Torsion of a helical rod

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Let us consider the massless homogeneous helical spring with the infinitesimal wire diameter d and wire length l. We denote the spring radius as R. Now we consider the curvilinear spring section of length dl. We draw radii from the spiral axis to the centers of the end cross-sections of this section. After the twisting deformation the second cross-section rotates relative to the first by angle dφ=Mdl/(GJ), where M is the twisting torque; G is the shear modulus; J is the torsion constant. At this rate second radius rotates relative to the first one by the same angle dφ and second end of the section is shifted along the axis of the spiral by a distance dz=Rdφcosα, where α is the current pitch angle of spring (helix). Then z=Mlcosα/(GJ), where l is the total rod length (we assume that this length remains constant during the torsion process). On the other hand, if the initial pitch angle (before twisting) close to zero, then z=l sinα. As a result, we have: M/(GJ)=sinα/(R cosα). This equation differs from that given in cin literature http://www.manuscriptsystem.com/Journal/articles.aspx?journalid=1108 article"Solving Geometrically Nonlinear Problem on Deformation of a Helical Spring through Variational Methods" (there is M/(GJ)=sinαcosα/R), but I can not find mistake in my derivation. Please help me with this problem
 
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  • #2
If you are twisting a helical spring, I believe the stress in the wire is essentially simple bending, not torsion. The wire of a helical spring subjected to compression parallel to its axis is stressed in torsion.
 

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