Slide 1

PCI/NSF/CPF PART 3: 1 of 34 NEES/EERI Webinar April 23 2012 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

Slide 2

PCI/NSF/CPF PART 3: 2 of 34 NEES/EERI Webinar April 23 2012 Design Methodology Documents

Slide 3

PCI/NSF/CPF PART 3: 3 of 34 NEES/EERI Webinar April 23 2012 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

Slide 4

PCI/NSF/CPF PART 3: 4 of 34 NEES/EERI Webinar April 23 2012 Diaphragm Design Example Example 1: 4-story Parking Garage - Knoxville (SDC C)

Slide 5

PCI/NSF/CPF PART 3: 5 of 34 NEES/EERI Webinar April 23 2012 Diaphragm Design Example Step 1: Determine the diaphragm seismic baseline design forces as per ASCE 7-05 Step 1: Baseline design force Seismic design parameters Design site:Knoxville, TN SDCC S s 0.58 S 1 0.147 Soil site classC F a 1.17 F v 1.65 S ms = F a S s 0.68 S m1 = F v S 1 0.24 S DS = 2/3 S ms 0.45 S D1 = 2/3 S m1 0.16 N-S: Intermediate Precast Shear Walls R=5, 0 =2.5, C d =4.5 E-W: Intermediate precast bearing wall R=4, 0 =2.5, C d =4

Slide 6

PCI/NSF/CPF PART 3: 6 of 34 NEES/EERI Webinar April 23 2012 Diaphragm Design Example Step 1: Baseline design force (cont) Diaphragm maximum design acceleration: C dia, max =max (F x /w x ) [Eqn.1] Diaphragm baseline design force F Dx = x C dia, max w x [Eqn.2] See Tables next page

Slide 7

PCI/NSF/CPF PART 3: 7 of 34 NEES/EERI Webinar April 23 2012 Diaphragm Design Example Step 1: Baseline design force (cont) h x (ft)W x (k)W x h x k C vx F x (k)C dia, max (1) x F Dx (k) (2) Roof47.55529*3009260.354820.087 1.0482 4th3762452624190.314200.087 0.68370 3rd26.562451857500.222970.087 0.68370 2nd1662451101730.131760.087 0.68370 Sum02426285926711375 h x (ft)W x (k)W x h x k C vx F x (k)C dia, max (1) x F Dx (k) (2) Roof47.55529*3009260.356020.109 1.0602 4th3762452624190.315250.109 0.68462 3rd26.562451857500.223720.109 0.68462 2nd1662451101730.132200.109 0.68462 Sum02426285926711719 N-S direction E-W direction * The top floor has less seismic mass due to ramp. Parking structure: x =1.0 at top floor and x =0.68 at lower floors. Baseline (unamplified) forces

Slide 8

PCI/NSF/CPF PART 3: 8 of 34 NEES/EERI Webinar April 23 2012 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.

Slide 9

PCI/NSF/CPF PART 3: 9 of 34 NEES/EERI Webinar April 23 2012 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 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): E =1.7 4 0.38 [1 0.04(3-4) 2 ] 1.05 (300/60-4) =2.9

Slide 10

PCI/NSF/CPF PART 3: 10 of 34 NEES/EERI Webinar April 23 2012 Diaphragm Design Example Step 6: Force Overstrength Factor Step 6: Determine Diaphragm Shear Overstrength Factor For elastic design (EDO), no further overstrength required.

Slide 11

PCI/NSF/CPF PART 3: 11 of 34 NEES/EERI Webinar April 23 2012 Diaphragm Design Example Diaphragm Design Forces Required for Different Available Options in Design Procedure Design Force Comparison Design example 1A: (EDO) Eqn. 3: E = 1.7 4 0.38 [1-0.04(3-4) 2 ] 1.05 (300/60-4) = 2.9 Eqn. 6: E = 1.0 Design example 1B: (BDO) Eqn. 4: D = 1.65 4 0.21 [1-0.03(3-4) 2 ] 1.05 (300/60-4) = 2.25 Eqn. 7: B = 1.42 AR -0.13 = 1.42 4 -0.13 = 1.19 Design example 1C: (RDO) Eqn. 5: R = 1.05 4 0.3 [1-0.03(2.5-4) 2 ] 1.05 (300/60-4) = 1.56 Eqn. 8: R = 1.92 AR -0.18 = 1.92 4 -0.18 = 1.5

Slide 12

PCI/NSF/CPF PART 3: 12 of 34 NEES/EERI Webinar April 23 2012 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: F dia = E F Dx = 2.9 482 =1398 kips > 0.2S DS Iw x = 498 kips Other Floors F dia = E F Dx = 2.9 370 = 1073 kips > 0.2S DS Iw x = 562 kips E-W direction: Top Floor F dia = E F Dx = 2.9 602 = 1747 kips > 0.2S DS Iw x = 498 kips Other Floors F dia = E F Dx = 2.9 462 = 1342 kips > 0.2S DS Iw x = 562 kips

Slide 13

PCI/NSF/CPF PART 3: 13 of 34 NEES/EERI Webinar April 23 2012 Diaphragm Design Example Step 8: Determine Diaphragm Internal Forces Step 8 makes use of PART 3: Analysis Techniques and Design Aids for Diaphragm Design The structure has a commonly-used configuration. Step 8: Diaphragm Internal Force 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 Select free-body diagram method.

Slide 14

PCI/NSF/CPF PART 3: 14 of 34 NEES/EERI Webinar April 23 2012 Diaphragm Design Analysis Techniques Step 8: Internal Force (cont) w V sw N lw N beam V beam N beam V beam x Distributed Load: Top floor: w = F Dx (3L L /2) Other floors: w = E F Dx /3L Reactions at Boundary: N lw = 0.15w N bea = 0.5(wL N lw L )/2 V SW = 0.5(wL N lw L )/2 V beam = VQ L beam / I Diaphragm Joint along Ramp, L beam < x L beam /2: N u = V beam V u = V sw wx N beam + N lw ( x L beam ) M u = xV sw wx 2 /2 N beam ( x 2L beam /3) + N lw ( x L beam ) 2 /2 Diaphragm Joint at End Flat, 0 x L beam : N u = xV beam /L beam V u = V sw wx xN beam /L beam M u = xV sw wx 2 /2 x 2 N beam /3L beam

Slide 15

PCI/NSF/CPF PART 3: 15 of 34 NEES/EERI Webinar April 23 2012 Diaphragm Design Analysis Techniques DIAPHRAGM DESIGN: SPREADSHEET PROGRAM Step 8: Internal Force (cont) Enter Site Information Enter Bldg Geometry Enter LFRS Factors

Slide 16

PCI/NSF/CPF PART 3: 16 of 34 NEES/EERI Webinar April 23 2012 Diaphragm Design Spreadsheet Program Step 8: Internal Force (cont) Calculates FBD Forces Generates diaphragm joint locations (Col D) based on span and panel width and calcs all internal forces N u, V u, M u

Slide 17

PCI/NSF/CPF PART 3: 17 of 34 NEES/EERI Webinar April 23 2012 Diaphragm Design Example Step 8: Internal Force (cont) The maximum internal forces N u, V u, M u represent the required strength at each diaphragm joint. These values calculated considering the effect of two orthogonal directions (transverse and longitudinal) independently.

Slide 18

PCI/NSF/CPF PART 3: 18 of 34 NEES/EERI Webinar April 23 2012 Diaphragm Design Example Step 9: Diaphragm Reinforcement Diaphragm reinforcement types selected must meet the Required Diaphragm Reinforcement Classification from Step 4. Step 9: Select Diaphragm Reinforcement Prequalified connectors will be used in this example. Select appropriate diaphragm reinforcement types from PART 2: Table 2A-1.

Slide 19

PCI/NSF/CPF PART 3: 19 of 34 NEES/EERI Webinar April 23 2012 Diaphragm Design Example Design example 1B: (BDO) Flat plate connector ( MDE ) Design example 1C: (RDO) Continuous Bars in Pour strip ( HDE ) Chord Details Meeting Requirements for Different Design Options Reinforcement Detail Comparison Design example 1A: (EDO) Dry chord connector ( LDE ) Increasing Deformation Capacity

Slide 20

PCI/NSF/CPF PART 3: 20 of 34 NEES/EERI Webinar April 23 2012 Diaphragm Design Analysis Techniques Topped Hairpin w/ Ductile Mesh Reinforcement Detail Comparison Web Details Meeting Requirements for Different Design Options LDEHDE JVI Vector Wire Mesh

Slide 21

PCI/NSF/CPF PART 3: 21 of 34 NEES/EERI Webinar April 23 2012 Diaphragm Design Analysis Techniques Straight Bar Connector LFRS-to-Diaphragm Connections Angled Plate Bar Connector Reinforcement Detail Comparison MDE Threaded Inserts, Dowel Bars in pour strip HDE

Slide 22

PCI/NSF/CPF PART 3: 22 of 34 NEES/EERI Webinar April 23 2012 Diaphragm Design Example Step 9: Diaphragm Reinforcement (cont) Determine Diaphragm Reinforcement Properties: The diaphragm reinforcement selected is prequalified. Thus, diaphragm reinforcement properties can be looked up in PART 2: Table 2A-1.

Slide 23

PCI/NSF/CPF PART 3: 23 of 34 NEES/EERI Webinar April 23 2012 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 (M u, N u and V u ) are from Step 8. The diaphragm joint nominal design strength values (M n, N n and V n ) are based on v n and t n from Step 9. Selection of a trial design is greatly facilitated through the use of spreadsheet methods. N n = t n V n = v n M n = t n y i

Slide 24

PCI/NSF/CPF PART 3: 24 of 34 NEES/EERI Webinar April 23 2012 Diaphragm Design Example OUPUT FROM SPREADSHEET DESIGN PROGRAM Step 10: Strength Design (cont) Automatically imports diaphragm internal forces calculated in Step 8 Enter trial chord and shear reinforcement at each joint

Slide 25

PCI/NSF/CPF PART 3: 25 of 34 NEES/EERI Webinar April 23 2012 Diaphragm Design Example Joint North/South flatRamp ChordJVIM-N-VChordJVIM-N-V Size##s (ft) Trans verse Longit udinalSize##s (ft) Trans verse Longit udinal 1#64124.80.520.32#68183.10.940.18 2#64124.80.790.65#68183.10.770.36 3#66124.80.820.70#68183.10.580.55 4#66163.50.970.92#67163.50.420.83 5#66163.50.820.83#67163.50.430.86 6#66163.50.960.75#67163.50.460.89 7#68134.40.880.54#67134.40.530.94 8#68134.40.950.50#67134.40.570.97 9#68134.41.020.47#67134.40.591.00 10#6978.80.990.47#6878.80.560.95 11#6978.81.020.48#6878.80.550.97 12#6978.81.030.48#6878.80.521.00 Final Design Summary Table: Top Floor Step 10: Strength Design (cont)

Slide 26

PCI/NSF/CPF PART 3: 26 of 34 NEES/EERI Webinar April 23 2012 Diaphragm Design Example Comparison of Simple Beam Method Design to FBD Method Step 10: Strength Design (cont) Simple Beam method produces higher internal force demand. Thus the FBD Method will produce a more economical design

Slide 27 o =2.5 OK Diaphragm collector reinforcement: Collectors designed to the shear tributary to the shear wall A s = V u / f y = 1.0 117/0.9/60 = 2.16 in 2 Select 5 # 6 at each end of structure">

PCI/NSF/CPF PART 3: 27 of 34 NEES/EERI Webinar April 23 2012 Diaphragm Design Example LFRS-to-Diaphragm Connection Wall length V u N u M u v n * t n * Req'd # Provide Anchorage design [ft] kips [kips] [k-ft] [kips] Per wall #4 angled bar Top 25 349 0 0 31.1 18.6 13.2 14 NS shear wall Others 25 268 0 0 31.1 18.6 10.2 11 Top 8 31 6.2 0 31.1 18.6 1.2 2 EW lite wall - S/N Flat Others 8 21 4.3 0 31.1 18.6 0.8 2 EW lite wall - Ramp All floors Provide flexible connector: 4"x3"x1/2"-5" angle plate with C-shape weld per wall Step 10: Strength Design (cont) Check E vE = 2.9 1.0=2.9> o =2.5 OK Diaphragm collector reinforcement: Collectors designed to the shear tributary to the shear wall A s = V u / f y = 1.0 117/0.9/60 = 2.16 in 2 Select 5 # 6 at each end of structure

Slide 28

PCI/NSF/CPF PART 3: 28 of 34 NEES/EERI Webinar April 23 2012 Diaphragm Design Example Step 11: Determine the diaphragm effective elastic modulus and shear modulus The diaphragm joint effective elastic Youngs modulus (E eff ) and effective shear modulus (G eff ) are calculated using an analytical procedure based on the stiffness (k v, k t ) of the selected diaphragm reinforcement. E eff and G eff were calculated at each joint in the spreadsheet during Step 10. An average value across the joints is recommended for use in the design. Joint Top FloorOther Floors North/South flatRampNorth/South flatRamp E eff G eff E eff G eff E eff G eff E eff G eff [ksi] Min7152131025222566142841191 Max11742811122327914238933267 Ave1002.9255.51068.8277.8768.3199.9882.3244.5 Des944.52471073.5274.5740190887229 Step 11: Diaphragm Stiffness

Slide 29

PCI/NSF/CPF PART 3: 29 of 34 NEES/EERI Webinar April 23 2012 Diaphragm Design Example Step 12: Check the diaphragm amplified gravity column drift SubFloor CC C d,dia C r,dia dia,el dia dia Diaphragm [in] [rad] N/S FlatTop1.09 0.590.7080.7750.0036 Others1.09 0.590.6370.6970.0033 RampTop1.09 0.590.9341.0210.0048 Others1.09 0.590.7460.8160.0038 Step 12: Drift Check The table shows the diaphragm amplified gravity column drift at the midspan column from the spreadsheet design program in PART 3. 9% increase from P- due to Flexible Diaphragm No further increase for inelastic diaphragm action (EDO) Elastic diaphragm midspan deflection based here on FBD (or computer structural analysis model) Amplified deflection Converted to drift Reduction factor for combining diaphragm and LFRS drifts (not needed in this example) The maximum diaphragm amplified gravity column drift < 0.01, OK

Slide 30

PCI/NSF/CPF PART 3: 30 of 34 NEES/EERI Webinar April 23 2012 Diaphragm Design Example Top floor Other floors Secondary reinforcement SDC C EDO Final Diaphragm Design

Slide 31

PCI/NSF/CPF PART 3: 31 of 34 NEES/EERI Webinar April 23 2012 Cost Comparison Chord reinforcementShear reinforcement 4-story parking garage structure SDC C, Knoxville Steel Comparison: Current vs. New

Slide 32

PCI/NSF/CPF PART 3: 32 of 34 NEES/EERI Webinar April 23 2012 Diaphragm Design Example Example 3: 8-story Moment Frame Office Seattle (SDC D) RDO Design Other Examples are Provided

Slide 33

PCI/NSF/CPF PART 3: 33 of 34 NEES/EERI Webinar April 23 2012 Diaphragm Design Example transverse loading Office Building Force Diagrams

Slide 34

PCI/NSF/CPF PART 3: 34 of 34 NEES/EERI Webinar April 23 2012 Diaphragm Design Example SDC D RDO Office Building Final Design