CELE Construction Management & Methods — Construction Methods, Equipment and OperationsStudy Notes

Equipment productivity is the measure of how much work (output in m³, tonnes, or units) a piece of equipment can accomplish per unit time. The fundamental equation is: **Output = (Capacity per cycle) × (Cycles per hour) × (Efficiency factor)** Each component matters: **Capacity per cycle:** The volume or tonnage moved in one operating cycle. For excavators and loaders, this is the bucket size (m³); for trucks, the bin capacity. Always account for a **bucket fill factor** (typically 0.85–1.0) because buckets rarely fill to theoretical maximum, especially in hard or cohesive soils. **Cycles per hour:** Calculated as: Cycles/hour = 3600 / (cycle time in seconds) The cycle time includes all sequential steps: dig/load, swing, dump, and return (for excavators); load, haul, dump, return (for truck fleets). **Efficiency factor:** This reflects real-world conditions: - Working minutes per hour (e.g., 50-minute hour means 50/60 = 0.833 efficiency due to breaks, meetings, delays). - Job conditions: poor site access, weather, operator skill, material properties. - Management efficiency: downtime for maintenance, coordination issues. In PRC exam problems, efficiency is often stated as a percentage or working-minute ratio. Always convert to decimal form before multiplying. **Example: Excavator Output Calculation** An excavator on a Manila reclamation project has: - Bucket size: 1.5 m³ - Cycle time: 30 seconds (dig, swing, dump, return) - Fill factor: 0.95 (soil is well-broken) - Working efficiency: 50 minutes per hour (typical for Philippine site conditions with heat, crew coordination) Find the hourly output. **Solution:** Step 1: Calculate cycles per hour. Cycles/hr = 3600 s/hr ÷ 30 s/cycle = 120 cycles/hr Step 2: Convert efficiency to decimal. Efficiency = 50 min ÷ 60 min = 0.833 Step 3: Apply the output formula. Output = 1.5 m³ × 120 cycles/hr × 0.95 (fill factor) × 0.833 (efficiency) Output = 1.5 × 120 × 0.95 × 0.833 Output = 142.5 m³/hr **Verification:** This is reasonable for a 1.5 m³ excavator. Smaller buckets (0.8 m³) produce ~75 m³/hr; larger buckets (2.5 m³) produce ~250 m³/hr at similar efficiency. **Common Mistake:** Students forget to include the fill factor or misinterpret efficiency. The fill factor is **multiplied** into output because buckets are typically 85–95 % full, not 100 %. **Practical Application (PRC Context):** On a DPWH highway project or private commercial development in the Philippines, you must estimate earthmoving time. If the excavation is 10,000 m³ and you have one 1.5 m³ excavator operating at 142.5 m³/hr, the time required is: Time = 10,000 m³ ÷ 142.5 m³/hr ≈ 70.2 hours ≈ 1.75 working days (at 40 hr/week). This feeds directly into project scheduling and cost estimates.

In earthmoving operations, loaders (excavators, wheel loaders) and haul trucks work together as a team. The goal is to **match the number of trucks to the loader's productivity** so that: - The loader is kept continuously busy (not waiting for empty trucks). - Trucks don't queue excessively (wasting fuel and labour). - The fleet operates at minimum total cost. **The Fleet Matching Formula:** Number of trucks = (Truck cycle time) / (Truck load time) where: - **Truck cycle time** = load time + haul time + dump time + return time (travel back empty). - **Truck load time** = time for the loader to fill the truck (in minutes). **Why This Works:** If a truck takes 24 minutes for a complete round trip and 4 minutes to load, the loader fills one truck every 4 minutes. To keep the loader busy continuously, you need 24 ÷ 4 = 6 trucks. While one truck is loading, the other five are hauling, dumping, and returning. **Important:** Always **round up** the result. If you calculate 5.5 trucks, use 6 trucks; if 6.8 trucks, use 7. You need enough trucks to prevent the loader from sitting idle. **Example: Truck Fleet Matching** On a Metro Manila construction site, a loader is used to fill dump trucks for a nearby landfill. Given: - Load time (loader cycle to fill one truck): 4 minutes - Haul time (to landfill): 10 minutes - Dump time: 2 minutes - Return time (empty): 8 minutes How many trucks are required to keep the loader continuously busy? **Solution:** Step 1: Calculate truck cycle time. Truck cycle = 4 + 10 + 2 + 8 = 24 minutes Step 2: Divide cycle time by load time. N = 24 min ÷ 4 min = 6 trucks Step 3: Conclusion. You need **exactly 6 trucks** to keep one loader continuously busy at this site. **Sensitivity Analysis:** If haul time increases (farther landfill), cycle time increases, requiring more trucks. If the loader becomes faster (experienced operator, better material), load time decreases, requiring fewer trucks. **Economic Optimization:** Trucks are expensive to purchase/lease. Adding one extra truck costs money but eliminates loader idle time. Conversely, removing one truck saves cost but causes the loader to wait. The formula gives the **breakeven point**. In practice, sites slightly overmatch (7 trucks instead of 6) to account for truck breakdowns or weather delays. **Common Mistake:** Confusing load time with cycle time. Load time is only the filling portion; cycle time includes all activities. **Practical Application (PRC Exam Context):** Highway projects, quarries, and reclamation sites commonly use fleet matching questions. You might see: - "A quarry loader has a 5-min load time. The haul distance is 15 km at 50 km/h average speed, dump is 1 min, return is 15 km at 60 km/h empty. How many trucks?" You must calculate: - Haul time = 15 km ÷ (50 km/h ÷ 60 min/h) = 18 minutes - Return time = 15 km ÷ (60 km/h ÷ 60 min/h) = 15 minutes - Truck cycle = 5 + 18 + 1 + 15 = 39 minutes - N = 39 ÷ 5 = 7.8 → round up to **8 trucks**

A critical concept in earthmoving is that soil volume changes depending on its state: - **Bank volume (Vbank):** in-place, undisturbed soil (how it exists before excavation). - **Loose volume (Vloose):** soil after excavation; it is fluffier and occupies more space due to swell (aeration). - **Compacted volume (Vcompacted):** soil after placement and compaction; it is denser and occupies less space than bank. **Why This Matters:** When you excavate 100 m³ of bank soil, you cannot haul 100 m³. You must haul more (due to swell), and when you place it as fill, it shrinks (due to compaction). Ignoring these factors leads to cost underestimation and scheduling errors—a common pitfall on the PRC exam. **Key Relationships:** **Swell Factor (during excavation):** Swell is typically 10–40 % depending on soil type. - Sandy soil: 10–15 % swell - Clay and cohesive soil: 20–35 % swell - Rocky or blasted material: 40–50 % swell V_loose = V_bank × (1 + swell factor) **Shrinkage Factor (during compaction):** Shrinkage is typically 5–15 % depending on soil type and compaction method. - Sandy soil: 5–10 % shrinkage - Clay and cohesive soil: 8–15 % shrinkage V_compacted = V_bank × (1 − shrinkage factor) Or, rearranged: V_bank = V_compacted / (1 − shrinkage factor) **Example 1: Excavation and Haul Volume** A building excavation in Metro Manila requires removing 5,000 m³ of bank soil (in-place). The soil is a medium clay with a 25 % swell factor. What volume must the trucks haul away? **Solution:** Step 1: Apply swell factor. V_loose = 5,000 × (1 + 0.25) = 5,000 × 1.25 = 6,250 m³ Step 2: Conclusion. Trucks must haul **6,250 m³** (not 5,000 m³). If you underestimate this, your truck count, time, and cost projections are all wrong. **Example 2: Fill Material Requirement (Shrinkage)** A developer needs 10,000 m³ of compacted fill for a residential project in Laguna. The borrow source soil has a 12 % shrinkage factor when compacted. How much bank volume must be excavated from the borrow pit? **Solution:** Step 1: Apply shrinkage relationship. V_bank = V_compacted / (1 − shrinkage) V_bank = 10,000 / (1 − 0.12) = 10,000 / 0.88 = 11,363.6 m³ Step 2: Conclusion. You must excavate **11,364 m³** of bank soil from the borrow pit to obtain 10,000 m³ of compacted fill. The 1,364 m³ loss is due to compaction settling. **Example 3: Combined Swell and Shrinkage** A large cut-and-fill operation has: - Cut (excavation): 8,000 m³ bank, with 20 % swell during haul - Fill requirement: 7,500 m³ compacted, with 10 % shrinkage Does the excavated material suffice for the fill? If not, how much extra borrow is needed? **Solution:** Step 1: Calculate loose volume from cut. V_loose from cut = 8,000 × (1 + 0.20) = 9,600 m³ Step 2: Calculate bank volume needed for fill. V_bank needed = 7,500 / (1 − 0.10) = 7,500 / 0.90 = 8,333.3 m³ Step 3: Compare. Excavated: 8,000 m³ bank Needed: 8,333.3 m³ bank Shortfall: 333.3 m³ bank You need an additional **333.3 m³ of bank borrow** from a borrow pit. The cost of this borrow (excavation, haul, placement) must be added to the project budget. **Practical Application (PRC Exam):** On a real DPWH or private project: 1. Read the excavation volume (bank). 2. Multiply by (1 + swell) to find haul volume and truck requirements. 3. Read the required compacted fill volume. 4. Divide by (1 − shrinkage) to find the bank borrow volume. 5. Compare; if borrow exceeds cut, calculate and cost the shortfall. **Common Mistakes:** - Applying swell and shrinkage in the wrong direction (loose volume should be > bank; compacted volume should be < bank). - Forgetting that swell and shrinkage are two separate phenomena at different stages. - Using wrong swell/shrinkage factors for soil type (always verify from geotechnical report).

Concrete production and placement is a critical construction operation requiring coordination among batching, transport (mixer trucks or pumps), placement, consolidation, and curing. Unlike earthmoving (continuous production), concreting is cyclical and must achieve quality standards per ACI 318 and NSCP 2015. **Key Stages of Concrete Operations:** **1. Batching (Mixing Plant)** - Raw materials (cement, sand, coarse aggregate, water, admixtures) are measured and mixed. - Output: fresh concrete (typically 6–12 m³ per batch, depending on mixer size). - Cycle time: 3–5 minutes per batch (charge, mix, discharge). - For a 10 m³ mixer at 4-min cycle: output ≈ (10 m³ ÷ 4 min) × 60 = 150 m³/hr. **2. Transport (Mixer Trucks or Pumps)** - **Ready-mix trucks (concrete mixers):** Maintain slump via rotating drum; typical capacity 6–8 m³; haul time limited to 90 min (per ASTM C94, adopted in NSCP). - **Concrete pumps:** Deliver concrete via pipeline to high-rise or congested sites; rate 30–80 m³/hr depending on mix and distance. **3. Placement** - Concrete is discharged into formwork (beams, slabs, columns). - Rate depends on element type, access, and crew size. - Typical rates: 2–5 m³/hr for columns, 3–8 m³/hr for slabs, 5–10 m³/hr for beams (with pump support). **4. Consolidation (Vibration)** - Internal (immersion) or external (form) vibration removes air voids and achieves full compaction. - Required per ACI 318-19 (equivalent to NSCP 2015 Chapter 4 on concrete construction) to avoid honeycombing and ensure design strength. - Typical time: 10–20 seconds per pour zone; equipment: internal vibrators (25–60 Hz), external vibrators for formwork. **5. Curing** - Concrete strength development requires moisture and temperature control (20–25 °C optimal). - Curing time: minimum 7 days for standard cement (per NSCP 2015, Section 4.3.2), 28 days for design strength determination. - In the Philippines, high ambient temperature and humidity aid curing but also require protection from rapid drying and rain during plastic phase. **Concrete Production Rate Example** A high-rise residential project in Manila requires continuous concrete placement for a floor slab (500 m³). You have: - A 10 m³ batching plant with 5-min batch cycle. - A concrete pump (capacity 60 m³/hr). - A placement crew (average rate 4 m³/hr per pour section). What is the bottleneck, and how long will the floor take to place? **Solution:** Step 1: Calculate batching capacity. Batching rate = (10 m³ ÷ 5 min) × 60 = 120 m³/hr Step 2: Identify bottleneck. - Batching: 120 m³/hr - Pumping: 60 m³/hr - Placement: 4 m³/hr (single section) or ~20 m³/hr (5 parallel sections) The **placement crew** (4 m³/hr single section) is the bottleneck. You need at least 60 ÷ 4 = 15 placement sections (crews) working in parallel, or 500 ÷ 60 ≈ 8.3 hours with perfect coordination. Step 3: Realistic time. With setup, cleanup, and transitions: approximately **10–12 working hours** to complete one floor slab. **Quality and Compliance (NSCP 2015 / ACI 318):** - Slump: must be maintained per design specs, re-tempered only once with water per ACI 318-19 Section 19.2.5.3. - Air content: for exposed concrete, 4–8 % per NSCP 2015; measured at plant and placement point (ASTM C231). - Temperature: concrete must be 10–32 °C at placement (hotter in cold climates, cooler in tropics like the Philippines). - Strength testing: cylinders cast at batching plant, cured and tested at 7 and 28 days per ASTM C39. **Formwork Pressure and Falsework Design** Fresh concrete exerts hydrostatic pressure on formwork: P = ρ × g × h (where ρ = 2400 kg/m³, g = 9.81 m/s², h = height of concrete) For example, a 3 m column filled with fresh concrete: P = 2400 × 9.81 × 3 = 70,632 Pa ≈ 70.6 kPa Formwork (wall forms, props, beams) must be designed to resist this pressure plus dynamic loads from concrete placement and vibration. Undersized formwork is a major source of construction failures—a common pitfall in the Philippines where temporary structures are sometimes compromised. Per NSCP 2015 Section 4.1, formwork must be designed for: - Vertical loads (concrete weight, construction loads, equipment). - Lateral loads (wind, bracing, vibration). - Safety factor: ≥1.5 on yield, ≥2.0 on ultimate (or per design code applicable). **Practical Application (PRC Exam Context):** 1. **Scheduling:** Estimate concrete volume and crew rates to determine placement time. 2. **Cost estimation:** Sum batching, transport (truck or pump rental), placement labour, and curing time. 3. **Quality:** Know slump, air content, and temperature requirements per NSCP 2015. 4. **Formwork design:** Verify that forms resist concrete pressure; a question might ask you to calculate pressure and check form adequacy.

Formwork (temporary mold) and falsework (temporary support structure) must safely carry fresh concrete loads and construction activities while being economical. Failures can be catastrophic—collapses cause deaths, injuries, and project delays. Understanding the design is essential for a competent engineer. **Key Design Loads:** **1. Vertical Loads:** - **Self-weight of formwork/props:** typically 5–10 kN/m² (steel frames) to 8–15 kN/m² (timber). - **Fresh concrete weight:** 2400 kg/m³ = 23.5 kN/m³. A 0.5 m thick slab = 11.75 kN/m². - **Construction loads (live load):** workers, tools, vibrators ≈ 2.4–4.8 kN/m² (per building codes, typically 1.0–2.0 kPa × 1.5 safety factor). - **Total load:** self-weight + concrete + live load. For the slab above: 8 + 11.75 + 4 ≈ 24 kN/m² total. **2. Lateral Loads:** - **Hydrostatic pressure (from fresh concrete):** P = 2400 × 9.81 × h (Pa). Maximum pressure typically at base. - **Bracing for stability:** walls, columns, and formwork systems must have diagonal or moment-resisting bracing. - **Impact/vibration:** dynamic loads from compaction equipment; amplify static load by factor 1.2–1.5. **3. Deflection Limits:** - NSCP 2015 (based on ACI 347-14, Formwork for Concrete) specifies maximum deflection: - Slabs: L/180 to L/240 (where L is span). - Beams: L/270 to L/360. - Failure to limit deflection can cause: - Concrete surface irregularity (poor finish). - Uneven slab (causing water ponding or aesthetic issues). - Crack initiation in hardened concrete due to stress concentration. **Design Approach (Simplified):** Assume a 4 m × 4 m slab, 0.5 m thick, supported on props. Design the horizontal walers (beams supporting formwork panels). **Given:** - Slab thickness: 0.5 m - Concrete density: 2400 kg/m³ - Waler span (support spacing): 1.5 m - Live load: 2.4 kN/m² (construction) - Formwork self-weight: 0.8 kN/m² **Step 1: Calculate load per unit length on waler.** Load on slab = (2400 kg/m³ × 0.5 m + 100 kg/m² + 250 kg/m²) × 9.81 / 1000 ≈ (1200 + 100 + 250) kg/m² × 0.00981 kN/kg = 1550 kg/m² × 0.00981 ≈ 15.2 kN/m² Since waler is spaced 1.5 m apart, load per unit length on waler: w = 15.2 kN/m² × 1.5 m = 22.8 kN/m **Step 2: Calculate bending moment for simply supported 4 m span.** M = w × L² / 8 = 22.8 × 4² / 8 = 22.8 × 16 / 8 = 45.6 kN·m **Step 3: Check deflection.** For a timber waler, I (second moment of inertia) and E (modulus) must be sufficient: δ = 5 w L⁴ / (384 E I) ≤ L / 180 Typical timber (2×10 beam, rough) has I ≈ 480 cm⁴, E ≈ 10 GPa. δ = 5 × 22.8 × 4⁴ / (384 × 10,000 × 480 × 10⁻⁸) ≈ 3.2 mm < 4000 / 180 ≈ 22 mm ✓ (adequate) **Step 4: Check shear and bending stress.** Section modulus S = 2 I / c (where c is distance to extreme fiber) For the 2×10 timber, S ≈ 96 cm³ Bending stress: f_b = M / S = 45.6 × 10⁶ / (96 × 10⁻⁶) ≈ 14.4 MPa Allowable stress (sawn timber, short-term loading): ≈ 16 MPa ✓ (adequate) **In Practice (Philippine Sites):** - **Timber formwork:** Common, economical, reusable 3–5 times. Risk: fungal decay in humid climate (protect with sealant). - **Steel formwork:** Higher initial cost, very reusable (10+ uses), faster assembly. Risk: rust (maintain paint or oiling). - **Prefab modular systems:** Reduce labour, improve safety, but higher rental cost. **Common Failures (PRC Exam Knowledge):** 1. **Undersized props:** Buckling under load. Risk: sudden collapse. 2. **Inadequate lateral bracing:** Formwork tilts or overturns. Cause: wind, vibration, or uneven loading. 3. **Hydrostatic pressure underestimated:** Formwork bulges or splits at mid-height. 4. **Premature stripping:** Concrete not strong enough (< 70 % design strength per NSCP). Formwork removed too soon → cracking. 5. **Poor base support:** Props settle or sink into soft ground; formwork tilts. **Curing Relation to Formwork Removal (NSCP 2015):** - Formwork removal depends on concrete strength and location: - **Vertical supports (props):** remove when concrete reaches ≥50 % design strength (typically 3–7 days depending on temperature and cement type). - **Horizontal formwork (soffit):** remove when concrete reaches ≥75–100 % design strength (typically 7–28 days, with reshoring for spans > 4–5 m). - **In the Philippines:** high temperature accelerates hydration → faster strength gain; can strip earlier (3 days in favorable conditions). **Practical Application (PRC Exam):** You may see a question like: "A 5 m span beam formwork must be designed. Given concrete pressure, construction loads, and material properties, verify that the proposed waler (2×12 timber, spaced 1.2 m) is adequate." Your approach: 1. Calculate loads per unit length on the waler. 2. Calculate maximum bending moment (M = w L² / 8 for simply supported). 3. Check bending stress: f_b = M / S ≤ F_b (allowable). 4. Check deflection: δ = 5 w L⁴ / (384 E I) ≤ L / 180. 5. Conclude: adequate or not. Knowing these relationships is essential for design competence and exam success.

Beyond excavators and loaders, larger-scale earthmoving uses bulldozers, scrapers, and draglines. Each has distinct productivity characteristics and applications. **Bulldozers (Dozers):** Dozers are tracked vehicles with a blade for pushing soil. They are most productive in: - Shallow cuts (< 2 m depth), where push distance is limited. - Soft or loose soils (less ripping force needed). - Onsite material (no haul trucks needed; cost per m³ is low). **Productivity formula (simplified):** Dozer output = (blade capacity in m³) × (number of dozes per hour) × (efficiency) - **Blade capacity:** Typically 2–8 m³ depending on dozer size (D4–D11 in CAT classification). - **Doze cycles per hour:** Estimated from cycle time = push distance / speed + return speed. - **Efficiency:** 0.70–0.85 for good conditions. **Example: Dozer Productivity** A D8 bulldozer (6 m³ blade) is used to level a reclamation site in Manila Bay. Conditions: - Average push distance: 50 m - Push speed (loaded): 3 m/s - Return speed (empty): 4 m/s - Working efficiency: 45 min/hr (heat, coordination) Calculate hourly output. **Solution:** Step 1: Calculate cycle time. Push time = 50 m / 3 m/s = 16.7 s Return time = 50 m / 4 m/s = 12.5 s Cycle time = 16.7 + 12.5 ≈ 29.2 s Step 2: Calculate cycles per hour. Cycles/hr = 3600 / 29.2 ≈ 123 cycles/hr Step 3: Apply efficiency. Eff = 45 / 60 = 0.75 Output = 6 m³ × 123 × 0.75 = 553.5 m³/hr This is high output because the dozer is not queued with trucks—it works independently. However, push distances > 100 m reduce efficiency significantly; at that point, scrapers or haul trucks become more economical. **Scrapers:** A scraper is a self-loading, self-hauling unit: it digs, loads, transports, and dumps. Most economical for medium haul distances (300–2000 m) and soft soils. **Productivity formula:** Scraper output = (struck capacity in m³) × (number of cycles per hour) × (fill factor) × (efficiency) - **Struck capacity:** 10–40 m³ depending on model. - **Cycle time:** load time + haul time + dump + return. - **Fill factor:** 0.80–0.95 (usually 0.90). - **Efficiency:** 0.75–0.85. **Example: Scraper Productivity** A CAT 631 scraper (20 m³ struck capacity) is used for a highway fill project. Conditions: - Load time (full blade at site): 2 min - Haul distance: 1.5 km at 40 km/h average speed - Return: 1.5 km at 50 km/h empty - Dump: 30 s - Efficiency: 50 min/hr (includes repositioning) Calculate hourly output. **Solution:** Step 1: Calculate cycle time. Load time = 2 min Haul time = 1.5 km / (40 km/h / 60 min/h) = 1.5 × 60 / 40 = 2.25 min Dump time = 0.5 min Return time = 1.5 km / (50 km/h / 60 min/h) = 1.5 × 60 / 50 = 1.8 min Cycle time = 2 + 2.25 + 0.5 + 1.8 = 6.55 min Step 2: Calculate cycles per hour. Cycles/hr = 60 / 6.55 ≈ 9.17 cycles/hr Step 3: Apply fill factor and efficiency. Output = 20 × 9.17 × 0.90 × (50/60) = 20 × 9.17 × 0.90 × 0.833 = 137.4 m³/hr Comparison: at 1.5 km haul, a scraper outputs 137 m³/hr; larger hauls favor the scraper (lower push-haul cycle) but increase cost. Shorter hauls (< 300 m) favor dozers. **Draglines:** Used for excavation in wet environments (rivers, harbors, borrow pits, quarries). Draglines have long reach (up to 50+ m) and are effective for dredging and deep cuts. **Productivity formula (same as excavator):** Dragline output = (bucket capacity) × (cycles/hr) × (fill factor) × (efficiency) - Bucket sizes: 4–40 m³ depending on machine. - Cycle time: 45–90 s (slower than excavator due to longer reach). - Fill factor: 0.80–0.95 (depending on soil cohesion and operator skill). **Example: Dragline Productivity** A dragline with 6 m³ bucket, 60 s cycle, 0.85 fill factor, 48-min efficiency (dredging operations) is used on a Laguna lake expansion project. Output = 6 × (3600/60) × 0.85 × (48/60) = 6 × 60 × 0.85 × 0.80 = 244.8 m³/hr **Practical Application (PRC Exam):** 1. **Dozer vs. Scraper:** Dozer wins for onsite or short-haul (< 300 m); scraper wins for 300–2000 m hauls. 2. **Scraper vs. Haul trucks:** Both compete for 300–2000 m hauls; scraper is self-loading (advantage), but trucks can work with different loaders (flexibility). 3. **Dragline vs. Excavator + trucks:** Dragline for dredging, soft soils, deep reaches; excavator + trucks for general excavation, hard soils, multiple haul destinations. **Cost per m³:** - Dozer: lowest for onsite fill (no haul cost), ~50–100 PHP/m³ (Philippine rates). - Scraper: medium for 500–1500 m hauls, ~100–200 PHP/m³. - Excavator + haul trucks: highest for long hauls or hard soils, ~150–300 PHP/m³ depending on haul distance. - Dragline: moderate for dredging, ~120–250 PHP/m³. (Note: Rates vary by location, equipment availability, operator wages, and fuel costs; these are rough estimates for educational context.) **Swell and Shrinkage Revisited:** When estimating haul volumes: - Output (in bank m³ from excavator/dragline) must be multiplied by swell factor to get loose haul volume. - When scrapers or dozers place fill, the volume decreases by shrinkage factor. - Example: Dragline excavates 500 m³ bank limestone (35 % swell); loose volume for haul = 500 × 1.35 = 675 m³. If scrapers place it as fill at 12 % shrinkage: compacted volume = (500 / 1.35) ÷ (1 − 0.12) ≈ 340 m³ (net loss of 160 m³ due to shrinkage exceeding swell in the accounting).