P
E
E
R
Keith Porter
Bridge Testbed Meeting 21 Oct 2003

DVs measure performance in terms of





 collapse closure repair duration speed limitations load limitations other?

Probabilistic relationship between what bridge inspectors see and what decision they make

What they would do, not should do
Cases with inconclusive safety evidence
Bridge category: “like Humboldt”


AASHTO-Caltrans girder bridge
Multi-span, single-column bent

Modest traffic demand
If successful, I-880, other categories later

How do post-earthquake inspectors characterize performance?
(What are the DVs?)
What are the possible outcomes of a post-earthquake inspection?
(What values can DV take on?)
What evidence do inspectors consider when making their performance evaluation?
(What are the DMs?)
How is damage evidence assessed to result in a decision?
(How is DV related to DM?)

Primary concern: collapse potential

DV
1
: collapsed, not collapsed
Post-earthquake, if not collapsed, inspectors have 2 alternatives: open or closed; if open, keep open?

DV
2
: inspector closure decision: if not collapsed, open or close?
Assessment protocol
1.
Inspectors report inspections to Caltrans EOC
2.


3.
4.
EOC compiles database of observed damage; bridge open or closed; recommended repair, repair cost estimate.
DV
2
DV
3
= “closed” subdivided into closed briefly, closed longer
: cost
Inspectors & traffic engineers decide which routes to open first. Can important be opened after shoring?
Construction engineers or design engineers design repairs

Settlement
Misalignment
Large roadway gaps
Physical evidence of structural distress

Permanent deformation



Plastic hinging
Fracture or buckling of flexural steel
X-cracking and other evidence of shear failure

2
Can bridge stand up to live load?
Likely to collapse in an aftershock?
If there is any question of the capacity of the bridge, it is closed
Repair vs. replacement: time is the deciding factor, not cost

Cost is a less-important DV
Expert system in development

If widely adopted, present results may become outdated
Mean time, how to encode DM-DV practice

2
DV
2
: inspector’s closure decision meaningful only for no collapse; doesn’t address cost
Decision

Damage ↓
No closure Close 1-3 days
Settlement of approach
Vertical offset at abutment
<1 in
1-3 in
3-6 in
>6 in
<1 in
1-3 in
3-6 in
>6 in
<1 in
1-3 in
3-6 in
>6 in
<1 in
1-3 in
3-6 in
>6 in
(Thanks, Eberhard, Conte,
Kunnath, Mahin, DesRoches)

Instructions





Provide summary info
Review the damage measures (2 blanks)
Consider the decision values (2 blanks)
Judge the max DM consistent with DV
2
Comment

Name
Agency or affiliation
Area of expertise

Geotech, design, inpect/maint, traffic
Bridge category (Humboldt)
Level of familiarity (1-5)

Settlement of approach
Vertical offset at abutment
Horizontal offset at abutment
Vertical offset at expansion jt
Horizontal offset at expansion jt
Max. beam or column flexural crack width
Max. beam or column shear crack width
Concrete beam or column spalling (y/n)
Beam or column rebar buckling, fracture, pullout (y/n)
Shear key or backwall shear cracking or spalling (y/n)

2
No closure
Close 1-3 days
Close > 3 days
Reduced speed
Not examining closure duration as continuum—not an issue for the judgment of the inspector

1st Tri-center Workshop on Earthquake
Loss Estimation Methodologies for
Transportation Systems; June 2003
15-20 DOT engineers from around US
Administered the survey in 2 of 3 breakouts
12 responses
6 self-rate as 4 or 5 on 1-5 scale
Is 6 adequate?

DM
<1 in
1-3 in
3-6 in
>6 in no response
DM1: Settlement of approach
No closure Close 1-3 days Close > 3 days Reduced speed
1 0 0 0
3
2
1
3
0
1
3
2
0
0
1
1
3
2
0
1
6 of 6 say, “If DM
1
> 6 in, then we would close at ≥ 1 day”
4 of 6 say, “If DM
1
> 3 in, then close ≥ 1 day”
1 of 6 say, “If DM
1
> 1 in, then close ≥ 1 day”
Let X = capacity to resist ≥ 1 day closure in terms of DM m s
X
X
= 3.67 in.
= 1.97 in. m s lnX lnX
= 1.17
= 0.50
1

1.00
0.75
0.50
0.25
LN( m
X
=3.7 in; s lnX
=0.5)
0.00
0 2 4 6 8
Approach settlement, in.
10

Table 1. Parameters of DM-DV relationships m
X s

X
X


 x
ˆ b
X

DM1: Settlement of approach, in.
Close at least briefly Close at least 3 days
3.67
1.97
0.54
3.23
0.50
5.50
3.14
0.57
4.78
0.53 m
X s


X
X


DM2: Vertical offset at abutment, in.
Close at least briefly

0.40

|
Parameters of 13 functions b
X
 values 0.3 ~ 0.7
i
 x

5.50

0.57
4.78
0.53 ln
  b
X





Problem:
Survey did not test vector DMs

How to combine P[DV | DM i
]?
Possibilities
1.
2.
Independent decisions?
P[DV ≥ dv j
|DM] = 1 – Π i
(1 – P[DV ≥ dv j
|DM i
])
Worst DM controls?
P[DV ≥ dv j
|DM] = max i
(P[DV ≥ dv j
|DM i
])
Must still account for correlation in DM
Check w/survey using sample vectors DM?

Larger survey: 2 nd round in ~Nov; web survey ~Dec
Revise DVs
 closed >0, >3, >30 days to regular traffic
 ditto, emergency vehicles
Rephrase questions: “What is the minimum DM causing this decision?”
Include pictures
Explore I-880 DM-DV as well
Test clarity of questions

DMs & their ranges
Ditto, DVs
Combination of p[DV
2
] values
Combination with other DVs

m
X
 s

X
X


ˆ b
X
 m
X
 s
X

X


ˆ b
X
 m
X
 s

X
X


ˆ b
X

DM2: Vertical offset at abutment, in.
Close at least briefly Close at least 3 days
3.75 5.50
1.50
0.40
3.14
0.57
3.48
0.39
4.78
0.53
DM3: Horizontal offset at abutment, in.
Close at least briefly Close at least 3 days
6.00 7.00
3.29
0.55
3.90
0.56
5.26
0.51
6.12
0.52
DM4: Vertical offset at expansion joint, in.
Close at least briefly Close at least 3 days
0.83 1.17
0.26
0.31
0.65
0.56
0.80
0.30
1.02
0.52 m
X
 s
X

X x
ˆ

 b
X

DM5: Horizontal offset at expansion joint, in.
Close at least briefly Close at least 3 days
1.20 1.33
0.45
0.37
0.60
0.45
1.12
0.36
1.22
0.43
DM6: Maximum beam or column flexural crack width, in.
Close at least briefly Close at least 3 days m
X s
X

X b x
ˆ
X




0.078
0.051
0.66
0.065
0.60
0.156
0.063
0.40
0.145
0.39 m
X s
X

X x
ˆ


 b
X

DM7: Maximum beam or column shear crack width, in.
Close at least briefly Close at least 3 days
0.063
0.048
0.77
0.049
0.69
0.104
0.067
0.64
0.088
0.59

P[DV = dv j
|DM] = 1 – P[DV ≥ dv
1
|DM] j=0
= P[DV ≥ dv j
|DM] – P[DV ≥ dv j+1
|DM] 1 ≤ j < n
= P[DV ≥ dv n
|DM] j = n
F
DV|DM
(DV = dv n
|dm) = S j=1..n
P[DV = dv j
|DM] u ~ U(0,1)
DV* = F -1
DV|DM
(u)