Page 368 - Shigley's Mechanical Engineering Design

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Fatigue Failure Resulting from Variable Loading 343
might be

a
M a c F

= K f or = K f
a a
I A
, expressible as
a
respectively. This makes the COV of , namely C σ a

2 2 1/2
= C + C
C σ a Kf F
again a function of variabilities. The COV of S e , namely C Se , is
1/2
2 2 2 2 2
C Se = C + C + C + C + C
ka kc kd kf Se
again, a function of variabilities. An example will be useful.
EXAMPLE 6–21 A strap to be made from a cold-drawn steel strip workpiece is to carry a fully reversed
axial load F = LN(1000, 120) lbf as shown in Fig. 6–39. Consideration of adjacent
parts established the geometry as shown in the figure, except for the thickness t. Make a
decision as to the magnitude of the design factor if the reliability goal is to be 0.999 95,
then make a decision as to the workpiece thickness t.

Solution Let us take each a priori decision and note the consequence:

A Priori Decision Consequence

Use 1018 CD steel S ¯ = 87.6 kpsi, C Sut = 0.0655
ut
Function:
Carry axial load C F = 0.12, C kc = 0.125
R ≥ 0.999 95 z = 3.891
F = 1000 lbf Machined surfaces C ka = 0.058
a
2 1/2
2
Hole critical C Kf = 0.10, C a (0.10 0.12 ) = 0.156
Ambient temperature C kd = 0
3 -in D. drill
8 Correlation method C S e = 0.138
2 1/2
2
2
Hole drilled C Se = (0.058 + 0.125 + 0.138 ) = 0.195
, 2 2
-C Se + C 0.195 + 0.156 2
-
2
3 in C n = . σ a = = 0.2467
4 2 1 + 0.156 2
1 + C
σ a
/ √ 0
&
2
n = exp − (−3.891) ln(1 + 0.2467 ) + ln 1 + 0.2467 2
¯
F = 1000 lbf
a = 2.65
Figure 6–39
These eight a priori decisions have quantified the mean design factor as ¯n = 2.65.
A strap with a thickness t is Proceeding deterministically hereafter we write
subjected to a fully reversed
¯
S e F ¯
axial load of 1000 lbf. σ = = K f
¯

a
Example 6–21 considers the ¯ n (w − d)t
thickness necessary to attain a from which
reliability of 0.999 95 against ¯ ¯
K f ¯nF
a fatigue failure. t = (1)
¯
(w − d)S e