Tensile Test of Slip Ring - Working out Tensile Strength

  • Thread starter Delvaurius
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All,

Through work, I've recently been tasked with investigating a very old tensile test technique, according to BS HC 403:1977 (now withdrawn).

Essentially, you take a ring. Anywhere along it you make a slit all the way through. You then pull that ring in a tensile machine, ensuring that the slit is perpendicular to the direction of force.

The standard then states, to work out the tensile strength, use the equation:

S = (3*P*((D/t)-1)) / b*t

S = Stress, P = Load, D = Outer Diameter, t = radial thickness, b = width.

Now I get the whole thing over (b*t) where b*t is the cross sectional area of the ring, meeting S=F/A.

How is Force = 3P((D/t) - 1) derived?

Thanks for any help!
 

Answers and Replies

  • #2
JBA
Science Advisor
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In your converted formula F = P; instead, you should be seeking the derivation of the original formula as stated.
 

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