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Outline

Outline. Introduce PCI/NSF/CPF DSDM Research Effort Review Key Behaviors of Precast Diaphragms and Design Philosophy Adopted Summarize DSDM Research Project Findings Present Precast Diaphragm Design Procedure Cover Precast Diaphragm Design Example Discuss Codification Efforts.

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Outline

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  1. Outline NEES/EERI Webinar April 23 2012 Introduce PCI/NSF/CPF DSDM Research Effort Review Key Behaviors of Precast Diaphragms and Design Philosophy Adopted Summarize DSDM Research Project Findings Present Precast Diaphragm Design Procedure Cover Precast Diaphragm Design Example Discuss Codification Efforts

  2. Design Methodology Documents NEES/EERI Webinar April 23 2012

  3. Design Methodology PART 4 The procedure will be demonstrated for the Elastic Design Option, but will be compared to the design forces and details of the other options NEES/EERI Webinar April 23 2012

  4. Diaphragm Design Example Example 1: 4-story Parking Garage - Knoxville (SDC C) NEES/EERI Webinar April 23 2012

  5. Diaphragm Design Example Step 1: Baseline design force Step 1: Determine the diaphragm seismic baseline design forces as per ASCE 7-05 • Seismic design parameters • Design site: Knoxville, TN • SDC C • Ss0.58 • S10.147 • Soil site class C • Fa 1.17 • Fv 1.65 • Sms= Fa Ss 0.68 • Sm1= Fv S1 0.24 • SDS= 2/3 Sms 0.45 • SD1= 2/3 Sm1 0.16 N-S: Intermediate Precast Shear WallsR=5, W0=2.5, Cd=4.5 E-W: Intermediate precast bearing wallR=4, W0=2.5, Cd=4 NEES/EERI Webinar April 23 2012

  6. Diaphragm Design Example Step 1: Baseline design force (con’t) • Diaphragm maximum design acceleration: Cdia, max=max (Fx/wx) [Eqn.1] • Diaphragm baseline design force FDx = axCdia, max wx [Eqn.2] See Tables next page NEES/EERI Webinar April 23 2012

  7. Baseline (unamplified) forces Parking structure: ax =1.0 at top floor and ax =0.68 at lower floors. Diaphragm Design Example Step 1: Baseline design force (con’t) N-S direction hx (ft) Wx (k) Wx hxk Cvx Fx (k) Cdia, max (1)ax FDx (k) (2) Roof 47.5 5529* 300926 0.35 482 0.087 1.0 482 4th 37 6245 262419 0.31 420 0.087 0.68 370 3rd 26.5 6245 185750 0.22 297 0.087 0.68 370 2nd 16 6245 110173 0.13 176 0.087 0.68 370 Sum 0 24262 859267 1 1375 E-W direction hx (ft) Wx (k) Wx hxk Cvx Fx (k) Cdia, max (1)ax FDx (k) (2) Roof 47.5 5529* 300926 0.35 602 0.109 1.0 602 4th 37 6245 262419 0.31 525 0.109 0.68 462 3rd 26.5 6245 185750 0.22 372 0.109 0.68 462 2nd 16 6245 110173 0.13 220 0.109 0.68 462 Sum 0 24262 859267 1 1719 NEES/EERI Webinar April 23 2012 * The top floor has less seismic mass due to ramp.

  8. Diaphragm Design Example Steps 2-4: Design Option and Classifications Step 2: Determine the Diaphragm Seismic Demand Level For SDC C: Low Step 3: Select Diaphragm Design Option For low seismic demand: Elastic design option (EDO) Step 4: Determine Required Diaphragm Reinforcement Classification For elastic design option: Low deformability element (LDE) Note: The Basic design option (BDO)and the Reduced design option (RDO) are also available to the designer, requiring improved details (MDE and HDE), but permitting lower design forces. NEES/EERI Webinar April 23 2012

  9. Diaphragm Design Example Step 5: Force Amplification Factor Step 5: Determine Diaphragm Force Amplification Factor • The entire diaphragm is treated as three individual sub-diaphragms for the diaphragm design (North, South and Ramp): • L = 300 ft • AR = 300/60 = 5 • Limit: 0.25 ≤ AR ≤ 4.0 Use AR = 4 in Eqns. 3-8 • n = 4 • L/60-AR = 1 YE=1.7  40.38 [1 – 0.04(3-4)2]  1.05(300/60-4)=2.9 NEES/EERI Webinar April 23 2012

  10. Diaphragm Design Example Step 6: Force Overstrength Factor Step 6: Determine Diaphragm Shear Overstrength Factor For elastic design (EDO), no further overstrength required. NEES/EERI Webinar April 23 2012

  11. Diaphragm Design Example Design Force Comparison Diaphragm Design Forces Required for Different Available Options in Design Procedure • Design example 1A: (EDO) • Eqn. 3: YE = 1.740.38[1-0.04(3-4)2]1.05(300/60-4) = 2.9 • Eqn. 6: WE = 1.0 • Design example 1B: (BDO) • Eqn. 4: YD = 1.6540.21[1-0.03(3-4)2]1.05(300/60-4) = 2.25 • Eqn. 7: WB = 1.42 AR-0.13 = 1.42  4-0.13 = 1.19 • Design example 1C: (RDO) • Eqn. 5: YR = 1.0540.3 [1-0.03(2.5-4)2]1.05(300/60-4) = 1.56 • Eqn. 8: WR = 1.92 AR-0.18 = 1.92  4-0.18 = 1.5 NEES/EERI Webinar April 23 2012

  12. Diaphragm Design Example Step 7: Diaphragm Design Force Step 7: Determine Diaphragm Design Force Continuing with EDO Design: Insert baseline diaphragm forces (Step 1) and diaphragm amplification factor (Step 5) into Equation 9 N-S direction: Top Floor: Fdia = YEFDx = 2.9482 =1398 kips > 0.2SDSIwx= 498 kips Other Floors Fdia = YEFDx = 2.9370 = 1073 kips > 0.2SDSIwx= 562 kips E-W direction: Top Floor Fdia = YEFDx = 2.9602 = 1747 kips > 0.2SDSIwx= 498 kips Other Floors Fdia = YEFDx = 2.9462 = 1342 kips > 0.2SDSIwx= 562 kips NEES/EERI Webinar April 23 2012

  13. Diaphragm Design Example Step 8: Diaphragm Internal Force Step 8: Determine Diaphragm Internal Forces The structure has a commonly-used configuration. Select free-body diagram method. Step 8 makes use of PART 3: Analysis Techniques andDesign Aids for Diaphragm Design • Part 3 is used here for: • existing free body diagrams created for common precast diaphragm layouts • a design spreadsheet program embedded with associated free body calculations NEES/EERI Webinar April 23 2012

  14. x w Vsw Vsw Vbeam Vbeam Nlw Nbeam Nbeam Diaphragm Design Analysis Techniques Step 8: Internal Force (con’t) Distributed Load: Top floor: w = YEFDx (3L − L’/2) Other floors: w = YEFDx /3L Diaphragm Joint at End Flat, 0≤ x ≤ Lbeam: Nu = xVbeam/Lbeam Vu = Vsw− wx − xNbeam/Lbeam Mu = xVsw− wx2/2− x2Nbeam/3Lbeam Diaphragm Joint along Ramp, Lbeam< x ≤ Lbeam/2: Nu = Vbeam Vu = Vsw− wx − Nbeam + Nlw( x − Lbeam) Mu = xVsw− wx2/2− Nbeam( x − 2Lbeam/3) + Nlw( x − Lbeam)2/2 Reactions at Boundary: Nlw = 0.15w Nbea = 0.5(wL − Nlw L’ )/2 VSW = 0.5(wL − Nlw L’ )/2 Vbeam= VQ Lbeam/ I NEES/EERI Webinar April 23 2012

  15. Enter Site Information Enter Bldg Geometry Enter LFRS Factors Diaphragm Design Analysis Techniques Step 8: Internal Force (con’t) DIAPHRAGM DESIGN: SPREADSHEET PROGRAM NEES/EERI Webinar April 23 2012

  16. Calculates FBD Forces Generates diaphragm joint locations (Col D) based on span and panel width and calcs all internal forces Nu , Vu , Mu Diaphragm Design Spreadsheet Program Step 8: Internal Force (con’t) NEES/EERI Webinar April 23 2012

  17. Diaphragm Design Example Step 8: Internal Force (con’t) The maximum internal forces Nu , Vu , Murepresent the required strength at each diaphragm joint. These values calculated considering the effect of two orthogonal directions (transverse and longitudinal) independently. NEES/EERI Webinar April 23 2012

  18. Diaphragm Design Example Step 9: Diaphragm Reinforcement Step 9: Select Diaphragm Reinforcement Diaphragm reinforcement types selected must meet the Required Diaphragm Reinforcement Classification from Step 4. Prequalified connectors will be used in this example. Select appropriate diaphragm reinforcement types from PART 2: Table 2A-1. NEES/EERI Webinar April 23 2012

  19. Increasing Deformation Capacity Diaphragm Design Example Reinforcement Detail Comparison Chord Details Meeting Requirements for Different Design Options Design example 1A: (EDO) Dry chord connector ( LDE ) Design example 1B: (BDO) Flat plate connector ( MDE ) Design example 1C: (RDO) Continuous Bars inPour strip ( HDE ) NEES/EERI Webinar April 23 2012

  20. Diaphragm Design Analysis Techniques Reinforcement Detail Comparison Web Details Meeting Requirements for Different Design Options JVI Vector Topped Hairpin w/ Ductile Mesh Wire Mesh HDE LDE HDE NEES/EERI Webinar April 23 2012

  21. Diaphragm Design Analysis Techniques Reinforcement Detail Comparison LFRS-to-Diaphragm Connections Angled Plate Bar Connector Straight Bar Connector MDE MDE Threaded Inserts, Dowel Bars in pour strip HDE NEES/EERI Webinar April 23 2012

  22. Diaphragm Design Example Step 9: Diaphragm Reinforcement (con’t) Determine Diaphragm Reinforcement Properties: The diaphragm reinforcement selected is prequalified. Thus, diaphragm reinforcement properties can be looked up in PART 2: Table 2A-1. NEES/EERI Webinar April 23 2012

  23. Nn= Stn Vn= Svn Mn= Stn yi Diaphragm Design Example Step 10: Diaphragm Strength Design Step 10: Design the Diaphragm Reinforcement at Joints Use the interaction equation (Eqn. 10) to determine the required diaphragm reinforcement at each joint: The diaphragm joint required strength values (Mu, Nu and Vu) are from Step 8. The diaphragm joint nominal design strength values (Mn, Nn and Vn) are based on vn and tn from Step 9. Selection of a trial design is greatly facilitated through the use of spreadsheet methods. NEES/EERI Webinar April 23 2012

  24. Enter trial chord and shear reinforcement at each joint Automatically imports diaphragm internal forces calculated in Step 8 Diaphragm Design Example Step 10: Strength Design (con’t) OUPUT FROM SPREADSHEET DESIGN PROGRAM NEES/EERI Webinar April 23 2012

  25. Diaphragm Design Example Step 10: Strength Design (con’t) Final Design Summary Table: Top Floor NEES/EERI Webinar April 23 2012

  26. Simple Beam method produces higher internal force demand. Thus the FBD Method will produce a more economical design Diaphragm Design Example Step 10: Strength Design (con’t) Comparison of Simple Beam Method Design to FBD Method NEES/EERI Webinar April 23 2012

  27. Wall length V N M v * t * Req'd # Provide u u u n n Anchorage design [ft] kips [kips] [k - ft] [kips] [kips] Per wall #4 angled bar NS shear Top 25 349 0 0 31.1 18.6 13.2 14 wall Others 25 268 0 0 31.1 18.6 10.2 11 31 6.2 EW lite wall Top 8 0 31.1 18.6 1.2 2 2 1 4.3 - S/N Flat Others 8 0 31.1 18.6 0.8 2 EW lite wall All floors Provide flexible connector: 4"x3"x1/2" - 5" angle plate with C - shape weld per wall - Ramp • Diaphragm collector reinforcement: • Collectors designed to the shear tributary to the shear wall • As=WVu/ffy= 1.0  117/0.9/60 = 2.16 in2 Select 5 # 6 at each end of structure Diaphragm Design Example Step 10: Strength Design (con’t) LFRS-to-Diaphragm Connection • Check YEWvE= 2.9  1.0=2.9> Wo=2.5 OK NEES/EERI Webinar April 23 2012

  28. Diaphragm Design Example Step 11: Diaphragm Stiffness Step 11: Determine the diaphragm effective elastic modulus and shear modulus The diaphragm joint effective elastic Young’s modulus (Eeff) and effective shear modulus (Geff ) are calculated using an analytical procedure based on the stiffness (kv , kt) of the selected diaphragm reinforcement. Eeff and Geff were calculated at each joint in the spreadsheet during Step 10. An average value across the joints is recommended for use in the design. NEES/EERI Webinar April 23 2012

  29. Amplified deflection Converted to drift Elastic diaphragm midspan deflection based here on FBD (or computer structural analysis model) No further increase for inelastic diaphragm action (EDO) 9% increase from P-D due to Flexible Diaphragm The maximum diaphragm amplified gravity column drift < 0.01, OK Reduction factor for combining diaphragm and LFRS drifts (not needed in this example) Diaphragm Design Example Step 12: Drift Check Step 12: Check the diaphragm amplified gravity column drift The table shows the diaphragm amplified gravity column drift at the midspan column from the spreadsheet design program in PART 3. NEES/EERI Webinar April 23 2012

  30. Diaphragm Design Example Final Diaphragm Design Top floor SDC C EDO Secondary reinforcement Other floors NEES/EERI Webinar April 23 2012

  31. Cost Comparison Steel Comparison: Current vs. New 4-story parking garage structure SDC C, Knoxville Chord reinforcement Shear reinforcement NEES/EERI Webinar April 23 2012

  32. Diaphragm Design Example Other Examples are Provided Example 3: 8-story Moment Frame Office – Seattle (SDC D) RDO Design NEES/EERI Webinar April 23 2012

  33. Diaphragm Design Example Office Building Force Diagrams transverse loading NEES/EERI Webinar April 23 2012

  34. Diaphragm Design Example Office Building Final Design SDC D RDO NEES/EERI Webinar April 23 2012

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