---
title: "Journal bearing design with film thickness check"
subject: "Machine Design I"
chapter: 7
key: "semester_6/machine_design_i:journal-bearing-minimum-film-thickness"
source: https://www.bctengineer.com/q/semester_6/machine_design_i/journal-bearing-minimum-film-thickness
site: BCT Engineering
---

# Journal bearing design with film thickness check

Machine Design I · Chapter 7: Lubrication and journal bearings

- Asked in: year not given, 8 marks
- Video: [Hydrodynamic journal bearing | Sommerfeld number | Design of journal Bearings](https://www.youtube.com/watch?v=_W8zA77b7zA) — The Mechanical Engineer
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/machine_design_i/journal-bearing-minimum-film-thickness
- Whole chapter: https://www.bctengineer.com/md/semester_6/machine_design_i/chapter-7

## Question

A full journal bearing has journal diameter 60 mm, length 60 mm, radial clearance 0.06 mm and runs at 1500 rpm under a load of 4.5 kN with oil of viscosity 0.02 Pa.s. For the operating Sommerfeld number, the design chart for l/d = 1 gives h0/c = 0.35, (r/c)f = 2.9 and Q/(r c N l) = 3.6, where N is in rev/s. Find the bearing pressure, Sommerfeld number, minimum film thickness, coefficient of friction, power loss and oil flow rate. Check the film thickness against Trumpler's criterion h0 >= 0.005 + 0.00004 d (mm).

## Short answer

Bearing pressure p = W/(ld) = 1.25 MPa. Sommerfeld number S = (r/c)² μN/p = 0.10. Minimum film thickness h₀ = 0.35c = 0.021 mm (>> Trumpler’s 0.0074 mm). Coefficient of friction f = (r/c)f/(r/c) = 0.0058. Power loss = 2πN f W r = 123 W. Oil flow Q = 3.6 r c N l = 9.7 cm³/s.

## Answer

Data: $d = 60$ mm, $l = 60$ mm, $r = 30$ mm, $c = 0.06$ mm, so $r/c = 500$; $N = 1500/60 = 25$ rev/s; $W = 4500$ N; $\mu = 0.02$ Pa.s.

#### Bearing pressure

$$
p = \frac{W}{ld} = \frac{4500}{0.06 \times 0.06} = 1.25\ \text{MPa}
$$

#### Sommerfeld number

$$
S = \left(\frac{r}{c}\right)^2\frac{\mu N}{p} = 500^2 \times \frac{0.02 \times 25}{1.25\times10^6} = 0.10
$$

#### Minimum film thickness

$$
h_0 = 0.35\,c = 0.35 \times 0.06 = 0.021\ \text{mm}
$$

#### Coefficient of friction

$$
f = \frac{2.9}{r/c} = \frac{2.9}{500} = 0.0058
$$

#### Friction torque and power loss

$$
\begin{aligned} T &= f W r = 0.0058 \times 4500 \times 0.03 = 0.783\ \text{N.m} \\ P_{loss} &= 2\pi N T = 2\pi \times 25 \times 0.783 = 123\ \text{W} \end{aligned}
$$

#### Oil flow

$$
\begin{aligned} Q &= 3.6\,(r\,c\,N\,l) = 3.6 \times 0.03 \times 0.06\times10^{-3} \times 25 \times 0.06 \\ &= 9.72\times10^{-6}\ \text{m}^3/\text{s} = 9.72\ \text{cm}^3/\text{s} \end{aligned}
$$

#### Trumpler's criterion

$$
h_{0,min} = 0.005 + 0.00004 \times 60 = 0.0074\ \text{mm}
$$

The actual $h_0 = 0.021$ mm is nearly three times this, so the film is thick enough to cover the surface roughness and small particles: the design is **safe**. (The oil temperature rise, from the heat generated, should still be checked.)

**Answer: p = 1.25 MPa; S = 0.10; h0 = 0.021 mm (greater than 0.0074 mm); f = 0.0058; power loss = 123 W; oil flow = 9.7 cm$^3$/s.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is the bearing pressure found for the full journal bearing (W = 4500 N, d = l = 60 mm)?
   A. p = W / (pi d l) = 0.80 MPa
   B. p = W / (l d) on the projected area = 1.25 MPa
   C. p = W / (pi d^2) = 0.40 MPa
   D. p = W d / l = 4.5 MPa
   Answer: B. Bearing pressure uses the projected area l x d: 4500 / (0.06 x 0.06) = 1.25 MPa. (Key point: Bearing pressure)

2. How is the minimum film thickness obtained from the chart value h0/c = 0.35?
   A. h0 = 0.35 x r = 10.5 mm
   B. h0 = 0.35 x c = 0.021 mm
   C. h0 = c / 0.35 = 0.171 mm
   D. h0 = 0.35 x d = 21 mm
   Answer: B. h0/c is a ratio to radial clearance, so h0 = 0.35 x 0.06 = 0.021 mm. (Key point: Minimum film thickness)

3. The chart gives (r/c) f = 2.9. What is the coefficient of friction?
   A. 2.9 x 500 = 1450
   B. 2.9 / 500 = 0.0058
   C. 2.9 / 25 = 0.116
   D. 2.9 x 0.06 = 0.174
   Answer: B. f = 2.9 / (r/c) = 2.9 / 500 = 0.0058, then T = f W r gives the torque and power loss. (Key point: Friction coefficient from chart)

4. Why is the design judged safe by Trumpler's criterion?
   A. h0 = 0.021 mm is nearly three times the required minimum 0.0074 mm, so it clears roughness and particles
   B. h0 is below the minimum, which keeps the oil film thin
   C. The Sommerfeld number is above 1
   D. The power loss is less than 100 W
   Answer: A. Minimum needed is 0.005 + 0.00004 x 60 = 0.0074 mm; the actual 0.021 mm is well above it. (Key point: Trumpler check)

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