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2015-042 }r \\ TOWN .OF QUEENSBURY 742 Bay Road,Queensbury,NY 12804-5902 (518)761-8201 Community Development-Building & Codes (518) 761-8256 CERTIFICATE OF- OCCUPANCY . Permit Number: P20150042 Date Issued: Tuesday, February 02,.2016 This is to certify that work requested to be done as shown-by Permit Number P20150042 has-been completed. Location: 34 BIG BOOM Rd Tax Map Number: 523400-309=014-0001-002-000-0000 Owner: NRG REAL ESTATE, LLC Applicant: NRG REAL ESTATE, LLC This structure maybe occupied as a: Commercial Industrial Unheated By Order of Town Board TOWN OF QUEENSBURY Issuance of this Certificate of Occupancy DOES NOT relieve the property4 owner of the responsibility for compliance with.Site Plan,Variance, or other:issues and conditions as a result of approvals by the Planning Board' 'Director of Building&Code Enforcement or Zoning Board of Appeals. TOWN OF QUEENSBURY 742 Bay.Road,Queensbury,NY 12804-5902 (518)761-8201 Community Development.-Building&Codes (518.) 761-8256 . BUILDING PERMIT Permit Number: P20150042 Application Number: A,20150042 Tax,Map No: 523400-300-014-0001-002-000-0000 Permission is hereby granted to: NRG REAL ESTATE. LLC For property located at: . 34 BIG BOOM Rd. in the Town of Queensbury,to-construct or place at the above location in accordance with application together with plot plans and other information hereto filed and approved and in with the NYS Uniform Building Codes and the Queensbury Zoning Ordinance.. Type of Construction Value Owner Address: NRG REAL ESTATE; LLC Commercial/Industrial Unheatec $45,000.00 PO BOX 98 Total value LATHAM,NY 1-21.10-0000 $as,000.00 Contractor or Builder's Name/Address Electrical Inspection Agency . Plans&Specifications 2015-042 AFSCO Fence cold storage building- 5000 sq. ft. $500:00 PERMIT FEE PAID=THIS PERMIT EXPIRES: Sunday,March 13,2016 (If a longer period is required,an application for an extension must.be made to the code.Enforcement Officer of the Town of Queensbury before the expiration date) Dated at the To n of ens ;� arch 13,2015 SIGNED BY �Q for the Town of Queensbuty. Director.of Building&Code Enforcement 4 i PRINCIPAL STRUCTURE APPLICATION Office Use Only DATE l— 2 — tC E 0 c TAX nlIAP IDJKB i►�K 2�1 Pe i� 2 P F e ZONING �--I FeeN OF QUEENS Pla # �� I`a0 !ILDING &CODEf' HISTORIC SITE Yes ✓ No _ on # SUBDIVISION NAME Lot# TOWN-BD.'RESOLUTION 86 2013: $850 RECREATION FEE FOR NEW DWELLING UNITS,INCLUDING SINGLE FAMILY DWELLINGS,DUPLEXE; OR TWO FAMILY DWELLINGS, MULTIPLE FAMILY DWELLINGS, APARTMENTS, CONDOMINIUMS, TOWNHOUSES, AND/OR MANUFACTURED MODULAR HOMES,BUT NOT MOBILE HOMES. THIS IS IN ADDITION TO THE PERMIT FEE. APPLICANT C . OWNER k1jkfr ADDRESS 3 -I \ �'a�"�' ADDRESS 1�1� PHONE/E-MAIL ti �� e � �" PHONE/sE-MAILS -FR-n Le 4 CONTRACTOR P- � COST OF CONSTRUCTION(ESTIMATED): $ (� ADDRESS: BUILDING ADDRESS: PHONE/E-MAIL \ ,, CONTACT PERSON FOR BUILDING&CODES COMPLIANCE: l PHONE M` L TYPE OF CONSTRUCTION Check all that apply Please indicate measurements as required below New Addition Alteration 151 floor sq.ft. 2nd floor sq-.ft. Total sq.ft. Height Single Family Two-Family Multi-Family (# Of units ) Townhouse i Business Office Retail - Mercantile Factory- Industrial Attached Garage (#of ) Other C'0\ S- tGtia 5 k boo ms :5 .33 1 Town of Queensbury•Building&Codes Principal Structure Application July 2014 If commercial or industrial indicate name of business Proposed use of building or addition `Z v Source of heat(circle one) Gas Oil Propane . Solar Other Fireplace: complete a separate application for`FuelBirnmg Appliances & Chimneys �c Are there structures not shown on plot plan? Are there easements on the property? U Site Information a. Dimensions or.acreage of lot b. Is this a corner lot? o c. Will the grade be changed as a result of construction Yes N/No d. Public water or Private well. ID e. Sewer or Private Septic System Value of all work to be performed.(labor or materials) $ L1§ r 00 0 DECLARATION: 1. 1 acknowledge no construction shall be commenced prior to issuance of a valid permit and will be completed within a 12 month period. 2. If work is not complete by the 1 year expiration date the permit may be renewed,subject to fees and department approval. 3. 1 certify that the application, plans and supporting materials are a true and complete statement/description of the work proposed,that all work will be performed in accordance with the NY State Building Codes, local building laws and ordinances, and in conformance with local zoning regulations. 4: 1 acknowledge that prior to occupying the facilities proposed, I or my agents will obtain a:certificate of occupancy. 5. 1 also understand that I a we are required to provide an as-built survey by a.licensed land surveyor of all newly constructed facilities prior:to:issuance of'a certificate of occupancy. I have read and agree to the above: PRINT NAME: K 1 \R ��\J PV_\S DATE. l—2� — IS SIGNATURE: DATE. FOR OFF7CE USE ONLY Operating Permit:Issued, —Yes No Occupancy Type Construction Classification Assembly Occupancy Limit Special Conditions 2 Town of Queensbury Building:&Codes Principal Structure Application July 2014, Z� Foundation Inspection Report Office No. (518)761-8256 Date Inspection reque v �1� L5 Queensbury Building&Code Enforcement Arrive: a part: m/pm _ ? 742 Bay Rd., Queensbury,NY 12804 Inspector's Initial NAME: -�FSC� P T#: 5--042 - LOCATION: Gl b t` INSPECT ON: 2015 IV TYPE OF STRUCTURE: �� v Ip Comments Y A, N N/A Z— Footings Piers Monolithic Slab Reinforcement in Place The contractor is responsible for providing protection from freezing for 48 hours following the placement of the concrete. Materials for this purpose on site. Foundation/Wallpour Reinforcement in Place Footing Dowels or Keyway in place Foundation Dampproofing Foundation Waterproofing Footing Drain Daylight or Sump Footing Drain Stone: 12 inch width 6 inches above footing 6 mil poly for wet areas under slab Backfill Approval. Plumbing Under Slab PVC/Cast/Copper Foundation Insulation Interior/Exterior R- Rough Grade 6 inch drop within 10 ft. L:\Building&Codes Forms\Building&Codes\Inspection Forms\Foundation Inspection Report.doc Last printed 12/9/2014 R Foundation Inspection Report v P P Office No. (518)761-8256 Date Inspect'on request received: 1 b 1 Queensbury Building&Code Enforcement Arrive: am/p Depart: am/pm 742 Bay Rd., Queensbury,NY 12804 Inspector's Initials: 1 NAME: PERMIT#: —O L LOCATION: G �aM INSPECT ON:Cj TYPE OF STRUCTURE: c _-A Comments Y N N/A Footings A Piers Monolithic Slab Reinforcement in Place �r` e The contractor is responsible for providing protection from freezing , �� Aer for 48 hours following the placement of the concrete. - Materials for this purpose on site. Foundation/Wallpour Reinforcement in Place Footing Dowels or Keyway in place Foundation Dampproofmg Foundation Waterproofing Footing Drain Daylight or Sump Footing Drain Stone: 12 inch width 6 inches above footing 6 mil poly for wet areas under slab Backfill Approval Plumbing Under Slab PVC/Cast/Copper Foundation Insulation Interior/Exterior R- Rough Grade 6 inch drop within 10 ft. L:\Building &Codes Forms\Building&Codes\Inspection Forms\Foundation Inspection Report.doc Last printed 12/9/2014 Town of Queensbury Building&Code Enforcement Office No. (518)761-8256 Commercial Final Inspection Report Inspection request received: Name: PIA�S�_10 Inspected on: 2 Location: t &O Arrive: a. .l p.m Permit No.: ti5` Inspector's Initials: (:7 COMMENTS Y N NA Chimney/"B"Vent/Direct Vent Location Plumbing Vent Through Roof 6"/Roof Complete Exterior Finish/Grade Complete 6"in 10'or Equivalent Interior/Exterior Guardrails 42 inch Platform/Decks Interior/Exterior Balusters 4 inch Spacing Platform/Decks Stair Handrail 34 inch—38 inch/Step Risers 7"/Treads 11" Vestibules For Exit doors>3000 sq.ft. All Doors 36 inch w/Lever Handles/Panic Hardware,if required Exits At Grade Or Platform 36"(w).x 44"(1)/Canopy or Equiv. Gas Valve Shut-off Exposed&Regulator(18")Above Grade �y � Floor Bathroom Watertight/Other Floors Okay NZ: Relief Valve,Heat Trap/Water Temperature 110 Degrees Maximum Boiler/Furnace Enclosure 1 hour or Fire Extinguishing System Fresh Air Supply for Occupancy/Ventilation Combustion } -� Low Water Shut Off For Boilers __ I Gas Oil Furnace Shut Off at Entrance to Furnace Area Stockroom/Storage/Receiving/Shipping Room(2 hr.), 1 '/z doors >10%>1000 sq.ft. %Hour Corridor Doors&Closers Firewalls/Fire Separation,2 Hour,3 Hour Complete/Fire Dampers/Fire Doors Ceiling Fire Stopping,3,000 sq.ft.Wood Frame Attic Access 30"x 20"x 30"(h),Crawl Space Access 18"x 24" Smoke Vents Or Fan,if required Elevator Operation and Signage/Shaft Sealed Handicapped Bathroom Grab Bars/Sinks/Toilets/Mirrors Handicapped Bath/Parking Lot Signage Public Toilet Room Handicapped Accessible Handicapped Service Counters,34 inch,Checkout 36 inch \U Handicapped Ramp/Handrails Continuous/12 inch Beyond[Both sides] Active Listening System and Signage Assembly Space Final Electrical/Flex Gas Piping Bonded l Site Plan/Variance required Final Survey,New Structure/Flood Plain certification,if req. As-built Septic System Layout Required or On File Building Number or Tenant Address on Building or Driveway 4" TT--(�"�/ Water Fountain or Cooler / ` _ Building Access All Sides by 20'/Drivable Surface 20'wide / Lj O Special Inspections I Engineer or Architect Approval �� Okay To Issue Temporary or Permanent C/O Q /= Okay To Issue C/C Commercial Final Inspection_11 2712 Inspection Form Town of Queensbury Fire Marshal O Periodic Inspection Date I i4 Time: _. 742 Bay Road,Queensbury NY 12804 o Re-Inspection 518 761 8206/518 761 8205 ®CO Inspection Permit#: Firp Marshals Representative MJ Palmer Business Name: C— Location: ? 1`31�Rcin W' GK Stillman Contact: SLIt-- Type of Inspection N/A Yes No EXITS: Exit Access FC 1014&FC1029 NOTES Exit Enclosure FC 1020&FC1029 Exit Discharge FC 1024&FC1029 Locks and latches FC1008& FC1029.2 Sign:Normal FC 1011 &FC1029 Sign:backup FC 1011.5.3 8.FC1029.7.5 AISLES: Main Aisle Width FC 1024/1025& FC1029.11 ,^^ Secondary Aisle Width FC 1025&FC1029.11 y FIRE EXTINGUISHER: Hung FC 906 Inspection of extinguisher FC 906 / EVAC Plan FC 404.2 TRUSS ID SIGNAGE FC 505.3 EMERGENCY LIGHTING: Interior FC 1006.3&FC1029.8 Exterior FC 1006.3 Clearance to Electrical FC 605.3 Electric Wiring Enclosed/Labeled FC 605.3.1 Combustibles in Equipment Rooms FC315.2.3 F.D.Si na e- FC 510 No Smoking Signs FC 310.3 Storage FC 315.2 Compressed Gas FC 3003 Vehicle Impact Protection FC 312.1 Interior Finishes FC 803-804 Smoke Detectors FC 907 CO detectors FC 610 Clearance to Sprinkler/Ceiling FC 315.2.1 18" / 24" EVAC SIGNS IN Rooms FC 404.6 R1 &112 Fuel Pump Warning Signs FC2205.6 Fuel Station Emer Procedures FC2204.3.5 Exterior Storage FC 315.3 REINSPECTION DUE APPROXIMATELY Vacant Buildings FC 311 Emergency Disconnect FC 2203.2 21 DAYS Insp OK NC DATE: OK NC SYSTEMS: FC 901.6 Date Generator Annual DATE: OK NC Hood Installation Elevator Semi Annual FIRE ALARM Annual DATE: OK NC HVAC Shutdown Sprinkler System Annual slr.- roc Sprinkler FDC -ire.M@f§h I Ing to a )MVI(00 Kitchen Suppression Semi Annual CO Ma b th Fuel Island Suppression Semi Annual dull 1 ap oval Hood Cleaning 3-6-Annual Knox Box:installed/checked FC506 ry Operating Permit, if required will be issued after Completion of Inspection ToFirt al Queens bury � v Town of w ieensbury Building&Code Enforcement ]� Offi e No. (518)761-8256 Commercial Final Inspection Report Inspection request received: 1 Zl Name: Inspected on: 1 �� Location: tQ ZpQn�j VC�. Arrive: - a. M. i Permit No.: Inspector's Initials: COMMENTS Y NA Chimney/"B"Vent/Direct Vent Location Plumbing Vent Through Roof 6"/Roof Complete Exterior Finish/Grade Complete 6"in 10'or Equivalent Interior/Exterior Guardrails 42 inch Platform/Decks Interior/Exterior Balusters 4 inch Spacing Platform/Decks Stair Handrail 34 inch—38 inch/Step Risers 7"/Treads 11" Vestibules For Exit doors>3000 sq.ft. All Doors 36 inch w/Lever Handles/Panic Hardware,if required t tz,v v � Exits At Grade Or Platform 36"(w)x 44"(1)/Canopy or Equiv. Gas Valve Shut-off Exposed&Regulator(18")Above Grade 1'k-o ez�;D\z- Floor Bathroom Watertight/Other Floors Okay Relief Valve,Heat Trap/Water Temperature 110 Degrees Maximum Boiler/Furnace Enclosure 1 hour or Fire Extinguishing System Fresh Air Supply for Occupancy/Ventilation Combustion Low Water Shut Off For Boilers Gas Furnace Shut Off Within 30 ft.or Within Line Of Site Oil Furnace Shut Off at Entrance to Furnace Area Stockroom/Storage/Receiving/Shipping Room(2 hr.), 1 'Y2 doors >10%>1000 sq.ft. %Hour Corridor Doors&Closers Firewalls/Fire Separation,2 Hour,3 Hour Complete/Fire Dampers/Fire Doors �^ Ceiling Fire Stopping,3,000 sq.ft.Wood Frame �� tw\_ Attic Access 30"x 20"x 30"(h),Crawl Space Access 18"x 24" Smoke Vents Or Fan,if required Elevator Operation and Signage/Shaft Sealed Handicapped Bathroom Grab Bars I Sinks/Toilets I Mirrors Handicapped Bath/Parking Lot Signage Public Toilet Room Handicapped Accessible Handicapped Service Counters,34 inch,Checkout 36 inch Handicapped Ramp/Handrails Continuous/12 inch Beyond[Both sides] Active Listening System and Signage Assembly Space Final Electrical I Flex Gas Piping Bonded Site Plan I Variance required Final Survey,New Structure/Flood Plain certification,if req. As-built Septic System Layout Required or On File Building Number or Tenant Address on Building or Driveway 4" Water Fountain or Cooler L� Building Access All Sides by 20'/Drivable Surface 20'wide Special Inspections I Engineer or Architect Approval Okay To Issue Temporary or Permanent CIO Okay To Issue C/C Commercial Final Inspection_11 2712 o SOFIA ENGINEERING PLLC 7 Lorna Lane Loudonville;NY 12211 Tel. 518.482.0067 Fax.518.482.0204 e.mail:vsofia@nycap.rr.com To: Town of Queensbury Building Department 742 Bay Road Queensbury,NY 12804 From: Enzo Sofia. PE 7 Lorna Lane Loudonville,NY 12211 Date: January 13, 2016 Re: Pre Engineered Building at AFSCO 34 Big Boom Road Building Manufacturer: Hansen Buildings Contractor: D&J Construction East LLC Dear Sir/Madam: I, Vincenzo Sofia, a register professional Engineer in the State of New York, do hereby certify that I the construction of the above referenced project manufactured by Hansen Buildings. To the best of my knowledge the work under permit has been completed in accordance with the approved drawings and specifications in so far as structural, fire, health and life safety are concerned. Should you need additional information,please do not hesitate to contact me.- Thank you very much for your corporation.' OF NEW Y �P C1NZ 0 CO erely zo Sofia,PE C9 0700� ROFESSioNP� FIRE MARSHAL'S OFFICE Town of Queensbury 742 Bay Road, Queensbury,!NY 12804 "Home of Natural ,Beauty ... A Good Place to Live PLAN REVIEW AFSCO Fence 2015 - 042 3/6/2015 The following comments are based on drawings reviewed: • Truss ID signage is required • Locks/latches to comply w/Chapter 10 NYSFC Michael J Palmer Fire Marshal 742 Bay Road Queensbury NY 12804 firemarshal@queensbury.net Fi r e M a r s h a l's Off i c e • P h o n e: 518-761-8206 F a x: 518-745-4437 {tremarsltal@giteensbunJ.net 7VIVMqueensbunt.net Post Frame Building Summary Prepared Form. Kyle Steven #15-0203SK Site Address: 34 Big Boom Rd; queensbury, NY Mailing Address: 34 Big Boom Rd; Queensburry, NY 12804 Phone: 518.-260-7103 Code: 2010 BCNYS Building Occupancy Classification: U Type of Construction:V-B Number of Stories:I.- Building Footprint Width: 50'.., Building Footprint.Length: 1.00' Square Footage (area contained by embedded poles): 5000 ftZ Total"roof area: 5591 ft2 Total wall area: 5097 ft2 .. Building Eave Height (Roof Above 0.0:Grade:at Si.dewall): 17' Roof..Style: GABLE: " Roof slope:4/.12. . Mean Roof Height:.21..17' Non-conditioned Floor Live Load: does not apply. Interior double trusses:37.802 psf using ps".. Ground Snow Load: (pg) = 60 psf Flat-Roof Snow Load (pf) 40.32 psf: Roof Dead Load: 3:3 psf Ceiling Dead Load:.1 psf: Snow Exposure Factor (Ce): 1 Snow load importance factor(IS)0.8 Thermal Factor(Ct)::.1:2. Slope Factor: 0.938,: Sloped Roof Snow Load (ps): 37.802 psf Roof Duration of Loads for Gravity/Live: 1.1.5.` � OF NE� 'Basic Wind Speed:,112 mph" Wind Importance.Factor(I,ti,): 0.87 AP ,ONN C Age�Q� Risk Category.: Wind Exposure: B Applicable Internal Pressure Coefficient: 0.18 Components and Cladding Design Wind Pressure: Zone 1: 16.304 Zone 2: -22.059 NFp 08992�:: Zo ne 3• -36 269 A V "Zone 4: -16.304_. FESSIOP Zone 5: -16.304. Duration of'Load for Wind: 1.6 Enclosed Structure l Seismic.Use:Group::B Spectral Response_Coefficients (Ss): 0.287g(Sj): 0.0799(SDS): 0.30047g (Sol): 0.1264g Site Class: D Basic seismic-force-resisting system: Light-framed walls sheathed with wood structural panels rated for shear resistance or steel sheets Design Base Shear: 1708.709 Analysis Procedure: Equivalent Lateral Force per 12.8. ASCE Soil Class: 4 Allowable Foundation Pressure: 2000 Lateral Bearing: 150 psf/f Frost Depth: 40" Concrete Slab on Grade: Yes Commercial Grade Insulated Steel Entry Doors have Steel Frame with an R-131008.1.6 Landings at doors. Landings shall have a width not less than the width of the stairway or the door, whichever is - greater. Doors in the fully open position shall not reduce a required dimension by more than 7 inches (178 mm). When'a landing serves an occupant load of 50 or more,-doors in any position shall not reduce the landing to.less than one-half its required width. Landings shall have a length measured in the direction of travel of not less than 44 inches (1118 mm). Exception: Landing length in the direction of travel in Groups R-3 and U and within individual units.' of Group R-2 need not exceed 36 inches (914 mm). - ESR 1980-Et ESR 2240 Micropro Pressure Treated Wood . ESR 2644-ACQ Pressure Treated Wood ESR 1721 Wolmanized Outdoor Preservative Treated Wood ESR 2523 Simpson Strong-Tie Products ESR 2549,Et ESR 3096 LU series ESR 2613'Et ESR 3096 H1 ESR 2105 LSTA and HST series . ESR-1893 Palram polycarbonate products Sidewa(l post deflection: 3.4" . For structural:roofing and siding made of formed metal sheets, the total load deflection shall not . exceed 1/60. per footnote a of IBC table.1604.3 l = 17' = 20y4,"..204'/3.4" = 60 >_ 1/60. .. '` u Later Longitudinal Sarface Net Pressure with Net Pressure with' GC t � r • +GCPi) ( GCPi).,, P (-GCP,) (-GCpi) �" .....,_ ., .. 0:5'(6 5.597 11.587 .0.4 3.66 9.65 a2^ - 0 69 . l4.4 44 -8.485 _.r0.b9:-14474 8.485� 3 0.469 -10.79 ` -4.8 -0.37 =9.15 4 0 415 -9:906 , -3.916 -0.29 -7.819 -1.83 . �1 E �4i0 78 °9.985 �15.975 :.0.61:7.154 �13 143 2E 1 07 °-20 796 ; 14 807 -1.07 20.796 -14 807 3E 3-14.197 '-8.207 0.53 -11.81.2 ;-5.823 4E 0.618;-13.278 ` 7.289 -0.43,-10.149 -4.159 0 45 10 481 4 492 --` 0.45 -10.481 `-4.492 S: 64.697 psf D: 4.3 psf All doors supplied by parties other than Hansen Pole Buildings must meet or exceed the stated minimum design wind loads Kyle Steven Land Owner: Neal Galvin 34 Big Boom Rd Warren county queensbury, NY 12804 Design per 2010 BCNYS. Gold Package Seismic B Risk Category I (IS= 0.8 1,= 0.87 le= 1 Ce 1 Ct= 1.2) Flat Roof Snow Load (pf) =40.32 psf Ground Snow Load (p.) =60 psf CS = 0-938' Roof Dead Load: 3.3 psf Ceiling Dead Load:.1 psf. Wind Load:.112 mph Exposure: B Soil: 2000/150 Building Span: 50' Building Length: 100' Eave Height: 17' Roof.Pitch: 4/12 SLFTR: 1.0541 Angle: 18.435" Concrete Floor: Yes . Overhangs: Main Building: Enclosed 12" Overhangson Front Endwall, Rear Endwall, Left Sidewall, Right Sidewall Truss Spacing front to back): 10'0" 10'0" 10'0" 10'O" 10'0'. .10-0-! 10,0" 10'0.. 10-0-- 10'0" P g ( ) , Front Column Spacing: 10'8.25", 8'0.5 12'6.5", 9'8:75", 9'0" Rear Column Spacing: 9'6", 9'8.75", 12'6:5 8'0.5", 10'8.25' 0'0" Horizontal 2x4 nailer required halfway between bottom of truss and top of truss at truss peak, 12'x 14' Door by others Overhead Door: Rear Endwall 19'from left corner on Main Building with 6x6 SYP . #2 - Header-2x6#2 12'x 14' Door by others Overhead Door: Front-Endwall 19'from left corner on Main Building with 6x6 SYP#2 - Header-2x6 #2 6'x 6'8"'Steel.Entry Door: Front Endwall 40'10"/16' from left corner on Main Building (40'815/16'`from corner BL to CL of post), rough opening size:.74V8x80%s, with poles on'either side reaching the top of . the door 6'x 6'8"'Steel Entry Door: Rear Endwall 2'10"/16'from left corner'on Main Building (2'815/16'from corner BL to CL of post), rough.opening size: 745/sx805/a, with poles on either side reaching the top of the door Location Pole Size EmbedmentDepth Diameter Thickness Left Sidewall 3ply 2x8 Glu-lam 40" 30" 18" 41/s" Right Sidewall 3ply 2x8 Glu-lam 40" 30" 18" 41/s" 6x6 SYP#2, 6x6 SYP#2, 6x6 SYP#2; Front Endwall 6x6 SYP#2 40" F . 51/2', 51/z", 51/z", 51/z" .. 6x6 SYP#2, 6x6 SYP#2, 6x6 SYP#2, Rear Endwall .' 6x6 SYP#2 . 40'. 18" 18 51/z", 51/2", 51/2 , 51/2'• _. Front Left 6x4 SYP#2 Corner 51/z' 40" 18 18 Front Right 6x4 SYP#2 40" 98" 18" .Corner 51/z" Rear Left Corner 6x4 SYP#Z 51/z 40" 18 • :- 18 {j Rear Right 6x4 SYP#2 I Corner 51/z" 40 18 18 `poles to be placed 8''above bottom of hole Roofing: 29g Steel Siding: 29g Steel Wall Girts: Number of Bay Girt Blocking. Number of.. #of Wall Girt Size Bays Size Spacing Size Nails Girts Left Sidewall 10 10'0'` 6x2#2 3T' 2x4 STD 3 . 4,. between Right Sidewall 10 10'0" 6x2 #2 37" [27x4STD. 3 4 between. Front Endwall 1 g0,; barn style 14" 2 9 2x6#2 between 1T/ o. Front.Endwall 1. 9'81/a" 8x2#2 1 2x4 STD 2 9 . . between Front Endwall 1 12'61/2" 6x2#2 between 2x4 STD 2 9 , Front Endwall 1 g p,/2 barn style 14" 2 9 2x6#2 between ____[ Front Endwall 1 10'81/4- 10x2#2 17 /10 2x4 STD 2 9` between Rear Endwall 1 10'81/4" 10x2#2 177/10 2x4 STD 2 9 between Rear Endwall 1 8'01/z" barn style 14" 2 9 2x6#2 between Rear Endwall 1 12MY' 6x2#2. 17 /10 2x4 STD 2 9 between 177/10' Rear Endwall 1 9'83/4 8x2#2 2x4 STD 2 . 9 between Rear Endwall 1 9'0" barn style 14" 2 9 __. 2x6 2` between 1 Front Endwall barn style 15%s" 1 around doors 1 TO" barn Std between 2 9 Rear Endwall. 1 3.001' barn style 15%s" 2 9 around doors 2x4 Std between ,..' Front endwall to be sheeted with 1 layer of OSB in bays 1, 2, 4, 5 Rear endwall to be sheeted with 1 layer of OSB in.bays 1, 2; 5,.6 Roof steel requires#12x11/4"sidelap stitch screws @ 9_3/8'o.c. within 25.039'of each endwall where.it's not sheated with OSB Roof Purlins within 9.87'of the peak at 24"o.c. with sizes of: (2)10'0" bays 2x8 #2, (8)10'0" bays 2x8#2.., Other Purlins at 24"o.c. with sizes of: (2)10'0" bays 2x8#2, (8)10'0".bays 2x6 #2 . Ridge.Purlins with sizes of: (10)10'0" bays 2x8#2 Eave Purlins with sizes of: (2)10'0" bays 2x8#2, (8)10'0" bays 2x6#2 Fascia: 2x8#2 34"from bottom of soffit to top of next.girt down on left sidewall; 2x4 eavelight girt below top girt . which is moved to be placed just above eavelight girt 34"from bottom of soffit to.top.of next girt.down.on right sidewall; 2x4 eavelight girt below top girt ., which is moved to-be.placed just above eavelight girt Eave Girt to First Column: 10d Interior Truss to Column: (1) 5/8" bolt 5" long and (9) 20d nails . Endwall Truss to Corner Column: (3) LSTA-12. 2x4 X bracing to endwall columns over T from corners Single 2x4 bracing (maximum 20'on center) interior bays Distance from-top of eave girt to top of vertical 2x4= 11.25" :. Lower Endwall Truss 7%8" INDEX Building Criteria Wind Calculations Method 2 - Analytical Procedure for Low-Rise Buildings Method 2 - Analytical Procedure Cladding Snow Calculations Live Load Calculations Dead Load Estimates Seismic Calculations Design Diaphragms Shearwalls Connection.Strengths Column Calculations Girt Calculations Purlin Calculations References . 9. 13 - 16 18 19 20 21 24 26 28 , 31 112 147 175 _ . BUILDING CRITERIA Width: 50 . Length: 100. Height: 17. Slope: 4/12 Maximum bay size= 10' roof area of bigger side: 2795.453 ftz mean roof height: 21.167 ft. 6: 18.435 estimated dead load: 39799.46 lbs. ' top chord.dead-load: 3.3 psf: bottom chord dead load: 1 psf Wind Speed: 112 mph. Wind Exposure: B Sloped Roof Snow Load:.37.802 psf Seismic Zone: B . Location Pole Size Embedment Diameter Thickness ' Depth. Left Sidewall 3ply 2x8 Glu-lama 40" 30 18 4'�8 3ply 2x8 Glu-lam Right Sidewall 40" 30'. 18 6x6 SYP#2,.6x6 SYP#2,6x6 SYP.#2, Front EndwaR 6x6 SYP#2 40" 181. 18" 5�/i', 5�/z'", 51/z", 5�/i' .' 6x6 SYP#2,' 6x6 SYP#2, 6x6 SYP#2, Rear Endwall 6x6 SYP#2 40" 5'/i', 51/z", 51/z", 5'/i'. i Front Left 6x4 SYP#2 '/i' 40"_ 18„ . 181, Corner 5 Front Right 6x4 SYP#2 40" 18". 18" Corner, 51/z' Rear Left Corner 6x4 SYP#2 40" 18" 18' 51/z' [Rear Right 6x4 SYP#2 40 18' 18" Corner 5'/z" *poles to be placed 87.above bottom of.hole page 12 LOAD CALCULATIONS WIND CALCULATIONS Wind Variables and Definitions y: adjustment factor for building height and exposure y= 1 interpolated from ASCE figure 6-2 based on meanh and exposure. Kn: topographic factor ' Kn= 1 assuming a flat site I: importance factor- factor that accounts for the degree of hazard to human life and damage to property = 0.87 ASCE table 6-1 . z: height above ground z= 30'with a minimum per ASCE table 6-2 based on wind exposure zg: nominal height of atmospheric boundary zg=1200'from ASCE table 6-2 based on wind exposure a: three second gust speed power law exponent a=7 from ASCE table 6-2 based on wind exposure c: turbulence intensityfactor c =0.3 from ASCE table 6-2 based on wind exposure z: equivalent height structure z= 30'from ASCE table 6-2 based on wind exposure. . l: integral length scale factor l= 320'from ASCE table 6-2 based on wind exposure E: integral length scale power law exponent E =0.333 from ASCE table 6-2 based on wind exposure Kd: wind directionality factor . Kd= 0.85 from ASCE table 6-4 Directionality Factor Kd has been calibrated with combinations of loads specified in Section 2. This factor shall only be applied when used in conjunction with.Load combinations specified in 2.3 and 2.4 wind speed = 112 mph page 13 Method 2-Analytical Procedure for Low-Rise Buildings Wind Calculations = ,00256*2.01 * (z / Zs)2/°* KZt* Kd*V2* 1 ' (GCPf- GCpj) simplified from ASCE equations 6-18, 6-15, and Table 6-3- Zone 1 GCpf for 1: product external pressure coefficient gust factor GCpf for 1 =0.516 from ASCE figure 6-10 based on roof angle . GCpj for 1: product internal pressure coefficient gust factor = GCpj for 1 = 0.18 using ASCE figure 6-5 determined.using worst case conditions Wind Calc 1 = :00256 lbs hrs�/ft.2mi�*2.01 * (30' / 1200')�/7 1 *0.85 * (112 mph * 0.87' (0.516 - 0.18) Wind Calc 1 = 11.587 psf Zone 2 GCPf for 2: product.external pressure coefficient gust factor. _GCpf for 2 = -0.69 GCPj for 2: product internal pressure coefficient gust.fact.or. GCpj for 2 = 0.18 ; wind talc 2 14.474 psf Zone.3 GCpf for 3: product externs l pressure coefficient gust factor GCpf for 3 =,-0.469 GCpj for 3: product internal pressure.coefficient gust factor.. GCP;for 3 = 0.18 ..wind talc 3 = =10.79 psf page 14 Zone 4 GCPf for 4: product external pressure coefficient gust factor GCpf for 4= -0.415 GCp;for 4: product internal pressure coefficient gust factor GCp;,for 4= 0.18 wind talc 4= -9.906 psf . Zones 5 and 6 . GCpf for 5 and 6: product external pressure coefficient gust factor GCpf for.5 and 6 = -0.45: GCp;•for 5 and 6: product internal pressure coefficient gust factor GCp;for 5 and 6 =0.18 . wind calc 5 10.481 psf Zone 1 E GCpf.for 1 E:-product external pressure coefficient gust factor GCpf for 1 E=..0.78 GCpf for 1 E: product internal pressure coefficient gust factor GCpf for l E= =0.18 wind talc 1 E= 15.975 psf d. .Zone 2E . GCPf for 2E: product external pressure coefficient gust factor GCPf for 2E_ -1.07 GCp;for 2E: product internal pressure coefficient gust factor GCp;for.2E=0.18 wind talc 2E_ -20.796 psf Zone 3E. GCpf for 3E product external pressure coefficient gust factor GCpf for 3E=.-0.673. GCp;for.3E: product internal pressure coefficient gust factor: - GCP;for.3E=0.18 wind talc 3E= 14.197 psf _. page 15 Zone 4E GCPf for 4E: product external pressure coefficient gust factor GCPf for 4E= -0.618 GCP;for 4E: product internal pressure coefficient gust factor GCP;for 4E = 0.18 wind talc 4E_ -13.278 psf. page 16 Method 2 -Analytical Procedure for Components and Cladding Wind Calculations = .00256*2.01 * (z L Zg)2/°* Kzt* Kd *V2* I * (GCP GCp;) simplified from ASCE . equations 6-22, 6-15, and Table 6-3 Zone 1 GCP: product external pressure coefficient gust factor GCP for 1 = -0.8 from ASCE figure 6-11 GCpi: product internal pressure coefficient gust factor GCP; =0.18 using ASCE figure 6-5 determined using worst case conditions Wind Calc for 1 = .00256 lbs hrsZ/ft.2mi2*2.01 * (30' / 1200')2"* 1 * 0.85 * (112 mph)2*0.87* ( 0.8 - Wind Calc for 1 716.304 psf Zone 2 GCP: product external pressure coefficient gust factor GCp for 2 = -1.2 from ASCE figure 6 11. GCp;: product internal pressure coefficient gust factor GCp;= 0.18 using ASCE figure 6-5 determined using worst case conditions Wind Catc for 2 = .00256 lbs hrsZ/ft.2mi2* 2.01 * (30' / 1200')1 * 1 *0.85 * (112 mph) *0.87,* (-1.2 0.18) Wind Calc for 2 = -22.959 psf. page 17 Zone 3 GCP: product external pressure coefficient gust factor GCP for 3 = -2 from ASCE figure 6-11 GCP;: product internal pressure coefficient gust factor GCP;= 0.18 using ASCE figure 6-5 determined using worst case.conditions Wind Cale for 3 = .00256 lbs hrs2/ft.�mi2*2.01 * (30' / 1200')217* 1 `0.85 * (112 mph)2*0.87:* (-2 - 0.18) Wind Calc for 3 = -36.269 psf Zone 4 GCP: product external pressure coefficient gust factor GCP for 4= -0.8 from ASCE figure 6-11 GCp;: product internal pressure coefficient gust factor GCp;=0.18 using.ASCE figure 6-5 determined using worst case conditions Wind,Calc for 4= .00256 lbs hrs2/ft.zmi2*2.01 * (30' / 1200')2"* 1 * 0.85 * (112 mph) *0:87* ( 0.8 0.18) Wind Calc for 4= -16.304 psf Zone 5 GCp: product external pressure coefficient gust factor GCP for 5 =.-0.8.from ASCE figure 6-11 GCp;: product internal pressure coefficient gust factor GCp;=0.18 using ASCE figure 6-5 determined using worst case conditions Wind Calc for 5 =..00256.1bs hrs�/ft.2mi2*2.01 * (30' / 1200')2"* 1.* 0.85 * (112 mph)2.*0.87* (-0.8 0.18) Wind.Calc for 5 = -16.304 Of page 18 SNOW CALCULATIONS Ce: exposure factor Ce= 1 ASCE table 7-2 based on exposure assuming at least partially exposed roof Ct: thermal factor. Ct= 1.2 ASCE table 7-3 based on heating condition of the building I: importance factor- factor that accounts for the degree of hazard to human life and damage to property 1 0.8 ASCE table 7-4 Pf= flat roof snow load * Ce*C�* I *.pg from ASCE Pf= :7 Pf= .7* 1 * 1.2 *0.8 *60psf Pf=40.32 psf using minimums from ASCE.7.3 CS= 1 -,((6 - 15) /:55) MBSM 1.3.5.5 CS=.1 ((18.435° -.15') / 55) C-: slope factor CS=0.938 P5= C5'`•pf from ASCE 7.4 P5=0.938*40.32 psf ps=.37.802 psf. unbalanced light load = :3 * ps ASCE 7.6 1, unbalanced light load =. 11.341-psf unbalanced heavy load= ps+ hd*y /f(S)ASCE 7.6.1 hd: drift height . hd= :43 * I * (Pg:f 10)'�.-1:5 . tenth of the roof upwind of the drift 1� =25' hd= .43 27.406 3 (60+.10) 1.5 hd=2.137' y: snow density. y= .13 * pg +.14 with a maximum of 30 pcf ASCE 73.1 y= .13 * 60 + 14 y=21.8pcf. : 5= 3 . -unbalanced heavy load =37.802 psf t.2.137'*21.8 pcf/.f(3) drift width = 8 *.f(S) * hd / 3 ASCE 7.6.1 drift width = 8 f(3) *2.137' /.3 . . . drift width = 9.87' unbalanced.heavy load = 64.697 psf using greater of.the calculation here or ps page 19 LIVE LOAD CALCULATIONS Lr= La* R, * R2 where 12 s Lr< 20 ASCE 4.8.2 Lo=20 psf.ASCE TABLE 4-1 , R, = 1.2 - .001 *At bounded at .6 and 1'ASCE 4'.8.2 At= 500 ft.2 R, = 1.2 - .001 * 500 Ri =0.7 R2= 1.2 .05* F bounded at .6 and 1 ASCE 4..8.2 F=4. R2= 1.2 .05*4 R2= 1 Lr truss 20.*0.7* 1 Lr truss= 14 psf R, = 1.2 - .001 *At bounded at .6 and 1 ASCE 4.8.1. At=20 ft.2 Ri = 1.2 .001 *20 Ri - 1 - - RZ= 1.2 -.M.* F bounded at .6 and 1 ASCE 4.8.1 F=4 R2= 1.2 .05*4 Rz= 1 Lr purlin =20*.1 * 1 Lr purlin 20 psf page 20 DEAD LOAD ESTIMATES Truss weight estimated at: 350 lbs Dead load of truss estimated at: 0.7 psf Dead load of.roof framing estimated at: 62.4 pcf* .43 lbs/ft.3, /.(1 + .43 lbs/ft.3* .009.* .19) * (1 + .0019) * 1.5" / 12 in. *7.25" / 12 in./ft. * 10, / (24" / 12 in. Dead load of roof framing estimated at: 0.989 psf Dead load of reflective insulation: .036 psf roof steel dead load = .63 psf steel American Building Components catalogue Estimated dead load of 2.355 psf is less than used top chord dead load of 3.3 psf Bottom chord bracing max weight per bay: 62.4 pcf* .43 lbs/ft.3 / (1.+ .43 lbs/ft.3* .009 * .19) * (1 + .0019) * 1.5" / 12 in./ft. *3.5" / 12 in./ft. * 10'*6 . Dead load of bottom chord bracing: 0:118 psf Estimated dead load of 0.118 psf is less than used bottom chord dead.Load of 1 psf ;. page 21 SEISMIC.CALCULATIONS S5:mapped spectral response acceleration short SS=0.287 g value obtained by county from MBSM'chapter IX Sj: mapped.spectral response acceleration long S, =0.079 g value obtained by county from MBSM chapter, IX _ Fa: short period site coefficient Fa= 1.5704 from ASCE table 11.4-1 using site class D since unknown . F,,: long period site coefficient Fv=.2.4 from ASCE.table 11.4-2 using site,class D.since unknown. Sos: design.spectral response acceleration short Sos=2* Fa*Ss / 3 from ASCE 11.4 Sos =2* 1.5704*0.287 g / 3 Sos = 0.3 g Sp,: design spectral response acceleration-long SDI=2*.F„* SI./ 3 from ASCE.11.4 SDI=2 .-2.4*.0.079 g f 3 . Soy = 0.126 g ; Seismic Zone; B from ASCE 11.6.; I: importance factor .I =.1 T: fundamental period: .T,=Ct.*h„"from ASCE 12.8.2.1 . T= .02 * 07'+50'L.2,.* (4 / 12))3i4 T=0.226 s Ra.`response`modification_coefficient. Ra=7 from ASCE table 12.2-1 for Light-framed walls sheathed with wood structural panels rated for shear resistance or steel sheets page 22 CS =SDS / (R / 1) from ASCE 12.8.1.1 CS = 0.12639999999999998 g / (7 / 1) CS = 0.043 with limitations according to ASCE 12.8.1.1 W: co+ .2* pf ASCE 12.7 QE: CS*W ASCE 1.2.8.1 QF=0.043 *29893.832 psf Qe= 1283.171 Eh: p * OE ASCE 12.4.2.1 Eh =.1 * 1283.171 lbs E,,: .2*SDS* D ASCE 12.4.2.2 E„= .2 *0.126 g *29885.768 lbs E„= 755.512 lbs. seismic force: En+ E ASCE 12.4.2 seismic force= 1283.171 lbs +755.512 lbs seismic force= 2038.661.lbs page 23 The governing load combination is D +.(W or .7E)ASCE 2.4.1 equation 5 In the previous equation, the wind load intensity (W= 10 psf) is greater than the earthquake load intensity (.7E=0.285 psf) therefore the wind load will be used to design the endwalls. page 24 DESIGN DIAPHRAGMS Background: The diaphragm panel systems have been tested for shear capacity using recommendations from ASTM E455-76 "Static Load Testing of Framed Floor or Roof Diaphragm Constructions for Buildings"and ASAE EP484"Diaphragm Design of Metal-Clad, Post-Frame Rectangular Buildings". Limitations: The relative impact on shear strength of each component from the diaphragm testing has not been defined. The characteristics of the metal cladding and fasteners specified herein meet or exceed the tested products. The characteristics of the framing lumber meet or exceed the Specific Gravity (G) of the tested product, Spruce-Pine-Fir, G =0.42. Diaphragm Testing Data Applicability:. Applicable.for.buildings that utilize the method of diaphragm design to resist design wind loads: The test data are not applicable for roofs with skylights Diaphragm Testing Results (Table.1): Maximum Shear Construction Intensity, I (lbslft) " Roof Wall Sections : Sections ^ .. .. _ 110 lbs/ft 160 lbs/ft Field Screws -110 x 1-1/2" next to each major rib (9" o.c. Top a Bottom Screws -#14 x 1-'/2"at BOTH sides of.each major rib Field Screws #10 x 1-'/z"next to each major rib`(9"o.c.) 164 lbs/ft 173 ibs/ft Top It Bottom Screws - #10'x 1=''/a"at BOTH sides of each major,rib ." St - through all overlaps Stitch Screws-#12 x 5/e'at 9 3/8 on center �. �TAB.� 06.3.1 F16 /ft for a roof. '/16"osb applied directly to framing with 8d common nails 2" on center at panel edges; 6" on center intermediate framing members. Per footnote a, G > 0.42 use base value of,230#" .92 . IBC TABLE 2306.4.1 '/�6"osb applied directly to framing with 8d common nails 2" on.center at 538.2 lbs/ft for a wall panel edges; 6"on center intermediate framing members. Per footnote a, G section z 0.42 use base value of 585# ' .92 page 25 When the wall columns are supported at floor Level so as to prevent rotation of the base of the column, the Maximum Roof Shear Intensity, I (lbs/ft) can be calculated by:_ . I = (3/ * (qw* H.1 * L) + (q,* H2* L)) / (2*W) / 1.33 Alumax Roof Diaphragm Brochure 1991 where qu,= design wall pressure (psf) qr= design roof pressure(psf) L= the building length (ft) W= the building width (ft) H1 = the outside wall height (not inside clear) H2 = the height from the eave,to the ridge Calculate the Maximum Roof Shear,Intensity, I (lbs/ft). This is the shear per foot of slope length that occurs at the end walls of the building. Where: ' `. 1 = (3/a * (21.492 psf,* 17'' 100') + (10 psf*8.333'*.100')) / (2* 50) /1.33 =220.347 lbs%ft- .. As I is over 164, we will need to sheet that portion of the roof with.bSB which has shear.forces greater: than 164 220.347 lbs%ft / 164.lbs/ft 1.344 > 1.=- " 100' / 1:344=74.428' 100' - 74.428' = 25.572' / 2 = 12.786' Roof steel requires'/16' OSB"within at least 12.786'of each endwalL As I is over 110, we will need.to stitch screw.that portion of the roof.which.has shear forces greater" than.110 220.347,lbs/ft / 110 lbs/ft=2.003.> 1- - .. .-100'./.2.003 =49:921' 100' -49.921, = 50.079' / 2=25.039' Roof steel requires#12x1?/a"sidelap stitch screws C 9-3/8'.o.c.. within 25.039'of each endwall where it's not sheated with OSB page.26 Front End Girt-Shearwall Girt Spacing to be a maximum of 42" o.c. . Vyl *W V =220.347 lbs/ft* 50' V= 11017.353 lbs SC = (W - DW) * R.. SC = (50' - 18)* 538.2 lbs/ft SC = 17222.4 lbs V < SC - endwall will have 7/16"osb applied directly to framing with 8d common nails 2" on center at panel edges; 6"on center intermediate framing members for at least 20.5'. Rear End Girt-Shearwall` Girt Spacing to be a.maximum of 42"o.c. V=220.347 lbs/ft* 50' - V= 11017.353 lbs SC =jW- DW) * R ; ', ..... SC = (50' -.18) *538.2 lbs/ft SC _. 17222.4 lbs . ..:.. V <_ SC :. endwall will have'J1e' osb applied directly to framing with 8d common nails 2'`on center at panel edges; 6'on center intermediate framing members for at least 20.5': SC =.(W._ DW) *R . S: section modulus S = b * d * d / 6NDS3.3.2 S = 13.75"*7.125"*7.125 / 6 S = 116.338 in.-' fb: bending stress.post fb= IMI /.S NDS 3.3.2_ ' fe 12247539.969in.lbs I / 116.338 in.' fb= 19319.071 psi page 27 . CM=wet service factor CM= 1 because the post is-protected by the building shell, concrete, or embedment from excessive moisture CF= 1 NDS Supplement . Cr: repetitive member factor Cr= 1.15 NDS 4.3 Cf,,: flat use factor Cfu =.1 NDS 4.3 Fe: allowable bending stress post Ft = Fb*Cb*CM*Ct* CL* CF*Cfu*Ci* Cr NDS 4.3., Fe.= 1500 psi * 1.6 * 1 * 1 * 1 * 1 1 * 1 * 1.15 . Fe=2760 psi fb_< Ft 2247539:969"psi <_2247648.047 psi Left Side Girt-ShearwalC Girt Spacing to be a maximum of 42"O.C. V=W*A: . V.=21".492348620749496 psf,*425' V=9134.248 1bs SC = (W- DW) * R SC = (100'- 0') * 160 lbs/ft SC= 16000 lbs V<_SC :. sidewall steel requires#10 x 1-''Y2".next"to.each major rib, (9" o.c.)Right Side.Girt ShearwatL. Girt Spacing to.be a maximum'of 42" o.c.:. V=W*A V=21.492348620749496 psf*425'. V= 9134.248 lbs SC (W- DW) * R SC = (100' - 0') * 160,lbs/ft SC =.16000 lbs V< SC :- sidewall steel requires#10 x 1.-I/i' next to each major rib(9"o.c.) page 28 Connection Strengths 10d 3" nails Girt Nails D = 0.1481, lR,: main member length 1m = 1:5 ls: side member length is= Is. . Fem: main member dowel bearing strength Fem= 5550 psi NDS table 11.3.2 Fes: side member dowel bearing strength Fes= 5550 psi NDS table 11.3.2 Fyb: dowel bending yield strength Fyb =.90000 psi NDS table 11 N Rd: reduction term Rd.=2.2 NDS table 11.3.1 B Z for Yield Mode Im= D* lm* Fem / Rd NDS table 11.3.1A . Z for Yield'Mode Im = 0.148"* 1.5"* 5550 psi / 2.2 Z for Yield Mode Im= 560:045 lbs = D is k Fes / Rd NDS table 11.3.1A Z for Yield Mode IS Z for Yield Mode IS= 0.148"* 1.5".* 5550 psi / 2.2 . Z for Yield Mode Is= 560.045 lbs page 29 Z for Yield Mode II = ki* D * 1,* Fe, / Rd NDS table 11.3.1A k, = (./(Re+2 * Re * (1 + Rt + Rtz) + Rt2* Re) Re* (1 + Rt)) / (1 + Re) NDS table 11.3.1A Re= Fem / Fe5 NDS table 11.3.1A Re= 5550 psi / 5550 psi Re= 1. Rt= h; / I NDS table 11.3.1A Rt= 1.5" / 1.5" Rt= 1 kt = (f(1 +2* 1z* (1 + 1 + 12) + 1z* 13) 1 * (1 + 1)) / (1 + 1) k' = 0.414 Z for Yield Mode II =0.414*.0.148"* 1.5"* 5550.psi / 2.2 Z for Yield Made II =23.1.978 lbs Z for Yield Mode III,, = Kz* D * lm* Fem / ((1 +2 * Re) * Rd) NDS table 11.3.1A k2= -1 +'I(2 * (1 +.Re) + (2 * Fyb* (1 +2*Re)* D2) / (3 * Fem.* Imz)) NDS table.11.3.1A k2= -1 +f(2* (1 + 1) + (2 * 90000 psi* (1 +2 * 1:) *0.148z) / (3 * 5550 psi.* 1.5"z)) k2= 1.077 Z for Yield Mode IIIm =.1..077*0.148"* 1.5" *.5550 psi /. ((1 +2 *1) *2.2) Z for Yield,Mode lllm =201.137 lbs Z for Yield Mode Ills=.k2.* D* lm.* Fem / {(2.+ Re) * Rd) NDS table 11.3.1A k3= 1 +1(2.° (.1 + Re).+ (2.* Fye* (2.+ Re).* D?), / (3 * Fem* l,z))NDS table„11 3 1A ks = -1.+I(2 * (1 + 1) + (2. *90000 psi * (2 + 1) *0.148'.2) / (3 *,5550 psi * 1.5,.2)) k3 = 1.077 Z for Yield Mode III,= 1.077*0.148" *1.5" 5550 psi / ((2 + 1) *2.2) Z for.Yield Mode III, = 201.137 lbs Z for Yield Mode IV= D.z / Rd.*f((2 * Fem*.Fyb) / (3-* (1 :+. Re))) NDS table 11.3AA Z for Yield.:Mode IV= 0.148"2 / 2.2*.f((2* 5550 psi *90000 psi) / (3- (1;+ 1))) Z for Yield Mode IV-=,128.472 lbs page 30 using worst case of Z modified by applicable factors NDS 10.3.1 Z'=Z*CD* CM*Ct*Cs *Ca* Ceg*Cd;* Ctn CD: load duration factor CD= 1 NDS 2.3.2 factor not applicable due to metal strength controlling CM: wet service factor CM= 1 NDS 10.3 Ct: temperature factor Ct= 1 NDS 10.3 Cg: group action factor Cg= 1 NDS-10.3.6 Co: group action factor Ce=.1 NDS 10.3 Ceg: end grain factor Ce2 = .67 for end nailed girts NDS.11.5.2 Cd;: diaphragm factor Cdi 1 NDS 10.3 Ct,,: toe-nail factor, Ctn = 1 NDS 10.3 Z'=:128.472.lbs*. 1 * 1 * 1..* 1 * 1 *.0.67* 1 * 1 Z' = 86:076 lbs page 31 COLUMN CALCULATIONS Sidewall Columns Step 1: Calculate the'roof diaphragm stiffness Ch, by using a stiffness.adjustment procedure.based upon . the total.number of sheet to purlin fasteners Aside from panel end fasteners, panel length is proportional to the number of fasteners_when..the pattern of sheet to purlin fasteners in the diaphragm is maintained for.the predicted.buildmg diaphragm where: Cl: stiffness of,test panel C� = .5 k P 7 D, a / b ASAE EP 484:1, Eqn. 4 referenced in Design of Commercial Post-Frame p.44 for a simple beam test'where page 32 P: panel strength P = Pultimate* .4 Puldmate: ultimate panel strength Paltimata = 9600 lbs Townsend, p.4 P = 9600 lbs*-.4 -P 38401bs _. .. D5: deflection at P DS= 0.265"Townsend, p.4 a: test panel frame spacing .. a = 144"Townsend p.4 b: test panel length parallel-to=corrugations . b.= 140"Townsend, p.4 „ . .. . CI: stiffness of test panel Cf = .5.* P / DS*a / b Skaggs Eqn. 4 -C, = .5 *3840 lbs / .265"* 144" / 140 C, = 7452.291 lbs/in. ' Cz: single slope roof stiffness Cz =CI * U / b *a / sf from Design of Commercial Post-Frame Buildings p.44 b': roof slope length b'.= 328.877" sf: frame spacing .. -sf= 120"using largest bay for most conservative values CZ = *7452.291 lbs/in. * 328.877" / 140"* 144" / 120"*cos2(18.435°) CZ='18906.837 lbs/in. ... Ch: roof stiffness Ch 2*C2*cos7(8) from Design of Commercial Post-Frame Buildings Ch =2* 18506.837 lbs/in. * cos2(18.435') . Ch = 37813.673 lbs/in. page 33 Step 2: Calculate k, the frame stiffness . where: k: frame stiffness k= 6.*E* I,/ (h2* (.7*d + h)) Skaggs Eqn. 1: E: modulus of elasticity Ci: incising factor Ci= 1 NDS4.3.8 E' = 1700000 psi * 1 E'= 1700000 psi is moment of inertia = b*d3 / 12 NDS Supplement 3.1 ' b: breadth b=4.125.. - d: depth d =7.125.. 1 =4.125"'*7.125"3 / 12 1 = 124.336 in.4 d: embedment depth d =40" k=6 *-1700000:psi ' 124.336 in.4/. (204 z*,(7*40"+204".)) k= 131.3561bs/in. , . - page 34 Step 3: Calculate R, the potential lateral restraining force of the roof diaphragm R: lateral restraining force Rwind = I sf* (h * (wall force - leeward force) * (2.8 * d + 3 * h) / (8 * (.7* d + h)) + hZ* (roof force - roof leeward force))I Skaggs Eqn. 2 Rwind = 1120"* (204"* (11.587 -9.906) psf/ 144 Ps'/psf* (2.8 *40"+ 3 *204" / 8 100 (10 - 0) psf/ 144 Ps'/psf)I Rwind =2258.591 lbs Rcombo= I sf* (h * .75 * (wall force - leeward force) * (2.8 *d + 3 * h) / (8 * (.7 * d + h)) + hz* .75 * (roof. force- roof leeward force))I Skaggs Eqn. 2 Rcombo= 1120"* (204"* (11.587- -9.906) psf / 144 Psi/psf* (2.8 *40:' + 3 *204") / (8* (.7*40" +204")) + 100 * .75 * (10 - 0).psf/ 144 Ps,/Psf)I Rcombo= 1693.943 lbs Rsnow= 1 sf* (h 0* (wall force - leeward force) * (2.8 *d +3 * h) / (8 *.(.7,* d + h)) +hz*0 * (roof force . - roof leeward force))I Skaggs Eqn. 2. .; Rsnow= 1120"* (204"* (11.587 - 9.906) psf./ 144 Ps'/psf.* (2.8 *40" + 3 *204") / (8 * (.7*40 +204")) + (100*0* (10 0) psf / 144 Ps,/psf))1 Rsnow=0 lbs . Step 4:Calculate the diaphragm factor mD Where: frames= 11:. ratio of frame to roof stiffness= 131.356 lbs/in../ 37813.673 lbs/in: ratio =0.003 mD: diaphragm factor mD = 0.955 Skaggs Table page 35 Step 5: Calculate Shear at top of windward post V: shear at the top of post. Vwind= .5 * (Rw'nd *mD +.sf* (h* (wall force - Leeward force) * (2.8 *d,+3 * h) ! (8 * (.7* d + h)),.- hz (roof force-roof Leeward force))) Skaggs Eqn. 3 Vwind= .5 * (2258.59-1.lbs* 0.955 +120"* (204"k (11.587 - 9,906) psf/.144 psil'Psf* (2:8:*40"+3 *204") / (8* (.7*40"+204")) - 100* (10 - 0) psf / 144 P'/psf)) Vwind = 1374,736 lbs Vcombo= .5 * (Rcombo* mD + sf*.(h *..75 * (wall force - leeward force) * (2.8,* d +3 * h) / (8 * (.7*&+ h)) hz* .75*. (roof force roof leeward force)))Skaggs Eqn.. 3 Vcombo= -5:. (1693.943 lbs * 0.955 + 120".*,(204"* :75 * (11.587 - -9.906).psf/ 144 P51/Psf:* (2:$*40"+ 3 204") / (8 * (.7*40"+204")) - 100* .75' (10 - 0) psf/ 144 Ps'/"psf)) Vcombo = 1031.052.lbs Vsnow='•5 * (Rsnow mD +sf.* (h*0* (wall force = Leeward force) * (2.8`* d +3.* h)" (8 * (.7*`d t h))_ hz` *0 * (roof force - roof leeward force))) Skaggs Eqn. 3 Vsnow= .5* (0 lbs`*0.955 + 120"* (204"*0* (11.587.= 9.906) psf / 144 PS'/psf*,(2.8 *40 + 3 *204°') 1 (8 (•7*40"+.204')) - 100-*0 * (10 - 0) psf/.144 P'/Psf)). Vsnow.=0 lbS A =V / k 0= 1374:736 lbs / 131.356 lbs/inch ; A = 3.4" page 36 Step 6: Calculate maximum post moments M, and M2 M,: moment at groundline Mlwind = h * (VWind - ((sf,* (wall force leeward force) * h) / 2)) Skaggs Eqn. 4 Mlwind=204"* (1374.736 lbs - ((120"* (11.587 - -9.906) psf / 144 Ps'/psf*204") / 2)) wi Mlnd = =92231.248 in_lbs . Mlcombo = h * (Vcombo - ((sf* .75 * (wall force - leeward force) * h) / 2)) Skaggs Eqn. 4 Mlcombo= 204"* (1031.052 lbs - ((120"* .75 * (11.587 - -9.906).psf / 144 Ps'/psf*204") / 2)) Mfcombo = -69173.436 in.lbs Mlsnow= h * (Vsnow " ((sf*0* (wall force - leeward force) * h) / 2)) Skaggs Eqn. 4 Mfsnow=204"`' (0 lbs - ((120"*0* (11.587 -9.906).psf / 144 Ps'/Psf*204") / 2)) Mlsnow=0 in.lbS M2: moment above groundline . M2wind=Vwind2 / (2 *sf* (wall force - leeward force)) * 12 Skaggs Eqn. 5 M2wind= (1374.736 lbs)2 / (2* 120".*,(11.587 -.-9.906),psf./ 144 Ps'/ Psf)' M2Mnd= 52760.12 in.lbs M2combo=Vcombo2 / (2'*sf* .75 * (wall force - Leeward force)) * 12 Skaggs Eqn: 5 M2comb6 = (1031.052 lbs)2 (2* 120''* .75 * (11,.587 - -9.906) psf /.144 Ps/psf) M2combo = 39570.09 in.lbs Mlsnow=Vsnow2 /.(2 *sf* 0 * (wall force - leeward force)) * 12 Skaggs Eqn. 5 M2snow= (0 Lbs)2 / (2 * 120"*0 * (11.587-.-9..906) psf./ 144 Ps'/Psf) M2sn6w=0 in.lbs page 37 Step 7: Calculate axial compression force in the post Pf: compressive force post Pfwind =sf*w /.2* ((roof force roof leeward force) + dead load) Pfw;nd= 120"* 300"* ((10 - 0) psf / 144 Ps;/Psf+4.3 psf /,144 P"/psf)+ Pfwind= 3575 lbs Pfcombo= sf*w / 2 * (roof force+ dead load) Pfcombo= 120"*300"* (35.851 psf./ 144 Psi/Psf+4.3 psf / 144 P"/Psf) +. Pfcombo= 10037.851 lbs, . Pfsnow=sf*w/ 2* (snow, roof force+ dead load) Pfsnow 120"* 300".* (37.802 psf / 144 Psi/Psf+4.3 psf/ 144 PS'/Psf) { Pfsnow= 10525.468 lbs Step 8`. Check post adequacy at groundline, . CM: wet service factor. C-M= 1'because"the post is protected by the building thell,'concrete,.or.embedment from excessive moisture CP: column'stability factor CP.= 1..NDS 37. CF: size factor: CF= 1 NDS Supplement.table 4D C;: incising factor C,=.1 NDS4.3.8'..: Co: load duration factor CD= 1.6 NDS 2.3:2 Ct: factor - Cc= 1.NDS43. W. Fc' allowable.compressive force post ; Fc= Fc*CD*CM*Ct*CF.*C;*CPNDS4:3 Fc'.= 1650 psi,*. 1.6 *.1 * 1 Fc'= 2640 psi page 38 fc: axial compressive force post fc= PfAnd / (b *d) from Design of Commercial Post-Frame Buildings p.48, fc= 3575 lbs /..(4.125"*7.125") fc = 121.637 psi fc= Pfcombo /.(b *d) from Design of Commercial Post-Frame Buildings p.48 . fc = 10037.851-lbs / (4.125"*7.12Y) fc = 341.532 psi fc = Pfsnow / (b * d) from Design of Commercial Post-Frame Buildings p.48 fc = 10525.468 lbs / (4.125"*7.125") fc=.358.123 psi CF= 1 NDS Supplement . C�: repetitve member factor Cr=-1.15 NDS.4:3 Cf�: flat Lite factor Cf„ 1.NDS 4.3.. Fe: allowable bending stress post Fb'=.Fe*. Cp*CM.*•Ct.*CL* CF*Cf,,*Ci * Cr NDS 43 Fb'= 1500 psi * 1.6 * 1 * 1 * 1 * 1 * 1 * 1 * 1.15 ,: Fe=2760 psi fb: bending stress post . fb= I Mtwind 1 ./ S NDS 3.3.2 %= 1-92231.248 in.lbs 1 / 34.901 fb=.2642.626 psi fb=.I Mtcombo I / S NDS 3.3 2 fb = 1-69173.436 in.lbs 1 / 34.901 fb= 1981.969 psi fb= IMtsnowl /S.NDS 3.3.2 fb= 10 in.lbs 1 / 34.901 fb= 0Psi page 39 (fcwind / Fc)Z+fbwind / Fb' < 1 (121.637 psi / 2640 psi)z +2642.626 psi / 2760 psi <_ 1. 0.96 <_ 1 (fccombo / Fc)z+ fbcombo / Fb < 1 (341.532 psi./ 2640 psi)z + 1981.969 psi 42760 psi <_ 1 0.735 < 1 (fcsnow / Fc)z+fbsnow / Fb :5 1 '(358.123 psi / 2640 psi)z+0 psi / 2760 psi < 1 0.018 <_ 1 :. this design works at,this point with 4.125"X7.125" posts stressed to 96% Step 9: Check Post Adequacy at Positive Moment Region F, = F,*"CP NDS'4.3 Cp: column stability factor * * z _ CP= (1 + (FCE../ F� )) /.(2 . c) '-f(((1 + (FEE / Ft*)) / (2 c)) . : (FEE / Fc ) / c) FcE= .3..* E / (le:/ d)z NDS 3.7.1 le= Ke* (NDS.Appendix G l=40.8" length of column between lateral supports Ke=0.8NDS3.7:1 le=0.8*.40.8" le.= 32.64" FcE=..822* 1700000 psi/(32.64" / 7.125")z . FcE= 66587.112 psi CP= (1 + (66587.112-psi 12640 psi)) / (2* ..8) r(((1 + (66587.112 psi / 2640 psi)) / (2* .8))2 66587.112 psij 2640psi) / 8 Cp'= 0.992: Fc for.Mz: allowable compressive force post F,for Mz=2618.562 psi page 40 fb: bending stress post fbMnd =M2wind / (b*,d *d / 6) fbvA d= 52760.12 in.lbs / (4.125"*7.125"*7.125" / 6) fbwind = 1511.692 psi fbcombo=M2combo / (b*d *d / 6) fbcombo = 39570.09 in.lbs./ (4.125"*:7.125."*7.125" / 6) fbcombo= 1133.769 psi fbsnow=M2snow / (b* d *d / 6) fb:now=0 in.Lbs / (4.125"*7.125"*,7.125 / 6) fbsnow= 0 psi (fcAnd / Fc')2+fbwiod / (Fb'* (1 ' fcwind / FcE)) 1 (121.637 psi./ 2618.562 psi)2.+.1511.692_psi / (2760,psi * (1 - 121.637 psi /:66587.112 psi)) < 1 0.551 < 1. (fccombo / Fc)2+ fbcombo / (Fb'*.0 fbcombo �.FcE)) < 1 (341.532 psi'/2618.562 psi)2+ 1133.769 psi / (2760-psi * (1 - 341.532 psi / 66587.112 psi)) < 1 0.43 < 1 (fcsnow / Fc)2+fbsnow / (Fb'* (1 " fcsnow FcE)) 1 (358.123 psi / 2618.562 psi)2+ 0 psi / (2760 psi * (1 358.123 psi_/ 66587.112 psi)) < 1 0.019 < 1 this design works at this point with 4.125"X7.125"posts page 41 ' Ground line deflection . Ag: deflection of the post at the ground line Ag=Ogr+Agm Agm: ground line deflection.of the post due to applied moment Agm= 3 *M*I-d2 / (4* E * 1) Anderson equation 34 M=92231.248 in.lbs d: depth . . d =40" . E= 1700000 psi = 124.336 in:4. . Agm = 3 *92231.248�in.lbs*:40"2 / (4* 1700000 psi * 124.336 in. Agm =0.523616" _ Agr: ground line deflection of the post due to shear load Agr=9* R*d3./ (16 * E* 1) Anderson equation,33_ R: shear toad applied to post at ground line. R= 1374.736 Agr=.9* 1374.736 lbs*40"3 / (16 ' 1.700000 psi,'124.336 in.4) Andersoh equation 33 -Agr_=0.23414 Og=0:523616"+0:23414' Ag= 0,757756. Ow 0.757756"< 1 4" w. Post Eccentricity: (fc Fc)2+ (fb' +.fc_* (6 el / d)-* (1 + .234* (f� / FcEt))) / (Fb, * (I.'- (fc /•FcE,))) +..�(fb2+fc* (6 *.eZ / b)_*. (1,+ .234* (f,./ FcE2) + .234* ((fbi +fc* (6,*elf d)) / FbE)Z)) ( (Fb2'* (1:- (fc FcE2).-.((fbj +fc* (6. d)) / FbE)2)) ` 1 page 42 el: eccentricity, measured parallel to wide face from centerline of column to centerline of axial load el =0.563" e2: eccentricity, measured parallel to narrow face from centerline of column to centerline of axial load e2= 0" FcEt = .822 Emin' / (le / b)2 Emi,;: reference adjusted modulus of elasticity NDS Supplement Emin= 580000 psi FcE1 = .822* 580000 psi / (32._64" / 7.125")2 FcEi =22717.956 psi FcE2= .822 ` Emia / (le / b)Z FcE2= .822* 580000 psi / (32.64" / 4.125")2 FcE2=7614.606 psi FbE= 1.2 * Emin'./ RbE2 RB: slenderness ratio of bending member with a max of 50 NDS 3.3.3 RB =1(32.64 *4.125" / 7.125,.2) RB = 1.629 .. FbE= 1.2* 580000 psi /"1.629�. FeE=262424.799 psi 2"+.(1511.692 psi + 121'.637 psi * (6 *"0.563" /4.125") * (1.+ .234 * (121.637 psi / 22717.956 psi))) / (2760 psi * (1 (121.6.37 psi / 22717.956 psi))) + (0 + 121.637 psi * (6 *0" / 7.125") *(1 +".234* (121.637 . psi / 7614.606 psi) +..234* ((1511,692 psi+ 121.637.psi * (6 *0.563" / 4.125")) / 262424.799 psi)?)) / 2760 psi * 1 121.637 psi'/7614.606psi) 2760 psi + 121.637 psi * 6 *0.563" / 4.125" l . 262424.799 psi)2)).< 1 0.58912 < 1 " Trusses are notched into.columns at least 3 Lateral Strength Assessment b =2.5' Pu,- / fL> P,,ASD ASAE 486.2 11.3.1 Pz.ASD=283.333"lbs p,,,Z= ultimate lateral soil resistance at depth z z= 3.333' .: page 43 p,,,Z= 3 *6,,Z* KP ASAE 486.2 11.3.1 a,,;Z= effective vertical stress G.=y*Z d,Z= 105* 3.333 cy, = 350 psf. K - 1 +sin / 1 _sin ASAE EP 486.2 11.2.1 co =soil friction angle co = 30' ASAE_EP486.2 table 1. KP=.(1_+sin(30)) / (1 sin(30)) KP = 3. p;,Z=3 * 350*3 P,,,Z=3150 psf f�= 1.4 / (:61.-..01 * tp) ASAE EP486:2:table 4 fL=4.516 3150 /4.516 >-281333 ASAE 486.2 11.3.1 :..Lateral Strength adequate,for 40"deep and 2.5'.diameter hole. Determine if.the diameter of the concrete column encasement is adequate A?fB* Paso / (qB.=:Y;*.dF) ASAE EP486:2 10:2 A= bearing area. A._-rr*rz A =.4.909 ft: bay weight=wall steel'weight +post weight+ girt weight+ roof weight+skirt board weight' bay weight (.66 * (sf (h + panel length))) + (62.4* .55 *.(b*d *.l)) + (62.4:* .55 *,(b,* d * l) * 6).+'.6 (BCDL+TCDL) +,(62.4*:.55 * (b * d * l) * n) bay weight-;(:66 psf* (10';* (17'+27.406))) + (62 4 pcf*.55 lbs/ft 3 * ((4.125'./ 12 '"/ft.) * (7.125 / 12 ".'/ft.)* 19.667')) t(62:4 pcf*.55 lbs/ft:3 * ((1. 5 / 12 '"/-);* (9.25"/;12 mft_) * 10') * 1).+ fi*>(1 + 3.3) * .10.*.27.406 + (62.4.pcf:*...55_lbs/ft.3 *. ((1:5" / 12 m/it.).* (1.25"1 12 inV t.)`10.)). bay weight=996.743 lbs Paso = (bay weight.+ Pf): PAso = (10525.468.+ 996.743)' - PAso = 11522.211 lbs qB =ultimate soil.bearing capacity fB ASD factor of safety for bearing strength assessment fB =-1.4 / (.58 - :01 * co) ASAE EP486.2 table 2 % 5. qB =.y* (:5 * B *CWj* Ny*S +dF* C * Nq* dq*sq) ASAEEP486:210.4.1 page 44 y= moist unit weight of soil y= 105 pcf ASAE EP486.2 table 1 dF= 3.333' CWi = 1 ASAE EP486.2 10.4 C,n= 1 ASAE EP486.2 10.4 NY= 22.4 ASAE EP486.2 table 6 - sY=0.6 ASAE EP486.2 table 6 Nq= 18.4 ASAE EP486.2 table 6 dq= 1 +2 *tan(cp) * (1 - sinV))z*tan- (dF / B)ASAE EP486.2 10.4.1 dq = 1 +2*0.577* 0 - 0.5) *tan-1(3.333 / 2.5) dq= 1.2676870718657418 . sq= 1.58 ASAE EP486.2 table 6 q6 = 105 * (.5 *2.5 * 1 *22.4*0.6 + 3.333 '.1 * 18.4* 1.2676870718657418 ' 1.58) qB= 14662.969493648296 psf 4.909 >_ 5 * 11522.211 / (14662.969493648296 - 105 *3.333) 4.909 >_4.025 ::'Diameter 2.5' adequate for foundation pressure page 45 COLUMN UPLIFT CALCULATIONS g*MF+ U / f6 ASAE EP486.2 12.3 . g gravitation acceleration constant g= 1 ASAE EP486.2 4 PASD = uplift+ dead load uplift force = roof-wind-force* roof area uplift force =.-14.474 psf*263.523 ft.z uplift force='-3814.317 lbs PASD = -2817.574 lbs MF=foundation mass. MF=C*n* B,2*t - :. C: concrete density. C = 150.pcf-.. B,,: collar width B„ =2.5' t: collar thickness or thickness of attached concretepq t 1.5' MF= 150*n*2.5z* 15;. MF=4392.352 lbs AP =cross-sectional area of post .A- = 0.344'*0.594'=0'.204 ft.z. U = ultimate uplift resistance due to soil:mass U =y*Au* (TT*.d6*.sF:* B;, *K„*tan /:2.+ B„z*rr / 4- ):ASAE EP486 2.12 5:1 SF'= 1 + 1:105 *-10 s �zsis d,;,/ Bb ASAE EP.486.2 12.5 1 SF=.1 .+:00001.105 *.302.811 *:1:833 / 2.5 K°_ .95 ASAE EP486 2 12 5 1 U = 105 * 1. 3,* (n* 1..833 *`1.117*.2 5 * .95 *tan(30) / 2 +2.52 83 *.Tr / 4 0.204).. . *.(p) ASAE EP486.2 table 5 f = 0'5922 1 *4392352.lbs t 1754.430313 1 900447 lbs / 0.5922 2:2817.574 Ibs 7354.916 >_ 2817.574 In summary a 4 125 X7.125".sidewall:column`, in an augered hole.2.5':in:diameter and 40"in.depth'is adequate. A.concrete bottom collar 1.5'.high mono-poured,with,a 8':footing below.the column is . required Checking Adequacy-of:concrete under pole'using formulas by Or. Jeff R. Filler; P.E.;-Moscow, ID FV= 2*f(F�).=2.*f(2500) = 100 psi fV=V / (bo.*d) ; bo=.4* (d +a) bo=4.*(7.125 + 8) bo= 60.5..in. fV 10525.468;1 (60.5 *:8) . f� .21.747 psi :21.747:_ 100 page 46 Front Endwall Columns Calculate maximum post moments M M: moment M=wind force*spacing * h2 / 8 h: distance from either grade or top of concrete slab,to bottom chord of end truss h = 188.358" - M= (9.65 - -7.819) psf / 144 in.2/ft.2* 112.375"* 188.358"2_/ 8 ' M= 60457.85 in.Lbs M is shared.by two endwalls so the load on one endwall can be cut in half. S: section modulus . S= b*d*d / 6 NDS 3.3.2 S = 5.5"* 5.5"* 5.5" / 6 S 27.729 in.3 fb: bending stress post fb= IMI / S NDS 3.3.2 J6= 130228.925in.lbs 1 / 27.729 in s fb= 1090.149,psi' . .Cm wet service factor CM= 1 because the post is protected by the building shelf, concrete, or embedment from excessive moisture CF= 1 NDS Supplement.' ; C,: repetitve member factor . Cr= 1 NDS 43 Cf,,: flat use factor _ - Cf,;= 1 NDS4.3. .. Fe: allowable bending stress post Fb'- Fb*Cp*CM*Ct* Ci*CF*Cfu*C;*C,ND543 Fb = 850psi * 1.6*.1 * 1 * 1 * 1 * 1.* 1 * 1 Fb= 1360 psi fb z Fe 1090.149 psi <_.1360 psi. _.. page 47 Calculate axial compression force in the post Pf: compressive force post Pf= sf*w / 2*-(snow roof force+dead load) Pf= 60"* 112.37499999999999"* (37.802 psf / 144 Ps'/Psf+4.3,-psf % 144 Ps'/Psf) + Pf= 1971.332 lbs CF: size factor CF= 1 NDS Supplement table 4D CP: column stability"factor. CP = (1 + (FcE /,F,:*)) / (2*c) --MO +.(FcE / F�*)) / (2*c))Z - (FcE / Fc*) / c) FcE'= .3 * E / (Le /.d)2NDS3.7.1 le= Ke* l NDS.Appendix G l=40.8" length of column between lateral supports Ke= 0.8.NDS3.7.1 1e=0.8 *40.8" le= 32.64" FcE= .822* 1200000 psi / (32.64. / 5.5")2.. FEE=28007.745.psi .. . CP= (1 + (28007.745 psi % NaN psi)) / (2* .8) I(((1:+ (28007.745 psi L NaN.psi)) /.(2 *...8))z (28007..745 psi / NaN psi)./..8) CP = NaN F` _.NaN psi fc: axial compressive force post f,= Pf / (b*d) from Design of Commercial Post-Frame Buildings p:48 fc= 1971.332 lbs./ (5.5"* 5.5") fc= 65.1.68..psi Calculate-Shear at'top of e, ndwall post V: shear.at the..top of post .. V= 5%8 (wall force - leeward force).*"spacing *"h V 5/8 (9.65- 7.819) psf/ 144 Ps'/P sf*:112.37499999.999.999"* 188.358 , V= 1604.8671bs fv: bending shear post f„ 3,* 1604.867 lbs / (2 k 5.5"* 5.57) ' fv=19.58 psi F,;: allowable shear.stress post Fv = F�*Cp*CM* Ct*CiNDS4.3 F, = 206.25 psi.* 1.6 * 1.* 1 * I F, =264 psi page 48 79.58 psi <_264 psi Lateral Strength Assessment b = 1.5' Pu,z / fL? PZ,ASD ASAE 486.2 11.3.1 Pz,ASD =283.333 lbs p,,,,= ultimate Lateral soil resistance at depth z . z= 3.333' P.,Z= 3 *dv Z* KP ASAE 486.2 11.3.1 c3,,,,=effective.vertical stress 6'v,Z= 105 * 3.333 cyv,Z= 350 psf KP = (1 +.sin(tp)) / (1 ` sin(o))ASAE EP 486.2 11 2 1 tp =soil friction angle tp.= 30° ASAE EP486.2 tablet KP = (1 +sin(30))./ (1 =.sin(30)) KP.= 3. , Pu.Z= 3 * 350*3 P.,i= 3150 psf fL= 1.4 / (.61 - .01 * cp) ASAE EP486.2 table 4 fL=4.51.6 - 3150 /4.516 >_2-83.333 ASAE 486.2 11.3.1 ' :. Lateral Strength for 40"deep and.1.T diameterhole Determine if the diameter:of the concrete column encasement is adequate. A> f * P ' B ASD (q8 " Y*dF) ASAE.EP486.2:10.2 A= bearing area A= rr* rz A'= 1.767 ft bay weight=wall steel weight'+ post weight:+girt weight+ roof weight.+skirt board weight bay weight= (.66 * (sf* (h + panel length))) + (62.4* .55 * (b_*d * l)) +:(62.4.* .5.5 *,(b*.d * l) n) + .6 *: (BCDL+TCDL).+ (62.4* .55 * (b*d * L)`*n) bay weight (.66 psf* (10'* (17'+ 50'))) + (62.4 pcf .55 lbs/ft 3 * ((5 5' /:12 '" /ft.) * (5.5" / 12 i°'/ft) 19.667)) + (62.4 pcf* .55.Lbs/ft.3* ((1.5",/ 12 `/ff.)* (9..25" / 12:"'/tt.) * 10') *.1)t:.6_* (1 + 3 3)..* 10 *: 27.406 + (62.4 pcf*.55 lbs/ft.3.* ((1.5" / 12 in./ft.):* (7:25" / 12.m. bay weight,= 1000.771 lbs PASD = (bay weight + Pf) PAS = (1769.994 + 1000.771) PASD =2770.765 lbs qB ultimate soil bearing capacity. page 49 fB =ASD factor of safety for bearing strength assessment fB = 1.4 / (.58 - .01 * cp) ASAE EP486.2 table 2 fB = 5 qB =Y* (•5 * B *Cw, * NY*SY+ dF* Cwz* Nq*dq*sq).ASAE,EP486.2 10.4.1 y= moist unit weight of soil y= 105 pcf ASAE E0486.2 table 1 dF.= 3.333' Cwi = 1 ASAE EP486.2 10.4 Cwz=.1 ASAE EP486.2 10.4 NY=22A ASAE EP486.2 table 6 sy=0.6 ASAE EP486.2 table 6 Nq= 18.4 ASAE EP486.2 table 6 dq= 1 +2*tan(�p)* (1 -sin(cp))z.*tan-1(dF/ B)ASAE.EP486.2.10.4.1 dq = 1 +2*0.577* (1 70 5)2*tan''(3.333 %.1.5) dq = 1.3313824270181827. Sq= 1.58 ASAE EP486.2.table 6 qB= 105 * (:5;* 1.5 * 1,*214.*0.6 +3.333 * 1:* 18.4*'1.33.1 M24270181827*,1.58) : qB =.14605.482471395411 psf - 1.767 >_ 5 *2770.765 /_(14605.482471395411 -.10.5 * 3.333) 1.767 >_0'.972 Diameter.1.5'adequate for foundation pressure page.50 COLUMN UPLIFT CALCULATIONS d =40" U: uplift resistance U = a* (.33 *ir* (((d - t) + .5 *w/ tan(0))3*fan(0)Z - .125 *w3 / tan(0)) -AP* (d.- t)) + .25 * C*rr*w� *t a:-soil density a =85,pcf C: concrete density C = 150pcf t, h: collar thickness or thickness of attached concrete t, h = 1.5' w: collar width w= 1.5' 0: soil friction angle 0 = 0.524 radians AP: post cross section area AP= 0.21 ft.Z U = 85 pcf" (.33_* 3.14159* (�(((32" /.12 ' /ft.) - 1.5') + .5 * 1.5 / tan(0.524°}}3*tan(0.524 }Z .125 1.5" / tan(0.524°)) - 0.21 ft...* ((32 / 12'"-/ft.) - 1.T)) + .25 *.150 pcf.*,3.14159 * 1.5'2* 1.5` U = 752.721..1bs . total uplift resistance = 1000.771 lbs + 752.721 lbs = 1753.492 lbs uplift force= roof_wind_force* roof area / DOL uplift force= 14.474 psf*.46'823 ft.Z 1.1:6 uplift force =.464.353.lbs Total uplift force of 464.353 lbs is less than total uplift resistance of.7.53.497 lbs The direct bearing capacity under vertical load depends on the bearing capacity of the soil against the end of the post and the frictional resistance between the post and the'soil, without producing.much . ; end bearing pressure at the butt. Interestingly, it has been determined that firm backfilling of the,hole with compacted soil, sand, soil concrete.or concrete is a more effective way of increasing vertical' force resistance than placing mats of footings of concrete beneath the butt ends. Concrete encasement of the post in the ground contact area enlarges the friction surface and can generally be credited with a wood to concrete bond strength of 30 psi. Hoyle Et Woeste, p.335-336 page 51 . Connection Strength = (breadth *2 + depth.*2) *collar thickness* 30 Connection Strength = (5.5"*2 + 5.5"*2) * 10"* 30 psi =.6600 lbs Connection Strength =6600 lbs > 464.353 lbs In summary a 5.5"X5.5"SYP endwall column, in'an.augered hole 1.5'in.diameter and 40"in depth is adequate. A concrete bottom collar 1.5'.high mono-poured with a 8."footing below.the column.is.. required page 52 Front Endwall Columns Calculate maximum post moments M M: moment M=wind force *spacing * h2 / 8 h: distance from either grade or top of concrete slab, to bottom chord of end truss h = 188.358" ,. M= (9.65 - -7.819) psf / 144 in.2/ft.1 * 123.5"* 188.358"2 / 8 M= 66443.11 in.lbs M is shared by two endwalls so the load on one endwail-can be cut in half. S: section modulus S= b* d *d / 6 NDS 3.3.2 S =5.5"* 5.5"* 5.5" / 6 .: S.=27.729 in.3 :. fb: bending stress post A= IMI / S NDS 3.3.2 fb 133221.555in.lbs1 / 27.729 in.' fb= 1198.073 psi Cm=wet service factor. CM = 1 because the post is protected-by the building shell, concrete;or embedment from excessive. moisture CF= 1 NDS Supplement C,: repetitve member factor' . . .. Cr= 1 NDS 4.3 Cf,,: flat use factor. Cf,, = 1 NDS4.3 ., Fb: allowable bending stress post Fe = Fe*Cp* CM*CC'CL* CF*Cfu* C;*C,NDS 4.3 Fe=.850psi * 1.6*1 * 1 * 1 *.1 * 1 *1 *.1 Fe.= 1360 psi fb < Fe 1198.073 psi :5 1360 psi . page 53 Calculate axial compression force in the post " Pf: compressive force post Pf= sf*w / 2 * (snow roof force + dead load) Pf= 60E-* 123.5 *:(37.802 psf / 144 P"/Psf+4.3 psf / 144 PS'/PSf) + Pf=2166.492 lbs CF: size factor CF= 1 NDS Supplement table 4D CP: column stability factor CP = (1 +,(FcE / Fr*)) / (2 * c) -f(((1 + (FEE/ Fc*)) /.(2 c))Z - (FEE`/ F�*) / c) FEE = .3 * E / (le / d)2 NDS 3.7.1 le.= Ke* l NDS Appendix G l=40.8" length of column between lateral supports . Ke= 0.8 NDS 3.7.1 1e.= 0.8 *40.8 le= 32.64'.' FEE_ .822* 1200000 psi / (32.64" / 5.5")z Fc,=28607.745 psi CP= (1 + (28007.745`psi / NaN psi)) /_(2.* .8) =f(((1 + (28007:745 psi / NaN psi)) / (2 * :8))z (28007:745.psi'/ NaN psi) / .8) CP= NaN F�= NaN psi f,: axial compressive force post. fc= Pf/ (b* d) from.Design of Commercial Post-Frame Buildings p.48 fC = 2166.492 lbs / (5.5"* 5 5") f�= 71.62 psi Calculate.Shear at top of endwall post V: shear at the top of post . V= 5/8 (wall force - leeward force) *spacing* h . V= 5/8.(9.65 7.819) psf /.144 P"/P5f*, 123.5"* 188.358" V.= 1763.747 lbs_ fv: bending shear post.. fv..= 3 *V_/ (2* b * d) f,= 3 * 1763.747 lbs/ (2* 5.5" * 5.5") f°="87.459 psi Fv: allowable shearstress post F,;= F *CD*CM*Ct.*Q NDS 4.3 Fv'= 206.25 psi'*'1.6.*.1 * 1..* 1 F,;=264 psi page 54 f„< F� 87.459 psi <_264 psi - Lateral Strength Assessment P.,Z / fL?PZ,ASD ASAE 486.2 11.3.1 PZ,ASD =283.333 lbs p,,,Z= ultimate lateral soil resistance at depth z z= 3.333' p.,Z= 3 *d,,Z* KP ASAE 486.2 11.3.1 d Z= effective vertical stress . (YV,Z=y.*Z GV,Z= 105 *3.333 6v,Z= 350 psf KP= (1 +sin(cp)) / (1 sin((p))ASAE EP 486.2 11 2.1 cp =soil friction angle cp= 30' ASAE EP486.2 table 1 Kp = (1.+ sin(30)) / (1 sin(30)) KP.= 3 Pu,z= 3 * 350 '3. R�,Z= 3150 psf fr= 1.4 / (.61 .01 * cp) ASAE EP486.2 table 4 fi-=4.516 3150 14.516 >_283.333 ASAE 486.2 11.3.1 :. Lateral Strength adequate for-40"deep.and I.T.diameter hole Determine if the diameter of-the concrete column encasement is adequate A? fB* PASD /.(qB - y* dF):ASAE EP486.2 10.2 A= bearing area . A= rr*r2 - A = 1.767 ft: bay weight "wall steel weight+post weight+girt weight+ roof weight+skirt.board weight bay weight= (.66 * (sf* (h + panel length))) + (62.4* .55.* (b * d * l)) + (62.4* .55* (b*d * lj,*.n) + .6 (BCDL+TCDL) +.-(62.4* .55 * (b.*d * l) * n) bay weight= (.66 psf* (10' * (17'+ 50'-))) + (62.4 pcf. .55 lbs./ft.3 * ((5 5,, % 12 in./, * (5.5`' / 12 '"'/ft) 19.667')).+ (62.4 pcf* .55 lbs/ft.3 * ((.1.5"/ 12 'n./ft.) * (9.25 % 12 in./tt:) * 10') *.1) + .6 * (1 + 3.3) * 10*. 27.406 + (62.4 pcf* .55 lbs/ft."* ((1.5 / 12 in. * (7.25" / 12 m./ft.) ' 10')) bay weight'= 1000.771 lbs PAsD = (bay weight + Pf) PASD = (1945.221 + 1000.771) PASD= 2945.992 lbs qB = ultimate soil bearing capacity page 55 fB =ASD factor of safety for bearing strength assessment fB = 1.4 / (.58 - .01 * cp) ASAE EP486.2 table 2 fB = 5 qB =y* (.5* B * Cwl * Ny*sY+ dF* CWZ* Nq*dq *sq) ASAE EP486.210.4.1 y= moist unit weight of soil y= 105 pcf ASAE EP486.2 table 1 dF= 3.333` Cw, = 1 ASAE EP486.2 10.4 - CwZ= 1 ASAE EP486.2 10.4 NY=22.4 ASAE EP486.2 table 6 sy= 0.6 ASAE EP486.2 table 6 Nq = 18.4 ASAE EP486.2 table 6 dq = 1 +2*_tan((p)* (1 sin�cp))Z*tan-'(dF / B) ASAE EP486.2 10.4.1 dq =-1 +2 *-0.577* (1 -0.5) *tan-'(3J33 / 1.5) dq = 1.3313824270181827 sq= 1.58 ASAE EP486.2 table 6 qB 105 * (.5 * 1.5 * 1 *22.4*0.6 + 3:333 * 1 * 18.4* 1:3313824270181827* 1.58) qB = 1460.5.482471395411 psf 1.767 >_ 5 *2945.992 / (1460.5.48247139.5411 105 *3.333). 1.767>_ 1.033 . Diameter 1.5'_adequate for foundation pressure .`` page 56 COLUMN UPLIFT CALCULATIONS d =40" U: uplift resistance U = a* (.33 *rr* (((d - t) +-.5 *w / tan(A))3 *tan(6)Z - 125 *w3 / tan(6)) 7AP* (d - t)) + .25 * C *Trfew� *t a: soil density a= 85 pcf . . C: concrete density - C = 150 pcf t, h: collar thickness or thickness of attached concrete t, h = 1.T: . w: collar width w= 1.5' 6: soil friction angle 0 =0.524 radians AP: post cross section area AP = 0:21 ft.2 ; U.= 85 pcf* (.33 *3.14159 * ((((32" /92 '"'/ft.) — 1.5') + .5 *.1.5' /'tan(0:524°))3 *tan(0.524 )Z -`A25 1.5'3 / tan(0.524°)) -_0.21.ft.2* ((32" / 12 /ft_) 1.5')) + .25 150 pcf* 3.14159 ` 1.5'2* 1.5' U = 752.721 lbs . total uplift resistance = 1000.771 lbs + 752.721 lbs= 1753.492 lbs uplift force= roof wind force.* roof area / DOL' uplift force 14.474 psf* 51.458 ft.Z % 1.6 uplift force=510.324 lbs. Total uplift force of 510.324 lbs'is less than total uplift resistance of-1753.492 lbs The direct bearing capacity under vertical load depends on the bearing capacity of the soil against the end of the post-and the frictional resistance between the post and the soil, without producing much end bearing,pressure at the butt. Interestingly, it.has been determined that firm backfilling of the hole with compacted soil, sand, soil concrete or concrete is a more effective way of increasing:,vertical force resistance than placing mats of footings of concrete beneath the butt.ends.Concrete encasement of the.post in the ground contact area enlarges.the friction surface and can generally be credited with a Wood,to concrete bond strength of 30 psi. Hoyle it Woeste, p.335-336 page 57. Connection Strength = (breadth *2 +depth *2) *collar thickness `30 Connection Strength = (5.5"*2 + 5.5"*2) * 10"* 30 psi = 6600 lbs Connection Strength= 6600.lbs > 510.324 lbs In summary a 5.5"X5.5"SYF endwall column, in an augered hole 1.S in diameter and 40"in depth is adequate.-A concrete bottom-collar 1.5'high mono-poured with a 8"footing below the column is . required . page 58 Front Endwall Columns Calculate maximum post moments M - M: moment M=wind force*spacing* hz / 8 h: distance from either grade or top of concrete slab, to bottom chord of end truss . h = 188.358" M= (9.65 - -7.819) psf / 144 in.z/ft.z` 133.625"* 188.358"2 / 8 M= 71890.369 in.lbs M is shared by two endwalls so the toad:on one endwall can be cut,in half. S: section modulus S= b*d *d / 6 NDS 3.3.2 S= 5.5"* 5.5"* 5.5" / 6 S =27,729 in.3. , fb: bending stress post fb= IML / S NDS 3.3.2 fb= 135945.184in.lbs 1 / 27.729 in.3 fb=.1296.295 psi CM Wet service factor CM= 1 because the post is protected by the building shell, concrete, or embedment from excessive moisture. CF= 1 NDS Supplement . Cr: repetitve member factor Cr= 1 NDS 4.3 Cfu: flat use factor G,,= 1 NDSA3 Fe: allowable bending stress post F,;= Fb*.CD*CM*Ct*CL* CF*Cf,, Ci* C,NDS 4.3 Fe= 850psi *.1.6 * 1 * 1 * 1 * 1 * 1 * 1 * 1 Fe=1360 psi fb < Fe . 1296.295 psi <_ 1360 psi page 59 Calculate axial compression force in the post Pf: compressive force post- Pf=sf*w / 2* (snow roof force+ dead load) " Pf=60"* 1-33.625"* (37.802 psf / 144 P"/PSf+4.3 psf./ 144�P51/P:f) Pf=2344.109 lbs . CF: size factor CF=. 1 NDS Supplement table 4D CP: column stability factor . CP = (1 + (FEE / F�*)) / (2 *c) -t(((1 + (FEE/. F�:*))./ (2 * o))Z - (FEE / Fc*) / c) FcE= .3 * E / (le / d). NDS3J.1 le= Ke.* l NDS Appendix G l=40.8" length of column between lateral.supports. Ke= 0.8 NDS 3.7.1 IL = 0.8 *40.8" 1e= 32.64"_ FcE .822* 1200000 psi / (32.64" / 5.5")2 FEE =:28007.745 psi. . CP = (1 + (28007.745 psi'/ NaN psi)) / (2 .8) - I(((1 +:(28007 745 psi/ NaN psi)) / (2*..8))2 (28007.745 psi / NaN psi) / .8) Cp= NaN Fc= NaN psi. fc: axial compressive force post. fc=Pf /:(b*d).from Design of.Commercial Post-Frame Buildings p:48 f,= 2344.109.1bs / (5.5"*.5.5') f�=77.491 Calculate Shear at top of eridwall post V: shear at the;top of post - . V= 5/8 (wall force -leeward force) spacing * h V= 5/8 (9:65 - 7.819) psf'/ 144 P"/" * 133.625"*-188.358 psf V= 1908:346 1bs .. fr: bending shear post f,= 3.* 1908.346.lbs / (2* 5 5'* 5.5") f„= 94.629 psi . F�::allowable shear stress post .• F,;= Fv* Co*.CM* Ci*C; NDS 4.3 . Fv' 206.25 psi * 1.6* 1 *1 * 1 . Fv =264 psi page 60 94.629 psi <_.264 psi Lateral Strength Assessment .. P.,= / fL? PZ,ASD ASAE 486.2 11.3.1 PZ,ASD=283.333 lbs p,,,Z=ultimate lateral soil resistance at depth z z= 3.333' p.,z= 3 * a,Z* Kp ASAE 486.2 11.3.1 a' Z=effective vertical stress ov"=y*z 6v,z= 105* 3.333 .- a'v,Z= 350 psf Kp= (1 + sin(cp)) / (1 sin(cp))ASAE EP 486.2 11.2.1 cp=soil friction angle to = 30 ASAE EP486.2 table 1 Kp = (1 +sin(30)) I (1 - sin(30)) KP= 3 = 3 350* 3 P,,,Z'= 3150 psf fC= 1.4 / (.61 - .01 * fp) ASAE EP486.2 table 4 fL=4.516 3150 / 4.516 _>283.333 ASAE 486.2.11.3.1 :. Lateral Strength adequate for 40"deep and 1.T diameter hole Determine if the diameter of the concrete column encasement is adequate: . A >-fe*PASD;/ (qB "Y*dF)ASAE EP486.2.10:2 A= bearing area, A= n* r2 . A=.1.767 ff. bay weight=wall steel weight+ post weight + girt weight+ roof weight+skirt board weight bay weight = (:66 * (sf* (h + panel length'))) + (62.4*..55 * (b *V 1))+ (62.4 * .55 * (b* d ' 1) * n) + :6 (BCDL+TCDL) + (62.4* .55* (b* d * l) * n). bay weight= (.66 psf* (10'* (17+ 50'))) -F (62.4 pcf* .55 lbs/ft.3 *((5 5"% 12' /ft) * (5.5 / 12'" /ft) *. 19.667)) + (62.4 pcf* .55 lbs/ft'.3* ((1..5" / 12 in. /ft.) * (9.25" / 12 /ft.) *10-) * 1) + .6 * 0 +:3.3)*.10 27.406 + (62.4 pcf* .551bs/ft.3.' ((1.5" / 12 `Vft.) * (7.25" 1.12 "/ft.) *:10-)) bay weight=-1000.771 lbs.. PASD = (bay weight + Pf) PASD• (2104.698 +.1000.771) PAso= 3105.469 lbs qB = ultimate soit.bearing capacity page 61 fB =ASD factor of safety for bearing strength assessment .fB= 1.4 / (.58 - .01 * cp) ASAE EP486.2 table 2 fB = 5 . qB =y* 0 * B * CW, * NY*sY+dF*Cw2 Nq*dq*sq) ASAE EP486.2 10.4.1 . y= moist unit weight of soil Y= 105 pcf_ASAE EP486.2 able 1 dF= 3.333' Cw1 = 1 ASAE EP486.2 10.4 Cw2= 1 ASAE EP486.2 10.4 NY=22.4 ASAE EP486.2 table 6 sY= 0.6 ASAE EP486.2 table 6 Nq = 18.4`ASAE EP486.2 table 6 dq= 1 + 2 *tan(cp).* (1 sin((p))Z* tan'(dF/ B) ASAE EP486.2 10.4.1 dq= 1 + 2*0.577* 0 - 0.5)2* tan-'(3.333 / 1.5) dq= 1.3313824270181,827 sq= 1.58 ASAE.EP486.2 table 6 qB = 105* (.5 * 1.5 - 1 *22.4* 0.6 +.3.333 * 1 * 18.4*.1:3313824270181827* 1.58); qB = 14605.482471395411 psf. 1.767_2: 5`3105.469 / (14605.482471395411.- 105* 3.333) 1.767_> 1.089 :: Diameter 1.5'.adequate for foundation pressure page 62 COLUMN UPLIFT CALCULATIONS d =40" U: uplift resistance - U = a.* (.33 * Tr * (((d t) + .5 *w / tan(6))3*tan(0)2 125 *w3 / tan(6)) AP* (d - t)) + .25 *C *rr *wz *t a: soil density a= 85 pcf' C: concrete density - C = 150 pcf. t, h: collar thickness or thickness of.attached concrete t, h = 1.5' . w: collar width w= 1.5' 0: soil friction angle 6 = 0.524 radians AP: post cross section area AP= 0.21 ft.z U = 85 pcf* (.33 *3.14159* (z(((32"/ 12 i /f�.) - 1.5') + .5.* 1:5' /-tan(0.524°))3*.tan(0.524 )z 925 1.5'- /tan(0.524°)) - 0.21 ft. ((32" / 12 /ft.) - 1.5')) + .25 150 pcf 3.14159 1.5' 1.5' - U = 752.7.21 lbs total uplift resistance =,1000.771 lbs +.752.721 lbs = 1753.492 lbs ' - ,uplift force = roof wind_force*.roof_area./ DOL uplift force = 14.474 psf* 55.677 ft.? /.1.6 uplift force =_552.162 lbs Total uplift force of.552.162 lbs is less than total uplift resistance of.1753.492 lbs The direct bearing capacity under vertical load depends on the bearing capacity of the soil against the end of the post and the frictional resistance between the post acid the soil,without producing much . end bearing-pressure at the butt. Interestingly,.it has been determined that firm backfilling of the.hole, with compacted soil, sand, soil concrete or concrete.is a more effective.way of increasing vertical force resistance than placing.mats of footings of concrete.beneath the butt ends. Concrete encasement of the post in.the ground contact area enlarges the friction surface and can generally be credited with a wood to concrete bond strength of 30 psi. Hoyle EtWoeste, p.335=336 page 63 Connection Strength= (breadth,*2 + depth *2) *collar thickness *30 Connection Strength = (5.5"*2 + 5.5"*2) * 10"*30 psi = 6600 lbs Connection Strength.= 6600 lbs > 552.162 lbs In summary a 5.5"X5.5"SYP endwall column, in an augered hole 1.5'.in diameter and 40"in depth is adequate:.A concrete bottom collar 1.5'high mono-poured with a 8"footing below the column is required. page 64 Front Endwall Columns , Calculate maximum post moments M M: moment M=wind force.*spacing* hz / 8 h: distance from either grade or top of concrete slab, to bottom chord of end truss h = 188.358" M= (9.65 7.819) psf / 144 in..2lft.1* 112.375"* 188.358"2 ! 8 M= 60457.85 in.lbs . M is shared by two endwalls so the load on one endwall can be cut in half. S: section modulus S= b* d" d / 6 NDS 3.3.2 S= 5.51,* 5.5"*.5.5",! 6..;; " S =27.729 in.3.. . fb: bending stress post %="IMI / S NDS 3.3.2. fb= 130228.925in.lbs 1 "/ 27.729 in.3 fb.= 1090:149,psi "CM=wet-service.factor CM= 1'because the post is protected by the building shell, concrete, or embedment from excessive moisture CF= 1 NDS Supplement C,: repetitve member factor.,:. C,= 1 NDS 4.3 .` Cf,,: flat.use factor . CfU= 1 NDS 4.3 Fe: allowable bending stress post F6= Fb*Cp*Cµ*Ct*CL* CF*Cfy Ci* C,NDS 43 :: - Fb= 850psi-* 1.6 * 1 * 1 * 1 * 1,*._1.* 1:* 1 Fe.= 1360 psi fe < Fe 1090.149 psi < 1360 psi page 65 Calculate axial compression force in the post Pf: compressive force post Pf=sf*w / 2* (snow roof force+.dead load) Pf:=-60 *.412.-37500U00000003"* (37.802 psf /.144 PS'/P5f+4.3 psf / 1.44 P5'/psf) + Pf= 1971.332 lbs CF: size factor CF= 1 NDS Supplement table 4D CP: column stability factor CP= 0 + (F�e / F�*)) / (2 *c) - r(((l + (FcE / Fr*)) / (2, c))Z (F�E,L.F,*) / c)' FEE •3 * E / (le / d)Z.NDS.3.7.1 le= Ke* l NDS Appendix G l=40.8" length of column between lateral supports Ke=0.8 NDS 3.7.1 . 1e= 32.64" FEE=".8.22 * 1200000 psi / (32.64" / 5.5"). Fc =28007.745 psi CP_=(1 + (28007.745 psi / NaN psi)) / (2* 8.) =. r(((l + (28007.745 psi% NaN psi)) %,(2.* .8))Z 4. (28007.745 psi / NaN,psi) '/ :8) Cp = NaN Fc'=.NaN psi =f,: axial"compressive force post f - P /. b *d from Design of Commercial,Post-Frame'_Buildin s 48 f ( ) g g .p fc= 1971.332 lbs /".(5:5".*5.5") f�= 65.168 psi.: Calculate Shear at top of endwall post V: shear at the top of post V.= 5/8 (wall.force." leeward force) *spacing * h V= 5/8 (9.65 -:-7.819) psf / 144 Ps'/psff*•112.37500000000003"* 188:358'. V=1604.867 Tbs f\" bending shear post f 3 * 1604.867,1bs / (2"* 5 5"* 5.5 ) fv 79.58 psi F�: allowable shear stress post F, = F�* Cp *CM*Cr*C; NDS4.3 Fv' =206.25 psi * 1.6* 1.* 1 * 1 F,;=264 psi page 66 f < F 79.58 psi < 264 psi Lateral Strength Assessment b = 1.5' P.,Z / fL? PZ,ASD ASAE 486.2 11.3.1 Pz,ASD=283.333.Lbs p,,,Z= ultimate lateral soil resistance at depth z z= 3.333' p..Z= 3 *G,,Z* KP ASAE 486.2 11.3.1 6ffv,Z=effective vertical stress =Y* 6'�,,Z z 6vZ= 105 * 3.333 6v,Z= 350 psf KP:= (1 +sin(cp)) / (1 -sin(cp))ASAE EP 486.2 11.2.1 co=soil friction angle (p= 30' ASAE EP486.2 table"1 KP = 0 +sin(30)) / (1 sin(30)) KP =3. 3 * 350*.3. P,,,Z= 3150 psf A= 1.4 / (.61 .01 * cp) ASAE EP486.2 table 4 f,=4.516. 3150 /4.516 >_283.333 ASAE 486.2 11.3..1 :. Lateral Strength adequate for 40"deep and 1.5'.diameter hole Determine if the diameter of the concrete column encasement is adequate A?%*PASD / (qB -.Y.* dF) ASAE EP486.2 10.2 - A= bearing area... A =n..*rz:...: A= 1.767 ft. : bay weight.=wall steel weight+ post weight+'girt weight +roof weight+.skirt board weight. bay weight= (.66 * (sf.* (h + panel length))) +.(62.4* .55 * (b_*d *,1)):+.(62.4.* .55* (b *d * L)*n) +. .6 (BCDL+TCDL) + (62.4.* ..55 * (b *d * l) * n) bay weight (.66 psf* (10'* (17' + 50',))) + (62.4 pcf* .55 lbs/ft 3 ` [ 12 'n /fr.) *'(5.5" / 12 in./ft) * . 19.667)) + (62.4 pcf* .55 lbs/ft.3*.((1.5" / 12 tn./ft.) * (9.25" / 12 /ft.) * 10') * 1) + .6 * (1 + 3.3) * 10*. 27.406 + (62.4 pcf* .55 lbs/ft.3 * ((1.5";/ 12'n-/ft.) * (7.25" /..12 m: 10')). bay weight = 1000.771 lbs PASD = (bay weight + Pf) PASD= (1769.994 + 1000.771) PASD =2770.765 lbs qB = ultimate soil bearing capacity . page 67 fB =ASD.factor of safety for bearing strength assessment fB = 1.4 /.(.58 - .01 * cp) ASAE EP486.2 table 2 fB = 5 qB =y* (.5 * B *CW, `NY*sY+ dF*Cw* Nq* dq*sq) ASAE EP486.210.4.1 y= moist unit weight of soil y= 105-pcf ASAE,EP486.2 table 1 dF= 3.333' Cw, = 1 ASAE EP486.2 10.4 CW2= 1 ASAE EP486.2 10.4 NY= 22.4 ASAE EP486.2 table 6 sY= 0.6,ASAE EP486:2 table 6 Nq= 1.8.4 ASAE EP486.2.table 6 dq = 1 +2*tan(tp).* 0 -sin((p))2* tan'(dF / B)ASAE EP486.2 10.4.1 dq = 1.+2*0.577.* (1 - 05)2*tan-'(3.333 / 1.5) dq = 1.3313824270181827 sq 1.58 ASAE EP486.2 table 6 qB = 105.,* (.5 * 1.5 *.1 *22.4*0.6.+ 3.333 * 1 * 18.4*.1.3313824270.181827* 1.58)'. . qB = 14605.482471395411 psf 1.767 2: 5 *2770.765 / (14605.48247139.5411 105 * 3.333) - 1.767>0:972 :. Diameter 1 'adequate for foundation pressure- page 68 COLUMN UPLIFT CALCULATIONS d =40" U: uplift resistance n.. U =a* (.33 *rr* (((d - t) + .5 *w / tan(6))3* tan(6)z - .125 *w3 / tan(6)) -AP* (d - t)) + .25 * C *rr *wz *t : a: soil density a= 85 pcf, C: concrete density C = 150 pcf t, h: collar thickness or thickness of attached concrete t, h = 1.5' '. w: collar width . w 6: soil friction angle .6 = 0.524 radians AP: post cross section.area AP=0.21 ft. U = 85 pcf*,(.33 *3.14154 * (�(((32 /:12 ' /ft.) 1.5') + .5:* 1.5' / tan(0.524°))3*tan(0.524 )z 125 1.5')) 25 150 cf 3.14159 1.5 1.5' 1.5'3 / tan(0.524`)) - 0.21 ft.. * ((32" % 12.'"�/tc.) ' + - *. p * * z* U = 752.721 lbs. total uplift resistance = 1000.771 lbs + 752.721. lbs = 1753.492.1bs. uplift force= roof_wind_force _roof area / DOL uplift force = 14.474 psf*46.823 ft." / 1.6 uplift force=464.353 lbs' Total uplift force of.464.353 lbs is less than total uplift resistance of 1753.492 lbs The direct bearing capacity under vertical load depends on the bearing capacity of the soil against,the end of the post and the frictional:resistance between the post and.the soil, without producing much end bearing pressure at the butt. Interestingly, it has been determined that firm backfilling of the We with compacted soil, sand, soil concrete.or concrete is a more effective way of increasing vertical force resistance than placing mats of footings of concrete beneath the butt.ends: Concrete encasement of the post in the ground contact area enlarges the friction surface and.can generally be:credited with a wood to concrete bond strength of 30 psi. Hoyle Et Woeste, p-335-336 page 69 Connection Strength = (breadth *2+ depth *2) * collar thickness *30 Connection Strength = (5.5"*2 + 5.5".*2) * 10"* 30 psi =6600 lbs Connection Strength = 6600 lbs >.464.353 lbs In summary a 5.5"X5.5"SYP endwall column; in an augered hole 1.5'in diameter and 40"in depth is adequate:A•-concrete bottom collar 1.5' high mono-poured with a 8 footing below the column is -. required page 70 Rear Endwall Columns Calculate maximum post moments M M: moment M=wind force*spacing* hZ / 8 h: distance from either grade or top of concrete slab, to bottom chord of end truss h = 188.358" M= (9.65 - -7.819) psf/ 144 in.Z/ft.z*.112.375".* 188.358"- % 8 M= 60457.85 in.lbs M is shared by two endwalls so the.load on one endwall can be cut in half. S: section modulus S = b* d * d / 6 NDS 3.3.2 S = 5.5"* 5.5"* 5.5 / 6 S = 27.729 in.3 fb: bending stress post fb= I M I ./ S NDS 3.3.2 f = 130228.925in.lbs 1 %27.729 in:3 b fb= 1090.1.49 psi . . CM=wet service factor CM= 1 because the post is protected by the building shell; concrete, or embedment from excessive moisture CF= 1 NDS Supplement Cr: repetitve member factor Cr= 1_NDS 4.3.. Cf,,: flat use factor Cf,,= 1 NDS4.3 Fb': allowable bending stress post Fb = Fb*Co*CM.*.Ct*CL*CF*Cf. *Ci * CrNDS-4.3 Fe= 850 psi *.1.6.* 1 ` 1 Fb.="1360 psi fb <_ Fe . 1090.149 psi s 1360 psi page 71 • Calculate axial compression force in.the post Pf: compressive force post - Pf=sf*w / 2* (snow roof force+dead load) Pf=60"* 112.37499999999999"* (37.802 psf / 144 P'V-Psf+4:3 psf/ 144-PS'1Psf) +. Pf=1971.332 lbs . CF: size factor CF= 1 NDS Supplement table 4D CP: column stability factor CP= 0 + (FrE / F�*)) / (2 *c) -f(((1 + (FcE /:Fc*)) !"(2*c))2 - (F,E / F,*) / c) FEE= .3 * E / (le / d)Z NDS 3.7.1 . le= Ke* l NDS Appendix G l=40.8" length of column.between lateral supports Ke= 0.8NDS3.7.1 . 1, 0.8.*40.8 le=32.64". FEE= .822 * 1200000 psi / (32.64" / 5.5") , FEE=28007.745 psi . f( 1 (28007745 psi / NaN 2 ZCP= (1 + (28007.745 psi NaN psi)). 8) - 28007.745:psi /.NaN psi)'/ 8 CP =-NaN F,'= NaN psi f,- axial compressive force post fc= Pf/.:(b * d) from Design of Commercial Post-Frame Buildings p.48 f`= 1971..332 lbs / (5.5"* 5.5")- f`= 65.168 psi Calculate Shear at.top of endwall post V: shear at.the top of post V=.5/8 (wall force = leeward force) *spacing" h V= 5/8.(9.65 -.-7.819) psf:/ 144 P"/PSf..* 1.12.37499999999999"* 188.358" " V= 1604.867 lbs f,: bending shear post f = 3 *V / (2.*b*d) fV= 3 .* 1.694:867.lbs / (2 * 5.5"* 5.5").. fv= 79.58 psi F,: allowable shear stress post Fv = Fv* Co *Cµ* Ct*Ci NDS 4.3 Fv =.206.25 psi * 1.6 1 * 1 *.1 F;=264 psi . . page 72 fV Fv 79.58 psi < 264 psi Lateral Strength Assessment b = 1.5' Pu,z / fL.? Pz,AID ASAE 486.2 11.3.1 PZ,ASD =283.333 lbs p,,,Z= ultimate lateral soil resistance at depth z z= 3.333' P,,,z= 3 *a,,z* KP ASAE 486.2 11.3.1 o,,Z=effective vertical stress 6v,z= 105 *3.333 6'v,z= 350 psf: KP = (1 + sin(cp)) / (1 - sin(cp))ASAE EP 486.2 11.2.1. cp=soil friction angle , -f .30 ASAE EP486.2 table 1 P - KP= (I +sin(30)) /.(1 sin(30)). KP = 3 puz= 3 ' 350*3. 3150 psf fL= 1.4 /•(.61. - .01 * cp) ASAE EP486.2 table 4 .. fL=4.516 3150 /4:516 >_.283.333 ASAE 486.2.11.3.1 Lateral Strength adequate for 40".deep and 1.5' diameter hole Determine if the diafrieter of the concrete column encasement is adequate A ? fB * PASD / (qB Y*dF) ASAE EP486.2 10.2 A =.bearing aread. A=n*rz A =1:767 ft.. - bay weight =wall steel weight+ post weight+girt weight + roof weight+skirtl,board weight bay weight= (:66 *.(sf* (h'+ panel length))).+ (62.4.* .55 * (b-.* d * l)).+ (62.4*.,55 * (b' d * l) *h) +.6 (BCDL+TCDL) + (62.4* .55 * (b *d * l) bay weight = (.66 psf* (10'* (17'+.50'))) + (62.4 pcf.* .55 lbs/ft 3*((5 5 /_12 in /ft.) * (5.5'` / 12 ' 19.667)) + (62.4 pcf* .55 lbs/ft.3* ((1:5 /.12 in-/ft.) * (9.25" / 12 in'/ft.) * 10') * 1.).+ .6 * 0 +.3.3):* 10 27.406 + (62.4,pcf* .55 lbs/ft.3-* ((1.5" /.12'"'/ft.) * (7.25" / 12' /ft.) * 10')) bay weight= 1000.771 lbs PASD._ (bay weight + Pf) :'PASD (1769.994 + 1000.771) PASD = 2770.765 lbs qB = ultimate soil bearing capacity W. page 73 fB =ASD factor of safety for bearing strength assessment '. fB = 1.4 / (.58 - .01 * cp) ASAE EP486.2-table 2 fB = S qB =Y* (•5 * B * Cwl * NY*sY+ dF* Cw2* Nq*dq*sq)ASAE EP486.2 10.4.1 y= moist unit weight of soil y= 105 pcf ASAE-EP48'6:2 table 1 ..:.:..; dF=1333' CWI = 1 ASAE-EP486.210.4 Cw2= 1 ASAE EP486.2 10.4 NY=22.4 ASAE EP486.2 table 6 sY= 0.6 ASAE EP486.2 table 6 . Nq= 18.4 ASAE EP486.2 table 6 dq =1 +2 *tan(cp) * (1 - sin(cp))2*tan�(dF / B) ASAE EP486.2 10.4.1 dq =.1 +2*0.577* (-1 - 0.5)2*tan'(3.333 / 1.5) dq= 1.3313824270181827 sq=1.58 ASAE EP486.2 table'6 qB= 105.* (.5 * 1.5 * 1 *22.4*0.6 + 3.333.* 1 * 1,8.4*.1.3313824270181827* 1.58) qB= 1.4605.482471395411 psf 1.767_>..5 *.2770J65 / (14605.482471395411 =.105 *..3.333) . 1.767 2t 0.972 :. Diameter.1.5'adequate.for foundation pressure page 74 COLUMN UPLIFT CALCULATIONS d =40" U: uplift resistance U = a* (.33 * Tr * (((d t) + .5 *w / tan(6))3 *tan(6)2 .125 *w3 /tan(6)) -AP* (d t)) + .25 *C *rr*wZ *t a: soil density a=85 pcf C: concrete density. : C = 150 pcf t, h: collar thickness or thickness of attached concrete , t, h = 1.5' w: collar width IN=.1.5' 6: soil friction angle 6 =0.524 radians AP: post.cross section area AP=0.21 ft.2 U=85:pcf* (.33 * 3.14159 *'((((32" 12 '"'/ft.) - 1.T) + .5 * 1:5' /tan(0.524°))3 *tan(0.524 )? 1.5'3 / tan(0.524`)) - 0.21 ft. * ((32" / 12 ".'/fc 1.5'z * 1 . . 5 U =752.721 lbs total uplift resistance = 1000.771 lbs + 752.721 lbs = 1753.492 lbs uplift.force = roof wind force* roof area / DOL uplift force= 14.474 psf*46.823 ft. ,/ 1.6. uplift force =464.353 lbs . Total uplift force of 464.353 lbs is less than total uplift resistance of 1753.492 lbs The direct.bearing capacity under vertical load depends on the bearing capacity of the soil against the end of the post and the frictional resistance between the post and the soil, without producing much end bearing pressure at the butt. Interestingly, it has been determined that firm backfilling of the hole with compacted soil,sand, soil concrete or concrete is a more effective way of.increasing vertical force resistance than placing mats of footings of concrete beneath the butt ends: Concrete encasement of the post in the ground contact area enlarges the friction surface and can generally be credited with., . a wood to concrete bond.strength of 30 psi. Hoyle It Woeste, p.335-336' ..: page 75 Connection Strength = (breadth *2 + depth *2) * collar.thickness.*30 Connection Strength =,(5:5" *'2 + 5.5" *2) * 10"* 30 psi = 6600 lbs Connection Strength =6600 lbs > 464.353 lbs . In summary a 5.5"X5.5"SYP endwall column; in an augered hole 1.5'in diameter and.40"in depth is adequate. A concrete bottom collar.1.5' high mono-poured with a^8?'-footing below the column is required, page 76 Rear Endwall Columns Calculate maximum post moments M M: moment M=wind force `spacing * hZ / 8 h: distance from either grade or top of concrete slab, to bottom chord of end truss h = 188.358" M= (9.65 - 7.819) psf / 144 in.z/ft.2* 133.625"*'188.358"2 / 8 M=71890.369 in.lbs M is shared by two endwalls so the load on one endwall can be cut in half. S: section modulus S = b * d * d / 6 NDS 3.3.2. S = 5.5"*5.5 * 5.5" / 6 S =27.729 in.3'. fb: bending stress post fb= I M 1 / S NDS 3.3.2 fb = 135945.184in.lbs 1 / 27.729 in.3. fb= 1296.295 psi .. CM=wet service factor CM= 1 because the post is protected by the building shell, concrete, or embedment from excessive moisture CF.= 1 NDS Supplement Cr: repetitve member factor. Cr= 1 NDS 4.3 Cf,,: flat.use factor. Cf,,-= 1 NDS 4.3.. Fe: allowable bending stress post Fti-= Fb*Co*CM*Ct* CL* CF*Cf.*Ci*CrNDS.4.3 F6= 850 psi * 1.6 * 1 *.1 1 * 1 *1 * 1 * 1 . Fb= 1360 psi fb < Fe 1296.295 psi :5 1360 psi . page 77. Calculate axial compression force in the post Pf: compressive force post Pf=sf*w / 2* (snow roof force + dead load) _._Pf= 60"*.133 625.,.*_(37.802 psf/ 144 Psf/Psf+4.3 psf/ 144 Ps,/P5f) + - Pf=2344.109 lbs CF: size factor CF= 1 NDS Supplement table 4D . CP: column stability factor CP = (1 +.(FcE / F�*)) / (2 *c) f(((1.+ (F,E / F,*)) / (2 * c))Z -.(FEE / Fc*) / c) FcE = .3 * E / (le / d.)Z NDS 3.7.1, le= Ke* l NDS Appendix G I =40.8" length of column between tateral supports Ke=.0.8 NDS 3.7.1 le.=32:64" FcE.= :822 )�.� 1200000 psi / (32,64" /.5.5' FEE=28007.745 psi CP = (1 +. (28007.745 psi / NaN psi)) / (2 * .8) .f(((1 + (28007.745 psi / NaN psi)) / (2 *`.8))Z 28007.745psi:/ NaNpsi) / 8 CP = NaN Fc' = NaN psi :. . fc: axial compressive force post. . fc Pf./ (b,.*d) from.Design of Commercial Post-Frame Buildings:p.48,. f, =2344.109 lbs % (5.5"* 55")... fc=.77.491t psi Calculate Shear at top of endwall post V,-,shear at the.top of post V= 5/8 (wall force leeward force) *spacing.* h. V= 5/8 (9:65 - 7.819) psf./ 144 PS'/Psf;* 133.625".* 188.358 V=.1908.346 lbs f,,: bending shear.post. fv= 3 *V / (2 * b *d) f„= 3 * 1908.346 lbs / (2.* 5.5"* 5.5") f„= 94.629 psi " F,;: allowable shear stress post F'= F * CD*Cµ* Ct* C1 NDS 4.3 Fv' =206.25 psi *-1.6 1 * 1 *.1 F,;=264 psi page 78 f,,< Fv' 94.629 psi <_264 psi Lateral Strength Assessment b = 1.5' - Pu,z / fL? Pz,ASD ASAE 486.2 11.3.1 N,ASD =283.333 [bs pu'z= ultimate lateral soil resistance at depth z _ z= 3.333' p.,z= 3 * o'V z* KP ASAE 486.2 11.3.1 o,z=effective vertical stress _Y* o'�,,z z ov,z= 105 * 3.333 o v.z= 350 psf Kp = (1 + sin(cp)) / (1 sin(cp))ASAE EP 486.2 11.2.1 co=.soil.friction angle . cp = 30' ASAE EP486.2 table 1 Kp.= (1 + sin(30)) / (1 sin(30)). KP = 3 3.* 350*3 . Pu,z= 3150 psf. fL= 1.4 / (.61 - .01 * fp) ASAE EP486.2 table 4 fL=4.516 3150 /4.516 _:'283.333 ASAE 486.2 11.3.1 Lateral Strength adequate for 40"deep and 1.5'diameter hole Determine if the diameter of the concrete column encasement is adequate . - A? fB*PASD./.(qB -Y* dF)ASAE EP486.2 10.2 . A=..bearing area - A= rr* rz A= 1.767 ft: bay weight=wall steel weight+ post weight+ girt weight+goof weight+ skirt board weight bay weight=(.66*-(sf* (h + panel length))) + (62.4* .55:* (b *d* [)) + (62.4.* .55 * (b"_d-* [) *n).+ .6:F: (BCDL+TCDL) + (62.4* .55 * (b *d * [) * n) bay.weight= (.66 psf* (10'* (17'+ 50'))).+ (62.4 pcf* .55 lbs/ft.3 * ((5.5" / 12 in./*.) * (5.5"1 11,"Ift).* 19.667)) + (62.4 pcf* .55 lbs/ft.3* ((1.5"./ 12 "Ift.) * (9.25" / 12 "-/ft.) * 10') * 1) +..6 * (1 + 3.3) *.10 27.406 + (62.4 pcf* .55 lbs/f -3 * ((1 5.. / 12.m-./ft.) *.(7.25'' /:12 m./ft.) * 10')) bay weight= 1000.771.1bs PASD = (bay weight+ Pf). PASD= (2104.698+1000.771) PASD = 3105.469 [bs qB =ultimate soil bearing capacity. page 79. fB =ASD factor of safety for bearing strength assessment fB = 1.4 /,(.58 - .01 * cp) ASAE EP486.2 table 2 fB = 5 qB =y* (•5 * B *Cw, * NY*sY+dF*CWZ* Nq*dq*sq) ASAE EP486.216.4.1 y=moist.unit weight of soil y= 105 pcf ASAE EP486.2.table 1 dF= 3.333'. Cw, = 1 ASAE EP486.2 10.4 CWz= 1 ASAE EP486.2 10.4 NY=22.4 ASAE EP486.2 table 6 sy= 0.6 ASAE EP486.2 table 6' Nq= 18.4 ASAE EP486.2 table 6 dq = 1 +2 *tan(to) * (1 -sin((p))z*tan"(dF / B) ASAE EP486.2 10.4.1 ' . dq = 1 + 2*0.577* 00.5)z*tan"'(3.333 / 1,5). dq= 1.3313824270181827 sq= 1.58 ASAE EP486.2 table 6 . qB=.105 * (.5 * 1.5 * 1 *22.4*0.6 +3.333 * 1"* 18:4* 1.3313824270181827* 1.58) qB = 14605.482471395411 psf,' 1.767>5*3105.469 /. (14605.482471395411 - 105.*,3.333) 1.767 >_ 1.089 Diameter 1.5' adequate for foundation pressure . page 80 COLUMN UPLIFT CALCULATIONS d =40" U: uplift resistance U =a* (.33 * rr * (((d - t) + .5 *w / tan(9))3*tan(6)z - .125 *w3 / tan(6)) - AP* (d - t)) + .25 *C * rr*.wz * t_ a: soil density a = 85 pcf. C: concrete density C= 150 pcf t, h: collar.,thickness or thickness of attached concrete t; h = 1.5' w: collar width w= 1.5' 6: soil friction angle 6 =0.524 radians AP: post cross section area . AP 0:21 ft.z U =85.pcf* (:33 *3.14159 * �(((32" / 12 ' /ft.) -.1.5').+.:5 ` 1.5' / tan(0.524°))3 *tan(0.524`)z -.125.* 1.T3 / tan(0.524°)) - 0,21 ft. *.((32" / 12 '"'/ft.) - 1.5')j + .25 150 pcf* 3.14159 * 1.52* 1.5' U = 752.72-1 lbs ` . total uplift resistance= 1000.771 lbs + 752.721 lbs= 1753.492 lbs uplift force:= roof_wind_force roof area % DOL. :: uplift force = 14.474 psf.�' 55.677 ft.? /.1.6 uplift force_= 552.162 lbs Total uplift force of 552.1621bs is less than total uplift resistance of 1753.492 lbs The direct bearing capacity under vertical load depends on the bearing capacity of the soil against the end of the post and the frictional resistance between the post and the soil, without producing much ' end bearing pressure at the butt. Interestingly, it has been determined that firm backfilling of the hole. with compacted soil, sand, soil concrete or concrete is a more effective way of increasing vertical . force resistance than placing mats.of footings of concrete beneath the butt ends. Concrete encasement of the post in the ground contact area enlarges the friction.surface and can generally be credited with a wood to concrete.bond strength of 30 psi. Hoyle Et Woeste, p.335-336 . . page 81 Connection Strength = (breadth*2 +depth *2) *collar thickness * 30 Connection Strength = (5.5"*2+ 5.5"*2).* 10"* 30 psi = 6600 lbs Connection Strength =6600 lbs > 552.162 lbs In summary a 5.5"X5.5" SYP endwall column, in an augered hole 1.5'in diameter and 40"in depth is adequate.`A:concrete bottom collar 1.5'high mono-poured with a.8"footing below the column is required page 82 Rear Endwall Columns Calculate maximum post moments M M: moment . M=wind force * spacing * hZ / 8 h: distance from either grade or top of concrete stab, to bottom chord of end truss h = 188.358" M= (9.65 - 7.819) psf / 144 in.z/ft.z* 123.5"* 188.358"i / 8 M=66443.11 in.lbs M is shared by two endwalls so the load on one endwall can be cut in half: S: section modulus S = b *d `d / 6 NDS 3.3.2 S= 5.5"* 5.5" * 5.5" / 6 S =27.729 in.3 fb: bending stress post fb.=.IM1 / S NDS 3.3.2 fb = 13.3221.555in.lbs I / 27.729 in s fb= 11.98.073 psi CM=wet service factor CM= 1 because the post is protected by.the building shell, concrete, or embedment from excessive moisture. CF= 1 NDS Supplement Cr: repetitve member factor. . Cr= 1 NDS 4.3 Cf,,: flat use factor Cfu =.1 NDS 4.3 . . :Fe allowable bending stress post Fe= Fb*Co*CM*Ct*CL* CF*Cfu*Ci* C,NDS4.3 Fb= 850psi * 1.6 * 1 * 1 *.1.*.1 * 1 * 1 *.1 Fb= 1360 psi: fb 5 Fe .: - 1198.073 psi <_ 1360 psi page 83' Calculate axial compression force in the post Pf: compressive force post Pf=sf*w / 2* (snow roof force+ dead load) - Pf=60"* 123..5"* (37.802.psf / 144 PS'/PSf+4.3 psf./ 144-?`/Psf) +' Pf=2166.492 lbs CF: size factor CF= 1 NDS Supplement table 4D CP: column stability,factor CP = 0 + (FcE /.Fc*)) / (2 'C) f(((1 + (FcE / Fc*)) / (2*CW.- (FcE / Fc*) / c) FEE_ .3.* E / (le'/ d)2 NDS 3.7.1 le= Ke* l NDS Appendix G l=40.8"length of column between lateral supports Ke= 0.8 NDS 3.7.4 le=32.64 FcE= .822 * 1.200000 psi / (32.64" / 5.5)2 FEE=28007.745 psi CP= (1 +.(28007,745 psi / NaN psi)).,/ (2* .8) - f(((1 + (28007.745 psi./ NaN psi)) % (2*".8))Z.-. (28007.745 psi / NaN psi) /.•8) CP =.NaN . F� = NaN psi fc: axial compressive force post ft= Pf/ (b*d) from Design of Commercial Post-Frame Buildings p.48. fc=2166:492 lbs / (5.5"" 5.5") f� =71.62 psi Calculate Shear at top.of endwall poste' • V: shear at the top of post V= 5/8.(wall force.-leeward force)`.*spacing.* h V=:5/8 (9.65 7.819) psf./ 144 PS'/Psf* 123.5"*188.358" V= 1763.747 lbs V. bending shear post f,,= 3 *.1763.747 lbs / (2 * 5.5"*5.5") fv=.87.459 psi Fv: allowable shear stress post. Fv = F„* CD*Cµ*.Cf* Ci NDS 4.3 F,;=206.25 psi * 1.6 *..1. * 1 *.1 Fv'=264 psi; page 84 fv < FV 87.459 psi <_ 264 psi Lateral Strength Assessment b = 1.5' Pu,z / fL 2: PZ,Aso ASAE 486.2 11.3.1 Pz,ASD =283.333 lbs p,,,Z= ultimate lateral soil resistance at depth z z=3.333' p.,Z= 3 *o'V,Z* KP ASAE 486.2 11.3.1 a',,,Z=effective vertical stress . CYV,Z=y*z o v,Z= 105 *'.3.333" 6v,Z= 350 psf Kp = (1 +sin(cp)) / (1 - sin(tp))ASAE EP 486.2 11.2.1 cp=soil friction angle . cp= 30° ASAE EP486.2 table 1,; Kp = (1 +sin(30)) / (1 sin(30)) Kp =.3 3 *350*.3 p6,Z= 3150 psf " fL=.1.4 / (.61 - .01 * cp) ASAE EP486.2 table 4 fL=4.516 3150 /4.516 >_283.333 ASAE 486.2 11.3.1 :. Lateral Strength adequate for 40"deep and 1.5'diameter"hole `. Determine if the-diameter of the concrete column encasement is adequate . A >_ fB* PAso./.(qa. Y*.dF)ASAE.EP486.2.10:2 A= bearing area. A= n* r� A= 1.767 ft: bay weight =wall steel weight + post weight +girt weight+roof weight+skirt board weight bay weight= (.66 * (sf*"(h +"panel length))).+ (62.4,* .55.* (b *d *J)) + (62.4* .55,* (b*d * t) ' n) + .6 (BCDL+TCDL) + (62.4.* .55 * (b *d * l) * n) bay weight= (.66 psf* (10'* (17'.+ 50'))) + (62.4 pcf* .55 lbs/ft.3* ((5.5" / 12 in /ft.) * (5 "5" /"12 ' /ft.) 19.66T)) + (62.4 pcf*.55 lbs/ft.3 * ((1.5" / 12 'n-/ft.) * (9.25" / 12 in'/ft,) * 10') * 1) +:.6:* (1 + 3.3)*.10 27.406 + (62.4 pcf* .55 lbs/ft.3* ((1.5""/ 12,"_/ft.) * (7,25" / 12 ,"_/ft.) *.10,)) bay weight= 1000.771 lbs _ Paso = (bay weight + pf)" PAsD.=.(1945.221 + 1000.771) PASS=2945.992 lbs ... qB = ultimate soil bearing capacity page 85 fB =ASD factor of safety for bearing strength assessment fB = 1.4 / (.58 - .01 * cp) ASAE EP486.2 table 2 fB = 5 qB =y* (.5* B * Cwi * NY*SY+dF*Cwz* Nq* dq*sq) ASAE,EP486.2 10.4.1 y= moist unit weight of soil y'=105..pcf ASAE=EP486.2 table 1 dF= 3.333' Cwj = 1 ASAE EP486.2 10.4 Cwz= 1 ASAE EP486.2 10.4 NY=22.4 ASAE EP486.2 table b sY=0.6 ASAE EP486.2 table,6 Nq=.18.4 ASAE EP486.2,table 6 dq = 1 +2*tan(cp) * (1 - sin((p))2*tan-'(dF./ B) ASAE EP486.2:10.4.1 dq = 1 +2.' 0.577* (1 -'0.5)2*tan-.'(3.333 / 1.5) dq = 1.3313824270181827 sq= 1.58 ASAE EP486.2 table 6 . qB = 105 * (.5 * 1.5.* 1 *22.4` 0.6 + 3.333 * 1.* 18.4 * 1.331382427018.1827* 1.58) qB.=.14605.482471395411 psf - 1.767 >_ 5*2945.992 / (14605.482471395411 105.* 3.333). - 1.767.>_ 1.033 Diameter 1.5'adequate.for foundation pressure, page 86 COLUMN UPLIFT CALCULATIONS d =40" U: uplift resistance U = a* (.33 * rr * (((d - t) + .5 *w / tan(0))3*tan(0)Z -. .125 *w3./ tan(6)) AP.* (d - t)) + .25 *C*IT `wz *t a: soil density a= 85 pcf C: concrete density C = 150 pcf' t, h: collar thickness or thickness of attached concrete: t, h = 1.5'. w: collar width w= 1.5' 0: soil friction angle . - 0 =0.524 radians AP: post cross section area AP=0.21 U = 85 pcf* (.33 * 3.14159 *.((((32" / 12' /ft,) 1 5') + :5 * 1.5' / fan(0:52'4°))3 *tan(0.524°)� 125 1:5'3./ tan(0.524°)) - 0.21 ft. * ((32" / 12 ' !ft.) - 1.5')) + .25 ` 150 pcf* 3.14159 * 1.5'2 * 1 5' U = 752.721 lbs total uplift resistance = 1000.771 lbs + 752:721-lbs = 1753.492 lbs uplift force =.roof_wind_force* roof area / DOL uplift force = 14.474 psf,* 51.458 ft.z / 1:6,:: - uplift force =.510.324 lbs Total uplift force of 510.324 lbs is less than total uplift resistance of 1753.492 lbs. The,direct bearing capacity under vertical load depends on.thebearing capacity of the soil against the end of the post and the frictional resistance between the post and the soil,.without producing much end bearing pressure at the butt. Interestingly, it has been determined that firm backfilling of the hole with compacted soil,'sand, soil concrete or concrete is a more effective way of increasing vertical force resistance than placing mats of footings of concrete beneath the butt ends. Concrete encasement of the post in the ground contact area enlarges the friction surface and can generally.be credited with a wood to concrete.bond strength of 30 psi. Hoyle Et Woeste, p.335-336 page 87 Connection Strength = (breadth *2 + depth *2) * collar thickness *30 Connection Strength = (5.5"*2+ 5.5" *2) * 10"* 30 psi = 6600 lbs .Connection Strength = 6600 lbs > 510.324 lbs In summary a 5.5"X5.5"SYP endwall column, in an augered hole 1.5'in diameter and 40"in depth is adequate. A concrete bottom collar 1.5' high mono--poured with'a 8"footing below the column is required page 88 Front Left Corner Columns Calculate maximum post moments M M: moment M=wind force*spacing * h2 / 8 h: distance from either grade or top of concrete stab, to bottom chord of end truss h = 188.358" M= (9.65 - -7.819) psf /.144 in.2/ft.2*75.25"* 188.358"2 /.8 M=40484.567 in.lbs M is shared by two endwalls so the.load on one endwall can be cut in half. S: section modulus S = b * d *d / 6 NDS 3.3.2 S= 3.5"*5.5"* 5.5" / 6 . S= 17.646 in:3 .. fb: bending stress post fb I M I / S NDS 3.3.2 - fb 120242.283i.n.lbs I / 17.646 in.3 fb.= 1147.142 psi CM=wet service factor CM= 1 because the post is protected by the building shell; concrete, or embedment from excessive moisture CF= 1 NDS Supplement Cr: repetitye member factor . Cr=.1 NDS 4.3 Cf,,: flat use factor Cf,,= 1 NDS.4.3 Fb:: allowable bending stress post F6= Fb* Co *CM*Ci*CL* CF*Cf,, * C;*C'NDS4.3 Fe=. 1000 psi * 1.6 1 *-1 * 1 * 1 * 1 * 1 * 1 .: Fe = 1600 psi fb < Fb 1.147.142 psi :5.1600 psi page 89 Calculate axial compression force in the post Pf: compressive force post Of= sf*w / 2 * (snow roof force+ dead load) Pf=60"*75•.25"*(37.802 psf / 144 P5;/Psf+4.3 psf / 144 P5;/Psf) + �. Of= 1320.069 lbs CF: size factor . CF= 1 NDS Supplement table 4D CP: column stability factor CP = (1 + (FrE / F�*)) / (2 *c) - f(((1 + (Fce/. Fc*)) /.(2*c))Z ' (FcE / F�*) / c) . . F,E= .3 * E / (le /.d)Z NDS 3.7.1 . le= Ke* l NDS Appendix.G . l =40.8" length of column between lateral supports Ke:= 0.8 NDS 3.7.1 le=32.64" FcE_ .822 * 1400000 psi / (32.64' / 5.5")z. Fc = 32675.702 psi CP = (1 + (32675.702 psi %2240 psi)) / (2* .'8). f(((1 + (32675.702 psi /, 2240 psi))./ (2 *. .8))2 (32675.702 psi'/ 2240 psi):/..8) CP = 0.986 c F2207.997 psi: fc: axial compressive force post f,=Of /- (b_*d) froni Design of.Commercial Post=Frame Buildings p.48 . fc= 1320.069.1bs / (3.5"* 5.5") f,= 68..575 psi Calculate_Shear.at top of endwall post V: shear at the top of post.. V= 5/8 (wall force - leeward force) *spacing.* h V= 5/8 (9.65 - 7.819) psf 1.144 P"/Psf*75.25'.'* 188.358". V=.1074.672 lbs f.,,: bending shear post` fv= 3 * 1074.672 lbs / (2 * 3;5"* 5.5") f,=.83.741.psi, Fv: allowable shear stress.post. F, = F„* CD*CM* Cf*C; NDS 4.3 F,;=218.75 psi * 1..6 *.1 * 1 * 1 F,;=280 psi page 90 fV� F„ 83.741 psi < 280 psi Lateral Strength Assessment b = 1.5' __.. PI,,z / fL? Pz,ASD ASAE 486.2 11.3.1 Pz,ASD =283.333 lbs ultimate lateral soil resistance at depth z z= 3.333' p,,,Z= 3 * 6�Z* KP ASAE 486.2 11.3.1 6V,Z=.effective vertical stress ,Z=Y*z 105 * 3.333 6'v,Z= 350 psf Kp = (1 +sin(cp)) 1 (1 - sin (tP)) ASAE EP 486.2 11.2.1.. .. . . -.. . . . . . . . cp =soil friction angle fP= 30' ASAE EP486.2 table 1 KP= (1 +sin(30)) / (1 sin(30)) . KP = 3 Pu;Z=3 *350 *3 pt,,Z= 3150 psf fL= 1.4 / (.61 - .01 * cp) ASAE EP486.2 table 4 fL=4.516 3150 /4.516 >_283.333 ASAE 486.2 11.3.1 :. Lateral Strength adequate for 40" deep and.1.5' diameter hole Determine if the diameter of the concrete column encasement is adequate A? fB* PASD / (qe Y*dF)ASAE EP486.2 10.2 . A = bearing area . . A- it*rz. A= 1.767.ft. bay weight=wall steel weight+ post weight+ girt weight+ roof weight+ skirt board weight bay weight= (.66 * (sf* (h + panel length))) + (62.4* .55 * (b,*d,* l)) +.(62.4 * .55 * (b * d * 1).* n)+ .6 (BCDL+TCDL) + (62.4* .55 * (b * d * l) *n) bay weight= (.66 psf* (5,'* (17' + 50'))) + (62.4 pcf*`.55 lbs/ft s * ((3 5" /.12 `-If.) * (5 5" / 12 in'%ft.) 19.667)) +.(62.4 pcf* .55 lbs/ft.3 * ((1.5" / 1.2 tn./tt.) x (9.25" /.12' /tc.) * T) * 1) + .6 * (1 +:3.3.).* 5 27.406 + (62.4 pcf* .55(bs/ft.3* ((1.5"./ 12 tn./ft.) * (7:25" / 12 /ft.) * 5)), bay weight= 519.72'lbs _ PAso = (bay weight + pf) PASD _ (592.623 + 519.72) PASD = 1112.343 lbs qB = ultimate soil bearing capacity page 91 fB =ASD factor of safety for bearing strength assessment fe=1.4 / (.58 - .01 * cp) ASAE EP486.2 table 2 fB = 5 qB =y* (.5 * B *CW, * NY*sY+ dF*Cw2* Nq* dq*sq) ASAE EP486.2 10.4.1 y=moist unit weight of:soil - y= 105 pcf ASAE.EP486.2 table 1 - dF= 3:333' Cwt = 1 ASAE EP486.210.4 Cw.,= 1 ASAE EP486.2 10.4 NY=22.4 ASAE EP486.2 table 6 sy=0.6 ASAE EP486.2 table 6 Nq= 18.4 ASAE EP486.2 table 6 dq = 1 +2*tan(cp) * (1 - sin( ))2* tan"'(dp / B) ASAE EP486.2 10.4.1 dq = 1 +2 *0.577* (1 - 0.5)2*tan-'(3.333 / 1.5) dq = 1.3313824270181827 sq = 1.58 ASAE EP486.2 table 6 q67105 * (.5 * 1.5 * 1 *22.4*0.6 + 3.333 * 1 * 18.4* 1.3313824270181827* 1.58) qs = 14605.482471395411 psf 1.767 >_ 5 * 1112.343 / (14605.482471395411 -105 *3.333) :. Diameter 1.5 adequate.for foundation pressure page 92 COLUMN UPLIFT CALCULATIONS d =40" U: uplift resistance - U = a* (.33 *n* (((d - t) + .5 *w / tan(6))3.*tan(6)2 -..125 *w3 / tan(6)) -AP* (d - t)) + .25,*C *TT wz * t a: soil density a= 85 pcf. C: concrete density C = 150pcf t, h: collar thickness or thickness of attached concrete t, h = 1.5' w: collar width 6: soil friction angle . 6 =0:524 radians ; AP:-post cross section area - AP=0.134 ft.?. U = 85 pcf* (.33 *3.14159 k ((((32" / 12 in'/ft.) - 1.5') + .5 * 1.5' / tan(0.524'))3 *tan(0.524 )Z 125 1.5'3 / tan(0.524°)) - 0:134 ft.z* ((32" / 12 '"'/ft.) 1.5')) + .25 * 150 pcf* 3.14159 * 1.5'2* 1.5' U = 760.296 lbs ' bay weight.=steel weight+ post weight+ girt.weight +roof weight +skirt board.Weight' bay weight=(.66.* (spacing* h + sf/ 2* roof length)) + (62.4* .55 * (b`d* l)) + (62.4* .55,* (b.*.d.* l) * n) + •6 *.(BCDL+TCDL) *sf % 2*w+(62.4* .55 *.(b d * l).*.n).. ; bay weight= (.66 psf. (6.270833333333333 *21.167 + 5' /2* (50' / 2 / cos(18.435°)))) + (62.4 pcf* .55 lbs/ft_3* ((3.5" / 12' /ft.)* (5.5" / 12 /ft.) * 19.667')) + (62.4 pcf*:.55 lbs/ft.3* ((1.5" / 12 ' /ft.) *. _(9.25 / 12 ' /ft)*6.270833333333333') * 1) +_ .6 *.(1 + 3.3)* 5 / 2 *6.270833333333333 + (62:4 pcf.* .55 lbs/ft.' k ((1.5" / 12 i /ft.) * (7.25" % 12.1 /ft.) *,6.270833333333333.')) bay weight=255.27 lbs total uplift resistance = 255.27 lbs +760.296 lbs= 1015.566.1bs page 93 uplift force.= roof—wind—force* roof area / DOL uplift force = 14.474 psf* 15.677 ft.2 / 1.6 uplift force = 155.473 lbs Total uplift force of 155.473 lbs is less than total uplift resistance of 1015.566 lbs The direct bearing capacity under vertical load"depends on the bearing capacity of the soil against the end of the Post-and the frictional resistance between the post and the soil, without producing much end bearing pressure at the butt. Interestingly, it.has been determined that firm backfilling of the hole with compacted.soil, sand, soil concrete or concrete is a more effective way of increasing vertical force resistance than placing mats of footings of concrete beneath the butt ends.Concrete encasement of the post in the ground.contact area enlarges the friction surface and.can generally be"credited-with a wood to-concrete bond strength of 30 psi. Hoyle Et Woeste, p.335-336 Connection Strength = (breadth.*2+ depth *2) *collar thickness*30 Connection Strength _ (3.5'*2 +5.5"*2):* 10"* 30 psi = 5400 lbs Connection Strength = 5400 lbs > 155.473. lbs . In summary a 3.5"X5.5"SYP endwall column, in an augered hole IS in.diameter and 4,0"in depth is adequate.A.concrete bottom collar 1.5'high mono-poured with a 87 footing below the column is required_ page 94 Front Right Corner Columns Calculate maximum post moments M M: moment M=wind force*-spacing* hZ / 8 h: distance from either grade or top of concrete slab, to bottom chord of end truss h = 188.358" : M= (9.65 7.819) psf/ 144 in.Z/ft.�*75.25".* 188.358"� / 8 M=40484.567 ibAbs M is shared by two endwalls.so the load on one endwall can be cut in half: S: section modulus . . S = b*d* d /'6 NDS 3.3.2 . S = 3.5"* 5.5".5.5"./ 6. S= 17.646 in.3 fb: bending stress post. . . fb 1.M1 / S NDS.3.3.2.. fb = 120242.283in.lbs I / 17.646 in.3 fb= 1147.142psi CM=wet service factor CM= 1 because the post is protected by the building shell; concrete, or embedment from excessive moisture CF= 1 NDS Supplement,. Cr:.repetitve member.factor Cr= 1 NDS 4.3 Cf,,: flat use factor Cf� = 1 NDS 4:3 . Fe: allowable bending stress post : Fb= Fb* Co*CM*.Ct*CL*CF*Cf„*C;*-Cr NDS 4.3. - Fe = 1000 psi *.1.6-* 1,*.1 * 1 * 1 * 1 * 1 *.1 Fb:= 1600 psi fb < Fb .. 1147.142 psi <_ 1600 psi page 95. Calculate.axial compression force in the post Pf: compressive force post Pf=sf*w / 2* (snow roof force + dead load) Pf= 60"* 75:25"* (37.802 psf / 144 PS'/Psf+43 psf 1144 P51 Pf= 1320.069 lbs CF: size factor CF= 1 NDS Supplement table 4D . Cp.: column stability factor CP = (1 + (FcE / Fc*)) / (2 *c) -f(((1 + (FcE / Fc*)) / (2*c))z- (FcE / Fc*) / c) FcE_ .3 *,E / (le./ d)Z NDS 3.7.1 le= Ke* l NDS Appendix G l =40.8" length of column between lateral supports Ke=0.8 NDS 3.7.1 1e= 0.-8`..40.8" le= 32.64" . FcE= .822 * 1400000 psi / (32.64" / 5.5")7. FcE= 32675.702 psi CP= (1 + (32675.702 psi / 2240 psi))./ (2* 8) -f(((1 + (32675.702.psi / 2240 psi)) / (2 * .8))z (32675.702 psi /..2240 psi) / .8) CP = 0.986 Ft= 2207.997 psi f • axial compressive for epost.: f� Pf/ (b.*d)from'.Design of Commercial Post Frame Buildings p,48 fc= 1320.069 lbs"/'(3.5"k 5.57) . fc.= 68.575 psi Calculate Shear at top of endwallpost V: shear at the top of post V= 5/8 (wall force- leeward force).* spacing* h V=.5/8 (9.65 - -7.819) psf /.144 P51/Psf.*:75.25"* 188.358" V= 1074.672 lbs'. f,,: bending shear post... fV= 3 *V / (2 * b *d) : . f„= 3 * 1074.672lbs % (2 * IT'*5 5") f�= 83.741 psi - _ Fv: allowable shear stress post F, = F„*.CD*CM* Ct' Q NDS 4.3 Fv'=:218.75.psi * 1.6 * 1 *.1 * 1 Fv 280 psi page 96 83.741 psi <_280 psi Lateral Strength Assessment. PwZ / fL' PZ,ASD ASAE 486.2 11.3.1 Pz,ASD=283.333 lbs - p,,,Z=ultimate lateral soil resistance at depth z z= 3.333' p.,Z= 3 *6,,,7* KP ASAE 486.2 11.3.1 o',,;Z=effective vertical stress aV'Z=Y*z a"',= 105 * 3.333 6,,x= 350 psf. p = (1 +.sin(0)) / (1 sin(fp))ASAE EP 486.2 11.2.1 fP = soil friction angle tp= 30 ASAE EP486.2 table 1 Kp=.(1 +.sin(30)) 7 (1 sin(30)) KP= 3 PU,r= 3.*,350 3 .. p,,,Z= 3150 psf fL= 1.4 / (._61 - .01 * cp) ASAE.EP486.2.table 4 fL=.4.516 3150 /4.516,>t 283.333 ASAE 486.2.11.3.1 :. Lateral.Strength adequate for.40"deep and 1.5'diameter hole Determine if the diameter of the concrete column encasement is adequate. A_fB*.PASD J (qB - Y* dF) ASAE EP486.2 10.2 A= bearing area.. A= rr*rZ A= 1.767 ft.. bay weight=wall steel weight.+ post weight+girt weight + roof weight+skirt board weight . bay weight= (.66* (sf* (h .+pane length))) + (62.4 .55* (b k.d*_l)) + (62.4*..55'.(b* d* t) *.n) +,.6* ' (BCDL+TCDL) + (62.4* .55.*-(b *d* l} *n} bay weight= (.66 psf* (5'*.(17'+ 50'))) + (62.4 pcf* 55.lbs/ft s * ((3.5 / 12'n'/f.) * (5.5' / 12,in./ft.) 1.9.667)) + (62.4 pcf*.:55 lbs/ft.3* ((1.5" / 12 in./ft.) *.(9.25" / 12 m./ft.) * 5')* 1) +..6* (1 + 3.3) k.5 27.406 +. (62.4.pcf .55 lbs/ft. .((1.5 / 12 /ft.) .(7:25' / 12 _ /ft.). 5')) bay weight=.519.72 lbs PASD= (bay weight+ Pf) PASD•=.(592.623 + 519.72) PASD = 1112.343 lbs qB = ultimate soil bearing.capacity page.97 fB =ASP factor of safety for bearing strength.assessment fe = 1.4 / (.58 - .01 * cp) ASAE EP486.2 table 2 fB = 5 . qB =Y* (•5 * B *Cw, * NY*sY+ dF* CwZ* Nq*dq*sq) ASAE EP486.210.4.1 y= moist unit weight of soil y--1'05 pcf ASAE EP486.2 table 1 dF= 3.333' Cw, ='1 ASAE EP486.2 10.4. . Cw2= 1 ASAE EP486.2 10.4 NY=22.4 ASAE EP486.2 table 6 sY= 0.6 ASAE EP486.2 table 6 Nq.= 18AASAE EP486.2 table 6 dq = 1 +2*tan(cp) * (1 •sin(W))Z* tan"OF./ B) ASAE EP486.2 10:4.1 dq = 1`+2 *0.577* (1 - 0.5)z*tan-1(3.333 / 1.5) dq = 1.3313824270181827. sq = 1.58 ASAE EP486.2 table 6 qB = 105 * (.5 * 1.5 1 *22.4*.0.6 +.3.333 *.1 * 18.41* 1.3313824270181827* 1.58) qB = 14605.482471395411 psf 1.767 >_ 5 * 1112.343 /.(14605.482471395411 - 105 *3.333) 1.767 >_0.39 .... :. Diameter 1.5'adequate for foundation pressure page 98 COLUMN UPLIFT CALCULATIONS d =40" U: uplift resistance U = a* (.33 *n * (((d t) + .5 *w / tan(0))3 *tan(A)z 125 *w3 / tan(0)) AP* (d t)) + .25 * C *Tf*wZ *.t . a: soil density a= 85 pcf C: concrete density C = 150 pcf t, h: collar thickness or thickness of attached concrete t, h =-1.5' • w: collar width w= 1.5' . 6: soil friction angle 0 =0.524 radians AP: post cross section area AP=0.134 ft.Z U = 85-pcf*:(`.33 * 3.14159 * ((((32" / 12'"'/ft•) - 1.5') +.5 *..:1.5•% tan(0.524°))3*tan(0.524°)2 .125 1.5'3 / tan(0.524°)).- 0.134 ft.Z* ((32"./ 12 ' /ft.) - 1.5')) + .25 *.150 pcf*3.14159 * 1.5" * 1 5' U =760.296 lbs bay weight= steel weight+ post weight+ girt weight +goof weight+ skirt board weight bay weight (:66 * (spacing*.h +sf / 2"..roof length)) + (62,4*•.55.. (b*d * l)) + (62A .55 * (b * d * n).+ .6 * (BCDL+TCDL) *sf / 2*w+(62.4* .55 * (b *-d * I).* n) bay weight= (.66 0sf* (6.270833333333333 *21.167+5' / 2* (56' / 2 / cos(18.435°)))) + (62.4 pcf* .55 lbs * ((3.5" / 12 `V t.) * (5.5" / 12 i /ft.) * 19.66T)) + (62.4 pcf* .55 lbs/ft,:3 * ((1.5" / 12 in/tt.) (9.25" / 12"Vft.) *6.270833333333333) * 1.).+ .6 * (1 + 3.3) * 5 / 2 * 6.270833333333333 + (62.4 pcf .55 lbs/ft:3* ((1.5" / 12 '. /ft.) * (7.25" /-12' /ft.) *6.270833333333333')) - bay weight =255:27 lbs total uplift resistance =255.27 lbs +760.296 lbs = 1015.566.1bs. page,99 uplift.force = roof—wind—force* roof area / DOL uplift force = 14.474 psf* 15.677 ft.2 / 1.6 uplift force = 155.473 Lbs Total uplift force of 155.473 lbs is less than total uplift resistance of 1015.566 lbs The direct bearing capacity undervertical Load depends on the bearing capacity of the soil against the . end of the post and.the frictional resistance between the='post and the soil, without producing much end bearing pressure at the butt. Interestingly, it has been determined that firm backfilling of the hole with compacted soil, sand, soil concrete or concrete.is a more effective way of increasing vertical force resistance than placing mats of footings of concrete beneath.the butt ends.Concrete encasement of the post in the ground contact area enlarges the friction surface and can generally be credited with a wood to concrete bond strength of 30 psi. Hoyle Ft Woeste, 0.335-336 Connection Strength =`(breadth *2 + depth*2)' collar thickness*30. Connection Strength (3.5".*2 + 5.5"*2) `10"* 30 psi =5400 lbs Connection Strength = 5400.lbs > 155.473 lbs In summary a 3.5"X5.5"SYP endwall:column, in an augered hole 1:5'in diameter and 40"`.in depth is .. Adequate. A concrete bottom collar 1:5' high.mono-poured with a 8"footing below the column is required page 100 Rear Left Corner Columns Calculate maximum post moments M M: moment . -.. M=wind force*.spacing * hZ / 8 h: distance from either grade or top of concrete slab, to bottom chord of end truss h=-188:358" M= (9.65 - 7.819) psf ! 144 in.z/ft.z*75.25"* 188.358"Z ! 8 M=40484.567 in.lbs M is shared by two endwalls so the load on one endwall can be cut in half.: . S: section modulus S= b*.d* d / 6 NDS 3.3.2` S= 3.5' 5.5* 5.5" f6 . 5 = 17.646in.a fb: bending stress post. fb= I I /.S NDS 3.3.2 : fb= I20242.283in:lbs1 f 17.646 in.3 . fb= 1147.142 psi CM.=wet service-factor r CM= 1 because the post is protected by the building shell, concrete, or embedment from excessive moisture CF= i NDS Supplement Cr: repetitve member factor: C,= 1. NDS 4.3 Cf,,: flat use factor Cf,, = 1 NDS4.3 Fe: allowable bending stress post Fe= F6* CD*Cµ*Ct*CL* CF* Cfj*Ci*C,NDS 4.3 . Fb= 1000.psi * 1.6 * 1 * 1 *,1 *.1 * 1 * 1 *1 .; . Fb= 1600 psi fb < Fe: 1147.142 psi <_ 1600 psi page 101 Calculate axial compression,force in the post Pf: compressive force post Pf=sf*'w / 2* (snow roof force+ dead load) _...-Pf=60".-*..75:25"*.(37.802 psf / 144 Ps/psf+4.3.psf / 144 as;/psf) + ::-J.:. ,.., — Pf= 1320.069 lbs CF: size factor CF= 1 NDS Supplement table 4D Cp: column stability factor Cp= (1 + (FEE / Fc*)) / (2 *c) -f(((1 + (Fce./ F,*)) / (2 * c))z - (FEE / Fc*) / c) FcE = .3 * E / (le / d)2 NDS 3.7.1 le = Ke* l NDS Appendix G l=.40.8"length of column between lateral supports Ke=0.8 NDS 3.7.1 le 0.8 *40.8".. le=— 32.64" . TEE .= .822_* 1400000 psi / (32:64" / 5.5")2 F,E= 32675.702 psi .Cp= (1 + (32675.702 psi / 2240 psi)) / (2* .8) -I(((1.+ (32675.702 psi / 2240 psi)) / (2 *:8)) , (32675.702 psi / 2240 psi) / .8) Cp= 0.986 Fc'= 2207.997 psi f,: axial compressive-force post f� = Pf% (b * d).from Design of Commercial Post-Frame.Buildings p.48 f,= 1.320.069 lbs / (3.5"* 5.5") f`= 68.575.psi Calculate Shear at_top."of endwall post. V: shear at the,top of post V= 5/8 (wall force leeward force) 'spacing * h V= 5/8 (9:65 :-7.819) psf / .144 ps;/Psf.*.75.25"* 188.358".. V= 1074.672 lbs fv: bending'shear post f = 3 *V / (2*b *d).. f = 3 * 1074.672 lbs / (2 * 3.5"* 5.5") f,= 83.741 psi Fv: allowable shear stress post F F; CD,*CM* Ct* C; NDS 4.3 F� 218.75 psi * 1.6 * 1 * 1 * 1 F,;= 280 psi page 102 f, F 83.741 psi <_280 psi Lateral Strength Assessment = . b= 1.5' P,,,z / f L? Pz,ASD ASAE 486.2 11.3.1 Pz,ASD = 283.333 lbs p,,,z= ultimate lateral soil resistance at depth z z= 3.333' Pu,z= 3 *6,,z* Kp ASAE 486.2 11.3.1. 6'V,z= effective vertical stress CYV,z=y*Z 6'v,7= 105*3.333 av,z= 350 psf . Kp = (1 + sin(cp)) /-(1 sin(cp))ASAE EP 486.2 11.2.1 cp =soil friction angle, cp= 30 ASAE EP486.2 table'1 KP = (1 +sin(30)) / (1 - sin(30)) Kp= 3. P.,z= 3150 psf f�= 1.4 / (.61 - .01 * cp) ASAE EP486.2 table 4 fL=4.516 3150 % 4.516 2 283-.333 ASAE.486.2.11:3.1 :. Lateral Strength adequate for 40"deep and 1.5' diameter.hole Determine if the diameter of the concrete column encasement is adequate ... A 2:fB * Ptso / (qB - Y*dF)ASAE EP486.2 10.2 A= bearing area . . A=rr*rz A= 1.767 ft; bay weight=wall steel'weight+ post weight+ girt weight+ roof weight+ skirt board,weight.. , bay weight = (.66 * (sf* (h +panel length))) + (62.4* .55 * (b..*d * l))'+„(62.4.* .55 *, (b * d * l).* n).+ .6 (BCDL+TCDL).+ (62.4* .55 * (b *d * l) *n) bay weight= (.66 psf* (T-* (17' + 50'))) + (62.4 pcf* .55-lbs/ft.'* ((3.5" / 12 'n./ft.) * (5.5" / 12'n' 19.667)) + (62.4 pcf* .55 lbs/ft.3* ((1.5" / 12 'n'/fc.) * (9.25" / 1.2 in./tt.) * T) *.,1) + .6 *..(1:t3.3) * 5 27.406 + (62.4 pcf* .55 lbs/ft.3.*.((1.5" / 12,n./ft.) * (7.25" / 12 'n./ft.) *.5')) bay weight,= 519.72 lbs Paso = (bay weight + pf) Paso= (592.623 + 519.72). Paso = 11,12.343 lbs qB =ultimate soil bearing capacity page 103 fB=ASD factor of safety for bearing strength assessment fB = 1.4 / (.58 - .01 * co) ASAE EP486.2 table 2 fB = 5 qB =y* (.5* B *Cwl *NY*sY+ dF* Cvn* Nq* dq*Sq).ASAE EP486.2 10.4.1 y= moist unit weight of soil y= 105 pcf ASAE EP486.2 table 1 - dF= 3.333. . CW, = 1 ASAE EP486.2 10.4 Cw,= 1 ASAE EP486.2 10.4 NY=22.4 ASAE EP486.2 table 6 sy= 0.6 ASAE EP486.2 table 6 Nq= 18.4 ASAE EP486.2 table 6 dq= 1 +2*tan * (1 - sin(cp))Z*.tan''(dF/ B) ASAE EP486.2 10.4.1 dq = 1 +2*0.577* 0 -0.5)2*tan-'(3.333 / 1.5). dq =.1.3313824270181827 sq= 1.58 ASAE EP486.2 table 6 . qB = 105 * (.5 * 1.5 * 1 *22.4*0.6 +.3.333 *:1 *. 18.4* 1.3313824270181827* 1.58) . qB= 14605.482471395411 psf. 1.767 >_ 5,* 1112.343 /,(.14605.482471395411 = 105*3.333) 1 J67:>_ 0.39 :. Diameter-.1.5' adequate for foundation_pressure page 104 COLUMN UPLIFT CALCULATIONS d =40" U- uplift resistance U =a*. (.33 * rr* (((d - t) + .5 *w / tan(6))3 * tan(6)2 - .125 *w3./ tan(6)) - AP* (d - t)) + .25 *C*ir*w2 *t a: soil density a=85 pcf C: concrete density C.= 150 pcf t, h: collar thickness or thickness of attached concrete t, h = 1.5' w: collar width w= 1.5' 8: soil friction angle - '8 = 0.524 radians AP: post cross section area AP 0.134 ft.2 U = 85 pcf* (.33 *3.14159 *.((((32 / 12 m'/ft_) - 1.T) + .5 *1.5' / tan(0.524°))3 *tan(0.524°)2 - .125 1.5'3 / tan(0.524°)) - 0.134 ft.2* ((32" / 12 'n./ft_) - 1.5')) + .25 * 150 pcf*3.14159 * 1.T2* 1.5' :._-- U = 760.296 lbs- bay weight=steel weight + post weight-+ girt weight + roof weight+skirt board weight bay weight= (.66 * (spacing* h +sf./ 2* roof length)) + (62.4* .55 *.(b*d * l))..+ (62-4* .55 * (b* d * n) + .6 * (BCDL.+TCDL) *sf / 2*w+(62.4* .55 * (b* d* l)* n) bay weight•= (.66 psf* (6.270833333333333 *21.167+ 5' / 2* (50''/ 2 / cos(18.435°)))) + (62.4 pcf* .55 lbs/ft.3 * ((3.5'.' / 12 m./ft.) * (5.5" / 12 inVft.) * 19.667)) + (62.4 pcf*_.55 lbs/ft.'* ((1.5" / 12'n./ft.) (9.25" / 12 m'/ft.) *6.270833333333333') * 1) + .6 *(1,:+ 3.3) * 5 / 2 * 6.270833333333333 + (62.4 pcf 55 lbs/ft.3* ((1.5,, / 12in./ft.) *.(7.25' / 12 '"'/ft.) *6.270833333333333' )) bay weight=255.27 lbs total uplift resistance =255,27 lbs +.760.296 lbs= 1015.566 lbs page 105 uplift force.= roof_wind_force* roof area / DOL . uplift force = 14.474 psf* 15.677 ft.z / 1.6 uplift force = 155.473 Lbs -Total uplift force of 155.473 lbs is less than total uplift resistance of 1015.566 Lbs The direct bearing capacity under vertical Load depends on the bearing capacity of the.soil against the ----end of thepost and the frictional resistance between the post and the soil, without producing much z. = end bearing pressure at the butt. Interestingly, it has been determined that firm backfilling of the hole with compacted soil,sand, soil concrete or concrete is a more effective way of increasing vertical.: .: force resistance than placing mats of footings of concrete beneath the butt ends. Concrete encasement' of the post in the ground contact area enlarges the friction surface and can generally be.creclited with. a wood to concrete bond strength of 30.psi. Hoyle Et Woeste, p.335-336 Connection.Strength = (breadth.*2 +depth *2) *collar thickness.* 30 Connection:Strength 3 5"*2 + 5.Y*2 * 10"* 30 psi = 5400 lbs. Connection Strength = 5400 Lbs > 155.473 lbs. In summary a 3.5"X5.5"SYP endwall column,in an augered hole 1.5'in diameter and 40"in depth is adequate.. A concrete bottom collar 1.5' high mono=poured with a 8"footing below the column..is required page 106.. . Rear Right Corner Columns Calculate maximum post moments M M: moment M=wind force * spacing* hZ / 8 h: distance from either grade or top of concrete slab, to bottom chord of end truss h = 188.358' M= (9.65 7.819) psf / 144 in.2/ft.z* 75.25"* 188.358"Z / 8 M=40484.567 in.lbs M is shared by two endwalls so the load on one endwall can be cut in half. S: section modulus S = b * d *d / 6 NDS 3.3.2 . S= IF * 5.5" * 5.5" / 6 S = 17.646in.3 fb: bending stress post, fb=. I M 1 / S NDS 3.3.2: fb = 120242.283in.lbs I / 17.646 in.3 fb= 1147.142 psi. CM=wet service factor CM= 1 because the post is protected by the building shell, concrete; or embedment from excessive moisture CF= 1 NDS'Supplement Cr: repetitve member factor 'Cr= 1 NDS4.3 Cf,,: flat use factor Cf,= 1 NDS 4.3. Fe: allowable be stress post Fe= Fb* Cp*CM* Ct*CL*CF*Cf.* C,* Cr NDS-4.3 Fe= 1000 psi * 1.6* 1 * 1 * 1 * 1 * 1 * 1- 1 : Fb 1600 psi , fb < Fb 1147.142.psi <_ 1600 psi . page107 Calculate axial compression force in the post Pf: compressive.force post Pf= sf*w / 2 * (snow roof force+ dead load) Pf.= 60"*75.25"* (37.802 psf / 144 PSI/ -+4.3-psf 1144'PS'/PSf)-+ Pf= 1320.069 tbs.- CF: size factor CF= 1 NDS Supplement table 4D. CP: column stability factor CP = (1 + (Fcj / Fc*)) / (2 *c) -f(((1 + (FcE / Fc*)) / (2* c))z - (FcE / Fc*.) / c) FcE= .3 * E /.(le / d)z NDS 3.7.1 le= Ke* l NDS Appendix G l =40.8"length of column between lateral.supports Ke= 0.8 NDS.3:7:1 .. p. le= 32.64" FcE= .822* 1400000 psi / (32.64" / 5.5„)zW. .::FcE.= 32675:702 psi CP = (I + (32675.702 psi /.2240 psi)) /:.(2.* .8) 'f(((1 + (32675.702.psi / 2240 psi))./,'(2 * .8))2 (32675.702 psi / 2240 psi)./ .8) . CP = 0.986 Fc =2207.997 psi fc:,axial compressive,force post fc.= Pf/ (b:*d) from Design of Commercial Post Frame Buildings p.48 f�.= 1320.069 lbs / (3.5 * 5.5') f,= 68.575 psi.,.: Calculate Shear at top of endwall post` V: shear at.the top of post'. V= 5/8 (wall'.force'- leeward force) *spacing * h V= 5/8 (9.65 - -7.819) psf./ 144 Psi/Psf*75.25* 188.358" V= 1074.672 tbs. . f,,: bending shear-post fV.= 3 * 1074.672.lbs / (2* 3:5"*5.5") fV= 83.741 psi F,;:.allowable shear stress post. Fv = FV*Co*CM*.Ct*.Ci NDS4.3 F,; 218.75 psi * 1.6 * 1 * 1 * 1 F,'=280.psi page 108 f F,' 83.741 psi < 280 psi Lateral Strength Assessment b = 1.5 - Pu,z / fL a Pz,ASD ASAE 486..2 11.3.1 Pz,ASD=283.333 lbs p,,,z= ultimate lateral soil resistance at depth z z= 3.333' Pu,z= 3 * d,,z.*KP ASAE 486.2 11.3.1 6v,z= effective vertical stress. 6�z=Y*z 6�z= 105 * 3.333 . 350 psf. KP = (1 +sin(cp)) / (1 - sin(cp))ASAE EP 486.2 11.2A cp=soil friction angle cp= 30° ASAE EP486.2 table 1 KP= (1 +sin(30))-_/ (1 sin(30)) KP= 3 Pu,z= 3 *350*3 , . Pu,z= 3150 psf fL= 1.4 / (.61 - .01 * fp) ASAE EP486.2 table 4 fL=4:516 3150 /4.516 >_283.333 ASAE 486.2 11.3.1. Lateral Strength adequate for 40"deep and 1.5'diameter hole Determine if the-diameter of the concrete column encasement is adequate * *dF ASAE EP486.2 10.2 A=.bearing area A= rr* rz A = 1.767 ft. bay weight=wall steel weight+ post weight+ girt weight+ roof weight+skirt board weight bay weight= (.66 * (sf* (h +panel length)}) + (62.4*. .55:* (b*d * 1)) + (62.4* .55 * (b * d * l) * n) + :6 *- (BCDL+TCDL) + (62.4* .55 * (b *d * 1) *.n) bay weight= (.66 psf* (5'*.(17'+ 50'))) + (62A'pcf* .55.1bs/ft.3 *: ((3 5" / 12 'n•/ft.) * (5.5 / 12'.n /ft),�; 19.667)) + (62.4 pcf* .55 lbslft.3 * ((1.5" / 12 /ftin.) * (9.25" / 12.in./ft:) * 5') * 1) + 6* (1 + 3.3).'.5 27.406 + (62.4 pcf* 55 lbs/ft.3 * ((1.5,, 1 12 in•/ft•} 7 25'. / 12... /ft.) : 5')} bay weight= 519.72 lbs `. PASD = (bay weight + Pf) PASD = (592.623 + 519.72):.:: PASD = 1112.343 this qB =ultimate soil bearing capacity, page 109 fB =ASD factor of safety for bearing strength assessment, fB = 1.4 / (.58. - .01 * cp) ASAE EP486.2 table 2 . fB = 5 qB =y* (.5 * B * Cwj * NY*SY+ dF* CW2*Nq*dq *sq) ASAE EP486..2 10A.1 y= moist unit weight of soil y 105 Pcf ASAE EP486.2 table 1 dF= 3.333' CW1 = 1 ASAE EP486.2 10.4 CW2= 1 ASAE EP486.2 10.4 NY=22.4 ASAE EP486.2 table 6. sY 0.6 ASAE EP486.2 table 6 Nq= 18.4 ASAE EP486.2 table 6 dq= 1 + 2*tan((p) * (1 - sin(ip))z*tan-'(dF / B) ASAE EP486.2 10.4.1 dq = 1 +2 *0.577* (1 0'5)z*tan-'(3.333 / 1.5) 'dq 1.3313824270181827. sq= 1.58 ASAE EP486.2 table 6 . qB= 105.* (.5 * 1.5 * 1 *.22.4*0.6 + 3.333 *1. ' 18:4* 1.3313824270181827* 1.58) qB = 14605.482471395411 psf. . 1.767> 5 * 1112.343 / (1.4605.482471395411 - 105 *3.333) 1.767> 0.39 ;. Diameter 1.5' adequate for foundation.pressure page 110 COLUMN UPLIFT CALCULATIONS d =40" U: uplift resistance U = a* (.33 *n* (((d t) + .5 *w / tan(8))3.* tan(6)2 125 *w3 / to.n(6)) .AP * (d t)) + .25 *C *rr*wZ * t a: soil density a= 85 pcf C: concrete density. C = 150 pcf t, h: collar thickness or thickness of attached-concrete t,.h = 1.5' w: collar width w 8: soil friction angle 0 =0.524 radians AP: post cross section area AP = 0.134 ft.Z . U = 85 pcf* (.33 *3.14159 * ((((32" / 12 i /ft) - 1.T) + .5 *.1:5' / tan(0.524°))3*tan(0.524 )z .125 tan(0.524`)) 0.134 ft.' * ((32"/ 12 '" 1.5'))+ .25 * 150 pcf* 3.14159 * 1.5'. * 1.5' U = 760.296 lbs bay weight = steel weight+,post weight + girt weight + roof weight.+skirt board weight bay weight= (.66 * (spacing*h +sf/.2* roof length)).+ (62.4*..55 *;.(b*.d * l)).+ (62.4* .55 . (b *d * l) * n) + .6 * (BCDL+TCDL) *sf/ 2*w+(62.4* .55 * (b "d * l) *n) bay weight = (.66.psf* (6.270833333333333 *21.167+ 5'../ 2 * (50' / 2 / cos(18.435°)))) + (62.4'.pcf* .55 lbs/ft.3 * ((3.5" / 12 'n./ft.) *.(5.5" / 12 in./ft.) * 19.667)) +.(62.4 pcf* .55 lbs/ft:3* ((1.5" / 12 '"./ft.) (9.25'' / 12' /ft,) *:6.270833333333333'),*.1.) + .6 * (1 + 3.3) *5:/ 2.* 6.270833333333333 + (62.4 pcf .55 lbs/ft.3 * ((1.5" / 12 'n.�ft.) * (7.25" / 12 ",/ft.) *6.270833333333333')) bay.weight.=.255.27lbs . total uplift resistance 255.27 lbs + 760.296'lbs =.1015.5661bs ; page 111 uplift force= roof—wind—force* roof area / DOL uplift force = 14.474 psf* 15.677 ft.z / 1.6 uplift force = 155.473 lbs Total uplift force of 155.473 lbs is less'-than total uplift resistance of 1015.566 lbs The direct bearing capacity under vertical load depends on the bearing capacity of the soil against the end of the post and the.frictional resistance-between the-post thepost and:the soil, without producing.much end bearing pressure at the butt. Interestingly, it has been determined that,firm backfilli,ng.of the hole with compacted soil, sand, soil concrete or concrete is-a more effective way of increasing vertical force resistance than placing mats of footings of concrete beneath the butt ends. Concrete encasement of the post in the ground contact area_enlarges'the.friction surface and can generally.be credited with a wood to concrete bond strength of 30 psi. Hoyle Et Woeste, p.335-336 Connection Strength = (breadth ' 2 +.depth *2) *collar thickness *-30 Connection Strength = (3.5"*2 + 5.5"*2)* 10"* 30 psi, 5400 lbs Connection Strength = 5400 lbs > 155.473,lbs In summary a 3.5"X5.5"SYP endwall column;in'an angered hole 1.T.in.diameter and 40";in"depth is. adequate. A concrete bottom collar 1.5'high mono-poured_with a8"footing below the column.is required page 112 GIRT CALCULATIONS Left Sidewall Girts wall girt size: 2"X6" spacing between girts = 37 girt span 115.875" supported by 2x4 blocking every 115.875" page 113 Fb: allowable girt pressure .Fe = Fb*CD' CM*Ct*CL* CF*Cfu * Ci* Cr NDS 4.3 Co: load duration factor Co = 1.6 NDS 4.3 CM: wet service factor CM= 1 because girts are protected from moisture by building envelope Ct: temperature.factor Ct = 1 NDS 4.3 Cf,,: flat use factor Cf,, = 1 NDS4.3 Ci: incising factor Ci = 1 'NDS 4.3 Emi,,: reference adjusted modulus of elasticity Emig =470000 psi NDS Supplement Cr: repetitive member factor Cr=. 1 NDS'4,3 1,,: laterally-unsupported span length lu = 115.875" le: effective length le= 1.63 * l„ NDS table 3.3.3 le=205.376.,. .. CF: size factor . _ CF= 1.3 NDS 4.3 Ct: beam stability factor CL= 1 NDS 3.3.3 Fb = 850 psi *:1..6*"1 * 1 * 1 * 1.:3 .1 * 1 * 1 Fe 1768 psi fb:.girt test pressure fb= 6 *wall—wind—force / 144*girtSpacing*span z / 8 /. (b *dz) NDS 3.3 fb.= 6 * 16:304 psf / 144 in:z/ft.z* 38.5"* 115.875"z / 8 /.'(1.5"* 5.-5`'z) fb=967.447 psi .967.447 <_.1768 stressed-to 55% :• 6"X2"#2 OK in bending page 114 F,;: allowable shear pressure F„= 135 NDS Supplement Table 4-A F,; = F * Cp*CM* Ct*C; NDS 4.3 F,;=.135 psi * 1.6* 1 * 1 * 1 F�= 216 psi NDS Supplement f,: shear girt pressure f = 3 * (wall_wind_force / 144* girtSpacing*span / 2) / (2 * b * d) NDS 3.4 f = 3 * (16.304 psf/ 144 in.Z/ft.Z*38.5"* 115.875" / 2) / (2* 1.5" * 5:5") f„=45.92 psi 45.92 < 216 stressed to 21%,.. 6"X2"#2 OK in shear Deflection Dail,,,,: allowable deflection l= 115.875' Aak,= 115.875" / 90 . Aauow 1.2875" - - - Amax: maxiinum deflection Amax= 5 *W*spacing *spa_n4 / 384 / E /I from'http://www.awc.org/pdf/DA6-BeamFormulas.odf p.4 E'. Modulus of Elasticity E= 1300000 psi NDS Supplement I: moment of inertia I ` b *.d3 / 12 . 1 = 1.5 * 5.5"3 / 12 I =20.796875.in.4 Amax- 5 *..11.413 psf / 144 P"../p,f k 38.,5"* 115.875"4 / 384,/ 1300000 psi / 20.796875:in.4 components and cladding.reduced by .7 per footnote f of IBC table 1604.3. Amax= 0.26495" < 1.2875" Right Sidewall Girts page 115 wall girt size: 2"W' spacing between girts = 37" - girt span = 115.875" supported by 2x4 blocking every 115:-875" —1 Fb: allowable girt pressure " Fe = Fb* Cp* Cv,* Ct* CL* CF*Cf,,* Ci* Cr NDS 4.3 Co: load duration factor :. Cp = 1.6 NDS 4.3 . Cµ: wet service factor CM.= 1 because girts are protected from moisture by.building envelope . Ct: temperature factor Ct=.1 NDS.4.3 Cf,,: flat use factor Cf„ = 1 NDS4.3 .. Ci: incising factor . Ci= 1 NDS4.3. ,. Emi,,: reference,adjusted.modulus of elasticity Emin =470000.psi NDS Supplement C: repetitive member factor Cr= 1 ND5.4.3 1,,: laterally unsupported span length 1„ :115.875". A,: effective length le= 1.63 * l,; NDS table 3.3.3 le=205.376" CF size factor CF= 1.3 NDS 4.3.''. CL: beam stability factor: CL= 1 NDS 3.3.3 Fb'= 850 psi * 1.6`1 * 1 * 1 * 1.3 * 1:* 1 * 1 Fb'.=..1768 psi page 116 fb: girt test pressure fb = 6*wall—wind—force / 144* girtSpacing*spanZ / 8 / (b * dZ) NDS 3.3 fb =6* 16.304 psf / 144 in.2/ft.Z*38.5"* 115.875"2 / 8 / (1.5"* 5.5"2) fb=967.447 psi 967.447 <_ 1768 stressed to 55% :- 6"X2"#2 OK in bending F,': allowable shear pressure F = 135 NDS Supplement Table 4-A F�= F,* Cp* CM*.Ct*Ci NDS4.3 F, = 135psi * 1.6 * 1 * 1 * 1 F, =216 psi NDS Supplement f,: shear girt pressure f = 3 * (Wall—wind—force / 144* girtSpacing *span /.2) / (2 * b'* d) NDS 3.4 f = 3 * (16.304.psf/ 144 in.2/ft:Z* 38.5"* 115.875." / 2) / (2* 1.5" 5.5") f„=45.92 psi 45.92 < 216 stressed to 21% :• 6"X2"#2.OK in shear Deflection . dauow: allowable deflection l = 115.875" Hallow= 115.875" / 90 Hallow= 1.28.75" page 117. A,,,,,: maximum deflection Amzx= 5 *W*spacing * span / 384 / E / I from http://www.awc.org/pdf/DA6-BeamForniulas.pdf p.4 E: Modulus of Elasticity E= 1300000 psi NDS Supplement I: moment of inertia = 1.5"*55s / 12 =20.796875 in.4 Amax= 5 * 11.413 psf / 144 PS'/Psf*38.5"* 115.875"4 / 384./ 1300000 psi_ / 20.796875 in.4 components and cladding reduced by .7 per footnote f of IBC table 1604.3 Amax=0.26495" < 1.2875" Front Endwall Gifts page 118 wall girt.size: 6"X2" spacing between girts= 13.7" girt span = 102.50000000000009" Fb: allowable girt pressure Fe = Fb*Co*-CM*Ct*CL* CF*Cf,* Ci* C, NDS 4.3 CD: load duration factor CD= 1.6 NDS 4.3 CM: wet service factor . CM= 1 because girts are protected from.moisture by building envelope Ct: temperature factor Ct= 1 NDS 4.3 Cf,,: flat use factor Cf,, = 1.15 NDS 4.3 . Ci: incising factor Ci = 1 NDS 4.3 Emig: reference adjusted modulus of elasticity Emi„ =470000 psi NDS Supplement Cr: repetitive member-factor: Cr'= 1.15 NDS 4.3 laterally unsupported span length l,;= 120 le: effective length le = 1.63*:lu NDS table 3.3.3 le= 200:1" CF: size factor CF= 1.3 NDS 4.3 CL: beam stability factor CL= 1 NDS 3.3.3. Fe 850 psi 1.6 1 . 1 .1 1.3 .1.15 .1 1.15 Fb'=2338.18 psi page 119 fb: girt test pressure fb = 6 *wall_wind_force / 144*girtSpacing*span, / 8 / (b dZ) NOS 3.3 fb= 6 * 16.304 psf / 144 in.z/ft.Z* 19.2"* 102.50000000000009"2 / 8 / (5.5"* 1.5"Z) fb= 1384.227 psi 1384.227 <_2338.18 stressed to 59% :• 6"X2"#2 OK in bending' F,;: allowable shear pressure F = 135 NDS Supplement Table 4-A F,,'F * Cp*CM* Ct*QNDS4.3 F, 135 psi * 1.6 *.1 1 * 1 F,;= 216 psi NOS Supplement. f,: shear gi.rt.pressure f„= 3 * (wal.l_wind_force / .144*girtSpacing*span,/* 2) / (2 * b*d) NOS 3.4 f = 3 * (16.304 psf / 144 in.Z%ft.z* 19.2"* 102.50000000000009" / 2)./ (2* 5.5" *.1.5") f =20.257 psi 20.257< 216 stressed to 9% • 6"X2"#2 OK in shear - Deflection. Aauow: allowable deflection l= 102.5" Aauow= 102.5 / 90 Aauow-= 1:1389" page 120 AmaX: maximum deflection Ama,,= 5 *W*spacing *span4 / 384 / E / I from http://www.awc.org/pdf/DA6-BeamFormulas.pdf p.4 E: Modulus of Elasticity E= 1300000 psi NDS Supplement I: moment of inertia I = b *.ds.% 12 1.50 1 =-1.546875 in.4 Amax= 5 * 11.413 psf/ 144 P`/Psf* 19.2 * 102.5"4 / 384`/ 1300000 psi / 1.546875 in.4 components and cladding reduced by .7 per footnote f of IBC table 1604.3 Amax= 1.08762" <_ 1.1389.. Front.Endwall Girts .. page 121 wall girt size: 8„X2" spacing between girts = 11.95" girt span = 111.25 Fb: allowable girt pressure Fe='Fb*CD*CM* Ct.*.CL*CF*Cf„*Ci*-Cr NDS 4.3 . CD: load duration factor CD = 1.6 NDS 4.3 CM: wet service factor. CM= 1.because girts are protected from.moisture by building envelope Cf: temperature factor. CL= 1 NDS 4.3 . Cf,,: flat Use factor: Cf„ = 1.NDS 4.31. Ci: incising factor Ci= 1 NDS-4.3 Emi,,: reference adjusted modulus of elasticity Emi„=470000 psi NDS Supplement Cr: repetitive member factor. Cr= 1.15 NDS 4.3 t,,: laterally unsupported span length . l = 120". le: effective length le= 1.63 * l„ NDS table 3.3.3 le =200.1" 6. CF: size factor'.. CF=:1.2.NDS 4.3 CL: beam.stability factor. CL= 1 NDS 3:3:3 Fe 850 psi *1.6 * 1 *.1 *.1 *.t2* 1.* 1 * 1.15 Fb = 1876.8 psi page 122 fb: girt test pressure fb= 6 *wall—wind—force / 144* girtSpacing*spanz / 8 / (b*d) NDS 3.3 fb = 6 * 16.304 psf / 144 in.Z/ft.z* 19.2"* 111.25"2 / 8 / (7.25"* 1.5"z) fb= 1237.042 psi 1237.042 < 1876.8 stressed to 66% :. 8"X2"#2 OK in bending F, : allowable shear pressure F„= 135 NDS Supplement Table 4-A F, = Fv* CD.*CM* Ct*Ci NDS 4.3 F,;= 135psi * 1.6* 1 * 1 * 1 F, =216 psi NDS Supplement f,,: shear girt pressure fv= I* (wall wind_force / 144* girtSpacing"span / 2) / (2 * b *d) NDS 3.4 fv= 3 * (16.304 psf/ 144 in.z/ft.z* 19.2".* 111.25- % 2) / (2 *7.25" *1.5") f = 16.679 psi 16.679 <_ 216 stressed`to 8% :. 8"X2"#2 OK in shear Deflection Aauow: allowable deflection l = 111.25" �auow= 111.25" / 90 Aauow=.1.2361 page 123 An,aX: maximum deflection Amax= 5'*W*spacing *span / 384 / E / I from http://www.awc.org/pdf/DA6-Beam Formulas.pdf pA E: Modulus of Elasticity E= 1300000.psi NDS Supplement I: moment of inertia I = b*d3. / 12 . . 7.25"* 1.5,s / 12 =2.0390625 in.4 Amax= 5 * 11.413 psf / 144 P"IPsf* 19.2" * 111.25"4 / 384 / 1300000 psi./ 2.0390625 in.4 components and cladding reduced by .7 per footnote f of IBC table 1604.3. Amax= 1.145 < 1.2361" Front Endwall Girts Wall girt.size: 2"X6".. spacing between girts= 17.7" girt span = 145" supported by 2x4 blocking every 145" page 124 Fb: allowable girt pressure Fe = Fb*Cp* CM* Ct*CL* CF*.Cf„ * Ci* Cr NDS 4.3 CD: load duration factor CD = 1.6 NDS 4.3 CM: wet service factor CM= 1 because girts are protected from moisture by building envelope Ct: temperature factor Ct = 1 NDS 4,3 CfU: flat use factor Cf„ = 1 NDS 4.3 Ci: incising factor Ci = 1 NDS 4.3 Emii,: reference.adjusted modulus of elasticity Emi„ =470000 psi NDS Supplement C,: repetitive member.factor Cr= 1.15 NDS 4.3 laterally unsupported span length l„ = 145" le: effective length - le= 1.63 * t„ NDS table 3.3.3 .. le-=252.85" CF: size factor CF= 1.3 NDS,4.3 CL: beam stability factor, CL= 1 NDS 3.3.3 . Fn:=850 psi * 1.6 *.1 *-1 * 1 * 1.3:* 1 * 1 *_1.15 .: Fb:=2033.2'psi fb: girt test pressure fb=.6 *wall_wind_foree / 144*girtSpacing*span / 8 / (b* dz) NDS 3.3 fb= 6 * 16.304 ' f / 144 in.z/ft.z* 19.2"* 145,2 / 8 / (1.5"* 5.5.,z) ' fb=755.482 psi 755.482 <_ 2033.2 stressed to 37% 6"X2' #2 OK in bending page 1.25 F,;: allowable shear pressure F„= 135 NDS Supplement Table 4-A F,;= F * Co *CM*Ct*Ci NDS 4.3 F,;= 135 psi * 1.6* 1 * 1.* 1 F, =216 psi NDS Supplement - f,: shear girt pressure f = 3 * (wall_wind_force / 144*'girtSpacing k span / 2) / (2* b d).NDS 3.4 f„=3 * (16.304 psf/ 144 in.z/ft.z* 19.2"* 145" L 2) / (2 * 1.5" * 5.5") f„=28.656 psi 28.1656 <_ 216 stressed to 13% :. 6"X2"#2 OK in shear Deflection.. . Aallowi allowable deflection l= 145 Aaiiow= 145"./ 90 Aall.= 1.6111" : A,,,ax: maximum deflection . Amax= 5 *W* spacing *span4-/ 384 / E / 1 from http://www.awc.org/pdf/DA6-BeamFormulas.pdf p.4 E: Modulus of Elasticity : E= 1300000 psi NDS Supplement I: moment of inertia 1'= b* &1-12 = 1.5":* 5.5"3 / 12 1 =20.796875 in.4 Amax= 5 * 11.413 psf / 144 Os. p,f.* 19.2"* 145"4 / 384 / 1300000 psi /'20.796875 in.4 components and cladding reduced by .7 per footnote f of IBC table.1604.3 Amaz= 0.32397" <.1-611.1" Front Endwall Girts wall girt size: 6"X2". . spacing between girts = 13.7 girt span = 91" - page 126 Fb: allowable girt pressure Fb'= Fb* Co *CM* CtkCL* CFkCfL *Ci* C, NDS 4.3 CD: load duration factor, CD= 1.6 NDS 4.3 CM: wet service.factor CM= 1.because girts are protected from moisture by building envelope Ct: temperature factor Ct.= 1 NDS 4.3 Cf;,: flat use factor Cf,, = 1.15 NDS 4.3 Ci: incising factor Ci = 1 NDS 4.3 Emi,,: reference.adjusted modulus of elasticity . Emir, =470000 psi NDS Supplement C,: repetitive member.factor Cr= 1.15 NDS 4.3: l,,:.laterally unsupported span length l„= 96`' le: effective length le= 1.63-* l„ NDS table 3.3.3 . le= 160:98' CF size factor. CF= 1.3 NDS 4.3 CL: beam stability factor CL= 1 NDS 3.3.3. . . -.. .. *. *' k * ..* * * * - Fe=.850 psi . 1.5 1 1 1 1.3. 1.15. 1. 1.15 Fe =2338.18 psi fb: girt.test pressure fb=6 *wall—wind—force / 144*girtSpacing*•span2 / 8 dz) NDS 3.3 fb= 6 * 16.304 psf / 144 in.Z/ft.Z*.19.2,,*91..2 / 8 / (5.5`'* 1.5"7p . fb=.1091.045 psi 1091.045 <_2338.18.stressed to 47% :. 6"X2"#2 OK in bending. page 127 F,;: allowable shear pressure F„= 135 NDS Supplement Table 4-A F, = F\,* CD*CM" Ct* Ci NDS 4.3 F,; = 135 psi* 1.6 * 1 * 1 * 1 F,;=216 psi NDS Supplement f,,: shear girt pressure f„=3 * (wall_wind_force / 144*girtSpacing*span / 2) / (2 * b *d) NDS 3.4 f„=3 * (16:304 psf / 144 in.Z/ft.z* 19.2"*91" / 2) / (2* 5.5" * 1.5") f = 17.984 psi 17.984 < 216 stressed to 8% 6"X2"#2 OK in shear Deflection Aallow: allowable deflection 1 = 91" Aaliow=91" / 90 kilm,= I-0111... Amax: maximum deflection ." Amax= 5,*W*.spacing*span4 / 384 / E / I from htto://Www.awc.org/pdf/`DA6-Beam.Fort-hutat.p.df p 4 . E: Modulus of Elasticity:. E= 1300000 psi NDS Supplement I: moment of inertia 1 = b * d3 / 12 . 1 = 5.5"*.1.5"112 4 1 = 1.546875 in. . Amax= 5* 11..413 psf./ 144 Psi/P5t*.19.2"*91"4 / 384 / 1300000 psi./ 1.546875 in components and: cladding reduced by .7.per footnote f of IBC table 1604.3 Amax=0.67569" <_.1.0111 Front Endwall Girts page 128 wall girt size: 10"X2" spacing between girts = 9.95" girt span = 122.75" Fb: allowable girt pressure Fb ---b*-CD* C *.Ct*".CL*-CF* Cfu* C;* Cr NDS 4.3 CD: toad duration factor Co= 1.6 NDS 4.3 CM: wet service factor CM= 1 because girts are protected from moisture by building envelope Ct: temperature factor Ct= 1 NDS 4.3. Cf,,: flat use factor Cf, = 1 NDS 4:3 Ci: incising factor Ci=1 NDS 4.3 ER,i,,: reference adjusted modulus of elasticity Emi„=470000 psi NDS Supplement Cr: repetitive member factor C,.= 1.15 NDS 4.3 1,,: laterally unsupported span length i„ = 144 le: effective•length le= 1.63 * lu NDS table.3.3.3 1e= 239..22'` . CF: size factor CF= 1..1 NDS 4.3 CL: beam stability factor CL_ 0 + (FK/ Fb*)) / 1.9 -f(((1 + (FeE/ Ft)*))./ .1.9)2 - FeE / Fb* / .95) FeE= 1.2 * Emiri''/ Rez RE, = r(le*d / bz) RB =f(239:22 * 1.5" 19.25"z) RB=2.648 FeE= 1.2*470000 psi / 2.0482 FbE= 134484:993 psi Fb*:.Fb*Cp*CM*Ct* CF*Ci*C,NDS 3.3.3 Fb= 850 psi.NDS Supplement Table 4-A Fb*= 850 psi `1.6 * 1 * 1 * 1.1 * 1 * 1.15 Fb*.= 1720 psi CL= (1 +.(134484.993 psi / 1720 psi.)) 11.9 -f(((1'+ (134484.993 psi / 1720.psi)) /:1.9)z - (134484.993 psi / 1720 psi) /-.95) CL= 0.999 . Fe= 50 psi * 1.6* 1 * 1.'*0.999* 1.1 * 1_* 1 * 1.15 Fe= 1719.287 psi page 129 fb: girt test pressure fb= 6 *wall_wind_force / 144*girtSpacing* span / 8 / (b* d2) NDS 3.3 fb =6* 16.304 psf / 144 in.2/ft.2* 19.2"* 122.75"2 / 8 / (9.25"* 1.5"2) fb= 1180.385 psi 1180.385 <_ 1719.:287 stressed to 69% :-.10"X2"#2 OK in bending. F,;: allowable shear pressure F = 135 NDS Supplement Table 4-A F, = F *Co*CM* Ct* Ci NDS 4.3 F,;= 135 psi *.1.6 * 1 * 1 *.1 F,; =216 psi NDS Supplement f,:,shear girt pressure . . f = 3 * (wall—wind—force / 144*girtSpacing-*span/ 2) / (2 * b *d) NDS 3.4 f„= 3 * (16,304 psf 1 144 in.2/ft:2* 19.2"-*122.75''./ 2) / (2.*9.25" * 1.5") f„= 14.424 psi 14.424 s 216.stressed to 7% • 10"X2"#2 OK in shear Deflection Aailow: allowable deflection l= 122.75". Aauow= 122.75" / 90 DaR.W= 1.3639'.' page 130 A,.,: maximum deflection 5 *W*spacing *span 4 / 384 / E / I from http://www.awc.org/pdf/DA6-BearhFormutas.pdf 0.4 E: Modulus of Elasticity E= 1300000 psi NDS Supplement I: moment of inertia 1 =b * d3 / 12 1 =9.25"* 1.5"3 / 12 =2.6015625 in.4 Amax= 5 * 11.413 psf / 144 Psi/p,f* 19.2"* 122.75"4 / 384 / 1300000 psi / 2.6015625 in.4 components and cladding reduced by .7 per footnote f of IBC table 1604.3 ARC= 1.33011" <_ 1.3639" Rear Endwall Girts wall girt size: 10"X2" spacing-between girts =9.95" girt span = 122.75" page 132 Fb: allowable girt pressure Fe = Fb*CD * CM*.Ct* CL* CF* Cf, *Ci* Cr NDS 4.3 CD: load duration factor CD =.1.6 NDS 4.3 CM: wet service factor CM= 1 because girts are protected from moisture by-building envelope Ct: temperature factor Ct= 1 NDS 4.3 CfU: flat use factor Cf„ = 1 NDS 4.3 _ Ci: incising factor Ci= 1 NDS 4.3 :Emi,,: reference adjusted modulus of elasticity Emi„ =470000 psi NDS Supplement Cr: repetitive member factor Cr= 1.15 NDS 4.3q. laterally unsupported span length l;,=.144''. le: effective length. le=1.63 * l„NDS table 3.3.3: le=239.22". CF: size factor CF= 1.1 NDS 4.3 CL: beam stability factor.' CL_ (1 + (Fbe / Fb*)) ..1.9 4(((1 + (FbE / Fb*))./:1.9)2.- Fbe-/ Fb* / .95) Fbe= 1.2.*Emig %RBz . Rs = r([' *,d % bz) Re=I(239-22"* 1.5' / 9:25 z) RB=2.048 Fbe= 1.2*470000 psi %.2.0482 :.FbE= 134484.993 psi... .. :. _ Fb*: Fb* CD*CM*Ct*CF*C; * C,NDS 3:3:3 . Fb =.850.psi NDS Supplement Table 4-A Fb*=850 Psi * 1.6* 1 * 1.* 1.1 *.1 * 1.15 Fb* = 1720 psi CL (1 + (134484.993 psi / 1720 psi_)) / 1.9 .f(((1 + (134484.993 psi / 1720 psi)) / 1.9}7 - (134484.993 - psi / 1720 psi) % .95-} C�=0.999 Fe=850psi * 1:6 *.1 "1.* 0-999 * 1.1 * 1 *.1 *1..15 Fb':= 1719.287 psi fb: girt test pressure fb= 6 *wall_wind_force / 144*girtSpacing*span z / 8 / (b ' dz) NDS 3.3 fb= 6 * 16.304 psf / 144.in.z/ft.z*.19:2 * 122.75"2 / 8./ (9.25" * 1.5"z) fb= 1180.385.psi 11.80.385 <_ 1719.287 stressed to 69% .:.107X2'`#2 OK in bending page 133 F,;: allowable shear pressure F„= 135 NDS Supplement Table 4-A F, = F„* Cp* Cµ* Ct*C; NDS 4.3 F, = 135psi * 1.6 * 1 * 1 * 1 F,;=.216 psi NDS Supplement. f,,: shear girt pressure f = 3 * (wall—wind—force /.144*girtSpacing*span / 2) / (2 * b *d) NDS 3.4 f = 3 * (16.304 psf / 144 in.2/ft.Z* 19.2 * 122.75" / 2) / (2*9.25" * 1.5"). f = 14.424 psi 14.424.5 216 stressed to 7% :•.10"X2"#2 OK in shear Deflection AauoW: allowable.deflection l= 122.75" Aauow= 122.75" / 90 AaUow= 1.3639". _ Amax maximum deflection Amax= 5 *W* spacing * span4 / 384 / E"/ I from-http://www:awc.org/pdf/DA6-Beam Formulas.pdf p.4. E: Modulus of Elasticity E= 1300000 psi NDS Supplement I: moment of inertia 1 = b *d3 / 12 1 =2.6015625 in -Amax=5 * 11.413 psf / 144 PS'/P5f* 19.2".* 1,22.75"4:/ 384 / 1300000 psi /..2.6015625 in.4 components and' cladding reduced by-.7 per.footnote f of IBC table-1604.3 Amax 1.33011 < 1.3639". Rear Endwall Girts page 135 wall girt size: 6"X2" . spacing between girts = 13.7" girt span 91 Fb: allowable.girt pressure ;r.. ,F _F - � b, - b*.Cp*CM-Ct* CL* CF*Cf„*_C;* Cr-NDS4:3 ." CD: load duration factor CI) = 1.6 NDS 4.3 CM: wet service factor CM= 1 because girts are protected from moisture by building envelope. Ct: temperature factor Ct 1 NDS 4.3. Cf,,: flat use factor Cf„ = 1.15 NDS 4.3 . C;: incising factor C;= 1 NDS 4.3 ' Em;,,: reference.adjusted modulus of elasticity. Em;,, =.470000 psi NDS Supplement C,: repetitive member factor C,= 1.15 NDS 4.3 laterally unsupported span length: - l„= 91' le: effective length le = 1.63 * lu NDS table 3.3.3 . le= 152.83" CF: size factor: CF=.1.3 NDS.4.3 CL: beam stability factor.*,. . CL= 1 NDS 3.3.3 < - Fb= 850 psi ` 1.6* 1 * 1 * 1 *.1.3 * 1.15.* 1 * 1.15 Fe=2338.18 psi . page 136 fb: girt test pressure fb= 6*wall_wind_force / 144* girtSpacing*span2 / 8 / (b * d2) NDS 3.3 fb = 6* 16.304 psf / 144 in.2/ft.2* 19.2"*91"- / 8 / (5.5"* 1.5"2) fb= 1091.045 psi 1091.045 <_2338.18 stressed to 47% • 6"X2"#2 OK in bending F�: allowable shear pressure F = 135 NDS Supplement Table 4-A F�= Fv.* CD,* CM* Ct*C; NDS4.3 F, = 135 psi * 1.6 * 1 * 1 * 1 F,;=216 psi NDS Supplement . fv: shear girt pressure f„= 3 * (wall—wind—force / 144*girtSpacing*span / 2) / (2* b *.d) NDS 3.4 fv= 3 * 06.304psf / 144in.2/ft.2* 19.2" *91" / 2) / (2* 5.5"` 1.5") ; f"= 17.984 psi.. 17.984 < 216 stressed to 8% 6"X2"#2.OK in shear Deflection Aaltow: allowable deflection on Hallow= 91 ./,90 Aauow 1.0111 page 137 Amax: maximum deflection AmaX= 5 *W spacing *span4 / 384 / E / I from http://www.awc.org/pdf/DA6-BeamFormulas.pdf p.4. E: Modulus of Elasticity E = 1300000 psi NDS Supplement I: moment of inertia. I = b * d3 / 12 1 = 5.5"* 15"}/ 12 1.546875 in.4 Amax= 5 * 11.413 psf / 144.P5l/P5f* 19.2..*91"4 / 384;/ 1300000 psi /.1.546875 in.4 components,and cladding reduced by .7 per:footnote f of IBC table 160, 4.3 Amax= 0.67569—:5 1.0111" Rear Endwall Girts wall girt site: 2"X6" spacing between girts.= 17 7".:. girt span= 145' supported by 25c4 blocking every .145.' page 138 Fb: allowable girt pressure Fe= Fb*CD*CM*Ct * CL* CF*Cf„*Ci* C,NDS 4.3 CD: load duration factor Cp= 1.6 NDS 4.3 CM: wet service factor CM= 1 because girts are'protected from moisture by building envelope, Ct: temperature factor Ct= 1 NDS 4.3 Cf,,: flat use factor Cfu = 1 NDS 4.3 Ci: incising factor Ci= 1 NDS.4.3 Emi,,: reference adjusted modulus of elasticity, Emi„=470000 psi NDS Supplement C,: repetitive member factor . Cr= 1.15 NDS 4.3 lu: laterally unsupported span length l„ = 145" le: effective length le= 1.63 * lu NDS table 3.3.3 le=252.85.".'. CF: size factor CF= 1.3 NDS 4.3 CL: beam stability factor " CL= 1 NDS 3`.3.3 Fe= 850psi * 1:6 * 1 * 1 * 1 * 1.3 * 1 * 1 1.15 . Fe=2033.2 psi fb: girt test pressure. fb = 6 *wall—wind—force / 144*girtSpacing span Z / 8 / (b* dz) NDS 3.3 fb=6 * 16.304 psf /.144 in.2/ft.2* 19.2,* 145,.2 / 8 /.(1.5" k 5.5"Z) .. fb=755.482 psi 755.482 < 2033.2 stressed to 37% :. 6"X2"#2 OK in bending page 139 F, : allowable shear pressure F„= 135 NDS Supplement Table 4-A Cm F,;= Fv* Co Fv' = 135 psi* 1.6 * 1 * 1 * 1 F�= 216 psi NDS Supplement : fv: shear girt pressure f;= 3 *.(wall_wind_force / 144*girtSpacing *span./ 2) / (2 *b *d) NDS 3.4 fv= 3 * (16.304 psf'/ 1.44 in.Z/ft.z* 19.2"*145" / 2) / (2 * 1 5"*.5.5") fv 28.656 psi 28.656 <_ 216 stressed to 13% :. 6"X2"#2 OK in shear Deflection Aatlow: allowable deflection l= 145" A.,,.,=.145" / 90 Aaliow= 1.6,111" Amax: maximum deflection Amax= 5 *W*spacing_*span., /.384 /E / I_from http://wWW.awc.org/pdf/`DA6-BeamFormuLas.pdf PA E: Modulus of Elasticity E= 1300000 psi NDS Supplement . I: moment of inertia 1 =.b * d3 /.12 1= 1.5"* 5.5"3 / 12 I'=20.796875 in 4 _ . "Amax= 5-* 11.413 psf /,.144 PS'/PSf*.19.2"* 145"4 / 384 / 1300000 psi'./ 20,796875 in.4 components and. : cladding reduced by .7 per:footnote f,of IBC table 1604 3 Amax= 0.32397" <_ 1.6111 Rear Endwall Girts wall girt size:'S..X2.. spacing between girts = 11.95" girt span 111.25' page 140 Fb: allowable girt pressure F6= Fb*Cp*CM*Ct* CL* CF* Cft*Ci* Cr NDS 4.3 .` CD: load duration factor Co= 1.6 NDS 4.3 CM: wet service factor CM= 1 because girts are protected from moisture by building envelope Ct: temperature factor Ct= 1 NDS 4.3 Cf,,: flat use factor Cfu = 1 NDS 4.3 Ci: incising factor' Ci = 1 NDS 4.3 Emi,,: reference adjusted modulus of elasticity Emir, =470000 psi NDS Supplement Cr: repetitive Member.factor C,= 1.15NDS4.3 laterally unsupported span length l„ = 120" I e: effective length . le= 1.63 * l„'NDS table 3.3.3 le =.200:1 CF: size factor CF= 1.2 NDS 4.3. . . CL: beam stability factor. CL= 1 NDS 3.3.3 Fe=850psi * 1.6 * 1 * 1 * 1.* 1.2* 1 *'1 * 1.15 Fb = 1876.8 psi fb: girt test pressure. fb= 6*wall_wind_force /144* girtSpacing*span? % 8 / (b*dz) NDS 3.3 fb=.6* 16.304 psf / 144.in.z/ft.z* - 19.2"* 111.. / 8 / (7.25 *1.5") fb= 1237.042,psi 1237.042 <1876.8 stressed to 66% •= 8"XT #2 OK:in bending ' page 141 F,': allowable shear pressure. F = 135 NDS Supplement Table 4-A F�= Fv* Cp* CM* Ct*CI NDS4.3 F,;= 135 psi * 1.6* 1 *.1 * 1 F,;=216 psi.NDS Supplement fl: shear girt pressure f„= 3 * (Wall—wind—force / 144*girtSpacing *span'/ 2) / (2* b*d) NDS 3.4 f,= 3 * (16.304 psf/ 144 in.2/ft.2* 19.2"* 111.25" / 2) / (2 *7.25"* 1.5") f = 16.679 psi 16.679 <_ 216 stressed to 8% =• 8"X2"#2 OK in.shear Deflection Aallow: allowable deflection Aauow 111.25 / 90 . . Aauow= 1:2361.. Amax:maximum deflection. Amax= 5*W:*spacing*span / 384 / E /,I from http://www.awc.brq/bdf`/bA6-BeamFormulas.pd,f.p'.4 E: Modulus.of Elasticity E = 1300000 psi NDS Supplement . I: moment of inertia = b*d3 /.12 . 1.=.7.25"* 1 5.3:/ 12 =2.0390625 in:4. Amax= 5 * 11..413 psf % 144 PS'7PSf* 19.2"* 111.25"4 /,38411300000 1300000 psi /.2.0390625 in.4.components and cladding reduced by .7 per footnote f of:IBC.table 1604.3 Amax 1.145".<.1.2.361.. Rear Endwall:Girts wall girt'size: 6"X2" spacing between gists 13.7,... girt span = 102.5" page 142 Fe: allowable girt pressure Fe= Fb*CD* CM*Cc* CL* CF*Cfu*C;* C,NDS 4.3 Co: load duration factor CD= 1.6 NDS 4.3 CM: wet service factor "CM= 1-because girts are protected from moisture by building envelope . Ct: temperature factor Ct= 1 NDS 4:3: Cf,,: flat use factor Cf,, = 1.15 NDS 4.3. C;: incising factor - C; = 1 NDS 4.3 Emin: reference adjusted modulus of elasticity Em;n =470000 psi NDS Supplement Cr: repetitive member factor Cr=1.15 NDS 4.3 laterally unsupported span length .- l„ = 120" [,,:'effective length le= 1.63 * L„-NDS table 3.3.3 le=.200.1" CF: size factor CF= 1.3 NDS 4.3. CL: beam stability factor CL= 1 NDS 3.3.3 Fb = 850 psi * *. 1.6 .1 1 * 1 * 1.3 1.15 1 1.15 * * * Fe =2338.18 psi fb: girt test pressure . .: fb = 6 *.wall_wind_force /:144 girtSpacing*spanz / 8 / (b*dz) NDS 3.3 . fb= 6 * 16.304 psf / 144 in.z/ft.z* 19.2"* 102.5"z /, 8 /_(5.5"* 1.5„z - fb= 1384.227 psi, 1384.227< 2338.18 stressed to 59% :. 6"X2"#2 OK in.bending page 143 F, : allowable shear pressure F = 135 NDS Supplement Table 4-A F, F *Co*CM* Ct*Ci NDS'4.3 F, = 135 psi * 1._6 * 1 * 1 * 1 F� =216 psi NDS Supplement f,: shear girt pressure: f„= 3 * (wall—wind—force 1144*girtSpacing*span / 2) / (2* b * d) NDS 3.4 f„= 3 * (16.304 psf /.144 in.2/ft.?* 19.2" * 102.5"./ 2) / (2* 5 5",* 1 5'.) fl,=20.257 psi 20.257< 216 stressed to.9% :. 6"X2"#2 OK in shear. Deflection. Aaliow: allowable deflection l.=.102.5" Aauow= 1.1389". Amax: maximum deflection Am,,* = 5 *W`spacing *span4 /-384 / E /.I.from http //www.awc.org/pdf/DA6-BeamFormulas.,pdf p 4 , E: Modulus.ofElasticity.' - ' E= 1300000 psi NDS Supplement- I: moment of.inertia 1 = b*.d3 / 12 * 1.5", f 12 1 = 1.546875 in:4 Amax= 5 * 11.413 psf_/ 144,P5lp,f* 19.2"* 102.5"7 / 384 / 1300000 psi /.1.546875 in 4 components and cladding reduced. by .7 perfootnote f of IBC table 1604 3., Amax 1.08762":< 1.1389" page 144 Wall Girt Connections connection pressure =wind force* girt spacing*sf/ 21 connection pressure= 16.304 psf / 144 PS'/psf* 38.5"* 115.875" / 2 connection pressure = 252.559 lbs girt nails = connection pressure / Z'with a min of 2 for safety-" - - girt nails =252.559./ 86.076-with a min of 2 for safety girt nails= 3 nails in each end connection pressure=-wind force* girt spacing *sf./ 2. . connection pressure= 16.304 psf / 144 P'/Psf* 38.5" * 115.875" / 2 connection pressure =252.559 lbs girt nails = connection pressure / Z'with a min of 2 for safety girt nails =252.55.9 / 86.076 with a min of 2 for safety,. girt nails = 3 nails in each end connection pressure=wind force P girt.spacing sf'/ 2 connection pressure.= 16.304 psf / 144 Ps'/Psf* 19.2"* 102.5" / 2 . connection pressure = 111.413 lbs girt nails=.connection pressure./ W'With a min of 2 for safety W' =46* 1.6* 1.* 1 *.1 * 1 W'=73.60000000000001 girt nails = 111:41341912603896 / (73.60000000000001.* 3) with a min of 2 for safety girt nails=2 nails in each end connection pressure =wind force* girt spacing *sf / 2 connection„pressure= 16.304 psf / 144 PS'%Psf*.19.2"* 111.25" / 2 connection pressure= 120.924 lbs girt nails =-connection pressure /.T with,a min of 2 for`safety girt nails— 120.9241 86.076 with a min of 2 for safety girt nails='2 nails in each end connection pressure =wind force*girt spacing*sf/ 2 Co*nnection pressure= 16.304 psf / 144 PS'/Psf' 14.2"* 145"_/2 ow connection,pressure =157.609 lbs girt nails connection pressure / Z'with a min of 2 for safety girt nails = 157.609 / 86.076 with.a min of 2 for safety girt nails =2.nails in.each end Page 145 connection pressure=wind force* girt spacing *sf /2' connection pressure= 16.304 psf / 144 Ps/psf* 19.2" *91" / 2 connection pressure = 98.913 lbs girt nails =connection pressure / W'with a min of 2 for safety W=W* CD * Cm*Ct*.Ceg* C W' =73.60000000000001 girt nails =98.91.3.3769801906 / (73.60000000000001 *3) with,a min of 2 for safety girt nails =2 nails in each end connection pressure =wind force*girt`spacing*sf / 2 connection pressure.= 16.304 psf / 144 PS1/psf* 19.2"* 122.75" / 2 connection pressure= 133.424 lbs girt nails =connection pressure / Z'with a min of 2 for safety ..girt nails ='133.424./ 86.076 with a rriin.of 2 for safety girt nails =2 nails in-each end connection pressure=wind force* girt.spacing*sf 12. connection pressure=-16.304 psf,/ 144 P51/Psf *. 2 connection pressure= 5.978-1bs girt nails =connection pressure./ Z'with a.min of 2.:for safety girt nails = 5.978-/ 86.076 with a min of 2 for_safety.. girt nails =.2..nails ineach end connection pressure =wind-force* girt spacing.,*-sf 2 connection pressure=,16.304 psf/ 144 Ps'/psf* 19.2"* 122.75"/ 2 . J. connection pressure = 133.424.lbs girt nails=.connection pressure / Z'with.a min of 2 for safety girt nails = 133.424/ 86.076 with a min of 2 for.safety girt nails 2 nails in each end : connection pressure=wind force# girt.spacing �'sf% 2 . connection,pressure = 16.304 psf 1 144 psi *19.2` *•91",/ 2 connection`pressure.= 98:9.13 lbs girt nails= connection_pressure / "r with a-min of 2.for safety W'=W* CD *CM.*Ct* Ceg*C W'=.46 * 1.6 * 1,*.1.* 1•..* 1 .. W'=73.60000000000001: . girt nails:=98.9133.769801906 1 (73.6000000000001 *3)with a min of,2 for'safety girt nails=2 nails in each end connection pressure=wind force* girt spacing *sf / 2 connection pressure 16.304 psf ( 144 PS'/P5f* 19.2".* 145" /.2 connection pressure = 157.609 lbs girt nails =.connection pressure / Z with a min of 2 for safety girt nails =`157.609 / 86.076 with a min of 2 for safety girt nails.=.2 nails in each end Page.146 connection pressure =wind force* girt spacing*sf / 2 connection pressure = 16.304 psf / 144 PS'/psf* 19.2"*,111.25" / 2. . connection pressure = 120.924 lbs girt nails =connection pressure / Z with a min of 2 for safety girt nails = 120.924 / 86.076 with a min of 2 for safety -girt nails =2 nails in each end connection pressure=wind force*girt spacing *sf / 2 connection.pressure 16.304 psf / 144 PS'/Psf 19.2'.* 102.5" / 2 . connection pressure = 111.413 lbs girt nails = connection pressure / W'with a min of 2 for safety , W' =W*Co *CM*Ct*Ce4 * C W'=46* 1.6 * 1 * 1 * 1 * 1 W':=73.60000000000001 girt nails= 111.41341.912603887 / (73.60000000000001 *3)with a min of 2 for safety girt nails=2 nails in each end page 147 PURLIN CALCULATIONS Purlinswithin Drift Zane drift width = 8 *I(S) * hd / 3 ASCE 7.6.1 - drift width = 8 *I(3) *2.137' / 3 drift width = 9.87' ROOF PURLIN DESIGN Assumptions: .4:12 roof slope (18.435° roof angle) Trusses spaced 10-ft. o.c. Purlin span 9J54t. purlin spacing =24 in. . roof steel dead load = .63 psf steel Ameri can.Building Components catalogue.. roof lumber dead load 62.4 pcf.*.43 lbs/ft.3 % (1.+..43 lbs/ft_3 `..009 *.19) *`(.1 +,.0019) ' 145"/ 12 in. �7.25"./.12'"'/fc;'*.(10'. T' 7 12'" Lft;) % 10' /_(24" /..12'"'/fc;) psf iri purlin weight based on :43 G ..:. NDS. roof OSB dead load = 62-.4 pcf k .5 lbs/ft.3 *7/16' / 12 in %ft. psf in OSB weight based on .5 G NDS ' total purlin dead load =2.757 psf' page 148 Check for gravity loads Bending Stresses Fb: allowable bending pressure Fe= Fb* Co*CM * Ct*CL* CF*Cf„*Ci*Cr CD: load duration factor Cp = 1.15 NDS 2.3.2 . CM: wet service factor CM= 1 because purlins are protected from moisture by roof Ct:.temperature factor Ct= 1 NDS 2.3.3 :. CL: beam stability factor CL= 1 NDS 4.4.1 CF: size factor CF= 1.2 NDS Supplement table 4A Cf,,: flat use factor. Cf„ = 1 NDS Supplement table 4A Ci: incising factor Ci = 1 NDS4.3..8 . Cr: repetitive member factor. . Cr= 1.15 NDS 4.3.9 Fb= 850 psi NDS Supplement Table 4-A Fe=850psi *.1:15 * 1 * 1.* 1 * 1.2* 1. *.1 * 1.15 Fb= 1348.95 psi... . fb: bending stress from snow/dead loads fb = (purlin_dead_load + S):*spacing,/ 12*cos(6) / 12* (sf* 12 - 3)2./ 8*6 / b / d2*cos(8) S = 64.697 psf using the appropriate load calculated above- - : fb= 67.453 psf*24" / 12 ' /ft::*cos(18.435) / 12 in./ft:.* (10'*.12-."-/f,.--'3")2 /;8 *.6 / 1.5" i..7.25"2*: cos(18.435) fb = 1317.524 psi:<_ 1348.95 psi; stressed to 97.7% page 149 Shear Stresses F,,: allowable shear pressure F,;= F * CD*CM* Ct*C; CD: load duration factor Co= 1.15 NDS 2.3.2 - F = 135 psi NDS Supplement Table 4-A F, = 135 psi * 1.15 * 1 * 1;* 1 F,;= 155.25 psi f,,: shear stress-from snow/dead loads f = (purlin_dead_load + S) *spacing / 12* cos(6 *rr / 180) / 12* (sf*.12= b =2*d),`/ 2*3 f„= 67.453.psf*24" / 12.'"'/ft. ".cos(18.435'-*3.14159 % 180) / 12 /f[. *.(10`* 12'"' 3" 2*7..75.) 2*3 / 2 / (1.5"*.7.25"): f„ 75.392 psi <_ 155.25 psi; stressed to 48.6% Deflection, 4ailow:allowable deflection Aauow= l /:150 IBC table 16043` l= 117". ,. Hallow= 117"% 150 Allow0.78 page 150 Am.: maximum deflection Amax= 5 * S *spacing *cos(6 *n / 180) * (sf* 12 - 3)4 / 3841 E / I from http://www.awc.org/pdf/DA6- BeamFormulas.pdf p.4 E: Modulus of Elasticity_ E= 1300000 psi NDS Supplement I: moment of inertia - 1 = b*d3 / 12 I = 1.5"* 7.25"3 / 12 =47.634765625 in.4 Amax= 5 *64.697 psf / 144 PS'/Psf*24"*cos(18.435° *3.14159 /.180)* (10'* 12 '"'/ft,.-.3")4 / 384 / 1300000 psi / 47.634765625.in.4. Amax=0-40306"< 0,78" Check for Wind (Uplift) Loads W. Bending Stresses Fe: allowable bending pressure Fb:=,Fb k CD.*CM Ct*CL-.* CF* Cf..*Ci k Cr CD: load duration factor,. CD= 1.6 NDS 2.3.2 Fe= 850*.Psi 1.6 ' 1 * 1 .*,.1'* 1.2 * 1 . 1 *1.15 Fb'= 1876.8 psi .' fb: bending stress from wind/dead loads fb= (purtin_dead_load +V,/) *spacing / 12*cos(6*rr./ 180) / 12* (sf" 12` b)z /.8.` 6 / .b / dz W=36.269 psf using the wind pressure.calculated.for roof cladding above fb'= 33.512 psf*24. / 12 1n'/ft. *cos(18.435° *3:14159 / 180):/ 12 '"' * (10'* 12'°;/ft..= 3")z/.8* 6 / 1.5" /:7.25' fb= 689.989 psi <_ 1876:8 psi page Shear Stresses F,,: allowable shear pressure F,,' F * Co*CM* Ct*C; CD: load duration factor - Co= 1.6NDS2.3.2 F,;= 135 psi * 1.6 * 1 * 1 *.1 F, =21.6 psi f,: shear stress from wind/dead loads f�_ (purlin_dead_load +W) spacing'/ 12*cos(6*Tf/.180) / 12*.(sf* 12 b) / 2*3 12 / (b* d) f = 33.512 psf*24 / 12",/ft..* cos(18.435' ' 3.14159 1 180) / 12 ' /fi. *.(10':* 12 in. 3") / 2*3:/ 2 / f�=42.756 psi <21.6 psi.: , Deflection Amy 5 * .7`W*spacing Y 12*cos(6 *tr 1180) / 12 (sf* 12 b)4 / 384% E / I from http.//www.awc:org/pdf/DA6-BeamFormulas.pdfp.4 components and cladding,reduced by 7 per footnote of IBC table.1604.3. Ama,= 5-* .7*36.269 psf*24" / 12'".Ift *cos(18 435° *3A4159 / 180)./ 12'"/ft:•* (10'* 12 '" /ft 3 )4 384 / 1300000 psi / 47.634765625 in.4 Amax 0.1582":< 0.78". Aauow=:117, / 150 Aallow 0;7$ page 152 . Amax: maximum deflection Amax= 5 * L*spacing*cos(0 *Tr /.180) * (sf* 12 - 3)4 / 384 / E / I from http://www.awc.org/pdf/DA6- BeamFormulas.pdf p.4 Amax= 5 *64.697 psf/ 144 P`/psf*24",*cos(18.435° *3.14159 / 180).* (10,* 12:in'/ft 3")4 / 384 / 1300000 psi / 47.634765625 in.4 _. Aauow= 117 120 Aauow=0.975 Amax: maximum deflection Amax= 5 * (L+ .5D) *spacing*cos(6 * rr / 180) *,(sf* 12 - 3)4 / 384 / E /1 from - http://www.awc.org/pdf/DA6-BeamFormu[as.pdf p.4 Amex=5 *21.378 psf / 144 Pr./psf*24"*cos(18.435°.*•3,14159 / 1.80) * (10-* 12'"/ft 3')4 / 384 1 A300000 psi % 47.634765625 in.4 ROOF.PUKIN.DESIGN Assumptions 4:12 roof slope'(18.435.° roof angle) , '; Trusses spaced 10 ft O.C. M. Purlin span 9:75=ft purlin spacing.=24.in roof steel dead load = .63 psf steel American Building; Components catalogue roof lumber dead.load;=.62 4 pcf* 43:lbs/ft : / (1 +:43 lbs/ft s * .009 .19) * (1 + 0019) * 1 S'''./ 12 in 6-*.5-*5"./" 12'"-/fi * (10 Y L 12 fin,/ft}/ 10'_/ (24' / 12 i"/ft) psf in purli.n weight based on. 43 G NDS' roof OB dead load =`62.4 pcf* ,5 lbs/ft 3 *7/16 / 12 j"Vft. psf in.056 weight.based 6n::5 G NDS;'..: total purlin dead load =2:518 psf page 153 Check for gravity loads Bending Stresses Fb: allowable bending pressure - Fb = Fb*CD*.CM*Ct*CL.* CF*Cfu«Ci«Cr CD: load duration factor CD= 1.15 NDS.2.3.2 CM: wet.service factor.- CM= 1 because purlins are protected from moisture by roof. Ct: temperature factor. Ct=.1 NDS III.: CL: beam stability.factor CL= I NDS 4.4.1. ' CF: size factor, CF 1:3 NDS Supplement'tab le-4A Cf;:,flat use factor Cf.. 1 NDS.;Supplement table4A F.Ci:incising factor.., :.. C;= 1'NDS 4.3.8 Cr: repetitive member factor'. Cr= 1.15.NDS 4.3.9 Fb=;850 psi:NDS.Supplement Table 'Fb'..=.850,psi'.1:15 « 1 ::*::1 ,. 1 «:1..3 « 1:' :1 *_1::15 Fb: 1461:363:psi. fb .;bending stress from snow/dead loads fb= (putlin dead load +-S) *spacing /:12*cos(6) / 12* (sf / 8*6 / b /:c «cos(8) S.= 37.802 psf using the..appropnate load calculated above fb=40.32,psf*24 ./..12 in./ft: *Cos(18.435):/ 12 m:�ft..* (10 *.12'"/ff:, 3 )Z.4:8 «6./:1..5 / 5 5 z« c6s(18.435) fe= 1368436 psi.<- 1461.363 psi; stressed.:to.93 6% { page 154 Shear Stresses : Fv: allowable shear pressure :F�= Fv* Cp*CM* Ct*Ci Co: load duration factor CD = 1.15 NDS 2.3.2 Fv= 135 psi NDS Supplement Table 4-A F, = 135psi * 1.15 * 1 * 1 * 1 - F,; 155.25 psi . fv: shear stress from snow/dead loads,..,` fv= (purlin_dead_load +S)*spacing /'12*cos(6'rr./`.180) / 12* (sf.*.12 b'- 2*d) / 2 ' 3 / 2/:'(b*d) fv=40.32 psf*24" / 12 in./tt. *cos(18.435°. *3.14159 / 189)./. 12in, ft. Y.-2 *.5.5 )./ 2_' . 3 / 2 / (1.5''.*5.5') fv= 61.433 psi < 155.25 psi; stressed to.39.69. Deflection AauoW :allowable deflection �. .<: Balm. l % 150.IBC table 1604.3 :. l= 117' dailow= 117" /.150 Dauow=0.78 : page 155 Ama,�:.maximum deflection _ - Amax= 5.*S * spacing*cos(6_*rr / 180) * (sf* 12 - 3)4/ 384./ E / I from http://www.awc.org/pdf/DA6- BeamFormulas.pdf p.4 E: Modulus of Elasticity E = 1300000 psi NDS Supplement. . I: moment of inertia = 1.5"*_5.5"3 / 12 = 20.796875 in.4.. . . Amy= 5 *37.802 psf / 144 PS'/' *,24''*.cos(18.435 *3.14159 /:180).*(10'•` 12 ' /ft --3")4 /.384./.. 1.300000 psi / 20.796875 ih.4 q. Amax= 0.53942" < 0.78" `. Check for Wind (Up lift),Loads Bending Stresses 4. Fb`. allowable bonding.Kessure F'_F.:*C *C *C *C `C :*:Gf„*G.*C� -. b b D M..,. .t L F Cp: load du ration.factor:: �.Co,= -&NDS 2.3.2 :. Fb'= 850",psi-1.,6 * 1 `.`1.* 1 *'1.3 * 1 *`1 * 1.15 Fb'..=.2033.2 psi. , fb. bending stress from wind/dead loads fb= (purlih dead_load t W) *spacing /,12*cos(6 *n'/ 180)•/ 12 �.(sf*12- b)./ 8 *6 / b./ dz W=.36.269 psf using th.e:wind pressure.calculated for roof cladding above, fb= 33.751 psf*24 / 12.'"/ft *,cos(18..435 *-IA 4159 / 180) / 12' /ft * (10 * 12, /ft 3 )z / 8 * 6 / 1:5"'/ 5:5,,z, fb:= 1207.468 psi.<_2033.2 psi '..: page 156 Shear Stresses - F,,: allowable shear pressure Fv'= Fv*Co.* CM*Ct*Ci Co: load duration factor CD= 1.6 NDS 2.3.2 F, = 135 psi * 1.6 * 1 * 1 * 1 F,; =216 psi • f,: shear stress:from wind/dead loads fv= (purlin_dead_load +W) *spacing / 12.*.cos(6* rr:/ 180) / 12* (sf* 12 b) / 2. 3 / 2 / (b* d) f,;= 33.751 psf*24" / 12 in./ft. * cos(18.435° #3.14159 / 180) / 12 In./ft..* (10,* 12.i"'/ft. = 3") / 2:*3:/2 _'.f„= 56.761.psis 215 psi Deflection An, 5 t .7*W`spacing / 12 cos(6:'�rr /.180) / 11* (sf*'12 b)4;/. 384 /.E / I from http.//www awc:org/pdf/DA67BeamForrhulas.pdf p 4, components and cladding reduced by. 7 per footnote f of IBC table 1604.3 �rnaz 5 *_,7*36.269 psf``24" / 12 i"'/ft. *cos(18 435 * 3.1,4159 / 180) /.12'" /ft ' (10 *:12'" /ft 3 )4 384 / 1300000 psi / 20.796875 in a Amax=.0.3623.< 0.78 117. 7.150 ; Aallow=0;78 page 157 .A,,:' m,,: maximum deflection I... . . Amax= 5 * L spacing * cos(0*n / 180)* (sf* 12 - 3)4 / 384 / E /I from http://www.awc.org/pdf/DA6- . BeamFormulas.pdf p.4 . Amax= 5 *37.802 psf / 144 Psi/psf*24"*cos(18.435°.* 3.14159 / 180)* (10'* 12' /ft. -_3")4 / 384 / , 1300000 psi / 20.796875.in.4 - - Aallow= 117" / 120 - . . . . . Aauow= 0.975 . . . . . - .. AmaX- maximum deflection .. Amax= 5 * (L+. .5D) *spacing*cos(6*rr /:180)'.* (sf,*. 12 3)4 / 384 / E/1 from http;%/www.awc:org/'pdf/DA6-BeamFormulas.pdf p.4 . - -.Om =5.*21.259 psf Y-1.44,R"/psf*24".*cos(18:435.' *3.14.159 /.180) * (10 *,12' /f� 3")4 / 384 / 1300000 psi / 20.796875 m,4 v. , _ ::.,1. Y � n ; page 158 RIDGE PURLIN DESIGN Assumptions: 4:12 roof slope (18.435° roof angle) Trusses spaced 10-ft. o.c. Purlin span.9.75-ft. purlin spacing =22 in: roof steel dead load = .63 psf steel American Building Components catalogue roof'lumber dead load = 62.4 pcf* .43 lbs/ft.3 / (1 + .43 lbs/ft.3 * :009-* '19) * (1 + .0019) * 1 5'_/ 12 ' /ft: "7.25'. /.12 i"A. * (10.- 3" / 12 1n'/ft.) / 10' / (22' / 12' /ft.) psf in purlin weight based,on .43 G NDS ` ... roof OSB dead load = 62.4 pcf* .5 lbs/ft.3 *.7/16' / IT in,/fc psf m.OSB weight based on .5.G NDS total puffin dead load =2.846 psf page 159 Check for gravity loads.. Bending Stresses Fb: allowable bending pressure ' Fe= Fe*CD *CM*Ct*.CL*CF*Cfu* Ci* Cr CD: load duration factor CD= 1.15 NDS 2.3.2 Cµ: wet service factor. : CM= 1 because pur(ins are protected from moisture by roof Ct: temperature factor': Ct= 1 NDS 2.3:3.:.. CL: beam stability factor CL=.1 NDS 4.4:1 CF: size factor.... CF=.1.2 NDS Supplement table 4A Cf*: flat use.:factor. Cf„ 1 NDS_Supplement table 4A C;: incising.factor. _ C; =.1 NDS 4.3.8 C • repetitive ,. .. . r- member factor. Cr= 1.15 NDS 4.3.9- Fe 850 psi NDS Supplement Table 4-A' Fe = 850 psi *.1 15* 1` 1 *.1.2" 4 :d * i 15 Fe=.1348:95"psi fe: bending stress from snow/dead loads fb.=,(purlin_dead_load:+:S) *spaang Y 12 'cos(0) / 12* (sf*.12 3)z/ 8*6 S =64.697 psf using the appropriate:load calculated above fb=67.543 psf:*.22" / 12 in./ft.:*.cos(18.435) /A2 `"./ft. * (10: 12'n..�tt.. ..3 )2 / 8 *6,;/. 1.5' / 7J5 :.( cos(18:435) fb=.1209.34 psi.:_ 1348:95 psi-stressed to,89.7% page 160 Shear Stresses F,,: allowable shear pressure Fv-= F„*CD*CM* Ct*C; CD: load duration factor CD = 1.15 NDS 2.3.2 F„= 135 psi NDS Supplement Table 4-A F, 135 psi * 1.15 * 1.* 1 * 1 . F"= 155.25.Psi fv: shear stress.from snow/dead Loads- f = (purlin_dead_load +S) *spacing / 12*cos(6*rr / 180) / 12 *.(sf* 12 - b- 2*.d) %".2 * 3 L 21 (b* d).` f� 67.543 psf*22"./.12 in'/ft..*cos(18.435°..*3.14159 /.180)"/".12 inIft. * (10'* 12'n' Y—* 2:*7.25"1 / 2 *3:/.2./ (1..5"*7.25")" f� 69.202 psi <_ 155.25 psi; stressed to"44:6% Deflection �auow,allowable.deflection �auow l_./.;150 IBC table.1604.3 - Aallow=:117" /:150 Aallow,=0.78 page 161 Lmax: maximum deflection Amax= 5.* S * spacing*cos(6 *rr / 180) * (sf* 12 - 3)4 / 384 / E / I frorn http://www.awc.org/pdf/DA6- BeamFormulas.pdf p.4 E: Modulus of Elasticity E= 1300000 psi NDS Supplement I: moment of inertia-: 1 = b* d3 / 12. . : = 1.5" *7.25"3 / 12•.. =47:634765625 in.4. Amax= 5 * 64:697 psf/ 144 Psi/Psf*22"*Cos(18.435°,*3.14159 / 180) * (10' .12 ' /ft.,7.37). / 384 / . 1300000 psi / 47.634765625.in.4 Amax=..0.36947" < 0.78" :. Check for.Wind (Uplift) Loads r . Bending Stresses Fb: allowable bending pressure *CD*.CM:".Ct*CL.# CF. fu*Ci*.0 Fb= Fbr Co; load duration factor, CD= 1.6 ND S 2.3.2 Fti=.850*psi1.6 *1 * 1 * 1 * 1.2 * 1;`.,1 : Fe= 1876.8 psi, fb .bendin'g stress from wind/dead loads fb.= (purlin_dead_load+.W)*spacing./12*cos(8* rr:/ 180) / 12* (sf* 12.- b)� / 8 * 6 b / dZ W�= 36.269 psf.using the wind..pressure calculated,fo.r roof cladding above fe= 33.423 Zpsf*.22 / 12 '"/ft; *cos(18.435 *3.14159 / 180) / 12 j /ft * (10 * 12:'"/ft - 3 )Z / fb= 030.793 psi:5;1876.8 psi page 162 Shear Stresses F,: allowable shear pressure F:= Fv* Co* CM* Ct* C CD: load duration factor Co = 1.6NDS2.3.2 F, = 135. psi * 1.6* 1 * 1 * 1 F,; =216 psi fv: shear stress from wind/dead loads f„= (purlin_dead_load +W) *spacing / 1.2*cos(6 *Tr 180) / 12 ` (sf* 12 - b) / 2 *3 / 2./ (b *d) "/f.'* ; i" * '"'/f /.2f,= 33.423 psf`22 1:2 cos(18.435 *3.14159 / 180) / 12 /ft, (10 * 12 3 / 2 (1.5"*7.25') fv= 39.088 psi <_216 psi Deflection46. Amy=5*..7*W`spacing / 1 T*cos(6 *,it/'180)./ 12" (sf* 12 b)4 1384 / E / I from http:.//wwW.awc.org/pdf/DA6-BeamFormulas.pdf p 4, components-and xladding reduced by 7. per, footnote f of IBC table 1604.3.: Amax 5 *..7* 36.269 psf*22":/ 12'n-/ft, k Cos(18 435° *3 141.59 / 180) / 12 '" 4 /ft *.(10 r. . . 384./ 1300000 psi / 47.634765625 4 in. .,,, Amax=0r145 :5:0:78" Aailnw= 11:T :/'-;150 Aalloiv=0.78 page 163 Amy: maximum deflection Amax= 5 * L*.spacing*cos(o *Tr / 180) * (sf* 12 - 3)4./ 384 / E / I from http://www.awc.org/pdf/DA6- BeamFormulas.pdf p.4 Amax= 5.-*64.697 psf / 144 PS'/PSf*22"*cos(18.435° *3:14159 / 180) * (10'* 12 ' 3")4 /,384 / - = 1300000,psi / 47.634765625 in.4. - AaUow= 117" /120.: AaRm,= 0.975" pm : maximum deflection Amax'=5 * (L+ .5D) *.spaang*cos(6 * rr / 180y* (sf* 12 - 3)4,/ 384/ El i from http://WwW.awc.6r�/pdf/DA6-BeamFormulas.pdf p:4 Amax= 5 *21..423 psf 1 144 P'/Psf*22"*cos(1$.435 *.3 14159 / 180) * (10'* 12.in-7ft- 3 ), / 384'% 1300000 psi /47.634765625 in.4. 0. J wr W. page 164 EAVE GIRT DESIGN Assumptions: 4:12 roof slope (18.435° roof angle) Trusses spaced 10-ft. o.c. Purlin span 9.75-ft. purlin spacing = 12 in. roof steel dead Load = .63 psf steel American Building Components catalogue roof lumber dead load =62.4 pcf* .43 lbs/ft.3 / (1 +.43 lbs/ft.3-:* .009* .19) * (1.+ .001.9) * 1 5'>% 12: .. i /,.,* 5.5"./ 12 '"'/ft, * (10'.- Y /.12 7Vit.) / 10' / (12"/ 12.'"../ft.) psf in purlin.weight_based on .43 G NDS roof OSB dead Load = 62.4 pcf* .5 lbs/ft.3: /16' /.12'.."�/fc. psf in OSB weight based on..5 G NDS total purlin dead load =3.268.psf page 165 Check for gravity loads Bending Stresses Fb: allowable bending pressure Fe = Fb*Cp*Cµ*Ct*CL*CF*Cfu*Ci*Cr. Co: load duration factor Cp= 1.15 NDS 2.3.2 CM: wet service,factor: CM= 1 because purlins are protected from moisture by roof. Ct: temperature factor. Ct= 1 NDS 2.3.3 . : CL: beam stability factor Cc= 1 NDS 4.4.1 . CF: size factor ' CF 1.3.NDS,Supplement table 4A - U. flat use.factor Cry=1 NDS Supplement table 4A Ci: incising'factor Ci= 1 NDS43.8 .. ,. Cr:'repetitive,member factor C�=...1.15NDS4.3.9 Fb 850 psi:,NDS_Supplement Table 4-A.. :'Ft = 850,psi*1.15* 1._k,1,* 1 * 1.3 *1 k,1,*.1.15 Fb= 1461:363 psi 1 fb: bending stress from snow/dead loads fb (purlin dead_load+:S) *spacing / 12*cos(6) / 42 (sf* 12 3)? / 8 *6 / b / dZ*cos(6) S = 37.802 psf using.the appropriate load calculated above' . %=41.07 psf k 12 / 12 in:/ft.*cos(18.435)'/ 12'" /ft *,(10 *'12 '" /fc ..3 ). / 8 *6 /:1 5 /.5 5 z*. cos(18.435) fb' 696.95 psi <_ 1461..363 psi;:stressed to 47.' page 166 Shear Stresses F,,: allowable shear pressure F-= F,,*Co*CM* Ct*C; CD:load duration factor Cp= 1.15NDS2.3.2_ F = 135 psi NDS Supplement Table 4-A F, 135 psi * 1.15 * 1 * 1 * 1 F, = 155.25 psi f,,: shear stress from snow/dead loads f„_.(purlin_dead_Load +:S).*spacing / 12 *cos(A" n / 180) / 12* (sf* 12 -b - 2 i`d) /`2 ` 3 / 2 (b* d) f„=.41.07 psf:* 12" / 12;in/ft. *cos(18.435° *3.14159 / 180),/ 12.n./f, *.(10'* 12 in. ` 3" - 2*5:5")./ 2*.:::. f�=.31'.288'psi _155.25 psi;. <_ tressed to 20.2%' Deflection. Aallow allowable deflection Aaiiow 1 Y 150 IBC table.1,604 3 Aauow.= 117",/.150 allow-0.78 page 167 . Amax: maximum deflection Amax= 5 * S*spacing *cos(6 *it / 180) * (st* 12 - 3)4 /.384 / E / I from http://www.awc.org/.pdf/DA6- Beam Formu Las.pdf p.4 E: Modulus of Elasticity E= 1300000 psi NDS Supplement I: moment of inertia - I = b * d3 /.12 1 = 1.5"* 5.5"3 / 12 =20.796875 in Amax= 5 *37.802 psf / 144.Ps'/psf* 12"*.cos(18.4357 * 3.141591 1.80) *(10':* 12' /tt.:=3")4 / 384,/ . 1300000 psi / 20.796875 in.4 max Check for Wind (Up lift)-:Loads. Bending Stresses " Fb:'allowable bending-pressure::. CD.*CM C*,*.Ctu* C{*C� Cb: load duration.factor Co=.1.6 NDS 2.3.2 ' Fe 8850* psi 1.6 * 1`:1 * 1.* 1.3 * 1:*:1'*.1 A 5 Fe"_"2033.2 psi fb: bending`stressfrom Wind/dead loads" fb = (purlin_dead_load +.W) ".spacing j 12`.cos(A *rr/ 1.80) / 12*,(st:*.12 b), / 8.*6 / b / d� W= 36.269 psf;using.the wind pressure calculated for roof.cladding above fb=.:33.001 psf ' 12 / 12 ' /tt, *cos(18,435 *3.14159 / 180)`/ 12 ' 10..* 12 /tt 3')Z / 8* 6 / fb.= 590.313 psi'k2033:2 psi page 168 Shear Stresses F,: allowable.shear pressure F�= F„* Cp* CM* Ct*C; Co: load duration factor Cp= 1.6NDS2.3.2 F,.= 135 psi * 1.6 * 1 * 1 * 1 F,;=216 psi f,,: shear stress from wind/dead loads f;,_ (purlin_dead_load+W) *spacing /.12 cos(0*_n / 1,80) / 12 * (sf* 12 - b) / 2 *3 /2 %(b * d) fl. 33.001 psf* 12" / 12 in./ft. *cos(18.435° ` 3.14159 / 180) / 12 /ft.« (10,* 12 m /ft. - 3")./ 2 *3 / 2 (1.5"* 5.5')' ._ f,;=27.75 pst<216 psi. .:. Deflection Amax 5 *.;7-*W. spacing / 12 os(A *'rr /_780) / 1.2 * (sf*12 b)4/:384 % E I I from http //www awt. or4/pdf/DA6-B6arnFormulas.odf p 4, components and cladding reduced:by 7 per footnote:f of.IBC table 1604.3_;., Amax 5*.7'.*36.269 psf* 12" /.12 '" /ft. *COs(18 435°,* 3.14159 / 180) / 12.'" /ft:. *.(10' *12 '" /ft 7.3 3841 1300000 psi / 20.79.6875 in.4 Omax=.0.181.1" <_ 0.78" allow:117 / 1,50 pallow=,0:78 page169 Amax: maximum deflection Amax= 5 * L*spacing*cos(6*Tr,/ 180) * (sf* 12 - 3)4 / 384 / E / I from http://www.awc.org/pdf/DA6- BeamFormulas.pdf p.4 . . Amax=5 *37.802 psf 1144 P'/Psf* 12",*cos(18.435° *3.14159 / 180)* (10'* 12 m,/ft. - 3")4 /384 / 1300000 psi / 20:796875 in.4. . Aailow= 117" / 120 dauow=0.975"Amax: maximum deflection Amax=:5 *.(L+ .5D);*.spacing-*cos(6 *:Tr,/ 180) * (sf* 12 -3)4 1;384/ E /.1 from http://www.awc.org/pdf/DA6-BeamFormulat.pdf p 4 Am az= 5 `21.634 psf / 144 Ps/Psf*12"*cos(18.435' * 3.14159 /.180) * (10'* 12 ' /ft. 3')4 / 384I .1300000.psi /'20.796875 in page 170 Purlin Connections LU-210 joist hanger strength = 850 lbs Simpson product specifications R.= roof force*purlin spacing sf „ R= 14.474psf*2.108'* 10' R= 305.145 lbs 305.145 lbs < 850 lbs so LU-210 joist hanger is adequate with 6 10d nails LU-210 joist hanger strength = 1345.lbs Simpson product specifications . . R= roof force' purlin spacing' sf R= 64.697 psf*2.108'*.10' R= 1363.922 lbs page 171 . Truss Connections End Trusses connection pressure="W/ 2* (sf / 2 + end overhang)* ((.6*design dead loads)..+ uplift—force + Hz? / W / W* uplift force) connection pressure 26'* (10' / 2 + V)* (.6 * (3.3 + 1) + -16.304 psf t 8.333" / 506 / 50'* 16.304 psf) connection pressure =2211.659 lbs end truss nails= connection pressure /Z'. end truss nails = 2211.659 lbs:/ 128.472 lbs . end truss nails =.,18 nails. Connection requires:more nails than..our standard allows; so we use LSTA 12 brackets LSTA 12 strength,=925 lbs from Simp5o.n product catalogue:. 2211.6592487621806 lbs<925 lbs*.3;so only 3 bracket per connection are.required �pt m �•a s x2 *_ f _ `• page 172 Interior Trusses connection pressure=W./ 2*sf* ((.6 * design dead loads) + Uplift force+ H7.2 / W / W* uplift force) connection pressure =26'* 10'* (.6 * (3.3 + 1) + -16.304 psf+ 8.667.'.2,/ 50'./ 50'*-16.304-psf) connection pressure=3695.707 lbs 1m = 3" l5 = 1.3333333333333333 Fye =45000 psi Ke = 1 + .25 * (6 / 90) Ke = 1 + .25 * (71.565 / 90). Ke = 1.199: Rd=4*Kjor Rd=4.795 for Rd= 3.6* Ke for I Rd 4.316 for.11 Rd= 3.2* Ke for.l Rd= 3.836 for III and IV Rd..=.1.199 NDS table'.11.3,1B' Z for Yield Mode Im =0.625"*3 *5550 psi'/ 4.795 Z for Yield Mode In.-.2170:154 tbsY Z for Yield Mode'15=,0:.625"_* 1.3333333333333333'`*5550 psi / 4:795 Z for Yield Mode 15 = 964.51.3 lbs Re.'= 5550 psi,/ 5550.psi wq Re = 1 Rt 3 / t.3333.333333333333' Rt 2.25 = kf= (f(1 +2 . 1z:.* (1 +2.25 +.2.25Z) + 2.252* 13) 1 * ('1 +2.25)) /. kf.=0.757. Z.for Yield Mode 11 =0.757*0.625"k 1.3333333333333333 *.5550 psi:/4.316 Z for Yield Mode LI = 810.801 psi kZ:=-1 +t(2. (1 + .) +:(2 45000 psi (1 +2. 1 ) 0.625 ) / (3 5550 psi 3 )) Z for Yield Mode II1 = 1:.169 0:625"*3"* 5550 psi./.((1 +2* 1),* 3:836) Z for.Yield Mode Ilim= 1056.893 lbs:. . page 173 k3 = -1 +f(2* 0 + 1) + (2 *45000 psi *:(2 +A *0.625"2) / (3 *,5550 psi * 1.3333333333333333'2)) k3 = 1.75 Z for Yield Mode IiIS= 1.75 * 0.625"* 1.3333333333333333"* 5550 psi L ((2 t 1) * 3.836) Z for Yield Mode IIIS= 703.337 lbs Z for Yield Mode IV= 0.625"2 / 3.836*f((2*5550.psi *45000 psi) / (3 * (1 + 1))) Z for Yield Mode IV= 929:091 lbs - using worst case of Z modified by applicable factors NDS 10.3.1 Z'=703.337lbs* 1.6 * 1 * 1 * 1 * 1 * 1 * 1 * t 1- 3,. : is= V. - Fye= 80000 psi . . KB = 1 + .25 *($theta;"/ 90) .25,* (71.565 / 90) Rd='10*.D'+ 5 Rd 10* 0:192 .5 Rd=2.42 N.DS-:table.11'.3.1 B Z for Yield Mode Im = 0.192"* 3"* 5550,psi"/-2.42 Z for.Yield`Mode Im - 1320,992 Ibs Z for Yield Models=b:192' 42 ,< Z for Yi6ld'Mode I5=:440.331.1bs Re 5550.psi / 5550 psi Rt=3., �y.. Rt= 3 ki'= (((1 +2 1 (1 +.3 + 3 )t.3 k� Z,for Yield Mode II = 1 *0.192"" 1".* 5550 psi,/ .42,; Z for Yield:Mode II 440.331 psi page 174 kz = -1 +,r(2 * (1 + 1) + (2.*80000 psi'* (1 +2 *.1 ) *0.192..z) / (3 *,5550 psi * 3,2)). . . kz = 1.029 . I. Z for Yield Mode IIIR, 1.029 *0.192"*3"*5550 psi / ((1 +2* 1) *2.42) . Z for Yield Mode Illm=453.235 lbs . . . - -. k3 = -1 +f(2 * (1 + 1) + (2*80000 psi * (2 + 1 ) *0.192"z) / (3 * 5550 psi * 1"z k3 = 1.251. ;�_ P . .. . Z for Yield Mode III, = 1.25 *0.192"* 1"* 5550 psi / ((2 + 1) *.2.42) . . Z for Yield Mode III, = 183.4.79 lbs , . . . Z for Yield Mode IV=0.192"z / 2.42_*I((2 * 5550 psi * 80000 psi) / (3* (1 + 1))), �:.. -. . . Z for Yield Mode IV= 185.318 lbs . using worst case of Z modified by applicable factors,NDS 10.3 1 , Z'= 183.479.lbs* 1'.6*.1 * 1 * 1:* 1 `A-* 1 * 1 :: I. . "connection.pressure=W./ 2.sf* ((.6*design dead;loads) +`.75 '`uplift_force +-H? /W'/ W uplift force) connection pressure =.. * 10'* (.6.*.(3.3 +.1) + 16:304 psf+ 8 667'2 / 50`/ 50 *:;16 304 psf) - connection pressure=.3695.707 lbs ':.. 20d.nails=.(connection pressure .boltstrength) / Z.with a:min of.2 for safety 20d nails =',(3695.707 lbs = 1125:339',lbs) /293.56655411.40858 lbs . 20d nails=9 nail through truss into:column :' ; _...._ . -- . . i CONSTRUCTION TECHNOLOGY INSPECTION & TESTING DIVISION, P.D.&T.S., INC. 4 William Street,Ballston Lake,New York 12019 - Phone: (518)399-1848 Fax: (518)399-1913 E C E MAR 13 2015 March 13,2015 TOWN OF QUEENSBURY BUILDING& CODES TOWN OF QUEENSBURY BUILDING DEPARTMENT 742 Bay Road Queensbury,New York 12804 FILE COPY Att'n: Building Code Enforcement Officials Re: COLD STORAGE BUILDING: 34 BIG BOOM ROAD, QUEENSBURY,NEW YORK Construction Materials Inspection and Testing Services Greetings, Per the request,Mr. Kyle Stevens,AFSCO Fence &Deck, we are pleased to inform you CONSTRUCTION TECHNOLOGY has been retained to provide the construction materials inspection and testing services required of the referenced project including but not limited to: Soil compaction and portland cement concrete inspection and testing. As such, we would like a moment to introduce our organization. CONSTRUCTION TECHNOLOGY is the construction materials inspection division of Product Development&Technical Services,Inc. Since our 1984 incorporation in New York State, we have been devoted to providing the highest standards of on time field inspection, testing and laboratory support to the construction industry. Our desire is a QC/QA involvement with projects committed to timely completion while satisfying the quality controls and assurances necessary to support today's strict design criteria. All too often materials inspections are viewed as a necessary evil or a requirement to be overlooked.Many times the laboratory brings this attitude on itself because of less than fully qualified personnel,tardy response time or uncompromising attitudes of technicians. Given the opportunity,we are confident we can provide the services that reflect our strong desire to be an asset to the construction process. Our facility and field technical force is supported by state of the art test equipment,including nuclear soil/asphalt density test equipment, automatic soil moisture/density(Proctor)apparatus, temperature/humidity controlled curing environments for concrete specimens, and all other necessary equipment to satisfy the requirements of the American Society for Testing and Materials (ASTM) standards C-1077 and E-329 as they relate to a commercial inspection agency. CONSTRUCTION TECHNOLOGY has submitted to outside auditing by having our personnel,equipment,facility and quality assurance plan evaluated by the Cement and Concrete Reference Laboratory (CCRL) of the National Standards Institute.A copy of the CCRL/NSI audit and finding is available upon request. March 13, 2015 TOWN OF QUEENSBURY BUILDING DEPARTMENT Re: COLD STORAGE BUILDING: 34 BIG BOOM ROAD, QUEENSBURY,NEW YORK Construction Materials Inspection and Testing Services Our staff ranges from consultant Professional Engineers,hydrologists and geologists to field technicians and materials inspectors with technical backgrounds of educational degrees and/or substantiated experience in a construction oriented discipline.Field inspectors also possess a combination of the following certifications: * Certification at various levels (AET-I through SET-IV)by the National Institute for Certification in the Engineering Technologies (NICET). *American Concrete Institute (ACI) Grade I Concrete Inspector. * American Concrete Institute (ACI)Certified Concrete Construction Special Inspector. *North East Transportation Training &Certification Program (NETTCP). Certified Soil and Aggregate Inspector *Troxler Nuclear Gauge Training Program for Radiation Safety Officers and Field Operators. * NYSDoT Certified Asphalt Density Inspector. *J&L Testing, Inc. Certification Program for Sanitary Landfill Inspectors. * J&L Testing, Inc. Certification Program for Quality Control and Inspection of Geosynthetic Systems for Use in Solid and Hazardous Waste Disposal Facilities. In addition to particular qualifications,field inspectors are familiar with all phases of construction materials testing and inspection including soils, concrete,reinforcement,masonry, structural steel,fireproofing and roof installation.This generally alleviates the need and expense of employing more than one inspector on a particular project. CONSTRUCTION TECHNOLOGY currently has the capabilities to provide inspection and testing services relating to many construction concerns including but not limited to: subsurface drilling and investigation, soils placement and compaction, foundation design recommendations,portland cement concrete,unit masonry,mortars, grouts,bituminous concrete, structural steel and fireproofing, as well as all associated laboratory services. In general, all forms of materials testing and inspection services, as they relate to the construction industry,may be obtained through CONSTRUCTION TECHNOLOGY. We appreciate your interest in our firm and our line of work. We look forward to completing-the construction materials inspection,testing and laboratory requirements for: 34 BIG BOOM ROAD,QUEENSBURY,NEW YORK. If there are any questions, or when we may be of assistance,please contact this office immediately. Respectfully, CONSTRUCTION TECHNOLOGY Tom Joshn, S.E.T., (NICET) Manager Technical Services CONSTRUCTION TECHNOLOGY INSPECTION & TESTING DIVISION, P.D.&T.S., INC. 4 William Street,Ballston Lake,New York 12019 Phone: (518)399-1848 Fax: (518)399-1913 f� j�- f D L5 yJ 15 2MAR 13 2015 March 13, 2015 TOWN OF QUEENSBURY AFSCO FENCE & DECK BUILDING& CODES 34 Big Boom Road Queensbury,New York 12804 Att'n: Mr. Kyle Stevens Re: COLD STORAGE BUILDING: 34 BIG BOOM ROAD,QUEENSBURY,NEW YORK Construction Materials Inspection and Testing Services Dear Mr. Stevens, CONSTRUCTION TECHNOLOGY is pleased to submit the following fee schedule and proposal to provide all necessary labor, equipment,tools, and supplies to complete the construction materials inspection, testing and laboratory evaluations required of the referenced project. From our conversation, we anticipate our services may include: I. FIELD SERVICES: * SOIL FILLS AND BACKFILLS: Services of a Senior Materials Technician to provide onsite soil compaction verification.Testing and.evaluation of rill and backfill materials shall be in accordance with ASTM D-1556: Sand Cone Method or ASTM D-2922: Nuclear Gauge Density Method. * CONCRETE STEEL REINFORCEMENT: Services of a Senior Materials Technician to provide onsite concrete reinforcing steel verification.Evaluation of reinforcement installations shall include bar size; welding,bending and fabrication techniques; lapped and tied connections,cleanliness, alignment, spacing and formwork clearances.Evaluation and recordation shall be in accordance with the American Concrete Institute (ACI) and the Concrete Reinforcing Steel Institute(CRSI). * PORTLAND CEMENT CONCRETE: Services of a Senior Materials Technician, certified as required in ASTM C-94,by the American Concrete Institute(ACI)to Grade I,Field Technician. Services shall include tests for slump, entrained air,temperatures,unit weight/yield and fabrication of test specimens.Recordation of placement shall be in accordance with applicable ASTM Standards and ACI Specifications. * Concrete test cylinder and construction materials pickup. Technician hourly rate for above listed services $ 31.50 Transportation per mile portal-to-portal 0.65 II.LABORATORY SERVICES: Soil Particle Size Analysis (Sieve Analysis),per sample $ 45.00 Soil Moisture/Density Relation (Proctor),per sample 85.00 Compression Test of Concrete Test Cylinders,per cylinder 7.50 1 March 13, 2015 AFSCO FENCE &DECK Re: COLD STORAGE BUILDING:34 BIG BOOM ROAD, QUEENSBURY,NEW YORK Construction Materials Inspection and Testing Services I1I.EQUIPMENT CHARGES: Nuclear Soil/Asphalt Density Gauge,per day $ 45.00 All listed fees INCLUDE administrative review and typed report distribution to our client and those parties designated by project specification. CONSTRUCTION TECHNOLOGY also provides other forms of testing and inspection services. Should there be requirements other than those listed,please allow us to modify this proposal or resubmit a suitable proposal for additional work. We appreciate your interest in our firm.We look forward to serving the construction materials inspection, testing and laboratory requirements of AFSCO FENCE &DECK. If there are any questions, or when we may be of assistance,please contact this office immediately. Respectfully, CONSTRUCTION TECHNOLOGY Tom Joslin,S.E.T. (NICET) Manager Technical Services SOFIA ENGINEERING PLLC 7 Lorna Lane Loudonville,NY 12211 Tel. 518.424.6599 e.mail:vsofa@nycap.rr.com To:Mr. John O'Brien CEO Town of Queensbury Building Department � rn* E B Y E 742 Bay Road V� Queensbury,NY 12804 MAR 13 2015 From: Enzo Sofia.PE 7 Lorna Lane TOWN OF QUEENBBURY Loudonville,NY 12211 BUILDING&CODES Date: September 23, 2014 Re:Neal Galvin new 100'x50 "SmartKits"Building 34 Big Boom Road Queensbury,NY 12804 Dear Sir or Madam: Please note that we will perform structural inspection during construction at the above project. Should you have any uestion let me know. S' ;erel Sofa, OON57RUCTION TECHNOLOGY 4 William Street,Ballston Lake,New York 12019 Phone:(518)399-1848 Pax:(518)399-1913 CLIENT: AFSCO FENCE&DECK REPORT NUMBER: 2 :PAGE#:1 34 BIG BOOM ROAD INSPECTION DATE: 04/22/15 QUEENSBURY,NEW YORK 12804 OUR FILE NUMBER: 1582.001 INSPECTOR&TEST SET: BRYAN CASAW #12 ATT'N: MR.KYLE STEVENS AMBIENT WEATHER: 50's:CLOUDY PROJECT: COLD STORAGE BUILDING:34 BIG BOOM ROAD,QUEENSBURY,NY. OUR FILE LOCATION: 154453 CONCRETE FIELD INSPECTION & COMPRESSION TEST RESULTS PLACEMENT LOCATION OF LOAD# 1': POLE BARN:POST INFILL:SOUTH,EAST&WEST WALL LINE• DELIVERED LOAD NUMBER: 1 TRUCK NUMBER/TICKET NUMBER: 4589/ 805244 YARDAGE-DELIVERED/SUBTOTAL: 8.00/ 8.00 TIME CONC. BATCHED/ARRIVED: 10.05/ 10.16 TIME PLACEMENT BEGAN/ENDED: 10.23/ 10.45 CONCRETE AGE(HOURS) (SPEC: MAX: 1.50) 0.67 SLUMP ON ARRIVAL (INCHES): 4.50 WATER ADDED ONSITE(GALLONS): WATER ADDED AT DISCRETION OF: SLUMP OF CONCRETE INTO PUMP: IF APPLICABLE: PLACEMENT SLUMP(INCH) (SPEC: 5.00) 4.50 . ENTRAINED AIR (%VOL) (SPEC: 4.00- 7.00) 8.501 UNIT WEIGHT (PCF) (SPEC: ) 147.53 CONCRETE TEMP. (F) (SPEC: 45 - 90) 66 NUMBER OF TEST SPECIMEN CAST: 6 LAB CYLINDER CONTROL NUMBERS 154453- 154458 DISCREPANCIES&REMARKS: 1:CONTRACTOR NOTIFIED:ENTRAINED AIR UNLESS NOTED ALL TESTING JAW: ASTM: C31, C138, C143, C172, C173, C231, C470, C567, C617,"C1064 CONCRETE SUPPLIER: .IOINTA-GALUSHA,LLC. CONCRETE TEST CYLINDER COMPRESSIVE RESULTS PER: ASTM:C39 DESIGN STRENGTH&FORMULA: 3000 P.S.I.@ 28 DAYS UNLESS OTHERWISE NOTED ALL CYLINDERS RECEIVED: 04/23/15 - CEMENT: LBS. CYLINDER TEST .TEST AGE ULTIMATE UNIT CEMENT: LBS. NUMBER . DATE DAYS APPLIED LOAD P.S.I. WATER: GAL. COARSE AGGREGATE#1: LBS. 154453 04/29/15 7 ' 43,400 3460 COARSE AGGREGATE#2: LBS. 154454 04/29/15 7 42,250 3360 COARSE AGGREGATE#3: LBS. 154455 05/20/15 28 55,100 4390 FINE AGGREGATE: LBS. 154456 05/20/15 28 54,650 4350 ADMIXTURE#1: OZS. 154457 05/20/15 28 53,906. 4290 ADMIXTURE#2: OZS. 15445.8 SPARE ADMIXTURE 43: OZS. AIR ENTRAINING AGENT: OZS. - REPORT DISTRIBUTION: 1:FILE 5: 2:SUPPLIER 6: 3: 7: 4: 8: RESPECTFULLY SUBMITTED, CONSTRUCTION TECHNOLOGY TOM JOSLIN,S.E.T. (NICET) MANAGER TECHNICAL SVCS. CONS'T UCTI®N TECHNOLOGY 4 William Street,Ballston Lake,New York 12019 Phone:(518)399-1848 Fax:(518)399-1913 CLIENT: AFSCO FENCE&DECK REPORT NUMBER: 1 :PAGE#:1 34 BIG BOOM ROAD INSPECTION DATE: 04/17/16 QUEENSBURY,NEW YORK 12804 OUR FILE NUMBER: 1582.001 ._INSPECTOR&TEST SET: PETE DUNHAM #9 ATT'N: MR.KYLESTEVENS AMBIENT WEATHER: 60's:CLOUDY PROJECT: COLD STORAGE BUILDING:34 BIG BOOM ROAD,QUEENSBURY,NY. OUR FILE LOCATION: 154289 CONCRETE FIELD INSPECTION & " COMPRES"SION . TEST RESULTS PLACEMENT LOCATION OF LOAD# 1 : POLE BARN:POST INFILL:NORTH WALL LINE DELIVERED LOAD NUMBER: 1 TRUCK NUMBER/TICKET NUMBER: 4560/ 805207 YARDAGE DELIVERED/SUBTOTAL: 2.00/ 2.00 TIME CONC. BATCHED/ARRIVED: 1.07/ 1.19 TIME PLACEMENT BEGAN/ENDED: 1.25/ 1.36 CONCRETE AGE(HOURS) (SPEC: MAX: 1.50) 0.48 SLUMP ON ARRIVAL (INCHES): 2.00 WATER ADDED ONSITE(GALLONS): WATER ADDED AT DISCRETION OF: SLUMP OF CONCRETE INTO PUMP: IF APPLICABLE: PLACEMENT SLUMP(INCH) (SPEC: - 5.00) 2.00 ENTRAINED AIR (%VOL) (SPEC: 4.00- 7.00) 4.10 UNIT WEIGHT (PCF) (SPEC: ) 152.79 CONCRETE TEMP. (F) (SPEC: 45 - 90) 71 NUMBER OF TEST SPECIMEN CAST: 6 LAB CYLINDER CONTROL NUMBERS 154289- 154294 DISCREPANCIES&REMARKS: UNLESS'NOTED ALL TESTING IAW: ASTM: C31, C138, C143, C172,.C173,.C231, C470, C567, C617, C1064 CONCRETE SUPPLIER: JOINTA-GALUSHA,LLC. CONCRETE TEST CYLINDER COMPRESSIVE RESULTS PER: ASTM:C39 DESIGN STRENGTH&FORMULA: 3000 P.S.I.@ 28 DAYS UNLESS OTHERWISE NOTED ALL CYLINDERS RECEIVED: 04/20/15 CEMENT: LBS. CYLINDER " TEST TEST AGE ULTIMATE UNIT. CEMENT: LBS. NUMBER DATE DAYS APPLIED LOAD P.S.I. WATER: GAL. COARSE AGGREGATE#1: LBS. 154289 04/24/15 7 40,300 3210 COARSE AGGREGATE#2: LBS. 154290 04/24/15 7 42,500 3380 COARSE AGGREGATE#3: LBS. 154291 05/15/15 28 58,750 4680 FINE AGGREGATE: LBS. 154292 05/15/15 28 57,100 4550 ADMIXTURE#1: OZS. 154293 05/15/15 28 56,550 4500 ADMIXTURE#2: OZS. 154294 SPARE ADMIXTURE#3: OZS. AIR ENTRAINING AGENT: OZS: REPORT DISTRIBUTION: 1:FILE 5: 2:SUPPLIER .6: 3: 7: 4: 8: RESPECTFULLY SUBMITTED, CONSTRUCTION TECHNOLOGY TOM JOSLIN,S.E.T. (NICET) MANAGER TECHNICAL SVCS.