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Consequences of pressure on helix

  1. Feb 8, 2007 #1
    Two engineers are disagreeing on the following issue.

    Can anyone assist in providing a 3rd opinion?

    The question relates to the force required to compress a simple helical compression spring.

    Assume a compression spring is resting on a work table in absolute pressure of approximately 14.7 PSI. The spring requires 50 pounds of force to compress the spring.

    Next, assume the same spring was submerged in 10,000 PSI hydraulic fluid. What would be the force required to compress the spring?

    One engineer has a simple math formula to solve the problem.....while another engineer has a more complex math formula to solve the problem.
  2. jcsd
  3. Feb 8, 2007 #2


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    Why would there be any difference? The pressure at all points of the wire spring are equal such that they cancel out and don't produce any resultant load on the spring.
  4. Feb 8, 2007 #3


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    Tell the engineers to look up the definition of a hydrostatic force.
  5. Feb 8, 2007 #4
    Would a difference be resulting from the effective area differences for each individual coil loop?

    If we look at a single helical coil of the spring, the outer diameter is obviously larger than the inner diameter. Hence, the outer diameter, or edge of the helical coil is physically in contact with more hydraulic fluid than the inner edge.

    As the spring is being compressed, the diameters of each helical coil is increasing. Since the outer edge is under more "resistance" pressure to enlarge versus the inner edge desire to expand, would this not require more force be added to compress the spring - to overcome the resist forces from the differences in area?
  6. Feb 8, 2007 #5


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    The only difference is, there is a compressive hydrostatic stress field in the spring. But 10,000 psi is way below the yield point of most "normal" spring materials, so the stress would have no effect unless you were planning to stretch the spring close to its elastic limit.
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