2080 Bhadra · 4 marks

SPT corrected N-value, submerged sand

Short answer

Observed N = 23+17 = 40 (first 150 mm ignored). Effective overburden stress (19−9.81)×6 = 55.14 kN/m². CN=0.77log10(2000/55.14)=1.201, N'=48.0. Dilatancy correction N''=15+0.5(48.0−15)=31.5, so corrected N-value ≈ 31.5.

Calculate the corrected N-value, if a standard penetration test was conducted 6 m below the ground level in a deposit of fully submerged fine sand. Numbers of blows were found to be 18, 23 and 17 for each 150 mm penetration. The average saturated unit weight of the soil is $19\ \text{kN/m}^3$ .

Method: overburden correction $N' = C_N N$ , $C_N = 0.77\log_{10}(2000/\sigma_v')$ , then dilatancy correction (saturated fine sand) $N'' = 15+\tfrac12(N'-15)$ for $N'>15$ .

Observed N: ignore the first 150 mm.

N = 23 + 17 = 40

Effective overburden (fully submerged, $\gamma_w = 9.81\ \text{kN/m}^3$ ):

\sigma_v' = (19 - 9.81)\times 6 = 55.14\ \text{kN/m}^2

Overburden correction:

\begin{aligned} C_N &= 0.77\log_{10}\frac{2000}{55.14} = 1.201 \\ N' &= 1.201\times 40 = 48.03 \end{aligned}

Dilatancy correction ( $N'>15$ , fine saturated sand):

N'' = 15 + \tfrac12(48.03-15) = 31.52

Answer: corrected N-value $\approx 31.5$ (about 32).

✓ Answer checked by Claude Sonnet 5.5

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