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  Precision ground stainless steel shafts in undersize, nominal and oversize dimensions, as well as cold-drawn precision low-carbon steel shafts are available from SDP, as featured in our catalog. A new line of case hardened steel (C1060), case hardened stainless steel (440C) and thru hardened stainless steel (416) is being offered for use with bronze and needle bearings.

(e) Combined Torsional and Bending Loads for Fluctuating Loads With Stress Reversal
In such cases fatigue failure governs the design. Various failure theories have been proposed (e.g. Soderberg, Mises—Hencky etc.). All of these result in formulas involving both the average loading and the reversing component of the loading. Agreement with experiment varies depending on the nature of the material and failure mode. The equations resulting from these several theories also involve yield-point stresses and endurance limits and the various considerations governing the applicability of a particular set of equations extends beyond the scope of this discussion. For a fuller treatment the reader is referred to the following literature:
(i) R.M. Phelan “Fundamentals of Mechanical Design”, Third Edition, McGraw- Hill, New York,
                         N.Y., 1970. Chapter 6.
(ii) J.E. Shigley: “Mechanical Engineering Design”, Third Edition, McGraw-I-fill, New York, N.Y.,
                           1977, Chapter 13.
(iii) M.F. Spotts: “Design of Machine Elements”, Third Edition, Prentice-Hall, Englewood Cliffs,
                         New Jersey, 1961, Chapter 3.

3.0 HOLLOW CYLINDRICAL SHAFTS

In the case of hollow cylindrical shafts under torsion and/or bending, we can proceed as follows. Suppose a hollow shaft has an outside diameter, d₀, and an inside diameter, d_j, To size such a shaft for torsional and/or bending loads, we can convert the design calculation to that for an equivalent solid shaft of the same material by noting that the stress is inversely proportional to the ratio of moment of inertia (of the shaft cross-section) to shaft radius.
Equating this ratio for both the hollow shaft and the equivalent solid shaft of diameter d_eq we have:

    p (d₀⁴ — d₁⁴) =  pd³ _eg.  
           32d₀                  32

Solving for d_eq, we have:
d_eq =  ( d₀⁴ — d₁⁴ )^1/3                                                                                                                                 (6)
                    d₀

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