#### Installation — business terrible - 1 part

September 8th, 2015

**3.5.1 **Calculation of the Displacement

The procedure to obtain the displacement from the expression of Kj as a function of the crack depth is, in fact, the reverse of that used in Section 2.1 to obtain Q from the expression of the compliance. To obtain the compliance (from which the displacement follows trivially), we couple (2.1.32) with Irwin’s relationship (2.2.22) to get

1 2Ща) _ Kj 2b da ‘ E’

This provides the basic equation to solve for the compliance, which may be simplified by using the general expression (2.3.11) for Kj:

tlC(a) _ 2

da bDE‘

This equation may be integrated between the limits for no crack, for which the compliance is Co, and an arbitrary crack length a:

where the second expression follows by setting da — D da. Thus, setting и — CP, the displacement can be written as

“ =

in which uo is the elastic displacement of the structure in the absence of crack and vc(a) is the additional displacement due to the crack.

Often the displacement is expressed in terms of the nominal strength art = CnP/bD instead of the load P. Making the substitution and taking into account the definitions of functions k(a) and fc(a) given in and below (2.3.11), one gets

(7дг

U~~K

The dimensionless function v(a)js given by

v(a) = v0 + vc(a), v0 =

in which, again, щ is the elastic displacement the structure would experiment if uncracked. Note that the equations are formally the same as before, except that the functions labeled by hats (which we always use with P) are replaced by functions without hats, representing variables expressed in terms of <7дг.

Example 3,5.1 Consider a large plate with a very short edge crack (a D) subjected to remote stress a (Fig. 3.5.1). The approximate expression of the stress intensity factor is Ki = 1Л215сгт/па, which may be rewritten as Kj = as/Dl. 1215,/тга. Therefore, taking art = a we have

к(а) — 1.1215т/тгсї

Thus, from the last equation of (3.5.7), we get

r a

vc(a) — 2 x 1.258tt / a’ da’ — 1.25 87m2

Jo

Figure 3.5.1 Single-edge cracked panel subjected to remote uniaxial stress.

Since the displacement for an uncracked panel is щ — uH/E, the second equation of (3.5.7) gives Vo — H/D. Therefore, the total displacement is

U=~&d{^ + l-2587fa2^ (3.5.10)

Identical results are obtained if calculations are done in terms of the resultant load P = abD. D

The foregoing expressions hold for a single-tipped crack, and a stands for the total crack length. For the center cracked panel, or more generally for internal cracks, where the total crack length is customarily represented by 2a, the matters become a little more complex in the case of loadings that are not symmetric with respect to the axis normal to the crack. In such a case, the energy release rates are different at one and the other tip. This means that in reality dU*/да = + Q~ where the superscripts + and — refer

to the right and left tips, respectively. Therefore, we must also distinguish the stress intensity factors Kf and KJ and their associated shape factors k+(a) and k~(a). A development strictly parallel to the previous one for a single-tipped crack leads to identical expressions for the displacement, except that function vc(a) is now given by

vc{a) = 2 J jfc+ (a’) + k~ (a’)j da’ (3.5.11)

and similarly for the functions with hats. This shows that, for symmetric loadings (k+ = k~ — k), the double-tip case reduces to the single-tip one just by replacing da by Ida. But for nonsymmetric loading, the expression is quite different.