4 meth deform

CIVIL-706 - displacement-based methodsEPFL-ENAC-SGC 2005 -1-

Doctoral School: StructuresCIVIL-706 Advanced Earthquake Engineering

Displacement-based methods


Content

• Link to force-based methods

• Assumptions

• Reinforced concrete: chord rotation

• Capacity curve

• Examples: RC shear walls and frames

• Unreinforced masonry

• Example: masonry shear walls


Link to force-based methods• Application of behavior factor q specified in

the codes SIA 262-266

• For existing buildings, non ductile behavior(q = 1.5 or q = 2.0 for reinforced concrete)

• Displacement-based method: more realisticapproach of real seismic behavior

• Allows to justify an higher value of q


Displacement-based Method (DbM)• Applicable to deformable structures

• Brittle failure modes excluded:- shear failure- bending with concrete failure before steel yielding- bending with steel failure without sufficient plastic deformation- failure of overlapping zones or anchorage zones


Displacement-based Method (DbM)• More realistic approach of real behavior

Joe’s

Beer!Beer!Food!Food!

lateral displacement

lateralforce

Joe’s

Beer!Beer!Food!Food!

small change inforce level

large change indeformation level


Method included in SIA 2018• Slides prepared by

Prof. Alessandro Dazio, IBK- ETH ZürichErdbebeningenieurwesen und Baudynamik


Reinforced concrete: chord rotation

V

V

M

θ

shear wall

M

M

θV

V

column

θ

Vl

Ml

Vr

Mr

beam


Reinforced concrete: chord rotation• Angle formed by the tg to element axis at

Mmax location and the chord joining this pointto the zero moment point

• Represents the element sollicitation(removal of rigid displacements)

• Verification by the comparison (demand-capacity) of chord rotations at the elementlevel


Chord rotation θ: definitions

rigid body rotation

V

V

M

θ

shear wall

M

M

θV

V

column

θ

Vl

Ml

Vr

Mr

beam

= chord= tangent to element axis= zero moment point


Yielding chord rotation θy

Fy

Vy

My

θy

Lv

moment

My φy

curvature Fy

Vy

My

Δy

vyvv

y

2vy

3vy

y LL3

L3

LEIM

EI3LF

!=""#="==$


Ultimate chord rotation θu

Fu

Vu

Mu

θu

Lv

moment

Mu φyφp

φu

curvature

Lpl

Lpl = plastic hinge

( )blsvstpl dfLaL !!+!= 022.008.0

hsp

hsp = strain penetration

hpl

hpl = plastic zone


Ultimate chord rotation θu

Fu

VuMu

Δyφyφp

φu

Lpl Lpl

θy

Lv

( ) )L5.0L(LLL plvplyuvyvuu !"!!#"#+$=$=%

Δp

Δu

θp

θu


Δ

F

Non-linear force-displacement relationship

F

V

M

θ

Lv

Δ

reality

Δy Δu

Fy

Fu

approximation

Δy = f(φy), Fy = f(My), Δu = f(φy, φu), Fu = f(Mu),


Bending moment-curvature relationship

φ’y

M’y

φy

Mn

φy = φ’y Mn / M’y

M

φ


Bending moment-curvature relationship• shear wall

N = -1099 kN

200

4000

36 Ø82 Ø20 2 Ø20



First yieldεs = εyεc = 0.002

φ’y

M’y

Nominal strengthεs = 0.015εc = 0.004

φy

Mn

Nominal yield

0

1

2

3

4

5

6

0 1 2 3 4 5 6 7curvature [10-3 m-1]

bend

ing

mom

ent

[MN

m] Ultimate

εs = εs,max, εc = εc,max

φu



0

1

2

3

4

5

6

0 1 2 3 4 5 6 7curvature [10-3 m-1]

bend

ing

mom

ent [

MN

m]

φ’y φy φu

Mn

M’y

1st approximation

modified approximation


Bending moment-curvature relationship• nominal yield curvature φy


Bending moment-curvature relationship• Material mechanical properties


Material mechanical properties

0

5

10

15

20

25

0 1 2 3 4 5strain [‰]

stre

ss [M

Pa]

a) concrete BH300, SIA 168/68

1) fcd acc. to SIA 262 2) fck acc. to SIA 2018 3) fck(t) acc. to SIA 2018

1)

2)

εcu

3)

fck(t)

0

100

200

300

400

500

600

0 10 20 30 40 50strain [‰]

stre

ss [M

Pa]

b) steel IIIa, SIA 168/68

1) fsd acc. to SIA 262 2) fsk approx.3) fsk type

1)

2)3)

εuk

ftk

fsk


Material mechanical properties• Characteristic values

• Partial factor for deformation capacity γD = 1.3

• Ultimate concrete deformationεcmax = εcu = 0.004 (in general)

• Peak steel strain εsmax = αεsu - in general: α = 1.0

- if Mu<2Mcr: α = 0.5- special cases: α = ?


Application of DbM• Non-linear element behavior

• Equivalent SDOF

• Force-displacement relationship of structure

• Capacity curve

• Seismic displacement demand

• Shear strength verification


Example: reinforced concrete

• 5 stories

• Building end of 60’

• 1 x direction « wall »

• 1 x direction « frame »

• Zone 3b, CS C, CO I

Main properties


Example RC: dimensions and structurem5=213t

m4=380t

m3=380t

m2=380t

380t

A

B

C

1 2 3 4 5

11 12 13 14 15

16 17 18 19 20

6

7 8 9

10

mtot = 1733t


Example RC: element dimensionsshear wall cross-section

beam at mid span beam at connection

column type A column type B


Example RC: shear walls direction

Fd

wStructure réelle

m1

m2

m3

m4

m5

Fd

w(MDOF)

kMDOF =Σkwalls

m*=1163t

Fd

h*=1

1.95

m

k*

w/Γ

(SDOF)


Example RC: shear walls direction• Force-displacement relationship

0

100

200

300

400

500

600

700

800

900

0.00 0.05 0.10 0.15 0.20horizontal displacement, w/Γ [m]

glo

bal h

oriz

onta

l for

ce F d

[kN

m]

shear walls

building

Fu=828kN

Fy=744kN

Fu=414kNFy= 372kN

wy/Γ=0.039m wu/Γ=0.119m


Example RC: shear walls direction• ADRS spectrum with capacity curve

0

1

2

3

4

5

0.00 0.05 0.10 0.15 0.20Sud, w/Γ [m]

S ad,

F d/m

* [m

/s2 ]

elastic design spectrum (Z3b, CO I, CS C)

T=1.55s

wd/Γ wu/ΓwR,d/Γcapacity curve


Example RC: frame direction• Force-displacement rel.: ∑frames A, B and C

• Torsion neglected

A

B

C


Example RC: frame direction• Modelisation

Elément traverse

Lpl Lpl

L

αEIg (var.)

Elément pilier

Lpl

Lpl

L αEIg


Example RC: frame direction• Column element type A

M

MV

VLpl/2

= plastic hinge M/θ (rigid-plastic with softening)

Lpl/2

N

N

0100200300400500

0 10 20 30 40curvature [10 -3 m-1]

bend

ing

mom

. [K

Nm

]

N=-1500kN

-1000kN-500kN

0kN

0100200300400500

0 5 10 15 plastic rotation [10-3]

bend

ing

mom

. [K

Nm

]-1000kN

N=0kN-500kN

-1500kN


Example RC: frame direction• Beam element

-400

-300

-200

-100

0

100

200

-100 -50 0 50 100 150 200

Courbure [10-3 m-1]

Mom

ent [

KN

m]

= Rotule plastique M/! (rigide-plastique avec adoucissement)

Type 2Type 8

Type 6

Type 4

V

M

V

M

Lpl/2 Lpl/2

-400-300-200-100

0100200

-20 0 20 40 60

Rotation plastique [10-3]

Mom

ent [

KN

m]

Type 2Type 8

Type 6

Type 4

2

Type 8

6

4

2

6

4Type 8


Example RC: frame direction• Force-displacement relationship

0

500

1000

1500

2000

2500

0.00 0.05 0.10 0.15 0.20 0.25 0.30

horizontal displacement Wd [m]

gl

obal

hor

izon

tal f

orce

Fd [

kN]

frame A

frame B

building = 2 x frame A + 1 x frame B

Fy=2032kN

wy=0.098m wu=0.182m

onset of plastichinge softening


Example RC: frame direction• Ultimate deformation of frame B

Fd

wd

= Rotule= Rotule (rupture atteinte)


Example RC: frame direction• ADRS spectrum with capacity curve

wu/ΓwR,d/Γ0

1

2

3

4

5

0.00 0.05 0.10 0.15 0.20Sud, wd/Γ [m]

S ad,

F d/m

* [m

/s2 ]

elastic design spectrum (Z3b, CO I, CS C)

T=1.52s

wd/Γ

capacity curve


Unreinforced masonry• Application of DbM more delicate


DbM: unreinforced masonry• Assumptions

(FEMA 356)

plastic deformationsconcentrated in someelements (piers)


DbM: unreinforced masonry• Limitation: out-of-plane failure excluded

Molise 2002EERI


Masonry: out-of-plane failure• Control with h/t ratio (FEMA 356)


Masonry: out-of-plane failure• Control with h/t ratio (SIA 2018)

≤ 17≤ 18≤ 19all other cases

≤ 18≤ 19≤ 20first story of multistorybuilding

≤ 17≤ 18≤ 18top story of multistorybuilding

Z3 / CO IZ3 / CO II

CO III

Z2 / CO IZ2 / CO II

Z1 / CO IZ1 / CO II

seismic zone / building class

limit values of ratioh/tw


Masonry: out-of-plane failure• More elaborated with stress fields

(SIA 266)


Masonry: deformation capacity• Compression depending (tests EPFZ)

V

δ

0

100

200

300

400

500

600

0.000 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008

Schiefstellung

Schu

bkra

ft [k

N]

W6

W4

W1

W7W2

drift

late

ral

forc

e [k

N]


Masonry: deformation capacity• Influence of compression

0

100

200

300

400

500

600

0.000 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008

Schiefstellung

Schu

bkra

ft [k

N]

W6

W4

W1

W7W2

large compression

low compression

drift

late

ral

forc

e [k

N]


Masonry: deformation capacity• In function of normal force (K. Lang, 2002)

based on test results

nu !" #$= 25.08.0

maximuminterstorey drift

0.0

0.2

0.4

0.6

0.8

1.0

0.0 0.5 1.0 1.5 2.0 2.5 3.0

Normalspannung [MPa]

Stoc

kwer

kssc

hief

stel

lung

[%]

compression stress [MPa]

i

nte

rsto

rey d

rift

[%

]


Strength of unreinforced masonry• In function of the inclination of σ2


Lateral strength of URM elements• Determination with stress fields

Principe:

superposition of avertical fieldand aninclined field


Lateral strength of URM elements• Determination with SIA 266

ydwvd ftlkV !!!=1

xd

dzw

N

Mll 1

12 !"=


Failure modes• Rocking


Failure modes• Shear


Lateral strength of URM elements• Simplified formulae (FEMA, EC8)

rocking (FEMA):

shear (FEMA):

with

0

,2

9,0h

lNV w

xdRRd!

!!=

wwdSRd tlV !!!= "67,0,

!!"

#$$%

&

'+''=

ww

xdmdd

tl

N(( 75,05,0

2/35,0 mmN

m

mkmd !=

"

##

!

"mk# 0,5 N /mm

2


Lateral strength of URM elements• Simplified formulae (FEMA, EC8)

rocking (EC8):

shear (EC8):VRd,S = fvd · t · lc

withfvd = fvd0 + 0,4 · Nxd / (lc · t)

fvd ≤ 0,065 · fb

!

VRd ,R =

lw" N

xd

2 " h0

" 1#1,15 " Nxd

lw" t

w" f

xd

$

% &

'

( )


Masonry: deformation capacity• According to EC8

rocking:

shear:

[%]8.00

wu

l

h!="

[%]4.0=u!


Unreinforced masonry buildings• Swiss typical building (Yverdon)


Example: unreinforced masonry• Simplified typology


Example: unreinforced masonry• Lateral strength in transversal direction

1440690total

624

216

96

with total couplingeffect

VRd [kN]

3671520lw = 6.0 m

117360lw = 5.0 m

22180lw = 2.0 m

without couplingeffect

VRd [kN]Nd [kN]


Example: unreinforced masonry• Displacement determination (Lang 2002)


Example: unreinforced masonry• Capacity curve in transversal direction

0

100

200

300

400

500

600

700

0 5 10 15 20 25top building displacement [mm]

horiz

onta

l str

engt

h [

kN]

wall 2m wall 5m wall 6m building y

capacity curve in y direction

wR


Example: unreinforced masonry• Without frame effect

displacement capacity:

target displacement:(equal displacement rule)

2063.0* TSaSw gdfudd !!!!== "

mm7.71042.015.16.00.1063.032

=!!!!!=

mmwR 16=

mmw

wd

RdR 3.12

3.1

16

,===

!

mmw

wdR

dR 1.935.1

3.12*

,

,==

!=


Masonry: lot of uncertainties• Real element deformation capacity?

• Effective coupling effect?

• Effective element stiffness (crack pattern)?

• Research efforts needed


Real behavior ?• Coupling effect (frame effect)

flexible floors(without coupling effect):

very stiff lintels(total coupling effect):

Htot

hst

lo

lo

hst

h0

Htot

hst

lo

lo

hst

h0

F2

F1

F2

F1

!

h0 "2

3 # htot

!

h0 "12 # hst


Masonry: experimental research• Master project Békir Omérovic, EPFL 2005

4 meth deform

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