SPECIFICATION FOR THE DESIGN OF - Transcon Steel

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46 Commentary on the Prescriptive Method for One and Two Family Dwellings - 2004 E. WALL FRAMING <strong>DESIGN</strong> EXAMPLES E1 Wall Stud Design Calculate the minimum size and thickness for an 8-foot wall stud located at the upper story of a 28-foot wide two-story house that is subjected to an 90 mph, Exposure Category B winds. The studs are spaced at 24” on center. The maximum ground snow load is 30 psf. The wall studs are laterally restrained at mid-height. All referenced equations and sections are to the Specification (AISI, 1999) unless noted. E1.1 Design Assumptions 28’ Building Width 2’ Roof Overhang 2’ Stud Spacing 33 ksi <strong>Steel</strong> Yield Strength 4:12 Roof Pitch 30 psf Ground Snow Load 0.7 Snow Reduction Factor 8’ 16 psf Minimum Roof Live Load 7 psf Roof Dead Load 5 psf Ceiling Dead Load 7 psf Soffit Dead Load 45’ Building Length 8’ Wall Height L/240 Live Load + Dead Load Deflection Limit 28’ 12 4 2’ E1.2 Design Loads Dead Load: Ceiling Dead Load = 5(28)/2 Roof Dead Load = 7(28)/2 Soffit Dead Load = 2 x 7 Total Dead Load (D) = 70 plf = 98 plf = 14 plf = 182 plf Live Loads: Roof Live Load (L r ) = 16(2 + 28/2) = 256 plf Roof Snow Load (S) = (0.7x30)(2 + 28/2) = 336 plf controls Wind Loads: Wind loads were calculated in accordance with ASCE 7 (ASCE, 1998) equations using MWFRS and components and cladding (refer to Table C2.4). MWFRS: ρ = q h [(GC pf - GC pi ) lb/ft 2 q h = 0.00256K z K zt K d V 2 I lb/ft 2 where K z = 0.70 (Table 6-5 at 30’ height above ground, Case 2, Exposure Category B)
Commentary on the Prescriptive Method for One and Two Family Dwellings - 2004 47 K zt = 1.0 (No topographic multiplier is considered) K d = 0.85 (Table 6-6) GC pf = 0.53 (Figure 6-4, 4:12 roof slope, Cases A, B, Building Surfaces 1, 4, 5 & 6) GC pi = ±0.18 (Table 6-7, enclosed buildings) V = 90 mph (3-second gust wind speed) I = 1 (Importance factor for residential buildings) q h = 0.00256(0.7)(1.0)(0.85)(90) 2 (1.0) = 12.34 lb/ft 2 ρ = 12.34(0.53 + 0.18) = 8.76 lb/ft 2 ASCE 7 Section 6.1.4.1 requires all enclosed building to be designed to a wind pressure not less than 10 psf. Therefore, ρ = 10 psf (for interior zone walls) Components and Cladding: Using Figure 6-5A of ASCE 7-98, the C&C wind pressure for this design example is calculated to be 13.9 psf and –15.1 psf. The higher pressure (15.1 psf) is selected to design the stud. E1.3 Load Combinations 1. 1.4D 2. 1.2D + 1.6L + 0.5(L r or S) 3. 1.2D + 0.5L + 1.6(L r or S) 4. 1.2D + 0.8W + 1.6(L r or S) (using MWFRS for W) 5. 1.2D + 1.6W + 0.5L + 0.5(L r or S) (using MWFRS for W) 6. 1.6W (C&C wind coefficients used to check bending) 7. 0.7W (C&C wind coefficients used to check deflections) The load combinations listed below will be checked: 1. 1.4D = 255 plf, No lateral load 2. 1.2D + 1.6L+ 0.5(L r or S) = 386 plf, No lateral load 3. 1.2D + 0.5L +1.6(L r or S) = 756 plf, No lateral load 4. 1.2D + 1.6(L r or S) = 756 plf, 0.8W = 0.8(10 x 2) = 16 plf 5. 1.2D + 0.5(L r or S) = 386 plf, 1.6W = 1.6(10 x 2) = 32 plf 6. 1.6W = 1.6(15.1x 2) = 48.3 plf Therefore, the controlling load combinations to be checked are: 3. 756 x 2 = 1,512 lb axial load and 0 lateral load 4. 756 x 2 = 1,512 lb axial load and 16 plf lateral load 5. 386 x 2 = 772 lb axial load and 32 plf lateral load 6. 0 axial load and 48.3 plf lateral load E1.4 Member Properties Select a 350S162-33 member and check its adequacy. The calculated capacities for this member in accordance with the Specification are: Nominal flexural strength*Φ = 7,974 in-lb
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SPECIFICATION FOR THE DESIGN OF - Transcon Steel

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