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STRUCTURAL ANALYSIS : Determining Structural Capacity. From Structural Analysis we have developed an understanding of all : Actions - Applied forces such as dead load, live load, wind load, seismic load. Reactions - Forces generated at the boundary conditions that maintain equilibrium.

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## STRUCTURAL ANALYSIS : Determining Structural Capacity

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**STRUCTURAL ANALYSIS :**Determining Structural Capacity**From Structural Analysis we have developed an understanding**of all : Actions - Applied forces such as dead load, live load, wind load, seismic load. Reactions - Forces generated at the boundary conditions that maintain equilibrium. Internal forces- Axial, shear and moment (P V M) in each structural element.**Determination of StructuralCapacity is based on each**element’s ability to perform under the applied actions, consequent reactions and internal forces without : Yielding - material deforming plastically (tension and/or stocky compression). Buckling- phenomenon of compression when a slender element loses stability. Deflecting Excessively- elastic defection that may cause damage to attached materials/finishes – bouncy floors.**stress**Plastic Range FY = yield stress Elastic Range deformation**(fA = P/Area of Section)**stress FY fA P1 Force on the spring generates an axial stress and elastic deformation deformation**stress**FY When Force is removed, the spring elastically returns to its original shape deformation**stress**fA FY A Larger Force may generate an axial stress sufficient to cause plastic deformation deformation P2**stress**fA FY When the larger force is removed, the plastic deformation remains (permanent offset) deformation**To be certain that the tension stress never reaches the**yield stress, Set an ALLOWABLE TENSILE STRESS : FTension = 0.60 FY stress FY fT deformation**If using A36 Steel : FY = 36 ksi**Allowable Tensile Stress (FT ): FT = 0.60 FY = 0.60 (36 ksi) = 21.6 ksi**fA stress**If using A36 Steel : FY = 36 ksi Allowable Tensile Stress : FT = 0.60 FY = 0.60 (36 ksi) = 21.6 ksi P = 5,000 lb or 5 kips fA = P/Area (actual axial stress = P/A) Aarea P force**FT stress**If using A36 Steel : FY = 36 ksi Allowable Tensile Stress : FT = 0.60 FY = 0.60 (36 ksi) = 21.6 ksi P = 5,000 lb or 5 kips fA = P/Area FT = Pmax /AreaRequired Areq Pmax**FT stress**If using A36 Steel : FY = 36 ksi Allowable Tensile Stress : FT = 0.60 FY = 0.60 (36 ksi) = 21.6 ksi P = 5,000 lb or 5 kips fA = P/Area FT = Pmax /AreaRequired AreaRequired = Pmax/FT Areq Pmax**21.6 ksi**If using A36 Steel : FY = 36 ksi Allowable Tensile Stress : FT = 0.60 FY = 0.60 (36 ksi) = 21.6 ksi P = 5,000 lb or 5 kips fA = P/Area FT = Pmax /AreaRequired AreaRequired = Pmax/FT = 5k / 21.6 ksi = .25 in2 Areq 5k**stress**Plastic Range FY = yield stress Elastic Range deformation fb = M/S S = Section Modulus**P1**stress FY Force on the BEAM generates an bending stress (tension and compression) and elastic deformation (fb = Mmax/S) fb deformation**stress**FY When Force is removed, the BEAM elastically returns to its original shape deformation**P2**stress fb FY A Larger Force may generate an bending stress sufficient to cause plastic deformation deformation**stress**fb FY When the larger force is removed, the plastic deformation remains. deformation**To be certain that the bending stress never reaches the**yield stress, Set an ALLOWABLE BENDING STRESS : Fbending = 0.60 FY stress FY Fb deformation**If using A36 Steel : FY = 36 ksi**Allowable Bending Stress (Fb) : Fb = 0.60 FY = 0.60 (36 ksi) = 21.6 ksi**If using A36 Steel : FY = 36 ksi**Allowable Bending Stress : Fb = 0.60 FY = 0.60 (36 ksi) = 21.6 ksi Mmax = 316 k-ft Mmax = 316 k-ft (12 in / ft) = 3792 k-in**If using A36 Steel : FY = 36 ksi**Allowable Bending Stress : Fb = 0.60 FY = 0.60 (36 ksi) = 21.6 ksi Mmax = 316 k-ft Mmax = 316 k-ft (12 in / ft) = 3792 k-in fb = M/S (actual bending stress = M/S)**If using A36 Steel : FY = 36 ksi**Allowable Bending Stress : Fb = 0.60 FY = 0.60 (36 ksi) = 21.6 ksi Mmax = 316 k-ft Mmax = 316 k-ft (12 in / ft) = 3792 k-in fb = M/S Fb = Mmax / SRequired**If using A36 Steel : FY = 36 ksi**Allowable Bending Stress : Fb = 0.60 FY = 0.60 (36 ksi) = 21.6 ksi Mmax = 316 k-ft Mmax = 316 k-ft (12 in / ft) = 3792 k-in fb = M/S Fb = Mmax / SRequired SRequired = Mmax / Fb**If using A36 Steel : FY = 36 ksi**Allowable Bending Stress : Fb = 0.60 FY = 0.60 (36 ksi) = 21.6 ksi Mmax = 316 k-ft Mmax = 316 k-ft (12 in / ft) = 3792 k-in fb = M/S Fb = Mmax / SRequired SRequired = Mmax / Fb = 3792 k-in / 21.6 ksi = 176 in3**If using A36 Steel : FY = 36 ksi**Mmax = 316 k-ft Mmax = 316 k-ft (12 in / ft) = 3792 k-in Allowable Bending Stress : Fb = 0.60 FY = 0.60 (36 ksi) = 21.6 ksi fb = M/S Fb = Mmax / SRequired SRequired = Mmax/Fb = 3792 k-in / 21.6 ksi = 176 in3 Use W24x76 : SX-X = 176in3**PC**Buckling is a compressive phenomenon that depends on : material ‘unbraced length’ of the compression element shape of the section compressive stress.**Allowable Compression Stress depends on ‘kl/r’**k = 1.0 l = 15 ft = 180 in assume r = 3.0 in.** kl/r = 60 Fc = 17.4 ksi ** we must always come back and verify this assumption ****If using A36 Steel : FY = 36 ksi**Pmax = 240 kips (typ. read this from your P diagram] Allowable Compression Stress (Fc) : FC = 17.4 ksi fC = P/Area FC = Pmax/AreaRequired AreaRequired = Pmax/FC = 240k / 17.4 ksi = 13.8 in2**W12x65 A = 19.1 in2**Ryy = 3.02 in**If using A36 Steel : FY = 36 ksi**Pmax = 240 kips Allowable Compression Stress : FC = 17.4 ksi fC = P/Area FC = Pmax/AreaRequired AreaRequired = Pmax/FC = 240k / 17.4 ksi = 13.8 in2 Use W12x65 Area = 19.1 in2 check actual stress: fC = P/A fC = 240 kips / 19.1 in2 = 12.6 ksi**Allowable Compression Stress depends on slenderness ratio =**kl/r**Slenderness Ratio = kl/r**k = coefficient which accounts for buckling shape for our project gravity columns, k=1.0 for moment frames see deformed shape**Slenderness Ratio = kl/r**l = unbraced length (inches)**Slenderness Ratio = kl/r**r = radius of gyration (inches) typical use ry (weak direction) rx > ry**Allowable Compression Stress (Fc)**slenderness ratio = kl/r assume r = 2 in., k = 1.0 lcolumn = 180 in kl/r = 90 use Table C-36 to determine Fc = 14.2 ksi**Pmax Column 2 = 238 kips**(assume columns are continuous from foundation to roof, total length 30 feet) FC = 14.2 ksi AreaRequired = Pmax/FC = 238k / 14.2 ksi = 16.8 in2

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