The observed standard penetration test value in a deposit of fully submerged sand was 45 at a depth of 6.5 m. The average effective unit weight of the soil is 9.69 kN/m^3. The other data given are (a) hammer efficiency = 0.8, (b) drill rod length correction factor = 0.9, and (c) borehole correction

Question

The observed standard penetration test value in a deposit of fully submerged sand was 45 at a depth of 6.5 m. The average effective unit weight of the soil is 9.69 [latex]kN / m ^{3}[/latex]. The other data given are (a) hammer efficiency = 0.8, (b) drill rod length correction factor = 0.9, and (c) borehole correction factor = 1.05. Determine the corrected SPT value for standard energy (a) [latex]R_{e s}=60[/latex] percent, and (b) [latex]R_{e s}=70 \%[/latex].

Accepted Answer

Per Eq (9.6), the equation for [latex]N _{60}[/latex] may be written as

[latex]N_{c o r}=C_{N} N E_{h} C_{d} C_{s} C_{b}[/latex] (9.6)

(i) [latex]N_{60} \quad=C_{N} N E_{h} C_{d} C_{s} C_{b}[/latex]

where N = observed SPT value

[latex]C_{N}[/latex] = overburden correction

Per Eq (9.5) we have

[latex]C_{N}=\left[\frac{95.76}{ ho_{o}^{\prime}} ight]^{1 / 2}[/latex] (9.5)

[latex]C_{N}=\frac{95.76}{p_{o}^{\prime}}[/latex]

where [latex]p_{0}^{\prime}[/latex] = effective overburden pressure

[latex]=6.5 imes 9.69=63 kN / m ^{2}[/latex]

Substituting for [latex]p_{0}^{\prime}[/latex],

[latex]C_{N}=\frac{95.76}{60}^{1 / 2}=1.233[/latex]

Substituting the known values, the corrected [latex]N _{60}[/latex] is

[latex]N _{60}=1.233 imes 45 imes 0.8 imes 0.9 imes 1.05=42[/latex]

For 70 percent standard energy

[latex]N_{70}=42 imes \frac{0.6}{0.7}=36[/latex]

Question 9.3: The observed standard penetration test value in a deposit of...

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