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