Page 393 - Shigley's Mechanical Engineering Design

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368 Mechanical Engineering Design
Combining these stresses in accordance with the distortion energy failure theory,
the von Mises stresses for rotating round, solid shafts, neglecting axial loads, are
given by
1/2
2 2

2
2 1/2

σ = (σ + 3τ ) = 32K f M a + 3 16K fs T a (7–5)
a a a 3 3
πd πd
1/2

2 2

2
2 1/2
σ = (σ + 3τ ) = 32K f M m + 3 16K fs T m (7–6)

m m m 3 3
πd πd
Note that the stress-concentration factors are sometimes considered optional for the
midrange components with ductile materials, because of the capacity of the ductile
material to yield locally at the discontinuity.
These equivalent alternating and midrange stresses can be evaluated using an
appropriate failure curve on the modified Goodman diagram (See Sec. 6–12, p. 303, and
Fig. 6–27). For example, the fatigue failure criteria for the modified Goodman line as
expressed previously in Eq. (6–46) is
1 σ a σ m
= +
n S e S ut
Substitution of σ and σ from Eqs. (7–5) and (7–6) results in

a m
1 16 1 2 2 1/2 1 2 2 1/2

= 4(K f M a ) + 3(K fs T a ) + 4(K f M m ) + 3(K fs T m )
n πd 3 S e S ut
For design purposes, it is also desirable to solve the equation for the diameter. This
results in
16n 1 2 2 1/2

d = 4(K f M a ) + 3(K fs T a )
π S e
1/3

1 2 2 1/2

+ 4(K f M m ) + 3(K fs T m )
S ut
Similar expressions can be obtained for any of the common failure criteria by sub-
stituting the von Mises stresses from Eqs. (7–5) and (7–6) into any of the failure
criteria expressed by Eqs. (6–45) through (6–48), p. 306. The resulting equations for
several of the commonly used failure curves are summarized below. The names
given to each set of equations identifies the significant failure theory, followed by a
fatigue failure locus name. For example, DE-Gerber indicates the stresses are com-
bined using the distortion energy (DE) theory, and the Gerber criteria is used for the
fatigue failure.
DE-Goodman
1 16 1 2 2 1/2 1 2 2 1/2

= 4(K f M a ) + 3(K fs T a ) + 4(K f M m ) + 3(K fs T m )
n πd 3 S e S ut
(7–7)
16n 1 2 2 1/2

d = 4(K f M a ) + 3(K fs T a )
π S e
1/3 (7–8)

1 2 2 1/2
+ 4(K f M m ) + 3(K fs T m )
S ut