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Home My WebLink About1602 Butler St Technical - BuildingTECHNICAL Permit 0 Address 6oZ 5 50 '.e-r 5 Project description 36' ).e, 13,4' h E•QivNel Date the permit vves---fi-Raled oG otl' Number of technical pages �Z 00N 1 of 31 *t. oF Post Frame Building Calculations Prepared By RAMON M RIBA, P E 4627 MANOR PARK DR NW ROCHESTER, MN 55901 TEL (507) 287 -0086 EFAX (707) 929 -1588 EMAIL RMRIBA @CHARTER NET Date Owner Address County City, State Building Span Building Length Eave Height Roof Pitch Avg Roof Height Design Seismic Design Category Wind Zone 100Mph Exposure C Enclosed Structure YES Roof Live Load 25 00 psf Roof Dead Load 3 3 psf Ceiling Load 1 0 psf Soil 2000/150 psf LOAD CALCULATIONS WIND LOAD Primary Frames and Systems Method 1 Normal Force Method Design wind pressures for main wind force resisting systems are given by Table 1609 6 2 1(1) p 322 These pressures are multiplied by the height /exposure coefficient in Table 1609 6 2 1(4) and the importance factor from Table 1604 5 The combined height exposure and gust factor coefficient shall be evaluated at the mean roof height for pressures on roofs and leeward walls ROOF INCREASE Ce @<15 (((DELTA Ce Ce @<15' 10) (MEAN ROOF height 15)) 1 21 (((1 29 1 21) 10) (16 67 15) WALLS 1 237 for mean roof height EXPIRES October 16, 2004 BILL SIPE 1602 S BUTLER ST PORT ANGELES, WA 32 ft 36 ft 14 ft 4 12 16 67 ft IBC -00 1 1 21 for eave height of windward wall Basic wind speed 100 mph Iw importance factor for wind 1 H1 and H2 are given by Figure 1 Building Schematic Wind pressures have positive signs for inward pressures and negative signs for outward pressures suction 3 922 psf -13 368 psf H2 5 33 1 1 1 1 H1 14 1 1 14 320 psf -12 384 psf 1 1 1 1 1 1 Figure 1 Building Schematic Wind perpendicular to the roof ridge where qWW design windward wall pressure psf qWR design windward roof pressure psf qLW design leeward wall pressure psf qLR design leeward roof pressure psf qWW 14 320 psf qWR 3 922 psf qLR -13 368 psf qLW -12 384 psf Earthquake Load Occupancy Category 4 Since the building is one story and has an occupancy category 4, we may use the simplified static lateral force procedure Not having any site specific soil condition data, a Site Class Definition D' is used Page 2 of 31 t Geographical location of building 1602 S BUTLER ST PORT ANGELES WA Building natural period T 0 020 (h) T 0 020 (16 67) 0 75 0 165 sec Since T TO 2 0 2 sec, Sa Fa Ss Since we have a framing system to support gravity loads and shear walls to carry lateral loads our building falls into the Building Frame Systems category In addition our building is a light -frame structure with shear walls consisting of metal cladding instead of plywood sheeting, so R 6 5 32) Determine the weight of the structure Wtotal WSideWall WEndWalls WRoof (Bender Woeste, p 13) WSideWall Length Eave Height psf 2 walls WSideWall 36 ft 14 ft 2 5 psf 2 walls 2520 00 lb WEndWall (Span Eave Height 2 walls) (Span Roof Height) psf WEndWall (32 ft 14 ft 2 walls) (32 ft 5 33 ft) 2 5 psf 1546 67 lb Wroof (Span Overhang) (Length Overhang) Deadload WRoof (32 ft 0 ft 0 ft) (36 ft 0 ft) 4 3 psf 5159 24 lb WTotal 2520 00 lb 1546 67 lb 5159 24 lb 9,226 lb Determine the seismic coefficient Page 3 of 31 For a site class D Fa 1 44 Fv 2 24 Ss 0 64 S1 0 315 Thus SDS (2/3) Fa Ss SDS (2/3) 1 44 0 64 0 614 SD1 (2/3) Fv S1 SD1 (2/3) 2 24 0 315 0 470 Cs 1 2 SDS R 1 2* 0 614 6 5= 0 113 For the simplified procedure, the base shear is calculated as follows V Cs W V 0 113 9 226 lb 1 046 lb Determine the Redundancy Factor p must be between 1 and 1 5 1< p 1 5 Eccentricity Analysis Accidental Eccentricity e 5 %L F Shearwall force due to eccentricity V Story Shear F 05V ri 55V) (10 /Lw)) V Page 4 of 31 ri 5 5/ Lw One story i 1 ri Rmax Ab ground floor area p 2 (20 /(Rmax sqr(Ab))) p 2 (20 (0 172 sqr(1152))) p -1 428 Then p 1 1 1 =1 5 OK WALL GIRT DESIGN Wall Girts (40 375 BETWEEN) 2 x 6 1650f -1 3E MSR 12' bays Tributary Area (48 in 12 in) 12 sft 48 sft From Table TABLE 1609 6 2 1(2) Wind pressure 25 600 psf Standard Girts BENDING P 25 600 psf W 25 29 psf (1 ft ^2 144 ^2) 41 875 in 7 35 pli Mmax wL ^2 8 16627 31 in -lbs SM ((b 2) d) 6 Page 5 of 31 (Bender Woeste, p 16) (Hoyle Woeste, p 381) (NDS 3 3 2, eqn 3 3 -4 p 16) SM 7 56 in ^3 Fb Mmax /SM 16627 31 in -lbs 7 56 in ^3 2198 65 psi Fb' Fb (CD CL Cfu CR) CD 1 6 CL (beam stability factor) 1 0 Cfu 1 Cr 1 15 p 28) FbMax 1650 psi 1 6 1 1* 1 15 2640 psi 2198 65 psi 2640 psi, Therefore wind load checks OK Therefore use 2 x 6 1650f -1 3E MSR with 2 10d common nails each end Roof Purlins (12' bays 2 x 6 HEM -FIR2 24 oc Assumptions Dead Load Calcs 0 06 2 0 12 psf ROOF PURLIN DESIGN 4 12 roof slope (18 435 degree roof angle) Trusses spaced 12 -ft o c Purlin span 11 875 -ft 24 in purlin spacing 1 43 psf dead load (NDS 3 3 3 1 p 16) (NDS Supp Table 4C, p 34) (NDS 4 3 4 Steel average thickness 0157 in (Vic West Steel) Steel weight 490 pcf (Lindley Whitaker, Table 6 -1, p 137) Therefore 490 pcf 0157 in 12 in (40 625 in /36 in) 723 psf Fiberglass blanket and Batt insulation 06 lb /ft ^2 /in (Manbeck, et al Table 6 1, p 132) The following formula shall be used to determine the density in lb /ft ^3 of wood (NDS Supp p 8) density 62 4 [G 1 G(0 009) (m c [1 m c 100] Page 6 of 31 (NDS 3 3 2, eqn 3 3 -2, p 16) (NDS Table 2 3 1 p 9) in which G specific gravity of wood 0 43 for Hem Fir(NDS Table 8A, p 56) m c moisture content of wood density 62 4 (0 43 (1 (0 43 0 009 0 19))) (1 (0 19 100)) lb /ft ^3 Lumber density per unit area lb /ft ^3 144 in ^2 0 187 lb(ft ^3 /in ^2) Unif distributed load (1 5 in 5 5 in 0 187 lb /(ft ^3 in ^2) (1 ft) (12 in 24 in) 5) 0 770 psf 1 613 psf 0 723 psf 0 12 psf 0 770 psf Dead Load is 1 613 psf 25 00 psf roof snow load Cd 1 15 for snow Page 7 of 31 BENDING SNOW LOAD Wt [25 00 psf (24 12 ft cos 18 435)] (1 613 psf 24 12 ft) (UBC Eqn 12 -12 p 2- 50 6619 04 plf Wt 50 6619 04 plf 12 in /ft 4 222 pli Solve for the bending moment and bending moment and bending stress about the x -x axis (Hoyle Woeste p 320) Mxx Wt cos L 2 8 Mxx ((4 222 pli cos 18 435) 141 7386 5 in 2) 8 9533 811 in -lb Sxx (1 5 in (5 5 in 2)) 6 7 56 in 3 (NDS Supp p 8) fbxx Mxx in -lb Sxx in 9533 811 in -lb 7 56 in (NDS Eqn 3 3 2, p 16) 1260 669 psi Fbxx' Fb (CD CL CF CR) Fb 850 psi CD 1 15 (NDS Table 2 3 1 p 5) (NDS Supp Table 4C, p 35) (UBC Section 2316 2 Amend 6 p 2 -291) CL 1 (NDS 3 3 3 3, p 16) CF 1 3 Cr 1 15 (NDS 4 3 4, p 28) Fbxx 850 psi 1 15 1 1 3 1 15 1461 36 psi 1260 669 psi 1461 36 psi therefore bending stress for snow load checks OK 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 Diaphragm Testing Results Table 1 Maximum Shear Intensity I (lbs /ft) Roof Wall Sections Sections 110 lbs /ft 160 lbs /ft Construction Page 8 of 31 Field Screws #10 x 1 -1/2 next to each major rib 9 o c Top Bottom Screws #14 x 1 -1/2 at BOTH sides of each major rib 164 lbs /ft 173 lbs /ft Field Screws #10 x 1 -1/2 next to each major rib 9 o c Top Bottom Screws #10 x 1 -1/2 at BOTH sides of each major rib Stitch Screws #12 x 5/8 at 9 -3/8 on center through all overlaps 1997 UBC Table 23- II -I -1 598 6 lbs /ft 7/16 osb applied directly to framing with 8d 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 common nails 2 on center at panel edges 6 on center intermediate framing members Per footnote number #1 SG 0 42 use Structural 1 base value of 730# 82 (Per footnote #4 panels applied long dimensions over studs use 15/32 values) 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 3/8 qww H1 L qlw H1 L qwr H2 L qlr H2 L 2 W (The Alumax Roof Diaphragm Brochure 1991) where qww design windward wall pressure (psf) qwr design windward roof pressure (psf) qlw design leeward wall pressure (psf) qlr design leeward roof pressure (psf) L the building length in feet W the building width in feet H1 the outside wall height not inside clear) H2 the height from the eave to the ridge Where I [3/8 ((14 320 psf 14 ft 36 ft) -12 384 psf 14 ft 36 ft)) ((3 922 psf 5 33 ft 36 ft) -13 368 psf 5 33 ft 36 ft))] (2 32 ft) I 100 56 lbs /ft The following minimum diaphragm requirements are needed Page 9 of 31 Field Screws #10 x 1 -1/2 next to each major rib 9" o c Top Bottom Screws #14 x 1 -1/2 at BOTH sides of each major rib Calculate the Maximum Endwall Shear Load Vendwall (lbs /ft), by multiplying the maximum roof shear intensity I, by the building end -wall width W (Bender Woeste, p A -30 A -33) Vendwall I W Vendwall 3 217 95 lbs 100 56 lbs /ft 32 ft Calculate the Allowable End -Wall Shear Capacity, SC BACKWALL SC W DW R DW the door or opening width in feet, or distance between end -wall structural posts surrounding the door, whichever is greater If more than one door is installed DW is the total of door widths R the allowable end -wall shear strength in lbs /ft from Table 1 Multiplying this value by the one -third increase in allowable stresses for all combinations including W or E, We end up with the following R 160 lb /ft 1 33 212 8 lb /ft SC back endwall 6,809 6 lbs (32 ft 0 ft) 212 8 lbs /ft SC W DW R R 160 lb /ft 1 33 212 8 lb /ft SC front endwall 851 2 lbs (32 ft 28 ft) 212 8 lbs /ft Checking by equation for adequacy Vendwall SC 3,217 95 lbs 5107 2 851 2 lbs R 160 lb /ft 1 33 212 8 lb /ft Column Calculations 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 building diaphragm This is taken from the Design of Commercial Post -Frame Buildings (1997) where 1 P a C1 2 DeltaS b for a simple beam test where P 0 4 Pultimate (ASAE EP484 1, Eqn 4, p 526) Page 10 of 31 Putimate ultimate strength of panel P 0 4 9600 P 3840 DeltaS deflection at P adjusted for sinking supports DeltaS 0 235 a Frame Spacing a 12 ft b test panel length parallel -to- corrugations b 11 667 ft C1 1 3840 12 ft 2 0 235 11 667 ft 01 8 403 41 lb /in stiffness of the test panel Calculate Stiffness of One Roof Slope b a C2 01 b Sf where C2 stiffness of diaphragm being predicted(lb /in 01 stiffness of test panel (lb /in b roof slope length(ft) b test panel length a width of test panel(ft) Sf frame spacing of diaphragm being predicted(ft) b 16 ft cos 18 435 Page 11 of 31 (Townsend p 4) (Townsend, p 4) (Anderson Bundy Ch 6) b' 16 87 ft a 12 ft Sf 12 ft C2 11,810 63 lb /in 8,403 41 lb /in 16 87 ft 12 ft) (12 ft 12 ft) Calculate Stiffness of Roof with Two Equal Sides The stiffness of the entire roof diaphragm with two equal sides can be calcuated by the following equation (Anderson Bundy) Ch 2 C2 cos ^2 roof slope where Ch roof diaphragm stiffness(lb /in Ch 2 11 810 63 lb /in Cos ^2 18 435) Ch 21 259 13 lb /in STEP 2 Calculate k the frame stiffness k (6 E I) (H1 ^2 ((0 7 d) H1) E modulus of elasticity of the post psi E' E CI CI 95 (NDS Table 2 3 1 1, p 11) E' 1300000 psi 95 1235000 psi (NDS Supp Table 4D, p 39) See Hem -Fir I moment of inertia of the post in ^4 I b x d"3 12 256 00 6 8 "3 12 H1 wall height in inches H1 168 in d embedment depth in inches d 40 00 in k 342 913 lb /in (6 1235000 psi 256 00 in'4) ((168 in ^2) 7 40 00 in 168 in)) STEP 3 Calculate R the lateral restraining force of the roof diaphragm (Bender Woeste, p 45) The lateral restraining force can be calculated as follows (Skaggs Eqn 2) For full Wind combinations R Sf ((Hi (qww qlw) ((2 8 d) (3 H1))) (8 ((0 7 d) H1)) H2 (qwr qlr)) (Skaggs Eqn 1) (NDS Supp R 12 ft ((14 ft (14 320 psf -12 384 psf) ((2 8 3 333 ft) (3 10 333 ft))) (8 7 3 000 ft) 10 333 ft)) 5 33 ft (3 922 psf -13 368 psf)) Page 12 of 31 p 8) R 2,869 00 lbs For 1/2 Wind combinations R Sf ((H1 ((qww /2) (qlw /2)) ((2 8 d) (3 H1))) (8 ((0 7 d) H1)) H2 ((qwr /2) (qlr /2))) R 12 ft ((14 ft (14 320/2 psf -12 384/2psf) ((2 8 3 333 ft) (3 10 333 ft))) (8 7 3 000 ft) 10 333 ft)) 5 33 ft (3 922/2 psf -13 368/2 psf)) R 1 434 50 lbs STEP 4 Calculate the Diaphragm Factor mD (Bender Woeste, p 45 -46) First we must determine the ratio of frame -to -roof diaphragm stiffness k Ch 0 0161 342 913 lb /in 21,259 13 lb /in (Bender Woeste, p 45) Frames 4 From Table 1 of Skaggs mD 0 96 STEP 5 Calculate Shear at Top of the Windward Post The Shear at the top of each post is calculated by the following (Skaggs Eqn 3) For full Wind combinations V 1 2 ((R mD) Sf ((H1 (qww qlw) ((2 8 d) (3 H1)) (8 ((0 7 d) H1)) H2 (qwr qlr))) V 1 2 2 869 00 lb 0 96 12 ft ((14 ft (14 320 psf -12 384 psf) ((2 8 3 333 ft) (3 14 ft))) (8 ((0 7 3 333 ft) 14 ft)) 5 33 ft (3 922 psf -13 368 psf))) V 916 42 lb For 1/2 Wind combinations V 1 2 ((R mD) Sf ((H1 ((qww /2) (qlw /2)) ((2 8 d) (3 H1))) (8 ((0 7 d) H1)) H2 ((qwr /2) (qlr /2)))) V 1 2 2 869 00 lb 0 96 12 ft ((14 ft (14 320/2 psf -12 384/2 psf) ((2 8 3 333 ft) (3 14 ft))) (8 ((0 7 3 333 ft) 14 ft)) 5 33 ft (3 922/2 psf -13 368/2 psf))) V 458 21 lb STEP 6 Calculate Maximum Post Moments M1 and M2 (Bender Woeste, p 46 -47) The moment at the groundline can be calculated as follows Page 13 of 31 k1 0 97 (the correction factor from Table 2 of Skaggs) For full Wind combinations M1 kl H1 (V ((Sf qww H1) 2)) (Skaggs Eqn 4) M1 0 97 14 ft 916 42 lb ((12 ft 14 320 psf 14 ft) 2)) 12 in M1 29 670 04 in -lb For 1/2 Wind combinations M1 kl H1 (V ((Sf (qww /2) H1) 2)) (Skaggs Eqn 4) M1 0 97 14 ft 458 21 lb ((12 ft (14 320 psf /2) 14 ft) 2)) 12 in M1 21 249 96 in -lb If V is positive the maximum positive moment moment above groundline can be calculated k2 1 13 (the correction factor from Table 3 of Skaggs) For full Wind combinations M2 k2 (V ^2 (2 Sf qww)) (Skaggs Eqn 5) M2 1 13 916 42 lb ^2 (2 12 ft 14 320 psf)) 12 in M2 2,825 68 in -lb For 1/2 Wind combinations M2 k2 (V ^2 (2 Sf (qww /2))) (Skaggs Eqn 5) M2 1 13 458 21 lb ^2 (2 12 ft (14 320 psf /2))) 12 in M2 1,412 84 in -lb STEP 7 Calculate Axial Compression Force in Post Sf is the frame spacing W width of building feet S 25 00 psf roof snow load DT 3 3 psf top chord dead load DB 1 psf bottom chord dead load D L W S/2 (IBC) Page 14 of 31 Pf ((Sf W) 2) ((S 2) DT DB (((3 qwr) qlr) 4)) (H2 (2 W)) ((R mD) (Sf H2 (qlr qwr))) Pf ((12 ft 32 ft) 2) ((25 00 psf 2) 3 3 psf 1 0 psf (((3 3 922 psf) -13 368 psf) 4)) (5 33 ft (2 32 ft)) 2,869 00 lb 0 96 (12 ft 5 33 ft -13 368 psf 3 922 psf))) 3,048 09 lb D L S W /2 (IBC Pf ((Sf W) 2) (S DT DB (((3 (qwr /2)) (qlr /2)) 4)) (H2 (2 W)) ((R mD) (Sf H2 ((qlr /2) (qwr /2)))) Pf ((12 ft 32 ft) 2) ((25 00 psf 2) 3 3 psf 1 0 psf (((3 3 922 /2 psf) -13 368 /2 psf) 4)) (5 33 ft (2 32 ft)) 2,869 00 lb 0 96 (12 ft 5 33 ft -13 368 /2 psf 3 922 /2 psf))) 5,750 00 lb Page 15 of 31 NOTE It was found that Deadload Snowload windload /2 controlled both the groundline moment adequacy and post adequacy in the positive moment region STEP 8 Check Post Adequacy at Groundline fc Pf A A b d A 6 in 8 in 48 00 in ^2 D L W S/2 (IBC) fc 63 50 psi 3 048 09 lb 48 00 in ^2 D L S W /2 (IBC) fc 119 79 psi 5,750 00 lb 48 00 in ^2 The allowable compressive stress in the post is given by Fc Fc (CD CM CF CP CI) (NDS Table 2 3 1 p 9) Fc 850 psi actual compression stress parallel -to -grain (NDS Supp Table 4D, p 39) See Hem -Fir CD the load duration factor for wind of 1 6 (NDS Table 2 3 2) 1997 UBC 2316 Table 2 3 2 footnote 2 provides, 1 6 may be used for members and nailed and bolted connections exhibiting Mode III or IV behavior, except that the increases for wind are not combined with the increase allowed in Section 1612 3 The 60- percent increase shall not apply to the allowable shear values in Tables 23 -II -H, 23- II,I -1 23- II -I -2, 23 -II J or in Section 2315 3 Since the post will be subjected to wet conditions at and below the groundline, we choose Page 16 of 31 CM (Wet service factor) 0 91 (NDS Supp Table 4D, p 37) CF 1 (NDS Supp Table 4D p 37) Since the column is laterally supported from all sides at the groundline, the column stability factor is CP 1 (NDS 3 7 1 1 p 22) CI 0 85 (NDS Table 2 3 1 1 p 11) F'c 1051 96 psi 850 psi 1 6 1 0 91 1 0 85 Using the full bending moment at groundline the bending stress in the post is given by Fbl Ml S S 64 00 in ^3 For full Wind combinations Fbl 729 37 psi 29,670 04 in -lb 64 00 in ^3 For 1/2 Wind combinations Fbl 364 68 psi 21 249 96 in -lb 64 00 in ^3 The allowable bending stress is given by Fb' Fb (CD CM CL CF CI CR) (NDS Table 2 3 1 p 9) Fb 975 psi (NDS Supp Table 4D p 39) See Hem -Fir CD 1 6 (IBC) Cm 1 (NDS Supp Table 4D p 37) Cfu 1 (NDS Supp Table 4D p 37) Cr 1 (NDS Supp Table 4C p 34) CL 1 (NDS 3 3 3 1, p 16) CF 1 (NDS Supp Table 4D, p 37) CI 0 85 (NDS Table 2 3 1 1, p 11) Fb 975 psi 1 6 1 1 1 0 85 1 1 1326 psi The interaction equation for checking bending about one axis and compression is CSI (Fc F c) ^2 (Fbl (F b (1 (Fc FcE)))) 1 0 (NDS Eqn 3 9 -3 p 23) For a laterally supported beam column that is restrained from buckling and twisting the interaction equation can be written as follows CSI1 (Fc F'C) ^2 Fbl F b =1 D L W S/2 (IBC) CSI1 63 50 psi 1051 96 psi) ^2) 729 37 psi 1326 psi) CSI1 0 004 0 55 0 55 1 D L S W /2 (IBC) CSI1 119 79 psi 1051 96 psi) ^2) 364 68 psi 1326 psi) CSI1 0 013 0 28 0 29 1 STEP 9 Check Post Adequacy at Positive Moment Region Fc Pf A A b d D L W S/2 (IBC) fc 63 50 psi 3 048 09 lb 48 00 in ^2 D L S W /2 (IBC) fc 119 79 psi 5 750 00 lb 48 00 in ^2 Page 17 of 31 The allowable compressive stress is given by Fc Fc (CD CP CM CI CT CF) (NDS Table 2 3 1 p 9) Fc 850 psi (NDS Supp Table 4D, p 39) CD 1 6 To calculate the column stability factor CP, we begin by determining the effective column length Le Ke 1 (NDS 3 7 1 2 p 22) Ke buckling length coefficient for compression members 0 8 for a propped cantilever le 126 8 in 0 8 158 5 in Fc FcE 119 79 psi 1474 79 psi Fc* Fc(CD CM Ct CF CI) Fc* 1156 psi 850 psi 1 6 1 0 1 0 85 Cp ((1 (FCE Fc (2 c)) SQR(((1 (FCE Fc (2 c)) ^2 ((FCE Fc* psi) c)) Cp ((1 (1474 79 psi 1156 psi)) (2 0 8)) SQR(((1 (1474 79 psi 1156 psi)) (2 0 8)) ^2 ((1474 79 psi 1156 psi) 0 8)) 0 768 Page 18 of 31 KcE 0 3 for visually graded lumber (NDS 3 7 1 5 p 22) c= 0 8 for sawn lumber (NDS 3 7 1 5 p 22) E E CI CI 0 95 (NDS Table 2 3 1 1, p 11) E 1300000 psi 0 95 1235000 psi (NDS Supp Table 4D p 39) See Hem -Fir FcE (KcE E) (Le d)"2 (NDS 3 7 1 5 p 22) FcE 1474 79 psi (0 3 1235000 psi) (158 5 in 8 in) ^2 To use Eqn 3 9 -3 in the NDS the values must satisfy the following equation (NDS 3 9 2, p 23) (NDS 3 7 1 5, p 22) (NDS 3 7 1 5 p 22) CI 0 85 (NDS Table 2 3 1 1, p 11) F c 804 0870551 psi 850 psi 1 6 0 768 0 85 1 Calculating the bending stress in the positive moment region For full Wind combinations Fb2 44 15 psi 2,825 68 in -lb 64 00 in ^3 For 1/2 Wind combinations Fb2 22 08 psi 1,412 84 in -lb 64 00 in ^3 The allowable stress is given by Fb Fb (CD CL CI CF CFU CM) (NDS Table 2 3 1, p 9) Fb 975 psi (NDS Supp Table 4D p 39) See Hem -Fir CD 1 6 Page 19 of 31 CI 0 85 (NDS Table 2 3 1 1, p 11) CL ((1 (FbE Fb 1 9) SQR((((1 (FbE Fb 1 9) ^2) ((FbE Fb 0 95)) (NDS Eqn 3 3 3 1 p 16) Lu Ke 1 (NDS 3 7 1 2, p 22) Lu 126 80 in 0 8 158 5 in le 233 31 in 1 84 126 80 in (NDS Table 3 3 3 p 17) RB SQR(Le d) b ^2 (NDS Eqn 3 3 -5 p 16) RB 0 99 SQR(( 233 31 in 8 in) 6 in ^2) FbE (KbE E) RB ^2 KbE 439 Euler buckling coefficient for beams FBE 554,137 10 psi (0 439 1235000 psi) 0 99 in ^2 Fb' Fb CD CI CM CF) Fb 1 326 00 psi 975 psi 1 6 0 85 1 1 CL 1 000 (NDS 3 3 3 1, p 16) CM 1 CF 1 (NDS Supp Table 4D, p 37) Cfu 1 (NDS Supp Table 4D p 37) Fbl' 1 325 84 psi 975 psi 1 6 1 000 0 85 1 1 1 1 The interaction applies again CSI2 (Fc F C) ^2 Fbl F b =1 (NDS Eqn 3 9 -3, p 23) D L W S/2 CSI2 0 11 =1 D L S W /2 CSI2 0 08 =1 Bond Strength Of Concrete -To -Wood (Hoyle Woeste, p 335 -336) 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 Page 20 of 31 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 Strength ((b in 2) (d in 2)) embed in 30 psi 33600 lb ((6 in 2) (8 in 2)) 40 00 in 30 psi 25 00 psf 3 3 psf 1 psf) 30 ft /2 5 14 12 ft 6465 6 lb 6465 6 lb 33600 lb Therefore no concrete footing pad is required below columns FOUNDATION UPLIFT From ASAE EP486 the following equation calculates the soil uplift resistance of a 32 004 in deep post with a concrete collar 26 in thick x 24 in diameter (Walker Woeste p 163) The presumed type 4 soil has a soil friction angle 26 degrees and density 85 lb /ft ^3 (Walker Woeste Appendix E Table 1 p 340) a soil density 85 lb /ft ^3 (Walker Woeste Appendix E Table 1 p 340) C concrete density 150 lb /ft ^3 (Walker Woeste p 163) G gravity acceleration constant 1 0 lbf /lb d embedment depth 40 00 in t collar thickness 31 in w collar width 24 in Phi soil friction angle 26 degrees (Walker Woeste, Append E, Table 1 p 340) Ap post cross sectional area 6 in 8 in 48 00 in ^2 h thickness of attached concrete (depth of collar) 31 in d d h then U soil and foundation uplift resistance (in lb) a 3 14 (d [w ^2 d w tan(Phi) d ^2 (3 tan(Phi)] d' Ap 85 3 14 ((9 00 [24 ^2 9 00 24 tan(26) 9 00 ^2 (3 tan(26) 9 00 48 5,254 67 lb Constant Value Uplift (TrussSpan Sidewall overhangs) Sidewall) 2) {PSCCBLoad [((3 QWR) QLR) /4] H2 2 /TrussSpan 2 (QLR QWR)} (Bender Woeste, p 56) -(32 ft 0 ft)* 12 ft) /2) (2 313 psf [((3 -3 922 psf) -13 368 psf) /4] 5 33 ft 2/ 32 ft 2 -13 368 psf (3 922 psf)1} 2003 44 lb 5,254 67 lb 2003 44 lb Therefore collar is sufficient to resist uplift The normal duration uplift (for truss -to- column 'connections) is Load 2003 44 lb 1 6 1252 15 lb BOLTS Design Values for Single Shear Connections (NDS Sec 8 2, p 53) Page 21 of 31 Page 22 of 31 Wood -to -Wood Connections (NDS, 8 2 1) Mode Im Z (D Tm Fem) (4 KTheta) 1760 (0 625 in 6 2250 psi) (4 1 1988) (NDS, Eq 8 2 -1) Mode Is Z (D Ts Fes) (4 KTheta) 789 (0 625 in 2 75 in 2200 psi) (4 1 1988) (NDS Eq 8 2 -2) Mode II Z (K1 D Ts Fes) (3 6 KTheta) 2138 (2 440 0 625 in 2 75 in 2200 psi) (3 6 1 1988) (NDS, Eq 8 2 -3) Mode IIIm Z (K2 D Tm Fem) (3 2 (1 (2 Re)) KTheta) 807 (1 118 0 625 in 6 in 2250 psi) (3 2 (1 (2 1 023)) 1 1988) (NDS Eq 8 2 -4) (Note when a mode N/A the mode is not applicable Mode IIIs Z (K3 D Ts Fes) (3 2 (2 Re) KTheta) 486 (1 457 0 625 in 3 in 2200 psi) (3 2 (2 1 023) 1 1988) (NDS Eq 8 2 -5) Mode IV Z (D (3 2 KTheta)) Sgr((2 Fem Fyb) (3 (1 Re))) 588 (0 625 in (3 2 1 1988)) Sgr((2 2250 psi 45000 psi) (3 (1 1 023))) (NDS, Eq 8 2 -6) Where Kl ((Sqr(Re ((2 (Re 2)) (1 Rt Rt 2)) ((Rt 2) (Re 3))))- (Re (1 Rt))) (1 Re) 2 440 ((Sqr(1 023 ((2 (1 023 2)) (1 2 18 2 18 2)) ((2 18 2) (2 18 3)))) (1 023 (1 2 18))) (1 1 023) (NDS 8 2 1) K2 -1 Sqr((2 (1 Re)) (((2 Fyb) (1 (2 Re)) D 2) (3 Fem (Tm 2)))) 1 118 -1 Sgr((2 (1 1 023 (((2 45000 psi) (1 (2 1 023 0 625 in 2) (3 2250 psi (6 2)))) (NDS 8 2 1) K3 -1 Sgr(((2 (1 Re)) Re) (((2 Fyb) (2 Re) D 2) (3 Fem (Ts 2)))) 1 457 -1 Sgr(((2 (1 1 023 1 023 (((2 45000 psi) (2 1 023 0 625 in 2) (3 2250 psi (2 75 2)))) (NDS 8 2 1) Re Fem Fes (NDS 8 2 1) 1 023 2250 psi 2200 psi Rt Tm Ts (NDS 8 2 1) 2 18 6 in 2 75 in Tm Thickness of main (thicker) member inches 6 in Ts Thickness of side (thinner) member, inches 2 75 in Fem Dowel bearing strength of the main (thicker) member, psi 2250 psi (See NDS Table 8A p 53, Using Spruce -Pine Fir) Fes Dowel bearing strength of the side (thinner) member, psi 2200 psi (See NDS Table 8A p 53 Using Spruce -Pine Fir) Fyb Bending yield strength of bolt, psi 45000 psi D nominal bolt diameter inches 0 625 in KTheta 1 (ThetaMax 360) 1 1988 1 (71 57 360) ThetaMax maximum angle of load to grain (0 Theta 90) for any member in a connection 71 57 therefore Z 486 lb Load 2003 44 lb 1 6 1252 15 lb 1252 15 lb 486 lb 2 577 bolts 3 bolts Page 23 of 31 Therefore use (3) 5/8 bolts thru (NDS eq 12 3 -2) Note Minimum edge end and on center spacings of 2 -1/2 must be maintained ALTERNATE The nominal nail or spike lateral design values, Z, shall be the lesser of Mode Is Z D ts Fes KD (NDS eq 12 3 -1) where D nail or spike diameter, inches 0 177 inch where ts thickness of side member in inches 2 75 in where Fes dowel bearing strength of side member 3350 psi for HF (NDS Table 12A) where KD 10D 0 5 for 0 17 in D 0 25 in KD 2 27 (10 0 177 0 177 in) 0 5 Z 718 33 lb 0 177 in 2 75 in 3350 psi 2 27 Mode IIIm Z k1 D p Fem KD (1 2 Re) where kl -1 SQR( 2 (1 Re) (2 Fyb (1 2 Re) D ^2 3 Fem P ^2)) where Re Fem Fes Where Fem dowel bearing strength of main member (member holding point) Page 24 of 31 3350 psi for HemFir (NDS Table 12A) Re 1 045 3500 3350 Where Fyb bending yield strength of nail of spike 115 000 psi for 0 177 diameter threaded hardened -steel nails (NDS Table 12 3C fn 2) where p penetration of nail or spike in main member 2 25 in K1 1 123 -1 SQR(2 (1 1 045) (2 115 000 psi (1 2 1 045) 0 177 in ^2 /(3 3500 psi in ^2))) Z 223 25 lb 1 123 0 177 in 2 in 3500 psi 2 27 (1 2 1 045) Mode IIIs Z k2 D ts Fem KD (2 Re) where k2 -1 SQR((2 (1 Re) Re) (2 Fyb 2 Re) D ^2 3 Fem is ^2)) K2 1 047 -1 SQR ((2 (1 1 045) 1 045) (2 115,000 psi (2 1 045) 0 177 in ^2 (3 3500 psi 2 75 in ^2))) Z 258 09 lb 1 047 0 177 in 2 75 in 3500 psi (2 27 (2 1 045) Mode IV Z (D ^2 KD) SQR((2 Fem Fyb) (3 (1 Re))) Z 158 10 lb (0 177 in ^2 2 27) SQR((2 3500 psi 115,000 psi) (3 (1 1 045))) Mode IV governs, therefore Z 158 lb Z' Z CD CM Ct Cd Ceg Cdi Ctn CD 1 6 CM 1 0 for wood used in continuously dry conditions (NDS 7 3 3) Ct 1 0 for wood which will not experience sustained exposure to temperatures over 100 degrees F Cd 1 00 p (12 D) 1 0 2 in (12 0 177 in) Ceg does not apply nails not driven in the end grain Cdi 1 0 for nails used in diaphragm construction Ctn does not apply, not a toe nailed connection NOTE For ease of application CD is applied to the load Z 158 10 lb 158 lb 1 0 1 0 1 00* 1 0 Calculate required number of nails Load 2003 44 lb 1 6 1252 15 lb 1252 15 lb 486 lb 158 10 lb /nail 4 846 nails 5 nails (NDS Eq 12 3 -4) (NDS Eq 12 3 -5) (NDS 12 3 5) (NDS 12 3 6) Page 25 of 31 Therefore, use (1) 5/8 in bolt through plus 5 40d threaded hardened nail According to NDS p 27 paragraph 7 1 1 1 'mixed fastener connections shall be based on tests or other analysis (see 1 1 1 4) In completing the previous design, we added a bolt allowable value to nail values Based upon our experience with truss uplift at post connections such designs have performed satisfactorily and thereby meet the requirements of the NDS Sidewall Embedment Determine the required column embedment h Base moment V h 2 70 in 29 670 04 ft -lb 916 42 lb) 12 inches S1 Column Depth Lateral Soil Bearing Pressure 2 3 S1 333 33 lb 3 333 ft 2000 lb -ft 2 3 The Lateral Soil Bearing values may be increased the amount of the designated value for each additional foot of depth to a maximum of 15 times the designated value Isolated poles for uses such as poles used to support buildings that are not adversely affected by a 1/2 -inch motion at ground surface due to short -term lateral loads may be designed using lateral bearing values equal to two times the tabulated values Additionally, an increase of one third shall be permitted when considering load combinations, including wind or earthquake loads, as permitted by 2 A (2 34 P) (S1 diameter) A 2 490 (2 34 916 42 in -lb) (333 33 lb 2 00) d (A 2) (1 Sqr(1 ((4 36 h) A)) d 4 501ft (2 490 2) (1 SQR(1 ((4 36 3 333 ft) 2 490 Therefore Depth (d) 3 333 ft is adequate to withstand lateral loads Determine the required diameter of the concrete column encasement The allowable foundation pressure may be increased 20 percent for each additional foot of width or depth (beyond one foot) to a maximum value of 3 times the designated value No increase for width is allowed for 1000 psf soil AFP 3,333 33 psf 2000 psf (1 ((3 333 ft 1 ft) 20 ((2 00 ft 1) 20 Load carrying capacity AFP PI r ^2 10 472 00 lbs 3 333 33 psf 3 1416 1 00 ft 2 Load on Footing (S DT DB) W 2 Sf 6,045 60 lbs (25 00 psf 3 3 psf 1 0 psf) 32 ft 2 (12 ft 2) 6,045 60 lbs 10 472 00 lbs Therefore Diameter 24 in Page 26 of 31 Embed 3 333 feet (40 00 in) deep in 2 58 foot (31 in) concrete backfill with 24 in diameter 6 x 8 Rough HEM -FIR1 Sidewall columns are required These columns are basically only subject to the wind load They are supported at least at the roof line, but we will also support them at the truss bottom chord line and use these braces for stabilizing the roof trusses against wind uplift They also are embedded below grade in pole footings Consider them as propped cantilevers (Barbera p 13) Endwall Design Column Spacings 12 12 Eave Height 14 ft (168 in) Overall heel height of roof truss at eave line 6 in Unsupported length of column 126 8 in BENDING ENDWALL WIND COLUMN DESIGN Page 27 of 31 P 14 320 psf W 14 320 psf (1 ft ^2 144 in ^2) ((12 ft 12 ft) 2 12 in) 8 077 pli Mmax wL ^2 8 8 077 pli 126 8 in ^2 8 16232 838 in -lbs SM (W (D 2)) 6 (NDS 3 3 2 2 eqn 3 3 -4, p 16) SM (6 in (6 in 2)) 6 SM 36 00 in A3 Fb Mmax SM 16232 838 in -lbs 36 00 in ^3 450 912 psi Fb' Fb (CD CL CF Cfu CI) (NDS Table 2 3 1, p 9) CD 1 6 (UBC Section 2316 2, Amend 6 fn 2 p 2 -291) CL (beam stability factor) 1 0 (NDS 3 3 3 1 p 16) CF 1 Page 28 of 31 Cfu 1 00 (NDS Supp Table 4D, p 39) CI 0 85 (NDS Table 2 3 1 1, p 11) Fb 575 psi 1 6 1 1 00 1 0 85 782 psi 450 912 psi 782 psi, therefore wind load checks ok 6 x 6 Rough Hem -Fir #2 Endwall columns are required Endwall Truss to Column Calculations Raw Nails PSCCBLoad Steel Weight CCB Weight Purlin Weight Truss Weight 0 723 0 12 0 767 0 7 PSCCBLoad 2 31 psf Qwr 3 922 psf Qlr -13 368 psf Sidewall 12 ft on center TrussSpan 32 Total DOL 1 6 Uplift (TrussSpan Sidewall overhangs) 2) Sidewall /2) Endwall overhang)) PSCCBLoad [((3 -QWR) QLR) /4] H2 ^2 /TrussSpan ^2 -QLR -QWR)} Uplift -(32 ft 0 ft) 2) ((12 ft /2) 0 ft)) {2 31 psf [((3 -3 922 psf) -13 368 psf) /4] 5 33 ft ^2/ 32 ft ^2 -13 368 psf -(3 922 psf)} 640 95 lb The nominal nail or spike lateral design values, Z shall be the lesser of Mode Is Z D ts Fes KD (NDS eq 12 3 -1) where D nail or spike diameter, inches 0 148 inch where ts thickness of side member in inches 1 5 in where Fes dowel bearing strength of side member 3500 psi for SPF (NDS Table 12A) where KD 2 20 for D 0 17 in Z 353 18 lb 0 148 in 1 5 in 3500 psi 2 20 Mode IIIm Z k1 D p Fem KD (1 2 Re) (NDS eq 12 3-2) where kl -1 SQR( 2 (1 Re) (2 Fyb (1 2 Re) D ^2 3 Fem P ^2)) where Re Fem Fes Where Fem dowel bearing strength of main member (member holding point) 3500 psi for HemFir (NDS Table 12A) Re 1 000 3500 3500 Where Fyb bending yield strength of nail of spike 115,000 psi for 0 148 diameter threaded hardened -steel nails (NDS Table 12 3C fn 2) where p penetration of nail or spike in main member 1 5 in K1 1 154 -1 SQR(2 (1 1 000) (2 115 000 psi (1 2 1 000) 0 148 in ^2 (3 3500 psi 1 5 in ^2))) Z 135 86 lb 1 154 0 148 in 1 5 in 3500 psi 2 20 (1 2 1 000) Mode IIIs Z k2 D is Fem KD (2 Re) where k2 -1 SQR((2 (1 Re) Re) (2 Fyb 2 Re) D ^2 3 Fem is ^2)) K2 1 154 -1 SQR ((2 (1 1 000) 1 000) (2 115 000 psi (2 1 000) 0 148 in ^2 (3 3500 psi 1 5 in ^2))) Z 135 86 lb 1 154 0 148 in 1 5 in 3500 psi 2 20 (2 1 000) Mode IV Z (D ^2 KD) SQR((2 Fem Fyb) (3 (1 Re))) (NDS Eq 12 3 -4) Z 115 32 lb (0 148 in ^2 2 20) SQR((2 3500 psi 115 000 psi) (3 (1 1 000))) Therefore Z 184 52 lb CM 1 0 for wood used in continuously dry conditions (NDS 7 3 3) Ct 1 0 for wood which will not experience sustained exposure to Page 29 of 31 Page 30 of 31 temperatures over 100 degrees F Cd 0 84 p (12 D) 1 0 1 5 in (12 0 148 in) (NDS Eq 12 3 -5) Ceg does not apply, nails not driven in the end grain Cdi 1 0 for nails not used in diaphragm construction Ctn does not apply, not a toe nailed connection Z 155 84 lb 100 89 lb 1 6 1 0 1 0 8446 1 1 Calculate required number of nails 400 60 lb 155 84 lb /nail 2 57 nails 3 nails Therefore, use 3 10d commons (NDS 12 3 5) (NDS 12 3 6) WARNING Added insulation, ceiling materials, lighting and HVAC equipment or other loads imposed upon the roof system BEYOND those stated above will reduce the overall capacity of the structure and its ability to resist the required design snow loads Sources Alumax Building Products 1992 Diaphragm Loading On Roofs and End Wall Sections Alumax Building Products Testing Laboratories Perris California American Forest and Paper Association 1997 ANSI /AF &PA NDS -1997 National Design Specification for Wood Construction AF &PA, Washington D C American Lumber Standards Committee (ALSC) December 1999 Woodwords, pg 41 Bender DA TD Skaggs, and F E Woeste 1990 Rigid Roof Design for Post -Frame Buildings Paper #90 -4525 ASAE St Joseph MI 49085 -9659 1991 Applied Engineering in Agriculture 7(6) 755 -760 1991 Frame Building Professional 3 (6) 4 -11 30 -33 Bender DA TD Skaggs and F E Woeste 1992 Design of Side -Wall Posts in Post -Frame Buildings Paper #92 -4541 ASAE St Joseph, MI 49085 -9659 referred to as Skaggs et al Also published in Frame Building News, March 1993 Applied Engineering in Agriculture 9(2) 253 -259 Hoyle, R J and F E Woeste 1989 Wood Technology in the Design of Structures Iowa State University Press Ames Iowa International Conference of Building Officials (ICBO) 1997 Uniform Building Code ICBO Whittier, CA National Forest Products Association (NFPA) 1991 National Design Specification (NDS) for Wood Construction Washington DC NFPA USP Full Line Catalog Construction Hardware 1996 United Steel Products Products Company, Inc Silver Metal Products, Inc Pollock DG DA Bender and K G Gebremedhin 1996 Designing for chord forces in post -frame roof diaphragms Frame Building News Page 31 of 31 8 (5) 40 -44 Taylor, Steven E 1996 Earthquake considerations in post -frame building design Frame Building News 9(1) 62 .1 Townsend Merl 1992 Diaphragm Loading on Roofs and End Wall Sections Alumax Building Products Inc Perris Valley, CA Uniform Building Code (UBC) 1997 International Conference of Building Officials (ICBO) Whittier CA Walker, John N Frank E Woeste 1992 Post -Frame Building Design American Society of Agricultural Engineers