1 Exercise Problems in PSHA Lecture-18
1
Exercise Problems in PSHA
Lecture-18
Example Problem 1
2
The hypothetical vertical fault segment shown in Fig.1 is represented as a quarter-circle. On the same graph, plot histograms of expected epicentral distance for motions at site A and site B assuming:(a)Earthquakes are equally likely to occur at any point on the fault segment.(b)Earthquakes are twice as likely to occur at the midpoint of the fault segment as at either end and the likelihood is linearly distributed between the midpoint and the ends.
Fault
Site B12km
Site A
30km
12km30km
NFig 1
3
Fault
Site B12km
Site A
30km
12km30km
N
Area =1
(a)Uniform distribution
0 90
1/90
P
θ
Solution
4
Area =1
(b)Non-uniform distribution
0 90
1/135
P
θ
2/135
Site A Histogram
Distance
Freq
uenc
y
30km
Solution
Uniform/Non-uniform distribution
5
Solution
Site B Histogram
Distance
Freq
uenc
y30km 45km
Uniform distribution
Site B Histogram
Distance
Freq
uenc
y
30km 45km
Non-uniform distribution
In a hypothetical seismically active region, earthquakes have been recorded over an 80-year period. Part of the record is instrumental, but part is not. Combining all available data, it appears that the earthquakes have been distributed as follows:
(a) Estimate the Gutenberg-Richter parameters for the region.(b) Neglecting earthquakes of magnitude less than 3, compute the probability that an
earthquake in the region will have a moment magnitude between 5.5 and 6.5.(c) Repeat Part (b) assuming that paleoseismic evidence indicates that the region is
not capable of producing earthquakes of moment magnitude greater than 6.5.
6
MOMENT MAGNITUDE
3-4
4-5
5-6
>6
1800
150
11
1
NUMBER OF EARTHQUAKES
Example Problem 2
t= 80 yearsMw>3, Nm = 1962, λm=24.525/year
log λm = 1.390
Mw>4, Nm = 162, λm=2.025/year
log λm= 0.307
Mw>5, Nm = 12, λm= 0.151/year
log λm= -0.824
Mw>6, Nm = 1, λm= 0.0125/year
log λm = -1.8867
Mw
3-4
4-5
5-6
>6
1800
150
11
1
N
Solution
8
M
logλM=4.6782-1.096x
logλ
M
Solution
(a) From the plot of M vs log M, Gutenberg-Richter parameters for the region are:
a = 4.6782b = 1.096
9
M
%.
][][
]|.[]|.[
]|..[
]|.[]|.[
]..[)(
).(..
001670
11
5556
35655
3
5556
7655
3550961303235609613032
00
0
00
0
ee
MMMPMMMP
MMP
ForM
MMMPMMMP
MMMPb
).(..
logλM=4.6782-1.096x
logλ
M
Solution
10
00170
11
56356553550961303235609613032
.
][][
].|..[)().(..).(..
ee
MMPc
Solution
logλM=4.6782-1.096x
logλ
M
11
The seismicity of a particular region is described by the Gutenberg-Ritcher recurrence law:
log λm=4.0-0.7M
(a) What is the probability that at least one earthquake of magnitude greater than 7.0 will occur in a 10-year period? In a 250-year period?
(b) What is the probability that exactly one earthquake of magnitude greater than 7.0 will occur in a 10-year period? In a 50 year period? In a 250-year period?
(c)Determine the earthquake magitude that would have a 10% probability of being exceeded at least once in a 50-year period?
Example Problem 3
• log λm =4.0- 0.7M
T=10years, P = 0.716
T=50years, P= 0.998
T=250years, P= 1
12
12590107
1
011
77040 .,
][][
!
.
m
t
tn
M
e
NPNP
n
e
m
For
t n]P[N (a)
Solution
13
(b) P[N=1]= λt e-λt
T=10years, P = 0.357
T=50years, P= 0.012
T=250years, P=7x10-13
597000210704
0021070
..
).log(
.
M
50
ln(0.9)t
P)-ln(1m
(c) P=1- e-λmt
Solution
Using given seismic hazard curve, estimate the probability of exceeding amax = 0.3g in a 50 year period and in 500 years period
• Combining uncertainties-probability computations
14
log
TR
amax
0.001
Example Problem 4
• Combining uncertainties-probability computations
• In a 50 year periodP = 1- e-λt
= 1- exp[-(0.001)(50)] = 0.049 = 4.9%
• In a 500 yr periodP= 0.393= 39.3%
15
log
TR
amax
=0.30g
0.001
Solution
• Combining uncertainties-probability computations
log
TR
amax =0.30g
0.001
Using seismic hazard curve, estimate the peak acceleration that has 10% probability of being exceeded in a 50 yr period.
• Combining uncertainties-probability computations
16
log
TR
amax =0.21g
0.0021
Example Problem 5
• Combining uncertainties-probability computations
• 10% in 50 yrs: λ = 0.0021
or
TR = 475 yrs
• Use seismic hazard curve to find amax value corresponding to
λ = 0.0021
17
log
TR
amax =0.21g
0.0021 475 yrs
Solution