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Question 6.10: Use the data from the example problem in Sec. 6.4.5, but ass...

Use the data from the example problem in Sec. 6.4.5, but assume that the soil type is eolian sand (i.e., soil type 4, see Fig. 6.12). Calculate the factor of safety against liquefaction.

6.12
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CSR =0.65 r_{d} \frac{\sigma_{v 0}}{\sigma_{v 0}^{\prime}} \frac{a_{ max }}{g}                Eq. (6.6)

= 0.65 (0.96) (1.35) (0.40) = 0.34

The eolian sand has about 6 percent fines (i.e., 6 percent passing No. 200 sieve). This is very close to the clean sand condition. Per Fig. 6.6, for \left(N_{1}\right)_{60}=7.7 and intersecting the clean sand curve, CRR = 0.09.

FS =\frac{ CRR }{ CSR }=\frac{0.09}{0.34}=0.26

(Note that since the soil is essentially a clean sand, the answer is identical to the example problem in Sec. 6.4.5.)

See Table 6.3.

TABLE 6.3 Summary of Answers for Probs. 6.3 to 6.11
Problem no. Soil type a_{\max } / g Earthquake magnitude \left(N_{1}\right)_{60} blows/ft; q_{c 1}, MPa ; V_{s 1}, m / s Cyclic stress ratio Cyclic resistance ratio FS = CRR/CSR
Section 6.4.5 Clean sand 0.40 7½ 7.7 blows/ft 0.34 0.09 0.26
Problem 6.3 Sand—15% fines 0.10 7½ 7.7 blows/ft 0.084 0.14 1.67
Problem 6.4 Clean sand 0.20 5¼ 7.7 blows/ft 0.17 0.14 0.82
Problem 6.5 Clean sand 0.40 7½ 5.8 MPa 0.34 0.09 0.26
Problem 6.6 Clean sand 0.40 7½ 185 m/s 0.34 0.16 0.47
Problem 6.7 Crushed limestone 0.40 7½ 5.0 MPa 0.34 0.18 0.53
Problem 6.8 Silty gravel 0.40 7½ 7.5 MPa 0.34 0.27 0.79
Problem 6.9 Clean gravelly sand 0.40 7½ 14 MPa 0.34 0.44 1.29
Problem 6.10 Eolian sand 0.40 7½ 7.7 blows/ft 0.34 0.09 0.26
Problem 6.11 Loess 0.40 7½ 7.7 blows/ft 0.34 0.18 0.53
Note: See App. E for solutions.
6.6

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