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the amount of k in torsion spring

the amount of k in torsion spring

the amount of k in torsion spring

(OP)
hi
i am looking for the amount of the spring constant in torsion spring

i have a little torsion spring used in a two-hole paper punch with the specifications written below, and i need to know the amount of the k of this little spring:


Diameter of spring wire: 1.5 mm
Mean coil diameter : 2.75 mm
Number of active coils : 4
Spring length : 4 cm


thanks !

RE: the amount of k in torsion spring

According to SMI Handbook 1991:

k = E d 4 / (10.8 Nt D)

Where
E = Modulus of elasticity (MPa | psi)
d = Diameter of wire (mm | in.)
Nt = Number of total coils
D = Mean coil diameter (mm | in.)
k = Rate (N·mm/rev. | lbf·in./rev.)

You can handle the expression to get k per degrees of deflection instead of per revolutions of deflection.

You have to check the index of the spring (C = D/d). You have C = 1.83 now: it's a low value. You have to achieve an index of 3.5 min. Or, for feasible (though very difficult) coiling, 3.0. However, ask your supplier (or check the documentation of your coiler machine.)

The number of total coils is the sum of the body coils and the equivalent to the effect of the spring arms. According to Shigley:
Nt = Nb + Na
Na = (L1 + L2) / (3 π D)

If the angle between both arms is different than 180°, the number of body coils has a fractional part.

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