How an O-Ring Functions (3)

The diametral squeeze applied to the constant volume of the O-Ring material will produce an increase in length of rubber across the groove. Expansion or swell of the rubber compound in the fluid or from heat will further increase this length of squeezed rubber. The groove must always be made sufficiently long to allow for the maximum expansion of the rubber, otherwise very high stresses will be set up on the installation. Normally the groove length will permit the O-ring to slide or to roll a small amount within the groove.

When break-out force is applied, the O-Ring will slide or roll in the direction of force applied until it contacts the end of the groove. From this point on, further pressure or force can only result in deformation of the O-Ring into tighter contact with the inner wall, the groove end and the outer wall.

The O-Ring will initially deform into a "D" shape. This normal deformation will increase the surface contact area to 70 to 80 percent of the initial cross-section. Thus, the contact area of sealing under pressure is roughly twice the area of contact of the original zero-pressure seal resulting from diametral squeeze. It will be apparent from the foregoing that the O-Ring will seal in either pressure direction.

Figure 5 shows the extreme case of deformation just before failure. Note that a small portion of the rubber material has been forced into the small clearance beyond the groove. Assuming the rubber has reached its limit of flow under pressure, further increase of force will result in failure by shear or extrusion as shown in Figure 6. The clearance allowed will bear a direct relation to the force causing failure.

Section 1 | Section 2 | Section 3