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Force Based DesignFundamentals

Ian BuckleC f CDirector Center for Civil Engineering Earthquake Research

University of Nevada, Reno

Learning OutcomesgExplain difference between elastic forces, actual forces and Modified Design Forcesactual forces and Modified Design ForcesDefine a plastic hingeE l i C it P t ti Phil hExplain Capacity Protection PhilosophyDescribe the Component Capacity/Demand R t fit M th d (M th d C)Retrofit Method (Method C)Describe how P-Δ effects are considered in d idesign

Force vs. Displacement Force Displ. RetroTopic Applicability

Elastic Response

rce FEQ

ΔEQ

eral

For FEQ

Late

Displacement - ΔΔEQ

p

Elastic Analysis & Response

Force vs. DisplacementTopic ApplicabilityForce Displ. Retrop

For moderate-large earthquakes, FEQ is a very large force (can be > structure weight) and it islarge force (can be > structure weight) and it is uneconomical to design an ordinary bridge to resist this force elasticallyInelastic behavior (damage) is therefore accepted, provided it

does not cause collapsedoes not cause collapseis ductile in nature (not brittle), and occurs in designated components (members)

Exceptions are made for special structuresExceptions are made for special structures where the higher cost of elastic (or almost elastic) behavior can be justified

Force vs. Displacement Force Displ. RetroTopic Applicability

orce

Pl ti hi

Fmechanism

Fmechanism

ater

al F

o

First-Order Rigid-Plastic Analysis

Plastic hinge

FLa

Displacement - Δ Plastic hinge

Fmechanism

First Order Rigid-Plastic analysis assumes all deformations take place at discrete regions, called plastic hinges, and that a sufficient number of plastic

First-Order Rigid Plastic Pier Response

d sc ete eg o s, ca ed p ast c ges, a d t at a su c e t u be o p ast chinges have formed to form a mechanism in the pier/bent.

Force vs. Displacement

c

Force Displ. RetroTopic Applicability

Elastic Response

rst-O

rder

gid-

Pla

stic

espo

nse

l For

ce Fir

Rig

Re

Late

ra IdealizedElasto-PlasticResponse

Displacement - Δ

Idealized Elasto-Plastic Response

Force vs. Displacement Force Displ. RetroTopic Applicability

orce FEQ

ΔEQActual Response

Fmax

ater

al F

o

2 3 4 Plastic Hinge

Fmax

La

Displacement Δ

1

Displacement - ΔActual Response Milestones

1 P d i ld P i t1 - Pseudo-yield Point2 – Maximum Plastic Deformations3 – Onset of Collapse4 - Collapsep

Force vs. Displacement Force Displ. RetroTopic Applicability

Elastic Response

FEQF

ΔEQ

Actual

l For

ce

FEQActual Response

Late

ra

Displacement - ΔΔEQ

p

Elastic vs. Actual Response

Force vs. Displacement

c

Force Displ. RetroTopic Applicability

Elastic Response

A t l Rrst-O

rder

gid-

Pla

stic

espo

nse

l For

ceActual ResponseFi

rR

igR

e

Late

ra IdealizedElasto-PlasticResponse

Displacement - Δ

Idealized Elasto-Plastic Response

Force vs. Displacement Force Displ. RetroTopic Applicability

Elastic Analysis

orce

First-OrderElastic-Plastic

ater

al F

o

Second-Order Elastic-Plastic (slender column response shown- requirements are

f )La

Displacement - Δ

in place to limit this type of behavior)

Elastic Analysis conducted using to linear methodsFirst-order elasto-plastic behavior includes reduced pier/bent stiffness and strength with progressive plastic hinge formation.

Pier Response with Higher Order Effects

Second-order elasto-plastic behavior traces formation of plastic hinges and includes geometric non-linearities (e.g. P-Δ)

Pier Response with Higher Order Effects

Force vs. DisplacementTopic ApplicabilityForce Displ. Retro

p

Member strengths:gNominal strength, SnDesign strength, Sd = φ Sn

h φ i th t th d ti f t <1 0where φ is the strength reduction factor <1.0 (see AASHTO LRFD Specifications)[Expected strength, Se = φe Sn[Expected strength, Se φe Snwhere, in absence of mill certificates, φe=1.2 for steel and 1.3 for concrete members] Over-strength, So = φo Snwhere φo is the over-strength factor >1.0 (1.7 – 2.7)

Force vs. Displacement Force Displ. RetroTopic Applicability

e D i f l f

Felastic

al F

orce Design for lesser of

Felastic or FoverstrengthFoverstrengthFmechanism

Overstrength Factor

Late

ra Force Displ. Retro1.7 1.7

Overstrength Factor

2.0

Displacement - Δ

Foverstrength = φo Fmechanism where φo = over-strength factor

Over-strength Forces

overstrength φo mechanism φo g

Force vs. Displacement Topic ApplicabilityForce Displ. Retro

cera

l For

c

μ = Δ

Late

r

Δ Δ

μ = ΔmaxΔyield

Displacement - Δ

Δyield Δmax

Displacement Ductility Factor - μ

Force vs. DisplacementTopic ApplicabilityForce Displ. Retrop

To calculate ΔmaxNonlinear time history analysis using software such as SAP2000, SEISAB-NL, ADINA, ABAQUS…Elastic analysis using either ULM SMSA MMSA THElastic analysis using either ULM, SMSA, MMSA, TH to find maximum elastic displacement ΔEQ and then assumeeither: Equal displacements, i.e. Δmax = ΔEQ

or: Equal work done during elasto-plastic response as during elastic response, and solve for Δmax

Force vs. DisplacementTopic ApplicabilityForce Displ. Retro

rce

Felastic

Then:Δ = ΔEQ

tera

l Fo

Felastic R and:

Δ R

FelasticDefine R =

Fmechanism

Fmechanism

Δmax ΔEQ

Lat

Δmax = ΔEQΔyield

μ = Δmax = RΔyield

(Displacement

mechanism

(Response Modification Factor)

Relationship between ductility and “R”

Displacement - Δ( pDuctility Factor)

p yEqual Displacement Assumption

Force vs. DisplacementTopic ApplicabilityForce Displ. Retro

rce

Felastic

Then:Δ = ΔEQ

tera

l Fo

Felastic R and:

Δ R

FelasticDefine R =

Fmechanism

Fmechanism

Δmax ΔEQ

Lat

Δmax = ΔEQΔyield

μ = Δmax = RΔyield

(Displacement

mechanism

(Response Modification Factor)

Relationship between ductility and “R”

Displacement - Δ( pDuctility Factor)

p yEqual Displacement Assumption

Force vs. DisplacementTopic ApplicabilityForce Displ. Retrop

Felastic Then:

Forc

eand:

Felastic R

FelasticDefine R =

Fmechanism F

Δmax= (R ΔEQ + Δyield) / 2

Late

ral and:

μ = Δmax ≈ (R2 + 1) / 2Δyield

(Displacement

Rmechanism

(Response Modification Factor)

Fmechanism

Displacement - ΔΔyield Δmax

Ductility Factor)ΔEQ

Relationship between ductility and “R” Equal Work Done Assumption

Force vs. DisplacementTopic ApplicabilityForce Displ. Retro

pFelastic Then:

Forc

eand:

Felastic R

FelasticDefine R =

Fmechanism F

Δmax= (R ΔEQ + Δyield) / 2

Late

ral and:

μ = Δmax ≈ (R2 + 1) / 2Δyield

(Displacement

Rmechanism

(Response Modification Factor)

Fmechanism

Displacement - ΔΔyield Δmax

Ductility Factor)ΔEQ

Relationship between ductility and “R” Equal Work Done Assumption

Force vs. Displacement Force Displ. RetroTopic Applicability

rce

μ = R = 1

ilityμ = Δmax ≈ R

Δyield

eral

For

μ = R = 2

ng D

uctiyield

Late

μ = R = 3

μ = R = 5

ncre

asin

ΔyieldΔmax

InStrength and Ductility Relationship

Displacement - Δyield

Substructure Type

Force vs. Displacement Force Displ. RetroTopic Applicability

rce

μ = R = 1

ilityμ = Δmax ≈ R

Δyield xibi

lity

ΔE

Q

eral

For

μ = R = 2

ng D

uctiyield

sing

flex

high

er Δ

Late

μ = R = 3

μ = R = 5

ncre

asin

Incr

eas

perio

d,

ΔyieldΔmax

In

NO

TE:

long

er

Strength and Ductility RelationshipDisplacement - Δ

yield

Substructure Type

Force vs. Displacement Force Displ. RetroTopic Applicability

rce

μ = R = 1

ilityμ = Δmax ≈ R

Δyield xibi

lity

ΔE

Q

eral

For

μ = R = 2

ng D

uctiyield

sing

flex

high

er Δ

Late

μ = R = 3

μ = R = 5

ncre

asin

Incr

eas

perio

d,

ΔyieldΔmax

In

NO

TE:

long

er

Strength and Ductility RelationshipDisplacement - Δ

ΔyieldSubstructure Type

Force vs. Displacement Force Displ. RetroTopic Applicability

e ct R

Felastic

F

AxialForce (Pn , Mn)

(Pn , Mn) w/ Over-strength

al F

orce

Effe

cof

R

Fmechanism = Fdesign

F

Foverstrength

ΦPn ΦMn

P Mign

Late

ra FelasticR

P,Melastic

P,M

des

Displacement - Δ MomentFelastic & P,Melastic force effect generated from Extreme Event 1 load combination

Elastic vs. Modified Design vs.Over-strength Forces

Capacity Protection Topic ApplicabilityForce Displ. Retrop y

The principle of Capacity Protection DesignThe principle of Capacity Protection Designis that yielding in one component (or member) will cap (or limit) the forces andmember) will cap (or limit) the forces and moments in an adjacent component (member)(member). i.e. reaching the elastic capacity (yield) of one member protects adjacent membersone member protects adjacent members from excessive forces

Capacity Protection Topic ApplicabilityForce Displ. Retrop y

F, Δ A B C D

Bridge Geometry

, A B C D

h1 h2

EF

2

Capacity Protection Topic ApplicabilityForce Displ. Retrop y

F, Δ F, ΔA B C D

Bending Moment Diagram

Mp

Bridge Geometry

1

2 4

, ,

VEB

A B C D

h1 h2 M

p

3EF

VFC

Two plastic hinges in BE,

2 Mp

VEB max = 2 Mp / h1

Capacity Protection Topic ApplicabilityForce Displ. Retrop y

F, Δ F, ΔA B C D

Bending Moment Diagram

Mp

Bridge Geometry

1

2 4

, ,

VEB

A B C D

h1 h2 M

p

3

EB

EF

VFC

Two plastic hinges in BE,

2 Mp

FVEB max = 2 Mp / h1

F

1

23 4

VFC

Δ

1

VEB

F = VEB + VFC

Capacity Protection Topic ApplicabilityForce Displ. Retrop y

F, Δ F, ΔA B C D

Bending Moment Diagram

Mp

Bridge Geometry

1

2 4

, ,

VEB

A B C D

h1 h2 M

p

3

EB

EF

VFCTwo plastic hinges in EB, V = 2 M / h

2 Mp

F VEB max = 2 Mp / h1F

1

23 4

VFCMaximum shear and moment in foundation at E is limited by yielding

Δ

1

VEB

F = VEB + VFC

foundation at E is limited by yielding in EB, i.e. the foundation is capacity protected by column yield

Capacity Protection Topic ApplicabilityForce Displ. Retrop y

F, Δ F, ΔA B C D

Bending Moment Diagram

Mp

Bridge Geometry

1

2 4

, ,

VEB

A B C D

h1 h2 M

p

3

EB

EF

VFCTwo plastic hinges in EB, V = 2 M / h

2 Mp

F VEB max = 2 Mp / h1F

1

23 4

VFCNOTE: For design Mp = φ Mn, but to calculate max value of V should

Δ

1

VEB

F = VEB + VFC

calculate max. value of VEB, should use the overstrength moment, i.e. Mp=φo Mn where (φo > 1)

Seismic Retrofit ProcessTopic ApplicabilityForce Displ. Retro

Three step process:Three step process:Screening and PrioritizationD t il d E l tiDetailed EvaluationRetrofit – strategies, approaches and measures

Evaluation MethodsTopic ApplicabilityForce Displ. Retro

FHWA Manual describes 5 basic methodsFHWA Manual describes 5 basic methods for evaluationMethods vary in rigor from ‘no analysis’ toMethods vary in rigor from no-analysis to ‘nonlinear dynamic time history analysis’S l ti / li ti d d thSelection / application depends on the Seismic Retrofit Category A – D for the b id ( i i h d d fbridge (seismic hazard and performance required), and its geometry (regularity).

Evaluation MethodsTopic ApplicabilityForce Displ. Retro

A and B: Default capacity methodA and B: Default capacity methodchecks capacity of components due to non-seismic loads against specified g pminima – no analysis is required (e.g. connections and support lengths)C: Capacity/demand method compares component capacities against forced d f l ti l idemands from an elastic analysis - on a component-by-component basis

Evaluation MethodsTopic ApplicabilityForce Displ. Retro

D: Capacity/spectrum method(s) useD: Capacity/spectrum method(s) use nonlinear capacity models for individual piers (or complete bridge), i.e. a pushover p ( p g ) pcurve, and displacement demands from an elastic analysis E: Nonlinear dynamic method explicitly models nonlinear behavior of components i ti hi t l i f l tin a time history analysis of complete bridge

Evaluation Method CTopic ApplicabilityForce Displ. Retro

Capacity / Demand Ratio MethodCapacity / Demand Ratio MethodForce demands are calculated by elastic methods, such as multi-modal spectral analysis method (i.e. without regard to any yielding that may occur). Elastic displacements also usedCapacities are obtained from combination ofCapacities are obtained from combination of theory (strength of materials) and engineering judgment, for each major componentj g , j pGives good results for bridges that behave elastically, or nearly so

Evaluation Method CTopic ApplicabilityForce Displ. Retro

Five step procedure (FHWA 2006):Five step procedure (FHWA 2006):1. Determine applicability of the method2 Determine capacity Q i for all components (i)2. Determine capacity, Qci for all components (i)

using theory and engineering judgement (e.g. Appendix D, FHWA Manual, 2006)pp , , )

3. Determine sum of non-seismic force and displacement demands, ΣQNSi for all NSicomponents for each load combination in LRFD design specification

Evaluation Method CTopic ApplicabilityForce Displ. Retro

Procedure continued:4. Determine seismic demand, QEQi on each

component by an elastic method of analysis

5. For each component determine capacity/demand ratio from:capacity/demand ratio from:

Qci – Σ QNSir =ri =

QEQi

Evaluation Method CTopic ApplicabilityForce Displ. Retro

If ri > 1.0 component has adequate capacity

If 0.5 < ri < 1.0 component may be acceptable i ywithout retrofitting, depending on

other deficiencies in member, if any, and f f ilconsequences of failure

If 0.5 > ri component has inadequate capacity and retrofitting is indicated

Evaluation Method CTopic ApplicabilityForce Displ. Retro

Note that for each component, several c/d ratios (ri) may need to be calculated. Example 1: Support lengths and bearingsp pp g g

Evaluation Method CTopic ApplicabilityForce Displ. Retro

Example 2: ColumnsFive (5) c/d ratios should be found for each column for the following: g

Anchorage length of longitudinal rebar, rca

Transverse confinement, rccTransverse confinement, rcc

Splice length, rcs

Shear force rrsplice

r con

finem

ent

Shear force, rcv

Bending moment, rec ranchorage

r

P-Δ Effects in Bridge Columns

Topic ApplicabilityForce Displ. Retro

ΔPPP

Bridge Columns

P

FF, Δ

PP

Fh Kpier

Kpier

Kpier h

Equilibrium in deformed state requires ΣMA = 0 i e F h + P Δ = (K i Δ) h

A

Equilibrium in deformed state requires ΣMA 0, i.e. F h + P Δ (Kpier Δ) h

Therefore F = (Kpier – P / h) Δ = K'pier Δ

P-Δ Effects in Elastic Bridge Columns

Topic ApplicabilityForce Displ. Retro

Elastic Bridge ColumnsElastic capacity curve for

Forcecolumn, slope = Kpier Elastic capacity curve

with P-Δ included, slope = K'pier

FLoss in capacity at displacement Δ, due to P-Δ effect

= PΔ / h

DisplacementΔ

P-Δ Effects in Yielding Bridge Columns

Topic ApplicabilityForce Displ. Retro

Yielding Bridge ColumnsElasto-plastic capacity curve for column; initial slope = Kpier, second slope = 0 Elasto-plastic capacity curve

Force

p pier p Elasto-plastic capacity curve with P-Δ included;initial slope = K'piersecond slope = - P/h

FLoss in capacity at displacement Δdue to P-Δ effect

= PΔ / h

DisplacementΔyield Δ

P-Δ Effects in Yielding Bridge Columns

Topic ApplicabilityForce Displ. Retro

Yielding Bridge ColumnsElasto-plastic capacity curve for column; initial slope = Kpier, second slope = 0 Elasto-plastic capacity curve

Force

p pier p Elasto-plastic capacity curve with P-Δ included;initial slope = K'piersecond slope = - P/h

FLoss in capacity at displacement Δdue to P-Δ effect

= PΔ / h

DisplacementΔyield Δ

P-Δ Effects in Yielding Bridge Columns

Topic ApplicabilityForce Displ. Retro

Yielding Bridge ColumnsAASHTO LRFD Specifications andAASHTO LRFD Specifications and FHWA Retrofit Manual require that if

PΔ / h > 0 25 Mp / hPΔ / h > 0.25 Mp / ha refined analysis must be undertaken that explicitly includes nonlinear geometricexplicitly includes nonlinear geometric effects (P-Δ)Encouraged to limit Δ such thatEncouraged to limit Δ such that

Δmax < 0.25 Mp / P

Learning OutcomesgExplain difference between elastic forces, actual forces and Modified Design Forcesactual forces and Modified Design ForcesDefine a plastic hingeE l i C it P t ti Phil hExplain Capacity Protection PhilosophyDescribe the Component Capacity/Demand R t fit M th d (M th d C)Retrofit Method (Method C)Describe how P-Δ effects are considered in d idesign

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