Q:

I have been trying my best to come up with a formula for circumferential force on an underlying part using a material that has been given shrink force characteristics. Consider the following:Using an Instron I have been able to determine that my oriented material has 8.5lbs/inch width of material shrink force. Regardless of the length the material, it will shrink 20% with 8.5lbs of force at 1 inch wide. What I would like to know is how this shrink force can be applied to a calculation, where the material is wrapped around a cylinder with a diameter of "x", and "w" pounds of initial tension. Once the shrink commences (heat application) I have struggled to figure out how the longitudinal shrink force relates to the other variables to determine an effective cumulative PSI on the substrate or underlying part.Any help is much appreciated!

- Drew (age 30)

Charlotte, North Carolina

- Drew (age 30)

Charlotte, North Carolina

A:

That's a nice physics problem. The tension force in the tape is the same as the dF/dL where F is the free energy and L is the length. You say T=(8.5 lbs/inch)*W where W is tape width, presumably assuming that the tape is not allowed to contract.

The pressure is dF/dV where V is volume. Now we have to make a physical assumption. Can the material squish out the ends of the cylinder? If so, squeezing it won't put it under extra pressure.If it's held in so it can't squish out, than dV=WdA where A is the cross-section area. A=L^{2}/4π. So dV=(WL/2π)dL.

The material is probably highly incompressible, so you get the maximum T. The added pressure is then dF/dV=(2π/WL)dF/dL=(2π/L)*8.5 lbs/inch. That's about 54 lbs/inch*L. If you measure L in inches, that gives pressure in convenient pounds per square inch. A 10" tape would give 5.4 psi.

Mike W.

*(published on 10/12/2017)*