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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
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