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

Eco, Structural Calculations, 2/23/2023, Pg. 10 of 25 Wind Analysis-Calculation of Shear Wall Loads (continued) Wall Location: Grid E Description: New Tributary Projected Area(ft) Calculations Height 12 ft P = 16.00 psf Wind Pressure from Wind Analysis page 1 Width 4 ft Fwind =P*A = 768 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= 768 lb Load resisted by shear walls Area, A 48 ftT PLFshear = Fs/Ls 256 plf Shear Wall Load(pounds per length foot) Resisting Shear Wall(ft) Shear Wall Case 11 Shear Wall, Lx 3 ft Case 1:Conventional O.K. PLF< 300 0 Case 2:SW4&HDU2's 300 <PLF< 685 0 Case 3:SW5 orSW8&HDU2's 685 <PLF< 980 0 Case 4:SW5&HDU8's 980 <PLF< 1400 0 Case S: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 Pres. L3 ft accordance with SPDWS Section 4.3. Shear wall aspect ration shall not exceed 3.5:1. Wall, LP 0 If requirements are not met portal framing methods can be used. Summary:Grid E new walls are Case 1,adequately braced by conventional sheathing.No special shearwalls or hold-downs required.The lateral resistance provided by conventional cheating 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. Wall Location: Grid F Description: New Tributary Projected Area(ft) Calculations Height 12 ft P = 16.00 psf Wind Pressure from Wind Analysis page 1 Width 6.5 ft Fwind =P*A = 1248 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= 1248 lb Load resisted by shear walls Area A 78 ft PLFshear =Fs/Ls 271 plf Shear Wall Load(pounds per length foot) Resisting Shear Wall(ft) Shear Wall Case 11 Shear Wall, Lx 2.3 ft Case 1:Conventional O.K. PLF< 300 2.3 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 pres. = E L Wall, LP 0 4.6 ft accordance with SPDWS Section 4.3. Shear wall aspect ration shall not exceed 3.5:1. s x If requirements are not met portal framing methods can be used. Summary:Grid F new walls are case 2,conventional sheathing does not adequately resist lateral loads.SW4 and HDU2's required.