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HomeMy WebLinkAboutSTRUCTURAL CALCULATIONry F�- U) z F- U) w M 5tructural Calculati,on SCANN20 BY St. Lucie County Job Number: 09-935 Date: May 07,2009 Job Name � Addre55: INTERNATIONAL AIRPORT E3U51NE55 PARK, 325 1 5t. Lucie Blvd., ffort Pierce, floragAA ENGINEER Of RECORD: AEC Services, Inc. bcen5e: #9277 Q5:#0011445 I G I G ALL15ON W001)5 LANE TAMPA, FL 33G 19 (513)G84-1 234 (813)G84-2GGO (f) WWW. KUN r-AIK, r.r. -ft #50738 I Design Wind Pressures For OPEN Structure Using the ASCE 7-05 Date: 5-7—'2009 With Roof Angle between 0 and 5 degrees Client: AEC Services Job: 09-935 40'X 75'Canopy 1616 Allison Woods Lane Tampa Florida BEST Industries Inc. —Pr6pared By: TOMAS PONCE P.E. SiteAddress: INTL. AIRPORT BUSINESS PARK 3251 St Lucie Blvd. Fort Pierce, FL Structure: Canopy Wind Speed (3-sec Gust) Importance Facto Exposure Cateno 2 Lile STEEL MOE V3:= 140 1:= 1.0 C hr E := 29000.ksi Structure Parameters: Symmetrical, No expected Torsion Canopy Heigh Total Height: Dead Load Live Load L:= 75-11 hfascia:= 7.16-ft bcolumn:= 15-ft H:= heolumn + hfascia DLr:= 8-psf LLr:= 30-psf W:= 40-ft Slope:= 0-deg If= 22.16 ft SQUARE Column Used: COL:= 4 Cohunn:= COL — S COL C:= C—SCOL X:= 71—SCOL DLcoi:= DL—SCOL AC:= A — S COL IX:= I—SCOL S:= S—S COL r:= r — S COL Z— Z — S COL mt�- n?l Column = "HSS 10" X 10" X 1/2" . Size Weigh X- Sec Area \ect Pro ert as 2 4\ 5 3 3 C= 10in DLco, = 62.3 Of AC = 17.2 in Ix = 25 = 51.2 in r = 3.86 in Z = 60.7 In. NumberofRows NumberperRow NumberofColumns ColumnSpacing Nrows:= 2 Neel:= 3 NC:= Nrows,Ncol OCL:= 26-ft NC = 6 OCW:= 16-ft L = 75ft W=40ft OCL = 26ft Side View hfascia = 7.16 ft heolumn = 15 ft OCW= 16ft End View Page I of 119 Determine the Velocity Pressure q Mean Roof Heigh Gust Effect Facto Topociraphic Facto Directionality Factor lI:=hcoIUMn h=15ft G:= 0.85 Kzt:= 1.0 Kd:= 0-85 Velocity Pressure Exposure Coefficients Table 6-2: Exposure C o:= 9.5 900-ft Z9 1:= 500-ft c:= 0.20 8:= -L 2�nin:= 15-ft 5.0 Table 6-3: Kz A, 2 z:= H z = 22.16 ft L15 - fit ) ct Q Kz:= if z < 15.ft' 2.01. 12.01.(-L) ( Kz = 0.922 Z9 ) Z9) - Velocity Pressure q 2 qh:= 0.00256.KZ-Kzt.Kd'V3 . I-psf qh = 39.3 psf Determine the Design Wind Loads on Open Buildings with Monoslope, Pitched, or Troughed Roofs MWFRS using ASCE 7-05 Section 6.5.13.2 Net Pressure Coefficient I Wini .Y=0 L 1 0.5 L 0.5 L =C-I r-�I h = 15ft 1 It 0 C. I Wind I Direction b fmm Windmrd Edge Wind Direction T-900 NJZM� Transverse Wind -j = 0 or y = 18, Longitudinal Wind Y = 90 1 .00 �w 0 := Slope Clear Wind Flow Roof Angle Load Case Transverse Wind Direction: Longitudinal Wind Direction: Horizontal Distance Windward Leeward Load Case from Windward Edge 0 = Odeg A CTANW:= 1.2 CTANL:= 0.3 A B B CTBNW:= -1-1 CTBNL:= -0-1 CIAN:= 4.8 ClBN:= 0.8 d:9 It h = 15ft C2AN:= -0.6 C2BN:= 0.5 C3AN:= -03 C3BN:= 0.3 h:5d:92-h 2-h=30ft d �: 2-h Page 2 of 19 Transverse Wind Downward and Uplift Forces on Roof Downward Pressure Uplift Pressure Windward Leeward Windward Leeward qh,G CTANW + qh.GCTANL qh-G CTBNW + qh-G CTBNL pTdown:' 2 PTuplift:� 2 pTdown = 25.1 psf pTuplift = -20.05psf pTdown = 25.06psf pTuplift = -20.05psf WIND ----=> Downward Force Uplift Force Frd:= pTdown-L-W Fruj:� pTuplift*L,W Flrd = 75.2 kip Fru, = -60.1 kip Longitudinal Wind Downward and Uplift Forces on Roof Distances of the Three Different Zones of MWFRS Measured From the Windward edge dl := h dI = 1511 d2:= 2-h - h d2 = 15ft d3:= L - 2-h 0 = 45ft Downward Pressure d!5 h h!9 d --� 2-h d �: 2-h pLIdown:= qh.GClBN pUdown:= qIl-G C213N P"down = 26.73psf pL2down = 16.7psf Downward Force FLdl:=pL'down.dl.W FLd2:=PL2down-d2-W FLd I = 16 kip FLO = 10 kip d �5 h pLIUL:= qh-GCIAN pLIUL = —26.73psf FLIUL:= pL'UL.dl.W FLIUL = —16 kip PL3down:= qh-GC3BN p0down = 10.02psf Total Downward Force FLd3:= pUdovm-O-W FL — FLdl + FLd2 + FLO FLd3 = 18 kip 44.1 kip Uplift Pressure h !� d:9 2-h pL2 UL;e PL2ULT20-Usof Uplift Force FL2UL:= pL2UL-d2-W FL2UL = -12 kip From 0 ft to dl = 15 ft From dl 15 ft to 2-h 30 ft pLIdown = 26.7 psf pL2down = 16.7 psf pLI UL = -26.7 psf pL2UL = -20 psf WIND 0011,10Y 1% %Mlf d �t 2-h pL3UL:= qh-GC3AN From 2-h=30ft to L=75ft PUdown = 1OPsf pL3UL = -10 Psf pL3[jL = -10.02psf FL3UL:= pL3UL-d3-W MUL = -18 kip Page 3 of 19 Determine the Design Wind Loads on Fascia MWFRS using ASCE 7-05 Section 6.5.12.4 Windward Fascia Leeward Fascia -GCWpn.:= 1.5 GCLpn:= -1-0 VrmdDkmdan deloff used to Develog External Pamet qW t AVINO wMa rorce Kesmung bysuaw ana Lmniponeras ana Combined Net Pressure an Windward Fascia PWf:= qh.Gcwpn PWf = 59psf Force in the Transverse Direction vrfascia:= (Pwf - PLf).hfascia-L FTfascia = 52.8kip Combined Net Pressure on Leeward Fascia PLf:= qh-GCLpn PLf = -39.3 psf Force In the Lonaltudinal Direction FLfascia �� (Pwf - a'w FL,. -20 ek, Total Lateral Force On Fascia Ffascia := FTfascia F. fascia = 52 1, p %ZOO# Determine the Design Wind Loads on Fasc I C&C using ASCE 7-05 Section 6.5.12.4.4 .............................................. . ..... . ..... ------ Use wall pressures from Figure 6-11A Assume Worst Case of 10 scift area Internal Pressure Positive Pressure Coefficient Negative Pressures From Zone 4 and 5 GCpi:= 0 No cavity Velocity Pressure qp:= qh was evaluated at H = 14 feet GC P-P := 1.0 Load Case A Front Side Back Side pFSA:= qp.GC P-P pBSA:= qp.GC pj pFSA = 39.3 psf pBSA = -55 psf GCP-A:= -1.1 GCpj:= -IA Load Case B Front Side Back Side pFSB:= qp-GC pj pBSB:= qp-GCp_p pFSB = -55 psf PHSB = 39.3 psf Page 4 of 19 Determine the Design Wind Loads on Columns MWFRS using ASCE 7-05 Section 6.5.12.4 ................... . ... . ..................... . ...... Column Dimension Force Coefficient Cf h/D Ratio Interpolate Figure 6-21 Values C = 0.83311 h/D 1 7 25 111COI X1 Y1 hcolumn Cf 1.3 1.4 2.0 C 18 Cf:= linterp (X1,Y1, C 14 25H 2 M*O Cf = 1.767 Wind Pressure Col Surface Area Wind Force on (1) Column Wind Force On all Columns PC:= qh'G'Cf Af:= C*hcojumn Fwcojumn:= qh,G,CrAf Fcolumn (Ncol + Nrows)'Fwcolumn PC = 59 PSI` Af = 12.5 ft 2 FwcoluFnn = 0-74 kip Fcolumn 3.7 kip Determine the Design Wind Loads on Roof Components and Cladding (C&C) using ASCE 7-05 Section 6.5.13.3 ................................................................................................................................. End Zone Distance numl:= min(O.10-W, 0.4-h) numl = 411 nuvn2:=max(O.04-W,3-ft) nurn2=3ft a:=if(numI<num2,num2,num1)a=4ft END:=2-a END=8ft L = 7511 a a=4ft a 2-a = Sit 2 3 a I a a=4ft L 2.a= 811 0 = 0 NET PRESSURE COEFFICIENTS Effective Area Zone 3 Zone 2 Zone I A:� a 2 a 2 = 16 W ClN3p:= 2.4 ClN3n:= -3.3 CIN2p:= 1-8 ClN2n:= -1.7 CN,p:= 1.2 * 2 -< A 5 4a 2 4-a 2= 64 ft 2 C2N3p:= 1.8 C2N3n:= -1.7 C2N2p:= 1.8 C2N2n:= -1.7 CNln:= -1.1 * 2 --� A C3N3p:= 1.2 C3N3n:= -1-1 C3N2p:= 1.2 C3N2n:= -1-1 Component and Cladding Pressure Equation qh = 39.3 psf Velocity Pressure at h Pec = qh.G'CN G = 0.85 Gust Factor CN Net Pressure Coefficient Page 5 of 19 G Components and Cladding Roof Pressure Comi3onent/Cladding Area Zone 3 Gravity/Uplift Zone 2 Gravity/Uplift Zone 1 Gravity/Upli 2 2 'For A <- pl—Z3—Pos,, = 80.2psf pl—Z2—Poscc = 60.1 psf pl—Zl—Posee = 40.1 psf —2 =-16ft pl—Z3—Negcc = —110.2psf pl—M—Nekc = —56.8psf pl—Zl—Negcc = —36.7psf 2 2 Pz--V�-poscc = 60.1 psf p2_Z2_Pos., = 60.1 psf pLzl—poscc � 40.1 psf For A < 4-a = 64ft pZ__Z3_Negcc = —56.8 psf pZ__Z2_Negcc = —56.8 psf p2_Zl—Nekc = —36.7psf 2 2 p3—Z3—Poscc � 40.1 psf p3_Z2_Poscc = 40.1 psf p3_Zl — Pos cc = 40.1 psf For A —> 4-a = 64ft p3LZ3LNegcc = —36.7psf p3_Z2_Negcc = —36.7psf p3LZ1—Negcc = —36.7 psf DecklngSlze W-16-in=S3.333ft 2 a Page 6 of 19 Gravity (Dead Loads) From The Canopy Deck Fascia Fascia:=: 3. lbf DLFascia:= Fascia.hfascia 2- - Outrigne Outrigger:= 3-lbf oCoutrig:= 32-in Angle Stiffeners Angle:= "L 2.511 X 2.5" X 3/16" Decking DLdeck:= DLr Moutrig:= Outrigger Ocoutrig DLFascia � 21A8 lbf . ft DLoutrig = 1.125 lbf ft lbf Mangle:= 3.06- ft Lbf DLdeck = 8 2 Column Loading (ASDI Tributory Area For Each Column LLr=30psf DLr=8psf Trib A W Trib�Acol = 520ft 2 - col:= OCL. 2 Dead Load on Each Column Live Load On Each Column Wind Load on Each Column Pe DL:= DLr-Trib-Ac0l Pe LL:= LLr-Trib-Ac0l Pc - WLI:= pTdown-Trib-Acol Pe-DL = 4.2 kip t Pc WL1 131 kip Pq LL = 15.6 kip r P _AJ c-WL2;= P co P MCC 2 Downward Uplift Page 7 of 19 Find Moment at Bottom of Column Due To Main Force Wind Ffasda�52*8kip Fe—olumn = 3-7kip Base Moment Ffascia*(H — blascia) + Fcolumn, healunin MB:= — 2 ) 2 MB 168ft-kip NC 6 NC Axial Force On A Single Column To Size Footing For Soil Bearing Condition (ASO Pc—DL � 4.16 kip Dead Load on Each Column PDL:= Pc—DL + DLeol, heolunin PDL = 5.095 kip Pc—LL = 15o6 kip Live Load On Each Column PLL:= Pc—LL PLL � 15.6 kip Wind Load On Each Column Pc—WLI � 13.029 kip Use Highest Magnitude for PWL:= PC—WLI PWL = 13.029 kip Pe—WL2 = —10A24 kip uplift and gravity P—asd—LC2:= PDL + PILL P�_asd_]LC2 = 20o7 kip P—asd—LC5:= PDL + PWL P�_asd—]LC5 = 18.1 kip P—asd—LC7:= 0.6oPDL + PWL P�_asd—LC7 = 16ol kip P—asd—LC7B:= 0.6-PDL — PWL F�_asd—LC7B=-10kip <==Critical Load cas ForFooting Windin I! ondition P—col—asd:= max(P�_asd—LC2, P—asd—LCS, P—asd—LC7) P_col—asd = 20.7 kip R.nrfinn.Rtrpccfnrii,rariAtThaRnconfFnr-hrniiimnfmm-RpmirpL MB P col asd fhend:= — + � fbend S AC AM Footing Size tl,b:= 2-ft Square Bsq:=7.5-ft Bsq=7.5ft' Depth:= 5-ft Axial Force On A Single Column LRFD Load Combinations DL:= (DLeWheolunin + PDL) LL:= PLL Pu—LCI:= IADL Pu—LC2:= 1.2DL + 1.6-LL Pu—LC4:= 1.2DL + 1.6-WL + LL Pu—LC6:= 0.9DL + 1.6-WL Pu—LC1 = 8.441 kip Pu—LC2 = 32.195 ldp Pu—LC4 = 43.682 kip Pu—LC6 = 26.273 kip Footing Dead Load 2 DLfooting:= 150"b—f -Bsq Depth ft 3 DLfbotlng = 42.188 kip WL:= PwL PU:= max(Pu—LC1,pu—LC2,pu—LC4,Pu—LC6) PU = 43.7 kip Page 8 of 19 Interaction Ecluations for HSS Steel Column Usinci LRFD Fy_gSS:= 42.ksi Column = "HSS 1011 X 1011 X 1/211 E = 290OOksi Ob:= 0-9 Oc:= 0-85 Column Data From Table 1-12 Slenderness Moment Of Inertia Section Modulus Radius of gyration Plastic Modulus A. = 18.5 Ix = 256in 4 S = 51.2 in 3 r = 3.86 in Z = 60.7 in 3 Check Compactness I p := 0.0714- E X p = 49.3 7, = 18.5 FY_HSS Noncompact Section ).r:= 0.309. E ;Lr = 213.4 1 = 18.5 FY_HSS Q := if 491r, 1.0, 0.0379-E Q=1 X pp := 0.448. E 71 pp = 309.3 a X = 18.5 (26 FY_HSS.X 3) FyjISS Critical Load !Lo�fFy H*S . y _ $S 12LC = 0.5,65 r.7c E Q.;L 2) 0.877 Fer -,,IQ 15 1.5, Q-(�.658 c)*FyjSS 2 Fy -HSS 21C Axial Load Capacl A6plied Fact"ed ead� tenderness Parameters pn:= �c.Fcr.AC Pu f43.% kx:= 1 ky:= 1 ...,3 %,_�A Opn = 537.3 kip Nominal Moment Capacity Mn HSS:= if X 2L P, FY_HSS.Z, (0.0207. E + 1�.F HSS'S11 I I ( X FyjSS ) Y- P u Opn - 0.081 < 0.2 ==> Use Equation 7.1.2 F I rom. LRFD HSS Design Interaction Equation Pu + MB = 0.92 < 1.0 OKII 2.�Pn h'Mn_HSS kx*hcolumn - = 46.6 200 OK r Mn­HSS = 212Aft-kip Page 9 of 19 5 Design of Isolated Column SQUARE Footing Code: ACI 318 - 05 Prepared By: Tomas Ponce PE Factored Load Steel Yield Bar Size Concrete Bearing Plate Pu = 43.7 kip ly := 60.ksi bar:= 5 fC:= 2500-psi BP:= 18-in Square Footing Size Depth of Bars Column Width Eccentric! Allowable Bearina Pressure B:= Bsq d:= Depth - 4-in c:= C MB qallow:= 2000-psf B = 7.5ft t:= Depth c= 10in P-col-asd cover:= 3-in e = 8.12 ft Bsq — 11.25in 8 Maximum Soil Pressure P-col_asd = 20.7 kip Maximum Allowable Load To Reach Soil Bearing Pressu P col sad 6*MB qmax:= � + — Bsq 2 Bsq 3 Reoulred Reinforcement Asreq:= 0.0018-B-d Asreq 9.07 in 2 qmax = 2757psf OR w Bsq MB ) P:= i.qaHow'(3-_ _ - -- Bsq 2 2 i�coLasd) P = 23.5 kip ]?�_col-asd = 20.7 kip ement Provided foot:= 30 Nfo t -�� = 15 @ 4" OC Top 2 and Bottom As:= Nfoot-BarNo5 As = 9.3 in 2 Page 10 of 19 K Desiqn of Isolated Column CIRCULAR Footin Code: ACI 318 - 02 Prepared By: Tomas Ponce PE Factored Load Steel Yield Bar Size Concrete Bearing Plate PU = 43.7 kip fy = 60 ksi bar= 5 f1c = 2500 psi BP = 18in Square Footinci Si Depth of Bars Column Width Eccentric! Allowable Bearinci Pressure Beir:= 8-ft d 56in c = 0.833 ft MB qa1low � 2000psf t 60 in c= 10in P—col—asd cover= 3 in Bcir e = 8.12 ft — = 121n 8 Axial Load On Column P_col_asd = 20.7 kip Maximum Allowable Load To Reach Soil Bearing Pressu 4-P col asd 32-MB P:= 1qa11ow'(3-.jc1r _ __!B__).Bcir qrnax:� � + — qroax = 3754psf 2 2 P ol asd) n-Bcir 2 u-Beir 3 OR _jc — P = 31.1 kip P�_col—asd = 20.7 kip Required Reinforcement Asreq:= 0.0018-Beir*d Asreq = 9.68 in 2 00 eincement Provided Nfoot:= 32 Nfoot — 16 @ 4" OC Top 2 and Bottom As:= Nfoot-BarNo 5 < As = 9.92 in 2 Page 11 of 19 Total Uplift Longitudinal Wind IJLLW:� [pLlUL-dl + pL2UL.dI + pL3UL-(L - 2.h)].W ULLW = -46.1 kip Transverse Wind- ULTW:= Frul Maximum Uplift UL max:= Min (ULLW,ULTW) ULmax = -60-1 kip Base Shear (Transverse) VT := Ffascia + Fcolumn VT = 56 kip Base Plate Number of Bolts D'mBolt:= 16-in N:= 4 Direct Shear Vbolt:= 1.1.6 N SOLT Bolt Diameter db.:= 5 --M 8 -in 4 -in 8 R 1 UU in VbOlt = 4kip j:= 1.. 5 Bolt Area 2 Ab := :�� 4 �0.31 OA4 Ali 0.6 n2 0.79 ,1.23 ULTW = -60.1 kip Uplift Per Column UL a UL:= 0.6.(DLfootino + NC Base Shear Per Column VT V:= — V = 9.4 kip NC Diagonal Diag:= Vf-2.DimBolt Moment About Base M:= 1.6-MB Tensile Force On Bolt M -UL Diag N bolt Vi Abj z F 'J_ := if 9.ksi - 2.5-f Vi :5 45-ksi, 59-ksi - 2.5.fv,,45-ks) > 45.0 ksl �Ft i := 0.75-Ft i 21.2) 283 OFt = 32:5 ksi 33.8 ,33.8) T N-Abj Must '113.1 B e 785 > ft = 57:7 ks! 44.2 ,28.3 ==> 1 - 1/4" Bolts are ok Diag = 1.9ft M = 3226 in -kip UL = 15.3 kip '= 139 kip idolnhk. RN( CIO '12.3 8*5 fV = 6.3 ksi 4.8 ,3.1 '28.3 37.7 Ft = 43A ksi 45 45 Page 12 of 19 P Check Thickness of Base Plate hcolumn � is ft OPR = 537.3 kip B:= D'mBolt N:= B B 16in in N — 0.95-C m 0.27111 2 n:= B — 0.8-C n 0.333 ft 2 2 n' := NFC n' 2in 5 1:= max(m,n,X-n') I = 0.333 It 2, P" t := I --_ min -[Z9 36 . ksi--B.N tmin = 0.411 in 3/4" Plate is good Pu = 43.682 kip pu = 537.271 kip 4-C-C Pu X:= X = 0.081 (2.C)2 �Pn 26:= 2.NFX X = 0.291 1 + NF, --x 7,-nl = 0.582in Page 13 of 19 Loads on Beam Bl: Equal Overhanas On Both Ends Unbraced Length for Beams Lb:=OCW Lb=16ft SpanBI:= Lb OHB1:= 12-ft Fy:= 50-ksi lbf 3 4 Beaml:= "W14 X 4811 �m -n-Bl:= 232-ft-kip DLB1:= 48-- ZK Bl:= 78.4-in IB1:= 485-in ft Tributary Area Uniform Loading From Decking Placed on Beam B2 lbf TribBl:= OCL WBIdeck:= DLdeck*(TribBI) wB1 deck= 208 ft Dead Load On Beam (Includinci Self Weigh Live Load Lbf bf DI, Bl:= DLBI + DLr-TribB, DL BI = 256 LL Bl:= LLr-TribB, LL BI = 780L- ft ft C&C Wind Force Pressure (Worst Case wcc:= P_V�_Poscc Wcc = 40.1 PSf wCC BI:= WCC TribBI wCC B1 = 1042A Lbf Apply Downward ft on beam B1 Moments Generated DL—BI.OHBI 2. NIDL_Bl:�' 2 Dead Load MDL_BI = 18.4ft-kip LL-BI.OHBI 2. MLL-Bl:= — 2 Live Load NILL-BI = 56.2ft. kip -'2. MW Bl:= WCC BFOHBI 2 Wind Load MW_B1 = 75ft-kip NIBI LC1:= 1.4-MDL BI -9 25.8ft-kip MBI-LC2:= 1.2'MDL_BI + 1.6-MLL-BI MBI-LC2= 112ft-kip MBI-LC4:= 1.2'MDL-BI + MLL-BI + 1.6.MW-Bl MBI-LC4 = 198A ft-kip MBI-LC5:= 1.2.M[DL­B1 + MLL-BI MBI-LC5 = 78.3 ft-kip MBI-LC6:= 0.9.MDL 111 + 1.6.MW B1 MBI-LC6= 136.7ft-kip MBI-LC7:= 0.9.MDL-Bl KB1_LC7 = 16.6 ft -kip Mu Bl:= max(MBI-LCI, M[BI_LC2, M[B1_LC4, MB1_LC5, MBI-LC6, MB1-LC7) Mu B1 = 198Aft-kip UnbracedLength Find Moment Capacity from AISC LRFD MANUAL Lb 16ft Beaml = "W14 X 48" Omn-Bl = 232ft-kip Zx B 1 78.4 in 3 Must Be > M[u_B1 = 52.9 in 3 SpanB1 = 16ft 1B, = 4851n4 E = 29000 ksi Fy 50 ksi 0.9-F Y Equivalent Loadin-q Check actual Deflection Deflection Limit — 'vBl:= NILL Bl'2 wB,.OHB1 3 3) 2fE 2 ABI:= .(4'O"BI'.SpanBI - SpanBI OHBI 24.E.IBI + 3.OHB, Alim:= 180 wBI � 780plf AB1 = 0.494in < Allm = 1.067 in Page 14 of 19 I Moment Demand in the Weak Axis Due to Wind Force On Fascia (Blaxiai Bendina Moment Arm for Loads from Beam Centerline . RBI := 1-0-ft RB1 = lit Tributary Area Uniform Loadinci From Decking Placed on Beam BI Live Load On Beam BI Tribweak:= 1-0-ft wBldeck:� DLdeck'(Tribweak) wB'deck = 8Lb-f LLB1:=LLr.(Tribweak) LLBI=30pif ft Dead Load On Beam (From Fascia) Live Load DL_Bl:=DLFaseia+DLr'Trlbweak DLBI=29.48 Lbf LL-Bl:= LLr.Tribweak LL_B1 = 30Lbf- ft ft C&C Wind Force Pressure (Worst Case) lbf Apply Downward WCC:=p3_Z3LPoscc WCC=40.lpsf WCC_B1:=WCC.Tribweak wCC-Bi=40.1 ft on beam B1 MDL-Bl:= RB1'(SpanB, + OHB1).DL_B1 MLL-Bl:= RB,,(SpanBI + OHBI)-LL-Bl MW_BI := RBI*(SpanB, + OHBI)'wCC_B1 Dead Load MDL_Bl = 0.8ft-kip MBI-LCI:= 1.4.MDL'-Bl Live Load MLL-BI = 0.8ft-kiP MBI-LC2:= 1.2.MDL-BI + 1-6,MLL-BI MBI-LC4:= 1.2.MDL-Bl + MLL-BI + 1.6.MW - B, MBI-LC5:= 1.2.MDL-BI + MLL-BI MBI-LC6:= 0'9'MDL-BI + 1.6'MW-Bl MBI-LC7:= 0'9'MDL-BI MM1_LC1 = 1.2ft-kip MZ1_LC2 = 2.3 ft. kip MB1-LC4 = 3.6ft-kip MBI-LC5= 1.8ft-kip MBI-LC6 = 2.5 ft -kip MBI-LC7 = 0.7ft-kip Myu-B I:= max (MB1_LC1, MBI-LC2, XB1_LC4, MBI-LCS, MBI-LC6, MBI-LC7) MYu_B1=3.6ft-ldp muy:= myu-Bl MuY = 3.6ft-kip interaction ifouation M ux muy , — = 0.91 �bmx �bmy Moment Demand about the Strong Axis MUX:= Mu_B1 Mux � 19SAft-kip 1.0 BEAM IS GOOD End Beam hold 100% of trIb Area No middle beam Beaml = "W14 X 48" Moment Capacities AISC LRFD Manual ObMx:= Wn-B1 ObMx = 232ft-kip �Wy:= 72-ft-kip Page IS of 19 Loads on Beam B2 Overhang on One End Unbraced Length for Beams LbB2:= 13-ft SpanB2:= 26-ft ORB2:= 12-ft Bcam2:= "W16 X 26" Omn-B2:= 78-ft-kip DLB2:= 26. lbf ft Tributary Area Uniform Loading From Decking Placed on Beam B2 Trib := 7.33-ft wB2 DL Trib wB2 58.64 Lbf B2 deck deck*( B2) deck ft ZZ-112:= 44.2-in 3 IB2:= 301-in 4 Live Load On Beam B2 LLB2:= LL,.(TribB2) LLB2 = 220Lbf- ft Dead Load On Beam (Including Self Weigh - Live Load DL-132 := DLB2 + DLr-TribB2 DL-B2 = 84.64plf LL-B2:= LLr.TribB2 LL-B2 = 219.9Lbf- ft C&C Wind Force Pressure (Worst Case) WCC:= p3_Z32osee WCC=40.lpsf wCCB2:=WCC-TribB2WCCB2=293.9 lbf Apply Downward ft on beam B2 Moments Generated Between Support := DEt_B2.x 2 2 ?120 MBSW 2-SpanB2 (SpanB2 OHB2 - SPanB2'X) IF Itt COP Use Graph to find Maximum Moment Location 61 1 1 1 1 1 1 1 1 1 i I 4 MBSW ft-kip 2 L L L 0 3 5 6 8 9 i'O i'l i2 13 ;4 �5 �6 i7 iS i9 i"O I i2 �3 24 25 x x:= 12.5-ft DL�W-x 2 2 MDL-W:� 2-SpaHB2 . (SpanB2 _ OHB2 _ SpanB2'x) MDL-112 = 4.2ft-kiP 1,L_B2.x 2 2 MLL-B2:= 2'SP`mB2 (SpanB2 _ OHB2 _ SpanB2'X) MLL-B2 = 10.9ft-kip wCC�M'x 2 2 MW-B2:= 2-SpanB2 (SPanB2 _ OHB2 _ SPanB2'X) MW-132 = 14.6ft-kip Page 16 of 19 Moments Generated From Overhang At Right Suppo MDL 132;� DI�1]12 OHB22) 2 MLL_B2:� LL�W . (OHB22) 2 'VCC B2 2) MW B2:= �' OHB2 2 Dead Load MDL_B2 - 6.1 ft -kip MDL­B2 = 6.1 ft -kip M[LL_ B2 = 15.8ft-kiP, MW B,2 = 211 ft-kip Live Load MLL-B2 15.il ft -kip <== THESE ARE LARGER Wind Load Mv�_]212 = 21.211 -kip USE ASCE 7-05 Strength Design Load Combinations (Section 2.3.21 To Find Maximum Moment Demand MB2-LCI:= 1.4.MDL-B2 MZ2_LCI = 8.5 ft. kip M32-LC2:= 1.2.MDL-B2 + 1.6,MLL-B2 MB2-LC2 = 32.6 ft-kip MB2-LC4:= 1.2'MDL-B2 + MLL_B2 + 1.6'MW-B2 MB2_LC4 = 57 ft.kii MB2-LC5:= 1.2'MDLB2 + MLL-B2 p .3ft-kip MB2-LC6:, 0.9.MDL B2 + 1.6,MW B2 goo . 5#i 9955 MB2-LC7:= 0.9.MDL-B2 MB2-LC7 5.5 ft -kip mu-B2:= max (MB2-LCl, MB2-LC2, XB2_LC4, MB2_LC5, MB2-LC6, MB2-LC7) Mu-B2=57ft-kip UnbracedLength Find Moment Capacity from AISC LRFD MANUAL LbB2 � 13 ft Beam2 = IIW16 X 2611 Wn-B2 = 7811-kip ZX B2 � 44.2 in 3 Must Be > Mu_B2 = 15.2 in 3 SPanB2 = 26ft 0.9.17Y Eaulvalent Loadin NILL 132,2 WB2:= — 2 OHB2 wB2 = 219.9 plf Check actual Deflection 'B2 = 301 in 4 E = 29000 ksi Fy = 50 ksi AB2:= WB2'OHB2. (4-O'HB2 2 -SpanB2- SpanB2 3 + 3.01IB2 3) 24-E-IB2 AB2 = 0.056 in 4 Aiim = 1.733 in Deflection Limit Aiim:= 22E 180 Page 17 of 19 Moment Demand in the Weak Axis Due to Wind Force On Fascia (Blaxial Bending Moment Arm for Loads from Beam Centerline RB2:= 2-11 RB2 = 211 Tributary Area Uniform Loading From Decking Placed on Beam B2 Live Load On Beam B2 Y Tribweak:� 0'ft w]32deck:� DLdeck'(Tribweak) wB2deck � 0 Lbf LLB2:= LLr.(TribwealLLB2 = 0 Ibf ft ft ,;4 Dead Load On Beam (From Fascia Live Load DL—B2:= DLFascia + DLr'Tribweak DL B2 = 21.48 jbf LL—B2:= LLr-Tribweak LL—B2 = 0 jbI ft ft C&C Wind Force'Pressure (Worst Case WCC:=p3_Z3_PosCC WCC=40.lpsf wCCB2:=WCC.TribwezWCC ­0 IN Apply Downward _B2 ft on beam B2 MDL­1[12:� RB2'(SPaHB2 + OHB2) -DL—B2 MLL—B2:= RB2*(SPanB2 + OHB2).LL—B2 MW—B2:= RB2*(SpanB2 + OHB2)'wCC—B2 Dead Load Live Load MDL_B2 = 1.611-kip MILL-132 = 0 ft'kP opy IndILE MW_B2 = Oft'k'P USE ASCE 7-05 Strength Design Load Combinations (Section 2.3.21 To Find Maximum Moment Demand MB2—LCI:= 1.4'MDL—B2 MB2_LC1 = 2.3 ft-kip MB2—LC2:= 1.2.MDL B2 + 1.6-MLL B2 MB2—LC2 = 2ft-kip MB2—LC4:= I.2.MDL­B2 + MLL—B2 + 1.6.MW—B2 MB2—LC4=,2ft-kip MB2—LCS:= I.2.MDL_B2 + MLL-132 MB2—LC5 = 2ft-kip MB2—LC6:= 0.9.MDL B2 + 1.6*MW B2 MB2—LC6 = 1.5 ft- kip f MB2—LC7:= 0'9'MDL—B2 XB2_LC7 = 1.5 ft-kip Myu—B2:= max (MB2_LCI, MB2—LC2, MB2—LC4, MB2—LC5, M[B2_LC6, MB2—LC7) Myu—B2=2.3ft-kip Moment Demand about the Strong Axis Moment Capacities Muy:= MYu_B2 MUX:= Mu B2 End Beam holds AISC LRFD Manual May = 2.3 ft -kip 100% Trib Area bmx:= Min — B2 Max = 57ft-kip as middle beam ObMx = 78ft-kip Interaction Equation M ux May = 0.85 1.0 BEAM IS GOOD �bMx �bNfy Beam2 = "W16 X 26" ObMy:= 19.6-ft-kip Page 18 of 19 I SUMMERY OF RESULTS DeckingSize W-16-in=53.333ft 2 Decking Must Resist MWFRS Pressure Gravity MRO pLidown � 26.727 psf pLIUL = —26.727 psf Decking Must Resist Component & Cladding Pressu Gravity MRO 2 2 For A < 4-a = 64ft pl—Z3—posce = 80.2 psf pl—Z3—Nekc = —110.2psf Fascia Must Resist Main Wind Pressure Pwf = 59 pSf Fascia Must Resist C&C Pressure Front Side Back Side pFSA = 39.3 psf PBSA = —55 psf COLUMNS Column = "HSS 10" X 10" X 1/2" Beams BeamI="W14X48" MainBearns Beam2 = "W16 X 26" Purlins (MUST BE LATERALLY BRACED AT MIDPOINTS) SQUARE FOOTING SIZE Depth Of Rebar Total Depth BSq = 7.5ft d = 56in Depth = 511 Reinforcinq Bars bar= 5 @ 4" OC Each Way Top and Bottom Anchor Bolts: (4) 1 1/4" ASTM A307 Bolts W' LONG Base Plate Thickness train = 0.41 in 3/4" Thick Base Plate Is OK ROUND FOOTING SIZE Bcir� 8ft Diameter Page 19 of 19