CIVL 228
–
Introduction to Structural Engineering, Spring 2015
38
Normal distribution
•
A statistical model to describe the probability of a distribution
4
2
0
2
4
6
8
10
0
0.05
0.1
0.15
0.2
0.25
0.3
X
m
s
s
s
s
Area under
m
±
2
s
= 95%
Area under
m
±
s
= 68%
CIVL 228
–
Introduction to Structural Engineering, Spring 2015
39
Safety index•A successful design will have membercapacitiesthatexceed the memberdemandsthroughout the life of thestructure. But how do we ensure that the capacityexceeds the demand considering the variability in both thecapacity and demand described above?
CIVL 228
–
Introduction to Structural Engineering, Spring 2015
40
Demand vs. Capacity•Normalized probability density function (pdf):1000100 200 300400 500600 7008000123456x 103DemandCapacityHow do we calculate the probability of failure?
CIVL 228
–
Introduction to Structural Engineering, Spring 2015
41
Safety index
•
pdf for the failure:
•
Lets assume both Capacity, C, and Demand, D, are
normal distribution.
•
Hence its probability can be modeled using the
standard “bell” curve.
•
Let X = CD
•
Hence, when X<0
D>C
the system failed.
•
Using basic probability theory
•
μ
X
= μ
C
 μ
D
• σ
X
= sqrt (
σ
C
2
+
σ
D
2
)
• β
X
= μ
X
/
σ
X
CIVL 228
–
Introduction to Structural Engineering, Spring 2015
42
Safety index
•
P(system failure) = P(X):
200 100
0
100 200
300 400
500 600
700
0
0.5
1
1.5
2
2.5
3
3.5
x 10
3
Capacity  Demand [kN], X
pdf []
Probability of failure [
m
X
= 246,
s
X
= 126]
β
X
σ
X
, where
β
X
= the safety index.
X = 0
X = μ
X
Area under the curve = probability of failure.
CIVL 228
–
Introduction to Structural Engineering, Spring 2015
43
•What is the safety index,bX?
/μ
CIVL 228
–
Introduction to Structural Engineering, Spring 2015
44
Safety index
•
Load and resistant factors:
• β
X
σ
X
= μ
X
•
μ
X
= μ
C
 μ
D
• σ
X
= sqrt (
σ
C
2
+
σ
D
2
)
β
X
(
σ
C
2
+
σ
D
2
)= μ
C
 μ
D
β
X
σ
C
2
μ
C
=  μ
D

β
X
σ
D
2
μ
C
(1 
β
X
σ
C
2
/μ
C
) = μ
D
(1 +
β
X
σ
D
2
/μ
D
)
•
To have a safe design:
•
f
should be < 1 and
g
should be > 1.
•
Hint:
σ
/μ = COV = coefficient of variance.
f
= resistant factor
g
= load factor
CIVL 228
–
Introduction to Structural Engineering, Spring 2015
45
Example of load factors
•
Load combinations (NBCC 2005):
Case
Load Combination
(2,3)
Principal Loads
Companion Loads
(9)
1
1.4D
─
2
(1.25D
(7)
or 0.9D
(4)
) + 1.5L
(5)
0.5S
(1)
or 0.4W
3
(1.25D
(7)
or 0.9D
(4)
) + 1.5S
0.5L
(1,6)
or 0.4W
4
(1.25D
(7)
or 0.9D
(4)
) + 1.4W
0.5L
(6)
or 0.5S
5
1.0D
(4)
+ 1.0E
(8)
0.5L
(1,6)
+ 0.25S
(1)
Check NBCC 2005 Section 4.1.3.2 for details.
CIVL 228
–
Introduction to Structural Engineering, Spring 2015
46
Example of resistant factors
•
Concrete (ACI):
•
Tension: 0.9
•
Compression: 0.75
•
Bending: 0.9
•
Shear: 0.85
•
Steel (AISC):
•
Tension yielding: 0.9
•
Tension fracture: 0.75
•
Compression: 0.85
•
Bending: 0.85
•
Shear: 0.9
•
Why different
f
for different loading cases?
•
Lower
f
higher
b
.
•
More confident in certain failure mode.
CIVL 228
–
Introduction to Structural Engineering, Spring 2015
47
Structural failure (snow load)
•
A factory building in North America
CIVL 228
–
Introduction to Structural Engineering, Spring 2015
48
Structural failure (wind load)
•
Hurricane Katrina, New Orleans (2005)
Levee fails
CIVL 228
–
Introduction to Structural Engineering, Spring 2015
49
Structural failure (Hurricane
–
wind load)
•
Hurricane Katrina, New Orleans (2005)
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 Winter '13
 YANG
 Civil Engineering, Structural Engineering, Probability theory, Structural load