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CIVIL-706 - displacement-based methodsEPFL-ENAC-SGC 2005 -1-
Doctoral School: StructuresCIVIL-706 Advanced Earthquake Engineering
Displacement-based methods
CIVIL-706 - displacement-based methodsEPFL-ENAC-SGC 2005 -2-
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
CIVIL-706 - displacement-based methodsEPFL-ENAC-SGC 2005 -3-
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
CIVIL-706 - displacement-based methodsEPFL-ENAC-SGC 2005 -4-
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
CIVIL-706 - displacement-based methodsEPFL-ENAC-SGC 2005 -5-
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
CIVIL-706 - displacement-based methodsEPFL-ENAC-SGC 2005 -6-
Method included in SIA 2018• Slides prepared by
Prof. Alessandro Dazio, IBK- ETH ZürichErdbebeningenieurwesen und Baudynamik
CIVIL-706 - displacement-based methodsEPFL-ENAC-SGC 2005 -7-
Reinforced concrete: chord rotation
V
V
M
θ
shear wall
M
M
θV
V
column
θ
Vl
Ml
Vr
Mr
beam
CIVIL-706 - displacement-based methodsEPFL-ENAC-SGC 2005 -8-
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
CIVIL-706 - displacement-based methodsEPFL-ENAC-SGC 2005 -9-
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
CIVIL-706 - displacement-based methodsEPFL-ENAC-SGC 2005 -10-
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
!=""#="==$
CIVIL-706 - displacement-based methodsEPFL-ENAC-SGC 2005 -11-
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
CIVIL-706 - displacement-based methodsEPFL-ENAC-SGC 2005 -12-
Ultimate chord rotation θu
Fu
VuMu
Δyφyφp
φu
Lpl Lpl
θy
Lv
( ) )L5.0L(LLL plvplyuvyvuu !"!!#"#+$=$=%
Δp
Δu
θp
θu
CIVIL-706 - displacement-based methodsEPFL-ENAC-SGC 2005 -13-
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),
CIVIL-706 - displacement-based methodsEPFL-ENAC-SGC 2005 -14-
Bending moment-curvature relationship
φ’y
M’y
φy
Mn
φy = φ’y Mn / M’y
M
φ
CIVIL-706 - displacement-based methodsEPFL-ENAC-SGC 2005 -15-
Bending moment-curvature relationship• shear wall
N = -1099 kN
200
4000
36 Ø82 Ø20 2 Ø20
CIVIL-706 - displacement-based methodsEPFL-ENAC-SGC 2005 -16-
Bending moment-curvature relationship
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
CIVIL-706 - displacement-based methodsEPFL-ENAC-SGC 2005 -17-
Bending moment-curvature relationship
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
CIVIL-706 - displacement-based methodsEPFL-ENAC-SGC 2005 -18-
Bending moment-curvature relationship• nominal yield curvature φy
CIVIL-706 - displacement-based methodsEPFL-ENAC-SGC 2005 -19-
Bending moment-curvature relationship• Material mechanical properties
CIVIL-706 - displacement-based methodsEPFL-ENAC-SGC 2005 -20-
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
CIVIL-706 - displacement-based methodsEPFL-ENAC-SGC 2005 -21-
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 = ?
CIVIL-706 - displacement-based methodsEPFL-ENAC-SGC 2005 -22-
Application of DbM• Non-linear element behavior
• Equivalent SDOF
• Force-displacement relationship of structure
• Capacity curve
• Seismic displacement demand
• Shear strength verification
CIVIL-706 - displacement-based methodsEPFL-ENAC-SGC 2005 -23-
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
CIVIL-706 - displacement-based methodsEPFL-ENAC-SGC 2005 -24-
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
CIVIL-706 - displacement-based methodsEPFL-ENAC-SGC 2005 -25-
Example RC: element dimensionsshear wall cross-section
beam at mid span beam at connection
column type A column type B
CIVIL-706 - displacement-based methodsEPFL-ENAC-SGC 2005 -26-
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)
CIVIL-706 - displacement-based methodsEPFL-ENAC-SGC 2005 -27-
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
CIVIL-706 - displacement-based methodsEPFL-ENAC-SGC 2005 -28-
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
CIVIL-706 - displacement-based methodsEPFL-ENAC-SGC 2005 -29-
Example RC: frame direction• Force-displacement rel.: ∑frames A, B and C
• Torsion neglected
A
B
C
CIVIL-706 - displacement-based methodsEPFL-ENAC-SGC 2005 -30-
Example RC: frame direction• Modelisation
Elément traverse
Lpl Lpl
L
αEIg (var.)
Elément pilier
Lpl
Lpl
L αEIg
CIVIL-706 - displacement-based methodsEPFL-ENAC-SGC 2005 -31-
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
CIVIL-706 - displacement-based methodsEPFL-ENAC-SGC 2005 -32-
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
CIVIL-706 - displacement-based methodsEPFL-ENAC-SGC 2005 -33-
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
CIVIL-706 - displacement-based methodsEPFL-ENAC-SGC 2005 -34-
Example RC: frame direction• Ultimate deformation of frame B
Fd
wd
= Rotule= Rotule (rupture atteinte)
CIVIL-706 - displacement-based methodsEPFL-ENAC-SGC 2005 -35-
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
CIVIL-706 - displacement-based methodsEPFL-ENAC-SGC 2005 -36-
Unreinforced masonry• Application of DbM more delicate
CIVIL-706 - displacement-based methodsEPFL-ENAC-SGC 2005 -37-
DbM: unreinforced masonry• Assumptions
(FEMA 356)
plastic deformationsconcentrated in someelements (piers)
CIVIL-706 - displacement-based methodsEPFL-ENAC-SGC 2005 -38-
DbM: unreinforced masonry• Limitation: out-of-plane failure excluded
Molise 2002EERI
CIVIL-706 - displacement-based methodsEPFL-ENAC-SGC 2005 -39-
Masonry: out-of-plane failure• Control with h/t ratio (FEMA 356)
CIVIL-706 - displacement-based methodsEPFL-ENAC-SGC 2005 -40-
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
CIVIL-706 - displacement-based methodsEPFL-ENAC-SGC 2005 -41-
Masonry: out-of-plane failure• More elaborated with stress fields
(SIA 266)
CIVIL-706 - displacement-based methodsEPFL-ENAC-SGC 2005 -42-
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]
CIVIL-706 - displacement-based methodsEPFL-ENAC-SGC 2005 -43-
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]
CIVIL-706 - displacement-based methodsEPFL-ENAC-SGC 2005 -44-
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
[%
]
CIVIL-706 - displacement-based methodsEPFL-ENAC-SGC 2005 -45-
Strength of unreinforced masonry• In function of the inclination of σ2
CIVIL-706 - displacement-based methodsEPFL-ENAC-SGC 2005 -46-
Lateral strength of URM elements• Determination with stress fields
Principe:
superposition of avertical fieldand aninclined field
CIVIL-706 - displacement-based methodsEPFL-ENAC-SGC 2005 -47-
Lateral strength of URM elements• Determination with stress fields
CIVIL-706 - displacement-based methodsEPFL-ENAC-SGC 2005 -48-
Lateral strength of URM elements• Determination with stress fields
CIVIL-706 - displacement-based methodsEPFL-ENAC-SGC 2005 -49-
Lateral strength of URM elements• Determination with SIA 266
ydwvd ftlkV !!!=1
xd
dzw
N
Mll 1
12 !"=
CIVIL-706 - displacement-based methodsEPFL-ENAC-SGC 2005 -50-
Lateral strength of URM elements• Determination with SIA 266
ydwvd ftlkV !!!=1
xd
dzw
N
Mll 1
12 !"=
CIVIL-706 - displacement-based methodsEPFL-ENAC-SGC 2005 -51-
Failure modes• Rocking
CIVIL-706 - displacement-based methodsEPFL-ENAC-SGC 2005 -52-
Failure modes• Shear
CIVIL-706 - displacement-based methodsEPFL-ENAC-SGC 2005 -53-
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
CIVIL-706 - displacement-based methodsEPFL-ENAC-SGC 2005 -54-
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
$
% &
'
( )
CIVIL-706 - displacement-based methodsEPFL-ENAC-SGC 2005 -55-
Masonry: deformation capacity• According to EC8
rocking:
shear:
[%]8.00
wu
l
h!="
[%]4.0=u!
CIVIL-706 - displacement-based methodsEPFL-ENAC-SGC 2005 -56-
Unreinforced masonry buildings• Swiss typical building (Yverdon)
CIVIL-706 - displacement-based methodsEPFL-ENAC-SGC 2005 -57-
Example: unreinforced masonry• Simplified typology
CIVIL-706 - displacement-based methodsEPFL-ENAC-SGC 2005 -58-
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]
CIVIL-706 - displacement-based methodsEPFL-ENAC-SGC 2005 -59-
Example: unreinforced masonry• Displacement determination (Lang 2002)
CIVIL-706 - displacement-based methodsEPFL-ENAC-SGC 2005 -60-
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
CIVIL-706 - displacement-based methodsEPFL-ENAC-SGC 2005 -61-
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*
,
,==
!=
CIVIL-706 - displacement-based methodsEPFL-ENAC-SGC 2005 -62-
Masonry: lot of uncertainties• Real element deformation capacity?
• Effective coupling effect?
• Effective element stiffness (crack pattern)?
• Research efforts needed
CIVIL-706 - displacement-based methodsEPFL-ENAC-SGC 2005 -63-
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
CIVIL-706 - displacement-based methodsEPFL-ENAC-SGC 2005 -64-
Masonry: experimental research• Master project Békir Omérovic, EPFL 2005
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