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Automation of Overhead Cranes

CHAPTER 4

Hoist, trolley and bridge: drives, brakes and automation

Electrical Control of Industrial Mechanisms

Master 1, Semester 3 · 2025–2026 · Prof. Z. TIR — University of El Oued

Photo: Methem, Public domain, Wikimedia Commons

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What you will be able to do

CHAPTER 4 · LEARNING OUTCOMES

1

Distinguish the three motions; follow the hoist torque through reeving and gearbox

2

Size a hoist motor, its braking resistor and its holding brakes

3

Explain brake-lift sequencing: torque before brake release

4

Compare legacy DC/thyristor and modern AC vector drives for crane duty

5

Compute and cancel load sway; specify skew, power-feed and operator provisions

Lectures, Fig. 4.1

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 4 · 2

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Meet the reference crane of the chapter

4.1 · REFERENCE CRANE

Quantity	Value	Remember it as
Rated load	8 t on the hook	a workshop bay crane
Hoisting speed	0.2 m/s full load, 0.6 m/s light	3:1 field weakening
Hoist drive	drum 0.35 m, 4 falls, gearbox 67, η 0.88	116 N·m at 1460 rpm
Hoist power	15.7 kW at the hook	17.8 kW at the motor (18.5 kW)
Trolley	2 t, 0.6 m/s in 3 s, 2 of 4 wheels driven	adhesion limit 0.74 m/s²

!

The same crane is used for the drive, the brakes, the resistor and the load-dependent speed

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 4 · 3

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Three motions, two kinds of load

4.1 · CRANE MOTIONS

Hoist: constant torque, overhauling — four quadrants, never lose the load
Trolley and bridge: inertia plus rolling friction, frequent starts and reversals
Hoist rated on continuous overhauling torque and zero-speed holding
Travel drives rated on peak accelerating torque; limited by wheel adhesion μgφ

Lectures, Fig. 4.1

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 4 · 4

Walk through Table 4.2 of the booklet orally. Hoist: torque set by the weight, essentially independent of speed, reversing role between raising (motoring) and lowering (generating): all four quadrants and zero-speed torque. Trolley and bridge: once rolling friction is overcome, the demand is the torque to accelerate a large mass; quadrants I and III when running, II and IV for a moment at every stop; duty dominated by starts, stops and reversals. The hoist torque is dominated by g, divided by the reeving ratio i (the number of falls per rope end) and the gear ratio, with the familiar efficiency exponent. The travel torque is dominated by d omega / dt — and the travel has a grip limit, like the rope traction of a lift: steel wheels on steel rails transmit at most mu, about 0.15 dry, times the weight on the driven wheels. Driving two wheels of four, a trolley cannot exceed 0.15 × 9.81 × 0.5 = 0.74 m/s², whatever its motor.

Ask self-check 4.1: does that limit depend on the load? (No: weight and inertia grow together.)

Timing: 5 minutes. Reference: Lectures section 4.1.1, Fig. 4.1, eqs 4.1 to 4.3, Table 4.2.

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Hoist versus travel drive sizing

4.1 · WORKED EXAMPLE

GIVEN

Reference crane. Hoist: 8 t at 0.2 m/s, 4 falls, drum radius 0.35 m, η = 0.88, motor 1460 rpm. Trolley: + 2 t carriage, 0.6 m/s in 3 s, rolling resistance 2 %.

FIND

What sizes each drive; the hoist gear ratio and motor torque

Photo: Syced, CC0, Wikimedia Commons

SOLUTION

Hoist: P = mgv = 15.7 kW; motor P/η ≈ 17.8 kW, continuously, both ways
Rope 4 × 0.2 = 0.8 m/s, drum 2.29 rad/s; k_gear ≈ 67; T = 116 N·m
Trolley rolling resistance: 0.02 × 10 000 × 9.81 = 1.96 kN, i.e. 1.18 kW at full speed
Trolley accelerating force: 10 000 × 0.6 / 3 = 2.0 kN, for 3 s per cycle
Trolley at 0.2 m/s² is well inside the 0.74 m/s² adhesion limit

✓

Check. 116 N·m × 152.9 rad/s = 17.8 kW, as from P/η. The trolley peak, (1.96 + 2.0) × 0.6 ≈ 2.4 kW, is a seventh of the hoist's.

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 4 · 5

Same load, two drives, two rules. The hoist is gravity-dominated: 15.7 kW of useful power, 17.8 kW at the motor, delivered continuously for the whole lift, in both directions. Follow the torque: four falls make the rope run at 0.8 m/s, the drum turns at 2.29 rad/s, a 67:1 gearbox brings the motor to 1460 rpm, and the motor torque is 8000 × 9.81 × 0.35 / (4 × 67 × 0.88) = 116 N·m. The trolley is inertia-dominated: steady rolling resistance only 1.96 kN, about 1.2 kW, but accelerating 10 t to 0.6 m/s in 3 s needs another 2 kN — as much again — demanded only for three seconds per cycle.

Conclusion from the booklet: sizing either drive by the other's rule gives the wrong machine.

Ask: if the trolley had to reach full speed in 1.5 s, what would change? (The accelerating force doubles to 4 kN; the peak torque rating, not the steady one, decides.)

Timing: 6 minutes. Reference: Lectures section 4.1.1, worked example; Tutorials Exercise 4.1.

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Same capacity, different crane: the load spectrum

4.1 · DUTY CLASSIFICATION

Lectures, Fig. 4.2

→

Damage ∝ (P/P_max)³: a half load does one eighth of the damage, so the heavy lifts dominate

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 4 · 6

Two cranes of the same rated load can need different drives, gearboxes and structures, because capacity says nothing about how often and how heavily the crane lifts. Two measures: the class of utilisation (total cycles in the design life) and the load spectrum (the distribution of loads). The spectrum factor weights each load level by the cube of its fraction of rated load — the fatigue law of steel and gearing. Panel (a): fraction of cycles at each load for three cranes. Panel (b): the same weighted by the cube. Light loads fill most cycles but cause almost no damage; for the two heavy spectra the top two load levels cause 85 % and 96 % of it, and even for the light crane the 7 % of lifts above 0.6 of rated cause almost half.

Timing: 5 minutes. Reference: Lectures section 4.1.2, eq. 4.4, Fig. 4.2.

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Duty group: a design input, not a label

4.1 · DUTY CLASSIFICATION

ISO 4301 / FEM groups M1–M8 (CMAA classes A–F)
Fixes structure fatigue life, gearbox service factor, rope safety factor
Motor rated on cyclic duration factor, typically 40 % CDF over 10 min
An M7 hoist may exceed 300 starts per hour: DOL is impossible, an inverter is needed

Photo: Syced, CC0, Wikimedia Commons

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 4 · 7

The duty group combines utilisation and spectrum: M3 for a maintenance hoist, M5 for a workshop crane, M6 for foundry and process cranes, M7 to M8 for ladle, grab and magnet cranes lifting near rated load on almost every cycle — like the container cranes in the photo, which work around the clock. For the control engineer, the duty group sets the motor's thermal basis: the cyclic duration factor, the S3 rating of Chapter 1, for example 40 % over a 10-minute cycle. And it sets the number of starts: above 300 starts per hour a direct-on-line cage motor would overheat on starting current alone, so the duty requires an inverter (or, historically, a wound rotor). End-effectors briefly: hook, lifting magnet with battery back-up so a power cut does not drop the load, and grab with two ropes.

Timing: 5 minutes. Reference: Lectures sections 4.1.2 and 4.1.3.

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Electrical drive and braking requirements

4.2

Every crane brakes twice: electrically to control the load, mechanically to hold it.

Photo: Williams Sparks, Puget Sound Naval Shipyard (HAER), Public domain, Wikimedia Commons

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 4 · 8

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Mechanical holding brake plus electrical braking

4.2 · BRAKING

Lectures, Fig. 4.3

Spring-applied brake on every motion: holds the load whenever power is absent
Electrical braking controls the load during motion
Dynamic: chopper and resistor across the DC link; regenerative: active front end
Stopping: drive to zero speed first, then brake, then current off

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 4 · 9

The mechanical brake is the elevator brake of Chapter 3: released electrically, applied by springs, fail-safe. It holds; it is not meant to stop a moving load. Electrical braking controls the load in motion and absorbs the energy when the load overhauls the motor. The block diagram: grid, rectifier, DC link, inverter, motor, hoist; when lowering, the generated energy flows back to the DC link and the chopper diverts it into the braking resistor — or, with an active front end, returns it to the network. A hoist that lowers heavy loads for a large part of its cycle should regenerate.

Ask: what happens if the chopper fails while a heavy load is lowered? (The bus voltage rises, the drive trips on overvoltage, and the brake must stop a moving load — the case it is not meant for.)

Timing: 4 minutes. Reference: Lectures section 4.2.1, Fig. 4.3.

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Braking resistor for an intermittent hoist

4.2 · WORKED EXAMPLE

GIVEN

The 17.8 kW hoist lowers its rated load for 12 s in every 90 s cycle; chopper at U_dc = 750 V, rated 40 A

FIND

The resistance and its peak and continuous ratings

SOLUTION

Lowering, losses subtract: P_gen = mgv·η = 15 700 × 0.88 ≈ 13.8 kW
R_max = 750² / 13 800 ≈ 41 Ω, R_min = 750/40 = 19 Ω; choose 36 Ω
Peak capability: 750² / 36 ≈ 15.6 kW, covering 13.8 kW with margin
Continuous: 13.8 × 12/90 ≈ 1.8 kW → a 2.5 kW resistor suffices

✓

Check. Energy per cycle 13.8 × 12 ≈ 166 kJ, over 90 s = 1.8 kW; and 36 < 41 Ω, the safe side of R_max.

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 4 · 10

Step 1 is the Chapter 1 pitfall in action: when lowering, the losses are paid by the load, so the power reaching the DC link is mgv times eta, not divided. Step 2 uses the two-sided bound of Chapter 3: R_max = 750 squared / 13 800, about 41 ohm, and R_min = 750/40 = 19 ohm; choose a standard 36 ohm, about 12 % below R_max. Why 750 V and not lower? The bus rests at 566 V and reaches 622 V when the mains is 10 % high; a lower threshold would burn grid energy in the resistor (Tutorials Exercise 4.3). Step 4 is the saving students miss: the continuous rating is set by the duty cycle. The energy dissipated equals the energy regenerated whatever the resistance, so 13.8 kW for 12 s every 90 s averages 1.8 kW. A 2.5 kW continuous resistor suffices, although it absorbs 13.8 kW at peak — a factor of seven in resistor size.

Ask: what if the crane lowers for 45 s per 90 s cycle? (6.9 kW continuous.)

Timing: 6 minutes. Reference: Lectures section 4.2.1, worked example; Tutorials Exercise 4.3.

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Sizing the holding brake — and why there are two

4.2 · HOLDING BRAKE

Photo: Williams Sparks, Puget Sound Naval Shipyard (HAER), Public domain, Wikimedia Commons

Lectures, Fig. 4.4

→

T_brake ≥ β·m·g·R·η / (i·k): η multiplies, because the load drives the gearbox. Reference crane, β = 1.5: 135 N·m on the motor shaft

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 4 · 11

The brake is sized on the torque the load applies at the brake shaft, times a factor beta: 1.5 for general hoisting, 2.0 or more for molten metal or high duty groups. It covers lining wear, thermal fade and oil contamination. Look where eta is: in the numerator. When the brake holds a hanging load, the load drives the mechanism, so gearbox friction helps the brake. Putting eta in the denominator, the motoring habit, overstates the brake torque — self-check 4.2. Reference crane, beta = 1.5: on the motor shaft 1.5 × 8000 × 9.81 × 0.35 × 0.88 / (4 × 67) = 135 N·m, just above the motor torque; on the drum, before the gearbox multiplies it, 9.1 kN m. The figure: two independent brakes, each able to hold the rated load, one on the motor shaft and one on the drum shaft. If the gearbox or a coupling breaks, B1 is left holding nothing, while B2 on the drum shaft still holds the load — separation, not duplication, removes the common failure (self-check 4.6). The photo shows real drum brakes on a large hoist.

Timing: 5 minutes. Reference: Lectures section 4.2.2, eq. 4.5, Fig. 4.4.

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Brake-lift sequencing: torque before release

4.2 · BRAKE-LIFT SEQUENCING

Lectures, Fig. 4.5

→

Torque proved (0.3 s) → brake released, opens in 0.2 s → motion. Stop: zero speed → brake applied → drive off 0.3 s later

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 4 · 12

Read the six events with the class. (1) Start command: drive enabled, torque built against the closed brake. (2) After 0.3 s torque is proved and brake release commanded. (3) The brake is mechanically open 0.2 s later and only then does the speed ramp begin. (4) Stop command, speed ramped down. (5) Zero speed: brake applied. (6) Drive disabled 0.3 s later, once the brake holds. The two delays look like wasted time and are not: the brake is mechanical and takes time to lift and to close. Holding torque is present whenever the brake is open.

Ask self-check 4.3: what if the brake is released first? (The load drops or rolls back before the drive holds it.)

Lab link: TP 4 Part A programs exactly this handshake as a GRAFCET on the PLC-200.

Timing: 5 minutes. Reference: Lectures section 4.2.2, Fig. 4.5.

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Automation and control technologies

4.3

DC performance from a cage motor — then stop the swing, keep the gantry square, know the load.

Photo: Syced, CC0, Wikimedia Commons

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 4 · 13

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From DC and wound-rotor drives to vector control

4.3 · DRIVE TECHNOLOGIES

Hoist needs: full torque at zero speed, smooth low speed, wide constant-power range
Legacy: Ward-Leonard, then thyristor converters on DC motors
Wound rotor: one resistor step = one curve = one speed, changing with load
Today: cage motor + encoder vector drive, often regenerative

Lectures, Fig. 1.5

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 4 · 14

The hoist's requirements — zero-speed torque, smooth low-speed spotting, and a wide constant-power range for light-load-fast, heavy-load-slow — historically justified DC drives: first Ward-Leonard, then the static thyristor converter, as in Chapter 3's mine hoists. Many cranes used wound-rotor motors instead. The figure is Chapter 1's: each contactor step in the rotor circuit selects one curve and so one running speed for a given load — which is why the crane's speed changed whenever the load did, and why low-speed operation burned energy in the resistors. Today the standard is the squirrel-cage motor with a closed-loop vector drive. Pitfall: zero-speed torque does not require a DC machine.

Timing: 4 minutes. Reference: Lectures sections 4.2.3, 4.3.1 and 4.3.2, Fig. 1.5; Tutorials Exercise 4.6.

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Field-oriented control: an AC motor behaving like DC

4.3 · VECTOR CONTROL

Lectures, Fig. 4.6

Measured currents are transformed into a frame turning with the rotor flux
i_d sets the flux, like a DC field current
i_q sets the torque, like a DC armature current
The encoder angle keeps the frame aligned: full torque at standstill

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 4 · 15

Trace the block diagram. The speed error goes through the speed PI to give the torque current reference i_q*. The flux reference i_d* is separate. Two current loops, each a PI. Their outputs are voltages in the rotating frame, transformed back to three phases and synthesised by the PWM inverter. The measured phase currents are transformed into the same frame using the rotor angle theta from the encoder. In that frame the currents are two steady quantities — i_d the flux, i_q the torque — exactly the field and armature currents of the DC motor of Chapter 1. That is how a rugged cage motor gets DC control quality, including full torque at zero speed.

Ask: why is the encoder essential for holding at standstill? (At zero speed there is no back-EMF to estimate the flux angle from.)

Timing: 6 minutes. Reference: Lectures section 4.3.2, Fig. 4.6.

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The load is a pendulum

4.3 · ANTI-SWAY

Lectures, Fig. 4.7

Rope length L sets the natural period
An uncompensated trolley acceleration excites the swing
A ramp lasting exactly one period T_sway leaves no swing
Anti-sway shapes the acceleration so the residual swing cancels
It is a control problem, not a mechanical one

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 4 · 16

The load hanging on the rope is a pendulum of period 2 pi root L over g. Accelerate the trolley and the rope lags; stop it and the load keeps swinging. The swing is excited by the acceleration profile the drive commands, so it can be removed in the controller — input shaping or feedback of the sway angle — with no mechanical addition. That is the fourth pitfall of the booklet. The residual-sway formula gives the size of the swing after a ramp, and a surprise: if the acceleration lasts exactly one sway period, nothing remains — the swing started at the beginning of the ramp is cancelled by the one started at its end.

Ask self-check 4.4: L = 6 m. (T = 2 pi root(6/9.81) = 4.9 s.)

Ask: does the period depend on the mass of the load? (No — only on the rope length.)

Timing: 4 minutes. Reference: Lectures section 4.3.3, eqs 4.6 and 4.7, Fig. 4.7.

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Anti-sway pulse timing

4.3 · WORKED EXAMPLE

GIVEN

A load hangs 9 m below the trolley; later the hoist reels in to 4 m

FIND

The timing of a two-pulse input shaper, and what happens when the rope shortens

SOLUTION

T_sway = 2π√(9/9.81) ≈ 6.02 s
Split the acceleration into two equal steps T_sway/2 ≈ 3.0 s apart
The second step launches a swing in antiphase with the first: they cancel
At 4 m the period is 4.01 s: a shaper tuned for 9 m no longer cancels

✓

Check. T_sway ∝ √L: 6.02 × √(4/9) = 4.01 s. Real controllers must track the rope length.

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 4 · 17

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Input shaping tested in a multibody simulation

4.3 · ANTI-SWAY

Lectures, Fig. 4.8

→

Unshaped: ±40 cm swing for ever (period 6.02 s). Shaped for 9 m: none. Same shaper on a 4 m rope: ±14 cm returns

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 4 · 18

Simscape Multibody simulation: a trolley with prescribed motion carrying a 5 t load. Panel (a): the speed command, unshaped (one 2 s acceleration to 0.5 m/s) and zero-vibration shaped (two half-steps 3.01 s apart). Panel (b): the horizontal sway. Unshaped, the 9 m pendulum swings plus or minus 40 cm indefinitely — the residual-sway formula gives 2 × 0.25 / 1.09 x sin(1.04) = 0.40 m, and the simulated period, 6.02 s, matches the formula. Shaped, the residual sway disappears completely. With the rope shortened to 4 m but the shaper still tuned for 9 m, 14 cm of sway returns — the numerical proof of why the controller must track rope length.

Ask: the shaped move arrives later than the unshaped one — by how much? (About half a period, 3 s.)

Timing: 4 minutes. Reference: Lectures section 4.3.3, Fig. 4.8.

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Master–slave and skew control on wide gantries

4.3 · SKEW CONTROL

Lectures, Fig. 4.9

Two legs, two drives: a small speed difference integrates into a skew
Wheel flanges bind: lateral forces several times the tractive effort
Each leg measures its own position; one speed reference is trimmed
A deliberately slow loop: seconds, not milliseconds

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 4 · 19

Master-slave control coordinates the two drives of a bridge so the girder stays square and the tractive effort is shared. On a wide gantry with independent motors on each leg, one leg running fractionally faster makes the portal rotate in plan — skew — the wheel flanges bind, and lateral forces can be several times the tractive effort. A speed difference integrates into a position error, so the loop must close on the difference of positions, not speeds. Each leg has an absolute position reference — a measuring wheel encoder, a laser range-finder or a barcode tape — and one speed reference is trimmed in proportion to the skew. The gain is high enough to stay inside the flange clearance, low enough not to fight the structure's elasticity.

Timing: 4 minutes. Reference: Lectures section 4.3.3, eq. 4.8, Fig. 4.9.

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Knowing the load: protection and productivity

4.3 · LOAD MONITORING

Load pin, strain-gauged link, or the vector drive's own torque estimate
Graded response: warn at 90 %, block hoisting (not lowering) at 100 %, trip at 110 %
Snag and slack-rope detection
Light loads hoisted faster: constant power above base speed, v = P/(mg)

Lectures, Fig. 4.10

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 4 · 20

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Power and control interfaces

4.4

Feeding a machine that moves, and commanding it safely.

Photo: Streetcrane, CC BY-SA 4.0, Wikimedia Commons

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 4 · 21

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Power feed and operator controls

4.4 · INTERFACES

Festoon: cable loops on trolleys — simple, limited speed and length
Conductor bars with collector shoes — long runways, heavy industry
Pendant or radio remote; maintained-contact emergency stop
Dead-man (hold-to-run); radio stops automatically on loss of signal

Photo: Streetcrane, CC BY-SA 4.0, Wikimedia Commons

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 4 · 22

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Common mistakes with crane drives

WATCH OUT

✗

MYTH

“A hoist drive can be sized from the lifting power alone.”

✓

WHAT IS TRUE

Lowering returns energy to the DC link: without a chopper or active front end sized for it, the bus trips — or the load runs away. Two separate calculations.

✗

MYTH

“The electromechanical brake stops the load.”

✓

WHAT IS TRUE

It holds and acts in emergencies. The drive decelerates; the brake closes only at zero speed. Braking a moving load wears it and is uncontrollable.

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 4 · 23

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Two more traps

WATCH OUT

✗

MYTH

“Zero-speed torque requires a DC machine.”

✓

WHAT IS TRUE

A closed-loop vector drive with an encoder holds full torque at zero speed on a standard cage motor — which is why AC replaced DC on cranes.

✗

MYTH

“Anti-sway is a mechanical problem.”

✓

WHAT IS TRUE

The swing is excited by the acceleration the drive commands. Shaping that profile suppresses it in the controller, with no mechanical addition.

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 4 · 24

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Quick check — try before you look

SELF-CHECK 4.1–4.6

4.1

A trolley drives 2 of its 4 wheels on dry rails (μ = 0.15): largest acceleration? Does it depend on the load?

4.2

Why does the efficiency multiply, not divide, the load torque in the holding-brake equation?

4.3

List the brake-lift sequence when starting a hoist. What if the brake is released first?

4.4

A load hangs 6 m below the trolley: sway period, and spacing of a two-pulse shaper?

4.5

The 8 t hoist of Fig. 4.10 lifts 4 t. How fast may it hoist?

4.6

Why are the two brakes of a hoist on opposite sides of the gearbox?

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 4 · 25

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Summary

CHAPTER 4 · SUMMARY

Hoist: overhauling, gravity-dominated, four quadrants; torque divided by reeving and gearbox
Trolley and bridge: inertia-dominated, limited by wheel adhesion
Duty group (load spectrum, cycles) sets fatigue life and motor thermal basis
Brake twice: spring brake to hold, drive (dynamic or regenerative) to control
Torque before brake release; brake after zero speed
Vector drives plus anti-sway, skew control and load monitoring

KEY EQUATIONS

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 4 · 26

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Practise this

CHAPTER 4 · NEXT STEPS

TD

Tutorials

Exercises 4.1–4.7: torque chain, duty group, resistor, brakes, sway, drives, load-dependent speed.

TP

Lab works

TP 4 — hoist: brake sequencing, overload and anti-sway on the PLC-200.

?

Before next week

Redo the self-check questions; then read Chapter 5 §5.1 on belt tensions.

Photo: Methem, Public domain, Wikimedia Commons

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 4 · 27

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Image credits

CHAPTER 4 · SOURCES

Figures and diagrams: Lectures booklet, Electrical Control of Industrial Mechanisms (computed in Python/Simscape or drawn in TikZ from the chapter's own equations).
Overhead cranes at Rajaville Jul2008 — Methem, Public domain, Wikimedia Commons, https://commons.wikimedia.org/wiki/File:Overhead_cranes_at_Rajaville_Jul2008.jpg
Container cranes in Tokyo Bay — Syced, CC0, Wikimedia Commons, https://commons.wikimedia.org/wiki/File:Container_cranes_in_Tokyo_Bay.jpg
View of main hoist wire rope drum and brakes, open contact boards are in view… — Williams Sparks, Puget Sound Naval Shipyard (HAER), Public domain, Wikimedia Commons, https://commons.wikimedia.org/wiki/File:View_of_main_hoist_wire_rope_drum_and_brakes,_open_contact_boards_are_in_view_at_the_far_right_wall_-_Puget_Sound_Naval_Shipyard,_Portal_Gantry_Crane_No._42,_Pier_5,_Farragut_Avenue,_HAER_WA-116-F-8.tif
MMD Derby009 — Streetcrane, CC BY-SA 4.0, Wikimedia Commons, https://commons.wikimedia.org/wiki/File:MMD_Derby009.jpg
Licence texts: creativecommons.org/licenses (CC BY, CC BY-SA); creativecommons.org/publicdomain/zero/1.0 (CC0). Photos are shown unmodified except for cropping and scaling.

Electrical Control of Industrial Mechanisms · Prof. Z. TIR — University of El Oued

Chapter 4 · 28