Page 358 - Mechanical Behavior of Materials

Section 8.5 Additional Topics on Application of K 359

√
Solution The stress intensity K = FS g πa must be below K Ic by a safety factor X K = 3.0.
Noting this and substituting the expression for S g from Fig. 8.12(c) gives

√
K Ic P √ 66 MPa m 55,000 N
K = = F πa, = F π(0.006 m)
X K bt 3.0 (40 mm)(t, mm)
where K Ic is from Table 8.1. Note that F = 1.12 with 10% accuracy up to α = 0.13. But α =
a/b = (6mm)/(40 mm) = 0.15 is beyond this limit, so F must be calculated by substituting this
α into the appropriate polynomial-type expression. The result is

0.857 + 0.265α
4
F = 0.265(1 − α) + = 1.283
(1 − α) 3/2
Substituting this F and solving gives t = 11.01 mm (Ans.).
However, we need to check that the safety factor is also met for yielding. The fully plastic
limit force, from Fig. A.16(d), is

" √ #
2
P o = btσ o −α + 2α − 2α + 1
" #
2
P o = (40 mm)(11.01 mm)(925 MPa) −0.15 + 2(0.15) − 2(0.15) + 1 = 290,400 N
where the yield strength is from Table 8.1. Hence, the safety factor against yielding is

P o 290.4kN

X = = = 5.28
o
P 55 kN
which exceeds the required value, so the preceding result of t = 11.01 mm is the final answer.

8.5 ADDITIONAL TOPICS ON APPLICATION OF K

Following the basic treatment of the previous section, it is useful to consider some additional topics
related to the application of K to design and analysis. These topics include some special cracked
member configurations, superposition for handling combined loading, cracks inclined to the stress
direction, and also leak-before-break for pressure vessels.

8.5.1 Cases of Special Interest for Practical Applications
The handbooks listed in the References contain a wide variety of additional useful cases. These
include not only additional situations of cracked plates and shafts, but also cracked tubes, discs,
stiffened panels, etc., including three-dimensional cases.