Question 3 of 6: Dumping Skips, Overwinding, Locked Coil Rope, Motor Pole Pairs and Shaft Hoist Design

Question 3: Dumping Skips, Overwinding, Locked Coil Rope, Motor Pole Pairs and Shaft Hoist Design (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

3.1.1 — Automatic dumping skip types

An automatic dumping skip discharges its load at surface (or at an intermediate dump level) without a separate manual unlatching step, actuated purely by the skip's own geometry riding fixed dump rails/rollers built into the headframe as it approaches the dump position. The three common designs are: (i) fixed-body bottom-dump — a rigid body whose hinged bottom door is tripped open by a cam/roller arrangement at the dump station, releasing the load downward through the door into the receiving bin, then closing automatically as the skip descends past the dump rollers; (ii) overturning (Kimberley-type) — the entire skip body pivots about trunnions as guide rollers on the body ride a curved fixed rail at the dump level, rotating the body through roughly 130–170° to tip the full contents out over the top, then rotating back upright to descend; and (iii) swing-out (side-dump) — the body is hinged near the top and swings outward/sideways, guided by a fixed cam track, to discharge to one side rather than rotating fully over. All three are "automatic" in the sense that the mechanical dump action is triggered purely by the skip's position in the headframe, needing no operator intervention beyond stopping the skip at the correct dump elevation.

3.1.2 — Overwinding and its prevention

Overwinding is the condition where the conveyance (skip or cage) travels past its intended stopping point at the top of the shaft and continues into the headframe, potentially striking the sheave wheel, running the rope off the drum, or crashing through the headframe structure — typically caused by hoist-operator error, a control or brake malfunction, or an over-speed condition the normal deceleration/creep-speed control fails to catch in time. It is prevented, in combination, by: (1) an electrical depth-indicator/limit-switch interlock (e.g. a Lilly controller or equivalent geared depth-tracking device on the drum shaft) that independently tracks conveyance position and automatically cuts power and applies the emergency brake if the conveyance passes a preset limit before reaching bank; and (2) a mechanical overtravel/catch device at the headframe (a keps, crash beam, or arrestor system) that physically arrests the conveyance if it travels beyond the electrical limit, giving a redundant, purely mechanical last line of defence independent of the control system.

3.1.3 — Locked coil rope, types and post-installation inspection

A locked coil rope is constructed with an outer layer (or layers) of specially shaped, interlocking wires — typically Z-shaped (full-lock) or half-round profile — that key together around the inner round-strand core to form a smooth, sealed, essentially valley-free outer surface, unlike an ordinary round-strand rope where the strand valleys remain exposed. This sealed surface strongly resists water and dirt ingress into the rope's interior and gives excellent abrasion resistance against the drum and sheaves, at the cost of greater stiffness and more specialised (and more critical) termination/socketing. Two types: (i) full-locked coil — the entire outer layer is Z-shaped locking wire, giving the smoothest, most fully sealed surface and the highest resistance to wear and ingress, used on the largest, deepest, highest-duty single or multi-rope shaft hoists; (ii) half-locked coil — an inner core of ordinary round strands with only the outermost layer in locking (typically half-round) profile, giving most of the sealed-surface wear benefit with somewhat more flexibility than a fully locked construction, used on medium-to-large installations where full locked-coil stiffness is not warranted.

After installation, a locked coil rope is examined for defects by a combination of: a close visual inspection along its full running length (walked or run slowly past an inspector) for external broken wires, corrosion, kinking, flattening, or crown-wire wear, concentrating on the high-duty zones (the rope sections that repeatedly cross the drum/sheave at the same point, and the section immediately above the rope socket/attachment); diameter measurement at regular intervals to detect abnormal reduction (internal wire loss/corrosion shows up as a diameter loss even when the outer surface still looks intact, which is exactly the failure mode a sealed locked-coil surface can otherwise hide from a purely visual check); and periodic non-destructive electromagnetic (magnetic flux leakage) testing, which passes the rope through a magnetising/sensing head that detects internal broken wires and cross-sectional loss the sealed outer surface conceals from a visual inspection alone — a check that is routinely mandated on a fixed interval for hoisting ropes under BC's Health, Safety and Reclamation Code for Mines precisely because locked coil's sealed surface makes it the rope construction where a purely visual inspection is least reliable.

3.1.4 — Motor pole pairs

In an AC induction or synchronous motor, "pairs of poles" refers to the number of north-south magnetic pole pairs built into the stator winding; the motor's synchronous speed is fixed by the supply frequency and this pole-pair count, $n_{sync} = \dfrac{120f}{\text{poles}} = \dfrac{60f}{\text{pole pairs}}$ (Part 3.2.10 applies exactly this relation with 8 pole pairs on a 60 Hz Canadian supply to get 450 rpm). For a DC motor, which has no fixed synchronous speed, "poles" instead refers to the number of field poles in the frame; a higher pole count still lets a DC motor develop the same torque at a lower armature/base speed for a given power rating, by the analogous mechanism of more, smaller torque-producing zones around the air gap. Advantage of a high pole-pair count (e.g. 8 pairs) on AC motors: it directly delivers a low synchronous speed (450 rpm here, versus 3600 rpm for a 1-pole-pair motor at 60 Hz) without needing external gearing, well matched to a large, low-speed hoist drum — though a gearbox is still used here to further reduce to the drum's own required rpm (Part 3.2.11). Advantage on DC motors: a higher pole count reduces the required armature speed for a given power and torque, which reduces commutator/brush surface speed and the resulting brush wear and commutation stress, letting the motor run its heavy, continuously reversing hoist duty with less maintenance than an equivalent low-pole, high-speed DC machine geared down externally.

3.2 — Shaft hoist design

Given.

Find. The rope diameter (3.2.1), rope weight (3.2.2), drum diameter (3.2.3), winds/hour and cycle time (3.2.4), the velocity–time diagram (3.2.5), steady-state hoisting velocity (3.2.6), maximum drum rpm (3.2.7), average linear and angular acceleration (3.2.8–3.2.9), motor speed and gearbox ratio (3.2.10–3.2.11), maximum static rope load (3.2.12), and the steady-state, acceleration and maximum horsepower (3.2.13–3.2.15).

Approach. Select the smallest available rope diameter that clears the assumed 7.5:1 static safety factor against its own weight plus the loaded skip (3.2.1–3.2.2); size the drum from the given D/d ratio (3.2.3); derive winds/hour directly from the required production rate and cycle time from the reciprocal (3.2.4); close the trapezoidal velocity–time profile against the shaft depth to solve for the steady-state hoisting velocity (3.2.5–3.2.6), from which every kinematic quantity (3.2.7–3.2.9) and, via the motor's synchronous speed, the gearbox ratio (3.2.10–3.2.11) follow; then combine the maximum static rope load (3.2.12) with the steady velocity and the average acceleration to get the three horsepower figures (3.2.13–3.2.15).

1. Part 3.2.1–3.2.2 — select the rope diameter and find its weight. For each candidate diameter, the breaking load and unit weight follow from the given formulas, and the maximum static load is the loaded skip (22 t) plus the full 450 m of suspended rope:

   Rope selection (breaking load / static load, tonnes)

   d (mm)	BL = 0.07625d² (t)	w = 0.00577d² (kg/m)	Rope wt over 450 m (t)	Static load, skip+rope (t)	Factor of safety
   47.6	172.8	13.07	5.88	27.88	6.20
   50.8	196.8	14.89	6.70	28.70	6.86
   54.0	222.3	16.83	7.57	29.57	7.52
   63.5	307.5	23.27	10.47	32.47	9.47

   The two smaller ropes fall short of the assumed 7.5:1 minimum; the 54.0 mm (2.125 in) rope is the smallest that clears it, at SF = 7.52. $$\boxed{d = 54.0\ \text{mm}, \qquad w = 0.00577(54.0)^2 = 16.83\ \text{kg/m}}$$
2. Part 3.2.3 — drum diameter. $$D_{drum} = 108\,d = 108 \times 0.0540 = \boxed{5.832\ \text{m}}$$
3. Part 3.2.4 — winds per hour and cycle time. Each wind delivers the 10 t payload, so at 500 t/hr: $$\text{winds/hr} = \dfrac{500}{10} = 50, \qquad t_{cycle} = \dfrac{3600\ \text{s}}{50} = \boxed{72\ \text{s}}$$
4. Part 3.2.5 — velocity–time diagram. The cycle is a trapezoid: linear acceleration for 12 s, constant velocity V for the remaining travel time, linear deceleration for 12 s, then the 10 s decking dwell at zero velocity before the next wind begins.
   Velocity–time profile for one 72 s hoisting cycle: 0–12 s accelerate, 12–50 s constant V = 8.5 m/s, 50–62 s decelerate, 62–72 s decking dwell.
5. Part 3.2.6 — steady-state hoisting velocity. The shaft depth equals the area under the trapezoid: accel and decel each cover $\tfrac12 V t_{acc}$, and the constant-speed portion covers $V(t_{cycle} - t_{acc} - t_{dec} - t_{deck})$: $$D_{shaft} = \tfrac12 V t_{acc} + \tfrac12 V t_{dec} + V(t_{cycle}-t_{acc}-t_{dec}-t_{deck}) = V(t_{cycle} - t_{acc} - t_{dec} - t_{deck} + t_{acc})$$ Substituting $t_{cycle}=72$, $t_{acc}=t_{dec}=12$, $t_{deck}=10$ simplifies the bracket to $(72 - 12 - 10) = 50$ s of effective travel time: $$V = \dfrac{425}{50} = \boxed{8.5\ \text{m/s}}$$ Check: accel + decel distance $=2\times\tfrac12(8.5)(12)=102$ m; constant-speed distance $=8.5\times(72-12-12-10)=8.5\times38=323$ m; total $=102+323=425$ m, matching the shaft depth exactly.
6. Part 3.2.7 — maximum drum rpm. At V = 8.5 m/s the drum surface speed equals V, so $$n_{drum} = \dfrac{V}{\pi D_{drum}} \times 60 = \dfrac{8.5}{\pi(5.832)}\times 60 = \boxed{27.84\ \text{rpm}}$$
7. Part 3.2.8 — average linear acceleration. $$a = \dfrac{V}{t_{acc}} = \dfrac{8.5}{12} = \boxed{0.708\ \text{m/s}^2}$$
8. Part 3.2.9 — average angular acceleration of the drum. $$\alpha = \dfrac{a}{D_{drum}/2} = \dfrac{0.708}{2.916} = \boxed{0.243\ \text{rad/s}^2}\ \ (\approx 2.32\ \text{rpm/s})$$
9. Part 3.2.10 — motor speed at steady state. With 8 pole pairs (16 poles, Part 3.1.4) on a 60 Hz Canadian supply, the motor's synchronous speed — and, to a first approximation, its steady running speed — is $$n_{motor} = \dfrac{120f}{\text{poles}} = \dfrac{120(60)}{16} = \boxed{450\ \text{rpm}}$$
10. Part 3.2.11 — gearbox ratio at steady state. $$\text{ratio} = \dfrac{n_{motor}}{n_{drum}} = \dfrac{450}{27.84} = \boxed{16.2:1}$$
11. Part 3.2.12 — maximum static load on the rope. The maximum static tension occurs with the loaded skip at the shaft bottom and the full 450 m of rope suspended, exactly the load used to select the rope in Step 1: $$F_{static} = (12+10+7.57)\ \text{t} = 29.57\ \text{t} = 29{,}570\ \text{kg} \times 9.81 = \boxed{290.1\ \text{kN}}$$
12. Part 3.2.13 — horsepower at steady state, HP(M)₃. At constant velocity the motor need only overcome gravity on the maximum static load found in Step 11 (the worst case, skip at the bottom of its travel): $$P_3 = F_{static}\,V = (290{,}100)(8.5) = 2{,}465.8\ \text{kW}$$ $$\boxed{HP(M)_3 = \dfrac{2{,}465.8}{0.7355} = 3{,}352.6\ \text{HP(M)}}$$
13. Part 3.2.14 — horsepower to accelerate the maximum static load, HP(M)₁. The inertial force needed to accelerate the full static mass at the average linear acceleration, evaluated at the instant it reaches full speed V (the point of maximum accelerating power): $$F_{inertia} = m\,a = (29{,}570\ \text{kg})(0.708\ \text{m/s}^2) = 20{,}946\ \text{N}$$ $$P_1 = F_{inertia}\,V = (20{,}946)(8.5) = 178.0\ \text{kW}$$ $$\boxed{HP(M)_1 = \dfrac{178.0}{0.7355} = 242.1\ \text{HP(M)}}$$
14. Part 3.2.15 — estimated maximum horsepower, HP(M)_max. The motor sees its greatest demand at the instant it is still accelerating the maximum static load and has just reached full speed — it must supply the steady-state (gravity) power of Step 12 and the inertial power of Step 13 simultaneously: $$P_{max} = P_1+P_3 = 178.0+2465.8 = 2{,}643.9\ \text{kW}$$ $$\boxed{HP(M)_{max} = HP(M)_1+HP(M)_3 = 242.1+3{,}352.6 = 3{,}594.6\ \text{HP(M)}}$$