How an O-Ring Functions (2)
Figure 3 shows the proper installation of an O-Ring seal. Notice that the clearance for the O-Ring is less than its free outer diameter, that the O-Ring cross-section is squeezed diametrically out-of-round even before the application of pressure. This ensures contact with the inner and outer walls of the passage under static conditions.
Figure 4 now shows the action of this properly installed O-Ring when pressure is applied. Since both the inner and outer walls are in firm contact with the O-Ring material, the pressure tends to force it along its groove. Engineered to deform, the rubber compound flows up to the passage, completely sealing it against leakage. The higher the pressure trying to leak past, the tighter the seal that is thus formed. Upon release of the pressure, the resiliency of the rubber compound results in the O-Ring returning to its natural round form, undamaged and ready for similar cycles.
By this fundamental explanation, the criteria of design are clearly visible. The initial "diametral squeeze" is vitally important. An initial "diametral squeeze" of 10% results in a flat sealing surface of about 40 to 45% of the initial cross-section area of the O-Ring AT ZERO PRESSURE. Thus, at zero or very low pressures, the natural resiliency of the rubber compound provides the seal. It follows that very low pressure sealing may be improved by increased "diametral squeeze" (but note that such increased squeeze may adversely affect dynamic sealing at higher pressures).
The "diametral squeeze" induces a frictional force between the O-Ring and the walls of the sealed passage that tend to hold the O-Ring in "neutral" position. Until the forces applied are sufficient to either overcome the frictional force or deform the rubber compound, the O-Ring will retain its initially deformed shape and will seal purely by diametral pressure.
Section 1 | Section 2 | Section 3