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PCI 6 th Edition

PCI 6 th Edition. Lateral Component Design. Presentation Outline. Architectural Components Earthquake Loading Shear Wall Systems Distribution of lateral loads Load bearing shear wall analysis Rigid diaphragm analysis. Architectural Components.

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PCI 6 th Edition

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  1. PCI 6th Edition Lateral Component Design

  2. Presentation Outline • Architectural Components • Earthquake Loading • Shear Wall Systems • Distribution of lateral loads • Load bearing shear wall analysis • Rigid diaphragm analysis

  3. Architectural Components • Must resist seismic forces and be attached to the SFRS • Exceptions • Seismic Design Category A • Seismic Design Category B with I=1.0 (other than parapets supported by bearing or shear walls).

  4. Seismic Design Force, Fp Where: ap = component amplification factorfrom Figure 3.10.10

  5. Seismic Design Force, Fp Where: Rp = component response modification factor from Figure 3.10.10

  6. Seismic Design Force, Fp Where: h = average roof height of structure SDS= Design, 5% damped, spectral response acceleration at short periods Wp = component weight z= height in structure at attachment point < h

  7. Cladding Seismic Load Example • Given: • A hospital building in Memphis, TN • Cladding panels are 7 ft tall by 28 ft long. A 6 ft high window is attached to the top of the panel, and an 8 ft high window is attached to the bottom. • Window weight = 10 psf • Site Class C

  8. Problem: Determine the seismic forces on the panel Assumptions Connections only resist load in direction assumed Vertical load resistance at bearing is 71/2” from exterior face of panel Lateral Load (x-direction) resistance is 41/2” from exterior face of the panel Element being consider is at top of building, z/h=1.0 Cladding Seismic Load Example

  9. Solution Steps Step 1 – Determine Component Factors Step 2 – Calculate Design Spectral Response Acceleration Step 3 – Calculate Seismic Force in terms of panel weight Step 4 – Check limits Step 5 – Calculate panel loading Step 6 – Determine connection forces Step 7 – Summarize connection forces

  10. Figure 3.10.10 Step 1 – Determine ap and Rp ap Rp

  11. Step 2 – Calculate the 5%-Damped Design Spectral Response Acceleration Where: SMS=FaSS Ss =1.5 From maps found in IBC 2003 Fa= 1.0 From figure 3.10.7

  12. Step 3 – Calculate Fp in Terms of Wp Wall Element: Body of Connections: Fasteners:

  13. Step 4 – Check Fp Limits Wall Element: Body of Connections: Fasteners:

  14. Step 5 – Panel Loading • Gravity Loading • Seismic Loading Parallel to Panel Face • Seismic or Wind Loading Perpendicular to Panel Face

  15. Step 5 – Panel Loading • Panel Weight • Area = 465.75 in2 • Wp=485(28)=13,580 lb • Seismic Design Force • Fp=0.48(13580)=6518 lb

  16. Step 5 – Panel Loading • Upper Window Weight • Height =6 ft • Wwindow=6(28)(10)=1680 lb • Seismic Design Force • Inward or Outward • Consider ½ of Window • Wp=3.0(10)=30 plf • Fp=0.48(30)=14.4 plf • 14.4(28)=403 lb • Wp=485(28)=13,580 lb • Seismic Design Force • Fp=0.48(13580)=6518 lb

  17. Step 5 – Panel Loading • Lower Window Weight • No weight on panel • Seismic Design Force • Inward or outward • Consider ½ of window • height=8 ft • Wp=4.0(10)=40 plf • Fp=0.48(30)=19.2 plf • 19.2(28)=538 lb

  18. Step 5 Loads to Connections

  19. Step 6Loads to Connections • Equivalent Load Eccentricity z=64,470/15,260=4.2 in • Dead Load to Connections • Vertical =15,260/2=7630 lb • Horizontal = 7630 (7.5-4.2)/32.5 =774.7/2=387 lb

  20. Step 6 – Loads to Connections

  21. Step 6 – Loads to Connections

  22. Step 6 – Loads to Connections • Center of equivalent seismic load from lower left • y=258,723/7459 y=34.7 in • z=41,973/7459 • z=5.6 in

  23. Step 6 – Seismic In-Out Loads • Equivalent Seismic Load • y=34.7 in • Fp=7459 lb • Moments about Rb • Rt=7459(34.7 -27.5)/32.5 • Rt=1652 lb • Force equilibrium • Rb=7459-1652 • Rb=5807 lb

  24. Step 6 – Wind Outward Loads

  25. Step 6 – Wind Outward Loads Fp • Center of equivalent wind load from lower left • y=267,540/6860 • y=39.0 in • Outward Wind Load • Fp=6,860 lb

  26. Step 6 – Wind Outward Loads • Moments about Rb • Rt=7459(39.0 -27.5)/32.5 • Rt=2427 lb • Force equilibrium • Rb=6860-2427 • Rb=4433 lb

  27. Step 6 – Wind Inward Loads • Outward Wind Reactions • Rt=2427 lb • Rb=4433 lb • Inward Wind Loads • Proportional to pressure • Rt=(11.3/12.9)2427 lb • Rt=2126 lb • Rb=(11.3/12.9)4433 lb • Rb=3883 lb

  28. Step 6 – Seismic Loads Normal to Surface • Load distribution (Based on Continuous Beam Model) • Center connections = .58 (Load) • End connections = 0.21 (Load)

  29. Step 6 – Seismic Loads Parallel to Face • Parallel load • =+ 7459 lb

  30. Step 6 – Seismic Loads Parallel to Face • Up-down load

  31. Step 6 – Seismic Loads Parallel to Face • In-out load

  32. Step 7 – Summary of Factored Loads • Load Factor of 1.2 Applied • Load Factor of 1.0 Applied • Load Factor of 1.6 Applied

  33. Distribution of Lateral Loads Shear Wall Systems • For Rigid diaphragms • Lateral Load Distributed based on total rigidity, r Where: r=1/D D=sum of flexural and shear deflections

  34. Distribution of Lateral Loads Shear Wall Systems • Neglect Flexural Stiffness Provided: • Rectangular walls • Consistent materials • Height to length ratio < 0.3 • Distribution based on • Cross-Sectional Area

  35. Distribution of Lateral Loads Shear Wall Systems • Neglect Shear Stiffness Provided: • Rectangular walls • Consistent materials • Height to length ratio > 3.0 • Distribution based on • Moment of Inertia

  36. Distribution of Lateral Loads Shear Wall Systems • Symmetrical Shear Walls Where: Fi = Force Resisted by individual shear wall ki=rigidity of wall i Sr=sum of all wall rigidities Vx=total lateral load

  37. Distribution of Lateral Loads “Polar Moment of Stiffness Method” • Unsymmetrical Shear Walls Force in the y-direction is distributed to a given wall at a given level due to an applied force in the y-direction at that level

  38. Distribution of Lateral Loads “Polar Moment of Stiffness Method” • Unsymmetrical Shear Walls Where: Vy = lateral force at level being considered Kx,Ky = rigidity in x and y directions of wall SKx, SKy = summation of rigidities of all walls T = Torsional Moment x = wall x-distance from the center of stiffness y = wall y-distance from the center of stiffness

  39. Distribution of Lateral Loads “Polar Moment of Stiffness Method” • Unsymmetrical Shear Walls Force in the x-direction is distributed to a given wall at a given level due to an applied force in the y-direction at that level.

  40. Distribution of Lateral Loads “Polar Moment of Stiffness Method” • Unsymmetrical Shear Walls Where: Vy=lateral force at level being considered Kx,Ky=rigidity in x and y directions of wall SKx, SKy=summation of rigidities of all walls T=Torsional Moment x=wall x-distance from the center of stiffness y=wall y-distance from the center of stiffness

  41. Unsymmetrical Shear Wall Example Given: • Walls are 8 ft high and 8 in thick

  42. Unsymmetrical Shear Wall Example Problem: • Determine the shear in each wall due to the wind load, w • Assumptions: • Floors and roofs are rigid diaphragms • Walls D and E are not connected to Wall B • Solution Method: • Neglect flexural stiffness h/L < 0.3 • Distribute load in proportion to wall length

  43. Solution Steps Step 1 – Determine lateral diaphragm torsion Step 2 – Determine shear wall stiffness Step 3 – Determine wall forces

  44. Step 1 – Determine Lateral Diaphragm Torsion • Total Lateral Load Vx=0.20 x 200 = 40 kips

  45. Step 1 – Determine Lateral Diaphragm Torsion • Center of Rigidity from left

  46. Step 1 – Determine Lateral Diaphragm Torsion • Center of Rigidity y=center of building

  47. Step 1 – Determine Lateral Diaphragm Torsion • Center of Lateral Load from left xload=200/2=100 ft • Torsional Moment MT=40(130.9-100)=1236 kip-ft

  48. Step 2 – Determine Shear Wall Stiffness • Polar Moment of Stiffness

  49. Step 3 – Determine Wall Forces • Shear in North-South Walls

  50. Step 3 – Determine Wall Forces • Shear in North-South Walls

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