Page 336 - Shigley's Mechanical Engineering Design

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Fatigue Failure Resulting from Variable Loading 311

2
2
(1) 33.9 (84) 2 S a 31.4
Answer S a = = 31.4 kpsi, S m = = = 31.4 kpsi
2
2
33.9 + (1) 84 2 r 1
To verify the fatigue factor of safety, n f = S a /σ a = 31.4/8.38 = 3.75.
As before, let us calculate r crit . From the third row second column panel of
Table 6–8,
2(84)33.9 2
S a = 2 2 = 23.5 kpsi, S m = S y − S a = 84 − 23.5 = 60.5kpsi
33.9 + 84
23.5
S a
r crit = = = 0.388
S m 60.5
which again is less than r = 1, verifying that fatigue occurs first with n f = 3.75.
The Gerber and the ASME-elliptic fatigue failure criteria are very close to each
other and are used interchangeably. The ANSI/ASME Standard B106.1M–1985 uses
ASME-elliptic for shafting.

EXAMPLE 6–11 A flat-leaf spring is used to retain an oscillating flat-faced follower in contact with a
plate cam. The follower range of motion is 2 in and fixed, so the alternating component
of force, bending moment, and stress is fixed, too. The spring is preloaded to adjust to
various cam speeds. The preload must be increased to prevent follower float or jump.
For lower speeds the preload should be decreased to obtain longer life of cam and
1
follower surfaces. The spring is a steel cantilever 32 in long, 2 in wide, and in thick,
4
as seen in Fig. 6–30a. The spring strengths are S ut = 150 kpsi, S y = 127 kpsi, and S e =
28 kpsi fully corrected. The total cam motion is 2 in. The designer wishes to preload
the spring by deflecting it 2 in for low speed and 5 in for high speed.
(a) Plot the Gerber-Langer failure lines with the load line.
(b) What are the strength factors of safety corresponding to 2 in and 5 in preload?
Solution We begin with preliminaries. The second area moment of the cantilever cross section is

bh 3 2(0.25) 3 4
I = = = 0.00260 in
12 12
Since, from Table A–9, beam 1, force F and deflection y in a cantilever are related by
3
F = 3EIy/l , then stress σ and deflection y are related by
Mc 32Fc 32(3EIy) c 96Ecy
σ = = = 3 = 3 = Ky
I I l I l
6
96Ec 96(30 · 10 )0.125 3
where K = 3 = 3 = 10.99(10 ) psi/in = 10.99 kpsi/in
l 32
Now the minimums and maximums of y and σ can be defined by
y min = δ y max = 2 + δ
σ min = Kδ σ max = K(2 + δ)