Question 6.15: Figure 6.15 shows the subsoil profile at a sewage disposal s...

Figure 6.15 shows the subsoil profile at a sewage disposal site in Niigata. Assume a level-ground site with the groundwater table at a depth of 0.4 m below ground surface, the medium to coarse sand has less than 5 percent fines, the total unit weight \gamma_{t} of the soil above the groundwater table is 18.3 kN/m³, and the buoyant unit weight \gamma_{b} of the soil below the groundwater table is 9.7 kN/m³.

The standard penetration data shown in Fig. 6.15 are uncorrected N values. Assume a hammer efficiency E_{m} of 0.6 and a boring diameter of 100 mm, and the length of drill rods is equal to the depth of the SPT test below ground surface. The earthquake conditions are a peak ground acceleration a_{\max } of 0.16g and a magnitude of 7.5. Using the standard penetration test data, determine the factor of safety against liquefaction versus depth.

6.15
The blue check mark means that this solution has been answered and checked by an expert. This guarantees that the final answer is accurate.
Learn more on how we answer questions.

See App. E for the solution and Fig. 6.16 for a plot of the factor of safety against liquefaction versus depth.

See page E.8.

Depth, m Cyclic stress ratio N value corrections CRR FS = CRR/CSR
\sigma_{v}, kPa \sigma_{v}^{\prime}, kPa \sigma_{v} / \sigma_{v}^{\prime} r_{d} CSR N value C_{r} N_{60} C_{N} \left(N_{1}\right)_{60}
1.2 22.9 15.1 1.52 0.99 0.16 4 0.75 3.0 2.57* 7.7 0.09 0.56
2.2 42.4 24.8 1.71 0.97 0.17 6 0.75 4.5 2.01* 9.0 0.1 0.59
3.2 61.9 34.5 1.79 0.96 0.18 5 0.75 3.8 1.70 6.5 0.07 0.39
4.2 81.5 44.2 1.84 0.95 0.18 8 0.85 6.8 1.50 10 0.11 0.61
5.2 101 53.9 1.87 0.94 0.18 7 0.85 6.0 1.36 8.2 0.09 0.50
6.2 120 63.6 1.89 0.93 0.18 13 0.95 12 1.25 15 0.16 0.89
7.2 140 73.3 1.91 0.91 0.18 36 0.95 34 1.17 40 >0.5 >2.8
8.2 159 83.0 1.92          0.90 0.18 24 0.95 23 1.09 25 0.29 1.61
9.2 179 92.7 1.93 0.89 0.18 35 0.95 33 1.04 34 >0.5 >2.8
10.2 199 102 1.95 0.88 0.18 30 1.00 30 0.99 30 0.50 2.78
11.2 218 112 1.95 0.87 0.18 28 1.00 28 0.94 26 0.30 1.67
12.2 238 122 1.95 0.85 0.17 32 1.00 32 0.91 29 0.45 2.65
13.2 257 131 1.96 0.84 0.17 16 1.00 16 0.87 14 0.16 0.94
14.2 277 141 1.96 0.83 0.17 28 1.00 28 0.84 24 0.28 1.65
15.2 296 151 1.96 0.82 0.17 27 1.00 27 0.81 22 0.25 1.47
16.2 316 161 1.96 0.81 0.17 23 1.00 23 0.79 18 0.20 1.18
17.2 335 170 1.97 0.79 0.16 39 1.00 38 0.77 29 0.45 2.81
18.2 355 180 1.97 0.78 0.16 32 1.00 32 0.75 24 0.28 1.75
19.2 374 190 1.97 0.77 0.16 47 1.00 47 0.73 34 >0.5 >3.1
Notes: Cyclic stress ratio:  a_{\max }=0.16 g, r_{d} from Eq. (6.7). N value corrections: E_{m}=0.6, C_{b}=1.0, C_{N} from Eq. (5.2). CRR from Fig. 6.6. *Suggested maximum values of C_{N} range from 1.7 to 2.0 (Youd and Idriss 1997, 2001).
r_{d}=1-0.012 z         (6.7)
\left(N_{1}\right)_{60}=C_{N} N_{60}=\left(100 / \sigma_{v 0}^{\prime}\right)^{0.5} N_{60}  (5.2)
6.16
6.6

Related Answered Questions

Question: 6.17

Verified Answer:

Per Fig. 6.16, the standard penetration test data ...
Question: 6.14

Verified Answer:

Per Fig. 6.14, the standard penetration test data ...