Rolling Thrust Bearings: Selection and Calculation: What Thrust Bearings Actually Do (And Why You Can't...

Context and scope

Every rotating shaft pushes somewhere. If you don't control where it pushes, it controls you.

That's the lesson the practitioner learned the hard way—standing in a flooded turbine hall at 2 a.m., watching maintenance crews tear apart a high-speed compressor because the wrong thrust bearing let a rotor walk 0.040 inches in the wrong direction. The repair cost six figures. The production loss cost seven.

This guide exists so you never become the practitioner.

Whether you're a mechanical engineer sizing your first thrust bearing, a maintenance professional troubleshooting a chronic failure, or a designer evaluating which bearing type best fits your next machine, this is your definitive reference. Every formula. Every design procedure. Every worked example. Every decision point—extracted from the most authoritative engineering handbooks and organized so you can act on it immediately.

What Thrust Bearings Actually Do (And Why You Can't Ignore Them)

Thrust bearings serve exactly two purposes:

• Absorb axial shaft loads — the forces that try to push a shaft lengthwise through its housing
• Position shafts axially — holding a rotor exactly where it needs to be, down to thousandths of an inch

That's it. But the consequences of getting this wrong cascade through every connected system: seals fail, impellers rub housings, gear meshes lose alignment, and catastrophic contact occurs between rotating and stationary parts.

The Four Major Thrust Bearing Types

Every hydrodynamic thrust bearing you'll encounter in practice falls into one of four categories. Each represents a different trade-off between cost, load capacity, alignment tolerance, and manufacturing complexity.

Notice something critical: the step, tapered land, and tilting pad bearings all share the same load rating range—200 psi normal, 500 psi maximum. So load capacity alone doesn't drive your selection. The real differentiators are alignment sensitivity, manufacturing cost, and size range.

The Hero's Journey: How the practitioner Got It Right the Second Time

Thrust Bearing Design Notation — Your Master Reference

Before you can design or evaluate any thrust bearing, you need to speak the language. These symbols appear in every design procedure that follows.

Keep this table bookmarked. You'll reference it constantly.

Subscript Convention: Subscript 1 = inside diameter. Subscript 2 = outside diameter. Subscript $i$ = inlet. Subscript $o$ = outlet.

Type 1: Flat Plate Thrust Bearing Design

When to Use It

The flat plate (parallel surface) thrust bearing is the most frequently used type of thrust bearing. It wins on simplicity and cost every time—but it pays for those advantages with the lowest load capacity of any design.

Use flat plate bearings when:

• Loads are light or occasional
• The bearing is primarily a positioning device, not a load carrier
• Cost and simplicity are top priorities
• The outside diameter is between 1.5× and 2.5× the inside diameter
• Unit loads stay below 75 psi normal, 150 psi maximum

Key General Parameters

• Maximum unit load: 75–100 psi
• $D_2 / D_1$ ratio: typically 1.5 to 2.5
• $K_g$: 0.8 (fraction of circumference occupied by pads)

The Complete Design Procedure

Each bearing section is wedge-shaped in reality. But for calculation purposes, you treat it as a rectangle with length $b$ (circumferential length along the pitch line) and width $a$ (radial width — the difference between external and internal radii).

Step 1 — Inside Diameter, $D_1$

Determined by shaft size and clearance. The bearing bore must clear the shaft with adequate running room.

Step 2 — Outside Diameter, $D_2$

$$D_2 = \left(\frac{4W}{\pi K_g p} + D_1^2\right)^{1/2}$$

Where:

• $W$ = applied load (lb)
• $K_g$ = fraction of circumference occupied by pads (usually 0.8)
• $p$ = bearing unit load (psi)

Step 3 — Radial Pad Width, $a$

$$a = \frac{D_2 - D_1}{2}$$

Step 4 — Pitch-Line Circumference, $B$

$$B = \pi(D_2 - a)$$

This is calculated at the pitch diameter, which lies at the midpoint of the pad's radial width.

Step 5 — Number of Pads, $i$

Assume an oil groove width $s$. If the pad length is assumed equal to its width (optimum geometry):

$$i_{approx} = \frac{B}{a + s}$$

Take $i$ as the nearest even number.

Step 6 — Length of Pad, $b$

$$b = \frac{B \times (i - s)}{i}$$

Or more precisely, once $i$ and $s$ are known:

$$b = \frac{B - is}{i}$$

Step 7 — Actual Unit Load, $p$

$$p = \frac{W}{i \cdot a \cdot b}$$

Verify this stays within the allowable range (< 75 psi normal, < 150 psi maximum).

Step 8 — Pitch-Line Velocity, $U$

$$U = \frac{B \cdot N}{12} \quad \text{(ft/min)}$$

Step 9 — Friction Power Loss, $P_f$

Friction power loss is difficult to calculate theoretically for flat plate bearings because there's no reliable method to determine the operating film thickness. However, a good approximation uses the $M$ value (horsepower loss per square inch of bearing surface) from empirical curves of $M$ vs. peripheral speed $U$.

$$P_f = i \cdot a \cdot b \cdot M$$

Engineering note: The $M$ value depends on both pitch-line velocity and unit load. At 4,000 ft/min and loads below 100 psi, typical values of $M$ are around 0.19 hp/in².

Step 10 — Oil Flow Required, $Q$

$$Q = \frac{42.4 \cdot P_f}{c \cdot \Delta t} \quad \text{(gpm)}$$

Where:

• $c$ = specific heat of oil (Btu/gal/°F)
• $\Delta t$ = temperature rise of the oil (°F)

Critical limit: A $\Delta t$ of 50°F is the acceptable maximum. Exceed this and you're degrading the oil and risking thermal runaway.

Step 11 — Film Flow, $Q_F$

$$Q_F = \frac{1.5 \times 10^5 \cdot i \cdot V \cdot h^3 \cdot p_s}{Z_2}$$

Where:

• $V$ = effective width-to-length ratio ($a/b$)
• $h$ = film thickness (use 0.002 inches as an approximation since $h$ cannot be calculated theoretically for flat plate bearings)
• $Z_2$ = oil viscosity at outlet temperature

Practical rule: It's desirable to have a minimum of one-half of the desired oil flow pass through the chamfer.

Step 12 — Required Flow Per Chamfer, $Q_c$

$$Q_c = \frac{Q}{i}$$

Step 13 — Kinetic Energy Correction Factor, $\xi$

Assume a chamfer length $l$ and enter the empirical curve with values of $Z_2 \cdot l$ and $Q_c$ to determine $\xi$.

Step 14 — Uncorrected Required Flow Per Chamfer, $Q_c^0$

$$Q_c^0 = \frac{Q_c}{\xi}$$

Step 15 — Depth of Chamfer, $g$

$$g = \left(\frac{Q_c^0 \cdot l \cdot Z_2}{4.74 \times 10^4 \cdot p_s}\right)^{1/4}$$

Worked Example: Flat Plate Thrust Bearing

Problem: Design a flat plate thrust bearing to carry 900 pounds at 4,000 rpm using SAE 10 oil with a specific heat of 3.5 Btu/gal/°F at 120°F and 30 psi inlet conditions. The shaft is 2.75 inches in diameter and the temperature rise must not exceed 40°F.

Step 1 — Inside diameter: $D_1 = 3$ inches (to clear shaft)

Step 2 — Outside diameter:

Assuming $p = 75$ psi from the load table:

$$D_2 = \left(\frac{4 \times 900}{\pi \times 0.8 \times 75} + 3^2\right)^{1/2} = 5.30 \text{ inches}$$

Use $D_2 = 5.5$ inches.

Step 3 — Radial pad width:

$$a = \frac{5.5 - 3}{2} = 1.25 \text{ inches}$$

Step 4 — Pitch-line circumference:

$$B = \pi \times 4.25 = 13.3 \text{ inches}$$

Step 5 — Number of pads:

Assume oil groove width $s = 3/16$ inch = 0.1875 inch.

$$i_{approx} = \frac{13.3}{1.25 + 0.1875} = 9.25$$

Take $i = 10$.

Step 6 — Length of pad:

$$b = \frac{13.3 - 10 \times 0.1875}{10} = 1.14 \text{ inches}$$

Step 7 — Actual unit load:

$$p = \frac{900}{10 \times 1.25 \times 1.14} = 63 \text{ psi} \quad \checkmark \text{ (below 75 psi)}$$

Step 8 — Pitch-line velocity:

$$U = \frac{13.3 \times 4000}{12} = 4{,}430 \text{ ft/min}$$

Step 9 — Friction power loss:

From empirical data at $U = 4{,}430$ ft/min and $p = 63$ psi: $M = 0.19$

$$P_f = 10 \times 1.25 \times 1.14 \times 0.19 = 2.7 \text{ hp}$$

Step 10 — Oil flow required:

Assuming $\Delta t = 40°F$ (the maximum allowable), the operating temperature becomes $120 + 40 = 160°F$, giving an oil viscosity $Z_2 = 9.6$ centipoises.

$$Q = \frac{42.4 \times 2.7}{3.5 \times 40} = 0.82 \text{ gpm}$$

Step 11 — Film flow:

$$Q_F = \frac{1.5 \times 10^5 \times 10 \times 1 \times 0.002^3 \times 30}{9.6} = 0.038 \text{ gpm}$$

Key finding: 0.038 gpm is a very small fraction of the required 0.82 gpm. The bulk of the flow must be carried through the chamfers.

Step 12 — Required flow per chamfer:

$$Q_c = \frac{0.82}{10} = 0.082 \text{ gpm}$$

Step 13 — Kinetic energy correction factor:

With chamfer length $l = 1/8$ inch: $Z_2 \cdot l = 9.6 \times 0.125 = 1.2$

From the empirical curve: $\xi = 0.44$

Step 14 — Uncorrected required flow:

$$Q_c^0 = \frac{0.082}{0.44} = 0.186 \text{ gpm}$$

Step 15 — Depth of chamfer:

$$g = \left(\frac{0.186 \times 0.125 \times 9.6}{4.74 \times 10^4 \times 30}\right)^{1/4} = 0.02 \text{ inches}$$

Final Flat Plate Design Summary

        Schematic — Flat Plate Thrust Bearing (Top View)
        ┌─────────────────────────────────────────┐
        │                                         │
        │      ┌──1──┐  ┌──2──┐  ┌──3──┐         │
        │     │ Pad  │  │ Pad  │  │ Pad  │        │
        │      └─────┘  └─────┘  └─────┘         │
        │   ┌──10─┐                  ┌──4──┐      │
        │  │ Pad  │    ○ Shaft      │ Pad  │     │
        │   └─────┘     Center       └─────┘      │
        │      ┌──9──┐  ┌──8──┐  ┌──7──┐         │
        │     │ Pad  │  │ Pad  │  │ Pad  │        │
        │      └─────┘  └─────┘  └─────┘         │
        │         ┌──6──┐  ┌──5──┐               │
        │        │ Pad  │  │ Pad  │              │
        │         └─────┘  └─────┘               │
        │                                         │
        │    Oil Grooves between each pad          │
        │    Chamfers at leading edges             │
        └─────────────────────────────────────────┘

Type 2: Step Thrust Bearing Design

When to Use It

The step bearing is the workhorse for small, high-volume applications. It accepts normal thrust loads (up to 200 psi normal, 500 psi maximum) and is inexpensive to produce. the practitioner filed this type under "the bearing you use when you need real load capacity but don't have the budget or space for something fancy."

The catch: Alignment sensitivity increases with size. Keep step bearings small or accept the alignment risk.

Key General Parameters (Optimum Proportions)

These are critical for proper step bearing geometry:

• $a = b$ (pad width equals pad length)
• $b_2 = 1.2 \cdot b_1$ (step length is 1.2× the land length)
• $e = 0.7h$ (step depth is 70% of film thickness)

Since $b_2 = 1.2 \cdot b_1$ and $b_2 + b_1 = b$, we get $b_2 = \frac{1.2b}{2.2}$

    Cross-Section — Step Thrust Bearing Pad

          ┌──── b ────────────────────┐
          │                           │
          │   b₂ (step)  │  b₁ (land) │
          │              │            │
    ──────┤              │            │──── Film surface
          │   (raised)   ▼   (flat)   │
          │         ┌─e──┐            │
    ──────┼─────────┘    └────────────┼──── Pad surface
          │                           │
          └───────────────────────────┘
          ◄──── Direction of motion U ────►

The Complete Step Bearing Design Procedure

Step 1 — Internal Diameter, $D_1$

Assume a diameter sufficient to clear the shaft.

Step 2 — External Diameter, $D_2$

$$D_2 = \left(\frac{4W}{\pi K_g p} + D_1^2\right)^{1/2}$$

Where $p$ is assumed from the thrust bearing load table (up to 200 psi normal).

Step 3 — Radial Pad Width, $a$

$$a = \frac{D_2 - D_1}{2}$$

Step 4 — Pitch-Line Circumference, $B$

$$B = \pi \cdot \frac{D_1 + D_2}{2}$$

Step 5 — Number of Pads, $i$

Assume an oil groove width $s$ (0.062 inch minimum). If pad length equals width:

$$i_{approx} = \frac{B}{a + s}$$

Take $i$ as the nearest even number. If a chamfer is found necessary later to increase oil flow, the groove width should be greater than the chamfer width.

Step 6 — Length of Pad, $b$

$$b = \frac{B}{i} - s$$

Step 7 — Pitch-Line Velocity, $U$

$$U = \frac{B \cdot N}{12} \quad \text{(ft/min)}$$

Step 8 — Film Thickness, $h$

$$h = \frac{2.09 \times 10^{-9} \cdot i \cdot a^3 \cdot U \cdot Z}{W} \quad \text{(inches)}$$

Step 9 — Depth of Step, $e$

$$e = 0.7h$$

Step 10 — Friction Power Loss, $P_f$

$$P_f = \frac{7.35 \times 10^{-13} \cdot i \cdot a^2 \cdot U^2 \cdot Z}{h} \quad \text{(hp)}$$

Step 11 — Pad Step Length, $b_2$

$$b_2 = \frac{1.2b}{2.2}$$

Step 12 — Hydrodynamic Oil Flow, $Q$

$$Q = 6.65 \times 10^{-4} \cdot i \cdot a \cdot h \cdot U \quad \text{(gpm)}$$

Step 13 — Temperature Rise, $\Delta t$

$$\Delta t = \frac{42.4 \cdot P_f}{c \cdot Q} \quad \text{(°F)}$$

If the temperature rise exceeds 50°F, chamfers can be added using the same procedure from Steps 12–15 of the flat plate design.

Worked Example: Step Thrust Bearing

Problem: Design a step thrust bearing for positioning a 7/8-inch diameter shaft operating with a 25-pound thrust load at 5,000 rpm. The oil has a viscosity of 25 centipoises at the operating temperature of 160°F and a specific heat of 3.4 Btu/gal/°F.

Step 1 — Internal diameter: $D_1 = 1$ inch (to clear shaft)

Step 2 — External diameter:

Because this is a positioning bearing with very low total load, the unit load will be negligible. Rather than using the formula, a convenient size is selected for desired proportions: $D_2 = 3$ inches.

Step 3 — Radial pad width:

$$a = \frac{3 - 1}{2} = 1 \text{ inch}$$

Step 4 — Pitch-line circumference:

$$B = \pi \cdot \frac{3 + 1}{2} = 6.28 \text{ inches}$$

Step 5 — Number of pads:

Assuming minimum groove width of 0.062 inch:

$$i_{approx} = \frac{6.28}{1 + 0.062} = 5.9$$

Take $i = 6$.

Step 6 — Length of pad:

$$b = \frac{6.28}{6} - 0.062 = 0.985 \text{ inches}$$

Step 7 — Pitch-line velocity:

$$U = \frac{6.28 \times 5{,}000}{12} = 2{,}620 \text{ ft/min}$$

Step 8 — Film thickness:

$$h = \frac{2.09 \times 10^{-9} \times 6 \times 1^3 \times 2{,}620 \times 25}{25} = 0.0057 \text{ inches}$$

5.7 mils of film thickness — excellent for a positioning bearing. This is thick enough to ensure complete hydrodynamic separation.

Step 9 — Depth of step:

$$e = 0.7 \times 0.0057 = 0.004 \text{ inches}$$

Step 10 — Friction power loss:

$$P_f = \frac{7.35 \times 10^{-13} \times 6 \times 1^2 \times 2{,}620^2 \times 25}{0.0057} = 0.133 \text{ hp}$$

Step 11 — Pad step length:

$$b_2 = \frac{1.2 \times 0.985}{2.2} = 0.537 \text{ inches}$$

Step 12 — Hydrodynamic oil flow:

$$Q = 6.65 \times 10^{-4} \times 6 \times 1 \times 0.0057 \times 2{,}620 = 0.060 \text{ gpm}$$

Step 13 — Temperature rise:

$$\Delta t = \frac{42.4 \times 0.133}{3.4 \times 0.060} = 28°F$$

Step Bearing Design Verdict

$\Delta t = 28°F$ is well within the 50°F maximum. No chamfers are needed. This is a clean, efficient design for a light positioning application.

Type 3: Tapered Land Thrust Bearing Design

When to Use It

This is where the practitioner's story gets personal. The tapered land thrust bearing is the first choice for large, heavily loaded applications — compressors, turbines, and industrial drives. It handles the same 200–500 psi load range as step and tilting pad bearings, but it does so with a machined taper that creates a converging oil wedge.

Strengths:

• High load capacity
• Can be used in larger sizes than step bearings
• Well-proven, standardized design

Weaknesses:

• More costly to manufacture than step bearings (the taper must be precision-machined)
• Requires good alignment as size increases — this is the vulnerability the practitioner missed
• Taper extends to 80% of pad length; remaining 20% is flat

Key General Parameters

• $b_2 = 0.8b$ (taper extends over 80% of pad length)
• $b_1 = 0.2b$ (flat land covers the remaining 20%)
• $K_g = 0.8$ or $0.9$

Taper Values — Critical Reference Table

These taper values ($\delta$) are determined by pad dimensions and represent the height difference between the leading and trailing edges of the tapered section.

Note: Taper values at the inner diameter ($\delta_1$) are always smaller than at the outer diameter ($\delta_2$) because the linear speed is lower at the ID. The taper compensates for the velocity gradient across the pad width.

    Cross-Section — Tapered Land Thrust Bearing Pad

          ┌──── b ─────────────────────────────┐
          │                                     │
          │    b₂ (tapered, 80%)  │  b₁ (flat)  │
          │                       │   (20%)     │
    ──────┤\                      │             │──── Film
          │  \   δ (taper)        │             │     surface
          │    \                  │             │
    ──────┼──────\────────────────┴─────────────┼──── Pad
          │       h₂               h₁           │     surface
          └─────────────────────────────────────┘
          ◄──── Direction of motion U ─────────►

The Complete Tapered Land Design Procedure

Step 1 — Inside Diameter, $D_1$: Determined by shaft size and clearance.

Step 2 — Outside Diameter, $D_2$:

$$D_2 = \left(\frac{4W}{\pi K_g P_a} + D_1^2\right)^{1/2}$$

Where $P_a$ = assumed unit load from the thrust bearing load table.

Step 3 — Radial Pad Width, $a$:

$$a = \frac{D_2 - D_1}{2}$$

Step 4 — Pitch-Line Circumference, $B$:

$$B = \pi \cdot \frac{D_1 + D_2}{2}$$

Step 5 — Number of Pads, $i$:

Assume an oil groove width $s$, pad length ≈ pad width:

$$i_{approx} = \frac{B}{a + s}$$

Take $i$ as the nearest even number.

Step 6 — Length of Pad, $b$:

$$b = \frac{B - is}{i}$$

Step 7 — Taper Values, $\delta_1$ and $\delta_2$:

Interpolate from the taper values table above based on actual pad dimensions.

Step 8 — Actual Bearing Unit Load, $p$:

$$p = \frac{W}{i \cdot a \cdot b}$$

Step 9 — Pitch-Line Velocity, $U$:

$$U = \frac{B \cdot N}{12} \quad \text{(ft/min)}$$

Step 10 — Oil Leakage Factor, $Y_L$:

Found from empirical curves of $Y_L$ vs. pad dimensions $a$ and $b$, or from:

$$Y_L = \frac{b}{1 / (\pi^2 b^2 + 12a^2)}$$

Practical note: For pads with $a = 5$ inches and $b \approx 5.78$ inches, $Y_L \approx 2.75$.

Step 11 — Film Thickness Factor, $K$:

$$K = \frac{5.75 \times 10^6 \times p}{U \cdot Y_L \cdot Z}$$

Step 12 — Minimum Film Thickness, $h$:

Using the value of $K$ and selected taper values, $h$ is found from empirical curves.

Target values:

• 0.001 inch for small bearings
• 0.002 inch for larger and high-speed bearings

Step 13 — Friction Power Loss, $P_f$:

Using the film thickness $h$, the power-loss coefficient $J$ is obtained from empirical curves. Then:

$$P_f = 8.79 \times 10^{-13} \cdot i \cdot a \cdot b \cdot J \cdot U^2 \cdot Z \quad \text{(hp)}$$

Step 14 — Required Oil Flow, $Q$:

$$Q = \frac{42.4 \cdot P_f}{c \cdot \Delta t} \quad \text{(gpm)}$$

Limit: $\Delta t = 50°F$ maximum.

Step 15 — Shape Factor, $Y_S$:

$$Y_S = \frac{8ab}{D_2^2 - D_1^2}$$

Step 16 — Oil Flow Factor, $Y_G$:

Found from empirical curves using $Y_S$ and $D_1/D_2$.

Step 17 — Actual Oil Film Flow, $Q_F$:

$$Q_F = \frac{8.9 \times 10^{-4} \cdot i \cdot \delta_2 \cdot D_2^3 \cdot N \cdot Y_G \cdot Y_S^2}{D_2 - D_1} \quad \text{(gpm)}$$

Step 18 — Flow Adequacy Check:

If $Q_F < Q$ (film flow is less than required flow), either increase the tapers or add chamfers using the flat plate bearing chamfer procedure (Steps 12–15 of flat plate design).

Worked Example: Tapered Land Thrust Bearing

Problem: Design a tapered land thrust bearing for 70,000 pounds at 3,600 rpm. Shaft diameter is 6.5 inches. Oil inlet temperature is 110°F at 20 psi. Maximum temperature rise of 50°F is acceptable, resulting in an outlet viscosity of 18 centipoises. Use $K_g = 0.9$ and $c = 3.5$ Btu/gal/°F.

Step 1 — Internal diameter: $D_1 = 7$ inches (to clear shaft).

Step 2 — External diameter:

Assume $P_a = 400$ psi:

$$D_2 = \left(\frac{4 \times 70{,}000}{3.14 \times 0.9 \times 400} + 7^2\right)^{1/2} = 17.2 \text{ inches}$$

Round to $D_2 = 17$ inches.

Step 3 — Radial pad width:

$$a = \frac{17 - 7}{2} = 5 \text{ inches}$$

Step 4 — Pitch-line circumference:

$$B = 3.14 \times \frac{17 + 7}{2} = 37.7 \text{ inches}$$

Step 5 — Number of pads:

Assume groove width $s = 0.5$ inch:

$$i_{approx} = \frac{37.7}{5 + 0.5} = 6.85$$

Take $i = 6$.

Step 6 — Length of pad:

$$b = \frac{37.7 - 6 \times 0.5}{6} = 5.78 \text{ inches}$$

Step 7 — Taper values:

Interpolating from the table for $a \times b \approx 5 \times 5.78$:

$$\delta_1 = 0.008 \text{ inch (at ID)}, \quad \delta_2 = 0.005 \text{ inch (at OD)}$$

Step 8 — Actual bearing unit load:

$$p = \frac{70{,}000}{6 \times 5 \times 5.78} = 404 \text{ psi}$$

Step 9 — Pitch-line velocity:

$$U = \frac{37.7 \times 3{,}600}{12} = 11{,}300 \text{ ft/min}$$

Step 10 — Oil leakage factor:

From empirical data: $Y_L = 2.75$

Step 11 — Film thickness factor:

$$K = \frac{5.75 \times 10^6 \times 404}{11{,}300 \times 2.75 \times 18} = 4{,}150$$

Step 12 — Minimum film thickness:

From the empirical curve using $K = 4{,}150$ and taper values $\delta_2 \times \delta_1 = 0.005 \times 0.008$:

$$h = 2.2 \text{ mils} = 0.0022 \text{ inches}$$

✅ This exceeds the 0.002-inch target for larger bearings. Good.

Step 13 — Friction power loss:

From the empirical curve, $J = 260$:

$$P_f = 8.79 \times 10^{-13} \times 6 \times 5 \times 5.78 \times 260 \times 11{,}300^2 \times 18 = 91 \text{ hp}$$

91 horsepower of friction loss. This is substantial but typical for a bearing handling 70,000 pounds at 11,300 ft/min.

Step 14 — Required oil flow:

$$Q = \frac{42.4 \times 91}{3.5 \times 50} = 22.0 \text{ gpm}$$

Step 15 — Shape factor:

$$Y_S = \frac{8 \times 5 \times 5.78}{17^2 - 7^2} = 0.963$$

Step 16 — Oil flow factor:

From empirical data with $Y_S = 0.963$ and $D_1/D_2 = 0.41$: $Y_G = 0.61$

Step 17 — Actual oil film flow:

$$Q_F = \frac{8.9 \times 10^{-4} \times 6 \times 0.005 \times 17^3 \times 3{,}600 \times 0.61 \times 0.963^2}{17 - 7} = 26.7 \text{ gpm}$$

The Critical Verdict

$Q_F = 26.7$ gpm exceeds $Q = 22.0$ gpm.

The film flow exceeds the required oil flow. No chamfers are necessary. The bearing can supply enough oil through its own hydrodynamic film action to maintain adequate cooling.

This is the ideal outcome. When film flow exceeds required flow, the bearing is thermally self-sufficient. If it had been the other way around ($Q_F < Q$), you'd need chamfers or increased taper to bridge the gap.

Tapered Land Design Summary

Type 4: Tilting Pad (Kingsbury) Thrust Bearing Design

When to Use It

This is the bearing the practitioner should have specified for that compressor train. The tilting pad — commonly called the Kingsbury bearing — is the premium solution for high-thrust applications.

Its defining advantage: the ability to absorb significant amounts of misalignment.

Each pad pivots independently on its own support point, allowing it to self-adjust its tilt angle and maintain a proper oil wedge even when the shaft isn't perfectly perpendicular to the bearing face. This is exactly what the tapered land bearing cannot do.

Trade-offs:

• Higher cost due to more complex construction
• More components (individual pads, pivots, retaining mechanisms)
• Larger axial envelope in some configurations

But for critical, high-value machinery where misalignment is possible? The tilting pad bearing is worth every unit of additional cost.

Pivot Location

The optimum pivot location is not at the center of the pad. It's offset toward the trailing edge (approximately 58% from the leading edge). This asymmetry creates the most effective converging oil wedge.

However: If shaft rotation in both directions is required, the pivot must be at the midpoint. This results in little or no detrimental effect on performance.

    Cross-Section — Tilting Pad Thrust Bearing

    ┌──── b ──────────────────────────┐
    │                                  │
    │          Oil film wedge          │
    │  ╲                          ╱    │
    │    ╲   hmin              ╱      │
    │      ╲                ╱         │
    │        ╲           ╱            │
    │          ╲      ╱               │
    │            ╲ ╱   ▲              │
    │             ╳    │ Pivot        │
    │           ╱  ╲   │ (at 0.58b)  │
    └─────────────────────────────────┘
    ◄──── Direction of motion U ─────►

    │←── 0.58b ──→│←── 0.42b ──→│

The Complete Tilting Pad Design Procedure

Step 1 — Inside Diameter, $D_1$: Determined by shaft size and clearance.

Step 2 — Outside Diameter, $D_2$:

$$D_2 = \left(\frac{4W}{\pi K_g p} + D_1^2\right)^{1/2}$$

Where $K_g = 0.8$ and $p$ = unit load from the load table.

Step 3 — Radial Pad Width, $a$:

$$a = \frac{D_2 - D_1}{2}$$

Step 4 — Pitch-Line Circumference, $B$:

$$B = \pi \cdot \frac{D_1 + D_2}{2}$$

Step 5 — Number of Pads, $i$:

$$i = \frac{B \cdot K_g}{a}$$

Select the nearest even number.

Step 6 — Length of Pad, $b$:

$$b \cong \frac{B \cdot K_g}{i}$$

Step 7 — Pitch-Line Velocity, $U$:

$$U = \frac{B \cdot N}{12} \quad \text{(ft/min)}$$

Step 8 — Bearing Unit Load, $p$:

$$p = \frac{W}{i \cdot a \cdot b}$$

Step 9 — Operating Number, $O$:

$$O = \frac{1.45 \times 10^{-7} \cdot Z_2 \cdot U}{5 \cdot p \cdot b}$$

Where $Z_2$ = viscosity at outlet temperature (inlet temperature plus assumed temperature rise).

Step 10 — Minimum Film Thickness, $h_{min}$:

Using the operating number $O$, the dimensionless film thickness $\alpha$ is found from empirical curves (plotted against $O$ for various $b/a$ ratios). Then:

$$h_{min} = \alpha \cdot b$$

Target values:

• 0.001 inch for small bearings
• 0.002 inch for larger and high-speed bearings

Step 11 — Coefficient of Friction, $f$:

Found from empirical curves of $f$ vs. $\alpha$ for various $b/a$ ratios.

Step 12 — Friction Power Loss, $P_f$:

$$P_f = \frac{f \cdot W \cdot U}{33{,}000} \quad \text{(hp)}$$

Step 13 — Actual Oil Flow, $Q$:

$$Q = 0.0591 \cdot \alpha \cdot i \cdot a \cdot b \cdot U \quad \text{(gpm)}$$

Step 14 — Temperature Rise, $\Delta t$:

$$\Delta t = \frac{0.0217 \cdot f \cdot p}{\alpha \cdot c} \quad \text{(°F)}$$

Maximum acceptable: 50°F. If exceeded, chamfers can be added per the flat plate bearing procedure.