of the roof
c= 2926 lbs/in-stiffness of actual roof diaphragm
4.Calculate the equivalent horizontal roof stiffness,g for the entire roof
cn 2xcx(cos28)x(b'/a) cn 5,089 lb/in a= 144 In.post spacing
5. Calculate the stiffness, k,of the post frame,which is the load required for the top of the frame a distance,d
For d=1",k=P=(6xdxEoxl )/L3
d= 1 in-deflection used to establish k Ip 108 in4-Momentof Inertia of post
Ep 1.10E+06 psi-Modulus of elasticity of post L= 132 in-Bending length of post
k= 310 lbs/in
6.Determine the side sway force,mD from tables based on k/g verses number of frames.
NF= 3 frames in building(including end walls) k/q,= 0.0609
mD= 0.97 =calculated stiffness of metal roof diaphragm
Since roof sheathing is wood, mD used for calculations is 1.00
Page 5 of 14
Post Wind Load Calculation
Determine the bending stress on the post from the wind load
Windward wall wind pressure= 7.76 psf
Leeward wall wind pressure= -6.33 psf
Total wind pressure= 14.09 psf
Total wall pressure to use= 14.09 psf(10 psf min.per code)
L= 132 in Bending length of the post
w= 14.09 p1i Distributed wind load on the post
Mpc= 15,344 lbf-in Moment as a propped cantilever(w x 12)/(2 x 8)
fb.1),= 426 psi Stress on the post from the distirbuted wall wind,=M /S,
R= 697 lbf Total side sway force=3 x w x(L/8)
mD= 1.00 Stiffness coefficient from diaphragm stiffness calculation,
or 1.0 if wood sheathing in roof
Q= 697 Ibf Side sway force resisted by the roof diaphragm=mD x R
WR= 14.1 pli The total distributed wind load resisted by the roof diaphragm=8 x((Q/(3 x L))
wpost= 0.00 pli The total distributed wind load NOT resisted by the roof diaphragm
for which the post must resist.Wosst=w-wR
Mgt= 0 Ibf-in The moment in the post as a simple cantilever
=wpost x((L2)I2)(This value is 0 if roof is a wood diaphragm)
ffant= 0 psi The fiber stress in the post from simple cantilever stress
=Mcan/(2 x Ss)(This value is 0 if roof is a wood diaphragm)
Mosst= 15,344 lbf-in The total moment in the post=(mD x +M�t
fist= 426 psi The total bending stress on the post=(mD x )+f ant
Page 6 of 14
Post Design
Determine the allowable bending and compression stresses for the eave wall posts per 2012 NDS
Nominal Design Values(allowable) Adjustment factors per Table 4.3.1
Fb: 575 psi-bending Co for snow 1.15 LDF for snow
Fc: 575 psi-compression CD for wind/seismic 1.6 LDF for wind/seismic
Co for post 1.0 Size factor for posts< 12"in depth
Final Design Values Cp= 0.83 Column stability factor per Section 3.7
Fb_design: 920 psi final allowable bending stress
Fc_design: 546 psi final allowable compression stress
Combined Bending And Compressive(CBAC)Post Loads by Load Case
Determine the maximum Combined Bending And Compressive stresses in the eave wall post per NDS 3.9.2
using applicable load cases from ASCE 7-16 Section 2.4.
Load Case 1 -Dead Load+Snow
Fb_desion: 920 psi Final allowable bending stress
Fc de5�: 546 psi Final allowable compression stress
Pdead= 1008 lbs Dead load
Ps no,.,= 3600 lbs Snow load
A= 36 sq-in Cross-sectional area of post FcE= 1,061 psi
fb= 0 ps1=0 fc= 128 psi=(Psi,+Pdead)/A
CBAC1= 0.06 =((fc/Fe_design)2)+((fb/(Fb_design(1-(fc/FcE))))))
Load Case 2-Dead Load+0.6Wind
Fb_design: 920 psi Final allowable bending stress
Fc_ibs;g,: 546 psi Final allowable compression stress
Pdead= 1008 lbs Dead load
Psno,,,= 3600 lbs Snow load
A= 36 sq-in Cross-sectional area of post FcE= 1,061 psi
fb= 256 Psi=0.6 x fbpost fc= 28 psi=Pdead/A
CBAC2= 0.29 =((fc/Fc_design)2)+((fb/(Fb_design(1-(fc/FcE))))))
Load Case 3-Dead Load+0.75(0.6W ind)+0.75Snow
Fb_design: 920 psi Final allowable bending stress
Fc_des : 546 psi Final allowable compression stress
Pdead= 1008 lbs Dead load
PST,o„,= 3600 lbs Snow load
A= 36 sq-in Cross-sectional area of post FcE= 1,061 psi
fb= 192 psi=.75 x(0.6 x fit) fc= 103 psi=((.75 x Ps )+Pdead)/A
CBAC3= 0.27 =((fc/Fcdesv)2)+((fb/(Fb_design(1-(fc/FcE))))))
Max. CBAC= 29% » Maximum post usage<100%OK
Page 7 oT 14
Post Embedment Calculation
Determine the minimum posthole diameter and embedment depth for the eave wall posts
per ASAE EP486.1
Since there is a slab,the post will be considered constrained at the top.
The backfill will be compacted gravel or sand full depth unless otherwise required for shear wall uplift.
Design Criteria:
Sy 1500 psf-vertical soil bearing capacity
S= 150 psf-lateral soil bearing capacity
M1= 767 ft-lbs-Moment at tap of one posthole
Va 349 lbs-Lateral load on post at top of posthole
Posthole dia.= 2 ft.
b= 0.71 ft-maximum width of post in soil
Ang 3.14 ft-area of footing
d= - ft-depth of footing to be determined below
Per Sections 4.2.2.1 and 4.2.2.2,allowable lateral soil bearing capacities may be increased
by 2 for isolated posts(spaced at least 3 ft.apart),and by 1.33 for wind loading
SLAT 318 psf-factored lateral soil bearing capacity
Minimum embedment depth required for lateral load,constrained at the top,
gravel backfill,per Section 6.5
dmm= [(4 x Mpost)!(SLAT x b))^113 dmin_r= 2.39 ft.-minimum depth requried for
lateral load
Allowable vertical soil bearing pressure for gravity loads
Sy= Sy x Aftg x(1+(0.2x(d-1))
Sy= 1500 psf-vertical soil bearing capacity
Aftg 3.14 ft-area of footing
d= minimum depth for vertical bearing requirements
Maximum vertical load on footing from gravity load Prnor 4,608 lbs-vertical load on footing
Posthole depth for this building= 4.00 ft-minimum depth to bottom of footing
Vertical capacity for footing Polio 7,540 lbs->Pfooting-OK
Page 8 of 14
Roof and Gable Wall Shear Loads and Diaphragm Design
Roof
Roof width= 22 ft.
Hrecr= 3.67 ft.
Total roof wind pres.,0.6 x P -2.02 psf(0.6 x Pr)
Total roof wind pressure to use= 4.80 psf-use 0 if Pr<0
Total wall wind pressure= 8.45 psf(0.6 x(qa -qir))
Total wall wind pressure to use= 9.60 psf-use 0.6 x 16=9.6 psf minimum
Diaphragm seismic load= 158 lbs-VRoor x 0.7
Diaphragm wind load= 730 lbs
Diaphragm load to use= 730 lbs-Wind load controls
Roof shear 33 plf
Sheathing= Composition Roof
Allowable shear- 230 plf> Roof shear-OK
Sheathing fastening= 8d nails 6 in.o.c.edges
12 in.o.c.field
Gable walls
Left Gable Wall
Left gable wall shear V„,gr„;t= 534 lbs-VLGw x 0.7
Left gable wall shear V.,,,d= 730 lbs-from Diaphragm wind load above
Diaphragm load to use= 730 lbs-Wind controls
Left Gable wall= 33 plf
Allowable shear- 113 plf>Wall shear-OK
Sheathing fastening= #9 screws at 9"o.c.
Right Gable Wall
Right gable wall shear V=;,,,,ic= 534 lbs-VRGw x 0.7
Right gable wall shear V(,i,d= 730 lbs-from diaphragm wind load above
Diaphragm load to use= 730 lbs-Wind controls
Right Gable wall= 38 plf
Allowable shear 113 plf>Wall shear-OK
Sheathing fastening= #9 screws at 9"o.c.
Page 9 or 14
Eave Wall Shear Loads and Diaphragm Design
Eave walls
Building Length= 24 ft.
Gable wall wind pressure= 9.60 psf-use 0.6 x 16=9.6 psf minimum
Diaphragm wind load= 548 lbs
Front Eave Wall
Front eave wall shear Vse151111C= 529 Ibs-VFEw x 0.7
Front eave wall shear V,,,,a= 548 lbs-from diaphragm wind load above
Diaphragm load to use= 548 lbs-Wind controls-see below*
Front eave wall= 0 plf-see below
Allowable shear= 113 plf>Wall shear-OK
Sheathing fastening= #9 screws at 9"o.c.
'Use post-bending calculation on following pages-use concrete backfill
Rear Eave Wall
Rear eave wall shear151,,,c= 529 Ibs-VRE,h,x 0.7
Rear eave wall shear Vvand= 548 lbs-from diaphragm wind load above
Diaphragm load to user 548 lbs-Wind controls
Rear eave wall= 23
Allowable shear 113 plf>Wall shear-OK
Sheathing fastening= #9 screws at 9"o.c.
Page 10 of 14
Post bending calculation
This calculation determines the adequacy of the posts to resist the shear load of the walls when there are no
adequate shear panels in the wall. The posts are modeled as simple cantilevers,and the load is applied to the
top of the post frame system and distributed throughout all of the posts in the wall as appropriate.
Eave wall with no shear panels(large openings)
Intermediate posts Corner posts
No.intermediate posts= 1 No.corner posts= 2
Intermediate post size= 6x8 Oriented perp. Corner post size= 6x6
Intermediate post grade= #2 H-F to eave wall Corner post grader #2 H-F
Post type= Rough-sawn Post type=Rough-sawn
loty't= 144 in4 icomeryost= 108 in4
Sctyost= 48 in3 Scomery°st= 36 Ins
Determine equivalent stiffness of posts based on post properties
%load distributed to each intermediate post= 40%
load distributed to each corner post= 30%
Bending height,h= 120 in
Total bending moment in frame from wind= 65,736 in-lbs
Total bending moment in frame from seismic= 105,767 in-lbs
Moment to use: 105,767 in-lbs-Seismic controls
fb;Myost= 881 psi-bending stress in
each intermediate post
Fb_intiskost= 920 psi-allowable bending stress-OK
fb_comer_post= 881 psi-bending stress in
each corner post
Fb_cortteryest= 920 psi-allowable bending stress-OK
Page it oft4
Purlin&Girt Calculations
Purlin Calculation
Roof Pitch: 4 /12
Roof Angle: 18.4
Greatest purlin span: 138 in
Purlin Sx: 7.56 in3
Live+dead load: 32 psf
Max.o.c.spacing: 24 in.o.c.
M: 12,044 in-lbf
fa: 1.593 psi
Fb allowable: 2,182 psi-per NDS Section 4 and Design Values for Wood Construction
Purlin usage: 73% OK
End reactions:
Snow load: 368 lbs If joist hanging,use LU26 joist hanger w/10d nails
or JB26 top-flange joist hanger w/10d nails
uplift: 220 lbs (2)16d nails each side of purlin block or joist hanger adequate
Girt Calculation
Greatest Bay Spacing: 12 ft.
O.C.Spacing: 24 in
Girt Sx: 7.56 Ina
Total wind pressure: 10.39 psf
w: 1.73 pli
Girt Span: 138 in
M: 4,122 lbf-in
fb: 545 psi
Fb allowable: 2,153 psi-per NDS Section 4 and Design Values for Wood Construction
Girt usage: 25% OK
Page 12 of14
Bearing Block Nails Between Commercial Girts
Calculate required number of nails and the correct o.c.spacings and bearing block size for the intermediate truss
bearing posts. Posts are assumed to be#2 HF; bearing blocks assumed to be#2 HF.
Total load
from one truss= 2,304 lbs
Allowable shear loads for nails(including increases)
16d box nail= 140 lbs
20d box nail= 169 lbs
For minimum bearing block length design using 2 rows of nails
-Use 2 vertical rows of nails,staggered
-Use 2"vertical spacing between nails
-Use 2"minimum end distances for top and bottom of block
16d nails 20d nails
Total nails required 17 I 14 f
Since commercial girts are used,the nails in the girt blocking will provide adequate nailing
Page 13 of 14
N.
Bearing Block Bolts In Double Shear
Calculate required number of bolts,and the correct bolts spacings and bearing block size for the intermediate truss
bearing posts. Posts are assumed to be#2 HF; bearing blocks assumed to be#2 HF.
Total load
from both trusses= 4,608 lbs
Bolt size= 3/4 "0(NOTE: Use 5/8"0,3/4"0 7/8"0 or 1"0 only)
Main member,f,,,= 6 in-post width
Post depth= 6 in-post depth
Side member(s),fs= 3 in-total for 2 side members
No.of fastener columns= 2
No.of bolts required,rb= 1.83 >> nb= 2 Bolt(s)in block
Truss bearing block= 2x6 Verify with truss engineering
Minimum block length, Lb= 11.25 in(no less than 12")
Minimum block width,Wa 5.5 in(<[(2 x de)+dos],
<truss bearing block)
Dimension Summary
det= 5 114 in(min)
deb= 3 in(min)
dr s= 3 in(min)
dchs= 1 118 in(min)
de= 1 1/8 in(min) I'
0 - -J----
Number of bolts,nb= 2 (min) d s
Block length,Lb= 11 1/4 in(min) Lb drys
Block width,Wb= 51/2 in(min) 0 d..5
deb
de Idyl de
—Wb NOTE: Number of bolts shown for
example only-use nb for actual design.
Page 14 0114