Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — May 2016 — 04-Geol-A6 Soil Mechanics. Three-hour, closed-book exam; a Casio or Sharp approved calculator, compass and ruler are permitted. Six questions constitute the complete 100-mark paper: Questions 1–5 are compulsory; Question 6 offers five 5-mark optional items of which the source asks for three.
Find. (a) $n$, (b) $\gamma_{moist}$, (c) $\gamma_{dry}$, (d) $S$, (e) mass of water to add to $10\text{ m}^3$ for full saturation.
Approach. Standard three-phase relations, assuming the void ratio stays constant as saturation increases (rigid skeleton) so only the air voids fill with water.
Check: full saturation is assumed to occur at constant $e$ (no volume change) — the added water simply displaces the remaining air in the voids.
Moist unit weight. $\gamma=\dfrac{G_s\gamma_w(1+w)}{1+e}=\dfrac{2.70\times9.81\times1.15}{1.70}=\boxed{17.9\text{ kN/m}^3}$.
Dry unit weight. $\gamma_d=\dfrac{\gamma}{1+w}=\dfrac{17.9}{1.15}=\boxed{15.6\text{ kN/m}^3}$.
Degree of saturation. $S=\dfrac{wG_s}{e}=\dfrac{0.15\times2.70}{0.70}=\boxed{57.9\%}$.
Water to add for full saturation. The air-void fraction of total volume is $n(1-S)=0.412\times(1-0.579)=0.1735$. For $V=10\text{ m}^3$: $\Delta V_w=0.1735\times10=1.735\text{ m}^3$, so $\boxed{\Delta m_w\approx1735\text{ kg}}$.
Quantity
Value
Porosity, $n$
41.2%
Moist unit weight, $\gamma$
17.9 kN/m³
Dry unit weight, $\gamma_d$
15.6 kN/m³
Degree of saturation, $S$
57.9%
Water to add (10 m³)
1735 kg (≈1.74 m³)
(2.2) Embankment borrow-pit volumes (part of 15 marks)
Find. (a) Borrow volume per m³ of embankment. (b) Water to add per m³ of embankment.
Approach. The mass of dry solids is conserved during hauling and compaction, so equate the dry weight delivered from the borrow pit to the dry weight required in place; any shortfall in water content is made up by adding water.
Borrow dry unit weight. $\gamma_{d,borrow}=\dfrac{\gamma_{borrow}}{1+w_{borrow}}=\dfrac{14}{1.045}=\boxed{13.40\text{ kN/m}^3}$.
Borrow volume for 1 m³ embankment. Dry weight needed $=\gamma_{d,target}\times1=16.5\text{ kN}$ (solids only, per m³ of finished embankment). $V_{borrow}=\dfrac{16.5}{\gamma_{d,borrow}}=\dfrac{16.5}{13.40}=\boxed{1.232\text{ m}^3}$.
Water carried from the borrow pit. $W_{w,initial}=(\gamma_{borrow}-\gamma_{d,borrow})\times V_{borrow}=(14-13.40)\times1.232=0.743\text{ kN}$.
Water required in place. $W_{w,target}=\gamma_{d,target}\times w_{target}=16.5\times0.19=3.135\text{ kN}$.
Water to add. $\Delta W_w=W_{w,target}-W_{w,initial}=3.135-0.743=\boxed{2.39\text{ kN/m}^3\text{ embankment}}$, i.e. $\approx\boxed{244\text{ kg (244 L)}}$ of water per m³.
