Hi Sheri,

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

fine blade.

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.

e.g.

http://www.ganoksin.com/gnkurl/ep81s6

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

treatment/microstructure

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.