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# STRUCTURAL ANALYSIS : Determining Structural Capacity

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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. 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

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

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

40. 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

41. 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