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Fatigue Failure Resulting from Variable Loading 289
here employs an equivalent diameter d e obtained by equating the volume of material
stressed at and above 95 percent of the maximum stress to the same volume in the
rotating-beam specimen. 16 It turns out that when these two volumes are equated,
the lengths cancel, and so we need only consider the areas. For a rotating round section,
the 95 percent stress area is the area in a ring having an outside diameter d and an inside
diameter of 0.95d. So, designating the 95 percent stress area A 0.95σ , we have
π 2 2 2
A 0.95σ = [d − (0.95d) ] = 0.0766d (6–22)
4
This equation is also valid for a rotating hollow round. For nonrotating solid or hollow
rounds, the 95 percent stress area is twice the area outside of two parallel chords hav-
ing a spacing of 0.95d, where d is the diameter. Using an exact computation, this is
A 0.95σ = 0.01046d 2 (6–23)
With d e in Eq. (6–22), setting Eqs. (6–22) and (6–23) equal to each other enables us to
solve for the effective diameter. This gives
d e = 0.370d (6–24)
as the effective size of a round corresponding to a nonrotating solid or hollow round.
A rectangular section of dimensions h × b has A 0.95σ = 0.05hb. Using the same
approach as before,
1/2
d e = 0.808(hb) (6–25)
Table 6–3 provides A 0.95σ areas of common structural shapes undergoing non-
rotating bending.
EXAMPLE 6–4 A steel shaft loaded in bending is 32 mm in diameter, abutting a filleted shoulder 38 mm
in diameter. The shaft material has a mean ultimate tensile strength of 690 MPa.
Estimate the Marin size factor k b if the shaft is used in
(a) A rotating mode.
(b) A nonrotating mode.
Solution (a) From Eq. (6–20)
d −0.107 32 −0.107
Answer k b = = = 0.858
7.62 7.62
(b) From Table 6–3,
d e = 0.37d = 0.37(32) = 11.84 mm
From Eq. (6–20),
−0.107
11.84
Answer k b = = 0.954
7.62
16 See R. Kuguel, “A Relation between Theoretical Stress-Concentration Factor and Fatigue Notch Factor
Deduced from the Concept of Highly Stressed Volume,” Proc. ASTM, vol. 61, 1961, pp. 732–748.