18-Env-A3 Geotechnical and Hydrogeological Engineering · Undated paper
Question 1 of 6: Phase Relationships in Natural Conditions
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — 18-Env-A3, Geotechnical and Hydrogeological Engineering — 3 hours, open book. Six questions, each 20 marks, equal value; all six are solved below.
Given. Moist (natural) unit weight, saturated unit weight and specific gravity of solids below.
Given data
Quantity
Value
Natural (moist) unit weight, $\gamma$
19.1 kN/m³
Saturated unit weight, $\gamma_{sat}$
20.1 kN/m³
Specific gravity of solids, $G_s$
2.5
Find. (a) void ratio, (b) porosity, (c) degree of saturation, (d) moisture content, (e) buoyant unit weight — all in the natural state.
Approach. The void ratio doesn't change between the natural and saturated states of the same soil skeleton, so back it out of $\gamma_{sat}=(G_s+e)\gamma_w/(1+e)$ first, then use it with $\gamma$ (natural) to close the rest of the phase-relationship system.
(c) Degree of saturation. The natural unit weight of a partly-saturated soil is $\gamma=(G_s+Se)\gamma_w/(1+e)$; solving for $S$: $$S = \dfrac{1}{e}\left(\dfrac{\gamma(1+e)}{\gamma_w} - G_s\right) = \dfrac{1}{0.430}\left(\dfrac{19.1\times1.430}{9.81}-2.5\right) = \boxed{66.1\%}$$
(d) Moisture content. From the identity $Se=wG_s$: $$w = \dfrac{Se}{G_s} = \dfrac{0.661\times0.430}{2.5} = \boxed{11.4\%}$$ Check: $\gamma_d=G_s\gamma_w/(1+e)=17.15\text{ kN/m}^3$, and $\gamma_d(1+w)=17.15\times1.114=19.1\text{ kN/m}^3$ — matches the given natural unit weight.
