Loading...
HomeMy WebLinkAbout1827 W 16th St Technical - Building TECHNICAL ._-,_.~._,',. - ..-,.~_..~,.,-". -\~3 ~ D <=6 / \" ~ \~~ s\ \'1J 1-1 \_ r\.-~ ~YlJAe.. \)Q:\~ ' ,. ~ ~' \ .\ <l~2\-Og \l y\11 ,~C^- ---.- --- -.1 LJANCE ENGINEERIN PROJECT No. 150408 Rev. 2 ~ o ~ 2J Ul \fJ POST FRAME BUILDING STRUCTURAL CALCULATION (This structure has been analyzed and designed for structural adequacy only.) BUILDING OWNER I LOCATION: Dale Seward FILE 1827 W 16th St ~ Port Angeles, WA 98363 S3 <t. CLIENT: , Sound Building Systems, Inc. - t:r 3546 Thorndyke Rd 1- Port Ludlow, WA 98365 RECEIVED ~ JUL 1 4 200B ENGINEER: CITY OF PORT ANGELES 12..:00 BUILDING DIVISION tJ ~ "o/'j ~- ~~.-( , ~XP!P'ES: '/1--:: "~i~lI""'''''''''\_"", ..l , Property of Alliance Engineering ar , authorized duplication prohibited. Copyright @ Alliance Eng. eerin of Oregon, Inc. 2700 Market Street N.E. Alliance Engineering of Oregon, Inc. Phone: (503) 589-1727 Salem, OR 97301 www.aeoregon.com Fax: (503) 589-1728 ,-- I 7/3/2008 150408R2 (Seward) 28x36x18.xmcd 1 POST FRAME BUILDING SUMMARY: This is a post-frame building with wooden trusses or rafters and preservately treated posts that are pressure treated for burial. Post size, post embedment depth. post hot.! di~~ter and backfill is given in the body of the calculation. The building will depend on the diaphragm action of the roof and wall sheathing for lateral stability. The posts will be modeled as propped cantilevers that are fixed at the base and propped by the deep beam action of the roof. The roof structure spans horizontally between the wall diaphragms where it is simply supported. The post frames will be assumed to act as a unit. Wind loads will be imposed on the windward and leeward sides of the building simultaneously. .If 1here.is .no concrete floor, the concrete backfill will provide lateral constraint in the windward and leeward direction. If a concrete floor is used, lateral restraint for the post will be provided at the ground line by the concrete floor. REFERENCES: 1. 2006 Edition of the InternationalBuikiin~ Code 2. ASCE 7..()S - Minimum Design Loads for Buildings and Other Structures American Society of Civil Engineers, 2006 3. 2005 Edition, National Design SpecnlCation (NOS) Supplement For Wood Construction, American Wood Counsel 7/3/2008 150408R2 (Seward) 28x36x18.xmcd 2 DESIGN INPUT VALUES: Building Dimensions Width of Building Wb1dg:= 24 ft 4,ldg := 30 ft Hb1dg:= 18 ft Length of Building Eave Height of Building Overhang:= 18 in Length of Eave Overhang Rpitch := 4 /12 Roof pitch Bay:= 10 ft Greatest nominal spacing between eavewall posts W gableopenings:= 16 ft Total width of openings in one gable wall Weaveopenings:= 0 ft Total width of openings in one eave wall DesiQn Loads for Buildina: Wind DesiQn Values: Fastest wind speed (3 second gust) V wind:= 100 MPH Wind Exposure: ExposlU'C:= "cn Roof Load DesiQn Values: Pg := 25 Pd:= 3 Pd2:= 0 Ibs Ground snow load Ibs Roof dead load Ibs Additional truss bottom chord dead load (if applicable) Seismic Deskin Values: Ss:= 145.6 Mapped spectral acceleration for short period Sl := 60.0 Mapped spectral acceleration for 1 second period In = 1.00 Importance factor Rs := 7 Response modification factor 7/3/2008 150408R2 (Seward) 28X?.A)x18.xmcd 3 DESIGN INPUT VALUES (Continued): Structural Members for BuildinQ: Post Properties: P width:= 6 P depth:= 8 in Post width y-:axis in Post depth x-axis POST SIZE (Solid rough-:sawn Hem-Fir post unless otherwise specified) Grade := "2" Grade of Post ( 2, 1, or SS" Select Structural) PurlinProoerties: P urlin _spacing:= 24 in GirtProoerties: Girt_spacing := 48 in Spurlin:= Sx26 Sgirt;= 2 Sy26 + Sx26 F purlin:= F bDF2dim Fgirt:= FbHF2dim Footinq and Post HoleDesKID Values: qsoil:= 1500 pst Assumed soil vertical bearing capacity Ssoil = 150 pst Assumed soH lateral bearing capacity <!jaJooting := 2 ft Main truss post footing diameter Slab and backfill infonnation Concrete_slab = "Required for post cOllStraint" Concrete backfill = "No" Backfill in main posts (GO TO LAST PAGE FOR SUMMARY OFRESUL TS) 7/3/2008 150408R2 (8evyard) 28x36x18.xmcd 4 SNOW lOAD ANALYSIS: Design per ASeE 7-05 For roof slopes greater than 5 degrees, and less than 70 d~grees. Pg = 25 psf Ground Snow Load (from above) Ce:= 1.0 Exposure factor Ct:= 1.0 Thermal Factor C. = 1.00 Roof slope factor I,. = 1.00 Importance factor Pt= Flat roof snow load, psf (see analysis below) Ps= Sloped roof snow load, psf (see analysis below) 1. Detennine Pf and Ps Pr:= .7-Ce-Ct-I,.-Pg pr= 17.5 psf Flat roof snow load Note: This is NOT the snow Ps = 17.5 psf Sloped (balanced) roof snow load load Ut Sbedttfor desf ign-See Psu a 0 om 0 page. 2. Detenninethe unbalanced snow load Ps := PrCs Wh1d!l W rid"~:= ----::.. =' J Wridgc = 12 ft Horizontal distance from eave to ridge Note: If Wridge < 20', use Method 1 to ~etermine unbalanced snow load, otherwise use Method 2 Method 1 Psul := I,..Pg Psul = 25 psf Unbalanced snow load for buildings with Wridge < 20' Method 2 The unbalanced snow load will occur from the ridge to a distance Is. and intensity, Psu2 as follows: h.t = 1.56 ft Height of drifted snow y = 17.25 pcf Snow density S = 3 ft Run in roof for a rise of 1 Is = 7.2 ft Distance of unbalanced snow from ridge (if applicable-see below) Psu2 = 33 psf Unbalanced snow load for buildings with Wridge> 20' Final unbalanced snow load Psu = 25 psf Final (roof) snow load used for desiQn of stmctural members and connections as required per Chapter 7 of ASCE 7-05 Application of snow load to buildina The snow load, Psu' was calculated using Method = I ,therefore the final roof snow load used for desiqn shall be Distributed = "across entire building width" If Method 2 is used. the remainder of roof shall be designed using no less than Ps = 20 pst snow load 7/312008 150408R2 (Seward) 28x36x18.xmcd 5 WIND ANALYSIS: Design per ASCE 7-05 Methoq 2 - Analytical Procedure V wind = 100 Basic Wind Speed k.J := .85 WlOd Directionality Factor kn = 1.0 Topographic Factor kz = 0.902 Wind Exposure Factor Iw = 1.00 Importance factor 2 %:= .00256.kz'kn'k.J. V wind .Iw Velocity Pressure qb = 19.63 pst Calculated Wind Pressures: Wmdward Eave Wall: Leeward Eave Wall: qww:= %.GCpfWw qlw := qh.GCpflw qww = 10.14 pst qlw = -8.15 pst Wmdward Gable Wall: Leeward Gable Wall: qwwg:= qh.GCpfwwg qlwg:= %.GCpflwg qwwg = 7.85 pst qlwg = -5.69 pst Wmdward Roof: Leeward Roof: qwr:= qh'GCpfWr <Ilr:= <Ib.GCpflr %- = -9.20 pst qwr = -13.54 pst Wall Elements: Roof Elements: qwe:= qh.GCpfw qwe = -20.02 pst qr:= ~.GCpfr q, = -29.05 pst Internal Wind Pressure (+/-): qj:= %.GCpj qj = 3.53 pst ~-- 7/3/2008 150408R2 (Seward) 28x"...6x18.xmcd 6 BUILDING MODEL: STEP 1: DETERMINE THE SHEAR STIFFNESS OF THE TE~T PANEL This procedure relies on tests conducted by the National Frame Builders Association. The test was conducted using 29 gauge ribbed steel panels. These ribbed steel panels are similar to Strongpanel,Norclad, and Delta-Rib which are in common use by builders in this area. The material and section properties for the test panels are thus reasonable and will be used throughout. The stiffness of the test panel was calculated to be: c"" 2166 Iblin STEP 2: CAlCULATED ROOF DIAPHRAGM STIFFNESS OF THE TEST PANEL c' = (E X t) I (2 X (1+V) X (gJp) + (K2/ (b' X t)^2)) Where: E = t= v= g/p ;; b'= 27 .5x1 0^6 psi (modulus of elasticity for steel) 0.017" (thickness of 29 gauge steel) 0.3 (Poisson's Ratio for steel) 1 .139 ratio of sheathing corrugation length to corrugation pitch 144" (12'-0" length attest pan el) STEP 2.1 This equation was set equal to the stiffness of the test panel (2166 Iblin) and the unknown value (K2) was solved for. K2 = 1215in4 sheet edge purlin fastening constant STEP 2.2: Use new building width to determine stiffness of new roof diaphragm ( Ch bn.:w:= Wb1dg.12 2 cos((':') ) K.,;= 1175 Ibf I ft t:= 0.017 in E> = 18.435 deg (Angle of roof pitch from horizontal) bnew = 152 in E := 27500000 E.! .c:= K"} C = 2404 Ibf / in 2.961 + ., (b""".tt STEP 2.3 & 2.4: Calculate the equivalent horizontal roof stiffness ( cJJ for the full roof: Since Ch is for the full roof, the roof length must be ratioed by the aspect ratio of the roof panel (b / a) where "aN is the truss spacing in inches. a := Bay" 12 ( )2 bn.:" cll := 2.c.cos (~) .- n a = 120 in Ch;;= 5474 lbf / in 71312008 150408R2 (8eward) 28JC>v6x18.xmcd 7 STEP 3: DETERMINE THE STIFFNESS OF THE POST FRAME (k): Since the connection between the posts and the rafters can be assumed to be a pinned joint, the model for the post frame can be assumed to be the sum of two cantilevers (the posts) that act in parallel. The stiffness of the post frame can be calculated from the amount of force required to deflect the system one inch. The spring constant (k) in pounds per inch of deflection results directly. k = 72 Ibffm STEP 4: DETERMINE THE TOTAL SIDE SWAY FORCE (R): Apply wind loads to the walls to determine the moment, fiber stress and end reaction at prop point R. Calculate Total Wind Pressure: qc:= if(qww - qlw ~ 10, lO,qww - ~w) qc = 18.29 psf '( '\ a q\\\\vosl:= q". 12.12) qwwposl = 15.24 pH .., Lposl_hndg .. MWind := q""wp,,sr _ X Mwind = 79289 in-Ibf Mwind t~'-ind := Sxea"epost fwind = 619 psi . LposI_lwdg R := .' .tlm'lJOsl" 8 R = 1166 Ibs STEP 5: DETERMINE THE RATIO OF THE FRAME STIFFNESS TO THE ROOF STIFFNESS: This ratio (kJ ctJ will be used to determine the side sway force modifiers. k - = 0.013 ch STEP 6: DETERMINE SIDE SWAY RESISTANCE FORCE: mD = 0.99 STEP 7: DETERMINE THE ROOF DIAPHRAGM SIDE SWAY RESISTANCE FORCE: Q:= mD.R Q = 1151 Ibf Since not all of the total side sway force (R) is resisted by the roof diaphragm, some translation will occur at the top of the post. The distributed load that is not resisted by the roof diaphragm will apply additional moment and fiber stress to the post. Mdfl = 4193 in-lbf fdfl = 33 psi Calculate the total moment and the total fiber stress in the post Mtat := mD. Mwind + ~ Mtot = 82434 in-Ibf ftot := mD.fwind + fdfl ftot = 644 psi 7/3/2008 150408R2 (Seward) 28x36x18.xmcd 8 MAIN POST DESIGN: Calculate allowable unit compression stress, Fcc' Fcl = 575 psi F c := F cl' 1.15 Fc = 661 psi Allowable compression stress including load factors ~_Dndg = 204 in Bending length of post ~ = 8 in Minimum unbraced dimension of post K.: := 0.8 c:= 0.8 Ewood = 400000 psi Ie:= K.:.I1IosUllldg Ie = i63.2in .882. Ewo<m FeE := (~r FcE = 848 CalculateCohlmn Stability FactQr. ep: ~( FeE J ( FeE J:2 . ..F'cE.'.l 1+- 1+- - . '_ Fe Fe 2 Cp ._ . . . ... - ..... ..0 .... ,-.. 2.c 2.c c Cp = 0.77 Fcc := Fc'Cp Fcc = 509 psi Allowable compression stress on the post Wroof = 28 psf Total roof loading Psnowpost = 3375 fbs Axial loading per post due to roof snowload Pdeadpost = 405 Ibs Axial loading per post due to roof dead load Fb := I-b}" 1.6 Fb = 920 psi Allowable bending stress per post including load factors 7/3/2008 150408R2 (Seward) 28x"..6x18.xmcd 9 Check Load Cases: Load Case 1: Dead Load + .75 * Wind Load +.75 * Snow Load fbI := .75ftot fbI = 483 psi Actual bending stress on post . '_ .75 P snoul"lSl + P d""dl")sl f.:.- 1\ost f~=61 psi Actual compression stress per post CCF ALl I := l(~)2 + F.:.: f"J ( t~) F". I - - \. F.:E J 1 CCFALI1 = 0.58 Load Case 2: Dead Load -+ Wind Load fbI := ftot fl.} = 644 psi (Actual bending stress on post) . P d"..dl"lSt t.: := Apost fe= 8 psi (Actual compression stress per post) r.., j CCFAU2= (:J + ('b' () Ft>. 1-- F.:E CCFALI2 = 0.71 Load Case 3: Dead Load + Snow Load fbI:= 0 fbI = 0 psi (Actual bending stress on post) , P snowpost + P deadpost fe := Apoo f~ = 79 psi (Actual compression stress per post) l 7 j f - f CCFALI3= (F:) +, ( bI ( I FI,I-- F.:E) CCFALI3 = 0.02 CCFALI = 0.71 Less than or equal to 1.00 thus OK 7/3/2008 150408R2 (Seward) 28x36x18.xmcd 10 SEISMIC CALCULATIONS: Design per A8CE 7.:.05 Ss = 145.6 Mapped spectral acceleration for short periods (from above) SI = 60 Mapped spectral acceleration for 1-second period (from above) IE = 1.0 Importance factor W = Dead load of building Rs = 7 Response modification factor (from above) 1. Determine the Seismic Design Category a. Calculate Sos and SDl For 8DS: For SD1: For Ss = 1.46 For SI = 0.60 Fa == 1.00 Fv == 1.50 SMS:== Ss.Fa S~H := SrFv SMS '" 1.46 Sr.u '" 0.9 (")\ Sps:== i )"S!\IS SDJ := (f }SHI Sns = 0.97 Sm = 0.60 Seismic _ Design_Category = ''0'' 2. Determinetlte building parameters Building dead load weight, W: W:== [(Wb1dg'4Idg),(pr.2)J + [(Wbldg'4Idg) + [2.(Wb1dg + 4Idg)' H;dg]'Pd W = 5076 Ibf Building area, Ab: At, ;= Lbldg' Wb1dg At, == 720 ft2 7/312008 150408R2 (Seward) 28x36x18.xmcd 11 3. Determine the shear force to be applied a. Determine the structural period, T Ta:= .02. (Hb1dg+ H.oof).75 T:= Ta b. Detemine the Seismic Response Coefficient, Cs: Cs is calculated as: '. ,_ SDS C." ,- , '- Rs T = 0.20 Cs2 = 0.139 IE But shall not be less than: C.l := .044.Sos.IE Csl = 0.043 But need not exceed: Cs3 := SDl Cs3 = 0.422 ('RS)' T- IE Cs::=' 0.139 c. Detemine the Seismic Base Shear: Vhase_shear:= Cs.W V base_shear = 704 Ibf 4. Detennine the seismic load on the building: Per ASCE 7-05 Section 12.3.4.1 & 12.3.4.2, for Seismic Design Category's A, S, andC, p =1.0; for Seismic Design Category 0, E, or F, p shall 1.3. Since Seismic_Design _Category = "D", p = 1.3 E:= P,Vbase_shear E = 915 Ibf Seismic load on building - ----, 7/3/2008 150408R2 (Seward) 28x36x18.xmcd 12 DETERMINE GABLE WALL SHEAR LOADS: 1. Determine the wind load on the eave wall to be resisted by the gable wall in shear: 4e == 18.3 pst Eave wall wind pressure from above . (O.37S.mD.l4.l<Ig.Lhklg.q,,) + (I4m,rLbl<lg'(L-oof) Veuye wmd:== - ~ Veave wind == 1827 Ibf 2. Determine the seismic load to be resisted by the gable wall in shear: .. E Vea,'c seISllllC:== - - ... Veave seismic == 458 Ibt 3. Determine the controlling load to be resisted by the gable wall in shean The controlling load = "Vcavc_wind" . Therefore, V gable shear = 1827 Ibf V gable_shear is the shear load that is transmitted through the roof diaphragm to each gable wall. Normalize the load to a per foot basis. Ygabl,,\\all :~ Whldg - 12 V gahl,,_ sh"ar v gablcwall = 152 pit The gable wall diaphragms can resist the shear loads as follows: v gablewall < 11 0 plf Vgablewall < 188 plf Use 29 gauge metal sheathing. Install per the Typical Screw Schedule as shown on the Standard Details drawing in the engineered drawing package. Use 29 gauge metal sheathing. Install per the Alternate Screw Schedule as shown on the Standard Details drawing in the engineered drawing package. Determine the lateral load that is transmitted to the gablewall with the -large openings that will be resisted by the gable wall posts in bending. Check the bending stress in these posts. OpcningJlCight:= 120 Mgablcwall :== V gablc_shcar" Opening_height fvlg;Jbl"wall Fxgahk\\a/l := 4'~fi8 Fxgablcwall :::: 857 psi Fxallow:== 1.6.575 Fxallow = 920 psi Since F xgablewall < F xallow this is ok. 7/312008 150408R2 (Seward) 28x36x18.xmcd 13 DETERMINE EAVE ViALL SHEAR LOADS: 1. Detennine the wind load on the gable wan to be resisted by the eave wall in shear: qg := if( qwwg - qlwg ~ 10, 10, qwwg - qlwJ qg = 13.5 psf Gable wall wind pressure H.-oof = 4 ft O.375.mD. Hhldl!' W"ldl!'ql! + O.5.II.-oof' W"ldl!'q!! V gable \\'ind:= ~ ~ ~ - ~ - 2 VgabJe_wind;::= 1407 Ibf 2. Determine the seismic toad to be resisted by the eave wall in shear: .. E V gable selsmIC:= - - 1 Vgable_seismic = 458 Ibf 3. Determine the controlling load to be n:-sisted by the eave wall in shear: The controlling load = "V gable_wind" . Therefore, V cave _ sbwr = 1407 Ibf Veave_shear is the shear load that is transmitted through the roof diaphragm to each eave wall. Normalize the load to a per foot basis. Veave shear veavewall:= T. w- '-1>ldg - eavcopeningll veavewall = 47 pit The eave wall diaphragms can resist the shear loads as follows: Veavewall < 110 pIt Use 29 gauge metal sheathing. Install per the Typical Screw Schedule as shown on the Standard Details drawing in the engineered drawing package. - -- -----, 7/3/2008 150408R2 (Se'Nard) 28x36x18.xmcd 14 EMBEDMENT FOR MAIN POST: Calculate the minimum required post embedment depth tor lateral loading tor the main posts. The backfill may be gravel, natural or concrete backfill as specified on page 3. Post_is = "constrained by a concrete slab" Concrete backfill = "No" (Input from page 3) Va = 1037 Ibf Lateral shear load at thegroundline Ma = 6869 ft-Ibt Moment at the groundline djaJooting = 2 ft. Main post footing diameter Ssoil = 150 pst Lateral capacity of soil Trial depth:: 1.5 ft.- The starting depth of the post hole depth. The final post hole depth is determined by iterating to a final depth, per ASAE EP486.1, as allowed per 2006 IBC. depth Jl<lS1 = 3 ft. This is the minimum required post embedment depth for lateral loading Gable walt uplift due to shear loading on gable wall shear panel: Calculate uplift pullout of the gable wall posts due to shear loads on the gable walls. Veave wind = 1827 Ibf Calculated from above Veaye _wind. Ht.ldg Cpo.<:l := Wh1dg - 11 Cpos! = 2741 Ibf This is the uplift load on one gable wall post Assume a dead load weight of roof and wall area to be 2.0 pst. The area of the roof and wall that will tend to keep the gable wall post in the ground will be as follows: B,,\ . Roof:= ---=-- Wh1d,z-2 ., . .. Roof = 240 Ibs Dead load of roof Gablc_"(tll := [Ht.IJ{(Wbldg - 12) + (HlOofOWbldg) + (Hbldg" 2o:-!ay)J2 Gable wall = 984 Ibf Dead load of gable wall <\.1llh.gabl..:]ootllg = 5.0 ft gable post embedment depth Posts:= (Hb1dg + <\:plIUlable Jooting)' W post Posts = 268 Ibs Weight of post dja .$olble Jooting = 2 ft Diameter of gable wall posthole footing Concrete backfill in the gable end posts is = "required" to resist gable wall panel uplift. Backfill = 21051bs Gable post backfill weight if gable end post hole is backfilled with concrete (0 if granular or native soil backfill. Concrete backfill mayor may not be required to resist gable wall panel uplift). Wttot := Gable_wall + Roof + Posts + Backfill Wttot = 3597 Ibf Total resistance for gable wall panel uplift. Since ~ is greater than the gable wall panel uplift, Cpost> the gable wall footing is adequate. ,- 7/3/2008 150408H2 (Se-.vard) 28x36x18.xmcd 15 FOOTING DESIGN FOR MAIN POST: Determine the footing size and depth for vertical bearing for the main posts. ( 2J' ~a Jooting Afooting:= 1t. 4 Atooting = 3.14 ft2 Footing area qsoil = 1500 psf S9i1 bearing capacity for f99ting ~a~tOoting = 2 ft Footing diameter P oslo.depth = 5 ft Minimum required post embedment depth Pfooting:= Acooting.qsoil.dfactor Pfooting = 8482 Ibf End bearing capacity of footing p snow = 3780 1M Total footing load Note that the end bearing capacity (Pfooting) is greater than the snow load (P snow)' This is OK. 7/3/2008 150408R2 (Seward) 28x36x18.xmcd 16 GIRT DESIGN: The girts will simple span between posts and loaded horizontally for wind. Calculate bending stress due to wind loading and determine the adequacy of the girts. Girt _SP.1Cing qw~girt := q"int! _ltirt. 12. 12 qwegirt = 7.85 pli Lgirt_span = 138 in Orientation = "Strongback" ., Lgirt _span'" M Ilirt : = q,,~ltirt. - - 8 , Mgirt tbllil1 := - - Sgirt Mgirt = 18687 in-Ibf fbgirt = 1600 psi Stress applied to the girt Determine the allowable member stress including load factors. LDFwind := 1.6 Cfugirt = U5 CFgirt = 1.00 Cc:= 1.15 F girt = 850 psi Fbgirt := LDFwind.Cfugirt.CFgirt,Cc.Fgirt Fbgirt = 1799 psi > fbgirt This is OK. PURLlN DESIGN: The purlins simply span between pairs of trusses or rafters. Determine the adequacy of the purl ins. Purlin = "2x6" Purlin_spacing = 24 in O.C. l:'pllrlin_~ = III in Bending length of purl in "'purlin = 4.43 pli Distributed snow load along top edge of purl in ') wpllrliu' Lpurlin --"p.,n- Mpurlin := :;: M.-Jin = 6818 in-lbf Bending moment in the purlin M "in ..... ~':1"'" lbpurlin := - Spurlin tb"unm = 902 psi Bending stress applied to the purlin Determine the allowable member stress including load factors LDFsnow:= 1.15 CFpurI;n = 1.30 Cc:= 1.15 Cfupurlin = 1.00 F purI;n = 900 psi F bpurlin := LDF snow,CFpurlin ,Cc,Cfupurlin' Fpurlin Fbpurlin = 1547 psi> fbpurlin This is OK I 7/312008 150408R2 (Seward) 28x"..6x18.xmcd 17 MAIN POST CORBEL BLOCK DESIGN: Determine the required number and size of bolts required in the main post corbel block. A$sum~ full snow load and dead load on the. roof. Allowable fastener shear capacities PbolU8 := 1590 Ibf Shear capacity for 5/8" dia. bolts Pbolt 34:= 2190 Ibf Shear capacity for 3/4" dia. bolts Pbolt 10 := 3600 Ibf Shear capacity for 1" dia. bolts PI6d:= 122 Ibf Shear capacity for 16d nails P20d:= 147 1M Shear capacity for 20d nails p snow = 3780 Ibf Combined snow and dead load on corbels If 5/8 dia. bolts ale used: Nbol\s58 = 2.1 Number of 5/8" dia. bolts required in the corbel block If 3/4 dia. bolts are used: Nbo1ts34 = 1.5 Number of 3/4" dia. bolts required in the corbel block If 1 dia. bolts are used: NbollslO = 0.9 Number of 1" dia. bolts required in the corbel block If 20d nails are to be used: N ail.2Od = i1.2 number of 20dnails required in each corbel block. If 16d nails are to be used: Nailsl6d = 135 number of 16d nails required in each corbel block. 7/312008 150408R2 (Seward) 28x?v6x18.xmcd 18 SUMMARY OF RESULTS: Building Dimensions Building Des",n Loads Wb1dg = 24 ft (Width of Building) Wind_speed = 100 MPH Growld_snow_load = 25 psf 4,ldg = 30 ft (Length of Building) Wind_exposure = "C" Roof snow load = 25 pst Roof dead load = 3 psf Hb1dg = 18 ft (Eave Height of Building) Seismic_Design_Category = "0" Overhang = 18 in (Length of Eave Overhang) Rpitch = 4 /12 (Roof pitch) FootinQ Details: Post Details Post size = "6x8" Postjs = "constrained by a concrete slab" Post_grade = "No.2 Hem-Fir" Usage;;::: 71 % (C9mbinedstre.ss usage !Jfpost) Shear Wall Details: Postdepth = 5.0 ft (Design Post Depth) !1jaJooting"" 2 ft(Design Footing Diameter) Footingusage = 45 % (Stress usage of footing) Vgablewall = 152 plf (Max. shear in gable wall) Veavew:lIl = 47 plf (Max. shear in eave wall) Girt Details: Girt_usage = 89 % (Stress usage of wall girt) Orientation = "Strongback" Purtio Details: Purlin_usage = 58 % (Stress usage of roof purlin for snow loading) Corbel Block Bolts: Nbolts58 = 2.1 Number of 5/8" dia. bolts required in the corbel block if used. NboIts34 = 1.5 Number of 314" dia. bolts required in the corbel block if used. NboltslO = 0.9 Number of 1" dia. bolts required in the corbel block if used. N ails20d = 11.2 Number of 20d nails required in each corbel block if used. Nailsl6d = 13.5 Number of 16d nails required in each corbel block if used. SPECIAL NOTE: The drawings attendant to this calculation shall not be modified by the builder unless authorized in writing by the engineer. No special inspections are required. No structural observation by the design engineer is required.