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Strong Motion Seismology EPS130

Homework 01Qingkai KONG


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Exercise 1

Problem A

The Gutenberg-Richter relationship for this catalog is: log(N) = 5.2112-0.8095M

Figure 1 Gutenberg-Richter relationship by using 1910-1998 catalog

Is the Gutenberg-Richter relationship a good model for yourdata?

Yes, we can see from the figure above, it fits the data very well.

How do the values you obtain compare with those typicallyobserved in global catalog searches

Generally, the b value is between 2/3 and 1. From the equation, we can see that the

b value for this data set is 0.8095, which falls into this range.

Completeness of the seismicity catalog is important to thisanalysis. How can you tell if the catalog is complete over themagnitude range used? Is your catalog complete?

log(N) = 5.2112-0.8095MR = 0.99629

0

0.5

11.5

2

2.5

3

2 3 4 5 6 7 L o g .

e v e n t s > = M

Magnitude

Gutenberg-Richter relationship


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Generally speaking, if the catalog is incomplete, most of the time, it is due to smallmagnitude events that are hard to detect. That is to say, if the catalog is incomplete,the small magnitude events will be not fit very well. But from the figure above, wecan see that, the small magnitude events (down to magnitude 3) fit quite well, so wecan say this catalog is complete.

Problem B

What is the magnitude of the 200-year event?

log(N) = 5.2112-0.8095M

N yr = N 89 yrs

89=

10 a10bM

89= 10 a

log8910bM

N yr = N 89 yrs

89=

10 5.21120.8095 M

89= 10 5.2112

0.8095 M log89

M =5.2112 log(89 N yr )

89For the 200-year event, let Nyr = 1/200, M is 6.87. Therefore, the magnitude of a200-year event is 6.9.

How does this magnitude compare with the size andrecurrence times of the events forecasted to occur?

Compare with the forecasted events in table 1, this one is similar to the events on

the Hayward fault, which have a recurrence time 210 years, and have a magnitude7.0. If we calculate the moment according to the following equation:log(M 0) = 1.5M+16.05

The moment is M0 is 2.51e26, which is slightly small than 3.4e26, but with the sameorder of 26.

Problem C

For the first (1910-1954) seismicity catalog, the obtained Gutenberg-Richterrelationship is

N yr=

N 45 yrs45

=10 a10

bM

45=

10alog45

10bM

=

103.8638 0.6549 M log45

According to this relationship, the magnitude for the 200-year earthquake is 6.89


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For the second (1955-1998) seismicity catalog, the obtained Gutenberg-Richterrelationship is:

N yr = N 44 yrs

44=

10 a10bM

44= 10 a

log4410bM

= 10 5.32490.8631 M log44


logN = 3.8638 - 0.6549MR = 0.87976

0

0.5

1

1.52

2.5

2 3 4 5 6 7 L o g .

e v e n t s > = M

Magnitude


logN =5.3249 -0.8631MR = 0.9869

0

0.5

1

1.5

2

2.5

3

2 3 4 5 6 7 L o g

e v e n t s > = M

Magnitude



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According to this relationship, the magnitude for the 200-year earthquake is 6.93.

Table 1 magnitude for the 200-year earthquake from different catalogs

1910-1954 Catalog 1955-1998 Catalog 1910-1998 Catalog

6.89 6.93 6.87

We can see that although the relation equation is different, but the estimation of themagnitude for the 200-year earthquake is almost the same, only slightly different.For the first half period, because there were not enough earthquakes whichmagnitude is above 5, so the fitting is not so good.

Exercise 2

Calculate the Nyr Gutenberg-Richter relationship for

Mendocino area.For the Mendocino region (1955-1998), the Gutenberg-Richter relationship is asfollows:

N yr = N 44 yrs

44=

10 a10bM

44= 10 a

log4410bM

= 10 5.79230.8808 M log44

Figure 4 Gutenberg-Richter relationship for Mendicino (1955-1998)

How much larger is the production rate (Nyr) of earthquakes

logN = 5.7923 - 0.8808MR = 0.99489

00.5

11.5

22.5

3

3.5

2 3 4 5 6 7 L o g .

e v e n t s > = M

Magnitude



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in Mendicino compared to the Bay Area? What accounts forthe higher rate of seismic production?

The production rate of earthquakes in Mendicino compared to the Bay Area is:

ratio = N yr , Mendicino N yr , BayArea

= 105.7923 0.8808 M log44

/105.3249 0.8631 M log44

= 100.4674 0.0177 M

We can see from the ratio above clearly that even M is small, the ratio is greater than

1, that is to say the Mendicino area have a higher rate. This is because Mendocino

lies in a Triple Junction, which is a geologic triple junction where San Andreas Fault

meets the Mendocino Fault and the Cascadia subduction zone, separating three

tectonic plates: the Pacific Plate, the North American Plate and the Gorda Plate.

While the bay area lies in a transform boundary.

Exercise 3It is an urban legend that if enough small earthquakes occurthey will release strain before it builds to the point of acatastrophic earthquake. Prove that this is not true using your1910-1998 seismicity catalog.

According to the equation which relates the magnitude with the released moment:log(M 0) = 1.5M+16.05

The total moment released by events in each magnitude bin is listed in table 2.

Table 2 moment released by events in each magnitude binMagnitude Bin Number of Events Moment Released (dyne cm)3.0 M < 4.0 614 9.06e+234.0 M < 5.0 78 4.16e+245.0 M < 6.0 10 2.25e+256.0 M < 7.0 2 1.07e+26

Sum 704 1.34e+26

So from the above table, we can calculate the total moment released by earthquakeswhich have magnitudes between [3,6) is 2.75e+25 dyne cm. However, the moment

released by the earthquakes that have magnitudes between [6,7) is 1.34e+26, is 4.87times larger than the 2.75e+25. Even on average, one 6+ earthquake will release5.35e+25 which is greater than that released by events [3,6). Therefore, it is not trueto say that enough small earthquakes occur will release strain before it builds to thepoint of a catastrophic earthquake.

Another way to look at this problem is that, according to the relationship betweenmagnitude and the moment,


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M 0

= 10 1.5 M + 16.05

If the magnitude increases 1 unit, then the energy released will be 31.62 timeslarger, as: M

0= 10 1.5( M + 1)+ 16.05 /10 1.5 M + 16.05 = 10 1.5 = 31.62

But according to Gutenberg-Richter relationship, N = 10 5.2112 0.8095 M

If the magnitude decreases 1 unit, then the number of the earthquakes will be 6.45times larger. N = 10 5.2112 0.8095 M /10 5.2112 0.8095( M 1) = 10 0.8095 = 6.45Because the b value is generally between 2/3 to 1. So the maxum N is when bequals 1, that is N = 10. Thus, if the magnitude increase 1 unit, then the number ofearthquakes will be 1/10 of the original magnitude. Therefore, the total momentreleased by the events which have 1 unit smaller magnitude earthquake wont havea significant impact on that released by the earthquakes 1 unit larger.

What fraction of the total seismic moment over the past 89years does this 6.6 event represent?

The seismic moment released by the magnitude-6.6 event is M

0= 10 1.5 6,6 + 16.05 = 8.91 e + 25

During 1910-1998, there were altogether 704 events occurred in Bay Area. The totalseismic moment is: 1.34e+26Therefore, the fraction is: (8.91e+25/1.34e+26)100%=66.5%

Exercise 4

Problem A) Using the search results plot the magnitude of theevents as a function of time. Describe the behavior of theaftershock sequence in terms of the aftershock frequency andmagnitude.

The plot of the magnitude of the events as a function of time is shown in figure 5, thered dot is the main shock, and the blue ones are the aftershocks. We can see clearlythat with time increases, the number of the aftershocks is decreasing.


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Figure 5 magnitudes after the main shock

Figure 6 number of aftershocks after the main shock

Figure 6 shows the number of aftershocks with time after the main shock. We canalso see from the above figure, that with time increases, the number of theaftershocks has a trend of decreasing.

0

1

2

3

4

5

6

7

0 48 96 144 192

M a g n

i t u d e

Hours after the mainshock

89

23 27 2110 10 13

169

4 8 5

0

20

40

60

80

100

12 24 36 48 60 72 84 96 108120 132 144 N u m

b e r o

f a f t e r s h o c k s

Hours after the mainshock


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Problem B) Tabulate the numbers of aftershocks in 24-hour binsand determine the C and P parameters. Discuss how well OmorisLaw fits the Morgan Hill earthquake data.

The number of aftershocks in 24-hour bins is in table 3 and figure 7.Table 3 number of aftershocks based on days

Day No. of aftershocks1 1122 483 204 295 136 13

Figure 7 number of aftershocks after the main shock based on days

The Omoris law is n t = C / (K + t )P . Assuming K equals 0, and by using the above

data we can calculate C = 107.04 and P= 1.209. From the following figure, we can seethat the Omoris law fits the Morgan Hill earthquake data very well.

112

48

20

29

13 13

0

20

40

60

80

100

120

1 2 3 4 5 6 N u m

b e r o


Days after the mainshock


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Figure 8 Omoris law

Problem C

The background rate of M>1 events before the earthquake was 1 per week, that is0.1429 event per day during one week. According to Omoris law, when t is 236, thenext 7 day average is 0.1423, which means after the mainshock, it will take 236 daysbefore the rate of earthquake occurrence in the area returns to the background rate.

References

Thorne Lay, Terry C. Wallace (1995) Modern Global Seismology. Academic Press.Lecture notes of EPS 130, 2012 spring, UC Berkeley

n = 107.04/t 1.209

0

20

40

60

80

100

120

0 2 4 6 8 N u m

b e r o


Days after the mainshock

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