Solution Manual for Introduction to Seismology, 3rd Edition

Preview (16 of 80 Pages)

100%

Purchase to unlock

Loading page image...

Chapter 1Exercise solutions1. The radii of the Earth, Moon, and Sun are 6,371 km, 1,738 km, and 6.951×105km, respectively. From Figures 1.1, 1.5, and 1.6, make a rough estimateof how long it takes aP-wave to traverse the diameter of each body.Crude estimates based on mean velocity follow:Earth: 2×6371 km / 11 km/s = 1160 s = 19.3 minMoon: 2×1738 km / 7.6 km/s = 460 s = 7.7 minSun: 2×7×105km / 250 km/s = 5600 s = 93 min2. ThePtoSvelocity ratio for most common rocks is about 1.7 (∼ √3). Whatsolid part of the Earth has a very differentP/Svelocity ratio? Hint: Look atFigure 1.1.The inner core, where theP/Svelocity ratio is about 3.3. Assume that theSvelocity perturbations plotted at 150 km depth in Figure1.7 extend throughout the uppermost 300 km of the mantle.Estimate howmany seconds earlier a vertically upgoingS-wave will arrive at a seismic stationin the middle of Canada, compared to a station in the eastern Pacific. Ignoreany topographic or crustal thickness differences between the sites; consideronly the integrated travel time difference through the upper mantle.Svelocity is about 4.5 km/s, so wave takes 300/4.5 = 67 s to go 300 km.Canada is about 1.4% fast; eastern Pacific is about 1.8% slow. The differenceis 3.2% or 0.032.Thus, S-wave will arrive earlier at central Canadian station by about (0.032)67=2.1 s. This may be an underestimate because the plot is saturated so theperturbations could exceed these values.4. Assuming that thePvelocity in the ocean is 1.5 km/s, estimate the minimumand maximum water depths shown in Figure 1.8. If the crustalPvelocity is 5km/s, what is the depth to the top of the magma chamber from the sea floor?1

Loading page image...

2CHAPTER 1.EXERCISE SOLUTIONSMinimum water depth = 3.5×1.5 / 2 = 2.62 kmMaximum water depth = 3.93×1.5 / 2 = 2.95 kmTop of magma chamber = (3.95 - 3.5)×5 / 2 = 1.1 km below sea floor5. Earthquake moment is defined asM0=μDA, whereμis the shear modulus,Dis the average displacement on the fault, andAis the fault area that slipped.Themoment magnitude,MW, is defined asMW=23[log10M0−9.1], wherethe momentM0is in N m.(a) The moment of the 2004 Sumatra-Andaman earthquake has been esti-mated to be about 1.0×1023N m. What moment magnitude does thiscorrespond to?Assuming that the fault is horizontal, crudely estimatethe slip area from the image shown in Figure1.9.Assuming that theshear modulusμ= 3.0×1010N/m2, then compute the average displace-ment on the fault.From the equation given, we haveMW=23[log10(1023)−9.1]=23(23−9.1) = 9.27The moment magnitude is9.27 .The slip area is about 200 km wide and 1200 km long, thusA= 240,000km2= 2.4×1011m2, and we haveD=M0μA=10233.0×1010×2.4×1011= 1007.2 = 14 mThe average displacement is about14 m .(b) A Hollywood director wants to make a movie about a devastating magni-tude 10 earthquake in California with 40 m of slip (displacement) alongthe San Andreas Fault.Given that the crust is about 35 km thick inCalifornia, how scientifically plausible is this scenario?Hint:you mayassume that the shear modulusμ= 3×1010N/m2We also need to know the length of the San Andreas Fault (SAF) thatruptures, which is not given, so some research is needed.The SAF isabout 1200 km long, which is also roughly the N-S extent of California.So 1200 km is a reasonable upper bound on the maximum SAF rupturelength. If we assume that the rupture extends throughout the crust (anassumption implied by the question wording, but which is probably anoverestimate for typical SAF earthquakes), then the fault area isA=1.2×106×3.5×104m2= 4.2×1010m. We then haveM0=μDA= 3×1010·40·4.2×1010≈5×1022N mThe moment magnitude isMW=23[log10(5×1022)−9.1]=23(22.7−9.1) = 9.07

Loading page image...

3The moment magnitude is about9.1 , which is much less than 10. Thusthe director’s idea for a magnitude 10 earthquake isnot plausibleforthe SAF in California.6. The 2004 Sumatra earthquake lasted about 10 minutes and radiated about2×1017joules of seismic energy.Compute the corresponding power outputin terrawatts (TW). Then do some research on the web and compare thispower to: (a) average rate of electricity consumption in the United States, (b)average dissipation rate of tidal energy in the world’s oceans, and (c) totalheat flow out of the Earth. Note that the total energy release (including heatgenerated on the fault, etc.) of the Sumatra earthquake may be significantlygreater than the seismically radiated energy. This is discussed in Chapter 9.A watt is one joule per second. 10 minutes is 600 s. Thus, the radiated powerduring the earthquake was 2×1017/600 = 3.33×1014watts. A terrawatt (TW)is 1012watts, sothe radiated power was about 330 TW . For comparison,(a) 2017 US power consumption was about 4000 TW hours (source Wikipedia).Thus the average power usage is 4000/(365·24) =0.46 TW .(b) Earth’s tidal energy dissipation is about3.7 TW , most of which occursin the oceans (see p. 106–108 of Stacey and Davis,Physics of the Earth, 2008).The Wikipedia article on tidal acceleration cites 3.3 TW.(c) Earth’s integrated heat flux is about44 TW(p. 337, Stacey and Davis,Physics of the Earth, 2008). The Wikipedia article on Earth’s energy budgetcites 47 TW from Davies and Davies, Earth’s surface heat flux,Solid Earth,1(1), 524, 2010.

Loading page image...

Chapter 2Exercise solutions1. Assume that the horizontal components of the 2-D stress tensor areτττ=[τxxτxyτyxτyy]=[−30−20−20−40]MPa(a) Compute the normal and shear stresses on a fault that strikes 10◦eastof north.cos 10◦= 0.9848, sin 10◦= 0.1736, thus fault parallel vectorˆf= (0.1736,0.9848), fault normal vectorˆn= (0.9848, -0.1736). The traction on thefault plane is given byt(ˆn) =τττˆn=[−30−20−20−40] [0.9848−0.1736]=[−26.07−12.75]MPa.andtN=t·ˆn= (−26.07,−12.75)·(0.9848,−0.1736) = -23.46MPatS=t·ˆf= (−26.07,−12.75)·(0.1736,0.9848) = -17.08MPaThe fault normal compression is 23.46 MPa.The shear stress is 17.08MPa.(b) Compute the principal stresses, and give the azimuths (in degrees east ofnorth) of the maximum and minimum compressional stress axes.λ1=−55.61 MPa,38◦E of Nλ2=−14.38 MPa, 128◦E of N(solution computed using Matlab)2. The principal stress axes for a 2-D geometry are oriented at N45◦E andN135◦E, corresponding to principal stresses of -15 and -10 MPa. What are the4 components of the 2-D stress tensor in a (x= east,y= north) coordinatesystem?The eigenvector matrix isN=[1/√21/√21/√2−1/√2]1

Loading page image...

2CHAPTER 2.EXERCISE SOLUTIONSNote thatN=NT=N−1. The principal (diagonalized) stress tensor isτττP=[−1500−10]which we can rotate to the stress tensor in E-N coordinates asτττ=NτττPNT=[−12.5−2.5−2.5−12.5]MPa3. Figure 2.5 shows a vertical-component seismogram of the 1989 Loma Prietaearthquake recorded in Finland.(a) Estimate the dominant period,T, of the surface wave from its first tencycles. Then compute the frequencyf= 1/T.There are about 8 peaks in 200 s so the period,T, is about25 s .Thefrequency,f, is 1/25 =0.04 Hz .(b) Make an estimate of themaximumsurface-wave strain recorded at thissite.Hints:1 micron = 10−6m, assume the Rayleigh surface wave phasevelocity at the dominant period is 3.9 km/s, remember that strain is∂uz/∂x,Table 3.1 may be helpful.The amplitudeA= 300 microns = 3×10−4m.The wavelength Λ =cT=(3.9 km/s)(25 s) = 97.5 km = about 100 km = 1×105m. The wavenumberk= 2π/Λ = 0.0628/km = 6.28×10−5/m. We can approximate displacementas a harmonic wave asux=Asin(kx). Strain =∂ux/∂x=kAcos(kx) so themaximum strain occurs when cos = 1 and isAk= (3×10−4)(6.28×10−5) =1.9×10−8.Note that this value is halved if we express this in terms of the strain tensor,i.e.,emax≈[010−810−80]4. Using Equations (2.4), (2.23), and (2.30), show that the principal stress axesalways coincide with the principal strain axes for isotropic media.In otherwords, show that ifxis an eigenvector ofe, then it is also an eigenvector ofτττ.We can show that the principal strain axes coincide if we prove that ifxis aneigenvector of the strain tensorethen it also must be an eigenvector of thestress tensorτ, that is ifeijxj=αxithenτijxj=α′xi

Loading page image...

3Using the isotropic stress/strain relationτij=λekkδij+ 2μeij(equation 2.29),we haveτijxj=λekkδijxj+ 2μeijxj=λekkxi+ 2μαxi= (λekk+ 2μα)xi=α′xi(2.1)whereα′=λekk+ 2μα. This completes the proof.5. From Equations (2.34) and (2.35) derive expressions for the Lam´e parametersin terms of the seismic velocities and density.μ=ρβ2λ=ρ(α2−2β2)6. Seismic observations ofSvelocity can be directly related to the shear modulusμ. However,Pvelocity is a function of both the shear and bulk moduli. Forthis reason, sometimes seismologists will compute thebulk sound speed, definedas:Vc=√κρ(2.2)which isolates the sensitivity to the bulk modulusκ.(a) Derive an equation forVcin terms of thePvelocity,α, and theSvelocity,β.We haveVc=√κρWe need to solve forκin terms ofαandβ. We have from thePandSvelocitydefinitions:λ+ 2μ=ρα2μ=ρβ2λ=ρ(α2−2β2)The bulk modulusκ=λ+ 2/3μand thusκ=ρ(α2−2ρβ2+ 2/3ρβ2)=ρ(α2−4/3β2)and thusVc=√α2−43β2

Loading page image...

4CHAPTER 2.EXERCISE SOLUTIONS(b) For the specific case of a Poisson solid, expressVcas a fraction of thePvelocity.For a Poisson solid,α=√3β, soβ2=13α2. Substituting into our result from(a), we haveVc=√α2−49α2=α√99−49 =√53α= 0.745α7. What is theP/Svelocity ratio for a rock with a Poisson’s ratio of 0.30?Starting with equation (2.36), we haveσ=(α/β)2−22(α/β)2−22σ(α/β)2−2σ=(α/β)2−22−2σ=(α/β)2−2σ(α/β)2= (α/β)2[1−2σ]αβ=√2−2σ1−2σαβ=√2−0.61−0.6 =√1.40.4 =√3.5 =1.87forσ= 0.308. A sample of granite in the laboratory is observed to have aPvelocity of 5.5km/s and a density of 2.6 Mg/m3. Assuming it is a Poisson solid, obtain valuesfor the Lam´e parameters, Young’s modulus, and the bulk modulus. Expressyour answers in pascals.For a Poisson solid,β=α/√3 = 3.1754 km/s forα= 5.5 km/s. From Exercise2.5, we haveλ=ρ(α2−2β2)=2.6(5.52−2(3.1754)2)λ=26.22 Mgm3km2s2106m2km2103kgMgλ=26.22×109kgm s2λ=26.22 GPaandμ=ρβ2= 2.6(3.1754)2μ=26.22 GPaNote thatλ=μas should be the case for a Poisson solid.The Young’smodulus is given byE=(3λ+ 2μ)μλ+μ= 2.5μsinceλ=μ

Loading page image...

5E=65.55 GPaThe bulk modulus is given byκ=λ+ 23μ= 53μκ=43.7 GPaSummarizing, we have:λ=μ= 26.22 GPA,E= 65.55 GPa,κ= 43.7 GPa .9. Using values from the PREM model (Appendix 1), compute values for thebulk modulus on both sides of (a) the core–mantle boundary (CMB) and (b)the inner-core boundary (ICB). Express your answers in pascals.The bulk modulus is given byκ=λ+ 23μFrom Exercise 2.5 we haveμ=ρβ2andλ=ρ(α2−2β2) and thusκ=ρ(α2−43β2)From the PREM model in the Appendix, we can thus construct a table withtheκvalues:VpVsdenkappaCMB top13.727.265.57657 GpaCMB bot8.0609.90643 GPaICB top10.36012.171306 GPaICB bot11.033.512.761344 GPa10. Figure 2.6 shows surface displacement rates as a function of distance from theSan Andreas Fault in California.(a) Consider this as a 2-D problem with thex-axis perpendicular to the faultand they-axis parallel to the fault. From these data, estimate the yearlystrain (e) and rotation (ΩΩΩ) tensors for a point on the fault. Express youranswers as 2×2 matrices.Each year,∂y/∂x≈43 mm / 50 km = (43×10−3m) / (50×103m) =0.86×10−6= 8.6×10−7/yr . This is the only non-zero partial derivativein theJmatrix. We thus haveJ=[0θ00]=e+Ω=[0θ/2θ/20]+[0θ/2−θ/20]whereθ=∂y/∂x= 8.6×10−7and thuse=[04.3×10−74.3×10−70]

Loading page image...

6CHAPTER 2.EXERCISE SOLUTIONSandΩ=[0−4.3×10−74.3×10−70](b) Assuming the crustal shear modulus is 27 GPa, compute the yearlychange in the stress tensor. Express your answer as a 2×2 matrix withappropriate units.τττ=[02.3×1042.3×1040]Pa(c) If the crustal shear modulus is 27 GPa, what is the shear stress acrossthe fault after 200 years, assuming zero initial shear stress?4.6×106Pa (4.6 MPa)(d) If large earthquakes occur every 200 years and release all of the distributedstrain by movement along the fault, what, if anything, can be inferredabout theabsolutelevel of shear stress?The absolute shear stress across the fault before the earthquake must beat least 4.6 MPa, but it could also be much bigger, depending upon howlarge the stress drop is relative to the absolute stress. The involves thefrictional properties of the fault, and is discussed in Chapter 9.(Note:this question is to get students to think or do research—the answer is notin the chapter)(e) What, if anything, can be learned about the fault from the observationthat most of the deformation occurs within a zone less than 50 km wide?Vertical strike-slip faults like the San Andreas are typically modeled asa locked zone (where the earthquakes occur) above a creeping zone atdepth. The width of the deformation zone is proportional in some senseto the depth of the locked zone. (Note: this question is to get studentsto think or do research—the answer is not in the chapter.)11. Do some research on the observed density of the Sun.Are the high soundvelocities in the Sun (see Fig. 1.6) compared to Earth’sPvelocities causedprimarily by low solar densities compared to the Earth, a higher bulk modulusor some combination of these factors?The average solar density is about 1.4 g/cc, compared to 4 to 5 g/cc in Earth’smantle. This would make solar velocities higher by about a factor of√3∼1.7.But the average solarPvelocity is about 50 times higher than in Earth, so themore important difference is that the bulk modulus,κ, is much higher for theSun. Students can make this more complicated by trying to take into accountthe velocity depth dependence, but I was just looking for a rough estimatebased on average properties.

Loading page image...

712. The University of California, San Diego, operates the Pi˜non Flat Observatory(PFO) in the mountains northeast of San Diego (near Anza).Instrumentsinclude high-quality strain meters for measuring crustal deformation.(a) Assume, at 5 km depth beneath PFO, the seismic velocities areα= 6km/s andβ= 3.5 km/s and the density isρ= 2.7 Mg/m3.Computevalues for the Lam´e parameters,λandμ, from these numbers. Expressyour answer in units of pascals.From the solution to Exercise 5, we haveλ=ρ(α2−2β2) = 2.7(6.02−2(3.5)2)λ=31.05 Mgm3km2s2106m2km2103kgMgλ=31.05×109kgm s2λ=31.05 GPaμ=ρβ2= 2.7(3.5)2μ=33.07 GPaSummarizing,λ= 31.05 GPa,μ= 33.07 GPa .(b) Following the 1992 Landers earthquake (MS= 7.3), located in south-ern California 80 km north of PFO (Fig. 2.7), the PFO strain metersmeasured a large static change in strain compared to values before theevent. Horizontal components of the strain tensor changed by the follow-ing amounts:e11=−0.26×10−6,e22= 0.92×10−6,e12=−0.69×10−6.In this notation 1 is east, 2 is north, and extension is positive. You mayassume that this strain change occurred instantaneously at the time ofthe event. Assuming these strain values are also accurate at depth, usethe result you obtained in part (a) to determine the change in stressdue to the Landers earthquake at 5 km, that is, compute the change inτ11,τ22, andτ12. Treat this as a two-dimensional problem by assumingthere is no strain in the vertical direction and no depth dependence ofthe strain.From (2.30), we haveτττ=[λtr[e] + 2μe112μe122μe21λtr[e] + 2μe22]Given the Landers strain changes ofe11=−0.26×10−6,e22= 0.92×10−6,e12=−0.69×10−6, then tr[e] = 0.66×10−6andτττ=[3300−45600−4560081300]Paorτ11= 3.3×103Pa,τ22= 8.13×104Pa, andτ12=−4.56×104Pa(c) Compute the orientations of the principal strain axes (horizontal) for theresponse at PFO to the Landers event. Express your answers as azimuths(degrees east of north).

Loading page image...

8CHAPTER 2.EXERCISE SOLUTIONSThe strain tensor is[−0.26−0.69−0.690.92]×10−6The solution to the eigenvalue problem for this matrix gives orientationsof principal strains asN65◦E and N155◦E (or N25◦W) .(d) A steady long-term change in strain at PFO has been observed to occur inwhich the changes in one year are:e11= 0.101×10−6,e22=−0.02×10−6,e12= 0.005×10−6. Note that the long-term strain change is close to sim-ple E–W extension. Assuming that this strain rate has occurred steadilyfor the last 1,000 years, from an initial state of zero stress, compute thecomponents of the stress tensor at 5 km depth.(Note:This is proba-bly not a very realistic assumption!) Don’t include the large hydrostaticcomponent of stress at 5 km depth.From (2.30), we haveτττ=[λtr[e] + 2μe112μe122μe21λtr[e] + 2μe22]Given the 1000 year strain changes ofe11= 0.101×10−3,e22=−0.02×10−3,e12= 0.005×10−3, then tr[e] = 0.081×10−3and usingλ= 31.05GPa andμ= 33.07 GPa from 2.6a we haveτττ=[9.200.330.331.19]MPa(e) Farmer Bob owns a 1 km2plot of land near PFO that he has fenced andsurveyed with great precision. How much land does Farmer Bob gain orlose each year? How much did he gain or lose as a result of the Landersearthquake? Express your answers in m2.Use tr[e] as a measure of area change. Each year, tr[e] = 0.081×10−6soFarmer Bob gains 0.081 m2of land each year .For Landers tr[e] = 0.66×10−6soFarmer Bob gained 0.66 m2of land following the Landers earthquake .(f) (COMPUTER) Write a computer program that computes the stress acrossvertical faults at azimuths between 0 and 170 degrees (east from north,at 10 degree increments). For the stress tensors that you calculated in (b)and (d), make a table that lists the fault azimuth and the correspond-ing shear stress and normal stress across the fault (for Landers theseare the stress changes, not absolute stresses). At what azimuths are themaximum shear stresses for each case?

Loading page image...

9(g) (COMPUTER) Several studies (e.g., Stein et al., 1992, 1994; Harris andSimpson, 1992; Harris et al., 1995; Stein, 1999; Harris, 2002) have mod-eled the spatial distribution of events following large earthquakes by as-suming that the likelihood of earthquake rupture along a fault is relatedto theCoulomb failure function(CFF). Ignoring the effect of pore fluidpressure, the change in CFF may be expressed as:∆CFF = ∆|τs|+μs∆τn,whereτsis the shear stress (traction),τnis the normal stress, andμsisthe coefficient of static friction (don’t confuse this with the shear mod-ulus!).Note that CFF increases as the shear stress increases, and asthe compressional stress on the fault is reduced (recall in our sign con-vention that extensional stresses are positive and compressional stressesare negative).Assume thatμs= 0.2 and modify your computer pro-gram to compute ∆CFF for each fault orientation. Make a table of theyearly change in ∆CFF due to the long-term strain change at each faultazimuth.(h) (COMPUTER) Now assume that the faults will fail when their long-term CFF reaches some critical threshold value. The change in time tothe next earthquake may be expressed as∆t= CFF1000+L−CFF1000CFFa,where CFFais the annual change in CFF, CFF1000is the thousand yearchange in CFF, and CFF1000+Lis the thousand year + Landers changein CFF (note that CFF1000+L6= CFF1000+ CFFL). Compute the effectof the Landers earthquake in terms of advancing or retarding the timeuntil the next earthquake for each fault orientation. Express your answerin years, using the sign convention of positive time for advancement ofthe next earthquake and negative time for retardation.

Loading page image...

10CHAPTER 2.EXERCISE SOLUTIONSResults for partsf–hare given in the following table:----Landers----Annual (*1000)azitau-stau-ntau-stau-nCFF-LCFF-adtL/a0-45643.53294.0330.79196.246302.32170.0-20.721.310-56239.421258.71679.68841.760491.13447.9-15.117.520-60052.041764.02825.98047.368404.84435.3-11.715.430-56621.462336.73631.36908.769088.85013.0-8.813.840-46361.580495.33998.75563.462460.55111.4-5.912.250-30509.794049.83883.84173.449319.64718.5-2.510.560-10977.9101365.23300.52906.631251.03881.92.48.1709877.9101559.12319.11915.830189.82702.311.211.28029542.394608.31058.01320.348464.01322.036.736.79045643.581351.0-330.81192.061913.7569.2-51.6108.810056239.463386.3-1679.61546.568916.61988.9-21.934.711060052.042881.0-2825.92341.068628.23294.0-15.620.812056621.422308.3-3631.33479.561083.14327.2-12.114.113046361.54149.6-3998.74824.947191.44963.7-9.29.514030509.6-9404.8-3883.86214.828628.75126.8-6.35.615010977.9-16720.2-3300.57481.67633.94796.8-3.01.6160-9877.9-16914.1-2319.18472.56495.14013.61.61.6170-29542.3-9963.3-1058.09068.027549.72871.69.69.6(i) No increase in seismicity (small earthquake activity) has been observednear PFO following the Landers event. Does this say anything about thevalidity of the threshold CFF model?Maybe! (another food for thought question to get students thinking, withno simple answer)

Loading page image...

Chapter 3Exercise solutions1. PeriodTis to angular frequencyωas wavelength Λ is to: (a) wavenumberk,(b) velocityc, (c) frequencyf, (d) timet, (e) none of the above.(a) wave number2. Figure 3.7 plots a harmonic plane wave att= 0, traveling in thexdirectionat 5 km/s. (a) Write down an equation for this wave using the sine functionthat describes displacement,u, as a function ofxandt.We have wavelength Λ = 8000 m, velocityc= 5000 m/s, thusf=c/Λ = 5/8Hz. We then haveω= 2πf= 5π/4 andk= 2π/Λ =π/4000. Thus we havefor a wave traveling in the +xdirection:u=0.04 sin[(π/4000)x−(5π/4)t]=0.04 sin[π(x4000−54t)]=0.04 sin[0.000785x−3.927t](t in seconds,x in meters)The correct signs of thexandtterms in this equation can be verified byinspecting the figure and considering the casest= 0 andx= 0. Alternatively,we could reverse the signs fortandxby adding a phase termπ, i.e.,u= 0.04 sin[5π4t−π4000x+π](b) Write the corresponding equation using the cosine function.u= 0.04 cos[0.000785x−3.927t−π/2](t in seconds,x in meters)(c) What is the displacement atx= 6 km andt= 30 s?Zero, since 6000/4000 - 5(30)/4 = 1.5 - 37.5 = -36, sin−36π= 0.(d) What is the maximum strain for this wave?1

Loading page image...

2CHAPTER 3.EXERCISE SOLUTIONSThe strain is∂u/∂x= (0.000785)(0.04) cos[0.000785x−3.927t] so the maxi-mum is (0.000785)(0.04) = 0.0000314 =3.14×10−5.If this is a shear wave, the 2-D strain tensor would have half this value as theoff-diagonal terms.3. Consider two types of monochromatic plane waves propagating in thexdi-rection in a uniform medium: (a)P-wave in whichux=Asin(ωt−kx), (b)S-wave with displacements in theydirection, i.e.,uy=Asin(ωt−kx). Foreach case, derive expressions for the nonzero components of the stress tensor.Refer to 2.17 to get the components of the strain tensor; then use 2.30 toobtain the stress components. Hint: Look at Example 3.4.1.(a)From (2.17) the only nonzero component of the strain tensor ise11=∂ux∂x=−kAcos(ωt−kx)Then from (2.29) we haveτ11=−(λ+ 2μ)kAcos(ωt−kx)τ22=τ33=−λkAcos(ωt−kx)(b)We are givenuy=Asin(ωt−kx) and are asked to solve for the nonzerocomponents of the stress tensor. From (2.17) the only nonzero components ofthe strain tensor aree12=e21=12∂uy∂x=−12kAcos(ωt−kx)Then from (2.30) we haveτ12=τ21=−μkAcos(ωt−kx)4. Assume harmonicP-waves are traveling through a solid withα= 10 km/s. Ifthe maximum strain is 10−8, what is the maximum particle displacement forwaves with periods of: (a) 1 s, (b) 10 s, (c) 100 s?For harmonicP-wave traveling inxdirection we haveux=Asin(ωt−kx)The only nonzero strain component ise11=∂ux∂x=−kAcos(ωt−kx)

Preview Mode

This document has 80 pages. Sign in to access the full document!

Report

Study Now!

Document Details

Related Documents

Exploring Coral Reefs Understanding Abiotic Factor

Building Topographic Maps Creating 3D Maps with Wa

Understanding the Three Types of Regions

The Lowest Mountain Unveiling Mount Wycheproof's G

Hunter McCarty Carbon Cycle Gizmo 2021 Bio

Understanding Bakersfield's Rainfall Patterns Prec

Student Exploration Water Cycle

Geosystems: An Introduction To Physical Geography, 10th Edition Test Bank

Geography: Realms, Regions, and Concepts, 16th Edition Test Bank

Solution Manual for Earth: An Introduction to Physical Geology, 12th Edition

Company

Explore

Study Tools