Chapter 17, Problem 17.9P

Principles of Geotechnical Enginee...

9th Edition
Braja M. Das + 1 other
ISBN: 9781305970939

Chapter
Section

Principles of Geotechnical Enginee...

9th Edition
Braja M. Das + 1 other
ISBN: 9781305970939
Textbook Problem

Refer to Figure 17.16. Estimate the variation of cone penetration resistance, qc, with depth, using Eq. (17.39) and values of c and a given by Kulhawy and Mayne (Table 17.7). Assume D50 = 0.46 mm.17.7 Refer to the boring log shown in Figure 17.16. Estimate the average drained friction angle, ϕ ′ , using the Kulhawy and Mayne correlation [Eq. (17.24)]. Assume pa ≈ 100 kN/m2.Figure 17.16

To determine

Find the variation of cone penetration resistance with various depth.

Explanation

Given information:

The atmospheric pressure is (pa) 100kN/m2.

Assume the grain size (D50) 0.46mm.

Show the depth and number of blows shown in table:

 Depth,(m) N60 1 6 2.5 8 4 11 5.5 13 7 14 8.5 16

Calculation:

Show the expression of correlation between qcandN60 using the Equation:

(qcpa)N60=cD50a (1)

Here, pa is atmospheric pressure, N60 is standard penetration resistance, c and a as developed from various studies.

For c and a value given by Kulhawy and Mayne(1990).

Refer Table 17.7 in section 17.11 “Cone penetration test” in the textbook.

 c a 5.44 0.26

Find the variation of cone penetration resistance with various depth 1m

Substitute 100kN/m2 for pa, 0.46mm for D50, 5.44 for c, 0.26 for a, and 6 for N60 in Equation (1).

(qc100)6=5.44(0.46)0.26(qc100)6=4.4454(qc100)=26.67qc=2,667kN/m2

Thus, the variation of cone penetration resistance at the depth 1m is 2,667kN/m2_.

Find the variation of cone penetration resistance with various depth 2.5m.

Substitute 100kN/m2 for pa, 0.46mm for D50, 5.44 for c, 0.26 for a, and 8 for N60 in Equation (1).

(qc100)6=5.44(0.46)0.26(qc100)8=4.4454(qc100)=35.56qc=3,556kN/m2

Thus, the variation of cone penetration resistance at the depth 2.5m is 3,556kN/m2_.

Find the variation of cone penetration resistance with various depth 4m.

Substitute 100kN/m2 for pa, 0.46mm for D50, 5.44 for c, 0.26 for a, and 11 for N60 in Equation (1).

(qc100)6=5.44(0.46)0.26(qc100)11=4

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