215 110TH PL SE 2024-11-08 - Laserfiche WebLink

Eco, Structural Calculations, 2/23/2023, Pg. 8 of 25 Wind Analysis - Calculation of Shear Wall Loads Wall Location: Grid A walls Description: Existing walls at garage door opening Tributary Projected Area(ft) Calculations Height 10 ft P = 16.00 psf Wind Pressure from Wind Analysis page 1 Width 12 ft Fwind =P*A = 1920 lb Calculate total wind force acting on wall Height 2(opt.) 0 Fp = Lp*300plf= 0 lb Load resisted by prescriptive/interior walls Width 2(opt.) 0 Fs = Fwind -Fp= 1920 lb Load resisted by shear walls Area, A 120 ft71 PLFshear = F I Ls 404 plf Shear Wall Load(pounds per length foot) Resisting Shear Wall(ft) Shear Wall Case 2 Shear Wall, Lx 3 ft Case 1:Conventional O.K. PLF< 300 1.75 ft Case 2:SW4&HDU2's 300 <PLF< 685 0 Case 3:SW5 or SW8&HDU2's 685 <PLF< 980 0 Case 4:SW5&HDU8's 980 <PLF< 1400 0 Case 5:Other sheathing methods required 1400 <PLF< 2435 0 Case 6:Shear capacity for double sided wall exceeded 2435 <PLF 0 0 Note:Shear nominal capacities taken from SDPWS Table 4.3 and adjusted in 4.75 ft accordance with SPDWS Section 4.3. Shear wall aspect ration shall not exceed 3.5:1. s — x Pres.Wall, LP 0 If requirements are not met portal framing methods can be used. Summary:Grid A existing walls are case 2,conventional sheathing does not adequately resist lateral loads. SW4 and HDU2's required. Wall Location: Grid B Walls Description: Existing Tributary Projected Area(ft) Calculations Height 12 ft P = 16.00 psf Wind Pressure from Wind Analysis page 1 Width 12 ft Fw ft,d =P*A = 2304 lb Calculate total wind force acting on wall Height 2(opt.) 0 Fp = Lp*300plf= 0 lb Load resisted by prescriptive/interior walls Width 2(opt.) 0 Fs = Fwind-Fp= 2304 lb Load resisted by shear walls Area A 144 ft PLFshear =Fs/Ls 121 plf Shear Wall Load(pounds per length foot) Resisting Shear Wall(ft) Shear Wall Case I1 Shear Wall, Lx 5 ft Case 1:Conventional O.K. PLF< 300 4 ft Case 2:SW4&HDU2's 300 <PLF< 685 7 ft Case 3:SW5 or SW8&HDU2's 685 <PLF< 980 3 ft Case 4:SW5&HDU8's 980 <PLF< 1400 0 Case 5:Other sheathing methods required 1400 <PLF< 2435 0 Case 6:Shear capacity for double sided wall exceeded 2435 <PLF 0 0 Note:Shear nominal capacities taken from SDPWS Table 4.3 and adjusted in 19 ft accordance with SPDWS Section 4.3. Shear wall aspect ration shall not exceed 3.5:1. es. s .Wall, LP 0 x If requirements are not met portal framing methods can be used. Pr Summary:Grid B existing walls are Case 1,adequately braced by conventional sheathing.No special shearwalls or hold-downs required.The lateral resistance provided by conventional sheating is structurally adequate to resist the tributary lateral loads imposed on them without additional lateral resistance provided by shearwalls and hold-downs/strap-ties,etc.