Underground cutting tools

Assume 8,000 kN net thrust across 40 engaged cutters, an illustrative load-distribution factor of 1.2, and a candidate allowable load of 267 kN.

Mean F = net thrust / N; estimated F = k × mean F; margin = allowable F − estimated F

1. 8,000 / 40 = 200 kN
2. 1.2 × 200 = 240 kN
3. 267 − 240 = 27 kN

267 kN is an example input, not a rating for products on this site. A historical Robbins sheet lists this order of load. Verify the distribution factor; total machine thrust is not net cutter thrust.

Assume a 3 m cutter track radius, 6 rpm cutterhead speed and a supplier-confirmed track-speed limit of 2 m/s.

v = 2πrn / 60; rpm limit = 60 × speed limit / (2πr)

1. 2π × 3 × 6 / 60 = 1.885 m/s
2. 60 × 2 / (2π × 3) = 6.366 rpm

Obtain speed limits from the cutter supplier. Track radius is the cutter position measured from the cutterhead centre, not the disc radius.

Assume a 432 mm disc on a 2.5 m track at 5 cutterhead rpm; illustrative disc-speed limit 80 rpm.

Disc rpm = 2 × track radius × cutterhead rpm / disc diameter (diameter in m)

1. 432 mm = 0.432 m
2. 2 × 2.5 × 5 / 0.432 = 57.870 rpm
3. 80 − 57.870 = 22.130 rpm

Assumes pure rolling with aligned rolling direction. Wear, tilt and contact geometry affect the result. 80 rpm is not a universal rating.

Assume a 2,800 mm radial span, 80 mm track spacing and 6 mm penetration per revolution.

S/p = spacing / penetration; intervals = W/S; initial track count = ceil(W/S)+1

1. 80 / 6 = 13.333
2. 2,800 / 80 = 35 intervals
3. 35 + 1 = 36 tracks

Track count is not total cutter count; centre/gauge arrangements, duplicate tracks and structural constraints need separate design. There is no universal optimum S/p.

Assume 6 mm/rev penetration, 5 rpm, a 20-hour daily window and 35% boring utilisation.

Net rate = 60pn / 1000; daily advance = net rate × hours × utilisation/100

1. 60 × 6 × 5 / 1,000 = 1.800 m/h
2. 1.8 × 20 × 35% = 12.600 m/day

Penetration cannot be assumed constant when rpm increases. Utilisation includes cutter changes, support and other stoppages.

Assume 1,000 kN·m torque, 5 rpm, 90% drivetrain efficiency and 660 kW available mechanical input.

Input power = 2πMn / (60η); margin = available input − required input

1. η = 90% = 0.9
2. 2π × 1,000 × 5 / (60 × 0.9) = 581.776 kW
3. 660 − 581.776 = 78.224 kW

Mechanical input power excludes motor electrical losses and auxiliaries. Low-speed operation also depends on the torque-speed envelope.

Assume 200 kN axial component, a 30° angle between resultant force and machine axis, and a 267 kN resultant limit.

Resultant F = axial F / cosθ; radial F = axial F × tanθ

1. 200 / cos30° = 230.940 kN
2. 200 × tan30° = 115.470 kN
3. 267 − 230.940 = 36.060 kN

Two-dimensional static resolution excludes rolling and other side loads. Confirm angle definitions and rated load directions with the supplier.

Assume 60 m head, density 1,000 kg/m³, an added 0.10 MPa allowance, and an illustrative 0.80 MPa assembly external-pressure rating.

Hydrostatic p = ρgh / 10⁶; required p = hydrostatic p + increment

1. 1,000 × 9.80665 × 60 / 10⁶ = 0.588399 MPa
2. 0.588399 + 0.10 = 0.688399 MPa
3. 0.80 − 0.688399 = 0.111601 MPa

This screens assembly external pressure, not seal differential pressure or compensation adequacy. Check slurry pressure, temperature and transients separately.

Assume a 250±0.5 mm housing, a 248±0.3 mm tool envelope and at least 1 mm total clearance.

Minimum gap = (housing width − housing tolerance) − (tool width + tool tolerance); margin = minimum gap − required gap

1. 250 − 0.5 = 249.5 mm
2. 248 + 0.3 = 248.3 mm
3. 249.5 − 248.3 − 1 = 0.2 mm

A single static dimension chain does not establish interchangeability. Check shaft ends, bolts, wedges, expansion and removal path separately.

At one position, radial wear increases from 8 to 12 mm over 200 m. Assume a 20 mm radial-wear limit and a 300 m next drive section.

Wear rate = (wear₂ − wear₁) / advance; remaining advance = (wear limit − wear₂) / wear rate

1. (12 − 8) / 200 = 0.020 mm/m
2. (20 − 12) / 0.020 = 400 m
3. 300 / 400 = 75%

Linear extrapolation assumes broadly stable conditions. Radial wear is not diameter loss. Flat wear, chipping, seizure or changing ground invalidate this forecast.

Assume 0.8 replacements/day, 25-day lead time, 30% buffer, 12 serviceable cutters on hand and 3 confirmed arrivals within the period.

Target stock = ceil(daily demand × lead time × (1+buffer)); order = max(0,target − stock − arrivals)

1. 0.8 × 25 = 20 cutters
2. ceil(20 × 1.30) = 26 cutters
3. max(0, 26 − 12 − 3) = 11 cutters

A deterministic planning model, not a statistical service-level model. Group by position/interface; exclude repair or inspection holds from usable stock.

Assume tooling cost 120,000 yuan, 10 hours of cutter-change downtime at 5,000 yuan/hour, and 500 m advance; comparison option 400 yuan/m.

Cost/m = (tooling cost + downtime × hourly cost) / advance; ratio = comparison cost / this cost

1. 120,000 + 10 × 5,000 = 170,000 yuan
2. 170,000 / 500 = 340 yuan/m
3. 400 / 340 = 1.176

All costs are hypothetical, not quotations. Align currency, advance, geology, residual value and downtime accounting between options; avoid double counting.

1. Thrust is concentrated at the cutter-rock contact.
2. A crushed zone forms under concentrated contact stress.
3. Cracks extend away from the contact region.
4. Schematic only: no force or crack-length prediction.

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1. Adjacent tracks have spacing S and penetration p.
2. Successive loading creates crushed zones and cracks.
3. Cracks may connect and release rock between tracks.
4. Match spacing to geology and cutting-test evidence.

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1. Cutter positions follow tracks of different radii.
2. Positions on one cutterhead share angular speed.
3. Track speed v equals 2 pi r n divided by 60.
4. Outer tracks run faster; check tool speed limits.

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1. The cutterhead carries the disc along its face track.
2. Contact causes the disc to rotate about its own axle.
3. Under pure rolling, track speed and diameter set disc rpm.
4. Seizure or slip invalidates the pure-rolling model.

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1. Define theta between resultant force and machine axis.
2. Resolve the force into axial and radial components.
3. At fixed axial force, larger angle needs greater resultant.
4. Rolling force and dynamic side loads are not included.

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1. The scraper edge enters ground as the cutterhead moves.
2. Carbide cuts while the steel body transfers load.
3. Cut material passes through openings to muck handling.
4. Check cutting direction, projection and mounting drawings.

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1. Uniform radial wear reduces effective ring diameter.
2. Flat wear can accompany seizure or uneven contact.
3. Chipping is local loss, not uniform average wear.
4. Record position, advance and condition before diagnosis.

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1. External hydrostatic pressure rises with water head.
2. Seal duty depends on pressure differential and temperature.
3. Where compensation exists, verify range and response.
4. External-pressure rating is not seal differential rating.

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1. Both housing and tool envelope have manufacturing tolerance.
2. Worst case combines minimum housing and maximum tool.
3. The difference must exceed the required installation gap.
4. Passing one clearance check does not prove interchangeability.

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1. Cutterhead torque overcomes cutting and other resistance.
2. Mechanical output power equals torque times angular speed.
3. Transmission losses make input exceed output power.
4. Also check the drive envelope and transient peaks.

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