Is the Stress in a Twisted Helical Spring Caused by Torsion or Bending?

In summary, the conversation discusses a massless, homogeneous helical spring with infinitesimal wire diameter and length. The spring radius is denoted as R and a curvilinear section of length dl is considered. The second cross-section rotates relative to the first by an angle dφ after twisting deformation, with M representing the twisting torque, G the shear modulus, and J the torsion constant. The second end of the section is shifted along the spiral axis by a distance dz and the resulting equation is M/(GJ)=sinα/(Rcosα). However, this differs from the equation given in the cited article, and the speaker is seeking help in finding the mistake in their derivation.
  • #1
reterty
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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.
 

1. What is the difference between torsion and bending stress in a twisted helical spring?

Torsion stress is caused by twisting or rotating forces, while bending stress is caused by bending or flexing forces. In a twisted helical spring, both torsion and bending stress can be present, but the dominant type of stress depends on the specific design and use of the spring.

2. How do I determine if the stress in a twisted helical spring is caused by torsion or bending?

The type of stress in a twisted helical spring can be determined by analyzing the design and loading conditions of the spring. Torsion stress is typically caused by a concentrated twisting force at the ends of the spring, while bending stress can be caused by a distributed load along the length of the spring.

3. What factors affect the amount of torsion or bending stress in a twisted helical spring?

The amount of torsion or bending stress in a twisted helical spring depends on several factors, including the material properties of the spring, the dimensions and shape of the spring, the magnitude and direction of the applied load, and the number of active coils in the spring.

4. How does the stress in a twisted helical spring affect its performance?

The stress in a twisted helical spring can affect its performance in several ways. Excessive stress can cause the spring to deform or fail, leading to a decrease in its load-carrying capacity. Additionally, different types of stress can cause different types of failure, so it is important to understand and manage the stress in a spring to ensure its optimal performance.

5. Can the stress in a twisted helical spring be reduced or controlled?

Yes, the stress in a twisted helical spring can be reduced or controlled through various design and material choices. For example, increasing the wire diameter or changing the spring's geometry can help distribute the load and reduce stress concentrations. Additionally, selecting a material with higher strength or fatigue resistance can help minimize the effects of stress on the spring's performance.

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