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SEISMIC FORCES

SEISMIC FORCES. seismic load. Seismic Load is generated by the inertia of the mass of the structure : V BASE Redistributed (based on relative height and weight) to each level as a ‘Point Load’ at the center of mass of the structure or element in question : F X V BASE w x h x

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SEISMIC FORCES

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  1. SEISMIC FORCES

  2. seismic load

  3. Seismic Load is generated by the inertia of the mass of the structure : VBASE Redistributed (based on relative height and weight) to each level as a ‘Point Load’ at the center of mass of the structure or element in question : FX VBASE wx hx S(w h) VBASE = (Cs)(W) ( VBASE ) Fx =

  4. Total Seismic Loading : VBASE = 0.3 W W = wroof + wsecond

  5. wroof

  6. wsecond flr

  7. W = wroof + wsecond flr

  8. VBASE

  9. Redistribute Total Seismic Load to each level based on relative height and weight Froof Fsecond flr VBASE (wx) (hx) S (w h) Fx =

  10. Load Flow to Lateral Resisting System : Distribution based on Relative Rigidity Assume Relative Rigidity : Single Bay MF : Rel Rigidity = 1 2 - Bay MF : Rel Rigidity = 2 3 - Bay MF : Rel Rigidity = 3

  11. Distribution based on Relative Rigidity : SR = 1+1+1+1 = 4 Px = ( Rx / SR ) (Ptotal) PMF1 = 1/4 Ptotal

  12. EXTERNAL FORCES

  13. Where are we going with all of this ???? global stability & load flow (Proj. 1) tension, compression, continuity equilibrium, forces on rigid bodies boundary conditions: fixed, pin, roller external forces applied to beams & columns categories of external loading: DL, LL, W, E, S, H (fluid pressure) internal forces in members: axial, shear, bending/flexure stresses: tension, compression, shear, bending stress stability, slenderness, and allowable compression stress member sizing for flexure member sizing for combined flexure and axial stress (Proj. 2) Trusses (Proj. 3)

  14. FORCES + REACTIONS ACTING UPON BODIES

  15. 200 lb ( + ) SM1 = 0 0= -200 lb(10 ft)+ RY2(15 ft) RY2(15 ft) = 2000 lb-ft RY2 = 133 lb ( +) SFY = 0 RY1 + RY2 - 200 lb = 0 RY1 + 133 lb- 200 lb = 0 RY1 = 67 lb ( +) SFX = 0 RX1 = 0 RX1 10 ft 5 ft RY2 RY1 200 lb 0 lb 10 ft 5 ft 67 lb 133 lb

  16. w = 880lb/ft RX1 RY1 RY2 24 ft

  17. resultant force - equivalent total load that is a result of a distributed line load resultant force = area loading diagram resultant force = w( L) RX1 RY1 RY2 24 ft

  18. resultant force - equivalent total load that is a result of a distributed line load resultant force = area loading diagram resultant force = w( L) = 880 lb/ft(24ft) = 21,120 lb RX1 12 ft 12 ft RY1 RY2 24 ft

  19. resultant force - equivalent total load that is a result of a distributed line load resultant force = area loading diagram resultant force = w( L) = 880 lb/ft(24ft) = 21,120 lb RX1 12 ft 12 ft RY1 RY2 24 ft ( + ) SM1 = 0

  20. resultant force - equivalent total load that is a result of a distributed line load resultant force = area loading diagram resultant force = w( L) = 880 lb/ft(24ft) = 21,120 lb RX1 12 ft 12 ft RY1 RY2 24 ft ( + ) SM1 = 0 –21,120 lb(12 ft)+ RY2(24 ft) = 0

  21. resultant force - equivalent total load that is a result of a distributed line load resultant force = area loading diagram resultant force = w( L) = 880 lb/ft(24ft) = 21,120 lb RX1 12 ft 12 ft RY1 RY2 24 ft ( + ) SM1 = 0 –21,120 lb(12 ft)+ RY2(24 ft) = 0 RY2(24 ft) = 253,440 lb-ft

  22. resultant force - equivalent total load that is a result of a distributed line load resultant force = area loading diagram resultant force = w( L) = 880 lb/ft(24ft) = 21,120 lb RX1 12 ft 12 ft RY1 RY2 24 ft ( + ) SM1 = 0 –21,120 lb(12 ft)+ RY2(24 ft) = 0 RY2(24 ft) = 253,440 lb-ft RY2 = 10,560 lb

  23. resultant force - equivalent total load that is a result of a distributed line load resultant force = area loading diagram resultant force = w( L) = 880 lb/ft(24ft) = 21,120 lb RX1 12 ft 12 ft RY1 RY2 24 ft ( + ) SM1 = 0 –21,120 lb(12 ft)+ RY2(24 ft) = 0 RY2(24 ft) = 253,440 lb-ft RY2 = 10,560 lb ( +) SFY = 0

  24. resultant force - equivalent total load that is a result of a distributed line load resultant force = area loading diagram resultant force = w( L) = 880 lb/ft(24ft) = 21,120 lb RX1 12 ft 12 ft RY1 RY2 24 ft ( + ) SM1 = 0 –21,120 lb(12 ft)+ RY2(24 ft) = 0 RY2(24 ft) = 253,440 lb-ft RY2 = 10,560 lb ( +) SFY = 0 RY1 + RY2 – 21,120 lb = 0

  25. resultant force - equivalent total load that is a result of a distributed line load resultant force = area loading diagram resultant force = w( L) = 880 lb/ft(24ft) = 21,120 lb RX1 12 ft 12 ft RY1 RY2 24 ft ( + ) SM1 = 0 –21,120 lb(12 ft)+ RY2(24 ft) = 0 RY2(24 ft) = 253,440 lb-ft RY2 = 10,560 lb ( +) SFY = 0 RY1 + RY2 – 21,120 lb = 0 RY1 + 10,560 lb– 21,120 lb = 0

  26. resultant force - equivalent total load that is a result of a distributed line load resultant force = area loading diagram resultant force = w( L) = 880 lb/ft(24ft) = 21,120 lb RX1 12 ft 12 ft RY1 RY2 24 ft ( + ) SM1 = 0 –21,120 lb(12 ft)+ RY2(24 ft) = 0 RY2(24 ft) = 253,440 lb-ft RY2 = 10,560 lb ( +) SFY = 0 RY1 + RY2 – 21,120 lb = 0 RY1 + 10,560 lb– 21,120 lb = 0 RY1 = 10,560 lb

  27. resultant force - equivalent total load that is a result of a distributed line load resultant force = area loading diagram resultant force = w( L) = 880 lb/ft(24ft) = 21,120 lb RX1 12 ft 12 ft RY1 RY2 24 ft ( + ) SM1 = 0 –21,120 lb(12 ft)+ RY2(24 ft) = 0 RY2(24 ft) = 253,440 lb-ft RY2 = 10,560 lb ( +) SFY = 0 RY1 + RY2 – 21,120 lb = 0 RY1 + 10,560 lb– 21,120 lb = 0 RY1 = 10,560 lb ( +) SFX = 0 RX1 = 0

  28. resultant force - equivalent total load that is a result of a distributed line load resultant force = area loading diagram resultant force = w( L) = 880 lb/ft(24ft) = 21,120 lb RX1 12 ft 12 ft RY1 RY2 24 ft ( + ) SM1 = 0 –21,120 lb(12 ft)+ RY2(24 ft) = 0 RY2(24 ft) = 253,440 lb-ft RY2 = 10,560 lb ( +) SFY = 0 RY1 + RY2 – 21,120 lb = 0 RY1 + 10,560 lb– 21,120 lb = 0 RY1 = 10,560 lb ( +) SFX = 0 RX1 = 0 w = 880 lb/ft 0 lb 24 ft 10,560 lb 10,560 lb

  29. SIGN CONVENTIONS (always confusing and very frustrating) External – for reactions & solving (Applied Loading & Support Reactions) + X pos. to right - X to left neg. + Y pos. up - Y down neg + Rotation pos. counter-clockwise - CW rot. neg. Internal – for PVM diagrams (Axial, Shear, and Moment inside members) Axial Tension (elongation) pos. | Axial Compression (shortening) neg. Shear Force (spin clockwise) pos. | Shear Force (spin CCW) neg. Bending Moment (smiling) pos. | Bending Moment (frowning) neg.

  30. STRUCTURAL ANALYSIS : PVM

  31. INTERNAL FORCES Axial (P) Shear (V) Moment (M)

  32. + + + P M V

  33. - - - P M V

  34. RULES FOR CREATING P DIAGRAMS 1. concentrated axial load | reaction = jump in the axial diagram 2. value of distributed axial loading = slope of axial diagram 3. sum of distributed axial loading = change in axial diagram

  35. - 0 + -10k -10k -10k -10k -20k -10k +20k +20k • 20k 0 • compression

  36. RULES FOR CREATING V M DIAGRAMS (3/6) 1. a concentrated load | reaction = a jump in the shear diagram 2. the value of loading diagram = the slope of shear diagram 3. the area of loading diagram = the change in shear diagram

  37. w = - 880 lb/ft 0 lb 24 ft 10,560 lb 10,560 lb P 0 0 Area of Loading Diagram -0.88k/ft * 24ft = -21.12k 10.56k + -21.12k = -10.56k +10.56k -880 plf = slope +10.56 k 0 V 0 +10.56 k -10.56k

  38. RULES FOR CREATING V M DIAGRAMS, Cont. (6/6) 4. a concentrated moment = a jump in the moment diagram 5. the value of shear diagram = the slope of moment diagram 6. the area of shear diagram = the change in moment diagram

  39. w = - 880 lb/ft 0 lb 24 ft 10,560 lb 10,560 lb P 0 0 0 0 Area of Loading Diagram -0.88k/ft * 24ft = -21.12k 10.56k + -21.12k = -10.56k +10.56k -880 plf = slope +10.56 k 0 V 0 +10.56 k zero slope -10.56k 63.36k’ -63.36 k-ft Slope initial = +10.56k +63.36 k-ft neg. slope pos. slope Area of Shear Diagram (10.56k )(12ft ) 0.5 = 63.36 k-ft M (-10.56k)(12ft)(0.5) = -63.36 k-ft

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