Question 6.8: 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 silty gravel (i.e., soil type 2, see Fig. 6.12) and at a depth of 3 m, q_{c 1}=7.5 MPa. Calculate the factor of safety against liquefaction.

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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
Silty gravel has more than 50 percent gravel-size particles with about 10 percent fines. Given plasticity characteristics (liquid limit = 23 and plasticity index = 3), use silty gravel curve in Fig. 6.9. For q_{c l}=7.5 MPa and intersecting silty gravel curve, CRR = 0.27.
FS =\frac{ CRR }{ CSR }=\frac{0.27}{0.34}=0.79See 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. |

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