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Unit 9 · Lesson 9-1
Overall Layout and Rotor Support Schemes
1h 35m1300 XP
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Overall Layout and Rotor Support Schemes

Design principles, support schemes and their notation, intermediate and thrust bearings, thrust balance, bearing speed and life

Three shafts, 1971
In the late 1960s Rolls-Royce designed its big new engine, the RB211, with three shafts where everyone else used two. It became the first production three-spool engine (the RB211). Rolls-Royce also gave it a fan of a new carbon-fibre composite, Hyfil, which shattered in May 1970 when a chicken carcass was fired into it at high speed in a bird-strike test. By January 1971 Rolls-Royce was insolvent; on 4 February it went into receivership, and the government nationalised it. The three shafts survived. But three shafts or two, every rotor has to be held — at the right places, by the right bearings, through frames that carry its loads out to the aircraft. This lesson is about where.
The course deck sets out the principles of an engine's overall design: on the premise of reliable operation, keep the structure simple, with few parts, a reasonable strength reserve and low weight; adopt proven technologies and balance the risks; learn from earlier engines; and keep reliability, maintainability and supportability in view throughout. The overall layout then decides where the rotors are supported, how loads travel to the mounts, how rotors and stators stay concentric, how the engine is assembled, lubricated and sealed — and what it weighs.
💡Support schemes and their notation
A rotor's support scheme says how many bearings (support points) it has and where. The deck writes it a-b-c: support points in front of the rotor's compressor, between its compressor and turbine, and behind its turbine. Single-spool engines use 2–4 support points, two-spool engines 4–7, three-spool engines 7–8. The five-point scheme — LP 0-2-1 with HP 1-1-0 or 1-0-1 — is the most typical for two-spool engines; the PW4000, CFM56 and V2500 all use it. Low-pressure rotors mostly use 1-1-1, 0-2-1 or 1-2-0, high-pressure rotors 1-0-1 or 1-1-0.
📐Six engines, six answers
From the course deck:
  • F100: LP 1-1-1, HP 1-1-0. The HP turbine is overhung and the combustor casing carries a bearing: four load-bearing frames, four oil chambers.
  • F110: LP 1-1-1, HP 1-0-1, the HP turbine's rear bearing riding on the LP shaft: three frames, three oil chambers.
  • PW4000: LP 0-2-1, HP 1-1-0. CFM56: LP 0-2-1, HP 1-0-1 with an intermediate bearing. Five points each.
  • GE90: five points on three frames. On the GE90-115B the No. 1 fan bearing became a roller with a 503 mm bore, up from 183 mm on the GE90-94B, and the ball bearing moved to No. 2.
  • RB211: eight points — LP 0-2-1, IP 1-2-0, HP 1-0-1 — with the No. 3 thrust bearing an intermediate bearing.
  • Trent XWB: with the airflow up from 1088 kg/s on the Trent 1000 to 1436 kg/s, the fan's ball bearing moved to No. 1, behind the fan disk, with four times the axial capacity.
💡Intermediate and thrust bearings
An intermediate bearing carries one rotor on another. It shortens the rotor, saves a load-bearing frame and its weight, and simplifies the layout — but it couples the rotors (the LP rotor's vibration reaches the HP rotor), demands a rigid LP coupling and precise machining, and is hard to lubricate, seal and assemble. Every rotor has exactly one thrust bearing — a ball bearing that locates it axially and carries its net axial force. It goes where it is cool: in front of the compressor, or just behind the fan when there are no inlet guide vanes.
Derivation
Bearing loads from statics
Two bearings at x1x_1 and x2x_2 carry point loads FiF_i at xix_i — weights, and unbalance forces m e Ω2m\,e\,\Omega^2. Force and moment balance give
R1+R2=∑Fi,R2=∑Fi xi−x1x2−x1.R_1 + R_2 = \sum F_i, \qquad R_2 = \sum F_i\,\frac{x_i - x_1}{x_2 - x_1}.
For a load between the bearings the fraction lies between 0 and 1. For an overhung load, outside them, it exceeds 1: the near bearing carries more than the load itself, and the far bearing is pulled the other way — a lever.
📐Straddled or overhung
An HP rotor at 15,000 rpm with a 120 kg compressor drum and a 90 kg turbine (Lab 1):
  • Balanced to ISO grade G2.5, its residual unbalance adds less than its weight: about 0.5 kN.
  • Lose 20 g from a turbine blade tip and the turbine throws out 17.3 kN. Straddled (1-0-1), the rear bearing sees 14.9 kN; overhung (1-1-0), 21.9 kN, and the front bearing is pulled the other way with 4.6 kN.
💡Thrust balance
The HP compressor's rotor is pushed forwards, the HP turbine's rotor rearwards. Their difference lands on the thrust bearing, and it changes with throttle. Balance pistons — sealed cavities on the rotor held at chosen pressures — trim it so that the bearing stays loaded in one direction at every operating point (a lightly loaded ball bearing skids) without exceeding its rating.
📐Sizing a balance piston
With the station pressures of Lesson 8-2's engine (Lab 2):
  • No piston: 73.1 kN forward at take-off, 22.3 kN at cruise, 17.2 kN at flight idle.
  • A cavity at compressor-delivery pressure, vented to compressor inlet pressure, keeps the load between 3 and 40 kN everywhere only if its outer radius lies between 99.9 and 110.1 mm — a 10 mm window.
  • At 105 mm: 30.2 kN at take-off, 9.2 kN at cruise, 5.7 kN at flight idle.
DN=d [mm]×n [rpm],L10=(CP)p×106 revolutions,Peq=(∑Pi p Ni∑Ni)1/p\mathrm{DN} = d\,[\mathrm{mm}] \times n\,[\mathrm{rpm}], \qquad L_{10} = \left(\frac{C}{P}\right)^{p} \times 10^6\ \text{revolutions}, \qquad P_{eq} = \left(\frac{\sum P_i^{\,p}\,N_i}{\sum N_i}\right)^{1/p}

Bearing speed (DN) and basic rating life: C is the dynamic load rating, P the load, p = 3 for ball and 10/3 for roller bearings; N_i are the revolutions run at each load.

💡Bearing speed and life
Engine bearings are rolling-element bearings — rollers for radial load, balls for thrust — with large diameters and small widths, working up to about 300 °C. High-speed engine bearings generally run above a DN of 1,000,000, so centrifugal and friction forces are large and they need good lubrication and cooling (course deck). Their basic rating life L10L_{10} is the number of revolutions that 90 % of a population survives, and it falls as the cube of the load: a bearing's life is decided by its heaviest hours.
📐DN and life
Lab 3:
  • DN: the GE90-115B's No. 1 fan roller, 503 mm at 2355 rpm, 1.18 million; this unit's HP thrust ball bearing, 140 mm at 15,000 rpm, 2.10 million.
  • Over a flight (2 minutes of take-off, 20 of climb, 5 hours of cruise, 40 minutes of descent and idle) with Lab 2's loads, the equivalent load is 10.97 kN; with a rating of 320 kN the L10L_{10} life is about 30,000 flight hours.
  • 10 % more load costs a quarter of that life.
A complex system that works is invariably found to have evolved from a simple system that worked.
– John Gall, General Systemantics (1975) (Wikiquote)
💡Three shafts or two
A third shaft lets each compressor turn nearer its own best speed and keeps each rotor short and stiff — at the price of a third rotor to carry, eight support points instead of five, and more bearing chambers to seal and feed. Lesson 10-2's geared fan attacks the same problem differently: two shafts, and a gearbox between the fan and its turbine.
💡Where you will use this
Lesson 9-2 asks how these rotors vibrate on their bearings, Lesson 9-3 how the loads continue through frames and mounts to the wing. The same statics, thrust balance and bearing-life arithmetic size every gearbox, pump and turbine you will meet.
📚Further reading
  • Rolls-Royce, The Jet Engine (5th ed., Rolls-Royce plc, 1986), on engine construction.
  • Stanley Hooker, Not Much of an Engineer (Airlife, 1984), an engineer's account of the RB211 crisis.
  • The RB211 on Wikipedia.
↓But there is a problem
The rotors are held. But a rotor spinning at thousands of revolutions per minute has speeds at which it wants to whirl — and the parts on it carry enormous energy. On 19 July 1989 a fan disk broke over Iowa.