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SPT corrected N-value, dense sand

Short answer

Observed N = 21+22 = 43 after seating drive. Overburden stress 340 kN/m², CN=0.77log10(2000/340)=0.593 gives N'=25.5. Dilatancy correction N''=15+0.5(N'−15) = 20.2, so corrected N-value ≈ 20.

A SPT was performed at a depth of 20 m in a dense sand deposit with a unit weight of 17 kN/m317\ \text{kN/m}^3. If the number of blows were found to be 18, 21 and 22 for each 150 mm penetration, what is the N-value corrected for overburden and dilatancy?

Method: Terzaghi-Peck corrections. First the overburden correction N′=CNNN' = C_N N with CN=0.77log⁡10(2000/σv′)C_N = 0.77\log_{10}(2000/\sigma_v'), then the dilatancy correction N′′=15+12(N′−15)N'' = 15 + \tfrac12(N'-15) for N′>15N' > 15.

Observed N: the first 150 mm is the seating drive, so

N=21+22=43N = 21 + 22 = 43

Overburden stress: no water table is given, so

σv′=γz=17×20=340 kN/m2\sigma_v' = \gamma z = 17 \times 20 = 340\ \text{kN/m}^2

Overburden correction:

CN=0.77log⁡102000340=0.593N′=0.593×43=25.48\begin{aligned} C_N &= 0.77\log_{10}\frac{2000}{340} = 0.593 \\ N' &= 0.593\times 43 = 25.48 \end{aligned}

Dilatancy correction (N′>15N' > 15):

N′′=15+12(25.48−15)=20.24N'' = 15 + \tfrac12(25.48-15) = 20.24

Answer: corrected N-value ≈20.2\approx 20.2, say 20. (If the dilatancy correction is applied first, 15+0.5(43−15)=2915+0.5(43-15)=29 and 29×0.593=17.229\times0.593=17.2; the order recommended by Terzaghi-Peck, overburden then dilatancy, is used here.)

✓ Answer checked by Claude Sonnet 5.5

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