Solids- Water –Air Relations and Index properties of soils

Soil changes its state with water content: solid, semi-solid, plastic and liquid, separated by the shrinkage limit ($w_s$), plastic limit ($w_p$) and liquid limit ($w_l$). The stress-strain behaviour differs in each state.

Stress-strain curves

 Stress
  |      ___ Solid (brittle, peak
  |    /\     then sudden drop)
  |   /  \______ Semi-solid
  |  /    ______ Plastic (yields,
  | /  __/       strain hardens
  |/ _/ ......... Liquid (no
  |/.-'           stress, flows)
  +-------------------- Strain

As water content rises, the peak strength drops and the strain at failure increases. The brittle curve has a sharp peak with a post-peak drop, the plastic curve rises gradually and levels off, and the liquid curve lies on the strain axis.

Index properties

Index properties are simple properties of soil that help to identify and classify it and indicate its probable engineering behaviour. They do not give design values directly.

• Soil grain properties: specific gravity, particle size and shape, grain size distribution.
• Soil mass properties: water content, unit weight, void ratio, porosity, relative density, consistency limits, consistency index, sensitivity, activity.

Engineering properties

These are the properties that control the behaviour of soil in design:

• Permeability, compressibility (consolidation), shear strength, compaction characteristics (MDD, OMC), bearing capacity, swelling.

Difference

Why index properties are determined

• To identify and classify soil quickly (USCS, IS, AASHTO).
• To judge the soil's probable behaviour (for example high $w_L$ means high compressibility).
• To compare soils from different sites and check uniformity along a site.
• To use empirical correlations for preliminary design (e.g. $C_c = 0.009(w_L - 10)$).
• To check quality control on fill.

For coarse-grained soils, grain size distribution is the most significant index property; for fine-grained soils, plasticity (consistency limits) is the most significant.

Tests for individual soil grains

• Specific gravity test (pycnometer or density bottle).
• Particle size analysis (sieve analysis for coarse grains, hydrometer or pipette analysis for fine grains).
• Particle shape and mineralogical tests (visual, microscope).

Tests for the soil mass as a whole

• Water content (oven drying, calcium carbide, sand bath, pycnometer).
• Unit weight / density (core cutter, sand replacement, water displacement method).
• Void ratio, porosity and degree of saturation (calculated from the above).
• Relative density test for sands (maximum and minimum density).
• Consistency limits (Atterberg limits): liquid limit (Casagrande/cone penetrometer), plastic limit, shrinkage limit.
• Sensitivity (unconfined compression) and field tests such as dilatancy and toughness.

A phase diagram shows soil as separate blocks of solids, water and air, with volumes on the left and weights on the right. Here $V_s, V_w, V_a$ are volumes of solids, water and air, $V_v = V_w + V_a$ is the void volume, and $W_s, W_w$ are the weights (air weight is taken as zero).

Partially saturated

  Volume              Weight
 +--------+  ---   +----------+
 |  Air   | Va     |    0     |
 |--------|  | Vv  |----------|
 | Water  | Vw     |   Ww     |
 |--------|  ---   |----------| W
 | Solids | Vs     |   Ws     |
 +--------+        +----------+
   V=Vs+Vw+Va         W=Ws+Ww

Fully saturated ($S = 100\%$, $V_a = 0$)

 +--------+  ---   +----------+
 | Water  | Vw=Vv  |   Ww     |
 |--------|  ---   |----------|
 | Solids | Vs     |   Ws     |
 +--------+        +----------+

Dry ($S = 0$, $V_w = 0$, $W_w = 0$)

 +--------+  ---   +----------+
 |  Air   | Va=Vv  |    0     |
 |--------|  ---   |----------|
 | Solids | Vs     | Ws = W   |
 +--------+        +----------+

The in-situ density is found by taking out a known volume of soil from the ground and weighing it; the water content gives the dry density, $\gamma_d = \gamma/(1+w)$. The method depends on the soil type.

1. Core cutter method (IS 2720 Part 29)

Used for soft to medium cohesive soils that stand without crumbling. A steel cylinder (usually 100 mm dia, 130 mm high) is driven into the ground with a dolly and rammer, dug out with the soil inside, trimmed and weighed. Volume = volume of the cutter, so $\gamma = W/V$.

2. Sand replacement method (IS 2720 Part 28)

Used for gravelly and sandy soils, hard clay, or where the core cutter cannot be used. A hole is dug, the soil is weighed, and the hole volume is found by filling it with calibrated dry sand (Ottawa/ standard sand) from a pouring cylinder.

$$V = \frac{W_{sand\ in\ hole}}{\gamma_{sand}}$$

3. Water displacement method

Used for soils with irregular holes. The hole is lined with a thin rubber balloon or plastic sheet and filled with a measured volume of water to find the volume.

4. Rubber balloon method

A balloon density apparatus measures the volume of the hole directly from the water pressed into the balloon.

5. Nuclear density gauge

Gives density and water content quickly by gamma-ray backscatter; used for rapid quality control of compaction.

6. Wax-coated lump (paraffin) method

For cohesive, hard lumps of irregular shape; the lump is coated with wax and its volume found by water displacement.

Selection: core cutter for soft fine-grained soil, sand replacement for coarse soils, and nuclear gauge for large-scale quick control.

Difference

Index tests

• Coarse-grained soils: sieve analysis (grain size distribution, $C_u$, $C_c$), specific gravity, relative density (maximum and minimum density), in-situ density, water content, particle shape.
• Fine-grained soils: hydrometer/pipette analysis, Atterberg limits (liquid limit, plastic limit, shrinkage limit), plasticity index, consistency index, activity, sensitivity, and field tests (dilatancy, dry strength, toughness).

Phase diagram

  Volume               Weight
 +---------+ ---    +-----------+
 |   Air   | Va     |     0     |
 |---------|  | Vv  |-----------|
 |  Water  | Vw     |    Ww     |
 |---------| ---    |-----------| W
 |  Solids | Vs     |    Ws     |
 +---------+        +-----------+
  V = Vs+Vw+Va        W = Ws+Ww

Here $V_v = V_w + V_a$ is the volume of voids; $W_w$ and $W_s$ are weights of water and solids.

Void ratio ($e$)

The ratio of the volume of voids to the volume of solids:

$$e = \frac{V_v}{V_s}$$

It is a ratio (no unit) and can be more than 1 (for example in soft clays).

Degree of saturation ($S$)

The ratio of the volume of water to the volume of voids, in percent:

$$S = \frac{V_w}{V_v} \times 100$$

$S = 0$ for dry soil and $S = 100\%$ for fully saturated soil.

Compaction reduces air voids by mechanical energy (no water leaves); consolidation reduces voids by squeezing water out of saturated soil under a sustained load. Solids volume $V_s$ is constant in both.

Dry soil before and after compaction

 Before            After (V reduced)
 +------+          +------+
 | Air  |          | Air  |  (less)
 |      |          +------+
 |      |          |Solids|  Vs same
 +------+          +------+
 |Solids|
 +------+

Air volume decreases, so the void ratio and total volume decrease; $W_s$ is unchanged.

Saturated soil before and after consolidation

 Before            After (V reduced)
 +------+          +------+
 |Water |          |Water |  (less)
 |      |          +------+
 |      |          |Solids|  Vs same
 +------+          +------+
 |Solids|
 +------+

Water is expelled from the voids, so $V_w$ and $V$ decrease, $S$ remains 100% and $W_w$ decreases.

In both cases the void ratio falls and dry unit weight rises.

Symbols: $W_s$ = weight of solids, $W_w$ = weight of water, $W = W_s + W_w$, $V_s$ = volume of solids, $V_w$ = volume of water, $V_a$ = volume of air, $V_v = V_w + V_a$, $V = V_s + V_w + V_a$.

Toughness index ($I_t$)

The ratio of plasticity index to flow index:

$$I_t = \frac{I_p}{I_f}$$

It measures the shear strength of soil at the plastic limit. Typical values are 0 to 3; low values mean the soil is friable and has low toughness.

Coefficient of curvature ($C_c$)

A shape parameter of the grain size curve:

$$C_c = \frac{D_{30}^2}{D_{10} \times D_{60}}$$

A well-graded gravel or sand has $C_c$ between 1 and 3.

Activity of soil ($A$)

The ratio of the plasticity index to the percent of clay-size particles (< 2 micron):

$$A = \frac{I_p}{\%\ \text{clay fraction}}$$

$A < 0.75$ inactive (kaolinite), 0.75 - 1.25 normal (illite), $> 1.25$ active (montmorillonite). It indicates the swelling potential.

Air content ($a_c$)

The ratio of volume of air to total volume of the soil:

$$a_c = \frac{V_a}{V}$$

Also $a_c = n(1 - S)$, where $n$ is porosity and $S$ is the degree of saturation.

Thixotropy

Thixotropy is the property of some clay soils to lose strength when disturbed (remoulded) at constant water content, and regain it with time on standing at rest. The regain is caused by reorientation of water molecules and particles in the adsorbed layer. Example: pile driving in clay softens the soil, then the strength returns after some days (pile "freeze").

Flow index ($I_f$)

The slope of the flow curve, i.e. the line of water content against log of number of blows in the liquid limit test. It is the change in water content per one log cycle of blows:

$$I_f = \frac{w_1 - w_2}{\log_{10}(N_2/N_1)}$$

A high flow index means the shear strength falls quickly with increase in water content.

Relative consistency

The relative consistency (consistency index) $I_c$ is the ratio of the difference between the liquid limit and the natural water content to the plasticity index:

$$I_c = \frac{w_L - w}{I_p}$$

$I_c = 1$ means the soil is at the plastic limit (stiff), and $I_c = 0$ means it is at the liquid limit (very soft).

Given

$w_L = 60\%$, $w_P = 30\%$, $w_S = 20\%$, $V_{LL} = 40$ cc, $V_{SL} = 23.5$ cc, $I_p = 60 - 30 = 30\%$.

Mass of solids

The soil is saturated at the liquid limit and at the shrinkage limit, and the volume of solids is the same. The volume difference is the volume of water lost between $w_L$ and $w_S$ (density of water 1 g/cc):

$$\begin{aligned} V_{LL} - V_{SL} &= \frac{(w_L - w_S) W_s}{\rho_w} \\ 40 - 23.5 &= (0.60 - 0.20) W_s \\ W_s &= \frac{16.5}{0.40} = 41.25\ \text{g} \end{aligned}$$

Specific gravity

At the shrinkage limit the soil is just saturated: $V_{SL} = V_s + w_S W_s$.

$$\begin{aligned} 23.5 &= V_s + 0.20 \times 41.25 = V_s + 8.25 \\ V_s &= 15.25\ \text{cc} \\ G &= \frac{W_s}{V_s \rho_w} = \frac{41.25}{15.25} = 2.70 \end{aligned}$$

Shrinkage ratio

$$R = \frac{W_s}{V_{SL}\,\rho_w} = \frac{41.25}{23.5} = 1.76$$

Answer: $G = 2.70$, $R = 1.76$.

Phase diagrams (per the sample)

 At liquid limit        At shrinkage limit
 V = 40 cc              V = 23.5 cc
 +--------+            +--------+
 | Water  | 24.75 cc   | Water  | 8.25 cc
 | 24.75 g|            | 8.25 g |
 |--------|            |--------|
 | Solids | 15.25 cc   | Solids | 15.25 cc
 | 41.25 g|            | 41.25 g|
 +--------+            +--------+
 W = 66.0 g             W = 49.5 g
 S = 100%, Va = 0       S = 100%, Va = 0

Weights of water: $0.60 \times 41.25 = 24.75$ g at the liquid limit and $0.20 \times 41.25 = 8.25$ g at the shrinkage limit. At both limits there is no air.

Given

In-situ: $\gamma = 18$ kN/m³, $w = 10\%$. Compacted fill: $\gamma_d = 19$ kN/m³, $w = 18\%$, volume = 1 m³. Take $\gamma_w = 9.81$ kN/m³ (it cancels in the answer).

Solution

Dry unit weight at the excavation site:

$$\gamma_{d,field} = \frac{\gamma}{1 + w} = \frac{18}{1.10} = 16.36\ \text{kN/m}^3$$

Weight of dry soil (solids) needed in 1 m³ of embankment:

$$W_s = \gamma_{d,fill} \times V = 19 \times 1 = 19\ \text{kN}$$

The solids weight is the same at the borrow site and in the fill, so the volume to excavate is:

$$V_{exc} = \frac{W_s}{\gamma_{d,field}} = \frac{19}{16.36} = 1.161\ \text{m}^3$$

Answer: about 1.16 m³ of soil must be excavated for each 1 m³ of compacted fill.

Assumption: no loss of soil during hauling and spreading.

Given

$M = 20$ kg, $V = 0.011$ m³, dry mass $M_s = 16.5$ kg, $G = 2.70$, $\rho_w = 1000$ kg/m³.

Solution

Mass of water: $M_w = 20 - 16.5 = 3.5$ kg.

Water content

$$w = \frac{M_w}{M_s} = \frac{3.5}{16.5} = 0.212 = 21.2\%$$

Dry density

$$\rho_d = \frac{M_s}{V} = \frac{16.5}{0.011} = 1500\ \text{kg/m}^3$$

Void ratio (from $\rho_d = \dfrac{G\rho_w}{1+e}$):

$$e = \frac{G\rho_w}{\rho_d} - 1 = \frac{2700}{1500} - 1 = 0.80$$

Degree of saturation

$$S = \frac{wG}{e} = \frac{0.2121 \times 2.70}{0.80} = 0.716 = 71.6\%$$

Porosity

$$n = \frac{e}{1+e} = \frac{0.80}{1.80} = 0.444 = 44.4\%$$

Answer: $w = 21.2\%$, $\rho_d = 1500$ kg/m³, $S = 71.6\%$, $n = 44.4\%$.

Given

$V = 20$ cc, $W = 38$ g, dry weight $W_s = 28$ g, $S = 100\%$.

Solution

Weight (and volume) of water: $W_w = 38 - 28 = 10$ g, so $V_w = 10$ cc (taking $\rho_w = 1$ g/cc).

As the clay is saturated, $V_v = V_w = 10$ cc, so:

$$\begin{aligned} V_s &= V - V_v = 20 - 10 = 10\ \text{cc} \\ e &= \frac{V_v}{V_s} = \frac{10}{10} = 1.0 \\ G &= \frac{W_s}{V_s\,\rho_w} = \frac{28}{10 \times 1} = 2.80 \end{aligned}$$

Check: $w = 10/28 = 35.7\%$, and $e = wG = 0.357 \times 2.8 = 1.0$ (agrees).

Answer: void ratio $e = 1.0$, specific gravity $G = 2.80$.

Given

$w = 15\%$, $G = 2.7$, $\gamma = 18$ kN/m³, specified $e < 0.75$. Take $\gamma_w = 9.81$ kN/m³.

Solution

Dry unit weight:

$$\gamma_d = \frac{\gamma}{1+w} = \frac{18}{1.15} = 15.65\ \text{kN/m}^3$$

Void ratio from $\gamma_d = \dfrac{G\gamma_w}{1+e}$:

$$e = \frac{G\gamma_w}{\gamma_d} - 1 = \frac{2.7 \times 9.81}{15.65} - 1 = 0.692$$

Since $e = 0.692 < 0.75$, the compaction meets the specification.

Decision: pass the bill for this base course work. (A check using the average of several samples is taken as given; individual samples should also be checked.)

Given

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

Volume and bulk unit weight

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

Water content

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

(i) Dry density

$$\gamma_d = \frac{\gamma}{1+w} = \frac{23.80}{1.0573} = 22.51\ \text{kN/m}^3$$ $$\rho_d = \frac{22.51}{9.81} = 2.294\ \text{Mg/m}^3 = 2294\ \text{kg/m}^3$$

(ii) Void ratio

$$e = \frac{G\gamma_w}{\gamma_d} - 1 = \frac{2.72 \times 9.81}{22.51} - 1 = 0.186$$

(iii) Degree of saturation

$$S = \frac{wG}{e} = \frac{0.0573 \times 2.72}{0.1856} = 0.840 = 84.0\%$$

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

Given

Container: $D = 0.2$ m, $H = 0.2$ m, so $V = \frac{\pi}{4}(0.2)^2(0.2) = 6.283 \times 10^{-3}$ m³. Dry sand $M_s = 10$ kg, $G = 2.65$, $e = 0.54$ (constant). $\rho_w = 1000$ kg/m³.

Assumption: the stated void ratio of 0.54 is used directly with the given mass of sand; the mass of solids fixes $V_s$, and the void volume is $V_v = eV_s$. (Using the container volume with $e = 0.54$ would give a slightly different mass of 10.8 kg, so the data are only approximately consistent.)

Volume of solids

$$V_s = \frac{M_s}{G\rho_w} = \frac{10}{2.65 \times 1000} = 3.774 \times 10^{-3}\ \text{m}^3$$

(i) Water to saturate the sand

$$V_v = eV_s = 0.54 \times 3.774 \times 10^{-3} = 2.038 \times 10^{-3}\ \text{m}^3$$

Water needed $= 2.038 \times 10^{-3} \times 1000 = 2.04$ kg (litres).

Water content at saturation:

$$w = \frac{M_w}{M_s} = \frac{2.038}{10} = 20.4\% \quad (\text{also } w = e/G = 0.54/2.65 = 20.4\%)$$

(ii) Water for 80% saturation

$$M_w = S\,e\,V_s\rho_w = 0.80 \times 2.038 = 1.63\ \text{kg}$$

So 1.63 litres of water must be added to the dry sand.

Answer: (i) 2.04 kg of water, $w = 20.4\%$; (ii) 1.63 kg (litres) of water.

Given

Borrow pit: $\gamma_d = 17$ kN/m³, $w = 10\%$. Embankment: $\gamma_d = 18$ kN/m³, $w = 15\%$, $V = 100$ m³. $\gamma_w = 9.81$ kN/m³.

Dry weight of soil needed

$$W_s = \gamma_{d,emb} \times V_{emb} = 18 \times 100 = 1800\ \text{kN}$$

Volume to be excavated

The same solids weight must come from the pit:

$$V_{pit} = \frac{W_s}{\gamma_{d,pit}} = \frac{1800}{17} = 105.9\ \text{m}^3$$

Water to be added

$$\begin{aligned} W_{w,\ pit} &= 0.10 \times 1800 = 180\ \text{kN} \\ W_{w,\ required} &= 0.15 \times 1800 = 270\ \text{kN} \\ W_{w,\ added} &= 270 - 180 = 90\ \text{kN} \end{aligned}$$

In volume: $\dfrac{90}{9.81} = 9.17$ m³ $\approx 9170$ litres.

Answer: excavate 105.9 m³ from the borrow pit; add 90 kN (about 9.17 m³ or 9170 litres) of water.

Principle

Compaction does not change the volume of solids. Only the void volume is reduced.

Given

Borrow pit: $V_1 = 1{,}00{,}000$ m³, $e_1 = 0.8$. Embankment: $e_2 = 0.6$.

Solution

$$V = V_s(1+e) \Rightarrow V_s = \frac{V_1}{1+e_1} = \frac{100000}{1.8} = 55\,555.6\ \text{m}^3$$

Volume of the compacted embankment:

$$V_2 = V_s(1+e_2) = 55\,555.6 \times 1.6 = 88\,888.9\ \text{m}^3$$

Answer: volume of embankment $\approx 88\,889$ m³ (about 11,111 m³ less than excavated).

Given

Embankment: $V = 1{,}00{,}000$ m³, $\gamma_d = 16$ kN/m³. Excavation site: $\gamma = 12$ kN/m³, $w = 15\%$, $G = 2.70$.

Dry weight needed in the embankment

$$W_s = 16 \times 100\,000 = 1.6 \times 10^{6}\ \text{kN}$$

Dry unit weight at the excavation site

$$\gamma_d = \frac{\gamma}{1+w} = \frac{12}{1.15} = 10.435\ \text{kN/m}^3$$

Volume to be excavated

$$V_{exc} = \frac{W_s}{\gamma_d} = \frac{1.6 \times 10^6}{10.435} = 1.533 \times 10^{5}\ \text{m}^3$$

Weight of soil to be excavated (moist)

$$W = \gamma \times V_{exc} = 12 \times 153\,333 = 1.84 \times 10^{6}\ \text{kN}$$

(check: $W_s(1+w) = 1.6 \times 10^6 \times 1.15 = 1.84 \times 10^6$ kN)

The specific gravity is not needed for this answer; it would only be used to find void ratios.

Answer: volume $\approx 1{,}53{,}333$ m³; weight $\approx 1.84 \times 10^{6}$ kN (dry weight $1.6 \times 10^{6}$ kN).

Given

$e_{max} = 1.25$, $e_{min} = 0.45$, $D_r = 40\%$, $G = 2.65$, $H = 3$ m, $\gamma_w = 9.81$ kN/m³.

1. Present state

$$D_r = \frac{e_{max} - e}{e_{max} - e_{min}} \Rightarrow e_0 = 1.25 - 0.40(0.80) = 0.93$$ $$\gamma_d = \frac{G\gamma_w}{1+e_0} = \frac{2.65 \times 9.81}{1.93} = 13.47\ \text{kN/m}^3\ (\rho_d \approx 1373\ \text{kg/m}^3)$$

2. After densifying to $D_r = 60\%$

$$e_1 = 1.25 - 0.60(0.80) = 0.77$$

Reduction in thickness (volume of solids constant, area constant):

$$\frac{\Delta H}{H} = \frac{e_0 - e_1}{1+e_0} \Rightarrow \Delta H = 3 \times \frac{0.93 - 0.77}{1.93} = 0.249\ \text{m}$$

New thickness $= 3 - 0.249 = 2.751$ m.

3. New unit weights

$$\gamma_d = \frac{2.65 \times 9.81}{1.77} = 14.69\ \text{kN/m}^3$$ $$\gamma_{sat} = \frac{(G+e_1)\gamma_w}{1+e_1} = \frac{(2.65+0.77) \times 9.81}{1.77} = 18.95\ \text{kN/m}^3$$

Answer: present $\gamma_d = 13.47$ kN/m³; thickness reduces by 0.249 m; new $\gamma_d = 14.69$ kN/m³ and $\gamma_{sat} = 18.95$ kN/m³.

Given

$G = 2.65$, $\gamma_d = 15$ kN/m³, $\gamma = 18$ kN/m³, $\gamma_w = 9.81$ kN/m³.

(i) Water content

$$w = \frac{\gamma}{\gamma_d} - 1 = \frac{18}{15} - 1 = 0.20 = 20\%$$

(ii) Degree of saturation

Void ratio:

$$e = \frac{G\gamma_w}{\gamma_d} - 1 = \frac{2.65 \times 9.81}{15} - 1 = 0.733$$ $$S = \frac{wG}{e} = \frac{0.20 \times 2.65}{0.733} = 0.723 = 72.3\%$$

(iii) Submerged unit weight

Imagine the soil saturated at the same void ratio:

$$\gamma_{sat} = \frac{(G+e)\gamma_w}{1+e} = \frac{(2.65+0.733) \times 9.81}{1.733} = 19.15\ \text{kN/m}^3$$ $$\gamma' = \gamma_{sat} - \gamma_w = 19.15 - 9.81 = 9.34\ \text{kN/m}^3$$

Answer: $w = 20\%$, $S = 72.3\%$, $\gamma' = 9.34$ kN/m³.

Given

Initial: $V_1 = 97$ cm³, $M_1 = 202$ g. Dry: $V_d = 87$ cm³, $M_s = 167$ g. $\rho_w = 1$ g/cm³.

(i) Initial water content

$$w_1 = \frac{M_1 - M_s}{M_s} = \frac{202 - 167}{167} = 0.2096 = 20.96\%$$

(ii) Specific gravity

The sample is saturated initially, so $V_1 = V_s + V_{w1}$:

$$V_{w1} = 35\ \text{cm}^3 \Rightarrow V_s = 97 - 35 = 62\ \text{cm}^3$$ $$G = \frac{M_s}{V_s\rho_w} = \frac{167}{62} = 2.69$$

(iii) Shrinkage limit

$$w_s = w_1 - \frac{(V_1 - V_d)\rho_w}{M_s} = 0.2096 - \frac{(97 - 87)}{167} = 0.2096 - 0.0599 = 0.1497$$

So $w_s = 15.0\%$.

Check using the dry volume: $V_d = V_s + w_s M_s = 62 + 0.1497 \times 167 = 87.0$ cm³ (agrees).

Answer: (i) $w = 20.96\%$, (ii) $G = 2.69$, (iii) shrinkage limit $= 15.0\%$.