Cantilevered Loads Simplicity ® Plain Bearings
Cantilevered Loads • Maximum 2:1 ratio
Balance the moments: f • s = L • P L / s = f / P Compute friction force: F = f • µ Note: Total friction force pushing up is 2 * F. To lock up the slide, the total friction force must be equal to
• 1x = bearing separation on same shaft • 2x = distance from shaft to load or force Example: If 2x equals 10” then 1x must be at least 5”
(or greater than) P. P = 2 • F = 2 • f • µ
Binding will occur if the 2:1 ratio is exceeded!! This principle is NOT load dependent. It is NOT due to edge loading. It is also NOT dependent on the driving force used. The bearings will bind whether hand or mechanically driven. This principle is a product of friction. Working through the following equation will explain why this is a product of friction: P = force being applied L = distance out from shaft that P is being applied
Substitute for P: L / s = f / ( 2 • f • µ) = 1 / ( 2 • µ ) = > L / s = 1 / ( 2 • µ ) Note: The forces drop out of the equation. Assume static coefficient of friction is .25 (µ = .25) then L / s = 2 That is the 2:1 ratio. There may be other factors that add to the braking effect, but the coefficient of friction is the main cause. Note: Proper lubrication can help to drop friction and extend the 2:1 ratio.
s = center to center spacing of bearings f = resultant force on bearings by shaft F = friction force on each bearing µ = coefficient of friction (about .25 when not moving)
2X MAX
L
P
Load or Force
f
S F
1X
f
F
pbclinear.com • Engineering Your Linear Motion Solutions 143
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