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Consequences of Strong Vertical Accelerations on Shear Behavior of Reinforced Concrete Bridge Columns
Khalid M. Mosalam, Professor, UC Berkeley
Hyerin Lee, Post-doctoral researcher, UC Berkeley
Selim Günay, Post-doctoral researcher, UC Berkeley
Pardeep Kumar, PhD student, UC Berkeley
Shakhzod Takhirov, Lab. Manager, UC Berkeley
Sashi Kunnath, Professor and Chair, UC Davis
Sponsors: Caltrans and PEER
5th Kwang-Hwa Forum, Tongji University, Shanghai, China, December 8-10, 2012
December 9, 2012
Outline
Motivation
Test Specimens and Test Sequence
Test Results: Evidence of Axial Force Effect on Shear Strength
Test Results vs. Code Estimations
Post-Test Simulation
Concluding Remarks
Further Developments: Laser Scanning & Repair
5th Kwang-Hwa Forum, Tongji University, Shanghai, China, December 8-10, 2012
5th Kwang-Hwa Forum, Tongji University, Shanghai, China, December 8-10, 2012
Motivation
Shear Failure
Various codes and guidelines
Consensus?
- Unexpected increase of shear force from P-M interaction
- Axial force or axial strain affects the shear strength of RC members
RC structures damaged by past earthquakes Papazoglou and Elnashai (1996)
Bridge Structures
- Do not have enough redundancy
- Columns are the most critical part
Shear Strength Shear Demand
i jV M M L
Test Specimens
¼ -scale Plumas-Arboga Overhead Bridge (prototype)
Two specimens (SP1 & SP2) with Aspect Ratio=3.5, D=20″, H=70″
Reinf. Ratio: Longitudinal=1.56%
Transverse=0.55% (SP1) > 0.47% by BDS
0.36% (SP2) < 0.47% by BDS
Weight of mass blocks = 85.6 kips (6.8% axial load ratio)
Periods: Lateral = 0.49 sec.
Rotational = 0.10 sec.
Vertical = 0.03 sec.
5th Kwang-Hwa Forum, Tongji University, Shanghai, China, December 8-10, 2012
Test Specimens
16-#5
hoops
#2@2” (SP 1)
#2@3” (SP 2)
D=20”
cover=1”
AA
B
B A-A B-B
24”
70”
18”
60” 84.85”
42.4” 30”
109”CC
C-C
16-#5
hoops
#2@2” (SP 1)
#2@3” (SP 2)
D=20”
cover=1”
16-#5
hoops
#2@2” (SP 1)
#2@3” (SP 2)
D=20”
cover=1”
AA
B
B
AA
B
B A-A B-B
24”
70”
18”
60” 84.85”
42.4” 30”
109”CC
C-C
5th Kwang-Hwa Forum, Tongji University, Shanghai, China, December 8-10, 2012
Test Sequence
Input: 1994 Northridge EQ at Pacoima Dam
2D (X and Z) and 1D (X) excitations
½ time-compressed
-2
-1
0
1
2
0 2 4 6 8
Time (sec)
(a) X direction
1.585g
-2
-1
0
1
2
0 2 4 6 8
Time (sec)
(b) Z direction
1.229g
Acc
ele
ration (
g)
Acc
ele
ration (
g)
-2
-1
0
1
2
0 2 4 6 8
Time (sec)
(a) X direction
1.585g
-2
-1
0
1
2
0 2 4 6 8
Time (sec)
(b) Z direction
1.229g
Acc
ele
ration (
g)
Acc
ele
ration (
g)
5th Kwang-Hwa Forum, Tongji University, Shanghai, China, December 8-10, 2012
Test Results: Evidence of Axial Force Effect on Shear Strength
Drift Ratio = 1.92~2.04% Drift Ratio = 1.82~2.34%
Shear force-lateral displacement relationships in the 125% tests
Decrease in stiffness
Shear degradation due to significant axial tension
-1
-0.5
0
0.5
1
-1.5 0 1.5 3
1st X+Z
X only
2nd X+Z
-1
-0.5
0
0.5
1
-1.5 0 1.5 3
1st X+Z
X only
2nd X+Z
(a) SP1 (b) SP2
Forc
e (
100 k
ips)
Drift (%)
4
2
0
-4
-2Forc
e (
100
kN
)
Forc
e (
100 k
ips)
Drift (%)
4
2
0
-4
-2
Forc
e (
100
kN
)
5th Kwang-Hwa Forum, Tongji University, Shanghai, China, December 8-10, 2012
Test Results: Evidence of Axial Force Effect on Shear Strength
Shear and axial force histories in the 125% tests
Shear degradation due to significant axial tension
-100
0
100
200
300
0.0 0.5 1.0 1.5 2.0
-100
0
100
200
300
0.0 0.5 1.0 1.5 2.0
-100
0
100
200
300
0.0 0.5 1.0 1.5 2.0
Axial Force(+ → Compression)
12
8
4
-4
0
0.0 0.5 1.0 1.5 2.0
3
2
1
0
-1
Forc
e (
100 k
ips)
Time (sec)
Forc
e (
100
kN
)
(a) SP2 1st X+Z (2-9)
-100
-50
0
50
100
0.0 0.5 1.0 1.5 2.0
77.4 kips
0.195 s
Shear Force4
2
0
-4
-2
0.5 1.0 1.5 2.0
1
0.5
0
-0.5
-1
Forc
e (
100 k
ips)
Time (sec)
Forc
e (
100
kN
)
0.0
-61.6 kips
56.7 kips
Axial Force (+ → Compression)
(b) SP2 X only (2-10)
-100
-50
0
50
100
0.0 0.5 1.0 1.5 2.0
80.9 kips
Shear Force
12
8
4
-4
0
0.0 0.5 1.0 1.5 2.0
3
2
1
0
-1
Forc
e (
100 k
ips)
Time (sec)
Forc
e (
100
kN
)
4
2
0
-4
-2
0.5 1.0 1.5 2.0
1
0.5
0
-0.5
-1
Forc
e (
100 k
ips)
Time (sec)
Forc
e (
100
kN
)0.0
Axial Force (+ → Compression)
(c) SP2 2nd X+Z (2-11)
-100
-50
0
50
100
0.0 0.5 1.0 1.5 2.0
67.0 kips
0.180 s
Shear Force
-63.3 kips
12
8
4
-4
0
0.0 0.5 1.0 1.5 2.0
3
2
1
0
-1
Forc
e (
100 k
ips)
Time (sec)
Forc
e (
100
kN
)
4
2
0
-4
-2
0.5 1.0 1.5 2.0
1
0.5
0
-0.5
-1
Forc
e (
100 k
ips)
Time (sec)
Forc
e (
100
kN
)
0.0
5th Kwang-Hwa Forum, Tongji University, Shanghai, China, December 8-10, 2012
Test Results: Evidence of Axial Force Effect on Shear Strength
0”
10”
20”
30”
40”
50”
60”
70”
60° 90° 180° 270° 0°
W S E N
60°
After 70%-scale (1-7) test After 70%-scale (2-7) test
0”
10”
20”
30”
40”
50”
60”
70”
60° 90° 180° 270° 0°
W S E N
60°
16-#5
hoops
#2@2” (SP 1)
#2@3” (SP 2)
D=20”
cover=1”
AA
B
B A-A B-B
24”
70”
18”
60” 84.85”
42.4” 30”
109”CC
C-C
16-#5
hoops
#2@2” (SP 1)
#2@3” (SP 2)
D=20”
cover=1”
16-#5
hoops
#2@2” (SP 1)
#2@3” (SP 2)
D=20”
cover=1”
AA
B
B
AA
B
B A-A B-B
24”
70”
18”
60” 84.85”
42.4” 30”
109”CC
C-C
N S
W
E SP1 SP2
5th Kwang-Hwa Forum, Tongji University, Shanghai, China, December 8-10, 2012
0”
10”
20”
30”
40”
50”
60”
70”
60° 90° 180° 270° 0°
W S E N
60°
Test Results: Evidence of Axial Force Effect on Shear Strength
After 125%-scale
“1st X+Z” test
5th Kwang-Hwa Forum, Tongji University, Shanghai, China, December 8-10, 2012
Crack Pattern of SP2
0”
10”
20”
30”
40”
50”
60”
70”
60° 90° 180° 270° 0°
W S E N
60°
Test Results: Evidence of Axial Force Effect on Shear Strength
After 125%-scale
“X only” test
5th Kwang-Hwa Forum, Tongji University, Shanghai, China, December 8-10, 2012
Crack Pattern of SP2
0”
10”
20”
30”
40”
50”
60”
70”
60° 90° 180° 270° 0°
W S E N
60°
Test Results: Evidence of Axial Force Effect on Shear Strength
After 125%-scale
“2nd X+Z” test
(a) First shear peak
N SN S N S
(b) Second shear peak (c) Third shear peak
lateraltranslation
rotation
Shear crack opening and closing at each shear peak during the 125%-scale tests
5th Kwang-Hwa Forum, Tongji University, Shanghai, China, December 8-10, 2012
Crack Pattern of SP2
Micro-cracks produced by tensile axial forces contribute to further widening of diagonal cracks under combined effect of vertical & horizontal excitations.
Test Results vs. Code Estimations
Code Estimations: Different design approaches and code equations
Influences of axial load, flexural ductility, and size of members & aggregates are not well agreed upon within different codes.
Three main approaches:
I. Axial force: ACI, Eurocode
II. Axial force + Ductility: SDC, Priestley et al.
III. Axial strain: AASHTO, CSA (based on MCFT)
5th Kwang-Hwa Forum, Tongji University, Shanghai, China, December 8-10, 2012
s
DfAV
yv
s
)8.0(
' 22 1 0.82000
c c
g
NV f D
A
' 22 1 0.8500
c c
g
NV f D
A
Circular member: Axial compression
Circular member: Axial tension
N: axial force (+ if compression, - if tension)
scn VVV
ACI (318-08) Eurocode (2004)
s
DfAV
yv
s
72.0
2
1.2 40 0.154
cc rd l cp
DV k
scn VVV
'0.25 0.7rd cf
1kcp
c
N
A
bwc dcDD 22
Approach I: Axial force affects the estimated shear strength
5th Kwang-Hwa Forum, Tongji University, Shanghai, China, December 8-10, 2012
Test Results vs. Code Estimations
0c
scn VVV
Priestley et al. (1996) Caltrans SDC (2010)
Approach II: Axial force and ductility affect the estimated shear strength.
'
cot2
v y
s
A f DV
s
'c c eV k f A
1: 0.29
1 3: 0.10 0.19(3 ) 2
3 7 : 0.05 0.05(7 ) 4
7 : 0.05
k
k
k
k
tan2
p
D cV P P
a
5th Kwang-Hwa Forum, Tongji University, Shanghai, China, December 8-10, 2012
'
v y
s
A f DV
s
c c eV A
' 'Factor1 Factor2 0.33c c cf f
' '0.25 Factor2 0.33c c cf f
0.025 Factor1 0.305 0.083 0.2512.5
s yh
d
f
Factor2 1 1.513.8
c
g
P
A
scn VVV
Inside PHZ
Outside PHZ
for axial tension shear strength increase by axial compression
Test Results vs. Code Estimations
s
dfAV
vyv
s
sincotcot
'0.083c c v vV f b d
scn VVV
Dbv ev dd 9.02
re
DDd
, : function of /f’c and
AASHTO (2010) CSA (2004)
s
dfAV
vyv
s
cot
'
c c v vV f b d
scn VVV
Dbv Ddv 72.0
'
,max 0.25n c v vV f b d
700029 o
zes1000
1300
15001
4.0
350.85
15
zze z
g
ss s
a
Approach III: Longitudinal strain at the centroid of the section affects the estimated shear strength.
300mmz vs or d
5th Kwang-Hwa Forum, Tongji University, Shanghai, China, December 8-10, 2012
: angle of inclination of transverse rebars
Test Results vs. Code Estimations
0
20
40
60
80
100
120
140
160
180
200
6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5
Forc
e (
kip
s)
Time (sec)
Shear Force ACI
Eurocode AASHTO
CSA SDC
0
20
40
60
80
100
120
140
160
180
200
6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5
Fo
rce
(kip
s)
Time (sec)
Shear Force ACI
Eurocode AASHTO
CSA SDC
Comparison of 1-10 and 1-11, 125% Northridge EQ 1-10: X only, 1-11: 2nd X+Z
X only (1-10) at h=10” 2nd X+Z (1-11) at h=10”
h=10”
h=60” Test Results vs. Code Estimates
5th Kwang-Hwa Forum, Tongji University, Shanghai, China, December 8-10, 2012
SP1
0
20
40
60
80
100
120
140
160
180
200
6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5
Fo
rce
(kip
s)
Time (sec)
Shear Force ACI
Eurocode AASHTO
CSA SDC
0
20
40
60
80
100
120
140
160
180
200
6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5
Forc
e (
kip
s)
Time (sec)
Shear Force ACI
Eurocode AASHTO
CSA SDC
Comparison of 1-10 and 1-11, 125% Northridge EQ 1-10: X only, 1-11: 2nd X+Z
h=10”
h=60” Test Results vs. Code Estimates
X only (1-10) at h=60” 2nd X+Z (1-11) at h=60”
Significant difference between AASHTO/CSA estimations at h=10” & 60” ACI, Eurocode/SDC are similar, but there are transient differences due to axial
tension or large displacement ductility.
5th Kwang-Hwa Forum, Tongji University, Shanghai, China, December 8-10, 2012
SP1
0
20
40
60
80
100
120
140
160
180
200
7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12
Forc
e (
kip
s)
Time (sec)
Shear Force ACI
Eurocode AASHTO
CSA SDC
0
20
40
60
80
100
120
140
160
180
200
7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12
Forc
e (
kip
s)
Time (sec)
Shear Force ACI
Eurocode AASHTO
CSA SDC
Comparison of 2-10 and 2-11, 125% Northridge EQ 2-10: X only, 2-11: 2nd X+Z
h=10”
h=60” Test Results vs. Code Estimates
X only (2-10) at h=10” 2nd X+Z (2-11) at h=10”
5th Kwang-Hwa Forum, Tongji University, Shanghai, China, December 8-10, 2012
SP2
0
20
40
60
80
100
120
140
160
180
200
7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12
Fo
rce
(kip
s)
Time (sec)
Shear Force ACI
Eurocode AASHTO
CSA SDC
0
20
40
60
80
100
120
140
160
180
200
7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12
Fo
rce
(kip
s)
Time (sec)
Shear Force ACI
Eurocode AASHTO
CSA SDC
Comparison of 2-10 and 2-11, 125% Northridge EQ 2-10: X only, 2-11: 2nd X+Z
h=10”
h=60” Test Results vs. Code Estimates
X only (2-10) at h=60” 2nd X+Z (2-11) at h=60”
A similar trend is observed, but AASHTO/CSA estimations are lower due to larger strains, especially at h=10”
5th Kwang-Hwa Forum, Tongji University, Shanghai, China, December 8-10, 2012
SP2
Post-Test Simulation
OpenSees ACI/SDC shear spring:
Based on the “Test Results vs. Code Estimations”, two methods chosen from Approaches I and II
Incorporates ACI & SDC code equations for shear capacity as a new material implemented into source code.
Within a zero-length element connected to a beam-column element, this material can be effectively used.
Displacement0
Forc
e
Vy
-Vy
Kelastic
rKelastic
0
Displacement0
Forc
e
Vy
-Vy
Kelastic
rKelastic
0Depends on axial load, … etc.
5th Kwang-Hwa Forum, Tongji University, Shanghai, China, December 8-10, 2012
Post-Test Simulation
OpenSees ACI/SDC shear spring: 1. Before the shear demand reaches the capacity (i.e. before yielding):
The yield force is updated at each integration time step.
2. At the time step where the demand reaches the capacity:
Yielding takes place & force-displacement relationship follows a post-yield behavior.
3. After 2: Yield force is not updated & kept constant unless the column experiences any axial tension for Caltrans SDC spring & a predetermined tension for ACI spring. The yield force is kept constant after this final modification.
5th Kwang-Hwa Forum, Tongji University, Shanghai, China, December 8-10, 2012
0
Forc
e
Displacement
0
Kelastic
0
Forc
e
Displacement
0
Kelastic
Vy
-Vy
0 Fo
rce
Displacement
0
Vy
-Vy rKelastic
Post-Test Simulation
Case A-1 Case A-2
Nodal Mass
Mx, My, Mz
Imx, Imy, Imz
Case B-1 Case B-2
BWH
Element70”
Shear Spring
Rigid
End Zone
Rigid
End Zone
Rotational Spring
(a) Beam With Hinges Element
Nodal Mass
NLBC
Elements
Shear Spring
Rigid
End Zone
Rigid
End Zone
Rotational Spring
(b) Nonlinear Beam Column Elements
Nodal Mass
15”
20”
20”
15”
Lp
Lp
Lp
Lp
Sectionat integration points
OpenSees model: BWH element (Model A) or NLBC elements (Model B)
5th Kwang-Hwa Forum, Tongji University, Shanghai, China, December 8-10, 2012
Post-Test Simulation
247 248 249 250 251-100
-50
0
50
100
time (sec)
sh
ea
r fo
rce
(kip
)
test data
B-2-SDC
247 248 249 250 251-100
-50
0
50
100
time (sec)
sh
ea
r fo
rce
(kip
)
test data
B-2-ACI
247 248 249 250 251-100
-50
0
50
100
time (sec)
sh
ea
r fo
rce
(kip
)
test data
A-2-SDC
247 248 249 250 251-100
-50
0
50
100
time (sec)
sh
ea
r fo
rce
(kip
)
test data
A-2-SDC
247 248 249 250 251-100
-50
0
50
100
time (sec)
sh
ea
r fo
rce
(kip
)
test data
A-1
247 248 249 250 251-100
-50
0
50
100
time (sec)
sh
ea
r fo
rce
(kip
)
test data
B-1
A-2-SDC and Run 2-11
A-2-ACI and Run 2-11
A-1 and Run 2-11
B-2-SDC and Run 2-11
B-2-ACI and Run 2-11
B-1 and Run 2-11
(c) 125% 2nd X+Z
Beam
wit
h H
ing
es E
lem
en
t (M
od
el A
)
No
nlin
ear
Beam
-Co
lum
n E
lem
en
ts (
Mo
del B
)
No
Shear
Spring
SDC
Shear
Spring
ACI
Shear
Spring
SP2 125%-scale “2nd X+Z” test data and simulation results
ACI
5th Kwang-Hwa Forum, Tongji University, Shanghai, China, December 8-10, 2012
Post-Test Simulation
247 248 249 250 251-100
-50
0
50
100
time (sec)
sh
ea
r fo
rce
(kip
)
test data
B-2-SDC
247 248 249 250 251-100
-50
0
50
100
time (sec)
sh
ea
r fo
rce
(kip
)
test data
B-2-ACI
247 248 249 250 251-100
-50
0
50
100
time (sec)
sh
ea
r fo
rce
(kip
)
test data
A-2-SDC
247 248 249 250 251-100
-50
0
50
100
time (sec)
sh
ea
r fo
rce
(kip
)
test data
A-2-SDC
247 248 249 250 251-100
-50
0
50
100
time (sec)
sh
ea
r fo
rce
(kip
)
test data
A-1
247 248 249 250 251-100
-50
0
50
100
time (sec)
sh
ea
r fo
rce
(kip
)
test data
B-1
A-2-SDC and Run 2-11
A-2-ACI and Run 2-11
A-1 and Run 2-11
B-2-SDC and Run 2-11
B-2-ACI and Run 2-11
B-1 and Run 2-11
(c) 125% 2nd X+Z
SP2 125%-scale “2nd X+Z” test data and simulation results from Model A
5th Kwang-Hwa Forum, Tongji University, Shanghai, China, December 8-10, 2012
247 248 249 250 251-100
-50
0
50
100
time (sec)
sh
ea
r fo
rce
(kip
)
test data
B-2-SDC
247 248 249 250 251-100
-50
0
50
100
time (sec)
sh
ea
r fo
rce
(kip
)
test data
B-2-ACI
247 248 249 250 251-100
-50
0
50
100
time (sec)
sh
ea
r fo
rce
(kip
)
test data
A-2-SDC
247 248 249 250 251-100
-50
0
50
100
time (sec)
sh
ea
r fo
rce
(kip
)
test data
A-2-SDC
247 248 249 250 251-100
-50
0
50
100
time (sec)
sh
ea
r fo
rce
(kip
)
test data
A-1
247 248 249 250 251-100
-50
0
50
100
time (sec)
sh
ea
r fo
rce
(kip
)
test data
B-1
A-2-SDC and Run 2-11
A-2-ACI and Run 2-11
A-1 and Run 2-11
B-2-SDC and Run 2-11
B-2-ACI and Run 2-11
B-1 and Run 2-11
(c) 125% 2nd X+Z
247 248 249 250 251-100
-50
0
50
100
time (sec)
sh
ea
r fo
rce
(kip
)
test data
B-2-SDC
247 248 249 250 251-100
-50
0
50
100
time (sec)
sh
ea
r fo
rce
(kip
)
test data
B-2-ACI
247 248 249 250 251-100
-50
0
50
100
time (sec)
sh
ea
r fo
rce
(kip
)
test data
A-2-SDC
247 248 249 250 251-100
-50
0
50
100
time (sec)
sh
ea
r fo
rce
(kip
)
test data
A-2-SDC
247 248 249 250 251-100
-50
0
50
100
time (sec)
sh
ea
r fo
rce
(kip
)
test data
A-1
247 248 249 250 251-100
-50
0
50
100
time (sec)
sh
ea
r fo
rce
(kip
)
test data
B-1
A-2-SDC and Run 2-11
A-2-ACI and Run 2-11
A-1 and Run 2-11
B-2-SDC and Run 2-11
B-2-ACI and Run 2-11
B-1 and Run 2-11
(c) 125% 2nd X+Z
ACI
Effect of OpenSees ACI/SDC shear spring Close resemblance in shear force responses
Notable difference in the inelastic response of the shear spring for the peak values.
SDC shear spring provides more conservative estimates than ACI spring does due to the different yielding patterns of the springs observed in the hysteresis.
Post-Test Simulation
5th Kwang-Hwa Forum, Tongji University, Shanghai, China, December 8-10, 2012
AC
I sh
ear
sp
rin
g h
yste
res
is
SD
C s
hear
sp
rin
g h
yste
res
is 125%
1st X+Z
2nd X+Z
X only
-1.5 -1 -0.5 0 0.5 1 1.5-100
-50
0
50
100
displacement (in)sh
ea
r fo
rce
(kip
)
A-2-ACI
-1.5 -1 -0.5 0 0.5 1 1.5-100
-50
0
50
100
displacement (in)
sh
ea
r fo
rce
(kip
)
A-2-ACI
-1.5 -1 -0.5 0 0.5 1 1.5-100
-50
0
50
100
displacement (in)
sh
ea
r fo
rce
(kip
)
A-2-ACI
-1.5 -1 -0.5 0 0.5 1 1.5-100
-50
0
50
100
displacement (in)
sh
ea
r fo
rce
(kip
)
A-2-SDC
-1.5 -1 -0.5 0 0.5 1 1.5-100
-50
0
50
100
displacement (in)
sh
ea
r fo
rce
(kip
)
A-2-SDC
-1.5 -1 -0.5 0 0.5 1 1.5-100
-50
0
50
100
displacement (in)
sh
ea
r fo
rce
(kip
)
A-2-SDC(a-1) 1st X+Z (a-2) 1st X+Z
(b-1) X only (b-2) X only
(c-1) 2nd X+Z (c-2) 2nd X+Z
ACI/SDC spring hysteresis obtained from Model A (SP2 125%-scale simulations)
Concluding Remarks
There exists an apparent evidence of shear strength reduction due to the presence of axial tension from:
The shear force measurements
The crack patterns
Shear strength reduction is mainly due to the degradation of concrete contribution to this strength.
ACI & SDC capture the shear strength degradation due to axial force.
Both approaches provide results on the conservative side with SDC predictions being more conservative. They can be enhanced by:
Modifying ACI to consider effect of ductility
Replacing the sharp tensile force effect in SDC to be more gradual
5th Kwang-Hwa Forum, Tongji University, Shanghai, China, December 8-10, 2012
Further Developments: Laser Scanning
5th Kwang-Hwa Forum, Tongji University, Shanghai, China, December 8-10, 2012
Leica ScanStation C10
Measured @10.8%
Measured @11.8%
Lab. (nees@berkeley)
Filed (Haiti)
Point Cloud: Specimen SP1, Undamaged Configuration
5th Kwang-Hwa Forum, Tongji University, Shanghai, China, December 8-10, 2012
Point Clouds at the Center of Column in Loading Direction: Raw Data
Undamaged Damaged
Footing
Column
Top Block
N S
W
E
5th Kwang-Hwa Forum, Tongji University, Shanghai, China, December 8-10, 2012
-15 -10 -5 0 5 10 15 20 25 30 35-40
-20
0
20
40
60
80
100
Longitudin
al coord
inate
, in
ch
Transversal coordinate, inch
Damaged configuration
Undamaged configuration
Footing
Column
Top Block
Steel Plate
-15 -10 -5 0 5 10 15 20 25 30 35-40
-20
0
20
40
60
80
100
Longitudin
al coord
inate
, in
ch
Transversal coordinate, inch
Undamaged configuration
Damaged configuration
-6.5 -6 -5.5 -5 -4.5 -4
0
10
20
30
40
50
60
Longitudin
al coord
inate
, in
ch
Transversal coordinate, inch
Damaged configuration
Undamaged configuration
Undam
aged
Dam
aged
Point Clouds at the Center of Column in Loading Direction: Processing
5th Kwang-Hwa Forum, Tongji University, Shanghai, China, December 8-10, 2012
0 0.1 0.2 0.30
10
20
30
40
50
60
70
Lo
ng
itu
din
al co
ord
ina
te, in
ch
Transversal coordinate, inch
Undamaged configuration
0 0.1 0.2 0.30
10
20
30
40
50
60
70
Lo
ng
itu
din
al co
ord
ina
te, in
ch
Residual Displacement, inch
0 0.1 0.2 0.30
10
20
30
40
50
60
70
Lo
ng
itu
din
al co
ord
ina
te, in
ch
Transversal coordinate, inch
Damaged configuration
0 0.1 0.2 0.30
10
20
30
40
50
60
70
Lo
ng
itu
din
al co
ord
ina
te, in
ch
Transversal coordinate, inch
Undamaged configuration
0 0.1 0.2 0.30
10
20
30
40
50
60
70
Lo
ng
itu
din
al co
ord
ina
te, in
ch
Residual Displacement, inch
0 0.1 0.2 0.30
10
20
30
40
50
60
70
Lo
ng
itu
din
al co
ord
ina
te, in
ch
Transversal coordinate, inch
Damaged configuration
Deformed Shape from Point Clouds
Points with squares are from averaging points in the cloud within 2 inches
0 0.1 0.2 0.30
10
20
30
40
50
60
70
Lo
ng
itu
din
al co
ord
ina
te, in
ch
Transversal coordinate, inch
Undamaged configuration
0 0.1 0.2 0.30
10
20
30
40
50
60
70
Lo
ng
itu
din
al co
ord
ina
te, in
ch
Residual Displacement, inch
0 0.1 0.2 0.30
10
20
30
40
50
60
70
Lo
ng
itu
din
al co
ord
ina
te, in
ch
Transversal coordinate, inch
Damaged configuration
Point Clouds at the Center of Column in Loading Direction: Interpretation
5th Kwang-Hwa Forum, Tongji University, Shanghai, China, December 8-10, 2012
0 0.1 0.2 0.30
10
20
30
40
50
60
70
Lo
ng
itu
din
al co
ord
ina
te, in
ch
Transversal coordinate, inch
Undamaged configuration
0 0.1 0.2 0.30
10
20
30
40
50
60
70
Lo
ng
itu
din
al co
ord
ina
te, in
ch
Residual Displacement, inch
0 0.1 0.2 0.30
10
20
30
40
50
60
70
Lo
ng
itu
din
al co
ord
ina
te, in
ch
Transversal coordinate, inch
Damaged configuration
h=55”
0.10” (Laser Scanner) 0.11” (Wire Potentiometer)
h=35”
0.15” (Laser Scanner) 0.18” (Wire Potentiometer)
h=15” 0.05” (Laser Scanner) 0.04” (Wire Potentiometer)
16.7%
9.1%
-25.0%
(W-L)/W×100
Deformed Shape from Point Clouds
Point Clouds at the Center of Column in Loading Direction: Interpretation
5th Kwang-Hwa Forum, Tongji University, Shanghai, China, December 8-10, 2012
SP1
After Before
SP2
After Before
5th Kwang-Hwa Forum, Tongji University, Shanghai, China, December 8-10, 2012
Further Developments: Repair
Grind Surface Remove Loose Concrete Mortar and Epoxy Gel Epoxy Injection
1 2 3 4
Thank You!
Questions? Comments?
5th Kwang-Hwa Forum, Tongji University, Shanghai, China, December 8-10, 2012
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