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22-Agric-A2 Soil Physics and Mechanics · May 2014

Question 3 of 7: Weight-Volume Relationships for a Sandy Loam Sample

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. 04-Agric-A2 Soil Physics & Mechanics, National Exams May 2014 — a three-hour open-book examination; any non-communicating calculator is permitted. The cover page states that five (5) questions constitute a complete exam paper and that only the first five as they appear in the answer book are marked, that each question is of equal value, and that some questions require a written answer whose clarity and organization matter for marks. All seven printed questions are worked here, because the set is a study resource rather than a timed attempt; on exam day a candidate submits only the first five, in order.

Reference texts. B.M. Das, Principles of Geotechnical Engineering, 9th ed. (weight-volume relationships, permeability, effective stress, shear strength, particle-size classification, flow to wells); R.F. Craig, Craig's Soil Mechanics, 9th ed. (effective stress, seepage, shear strength).

Question 3: Weight-Volume Relationships for a Sandy Loam Sample (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

CharacteristicValue
Sample volume, V1200 cm³
Mass, smaller of the two printed values2.005 kg
Mass, larger of the two printed values2.915 kg
Particle (grain) density, Gs2.63 g/cm³

Find. Void ratio and porosity (a); wet and dry density (b); degree of saturation (c); and whether the resulting numbers are consistent for a sandy loam (d).

Check

The source table labels the smaller mass (2.005 kg) "Original Mass" and the larger (2.915 kg) "Mass after drying" — but oven-drying only removes water, so the post-drying mass can never exceed the original mass. The only physically consistent reading is that the two labels are reversed: Ms (dry) = 2.005 kg and Mt (original, wet) = 2.915 kg. This solution proceeds on that basis; part (d) shows that even the corrected pairing is not fully self-consistent, which is the intended teaching point of that sub-part.

Approach. Get the solids volume from the dry mass and Gs, then the void volume by difference from the total sample volume; void ratio and porosity follow directly, wet/dry density from the two masses over the same total volume, and degree of saturation from the water volume implied by the mass difference.

  1. a) Void ratio and porosity. With $M_s = 2005\ \text{g}$, $G_s = 2.63$, and $\rho_w = 1\ \text{g/cm}^3$, $$V_s = \frac{M_s}{G_s} = \frac{2005}{2.63} = 762.4\ \text{cm}^3, \qquad V_v = V - V_s = 1200 - 762.4 = 437.6\ \text{cm}^3$$ $$e = \frac{V_v}{V_s} = \frac{437.6}{762.4} = \boxed{0.574}, \qquad n = \frac{V_v}{V} = \frac{437.6}{1200} = \boxed{0.365\ (36.5\%)}$$
  2. b) Wet and dry densities. Both use the same total volume V = 1200 cm³: $$\rho_{\text{wet}} = \frac{M_t}{V} = \frac{2915}{1200} = \boxed{2.429\ \text{g/cm}^3\ (2429\ \text{kg/m}^3)}, \qquad \rho_{\text{dry}} = \frac{M_s}{V} = \frac{2005}{1200} = \boxed{1.671\ \text{g/cm}^3\ (1671\ \text{kg/m}^3)}$$
  3. c) Degree of saturation. The mass difference is water, and with $\rho_w = 1\ \text{g/cm}^3$ that mass converts directly to a volume: $$M_w = M_t - M_s = 2915 - 2005 = 910\ \text{g} \ \Rightarrow\ V_w = 910\ \text{cm}^3$$ $$S = \frac{V_w}{V_v}\times100\% = \frac{910}{437.6}\times100\% = \boxed{208\%}$$

d) Are these values appropriate for a sandy loam? Partly. The void ratio (0.574) and porosity (36.5%), and the wet/dry densities, all fall squarely within the normal range for a sandy loam (e typically 0.4–0.7, n roughly 30–45%), so the corrected mass assignment in the Verify callout above is at least internally plausible for this soil type. The degree of saturation, however, computes to about 208% — physically impossible, since a sample cannot hold more water than the volume of its own voids ($S \le 100\%$ always). This signals that the data set still contains an error beyond the simple label swap already corrected: most likely a further error in one of the two masses, the particle density, or the sample volume. Practically, a candidate would flag the void ratio, porosity and densities as usable design values (they are self-consistent and typical) while noting that the saturation figure cannot be trusted and the sample should be re-weighed or re-measured before any water-content-dependent design decision is made.

QuantityValue
Void ratio, e0.574
Porosity, n36.5%
Wet (bulk) density2.429 g/cm³ (2429 kg/m³)
Dry density1.671 g/cm³ (1671 kg/m³)
Degree of saturation, S≈208% — physically impossible; data flagged