Page 307 - Shigley's Mechanical Engineering Design

bud29281_ch06_265-357.qxd 12/02/2009 6:49 pm Page 282 pinnacle s-171:Desktop Folder:Temp Work:Don't Delete (Jobs):MHDQ196/Budynas:

282 Mechanical Engineering Design
From Fig. 5–27, we compute the ratio a f /h as

a f 0.1278
= = 0.256
h 0.5
Thus a f /h varies from near zero to approximately 0.256. From Fig. 5–27, for this range
β is nearly constant at approximately 1.07. We will assume it to be so, and re-evaluate
a f as
2

1 73

a f = = 0.112 in
π 1.07(115.2)
Thus, from Eq. (6–6), the estimated remaining life is
1 a f da 1 0.112 da
Answer N f = √ = √
C (β σ πa) m 3.8(10 −11 ) 0.004 [1.07(115.2) πa] 3
a i
3
0.112
5.047(10 )
3
=− √
= 64.7 (10 ) cycles
a
0.004

6–7 The Endurance Limit

The determination of endurance limits by fatigue testing is now routine, though a lengthy
procedure. Generally, stress testing is preferred to strain testing for endurance limits.
For preliminary and prototype design and for some failure analysis as well, a quick
method of estimating endurance limits is needed. There are great quantities of data in
the literature on the results of rotating-beam tests and simple tension tests of specimens
taken from the same bar or ingot. By plotting these as in Fig. 6–17, it is possible to see
whether there is any correlation between the two sets of results. The graph appears to
suggest that the endurance limit ranges from about 40 to 60 percent of the tensile
strength for steels up to about 210 kpsi (1450 MPa). Beginning at about S ut = 210 kpsi
(1450 MPa), the scatter appears to increase, but the trend seems to level off, as sug-

gested by the dashed horizontal line at S = 105 kpsi.
e
We wish now to present a method for estimating endurance limits. Note that esti-
mates obtained from quantities of data obtained from many sources probably have a
large spread and might deviate significantly from the results of actual laboratory tests of
the mechanical properties of specimens obtained through strict purchase-order specifi-
cations. Since the area of uncertainty is greater, compensation must be made by employ-
ing larger design factors than would be used for static design.
For steels, simplifying our observation of Fig. 6–17, we will estimate the endurance
limit as
⎧
0.5S ut S ut ≤ 200 kpsi (1400 MPa)
⎨
(6–8)
S = 100 kpsi S ut > 200 kpsi
e
⎩
700 MPa S ut > 1400 MPa

where S ut is the minimum tensile strength. The prime mark on S in this equation refers
e
to the rotating-beam specimen itself. We wish to reserve the unprimed symbol S e for the
endurance limit of an actual machine element subjected to any kind of loading. Soon
we shall learn that the two strengths may be quite different.