Calculating Torsion Spring Constant | 20 Turns, 6mm Mean Dia, ASTM A228 Material

[evo_vil]

Homework Statement

I need to calculate the spring constant of a torsion spring.

Turns : 20
Mean Diameter : 6mm
Wire Diameter : 2mm
Material : ASTM A228
Arm Length : will be negligible, as they will be fixed
Orientation : Arms inline

Homework Equations

k=\frac{Ed^4}{10.8DN}

The Attempt at a Solution

Subbing values in i get a spring constant of 1.944 Nm/revolution

Does this look correct to you?
Am i using the right equation?

If the one arm is fixed and the other is rotated (with arm length 25mm) 1/4 revolution does this result in 1.944/4*0.025 = 0.01215 N or 1.944/4/0.025 = 19.44 N at the end of the arm?

 

[drmohr]

I'm not exactly sure what modulus you used but ASTM A228 is music wire so E≈207GPa. With that value the spring constant I found was:

\kappa = \frac{Ed^4}{10.8DN} = \frac{(207GPa)(2mm)^4}{10.8(6mm)(20)}=2.556Nm/rev

As for your question, to calculate the force the spring will apply you need to multiply the spring constant by the deflection and then divide the resulting torque by the moment arm (arm length). Therefore if the spring is deflected a quarter turn and the free arm is 25mm long:

F=\frac{\kappa \delta}{L}=(2.556Nm/rev)*(0.25rev)/(0.025m)=25.56N

Hope that helps.

 

[evo_vil]

Yes that help alot... Thank you... I was being stupid at the end and not thinking about the units... Newton meters divided by meters provides Newtons... Duh... After redoing the number i get your value... Thanks

(sorry bout the formating, done on my phone)