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1656_C009.fm Page 390 Monday, May 23, 2005 3:58 PM
390 Fracture Mechanics: Fundamentals and Applications
FIGURE 9.6 Internal and external axial surface flaws in a pressurized cylinder.
That is, we have superimposed K solutions for power-law loading with n = 0, 1, 2, 3, 4 to obtain
I
the solution for the polynomial distribution. By comparing each term in the above expression with
Equation (9.6), we see that
p = σ n a n
t
n
Consider the example of a pressurized cylinder with an internal axial surface flaw, as illustrated
in Figure 9.6. In the absence of the crack, the hoop stress in a thick wall pressure vessel is as follows:
pR 2 R 2
σ θθ = R o 2 − R i i 2 1 + r o (9.8)
where p is the internal pressure and the other terms are defined in Figure 9.6. If we define the
origin at the inner wall (x = r – R) and perform a Taylor series expansion about x = 0, Equation (9.8)
i
becomes
pR 2 R 2 x x 2 x 3 x 4
σ θθ = R o 2 − R o i 2 1 + R o i − 2 R i + 3 R i − 4 R i + 5 R i + 0 ( ≤ i ≤ xR / 1) (9.9)
where x is in the radial direction with the origin at R . The first five terms of this expansion give
i
the desired fourth-order polynomial. An alternate approach would be to curve-fit a polynomial to

