I also am not sure exactly why you want to calculate the amount of
springback in the rings. Is it because you want to get all the rings
precisely the same diameter? this will happen already as long as the
rings are evenly wound onto the mandrel and the rings are cut with a
when you close (bend) the ring and make it flush, it will close the
gap created by both the spring back and the saw cut. In terms of the
final inner diameter, because you are bending the ring to close the
gap, it will 1) reverse any spring back and 2) account for the saw
cut. In effect the only factor you need to worry about is the width
of the saw cut on the inner circumference.
(presuming you are not intending to join the links…)
In terms of materials engineering maths, continuum mechanics (the
maths describing how to calculate final shape after deformation) was
the single hardest subject I ever attempted.
You need to know the yield point, modulus and deformation hardening
rate (which depends on composition microstructure and heat treatment)
for even a linear approximation.
if you don’t need micron accuracy, you can do better, but only if
you know the yield point. Ignoring hardening, the amount of spring
back is determined from just the modulus and the yield point.
modulus = stress/strain
spring back strain = yield stress/youngs modulus
for “silver” from the previous link, Youngs Modulus for silver = 71
GPa - this will stay almost constant for alloys that are mostly
silver, regardless of heat treatment
Yeild Stress = 54 MPa for Silver, 126 for sterling silver, 136 for
ASA: this is the factor that changes with composition and heat
lets go with sterling:
strain = 126/71,000 = 1.77 x 10^-3
so for a linear length of sterling silver wire 1 m long, stretched
past it’s yield point, when it is cut it will spring back 1.77mm.