---
title: "SPT corrected N-value, submerged sand"
subject: "Foundation Engineering"
chapter: 2
key: "semester_5/foundation_engineering:spt-corrected-6m-submerged"
source: https://www.bctengineer.com/q/semester_5/foundation_engineering/spt-corrected-6m-submerged
site: BCT Engineering
---

# SPT corrected N-value, submerged sand

Foundation Engineering · Chapter 2: Soil Exploration

- Asked in: 2080 Bhadra, 4 marks
- Video: [Correcting SPT N values to Hammer Energy and Overburden Pressure | Formulas and Procedure](https://www.youtube.com/watch?v=NwQ24-zlxyU) — Soil Mechanics and Engineering Geology
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/foundation_engineering/spt-corrected-6m-submerged
- Whole chapter: https://www.bctengineer.com/md/semester_5/foundation_engineering/chapter-2

## Question

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 $19\ \text{kN/m}^3$.

## 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.

## Answer

**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).**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. From blows 18, 23, 17, observed N is
   A. 58
   B. 41
   C. 40
   D. 35
   Answer: C. 23 + 17 = 40; first 150 mm ignored. (Key point: Observed N)

2. For fully submerged soil, the effective stress at 6 m is
   A. 19 x 6 = 114 kN/m^2
   B. 9.81 x 6 = 58.9 kN/m^2
   C. (19 - 9.81) x 6 = 55.14 kN/m^2
   D. (19 + 9.81) x 6
   Answer: C. Subtract the unit weight of water for submerged soil. (Key point: Effective stress)

3. N' = CN x N with CN = 1.201 gives about
   A. 40
   B. 48
   C. 31
   D. 15
   Answer: B. 1.201 x 40 = 48.03. (Key point: Overburden corrected)

4. The final corrected value after dilatancy is about
   A. 31.5
   B. 48
   C. 40
   D. 15
   Answer: A. 15 + 0.5(48.03 - 15) = 31.5. (Key point: Final)

## Related in Foundation Engineering

- [SPT corrected N-value, dense sand](https://www.bctengineer.com/question.md?key=semester_5%2Ffoundation_engineering%3Aspt-dilatancy-20m) — Foundation Engineering, chapter 2
- [Standard penetration test and corrections](https://www.bctengineer.com/question.md?key=semester_5%2Ffoundation_engineering%3Aspt-procedure-corrections) — Foundation Engineering, chapter 2, asked 3×

## Related in other subjects

- [Two-layer sand, plot stresses (bulk 18)](https://www.bctengineer.com/question.md?key=semester_4%2Fsoil_mechanics%3Atwo-layer-sand-stress-plot) — Soil Mechanics, chapter 6
- [Flow through partially and fully submerged orifice](https://www.bctengineer.com/question.md?key=semester_3%2Ffluid_mechanics%3Asubmerged-orifice-flow-expression) — Fluid Mechanics, chapter 6