Loading...
The URL can be used to link to this page
Your browser does not support the video tag.
REVIEWED structural calcs
REVISION Ken Huffman Mar 1 2021 DS E CITY OFDMONDS 10620 235th Pl. S.W. DEVELOPMENT SERVICES Edmonds, WA DEPARTMENTMENT Structural Calculations BLD2018-1592 Revision 5 DVD7 Job No. - 18287 Drawings provided by Ken Huffman 03/08/2021 Index Description Design Criteria Vertical Design Foundation Design Drawings DVD7 Page 1 3 14 S1.0 REVIEWED BY CITY OF EDMONDS DVD7 Engineering, Inc. Structural Engineering (360) 933-1345 Job No.: 18287 — Rev. 5 DVDT Date: 3/5/2021 Title: Ken Huffinan Loc.: 10620 235th Pl. S.W., Edmonds DVDT Engineering, Inc. Structural Engineering Description: Ken Huffman has asked for structural engineering calculations for a residential house. Page: 1 of 14 Scope: Perform the structural calculations necessary to build the residence according to code, and obtain a building permit. Building Department: The calculations and drawings will be submitted to The City of Edmonds Building Services. References: 2015 IBC, ASCE 7-10, 2015 NDS Loads: Roof Load Ri=2 RDL = Roof Dead Load = 15 psf DF1 = Roof Dead Load Factor = ' sgrt((Rl)2 + (12)2) _ (I x sgrt((2 )2 + (12)2) = 1.01 TO Rcc = Roof Live Load = 20 x1.0x(1.2-O.05xR1) = 20 x1.0x(1.2-0.05x2) = 22 psf Snow Load Pg = Ground Snow Load = 25 psf Ce = Exposure Factor = 1.0 Ct = Thermal Factor = 1.1 I = Importance Factor = 1.0 Pf= Flat -Roof Snow Load = 0.7 Ce CtI Pg = (0.7) (1.0)(1.1) (1.0) (25) = 19.25 psf RsL = Design Roof Snow Load = 25 psf Wall Load WDL = Wall Dead Load = 12 psf Floor Load FDL = Floor Dead Load = 13 psf; FLL = Floor Live Load = 40 psf; Seismic Load Equivalent Lateral Force Procedure, Fa = 1.1; Ss = 1.0 Fv = 1.7; Si = 0.35 SMs = Fa Ss = (1.1)(1.0) = 1.1 SM1 = FvS1 = (1.7)(0.35) = 0.595 SDs=2XSMs=3x1.1 =0.733 SD1=2XSM1=3x0.595 =0.397 FDLD = Deck Dead Load = 9 psf FLLD = Deck Live Load = 60 psf Wood shear wall system, Seismic Design Cat. D, Category II, Site Class D R = 6.5; Io = Occupancy Importance Factor = 1.0 Wind Load - Enclosed, Simple Diaphragm Building Vuit = 3-second gust wind speed = 110 mph; KZt = 1.0 Allowable Soil Load GAR = Allowable Soil Load = 2000 psf Concrete f, = 2500 psi fy4 = 40,000 psi - Grade 40 rebar; Exposure B fy6 = 60,000 psi - Grade 60 rebar Job No.: 18287 — Rev. 5 DVDT Date: 3/5/2021 Title: Ken Huffinan Loc.: 10620 235th Pl. S.W., Edmonds DVDT Engineering, Inc. Structural Engineering Page: 2 of 14 Building Information h, = Ridge Height = 18.83 ft hem = Main Floor Plate Height = 7.63 ft heu = Upper Floor Plate Height = 16.08 ft hmr = Mean Roof Height = z x (h,. + h,u) = 2 (18.83 + 16.08) = 17.5 ft LT = Transverse horizontal dimension = 60 ft; LL = Longitudinal horizontal dimension = 56 ft W= Living Room Addition Weight = 233xRDL+35x4x WDL = 233x15+35x4x12 = 5,175 lbs Cs = Seismic Coefficient = SDs = °'�'S 3 = 0.113 lbs T, 1.0 Vs= Seismic base shear = 1.3xCsx Wx0.7 = 1.34.113x51754.7 = 532 lbs Wind Load E,G E,G F,H F,H 1 1 1 1 1 1 B,D A,C A,C Al = 21.6, Bi = -9.0, C1 = 14.4, Di = -5.2, El = -23.1; F1 = -14.1; Gi = -16.0, Hi =-10.8,• EOH1 = -32.3, GoHi = -25.3 al=0.4xhmr=0.4x17.5=7ft a2 = 0.lxmin(LT; LL) = 0.lxmin(60; 56) = 5.6 ft a,,,i„ = Minimum = 3 ft ?, = Height and Exposure Adjustment Factor (Figure 6-2) = 1.0 Le = End zone length = 2 max(a,,,i,,;min(a1;a2)) = 2xmax(3;min(7;5.6)) = 11.2 ft The shear walls and lateral strength of the building is acceptable without change. Job No.: 18287 — Rev. 5 DVDT Date: 3/5/2021 Title: Ken Huffman DVDT Engineering, Inc. Loc.: 10620 2351h Pl. S.W., Edmonds Structural Engineering Page: 3 of 14 Vertical Load Roof Ledger Connection Lsp. = Length of span = 12 ft Lft = Floor Tributary area = 0 ft LRt = Roof Tributary area = 2 ft hW = Wall Height = 0 ft WD = Dead Load = LftxFDL +LRtxRDLxDF1 +hwxWDL = Ox13 +2x15x1.01 +Ox12 = 30.3 plf WL = Live Load = LftxFLL +LRtxRsL= Ox40 +2x25 = 50 plf W = Total Load = WD + WL = 30.3 + 50 = 80.3 plf VMaX = 2 x W Lspan x l= 2 (80.3) (12) x 2= 241 plf LRt CD = Duration = 1.15 Zs = 200 lbs per 1/4"0 1/2" Simpson SDS screw ss = Screw Spacing = 1.33 ft ZS' = s 2 x Zs CD = 1.ss (2) (200) (1.15) = 3461bs/ft Therefore: Use (2)1/4"0 1/2" Simpson SDS screws @ 16" o.c. Job No.: 18287 - Rev. 5 DVD7 Date: 3/5/2021 Title: Ken Huffman DVDT Engineering, Inc. Loc.: 10620 235th Pl. S.W., Edmonds Structural Engineering RO1 - 5 1/2" x 18" GLB Ridge Beam b = Width = 5.5 in d = Depth = 18 in lb = Bearing Length (assumed) = 3.5 in = 0.292 It ws = Self Weight = 35 x 12 x b x 12 x d = 35 x' x 5.5 x l x 18 = 24.1 lb/ft Lsp. = Length of span = 23.5 ft Lft = Floor Tributary area = 0 It Let = Roof Tributary area = z x 24 = 12 ft hW = Wall Height = 0 ft Page: 4 of 14 WD = Dead Load = ws+LftxFDL +LRtxRDLxDF1 +hwxWDL = 24.1+Ox13 +12x15x1.01 +Ox12 = 206 plf WL = Live Load = LftxFLL +LRtxRsL= Ox40 +12x25 = 300 plf W = Total Load = WD + WL = 206 + 300 = 506 plf P = From R04 = 633 lbs Forces 0.633 0.506 loading 36.9 23.5' 29.4 6.11 6.42 bending 6.11 deflection shear -0.765" 6.42 Allowable Forces - 24F-V4 DF S,, = b b d2 = (1)(5.5)(18)2 = 297 in3 L = iz b d3 = (1 (5.5) (18)3 = 2,673 in4 A = b d = (5.5)(18) = 99 in2 E = 1800000 psi CD = 1.15 21 0.1 1z o.1 s.1zs Go 1 21 0.1 12o.1 .1 Cv = NDS 5.3.6 =min 1.0; ) (12) ( ) =min (1.0; (-) (-) (5.125)0 ) = 0.943 pan d b 23.5 18 5.5 F&,,p = Design compression stress perpendicular to grain = 650 psi RAu=&Perpxbxlbx12 = 650x5.5xO.292x12 = 12.5271bs Fv = Design shear stress = 265 psi VAu = F, CDxA = (265)(1.15)x99 = 30,170 lbs Fb = Design bending stress = 2400 psi Man = FbCDCV (12) = (2400) (1.15) (0. 943) (197) = 64,416 ft-lb Deflection anX12 _ 23.5x12 Allowable Total Deflection = L,,,p= 1.175 in 240 240 Therefore use a 5 1/2" x 18" GLB for the beam. OK Job No.: 18287 - Rev. 5 DVD7 Date: 3/5/2021 Title: Ken Huffman DVDT Engineering, Inc. Loc.: 10620 235th Pl. S.W., Edmonds Structural Engineering R02 - 5 1/2" x 18" GLB Roof Beam b = Width = 5.5 in d = Depth = 18 in lb = Bearing Length (assumed) = 6 in = 0.5 ft ws=Self Weight =35x12xbx12xd=35x12 Lsp. = Length of span = 12.5 ft Lft = Floor Tributary area = z x 3.5 = 1.75 ft x 5.5 x 1, x 18 = 24.1 lb/ft Lat = Roof Tributary area = Z x 24.67 + 1.5 = 13.8 ft hW = Wall Height = 3 ft Page: 5 of 14 WD = Dead Load = ws+LftxFDL +LatxRDLxDF1 +hwxWDL = 24.1+1.75x13 +13.8x15x1.01 +3x12 = 292 plf WL = Live Load = LftxFLL +LatxRsL= 1.75x40 +13.8x25 = 416 plf W = Total Load = WD + WL = 292 + 416 = 708 plf P = From R01+R04 = 6420+633 = 7,053 lbs Forces 7.05 0.708 16 loading 12.5' 6.56 bending 4.76 11.1 4.76 deflection shear -0.095 6" -11.1 Allowable Forces - 24F-V4 DF S,, = e b d2 = (1) (5.5) (18)2 = 297 in3 IX = Z b d3 = (Z) (5.5) (18)3 = 2,673 in4 A = b d = (5.5)(18) = 99 in2 E = 1800000 psi CD = 1.15 21 0.1 12 ° 1 s.12s 01 21 0.1 12 0.1 s.12s 0.1 Cv = NDS 5.3.6 = min 1.0; (L ) (-) ( ) = min (1.0; ((-) ( ) ) = 1.0 span d b 12.5 18 S.S Fepe,p = Design compression stress perpendicular to grain = 650 psi Ran = Fcre�pxbxlbxl2 = 650x5.5x0.5x12 = 21,4501bs F, = Design shear stress = 265 psi VAu = Fv CDxA = (265)(1.15)x99 = 30,170 lbs Fb = Design bending stress = 2400 psi MAu = FbCDCV (1z) _ (2400) (1.15) (1.0) (211 = 68.310 ft-lb Deflection Allowable Total Deflection = L,,panx12 _ 12.Sx12= 0.625 in 240 240 Therefore use a 5 1/2" x 18" GLB for the header. OK Job No.: 18287 - Rev. 5 DVD7 Date: 3/5/2021 Title: Ken Huffman DVDT Engineering, Inc. Loc.: 10620 2351h Pl. S.W., Edmonds Structural Engineering Page: 6 of 14 R03 - 5 1/2" x 10 1/211 GLB Roof Beam b = Width = 5.5 in; d = Depth = 10.5 in lb = Bearing Length (assumed) = 3.5 in = 0.292 ft ws=Self Weight =35x12xbx12xd=35x12 x5.5x12 x10.5=14.0lb/ft L,p. = Length of span = 10.5 ft Lft = Floor Tributary area = 0 ft Lat = Roof Tributary area = Z x 12 + 4 = 10 ft hW = Wall Height = 0 ft WD = Dead Load = ws+LftxFDL +LatxRDLXDF1 +hWxWDL = 14.O+Ox13 +10x15x1.01 +Ox12 = 166 plf WL = Live Load = LftxFLL +LatxRsL= Ox40 +10x25 = 250 plf W = Total Load = WD + WL = 166 + 250 = 416 plf Lft = Floor Tributary area = z x 1.33 = 0.665 ft Lat = Roof Tributary area = Z (2 + 4) = 3 ft hW = Wall Height = 3 ft WD = Dead Load = ws+LftxFDL +LatxRDLxDF1 +hWxWDL = 14.0+0.665x13 +3x15x1.01 +3x12 = 104 plf WL = Live Load = LftxFLL +LatxRsL= 0.665x4O +3x25 = 102 plf W = Total Load = WD + WL = 104 + 102 = 206 plf P = From R02 = 4,6901bs 4.69 Forces 0.416 0.206 loading 14.6 10' 3.59 4.52 bending 3.59 0.889 deflection shear -0.24" Allowable Forces - 24F-V4 DF -4.52 SX=1bd2=Q (5.5)(10.5)2 = 101 in3 IX = 12 b d3 = (11) (5.5) (10.5)3 = 531 in4 A = b d = (5.5)(10.5) = 57.75 in2 E = 1800000 psi; CD = 1.15 21 0.1 12l ° 1 s.12s 0 1 21 l0.1 12 0.1 5.125l 0.1 Cv = NDS 5.3.6 = min 1.0; (L ) (-I ( ) = min (1.0; ( (-) ( I ) = 1.0 span 5 10.5 5.5 Fcpe,p = Design compression stress perpendicular to grain = 650 psi Rmi=FcPe�pxbxlbxl2 = 65Ox5.5xO.292x12 = 12.5271bs F, = Design shear stress = 265 psi VAu = Fv CDxA = (265)(1.15)x57.75 = 17,5991bs Fb = Design bending stress = 2400 psi Man = FbCDCv (iz) = (2400) (1.15) (1. 0) (01 = 23.244 ft-lb12) Deflection Allowable Total Deflection = L,,panx12 _ 10.5x12= 0.525 in OK 240 240 Replace the existing beam with a 5 1/2" x 10 1/2" GLB for the beam. Job No.: 18287 - Rev. 5 DVDT Date: 3/5/2021 Title: Ken Huffinan Loc.: 10620 235th Pl. S.W., Edmonds DVDT Engineering, Inc. Structural Engineering R04 - 3 1/2" x 10 1/2" GLB Ceiling Joist Beam b = Width = 3.5 in d = Depth = 10.5 in lb = Bearing Length (assumed) = 1.5 in = 0.125 ft ws=Self Weight =35x12xbx12xd=35x z x3.5x Z x10.5=8.93lb/ft Lsp. = Length of span = 12.5 ft Page: 7 of 14 Lft = Floor Tributary area = 0 ft Lat = Roof Tributary area = Z x 24 = 12 ft hW = Wall Height = 0 ft WD = Dead Load = ws+LftxFDL +LatxRDLxDF1 +hwxWDL = 8.93+Ox13 +12x15x1.01 +Ox12 = 191 plf WL = Live Load = LftxFLL +LatxRsL= Ox40 +12x25 = 300 plf W = Total Load = WD + WL = 191 + 300 = 491 plf Forces VMax = z x W Lspan = z (491) (12.5) = 3,067 lbs Mm,,x = $ X W X L2pan = $ X 491 x 12.52 = 9.585 ft•lbs Allowable Forces - 24F-V4 DF SX = e b d2 (10.5)2 = 64.3 in3 IX = z b d3 = (2) (3.5) (10.5)3 = 338 in4 A = b d = (3.5)(10.5) = 36.75 in2 E = 1800000 psi; CD = 1.15 21 0.1 12 0.1 5.125 01 21 0.1 12 0.1 5.125 0.1 Cv = NDS 5.3.6 =min 1.0; ( ) (12) ( ) =min (1.0; (-) (-) ( ) ) = 1.0 L,span d b 12.5 10.5 3.5 F&,,p = Design compression stress perpendicular to grain = 650 psi RAu = &Perpxhxlbx12 = 65Ox3.5xO.125x12 = 3.4131bs F, = Design shear stress = 265 psi VAn = Fv CDxA = (265)(1.15)x36.75 = 11,2001bs Fb = Design bending stress = 2400 psi MAn = FbCDCV (12) = (2400) (1.15) (1.0) (66 .3) = 14.792 ft•lbs Deflection SXWLx12x(LspanX12)4 _ Sx300x12 (12.Sx12)4 AL - 0.271 In 384EIx (384)(1800000)(338) LSpanxl2 12.5x12 Allowable Live Load Deflection = 360 360 = 0.417 in OK SxWx12x(Lspanxl2)4 _ Sx491x12 (12.Sx12)4 QN/aX = 384 E Ix (384)(1800000)(338) - 0.443 In Ls anx12 _ 12.Sx12 Allowable Total Deflection = p240 240 = 0.625 in OK Therefore use a 3 1/2" x 101/2" GLB for the beam. Job No.: 18287 - Rev. 5 DVDT Date: 3/5/2021 Title: Ken Huffinan Loc.: 10620 235th Pl. S.W., Edmonds DVDT Engineering, Inc. Structural Engineering R05 - 4x10 #2 DF Living Room Slider Door Header b = Width = 3.5 in d = Depth = 9.25 in lb = Bearing Length (assumed) = 1.5 in = 0.125 ft ws=Self Weight =35x12xbx12xd =35x12 x3.5x12 x9.25=7.87lb/ft Lsp. = Length of span = 8.5 ft Lft = Floor Tributary area = 0 ft Let = Roof Tributary area = z x 12 + 1.5 = 7.5 ft hW = Wall Height = 0 ft Page: 8 of 14 WD = Dead Load = ws+LftxFDL +LRtxRDLxDF1 +hwxWDL = 7.87+Ox13 +7.5x15x1.01 +Ox12 = 121 plf WL = Live Load = LftxFLL +LRtxRsL= Ox40 +7.5x25 = 187.5 plf W = Total Load = WD + WL = 121 + 187.5 = 309 plf Forces Vmw, = a x W Lspan = z (309) (8.5) = 1.313 lbs MMax = 8 x W x Lspan = 8 x 309 x 8.52 = 2,791 ft•lbs Allowable Forces - #2 DF SX = e b d2 = (6) (3.5) (9.25)2 = 49.9 in3 IX = 2 b d3 = (Z) (3.5) (9.25)3 = 231 in4 A = b d = (3.5)(9.25) = 32.4 in2 E = 1600000 psi CD = Duration = 1.0 CF = NDS Table 4A = 1.3 F&,,p = Design compression stress perpendicular to grain = 625 psi RAu = FcPerpxbxlbxl2 = 625x3.5xO.125x12 = 3,2811bs F, = Design shear stress = 180 psi VAu = F, CDxA = (180) (1) x32.4 = 5,8281bs Fb = Design bending stress = 900 psi MAu = FbCDCF (12) = (900) (1) (1.3) (49.9 4 29) = 4.866 ft•lbs Deflection AL _ 5xWLx12x(LspanX12)' _ Sx187.5x12 (8.Sx12)4 = 0.0596 in 384EIx (384)(1600000)(231) anx12 8.5x12 _ Allowable Live Load Deflection = Ls p36o 360 = 0.283 in OK SxWx12x(Lspanxl2)' SX309X12 (8.Sx12)4 QMax = _ = 0.0983 in 384EIx (384)(1600000)(231) anx12 _ 8.5x12 Allowable Total Deflection = Ls p- 240 = 0.425 in OK 240 Therefore use a 4x8 #2 DF for the typical headers. Job No.: 18287 - Rev. 5 D v D % Title: Ken Huffman DVDT Engineering, Inc. Loc.: 10620 2351h Pl. S.W., Edmonds Structural Engineering R06 - 2x8 #2 DF Rafters-12'-0" Span - 16" ox. b = Width = 1.5 in d = Depth = 7.25 in lb = Bearing Length (assumed) = 1.5 in = 0.125 ft ws=Self Weight =35x12xbx12xd =35x12 x1.5x12 x7.25=2.64lb/ft Lsp. = Length of span = 12 ft Lft = Floor Tributary area = 0 ft Lac = Roof Tributary area = 1.33 ft hW = Wall Height = 0 ft WD = Dead Load = LftxFDL +LRtxRDLxDFI +hWxWDL = Ox13 +1.33x15x1.01 +0x12 = 20.1 plf WL = Live Load = LftxFLL +LRtxRsL= Ox4O +1.33x25 = 33.25 plf W = Total Load = WD + WL = 20.1 + 33.25 = 53.4 plf Forces VMax = a X W Lspan = z (53.4) (12) = 3201bs MMax = $ X W X L2pan = a X 53.4 X 122 = 961 ft-lbs Allowable Forces - #2 DF S. = e b d2 = (e) (1.5) (7.25)2 = 13.1 in3 Ix =12 b d3 = \lz) (1.5) (7.25)3 = 47.6 in4 A = b d = (1.5)(7.25) = 10.9 in2 E = 1600000 psi CD = Duration = 1.15 CF = NDS Table 4A = 1.2 CR = Repetitive = 1.15 Fcpe,p = Design compression stress perpendicular to grain = 625 psi RAu=FcPerpxbxlbxl2 = 625x1.5xO.125x12 = 1.4061bs Fv = Design shear stress = 180 psi VAu = Fv CDxA = (180)(1.15)x10.9 = 2.251 lbs Fb = Design bending stress = 900 psi MAu = FbCDCFCR (iz) = (900) (1.15) (1.2) (1.15) (1iz) = 1.559 ft•lbs Deflection AL _ SxWtxlyx(Lspanxl2)4 _ 5x33.25x12 (12xi2)4 _ 0.204 in 384EIx (384)(1600000)(47.6) a,XAllowable Live Load Deflection = Lsp 360 60 _ 1360 = 0.4 in SxWx12x(LspanX12)4 _ 5x53.4x12 0.327 In (12x12)4 Q_ MaX = 384 E Ix (384)(1600000)(47.6) Allowable Total Deflection = Lspanx12 _ 12x12 = 0.6 in 240 240 Therefore use 2x8 #2 DF @ 16" for the rafters. Date: 3/5/2021 Page: 9 of 14 OK OK Job No.: 18287 — Rev. 5 DVDT Date: 3/5/2021 Title: Ken Huffman DVDT Engineering, Inc. Loc.: 10620 2351h Pl. S.W., Edmonds Structural Engineering Page: 10 of 14 R07 - 2x6 #2 SPF Rafters - 12'-0" Span - 12" ox. b = Width = 1.5 in d = Depth = 5.5 in lb = Bearing Length (assumed) = 1.5 in = 0.125 ft ws=Self Weight =35x12xbx12xd =35x12 x1.5x12 x5.5=2.01lb/ft Lsp. = Length of span = 12 ft Lft = Floor Tributary area = 0 ft Lac = Roof Tributary area = 1.0 ft hW = Wall Height = 0 ft WD = Dead Load = LftxFDL +LRtxRDLxDF1 +hWxWDL = Ox13 +1x15x1.01 +Ox12 = 15.15 Of WL = Live Load = LftxFLL +LRtxRsL= 0x40 + 1 x 2 5 = 25 plf W = Total Load = WD + WL = 15.15 + 25 = 40.15 plf Forces VMax = a x W Lspan = z (40.15) (12) = 241 lbs MMax = $ x W x L2pan = a x 40.15 x 122 = 723 ft-lbs Allowable Forces - #2 SPF S. =1 b d2 = Q (1.5) (5.5)2 = 7.56 in3 Ix =12 b d3 = (12) (1.5) (5.5)3 = 20.8 in4 A = b d = (1.5)(5.5) = 8.25 in2 E = 1600000 psi CD = Duration = 1.15 CF = NDS Table 4A = 1.3 CR = Repetitive = 1.15 Fcpe,p = Design compression stress perpendicular to grain = 335 psi RAu = FcPe,pxbxlbxl2 = 335x1.5xO.125x12 = 7541bs Fv = Design shear stress = 135 psi VAu = Fv CDxA = (135)(1.15)x8.25 = 1,281 lbs Fb = Design bending stress = 775 psi MAu = FbCDCFCR (izl = (775) (1.15) (1.3) (1.15) ( ") = 840 ft•lbs12 Deflection AL _ SxWLxlyx(Lspanxl2)4 _ 5x2sx12 (12x12)4 = 0.350 in 384EIx (384)(1600000)(20.8) a,XAllowable Live Load Deflection = Lsp 360 60 _ 1360 = 0.4 in OK sxwxl2x(LspanX12)4 _ sx40.1sx12 (12x12)4 = 0.563 in QMaX =. 384 E Ix (384)(1600000)(20.8) Allowable Total Deflection = Lspanx12 = 12x12 = 0.6 in OK 240 240 Therefore use 2x6 #2 SPF @ 12" for the rafters. Job No.: 18287 - Rev. 5 DVDT Date: 3/5/2021 Title: Ken Huffinan Loc.: 10620 235th Pl. S.W., Edmonds 4x8 #2 DF Typical Headers DVDT Engineering, Inc. Structural Engineering b = Width = 3.5 in d = Depth = 7.25 in lb = Bearing Length (assumed) = 1.5 in = 0.125 It ws=Self Weight=35x12xbx12xd =35xl Lsp. = Length of span = 6.5 ft Lft = Floor Tributary area = 0 ft Let = Roof Tributary area = z x 12 + 1.5 = 7.5 ft hW = Wall Height = 0 ft x 3.5 x 1 x 7.25 = 6.17 lb/ft Page: 11 of 14 WD = Dead Load = ws+LftxFDL +LRtxRDLxDF1 +hwxWDL = 6.17+Ox13 +7.5x15x1.01 +Ox12 = 120 plf WL = Live Load = LftxFLL +LRtxRsL= Ox4O +7.5x25 = 187.5 plf W = Total Load = WD + WL = 120 + 187.5 = 307 plf Forces Vmw, = a x W Lspan = z (307) (6.5) = 9981bs MMax = 8 x W x Lspan = 8 x 307 x 6.52 = 1,621 ft•lbs Allowable Forces - #2 DF SX = e b d2 = (6) (3.5) (7.25)2 = 30.7 in3 IX = Z b d3 = (Z) (3.5) (7.25)3 = 111 in4 A = b d = (3.5)(7.25) = 25.4 in2 E = 1600000 psi CD = Duration = 1.0 CF = NDS Table 4A = 1.3 F&,,p = Design compression stress perpendicular to grain = 625 psi RAu = FcPerpxbxlbxl2 = 625x3.5xO.125x12 = 3,2811bs F, = Design shear stress = 180 psi VAu = F, CDxA = (180)(1)x25.4 = 4.5681bs Fb = Design bending stress = 900 psi MAu = FbCDCF (12) = (900) (1) (1.3) (3 1) = 2.989 ft-lbs Deflection 5xWLxi2x(LspanX12)' _ Sx187.5x12 (6.SX12)4 AL 0.04241n 384EIx (384)(1600000)(111) - Allowable Live Load Deflection = LspanX12 = 6.5x12 = 0.217 in OK 360 360 SXWxlZx(LspanX12)4 Sx307x12 (6.Sx12)4 QMax = _ = 0.0694 in 384EIx (384)(1600000)(111) anx12 _ 6.Sx12 Allowable Total Deflection = Ls p- 240 = 0.325 in OK 240 Therefore use a 4x8 #2 DF for the typical headers. Job No.: 18287 — Rev. 5 DVD7 Date: 3/5/2021 Title: Ken Huffman DVDT Engineering, Inc. Loc.: 10620 235th Pl. S.W., Edmonds Structural Engineering BO1 - 5 1/4" x 9 1/4" PSL Floor Beam b = Width = 5.25 in d = Depth = 9.25 in lb = Bearing Length (assumed) = 1.5 in = 0.125 ft ws=Self Weight =45x12xbx12xd=45x12 x5.25x12 x9.25=15.2lb/ft Lsp. = Length of span = 9.5 ft Lft = Floor Tributary area = z x 24 = 12 ft Let = Roof Tributary area = 0 ft hW = Wall Height = 0 ft Page: 12 of 14 WD = Dead Load = ws+LftxFDL +LRtxRDLxDF1 +hwxWDL = 15.2+12x13 +Ox15x1.01 +Ox12 = 171 plf WL = Live Load = LftxFLL +LRtxRsL= 12x40 +Ox25 = 480 plf W = Total Load = WD + WL = 171 + 480 = 651 plf P = From B06 = Z970 lbs Forces 2.97 0.651 loading 10.2 8.65 9.5' 3.64 5.52 bending 3.64 deflection shear -0.231" -5.52 Allowable Forces - 2.0E Parallam from Trus Joist Tables S,, = e b d2 = (6) (5.25) (9.25)2 = 74.9 in3 IX = Z b d3 = (Z) (5.25) (9.25)3 = 346 in4 A = b d = (5.25)(9.25) = 48.6 in2 E = 1600000 psi CD = Duration =(1.15 CF= (a)o.11l _ \9zs)o.111 _ 1.03 F&,,p = Design compression stress perpendicular to grain = 750 psi RAu = &Perpxbxlbx12 = 750x5.25xO.125x12 = 5.9061bs F, = Design shear stress = 190 psi VAu = F, CDxA = (190)(1.15)x48.6 = 10,6111bs Fb = Design bending stress = 2900 psi MAu = FbCDCF (12) = (2900) (1.15) (1.03) \ 129) = 21.417 ft•lbs Deflection Allowable Total Deflection = L,panx12 _ 13x12 = 0.65 in 240 240 Therefore a 5 1/4" x 9 1/4" parallam is OK for this beam. OK Job No.: 18287 — Rev. 5 DVDT Date: 3/5/2021 Title: Ken Huffman DVDT Engineering, Inc. Loc.: 10620 235th Pl. S.W., Edmonds Structural Engineering B02 - (2) WO #2 DF Floor Beam b = Width = 3 in d = Depth = 9.25 in lb = Bearing Length (assumed) = 1.5 in = 0.125 ft ws = Self Weight = 35 x 1 x b x 1 x d= 35 x z x 3 x lZ x 9.25 = 6.74 lb/ft Lsp. = Length of span = 4 ft Page: 13 of 14 Lft = Floor Tributary area = z x (9.25 + 14.75) =12 ft Let = Roof Tributary area = 0 ft hW = Wall Height = 0 ft WD = Dead Load = ws+LftxFDL +LRtxRDLxDF1 +hwxWDL = 6.74+12x13 +Ox15x1.01 +Ox12 = 163 plf WL = Live Load = LftxFLL +LRtxRsL= 12x40 +Ox25 = 480 plf W = Total Load = WD + WL = 163 + 480 = 643 plf Forces VMax = a x W Lspan = a (643) (4) = 1.285 lbs MMax = $ x W x Lspan = e x 643 x 42 = 1.285 ft-lbs Allowable Forces - #2 DF Sx = e b d2 = (6) (3) (9.25)2 = 42.8 in3 Ix = Z b d3 = (Z) (3) (9.25)3 = 198 in4 A = b d = (3)(9.25) = 27.75 in2 E = 1600000 psi CD = Duration = 1.0 CF = NDS Table 4A = 1.1 F&,,p = Design compression stress perpendicular to grain = 625 psi RAu = FcPerpxbxlbxl2 = 625x3xO.125x12 = 2,8131bs F, = Design shear stress = 180 psi VAu = F, CDxA = (180)(1)x27.75 = 4.9951bs Fb = Design bending stress = 900 psi MAu = FbCDCF (12) _ (900) (1) (1.1) (12) = 3,531 ftlbs Deflection SxWLx12x(LspanX12)4 Sx480x i (4x12)4 AL = = 12 = 0.00873 in 384EIx (384)(1600000)(198) anx12 4x12 Allowable Live Load Deflection = Ls p360 360 = 0.133 in Sxwx 1-x(LspanX12)' Sx643x1z (4x12)4 QMQX = - 0.0117 In 384 E Ix (384)(1600000)(198) anx12 _ 4x12 Allowable Total Deflection = Ls p240 240 = 0.2 in Therefore use (2) 2x10 #2 DF for the beam with LUS210-2 hanger. Sister the beam with (2) rows 16d (0.128"0") nails @ 12" o.c. OK M Job No.: 18287 — Rev. 5 DVD7 Date: 3/5/2021 Title: Ken Huffman DVDT Engineering, Inc. Loc.: 10620 235th Pl. S.W., Edmonds Structural Engineering Page: 14 of 14 B03 - (3) WO #2 DF Floor Beam b = Width = 4.5 in d = Depth = 9.25 in lb = Bearing Length (assumed) = 1.5 in = 0.125 It ws=Self Weight=35x12xbx12xd=35x Z x4.5x12 x9.25=10.11b/ft Lsp. = Length of span = 6 ft Lfc = Floor Tributary area = z x 1.33 = 0.665 ft Lac = Roof Tributary area = Z (24.67 + 12) + 2 = 20.3 ft hW = Wall Height = 8 ft WD = Dead Load = ws+LftxFDL +LatxRDLxDF1 +h,vxWDL = 10.1+0.665x13 +20.3x15x1.01 +8x12 = 423 plf WL = Live Load = LftxFLL +LacxRsL= 0.665x40 +20.3x25 = 535 plf W = Total Load = WD + WL = 423 + 535 = 958 plf P = From B02 = 1,2851bs Forces 1.29 0.958 loading 4.64 6' 1.91 4.05 2.98 bending 4.05 FL�� deflection shear -0.064 " 2.98 Allowable Forces - #2 DF S,, = e b d2 = (6) (4.5) (9.25)2 = 64.2 in3 L = iz b d3 = \TO (4.5) (9.25)3 = 297 in4 A = b d = (4.5)(9.25) = 41.6 in2 E = 1600000 psi CD = Duration = 1.15 CF = NDS Table 4A = 1.1 Fcpe,p = Design compression stress perpendicular to grain = 625 psi RAu = Fcpe�pxbxlbxl2 = 625x4.5xO.125x12 = 4.2191bs F, = Design shear stress = 180 psi VAu = Fv CDXA = (180)(1.15)x41.6 = 8,616 lbs Fb = Design bending stress = 900 psi MAII = FbCpCp (12) = (900) (1.15) (1.1) (6122) = 6.088 ft-lb Deflection Allowable Total Deflection = LS p240anx12 = 6x12 240 = 0.3 in OK Therefore use (3) 2x10 #2 DF for the beam. Sister the beam with (3) rows 16d (0.128"0") nails @ 12" o.c. each side. Job No.: 18287 - Rev. 5 DVDT Date: 3/5/2021 Title: Ken Huffman DVDT Engineering, Inc. Loc.: 10620 235th Pl. S.W., Edmonds Structural Engineering B04 - 2x6 P.T. #2 HF Deck Joists b = Width = 1.5 in d = Depth = 5.5 in lb = Bearing Length (assumed) = 1.5 in = 0.125 It ws=Self Weight=35x12xbx12xd =35x12 x1.5xl x5.5=2.011b/ft Lsp. = Length of span = 5.25 ft Lft = Floor Tributary area = 1.33 ft WD = Dead Load = LftxFDLD = 1.33x9 = 12.0 plf WL = Live Load = LftxFLLD = 1.33x60 = 79.8 plf W = Total Load = WD + WL = 12.0 + 79.8 = 91.8 plf Forces VMax = a x W Lspan = z (91.8) (5.25) = 241 lbs MMax=8xWxLspan =$ x91.8x5.252=316ft•lbs Allowable Forces - #2 HF Sx=1bd2=Q (1.5)(5.5)2=7.56in3 Ix = Z b d3 = (Z) (1.5) (5.5)3 = 20.8 in4 A = b d = (1.5)(5.5) = 8.25 in2 E = 1300000 psi CD = Duration = 1.0 CF = NDS Table 4A = 1.3 C, = Repetitive = 1.15 Fepe,p = Design compression stress perpendicular to grain = 405 psi Ran = F�pe pxbxlbxl2 = 405xl.5xO.125xl2 = 9111bs Fv = Design shear stress = 150 psi Van = F, CDxA = (150)(1)x8.25 = 1,2381bs Fb = Design bending stress = 850 psi Ma(( = FbCDCF C,. (12) _ (850) (1) (1.3) (1.15) \ iz61 = 801 ft-lbs Deflection SxWLX 1 x(Lspanxl2)4 AL = 384 E Ix Sx79.8x' (5.25x12)4 (384)(1300000)(20.8) - 0.0504 In Lspa„x12 5.25x12 Allowable Live Load Deflection = 360 - 360 = 0.175 in )4-sx91.8xSXWX „12 Amax = 1-2p(5.25x12)4 = 0.0580 in 384 E Ix (384)(1300000)(20.8) a„x12 _ s.25x12 Allowable Total Deflection = Lsp= 0.2625 in 240 240 Therefore use 2x6 P.T. #2 HF deck joists @ 16" o.c. Deck Ledger Connection V = Shear = VMaX - = (241) (16 12 = 181 lbs/ft Zs = 340 lbs per 1/2"x5" Simpson Titan HD screw ss = Screw Spacing = 1.33 ft ZS'= s 1 x Zs CD = 1 33 (1) (340) (1) = 256 lbs/ft s Therefore: Use (1) 1/2"x5" Simpson Titan HD screw @ 16" o.c. Page: 15 of 14 OK OK Job No.: 18287 - Rev. 5 DVD7 Date: 3/5/2021 Title: Ken Huffman DVDT Engineering, Inc. Loc.: 10620 235th Pl. S.W., Edmonds Structural Engineering B05 - 4x6 P.T. #2 HF Deck Beam b = Width = 3.5 in d = Depth = 5.5 in lb = Bearing Length (assumed) = 1.5 in = 0.125 ft ws = Self Weight = 35 x Z x b x Z x d = (35) (Z) (3.5) (Z) (9.25) = 7.87 lb/ft Lsp. = Length of span = 8 ft Lc = Length of cantilever = 2 ft Lft = Floor Tributary area = Z x 5.25 + 1 = 3.625 ft Lat = Roof Tributary area = 0 ft hW = Wall Height = 0 ft Page: 16 of 14 WD = Dead Load = ws+LftxFDLD +LatxRDLxDF1 +hwxWDL = 7.87+3.625x9 +Ox15x1.01 +Ox12 = 40.5 plf WL = Live Load = LftxFLLD +LatxRsL = 3.625x60 +Ox25 = 217.5 plf W = Total Load = WD + WL = 40.5 + 217.5 = 258 plf Forces 0.258 loading 2' 8' 2' 1.55 1.55 1.03 0.516 shear -0.516 1.03 Allowable Stresses - #2 HF S,, = 6 b d2 = (6) (3.5) (5.5)2 = 17.6 in3 IX = Z b d3 = (Z) (3.5) (5.5)3 = 48.5 in4 A = b d = (3.5)(5.5) = 19.25 in2 E = 1300000 psi CD = Duration = 1.0 CF = NDS Table 4A = 1.3 1.55 bending 0.516 -0.516 0.174" deflection 0.264" Fcpe,p = Design compression stress perpendicular to grain = 405 psi RAtt = &Pe,Pxbx1bx12 = 405x3.5xO.125xl2 = 2.1261bs Fv = Design shear stress = 150 psi VAu = F, CDxA = (150)(1)x19.25 = 2,8881bs Fb = Design bending stress = 850 psi MAn = FbCDCF `i2) = (850) (1) (1. 3) (12) = 1.625 ft-lb Deflection Allowable Total Deflection = LS P- 2anx12 _ 8x12 40 = 0.4 in OK 240 Allowable Cantilever Deflection = 24zx240 = zx2x 240 - = 0.2 in OK Therefore use a 4x6 P.T. #2 HF for this beam. Job No.: 18287 - Rev. 5 DVD7 Date: 3/5/2021 Title: Ken Huffman DVDT Engineering, Inc. Loc.: 10620 235th Pl. S.W., Edmonds Structural Engineering B06 - 5 1/4" x 9 1/4" PSL Floor Beam b = Width = 5.5 in d = Depth = 9.25 in lb = Bearing Length (assumed) = 1.5 in = 0.125 ft ws=Self Weight =45x12xbx12xd=45x12 x5.5x12 x9.25=15.9lb/ft Lsp. = Length of span = 14.75 ft Lfc = Floor Tributary area = 1.33 ft Lac = Roof Tributary area = 0 ft hW = Wall Height = 0 ft Page: 17 of 14 WD = Dead Load = ws+LftxFDL +LRcxRDLxDF1 +hwxWDL = 15.9+1.33x13 +Ox15x1.01 +Ox12 = 33.2 plf WL = Live Load = LftxFLL +LRcxRsL= 1.33x40 +Ox25 = 53.2 plf W = Total Load = WD + WL = 33.2 + 53.2 = 86.4 plf P = From R04 = 3,0671bs Forces 3.07 0.086 9.88 loading 14.75' 2.97 1.36 bending 2.97 deflection shear-0.451" -0.395 -1.36 Allowable Forces - 2.0E Parallam from Trus Joist Tables S. = b b d2 = (6) (5.5) (9.25)2 = 78.4 in3 IX = 12 b d3 = \lz� (5.5) (9.25)3 = 363 in4 A = b d = (5.5)(9.25) = 50.9 in2 E = 1600000 psi CD = Duration = 1.0 12 0.111 12 0.111 CF =min (1.0; ( d ) ) =min (1.0; (9.25) ) = 1.0 F,pe,p = Design compression stress perpendicular to grain = 750 psi RAu = FePe pxbxlbxl2 = 75Ox5.5xO.125x12 = 6.1881bs F, = Design shear stress = 190 psi VAu = F, CDxA = (190)(1)x50.9 = 9.671 lbs Fb = Design bending stress = 2900 psi - (2900) (1) (1) (24) = 18.947 ft lb MAu = FbCDCF (12) s Deflection Allowable Total Deflection = LS p= 0.7375 in 240 240anx12 _ 14.75x12 OK Therefore a 5 1/4" x 9 1/4" parallam is OK for this beam with HGUS5.50/10 hanger. Job No.: 18287 — Rev. 5 DVDT Date: 3/5/2021 Title: Ken Huffman DVDT Engineering, Inc. Loc.: 10620 235th Pl. S.W., Edmonds Structural Engineering DUCTOI - (2) WO #2 DF Floor Beam b = Width = 3 in d = Depth = 9.25 in lb = Bearing Length (assumed) = 1.5 in = 0.125 ft ws = Self Weight = 35 x l x b x l x d= 35 x l x 3 x' x 9.25 = 6.74 lb/ft Lsp. = Length of span = 6 ft Lft = Floor Tributary area = z x 13.5 = 6.75 ft Let = Roof Tributary area = 0 ft hW = Wall Height = 0 ft Page: 18 of 14 WD = Dead Load = ws+LftxFDL +LRtxRDLxDF1 +hwxWDL = 6.74+6.75x13 +Ox15x1.01 +Ox12 = 94.5 plf WL = Live Load = LftxFLL +LRtxRsL= 6.75x40 +Ox25 = 270 plf W = Total Load = WD + WL = 94.5 + 270 = 364 plf Forces VMax = a x W Lspan = z (364) (6) = 1.092 lbs MMax = $ x W x Lspan = $ x 364 x 62 = 1.638 ft-lbs Allowable Forces - #2 DF S,, = e b d2 = (6) (3) (9.25)2 = 42.8 in3 IX = Z b d3 = (Z) (3)(9.2 = 198 in4 A = b d = (3)(9.25) = 27.75 in2 E = 1600000 psi CD = Duration = 1.0 CF = NDS Table 4A = 1.1 F&,,p = Design compression stress perpendicular to grain = 625 psi RAu = FcPerpxbxlbxl2 = 625x3xO.125x12 = 2,8131bs F, = Design shear stress = 180 psi VAu = F, CDxA = (180)(1)x27.75 = 4.995 lbs Fb = Design bending stress = 900 psi MAu = FbCDCF (12) = (900) (1) (1.1) (12) = 3.529 ft-lb nefleetion SxWLX 1 x(LspanX12)' Sx270x 1 (6x12)4 AL = = 12 = 0.0249 in 384EIx (384)(1600000)(198) Allowable Give Load Deflection = Ls p360 360anx12 6x12 = 0.2 in OK SxWx1-x(LspanX12)4 Sx364x1z (6x12)4 QMax = _ = 0.0335 in 384EIx (384)(1600000)(198) Allowable Total Deflection = Ls p240 240anx12 _ 6x12 = 0.3 in OK Therefore use (2) 2x10 #2 DF for the beam with LUS210-2 hanger. Sister the beam with (2) rows 16d (0.128"0") nails @ 12" o.c. Job No.: 18287 - Rev. 5 DVD7 Date: 3/5/2021 Title: Ken Huffman DVDT Engineering, Inc. Loc.: 10620 235th Pl. S.W., Edmonds Structural Engineering DUCT02 - (3) WO #2 DF Floor Beam b = Width = 4.5 in d = Depth = 9.25 in lb = Bearing Length (assumed) = 1.5 in = 0.125 It ws=Self Weight=35x12xbx12xd=35x Z x4.5x12 x9.25=10.11b/ft Lsp. = Length of span = 14.5 ft Lft = Floor Tributary area = 1.33 ft Lat = Roof Tributary area = 0 It hW = Wall Height = 0 ft Page: 19 of 14 WD = Dead Load = ws+LftxFDL +LatxRDLxDF1 +hwxWDL = 10.1+1.33x13 +Ox15x1.01 +Ox12 = 27.4 plf WL = Live Load = LftxFLL +LatxRsL= 1.33x40 +Ox25 = 53.2 plf W = Total Load = WD + WL = 27.4 + 53.2 = 80.6 plf P = From B02 = 1,2851bs Forces 1.29 0.081 loading 3.2 �f T 14.5' 2.52 1.74 0.72 bending 1.74 FL-- deflection shear -0.261" 0.72 Allowable Forces - #2 DF S,, = e b d2 = (6) (4.5) (9.25)2 = 64.2 in3 IX = i2 b d3 = \iz� (4.5) (9.25)3 = 297 in4 A = b d = (4.5)(9.25) = 41.6 in2 E = 1600000 psi CD = Duration = 1.15 CF = NDS Table 4A = 1.1 F,pe,p = Design compression stress perpendicular to grain = 625 psi RAn = &Pe,Pxbxlex12 = 625x4.5x0.125x12 = 4,2191bs F = Design shear stress = 180 psi VAn = F„ CDxA = (180)(1.15)x41.6 = 8,616 lbs Fb = Design bending stress = 900 psi Man = FbCDCF (12) Lx (900) (1.15) (1.1) (61/ = 6.088 ft-Ibs Deflection Allowable Total Deflection = LS P= 240 6x12 = 0.3 in 240 anx12 Therefore use (3) 2x10 #2 DF for the beam with LUS210-3 hanger. Sister the beam with (3) rows 16d (0.128"0") nails @ 12" o.c. each side. OK Job No.: 18287 — Rev. 5 DVDT Date: 3/5/2021 Title: Ken Huffman DVDT Engineering, Inc. Loc.: 10620 235th Pl. S.W., Edmonds Structural Engineering Page: 20 of 14 Foundation Typical Bearing Point Footing P = 6500 lbs b = Pad Width = 1.33 ft L = Pad Length = 3.5 ft a = Soil Bearing Pressure = n LL (1.3 �(3.$) = 1,396 psf Therefore the 16" wide footing is acceptable for loads under 6,500 lbs. F1 - Bearing Point Footing P = From B01 = 5,520 lbs b = Pad Width = 1.75 ft L = Pad Length = 1.75 ft 6 = Soil Bearing Pressure = b (1.75)(1.75) = 1,802 psf Therefore a 21" wide footing footing is acceptable.