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PCI 6th Edition

Presentation Outline. Architectural ComponentsEarthquake LoadingShear Wall SystemsDistribution of lateral loadsLoad bearing shear wall analysisRigid diaphragm analysis. Architectural Components. Must resist seismic forces and be attached to the SFRSExceptionsSeismic Design Category ASeismic Design Category B with I=1.0 (other than parapets supported by bearing or shear walls)..

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PCI 6th 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

    5. Seismic Design Force, Fp

    6. Seismic Design Force, Fp

    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. Cladding Seismic Load Example 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

    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. Step 1 – Determine ap and Rp Figure 3.10.10 Both the wall element and the body of the connection require the same modification factors. Fasteners have modifications factors that create a larger seismic design force. Both the wall element and the body of the connection require the same modification factors. Fasteners have modifications factors that create a larger seismic design force.

    11. Step 2 – Calculate the 5%-Damped Design Spectral Response Acceleration

    12. Step 3 – Calculate Fp in Terms of Wp Space bar enters individual forces Space bar enters individual forces

    13. Step 4 – Check Fp Limits Space bar enters individual forces and check mark Space bar enters individual forces and check mark

    14. Step 5 – Panel Loading

    15. Step 5 – Panel Loading

    16. Step 5 – Panel Loading

    17. Step 5 – Panel Loading

    18. Step 5 Loads to Connections Space bar enters force arrows Space bar enters force arrows

    19. Step 6 Loads to Connections Space bar enters text and enters and exits appropriate force arrows. Space bar enters text and enters and exits appropriate force arrows.

    20. Step 6 – Loads to Connections Space bar enters and exists force arrows Space bar enters and exists force arrows

    21. Step 6 – Loads to Connections

    22. Step 6 – Loads to Connections

    23. Step 6 – Seismic In-Out Loads

    24. Step 6 – Wind Outward Loads

    25. Step 6 – Wind Outward Loads Space exits force components and enters equivalent load Space exits force components and enters equivalent load

    26. Step 6 – Wind Outward Loads

    27. Step 6 – Wind Inward Loads

    28. Step 6 – Seismic Loads Normal to Surface

    29. Step 6 – Seismic Loads Parallel to Face

    30. Step 6 – Seismic Loads Parallel to Face

    31. Step 6 – Seismic Loads Parallel to Face

    32. Step 7 – Summary of Factored Loads Space bar zooms table Space bar zooms table

    33. Distribution of Lateral Loads Shear Wall Systems For Rigid diaphragms Lateral Load Distributed based on total rigidity, r

    34. Distribution of Lateral Loads Shear Wall Systems

    35. Distribution of Lateral Loads Shear Wall Systems

    36. Distribution of Lateral Loads Shear Wall Systems Symmetrical Shear Walls

    37. Distribution of Lateral Loads “Polar Moment of Stiffness Method” Unsymmetrical Shear Walls

    38. Unsymmetrical Shear Walls Distribution of Lateral Loads “Polar Moment of Stiffness Method”

    39. Unsymmetrical Shear Walls Distribution of Lateral Loads “Polar Moment of Stiffness Method”

    40. Unsymmetrical Shear Walls Distribution of Lateral Loads “Polar Moment of Stiffness Method”

    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 Walls D and E are placed symmetrically about the center of the building in the north-south directionWalls D and E are placed symmetrically about the center of the building in the north-south direction

    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

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

    52. Step 3 – Determine Wall Forces Shear in East-West Walls

    53. Load Bearing Shear Wall Example Given:

    54. Load Bearing Shear Wall Example Given Continued: Three level parking structure Seismic Design Controls Symmetrically placed shear walls Corner Stairwells are not part of the SFRS

    55. Load Bearing Shear Wall Example Problem: Determine the tension steel requirements for the load bearing shear walls in the north-south direction required to resist seismic loading

    56. Load Bearing Shear Wall Example Solution Method: Accidental torsion must be included in the analysis The torsion is assumed to be resisted by the walls perpendicular to the direction of the applied lateral force

    57. Solution Steps Step 1 – Calculate force on wall Step 2 – Calculate overturning moment Step 3 – Calculate dead load Step 4 – Calculate net tension force Step 5 – Calculate steel requirements

    58. Step 1 – Calculate Force in Shear Wall Accidental Eccentricity=0.05(264)=13.2 ft Force in two walls

    59. Step 1 – Calculate Force in Shear Wall Force at each level Level 3 F1W=0.500(270)=135 kips Level 2 F1W=0.333(270)= 90 kips Level 1 F1W=0.167(270)= 45 kips

    60. Step 2 – Calculate Overturning Moment Force at each level Level 3 F1W=0.500(270)=135 kips Level 2 F1W=0.333(270)= 90 kips Level 1 F1W=0.167(270)= 45 kips Overturning moment, MOT MOT=135(31.5)+90(21)+45(10.5) MOT=6615 kip-ft

    61. Step 3 – Calculate Dead Load Load on each Wall Dead Load = .110 ksf (all components) Supported Area = (60)(21)=1260 ft2 Wwall=1260(.110)=138.6 kips Total Load Wtotal=3(138.6)=415.8~416 kips

    62. Step 4 – Calculate Tension Force Governing load Combination U=[0.9-0.2(0.24)]D+1.0E Eq. 3.2.6.7a U=0.85D+1.0E Tension Force

    63. Step 5 – Reinforcement Requirements Tension Steel, As Reinforcement Details Use 4 - #8 bars = 3.17 in2 Locate 2 ft from each end

    64. Rigid Diaphragm Analysis Example Given:

    65. Rigid Diaphragm Analysis Example Given Continued: Three level parking structure (ramp at middle bay) Seismic Design Controls Seismic Design Category C Corner Stairwells are not part of the SFRS

    66. Rigid Diaphragm Analysis Example Problem: Part A Determine diaphragm reinforcement required for moment design Part B Determine the diaphragm reinforcement required for shear design

    67. Solution Steps Step 1 – Determine diaphragm force Step 2 – Determine force distribution Step 3 – Determine statics model Step 4 – Determine design forces Step 5 – Diaphragm moment design Step 6 – Diaphragm shear design

    68. Step 1 – Diaphragm Force, Fp Fp, Eq. 3.8.3.1 Fp = 0.2·IE·SDS·Wp + Vpx but not less than any force in the lateral force distribution table

    69. Step 1 – Diaphragm Force, Fp Fp, Eq. 3.8.3.1 Fp =(1.0)(0.24)(5227)+0.0=251 kips Fp=471 kips

    70. Step 2 – Diaphragm Force, Fp, Distribution Assume the forces are uniformly distributed Total Uniform Load, w Distribute the force equally to the three bays

    71. Step 3 – Diaphragm Model Ramp Model

    72. Step 3 – Diaphragm Model Flat Area Model

    73. Step 3 – Diaphragm Model Flat Area Model Half of the load of the center bay is assumed to be taken by each of the north and south bays w2=0.59+0.59/2=0.89 kip/ft Stress reduction due to cantilevers is neglected. Positive Moment design is based on ramp moment

    74. Step 4 – Design Forces Ultimate Positive Moment, +Mu Ultimate Negative Moment Ultimate Shear

    75. Step 5 – Diaphragm Moment Design Assuming a 58 ft moment arm Tu=2390/58=41 kips Required Reinforcement, As Tensile force may be resisted by: Field placed reinforcing bars Welding erection material to embedded plates

    76. Step 6 – Diaphragm Shear Design Force to be transferred to each wall Each wall is connected to the diaphragm, 10 ft Shear/ft=Vwall/10=66.625/10=6.625 klf Providing connections at 5 ft centers Vconnection=6.625(5)=33.125 kips/connection Error on page 3-53 – Diaphragm Shear Design section the 24 should be 27kips from the previous page (3-52) Error on page 3-53 – Diaphragm Shear Design section the 24 should be 27kips from the previous page (3-52)

    77. Step 6 – Diaphragm Shear Design Force to be transferred between Tees For the first interior Tee Vtransfer=Vu-(10)0.59=47.1 kips Shear/ft=Vtransfer/60=47.1/60=0.79 klf Providing Connections at 5 ft centers Vconnection=0.79(5)=4 kips

    78. Questions?

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