Reinforced Concrete Beam - Column Joint: Macroscopic Super...

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Page 1: Reinforced Concrete Beam - Column Joint: Macroscopic Super ...nilanjan/RCBC-Jnt-SuperEle-model.pdf · Reinforced Concrete Beam - Column Joint: Macroscopic Super-element models-Nilanjan

Reinforced Concrete Beam-Column Joint: Macroscopic

Super-element models

-Nilanjan Mitra

(work performed as a PhD student while at University of Washington between 2001-2006)

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Need for the study

Reinforced concrete beam column jointssubjected to earthquake loading

Experimental Investigation@ UW

I-280 Freeway, San Francisco, CAfollowing Loma Prieta Earthquake in 1989

Courtesy: NISEE, Univ. of California, Berkeley.

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Loading in a joint region

Earthquake Loading of Beam-Column Joint

compression resultant (concrete and steel)

shear resultant (concrete)

Earthquake Loading of Beam-Column Joint

compression resultant (concrete and steel)

compression resultant (concrete and steel)

shear resultant (concrete)

shear resultant (concrete) tension resultant (steel)

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anchorage bond stress acting on joint core concrete

compression force carried by joint core concrete

Internal load distribution in a joint

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Macroscopic beam-column joint element models

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Macroscopic beam-column joint element models

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Macroscopic beam-column joint element models

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shear panel

external node

internal node

rigid externalinterface plane

shown with finite widthto facilitate discussion

beam element

zero-width region

interface-shear spring

bar-slip spring

zero-length

zero-length

elem

ent

colu

mn

Proposed Beam-column super-element model• 4-noded 12-dof element• 8 bar-slip springs to simulate

anchorage failure• 4 interface-shear springs to simulate

shear transfer failure at joint interface• 1 shear-panel to simulate inelastic

action of shear within joint core

Note: The location of the bar-slipsprings is at the centroid of thetension-compression couple at nominalstrength of the beams.

[Mitra & Lowes; J. Structural Eng. ASCE, 2007: 133 (1): 105-20 ]

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Joint element formulation: Kinematics

External, Internal and Component deformation

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Joint element formulation: Equilibrium

External, Internal and Component forces

Solution of element state achieved by an iterative procedure and requires solving for zero reaction in the 4 internal degrees of freedom

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Characterized by Response envelope Unload reload path Damage rules

Hysteretic one dimensional material model

deformation

load

(ePd1,ePf1)

(ePd4,ePf4)

(ePd3,ePf3)(ePd2,ePf2)

(eNd3,eNf3)(eNd2,eNf2)

(*,uForceP.ePf3)

(dmin,f(dmin))

(dmax,f(dmax))

(rDispP.dmax,rForceP.f(dmax))

(rDispN.dmin,rForceN.f(dmin))

(*,uForceN.eNf3)

(eNd1,eNf1)

(eNd4,eNf4)

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Damage simulation in material model

-0.015 -0.01 -0.005 0 0.005 0.01 0.015-8

-6

-4

-2

0

2

4

6

8

deformation

load

without damagewith unloading stiffness damage

-0.015 -0.01 -0.005 0 0.005 0.01 0.015-8

-6

-4

-2

0

2

4

6

8

deformationlo

ad

without damagewith reloading stiffness damage

-0.015 -0.01 -0.005 0 0.005 0.01 0.015-8

-6

-4

-2

0

2

4

6

8

deformation

load

without damagewith strength damage

( ) ( )( )3 4

1 max 2i dα αδ α α χ= +

max minmax

max min

max. ,i id dd

def def

=

( ).f No of load cyclesχ =

( )f Accumulated Energyχ =

( )0 1 ki ik k δ= −

( ) ( ) ( )max max 01 f

iif f δ= −

( ) ( ) ( )max max 01 d

iid d δ= +

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Damage simulation in material model

load historyi

monotonic

monotonic load history

dEE

EgE dE

χ = =

-0.015 -0.01 -0.005 0 0.005 0.01 0.015-8

-6

-4

-2

0

2

4

6

8

deformation

load

without damagewith all 3 damage rules

-0.015 -0.01 -0.005 0 0.005 0.01 0.015-5

-4

-3

-2

-1

0

1

2

3

4

5

deformation

load

with all 3 damages (Energy)with all 3 damages (Cyclic)

max4duu

χ =∑

Energy criterion

No. of load cycle criterion: rain-flow-counting algorithm

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Shear-panel calibration

column

shear panel

• Shear panel envelope calibration• MCFT• Diagonal compression strut

• Compression envelope reduction

• Determination of hysteretic model parameters

Typical response envelope

Observed Simulated

Spec

imen

SE8

(S

teve

ns e

t al.

1987

)

-0.012 -0.008 -0.004 0 0.004 0.008 0.012-10

-8

-6

-4

-2

0

2

4

6

8

10

Shear strain

Shea

r st

ress

(MPa

)

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Shear panel envelope calibration using proposedDiagonal compression strut mechanism

_ cosc strut strut strutstrut

jnt

f ww

ατ

⋅ ⋅=

• Mander et al. (1988) concrete

• Column longitudinal and joint hoop

steel confine the strut.

• Reduction in concrete to account for

perpendicular tensile stress to the strut

cyclic loading.

• Strut force is converted to panel shear

stress as

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2_

_3.62 2.82 1 0.39

0.45 0.39

c strut t t t

c Mander cc cc cc

t

cc

ff

ε ε εε ε ε

εε

= − + ∀ <

= ∀ ≥

Proposed concrete compression envelope reduction

0 1 2 3 4 5 60

0.2

0.4

0.6

0.8

1

εt / εcc

f c_ob

s / f c_

Man

der

Data with ρj > 0

Data with ρj = 0

Vecchio 1986Stevens 1991Hsu 1995Noguchi 1992Proposed eq. for ρ

j > 0

Proposed eq. for ρj = 0

2_

_0.36 0.60 1 0.83

0.75 0.83

c strut t t t

c Mander cc cc cc

t

cc

ff

ε ε εε ε ε

εε

= − + ∀ <

= ∀ ≥

0jρ =0jρ >Eq. for Eq. for

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Comparison of MCFT and Diagonal Compression Strut model in shear-panel envelope calibration

0 0.5 1 1.5 2 2.50

0.5

1

1.5

2

2.5

φ

τ mcf

t_cy

clic

/ τ m

ax

0.55JFBYJFBY

0 0.5 1 1.5 2 2.50

0.5

1

1.5

2

φ

τ diag

onal

_stru

t / τ

max

JFBYJFBY

[Lowes, Altoontash and Mitra, J. Structural Eng. ASCE, 2005: 131 (6) ]

Transverse steel contribution to shear stress

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Bar slip material model calibration

column

Bar-slip spring• Mechanistic model :- envelope

• Hysteretic model calibration

• Strength deterioration model

-2 0 2 4 6 8 10 12 14 16-1000

-500

0

500

1000

slip (mm)

bar-

spri

ng fo

rce

(kN

)

Typical response envelope

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Bar slip mechanistic model

Assumptions for anchorage response of bond within the joint region:

• Bond stress uniform for elastic reinforcement, piecewise uniform for reinforcement

loaded beyond yield

• Slip is the relative movement of reinforcement bar with respect to the joint perimeter

• Slip is a function of strain distribution in the joint

• Bar exhibits zero slip at zero bar stress

2

0

2fsl

fsE b Eslip s y

b b

ldd x dx f fA E E d

τ π τ⋅= = ∀ <

⋅∫

( )0

e ye

e

l llyE b Y b

slip eb b hl

fd dd x dx x l dxA E E A E

τ π τ π+ ⋅ ⋅

= + + − ⋅ ⋅ ∫ ∫

22

2 2y y yeE Ys y

b b

f l ll f fE d E E dτ τ

= + + ∀ ≥Mechanistic model

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Strength deterioration– Is activated once slip exceeds the slip level corresponding to ultimate stress

in the reinforcing bars.– Is observed upon reloading, with the result that bar-slip springs always

exhibit positive tangent stiffness.

0 5 10 15 200

1

2

3

4

5

specimen number

max

imum

slip

/

slip

with

anc

hora

ge le

ngth

equ

al to

join

t wid

th

BYJFBY

0 5 10 15 200

5

10

15

20

specimen number

Sim

ulat

ed m

axim

um b

ar-s

lip

BYJFBY

( )max, lim max,f f

i i ult i ultd d d dδ α δ= − ≤ ∀ ≥

Strength deterioration calibration for bar-slip spring

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Steps for calibrating the joint model• Calculate moment curvature of beams and columns• From moment curvature analysis determine

• moment associated with first yield of the reinforcing bar• tension-compression couple distance at nominal yield strength• neutral axis depth at nominal yield strength

• Define joint elements parameters using joint geometry and tension-compression couple distance

• Determine concrete compression strut response• Mander model for concrete• Concrete strength reduction eq. proposed to account for perpendicular cracks

and cyclic loading• Hysteretic parameters defined for shear panel

• Determine bar-slip response• Mechanistic model for bond• Hysteretic parameters defined for bar-slip model

• Interface slip-springs are defined to be stiff and elastic

Page 22: Reinforced Concrete Beam - Column Joint: Macroscopic Super ...nilanjan/RCBC-Jnt-SuperEle-model.pdf · Reinforced Concrete Beam - Column Joint: Macroscopic Super-element models-Nilanjan

Concrete Stress-Strain(Compressive only, no tensile strength)

Reinforcing Steel Stress-Strain

Beam-Column Elements:Force based lumped plasticity element

Plastic Hinge region

Elastic region

Fiber discretisation

joint element

plastic hinge length

column axial load applied under load control

beam-column element

lateral load applied under displacement

control

Model simulationL

ab te

st

OpenSees Model

Page 23: Reinforced Concrete Beam - Column Joint: Macroscopic Super ...nilanjan/RCBC-Jnt-SuperEle-model.pdf · Reinforced Concrete Beam - Column Joint: Macroscopic Super-element models-Nilanjan

Validation study

-6 -4 -2 0 2 4 6-300

-200

-100

0

100

200

300

Drift (%)

Col

umn

shea

r (kN

)

-6 -4 -2 0 2 4 6-300

-200

-100

0

100

200

300

Drift (%)

Col

umn

shea

r (kN

)

-6 -4 -2 0 2 4 6-300

-200

-100

0

100

200

300

Drift (%)

Col

umn

shea

r (kN

)Specimen OSJ10:

Page 24: Reinforced Concrete Beam - Column Joint: Macroscopic Super ...nilanjan/RCBC-Jnt-SuperEle-model.pdf · Reinforced Concrete Beam - Column Joint: Macroscopic Super-element models-Nilanjan

Validation study discussion & conclusion• Failure mechanism

– For joints exhibiting JF (joint failure prior to beam yielding), 82% accurate.

– For joints exhibiting BYJF (beam yielding followed by joint failure), 89% accurate.

– For joints exhibiting BY (beam yielding), 94% accurate.• Initial and unloading stiffness

– For all joints, mean of simulated to observed ranges from 1.03 to 1.06 with an average C.O.V. = 0.15.

• Post-yield tangent stiffness– For joints that exhibit BYJF, mean ratio of simulated to observed is 1.0

with a C.O.V. = 0.22.• Maximum strength

– For all joints, mean of simulated to observed is 1.03 with a C.O.V. = 0.17.

• Drift at maximum strength– For all joints, mean of simulated to observed is 1.12 with a C.O.V. =

0.27.• Strength at final drift level

– For all joints, strength for final drift cycle is 1.04 with a C.O.V = 0.2.• Pinching ratio (ratio of strength at zero drift to maximum strength)

– For all joints, pinching ratio is 1.04 with a C.O.V = 0.12.