Page 343 - Shigley's Mechanical Engineering Design

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318 Mechanical Engineering Design
multiple stresses on a stress element of a ductile material into a single equivalent von
Mises stress. The same approach will be used here.
The first step is to generate two stress elements—one for the alternating stresses and
one for the midrange stresses. Apply the appropriate fatigue stress-concentration factors
to each of the stresses; i.e., apply (K f ) bending for the bending stresses, (K fs ) torsion for the
torsional stresses, and (K f ) axial for the axial stresses. Next, calculate an equivalent von
Mises stress for each of these two stress elements, σ and σ . Finally, select a fatigue

a
m
failure criterion (modified Goodman, Gerber, ASME-elliptic, or Soderberg) to complete
the fatigue analysis. For the endurance limit, S e , use the endurance limit modifiers,
k a , k b , and k c , for bending. The torsional load factor, k c = 0.59 should not be applied as it
is already accounted for in the von Mises stress calculation (see footnote 17 on p. 290). The
load factor for the axial load can be accounted for by dividing the alternating axial stress
by the axial load factor of 0.85. For example, consider the common case of a shaft with
bending stresses, torsional shear stresses, and axial stresses. For this case, the von Mises
2 2 1/2
. Considering that the bending, torsional, and
stress is of the form σ = σ x + 3τ xy
axial stresses have alternating and midrange components, the von Mises stresses for the
two stress elements can be written as
1/2
2

(σ a ) axial 2

σ = (K f ) bending (σ a ) bending + (K f ) axial + 3 (K fs ) torsion (τ a ) torsion
a
0.85
(6–55)
1/2

2 2

σ = (K f ) bending (σ m ) bending + (K f ) axial (σ m ) axial + 3 (K fs ) torsion (τ m ) torsion
m
(6–56)
For first-cycle localized yielding, the maximum von Mises stress is calculated. This
would be done by first adding the axial and bending alternating and midrange stresses to
obtain σ max and adding the alternating and midrange shear stresses to obtain τ max . Then
substitute σ max and τ max into the equation for the von Mises stress. A simpler and more con-
.

servative method is to add Eq. (6–55) and Eq. (6–56). That is, let σ max = σ + σ .

a
m
If the stress components are not in phase but have the same frequency, the maxima
can be found by expressing each component in trigonometric terms, using phase angles,
and then finding the sum. If two or more stress components have differing frequencies,
the problem is difficult; one solution is to assume that the two (or more) components
often reach an in-phase condition, so that their magnitudes are additive.
EXAMPLE 6–14 A rotating shaft is made of 42- × 4-mm AISI 1018 cold-drawn steel tubing and has a
6-mm-diameter hole drilled transversely through it. Estimate the factor of safety guard-
ing against fatigue and static failures using the Gerber and Langer failure criteria for the
following loading conditions:
(a) The shaft is subjected to a completely reversed torque of 120 N · m in phase with a
completely reversed bending moment of 150 N · m.
(b) The shaft is subjected to a pulsating torque fluctuating from 20 to 160 N · m and a
steady bending moment of 150 N · m.
Solution Here we follow the procedure of estimating the strengths and then the stresses, followed
by relating the two.