Calculation of the Crack Opening Profile

To obtain the crack opening profile from K[ expressions, we need to know the stress intensity factor produced by a pair of point loads at an arbitrary position x along the crack. Let Px be the magnitude of the loads located at x (Fig 3.5.4). The displacement conjugate of the pair Px is the crack opening at point x, w(x). The stress intensity factor may be written as in (3.1.10), where now it is essential to remark that when the point loads are applied on the uncracked part of the crack plane (i. c., when x > a) they do not generate any stress intensity factor at the crack tip, so that we can write

kc(ce, x/D) = 0 for x > a (or x/D > a) (3.5.38)

Then we proceed as we did in calculating the CMOD. We first write the displacements as

и = C(a) P + Cx(a) Px (3.5.39)

w(x) = Cx(a) P + C? x(a) Px (3.5.40)

The compliance Cx = Ci is obtained from (3.5.18) with k (a) = k(a), fc2(a) = ka(a, x/D), and

C120 =

2 ra/D „

Cx{a)~-^— j k(a)kG{a, x/D)da (3.5.41)

Now, taking into account (3.5.38) and (3.5.40), the crack opening profile is obtained as P fa –

ui(x, a) = — vx(a) , vx(a)=2 k(a)ka{a’,x/D)da’ (3.5.42)

bb Jx/D

where it is understood that this equation is valid for x < a, and that, obviously, w(x, a) = 0 for x > a. The foregoing equation can be rewritten in terms of сгдг as

w(x, a) – ~rDvx(a) , vx{a)-2 [ k^^kai^ ,x/D)da’ (3.5.43)

& Jx/D

where, again, &(a) is defined in (2.3.11).

For an internal crack for which a is the half crack length, the previous adjustment for two crack tips must be performed again. The result is identical to (3.5.43) except that v(a) is now replaced by

vx(a) = 2 f [k+(a)k^(a,,x/D) 4- k~ (а)к^(а/ ^x/D)] da’ (3.5.44)

Jx/D

Example 3.5.4 Consider the center cracked panel of Fig. 2.1.1 subjected to remote uniaxial stress a, and assume the crack to be very small relative to the dimensions of the panel. To obtain the crack opening profile we use the stress intensity factor for a pair of point loads on the crack faces introduced in Example 3.1.7 (Fig. 3.1.6c), with kG(a, x/D) given Eq. (3.1.13). Because of the symmetry, the shape factor for the tip on the left is к/, (a, x/D) — k/,(a, – x/D). Introducing this and k+(a) = k~(a) = фа into (3.5.44), and substituting the result in the first of (3.5.43), we get

u>(x, a) = [ 4a -=da’ = ~ z2 (3.5.45)

E Jx/D ^a,2.. (xjDy E

which does coincide with the solution obtained by the complete elastic analysis of the problem; see Section

2.2.1. 0