Seismic Design Tools

Seismic Pushover Analysis Using AASHTO Guide Specifications for

LRFD Seismic Bridge Design

Elmer E. Marx, Alaska Department of Transportation and Public Facilities Michael Keever, California Department of Transportation

1 2015 TRB Pushover Webinar

Learning Objective

• Given the plans for a conventional bridge pier, calculate the lateral force-displacement (pushover) response of the system necessary for the seismic design of bridges in accordance with the AASHTO Guide Specifications for LRFD Seismic Bridge Design (SGS).

• Focus is pushover analysis not modeling, seismic displacement demand or capacity design

2

Overview

• Flexural mechanics refresher

• Generation of Moment-Curvature ( M - φ )

• Material models and failure strains

• Hand checking M - φ

• Analytical plastic hinge length, Lp

• Force-Displacement pushover example

• Hand checking force-displacement

• Design Example – Two circular column bent

3

Why Pushover?

• SGS is primarily a displacement-based approach

• More rational approach than force-based method of AASHTO LRFD

• Seismic Design Category D (SDC D) requires pushover analysis

4

Why Pushover?

• Need to verify that the seismic displacement demand, ∆L

D, exceed the displacement capacity, ∆LC

• Results can be used to help create the seismic model and subsequent displacement demands

• Provides for a better understanding of bridge response and behavior

5

Flexural Mechanics

• Relationship between force, shear, moment, curvature, slope and deflection – Integration

– Moment area

– Energy methods

– Stiffness methods

6

Force – Displacement Relationship

∫ ⋅= dxPV

∫ ⋅= dxVM

loadP =

7


∫ ⋅= dxVM

EIM

=ϕ

8


IEM⋅

=ϕ

∫ ⋅= dxϕθ

∫ ⋅=∆ dxθ9

Example - Cantilever

P PV =

LPM o ⋅=IELP

IEM o

o ⋅⋅

=⋅

=ϕ

2Lo

L⋅

=ϕθ

3

2LoL

⋅=∆

ϕ

L

10

Compatibility

• Bernoulli-Euler assumption – sections that are plane (linear) before bending remain plane after bending

• Perfect bond between the steel reinforcing bars and surrounding concrete

• Material (constitutive σ-ε) models are representative of actual material response (confinement, strain hardening, buckling, spalling, cracking, creep, strain-rate, shear, etc.)

11

Moment – Curvature ( M - φ )

12


• For small angles, curvature (φ ) may be calculated as:

φ = εc / c = εt / (D – c)

φ = strain / distance from neutral axis 13


• From equilibrium:

P = FSC + FCC + FCU – FST

• Summing moments about the neutral axis

M = CGSC*FSC+CGCC*FCC+CGCU*FCU+CGST*FST+P*(D/2 – c) 14

15

Generation of Moment-Curvature Relationship


SGS 2011

Predicted Response

Idealized Bilinear Response used in SGS

16

Idealized M – φ Response

• Used for design purposes

• First yield Moment and Curvature, My and φy

• Effective stiffness, Ece* Ieff = My / φy

• Idealized yield Moment and Curvature, Mp and φyi

• Expected nominal moment, Mne at εc = 0.003

• Balance the area above and below the curve

17

SGS Material Models

• Concrete: Mander et al model for unconfined and confined condition

• Reinforcing steel: elastic - perfectly plastic with strain hardening

• Other: prestressing steel (see SGS)

• Any function / model that accurately captures the material σ-ε relationship is acceptable

18

• Stress-strain relationship for concrete where:

SGS Material Models - Concrete

rcc

c xrxrff+−

=1

'

−−+= 254.1294.71254.2 '

'

'

'''

ce

l

ce

lcecc f

ff

fff

cc

cxεε

=

secEEEr

ce

ce

−=

Mander, Priestley, Park 1988 19


−+= 151002.0 '

'

ce

cccc f

fε

'1900 cece fE =

cc

ccfEε

'

sec =

(continue)


• Confinement of circular columns


sDAsp

s '

4=ρ

llel ffKf 95.0' ≈=

22

'yhsyhsp

l

fsDfA

fρ

==


• Confinement of rectangular columns


( )2

)(8.02

' yhyxyhyxel

ffKf

ρρρρ +≈

+≈

c

sxx sH

A=ρ

c

syy sB

A=ρ


SGS Material Models - Concrete εc = strain in concrete (IN/IN) fc = stress in concrete corresponding to strain εc (KSI) f’ce = expected nominal compressive strength (KSI) fyh = nominal yield stress of transverse reinforcing (KSI) εR

su = reduced ultimate tensile strain of transverse bars (IN/IN Asx = total area of transverse bars in the “x” axis (IN2) Asy = total area of transverse bars in the “y” axis (IN2) Asp = area of hoop /spiral bar for circular column (IN2) Hc = confined core dimension in the “y” axis (IN) Bc = confined core dimension in the “x” axis (IN) D’ = diameter of spiral or hoop for circular column (IN) s = pitch of spiral or spacing of hoops or ties (IN)



Chen 2003 24

• Effective confinement factor for circular columns

• Effective confinement factor for rectangular columns


80.0~1

2'1

2'1

6)(1

1

2'

cc

cc

N

i cd

i

e

ds

bs

dbw

Kρ−

−

−

−

=∑

=

95.0~1

'2'1

cc

n

eDs

Kρ−

−

=

Chen 2003 25

SGS Material Models - Concrete s’ = clear distance between transverse bars = s – dbh (IN) dbh = diameter of transverse reinforcing bar (IN) D’ = centerline diameter of hoop or spiral (IN) ρcc = Ast / Acc

Ast = total area of longitudinal reinforcement (IN2) Acc = area of confined concrete core (IN2) n = 1 for continuous spiral n = 2 for individual hoops wi’ = clear distance between adjacent tied longitudinal bars(IN) bd = confined core dimension in the longer direction(IN) dc = confined core dimension in the shorter direction (IN) N = number of spaces between longitudinal bars

Chen 2003 26

SGS Material Models - Concrete • Stress-strain curves for concrete

• Recommend εcu < 0.02 for design purposes

Spalling strain, εsp, not to exceed 0.005

Confined concrete crushing strain, εcu, critical component

εcu

27 SGS 2011

SGS Failure Strain - Concrete

• Confined concrete crushing strain limit, εcu

where:

ρs = transverse reinforcement ratio (ρx+ρy for rect.)

fyh = nominal yield stress of transverse steel

εsu = rupture strain of transverse steel ~ 0.09 to 0.12

f’cc = confined concrete compressive stress

025.04.1

004.0 ' <+=cc

suyhscu f

f ερεimportant


SGS Material Models - Steel

Kowalsky 2013 and SGS 29

SGS Material Models - Steel

• Stress-strain relationship for reinforcing steel

yes εε ≤

shsye εεε ≤≤

Rsussh εεε ≤≤

sss Ef ε=

yes ff =

−−

−−=2

)(1shsu

ssuyeueues ffff

εεεε

Priestley, Calvi, Kowalsky 2007 30

Elastic

Plastic

Strain hardening

SGS Failure Strains - Steel • Reinforcing steel failure strain, εR

su

Property Notation Bar Size ASTM A706 ASTM A615 Grade 60

Specified minimum yield stress (ksi) fy #3 - #18 60 60

Expected yield stress (ksi) fye #3 - #18 68 68

Expected tensile strength (ksi) fue #3 - #18 95 95

Expected yield strain εye #3 - #18 0.0023 0.0023

Onset of strain hardening #3 - #8 0.0150 0.0150

#9 0.0125 0.0125

εsh #10 - #11 0.0115 0.0115

#14 0.0075 0.0075

#18 0.0050 0.0050

Reduced ultimate tensile strain #4 - #10 0.090 0.060

#11 - #18 0.060 0.040

Ultimate tensile strain εsu #4 - #10 0.120 0.090

#11 - #18 0.090 0.060

Rsuε

SGS 2011 31

SGS Material Models - Steel • ASTM A 706 Grade 60 v. ASTM A 615 Grade 60

ASTM 2011, 2012 32

Quasi-static v. Cyclic Discussion • Quasi-static constitutive models for cyclic response

Kowalsky et al. 2010 33

• Find the M-φ for the column shown below

• ASTM A 706 Grade 60 f’c = 4 KSI

Moment – Curvature Example

D = 48 IN

L =H

o = 2

0 FT

#5 hoop @ 4IN

20 #11

2 IN clr.

940 KIP

34

Moment – Curvature Example • Diameter, D = 48 IN • Gross Area, Ag = π*D2/4 = 1810 IN^2 • 20 #11 Ast = 20*1.56 = 31.2 IN^2 • ρl = Ast / Ag = 0.01724 = 1.724% • #5 hoop (dsp = 0.625 IN and Asp = 0.31 IN^2) • Hoop spacing = 4 IN pitch (s = 4 IN) • Clear cover, cov = 2 IN over transverse bars • Core diameter, D’ = D – 2*cov – dsp = 43.375 IN • ρs = 4*Asp / (D’ * s) = 0.00715 = 0.715% • ASTM A 706 Grade 60 so fye = 68 KSI, fue = 95 KSI • f’c = 4 KSI so f’ce = 5.2 KSI • P = 940 K Axial Load Ratio, ALR = P / (f’ce*Ag) = 0.1

35


• Confined concrete crushing strain limit, εcu where: ρs = 4*Asp/(D’*s) = 0.00715 f’l = Ke*ρs*fyh/2 ~ 0.95*0.00715*60/2 = 0.204 KSI fyh = 60 KSI (use nominal for horizontal steel) εsu = εR

su = 0.09 for #5 hoops

• Spreadsheet demo then check with commercial

01232.04.1

004.0 ' =+=cc

suyhscu f

f ερε

49.6254.1294.71254.2 '

'

'

''' =

−−+=

ce

l

ce

lcecc f

ff

fff

36

M – φ Spreadsheet Example

37

Moment – Curvature Example • Spreadsheet summary results for ALR = 0.1

38

Idealized Moment – Curvature • Elastic-perfectly plastic relationship is defined as:

Mi = idealized moment values = EceIeff*φ < Mp Mp = idealized plastic moment (solving for this) φ = actual calculated curvature from M-φ results EceIeff = effective stiffness at first yield = My / φy ∆(Mφ) = actual - idealized = (∆M)*(∆φ) Σ ∆(Mφ) = sum of difference = 0 by changing Mp

• Spreadsheet demo then check with commercial 39

Idealized Moment – Curvature

40

Moment – Curvature Example • Idealized M - φ relationship using more data points

41


42


P = 940 K φyi = 0.0001109 1/IN 2% φu = 0.001073 1/IN 0.2% Mp = 50460 K-IN 2% EceIeff = 454.85E6 K-IN2 0.3%

Spreadsheet Comparison φyi = 0.0001132 1/IN φu = 0.001075 1/IN Mp = 51626 K-IN EceIeff = 456.3E6 K-IN2

43

Approximate Methods

• the use of curvature ( φ ) in design is uncommon

• don’t have a good “feel” for curvature values

• simple method to check the computer results

• can help with iterative processes

• use these approximations with due skepticism

44

Approximate Methods - φyi

• For a rough check of conventional circular reinforced concrete column sections: φyi ~ 2.25*εye/12Bo ~ 1/2300Bo

φyi ~ 2.25*εye/D ~ 1/190D where: φyi = idealized yield curvature (1/IN)

Bo = column diameter (FT) D = column diameter (IN) εye = expected yield strain ~ 0.002345 (IN/IN)

45 Priestley, Calvi, Kowalsky 2007

Approximate Methods - φyi

• For a rough check of conventional rectangular reinforced concrete column sections: φyi ~ 2.1*εye/12Bo ~ 1/2500Bo where: φyi = idealized yield curvature (1/IN)

Bo = column width in loaded direction (FT) εye = expected yield strain ~ 0.002345 (IN/IN)

46 Priestley, Calvi, Kowalsky 2007

Approximate Methods - φu

• And for a very rough check of conventional circular or rectangular reinforced concrete column sections with good confinement: φu = min (εcu/cc , εsu

R/d-c) ~ εsuR /12Bo

where: φu = ultimate curvature (1/IN)

εcu = ultimate confined concrete strain (1/IN) εsu

R = reduced ultimate tensile strain (IN/IN)

cc = neutral axis to edge of confined core (IN) d-c = neutral axis to extreme tension bar (IN) Bo = column diameter / width (FT)

47

Effective Stiffness - Ece*Ieff

Ece*Ieff = My / φy

where: Ece = Expected modulus of elasticity for concrete My = moment at first yield (expected materials at first yield) φy = curvature at first yield (expected materials at first yield) So,

Ieff = My /( φy *Ece) And typically,

0.7 < Ieff / Ig < 0.3

48

Approximate Methods – Ieff / Ig

Ieff/Ig ~ 0.2 + 0.1*ρl + 0.5*(P/f’ce*Ag)

where: ρl = longitudinal reinforcement ratio in % = Ast / Ag * 100 < 3% practical limit P = axial load – keep below 0.2*Ag*f’ce

SGS 2011 49

Approximate Methods - Mp

Mp ~ D3 * [0.05 + 0.2*ρl + P/(f’ce*Ag)] where: Mp = idealized plastic moment for circular section (K-IN) D = column diameter (IN) > 30 IN ρl = longitudinal reinforcement ratio in % = Ast / Ag * 100 < 4% but 3% is practical limit P = axial load on column (K) < 0.2 * f’ce* Ag f’ce = expected concrete strength (KSI) Ag = gross area of column (IN^2)

• Assumes ASTM A 706 Grade 60 reinforcing steel and f’ce = 5.2 50

Approximate Methods - Mp

Mp ~ b*d2 * [0.15 + 0.25*ρl + 1.5*P/(f’ce*Ag)] where: Mp = idealized plastic moment for rectangular section (K-IN) d = column depth in direction of loading (IN) > 30 IN b = column width (IN) > 30 IN ρl = Ast / Ag * 100 = longitudinal reinforcement ratio in % P = axial load on column (K) < 0.2 * f’ce* Ag f’ce = expected concrete strength (KSI) Ag = b * d = gross area of column (IN^2)

• Assumes ASTM A 706 Grade 60 reinforcing steel and f’ce = 5.2 51

Moment – Curvature Check

• Diameter, D = 48 IN • Ig = D4π/64 = 260576 IN4

• ρl = Ast / Ag = 0.01724 = 1.724% • Axial Load Ratio, ALR = P / (f’ce*Ag) = 0.1 [P ~ 940 K] • εye = fye/Es = 68/29000 = 0.002345 εR

su = 0.06 (#11)

φyi ~2.25*0.00234/48 = 0.000110 1/IN v. 0.0001132 1/IN φu ~ 0.06/48 = 0.00125 1/IN v. 0.001075 1/IN Ieff ~ (0.2+.1*1.724+0.5*0.1)*260576 = 0.42*260576 Ieff ~ 109963 IN4 v. 105307 IN4 4% Mp ~ 483(0.05+0.2*1.724+0.1) Mp ~ 54713 K-IN v. 51626 K-IN 6%

52

Moment – Curvature Check

Predicted Response

Approximate Method

Idealized Response

XTRACT™ Response

53

• Calculate deflections using M-φ result

• Elastic deformation component, ∆yi

• Plastic deformation component, ∆p

• Foundation deformation component – important

but not specifically addressed in this workshop

• Conservative to neglect shear deformations

Force - Displacement

54


Caltrans SDC Version 1.7 55

Analytical Plastic Hinge Length - Lp

• Approximation used to simplify analysis

• Converts (integrates) curvature to rotation

• Includes a moment gradient part (integration), tension shift (diagonal cracking) and a strain penetration part (yielding into cap, footing or shaft)

• Calibrated to the failure condition only and modification may be needed for full strain-displacement response 56

Analytical Plastic Hinge Length - Lp

Lp = k * L + Lsp > 2 * Lsp

Lp = 0.08 * L + 0.15 * fye * dbl > 0.3 * fye * dbl

where:

k = 0.2*(fue/fye – 1) ≤ 0.08 L = length of column from point of maximum moment to the point of moment contraflexure (IN) Lsp = strain penetration component = 0.15*fye*dbl (IN) fye = expected yield stress of longitudinal bars (KSI) fue = expected tensile strength (KSI) dbl = diameter of longitudinal column bars (IN)

57 SGS 2011

Predicted Force – Displacement

∆y ~ 1/3*φy*(L+Lsp)2

∆(M, φ) ~ ∆y*(M/My)+ (φ - φy )*Lp*(L-Lp/2)

where: φy = curvature at first yield L = column height Lp = analytical plastic hinge length Lsp = strain penetration φ = curvature at point of interest My = moment at first yield M = moment associated with φ F = force associated with ∆ = M / L


Idealized Force – Displacement

∆yi ~ 1/3*φyi*(L+Lsp)2

∆LC = ∆yi + ∆p ~ ∆yi + (φu - φyi )*Lp*(L-Lp/2)

where: φyi = idealized yield curvature ∆yi = idealized yield displacement ∆p = plastic displacement capacity L = column height Lp = analytical plastic hinge length Lsp = strain penetration φu = ultimate curvature ∆L

C = ultimate displacement Fp = plastic force = Mp / L


Idealized Force - Displacement • So the deformation values for the example problem are:

∆yi = 1/3*0.0001132*(240+14.38)2 = 2.44 IN

∆LC = 2.44 + (0.001075 – 0.0001132 )*33.58*(240-33.58/2) = 9.62 IN

where: φyi = 0.0001132 1/IN L = 20 FT = 240 IN Lsp = 0.15*68*1.41 = 14.38 IN Lp = 0.08*240 + 14.38 = 33.58 IN > 28.8 IN φu = 0.001075 1/IN Mp = Myi = Mu = 51626 K-IN Fp = Mp / L = 51626 / 240 = 215 KIP µD = ∆L

C / ∆yi = 9.62 / 2.44 = 3.94

60

• The idealized response for the single column example


Approximate Mp, φyi and φu

Predicted Response from Spreadsheet calculated M-φ

Idealized Spreadsheet Mp, φyi and φu

61

Approximate Methods

• As with M-φ, we prefer a simplified method to check the computer results

• simple method to check the computer results

• could use the closed-form AASHTO equations to start but they are developed for specific target ductility / strain limits

• use these approximations with due skepticism

62

Approximate Methods - ∆yi

• For a rough check of conventional circular reinforced concrete column sections: ∆yi ~ 1/3*φyi*(12*L+0.15*fye*db)2 ~ L2 / 42Bo

where: ∆yi = idealized yield displacement (IN)

L = contraflexure to plastic hinge distance (FT) db = diameter of longitudinal column bar (IN) φyi = idealized yield curvature ~ 2.25*εye/Bo (1/IN)

Bo = column diameter (FT) fye = expected yield stress (KSI)

63

Approximate Methods - ∆yi

• For a rough check of conventional rectangular reinforced concrete column sections: ∆yi ~ 1/3*φyi*(12*L+0.15*fye*db)2 ~ L2 / 45Bo

where: ∆yi = idealized yield displacement (IN)

L = contraflexure to plastic hinge distance (FT) db = diameter of longitudinal column bar (IN) φyi = idealized yield curvature ~ 2.1*εye/12Bo (1/IN)

Bo = column width in direction of loading (FT) fye = expected yield stress (KSI)

64

Approximate Methods - ∆cL

• And for a very rough check of conventional reinforced concrete columns with L/Bo > 4 : ∆L

C ~ ∆yi + (φu - φyi )*Lp*(12*L-Lp/2) ~ L2 / 10Bo

where: ∆L

C = local displacement capacity (IN) L = contraflexure to plastic hinge distance (FT) Lp = analytical plastic hinge length (IN) φu = ultimate curvature ~ εsu

R /12Bo (1/IN) φyi = idealized yield curvature (1/IN)

Bo = column diameter / width (FT) 65

Force-Displacement Check

• Diameter, D = 48 IN => Bo = 4 FT

• Height, L = Ho = 20 FT

• Mp ~ 54713K-IN/12 = 4560 K-FT (see previous check)

∆yi ~ L2 / 42Bo = 202/(42*4) = 2.4 IN v. 2.4 IN ∆L

C ~ L2 / 10Bo = 202/(10*4) = 10 IN v. 9.6 IN

Fp = Mp / L = 4560/20 = 228 KIP v. 215 KIP

66

• Idealized response for the single column example


Approximate Mp, ∆yi and ∆L

C

Spreadsheet calculated Mp, φyi and φu

67

What about Double Curvature? • Effective column height, L, is taken from the maximum

moment location to the contraflexure point • Then add the displacement results for each part

L1

L2

SGS 2011 68

What about P-∆ Effects?

Caltrans SDC Version 1.7

From equilibrium and summing moments about the base of the column: Mp = Fp * Ho + P * ∆

69

• The moment left to resist lateral forces becomes,

M’p = Mp – P * ∆

F’P = M’P / Ho

• But when using an idealized elastic perfectly-plastic force-displacement relationship check (SGS method)

Pdl * ∆r ≤ 0.25 * Mp


70

What about P-∆ Effects? • Adjusted response for the single column example

Unadjusted for P-∆

Adjusted for P-∆

71

• Rearranging the P-∆ limit the maximum permissible deflection for the single column example

∆r ≤ 0.25 * Mp / Pdl

∆r ≤ 0.25 * 51626 K-IN / 940 K = 13.7 IN

• In this case, the P-∆ limit is greater than the calculated ∆L

C value

• With more confinement, the P-∆ limit may govern


72

Local Ductility v. Global Ductility

• Use local member displacements

∆DL < ∆C

L

µD = ∆DL / ∆yi

where: ∆D

L = Local member deformation demand ∆C

L = Local member deformation capacity µD = local member displacement ductility demand

∆yi = idealized yield deformation 73

Local Ductility v. Global Ductility


• Must remove the deformation components associated with non-column deformation demands

Design Example

D = 48 IN

Ho =

24

FT

#5 hoop @ 4IN

20 #11

2 IN clr. 940 KIP 940 KIP

Z = 32 FT

d = 4 FT

Rigid cap and footings

75

Design Example • Assume that the point of contraflexure is at column mid-height

L1 =L = Ho / 2

L2 = L = Ho / 2


Design Example – Predict Response • Use approximate methods to predict expected response • Diameter, D = 48 IN => Bo = 4 FT • Height, Ho = 24 FT so for the transverse direction, L = 12 FT • P = PDL ± PEQ • PEQ = (Mp-L + Mp-R) * (1 + d / Ho) / Z why? • Use P = PDL = 940 K for first iteration

• Mp ~ 54720K-IN/12 = 4560 K-FT (see previous calculations)

∆yi ~ 2 * L2/42Bo = 2 * 122/(42*4) = 1.71 IN ∆L

C ~ 2 * L2/10Bo = 2 * 122/(10*4) = 7.2 IN

Fp = (2*Mp-L + 2*Mp-R) / Ho ~ 4 * Mp /24 = 760 KIP why? 77

Design Example - FBD

d /2

L =

Ho /

2

Moment Diagram – slope of line is the shear – shear in cap beam is axial force in column

Mp-L Mp-R

Mp-L Mp-R

Mp-L*(1 + d/Ho) Mp-R*(1 + d/Ho) Fp

78

Moment contraflexure

Design Example – Predict Response • Perform a second iteration to verify initial assumptions • PEQ = (Mp-L+Mp-R)*(1+d/Ho)/ Z ~ 2*54720*1.167 /12/32 ~ 332 K

• P = PDL ± PEQ = 940 -/+ 332 = 608K and 1272K [compression]

• Mp-L ~ (0.05+0.2*1.724+0.0643) * 483 = 50808 K-IN

• Mp-R ~ (0.05+0.2*1.724+0.130) * 483 = 58612 K-IN

• PEQ = (Mp-L + Mp-R)*(1 + d / Ho) / Z • PEQ = (50808+58612)(1.167)/32/12 = 332 K

• Fp = (2 * Mp-L + 2 * Mp-R) / Ho = 760 KIP

79

Design Example - Process

• Now use the refined analysis to determine the force-displacement response of the pier

• First calculation idealized M-φ

• Then calculate Lsp and Lp

• Then calculate the force-displacement for each column

• Then add the results of each column to find the total pier

response

80

Design Example – Refined Analysis

81

Design Example – Refined Analysis • Verify that the axial forces are reasonably close to the initial

estimate PEQ = (Mp-L + Mp-R) * (1 + d / Ho) / Z PEQ = (48172+55174)*1.167 /(12*32) ~ 314 K v. 332k P = PDL - PEQ = 940 - 314 = 626K v. 608K – close enough @ 3% P = PDL + PEQ = 940 + 314 = 1254K v. 1272K – close enough @ 2% • Use the initial values since they are close 82

Design Example – Hinge Formation

• Order of hinge formation and hinge failure

83

Design Example – Displacement

• For the left / trailing side column:

∆yi-L = 1/3*0.0001106*(144+14.38)2 = 0.92 IN

∆LC-L = 0.92 +(0.0011844– 0.0001106 )*28.8*(144-28.8/2) = 4.92 IN

where: φyi = 0.0001106/IN (from M-φ analysis) L = Ho / 2 = 24 FT / 2 * 12 = 144 IN Lsp = 0.15*68*1.41 = 14.38 IN Lp = 0.08*144 + 14.38 = 25.9 IN < 2*Lsp = 28.8 IN φu = 0.0011844 1/IN (from M-φ analysis) Mp = 48172 K-IN (from M-φ analysis) Fp-L = Mp / L = 48172 / 144 = 335 KIP

84

Design Example – Displacement

• For the right / right side column:

∆yi-R = 1/3*0.00011231*(144+14.38)2 = 0.94 IN

∆LC-R = 0.94 +(0.0009898 – 0.00011231)*28.8*(144-28.8/2) = 4.21 IN

where: φyi = 0.00011231/IN (from M-φ analysis) L = Ho / 2 = 24 FT / 2 * 12 = 144 IN Lsp = 0.15*68*1.41 = 14.38 IN Lp = 0.08*144 + 14.38 = 25.9 IN < 2*Lsp = 28.8 IN φu = 0.0009898 1/IN (from M-φ analysis) Mp = 55174 K-IN (from M-φ analysis) Fp-R = Mp / L = 55174 / 144 = 383 KIP

85

Design Example Left Column Right Column

Axial compression, P 608 1272

Effective stiffness, (EI)eff 4.36E9 4.91E9

Plastic moment, Mp 48172 55174

Idealized yield curvature, φyi 0.00011060 0.00011231

Ultimate curvature, φu 0.0011844 0.0009898

Plastic curvature, φp 0.001074 0.000877

Analytical Plastic Hinge Length, Lp 28.8 28.8

∆yi (per half of column height) 0.92 0.94

∆p (per half of column height) 4.00 3.27

∆u (per half of column height) 4.92 4.21

Plastic shear, Fp 335 383 86

Design Example – Combined Response

• Total effective lateral force Fp = Fp-L + Fp-R = 335K + 383K = 718 K

• Idealized yield displacement ∆yi ~ (2*∆yi-L + 2*∆yi-R) / 2 = 0.92 + 0.94 = 1.86 IN

• Ultimate displacement (first hinge failure on right) ∆L

c = 2 * ∆Lc-R = 2 * 4.21 IN = 8.42 IN

87

Design Example – Idealized Response

Combined

Left / Trailing Column

Right / Leading Column

88

Design Example – Refined Response • Idealized force displacement for the two column pier

Approximate Mp, ∆yi and ∆L

C

Spreadsheet Idealized Mp, φyi and φu

89

• The maximum deflection based on P-∆ limits

∆r ≤ 0.25 * 48172 K-IN / 940 K = 12.8 IN (per half)

• And the displacement ductility capacity µD < ∆L

c /∆yi = 8.42 / 1.86 = 4.5

Design Example – Verification

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Capacity Design

• Designer dictates the bridge response (e.g., column hinge response mechanism – Type I)

• Preclude all failure modes that would prevent the formation of the predicted force-deflection response including:

– Shear failure inside and outside the plastic hinge region (not the same as analytical plastic hinge length, Lp)

– Column-cap and column-footing joint failure – Cap beam, footing or shaft moment, shear, and axial

overload / hinging – Footing overturning, sliding, uplift and rocking failure – Any other failure that occurs prior to reaching ∆L

C

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Thank you - Questions

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Seismic Design Tools

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