Geology and Construction Materials

Reserve

A reserve is that part of a resource which has been identified (measured) in quantity and quality and can be extracted economically, legally and with existing technology at the time of estimate. Hence reserve is a subset of resource. For construction materials, reserve = the volume or tonnage of usable aggregate/sand/gravel of acceptable quality that can actually be quarried.

Reserve of aggregate from bedrock

1. Select the site: by topographic and geological maps and field survey; identify the rock type, outcrop and quality (strength, durability, specific gravity, abrasion).
2. Determine the quarry area: map the boundary (strike length $\times$ dip length or plan area $A$), excluding settlements, forests, rivers and roads.
3. Determine the attitude and thickness: measure strike and dip from outcrops, and the vertical thickness $t_v$ from boreholes or section lines; compute the true thickness

$$t = t_v\cos\delta$$

4. Compute the volume: $V = A \times t$ (or plan area $\times\ t_v$).
5. Deduct losses: overburden, weathered rock, waste bands, unusable rock, and set-back zones; apply a recovery factor (e.g. 0.7 to 0.9).
6. Compute tonnage: mass = $V\times$ density (bulk density of rock, about 2.6 to 2.7 t/m³).
7. Classify: as measured, indicated or inferred according to spacing of data.

Reserve of sand, gravel and boulders (unconsolidated deposits)

1. Locate river terraces, floodplains, fans and moraines from topographic maps, aerial photos and geomorphology.
2. Survey the deposit area (grid or cross-sections) and dig test pits, trenches or drill holes at regular spacing to measure thickness and depth to the water table.
3. Sample for grading, fines, organic and clay content, and quality tests.
4. Calculate $V =$ area $\times$ average workable thickness (or by cross-section/prismoidal method), deducting overburden and the portion below the groundwater level.
5. Multiply by bulk density to get tonnage, and apply the percentage of oversize and fines.

Example: area 150 m x 150 m, thickness 16 m: $V = 360{,}000\ \text{m}^3$.

Construction materials (aggregate, sand, gravel, boulders, clay, limestone) must be found close to the project, in sufficient quantity and quality. Maps are the first tools in this search.

1. Topographical (and aerial) map

• Shows contours, rivers, terraces, slopes, roads, villages and land use, helping to locate accessible sites and estimate haul distance and area.
• Shows landforms such as river terraces, fans and floodplains, which are likely sources of sand and gravel.
• Aerial photographs and satellite images show vegetation, outcrops, river channels, landslides and tonal contrasts that indicate rock or soil types, and allow measurement of area.
• Contours help to estimate overburden thickness, slope for quarry benches, and drainage.

2. Geological map

• Shows the type, age and distribution of rock formations, and the attitude (strike and dip) of beds, faults, folds and contacts.
• Helps to choose the formation suitable for aggregate (quartzite, limestone, granite, gneiss), cement raw materials (limestone, clay) or building stone, and to estimate thickness and extent, thus reserve.
• Shows weak, shattered and weathered zones (faults, shear zones) to be avoided.

3. Engineering geological map

• Shows rock/soil engineering properties, weathering grade, joint spacing, rock mass classes, slope stability, groundwater and geohazards.
• Allows direct selection of the best site considering quality, safety of quarry slopes, stability and environment.
• Shows areas to be avoided (landslides, active gullies, flood zones) and areas for waste disposal.

Together the maps narrow the search from the regional scale to the quarry or borrow site, which is then confirmed by field mapping, test pits and laboratory tests.

Geomorphology is the study of landforms and the processes shaping them. Because each landform has a typical material, it directs the search for construction materials.

Major applications

1. River deposits: active channels, bars and floodplains give sand and gravel; river terraces give thick, dry, well-graded sand, gravel and boulders above flood level, suitable for exploitation.
2. Alluvial fans and debris cones: coarse material (boulders, cobbles) at the mouths of tributaries.
3. Glacial and periglacial landforms: moraines and outwash plains in high Himalaya give mixed boulders and gravel.
4. Colluvial and talus slopes: angular rock fragments at the foot of cliffs for rip-rap and rock fill, but variable in quality.
5. Hill ridges, scarps and outcrops: show rock exposed at surface with little overburden, ideal for quarries; the slope and relief indicate quarry bench design.
6. Weathering mantle and residual soil: thick residual soil gives clay and earth fill; fresh rock is shallow where erosion has been strong.
7. Drainage pattern and slope: show lithology, fractures and structure and help in locating accessible, safe sites away from landslides and floods.
8. Hazard recognition: old landslides, active gullies and flood plains are identified from landforms so that unsafe sites are avoided.

Thus geomorphological mapping (with aerial photos) cuts the cost and time of exploration.

Resource

A resource is the total concentration of naturally occurring material (rock, sand, minerals, fuels) in or on the earth's crust in such form and quantity that its economic extraction is currently or potentially feasible. It includes both discovered and undiscovered material.

Reserve versus resource

Total resource versus undiscovered resource

A borrow area is a place from which soil, sand, gravel or rock is excavated for use in construction (embankments, concrete, road base).

Requirements for selection

1. Quality: suitable grading, strength, durability and soundness; free of organic matter, excess clay, salts and harmful minerals; passes the tests required for use (embankment fill, concrete aggregate, etc.).
2. Quantity: adequate volume (usually 1.5 to 2 times the requirement) after deducting overburden, waste and the portion under water.
3. Location: close to the site (short haul distance), with easy access by road and low transport cost.
4. Ease of excavation: thin overburden, workable slopes, little blasting, and above water table.
5. Stability and safety: away from landslide, flood and hazard zones; stable excavated slopes.
6. Legal and social: land ownership and clearance, government permission, compensation and acceptable to the community.
7. Environment: low damage to forests, farmland, water sources and scenery; the area can be rehabilitated; no effect on river morphology or structures such as bridges.
8. Drainage and water: no water logging, and water available for processing.
9. Economics: low cost of excavation, processing and royalty.

Types of resource

Each can be economic (reserve) or sub-economic. For construction materials the sources are also grouped as rock quarries (bedrock aggregate), river-bed and terrace deposits (sand and gravel), and soil borrow pits.

Principle: reserve (volume) of a bed = area $\times$ true thickness, where the true thickness is obtained from the vertical thickness by $t = t_v\cos\delta$.

Given

Vertical thickness $t_v = 300$ m, attitude S72°W/43° (dip $\delta = 43^\circ$), area $A = 227\ \text{km}^2$.

Calculation

$$\begin{aligned} t &= 300\cos 43^\circ = 300 \times 0.7314 = 219.41\ \text{m} = 0.2194\ \text{km} \\ V &= A \times t = 227 \times 0.2194 = 49.81\ \text{km}^3 \end{aligned}$$ $$V = 49.81\ \text{km}^3 = 4.98 \times 10^{10}\ \text{m}^3$$

Answer: reserve of sandstone aggregate = 49.8 km³ (4.98 × 10¹⁰ m³).

(If the 227 km² is the plan area on the map, the volume under it is $A\times t_v = 227 \times 0.3 = 68.1\ \text{km}^3$.)

Principle: reserve = area $\times$ true thickness, with $t = t_v\cos\delta$.

Given

$t_v = 300$ m, attitude S72°W/43° (dip $43^\circ$), area $A = 437\ \text{km}^2$.

Calculation

$$\begin{aligned} t &= 300\cos 43^\circ = 219.41\ \text{m} = 0.2194\ \text{km} \\ V &= 437 \times 0.2194 = 95.88\ \text{km}^3 = 9.59 \times 10^{10}\ \text{m}^3 \end{aligned}$$

Answer: reserve of quartzite aggregate = 95.9 km³ (9.59 × 10¹⁰ m³).

(If 437 km² is the plan area, the volume under it is $437 \times 0.3 = 131.1\ \text{km}^3$.)

Principle: the area of the quarry on the bed surface = strike length $\times$ dip length; reserve = area $\times$ true thickness, with $t = t_v\cos\delta$.

Given

Attitude N42°W/34° (dip $34^\circ$), vertical thickness (top to bottom) $t_v = 80$ m, strike length 2.5 km, dip length 0.5 km (taken along the dip of the bed).

Calculation

$$\begin{aligned} t &= 80\cos 34^\circ = 66.32\ \text{m} \\ A &= 2500 \times 500 = 1.25 \times 10^6\ \text{m}^2 \\ V &= 1.25 \times 10^6 \times 66.32 = 8.29 \times 10^7\ \text{m}^3 \end{aligned}$$

Answer: reserve of aggregate = 8.29 × 10⁷ m³ (about 82.9 million m³ = 0.083 km³).

Principle: reserve = area on the bed $\times$ true thickness, with $t = t_v\cos\delta$.

Given

Attitude N44°W/36° (dip $36^\circ$), vertical thickness $t_v = 82$ m, strike length 2.7 km, dip length 0.62 km (along the dip of the bed).

Calculation

$$\begin{aligned} t &= 82\cos 36^\circ = 66.34\ \text{m} \\ A &= 2700 \times 620 = 1.674 \times 10^6\ \text{m}^2 \\ V &= 1.674 \times 10^6 \times 66.34 = 1.11 \times 10^8\ \text{m}^3 \end{aligned}$$

Answer: reserve of aggregate = 1.11 × 10⁸ m³ (about 111 million m³ = 0.111 km³).

Step 1: true dip from the two apparent dips

For a direction of bearing $\beta$ the gradient of the bed satisfies $g_E\sin\beta + g_N\cos\beta = \tan\delta_a$ ($g_E$, $g_N$ = east and north components of the true gradient).

$$\begin{aligned} g_E\sin 20^\circ + g_N\cos 20^\circ &= \tfrac{1}{9} = 0.11111 \\ g_E\sin 65^\circ + g_N\cos 65^\circ &= \tfrac{1}{12} = 0.08333 \end{aligned}$$

Solving: $g_E = 0.04434$, $g_N = 0.10211$.

$$\tan\delta = \sqrt{0.04434^2 + 0.10211^2} = 0.1113 \Rightarrow \delta = 6.35^\circ\ (1:8.98)$$

Dip direction $= \arctan(0.04434/0.10211) = 23.5^\circ$ (N23.5°E).

Step 2: true thickness

$$t = t_v\cos\delta = 105\cos 6.35^\circ = 104.36\ \text{m}$$

Step 3: volume (reserve)

Area of the terrain $= 2.7\ \text{km} \times 3.1\ \text{km} = 8.37\ \text{km}^2 = 8.37\times 10^6\ \text{m}^2$

$$V = A \times t = 8.37\times 10^6 \times 104.36 = 8.73 \times 10^8\ \text{m}^3$$

Answer: true dip = 6.35° (1:9.0) towards N23.5°E; true thickness = 104.4 m; reserve of quartzite = 8.73 × 10⁸ m³ (0.873 km³).

Step 1: true dip from the two apparent dips

For a direction of bearing $\beta$ the gradient of the bed satisfies $g_E\sin\beta + g_N\cos\beta = \tan\delta_a$ ($g_E$, $g_N$ = east and north components of the true gradient).

$$\begin{aligned} g_E\sin 20^\circ + g_N\cos 20^\circ &= \tfrac{1}{9} = 0.11111 \\ g_E\sin 65^\circ + g_N\cos 65^\circ &= \tfrac{1}{12} = 0.08333 \end{aligned}$$

Solving: $g_E = 0.04434$, $g_N = 0.10211$.

$$\tan\delta = \sqrt{0.04434^2 + 0.10211^2} = 0.1113 \Rightarrow \delta = 6.35^\circ\ (1:8.98)$$

Dip direction $= \arctan(0.04434/0.10211) = 23.5^\circ$ (N23.5°E).

Step 2: true thickness

$$t = t_v\cos\delta = 105\cos 6.35^\circ = 104.36\ \text{m}$$

Step 3: volume (reserve)

Area $= 4.6\ \text{km}^2 = 4.6\times 10^6\ \text{m}^2$

$$V = A \times t = 4.6\times 10^6 \times 104.36 = 4.80 \times 10^8\ \text{m}^3$$

Answer: true dip = 6.35° (1:9.0) towards N23.5°E; true thickness = 104.4 m; reserve = 4.80 × 10⁸ m³ (0.480 km³).

Principle: reserve = area $\times$ true thickness, with $t = t_v\cos\delta$.

Given

$t_v = 400$ m, attitude S72°W/48° SE (dip $\delta = 48^\circ$), area $A = 537\ \text{km}^2$.

Calculation

$$\begin{aligned} t &= 400\cos 48^\circ = 400 \times 0.6691 = 267.65\ \text{m} = 0.2677\ \text{km} \\ V &= 537 \times 0.2677 = 143.73\ \text{km}^3 = 1.44 \times 10^{11}\ \text{m}^3 \end{aligned}$$

Answer: reserve of quartzite = 143.7 km³ (1.44 × 10¹¹ m³).

(If 537 km² is the plan area, the volume under it is $537 \times 0.4 = 214.8\ \text{km}^3$.)

Principle: the apparent (vertical) thickness measured in a borehole is converted to true thickness by $t = t_v\cos\delta$; reserve = area $\times$ true thickness.

Given

Dip $\delta = 30^\circ$, apparent thickness $t_v = 20$ m, area $A = 1128.5\ \text{m}^2$.

Calculation

$$\begin{aligned} t &= 20\cos 30^\circ = 20 \times 0.8660 = 17.32\ \text{m} \\ V &= 1128.5 \times 17.32 = 19{,}546\ \text{m}^3 \end{aligned}$$

Answer: reserve of limestone = 19,546 m³ (about 1.95 × 10⁴ m³).

(If the area is the plan area, the volume is $1128.5 \times 20 = 22{,}570\ \text{m}^3$.)

Step 1: true dip from the two apparent dips

Apparent dips: 1:16 along S15°E (bearing 165°) and 1:22 along S27°W (bearing 207°).

$$\begin{aligned} g_E\sin 165^\circ + g_N\cos 165^\circ &= \tfrac{1}{16} = 0.06250 \\ g_E\sin 207^\circ + g_N\cos 207^\circ &= \tfrac{1}{22} = 0.04545 \end{aligned}$$

Solving: $g_E = 0.01761$, $g_N = -0.05999$.

$$\tan\delta = \sqrt{0.01761^2 + 0.05999^2} = 0.06252 \Rightarrow \delta = 3.58^\circ\ (1:16.0)$$

Dip direction $= 163.6^\circ$ (S16.4°E).

Step 2: true thickness

$$t = 215\cos 3.58^\circ = 214.58\ \text{m}$$

Step 3: volume

The extent of the rock is read as 3.27 km $\times$ 72.05 m (length $\times$ width in plan):

$$A = 3270 \times 72.05 = 2.356\times 10^5\ \text{m}^2$$ $$V = A \times t = 2.356\times 10^5 \times 214.58 = 5.06 \times 10^7\ \text{m}^3$$

Step 4: mass (reserve in tonnes)

$$\text{Mass} = V \times \rho = 5.056\times 10^7 \times 2.75\times 10^3 = 1.39 \times 10^{11}\ \text{kg} = 1.39 \times 10^{8}\ \text{tonnes}$$

Answer: true dip = 3.58° (1:16) towards S16.4°E; true thickness = 214.6 m; volume = 5.06 × 10⁷ m³; reserve of limestone = 1.39 × 10¹¹ kg (about 139 million tonnes).

Principle: for a horizontal deposit the thickness is the true thickness, so reserve (volume) = area $\times$ thickness; mass = volume $\times$ density.

Given

Thickness $t = 16$ m; area $= 150 \times 150 = 22{,}500\ \text{m}^2$; density $\rho = 2870\ \text{kg/m}^3$.

Calculation

$$\begin{aligned} V &= 22{,}500 \times 16 = 360{,}000\ \text{m}^3 \\ M &= V\rho = 360{,}000 \times 2870 = 1.0332\times 10^{9}\ \text{kg} \\ &= 1{,}033{,}200\ \text{tonnes} \end{aligned}$$

(1 tonne = 1000 kg.)

Answer: reserve of sand = 3.6 × 10⁵ m³, i.e. about 1.03 × 10⁶ tonnes (1,033,200 t).

Principle: the vertical thickness $t_v$ (as in a vertical hole) is converted to the true thickness (perpendicular to bedding) by

$$t = t_v\cos\delta$$

where $\delta$ is the dip of the bed. The strike (N75°E) and dip direction (NW) do not affect the thickness; only the dip amount, $50^\circ$, is needed.

Calculation

$$\begin{aligned} t &= 600\cos 50^\circ \\ &= 600 \times 0.6428 \\ &= 385.67\ \text{m} \end{aligned}$$

Answer: true thickness of the limestone = 385.7 m.

Volume under 550 km² (for reserve)

$$V = 550\ \text{km}^2 \times 0.38567\ \text{km} = 212.1\ \text{km}^3 \approx 2.12\times 10^{11}\ \text{m}^3$$

(taking the 550 km² as the area measured on the bed).