Page 339 - Shigley's Mechanical Engineering Design

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314 Mechanical Engineering Design
This completely reversed stress can be obtained by replacing S e with σ rev in Eq. (6–46)
for the modified Goodman line resulting in
σ a 40
σ rev = = = 53.3 kpsi
σ m 20
1 − 1 −
S ut 80
From the material properties, Eqs. (6–14) to (6–16), p. 285, give
( fS ut ) 2 [0.9(80)] 2
a = = = 129.6 kpsi
S e 40
1 fS ut 1 0.9(80)
b =− log =− log =−0.0851
3 S e 3 40
1/b −1/0.0851
σ rev σ rev

N = = (1)
a 129.6
Substituting σ rev into Eq. (1) yields
53.3 −1/0.0851
.
4
Answer N = = 3.4(10 ) cycles
129.6
(b) For Gerber, similar to part (a), from Eq. (6–47),
40
σ a
σ rev = = = 42.7 kpsi
2 2
σ m 20
1 − 1 −
S ut 80
Again, from Eq. (1),
42.7 −1/0.0851
.
Answer N = = 4.6(10 ) cycles
5
129.6
Comparing the answers, we see a large difference in the results. Again, the modified
Goodman criterion is conservative as compared to Gerber for which the moderate dif-
ference in S f is then magnified by a logarithmic S, N relationship.

For many brittle materials, the first quadrant fatigue failure criteria follows a con-
cave upward Smith-Dolan locus represented by
S a 1 − S m /S ut
= (6–50)
S e 1 + S m /S ut
or as a design equation,
nσ a 1 − nσ m /S ut
= (6–51)
S e 1 + nσ m /S ut
For a radial load line of slope r, we substitute S a /r for S m in Eq. (6–50) and solve for
S a , obtaining the intersect

rS ut + S e 4rS ut S e
S a = −1 + 1 + 2 (6–52)
2 (rS ut + S e )
The fatigue diagram for a brittle material differs markedly from that of a ductile material
because:
• Yielding is not involved since the material may not have a yield strength.
• Characteristically, the compressive ultimate strength exceeds the ultimate tensile
strength severalfold.