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Question 6.9: 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 gravelly sand (i.e., soil type 3, see Fig. 6.12) and at a depth of 3 m, q_{c 1}=14 MPa. 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 gravelly sand does not have any fines (hence clean soil). The soil type is approximately midway between clean sand and clean gravel. In Fig. 6.9, using q_{c l}=14 MPa and at a midpoint between the clean sand and clean gravel curves, CRR = 0.44.

FS =\frac{ CRR }{ CSR }=\frac{0.44}{0.34}=1.29

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.
6.9

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