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Specifications (30) JUN 1 ? 2019 CITY Or _.ILD1NG Divi :O :9 South Valley Engineering 4742 Liberty Rd. S #151 • Salem, OR. 97302 Ph. (503) 302-7020 • Fax (888) 535-6341 www.southvalleyengineering.com Project No. 11905067 Calculations for Margie Johnson 7015 SW Oak St. Tigard, OR. 97223 Date 6/7/2019 Engineer •oP / el , "fp ci OREGON . sem,, ✓G oti 4 FA <y92- 60 6$'64' R. 1.10 RENEWS: 6/30/19 POST FRAME BUILDING SUMMARY SHEET Owner:Margie Johnson Date:61712019 Building location:7015 SW Oak St_ Tigard,OR_97223 Project No.: 11905067 Building Description:Private shop Building Codes:2014 OSSC,ASCE 7-10 Building dimensions: Environmental information: Width: 16 ft_ Wind speed: 120 MPH Length: 32 ft_ Wind exposure: B Height: 13 ft_ Seismic design category D Eave overhang 0 ft_ S,: 0.98 Gable overhang 0 ft_ S1: 0.42 Roof pitch: 3 /12 Ground snow load: 25 psf_ Greatest bay spacing: 10 ft_ Design Snow Load, 25 psf. Greatest post tributary width: 11 ft_ Roof dead load 3 psf.(incl.ceiling load if any) Concrete Slab: No Soil bearing capacity 1,500 psf. Risk Category: II Per Table 1.5-1 ASCE 7-10 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: no slab,concrete Post Constraintibackfill. no slab,concrete backfill backfill *Rough Sawn *Rough Sawn **To bottom of footing **To bottom of footing Purlin&girt information: Purlins Girts Size: 2x6 Size&orientation 2x6 Commercial Grade: #2 D-F Grade: #2 D-F Spacing: 24 in.o.c. Spacing: 24 in_o.c. Sheathing information: Roof:29 ga_Metal only Walls: All walls are 29 ga_metal only Page 1of13 Snow Load Calculations Snow load calculations per ASCE 7-10 Chapter 7 Pg: 25 psf-Ground Snow Loac Ce: 1.0 Exposure Factor from ASCE Table 7-2 Ci: 1.2 Thermal Factor from ASCE Table 7-3 Is: 1.0 Importance Factor from ASCE Table 1.5-2 Flat Roof Snow Load,p=0.7 x pg x Ce x Ct x I, pi: 21 psf-Flat Roof Snow Load C8: 0.95 Figure 7-2 based on Ct,roof slope and surface pg: 20.0 psf-Sloped roof snow load Pdesign: 25 psf-Design Snow Load Pape 2 of 13 Wind Pressure Calculations Wind calculations per ASCE 7-10 Chapter 28 Part 1:Enclosed and Partially Enclosed Low Rise Buildings Roof Pitch: 3 /12 Design Wind Speed,V: 120 MPH Eave Height: 13 ft. Wind Exposure: B Risk Category It Velocity pressures q &q,,per equation 28.3-1: qZ=0.00256xIcxKKxKdxV2 at eave height z q,,=0.00256x1t„xKZtxKdxV2 at mean roof height h Angle: 14.04° KZ: 0.62 Velocity pressure coefficient at cave ht,z from Table 27.3-1 Kh: 0/0 Velocity pressure coefficient at roof ht_h from Table 27.3-1 Kn: 1.0 Topographic effrect-assume no ridges or escarpments Kd: 0.85 Wind Directionality Factor,Table 26.6-1 Velocity Pressures:c}= 19.43 psf q,,= 21.93 psf Determine Velocity Pressure Coefficients&Wind Pressures per ASCE 7-10 Figure 28.4-1 for MWFRS MWFRS 1. Windward Eave Wall Pressure 2_ Leeward Eave Wall: GCp,r 0.48 GC : -0.37 q,,,,,: 9.29 psf qn,,,: -7.27 psf 3. Windward Eave Roof Pressure 4. Leeward Eave Roof: GCpr,,,,r: -0.69 GCpnr: -0.44 qtr: -13.40 psf qh: -8.48 psf 5. Windward Gable Wall: 6. Leeward Gable Wall: GCS 0.40 Cpl -0.29 qt.: 7.77 psf qt.: -5.63 psf Components&Cladding GCp;: 0.18 Internal pressure per Figure 26.11-1 7. Roof elements GCpr: -0.84 qef: 22.30 psf Roof elements per Figure 30.4-2B 8. Wall elements: GCpw: -1.01 ger: 26.00 psf (Wall elements per Figure 30.4-1 Page 3 of 13 Seismic Design Parameters Calculate seismic building loads from ASCE 7-10 Section 1214.8 Seismic Parameters Sg 0.98 S1= 0.42 Fa 1.11 FY 1.58 per Tables 11.4-1 &11.4-2 SMS= 1.08 SMI= 0.67 Calculated per Section 11.4.3 SDs= 0.72 SDI= 0.45 Calculated per Section 11.4A Seismic Design Category= D From Section 11.6 F= 1.0 for 1 story building Response Mod.Factor R: Roof: 2.5 From Table 12.14-1,Section B-24 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= 16 ft. Building length= 32 ft. Building height= 13 ft. Roof area= 512 sf Gable wall area= 224 sf Eave wall area= 416 sf Roof+ceiling DL= 3 psf Snow LL(if appliable)= 0 psf Roof W= 1,536 lbs Loft(yin): n Loft dead load: N/A psf Full or partial loft: N/A Wall Areas Building dead loads Loft dead loads Left gable wall: 224 SF Left gable wall: 3 psf Left gable wall: 0 lbs Right gable wall: 224 SF Right gable wall: 3 psf Right gable wall: 0 lbs Front eave wall: 416 SF Front eave wall: 3 psf Front eave wall 0 lbs Rear eave wall: 416 SF Rear eave wall 3 psf Rear eave wall: 0 lbs Calculate Seismic Base Shear,V per Section 12.14.8 V=[(FxSDs)/R]xW (Eqn.12.14-11) Total dead loads.W find roof.loft) Roof: 768 lbs Vr,,01= 222 lbs base shear for roof diaphragm Left gable wall: 1,104 lbs VLGW 319 lbs base shear for wall diaphragm Right gable wall: 1,104 lbs VRcw= 319 lbs base shear for wall diaphragm Front eave wall: 1,392 lbs VFE 402 lbs base shear for wall diaphragm Rear eave wall: 1,392 lbs VREw= 402 lbs base shear for wall diaphragm Page 4 of 13 Diaphragm 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)+(Ic/(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 l is known. Solving for K2 yields: K2=[((Ext)x(bxff))/c]-[2x(1+u)x(bxt)2x(g/p) 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= 3 /12 Building width= 16 ft 0= 14.04 °roof angle b'= 98.95 in=length of steel roof panel at the given angle for 1/2 of the roof c= 1493 Ibsfin-stiffness of actual roof diaphragm 4.Calculate the equivalent horizontal roof stiffness,,pfor the entire roof ch=2xcx(cos20)x(b7a) ch= 2,107 lb/in a= 132 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=(6xdxEpx(p)/L3 d= 1 in-deflection used to establish k 1p= 108 in4-Momentof inertia of post EP= 1.10E+06 psi-Modulus of elasticity of post L= 144 in-Bending length of post k= 239 lbs/in 6.Determine the side sway force,mD from tables based on k/cverses number of frames_ NF= 4 frames in building(including end walls) k/ch= 0.1133 mD= 0.91 =calculated stiffness of metal roof diaphragm Since roof sheathing is metal, mD used for calculations is 0.91 Page 5 of 13 Post Wind Load Calculation Determine the bending stress on the post from the wind load Windward wall wind pressure= 9.29 psf Leeward wall wind pressure= -7.27 psf Total wind pressure= 16.56 psf Total wall pressure to use= 16.56 psf(10 psf min_per code) L= 144 in Bending length of the post w= 15.18 phi Distributed wind load on the post Moo= 19,679 lbf-in Moment as a propped cantilever(w xL)1(2 x 8) fes= 547 psi Stress on the post from the distirbuted wall wind,= /S. R= 820 lbf Total side sway force=3 x w x(L/8) mD= 0.91 Stiffness coefficient from diaphragm stiffness calculation, or 1.0 if wood sheathing in roof 0= 746 lbf Side sway force resisted by the roof diaphragm=mD x R WR= 13.8 pli The total distributed wind load resisted by the roof diaphragm=8 x((Q/(3 x L)) wit= 1.37 phi The total distributed wind load NOT resisted by the roof diaphragm for which the post must resist.Wpost=w-WR Mat= 14,169 lbf-in The moment in the post as a simple cantilever =Noe x((L2)/2)(This value is 0 if roof is a wood diaphragm) foots= 197 psi The fiber stress in the post from simple cantilever stress =Mos„t/(2 x Sx)(This value is 0 if roof is a wood diaphragm) Mme= 32,077 lbf-in The total moment in the post=(mD x l4)+Kant fit= 694 psi The total bending stress on the post=(mD x, )+foont Page 6 of 13 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 C©for post 1.0 Size factor for posts<12"in depth Final Design Values Cp= 0/8 Column stability factor per Section 3.7 Fb design: 920 psi final allowable bending stress F=design: 518 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 7-10 Section 2A. Load Case 1 -Dead Load+Snow Fb Win- 920 psi Final allowable bending stress Fc desgn: 518 psi Final allowable compression stress Pdesd= 264 lbs Dead load PSflOff 2200 lbs Snow load A= 36 sq-in Cross-sectional area of post FcE= 892 psi fb= 0 psi=0 fC 68 psi=(P.+PdeadYA CBAC1= 0.02 =((fc/F design)2)+((fb/(Fb desig�1-(fc1FcE)))))) Load Case 2-Dead Load+0.6Wind Fb_d gn: 920 psi Final allowable bending stress Fcgn: 518 psi Final allowable compression stress P de.d= 264 lbs Dead load P snow= 2200 lbs Snow load A= 36 sq-in Cross-sectional area of post FcE= 892 psi fb= 417 psi=0.6 x t_post fc= 7 psi=Pdead/A CBAC2= 0.46 =((fc/Fc_degim)2)+((fbi(F:,desig41-(fc/F )))))) Load Case 3-Dead Load+0.75(0.6Wind)+0.75Snow Fb design: 920 psi Final allowable bending stress Fcdesign: 518 psi Final allowable compression stress Plead 264 lbs Dead load P al 2200 lbs Snow load A= 36 sq-in Cross-sectional area of post FcE= 892 psi fb= 312 psi=.75 x(0.6 x(,post) f.= 53 psi=((.75 x P ,)+Pd )/A CBAC3= 0.37 =((fc/Fc design)2)+((fb/(Fb_design(1-(fJFth)))))) Max.CBAC= 46% >> Maximum post usage<100%OK Page 7 0113 Roof and Gable Wall Shear Loads and Diaphragm Desitin Roof Roof width= 16 ft. Fly= 2.00 ft_ Total roof wind pres.,0.6 x P= -2.96 psf(0.6 x Pr) Total roof wind pressure to use= 4.80 psf-use 0 if Pr<0 Total wall wind pressure= 9.94 psf(0.6 x(qM,M-qir)) Total wall wind pressure to use= 9.94 psf-use 0.6 x 16=9.6 psf minimum Diaphragm seismic load= 155 lbs-VRo x 0.7 Diaphragm wind load= 859 lbs Diaphragm load to used 859 lbs-Wind load controls Roof shear= 54 plf Sheathing=29 ga.Metal only Allowable shear= 113 plf>Roof shear-OK Sheathing fastening=#10 screws at 9"o_c_ Gable walls Left Gable Wall Left gable wall shear C18,,,;c= 223 lbs-V10w x 0.7 Left gable wall shear Wend= 859 lbs-from Diaphragm wind load above Diaphragm load to use= 859 plf-Wind controls-see below* Left Gable wall= 0 plf-see below* Allowable shear= 113 plf>Wall shear-OK Sheathing fastening= #10 screws at 9"o.c_ *Use post-bending calculation on following pages-use concrete backfill Right Gable Wall Right gable wall shear 223 lbs-VRcw x 0.7 Right gable wall shear'L d= 859 lbs-from diaphragm wind load above Diaphragm load to use= 859 plf-Wind controls Right Gable wall= 54 plf Allowable shear= 113 plf>Wall shear-OK Sheathing fastening= #10 screws at 9"o_c_ Backfill posthole with concrete as follows: Posthole diameter= 2 ft Posthole depth= 4.0 ft Page 9of13 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. Gable wall with no shear panels(large openings) Intermediate posts Corner posts No.intermediate posts= 2 No_corner posts= 2 Intermediate post size= 6x6 Corner post size= 6x6 Intermediate post grade= #2 H-F Comer post grade= #2 H-F Post type= Rough-sawn Post type=Rough-sawn knc�= 108 i " nlearner_pogr= 108 inA Sint_past= 36 in3 Scomer_post= 36 in3 Determine equivalent stiffness of posts based on post properties %load distributed to each intermediate post= 25% %load distributed to each corner post= 25% Bending height,h= 144 in Total bending moment in frame from wind= 123,705 in-lbs Total bending moment in frame from seismic= 53,563 in-lbs Moment to use: 123,705 in-lbs-Wind controls fb nice= 859 psi-bending stress in each intermediate post Fb_int t 920 psi-allowable bending stress-OK fb comer_pos= 859 psi-bending stress in each corner post Fb swner_post= 920 psi-allowable bending stress-OK Page 10 0113 Eave Wall Shear Loads and Diaphragm Design Eave walls Building Length= 32 ft. Gable wall wind pressure= 9.60 psf-use 0.6 x 16=9.6 psf minimum Diaphragm wind load= 367 lbs Front Eave Wall Front eave wall shear\Lismic= 281 lbs-VFE,r,x 0.7 Front eave wall shear Vvind= 367 lbs-from diaphragm wind load above Diaphragm load to use= 367 plf-Wind controls Front eave wall= 11 plf Allowable shear= 113 plf>Wall shear-OK Sheathing fastening= #10 screws at 9"o_c. Backfill posthole with concrete as follows: Posthole diameter= 2 ft Posthole depth= 4.0 ft Rear Eave Wall Rear eave wall shear\e;9,,,;,= 281 lbs-Vpw x 0.7 Rear eave wall shear „;,,d= 367 lbs-from diaphragm wind load above Diaphragm load to use= 367 plf-Wind controls Rear eave wall= 13 plf Allowable shear= 113 plf>Wall shear-OK Sheathing fastening= #10 screws at 9"o_c_ Backfill posthole with concrete as follows: Posthole diameter= 2 ft Posthole depth= 4.0 ft Page 11 of 13 Purlin&Girt Calculations Purlin Calculation Roof Pitch: 3 112 Roof Angle: 14.0 Greatest purlin span: 114 in Purlin SX: 7.56 in3 Live+dead load: 28 psf Max_o.c.spacing: 24 in.o_c. M: 7,355 in-lbf fb: 973 psi Fb allowable: 1,547 psi per NDS Section 4 and Design Values for Wood Construction Purlin usage: 63% OK End reactions: Snow load: 266 lbs If joist hanging,use LU26 joist hanger wl 10d nails or JB26 top-flange joist hanger WI 10d nails uplift: 294 lbs (2)16d nails each side of purlin block or joist hanger adequate Girt Calculation Greatest Bay Spacing: 10 ft. O.C.Spacing: 24 in Girt SX: 7.56 in3 Total wind pressure: 15.60 psf w: 2.60 pli Girt Span: 114 in M: 4,224 lbf-in fb: 559 psi Fb allowable: 2,153 psi-per NDS Section 4 and Design Values for Wood Construction Girt usage: 26% OK Page 12 of 13 Bearing Block Nails Between Commercial Girls 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 HE Total load from one truss= 1,120 lbs Allowable shear loads for nails includin+ 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 8 I 7 Since commercial girts are used,the nails in the girt blocking will provide adequate nailing Page 13 ot 13