Page 394 - Shigley's Mechanical Engineering Design

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Shafts and Shaft Components 369
DE-Gerber
⎧ ⎫
2 1/2
1 8A ⎨ 2BS e ⎬
= 1 + 1 + (7–9)
3
n πd S e ⎩ AS ut ⎭
1/3
⎛ ⎧ ⎫⎞
2 1/2
8nA ⎨ 2BS e ⎬
d = ⎝ 1 + 1 + ⎠ (7–10)
πS e ⎩ AS ut ⎭
where

2
A = 4(K f M a ) + 3(K fs T a ) 2

2
B = 4(K f M m ) + 3(K fs T m ) 2
DE-ASME Elliptic
1/2
2 2 2 2
1 16 K f M a K fs T a K f M m K fs T m
= 4 + 3 + 4 + 3
n πd 3 S e S e S y S y
(7–11)
1/3
⎧ ⎫
2 2 2 2 1/2
16n K f M a K fs T a K f M m K fs T m
⎨ ⎬
d = 4 + 3 + 4 + 3
⎩ π S e S e S y S y ⎭
(7–12)
DE-Soderberg
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 yt
(7–13)

16n 1 2 2 1/2

d = 4(K f M a ) + 3(K fs T a )
π S e
(7–14)
1/3

1 2 2 1/2

+ 4(K f M m ) + 3(K fs T m )
S yt
For a rotating shaft with constant bending and torsion, the bending stress is com-
pletely reversed and the torsion is steady. Equations (7–7) through (7–14) can be sim-
plified by setting M m and T a equal to 0, which simply drops out some of the terms.
Note that in an analysis situation in which the diameter is known and the factor of
safety is desired, as an alternative to using the specialized equations above, it is always
still valid to calculate the alternating and mid-range stresses using Eqs. (7–5) and (7–6),
and substitute them into one of the equations for the failure criteria, Eqs. (6–45) through
(6–48), and solve directly for n. In a design situation, however, having the equations
pre-solved for diameter is quite helpful.
It is always necessary to consider the possibility of static failure in the first load cycle.
The Soderberg criteria inherently guards against yielding, as can be seen by noting that
its failure curve is conservatively within the yield (Langer) line on Fig. 6–27, p. 305. The
ASME Elliptic also takes yielding into account, but is not entirely conservative