Question 4 of 6: Journal Bearing Sized by No-Load (Petroff) Friction
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — December 2013 · 07-Mec-B1 (22-Mec-B1) Advanced Machine Design.
Open-book, 3 hours, 100 marks. The paper requires all of Part I (Problems 1 and 2)
plus any three of the four Part II problems (3–6). All six problems are solved in full below.
Reference texts. R. G. Budynas & J. K. Nisbett, Shigley’s
Mechanical Engineering Design, 10th ed. (shafts & critical speed §7; bolted joints
§8; lubrication & journal bearings §12; brakes §16); R. C. Juvinall &
K. M. Marshek, Fundamentals of Machine Component Design (bearings, brakes, impact);
R. C. Hibbeler, Mechanics of Materials (bending, impact loading).
Note on scope. The exam instructs the candidate to attempt Part I
plus three of Part II. This document answers every problem so that the paper functions as a complete
study set; on exam day a candidate would submit Problems 1, 2 and any three of 3–6.
Given. Full journal bearing, speed $N = 250$ rpm; lubricant ISO VG100
(SAE 30); $L = 1.2\,D$; diametral clearance $c_d = 0.0045\,D$ (radial $c_r = 0.00225\,D$);
no-load power loss limit $P_f \le 2.5\times10^{-4}$ hp $= 0.186$ W.
Find. The maximum journal diameter and the corresponding allowable oil-operating
temperature.
Approach. At no external load the friction is Petroff friction; write the friction power
in terms of $D$ using $L=1.2D$ and $c_r=0.00225D$, which reduces the limit to a fixed product
$\mu D^3$. To make $D$ as large as possible the viscosity must be as small as allowed, i.e. the oil is
run at its highest safe temperature; the Walther chart then fixes $\mu$ and hence $D$.
Petroff friction power. The Petroff no-load friction torque is
$T_f = \dfrac{4\pi^{2}\mu N_s r^{3}L}{c_r}$ and the power is $P_f = T_f\,(2\pi N_s)$, with
$N_s = 250/60 = 4.17\ \text{rev/s}$.
Collapse to $\mu D^3$. Substituting $r=D/2$, $L=1.2D$ and $c_r=0.00225D$,
every geometric term scales with $D$ so
$$P_f = K\,\mu\,D^{3},\qquad K = \dfrac{4\pi^{2}N_s(2\pi N_s)(D/2)^3(1.2D)}{0.00225D}\Big/D^{3}.$$
Imposing $P_f = 0.186$ W gives the design constraint
$$\boxed{\mu D^{3} = 6.49\times10^{-7}\ \text{Pa}\cdot\text{s}\cdot\text{m}^3}.$$
Minimise viscosity → run at the thermal limit. Since $\mu D^3$ is fixed, the
largest $D$ comes from the smallest safe $\mu$, i.e. the highest safe oil temperature. For a mineral
oil the practical continuous limit is about $\boxed{T \approx 70\,{}^{\circ}\text{C}}$ (above this,
oxidation and film loss accelerate).
Viscosity at 70 °C (Walther / ASTM D341). For ISO VG100
($\nu = 100$ cSt at 40 °C, $11.4$ cSt at 100 °C) the Walther fit gives
$\nu(70\,{}^{\circ}\text{C}) = 27.7\ \text{cSt}$; with density $\rho \approx 854\ \text{kg/m}^3$,
$\mu = \rho\nu = 0.0236\ \text{Pa}\cdot\text{s}$.
Check: the answer hinges on the oil’s allowable temperature.
70 °C is the standard continuous limit for a mineral SAE 30 oil; a higher rated synthetic
(lower $\mu$) would allow a larger journal, and a cooler design a smaller one. The bearing must also be
checked against a real applied load ($p$, Sommerfeld number) — here only the no-load friction governs.
