Slide9 Displ Method 16 - EPFL · CIVIL 706 - displacement-based methods EDCE-EPFL-ENAC-SGC 2016 -32- Example RC: frame direction •!Beam element -400-300-200-100 0 100 200-100 -50

CIVIL 706 - displacement-based methods EDCE-EPFL-ENAC-SGC 2016 -1-

EDCE: Civil and Environmental Engineering CIVIL 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 realistic approach 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! Food!

lateral displacement

lateral force

Joe s

Beer! Food!

small change in force level

large change in deformation level


Method included in SIA 2018 •! Slides prepared by

Prof. Alessandro Dazio, IBK- ETH Zürich

i b k Erdbebeningenieurwesen und Baudynamik


Reinforced concrete: chord rotation

V

V

M

!

shear wall

M

M V

V

column

!

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 point to the zero moment point

•! Represents the element sollicitation (removal of rigid displacements)

•! Verification by the comparison (demand-capacity) of chord rotations at the element level


Chord rotation !: definitions

rigid body rotation

V

V

M

!

shear wall

M

M V

V

column

!

Vl

Ml

Vr

Mr

beam

= chord = tangent to element axis = zero moment point


Yielding chord rotation !y

Fy

Vy

My

!y

Lv

moment

My "y

curvature Fy

Vy

My

#y

vyvv

y

2vy

3vy

y LL3L

3L

EIM

EI3LF

!=""#="==$


Ultimate chord rotation !u

Fu

Vu

Mu

!u

Lv

moment

Mu "y "p

"u

curvature

Lpl

Lpl = plastic hinge

( )blsvstpl dfLaL !!+!= 022.008.0

hsp

hsp = strain penetration

hpl

hpl = plastic zone


Ultimate chord rotation !u

Fu

Vu Mu

#y "y "p

"u

Lpl Lpl

!y

Lv

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

#p

#u

!p

!u


#

F

Non-linear force-displacement relationship

F

V

M

!

Lv

#

reality

#y #u

Fy

Fu

approximation

#y = f("y), Fy = f(My), #u = f("y, "u), Fu = f(Mu),


Bending moment-curvature relationship

"’y

M’y

"y

Mn

"y = "’y Mn / M’y

M

"


Bending moment-curvature relationship •! shear wall

N = -1099 kN

200

4000

36 Ø82 Ø20 2 Ø20



First yield $s = $y $c = 0.002

"’y

M’y

Nominal strength $s = 0.015 $c = 0.004

"y

Mn Nominal yield

0

1

2

3

4

5

6

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

bend

ing

mom

ent

[MN

m] Ultimate

$s = $s,max, $c = $c,max

"u



0

1

2

3

4

5

6

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

bend

ing

mom

ent [

MN

m]

"’y "y "u

Mn

M’y

1st approximation

modified approximation


Bending moment-curvature relationship •! nominal yield curvature "y


Bending moment-curvature relationship •! Material mechanical properties


Material mechanical properties

0

5

10

15

20

25

0 1 2 3 4 5 strain [‰]

stre

ss [M

Pa]

a) concrete BH300, SIA 168/68

1) f cd acc. to SIA 262 2) f ck acc. to SIA 2018 3) f ck (t) acc. to SIA 2018

1)

2)

$ cu

3)

f ck (t)

0

100

200

300

400

500

600

0 10 20 30 40 50 strain [‰]

stre

ss [M

Pa]

b) steel IIIa, SIA 168/68

1) f sd acc. to SIA 262 2) f sk approx. 3) f sk type

1)

2) 3)

$ uk

f tk

f sk


Material mechanical properties •! Characteristic values

•! Partial factor for deformation capacity %D = 1.3

•! Ultimate concrete deformation $cmax = $cu = 0.004 (in general)

•! Peak steel strain $smax = &$su - in general: & = 1.0 - if Mu<2Mcr: & = 0.5 - special cases: & = ?


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 structure m5=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 dimensions shear wall cross-section

beam at mid span beam at connection

column type A column type B


Example RC: shear walls direction

Fd

wStructure réelle

m 1

m 2

m 3

m 4

m 5

F d

w (MDOF)

k MDOF = ' k walls

m*=1163t

F d

h*=1

1.95

m

k*

w/ (

(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 0.20 horizontal displacement, w/ ( [m]

glo

bal h

oriz

onta

l for

ce F

d [kN

m]

shear walls

building

F u =828kN F y =744kN

F u =414kN F y = 372kN

w y / ( =0.039m w u / ( =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 0.20 S ud , w / ( [m]

S ad ,

F d /m

* [m

/s 2 ]

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

T=1.55s

w d / ( w u / ( w R,d / ( capacity 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

!EIg (var.)

Elément pilier

Lpl

Lpl

L !EIg


Example RC: frame direction •! Column element type A

M

M V

V L pl /2

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

L pl /2

N

N

0 100 200 300 400 500

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

bend

ing

mom

. [K

Nm

]

N=-1500kN -1000kN -500kN

0kN

0 100 200 300 400 500

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

4

Type 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 W d [m]

gl

obal

hor

izon

tal f

orce

F d [

kN]

frame A

frame B

building = 2 x frame A + 1 x frame B F y =2032kN

w y =0.098m w u =0.182m

onset of plastic hinge 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

w u / ( w R,d / ( 0

1

2

3

4

5

0.00 0.05 0.10 0.15 0.20 S ud , w d / ( [m]

S ad ,

F d /m

* [m

/s 2 ]

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

T=1.52s

w d / (

capacity curve


Unreinforced masonry

Yverdon-Les-Bains

Vernier D

örfli

ngen

dou

ane

O

berriet


DbM: unreinforced masonry •! Assumptions

(FEMA 356) plastic deformations concentrated in some elements (piers)


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


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


Masonry: deformation capacity •! Compression depending (tests EPFZ)

V

)

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

W7 W2

drift

late

ral f

orce

[kN

]


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

W7 W2

large compression

low compression

drift

late

ral f

orce

[kN

]


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

based on test results

nu !" #$= 25.08.0

maximum interstorey 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]

Sto

ckw

erks

schi

efst

ellu

ng [%

]

compression stress [MPa]

in

ters

tore

y dr

ift [%

]


Strength of unreinforced masonry •! In function of the inclination of *2


Lateral strength of URM elements •! Determination with stress fields

Principle: superposition of a

vertical field

and an

inclined field





– équilibre: ! !Nv + Nn = !! ! ! !Nv·e1v + Nn·e1n = M1

! ! ! !Nv·e2v + Nn·e2n = M2 = M1 + V·h

– limitation des sollicitations:

f! = ! fmy l2v·t·(cos!)Nv

2

fv = ! (fmx " fmy)l2n·tNn

tan! ! tan#

" "l2v = lw " 2·e2v

avec

" "l2n = lw " 2·e2n

– résistance: ! !V = Nv·tan!


Failure modes •! Rocking


Failure modes •! Shear


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

rocking (EC8): shear (EC8):

VRd,S = fvd · t · lc

with fvd = fvd0 + 0,4 · Nxd / (lc · t)

fvd ! 0,065 · fb

!

VRd ,R =lw " Nxd

2 " h0" 1#1,15 " Nxd

lw " tw " f xd

$

% &

'

( )


Masonry: deformation capacity •! According to EC8

rocking: shear:

[%]8.0 0

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

Nd [kN] without coupling

effect VRd [kN]

with total coupling effect

VRd [kN]

lw = 2.0 m 180 22 96

lw = 5.0 m 360 117 216

lw = 6.0 m 1520 367 624

total 690 1440


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 25 top building displacement [mm]

horiz

onta

l str

engt

h [

kN]

wall 2m wall 5m wall 6m building y

capacity curve in y direction

wR


Unreinforced masonry buildings •! Swiss typical building (“in line” building) RC rigid floors, diaphragm effect Very regular, no torsion 3 shear walls, long. direction Zone Z1, agd = 0.6 m/s2 Soil E, S = 1.40 Hstorey = 2.85 m Mstorey = 470 t VM,base = 880 kN Top, wy = 20 mm Top, wu = 35 mm


Unreinforced masonry buildings •! Swiss typical building (“in line” building)


Unreinforced masonry buildings •! Swiss typical building (“in line” building)

!

"!!

#!!

$!!

%!!

&!!

'!!

(!!

)!!

*!!

"!!!

! & "! "& #! #& $! $& %!

+,-./0121345/657822145922:

;,7<74/3015=5./5>/7159?@:

!"!

!"#

$"!

$"#

%"!

%"#

! $! %! &! '! #! (!

)*+,*-Γ .//0

)1+,2/-/3,./-4% 0


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

Documents

Slide9 Displ Method 16 - EPFL · CIVIL 706 - displacement-based methods EDCE-EPFL-ENAC-SGC 2016 -32- Example RC: frame direction •!Beam element -400-300-200-100 0 100 200-100 -50