6.2 Construction Equipment, Formwork, and Construction Methods

6.2 Construction Equipment, Formwork, and Construction Methods

Construction equipment and methods form the last area of the AMSTHC table of specifications, Construction Equipment and Methods. Its five competencies cover ground engineering and excavating equipment; concrete production and road-pavement equipment; bridge construction methods; production of excavating, lifting, loading and hauling equipment; and compressed air and water systems. This section covers all five, together with formwork pressure, a frequent construction problem. Safety regulation is treated separately in the construction safety and health section.

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Heavy Equipment Productivity & Earthmoving Mechanics

Earthmoving operations involve excavating, loading, hauling, dumping, and compacting soil or blasted rock. Soil volume changes across these phases:

• Bank Measure ($V_B$): Soil in its natural, undisturbed in-situ state.
• Loose Measure ($V_L$): Soil after excavation, which expands in volume due to voids (Swell): $\text{Swell (\%)} = \left( \frac{V_L - V_B}{V_B} \right) \times 100\%$ $\text{Load Factor } (L_F) = \frac{V_B}{V_L} = \frac{1}{1 + \text{Swell}}$
• Compacted Measure ($V_C$): Soil after mechanical compaction, which is less than bank volume (Shrinkage): $\text{Shrinkage (\%)} = \left( \frac{V_B - V_C}{V_B} \right) \times 100\%$

1. Hydraulic Excavator & Backhoe Productivity

Hourly production of an excavator loading haul units is computed by:

$Q = \frac{q \times 3600 \times E}{t_c}$

Where:

• $Q$ = hourly production in loose cubic meters ($LCBM/hr$) or bank cubic meters ($BCBM/hr$)
• $q$ = bucket payload per cycle ($q = q_{\text{rated}} \times K$, where $K$ is the bucket fill factor, e.g., $0.85$ for blasted rock to $1.10$ for moist loam)
• $t_c$ = total cycle time in seconds (dig + swing loaded + dump + swing empty, typically $15\text{ to }30\text{ s}$)
• $E$ = operational efficiency factor (e.g., $50\text{ min/hr} = 50/60 = 0.833$)

2. Bulldozer Productivity

Bulldozer output is governed by blade capacity and pushing distance:

$Q = \frac{q_{\text{blade}} \times 60 \times E}{T_{\text{cycle}}}$

Where blade capacity $q_{\text{blade}} \approx 0.5 W H^2$ ($W$ = blade width, $H$ = blade height) and $T_{\text{cycle}} = \frac{D}{v_{\text{forward}}} + \frac{D}{v_{\text{reverse}}} + t_{\text{shift}}$.

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Haul Fleet Matching & Loader-Truck Balancing

A critical objective in earthmoving economics is matching the number of haul trucks to the primary loading unit (excavator or wheel loader) to eliminate machine idle time.

Mathematical Formulation

1. Number of Loader Passes per Truck ($n$): $n = \frac{\text{Truck Capacity } (C_T)}{\text{Loader Bucket Payload } (q)}$ (Round to the nearest whole integer or load limit).
2. Loading Time per Truck ($t_l$): $t_l = n \times t_c$ Where $t_c$ is the excavator cycle time.
3. Total Round-Trip Truck Cycle Time ($T_{\text{truck}}$): $T_{\text{truck}} = t_l + t_h + t_d + t_r$ Where $t_h$ is haul time, $t_d$ is dump and maneuver time, and $t_r$ is empty return time.
4. Balanced Fleet Size ($N_{\text{bal}}$): $N_{\text{bal}} = \frac{T_{\text{truck}}}{t_l} = \frac{t_l + t_h + t_d + t_r}{t_l} = 1 + \frac{t_h + t_d + t_r}{t_l}$

Operational Implications of Fleet Size ($N$)

• If $N < N_{\text{bal}}$ (Under-trucked): The loader is underutilized and must wait for returning trucks. The truck fleet governs production: $Q_{\text{system}} = N \times \left( \frac{60}{T_{\text{truck}}} \right) \times C_T \times E$
• If $N > N_{\text{bal}}$ (Over-trucked): The loader operates at 100% capacity, but haul trucks form queues at the loading pit. The loader governs production: $Q_{\text{system}} = \left( \frac{60}{t_l} \right) \times C_T \times E$

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Soil Compaction Equipment & Roller Selection

Specifications commonly require a relative compaction of about 95% to 100% of a laboratory maximum dry density. Meeting it requires matching roller mechanics to soil type:

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Lateral Pressure of Fresh Concrete on Formwork (ACI 347)

Freshly poured concrete initially behaves as a dense hydrostatic fluid. As cement hydration initiates, internal shearing resistance develops, causing lateral pressure to peak and then stabilize.

According to ACI 347 (Guide to Formwork for Concrete), the lateral design pressure depends on placement rate $R$, concrete temperature $T$, and admixture chemistry:

1. Maximum Hydrostatic Pressure (Envelope Limit)

At any point, lateral pressure cannot exceed full liquid head:

$p_{\max} \le \gamma \cdot h$

Where $\gamma$ is concrete unit weight (typically $23.6\text{ to }24.0\text{ kN/m}^3$) and $h$ is total formwork height in meters.

2. Formwork Design Formulas for Walls

• For placement rate $R \le 2.1\text{ m/hr}$ and vertical placement height $h \le 4.2\text{ m}$: $p_{\max} = C_w C_c \left[ 7.2 + \frac{785 R}{T + 17.8} \right] \text{ kPa}$
• For placement rate $2.1 < R \le 4.5\text{ m/hr}$ or all walls where $h > 4.2\text{ m}$: $p_{\max} = C_w C_c \left[ 7.2 + \frac{1156}{T + 17.8} + \frac{244 R}{T + 17.8} \right] \text{ kPa}$
• Mandatory Limits: Under all conditions, $p_{\max}$ must satisfy: $30 C_w \le p_{\max} \le \gamma h$ Where:
  • $p_{\max}$ = maximum lateral pressure (kPa)
  • $R$ = rate of concrete placement (m/hr)
  • $T$ = concrete mix temperature ($^\circ\text{C}$)
  • $C_w$ = unit weight coefficient ($1.0$ for normal-weight concrete $23.0\text{ to }24.0\text{ kN/m}^3$)
  • $C_c$ = chemistry coefficient ($1.0$ for Type I cement without retarding admixtures; $1.2\text{ to }1.4$ for slag or fly ash blends)

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Lifting Equipment and Crane Capacity

Cranes are selected from the manufacturer's load chart, which gives rated capacity as a function of boom length and operating radius. The radius is the horizontal distance from the center of rotation to the load hook.

• Rated capacity falls quickly as radius increases, because the tipping moment is load × radius. The chart may also be governed by the structural strength of the boom at short radii.
• Gross load includes the load itself plus the hook block, slings, spreader beams and other rigging.
• Industry practice limits tipping-governed ratings to a fraction of the actual tipping load. For mobile cranes on outriggers this is commonly 85%; for crawler cranes it is commonly 75%.
• Tower cranes are used for high-rise building work. Lifting cycles are estimated like loader cycles: hook time, swing, landing and return.

Production example. A tower crane places concrete buckets of $1.0\text{ m}^3$ on a 6-minute cycle at 50 working minutes per hour. Hourly placement is $(50/6)(1.0) = 8.3\text{ m}^3/\text{h}$.

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Ground Engineering Methods

Excavations below the water table, or in weak soils, need groundwater control, excavation support, or ground improvement before structural work can start.

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Concrete Production and Pavement Construction Equipment

Transit mixer example. A plant loads a $7\text{ m}^3$ mixer in 6 minutes. The round trip, including discharge at the pour, takes 54 minutes. Trucks needed for continuous placing: $54/6 = 9$, so the plant can deliver $(60/6)(7) = 70\text{ m}^3/\text{h}$ if 9 trucks are assigned.

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Bridge Construction Methods

Substructure work uses cofferdams for footings in water, together with bored piles (drilled shafts) or driven piles for foundations.

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Compressed Air and Water Systems

Compressed air powers rock drills, jackhammers, sandblasters, pneumatic tools and grouting equipment. Plan for these points:

• Capacity. Compressor output is rated as free air delivered, in $\text{m}^3/\text{min}$ at atmospheric conditions. Total demand equals the sum of each tool's air consumption times a use factor, the fraction of time tools actually run. Add an allowance for leakage.
• Pressure. Most pneumatic tools operate at about $600$ to $700\text{ kPa}$ gauge. Pressure drop in long hoses and pipes must be limited by choosing adequate diameters.
• Altitude. At higher elevations the air is thinner, so a compressor delivers less mass of air per minute, and tool demand expressed as free air rises.
• Receivers smooth out demand peaks and allow moisture to condense and be drained.

Water systems on site serve dewatering and supply for mixing, curing and dust control. Pump power follows the same energy principle used in hydraulics:

$P_{\text{input}} = \frac{\gamma Q H}{\eta}$

Example. A dewatering pump lifts $0.020\text{ m}^3/\text{s}$ against a total dynamic head of $18\text{ m}$ at $\eta = 0.60$. The required input is $9.81(0.020)(18)/0.60 = 5.89\text{ kW}$.

Step-by-Step Worked Problem Examples

Worked Example 1: Earthmoving Fleet Production Matching

Problem: A mass excavation operation utilizes a hydraulic backhoe with a $1.6\text{ m}^3$ struck bucket capacity. The soil has a swell of $25\%$ and the bucket fill factor is $K = 0.95$. The backhoe has a cycle time $t_c = 24\text{ seconds}$. A fleet of dump trucks with an individual loose capacity of $12.0\text{ m}^3$ transports the soil. The haul cycle parameters are:

• Haul time to spoil site: $t_h = 8.0\text{ minutes}$
• Dump and maneuver time: $t_d = 1.5\text{ minutes}$
• Empty return time: $t_r = 6.5\text{ minutes}$
• Overall job efficiency: $E = 50\text{ min/hr} = 0.833$

Calculate:

1. The loader bucket payload and the number of passes required to load one truck.
2. The truck loading time $t_l$ and total truck round-trip cycle time $T_{\text{truck}}$.
3. The balanced fleet size $N_{\text{bal}}$.
4. The hourly production rate in Bank Cubic Meters ($BCBM/hr$) if 6 trucks are deployed versus 8 trucks.

Solution:

1. Bucket Payload and Passes: $q_{\text{loose}} = q_{\text{struck}} \times K = 1.6 \times 0.95 = 1.52\text{ m}^3\text{ (loose)}$ $\text{Passes } n = \frac{C_T}{q_{\text{loose}}} = \frac{12.0}{1.52} = 7.89 \implies \mathbf{8\text{ passes}}$ $\text{Actual truck load} = 8 \times 1.52 = 12.16\text{ m}^3\text{ (loose)}$
2. Loading Time and Truck Cycle: $t_l = n \times t_c = 8 \times 24\text{ s} = 192\text{ seconds} = \mathbf{3.20\text{ minutes}}$ $T_{\text{truck}} = t_l + t_h + t_d + t_r = 3.20 + 8.0 + 1.5 + 6.5 = \mathbf{19.20\text{ minutes}}$
3. Balanced Fleet Size: $N_{\text{bal}} = \frac{T_{\text{truck}}}{t_l} = \frac{19.20}{3.20} = \mathbf{6.0\text{ trucks}}$ Exactly 6 trucks create perfect fleet equilibrium with zero loader or truck waiting time.
4. Hourly Production Comparison:
   • Load Factor: $L_F = \frac{1}{1 + 0.25} = 0.80$
   • Case A: 6 Trucks Deployed ($N = 6 = N_{\text{bal}}$): Loader is 100% utilized. Loading frequency = $\frac{60}{t_l} = \frac{60}{3.20} = 18.75\text{ trucks/hr}$. $Q_{\text{loose}} = 18.75 \times 12.16 \times 0.833 = 189.92\text{ LCBM/hr}$ $Q_{\text{bank}} = Q_{\text{loose}} \times L_F = 189.92 \times 0.80 = \mathbf{151.94\text{ BCBM/hr}}$
   • Case B: 8 Trucks Deployed ($N = 8 > N_{\text{bal}}$): Because the loader was already at 100% capacity with 6 trucks, adding 2 more trucks merely causes trucks to queue at the shovel. The production rate remains governed by the loader at $151.94\text{ BCBM/hr}$.

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Worked Example 2: Formwork Maximum Lateral Pressure (ACI 347)

Problem: A reinforced concrete retaining wall formwork with height $H = 4.0\text{ m}$ is being cast using normal-weight concrete ($\gamma = 24.0\text{ kN/m}^3$) with Type I Portland cement and no admixtures ($C_w = 1.0, C_c = 1.0$). Concrete is placed at a rate $R = 2.0\text{ m/hr}$ and the mix temperature is $T = 30^\circ\text{C}$.

1. Calculate the maximum lateral pressure $p_{\max}$ on the formwork.
2. Check the minimum and maximum pressure limits per ACI 347.

Solution:

1. Check Governing ACI Formula: Since $R = 2.0\text{ m/hr} \le 2.1\text{ m/hr}$ and $H = 4.0\text{ m} \le 4.2\text{ m}$, use: $p_{\max} = C_w C_c \left[ 7.2 + \frac{785 R}{T + 17.8} \right]$ $p_{\max} = (1.0)(1.0) \left[ 7.2 + \frac{785 \times 2.0}{30 + 17.8} \right] = 7.2 + \frac{1570}{47.8} = 7.2 + 32.85 = \mathbf{40.05\text{ kPa}}$
2. Check Code Boundary Conditions:
   • Minimum Pressure Limit: $p_{\min} = 30 C_w = 30(1.0) = 30\text{ kPa} \implies 40.05\text{ kPa} > 30\text{ kPa} \quad (\text{Satisfied})$
   • Maximum Hydrostatic Limit: $p_{\text{hydro}} = \gamma \cdot H = 24.0 \times 4.0 = 96.0\text{ kPa} \implies 40.05\text{ kPa} < 96.0\text{ kPa} \quad (\text{Satisfied})$ Design Lateral Pressure: $p_{\text{design}} = \mathbf{40.05\text{ kPa}}$.

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