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

Click For Summary
SUMMARY

The calculation of the torsion spring constant for a spring with 20 turns, a mean diameter of 6mm, and made from ASTM A228 material yields a spring constant of 2.556 Nm/revolution. The formula used is k = (Ed^4) / (10.8DN), where E is the modulus of elasticity (approximately 207 GPa for ASTM A228), d is the wire diameter (2mm), D is the mean diameter (6mm), and N is the number of turns (20). The resulting force applied by the spring when deflected a quarter turn with a 25mm arm length is calculated to be 25.56 N. Proper unit conversion is crucial for accurate results.

PREREQUISITES
  • Understanding of torsion spring mechanics
  • Familiarity with ASTM A228 material properties
  • Knowledge of the modulus of elasticity (E) and its application in spring calculations
  • Proficiency in unit conversions for torque and force calculations
NEXT STEPS
  • Research the effects of varying the number of turns on torsion spring constants
  • Explore advanced calculations for torsion springs using finite element analysis (FEA) tools
  • Learn about the impact of different materials on spring performance, specifically comparing ASTM A228 with other alloys
  • Investigate practical applications of torsion springs in mechanical design and engineering
USEFUL FOR

Mechanical engineers, students studying mechanics of materials, and professionals involved in spring design and analysis will benefit from this discussion.

evo_vil
Messages
10
Reaction score
0

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



[itex]k=\frac{Ed^4}{10.8DN}[/itex]

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?
 
Last edited:
Physics news on Phys.org
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:

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

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:

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

Hope that helps.
 
Yes that help a lot... 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)
 

Similar threads

Undergrad Finding the spring constant of a torsion spring.
  • · Replies 2 ·
Replies
2
Views
8K
Mechanics of material shaft coupling
  • · Replies 3 ·
Replies
3
Views
3K