By Sastry Ganti
Senior Engineer
Formeta Technologies
Wheeling, Ill.

Edited by Kenneth Korane

Off-center loads can tilt and lock up a shaft inside the bearing.

Off-center loads can tilt and lock up a shaft inside the bearing.


Experienced engineers and production personnel are generally well versed in the basics of slide bearings. There are plenty of resources available to assist with the choices necessary for a successful design. For instance, designers must weigh the benefits of plain versus ball bearings, ensure a rigid structure, calculate wear and life of the elements, determine the span of dual bearings, and select the right materials and lubricants.

All this is fine when forces applied to the shaft are perfectly centered. Rarely, however, do engineers consider the possibility that the system might see eccentric loads that lock up the guiding elements. Rules of thumb abound, such as make the guide 1.5 times the shaft diameter or, if room permits, use two bearings spaced as far apart as possible. And make the shaft as rigid as possible.

But most engineering textbooks and design manuals do not address the conditions that cause an eccentrically loaded slide bushing to lock up. The following calculations help provide the answer.

Let's look at a typical plain bearing or bushing containing a sliding shaft. Assume that the elements are fairly rigid in relation to applied loads, so deflections are minimal. The accompanying graphic shows an exaggerated sliding shaft tilted inside the bearing due to an eccentrically applied load. Motion is imminent in the P direction, or the bearing and shaft are locked.

Begin the analysis by balancing vertical forces,

P = 2µFh + W.

Taking moments of all forces about Q results in P(L – (H⁄2)) – FhV = 0.

Solving for the horizontal force,

Substituting Fh in the first equation results in

Finally, the interrelated equation for determining motion or lock-up is

From this we can check various operating conditions.

  1. P perfectly lines up with W, thus L = H⁄2. Therefore, PW = 1⁄(1-0) = 1, P = W and friction, however large, theoretically does not play a role.
  2. The bearing is frictionless, µ = 0. The denominator = 1 and P = W.
  3. Although L can have any value, assume L = H. Then the denominator is 1 – ( µHV). The goal is for µHV to approach 0, so the denominator is as close to 1 as possible. Thus, the bearing design should use low-friction material, minimizing µ; have H as small as possible without compromising rigidity; and V as large as proportions and space allow.
  4. If the denominator becomes zero or negative, PW does not have a meaningful solution, which indicates lock-up conditions. Thus, whether or not a sliding bearing will lock up depends only on friction µ and bearing proportions HV. The shaft-to-bushing clearance does not really enter into consideration, of course provided that shaft and bushing diameters are not the same.
Nomenclature
Fh= Horizontal force
H = Bearing diameter
L = Distance from bearing edge to force P.
P = Eccentrically applied reaction force causing sliding or lock-up.
Q = Point on shaft centerline at top of bearing
V = Bearing length
W = Weight and other forces acting centrally on sliding shaft.
µ = Coefficient of friction