Outline

NEES/EERI Webinar

April 23 2012PCI/NSF/CPFPART 3:1 of 34

OutlineOutline

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

NEES/EERI Webinar


Design Methodology Documents

Seismic Design Methodology Document for Precast Concrete Diaphragms

PART 4: DESIGN EXAMPLES

NEES/EERI Webinar


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


Diaphragm Design Example

Example 1: 4-story Parking Garage - Knoxville (SDC C)

Lbeam=48' L'=204' 48'

L=300'

60'

d=60'

60'

a=180'

25'8'

Lite wall

Shear wall

b=12'

Ramp Span Transverse

Longitudinal

North

South

RampLanding

Joint #: 1 2 3 4 5 6 7 8 9 10 11 12

DT-IT Joint

16'

10'-6"

10'-6"

10'-6"

47'-6"

204'

NEES/EERI Webinar



Step 1: Determine the diaphragm seismic baseline design forces as per ASCE 7-05

Step 1: Baseline design force

Seismic design parametersDesign site: Knoxville, TNSDC CSs 0.58S1 0.147Soil site class C

Fa 1.17Fv 1.65

Sms= Fa Ss 0.68Sm1= Fv S1 0.24SDS= 2/3 Sms 0.45SD1= 2/3 Sm1 0.16

N-S: Intermediate Precast Shear Walls R=5, 0=2.5, Cd=4.5

E-W: Intermediate precast bearing wall R=4, 0=2.5, Cd=4

NEES/EERI Webinar



Seismic response coefficient Cs

362.05.4702.0 4/34/3 nta HCT sec; 571.0362.058.1 auTCT sec

Step 1: Baseline design force (con’t)

N-S direction:

09.0/

E

DSS IR

SC , 01.0min, SC , 057.0

571.05

16.0

)/(1

max,

TIR

SC

E

DS

controls

E-W direction:

11.0/

E

DSS IR

SC , 01.0min, SC , 071.0

571.04

16.0

)/(1

max,

TIR

SC

E

DS

controls

Diaphragm maximum design acceleration: Cdia, max=max (Fx/wx) [Eqn.1]

Diaphragm baseline design force FDx = x Cdia, max wx [Eqn.2]

See Tables next page

Seismic Base Shear: WCV s [Eqn.12.8-1 (ASCE 7 2005)]

Lateral Seismic Force: VCF vxx [Eqn.12.8-11 (ASCE 7 2005)]

where

n

i

kii

kxxvx hwhwC

1

/ [Eqn.12.8-12 (ASCE 7 2005)]

Seismic Base Shear: WCV s [Eqn.12.8-1 (ASCE 7 2005)]

Lateral Seismic Force: VCF vxx [Eqn.12.8-11 (ASCE 7 2005)]

where

n

i

kii

kxxvx hwhwC

1

/ [Eqn.12.8-12 (ASCE 7 2005)]

NEES/EERI Webinar


Diaphragm Design ExampleStep 1: Baseline design force (con’t)

hx (ft) Wx (k) Wx hxk Cvx Fx (k) Cdia, max

(1) x 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

hx (ft) Wx (k) Wx hxk Cvx Fx (k) Cdia, max

(1) x 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

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

NEES/EERI Webinar



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


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 ExampleStep 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):

Eqn. 3: )60/(238.0 05.1])3(04.01[7.1 ARLE ARn

9.205.1])43(04.01[47.1 )460/300(238.0 E

E=1.7 4 0.38 [1 – 0.04(3-4)2] 1.05(300/60-4) =2.9

NEES/EERI Webinar


Diaphragm Design ExampleStep 6: Force Overstrength Factor

Step 6: Determine Diaphragm Shear Overstrength Factor

For elastic design (EDO), no further overstrength required.

NEES/EERI Webinar



Diaphragm Design Forces Required for Different Available Options in Design Procedure

Design Force Comparison

Design example 1A: (EDO)

Eqn. 3: E = 1.740.38[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 = 1.6540.21[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 = 1.0540.3 [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

NEES/EERI Webinar


Diaphragm Design ExampleStep 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 = EFDx = 2.9482 =1398 kips > 0.2SDSIwx= 498 kips

Other FloorsFdia = EFDx = 2.9370 = 1073 kips > 0.2SDSIwx= 562 kips

E-W direction:

Top FloorFdia = EFDx = 2.9602 = 1747 kips > 0.2SDSIwx= 498 kips

Other FloorsFdia = EFDx = 2.9462 = 1342 kips > 0.2SDSIwx= 562 kips

NEES/EERI Webinar



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

DSDM PROJECT: UA, UCSD,LU

Seismic Design Methodology Document for Precast Concrete Diaphragms

PART 3: ANALYSIS TECHNIQUES

AND DESIGN AIDS PROGRAM

for Precast 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

Select free-body diagram method.

NEES/EERI Webinar


Diaphragm Design Analysis Techniques

Step 8: Internal Force (con’t)

Lbeam L'

L

d

a-2d

d

a

Lite wall

Shear wall

Ramp Span Transverse (NS)

Longitudinal (EW)

North

South

RampLanding

Lbeam

b

LLbeam

L'Lbeam

d

w

Vsw Vsw

NlwNbeam

VbeamNbeam

Vbeam

x

Distributed Load: Top floor: w = FDx (3L − L’/2) Other floors: w = EFDx /3L

Reactions at Boundary: Nlw = 0.15w Nbea = 0.5(wL − Nlw L’ )/2 VSW = 0.5(wL − Nlw L’ )/2 Vbeam= VQ Lbeam / I

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

Diaphragm Joint at End Flat, 0≤ x ≤ Lbeam:

Nu = xVbeam/Lbeam

Vu = Vsw− wx − xNbeam/Lbeam

Mu = xVsw− wx2/2 − x2Nbeam/3Lbeam

NEES/EERI Webinar



DIAPHRAGM DESIGN: SPREADSHEET PROGRAM


Enter Site Information

Enter Bldg Geometry

Enter LFRS Factors

NEES/EERI Webinar


Diaphragm Design Spreadsheet Program


Calculates FBD Forces

Generates diaphragm joint locations (Col D) based on span and panel width and calcs all internal forces Nu , Vu , Mu

Lseam=48' L'=204' 48'

L=300'

60'

d=60'

60'

a=180'

25'8'

Lite wall

Shear wall

b=12'

Ramp Span Transverse

Longitudinal

North

South

RampLanding

Joint #: 1 2 3 4 5 6 7 8 9 10 11 12

NEES/EERI Webinar




The maximum internal forces Nu , Vu , Mu represent the required strength at each diaphragm joint. These values calculated considering the effect of two

orthogonal directions (transverse and longitudinal) independently.

NEES/EERI Webinar


Diaphragm Design ExampleStep 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.

0.121.68216.8HDE0.60.04354.21260Ductile mesh Gr.1018

0.124.2382HDE0.70.0568601234Pour strip chord Gr.60

0.124.2382MDE0.30.071601018Dry chord w/ flat plate Gr. 60

0.124.2382LDE0.10.071601018Dry chord Gr.60

[in][ksi][ksi/in][in][in][ksi][ksi/in]

vyvykvClassificationtutytykt

ShearTension

Reinforcing bars

0.121.68216.8HDE0.60.04354.21260Ductile mesh Gr.1018

0.124.2382HDE0.70.0568601234Pour strip chord Gr.60

0.124.2382MDE0.30.071601018Dry chord w/ flat plate Gr. 60

0.124.2382LDE0.10.071601018Dry chord Gr.60

[in][ksi][ksi/in][in][in][ksi][ksi/in]


ShearTension

Reinforcing bars

0.0517.1372MDE0.30.04110.22300Angled bar (#3)

0.118.1181HDE0.60.0439209Hairpin (#4)

0.08218.1226HDE0.60.0663.155JVI

[in][k][k/in][in][in][k][k/in]


ShearTension

Connectors

0.0517.1372MDE0.30.04110.22300Angled bar (#3)

0.118.1181HDE0.60.0439209Hairpin (#4)

0.08218.1226HDE0.60.0663.155JVI

[in][k][k/in][in][in][k][k/in]


ShearTension

Connectors

NEES/EERI Webinar



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

NEES/EERI Webinar



Topped Hairpin w/ Ductile Mesh


Web Details Meeting Requirements for Different Design Options

LDE HDE HDE

JVI VectorWire Mesh

NEES/EERI Webinar



Straight Bar Connector

LFRS-to-Diaphragm Connections

Angled Plate Bar Connector


MDE MDE

Threaded Inserts, Dowel Bars in pour strip

HDE

NEES/EERI Webinar


Diaphragm Design ExampleStep 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.

kt tn

ty

tu Classification kv vn

vy

[k/in] [kips] [in] [in] [k/in] [kips] [in]

JVI 55 3.1 0.066 0.6 HDE 226 18.1 0.082

Hairpin (#4) 209 9 0.043 0.6 HDE 181 18.1 0.1

Angled bar (#3) 300 10.22 0.041 0.3 MDE 372 17.1 0.05

ConnectorsTension Shear

kt /As tn /As

ty

tu Classification kv /Av vn /Av

vy

[k/in/in2] [ksi] [in] [in] [k/in/in2] [ksi] [in]

Dry chord Gr.60 1018 60 0.071 0.1 LDE 382 24.2 0.1

Dry chord w/ flat plate Gr. 60 1018 60 0.071 0.3 MDE 382 24.2 0.1

Pour strip chord Gr.60 1234 60 0.0568 0.7 HDE 382 24.2 0.1

Ductile mesh Gr.1018 1260 54.2 0.043 0.6 HDE 216.8 21.68 0.1

Reinforcing Bars

Tension Shear

NEES/EERI Webinar


Diaphragm Design ExampleStep 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:

0.122

nv

uv

nf

u

nf

u

V

V

N

N

M

M

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.

Nn = tn Vn = vn Mn = tn yi

NEES/EERI Webinar



OUPUT FROM SPREADSHEET DESIGN PROGRAM

0.122

nv

uv

nf

u

nf

u

V

V

N

N

M

M

Step 10: Strength Design (con’t)

Automatically imports diaphragm internal forces calculated in

Step 8

Enter trial chord and shear reinforcement at

each joint

NEES/EERI Webinar



Joint

North/South flat Ramp

Chord JVI M-N-V Chord JVI M-N-V

Size # # s (ft)Transverse

Longitudinal Size # # s (ft)

Transverse

Longitudinal

1 #6 4 12 4.8 0.52 0.32 #6 8 18 3.1 0.94 0.18

2 #6 4 12 4.8 0.79 0.65 #6 8 18 3.1 0.77 0.36

3 #6 6 12 4.8 0.82 0.70 #6 8 18 3.1 0.58 0.55

4 #6 6 16 3.5 0.97 0.92 #6 7 16 3.5 0.42 0.83

5 #6 6 16 3.5 0.82 0.83 #6 7 16 3.5 0.43 0.86

6 #6 6 16 3.5 0.96 0.75 #6 7 16 3.5 0.46 0.89

7 #6 8 13 4.4 0.88 0.54 #6 7 13 4.4 0.53 0.94

8 #6 8 13 4.4 0.95 0.50 #6 7 13 4.4 0.57 0.97

9 #6 8 13 4.4 1.02 0.47 #6 7 13 4.4 0.59 1.00

10 #6 9 7 8.8 0.99 0.47 #6 8 7 8.8 0.56 0.95

11 #6 9 7 8.8 1.02 0.48 #6 8 7 8.8 0.55 0.97

12 #6 9 7 8.8 1.03 0.48 #6 8 7 8.8 0.52 1.00

Final Design Summary Table: Top Floor


NEES/EERI Webinar



Comparison of Simple Beam Method Design to FBD Method


0

50

100

150

200

250

0 100 200 300

x (ft)

N (

kips

)

Simple Beam

DSDM FBD

-300

-200

-100

0

100

200

300

0 100 200 300

x (ft)V

(ki

ps)

Simple Beam

DSDM FBD

-5000

0

5000

10000

15000

20000

0 100 200 300

x (ft)

M (

k-ft

)

Simple Beam

DSDM FBD

00.20.40.60.8

11.21.41.61.8

2

0 100 200 300x (ft)

M-N

-V D

eman

d

Simple Beam

DSDM FBD

Simple Beam method produces higher internal force demand. Thus the FBD Method will produce a more economical design

NEES/EERI Webinar



LFRS-to-Diaphragm Connection

Wall length Vu Nu Mu vn * tn * Req'd # Provide Anchorage design

[ft] kips [kips] [k-ft] [kips] [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


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

As=Vu/fy= 1.0 117/0.9/60 = 2.16 in2

Select 5 # 6 at each end of structure

NEES/EERI Webinar



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.

Joint

Top Floor Other Floors

North/South flat Ramp North/South flat Ramp

Eeff Geff Eeff Geff Eeff Geff Eeff Geff

[ksi] [ksi] [ksi] [ksi] [ksi] [ksi] [ksi] [ksi]

Min 715 213 1025 222 566 142 841 191

Max 1174 281 1122 327 914 238 933 267

Ave 1002.9 255.5 1068.8 277.8 768.3 199.9 882.3 244.5

Des 944.5 247 1073.5 274.5 740 190 887 229

Step 11: Diaphragm Stiffness

NEES/EERI Webinar



Step 12: Check the diaphragm amplified gravity column drift

Sub Floor

C Cd,dia Cr,dia

dia,el dia dia

Diaphragm 　 [in] [in] [rad]

N/S Flat Top 1.09 1.09 0.59 0.708 0.775 0.0036

　 Others 1.09 1.09 0.59 0.637 0.697 0.0033

Ramp Top 1.09 1.09 0.59 0.934 1.021 0.0048

　 Others 1.09 1.09 0.59 0.746 0.816 0.0038

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

NEES/EERI Webinar



Top floor

Other floors

Secondary reinforcement

SDC C EDO8#6 4#6

8#6 8#6

16 JVI12 JVI 7 JVI

18 JVI

Cut-off 1#6 Cut-off 2#6

Downward to3rd floor

7 JVI16 JVI

9#6

North/South

Ramp

Cut-off 2#6

6#613 JVI

13 JVI 7#6

Cut-off 1#6

8#6

5#6 3#6

6#6 5#6

7 JVI 7 JVI

16 JVI

Cut-off 1#6 Cut-off 1#6

Downward tolower floor

7 JVI16 JVI

6#6

North/South

Ramp

Cut-off 1#6

4#613 JVI

13 JVI 5#6

Cut-off 1#6

Final Diaphragm Design

2#4 DT9A to flat12" Angle w/ C-shape weld to ramp

#4 angled bar connector:Top floor: 4 per panelOther floors: 3 per panel

60'

61'

60'

181'

25'

300'

#4 angled bar connector:Top floor: 14 per wallOther floors: 11 per wall

5 # 6 collectorreinforcement

NEES/EERI Webinar


Cost Comparison

Chord reinforcement

Shear reinforcement

4-story parking garage structure

SDC C, Knoxville

0

0.1

0.2

0.3

Current

EDOBDO

RDO

W/ bar cut-off

w/o bar cut-off

Cho

rd w

eigh

t (lb

/sq.

ft)

0

500

1000

1500

2000

2500

3000

3500

Current

EDOBDO

RDO

# of

she

ar c

onne

ctor

Steel Comparison: Current vs. New

NEES/EERI Webinar



Example 3: 8-story Moment Frame Office – Seattle (SDC D)

230'

147'

170'

South

North

West East

24.5'

24.5'

20'30'Joint # 1 2 3 4 5 6 7 8 9 10 11

30' 20' 20' 20' 20' 30'30' 30' 30'

230'

15'

106'

13'

15'

13'

13'

13'

13'

13'

13'

RDO Design

Other Examples are Provided

NEES/EERI Webinar



transverse loading

w=Fpx/L

Vint=VQb/I

RLFRS RLFRSLsupportLfree

x

d

Joint Shear Force

-600

-400

-200

0

200

400

600

0 50 100 150 200 250

x (ft)

V (

kips

)

Top floor

Other floors

Joint Moment

-5000

0

5000

10000

15000

20000

0 50 100 150 200 250

x (ft)

Mom

ent (

k-ft

)

Top floor

Other floors

Office Building Force Diagrams

NEES/EERI Webinar



SDC D RDO

2#3 collector

Hairpin Connector Ductile mesh

W2.9xW2.9 mesh

Hairpin

15@10'

[email protected]'

[email protected]'

15 6'3' 1' 6'

3#5 chord

#9 @ 3.9'

4#9 per panel

18

Office Building Final Design

Outline

Documents

Transcript of Outline