lecture18 psha problems

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


• 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


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.


16

log

TR

amax =0.21g

0.0021

Example Problem 5


• 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

lecture18 psha problems

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