| K = A | cos³ α | + Cτ sin α. |
| r |
The thrust across any section is R sin α parallel to the tangent to the helix, and the shearing stress-resultant is R cos α at right angles to the osculating plane.
When the twist is such that, if the rod were simply unbent, it would also be untwisted, τ is (sin α cos α) / r, and then, restoring the values of A and C, we have
| R = | Eπc4 | σ | sin α cos² α, | |
| 4r² | 1 + σ |
| K = | Eπc4 | 1 + σ cos² α | cos α. | |
| 4r | 1 + σ |
65. The theory of spiral springs affords an application of these results. The stress-couples called into play when a naturally helical spring (α, r) is held in the form of a helix (α′, r′), are equal to the differences between those called into play when a straight rod of the same material and section is held in the first form, and those called into play when it is held in the second form.
Thus the torsional couple is
| C ( | sin α′ cos α′ | − | sin α cos α | ). |
| r′ | r |
and the flexural couple is
| A ( | cos² α′ | − | cos² α | ). |
| r′ | r |