1 / 95

STRUCTURAL ANALYSIS : Determining Structural Capacity

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.

vita
Télécharger la présentation

STRUCTURAL ANALYSIS : Determining Structural Capacity

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. STRUCTURAL ANALYSIS : Determining Structural Capacity

  2. 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.

  3. 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.

  4. TENSILE YIELDING and ALLOWABLE STRESS :

  5. stress Plastic Range FY = yield stress Elastic Range deformation

  6. (fA = P/Area of Section) stress FY fA P1 Force on the spring generates an axial stress and elastic deformation deformation

  7. stress FY When Force is removed, the spring elastically returns to its original shape deformation

  8. stress fA FY A Larger Force may generate an axial stress sufficient to cause plastic deformation deformation P2

  9. stress fA FY When the larger force is removed, the plastic deformation remains (permanent offset) deformation

  10. 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

  11. If using A36 Steel : FY = 36 ksi Allowable Tensile Stress (FT ): FT = 0.60 FY = 0.60 (36 ksi) = 21.6 ksi

  12. 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

  13. 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

  14. 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

  15. 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

  16. FLEXURAL YIELDING and ALLOWABLE BENDING STRESS :

  17. stress Plastic Range FY = yield stress Elastic Range deformation fb = M/S S = Section Modulus

  18. P1 stress FY Force on the BEAM generates an bending stress (tension and compression) and elastic deformation (fb = Mmax/S) fb deformation

  19. stress FY When Force is removed, the BEAM elastically returns to its original shape deformation

  20. P2 stress fb FY A Larger Force may generate an bending stress sufficient to cause plastic deformation deformation

  21. stress fb FY When the larger force is removed, the plastic deformation remains. deformation

  22. 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

  23. If using A36 Steel : FY = 36 ksi Allowable Bending Stress (Fb) : Fb = 0.60 FY = 0.60 (36 ksi) = 21.6 ksi

  24. 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

  25. 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)

  26. 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

  27. 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

  28. 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

  29. 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

  30. BUCKLING and ALLOWABLE COMPRESSION STRESS :

  31. PC Buckling is a compressive phenomenon that depends on : material ‘unbraced length’ of the compression element shape of the section compressive stress.

  32. 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 **

  33. 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

  34. W12x65 A = 19.1 in2 Ryy = 3.02 in

  35. 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

  36. BUCKLING and ALLOWABLE COMPRESSION STRESS :

  37. Allowable Compression Stress depends on slenderness ratio = kl/r

  38. 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

  39. Slenderness Ratio = kl/r l = unbraced length (inches)

  40. Slenderness Ratio = kl/r r = radius of gyration (inches) typical use ry (weak direction) rx > ry

  41. 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

  42. 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

More Related