Relating Seismic Design Values to the Richter Scale 2

[jdblake]

I have a client asking me what the comparable earthquake magnitude would be (Richter Scale) for the design loads that were calculated. I’m in Portland Oregon and the typical Sds is around 0.70 and Sd1 is around 0.40. Is there anything that relates the two together?

 

[TERIO]

Take a look at TID-7024, "Nuclear Reactor and Earthquakes".

http://www.osti.gov/energycitations/product.biblio.jsp?osti_id=4620608

In the first chapter there are some charts/equations that roughly relate ground acceleration to Richter Scale magnitude, which may help you.

 

[JAE]

The "Richter" scale is not the current measure of magnitude that is used today.

 

[jdblake]

It may not be the current "tool" for measurement, but it is what is relatable to the client.

 

[dgillette]

If it's what you're going to explain to the client, you'll have to explain it as "the equivalent of a 6.0 20 miles away, or a 6.5 35 miles away, etc.," since magnitude is a measure of the amount of energy released, not the intensity of shaking at the site.

On that note, JAE is correct. Generally they use moment magnitude, M-sub-w, nowadays, which is more directly related to energy than the Richter magnitude. Richter M is based on the behavior of a specific type of instrument at some specific distance, so it is a very indirect measure of energy. For a client, you could call moment mag an improved Richter scale - not exactly true, but not a lie.

 

[Qshake]

Agree with both JAE and dgillette. unfortunately, reporters, you know those scientific types, are hooked on the Richter scale. That's all that comes out of their mouths.

As a footnote, the Richter is really only good in California and it is a Wood-Anderson seismograph within a limited distance from the epicenter.

Regards,
Qshake

Eng-Tips Forums:Real Solutions for Real Problems Really Quick.

 

[cvg]

regardless of the amount of energy released by the earthquake or the estimated Richter or Mercalli intensity, your site may experience more or less intense "shaking" related to the local geologic conditions such as depth to bedrock, depth to water table and type of soil. Say you have two sites at equal distances from the epicenter. One is near the ocean, the other is sitting on bedrock in the foothills. The intensity of the shaking will be completely different for each site. This concept should be part of the conversation with the owner. In other words, the charts and equations should be taken with several grains of salt.

 

[KeySol]

I agree with everyone else so far. The best way to explain it to your client is to explain to them the Code does not use a specific magnitude earthquake for design. The code uses historical data as well as local fault information to determine an amount of acceleration that your building will experience. There is a 2% chance in 50 years your building will experience that much acceleration.

Regardless, if your client has no idea about anything to do with seismic design and asks a question like that. Think long and hard about it, and then make something up.

 

[bdSE]

It may also be helpful to relate to the client the Return Period associated with a 2% in 50 year event (some USGS values provide 10% in 50, not 2% in 50). If one of those earthquakes occurred today, it is not expected to exceed that earthquake again for another 2500 years. Equation included below.

RP = T / [r(1+0.5r)]
r = Probability of exceedance.
T = Exposure time
RP = 50 / [10%(1+0.5(10%))]
RP = 475 years (aka 500 year EQ)
Likewise the 2% in 50 years design is known as the 2500 year EQ.

You'll want to check to see which load probabilities were used in your seismic analysis before giving the 2500 year value a shot.

 

[RonRoberts]

Take a look at:

Boore, D. M., W. B. Joyner, and T. E. Fumal (1993). Estimation of Response Spectra and Peak Accelerations from Western North American Earthquakes: An Interim Report, U.S. Geological Survey Open-File Report 93-509, 72 pp.

and subsequent work/comments by Boore in:

Seismological Research Letters; May/June 2005; v. 76; no. 3; p. 368-369; DOI: 10.1785/gssrl.76.3.368

 

[dgillette]

"If one of those earthquakes occurred today, it is not expected to exceed that earthquake again for another 2500 years."

This is a common misunderstanding that I can't let go by, lest it be propagated further. The term "return period" misleads a lot of people. It's true that SOME faults show identifiable periodic behavior (including some around the island of Hispaniola, in the news recently), but that is not the general rule, and there are often other potential sources of strong motion in addition to that one particular fault that might have moved this morning. Seismologists typically estimate exeedance probability from historic seismicity using a Poisson model, in which each year is essentially considered an independent trial, so occurrence of the so-called "2500-year earthquake" in 2010 doesn't get us off the hook for larger earthquakes in the following 2500 years. By the Poisson model, it's just as likely in 2011 regardless of whether it occurred in 2010. "Return period" is really just the unfortunate shorthand term used for "reciprocal of annual probability of exceedance." It is more correct to say "The probability of exceedance of this level of shaking is 1/2500 in any given year."

The probability of exceeding the so-called 2500-year EQ in 50 years is

1-(1-1/2500)^50 = 0.0198,

or running it backward, the earthquake with 2% prob of exceedence in 50 years has annual probability equal to

Pa = 1 - (1-0.02)^(1/50) = 4.04x10^-4 = 1/2475

The probability of the "2500-year earthquake" being exceeded in the next 2500 years is

1-(1-1/2500)^2500 = 0.63

Exceedance of the 1/2500 earthquake is at least slightly probable for any period exceeding about 1735 years:

1-(1-1/2500)^1735 > 0.5.

I'd like to see "return period" struck from the dictionary, but it's probably too late for that.