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Core cutter field density numerical

Short answer

Dry density = 22.51 kN/m³ (2294 kg/m³), void ratio = 0.186, degree of saturation = 84.0%.

A field density test was conducted by core cutter method and the following data was obtained. Weight of empty core-cutter = 22.80 N Weight of soil and core-cutter = 50.05 N Inside diameter of the core-cutter = 90.0 mm Height of core-cutter = 180.0 mm Weight of wet sample for moisture determination = 0.5405 N Weight of oven-dry sample = 0.5112 N Specific gravity of soil grains = 2.72 Determine (i) dry density, (ii) void ratio, and (iii) degree of saturation.

Given

Empty cutter 22.80 N; soil + cutter 50.05 N; D=90D = 90 mm; H=180H = 180 mm; G=2.72G = 2.72; moisture sample 0.5405 N wet and 0.5112 N dry. γw=9.81\gamma_w = 9.81 kN/m³.

Volume and bulk unit weight

Wsoil=50.05−22.80=27.25 NV=π4(0.09)2(0.18)=1.1451×10−3 m3γ=27.251.1451×10−3=23 797 N/m3=23.80 kN/m3\begin{aligned} W_{soil} &= 50.05 - 22.80 = 27.25\ \text{N} \\ V &= \frac{\pi}{4}(0.09)^2(0.18) = 1.1451 \times 10^{-3}\ \text{m}^3 \\ \gamma &= \frac{27.25}{1.1451 \times 10^{-3}} = 23\,797\ \text{N/m}^3 = 23.80\ \text{kN/m}^3 \end{aligned}

Water content

w=0.5405−0.51120.5112=0.0573=5.73%w = \frac{0.5405 - 0.5112}{0.5112} = 0.0573 = 5.73\%

(i) Dry density

γd=γ1+w=23.801.0573=22.51 kN/m3\gamma_d = \frac{\gamma}{1+w} = \frac{23.80}{1.0573} = 22.51\ \text{kN/m}^3 ρd=22.519.81=2.294 Mg/m3=2294 kg/m3\rho_d = \frac{22.51}{9.81} = 2.294\ \text{Mg/m}^3 = 2294\ \text{kg/m}^3

(ii) Void ratio

e=Gγwγd−1=2.72×9.8122.51−1=0.186e = \frac{G\gamma_w}{\gamma_d} - 1 = \frac{2.72 \times 9.81}{22.51} - 1 = 0.186

(iii) Degree of saturation

S=wGe=0.0573×2.720.1856=0.840=84.0%S = \frac{wG}{e} = \frac{0.0573 \times 2.72}{0.1856} = 0.840 = 84.0\%

Answer: γd=22.51\gamma_d = 22.51 kN/m³ (ρd=2294\rho_d = 2294 kg/m³), e=0.186e = 0.186, S=84.0%S = 84.0\%.

✓ Answer checked by Claude Opus 5.5 (high effort)

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