Exercise Problems in LiquefactionLecture-33*

Exercise Problem -1A rounded sand subjected to a series of monotonic triaxial compression tests exhibits the steady-state line(SSL) shown in the next slide. Assuming that the sand can mobilize a friction angle of 330 in the steady state, estimate the steady state strength that would be mobilized by the following test specimens:

*Initial void ratioTest ConditionDrainedUndrainedDrainedUndrained0.750.750.600.60

*1010010000.50.60.70.81Effective confining pressure-kPaVoid ratio, eFig :1Exercise Problem -1

SolutionFrom graph, steady state 3C is read as

For granular soils,

Steady state shear strength*

*Solution

Consider an anisotropically consolidated direct simple shear test specimen with the initial conditions shown below. Show graphically how the pore pressure ratio at the initiation of liquefaction, ru varies with the amplitude of the cyclic shear stress.*50100150p-kPa0255075q-kPaExercise Problem -2

Excess pore water pressure uexcess decreases with cyclic stress amplitude

where is the effective confining pressureru also decreases with increase in cyclic shear stress amplitude.

*50100150p-kPa0255075q-kPauexcessuexcessSolution

A 10m thick deposit of loose sand is saturated below a depth of 3m. The soil below the water table is highly susceptible to liquefaction. Estimate the ground surface acceleration that would be required to produce sand boils in a M=7.7 earthquake.

H1 = 3mH2 =7m*Exercise Problem -3

*Figure: Thickness of the overlying layer required to prevent level ground liquefaction related damage (After Ishihara, 1985). From the figure, ground surface acceleration that would be required to produce sand boils for H1 = 3m and H2 = 7m is 0.2 g.SolutionSource: Kramer (1996)

Estimate the variation of uniform cyclic shear stress amplitude with depth for the upper 50ft of soil (assume g=120pcf) at the Gilroy No. 2 (soil) station in the Loma Prieta earthquake. Estimate the number of equivalent uniform stress cycles.

*Exercise Problem -4Figure: Responses at Gilroy stations for Loma Prieta EQSource: Kramer (1996)

Gilroy No. 2 earthquake parameters: M=7.1, amax = 0.332g

g = 120 pcf.

cyc= 0.65 0.332 120 z rd *Solution

z (ft)rd cyc (psf)01.0050.99128100.98254.....

*SolutionFigure: Variation of cyclic shear stress amplitude with depth

z (ft)rd cyc (psf)01.0050.99128100.98254.....

*Figure: Number of equivalent uniform stress cycles (Neq) for earthquakes of different magnitude (After Seed et al., 1975)SolutionFrom figure, for M=7.1, Neq =10 (Taking Mean)

A level deposit of saturated clean sand has an average (N1)60 value of 18 and an average dry unit weight of 105 pcf. Plot the variation of cyclic shear stress required to produce liquefaction in M= 5.5, 6.5, 7.5 and 8.5 earthquakes.*Exercise Problem -5

(N1)60 = 18;gd = 105pcf

For saturated sands, assume specific gravity G= 2.6.

*Solution

*Fig: Relationship between cyclic stress ratio and (N1)60 for Mw = 7.5 earthquakesSolution

From figure, for (N1)60 = 18, CSRM=7.5 =0.2For other magnitudes, magnitude correction factors should be applied.

SolutionTable: Magnitude Correction Factors*

MagnitudeCSRM/CSRM=7.51.5061.321.131.000.89

*SolutionTable: Cyclic stress Ratios

MagnitudeR = CSRM/CSRM=7.5CSRM = R0.25.51.440.2886.51.200.247.51.000.208.50.900.18

*Solution

Depth z ft)

M = 5.5M = 6.5M = 7.5M = 8.5

00000593.16877.64064.70058.2310186.336155.280129.400116.4620372.672310.560258.800232.9230559.008465.840388.200349.3840745.344621.120517.600465.8450931.680776.400647.000465.84

*Solution

A level deposit of saturated clean sand has an average (N1)60 value of 18 and an average dry unit weight of 105pcf. Plot the variation of cyclic shear stress required to produce liquefaction in M= 5.5, 6.5, 7.5 and 8.5 earthquakes. Assume that the sand has 15% fines.

*Exercise Problem -6

**Fig: Relationship between cyclic stress ratio and (N1)60 for Mw = 7.5 earthquakesSolutionSource: Kramer (1996)

From figure, for (N1)60 = 18, with 15% fines as shown in figure (previous slide) CSRM=7.5 =0.27For other magnitudes, magnitude correction factors should be applied.

SolutionTable: Magnitude Correction Factors*

MagnitudeCSRM/CSRM=7.51.5061.321.131.000.89

*SolutionTable: Cyclic stress Ratios

MagnitudeR = CSRM/CSRM=7.5CSRM = R0.275.51.440.3886.51.200.327.51.000.278.50.900.24

*Solution

Depth z ft)

M = 5.5M = 6.5M = 7.5M = 8.5

000005125.7768104.81487.34578.610510251.5536209.628174.69157.22120503.1072419.256349.38314.44230754.6608628.884524.07471.663401006.214838.512698.76628.884501257.7681048.14873.45786.105

*Solution

Prior to the 1989 Loma Prieta earthquake, a site investigation at the Hunters Point Navy Base in San Francisco showed 13 to 15m of clean, hydraulically filled sand with D50= 0.29mm. The water table was at a depth of about 2.5m.CPT tests produced the following average tip resistances. The Loma Prieta earthquake produced peak ground accelerations on the order of 0.15g to 0.20g at similar sites in the Bay Area. Assuming that the sand has an average dry density of 1.6Mg/m3, compute and plot the factor of safety against liquefaction that would have been expected in the Loma Prieta earthquake.

*Depth Interval(m)Average qc(MPa)3-44-55-66-77-88-99-1010-116.55.75.24.95.16.36.97.6Exercise Problem -7

*Solutiond = 1.6Mg/m3gd= 15.7kN/m3

D50 of sand 0.29 mmDepth of water table 2.5 mMagnitude of earthquake 7.1Peak ground accelerarationCase I0.15 gCase II0.2 gMax dry density (d)1.6 Mg/m3 =15.696 kN/m3Specific gravity G (assumed)2.6Unit weight of water (w)9.81 kPavoid ratio (e )0.625Saturated unit weight (sat)19.4690 kN/m3

*Fig: Relationship between cyclic stress ratio and normalized cone resistance (Mitchell and Tseng, 1990)Solution

*Solution

Depth interval (m)Avg depth(m)rdAvg qc(MPa)v0' (kPa)CRR7.5CRR7.1/CRR7.5CRR 7.1v0 (kPa)CSR 0.15gCSR 0.2gFOS at 0.15gFOS at 0.2g3-43.50.986.548.900.131.070.1358.710.1150.1531.170.8744-54.50.975.758.560.111.070.1278.180.1260.1680.930.6995-65.50.965.268.220.111.070.1197.650.1340.1790.840.6296-76.50.954.977.880.101.070.11117.120.1390.1860.770.5767-87.50.945.187.540.101.070.11136.590.1430.1910.770.5788-98.50.936.397.190.121.070.13156.050.1460.1940.900.6729-109.50.926.9106.850.131.070.14175.520.1470.1960.940.70310-1110.50.907.6116.510.141.070.15194.990.1470.1961.010.760

*Solution

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