From the Rolls-Royce experimental archive: a quarter of a million communications from Rolls-Royce, 1906 to 1960's. Documents from the Sir Henry Royce Memorial Foundation (SHRMF).
Notes on the design of helical compression springs.
| Identifier | ExFiles\Box 168\3\ img319 | |
| Date | 26th May 1943 | |
| 6119 Rm{William Robotham - Chief Engineer}/BC.1/BM. 26.5.43. NOTES ON THE DESIGN OF HELICAL COMPRESSION SPRINGS. The application of a compressive load P along the axis of a helical spring of mean coil radius R and helix angle ϕ produces at any section a shear force P cos ϕ, a compressive load P sin ϕ, a torque P R cos ϕ and a bending moment P R sin ϕ. Provided the helix angle ϕ is fairly small (and this will always be the case when the stresses are high, for then the spring will be nearly closed) the direct stresses due to the bending moment are insignificant in comparison with the shear stresses due to torsion. Accordingly only the latter need be considered. (a) Circular Section. (1) Basic Formulae. The usual elementary formulas for the stress in and stiffness of circular section materials are:- q = 8PD / πd³ ... (1) and S = Gd⁴ / 8nD³ ... (2) where P = the applied load. q = the maximum torsional stress. G = Modulus of rigidity. D = Mean coil diameter (=2R). d = diameter of cross-section of the material S = Stiffness of spring rate. n = effective number of coils (total number of coils less one or two, according to the degree to which the end coils are close coiled to enable the ends of the finished springs to be squared off). (2) Effect of curvature on stress. These formulas are based on the torque-stress and torque-strain relations for a straight bar under torsion; but owing to the curvature of the axis of the material, at the inside of the coil the twist takes place over a shorter length of material than at the outside of the coil. As a result at the inside of the coil the shear stress is greater than it would be in a straight bar and the value of "q" given by equation (1) must be corrected by multiplying by a factor K (greater than 1) the value of K depending only on the spring index D/d = C. Recommended formula for K K = (C + 0.2) / (C - 1). Wahl's factor is slightly more correct and equals (4C - 1) / (4C - 4) + 0.615 / C. Hence, q = 8PD / (πd³) . K = 2.55 PD / d³ x (C + 0.2) / (C - 1) | ||